Unformatted text preview: SOLUTION MANUAL SI UNIT PROBLEMS CHAPTER 2 SONNTAG • BORGNAKKE • VAN WYLEN FUNDAMENTALS of Thermodynamics Sixth Edition CONTENT SUBSECTION Correspondence table Concept-Study Guide Problems Properties and Units Force and Energy Specific Volume Pressure Manometers and Barometers Temperature Review Problems PROB NO. 1-22 23-26 27-37 38-43 44-57 58-76 77-80 81-86 Sonntag, Borgnakke and van Wylen Correspondence table CHAPTER 2 6th edition Sonntag/Borgnakke/Wylen The correspondence between the problem set in this sixth edition versus the problem set in the 5'th edition text. Problems that are new are marked new and those that are only slightly altered are marked as modified (mod). Study guide problems 2.1-2.22 and 2.23-2.26 are all new problems. New 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 5th Ed. 1 new 2 new 3 new 5 6 7 9 10 12 new new new 11 13 new 18 14 New 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 5th Ed. new 16 17 new new 19 new 34 29 new 28 mod new 20 26 new 21 new new 15 new New 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 5th Ed. 24 new new 23 new 30 32 33 new 37 27 new 38 new 31 new 22 35 36 new English Unit Problems New 5th Ed. SI New 5th Ed. SI 87 new 97 43E 43 88 new 11 98 new 50 89 new 12 99 new 53 90 new 19 100 45E 70 91 new 20 101 46E 45 92 new 24 102 new 82 93 39E 33 103 48E 55 94 40E 104 new 80 95 new 47 105 47E 77 96 42E 42 Design and Open ended problems 106-116 are from 5th edition problems 2.502.60 Sonntag, Borgnakke and van Wylen Concept-Study Guide Problems 2.1 Make a control volume around the turbine in the steam power plant in Fig. 1.1 and list the flows of mass and energy that are there. Solution: We see hot high pressure steam flowing in at state 1 from the steam drum through a flow control (not shown). The steam leaves at a lower pressure to the condenser (heat exchanger) at state 2. A rotating shaft gives a rate of energy (power) to the electric generator set. 1 WT 2 Sonntag, Borgnakke and van Wylen 2.2 Make a control volume around the whole power plant in Figure 1.2 and with the help of Fig. 1.1 list what flows of mass and energy are in or out and any storage of energy. Make sure you know what is inside and what is outside your chosen C.V. Solution: Smoke stack Boiler building Coal conveyor system Storage gypsum cb flue gas Coal storage Turbine house Dock Combustion air Flue gas Underground Welectrical power cable District heating Cold return Hot water m m m Storage for later m transport out: Gypsum, fly ash, slag m Coal Sonntag, Borgnakke and van Wylen 2.3 Make a control volume that includes the steam flow around in the main turbine loop in the nuclear propulsion system in Fig.1.3. Identify mass flows (hot or cold) and energy transfers that enter or leave the C.V. Solution: 1 Hot steam from generator 1 Electric power gen. cb Welectrical WT 3 2 5 Condensate to steam gen. cold 4 7 6 Cooling by seawater The electrical power also leaves the C.V. to be used for lights, instruments and to charge the batteries. Sonntag, Borgnakke and van Wylen 2.4 Take a control volume around your kitchen refrigerator and indicate where the components shown in Figure 1.6 are located and show all flows of energy transfer. Solution: The valve and the cold line, the evaporator, is inside close to the inside wall and usually a small blower distributes cold air from the freezer box to the refrigerator room. Q leak . The black grille in the back or at the bottom is the condenser that gives heat to the room air. Q . W cb The compressor sits at the bottom. Sonntag, Borgnakke and van Wylen 2.5 An electric dip heater is put into a cup of water and heats it from 20oC to 80oC. Show the energy flow(s) and storage and explain what changes. Solution: Electric power is converted in the heater element (an electric resistor) so it becomes hot and gives energy by heat transfer to the water. The water heats up and thus stores energy and as it is warmer than the cup material it heats the cup which also stores some energy. The cup being warmer than the air gives a smaller amount of energy (a rate) to the air as a heat loss. Welectric C B Q loss Sonntag, Borgnakke and van Wylen 2.6 Separate the list P, F, V, v, ρ, T, a, m, L, t and V into intensive, extensive and nonproperties. Solution: Intensive properties are independent upon mass: P, v, ρ, T Extensive properties scales with mass: V, m Non-properties: F, a, L, t, V Comment: You could claim that acceleration a and velocity V are physical properties for the dynamic motion of the mass, but not thermal properties. Sonntag, Borgnakke and van Wylen 2.7 An escalator brings four people of total 300 kg, 25 m up in a building. Explain what happens with respect to energy transfer and stored energy. Solution: The four people (300 kg) have their potential energy raised, which is how the energy is stored. The energy is supplied as electrical power to the motor that pulls the escalator with a cable. Sonntag, Borgnakke and van Wylen 2.8 Water in nature exist in different phases like solid, liquid and vapor (gas). Indicate the relative magnitude of density and specific volume for the three phases. Solution: Values are indicated in Figure 2.7 as density for common substances. More accurate values are found in Tables A.3, A.4 and A.5 Water as solid (ice) has density of around 900 kg/m3 Water as liquid has density of around 1000 kg/m3 Water as vapor has density of around 1 kg/m3 (sensitive to P and T) Sonntag, Borgnakke and van Wylen 2.9 Is density a unique measure of mass distribution in a volume? Does it vary? If so, on what kind of scale (distance)? Solution: Density is an average of mass per unit volume and we sense if it is not evenly distributed by holding a mass that is more heavy in one side than the other. Through the volume of the same substance (say air in a room) density varies only little from one location to another on scales of meter, cm or mm. If the volume you look at has different substances (air and the furniture in the room) then it can change abruptly as you look at a small volume of air next to a volume of hardwood. Finally if we look at very small scales on the order of the size of atoms the density can vary infinitely, since the mass (electrons, neutrons and positrons) occupy very little volume relative to all the empty space between them. Sonntag, Borgnakke and van Wylen 2.10 Density of fibers, rock wool insulation, foams and cotton is fairly low. Why is that? Solution: All these materials consists of some solid substance and mainly air or other gas. The volume of fibers (clothes) and rockwool that is solid substance is low relative to the total volume that includes air. The overall density is m msolid + mair ρ=V= V solid + Vair where most of the mass is the solid and most of the volume is air. If you talk about the density of the solid only, it is high. Sonntag, Borgnakke and van Wylen 2.11 How much mass is there approximately in 1 L of mercury (Hg)? Atmospheric air? Solution: A volume of 1 L equals 0.001 m3, see Table A.1. From Figure 2.7 the density is in the range of 10 000 kg/m3 so we get m = ρV = 10 000 kg/m3 × 0.001 m3 = 10 kg A more accurate value from Table A.4 is ρ = 13 580 kg/m3. For the air we see in Figure 2.7 that density is about 1 kg/m3 so we get m = ρV = 1 kg/m3 × 0.001 m3 = 0.001 kg A more accurate value from Table A.5 is ρ = 1.17 kg/m3 at 100 kPa, 25oC. Sonntag, Borgnakke and van Wylen 2.12 Can you carry 1 m3 of liquid water? Solution: The density of liquid water is about 1000 kg/m3 from Figure 2.7, see also Table A.3. Therefore the mass in one cubic meter is m = ρV = 1000 kg/m3 × 1 m3 = 1000 kg and we can not carry that in the standard gravitational field. 2.13 A manometer shows a pressure difference of 1 m of liquid mercury. Find ∆P in kPa. Solution: Hg : L = 1 m; ρ = 13 580 kg/m3 from Table A.4 (or read Fig 2.7) The pressure difference ∆P balances the column of height L so from Eq.2.2 ∆P = ρ g L = 13 580 kg/m3 × 9.80665 m/s2 × 1.0 m × 10-3 kPa/Pa = 133.2 kPa Sonntag, Borgnakke and van Wylen 2.14 You dive 5 m down in the ocean. What is the absolute pressure there? Solution: The pressure difference for a column is from Eq.2.2 and the density of water is from Table A.4. ∆P = ρgH = 997 kg/m3 × 9.81 m/s2 × 5 m = 48 903 Pa = 48.903 kPa Pocean= P0 + ∆P = 101.325 + 48.903 = 150 kPa Sonntag, Borgnakke and van Wylen 2.15 What pressure difference does a 10 m column of atmospheric air show? Solution: The pressure difference for a column is from Eq.2.2 ∆P = ρgH So we need density of air from Fig.2.7, ρ = 1.2 kg/m3 ∆P = 1.2 kg/m3 × 9.81 ms-2 × 10 m = 117.7 Pa = 0.12 kPa Sonntag, Borgnakke and van Wylen 2.16 The pressure at the bottom of a swimming pool is evenly distributed. Suppose we look at a cast iron plate of 7272 kg lying on the ground with an area of 100 m2. What is the average pressure below that? Is it just as evenly distributed? Solution: The pressure is force per unit area from page 25: P = F/A = mg/A = 7272 kg × (9.81 m/s2) / 100 m2 = 713.4 Pa The iron plate being cast can be reasonable plane and flat, but it is stiff and rigid. However, the ground is usually uneven so the contact between the plate and the ground is made over an area much smaller than the 100 m2. Thus the local pressure at the contact locations is much larger than the quoted value above. The pressure at the bottom of the swimming pool is very even due to the ability of the fluid (water) to have full contact with the bottom by deforming itself. This is the main difference between a fluid behavior and a solid behavior. Iron plate Ground Sonntag, Borgnakke and van Wylen 2.17 A laboratory room keeps a vacuum of 0.1 kPa. What net force does that put on the door of size 2 m by 1 m? Solution: The net force on the door is the difference between the forces on the two sides as the pressure times the area F = Poutside A – Pinside A = ∆P A = 0.1 kPa × 2 m × 1 m = 200 N Remember that kPa is kN/m2. Pabs = Po - ∆P ∆P = 0.1 kPa Sonntag, Borgnakke and van Wylen 2.18 A tornado rips off a 100 m2 roof with a mass of 1000 kg. What is the minimum vacuum pressure needed to do that if we neglect the anchoring forces? Solution: The net force on the roof is the difference between the forces on the two sides as the pressure times the area F = Pinside A – PoutsideA = ∆P A That force must overcome the gravitation mg, so the balance is ∆P A = mg ∆P = mg/A = (1000 kg × 9.807 m/s2 )/100 m2 = 98 Pa = 0.098 kPa Remember that kPa is kN/m2. Sonntag, Borgnakke and van Wylen 2.19 What is a temperature of –5oC in degrees Kelvin? Solution: The offset from Celsius to Kelvin is 273.15 K, so we get TK = TC + 273.15 = -5 + 273.15 = 268.15 K Sonntag, Borgnakke and van Wylen 2.20 What is the smallest temperature in degrees Celsuis you can have? Kelvin? Solution: The lowest temperature is absolute zero which is at zero degrees Kelvin at which point the temperature in Celsius is negative TK = 0 K = −273.15 oC Sonntag, Borgnakke and van Wylen 2.21 Density of liquid water is ρ = 1008 – T/2 [kg/m3] with T in oC. If the temperature increases 10oC how much deeper does a 1 m layer of water become? Solution: The density change for a change in temperature of 10oC becomes ∆ρ = – ∆T/2 = –5 kg/m3 from an ambient density of ρ = 1008 – T/2 = 1008 – 25/2 = 995.5 kg/m3 Assume the area is the same and the mass is the same m = ρV = ρAH, then we have ∆m = 0 = V∆ρ + ρ∆V ⇒ ∆V = - V∆ρ/ρ and the change in the height is ∆V H∆V -H∆ρ -1 × (-5) ∆H = A = V = = 995.5 = 0.005 m ρ barely measurable. Sonntag, Borgnakke and van Wylen 2.22 Convert the formula for water density in problem 21 to be for T in degrees Kelvin. Solution: ρ = 1008 – TC/2 [kg/m3] We need to express degrees Celsius in degrees Kelvin TC = TK – 273.15 and substitute into formula ρ = 1008 – TC/2 = 1008 – (TK – 273.15)/2 = 1144.6 – TK/2 Sonntag, Borgnakke and van Wylen Properties and units 2.23 A steel cylinder of mass 2 kg contains 4 L of liquid water at 25oC at 200 kPa. Find the total mass and volume of the system. List two extensive and three intensive properties of the water Solution: Density of steel in Table A.3: ρ = 7820 kg/m3 Volume of steel: V = m/ρ = 2 kg = 0.000 256 m3 7820 kg/m3 Density of water in Table A.4: ρ = 997 kg/m3 Mass of water: m = ρV = 997 kg/m3 ×0.004 m3 = 3.988 kg Total mass: m = msteel + mwater = 2 + 3.988 = 5.988 kg Total volume: V = Vsteel + Vwater = 0.000 256 + 0.004 = 0.004 256 m3 = 4.26 L Sonntag, Borgnakke and van Wylen 2.24 An apple “weighs” 80 g and has a volume of 100 cm3 in a refrigerator at 8oC. What is the apple density? List three intensive and two extensive properties of the apple. Solution: m 0.08 kg kg ρ = V = 0.0001 3 = 800 m3 m Intensive kg ; m3 T = 8°C; ρ = 800 v= 1 m3 = 0.001 25 kg ρ P = 101 kPa Extensive m = 80 g = 0.08 kg V =100 cm3 = 0.1 L = 0.0001 m3 Sonntag, Borgnakke and van Wylen 2.25 One kilopond (1 kp) is the weight of 1 kg in the standard gravitational field. How many Newtons (N) is that? F = ma = mg 1 kp = 1 kg × 9.807 m/s2 = 9.807 N Sonntag, Borgnakke and van Wylen 2.26 A pressurized steel bottle is charged with 5 kg of oxygen gas and 7 kg of nitrogen gas. How many kmoles are in the bottle? Table A2 : MO2 = 31.999 ; MN2 = 28.013 5 nO2 = mO2 / MO2 = 31.999 = 0.15625 kmol 7 nO2 = mN2 / MN2 = 28.013 = 0.24988 kmol ntot = nO2 + nN2 = 0.15625 + 0.24988 = 0.406 kmol Sonntag, Borgnakke and van Wylen Force and Energy 2.27 The “standard” acceleration (at sea level and 45° latitude) due to gravity is 9.80665 m/s2. What is the force needed to hold a mass of 2 kg at rest in this gravitational field ? How much mass can a force of 1 N support ? Solution: ma = 0 = ∑ F = F - mg F F = mg = 2 × 9.80665 = 19.613 N F = mg => m = F/g = 1 / 9.80665 = 0.102 kg m g Sonntag, Borgnakke and van Wylen 2.28 A force of 125 N is applied to a mass of 12 kg in addition to the standard gravitation. If the direction of the force is vertical up find the acceleration of the mass. Solution: Fup = ma = F – mg F – mg F 125 a= m = m – g = 12 – 9.807 = 0.61 ms-2 x F m g Sonntag, Borgnakke and van Wylen 2.29 A model car rolls down an incline with a slope so the gravitational “pull” in the direction of motion is one third of the standard gravitational force (see Problem 2.1). If the car has a mass of 0.45 kg find the acceleration. Solution: ma = ∑ F = mg / 3 a = mg / 3m = g/3 = 9.80665 / 3 = 3.27 m/s2 g This acceleration does not depend on the mass of the model car. Sonntag, Borgnakke and van Wylen 2.30 When you move up from the surface of the earth the gravitation is reduced as g = 9.807 − 3.32 × 10-6 z, with z as the elevation in meters. How many percent is the weight of an airplane reduced when it cruises at 11 000 m? Solution: go= 9.807 ms-2 gH = 9.807 – 3.32 × 10-6 × 11 000 = 9.7705 ms-2 Wo = m g o ; WH = m g H 9.7705 WH/Wo = gH/go = 9.807 = 0.9963 Reduction = 1 – 0.9963 = 0.0037 or 0.37% i.e. we can neglect that for most application Sonntag, Borgnakke and van Wylen 2.31 A car drives at 60 km/h and is brought to a full stop with constant deceleration in 5 seconds. If the total car and driver mass is 1075 kg find the necessary force. Solution: Acceleration is the time rate of change of velocity. dV 60 × 1000 a = dt = = 3.333 m/s2 3600 × 5 ma = ∑ F ; Fnet = ma = 1075 kg × 3.333 m/s2 = 3583 N Sonntag, Borgnakke and van Wylen 2.32 A car of mass 1775 kg travels with a velocity of 100 km/h. Find the kinetic energy. How high should it be lifted in the standard gravitational field to have a potential energy that equals the kinetic energy? Solution: Standard kinetic energy of the mass is 100 × 10002 KIN = ½ m V2 = ½ × 1775 kg × 3600 m2/s2 = ½ × 1775 × 27.778 Nm = 684 800 J = 684.8 kJ Standard potential energy is POT = mgh h = ½ m V2 / mg = 684 800 = 39.3 m 1775 × 9.807 Sonntag, Borgnakke and van Wylen 2.33 A 1200-kg car moving at 20 km/h is accelerated at a constant rate of 4 m/s2 up to a speed of 75 km/h. What are the force and total time required? Solution: dV ∆V a = dt = => ∆t (75 − 20) 1000 ∆V ∆t = a = = 3.82 sec 3600 × 5 F = ma = 1200 kg × 4 m/s2 = 4800 N Sonntag, Borgnakke and van Wylen 2.34 A steel plate of 950 kg accelerates from rest with 3 m/s2 for a period of 10s. What force is needed and what is the final velocity? Solution: Constant acceleration can be integrated to get velocity. dV a = dt => ∫ dV = ∫ a dt => ∆V = a ∆t ∆V = a ∆t = 3 m/s2 × 10 s = 30 m/s => V = 30 m/s F = ma = 950 kg × 3 m/s2 = 2850 N F Sonntag, Borgnakke and van Wylen 2.35 A 15 kg steel container has 1.75 kilomoles of liquid propane inside. A force of 2 kN now accelerates this system. What is the acceleration? Solution: The molecular weight for propane is M = 44.094 from Table A.2. The force must accelerate both the container mass and the propane mass. m = msteel + mpropane = 15 + (1.75 × 44.094) = 92.165 kg ma = ∑ F ⇒ a = ∑ F / m 2000 N a = 92.165 kg = 21.7 m/s2 Sonntag, Borgnakke and van Wylen 2.36 A bucket of concrete of total mass 200 kg is raised by a crane with an acceleration of 2 m/s2 relative to the ground at a location where the local gravitational acceleration is 9.5 m/s2. Find the required force. Solution: Fup F = ma = Fup − mg Fup = ma + mg = 200 ( 2 + 9.5 ) = 2300 N g Sonntag, Borgnakke and van Wylen 2.37 On the moon the gravitational acceleration is approximately one-sixth that on the surface of the earth. A 5-kg mass is “weighed” with a beam balance on the surface on the moon. What is the expected reading? If this mass is weighed with a spring scale that reads correctly for standard gravity on earth (see Problem 2.1), what is the reading? Solution: Moon gravitation is: g = gearth/6 m Beam Balance Reading is 5 kg This is mass comparison m m Spring Balance Reading is in kg units Force comparison length ∝ F ∝ g 5 Reading will be kg 6 Sonntag, Borgnakke and van Wylen Specific Volume 2.38 A 5 m3 container is filled with 900 kg of granite (density 2400 kg/m3 ) and the rest of the volume is air with density 1.15 kg/m3. Find the mass of air and the overall (average) specific volume. Solution: mair = ρ V = ρair ( Vtot − mgranite ) ρ 900 = 1.15 [ 5 - 2400 ] = 1.15 × 4.625 = 5.32 kg V 5 v = m = 900 + 5.32 = 0.005 52 m3/kg Comment: Because the air and the granite are not mixed or evenly distributed in the container the overall specific volume or density does not have much meaning. Sonntag, Borgnakke and van Wylen 2.39 A tank has two rooms separated by a membrane. Room A has 1 kg air and volume 0.5 m3, room B has 0.75 m3 air with density 0.8 kg/m3. The membrane is broken and the air comes to a uniform state. Find the final density of the air. Solution: Density is mass per unit volume m = mA + mB = mA + ρBVB = 1 + 0.8 × 0.75 = 1.6 kg V = VA + VB = 0.5 + 0.75 = 1.25 m3 m 1.6 ρ = V = 1.25 = 1.28 kg/m3 A B cb Sonntag, Borgnakke and van Wylen 2.40 A 1 m3 container is filled with 400 kg of granite stone, 200 kg dry sand and 0.2 m3 of liquid 25°C water. Use properties from tables A.3 and A.4. Find the average specific volume and density of the masses when you exclude air mass and volume. Solution: Specific volume and density are ratios of total mass and total volume. mliq = Vliq/vliq = Vliq ρliq = 0.2 × 997 = 199.4 kg mTOT = mstone + msand + mliq = 400 + 200 + 199.4 = 799.4 kg Vstone = mv = m/ρ = 400/ 2750 = 0.1455 m3 Vsand = mv = m/ρ = 200/ 1500 = 0.1333 m3 VTOT = Vstone + Vsand + Vliq = 0.1455 + 0.1333 + 0.2 = 0.4788 m3 v = VTOT / mTOT = 0.4788/799.4 = 0.000599 m3/kg ρ = 1/v = mTOT/VTOT = 799.4/0.4788 = 1669.6 kg/m3 Sonntag, Borgnakke and van Wylen 2.41 A 1 m3 container is filled with 400 kg of granite stone, 200 kg dry sand and 0.2 m3 of liquid 25°C water. Use properties from tables A.3 and A.4 and use air density of 1.1 kg/m3. Find the average specific volume and density of the 1 m3 volume. Solution: Specific volume and density are ratios of total mass and total volume. Vstone = mv = m/ρ = 400/ 2750 = 0.1455 m3 Vsand = mv = m/ρ = 200/ 1500 = 0.1333 m3 Vair = VTOT − Vstone − Vsand − Vliq = 1− 0.1455 − 0.1333 − 0.2 = 0.5212 m3 mair = Vair/vair = Vair ρair = 0.5212 × 1.1 = 0.573 kg mliq = Vliq/vliq = Vliq ρliq = 0.2 × 997 = 199.4 kg mTOT = mstone + msand + mliq + mair = 400 + 200 + 199.4 + 0.573 ≈ 800 kg v = VTOT / mTOT = 1/800 = 0.00125 m3/kg ρ = 1/v = mTOT/VTOT = 800/1 = 800 kg/m3 Sonntag, Borgnakke and van Wylen 2.42 One kilogram of diatomic oxygen (O2 molecular weight 32) is contained in a 500L tank. Find the specific volume on both a mass and mole basis (v and v ). Solution: From the definition of the specific volume V 0.5 v = m = 1 = 0.5 m3/kg V V 3 v= = n m/M = M v = 32 × 0.5 = 16 m /kmol Sonntag, Borgnakke and van Wylen 2.43 A 15-kg steel gas tank holds 300 L of liquid gasoline, having a density of 800 kg/m3. If the system is decelerated with 6 m/s2 what is the needed force? Solution: m = mtank + mgasoline = 15 kg + 0.3 m3 × 800 kg/m3 = 255 kg F = ma = 255 kg × 6 m/s2 = 1530 N cb Sonntag, Borgnakke and van Wylen Pressure 2.44 A hydraulic lift has a maximum fluid pressure of 500 kPa. What should the piston-cylinder diameter be so it can lift a mass of 850 kg? Solution: With the piston at rest the static force balance is F↑ = P A = F↓ = mg A = π r2 = π D2/4 PA = P π D2/4 = mg D=2 mg =2 Pπ ⇒ D2 = 4mg Pπ 850 × 9.807 = 0.146 m 500 π × 1000 Sonntag, Borgnakke and van Wylen 2.45 A piston/cylinder with cross sectional area of 0.01 m2 has a piston mass of 100 kg resting on the stops, as shown in Fig. P2.45. With an outside atmospheric pressure of 100 kPa, what should the water pressure be to lift the piston? Solution: The force acting down on the piston comes from gravitation and the outside atmospheric pressure acting over the top surface. Force balance: F↑ = F↓ = PA = mpg + P0A Now solve for P (divide by 1000 to convert to kPa for 2nd term) mpg 100 × 9.80665 P = P0 + A = 100 kPa + 0.01 × 1000 = 100 kPa + 98.07 kPa = 198 kPa cb Water Sonntag, Borgnakke and van Wylen 2.46 A vertical hydraulic cylinder has a 125-mm diameter piston with hydraulic fluid inside the cylinder and an ambient pressure of 1 bar. Assuming standard gravity, find the piston mass that will create a pressure inside of 1500 kPa. Solution: Force balance: Po F↑ = PA = F↓ = P0A + mpg; P0 = 1 bar = 100 kPa cb A = (π/4) D2 = (π/4) × 0.1252 = 0.01227 m2 A 0.01227 mp = (P − P0) g = ( 1500 − 100 ) × 1000 × 9.80665 = 1752 kg g Sonntag, Borgnakke and van Wylen 2.47 A valve in a cylinder has a cross sectional area of 11 cm2 with a pressure of 735 kPa inside the cylinder and 99 kPa outside. How large a force is needed to open the valve? Fnet = PinA – PoutA = (735 – 99) kPa × 11 cm2 Pcyl = 6996 kPa cm2 = 6996 × = 700 N kN -4 2 2 × 10 m m cb Sonntag, Borgnakke and van Wylen 2.48 A cannon-ball of 5 kg acts as a piston in a cylinder of 0.15 m diameter. As the gun-powder is burned a pressure of 7 MPa is created in the gas behind the ball. What is the acceleration of the ball if the cylinder (cannon) is pointing horizontally? Solution: The cannon ball has 101 kPa on the side facing the atmosphere. ma = F = P1 × A − P0 × A = (P1 − P0 ) × A 2 2 = (7000 – 101) kPa × π ( 0.15 /4 ) m = 121.9 kN F 121.9 kN a = m = 5 kg = 24 380 m/s2 Sonntag, Borgnakke and van Wylen 2.49 Repeat the previous problem for a cylinder (cannon) pointing 40 degrees up relative to the horizontal direction. Solution: ma = F = ( P1 - P0 ) A - mg sin 400 2 ma = (7000 - 101 ) kPa × π × ( 0.152 / 4 ) m - 5 × 9.807 × 0.6428 N = 121.9 kN - 31.52 N = 121.87 kN F 121.87 kN a= m = = 24 374 m/s2 5 kg Sonntag, Borgnakke and van Wylen 2.50 A large exhaust fan in a laboratory room keeps the pressure inside at 10 cm water relative vacuum to the hallway. What is the net force on the door measuring 1.9 m by 1.1 m? Solution: The net force on the door is the difference between the forces on the two sides as the pressure times the area F = Poutside A – Pinside A = ∆P × A = 10 cm H2O × 1.9 m × 1.1 m = 0.10 × 9.80638 kPa × 2.09 m2 = 2049 N Table A.1: 1 m H2O is 9.80638 kPa and kPa is kN/m2. Sonntag, Borgnakke and van Wylen 2.51 What is the pressure at the bottom of a 5 m tall column of fluid with atmospheric pressure 101 kPa on the top surface if the fluid is a) water at 20°C b) glycerine 25°C or c) light oil Solution: ρH2O = 997 kg/m3; ρGlyc = 1260 kg/m3; ∆P = ρgh Table A.4: P = Ptop + ∆P a) ∆P = ρgh = 997× 9.807× 5 = 48887.9 Pa P = 101 + 48.99 = 149.9 kPa b) ∆P = ρgh = 1260× 9.807× 5 = 61784 Pa P = 101 + 61.8 = 162.8 kPa c) ∆P = ρgh = 910× 9.807× 5 = 44622 Pa P = 101 + 44.6 = 145.6 kPa ρOil = 910 kg/m3 Sonntag, Borgnakke and van Wylen 2.52 The hydraulic lift in an auto-repair shop has a cylinder diameter of 0.2 m. To what pressure should the hydraulic fluid be pumped to lift 40 kg of piston/arms and 700 kg of a car? Solution: Force acting on the mass by the gravitational field F↓ = ma = mg = 740 × 9.80665 = 7256.9 N Force balance: F↑ = ( P - P0 ) A = F↓ => P = P0 + F↓ / A A = π D2 (1 / 4) = 0.031416 m2 P = 101 + 7256.9 / (0.031416 × 1000) = 332 kPa Sonntag, Borgnakke and van Wylen 2.53 A 2.5 m tall steel cylinder has a cross sectional area of 1.5 m2. At the bottom with a height of 0.5 m is liquid water on top of which is a 1 m high layer of gasoline. The gasoline surface is exposed to atmospheric air at 101 kPa. What is the highest pressure in the water? Solution: The pressure in the fluid goes up with the depth as P = Ptop + ∆P = Ptop + ρgh and since we have two fluid layers we get P = Ptop + [(ρh)gasoline + (ρh)water]g The densities from Table A.4 are: ρgasoline = 750 kg/m3; Air 1m 0.5 m ρwater = 997 kg/m3 9.807 P = 101 + [750 × 1 + 997 × 0.5] 1000 = 113.2 kPa Gasoline Water Sonntag, Borgnakke and van Wylen 2.54 At the beach, atmospheric pressure is 1025 mbar. You dive 15 m down in the ocean and you later climb a hill up to 250 m elevation. Assume the density of water is about 1000 kg/m3 and the density of air is 1.18 kg/m3. What pressure do you feel at each place? Solution: ∆P = ρgh Pocean= P0 + ∆P = 1025 × 100 + 1000 × 9.81 × 15 = 2.4965 × 105 Pa = 250 kPa Phill = P0 - ∆P = 1025 × 100 - 1.18 × 9.81 × 250 = 0.99606 × 105 Pa = 99.61 kPa Sonntag, Borgnakke and van Wylen 2.55 A piston, mp= 5 kg, is fitted in a cylinder, A = 15 cm2, that contains a gas. The setup is in a centrifuge that creates an acceleration of 25 m/s2 in the direction of piston motion towards the gas. Assuming standard atmospheric pressure outside the cylinder, find the gas pressure. Solution: Force balance: mpg P = P0 + A = 101.325 + = 184.7 kPa Po F↑ = F↓ = P0A + mpg = PA g 5 × 25 kPa kg m/s2 1000 × 0.0015 Pa m2 gas Sonntag, Borgnakke and van Wylen 2.56 A steel tank of cross sectional area 3 m2 and 16 m tall weighs 10 000 kg and it is open at the top. We want to float it in the ocean so it sticks 10 m straight down by pouring concrete into the bottom of it. How much concrete should I put in? Solution: The force up on the tank is from the water pressure at the bottom times its area. The force down is the gravitation times mass and the atmospheric pressure. F↑ = PA = (ρoceangh + P0)A F↓ = (mtank + mconcrete)g + P0A Air Ocean 10 m Concrete The force balance becomes F↑ = F↓ = (ρoceangh + P0)A = (mtank + mconcrete)g + P0A Solve for the mass of concrete mconcrete = (ρoceanhA - mtank) = 997 × 10 × 3 – 10 000 = 19 910 kg Notice: The first term is the mass of the displaced ocean water. The net force up is the weight (mg) of this mass called bouyancy, P0 cancel. Sonntag, Borgnakke and van Wylen 2.57 Liquid water with density ρ is filled on top of a thin piston in a cylinder with cross-sectional area A and total height H. Air is let in under the piston so it pushes up, spilling the water over the edge. Deduce the formula for the air pressure as a function of the piston elevation from the bottom, h. Solution: Force balance Piston: F↑ = F↓ P0 H h PA = P0A + mH Og 2 P = P0 + mH Og/A 2 P = P0 + (H − h)ρg P P0 h, V air Sonntag, Borgnakke and van Wylen Manometers and Barometers 2.58 The density of atmospheric air is about 1.15 kg/m3, which we assume is constant. How large an absolute pressure will a pilot see when flying 1500 m above ground level where the pressure is 101 kPa. Solution: Assume g and ρ are constant then the pressure difference to carry a column of height 1500 m is from Fig.2.10 ∆P = ρgh = 1.15 kg/m3 × 9.807 ms-2 × 1500 m = 16 917 Pa = 16.9 kPa The pressure on top of the column of air is then P = P0 – ∆P = 101 – 16.9 = 84.1 kPa Sonntag, Borgnakke and van Wylen 2.59 A differential pressure gauge mounted on a vessel shows 1.25 MPa and a local barometer gives atmospheric pressure as 0.96 bar. Find the absolute pressure inside the vessel. Solution: Convert all pressures to units of kPa. Pgauge = 1.25 MPa = 1250 kPa; P0 = 0.96 bar = 96 kPa P = Pgauge + P0 = 1250 + 96 = 1346 kPa Sonntag, Borgnakke and van Wylen 2.60 Two vertical cylindrical storage tanks are full of liquid water, density 1000 kg/m3, the top open to the atmoshere. One is 10 m tall, 2 m diameter, the other is 2.5 m tall with diameter 4 m. What is the total force from the bottom of each tank to the water and what is the pressure at the bottom of each tank? Solution: VA = H × πD2 × (1 / 4) = 10 × π × 22 × ( 1 / 4) = 31.416 m3 VB = H × πD2 × (1 / 4) = 2.5 × π × 42 × ( 1 / 4) = 31.416 m3 Tanks have the same volume, so same mass of water gives gravitational force F = mg = ρ V g = 1000 × 31.416 × 9.80665 = 308 086 N this is the force the legs have to supply (assuming Po below the bottom). Tanks have total force up from bottom as Ftot A = F + PoA = 308 086 + 101325 × 3.1416 = 626 408 N Ftot B = F + PoA = 308 086 + 101325 × 12.5664 = 1 581 374 N Pbot = Po + ρ H g Pbot A = 101 + (1000 × 10 × 9.80665 / 1000) = 199 kPa Pbot B = 101 + (1000 × 2.5 × 9.80665 / 1000) = 125.5 kPa A Po g m B cb m Po Sonntag, Borgnakke and van Wylen 2.61 Blue manometer fluid of density 925 kg/m3 shows a column height difference of 6 cm vacuum with one end attached to a pipe and the other open to P0 = 101 kPa. What is the absolute pressure in the pipe? Solution: Since the manometer shows a vacuum we have Pipe Po PPIPE = P0 - ∆P ∆P = ρgh = 925 × 9.807 × 0.06 = 544.3 Pa = 0.544 kPa PPIPE = 101 – 0.544 = 100.46 kPa cb Sonntag, Borgnakke and van Wylen 2.62 The absolute pressure in a tank is 85 kPa and the local ambient absolute pressure is 97 kPa. If a U-tube with mercury, density 13550 kg/m3, is attached to the tank to measure the vacuum, what column height difference would it show? Solution: ∆P = P0 - Ptank = ρg H H = ( P0 - Ptank ) / ρg = [(97 - 85 ) × 1000 ] / (13550 × 9.80665) = 0.090 m = 90 mm H Sonntag, Borgnakke and van Wylen 2.63 The pressure gauge on an air tank shows 75 kPa when the diver is 10 m down in the ocean. At what depth will the gauge pressure be zero? What does that mean? Ocean H20 pressure at 10 m depth is P H20 = Po + ρLg = 101.3 + 997 × 10 × 9.80665 = 199 kPa 1000 Air Pressure (absolute) in tank Ptank = 199 + 75 = 274 kPa Tank Pressure (gauge) reads zero at H20 local pressure 274 = 101.3 + 997 × 9.80665 L 1000 L = 17.66 m At this depth you will have to suck the air in, it can no longer push itself through a valve. Sonntag, Borgnakke and van Wylen 2.64 A submarine maintains 101 kPa inside it and it dives 240 m down in the ocean having an average density of 1030 kg/m3. What is the pressure difference between the inside and the outside of the submarine hull? Solution: Assume the atmosphere over the ocean is at 101 kPa, then ∆P is from the 240 m column water. ∆P = ρLg = (1030 kg/m3 × 240 m × 9.807 m/s2) / 1000 = 2424 kPa Sonntag, Borgnakke and van Wylen 2.65 A barometer to measure absolute pressure shows a mercury column height of 725 mm. The temperature is such that the density of the mercury is 13 550 kg/m3. Find the ambient pressure. Solution: Hg : L = 725 mm = 0.725 m; ρ = 13 550 kg/m3 The external pressure P balances the column of height L so from Fig.2.10 P = ρ L g = 13 550 kg/m3 × 9.80665 m/s2 × 0.725 m × 10-3 kPa/Pa = 96.34 kPa Sonntag, Borgnakke and van Wylen 2.66 An absolute pressure gauge attached to a steel cylinder shows 135 kPa. We want to attach a manometer using liquid water a day that Patm = 101 kPa. How high a fluid level difference must we plan for? Solution: Since the manometer shows a pressure difference we have ∆P = PCYL - Patm = ρ L g L = ∆P / ρg = (135 – 101) kPa 1000 Pa -3 × 10 × 9.807 m/s2 kPa 997 kg m = 3.467 m H Sonntag, Borgnakke and van Wylen 2.67 The difference in height between the columns of a manometer is 200 mm with a fluid of density 900 kg/m3. What is the pressure difference? What is the height difference if the same pressure difference is measured using mercury, density 13600 kg/ m3, as manometer fluid? Solution: ∆P = ρ1gh1 = 900 kg/m3 × 9.807 m/s2 × 0.2 m = 1765.26 Pa = 1.77 kPa 900 hHg = ∆P/ (ρhg g) = (ρ1 gh1) / (ρhg g) = 13600 × 0.2 = 0.0132 m= 13.2 mm Sonntag, Borgnakke and van Wylen 2.68 An exploration submarine should be able to go 4000 m down in the ocean. If the ocean density is 1020 kg/m3 what is the maximum pressure on the submarine hull? Solution: ∆P = ρLg = (1020 kg/m3 × 4000 m × 9.807 m/s2) / 1000 = 40 012 kPa ≈ 40 MPa Sonntag, Borgnakke and van Wylen 2.69 Assume we use a pressure gauge to measure the air pressure at street level and at the roof of a tall building. If the pressure difference can be determined with an accuracy of 1 mbar (0.001 bar) what uncertainty in the height estimate does that corresponds to? Solution: ρair = 1.169 kg/m3 from Table A.5 ∆P = 0.001 bar = 100 Pa L= ∆P 100 = = 8.72 m ρg 1.169 × 9.807 Sonntag, Borgnakke and van Wylen 2.70 A U-tube manometer filled with water, density 1000 kg/m3, shows a height difference of 25 cm. What is the gauge pressure? If the right branch is tilted to make an angle of 30° with the horizontal, as shown in Fig. P2.70, what should the length of the column in the tilted tube be relative to the U-tube? Solution: Same height in the two sides in the direction of g. ∆P = F/A = mg/A = Vρg/A = hρg = 0.25 × 1000 × 9.807 = 2452.5 Pa = 2.45 kPa H h 30o h = H × sin 30° ⇒ H = h/sin 30° = 2h = 50 cm Sonntag, Borgnakke and van Wylen 2.71 A barometer measures 760 mmHg at street level and 735 mmHg on top of a building. How tall is the building if we assume air density of 1.15 kg/m3? Solution: ∆P = ρgH H = ∆P/ρg = 760 – 735 mmHg 133.32 Pa = 295 m 1.15 × 9.807 kg/m2s2 mmHg Sonntag, Borgnakke and van Wylen 2.72 A piece of experimental apparatus is located where g = 9.5 m/s2 and the temperature is 5°C. An air flow inside the apparatus is determined by measuring the pressure drop across an orifice with a mercury manometer (see Problem 2.77 for density) showing a height difference of 200 mm. What is the pressure drop in kPa? Solution: ∆P = ρgh ; ρHg = 13600 kg/m3 ∆P = 13 600 kg/m3 × 9.5 m/s2 × 0.2 m = 25840 Pa = 25.84 kPa Air g Sonntag, Borgnakke and van Wylen 2.73 Two piston/cylinder arrangements, A and B, have their gas chambers connected by a pipe. Cross-sectional areas are AA = 75 cm2 and AB = 25 cm2 with the piston mass in A being mA = 25 kg. Outside pressure is 100 kPa and standard gravitation. Find the mass mB so that none of the pistons have to rest on the bottom. Solution: Po P o cb Force balance for both pistons: A: mPAg + P0AA = PAA B: F↑ = F↓ mPBg + P0AB = PAB Same P in A and B gives no flow between them. mPAg mPBg + P0 = A + P0 A A B => mPB = mPA AA/ AB = 25 × 25/75 = 8.33 kg Sonntag, Borgnakke and van Wylen 2.74 Two hydraulic piston/cylinders are of same size and setup as in Problem 2.73, but with negligible piston masses. A single point force of 250 N presses down on piston A. Find the needed extra force on piston B so that none of the pistons have to move. Solution: AA = 75 cm2 ; FA Po AB = 25 cm2 FB P o No motion in connecting pipe: PA = PB A B cb Forces on pistons balance PA = P0 + FA / AA = PB = P0 + FB / AB AB 25 FB = FA × A = 250 × 75 = 83.33 N A Sonntag, Borgnakke and van Wylen 2.75 A pipe flowing light oil has a manometer attached as shown in Fig. P2.75. What is the absolute pressure in the pipe flow? Solution: Table A.3: ρoil = 910 kg/m3; ρwater = 997 kg/m3 PBOT = P0 + ρwater g Htot = P0 + 997 × 9.807 × 0.8 = Po + 7822 Pa PPIPE = PBOT – ρwater g H1 – ρoil g H2 = PBOT – 997 × 9.807 × 0.1 – 910 × 9.807 × 0.2 = PBOT – 977.7 Pa – 1784.9 Pa PPIPE = Po + (7822 – 977.7 – 1784.9) Pa = Po + 5059.4 Pa = 101.325 + 5.06 = 106.4 kPa Sonntag, Borgnakke and van Wylen 2.76 Two cylinders are filled with liquid water, ρ = 1000 kg/m3, and connected by a line with a closed valve. A has 100 kg and B has 500 kg of water, their crosssectional areas are AA = 0.1 m2 and AB = 0.25 m2 and the height h is 1 m. Find the pressure on each side of the valve. The valve is opened and water flows to an equilibrium. Find the final pressure at the valve location. Solution: VA = vH OmA = mA/ρ = 0.1 = AAhA 2 => hA = 1 m VB = vH OmB = mB/ρ = 0.5 = ABhB 2 => hB = 2 m PVB = P0 + ρg(hB+H) = 101325 + 1000 × 9.81 × 3 = 130 755 Pa PVA = P0 + ρghA = 101325 + 1000 × 9.81 × 1 = 111 135 Pa Equilibrium: same height over valve in both Vtot = VA + VB = h2AA + (h2 - H)AB ⇒ h2 = hAAA + (hB+H)AB AA + AB = 2.43 m PV2 = P0 + ρgh2 = 101.325 + (1000 × 9.81 × 2.43)/1000 = 125.2 kPa Sonntag, Borgnakke and van Wylen Temperature 2.77 The density of mercury changes approximately linearly with temperature as ρHg = 13595 − 2.5 T kg/ m3 T in Celsius so the same pressure difference will result in a manometer reading that is influenced by temperature. If a pressure difference of 100 kPa is measured in the summer at 35°C and in the winter at −15°C, what is the difference in column height between the two measurements? Solution: The manometer reading h relates to the pressure difference as ∆P ∆P = ρ L g ⇒ L = ρg The manometer fluid density from the given formula gives ρsu = 13595 − 2.5 × 35 = 13507.5 kg/m3 ρw = 13595 − 2.5 × (−15) = 13632.5 kg/m3 The two different heights that we will measure become kPa (Pa/kPa) 100 × 103 = 0.7549 m Lsu = 13507.5 × 9.807 (kg/m3) m/s2 100 × 103 kPa (Pa/kPa) Lw = = 0.7480 m 13632.5 × 9.807 (kg/m3) m/s2 ∆L = Lsu - Lw = 0.0069 m = 6.9 mm Sonntag, Borgnakke and van Wylen 2.78 A mercury thermometer measures temperature by measuring the volume expansion of a fixed mass of liquid Hg due to a change in the density, see problem 2.35. Find the relative change (%) in volume for a change in temperature from 10°C to 20°C. Solution: From 10°C to 20°C At 10°C : ρHg = 13595 – 2.5 × 10 = 13570 kg/m3 At 20°C : ρHg = 13595 – 2.5 × 20 = 13545 kg/m3 The volume from the mass and density is: Relative Change = = V = m/ρ V20– V10 (m/ρ20) - (m/ρ10) = V10 m/ρ10 ρ10 13570 – 1 = 13545 – 1 = 0.0018 (0.18%) ρ20 Sonntag, Borgnakke and van Wylen 2.79 Using the freezing and boiling point temperatures for water in both Celsius and Fahrenheit scales, develop a conversion formula between the scales. Find the conversion formula between Kelvin and Rankine temperature scales. Solution: TFreezing = 0 oC = 32 F; TBoiling = 100 oC = 212 F ∆T = 100 oC = 180 F ⇒ ToC = (TF - 32)/1.8 or TF = 1.8 ToC + 32 For the absolute K & R scales both are zero at absolute zero. TR = 1.8 × TK Sonntag, Borgnakke and van Wylen 2.80 The atmosphere becomes colder at higher elevation. As an average the standard atmospheric absolute temperature can be expressed as Tatm = 288 - 6.5 × 10−3 z, where z is the elevation in meters. How cold is it outside an airplane cruising at 12 000 m expressed in Kelvin and in Celsius? Solution: For an elevation of z = 12 000 m we get Tatm = 288 - 6.5 × 10−3 z = 210 K To express that in degrees Celsius we get T = T – 273.15 = −63.15oC C Sonntag, Borgnakke and van Wylen Review Problems 2.81 Repeat problem 2.72 if the flow inside the apparatus is liquid water, ρ ≅ 1000 kg/m3, instead of air. Find the pressure difference between the two holes flush with the bottom of the channel. You cannot neglect the two unequal water columns. Solution: P1 . H h1 Balance forces in the manometer: P2 · (H - h2) - (H - h1) = ∆hHg = h1 - h2 h2 P1A + ρH Oh1gA + ρHg(H - h1)gA 2 = P2A + ρH Oh2gA + ρHg(H - h2)gA 2 ⇒ P1 - P2 = ρH O(h2 - h1)g + ρHg(h1 - h2)g 2 P1 - P2 = ρHg∆hHgg - ρH O∆hHgg = 13600 × 0.2 × 9.5 - 1000 × 0.2 × 9.5 2 = 25840 - 1900 = 23940 Pa = 23.94 kPa Sonntag, Borgnakke and van Wylen 2.82 The main waterline into a tall building has a pressure of 600 kPa at 5 m elevation below ground level. How much extra pressure does a pump need to add to ensure a water line pressure of 200 kPa at the top floor 150 m above ground? Solution: The pump exit pressure must balance the top pressure plus the column ∆P. The pump inlet pressure provides part of the absolute pressure. Pafter pump = Ptop + ∆P ∆P = ρgh = 997 kg/m3 × 9.807 m/s2 × (150 + 5) m = 1 515 525 Pa = 1516 kPa Pafter pump = 200 + 1516 = 1716 kPa ∆Ppump = 1716 – 600 = 1116 kPa Sonntag, Borgnakke and van Wylen 2.83 A 5-kg piston in a cylinder with diameter of 100 mm is loaded with a linear spring and the outside atmospheric pressure of 100 kPa. The spring exerts no force on the piston when it is at the bottom of the cylinder and for the state shown, the pressure is 400 kPa with volume 0.4 L. The valve is opened to let some air in, causing the piston to rise 2 cm. Find the new pressure. Solution: A linear spring has a force linear proportional to displacement. F = k x, so the equilibrium pressure then varies linearly with volume: P = a + bV, with an intersect a and a slope b = dP/dV. Look at the balancing pressure at zero volume (V -> 0) when there is no spring force F = PA = PoA + mpg and the initial state. These two points determine the straight line shown in the P-V diagram. Piston area = AP = (π/4) × 0.12 = 0.00785 m2 mpg 5 × 9.80665 a = P0 + A = 100 kPa + 0.00785 Pa p = 106.2 kPa intersect for zero volume. P P2 2 400 1 V2 = 0.4 + 0.00785 × 20 = 0.557 L 106.2 dP P2 = P1 + dV ∆V V 0 0.4 0.557 (400-106.2) 0.4 - 0 (0.557 - 0.4) = 515.3 kPa = 400 + Sonntag, Borgnakke and van Wylen 2.84 In the city water tower, water is pumped up to a level 25 m above ground in a pressurized tank with air at 125 kPa over the water surface. This is illustrated in Fig. P2.84. Assuming the water density is 1000 kg/m3 and standard gravity, find the pressure required to pump more water in at ground level. Solution: ∆P = ρ L g = 1000 kg/m3 × 25 m × 9.807 m/s2 = 245 175 Pa = 245.2 kPa Pbottom = Ptop + ∆P = 125 + 245.2 = 370 kPa cb Sonntag, Borgnakke and van Wylen 2.85 Two cylinders are connected by a piston as shown in Fig. P2.85. Cylinder A is used as a hydraulic lift and pumped up to 500 kPa. The piston mass is 25 kg and there is standard gravity. What is the gas pressure in cylinder B? Solution: Force balance for the piston: PBAB + mpg + P0(AA - AB) = PAAA AA = (π/4)0.12 = 0.00785 m2; AB = (π/4)0.0252 = 0.000 491 m2 PBAB = PAAA - mpg - P0(AA - AB) = 500× 0.00785 - (25 × 9.807/1000) - 100 (0.00785 - 0.000 491) = 2.944 kN PB = 2.944/0.000 491 = 5996 kPa = 6.0 MPa B GAS P o cb A Oil P Sonntag, Borgnakke and van Wylen 2.86 A dam retains a lake 6 m deep. To construct a gate in the dam we need to know the net horizontal force on a 5 m wide and 6 m tall port section that then replaces a 5 m section of the dam. Find the net horizontal force from the water on one side and air on the other side of the port. Solution: Pbot = P0 + ∆P ∆P = ρgh = 997× 9.807× 6 = 58 665 Pa = 58.66 kPa Neglect ∆P in air Fnet = Fright – Fleft = Pavg A - P0A Pavg = P0 + 0.5 ∆P Since a linear pressure variation with depth. Fnet = (P0 + 0.5 ∆P)A - P0A = 0.5 ∆P A = 0.5 × 58.66 × 5 × 6 = 880 kN Fle ft Frig h t SOLUTION MANUAL ENGLISH UNIT PROBLEMS CHAPTER 3 SONNTAG • BORGNAKKE • VAN WYLEN FUNDAMENTALS of Thermodynamics Sixth Edition Sonntag, Borgnakke and van Wylen CONTENT CHAPTER 3 SUBSECTION PROB NO. Correspondence table Study guide problems Phase diagrams, triple and critical points General tables Ideal gas Compressibility factor Review problems Linear interpolation Computer tables 1-20 21-28 29-63 64-79 79-89 90-115 116-121 122-127 English unit problems 128-158 Sonntag, Borgnakke and van Wylen Correspondence Table CHAPTER 3 6th edition Sonntag/Borgnakke/Wylen The set of problems have a correspondence to the 5th edition Fundamentals of Thermodynamics as: Problems 3.1 through 3.20 are all new New 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 5th new 2 1 3 new 4 28 mod new 23 28 mod 24 new new new 29 new new 27 mod new 37 41 new new new new 36 new 58 35 42 new 43 new 40 44 New 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 5th 46 48 39 mod 57 51 new new 5 new 22 6 new 8 new 10 13 new 25 new new new 17 14 19 33 new new new new 20 new 21 18 26 mod 16 mod 30 mod New 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 5th 30 mod 31 mod 32 new 60 55 new 59 53 54 50 49 45 56 9 52 7 47 11 12 16 38 34 new new new new new new new new new new 86 87 Sonntag, Borgnakke and van Wylen The English unit problem correspondence is New 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 5th Ed. new new new new new new 61E 68E a-c 68E d-f new 70E 73E 74E new 76E SI 5 7 9 11 17 23 27 30 30 40 36 47 41 44 51 New 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 5th Ed. 77E new 79E 62E new 69E c+d 70E d 72E 64E new 81E new 71E 80E 83E 65E 66E SI 53 62 58 69 65 81 113 74 49 99 95 61 106 89 - The Computer, design and open-ended problem correspondence is New 159 160 161 162 5th new new 88 89 New 163 164 165 166 5th 90 91 92 93 New 167 168 5th 94 95 mod indicates a modification from the previous problem that changes the solution but otherwise is the same type problem. Sonntag, Borgnakke and van Wylen Concept-Study Guide Problems Sonntag, Borgnakke and van Wylen 3.1 What is the lowest temperature (approximately) at which water can be liquid? ln P Look at the phase diagram in Fig. 3.7. At the border between ice I, ice III and the liquid region is a triple point which is the lowest T where you can have liquid. From the figure it is estimated to be about 255 K i.e. at -18oC. lowest T liquid L S CR.P. V T ≈ 255 K ≈ - 18°C T 3.2 What is the percent change in volume as liquid water freezes? Mention some effects in nature and for our households the volume change can have. The density of water in the different phases can be found in Tables A.3 and A.4 and in Table B.1. From Table B.1.1 vf = 0.00100 m3/kg From Table B.1.5 Percent change: 100 vi = 0.0010908 m3/kg vi – vf 0.0010908 – 0.001 = 100 × = 9.1 % increase 0.001 vf Liquid water that seeps into cracks or other confined spaces and then freezes will expand and widen the cracks. This is what destroys any pourous material exposed to the weather on buildings, roads and mountains. 3.3 When you skate on ice a thin liquid film forms under the skate; how can that be? The ice is at some temperature below the freezing temperature for the atmospheric pressure of 100 kPa = 0.1 MPa and thus to the left of the fusion line in the solid ice I region of Fig. 3.7. As the skate comes over the ice the pressure is increased dramatically right under the blade so it brings the state straight up in the diagram crossing the fusion line and brings it into a liquid state at same temperature. The very thin liquid film under the skate changes the friction to be viscous rather than a solid to solid contact friction. Friction is thus significantly reduced. Sonntag, Borgnakke and van Wylen 3.4 An external water tap has the valve activated by a long spindle so the closing mechanism is located well inside the wall. Why is that? Solution: By having the spindle inside the wall the coldest location with water when the valve is closed is kept at a temperature above the freezing point. If the valve spindle was outside there would be some amount of water that could freeze while it is trapped inside the pipe section potentially rupturing the pipe. 3.5 Some tools should be cleaned in water at a least 150oC. How high a P is needed? Solution: If I need liquid water at 150oC I must have a pressure that is at least the saturation pressure for this temperature. Table B.1.1: 150oC Psat = 475.9 kPa. 3.6 Are the pressures in the tables absolute or gauge pressures? Solution: The behavior of a pure substance depends on the absolute pressure, so P in the tables is absolute. 3.7 If I have 1 L ammonia at room pressure and temperature (100 kPa, 20oC) how much mass is that? Ammonia Tables B.2: B.2.1 Psat = 857.5 kPa at 20oC so superheated vapor. B.2.2 v = 1.4153 m3/kg under subheading 100 kPa 3 V 0.001 m m= v = = 0.000 706 kg = 0.706 g 1.4153 m3/kg Sonntag, Borgnakke and van Wylen 3.8 How much is the change in liquid specific volume for water at 20oC as you move up from state i towards state j in figure 3.12 reaching 15 000 kPa? State “i”, here “a”, is saturated liquid and up is then compressed liquid states a Table B.1.1: vf = 0.001 002 m3/kg at 2.34 kPa Table B.1.4: vf = 0.001 002 m3/kg at c Table B.1.4: d Table B.1.4: e Table B.1.4: f Table B.1.4: vf = 0.001 001 m3/kg at 2000 kPa vf = 0.001 000 m3/kg at 5000 kPa vf = 0.000 995 m3/kg at 15 000 kPa vf = 0.000 980 m3/kg at 50 000 kPa b 500 kPa Notice how small the changes in v are for very large changes in P. P T f e d c b a f-a o T = 20 C v v P L f T S C.P. V a v Sonntag, Borgnakke and van Wylen 3.9 For water at 100 kPa with a quality of 10% find the volume fraction of vapor. This is a two-phase state at a given pressure: Table B.1.2: vf = 0.001 043 m3/kg, vg = 1.6940 m3/kg From the definition of quality we get the masses from total mass, m, as mf = (1 – x) m, mg = x m The volumes are Vf = mf vf = (1 – x) m vf, Vg = mg vg = x m vg So the volume fraction of vapor is Vg Vg x m vg Fraction = V = V + V = x m v + (1 – x)m v g f g f = 0.1694 0.1 × 1.694 = 0.17034 = 0.9945 0.1 × 1.694 + 0.9 × 0.001043 Notice that the liquid volume is only about 0.5% of the total. We could also have found the overall v = vf + xvfg and then V = m v. Sonntag, Borgnakke and van Wylen 3.10 Sketch two constant-pressure curves (500 kPa and 30 000 kPa) in a T-v diagram and indicate on the curves where in the water tables you see the properties. MPa P 30 0.5 B 1 4 B.1.3 T C.P. B 1 4 B.1.3 30 MPa B.1.3 500 kPa B.1.1 B.1.3 B.1.2 B.1.5 v B.1.5 v The 30 MPa line in Table B.1.4 starts at 0oC and table ends at 380oC, the line is continued in Table B.1.3 starting at 375oC and table ends at 1300oC. The 500 kPa line in Table B.1.4 starts at 0.01oC and table ends at the saturated liquid state (151.86oC). The line is continued in Table B.1.3 starting at the saturated vapor state (151.86oC) continuing up to 1300oC. Sonntag, Borgnakke and van Wylen 3.11 Locate the state of ammonia at 200 kPa, -10oC. Indicate in both the P-v and the T-v diagrams the location of the nearest states listed in the printed table B.2 P T C.P. C.P. 200 kPa 290.9 200 0 -10 -18.9 -10 C -18.9 C T v 150 kPa v 3.12 Why are most of the compressed liquid or solid regions not included in the printed tables? For the compressed liquid and the solid phases the specific volume and thus density is nearly constant. These surfaces are very steep nearly constant v and there is then no reason to fill up a table with the same value of v for different P and T. Sonntag, Borgnakke and van Wylen 3.13 Water at 120oC with a quality of 25% has its temperature raised 20oC in a constant volume process. What is the new quality and pressure? Solution: State 1 from Table B.1.1 at 120oC v = vf + x vfg = 0.001060 + 0.25 × 0.8908 = 0.22376 m3/kg State 2 has same v at 140oC also from Table B.1.1 v - vf 0.22376 - 0.00108 x= v = = 0.4385 0.50777 fg P = Psat = 361.3 kPa P C.P. 361.3 198.5 140 C 120 C T C.P. 140 120 T v v Sonntag, Borgnakke and van Wylen 3.14 Water at 200 kPa with a quality of 25% has its temperature raised 20oC in a constant pressure process. What is the new quality and volume? Solution: State 1 from Table B.1.2 at 200 kPa v = vf + x vfg = 0.001061 + 0.25 × 0.88467 = 0.22223 m3/kg State 2 has same P from Table B.1.2 at 200 kPa T = T + 20 = 120.23 + 20 = 140.23oC 2 sat so state is superheated vapor x = undefined 20 v = 0.88573 + (0.95964 – 0.88573)150 - 120.23 = 0.9354 m3/kg P C.P. T C.P. 200 kPa 140 C 200 120.2 C 140 120 T v v 3.15 Why is it not typical to find tables for Ar, He, Ne or air like an Appendix B table? The temperature at which these substances are close to the two-phase region is very low. For technical applications with temperatures around atmospheric or higher they are ideal gases. Look in Table A.2 and we can see the critical temperatures as Ar : 150.8 K He: 5.19 K Ne: 44.4 K It requires a special refrigerator in a laboratory to bring a substance down to these cryogenic temperatures. Sonntag, Borgnakke and van Wylen 3.16 What is the relative (%) change in P if we double the absolute temperature of an ideal gas keeping mass and volume constant? Repeat if we double V having m, T constant. Ideal gas law: PV = mRT State 2: P2V = mRT2 = mR2T1 = 2P1V ⇒ P2 = 2P1 Relative change = ∆P/P1 = P1/P1 = 1 = 100% State 3: P3V3 = mRT1 = P1V1 ⇒ P3 = P1V1/V3 = P1/2 Relative change = ∆P/P1 = -P1/2P1 = -0.5 = -50% P T 2 2 1 3 T2 T1 1 V 3 V Sonntag, Borgnakke and van Wylen 3.17 Calculate the ideal gas constant for argon and hydrogen based on table A.2 and verify the value with Table A.5 The gas constant for a substance can be found from the universal gas constant from the front inside cover and the molecular weight from Table A.2 _ R 8.3145 Argon: R = M = 39.948 = 0.2081 kJ/kg K _ R 8.3145 Hydrogen: R = M = 2.016 = 4.1243 kJ/kg K Sonntag, Borgnakke and van Wylen 3.18 How close to ideal gas behavior (find Z) is ammonia at saturated vapor, 100 kPa? How about saturated vapor at 2000 kPa? v1 = 1.1381 m3/kg, Table B.2.2: T1 = -33.6oC, v2 = 0.06444 m3/kg, T2 = 49.37oC, Table A.5: Pv = ZRT P2 = 2000 kPa R = 0.4882 kJ/kg K Extended gas law: P1 = 100 kPa so we can calculate Z from this P1v1 100 × 1.1381 Z1 = RT = = 0.973 1 0.4882 × (273.15 - 33.6) P2v2 2000 × 0.06444 Z2 = RT = = 0.8185 2 0.4882 × (273.15 + 49.37) So state 1 is close to ideal gas and state 2 is not so close. Z 1 Tr= 2.0 2 Tr = 1.2 Tr = 0.7 Tr = 0.7 0.1 1 ln Pr Sonntag, Borgnakke and van Wylen 3.19 Find the volume of 2 kg of ethylene at 270 K, 2500 kPa using Z from Fig. D.1 Ethylene Table A.2: Table A.5: Tc = 282.4 K, Pc = 5.04 MPa R = 0.2964 kJ/kg K The reduced temperature and pressure are: T 270 Tr = T = 282.4 = 0.956, c P 2.5 Pr = P = 5.04 = 0.496 c Enter the chart with these coordinates and read: V= Z = 0.76 mZRT 2 × 0.76 × 0.2964 × 270 = 0.0487 m3 P= 2500 Z Tr= 2.0 Tr = 1.2 Tr = 0.96 Tr = 0.7 Tr = 0.7 0.1 0.5 1 ln Pr Sonntag, Borgnakke and van Wylen 3.20 With Tr = 0.85 and a quality of 0.6 find the compressibility factor using Fig. D.1 For the saturated states we will use Table D.4 instead of the figure. There we can see at Tr = 0.85 Zf = 0.062, Zg = 0.747 Z = (1 – x) Zf + xZg = (1 – 0.6) 0.062 + 0.6 × 0.747 = 0.473 Sonntag, Borgnakke and van Wylen Phase Diagrams, Triple and Critical Points 3.21 Modern extraction techniques can be based on dissolving material in supercritical fluids such as carbon dioxide. How high are pressure and density of carbon dioxide when the pressure and temperature are around the critical point. Repeat for ethyl alcohol. Solution: CO2 : Table A.2: Pc = 7.38 MPa, Tc = 304 K, vc = 0.00212 m3/kg ρc = 1/vc = 1/0.00212 = 472 kg/m3 C2H5OH: Table A.2: Pc = 6.14 MPa, Tc = 514 K, vc = 0.00363 m3/kg ρc = 1/vc = 1/0.00363 = 275 kg/m3 Sonntag, Borgnakke and van Wylen 3.22 Find the lowest temperature at which it is possible to have water in the liquid phase. At what pressure must the liquid exist? Solution: ln P There is no liquid at lower temperatures than on the fusion line, see Fig. 3.6, saturated ice III to liquid phase boundary is at T ≈ 263K ≈ - 10°C and P ≈ 2100 MPa lowest T liquid L S CR.P. V T Sonntag, Borgnakke and van Wylen 3.23 Water at 27°C can exist in different phases dependent upon the pressure. Give the approximate pressure range in kPa for water being in each one of the three phases vapor, liquid or solid. Solution: ln P The phases can be seen in Fig. 3.6, a sketch of which is shown to the right. T = 27 °C = 300 Κ From Fig. 3.6: P ≈ 4 × 10−3 MPa = 4 kPa, S L S CR.P. V VL PLS = 103 MPa 0<P< 4 kPa 0.004 MPa < P < 1000 MPa P > 1000 MPa T VAPOR LIQUID SOLID(ICE) Sonntag, Borgnakke and van Wylen 3.24 What is the lowest temperature in Kelvins for which you can see metal as a liquid if the metal is a. silver b. copper Solution: Assume the two substances have a phase diagram similar to Fig. 3.6, then we can see the triple point data in Table 3.2 Ta = 961oC = 1234 K Tb = 1083oC = 1356 K Sonntag, Borgnakke and van Wylen 3.25 If density of ice is 920 kg/m3, find the pressure at the bottom of a 1000 m thick ice cap on the north pole. What is the melting temperature at that pressure? Solution: ρICE = 920 kg/m3 ∆P = ρgH = 920 kg/m3 × 9.80665 m/s2 × 1000 = 9022 118 Pa P = Po + ∆P = 101.325 + 9022 = 9123 kPa See figure 3.6 liquid solid interphase => TLS = −1°C Sonntag, Borgnakke and van Wylen 3.26 Dry ice is the name of solid carbon dioxide. How cold must it be at atmospheric (100 kPa) pressure? If it is heated at 100 kPa what eventually happens? Solution: The phase boundaries are shown in Figure 3.6 At 100 kPa the carbon dioxide is solid if T < 190 K It goes directly to a vapor state without becoming a liquid hence its name. ln P The 100 kPa is below the triple point. S 100 kPa L V T Sonntag, Borgnakke and van Wylen 3.27 A substance is at 2 MPa, 17°C in a rigid tank. Using only the critical properties can the phase of the mass be determined if the substance is nitrogen, water or propane ? Solution: Find state relative to critical point properties which are from Table A.2: a) Nitrogen N2 : 3.39 MPa 126.2 K b) Water H2O : 22.12 MPa c) Propane C3H8 : 4.25 MPa 647.3 K 369.8 K State is at 17 °C = 290 K and 2 MPa < Pc for all cases: N2 : T >> Tc Superheated vapor P < Pc H2O : T << Tc ; P << Pc you cannot say. C3H8 : T < Tc ; P < Pc you cannot say ln P Liquid b c Cr.P. a Vapor T Sonntag, Borgnakke and van Wylen 3.28 Give the phase for the following states. Solution: a. CO2 T = 267°C b. Air superheated vapor assume ideal gas Table A.5 T = 20°C P = 200 kPa Table A.2 c. NH3 superheated vapor assume ideal gas Table A.5 T = 170°C P = 600 kPa Table B.2.2 or A.2 T > Tc => P = 0.5 MPa Table A.2 superheated vapor P C.P. T a, b, c a,b,c P = const. T v v Sonntag, Borgnakke and van Wylen 3.29 Determine the phase of the substance at the given state using Appendix B tables a) Water 100°C, 500 kPa b) Ammonia -10°C, 150 kPa c) R-12 0°C, 350 kPa Solution: a) From Table B.1.1 Psat(100°C) = 101.3 kPa 500 kPa > Psat then it is compressed liquid OR from Table B.1.2 Tsat(500 kPa) = 152°C 100°C < Tsat then it is subcooled liquid = compressed liquid b) Ammonia NH3 : Table B.2.1: P < Psat(-10 °C) = 291 kPa Superheated vapor c) R-12 Table B.3.1: P > Psat(0 °C) = 309 kPa Compressed liquid. ln P The S-L fusion line goes slightly to the left for water. It tilts slightly to the right for most other substances. L a, c S Cr.P. b Vapor T Sonntag, Borgnakke and van Wylen 3.30 Determine whether water at each of the following states is a compressed liquid, a superheated vapor, or a mixture of saturated liquid and vapor. a. P = 10 MPa, v = 0.003 m3/kg b. 1 MPa, 190°C 3/kg c. 200°C, 0.1 m d. 10 kPa, 10°C Solution: For all states start search in table B.1.1 (if T given) or B.1.2 (if P given) a. P = 10 MPa, v = 0.003 m3/kg so look in B.1.2 at 10 MPa vf = 0.001452; vg = 0.01803 m3/kg, => v f < v < vg => so mixture of liquid and vapor. b. 1 MPa, 190°C : Only one of the two look-ups is needed B.1.1: P < Psat = 1254.4 kPa so it is superheated vapor B.1.2: T > Tsat = 179.91°C so it is superheated vapor c. 200°C, 0.1 m3/kg: look in B.1.1 vf = 0.001156 m3/kg ; vg = 0.12736 m3/kg, => so mixture of liquid and vapor. => v f < v < vg d. 10 kPa, 10°C : From B.1.1: From B.1.2: Only one of the two look-ups is needed P > Pg = 1.2276 kPa so compressed liquid T < Tsat = 45.8 °C so compressed liquid P C.P. States shown are placed relative to the two-phase region, not to each other. T C.P. P = const. d b b a c T d v a c v Sonntag, Borgnakke and van Wylen 3.31 Give the phase for the following states. Solution: a. H2O T = 275°C P = 5 MPa Table B.1.1 or B.1.2 B.1.1 Psat = 5.94 MPa B.1.2 Tsat = 264°C b. H2O => superheated vapor => superheated vapor T = −2°C Table B.1.1 P = 100 kPa T < Ttriple point Table B.1.5 at −2°C Psat = 0.518 kPa since P > Psat => compressed solid P C.P. States shown are placed relative to the two-phase region, not to each other. T a v b P L C.P. a S V T b v a P = const. T b Note state b in P-v, see in 3-D figure, is up on the solid face. C.P. v Sonntag, Borgnakke and van Wylen 3.32 Determine whether refrigerant R-22 in each of the following states is a compressed liquid, a superheated vapor, or a mixture of saturated liquid and vapor. Solution: All cases are seen in Table B.4.1 a. 50°C, 0.05 m3/kg From table B.4.1 at 50°C vg = 0.01167 m3/kg since v > vg we have superheated vapor b. 1.0 MPa, 20°C From table B.4.1 at 20°C Pg = 909.9 kPa since P > Pg we have compressed liquid c. 0.1 MPa, 0.1 m3/kg From table B.4.1 at 0.1 MPa (use 101 kPa) vf = 0.0007 and vg = 0.2126 m3/kg as vf < v < vg we have a mixture of liquid & vapor d −20°C, 200 kPa superheated vapor, P < Pg = 244.8 kPa at -20°C P C.P. States shown are placed relative to the two-phase region, not to each other. T C.P. P = const. a b c d T v b c a d v Sonntag, Borgnakke and van Wylen General Tables 3.33 Fill out the following table for substance water: Solution: P [kPa] T [ oC] v [m3/kg] a) 500 20 0.001002 b) 500 151.86 0.20 c) 1400 200 0.14302 d) 8581 300 0.01762 x Undefined 0.532 Undefined 0.8 a) Table B.1.1 P > Psat so it is compressed liquid => Table B.1.4 b) Table B.1.2 vf < v < vg so two phase L + V v - vf x = v = (0.2 – 0.001093) / 0.3738 = 0.532 fg T = Tsat = 151.86oC c) Only one of the two look-up is needed Table B.1.1 200oC P < Psat = => superheated vapor T > Tsat = 195oC Table B.1.2 1400 kPa Table B.1.3 subtable for 1400 kPa gives the state properties d) Table B.1.1 since quality is given it is two-phase v = vf + x × vfg = 0.001404 + 0.8 × 0.02027 = 0.01762 m3/kg 3.34 Place the four states a-d listed in Problem 3.33 as labeled dots in a sketch of the P-v and T-v diagrams. Solution: 8581 1400 500 P C.P. d T 300 200 c T a 152 20 b v C.P. d a P = const. c b v Sonntag, Borgnakke and van Wylen 3.35 Determine the phase and the specific volume for ammonia at these states using the Appendix B table. a. –10oC, 150 kPa b. 20oC, 100 kPa c. 60oC, quality 25% Solution: Ammonia, NH3, properties from Table B.2 a) Table B.2.1: P < Psat(-10 °C) = 291 kPa Superheated vapor B.2.2 v = 0.8336 m3/kg Table B.2.1 at given T: Psat = 847.5 kPa b) Superheated vapor B.2.2 so P < Psat v = 1.4153 m3/kg c) Table B.2.1 enter with T (this is two-phase L + V) v = vf + x vfg = 0.001834 + x × 0.04697 = 0.01358 m3/kg Sonntag, Borgnakke and van Wylen 3.36 Give the phase and the specific volume. Solution: a. R-22 T = −25°C P = 100 kPa Table B.4.1 at given T: Psat = 201 kPa sup. vap. B.4.2 b. R-22 T = −25°C so P < Psat => v ≅ (0.22675 + 0.23706)/2 = 0.2319 m3/kg P = 300 kPa Table B.4.1 at given T: Psat = 201 kPa compr. liq. as P > Psat so 3 v ≅ vf = 0.000733 m /kg c. R-12 T = 5°C P = 200 kPa Table B.3.1 at given T: Psat = 362.6 kPa sup. vap. B.3.2 v ≅ (0.08861 + 0.09255)/2 = 0.09058 m3/kg P C.P. States shown are placed relative to the two-phase region, not to each other. so P < Psat b T C.P. P = const. a, c b T v a, c v Sonntag, Borgnakke and van Wylen 3.37 Fill out the following table for substance ammonia: Solution: P [kPa] T [ oC] v [m3/kg] a) 1200 50 0.1185 b) 2033 50 0.0326 a) b) x Undefined 0.5 B.2.1 v > vg => superheated vapor Look in B.2.2 B.2.1 P = Psat = 2033 kPa v = vf + x vfg = 0.001777 + 0.5 × 0.06159 = 0.0326 m3/kg 3.38 Place the two states a-b listed in Problem 3.37 as labeled dots in a sketch of the Pv and T-v diagrams. Solution: P C.P. 2033 1200 b T C.P. P = const. a T 50 v b a v Sonntag, Borgnakke and van Wylen 3.39 Calculate the following specific volumes a. R-134a: 50°C, 80% quality b. Water c. Nitrogen 4 MPa, 90% quality 120 K, 60% quality Solution: All states are two-phase with quality given. The overall specific volume is given by Eq.3.1 or 3.2 v = vf + x vfg = (1-x)vf + x vg a. R-134a: 50°C, 80% quality in Table B.5.1 v = 0.000908 + x × 0.01422 = 0.01228 m3/kg b. Water 4 MPa, 90% quality in Table B.1.2 v = 0.001252(1-x) + x × 0.04978 = 0.04493 m3/kg c. Nitrogen 120 K, 60% quality in Table B.6.1 v = 0.001915 + x × 0.00608 = 0.005563 m3/kg Sonntag, Borgnakke and van Wylen 3.40 Give the phase and the missing property of P, T, v and x. a. R-134a T = -20oC, P = 150 kPa b. R-134a P = 300 kPa, v = 0.072 m3/kg c. CH4 T = 155 K, v = 0.04 m3/kg d. T = 350 K, v = 0.25 m3/kg CH4 Solution: a) B.5.1 P > Psat = 133.7 kPa ⇒ compressed liquid v ~ vf = 0.000738 m3/kg x = undefined b) B.5.2 v > vg at 300 kPa T = 10 + (20-10) ( ⇒ superheated vapor 0.072 - 0.07111 0.07441 - 0.07111 = 12.7°C ) x = undefined c) B.7.1 v > vg = 0.04892 m3/kg 2-phase v - vf 0.04-0.002877 x= v = = 0.806 0.04605 fg P = Psat = 1295.6 kPa d) B.7.1 T > Tc and v >> vc ⇒ superheated vapor B.7.2 located between 600 & 800 kPa 0.25-0.30067 P = 600 + 200 0.2251-0.30067 = 734 kPa P T c a b C.P. c d T d b P = const. a v v Sonntag, Borgnakke and van Wylen 3.41 A sealed rigid vessel has volume of 1 m3 and contains 2 kg of water at 100°C. The vessel is now heated. If a safety pressure valve is installed, at what pressure should the valve be set to have a maximum temperature of 200°C? Solution: Process: v = V/m = constant State 1: v1 = 1/2 = 0.5 m3/kg T C.P. from Table B.1.1 it is 2-phase State 2: 200°C, 0.5 m3/kg Table B.1.3 between 400 and 500 kPa so interpolate 0.5-0.53422 P ≅ 400 + 0.42492-0.53422 × (500-400) = 431.3 kPa 500 kPa 400 kPa 100 C v Sonntag, Borgnakke and van Wylen 3.42 Saturated liquid water at 60°C is put under pressure to decrease the volume by 1% keeping the temperature constant. To what pressure should it be compressed? Solution: State 1: T = 60°C , x = 0.0; Table B.1.1: v = 0.001017 m3/kg Process: T = constant = 60°C State 2: T, v = 0.99 × vf (60°C) = 0.99×0.001017 = 0.0010068 m3/kg Between 20 & 30 MPa in Table B.1.4, P 2 30 MPa T C.P. 2 1 v P ≅ 23.8 MPa 20 MPa 1 v Sonntag, Borgnakke and van Wylen 3.43 Saturated water vapor at 200 kPa is in a constant pressure piston cylinder. At this state the piston is 0.1 m from the cylinder bottom. How much is this distance if the temperature is changed to a) 200 oC and b) 100 oC. Solution: State 1: (200 kPa, x = 1) in B.1.2: v1 = vg (200 kPa) = 0.8857 m3/kg State a: (200 kPa, 200 oC) B.1.3: va = 1.083 m3/kg State b: (200 kPa, 100 oC) B.1.1: vb = 0.001044 m3/kg As the piston height is proportional to the volume we get ha = h1 (va /v1) = 0.1 × (1.0803 / 0.8857) = 0.12 m hb = h1 (vb / v1) = 0.1 × (0.001044 / 0.8857) = 0.00011 m P C.P. T C.P. P = 200 kPa 200 b 200 1a 120 100 b T v a 1 v Sonntag, Borgnakke and van Wylen 3.44 You want a pot of water to boil at 105oC. How heavy a lid should you put on the 15 cm diameter pot when Patm = 101 kPa? Solution: Table B.1.1 at 105oC : Psat = 120.8 kPa π π A = 4 D2 = 4 0.152 = 0.01767 m2 Fnet = (Psat –Patm) A = (120.8 - 101) kPa × 0.01767 m2 = 0.3498 kN = 350 N Fnet = mlid g 350 mlid = Fnet/g = 9.807 = 35.7 kg Some lids are clamped on, the problem deals with one that stays on due to its weight. Sonntag, Borgnakke and van Wylen 3.45 In your refrigerator the working substance evaporates from liquid to vapor at -20 oC inside a pipe around the cold section. Outside (on the back or below) is a black grille inside which the working substance condenses from vapor to liquid at +40 oC. For each location find the pressure and the change in specific volume (v) if a) the substance is R-12 b) the substance is ammonia Solution: The properties come from the saturated tables where each phase change takes place at constant pressure and constant temperature. Substance TABLE T Psat , kPa ∆v = vfg R-12 B.3.1 961 0.017 R-12 B.3.1 40 oC -20 oC 151 0.108 Ammonia B.2.1 1555 0.0814 Ammonia B.2.1 40 oC -20 oC 190 0.622 P C.P. T 40 40C 4 1 3 T -20 2 v C.P. 4 1 3 2 v Sonntag, Borgnakke and van Wylen 3.46 In your refrigerator the working substance evaporates from liquid to vapor at -20 oC inside a pipe around the cold section. Outside (on the back or below) is a black grille inside which the working substance condenses from vapor to liquid at +40 oC. For each location find the pressure and the change in specific volume (v) if: a) the substance is R-134a b) the substance is R-22 Solution: The properties come from the saturated tables where each phase change takes place at constant pressure and constant temperature. Substance TABLE T Psat , kPa ∆v = vfg R-134a B.5.1 1017 0.019 R-134a B.5.1 40 oC -20 oC 134 0.146 B.4.1 40 oC 1534 0.0143 B.4.1 -20 oC 245 0.092 R-22 R-22 P C.P. T 40 40C 4 1 3 T -20 2 v C.P. 4 1 3 2 v Sonntag, Borgnakke and van Wylen 3.47 A water storage tank contains liquid and vapor in equilibrium at 110°C. The distance from the bottom of the tank to the liquid level is 8 m. What is the absolute pressure at the bottom of the tank? Solution: Saturated conditions from Table B.1.1: Psat = 143.3 kPa vf = 0.001052 m3/kg ; gh 9.807 × 8 ∆P = v = 0.001052 = 74 578 Pa = 74.578 kPa f Pbottom = Ptop + ∆P = 143.3 + 74.578 = 217.88 kPa H Sonntag, Borgnakke and van Wylen 3.48 Saturated water vapor at 200 kPa is in a constant pressure piston cylinder. At this state the piston is 0.1 m from the cylinder bottom. How much is this distance and the temperature if the water is cooled to occupy half the original volume? Solution: Process: B 1.2 v1 = vg (200 kPa) = 0.8857 m3/kg, P = constant = 200 kPa State 2: P, v2 = v1/2 = 0.44285 m3/kg State 1: T1 = 120.2°C v2 < vg so two phase T2 = Tsat = 120.2°C Height is proportional to volume h2 = h1 × v2/v1 = 0.1 × 0.5 = 0.05m Table B.1.2 P C.P. 200 2 T C.P. P = 200 kPa 1 120 T v 2 1 v Sonntag, Borgnakke and van Wylen 3.49 Two tanks are connected as shown in Fig. P3.49, both containing water. Tank A is at 200 kPa, v = 0.5 m3/kg, V = 1 m3 and tank B contains 3.5 kg at 0.5 MPa, 400°C. A The valve is now opened and the two come to a uniform state. Find the final specific volume. Solution: Control volume: both tanks. Constant total volume and mass process. A State A1: (P, v) B sup. vapor mA = VA/vA = 1/0.5 = 2 kg State B1: (P, T) Table B.1.3 vB = 0.6173 m3/kg ⇒ VB = mBvB = 3.5 × 0.6173 = 2.1606 m3 Final state: mtot = mA + mB = 5.5 kg Vtot = VA + VB = 3.1606 m3 v2 = Vtot/mtot = 0.5746 m3/kg Sonntag, Borgnakke and van Wylen 3.50 Determine the mass of methane gas stored in a 2 m3 tank at −30°C, 3 MPa. Estimate the percent error in the mass determination if the ideal gas model is used. Solution: Methane Table B.7.1 at −30°C = 243.15 K > Tc = 190.6 K, so superheated vapor in Table B.7.2. Linear interpolation between 225 and 250 K. 243.15-225 ⇒ v ≅ 0.03333 + 250-225 ×(0.03896 - 0.03333) = 0.03742 m3/kg m = V/v = 2/0.03742 = 53.45 kg Ideal gas assumption v = RT/P = 0.51835 × 243.15/3000 = 0.042 m3/kg m = V/v = 2/0.042 = 47.62 kg Error: 5.83 kg 10.9% too small Sonntag, Borgnakke and van Wylen 3.51 Saturated water vapor at 60°C has its pressure decreased to increase the volume by 10% keeping the temperature constant. To what pressure should it be expanded? Solution: Initial state: v = 7.6707 m3/kg from table B.1.1 Final state: v = 1.10 × vg = 1.1 × 7.6707 = 8.4378 m3/kg Interpolate at 60°C between saturated (P = 19.94 kPa) and superheated vapor P = 10 kPa in Tables B.1.1 and B.1.3 8.4378 − 7.6707 P ≅ 19.941 + (10 − 19.941) = 18.9 kPa 15.3345 − 7.6707 P C.P. T P = 10 kPa o 60 C 10 kPa C.P. T v v Comment: T,v ⇒ P = 18 kPa (software) v is not linear in P, more like 1/P, so the linear interpolation in P is not very accurate. Sonntag, Borgnakke and van Wylen 3.52 Saturated water vapor at 200 kPa is in a constant pressure piston cylinder. At this state the piston is 0.1 m from the cylinder bottom. How much is this distance and the temperature if the water is heated to occupy twice the original volume? Solution: From B.1.2, v1 = 0.8857 m3/kg P2 = P1, v2 = 2v1 = 2 × 0.8857 = 1.7714 m3/kg Since the cross sectional area is constant the height is proportional to volume h2 = h1 v2/v1 = 2h1 = 0.2 m 2: From B.1.3., Interpolate for the temperature 1.7714 – 1.5493 T2 = 400 + 100 1.78139 – 1.5493 ≈ 496°C Sonntag, Borgnakke and van Wylen 3.53 A boiler feed pump delivers 0.05 m3/s of water at 240°C, 20 MPa. What is the mass flowrate (kg/s)? What would be the percent error if the properties of saturated liquid at 240°C were used in the calculation? What if the properties of saturated liquid at 20 MPa were used? Solution: State 1: (T, P) compressed liquid seen in B.1.4: v = 0.001205 m3/kg . . m = V/v = 0.05/0.001205 = 41.5 kg/s . vf (240°C) = 0.001229 m3/kg ⇒ m = 40.68 kg/s error 2% . vf (20 MPa) = 0.002036 m3/kg ⇒ m = 24.56 kg/s error 41% P C.P. 20 MPa T C.P. P = 20 MPa 240 o 240 C v v The constant T line is nearly vertical for the liquid phase in the P-v diagram. The state is at so high P, T that the saturated liquid line is not extremely steep. Sonntag, Borgnakke and van Wylen 3.54 Saturated vapor R-134a at 50oC changes volume at constant temperature. Find the new pressure, and quality if saturated, if the volume doubles. Repeat the question for the case the volume is reduced to half the original volume. Solution: v1 = vg = 0.01512 m3/kg, 1: (T, x) B.4.1: 2: v2 = 2v1 = 0.03024 m3/kg superheated vapor Interpolate between 600 kPa and 800 kPa 0.03024 – 0.03974 P2 = 600 + 200 × 0.02861 – 0.03974 = 771 kPa 3: v3 = v1/2 = 0.00756 m3/kg < vg : two phase v3 - vf 0.00756 – 0.000908 = = 0.4678 x3 = v 0.01422 fg P3 = Psat = 1318 kPa P C.P. T P1 = Psat = 1318 Kpa C.P. P = 1318 kPa 1318 3 1 2 50 T v 3 1 2 v Sonntag, Borgnakke and van Wylen 3.55 A storage tank holds methane at 120 K, with a quality of 25 %, and it warms up by 5°C per hour due to a failure in the refrigeration system. How long time will it take before the methane becomes single phase and what is the pressure then? Solution: Use Table B.7.1 Assume rigid tank v = constant = v1 v1 = 0.002439 + 0.25×0.30367 = 0.078366 m3/kg v1 > vc = 0.00615 m3/kg All single phase when v = vg => T ≅ 145 K We then also see that ∆t = ∆T/(5°C/h) ≅ (145 – 120 ) / 5 = 5 hours P = Psat= 824 kPa Sonntag, Borgnakke and van Wylen 3.56 A glass jar is filled with saturated water at 500 kPa, quality 25%, and a tight lid is put on. Now it is cooled to −10°C. What is the mass fraction of solid at this temperature? Solution: Constant volume and mass ⇒ v1 = v2 = V/m From Table B.1.2: v1 = 0.001093 + 0.25 × 0.3738 = 0.094543 v2 = 0.0010891 + x2 × 446.756 = v1 = 0.094543 From Table B.1.5: ⇒ x2 = 0.0002 mass fraction vapor or 99.98 % xsolid =1 - x2 = 0.9998 P C.P. T C.P. 1 1 T 2 v 2 v P L T C.P. 1 S 2 v V Sonntag, Borgnakke and van Wylen 3.57 Saturated (liquid + vapor) ammonia at 60°C is contained in a rigid steel tank. It is used in an experiment, where it should pass through the critical point when the system is heated. What should the initial mass fraction of liquid be? Solution: Process: Constant mass and volume, From table B.2.1: v=C T Crit. point v2 = vc = 0.004255 m3/kg v1 = 0.001834 + x1 × 0.04697 = 0.004255 => x1 = 0.01515 liquid mass fraction = 1 - x1 = 0.948 1 60 C v Sonntag, Borgnakke and van Wylen 3.58 A steel tank contains 6 kg of propane (liquid + vapor) at 20°C with a volume of 0.015 m3. The tank is now slowly heated. Will the liquid level inside eventually rise to the top or drop to the bottom of the tank? What if the initial mass is 1 kg instead of 6 kg? Solution: V 0.015 m3 v2 = v1 = m = 6 kg = 0.0025 m3/kg Constant volume and mass T A.2: vc = 0.00454 m3/kg > v1 eventually reaches sat. liquid. ⇒ level rises to top C.P. Liq. Vapor a b 20°C v vc If m = 1 kg ⇒ v1 = 0.015 m3/kg > vc then it will reach saturated vapor. ⇒ level falls Sonntag, Borgnakke and van Wylen 3.59 A 400-m3 storage tank is being constructed to hold LNG, liquified natural gas, which may be assumed to be essentially pure methane. If the tank is to contain 90% liquid and 10% vapor, by volume, at 100 kPa, what mass of LNG (kg) will the tank hold? What is the quality in the tank? Solution: CH4 is in the section B tables. From Table B.7.1: vf ≅ 0.002366 m3/kg, (interpolated) From Table B.7.2: vg ≅ 0.55665 m3/kg (first entry 100 kPa) Vliq 0.9 × 400 Vvap 0.1 × 400 mliq = v = 0.002366 = 152 155.5 kg; mvap = v = 0.55665 = 71.86 kg f g mtot = 152 227 kg, x = mvap / mtot = 4.72 × 10-4 Sonntag, Borgnakke and van Wylen 3.60 A sealed rigid vessel of 2 m3 contains a saturated mixture of liquid and vapor R134a at 10°C. If it is heated to 50°C, the liquid phase disappears. Find the pressure at 50°C and the initial mass of the liquid. Solution: Process: constant volume and constant mass. P State 2 is saturated vapor, from table B.5.1 P2 = Psat(50°C) = 1.318 MPa State 1: same specific volume as state 2 v = v = 0.015124 m3/kg 2 1 1 v 2 v1 = 0.000794 + x1 × 0.048658 m = V/v1 = 2/0.015124 = 132.24 kg; ⇒ x1 = 0.2945 mliq = (1 - x1)m = 93.295 kg Sonntag, Borgnakke and van Wylen 3.61 A pressure cooker (closed tank) contains water at 100°C with the liquid volume being 1/10 of the vapor volume. It is heated until the pressure reaches 2.0 MPa. Find the final temperature. Has the final state more or less vapor than the initial state? Solution: State 1: Vf = mf vf = Vg/10 = mgvg/10 ; vf = 0.001044 m3/kg, vg = 1.6729 m3/kg mg 10 mfvf / vg 10 vf 0.01044 x1 = m + m = m + 10 m v / v = 10 v + v = 0.01044 + 1.6729 = 0.0062 g f f ff g f g Table B.1.1: v1 = 0.001044 + 0.0062×1.67185 = 0.01141 m3/kg State 2: v2 = v1 = 0.01141 m3/kg < vg(2MPa) from B.1.2 so two-phase P At state 2: v2 = vf + x2 vfg 0.01141 = 0.001177 + x2 × 0.09845 2 1 => x2 = 0.104 More vapor at final state T2 = Tsat(2MPa) = 212.4°C v Sonntag, Borgnakke and van Wylen 3.62 A pressure cooker has the lid screwed on tight. A small opening with A = 5 mm2 is covered with a petcock that can be lifted to let steam escape. How much mass should the petcock have to allow boiling at 120oC with an outside atmosphere at 101.3 kPa? Table B.1.1.: Psat = 198.5 kPa F = mg = ∆P × A m = ∆P × A/g (198.5-101.3)×1000×5×10-6 = 9.807 = 0.0496 kg = 50 g Sonntag, Borgnakke and van Wylen 3.63 Ammonia at 10 oC and mass 0.1 kg is in a piston cylinder with an initial volume of 1 m3. The piston initially resting on the stops has a mass such that a pressure of 900 kPa will float it. Now the ammonia is slowly heated to 50oC. Find the final pressure and volume. Solution: C.V. Ammonia, constant mass. Process: V = constant unless P = Pfloat P V1 State 1: T = 10 oC, v1 = m = 10 = 0.1 m3/kg From Table B.2.1 vf < v < vg v - vf 0.1 - 0.0016 x1 = v = 0.20381 = 0.4828 fg 1a P1 1 State 1a: P = 900 kPa, v = v1 = 0.1 m3/kg < vg at 900 kPa This state is two-phase T1a = 21.52oC Since T2 > T1a then v2 > v1a State 2: 50oC and on line(s) means P2 = 900 kPa which is superheated vapor. Table B.2.2 : v2 = 0.16263 m3/kg V2 = mv2 = 1.6263 m3 2 P2 V Sonntag, Borgnakke and van Wylen Ideal Gas Law 3.64 A cylinder fitted with a frictionless piston contains butane at 25°C, 500 kPa. Can the butane reasonably be assumed to behave as an ideal gas at this state ? Solution Butane 25°C, 500 kPa, Table A.2: Tc = 425 K; Pc = 3.8 MPa 25 + 273 0.5 Tr = 425 = 0.701; Pr = 3.8 = 0.13 Look at generalized chart in Figure D.1 Actual Pr > Pr, sat = 0.1 => liquid!! not a gas The pressure should be less than 380 kPa to have a gas at that T. Sonntag, Borgnakke and van Wylen 3.65 A spherical helium balloon of 10 m in diameter is at ambient T and P, 15oC and 100 kPa. How much helium does it contain? It can lift a total mass that equals the mass of displaced atmospheric air. How much mass of the balloon fabric and cage can then be lifted? π π V = 6 D3 = 6 103 = 523.6 m3 V PV mHe = ρV = v = RT 100 × 523.6 = = 87.5 kg 2.0771 × 288 PV 100 × 523.6 mair = RT = = 633 kg 0.287 × 288 mlift = mair – mHe = 633-87.5 = 545.5 kg Sonntag, Borgnakke and van Wylen 3.66 Is it reasonable to assume that at the given states the substance behaves as an ideal gas? Solution: a) Oxygen, O2 at 30°C, 3 MPa Ideal Gas ( T » Tc = 155 K from A.2) b) Methane, CH4 at 30°C, 3 MPa Ideal Gas ( T » Tc = 190 K from A.2) c) Water, H2O at 30°C, 3 MPa NO compressed liquid P > Psat (B.1.1) d) R-134a e) R-134a at 30°C, 3 MPa NO compressed liquid P > Psat (B.5.1) at 30°C, 100 kPa Ideal Gas P is low < Psat (B.5.1) ln P c, d Liq. e Cr.P. a, b Vapor T Sonntag, Borgnakke and van Wylen 3.67 A 1-m3 tank is filled with a gas at room temperature 20°C and pressure 100 kPa. How much mass is there if the gas is a) air, b) neon or c) propane ? Solution: Use Table A.2 to compare T and P to the critical T and P with T = 20°C = 293.15 K ; P = 100 kPa << Pc for all Air : T >> TC,N2; TC,O2 = 154.6 K so ideal gas; R= 0.287 kJ/kg K Neon: T >> Tc = 44.4 K so ideal gas; R = 0.41195 kJ/kg K Propane: T < Tc = 370 K, but P << Pc = 4.25 MPa so gas R = 0.18855 kJ/kg K All states are ideal gas states so the ideal gas law applies PV = mRT PV a) m = RT = 100 × 1 = 1.189 kg 0.287 × 293.15 b) m = 100 × 1 = 0.828 kg 0.41195 × 293.15 c) m = 100 × 1 = 1.809 kg 0.18855 × 293.15 Sonntag, Borgnakke and van Wylen 3.68 A rigid tank of 1 m3 contains nitrogen gas at 600 kPa, 400 K. By mistake someone lets 0.5 kg flow out. If the final temperature is 375 K what is then the final pressure? Solution: PV m = RT = 600 × 1 = 5.054 kg 0.2968 × 400 m2 = m - 0.5 = 4.554 kg m2RT2 4.554 × 0.2968 × 375 P2 = V = = 506.9 kPa 1 Sonntag, Borgnakke and van Wylen 3.69 A cylindrical gas tank 1 m long, inside diameter of 20 cm, is evacuated and then filled with carbon dioxide gas at 25°C. To what pressure should it be charged if there should be 1.2 kg of carbon dioxide? Solution: Assume CO2 is an ideal gas, table A.5: R = 0.1889 kJ/kg K π Vcyl = A × L = 4(0.2)2 × 1 = 0.031416 m3 P V = mRT ⇒P= => mRT P= V 1.2 kg × 0.1889 kJ/kg Κ × (273.15 + 25) K = 2152 kPa 0.031416 m3 Sonntag, Borgnakke and van Wylen 3.70 A glass is cleaned in 45oC hot water and placed on the table bottom up. The room air at 20oC that was trapped in the glass gets heated up to 40oC and some of it leaks out so the net resulting pressure inside is 2 kPa above ambient pressure of 101 kPa. Now the glass and the air inside cools down to room temperature. What is the pressure inside the glass? Solution: 1 air: 40oC, 103 kPa 2 air: 20oC, ? AIR Constant Volume: V1 = V2, Constant Mass m1 = m2 Ideal Gas P1V1 = m1RT1 and P2V2 = m1RT2 Take Ratio T1 20 + 273 P2 = P1 T = 103 × 40 + 273 = 96.4 kPa 2 Slight amount of liquid water seals to table top Sonntag, Borgnakke and van Wylen 3.71 A hollow metal sphere of 150-mm inside diameter is weighed on a precision beam balance when evacuated and again after being filled to 875 kPa with an unknown gas. The difference in mass is 0.0025 kg, and the temperature is 25°C. What is the gas, assuming it is a pure substance listed in Table A.5 ? Solution: π Assume an ideal gas with total volume: V = 6(0.15)3 = 0.001767 m3 _ mRT 0.0025 × 8.3145 × 298.2 M = PV = = 4.009 ≈ MHe 875 × 0.001767 => Helium Gas Sonntag, Borgnakke and van Wylen 3.72 A vacuum pump is used to evacuate a chamber where some specimens are dried at 50°C. The pump rate of volume displacement is 0.5 m3/s with an inlet pressure of 0.1 kPa and temperature 50°C. How much water vapor has been removed over a 30min period? Solution: Use ideal gas since P << lowest P in steam tables. From table A.5 we get R = 0.46152 kJ/kg K . . .. m = m ∆t with mass flow rate as: m= V/v = PV/RT . 0.1 × 0.5 × 30×60 ⇒ m = PV∆t/RT = = 0.603 kg (0.46152 × 323.15) (ideal gas) Sonntag, Borgnakke and van Wylen 3.73 A 1 m3 rigid tank has propane at 100 kPa, 300 K and connected by a valve to another tank of 0.5 m3 with propane at 250 kPa, 400 K. The valve is opened and the two tanks come to a uniform state at 325 K. What is the final pressure? Solution: Propane is an ideal gas (P << Pc) with R = 0.1886 kJ/kgK from Tbl. A.5 PAVA 100 × 1 mA = RT = = 1.7674 kg 0.1886 × 300 A PBVB 250 × 0.5 m = RT = = 1.6564 kg 0.1886 × 400 B V2 = VA + VB = 1.5 m3 m2 = mA + mB = 3.4243 kg m2RT2 3.4243 × 0.1886 × 325 = = 139.9 kPa P2 = V 1.5 2 Sonntag, Borgnakke and van Wylen 3.74 Verify the accuracy of the ideal gas model when it is used to calculate specific volume for saturated water vapor as shown in Fig. 3.9. Do the calculation for 10 kPa and 1 MPa. Solution: Look at the two states assuming ideal gas and then the steam tables. Ideal gas: v = RT/P => v1 = 0.46152 × (45.81 + 273.15)/10 = 14.72 m3/kg v2 = 0.46152 × (179.91 + 273.15)/1000 = 0.209 m3/kg Real gas: Table B.1.2: v1 = 14.647 m3/kg so error = 0.3 % 3/kg so error = 7.49 % v2 = 0.19444 m Sonntag, Borgnakke and van Wylen 3.75 Assume we have 3 states of saturated vapor R-134a at +40 oC, 0 oC and -40 oC. Calculate the specific volume at the set of temperatures and corresponding saturated pressure assuming ideal gas behavior. Find the percent relative error = 100(v - vg)/vg with vg from the saturated R-134a table. Solution: R-134a. Table values from Table B.5.1 Psat, vg(T) Ideal gas constant from Table A.5: RR-134a = 0.08149 kJ/kg K T Psat , kPa vg vID.G. = RT / Psat error % -40 oC 51.8 0.35696 0.36678 2.75 0 oC 294 0.06919 0.07571 9.4 40 oC 1017 0.02002 0.02509 25.3 T P 3 3 2 2 1 1 v v Sonntag, Borgnakke and van Wylen 3.76 Do Problem 3.75, but for the substance R-12. Solution: R-12. Table values from Table B.3.1 Psat, vg(T) Ideal gas constant from Table A.5: RR-12 = 0.08149 kJ/kg K T Psat , kPa vg vID.G. = RT / Psat error % -40 oC 64.2 0.24191 0.2497 3.2 0 oC 308.6 0.05539 0.06086 9.9 40 oC 960.7 0.01817 0.02241 23.4 T P 3 3 2 2 1 1 v v Sonntag, Borgnakke and van Wylen 3.77 Do Problem 3.75, but for the substance ammonia. Solution: NH3. Table values from Table B.2.1 Psat, vg(T) Ideal gas constant from Table A.5: Rammonia = 0.4882 kJ/kg K T Psat , kPa vg vID.G. = RT / Psat error % -40 oC 71.7 1.5526 1.5875 2.25 0 oC 429.6 0.28929 0.3104 7.3 40 oC 1555 0.08313 0.09832 18.3 T P 3 3 2 2 1 1 v v Sonntag, Borgnakke and van Wylen 3.78 Air in an automobile tire is initially at −10°C and 190 kPa. After the automobile is driven awhile, the temperature gets up to 10°C. Find the new pressure. You must make one assumption on your own. Solution: Assume constant volume and that air is an ideal gas P2 = P1 × T2/T1 283.15 = 190 × 263.15 = 204.4 kPa Sonntag, Borgnakke and van Wylen 3.79 An initially deflated and flat balloon is connected by a valve to a 12 m3 storage tank containing helium gas at 2 MPa and ambient temperature, 20°C. The valve is opened and the balloon is inflated at constant pressure, Po = 100 kPa, equal to ambient pressure, until it becomes spherical at D1 = 1 m. If the balloon is larger than this, the balloon material is stretched giving a pressure inside as D1 D1 P = P0 + C 1 − D D The balloon is inflated to a final diameter of 4 m, at which point the pressure inside is 400 kPa. The temperature remains constant at 20°C. What is the maximum pressure inside the balloon at any time during this inflation process? What is the pressure inside the helium storage tank at this time? Solution: At the end of the process we have D = 4 m so we can get the constant C as 11 P = 400 = P0 + C ( 1 – 4 ) 4 = 100 + C × 3/16 => C = 1600 –1 –1 X = D / D1 The pressure is: P = 100 + 1600 ( 1 – X ) X ; dP –2 –3 + 2 X ) / D1 = 0 dD = C ( - X –2 –3 X=2 => - X + 2 X = 0 => π3 3 at max P => D = 2D1 = 2 m; V = 6 D = 4.18 m 11 Pmax = 100 + 1600 ( 1 - 2 ) 2 = 500 kPa Differentiate to find max: PV Helium is ideal gas A.5: m = RT = 500 × 4.189 = 3.44 kg 2.0771 × 293.15 2000 × 12 = 39.416 kg 2.0771 × 293.15 mTANK, 2 = 39.416 – 3.44 = 35.976 kg PT2 = mTANK, 2 RT/V = ( mTANK, 1 / mTANK, 2 ) × P1 = 1825.5 kPa PV mTANK, 1 = RT = Sonntag, Borgnakke and van Wylen Compressibility Factor 3.80 Argon is kept in a rigid 5 m3 tank at −30°C, 3 MPa. Determine the mass using the compressibility factor. What is the error (%) if the ideal gas model is used? Solution: No Argon table, so we use generalized chart Fig. D.1 Tr = 243.15/150.8 = 1.612, Pr = 3000/4870 = 0.616 => PV m = ZRT = 3000 × 5 = 308.75 kg 0.96 × 0.2081 × 243.2 Ideal gas Z = 1 PV m = RT = 296.4 kg 4% error Z ≅ 0.96 Sonntag, Borgnakke and van Wylen 3.81 What is the percent error in specific volume if the ideal gas model is used to represent the behavior of superheated ammonia at 40°C, 500 kPa? What if the generalized compressibility chart, Fig. D.1, is used instead? Solution: NH3 T = 40°C = 313.15 K, Tc = 405.5 K, Pc = 11.35 MPa from Table A.1 v = 0.2923 m3/kg RT 0.48819 × 313 = 0.3056 m3/kg ⇒ 4.5% error Ideal gas: v = P = 500 313.15 0.5 Figure D.1: Tr = 405.5 = 0.772, Pr = 11.35 = 0.044 ⇒ Z = 0.97 ZRT v = P = 0.2964 m3/kg ⇒ 1.4% error Table B.2.2: Sonntag, Borgnakke and van Wylen 3.82 A new refrigerant R-125 is stored as a liquid at -20 oC with a small amount of vapor. For a total of 1.5 kg R-125 find the pressure and the volume. Solution: As there is no section B table use compressibility chart. Table A.2: R-125 Tc = 339.2 K Pc = 3.62 MPa Tr = T / Tc = 253.15 / 339.2 = 0.746 We can read from Figure D.1 or a little more accurately interpolate from table D.4 entries: Pr sat = 0.16 ; Zg = 0.86 ; Zf = 0.029 P = Pr sat Pc = 0.16 × 3620 = 579 kPa PVliq = Zf mliq RT = 0.029 × 1.5 × 0.06927 × 253.15 / 579 = 0.0013 m3 Z sat vapor Tr= 2.0 Tr = 0.7 Tr = 0.7 sat liq. 0.1 1 ln Pr Sonntag, Borgnakke and van Wylen 3.83 Many substances that normally do not mix well do so easily under supercritical pressures. A mass of 125 kg ethylene at 7.5 MPa, 296.5 K is stored for such a process. How much volume does it occupy? Solution: There is no section B table for ethylene so use compressibility chart. Table A.2: Ethylene Tc = 282.4 K Pc = 5.04 MPa Tr = T/Tc = 296.5 / 282.4 = 1.05 ; Pr = P/Pc = 7.5 / 5.04 = 1.49 Z = 0.32 from Figure D.1 V = mZRT / P = 125 × 0.32 × 0.2964 × 296.5 / 7500 = 0.469 m3 Z Tr= 2.0 Tr = 1.05 Tr = 0.7 Tr = 0.7 0.1 1 ln Pr Sonntag, Borgnakke and van Wylen 3.84 Carbon dioxide at 330 K is pumped at a very high pressure, 10 MPa, into an oilwell. As it penetrates the rock/oil the oil viscosity is lowered so it flows out easily. For this process we need to know the density of the carbon dioxide being pumped. Solution: There is not a B section table so use compressibility chart Table A.2 CO2: Tc = 304.1 K Pc = 7.38 MPa Tr = T/Tc = 330/304.1 = 1.085 Pr = P/Pc = 10/7.38 = 1.355 From Figure D.1: Z ≈ 0.45 ρ = 1/v = P / ZRT = 10000/(0.45 × 0.1889 × 330) = 356 kg/m3 Z Tr= 2.0 Tr = 1.1 Tr = 0.7 Tr = 0.7 0.1 1 ln Pr Sonntag, Borgnakke and van Wylen 3.85 To plan a commercial refrigeration system using R-123 we would like to know how much more volume saturated vapor R-123 occupies per kg at -30 oC compared to the saturated liquid state. Solution: For R-123 there is no section B table printed. We will use compressibility chart. From Table A.2 Tc = 456.9 K ; Pc = 3.66 MPa ; M = 152.93 Tr = T/Tc = 243/456.9 = 0.53 _ R = R/M = 8.31451 / 152.93 = 0.0544 The value of Tr is below the range in Fig. D.1 so use the table D.4 Table D.4, Zg = 0.979 Zf = 0.00222 Zfg = 0.979 − 0.0022 = 0.9768; Pr = Pr sat = 0.0116 P = Pr × Pc = 42.5 vfg = Zfg RT/P = 0.9768 × 0.0544 × 243 / 42.5 = 0.304 m3/kg Sonntag, Borgnakke and van Wylen 3.86 A bottle with a volume of 0.1 m3 contains butane with a quality of 75% and a temperature of 300 K. Estimate the total butane mass in the bottle using the generalized compressibility chart. Solution: We need to find the property v the mass is: m = V/v so find v given T1 and x as : v = vf + x vfg Table A.2: Butane Tc = 425.2 K Tr = 300/425.2 = 0.705 => From Fig. D.1 or table D.4: Z Pc = 3.8 MPa = 3800 kPa Zf ≈ 0.02; g Zg ≈ 0.9; Tr= 2.0 Tr = 0.7 Tr = 0.7 f 0.1 1 ln Pr P = Psat = Pr sat × Pc = 0.1× 3.80 ×1000 = 380 kPa vf = ZfRT/P = 0.02 × 0.14304 × 300/380 = 0.00226 m3/kg vg = ZgRT/P = 0.9 × 0.14304 × 300/380 = 0.1016 m3/kg v = 0.00226 + 0.75 × (0.1016 – 0.00226) = 0.076765 m3/kg V 0.1 m = v = 0.076765 = 1.303 kg Pr sat = 0.1 Sonntag, Borgnakke and van Wylen 3.87 Refrigerant R-32 is at -10 oC with a quality of 15%. Find the pressure and specific volume. Solution: For R-32 there is no section B table printed. We will use compressibility chart. From Table A.2: Tc = 351.3 K ; Pc = 5.78 MPa ; From Table A.5: R = 0.1598 kJ/kg K Tr = T/Tc = 263/351.3 = 0.749 From Table D.4 or Figure D.1, Zf ≈ 0.029 ; Zg ≈ 0.86 ; P = Pr sat Pc = 0.16 × 5780 = 925 kPa v = vf + x vfg = (Zf + x × Zfg) RT/P = [0.029 + 0.15 × (0.86 – 0.029)] × 0.1598 × 263 / 925 = 0.007 m3/kg Z Tr= 2.0 Tr = 0.7 0.1 Tr = 0.7 1 ln Pr Pr sat ≈ 0.16 Sonntag, Borgnakke and van Wylen 3.88 A mass of 2 kg of acetylene is in a 0.045 m3 rigid container at a pressure of 4.3 MPa. Use the generalized charts to estimate the temperature. (This becomes trial and error). Solution: Table A.2, A.5: Pr = 4.3/6.14 = 0.70; Tc = 308.3 K; R = 0.3193 kJ/kg K v = V/m = 0.045/2 = 0.0225 m3/kg ZRT State given by (P, v) v= P Since Z is a function of the state Fig. D.1 and thus T, we have trial and error. Try sat. vapor at Pr = 0.7 => Fig. D.1: Zg = 0.59; Tr = 0.94 vg = 0.59 × 0.3193 × 0.94 × 308.3/4300 = 0.0127 m3/kg too small Tr = 1 => Z = 0.7 => v = Tr = 1.2 => Z = 0.86 => v = 0.7 × 0.3193 × 1 × 308.3 = 0.016 m3/kg 4300 0.86 × 0.3193 × 1.2 × 308.3 = 0.0236 m3/kg 4300 Interpolate to get: Tr ≈ 1.17 T ≈ 361 K Sonntag, Borgnakke and van Wylen 3.89 A substance is at 2 MPa, 17°C in a 0.25-m3 rigid tank. Estimate the mass from the compressibility factor if the substance is a) air, b) butane or c) propane. Solution: Figure D.1 for compressibility Z and table A.2 for critical properties. Pr = P/Pc and Tr = T/Tc Air is a mixture so we will estimate from the major component. Nitrogen Pr = 2/3.39 = 0.59; Tr = 290/126.2 = 2.3; Z ≈ 0.98 m = PV/ZRT = 2000 × 0.25/(0.98 × 0.2968 × 290) = 5.928 kg Butane Pr = 2/3.80 = 0.526; Tr = 290/425.2 = 0.682; Z ≈ 0.085 m = PV/ZRT = 2000 × 0.25/(0.085 × 0.14304 × 290) = 141.8 kg Propane Pr = 2/4.25 = 0.47; Tr = 290/369.8 = 0.784; Z ≈ 0.08 m = PV/ZRT = 2000 × 0.25/(0.08 × 0.18855 × 290) = 114.3 kg Z a Tr= 2.0 Tr = 0.7 Tr = 0.7 cb 0.1 1 ln Pr Sonntag, Borgnakke and van Wylen Review Problems 3.90 Determine the quality (if saturated) or temperature (if superheated) of the following substances at the given two states: Solution: a) Water, H2O, use Table B.1.1 or B.1.2 1) 120°C, 1 m3/kg v > vg superheated vapor, T = 120 °C 2) 10 MPa, 0.01 m3/kg => two-phase v < vg x = ( 0.01 – 0.001452 ) / 0.01657 = 0.516 b) Nitrogen, N2, table B.6 1) 1 MPa, 0.03 m3/kg => superheated vapor since v > vg Interpolate between sat. vapor and superheated vapor B.6.2: 0.03−0.02416 T ≅ 103.73 + (120-103.73) × = 117 K 0.03117−0.02416 2) 100 K, 0.03 m3/kg => sat. liquid + vapor as two-phase v < vg v = 0.03 = 0.001452 + x × 0.029764 ⇒ x = 0.959 P C.P. States shown are placed relative to the two-phase region, not to each other. a2 b2 T C.P. P = const. a1, b1 a2 a1, b1 b2 T v v Sonntag, Borgnakke and van Wylen 3.91 Fill out the following table for substance ammonia: Solution: P [kPa] T [ oC] v [m3/kg] a) 400 -10 0.001534 b) 855 20 0.15 x Undefined 1.0 B.2.1 P > Psat(-10oC) = 291 kPa => compressed liquid v ≅ vf = 0.001534 m3/kg B.2.1 search along the vg values a) b) P C.P. T C.P. P = const. 1200 400 b a b 20 T -10 v a v Sonntag, Borgnakke and van Wylen 3.92 Find the phase, quality x if applicable and the missing property P or T. Solution: a. H2O T = 120°C v = 0.5 m3/kg v < vg = 0.89186 Table B.1.1 at given T: b. H2O sat. liq. + vap. P = Psat = 198.5 kPa, x = (v - vf)/vfg = (0.5 - 0.00106)/0.8908 = 0.56 P = 100 kPa v = 1.8 m3/kg v > vg = 1.694 sup. vap., interpolate in Table B.1.3 1.8 − 1.694 T= (150 – 99.62) + 99.62 = 121.65 °C 1.93636 − 1.694 Table B.1.2 at given P: c. H2O T = 263 K v = 0.2 m3/kg Table B.1.5 at given T = -10 °C: v < vg = 466.757 sat. solid + vap., P = Psat = 0.26 kPa, x = (v - vi)/vig = (200 - 0.001)/466.756 = 0.4285 P C.P. States shown are placed relative to the two-phase region, not to each other. T C.P. P = const. b a a T c v c b v Sonntag, Borgnakke and van Wylen 3.93 Find the phase, quality x if applicable and the missing property P or T. Solution: a. NH3 P = 800 kPa v = 0.2 m3/kg; Superheated Vapor (v > vg at 800 kPa) Table B 2.2 interpolate between 70°C and 80°C T = 71.4°C b. NH3 T = 20°C v = 0.1 m3/kg v < vg = 0.14922 sat. liq. + vap. , P = Psat = 857.5 kPa, x = (v - vf)/vfg = (0.1 - 0.00164)/0.14758 = 0.666 Table B.2.1 at given T: P C.P. T a C.P. a P = const. b b T v v Sonntag, Borgnakke and van Wylen 3.94 Give the phase and the missing properties of P, T, v and x. Solution: a. R-22 T = 10°C Table B.4.1 v = 0.01 m3/kg v < vg = 0.03471 m3/kg sat. liq. + vap. P = Psat = 680.7 kPa, x = (v - vf)/vfg = (0.01 - 0.0008)/0.03391 = 0.2713 b. H2O T = 350°C v = 0.2 m3/kg Table B.1.1 at given T: c. R-12 sup. vap. T = - 5 °C v > vg = 0.00881 P ≅ 1.40 MPa, x = undefined P = 200 kPa sup. vap. (P < Pg at -5°C) Table B 3.2: v = 0.08354 m3/kg at –12.5°C v = 0.08861 m3/kg at 0°C => v = 0.08658 m3/kg at -5°C d. R-134a P = 294 kPa, v = 0.05 m3/kg Table B.5.1: v < vg = 0.06919 m3/kg two-phase T = Tsat = 0°C x = (v - vf)/vfg = (0.05 - 0.000773)/0.06842 = 0.7195 P C.P. States shown are placed relative to the two-phase region, not to each other. T c c C.P. P = const. b a, d a, d T v b v Sonntag, Borgnakke and van Wylen 3.95 Give the phase and the missing properties of P, T, v and x. These may be a little more difficult if the appendix tables are used instead of the software. Solution: a) R-22 at T = 10°C, v = 0.036 m3/kg: Table B.4.1 v > vg at 10°C => sup. vap. Table B.4.2 interpolate between sat. and sup. both at 10°C 0.036-0.03471 P = 680.7 + (600 - 680.7) 0.04018-0.03471 = 661.7 kPa b) H2O v = 0.2 m3/kg , x = 0.5: Table B.1.1 v = (1-x) vf + x vg => vf + vg = 0.4 m3/kg since vf is so small we find it approximately where vg = 0.4 m3/kg. sat. liq. + vap. vf + vg = 0.39387 at 150°C, vf + vg = 0.4474 at 145°C. An interpolation gives T ≅ 149.4°C, P ≅ 468.2 kPa c) H2O T = 60°C, v = 0.001016 m3/kg: Table B.1.1 v < vf = 0.001017 => compr. liq. see Table B.1.4 v = 0.001015 at 5 MPa so P ≅ 0.5(5000 + 19.9) = 2.51 MPa d) NH3 T = 30°C, P = 60 kPa : Table B.2.1 P < Psat => sup. vapor interpolate in Table B.2.2 60 - 50 v = 2.94578 + (1.95906 - 2.94578) 75 - 50 = 2.551 m3/kg v is not linearly proportional to P (more like 1/P) so the computer table gives a more accurate value of 2.45 m3/kg e) R-134a v = 0.005 m3/kg , x = 0.5: sat. liq. + vap. Table B.5.1 v = (1-x) vf + x vg => vf + vg = 0.01 m3/kg vf + vg = 0.010946 at 65°C, vf + vg = 0.009665 at 70°C. T ≅ 68.7°C, P = 2.06 MPa An interpolation gives: P C.P. States shown are placed relative to the two-phase region, not to each other. c T a b, e C.P. d T c v b, e P = const. a d v Sonntag, Borgnakke and van Wylen 3.96 A 5 m long vertical tube of cross sectional area 200 cm2 is placed in a water fountain. It is filled with 15oC water, the bottom closed and the top open to the 100 kPa atmosphere. a) How much water is in the tube? b) What is the pressure at the bottom of the tube Solution: State 1: slightly compressed liquid from Table B.1.1 Mass: m = ρ V = V/v = AH/v = 200 × 10−4 × 5/0.001001 = 99.9 kg ∆P = ρ gH = gH/v = 9.80665 × 5/0.001001 = 48 984 Pa = 48.98 kPa Ptot = Ptop + ∆P = 149 kPa Sonntag, Borgnakke and van Wylen 3.97 Consider two tanks, A and B, connected by a valve, as shown in Fig. P3.97. Each has a volume of 200 L and tank A has R-12 at 25°C, 10% liquid and 90% vapor by volume, while tank B is evacuated. The valve is now opened and saturated vapor flows from A to B until the pressure in B has reached that in A, at which point the valve is closed. This process occurs slowly such that all temperatures stay at 25°C throughout the process. How much has the quality changed in tank A during the process? Solution: A B vacuum vf = 0.000763 m3/kg, vg = 0.026854 m3/kg Vliq1 Vvap1 0.1 × 0.2 0.9 × 0.2 + = + mA1 = v f 25°C vg 25°C 0.000763 0.026854 State A1: Table B.3.1 = 26.212 + 6.703 = 32.915 kg 6.703 xA1 = 32.915 = 0.2036 ; State B2: Assume A still two-phase so saturated P for given T VB 0.2 = 0.26854 = 7.448 kg mB2 = v g 25°C State A2: mass left is mA2 = 32.915 - 7.448 = 25.467 kg 0.2 vA2 = 25.467 = 0.007853 = 0.000763 + xA2 × 0.026091 xA2 = 0.2718 ∆x = 6.82% Sonntag, Borgnakke and van Wylen 3.98 A spring-loaded piston/cylinder contains water at 500°C, 3 MPa. The setup is such that pressure is proportional to volume, P = CV. It is now cooled until the water becomes saturated vapor. Sketch the P-v diagram and find the final pressure. Solution: State 1: Table B.1.3: v1 = 0.11619 m3/kg Process: m is constant and P = C0V = C0m v = C v P = Cv ⇒ C = P1/v1 = 3000/0.11619 = 25820 kPa kg/m3 State 2: x2 = 1 & P2 = Cv2 (on process line) P Trial & error on T2sat or P2sat: 1 Here from B.1.2: at 2 MPa vg = 0.09963 ⇒ C = 20074 (low) 2 vg = 0.07998 ⇒ C = 31258 (high) 2.25 MPa vg = 0.08875 ⇒ C = 25352 (low) 2.5 MPa v Interpolate to get the right C ⇒ P2 = 2270 kPa Sonntag, Borgnakke and van Wylen 3.99 A 1 m3 rigid tank has air at 1500 kPa and ambient 300 K connected by a valve to a piston cylinder. The piston of area 0.1 m2 requires 250 kPa below it to float. The valve is opened and the piston moves slowly 2 m up and the valve is closed. During the process air temperature remains at 300 K. What is the final pressure in the tank? PAVA 1500×1 mA = RT = = 17.422 kg 0.287×300 A ∆VA ∆VBPB 0.1×2×250 mB2 - mB1 = v = RT = = 0.581 kg 0.287×300 B mA2 = mA – (mB2 - mB1) = 17.422 – 0.581 = 16.841 kg PA2 = mA2RT 16.841×0.287×300 = 1450 kPa 1 VA = Sonntag, Borgnakke and van Wylen 3.100 A tank contains 2 kg of nitrogen at 100 K with a quality of 50%. Through a volume flowmeter and valve, 0.5 kg is now removed while the temperature remains constant. Find the final state inside the tank and the volume of nitrogen removed if the valve/meter is located at a. The top of the tank b. The bottom of the tank Solution Table B.6.1: v1 = 0.001452 + x1 × 0.029764 = 0.016334 m3/kg Vtank = m1v1 = 0.0327 m3 m2 = m1 - 0.5 = 1.5 kg v2 = Vtank/m2 = 0.0218 < vg(T) 0.0218-0.001452 x2 = 0.031216-0.001452 = 0.6836 Top: flow out is sat. vap. vg = 0.031216 m3/kg, Vout = moutvg = 0.0156 m3 Bottom: flow out is sat. liq. vf = 0.001452 Vout = moutvf = 0.000726 m3 Sonntag, Borgnakke and van Wylen 3.101 A piston/cylinder arrangement is loaded with a linear spring and the outside atmosphere. It contains water at 5 MPa, 400°C with the volume being 0.1 m3. If the piston is at the bottom, the spring exerts a force such that Plift = 200 kPa. The system now cools until the pressure reaches 1200 kPa. Find the mass of water, the final state (T2, v2) and plot the P–v diagram for the process. Solution: P 1: Table B.1.3 m = V/v1 = 0.1/0.05781 = 1.73 kg 1 5000 Straight line: P = Pa + C × v P2 - Pa v2 = v1 P - P = 0.01204 m3/kg 1 a 2 1200 200 ⇒ v1= 0.05781 m3/kg a v 0 ? 0.05781 v2 < vg(1200 kPa) so two-phase T2 = 188°C ⇒ x2 = (v2 - 0.001139)/0.1622 = 0.0672 Sonntag, Borgnakke and van Wylen 3.102 Water in a piston/cylinder is at 90°C, 100 kPa, and the piston loading is such that pressure is proportional to volume, P = CV. Heat is now added until the temperature reaches 200°C. Find the final pressure and also the quality if in the two-phase region. Solution: Final state: 200°C , on process line P = CV P State 1: Table B.1.1: v1 = 0.001036 m3/kg 2 P2 = P1v2/v1 from process equation Check state 2 in Table B.1.1 1 v If v2 = vg(T2) vg(T2) = 0.12736; Pg(T2) = 1.5538 MPa ⇒ P2 = 12.3 MPa > Pg not OK If sat. P2 = Pg(T2) = 1553.8 kPa ⇒ v2 = 0.0161 m3kg < vg sat. OK, P2 = 1553.8 kPa, x2 = (0.0161 - 0.001156) / 0.1262 = 0.118 Sonntag, Borgnakke and van Wylen 3.103 A container with liquid nitrogen at 100 K has a cross sectional area of 0.5 m2. Due to heat transfer, some of the liquid evaporates and in one hour the liquid level drops 30 mm. The vapor leaving the container passes through a valve and a heater and exits at 500 kPa, 260 K. Calculate the volume rate of flow of nitrogen gas exiting the heater. Solution: Properties from table B.6.1 for volume change, exit flow from table B.6.2: ∆V = A × ∆h = 0.5 × 0.03 = 0.015 m3 ∆mliq = -∆V/vf = -0.015/0.001452 = -10.3306 kg ∆mvap = ∆V/vg = 0.015/0.0312 = 0.4808 kg mout = 10.3306 - 0.4808 = 9.85 kg 3 vexit = 0.15385 m /kg . . 3 V = mvexit = (9.85 / 1 h)× 0.15385 m /kg 3 = 1.5015 m /h = 0.02526 m3/min Sonntag, Borgnakke and van Wylen 3.104 A cylinder containing ammonia is fitted with a piston restrained by an external force that is proportional to cylinder volume squared. Initial conditions are 10°C, 90% quality and a volume of 5 L. A valve on the cylinder is opened and additional ammonia flows into the cylinder until the mass inside has doubled. If at this point the pressure is 1.2 MPa, what is the final temperature? Solution: State 1 Table B.2.1: v1 = 0.0016 + 0.9(0.205525 - 0.0016) = 0.18513 m3/kg P1 = 615 kPa; V1 = 5 L = 0.005 m3 m1 = V/v = 0.005/0.18513 = 0.027 kg State 2: P2 = 1.2 MPa, Flow in so: m2 = 2 m1 = 0.054 kg Process: Piston Fext = KV2 = PA => P = CV2 => P2 = P1 (V2/V1)2 From the process equation we then get: 1200 1/2 1/2 V2 = V1 (P2/P1) = 0.005 ( 615 ) = 0.006984 m3 0.006984 v2 = V/m = 0.054 = 0.12934 m3/kg At P2, v2: T2 = 70.9°C Sonntag, Borgnakke and van Wylen 3.105 A cylinder/piston arrangement contains water at 105°C, 85% quality with a volume of 1 L. The system is heated, causing the piston to rise and encounter a linear spring as shown in Fig. P3.105. At this point the volume is 1.5 L, piston diameter is 150 mm, and the spring constant is 100 N/mm. The heating continues, so the piston compresses the spring. What is the cylinder temperature when the pressure reaches 200 kPa? Solution: P1 = 120.8 kPa, v1 = vf + x vfg = 0.001047 + 0.85*1.41831 = 1.20661 P 0.001 m = V1/ v1 = 1.20661 = 8.288×10-4 kg 200 v2 = v1 (V2 / V1) = 1.20661× 1.5 = 1.8099 & P = P1 = 120.8 kPa ( T2 = 203.5°C ) 1 2 v P3 = P2 + (ks/Ap2) m(v3-v2) linear spring 1 1.5 Ap = (π/4) × 0.152 = 0.01767 m2 ; ks = 100 kN/m (matches P in kPa) 200 = 120.8 + (100/0.01767 2 ) × 8.288×10-4(v -1.8099) 3 200 = 120.8 + 265.446 (v3 – 1.8099) => 3 3 v3 = 2.1083 m /kg T3 ≅ 600 + 100 × (2.1083 – 2.01297)/(2.2443-2.01297) ≅ 641°C liters Sonntag, Borgnakke and van Wylen 3.106 Refrigerant-12 in a piston/cylinder arrangement is initially at 50°C, x = 1. It is then expanded in a process so that P = Cv−1 to a pressure of 100 kPa. Find the final temperature and specific volume. Solution: P1 = 1219.3 kPa, v1 = 0.01417 m3/kg State 1: 50°C, x = 1 Table B.3.1: Process: Pv = C = P1v1; => P2 = C/v2= P1v1/v2 State 2: 100 kPa and v2 = v1P1/P2 = 0.1728 m3/kg T2 ≅ -13.2°C from Table B.3.2 Notice T not constant T P 1 1 2 2 v v Sonntag, Borgnakke and van Wylen 3.107 A 1-m3 rigid tank with air at 1 MPa, 400 K is connected to an air line as shown in Fig. P3.107. The valve is opened and air flows into the tank until the pressure reaches 5 MPa, at which point the valve is closed and the temperature inside is 450K. a. What is the mass of air in the tank before and after the process? b. The tank eventually cools to room temperature, 300 K. What is the pressure inside the tank then? Solution: P, T known at both states and assume the air behaves as an ideal gas. P1V 1000 × 1 mair1 = RT = = 8.711 kg 0.287 × 400 1 P2V 5000 × 1 mair2 = RT = = 38.715 kg 2 0.287 × 450 Process 2 → 3 is constant V, constant mass cooling to T3 P3 = P2 × (T3/T2) = 5000 × (300/450) = 3.33 MPa Sonntag, Borgnakke and van Wylen 3.108 Ammonia in a piston/cylinder arrangement is at 700 kPa, 80°C. It is now cooled at constant pressure to saturated vapor (state 2) at which point the piston is locked with a pin. The cooling continues to −10°C (state 3). Show the processes 1 to 2 and 2 to 3 on both a P–v and T–v diagram. Solution: State 1: T, P from table B.2.2 this is superheated vapor. State 2: T, x from table B.2.1 State 3: T, v two-phase T P 700 2 80 1 1 14 290 2 -10 3 3 v v Sonntag, Borgnakke and van Wylen 3.109 A cylinder has a thick piston initially held by a pin as shown in Fig. P3.109. The cylinder contains carbon dioxide at 200 kPa and ambient temperature of 290 K. The metal piston has a density of 8000 kg/m3 and the atmospheric pressure is 101 kPa. The pin is now removed, allowing the piston to move and after a while the gas returns to ambient temperature. Is the piston against the stops? Solution: Force balance on piston determines equilibrium float pressure. Piston mp = Ap × l × ρ ρpiston = 8000 kg/m3 mpg Ap × 0.1 × 9.807 × 8000 Pext on CO = P0 + A = 101 + = 108.8 kPa Ap × 1000 p 2 Pin released, as P1 > Pext piston moves up, T2 = To & if piston at stops, then V2 = V1 × Η2/Η1 = V1 × 150 / 100 Ideal gas with T2 = T1 then gives 100 ⇒ P2 = P1 × V1 / V2 = 200 × 150 = 133 kPa > Pext ⇒ piston is at stops, and P2 = 133 kPa Sonntag, Borgnakke and van Wylen 3.110 For a certain experiment, R-22 vapor is contained in a sealed glass tube at 20°C. It is desired to know the pressure at this condition, but there is no means of measuring it, since the tube is sealed. However, if the tube is cooled to −20°C small droplets of liquid are observed on the glass walls. What is the initial pressure? Solution: Control volume: R-22 fixed volume (V) & mass (m) at 20°C Process: cool to -20°C at constant v, so we assume saturated vapor State 2: v2 = vg at -20°C = 0.092843 m3/kg State 1: 20°C, v1 = v2 = 0.092843 m3/kg interpolate between 250 and 300 kPa in Table B.4.2 => P1 = 291 kPa T P 300 kPa 250 kPa 1 20 C -20 C 2 300 kPa 20 C -20 C v P1 250 kPa 2 v Sonntag, Borgnakke and van Wylen 3.111 A piston/cylinder arrangement, shown in Fig. P3.111, contains air at 250 kPa, 300°C. The 50-kg piston has a diameter of 0.1 m and initially pushes against the stops. The atmosphere is at 100 kPa and 20°C. The cylinder now cools as heat is transferred to the ambient. a. At what temperature does the piston begin to move down? b. How far has the piston dropped when the temperature reaches ambient? Solution: π Ap = 4 × 0.12 = 0.00785 m2 Balance forces when piston floats: mpg 50 × 9.807 Pfloat = Po + A = 100 + 0.00785 × 1000 p = 162.5 kPa = P2 = P3 P Piston 3 P2 1 2 V V stop To find temperature at 2 assume ideal gas: P2 162.5 T2 = T1 × P = 573.15 × 250 = 372.5 K 1 b) Process 2 -> 3 is constant pressure as piston floats to T3 = To = 293.15 K V2 = V1 = Ap × H = 0.00785 × 0.25 = 0.00196 m3 = 1.96 L Ideal gas and P2 = P3 => T3 293.15 V3 = V2 × T = 1.96 × 372.5 = 1.54 L 2 ∆H = (V2 -V3)/A = (1.96-1.54) × 0.001/0.00785 = 0.053 m = 5.3 cm Sonntag, Borgnakke and van Wylen 3.112 Air in a tank is at 1 MPa and room temperature of 20°C. It is used to fill an initially empty balloon to a pressure of 200 kPa, at which point the radius is 2 m and the temperature is 20°C. Assume the pressure in the balloon is linearly proportional to its radius and that the air in the tank also remains at 20°C throughout the process. Find the mass of air in the balloon and the minimum required volume of the tank. Solution: Assume air is an ideal gas. Balloon final state: V2 = (4/3) π r3 = (4/3) π 23 = 33.51 m3 P2V2 200× 33.51 m2bal = RT = = 79.66 kg 0.287 × 293.15 2 Tank must have P2 ≥ 200 kPa => m2 tank ≥ P2 VTANK /RT2 Initial mass must be enough: m1 = m2bal + m2 tank = P1V1 / R T1 P1VTANK / R T1 = m2bal + P2VTANK / RT2 => RTm2bal 0.287 × 293.15 × 79.66 VTANK = P - P = = 8.377 m3 1000 – 200 1 2 Sonntag, Borgnakke and van Wylen 3.113 A cylinder is fitted with a 10-cm-diameter piston that is restrained by a linear spring (force proportional to distance) as shown in Fig. P3.113. The spring force constant is 80 kN/m and the piston initially rests on the stops, with a cylinder volume of 1 L. The valve to the air line is opened and the piston begins to rise when the cylinder pressure is 150 kPa. When the valve is closed, the cylinder volume is 1.5 L and the temperature is 80°C. What mass of air is inside the cylinder? Solution: π Fs = ks∆x = ks ∆V/Ap ; V1 = 1 L = 0.001 m3, Ap = 4 0.12 = 0.007854 m2 State 2: V3 = 1.5 L = 0.0015 m3; T3 = 80°C = 353.15 K The pressure varies linearly with volume seen from a force balance as: PAp = P0 Ap + mp g + ks(V - V0)/Ap Between the states 1 and 2 only volume varies so: ks(V3-V2) 80×103(0.0015 - 0.001) P3 = P2 + = 150 + Ap2 0.0078542 × 1000 = 798.5 kPa P3V3 798.5 × 0.0015 m = RT = = 0.012 kg 0.287 × 353.15 3 P 3 2 1 v Sonntag, Borgnakke and van Wylen 3.114 A 500-L tank stores 100 kg of nitrogen gas at 150 K. To design the tank the pressure must be estimated and three different methods are suggested. Which is the most accurate, and how different in percent are the other two? a. Nitrogen tables, Table B.6 b. Ideal gas c. Generalized compressibility chart, Fig. D.1 Solution: State 1: 150 K, v = V/m = 0.5/100 = 0.005 m3/kg a) Table B.6, interpolate between 3 & 6 MPa with both at 150 K: 3 MPa : v = 0.01194 m3/kg, 6 MPa : v = 0.0042485 m3/kg P= 3 + (0.005-0.01194)×(6-3)/(0.0042485-0.01194) = 5.707 MPa RT 0.2968 × 150 b) Ideal gas table A.5: P= v = = 8.904 MPa 0.005 c) Table A.2 Tc = 126.2 K, Pc = 3.39 MPa so Tr = 150/126.2 = 1.189 Z is a function of P so it becomes trial and error. Start with P = 5.7 MPa ZRT Pr ≅ 1.68 ⇒ Z = 0.60 ⇒ P = v = 5342 kPa Now repeat finding the proper Z value. ⇒ Pr = 1.58 ⇒ Z = 0.62 ⇒ P = 5520 kPa OK Z Tr= 2.0 Tr = 1.2 Tr = 0.7 Tr = 0.7 0.1 1 ln Pr ANSWER: a) is the most accurate with others off by b) 60% c) 1% Sonntag, Borgnakke and van Wylen 3.115 What is the percent error in pressure if the ideal gas model is used to represent the behavior of superheated vapor R-22 at 50°C, 0.03082 m3/kg? What if the generalized compressibility chart, Fig. D.1, is used instead (iterations needed)? Solution: Real gas behavior: P = 900 kPa from Table B.4.2 _ Ideal gas constant: R = R/M = 8.31451/86.47 = 0.096155 kJ/kg K P = RT/v = 0.096155 × (273.15 + 50) / 0.03082 = 1008 kPa which is 12% too high Generalized chart Fig D.1 and critical properties from A.2: Tr = 323.2/363.3 = 0.875; Pc = 4970 kPa Assume P = 900 kPa => Pr = 0.181 => Z ≅ 0.905 v = ZRT/P = 0.905 × 0.096155 × 323.15 / 900 = 0.03125 too high Assume P = 950 kPa => Pr = 0.191 => Z ≅ 0.9 v = ZRT/P = 0.9 × 0.096155 × 323.15 / 950 = 0.029473 too low 0.03082 − 0.029437 = 938 kPa 4.2 % high P ≅ 900 + ( 950 − 900 ) × 0.03125 − 0.029437 Sonntag, Borgnakke and van Wylen Linear Interpolation 3.116 Find the pressure and temperature for saturated vapor R-12 with v = 0.1 m3/kg Solution: Table B.3.1 Look at the saturated vapor column vg and it is found between −20° C and −15°C. We must then do a linear interpolation between these values. T = −20 + [ –15 – (–20)] 0.1 − 0.10885 0.09101 − 0.10885 = −20 + 5 × 0.4961 = −17.5°C P = 150.9 + (182.6 – 150.9) × 0.4961 = 166.6 kPa T P 2 2 -15 182.6 1 150.9 1 -20 v 0.09101 0.10885 0.1 v 0.09101 0.10885 0.1 To understand the interpolation equation look at the smaller and larger triangles formed in the figure. The ratio of the side of the small triangle in v as (0.10885 - 0.1) to the side of the large triangle (0.10885 - 0.09101) is equal to 0.4961. This fraction of the total ∆P = 182.6 - 150.9 or ∆T = -15 -(-20) is added to the lower value to get the desired interpolated result. Sonntag, Borgnakke and van Wylen 3.117 Use a linear interpolation to estimate properties of ammonia to fill out the table below P [kPa] T [ °C] v [m3/kg] x a) 550 0.75 b) 80 20 c) 10 0.4 Solution: a) Find the pressures in Table B.2.1 that brackets the given pressure. 550 − 515.9 T = 5 + (10 – 5) = 5 + 5 × 0.341 = 6.7 °C 615.2 − 515.9 vf = 0.001583 + (0.0016 – 0.001583) 0.341 = 0.001589 m3/kg vg = 0.24299 + (0.20541 – 0.24299) 0.341 = 0.230175 m3/kg v = vf + xvfg = 0.001589 + 0.75(0.230175 – 0.001589) b) c) = 0.1729 m3/kg Interpolate between 50 and 100 kPa to get properties at 80 kPa 80 − 50 v = 2.8466 + (1.4153 – 2.8466) 100 − 50 = 2.8466 + ( − 1.4313) × 0.6 = 1.9878 m3/kg x: Undefined Table B.2.1: v > vg so the it is superheated vapor. Table B.2.2 locate state between 300 and 400 kPa. 0.4 - 0.44251 P = 300 + (400 – 300) 0.32701 − 0.44251 = 300 + 100 × 0.368 = 336.8 kPa x: Undefined Sonntag, Borgnakke and van Wylen 3.118 Use a linear interpolation to estimate Tsat at 900 kPa for nitrogen. Sketch by hand the curve Psat(T) by using a few table entries around 900 kPa from table B.6.1. Is your linear interpolation over or below the actual curve? Solution: The 900 kPa in Table B.6.1 is located between 100 and 105 K. 900 − 779.2 1084.6 − 779.2 = 100 + 5 × 0.3955 = 102 K T = 100 + (105 – 100) The actual curve has a positive second derivative (it curves up) so T is slightly underestimated by use of the chord between the 100 K and the 105 K points, as the chord is above the curve. P 1467.6 1084.6 900 779.2 T 100 105 110 Sonntag, Borgnakke and van Wylen 3.119 Use a double linear interpolation to find the pressure for superheated R-134a at 13°C with v = 0.3 m3/kg. Solution: Table B.5.2: Superheated vapor At 10°C, 0.3 m3/kg 0.3 - 0.45608 P = 50 + (100 – 50) × 0.22527 - 0.45608 = 83.8 kPa At 20°C, 0.3 m3/kg 0.3 - 0.47287 P = 50 + (100 – 50) × 0.23392 - 0.47287 = 86.2 kPa Interpolating at 13°C, P = 83.8 + (3/10) × (86.2 − 83.8) = 84.5 kPa This could also be interpolated as following: At 13°C, 50 kPa, v = 0.45608 + (3/10) × 0.0168 = 0.4611 m3/kg At 13°C, 100 kPa, v = 0.22527 + (3/10) × 0.0087 = 0.2279 m3/kg Interpolating at 0.3 m3/kg. 0.1611 P= 50 + (100 – 50) × 0.2332 = 84.5 kPa Sonntag, Borgnakke and van Wylen 3.120 Find the specific volume of ammonia at 140 kPa and 0°C. Solution: The state is superheated vapor in Table B.2.2 between 100 and 150 kPa. 140 − 100 v = 1.3136 + (0.8689 – 1.3136) 150 − 100 = 1.3136 + ( − 0.4447) × 0.8 = 0.9578 m3/kg 3.121 Find the pressure of water at 200°C and specific volume of 1.5 m3/kg. Solution: Table B.1.1: Table B.1.3: v > vg so that it is superheated vapor. Between 100 kPa and 200 kPa 1.5 − 2.17226 P = 100 + (200 – 100) = 161.6 kPa 1.08034 − 2.17226 Sonntag, Borgnakke and van Wylen Computer Tables 3.122 Use the computer software to find the properties for water at the 4 states in Problem 3.33 Start the software, click the tab for water as the substance, and click the small calculator icon. Select the proper CASE for the given properties. a) b) c) d) CASE 1 (T, P) 5 (P, v) 1 (T, P) 4 (T, x) RESULT Compressed liquid, x = undefined, v = 0.001002 Two-phase, T = 151.9°C, x = 0.5321 Sup. vapor, x = undefined, v = 0.143 m3/kg P = Psat = 8581 kPa, v = 0.01762 m3/kg Sonntag, Borgnakke and van Wylen 3.123 Use the computer software to find the properties for ammonia at the 2 states listed in Problem 3.37 Start the software, click the tab for cryogenic substances, and click the tab for the substance ammonia. Then click the small calculator icon and select the proper CASE for the given properties. a) b) c) d) CASE RESULT 2 (T, v) Sup. vapor, x = undefined, P = 1200 kPa 4 (T, x) Two-phase, P = 2033 kPa, v = 0.03257 m3/kg 1 (T, P) Compressed liquid, x = undefined, v = 0.001534 m3/kg No (v, x) entry so use 4 (T, x) OR 8 (P, x) several times T = 19.84°C, P = 853.1 kPa T = 19.83°C, P = 852.9 kPa Sonntag, Borgnakke and van Wylen 3.124 Use the computer software to find the properties for ammonia at the 3 states listed in Problem 3.117 Start the software, click the tab for cryogenic substances, select ammonia and click the small calculator icon. Select the proper CASE for the given properties. a) b) c) CASE 8 (P, x) 1 (T, P) 2 (T, v) RESULT T = 6.795°C, v = 0.1719 m3/kg Sup. vapor, x = undefined, v = 1.773 m3/kg Sup. vapor, x = undefined, P = 330.4 kPa Sonntag, Borgnakke and van Wylen 3.125 Find the value of the saturated temperature for nitrogen by linear interpolation in table B.6.1 for a pressure of 900 kPa. Compare this to the value given by the computer software. The 900 kPa in Table B.6.1 is located between 100 and 105 K. 900 − 779.2 1084.6 − 779.2 = 100 + 5 × 0.3955 = 101.98 K The actual curve has a positive second derivative (it curves up) so T is slightly underestimated by use of the chord between the 100 K and the 105 K points, as the chord is above the curve. From the computer software: CASE: 8 (P,x) T = -171°C = 102.15 K So we notice that the curvature has only a minor effect. T = 100 + (105 – 100) P 1467.6 1084.6 900 779.2 T 100 105 110 Sonntag, Borgnakke and van Wylen 3.126 Write a computer program that lists the states P, T, and v along the process curve in Problem 3.111 State 1: 250 kPa, 300°C = 573 K State 2: 162.5 kPa, 372.5 K State 3: 162.5 kPa, 293 K Since we have an ideal gas the relations among the pressure, temperature and the volume are very simple. The process curves are shown in the figure below. 1 T T2 3 T3 P 2 P/mR 3 V V3 V stop 1 P2 2 V stop V Sonntag, Borgnakke and van Wylen 3.127 Use the computer software to sketch the variation of pressure with temperature in Problem 3.41. Extend the curve a little into the single-phase region. P was found for a number of temperatures. A small table of (P, T) values were entered into a spreadsheet and a graph made as shown below. The superheated vapor region is reached at about 140°C and the graph shows a small kink at that point. 430 380 330 280 P 230 180 130 80 100 110 120 130 T 140 150 160 CHAPTER 4 SI UNIT PROBLEMS SOLUTION MANUAL SONNTAG • BORGNAKKE • VAN WYLEN FUNDAMENTALS of Thermodynamics Sixth Edition Sonntag, Borgnakke and van Wylen CONTENT SUBSECTION Correspondence table Concept problems Force displacement work Boundary work: simple one-step process Polytropic process Boundary work: multistep process Other types of work and general concepts Rates of work Heat transfer rates Review problems English unit concept problems PROB NO. 1-19 20-30 31-46 47-58 59-70 71-81 82-94 95-105 106-116 117-122 English unit problems 123-143 Sonntag, Borgnakke and van Wylen th CHAPTER 4 6 ed. CORRESPONDANCE TABLE The new problem set relative to the problems in the fifth edition. New 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 5th 1 2mod new New New 3 4 new New new New New 18 27 new new 5 new New 13 new new New New New 22 45 mod 8 12 14 New New New New 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 5th new 19 20 33 mod 37 36 15 30 6 New 32 7 9 34 10 New New 26 39 New 40 New New New New 58 59 60 61 New New New New New 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 5th new new new 43 new New new new New 47 HT 48 HT 49 HT 50 HT mod 51 HT mod 52 HT 53 HT 54 HT 55 HT 56 HT 57 HT 31 mod 11 16 17 23 21 mod 28 29 24 44 35 Sonntag, Borgnakke and van Wylen The English unit problem set is New 117 118 119 120 121 122 123 124 125 5th new new new new new new new 68 64 New 126 127 128 129 130 131 132 133 134 5th New new 62 67 70 new 66 65 75 New 135 136 137 138 139 140 141 142 143 5th 69 73 72 76 63 new 77 78 79 The computer, design and open-ended problem set is: New 144 145 146 147 5th 80 81 82 83 New 148 149 150 151 5th 84 85 86 87 New 152 153 5th 88 89 Sonntag, Borgnakke and van Wylen Concept-Study Guide Problems 4.1 The electric company charges the customers per kW-hour. What is that in SI units? Solution: The unit kW-hour is a rate multiplied with time. For the standard SI units the rate of energy is in W and the time is in seconds. The integration in Eq.4.21 becomes min s 1 kW- hour = 1000 W × 60 hour hour × 60 min = 3 600 000 Ws = 3 600 000 J = 3.6 MJ Sonntag, Borgnakke and van Wylen 4.2 A car engine is rated at 160 hp. What is the power in SI units? Solution: The horsepower is an older unit for power usually used for car engines. The conversion to standard SI units is given in Table A.1 1 hp = 0.7355 kW = 735.5 W 1 hp = 0.7457 kW for the UK horsepower 160 hp = 160 × 745.7 W = 119 312 W = 119.3 kW Sonntag, Borgnakke and van Wylen 4.3 A 1200 hp dragster engine has a drive shaft rotating at 2000 RPM. How much torque is on the shaft? Power is force times rate of displacement as in Eq.4.2 . . Power, rate of work W = FV=PV=Tω We need to convert the RPM to a value for angular velocity ω 2π 2π rad ω = RPM × 60 s = 2000 × 60 s = 209.44 s We need power in watts: 1 hp = 0.7355 kW = 735.5 W . 1200 hp × 735.5 W/hp T=W/ω= = 4214 Ws = 4214 Nm 209.44 rad/s Sonntag, Borgnakke and van Wylen 4.4 A 1200 hp dragster engine drives the car with a speed of 100 km/h. How much force is between the tires and the road? Power is force times rate of displacement as in Eq.4.2 . . Power, rate of work W = FV=PV=Tω We need the velocity in m/s: V = 100 × 1000 / 3600 = 27.78 m/s We need power in watts: 1 hp = 0.7355 kW = 735.5 W . 1200 × 735.5 W Nm/s F =W/V= 27.78 m/s = 31 771 m/s = 31 771 N = 31.8 kN 4.5 Two hydraulic piston/cylinders are connected through a hydraulic line so they have roughly the same pressure. If they have diameters of D1 and D2 = 2D1 respectively, what can you say about the piston forces F1 and F2? For each cylinder we have the total force as: F = PAcyl = P π D2/4 2 F1 = PAcyl 1 = P π D1/4 2 2 F2 = PAcyl 2 = P π D2/4 = P π 4 D1/4 = 4 F1 F1 F2 2 1 cb The forces are the total force acting up due to the cylinder pressure. There must be other forces on each piston to have a force balance so the pistons do not move. Sonntag, Borgnakke and van Wylen 4.6 Normally pistons have a flat head, but in diesel engines pistons can have bowls in them and protruding ridges. Does this geometry influence the work term? The shape of the surface does not influence the displacement dV = An dx where An is the area projected to the plane normal to the direction of motion. An = Acyl = π D2/4 Work is dW = F dx = P dV = P An dx = P Acyl dx and thus unaffected by the surface shape. x Semi-spherical head is made to make room for larger valves. Ridge Bowl Piston normal plane Sonntag, Borgnakke and van Wylen 4.7 What is roughly the relative magnitude of the work in the process 1-2c versus the process 1-2a shown in figure 4.8? By visual inspection the area below the curve 1-2c is roughly 50% of the rectangular area below the curve 1-2a. To see this better draw a straight line from state 1 to point f on the axis. This curve has exactly 50% of the area below it. 4.8 A hydraulic cylinder of area 0.01 m2 must push a 1000 kg arm and shovel 0.5 m straight up. What pressure is needed and how much work is done? F = mg = 1000 kg × 9.81 m/s2 = 9810 N = PA P = F/A = 9810 N/ 0.01 m2 = 981 000 Pa = 981 kPa W = ⌠F dx = F ∆x = 9810 N × 0.5 m = 4905 J ⌡ Sonntag, Borgnakke and van Wylen 4.9 A work of 2.5 kJ must be delivered on a rod from a pneumatic piston/cylinder where the air pressure is limited to 500 kPa. What diameter cylinder should I have to restrict the rod motion to maximum 0.5 m? π W = ⌠F dx = ⌠P dV = ⌠PA dx = PA ∆x = P 4 D2 ∆x ⌡ ⌡ ⌡ D= 4W = πP∆x 4 × 2.5 kJ = 0.113 m π × 500 kPa × 0.5 m 4.10 Helium gas expands from 125 kPa, 350 K and 0.25 m3 to 100 kPa in a polytropic process with n = 1.667. Is the work positive, negative or zero? W = ⌠P dV ⌡ The boundary work is: P drops but does V go up or down? The process equation is: PVn = C so we can solve for P to show it in a P-V diagram P = CV-n as n = 1.667 the curve drops as V goes up we see V2 > V1 giving dV > 0 and the work is then positive. P 1 2 W V Sonntag, Borgnakke and van Wylen 4.11 An ideal gas goes through an expansion process where the volume doubles. Which process will lead to the larger work output: an isothermal process or a polytropic process with n = 1.25? The process equation is: PVn = C The polytropic process with n = 1.25 drops the pressure faster than the isothermal process with n = 1 and the area below the curve is then smaller. P 1 n=1 2 W V 4.12 Show how the polytropic exponent n can be evaluated if you know the end state properties, (P1, V1) and (P2, V2). Polytropic process: PVn = C n n Both states must be on the process line: P2V2 = C = P1V1 Take the ratio to get: P1 V2 P2 = V1 n and then take ln of the ratio n P1 V2 V2 l n P = l n V = n ln V 1 1 2 now solve for the exponent n P1 V2 n = l n P ln V 2 1 / Sonntag, Borgnakke and van Wylen 4.13 A drag force on an object moving through a medium (like a car through air or a submarine through water) is Fd = 0.225 A ρV2. Verify the unit becomes Newton. Solution: Fd = 0.225 A ρV2 Units = m2 × ( kg/m3 ) × ( m2/ s2 ) = kg m / s2 = N 4.14 A force of 1.2 kN moves a truck with 60 km/h up a hill. What is the power? Solution: . W = F V = 1.2 kN × 60 (km/h) 3 3 × 60 × 10 Nm = 1.2 × 10 3600 = 20 000 W = 20 kW s Sonntag, Borgnakke and van Wylen 4.15 Electric power is volts times ampere (P = V i). When a car battery at 12 V is charged with 6 amp for 3 hours how much energy is delivered? Solution: . . W = ⌠ W dt = W ∆t = V i ∆t ⌡ = 12 V × 6 Amp × 3 × 3600 s = 777 600 J = 777.6 kJ Remark: Volt times ampere is also watts, 1 W = 1 V × 1 Amp. 4.16 Torque and energy and work have the same units (N m). Explain the difference. Solution: Work = force × displacement, so units are N × m. Energy in transfer Energy is stored, could be from work input 1 J = 1 N m Torque = force × arm static, no displacement needed Sonntag, Borgnakke and van Wylen 4.17 Find the rate of conduction heat transfer through a 1.5 cm thick hardwood board, k = 0.16 W/m K, with a temperature difference between the two sides of 20oC. One dimensional heat transfer by conduction, we do not know the area so we can find the flux (heat transfer per unit area W/m2). .. ∆T W 20 K q = Q/A = k = 0.16 m K × 0.015 m = 213 W/m2 ∆x 4.18 A 2 m2 window has a surface temperature of 15oC and the outside wind is blowing air at 2oC across it with a convection heat transfer coefficient of h = 125 W/m2K. What is the total heat transfer loss? Solution: . Q = h A ∆T = 125 W/m2K × 2 m2 × (15 – 2) K = 3250 W as a rate of heat transfer out. o 15 C o 2C Sonntag, Borgnakke and van Wylen 4.19 A radiant heating lamp has a surface temperature of 1000 K with ε = 0.8. How large a surface area is needed to provide 250 W of radiation heat transfer? Radiation heat transfer. We do not know the ambient so let us find the area for an emitted radiation of 250 W from the surface . Q = εσAT4 . 250 Q A= 4 = 0.8 × 5.67 × 10-8 × 10004 εσT = 0.0055 m2 Sonntag, Borgnakke and van Wylen Force displacement work 4.20 A piston of mass 2 kg is lowered 0.5 m in the standard gravitational field. Find the required force and work involved in the process. Solution: F = ma = 2 kg × 9.80665 m/s2 = 19.61 N W= ∫ F dx = F ∫ dx = F ∆x = 19.61 N × 0.5 m = 9.805 J 4.21 An escalator raises a 100 kg bucket of sand 10 m in 1 minute. Determine the total amount of work done during the process. Solution: The work is a force with a displacement and force is constant: F = mg W= ∫ F dx = F ∫ dx = F ∆x = 100 kg × 9.80665 m/s2 × 10 m = 9807 J Sonntag, Borgnakke and van Wylen 4.22 A bulldozer pushes 500 kg of dirt 100 m with a force of 1500 N. It then lifts the dirt 3 m up to put it in a dump truck. How much work did it do in each situation? Solution: W = ∫ F dx = F ∆x = 1500 N × 100 m = 150 000 J = 150 kJ W = ∫ F dz = ∫ mg dz = mg ∆Z = 500 kg × 9.807 m/s2 × 3 m = 14 710 J = 14.7 kJ Sonntag, Borgnakke and van Wylen 4.23 A hydraulic cylinder has a piston of cross sectional area 25 cm2 and a fluid pressure of 2 MPa. If the piston is moved 0.25 m how much work is done? Solution: The work is a force with a displacement and force is constant: F = PA W= ∫ F dx = ∫ PA dx = PA ∆x = 2000 kPa × 25 × 10-4 m2 × 0.25 m = 1.25 kJ Units: kPa m2 m = kN m-2 m2 m = kN m = kJ Sonntag, Borgnakke and van Wylen 4.24 Two hydraulic cylinders maintain a pressure of 1200 kPa. One has a cross sectional area of 0.01 m2 the other 0.03 m2. To deliver a work of 1 kJ to the piston how large a displacement (V) and piston motion H is needed for each cylinder? Neglect Patm. Solution: W = ∫ F dx = ∫ P dV = ∫ PA dx = PA* H = P∆V W 1 kJ ∆V = P = 1200 kPa = 0.000 833 m3 Both cases the height is H = ∆V/A 0.000833 H1 = 0.01 = 0.0833 m 0.000833 H2 = 0.03 = 0.0278 m F1 F2 2 1 cb Sonntag, Borgnakke and van Wylen 4.25 A linear spring, F = ks(x − x0), with spring constant ks = 500 N/m, is stretched until it is 100 mm longer. Find the required force and work input. Solution: F = ks(x - x0) = 500 × 0.1 = 50 N W = ∫ F dx = ⌠ ks(x - x0)d(x - x0) = ks(x - x0)2/2 ⌡ N = 500 m × (0.12/2) m2 = 2.5 J Sonntag, Borgnakke and van Wylen 4.26 n A nonlinear spring has the force versus displacement relation of F = kns(x − x0) . If the spring end is moved to x1 from the relaxed state, determine the formula for the required work. Solution: In this case we know F as a function of x and can integrate kns W = ⌠Fdx = ⌠ kns(x - xo)n d(x - xo) = n + 1 (x1 - xo)n+1 ⌡ ⌡ Sonntag, Borgnakke and van Wylen 4.27 The rolling resistance of a car depends on its weight as: F = 0.006 mg. How long will a car of 1400 kg drive for a work input of 25 kJ? Solution: Work is force times distance so assuming a constant force we get W = ⌠ F dx = F x = 0.006 mgx ⌡ Solve for x W x = 0.006 mg = 25 kJ = 303.5 m 0.006 × 1400 kg × 9.807 m/s2 Sonntag, Borgnakke and van Wylen 4.28 A car drives for half an hour at constant speed and uses 30 MJ over a distance of 40 km. What was the traction force to the road and its speed? Solution: We need to relate the work to the force and distance W = ⌡F dx = F x ⌠ W 30 000 000 J F = x = 40 000 m = 750 N L 40 km km 1000 m V = t = 0.5 h = 80 h = 80 3600 s = 22.2 ms−1 Sonntag, Borgnakke and van Wylen 4.29 The air drag force on a car is 0.225 A ρV2. Assume air at 290 K, 100 kPa and a car frontal area of 4 m2 driving at 90 km/h. How much energy is used to overcome the air drag driving for 30 minutes? 1P 100 kg ρ = v = RT = = 1.2015 3 0.287 ×290 m km 1000 m V = 90 h = 90 × 3600 s = 25 m/s ∆x = V ∆t = 25 × 30 × 60 = 45 000 m F = 0.225 A ρV2 = 0.225 ×4 ×1.2015 ×252 2 2 kg × m = 676 N = 675.8 m 3 m s2 W = F ∆x = 676 N × 45 000 m = 30 420 000 J = 30.42 MJ Sonntag, Borgnakke and van Wylen 4.30 Two hydraulic piston/cylinders are connected with a line. The master cylinder has an area of 5 cm2 creating a pressure of 1000 kPa. The slave cylinder has an area of 3 cm2. If 25 J is the work input to the master cylinder what is the force and displacement of each piston and the work out put of the slave cylinder piston? Solution: W = ∫ Fx dx = ∫ P dv = ∫ P A dx = P A ∆x W 25 ∆xmaster = PA = = 0.05 m 1000×5×10-4 A∆x = ∆V = 5 ×10-4× 0.05 = 2.5 ×10-5 m = ∆Vslave = A ∆x ∆xslave = ∆V/A = 2.5 × 10-5 / 3 ×10-4 = 0.0083 33 m Fmaster = P A = 1000× 5 ×10-4 ×103 = 500 N Fslave = P A = 1000 ×103× 3 ×10-4 = 300 N Wslave = F ∆x = 300 × 0.08333 = 25 J Master Slave Sonntag, Borgnakke and van Wylen Boundary work simple 1 step process 4.31 A constant pressure piston cylinder contains 0.2 kg water as saturated vapor at 400 kPa. It is now cooled so the water occupies half the original volume. Find the work in the process. Solution: Table B.1.2 v1= 0.4625 m3/kg v2 = v1/ 2 = 0.23125 m3/kg V1 = mv1 = 0.0925 m3 V2 = V1 / 2 = 0.04625 m3 Process: P = C so the work term integral is W = ∫ PdV = P(V2-V1) = 400 kPa × (0.04625 – 0.0925) m3 = -18.5 kJ P C.P. T C.P. P = 400 kPa 400 2 1 144 T 2 cb v 1 v Sonntag, Borgnakke and van Wylen 4.32 A steam radiator in a room at 25°C has saturated water vapor at 110 kPa flowing through it, when the inlet and exit valves are closed. What is the pressure and the quality of the water, when it has cooled to 25oC? How much work is done? Solution: Control volume radiator. After the valve is closed no more flow, constant volume and mass. 1: x1 = 1, P1 = 110 kPa ⇒ v1 = vg = 1.566 m3/kg from Table B.1.2 2: T2 = 25oC, ? Process: v2 = v1 = 1.566 m3/kg = [0.001003 + x2 × 43.359] m3/kg x2 = State 2 : T2 , x2 1.566 – 0.001003 = 0.0361 43.359 From Table B.1.1 1W2 = ⌠PdV = 0 ⌡ P2 = Psat = 3.169 kPa Sonntag, Borgnakke and van Wylen 4.33 A 400-L tank A, see figure P4.33, contains argon gas at 250 kPa, 30oC. Cylinder B, having a frictionless piston of such mass that a pressure of 150 kPa will float it, is initially empty. The valve is opened and argon flows into B and eventually reaches a uniform state of 150 kPa, 30oC throughout. What is the work done by the argon? Solution: Take C.V. as all the argon in both A and B. Boundary movement work done in cylinder B against constant external pressure of 150 kPa. Argon is an ideal gas, so write out that the mass and temperature at state 1 and 2 are the same PA1VA = mARTA1 = mART2 = P2( VA + VB2) => VB2 = 2 250 × 0.4 - 0.4 = 0.2667 m3 150 3 ⌠ 1W2 = ⌡ PextdV = Pext(VB2 - VB1) = 150 kPa (0.2667 - 0) m = 40 kJ 1 Sonntag, Borgnakke and van Wylen 4.34 A piston cylinder contains air at 600 kPa, 290 K and a volume of 0.01 m3. A constant pressure process gives 54 kJ of work out. Find the final volume and temperature of the air. Solution: W = ∫ P dV = P∆V 54 ∆V = W/P = 600 = 0.09 m3 V2 = V1 + ∆V = 0.01 + 0.09 = 0.1 m3 Assuming ideal gas, PV = mRT, then we have P2 V2 P2 V2 V2 0.1 T2 = mR = P V T1= V T1 = 0.01 290 = 2900 K 11 1 Sonntag, Borgnakke and van Wylen 4.35 Saturated water vapor at 200 kPa is in a constant pressure piston cylinder. At this state the piston is 0.1 m from the cylinder bottom and cylinder area is 0.25 m2. The temperature is then changed to 200oC. Find the work in the process. Solution: State 1 from B.1.2 (P, x): v1 = vg = 0.8857 m3/kg State 2 from B.1.3 (P, T): v2 = 1.0803 m3/kg (also in B.1.3) Since the mass and the cross sectional area is the same we get v2 1.0803 h2 = v × h1 = 0.8857 × 0.1 = 0.122 m 1 Process: P = C so the work integral is W = ∫ PdV = P(V2 - V1) = PA (h2 - h1) W = 200 kPa × 0.25 m2 × (0.122 − 0.1) m = 1.1 kJ P C.P. T C.P. P = 200 kPa 200 1 200 2 2 120 T 1 cb v v Sonntag, Borgnakke and van Wylen 4.36 A cylinder fitted with a frictionless piston contains 5 kg of superheated refrigerant R-134a vapor at 1000 kPa, 140°C. The setup is cooled at constant pressure until the R-134a reaches a quality of 25%. Calculate the work done in the process. Solution: Constant pressure process boundary work. State properties from Table B.5.2 State 1: v = 0.03150 m3/kg , State 2: v = 0.000871 + 0.25 × 0.01956 = 0.00576 m3/kg Interpolated to be at 1000 kPa, numbers at 1017 kPa could have been used in which case: 1W2 = v = 0.00566 m3/kg ∫ P dV = P (V2-V1) = mP (v2-v1) = 5 × 1000 (0.00576 - 0.03150) = -128.7 kJ P C.P. T C.P. P = 1000 kPa 1000 2 1 140 1 39 T 2 cb v v Sonntag, Borgnakke and van Wylen 4.37 Find the specific work in Problem 3.54 for the case the volume is reduced. Saturated vapor R-134a at 50oC changes volume at constant temperature. Find the new pressure, and quality if saturated, if the volume doubles. Repeat the question for the case the volume is reduced to half the original volume. Solution: R-134a 50oC Table B.4.1: 1W2 v1 = vg = 0.01512 m3/kg, v2 = v1 / 2 = 0.00756 m3/kg = ∫ PdV = 1318.1 kPa (0.00756 – 0.01512) m3/kg = -9.96 kJ/kg P C.P. T C.P. P = 1318 kPa 1318 2 1 50 T cb v 2 1 v Sonntag, Borgnakke and van Wylen 4.38 A piston/cylinder has 5 m of liquid 20oC water on top of the piston (m = 0) with cross-sectional area of 0.1 m2, see Fig. P2.57. Air is let in under the piston that rises and pushes the water out over the top edge. Find the necessary work to push all the water out and plot the process in a P-V diagram. Solution: P1 = Po + ρgH = 101.32 + 997 × 9.807 × 5 / 1000 = 150.2 kPa ∆V = H × A = 5 × 0.1 = 0.5 m3 1W2 = AREA = ∫ P dV = ½ (P1 + Po )(Vmax -V1) = ½ (150.2 + 101.32) kPa × 0.5 m3 = 62.88 kJ Po P H2O cb P1 1 2 P0 V Air V1 Vmax Sonntag, Borgnakke and van Wylen 4.39 Air in a spring loaded piston/cylinder has a pressure that is linear with volume, P = A + BV. With an initial state of P = 150 kPa, V = 1 L and a final state of 800 kPa and volume 1.5 L it is similar to the setup in Problem 3.113. Find the work done by the air. Solution: Knowing the process equation: P = A + BV giving a linear variation of pressure versus volume the straight line in the P-V diagram is fixed by the two points as state 1 and state 2. The work as the integral of PdV equals the area under the process curve in the P-V diagram. State 1: P1 = 150 kPa V1 = 1 L = 0.001 m3 P State 2: P2 = 800 kPa V2 = 1.5 L = 0.0015 m3 Process: P = A + BV ⇒ linear in V 2 P1 + P2 W2 = ⌠ PdV = (V2 - V1) 1 ⌡ 2 2 1 ( ) 1 W 1 = 2 (150 + 800) kPa (1.5 - 1) × 0.001 m3 = 0.2375 kJ V Sonntag, Borgnakke and van Wylen 4.40 Find the specific work in Problem 3.43. Saturated water vapor at 200 kPa is in a constant pressure piston cylinder. At this state the piston is 0.1 m from the cylinder bottom. How much is this distance if the temperature is changed to a) 200 oC and b) 100 oC. Solution: Process: P=C ⇒ w = ∫ Pdv = P1(v – v1) State 1: (200 kPa, x = 1) in B.1.2: v1 = vg (200 kPa) = 0.8857 m3/kg CASE a) State a: (200 kPa, 200oC) B.1.3: 1wa va = 1.083 m3/kg = ∫ Pdv = 200(1.0803 – 0.8857) = 38.92 kJ/kg CASE b) State b: (200 kPa, 100oC) B.1.1: 1Wb vb ≈ vf = 0.001044 m3/kg = ∫ PdV = 200(0.001044 – 0.8857) = -176.9 kJ/kg P C.P. T C.P. P = 200 kPa 200 200 1a b 120 100 b T bW1 cb a 1W v a 1 v Sonntag, Borgnakke and van Wylen 4.41 A piston/cylinder contains 1 kg water at 20oC with volume 0.1 m3. By mistake someone locks the piston preventing it from moving while we heat the water to saturated vapor. Find the final temperature, volume and the process work. Solution 1: v1 = V/m = 0.1 m3/1 kg = 0.1 m3/kg 2: Constant volume: v2 = vg = v1 V2 = V1 = 0.1 m3 1W2 = ∫ P dV = 0 0.1 - 0.10324 T2 = Tsat = 210 + 5 0.09361 - 0.10324 = 211.7°C Sonntag, Borgnakke and van Wylen 4.42 A piston cylinder contains 1 kg of liquid water at 20oC and 300 kPa. There is a linear spring mounted on the piston such that when the water is heated the pressure reaches 3 MPa with a volume of 0.1 m3. a) Find the final temperature b) Plot the process in a P-v diagram. c) Find the work in the process. Solution: Take CV as the water. This is a constant mass: m2 = m1 = m ; State 1: Compressed liquid, take saturated liquid at same temperature. B.1.1: v1 = vf(20) = 0.001002 m3/kg, State 2: v2 = V2/m = 0.1/1 = 0.1 m3/kg and P = 3000 kPa from B.1.3 => Superheated vapor close to T = 400oC Interpolate: T2 = 404oC Work is done while piston moves at linearly varying pressure, so we get: 1 1W2 = ∫ P dV = area = Pavg (V2 − V1) = 2 (P1 + P2)(V2 - V1) = 0.5 (300 + 3000)(0.1 − 0.001) = 163.35 kJ P C.P. 300 1 T 2 C.P. 2 300 kPa T 20 v 1 v Sonntag, Borgnakke and van Wylen 4.43 A piston cylinder contains 3 kg of air at 20oC and 300 kPa. It is now heated up in a constant pressure process to 600 K. a) Find the final volume b) Plot the process path in a P-v diagram c) Find the work in the process. Solution: Ideal gas PV = mRT State 1: T1, P1 ideal gas so P1V1 = mRT1 V1 = mR T1 / P1 = 3 × 0.287 × 293.15/300 = 0.8413 m3 State 2: T2, P2 = P1 and ideal gas so P2V2 = mRT2 V2 = mR T2 / P2 = 3 × 0.287 × 600/300 = 1.722 m3 W2 1 = ⌠ PdV = P (V2 - V1) = 300 (1.722 – 0.8413) = 264.2 kJ ⌡ P 300 T 2 1 T1 2 600 300 kPa T2 293 v 1 v Sonntag, Borgnakke and van Wylen 4.44 A piston cylinder contains 0.5 kg air at 500 kPa, 500 K. The air expands in a process so P is linearly decreasing with volume to a final state of 100 kPa, 300 K. Find the work in the process. Solution: Process: P = A + BV (linear in V, decreasing means B is negative) 1 W2 = ⌠ PdV = AREA = 2 (P1 + P2)(V2 - V1) ⌡ From the process: 1 V1 = mR T1/ P1 = 0.5 × 0.287 × (500/500) = 0.1435 m3 V2 = mR T2/ P2 = 0.5 × 0.287 × (300/100) = 0.4305 m3 1 W2 = 2 × (500 + 100) kPa × (0.4305 - 0.1435) m3 = 86.1 kJ 1 P T 1 500 500 1 T1 T2 2 100 cb 300 v 2 v Sonntag, Borgnakke and van Wylen 4.45 Consider the nonequilibrium process described in Problem 3.109. Determine the work done by the carbon dioxide in the cylinder during the process. A cylinder has a thick piston initially held by a pin as shown in Fig. P3.109. The cylinder contains carbon dioxide at 200 kPa and ambient temperature of 290 K. The metal piston has a density of 8000 kg/m3 and the atmospheric pressure is 101 kPa. The pin is now removed, allowing the piston to move and after a while the gas returns to ambient temperature. Is the piston against the stops? Solution: Knowing the process (P vs. V) and the states 1 and 2 we can find W. If piston floats or moves: P = Plift = Po + ρHg = 101.3 + 8000 × 0.1 × 9.807 / 1000 = 108.8 kPa Assume the piston is at the stops (since P1 > Plift piston would move) V2 = V1 × 150 / 100 = (π/4) 0.12 × 0.1× 1.5 = 0.000785× 1.5 = 0.001 1775 m3 For max volume we must have P > Plift so check using ideal gas and constant T process: P2 = P1 V1/ V2 = 200/1.5 = 133 kPa > Plift and piston is at stops. 1W2 = ∫ Plift dV = Plift (V2 -V1) = 108.8 (0.0011775 - 0.000785) = 0.0427 kJ Remark: The work is determined by the equilibrium pressure, Plift, and not the instantaneous pressure that will accelerate the piston (give it kinetic energy). We need to consider the quasi-equilibrium process to get W. Sonntag, Borgnakke and van Wylen 4.46 Consider the problem of inflating the helium balloon, as described in problem 3.79. For a control volume that consists of the helium inside the balloon determine the work done during the filling process when the diameter changes from 1 m to 4 m. Solution : Inflation at constant P = P0 = 100 kPa to D1 = 1 m, then P = P0 + C ( D* -1 - D* -2 ), D* = D / D1, to D2 = 4 m, P2 = 400 kPa, from which we find the constant C as: 400 = 100 + C[ (1/4) - (1/4)2 ] => C = 1600 kPa π The volumes are: V = 6 D3 => V1 = 0.5236 m3; V2 = 33.51 m3 2 WCV = ⌠ PdV ⌡ 1 2 = P0(V2 - V1) + ⌠ C(D* -1 - D* -2)dV ⌡ 1 π V = 6 D3, π π dV = 2 D2 dD = 2 D13 D* 2 dD* D2*=4 ⇒ WCV = P0(V2 - V1) + 3CV1 ⌠ ⌡ (D*-1)dD* D1*=1 4 D2* 2 - D1* 2 * * = P0(V2 - V1) + 3CV1[ - (D2 - D1 )] 2 1 16-1 = 100 × (33.51 – 0.5236) + 3 × 1600 × 0.5236 [ 2 – (4–1)] = 14 608 kJ Sonntag, Borgnakke and van Wylen Polytropic process 4.47 Consider a mass going through a polytropic process where pressure is directly proportional to volume (n = − 1). The process start with P = 0, V = 0 and ends with P = 600 kPa, V = 0.01 m3. The physical setup could be as in Problem 2.22. Find the boundary work done by the mass. Solution: The setup has a pressure that varies linear with volume going through the initial and the final state points. The work is the area below the process curve. P W = ⌠ PdV = AREA ⌡ 600 1 = 2 (P1 + P2)(V2 - V1) W 0 0 0.01 1 V = 2 (P2 + 0)( V2 - 0) 1 1 = 2 P2 V2 = 2 × 600 × 0.01 = 3 kJ Sonntag, Borgnakke and van Wylen 4.48 The piston/cylinder shown in Fig. P4.48 contains carbon dioxide at 300 kPa, 100°C with a volume of 0.2 m3. Mass is added at such a rate that the gas compresses according to the relation PV1.2 = constant to a final temperature of 200°C. Determine the work done during the process. Solution: From Eq. 4.4 for the polytopic process PVn = const ( n = 1 ) / 2 P2V2 - P1V1 W2 = ⌠ PdV = 1 ⌡ 1-n 1 Assuming ideal gas, PV = mRT mR(T2 - T1) , 1W2 = 1-n P1V1 300 × 0.2 kPa m3 But mR = T = 373.15 K = 0.1608 kJ/K 1 1W2 = 0.1608(473.2 - 373.2) kJ K 1 - 1.2 K = -80.4 kJ Sonntag, Borgnakke and van Wylen 4.49 A gas initially at 1 MPa, 500°C is contained in a piston and cylinder arrangement with an initial volume of 0.1 m3. The gas is then slowly expanded according to the relation PV = constant until a final pressure of 100 kPa is reached. Determine the work for this process. Solution: By knowing the process and the states 1 and 2 we can find the relation between the pressure and the volume so the work integral can be performed. Process: PV = C ⇒ V2 = P1V1/P2 = 1000 × 0.1/100 = 1 m3 For this process work is integrated to Eq.4.5 1W2 ⌠ = ∫ P dV = ⌡ CV-1dV = C ln(V2/V1) V2 1W2 = P1V1 ln V = 1000 × 0.1 ln (1/0.1) 1 = 230.3 kJ P 1 2 W V Sonntag, Borgnakke and van Wylen 4.50 Helium gas expands from 125 kPa, 350 K and 0.25 m3 to 100 kPa in a polytropic process with n = 1.667. How much work does it give out? Solution: Process equation: n n n PV = constant = P1V1 = P2V2 Solve for the volume at state 2 V2 = V1 (P1/P2) 1/n 1250.6 = 0.25 × 100 = 0.2852 m3 Work from Eq.4.4 P2V2- P1 V1 100× 0.2852 - 125× 0.25 W2 = = kPa m3 = 4.09 kJ 1 1 - 1.667 1-n Sonntag, Borgnakke and van Wylen 4.51 Air goes through a polytropic process from 125 kPa, 325 K to 300 kPa and 500 K. Find the polytropic exponent n and the specific work in the process. Solution: n n Process: Pvn = Const = P1v1 = P2 v2 Ideal gas Pv = RT so RT 0.287 × 325 = 0.7462 m3/kg v1 = P = 125 RT 0.287 × 500 v2 = P = = 0.47833 m3/kg 300 From the process equation (P2/ P1) = (v1/ v2) n => ln(P2/ P1) = n ln(v1/ v2) ln 2.4 n = ln(P2/ P1) / ln(v1/ v2) = ln 1.56 = 1.969 The work is now from Eq.4.4 per unit mass P2v2-P1v1 R(T2 - T1) 0.287(500 - 325) = = = -51.8 kJ/kg 1w2 = 1-n 1-n 1-1.969 Sonntag, Borgnakke and van Wylen 4.52 A piston cylinder contains 0.1 kg air at 100 kPa, 400 K which goes through a polytropic compression process with n = 1.3 to a pressure of 300 kPa. How much work has the air done in the process? Solution: Process: Pvn = Const. T2 = T1 ( P2 V2 / P1V1) = T1 ( P2 / P1)(P1 / P2 ) 1/n (1 - 1/1.3) = 400 × (300/100) = 515.4 K Work term is already integrated giving Eq.4.4 1 mR = (P V - P1V1) = (T -T ) 1W2 1−n 2 2 1−n 2 1 = 0.2 × 0.287 × (515.4-400) = -477 kJ 1 − 1.3 P2 1 W n=1 V Since Ideal gas, Sonntag, Borgnakke and van Wylen 4.53 A balloon behaves so the pressure is P = C2 V1/3, C2 = 100 kPa/m. The balloon is blown up with air from a starting volume of 1 m3 to a volume of 3 m3. Find the final mass of air assuming it is at 25oC and the work done by the air. Solution: The process is polytropic with exponent n = -1/3. P P1 = C2 V1/3 = 100 × 11/3 = 100 kPa P2 = C2 V1/3 = 100 × 31/3 = 144.22 kPa 1W2 = ∫ P dV = P2V2 - P1V1 1-n (Equation 4.4) 144.22 × 3 - 100 × 1 = 249.5 kJ 1 - (-1/3) P2V2 144.22 × 3 = 5.056 kg m2 = RT = 0.287 × 298 2 = 2 1 W V Sonntag, Borgnakke and van Wylen 4.54 A balloon behaves such that the pressure inside is proportional to the diameter squared. It contains 2 kg of ammonia at 0°C, 60% quality. The balloon and ammonia are now heated so that a final pressure of 600 kPa is reached. Considering the ammonia as a control mass, find the amount of work done in the process. Solution: Process : P ∝ D2, with V ∝ D3 this implies P ∝ D2 ∝ V2/3 so PV -2/3 = constant, which is a polytropic process, n = −2/3 From table B.2.1: V1 = mv1 = 2(0.001566 + 0.6 × 0.28783) = 0.3485 m3 P23/2 600 3/2 V2 = V1 P = 0.3485 429.3 = 0.5758 m3 1 P2V2 - P1V1 (Equation 4.4) 1W2 = ∫ P dV = 1-n = 600 × 0.5758 - 429.3 × 0.3485 = 117.5 kJ 1 - (-2/3) P 2 1 W V Sonntag, Borgnakke and van Wylen 4.55 Consider a piston cylinder with 0.5 kg of R-134a as saturated vapor at -10°C. It is now compressed to a pressure of 500 kPa in a polytropic process with n = 1.5. Find the final volume and temperature, and determine the work done during the process. Solution: Take CV as the R-134a which is a control mass. Process: Pv 1.5 = constant m2 = m1 = m until P = 500 kPa 1: (T, x) v1 = 0.09921 m3/kg, P = Psat = 201.7 kPa from Table B.5.1 (1/1.5) 2/3 2: (P, process) v2 = v1 (P1/P2) = 0.09921× (201.7/500) = 0.05416 Given (P, v) at state 2 from B.5.2 it is superheated vapor at T2 = 79°C Process gives P = C v 1W2 = -1.5 , which is integrated for the work term, Eq.(4.4) ∫ P dV = 1 -m (P2v2 - P1v1) 1.5 2 = - 0.5 × (500 × 0.05416 - 201.7 × 0.09921) = -7.07 kJ Sonntag, Borgnakke and van Wylen 4.56 Consider the process described in Problem 3.98. With 1 kg water as a control mass, determine the boundary work during the process. A spring-loaded piston/cylinder contains water at 500°C, 3 MPa. The setup is such that pressure is proportional to volume, P = CV. It is now cooled until the water becomes saturated vapor. Sketch the P-v diagram and find the final pressure. Solution : State 1: Table B.1.3: v1 = 0.11619 m3/kg Process: m is constant and P = C0V = C0m v = C v P = Cv ⇒ C = P1/v1 = 3000/0.11619 = 25820 kPa kg/m3 State 2: x2 = 1 & P2 = Cv2 (on process line) P Trial & error on T2sat or P2sat: 1 Here from B.1.2: 2 at 2 MPa vg = 0.09963 ⇒ C = P/vg = 20074 (low) 2.5 MPa vg = 0.07998 ⇒ C = P/vg = 31258 (high) C v 2.25 MPa vg = 0.08875 ⇒ C = P/vg = 25352 (low) Now interpolate to match the right slope C: P2 = 2270 kPa, v2 = P2/C = 2270/25820 = 0.0879 m3/kg P is linear in V so the work becomes (area in P-v diagram) 1 1w2 = ∫ P dv = 2(P1 + P2)(v2 - v1) 1 = 2 (3000 + 2270)(0.0879 - 0.11619) = - 74.5 kJ/kg Sonntag, Borgnakke and van Wylen 4.57 Find the work for Problem 3.106. Refrigerant-12 in a piston/cylinder arrangement is initially at 50°C, x = 1. It is then expanded in a process so that P = Cv−1 to a pressure of 100 kPa. Find the final temperature and specific volume. Solution: Knowing the process (P versus V) and states 1 and 2 allows calculation of W. State 1: 50°C, x=1 Table B.3.1: P1 = 1219.3 kPa, v1 = 0.01417 m3/kg Process: v2 P = Cv-1 ⇒ 1w2 = ∫ P dv = C ln v 1 State 2: 100 kPa and on process curve: same as Eq.4.5 v2 = v1P1/P2 = 0.1728 m3/kg From table B.3.2 T = - 13.2°C The constant C for the work term is P1v1 so per unit mass we get v2 0.1728 w2 = P1v1 ln v = 1219.3 × 0.01417 × ln 0.01417 = 43.2 kJ/kg 1 1 T P 1 1 2 2 v Notice T is not constant. It is not an ideal gas in this range. v Sonntag, Borgnakke and van Wylen 4.58 A piston/cylinder contains water at 500°C, 3 MPa. It is cooled in a polytropic process to 200°C, 1 MPa. Find the polytropic exponent and the specific work in the process. Solution: Polytropic process: Pvn = C n n Both states must be on the process line: P2v2 = C = P1v1 Take the ratio to get: P1 v2 P2 = v1 n n and then take ln of the ratio: P1 v2 v2 ln P = ln v = n ln v 2 1 1 now solve for the exponent n P1 v2 1.0986 n = ln P ln v = 0.57246 = 1.919 2 1 / 1w2 = ∫ P dv = = P2v2 - P1v1 1-n (Equation 4.4) 1000 × 0.20596 - 3000 × 0.11619 = 155.2 kJ 1 - 1.919 Sonntag, Borgnakke and van Wylen 4.59 Consider a two-part process with an expansion from 0.1 to 0.2 m3 at a constant pressure of 150 kPa followed by an expansion from 0.2 to 0.4 m3 with a linearly rising pressure from 150 kPa ending at 300 kPa. Show the process in a P-V diagram and find the boundary work. Solution: By knowing the pressure versus volume variation the work is found. If we plot the pressure versus the volume we see the work as the area below the process curve. P 3 300 150 1 2 V 0.1 0.2 0.4 2 3 W3 = 1W2 + 2W3 = ⌠ PdV + ⌠ PdV 1 ⌡ ⌡ 1 2 1 = P1 (V2 – V1) + 2 (P2 + P3)(V3-V2) 1 = 150 (0.2-1.0) + 2 (150 + 300) (0.4 - 0.2) = 15 + 45 = 60 kJ Sonntag, Borgnakke and van Wylen 4.60 A cylinder containing 1 kg of ammonia has an externally loaded piston. Initially the ammonia is at 2 MPa, 180°C and is now cooled to saturated vapor at 40°C, and then further cooled to 20°C, at which point the quality is 50%. Find the total work for the process, assuming a piecewise linear variation of P versus V. Solution: P 1 cb 2000 2 1555 3 857 State 1: (T, P) 180 o C 40 oC o 20 C Table B.2.2 v1 = 0.10571 m3/kg State 2: (T, x) Table B.2.1 sat. vap. P2 = 1555 kPa, v2 = 0.08313 m3/kg v State 3: (T, x) P3 = 857 kPa, v3 = (0.001638 + 0.14922)/2 = 0.07543 m3/kg Sum the the work as two integrals each evaluated by the area in the P-v diagram. 3 ⌠ 1W3 = ⌡ PdV ≈ ( 1 = P2 + P3 P1 + P2 ) m(v2 - v1) + ( ) m(v3 - v2) 2 2 1555 + 857 2000 + 1555 1(0.08313 - 0.10571) + 1(0.07543 - 0.08313) 2 2 = -49.4 kJ Sonntag, Borgnakke and van Wylen 4.61 A piston/cylinder arrangement shown in Fig. P4.61 initially contains air at 150 kPa, 400°C. The setup is allowed to cool to the ambient temperature of 20°C. a. Is the piston resting on the stops in the final state? What is the final pressure in the cylinder? b. What is the specific work done by the air during this process? Solution: State 1: State 2: P1 = 150 kPa, T1 = 400°C = 673.2 K T2 = T0 = 20°C = 293.2 K For all states air behave as an ideal gas. a) If piston at stops at 2, V2 = V1/2 and pressure less than Plift = P1 V1 T2 293.2 ⇒ P2 = P1 × V × T = 150 × 2 × 673.2 = 130.7 kPa < P1 2 1 ⇒ Piston is resting on stops at state 2. b) Work done while piston is moving at constant Pext = P1. 1W2 = ∫ Pext dV = P1 (V2 - V1) 1 1 1 V2 = 2 V1 = 2 m RT1/P1 ; 1 1w2 = 1W2/m = RT1 (2 - 1 ) = -2 × 0.287 × 673.2 = -96.6 kJ/kg P 1a T 1 T1 P1 T1a P2 2 V T2 1 1a 2 V Sonntag, Borgnakke and van Wylen 4.62 A piston cylinder has 1.5 kg of air at 300 K and 150 kPa. It is now heated up in a two step process. First constant volume to 1000 K (state 2) then followed by a constant pressure process to 1500 K, state 3. Find the final volume and the work in the process. Solution: P The two processes are: 1 -> 2: Constant volume V2 = V1 2 -> 3: Constant pressure P3 = P2 2 P2 P1 3 1 V Use ideal gas approximation for air. State 1: T, P => State 2: V2 = V1 V1 = mRT1/P1 = 1.5×0.287×300/150 = 0.861 m3 => P2 = P1 (T2/T1) = 150×1000/300 = 500 kPa => V3 = V2 (T3/T2) = 0.861×1500/1000 = 1.2915 m3 State 3: P3 = P2 We find the work by summing along the process path. 1W3 = 1W2 + 2W3 = 2W3 = P3(V3 - V2) = 500(1.2915 - 0.861) = 215.3 kJ Sonntag, Borgnakke and van Wylen 4.63 A piston/cylinder assembly (Fig. P4.63) has 1 kg of R-134a at state 1 with 110°C, 600 kPa, and is then brought to saturated vapor, state 2, by cooling while the piston is locked with a pin. Now the piston is balanced with an additional constant force and the pin is removed. The cooling continues to a state 3 where the R-134a is saturated liquid. Show the processes in a P-V diagram and find the work in each of the two steps, 1 to 2 and 2 to 3. Solution : CV R-134a This is a control mass. Properties from table B.5.1 and 5.2 State 1: (T,P) B.5.2 => v = 0.04943 m3/kg State 2: given by fixed volume v2 = v1 and x2 = 1.0 v2 = v1 = vg = 0.04943 m3/kg State 3 reached at constant P (F = constant) so from B.5.1 => T = 10°C v3 = vf = 0.000794 m3/kg P 1 3 cb 2 V Since no volume change from 1 to 2 => 1W2 = 0 Constant pressure 2W3 = ∫P dV = P(V3 -V2) = mP(v3 -v2) = 415.8 (0.000794 - 0.04943) 1 = -20.22 kJ Sonntag, Borgnakke and van Wylen 4.64 The refrigerant R-22 is contained in a piston/cylinder as shown in Fig. P4.64, where the volume is 11 L when the piston hits the stops. The initial state is −30°C, 150 kPa with a volume of 10 L. This system is brought indoors and warms up to 15°C. a. Is the piston at the stops in the final state? b. Find the work done by the R-22 during this process. Solution: Initially piston floats, V < Vstop so the piston moves at constant Pext = P1 until it reaches the stops or 15°C, whichever is first. a) From Table B.4.2: v1 = 0.1487 m3/kg, 0.010 m = V/v = 0.1487 = 0.06725 kg Po P 2 mp P1 1 1a R-22 V V stop Check the temperature at state 1a: P1a = 150 kPa, v = Vstop/m. 0.011 v1a = V/m = 0.06725 = 0.16357 m3/kg => T1a = -9°C & T2 = 15°C Since T2 > T1a then it follows that P2 > P1 and the piston is against stop. b) Work done at constant Pext = P1. 1W2 = ∫ Pext dV = Pext(V2 - V1) = 150(0.011 - 0.010) = 0.15 kJ Sonntag, Borgnakke and van Wylen 4.65 A piston/cylinder contains 50 kg of water at 200 kPa with a volume of 0.1 m3. Stops in the cylinder restricts the enclosed volume to 0.5 m3, similar to the setup in Problem 4.7. The water is now heated to 200°C. Find the final pressure, volume and the work done by the water. Solution: Initially the piston floats so the equilibrium lift pressure is 200 kPa 1: 200 kPa, v1= 0.1/50 = 0.002 m3/kg, 2: 200°C, on line Check state 1a: vstop = 0.5/50 = 0.01 m3/kg P 2 P1 1 1a V V stop => Table B.1.2: 200 kPa , vf < vstop < vg State 1a is two phase at 200 kPa and Tstop ≈ 120.2 °C so as T2 > Tstop the state is higher up in the P-V diagram with v2 = vstop < vg = 0.127 m3/kg (at 200°C) State 2 two phase => P2 = Psat(T2) = 1.554 MPa, V2 = Vstop = 0.5 m3 1W2 = 1Wstop = 200 (0.5 – 0.1) = 80 kJ Sonntag, Borgnakke and van Wylen 4.66 Find the work for Problem 3.108. Ammonia in a piston/cylinder arrangement is at 700 kPa, 80°C. It is now cooled at constant pressure to saturated vapor (state 2) at which point the piston is locked with a pin. The cooling continues to −10°C (state 3). Show the processes 1 to 2 and 2 to 3 on both a P–v and T–v diagram. Solution : T P 2 700 1 80 1 2 14 290 3 cb v -10 3 v 2 W3 = 1W2 + 2W3 = ⌠ PdV = P1(V2 - V1) = mP1(v2 - v1) 1 ⌡ 1 Since constant volume from 2 to 3, see P-v diagram. From table B.2 v1 = 0.2367 m3/kg, P1 = 700 kPa, v2 = vg = 0.1815 m3/kg 1w3 = P1(v2- v1) = 700 × (0.1815 - 0.2367) = -38.64 kJ/kg Sonntag, Borgnakke and van Wylen 4.67 A piston/cylinder contains 1 kg of liquid water at 20°C and 300 kPa. Initially the piston floats, similar to the setup in Problem 4.64, with a maximum enclosed volume of 0.002 m3 if the piston touches the stops. Now heat is added so a final pressure of 600 kPa is reached. Find the final volume and the work in the process. Solution: Take CV as the water which is a control mass: m2 = m1 = m ; Table B.1.1: 20°C => Psat = 2.34 kPa State 1: Compressed liquid v = vf(20) = 0.001002 m3/kg State 1a: vstop = 0.002 m3/kg , 300 kPa State 2: Since P2 = 600 kPa > Plift then piston is pressed against the stops v2 = vstop = 0.002 m3/kg and V = 0.002 m3 For the given P : vf < v < vg so 2-phase T = Tsat = 158.85 °C Work is done while piston moves at Plift = constant = 300 kPa so we get 1W2 = ∫ P dV = m Plift(v2 -v1) = 1 × 300(0.002 − 0.001002) = 0.30 kJ P P o cb H2O 1 2 V Sonntag, Borgnakke and van Wylen 4.68 10 kg of water in a piston cylinder arrangement exists as saturated liquid/vapor at 100 kPa, with a quality of 50%. It is now heated so the volume triples. The mass of the piston is such that a cylinder pressure of 200 kPa will float it. a) Find the final temperature and volume of the water. b) Find the work given out by the water. Solution: Take CV as the water m2 = m1 = m; Process: v = constant until P = Plift then P is constant. State 1: v1 = vf + x vfg = 0.001043 + 0.5 × 1.69296 = 0.8475 m3/kg State 2: v2, P2 ≤ Plift => v2 = 3 × 0.8475 = 2.5425 m3/kg; T2 = 829°C ; V2 = m v2 = 25.425 m3 1W2 = ∫ P dV = Plift × (V2 - V1) = 200 kPa × 10 kg × (2.5425 – 0.8475) m3/kg = 3390 kJ P Po 2 P2 cb H2O P1 1 V Sonntag, Borgnakke and van Wylen 4.69 Find the work in Problem 3.43. Ammonia at 10oC with a mass of 10 kg is in a piston cylinder arrangement with an initial volume of 1 m3. The piston initially resting on the stops has a mass such that a pressure of 900 kPa will float it. The ammonia is now slowly heated to 50oC. Find the work in the process. C.V. Ammonia, constant mass. Process: V = constant unless P = Pfloat P V1 State 1: T = 10oC, v1 = m = 10 = 0.1 m3/kg From Table B.2.1 1a 2 P2 vf < v < vg x1 = (v - vf)/vfg = (0.1 - 0.0016)/0.20381 P1 = 0.4828 1 V cb State 1a: P = 900 kPa, v = v1 = 0.1 < vg at 900 kPa This state is two-phase T1a = 21.52oC Since T2 > T1a then v2 > v1a State 2: 50oC and on line(s) means 900 kPa which is superheated vapor. From Table B.2.2 linear interpolation between 800 and 1000 kPa: v2 = 0.1648 m3/kg, 1W2 = V2 = mv2 = 1.648 m3 ∫ P dV = Pfloat (V2 - V1) = 900 (1.648 - 1.0) = 583.2 kJ Sonntag, Borgnakke and van Wylen 4.70 A piston cylinder setup similar to Problem 4.68 contains 0.1 kg saturated liquid and vapor water at 100 kPa with quality 25%. The mass of the piston is such that a pressure of 500 kPa will float it. The water is heated to 300°C. Find the final pressure, volume and the work, 1W2. Solution: P Take CV as the water: m2 = m1 = m 1a 2 Process: v = constant until P = Plift To locate state 1: Table B.1.2 v1 = 0.001043 + 0.25×1.69296 = 0.42428 m3/kg 1a: v1a = v1 = 0.42428 m3/kg > vg at 500 kPa Plift P1 1 cb V so state 1a is Sup.Vapor T1a = 200°C State 2 is 300°C so heating continues after state 1a to 2 at constant P => 2: T2, P2 = Plift => Tbl B.1.3 v2 =0.52256 m3/kg ; V2 = mv2 = 0.05226 m3 1W2 = Plift (V2 - V1) = 500(0.05226 - 0.04243) = 4.91 kJ Sonntag, Borgnakke and van Wylen Other types of work and general concepts 4.71 A 0.5-m-long steel rod with a 1-cm diameter is stretched in a tensile test. What is the required work to obtain a relative strain of 0.1%? The modulus of elasticity of steel is 2 × 108 kPa. Solution : −1W2 = −1W2 = AEL0 2 (e)2, π A = 4 (0.01)2 = 78.54 × 10-6 m2 78.54×10-6 × 2×108 × 0.5 (10-3)2 = 3.93 J 2 Sonntag, Borgnakke and van Wylen 4.72 A copper wire of diameter 2 mm is 10 m long and stretched out between two posts. The normal stress (pressure) σ = E(L – Lo)/Lo , depends on the length L versus the unstretched length Lo and Young’s modulus E = 1.1 × 106 kPa. The force is F = Aσ and measured to be 110 N. How much longer is the wire and how much work was put in? Solution: F = As = A E ∆L/ Lo and ∆L = FLo /AE π π A = 4D2 = 4 × 0.0022 = 3.142 ×10-6 m2 ∆L = 110 ×10 = 0.318 m 3.142×10-6 ×1.1 ×106 ×103 x 1W2 = ∫ F dx = ∫ A s dx = ∫ AE L dx o AE = L ½ x2 o = where x = L - Lo 3.142×10 -6 ×1.1 ×106 ×103 × ½ × 0.3182 = 17.47 J 10 Sonntag, Borgnakke and van Wylen 4.73 A film of ethanol at 20°C has a surface tension of 22.3 mN/m and is maintained on a wire frame as shown in Fig. P4.73. Consider the film with two surfaces as a control mass and find the work done when the wire is moved 10 mm to make the film 20 × 40 mm. Solution : Assume a free surface on both sides of the frame, i.e., there are two surfaces 20 × 30 mm W = −⌠ S dA = −22.3×10-3 × 2(800 − 600)×10-6 ⌡ = −8.92×10-6 J = -8.92 µJ Sonntag, Borgnakke and van Wylen 4.74 Assume a balloon material with a constant surface tension of S = 2 N/m. What is the work required to stretch a spherical balloon up to a radius of r = 0.5 m? Neglect any effect from atmospheric pressure. Assume the initial area is small, and that we have 2 surfaces inside and out W = -∫ S dA = -S (A2 − A1) 2 = - S(A2) = -S( 2× π D2 ) = -2 × 2 × π × 1 = -12.57 J Win = -W = 12.57 J Sonntag, Borgnakke and van Wylen 4.75 A soap bubble has a surface tension of S = 3 × 10-4 N/cm as it sits flat on a rigid ring of diameter 5 cm. You now blow on the film to create a half sphere surface of diameter 5 cm. How much work was done? 1W2 = ∫ F dx = ∫ S dA = S ∆A π π = 2 × S × ( 2 D2 - 4 D 2) π = 2 × 3 × 10-4 × 100 × 2 0.052 ( 1- 0.5 ) = 1.18 × 10-4 J Notice the bubble has 2 surfaces. π A1 = 4 D 2 , A2 = ½ π D2 Sonntag, Borgnakke and van Wylen 4.76 Assume we fill a spherical balloon from a bottle of helium gas. The helium gas ∫ PdV that stretches the balloon material ∫ S dA and pushes back the atmosphere ∫ Po dV. Write the incremental balance for dWhelium = dWstretch + provides work dWatm to establish the connection between the helium pressure , the surface tension S and Po as a function of radius. WHe = ∫ P dV = ∫ S dA + ∫ Po dV dWHe = P dV = S dA + Po dV π π dV = d ( 6 D3 ) = 6 × 3D2 dD dA = d ( 2 × π × D2) = 2π (2D) dD π π P 2 D2 dD = S (4π)D dD + Po2 D2 dD PHe = Po + 8 S D Sonntag, Borgnakke and van Wylen 4.77 A sheet of rubber is stretched out over a ring of radius 0.25 m. I pour liquid water at 20oC on it so the rubber forms a half sphere (cup). Neglect the rubber mass and find the surface tension near the ring? Solution: F ↑ = F ↓ ; F ↑ = SL The length is the perimeter, 2πr, and there is two surfaces 1 2 S × 2 × 2πr = mH2o g = ρH2o Vg = ρH2o× 12 π (2r) 3g = ρH2o× π 3 r 3 1 1 S = ρH2o 6 r2 g = 997 × 6 × 0.252 × 9.81 = 101.9 N/m Sonntag, Borgnakke and van Wylen 4.78 Consider a window-mounted air conditioning unit used in the summer to cool incoming air. Examine the system boundaries for rates of work and heat transfer, including signs. Solution : Air-conditioner unit, steady operation with no change of temperature of AC unit. Cool side Inside 25°C 15°C Hot side Outside C 30°C 37°C - electrical work (power) input operates unit, +Q rate of heat transfer from the room, a larger -Q rate of heat transfer (sum of the other two energy rates) out to the outside air. Sonntag, Borgnakke and van Wylen 4.79 Consider a hot-air heating system for a home. Examine the following systems for heat transfer. a) The combustion chamber and combustion gas side of the heat transfer area. b) The furnace as a whole, including the hot- and cold-air ducts and chimney. Solution: a) Fuel and air enter, warm products of the combustion exit, large -Q to the air in the duct system, small -Q loss directly to the room. b) Fuel and air enter, warm products exit through the chimney, cool air into the cold air return duct, warm air exit hot-air duct to heat the house. Small heat transfer losses from furnace, chimney and ductwork to the house. Sonntag, Borgnakke and van Wylen 4.80 Consider a household refrigerator that has just been filled up with roomtemperature food. Define a control volume (mass) and examine its boundaries for rates of work and heat transfer, including sign. a. Immediately after the food is placed in the refrigerator b. After a long period of time has elapsed and the food is cold Solution: I. C.V. Food. a) short term.: -Q from warm food to cold refrigerator air. Food cools. b) Long term: -Q goes to zero after food has reached refrigerator T. II. C.V. refrigerator space, not food, not refrigerator system a) short term: +Q from the warm food, +Q from heat leak from room into cold space. -Q (sum of both) to refrigeration system. If not equal the refrigerator space initially warms slightly and then cools down to preset T. b) long term: small -Q heat leak balanced by -Q to refrigeration system. Note: For refrigeration system CV any Q in from refrigerator space plus electrical W input to operate system, sum of which is Q rejected to the room. Sonntag, Borgnakke and van Wylen 4.81 A room is heated with an electric space heater on a winter day. Examine the following control volumes, regarding heat transfer and work , including sign. a) The space heater. b) Room c) The space heater and the room together Solution: a) The space heater. Electrical work (power) input, and equal (after system warm up) Q out to the room. b) Room Q input from the heater balances Q loss to the outside, for steady (no temperature change) operation. c) The space heater and the room together Electrical work input balances Q loss to the outside, for steady operation. Sonntag, Borgnakke and van Wylen Rates of work 4.82 An escalator raises a 100 kg bucket of sand 10 m in 1 minute. Determine the rate of work done during the process. Solution: The work is a force with a displacement and force is constant: F = mg W= ∫ F dx = F ∫ dx = F ∆x = 100 kg × 9.80665 m/s2 × 10 m = 9807 J The rate of work is work per unit time . W 9807 J W = = 60 s = 163 W ∆t Sonntag, Borgnakke and van Wylen 4.83 A car uses 25 hp to drive at a horizontal level at constant 100 km/h. What is the traction force between the tires and the road? Solution: We need to relate the rate of work to the force and velocity dW . dx dW = F dx => dt = W = F dt = FV . F=W/V . W = 25 hp = 25 × 0.7355 kW = 18.39 kW 1000 V = 100 × 3600 = 27.78 m/s . F = W / V = (18.39 / 27.78) kN = 0.66 kN Units: kW / (ms−1) = kW s m−1 = kJ s−1s m−1 = kN m m−1 = kN Sonntag, Borgnakke and van Wylen 4.84 A piston/cylinder of cross sectional area 0.01 m2 maintains constant pressure. It contains 1 kg water with a quality of 5% at 150oC. If we heat so 1 g/s liquid turns into vapor what is the rate of work out? Vvapor = mvapor vg , Vliq = mliq vf mtot = constant = mvapor mliq Vtot = Vvapor + Vliq . . . = 0 = mvapor + mliq mtot ⇒ . . mliq = -mvapor . . . . . Vtot = Vvapor + Vliq = mvaporvg + mliqvf . . = mvapor (vg- vf ) = mvapor vfg . . . W = PV = P mvapor vfg = 475.9 × 0.001 × 0.39169 = 0.1864 kW = 186 W Sonntag, Borgnakke and van Wylen 4.85 A crane lifts a bucket of cement with a total mass of 450 kg vertically up with a constant velocity of 2 m/s. Find the rate of work needed to do that. Solution: Rate of work is force times rate of displacement. The force is due to gravity (a = 0) alone. . W = FV = mg × V = 450 kg × 9.807 ms−2 × 2 ms−1 = 8826 J/s . W = 8.83 kW Sonntag, Borgnakke and van Wylen 4.86 Consider the car with the rolling resistance as in problem 4.27. How fast can it drive using 30 hp? F = 0.006 mg . Power = F × V = 30 hp = W . . W 30 ×0.7457 ×1000 V = W / F = 0.006 mg = = 271.5 m/s 0.006 ×1200 ×9.81 Comment : This is a very high velocity, the rolling resistance is low relative to the air resistance. Sonntag, Borgnakke and van Wylen 4.87 Consider the car with the air drag force as in problem 4.29. How fast can it drive using 30 hp? 1P 100 kg ρ = v = RT = = 1.2015 3 and A = 4 m2 0.287 ×290 m Fdrag = 0.225 A ρ V2 . Power for drag force: Wdrag = 30 hp × 0.7457 = 22.371 kW . Wdrag = Fdrag V = 0.225 × 4 × 1.2015 × V3 . V3 = Wdrag /(0.225 × 4 × 1.2015) = 20 688 Drag force: 3600 V = 27.452 m/s = 27.452 × 1000 = 98.8 km/h Sonntag, Borgnakke and van Wylen 4.88 Consider a 1400 kg car having the rolling resistance as in problem 4.27 and air resistance as in problem 4.29. How fast can it drive using 30 hp? Ftot = Frolling + Fair = 0.006 mg + 0.225 AρV2 m = 1400 kg , A = 4 m2 ρ = P/RT = 1.2015 kg/m3 . W = FV = 0.006 mgV + 0.225 ρAV3 Nonlinear in V so solve by trial and error. . W = 30 hp = 30 × 0.7355 kW = 22.06 kW = 0.0006 × 1400 × 9.807 V + 0.225 × 1.2015 × 4 V3 = 82.379V + 1.08135 V3 . V = 25 m/s ⇒ W = 18 956 W . V = 26 m/s W = 21 148 W . V = 27 m/s W = 23508 W Linear interpolation V = 26.4 m/s = 95 km/h Sonntag, Borgnakke and van Wylen 4.89 A battery is well insulated while being charged by 12.3 V at a current of 6 A. Take the battery as a control mass and find the instantaneous rate of work and the total work done over 4 hours. Solution : Battery thermally insulated ⇒ Q = 0 For constant voltage E and current i, Power = E i = 12.3 × 6 = 73.8 W [Units V × A = W] W = ∫ power dt = power ∆t = 73.8 × 4 × 60 × 60 = 1 062 720 J = 1062.7 kJ Sonntag, Borgnakke and van Wylen 4.90 A current of 10 amp runs through a resistor with a resistance of 15 ohms. Find the rate of work that heats the resistor up. Solution: . W = power = E i = R i2 = 15 × 10 × 10 = 1500 W R Sonntag, Borgnakke and van Wylen 4.91 A pressure of 650 kPa pushes a piston of diameter 0.25 m with V = 5 m/s. What is the volume displacement rate, the force and the transmitted power? π A = 4 D2 = 0.049087 m2 . V = AV = 0049087 m2 × 5 m/s = 0.2454 m3/s . . W = power = F V = P V = 650 kPa × 0.2454 m3/s = 159.5 kW P V Sonntag, Borgnakke and van Wylen 4.92 Assume the process in Problem 4.37 takes place with a constant rate of change in volume over 2 minutes. Show the power (rate of work) as a function of time. Solution: W = ∫ P dV since 2 min = 120 secs . W = P (∆V / ∆t) (∆V / ∆t) = 0.3 / 120 = 0.0025 m3/s P kW 3 300 150 1 W 3 0.75 2 0.375 1 2 V 0.1 0.2 0.4 V 0.1 0.2 0.4 Sonntag, Borgnakke and van Wylen 4.93 Air at a constant pressure in a piston cylinder is at 300 kPa, 300 K and a volume of 0.1 m3. It is heated to 600 K over 30 seconds in a process with constant piston velocity. Find the power delivered to the piston. Solution: . . Process: P = constant : dW = P dV => W = PV V2 = V1× (T2/T1) = 0.1 × (600/300) = 0.2 . W = P (∆V / ∆t) = 300 × (0.2-0.1)/30 = 1 kW Sonntag, Borgnakke and van Wylen 4.94 A torque of 650 Nm rotates a shaft of diameter 0.25 m with ω = 50 rad/s. What are the shaft surface speed and the transmitted power? Solution: V = ωr = ωD/2 = 50 × 0.25 / 2 = 6.25 m/s Power = Tω = 650 × 50 Nm/s = 32 500 W = 32.5 kW Sonntag, Borgnakke and van Wylen Heat Transfer rates 4.95 The sun shines on a 150 m2 road surface so it is at 45°C. Below the 5 cm thick asphalt, average conductivity of 0.06 W/m K, is a layer of compacted rubbles at a temperature of 15°C. Find the rate of heat transfer to the rubbles. Solution : This is steady one dimensional conduction through the asphalt layer. . ∆T Q=k A ∆x 45-15 = 0.06 × 150 × 0.05 = 5400 W Sonntag, Borgnakke and van Wylen 4.96 A pot of steel, conductivity 50 W/m K, with a 5 mm thick bottom is filled with 15°C liquid water. The pot has a diameter of 20 cm and is now placed on an electric stove that delivers 250 W as heat transfer. Find the temperature on the outer pot bottom surface assuming the inner surface is at 15°C. Solution : Steady conduction through the bottom of the steel pot. Assume the inside surface is at the liquid water temperature. . . ∆T Q=k A ⇒ ∆Τ = Q ∆x / kΑ ∆x π ∆T = 250 × 0.005/(50 × 4 × 0.22) = 0.796 T = 15 + 0.796 ≅ 15.8°C cb Sonntag, Borgnakke and van Wylen 4.97 A water-heater is covered up with insulation boards over a total surface area of 3 m2. The inside board surface is at 75°C and the outside surface is at 20°C and the board material has a conductivity of 0.08 W/m K. How thick a board should it be to limit the heat transfer loss to 200 W ? Solution : Steady state conduction through a single layer board. . . ∆T Q cond = k A ⇒ ∆x = k Α ∆Τ/Q ∆x ∆x = 0.08 × 3 × 75 − 20 200 = 0.066 m Sonntag, Borgnakke and van Wylen 4.98 You drive a car on a winter day with the atmospheric air at −15°C and you keep the outside front windshield surface temperature at +2°C by blowing hot air on the inside surface. If the windshield is 0.5 m2 and the outside convection coefficient is 250 W/m2K find the rate of energy loos through the front windshield. For that heat transfer rate and a 5 mm thick glass with k = 1.25 W/m K what is then the inside windshield surface temperature? Solution : The heat transfer from the inside must match the loss on the outer surface to give a steady state (frost free) outside surface temperature. . Q conv = h A ∆Τ = 250 × 0.5 × [2 − ( −15)] = 250 × 0.5 × 17 = 2125 W This is a substantial amount of power. . . ∆T Q Q cond = k A ⇒ ∆Τ = kA ∆x ∆x ∆Τ = 2125 W 0.005 m = 17 K 1.25 W/mK × 0.5 m2 Tin = Tout + ∆T = 2 + 17 = 19°C o -15 C o Windshield 2C T=? Warm air Sonntag, Borgnakke and van Wylen 4.99 A large condenser (heat exchanger) in a power plant must transfer a total of 100 MW from steam running in a pipe to sea water being pumped through the heat exchanger. Assume the wall separating the steam and seawater is 4 mm of steel, conductivity 15 W/m K and that a maximum of 5°C difference between the two fluids is allowed in the design. Find the required minimum area for the heat transfer neglecting any convective heat transfer in the flows. Solution : Steady conduction through the 4 mm steel wall. . . ∆T Q=k A ⇒ Α = Q ∆x / k∆Τ ∆x A = 100 × 10 × 0.004 / (15 × 5) = 480 m2 6 Condensing water Sea water cb Sonntag, Borgnakke and van Wylen 4.100 The black grille on the back of a refrigerator has a surface temperature of 35°C with a total surface area of 1 m2. Heat transfer to the room air at 20°C takes place with an average convective heat transfer coefficient of 15 W/m2 K. How much energy can be removed during 15 minutes of operation? Solution : . . Q = hA ∆T; Q = Q ∆t = hA ∆T ∆t Q = 15 × 1 × (35-20) × 15 × 60 = 202500 J = 202.5 kJ Sonntag, Borgnakke and van Wylen 4.101 Due to a faulty door contact the small light bulb (25 W) inside a refrigerator is kept on and limited insulation lets 50 W of energy from the outside seep into the refrigerated space. How much of a temperature difference to the ambient at 20°C must the refrigerator have in its heat exchanger with an area of 1 m2 and an average heat transfer coefficient of 15 W/m2 K to reject the leaks of energy. Solution : . Q tot = 25 + 50 = 75 W to go out . Q = hA∆T = 15 × 1 × ∆T = 75 . ∆T = Q / hA = 75/(15×1) = 5 °C OR T must be at least 25 °C Sonntag, Borgnakke and van Wylen 4.102 The brake shoe and steel drum on a car continuously absorbs 25 W as the car slows down. Assume a total outside surface area of 0.1 m2 with a convective heat transfer coefficient of 10 W/m2 K to the air at 20°C. How hot does the outside brake and drum surface become when steady conditions are reached? Solution : . . Q = hA∆Τ ⇒ ∆Τ = Q / hA ∆T = ( ΤBRAKE − 20 ) = 25/(10 × 0.1) = 25 °C TBRAKE = 20 + 25 = 45°C Sonntag, Borgnakke and van Wylen 4.103 A wall surface on a house is at 30°C with an emissivity of ε = 0.7. The surrounding ambient to the house is at 15°C, average emissivity of 0.9. Find the rate of radiation energy from each of those surfaces per unit area. Solution : . –8 Q /A = εσAT4, σ =5.67 × 10 . -8 a) Q/A = 0.7 × 5.67 × 10 × ( 273.15 + 30)4 = 335 W/m2 . -8 4 b) Q/A = 0.9 × 5.67 × 10 × 288.15 = 352 W/m2 Sonntag, Borgnakke and van Wylen 4.104 A log of burning wood in the fireplace has a surface temperature of 450°C. Assume the emissivity is 1 (perfect black body) and find the radiant emission of energy per unit surface area. Solution : . Q /A = 1 × σ T4 –8 4 = 5.67 × 10 × ( 273.15 + 450) = 15505 W/m2 = 15.5 kW/m2 Sonntag, Borgnakke and van Wylen 4.105 A radiant heat lamp is a rod, 0.5 m long and 0.5 cm in diameter, through which 400 W of electric energy is deposited. Assume the surface has an emissivity of 0.9 and neglect incoming radiation. What will the rod surface temperature be ? Solution : For constant surface temperature outgoing power equals electric power. . . 4 Qrad = εσAT = Qel ⇒ . 4 –8 T = Qel / εσA = 400 / (0.9 × 5.67 ×10 × 0.5 × π × 0.005) 11 4 = 9.9803 ×10 K ⇒ T ≅ 1000 K OR 725 °C Sonntag, Borgnakke and van Wylen Review Problems 4.106 A vertical cylinder (Fig. P4.106) has a 61.18-kg piston locked with a pin trapping 10 L of R-22 at 10°C, 90% quality inside. Atmospheric pressure is 100 kPa, and the cylinder cross-sectional area is 0.006 m2. The pin is removed, allowing the piston to move and come to rest with a final temperature of 10°C for the R-22. Find the final pressure, final volume and the work done by the R-22. Solution: Po State 1: (T, x) from table B.4.1 v1 = 0.0008 + 0.9 × 0.03391 = 0.03132 m3/kg mp m = V1/v1 = 0.010/0.03132 = 0.319 kg R-22 g Force balance on piston gives the equilibrium pressure 61.18 × 9.807 P2 = P0 + mPg/ AP = 100 + = 200 kPa 0.006 × 1000 State 2: (T,P) in Table B.4.2 v2 = 0.13129 m3/kg V2 = mv2 = 0.319 kg × 0.13129 m3/kg = 0.04188 m3 = 41.88 L 1W2 = ⌠Pequil dV = P2(V2-V1) = 200 kPa (0.04188- 0.010) m3 = 6.38 kJ ⌡ Sonntag, Borgnakke and van Wylen 4.107 A piston/cylinder contains butane, C4H10, at 300°C, 100 kPa with a volume of 0.02 m3. The gas is now compressed slowly in an isothermal process to 300 kPa. a. Show that it is reasonable to assume that butane behaves as an ideal gas during this process. b. Determine the work done by the butane during the process. Solution: a) T 573.15 Tr1 = T = 425.2 = 1.35; c P 100 Pr1 = P = 3800 = 0.026 c From the generalized chart in figure D.1 Z1 = 0.99 T 573.15 P 300 Tr2 = T = 425.2 = 1.35; Pr2 = P = 3800 = 0.079 c c From the generalized chart in figure D.1 Z2 = 0.98 Ideal gas model is adequate for both states. b) Ideal gas T = constant ⇒ PV = mRT = constant P1 100 W = ⌠P dV = P1V1 ln P = 100 × 0.02 × ln 300 = -2.2 kJ ⌡ 2 Sonntag, Borgnakke and van Wylen 4.108 A cylinder fitted with a piston contains propane gas at 100 kPa, 300 K with a volume of 0.2 m3. The gas is now slowly compressed according to the relation PV1.1 = constant to a final temperature of 340 K. Justify the use of the ideal gas model. Find the final pressure and the work done during the process. Solution: The process equation and T determines state 2. Use ideal gas law to say 1.1 T2 n 340 0.1 n-1 = 100 ( P2 = P1 ( T ) = 396 kPa 300 ) 1 P1 1/n 100 1/1.1 V2 = V1 ( P ) = 0.2 ( 396 ) = 0.0572 m3 2 For propane Table A.2: Tc = 370 K, Pc = 4260 kPa, Figure D.1 gives Z. Tr1 = 0.81, Pr1 = 0.023 => Z1 = 0.98 Tr2 = 0.92, Pr2 = 0.093 => Z2 = 0.95 Ideal gas model OK for both states, minor corrections could be used. The work is integrated to give Eq.4.4 1W2 = ∫ P dV = P2V2-P1V1 (396 × 0.0572) - (100 × 0.2) = = -26.7 kJ 1-n 1 - 1.1 Sonntag, Borgnakke and van Wylen 4.109 The gas space above the water in a closed storage tank contains nitrogen at 25°C, 100 kPa. Total tank volume is 4 m3, and there is 500 kg of water at 25°C. An additional 500 kg water is now forced into the tank. Assuming constant temperature throughout, find the final pressure of the nitrogen and the work done on the nitrogen in this process. Solution: The water is compressed liquid and in the process the pressure goes up so the water stays as liquid. Incompressible so the specific volume does not change. The nitrogen is an ideal gas and thus highly compressible. State 1: VH O 1 = 500 × 0.001003 2 VN 1 = 4.0 - 0.5015 2 State 2: Process: = 0.5015 m3 = 3.4985 m3 VN 2 = 4.0 - 2 × 0.5015 = 2.997 m3 2 T = C gives P1V1 = mRT = P2V2 3.4985 PN 2 = 100 × 2.997 = 116.7 kPa 2 Constant temperature gives P = mRT/V i.e. pressure inverse in V for which the work term is integrated to give Eq.4.5 2 Wby N = ⌠ PN dVN = P1V1 ln(V2/V1) 2⌡ 2 2 1 2.997 = 100 × 3.4985 × ln 3.4985 = -54.1 kJ Sonntag, Borgnakke and van Wylen 4.110 Two kilograms of water is contained in a piston/cylinder (Fig. P4.110) with a massless piston loaded with a linear spring and the outside atmosphere. Initially the spring force is zero and P1 = Po = 100 kPa with a volume of 0.2 m3. If the piston just hits the upper stops the volume is 0.8 m3 and T = 600°C. Heat is now added until the pressure reaches 1.2 MPa. Find the final temperature, show the P– V diagram and find the work done during the process. Solution: , P State 1: v1 = V/m = 0.2 / 2 = 0.1 m3/kg 2 , 3 Process: 1 → 2 → 3 or 1 → 3’ 3 State at stops: 2 or 2’ 2 v2 = Vstop/m = 0.4 m3/kg & T2 = 600°C P1 Table B.1.3 ⇒ Pstop = 1 MPa < P3 1 V stop V1 V since Pstop < P3 the process is as 1 → 2 → 3 State 3: P3 = 1.2 MPa, v3 = v2 = 0.4 m3/kg ⇒ T3 ≅ 770°C 1 1 W13 = W12 + W23 = 2(P1 + P2)(V2 - V1) + 0 = 2(100 + 1000)(0.8 - 0.2) = 330 kJ Sonntag, Borgnakke and van Wylen 4.111 A cylinder having an initial volume of 3 m3 contains 0.1 kg of water at 40°C. The water is then compressed in an isothermal quasi-equilibrium process until it has a quality of 50%. Calculate the work done in the process splitting it into two steps. Assume the water vapor is an ideal gas during the first step of the process. Solution: C.V. Water State 2: (40°C, x = 1) Tbl B.1.1 => PG = 7.384 kPa, vG = 19.52 v1 = V1/m = 3 / 0.1 = 30 m3/kg State 1: ( > vG ) so H2O ~ ideal gas from 1-2 so since constant T vG 19.52 P1 = PG v = 7.384 × 30 = 4.8 kPa 1 V2 = mv2 = 0.1 × 19.52 = 1.952 m3 P C.P. T C.P. Psat 3 7.38 2 1 T 40 3 2 v Process T = C: P1 1 v and ideal gas gives work from Eq.4.5 2 V2 1.952 ⌠ 1W2 = ⌡ PdV = P1V1ln V = 4.8 × 3.0 × ln 3 = −6.19 kJ 1 1 v3 = 0.001008 + 0.5 × 19.519 = 9.7605 => V3 = mv3 = 0.976 m3 P = C = Pg: This gives a work term as 3 2W3 = ⌠ PdV = Pg (V3−V2) = 7.384(0.976 - 1.952) = −7.21 kJ ⌡ 2 Total work: 1W3 = 1W2 + 2W3 = − 6.19 − 7.21 = -13.4 kJ Sonntag, Borgnakke and van Wylen 4.112 Air at 200 kPa, 30°C is contained in a cylinder/piston arrangement with initial volume 0.1 m3 . The inside pressure balances ambient pressure of 100 kPa plus an externally imposed force that is proportional to V0.5. Now heat is transferred to the system to a final pressure of 225 kPa. Find the final temperature and the work done in the process. Solution: C.V. Air. This is a control mass. Use initial state and process to find T2 P1 = P0 + CV1/2; 200 = 100 + C(0.1)1/2, 225 = 100 + CV21/2 ⇒ V2 = 0.156 m3 P1V1 P2V2 = mRT2 = T T2 1 C = 316.23 => ⇒ T2 = (P2V2 / P1V1) T1 = 225 × 0.156 ×303.15 / (200 ×0.1) = 532 K = 258.9°C W12 = ∫ P dV = ∫ (P0 + CV1/2) dV 2 = P0 (V2 - V1) + C × 3 × (V23/2 - V13/2) 2 = 100 (0.156 – 0.1) + 316.23 × 3 × (0.1563/2 – 0.13/2) = 5.6 + 6.32 = 11.9 kJ Sonntag, Borgnakke and van Wylen 4.113 A spring-loaded piston/cylinder arrangement contains R-134a at 20°C, 24% quality with a volume 50 L. The setup is heated and thus expands, moving the piston. It is noted that when the last drop of liquid disappears the temperature is 40°C. The heating is stopped when T = 130°C. Verify the final pressure is about 1200 kPa by iteration and find the work done in the process. Solution: C.V. R-134a. This is a control mass. State 1: Table B.5.1 => v1 = 0.000817 + 0.24*0.03524 = 0.009274 P 3 P3 P1 = 572.8 kPa, P1 m = V/ v1 = 0.050 / 0.009274 = 5.391 kg 2 P2 Process: Linear Spring P = A + Bv 1 v State 2: x2 = 1, T2 ⇒ P2 = 1.017 MPa, v2 = 0.02002 m3/kg Now we have fixed two points on the process line so for final state 3: P2 - P1 P3 = P1 + v - v (v3 - v1) = RHS Relation between P3 and v3 2 1 State 3: T3 and on process line ⇒ iterate on P3 given T3 at P3 = 1.2 MPa => v3 = 0.02504 => P3 - RHS = -0.0247 at P3 = 1.4 MPa => v3 = 0.02112 => P3 - RHS = 0.3376 Linear interpolation gives : 0.0247 P3 ≅ 1200 + 0.3376 + 0.0247 (1400-1200) = 1214 kPa 0.0247 v3 = 0.02504 + 0.3376 + 0.0247 (0.02112-0.02504) = 0.02478 m3/kg W13 = ∫ P dV = 2 (P1 + P3)(V3 - V1) = 2 (P1 + P3) m (v3 - v1) 1 1 1 = 2 5.391(572.8 + 1214)(0.02478 - 0.009274) = 74.7 kJ Sonntag, Borgnakke and van Wylen 4.114 A piston/cylinder (Fig. P4.114) contains 1 kg of water at 20°C with a volume of 0.1 m3. Initially the piston rests on some stops with the top surface open to the atmosphere, Po and a mass so a water pressure of 400 kPa will lift it. To what temperature should the water be heated to lift the piston? If it is heated to saturated vapor find the final temperature, volume and the work, 1W2. Solution: (a) State to reach lift pressure of P = 400 kPa, v = V/m = 0.1 m3/kg Table B.1.2: vf < v < vg = 0.4625 m3/kg => T = T sat = 143.63°C (b) State 2 is saturated vapor at 400 kPa since state 1a is two-phase. P P o H2O 1a 2 1 V v2 = vg = 0.4625 m3/kg , V2 = m v2 = 0.4625 m3, Pressure is constant as volume increase beyond initial volume. 1W2 = ∫ P dV = P (V2 - V1) = Plift (V2 – V1) = 400 (0.4625 – 0.1) = 145 kJ Sonntag, Borgnakke and van Wylen 4.115 Two springs with same spring constant are installed in a massless piston/cylinder with the outside air at 100 kPa. If the piston is at the bottom, both springs are relaxed and the second spring comes in contact with the piston at V = 2 m3. The cylinder (Fig. P4.115) contains ammonia initially at −2°C, x = 0.13, V = 1 m3, which is then heated until the pressure finally reaches 1200 kPa. At what pressure will the piston touch the second spring? Find the final temperature and the total work done by the ammonia. Solution : P State 1: P = 399.7 kPa Table B.2.1 v = 0.00156 + 0.13×0.3106 = 0.0419 3 At bottom state 0: 0 m3, 100 kPa 2 0 1 cb 0 2W3 1W2 P0 1 2 V State 2: V = 2 m3 and on line 0-1-2 Final state 3: 1200 kPa, on line segment 2. V3 Slope of line 0-1-2: ∆P/ ∆V = (P1 - P0)/∆V = (399.7-100)/1 = 299.7 kPa/ m3 P2 = P1 + (V2 - V1)∆P/∆V = 399.7 + (2-1)×299.7 = 699.4 kPa State 3: Last line segment has twice the slope. P3 = P2 + (V3 - V2)2∆P/∆V ⇒ V3 = V2+ (P3 - P2)/(2∆P/∆V) V3 = 2 + (1200-699.4)/599.4 = 2.835 m3 v3 = v1V3/V1 = 0.0419×2.835/1 = 0.1188 1 ⇒ T = 51°C 1 1W3 = 1W2 + 2W3 = 2 (P1 + P2)(V2 - V1) + 2 (P3 + P2)(V3 - V2) = 549.6 + 793.0 = 1342.6 kJ Sonntag, Borgnakke and van Wylen 4.116 Find the work for Problem 3.101. A piston/cylinder arrangement is loaded with a linear spring and the outside atmosphere. It contains water at 5 MPa, 400°C with the volume being 0.1 m3. If the piston is at the bottom, the spring exerts a force such that Plift = 200 kPa. The system now cools until the pressure reaches 1200 kPa. Find the mass of water, the final state (T2, v2) and plot the P–v diagram for the process. Solution : P 1: 5 MPa, 400°C ⇒ v1= 0.05781 m3/kg m = V/v1 = 0.1/0.05781 = 1.73 kg 1 5000 Straight line: P = Pa + Cv P2 - Pa v2 = v1 P - P = 0.01204 m3/kg 1 a 2 1200 200 a v 0 ? v2 < vg(1200 kPa) so two-phase T2 = 188°C 0.05781 v2 - 0.001139 = 0.0672 0.1622 The P-V coordinates for the two states are then: ⇒ x2 = P1 = 5 MPa, V1 = 0.1 m3, P2 = 1200 kPa, V2 = mv2 = 0.02083 m3 P vs. V is linear so 1 1W2 = ⌠PdV = 2 (P1 + P2)(V2 - V1) ⌡ 1 = 2 (5000 + 1200)(0.02083 - 0.1) = -245.4 kJ SOLUTION MANUAL SI UNIT PROBLEMS CHAPTER 5 SONNTAG • BORGNAKKE • VAN WYLEN FUNDAMENTALS of Thermodynamics Sixth Edition Sonntag, Borgnakke and van Wylen CONTENT SUBSECTION PROB NO. Correspondence table Concept-Study Guide Problems Kinetic and potential energy Properties (u,h) from general tables Energy equation: simple process Energy eqaution: multistep process Energy equation: solids and liquids Properties (u, h, Cv, Cp), ideal gas Energy equation: ideal gas Energy equation: polytropic process Energy equation in rate form Review Problems 1-19 20-27 28-34 35-60 61-73 74-81 82-88 89-102 103-115 116-125 126-138 Sonntag, Borgnakke and van Wylen CHAPTER 5 CORRESPONDENCE TABLE The correspondence between this problem set and 5th edition chapter 5 problem set. Study guide problems 5.1-5.19 are all new New 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 5th 1 4 2mod 3 new 5 new new 6 mod new 7 mod new 8 mod 9 mod new 10 mod new 12 14 11 new 13 15 21 new new new 26 41 new New 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 5th 28 new 17 new 27 51 53 40 37 44 42 new 38 39 20 23 mod 43 24 45 new new 49 mod 55 36 new 58 60 new 59 61 New 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 5th new new new new new 67 mod new 68 mod 62 72 mod 63 new new 79 new 64 new 65 new new new 69 new new 74 76 new 66 new 46 New 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 5th new 84 77 30 54 82 new 89 87 new 90 new 86 new new new 22 29 57 35 31 32 48 56 18 new 83 new 85 Sonntag, Borgnakke and van Wylen The english unit problem set corresponds to the 5th edition as New 139 140 141 142 143 144 145 146 147 148 149 150 5th new new new new new new new 102 103 104 mod 105 mod 104 mod New 151 152 153 154 155 156 157 158 159 160 161 162 5th 107 108 106 new 112 115 111 110 109 113 114 118 New 163 164 165 166 167 168 169 170 171 172 173 174 5th 124 119 new 120 new 122 121 new 125 130 129 123 New 175 176 177 178 179 180 181 182 5th 127 new 131 132 135 new 136 134 Sonntag, Borgnakke and van Wylen Concept-Study Guide Problems 5.1 What is 1 cal in SI units and what is the name given to 1 N-m? Look in the conversion factor table A.1 under energy: 1 cal (Int.) = 4.1868 J = 4.1868 Nm = 4.1868 kg m2/s2 This was historically defined as the heat transfer needed to bring 1 g of liquid water from 14.5oC to 15.5oC, notice the value of the heat capacity of water in Table A.4 1 N-m = 1 J or Force times displacement = energy = Joule 5.2 In a complete cycle what is the net change in energy and in volume? For a complete cycle the substance has no change in energy and therefore no storage, so the net change in energy is zero. For a complete cycle the substance returns to its beginning state, so it has no change in specific volume and therefore no change in total volume. 5.3 Why do we write ∆E or E2 – E1 whereas we write 1Q2 and 1W2? ∆E or E2 – E1 is the change from state 1 to state 2 and depends only on states 1 and 2 not upon the process between 1 and 2. 1Q2 and 1W2 are amounts of energy transferred during the process between 1 and 2 and depend on the process path. Sonntag, Borgnakke and van Wylen 5.4 When you wind a spring up in a toy or stretch a rubber band what happens in terms of work, energy and heat transfer? Later when they are released, what happens then? In both processes work is put into the device and the energy is stored as potential energy. If the spring or rubber is inelastic some of the work input goes into internal energy (it becomes warmer) and not its potential energy and being warmer than the ambient air it cools slowly to ambient temperature. When the spring or rubber band is released the potential energy is transferred back into work given to the system connected to the end of the spring or rubber band. If nothing is connected the energy goes into kinetic energy and the motion is then dampened as the energy is transformed into internal energy. 5.5 Explain in words what happens with the energy terms for the stone in Example 5.2. What would happen if it were a bouncing ball falling to a hard surface? In the beginning all the energy is potential energy associated with the gravitational force. As the stone falls the potential energy is turned into kinetic energy and in the impact the kinetic energy is turned into internal energy of the stone and the water. Finally the higher temperature of the stone and water causes a heat transfer to the ambient until ambient temperature is reached. With a hard ball instead of the stone the impact would be close to elastic transforming the kinetic energy into potential energy (the material acts as a spring) that is then turned into kinetic energy again as the ball bounces back up. Then the ball rises up transforming the kinetic energy into potential energy (mgZ) until zero velocity is reached and it starts to fall down again. The collision with the floor is not perfectly elastic so the ball does not rise exactly up to the original height loosing a little energy into internal energy (higher temperature due to internal friction) with every bounce and finally the motion will die out. All the energy eventually is lost by heat transfer to the ambient or sits in lasting deformation (internal energy) of the substance. Sonntag, Borgnakke and van Wylen 5.6 Make a list of at least 5 systems that store energy, explaining which form of energy. A spring that is compressed. Potential energy (1/2)kx2 A battery that is charged. Electrical potential energy. V Amp h A raised mass (could be water pumped up higher) Potential energy mgH A cylinder with compressed air. Potential (internal) energy like a spring. A tank with hot water. Internal energy mu A fly-wheel. Kinetic energy (rotation) (1/2)Iω2 A mass in motion. Kinetic energy (1/2)mV2 5.7 A 1200 kg car is accelerated from 30 to 50 km/h in 5 s. How much work is that? If you continue from 50 to 70 km/h in 5 s is that the same? The work input is the increase in kinetic energy. 2 2 E2 – E1 = (1/2)m[V2 - V1] = 1W2 km2 = 0.5 × 1200 kg [502 – 302] h 1000 m2 = 600 [ 2500 – 900 ] kg 3600 s = 74 074 J = 74.1 kJ The second set of conditions does not become the same 2 2 1000 m2 E2 – E1 = (1/2)m[V2 - V1] = 600 [ 702 – 502 ] kg 3600 s = 111 kJ Sonntag, Borgnakke and van Wylen 5.8 A crane use 2 kW to raise a 100 kg box 20 m. How much time does it take? . L Power = W = FV = mgV = mg t mgL 100 kg 9.807 m/s2 20 m = 9.81 s t= . = 2000 W W 5.9 Saturated water vapor has a maximum for u and h at around 235oC. Is it similar for other substances? Look at the various substances listed in appendix B. Everyone has a maximum u and h somewhere along the saturated vapor line at different T for each substance. This means the constant u and h curves are different from the constant T curves and some of them cross over the saturated vapor line twice, see sketch below. P C.P. Constant h lines are similar to the constant u line shown. T u=C C.P. P=C T u=C v v Notice the constant u(h) line becomes parallel to the constant T lines in the superheated vapor region for low P where it is an ideal gas. In the T-v diagram the constant u (h) line becomes horizontal. Sonntag, Borgnakke and van Wylen 5.10 A pot of water is boiling on a stove supplying 325 W to the water. What is the rate of mass (kg/s) vaporizing assuming a constant pressure process? To answer this we must assume all the power goes into the water and that the process takes place at atmospheric pressure 101 kPa, so T = 100oC. Energy equation dQ = dE + dW = dU + PdV = dH = hfg dm dQ dm = hfg dt dt . 325 W dm Q = h = 2257 kJ/kg = 0.144 g/s dt fg The volume rate of increase is dV dm 3 dt = dt vfg = 0.144 g/s × 1.67185 m /kg = 0.24 × 10-3 m3/s = 0.24 L/s 5.11 A constant mass goes through a process where 100 W of heat transfer comes in and 100 W of work leaves. Does the mass change state? Yes it does. As work leaves a control mass its volume must go up, v increases As heat transfer comes in at a rate equal to the work out means u is constant if there are no changes in kinetic or potential energy. Sonntag, Borgnakke and van Wylen 5.12 I have 2 kg of liquid water at 20oC, 100 kPa. I now add 20 kJ of energy at a constant pressure. How hot does it get if it is heated? How fast does it move if it is pushed by a constant horizontal force? How high does it go if it is raised straight up? a) Heat at 100 kPa. Energy equation: E2 – E1 = 1Q2 – 1W2 = 1Q2 – P(V2 – V1) = H2 – H1= m(h2 – h1) h2 = h1 + 1Q2/m = 83.94 + 20/2 = 94.04 kJ/kg Back interpolate in Table B.1.1: T2 = 22.5oC (We could also have used ∆T = 1Q2/mC = 20 / (2*4.18) = 2.4oC) b) Push at constant P. It gains kinetic energy. 2 0.5 m V2 = 1W2 V2 = 2 1W2/m = 2 × 20 × 1000 J/2 kg = 141.4 m/s c) Raised in gravitational field m g Z2 = 1W2 Z2 = 1W2/m g = 20 000 J = 1019 m 2 kg × 9.807 m/s2 Sonntag, Borgnakke and van Wylen 5.13 Water is heated from 100 kPa, 20oC to 1000 kPa, 200oC. In one case pressure is raised at T = C, then T is raised at P = C. In a second case the opposite order is done. Does that make a difference for 1Q2 and 1W2? Yes it does. Both 1Q2 and 1W2 are process dependent. We can illustrate the work term in a P-v diagram. P Cr.P. L S 1000 100 a 2 1 V T 20 200 P a 1000 100 1 T 2 1553 kPa 1000 200 200 C b 20 C C.P. 180 C v 20 2 a b 100 1 v In one case the process proceeds from 1 to state “a” along constant T then from “a” to state 2 along constant P. The other case proceeds from 1 to state “b” along constant P and then from “b” to state 2 along constant T. Sonntag, Borgnakke and van Wylen 5.14 Two kg water at 120oC with a quality of 25% has its temperature raised 20oC in a constant volume process. What are the new quality and specific internal energy? Solution: State 1 from Table B.1.1 at 120oC v = vf + x vfg = 0.001060 + 0.25 × 0.8908 = 0.22376 m3/kg State 2 has same v at 140oC also from Table B.1.1 v - vf 0.22376 - 0.00108 x= v = = 0.4385 0.50777 fg u = uf + x ufg = 588.72 + 0.4385 × 1961.3 = 1448.8 kJ/kg P C.P. 361.3 198.5 140 C 120 C T C.P. 140 120 T v v Sonntag, Borgnakke and van Wylen 5.15 Two kg water at 200 kPa with a quality of 25% has its temperature raised 20oC in a constant pressure process. What is the change in enthalpy? Solution: State 1 from Table B.1.2 at 200 kPa h = hf + x hfg = 504.68 + 0.25 × 2201.96 = 1055.2 kJ/kg State 2 has same P from Table B.1.2 at 200 kPa T = T + 20 = 120.23 + 20 = 140.23oC 2 sat so state 2 is superheated vapor (x = undefined) from Table B.1.3 20 h2 = 2706.63 + (2768.8 – 2706.63)150 - 120.23 = 2748.4 kJ/kg h2 – h1 = 2748.4 – 1055.2 = 1693.2 kJ/kg P C.P. T C.P. 200 kPa 140 C 200 120.2 C 140 120 T v v 5.16 You heat a gas 10 K at P = C. Which one in table A.5 requires most energy? Why? A constant pressure process in a control mass gives (recall Eq.5.29) 1q2 = u2 − u1 + 1w2 = h2 − h1 ≈ Cp ∆T The one with the highest specific heat is hydrogen, H2. The hydrogen has the smallest mass but the same kinetic energy per mol as other molecules and thus the most energy per unit mass is needed to increase the temperature. Sonntag, Borgnakke and van Wylen 5.17 Air is heated from 300 to 350 K at V = C. Find 1q2? What if from 1300 to 1350 K? Process: V = C Energy Eq.: 1W2 = Ø u2 − u1 = 1q2 – 0 1q2 = u2 − u1 Read the u-values from Table A.7.1 a) 1q2 = u2 − u1 = 250.32 – 214.36 = 36.0 kJ/kg b) 1q2 = u2 − u1 = 1067.94 – 1022.75 = 45.2 kJ/kg case a) Cv ≈ 36/50 = 0.72 kJ/kg K , see A.5 case b) Cv ≈ 45.2/50 = 0.904 kJ/kg K (25 % higher) 5.18 A mass of 3 kg nitrogen gas at 2000 K, V = C, cools with 500 W. What is dT/dt? Process: V=C 1W2= 0 . dE dU dU dT . = dt = m dt = mCv dt = Q – W = Q = -500 W dt du ∆u u2100 - u1900 1819.08 - 1621.66 Cv 2000 = dT = = = = 0.987 kJ/kg K 200 ∆T 2100-1900 . dT Q -500 W K = mC = = -0.17 dt s v 3 × 0.987 kJ/K Remark: Specific heat from Table A.5 has Cv 300 = 0.745 kJ/kg K which is nearly 25% lower and thus would over-estimate the rate with 25%. Sonntag, Borgnakke and van Wylen 5.19 A drag force on a car, with frontal area A = 2 m2, driving at 80 km/h in air at 20oC is Fd = 0.225 A ρairV2. How much power is needed and what is the traction force? . W = FV km 1000 V = 80 h = 80 × 3600 ms-1 = 22.22 ms-1 P 101 ρAIR = RT = = 1.20 kg/m3 0.287 × 293 Fd = 0.225 AρV2 = 0.225 × 2 × 1.2 × 22.222 = 266.61 N . W = FV = 266.61 N × 22.22 m/s = 5924 W = 5.92 kW Sonntag, Borgnakke and van Wylen Kinetic and Potential Energy 5.20 A hydraulic hoist raises a 1750 kg car 1.8 m in an auto repair shop. The hydraulic pump has a constant pressure of 800 kPa on its piston. What is the increase in potential energy of the car and how much volume should the pump displace to deliver that amount of work? Solution: C.V. Car. No change in kinetic or internal energy of the car, neglect hoist mass. E2 – E1 = PE2 - PE1 = mg (Z2 – Z1) = 1750 × 9.80665 × 1.8 = 30 891 J The increase in potential energy is work into car from pump at constant P. W = E2 – E1 = ∫ P dV = P ∆V ∆V = ⇒ E2 – E1 30891 = 800 × 1000 = 0.0386 m3 P Sonntag, Borgnakke and van Wylen 5.21 A piston motion moves a 25 kg hammerhead vertically down 1 m from rest to a velocity of 50 m/s in a stamping machine. What is the change in total energy of the hammerhead? Solution: C.V. Hammerhead The hammerhead does not change internal energy (i.e. same P, T), but it does have a change in kinetic and potential energy. E2 – E1 = m(u2 – u1) + m[(1/2)V2 2 – 0] + mg (h2 - 0) = 0 + 25 × (1/2) × 502 + 25 × 9.80665 × (-1) = 31250 – 245.17 = 31005 J = 31 kJ Sonntag, Borgnakke and van Wylen 5.22 Airplane takeoff from an aircraft carrier is assisted by a steam driven piston/cylinder device with an average pressure of 1250 kPa. A 17500 kg airplane should be accelerated from zero to a speed of 30 m/s with 30% of the energy coming from the steam piston. Find the needed piston displacement volume. Solution: C.V. Airplane. No change in internal or potential energy; only kinetic energy is changed. 2 E2 – E1 = m (1/2) (V2 - 0) = 17500 × (1/2) × 302 = 7875 000 J = 7875 kJ The work supplied by the piston is 30% of the energy increase. W = ∫ P dV = Pavg ∆V = 0.30 (E2 – E1) = 0.30 × 7875 = 2362.5 kJ W 2362.5 ∆V = P = 1250 = 1.89 m3 avg Sonntag, Borgnakke and van Wylen 5.23 Solve Problem 5.22, but assume the steam pressure in the cylinder starts at 1000 kPa, dropping linearly with volume to reach 100 kPa at the end of the process. Solution: C.V. Airplane. P E2 – E1 = m (1/2) (V22 - 0) = 3500 × (1/2) × 302 = 1575000 J = 1575 kJ W = 0.25(E2 – E1) = 0.25 × 1575 = 393.75 kJ W = ∫ P dV = (1/2)(Pbeg + Pend) ∆V W 2362.5 ∆V = P = 1/2(1000 + 100) = 4.29 m3 avg 1000 100 1 W 2 V Sonntag, Borgnakke and van Wylen 5.24 A 1200 kg car accelerates from zero to 100 km/h over a distance of 400 m. The road at the end of the 400 m is at 10 m higher elevation. What is the total increase in the car kinetic and potential energy? Solution: 2 2 ∆KE = ½ m (V2 - V1) V2 = 100 km/h = 100 × 1000 m/s 3600 = 27.78 m/s ∆KE = ½ ×1200 kg × (27.782 – 02) (m/s)2 = 463 037 J = 463 kJ ∆PE = mg(Z2 – Z1) = 1200 kg × 9.807 m/s2 ( 10 - 0 ) m = 117684 J = 117.7 kJ Sonntag, Borgnakke and van Wylen 5.25 A 25 kg piston is above a gas in a long vertical cylinder. Now the piston is released from rest and accelerates up in the cylinder reaching the end 5 m higher at a velocity of 25 m/s. The gas pressure drops during the process so the average is 600 kPa with an outside atmosphere at 100 kPa. Neglect the change in gas kinetic and potential energy, and find the needed change in the gas volume. Solution: C.V. Piston (E2 – E1)PIST. = m(u2 – u1) + m[(1/2)V2 2 – 0] + mg (h2 – 0) = 0 + 25 × (1/2) × 252 + 25 × 9.80665 × 5 = 7812.5 + 1225.8 = 9038.3 J = 9.038 kJ Energy equation for the piston is: E2 – E1 = Wgas - Watm = Pavg ∆Vgas – Po ∆Vgas (remark ∆Vatm = – ∆Vgas so the two work terms are of opposite sign) ∆Vgas = 9.038/(600 – 100) = 0.018 m3 V Po g P H Pavg 1 2 V Sonntag, Borgnakke and van Wylen 5.26 The rolling resistance of a car depends on its weight as: F = 0.006 mg. How far will a car of 1200 kg roll if the gear is put in neutral when it drives at 90 km/h on a level road without air resistance? Solution: The car decreases its kinetic energy to zero due to the force (constant) acting over the distance. 2 2 m (1/2V2 –1/2V1) = -1W2 = -∫ F dx = -FL km 90 ×1000 V1 = 90 h = 3600 ms-1 = 25 ms-1 V2 = 0, 2 -1/2 mV1 = -FL = - 0.006 mgL 2 0.5 V1 0.5×252 m2/s2 = 5311 m L = 0.0006g = 0.006×9.807 m/s2 Remark: Over 5 km! The air resistance is much higher than the rolling resistance so this is not a realistic number by itself. Sonntag, Borgnakke and van Wylen 5.27 A mass of 5 kg is tied to an elastic cord, 5 m long, and dropped from a tall bridge. Assume the cord, once straight, acts as a spring with k = 100 N/m. Find the velocity of the mass when the cord is straight (5 m down). At what level does the mass come to rest after bouncing up and down? Solution: Let us assume we can neglect the cord mass and motion. 1: V1 = 0, Z1= 0 3: V3 = 0, Z3= -L , Fup = mg = ks ∆L 1 2: 2 : V2, Z2= -5m 2 2 ½ mV1 + mg Z1 = ½ V2 + mgZ2 Divide by mass and left hand side is zero so 2 ½ V2 + g Z2 = 0 V2 = (-2g Z2)1/2 = ( -2 ×9.807 × (-5)) 1/2 = 9.9 m/s State 3: m is at rest so Fup = Fdown ks ∆L = mg mg 5 ×9.807 kg ms-2 ∆L = k = 100 = 0.49 m s Nm-1 L = Lo + ∆L = 5 + 0.49 = 5.49 m So: Z2 = -L = - 5.49 m BRIDGE m V Sonntag, Borgnakke and van Wylen Properties (u, h) from General Tables 5.28 Find the missing properties. a. H2O T = 250°C, v = 0.02 m3/kg P=? u=? b. N2 T = 120 K, P = 0.8 MPa x=? h=? c. H2O T = −2°C, P = 100 kPa u=? v=? d. R-134a Solution: P = 200 kPa, v = 0.12 m3/kg u=? T=? a) Table B.1.1 at 250°C: ⇒ vf < v < vg P = Psat = 3973 kPa x = (v - vf)/ vfg = (0.02 – 0.001251)/0.04887 = 0.38365 u = uf + x ufg = 1080.37 + 0.38365 × 1522.0 = 1664.28 kJ/kg b) Table B.6.1 P is lower than Psat so it is super heated vapor => x = undefined Table B.6.2: and we find the state in Table B.6.2 h = 114.02 kJ/kg c) Table B.1.1 : T < Ttriple point => B.1.5: P > Psat so compressed solid u ≅ ui = -337.62 kJ/kg v ≅ vi = 1.09×10-3 m3/kg approximate compressed solid with saturated solid properties at same T. d) Table B.5.1 v > vg superheated vapor => Table B.5.2. T ~ 32.5°C = 30 + (40 – 30) × (0.12 – 0.11889)/(0.12335 - 0.11889) u = 403.1 + (411.04 – 403.1) × 0.24888 = 405.07 kJ/kg P C.P. L S T a V P C.P. v C.P. d a c b P=C b b d c T a T v c d v Sonntag, Borgnakke and van Wylen 5.29 Find the missing properties of T, P, v, u, h and x if applicable and plot the location of the three states as points in the T-v and the P-v diagrams a. Water at 5000 kPa, u = 800 kJ/kg b. Water at 5000 kPa, v = 0.06 m3/kg c. R-134a at 35oC, v = 0.01 m3/kg Solution: a) Look in Table B.1.2 at 5000 kPa u < uf = 1147.78 => compressed liquid between 180 oC and 200 oC Table B.1.4: 800 - 759.62 T = 180 + (200 - 180) 848.08 - 759.62 = 180 + 20*0.4567 = 189.1 C v = 0.001124 + 0.4567 (0.001153 - 0.001124) = 0.001137 b) Look in Table B.1.2 at 5000 kPa v > vg = 0.03944 => superheated vapor between 400 oC and 450 oC. Table B.1.3: T = 400 + 50*(0.06 - 0.05781)/(0.0633 - 0.05781) = 400 + 50*0.3989 = 419.95 oC h = 3195.64 + 0.3989 *(3316.15 - 3195.64) = 3243.71 c) B.5.1: v f < v < vg => 2-phase, P = Psat = 887.6 kPa, x = (v - vf ) / vfg = (0.01 - 0.000857)/0.02224 = 0.4111 u = uf + x ufg = 248.34 + 0.4111*148.68 = 309.46 kJ/kg P C.P. States shown are placed relative to the two-phase region, not to each other. T C.P. P = const. a b b T a c v c v Sonntag, Borgnakke and van Wylen 5.30 Find the missing properties and give the phase of the ammonia, NH3. a. T = 65oC, P = 600 kPa u=? v=? b. T = 20oC, P = 100 kPa u=? v=? x=? c. T = 50oC, v = 0.1185 m3/kg u=? P=? x=? Solution: a) Table B.2.1 P < Psat => superheated vapor Table B.2.2: v = 0.5 × 0.25981 + 0.5 × 0.26888 = 0.2645 m3/kg u = 0.5 × 1425.7 + 0.5 × 1444.3 = 1435 kJ/kg b) Table B.2.1: P < Psat => x = undefined, superheated vapor, from B.2.2: v = 1.4153 m3/kg ; u = 1374.5 kJ/kg c) Sup. vap. ( v > vg) Table B.2.2. P = 1200 kPa, x = undefined u = 1383 kJ/kg P C.P. States shown are placed relative to the two-phase region, not to each other. T c C.P. c a T b v 1200 kPa 600 kPa a b v Sonntag, Borgnakke and van Wylen 5.31 Find the phase and missing properties of P, T, v, u, and x. a. Water at 5000 kPa, u = 1000 kJ/kg (Table B.1 reference) b. R-134a at 20oC, u = 300 kJ/kg c. Nitrogen at 250 K, 200 kPa Show also the three states as labeled dots in a T-v diagram with correct position relative to the two-phase region. Solution: a) Compressed liquid: B.1.4 interpolate between 220oC and 240oC. T = 233.3oC, v = 0.001213 m3/kg, x = undefined b) Table B.5.1: u < ug => two-phase liquid and vapor x = (u - uf)/ufg = (300 - 227.03)/162.16 = 0.449988 = 0.45 v = 0.000817 + 0.45*0.03524 = 0.01667 m3/kg c) Table B.6.1: T > Tsat (200 kPa) so superheated vapor in Table B.6.2 x = undefined v = 0.5(0.35546 + 0.38535) = 0.3704 m3/kg, u = 0.5(177.23 + 192.14) = 184.7 kJ/kg P C.P. States shown are placed relative to the two-phase region, not to each other. T a a b C.P. P = const. b c T v c v Sonntag, Borgnakke and van Wylen 5.32 Find the missing properties and give the phase of the substance a. H2O T = 120°C, v = 0.5 m3/kg u=? P=? x=? b. H2O T = 100°C, P = 10 MPa u=? x=? v=? c. N2 T = 200 K, P = 200 kPa v=? u=? d. NH3 T = 100°C, v = 0.1 m3/kg P=? x=? e. N2 T = 100 K, x = 0.75 v=? u=? Solution: a) Table B.1.1: vf < v < vg => L+V mixture, P = 198.5 kPa, x = (0.5 - 0.00106)/0.8908 = 0.56, u = 503.48 + 0.56 × 2025.76 = 1637.9 kJ/kg b) Table B.1.4: compressed liquid, v = 0.001039 m3/kg, u = 416.1 kJ/kg c) Table B.6.2: 200 K, 200 kPa v = 0.29551 m3/kg ; d) Table B.2.1: v > vg u = 147.37 kJ/kg => superheated vapor, x = undefined 0.1 - 0.10539 B.2.2: P = 1600 + 400 × 0.08248-0.10539 = 1694 kPa e) Table B.6.1: 100 K, x = 0.75 v = 0.001452 + 0.75 × 0.02975 = 0.023765 m3/kg u = -74.33 + 0.75 ×137.5 = 28.8 kJ/kg P C.P. States shown are placed relative to the two-phase region, not to each other. T c C.P. > P = const. b c a d b T a d e e v v Sonntag, Borgnakke and van Wylen 5.33 Find the missing properties among (T, P, v, u, h and x if applicable) and give the phase of the substance and indicate the states relative to the two-phase region in both a T-v and a P-v diagram. a. R-12 P = 500 kPa, h = 230 kJ/kg b. R-22 T = 10oC, u = 200 kJ/kg c. R-134a T = 40oC, h = 400 kJ/kg Solution: a) Table B.3.2: h > hg = > superheated vapor, look in section 500 kPa and interpolate T = 68.06°C, v = 0.04387 m3/kg, u = 208.07 kJ/kg b) Table B.4.1: u < ug => L+V mixture, P = 680.7 kPa u - uf 200 - 55.92 x = u = 173.87 = 0.8287, fg v = 0.0008 + 0.8287 × 0.03391 = 0.0289 m3/kg, h = 56.46 + 0.8287 × 196.96 = 219.7 kJ/kg c) Table B.5.1: h < hg => two-phase L + V, look in B.5.1 at 40°C: h - hf 400 - 256.5 x = h = 163.3 = 0.87875 fg P = Psat = 1017 kPa, v = 0.000 873 + 0.87875 × 0.01915 = 0.0177 m3/kg u = 255.7 + 0.87875 × 143.8 = 382.1 kJ/kg P C.P. States shown are placed relative to the two-phase region, not to each other. T C.P. P=C a b, c b, c T v a v Sonntag, Borgnakke and van Wylen 5.34 Saturated liquid water at 20oC is compressed to a higher pressure with constant temperature. Find the changes in u and h from the initial state when the final pressure is a) 500 kPa, b) 2000 kPa, c) 20 000 kPa Solution: State 1 is located in Table B.1.1 and the states a-c are from Table B.1.4 State u [kJ/kg] h [kJ/kg] ∆u = u - u1 ∆h = h - h1 ∆(Pv) 1 a b c 83.94 83.91 83.82 82.75 83.94 84.41 85.82 102.61 -0.03 -0.12 -1.19 0.47 1.88 18.67 0.5 2 20 For these states u stays nearly constant, dropping slightly as P goes up. h varies with Pv changes. T P c b a 1 c,b,a,1 o T = 20 C v v P L T c b a S C.P. V 1 cb v Sonntag, Borgnakke and van Wylen Energy Equation: Simple Process 5.35 A 100-L rigid tank contains nitrogen (N2) at 900 K, 3 MPa. The tank is now cooled to 100 K. What are the work and heat transfer for this process? Solution: C.V.: Nitrogen in tank. Energy Eq.5.11: m2 = m1 ; m(u2 - u1) = 1Q2 - 1W2 Process: V = constant, v2 = v1 = V/m => / 1W2 = 0 Table B.6.2: State 1: v1 = 0.0900 m3/kg => m = V/v1 = 1.111 kg u1 = 691.7 kJ/kg State 2: 100 K, v2 = v1 = V/m, look in Table B.6.2 at 100 K 200 kPa: v = 0.1425 m3/kg; u = 71.7 kJ/kg 400 kPa: v = 0.0681 m3/kg; u = 69.3 kJ/kg so a linear interpolation gives: P2 = 200 + 200 (0.09 – 0.1425)/(0.0681 – 0.1425) = 341 kPa 0.09 – 0.1425 u2 = 71.7 + (69.3 – 71.7) 0.0681 – 0.1425 = 70.0 kJ/kg, 1Q2 = m(u2 - u1) = 1.111 (70.0 – 691.7) = −690.7 kJ Sonntag, Borgnakke and van Wylen 5.36 A rigid container has 0.75 kg water at 300oC, 1200 kPa. The water is now cooled to a final pressure of 300 kPa. Find the final temperature, the work and the heat transfer in the process. Solution: C.V. Water. Constant mass so this is a control mass Energy Eq.: Process eq.: V = constant. (rigid) P U2 - U1 = 1Q2 - 1W2 1200 1 300 2 1W2 = ∫ P dV = 0 => o State 1: 300 C, 1200 kPa => superheated vapor Table B.1.3 v = 0.21382 m3/kg, v u = 2789.22 kJ/kg State 2: 300 kPa and v2 = v1 from Table B.1.2 v2 < vg T2 = Tsat = 133.55oC v2 - vf 0.21382 - 0.001073 = = 0.35179 x2 = v 0.60475 fg u2 = uf + x2 ufg = 561.13 + x2 1982.43 = 1258.5 kJ/kg 1Q2 = m(u2 - u1) + 1W2 = m(u2 - u1) = 0.75 (1258.5 - 2789.22) = -1148 kJ two-phase Sonntag, Borgnakke and van Wylen 5.37 A cylinder fitted with a frictionless piston contains 2 kg of superheated refrigerant R134a vapor at 350 kPa, 100oC. The cylinder is now cooled so the R-134a remains at constant pressure until it reaches a quality of 75%. Calculate the heat transfer in the process. Solution: C.V.: R-134a Energy Eq.5.11 m2 = m1 = m; m(u2 - u1) = 1Q2 - 1W2 Process: P = const. ⇒ 1W2 = ⌡PdV = P∆V = P(V2 - V1) = Pm(v2 - v1) ⌠ P T 2 1 1 2 V V State 1: Table B.5.2 h1 = (490.48 + 489.52)/2 = 490 kJ/kg State 2: Table B.5.1 h2 = 206.75 + 0.75 ×194.57 = 352.7 kJ/kg (350.9 kPa) 1Q2 = m(u2 - u1) + 1W2 = m(u2 - u1) + Pm(v2 - v1) = m(h2 - h1) 1Q2 = 2 × (352.7 - 490) = -274.6 kJ Sonntag, Borgnakke and van Wylen 5.38 Ammonia at 0°C, quality 60% is contained in a rigid 200-L tank. The tank and ammonia is now heated to a final pressure of 1 MPa. Determine the heat transfer for the process. Solution: C.V.: NH3 P 2 1 V Continuity Eq.: m2 = m1 = m ; Energy Eq.5.11: m(u2 - u1) = 1Q2 - 1W2 Process: Constant volume ⇒ v2 = v1 & 1W2 = 0 State 1: Table B.2.1 two-phase state. v1 = 0.001566 + x1 × 0.28783 = 0.17426 m3/kg u1 = 179.69 + 0.6 × 1138.3 = 862.67 kJ/kg m = V/v1 = 0.2/0.17426 = 1.148 kg State 2: P2 , v2 = v1 superheated vapor Table B.2.2 ⇒ T2 ≅ 100°C, u2 ≅ 1490.5 kJ/kg So solve for heat transfer in the energy equation 1Q2 = m(u2 - u1) = 1.148(1490.5 - 862.67) = 720.75 kJ Sonntag, Borgnakke and van Wylen 5.39 Water in a 150-L closed, rigid tank is at 100°C, 90% quality. The tank is then cooled to −10°C. Calculate the heat transfer during the process. Solution: C.V.: Water in tank. m2 = m1 ; Energy Eq.5.11: m(u2 - u1) = 1Q2 - 1W2 Process: V = constant, v2 = v1, 1W2 = 0 State 1: Two-phase L + V look in Table B.1.1 v1 = 0.001044 + 0.9 × 1.6719 = 1.5057 m3/kg u1 = 418.94 + 0.9 × 2087.6 = 2297.8 kJ/kg ⇒ mix of saturated solid + vapor Table B.1.5 State 2: T2, v2 = v1 v2 = 1.5057 = 0.0010891 + x2 × 466.7 => x2 = 0.003224 u2 = -354.09 + 0.003224 × 2715.5 = -345.34 kJ/kg m = V/v1 = 0.15/1.5057 = 0.09962 kg 1Q2 = m(u2 - u1) = 0.09962(-345.34 - 2297.8) = -263.3 kJ P C.P. T C.P. P = const. 1 1 T 2 v 2 P C.P. 1 L T V S L+V 2 S+V v v Sonntag, Borgnakke and van Wylen 5.40 A piston/cylinder contains 1 kg water at 20oC with volume 0.1 m3. By mistake someone locks the piston preventing it from moving while we heat the water to saturated vapor. Find the final temperature and the amount of heat transfer in the process. Solution: C.V. Water. This is a control mass Energy Eq.: m (u2 − u1 ) = 1Q2 − 1W2 Process : State 1: 1W2 = 0 T, v1 = V1/m = 0.1 m3/kg > vf so two-phase V = constant v1 - vf 0.1-0.001002 x1 = v = 57.7887 = 0.0017131 fg u1 = uf + x1 ufg = 83.94 + x1 × 2318.98 = 87.913 kJ/kg State 2: v2 = v1 = 0.1 & x2 =1 found in Table B.1.1 between 210°C and 215° C 0.1-0.10441 T2 = 210 + 5 × 0.09479-0.10441 = 210 + 5 × 0.4584 = 212.3°C u2 = 2599.44 + 0.4584 (2601.06 – 2599.44) = 2600.2 kJ/kg From the energy equation 1Q2 = m(u2 − u1) = 1( 2600.2 – 87.913) = 2512.3 kJ P T 2 2 1 1 V V Sonntag, Borgnakke and van Wylen 5.41 A test cylinder with constant volume of 0.1 L contains water at the critical point. It now cools down to room temperature of 20°C. Calculate the heat transfer from the water. Solution: C.V.: Water P m2 = m1 = m ; 1 Energy Eq.5.11: m(u2 - u1) = 1Q2 - 1W2 Process: Constant volume ⇒ v2 = v1 Properties from Table B.1.1 2 State 1: v1 = vc = 0.003155 m3/kg, u1 = 2029.6 kJ/kg m = V/v1 = 0.0317 kg State 2: T2, v2 = v1 = 0.001002 + x2 × 57.79 x2 = 3.7×10-5, u2 = 83.95 + x2 × 2319 = 84.04 kJ/kg Constant volume => / 1W2 = 0 1Q2 = m(u2 - u1) = 0.0317(84.04 - 2029.6) = -61.7 kJ v Sonntag, Borgnakke and van Wylen 5.42 A 10-L rigid tank contains R-22 at −10°C, 80% quality. A 10-A electric current (from a 6-V battery) is passed through a resistor inside the tank for 10 min, after which the R-22 temperature is 40°C. What was the heat transfer to or from the tank during this process? Solution: C.V. R-22 in tank. Control mass at constant V. Continuity Eq.: m2 = m1 = m ; Energy Eq.: P 2 m(u2 - u1) = 1Q2 - 1W2 Constant V ⇒ v2 = v1 => no boundary work, but electrical work Process: 1 V State 1 from table B.4.1 v1 = 0.000759 + 0.8 × 0.06458 = 0.05242 m3/kg u1 = 32.74 + 0.8 × 190.25 = 184.9 kJ/kg m = V/v = 0.010/0.05242 = 0.1908 kg State 2: Table B.4.2 at 40°C and v2 = v1 = 0.05242 m3/kg => sup.vapor, so use linear interpolation to get P2 = 500 + 100 × (0.05242 – 0.05636)/(0.04628 – 0.05636) = 535 kPa, u2 = 250.51 + 0.35× (249.48 – 250.51) = 250.2 kJ/kg 1W2 elec = –power × ∆t = –Amp × volts × ∆t = – 10 × 6 × 10 × 60 = –36 kJ 1000 1Q2 = m(u2 – u1) + 1W2 = 0.1908 ( 250.2 – 184.9) – 36 = –23.5 kJ Sonntag, Borgnakke and van Wylen 5.43 A piston/cylinder contains 50 kg of water at 200 kPa with a volume of 0.1 m3. Stops in the cylinder are placed to restrict the enclosed volume to a maximum of 0.5 m3. The water is now heated until the piston reaches the stops. Find the necessary heat transfer. Solution: C.V. H2O m = constant Energy Eq.5.11: m(e2 – e1) = m(u2 – u1) = 1Q2 - 1W2 Process : P = constant (forces on piston constant) ⇒ 1W2 = ∫ P dV = P1 (V2 – V1) P 1 2 0.1 0.5 V Properties from Table B.1.1 State 1 : v1 = 0.1/50 = 0.002 m3/kg => 2-phase as v1 < vg v1 – vf 0.002 – 0.001061 x= = 0.001061 0.88467 vfg = h = 504.68 + 0.001061 × 2201.96 = 507.02 kJ/kg State 2 : v2= 0.5/50 = 0.01 m3/kg also 2-phase same P v2 – vf 0.01 – 0.001061 = = 0.01010 x2 = v 0.88467 fg h2 = 504.68 + 0.01010 × 2201.96 = 526.92 kJ/kg Find the heat transfer from the energy equation as 1Q2 = m(u2 – u1) + 1W2 = m(h2 – h1) 1Q2 = 50 kg × (526.92 – 507.02) kJ/kg = 995 kJ [ Notice that 1W2 = P1 (V2 – V1) = 200 × (0.5 – 0.1) = 80 kJ ] Sonntag, Borgnakke and van Wylen 5.44 A constant pressure piston/cylinder assembly contains 0.2 kg water as saturated vapor at 400 kPa. It is now cooled so the water occupies half the original volume. Find the heat transfer in the process. Solution: C.V. Water. This is a control mass. Energy Eq.5.11: m(u2 – u1) = 1Q2 – 1W2 Process: P = constant => 1W2 = Pm(v2 – v1) So solve for the heat transfer: 1Q2 = m(u2 - u1) + 1W2 = m(u2 - u1) + Pm(v2 - v1) = m(h2 - h1) State 1: Table B.1.2 v1 = 0.46246 m3/kg; h1 = 2738.53 kJ/kg State 2: v2 = v1 / 2 = 0.23123 = vf + x vfg from Table B.1.2 x2 = (v2 – vf) / vfg = (0.23123 – 0.001084) / 0.46138 = 0.4988 h2 = hf + x2 hfg = 604.73 + 0.4988 × 2133.81 = 1669.07 kJ/kg 1Q2 = 0.2 (1669.07 – 2738.53) = –213.9 KJ Sonntag, Borgnakke and van Wylen 5.45 Two kg water at 120oC with a quality of 25% has its temperature raised 20oC in a constant volume process as in Fig. P5.45. What are the heat transfer and work in the process? Solution: C.V. Water. This is a control mass Energy Eq.: m (u2 − u1 ) = 1Q2 − 1W2 Process : V = constant 1W2 = State 1: ∫ P dV = 0 T, x1 from Table B.1.1 v1 = vf + x1 vfg = 0.00106 + 0.25 × 0.8908 = 0.22376 m3/kg u1 = uf + x1 ufg = 503.48 + 0.25 × 2025.76 = 1009.92 kJ/kg State 2: T2, v2 = v1< vg2 = 0.50885 m3/kg so two-phase v2 - vf2 0.22376 - 0.00108 = = 0.43855 x2 = v 0.50777 fg2 u2 = uf2 + x2 ufg2 = 588.72 + x2 ×1961.3 = 1448.84 kJ/kg From the energy equation 1Q2 = m(u2 − u1) = 2 ( 1448.84 – 1009.92 ) = 877.8 kJ P C.P. 361.3 198.5 140 C 120 C T C.P. 140 120 T v v Sonntag, Borgnakke and van Wylen 5.46 A 25 kg mass moves with 25 m/s. Now a brake system brings the mass to a complete stop with a constant deceleration over a period of 5 seconds. The brake energy is absorbed by 0.5 kg water initially at 20oC, 100 kPa. Assume the mass is at constant P and T. Find the energy the brake removes from the mass and the temperature increase of the water, assuming P = C. Solution: C.V. The mass in motion. 2 2 E2 - E1= ∆E = 0.5 mV = 0.5 × 25 × 25 /1000 = 7.8125 kJ C.V. The mass of water. m(u2 - u1) H2O = ∆E = 7.8125 kJ => u2 - u1 = 7.8125 / 0.5 = 15.63 u2 = u1 + 15.63 = 83.94 + 15.63 = 99.565 kJ/kg Assume u2 = uf then from Table B.1.1: T2 ≅ 23.7oC, ∆T = 3.7oC We could have used u2 - u1 = C∆T with C from Table A.4: C = 4.18 kJ/kg K giving ∆T = 15.63/4.18 = 3.7oC. Sonntag, Borgnakke and van Wylen 5.47 An insulated cylinder fitted with a piston contains R-12 at 25°C with a quality of 90% and a volume of 45 L. The piston is allowed to move, and the R-12 expands until it exists as saturated vapor. During this process the R-12 does 7.0 kJ of work against the piston. Determine the final temperature, assuming the process is adiabatic. Solution: Take CV as the R-12. Continuity Eq.: m2 = m1 = m ; m(u2 − u1) = 1Q2 - 1W2 Energy Eq.5.11: State 1: (T, x) Tabel B.3.1 => v1 = 0.000763 + 0.9 × 0.02609 = 0.024244 m3/kg m = V1/v1 = 0.045/0.024244 = 1.856 kg u1 = 59.21 + 0.9 × 121.03 = 168.137 kJ/kg State 2: (x = 1, ?) We need one property information. Apply now the energy equation with known work and adiabatic so / 1Q2 = 0 = m(u2 - u1) + 1W2 = 1.856 × (u2 - 168.137) + 7.0 u2 = 164.365 kJ/kg = ug at T2 => Table B.3.1 gives ug at different temperatures: T2 ≅ -15°C T P 1 1 2 v 2 v Sonntag, Borgnakke and van Wylen 5.48 A water-filled reactor with volume of 1 m3 is at 20 MPa, 360°C and placed inside a containment room as shown in Fig. P5.48. The room is well insulated and initially evacuated. Due to a failure, the reactor ruptures and the water fills the containment room. Find the minimum room volume so the final pressure does not exceed 200 kPa. Solution: Solution: C.V.: Containment room and reactor. Mass: m2 = m1 = Vreactor/v1 = 1/0.001823 = 548.5 kg Energy: m(u2 - u1) = 1Q2 - 1W2 = 0 - 0 = 0 v1 = 0.001823 m3/kg, u1 = 1702.8 kJ/kg Energy equation then gives u2 = u1 = 1702.8 kJ/kg State 1: Table B.1.4 P2 = 200 kPa, u2 < ug State 2: => Two-phase Table B.1.2 x2 = (u2 - uf)/ ufg = (1702.8 – 504.47)/2025.02 = 0.59176 v2 = 0.001061 + 0.59176 × 0.88467 = 0.52457 m3/kg V2 = m2 v2 = 548.5 ×0.52457 = 287.7 m3 T P 1 1 2 2 200 v P C.P. 1 L T 200 kPa 2 v 200 kPa u = const v Sonntag, Borgnakke and van Wylen 5.49 A piston/cylinder arrangement contains water of quality x = 0.7 in the initial volume of 0.1 m3, where the piston applies a constant pressure of 200 kPa. The system is now heated to a final temperature of 200°C. Determine the work and the heat transfer in the process. Take CV as the water. Continuity Eq.: m2 = m1 = m ; Energy Eq.5.11: m(u2 − u1) = 1Q2 - 1W2 Process: P = constant ⇒ 1W2 = PdV = Pm(v2 - v1) ⌡ State 1: Table B.1.2 T1 = Tsat at 200 kPa = 120.23°C v1 = vf + xvfg = 0.001061 + 0.7 × 0.88467 = 0.62033 m3 h1 = hf + xhfg = 504.68 + 0.7 × 2201.96 = 2046.05 kJ/kg Total mass can be determined from the initial condition, m = V1/v1 = 0.1/0.62033 = 0.1612 kg T2 = 200°C, P2 = 200 kPa (Table B.1.3) gives v2 = 1.08034 m3/kg h2 = 2870.46 kJ/kg (Table B.1.3) V2 = mv2 = 0.1612 kg × 1.08034 m3/kg = 0.174 m3 Substitute the work into the energy equation 1Q2 = U2 − U1 + 1W2 = m ( u2 – u1 + Pv2 – Pv1) = m(h2 − h1) 1Q2= 0.1612 kg × (2870.46−2046.05) kJ/kg = 132.9 kJ (heat added to system). P T 1 2 2 1 V V Sonntag, Borgnakke and van Wylen 5.50 A piston/cylinder arrangement has the piston loaded with outside atmospheric pressure and the piston mass to a pressure of 150 kPa, shown in Fig. P5.50. It contains water at −2°C, which is then heated until the water becomes saturated vapor. Find the final temperature and specific work and heat transfer for the process. Solution: C.V. Water in the piston cylinder. Continuity: m2 = m1, Energy Eq. per unit mass: u2 - u1 = 1q2 - 1w2 2 Process: P = constant = P1, => ⌡ 1w2 = ⌠ P dv = P1(v2 - v1) 1 State 1: T1 , P1 => Table B.1.5 compressed solid, take as saturated solid. v1 = 1.09×10-3 m3/kg, u1 = -337.62 kJ/kg State 2: x = 1, P2 = P1 = 150 kPa due to process => Table B.1.2 v2 = vg(P2) = 1.1593 m3/kg, T2 = 111.4°C ; u2 = 2519.7 kJ/kg From the process equation -3 1w2 = P1(v2 -v1) = 150(1.1593 -1.09×10 ) = 173.7 kJ/kg From the energy equation 1q2 = u2 - u1 + 1w2 = 2519.7 - (-337.62) + 173.7 = 3031 kJ/kg P L C.P. S T 1 V L+V S+V P C.P. 1 T P=C 2 2 2 v v C.P. 1 v Sonntag, Borgnakke and van Wylen 5.51 A piston/cylinder assembly contains 1 kg of liquid water at 20oC and 300 kPa. There is a linear spring mounted on the piston such that when the water is heated the pressure reaches 1 MPa with a volume of 0.1 m3. Find the final temperature and the heat transfer in the process. Solution: Take CV as the water. m2 = m1 = m ; m(u2 − u1) = 1Q2 - 1W2 State 1: Compressed liquid, take saturated liquid at same temperature. v1 = vf(20) = 0.001002 m3/kg, u1 = uf = 83.94 kJ/kg State 2: v2 = V2/m = 0.1/1 = 0.1 m3/kg and P = 1000 kPa so T2 = Tsat = 179.9°C => Two phase as v2 < vg x2 = (v2 - vf) /vfg = (0.1 - 0.001127)/0.19332 = 0.51145 u2 = uf + x2 ufg = 780.08 + 0.51147 × 1806.32 = 1703.96 kJ/kg Work is done while piston moves at linearly varying pressure, so we get 1W2 = ∫ P dV = area = Pavg (V2 − V1) = 0.5 × (300 + 1000)(0.1 − 0.001) = 64.35 kJ Heat transfer is found from the energy equation 1Q2 = m(u2 − u1) + 1W2 = 1 × (1703.96 - 83.94) + 64.35 = 1684 kJ P 2 P2 P 1 1 cb v Sonntag, Borgnakke and van Wylen 5.52 A closed steel bottle contains ammonia at −20°C, x = 20% and the volume is 0.05 m3. It has a safety valve that opens at a pressure of 1.4 MPa. By accident, the bottle is heated until the safety valve opens. Find the temperature and heat transfer when the valve first opens. Solution: C.V.: NH3 : m2 = m1 = m ; Energy Eq.5.11: m(u2 - u1) = 1Q2 - 1W2 P Process: constant volume process ⇒ 1W2 = 0 State 1: (T, x) Table B.2.1 v1 = 0.001504 + 0.2 × 0.62184 = 0.1259 m3/kg => 2 1 m = V/v1 = 0.05/0.1259 = 0.397 kg u1 = 88.76 + 0.2 × 1210.7 = 330.9 kJ/kg State 2: P2 , v2 = v1 => superheated vapor, interpolate in Table B.2.2: T ≅ 110°C = 100 + 20(0.1259 – 0.12172)/(0.12986 – 0.12172), u2 = 1481 + (1520.7 – 1481) × 0.51 = 1501.25 kJ/kg 1Q2 = m(u2 - u1) = 0.397(1501.25 – 330.9) = 464.6 kJ V Sonntag, Borgnakke and van Wylen 5.53 Two kg water at 200 kPa with a quality of 25% has its temperature raised 20oC in a constant pressure process. What are the heat transfer and work in the process? C.V. Water. This is a control mass Energy Eq.: m (u2 − u1 ) = 1Q2 − 1W2 Process : P = constant 1W2 = ∫ P dV = mP (v2 − v1) State 1: Two-phase given P,x so use Table B.1.2 v1 = 0.001061 + 0.25 × 0.88467 = 0.22223 m3/kg u1 = 504047 + 0.25 × 2025.02 = 1010.725 kJ/kg T = T + 20 = 120.23 + 20 = 140.23 State 2 is superheated vapor 20 v2 = 0.88573 + 150-120.23 × (0.95964 – 0.88573 ) = 0.9354 m3/kg 20 u2 = 2529.49 + 150-120.23 (2576.87- 2529.49) = 2561.32 kJ/kg From the process equation we get 1W2 = mP (v2 − v1) = 2 × 200 ( 0.9354 - 0.22223) = 285.3 kJ From the energy equation 1Q2 = m (u2 − u1) + 1W2 = 2 ( 2561.32 – 1010.725 ) + 285.3 = 3101.2 + 285.27 = 3386.5 kJ P T 1 2 2 1 V V Sonntag, Borgnakke and van Wylen 5.54 Two kilograms of nitrogen at 100 K, x = 0.5 is heated in a constant pressure process to 300 K in a piston/cylinder arrangement. Find the initial and final volumes and the total heat transfer required. Solution: Take CV as the nitrogen. Continuity Eq.: m2 = m1 = m ; m(u2 − u1) = 1Q2 - 1W2 Energy Eq.5.11: Process: P = constant ⇒ 1W2 = ⌡PdV = Pm(v2 - v1) ⌠ State 1: Table B.6.1 v1 = 0.001452 + 0.5 × 0.02975 = 0.01633 m3/kg, V1 = 0.0327 m3 h1 = -73.20 + 0.5 × 160.68 = 7.14 kJ/kg State 2: (P = 779.2 kPa , 300 K) => sup. vapor interpolate in Table B.6.2 v2 = 0.14824 + (0.11115-0.14824)× 179.2/200 = 0.115 m3/kg, V2 = 0.23 m3 h2 = 310.06 + (309.62-310.06) × 179.2/200 = 309.66 kJ/kg Now solve for the heat transfer from the energy equation 1Q2 = m(u2 - u1) + 1W2 = m(h2 - h1) = 2 × (309.66 - 7.14) = 605 kJ P T 1 2 2 1 V V Sonntag, Borgnakke and van Wylen 5.55 A 1-L capsule of water at 700 kPa, 150°C is placed in a larger insulated and otherwise evacuated vessel. The capsule breaks and its contents fill the entire volume. If the final pressure should not exceed 125 kPa, what should the vessel volume be? Solution: C.V. Larger vessel. Continuity: m2 = m1 = m = V/v1 = 0.916 kg Process: expansion with 1Q2 = 0 , 1W2 = 0 / / Energy: m(u2 - u1) = 1Q2 - 1W2 = 0 ⇒ u2 = u1 / State 1: v1 ≅ vf = 0.001091 m3/kg; State 2: P2 , u2 ⇒ x2 = u1 ≅ uf = 631.66 kJ/kg 631.66 – 444.16 = 0.09061 2069.3 v2 = 0.001048 + 0.09061 × 1.37385 = 0.1255 m3/kg V2 = mv2 = 0.916 × 0.1255 = 0.115 m3 = 115 L T P 1 1 2 2 200 v P C.P. 1 L T 200 kPa 2 v 200 kPa u = const v Sonntag, Borgnakke and van Wylen 5.56 Superheated refrigerant R-134a at 20°C, 0.5 MPa is cooled in a piston/cylinder arrangement at constant temperature to a final two-phase state with quality of 50%. The refrigerant mass is 5 kg, and during this process 500 kJ of heat is removed. Find the initial and final volumes and the necessary work. Solution: C.V. R-134a, this is a control mass. Continuity: m2 = m1 = m ; Energy Eq.5.11: m(u2 -u1) = 1Q2 - 1W2 = -500 - 1W2 State 1: T1 , P1 Table B.5.2, v1 = 0.04226 m3/kg ; u1 = 390.52 kJ/kg => V1 = mv1 = 0.211 m3 State 2: T2 , x2 ⇒ Table B.5.1 u2 = 227.03 + 0.5 × 162.16 = 308.11 kJ/kg, v2 = 0.000817 + 0.5 × 0.03524 = 0.018437 m3/kg => V2 = mv2 = 0.0922 m3 1W2 = -500 - m(u2 - u1) = -500 - 5 × (308.11 - 390.52) = -87.9 kJ T P 2 2 1 1 v v Sonntag, Borgnakke and van Wylen 5.57 A cylinder having a piston restrained by a linear spring (of spring constant 15 kN/m) contains 0.5 kg of saturated vapor water at 120°C, as shown in Fig. P5.57. Heat is transferred to the water, causing the piston to rise. If the piston cross-sectional area is 0.05 m2, and the pressure varies linearly with volume until a final pressure of 500 kPa is reached. Find the final temperature in the cylinder and the heat transfer for the process. Solution: C.V. Water in cylinder. Continuity: m2 = m1 = m ; Energy Eq.5.11: m(u2 - u1) = 1Q2 - 1W2 State 1: (T, x) Table B.1.1 => Process: State 2: P2 = P1 + v1 = 0.89186 m3/kg, ksm u1 = 2529.2 kJ/kg 15 × 0.5 (v - v ) = 198.5 + (v - 0.89186) Ap2 2 1 (0.05)2 2 P2 = 500 kPa and on the process curve (see above equation). v2 = 0.89186 + (500 - 198.5) × (0.052/7.5) = 0.9924 m3/kg => (P, v) Table B.1.3 => T2 = 803°C; u2 = 3668 kJ/kg P1 + P2 W12 = ⌠ PdV = 2 m(v2 - v1) ⌡ 198.5 + 500 = × 0.5 × (0.9924 - 0.89186) = 17.56 kJ 2 1Q2 = m(u2 - u1) + 1W2 = 0.5 × (3668 - 2529.2) + 17.56 = 587 kJ T P 2 1 ksm 2 A2 p 1 v v Sonntag, Borgnakke and van Wylen 5.58 A rigid tank is divided into two rooms by a membrane, both containing water, shown in Fig. P5.58. Room A is at 200 kPa, v = 0.5 m3/kg, VA = 1 m3, and room B contains 3.5 kg at 0.5 MPa, 400°C. The membrane now ruptures and heat transfer takes place so the water comes to a uniform state at 100°C. Find the heat transfer during the process. Solution: A C.V.: Both rooms A and B in tank. Continuity Eq.: m2 = mA1 + mB1 ; Energy Eq.: B m2u2 - mA1uA1 - mB1uB1 = 1Q2 - 1W2 State 1A: (P, v) Table B.1.2, mA1 = VA/vA1 = 1/0.5 = 2 kg v – vf 0.5 - 0.001061 = = 0.564 xA1 = v 0.88467 fg uA1 = uf + x ufg = 504.47 + 0.564 × 2025.02 = 1646.6 kJ/kg State 1B: Table B.1.3, vB1 = 0.6173, uB1 = 2963.2, VB = mB1vB1 = 2.16 m3 Process constant total volume: m2 = mA1 + mB1 = 5.5 kg State 2: T2 , v2 ⇒ Table B.1.1 x2 = Vtot = VA + VB = 3.16 m3 and 1W2 = 0 / => v2 = Vtot/m2 = 0.5746 m3/kg two-phase as v2 < vg v2 – vf 0.5746 – 0.001044 = = 0.343 , 1.67185 vfg u2 = uf + x ufg = 418.91 + 0.343 × 2087.58= 1134.95 kJ/kg Heat transfer is from the energy equation 1Q2 = m2u2 - mA1uA1 - mB1uB1 = -7421 kJ Sonntag, Borgnakke and van Wylen 5.59 A 10-m high open cylinder, Acyl = 0.1 m2, contains 20°C water above and 2 kg of 20°C water below a 198.5-kg thin insulated floating piston, shown in Fig. P5.59. Assume standard g, Po. Now heat is added to the water below the piston so that it expands, pushing the piston up, causing the water on top to spill over the edge. This process continues until the piston reaches the top of the cylinder. Find the final state of the water below the piston (T, P, v) and the heat added during the process. Solution: C.V. Water below the piston. Piston force balance at initial state: F↑ = F↓ = PAA = mpg + mBg + P0A State 1A,B: Comp. Liq. ⇒ v ≅ vf = 0.001002 m3/kg; VA1 = mAvA1 = 0.002 m3; mass above the piston mtot = Vtot/v = 1/0.001002 = 998 kg mB1 = mtot - mA = 996 kg PA1 = P0 + (mp + mB)g/A = 101.325 + State 2A: u1A = 83.95 kJ/kg (198.5+996)*9.807 = 218.5 kPa 0.1*1000 mpg PA2 = P0 + A = 120.8 kPa ; vA2 = Vtot/ mA= 0.5 m3/kg xA2 = (0.5 - 0.001047)/1.4183 = 0.352 ; T2 = 105°C uA2 = 440.0 + 0.352 × 2072.34 = 1169.5 kJ/kg Continuity eq. in A: mA2 = mA1 P Energy: mA(u2 - u1) = 1Q2 - 1W2 Process: 1 P linear in V as mB is linear with V 1 1W2 = ⌠PdV = 2(218.5 + 120.82)(1 - 0.002) ⌡ = 169.32 kJ 1Q2 = mA(u2 - u1) + 1W2 = 2170.1 + 169.3 = 2340.4 kJ W 2 cb V Sonntag, Borgnakke and van Wylen 5.60 Assume the same setup as in Problem 5.48, but the room has a volume of 100 m3. Show that the final state is two-phase and find the final pressure by trial and error. C.V.: Containment room and reactor. Mass: m2 = m1 = Vreactor/v1 = 1/0.001823 = 548.5 kg Energy: m(u2 - u1) = 1Q2 - 1W2 = 0 - 0 = 0 ⇒ u2 = u1 = 1702.8 kJ/kg Total volume and mass => v2 = Vroom/m2 = 0.1823 m3/kg State 2: u2 , v2 Table B.1.1 see Figure. Note that in the vicinity of v = 0.1823 m3/kg crossing the saturated vapor line the internal energy is about 2585 kJ/kg. However, at the actual state 2, u = 1702.8 kJ/kg. Therefore state 2 must be in the two-phase region. T Trial & error v = vf + xvfg ; u = uf + xufg v2 - vf ⇒ u2 = 1702.8 = uf + v ufg fg 1060 kPa 1060 kPa u=2585 Compute RHS for a guessed pressure P2: sat vap 0.184 v P2 = 600 kPa: RHS = 669.88 + 0.1823-0.001101 × 1897.52 = 1762.9 0.31457 too large P2 = 550 kPa: RHS = 655.30 + 0.1823-0.001097 × 1909.17 = 1668.1 0.34159 too small Linear interpolation to match u = 1702.8 gives P2 ≅ 568.5 kPa Sonntag, Borgnakke and van Wylen Energy Equation: Multistep Solution 5.61 10 kg of water in a piston cylinder arrangement exists as saturated liquid/vapor at 100 kPa, with a quality of 50%. It is now heated so the volume triples. The mass of the piston is such that a cylinder pressure of 200 kPa will float it, as in Fig. 4.68. Find the final temperature and the heat transfer in the process. Solution: Take CV as the water. m2 = m1 = m ; m(u2 − u1) = 1Q2 − 1W2 Process: v = constant until P = Plift , then P is constant. State 1: Two-phase so look in Table B.1.2 at 100 kPa u1 = 417.33 + 0.5 × 2088.72 = 1461.7 kJ/kg, v1 = 0.001043 + 0.5 × 1.69296 = 0.8475 m3/kg State 2: v2, P2 ≤ Plift => v2 = 3 × 0.8475 = 2.5425 m3/kg ; Interpolate: T2 = 829°C, u2 = 3718.76 kJ/kg => V2 = mv2 = 25.425 m3 1W2 = Plift(V2 −V1) = 200 × 10 (2.5425 − 0.8475) = 3390 kJ 1Q2 = m(u2 − u1) + 1W2 = 10×(3718.76 − 1461.7) + 3390 = 25 961 kJ P Po 2 P2 cb H2O P1 1 V cb Sonntag, Borgnakke and van Wylen 5.62 Two tanks are connected by a valve and line as shown in Fig. P5.62. The volumes are both 1 m3 with R-134a at 20°C, quality 15% in A and tank B is evacuated. The valve is opened and saturated vapor flows from A into B until the pressures become equal. The process occurs slowly enough that all temperatures stay at 20°C during the process. Find the total heat transfer to the R-134a during the process. Solution: C.V.: A + B State 1A: vA1 = 0.000817 + 0.15 × 0.03524 = 0.006103 m3/kg uA1 = 227.03 + 0.15 × 162.16 = 251.35 kJ/kg mA1 = VA/vA1 = 163.854 kg Process: Constant temperature and constant total volume. m2 = mA1 ; V2 = VA + VB = 2 m3 ; v2 = V2/m2 = 0.012206 m3/kg 1W2 = ∫ P dV = 0 State 2: T2 , v2 ⇒ x2 = (0.012206 – 0.000817)/0.03524 = 0.3232 u2 = 227.03 + 0.3232 × 162.16 = 279.44 kJ/kg 1Q2 = m2u2 - mA1uA1 - mB1uB1 + 1W2 = m2(u2 - uA1) = 163.854 × (279.44 - 251.35) = 4603 kJ A B Sonntag, Borgnakke and van Wylen 5.63 Consider the same system as in the previous problem. Let the valve be opened and transfer enough heat to both tanks so all the liquid disappears. Find the necessary heat transfer. Solution: C.V. A + B, so this is a control mass. State 1A: vA1 = 0.000817 + 0.15 × 0.03524 = 0.006 103 m3/kg uA1 = 227.03 + 0.15 × 162.16 = 251.35 kJ/kg mA1 = VA/vA1 = 163.854 kg Process: Constant temperature and total volume. m2 = mA1 ; V2 = VA + VB = 2 m3 ; v2 = V2/m2 = 0.012 206 m3/kg State 2: x2 = 100%, v2 = 0.012206 ⇒ T2 = 55 + 5 × (0.012206 – 0.01316)/(0.01146 – 0.01316) = 57.8°C u2 = 406.01 + 0.56 × (407.85 – 406.01) = 407.04 kJ/kg 1Q2 = m2(u2 - uA1) = 163.854 × (407.04 - 251.35) = 25 510 kJ A B Sonntag, Borgnakke and van Wylen 5.64 A vertical cylinder fitted with a piston contains 5 kg of R-22 at 10°C, shown in Fig. P5.64. Heat is transferred to the system, causing the piston to rise until it reaches a set of stops at which point the volume has doubled. Additional heat is transferred until the temperature inside reaches 50°C, at which point the pressure inside the cylinder is 1.3 MPa. a. What is the quality at the initial state? b. Calculate the heat transfer for the overall process. Solution: C.V. R-22. Control mass goes through process: 1 -> 2 -> 3 As piston floats pressure is constant (1 -> 2) and the volume is constant for the second part (2 -> 3). So we have: v3 = v2 = 2 × v1 State 3: Table B.4.2 (P,T) v3 = 0.02015 m3/kg, u3 = 248.4 kJ/kg P 3 P o cb R-22 1 2 V So we can then determine state 1 and 2 Table B.4.1: v1 = 0.010075 = 0.0008 + x1 × 0.03391 => x1 = 0.2735 b) u1 = 55.92 + 0.2735 × 173.87 = 103.5 kJ/kg State 2: v2 = 0.02015 m3/kg, P2 = P1 = 681 kPa this is still 2-phase. 2 ⌡ 1W3 = 1W2 = ⌠ PdV = P1(V2 - V1) = 681 × 5 (0.02 - 0.01) = 34.1 kJ 1 1Q3 = m(u3-u1) + 1W3 = 5(248.4 - 103.5) + 34.1 = 758.6 kJ Sonntag, Borgnakke and van Wylen 5.65 Find the heat transfer in Problem 4.67. A piston/cylinder contains 1 kg of liquid water at 20°C and 300 kPa. Initially the piston floats, similar to the setup in Problem 4.64, with a maximum enclosed volume of 0.002 m3 if the piston touches the stops. Now heat is added so a final pressure of 600 kPa is reached. Find the final volume and the work in the process. Solution: Take CV as the water. Properties from table B.1 m2 = m1 = m ; m(u2 - u1) = 1Q2 - 1W2 State 1: Compressed liq. v = vf (20) = 0.001002 m3/kg, u = uf = 83.94 kJ/kg State 2: Since P > Plift then v = vstop = 0.002 and P = 600 kPa For the given P : vf < v < vg so 2-phase T = Tsat = 158.85 °C v = 0.002 = 0.001101 + x × (0.3157-0.001101) => x = 0.002858 u = 669.88 + 0.002858 ×1897.5 = 675.3 kJ/kg Work is done while piston moves at Plift= constant = 300 kPa so we get 1W2 = ∫ P dV = m Plift (v2 -v1) = 1×300(0.002 - 0.001002) = 0.299 kJ Heat transfer is found from energy equation 1Q2 = m(u2 - u1) + 1W2 = 1(675.3 - 83.94) + 0.299 = 591.66 kJ P P o cb H2O 1 2 V Sonntag, Borgnakke and van Wylen 5.66 Refrigerant-12 is contained in a piston/cylinder arrangement at 2 MPa, 150°C with a massless piston against the stops, at which point V = 0.5 m3. The side above the piston is connected by an open valve to an air line at 10°C, 450 kPa, shown in Fig. P5.66. The whole setup now cools to the surrounding temperature of 10°C. Find the heat transfer and show the process in a P–v diagram. C.V.: R-12. Control mass. Continuity: m = constant, Energy Eq.5.11: m(u2 - u1) = 1Q2 - 1W2 Air line F↓ = F↑ = P A = PairA + Fstop Process: if V < Vstop ⇒ Fstop = 0 / This is illustrated in the P-v diagram shown below. R-22 v1 = 0.01265 m3/kg, u1 = 252.1 kJ/kg State 1: ⇒ m = V/v = 39.523 kg State 2: T2 and on line ⇒ compressed liquid, see figure below. v2 ≅ vf = 0.000733 m3/kg ⇒ V2 = 0.02897 m3; u2 = uf = 45.06 kJ/kg 1W2 = ⌠PdV = Plift(V2 - V1) = 450 (0.02897 - 0.5) = -212.0 kJ ; ⌡ Energy eq. ⇒ 1Q2 = 39.526 (45.06 - 252.1) - 212 = -8395 kJ P T 150 ~73 1 2 MPa P = 2 MPa 1 T = 10 P = 450 kPa v 450 kPa 2 11.96 10 2 v Sonntag, Borgnakke and van Wylen 5.67 Find the heat transfer in Problem 4.114. A piston/cylinder (Fig. P4.114) contains 1 kg of water at 20°C with a volume of 0.1 m3. Initially the piston rests on some stops with the top surface open to the atmosphere, Po and a mass so a water pressure of 400 kPa will lift it. To what temperature should the water be heated to lift the piston? If it is heated to saturated vapor find the final temperature, volume and the work, 1W2. Solution: C.V. Water. This is a control mass. m2 = m1 = m ; m(u2 - u1) = 1Q2 - 1W2 P P o 1a 2 1 H2O V State 1: 20 C, v1 = V/m = 0.1/1 = 0.1 m3/kg x = (0.1 - 0.001002)/57.789 = 0.001713 u1 = 83.94 + 0.001713 × 2318.98 = 87.92 kJ/kg To find state 2 check on state 1a: P = 400 kPa, v = v1 = 0.1 m3/kg Table B.1.2: vf < v < vg = 0.4625 m3/kg State 2 is saturated vapor at 400 kPa since state 1a is two-phase. v2 = vg = 0.4625 m3/kg , V2 = m v2 = 0.4625 m3, u2 = ug= 2553.6 kJ/kg Pressure is constant as volume increase beyond initial volume. 1W2 = ∫ P dV = P (V2 - V1) = Plift (V2 – V1) = 400 (0.4625 – 0.1) = 145 kJ 1Q2 = m(u2 - u1) + 1W2 = 1 (2553.6 – 87.92) + 145 = 2610.7 kJ Sonntag, Borgnakke and van Wylen 5.68 A rigid container has two rooms filled with water, each 1 m3 separated by a wall. Room A has P = 200 kPa with a quality x = 0.80. Room B has P = 2 MPa and T = 400°C. The partition wall is removed and the water comes to a uniform state, which after a while due to heat transfer has a temperature of 200°C. Find the final pressure and the heat transfer in the process. Solution: C.V. A + B. Constant total mass and constant total volume. Continuity: m2 – mA1– mB1= 0 ; V2= VA+ VB= 2 m3 Energy Eq.5.11: U2 – U1 = m2u2 – mA1uA1 – mA1uA1 = 1Q2 – 1W2 = 1Q2 Process: V = VA + VB = constant State 1A: Table B.1.2 => 1W2 = 0 uA1= 504.47 + 0.8 × 2025.02 = 2124.47 kJ/kg, vA1= 0.001061 + 0.8 × 0.88467 = 0.70877 m3/kg State 1B: Table B.1.3 u B1= 2945.2, mA1= 1/vA1= 1.411 kg vB1= 0.1512 mB1= 1/vB1= 6.614 kg State 2: T2, v2 = V2/m 2= 2/(1.411 + 6.614) = 0.24924 m3/kg Table B.1.3 superheated vapor. 800 kPa < P2 < 1 MPa Interpolate to get the proper v2 0.24924-0.2608 P2 ≅ 800 + 0.20596-0.2608 × 200 = 842 kPa u2 ≅ 2628.8 kJ/kg From the energy equation 1Q2 = 8.025 × 2628.8 – 1.411 × 2124.47 – 6.614 × 2945.2 = - 1381 kJ P PB1 A Q B1 2 B PA1 A1 v Sonntag, Borgnakke and van Wylen 5.69 The cylinder volume below the constant loaded piston has two compartments A and B filled with water. A has 0.5 kg at 200 kPa, 150oC and B has 400 kPa with a quality of 50% and a volume of 0.1 m3. The valve is opened and heat is transferred so the water comes to a uniform state with a total volume of 1.006 m3. a) Find the total mass of water and the total initial volume. b) Find the work in the process c) Find the process heat transfer. Solution: Take the water in A and B as CV. Continuity: m2 - m1A - m1B = 0 Energy: m2u2 - m1Au1A - m1Bu1B = 1Q2 - 1W2 Process: P = constant = P1A if piston floats (VA positive) i.e. if V2 > VB = 0.1 m3 State A1: Sup. vap. Table B.1.3 v = 0.95964 m3/kg, u = 2576.9 kJ/kg => V = mv = 0.5 × 0.95964 = 0.47982 State B1: Table B.1.2 v = (1-x) × 0.001084 + x × 0.4625 = 0.2318 m3/kg => m = V/v = 0.4314 kg u = 604.29 + 0.5 × 1949.3 = 1578.9 kJ/kg State 2: 200 kPa, v2 = V2/m = 1.006/0.9314 = 1.0801 m3/kg Table B.1.3 => close to T2 = 200oC and u2 = 2654.4 kJ/kg So now V1 = 0.47982 + 0.1 = 0.5798 m3, m1 = 0.5 + 0.4314 = 0.9314 kg Since volume at state 2 is larger than initial volume piston goes up and the pressure then is constant (200 kPa which floats piston). 1W2 = ∫ P dV = Plift (V2 - V1) = 200 (1.006 - 0.57982) = 85.24 kJ 1Q2 = m2u2 - m1Au1A - m1Bu1B + 1W2 = 0.9314 × 2654.4 - 0.5 × 2576.9 - 0.4314 × 1578.9 + 85.24 = 588 kJ Sonntag, Borgnakke and van Wylen 5.70 A rigid tank A of volume 0.6 m3 contains 3 kg water at 120oC and the rigid tank B is 0.4 m3 with water at 600 kPa, 200oC. They are connected to a piston cylinder initially empty with closed valves. The pressure in the cylinder should be 800 kPa to float the piston. Now the valves are slowly opened and heat is transferred so the water reaches a uniform state at 250oC with the valves open. Find the final volume and pressure and the work and heat transfer in the process. C.V.: A + B + C. Only work in C, total mass constant. m2 - m1 = 0 => C m2 = mA1 + mB1 1W2 = B A U2 - U1 = 1Q2 - 1W2 ; ∫ PdV = Plift (V2 - V1) 1A: v = 0.6/3 = 0.2 m3/kg => xA1 = (0.2 - 0.00106)/0.8908 = 0.223327 u = 503.48 + 0.223327 × 2025.76 = 955.89 kJ/kg 3/kg => m = 0.4/0.35202 = 1.1363 kg ; u = 2638.91 kJ/kg 1B: v = 0.35202 m B1 m2 = 3 + 1.1363 = 4.1363 kg and P V2 = VA+ VB + VC = 1 + VC Locate state 2: Must be on P-V lines shown State 1a: 800 kPa, V +V v1a = Am B = 0.24176 m3/kg 800 kPa, v1a => T = 173°C Assume 800 kPa: 250°C => 1a 2 P2 too low. v = 0.29314 m3/kg > v1a OK Final state is : 800 kPa; 250°C => u2 = 2715.46 kJ/kg W = 800(0.29314 - 0.24176) × 4.1363 = 800 × (1.2125 - 1) = 170 kJ Q = m2u2 - m1u1 + 1W2 = m2u2 - mA1uA1 - mB1uB1 + 1W2 = 4.1363 × 2715.46 - 3 × 955.89 - 1.1363 × 2638.91 + 170 = 11 232 - 2867.7 - 2998.6 + 170 = 5536 kJ V Sonntag, Borgnakke and van Wylen 5.71 Calculate the heat transfer for the process described in Problem 4.60. A cylinder containing 1 kg of ammonia has an externally loaded piston. Initially the ammonia is at 2 MPa, 180°C and is now cooled to saturated vapor at 40°C, and then further cooled to 20°C, at which point the quality is 50%. Find the total work for the process, assuming a piecewise linear variation of P versus V. Solution: C.V. Ammonia going through process 1 - 2 - 3. Control mass. Continuity: m = constant, Energy Eq.5.11: m(u3 - u1) = 1Q3 - 1W3 Process: P is piecewise linear in V State 1: (T, P) Table B.2.2: v1 = 0.10571 m3/kg, u1 = 1630.7 kJ/kg State 2: (T, x) Table B.2.1 sat. vap. P2 = 1555 kPa, v2 = 0.08313 m3/kg P 1 2000 2 1555 857 o 180 C o 40 C 3 o 20 C v State 3: (T, x) P3 = 857 kPa, v3 = (0.001638+0.14922)/2 = 0.07543 u3 = (272.89 + 1332.2)/2 = 802.7 kJ/kg Process: piecewise linear P versus V, see diagram. Work is area as: 3 W13 = ⌠ PdV ≈ ( ⌡ 1 = P2 + P3 P1 + P2 ) m(v2 - v1) + ( 2 ) m(v3 - v2) 2 2000 + 1555 1555 + 857 1(0.08313 - 0.10571) + 1(0.07543 - 0.08313) 2 2 = -49.4 kJ From the energy equation, we get the heat transfer as: 1Q3 = m(u3 - u1) + 1W3 = 1× (802.7 - 1630.7) - 49.4 = -877.4 kJ Sonntag, Borgnakke and van Wylen 5.72 Calculate the heat transfer for the process described in Problem 4.70. A piston cylinder setup similar to Problem 4.24 contains 0.1 kg saturated liquid and vapor water at 100 kPa with quality 25%. The mass of the piston is such that a pressure of 500 kPa will float it. The water is heated to 300°C. Find the final pressure, volume and the work, 1W2. Solution: P Take CV as the water: m2 = m1 = m Energy Eq.5.11: Process: m(u2 - u1) = 1Q2 - 1W2 v = constant until P = Plift To locate state 1: Table B.1.2 1a Plift P1 v1 = 0.001043 + 0.25×1.69296 = 0.42428 m3/kg 2 1 cb V u1 = 417.33 + 0.25×2088.7 = 939.5 kJ/kg State 1a: 500 kPa, v1a = v1 = 0.42428 > vg at 500 kPa, so state 1a is superheated vapor Table B.1.3 T1a = 200°C State 2 is 300°C so heating continues after state 1a to 2 at constant P = 500 kPa. 2: T2, P2 = Plift => Tbl B.1.3 v2 =0.52256 m3/kg; u2 = 2802.9 kJ/kg From the process, see also area in P-V diagram 1W2 = Plift m(v2 - v1) = 500 × 0.1 (0.5226 - 0.4243) = 4.91 kJ From the energy equation 1Q2 = m(u2 - u1) + 1W2 = 0.1(2802.9 - 939.5) + 4.91 = 191.25 kJ Sonntag, Borgnakke and van Wylen 5.73 A cylinder/piston arrangement contains 5 kg of water at 100°C with x = 20% and the piston, mP = 75 kg, resting on some stops, similar to Fig. P5.73. The outside pressure is 100 kPa, and the cylinder area is A = 24.5 cm2. Heat is now added until the water reaches a saturated vapor state. Find the initial volume, final pressure, work, and heat transfer terms and show the P–v diagram. Solution: C.V. The 5 kg water. Continuty: m2 = m1 = m ; Energy: m(u2 - u1) = 1Q2 - 1W2 Process: V = constant if P < Plift otherwise P = Plift see P-v diagram. 75 × 9.807 P3 = P2 = Plift = P0 + mp g / Ap = 100 + 0.00245 × 1000 = 400 kPa P P o 2 3 o 143 C cb H2O o cb 1 100 C v State 1: (T,x) Table B.1.1 v1 = 0.001044 + 0.2 × 1.6719, V1 = mv1 = 5 × 0.3354 = 1.677 m3 u1 = 418.91 + 0.2 × 2087.58 = 836.4 kJ/kg State 3: (P, x = 1) Table B.1.2 => v3 = 0.4625 > v1, u3 = 2553.6 kJ/kg Work is seen in the P-V diagram (if volume changes then P = Plift) 1W3 = 2W3 = Pextm(v3 - v2) = 400 × 5(0.46246 - 0.3354) = 254.1 kJ Heat transfer is from the energy equation 1Q3 = 5 (2553.6 - 836.4) + 254.1 = 8840 kJ Sonntag, Borgnakke and van Wylen Energy Equation: Solids and Liquids 5.74 Because a hot water supply must also heat some pipe mass as it is turned on so it does not come out hot right away. Assume 80oC liquid water at 100 kPa is cooled to 45oC as it heats 15 kg of copper pipe from 20 to 45oC. How much mass (kg) of water is needed? Solution: C.V. Water and copper pipe. No external heat transfer, no work. Energy Eq.5.11: U2 – U1 = ∆Ucu + ∆UH2O = 0 – 0 From Eq.5.18 and Table A.3: kJ ∆Ucu = mC ∆Τ = 15 kg × 0.42 kg K × (45 – 20) K = 157.5 kJ From the energy equation mH2O = - ∆Ucu / ∆uH2O 157.5 mH2O = ∆Ucu / CH2O(- ∆ΤH2O) = = 1.076 kg 4.18 × 35 or using Table B.1.1 for water 157.5 mH2O = ∆Ucu / ( u1- u2) = 334.84 – 188.41 = 1.076 kg Cu pipe Water The real problem involves a flow and is not analyzed by this simple process. Sonntag, Borgnakke and van Wylen 5.75 A house is being designed to use a thick concrete floor mass as thermal storage material for solar energy heating. The concrete is 30 cm thick and the area exposed to the sun during the daytime is 4 m × 6 m. It is expected that this mass will undergo an average temperature rise of about 3°C during the day. How much energy will be available for heating during the nighttime hours? Solution: C.V. The mass of concrete. Concrete is a solid with some properties listed in Table A.3 V = 4 × 6 × 0.3 = 7.2 m3 ; m = ρV = 2200 kg/m3 × 7.2 m3 = 15 840 kg From Eq.5.18 and C from table A.3 kJ ∆U = m C ∆T = 15840 kg × 0.88 kg K × 3 K = 41818 kJ = 41.82 MJ Sonntag, Borgnakke and van Wylen 5.76 A copper block of volume 1 L is heat treated at 500°C and now cooled in a 200-L oil bath initially at 20°C, shown in Fig. P5.76. Assuming no heat transfer with the surroundings, what is the final temperature? Solution: C.V. Copper block and the oil bath. Also assume no change in volume so the work will be zero. Energy Eq.: U2 - U1 = mmet(u2 - u1)met + moil(u2 - u1)oil = 1Q2 - 1W2 = 0 Properties from Table A.3 and A.4 mmet = Vρ = 0.001 m3 × 8300 kg/m3 = 8.3 kg, moil = Vρ = 0.2 m3 × 910 kg/m3 = 182 kg Solid and liquid Eq.5.17: ∆u ≅ Cv ∆T, Table A.3 and A.4: kJ kJ Cv met = 0.42 kg K, Cv oil = 1.8 kg K The energy equation for the C.V. becomes mmetCv met(T2 − T1,met) + moilCv oil(T2 − T1,oil) = 0 8.3 × 0.42(T2 − 500) + 182 × 1.8 (T2 − 20) = 0 331.09 T2 – 1743 – 6552 = 0 ⇒ T2 = 25 °C Sonntag, Borgnakke and van Wylen 5.77 A 1 kg steel pot contains 1 kg liquid water both at 15oC. It is now put on the stove where it is heated to the boiling point of the water. Neglect any air being heated and find the total amount of energy needed. Solution: Energy Eq.: U2 − U1= 1Q2 − 1W2 The steel does not change volume and the change for the liquid is minimal, so 1W2 ≅ 0. State 2: T2 = Tsat (1atm) = 100oC Tbl B.1.1 : u1 = 62.98 kJ/kg, u2 = 418.91 kJ/kg Tbl A.3 : Cst = 0.46 kJ/kg K Solve for the heat transfer from the energy equation 1Q2 = U2 − U1 = mst (u2 − u1)st + mH2O (u2 − u1)H2O = mstCst (T2 – T1) + mH2O (u2 − u1)H2O kJ 1Q2 = 1 kg × 0.46 kg K ×(100 – 15) K + 1 kg ×(418.91 – 62.98) kJ/kg = 39.1 + 355.93 = 395 kJ Sonntag, Borgnakke and van Wylen 5.78 A car with mass 1275 kg drives at 60 km/h when the brakes are applied quickly to decrease its speed to 20 km/h. Assume the brake pads are 0.5 kg mass with heat capacity of 1.1 kJ/kg K and the brake discs/drums are 4.0 kg steel. Further assume both masses are heated uniformly. Find the temperature increase in the brake assembly. Solution: C.V. Car. Car loses kinetic energy and brake system gains internal u. No heat transfer (short time) and no work term. m = constant; Energy Eq.5.11: 1 2 2 E2 - E1 = 0 - 0 = mcar 2(V2 − V1) + mbrake(u2 − u1) The brake system mass is two different kinds so split it, also use Cv from Table A.3 since we do not have a u table for steel or brake pad material. 2 1000 msteel Cv ∆T + mpad Cv ∆T = mcar 0.5 (602 − 202) 3600 m2/s2 kJ (4 × 0.46 + 0.5 × 1.1) K ∆T = 1275 kg × 0.5 × (3200 × 0.077 16) m2/s2 = 157 406 J = 157.4 kJ => ∆T = 65.9 °C Sonntag, Borgnakke and van Wylen 5.79 Saturated, x = 1%, water at 25°C is contained in a hollow spherical aluminum vessel with inside diameter of 0.5 m and a 1-cm thick wall. The vessel is heated until the water inside is saturated vapor. Considering the vessel and water together as a control mass, calculate the heat transfer for the process. Solution: C.V. Vessel and water. This is a control mass of constant volume. Continuity Eq.: m2 = m1 Energy Eq.5.11: Process: U2 - U1 = 1Q2 - 1W2 = 1Q2 V = constant => 1W2 = 0 used above State 1: v1 = 0.001003 + 0.01 × 43.359 = 0.4346 m3/kg u1 = 104.88 + 0.01 × 2304.9 = 127.9 kJ/kg State 2: x2 = 1 and constant volume so v2 = v1 = V/m vg T2 = v1 = 0.4346 => T2 = 146.1°C; u2 = uG2 = 2555.9 0.06545 π VINSIDE = 6 (0.5)3 = 0.06545 m3 ; mH2O = 0.4346 = 0.1506 kg π Valu = 6((0.52)3 - (0.5)3) = 0.00817 m3 malu = ρaluValu = 2700 × 0.00817 = 22.065 kg From the energy equation 1Q2 = U2 - U1 = mH2O(u2 - u1)H2O + maluCv alu(T2 - T1) = 0.1506(2555.9 - 127.9) + 22.065 × 0.9(146.1 - 25) = 2770.6 kJ Sonntag, Borgnakke and van Wylen 5.80 A 25 kg steel tank initially at –10oC is filled up with 100 kg of milk (assume properties as water) at 30oC. The milk and the steel come to a uniform temperature of +5 oC in a storage room. How much heat transfer is needed for this process? Solution: C.V. Steel + Milk. This is a control mass. Energy Eq.5.11: U2 − U1 = 1Q2 − 1W2 = 1Q2 Process: V = constant, so there is no work 1W2 = 0. Use Eq.5.18 and values from A.3 and A.4 to evaluate changes in u 1Q2 = msteel (u2 - u1)steel + mmilk(u2 - u1)milk kJ kJ = 25 kg × 0.466 kg K × [5 − (−10)] Κ + 100 kg ×4.18 kg K × (5 − 30) Κ = 172.5 − 10450 = −10277 kJ Sonntag, Borgnakke and van Wylen 5.81 An engine consists of a 100 kg cast iron block with a 20 kg aluminum head, 20 kg steel parts, 5 kg engine oil and 6 kg glycerine (antifreeze). Everything begins at 5oC and as the engine starts we want to know how hot it becomes if it absorbs a net of 7000 kJ before it reaches a steady uniform temperature. Energy Eq.: U2 − U1= 1Q2 − 1W2 Process: The steel does not change volume and the change for the liquid is minimal, so 1W2 ≅ 0. So sum over the various parts of the left hand side in the energy equation mFe (u2 − u1) + mAl (u2 − u1)Al + mst (u − u1)st + moil (u2 − u1)oil + mgly (u2 − u1)gly = 1Q2 Tbl A.3 : CFe = 0.42 , CAl = 0.9, Cst = 0.46 all units of kJ/kg K Tbl A.4 : Coil = 1.9 , Cgly = 2.42 all units of kJ/kg K So now we factor out T2 –T1 as u2 − u1 = C(T2 –T1) for each term [ mFeCFe + mAlCAl + mstCst+ moilCoil + mglyCgly ] (T2 –T1) = 1Q2 T2 –T1 = 1Q2 / Σmi Ci 7000 100× 0.42 + 20× 0.9 + 20× 0.46 + 5 ×1.9 + 6 ×2.42 7000 = 93.22 = 75 K = T2 = T1 + 75 = 5 + 75 = 80oC Air intake filter Shaft power Exhaust flow Coolant flow Fan Radiator Atm. air Sonntag, Borgnakke and van Wylen Properties (u, h, Cv and Cp), Ideal Gas 5.82 Use the ideal gas air table A.7 to evaluate the heat capacity Cp at 300 K as a slope of the curve h(T) by ∆h/∆T. How much larger is it at 1000 K and 1500 K. Solution : From Eq.5.24: dh ∆h h320 - h290 = = 1.005 kJ/kg K Cp = dT = ∆T 320 - 290 1000K Cp = ∆h h1050 - h950 1103.48 - 989.44 = = = 1.140 kJ/kg K 100 ∆T 1050 - 950 1500K Cp = ∆h h1550 - h1450 1696.45 - 1575.4 = = = 1.21 kJ/kg K 100 ∆T 1550 - 1450 Notice an increase of 14%, 21% respectively. h C p 1500 Cp 300 300 1000 1500 T Sonntag, Borgnakke and van Wylen 5.83 We want to find the change in u for carbon dioxide between 600 K and 1200 K. a) Find it from a constant Cvo from table A.5 b) Find it from a Cvo evaluated from equation in A.6 at the average T. c) Find it from the values of u listed in table A.8 Solution : a) ∆u ≅ Cvo ∆T = 0.653 × (1200 – 600) = 391.8 kJ/kg b) 1 Tavg = 2 (1200 + 600) = 900, T 900 θ = 1000 = 1000 = 0.9 Cpo = 0.45 + 1.67 × 0.9 - 1.27 × 0.92 + 0.39 × 0.93 = 1.2086 kJ/kg K Cvo = Cpo – R = 1.2086 – 0.1889 = 1.0197 kJ/kg K ∆u = 1.0197 × (1200 – 600) = 611.8 kJ/kg c) ∆u = 996.64 – 392.72 = 603.92 kJ/kg u u1200 u600 T 300 600 1200 Sonntag, Borgnakke and van Wylen 5.84 We want to find the change in u for oxygen gas between 600 K and 1200 K. a) Find it from a constant Cvo from table A.5 b) Find it from a Cvo evaluated from equation in A.6 at the average T. c) Find it from the values of u listed in table A.8 Solution: a) ∆u ≅ Cvo ∆T = 0.662 × (1200 − 600) = 397.2 kJ/kg b) 1 Tavg = 2 (1200 + 600) = 900 K, T 900 θ = 1000 = 1000 = 0.9 Cpo = 0.88 − 0.0001 × 0.9 + 0.54 × 0.92 − 0.33 × 0.93 = 1.0767 Cvo = Cpo − R = 1.0767 − 0.2598 = 0.8169 kJ/kg K ∆u = 0.8169 × (1200 − 600)= 490.1 kJ/kg c) ∆u = 889.72 − 404.46 = 485.3 kJ/kg u u1200 u600 T 300 600 1200 Sonntag, Borgnakke and van Wylen 5.85 Water at 20°C, 100 kPa, is brought to 200 kPa, 1500°C. Find the change in the specific internal energy, using the water table and the ideal gas water table in combination. Solution: State 1: Table B.1.1 u1 ≅ uf = 83.95 kJ/kg State 2: Highest T in Table B.1.3 is 1300°C Using a ∆u from the ideal gas tables, A.8, we get u1500 = 3139 kJ/kg u1300 = 2690.72 kJ/kg u1500 - u1300 = 448.26 kJ/kg We now add the ideal gas change at low P to the steam tables, B.1.3, ux = 4683.23 kJ/kg as the reference. u2 - u1 = (u2 - ux)ID.G. + (ux - u1) = 448.28 + 4683.23 - 83.95 = 5048 kJ/kg Sonntag, Borgnakke and van Wylen 5.86 We want to find the increase in temperature of nitrogen gas at 1200 K when the specific internal energy is increased with 40 kJ/kg. a) Find it from a constant Cvo from table A.5 b) Find it from a Cvo evaluated from equation in A.6 at 1200 K. c) Find it from the values of u listed in table A.8 Solution : ∆u = ∆uA.8 ≅ Cv avg ∆T ≅ Cvo ∆T a) 40 ∆T = ∆u / Cvo = 0.745 = 53.69°C b) θ = 1200 / 1000 =1.2 Cpo = 1.11 – 0.48 × 1.2 + 0.96 × 1.22 – 0.42 × 1.2 3 = 1.1906 kJ/kg K Cvo = Cpo – R = 1.1906 – 0.2968 = 0.8938 kJ/kg K ∆T = ∆u / Cvo = 40 / 0.8938 = 44.75°C c) u = u1 + ∆u = 957 + 40 = 997 kJ/kg less than 1300 K so linear interpolation. 1300 – 1200 ∆T = 1048.46 – 957 × 40 = 43.73°C Cvo ≅ (1048.46 – 957) / 100 = 0.915 kJ/kg K So the formula in A.6 is accurate within 2.3%. Sonntag, Borgnakke and van Wylen 5.87 For an application the change in enthalpy of carbon dioxide from 30 to 1500°C at 100 kPa is needed. Consider the following methods and indicate the most accurate one. a. Constant specific heat, value from Table A.5. b. Constant specific heat, value at average temperature from the equation in Table A.6. c. Variable specific heat, integrating the equation in Table A.6. d. Enthalpy from ideal gas tables in Table A.8. Solution: a) ∆h = Cpo∆T = 0.842 (1500 - 30) = 1237.7 kJ/kg b) Tave = 2 (30 + 1500) + 273.15 = 1038.15 K; θ = T/1000 = 1.0382 1 Table A.6 ⇒ Cpo =1.2513 ∆h = Cpo,ave ∆T = 1.2513 × 1470 = 1839 kJ/kg c) For the entry to Table A.6: θ2 = 1.77315 ; θ1 = 0.30315 ∆h = h2- h1 = ∫ Cpo dT 1 = [0.45 (θ2 - θ1) + 1.67 × 2 (θ22 - θ12) 1 1 4 4 –1.27 × 3 (θ23 - θ13) + 0.39× 4 (θ2 - θ1 )] = 1762.76 kJ/kg d) ∆h = 1981.35 – 217.12 = 1764.2 kJ/kg The result in d) is best, very similar to c). For large ∆T or small ∆T at high Tavg, a) is very poor. Sonntag, Borgnakke and van Wylen 5.88 An ideal gas is heated from 500 to 1500 K. Find the change in enthalpy using constant specific heat from Table A.5 (room temperature value) and discuss the accuracy of the result if the gas is a. Argon b. Oxygen c. Carbon dioxide Solution: T1 = 500 K, T2 = 1500 K, ∆h = CP0(T2-T1) a) Ar : ∆h = 0.520(1500-500) = 520 kJ/kg Monatomic inert gas very good approximation. b) O2 : ∆h = 0.922(1500-500) = 922 kJ/kg Diatomic gas approximation is OK with some error. c) CO2: ∆h = 0.842(1500-500) = 842 kJ/kg Polyatomic gas heat capacity changes, see figure 5.11 See also appendix C for more explanation. Sonntag, Borgnakke and van Wylen Energy Equation: Ideal Gas 5.89 A 250 L rigid tank contains methane at 500 K, 1500 kPa. It is now cooled down to 300 K. Find the mass of methane and the heat transfer using a) ideal gas and b) the methane tables. Solution: a) Assume ideal gas, P2 = P1 × (Τ2 / Τ1) = 1500 × 300 / 500 = 900 kPa 1500 × 0.25 m = P1V/RT1 = 0.5183 × 500 = 1.447 kg Use specific heat from Table A.5 u2 - u1 = Cv (T2 – T1) = 1.736 (300 – 500) = –347.2 kJ/kg 1Q2 = m(u2 - u1) = 1.447(-347.2) = –502.4 kJ b) Using the methane Table B.7, v1 = 0.17273 m3/kg, u1 = 872.37 kJ/kg m = V/v1 = 0.25/0.17273 = 1.4473 kg State 2: v2 = v1 and 300 K is found between 800 and 1000 kPa 0.17273 – 0.19172 u2 = 467.36 + (465.91 – 467.36) 0.15285 – 0.19172 = 466.65 kJ/kg 1Q2 = 1.4473 (466.65 – 872.37) = –587.2 kJ Sonntag, Borgnakke and van Wylen 5.90 A rigid insulated tank is separated into two rooms by a stiff plate. Room A of 0.5 m3 contains air at 250 kPa, 300 K and room B of 1 m3 has air at 150 kPa, 1000 K. The plate is removed and the air comes to a uniform state without any heat transfer. Find the final pressure and temperature. Solution: C.V. Total tank. Control mass of constant volume. Mass and volume: m2 = mA + mB; V = VA + VB = 1.5 m3 Energy Eq.: U2 – U1 = m2 u2 – mAuA1 – mBuB1 = Q – W = 0 Process Eq.: V = constant ⇒ W = 0; Ideal gas at 1: mA = PA1VA/RTA1 = 250 × 0.5/(0.287 × 300) = 1.452 kg Insulated ⇒ Q = 0 u A1= 214.364 kJ/kg from Table A.7 Ideal gas at 2: mB = PB1VB/RT B1= 150 × 1/(0.287 × 1000) = 0.523 kg u B1= 759.189 kJ/kg from Table A.7 m2 = mA + mB = 1.975 kg u2 = mAuA1 + mBuB1 1.452 × 214.364 + 0.523 × 759.189 = = 358.64 kJ/kg 1.975 m2 => Table A.7.1: T2 = 498.4 K P2 = m2 RT2 /V = 1.975 × 0.287 × 498.4/1.5 = 188.3 kPa A B cb Sonntag, Borgnakke and van Wylen 5.91 A rigid container has 2 kg of carbon dioxide gas at 100 kPa, 1200 K that is heated to 1400 K. Solve for the heat transfer using a. the heat capacity from Table A.5 and b. properties from Table A.8 Solution: C.V. Carbon dioxide, which is a control mass. Energy Eq.5.11: U2 – U1 = m (u2- u1) = 1Q2 − 1W2 ∆V = 0 ⇒ 1W2 = 0 a) For constant heat capacity we have: u2- u1 = Cvo (T2- T1) so Process: 1Q2 ≅ mCvo (T2- T1) = 2 × 0.653 × (1400 –1200) = 261.2 kJ b) Taking the u values from Table A.8 we get 1Q2 = m (u2- u1) = 2 × (1218.38 – 996.64) = 443.5 kJ Sonntag, Borgnakke and van Wylen 5.92 Do the previous problem for nitrogen, N2, gas. A rigid container has 2 kg of carbon dioxide gas at 100 kPa, 1200 K that is heated to 1400 K. Solve for the heat transfer using a. the heat capacity from Table A.5 and b. properties from Table A.8 Solution: C.V. Nitrogen gas, which is a control mass. Energy Eq.5.11: Process: U2 – U1 = m (u2- u1) = 1Q2 − 1W2 ∆V = 0 ⇒ 1W2 = 0 a) For constant heat capacity we have: u2- u1 = Cvo (T2 - T1) so 1Q2 ≅ mCvo (T2- T1) = 2 × 0.745 × (1400 – 1200) = 298 kJ b) Taking the u values from Table A.8, we get 1Q2 = m (u2- u1) = 2 × (1141.35 – 957) = 368.7 kJ Sonntag, Borgnakke and van Wylen 5.93 A 10-m high cylinder, cross-sectional area 0.1 m2, has a massless piston at the bottom with water at 20°C on top of it, shown in Fig. P5.93. Air at 300 K, volume 0.3 m3, under the piston is heated so that the piston moves up, spilling the water out over the side. Find the total heat transfer to the air when all the water has been pushed out. Solution: Po P H2O cb P1 1 2 P0 V air V1 Vmax The water on top is compressed liquid and has volume and mass VH2O = Vtot - Vair = 10 × 0.1 - 0.3 = 0.7 m3 mH2O = VH2O/vf = 0.7 / 0.001002 = 698.6 kg The initial air pressure is then 698.6 × 9.807 P1 = P0 + mH2Og/A = 101.325 + 0.1 × 1000 = 169.84 kPa 169.84 × 0.3 and then mair = PV/RT = 0.287 × 300 = 0.592 kg State 2: No liquid water over the piston so P2 = P0 + 0 = 101.325 kPa, / State 2: P2, V2 ⇒ V2 = 10×0.1 = 1 m3 T1P2V2 300×101.325×1 T2 = P V = 169.84×0.3 = 596.59 K 11 The process line shows the work as an area 1 1 ⌠ 1W2 = ⌡PdV = 2 (P1 + P2)(V2 - V1) = 2 (169.84 + 101.325)(1 - 0.3) = 94.91 kJ The energy equation solved for the heat transfer becomes 1Q2 = m(u2 - u1) + 1W2 ≅ mCv(T2 - T1) + 1W2 = 0.592 × 0.717 × (596.59 - 300) + 94.91 = 220.7 kJ Remark: we could have used u values from Table A.7: u2 - u1 = 432.5 - 214.36 = 218.14 kJ/kg versus 212.5 kJ/kg with Cv. Sonntag, Borgnakke and van Wylen 5.94 Find the heat transfer in Problem 4.43. A piston cylinder contains 3 kg of air at 20oC and 300 kPa. It is now heated up in a constant pressure process to 600 K. Solution: Ideal gas PV = mRT State 1: T1, P1 State 2: T2, P2 = P1 V2 = mR T2 / P2 = 3×0.287×600 / 300 = 1.722 m3 P2V2 = mRT2 Process: P = constant, W2 1 = ⌠ PdV = P (V2 - V1) = 300 (1.722 – 0.8413) = 264.2 kJ ⌡ Energy equation becomes U2 - U1 = 1Q2 - 1W2 = m(u2 - u1) Q = U2 - U1 + 1W2 = 3(435.097 – 209.45) + 264.2 = 941 kJ 12 P 300 T 2 1 T1 2 600 300 kPa T2 293 v 1 v Sonntag, Borgnakke and van Wylen 5.95 An insulated cylinder is divided into two parts of 1 m3 each by an initially locked piston, as shown in Fig. P5.95. Side A has air at 200 kPa, 300 K, and side B has air at 1.0 MPa, 1000 K. The piston is now unlocked so it is free to move, and it conducts heat so the air comes to a uniform temperature TA = TB. Find the mass in both A and B, and the final T and P. C.V. A + B Force balance on piston: PAA = PBA So the final state in A and B is the same. State 1A: Table A.7 uA1 = 214.364 kJ/kg, mA = PA1VA1/RTA1 = 200 × 1/(0.287 × 300) = 2.323 kg State 1B: Table A.7 uB1 = 759.189 kJ/kg, mB = PB1VB1/RTB1 = 1000 × 1/(0.287 × 1000) = 3.484 kg For chosen C.V. 1Q2 = 0 , 1W2 = 0 so the energy equation becomes mA(u2 - u1)A + mB(u2 - u1)B = 0 (mA + mB)u2 = mAuA1 + mBuB1 = 2.323 × 214.364 + 3.484 × 759.189 = 3143 kJ u2 = 3143/(3.484 + 2.323) = 541.24 kJ/kg From interpolation in Table A.7: ⇒ T2 = 736 K kJ P = (mA + mB)RT2/Vtot = 5.807 kg × 0.287 kg K × 736 K/ 2 m3 = 613 kPa A B Sonntag, Borgnakke and van Wylen 5.96 A piston cylinder contains air at 600 kPa, 290 K and a volume of 0.01 m3. A constant pressure process gives 54 kJ of work out. Find the final temperature of the air and the heat transfer input. Solution: C.V AIR control mass Continuity Eq.: m2 – m1 = 0 m (u2 − u1) = 1Q2 - 1W2 Energy Eq.: Process: 1 : P1 , T1,V1 P=C so 1W2 = ∫ P dV = P(V2 V1) 2 : P1 = P2 , ? m1 = P1V1/RT1 = 600 ×0.01 / 0.287 ×290 = 0.0721 kg 1W2 = P(V2 – V1) = 54 kJ V2 – V1 = 1W2 / P = 54 kJ / 600 kPa = 0.09 m3 V2 = V1 + 1W2 / P = 0.01 + 0.09 = 0.10 m3 Ideal gas law : P2V2 = mRT2 P2V2 0.10 T2 = P2V2 / mR = P V T1 = 0.01 × 290 = 2900 K 11 Energy equation with u’s from table A.7.1 1Q2 = m (u2 − u1 ) + 1W2 = 0.0721 ( 2563.8 – 207.2 ) + 54 = 223.9 kJ Sonntag, Borgnakke and van Wylen 5.97 A cylinder with a piston restrained by a linear spring contains 2 kg of carbon dioxide at 500 kPa, 400°C. It is cooled to 40°C, at which point the pressure is 300 kPa. Calculate the heat transfer for the process. Solution: C.V. The carbon dioxide, which is a control mass. Continuity Eq.: m2 – m1 = 0 Energy Eq.: m (u2 − u1) = 1Q2 - 1W2 Process Eq.: P = A + BV (linear spring) 1 1W2 = ⌠PdV = 2(P1 + P2)(V2 - V1) ⌡ Equation of state: PV = mRT (ideal gas) State 1: V1 = mRT1/P1 = 2 × 0.18892 × 673.15 /500 = 0.5087 m3 State 2: V2 = mRT2/P2 = 2 × 0.18892 × 313.15 /300 = 0.3944 m3 1 1W2 = 2(500 + 300)(0.3944 - 0.5087) = -45.72 kJ To evaluate u2 - u1 we will use the specific heat at the average temperature. From Figure 5.11: Cpo(Tavg) = 45/44 = 1.023 ⇒ Cvo = 0.83 = Cpo - R For comparison the value from Table A.5 at 300 K is Cvo = 0.653 kJ/kg K 1Q2 = m(u2 - u1) + 1W2 = mCvo(T2 - T1) + 1W2 = 2 × 0.83(40 - 400) - 45.72 = -643.3 kJ P 2 CO 2 1 v Remark: We could also have used the ideal gas table in A.8 to get u2 - u1. Sonntag, Borgnakke and van Wylen 5.98 Water at 100 kPa, 400 K is heated electrically adding 700 kJ/kg in a constant pressure process. Find the final temperature using a) The water tables B.1 b) The ideal gas tables A.8 c) Constant specific heat from A.5 Solution : Energy Eq.5.11: Process: u2 - u1 = 1q2 - 1w2 P = constant => 1w2 = P ( v2 - v1 ) Substitute this into the energy equation to get 1q2 = h2 - h1 Table B.1: h1 ≅ 2675.46 + 126.85 - 99.62 150 - 99.62 × (2776.38 –2675.46) = 2730.0 kJ/kg h2 = h1 + 1q2 = 2730 + 700 = 3430 kJ/kg 3430 - 3278.11 T2 = 400 + ( 500 – 400 ) × 3488.09 - 3278.11 = 472.3°C Table A.8: h2 = h1 + 1q2 = 742.4 + 700 = 1442.4 kJ/kg 1442.4 - 1338.56 T2 = 700 + (750 – 700 ) × 1443.43 - 1338.56 = 749.5 K = 476.3°C Table A.5 h2 - h1 ≅ Cpo ( T2 - T1 ) T2 = T1 + 1q2 / Cpo = 400 + 700 / 1.872 = 773.9K = 500.8°C Sonntag, Borgnakke and van Wylen 5.99 A piston/cylinder has 0.5 kg air at 2000 kPa, 1000 K as shown. The cylinder has stops so Vmin = 0.03 m3. The air now cools to 400 K by heat transfer to the ambient. Find the final volume and pressure of the air (does it hit the stops?) and the work and heat transfer in the process. Solution: We recognize this is a possible two-step process, one of constant P and one of constant V. This behavior is dictated by the construction of the device. Continuity Eq.: m2 – m1 = 0 Energy Eq.5.11: m(u2 - u1) = 1Q2 - 1W2 P = constant = F/A = P1 if V > Vmin Process: V = constant = V1a = Vmin State 1: (P, T) if P < P1 V1 = mRT1/P1 = 0.5 × 0.287 × 1000/2000 = 0.07175 m3 The only possible P-V combinations for this system is shown in the diagram so both state 1 and 2 must be on the two lines. For state 2 we need to know if it is on the horizontal P line segment or the vertical V segment. Let us check state 1a: State 1a: P1a = P1, V1a = Vmin V1a 0.03 Ideal gas so T1a = T1 V = 1000 × 0.07175 = 418 K 1 We see that T2 < T1a and state 2 must have V2 = V1a = Vmin = 0.03 m3. T2 V1 400 0.07175 P2 = P1× T × V = 2000 × 1000 × 0.03 = 1913.3 kPa 1 2 The work is the area under the process curve in the P-V diagram 3 2 1W2 = ⌠1 P dV = P1 (V1a – V1) = 2000 kPa (0.03 – 0.07175) m = – 83.5 kJ ⌡ Now the heat transfer is found from the energy equation, u’s from Table A.7.1, 1Q2 = m(u2 - u1) + 1W2 = 0.5 (286.49 - 759.19) – 83.5 = -319.85 kJ P 1a T 1 T1 P1 T1a P2 2 V T2 1 1a 2 V Sonntag, Borgnakke and van Wylen 5.100 A spring loaded piston/cylinder contains 1.5 kg of air at 27C and 160 kPa. It is now heated to 900 K in a process where the pressure is linear in volume to a final volume of twice the initial volume. Plot the process in a P-v diagram and find the work and heat transfer. Take CV as the air. m2 = m1 = m ; m(u2 -u1) = 1Q2 - 1W2 Process: P = A + BV => 1W2 = ∫ P dV = area = 0.5(P1 + P2)(V2 -V1) State 1: Ideal gas. V1 = mRT1/P1 = 1.5× 0.287 × 300/160 = 0.8072 m3 Table A.7 u1 = u(300) = 214.36 kJ/kg State 2: P2V2 = mRT2 so ratio it to the initial state properties P2V2 /P1V1 = P22 /P1 = mRT2 /mRT1 = T2 /T1 => P2 = P1 (T2 /T1 )(1/2) = 160 × (900/300) × (1/2) = 240 kPa Work is done while piston moves at linearly varying pressure, so we get 3 1W2 = 0.5(P1 + P2)(V2 -V1) = 0.5×(160 + 240) kPa × 0.8072 m = 161.4 kJ Heat transfer is found from energy equation 1Q2 = m(u2 - u1) + 1W2 = 1.5×(674.824 - 214.36) + 161.4 = 852.1 kJ P T 2 1 2 1 W V V Sonntag, Borgnakke and van Wylen 5.101 Air in a piston/cylinder at 200 kPa, 600 K, is expanded in a constant-pressure process to twice the initial volume (state 2), shown in Fig. P5.101. The piston is then locked with a pin and heat is transferred to a final temperature of 600 K. Find P, T, and h for states 2 and 3, and find the work and heat transfer in both processes. Solution: C.V. Air. Control mass m2 = m3 = m1 Energy Eq.5.11: Process 1 to 2: u2 - u1 = 1q2 - 1w2 ; P = constant 1w2 = => ∫ P dv = P1(v2 -v1) = R(T2 -T1) Ideal gas Pv = RT ⇒ T2 = T1v2/v1 = 2T1 = 1200 K P2 = P1 = 200 kPa, Table A.7 1w2 = RT1 = 172.2 kJ/kg h2 = 1277.8 kJ/kg, h3 = h1 = 607.3 kJ/kg 1q2 = u2 - u1 + 1w2 = h2 - h1 = 1277.8 - 607.3 = 670.5 kJ/kg Process 2→3: v3 = v2 = 2v1 ⇒ 2w3 = 0, P3 = P2T3/T2 = P1T1/2T1 = P1/2 = 100 kPa 2q3 = u3 - u2 = 435.1 - 933.4 = -498.3 kJ/kg P o 200 cb Air 100 P 1 T 2 2 1200 3 600 v 1 3 v Sonntag, Borgnakke and van Wylen 5.102 A vertical piston/cylinder has a linear spring mounted as shown so at zero cylinder volume a balancing pressure inside is zero. The cylinder contains 0.25 kg air at 500 kPa, 27oC. Heat is now added so the volume doubles. a) Show the process path in a P-V diagram b) Find the final pressure and temperature. c) Find the work and heat transfer. Solution: Take CV around the air. This is a control mass. Continuity: m2 = m1 = m ; Energy Eq.5.11: m(u2 -u1) = 1Q2 - 1W2 Process: P linear in V so, P = A + BV, since V = 0 => P = 0 => A = 0 now: P = BV; B = P1/V1 State 1: P, T Ideal gas : mRT 0.25 × 0.287 × 300 V= P = 500 b) a) = 0.04305 m3 State 2: V2 = 2 V1 ; ? must be on line in P-V diagram, this substitutes for the question mark only one state is on the line with that value of V2 P 2 P2 1 P1 0 V 0 V1 2V1 P2 = BV2 = (P1/V1)V2 = 2P1 = 1000 kPa. PV 2P12V1 4P1V1 T2 = mR = mR = mR = 4 T1 = 1200 K c) The work is boundary work and thus seen as area in the P-V diagram: 1W2 = ∫ P dV = 0.5(P1 + P2 )( 2V1 − V1) = 0.5(500 + 1000) 0.04305 = 32.3 kJ 1Q2 = m(u2 − u1) + 1W2 = 0.25(933.4 - 214.4) + 32.3 = 212 kJ Internal energy u was taken from air table A.7. If constant Cv were used then (u2 − u1) = 0.717 (1200 - 300) = 645.3 kJ/kg (versus 719 above) Sonntag, Borgnakke and van Wylen Energy Equation: Polytropic Process 5.103 A piston cylinder contains 0.1 kg air at 300 K and 100 kPa. The air is now slowly compressed in an isothermal (T = C) process to a final pressure of 250 kPa. Show the process in a P-V diagram and find both the work and heat transfer in the process. Solution : Process: ⇒ T = C & ideal gas PV = mRT = constant V2 P1 mRT W2 = ∫ PdV = ⌠ V dV = mRT ln V = mRT ln P 1 1 2 ⌡ = 0.1 × 0.287 × 300 ln (100 / 250 ) = -7.89 kJ since T1 = T2 ⇒ u2 = u1 The energy equation thus becomes 1Q2 = m × (u2 - u1 ) + 1W2 = 1W2 = -7.89 kJ P = C v -1 P T T=C 2 2 1 1 v v Sonntag, Borgnakke and van Wylen 5.104 Oxygen at 300 kPa, 100°C is in a piston/cylinder arrangement with a volume of 0.1 m3. It is now compressed in a polytropic process with exponent, n = 1.2, to a final temperature of 200°C. Calculate the heat transfer for the process. Solution: Continuty: m2 = m1 m(u2 − u1) = 1Q2 − 1W2 Energy Eq.5.11: State 1: T1 , P1 & ideal gas, small change in T, so use Table A.5 P1V1 300 × 0.1 m3 ⇒ m = RT = 0.25983 × 373.15 = 0.309 kg 1 Process: PVn = constant 1 mR 1W2 = 1-n (P2V2 - P1V1) = 1-n (T2 - T1) = 0.309 × 0.25983 (200 - 100) 1 - 1.2 = -40.2 kJ 1Q2 = m(u2 - u1) + 1W2 ≅ mCv(T2 - T1) + 1W2 = 0.3094 × 0.662 (200 - 100) - 40.2 = -19.7 kJ P = C v -1.2 P 2 T2 T T=Cv -0.2 2 1 T1 v 1 v Sonntag, Borgnakke and van Wylen 5.105 A piston/cylinder contains 0.001 m3 air at 300 K, 150 kPa. The air is now compressed in a process in which P V1.25 = C to a final pressure of 600 kPa. Find the work performed by the air and the heat transfer. Solution: C.V. Air. This is a control mass, values from Table A.5 are used. Continuty: m2 = m1 Energy Eq.5.11: Process : m(u2 − u1) = 1Q2 − 1W2 PV1.25 = const. V2 = V1 ( P1/P2 )1.25= 0.00033 m3 State 2: 600 × 0.00033 T2 = T1 P2V2/(P1V1) = 300 150 × 0.001 = 395.85 K 1 1 1W2 = n-1(P2 V2 – P1V1) = n-1 (600 × 0.00033 – 150 × 0.001) = - 0.192 kJ P1V1 1Q2 = m(u2 – u1) + 1W2 = RT Cv (T2 – T1) + 1W2 1 = 0.001742 × 0.717× 95.85 – 0.192 = - 0.072 kJ Sonntag, Borgnakke and van Wylen 5.106 Helium gas expands from 125 kPa, 350 K and 0.25 m3 to 100 kPa in a polytropic process with n = 1.667. How much heat transfer is involved? Solution: C.V. Helium gas, this is a control mass. Energy equation: m(u2 – u1) = 1Q2 – 1W2 n n n Process equation: PV = constant = P1V1 = P2V2 Ideal gas (A.5): m = PV/RT = 125 × 0.25 = 0.043 kg 2.0771 × 350 Solve for the volume at state 2 1250.6 = 0.25 × 100 = 0.2852 m3 100 × 0.2852 T2 = T1 P2V2/(P1V1) = 350 125 × 0.25 = 319.4 K Work from Eq.4.4 V2 = V1 (P1/P2) 1W2 = 1/n P2V2- P1 V1 100× 0.2852 - 125× 0.25 = kPa m3 = 4.09 kJ 1-n 1 - 1.667 Use specific heat from Table A.5 to evaluate u2 – u1, Cv = 3.116 kJ/kg K 1Q2 = m(u2 – u1) + 1W2 = m Cv (T2 – T1) + 1W2 = 0.043 × 3.116 × (319.4 – 350) + 4.09 = -0.01 kJ Sonntag, Borgnakke and van Wylen 5.107 A piston/cylinder in a car contains 0.2 L of air at 90 kPa, 20°C, shown in Fig. P5.107. The air is compressed in a quasi-equilibrium polytropic process with polytropic exponent n = 1.25 to a final volume six times smaller. Determine the final pressure, temperature, and the heat transfer for the process. Solution: C.V. Air. This is a control mass going through a polytropic process. Continuty: m2 = m1 m(u2 − u1) = 1Q2 − 1W2 Energy Eq.5.11: Process: Pvn = const. 1.25 P1v1n = P2v2n ⇒ P2 = P1(v1/v2)n = 90 × 6 = 845.15 kPa Substance ideal gas: Pv = RT T2 = T1(P2v2/P1v1) = 293.15(845.15/90 × 6) = 458.8 K P 2 -1.25 T -0.25 P=Cv 2 T=Cv 1 1 v v PV 90 × 0.2×10-3 m = RT = 0.287 × 293.15 = 2.14×10-4 kg The work is integrated as in Eq.4.4 1 R ⌠ 1w2 = ⌡Pdv = 1 - n (P2v2 - P1v1) = 1 - n (T2 - T1) 0.287 = 1 - 1.25(458.8 - 293.15) = -190.17 kJ/kg The energy equation with values of u from Table A.7 is 1q2 = u2 - u1 + 1w2 = 329.4 - 208.03 – 190.17 = -68.8 kJ/kg 1Q2 = m 1q2 = -0.0147 kJ (i.e a heat loss) Sonntag, Borgnakke and van Wylen 5.108 A piston/cylinder has nitrogen gas at 750 K and 1500 kPa. Now it is expanded in a polytropic process with n = 1.2 to P = 750 kPa. Find the final temperature, the specific work and specific heat transfer in the process. C.V. Nitrogen. This is a control mass going through a polytropic process. Continuty: m2 = m1 Energy Eq.5.11: m(u2 − u1) = 1Q2 − 1W2 Process: Substance ideal gas: Pvn = constant Pv = RT 0.2 n-1 n = 750 750 1.2 = 750 × 0.8909 = 668 K T2 = T1 (P2/P1) 1500 The work is integrated as in Eq.4.4 1 R 1w2 = ⌠Pdv = 1 - n (P2v2 - P1v1) = 1 - n (T2 - T1) ⌡ 0.2968 = 1 - 1.2 (668 - 750) = 121.7 kJ/kg The energy equation with values of u from Table A.8 is 1q2 = u2 - u1 + 1w2 = 502.8 - 568.45 + 121.7 = 56.0 kJ/kg If constant specific heat is used from Table A.5 1q2 = C(T2 - T1) + 1w2 = 0.745(668 – 750) + 121.7 = 60.6 kJ/kg Sonntag, Borgnakke and van Wylen 5.109 A piston/cylinder arrangement of initial volume 0.025 m3 contains saturated water vapor at 180°C. The steam now expands in a polytropic process with exponent n = 1 to a final pressure of 200 kPa, while it does work against the piston. Determine the heat transfer in this process. Solution: C.V. Water. This is a control mass. State 1: Table B.1.1 P = 1002.2 kPa, v1 = 0.19405 m3/kg, u1 = 2583.7 kJ/kg , m = V/v1 = 0.025/0.19405 = 0.129 kg Process: Pv = const. = P1v1 = P2v2 ; polytropic process n = 1. ⇒ v2 = v1P1/P2 = 0.19405 × 1002.1/200 = 0.9723 m3/kg State 2: P2, v2 ⇒ Table B.1.3 T2 ≅ 155°C , u2 = 2585 kJ/kg v2 0.9723 W2 = ⌠PdV = P1V1 ln v = 1002.2 × 0.025 ln 0.19405 = 40.37 kJ ⌡ 1 1 1Q2 = m(u2 - u1) + 1W2 = 0.129(2585 - 2583.7) + 40.37 = 40.54 kJ P Sat vapor line 1 T P = C v -1 T=C 2 1 v Notice T drops, it is not an ideal gas. 2 v Sonntag, Borgnakke and van Wylen 5.110 Air is expanded from 400 kPa, 600 K in a polytropic process to 150 kPa, 400 K in a piston cylinder arrangement. Find the polytropic exponent n and the work and heat transfer per kg air using constant heat capacity from A.5. Solution: Process: P V n = P V n 11 22 Ideal gas: PV = RT ⇒ V = RΤ/ P P1 ln P = ln (V2 / V1)n = n ln (V2 / V1) = n ln 2 P1 n = ln P / l n 2 P T T P 2 [ P2 × T1 ] 1 [ P1 × T2 ] = ln 400 / ln [ 400 × 400 ] 150 600 150 2 1 = 1.7047 The work integral is from Eq.4.4 R 0.287 ⌡ 1W2 = ⌠PdV = 1 − n (T2 – T1) = −0.7047 (400 – 600) = 81.45 kJ/kg Energy equation from Eq.5.11 1q2 = u2 - u1 + 1w2 = Cv(T2 - T1) + 1w2 = 0.717 (400-600) + 81.45 = -61.95 kJ/kg Sonntag, Borgnakke and van Wylen 5.111 A piston/cylinder has 1 kg propane gas at 700 kPa, 40°C. The piston cross-sectional area is 0.5 m2, and the total external force restraining the piston is directly proportional to the cylinder volume squared. Heat is transferred to the propane until its temperature reaches 700°C. Determine the final pressure inside the cylinder, the work done by the propane, and the heat transfer during the process. Solution: C.V. The 1 kg of propane. Energy Eq.5.11: m(u2 - u1) = 1Q2 - 1W2 PV-2 = constant, Process: P = Pext = CV2 ⇒ polytropic n = -2 Ideal gas: PV = mRT, and process yields n 700+273.152/3 n-1 = 700 P2 = P1(T2/T1) 40+273.15 = 1490.7 kPa The work is integrated as Eq.4.4 2 P2V2 - P1V1 mR(T2 - T1) W2 = ⌠ PdV = = 1 ⌡ 1-n 1-n 1 = 1× 0.18855 × (700 – 40) = 41.48 kJ 1– (–2) The energy equation with specific heat from Table A.5 becomes 1Q2 = m(u2 - u1) + 1W2 = mCv(T2 - T1) + 1W2 = 1 × 1.490 × (700 - 40) + 41.48 = 1024.9 kJ P P=CV T 2 3 T=CV 2 2 1 V 1 V Sonntag, Borgnakke and van Wylen 5.112 An air pistol contains compressed air in a small cylinder, shown in Fig. P5.112. Assume that the volume is 1 cm3, pressure is 1 MPa, and the temperature is 27°C when armed. A bullet, m = 15 g, acts as a piston initially held by a pin (trigger); when released, the air expands in an isothermal process (T = constant). If the air pressure is 0.1 MPa in the cylinder as the bullet leaves the gun, find a. The final volume and the mass of air. b. The work done by the air and work done on the atmosphere. c. The work to the bullet and the bullet exit velocity. Solution: C.V. Air. Air ideal gas: mair = P1V1/RT1 = 1000 × 10-6/(0.287 × 300) = 1.17×10-5 kg Process: PV = const = P1V1 = P2V2 ⇒ V2 = V1P1/P2 = 10 cm3 ⌠P1V1 1W2 = ⌠PdV = V dV = P1V1 ln (V2/V1) = 2.303 J ⌡ ⌡ -6 1W2,ATM = P0(V2 - V1) = 101 × (10 − 1) × 10 kJ = 0.909 J 1 Wbullet = 1W2 - 1W2,ATM = 1.394 J = 2 mbullet(Vexit)2 Vexit = (2Wbullet/mB)1/2 = (2 × 1.394/0.015)1/2 = 13.63 m/s Sonntag, Borgnakke and van Wylen 5.113 A spherical balloon contains 2 kg of R-22 at 0°C, 30% quality. This system is heated until the pressure in the balloon reaches 600 kPa. For this process, it can be assumed that the pressure in the balloon is directly proportional to the balloon diameter. How does pressure vary with volume and what is the heat transfer for the process? Solution: C.V. R-22 which is a control mass. m2 = m1 = m ; Energy Eq.5.11: m(u2 - u1) = 1Q2 - 1W2 State 1: 0°C, x = 0.3. Table B.4.1 gives P1 = 497.6 kPa v1 = 0.000778 + 0.3 × 0.04636 = 0.014686 m3/kg u1 = 44.2 + 0.3 × 182.3 = 98.9 kJ/kg Process: P ∝ D, V ∝ D3 => PV -1/3 = constant, polytropic n = -1/3. => V2 = mv2 = V1 ( P2 /P1 )3 = mv1 ( P2 /P1 )3 v2 = v1 ( P2 /P1 )3 = 0.014686 × (600 / 497.6)3 = 0.02575 m3/kg State 2: P2 = 600 kPa, process : v2 = 0.02575 → Table B.4.1 x2 = 0.647, u2 = 165.8 kJ/kg 1W2 = ∫ P dV = P2V2 - P1V1 600 × 0.05137 - 498 × 0.02937 = = 12.1 kJ 1 - (-1/3) 1-n 1Q2 = m(u2- u1) + 1W2 = 2(165.8 - 98.9) + 12.1 = 145.9 kJ Sonntag, Borgnakke and van Wylen 5.114 Calculate the heat transfer for the process described in Problem 4.55. Consider a piston cylinder with 0.5 kg of R-134a as saturated vapor at -10°C. It is now compressed to a pressure of 500 kPa in a polytropic process with n = 1.5. Find the final volume and temperature, and determine the work done during the process. Solution: Take CV as the R-134a which is a control mass Continuity: m2 = m1 = m ; Energy: m(u2 -u1) = 1Q2 - 1W2 Process: 1: (T, x) 1.5 Pv = constant. Polytropic process with n = 1.5 P = Psat = 201.7 kPa from Table B.5.1 v1 = 0.09921 m3/kg, u1 = 372.27 kJ/kg 2: (P, process) v2 = v1 (P1/P2) (1/1.5) = 0.09921× (201.7/500) 0.667 = 0.05416 => Table B.5.2 superheated vapor, T2 = 79°C, u2 = 440.9 kJ/kg Process gives P = C v (-1.5) , which is integrated for the work term, Eq.4.4 1W2 = ∫ P dV = m(P2v2 - P1v1)/(1-1.5) = -2×0.5× (500×0.05416 - 201.7×0.09921) = -7.07 kJ 1Q2 = m(u2 -u1) + 1W2 = 0.5(440.9 - 372.27) + (-7.07) = 27.25 kJ Sonntag, Borgnakke and van Wylen 5.115 A piston/cylinder setup contains argon gas at 140 kPa, 10°C, and the volume is 100 L. The gas is compressed in a polytropic process to 700 kPa, 280°C. Calculate the heat transfer during the process. Solution: Find the final volume, then knowing P1, V1, P2, V2 the polytropic exponent can be determined. Argon is an ideal monatomic gas (Cv is constant). P1 T2 140 553.15 V2 = V1 × P T = 0.1 × 700 283.15 = 0.0391 m3 21 P1V1n = P2V2n ⌠ 1W2 = ⌡PdV = ⇒ P2 V1 1.6094 n = ln (P ) / ln (V ) = 0.939 = 1.714 1 2 P2V2 -P1V1 700×0.0391 - 140×0.1 = = -18.73 kJ 1-n 1 - 1.714 m = P1V1/RT1 = 140 × 0.1/(0.20813 × 283.15) = 0.2376 kg 1Q2 = m(u2 - u1) + 1W2 = mCv(T2 - T1) + 1W2 = 0.2376 × 0.3122 (280 - 10) - 18.73 = 1.3 kJ Sonntag, Borgnakke and van Wylen Energy Equation in Rate Form 5.116 A crane lifts a load of 450 kg vertically up with a power input of 1 kW. How fast can the crane lift the load? Solution : Power is force times rate of displacement . W = F⋅V = mg⋅V . W 1000 W V = mg = 450 × 9.806 N = 0.227 m/s Sonntag, Borgnakke and van Wylen 5.117 A computer in a closed room of volume 200 m3 dissipates energy at a rate of 10 kW. The room has 50 kg wood, 25 kg steel and air, with all material at 300 K, 100 kPa. Assuming all the mass heats up uniformly, how long will it take to increase the temperature 10°C? Solution: C.V. Air, wood and steel. m2 = m1 ; no work . Energy Eq.5.11: U2 - U1 = 1Q2 = Q∆t The total volume is nearly all air, but we can find volume of the solids. Vwood = m/ρ = 50/510 = 0.098 m3 ; Vsteel = 25/7820 = 0.003 m3 Vair = 200 - 0.098 - 0.003 = 199.899 m3 mair = PV/RT = 101.325 × 199.899/(0.287 × 300) = 235.25 kg We do not have a u table for steel or wood so use heat capacity from A.3. ∆U = [mair Cv + mwood Cv + msteel Cv ]∆T = (235.25 × 0.717 + 50 × 1.38 + 25 × 0.46) 10 . = 1686.7 + 690 +115 = 2492 kJ = Q × ∆t = 10 kW × ∆t => ∆t = 2492/10 = 249.2 sec = 4.2 minutes Sonntag, Borgnakke and van Wylen 5.118 The rate of heat transfer to the surroundings from a person at rest is about 400 kJ/h. Suppose that the ventilation system fails in an auditorium containing 100 people. Assume the energy goes into the air of volume 1500 m3 initially at 300 K and 101 kPa. Find the rate (degrees per minute) of the air temperature change. Solution: . . Q = n q = 100× 400 = 40000 kJ/h = 666.7 kJ/min dEair dTair . = Q = mairCv dt dt mair = PV/RT = 101 × 1500 / 0.287 × 300 = 1759.6 kg dTair . dt = Q /mCv = 666.7 / (1759.6 × 0.717) = 0.53°C/min Sonntag, Borgnakke and van Wylen 5.119 A piston/cylinder of cross sectional area 0.01 m2 maintains constant pressure. It contains 1 kg water with a quality of 5% at 150oC. If we heat so 1 g/s liquid turns into vapor what is the rate of heat transfer needed? Solution: Control volume the water. Continuity Eq.: mtot = constant = mvapor + mliq . . . on a rate form: mtot = 0 = mvapor + mliq ⇒ Vvapor = mvapor vg , Vliq = mliq vf Vtot = Vvapor + Vliq . . . . . Vtot = Vvapor + Vliq = mvaporvg + mliqvf . . = mvapor (vg- vf ) = mvapor vfg . . . W = PV = P mvapor vfg = 475.9 × 0.001 × 0.39169 = 0.1864 kW = 186 W . . mliq = -mvapor Sonntag, Borgnakke and van Wylen 5.120 The heaters in a spacecraft suddenly fail. Heat is lost by radiation at the rate of 100 kJ/h, and the electric instruments generate 75 kJ/h. Initially, the air is at 100 kPa, 25°C with a volume of 10 m3. How long will it take to reach an air temperature of −20°C? Solution: C.M. Air . Q el C.V. . Qrad dM Continuity Eq: dt = 0 . dE . Energy Eq: dt = Qel - Qrad . W=0 . KE = 0 . PE = 0 .. . . . . E = U = Qel - Qrad = Qnet ⇒ U2 - U1 = m(u2 - u1) = Qnet(t2 - t1) P1V1 100 ×10 Ideal gas: m = RT = 0.287 × 298.15 = 11.688 kg 1 u2 - u1 = Cv0(T2 - T1) = 0.717 (-20 - 25) = -32.26 kJ/kg . t2 - t1 = mCv0(T2-T1)/Qnet = 11.688 × (−32.26)/(-25) = 15.08 h Sonntag, Borgnakke and van Wylen 5.121 A steam generating unit heats saturated liquid water at constant pressure of 200 kPa in a piston cylinder. If 1.5 kW of power is added by heat transfer find the rate (kg/s) of saturated vapor that is made. Solution: Energy equation on a rate form making saturated vapor from saturated liquid . . . . .. . . . U = (mu) = m∆u = Q - W = Q - P V = Q - Pm∆v . . . . m(∆u + ∆vP ) = Q = m∆h = mhfg . . m = Q/ hfg = 1500 / 2201.96 = 0.681 kg/s Sonntag, Borgnakke and van Wylen 5.122 A small elevator is being designed for a construction site. It is expected to carry four 75-kg workers to the top of a 100-m tall building in less than 2 min. The elevator cage will have a counterweight to balance its mass. What is the smallest size (power) electric motor that can drive this unit? Solution: m = 4 × 75 = 300 kg ; ∆Z = 100 m ; ∆t = 2 minutes . . ∆Z 300 × 9.807 × 100 -W = ∆PE = mg = 1000 × 2 × 60 = 2.45 kW ∆t Sonntag, Borgnakke and van Wylen 5.123 As fresh poured concrete hardens, the chemical transformation releases energy at a rate of 2 W/kg. Assume the center of a poured layer does not have any heat loss and that it has an average heat capacity of 0.9 kJ/kg K. Find the temperature rise during 1 hour of the hardening (curing) process. Solution: . . .. . U = (mu) = mCvT = Q = mq .. T = q/Cv = 2×10-3 / 0.9 = 2.222 × 10-3 °C/sec . ∆T = T∆t = 2.222 × 10-3 × 3600 = 8 °C Sonntag, Borgnakke and van Wylen 5.124 A 100 Watt heater is used to melt 2 kg of solid ice at −10oC to liquid at +5oC at a constant pressure of 150 kPa. a) Find the change in the total volume of the water. b) Find the energy the heater must provide to the water. c) Find the time the process will take assuming uniform T in the water. Solution: Take CV as the 2 kg of water. m2 = m1 = m ; m(u2 − u1) = 1Q2 − 1W2 Energy Eq.5.11 State 1: Compressed solid, take sat. solid at same temperature. v = vi(−10) = 0.0010891 m3/kg, h = hi = −354.09 kJ/kg State 2: Compressed liquid, take sat. liquid at same temperature v = vf = 0.001, h = hf = 20.98 kJ/kg Change in volume: V2 − V1 = m(v2 − v1) = 2(0.001 − 0.0010891) = 0.000178 m3 Work is done while piston moves at constant pressure, so we get 1W2 = ∫ P dV = area = P(V2 − V1) = -150 × 0.000178 = −0.027 kJ = −27 J Heat transfer is found from energy equation 1Q2 = m(u2 − u1) + 1W2 = m(h2 − h1) = 2 × [20.98 − (−354.09)] = 750 kJ The elapsed time is found from the heat transfer and the rate of heat transfer . t = 1Q2/Q = (750/100) 1000 = 7500 s = 125 min = 2 h 5 min P L C.P. S T 1 V L+V S+V 2 P C.P. 1 T P=C 2 2 v v C.P. 1 v Sonntag, Borgnakke and van Wylen 5.125 Water is in a piston cylinder maintaining constant P at 700 kPa, quality 90% with a volume of 0.1 m3. A heater is turned on heating the water with 2.5 kW. What is the rate of mass (kg/s) vaporizing? Solution: Control volume water. Continuity Eq.: mtot = constant = mvapor + mliq . . . . . on a rate form: mtot = 0 = mvapor + mliq ⇒ mliq = -mvapor . . . .. . Energy equation: U = Q - W = mvapor ufg = Q - P mvapor vfg . Rearrange to solve for mvapor . . . mvapor (ufg + Pvfg) = mvapor hfg = Q . . 2.5 kW mvapor = Q/hfg = 2066.3 kJ/kg = 0.0012 kg/s Sonntag, Borgnakke and van Wylen Review Problems 5.126 Ten kilograms of water in a piston/cylinder setup with constant pressure is at 450°C and a volume of 0.633 m3. It is now cooled to 20°C. Show the P–v diagram and find the work and heat transfer for the process. Solution: C.V. The 10 kg water. Energy Eq.5.11: m(u2 - u1) = 1Q2 − 1W2 Process: ⇒ P=C 1W2 = mP(v2 -v1) State 1: (T, v1 = 0.633/10 = 0.0633 m3/kg) P1 = 5 MPa, h1 = 3316.2 kJ/kg State 2: (P = P = 5 MPa, 20°C) ⇒ Table B.1.4 v2 = 0.000 999 5 m3/kg ; P 2 Table B.1.3 h2 = 88.65 kJ/kg T 1 1 5 MPa v 2 The work from the process equation is found as 1W2 = 10 × 5000 ×(0.0009995 - 0.0633) = -3115 kJ The heat transfer from the energy equation is 1Q2 = m(u2 - u1) + 1W2 = m(h2 - h1) 1Q2 = 10 ×(88.65 - 3316.2) = -32276 kJ v Sonntag, Borgnakke and van Wylen 5.127 Consider the system shown in Fig. P5.127. Tank A has a volume of 100 L and contains saturated vapor R-134a at 30°C. When the valve is cracked open, R-134a flows slowly into cylinder B. The piston mass requires a pressure of 200 kPa in cylinder B to raise the piston. The process ends when the pressure in tank A has fallen to 200 kPa. During this process heat is exchanged with the surroundings such that the R-134a always remains at 30°C. Calculate the heat transfer for the process. Solution: C.V. The R-134a. This is a control mass. Continuity Eq.: m2 = m1 = m ; Energy Eq.5.11: Process in B: m(u2 − u1) = 1Q2 - 1W2 If VB > 0 then P = Pfloat (piston must move) ⇒ 1W2 = ∫ Pfloat dV = Pfloatm(v2 - v1) Work done in B against constant external force (equilibrium P in cyl. B) State 1: 30°C, x = 1. Table B.5.1: v1 = 0.02671 m3/kg, u1 = 394.48 kJ/kg m = V/v1 = 0.1 / 0.02671 = 3.744 kg State 2: 30°C, 200 kPa superheated vapor Table B.5.2 v2 = 0.11889 m3/kg, u2 = 403.1 kJ/kg From the process equation 1W2 = Pfloatm(v2 - v1) = 200×3.744×(0.11889 - 0.02671) = 69.02 kJ From the energy equation 1Q2 = m(u2 - u1) + 1W2 = 3.744 ×(403.1 - 394.48) + 69.02 = 101.3 kJ Sonntag, Borgnakke and van Wylen 5.128 Ammonia, NH3, is contained in a sealed rigid tank at 0°C, x = 50% and is then heated to 100°C. Find the final state P2, u2 and the specific work and heat transfer. Solution: Continuity Eq.: m2 = m1 ; Energy Eq.5.11: E2 - E1 = 1Q2 ; / (1W2 = 0) Process: V2 = V1 ⇒ v2 = v1 = 0.001566 + 0.5 × 0.28783 = 0.14538 m3/kg v2 & T2 ⇒ between 1000 kPa and 1200 kPa 0.14538 – 0.17389 P2 = 1000 + 200 0.14347 – 0.17389 = 1187 kPa Table B.2.2: P u2 = 1490.5 + (1485.8 – 1490.5) × 0.935 2 = 1485.83 kJ/kg u1 = 179.69 + 0.5 × 1138.3 = 748.84 kJ/kg 1 V Process equation gives no displacement: 1w2 = 0 ; The energy equation then gives the heat transfer as 1q2 = u2 - u1 = 1485.83 – 748.84 = 737 kJ/kg Sonntag, Borgnakke and van Wylen 5.129 A piston/cylinder contains 1 kg of ammonia at 20°C with a volume of 0.1 m3, shown in Fig. P5.129. Initially the piston rests on some stops with the top surface open to the atmosphere, Po, so a pressure of 1400 kPa is required to lift it. To what temperature should the ammonia be heated to lift the piston? If it is heated to saturated vapor find the final temperature, volume, and the heat transfer. Solution: C.V. Ammonia which is a control mass. m2 = m1 = m ; m(u2 -u1) = 1Q2 - 1W2 State 1: 20°C; v1 = 0.10 < vg ⇒ x1 = (0.1 – 0.001638)/0.14758 = 0.6665 u1 = uf + x1 ufg = 272.89 + 0.6665 ×1059.3 = 978.9 kJ/kg Process: Piston starts to lift at state 1a (Plift, v1) State 1a: 1400 kPa, v1 Table B.2.2 (superheated vapor) 0.1 – 0.09942 Ta = 50 + (60 – 50) 0.10423 – 0.09942 = 51.2 °C T P 1400 1a 1a 1200 857 1 2 2 v 1 State 2: x = 1.0, v2 = v1 => V2 = mv2 = 0.1 m3 T2 = 30 + (0.1 – 0.11049) × 5/(0.09397 – 0.11049) = 33.2 °C u2 = 1338.7 kJ/kg; 1W2 = 0; 1Q2 = m1q2 = m(u2 – u1) = 1 (1338.7 – 978.9) = 359.8 kJ/kg v Sonntag, Borgnakke and van Wylen 5.130 A piston held by a pin in an insulated cylinder, shown in Fig. P5.130, contains 2 kg water at 100°C, quality 98%. The piston has a mass of 102 kg, with cross-sectional area of 100 cm2, and the ambient pressure is 100 kPa. The pin is released, which allows the piston to move. Determine the final state of the water, assuming the process to be adiabatic. Solution: C.V. The water. This is a control mass. Continuity Eq.: m2 = m1 = m ; Energy Eq.5.11: m(u2 − u1) = 1Q2 - 1W2 Process in cylinder: P = Pfloat (if piston not supported by pin) P2 = Pfloat = P0 + mpg/A = 100 + 102 × 9.807 = 200 kPa 100×10-4 × 103 We thus need one more property for state 2 and we have one equation namely the energy equation. From the equilibrium pressure the work becomes 1W2 = ∫ Pfloat dV = P2 m(v2 - v1) With this work the energy equation gives per unit mass u2 − u1 = 1q2 - 1w2 = 0 - P2(v2 - v1) or with rearrangement to have the unknowns on the left hand side u2 + P2v2 = h2 = u1 + P2v1 h2 = u1 + P2v1 = 2464.8 + 200 × 1.6395 = 2792.7 kJ/kg State 2: (P2 , h2) Table B.1.3 => T2 ≅ 161.75°C Sonntag, Borgnakke and van Wylen 5.131 A piston/cylinder arrangement has a linear spring and the outside atmosphere acting on the piston, shown in Fig. P5.131. It contains water at 3 MPa, 400°C with the volume being 0.1 m3. If the piston is at the bottom, the spring exerts a force such that a pressure of 200 kPa inside is required to balance the forces. The system now cools until the pressure reaches 1 MPa. Find the heat transfer for the process. Solution: C.V. Water. Continuity Eq.: m2 = m1 = m ; m(u2 − u1) = 1Q2 - 1W2 Energy Eq.5.11: P State 1: Table B.1.3 1 3 MPa v1 = 0.09936 m3/kg, u1 = 2932.8 kJ/kg m = V/v1 = 0.1/0.09936 = 1.006 kg 2 1 MPa 200 kPa V, v 0 v2 = v2 v1 Process: Linear spring so P linear in v. P = P0 + (P1 - P0)v/v1 (P2 - P0)v1 (1000 - 200)0.09936 = 0.02839 m3/kg 3000 - 200 P1 - P0 = State 2: P2 , v2 ⇒ x2 = (v2 - 0.001127)/0.19332 = 0.141, T2 = 179.91°C, u2 = 761.62 + x2 × 1821.97 = 1018.58 kJ/kg 1 Process => 1W2 = ⌠PdV = 2 m(P1 + P2)(v2 - v1) ⌡ 1 = 2 1.006 (3000 + 1000)(0.02839 -0.09936) = -142.79 kJ Heat transfer from the energy equation 1Q2 = m(u2 - u1) + 1W2 = 1.006(1018.58 - 2932.8) - 142.79 = -2068.5 kJ Sonntag, Borgnakke and van Wylen 5.132 Consider the piston/cylinder arrangement shown in Fig. P5.132. A frictionless piston is free to move between two sets of stops. When the piston rests on the lower stops, the enclosed volume is 400 L. When the piston reaches the upper stops, the volume is 600 L. The cylinder initially contains water at 100 kPa, 20% quality. It is heated until the water eventually exists as saturated vapor. The mass of the piston requires 300 kPa pressure to move it against the outside ambient pressure. Determine the final pressure in the cylinder, the heat transfer and the work for the overall process. Solution: C.V. Water. Check to see if piston reaches upper stops. m(u4 - u1) = 1Q4 − 1W4 Energy Eq.5.11: Process: If P < 300 kPa then V = 400 L, line 2-1 and below If P > 300 kPa then V = 600 L, line 3-4 and above If P = 300 kPa then 400 L < V < 600 L line 2-3 These three lines are shown in the P-V diagram below and is dictated by the motion of the piston (force balance). 0.4 State 1: v1 = 0.001043 + 0.2×1.693 = 0.33964; m = V1/v1 = 0.33964 = 1.178 kg u1 = 417.36 + 0.2 × 2088.7 = 835.1 kJ/kg 0.6 State 3: v3 = 1.178 = 0.5095 < vG = 0.6058 at P3 = 300 kPa ⇒ Piston does reach upper stops to reach sat. vapor. State 4: v4 = v3 = 0.5095 m3/kg = vG at P4 => P4 = 361 kPa, From Table B.1.2 u4 = 2550.0 kJ/kg 1W4 = 1W2 + 2W3 + 3W4 = 0 + 2W3 + 0 1W4 = P2(V3 - V2) = 300 × (0.6 - 0.4) = 60 kJ 1Q4 = m(u4 - u1) + 1W4 = 1.178(2550.0 - 835.1) + 60 = 2080 kJ T 4 P2= P3 = 300 2 3 1 P4 P1 Water v cb Sonntag, Borgnakke and van Wylen 5.133 A piston/cylinder, shown in Fig. P5.133, contains R-12 at − 30°C, x = 20%. The volume is 0.2 m3. It is known that Vstop = 0.4 m3, and if the piston sits at the bottom, the spring force balances the other loads on the piston. It is now heated up to 20°C. Find the mass of the fluid and show the P–v diagram. Find the work and heat transfer. Solution: C.V. R-12, this is a control mass. Properties in Table B.3 Continuity Eq.: m2 = m1 Energy Eq.5.11: Process: E2 - E1 = m(u2 - u1) = 1Q2 - 1W2 P = A + BV, V < 0.4 m3, A = 0 (at V = 0, P = 0) State 1: v1 = 0.000672 + 0.2 × 0.1587 = 0.0324 m3/kg u1 = 8.79 + 0.2 × 149.4 = 38.67 kJ/kg m = m1 = = V1/v1 = 6.17 kg P 2 System: on line T ≅ -5°C 2P 1 V ≤ Vstop; 1 P1 Pstop = 2P1 =200 kPa State stop: (P,v) ⇒ Tstop ≅ -12°C 0 T stop ≅ -12.5°C V 0 0.2 0.4 TWO-PHASE STATE Since T2 > Tstop ⇒ v2 = vstop = 0.0648 m3/kg 2: (T2 , v2) Table B.3.2: Interpolate between 200 and 400 kPa P2 = 292.3 kPa ; u2 = 181.9 kJ/kg From the process curve, see also area in P-V diagram, the work is 1 1 1W2 = ⌠PdV = 2 (P1 + Pstop)(Vstop - V1) = 2 (100 + 200)0.2 = 30 kJ ⌡ From the energy equation 1Q2 = m(u2 - u1) + 1W2 = 913.5 kJ Sonntag, Borgnakke and van Wylen 5.134 A piston/cylinder arrangement B is connected to a 1-m3 tank A by a line and valve, shown in Fig. P5.134. Initially both contain water, with A at 100 kPa, saturated vapor and B at 400°C, 300 kPa, 1 m3. The valve is now opened and, the water in both A and B comes to a uniform state. a. Find the initial mass in A and B. b. If the process results in T2 = 200°C, find the heat transfer and work. Solution: C.V.: A + B. This is a control mass. Continuity equation: m2 - (mA1 + mB1) = 0 ; Energy: m2u2 - mA1uA1 - mB1uB1 = 1Q2 - 1W2 System: if VB ≥ 0 piston floats ⇒ PB = PB1 = const. if VB = 0 then P2 < PB1 and v = VA/mtot see P-V diagram ⌠ 1W2 = ⌡PBdVB = PB1(V2 - V1)B = PB1(V2 - V1)tot State A1: Table B.1.1, x = 1 vA1 = 1.694 m3/kg, uA1 = 2506.1 kJ/kg mA1 = VA/vA1 = 0.5903 kg P a 2 PB1 State B1: Table B.1.2 sup. vapor vB1 = 1.0315 m3/kg, uB1 = 2965.5 kJ/kg mB1 = VB1/vB1 = 0.9695 kg m2 = mTOT = 1.56 kg * At (T2 , PB1) v2 = 0.7163 > va = VA/mtot = 0.641 so VB2 > 0 so now state 2: P2 = PB1 = 300 kPa, T2 = 200 °C => u2 = 2650.7 kJ/kg and V2 = m2 v2 = 1.56 × 0.7163 = 1.117 m3 (we could also have checked Ta at: 300 kPa, 0.641 m3/kg => T = 155 °C) 1W2 = PB1(V2 - V1) = -264.82 kJ 1Q2 = m2u2 - mA1uA1 - mB1uB1 + 1W2 = -484.7 kJ V2 Sonntag, Borgnakke and van Wylen 5.135 A small flexible bag contains 0.1 kg ammonia at –10oC and 300 kPa. The bag material is such that the pressure inside varies linear with volume. The bag is left in the sun with with an incident radiation of 75 W, loosing energy with an average 25 W to the ambient ground and air. After a while the bag is heated to 30oC at which time the pressure is 1000 kPa. Find the work and heat transfer in the process and the elapsed time. Solution: Take CV as the Ammonia, constant mass. Continuity Eq.: m2 = m1 = m ; Energy Eq.5.11: m(u2 − u1) = 1Q2 – 1W2 Process: P = A + BV (linear in V) State 1: Compressed liquid P > Psat, take saturated liquid at same temperature. v1 = vf(20) = 0.001002 m3/kg, State 2: Table B.2.1 at 30oC : u1 = uf = 133.96 kJ/kg P < Psat so superheated vapor v2 = 0.13206 m3/kg, u2 = 1347.1 kJ/kg, V2 = mv2 = 0.0132 m3 Work is done while piston moves at increacing pressure, so we get 1W2 = ½(300 + 1000)*0.1(0.13206 – 0.001534) = 8.484 kJ Heat transfer is found from the energy equation 1Q2 = m(u2 – u1) + 1W2 = 0.1 (1347.1 – 133.96) + 8.484 = 121.314 + 8.484 = 129.8 kJ P C.P. NH3 1000 T 2 300 -10 v . Qnet = 75 – 25 = 50 Watts . 129800 t = 1Q2 / Qnet = 50 = 2596 s = 43.3 min 2 30 1 C.P. 1 v Sonntag, Borgnakke and van Wylen 5.136 Water at 150°C, quality 50% is contained in a cylinder/piston arrangement with initial volume 0.05 m3. The loading of the piston is such that the inside pressure is linear with the square root of volume as P = 100 + CV 0.5 kPa. Now heat is transferred to the cylinder to a final pressure of 600 kPa. Find the heat transfer in the process. Continuty: m2 = m1 m(u2 − u1) = 1Q2 − 1W2 Energy: State 1: v1 = 0.1969, u1 = 1595.6 kJ/kg ⇒ m = V/v1 = 0.254 kg Process equation ⇒ P1 - 100 = CV11/2 so (V2/V1)1/2 = (P2 - 100)/(P1 - 100) 2 P2 - 1002 500 V2 = V1 × P - 100 = 0.05 × 475.8 - 100 = 0.0885 1 2 1/2)dV = 100×(V - V ) + C(V 1.5 - V 1.5) ⌠ 1W2 = ⌠PdV = ⌡(100 + CV 2 1 2 1 ⌡ 3 = 100(V2 - V1)(1 - 2/3) + (2/3)(P2V2 - P1V1) 1W2 = 100 (0.0885-0.05)/3 + 2 (600 × 0.0885-475.8 × 0.05)/3 = 20.82 kJ State 2: P2, v2 = V2/m = 0.3484 ⇒ u2 = 2631.9 kJ/kg, 1Q2 = 0.254 × (2631.9 - 1595.6) + 20.82 = 284 kJ P 1 100 P = 100 + C V 1/2 2 V T2 ≅ 196°C Sonntag, Borgnakke and van Wylen 5.137 A 1 m3 tank containing air at 25oC and 500 kPa is connected through a valve to another tank containing 4 kg of air at 60oC and 200 kPa. Now the valve is opened and the entire system reaches thermal equilibrium with the surroundings at 20oC. Assume constant specific heat at 25oC and determine the final pressure and the heat transfer. Control volume all the air. Assume air is an ideal gas. Continuity Eq.: m2 – mA1 – mB1 = 0 Energy Eq.: U2 − U1 = m2u2 – mA1uA1 – mB1uB1 = 1Q2 - 1W2 Process Eq.: V = constant ⇒ 1W2 = 0 State 1: PA1VA1 (500 kPa)(1m3) mA1 = RT = (0.287 kJ/kgK)(298.2 K) = 5.84 kg A1 VB1 = mB1RTB1 (4 kg)(0.287 kJ/kgK)(333.2 K) = 1.91 m3 PB1 = (200 kN/m2) State 2: T2 = 20°C, v2 = V2/m2 m2 = mA1 + mB1 = 4 + 5.84 = 9.84 kg V2 = VA1 + VB1 = 1 + 1.91 = 2.91 m3 P2 = m2RT2 (9.84 kg)(0.287 kJ/kgK)(293.2 K) = 284.5 kPa V2 = 2.91 m3 Energy Eq.5.5 or 5.11: 1Q2 = U2 − U1 = m2u2 – mA1uA1 – mB1uB1 = mA1(u2 – uA1) + mB1(u2 – uB1) = mA1Cv0(T2 – TA1) + mB1Cv0(T2 – TB1) = 5.84 × 0.717 (20 – 25) + 4 × 0.717 (20 – 60) = −135.6 kJ The air gave energy out. A B Sonntag, Borgnakke and van Wylen 5.138 A closed cylinder is divided into two rooms by a frictionless piston held in place by a pin, as shown in Fig. P5.138. Room A has 10 L air at 100 kPa, 30°C, and room B has 300 L saturated water vapor at 30°C. The pin is pulled, releasing the piston, and both rooms come to equilibrium at 30°C and as the water is compressed it becomes twophase. Considering a control mass of the air and water, determine the work done by the system and the heat transfer to the cylinder. Solution: C.V. A + B, control mass of constant total volume. Energy equation: mA(u2 – u1)A + mB(uB2 – uB1) = 1Q2 – 1W2 Process equation: V = C ⇒ 1W2 = 0 T = C ⇒ (u2 – u1)A = 0 (ideal gas) The pressure on both sides of the piston must be the same at state 2. Since two-phase: P2 = Pg H2O at 30°C = PA2 = PB2 = 4.246 kPa Air, I.G.: → VA2 = PA1VA1 = mARAT = PA2VA2 = Pg H2O at 30°C VA2 100 × 0.01 3 3 4.246 m = 0.2355 m Now the water volume is the rest of the total volume VB2 = VA1 + VB1 - VA2 = 0.30 + 0.01 - 0.2355 = 0.0745 m3 VB1 0.3 mB = v = 32.89 = 9.121×10-3 kg => B1 vB2 = 8.166 m3/kg 8.166 = 0.001004 + xB2 × (32.89 - 0.001) ⇒ xB2 = 0.2483 uB2 = 125.78 + 0.2483 × 2290.8 = 694.5 kJ/kg, uB1 = 2416.6 kJ/kg Q = m (u – u ) = 9.121×10-3(694.5 - 2416.6) = -15.7 kJ 12 B B2 B1 A B SOLUTION MANUAL SI UNIT PROBLEMS CHAPTER 6 SONNTAG • BORGNAKKE • VAN WYLEN FUNDAMENTALS of Thermodynamics Sixth Edition Sonntag, Borgnakke and van Wylen CONTENT SUBSECTION PROB NO. Correspondence table Concept-Study guide problems Continuity equation and flow rates Single flow, single-device processes Nozzles, diffusers Throttle flow Turbines, expanders Compressors, fans Heaters, coolers Pumps, pipe and channel flows Multiple flow, single-device processes Turbines, compressors, expanders Heat exchangers Mixing processes Multiple devices, cycle processes Transient processes Review Problems Heat Transfer Problems 78-83 84-90 91-98 99-107 108-123 124-134 135-138 English Unit Problems 139-175 1- 21 22-29 30-39 40-47 48-54 55-62 63-70 71-77 Sonntag, Borgnakke and van Wylen Correspondence List This chapter 6 homework problem set corresponds to the 5th edition chapter 6 as follows. Problems 1-21 are all new. New 5th 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 1 2 new 4 5 new 6 new 18 new 19 new new new new 20 new 22 23 new new 24 new 26 mod new 25 new 30 New 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 5th new 33 new 32 new new 36 new new 38 mod new new new 9 new new 10 12 new new new new 35 new 11 mod new new new 29 31 New 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 5th 37 mod 34 41 new 14 new 13 new new new new new 8 new 16 new 27 new new 39a 39b 40a 42 43 new 46 47 new 58 mod 53 New 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 5th new 49 new 48 51 new new new 50 52 new 57 59 new 15 new new new 44 45 55 56 62 63 65 new new new new Sonntag, Borgnakke and van Wylen CONCEPT-STUDY GUIDE PROBLEMS 6.1 A mass flow rate into a control volume requires a normal velocity component. Why? The tangential velocity component does not bring any substance across the control volume surface as it flows parallel to it, the normal component of velocity brings substance in or out of the control volume according to its sign. The normal component must be into the control volume to bring mass in, just like when you enter a bus (it does not help that you run parallel with the bus side). V Vtangential Vnormal 6.2 . A temperature difference drives a heat transfer. Does a similar concept apply to m? Yes. A pressure difference drives the flow. The fluid is accelerated in the direction of a lower pressure as it is being pushed harder behind it than in front of it. This also means a higher pressure in front can decelerate the flow to a lower velocity which happens at a stagnation on a wall. F1 = P1 A F2 = P 2 A 6.3 Can a steady state device have boundary work? No. Any change in size of the control volume would require either a change in mass inside or a change in state inside, neither of which is possible in a steady-state process. Sonntag, Borgnakke and van Wylen 6.4 . . Can you say something about changes in m and V through a steady flow device? The continuity equation expresses the conservation of mass, so the total . amount of m entering must be equal to the total amount leaving. For a single flow device the mass flow rate is constant through it, so you have the same mass flow rate across any total cross-section of the device from the inlet to the exit. The volume flow rate is related to the mass flow rate as . . V=vm so it can vary if the state changes (then v changes) for a constant mass flow rate. . This also means that the velocity can change (influenced by the area as V = VA) and the flow can experience an acceleration (like in a nozzle) or a deceleration (as in a diffuser). Sonntag, Borgnakke and van Wylen 6.5 How does a nozzle or sprayhead generate kinetic energy? By accelerating the fluid from a high pressure towards the lower pressure, which is outside the nozzle. The higher pressure pushes harder than the lower pressure so there is a net force on any mass element to accelerate it. 6.6 Liquid water at 15oC flows out of a nozzle straight up 15 m. What is nozzle Vexit? 1 2 1 2 Energy Eq.6.13: hexit + 2 Vexit + gHexit = h2 + 2 V2 + gH2 If the water can flow 15 m up it has specific potential energy of gH2 which must 2 equal the specific kinetic energy out of the nozzle Vexit/2. The water does not change P or T so h is the same. 2 Vexit/2 = g(H2 – Hexit) = gH Vexit = 2gH = => 2 × 9.807 × 15 m2/s2 = 17.15 m/s 6.7 What is the difference between a nozzle flow and a throttle process? In both processes a flow moves from a higher to a lower pressure. In the nozzle the pressure drop generates kinetic energy, whereas that does not take place in the throttle process. The pressure drop in the throttle is due to a flow restriction and represents a loss. Sonntag, Borgnakke and van Wylen 6.8 If you throttle a saturated liquid what happens to the fluid state? If it is an ideal gas? The throttle process is approximated as a constant enthalpy process. Changing the state from saturated liquid to a lower pressure with the same h gives a two-phase state so some of the liquid will vaporize and it becomes colder. 1 2 P 1 2 h=C T h=C v If the same process happens in an ideal gas then same h gives the same temperature (h a function of T only) at the lower pressure. 6.9 R-134a at 30oC, 800 kPa is throttled so it becomes cold at –10oC. What is exit P? State 1 is slightly compressed liquid so Table B.5.1: h = hf = 241.79 kJ/kg At the lower temperature it becomes two-phase since the throttle flow has constant h and at –10oC: hg = 392.28 kJ/kg P = Psat = 210.7 kPa 6.10 Air at 500 K, 500 kPa is expanded to 100 kPa in two steady flow cases. Case one is a throttle and case two is a turbine. Which has the highest exit T? Why? 1. Throttle. In the throttle flow no work is taken out, no kinetic energy is generated and we assume no heat transfer takes place and no potential energy change. The energy equation becomes constant h, which gives constant T since it is an ideal gas. 2. Turbine. In the turbine work is taken out on a shaft so the fluid expands and P and T drops. Sonntag, Borgnakke and van Wylen 6.11 A turbine at the bottom of a dam has a flow of liquid water through it. How does that produce power? Which terms in the energy equation are important? The water at the bottom of the dam in the turbine inlet is at a high pressure. It runs through a nozzle generating kinetic energy as the pressure drops. This high kinetic energy flow impacts a set of rotating blades or buckets which converts the kinetic energy to power on the shaft so the flow leaves at low pressure and low velocity. Lake DAM T H Sonntag, Borgnakke and van Wylen 6.12 A windmill takes a fraction of the wind kinetic energy out as power on a shaft. In what manner does the temperature and wind velocity influence the power? Hint: write the power as mass flow rate times specific work. The work as a fraction f of the flow of kinetic energy becomes . . .12 12 W = mw = m f 2 Vin = ρAVin f 2 Vin so the power is proportional to the velocity cubed. The temperature enters through the density, so assuming air as ideal gas ρ = 1/v = P/RT and the power is inversely proportional to temperature. Sonntag, Borgnakke and van Wylen 6.13 If you compress air the temperature goes up, why? When the hot air, high P flows in long pipes it eventually cools to ambient T. How does that change the flow? As the air is compressed, volume decreases so work is done on a mass element, its energy and hence temperature goes up. If it flows at nearly constant P and cools its density increases (v decreases) so it slows down . for same mass flow rate ( m = ρAV ) and flow area. 6.14 In a boiler you vaporize some liquid water at 100 kPa flowing at 1 m/s. What is the velocity of the saturated vapor at 100 kPa if the pipe size is the same? Can the flow then be constant P? The continuity equation with average values is written . . . mi = me = m = ρAV = AV/v = AVi/vi = AVe/ve From Table B.1.2 at 100 kPa we get vf = 0.001043 m3/kg; vg = 1.694 m3/kg 1.694 Ve = Vi ve/vi = 1 0.001043 = 1624 m/s To accelerate the flow up to that speed you need a large force ( ∆PA ) so a large pressure drop is needed. Pe < Pi Pi cb Sonntag, Borgnakke and van Wylen 6.15 A mixing chamber has all flows at the same P, neglecting losses. A heat exchanger has separate flows exchanging energy, but they do not mix. Why have both kinds? You might allow mixing when you can use the resulting output mixture, say it is the same substance. You may also allow it if you definitely want the outgoing mixture, like water out of a faucet where you mix hot and cold water. Even if it is different substances it may be desirable, say you add water to dry air to make it more moist, typical for a winter time air-conditioning set-up. In other cases it is different substances that flow at different pressures with one flow heating or cooling the other flow. This could be hot combustion gases heating a flow of water or a primary fluid flow around a nuclear reactor heating a transfer fluid flow. Here the fluid being heated should stay pure so it does not absorb gases or radioactive particles and becomes contaminated. Even when the two flows have the same substance there may be a reason to keep them at separate pressures. 1 2 1 MIXING 2 cb CHAMBER 3 4 3 cb Sonntag, Borgnakke and van Wylen 6.16 In a co-flowing (same direction) heat exchanger 1 kg/s air at 500 K flows into one channel and 2 kg/s air flows into the neighboring channel at 300 K. If it is infinitely long what is the exit temperature? Sketch the variation of T in the two flows. .. C.V. mixing section (no W, Q) Continuity Eq.: . . . . m1 = m3 and m2 = m4 Energy Eq.6.10: . . . . m1h1 + m2h2 = m1h3 + m2h4 Same exit T: . . . . h3 = h4 = [m1h1 + m2h2] / [m1 + m2] Using conctant specific heat T3 = T4 = . m1 . . m1 + m2 T1 + . m2 1 2 T2 = 3 × 500 + 3 × 300 = 367 K . . m1 + m2 T 3 1 x 4 2 cb 500 300 T1 T2 x Sonntag, Borgnakke and van Wylen 6.17 Air at 600 K flows with 3 kg/s into a heat exchanger and out at 100oC. How much (kg/s) water coming in at 100 kPa, 20oC can the air heat to the boiling point? C.V. Total heat exchanger. The flows are not mixed so the two flowrates are constant through the device. No external heat transfer and no work. . . . . Energy Eq.6.10: mairhair in + mwaterhwater in = mairhair out + mwaterhwater out . . mair[hair in - hair out] = mwater[hwater out – hwater in] Table B.1.2: hwater out – hwater in = 2675.46 – 83.94 = 2591.5 kJ/kg Table A.7.1: hair in - hair out = 607.32 – 374.14 = 233.18 kJ/kg Solve for the flow rate of water from the energy equation hair in - hair out 233.18 . . mwater = mair h - hwater in = 3 × 2591.5 = 0.27 kg/s water out Air out Air in cb Sonntag, Borgnakke and van Wylen 6.18 Steam at 500 kPa, 300oC is used to heat cold water at 15oC to 75oC for domestic hot water supply. How much steam per kg liquid water is needed if the steam should not condense? Solution: C.V. Each line separately. No work but there is heat transfer out of the steam flow and into the liquid water flow. . . . . . Water line energy Eq.: mliqhi + Q = mliqhe ⇒ Q = mliq(he – hi) For the liquid water look in Table B.1.1 ∆hliq = he – hi = 313.91 – 62.98 = 250.93 kJ/kg ( ≅ Cp ∆T = 4.18 (75 – 15) = 250.8 kJ/kg ) Steam line energy has the same heat transfer but it goes out . . . . . Steam Energy Eq.: msteamhi = Q + msteamhe ⇒ Q = msteam(hi – he) For the steam look in Table B.1.3 at 500 kPa ∆hsteam = hi – he = 3064.2 – 2748.67 = 315.53 kJ/kg Now the heat transfer for the steam is substituted into the energy equation for the water to give . . 250.93 msteam / mliq = ∆hliq / ∆hsteam = 315.53 = 0.795 Hot water out Steam in Steam out cb Cold water in Sonntag, Borgnakke and van Wylen 6.19 Air at 20 m/s, 260 K, 75 kPa with 5 kg/s flows into a jet engine and it flows out at 500 m/s, 800 K, 75 kPa. What is the change (power) in flow of kinetic energy? . .1 2 2 m ∆KE = m 2 (Ve – Vi ) 1 1 = 5 kg/s × 2 (5002 – 202) (m/s)2 1000 (kW/W) = 624 kW cb 6.20 An initially empty cylinder is filled with air from 20oC, 100 kPa until it is full. Assuming no heat transfer is the final temperature larger, equal to or smaller than 20oC? Does the final T depend on the size of the cylinder? This is a transient problem with no heat transfer and no work. The balance equations for the tank as C.V. become Continuity Eq.: m2 – 0 = mi Energy Eq.: m2u2 – 0 = mihi + Q – W = mihi + 0 – 0 Final state: u2 = hi & T2 > Ti and it does not depend on V P2 = Pi Sonntag, Borgnakke and van Wylen 6.21 A cylinder has 0.1 kg air at 25oC, 200 kPa with a 5 kg piston on top. A valve at the bottom is opened to let the air out and the piston drops 0.25 m towards the bottom. What is the work involved in this process? What happens to the energy? If we neglect acceleration of piston then P = C = Pequilibrium W = P ∆V To get the volume change from the height we need the cylinder area. The force balance on the piston gives mpg mpg 5 × 9.807 P = Po + A ⇒ A = P - P = 100 × 1000 = 0.000 49 m2 o ∆V = - AH = -0.000 49 × 0.25 = -0.000 1225 m3 W = P ∆V = 200 kPa × (-0.000 1225) m3 = -0.0245 kJ The air that remains inside has not changed state and therefore not energy. The work leaves as flow work Pv ∆m. m g cb AIR Pcyl e Sonntag, Borgnakke and van Wylen Continuity equation and flow rates 6.22 Air at 35°C, 105 kPa, flows in a 100 mm × 150 mm rectangular duct in a heating system. The volumetric flow rate is 0.015 m3/s. What is the velocity of the air flowing in the duct and what is the mass flow rate? Solution: Assume a constant velocity across the duct area with A = 100 × 150 ×10-6 m2 = 0.015 m2 and the volumetric flow rate from Eq.6.3, . . V = mv = AV . V 0.015 m3/s = 1.0 m/s V=A= 0.015 m2 Ideal gas so note: RT 0.287 × 308.2 v= P = = 0.8424 m3/kg 105 . . V 0.015 m = v = 0.8424 = 0.0178 kg/s Sonntag, Borgnakke and van Wylen 6.23 A boiler receives a constant flow of 5000 kg/h liquid water at 5 MPa, 20°C and it heats the flow such that the exit state is 450°C with a pressure of 4.5 MPa. Determine the necessary minimum pipe flow area in both the inlet and exit pipe(s) if there should be no velocities larger than 20 m/s. Solution: Mass flow rate from Eq.6.3, both V ≤ 20 m/s . . 1 mi = me = (AV/v) i = (AV/v) e = 5000 3600 kg/s Table B.1.4 vi = 0.001 m3/kg, Table B.1.3 ve = (0.08003 + 0.00633)/2 = 0.07166 m3/kg, . 5000 Ai ≥ vi m/Vi = 0.001× 3600 / 20 = 6.94 × 10-5 m2 = 0.69 cm2 . 5000 Ae ≥ ve m/Ve = 0.07166 × 3600 / 20 = 4.98 × 10-3 m2 = 50 cm2 vapor Inlet liquid i Super heater Q cb Q boiler e Exit Superheated vapor Sonntag, Borgnakke and van Wylen 6.24 An empty bathtub has its drain closed and is being filled with water from the faucet at a rate of 10 kg/min. After 10 minutes the drain is opened and 4 kg/min flows out and at the same time the inlet flow is reduced to 2 kg/min. Plot the mass of the water in the bathtub versus time and determine the time from the very beginning when the tub will be empty. Solution: During the first 10 minutes we have dmcv . . = mi = 10 kg/min , ∆m = m ∆t1 = 10 × 10 = 100 kg dt So we end up with 100 kg after 10 min. For the remaining period we have dmcv . . = mi - me= 2 – 4 = -2 kg/min dt . ∆m ∆m2 = mnet ∆t2 ∆t2 = . = -100/-2 = 50 min. mnet So it will take an additional 50 min. to empty ∆ttot = ∆t1 + ∆t2 = 10 + 50 = 60 min. . m m kg 100 0 10 t 0 10 20 min 0 -2 t 0 10 min Sonntag, Borgnakke and van Wylen 6.25 Nitrogen gas flowing in a 50-mm diameter pipe at 15°C, 200 kPa, at the rate of 0.05 kg/s, encounters a partially closed valve. If there is a pressure drop of 30 kPa across the valve and essentially no temperature change, what are the velocities upstream and downstream of the valve? Solution: Same inlet and exit area: π A = 4 (0.050)2 = 0.001963 m2 RTi 0.2968 × 288.2 Ideal gas: vi = P = = 0.4277 m3/kg 200 i From Eq.6.3, . mvi 0.05 × 0.4277 Vi = A = 0.001963 = 10.9 m/s RTe 0.2968 × 288.2 Ideal gas: ve = P = = 0.5032 m3/kg 170 e . mve 0.05 × 0.5032 Ve = A = 0.001963 = 12.8 m/s Sonntag, Borgnakke and van Wylen 6.26 Saturated vapor R-134a leaves the evaporator in a heat pump system at 10°C, with a steady mass flow rate of 0.1 kg/s. What is the smallest diameter tubing that can be used at this location if the velocity of the refrigerant is not to exceed 7 m/s? Solution: Mass flow rate Eq.6.3: . . m = V/v = AV/v Exit state Table B.5.1: (T = 10°C, x =1) => v = vg = 0.04945 m3/kg . The minimum area is associated with the maximum velocity for given m . mvg 0.1 kg/s × 0.04945 m3/kg π2 = = 0.000706 m2 = 4 DMIN AMIN = 7 m/s VMAX DMIN = 0.03 m = 30 mm Exit cb Sonntag, Borgnakke and van Wylen 6.27 A hot air home heating system takes 0.25 m3/s air at 100 kPa, 17oC into a furnace and heats it to 52oC and delivers the flow to a square duct 0.2 m by 0.2 m at 110 kPa. What is the velocity in the duct? Solution: . The inflate flow is given by a mi . . . Continuity Eq.: mi = Vi / vi = me = AeVe/ve RTi 0.287 × 290 m3 = 0.8323 kg Ideal gas: vi = P = 100 i RTe 0.287 × (52 + 273) ve = P = 110 e = 0.8479 m3/ kg . . mi = Vi/vi = 0.25/0.8323 = 0.30 kg/s . 0.3 × 0.8479 m3/s Ve = m ve/ Ae = = 6.36 m/s 0.2 × 0.2 m2 Sonntag, Borgnakke and van Wylen 6.28 Steam at 3 MPa, 400°C enters a turbine with a volume flow rate of 5 m3/s. An extraction of 15% of the inlet mass flow rate exits at 600 kPa, 200°C. The rest exits the turbine at 20 kPa with a quality of 90%, and a velocity of 20 m/s. Determine the volume flow rate of the extraction flow and the diameter of the final exit pipe. Solution: . . mi = V/v = 5/0.09936 = 50.32 kg/ s Inlet flow : Extraction flow : . . me = 0.15 mi = 7.55 kg/ s; (Table B.1.3) v = 0.35202 m3/kg . . Vex = mev = 7.55 × 0.35202 = 2.658 m3/ s . . Exit flow : m = 0.85 mi = 42.77 kg /s v = 0.001017 + 0.9 × 7.64835 = 6.8845 m3/kg Table B.1.2 . m = AV/v ⇒ . A = (π/4) D2 = m v/V = 42.77 × 6.8845/20 = 14.723 m2 ⇒ D = 4.33 m Inlet flow 1 2 Extraction flow WT HP section 3 Exit flow LP section Sonntag, Borgnakke and van Wylen 6.29 A household fan of diameter 0.75 m takes air in at 98 kPa, 22oC and delivers it at 105 kPa, 23oC with a velocity of 1.5 m/s. What are the mass flow rate (kg/s), the inlet velocity and the outgoing volume flow rate in m3/s? Solution: Continuity Eq. Ideal gas . . mi = me = AV/ v v = RT/P π π Area : A = 4 D 2 = 4× 0.752 = 0.442 m2 . Ve = AVe = 0.442 ×1.5 = 0.6627 m3/s RTe 0.287 × (23 + 273) = 0.8091 m3/kg ve = P = 105 e . . mi = Ve/ve = 0.6627/0.8091 = 0.819 kg/s . AVi /vi = mi = AVe / ve RTi 0.287 × (22 + 273) Vi = Ve × (vi / ve) = Ve × P v = 1.5 × = 1.6 m/s 98 × 0.8091 ie Sonntag, Borgnakke and van Wylen Single flow single device processes Nozzles, diffusers 6.30 Nitrogen gas flows into a convergent nozzle at 200 kPa, 400 K and very low velocity. It flows out of the nozzle at 100 kPa, 330 K. If the nozzle is insulated find the exit velocity. Solution: C.V. Nozzle steady state one inlet and exit flow, insulated so it is adiabatic. Inlet Exit Low V Hi V Low P, A Hi P, A Energy Eq.6.13: cb 1 2 h1 + ∅ = h2 + 2 V2 2 V2 = 2 ( h1 - h2 ) ≅ 2 CPN2 (T1 – T2 ) = 2 × 1.042 (400 – 330) = 145.88 kJ/kg = 145 880 J/kg ⇒ V2 = 381.9 m/s Sonntag, Borgnakke and van Wylen 6.31 A nozzle receives 0.1 kg/s steam at 1 MPa, 400oC with negligible kinetic energy. The exit is at 500 kPa, 350oC and the flow is adiabatic. Find the nozzle exit velocity and the exit area. Solution: 1 2 1 2 Energy Eq.6.13: h1+ 2 V1 + gZ1 = h2 + 2 V2 + gZ2 Process: Z1 = Z2 State 1: V1 = 0 , Table B.1.3 State 2: Table B.1.3 h1 = 3263.88 kJ/kg h2 = 3167.65 kJ/kg Then from the energy equation 12 2 V2 = h1 – h2 = 3263.88 – 3167.65 = 96.23 kJ/kg V2 = 2(h1 - h2) = 2 × 96.23 × 1000 = 438.7 m/s The mass flow rate from Eq.6.3 . m = ρAV = AV/v . 2 2 A = mv/V = 0.1 × 0.57012 / 438.7 = 0.00013 m = 1.3 cm Inlet Exit Low V Hi V Low P, A Hi P, A cb Sonntag, Borgnakke and van Wylen 6.32 Superheated vapor ammonia enters an insulated nozzle at 20°C, 800 kPa, shown in Fig. P6.32, with a low velocity and at the steady rate of 0.01 kg/s. The ammonia exits at 300 kPa with a velocity of 450 m/s. Determine the temperature (or quality, if saturated) and the exit area of the nozzle. Solution: C.V. Nozzle, steady state, 1 inlet and 1 exit flow, insulated so no heat transfer. 2 2 Energy Eq.6.13: q + hi + Vi /2 = he + Ve /2, Process: q = 0, Vi = 0 Table B.2.2: hi = 1464.9 = he + 4502/(2×1000) Table B.2.1: Pe = 300 kPa ⇒ he = 1363.6 kJ/kg Sat. state at −9.2°C : he = 1363.6 = 138.0 + xe × 1293.8, => ve = 0.001536 + xe × 0.4064 = 0.3864 m3/kg xe = 0.947, . Ae = meve/Ve = 0.01 × 0.3864 / 450 = 8.56 × 10-6 m2 Inlet Exit Low V Hi V Low P, A Hi P, A cb Sonntag, Borgnakke and van Wylen 6.33 In a jet engine a flow of air at 1000 K, 200 kPa and 30 m/s enters a nozzle, as shown in Fig. P6.33, where the air exits at 850 K, 90 kPa. What is the exit velocity assuming no heat loss? Solution: C.V. nozzle. No work, no heat transfer . . . Continuity mi = me = m . . Energy : m (hi + ½Vi2) = m(he+ ½Ve2) Due to high T take h from table A.7.1 ½Ve2 = ½ Vi2 + hi - he 1 = 2000 (30)2 + 1046.22 – 877.4 = 0.45 + 168.82 = 169.27 kJ/kg Ve = (2000 × 169.27)1/2 = 581.8 m/s Sonntag, Borgnakke and van Wylen 6.34 In a jet engine a flow of air at 1000 K, 200 kPa and 40 m/s enters a nozzle where the air exits at 500 m/s, 90 kPa. What is the exit temperature assuming no heat loss? Solution: C.V. nozzle, no work, no heat transfer . . . Continuity mi= me = m . . Energy : m (hi + ½Vi2) = m(he+ ½Ve2) Due to the high T we take the h value from Table A.7.1 he = hi + ½ Vi2 - ½Ve2 = 1046.22 + 0.5 × (402 – 5002) (1/1000) = 1046.22 – 124.2 = 922.02 kJ/kg Interpolation in Table A.7.1 922.02 - 877.4 Te = 850 + 50 933.15 - 877.4 = 890 K 40 m/s 200 kPa 500 m/s 90 kPa Sonntag, Borgnakke and van Wylen 6.35 A sluice gate dams water up 5 m. There is a small hole at the bottom of the gate so liquid water at 20oC comes out of a 1 cm diameter hole. Neglect any changes in internal energy and find the exit velocity and mass flow rate. Solution: 1 2 1 2 Energy Eq.6.13: h1+ 2 V1 + gZ1 = h2 + 2 V2 + gZ2 Process: h1 = h2 both at P = 1 atm Z1 = Z2 + 5 m V1 = 0 Water 12 2 V2 = g (Z1 − Z2) 2g(Z1 - Z2) = 2 × 9.806 × 5 = 9.902 m/s π . m = ρΑV = AV/v = 4 D2 × (V2/v) V2 = π = 4 × (0.01)2 × (9.902 / 0.001002) = 0.776 kg/s 5m Sonntag, Borgnakke and van Wylen 6.36 A diffuser, shown in Fig. P6.36, has air entering at 100 kPa, 300 K, with a velocity of 200 m/s. The inlet cross-sectional area of the diffuser is 100 mm2. At the exit, the area is 860 mm2, and the exit velocity is 20 m/s. Determine the exit pressure and temperature of the air. Solution: Continuity Eq.6.3: . . mi = AiVi/vi = me = AeVe/ve, Energy Eq.(per unit mass flow)6.13: 1 1 1 hi + 2Vi2 = he + 2Ve2 1 he - hi = 2 ×2002/1000 − 2 ×202/1000 = 19.8 kJ/kg Te = Ti + (he - hi)/Cp = 300 + 19.8/1.004 = 319.72 K Now use the continuity equation and the ideal gas law AeVe AeVe ve = vi = (RTi/Pi) A V = RTe/Pe AiVi i i Te AiVi 319.72 100 × 200 Pe = Pi T = 123.92 kPa = 100 300 860 × 20 i AeVe Inlet Hi V Low P, A Exit Low V Hi P, A Sonntag, Borgnakke and van Wylen 6.37 A diffuser receives an ideal gas flow at 100 kPa, 300 K with a velocity of 250 m/s and the exit velocity is 25 m/s. Determine the exit temperature if the gas is argon, helium or nitrogen. Solution: C.V. Diffuser: . . mi = me Energy Eq.6.13: hi + 2 Vi = 2 Ve + he ⇒ he = hi + 2 Vi - 2Ve 1 & assume no heat transfer ⇒ 2 1 1 2 1 2 2 2 12 1 he – hi ≈ Cp ( Te – Ti ) = 2 ( Vi - Ve ) = 2 ( 2502 – 252 ) = 30937.5 J/kg = 30.938 kJ/kg Specific heats for ideal gases are from table A.5 30.938 ∆T = 0.52 = 59.5 Argon Cp = 0.52 kJ/kg K; Helium 30.938 Cp = 5.913 kJ/kg K; ∆T = 5.193 = 5.96 Te = 306 K Nitrogen 30.938 Cp = 1.042 kJ/kg K; ∆T = 1.042 = 29.7 Te = 330 K Inlet Hi V Low P, A Te = 359.5 K Exit Low V Hi P, A Sonntag, Borgnakke and van Wylen 6.38 Air flows into a diffuser at 300 m/s, 300 K and 100 kPa. At the exit the velocity is very small but the pressure is high. Find the exit temperature assuming zero heat transfer. Solution: Energy Eq.: Process: 1 2 1 2 h1 + 2 V1 + gZ1 = h2 + 2 V2 + gZ2 Z1 = Z2 and V2 = 0 1 2 h2 = h1 + 2 V1 1 2 T2 = T1 + 2 × (V1 / Cp) 1 = 300 + 2 × 3002 / (1.004 × 1000) = 344.8K Inlet Hi V Low P, A Exit Low V Hi P, A Sonntag, Borgnakke and van Wylen 6.39 The front of a jet engine acts as a diffuser receiving air at 900 km/h, -5°C, 50 kPa, bringing it to 80 m/s relative to the engine before entering the compressor. If the flow area is reduced to 80% of the inlet area find the temperature and pressure in the compressor inlet. Solution: C.V. Diffuser, Steady state, 1 inlet, 1 exit flow, no q, no w. Continuity Eq.6.3: . . mi = me = (AV/v) . .12 12 Energy Eq.6.12: m ( hi + 2 Vi ) = m ( 2 Ve + he ) 1 2 1 2 1 900 × 1000 2 he – hi = Cp ( Te – Ti ) = 2 Vi - 2 Ve = 2 ( 3600 ) 1 − 2 (80)2 = 28050 J/kg = 28.05 kJ/kg ∆T = 28.05/1.004 = 27.9 ⇒ Te = −5 + 27.9 = 22.9°C Now use the continuity eq.: AiVi /vi = AeVe /ve ⇒ ve = v i × Ideal gas: AeVe ve = v i AiVi 0.8 × 80 = vi × 0.256 1 × 250 Pv = RT => ve = RTe/Pe = RT i × 0.256/Pi Pe = Pi (Te/T i)/0.256 = 50 × 296/268 × 0.256 = 215.7 kPa Fan Sonntag, Borgnakke and van Wylen Throttle flow 6.40 Helium is throttled from 1.2 MPa, 20°C, to a pressure of 100 kPa. The diameter of the exit pipe is so much larger than the inlet pipe that the inlet and exit velocities are equal. Find the exit temperature of the helium and the ratio of the pipe diameters. Solution: C.V. Throttle. Steady state, Process with: q = w = 0; and Energy Eq.6.13: hi = he, . AV m = RT/P Vi = Ve, Zi = Ze Ideal gas => . But m, V, T are constant ⇒ => Ti = Te = 20°C PiAi = PeAe De Pi 1/2 1.21/2 = 0.1 = 3.464 Di = Pe Sonntag, Borgnakke and van Wylen 6.41 Saturated vapor R-134a at 500 kPa is throttled to 200 kPa in a steady flow through a valve. The kinetic energy in the inlet and exit flow is the same. What is the exit temperature? Solution: Steady throttle flow . . . Continuity mi = me = m 1 2 1 2 Energy Eq.6.13: h1 + 2 V1 + gZ1 = h2 + 2 V2 + gZ2 and V2 = V1 Process: Z1 = Z2 ⇒ h2 = h1 = 407.45 kJ/kg from Table B.5.2 ⇒ superheated vapor State 2: P2 & h2 Interpolate between 0oC and 10oC in table B.5.2 in the 200 kPa subtable 407.45 – 400.91 T2 = 0 + 10 409.5 – 400.91 = 7.6oC i e T 500 kPa cb i h=C 200 e v Sonntag, Borgnakke and van Wylen 6.42 Saturated liquid R-12 at 25oC is throttled to 150.9 kPa in your refrigerator. What is the exit temperature? Find the percent increase in the volume flow rate. Solution: Steady throttle flow. Assume no heat transfer and no change in kinetic or potential energy. he = hi = hf 25oC = 59.70 kJ/kg = hf e + xe hfg e at 150.70 kPa From table B.3.1 we get Te = Tsat ( 150.9 kPa ) = -20oC he – hf e 59.7 – 17.82 xe = h = 160.92 = 0.26025 fg e ve = vf + xe vfg = 0.000685 + xe 0.10816 = 0.0288336 m3/kg vi = vf 25oC = 0.000763 m3/kg . . V = mv so the ratio becomes . . Ve mve ve 0.0288336 = . = v = 0.000763 = 37.79 . i mvi Vi So the increase is 36.79 times or 3679 % e i cb T i e h=C v Sonntag, Borgnakke and van Wylen 6.43 Water flowing in a line at 400 kPa, saturated vapor, is taken out through a valve to 100 kPa. What is the temperature as it leaves the valve assuming no changes in kinetic energy and no heat transfer? Solution: C.V. Valve. Steady state, single inlet and exit flow . . m1 = m2 Continuity Eq.: Energy Eq.6.12: 1 2 . . . . m1h1 + Q = m2h2 + W Process: Throttling . Small surface area: Q = 0; No shaft: . W=0 Table B.1.2: h2 = h1 = 2738.6 kJ/kg ⇒ T2 = 131.1°C Sonntag, Borgnakke and van Wylen 6.44 Liquid water at 180oC, 2000 kPa is throttled into a flash evaporator chamber having a pressure of 500 kPa. Neglect any change in the kinetic energy. What is the fraction of liquid and vapor in the chamber? Solution: 1 2 1 2 Energy Eq.6.13: h1 + 2 V1 + gZ1 = h2 + 2 V2 + gZ2 and V2 = V1 Process: Z1 = Z2 ⇒ h2 = h1 = 763.71 kJ/kg from Table B.1.4 ⇒ 2 – phase State 2: P2 & h2 h2 = hf + x2 hfg 763.71 – 640.21 x2 = (h2 - hf ) / hfg= = 0.0586 2108.47 Fraction of Vapor: x2 = 0.0586 Liquid: 1 - x2 = 0.941 (5.86 %) (94.1 %) Two-phase out of the valve. The liquid drops to the bottom. Sonntag, Borgnakke and van Wylen 6.45 Water at 1.5 MPa, 150°C, is throttled adiabatically through a valve to 200 kPa. The inlet velocity is 5 m/s, and the inlet and exit pipe diameters are the same. Determine the state (neglecting kinetic energy in the energy equation) and the velocity of the water at the exit. Solution: . m = const, CV: valve. A = const ⇒ Ve = Vi(ve/vi) Energy Eq.6.13: 1 2 ve2 (he - hi) + 2 Vi v - 1 = 0 i Now neglect the kinetic energy terms (relatively small) from table B.1.1 we have the compressed liquid approximated with saturated liquid same T 1 2 1 2 hi + 2 Vi = 2 Ve + he or he = hi = 632.18 kJ/kg ; Table B.1.2: vi = 0.001090 m3/kg he = 504.68 + xe × 2201.96, Substituting and solving, xe = 0.0579 ve = 0.001061 + xe × 0.88467 = 0.052286 m3/kg Ve = Vi(ve/vi) = 5 m/s (0.052286 / 0.00109) = 240 m/s Sonntag, Borgnakke and van Wylen 6.46 R-134a is throttled in a line flowing at 25oC, 750 kPa with negligible kinetic energy to a pressure of 165 kPa. Find the exit temperature and the ratio of exit pipe diameter to that of the inlet pipe (Dex/Din) so the velocity stays constant. Solution: Energy Eq.6.13: Process: 1 2 1 2 h1 + 2 V1 + gZ1 = h2 + 2 V2 + gZ2 and V2 = V1 Z1 = Z2 State 1, Table B.5.1: h1 = 234.59 kJ/kg, v1 = vf = 0.000829 m3/kg Use energy eq.: ⇒ h2 = h1 = 234.59 kJ/kg ⇒ 2 – phase and T2 = Tsat (165 kPa) = -15°C State 2: P2 & h2 h2 = hf + x2 hfg = 234.59 kJ/kg x2 = (h2 - hf ) / hfg= (234.59 – 180.19) / 209 = 0.2603 v2 = vf + x2 × vfg = 0.000746 + 0.2603 × 0.11932 = 0.0318 m3/kg Now the continuity equation with V2 = V1 gives, from Eq.6.3, . m = ρΑV = AV/v = A1V1/v1 = (A2 V1) / v2 (A2 / A1) = v2 / v1 = (D2 / D1) (D2/D1) = (v2 / v1) 0.5 2 = (0.0318 / 0.000829) 0.5 = 6.19 Sonntag, Borgnakke and van Wylen 6.47 Methane at 3 MPa, 300 K is throttled through a valve to 100 kPa. Calculate the exit temperature assuming no changes in the kinetic energy and ideal gas behavior. Repeat the answer for real-gas behavior. C.V. Throttle (valve, restriction), Steady flow, 1 inlet and exit, no q, w Energy Eq.: Real gas : hi = he => Ideal gas Ti = Te = 300 K hi = he = 598.71 Table B.7 Pe = 0.1 MPa Te = 13.85°C ( = 287 K) Sonntag, Borgnakke and van Wylen Turbines, Expanders 6.48 A steam turbine has an inlet of 2 kg/s water at 1000 kPa, 350oC and velocity of 15 m/s. The exit is at 100 kPa, x = 1 and very low velocity. Find the specific work and the power produced. Solution: 1 2 1 2 Energy Eq.6.13: h1 + 2 V1 + gZ1 = h2 + 2 V2 + gZ2 + wT and V2 = 0 Process: Z1 = Z2 Table B.1.3: h1 = 3157.65 kJ/kg, h2 = 2675.46 kJ/kg 1 2 2 wT = h1 + 2 V1 – h2 = 3157.65 + 15 / 2000 – 2675.46 = 482.3 kJ/kg . . WT = m × wT = 2 × 482.3 = 964.6 kW 1 WT 2 Sonntag, Borgnakke and van Wylen 6.49 A small, high-speed turbine operating on compressed air produces a power output of 100 W. The inlet state is 400 kPa, 50°C, and the exit state is 150 kPa, −30°C. Assuming the velocities to be low and the process to be adiabatic, find the required mass flow rate of air through the turbine. Solution: C.V. Turbine, no heat transfer, no ∆KE, no ∆PE Energy Eq.6.13: hin = hex + wT Ideal gas so use constant specific heat from Table A.5 wT = hin - hex ≅ Cp(Tin - Tex) = 1.004 (kJ/kg K) [50 - (-30)] K = 80.3 kJ/kg . . W = mwT ⇒ . . m = W/wT = 0.1/80.3 = 0.00125 kg/s The dentist’s drill has a small air flow and is not really adiabatic. Sonntag, Borgnakke and van Wylen 6.50 A liquid water turbine receives 2 kg/s water at 2000 kPa, 20oC and velocity of 15 m/s. The exit is at 100 kPa, 20oC and very low velocity. Find the specific work and the power produced. Solution: 1 2 1 2 Energy Eq.6.13: h1 + 2 V1 + gZ1 = h2 + 2 V2 + gZ2 + wT V2 = 0 Process: Z1 = Z2 and State 1: Table B.1.4 h1 = 85.82 kJ/kg State 2: Table B.1.1 h2 = 83.94 (which is at 2.3 kPa so we should add ∆Pv = 97.7 × 0.001 to this) 1 2 2 wT = h1 + 2 V1 − h2 = 85.82 + 15 /2000 – 83.94 = 1.99 kJ/kg . . WT = m × wT = 2 × 1.9925 = 3.985 kW Notice how insignificant the specific kinetic energy is. Sonntag, Borgnakke and van Wylen 6.51 Hoover Dam across the Colorado River dams up Lake Mead 200 m higher than the river downstream. The electric generators driven by water-powered turbines deliver 1300 MW of power. If the water is 17.5°C, find the minimum amount of water running through the turbines. Solution: C.V.: H2O pipe + turbines, Lake Mead DAM H T Continuity: . . min = mex; Energy Eq.6.13: Water states: (h+ V2/2 + gz)in = (h+ V2/2 + gz)ex + wT hin ≅ hex ; vin ≅ vex Now the specific turbine work becomes wT = gzin - gzex = 9.807 × 200/1000 = 1.961 kJ/kg . . 1300×103 kW m = WT/wT = 1.961 kJ/kg = 6.63 ×105 kg/s . . V = mv = 6.63 ×105 × 0.001001 = 664 m3/s Sonntag, Borgnakke and van Wylen 6.52 A windmill with rotor diameter of 30 m takes 40% of the kinetic energy out as shaft work on a day with 20oC and wind speed of 30 km/h. What power is produced? Solution: Continuity Eq. . . . mi = me = m . . . m (hi + ½Vi2 + gZi) = m(he+ ½Ve2 + gZe) + W . . Process information: W = m½Vi2 × 0.4 Energy . m = ρAV =AVi /vi π π A = 4 D 2 = 4 302 = 706.85 m2 0.287 × 293 vi = RTi/Pi = = 0.8301 m3/kg 101.3 30 × 1000 Vi = 30 km/h = 3600 = 8.3333 m/s . 706.85 × 8.3333 m = AVi /vi = = 7096 kg/s 0.8301 ½ Vi2 = ½ 8.33332 m2/s2 = 34.722 J/kg . . W = 0.4 m½ Vi2 = 0.4 ×7096 × 34.722 = 98 555 W = 98.56 kW Sonntag, Borgnakke and van Wylen 6.53 A small turbine, shown in Fig. P 6.53, is operated at part load by throttling a 0.25 kg/s steam supply at 1.4 MPa, 250°C down to 1.1 MPa before it enters the turbine and the exhaust is at 10 kPa. If the turbine produces 110 kW, find the exhaust temperature (and quality if saturated). Solution: C.V. Throttle, Steady, q = 0 and w = 0. No change in kinetic or potential energy. The energy equation then reduces to Energy Eq.6.13: h1 = h2 = 2927.2 kJ/kg from Table B.1.3 110 C.V. Turbine, Steady, no heat transfer, specific work: w = 0.25 = 440 kJ/kg Energy Eq.: h1 = h2 = h3 + w = 2927.2 kJ/kg ⇒ h3 = 2927.2 - 440 = 2487.2 kJ/kg State 3: (P, h) Table B.1.2 h < hg 2487.2 = 191.83 + x3 × 2392.8 ⇒ T = 45.8°C , x3 = 0.959 T 1 2 3 v Sonntag, Borgnakke and van Wylen 6.54 A small expander (a turbine with heat transfer) has 0.05 kg/s helium entering at 1000 kPa, 550 K and it leaves at 250 kPa, 300 K. The power output on the shaft is measured to 55 kW. Find the rate of heat transfer neglecting kinetic energies. Solution: C.V. Expander. Steady operation Cont. . . . . mhi + Q = mhe + W Q . . . mi= me = m Energy . i . . . Q = m (he-hi) + W Use heat capacity from tbl A.5: Cp He = 5.193 kJ/kg K . . . Q = mCp (Te-Ti) + W = 0.05× 5.193 (300 - 550) + 55 = - 64.91 + 55 = -9.9 kW WT cb e Sonntag, Borgnakke and van Wylen Compressors, fans 6.55 A compressor in a commercial refrigerator receives R-22 at -25oC, x = 1. The exit is at 800 kPa, 40oC. Neglect kinetic energies and find the specific work. Solution: i C.V. Compressor, steady state, single inlet and exit flow. For this device we also assume no heat transfer and Z1 = Z2 From Table B.4.1 : cb -WC h1 = 239.92 kJ/kg From Table B.4.2 : e h2 = 274.24 kJ/kg Energy Eq.6.13 reduces to wc = h1 - h2 = (239.92 – 274.24) = -34.3 kJ/kg Sonntag, Borgnakke and van Wylen 6.56 The compressor of a large gas turbine receives air from the ambient at 95 kPa, 20°C, with a low velocity. At the compressor discharge, air exits at 1.52 MPa, 430°C, with velocity of 90 m/s. The power input to the compressor is 5000 kW. Determine the mass flow rate of air through the unit. Solution: C.V. Compressor, steady state, single inlet and exit flow. Energy Eq.6.13: q + hi + Vi2/2 = he + Ve2/2 + w Here we assume q ≅ 0 and Vi ≅ 0 so using constant CPo from A.5 (90)2 -w = CPo(Te - Ti) + Ve2/2 = 1.004(430 - 20) + 2 × 1000 = 415.5 kJ/kg Notice the kinetic energy is 1% of the work and can be neglected in most cases. The mass flow rate is then from the power and the specific work . Wc . 5000 m = -w = 415.5 = 12.0 kg/s Sonntag, Borgnakke and van Wylen 6.57 A compressor brings R-134a from 150 kPa, -10oC to 1200 kPa, 50oC. It is water cooled with a heat loss estimated as 40 kW and the shaft work input is measured to be 150 kW. How much is the mass flow rate through the compressor? Solution: C.V Compressor. Steady flow. Neglect kinetic and potential energies. . . . . Energy : m hi + Q = mhe + W .. . m = (Q - W)/(he - hi) 1 2 Compressor -Wc Look in table B.5.2 hi = 393.84 kJ/kg, he = 426.84 kJ/kg -40 – (-150) . m = 426.84 – 393.84 = 3.333 kg/s Q cool Sonntag, Borgnakke and van Wylen 6.58 An ordinary portable fan blows 0.2 kg/s room air with a velocity of 18 m/s. What is the minimum power electric motor that can drive it? Hint: Are there any changes in P or T? Solution: C.V. Fan plus space out to near stagnant inlet room air. Energy Eq.6.13: q + hi + Vi2/2 = he + Ve2/2 + w Here q ≅ 0, Vi ≅ 0 and hi = he same P and T −w = Ve2/2 = 182/2000 = 0.162 kJ/kg . . −W = −mw = 0.2 kg/s × 0.162 kJ/kg = 0.032 kW Sonntag, Borgnakke and van Wylen 6.59 An air compressor takes in air at 100 kPa, 17°C and delivers it at 1 MPa, 600 K to a constant-pressure cooler, which it exits at 300 K. Find the specific compressor work and the specific heat transfer in the cooler. Solution C.V. air compressor q = 0 . . Continuity Eq.: m2 = m1 Energy Eq.6.13: h1 + wc = h2 1 2 Q cool Compressor -W c Compressor section Cooler section Table A.7: wc in = h2 - h1 = 607.02 - 290.17 = 316.85 kJ/kg C.V. cooler w = 0 / Continuity Eq.: Energy Eq.6.13: . . m3 = m1 h2 = qout + h3 qout = h2 - h3 = 607.02 - 300.19 = 306.83 kJ/kg 3 Sonntag, Borgnakke and van Wylen 6.60 A 4 kg/s steady flow of ammonia runs through a device where it goes through a polytropic process. The inlet state is 150 kPa, -20oC and the exit state is 400 kPa, 80oC, where all kinetic and potential energies can be neglected. The specific work input has been found to be given as [n/(n-1)] ∆(Pv). a) Find the polytropic exponent n b) Find the specific work and the specific heat transfer. Solution: C.V. Steady state device. Single inlet and single exit flows. 1 2 1 2 Energy Eq.6.13: h1 + 2 V1 + gZ1 + q = h2 + 2 V2 + gZ2 + w Process: Pvn = constant State 1: Table B.2.2 and Z1 = Z2 , V1 = V2 = 0 v1 = 0.79774, h1 = 1422.9 State 2: Table B.2.2 v2 = 0.4216, h2 = 1636.7 From the polytropic process equation and the two states we can find the exponent n: P2 v1 400 0.79774 n = ln P / ln v = ln 150 / ln 0.4216 = 1.538 1 2 Before we can do the heat transfer we need the work term w=− n (P v – P1v1) = -2.8587(400×0.4216 – 150×0.79774) n−1 2 2 = −140.0 kJ/kg q = h2 + w − h1 = 1636.7 – 140.0 – 1422.9 = 73.8 kJ/kg Sonntag, Borgnakke and van Wylen 6.61 An exhaust fan in a building should be able to move 2.5 kg/s air at 98 kPa, 20oC through a 0.4 m diameter vent hole. How high a velocity must it generate and how much power is required to do that? Solution: C.V. Fan and vent hole. Steady state with uniform velocity out. Continuity Eq.: . m = constant = ρΑV = AV / v =AVP/RT Ideal gas : Pv = RT, π and area is A = 4 D2 Now the velocity is found . π π V = m RT/(4 D2P) = 2.5 × 0.287 × 293.15 / ( 4 × 0.42 × 98) = 17.1 m/s The kinetic energy out is 12 1 2 2 V2 = 2 × 17.1 / 1000 = 0.146 kJ/kg which is provided by the work (only two terms in energy equation that does not cancel, we assume V1 = 0) . .12 Win = m 2 V2 = 2.5 × 0.146 = 0.366 kW Sonntag, Borgnakke and van Wylen 6.62 How much power is needed to run the fan in Problem 6.29? A household fan of diameter 0.75 m takes air in at 98 kPa, 22oC and delivers it at 105 kPa, 23oC with a velocity of 1.5 m/s. What are the mass flow rate (kg/s), the inlet velocity and the outgoing volume flow rate in m3/s? Solution: . . Continuity Eq. mi = me = AV/ v Ideal gas v = RT/P π π Area : A = 4 D2 = 4× 0.752 = 0.442 m2 . Ve = AVe = 0.442 ×1.5 = 0.6627 m3/s RTe 0.287 × 296 = 0.8091m3/kg ve = P = 105 e . . mi = Ve/ve = 0.6627/0.8091 = 0.819 kg/s . AVi /vi = mi = AVe / ve Vi = Ve × (vi / ve) = Ve × (RTi)/(Pive) = 1.5 × 0.287 × (22 + 273) = 1.6 m/s 98 × 0.8091 . . . m (hi + ½Vi2) = m(he+ ½Ve2) +W . . . W = m(hi + ½Vi2 – he – ½Ve2 ) = m [Cp (Ti-Te) + ½ Vi2 – ½Ve2 ] 1.62 - 1.52 = 0.819 [ 1.004 (-1) + 2000 ] = 0.819 [ -1.004 + 0.000155] = - 0.81 kW Sonntag, Borgnakke and van Wylen Heaters/Coolers 6.63 Carbon dioxide enters a steady-state, steady-flow heater at 300 kPa, 15oC, and exits at 275 kPa, 1200oC, as shown in Fig. P6.63. Changes in kinetic and potential energies are negligible. Calculate the required heat transfer per kilogram of carbon dioxide flowing through the heater. Solution: C.V. Heater Steady state single inlet and exit flow. Energy Eq.6.13: q + h i = he e i Q Table A.8: q = he - hi = 1579.2 – 204.6 = 1374.6 kJ/kg (If we use Cp0 from A.5 then q ≅ 0.842(1200 - 15) = 997.8 kJ/kg) Too large ∆T, Tave to use Cp0 at room temperature. Sonntag, Borgnakke and van Wylen 6.64 A condenser (cooler) receives 0.05 kg/s R-22 at 800 kPa, 40oC and cools it to 15o C. There is a small pressure drop so the exit state is saturated liquid. What cooling capacity (kW) must the condenser have? Solution: C.V. R-22 condenser. Steady state single flow, heat transfer out and no work. . . . m h1 = m h2 + Qout Energy Eq.6.12: Inlet state: Table B.4.2 h1 = 274.24 kJ/kg, Exit state: Table B.4.1 h2 = 62.52 kJ/kg Process: Neglect kinetic and potential energy changes. Cooling capacity is taken as the heat transfer out i.e. positive out so . . Qout = m ( h1- h2) = 0.05 kg/s (274.24 – 62.52) kJ/kg = 10.586 kW = 10.6 kW 1 Q cool 2 Sonntag, Borgnakke and van Wylen 6.65 A chiller cools liquid water for air-conditioning purposes. Assume 2.5 kg/s water at 20oC, 100 kPa is cooled to 5oC in a chiller. How much heat transfer (kW) is needed? Solution: C.V. Chiller. Steady state single flow with heat transfer. Neglect changes in kinetic and potential energy and no work term. Energy Eq.6.13: qout = hi – he Properties from Table B.1.1: hi = 83.94 kJ/kg and he = 20.98 kJ/kg Now the energy equation gives qout = 83.94 – 20.98 = 62.96 kJ/kg . . Qout = m qout = 2.5 × 62.96 = 157.4 kW Alternative property treatment since single phase and small ∆T If we take constant heat capacity for the liquid from Table A.4 qout = hi – he ≅ Cp (Ti - Te ) = 4.18 (20 – 5) = 62.7 kJ/kg . . Qout = m qout = 2.5 × 62.7 = 156.75 kW 1 Q cool 2 Sonntag, Borgnakke and van Wylen 6.66 Saturated liquid nitrogen at 500 kPa enters a boiler at a rate of 0.005 kg/s and exits as saturated vapor. It then flows into a super heater also at 500 kPa where it exits at 500 kPa, 275 K. Find the rate of heat transfer in the boiler and the super heater. Solution: C.V.: boiler steady single inlet and exit flow, neglect KE, PE energies in flow Continuity Eq.: . . . m1 = m2 = m3 vapor 1 2 Super heater Q cb T P 3 500 123 Q boiler 3 12 v Table B.6.1: h1 = -87.095 kJ/kg, h2 = 86.15 kJ/kg, Table B.6.2: h3 = 284.06 kJ/kg Energy Eq.6.13: qboiler = h2 – h1 = 86.15 - (- 87.095) = 173.25 kJ/kg . . Qboiler = m1qboiler = 0.005 × 173.25 = 0.866 kW C.V. Superheater (same approximations as for boiler) Energy Eq.6.13: qsup heater = h3 – h2 = 284.06 – 86.15 = 197.9 kJ/kg . . Qsup heater = m2qsup heater = 0.005 × 197.9 = 0.99 kW v Sonntag, Borgnakke and van Wylen 6.67 In a steam generator, compressed liquid water at 10 MPa, 30°C, enters a 30-mm diameter tube at the rate of 3 L/s. Steam at 9 MPa, 400°C exits the tube. Find the rate of heat transfer to the water. Solution: C.V. Steam generator. Steady state single inlet and exit flow. Constant diameter tube: Table B.1.4 π Ai = Ae = 4 (0.03)2 = 0.0007068 m2 . . m = Vi/vi = 0.003/0.0010003 = 3.0 kg/s . Vi = Vi/Ai = 0.003/0.0007068 = 4.24 m/s Exit state properties from Table B.1.3 Ve = Vi × ve/vi = 4.24 × 0.02993/0.0010003 = 126.86 m/s The energy equation Eq.6.12 is solved for the heat transfer as . . Q = m (he - hi) + Ve2 - Vi2 /2 126.862 - 4.242 = 3.0 3117.8 - 134.86 + 2 × 1000 = 8973 kW ( ) Steam exit gas out Typically hot combustion gas in cb liquid water in Sonntag, Borgnakke and van Wylen 6.68 The air conditioner in a house or a car has a cooler that brings atmospheric air from 30oC to 10oC both states at 101 kPa. If the flow rate is 0.5 kg/s find the rate of heat transfer. Solution: CV. Cooler. Steady state single flow with heat transfer. Neglect changes in kinetic and potential energy and no work term. Energy Eq.6.13: qout = hi – he Use constant heat capacity from Table A.5 (T is around 300 K) qout = hi − he = Cp (Ti − Te) kJ = 1.004 kg K × (30 – 10) K = 20.1 kJ/kg . Qout . = m qout = 0.5 × 20.1 = 10 kW Sonntag, Borgnakke and van Wylen 6.69 A flow of liquid glycerine flows around an engine, cooling it as it absorbs energy. The glycerine enters the engine at 60oC and receives 9 kW of heat transfer. What is the required mass flow rate if the glycerine should come out at maximum 95o C? Solution: C.V. Liquid flow (glycerine is the coolant), steady flow. no work. . . . Energy Eq.: mhi + Q = mhe . . . Q m = Q/( he - hi) = C (T - T ) gly e i From table A.4 Cgly = 2.42 kJ/kg-K . 9 m = 2.42 (95 – 60) = 0.106 kg/s Air intake filter Shaft power Fan Radiator Atm. air cb Exhaust flow Coolant flow Sonntag, Borgnakke and van Wylen 6.70 A cryogenic fluid as liquid nitrogen at 90 K, 400 kPa flows into a probe used in cryogenic surgery. In the return line the nitrogen is then at 160 K, 400 kPa. Find the specific heat transfer to the nitrogen. If the return line has a cross sectional area 100 times larger than the inlet line what is the ratio of the return velocity to the inlet velocity? Solution: C.V line with nitrogen. No kinetic or potential energy changes Continuity Eq.: . . . m = constant = me = mi = AeVe/ve = AiVi/vi Energy Eq.6.13: q = he − hi State i, Table B.6.1: hi = -95.58 kJ/kg, vi = 0.001343 m3/kg State e, Table B.6.2: he = 162.96 kJ/kg, ve = 0.11647 m3/kg From the energy equation q = he − hi = 162.96 – (-95.58) = 258.5 kJ/kg From the continuity equation 1 0.11647 Ve/Vi = Ai/Ae (ve/vi) = 100 0.001343 = 0.867 Sonntag, Borgnakke and van Wylen Pumps, pipe and channel flows 6.71 A small stream with 20oC water runs out over a cliff creating a 100 m tall waterfall. Estimate the downstream temperature when you neglect the horizontal flow velocities upstream and downstream from the waterfall. How fast was the water dropping just before it splashed into the pool at the bottom of the waterfall? Solution: . . CV. Waterfall, steady state. Assume no Q nor W 1 Energy Eq.6.13: h + 2V2 + gZ = const. State 1: At the top zero velocity Z1 = 100 m State 2: At the bottom just before impact, Z2 = 0 State 3: At the bottom after impact in the pool. 12 h1 + 0 + gZ1 = h2 + 2 V2 + 0 = h3 + 0 + 0 Properties: h1 ≅ h2 same T, P 12 => 2 V2 = gZ1 V2 = 2gZ1 = 2 × 9.806 × 100 = 44.3 m/s Energy equation from state 1 to state 3 h3 = h1 + gZ1 use ∆h = Cp ∆T with value from Table A.4 (liquid water) T3 = T1 + gZ1 / Cp = 20 + 9.806 × 100 /4180 = 20.23 °C Sonntag, Borgnakke and van Wylen 6.72 A small water pump is used in an irrigation system. The pump takes water in from a river at 10oC, 100 kPa at a rate of 5 kg/s. The exit line enters a pipe that goes up to an elevation 20 m above the pump and river, where the water runs into an open channel. Assume the process is adiabatic and that the water stays at 10oC. Find the required pump work. Solution: C.V. pump + pipe. Steady state , 1 inlet, 1 exit flow. Assume same velocity in and out, no heat transfer. Continuity Eq.: . . . min = mex = m Energy Eq.6.12: e . m(hin + (1/2)Vin2 + gzin) = . . m(hex + (1/2) Vex2 + gzex) + W States: hin = hex same (T, P) H i cb . . W = m g(zin - zex) = 5 × 9.807 × (0 - 20)/1000 = −0.98 kW I.E. 0.98 kW required input Sonntag, Borgnakke and van Wylen 6.73 A steam pipe for a 300-m tall building receives superheated steam at 200 kPa at ground level. At the top floor the pressure is 125 kPa and the heat loss in the pipe is 110 kJ/kg. What should the inlet temperature be so that no water will condense inside the pipe? Solution: C.V. Pipe from 0 to 300 m, no ∆KE, steady state, single inlet and exit flow. Neglect any changes in kinetic energy. Energy Eq.6.13: q + hi = he + gZe No condensation means: Table B.1.2, hi = he + gZe - q = 2685.4 + he = hg at 125 kPa = 2685.4 kJ/kg 9.807 × 300 - (-110) = 2810.1 kJ/kg 1000 At 200 kPa: T ~ 170oC Table B.1.3 Sonntag, Borgnakke and van Wylen 6.74 The main waterline into a tall building has a pressure of 600 kPa at 5 m below ground level. A pump brings the pressure up so the water can be delivered at 200 kPa at the top floor 150 m above ground level. Assume a flow rate of 10 kg/s liquid water at 10oC and neglect any difference in kinetic energy and internal energy u. Find the pump work. Solution: C.V. Pipe from inlet at -5 m up to exit at +150 m, 200 kPa. 1 1 Energy Eq.6.13: hi + 2Vi2 + gZi = he + 2Ve2 + gZe + w With the same u the difference in h’s are the Pv terms 1 w = hi - he + 2 (Vi2 - Ve2) + g (Zi- Ze) = Pivi – Peve + g (Zi – Ze) = 600 × 0.001 – 200 × 0.001 + 9.806 × (-5-150)/1000 = 0.4 – 1.52 = -1.12 kJ/kg . . W = mw = 10 × (-1.12) = -11.2 kW Sonntag, Borgnakke and van Wylen 6.75 Consider a water pump that receives liquid water at 15oC, 100 kPa and delivers it to a same diameter short pipe having a nozzle with exit diameter of 1 cm (0.01 m) to the atmosphere 100 kPa. Neglect the kinetic energy in the pipes and assume constant u for the water. Find the exit velocity and the mass flow rate if the pump draws a power of 1 kW. Solution: π2π . . Continuity Eq.: mi = me = AV/v ; A = 4 De = 4 × 0.01 2 = 7.854× 10 −5 12 12 Energy Eq.6.13: hi + 2Vi + gZi = he + 2Ve + gZe + w Properties: hi = ui + Pivi = he = ue + Peve ; Pi = Pe ; vi = ve . 13 12 .1 2 ⇒ −W = m (2 Ve ) = A × 2 Ve /ve w = − 2 Ve . −2 W ve 1/3 2 × 1000 × 0.001001 1/3 Ve = = = 29.43 m/s A 7.854×10 −5 ( ) ( ) . m = AVe/ve = 7.854× 10 −5 × 29.43 / 0.001001 = 2.31 kg/s Sonntag, Borgnakke and van Wylen 6.76 A cutting tool uses a nozzle that generates a high speed jet of liquid water. Assume an exit velocity of 1000 m/s of 20oC liquid water with a jet diameter of 2 mm (0.002 m). How much mass flow rate is this? What size (power) pump is needed to generate this from a steady supply of 20oC liquid water at 200 kPa? Solution: C.V. Nozzle. Steady state, single flow. Continuity equation with a uniform velocity across A π π . m = AV/v = 4 D2 V / v = 4 0.0022 × 1000 / 0.001002 = 3.135 kg/s Assume Zi = Ze = Ø, ue = ui and Vi = 0 Pe = 100 kPa (atmospheric) 12 Energy Eq.6.13: hi + Ø + Ø = he + 2Ve + Ø + w 12 12 w = hi − he − 2Ve = ui − ue + Pi vi − Pe ve − 2Ve 12 = (Pi - Pe) vi − 2Ve = 0.001002 × (200 – 100) – 0.5 × (10002 / 1000) = 0.1002 – 500 ≅ −500 kJ/kg . . W = mw = 3.135 (-500) = −1567.5 kW Sonntag, Borgnakke and van Wylen 6.77 A pipe flows water at 15oC from one building to another. In the winter time the pipe loses an estimated 500 W of heat transfer. What is the minimum required mass flow rate that will ensure that the water does not freeze (i.e. reach 0oC)? Solution: . . . Energy Eq.: mhi + Q = mhe Assume saturated liquid at given T from table B.1.1 . . -500 × 10-3 0.5 Q m = h - h = 0 - 62.98 = 62.98 = 0.007 94 kg/s e i . -Q 1 2 Sonntag, Borgnakke and van Wylen Multiple flow single device processes Turbines, Compressors, Expanders 6.78 A steam turbine receives water at 15 MPa, 600°C at a rate of 100 kg/s, shown in Fig. P6.78. In the middle section 20 kg/s is withdrawn at 2 MPa, 350°C, and the rest exits the turbine at 75 kPa, and 95% quality. Assuming no heat transfer and no changes in kinetic energy, find the total turbine power output. Solution: C.V. Turbine Steady state, 1 inlet and 2 exit flows. . . . Continuity Eq.6.9: m1 = m2 + m3 ; Energy Eq.6.10: Table B.1.3 => . . . m3 = m1 - m2 = 80 kg/s . . . . m1h1 = WT + m2h2 + m3h3 h1 = 3582.3 kJ/kg, 1 h2 = 3137 kJ/kg 2 Table B.1.2 : h3 = hf + x3hfg = 384.3 + 0.95×2278.6 = 2549.1 kJ/kg From the energy equation, Eq.6.10 => . . . . WT = m1h1 − m2h2 − m3h3 = 91.565 MW WT 3 Sonntag, Borgnakke and van Wylen 6.79 A steam turbine receives steam from two boilers. One flow is 5 kg/s at 3 MPa, 700°C and the other flow is 15 kg/s at 800 kPa, 500°C. The exit state is 10 kPa, with a quality of 96%. Find the total power out of the adiabatic turbine. Solution: . C.V. whole turbine steady, 2 inlets, 1 exit, no heat transfer Q = 0 Continuity Eq.6.9: Energy Eq.6.10: . . . m1 + m2 = m 3 = 5 + 15 = 20 kg/s . . . . m1h1 + m2h2 = m3h3 + WT Table B.1.3: h1 = 3911.7 kJ/kg, h2 = 3480.6 kJ/kg 1 2 Table B.1.2: h3 = 191.8 + 0.96 × 2392.8 = 2488.9 kJ/kg WT 3 . WT = 5 × 3911.7 + 15 × 3480.6 – 20 × 2488.9 = 21990 kW = 22 MW Sonntag, Borgnakke and van Wylen 6.80 Two steady flows of air enters a control volume, shown in Fig. P6.80. One is 0.025 kg/s flow at 350 kPa, 150°C, state 1, and the other enters at 450 kPa, 15°C, both flows with low velocity. A single flow of air exits at 100 kPa, −40°C, state 3. The control volume rejects 1 kW heat to the surroundings and produces 4 kW of power. Neglect kinetic energies and determine the mass flow rate at state 2. Solution: . 1 W C.V. Steady device with two inlet and one Engine 3 exit flows, we neglect kinetic energies. Notice here the Q is rejected so it goes out. . 2 Q loss Continuity Eq.6.9: . . . . m1 + m2 = m3 = 0.025 + m2 Energy Eq.6.10: . . . . . m1h1 + m2h2 = m3h3 + WCV + Qloss Substitute the work and heat transfer into the energy equation and use constant heat capacity . 0.025 × 1.004 × 423.2 + m2 × 1.004 × 288.2 . = (0.025 + m2) 1.004 × 233.2 + 4.0 + 1.0 . Now solve for m2. 4.0 + 1.0 + 0.025 × 1.004 × (233.2 – 423.2) . m2 = 1.004 (288.2 - 233.2) . Solving, m2 = 0.0042 kg/s Sonntag, Borgnakke and van Wylen 6.81 A large expansion engine has two low velocity flows of water entering. High pressure steam enters at point 1 with 2.0 kg/s at 2 MPa, 500°C and 0.5 kg/s cooling water at 120 kPa, 30°C enters at point 2. A single flow exits at point 3 with 150 kPa, 80% quality, through a 0.15 m diameter exhaust pipe. There is a heat loss of 300 kW. Find the exhaust velocity and the power output of the engine. Solution: C.V. : Engine (Steady state) . . Constant rates of flow, Qloss and W State 1: State 2: Table B.1.3: h1 = 3467.6 kJ/kg Table B.1.1: h2 = 125.77 kJ/kg . W 1 Engine 2 3 . Q loss h3 = 467.1 + 0.8 × 2226.5 = 2248.3 kJ/kg v3 = 0.00105 + 0.8 × 1.15825 = 0.92765 m3/kg . . . Continuity Eq.6.9: m1+ m2 = m3 = 2 + 0.5= 2.5 kg/s = (AV/v) = (π/4)D2V/v Energy Eq.6.10: . . . . . m1h1 + m2h2 = m3(h3 + 0.5 V2) + Qloss + W . π V = m3v3 / [4 D2] = 2.5 × 0.92765 / (0.7854 × 0.152 ) = 131.2 m/s 0.5 V2 = 0.5 × 131.22 /1000 = 8.6 kJ/kg ( remember units factor 1000) . W = 2 ×3467.6 + 0.5 ×125.77 – 2.5 (2248.3 + 8.6) – 300 = 1056 kW Sonntag, Borgnakke and van Wylen 6.82 Cogeneration is often used where a steam supply is needed for industrial process energy. Assume a supply of 5 kg/s steam at 0.5 MPa is needed. Rather than generating this from a pump and boiler, the setup in Fig. P6.82 is used so the supply is extracted from the high-pressure turbine. Find the power the turbine now cogenerates in this process. Solution: . C.V. Turbine, steady state, 1 inlet and 2 exit flows, assume adiabatic, QCV = 0 Continuity Eq.6.9: . . . m1 = m2 + m3 Energy Eq.6.10: . . . . . QCV + m1h1 = m2h2 + m3h3 + WT ; Supply state 1: 20 kg/s at 10 MPa, 500°C 1 2 Process steam 2: 5 kg/s, 0.5 MPa, 155°C, Exit state 3: 20 kPa, x = 0.9 Table B.1.3: h1 = 3373.7, h2 = 2755.9 kJ/kg, Table B.1.2: h3 = 251.4 + 0.9 × 2358.3 WT 3 HP = 2373.9 kJ/kg . WT = 20 × 3373.7 − 5 × 2755.9 − 15 × 2373.9 = 18.084 MW LP Sonntag, Borgnakke and van Wylen 6.83 A compressor receives 0.1 kg/s R-134a at 150 kPa, -10oC and delivers it at 1000 kPa, 40oC. The power input is measured to be 3 kW. The compressor has heat transfer to air at 100 kPa coming in at 20oC and leaving at 25oC. How much is the mass flow rate of air? Solution: C.V. Compressor, steady state, single inlet and exit flow. For this device we also have an air flow outside the compressor housing no changes in kenetic or potential energy. 1 Air 4 2 cb -WC 3 Air . . m2 = m1 . . . . . Energy Eq. 6.12: m1h1 + Win + mairh3 = m2h2 + mairh4 Ideal gas for air and constant heat capacity: h4 - h3 ~ Cp air (T4 –T3) Continuity Eq.: . . . mair = [m1 (h1 –h2) + Win ] / Cp air (T4 –T3) = 0.1 ( 393.84 – 420.25) + 3 0.359 =5 1.004 (25-20) = 0.0715 kg/s Sonntag, Borgnakke and van Wylen Heat Exchangers 6.84 A condenser (heat exchanger) brings 1 kg/s water flow at 10 kPa from 300°C to saturated liquid at 10 kPa, as shown in Fig. P6.84. The cooling is done by lake water at 20°C that returns to the lake at 30°C. For an insulated condenser, find the flow rate of cooling water. Solution: C.V. Heat exchanger Energy Eq.6.10: 1 kg/s . . . . mcoolh20 + mH Oh300 = mcoolh30 + mH Ohf, 10 kPa 2 2 300°C sat. liq. 30°C 20°C . m cool Table B.1.1: h20 = 83.96 kJ/kg , Table B.1.3: h300, 10kPa = 3076.5 kJ/kg, B.1.2: h30 = 125.79 kJ/kg hf, 10 kPa = 191.83 kJ/kg h300 - hf, 10kPa . 3076.5 - 191.83 . mcool = mH2O h - h = 1 × 125.79 - 83.96 = 69 kg/s 30 20 Sonntag, Borgnakke and van Wylen 6.85 A cooler in an air conditioner brings 0.5 kg/s air at 35oC to 5oC, both at 101 kPa and it then mix the output with a flow of 0.25 kg/s air at 20oC, 101 kPa sending the combined flow into a duct. Find the total heat transfer in the cooler and the temperature in the duct flow. Solution: Q cool 1 3 2 Cooler section 4 Mixing section . C.V. Cooler section (no W) Energy Eq.6.12: . . . mh1 = mh2 + Qcool . . . Qcool = m(h1 - h2) = m Cp (T1 - T2) = 0.5 × 1.004 × (35-5) = 15.06 kW .. C.V. mixing section (no W, Q) Continuity Eq.: . . . m2 + m3 = m4 Energy Eq.6.10: . . . m2h2 + m3h3 = m4h4 . . . m4 = m2 + m3 = 0.5 + 0.25 = 0.75 kg/s . . . . . m4h4 = (m2 + m3)h4 = m2h2 + m3h3 . . m2 (h4 - h2) + m3 (h4 - h3) = Ø . . m2 Cp (T4 - T2) + m3 Cp (T4 - T3) = Ø . . . . T4 = (m2 / m4) T2 + (m3 / m4) T3 = 5(0.5/0.75) + 20(0.25/0.75) = 10°C Sonntag, Borgnakke and van Wylen 6.86 A heat exchanger, shown in Fig. P6.86, is used to cool an air flow from 800 K to 360 K, both states at 1 MPa. The coolant is a water flow at 15°C, 0.1 MPa. If the . . water leaves as saturated vapor, find the ratio of the flow rates mH2O/mair Solution: C.V. Heat exchanger, steady flow 1 inlet and 1 exit for air and water each. The two flows exchange energy with no heat transfer to/from the outside. Continuity Eqs.: 4 1 air Each line has a constant flow rate through it. . . . . Energy Eq.6.10: mairh1 + mH2Oh3 = mairh2 + mH2Oh4 Process: Each line has a constant pressure. Air states, Table A.7.1: h1 = 822.20 kJ/kg, h2 = 360.86 kJ/kg Water states, Table B.1.1: Table B.1.2: h3 = 62.98 kJ/kg (at 15°C), h4 = 2675.5 kJ/kg (at 100 kPa) h1 - h2 822.20 - 360.86 . . mH2O/mair = h - h = 2675.5 - 62.99 = 0.1766 4 3 2 3 water Sonntag, Borgnakke and van Wylen 6.87 A superheater brings 2.5 kg/s saturated water vapor at 2 MPa to 450oC. The energy is provided by hot air at 1200 K flowing outside the steam tube in the opposite direction as the water, which is a counter flowing heat exchanger. Find the smallest possible mass flow rate of the air so the air exit temperature is 20oC larger than the incoming water temperature (so it can heat it). Solution: C.V. Superheater. Steady state with no 4 . . external Q or any W the two flows exchanges energy inside the box. Neglect 1 air kinetic and potential energies at all states. 2 3 water . . . . Energy Eq.6.10: mH2O h3 + mair h1 = mH2O h4 + mair h2 Process: Constant pressure in each line. State 1: Table B.1.2 T3 = 212.42°C, h3 = 2799.51 kJ/kg State 2: Table B.1.3 h4 = 3357.48 kJ/kg State 3: Table A.7 h1 = 1277.81 kJ/kg T2 = T3 + 20 = 232.42°C = 505.57 K State 4: A.7 : 5.57 h2 = 503.36 + 20 (523.98 – 503.36) = 509.1 kJ/kg From the energy equation we get . . mair / mH2O = (h4 - h3)/(h1 - h2) = 2.5 (3357.48 – 2799.51) / (1277.81 – 509.1) = 1.815 kg/s Sonntag, Borgnakke and van Wylen 6.88 An automotive radiator has glycerine at 95oC enter and return at 55oC as shown in Fig. P6.88. Air flows in at 20oC and leaves at 25oC. If the radiator should transfer 25 kW what is the mass flow rate of the glycerine and what is the volume flow rate of air in at 100 kPa? Solution: If we take a control volume around the whole radiator then there is no external heat transfer - it is all between the glycerin and the air. So we take a control volume around each flow separately. Glycerine: Table A.4: Air Table A.5: . . . mhi + (-Q) = mhe . . . -Q -Q -25 mgly = h - h = C (T -T ) = 2.42(55 - 95) = 0.258 kg/s e i gly e i . . . mhi+ Q = mhe . . . Q Q 25 mair = h - h = C (T -T ) = 1.004(25 - 20) = 4.98 kg/s e i air e i RTi 0.287 × 293 vi = P = = 0.8409 m3/kg 100 i . . V = mvi ; . . Vair = mvi = 4.98 × 0.8409 = 4.19 m3/s Air intake filter Shaft power Exhaust flow cb o 95 C o Coolant flow 55 C Atm. air Sonntag, Borgnakke and van Wylen 6.89 A two fluid heat exchanger has 2 kg/s liquid ammonia at 20oC, 1003 kPa entering state 3 and exiting at state 4. It is heated by a flow of 1 kg/s nitrogen at 1500 K, state 1, leaving at 600 K, state 2 similar to Fig. P6.86. Find the total rate of heat transfer inside the heat exchanger. Sketch the temperature versus distance for the ammonia and find state 4 (T, v) of the ammonia. Solution: . . CV: Nitrogen flow line, steady rates of flow, Q out and W = 0 . . . . . Continiuty: m1 = m2 = 1 kg/s ; Energy Eq: m1h1 = m2h2 + Qout Tbl. A.8: h1 = 1680.7 kJ/kg; h2 = 627.24 kJ/kg . . Qout = m1(h1 - h2) = 1 (1680.7 – 627.24) = 1053.5 kW If Tbl A.5 is used: Cp = 1.042 kJ/kg K . . Qout = m1 Cp (T1 - T2) = 1×1.042 (1500 - 600) = 937.8 kW . CV The whole heat exchanger: No external Q, constant pressure in each line. . . . . . . => h4 = h3 + m1(h1 - h2)/m3 m1h1 + m3h3 = m1h2 + m3h4 h4 = 274.3 + 1053.5 /2 = 801 kJ/kg < hg => 2-phase x4 = (h4 - hf)/ hfg = (801 - 298.25) / 1165.2 = 0.43147 v4 = vf + x4 vfg = 0.001658 + 0.43147×0.12647 = 0.05623 m3/kg T4 = T3a = 25oC This is the boiling temperature for 1003 kPa. T 298 293 3a 4 4 1 N2 3 x 2 3 NH 3 Sonntag, Borgnakke and van Wylen 6.90 A copper wire has been heat treated to 1000 K and is now pulled into a cooling chamber that has 1.5 kg/s air coming in at 20oC; the air leaves the other end at 60oC. If the wire moves 0.25 kg/s copper, how hot is the copper as it comes out? Solution: C.V. Total chamber, no external heat transfer . . . . Energy eq.: mcu h icu + mair hi air = mcu he cu + mair he air . . mcu ( he – hi )cu = mair( hi – he )air . . mcu Ccu ( Te – Ti )cu = mair Cp air( Te – Ti )air Heat capacities from A.3 for copper and A.5 for air . mairCp air 1.5 ×1.004 ( Te – Ti )cu = ( Te – Ti )air = (20 - 60) = - 573.7 K . 0.25 × 0.42 mcuCcu Te = Ti – 573.7 = 1000 - 573.7 = 426.3 K Air Air Cu Sonntag, Borgnakke and van Wylen Mixing processes 6.91 An open feedwater heater in a powerplant heats 4 kg/s water at 45oC, 100 kPa by mixing it with steam from the turbine at 100 kPa, 250oC. Assume the exit flow is saturated liquid at the given pressure and find the mass flow rate from the turbine. Solution: C.V. Feedwater heater. 1 . . No external Q or W MIXING 2 3 CHAMBER cb Continuity Eq.6.9: Energy Eq.6.10: State 1: State 2: State 3: . . . m1 + m2 = m3 . . . . . m1h1 + m2h2 = m3h3 = (m1+ m2)h3 Table B.1.1 Table B.1.3 Table B.1.2 h = hf = 188.42 kJ/kg at 45oC h2 = 2974.33 kJ/kg h3 = hf = 417.44 kJ/kg at 100 kPa 188.42 – 417.44 . . h1 - h3 m2 = m1× h - h = 4 × 417.44 – 2974.33 = 0.358 kg/s 3 2 P T 3 1 2 100 kPa 2 13 v v Sonntag, Borgnakke and van Wylen 6.92 A desuperheater mixes superheated water vapor with liquid water in a ratio that produces saturated water vapor as output without any external heat transfer. A flow of 0.5 kg/s superheated vapor at 5 MPa, 400°C and a flow of liquid water at 5 MPa, 40°C enter a desuperheater. If saturated water vapor at 4.5 MPa is produced, determine the flow rate of the liquid water. Solution: LIQ 2 3 Sat. vapor VAP 1 Continuity Eq.: Energy Eq.6.10: Table B.1 . QCV = 0 . . . m1 + m2 = m3 . . . m1h1 + m2h2 = m3h3 . . 0.5 × 3195.7 + m2 × 171.97 = (0.5 + m2) 2797.9 . => m2 = 0.0757 kg/s Sonntag, Borgnakke and van Wylen 6.93 Two air flows are combined to a single flow. Flow one is 1 m3/s at 20oC and the other is 2 m3/s at 200oC both at 100 kPa. They mix without any heat transfer to produce an exit flow at 100 kPa. Neglect kinetic energies and find the exit temperature and volume flow rate. Solution: . . . 2 Cont. mi = me = m 1 . . . 3 Energy m1h1 + m2h2 = m3h3 . . = (m1 + m2)h3 Mixing section . . m1 (h3 -h1) + m2 (h3 -h2) = 0 . . m1Cp ( T3-T1) + m2Cp( T3-T2) = 0 .. .. T3 = (mi/m3)/T1 + (m2/m3)T2 We need to find the mass flow rates v1 = RT1/P1 = (0.287 × 293)/100 = 0.8409 m3/kg v2 = RT2/P2 = (0.287 × 473)/100 = 1.3575 m3/kg . . V1 V2 . . 1 kg 2 kg m1 = v = 0.8409 = 1.1892 s , m2 = v = 1.3575 = 1.4733 s 1 2 . . . m3 = m1+ m2 = 2.6625 kg/s 1.1892 1.4733 T3 = 2.6625 × 20 + 2.6625 × 200 = 119.6o C RT3 0.287 (119.6 + 273) = 1.1268 m3/kg v3 = P = 100 3 . . V3 = m3 v3 = 2.6625 × 1.1268 = 3.0 m3/s Sonntag, Borgnakke and van Wylen 6.94 A mixing chamber with heat transfer receives 2 kg/s of R-22 at 1 MPa, 40°C in one line and 1 kg/s of R-22 at 30°C, quality 50% in a line with a valve. The outgoing flow is at 1 MPa, 60°C. Find the rate of heat transfer to the mixing chamber. Solution: C.V. Mixing chamber. Steady with 2 flows in and 1 out, heat transfer in. 1 P 3 Mixer Heater 2 13 . Q 2 Continuity Eq.6.9: Energy Eq.6.10: v . . . m1 + m2 = m3 ; => . m3 = 2 + 1 = 3 kg/s . . . . m1h1 + m2h2 + Q = m3h3 Properties: Table B.4.2: Table B.4.1: h1 = 271.04 kJ/kg, h3 = 286.97 kJ/kg h2 = 81.25 + 0.5 × 177.87 = 170.18 kJ/kg Energy equation then gives the heat transfer as . Q = 3 × 286.973 – 2 × 271.04 – 1 × 170.18 = 148.66 kW Sonntag, Borgnakke and van Wylen 6.95 Two flows are mixed to form a single flow. Flow at state 1 is 1.5 kg/s water at 400 kPa, 200oC and flow at state 2 is 500 kPa, 100oC. Which mass flow rate at state 2 will produce an exit T3 = 150oC if the exit pressure is kept at 300 kPa? Solution: C.V. Mixing chamber and valves. Steady state no heat transfer or work terms. . . . Continuity Eq.6.9: m1 + m2 = m3 . . . . . Energy Eq.6.10: m1h1 + m2h2 = m3h3 = (m1+ m2)h3 P 1 2 MIXING 2 3 1 3 CHAMBER Properties Table B.1.3: Table B.1.4: h1 = 2860.51 kJ/kg; v h3 = 2760.95 kJ/kg h2 = 419.32 kJ/kg 2860.51 – 2760.95 . . h1 - h3 m2 = m1× h - h = 1.5 × 2760.95 – 419.32 = 0.0638 kg/s 3 2 Sonntag, Borgnakke and van Wylen 6.96 An insulated mixing chamber receives 2 kg/s R-134a at 1 MPa, 100°C in a line with low velocity. Another line with R-134a as saturated liquid 60°C flows through a valve to the mixing chamber at 1 MPa after the valve. The exit flow is saturated vapor at 1 MPa flowing at 20 m/s. Find the flow rate for the second line. Solution: C.V. Mixing chamber. Steady state, 2 inlets and 1 exit flow. Insulated q = 0, No shaft or boundary motion w = 0. Continuity Eq.6.9: Energy Eq.6.10: . . . m1 + m2 = m3 ; . . . 12 m1h1 + m2h2 = m3( h3 + 2 V3 ) . . 12 12 m2 (h2 – h3 – 2 V3 ) = m1 ( h3 + 2 V3 – h1 ) 1: Table B.5.2: 1 MPa, 100°C, h1 = 483.36 kJ/kg 2: Table B.5.1: x = ∅, 60°C, 3: Table B.5.1: x = 1, 1 MPa, 20 m/s, h2 = 287.79 kJ/kg h3 = 419.54 kJ/kg . Now solve the energy equation for m2 2 . 1 12 1 20 m2 = 2 × [419.54 + 2 20 × 1000 – 483.36] / [287.79 – 419.54 – 2 1000] = 2 × ( -63.82 + 0.2) / ( -131.75 - 0.2) = 0.964 kg/s Notice how kinetic energy was insignificant. P 1 MIXING 2 cb CHAMBER 3 2 31 v Sonntag, Borgnakke and van Wylen 6.97 To keep a jet engine cool some intake air bypasses the combustion chamber. Assume 2 kg/s hot air at 2000 K, 500 kPa is mixed with 1.5 kg/s air 500 K, 500 kPa without any external heat transfer. Find the exit temperature by using constant heat capacity from Table A.5. Solution: C.V. Mixing Section Continuity Eq.6.9: . . . m1 + m2 = m3 => Energy Eq.6.10: . . . m1h1 + m2h2 = m3h3 . m3 = 2 + 1.5 = 3.5 kg/s . . . h3 = (m1h1 + m2h2) / m3 ; For a constant specific heat divide the equation for h3 with Cp to get . . m1 m2 2 1.5 T3 = T1 + T2 = 3.5 2000 + 3.5 500 = 1357 K . . m3 m3 2 1 3 Mixing section Sonntag, Borgnakke and van Wylen 6.98 To keep a jet engine cool some intake air bypasses the combustion chamber. Assume 2 kg/s hot air at 2000 K, 500 kPa is mixed with 1.5 kg/s air 500 K, 500 kPa without any external heat transfer. Find the exit temperature by using values from Table A.7. Solution: C.V. Mixing Section Continuity Eq.6.9: . . . m1 + m2 = m3 => Energy Eq.6.10: . . . m1h1 + m2h2 = m3h3 . m3 = 2 + 1.5 = 3.5 kg/s . . . h3 = (m1h1 + m2h2) / m3 ; Using A.7 we look up the h at states 1 and 2 to calculate h3 . . m1 m2 2 1.5 h3 = h1 + h2 = 3.5 2251.58 + 3.5 503.36 = 1502 kJ/kg . . m3 m3 Now we can backinterpolate to find at what temperature do we have that h 1502 – 1455.43 T3 = 1350 + 50 1515.27 – 1455.43 = 1389 K This procedure is the most accurate. 2 1 3 Mixing section Sonntag, Borgnakke and van Wylen Multiple Devices, Cycle Processes 6.99 The following data are for a simple steam power plant as shown in Fig. P6.99. State 1 2 3 4 5 6 7 P MPa 6.2 6.1 5.9 5.7 5.5 0.01 0.009 45 175 500 490 40 T °C h kJ/kg - 194 744 3426 3404 - 168 State 6 has x6 = 0.92, and velocity of 200 m/s. The rate of steam flow is 25 kg/s, with 300 kW power input to the pump. Piping diameters are 200 mm from steam generator to the turbine and 75 mm from the condenser to the steam generator. Determine the velocity at state 5 and the power output of the turbine. Solution: Turbine A5 = (π/4)(0.2)2 = 0.031 42 m2 . V5 = mv5/A5 = 25 × 0.061 63 / 0.031 42 = 49 m/s h6 = 191.83 + 0.92 × 2392.8 = 2393.2 kJ/kg 1 2 2 wT = h5 - h6 + 2 ( V5 - V6 ) = 3404 - 2393.2 + (492 - 2002 )/(2 × 1000) = 992 kJ/kg . . WT = mwT = 25 × 992 = 24 800 kW Remark: Notice the kinetic energy change is small relative to enthalpy change. Sonntag, Borgnakke and van Wylen 6.100 For the same steam power plant as shown in Fig. P6.99 and Problem 6.99, assume the cooling water comes from a lake at 15°C and is returned at 25°C. Determine the rate of heat transfer in the condenser and the mass flow rate of cooling water from the lake. Solution: Condenser A7 = (π/4)(0.075)2 = 0.004 418 m2, v7 = 0.001 008 m3/kg . V7 = mv7/A7 = 25 × 0.001 008 / 0.004 418 = 5.7 m/s h6 = 191.83 + 0.92 × 2392.8 = 2393.2 kJ/kg 1 2 2 qCOND = h7 - h6 + 2 ( V7 - V6 ) = 168 − 2393.2 + (5.72 − 2002 )/(2×1000) = −2245.2 kJ/kg . QCOND = 25 × (−2245.2) = −56 130 kW This rate of heat transfer is carried away by the cooling water so . . −QCOND = mH2O(hout − hin)H2O = 56 130 kW => . 56 130 mH2O = 104.9 - 63.0 = 1339.6 kg/s Sonntag, Borgnakke and van Wylen 6.101 For the same steam power plant as shown in Fig. P6.99 and Problem 6.99, determine the rate of heat transfer in the economizer, which is a low temperature heat exchanger. Find also the rate of heat transfer needed in the steam generator. Solution: Economizer 2 A7 = πD7/4 = 0.004 418 m2, v7 = 0.001 008 m3/kg . V2 = V7 = mv7/A7 = 25 × 0.001 008/0.004 418 = 5.7 m/s, V3 = (v3/v2)V2 = (0.001 118 / 0.001 008) 5.7 = 6.3 m/s ≈ V2 so kinetic energy change unimportant qECON = h3 - h2 = 744 - 194 = 550.0 kJ/kg . . QECON = mqECON = 25 (550.0) = 13 750 kW Generator 2 A4 = πD4/4 = 0.031 42 m2, v4 = 0.060 23 m3/kg . V4 = mv4/A4 = 25 × 0.060 23/0.031 42 = 47.9 m/s qGEN = 3426 - 744 + (47.92 - 6.32)/(2×1000) = 2683 kJ/kg . . QGEN = mqGEN = 25 × (2683) = 67 075 kW Sonntag, Borgnakke and van Wylen 6.102 A somewhat simplified flow diagram for a nuclear power plant shown in Fig. 1.4 is given in Fig. P6.102. Mass flow rates and the various states in the cycle are shown in the accompanying table. The cycle includes a number of heaters in which heat is transferred from steam, taken out of the turbine at some intermediate pressure, to liquid water pumped from the condenser on its way to the steam drum. The heat exchanger in the reactor supplies 157 MW, and it may be assumed that there is no heat transfer in the turbines. a. Assume the moisture separator has no heat transfer between the two turbinesections, determine the enthalpy and quality (h4, x4). b. Determine the power output of the low-pressure turbine. c. Determine the power output of the high-pressure turbine. d. Find the ratio of the total power output of the two turbines to the total power delivered by the reactor. 3 Solution: 2 5 moisture 4 separator WHP 17 12 9 WLP 8 a) Moisture Separator, steady state, no heat transfer, no work . . . . . . Energy: m3h3 = m4h4 + m9h9 ; Mass: m3 = m4 + m9, 62.874 × 2517 = 58.212 × h4 + 4.662 × 558 ⇒ h4 = 2673.9 kJ/kg h4 = 2673.9 = 566.18 + x4 × 2160.6 => x4 = 0.9755 b) Low Pressure Turbine, steady state no heat transfer . . . . Energy Eq.: m4h4 = m5h5 + m8h8+ WCV(LP) . . . . WCV(LP) = m4h4 - m5h5 - m8h8 = 58.212 × 2673.9 - 55.44 × 2279 - 2.772 × 2459 = 22 489 kW = 22.489 MW c) High Pressure Turbine, steady state no heat transfer . . . . . Energy Eq.: m2h2 = m3h3 + m12h12 + m17h17 + WCV(HP) . . . . . WCV(HP) = m2h2 - m3h3 - m12h12 - m17h17 = 75.6 × 2765 - 62.874 × 2517 - 8.064 × 2517 - 4.662 × 2593 d) = 18 394 kW = 18.394 MW . . . η = (WHP + WLP)/QREACT = 40.883/157 = 0.26 Sonntag, Borgnakke and van Wylen 6.103 Consider the powerplant as described in the previous problem. a.Determine the quality of the steam leaving the reactor. b.What is the power to the pump that feeds water to the reactor? Solution: . . . m20 = m21; QCV = 157 MW . . . Energy Eq.6.12: QCV + m20h20 = m21h21 21 a) Reactor: Cont.: Q 19 157 000 + 1386 × 1221 = 1386 × h21 20 h21 = 1334.3 = 1282.4 + x21 × 1458.3 => x21 = 0.0349 b) C.V. Reactor feedwater pump . . Cont. m19 = m20 Table B.1: Energy Eq.6.12: . . . m19h19 = m19h20 + WCv,P h19 = h(277°C, 7240 kPa) = 1220 kJ/kg, h20 = 1221 kJ/kg . . WCv,P = m19(h19 - h20) = 1386(1220 - 1221) = -1386 kW Sonntag, Borgnakke and van Wylen 6.104 A gas turbine setup to produce power during peak demand is shown in Fig. P6.104. The turbine provides power to the air compressor and the electric generator. If the electric generator should provide 5 MW what is the needed air flow at state 1 and the combustion heat transfer between state 2 and 3? Solution: 1: 90 kPa, 290 K ; 2: 900 kPa, 560 K ; 3: 900 kPa, 1400 K 4: 100 kPa, 850 K ; wc in = h2 – h1 = 565.47 – 290.43 = 275.04 kJ/kg wTout = h3 - h4 = 1515.27 – 877.4 = 637.87 kJ/kg q H = h3 – h2 = 1515.27 – 565.47 = 949.8 kJ/kg . . . Wel = mwT – mwc . . 5000 m = Wel / ( wT - wc ) = 637.87 - 275.04 = 13.78 kg/s . . QH = mqH = 13.78 × 949.8 = 13 088 kW = 13.1 MW Sonntag, Borgnakke and van Wylen 6.105 A proposal is made to use a geothermal supply of hot water to operate a steam turbine, as shown in Fig. P6.105. The high-pressure water at 1.5 MPa, 180°C, is throttled into a flash evaporator chamber, which forms liquid and vapor at a lower pressure of 400 kPa. The liquid is discarded while the saturated vapor feeds the turbine and exits at 10 kPa, 90% quality. If the turbine should produce 1 MW, find the required mass flow rate of hot geothermal water in kilograms per hour. Solution: Separation of phases in flash-evaporator constant h in the valve flow so Table B.1.3: h1 = 763.5 kJ/kg 1 FLASH EVAP. h1 = 763.5 = 604.74 + x × 2133.8 4 .. ⇒ x = 0.07439 = m2/m1 Table B.1.2: H2O 2 Sat. vap. Sat. liq. out h2 = 2738.6 kJ/kg; . W Turb 3 h3 = 191.83 + 0.9 × 2392.8 = 2345.4 kJ/kg Energy Eq.6.12 for the turbine . . W = m2(h2 - h3) => 1000 . m2 = 2738.6 - 2345.4 = 2.543 kg/s . . ⇒ m1 = m2/x = 34.19 kg/s = 123 075 kg/h Sonntag, Borgnakke and van Wylen 6.106 A R-12 heat pump cycle shown in Fig. P6.71 has a R-12 flow rate of 0.05 kg/s with 4 kW into the compressor. The following data are given State 1 2 3 4 5 6 P kPa 1250 1230 1200 320 300 290 120 110 45 0 5 T °C h kJ/kg 260 253 79.7 188 191 Calculate the heat transfer from the compressor, the heat transfer from the R-12 in the condenser and the heat transfer to the R-12 in the evaporator. Solution: CV: Compressor . . . QCOMP = m(h1 - he) + WCOMP = 0.05 (260 - 191) - 4.0 = -0.55 kW CV: Condenser . . QCOND = m (h3-h2) = 0.05 (79.7 - 253) = -8.665 kW CV: Evaporator h4 = h3 = 79.7 kJ/kg (from valve) . . QEVAP = m (h5- h4) = 0.05 (188 - 79.7) = 5.42 kW Sonntag, Borgnakke and van Wylen 6.107 A modern jet engine has a temperature after combustion of about 1500 K at 3200 kPa as it enters the turbine setion, see state 3 Fig. P.6.107. The compressor inlet is 80 kPa, 260 K state 1 and outlet state 2 is 3300 kPa, 780 K; the turbine outlet state 4 into the nozzle is 400 kPa, 900 K and nozzle exit state 5 at 80 kPa, 640 K. Neglect any heat transfer and neglect kinetic energy except out of the nozzle. Find the compressor and turbine specific work terms and the nozzle exit velocity. Solution: The compressor, turbine and nozzle are all steady state single flow devices and they are adiabatic. We will use air properties from table A.7.1: h1 = 260.32, h2 = 800.28, h3 = 1635.80, h4 = 933.15, h5 = 649.53 kJ/kg Energy equation for the compressor gives wc in = h2 – h1 = 800.28 – 260.32 = 539.36 kJ/kg Energy equation for the turbine gives wT = h3 – h4 = 1635.80 – 933.15 = 702.65 kJ/kg Energy equation for the nozzle gives 2 h4 = h5 + ½ V5 2 ½ V5 = h4 - h5 = 933.15 – 649.53 = 283.62 kJ/kg V5 = [2( h4 – h5) ] 1/2 = ( 2× 283.62 ×1000 ) 1/2 = 753 m/s Sonntag, Borgnakke and van Wylen Transient processes 6.108 A 1-m3, 40-kg rigid steel tank contains air at 500 kPa, and both tank and air are at 20°C. The tank is connected to a line flowing air at 2 MPa, 20°C. The valve is opened, allowing air to flow into the tank until the pressure reaches 1.5 MPa and is then closed. Assume the air and tank are always at the same temperature and the final temperature is 35°C. Find the final air mass and the heat transfer. Solution: Control volume: Air and the steel tank. Continuity Eq.6.15: Energy Eq.6.16: m2 - m1 = mi (m2u2 - m1u1)AIR + mST(u2 - u1)ST = mihi + 1Q2 P1V 500 × 1 m1 AIR = RT = 0.287 × 293.2 = 5.94 kg 1 P2V 1500 × 1 m2 AIR = RT = 0.287 × 308.2 = 16.96 kg 2 mi = (m2 - m1)AIR = 16.96 - 5.94 = 11.02 kg The energy equation now gives 1Q2 = (m2u2 - m1u1)AIR + mST(u2 - u1)ST - mihi = m1(u2 - u1) + mi(u2 - ui - RTi) + mSTCST(T2 – T1) ≅ m1Cv(T2 – T1) + mi[Cv(T2 – Ti) - RTi] + mSTCST(T2 – T1) = 5.94 × 0.717(35 – 20) + 11.02[0.717(35 – 20) – 0.287× 293.2] + 40 × 0.46(35 – 20) = 63.885 – 808.795 + 276 = – 468.9 kJ Sonntag, Borgnakke and van Wylen 6.109 An evacuated 150-L tank is connected to a line flowing air at room temperature, 25°C, and 8 MPa pressure. The valve is opened allowing air to flow into the tank until the pressure inside is 6 MPa. At this point the valve is closed. This filling process occurs rapidly and is essentially adiabatic. The tank is then placed in storage where it eventually returns to room temperature. What is the final pressure? Solution: C.V. Tank: Continuity Eq.6.15: mi = m2 Energy Eq.6.16: mihi = m2u2 => u 2 = hi Use constant specific heat CPo from table A.5 then energy equation: T2 = (CP/CV) Ti = kTi= 1.4 × 298.2 = 417.5 K Process: constant volume cooling to T3: P3 = P2 × T3/T2 = 6.0 × 298.15/ 417.5 = 4.29 MPa line Sonntag, Borgnakke and van Wylen 6.110 An initially empty bottle is filled with water from a line at 0.8 MPa, 350oC. Assume no heat transfer and that the bottle is closed when the pressure reaches the line pressure. If the final mass is 0.75 kg find the final temperature and the volume of the bottle. Solution; C.V. Bottle, transient process with no heat transfer or work. Continuity Eq.6.15: m2 - m1 = min ; Energy Eq.6.16: m2u2 – m1u1 = - min hin State 1: m1 = 0 => m2 = min and Line state: Table B.1.3: hin = 3161.68 kJ/kg State 2: P2 = Pline = 800 kPa, u2 = 3161.68 kJ/kg T2 = 520oC and v2 = 0.4554 m3/kg V2 = m2v2 = 0.75 × 0.4554 = 0.342 m3 line u2 = hin from Table B.1.3 Sonntag, Borgnakke and van Wylen 6.111 A 25-L tank, shown in Fig. P6.111, that is initially evacuated is connected by a valve to an air supply line flowing air at 20°C, 800 kPa. The valve is opened, and air flows into the tank until the pressure reaches 600 kPa.Determine the final temperature and mass inside the tank, assuming the process is adiabatic. Develop an expression for the relation between the line temperature and the final temperature using constant specific heats. Solution: C.V. Tank: Continuity Eq.6.15: m2 = mi Energy Eq.6.16: Table A.7: m2u2 = mihi TANK u2 = hi = 293.64 kJ/kg ⇒ T2 = 410.0 K P2V 600 × 0.025 m2 = RT = 0.287 × 410 = 0.1275 kg 2 Assuming constant specific heat, hi = ui + RTi = u2 , RTi = u2 - ui = Cvo(T2 - Ti) CPo CvoT2 = ( Cvo + R )Ti = CPoTi , T2 = C Ti = kTi vo For Ti = 293.2K & constant CPo, T2 = 1.40×293.2 = 410.5 K Sonntag, Borgnakke and van Wylen 6.112 Helium in a steel tank is at 250 kPa, 300 K with a volume of 0.1 m3. It is used to fill a balloon. When the tank pressure drops to 150 kPa the flow of helium stops by itself. If all the helium still is at 300 K how big a balloon did I get? Assume the pressure in the balloon varies linearly with volume from 100 kPa (V = 0) to the final 150 kPa. How much heat transfer did take place? Solution: Take a C.V. of all the helium. This is a control mass, the tank mass changes density and pressure. Energy Eq.: U2 – U1 = 1Q2 - 1W2 Process Eq.: P = 100 + CV State 1: P1, T1, V1 State 2: P2, T2, V2 = ? cb c i r c u s t h e r m o Ideal gas: P2 V2 = mRT2 = mRT1 = P1V1 V2 = V1(P1/P2) = 0.1× (250/150) = 0.16667 m3 Vbal = V2 – V1 = 0.16667 – 0.1 = 0.06667 m3 1W2 = ∫ P dV = AREA = ½ ( P1 + P2 )( V2 –V1 ) = ½( 250 + 150) × 0.06667 = 13.334 kJ U2 – U1 = 1Q2 - 1W2 = m (u2 – u1) = mCv ( T2 –T1 ) = 0 so 1Q2 = 1W2 = 13.334 kJ Remark: The process is transient, but you only see the flow mass if you select the tank or the balloon as a control volume. That analysis leads to more terms that must be elliminated between the tank control volume and the balloon control volume. Sonntag, Borgnakke and van Wylen 6.113 A rigid 100-L tank contains air at 1 MPa, 200°C. A valve on the tank is now opened and air flows out until the pressure drops to 100 kPa. During this process, heat is transferred from a heat source at 200°C, such that when the valve is closed, the temperature inside the tank is 50°C. What is the heat transfer? Solution: 1 : 1 MPa, 200°C, m1 = P1V1/RT1 = 1000 × 0.1/(0.287 × 473.1) = 0.736 kg 2 : 100 kPa, 50°C, m2 = P2V2/RT2 = 100 × 0.1/(0.287 × 323.1) = 0.1078 kg Continuity Eq.6.15: mex = m1 – m2 = 0.628 kg, Energy Eq.6.16: m2u2 – m1u1 = - mex hex + 1Q2 Table A.7: u1 = 340.0 kJ/kg, u2 = 231.0 kJ/kg, he ave = (h1 + h2)/2 = (475.8 + 323.75)/2 = 399.8 kJ/kg 1Q2 = 0.1078 × 231.0 – 0.736 × 340.0 + 0.628 × 399.8 = +25.7 kJ Sonntag, Borgnakke and van Wylen 6.114 A 1-m3 tank contains ammonia at 150 kPa, 25°C. The tank is attached to a line flowing ammonia at 1200 kPa, 60°C. The valve is opened, and mass flows in until the tank is half full of liquid, by volume at 25°C. Calculate the heat transferred from the tank during this process. Solution: C.V. Tank. Transient process as flow comes in. State 1 Table B.2.2 interpolate between 20 °C and 30°C: v1 = 0.9552 m3/kg; u1 = 1380.6 kJ/kg m1 = V/v1 = 1/0.9552 = 1.047 kg State 2: 0.5 m3 liquid and 0.5 m3 vapor from Table B.2.1 at 25°C vf = 0.001658 m3/kg; vg = 0.12813 m3/kg mLIQ2 = 0.5/0.001658 = 301.57 kg, mVAP2 = 0.5/0.12813 = 3.902 kg m2 = 305.47 kg, x2 = mVAP2/m2 = 0.01277, From continuity equation mi = m2 - m1 = 304.42 kg Table B.2.1: u2 = 296.6 + 0.01277 × 1038.4 = 309.9 kJ/kg State inlet: Table B.2.2 hi = 1553.3 kJ/kg Energy Eq.6.16: QCV + mihi = m2u2 - m1u1 QCV = 305.47 × 309.9 - 1.047 × 1380.6 - 304.42 × 1553.3 = -379 636 kJ line Q Sonntag, Borgnakke and van Wylen 6.115 An empty cannister of volume 1 L is filled with R-134a from a line flowing saturated liquid R-134a at 0oC. The filling is done quickly so it is adiabatic. How much mass of R-134a is there after filling? The cannister is placed on a storage shelf where it slowly heats up to room temperature 20oC. What is the final pressure? C.V. cannister, no work and no heat transfer. Continuity Eq.6.15: m2 = mi Energy Eq.6.16: Table B.5.1: m2u2 – 0 = mihi = mihline hline = 200.0 kJ/kg, Pline = 294 kPa From the energy equation we get u2 = hline = 200 kJ/kg > uf = 199.77 kJ/kg State 2 is two-phase P2 = Pline = 294 kPa and T2 = 0°C u2 - uf 200 – 199.77 x2 = u = = 0.00129 178.24 fg v2 = 0.000773 + x2 0.06842 = 0.000861 m3/kg m2 = V/v2 = 0.01/0.000861 = 11.61 kg At 20°C: vf = 0.000817 m3/kg < v2 so still two-phase P = Psat = 572.8 kPa Sonntag, Borgnakke and van Wylen 6.116 A piston cylinder contains 1 kg water at 20oC with a constant load on the piston such that the pressure is 250 kPa. A nozzle in a line to the cylinder is opened to enable a flow to the outside atmosphere at 100 kPa. The process continues to half the mass has flowed out and there is no heat transfer. Assume constant water temperature and find the exit velocity and total work done in the process. Solution: C.V. The cylinder and the nozzle. Continuity Eq.6.15: m2 - m1 = − me Energy Eq.6.16: m2u2 - m1u1 = - me(he + 2Ve ) - 1W2 Process: P=C 12 => 1W2 = ⌠P dV = P(V2 - V1) ⌡ State 1: Table B.1.1, 20oC => v1 = 0.001002, u1 = 83.94 kJ/kg State 2: Table B.1.1, 20oC => v2 = v1, u2 = u1; m2 = m1/2 = 0.5 kg => V2 = V1/2 1W2 = P(V2 - V1) = 250 (0.5 - 1) 0.001002 = -0.125 kJ Exit state: Table B.1.1, 20oC => he = 83.94 kJ/kg Solve for the kinetic energy in the energy equation 12 2Ve = [m1u1 - m2u2 - mehe - 1W2]/me = [1 × 83.94 - 0.5 × 83.94 - 0.5 × 83.94 + 0.125] / 0.5 = 0.125/0.5 = 0.25 kJ/kg V= 2 × 0.25 × 1000 = 22.36 m/s All the work ended up as kinetic energy in the exit flow. F cb AIR Pcyl e Sonntag, Borgnakke and van Wylen 6.117 A 200 liter tank initially contains water at 100 kPa and a quality of 1%. Heat is transferred to the water thereby raising its pressure and temperature. At a pressure of 2 MPa a safety valve opens and saturated vapor at 2 MPa flows out. The process continues, maintaining 2 MPa inside until the quality in the tank is 90%, then stops. Determine the total mass of water that flowed out and the total heat transfer. Solution: C.V. Tank, no work but heat transfer in and flow out. Denoting State 1: initial state, State 2: valve opens, State 3: final state. Continuity Eq.: Energy Eq.: m3 − m1 = − me m3u3 − m1u1 = − mehe + 1Q3 State 1 Table B.1.2: sat vap e . Qcv v1 = vf + x1vfg = 0.001043 + 0.01×1.69296 = 0.01797 m3/kg u1 = uf + x1ufg = 417.33 + 0.01×2088.72 = 438.22 kJ/kg m1 = V/v1 = 0.2 m3/(0.01797 m3/kg) = 11.13 kg State 3 (2MPa): v3 = vf + x3vfg = 0.001177 + 0.9×0.09845 = 0.8978 m3/kg u3 = uf + x3ufg = 906.42 + 0.9×1693.84 = 2430.88 kJ/kg Exit state (2MPa): m3 = V/v3 = 0.2 m3/(0.08978 m3/kg) = 2.23 kg he = hg = 2799.51 kJ/kg Hence: me = m1 − m3 = 11.13 kg − 2.23 kg = 8.90 kg Applying the 1st law between state 1 and state 3 1Q3 = m3u3 − m1u1 + mehe = 2.23 × 2430.88 − 11.13 × 438.22 + 8.90 × 2799.51 = 25 459 kJ = 25.46 MJ Sonntag, Borgnakke and van Wylen 6.118 A 100-L rigid tank contains carbon dioxide gas at 1 MPa, 300 K. A valve is cracked open, and carbon dioxide escapes slowly until the tank pressure has dropped to 500 kPa. At this point the valve is closed. The gas remaining inside the tank may be assumed to have undergone a polytropic expansion, with polytropic exponent n = 1.15. Find the final mass inside and the heat transferred to the tank during the process. Solution: Ideal gas law and value from table A.5 P1V 1000 × 0.1 m1 = RT = 0.18892 × 300 = 1.764 kg 1 Polytropic process and ideal gas law gives P2(n-1)/n 500 (0.15/1.15) T2 = T1 P = 300 1000 = 274 K 1 P2V 500 × 0.1 m2 = RT = 0.18892 × 274 = 0.966 kg 2 Energy Eq.6.16: QCV = m2u2 - m1u1 + mehe avg = m2CvoT2 - m1CvoT1 + (m1 - m2)CPo(T1 + T2)/2 = 0.966 × 0.6529 × 274 - 1.764 × 0.6529 × 300 + (1.764 - 0.966) × 0.8418 ×(300 + 274)/2 = +20.1 kJ cb Sonntag, Borgnakke and van Wylen 6.119 A nitrogen line, 300 K and 0.5 MPa, shown in Fig. P6.119, is connected to a turbine that exhausts to a closed initially empty tank of 50 m3. The turbine operates to a tank pressure of 0.5 MPa, at which point the temperature is 250 K. Assuming the entire process is adiabatic, determine the turbine work. Solution: C.V. turbine & tank ⇒ Transient process Conservation of mass Eq.6.15: mi = m2 ⇒ m Energy Eq.6.16: Table B.6.2: mihi = m2u2 + WCV ; WCV = m(hi - u2) Pi = 0.5 MPa, Ti = 300 K, Nitrogen; hi = 310.28 kJ/kg 2: P2 = 0.5 MPa, T2 = 250 K, u2 = 183.89 kJ/kg, v2 = 0.154 m3/kg m2 = V/v2 = 50/0.154 = 324.7 kg WCV = 324.7 (310.28 - 183.89) = 41 039 kJ = 41.04 MJ . 1 W Turb 2 TANK We could with good accuracy have solved using ideal gas and Table A.5 Sonntag, Borgnakke and van Wylen 6.120 A 2 m tall cylinder has a small hole in the bottom. It is filled with liquid water 1 m high, on top of which is 1 m high air column at atmospheric pressure of 100 kPa. As the liquid water near the hole has a higher P than 100 kPa it runs out. Assume a slow process with constant T. Will the flow ever stop? When? New fig. Solution: Pbot = Pair + ρgLliq For the air PV = mRT Pair = mRT/Vair ; Vair = A Lair = A ( H-Lliq ) maRaTa Pa1Va1 Pa1La1 Pbot = A(H-L ) + ρfg Lf = A(H-L ) + ρliq gLf = H-L + ρliq gLf ≥ liq liq f Po Solve for Lliq ; ρliq= 1/(vf) = 1/0.0021002 = 998 kg/m3 Pa1 La1 + ρg Lf ( H – Lf ) ≥ P ( H – Lf ) (ρgH + Po ) Lf – ρgL2f = Po H + Pa1 La1 ≥ 0 Put in numbers and solve quadratic eq. L2f – ( H +(Po/ρg) ) Lf + PoH-Pa1La1 =∅ ρg L2f – 12.217 Lf + 10.217 = 0 100 kPa m3 s3 (Po/ρg) = = 10.217 m 998 ×9.807 kg m PoH+Pa1La1 100 (2-1) = = 10.217 m ρg 998×9.807 Lf = 1m Air 1m Water 12.217 12.2173 12.217 1/2 -4] = 6.1085 × 5.2055 2 ×[ 4 => 11.314 or 0.903 m Verify La1 1 Pa2 = Pa1. H-L = 100 2 - 0.903 = 91.158 kPa f ρgLf = 998 × 9.807 × 0.903 = 8838 Pa = 8.838 kPa Pbot = Pa2 + ρgLf = 91.158 + 8.838 = 99.996 kPa OK Sonntag, Borgnakke and van Wylen 6.121 A 2-m3 insulated vessel, shown in Fig. P6.121, contains saturated vapor steam at 4 MPa. A valve on the top of the tank is opened, and steam is allowed to escape. During the process any liquid formed collects at the bottom of the vessel, so that only saturated vapor exits. Calculate the total mass that has escaped when the pressure inside reaches 1 MPa. Solution: C.V. Vessel: Mass flows out. Continuity Eq.6.15: me = m1 - m2 Energy Eq.6.16: m2u2 - m1u1 = - (m1-m2)he or m2(he-u2) = m1(he-u1) Average exit enthalpy State 1: he ≈ (hG1+hG2)/2 = (2801.4+2778.1)/2 = 2789.8 m1 = V/v1 = 40.177 kg, m2 = V/v2 Energy equation 2 ⇒ v (2789.8-u2) = 40.177(2789.8-2602.3) = 7533.19 2 But v2 = .001 127 + .193 313 x2 and Substituting and solving, u2 = 761.7 + 1822 x2 x2 = 0.7936 ⇒ m2 = V/v2 = 12.94 kg, me = 27.24 kg Sat. vapor out Vapor Liquid cb Sonntag, Borgnakke and van Wylen 6.122 A 750-L rigid tank, shown in Fig. P6.122, initially contains water at 250°C, 50% liquid and 50% vapor, by volume. A valve at the bottom of the tank is opened, and liquid is slowly withdrawn. Heat transfer takes place such that the temperature remains constant. Find the amount of heat transfer required to the state where half the initial mass is withdrawn. Solution: C.V. vessel Continuity Eq.6.15: m2 − m1 = − me Energy Eq.6.16: m2u2 − m1u1 = QCV − mehe State 1: 0.375 mLIQ1 = 0.001251 = 299.76 kg; 0.375 mVAP1 = 0.05013 = 7.48 kg m1u1 = 299.76 × 1080.37 + 7.48 × 2602.4 = 343 318 kJ m1 = 307.24 kg; State 2: me = m2 = 153.62 kg 0.75 v2 = 153.62 = 0.004882 = 0.001251 + x2 × 0.04888 x2 = 0.07428 ; u2 = 1080.37 + 0.07428 × 1522 = 1193.45 kJ/kg Exit state: he = hf = 1085.34 kJ/kg Energy equation now gives the heat transfer as QCV = m2u2 - m1u1 + mehe = 153.62 × 1193.45 – 343 318 + 153.62 × 1085.34 = 6750 kJ Vapor Liquid Sat. liq. out Sonntag, Borgnakke and van Wylen 6.123 Consider the previous problem but let the line and valve be located in the top of the tank. Now saturated vapor is slowly withdrawn while heat transfer keeps the temperature inside constant. Find the heat transfer required to reach a state where half the original mass is withdrawn. Solution: C.V. vessel Continuity Eq.6.15: m2 − m1 = − me Energy Eq.6.16: m2u2 − m1u1 = QCV − mehe State 1: 0.375 mLIQ1 = 0.001251 = 299.76 kg; 0.375 mVAP1 = 0.05013 = 7.48 kg m1u1 = 299.76 × 1080.37 + 7.48 × 2602.4 = 343 318 kJ m1 = 307.24 kg; State 2: me = m2 = 153.62 kg 0.75 v2 = 153.62 = 0.004882 = 0.001251 + x2 × 0.04888 x2 = 0.07428 ; u2 = 1080.37 + 0.07428 × 1522 = 1193.45 kJ/kg Exit state: he = hg = 2801.52 kJ/kg Energy equation now gives the heat transfer as QCV = m2u2 - m1u1 + mehe = 153.62 × 1193.45 – 343 318 + 153.62 × 2801.52 = 270 389 kJ Sat. vapor out Vapor Liquid Sonntag, Borgnakke and van Wylen Review Problems 6.124 Two kg of water at 500 kPa, 20oC is heated in a constant pressure process to 1700oC. Find the best estimate for the heat transfer. Solution: C.V. Heater; steady state 1 inlet and exit, no work term, no ∆KE, ∆PE . . . . Continuity Eq.: min = mex = m, Energy Eq.6.13: q + hin = hex ⇒ q = hex - hin steam tables only go up to 1300oC so use an intermediate state at lowest pressure (closest to ideal gas) hx(1300oC, 10 kPa) from Table B.1.3 and table A.8 for the high T change ∆h hex - hin = (hex - hx) + (hx - hin) = (71 423 – 51 629)/18.015 + 5409.7 - 83.96 = 6424.5 kJ/kg Q = m(hex - hin) = 2 × 6424.5 = 12 849 kJ Sonntag, Borgnakke and van Wylen 6.125 In a glass factory a 2 m wide sheet of glass at 1500 K comes out of the final rollers that fix the thickness at 5 mm with a speed of 0.5 m/s. Cooling air in the amount of 20 kg/s comes in at 17oC from a slot 2 m wide and flows parallel with the glass. Suppose this setup is very long so the glass and air comes to nearly the same temperature (a co-flowing heat exchanger) what is the exit temperature? Solution: Energy Eq.: . . . . mglasshglass 1 + mairhair 2 = mglasshglass 3 + mairhair 4 . . mglass = ρV = ρAV = 2500× 2 ×0.005× 0.5 = 12.5 kg/s . . mglassCglass ( T3 – T1 ) + mair CPa ( T4 – T2 ) = ∅ T4 = T3 , Cglass = 0.80 kJ/kg K, CPa = 1.004 kJ/kg K . . mglassCglass T1 + mairCPa T2 12.5×0.80×1500 + 20×1.004×290 T3 = = . . 12.5×0.80 + 20×1.004 mglassCglass + mairCPa = 692.3 K We could use table A.7.1 for air, but then it will be trial and error 2 Air 1 4 3 Sonntag, Borgnakke and van Wylen 6.126 Assume a setup similar to the previous problem but the air flows in the opposite direction of the glass, it comes in where the glass goes out. How much air flow at 17oC is required to cool the glass to 450 K assuming the air must be at least 120 K cooler than the glass at any location? Solution: . . . . Energy Eq.: m1h1 + m4h4 = m3h3 + m2h2 T4 = 290 K and T3 = 450 K . . mglass = ρV = ρAV = 2500× 2 ×0.005× 0.5 = 12.5 kg/s T2 ≤ T1 – 120 K = 1380 K . . h1-h3 . . m = m4 = m2 = m1 h -h 24 Let us check the limit and since T is high use table A.7.1 for air. h4 = 290.43 kJ/kg, h2 = 1491.33 kJ/kg . . . h1-h3 . Cglass(T1-T3) . m = m4 = m2 = m1 h -h = m1 h -h 2 4 2 4 0.8 (1500-450) . m = 12.5 1491.33 – 290.43 = 8.743 kg/s 2 Air 1 4 3 Sonntag, Borgnakke and van Wylen 6.127 Three air flows all at 200 kPa are connected to the same exit duct and mix without external heat transfer. Flow one has 1 kg/s at 400 K, flow two has 3 kg/s at 290 K and flow three has 2 kg/s at 700 K. Neglect kinetic energies and find the volume flow rate in the exit flow. Solution: . . . . Continuity Eq. m1+ m2 + m3 = m4h4 . . . . Energy Eq.: m1h1 + m2h2 = m3h3 + m4h4 . . V4 = m v4 . . . m1 m2 m3 1 3 2 h4 = h1 + h2+ h3 = 6 × 401.3 + 6 × 290.43 + 6 × 713.56 . . . m4 m4 m4 = 449.95 kJ/kg 449.95 - 441.93 T4 = 440 + 20 462.34 - 441.93 = 447.86 K v4 = RT4 /P4 = 0.287×447.86/200 = 0.643 m3/kg . . V4 = m4v4 = 6 × 0.643 = 3.858 m3/s 1 2 4 3 Sonntag, Borgnakke and van Wylen 6.128 Consider the power plant as described in Problem 6.102. a. Determine the temperature of the water leaving the intermediate pressure heater, T , assuming no heat transfer to the surroundings. 13 b. Determine the pump work, between states 13 and 16. Solution: a) Intermediate Pressure Heater Energy Eq.6.10: . . . . . m11h11 + m12h12 + m15h15 = m13h13 + m14h14 75.6×284.6 + 8.064×2517 + 4.662×584 = 75.6×h13 + 12.726×349 h13 = 530.35 → T13 = 126.3°C b) The high pressure pump Energy Eq.6.12: . . . m13h13 = m16h16 + WCv,P . . WCv,P = m13(h13 - h16) = 75.6(530.35 - 565) = -2620 kW Sonntag, Borgnakke and van Wylen 6.129 Consider the powerplant as described in Problem 6.102. a. Find the power removed in the condenser by the cooling water (not shown). b. Find the power to the condensate pump. c. Do the energy terms balance for the low pressure heater or is there a heat transfer not shown? Solution: a) Condenser: Energy Eq.6.10: . . . . QCV + m5h5 + m10h10 = m6h6 . QCV + 55.44 × 2279 + 20.16 × 142.51 = 75.6 × 138.3 . QCV = -118 765 kW = -118.77 MW b) The condensate pump . . WCv,P = m6(h6 - h7) = 75.6(138.31 - 140) = -127.8 kW c) Low pressure heater Assume no heat transfer . . . . . . m14h14 + m8h8 + m7h7 + m9h9 = m10h10 + m11h11 LHS = 12.726×349 + 2.772×2459 + 75.6×140 + 4.662×558 = 24 443 kW RHS = (12.726 + 2.772 + 4.662) × 142.51 + 75.6 × 284.87 = 24 409 kW A slight imbalance, but OK. Sonntag, Borgnakke and van Wylen 6.130 A 500-L insulated tank contains air at 40°C, 2 MPa. A valve on the tank is opened, and air escapes until half the original mass is gone, at which point the valve is closed. What is the pressure inside then? Solution: 2000 × 0.5 m1 = P1V/RT1 = 0.287 × 313.2 = 11.125 kg State 1: ideal gas Continuity eq.6.15: Energy Eq.6.16: me = m1 - m2, m2 = m1/2 ⇒ me = m2 = 5.5625 kg m2u2 - m1u1 = - mehe AV Substitute constant specific heat from table A.5 and evaluate the exit enthalpy as the average between the beginning and the end values 5.5625×0.717 T2 - 11.125×0.717×313.2 = - 5.5625×1.004 (313.2 + T2)/2 Solving, T2 = 239.4 K P2 = m2RT2 5.5625 × 0.287 × 239.4 = 764 kPa V= 0.5 cb Sonntag, Borgnakke and van Wylen 6.131 A steam engine based on a turbine is shown in Fig. P6.131. The boiler tank has a volume of 100 L and initially contains saturated liquid with a very small amount of vapor at 100 kPa. Heat is now added by the burner, and the pressure regulator does not open before the boiler pressure reaches 700 kPa, which it keeps constant. The saturated vapor enters the turbine at 700 kPa and is discharged to the atmosphere as saturated vapor at 100 kPa. The burner is turned off when no more liquid is present in the as boiler. Find the total turbine work and the total heat transfer to the boiler for this process. Solution: C.V. Boiler tank. Heat transfer, no work and flow out. Continuity Eq.6.15: m2 - m1 = − me Energy Eq.6.16: m2u2 - m1u1 = QCV - mehe State 1: Table B.1.1, 100 kPa => v1 = 0.001 043, u1 = 417.36 kJ/kg => m1 = V/v1 = 0.1/0.001 043 = 95.877 kg State 2: Table B.1.1, 700 kPa => v2 = vg = 0.2729, u2 = 2572.5 kJ/kg => m2 = V/vg = 0.1/0.2729 = 0.366 kg, Exit state: Table B.1.1, 700 kPa => he = 2763.5 kJ/kg From continuity eq.: me = m1 - m2 = 95.511 kg QCV = m2u2 - m1u1 + mehe = 0.366 × 2572.5 - 95.877 × 417.36 + 95.511 × 2763.5 = 224 871 kJ = 224.9 MJ C.V. Turbine, steady state, inlet state is boiler tank exit state. Turbine exit state: Table B.1.1, 100 kPa => he = 2675.5 kJ/kg Wturb = me (hin- hex) = 95.511 × (2763.5 - 2675.5) = 8405 kJ W cb Sonntag, Borgnakke and van Wylen 6.132 An insulated spring-loaded piston/cylinder, shown in Fig. P6.132, is connected to an air line flowing air at 600 kPa, 700 K by a valve. Initially the cylinder is empty and the spring force is zero. The valve is then opened until the cylinder pressure reaches 300 kPa. By noting that u2 = uline + CV(T2 − Tline) and hline − uline = RTline find an expression for T2 as a function of P2, Po, Tline. With P = 100 kPa, find T2. Solution: C.V. Air in cylinder, insulated so 1Q2 = 0 Continuity Eq.6.15: m2 - m1 = min Energy Eq.6.16: m2u2 - m1u1 = minhline - 1W2 1 m1 = 0 ⇒ min = m2 ; m2u2 = m2hline - 2 (P0 + P2)m2v2 ⇒ 1 u2 + 2 (P0 + P2)v2 = hline Use constant specific heat in the energy equation 1 Cv(T2 - Tline) + uline + 2 (P0 + P2)RT2/P2 = hline 1 P0 + P2 RT2 = (R + Cv)Tline Cv + 2 P 2 R + Cv Tline ; Cv/R = 1/(k-1) , k = 1.4 with #'s: T2 = 2 R + Cv 3 k-1+1 3k T2 = 2 Tline = 2k + 1 Tline = 1.105 Tline = 773.7 K 2 3k-3+1 Sonntag, Borgnakke and van Wylen 6.133 A mass-loaded piston/cylinder, shown in Fig. P6.133, containing air is at 300 kPa, 17°C with a volume of 0.25 m3, while at the stops V = 1 m3. An air line, 500 kPa, 600 K, is connected by a valve that is then opened until a final inside pressure of 400 kPa is reached, at which point T = 350 K. Find the air mass that enters, the work, and heat transfer. Solution: C.V. Cylinder volume. Continuity Eq.6.15: m2 - m1 = min Energy Eq.6.16: m2u2 - m1u1 = minhline + QCV - 1W2 Process: P1 is constant to stops, then constant V to state 2 at P2 P1V 300 × 0.25 m1 = RT = 0.287 × 290.2 = 0.90 kg State 1: P1, T1 1 State 2: Open to P2 = 400 kPa, T2 = 350 K 400 × 1 m2 = 0.287 × 350 = 3.982 kg AIR mi = 3.982 - 0.90 = 3.082 kg Only work while constant P 1W2 = P1(V2 - V1) = 300(1 - 0.25) = 225 kJ Energy Eq.: QCV + mihi = m2u2 - m1u1 + 1W2 QCV = 3.982 × 0.717 × 350 - 0.90 × 0.717 × 290.2 + 225 - 3.082 × 1.004 × 600 = -819.2 kJ We could also have used the air tables A.7.1 for the u’s and hi. Sonntag, Borgnakke and van Wylen 6.134 A 2-m3 storage tank contains 95% liquid and 5% vapor by volume of liquified natural gas (LNG) at 160 K, as shown in Fig. P6.65. It may be assumed that LNG has the same properties as pure methane. Heat is transferred to the tank and saturated vapor at 160 K flows into the a steady flow heater which it leaves at 300 K. The process continues until all the liquid in the storage tank is gone. Calculate the total amount of heat transfer to the tank and the total amount of heat transferred to the heater. Solution: CV: Tank, flow out, transient. Continuity Eq.: m2 - m1 = -me Energy Eq.: QTank = m2u2 - m1u1 + mehe At 160 K, from Table B.7: 0.95 × 2 mf = Vf /vf = 0.00297 = 639.73 kg , m1 = 642.271 kg, VAPOR Qheater LIQUID Q tank 0.05 × 2 mg = Vg/vg = 0.03935 = 2.541 kg m2 = V/vg2 = 2/0.03935 = 50.826 kg m1u1 = 639.73(-106.35) + 2.541(207.7) = -67507 kJ me = m1 - m2 = 591.445 kg QTank = 50.826 × 207.7 - (-67 507) + 591.445 × 270.3 = +237 931 kJ CV: Heater, steady flow, P = PG 160 K = 1593 kPa QHeater = me Tank(he - hi)Heater = 591.445(612.9 - 270.3) = 202 629 kJ Sonntag, Borgnakke and van Wylen Heat transfer problems 6.135 Liquid water at 80oC flows with 0.2 kg/s inside a square duct, side 2 cm insulated with a 1 cm thick layer of foam k = 0.1 W/m K. If the outside foam surface is at 25oC how much has the water temperature dropped for 10 m length of duct? Neglect the duct material and any corner effects (A = 4sL). Solution: Conduction heat transfer . dT ∆T Qout = kA dx = k 4 sL = 0.1× 4×0.02×10×(80-25)/0.01 = 440 W ∆x . . . Energy equation: m1h1= mhe + Qout .. he– hi = -Q/m = - (440/0.2) = -2200 J/kg = - 2.2 kJ/kg he= hi -2.2 kJ/kg = 334.88 – 2.2 = 332.68 kJ/kg 2.2 Te = 80 – 334.88-313.91 5 = 79.48oC ∆T = 0.52oC We could also have used he– hi = Cp∆T s cb L Sonntag, Borgnakke and van Wylen 6.136 A counter-flowing heat exchanger conserves energy by heating cold outside fresh air at 10oC with the outgoing combustion gas (air) at 100oC. Assume both flows are 1 kg/s and the temperature difference between the flows at any point is 50oC. What is the incoming fresh air temperature after the heat exchanger? What is the equivalent (single) convective heat transfer coefficient between the flows if the interface area is 2 m2? Solution: The outside fresh air is heated up to T4 = 50oC (100 – 50), the heat transfer needed is . . . kJ Q = m∆h = mCp∆T = 1 kg/s × 1.004 kg K × (50 – 10) K = 40 kW This heat transfer takes place with a temperature difference of 50oC throughout . Q = h A ∆T ⇒ . Q 40 000 W W h= = 2 K = 400 m2 K A∆T 2 × 50 m Often the flows may be concentric as a smaller pipe inside a larger pipe. Hot gas 1 Heat transfer 4 Wall 2 3 fresh air Sonntag, Borgnakke and van Wylen 6.137 Saturated liquid water at 1000 kPa flows at 2 kg/s inside a 10 cm outer diameter steel pipe and outside of the pipe is a flow of hot gases at 1000 K with a convection coefficient of h = 150 W/m2 K. Neglect any ∆T in the steel and any inside convection h and find the length of pipe needed to bring the water to saturated vapor. Solution: Energy Eq. water: Table B.1.2: . . . Q = m (he – hi) = m hfg hfg = 2015.29 kJ/kg, T = Tsat = 179.91oC = 453.1 K The energy is transferred by heat transfer so . Q = h A ∆T = h πD L ∆T Equate the two expressions for the heat transfer and solve for the length L . . mhfg Q 2 × 2015.29 × 1000 L= = = h πD ∆T h πD ∆T 150 × π × 0.1 ×(1000 - 453.1) = 156.4 m L Sonntag, Borgnakke and van Wylen 6.138 A flow of 1000 K, 100 kPa air with 0.5 kg/s in a furnace flows over a steel plate of surface temperature 400 K. The flow is such that the convective heat transfer coefficient is h = 125 W/m2 K. How much of a surface area does the air have to flow over to exit with a temperature of 800 K? How about 600 K? Solution: Convection heat transfer . Q = hA ∆T Inlet: ∆Ti = 1000 - 400 = 600 K a) Exit: ∆Te = 800 - 400 = 400 K, so we can use an average of ∆T = 500 K for heat transfer . . Q = ma (hi – he) = 0.5(1046.22 – 822.2) =112 kW . Q 112 × 1000 A= = = 1.79 m2 h ∆T 125 × 500 b) . . Q = ma (hi – he) = 0.5 (1046.22 – 607.32) = 219.45 kW Exit: ∆Tout = 600 - 400 = 200 K, so we have an average of ∆T = 400 K for heat transfer . Q 219.45 × 1000 A= = = 4.39 m2 h∆T 125 × 400 1000 K air 400 K T exit Q from air to steel SOLUTION MANUAL SI UNIT PROBLEMS CHAPTER 8 SONNTAG • BORGNAKKE • VAN WYLEN FUNDAMENTALS of Thermodynamics Sixth Edition Sonntag, Borgnakke and van Wylen CONTENT SUBSECTION PROB NO. Correspondence table Concept-Study Guide problems 1-20 Inequality of Clausius 21-25 Entropy of a pure substance 26-30 Reversible processes 31-60 Entropy generation 61-74 Entropy of a liquid or solid 75-86 Entropy of ideal gases 87-105 Polytropic processes 106-119 Rates or fluxes of entropy 120-125 Review problems 126-139 Problem solution repeated, but using the Pr and vr functions in Table A.7.2: 88, 95, 107, 112 Sonntag, Borgnakke and van Wylen Correspondence table CHAPTER 8 6th edition Sonntag/Borgnakke/Wylen The correspondence between this problem set and the 5th edition chapter 8 problem set. The study guide problems 8.1-8.20 are all new New 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 Old new new new new new new 2 new new new 3 4mod 5 6 7 8 new new 9 10 11 12 13 new 15 mod new 16 new 18 mod 20 New 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 Old 22 mod new new new new new 17 new 19 24 new new 29 new 27 28 new 30 new 32 new 36 37 new new 41 31 40 new new New 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 Old 42 43 new new new 44 45 new 47 51 new new 49 50 53 54 new 52 48 new 56 57 new 59 60 61 46 67 63 65 New 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 Old 64 new 62 new 69 70 new 71 66 new 26 mod new new new new 25 18 34 23 33 14 35 39 new 58 68 74 78 82 Sonntag, Borgnakke and van Wylen Concept-Study Guide Problems 8.1 Does Clausius say anything about the sign for o dQ ? ∫ No. The total (net) heat transfer can be coming in like in a heat engine (Wout = QH – QL) in which case it is positive. It can also be net going out like in a refrigerator or heat pump (Win = QH – QL) in which case the sign is negative. Finally if you look at a transmission gearbox there could be no heat transfer (first approximation) in which case the integral is zero. 8.2 When a substance has completed a cycle v, u, h and s are unchanged. Did anything happen? Explain. Yes. During various parts of the cycle work and heat transfer may be transferred. That happens at different P and T. The net work out equals the net heat transfer in (energy conservation) so dependent upon the sign it is a heat engine or a heat pump (refrigerator). The net effect is thus a conversion of energy from one storage location to another and it may also change nature (some Q got changed to W or the opposite) 8.3 Assume a heat engine with a given QH. Can you say anything about QL if the engine is reversible? If it is irreversible? For a reversible heat engine it must be that: QQ o dQ = 0 = H - L ∫T TH TL or integrals if T not constant So as TL is lower than TH then QL must be correspondingly lower than QH to obtain the net zero integral. For an irreversible heat engine we have QQ o dQ = H - L < 0 ∫T T T H L This means that QL is larger than before (given QH and the T’s). The irreversible heat engine rejects more energy and thus gives less out as work. Sonntag, Borgnakke and van Wylen 8.4 How can you change s of a substance going through a reversible process? From the definition of entropy dq ds = T for a reversible process. Thus only heat transfer gives a change in s, expansion/compression involving work does not give such a contribution. 8.5 Does the statement of Clausius require a constant T for the heat transfer as in a Carnot cycle? No. The statement for a cycle involves an integral of dQ/T so T can vary, which it does during most processes in actual devices. This just means that you cannot that easily get a closed expression for the integral. 8.6 A reversible process adds heat to a substance. If T is varying does that influence the change in s? Yes. dq Reversible: ds = T So if T goes up it means that s changes less per unit of dq, and the opposite if T decreases then s changes more per unit of dq. 8.7 Water at 100 kPa, 150oC receives 75 kJ/kg in a reversible process by heat transfer. Which process changes s the most: constant T, constant v or constant P? dq ds = T Look at the constant property lines in a T-s diagram, Fig. 8.5. The constant v line has a higher slope than the constant P line also at positive slope. Thus both the constant P and v processes have an increase in T. As T goes up the change in s is smaller. The constant T (isothermal) process therefore changes s the most. Sonntag, Borgnakke and van Wylen 8.8 A substance has heat transfer out. Can you say anything about changes in s if the process is reversible? If it is irreversible? Reversible: dq ds = T < 0 Irreversible: dq ds = T + dsgen = ? since dq < 0 dq < 0 but dsgen > 0 You cannot say, ds depends on the magnitude of dq/T versus dsgen Sonntag, Borgnakke and van Wylen 8.9 A substance is compressed adiabaticly so P and T go up. Does that change s? dq If the process is reversible then s is constant, ds = T = 0 If the process is irreversible then s goes up, dq ds = T + dsgen = dsgen > 0 8.10 Saturated water vapor at 200 kPa is compressed to 600 kPa in a reversible adiabatic process. Find the new v and T. Process adiabatic: dq = 0 Process reversible: dsgen = 0 Change in s: ds = dq/T + dsgen = 0 + 0 = 0 thus s is constant Table B.1.3: T1 = 120.23oC, v1 = 0.88573 m3/kg, s1 = 7.1271 kJ/kg K Table B.1.3 at 600 kPa and s = s1 = 7.1271 kJ/kg-K 7.1271 – 6.9665 T = 200 + 50 7.1816 – 6.9665 = 200 + 50 × 0.74663 = 237.3oC v = 0.35202 + (0.39383 – 0.35202) × 0.74663 = 0.38324 m3/kg Sonntag, Borgnakke and van Wylen 8.11 A computer chip dissipates 2 kJ of electric work over time and rejects that as heat transfer from its 50oC surface to 25oC air. How much entropy is generated in the chip? How much if any is generated outside the chip? C.V.1 Chip with surface at 50oC, we assume chip state is constant. Energy: U2 – U1 = 0 = 1Q2 – 1W2 = Welectrical in - Qout 1 Qout 1 + 1S2 gen1 Entropy: S2 – S1 = 0 = - T surf Qout 1 Welectrical in 2 kJ = = 323.15 K = 6.19 J/K 1S2 gen1 = T Tsurf surf C.V.2 From chip surface at 50oC to air at 25oC, assume constant state. Energy: U2 – U1 = 0 = 1Q2 – 1W2 = Qout 1 - Qout 2 Qout1 Qout2 Entropy: S2 – S1 = 0 = T - T + 1S2 gen2 surf air Qout2 Qout1 2 kJ 2 kJ -T = 298.15 K - 323.15 K = 0.519 J/K 1S2 gen2 = T air surf o 25 C air air flow cb Q o 50 C Sonntag, Borgnakke and van Wylen 8.12 A car uses an average power of 25 hp for a one hour round trip. With a thermal efficiency of 35% how much fuel energy was used? What happened to all the energy? What change in entropy took place if we assume ambient at 20oC? Since it is a round trip, there are no changes in storage of energy for the car after it has cooled down again. All the energy is given out to the ambient in the form of exhaust flow (hot air) and heat transfer from the radiator and underhood air flow. ⌠. E = ⌡ W dt = 25 hp × 0.7457 (kW/hp) × 3600 s = 67 113 kJ = η Q Q = E / η = 67 113 / 0.35 = 191 751 kJ ∆S = Q / T = 191 751 / 293.15 = 654.1 kJ/K All the energy ends up in the ambient at the ambient temperature. 8.13 A liquid is compressed in a reversible adiabatic process. What is the change in T? dq If the process is reversible then s is constant, ds = T = 0 Change in s for a liquid (an incompressible substance) is Eq. 8.19 C ds = T dT From this it follows that if ds = 0 then T is constant. Sonntag, Borgnakke and van Wylen 8.14 Two 5 kg blocks of steel, one at 250oC the other at 25oC, come in thermal contact. Find the final temperature and the total entropy generation in the process? C.V. Both blocks, no external heat transfer, C from Table A.3. Energy Eq.: U2 – U1 = mA(u2 – u1)A + mB(u2 – u1)B = 0 – 0 = mAC(T2 – TA1) + mBC(T2 – TB1) T2 = mATA1 + mBTB1 1 1 = 2 TA1 + 2 TB1 = 137.5oC mA + mB Entropy Eq.: S2 – S1 = mA(s2 – s1)A + mB(s2 – s1)B = 1S2 gen T2 T2 + mBC ln T 1S2 gen = mAC ln T A1 B1 137.5 + 273.15 137.5 + 273.15 = 5 × 0.46 ln 250 + 273.15 + 5 × 0.46 ln 298.15 = -0.5569 + 0.7363 = 0.1794 kJ/K B A Heat transfer over a finite temperature difference is an irreversible process Sonntag, Borgnakke and van Wylen 8.15 One kg of air at 300 K is mixed with one kg air at 400 K in a process at a constant 100 kPa and Q = 0. Find the final T and the entropy generation in the process. C.V. All the air. Energy Eq.: U2 – U1 = 0 – W Entropy Eq.: S2 – S1 = 0 + 1S2 gen Process Eq.: P = C; W = P(V2 – V1) Substitute W into energy Eq. U2 – U1 + W = U2 – U1 + P(V2 – V1) = H2 – H1 = 0 Due to the low T let us use constant specific heat H2 – H1 = mA(h2 – h1)A + mB(h2 – h1)B = mACp(T2 – TA1) + mBCp(T2 – TB1) = 0 mATA1 + mBTB1 1 1 = 2 TA1 + 2 TB1 = 350 K mA + mB Entropy change is from Eq. 8.25 with no change in P T2 T2 + mBCp ln T 1S2 gen = S2 – S1 = mACp ln T A1 B1 350 350 = 1 × 1.004 ln 300 + 1 × 1.004 ln 400 = 0.15477 - 0.13407 = 0.0207 kJ/K T2 = Remark: If you check, the volume does not change and there is no work. Sonntag, Borgnakke and van Wylen 8.16 One kg of air at 100 kPa is mixed with one kg air at 200 kPa, both at 300 K, in a rigid insulated tank. Find the final state (P, T) and the entropy generation in the process. C.V. All the air. Energy Eq.: U2 – U1 = 0 – 0 Entropy Eq.: S2 – S1 = 0 + 1S2 gen Process Eqs.: V = C; W = 0, Q = 0 States A1, B1: uA1 = uB1 cb VA = mART1/PA1; VB = mBRT1/PB1 U2 – U1 = m2u2 – mAuA1 – mBuB1 = 0 ⇒ u2 = (uA1 + uB1)/2 = uA1 State 2: T2 = T1 = 300 K (from u2); m2 = mA + mB = 2 kg; V2 = m2RT1/P2 = VA + VB = mART1/PA1 + mBRT1/PB1 Divide with mART1 and get 1 1 2/P2 = 1/PA1 + 1/PB1 = 100 + 200 = 0.015 kPa-1 ⇒ P2 = 133.3 kPa Entropy change from Eq. 8.25 with the same T, so only P changes P2 P2 – mBR ln P 1S2 gen = S2 – S1 = –mAR ln P A1 B1 133.3 133.3 = – 1 × 0.287 [ ln 100 + ln 200 ] = –0.287 (0.2874 – 0.4057) = 0.034 kJ/K Sonntag, Borgnakke and van Wylen 8.17 An ideal gas goes through a constant T reversible heat addition process. How do the properties (v, u, h, s, P) change (up, down or constant)? Ideal gas: u(T), h(T) so they are both constant Eq. 8.11 gives: ds = dq/T + dsgen = dq/T + 0 > 0 Eq. 8.21 gives: ds = (R/v) dv so v increases Eq. 8.23 gives: ds = -(R/P) dP so P decreases so s goes up by q/T T P 1 2 2 1 T q v s Sonntag, Borgnakke and van Wylen 8.18 Carbon dioxide is compressed to a smaller volume in a polytropic process with n = 1.2. How do the properties (u, h, s, P, T) change (up, down or constant)? For carbon dioxide Table A.5 k = 1.4 so we have n < k and the process curve can be recognized in Figure 8.18. From this we see a smaller volume means moving to the left in the P-v diagram and thus also up. P up, T up and s down. As T is up so is h and u. T P 2 1T n = 1.2 v 2 1 q s Sonntag, Borgnakke and van Wylen 8.19 Hot combustion air at 1500 K expands in a polytropic process to a volume 6 times as large with n = 1.5. Find the specific boundary work and the specific heat transfer. Energy Eq.: u2 – u1 = 1q2 - 1w2 1 R Reversible work Eq. 8.38: 1w2 = 1-n (P2v2 – P1v1) = 1-n (T2 – T1) Process Eq: n Pv = C; T2 = T1 (v1/v2) n-1 10.5 = 1500 6 = 612.4 K Properties from Table A.7.1: u1 = 444.6 kJ/kg, u2 = 1205.25 kJ/kg 0.287 1w2 = 1 - 1.5 (612.4 – 1500) = 509.5 kJ/kg 1q2 = u2 – u1 + 1w2 = 1205.25 – 444.6 + 509.5 = 1270 kJ/kg Sonntag, Borgnakke and van Wylen 8.20 A window receives 200 W of heat transfer at the inside surface of 20oC and transmits the 200 W from its outside surface at 2oC continuing to ambient air at – 5oC. Find the flux of entropy at all three surfaces and the window’s rate of entropy generation. Flux of entropy: . .Q S=T Window Inside Outside . 200 W Sinside = 293.15 K = 0.682 W/K . 200 W Swin = 275.15 K = 0.727 W/K . 200 W Samb = 268.15 K = 0.746 W/K o 20 C o 2C o -5 C . . . Window only: Sgen win = Swin – Sinside = 0.727 – 0.682 = 0.045 W/K If you want to include the generation in the outside air boundary layer where T changes from 2oC to the ambient –5oC then it is . . . Sgen tot = Samb – Sinside = 0.746 – 0.682 = 0.064 W/K Sonntag, Borgnakke and van Wylen Inequality of Clausius 8.21 Consider the steam power plant in Example 6.9 and assume an average T in the line between 1 and 2. Show that this cycle satisfies the inequality of Clausius. Solution: ⌠dQ ≤ 0 T ⌡ For this problem we have three heat transfer terms: qb = 2831 kJ/kg, qloss = 21 kJ/kg, qc = 2173.3 kJ/kg Show Clausius: ⌠dq = qb – qloss – qc T T Tavg 1-2 Tc ⌡ b 2831 21 2173.3 = 573 − 568 − 318 = –1.93 kJ/kg K < 0 OK Sonntag, Borgnakke and van Wylen 8.22 Assume the heat engine in Problem 7.25 has a high temperature of 1200 K and a low temperature of 400 K. What does the inequality of Clausius say about each of the four cases? Solution: ⌠. dQ 6 4 Cases a) T = 1200 – 400 = – 0.005 kW/K < 0 ⌡ OK ⌠. dQ 6 0 b) T = 1200 – 400 = 0.005 kW/K > 0 Impossible ⌡ ⌠. dQ 6 2 c) T = 1200 – 400 = 0 kW/K ⌡ Possible if reversible ⌠. dQ 6 6 d) T = 1200 – 400 = – 0.001 kW/K < 0 ⌡ TH = 1200 K QH HE W cb QL TL = 400 K OK Sonntag, Borgnakke and van Wylen 8.23 Let the steam power plant in Problem 7.26 have 700oC in the boiler and 40oC during the heat rejection in the condenser. Does that satisfy the inequality of Clausius? Repeat the question for the cycle operated in reverse as a refrigerator. Solution: . QH = 1 MW . QL = 0.58 MW ⌠. dQ 1000 580 T = 973 – 313 = –0.82 kW/K < 0 ⌡ Refrigerator ⌠. dQ 580 1000 T = 313 – 973 = 0.82 > 0 ⌡ OK Cannot be possible QH WT WP, in . QL Sonntag, Borgnakke and van Wylen 8.24 A heat engine receives 6 kW from a 250oC source and rejects heat at 30oC. Examine each of three cases with respect to the inequality of Clausius. . . a . W = 6 kW b. W = 0 kW c. Carnot cycle Solution: TH = 250 + 273 = 523 K ; TL = 30 + 273 = 303 K ⌠. dQ 6000 0 Case a) T = 523 – 303 = 11.47 kW/K > 0 Impossible ⌡ ⌠. dQ 6000 6000 b) T = 523 – 303 = –8.33 kW/K < 0 ⌡ . ⌠. dQ 6000 QL c) T = 0 = 523 – 303 ⌡ . 303 QL = 523 × 6 = 3.476 kW . . . W = QH – QL = 2.529 kW ⇒ OK Sonntag, Borgnakke and van Wylen 8.25 Examine the heat engine given in Problem 7.50 to see if it satisfies the inequality of Clausius. Solution: QH = 325 kJ QL = 125 kJ at TH = 1000 K at TL = 400 K ⌠ dQ = 325 – 125 = 0.0125 kJ/K > 0 T 1000 400 ⌡ Impossible TH = 1000 K QH = 325 kJ HE W = 200 kJ QL = 125 kJ cb TL = 400 K Sonntag, Borgnakke and van Wylen Entropy of a pure substance 8.26 Find the entropy for the following water states and indicate each state on a T-s diagram relative to the two-phase region. a. 250oC, v = 0.02 m3/kg b. 250oC, 2000 kPa c. –2oC, 100 kPa d. 20oC, 100 kPa e. 20oC, 10 000 kPa Solution: 0.02 - 0.001251 a) Table B.1.1: x = = 0.38365 0.04887 s = sf + x sfg = 2.7927 + 0.38365 × 3.2802 = 4.05 kJ/kg K b) Table B.1.3: s = 6.5452 kJ/kg K c) Table B.1.5: s = –1.2369 kJ/kg K d) Table B.1.1: s = 0.2966 kJ/kg K e) Table B.1.4 s = 0.2945 kJ/kg K T P e a b a e d c v c b d s Sonntag, Borgnakke and van Wylen 8.27 Find the missing properties and give the phase of the substance a. H2O s = 7.70 kJ/kg K, P = 25 kPa h = ? T = ? x = ? b. H2O u = 3400 kJ/kg, P = 10 MPa T = ? x = ? s = ? c. R-12 T = 0°C, P = 200 kPa s=?x=? d. R-134a T = −10°C, x = 0.45 v=?s=? e. NH3 T = 20°C, s = 5.50 kJ/kg K u = ? x = ? Solution: a) Table B.1.1 T = Tsat(P) = 64.97°C x = (s – sf)/sfg = 7.70 - 0.893 6.9383 = 0.981 h = 271.9 + 0.981 × 2346.3 = 2573.8 kJ/kg b) Table B.1.2 u > ug => Sup.vap Table B.1.3, x = undefined T ≅ 682°C , s ≅ 7.1223 kJ/kg K c) Table B.3.2, superheated vapor, x = undefined, s = 0.7325 kJ/kg K v = vf + xvfg = 0.000755 + 0.45 × 0.098454 = 0.04506 m3/kg d) Table B.5.1 s = sf + xsfg = 0.9507 + 0.45 × 0.7812 = 1.3022 kJ/kg K e) Table B.2.1, s > sg => Sup.vap. Table B.2.2, x = undefined u = h–Pv = 1492.8 – 439.18 × 0.3100 = 1356.7 kJ/kg P T b b c, e da da T v P c, e s Sonntag, Borgnakke and van Wylen 8.28 Saturated liquid water at 20oC is compressed to a higher pressure with constant temperature. Find the changes in u and s when the final pressure is a. 500 kPa b. 2000 kPa c. 20 000 kPa Solution: kJ/kg kJ/kg K B.1.1: u1 = 83.94 s1 = 0.2966 B.1.4: ua = 83.91 sa = 0.2965 ∆u = –0.03 ∆s = –0.0001 B.1.4: ub = 83.82 sb = 0.2962 ∆u = –0.12 ∆s = –0.0004 B.1.4: uc = 82.75 sc = 0.2922 ∆u = –1.19 ∆s = –0.0044 Nearly constant u and s, incompressible media T P c b c,b,a,1 a 1 v s Sonntag, Borgnakke and van Wylen 8.29 Saturated vapor water at 150oC is expanded to a lower pressure with constant temperature. Find the changes in u and s when the final pressure is a. 100 kPa b. 50 kPa c. 10 kPa Solution: Table B.1.1 for the first state then B.1.3 for the a, b and c states. kJ/kg kJ/kg K u1= 2559.54 s1= 6.8378 ua = 2582.75 sa = 7.6133 ∆u = 23.21 ∆s = 0.7755 ub = 2585.61 sb = 7.94 ∆u = 26.07 ∆s = 1.1022 uc = 2587.86 sc = 8.6881 ∆u = 28.32 ∆s = 1.8503 T P 1 ab 1 c v 476 kPa 100 50 10 ab c s Sonntag, Borgnakke and van Wylen 8.30 Determine the missing property among P, T, s, x for the following states: a. Ammonia 25oC, v = 0.10 m3/kg b. Ammonia 1000 kPa, s = 5.2 kJ/kg K c. R-134a 5oC, s = 1.7 kJ/kg K d. R-134a 50oC, s = 1.9 kJ/kg K e. R-22 100 kPa, v = 0.3 m3/kg Solution: P kPa 1003 1000 350.9 232.3 100 T oC 25 42.53 5 50 42.6 a) b) c) d) e) Table B2.1 B2.2 B5.1 B5.2 B4.2 a) x = (0.1 – 0.001658)/0.12647 = 0.7776 s kJ/kg K 4.1601 5.2 1.7 1.9 1.1975 x 0.7776 ----0.96598 --------- s = sf + x sfg = 1.121 + x × 3.9083 = 4.1601 kJ/kg K b) T = 40 + 10 × (5.2 – 5.1778)/(5.2654 – 5.1778) = 42.53oC superheated vapor so x is undefined c) x = (1.7 – 1.0243)/0.6995 = 0.96598 P = Psat = 350.9 kPa d) superheated vapor between 200 and 300 kPa P = 200 + 100 × (1.9 – 1.9117)/(1.8755 – 1.9117) = 232.3 kPa e) T = 40 + 10 × (0.3 – 0.29739)/(0.30729 – 0.29739) = 42.636oC s = 1.1919 + 0.2636 × (1.2132 – 1.1919) =1.1975 kJ/kg K T P a b b e c a d e d c v s Sonntag, Borgnakke and van Wylen Reversible processes 8.31 Consider a Carnot-cycle heat engine with water as the working fluid. The heat transfer to the water occurs at 300°C, during which process the water changes from saturated liquid to saturated vapor. The heat is rejected from the water at 40°C. Show the cycle on a T–s diagram and find the quality of the water at the beginning and end of the heat rejection process. Determine the net work output per kilogram of water and the cycle thermal efficiency. Solution: From the definition of the Carnot cycle, two constant s and two constant T processes. T 300 40 2 1 4 From table B.1.1 State 2 is saturated vapor so s3 = s2 = 5.7044 kJ/kg K = 0.5724 + x3(7.6845) 3 x3 = 0.6678 s State 1 is saturated liquid so s4 = s1 = 3.2533 kJ/kg K = 0.5724 + x4(7.6845) x4 = 0.3489 wNET TH – TL 260 ηTH = q = TH = 573.2 = 0.4536 H qH = TH(s2 – s1) = 573.2 K (5.7044 – 3.2533) kJ/kg K = 1405.0 kJ/kg wNET = ηTH × qH = 637.3 kJ/kg Sonntag, Borgnakke and van Wylen 8.32 In a Carnot engine with ammonia as the working fluid, the high temperature is 60°C and as QH is received, the ammonia changes from saturated liquid to saturated vapor. The ammonia pressure at the low temperature is 190 kPa. Find TL, the cycle thermal efficiency, the heat added per kilogram, and the entropy, s, at the beginning of the heat rejection process. Solution: Constant T ⇒ constant P from 1 to 2, Table B.2.1 T 1 qH = ∫ Tds = T (s2 – s1) = T sfg 2 = h2 – h1 = hfg = 997.0 kJ/kg s 43 States 3 & 4 are two-phase, Table B.2.1 ⇒ TL = T3 = T4 = Tsat(P) = –20°C TL 253.2 ηcycle = 1 – T = 1 – 333.2 = 0.24 H Table B.2.1: s3 = s2 = sg(60°C) = 4.6577 kJ/kg K Sonntag, Borgnakke and van Wylen 8.33 Water is used as the working fluid in a Carnot cycle heat engine, where it changes from saturated liquid to saturated vapor at 200°C as heat is added. Heat is rejected in a constant pressure process (also constant T) at 20 kPa. The heat engine powers a Carnot cycle refrigerator that operates between –15°C and +20°C. Find the heat added to the water per kg water. How much heat should be added to the water in the heat engine so the refrigerator can remove 1 kJ from the cold space? Solution: Carnot cycle heat engine: Constant T ⇒ constant P from 1 to 2, Table B.2.1 T 1 qH = ∫ Tds = T (s2 – s1) = T sfg = hfg 2 s 43 = 473.15 (4.1014) = 1940 kJ/kg States 3 & 4 are two-phase, Table B.2.1 ⇒ TL = T3 = T4 = Tsat(P) = 60.06oC Carnot cycle refrigerator (TL and TH are different from above): TL QL 273 – 15 258 βref = W = T – T = 20 – (–15) = 35 = 7.37 H L W= QL 1 = 7.37 = 0.136 kJ β The needed work comes from the heat engine TL 333 W = ηHE QH H2O ; ηHE = 1 – T = 1 – 473 = 0.296 H QH H2O = W 0.136 = = 0.46 kJ ηHE 0.296 Sonntag, Borgnakke and van Wylen 8.34 Consider a Carnot-cycle heat pump with R-22 as the working fluid. Heat is rejected from the R-22 at 40°C, during which process the R-22 changes from saturated vapor to saturated liquid. The heat is transferred to the R-22 at 0°C. a. Show the cycle on a T–s diagram. b. Find the quality of the R-22 at the beginning and end of the isothermal heat addition process at 0°C. c. Determine the coefficient of performance for the cycle. Solution: a) T 40 0 b) From Table B.4.1, state 3 is saturated liquid 2 3 1 4 s s4 = s3 = 0.3417 kJ/kg K = 0.1751 + x4(0.7518) => x4 = 0.2216 State 2 is saturated vapor so from Table B.4.1 s1 = s2 = 0.8746 kJ/kg K = 0.1751 + x1(0.7518) => c) x1 = 0.9304 qH TH 313.2 β′ = w = T – T = 40 = 7.83 H L IN Sonntag, Borgnakke and van Wylen 8.35 Do Problem 8.34 using refrigerant R-134a instead of R-22. Consider a Carnot-cycle heat pump with R-22 as the working fluid. Heat is rejected from the R-22 at 40°C, during which process the R-22 changes from saturated vapor to saturated liquid. The heat is transferred to the R-22 at 0°C. a. Show the cycle on a T–s diagram. b. Find the quality of the R-22 at the beginning and end of the isothermal heat addition process at 0°C. c. Determine the coefficient of performance for the cycle. Solution: a) T 40 0 b) From Table B.5.1, state 3 is saturated liquid 2 3 4 s4 = s3 = 1.1909 kJ/kg K = 1.00 + x4(0.7262) 1 => s x4 = 0.2629 State 2 is saturated vapor so from Table B.5.1 s1 = s2 = 1.7123 kJ/kg K = 1.00 + x1(0.7262) => c) x1 = 0.9809 qH TH 313.2 β′ = w = T – T = 40 = 7.83 H L IN Sonntag, Borgnakke and van Wylen 8.36 Water at 200 kPa, x = 1.0 is compressed in a piston/cylinder to 1 MPa, 250°C in a reversible process. Find the sign for the work and the sign for the heat transfer. Solution: State 1: Table B.1.1: v1 = 0.8857 m3/kg; u1 = 2529.5 kJ/kg; State 2: Table B.1.3: s1 = 7.1271 kJ/kg K v2 = 0.23268 m3/kg; u2 = 2709.9 kJ/kg; s2 = 6.9246 kJ/kg K v2 < v 1 => 1w2 = ∫ P dv < 0 s2 < s1 => 1q2 = ∫ T ds < 0 P T 2 2 1 1 v s Sonntag, Borgnakke and van Wylen 8.37 Water at 200 kPa, x = 1.0 is compressed in a piston/cylinder to 1 MPa, 350oC in a reversible process. Find the sign for the work and the sign for the heat transfer. Solution: 1w2 = ∫ P dv so sign dv 1q2 = ∫ T ds so sign ds B1.2 v1 = 0.88573 m3/kg s1 = 7.1271 kJ/kg K B1.3 v2 = 0.28247 m3/kg s2 = 7.301 kJ/kg K dv < 0 => w is negative ds > 0 => q is positive P T 2 2 1 1 v s Sonntag, Borgnakke and van Wylen 8.38 Ammonia at 1 MPa, 50oC is expanded in a piston/cylinder to 500 kPa, 20oC in a reversible process. Find the sign for both the work and the heat transfer. Solution: 1w2 = ∫ P dv so sign dv 1q2 = ∫ T ds so sign ds B.2.2 v1 = 0.14499 m3/kg s1 = 5.2654 kJ/kg K B.2.2 v2 = 0.26949 m3/kg s2 = 5.4244 kJ/kg K dv > 0 => w is positive ds > 0 => q is positive P T 1 1 2 2 v s Sonntag, Borgnakke and van Wylen 8.39 One kilogram of ammonia in a piston/cylinder at 50°C, 1000 kPa is expanded in a reversible isothermal process to 100 kPa. Find the work and heat transfer for this process. Solution: C.V.: NH3 This is a control mass m2 = m1 with a reversible process Energy Eq.5.11: m(u2 – u1) = 1Q2 – 1W2 Entropy Eq.8.3: m(s2 – s1) = ⌡ (1/T) dQ = 1Q2/T ⌠ Rev.: 1W2 = ⌠ PdV ⌡ ( = since reversible) P 1Q2 = ⌠ Tmds = mT(s2 – s1) ⌡ T 1 From Table B.2.2 2 1 T v State 1: u1 = 1391.3 kJ/kg; s1 = 5.265 kJ/kg K State 2: u2 = 1424.7 kJ/kg; s2 = 6.494 kJ/kg K; v2 = 1.5658 m3/kg; h2 = 1581.2 kJ/kg 1Q2 = 1 kg (273 + 50) K (6.494 – 5.265) kJ/kg K = 396.967 kJ 1W2 = 1Q2 – m(u2 – u1) = 363.75 kJ 2 s Sonntag, Borgnakke and van Wylen 8.40 One kilogram of ammonia in a piston/cylinder at 50°C, 1000 kPa is expanded in a reversible isobaric process to 140°C. Find the work and heat transfer for this process. Solution: Control mass. m(u2 - u1) = 1Q2 - 1W2 P 2 T 1 2 1 Process: P = constant ⇒ 1W2 = mP(v2 - v1) v State 1: Table B.2.2 v1 = 0.145 m3/kg, u1 = 1391.3 kJ/kg State 2: Table B.2.2 v2 = 0.1955 m3/kg, u2 = 1566.7 kJ/kg 1W2 = 1 × 1000(0.1955 - 0.145) = 50.5 kJ 1Q2 = m(u2 - u1) + 1W2 = 1 × (1566.7 - 1391.3) + 50.5 = 225.9 kJ s Sonntag, Borgnakke and van Wylen 8.41 One kilogram of ammonia in a piston/cylinder at 50°C, 1000 kPa is expanded in a reversible adiabatic process to 100 kPa. Find the work and heat transfer for this process. Solution: Control mass: Energy Eq.5.11: m(u2 - u1) = 1Q2 - 1W2 Entropy Eq.8.3: m(s2 - s1) = ∫ dQ/T ( = since reversible) ⇒ s2 = s 1 State 1: (P, T) Table B.2.2, u1 = 1391.3 kJ/kg, s1 = 5.2654 kJ/kg K Process: 1Q2 = 0 State 2: P2 , s2 ⇒ 2 phase Table B.2.1 Interpolate: sg2 = 5.8404 kJ/kg K, sf = 0.1192 kJ/kg K s - sf 5.2654 − 0.1192 x2 = s = = 0.90, 5.7212 fg u2 = uf + x2 ufg = 27.66 + 0.9×1257.0 = 1158.9 kJ/kg 1W2 = 1 × (1391.3 - 1158.9) = 232.4 kJ P T 1 1 2 2 v s Sonntag, Borgnakke and van Wylen 8.42 A cylinder fitted with a piston contains ammonia at 50°C, 20% quality with a volume of 1 L. The ammonia expands slowly, and during this process heat is transferred to maintain a constant temperature. The process continues until all the liquid is gone. Determine the work and heat transfer for this process. Solution: C.V. Ammonia in the cylinder. Table B.2.1: T1 = 50°C, x1 = 0.20, V1 = 1 L NH 3 v1 = 0.001777 + 0.2 ×0.06159 = 0.014095 m3/kg s1 = 1.5121 + 0.2 × 3.2493 = 2.1620 kJ/kg K T m = V1/v1 = 0.001/0.014095 = 0.071 kg v2 = vg = 0.06336 m3/kg, o 50 C 1 2 s Process: T = constant to x2 = 1.0, s2 = sg = 4.7613 kJ/kg K P = constant = 2.033 MPa From the constant pressure process 1W2 = ⌠PdV = Pm(v2 - v1) = 2033 × 0.071 × (0.06336 - 0.014095) = 7.11 kJ ⌡ From the second law Eq.8.3 with constant T 1Q2 = ⌠TdS = Tm(s2 - s1) = 323.2 × 0.071(4.7613 - 2.1620) = 59.65 kJ ⌡ or 1Q2 = m(u2 - u1) + 1W2 = m(h2 - h1) h1 = 421.48 + 0.2 × 1050.01 = 631.48 kJ/kg, 1Q2 = 0.071(1471.49 - 631.48) = 59.65 kJ h2 = 1471.49 kJ/kg Sonntag, Borgnakke and van Wylen 8.43 An insulated cylinder fitted with a piston contains 0.1 kg of water at 100°C, 90% quality. The piston is moved, compressing the water until it reaches a pressure of 1.2 MPa. How much work is required in the process? Solution: C.V. Water in cylinder. Energy Eq.5.11: 1Q2 = 0 = m(u2 - u1) + 1W2 Entropy Eq.8.3: m(s2 − s1) = ∫ dQ/T = 0 P State 1: 100°C, x1 = 0.90: Table B.1.1, s1 = 1.3068 + 0.90×6.048 (assume reversible) T 2 2 1 1 v = 6.7500 kJ/kg K u1 = 418.91 + 0.9 × 2087.58 = 2297.7 kJ/kg State 2: Given by (P, s) B.1.3 s2 = s1 = 6.7500 P2 = 1.2 MPa T2 = 232.3°C ⇒ u = 2672.9 2 1W2 = -m(u2 – u1) = -0.1(2672.9 - 2297.7) = -37.5 kJ s Sonntag, Borgnakke and van Wylen 8.44 Compression and heat transfer brings R-134a in a piston/cylinder from 500 kPa, 50oC to saturated vapor in an isothermal process. Find the specific heat transfer and the specific work. Solution: m = constant Energy Eq.5.11: u2 − u1 = 1q2 − 1w2 Entropy Eq.8.3: s2 - s1= ∫ dq/T = 1q2 /T Process: T = C and assume reversible State 1: Table B.5.2: u1 = 415.91 kJ/kg, s1 = 1.827 kJ/kg K ⇒ 1q2 = T (s2 - s1) T P 2 1 2 T State 2: Table B.5.1 u2 = 403.98 kJ/kg, v s2 = 1.7088 kJ/kg K 1q2 = (273 + 50) × (1.7088 – 1.827) = -38.18 kJ/kg 1w2 = 1q2 + u1 - u2 = -38.18 + 415.91 – 403.98 = -26.25 kJ/kg 1 s Sonntag, Borgnakke and van Wylen 8.45 One kilogram of water at 300°C expands against a piston in a cylinder until it reaches ambient pressure, 100 kPa, at which point the water has a quality of 90.2%. It may be assumed that the expansion is reversible and adiabatic. What was the initial pressure in the cylinder and how much work is done by the water? Solution: C.V. Water. Process: Rev., Q = 0 Energy Eq.5.11: m(u2 − u1) = 1Q2 − 1W2 = − 1W2 Entropy Eq.8.3: Process: m(s2 − s1) = ∫ dQ/T Adiabatic Q = 0 and reversible => s 2 = s1 State 2: P2 = 100 kPa, x2 = 0.902 from Table B.1.2 s2 = 1.3026 + 0.902 × 6.0568 = 6.7658 kJ/kg K u2 = 417.36 + 0.902 × 2088.7 = 2301.4 kJ/kg State 1 At T1 = 300°C, s1 = 6.7658 Find it in Table B.1.3 ⇒ P1 = 2000 kPa, u1 = 2772.6 kJ/kg From the energy equation 1W2 = m(u1 - u2) = 1(2772.6 – 2301.4) = 471.2 kJ T P 1 1 T1 2 2 v s Sonntag, Borgnakke and van Wylen 8.46 Water in a piston/cylinder at 400oC, 2000 kPa is expanded in a reversible adiabatic process. The specific work is measured to be 415.72 kJ/kg out. Find the final P and T and show the P-v and the T-s diagram for the process. Solution: C.V. Water, which is a control mass. Adiabatic so: 1q2 = 0 Energy Eq.5.11: u2 − u1 = 1q2 − 1w2 = -1w2 s2 - s1= ∫ dq/T = 0 Entropy Eq.8.3: State 1: Table B.1.3 State 2: (= since reversible) u1 = 2945.21 kJ/kg; s1 = 7.127 kJ/kg K (s, u): u2 = u1 - 1w2 = 2529.29 – 415.72 = 2529.49 kJ/kg => sat. vapor 200 kPa, T = 120.23°C T P 1 1 2 2 v s Sonntag, Borgnakke and van Wylen 8.47 A piston/cylinder has 2 kg ammonia at 50°C, 100 kPa which is compressed to 1000 kPa. The process happens so slowly that the temperature is constant. Find the heat transfer and work for the process assuming it to be reversible. Solution: CV : NH3 Control Mass Energy Eq.5.11: m(u2 − u1) = 1Q2 − 1W2 ; Entropy Eq.8.3: m(s2 − s1) = ∫ dQ/T Process: T = constant and assume reversible process 1: (T,P), Table B.2.2: v1 = 1.5658 m3/kg, u1 = 1424.7 kJ/kg, s1 = 6.4943 kJ/kg K 2: (T,P), Table B.2.2: v2 = 0.1450 m3/kg, u2 = 1391.3 kJ/kg, s2 = 5.2654 kJ/kg K P T 1 2 1 2 T v s From the entropy equation (2nd law) 1Q2 = mT(s2 − s1) = 2 × 323.15 (5.2654 - 6.4943) = -794.2 kJ From the energy equation 1W2 = 1Q2 - m(u2 - u1) = -794.24 - 2(1391.3 - 1424.62) = -727.6 kJ Sonntag, Borgnakke and van Wylen 8.48 A piston cylinder has R-134a at –20oC, 100 kPa which is compressed to 500 kPa in a reversible adiabatic process. Find the final temperature and the specific work. Solution: C.V. R-134a, Control mass of unknown size, adiabatic 1q2 = 0 Energy Eq.5.11: u2 − u1 = 1q2 − 1w2 = - 1w2 s2 − s1 = ∫ dq/T Entropy Eq.8.3: Process: Adiabatic and reversible => s 2 = s1 State 1: (T, P) B.5.2 u1 = 367.36 kJ/kg, State 2: (P, s) P2 = 500 kPa, B.5.2 s1 = 1.7665 kJ/kg K s2 = s1 = 1.7665 kJ/kg K very close at 30oC u2 = 398.99 kJ/kg 1w2 = u2 - u1 = 367.36 – 398.99 = -31.63 kJ/kg P T 2 2 1 1 v s Sonntag, Borgnakke and van Wylen 8.49 A closed tank, V = 10 L, containing 5 kg of water initially at 25°C, is heated to 175°C by a heat pump that is receiving heat from the surroundings at 25°C. Assume that this process is reversible. Find the heat transfer to the water and the change in entropy. Solution: C.V.: Water from state 1 to state 2. Energy Eq.5.11: m(u2 − u1) = 1Q2 − 1W2 Entropy Eq.8.3: m(s2 − s1) = ∫ dQ/T Process: constant volume (reversible isometric) so State 1: v1 = V/m = 0.002 1W2 = 0 from Table B.1.1 x1 = (0.002 - 0.001003)/43.358 = 0.000023 u1 = 104.86 + 0.000023 × 2304.9 = 104.93 kJ/kg s1 = 0.3673 + 0.000023 × 8.1905 = 0.36759 kJ/kg K Continuity eq. (same mass) and V = C fixes v2 State 2: T2, v2 = v1 so from Table B.1.1 x2 = (0.002 - 0.001121)/0.21568 = 0.004075 u2 = 740.16 + 0.004075 × 1840.03 = 747.67 kJ/kg s2 = 2.0909 + 0.004075 × 4.5347 = 2.1094 kJ/kg K Energy eq. has W = 0, thus provides heat transfer as 1Q2 = m(u2 - u1) = 3213.7 kJ The entropy change becomes m(s2 - s1) = 5(2.1094 – 0.36759) = 8.709 kJ/K P T 2 1 2 T 1 v s Notice we do not perform the integration ∫ dQ/T to find change in s as the equation for the dQ as a function of T is not known. Sonntag, Borgnakke and van Wylen 8.50 A cylinder containing R-134a at 10°C, 150 kPa, has an initial volume of 20 L. A piston compresses the R-134a in a reversible, isothermal process until it reaches the saturated vapor state. Calculate the required work and heat transfer to accomplish this process. Solution: C.V. R-134a. Cont.Eq.: m2 = m1 = m ; Energy Eq.:5.11 Entropy Eq.8.3: m(u2 − u1) = 1Q2 − 1W2 m(s2 − s1) = ∫ dQ/T Process: T = constant, reversible State 1: (T, P) Table B.5.2 u1 = 388.36 kJ/kg, s1 = 1.822 kJ/kg K m = V/v1 = 0.02/0.148283 = 0.1349 kg State 2: (10°C, sat. vapor) Table B.5.1 u2 = 383.67 kJ/kg, T P 2 1 s2 = 1.7218 kJ/kg K 2 1 T v As T is constant we can find Q by integration as 1Q2 = ⌠Tds = mT(s2 - s1) = 0.1349 × 283.15 × (1.7218 - 1.822) = -3.83 kJ ⌡ The work is then from the energy equation 1W2 = m(u1 - u2) + 1Q2 = 0.1349 × (388.36 - 383.67) - 3.83 = -3.197 kJ s Sonntag, Borgnakke and van Wylen 8.51 A heavily-insulated cylinder fitted with a frictionless piston, as shown in Fig. P8.51 contains ammonia at 5°C, 92.9% quality, at which point the volume is 200 L. The external force on the piston is now increased slowly, compressing the ammonia until its temperature reaches 50°C. How much work is done by the ammonia during this process? Solution: C.V. ammonia in cylinder, insulated so assume adiabatic Q = 0. Cont.Eq.: m2 = m1 = m ; m(u2 − u1) = 1Q2 − 1W2 Energy Eq.5.11: m(s2 − s1) = ∫ dQ/T Entropy Eq.8.3: State 1: T1 = 5oC, x1 = 0.929, V1 = 200 L = 0.2 m3 Table B.2.1 saturated vapor, P1 = Pg = 515.9 kPa v1 = vf + x1vfg = 0.001583 + 0.929 × 0.2414 = 0.2258 m3/kg, u1 = uf + x1ufg = 202.8 + 0.929 × 1119.2 = 1242.5 kJ/kg s1 = sf + x1sfg = 0.7951 + 0.929 × 4.44715 = 4.9491 kJ/kg K, m1 = V1/v1 = 0.2 / 0.2258 = 0.886 kg Process: 1 2 Adiabatic 1Q2 = 0 & Reversible => s1 = s2 State 2: T2 = 50oC, s2 = s1 = 4.9491 kJ/kg K superheated vapor, interpolate in Table B.2.2 P2 = 1600 kPa, u2 = 1364.9 kJ/kg => T P 2 2 T1 1 1 v Energy equation gives the work as 1W2 = m(u1 - u2) = 0.886 ( 1242.5 – 1364.9) = −108.4 kJ s Sonntag, Borgnakke and van Wylen 8.52 A piston/cylinder has 2 kg water at 1000 kPa, 250°C which is now cooled with a constant loading on the piston. This isobaric process ends when the water has reached a state of saturated liquid. Find the work and heat transfer and sketch the process in both a P-v and a T-s diagram. Solution: C.V. H2O Energy Eq.5.11: m(u2 − u1) = 1Q2 − 1W2 Entropy Eq.8.3: m(s2 − s1) = ∫ dQ/T Process: P = C => W = ∫ P dV = P(V2 − V1) v1= 0.23268 m3/kg, s1= 6.9246 kJ/kg K, u1 = 2709.91 kJ/kg State 1: B.1.3 v2 = 0.001127 m3/kg, s2 = 2.1386 kJ/kg K, u2 = 761.67 kJ/kg From the process equation State 2: B.1.2 1W2 = m P (v2 − v1) = 2 × 1000 (0.001127 – 0.23268) = -463.1 kJ From the energy equation we get 1Q2 = m(u2 − u1) + 1W2 = 2 (761.67 – 2709.91) – 463.1 = -4359.6 kJ P 2 T 1 1 2 v s Sonntag, Borgnakke and van Wylen 8.53 Water at 1000 kPa, 250°C is brought to saturated vapor in a piston/cylinder with an isothermal process. Find the specific work and heat transfer. Estimate the specific work from the area in the P-v diagram and compare it to the correct value. Solution: Continuity Eq.: m2 = m1 = m ; m(u2 − u1) = 1Q2 − 1W2 Energy Eq.:5.11 Entropy Eq.8.3: m(s2 − s1) = ∫ dQ/T Process: T = constant, reversible State 1: Table B.1.3: v1 = 0.23268 m3/kg; State 2: (T, x) Table B.1.1 u1 = 2709.91 kJ/kg; s1 = 6.9246 kJ/kg K P2 = 3973 kPa v2 = 0.05013 m3/kg, u2 = 2602.37 kJ/kg, s2 = 6.0729 kJ/kg K T P 2 1 2 1 T v s From the entropy equation 1q2 = ∫ T ds = T(s2 − s1) = (250 + 273) (6.0729 – 6.9246) = -445.6 kJ/kg From the energy equation 1w2 = 1q2 + u1 − u2 = -445.6 + 2709.91 – 2602.37 = -338 kJ/kg Estimation of the work term from the area in the P-v diagram 1 1 1w2 area ≅ 2 (P1+P2)(v2 − v1) = 2(1000 + 3973)(0.05013 – 0.23268) = –454 kJ/kg Not extremely accurate estimate; P-v curve not linear more like Pv = constant as curve has positive curvature the linear variation over-estimates area. Sonntag, Borgnakke and van Wylen 8.54 Water at 1000 kPa, 250°C is brought to saturated vapor in a rigid container, shown in Fig. P8.54. Find the final T and the specific heat transfer in this isometric process. Solution: Energy Eq.5.11: u2 − u1 = 1q2 - 1w2 Entropy Eq.8.3: s2 − s1 = ∫ dq/T Process: v = constant => u1 = 2709.91 kJ/kg, State 1: (T, P) Table B.1.3 State 2: x = 1 and v2 = v1 1w2 = 0 v1 = 0.23268 m3/kg so from Table B.1.1 we see P2 ≅ 800 kPa T2 = 170 + 5 × (0.23268 – 0.24283)/(0.2168 – 0.24283) = 170 + 5 × 0.38993 = 171.95°C u2 = 2576.46 + 0.38993 × (2580.19 – 2576.46) = 2577.9 kJ/kg From the energy equation 1q2 = u2 − u1 = 2577.9 – 2709.91 = −132 kJ/kg T P 1 1 v=C 2 2 v s Notice to get 1q2 = ∫ T ds we must know the function T(s) which we do not readily have for this process. Sonntag, Borgnakke and van Wylen 8.55 Estimate the specific heat transfer from the area in the T-s diagram and compare it to the correct value for the states and process in Problem 8.54. Solution: Energy Eq.5.11: u2 − u1 = 1q2 - 1w2 Entropy Eq.8.3: s2 − s1 = ∫ dq/T Process: v = constant => 1w2 = 0 State 1: (T, P) Table B.1.3 u1 = 2709.91 kJ/kg, v1 = 0.23268 m3/kg, s1 = 6.9246 kJ/kg K State 2: x = 1 and v2 = v1 so from Table B.1.1 we see P2 ≅ 800 kPa T2 = 170 + 5 × (0.23268 – 0.24283)/(0.2168 – 0.24283) = 170 + 5 × 0.38993 = 171.95°C u2 = 2576.46 + 0.38993 × (2580.19 – 2576.46) = 2577.9 kJ/kg s2 = 6.6663 + 0.38993 (6.6256 – 6.6663) = 6.6504 kJ/kg K From the energy equation 1q2 actual = u2 − u1 = 2577.9 – 2709.91 = −132 kJ/kg Assume a linear variation of T versus s. 1q2 = ∫ T ds = area ≅ 1 2 (T1 + T2)(s2 − s1) 1 = 2 (171.95 + (2 × 273.15) + 250)(6.6504 – 6.9246) = -132.74 kJ/kg very close i.e. the v = C curve is close to a straight line in the T-s diagram. Look at the constant v curves in Fig. E.1. In the two-phase region they curve slightly and more so in the region above the critical point. T P v=C 1 1 2 2 v s Sonntag, Borgnakke and van Wylen 8.56 Water at 1000 kPa, 250°C is brought to saturated vapor in a piston/cylinder with an isobaric process. Find the specific work and heat transfer. Estimate the specific heat transfer from the area in the T-s diagram and compare it to the correct value. Solution: C.V. H2O Energy Eq.5.11: u2 − u1 = 1q2 − 1w2 Entropy Eq.8.3: s2 − s1 = ∫ dq/T Process: P=C 1: B1.3 v1= 0.23268 m3/kg, 2: B1.3 v2 = 0.19444 m3/kg, s2 = 6.5864 kJ/kg K, u2 = 2583.64 kJ/kg, T2 = 179.91°C w = ∫ P dv = P(v2 − v1) => s1= 6.9246 kJ/kgK, u1 = 2709.91 kJ/kg From the process equation 1w2 = P (v2 − v1) = 1000 (0.1944 – 0.23268) = -38.28 kJ/kg From the energy equation 1q2 = u2 − u1 + 1w2 = 2583.64 – 2709.91 – 38.28 = -164.55 kJ/kg Now estimate the heat transfer from the T-s diagram. 1 1q2 = ∫ T ds = AREA ≅ 2 (T1 + T2)(s2 − s1) 1 = 2 (250 + 179.91 + 2 × 273.15)(6.5864 – 6.9246) = 488.105 × (-0.3382) = -165.1 kJ/kg very close approximation. The P = C curve in the T-s diagram is nearly a straight line. Look at the constant P curves on Fig.E.1. Up over the critical point they curve significantly. P T 2 1 1 2 v s Sonntag, Borgnakke and van Wylen 8.57 A heavily insulated cylinder/piston contains ammonia at 1200 kPa, 60°C. The piston is moved, expanding the ammonia in a reversible process until the temperature is −20°C. During the process 600 kJ of work is given out by the ammonia. What was the initial volume of the cylinder? C.V. ammonia. Control mass with no heat transfer. v1 = 0.1238 m3/kg, s1 = 5.2357 kJ/kg K State 1: Table B.2.2 u1 = h - Pv = 1553.3 - 1200×0.1238 = 1404.9 kJ/kg m(s2 − s1) = ∫ dQ/T + 1S2 gen Entropy Eq.: Process: reversible (1S2 gen = 0) and adiabatic (dQ = 0) P => T 1 1 2 2 v s State 2: T2, s2 ⇒ x2 = (5.2357 - 0.3657)/5.2498 = 0.928 u2 = 88.76 + 0.928×1210.7 = 1211.95 kJ/kg 1Q2 = 0 = m(u2 - u1) + 1W2 = m(1211.95 - 1404.9) + 600 ⇒ m = 3.110 kg V1 = mv1 = 3.11 × 0.1238 = 0.385 m3 s 2 = s1 Sonntag, Borgnakke and van Wylen 8.58 Water at 1000 kPa, 250°C is brought to saturated vapor in a piston/cylinder with an adiabatic process. Find the final T and the specific work. Estimate the specific work from the area in the P-v diagram and compare it to the correct value. Solution: C.V. Water, which is a control mass with unknown size. Energy Eq.5.11: u2 – u1 = 0 – 1w2 Entropy Eq.8.3: s2 – s1 = ∫ dq/T = 0 Process: Adiabatic State 1: Table B.1.3 v1 = 0.23268 m3/kg, u1 = 2709.91 kJ/kg, s1 = 6.9246 kJ/kg K State 2: Table B.1.1 x = 1 and s2 = s1 = 6.9246 kJ/kg K 1q2 = 0 and reversible as used above => T2 ≅ 140.56°C, P2 ≅ 367.34 kPa, v2 = 0.50187 m3/kg, u2 ≅ 2550.56 kJ/kg From the energy equation 1w2 = u1 – u2 = 2709.91 – 2550.56 = 159.35 kJ/kg Now estimate the work term from the area in the P-v diagram 1 1w2 ≅ 2 (P1 + P2)(v2 − v1) 1 = 2 (1000 + 367.34)(0.50187 – 0.23268) = 184 kJ/kg The s = constant curve is not a straight line in the the P-v diagram, notice the straight line overestimates the area slightly. T P 1 1 2 2 v s Sonntag, Borgnakke and van Wylen 8.59 A rigid, insulated vessel contains superheated vapor steam at 3 MPa, 400°C. A valve on the vessel is opened, allowing steam to escape. The overall process is irreversible, but the steam remaining inside the vessel goes through a reversible adiabatic expansion. Determine the fraction of steam that has escaped, when the final state inside is saturated vapor. C.V.: steam remaining inside tank. Rev. & Adiabatic (inside only) Cont.Eq.: m2 = m1 = m ; Entropy Eq.: P Energy Eq.: m(u2 − u1) = 1Q2 − 1W2 m(s2 − s1) = ∫ dQ/T + 1S2 gen T 1 1 C.V. m2 2 2 v Rev ( 1S2 gen = 0) Adiabatic ( Q = 0) => s s2 = s1 = 6.9212 = sG at T2 ⇒ T2 = 141°C, v2 = vg at T2 = 0.4972 m3/kg m2 v1 me m1-m2 0.09936 = m = 1 - m = 1 - v = 1 - 0.4972 = 0.80 m1 1 1 2 Sonntag, Borgnakke and van Wylen 8.60 A piston/cylinder contains 2 kg water at 200°C, 10 MPa. The piston is slowly moved to expand the water in an isothermal process to a pressure of 200 kPa. Any heat transfer takes place with an ambient at 200°C and the whole process may be assumed reversible. Sketch the process in a P-V diagram and calculate both the heat transfer and the total work. Solution: C.V. Water. Energy Eq.5.11: m(u2 − u1) = 1Q2 − 1W2 Entropy Eq.8.3: m(s2 − s1) = ∫ dQ/T = 1Q2 / T Process: T = C and reversible as used in entropy equation State 1: Table B.1.4 : v1 = 0.001148 m3/kg, u1 = 844.49 kJ/kg, s1 = 2.3178 kJ/kg K, V1 = mv1 = 0.0023 m3 State 2: Table B.1.3 : v2 = 1.08034 m3/kg, u2 = 2654.4 kJ/kg s2 = 7.5066 kJ/kg K V2 = mv2 = 2.1607 m3, T P 1 2 1 2 T v s From the entropy equation and the process equation 1Q2 = mT(s2 − s1) = 2 × 473.15 (7.5066 - 2.3178) = 4910 kJ From the energy equation 1W2 = 1Q2 - m(u2 - u1) = 1290.3 kJ Sonntag, Borgnakke and van Wylen Entropy generation 8.61 One kg water at 500oC and 1 kg saturated water vapor both at 200 kPa are mixed in a constant pressure and adiabatic process. Find the final temperature and the entropy generation for the process. Solution: Continuity Eq.: m2 − mA – mB = 0 Energy Eq.5.11: m2u2 − mAuA – mBuB = –1W2 Entropy Eq.8.14: m2s2 − mAsA – mBsB = ∫ dQ/T + 1S2 gen Process: P = Constant => 1W2 = ∫ PdV = P(V2 - V1) Q=0 Substitute the work term into the energy equation and rearrange to get m2u2 + P2V2 = m2h2 = mAuA + mBuB+ PV1 = mAhA + mBhB where the last rewrite used PV1 = PVA + PVB. State A1: Table B.1.3 hA= 3487.03 kJ/kg, sA= 8.5132 kJ/kg K State B1: Table B.1.2 hB = 2706.63 kJ/kg, sB= 7.1271 kJ/kg K Energy equation gives: mA mB 1 1 h2 = m hA + m hB = 2 3487.03 + 2 2706.63 = 3096.83 2 2 P2, h2 = 3096.83 kJ/kg => s2 = 7.9328 kJ/kg K; With the zero heat transfer we have State 2: T2 = 312.2°C 1S2 gen = m2s2 − mAsA – mBsB = 2 × 7.9328 – 1 × 8.5132 – 1 × 7.1271 = 0.225 kJ/K Sonntag, Borgnakke and van Wylen 8.62 The unrestrained expansion of the reactor water in Problem 5.48 has a final state in the two-phase region. Find the entropy generated in the process. A water-filled reactor with volume of 1 m3 is at 20 MPa, 360°C and placed inside a containment room as shown in Fig. P5.48. The room is well insulated and initially evacuated. Due to a failure, the reactor ruptures and the water fills the containment room. Find the minimum room volume so the final pressure does not exceed 200 kPa. Solution: C.V.: Containment room and reactor. Mass: m2 = m1 = Vreactor/v1 = 1/0.001823 = 548.5 kg Energy Eq.5.11: m(u2 - u1) = 1Q2 - 1W2 = 0 - 0 = 0 m(s2 – s1) = ∫ dQ/T + 1S2 gen State 1: (T, P) Table B.1.4 u1 = 1702.8 kJ/kg, s1 = 3.877 Entropy Eq.8.14: Energy equation implies State 2: u2 = u1 = 1702.8 kJ/kg P2 = 200 kPa, u2 < ug => Two-phase Table B.1.2 x2 = (u2 - uf)/ ufg = (1702.8 – 504.47)/2025.02 = 0.59176 v2 = 0.001061 + 0.59176 × 0.88467 = 0.52457 m3/kg s2 = sf + x2sfg = 1.53 + 0.59176 × 5.597 = 4.8421 kJ/kg K V2 = m2 v2 = 548.5 ×0.52457 = 287.7 m3 From the entropy equation the generation is 1S2 gen = m(s2 – s1) = 548.5 (4.8421 – 3.877) = 529.4 kJ/K T P 1 1 200 200 kPa 2 v 2 u = const s Entropy is generated due to the unrestrained expansion. No work was taken out as the volume goes up. Sonntag, Borgnakke and van Wylen 8.63 A mass and atmosphere loaded piston/cylinder contains 2 kg of water at 5 MPa, 100°C. Heat is added from a reservoir at 700°C to the water until it reaches 700°C. Find the work, heat transfer, and total entropy production for the system and surroundings. Solution: C.V. Water out to surroundings at 700°C. This is a control mass. Energy Eq.5.11: U2 - U1 = 1Q2 - 1W2 Entropy Eq.8.14: m(s2 - s1) = ⌠dQ/T + 1S2 gen = 1Q2/Tres + 1S2 gen ⌡ Process: P = constant so State 1: Table B.1.4: 1W2 = P(V2 - V1) = mP(v2 - v1) h1 = 422.72 kJ/kg, u1 = 417.52 kJ/kg, s1 = 1.303 kJ/kg K, v1 = 0.00104 m3/kg State 2: Table B.1.3: h2 = 3900.1 kJ/kg, u2 = 3457.6 kJ/kg, s2 = 7.5122 kJ/kg K, v2 = 0.08849 m3/kg P 1 T 2 2 1 v s Work is found from the process (area in P-V diagram) 1W2 = mP(v2 - v1) = 2 × 5000(0.08849 – 0.00104) = 874.6 kJ The heat transfer from the energy equation is 1Q2 = U2 - U1 + 1W2 = m(u2 - u1) + mP(v2 - v1) = m(h2 - h1) 1Q2 = 2(3900.1 - 422.72) = 6954.76 kJ Entropy generation from entropy equation (or Eq.8.18) 1S2 gen = m(s2 - s1) - 1Q2/Tres = 2(7.5122 - 1.303) - 6954/973 = 5.27 kJ/K Sonntag, Borgnakke and van Wylen 8.64 Ammonia is contained in a rigid sealed tank unknown quality at 0oC. When heated in boiling water to 100oC its pressure reaches 1200 kPa. Find the initial quality, the heat transfer to the ammonia and the total entropy generation. Solution: C.V. Ammonia, which is a control mass of constant volume. Energy Eq.5.11: u2 - u1 = 1q2 - 1w2 Entropy Eq.8.14: s2 – s1 = ∫ dq/T + 1s2 gen State 2: 1200 kPa, 100oC => s2 = 5.5325 kJ/kg K, State 1: v1 = v2 => Table B.2.2 v2 = 0.14347 m3/kg, u2 = 1485.8 kJ/kg Table B.2.1 x1 = (0.14347 – 0.001566)/0.28763 = 0.49336 u1 = 741.28 kJ/kg, s1 = 0.7114 + x1 × 4.6195 = 2.9905 kJ/kg K Process: V = constant => 1w2 = 0 1q2 = (u2 - u1) = 1485.8 – 741.28 = 744.52 kJ/kg To get the total entropy generation take the C.V out to the water at 100oC. 1s2 gen = s2 – s1 - 1q2/T = 5.5325 – 2.9905 – 744.52/373.15 = 0.547 kJ/kg K T P 2 2 v=C 1 1 v s Sonntag, Borgnakke and van Wylen 8.65 An insulated cylinder/piston contains R-134a at 1 MPa, 50°C, with a volume of 100 L. The R-134a expands, moving the piston until the pressure in the cylinder has dropped to 100 kPa. It is claimed that the R-134a does 190 kJ of work against the piston during the process. Is that possible? Solution: C.V. R-134a in cylinder. Insulated so assume Q = 0. v1 = 0.02185 m3/kg, u1 = 409.39 kJ/kg, s1 = 1.7494 kJ/kg K, m = V1/v1 = 0.1/0.02185 = 4.577 kg State 1: Table B.5.2, Energy Eq.5.11: m(u2 - u1) = 1Q2 - 1W2 = 0 - 190 ⇒ / u2 = u1 − 1W2/m = 367.89 kJ/kg State 2: P2 , u2 ⇒ Table B.5.2: T2 = -19.25°C ; s2 = 1.7689 kJ/kg K m(s2 - s1) = ⌡dQ/T + 1S2,gen = 1S2,gen ⌠ Entropy Eq.8.14: 1S2,gen = m(s2 - s1) = 0.0893 kJ/K This is possible since 1S2,gen > 0 / P s=C 1 T 1 2 2 v s Sonntag, Borgnakke and van Wylen 8.66 A piece of hot metal should be cooled rapidly (quenched) to 25°C, which requires removal of 1000 kJ from the metal. The cold space that absorbs the energy could be one of three possibilities: (1) Submerge the metal into a bath of liquid water and ice, thus melting the ice. (2) Let saturated liquid R-22 at −20°C absorb the energy so that it becomes saturated vapor. (3) Absorb the energy by vaporizing liquid nitrogen at 101.3 kPa pressure. a. Calculate the change of entropy of the cooling media for each of the three cases. b. Discuss the significance of the results. Solution: a) Melting or boiling at const P & T 1Q2 = m(u2 - u1) + Pm(v2 - v1) = m(h2 - h1) 1000 1) Ice melting at 0°C , Table B.1.5: m = 1Q2 /hig = 333.41 = 2.9993 kg ∆SH 2O = msig = 2.9993(1.221) = 3.662 kJ/K 1000 2) R-22 boiling at -20°C, Table B.4.1: m = 1Q2 /hfg = 220.327 = 4.539 kg ∆SR-22 = msfg = 4.539(0.8703) = 3.950 kJ/K 1000 3) N2 boiling at 101.3 kPa, Table B.6.1: m = 1Q2 /hfg = 198.842 = 5.029 kg ∆SN = msfg = 5.029(2.5708) = 12.929 kJ/K 2 b) The larger the ∆(1/T) through which the Q is transferred, the larger the ∆S. Sonntag, Borgnakke and van Wylen 8.67 A piston cylinder has 2.5 kg ammonia at 50 kPa, -20oC. Now it is heated to 50oC at constant pressure through the bottom of the cylinder from external hot gas at 200oC. Find the heat transfer to the ammonia and the total entropy generation. Solution: C.V. Ammonia plus space out to the hot gas. Energy Eq.5.11: m(u2 - u1) = 1Q2 - 1W2 Entropy Eq.8.14: m(s2 - s1) = ⌠dQ/T + 1S2,gen = 1Q2/ Tgas + 1S2 gen ⌡ Process: P=C State 1: Table B.2.2 => 1W2 = Pm(v2 - v1) v1 = 2.4463 m3/kg, h1 = 1434.6 kJ/kg, s1 = 6.3187 kJ/kg K State 2: Table B.2.2 v2 = 3.1435 m3/kg, h2 = 1583.5 kJ/kg, s2 = 6.8379 kJ/kg K Substitute the work into the energy equation and solve for the heat transfer 1Q2 = m(h2 - h1) = 2.5 (1583.5 – 1434.6) = 372.25 kJ 1S2 gen = m(s2 – s1) - 1Q2/Tgas = 2.5 (6.8379 – 6.3187) – 372.25/473.15 = 0.511 kJ/K P T 1 2 2 1 v s Remark: This is an internally reversible- externally irreversible process. The s is generated in the space between the 200oC gas and the ammonia. If there are any ∆T in the ammonia then it is also internally irreversible. Sonntag, Borgnakke and van Wylen 8.68 A cylinder fitted with a movable piston contains water at 3 MPa, 50% quality, at which point the volume is 20 L. The water now expands to 1.2 MPa as a result of receiving 600 kJ of heat from a large source at 300°C. It is claimed that the water does 124 kJ of work during this process. Is this possible? Solution: C.V.: H2O in Cylinder State 1: 3 MPa, x1 = 0.5, Table B.1.2: T1 = 233.9oC v1 = vf + x1vfg = 0.001216 + 0.5×0.06546 = 0.033948 m3/kg u1 = uf + x1ufg = 1804.5 kJ/kg, s1 = sf + x1sfg = 4.4162 kJ/kg-K m1 = V1/v1 = 0.02 / 0.033948 = 0.589 kg 1st Law: 1 2, m(u2 − u1) = 1Q2 − 1W2 ; 1Q2 = 600 kJ, 1W2 = 124 kJ ? Now solve for u2 u2 = 1804.5 + (600 - 124)/0.589 = 2612.6 kJ/kg State 2: P2 = 1.2 MPa : u2 = 2612.6 kJ/kg Table B.1.3 T2 ≅ 200oC, s2 = 6.5898 kJ/kgK 2nd Law Eq.8.18: Qcv ∆Snet = m(s2 − s1) - T ; H TH = 300oC, QCV = 1Q2 600 ∆Snet = 0.589 (6.5898 – 4.4162) – 300 + 273 = 0.2335 kJ/K > 0; Process is possible T P 1 2 2 1 T1 v s Sonntag, Borgnakke and van Wylen 8.69 A piston cylinder loaded so it gives constant pressure has 0.75 kg saturated vapor water at 200 kPa. It is now cooled so the volume becomes half the initial volume by heat transfer to the ambient at 20oC. Find the work, the heat transfer and the total entropy generation. Solution: Continuity Eq.: m2 − m1 = 0 Energy Eq.5.11: m(u2 − u1) = 1Q2 − 1W2 Entropy Eq.8.14: m(s2 − s1) = ∫ dQ/T + 1S2 gen Process: P=C => 1W2 = ∫ PdV = mP(v2 − v1) 1Q2 = m(u2 − u1) + 1W2 = m(h2 − h1) State 1: v1 = 0.88573 m3/kg, h1 = 2706.63 kJ/kg, s1 = 7.1271 kJ/kg K State 2: P2, v2 = v1/2 = 0.444286 m3/kg => Table B.1.2 x2 = (0.444286 − 0.001061)/0.88467 = 0.501 h2 = 504.68 + x2× 2201.96 = 1607.86 kJ/kg s2 = 1.53+ x2× 5.5970 = 4.3341 kJ/kg K 1W2 = 0.75 × 200(0.444286 − 0.88573) = -66.22 kJ 1Q2 = 0.75(1607.86 − 2706.63) = -824.1 kJ 1S2 gen = m(s2 - s1) − 1Q2/T = 0.75(4.3341 – 7.1271) – (-824.1/293.15) = -2.09475 + 2.81119 = 0.716 kJ/K Notice: The process is externally irreversible (T receiving Q is not T1) P T 2 2 1 v 1 s Sonntag, Borgnakke and van Wylen 8.70 A piston/cylinder contains 1 kg water at 150 kPa, 20°C. The piston is loaded so pressure is linear in volume. Heat is added from a 600°C source until the water is at 1 MPa, 500°C. Find the heat transfer and the total change in entropy. Solution: CV H2O out to the source, both 1Q2 and 1W2 Energy Eq.5.11: m(u2 − u1) = 1Q2 − 1W2 Entropy Eq.8.14: m(s2 - s1) = 1Q2 / TSOURCE + 1S2 gen Process: => P = A + BV 1W2 = ∫ P dV = ½ (P1 + P2 ) (V2 - V1) State 1: B.1.1 Compressed liquid use saturated liquid at same T: v1 = 0.001002 m3/kg; u1 = 83.94 kJ/kg; s1 = 0.2966 kJ/kg K State 2: Table B.1.3 sup. vap. P v2 = 0.35411 m3/kg u2 = 3124.3 kJ/kg; T 2 2 1 s2 = 7.7621 kJ/kg K 1 v 1W2 = ½ (1000 + 150) 1 (0.35411 - 0.001002) = 203 kJ 1Q2 = 1(3124.3 - 83.94) + 203 = 3243.4 kJ m(s2 - s1) = 1(7.7621 - 0.2968) = 7.4655 kJ/K 1Q2 / Tsource = 3.7146 kJ/K (for source Q = -1Q2 recall Eq.8.18) 1S2 gen = m(s2 - s1) − 1Q2 / TSOURCE = ∆Stotal = ∆SH2O + ∆Ssource = 7.4655 - 3.7146 = 3.751 kJ/K Remark: This is an external irreversible process (delta T to the source) P2 P1 s Sonntag, Borgnakke and van Wylen 8.71 A piston/cylinder has ammonia at 2000 kPa, 80oC with a volume of 0.1 m3. The piston is loaded with a linear spring and outside ambient is at 20oC, shown in Fig. P8.71. The ammonia now cools down to 20oC at which point it has a quality of 10%. Find the work, the heat transfer and the total entropy generation in the process. CV Ammonia out to the ambient, both 1Q2 and 1W2 Energy Eq.5.11: m(u2 − u1) = 1Q2 − 1W2 Entropy Eq.8.14: m(s2 - s1) = 1Q2 / Tambient + 1S2 gen Process: => P = A + BV 1W2 = ∫ P dV = ½ m(P1 + P2 ) (v2 - v1) State 1: Table B.2.2 v1 = 0.07595 m3/kg, u1 = 1421.6 kJ/kg, s1 = 5.0707 kJ/kg K m = V1/v1 = 0.1/0.07595 = 1.31665 kg State 2: Table B.2.1 v2= 0.001638 + 0.1×0.14758 = 0.016396 m3/kg u2 = 272.89 + 0.1×1059.3 =378.82 kJ/kg s2 = 1.0408 + 0.1×4.0452 = 1.44532 kJ/kg K 1W2 = ½ m(P1 + P2 )( v2 - v1) = ½ ×1.31665 (2000 + 857.5)( 0.016396 – 0.07595) = - 112 kJ Q2 = m(u2 − u1) + 1W2 = 1.31665 (378.82 – 1421.6) –112 1 = - 1484.98 kJ 1S2 gen = m(s2 − s1) – (1Q2/ Tamb) –1484.98 = 1.31665 (1.44532 – 5.0707) – 293.15 = – 4.77336 + 5.0656 = 0.292 kJ/k P T 1 1 2 P2 2 v s Sonntag, Borgnakke and van Wylen 8.72 A cylinder/piston contains water at 200 kPa, 200°C with a volume of 20 L. The piston is moved slowly, compressing the water to a pressure of 800 kPa. The loading on the piston is such that the product PV is a constant. Assuming that the room temperature is 20°C, show that this process does not violate the second law. Solution: C.V.: Water + cylinder out to room at 20°C Energy Eq.5.11: m(u2 - u1) = 1Q2 − 1W2 Entropy Eq.8.14: m(s2 - s1) = 1Q2 / Troom + 1S2 gen Process: PV = constant = Pmv ⇒ v2 = P1v1/P2 1w2 = ⌠Pdv = P1v1 ln(v2/v1) ⌡ v1 = 1.0803 m3/kg, u1 = 2654.4 kJ/kg, s1 = 7.5066 kJ/kg K State 1: Table B.1.3, State 2: P2 , v2 = P1v1/P2 = 200 × 1.0803/800 = 0.2701 m3/kg Table B.1.3: u2 = 2655.0 kJ/kg , s2 = 6.8822 kJ/kg K 0.2701 1w2 = 200 × 1.0803 ln1.0803 = -299.5 kJ/kg 1q2 = u2 - u1 + 1w2 = 2655.0 - 2654.4 - 299.5 = -298.9 kJ/kg 1q2 1s2,gen = s2 - s1 - T room 298.9 = 6.8822 - 7.5066 + 293.15 = 0.395 kJ/kg K > 0 satisfy 2nd law. Sonntag, Borgnakke and van Wylen 8.73 One kilogram of ammonia (NH3) is contained in a spring-loaded piston/cylinder, Fig. P8.73, as saturated liquid at −20°C. Heat is added from a reservoir at 100°C until a final condition of 800 kPa, 70°C is reached. Find the work, heat transfer, and entropy generation, assuming the process is internally reversible. Solution: C.V. = NH3 out to the reservoir. Continuity Eq.: m2 = m1 = m Energy Eq.5.11: E2 - E1 = m(u2 - u1) = 1Q2 - 1W2 Entropy Eq.8.14: S2 - S1 = ⌠dQ/T + 1S2,gen = 1Q2/Tres + 1S2,gen ⌡ Process: linear in V P = A + BV => 1 1 1W2 = ⌠PdV = 2 (P1 + P2)(V2 - V1) = 2 (P1 + P2)m(v2 - v1) ⌡ State 1: Table B.2.1 P P1 = 190.08 kPa, v1 = 0.001504 m3/kg u1 = 88.76 kJ/kg, T 2 2 1 1 s1 = 0.3657 kJ/kg K P2 v State 2: Table B.2.2 sup. vapor v2 = 0.199 m3/kg, u2 = 1438.3 kJ/kg, s2 = 5.5513 kJ/kg K 1 1W2 = 2(190.08 + 800)1(0.1990 - 0.001504) = 97.768 kJ 1Q2 = m(u2 - u1) + 1W2 = 1(1438.3 - 88.76) + 97.768 = 1447.3 kJ 1447.3 1S2,gen = m(s2 - s1) - 1Q2/Tres = 1(5.5513 - 0.3657) - 373.15 = 1.307 kJ/K s Sonntag, Borgnakke and van Wylen 8.74 A piston/cylinder device keeping a constant pressure has 1 kg water at 20oC and 1 kg of water at 100oC both at 500 kPa separated by a thin membrane. The membrane is broken and the water comes to a uniform state with no external heat transfer. Find the final temperature and the entropy generation for the process. Solution: m2 − mA – mB = 0 Continuity Eq.: Energy Eq.5.11: m2u2 − mAuA – mBuB = –1W2 Entropy Eq.8.14: m2s2 − mAsA – mBsB = ∫ dQ/T + 1S2 gen Process: P = Constant => 1W2 = ∫ PdV = P(V2 - V1) Q=0 Substitute the work term into the energy equation and rearrange to get m2u2 + P2V2 = m2h2 = mAuA + mBuB+ PV1 = mAhA + mBhB where the last rewrite used PV1 = PVA + PVB. State A1: Table B.1.4 hA= 84.41 kJ/kg sA= 0.2965 kJ/kg K State B1: Table B.1.4 hB = 419.32 kJ/kg sB= 1.3065 kJ/kg K Energy equation gives: mA mB 1 1 h2 = m hA + m hB = 2 84.41 + 2 419.32 = 251.865 kJ/kg 2 2 State 2: h2 = 251.865 kJ/kg & P2 = 500 kPa from Table B.1.4 T2 = 60.085°C, s2 = 0.83184 kJ/kg K With the zero heat transfer we have 1S2 gen = m2s2 − mAsA – mBsB = 2 × 0.83184 – 1 × 0.2965 – 1 × 1.3065 = 0.0607 kJ/K Water 20 C Water 100 C F cb Sonntag, Borgnakke and van Wylen Entropy of a liquid or a solid 8.75 A piston cylinder has constant pressure of 2000 kPa with water at 20oC. It is now heated up to 100oC. Find the heat transfer and the entropy change using the steam tables. Repeat the calculation using constant heat capacity and incompressibility. Solution: C.V. Water. Constant pressure heating. Energy Eq.5.11: u2 - u1 = 1q2 − 1w2 Entropy Eq.8.3: s2 - s1 = 1q2 / TSOURCE + 1s2 gen Process: P = P1 => 1w2 = P(v2 - v1) The energy equation then gives the heat transfer as 1q2= u2 - u1 + 1w2 = h2 - h1 Steam Tables B.1.4: h1 = 85.82 kJ/kg; s1= 0.2962 kJ/kg K h2 = 420.45 kJ/kg; s2 = 1.3053 kJ/kg K 1q2= h2 - h1= -85.82 + 420.45 = 334.63 kJ/kg s2 - s1= 1.3053 – 0.2962 = 1.0091 kJ/kg K Now using values from Table A.4: Liquid water Cp = 4.18 kJ/kg K h2 - h1 ≅ Cp(T2 – T1) = 4.18 × 80 = 334.4 kJ/kg 373.15 s2 - s1 ≅ Cp ln(T2/T1) = 4.18 ln 293.15 = 1.0086 kJ/kg K Approximations are very good Sonntag, Borgnakke and van Wylen 8.76 A large slab of concrete, 5 × 8 × 0.3 m, is used as a thermal storage mass in a solar-heated house. If the slab cools overnight from 23°C to 18°C in an 18°C house, what is the net entropy change associated with this process? Solution: C.V.: Control mass concrete. V = 5 × 8 × 0.3 = 12 m3 m = ρV = 2200 × 12 = 26 400 kg Energy Eq.: Entropy Eq.: Process: m(u2 - u1) = 1Q2 - 1W2 1Q2 m(s2 - s1) = T + 1S2 gen 0 V = constant so 1W2 = 0 Use heat capacity (Table A.3) for change in u of the slab 1Q2 = mC∆T = 26400 × 0.88(-5) = -116 160 kJ Eq.8.18 provides the equivalent of total entropy generation: T2 291.2 ∆SSYST = m(s2 - s1) = mC ln T = 26400 × 0.88 ln 296.2 = -395.5 kJ/K 1 -1Q2 +116 160 ∆SSURR = T = 291.2 = +398.9 kJ/K 0 ∆SNET = -395.5 + 398.9 = +3.4 kJ/K 1Q2 = m(s2 - s1) − T = 1S2 gen 0 Sonntag, Borgnakke and van Wylen 8.77 A 4 L jug of milk at 25°C is placed in your refrigerator where it is cooled down to the refrigerators inside constant temperature of 5°C. Assume the milk has the property of liquid water and find the entropy generated in the cooling process. Solution: C.V. Jug of milk. Control mass at constant pressure. Continuity Eq.: m2 = m1 = m ; Energy Eq.5.11: m(u2 − u1) = 1Q2 − 1W2 Entropy Eq.8.14: m(s2 − s1) = ∫ dQ/T + 1S2 gen State 1: Table B.1.1: v1 ≅ vf = 0.001003 m3/kg, h = hf = 104.87 kJ/kg; sf = 0.3673 kJ/kg K State 2: Table B.1.1: h = hf = 20.98 kJ/kg, s = sf = 0.0761 kJ/kg K Process: P = constant = 101 kPa => 1W2 = mP(v2 - v1) m = V/v1 = 0.004 / 0.001003 = 3.988 kg Substitute the work into the energy equation and solve for the heat transfer 1Q2 = m(h2 − h1) = 3.988 (20.98 - 104.87) = -3.988 × 83.89 = -334.55 kJ The entropy equation gives the generation as 1S2 gen = m(s2 − s1) − 1Q2/Trefrig = 3.988 (0.0761 − 0.3673) − (−334.55 / 278.15) = − 1.1613 + 1.2028 = 0.0415 kJ/K o 5C MILK cb AIR Sonntag, Borgnakke and van Wylen 8.78 A foundry form box with 25 kg of 200°C hot sand is dumped into a bucket with 50 L water at 15°C. Assuming no heat transfer with the surroundings and no boiling away of liquid water, calculate the net entropy change for the process. Solution: C.V. Sand and water, constant pressure process msand(u2 - u1)sand + mH O(u2 - u1)H O = -P(V2 - V1) 2 2 ⇒ msand∆hsand + mH O∆hH O = 0 2 2 For this problem we could also have said that the work is nearly zero as the solid sand and the liquid water will not change volume to any measurable extent. Now we get changes in u's instead of h's. For these phases CV = CP = C which is a consequence of the incompressibility. Now the energy equation becomes msandC∆Tsand + mH OCH O∆TH O = 0 2 2 2 -3 25 × 0.8×(T2 - 200) + (50×10 /0.001001) × 4.184 × (T2 - 15) = 0 T2 = 31.2°C 304.3 304.3 ∆S = 25 × 0.8 ln473.15 + 49.95 × 4.184 ln288.15 = 2.57 kJ/K Box holds the sand for form of the cast part Sonntag, Borgnakke and van Wylen 8.79 A 5-kg steel container is cured at 500oC. An amount of liquid water at 15oC, 100 kPa is added to the container so a final uniform temperature of the steel and the water becomes 75oC. Neglect any water that might evaporate during the process and any air in the container. How much water should be added and how much entropy was generated? Heat steel m(u2 − u1) = 1Q2 = mC (T2 + T4 ) 1Q2 = 5(0.46)(500-20) = 1104 kJ mH2O( u3-u2)H2O + mst( u3-u2) = 0 mH2o( 313.87 – 62.98) + mstC ( T3-T2) = 0 mH2O 250.89 + 5 × 0.46 × (75 - 500) = 0 mH2O = 977.5/250.89 = 3.896 kg mH2O ( s3-s2) + mst( s3 - s2) = ∅ + 2S3 gen 3.986 (1.0154 – 0.2245) + 5 × 0.46 ln 75+273 773 = 2S3 gen 2S3 gen = 3.0813 – 1.8356 = 1.246 kJ/K cb Sonntag, Borgnakke and van Wylen 8.80 A pan in an autoshop contains 5 L of engine oil at 20oC, 100 kPa. Now 2 L of hot 100oC oil is mixed into the pan. Neglect any work term and find the final temperature and the entropy generation. Solution: Since we have no information about the oil density, we assume the same for both from Table A.4: ρ = 885 kg/m3 Energy Eq.: m2u2 – mAuA – mBuB ≅ 0 – 0 ∆u ≅ Cv∆T so same Cv = 1.9 kJ/kg K for all oil states. mA mB 5 2 T2 = m TA + m TB = 7 × 20 + 7 × 100 = 42.868oC = 316.02 K 2 2 S2 - S1 = m2s2 − mAsA – mBsB = mA(s2 – sA) + mB(s2 – sB) 316.02 316.02 = 0.005 × 885 × 1.9 ln 293.15 + 0.002 × 885 × 1.9 ln 373.15 = 0.6316 – 0.5588 = + 0.0728 kJ/K Oils shown before mixed to final uniform state. Sonntag, Borgnakke and van Wylen 8.81 Find the total work the heat engine can give out as it receives energy from the rock bed as described in Problem 7.61 (see Fig.P 8.81). Hint: write the entropy balance equation for the control volume that is the combination of the rock bed and the heat engine. Solution: To get the work we must integrate over the process or do the 2nd law for a control volume around the whole setup out to T0 C.V. Heat engine plus rock bed out to T0. W and QL goes out. W C.V. HE Q H QL Energy Eq.5.11: (U2 − U1)rock = – QL – W QL T2 Entropy Eq.8.3: (S2 − S1)rock = − T = mC ln ( T ) 0 1 290 = 5500 × 0.89 ln 400 = −1574.15 kJ/K QL = −T0 (S2 − S1)rock = −290 (−1574.15) = 456 504 kJ The energy drop of the rock −(U2 − U1)rock equals QH into heat engine (U2−U1)rock = mC (T2−T1) = 5500 ×0.89 (290 − 400) = −538 450 kJ W = −(U2 − U1)rock − QL = 538450 − 456504 = 81 946 kJ Sonntag, Borgnakke and van Wylen 8.82 Two kg of liquid lead initially at 500°C are poured into a form. It then cools at constant pressure down to room temperature of 20°C as heat is transferred to the room. The melting point of lead is 327°C and the enthalpy change between the phases, hif , is 24.6 kJ/kg. The specific heats are in Tables A.3 and A.4. Calculate the net entropy change for this process. Solution: C.V. Lead, constant pressure process mPb(u2 - u1)Pb = 1Q2 - P(V2 - V1) We need to find changes in enthalpy (u + Pv) for each phase separately and then add the enthalpy change for the phase change. Consider the process in several steps: Cooling liquid to the melting temperature Solidification of the liquid to solid Cooling of the solid to the final temperature 1Q2 = mPb(h2 - h1) = mPb(h2 - h327,sol - hif + h327,f - h500) = 2 × (0.138 × (20 - 327) - 24.6 + 0.155 × (327 - 500)) = -84.732 - 49.2 - 53.63 = -187.56 kJ ∆SCV = mPb[Cp solln(T2/600) - (hif/600) + CP liqln(600/T1)] 293.15 24.6 600 = 2 × [0.138 ln 600 - 600 + 0.155 ln 773.15 ] = -0.358 kJ/K ∆SSUR = -1Q2/T0 = 187.56/293.15 = 0.64 kJ/K The net entropy change from Eq.8.18 is equivalent to total entropy generation ∆Snet = ∆SCV + ∆SSUR = 0.282 kJ/K Sonntag, Borgnakke and van Wylen 8.83 A 12 kg steel container has 0.2 kg superheated water vapor at 1000 kPa, both at 200oC. The total mass is now cooled to ambient temperature 30oC. How much heat transfer was taken out and what is the total entropy generation? Solution: C.V.: Steel and the water, control mass of constant volume. Energty Eq.5.11: U2 - U1 = 1Q2 - 1W2 Process: V = constant => 1W2 = 0 State 1: H20 Table B.1.3: u1 = 2621.9 kJ/kg, v1 = 0.20596 m3/kg, s1 = 6.6939 kJ/kg K State 2: H20: => from Table B.1.1 T2 , v2 = v1 v - vf 0.20596 – 0.001004 x2 = v = = 0.006231 32.8922 fg u2 = 125.77 + x2 × 2290.81 = 140.04 kJ/kg s2 = 0.4369 + x2 × 8.0164 = 0.48685 kJ/kg K 1Q2 = m(u2 − u1) = msteelCsteel (T2 – T1 ) + mH2O (u2 - u1) H2O = 12 × 0.42 (30 – 200) + 0.2 (140.04 –262.19) = -1353.2 kJ Entropy generation from Eq.8.18 1Q2 S2 gen = m2 s2 - m1s1 − T 1 amb T2 1Q2 = msteelCsteel ln ( T ) + mH2O (s2- s1)H2O − T 1 amb 303.15 −1353.2 =12 × 0.42 ln ( 473.15 ) + 0.2(0.48685 – 6.6939) – ( 303.15 ) = -2.2437 – 1.2414 + 4.4638 = 0.9787 kJ/K Sonntag, Borgnakke and van Wylen 8.84 A 5 kg aluminum radiator holds 2 kg of liquid R-134a both at –10oC. The setup is brought indoors and heated with 220 kJ from a heat source at 100oC. Find the total entropy generation for the process assuming the R-134a remains a liquid. Solution: C.V. The aluminum radiator and the R-134a. Energy Eq.5.11: m2u2 – m1u1 = 1Q2 – 0 Process: No change in volume so no work as used above. The energy equation now becomes (summing over the mass) mal (u2 - u1)al + mR134a (u2 - u1)R134a = 1Q2 Use specific heat from Table A.3 and A.4 malCal (T2 - T1) + m R134aC R134a ln (T2 - T1) = 1Q2 T2 - T1 = 1Q2 / [malCal + m R134aC R134a ] = 220 / [5 × 0.9 + 2 × 1.43] = 29.89oC T2 = -10 + 29.89 = 19.89oC Entropy generation from Eq.8.18 1S2 gen = m(s2 - s1)- 1Q2/T 1Q2 = malCal ln (T2/T1) + m R134aC R134a ln (T2/T1) − T amb = (5 × 0.9 + 2 × 1.43) ln (19.89 + 273.15) 220 -10 + 273.15 – 373.15 = 0.7918 – 0.5896 = 0.202 kJ/K 1Q 2 o 100 C Sonntag, Borgnakke and van Wylen 8.85 A piston/cylinder of total 1 kg steel contains 0.5 kg ammonia at 1600 kPa both masses at 120oC. Some stops are placed so a minimum volume is 0.02 m3, shown in Fig. P8.85. Now the whole system is cooled down to 30oC by heat transfer to the ambient at 20oC, and during the process the steel keeps same temperature as the ammonia. Find the work, the heat transfer and the total entropy generation in the process. 1 : v1 = 0.11265 m3/kg, u1 = 1516.6 kJ/kg, s1 = 5.5018 kJ/kg K V1 = mv1 = 0.05634 m3 Stop 1a: vstop = V/m = 0.02/0.5 = 0.04 m3/kg Pstop = P1 ⇒ T ~ 42oC (saturated) 2: 30oC < Tstop so v2 = vstop = 0.04 m3/kg v2-vf 0.04 - 0.00168 x2 = v = = 0.35217 0.10881 fg u2 = 320.46 + x2 ×1016.9 = 678.58 kJ/kg s2 = 1.2005 + x2 × 3.7734 = 2.5294 kJ/kg K 1W2= ∫ P dV = P1m (v2-v1) = 1600 × 0.5 (0.004 – 0.11268) = - 58.14 kJ 1Q2 = m (u2 - u1) + mst(u2 - u1) + 1W2 = 0.5( 678.58 – 1516.6 ) + 1×0.46(30 – 120) – 58.14 = -419.01 – 41.4 – 58.14 = –518.55 kJ 1S2 gen= m(s2 − s1) + mst (s2 − s1) – 1Q2/Tamb 273+30 -518.5 = 0.5 (2.5294 – 5.5018) + 1×0.46 ln 273+120 – 293.15 = - 1.4862 – 0.1196 + 1.6277 = 0.02186 kJ/K P 1a P o T 1 42 30 2 v 1a 1 2 s NH 3 T o Sonntag, Borgnakke and van Wylen 8.86 A hollow steel sphere with a 0.5-m inside diameter and a 2-mm thick wall contains water at 2 MPa, 250°C. The system (steel plus water) cools to the ambient temperature, 30°C. Calculate the net entropy change of the system and surroundings for this process. C.V.: Steel + water. This is a control mass. Energy Eq.: U2 – U1 = 1Q2 - 1W2 = mH O(u2 – u1) + msteel(u2 – u1) 2 Process: V = constant => 1W2 = 0 msteel = (ρV)steel = 8050 × (π/6)[(0.504)3 - (0.5)3] = 12.746 kg VH2O = (π/6)(0.5)3, mH2O = V/v = 6.545×10-2/0.11144 = 0.587 kg v2 = v1 = 0.11144 = 0.001004 + x2 × 32.889 => x2 = 3.358×10-3 u2 = 125.78 + 3.358×10-3 × 2290.8 = 133.5 kJ/kg s2 = 0.4639 + 3.358×10-3 × 8.0164 = 0.4638 kJ/kg K 1Q2 = mH2O(u2 – u1) + msteel(u2 – u1) = 0.587(133.5 - 2679.6) + 12.746 × 0.48(30 - 250) = -1494.6 + (-1346) = -2840.6 kJ ∆STOT = ∆SSTEEL + ∆SH2O = 12.746 × 0.48 ln (303.15 / 523.15) + 0.587(0.4638 - 6.545) = -6.908 kJ/K ∆SSURR = -1Q2/T0 = +2840.6/303.2 = +9.370 kJ/K ∆SNET = -6.908 + 9.370 = +2.462 kJ/K Steel Water Ambient Sonntag, Borgnakke and van Wylen Entropy of ideal gases 8.87 A mass of 1 kg of air contained in a cylinder at 1.5 MPa, 1000 K, expands in a reversible isothermal process to a volume 10 times larger. Calculate the heat transfer during the process and the change of entropy of the air. Solution: C.V. Air, control mass. Energy Eq. 5.11: m(u2 - u1) = 1Q2 - 1W2 = 0 Process: T = constant so with ideal gas T P 1 1 2 P1 2 => u 2 = u1 P2 v s From the process equation and ideal gas law PV = mRT = constant we can calculate the work term as in Eq.4.5 1Q2 = 1W2 = ⌠PdV = P1V1 ln (V2/V1) = mRT1 ln (V2/V1) ⌡ = 1 × 0.287 × 1000 ln (10) = 660.84 kJ The change of entropy from Eq.8.3 is ∆Sair = m(s2 - s1) = 1Q2/T = 660.84/1000 = 0.661 kJ/K If instead we use Eq.8.26 we would get T2 v2 ∆Sair = m(s2 - s1) = m(Cvo ln T + R ln v ) 1 1 = 1 [ 0 + 0.287 ln(10) ] = 0.661 kJ/K consistent with the above result. Sonntag, Borgnakke and van Wylen 8.88 A piston/cylinder setup contains air at 100 kPa, 400 K which is compressed to a final pressure of 1000 kPa. Consider two different processes (i) a reversible adiabatic process and (ii) a reversible isothermal process. Show both processes in P-v and a T-s diagram. Find the final temperature and the specific work for both processes. Solution: C.V. Air, control mass of unknown size and mass. Energy Eq.5.11: u2 – u1 = 1q2 – 1w2 Entropy Eq.8.14: s2 – s1 = ∫ dq/T + 1s2 gen Process: Reversible 1s2 gen = 0 i) dq = 0 so ii) T=C 1q2 = 0 so ∫ dq/T = 1q2/T i) For this process the entropy equation reduces to: s2 – s1 = 0 + 0 so we have constant s, an isentropic process. The relation for an ideal gas, constant s and k becomes Eq.8.32 0.4 k-1 k 0.28575 1000 T2 = T1( P2 / P1) = 400 100 1.4 = 400 × 10 = 772 K From the energy equation we get the work term 1w2 = u1 – u2 = Cv(T1 – T2) = 0.717(400 – 772) = -266.7 kJ/kg ii) For this process T2 = T1 so since ideal gas we get ° ° => Energy Eq.: u2 = u1 also sT2 = sT1 1w2 = 1q2 Now from the entropy equation we solve for 1q2 P2 P2 ° ° 1w2 = 1q2 = T(s2 – s1) = T[sT2 – sT1 – R ln P ] = −RT ln P 1 1 = − 0.287 × 400 ln 10 = −264 kJ/kg T P 2ii P2 2i 2i P1 2ii 1 1 v s Sonntag, Borgnakke and van Wylen 8.89 Consider a Carnot-cycle heat pump having 1 kg of nitrogen gas in a cylinder/piston arrangement. This heat pump operates between reservoirs at 300 K and 400 K. At the beginning of the low-temperature heat addition, the pressure is 1 MPa. During this process the volume triples. Analyze each of the four processes in the cycle and determine a. The pressure, volume, and temperature at each point b. The work and heat transfer for each process Solution: T 4 T1 = T2 = 300 K, T3 = T4 = 400 K, P1 = 1 MPa, V2 = 3 × V1 a) P2V2 = P1V1 => P2 = P1/3 = 0.3333 MPa mRT1 1 × 0.2968 × 300 V1 = P = = 0.08904 m3 1000 1 3 N2 2 1 V2 = 0.26712 m3 s P3 = P2(T3/T2) k k-1 4003.5 = 0.3333300 = 0.9123 MPa P2 T3 0.3333 400 V3 = V2 × P × T = 0.26712 × 0.9123 × 300 = 0.1302 m3 3 P4 = 2 k k-1 P1(T3/T1) 4003.5 = 1300 = 2.73707 MPa P1 T4 1 400 V4 = V1 × P × T = 0.08904 × 2.737 × 300 = 0.04337 m3 4 b) 1 1W2 = 1Q2 = mRT1 ln (P1/P2) = 1 × 0.2968 × 300 ln(1/0.333) = 97.82 kJ 3W4 = 3Q4 = mRT3 ln(P3/P4) = 1 × 0.2968 × 400 ln(0.9123/2.737) = -130.43 kJ 2W3 = -mCV0(T3 - T2) = -1 × 0.7448(400 - 300) = -74.48 kJ 4W1 = -mCV0(T1 - T4) = -1 × 0.7448(300 - 400) = +74.48 kJ 2Q3 = 0, 4Q1 = 0 Sonntag, Borgnakke and van Wylen 8.90 Consider a small air pistol with a cylinder volume of 1 cm3 at 250 kPa, 27°C. The bullet acts as a piston initially held by a trigger. The bullet is released so the air expands in an adiabatic process. If the pressure should be 100 kPa as the bullet leaves the cylinder find the final volume and the work done by the air. Solution: C.V. Air. Assume a reversible, adiabatic process. Energy Eq.5.11: u2 - u1 = 0 − 1w2 ; Entropy Eq.8.14: s2 - s1 = ∫ dq/T + 1s2 gen = 0 / State 1: State 2: (T1,P1) (P2, ?) So we realize that one piece of information is needed to get state 2. Process: Adiabatic 1q2 = 0 Reversible 1s2 gen = 0 With these two terms zero we have a zero for the entropy change. So this is a constant s (isentropic) expansion process giving s2 = s1. From Eq.8.32 k-1 k 0.4 0.28575 100 T2 = T1( P2 / P1) = 300 250 1.4 = 300 × 0.4 = 230.9 K The ideal gas law PV = mRT at both states leads to V2 = V1 P1 T2/P2 T1 = 1 × 250 × 230.9/100 × 300 = 1.92 cm3 The work term is from Eq.8.38 or Eq.4.4 with polytropic exponent n = k 1 1 -6 1W2 = 1 - k (P2V2 - P1V1) = 1 - 1.4 (100 × 1.92 - 250 × 1) ×10 = 0.145 J Sonntag, Borgnakke and van Wylen 8.91 Oxygen gas in a piston cylinder at 300 K, 100 kPa with volume 0.1 m3 is compressed in a reversible adiabatic process to a final temperature of 700 K. Find the final pressure and volume using Table A.5. Solution: C.V. Air. Assume a reversible, adiabatic process. Energy Eq.5.11: u2 - u1 = 0 − 1w2 ; Entropy Eq.8.14: s2 - s1 = ∫ dq/T + 1s2 gen = 0 Process: Adiabatic 1q2 = 0 Reversible 1s2 gen = 0 Properties: Table A.5: k = 1.393 With these two terms zero we have a zero for the entropy change. So this is a constant s (isentropic) expansion process. From Eq.8.32 P2 = P1( T2 / T1) k k-1 = 2015 kPa Using the ideal gas law to eliminate P from this equation leads to Eq.8.33 V2 = V1( T2 / T1) 1 1-k T P 2 1 700 = 0.1 × 300 1−1.393 = 0.0116 m3 P2 2 P1 1 1 v s Sonntag, Borgnakke and van Wylen 8.92 Oxygen gas in a piston cylinder at 300 K, 100 kPa with volume 0.1 m3 is compressed in a reversible adiabatic process to a final temperature of 700 K. Find the final pressure and volume using constant heat capacity from Table A.8. Solution: C.V. Air. Assume a reversible, adiabatic process. Energy Eq.5.11: u2 - u1 = 0 − 1w2 ; Entropy Eq.8.14: s2 - s1 = ∫ dq/T + 1s2 gen = 0 / Process: Adiabatic 1q2 = 0 Reversible 1s2 gen = 0 With these two terms zero we have a zero for the entropy change. So this is a constant s (isentropic) expansion process. From Eq.8.28 P2 o o sT2 – sT1 = R ln P 1 Properties: o Table A.8: o sT1 = 6.4168, sT2 = 7.2336 kJ/kg K P2 7.2336 – 6.4168 o o = exp [(sT2 – sT1)/R] = exp( ) = 23.1955 0.2598 P1 P2 = 100 × 23.1955 = 2320 kPa Ideal gas law: P1V1 = mRT1 and P2V2 = mRT2 Take the ratio of these so mR drops out to give 700 100 V2 = V1 × (T2 / T1) × (P1 / P2) = 0.1 × (300) × (2320) = 0.01 m3 T P 2 P2 2 P1 1 1 v s Sonntag, Borgnakke and van Wylen 8.93 A handheld pump for a bicycle has a volume of 25 cm3 when fully extended. You now press the plunger (piston) in while holding your thumb over the exit hole so that an air pressure of 300 kPa is obtained. The outside atmosphere is at P0, T0. Consider two cases: (1) it is done quickly (∼1 s), and (2) it is done very slowly (∼1 h). a. State assumptions about the process for each case. b. Find the final volume and temperature for both cases. Solution: C.V. Air in pump. Assume that both cases result in a reversible process. State 1: P0, T0 State 2: 300 kPa, ? One piece of information must resolve the ? for a state 2 property. Case I) Quickly means no time for heat transfer Q = 0, so a reversible adiabatic compression. u2 - u1 = -1w2 ; s2 - s1 = ∫ dq/T + 1s2 gen = 0 / With constant s and constant heat capacity we use Eq.8.32 0.4 k-1 k 300 T2 = T1( P2 / P1) = 298 101.325 1.4 = 405.3 K Use ideal gas law PV = mRT at both states so ratio gives => Case II) V2 = P1V1T2/T1P2 = 11.48 cm3 Slowly, time for heat transfer so T = constant = T0. The process is then a reversible isothermal compression. T2 = T0 = 298 K T P 2ii => P2 2i V2 = V1P1/P2 = 8.44 cm3 2i P1 2ii 1 1 v s Sonntag, Borgnakke and van Wylen 8.94 An insulated cylinder/piston contains carbon dioxide gas at 120 kPa, 400 K. The gas is compressed to 2.5 MPa in a reversible adiabatic process. Calculate the final temperature and the work per unit mass, assuming a. Variable specific heat, Table A.8 b. Constant specific heat, value from Table A.5 c. Constant specific heat, value at an intermediate temperature from Table A.6 Solution: C.V. Air, a control mass undergoing a reversible, adiabatic process. Energy Eq.5.11: u2 - u1 = 0 − 1w2 ; Entropy Eq.8.14: s2 - s1 = ∫ dq/T + 1s2 gen = 0 / Process: Adiabatic 1q2 = 0 Reversible 1s2 gen = 0 State 1: (400 K, 120 kPa) State 2: (2500 kPa, ?) With two terms zero in the entropy equation we have a zero for the entropy change. So this is a constant s (isentropic) expansion process, s2 = s1. a) Table A.8 for CO2 and Eq.8.28 s2 - s1 = 0 = s° – s° − R ln(P2/P1) T2 T1 ° sT2 = s° + R ln(P2/P1) = 5.1196 + 0.1889 ln(2500/120) = 5.6932 T1 Now interpolate in A.8 to find T2 T2 = 650 + 50 (5.6932 – 5.6151)/(5.6976 – 5.6151) = 697.3 K 1w2 = -(u2 - u1) = –(481.5 – 228.19) = –253.3 kJ/kg b) Table A.5: k = 1.289, CVo = 0.653 kJ/kg K and now Eq.8.32 P2 k-1 2.5 0.224 T2 = T1 P k = 400 0.12 = 789.7 K 1 1w2 = -CVo(T2-T1) = -0.653 (789.7 - 400) = -254.5 kJ/kg c) Find a heat capacity at an average temperature from Table A.6. Estimate T2 ~ 700 K giving TAVE ~ 550 K => θ = 0.55 CPo = 0.45 + 1.67 ×0.55 – 1.27 ×0.552 + 0.39 ×0.553 = 1.049 kJ/kg K CVo = CPo – R = 1.049 – 0.1889 = 0.8601, k = CPo/CVo = 1.2196 Eq.8.32: P2 k-1 2.5 0.18006 T2 = T1 P k = 400 0.12 = 691 K 1 1w2 = –CVo(T2–T1) = –0.8601 (691 – 400) = –250.3 kJ/kg Sonntag, Borgnakke and van Wylen 8.95 A piston/cylinder, shown in Fig. P8.95, contains air at 1380 K, 15 MPa, with V1 = 10 cm3, Acyl = 5 cm2. The piston is released, and just before the piston exits the end of the cylinder the pressure inside is 200 kPa. If the cylinder is insulated, what is its length? How much work is done by the air inside? Solution: C.V. Air, Cylinder is insulated so adiabatic, Q = 0. Continuity Eq.: m2 = m1 = m, Energy Eq.5.11: m(u2 - u1) = 1Q2 - 1W2 = - 1W2 Entropy Eq.8.14: m(s2 - s1) = ∫ dQ/T + 1S2 gen = 0 + 1S2 gen State 1: (T1, P1) State 2: (P2, ?) So one piece of information is needed for the ?, assume reversible process. 1S2 gen = 0 => s2 - s1 = 0 State 1: Table A.7: u1 = 1095.2 kJ/kg, m = P1V1/RT1 = o sT1 = 8.5115 kJ/kg K 15000 × 10×10-6 = 0.000379 kg 0.287 × 1380 State 2: P2 and from Entropy eq.: s2 = s1 so from Eq.8.28 P2 200 s° = s° + R ln P = 8.5115 + 0.287 ln(15000) = 7.2724 kJ/kg K T2 T1 1 Now interpolate in Table A.7 to get T2 T2 = 440 + 20 (7.2724 – 7.25607)/(7.30142 – 7.25607) = 447.2 K u2 = 315.64 + (330.31 – 315.64) 0.36 = 320.92 kJ/kg T2 P1 10 × 447.2 × 15000 V2 = V1 T P = = 243 cm3 1380 × 200 12 ⇒ L2 = V2 /Acyl = 243/5 = 48.6 cm 1w2 = u1 - u2 = 774.3 kJ/kg, 1W2 = m1w2 = 0.2935 kJ Sonntag, Borgnakke and van Wylen 8.96 Two rigid tanks, shown in Fig. P8.96, each contain 10 kg N2 gas at 1000 K, 500 kPa. They are now thermally connected to a reversible heat pump, which heats one and cools the other with no heat transfer to the surroundings. When one tank is heated to 1500 K the process stops. Find the final (P, T ) in both tanks and the work input to the heat pump, assuming constant heat capacities. Solution: Control volume of hot tank B, Process = constant volume & mass so no work Energy equation Eq.5.11 and specific heat in Eq.5.20 gives U2 - U1 ≅ mCv(T2 - T1) = 1Q2 = 10 × 0.7448 × 500 = 3724 kJ P2 = P1T2/T1 = 1.5(P1) = 750 kPa A 1Q 3 WHE H.P. 1 -> 3 1Q 2 B 1 -> 2 State: 1 = initial, 2 = final hot 3 = final cold To fix temperature in cold tank, C.V.: total For this CV only WHP cross the control surface no heat transfer. The entropy equation Eq.8.14 for a reversible process becomes (S2 - S1)tot = 0 = mhot(s2 - s1) + mcold(s3 - s1) Use specific heats to evaluate the changes in s from Eq.8.25 and division by m Cp,hot ln(T2 / T1) − R ln(P2 / P1) + Cp,coldln(T3 / T1) − R ln(P3 / P1) = 0 / P3 = P1T3/T1 and P2 = P1T2/T1 Now everything is in terms of T and Cp = Cv + R, so Cv,hotln(T2/T1) + Cv,coldln(T3/T1) = 0 same Cv: T3 = T1(T1/T2) = 667 K, P3 = 333 kPa Qcold = - 1Q3 = mCv(T3 - T1) = -2480 kJ, WHP = 1Q2 + Qcold = 1Q2 - 1Q3 = 1244 kJ Sonntag, Borgnakke and van Wylen 8.97 A spring loaded piston cylinder contains 1.5 kg air at 27oC and 160 kPa. It is now heated in a process where pressure is linear in volume, P = A + BV, to twice the initial volume where it reaches 900 K. Find the work, the heat transfer and the total entropy generation assuming a source at 900 K. Solution: C.V. Air out to the 900 K source. Since air T is lower than the source temperature we know that this is an irreversible process. Continuity Eq.: m2 = m1 = m, Energy Eq.5.11: m(u2 – u1) = 1Q2 – 1W2 Entropy Eq.8.14: m(s2 – s1) = ∫ dQ/T + 1S2 gen = 1Q2/TSOURCE + 1S2 gen Process: State 1: (T1, P1) P = A + BV Table A.7 u1 = 214.36 kJ/kg V1 = mRT1/ P1 = (1.5 × 0.287 ×300) / 160 = 0.8072 m3 State 2: (T2, v2 = 2 v1) Table A.7 u2 = 674.824 kJ/kg P2 = RT2/ v2 = RT2/2v1 = T2 P1/ 2T1= P1 T2/2 T1 = 160 × 900 / 2 × 300 = 240 kPa From the process equation we can express the work as 1W2 = ∫ PdV = 0.5 × (P1 + P2) (V2 - V1) = 0.5 × (P1 + P2) V1 = 0.5 × (160 + 240) 0.8072 = 161.4 kJ 1Q2 = 1.5 × ( 674.824 – 214.36) + 161.4 = 852.1 kJ Change in s from Eq.8.28 and Table A.7 values P2 o o S2 gen = m(sT2 – sT1 – R ln P ) – 1Q2/TSOURCE 1 1 240 852.1 = 1.5 × [8.0158 – 6.8693 – 0.287 ln ( 160 )] – ( 900 ) = 1.545 – 0.947 = 0.598 kJ/K P T 2 900 1 1 300 v 2 P1 s Sonntag, Borgnakke and van Wylen 8.98 A rigid storage tank of 1.5 m3 contains 1 kg argon at 30°C. Heat is then transferred to the argon from a furnace operating at 1300°C until the specific entropy of the argon has increased by 0.343 kJ/kg K. Find the total heat transfer and the entropy generated in the process. Solution: C.V. Argon out to 1300°C. Control mass. , m = 1 kg Argon is an ideal gas with constant heat capacity. Energy Eq.5.11: m (u2 - u1 ) = m Cv (T2 - T1) = 1Q2 - 1W2 Entropy Eq.8.14: m(s2 − s1) = 1Q2/Tres + 1S2 gen tot Process: V = constant => also 1W2 = 0 Properties: Table A.5 R = 0.20813, Cv = 0.312 kJ/kg K State 1: (T1, v1= V/m ) v 2 = v1 P1 = mRT1/V = 42.063 kPa State 2: s2 = s1 + 0.343, and change in s from Eq.8.28 or Eq.8.26 s2 - s1 = Cp ln (T2 / T1 ) - R ln (T2 / T1 ) = Cv ln (T2 / T1 ) s2 - s 1 0.343 T2 / T1 = exp[ C ] = exp[0.312] = exp(1.09936) = 3.0 v Pv = RT => (P2 / P1) (v2 / v1) = T2 / T1 = P2 / P1 T2 = 3.0 × T1 = 909.45 K, P2 = 3.0 × P1 = 126.189 kPa T P 2 v=C 2 1 P1 1 v s Heat transfer from energy equation 1Q2 = 1 × 0.312 (909.45 − 303.15) = 189.2 kJ Entropy generation from entropy equation (2nd law) 1S2 gen tot = m(s2 − s1) − 1Q2/Tres = 1 × 0.343 − 189.2 / (1300 + 273) = 0.223 kJ/K Sonntag, Borgnakke and van Wylen 8.99 A rigid tank contains 2 kg of air at 200 kPa and ambient temperature, 20°C. An electric current now passes through a resistor inside the tank. After a total of 100 kJ of electrical work has crossed the boundary, the air temperature inside is 80°C. Is this possible? Solution: C.V.: Air in tank out to ambient; Energy Eq.5.11: m(u2 − u1) = 1Q2 − 1W2 , Entropy Eq.8.14, 8.18: Process: 1W2 = −100 kJ m(s2 – s1) = ∫ dQ/T + 1S2 gen = 1Q2/Tamb + 1S2 gen Constant volume and mass so v2 = v1 State 1: T1 = 20oC, P1 = 200 kPa, m1 = 2 kg State 2: T2 = 80oC, v2 = v1 Ideal gas, Table A.5: R = 0.287 kJ/kg-K, Cv = 0.717 kJ/kg-K Assume constant specific heat then energy equation gives 1Q2 = mCv(T2 − T1) + 1W2 = 2 × 0.717(80 – 20) – 100 = −14.0 kJ Change in s from Eq.8.26 (since second term drops out) v2 s2 - s1 = Cv ln (T2/T1) + Rln v ; 1 v2 v2 = v1, ln v = 0 1 s2 - s1 = Cvln (T2/T1) = 0.1336 kJ/kg-K Now Eq.8.18 or from Eq.8.14 14 1S2 gen = m(s2 – s1) – 1Q2/Tamb = 2 × 0.1336 + 293 = 0.315 kJ/K ≥ 0, Process is Possible T2 Note: P2 = P1 T in Eq.8.28 1 same answer as Eq.8.26. T2 P2 s2 – s1 = Cp lnT - R ln P , results in the 1 T P + _ 1 2 v=C 2 1 P1 1 v s Sonntag, Borgnakke and van Wylen 8.100 Argon in a light bulb is at 90 kPa and heated from 20oC to 60oC with electrical power. Do not consider any radiation, nor the glass mass. Find the total entropy generation per unit mass of argon. Solution: C.V. Argon gas. Neglect any heat transfer. Energy Eq.5.11: m(u2 - u1) = 1W2 electrical in Entropy Eq.8.14: s2 - s1 = ∫ dq/T + 1s2 gen = 1s2 gen Process: v = constant and ideal gas => P2/ P1 = T2/T1 Evaluate changes in s from Eq.8.26 or 8.28 1s2 gen = s2 - s1 = Cp ln (T2/T1) – R ln (P2/ P1) = Cp ln (T2/T1) – R ln (T2/ T1) = Cv ln(T2/T1) Eq.8.28 Eq.8.26 = 0.312 ln [ (60 + 273)/(20 + 273) ] = 0.04 kJ/kg K Since there was no heat transfer but work input all the change in s is generated by the process (irreversible conversion of W to internal energy) Sonntag, Borgnakke and van Wylen 8.101 We wish to obtain a supply of cold helium gas by applying the following technique. Helium contained in a cylinder at ambient conditions, 100 kPa, 20°C, is compressed in a reversible isothermal process to 600 kPa, after which the gas is expanded back to 100 kPa in a reversible adiabatic process. a. Show the process on a T–s diagram. b. Calculate the final temperature and the net work per kilogram of helium. Solution: a) T T1 = T2 P 2 2 P =P 1 3 1 3 s P 2 600 100 3 1 v s2 = s3 b) The adiabatic reversible expansion gives constant s from the entropy equation Eq.8.14. With ideal gas and constant specific heat this gives relation in Eq.8.32 T3 = T2(P3/P2) k-1 k = 293.15 (100/600)0.4 = 143.15 K The net work is summed up over the two processes. The isothermal process has work as Eq.8.41 1w2 = -RT1 ln(P2/P1) = -2.0771 × 293.15 × ln(600/100) = -1091.0 kJ/kg The adiabatic process has a work term from energy equation with no q 2w3 = CVo(T2-T3) = 3.116 (293.15 - 143.15) = +467.4 kJ/kg The net work is the sum wNET = -1091.0 + 467.4 = -623.6 kJ/kg Sonntag, Borgnakke and van Wylen 8.102 A 1-m3 insulated, rigid tank contains air at 800 kPa, 25°C. A valve on the tank is opened, and the pressure inside quickly drops to 150 kPa, at which point the valve is closed. Assuming that the air remaining inside has undergone a reversible adiabatic expansion, calculate the mass withdrawn during the process. Solution: C.V.: Air remaining inside tank, m2. Cont.Eq.: m2 = m ; Energy Eq.5.11: m(u2 − u1) = 1Q2 − 1W2 Entropy Eq.8.14: m(s2 − s1) = ∫ dQ/T + 1S2 gen Process: P adiabatic 1Q2 = 0 and reversible 1S2 gen = 0 T 1 1 2 2 v C.V. m2 s Entropy eq. then gives s2 = s1 and ideal gas gives the relation in Eq.8.32 T2 = T1(P2/P1 k-1 )k = 298.2(150/800)0.286 = 184.8 K m1 = P1V/RT1 = (800 × 1)/(0.287 × 298.2) = 9.35 kg m2 = P2V/RT2 = (150 × 1)/(0.287 × 184.8) = 2.83 kg me = m1 - m2 = 6.52 kg Sonntag, Borgnakke and van Wylen 8.103 Nitrogen at 200oC, 300 kPa is in a piston cylinder, volume 5 L, with the piston locked with a pin. The forces on the piston require a pressure inside of 200 kPa to balance it without the pin. The pin is removed and the piston quickly comes to its equilibrium position without any heat transfer. Find the final P, T and V and the entropy generation due to this partly unrestrained expansion. Solution: C.V. Nitrogen gas. Energy Eq.5.11: m(u2 - u1) = 1Q2 - 1W2 = - ∫ Peq dV = -P2 (V2 - V1) Entropy Eq.8.14: Process: m(s2 - s1) = 0 + 1S2 gen 1Q2 = 0 (already used), State 1: 200 °C, 300 kPa P = Peq after pin is out. State 2: P2 = Peq = 200 kPa m = P1V1/RT1 = 300 × 0.005 / 0.2968 × 473.15 = 0.01068 kg The energy equation becomes mu2 + P2V2 = mu1 + P2V1 = mh2 => h2 = u1 + P2V1/m = u1 + P2V1 RT1 /P1V1 = u1 + (P2/P1) RT1 Solve using constant Cp, Cv Cp T2 = Cv T1 + (P2/P1) RT1 T2 = T1 [Cv + (P2/P1) R] / Cp = 473.15 [0.745 + (200 / 300) × 0.2368] / 1.042 = 428.13 K 428.13 300 V2 = V1( T2 / T1) × ( P1/P2 ) = 0.005 × 473.15 × 200 = 0.00679 m3 1S2 gen= m(s2 - s1) ≅ m[Cp ln (T2/T1) – R ln (P2/ P1)] = P1V1 /RT1 [Cp ln (T2/T1) – R ln (P2/ P1)] = 0.01068 [1.042 × ln (428.13/473.15) – 0.2968 × ln (200 / 300)] = 0.000173 kJ/K Sonntag, Borgnakke and van Wylen 8.104 A rigid container with volume 200 L is divided into two equal volumes by a partition, shown in Fig. P8.104. Both sides contain nitrogen, one side is at 2 MPa, 200°C, and the other at 200 kPa, 100°C. The partition ruptures, and the nitrogen comes to a uniform state at 70°C. Assume the temperature of the surroundings is 20°C, determine the work done and the net entropy change for the process. Solution: C.V. : A + B no change in volume. 1W2 = 0 mA1 = PA1VA1/RTA1 = (2000 × 0.1)/(0.2968 × 473.2) = 1.424 kg mB1 = PB1VB1/RTB1 = (200 × 0.1)/(0.2968 × 373.2) = 0.1806 kg P2 = mTOTRT2/VTOT = (1.6046 × 0.2968 × 343.2)/0.2 = 817 kPa From Eq.8.25 343.2 817 ∆SSYST = 1.424[1.042 ln 473.2 - 0.2968 ln 2000] 343.2 817 + 0.1806[1.042 ln 373.2 - 0.2968 ln 200] = -0.1894 kJ/K 1Q2 = U2 - U1 = 1.424 × 0.745(70 - 200) + 0.1806 × 0.745(70 - 100) = -141.95 kJ From Eq.8.18 ∆SSURR = - 1Q2/T0 = 141.95/293.2 = +0.4841 kJ/K ∆SNET = -0.1894 + 0.4841 = +0.2947 kJ/K Sonntag, Borgnakke and van Wylen 8.105 Nitrogen at 600 kPa, 127°C is in a 0.5 m3 insulated tank connected to a pipe with a valve to a second insulated initially empty tank of volume 0.5 m3, shown in Fig. P8.105. The valve is opened and the nitrogen fills both tanks at a uniform state. Find the final pressure and temperature and the entropy generation this process causes. Why is the process irreversible? Solution: CV Both tanks + pipe + valve Insulated : Q = 0 Rigid: W = 0 Energy Eq.5.11: m(u2 - u1) = 0 - 0 => u2 = u1 = ua1 m(s2 − s1) = ∫ dQ/T + 1S2 gen = 1S2 gen Entropy Eq.8.14: (dQ = 0) 1: P1 , T1 , Va => m = PV/RT = (600 × 0.5)/ (0.2968 × 400) = 2.527 2: V2 = Va + Vb ; uniform state v2 = V2 / m ; u2 = ua1 T P 1 1 P1 2 P2 s v Ideal gas u (T) => u2 = ua1 2 => T2 = Ta1 = 400 K P2 = mR T2 / V2 = (V1 / V2 ) P1 = ½ × 600 = 300 kPa From entropy equation and Eq.8.28 for entropy change Sgen = m(s2 − s1) = m[sT2 − sT1 − R ln(P2 / P1)] = m [0 - R ln (P2 / P1 )] = -2.527 × 0.2968 ln ½ = 0.52 kJ/K Irreversible due to unrestrained expansion in valve P ↓ but no work out. Sonntag, Borgnakke and van Wylen Polytropic processes 8.106 Neon at 400 kPa, 20°C is brought to 100°C in a polytropic process with n = 1.4. Give the sign for the heat transfer and work terms and explain. Solution: P Neon Table A.5 k = γ = 1.667 so n < k Cv = 0.618, R = 0.412 T 2 2 1 1 T=C v From definition Eq.8.2 From work term s ds = dq/T so dq = T ds dw = P dv From figures: v goes down so work in ( W < 0); s goes down so Q out ( Q < 0) We can also calculate the actual specific work from Eq.8.38 and heat transfer from the energy equation as: 1w2 = [R/(1-n)](T2 - T1) = -82.39 kJ/kg u2 - u1 = Cv(T2 - T1) = 49.432, 1q2 = ∆u + 1w2 = -32.958 1W2 Negative and 1Q2 Negative Sonntag, Borgnakke and van Wylen 8.107 A mass of 1 kg of air contained in a cylinder at 1.5 MPa, 1000 K, expands in a reversible adiabatic process to 100 kPa. Calculate the final temperature and the work done during the process, using a. Constant specific heat, value from Table A.5 b. The ideal gas tables, Table A.7 Solution: C.V. Air. Continuity Eq.: m2 = m1 = m ; Energy Eq.5.11: m(u2 − u1) = 1Q2 − 1W2 Entropy Eq.8.14: m(s2 − s1) = ∫ dQ/T + 1S2 gen Process: 1Q2 = 0, 1S2 gen = 0 => s2 = s1 a) Using constant Cp from Table A.5 gives the power relation Eq.8.32. k-1 k 0.10.286 = 10001.5 = 460.9 K 1W2 = -(U2 - U1) = mCVo(T1 - T2) T2 = T1(P2/P1) = 1 × 0.717(1000 - 460.9) = 386.5 kJ b) Use the standard entropy function that includes variable heat capacity from A.7.1 and Eq.8.28 P2 P2 o o o o s2 – s1 = sT2 – sT1 – R ln P = 0 ⇒ sT2 = sT1 + R ln P 1 1 o sT2 = 8.13493 + 0.287 ln(100/1500) = 7.35772 Interpolation gives T2 = 486 K and u2 = 349.5 kJ/kg 1W2 = m(u1 - u2) = 1(759.2 - 349.5) = 409.7 kJ Sonntag, Borgnakke and van Wylen 8.108 An ideal gas having a constant specific heat undergoes a reversible polytropic expansion with exponent, n = 1.4. If the gas is carbon dioxide will the heat transfer for this process be positive, negative, or zero? Solution: T n=k n>k 1 P = const n<k 2 CO2: Table A.5 k = 1.289 < n Since n > k and P2 < P1 it follows that s2 < s1 and thus Q flows out. / 1Q2 < 0 s Sonntag, Borgnakke and van Wylen 8.109 A cylinder/piston contains 1 kg methane gas at 100 kPa, 20°C. The gas is compressed reversibly to a pressure of 800 kPa. Calculate the work required if the process is a. Adiabatic b. Isothermal c. Polytropic, with exponent n = 1.15 Solution: C.V. Methane gas of constant mass m2 = m1 = m and reversible process. Energy Eq.5.11: m(u2 − u1) = 1Q2 − 1W2 Entropy Eq.8.14: m(s2 − s1) = ∫ dQ/T + 1S2 gen = ∫ dQ/T a) Process: 1Q2 = 0 => s 2 = s 1 thus isentropic process s = const and ideal gas gives relation in Eq.8.32 T2 = T1 (P2/P1) k-1 k 800 0.230 = 293.2 100 = 473.0 K 1W2 = -mCV0(T2 - T1) = -1 × 1.7354 (473.0 - 293.2) = -312.0 kJ b) Process: T = constant. For ideal gas then u2 = u1 and s° = s° T2 T1 Energy eq. gives 1W2 = 1Q2 and ∫ dQ/T = 1Q2/T with the entropy change found from Eq.8.28 => 1W2 = 1Q2 = mT(s2 - s1) = -mRT ln(P2/P1) = -0.51835× 293.2 ln(800/100) = -316.0 kJ c) Pvn = constant with n = 1.15 ; The T-P relation is given in Eq.8.37 Process: n -1 800 0.130 T2 = T1 (P2/P1) n = 293.2 100 = 384.2 K and the work term is given by Eq.8.38 1W2 = ∫ mP dv = m(P2v2 - P1v1)/(1 - n) = mR (T2 - T1)/(1 - n) = 1× 0.51835(384.2 - 293.2) = -314.5 kJ 1 - 1.15 Sonntag, Borgnakke and van Wylen 8.110 Helium in a piston/cylinder at 20°C, 100 kPa is brought to 400 K in a reversible polytropic process with exponent n = 1.25. You may assume helium is an ideal gas with constant specific heat. Find the final pressure and both the specific heat transfer and specific work. Solution: C.V. Helium Continuity Eq.: m2 = m1 = m ; Energy Eq.5.11: m(u2 − u1) = 1Q2 − 1W2 => Tvn-1 = C Process: Pvn = C & Pv = RT Table A.5: Cv = 3.116 kJ/kg K, R = 2.0771 kJ/kg K From the process equation and T1 = 293.15, T2 = 400 K T1 vn-1 = T2 vn-1 => P2 / P1 = (v1 / v2)n = 4.73 v2 / v1 = (T1 / T2 )1/n-1 = 0.2885 => P2 = 473 kPa The work is from Eq.8.38 per unit mass 1-n - v 1-n ) -n 1w2 = ∫ P dv = ∫ C v dv = [ C / (1-n) ] × ( v2 1 1 R = 1-n (P2 v2 - P1 v1) = 1-n (T2 - T1) = -887.7 kJ/kg The heat transfer follows from the energy equation 1q2 = u2 - u1 + 1w2 = Cv (T2 - T1 ) + (- 887.7) = -554.8 kJ/kg Sonntag, Borgnakke and van Wylen 8.111 The power stroke in an internal combustion engine can be approximated with a polytropic expansion. Consider air in a cylinder volume of 0.2 L at 7 MPa, 1800 K, shown in Fig. P8.111. It now expands in a reversible polytropic process with exponent, n = 1.5, through a volume ratio of 8:1. Show this process on P–v and T– s diagrams, and calculate the work and heat transfer for the process. Solution: C.V. Air of constant mass m2 = m1 = m. m(u2 − u1) = 1Q2 − 1W2 Energy Eq.5.11: m(s2 − s1) = ∫ dQ/T + 1S2 gen = ∫ dQ/T 1.50 = constant, PV V2/V1 = 8 Entropy Eq.8.14: Process: State 1: P1 = 7 MPa, T1 = 1800 K, V1 = 0.2 L State 2: P1V1 7000 × 0.2 × 10-3 m1 = RT = = 2.71×10-3 kg 0.287 × 1800 1 (v = V2/m, ?) Must be on process curve so Eq.8.37 gives Table A.7: T2 = T1 (V1/V2)n-1 = 1800 (1/8)0.5 = 636.4 K u1 = 1486.331 kJ/kg and interpolate u2 = 463.05 kJ/kg P T 1 1 Notice: n = 1.5, k = 1.4 2 V 2 S Work from the process expressed in Eq.8.38 1W2 = ⌠ PdV = mR(T2 - T1)/(1 - n) ⌡ = 2.71×10-3 × 0.287(636.4 - 1800) = 1.81 kJ 1 - 1.5 Heat transfer from the energy equation 1Q2 = m(u2 - u1) + 1W2 = 2.71×10-3 × (463.05 - 1486.331) + 1.81 = -0.963 kJ n>k Sonntag, Borgnakke and van Wylen 8.112 A piston/cylinder contains air at 300 K, 100 kPa. It is now compressed in a reversible adiabatic process to a volume 7 times as small. Use constant heat capacity and find the final pressure and temperature, the specific work and specific heat transfer for the process. Solution: Expansion ratio: v2/ v1 = 1/7 Process eq.: Rev. adiabatic and ideal gas gives P2 /P1 = (v2/v1)-k = 71.4 = 15.245 Pvn = C, with n = k P2 = P1 (71.4) = 100 × 15.245 = 1524.5 kPa T2 = T1 (v1/v2)k-1 = 300 × 70.4 = 653.4 K 1q2 = 0 kJ/kg Polytropic process work term from Eq.8.38 R 0.287 1w2 = 1 - k (T2 –T1) = -0.4 (653.4 – 300) = -253.6 kJ/kg Notice: Cv = R/(k-1) so the work term is also the change in u consistent with the energy equation. Sonntag, Borgnakke and van Wylen 8.113 A cylinder/piston contains carbon dioxide at 1 MPa, 300°C with a volume of 200 L. The total external force acting on the piston is proportional to V 3. This system is allowed to cool to room temperature, 20°C. What is the total entropy generation for the process? Solution: C.V. Carbon dioxide gas of constant mass m2 = m1 = m out to ambient. Energy Eq.5.11: m(u2 − u1) = 1Q2 − 1W2 m(s2 − s1) = ∫ dQ/T + 1S2 gen = 1Q2/Tamb + 1S2 gen Entropy Eq.8.14,18: Process: P = CV 3 or PV-3 = constant, which is polytropic with n = -3 State 1: (T, P) => m = P1V1/RT1 = 1000 × 0.2 = 1.847 kg 0.18892 × 573.2 State 2: (T, ?) state must be on process curve. This and ideal gas leads to Eq.8.37 ⇒ n n-1 = 1000(293.2/573.2)3/4 = 604.8 kPa P2 = P1(T2/T1) 1 V2 = V1(T1/T2)n-1 = 0.16914 m3 1W2 =⌠ PdV = (P2V2 - P1V1)/(1-n) ⌡ = [604.8 × 0.16914 - 1000 × 0.2] / [1 - (-3)] = -24.4 kJ 1Q2 = m(u2 − u1) + 1W2 = 1.847 × 0.653 (20 - 300) - 24.4 = -362.1 kJ From Eq.8.25 293.2 604.8 m(s2 − s1) = 1.847[0.842 ln 573.2 - 0.18892 ln 1000 ] = 1.847[-0.4694] = -0.87 kJ/K ∆SSURR = − 1Q2/Tamb = +362.1 / 293.2 = +1.235 kJ/K From Eq.8.18 1S2 gen = m(s2 − s1) − 1Q2/Tamb = ∆SNET = ∆SCO2 + ∆SSURR = −0.87 + 1.235 = +0.365 kJ/K P 1 1000 605 T 2 300 v 20 Notice: n = -3, k = 1.3 1 2 s n<k Sonntag, Borgnakke and van Wylen 8.114 A device brings 2 kg of ammonia from 150 kPa, -20oC to 400 kPa, 80oC in a polytropic process. Find the polytropic exponent, n, the work and the heat transfer. Find the total entropy generated assuming a source at 100oC. Solution: C.V. Ammonia of constant mass m2 = m1 = m out to source. m(u2 − u1) = 1Q2 − 1W2 Energy Eq.5.11: m(s2 − s1) = ∫ dQ/T + 1S2 gen = 1Q2/T + 1S2 gen Entropy Eq.8.14, 8.18: Process: P1v1n = P2v2n Eq. (8.36) State 1: Table B.2.2 v1 = 0.79774 m3/kg, s1 = 5.7465 kJ/kg K, u1 = 1303.3 kJ/kg State 2: Table B.2.2 v2 = 0.4216 m3/kg, s2 = 5.9907 kJ/kg K, u2 = 1468.0 kJ/kg ln (P2/P1) = ln (v2/v1)n = n × ln (v2/v1) 480 0.4216 ln ( 150 ) = n × ln ( 0.79774 ) = 0.98083 = n × 0.63773 ⇒ n = 1.538 The work term is integration of PdV as done in text leading to Eq.8.38 m 1W2 = 1 − n ( P2v2 - P1v1) = 2 × ( 400 × 0.4216 – 150 × 0.79774) = –182.08 kJ 1 − 1.538 Notice we did not use Pv = RT as we used the ammonia tables. 1Q2 = m(u2 - u1) + 1W2 = 2 (1468 – 1303.3) – 182.08 = 147.3 kJ From Eq.8.18 147.3 1S2 gen = m(s2 – s1) - 1Q2/T = 2 (5.9907 – 5.7465 ) – 373.15 = 0.0936 kJ/K P 400 150 T 2 80 Notice: n = 1.54, k = 1.3 2 1 v -20 1 s n>k Sonntag, Borgnakke and van Wylen 8.115 A cylinder/piston contains 100 L of air at 110 kPa, 25°C. The air is compressed in a reversible polytropic process to a final state of 800 kPa, 200°C. Assume the heat transfer is with the ambient at 25°C and determine the polytropic exponent n and the final volume of the air. Find the work done by the air, the heat transfer and the total entropy generation for the process. Solution: C.V. Air of constant mass m2 = m1 = m out to ambient. Energy Eq.5.11: m(u2 − u1) = 1Q2 − 1W2 Entropy Eq.8.14,18: Process: m(s2 − s1) = ∫ dQ/T + 1S2 gen = 1Q2/T0 + 1S2 gen Pv1n = P2v2n State 1: (T1, P1) Eq.8.36 State 2: (T2, P2) Thus the unknown is the exponent n. m = P1V1 /(RT1) = 110 × 0.1/(0.287 × 298.15) = 0.1286 kg The relation from the process and ideal gas is in Eq.8.37 n -1 T2/T1 = (P2/P1) n n -1 473.15 800 => 298.15 = 110 n ⇒ n-1 n = 0.2328 1 110 n = 1.3034, V2 = V1(P1/P2)n = 0.1 800 0.7672 = 0.02182 m3 The work is from Eq.8.38 P2V2 - P1V1 800 × 0.02182 - 110 × 0.1 = = -21.28 kJ ⌠ 1W2 = ⌡PdV = 1-n 1 - 1.3034 Heat transfer from the energy equation 1Q2 = mCv(T2 - T1) + 1W2 = 0.1286 × 0.717 × (200 - 25) - 21.28 = -5.144 kJ Entropy change from Eq.8.25 s2 - s1 = CP0ln(T2/T1) - R ln(P2/P1) kJ 473.15 800 = 1.004 ln 298.15 - 0.287 ln 110 = -0.106 kg K From the entropy equation (also Eq.8.18) 1S2,gen = m(s2 - s1) - 1Q2/T0 = 0.1286 × (-0.106) + (5.144/298.15) = 0.00362 kJ/K Sonntag, Borgnakke and van Wylen 8.116 A mass of 2 kg ethane gas at 500 kPa, 100°C, undergoes a reversible polytropic expansion with exponent, n = 1.3, to a final temperature of the ambient, 20°C. Calculate the total entropy generation for the process if the heat is exchanged with the ambient. Solution: C.V. Ethane gas of constant mass m2 = m1 = m out to ambient. Energy Eq.5.11: m(u2 − u1) = 1Q2 − 1W2 Entropy Eq.8.14,18: Process: m(s2 − s1) = ∫ dQ/T + 1S2 gen = 1Q2/Tamb + 1S2 gen Pv1n = P2v2n Eq.8.36 State 1: (T1, P1) State 2: (T2, ?) on process curve n 4.333 n-1 = 500 293.2 P2 = P1 (T2/T1) = 175.8 kPa 373.2 Work is integrated in Eq.8.38 2 1w2 = ⌠ Pdv = ⌡ 1 P2v2-P1v1 R(T2-T1) 0.2765(293.2-373.2) = 1-n = = +73.7 kJ/kg 1-n 1-1.30 Heat transfer is from the energy equation 1q2 = CV0(T2 - T1) + 1w2 = 1.49(293.2 - 373.2) + 73.7 = −45.5 kJ/kg Entropy change from Eq.8.25 s2 - s1 = CP0 ln (T2/T1) - R ln(P2/P1) 293.2 175.8 = 1.766 ln 373.2 − 0.2765 ln 500 = −0.1371 kJ/kg K m(s2 − s1) = ∆SSYST = 2(−0.1371) = −0.2742 kJ/K ∆SSURR = −1Q2/T0 = + 2 × 45.5/293.2 = +0.3104 kJ/K Generation from entropy equation or Eq.8.18 1S2 gen = m(s2 − s1) − 1Q2/Tamb = ∆SNET = ∆SSYST + ∆SSURR = −0.2742 + 0.3104 = +0.0362 kJ/K P 500 176 T 1 100 2 v 20 1 2 Notice: n = 1.3, k = 1.186 s n>k Sonntag, Borgnakke and van Wylen 8.117 A piston/cylinder contains air at 300 K, 100 kPa. A reversible polytropic process with n = 1.3 brings the air to 500 K. Any heat transfer if it comes in is from a 325oC reservoir and if it goes out it is to the ambient at 300 K. Sketch the process in a P-v and a T-s diagram. Find the specific work and specific heat transfer in the process. Find the specific entropy generation (external to the air) in the process. Solution: Process : Pvn = C P2v2-P1v1 R = ( T -T ) 1w2 = ∫ P dv = 1-n 1 - n 2 1 0.287 = 1 - 1.3 (500 - 300) = -191.3 kJ/kg Energy equation 1q2 = u2 – u1 +1w2 = Cv ( T2 –T1 ) + 1w2 = 0.717 (500 - 300) – 191.3 = -47.93 kJ/kg The 1q2 is negative and thus goes out. Entropy is generated between the air and ambient. s2 - s1 = 1q2/Tamb+ 1s2 gen 1s2 gen = s2 - s1 – 1q2/Tamb = Cp ln (T2/T1) – R ln (P2/P1) - 1q2/Tamb P2/P1 = (T2 /T1) n/(n-1) = (500/300) 1.3/0.3 = 9.148 500 1s2 gen = 1.004 ln (300) – 0.287 ln 9.148 – ( – 47.93 300 ) = 0.51287 – 0.635285 + 0.15977 = 0.03736 kJ/kg K P 915 100 T 2 1 500 v 300 Notice: n = 1.3, k = 1.4 2 1 s n<k Sonntag, Borgnakke and van Wylen 8.118 A cylinder/piston contains saturated vapor R-22 at 10°C; the volume is 10 L. The R-22 is compressed to 2 MPa, 60°C in a reversible (internally) polytropic process. If all the heat transfer during the process is with the ambient at 10°C, calculate the net entropy change. Solution: C.V. R-22 of constant mass m2 = m1 = m out to ambient. Energy Eq.5.11: m(u2 − u1) = 1Q2 − 1W2 Entropy Eq.8.14,18: Process: m(s2 − s1) = ∫ dQ/T + 1S2 gen = 1Q2/Tamb + 1S2 gen P1v1n = P2v2n Eq.8.36 State 1: (T1, x1) Table B.4.1 P1 = 0.681 MPa, v1 = 0.03471 m = V1/v1 = 0.01/0.03471 = 0.288 kg State 2: (T2, P2) Table B.4.2 v2 = 0.01214 m3/kg ( ) 0.03471 n 2.0 P2/P1 = 0.681 = 0.01214 => n = 1.0255 From process eq. The work is from Eq.8.38 P2v2 - P1v1 2000 × 0.01214 - 681 × 0.03471 = 0.288 1W2 = ⌠ PdV = m ⌡ 1-n 1 - 1.0255 = −7.26 kJ Heat transfer from energy equation 1Q2 = m(u2 − u1) + 1W2 = 0.288(247.3 − 229.8) − 7.26 = −2.22 kJ ∆SSYST = m(s2 − s1) = 0.288(0.8873 − 0.9129) = −0.00737 kJ/K ∆SSURR = − 1Q2/T0 = +2.22/283.2 = +0.00784 kJ/K Generation is from entropy equation or Eq.8.18 1S2 gen = m(s2 − s1) − 1Q2/Tamb = ∆SNET = ∆SSYST + ∆SSURR = −0.00737 + 0.00784 = +0.00047 kJ/K P 2000 681 L+V T 2 60 1 v 10 Notice: n = 1.03, k = 1.17 2 1 s n<k Sonntag, Borgnakke and van Wylen 8.119 A cylinder/piston contains air at ambient conditions, 100 kPa and 20°C with a volume of 0.3 m3. The air is compressed to 800 kPa in a reversible polytropic process with exponent, n = 1.2, after which it is expanded back to 100 kPa in a reversible adiabatic process. a. Show the two processes in P–v and T–s diagrams. b. Determine the final temperature and the net work. Solution: a) P T 2 P2 2 1 3 P1 1 v 3 m = P1V1/RT1 100 × 0.3 = 0.287 × 293.2 = 0.3565 kg s b) The process equation is expressed in Eq.8.37 n -1 8000.167 T2 = T1(P2/P1) n = 293.2 100 = 414.9 K The work is from Eq.8.38 2 1w2 = ⌠ Pdv = ⌡ 1 P2v2-P1v1 R(T2-T1) 0.287(414.9-293.2) = 1-n = = -174.6 kJ/kg 1-n 1-1.20 Isentropic relation is from Eq.8.32 k-1 1000.286 = 228.9 K T3 = T2 (P3/P2) k = 414.9 800 With zero heat transfer the energy equation gives the work 2w3 = CV0(T2 - T3) = 0.717(414.9 - 228.9) = +133.3 kJ/kg wNET = 0.3565(-174.6 + 133.3) = -14.7 kJ Sonntag, Borgnakke and van Wylen Rates or fluxes of entropy 8.120 A reversible heat pump uses 1 kW of power input to heat a 25oC room, drawing energy from the outside at 15oC. Assuming every process is reversible, what are the total rates of entropy into the heat pump from the outside and from the heat pump to the room? Solution: C.V.TOT. . . . Energy Eq.: QL+ W = QH . . QL QH . . TL Entropy Eq.: T - T = 0 ⇒ QL = QH T L H H . TL . . QH T + W = QH H ⇒ TH . . QH = T - T W H L . QH . 1 1 = T - T W = 25 - 15 (1) = 0.1 kW/K TH H L . . QL QH TL = TH = 0.1 kW/K W QL o 15 C HP QH o 25 C Sonntag, Borgnakke and van Wylen 8.121 An amount of power, say 1000 kW, comes from a furnace at 800°C going into water vapor at 400°C. From the water the power goes to a solid metal at 200°C and then into some air at 70°C. For each location calculate the flux of s through a . surface as (Q/T). What makes the flux larger and larger? Solution: 3 T1 = > T 2 = > T3 = > T4 furnace vapor metal air Flux of s: . Fs = Q/T 1 FURNACE 2 4 AIR FLOW with T as absolute temperature. Fs1 = 1000/1073.15 = 0.932 kW/K, Fs2 = 1000/673.15 = 1.486 kW/K Fs3 = 1000/473.15 = 2.11 kW/K, Fs4 = 1000/343.15 = 2.91 kW/K T T amb 800 1073 400 673 200 476 70 646 ( °C) K Q/T 0.932 1.486 2.114 2.915 kW/K 1S2 gen for every change in T Q over ∆T is an irreversible process Sonntag, Borgnakke and van Wylen 8.122 Room air at 23oC is heated by a 2000 W space heater with a surface filament temperature of 700 K, shown in Fig. P8.122. The room at steady state looses the power to the outside which is at 7oC. Find the rate(s) of entropy generation and specify where it is made. Solution: For any C.V at steady state. The entropy equation as a rate form is Eq.8.43 dSc.v. . . = 0 = ∫ dQ/T + Sgen dt C.V. Heater Element . . Sgen = –∫ dQ/T = -(-2000/700) = 2.857 W/K C.V. Space between heater 700 K and room 23°C . . Sgen = –∫ dQ/T = (-2000 / 700) – [-2000 / (23+273)] = 3.9 W/K C.V. Wall between 23°C inside and 7°C outside . . Sgen = –∫ dQ/T = [-2000 / (23+273)] – [2000 / (7 + 273)] = 0.389 W/K . Notice biggest Sgen is for the largest change in 1/T. Sonntag, Borgnakke and van Wylen 8.123 A small halogen light bulb receives an electrical power of 50 W. The small filament is at 1000 K and gives out 20% of the power as light and the rest as heat transfer to the gas, which is at 500 K; the glass is at 400 K. All the power is absorbed by the room walls at 25oC. Find the rate of generation of entropy in the filament, in the total bulb including glass and the total room including bulb. Solution: . Radiation Wel = 50 W g leads a . Conduction s Q = 10 W RAD glass . QCOND = 40 W We will assume steady state and no storage in the bulb, air or room walls. C.V. Filament steady-state Energy Eq.5.31: Entropy Eq.8.43: . . . dEc.v./dt = 0 = Wel – QRAD – QCOND . . QRAD QCOND . –T + Sgen dSc.v./dt = 0 = – T FILA FILA . . . . 50 Sgen = (QRAD + QCOND)/TFILA = Wel/TFILA = 1000 = 0.05 W/K C.V. Bulb including glass . QRAD leaves at 1000 K . QCOND leaves at 400 K . . Sgen = ∫ dQ/T = -(-10/1000) – (-40/400) = 0.11 W/K C.V. Total room. All energy leaves at 25°C Eq.5.31: Eq.8.43: . . . dEc.v./dt = 0 = Wel – QRAD – QCOND . QTOT . + Sgen dSc.v./dt = 0 = – T WALL . QTOT . Sgen = T = 50/(25+273) = 0.168 W/K WALL Sonntag, Borgnakke and van Wylen 8.124 A farmer runs a heat pump using 2 kW of power input. It keeps a chicken hatchery at a constant 30oC while the room loses 10 kW to the colder outside ambient at 10oC. What is the rate of entropy generated in the heat pump? What is the rate of entropy generated in the heat loss process? Solution: C.V. Hatchery, steady state. To have steady state at 30oC for the hatchery . . . . Energy Eq.: 0 = QH − QLoss ⇒ QH= QLoss = 10 kW C.V. Heat pump, steady state . . . Energy eq.: 0 = QL + W − QH . . . QL = QH − W = 8 kW ⇒ . . QL QH . Entropy Eq.: 0 = T − T + Sgen HP L H . . QH QL . 10 8 Sgen HP = T − T = 273 + 30 − 273 + 10 = 0.00473 kW/K H L C.V. From hatchery at 30oC to the ambient 10oC. This is typically the walls and . the outer thin boundary layer of air. Through this goes QLoss. . . QLoss QLoss . Entropy Eq.: 0 = T − T + Sgen walls H amb . . QLoss QLoss 10 . 10 − T = 283 − 303 = 0.00233 kW/K Sgen walls = T amb H W = 2 kW QL Q leak QH HP cb Sonntag, Borgnakke and van Wylen 8.125 The automatic transmission in a car receives 25 kW shaft work and gives out 24 kW to the drive shaft. The balance is dissipated in the hydraulic fluid and metal casing, all at 45oC, which in turn transmits it to the outer atmosphere at 20oC. What is the rate of entropy generation inside the transmission unit? What is it outside the unit? Solution: 1 kW C.V. Total unit. Steady state and surface at 45oC 24 kW 25 kW Energy Eq: . . . 0 = Win - Wout - Qout . Qout . Entropy Eq.: 0 = - T + Sgen oil . . . From energy Eq.: Qout = Win - Wout = 25 – 24 = 1 kW . Qout . 1 kW From entropy Eq.: Sgen = T = 273.15 + 45 K = 3.1 W/K oil C.V. From surface at 45oC to atm. at 20oC. . . Qout Qout . Entropy Eq.: 0 = T - T + Sgen outside oil amb . . 1 1 1 1 Sgen outside = Qout [T - T ] = 1 kW [293 - 318] = 0.268 W/K amb oil Sonntag, Borgnakke and van Wylen Review problems 8.126 An insulated cylinder/piston has an initial volume of 0.15 m3 and contains steam at 400 kPa, 200oC. The steam is expanded adiabaticly, and the work output is measured very carefully to be 30 kJ. It is claimed that the final state of the water is in the two-phase (liquid and vapor) region. What is your evaluation of the claim? Solution: C.V. Water. Energy Eq.5.11: m(u2 − u1) = 1Q2 − 1W2 Entropy Eq.8.3: m(s2 − s1) = ∫ dQ/T Process: 1Q2 = 0 and reversible State 1: (T, P) Table B.1.3 v1 = 0.5342, u1 = 2646.8, s1 = 7.1706 kJ/kg K T 1 o 130 C P1 V1 0.15 m = v = 0.5342 = 0.2808 kg 1 u = 2540 7.0259 s With the assumed reversible process we have from entropy equation s2 = s1 = 7.1706 kJ/kg K and from the energy equation 30 u2 = u1 − 1W2/m = 2646.8 - 0.2808 = 2540.0 kJ/kg State 2 given by (u, s) check Table B.1.1: sG (at uG = 2540) = 7.0259 < s1 ⇒ State 2 must be in superheated vapor region. Sonntag, Borgnakke and van Wylen 8.127 A closed tank, V = 10 L, containing 5 kg of water initially at 25°C, is heated to 175°C by a heat pump that is receiving heat from the surroundings at 25°C. Assume that this process is reversible. Find the heat transfer to the water and the work input to the heat pump. C.V.: Water from state 1 to state 2. Process: constant volume (reversible isometric) 1: v1 = V/m = 0.002 ⇒ x1 = (0.002 - 0.001003)/43.358 = 0.000023 u1 = 104.86 + 0.000023×2304.9 = 104.93 kJ/kg s1 = 0.3673 + 0.000023×8.1905 = 0.36759 kJ/kg K Continuity eq. (same mass) and V = C fixes v2 2: T2, v2 = v1 ⇒ P 2 T 1 v x2 = (0.002 - 0.001121)/0.21568 = 0.004075 u2 = 740.16 + 0.004075×1840.03 = 747.67 kJ/kg s2 = 2.0909 + 0.004075×4.5347 = 2.1094 kJ/kg K Energy eq. has W = 0, thus provides heat transfer as 1Q2 = m(u2 - u1) = 3213.7 kJ T 2 1 s Entropy equation for the total (tank plus heat pump) control volume gives for a reversible process: ⇒ QL = mT0(s2 - s1) = 2596.6 kJ m(s2 - s1) = QL/T0 and then the energy equation for the heat pump gives WHP = 1Q2 - QL = 617.1 kJ 1Q2 HP WHP Water QL T amb Sonntag, Borgnakke and van Wylen 8.128 Two tanks contain steam, and they are both connected to a piston/cylinder as shown in Fig. P8.128. Initially the piston is at the bottom and the mass of the piston is such that a pressure of 1.4 MPa below it will be able to lift it. Steam in A is 4 kg at 7 MPa, 700°C and B has 2 kg at 3 MPa, 350°C. The two valves are opened, and the water comes to a uniform state. Find the final temperature and the total entropy generation, assuming no heat transfer. Solution: Control mass: All water mA + mB. Continuity Eq.: m2 = mA + mB = 6 kg Energy Eq.5.11: m2u2 - mAuA1 - mBuB1 = 1Q2 - 1W2 = - 1W2 Entropy Eq.8.14: m2s2 - mAsA1 - mBsB1 = 1S2 gen B.1.3: vA1 = 0.06283, uA1 = 3448.5, sA1 = 7.3476 , VA = 0.2513 m3 B.1.3: vB1 = 0.09053, uB1 = 2843.7, sB1 = 6.7428, VB = 0.1811 m3 T P 1400 2 A1 B1 2 V s The only possible P, V combinations for state 2 are on the two lines. Assume V2 > VA + VB ⇒ P2 = Plift , 1W2 = P2(V2 - VA - VB) Substitute into energy equation: m2h2 = mAuA1 + mBuB1 + P2(VA + VB) = 4 × 3448.5 + 2 × 2843.7 + 1400 × 0.4324 State 2: h2 = 3347.8 kJ/kg, P2 = 1400 kPa, v2 = 0.2323, s2 = 7.433 T2 = 441.9 °C, Check assumption: V2 = m2v2 = 1.394 m3 > VA + VB OK. 1S2 gen = 6 × 7.433 - 4 ×7.3476 - 2 × 6.7428 = 1.722 kJ/K Sonntag, Borgnakke and van Wylen 8.129 A piston/cylinder with constant loading of piston contains 1 L water at 400 kPa, quality 15%. It has some stops mounted so the maximum possible volume is 11 L. A reversible heat pump extracting heat from the ambient at 300 K, 100 kPa heats the water to 300°C. Find the total work and heat transfer for the water and the work input to the heat pump. Solution: Take CV around the water and check possible P-V combinations. State 1: v1 = 0.001084 + 0.15×0.46138 = 0.07029 m3/kg u1 = 604.29 + 0.15 × 1949.26 = 896.68 kJ/kg s1 = 1.7766 + 0.15 × 5.1193 = 2.5445 kJ/kg K m1 = V1/v1 = 0.001/0.07029 = 0.0142 kg P 1Q2 HP WHP 1 2 a water QL V T amb T State a: v = 11 v1 = 0.77319 m3/kg, v=C 1 2 a 400 kPa => Sup. vapor Ta = 400oC > T2 s State 2: Since T2 < Ta then piston is not at stops but floating so P2 = 400 kPa. (T, P) => v2 = 0.65484 m3/kg => V2 = (v2/v1) × V1 = 9.316 L 1W2 = ∫ P dV = P(V2 - V1) = 400 (9.316 - 1) × 0.001 = 3.33 kJ 1Q2 = m(u2 − u1) + 1W2 = 0.0142 (2804.8 - 896.68) + 3.33 = 30.43 kJ Take CV as water plus the heat pump out to the ambient. m(s2 − s1) = QL/To => QL = mTo (s2 − s1) = 300×0.0142 (7.5661 - 2.5445) = 21.39 kJ WHP = 1Q2 - QL = 9.04 kJ Sonntag, Borgnakke and van Wylen 8.130 Water in a piston/cylinder is at 1 MPa, 500°C. There are two stops, a lower one at which Vmin = 1 m3 and an upper one at Vmax = 3 m3. The piston is loaded with a mass and outside atmosphere such that it floats when the pressure is 500 kPa. This setup is now cooled to 100°C by rejecting heat to the surroundings at 20°C. Find the total entropy generated in the process. C.V. Water. Initial state: Table B.1.3: v1 = 0.35411 m3/kg, u1 = 3124.3, s1 = 7.7621 m =V/v1 = 3/0.35411 = 8.472 kg T P 1000 500 v=C 1 1 2 2 s v Final state: 100°C and on line in P-V diagram. Notice the following: vg(500 kPa) = 0.3749 > v1, v1 = vg(154°C) Tsat(500 kPa) = 152°C > T2 , so now piston hits bottom stops. State 2: v2 = vbot = Vbot/m = 0.118 m3/kg, x2 = (0.118 − 0.001044)/1.67185 = 0.0699, u2 = 418.91 + 0.0699×2087.58 = 564.98 kJ/kg, s2 = 1.3068 + 0.0699×6.048 = 1.73 kJ/kg K Now we can do the work and then the heat transfer from the energy equation 1W2 = ⌠PdV = 500(V2 - V1) = -1000 kJ (1w2 = -118 kJ/kg) ⌡ 1Q2 = m(u2 - u1) + 1W2 = -22683.4 kJ (1q2 = -2677.5 kJ/kg) Take C.V. total out to where we have 20°C: m(s2 - s1) = 1Q2/T0 + Sgen ⇒ Sgen = m(s2 - s1) − 1Q2/T0 = 8.472 (1.73 - 7.7621) + 22683 / 293.15 = 26.27 kJ/K ( = ∆Swater + ∆Ssur ) Sonntag, Borgnakke and van Wylen 8.131 A cylinder fitted with a frictionless piston contains water. A constant hydraulic pressure on the back face of the piston maintains a cylinder pressure of 10 MPa. Initially, the water is at 700°C, and the volume is 100 L. The water is now cooled and condensed to saturated liquid. The heat released during this process is the Q supply to a cyclic heat engine that in turn rejects heat to the ambient at 30°C. If the overall process is reversible, what is the net work output of the heat engine? C.V.: H2O, 1 3, this is a control mass: Continuity Eq.: m1 = m3 = m P 2 3 1 m(u3-u1) = 1Q3 − 1W3; Energy Eq.: Process: P = C => 1W3 = ∫ P dV = Pm(v3-v1) v o State 1: 700 C, 10 MPa, V1 = 100 L Table B.1.4 v1 = 0.04358 m3/kg => m = m1 = V1/v1 = 2.295 kg h1 = 3870.5 kJ/kg, s1 = 7.1687 kJ/kg K State 3: P3 = P1 = 10 MPa, x3 = 0 T 3 1 2 Table B.1.2 h3 =hf = 1407.5 kJ/Kg, s3 = sf = 3.3595 kJ/Kg K s 1Q3 = m(u3-u1) + Pm(v3 - v1) = m(h3 - h1) = -5652.6 kJ Heat transfer to the heat engine: QH = -1Q3 = 5652.6 kJ Take control volume as total water and heat engine. Process: Rev., ∆Snet = 0 ; H.E. TL = 30oC 2nd Law: ∆Snet = m(s3 - s1) - Qcv/TL ; Qcv = To m(s3 - s1) = -2650.6 kJ => -1Q 3 QL = -Qcv = 2650.6 kJ Wnet = WHE = QH - QL = 3002 kJ QL Tamb WHE Sonntag, Borgnakke and van Wylen 8.132 A cylinder/piston contains 3 kg of water at 500 kPa, 600°C. The piston has a cross-sectional area of 0.1 m2 and is restrained by a linear spring with spring constant 10 kN/m. The setup is allowed to cool down to room temperature due to heat transfer to the room at 20°C. Calculate the total (water and surroundings) change in entropy for the process. State 1: Table B.1.3, v1 = 0.8041, u1 = 3299.6 , s1 = 7.3522 State 2: T2 & on line in P-V diagram. P P = P1 + (ks/A2 l)(V - V1) cy Assume state 2 is two-phase, P2 = Psat(T2) = 2.339 kPa v2 = v1 + (P2 - P1)A2 l/mks cy 1 2 v v2 = 0.8041 + (2.339 - 500)0.01/(3 × 10) = 0.6382 = vf + x2vfg x2 = (0.6382 - 0.001002)/57.7887 = 0.011, u2 = 109.46, s2 = 0.3887 1 1W2 = 2 (P1 + P2)m × (v2 - v1) 1 = 2 (500 + 2.339) × 3 × (0.6382 - 0.8041) = -125 kJ 1Q2 = m(u2 - u1) + 1W2 = 3(109.46 - 3299.6) - 125 = -9695.4 kJ ∆Stot = Sgen,tot = m(s2 - s1) - 1Q2/Troom = 3(0.3887 - 7.3522) + 9695.4/293.15 = 12.18 kJ/K Sonntag, Borgnakke and van Wylen 8.133 An insulated cylinder fitted with a frictionless piston contains saturated vapor R12 at ambient temperature, 20°C. The initial volume is 10 L. The R-12 is now expanded to a temperature of -30°C. The insulation is then removed from the cylinder, allowing it to warm at constant pressure to ambient temperature. Calculate the net work and the net entropy change for the overall process. C.V.: R-12 State 1: T1 = 20oC, V1 = 10 L = 0.01 m3, Sat. Vapor P1 = Pg = 567 kPa, v1 = vg = 0.03078 m3/kg, u1 = ug = 178.32 kJ/kg, x1 = 1.0 m1 =V1/v1 = 0.325 kg s1 = sg = 0.68841 kJ/kg-K State 2: T2 = -30oC Assume 1 2 Adiabatic & Reversible: s2 = s1 = 0.68841 kJ/kg-K s2 = sf + x2sfg; => x2 = 0.95789, P2 = Pg = 100.4 kPa v2 = vf + x2vfg = 0.15269 m3/kg, h2 = hf + x2hfg = 167.23 kJ/kg u2 = h2 - P2v2 = 151.96 kJ/kg State 3: T3 = 20oC, P3 = P2 = 100.41 kPa v3 = 0.19728 m3/kg, h3 = 203.86 kJ/kg, s3 = 0.82812 kJ/kg-K 1st Law: 1 2, 1Q2 = m(u2 − u1) + 1W2 ; 1Q2 = 0 1W2 = m(u1 - u2) = 8.57 kJ 2 3: Process: P = constant => 2W3 = ∫ Pm dv = Pm(v3 - v2) = 1.45 kJ WTOT = 1W2 + 2W3 = 8.57 + 1.45 = 10.02 kJ b) 1st Law: 2 3 2Q3 = m(u3 - u2) + 2W3; 2W3 = Pm(v3-v2) 2Q3 = m(u3 - u2) + Pm(v3 - v2) = m(h3 - h2) = 11.90 kJ 2nd Law: 1 3: To = 20oC, QCV = 1Q2 + 2Q3; 1Q2 = 0 ∆Snet = m(s3 - s1) - QCV/To = 0.0048 kJ/K P T 1 2 1 3 3 2 v s Sonntag, Borgnakke and van Wylen 8.134 A piston/cylinder assembly contains 2 kg of liquid water at 20°C, 100 kPa and it is now heated to 300°C by a source at 500°C. A pressure of 1000 kPa will lift the piston off the lower stops. Find the final volume, work, heat transfer and total entropy generation. Solution: C.V. Water out to source at 500°C. This is a control mass. Energy Eq.5.11: m(u2 - u1) = 1Q2 − 1W2 Entropy Eq.8.14: m(s2 - s1) = 1Q2 / TSOURCE + 1S2 gen Process: V = V1 if P < PLIFT or P = PLIFT if V > V1 Any state of this system must be on the two lines shown in the P-v diagram. Initial state: Table B.1.1: v1 = 0.001002, u1 = 83.94, = 0.2966 V1 = mv1 = 2 × 0.001002 = 0.002 m3 Final state: 300°C and on line in P-V diagram. Now check at state 1a. State 1a: v1a = v1, P = 1000 kPa => compressed liquid T1a < 180°C As final state is at 300°C higher than T1a we must be further out so State 2: 1000 kPa, 300°C => Superheated vapor in Table B.1.3 v2 = 0.25794, u2 = 2793.2, s2 = 7.1228 V2 = mv2 = 2 × 0.25794 = 0.51588 m3 1W2 = ⌠PdV = P2(V2 - V1) = 1000 (0.51588 – 0.002) = 513.9 kJ ⌡ 1Q2 = m(u2 - u1) + 1W2 = 2(2793.2 – 83.94) + 513.9 = 5932 kJ 5932 1S2 gen = m(s2 - s1) − 1Q2/TSOURCE = 2 (7.1228 - 0.2966) − 773.15 = 13.652 – 7.673 = 5.98 kJ/K ( = ∆Swater + ∆Ssur ) T P water 1Q2 500 C 1a 2 1a 1 1 v 2P 2 P1 s Sonntag, Borgnakke and van Wylen 8.135 An uninsulated cylinder fitted with a piston contains air at 500 kPa, 200°C, at which point the volume is 10 L. The external force on the piston is now varied in such a manner that the air expands to 150 kPa, 25 L volume. It is claimed that in this process the air produces 70% of the work that would have resulted from a reversible, adiabatic expansion from the same initial pressure and temperature to the same final pressure. Room temperature is 20°C. a) What is the amount of work claimed? b) Is this claim possible? Solution: C.V.: Air; R = 0.287 kJ/kg-K, Cp = 1.004 kJ/kg K, Cv = 0.717 kJ/kg K State 1: T1 = 200oC, P1 = 500 kPa, V1 = 10 L = 0.01 m3; m1 = V1/v1 = P1V1/RT1 = 0.0368 kg State 2: P2 = 150 kPa, V2 = 25 L = 0.025 m3 ηs = 70%; Actual Work is 70% of Isentropic Work a) Assume Reversible and Adiabatic Process; s1 = s2s k-1 P2 T2s = T1P k = 473.15 (150 / 500) = 335.4 K 1 1st Law: 1Q2s = m(u2s - u1) + 1W2s; 1Q2s = 0 Assume constant specific heat 1W2 s = mCv(T1 - T2s) = 3.63 kJ 1W2 ac = 0.7×1W2 s = 2.54 kJ b) Use Ideal Gas Law; T2 ac = T1P2V2 / P1V1 = 354.9 K 1st Law: 1Q2 ac = mCv(T2 ac - T1) + 1W2 ac = -0.58 kJ Qcv 2nd Law: ∆Snet = m(s2 − s1) - T ; o QCV = 1Q2 ac, To = 20oC T2 P2 s2 − s1 = Cp ln T - R ln P = 0.0569 kJ/kg-K 1 1 ∆Snet = 0.00406 kJ/K > 0 ; Process is Possible Sonntag, Borgnakke and van Wylen 8.136 A cylinder fitted with a piston contains 0.5 kg of R-134a at 60°C, with a quality of 50 percent. The R-134a now expands in an internally reversible polytropic process to ambient temperature, 20°C at which point the quality is 100 percent. Any heat transfer is with a constant-temperature source, which is at 60°C. Find the polytropic exponent n and show that this process satisfies the second law of thermodynamics. Solution: C.V.: R-134a, Internally Reversible, Polytropic Expansion: PVn = Const. Cont.Eq.: m2 = m1 = m ; Entropy Eq.: m(u2 − u1) = 1Q2 − 1W2 Energy Eq.: m(s2 − s1) = ∫ dQ/T + 1S2 gen State 1: T1 = 60oC, x1 = 0.5, Table B.5.1: P1 = Pg = 1681.8 kPa, v1 = vf + x1vfg = 0.000951 + 0.5×0.010511 = 0.006207 m3/kg s1 = sf + x1sfg = 1.2857 + 0.5×0.4182 = 1.4948 kJ/kg K, u1 = uf + x1ufg = 286.19 + 0.5×121.66 = 347.1 kJ/kg State 2: T2 = 20oC, x2 = 1.0, P2 = Pg = 572.8 kPa, Table B.5.1 v2 = vg = 0.03606 m3/kg, s2 = sg = 1.7183 kJ/kg-K u2 = ug = 389.19 kJ/kg Process: P1 v2n PVn = Const. => P = v => 2 1 1W2 = ∫ PdV = P1 v2 n = ln P / ln v = 0.6122 2 1 P2V2 - P1V1 1-n = 0.5(572.8 × 0.03606 - 1681.8 × 0.006207)/(1 - 0.6122) = 13.2 kJ 2nd Law for C.V.: R-134a plus wall out to source: QH Check ∆Snet > 0 ∆Snet = m(s2 − s1) − T ; H QH = 1Q2 = m(u2 − u1) + 1W2 = 34.2 kJ ∆Snet = 0.5(1.7183 - 1.4948) - 34.2/333.15 = 0.0092 kJ/K, ∆Snet > 0 Process Satisfies 2nd Law Sonntag, Borgnakke and van Wylen 8.137 A cylinder with a linear spring-loaded piston contains carbon dioxide gas at 2 MPa with a volume of 50 L. The device is of aluminum and has a mass of 4 kg. Everything (Al and gas) is initially at 200°C. By heat transfer the whole system cools to the ambient temperature of 25°C, at which point the gas pressure is 1.5 MPa. Find the total entropy generation for the process. CO2: m = P1V1/RT1 = 2000 × 0.05/(0.18892 × 473.2) = 1.1186 kg V2 = V1(P1/P2)(T2/T1) = 0.05(2/1.5)(298.2/473.2) = 0.042 m3 ⌠ 1W2 CO2 =⌡ PdV = P1+P2 2000+1500 (0.042 - 0.050) = -14.0 kJ 2 2 (V2 - V1) = 1Q2 CO2 = mCV0(T2-T1) + 1W2 = 1.1186×0.6529(25-200)-14.0 = -141.81 kJ 1Q2 Al = mC(T2 - T1) = 4 × 0.90(25 - 200) = -630 kJ System: CO2 + Al 1Q2 = -141.81 - 630 = -771.81 kJ ∆SSYST = mCO2(s2 - s1)CO2 + mAL(s2 - s1)AL [ 298.2 1.5 = 1.1186 0.8418 ln 473.2 - 0.18892 ln 2.0 + 4 × 0.9 ln(298.2/473.2) = -0.37407 - 1.6623 = -2.0364 kJ/K ∆SSURR = -(1Q2/T0) = + (771.81/298.15) = +2.5887 kJ/K ∆SNET = -2.0364 + 2.5887 = +0.552 kJ/K Al Q CO 2 Tamb Sonntag, Borgnakke and van Wylen 8.138 A vertical cylinder/piston contains R–22 at −20°C, 70% quality, and the volume is 50 L, shown in Fig. P8.138. This cylinder is brought into a 20°C room, and an electric current of 10 A is passed through a resistor inside the cylinder. The voltage drop across the resistor is 12 V. It is claimed that after 30 min the temperature inside the cylinder is 40°C. Is this possible? C.V. The R-22 out to the surroundings, i.e. include walls. Energy Eq.5.11: m(u2 – u1) = 1Q2 – 1W2 Entropy Eq.8.14: Process: m(s2 – s1) = ∫ dQ/T + 1S2 gen = 1Q2/Tamb + 1S2 gen Constant pressure P1 = P2 = 245 kPa v1 = 0.06521 m3/kg, h1 = 176 kJ/kg, s1 = 0.6982 kJ/kg K m = V1/v1 = 0.05/0.06521 = 0.767 kg State 2: Table B.4.2 Interpolate between 200 and 300 kPa h2 = 282.2 kJ/kg, s2 = 1.1033 kJ/kg K Electrical work: WELEC = -Ei ∆t = -12 × 10 × 30 × 60/1000 = -216 kJ Total work: 1W2 = Pm(v2 v1) + WELEC Now substitute into energy equation and solve for Q 1Q2 = m(u2 - u1) + Pm(v2 – v1) + WELEC = m(h2 - h1) + WELEC State 1: Table B.4.1 = 0.767(282.2 - 176.0) - 216 = -134.5 kJ Solve for the entropy generation from entropy equation 1S2 gen = m(s2 – s1) - 1Q2/Tamb 134.5 = 0.767 (1.1033 - 0.6982) + 293.15 = 0.3093 + 0.4587 = +0.768 kJ/K Claim is OK. Sonntag, Borgnakke and van Wylen 8.139 A gas in a rigid vessel is at ambient temperature and at a pressure, P1, slightly higher than ambient pressure, P0. A valve on the vessel is opened, so gas escapes and the pressure drops quickly to ambient pressure. The valve is closed and after a long time the remaining gas returns to ambient temperature at which point the pressure is P2. Develop an expression that allows a determination of the ratio of specific heats, k, in terms of the pressures. C.V.: air remaining in tank, First part of the process is an isentropic expansion s = constant. P1, T0 → P0, Tx Tx/T0 =(P0/P1) k-1 k Second part of the process is a const. vol. heat transfer. P0, Tx → P2, T0 P0 Tx P0 P0 = T ⇒ P = P P2 0 2 1 k-1 k → k= ln (P1/P0) ln (P1/P2) Sonntag, Borgnakke and van Wylen Solutions using the Pr and vr functions in Table A.7.2 8.88 A piston/cylinder setup contains air at 100 kPa, 400 K which is compressed to a final pressure of 1000 kPa. Consider two different processes (i) a reversible adiabatic process and (ii) a reversible isothermal process. Show both processes in P-v and a T-s diagram. Find the final temperature and the specific work for both processes. Solution: C.V. Air, control mass of unknown size and mass. Energy Eq.5.11: u2 – u1 = 1q2 – 1w2 Entropy Eq.8.14: s2 – s1 = ∫ dq/T + 1s2 gen Process: Reversible 1s2 gen = 0 i) dq = 0 so ii) T=C 1q2 = 0 so ∫ dq/T = 1q2/T i) For this process the entropy equation reduces to: s2 – s1 = 0 + 0 so we have constant s, an isentropic process. The relation for air from table A.7.2, constant s becomes Pr2 = Pr1( P2 / P1) = 3.06119 × 10 = 30.6119 u2 = 555.24 kJ/kg From A.7.2 => T2 = 753.6 K and From the energy equation we get the work term 1w2 = u1 – u2 = 286.5 - 555.2 = -268.7 kJ/kg ii) For this process T2 = T1 so since ideal gas we get o o => Energy Eq.: 1w2 = 1q2 u2 = u1 also sT2 = sT1 Now from the entropy equation we solve for 1q2 P2 P2 o o w2 = 1q2 = T(s2 – s1) = T[sT2 – sT1 – R ln P ] = −RT ln P 1 1 1 = − 0.287 × 400 ln 10 = −264 kJ/kg T P 2ii P2 2i 2i P1 2ii 1 1 v s Sonntag, Borgnakke and van Wylen 8.95 A piston/cylinder, shown in Fig. P8.95, contains air at 1380 K, 15 MPa, with V1 = 10 cm3, Acyl = 5 cm2. The piston is released, and just before the piston exits the end of the cylinder the pressure inside is 200 kPa. If the cylinder is insulated, what is its length? How much work is done by the air inside? Solution: C.V. Air, Cylinder is insulated so adiabatic, Q = 0. Continuity Eq.: m2 = m1 = m, Energy Eq.5.11: m(u2 - u1) = 1Q2 - 1W2 = - 1W2 Entropy Eq.8.14: m(s2 - s1) = ∫ dQ/T + 1S2 gen = 0 + 1S2 gen State 1: (T1, P1) State 2: (P2, ?) So one piece of information is needed for the ?, assume reversible process. 1S2 gen = 0 => s2 - s1 = 0 State 1: Table A.7.1: u1 = 1095.2 kJ/kg, Table A.7.2: Pr1 = 340.53 , vr1 = 2.7024 m = P1V1/RT1 = 15000 × 10×10-6 = 0.000379 kg 0.287 × 1380 State 2: P2 and from Entropy eq.: s2 = s1 => Pr2 = Pr1P2/P1 = 340.53×200/15000 = 4.5404 Interpolate in A.7.2 to match the Pr2 value T2 = 447 K, u2 = 320.85 kJ/kg, vr2 = 65.67 ⇒ V2 = V1vr2/vr1 = 10 × 65.67 / 2.7024 = 243 cm3 ⇒ L2 = V2 /Acyl = 243/5 = 48.6 cm ⇒ 1w2 = u1 - u2 = 774.4 kJ/kg, 1W2 = m1w2 = 0.2935 kJ We could also have done V2 = V1 (T2P1/T1P2) from ideal gas law and thus did not need the vr function for this problem Sonntag, Borgnakke and van Wylen 8.107 A mass of 1 kg of air contained in a cylinder at 1.5 MPa, 1000 K, expands in a reversible adiabatic process to 100 kPa. Calculate the final temperature and the work done during the process, using a. Constant specific heat, value from Table A.5 b. The ideal gas tables, Table A.7 Solution: C.V. Air. Continuity Eq.: m2 = m1 = m ; Energy Eq.5.11: m(u2 − u1) = 1Q2 − 1W2 Entropy Eq.8.14: m(s2 − s1) = ∫ dQ/T + 1S2 gen Process: 1Q2 = 0, 1S2 gen = 0 => s2 = s1 a) Using constant Cp from Table A.5 gives the power relation Eq.8.32. k-1 k 0.10.286 = 10001.5 = 460.9 K 1W2 = -(U2 - U1) = mCVo(T1 - T2) T2 = T1(P2/P1) = 1 × 0.717(1000 - 460.9) = 386.5 kJ b) Use the tabulated reduced pressure function that includes variable heat capacity from A.7.2 0.1 Pr2 = Pr1 × P2/P1 = 91.65 × 1.5 = 6.11 Interpolation gives T2 = 486 K and u2 = 349.4 kJ/kg 1W2 = m(u1 - u2) = 1(759.2 - 349.4) = 409.8 kJ Sonntag, Borgnakke and van Wylen 8.112 A piston/cylinder contains air at 300 K, 100 kPa. It is now compressed in a reversible adiabatic process to a volume 7 times as small. Use constant heat capacity and find the final pressure and temperature, the specific work and specific heat transfer for the process. Solution: Here we use the vr function from Table A.7.2 Expansion ratio: v2/ v1 = 1/7 Process eq.: Rev. adiabatic and ideal gas gives Pvn = C, with n = k vr1 = 179.49 => vr2 = vr1 v2/ v1 = 179.49/7 = 25.641 Table A.7.2: Interpolate T2 = 640.7 K P2 = P1× (T2 / T1) × (v1/v2) = 100 × (640.7/300) × 7 = 1495 kPa Adiabatic: 1q2 = 0 kJ/kg Polytropic process work term from Eq.8.38 1w2 = -(u2 – u1) = -(466.37 – 214.36) = -252.0 kJ/kg Notice: Here the solution is done with variable heat capacity.. SOLUTION MANUAL SI UNIT PROBLEMS CHAPTER 9 SONNTAG • BORGNAKKE • VAN WYLEN FUNDAMENTALS of Thermodynamics Sixth Edition Sonntag, Borgnakke and van Wylen CONTENT SUBSECTION Correspondence table Concept-Study Guide Problems Steady State Reversible Processes Single Flow Steady State Processes Multiple Devices and Cycles Steady State Irreversible Processes Transient Processes Reversible Shaft Work, Bernoulli Equation Device efficiency Review Problems Problems resolved with Pr and vr from Table A.7.2: 28, 32, 34, 69, 89, 127 PROB NO. 1-20 21-36 37-46 47-62 63-75 76-94 95-116 117-133 Sonntag, Borgnakke and van Wylen Correspondance Table CHAPTER 9 6th edition The correspondence between the new problem set and the previous 5th edition chapter 9 problem sets. New 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 5th new 1 new 2 new 3 10 new 4 14 new new new 21 new 15 5 6 16 new 20 22 mod 24 70 mod 73 mod 80 mod 8 17 new new new 12 new 50 new 19 13 new New 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 5th 18 23 new new new new 26 30 new new 31 mod new 33 new 27 29 new 40 38 41 39 42 new 43 44 46 48 45 new new new 47 new new new new new 51 New 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 5th 52 53 new new 55 57 mod 54 new 61 63 78 56 58 74 75 new 60 65 new 67 9 11 28 82a 25 36 37 78 80 84 89 90 32 34 mod 35 49 62 Sonntag, Borgnakke and van Wylen Concept-Study Guide Problems 9.1 In a steady state single flow s is either constant or it increases. Is that true? Solution: No. e dq se = si + ⌠ T + sgen ⌡i Entropy can only go up or stay constant due to sgen, but it can go up or down due to the heat transfer which can be positive or negative. So if the heat transfer is large enough it can overpower any entropy generation and drive s up or down. Steady state single flow: 9.2 Which process will make the previous statement true? Solution: If the process is said to be adiabatic then: Steady state adiabatic single flow: se = si + sgen ≥ si 9.3 A reversible adiabatic flow of liquid water in a pump has increasing P. How about T? Solution: e dq Steady state single flow: se = si + ⌠ T + sgen = si + 0 + 0 ⌡i Adiabatic (dq = 0) means integral vanishes and reversible means sgen = 0, so s is constant. Properties for liquid (incompressible) gives Eq.8.19 C ds = T dT then constant s gives constant T. Sonntag, Borgnakke and van Wylen 9.4 A reversible adiabatic flow of air in a compressor has increasing P. How about T? Solution: e dq se = si + ⌠ T + sgen = si + 0 + 0 ⌡i so s is constant. Properties for an ideal gas gives Eq.8.23 and for constant specific heat we get Eq.8.29. A higher P means a higher T, which is also the case for a variable specific heat, recall Eq.8.28 for the standard entropy. Steady state single flow: 9.5 An irreversible adiabatic flow of liquid water in a pump has higher P. How about T? Solution: e dq Steady state single flow: se = si + ⌠ T + sgen = si + 0 + sgen ⌡i so s is increasing. Properties for liquid (incompressible) gives Eq.8.19 where an increase in s gives an increasse in T. 9.6 A compressor receives R-134a at –10oC, 200 kPa with an exit of 1200 kPa, 50oC. What can you say about the process? Solution: Properties for R-134a are found in Table B.5 Inlet state: si = 1.7328 kJ/kg K Exit state: se = 1.7237 kJ/kg K e dq Steady state single flow: se = si + ⌠ T + sgen ⌡i Since s decreases slightly and the generation term can only be positive, it must be that the heat transfer is negative (out) so the integral gives a contribution that is smaller than -sgen. Sonntag, Borgnakke and van Wylen 9.7 An air compressor has a significant heat transfer out. See Example 9.4 for how high T becomes if no heat transfer. Is that good or should it be insulated? Solution: A lower T at a given pressure P means the specific volume is smaller, Ideal gas: Pv = RT ; Shaft work: w = -∫ v dP This gives a smaller work input which is good. 9.8 A large condenser in a steam power plant dumps 15 MW at 45oC with an ambient at 25oC. What is the entropy generation rate? Solution: This process transfers heat over a finite temperature difference between the water inside the condenser and the outside ambient (cooling water from the sea, lake or river or atmospheric air) C.V. The wall that separates the inside 45oC water from the ambient at 25oC. Condensing water Sea water Entropy Eq. 9.1 for steady state operation: cb 45oC dS dt = 0 = ∑ . . . . Q. Q Q + Sgen = T − T + Sgen T 45 25 . 15 MW 15 MW kW Sgen = 25 + 273 K − 45 + 273 K = 3.17 K 25oC Sonntag, Borgnakke and van Wylen 9.9 Air at 1000 kPa, 300 K is throttled to 500 kPa. What is the specific entropy generation? Solution: C.V. Throttle, single flow, steady state. We neglect kinetic and potential energies and there are no heat transfer and shaft work terms. Energy Eq. 6.13: hi = he ⇒ Ti = Te (ideal gas) e dq Entropy Eq. 9.9: se = si + ⌠ T + sgen = si + sgen ⌡i Pe Pe e dT Change in s Eq.8.24: se − si = ⌠ Cp T − R ln P = − R ln P ⌡i i i kJ 500 sgen = se − si = − 0.287 ln 1000 = 0.2 kg K 9.10 Friction in a pipe flow causes a slight pressure decrease and a slight temperature increase. How does that affect entropy? Solution: The friction converts flow work (P drops) into internal energy (T up if single phase). This is an irreversible process and s increases. If liquid: Eq. 8.19: C ds = T dT so s follows T dT dP If ideal gas Eq. 8.23: ds = Cp T − R P (both terms increase) 9.11 A flow of water at some velocity out of a nozzle is used to wash a car. The water then falls to the ground. What happens to the water state in terms of V, T and s? let us follow the water flow. It starts out with kinetic and potential energy of some magnitude at a compressed liquid state P, T. As the water splashes onto the car it looses its kinetic energy (it turns in to internal energy so T goes up by a very small amount). As it drops to the ground it then looses all the potential energy which goes into internal energy. Both of theses processes are irreversible so s goes up. If the water has a temperature different from the ambient then there will also be some heat transfer to or from the water which will affect both T and s. Sonntag, Borgnakke and van Wylen 9.12 The shaft work in a pump to increase the pressure is small compared to the shaft work in an air compressor for the same pressure increase. Why? The reversible work is given by Eq. 9.14 or 9.18 if no kinetic or potential energy changes w = −∫ v dP The liquid has a very small value for v compared to a large value for a gas. 9.13 If the pressure in a flow is constant, can you have shaft work? The reversible work is given by Eq.9.14 2 2 w = −∫ v dP + (Vi – Ve ) + g (Zi – Ze) For a constant pressure the first term drops out but the other two remains. Kinetic energy changes can give work out (windmill) and potential energy changes can give work out (a dam). 9.14 A pump has a 2 kW motor. How much liquid water at 15oC can I pump to 250 kPa from 100 kPa? Incompressible flow (liquid water) and we assume reversible. Then the shaftwork is from Eq.9.18 w = −∫ v dP = −v ∆P = −0.001 m3/kg (250 – 100) kPa = − 0.15 kJ/kg . 2 .W m = -w = 0.15 = 13.3 kg/s Sonntag, Borgnakke and van Wylen 9.15 Liquid water is sprayed into the hot gases before they enter the turbine section of a large gasturbine power plant. It is claimed that the larger mass flow rate produces more work. Is that the reason? No. More mass through the turbine does give more work, but the added mass is only a few percent. As the liquid vaporises the specific volume increases dramatically which gives a much larger volume flow throught the turbine and that gives more work output. . . . . . W = mw = −m∫ v dP = −∫ mv dP = −∫ V dP This should be seen relative to the small work required to bring the liquid water up to the higher turbine inlet pressure from the source of water (presumably atmospheric pressure). 9.16 A polytropic flow process with n = 0 might be which device? As the polytropic process is Pvn = C, then n = 0 is a constant pressure process. This can be a pipe flow, a heat exchanger flow (heater or cooler) or a boiler. 9.17 A steam turbine inlet is at 1200 kPa, 500oC. The exit is at 200 kPa. What is the lowest possible exit temperature? Which efficiency does that correspond to? We would expect the lowest possible exit temperature when the maximum amount of work is taken out. This happens in a reversible process so if we assume it is adiabatic this becomes an isentropic process. Exit: 200 kPa, s = sin = 7.6758 kJ/kg K ⇒ T = 241.9oC The efficiency from Eq.9.27 measures the turbine relative to an isentropic turbine, so the efficiency will be 100%. Sonntag, Borgnakke and van Wylen 9.18 A steam turbine inlet is at 1200 kPa, 500oC. The exit is at 200 kPa. What is the highest possible exit temperature? Which efficiency does that correspond to? The highest possible exit temperature would be if we did not get any work out, i.e. the turbine broke down. Now we have a throttle process with constant h assuming we do not have a significant exit velocity. Exit: 200 kPa, h = hin = 3476.28 kJ/kg ⇒ T = 495oC w η=w =0 Efficiency: s T P i i h=C e e v s Remark: Since process is irreversible there is no area under curve in T-s diagram that correspond to a q, nor is there any area in the P-v diagram corresponding to a shaft work. Sonntag, Borgnakke and van Wylen 9.19 A steam turbine inlet is at 1200 kPa, 500oC. The exit is at 200 kPa, 275oC. What is the isentropic efficiency? Inlet: hin = 3476.28 kJ/kg, sin = 7.6758 kJ/kg K Exit: hex = 3021.4 kJ/kg, sex = 7.8006 kJ/kg K Ideal Exit: 200 kPa, s = sin = 7.6758 kJ/kg K ⇒ hs = 2954.7 kJ/kg wac = hin - hex = 3476.28 – 3021.4 = 454.9 kJ/kg ws = hin - hs = 3476.28 – 2954.7 = 521.6 kJ/kg wac 454.9 η = w = 521.6 = 0.872 s T P 1200 kPa i i es e ac es e ac v 9.20 200 kPa s The exit velocity of a nozzle is 500 m/s. If ηnozzle = 0.88 what is the ideal exit velocity? The nozzle efficiency is given by Eq. 9.30 and since we have the actual exit velocity we get 2 2 Ve s = Vac/ηnozzle ⇒ Ve s = Vac/ ηnozzle = 500 / 0.88 = 533 m/s Sonntag, Borgnakke and van Wylen Steady state reversible processes single flow 9.21 A first stage in a turbine receives steam at 10 MPa, 800°C with an exit pressure of 800 kPa. Assume the stage is adiabatic and negelect kinetic energies. Find the exit temperature and the specific work. Solution: i e C.V. Stage 1 of turbine. The stage is adiabatic so q = 0 and we will assume reversible so sgen = 0 WT Energy Eq.6.13: wT = hi - he se = si + ∫ dq/T + sgen = si + 0 + 0 Entropy Eq.9.8: Inlet state: B.1.3: hi = 4114.9 kJ/kg, si = 7.4077 kJ/kg K Exit state: 800 kPa, s = si Table B.1.3 ⇒ T ≅ 349.7°C, he = 3161 kJ/kg wT = 4114.9 – 3161 = 953.9 kJ/kg T P i i es 10 MPa 800 kPa es v s Sonntag, Borgnakke and van Wylen 9.22 Steam enters a turbine at 3 MPa, 450°C, expands in a reversible adiabatic process and exhausts at 10 kPa. Changes in kinetic and potential energies between the inlet and the exit of the turbine are small. The power output of the turbine is 800 kW. What is the mass flow rate of steam through the turbine? Solution: . C.V. Turbine, Steady single inlet and exit flows. Adiabatic: Q = 0. Continuity Eq.6.11: . . . mhi = mhe + WT, Energy Eq.6.12: Entropy Eq.9.8: . . . mi = me = m, . ( Reversible Sgen = 0 ) . . msi + 0 = mse / P Explanation for the work term is in Sect. 9.3, Eq.9.18 T 1 1 2 2 v Inlet state: Table B.1.3 hi = 3344 kJ/kg, si = 7.0833 kJ/kg K Exit state: Pe , se = si ⇒ Table B.1.2 saturated as se < sg xe = (7.0833 - 0.6492)/7.501 = 0.8578, he = 191.81 + 0.8578 × 2392.82 = 2244.4 kJ/kg . . . m = WT/wT = WT/(hi - he) = 800/(3344 - 2244.4) = 0.728 kg/s s Sonntag, Borgnakke and van Wylen 9.23 A reversible adiabatic compressor receives 0.05 kg/s saturated vapor R-22 at 200 kPa and has an exit presure of 800 kPa. Neglect kinetic energies and find the exit temperature and the minimum power needed to drive the unit. Solution: . C.V. Compressor, Steady single inlet and exit flows. Adiabatic: Q = 0. Continuity Eq.6.11: Energy Eq.6.12: Entropy Eq.9.8: Inlet state: B 4.2.: . . . mi = me = m, . . . mhi = mhe + WC, . . msi + 0 = mse / . ( Reversible Sgen = 0 ) hi = 239.87 kJ/kg, si = 0.9688 kJ/kg K Exit state: Pe , se = si ⇒ Table B.4.2 he = 274.24 kJ/kg, Te ≅ 40°C –wc = he - hi = 274.24 – 239.87 = 34.37 kJ/kg . . – Wc = Power In = –wcm = 34.37 × 0.05 = 1.72 kW P Explanation for the work term is in Sect. 9.3, Eq.9.18 T 2 2 1 1 v s Sonntag, Borgnakke and van Wylen 9.24 In a heat pump that uses R-134a as the working fluid, the R-134a enters the compressor at 150 kPa, −10°C at a rate of 0.1 kg/s. In the compressor the R-134a is compressed in an adiabatic process to 1 MPa. Calculate the power input required to the compressor, assuming the process to be reversible. Solution: . C.V. Compressor, Steady single inlet and exit flows. Adiabatic: Q = 0. Continuity Eq.6.11: Energy Eq.6.12: Entropy Eq.9.8: . . . m1 = m2 = m, . . . mh1 = mh2 + WC, . ( Reversible Sgen = 0 ) h1 = 393.84 kJ/kg, s1 = 1.7606 kJ/kg K . . ms1 + 0 = ms2 / Inlet state: Table B.5.2 Exit state: P2 = 1 MPa & s2 = s1 ⇒ h2 = 434.9 kJ/kg . . . Wc = mwc = m(h1 - h2) = 0.1 × (393.84 - 434.9) = -4.1 kW P Explanation for the work term is in Sect. 9.3 Eq.9.18 T 2 2 1 1 v s Sonntag, Borgnakke and van Wylen 9.25 A boiler section boils 3 kg/s saturated liquid water at 2000 kPa to saturated vapor in a reversible constant pressure process. Assume you do not know that there is no work. Prove that there is no shaftwork using the first and second laws of thermodynamics. Solution: C.V. Boiler. Steady, single inlet and single exit flows. Energy Eq.6.13: hi + q = w + he; Entropy Eq.9.8: si + q/T = se States: Table B.1.2, T = Tsat = 212.42°C = 485.57 K hi = hf = 908.77 kJ/kg, si = 2.4473 kJ/kg K he = hg = 2799.51 kJ/kg, se = 6.3408 kJ/kg K q = T(se – si) = 485.57(6.3408 – 2.4473) = 1890.6 kJ/kg w = hi + q – he = 908.77 + 1890.6 – 2799.51 = -0.1 kJ/kg It should be zero (non-zero due to round off in values of s, h and Tsat). cb Often it is a long pipe and not a chamber Sonntag, Borgnakke and van Wylen 9.26 Consider the design of a nozzle in which nitrogen gas flowing in a pipe at 500 kPa, 200°C, and at a velocity of 10 m/s, is to be expanded to produce a velocity of 300 m/s. Determine the exit pressure and cross-sectional area of the nozzle if the mass flow rate is 0.15 kg/s, and the expansion is reversible and adiabatic. Solution: C.V. Nozzle. Steady flow, no work out and no heat transfer. 2 2 Energy Eq.6.13: hi + Vi /2 = he + Ve/2 Entropy Eq.9.8: si + ⌠ dq/T + sgen = si + 0 + 0 = se ⌡ Properties Ideal gas Table A.5: kJ kJ CPo = 1.042 kg K, R = 0.2968 kg K, k = 1.40 he - hi = CPo(Te - Ti) = 1.042(Te - 473.2) = (102 - 3002)/(2×1000) Solving for exit T: Process: si = se => Te = 430 K, For ideal gas expressed in Eq.8.32 k 430 3.5 Pe = Pi(Te/Ti)k-1 = 500473.2 = 357.6 kPa ve = RTe/Pe = (0.2968 × 430)/357.6 = 0.35689 m3/kg 0.15 × 0.35689 . Ae = mve/Ve = = 1.78 ×10-4 m2 300 P Inlet Exit Vi Ve cb T i i e e v s Sonntag, Borgnakke and van Wylen 9.27 Atmospheric air at -45°C, 60 kPa enters the front diffuser of a jet engine with a velocity of 900 km/h and frontal area of 1 m2. After the adiabatic diffuser the velocity is 20 m/s. Find the diffuser exit temperature and the maximum pressure possible. Solution: C.V. Diffuser, Steady single inlet and exit flow, no work or heat transfer. Energy Eq.6.13: Entropy Eq.9.8: 2 2 hi + Vi /2 = he + Ve/2, and he − hi = Cp(Te − Ti) si + ∫ dq/T + sgen = si + 0 + 0 = se (Reversible, adiabatic) Heat capacity and ratio of specific heats from Table A.5: kJ CPo = 1.004 kg K, k = 1.4, the energy equation then gives: 1.004[ Te - (-45)] = 0.5[(900×1000/3600)2 - 202 ]/1000 = 31.05 kJ/kg => Te = −14.05 °C = 259.1 K Constant s for an ideal gas is expressed in Eq.8.32: k Pe = Pi (Te/Ti)k-1 = 60 (259.1/228.1)3.5 = 93.6 kPa P T 2 2 Fan 1 2 1 1 v s Sonntag, Borgnakke and van Wylen 9.28 A compressor receives air at 290 K, 100 kPa and a shaft work of 5.5 kW from a gasoline engine. It should deliver a mass flow rate of 0.01 kg/s air to a pipeline. Find the maximum possible exit pressure of the compressor. Solution: . C.V. Compressor, Steady single inlet and exit flows. Adiabatic: Q = 0. Continuity Eq.6.11: Energy Eq.6.12: Entropy Eq.9.8: . . . mi = me = m, . . . mhi = mhe + WC, . . . msi + Sgen = mse . ( Reversible Sgen = 0 ) . . .. Wc = mwc => -wc = -W/m = 5.5/0.01 = 550 kJ/kg Use constant specific heat from Table A.5, CPo = 1.004, k = 1.4 he = hi + 550 => Te = Ti + 550/1.004 Te = 290 + 550/1.004 = 837.81 K si = s e k => Pe = Pi (Te/Ti)k-1 Eq.8.32 Pe = 100 × (837.81/290)3.5 = 4098 kPa P i T e ∆ h = 550 kJ/kg e i i v s e -WC Sonntag, Borgnakke and van Wylen 9.29 A compressor is surrounded by cold R-134a so it works as an isothermal compressor. The inlet state is 0°C, 100 kPa and the exit state is saturated vapor. Find the specific heat transfer and specific work. Solution: C.V. Compressor. Steady, single inlet and single exit flows. Energy Eq.6.13: hi + q = w + he; Entropy Eq.9.8: si + q/T = se Inlet state: Table B.5.2, hi = 403.4 kJ/kg, si = 1.8281 kJ/kg K Exit state: Table B.5.1, he = 398.36 kJ/kg, se = 1.7262 kJ/kg K q = T(se – si) = 273.15(1.7262 – 1.8281) = - 27.83 kJ/kg w = 403.4 + (-27.83) – 398.36 = -22.8 kJ/kg P Explanation for the work term is in Sect. 9.3 Eqs. 9.16 and 9.18 T e e i i v s Sonntag, Borgnakke and van Wylen 9.30 A diffuser is a steady-state device in which a fluid flowing at high velocity is decelerated such that the pressure increases in the process. Air at 120 kPa, 30°C enters a diffuser with velocity 200 m/s and exits with a velocity of 20 m/s. Assuming the process is reversible and adiabatic what are the exit pressure and temperature of the air? Solution: C.V. Diffuser, Steady single inlet and exit flow, no work or heat transfer. Energy Eq.6.13: Entropy Eq.9.8: 2 2 hi + Vi /2 = he + Ve /2, => he - hi = CPo(Te - Ti) si + ∫ dq/T + sgen = si + 0 + 0 = se (Reversible, adiabatic) kJ Use constant specific heat from Table A.5, CPo = 1.004 kg K, k = 1.4 Energy equation then gives: CPo(Te - Ti) = 1.004(Te - 303.2) = (2002 - 202)/(2×1000) => Te = 322.9 K The isentropic process (se = si) gives Eq.8.32 k Pe = Pi(Te/Ti)k-1 = 120(322.9/303.2)3.5 = 149.6 kPa P T e i Inlet e i v s Exit Hi V Low P, A Low V Hi P, A Sonntag, Borgnakke and van Wylen 9.31 The exit nozzle in a jet engine receives air at 1200 K, 150 kPa with neglible kinetic energy. The exit pressure is 80 kPa and the process is reversible and adiabatic. Use constant heat capacity at 300 K to find the exit velocity. Solution: C.V. Nozzle, Steady single inlet and exit flow, no work or heat transfer. 2 Energy Eq.6.13: hi = he + Ve /2 Entropy Eq.9.8: ( Zi = Ze ) se = si + ∫ dq/T + sgen = si + 0 + 0 kJ Use constant specific heat from Table A.5, CPo = 1.004 kg K, k = 1.4 The isentropic process (se = si) gives Eq.8.32 => Te = Ti( Pe/Pi) k-1 k = 1200 (80/150) 0.2857 = 1002.7 K The energy equation becomes 2 Ve /2 = hi - he ≅ CP( Ti - Te) Ve = 2 CP( Ti - Te) = P T i Hi P v s Low P Low V i e e 2×1.004(1200-1002.7) × 1000 = 629.4 m/s Hi V Sonntag, Borgnakke and van Wylen 9.32 Do the previous problem using the air tables in A.7 The exit nozzle in a jet engine receives air at 1200 K, 150 kPa with neglible kinetic energy. The exit pressure is 80 kPa and the process is reversible and adiabatic. Use constant heat capacity at 300 K to find the exit velocity. Solution: C.V. Nozzle, Steady single inlet and exit flow, no work or heat transfer. 2 Energy Eq.6.13: hi = he + Ve /2 Entropy Eq.9.8: Process: se = si + ∫ dq/T + sgen = si + 0 + 0 q = 0, Inlet state: ( Zi = Ze ) sgen = 0 as used above leads to se = si o hi = 1277.8 kJ/kg, sTi = 8.3460 kJ/kg K The constant s is rewritten from Eq.8.28 as o o sTe = sTi + R ln(Pe / Pi) = 8.3460 + 0.287 ln (80/150) = 8.1656 Interpolate in A.7 => 8.1656 – 8.1349 Te = 1000 + 50 8.1908 – 8.1349 = 1027.46 K 8.1656 – 8.1349 he = 1046.2 + (1103.5 – 1046.3) × 8.1908 – 8.1349 = 1077.7 2 From the energy equation we have Ve /2 = hi - he , so then Ve = 2 (hi - he) = P T i 2(1277.8 - 1077.7) × 1000 = 632.6 m/s Hi P e e v s Low P Low V i Hi V Sonntag, Borgnakke and van Wylen 9.33 An expander receives 0.5 kg/s air at 2000 kPa, 300 K with an exit state of 400 kPa, 300 K. Assume the process is reversible and isothermal. Find the rates of heat transfer and work neglecting kinetic and potential energy changes. Solution: C.V. Expander, single steady flow. . . . . mhi + Q = mhe + W . . . . Entropy Eq.: msi + Q/T + msgen = mse Process: T is constant and sgen = 0 Ideal gas and isothermal gives a change in entropy by Eq. 8.24, so we can solve for the heat transfer Pe . . . Q = Tm(se – si) = –mRT ln P i 400 = - 0.5 × 300 × 0.287 × ln 2000 = 69.3 kW From the energy equation we get . . . . W = m(hi – he) + Q = Q = 69.3 kW Energy Eq.: P i T i i e Q e e v s Wexp Sonntag, Borgnakke and van Wylen 9.34 Air enters a turbine at 800 kPa, 1200 K, and expands in a reversible adiabatic process to 100 kPa. Calculate the exit temperature and the work output per kilogram of air, using a. The ideal gas tables, Table A.7 b. Constant specific heat, value at 300 K from table A.5 Solution: air i C.V. Air turbine. Adiabatic: q = 0, reversible: sgen = 0 . W Turbine Energy Eq.6.13: Entropy Eq.9.8: e a) Table A.7: wT = hi − he , s e = si o hi = 1277.8 kJ/kg, sTi = 8.34596 kJ/kg K The constant s process is written from Eq.8.28 as Pe o o 100 ⇒ sTe = sTi + R ln( P ) = 8.34596 + 0.287 ln800 = 7.7492 kJ/kg K i ⇒ Te = 706 K, he = 719.9 kJ/kg w = hi - he = 557.9 kJ/kg Interpolate in A.7.1 b) Table A.5: CPo = 1.004 kJ/kg K, R = 0.287 kJ/kg K, k = 1.4, then from Eq.8.32 Te = Ti (Pe/Pi) k-1 k 1000.286 = 1200 800 = 662.1 K w = CPo(Ti - Te) = 1.004(1200 - 662.1) = 539.8 kJ/kg Sonntag, Borgnakke and van Wylen 9.35 A flow of 2 kg/s saturated vapor R-22 at 500 kPa is heated at constant pressure to 60oC. The heat is supplied by a heat pump that receives heat from the ambient at 300 K and work input, shown in Fig. P9.35. Assume everything is reversible and find the rate of work input. Solution: C.V. Heat exchanger . . m1 = m2 ; Continuity Eq.: Energy Eq.: . . . m1h1 + QH = m1h2 Table B.4.2: h1 = 250 kJ/kg, s1 = 0.9267 kJ/kg K 1 QH 2 W HP QL h2 = 293.22 kJ/kg, s2 = 1.0696 kJ/kg K TL . Notice we can find QH but the temperature TH is not constant making it difficult to evaluate the COP of the heat pump. C.V. Total setup and assume everything is reversible and steady state. Energy Eq.: Entropy Eq.: . . . . m1h1 + QL + W = m1h2 . . . m1s1 + QL/TL + 0 = m1s2 (TL is constant, sgen = 0) . . QL = m1TL [s2 - s1] = 2 × 300 [1.0696 – 0.9267] = 85.74 kW . . . W = m1[h2 - h1] - QL = 2 (293.22 – 250) – 85.74 = 0.7 kW Sonntag, Borgnakke and van Wylen 9.36 A reversible steady state device receives a flow of 1 kg/s air at 400 K, 450 kPa and the air leaves at 600 K, 100 kPa. Heat transfer of 800 kW is added from a 1000 K reservoir, 100 kW rejected at 350 K and some heat transfer takes place at 500 K. Find the heat transferred at 500 K and the rate of work produced. Solution: C.V. Device, single inlet and exit flows. Energy equation, Eq.6.12: T3 . . . . . . mh1 + Q3 - Q4 + Q5 = mh2 + W Entropy equation with zero generation, Eq.9.8: T4 Q3 Q4 1 2 . . . . . ms1 + Q3/T3 - Q4/T4+ Q5 /T5 = ms2 Q5 W 500 K Solve for the unknown heat transfer using Table A.7.1 and Eq. 8.28 for change in s T5 . . . T5 . Q5 = T5 [s2 - s1]m + T Q4 - T Q3 4 3 100 500 500 = 500 ×1 (7.5764 – 7.1593 – 0.287 ln 450 ) + 350 ×100 - 1000 × 800 = 424.4 + 142.8 – 400 = 167.2 kW Now the work from the energy equation is . W = 1 × (401.3 – 607.3) + 800 – 100 + 167.2 = 661.2 kW Sonntag, Borgnakke and van Wylen Steady state processes multiple devices and cycles 9.37 Air at 100 kPa, 17°C is compressed to 400 kPa after which it is expanded through a nozzle back to the atmosphere. The compressor and the nozzle are both reversible and adiabatic and kinetic energy in and out of the compressor can be neglected. Find the compressor work and its exit temperature and find the nozzle exit velocity. Solution: 1 T 2 2 -W Separate control volumes around compressor and nozzle. For ideal compressor we have inlet : 1 and exit : 2 P2 P1 1 s 1=3 Energy Eq.6.13: Entropy Eq.9.8: Adiabatic : q = 0. Reversible: sgen = 0 h1 + 0 = wC + h2; s1 + 0/T + 0 = s2 - wC = h2 - h1 , s2 = s 1 Properties Table A.5 air: CPo = 1.004 kJ/kg K, R = 0.287 kJ/kg K, k = 1.4 Process gives constant s (isentropic) which with constant CPo gives Eq.8.32 => ⇒ k-1 T2 = T1( P2/P1) k = 290 (400/100) 0.2857 = 430.9 K −wC = CPo(T2 – T1) = 1.004 (430.9 – 290) = 141.46 kJ/kg The ideal nozzle then expands back down to P1 (constant s) so state 3 equals state 1. The energy equation has no work but kinetic energy and gives: 12 2V = h2 - h1 = -wC = 141 460 J/kg ⇒ V3 = 2×141460 = 531.9 m/s (remember conversion to J) Sonntag, Borgnakke and van Wylen 9.38 A small turbine delivers 150 kW and is supplied with steam at 700°C, 2 MPa. The exhaust passes through a heat exchanger where the pressure is 10 kPa and exits as saturated liquid. The turbine is reversible and adiabatic. Find the specific turbine work, and the heat transfer in the heat exchanger. Solution: 1 2 3 Continuity Eq.6.11: Steady . . . . m1 = m2 = m3 = m -Q WT Turbine: Energy Eq.6.13: wT = h1 − h2 Entropy Eq.9.8: s2 = s1 + sT gen Inlet state: Table B.1.3 h1 = 3917.45 kJ/kg, s1 = 7.9487 kJ/kg K Ideal turbine sT gen = 0, s2 = s1 = 7.9487 = sf2 + x sfg2 State 3: P = 10 kPa, s2 < sg => saturated 2-phase in Table B.1.2 ⇒ x2,s = (s1 - sf2)/sfg2 = (7.9487 - 0.6492)/7.501 = 0.9731 ⇒ h2,s = hf2 + x hfg2 = 191.8 + 0.9731× 2392.8 = 2520.35 kJ/kg wT,s = h1 − h2,s = 1397.05 kJ/kg . . m = W / wT,s = 150 / 1397 = 0.1074 kg/s Heat exchanger: Energy Eq.6.13: q = h3 − h2 , Entropy Eq.9.8: s3 = s2 + ⌠ dq/T + sHe gen ⌡ q = h3 − h2,s = 191.83 - 2520.35 = -2328.5 kJ/kg . . Q = m q = 0.1074 × (-2328.5) = - 250 kW P Explanation for the work term is in Sect. 9.3, Eq.9.18 T 1 3 1 3 2 v 2 s Sonntag, Borgnakke and van Wylen 9.39 One technique for operating a steam turbine in part-load power output is to throttle the steam to a lower pressure before it enters the turbine, as shown in Fig. P9.39. The steamline conditions are 2 MPa, 400°C, and the turbine exhaust pressure is fixed at 10 kPa. Assuming the expansion inside the turbine to be reversible and adiabatic, determine a. The full-load specific work output of the turbine b. The pressure the steam must be throttled to for 80% of full-load output c. Show both processes in a T–s diagram. Solution: a) C.V Turbine. Full load reversible and adiabatic Entropy Eq.9.8 reduces to constant s so from Table B.1.3 and B.1.2 s3 = s1 = 7.1271 = 0.6493 + x3a × 7.5009 => x3a = 0.8636 h3a = 191.83 + 0.8636 × 2392.8 = 2258.3 kJ/kg Energy Eq.6.13 for turbine 1w3a = h1 - h3a = 3247.6 - 2258.3 = 989.3 kJ/kg b) The energy equation for the part load operation and notice that we have constant h in the throttle process. wT = 0.80 × 989.3 = 791.4 = 3247.6 - h3b h3b = 2456.2 = 191.83 + x3b × 2392.8 => x3b = 0.9463 s3b = 0.6492 + 0.9463 × 7.501 = 7.7474 kJ/kg s2b = s3b = 7.7474 P2b = 510 kPa → h2b = h1 = 3247.6 & T2b = 388.4°C c) T 1= 2a 2b h=C 1 3a 3b 2 3 WT s Sonntag, Borgnakke and van Wylen 9.40 Two flows of air both at 200 kPa, one has 1 kg/s at 400 K and the other has 2 kg/s at 290 K. The two lines exchange energy through a number of ideal heat engines taking energy from the hot line and rejecting it to the colder line. The two flows then leave at the same temperature. Assume the whole setup is reversible and find the exit temperature and the total power out of the heat engines. Solution: 1 HE W HE W HE W QL QL QL 2 3 QH QH QH 4 C.V. Total setup . . . . . Energy Eq.6.10: m1h1 + m2h2 = m1h3 + m2h4 + WTOT Entropy Eq.9.7: Process: . . . . . . m1s1 + m2s2 + Sgen + ∫ dQ/T = m1s3 + m2s4 . Sgen = 0 Reversible . Adiabatic Q = 0 Assume the exit flow has the same pressure as the inlet flow then the pressure part of the entropy cancels out and we have Exit same T, P => h3 = h4 = he; s3 = s4 = se . . . . m1h1 + m2h2 = mTOThe + WTOT . . . m1s1 + m2s2 = mTOTse se = · m1 · mTOT Table A.7: s1 + · m2 · mTOT 1 2 s2 = 3 × 7.1593 + 3 × 6.8352 = 6.9432 => Te ≅ 323 K; he = 323.6 . . . WTOT = m1(h1 - he) + m2 (h2 - he) = 1(401.3 – 323.6) + 2(290.43 – 323.6) =11.36 kW Note: The solution using constant heat capacity writes the entropy equation using Eq.8.25, the pressure terms cancel out so we get 1 2 3 Cp ln(Te/T1) + 3 Cp ln(Te/T2) = 0 => lnTe = (lnT1 + 2 lnT2)/3 Sonntag, Borgnakke and van Wylen 9.41 A certain industrial process requires a steady supply of saturated vapor steam at 200 kPa, at a rate of 0.5 kg/s. Also required is a steady supply of compressed air at 500 kPa, at a rate of 0.1 kg/s. Both are to be supplied by the process shown in Fig. P9.41. Steam is expanded in a turbine to supply the power needed to drive the air compressor, and the exhaust steam exits the turbine at the desired state. Air into the compressor is at the ambient conditions, 100 kPa, 20°C. Give the required steam inlet pressure and temperature, assuming that both the turbine and the compressor are reversible and adiabatic. Solution: 4 2 Steam turbine Compressor: s4 = s3 => 3 1 C.V. Each device. Steady flow. Both adiabatic (q = 0) and reversible (sgen = 0). T4 = T3(P4/P3 k-1 )k Air compressor 5000.286 = 293.2100 = 464.6 K . . WC = m3(h3 - h4) = 0.1 × 1.004(293.2 - 464.6) = -17.2 kW Turbine: . . Energy: WT = +17.2 kW = m1(h1 - h2); Entropy: s2 = s1 Table B.1.2: P2 = 200 kPa, x2 = 1 => h2 = 2706.6 kJ/kg, s2 = 7.1271 h1 = 2706.6 + 17.2/0.5 = 2741.0 kJ/kg s1 = s2 = 7.1271 kJ/kg K At h1, s1 → P1 = 242 kPa T1 = 138.3°C Sonntag, Borgnakke and van Wylen 9.42 Consider a steam turbine power plant operating near critical pressure, as shown in Fig. P9.42. As a first approximation, it may be assumed that the turbine and the pump processes are reversible and adiabatic. Neglecting any changes in kinetic and potential energies, calculate a. The specific turbine work output and the turbine exit state b. The pump work input and enthalpy at the pump exit state c. The thermal efficiency of the cycle Solution: QH P1 = P4 = 20 MPa T1 = 700 °C P2 = P3 = 20 kPa T3 = 40 °C 1 4 WT WP, in 3 . QL 2 a) State 1: (P, T) Table B.1.3 C.V. Turbine. Entropy Eq.9.8: Table B.1.2 h1 = 3809.1 kJ/kg, s1 = 6.7993 kJ/kg K s2 = s1 = 6.7993 kJ/kg K s2 = 0.8319 + x2 × 7.0766 => x2 = 0.8433 h2 = 251.4 + 0.8433× 2358.33 = 2240.1 Energy Eq.6.13: wT = h1 - h2 = 1569 kJ/kg b) State 3: (P, T) Compressed liquid, take sat. liq. Table B.1.1 h3 = 167.5 kJ/kg, v3 = 0.001008 m3/kg Property relation in Eq.9.13 gives work from Eq.9.18 as wP = - v3( P4 - P3) = -0.001008(20000 – 20) = -20.1 kJ/kg h4 = h3 - wP = 167.5 + 20.1 = 187.6 kJ/kg c) The heat transfer in the boiler is from energy Eq.6.13 qboiler = h1 - h4 = 3809.1 – 187.6 = 3621.5 kJ/kg wnet = 1569 – 20.1 = 1548.9 kJ/kg 1548.9 ηTH = wnet/qboiler = 3621.5 = 0.428 Sonntag, Borgnakke and van Wylen 9.43 A turbo charger boosts the inlet air pressure to an automobile engine. It consists of an exhaust gas driven turbine directly connected to an air compressor, as shown in Fig. P9.43. For a certain engine load the conditions are given in the figure. Assume that both the turbine and the compressor are reversible and adiabatic having also the same mass flow rate. Calculate the turbine exit temperature and power output. Find also the compressor exit pressure and temperature. Solution: CV: Turbine, Steady single inlet and exit flows, Engine Process: adiabatic: q = 0, reversible: sgen = 0 EnergyEq.6.13: 3 2 s4 = s 3 4 wT = h3 − h4 , Entropy Eq.9.8: W 1 Compressor Turbine The property relation for ideal gas gives Eq.8.32, k from Table A.5 k-1 1000.286 = 793.2 K s4 = s3 → T4 = T3(P4/P3) k = 923.2 170 The energy equation is evaluated with specific heat from Table A.5 wT = h3 − h4 = CP0(T3 - T4) = 1.004(923.2 - 793.2) = 130.5 kJ/kg . . WT = mwT = 13.05 kW C.V. Compressor, steady 1 inlet and 1 exit, same flow rate as turbine. Energy Eq.6.13: Entropy Eq.9.8: -wC = h2 − h1 , s2 = s 1 Express the energy equation for the shaft and compressor having the turbine power as input with the same mass flow rate so we get -wC = wT = 130.5 = CP0(T2 - T1) = 1.004(T2 - 303.2) T2 = 433.2 K The property relation for s2 = s1 is Eq.8.32 and inverted as k 433.23.5 P2 = P1(T2/T1)k-1 = 100303.2 = 348.7 kPa Sonntag, Borgnakke and van Wylen 9.44 A two-stage compressor having an interstage cooler takes in air, 300 K, 100 kPa, and compresses it to 2 MPa, as shown in Fig. P9.44. The cooler then cools the air to 340 K, after which it enters the second stage, which has an exit pressure of 15.74 MPa. Both stages are adiabatic, and reversible. Find q in the cooler, total specific work, and compare this to the work required with no intercooler. Solution: 2 1 · -W 1 intercooler C1 4 3 · Q C2 · -W2 C.V.: Stage 1 air, Steady flow Process: adibatic: q = 0, reversible: sgen = 0 Energy Eq.6.13: -wC1 = h2 − h1 , Entropy Eq.9.8: s 2 = s1 Assume constant CP0 = 1.004 from A.5 and isentropic leads to Eq.8.32 k-1 0.286 T2 = T1(P2/P1) k = 300(2000/100) = 706.7 K wC1 = h1 - h2 = CP0(T1 - T2) = 1.004(300 – 706.7) = -408.3 kJ/kg C.V. Intercooler, no work and no changes in kinetic or potential energy. q23 = h3 - h2 = CP0(T3 - T2) = 1.004(340 – 706.7) = -368.2 kJ/kg C.V. Stage 2. Analysis the same as stage 1. So from Eq.8.32 k-1 0.286 T4 = T3(P4/P3) k = 340(15.74/2) = 613.4 K wC2 = h3 - h4 = CP0(T3 - T4) = 1.004(340 – 613.4) = -274.5 kJ/kg Same flow rate through both stages so the total work is the sum of the two wcomp = wC1 + wC2 = –408.3 – 274.5 = –682.8 kJ/kg For no intercooler (P2 = 15.74 MPa) same analysis as stage 1. So Eq.8.32 T2 = 300(15740/100) 0.286 = 1274.9 K wcomp = 1.004(300 – 1274.9) = –978.8 kJ/kg Sonntag, Borgnakke and van Wylen 9.45 A heat-powered portable air compressor consists of three components: (a) an adiabatic compressor; (b) a constant pressure heater (heat supplied from an outside source); and (c) an adiabatic turbine. Ambient air enters the compressor at 100 kPa, 300 K, and is compressed to 600 kPa. All of the power from the turbine goes into the compressor, and the turbine exhaust is the supply of compressed air. If this pressure is required to be 200 kPa, what must the temperature be at the exit of the heater? Solution: 2 Heater T P2 = 600 kPa, P4 = 200 kPa Adiabatic and reversible compressor: Process: q = 0 and sgen = 0 4 Energy Eq.6.13: h − wc = h2 Entropy Eq.9.8: s 2 = s1 3 qH C 1 For constant specific heat the isentropic relation becomes Eq.8.32 k-1 P2 T2 = T1P k = 300(6)0.2857 = 500.8 K 1 −wc = CP0(T2 - T1) = 1.004(500.8 − 300) = 201.5 kJ/kg q = 0 and sgen = 0 Energy Eq.6.13: h3 = wT + h4 ; Entropy Eq.9.8: s4 = s3 For constant specific heat the isentropic relation becomes Eq.8.32 Adiabatic and reversible turbine: k-1 T4 = T3(P4/P3) k = T3(200/600)0.2857 = 0.7304 T3 −wc = wT = CP0(T3 − T4) Energy Eq. for shaft: 201.5 = 1.004 T3(1 − 0.7304) => T3 = 744.4 K P 23 T 3 600 kPa 2 4 1 300 v 4 1 200 kPa 100 kPa s Sonntag, Borgnakke and van Wylen 9.46 A certain industrial process requires a steady 0.5 kg/s supply of compressed air at 500 kPa, at a maximum temperature of 30°C. This air is to be supplied by installing a compressor and aftercooler. Local ambient conditions are 100 kPa, 20°C. Using an reversible compressor, determine the power required to drive the compressor and the rate of heat rejection in the aftercooler. Solution: Air Table A.5: R = 0.287 kJ/kg-K, Cp = 1.004 kJ/kg K, k = 1.4 . State 1: T1 = To = 20oC, P1 = Po = 100 kPa, m = 0.5 kg/s State 2: P2 = P3 = 500 kPa State 3: T3 = 30oC, P3 = 500 kPa Compressor: Assume Isentropic (adiabatic q = 0 and reversible sgen = 0 ) From entropy equation Eq.9.8 this gives constant s which is expressed for an ideal gas in Eq.8.32 k-1 T2 = T1 (P2/P1) k = 293.15 (500/100)0.2857 = 464.6 K 1st Law Eq.6.13: qc + h1 = h2 + wc; q c = 0, assume constant specific heat from Table A.5 wc = Cp(T1 - T2) = -172.0 kJ/kg . . WC = mwC = -86 kW Aftercooler Energy Eq.6.13: q + h2 = h3 + w; w = 0, assume constant specific heat q = Cp(T3 - T2) = -205 kJ/kg, 1 2 . . Q = mq = -102.5 kW Q cool Compressor -W c Compressor section Aftercooler section 3 Sonntag, Borgnakke and van Wylen Steady state irreversible processes 9.47 Analyze the steam turbine described in Problem 6.78. Is it possible? Solution: 1 C.V. Turbine. Steady flow and adiabatic. 2 . . . Continuity Eq.6.9: m1 = m2 + m3 ; Energy Eq.6.10: . . . . m1h1 = m2h2 + m3h3 + W Entropy Eq.9.7: WT . . . . m1s1 + Sgen = m2s2 + m3s3 3 States from Table B.1.3: s1 = 6.6775, s2 = 6.9562, s3 = 7.14413 kJ/kg K . Sgen = 20×6.9562 + 80×7.14413 - 100×6.6775 = 42.9 kW/K Since it is positive => possible. Notice the entropy is increasing through turbine: s1 < s2 < s3 >0 Sonntag, Borgnakke and van Wylen 9.48 Carbon dioxide at 300 K, 200 kPa is brought through a steady device where it is heated to 500 K by a 600 K reservoir in a constant pressure process. Find the specific work, specific heat transfer and specific entropy generation. Solution: C.V. Heater and walls out to the source. Steady single inlet and exit flows. Since the pressure is constant and there are no changes in kinetic or potential energy between the inlet and exit flows the work is zero. w=0 Continuity Eq.6.11: . . . mi = me = m Energy Eq.6.13: hi + q = he Entropy Eq.9.8, 9.23: si + ∫ dq/T + sgen = se = si + q/Tsource + sgen Properties are from Table A.8 so the energy equation gives q = he - hi = 401.52 – 214.38 = 187.1 kJ/kg From the entropy equation sgen = se - si - q/Tsource = (5.3375 – 4.8631) - 187.1/600 = 0.4744 - 0.3118 = 0.1626 kJ/kg K P 600 K 1 Q 1 2 T 2 600 500 T2 T1 300 v 2 1 s Sonntag, Borgnakke and van Wylen 9.49 Consider the steam turbine in Example 6.6. Is this a reversible process? Solution: At the given states Table B.1.3: si = 6.9552 kJ/kg K; se = 7.3593 kJ/kg K Do the second law for the turbine, Eq.9.8 . . . . mese = misi + ∫ dQ/T + Sgen se = si + ∫ dq/T + sgen sgen = se - si - ∫ dq/T = 7.3593 – 6.9552 – (negative) > 0 Entropy goes up even if q goes out. This is an irreversible process. T P 2 MPa i i e ac 100 kPa e ac v s Sonntag, Borgnakke and van Wylen 9.50 The throttle process described in Example 6.5 is an irreversible process. Find the entropy generation per kg ammonia in the throttling process. Solution: The process is adiabatic and irreversible. The consideration with the energy given in the example resulted in a constant h and two-phase exit flow. Table B.2.1: si = 1.2792 kJ/kg K Table B.2.1: se = sf + xe sfg = 0.5408 + 0.1638 × 4.9265 = 1.34776 kJ/kg K We assumed no heat transfer so the entropy equation Eq.9.8 gives sgen = se - si - ∫ dq/T = 1.34776 – 1.2792 – 0 = 0.0686 kJ/kg K T 1 1.5 MPa 2 i e h=C 291 kPa s Sonntag, Borgnakke and van Wylen 9.51 A geothermal supply of hot water at 500 kPa, 150°C is fed to an insulated flash evaporator at the rate of 1.5 kg/s. A stream of saturated liquid at 200 kPa is drained from the bottom of the chamber and a stream of saturated vapor at 200 kPa is drawn from the top and fed to a turbine. Find the rate of entropy generation in the flash evaporator. Solution: . . . Continuity Eq.6.9: m1 = m2 + m3 . . . Energy Eq.6.10: m1h1 = m2h2 + m3h3 . . . . . Entropy Eq.9.7: m1s1 + Sgen + ∫ dQ/T = m2s2 + m3s3 . Process: Q = 0, irreversible (throttle) 3 1 Two-phase out of the valve. The liquid drops to the bottom. 2 B.1.1 h1 = 632.18 kJ/kg, s1 = 1.8417 kJ/kg K B.1.2 h3 = 2706.63 kJ/kg, s3 = 7.1271 kJ/kg K, h2 = 504.68 kJ/kg, s2 = 1.53 kJ/kg K From the energy equation we solve for the flow rate . . m3 = m1(h1 - h2)/(h3 - h2) = 1.5 × 0.0579 = 0.08685 kg/s Continuity equation gives . . . m2 = m1 - m2 = 1.41315 kg/s Entropy equation now leads to . . . . Sgen = m2s2 + m3s3 - m1s1 = 1.41315 × 1.53 + 0.08685 × 7.127 – 1.5 × 1.8417 = 0.017 kW/K T P 500 kPa 200 kPa 1 2 1 3 2 v 3 s Sonntag, Borgnakke and van Wylen 9.52 Two flowstreams of water, one at 0.6 MPa, saturated vapor, and the other at 0.6 MPa, 600°C, mix adiabatically in a steady flow process to produce a single flow out at 0.6 MPa, 400°C. Find the total entropy generation for this process. Solution: 1: B.1.2 2: B.1.3 3: B.1.3 h1 = 2756.8 kJ/kg, s1 = 6.760 kJ/kg K h2 = 3700.9 kJ/kg, s2 = 8.2674 kJ/kg K h3 = 3270.3 kJ/kg, s3 = 7.7078 kJ/kg K Continuity Eq.6.9: . . . m3 = m1 + m2, Energy Eq.6.10: . . . m3h3 = m1h1 + m2h2 .. => m1/m3 = (h3 – h2) / (h1 – h2) = 0.456 Entropy Eq.9.7: . . . . m3s3 = m1s1 + m2s2 + Sgen => . . .. .. Sgen/m3 = s3 – (m1/m3) s1 – (m2/m3) s2 = 7.7078 – 0.456×6.760 – 0.544×8.2674 = 0.128 kJ/kg K T 1 2 Mixing chamber 600 kPa 3 1 3 2 The mixing process generates entropy. The two inlet flows could have exchanged energy (they have different T) through some heat engines and produced work, the process failed to do that, thus irreversible. s Sonntag, Borgnakke and van Wylen 9.53 A condenser in a power plant receives 5 kg/s steam at 15 kPa, quality 90% and rejects the heat to cooling water with an average temperature of 17°C. Find the power given to the cooling water in this constant pressure process and the total rate of enropy generation when condenser exit is saturated liquid. Solution: C.V. Condenser. Steady state with no shaft work term. Energy Eq.6.12: . . . m hi + Q = mhe Entropy Eq.9.8: . . . . m si + Q/T + Sgen = m se Properties are from Table B.1.2 hi = 225.91 + 0.9 × 2373.14 = 2361.74 kJ/kg , he= 225.91 kJ/kg si = 0.7548 + 0.9 × 7.2536 = 7.283 kJ/kg K, se = 0.7548 kJ/kg K . . . Qout = –Q = m (hi – he) = 5(2361.74 – 225.91) = 10679 kW . . . Sgen = m (se – si) + Qout/T = 5(0.7548 – 7.283) + 10679/(273 + 17) = –32.641 + 36.824 = 4.183 kW/K Sonntag, Borgnakke and van Wylen 9.54 A mixing chamber receives 5 kg/min ammonia as saturated liquid at −20°C from one line and ammonia at 40°C, 250 kPa from another line through a valve. The chamber also receives 325 kJ/min energy as heat transferred from a 40°C reservoir. This should produce saturated ammonia vapor at −20°C in the exit line. What is the mass flow rate in the second line and what is the total entropy generation in the process? Solution: CV: Mixing chamber out to reservoir Continuity Eq.6.9: . . . m1 + m2 = m3 Energy Eq.6.10: . . . . m1h1 + m2h2 + Q = m3h3 Entropy Eq.9.7: . . . . . m1s1 + m2s2 + Q/Tres + Sgen = m3s3 1 2 3 MIXING CHAMBER P 2 . Q 1 3 From the energy equation: . . . m2 = [(m1(h1 - h3) + Q]/(h3 - h2) = [5 × (89.05 - 1418.05) + 325] / (1418.05 - 1551.7) . = 47.288 kg/min ⇒ m3 = 52.288 kg/min . . . . . Sgen = m3s3 – m1s1 – m2s2 – Q/Tres = 52.288 × 5.6158 – 5 × 0.3657 − 47.288 × 5.9599 − 325/313.15 = 8.94 kJ/K min v Sonntag, Borgnakke and van Wylen 9.55 A heat exchanger that follows a compressor receives 0.1 kg/s air at 1000 kPa, 500 K and cools it in a constant pressure process to 320 K. The heat is absorbed by ambient ait at 300 K. Find the total rate of entropy generation. Solution: C.V. Heat exchanger to ambient, steady constant pressure so no work. Energy Eq.6.12: . . . mhi = mhe + Qout Entropy Eq.9.8, 9.23: . . . . msi + Sgen = mse + Qout/T Using Table A.5 and Eq.8.25 for change in s . . . Qout = m(hi – he) = mCPo(Ti – Te) = 0.1 × 1.004(500 – 320) = 18.07 kW . . . . . Sgen = m(se – si) + Qout/T = mCPo ln( Te/Ti ) + Qout/T = 0.1 × 1.004 ln( 320/500) + 18.07/300 = 0.0154 kW/K Using Table A.7.1 and Eq. 8.28 for change in entropy h500 = 503.36 kJ/kg, h320 = 320.58 kJ/kg; sT500 = 7.38692 kJ/kg K, sT320 = 6.93413 kJ/kg K . . Qout = m(hi – he) = 0.1 (503.36 – 320.58) = 18.28 kW . . . Sgen = m(se – si) + Qout/T = 0.1(6.93413 – 7.38692) + 18.28/300 = 0.0156 kW/K Sonntag, Borgnakke and van Wylen 9.56 Air at 327°C, 400 kPa with a volume flow 1 m3/s runs through an adiabatic turbine with exhaust pressure of 100 kPa. Neglect kinetic energies and use constant specific heats. Find the lowest and highest possible exit temperature. For each case find also the rate of work and the rate of entropy generation. Solution: C.V Turbine. Steady single inlet and exit flows, q = 0. vi= RTi/ Pi = 0.287 × 600/400 = 0.4305 m3/kg Inlet state: (T, P) . . m = V/vi = 1/0.4305 = 2.323 kg/s The lowest exit T is for maximum work out i.e. reversible case Process: Reversible and adiabatic => constant s from Eq.9.8 Eq.8.32: k-1 Te = Ti(Pe/Pi) k = 600 × (100/400) 0.2857 = 403.8 K ⇒ w = hi - he = CPo(Ti - Te) = 1.004 × ( 600 – 403.8) = 197 kJ/kg . . WT = mw = 2.323 × 197 = 457.6 kW and . Sgen = 0 Highest exit T occurs when there is no work out, throttling q = ∅; w = ∅ ⇒ hi - he = 0 ⇒ Te = Ti = 600 K Pe . . . 100 Sgen = m (se - si) = - mR ln P = -2.323 × 0.287 ln 400 = 0.924 kW/K i Sonntag, Borgnakke and van Wylen 9.57 In a heat-driven refrigerator with ammonia as the working fluid, a turbine with inlet conditions of 2.0 MPa, 70°C is used to drive a compressor with inlet saturated vapor at −20°C. The exhausts, both at 1.2 MPa, are then mixed together. The ratio of the mass flow rate to the turbine to the total exit flow was measured to be 0.62. Can this be true? Solution: Assume the compressor and the turbine are both adiabatic. C.V. Total: Compressor Continuity Eq.6.11: Energy Eq.6.10: . . . m5 = m1 + m3 . . . m5h5 = m1h1 + m3h3 . . . . Entropy: m5s5 = m1s1 + m3s3 + SC.V.,gen 1 Turbine 4 3 5 2 . . s5 = ys1 + (1-y)s3 + SC.V.,gen/m5 Assume .. y = m1/m5 = 0.62 State 1: Table B.2.2 State 3: Table B.2.1 h1 = 1542.7 kJ/kg, s1 = 4.982 kJ/kg K, h3 = 1418.1 kJ/kg, s3 = 5.616 kJ/kg K Solve for exit state 5 in the energy equation h5 = yh1 + (1-y)h3 = 0.62 × 1542.7 + (1 - 0.62)1418.1 = 1495.4 kJ/kg State 5: h5 = 1495.4 kJ/kg, P5 = 1200 kPa ⇒ s5 = 5.056 kJ/kg K Now check the 2nd law, entropy generation . . ⇒ SC.V.,gen/m5 = s5 - ys1 - (1-y)s3 = -0.1669 Impossible The problem could also have been solved assuming a reversible process and then find the needed flow rate ratio y. Then y would have been found larger than 0.62 so the stated process can not be true. Sonntag, Borgnakke and van Wylen 9.58 Two flows of air both at 200 kPa; one has 1 kg/s at 400 K and the other has 2 kg/s at 290 K. The two flows are mixed together in an insulated box to produce a single exit flow at 200 kPa. Find the exit temperature and the total rate of entropy generation. Solution: 2 Continuity Eq.6.9: . . . m1 + m2 = m3 = 1 + 2 = 3 kg/s 1 3 Energy Eq.6.10: . . . m1h1 + m2h2 = m3h3 Entropy Eq.9.7: . . . . m1s1 + m2s2 + Sgen = m3s3 Using constant specific heats from A.5 and Eq.8.25 for s change. . Divide the energy equation with m3CPo .. .. 1 2 T3 = (m1/m3)T1 + (m2/m3)T2 = 3 × 400 + 3 × 290 = 326.67 K . . . Sgen = m1(s3 - s1) + m2(s3 - s2) = 1 × 1.004 ln (326.67/400) + 2 × 1.004 ln (326.67/290) = 0.0358 kW/K Using A.7.1 and Eq.8.28 for change in s. .. .. 1 2 h3 = (m1/m3)h1 + (m2/m3)h2 = 3 × 401.3 + 3 × 290.43 = 327.39 kJ/kg From A.7.1: T3 = 326.77 K sT3 = 6.9548 kJ/kg K . Sgen = 1(6.9548 – 7.15926) + 2(6.9548 – 6.83521) = 0.0347 kW/K The pressure correction part of the entropy terms cancel out as all three states have the same pressure. Sonntag, Borgnakke and van Wylen 9.59 One type of feedwater heater for preheating the water before entering a boiler operates on the principle of mixing the water with steam that has been bled from the turbine. For the states as shown in Fig. P9.59, calculate the rate of net entropy increase for the process, assuming the process to be steady flow and adiabatic. Solution: CV: Feedwater heater, Steady flow, no external heat transfer. Continuity Eq.6.9: . . . m1 + m2 = m3 Energy Eq.6.10: . . . . m1h1 + (m3 - m1)h2 = m3h3 Properties: All states are given by (P,T) table B.1.1 and B.1.3 h1 = 168.42, h2 = 2828 , h3 = 675.8 s1 = 0.572, all kJ/kg s2 = 6.694 , s3 = 1.9422 all kJ/kg K T 1 FEED WATER HEATER 2 3 2 1 MPa 3 1 Solve for the flow rate from the energy equation . m3(h3 - h2) 4(675.8 - 2828) . m1 = (h - h ) = (168.42 - 2828) = 3.237 kg/s 1 ⇒ 2 . m2 = 4 - 3.237 = 0.763 kg/s . The second law for steady flow, SCV = 0, and no heat transfer, Eq.9.7: . . . . . SC.V.,gen = SSURR = m3s3 - m1s1 - m2s2 = 4(1.9422) - 3.237(0.572) - 0.763(6.694) = 0.8097 kJ/K s s Sonntag, Borgnakke and van Wylen 9.60 A supply of 5 kg/s ammonia at 500 kPa, 20°C is needed. Two sources are available one is saturated liquid at 20°C and the other is at 500 kPa and 140°C. Flows from the two sources are fed through valves to an insulated mixing chamber, which then produces the desired output state. Find the two source mass flow rates and the total rate of entropy generation by this setup. Solution: C.V. mixing chamber + valve. Steady, no heat transfer, no work. Continuity Eq.6.9: . . . m1 + m2 = m3; Energy Eq.6.10: . . . m1 h1 + m2h2 = m3h3 Entropy Eq.9.7: . . . . m1 s1 + m2s2 + Sgen = m3s3 T 1 2 MIXING 2 1 3 3 CHAMBER s State 1: Table B.2.1 h1 = 273.4 kJ/kg, s1= 1.0408 kJ/kg K State 2: Table B.2.2 h2 = 1773.8 kJ/kg, s2 = 6.2422 kJ/kg K State 3: Table B.2.2 h3= 1488.3 kJ/kg, s3= 5.4244 kJ/kg K As all states are known the energy equation establishes the ratio of mass flow rates and the entropy equation provides the entropy generation. . . . . m1 h1 +( m3 - m2)h2 = m3h3 => . . h3 - h2 m1 = m3 h - h = 0.952 kg/s 1 2 . . . m2 = m3 - m1 = 4.05 kg/s . Sgen= 5 × 5.4244 – 0.95 ×1.0408 – 4.05 × 6.2422 = 0.852 kW/K Sonntag, Borgnakke and van Wylen 9.61 A counter flowing heat exchanger has one line with 2 kg/s at 125 kPa, 1000 K entering and the air is leaving at 100 kPa, 400 K. The other line has 0.5 kg/s water coming in at 200 kPa, 20°C and leaving at 200 kPa. What is the exit temperature of the water and the total rate of entropy generation? Solution: C.V. Heat exchanger, steady flow 1 inlet and 1 exit for air and water each. The two flows exchange energy with no heat transfer to/from the outside. 4 2 1 air 3 water . . Energy Eq.6.10: mAIR∆hAIR = mH2O∆hH2O From A.7: h1 - h2 = 1046.22 – 401.3 = 644.92 kJ/kg From B.1.2 h3 = 83.94 kJ/kg; s3 = 0.2966 kJ/kg K . . h4 - h3 = (mAIR/mH2O)(h1 - h2) = (2/0.5)644.92 = 2579.68 kJ/kg h4 = h3 + 2579.68 = 2663.62 kJ/kg < hg at 200 kPa T4 = Tsat = 120.23°C, x4 = (2663.62 – 504.68)/2201.96 = 0.9805, s4 = 1.53 + x4 5.597 = 7.01786 kJ/kg K From entropy Eq.9.7 . . . Sgen = mH2O (s4 - s3) + mAIR(s2 - s1) = 0.5(7.01786 – 0.2966) + 2(7.1593 – 8.1349 – 0.287 ln (100/125)) = 3.3606 – 1.823 = 1.54 kW/K Sonntag, Borgnakke and van Wylen 9.62 A coflowing (same direction) heat exchanger has one line with 0.25 kg/s oxygen at 17°C, 200 kPa entering and the other line has 0.6 kg/s nitrogen at 150 kPa, 500 K entering. The heat exchanger is very long so the two flows exit at the same temperature. Use constant heat capacities and find the exit temperature and the total rate of entropy generation. Solution: 4 C.V. Heat exchanger, steady 2 flows in and two flows out. 1 2 3 . . . . Energy Eq.6.10: mO2h1 + mN2h3 = mO2h2 + mN2h4 Same exit temperature so T4 = T2 with values from Table A.5 . . . . mO2CP O2T1 + mN2CP N2T3 = (mO2CP O2 + mN2CP N2)T2 T2 = 0.25 × 0.922× 290 + 0.6 × 1.042 × 500 379.45 = 0.8557 0.25 × 0.922 + 0.6 × 1.042 = 443.4 K Entropy Eq.9.7 gives for the generation . . . Sgen = mO2(s2 - s1) + mN2(s4 - s3) . . = mO2CP ln (T2/T1) + mN2CP ln (T4/T3) = 0.25 × 0.922 ln (443.4/290) + 0.6 × 1.042 ln (443.4/500) = 0.0979 – 0.0751 = 0.0228 kW/K Sonntag, Borgnakke and van Wylen Transient processes 9.63 Calculate the specific entropy generated in the filling process given in Example 6.11. Solution: C.V. Cannister filling process where: 1Q2 = 0 ; 1W2 = 0 ; m1 = 0 Continuity Eq.6.15: m2 - 0 = min ; Energy Eq.6.16: m2u2 - 0 = minhline + 0 + 0 ⇒ u2 = hline Entropy Eq.9.12: m2s2 - 0 = minsline + 0 + 1S2 gen Inlet state : 1.4 MPa, 300°C, hi = 3040.4 kJ/kg, si = 6.9533 kJ/kg K final state: 1.4 MPa, u2 = hi = 3040.4 kJ/kg => T2 = 452°C, s2 = 7.45896 kJ/kg K 1S2 gen = m2(s2 - si) 1s2 gen = s2 - si = 7.45896 – 6.9533 = 0.506 kJ/kg K T line 2 s Sonntag, Borgnakke and van Wylen 9.64 Calculate the total entropy generated in the filling process given in Example 6.12. Solution: Since the solution to the problem is done in the example we will just add the second law analysis to that. Initial state: Table B.1.2: s1 = 6.9404 kJ/kg K 42 kJ Final state: Table B.1.3: s2 = 6.9533 + 50 × (7.1359 – 6.9533) = 7.1067 kg K Inlet state: Table B.1.3: Entropy Eq.9.12: si = 6.9533 kJ/kg K m2s2 − m1s1 = misi + 1S2 gen Now solve for the generation 1S2 gen = m2s2 − m1s1 - misi = 2.026 × 7.1067 – 0.763 × 6.9404 – 1.263 × 6.9533 = 0.32 kJ/K > 0 Sonntag, Borgnakke and van Wylen 9.65 An initially empty 0.1 m3 cannister is filled with R-12 from a line flowing saturated liquid at −5°C. This is done quickly such that the process is adiabatic. Find the final mass, liquid and vapor volumes, if any, in the cannister. Is the process reversible? Solution: C.V. Cannister filling process where: 1Q2 = 0 ; 1W2 = 0 ; m1 = 0 / / / Continuity Eq.6.15: m2 - 0 = min ; / Energy Eq.6.16: m2u2 - 0 = minhline + 0 + 0 ⇒ u2 = hline / // 2: P2 = PL ; u2 = hL ⇒ 2 phase u2 > uf ; Table B.3.1: u2 = uf + x2ufg uf = 31.26 ; ufg = 137.16 ; hf = 31.45 all kJ/kg x2 = (31.45 -31.26)/137.16 = 0.001385 ⇒ v2 = vf + x2vfg = 0.000708 + 0.001385×0.06426 = 0.000797 m3/kg ⇒ m2 = V/v2 = 125.47 kg ; mf = 125.296 kg; mg = 0.174 kg Vf = mfvf = 0.0887 m3; Vg = mgvg = 0.0113 m3 Process is irreversible (throttling) s2 > sf T line line 2 s Sonntag, Borgnakke and van Wylen 9.66 A 1-m3 rigid tank contains 100 kg R-22 at ambient temperature, 15°C. A valve on top of the tank is opened, and saturated vapor is throttled to ambient pressure, 100 kPa, and flows to a collector system. During the process the temperature inside the tank remains at 15°C. The valve is closed when no more liquid remains inside. Calculate the heat transfer to the tank and total entropy generation in the process. Solution: C.V. Tank out to surroundings. Rigid tank so no work term. m2 - m1 = − me ; Continuity Eq.6.15: Energy Eq.6.16: m2u2 - m1u1 = QCV − mehe Entropy Eq.9.12: m2s2 - m1s1 = QCV/TSUR − mese + Sgen State 1: Table B.3.1, x1 = 0.3149, v1 = V1/m1 = 1/100 = 0.000812 + x1 0.02918 u1 = 61.88 + 0.3149 × 169.47 = 115.25 kJ/kg s1 = 0.2382 + 0.3149 × 0.668 = 0.44855; he = hg = 255.0 kJ/kg State 2: v2 = vg = 0.02999, u2 = ug = 231.35, s2 = 0.9062 kJ/kg K Exit state: he = 255.0, Pe = 100 kPa → Te = -4.7C, se = 1.0917 m2 = 1/0.02999 = 33.34 kg; me = 100 - 33.34 = 66.66 kg QCV = m2u2 - m1u1 + mehe = 33.34×231.35 - 100×115.25 + 66.66×255 = 13 186 kJ ∆SCV = m2s2 - m1s1 = 33.34(0.9062) - 100(0.44855) = -14.642 ∆SSUR = − QCV/TSUR + mese = -13186/288.2 + 66.66(1.0917) = +27.012 Sgen = ∆SNET = -14.642 + 27.012 = +12.37 kJ/K sat vap P e 789 Qcv h=C e 12 v T P=C 12 e s Sonntag, Borgnakke and van Wylen 9.67 Air in a tank is at 300 kPa, 400 K with a volume of 2 m3. A valve on the tank is opened to let some air escape to the ambient to a final pressure inside of 200 kPa. Find the final temperature and mass assuming a reversible adiabatic process for the air remaining inside the tank. Solution: C.V. Total tank. Continuity Eq.6.15: m2 – m1 = –mex Energy Eq.6.16: m2u2 – m1u1 = –mexhex + 1Q2 - 1W2 Entropy Eq.9.12: m2s2 – m1s1 = –mexsex + ∫ dQ/T + 1S2 gen Process: Adiabatic 1Q2 = 0; rigid tank 1W2 = 0 This has too many unknowns (we do not know state 2). C.V. m2 the mass that remains in the tank. This is a control mass. Energy Eq.5.11: m2(u2 – u1) = 1Q2 - 1W2 Entropy Eq.8.14: m2(s2 – s1) = Process: ∫ dQ/T + 1S2 gen Adiabatic 1Q2 = 0; Reversible 1S2 gen = 0 ⇒ s2 = s1 Ideal gas and process Eq.8.32 k-1 P2 T2 = T1P k = 400(200/300)0.2857 = 356.25 K 1 P2V 200 × 2 m2 = RT = = 3.912 kg 0.287 × 356.25 2 Notice that the work term is not zero for mass m2. The work goes into pushing the mass mex out. cb m2 Sonntag, Borgnakke and van Wylen 9.68 An empty cannister of 0.002 m3 is filled with R-134a from a line flowing saturated liquid R-134a at 0°C. The filling is done quickly so it is adiabatic. Find the final mass in the cannister and the total entropy generation. Solution: C.V. Cannister filling process where: 1Q2 = 0 ; 1W2 = 0 ; m1 = 0 / / / Continuity Eq.6.15: m2 - 0 = min ; / Energy Eq.6.16: m2u2 - 0 = minhline + 0 + 0 ⇒ u2 = hline / // Entropy Eq.9.12: m2s2 - 0 = minsline + 0 + 1S2 gen / / Inlet state: Table B.5.1 State 2: P2 = Pline hline = 200 kJ/kg, and sline = 1.0 kJ/kg K u2 = hline = 200 kJ/kg > uf x2 = (200 – 199.77) / 178.24 = 0.00129 v2 = 0.000773 + x2 0.06842 = 0.000861 m3/kg s2 = 1.0 + x2 0.7262 = 1.000937 kJ/kg K m2 = V / v2 = 0.002/0.000861 = 2.323 kg 1S2 gen = m2(s2 - sline) = 2.323 (1.00094 – 1) = 0.0109 kJ/K T line 2 s Sonntag, Borgnakke and van Wylen 9.69 An old abandoned saltmine, 100 000 m3 in volume, contains air at 290 K, 100 kPa. The mine is used for energy storage so the local power plant pumps it up to 2.1 MPa using outside air at 290 K, 100 kPa. Assume the pump is ideal and the process is adiabatic. Find the final mass and temperature of the air and the required pump work. Solution: C.V. The mine volume and the pump Continuity Eq.6.15: m2 - m1 = min Energy Eq.6.16: m2u2 - m1u1 = 1Q2 - 1W2 + minhin Entropy Eq.9.12: m2s2 - m1s1 = ⌠dQ/T + 1S2 gen + minsin ⌡ Process: Adiabatic 1Q2 = 0 , Process ideal 1S2 gen = 0 , s1 = sin ⇒ m2s2 = m1s1 + minsin = (m1 + min)s1 = m2s1 ⇒ s2 = s1 Constant s ⇒ o o sT2 = sTi + R ln(P2 / Pin) Eq.8.28 o sT2 = 6.83521 + 0.287 ln( 21 ) = 7.7090 kJ/kg K ⇒ T2 = 680 K , u2 = 496.94 kJ/kg A.7 m1 = P1V1/RT1 = 100×105/(0.287 × 290) = 1.20149 × 105 kg m2 = P2V2/RT2 = 100 × 21×105/(0.287 × 680) = 10.760 × 105 kg ⇒ min = 9.5585×105 kg 1W2 = minhin + m1u1 - m2u2 = min(290.43) + m1(207.19) - m2(496.94) = -2.322 × 108 kJ P s=C 2 T T2 400 290 1, i v 2 100 kPa 1, i s Sonntag, Borgnakke and van Wylen 9.70 Air in a tank is at 300 kPa, 400 K with a volume of 2 m3. A valve on the tank is opened to let some air escape to the ambient to a final pressure inside of 200 kPa. At the same time the tank is heated so the air remaining has a constant temperature. What is the mass average value of the s leaving assuming this is an internally reversible process? Solution: C.V. Tank, emptying process with heat transfer. Continuity Eq.6.15: m2 - m1 = -me Energy Eq.6.16: m2u2 - m1u1 = -mehe + 1Q2 Entropy Eq.9.12: m2s2 - m1s1 = -mese + 1Q2/T + 0 Process: T2 = T1 Reversible State 1: Ideal gas => 1S2 gen = 0 1Q2 in at 400 K m1 = P1V/RT1 = 300 × 2/0.287 × 400 = 5.2265 kg State 2: 200 kPa, 400 K m2 = P2V/RT2 = 200 × 2/0.287 × 400 = 3.4843 kg => me = 1.7422 kg From the energy equation: 1Q2 = m2u2 - m1u1 + mehe = 3.4843 × 286.49 – 5.2265 × 286.49 + 1.7422 × 401.3 = 1.7422(401.3 – 286.49) = 200 kJ mese = m1s1- m2s2 + 1Q2/T = 5.2265[7.15926 – 0.287 ln (300/100)] – 3.4843[7.15926 – 0.287 ln (200/100)] + (200/400) mese = 35.770 – 24.252 + 0.5 = 12.018 kJ/K se = 12.018/1.7422 = 6.89817 = 6.8982 kJ/kg K Note that the exit state e in this process is for the air before it is throttled across the discharge valve. The throttling process from the tank pressure to ambient pressure is a highly irreversible process. Sonntag, Borgnakke and van Wylen 9.71 An insulated 2 m3 tank is to be charged with R-134a from a line flowing the refrigerant at 3 MPa. The tank is initially evacuated, and the valve is closed when the pressure inside the tank reaches 3 MPa. The line is supplied by an insulated compressor that takes in R-134a at 5°C, quality of 96.5 %, and compresses it to 3 MPa in a reversible process. Calculate the total work input to the compressor to charge the tank. Solution: C.V.: Compressor, R-134a. Steady 1 inlet and 1 exit flow, no heat transfer. 1st Law Eq.6.13: Entropy Eq.9.8: qc + h1 = h1 = h2 + wc s1 + ∫ dq/T + sgen = s1 + 0 = s2 inlet: T1 = 5oC, x1 = 0.965 use Table B.5.1 s1 = sf + x1sfg = 1.0243 + 0.965×0.6995 = 1.6993 kJ/kg K, h1 = hf + x1hfg = 206.8 + 0.965×194.6 = 394.6 kJ/kg exit: P2 = 3 MPa From the entropy eq.: s2 = s1 = 1.6993 kJ/kg K; T2 = 90oC, h2 = 436.2 kJ/kg wc = h1 - h2 = -41.6 kJ/kg C.V.: Tank; VT = 2 m3, PT = 3 MPa 1st Law Eq.6.16: Q + mihi = m2u2 - m1u1 + mehe + W; Process and states have: Q = 0, W = 0, me = 0, m1 = 0, m2 = mi u2 = hi = 436.2 kJ/kg Final state: PT = 3 MPa, u2 = 436.2 kJ/kg TT = 101.9oC, vT = 0.006783 m3/kg mT = VT/vT = 294.84 kg; The work term is from the specific compressor work and the total mass -Wc = mT(-wc) = 12 295 kJ Sonntag, Borgnakke and van Wylen 9.72 An 0.2 m3 initially empty container is filled with water from a line at 500 kPa, 200°C until there is no more flow. Assume the process is adiabatic and find the final mass, final temperature and the total entropy generation. Solution: C.V. The container volume and any valve out to line. Continuity Eq.6.15: m2 - m1 = m2 = mi Energy Eq.6.16: m2u2 - m1u1 = m2u2 = 1Q2 - 1W2 + mihi = mihi Entropy Eq.9.12: m2s2 - m1s1 = m2s2 = ⌠dQ/T + 1S2 gen + misi ⌡ Process: Adiabatic 1Q2 = 0 , Rigid State i: hi = 2855.37 kJ/kg; State 2: 1W2 = 0 Flow stops P2 = Pline si = 7.0592 kJ/kg K 500 kPa, u2 = hi = 2855.37 kJ/kg T2 ≅ 332.9°C , s2 = 7.5737 kJ/kg, => Table B.1.3 v2 = 0.55387 m3/kg m2 = V/v2 = 0.2/0.55387 = 0.361 kg From the entropy equation 1S2 gen = m2s2 - m2si = 0.361(7.5737 – 7.0592) = 0.186 kJ/K T line 2 s Sonntag, Borgnakke and van Wylen 9.73 Air from a line at 12 MPa, 15°C, flows into a 500-L rigid tank that initially contained air at ambient conditions, 100 kPa, 15°C. The process occurs rapidly and is essentially adiabatic. The valve is closed when the pressure inside reaches some value, P2. The tank eventually cools to room temperature, at which time the pressure inside is 5 MPa. What is the pressure P2? What is the net entropy change for the overall process? Solution: CV: Tank. Mass flows in, so this is transient. Find the mass first m1 = P1V/RT1 = 100 × 0.5 = 0.604 kg 0.287 × 288.2 T Fill to P2, then cool to T3 = 15°C, P3 = 5 MPa Mass: 21 12 MPa m3 = m2 = P3V/RT3 = v=C line 5 MPa 5000 × 0.5 = 30.225 kg 0.287 × 288.2 3 100 kPa s mi = m2 - m1 = 30.225 - 0.604 = 29.621 kg In the process 1-2 heat transfer = 0 1st law Eq.6.16: mihi = m2u2 - m1u1 ; T2 = miCP0Ti = m2CV0T2 - m1CV0T1 (29.621×1.004 + 0.604×0.717)×288.2 = 401.2 K 30.225 × 0.717 P2 = m2RT2/V = (30.225 × 0.287 × 401.2)/0.5 = 6.960 MPa Consider now the total process from the start to the finish at state 3. Energy Eq.6.16: QCV + mihi = m2u3 - m1u1 = m2h3 - m1h1 - (P3 - P1)V But, since Ti = T3 = T1, mihi = m2h3 - m1h1 ⇒ QCV = -(P3 - P1)V = -(5000 - 100)0.5 = -2450 kJ From Eqs.9.24-9.26 ∆SNET = m3s3 - m1s1 - misi - QCV/T0 = m3(s3 - si) - m1(s1 - si) - QCV/T0 [ [ 5 0.1 = 30.225 0-0.287 ln 12 - 0.604 0-0.287 ln 12 + (2450 / 288.2) = 15.265 kJ/K Sonntag, Borgnakke and van Wylen 9.74 An initially empty canister of volume 0.2 m3 is filled with carbon dioxide from a line at 1000 kPa, 500 K. Assume the process is adiabatic and the flow continues until it stops by itself. Use constant heat capacity to solve for the final mass and temperature of the carbon dioxide in the canister and the total entropy generated by the process. Solution: C.V. Cannister + valve out to line. No boundary/shaft work, m1 = 0; Q = 0. Continuity Eq.6.15: m2 − 0 = mi Energy Eq.6.16: m2 u2 − 0 = mi hi Entropy Eq.9.12: m2s2 − 0 = misi + 1S2 gen State 2: P2 = Pi and u2 = hi = hline = h2 − RT2 To reduce or eliminate guess use: Energy Eq. becomes: (ideal gas) h2 − hline = CPo(T2 − Tline) CPo(T2 − Tline) − RT2 = 0 T2 = Tline CPo/(CPo − R) = Tline CPo/CVo = k Tline kJ Use A.5: CP = 0.842 kg K, k = 1.289 => T2 = 1.289×500 = 644.5 K m2 = P2V/RT2 = 1000×0.2/(0.1889×644.5) = 1.643 kg 1S2 gen = m2 (s2 − si) = m2[ CP ln(T2 / Tline) − R ln(P2 / Pline)] = 1.644[0.842×ln(1.289) - 0] = 0.351 kJ/K kJ If we use A.8 at 550 K: CP = 1.045 kg K, k = 1.22 => T2 = 610 K, m2 = 1.735 kg P CO2 1 T 2 2 T2 Tline 500 v line s Sonntag, Borgnakke and van Wylen 9.75 A cook filled a pressure cooker with 3 kg water at 20°C and a small amount of air and forgot about it. The pressure cooker has a vent valve so if P > 200 kPa steam escapes to maintain a pressure of 200 kPa. How much entropy was generated in the throttling of the steam through the vent to 100 kPa when half the original mass has escaped? Solution: The pressure cooker goes through a transient process as it heats water up to the boiling temperature at 200 kPa then heats more as saturated vapor at 200 kPa escapes. The throttling process is steady state as it flows from saturated vapor at 200 kPa to 100 kPa which we assume is a constant h process. C.V. Pressure cooker, no work. Continuity Eq.6.15: m2 − m1 = −me Energy Eq.6.16: m2 u2 − m1u1 = −me he + 1Q2 Entropy Eq.9.12: m2s2 − m1s1 = −me se + ∫ dQ/T + 1S2 gen State 1: v1 = vf = 0.001002 m3/kg V = m1v1 = 0.003006 m3 State 2: m2 = m1/2 = 1.5 kg, v2 = V/m2 = 2v1, P2 = 200 kPa Exit: he = hg = 2706.63 kJ/kg, se = sg = 7.1271 kJ/kg K So we can find the needed heat transfer and entropy generation if we know the C.V. surface temperature T. If we assume T for water then 1S2 gen = 0, which is an internally reversible externally irreversible process, there is a ∆T between the water and the source. C.V. Valve, steady flow from state e (200 kPa) to state 3 (at 100 kPa). Energy Eq.: h3 = he Entropy Eq.: s3 = se + es3 gen State 3: 100 kPa, h3 = 2706.63 kJ/kg generation in valve (throttle) Table B.1.3 ⇒ 2706.63 - 2675.46 T3 = 99.62 + (150-99.62) 2776.38 - 2675.46 = 115.2°C s3 = 7.3593 + (7.6133 – 7.3593) 0.30886 = 7.4378 kJ/kg K eS3 gen = me(s3 – se) = 1.5 (7.4378 – 7.1271) = 0.466 kJ/K Sonntag, Borgnakke and van Wylen Reversible shaft work, Bernoulli equation 9.76 A large storage tank contains saturated liquid nitrogen at ambient pressure, 100 kPa; it is to be pumped to 500 kPa and fed to a pipeline at the rate of 0.5 kg/s. How much power input is required for the pump, assuming it to be reversible? Solution: C.V. Pump, liquid is assumed to be incompressible. Table B.6.1 at Pi = 101.3 kPa , vFi = 0.00124 m3/kg Eq.9.18 wPUMP = - wcv = ⌠vdP ≈ vFi(Pe - Pi) ⌡ = 0.00124(500 - 101) = 0.494 kJ/kg liquid i nitrogen . . WPUMP = mwPUMP = 0.5 kg/s (0.494 kJ/kg) = 0.247 kW e Sonntag, Borgnakke and van Wylen 9.77 Liquid water at ambient conditions, 100 kPa, 25°C, enters a pump at the rate of 0.5 kg/s. Power input to the pump is 3 kW. Assuming the pump process to be reversible, determine the pump exit pressure and temperature. Solution: C.V. Pump. Steady single inlet and exit flow with no heat transfer. .. Energy Eq.6.13: w = hi − he = W/m = -3/0.5 = - 6.0 kJ/kg Using also incompressible media we can use Eq.9.18 w = − ⌡vdP ≈ −vi(Pe − Pi) = −0.001003(Pe − 100) ⌠ from which we can solve for the exit pressure Pe = 100 + 6.0/0.001003 = 6082 kPa = 6.082 MPa e Pump . -W . -W = 3 kW, Pi = 100 kPa . Ti = 25°C , m = 0.5 kg/s i Energy Eq.: he = hi − w = 104.87 − (−6) = 110.87 kJ/kg Use Table B.1.4 at 5 MPa => Te = 25.3°C Remark: If we use the software we get: si = 0.36736 = se → Te = 25.1°C At se & Pe Sonntag, Borgnakke and van Wylen 9.78 A small dam has a pipe carrying liquid water at 150 kPa, 20°C with a flow rate of 2000 kg/s in a 0.5 m diameter pipe. The pipe runs to the bottom of the dam 15 m lower into a turbine with pipe diameter 0.35 m. Assume no friction or heat transfer in the pipe and find the pressure of the turbine inlet. If the turbine exhausts to 100 kPa with negligible kinetic energy what is the rate of work? Solution: C.V. Pipe. Steady flow no work, no heat transfer. 1 DAM 2 States: compressed liquid B.1.1 Turbine 3 v2 ≈ v1 ≈ vf = 0.001002 m3/kg . m = ρ AV = AV/v Continuity Eq.6.3: . π V1 = mv1 /A1 = 2000 × 0.001002 / ( 4 0.52 ) = 10.2 m s-1 . π V2 = mv2 /A2 = 2000 × 0.001002 / ( 4 0.352) = 20.83 m s-1 From Bernoulli Eq.9.17 for the pipe (incompressible substance): 2 1 2 v(P2 − P1) + 2 (V2 − V1) + g (Z2 – Z1 ) = ∅ ⇒ 2 1 2 P2 = P1 + [2 (V1 − V2) + g (Z1 – Z2)]/v 1 1 = 150 + [2×10.22 - 2× 20.832 + 9.80665 × 15]/(1000 × 0.001002) = 150 – 17.8 = 132.2 kPa Note that the pressure at the bottom should be higher due to the elevation difference but lower due to the acceleration. Now apply the energy equation Eq.9.14 for the total control volume 1 2 2 w = – ∫ v dP + 2 (V1 − V3) + g(Z1 – Z3 ) 1 = - 0.001002 (100 – 150) + [2×10.22 + 9.80665 × 15] /1000 = 0.25 kJ/kg . . W = mw = 2000 ×0.25 = 500 kW Sonntag, Borgnakke and van Wylen 9.79 A firefighter on a ladder 25 m above ground should be able to spray water an additional 10 m up with the hose nozzle of exit diameter 2.5 cm. Assume a water pump on the ground and a reversible flow (hose, nozzle included) and find the minimum required power. Solution: C.V.: pump + hose + water column, total height difference 35 m. Here V is velocity, not volume. Continuity Eq.6.3, 6.11: Energy Eq.6.12: Process: . . min = mex = (ρAV)nozzle . . . m(-wp) + m(h + V2/2 + gz)in = m(h + V2/2 + gz)ex hin ≅ hex , Vin ≅ Vex = 0 , zex - zin = 35 m , ρ = 1/v ≅ 1/vf -wp = g(zex - zin) = 9.81×(35 - 0) = 343.2 J/kg The velocity in the exit nozzle is such that it can rise 10 m. Make that column a C.V. for which Bernoulli Eq.9.17 is: 1 gznoz + 2V2noz = gzex + 0 Vnoz = = 2g(zex - znoz) 10 m 35 m 2 × 9.81 × 10 = 14 m/s . π D2 m = v 2 Vnoz = ( π/4) 0.0252 × 14 / 0.001 = 6.873 kg/s f . . -Wp = -mwp = 6.873 kg/s × 343.2 J/kg = 2.36 kW Sonntag, Borgnakke and van Wylen 9.80 A small pump is driven by a 2 kW motor with liquid water at 150 kPa, 10°C entering. Find the maximum water flow rate you can get with an exit pressure of 1 MPa and negligible kinetic energies. The exit flow goes through a small hole in a spray nozzle out to the atmosphere at 100 kPa. Find the spray velocity. Solution: C.V. Pump. Liquid water is incompressible so work from Eq.9.18 . . . W = mw = -mv(Pe - Pi) ⇒ .. m= W/ [-v(Pe - Pi) ] = -2/[-0.001003 ( 1000 – 150) ] = 2.35 kg/s C.V Nozzle. No work, no heat transfer, v ≈ constant => Bernoulli Eq.9.17 12 2Vex = v∆P = 0.001 ( 1000 – 100) = 0.9 kJ/kg = 900 J/kg Vex = 2 × 900 J/kg = 42.4 m s -1 Sonntag, Borgnakke and van Wylen 9.81 A garden water hose has liquid water at 200 kPa, 15°C. How high a velocity can be generated in a small ideal nozzle? If you direct the water spray straight up how high will it go? Solution: Liquid water is incompressible and we will assume process is reversible. 1 Bernoulli’s Eq. across the nozzle Eq.9.17: v∆P = ∆(2 V2) 2×0.001001 × (200-101) × 1000 = 14.08 m/s 1 Bernoulli’s Eq.9.17 for the column: ∆(2 V2) = ∆gZ V= 2v∆P = 1 ∆Z = ∆(2 V2)/g = v∆P/g = 0.001001 × (200 – 101) × 1000/9.807 = 10.1 m Sonntag, Borgnakke and van Wylen 9.82 Saturated R-134a at -10°C is pumped/compressed to a pressure of 1.0 MPa at the rate of 0.5 kg/s in a reversible adiabatic process. Calculate the power required and the exit temperature for the two cases of inlet state of the R-134a: a) quality of 100 %. b) quality of 0 %. Solution: . C.V.: Pump/Compressor, m = 0.5 kg/s, R-134a T1 = -10oC, x1 = 1.0 Saturated vapor a) State 1: Table B.5.1, P1 = Pg = 202 kPa, h1 = hg = 392.3 kJ/kg, s1 = sg = 1.7319 kJ/kg K Assume Compressor is isentropic, s2 = s1 = 1.7319 kJ/kg-K h2 = 425.7 kJ/kg, T2 = 45oC 1st Law Eq.6.13: qc + h1 = h2 + wc; wcs = h1 - h2 = -33.4 kJ/kg; b) State 1: T1 = -10oC, x1 = 0 => qc = 0 . . WC = mwC = -16.7 kW Saturated liquid. This is a pump. P1 = 202 kPa, h1 = hf = 186.72 kJ/kg, v1 = vf = 0.000755 m3/kg 1st Law Eq.6.13: qp + h1 = h2 + wp; qp = 0 Assume Pump is isentropic and the liquid is incompressible, Eq.9.18: wps = - ∫ v dP = -v1(P2 - P1) = -0.6 kJ/kg h2 = h1 - wp = 186.72 - ( - 0.6) = 187.3 kJ/kg, P2 = 1 MPa Assume State 2 is approximately a saturated liquid => T2 ≅ -9.6oC . . WP = mwP = -0.3 kW P 2b 1b T 2b 2a 1a 1b v 2a 1a s Sonntag, Borgnakke and van Wylen 9.83 A small water pump on ground level has an inlet pipe down into a well at a depth H with the water at 100 kPa, 15°C. The pump delivers water at 400 kPa to a building. The absolute pressure of the water must be at least twice the saturation pressure to avoid cavitation. What is the maximum depth this setup will allow? Solution: C.V. Pipe in well, no work, no heat transfer From Table B.1.1 e P inlet pump ≥ 2 Psat, 15C = 2×1.705 = 3.41 kPa Process: Assume ∆ KE ≈ ∅ , Bernoulli Eq.9.17: v ∆P + g H = 0 => v ≈ constant. => H i 1000 × 0.001001 ( 3.41 – 100) + 9.80665 × H = 0 ⇒ H = 9.86 m Since flow has some kinetic energy and there are losses in the pipe the height is overestimated. Also the start transient would generate a very low inlet pressure (it moves flow by suction) Sonntag, Borgnakke and van Wylen 9.84 A small pump takes in water at 20°C, 100 kPa and pumps it to 2.5 MPa at a flow rate of 100 kg/min. Find the required pump power input. Solution: C.V. Pump. Assume reversible pump and incompressible flow. This leads to the work in Eq.9.18 wp = -⌠vdP = -vi(Pe - Pi) = -0.001002(2500 - 100) = -2.4 kJ/kg ⌡ . . 100 kg/min Wp = mwp = 60 sec/min (-2.4 kJ/kg) = -4.0 kW Sonntag, Borgnakke and van Wylen 9.85 A pump/compressor pumps a substance from 100 kPa, 10°C to 1 MPa in a reversible adiabatic process. The exit pipe has a small crack, so that a small amount leaks to the atmosphere at 100 kPa. If the substance is (a) water, (b) R-12, find the temperature after compression and the temperature of the leak flow as it enters the atmosphere neglecting kinetic energies. Solution: C.V.: Compressor, reversible adiabatic 2 Eq.6.13: 3 C h1 − wc = h2 ; Eq.9.8: s1 = s2 State 2: P2, s2 = s1 . -Wc C.V.: Crack (Steady throttling process) Eq.6.13: h3 = h2 ; Eq.9.8: s3 = s2 + sgen 1 State 3: P3, h3 = h2 a) Water 1: compressed liquid, Table B.1.1 −wc = + ⌡vdP = vf1(P2 − P1) = 0.001 × (1000 − 100) = 0.9 kJ/kg ⌠ h2 = h1 − wc = 41.99 + 0.9 = 42.89 kJ/kg => T2 = 10.2°C P3 , h3 ⇒ compressed liquid at ~10.2°C P States 1 and 3 are at the same 100 kPa, and same v. You cannot separate them in the P-v fig. T 2 2 1, 3 13 v b) R-12 1: s superheated vapor, Table B.3.2, s1 = 0.8070 kJ/kg K s2 = s1 & P2 ⇒ T2 = 98.5°C , h2 = 246.51 kJ/kg −wc = h2 − h1 = 246.51 - 197.77 = 48.74 kJ/kg P3 , h3 ⇒ T3 = 86.8°C P T 2 2 1 100 kPa 3 1 v 3 h=C s Sonntag, Borgnakke and van Wylen 9.86 Atmospheric air at 100 kPa, 17°C blows at 60 km/h towards the side of a building. Assume the air is nearly incompressible find the pressure and the temperature at the stagnation point (zero velocity) on the wall. Solution: C.V. A stream line of flow from the freestream to the wall. Eq.9.17: V 122 v(Pe-Pi) + 2 (Ve -Vi ) + g(Ze - Zi) = 0 km m 1h Vi = 60 h × 1000 km × 3600 s = 16.667 m/s RTi 0.287 × 290.15 m3 v= P = = 0.8323 kg 100 i 12 16.6672 ∆P = 2v Vi = = 0.17 kPa 0.8323 × 2000 Pe = Pi + ∆ P = 100.17 kPa Then Eq.8.32 for an isentropic process: Te = Ti (Pe/Pi)0.286 = 290.15 × 1.0005 = 290.3 K Very small effect due to low velocity and air is light (large specific volume) Sonntag, Borgnakke and van Wylen 9.87 You drive on the highway with 120 km/h on a day with 17°C, 100 kPa atmosphere. When you put your hand out of the window flat against the wind you feel the force from the air stagnating, i.e. it comes to relative zero velocity on your skin. Assume the air is nearly incompressible and find the air temperature and pressure right on your hand. Solution: 12 Energy Eq.6.13: 2 V + ho = hst 1 1 Tst = To + 2 V2/Cp = 17 + 2 [(120×1000)/3600]2 × (1/1004) = 17 + 555.5/1004 = 17.6°C v = RTo/Po = 0.287 × 290/100 = 0.8323 m3/kg From Bernoulli Eq.9.17: 1 v∆P = 2 V2 1 Pst = Po + 2 V2/v = 100 + 555.5/(0.8323 × 1000) = 100.67 kPa Sonntag, Borgnakke and van Wylen 9.88 An air flow at 100 kPa, 290 K, 200 m/s is directed towards a wall. At the wall the flow stagnates (comes to zero velocity) without any heat transfer. Find the stagnation pressure a) assuming incompressible flow b) assume an adiabatic compression. Hint: T comes from the energy equation. Solution: v = RTo/Po = 0.287 × 290/100 = 0.8323 m3/kg Ideal gas: 121 2 2 V = 2 (200 /1000) = 20 kJ/kg Kinetic energy: a) Reversible and incompressible gives Bernoulli Eq.9.17: 1 ∆P = 2 V2/v = 20/0.8323 0 = 24 kPa Pst = Po + ∆P = 124 kPa St b) adiabatic compression 1 Energy Eq.6.13: 2 V2 + ho = hst 1 hst - ho = 2 V2 = Cp∆T 1 ∆T = 2 V2/Cp = 20/1.004 = 19.92°C => Tst = 290 + 19.92 = 309.92 K Entropy Eq.9.8 assume also reversible process: so + sgen + ⌠(1/T) dq = sst ⌡ as dq = 0 and sgen = 0 then it follows that s = constant This relation gives Eq.8.32: k Tst Pst = Po T k-1 = 100 × (309.92/290)3.5 = 126 kPa o cb Sonntag, Borgnakke and van Wylen 9.89 Calculate the air temperature and pressure at the stagnation point right in front of a meteorite entering the atmosphere (-50 °C, 50 kPa) with a velocity of 2000 m/s. Do this assuming air is incompressible at the given state and repeat for air being a compressible substance going through an adiabatic compression. Solution: 121 2 Kinetic energy: 2 V = 2 (2000) /1000 = 2000 kJ/kg Ideal gas: vatm = RT/P = 0.287 × 223/50 =1.28 m3/kg a) incompressible 1 ∆h = 2 V2 = 2000 kJ/kg Energy Eq.6.13: If A.5 ∆T = ∆h/Cp = 1992 K unreasonable, too high for that Cp Use A.7: 1 hst = ho + 2 V2 = 223.22 + 2000 = 2223.3 kJ/kg Tst = 1977 K Bernoulli (incompressible) Eq.9.17: 1 ∆P = Pst - Po = 2 V2/v = 2000/1.28 = 1562.5 kPa Pst = 1562.5 + 50 = 1612.5 kPa b) compressible Tst = 1977 K the same energy equation. From A.7.1: o sT st = 8.9517 kJ/kg K; o sT o = 6.5712 kJ/kg K Eq.8.28: o o Pst = Po × e(sT st - sT o)/R = 50 × exp [ 8.9517 - 6.5712 0.287 = 200 075 kPa Notice that this is highly compressible, v is not constant. Sonntag, Borgnakke and van Wylen 9.90 Helium gas enters a steady-flow expander at 800 kPa, 300°C, and exits at 120 kPa. The mass flow rate is 0.2 kg/s, and the expansion process can be considered as a reversible polytropic process with exponent, n = 1.3. Calculate the power output of the expander. Solution: i Q CV: expander, reversible polytropic process. From Eq.8.37: e Pe Te = Ti P i Wexp n-1 n 0.3 120 1.3 = 573.2 800 = 370 K Work evaluated from Eq.9.19 nR -1.3 × 2.07703 w = − ⌠vdP = − n-1 (Te - Ti) = (370 - 573.2) ⌡ 0.3 = 1828.9 kJ/kg . . W = mw = 0.2 × 1828.9 = 365.8 kW P T i n = k = 1.667 i n=1 e n = 1.3 v n=1 e n = 1.3 s Sonntag, Borgnakke and van Wylen 9.91 Air at 100 kPa, 300 K, flows through a device at steady state with the exit at 1000 K during which it went through a polytropic process with n = 1.3. Find the exit pressure, the specific work and heat transfer. Solution: C.V. Steady state device, single inlet and single exit flow. Energy Eq.6.13: h1 + q = h2 + w Neglect kinetic, potential energies Entropy Eq.9.8: Te = 1000 K; s1 + ∫ dq/T + sgen = s2 Ti = 300 K; Pi = 100 kPa n n-1 1.3 0.3 = 18 442 kPa Process Eq.8.37: Pe = Pi (Te/ Ti) = 100 (1000/300) and the process leads to Eq.9.19 for the work term n w= R (Te - Ti) = (1.3/-0.3) × 0.287 × (1000 - 300) n-1 = – 849.3 kJ/kg q = he - hi + w = 1046.2 – 300.5 – 849.3 = -103.6 kJ/kg P T e e n=1 i n = k = 1.4 n = 1.3 v n=1 i n = 1.3 s Notice: dP > 0 so dw <0 ds < 0 so dq < 0 Sonntag, Borgnakke and van Wylen 9.92 A flow of 4 kg/s ammonia goes through a device in a polytropic process with an inlet state of 150 kPa, -20°C and an exit state of 400 kPa, 80°C. Find the polytropic exponent n, the specific work and heat transfer. Solution: C.V. Steady state device, single inlet and single exit flow. Energy Eq.6.13: h1 + q = h2 + w Neglect kinetic, potential energies s1 + ∫ dq/T + sgen = s2 Entropy Eq.9.8: Process Eq.8.37: P1v1n = P2v2n: State 1: Table B.2.2 v1= 0.79774, s1= 5.7465 kJ/kg K, h1= 1422.9 kJ/kg State 2: Table B.2.2 v2= 0.4216, s2= 5.9907 kJ/kg K, h2= 1636.7 kJ/kg ln (P2/P1) = n ln (v1/ v2) => 0.98083 = n × 0.63772 n = ln (P2/P1) / ln (v1/ v2) = 1.538 From the process and the integration of v dP gives Eq.9.19. n wshaft = – n–1 (P2v2 – P1v1) = -2.8587 (168.64 –119.66) = -140.0 kJ/kg q = h2+ w - h1 = 1636.7 – 1422.9 – 140 = 73.8 kJ/kg P T 2 n = k = 1.3 2 n = 1.54 n=1 1 n=1 n = 1.54 v 1 s Notice: dP > 0 so dw <0 ds > 0 so dq > 0 Sonntag, Borgnakke and van Wylen 9.93 Carbon dioxide flows through a device entering at 300 K, 200 kPa and leaving at 500 K. The process is steady state polytropic with n = 3.8 and heat transfer comes from a 600 K source. Find the specific work, specific heat transfer and the specific entropy generation due to this process. Solution: C.V. Steady state device, single inlet and single exit flow. Energy Eq.6.13: hi + q = he + w Neglect kinetic, potential energies si + ∫ dq/T + sgen = se Entropy Eq.9.8: Process Eq.8.37: Pe = Pi (Te/ Ti) n n-1 = 200(500/300) 3.8 2.8 = 400 kPa and the process leads to Eq.9.19 for the work term n 3.8 w = -n-1 R (Te - Ti) = -2.8 × 0.1889 × (500 - 300) = -51.3 kJ/kg Energy equation gives q = he - hi + w = 401.52 – 214.38 – 51.3 = 135.8 kJ/kg Entropy equation gives (CV out to source) o o sgen = se – si – q/Tsource = sTe − sTi − R ln(Pe / Pi) – q/Tsource = 5.3375 – 4.8631 – 0.1889 ln (400/200) – (135.8/600) = 0.117 kJ/kg K P T e n=1 i n = 3.8 v Notice: dP > 0 so dw <0 n = k = 1.29 e n = 3.8 n=1 i s ds > 0 so dq > 0 Notice process is externally irreversible, ∆T between source and CO2 Sonntag, Borgnakke and van Wylen 9.94 An expansion in a gas turbine can be approximated with a polytropic process with exponent n = 1.25. The inlet air is at 1200 K, 800 kPa and the exit pressure is 125 kPa with a mass flow rate of 0.75 kg/s. Find the turbine heat transfer and power output. Solution: C.V. Steady state device, single inlet and single exit flow. Energy Eq.6.13: hi + q = he + w Neglect kinetic, potential energies si + ∫ dq/T + sgen = se Entropy Eq.9.8: Process Eq.8.37: Te = Ti (Pe/ Pi) n-1 n = 1200 (125/800) 0.25 1.25 = 827.84 K so the exit enthalpy is from Table A.7.1 27.84 he = 822.2 + 50 (877.4 – 822.2) = 852.94 kJ/kg The process leads to Eq.9.19 for the work term . . . nR 1.25 × 0.287 W = mw = -mn-1 (Te - Ti) = -0.75 × (827.84 - 1200) 0.25 = 400.5 kW Energy equation gives . . . . Q = mq = m(he - hi) + W = 0.75(852.94 – 1277.81) + 400.5 = -318.65 + 400.5 = 81.9 kW T P i i n=1 e n = 1.25 v Notice: dP < 0 so dw > 0 n = k = 1.4 n=1 e n = 1.25 s ds > 0 so dq > 0 Notice this process has some heat transfer in during expansion which is unusual. The typical process would have n = 1.5 with a heat loss. Sonntag, Borgnakke and van Wylen Device efficiency 9.95 Find the isentropic efficiency of the R-134a compressor in Example 6.10 Solution: State 1: Table B.5.2 h1 = 387.2 kJ/kg; s1 = 1.7665 kJ/kg K State 2ac: h2 = 435.1 kJ/kg State 2s: s = 1.7665 kJ/kg K, 800 kPa => h = 431.8 kJ/kg; T = 46.8°C -wc s = h2s - h 1 = 431.8 – 387.2 = 44.6 kJ/kg -wac = 5/0.1 = 50 kJ/kg η = wc s/ wac = 44.6/50 = 0.89 T P e s e ac e ac es i v s Sonntag, Borgnakke and van Wylen 9.96 A compressor is used to bring saturated water vapor at 1 MPa up to 17.5 MPa, where the actual exit temperature is 650°C. Find the isentropic compressor efficiency and the entropy generation. Solution: C.V. Compressor. Assume adiabatic and neglect kinetic energies. Energy Eq.6.13: w = h1 - h 2 Entropy Eq.9.8: s2 = s1 + sgen We have two different cases, the ideal and the actual compressor. States: 1: B.1.2 h1 = 2778.1 kJ/kg, s1 = 6.5865 kJ/kg K 2ac: B.1.3 h2,AC = 3693.9 kJ/kg, 2s: B.1.3 (P, s = s1) IDEAL: -wc,s = h2,s - h1 = 782 kJ/kg Definition Eq.9.28: s2,AC = 6.7357 kJ/kg K h2,s = 3560.1 kJ/kg ACTUAL: -wC,AC = h2,AC - h1 = 915.8 kJ/kg ηc = wc,s/wc,AC = 0.8539 ~ 85% Entropy Eq.9.8: sgen = s2 ac - s1 = 6.7357 - 6.5865 = 0.1492 kJ/kg K Sonntag, Borgnakke and van Wylen 9.97 Liquid water enters a pump at 15°C, 100 kPa, and exits at a pressure of 5 MPa. If the isentropic efficiency of the pump is 75%, determine the enthalpy (steam table reference) of the water at the pump exit. Solution: . CV: pump QCV ≈ 0, ∆KE ≈ 0, ∆PE ≈ 0 2nd law, reversible (ideal) process: ses = si ⇒ Eq.9.18 for work term. es ws = - ⌠vdP ≈ -vi(Pe - Pi) = -0.001001(5000 - 100) = -4.905 kJ/kg ⌡ i Real process Eq.9.28: Energy Eq.6.13: w = ws/ηs = -4.905/0.75 = -6.54 kJ/kg he = hi - w = 62.99 + 6.54 = 69.53 kJ/kg Sonntag, Borgnakke and van Wylen 9.98 A centrifugal compressor takes in ambient air at 100 kPa, 15°C, and discharges it at 450 kPa. The compressor has an isentropic efficiency of 80%. What is your best estimate for the discharge temperature? Solution: C.V. Compressor. Assume adiabatic, no kinetic energy is important. Energy Eq.6.13: w = h1 - h 2 Entropy Eq.9.8: s2 = s1 + sgen We have two different cases, the ideal and the actual compressor. We will solve using constant specific heat. State 2 for the ideal, sgen = 0 so s2 = s1 and it becomes: k-1 P2 k 0.2857 = 288.15 (450 / 100) = 442.83 K Eq.8.32: T2s = T1 P 1 ws = h1 - h2s = Cp (T1 - T2s) = 1.004 (288.15 - 442.83) = -155.299 The actual work from definition Eq.9.28 and then energy equation: wac = -155.299 / 0.8 = -194.12 kJ/kg = h1 - h2 = Cp(T1 - T2) ⇒ T2 = T1 - wac / Cp = 288.15 + 194.12/1.004 = 481.5 K -----------------------------------------------------------------------------Solving using Table A.7.1 instead will give State 1: Table A.7.1: o sT1 = 6.82869 kJ/kg K Now constant s for the ideal is done with Eq.8.28 P2 450 o o sT2s = sT1 + R ln(P ) = 6.82869 + 0.287 ln(100) = 7.26036 kJ/kg K 1 From A.7.1: T2s = 442.1 K and h2s = 443.86 kJ/kg ws = h1 - h2s = 288.57 - 443.86 = -155.29 kJ/kg The actual work from definition Eq.9.28 and then energy equation: wac = -155.29/0.8 = -194.11 kJ/kg ⇒ h2 = 194.11 + 288.57 = 482.68, Table A.7.1: T2 = 480 K Sonntag, Borgnakke and van Wylen 9.99 An emergency drain pump should be able to pump 0.1 m3/s liquid water at 15°C, 10 m vertically up delivering it with a velocity of 20 m/s. It is estimated that the pump, pipe and nozzle have a combined isentropic efficiency expressed for the pump as 60%. How much power is needed to drive the pump? Solution: C.V. Pump, pipe and nozzle together. Steady flow, no heat transfer. Consider the ideal case first (it is the reference for the efficiency). Energy Eq.6.12: . . . mi(hi + V2i/2 + gZi) + Win = me(he + V2e/2 + gZe) Solve for work and use reversible process Eq.9.13 . . Wins = m [he - hi + (V2e -V2i)/2 + g(Ze - Zi)] . = m[( Pe -Pi)v + V2e/2 + g∆Z] . . m = V/v = 0.1/0.001001 = 99.9 kg/s . Wins = 99.9[0 + (202/2) × (1/1000) + 9.807 × (10/1000)] = 99.9(0.2 + 0.09807) = 29.8 kW With the estimated efficiency the actual work, Eq.9.28 is . . Winactual = Wins/η = 29.8/0.6 = 49.7 kW = 50 kW Sonntag, Borgnakke and van Wylen 9.100 A pump receives water at 100 kPa, 15°C and a power input of 1.5 kW. The pump has an isentropic efficiency of 75% and it should flow 1.2 kg/s delivered at 30 m/s exit velocity. How high an exit pressure can the pump produce? Solution: CV Pump. We will assume the ideal and actual pumps have same exit pressure, then we can analyse the ideal pump. Specific work: wac = 1.5/1.2 = 1.25 kJ/kg Ideal work Eq.9.28: ws = η wac = 0.75 × 1.25 = 0.9375 kJ/kg As the water is incompressible (liquid) we get Energy Eq.9.14: ws = (Pe - Pi)v + V2e/2 = (Pe - Pi)0.001001 + (302/2)/1000 = (Pe - Pi)0.001001 + 0.45 Solve for the pressure difference Pe - Pi = (ws – 0.45)/0.001001 = 487 kPa Pe = 587 kPa Sonntag, Borgnakke and van Wylen 9.101 A small air turbine with an isentropic efficiency of 80% should produce 270 kJ/kg of work. The inlet temperature is 1000 K and it exhausts to the atmosphere. Find the required inlet pressure and the exhaust temperature. Solution: C.V. Turbine actual energy Eq.6.13: w = hi - he,ac = 270 kJ/kg Table A.7: hi = 1046.22 ⇒ he,ac = 776.22 kJ/kg, Te = 757.9 K C.V. Ideal turbine, Eq.9.27 and energy Eq.6.13: ws = w/ηs = 270/0.8 = 337.5 = hi - he,s ⇒ he,s = 708.72 kJ/kg From Table A.7: Te,s = 695.5 K Entropy Eq.9.8: si = se,s adiabatic and reversible To relate the entropy to the pressure use Eq.8.28 inverted and standard entropy from Table A.7.1: o o Pe/Pi = exp[ (sTe − sTi ) / R ] = exp[(7.733 - 8.13493)/0.287] = 0.2465 Pi = Pe / 0.2465 = 101.3/0.2465 = 411 kPa T P i i Pi Pe e, ac e, s s = C e, s e, ac v s If constant heat capacity were used Te = Ti - w/Cp = 1000 - 270/1.004 = 731 K C.V. Ideal turbine, Eq.9.27 and energy Eq.6.13: ws = w/ηs = 270/0.8 = 337.5 kJ/kg = hi - he,s = Cp(Ti - Te,s) Te,s = Ti - ws/Cp = 1000 - 337.5/1.004 = 663.8 K Eq.9.8 (adibatic and reversible) gives constant s and relation is Eq.8.32 P /P = (T /T )k/(k-1) ⇒ P = 101.3 (1000/663.8)3.5 = 425 kPa ei e i i Sonntag, Borgnakke and van Wylen 9.102 Repeat Problem 9.42 assuming the turbine and the pump each have an isentropic efficiency of 85%. Solution: QH P1 = P4 = 20 MPa T1 = 700 °C P2 = P3 = 20 kPa T3 = 40 °C ηP = ηT = 85% 1 4 WT WP, in 3 . QL 2 a) State 1: (P, T) Table B.1.3 h1 = 3809.1 kJ/kg, s1 = 6.7993 kJ/kg K C.V. Turbine. First we do the ideal, then the actual. Entropy Eq.9.8: s2 = s1 = 6.7993 kJ/kg K Table B.1.2 s2 = 0.8319 + x2 × 7.0766 => x2 = 0.8433 h2 s = 251.4 + 0.8433 × 2358.33 = 2240.1 kJ/kg Energy Eq.6.13: wT s = h1 - h2 s = 1569 kJ/kg wT AC = ηTwT s = 1333.65 = h1 - h2 AC h2 AC=h1 - wT AC = 2475.45 kJ/kg; x2,AC = (2475.45 - 251.4)/2358.3 = 0.943 , T2,AC=60.06°C b) State 3: (P, T) Compressed liquid, take sat. liq. Table B.1.1 h3 = 167.54 kJ/kg, v3 = 0.001008 m3/kg wP s = - v3( P4 - P3) = -0.001008(20000 – 20) = -20.1 kJ/kg -wP,AC = -wP,s/ηρ = 20.1/0.85 = 23.7 = h4,AC - h3 h4,AC = 191.2 T4,AC ≅ 45.7°C c) The heat transfer in the boiler is from energy Eq.6.13 qboiler = h1 - h4 = 3809.1 – 191.2 = 3617.9 kJ/kg wnet = 1333.65 – 23.7 = 1310 kJ/kg 1310 ηTH = wnet/qboiler = 3617.9 = 0.362 Sonntag, Borgnakke and van Wylen 9.103 Repeat Problem 9.41 assuming the steam turbine and the air compressor each have an isentropic efficiency of 80%. A certain industrial process requires a steady supply of saturated vapor steam at 200 kPa, at a rate of 0.5 kg/s. Also required is a steady supply of compressed air at 500 kPa, at a rate of 0.1 kg/s. Both are to be supplied by the process shown in Fig. P9.41. Steam is expanded in a turbine to supply the power needed to drive the air compressor, and the exhaust steam exits the turbine at the desired state. Air into the compressor is at the ambient conditions, 100 kPa, 20°C. Give the required steam inlet pressure and temperature, assuming that both the turbine and the compressor are reversible and adiabatic. Solution: 4 2 Steam turbine Air Eq.8.32, T4s = T3(P4/P3) 3 1 C.V. Each device. Steady flow. Both adiabatic (q = 0) and actual devices (sgen > 0) given by ηsT and ηsc. k-1 k Air compressor 5000.286 = 293.2100 = 464.6 K . . WCs = m3(h3 - h4s) = 0.1 × 1.004(293.2 - 464.6) = -17.21 kW . . . WCs = m3(h3 - h4) = WCs /ηsc = -17.2/0.80 = -21.5 kW Now the actual turbine must supply the actual compressor work. The actual state 2 is given so we must work backwards to state 1. . . WT = +21.5 kW = m1(h1 - h2) = 0.5(h1 - 2706.6) ⇒ h1 = 2749.6 kJ/kg Also, ηsT = 0.80 = (h1 - h2)/(h1 - h2s) = 43/(2749.6 - h2s) ⇒ h2s = 2695.8 kJ/kg 2695.8 = 504.7 + x2s(2706.6 - 504.7) => x2s = 0.9951 s2s = 1.5301 + 0.9951(7.1271 - 1.5301) = 7.0996 kJ/kg K (s1 = s2s, h1) → P1 = 269 kPa, T1 = 143.5°C Sonntag, Borgnakke and van Wylen 9.104 Steam enters a turbine at 300°C, 600 kPa and exhausts as saturated vapor at 20 kPa. What is the isentropic efficiency? Solution: C.V. Turbine. Steady single inlet and exit flow. To get the efficiency we must compare the actual turbine to the ideal one (the reference). Energy Eq.6.13: wT = h1 - h2 ; Entropy Eq.9.8: s2s = s1 + sgen = s1 Process: Ideal sgen = 0 State 1: Table B.1.3 h1 = 3061.63 kJ/kg, s1 = 7.3723 kJ/kg K State 2s: 20 kPa, s2s = s1 = 7.3723 kJ/kg K < sg so two-phase s - sf 7.3723 - 0.8319 x2s = s = = 0.92423 7.0766 fg h2s = hf + x2s hfg = 251.38 + x2s × 2358.33 = 2431.0 kJ/kg wTs = h1 - h2s = 3061.63 – 2431.0 = 630.61 kJ/kg State 2ac: Table B.1.2 h2ac = 2609.7 kJ/kg, s2ac = 7.9085 kJ/kg K Now we can consider the actual turbine from energy Eq.6.13: T wac = h1 - h2ac = 3061.63 – 2609.7 = 451.93 Then the efficiency from Eq. 9.27 T ηT = wac / wTs = 451.93/630.61 = 0.717 T 1 P1 2s 2ac P2 s Sonntag, Borgnakke and van Wylen 9.105 A turbine receives air at 1500 K, 1000 kPa and expands it to 100 kPa. The turbine has an isentropic efficiency of 85%. Find the actual turbine exit air temperature and the specific entropy increase in the actual turbine. Solution: C.V. Turbine. steady single inlet and exit flow. To analyze the actual turbine we must first do the ideal one (the reference). Energy Eq.6.13: wT = h1 - h2 ; Entropy Eq.9.8: s2 = s1 + sgen = s1 Entropy change in Eq.8.28 and Table A.7.1: o o sT2 = sT1 + R ln(P2/P1) = 8.61208 + 0.287 ln(100/1000) = 7.95124 Interpolate in A.7 => T2s = 849.2, h2s = 876.56 => wT = 1635.8 - 876.56 = 759.24 kJ/kg Now we can consider the actual turbine from Eq.9.27 and Eq.6.13: T wac = ηT wT = 0.85 × 759.24 = 645.35 = h1 - h2ac => T h2ac = h1 - wac = 990.45 => T2ac = 951 K The entropy balance equation is solved for the generation term sgen = s2ac - s1 = 8.078 - 8.6121 - 0.287 ln(100/1000) = 0.1268 kJ/kg K T 1 P1 2s 2ac P2 s Sonntag, Borgnakke and van Wylen 9.106 The small turbine in Problem 9.38 was ideal. Assume instead the isentropic turbine efficiency is 88%. Find the actual specific turbine work and the entropy generated in the turbine. Solution: Continuity Eq.6.11: (Steady) 1 2 3 . . . . m1 = m2 = m3 = m Turbine: Energy Eq.6.13: Entropy Eq.9.8: s2 = s1 + sT gen Inlet state: Table B.1.3 h1 = 3917.45 kJ/kg, Ideal turbine Q out WT wT = h1 − h2 s1 = 7.9487 kJ/kg K sT gen = 0, s2 = s1 = 7.9487 = sf2 + x sfg2 State 2: P = 10 kPa, s2 < sg => saturated 2-phase in Table B.1.2 ⇒ x2,s = (s1 - sf2)/sfg2 = (7.9487 - 0.6492)/7.501 = 0.9731 ⇒ h2,s = hf2 + x×hfg2 = 191.8 + 0.9731×2392.8 = 2520.35 kJ/kg wT,s = h1 − h2,s = 1397.05 kJ/kg P Explanation for the reversible work term is in sect. 9.3 Eq.9.18 T 1 1 2ac 2ac 3 3 2s v wT,AC = η × wT,s = 1229.9 kJ/kg = h1 - h2,AC ⇒ h2,AC = h1 - wT,AC = 2687.5 kJ/kg ⇒ T2,AC = 100°C , s2,AC = 8.4479 kJ/kg-K sT gen = s2,AC - s1 = 0.4992 kJ/kg K 2s s Sonntag, Borgnakke and van Wylen 9.107 Air enters an insulated turbine at 50°C, and exits the turbine at - 30°C, 100 kPa. The isentropic turbine efficiency is 70% and the inlet volumetric flow rate is 20 L/s. What is the turbine inlet pressure and the turbine power output? Solution: C.V.: Turbine, ηs = 0.7, Insulated Air table A.5: Cp = 1.004 kJ/kg K, R = 0.287 kJ/kg K, k = 1.4 . Inlet: Ti = 50oC, Vi = 20 L/s = 0.02 m3/s ; . . m = PV/RT = 100 × 0.02/(0.287 × 323.15) = 0.099 kg/s Exit (actual): Te = -30oC, Pe = 100 kPa 1st Law Steady state Eq.6.13: qT + hi = he + wT; qT = 0 Assume Constant Specific Heat wT = hi - he = Cp(Ti - Te) = 80.3 kJ/kg wTs = w/η = 114.7 kJ/kg, wTs = Cp(Ti - Tes) Solve for Tes = 208.9 K Isentropic Process Eq.8.32: k k-1 => Pe = Pi (Te / Ti) . . WT = mwT = 0.099 × 80.3 = 7.98 kW Pi = 461 kPa Sonntag, Borgnakke and van Wylen 9.108 Carbon dioxide, CO2, enters an adiabatic compressor at 100 kPa, 300 K, and exits at 1000 kPa, 520 K. Find the compressor efficiency and the entropy generation for the process. Solution: C.V. Ideal compressor. We will assume constant heat capacity. Energy Eq.6.13: wc = h1 - h2, k-1 P2 10000.2242 Entropy Eq.9.8, 8.32: s2 = s1 : T2s = T1P k = 300 100 = 502.7 K 1 wcs = Cp(T1 - T2s) = 0.842(300-502.7) = -170.67 kJ/kg C.V. Actual compressor wcac = Cp(T1 - T2ac) = 0.842(300 - 520) = -185.2 kJ/kg ηc = wcs/wcac = -170.67/(-185.2) = 0.92 Use Eq.8.25 for the change in entropy sgen = s2ac - s1 = Cp ln (T2ac/T1) - R ln (P2/P1) = 0.842 ln(520 / 300) - 0.1889 ln(1000 / 100) = 0.028 kJ/kg K P e, s T e, s e, ac Pe e, ac Pi i i s =C v s Constant heat capacity is not the best approximation. It would be more accurate to use Table A.8. Sonntag, Borgnakke and van Wylen 9.109 Air enters an insulated compressor at ambient conditions, 100 kPa, 20°C, at the rate of 0.1 kg/s and exits at 200°C. The isentropic efficiency of the compressor is 70%. What is the exit pressure? How much power is required to drive the compressor? Assume the ideal and actual compressor has the same exit pressure. Solution: C.V. Compressor: P1, T1, Te(real), ηs COMP known, assume constant CP0 Energy Eq.6.13 for real: -w = CP0(Te - Ti) = 1.004(200 - 20) = 180.72 Ideal -ws = -w × ηs = 180.72 × 0.70 = 126.5 Energy Eq.6.13 for ideal: 126.5 = CP0(Tes - Ti) = 1.004(Tes - 293.2), Tes = 419.2 K Constant entropy for ideal as in Eq.8.32: k Pe = Pi(Tes/Ti)k-1 = 100(419.2/293.20)3.5 = 349 kPa . . -WREAL = m(-w) = 0.1 × 180.72 = 18.07 kW P e, s T e, s e, ac Pe e, ac Pi i i s =C v s Sonntag, Borgnakke and van Wylen 9.110 Assume an actual compressor has the same exit pressure and specific heat transfer as the ideal isothermal compressor in Problem 9.8 with an isothermal efficiency of 80%. Find the specific work and exit temperature for the actual compressor. Solution: C.V. Compressor. Steady, single inlet and single exit flows. Energy Eq.6.13: hi + q = w + he; Entropy Eq.9.8: si + q/T = se Inlet state: Table B.5.2, hi = 403.4 kJ/kg, si = 1.8281 kJ/kg K Exit state: Table B.5.1, he = 398.36 kJ/kg, se = 1.7262 kJ/kg K q = T(se – si) = 273.15(1.7262 – 1.8281) = - 27.83 kJ/kg w = 403.4 + (-27.83) – 398.36 = -22.8 kJ/kg From Eq.9.29 for a cooled compressor wac = wT /η = - 22.8/0.8 = 28.5 kJ/kg Now the energy equation gives he= hi + q – wac = 403.4 + (-27.83) + 28.5= 404.07 Te ac ≈ 6°C Pe = 294 kPa P Explanation for the reversible work term is in Sect. 9.3 Eqs. 9.16 and 9.18 e,s T e,ac e,s e,ac i i v s Sonntag, Borgnakke and van Wylen 9.111 A water-cooled air compressor takes air in at 20°C, 90 kPa and compresses it to 500 kPa. The isothermal efficiency is 80% and the actual compressor has the same heat transfer as the ideal one. Find the specific compressor work and the exit temperature. Solution: Ideal isothermal compressor exit 500 kPa, 20°C Reversible process: dq = T ds => q = T(se – si) o o q = T(se – si) = T[sTe − sT1 − R ln(Pe / Pi)] = - RT ln (Pe / Pi) = - 0.287 × 293.15 ln (500/90) = - 144.3 kJ/kg As same temperature for the ideal compressor w = q = -144.3 kJ/kg => he = hi ⇒ wac = w /η = - 180.3 kJ/kg, qac = q Now for the actual compressor energy equation becomes qac + hi = he ac + wac ⇒ he ac - hi = qac - wac = - 144.3 – (-180.3) = 36 kJ/kg ≈ Cp (Te ac - Ti) Te ac = Ti + 36/1.004 = 55.9°C Sonntag, Borgnakke and van Wylen 9.112 A nozzle in a high pressure liquid water sprayer has an area of 0.5 cm2. It receives water at 250 kPa, 20°C and the exit pressure is 100 kPa. Neglect the inlet kinetic energy and assume a nozzle isentropic efficiency of 85%. Find the ideal nozzle exit velocity and the actual nozzle mass flow rate. Solution: C.V. Nozzle. Liquid water is incompressible v ≈ constant, no work, no heat transfer => Bernoulli Eq.9.17 12 2Vex – 0 = v(Pi - Pe) = 0.001002 ( 250 – 100) = 0.1503 kJ/kg Vex = 2 × 0.1503 × 1000 J/kg = 17.34 m s -1 This was the ideal nozzle now we can do the actual nozzle, Eq. 9.30 12 12 Vex ac = η 2Vex = 0.85 × 0.1503 = 0.12776 kJ/kg 2 Vex ac = 2 × 0.12776 × 1000 J/kg = 15.99 m s -1 . m= ρAVex ac = AVex ac/v = 0.5 × 10-4 × 15.99 / 0.001002 = 0.798 kg/s Sonntag, Borgnakke and van Wylen 9.113 A nozzle is required to produce a flow of air at 200 m/s at 20°C, 100 kPa. It is estimated that the nozzle has an isentropic efficiency of 92%. What nozzle inlet pressure and temperature is required assuming the inlet kinetic energy is negligible? Solution: C.V. Air nozzle: Pe, Te(real), Ve(real), ηs(real) 2 For the real process: hi = he + Ve /2 or 2 Ti = Te + Ve /2CP0 = 293.2 + 2002/2 × 1000 × 1.004 = 313.1 K For the ideal process, from Eq.9.30: 2 2 Ves/2 = Ve /2ηs = 2002/2 × 1000 × 0.92 = 21.74 kJ/kg and 2 hi = hes + (Ves/2) 2 Tes = Ti - Ves/(2CP0) = 313.1 - 21.74/1.004 = 291.4 K The constant s relation in Eq.8.32 gives ⇒ Pi = k k-1 Pe (Ti/Tes) 313.13.50 = 100291.4 = 128.6 kPa Sonntag, Borgnakke and van Wylen 9.114 Redo Problem 9.79 if the water pump has an isentropic efficiency of 85% (hose, nozzle included). Solution: C.V.: pump + hose + water column, height difference 35 m. V is velocity. Continuity Eq.6.11: Energy Eq.6.12: . . min = mex = (ρAV)nozzle; . . . m(-wp) + m(h + V2/2 + gz)in = m(h + V2/2 + gz)ex hin ≅ hex , Vin ≅ Vex = 0 , zex - zin = 35 m , ρ = 1/v ≅ 1/vf Process: 10 m 35 m -wp = g(zex - zin) = 9.80665(35 - 0) = 343.2 J/kg The velocity in nozzle is such that it can rise 10 m, so make that column C.V. 1 gznoz + 2V2noz = gzex + 0 ⇒ Vnoz = 2g(zex - znoz) = 2 × 9.81 × 10 = 14 m/s . m = (π/vf) (D2/4) Vnoz = ( π/4) 0.0252 × 14 / 0.001 = 6.873 kg/s ; . . -Wp = m(-wp)/η = 6.872 × 0.343/0.85 = 2.77 kW Sonntag, Borgnakke and van Wylen 9.115 Find the isentropic efficiency of the nozzle in example 6.4. Solution: C.V. adiabatic nozzle with known inlet state and velocity. Inlet state: B.1.3 hi = 2850.1 kJ/kg; si = 6.9665 kJ/kg K Process ideal: adiabatic and reversible Eq.9.8 gives constant s ideal exit, (150 kPa, s) ; xes = (6.9665 – 1.4335)/5.7897 = 0.9557 hes = hf + xes hfg = 2594.9 kJ/kg 2 2 Ves/2 = hi - hes + Vi /2 = 2850.1 – 2594.9 + (502)/2000 = 256.45 kJ/kg Ves = 716.2 m/s From Eq.9.30, 2 2 ηnoz.= (Ve /2)/( Ves/2) = 180/256.45 = 0.70 Sonntag, Borgnakke and van Wylen 9.116 Air flows into an insulated nozzle at 1 MPa, 1200 K with 15 m/s and mass flow rate of 2 kg/s. It expands to 650 kPa and exit temperature is 1100 K. Find the exit velocity, and the nozzle efficiency. Solution: C.V. Nozzle. Steady 1 inlet and 1 exit flows, no heat transfer, no work. 2 2 Energy Eq.6.13: hi + (1/2)Vi = he + (1/2)Ve Entropy Eq.9.8: si + sgen = se Ideal nozzle sgen = 0 and assume same exit pressure as actual nozzle. Instead of using the standard entropy from Table A.7 and Eq.8.28 let us use a constant heat capacity at the average T and Eq.8.32. First from A.7.1 1277.81 - 1161.18 Cp 1150 = 1200 - 1100 = 1.166 kJ/kg K; Cv = Cp 1150 - R = 1.166 - 0.287 = 0.8793, k = Cp 1150/Cv = 1.326 Notice how they differ from Table A.5 values. k-1 650 0.24585 Te s = Ti (Pe/Pi) k = 1200 1000 = 1079.4 K 12 12 1 2 2 Ve s = 2 Vi + C(Ti - Te s) = 2 ×15 + 1.166(1200 – 1079.4) × 1000 ⇒ = 112.5 + 140619.6 = 140732 J/kg Ve s = 530.5 m/s Actual nozzle with given exit temperature 12 12 2Ve ac = 2Vi + hi - he ac = 112.5 + 1.166(1200 – 1100) × 1000 = 116712.5 J/kg ⇒ Ve ac = 483 m/s 12 12 12 12 η noz = (2Ve ac - 2Vi )/ (2Ve s - 2Vi ) = 116600 = (hi - he, AC)/(hi - he, s) = 140619.6 = 0.829 Sonntag, Borgnakke and van Wylen Review Problems 9.117 A coflowing heat exchanger has one line with 2 kg/s saturated water vapor at 100 kPa entering. The other line is 1 kg/s air at 200 kPa, 1200 K. The heat exchanger is very long so the two flows exit at the same temperature. Find the exit temperature by trial and error. Calculate the rate of entropy generation. Solution: 4 C.V. Heat exchanger, steady 2 flows in and two flows out. No W, no external Q 1 2 3 Flows: . . . m1 = m2 = mH2O; . . . m3 = m4 = mair Energy: . . mH2O (h2 - h1) = mair (h3 - h4) State 1: Table B.1.2 h1 = 2675.5 kJ/kg State 2: 100 kPa, T2 State 3: Table A.7 h3 = 1277.8 kJ/kg, State 4: 200 kPa, T2 Only one unknown T2 and one equation the energy equation: 2( h2 - 2675.5) = 1(1277.8 - h4) => 2h2 + h4 = 6628.8 kW At 500 K: h2 = 2902.0, h4 = 503.36 => LHS = 6307 too small At 700 K: h2 = 3334.8, h4 = 713.56 => LHS = 7383 too large Linear interpolation T2 = 560 K, h2 = 3048.3, h4 = 565.47 => LHS = 6662 Final states are with T2 = 554.4 K = 281 °C H2O: Table B.1.3, AIR: Table A.7, h2 = 3036.8 kJ/kg, s2 = 8.1473, s1 = 7.3593 kJ/kg K h4 = 559.65 kJ/kg, sT4 = 7.4936, sT3 = 8.3460 kJ/kg K The entropy balance equation, Eq.9.7, is solved for the generation term: . . . Sgen = mH2O (s2 - s1) + mair (s4- s3) = 2(8.1473 - 7.3593) +1 (7.4936 - 8.3460) = 0.724 kW/K No pressure correction is needed as the air pressure for 4 and 3 is the same. Sonntag, Borgnakke and van Wylen 9.118 A vortex tube has an air inlet flow at 20°C, 200 kPa and two exit flows of 100 kPa, one at 0°C and the other at 40°C. The tube has no external heat transfer and no work and all the flows are steady and have negligible kinetic energy. Find the fraction of the inlet flow that comes out at 0°C. Is this setup possible? Solution: C.V. The vortex tube. Steady, single inlet and two exit flows. No q or w. Continuity Eq.: Entropy: . . . m1 = m2 + m3 ; Energy: . . . m1h1 = m2h2 + m3h3 . . . . m1s1 + Sgen = m2s2 + m3s3 States all given by temperature and pressure. Use constant heat capacity to .. evaluate changes in h and s. Solve for x = m2/m1 from the energy equation .. m3/m1 = 1 - x; h1 = x h2 + (1-x) h3 => x = (h1 - h3)/(h2 - h3) = (T1 - T3)/(T2 - T3) = (20−40)/(0−40) = 0.5 Evaluate the entropy generation . . Sgen/m1 = x s2 + (1-x)s3 - s1 = 0.5(s2 - s1 ) + 0.5(s3 - s1 ) = 0.5 [Cp ln(T2 / T1) − R ln(P2 / P1)] + 0.5[Cp ln(T3 / T1) − R ln(P3/ P1)] 273.15 100 = 0.5 [1.004 ln( 293.15 ) - 0.287 ln( 200 )] 313.15 100 + 0.5 [1.004 ln( 293.15 ) - 0.287 ln( 200 )] = 0.1966 kJ/kg K >0 So this is possible. Sonntag, Borgnakke and van Wylen 9.119 An initially empty spring-loaded piston/cylinder requires 100 kPa to float the piston. A compressor with a line and valve now charges the cylinder with water to a final pressure of 1.4 MPa at which point the volume is 0.6 m3, state 2. The inlet condition to the reversible adiabatic compressor is saturated vapor at 100 kPa. After charging the valve is closed and the water eventually cools to room temperature, 20°C, state 3. Find the final mass of water, the piston work from 1 to 2, the required compressor work, and the final pressure, P3. Solution: in Process 1→2: transient, adiabatic. for C.V. compressor + cylinder Assume process is reversible × -Wc ⇒ Continuity: m2 - 0 = min , Entropy Eq.: Energy: m2s2 - 0 = minsin + 0 / Inlet state: Table B.1.2, m2u2 - 0 = (minhin) - Wc - 1W2 / ⇒ s2 = sin hin = 2675.5 kJ/kg, sin = 7.3594 kJ/kg K 1 1 / 1W2 = ⌠PdV = 2 (Pfloat+ P2)(V2 - 0) = 2 (100+1400)0.6 = 450 kJ ⌡ State 2: P2 , s2 = sin Table B.1.3 ⇒ v2 = 0.2243, u2 = 2984.4 kJ/kg m2 = V2/v2 = 0.6/0.2243 = 2.675 kg Wc = minhin - m2u2 - 1W2 = 2.675 × (2675.5 - 2984.4) - 450 = -1276.3 kJ P 1400 2 Assume 2-phase ⇒ P3 = Psat(20°C) = 2.339 kPa 3 100 0 State 3 must be on line & 20°C 0.6 V less than Pfloat so compressed liquid Table B.1.1: v3 ≅ vf(20°C) = 0.001002 ⇒ V3 = m3v3 = 0.00268 m3 On line: P3 = 100 + (1400 - 100) × 0.00268/0.6 = 105.8 kPa Sonntag, Borgnakke and van Wylen 9.120 In a heat-powered refrigerator, a turbine is used to drive the compressor using the same working fluid. Consider the combination shown in Fig. P9.120 where the turbine produces just enough power to drive the compressor and the two exit flows are mixed together. List any assumptions made and find the ratio of mass .. flow rates m3/m1 and T5 (x5 if in two-phase region) if the turbine and the compressor are reversible and adiabatic Solution: CV: compressor s2S = s1 = 0.7082 kJ/kg K → T2S = 52.6°C wSC = h1 - h2S = 178.61 - 212.164 = -33.554 kJ/kg CV: turbine s4S = s3 = 0.6444 = 0.2767 + x4S × 0.4049 => x4S = 0.9081 h4S = 76.155 + 0.9081 × 127.427 = 191.875 kJ/kg wST = h3 - h4S = 209.843 - 191.875 = 17.968 kJ/kg wSC 33.554 . .. . As wTURB = -wCOMP , m3/m1 = - w = 17.968 = 1.867 ST CV: mixing portion . . . . m1h2S + m3h4S = (m1 + m3)h5 1 × 212.164 + 1.867 × 191.875 = 2.867 h5 ⇒ h5 = 198.980 = 76.155 + x5 × 127.427 => x5 = 0.9639 Sonntag, Borgnakke and van Wylen 9.121 A stream of ammonia enters a steady flow device at 100 kPa, 50°C, at the rate of 1 kg/s. Two streams exit the device at equal mass flow rates; one is at 200 kPa, 50°C, and the other as saturated liquid at 10°C. It is claimed that the device operates in a room at 25°C on an electrical power input of 250 kW. Is this possible? Solution: 1 Control volume: Steady device out 2 to ambient 25°C. Steady cb . Q device . Wel Energy Eq.6.10: . . . . . m1h1 + Q + Wel = m2h2 + m3h3 Entropy Eq.9.7: . . . . . m1s1 + Q/Troom + Sgen = m2s2 + m3s3 State 1: Table B.2.2, h1 = 1581.2 kJ/kg, s1 = 6.4943 kJ/kg K State 2: Table B.2.2 h2 = 1576.6 kJ/kg, s2 = 6.1453 kJ/kg K State 3: Table B.2.1 h3 = 226.97 kJ/kg, s3 = 0.8779 kJ/kg K From the energy equation . Q = 0.5 × 1576.6 + 0.5 × 226.97 - 1 × 1581.2 - 250 = -929.4 kW From the entropy equation . Sgen = 0.5×6.1453 + 0.5 × 0.8779 - 1 × 6.4943 - (-929.4)/298.15 = 0.1345 kW/K > 0 / . since Sgen > 0 this is possible / 3 Sonntag, Borgnakke and van Wylen 9.122 A frictionless piston/cylinder is loaded with a linear spring, spring constant 100 kN/m and the piston cross-sectional area is 0.1 m2. The cylinder initial volume of 20 L contains air at 200 kPa and ambient temperature, 10°C. The cylinder has a set of stops that prevent its volume from exceeding 50 L. A valve connects to a line flowing air at 800 kPa, 50°C. The valve is now opened, allowing air to flow in until the cylinder pressure reaches 800 kPa, at which point the temperature inside the cylinder is 80°C. The valve is then closed and the process ends. a) Is the piston at the stops at the final state? b) Taking the inside of the cylinder as a control volume, calculate the heat transfer during the process. c) Calculate the net entropy change for this process. line x 800 500 200 P To = 10oC = 283.15 K V Ap = 0.1 m2, Vstop = 50 L 20 50 Air from Table A.5: R = 0.287, Cp = 1.004, Cv = 0.717 kJ/kg-K State 1: T1 = 10oC, P1 = 200 kPa, V1 = 20 L = 0.02 m3, m1 = P1V1/RT1 = 200×0.02/(0.287×283.15) = 0.0492 kg State 2: T2 = 80oC, P2 = 800 kPa, Inlet: Ti = 50oC, Pi = 800 kPa ks a) Pstop = P1 + 2 (Vstop - V1) = 500 kPa, P2 > Pstop Ap Piston hits stops V2 = Vstop = 50 L, m2 = PV/RT = 0.3946 kg b) 1st Law: 1Q2 + mihi = m2u2 - m1u1 + mehe + 1W2; me = 0, mi = m2 - m1 1W2 = ∫ P dV = (P1 + Pstop)(Vstop - V1)/2 = 10.5 kJ Assume constant specific heat 1Q2 = m2CvT2 - m1CvT1 - (m2 - m1) CpTi + 1W2 = -11.6 kJ Qcv c) 2nd Law: ∆Snet = m2s2 - m1s1 - misi - T o Qcv ∆Snet = m2(s2 - si) - m1(s1 - si) - T o s2 - si = Cp ln(T2 / Ti) − R ln(P2 / Pi) = 0.08907 kJ/kg-K s1 - si = Cp ln(T1 / Ti) − R ln(P1 / Pi )= 0.26529 kJ/kg-K ∆Snet = 0.063 kJ/K (P2 = Pi) Sonntag, Borgnakke and van Wylen 9.123 An insulated piston/cylinder contains R-22 at 20°C, 85% quality, at a cylinder volume of 50 L. A valve at the closed end of the cylinder is connected to a line flowing R-22 at 2 MPa, 60°C. The valve is now opened, allowing R-22 to flow in, and at the same time the external force on the piston is decreased, and the piston moves. When the valve is closed, the cylinder contents are at 800 kPa, 20°C, and a positive work of 50 kJ has been done against the external force. What is the final volume of the cylinder? Does this process violate the second law of thermodynamics? Solution: C.V. Cylinder volume. A transient problem. Continuity Eq.: m2 - m1 = mi Energy Eq.: m2u2 - m1u1 = 1Q2 + mihi - 1W2 Entropy Eq.: m2s2 - m1s1 = 1Q2/T + misi + 1S2 gen Process: 1Q2 = 0, 1W2 = 50 kJ State 1: T1 = 20oC, x1 = 0.85, V1 = 50 L = 0.05 m3 P1 = Pg = 909.9 kPa, u1 = uf + x1ufg = 208.1 kJ/kg v1 = vf + x1vfg = 0.000824 + 0.85×0.02518 = 0.022226 m3/kg, s1 = sf + x1sfg = 0.259 + 0.85×0.6407 = 0.8036 kJ/kg K m1 = V1/v1 = 2.25 kg State 2: T2 = 20oC, P2 = 800 kPa, superheated, v2 = 0.03037 m3/kg, u2 = 234.44 kJ/kg, s2 = 0.9179 kJ/kg K Inlet: Ti = 60oC, Pi = 2 MPa, hi = 271.56 kJ/kg, si = 0.8873 kJ/kg K Solve for the mass m2 from the energy equation (the only unknown) m2 = [m1u1 - 1W2 - m1hi] / [u2 - hi] = 2.25 × 208.1 – 50 – 2.25 × 271.56 = 5.194 kg 234.44 – 271.56 V2 = m2v2 = 0.158 m3 Now check the second law 1S2 gen = m2s2 - m1s1 - 1Q2/T - misi = 5.194 ×0.9179 – 2.25 × 0.8036 – 0 – (5.194 – 2.25) 0.8873 = 0.347 kJ/K > 0, Satisfies 2nd Law Sonntag, Borgnakke and van Wylen 9.124 Air enters an insulated turbine at 50°C, and exits the turbine at - 30°C, 100 kPa. The isentropic turbine efficiency is 70% and the inlet volumetric flow rate is 20 L/s. What is the turbine inlet pressure and the turbine power output? C.V.: Turbine, ηs = 0.7, Insulated Air: Cp = 1.004 kJ/kg-K, R = 0.287 kJ/kg-K, k = 1.4 . Inlet: Ti = 50oC, Vi = 20 L/s = 0.02 m3/s Exit: Te = -30oC, Pe = 100 kPa a) 1st Law steady flow: q + hi = he + wT; q = 0 Assume Constant Specific Heat wT = hi - he = Cp(Ti - Te) = 80.3 kJ/kg wTs = w/η = 114.7 kJ/kg, wTs = Cp(Ti - Tes) Solve for Tes = 208.9 K k Isentropic Process: Pe = Pi (Te / Ti)k-1 => Pi = 461 kPa . . . . b) WT = mwT; m = PV/RT = 0.099 kg/s . => WT = 7.98 kW Sonntag, Borgnakke and van Wylen 9.125 A certain industrial process requires a steady 0.5 kg/s supply of compressed air at 500 kPa, at a maximum temperature of 30°C. This air is to be supplied by installing a compressor and aftercooler, see Fig. P9.46. Local ambient conditions are 100 kPa, 20°C. Using an isentropic compressor efficiency of 80%, determine the power required to drive the compressor and the rate of heat rejection in the aftercooler. Air: R = 0.287 kJ/kg-K, Cp = 1.004 kJ/kg-K, k = 1.4 . State 1: T1 = To = 20oC, P1 = Po = 100 kPa, m = 0.5 kg/s State 2: P2 = P3 = 500 kPa State 3: T3 = 30oC, P3 = 500 kPa Assume ηs = 80 % (Any value between 70%-90% is OK) Compressor: Assume Isentropic k-1 T2s = T1 (P2/P1) k , T2s = 464.6 K 1st Law: qc + h1 = h2 + wc; qc = 0, assume constant specific heat wcs = Cp(T1 - T2s) = -172.0 kJ/kg . . ηs = wcs/wc, wc = wcs/ηs = -215, WC = mwC = -107.5 kW wc = Cp(T1 - T2), solve for T2 = 507.5 K Aftercooler: 1st Law: q + h2 = h3 + w; w = 0, assume constant specific heat q = Cp(T3 - T2) = 205 kJ/kg, . . Q = mq = -102.5 kW Sonntag, Borgnakke and van Wylen 9.126 Consider the scheme shown in Fig. P9.126 for producing fresh water from salt water. The conditions are as shown in the figure. Assume that the properties of salt water are the same as for pure water, and that the pump is reversible and adiabatic. .. a. Determine the ratio (m7/m1), the fraction of salt water purified. b. Determine the input quantities, wP and qH. c. Make a second law analysis of the overall system. C.V. Flash evaporator: Steady flow, no external q, no work. Energy Eq.: . . . . m1h4 = (m1 - m7)h5 + m7h6 Table B.1.1 .. .. or 632.4 = (1 - (m7/m1)) 417.46 + (m7/m1) 2675.5 .. ⇒ m7/m1 = 0.0952 C.V. Pump steady flow, incompressible liq.: wP = -⌠vdP ≈ -v1(P2 - P1) = - 0.001001(700 - 100) = -0.6 kJ/kg ⌡ h2 = h1 - wP = 62.99 + 0.6 = 63.6 kJ/kg C.V. Heat exchanger: .. .. h2 + (m7/m1)h6 = h3 + (m7/m1)h7 63.6 + 0.0952 × 2675.5 = h3 + 0.0952 × 146.68 => h3 = 304.3 kJ/kg C.V. Heater: qH = h4 - h3 = 632.4 - 304.3 = 328.1 kJ/kg CV: entire unit, entropy equation per unit mass flow rate at state 1 .. .. SC.V.,gen = - qH/TH + (1 - (m7/m1))s5 +(m7/m1)s7 - s1 = (-328.1/473.15) + 0.9048 × 1.3026 + 0.0952 × 0.5053 - 0.2245 = 0.3088 kJ/K kg m1 Sonntag, Borgnakke and van Wylen 9.127 Supercharging of an engine is used to increase the inlet air density so that more fuel can be added, the result of which is an increased power output. Assume that ambient air, 100 kPa and 27°C, enters the supercharger at a rate of 250 L/s. The supercharger (compressor) has an isentropic efficiency of 75%, and uses 20 kW of power input. Assume that the ideal and actual compressor have the same exit pressure. Find the ideal specific work and verify that the exit pressure is 175 kPa. Find the percent increase in air density entering the engine due to the supercharger and the entropy generation. ex . -Wc C.V.: Air in compressor (steady flow) in . . . . Energy: mhin - W = mhex Assume: Q = 0 ⇐ . . . . Cont: min = mex = m = V/vin = 0.29 kg/s . . . Entropy: msin + Sgen = msex RTin o vin = P = 0.8614 m3/kg, sTi = 6.86975 kJ/kg K, hin = 300.62 kJ/kg in . . ηc = wC s/wC ac => -WS = -WAC × ηc = 15 kW .. -wC s = -WS/m = 51.724 kJ/kg, Table A.7: -wC ac = 68.966 kJ/kg hex s = hin - wC s = 300.62 + 51.724 = 352.3 kJ/kg o ⇒ Tex s = 351.5 K, sTe = 7.02830 kJ/kg K o o Pex = Pin × e(sT ex - sT in)/R = 100 × exp [ 7.0283 - 6.86975 0.287 = 173.75 kPa The actual exit state is hex ac = hin - wC ac = 369.6 kJ/kg ⇒ Tex ac = 368.6 K o vex = RTex/Pex = 0.6088 m3/kg, sTex ac = 7.0764 ρex/ρin = vin/vex = 0.8614/0.6088 = 1.415 or 41.5% increase 173.75 sgen = sex - sin = 7.0764 - 6.86975 - 0.287 ln( 100 ) = 0.0481 kJ/kg K Sonntag, Borgnakke and van Wylen 9.128 A jet-ejector pump, shown schematically in Fig. P9.128, is a device in which a low-pressure (secondary) fluid is compressed by entrainment in a high-velocity (primary) fluid stream. The compression results from the deceleration in a diffuser. For purposes of analysis this can be considered as equivalent to the turbine-compressor unit shown in Fig. P9.120 with the states 1, 3, and 5 corresponding to those in Fig. P9.128. Consider a steam jet-pump with state 1 as saturated vapor at 35 kPa; state 3 is 300 kPa, 150°C; and the discharge pressure, P5, is 100 kPa. .. a. Calculate the ideal mass flow ratio, m1/m3. .. .. b. The efficiency of a jet pump is defined as η = (m1/m3)actual / (m1/m3)ideal for the same inlet conditions and discharge pressure. Determine the discharge temperature of the jet pump if its efficiency is 10%. a) ideal processes (isen. comp. & exp.) expands 3-4s comp 1-2s then mix at const. P s4s = s3 = 7.0778 = 1.3026 + x4s × 6.0568 => x4s = 0.9535 h4s = 417.46 + 0.9535 × 2258.0 = 2570.5 kJ/kg s2s = s1 = 7.7193 → T2s = 174°C & h2s = 2823.8 kJ/kg . . m1(h2s - h1) = m3(h3 - h4s) 2761.0 - 2570.5 .. ⇒ (m1/m3)IDEAL = 2823.8 - 2631.1 = 0.9886 b) real processes with jet pump eff. = 0.10 .. ⇒ (m1/m3)ACTUAL = 0.10 × 0.9886 = 0.09886 . . . . 1st law m1h1 + m3h3 = (m1 + m3)h5 0.09886 × 2631.1 + 1 × 2761.0 = 1.09896 h5 State 5: h5 = 2749.3 kJ/kg, P5 = 100 kPa => T5 = 136.5 oC Sonntag, Borgnakke and van Wylen 9.129 A rigid steel bottle, V = 0.25 m3, contains air at 100 kPa, 300 K. The bottle is now charged with air from a line at 260 K, 6 MPa to a bottle pressure of 5 MPa, state 2, and the valve is closed. Assume that the process is adiabatic, and the charge always is uniform. In storage, the bottle slowly returns to room temperature at 300 K, state 3. Find the final mass, the temperature T2, the final pressure P3, the heat transfer 1Q3 and the total entropy generation. C.V. Bottle. Flow in, no work, no heat transfer. Continuity Eq.6.15: m2 - m1 = min ; Energy Eq.6.16: State 1 and inlet: m2u2 - m1u1 = minhin Table A.7, u1 = 214.36 kJ/kg, hin = 260.32 kJ/kg m1 = P1V/RT1 = (100 × 0.25)/(0.287 × 300) = 0.290 kg m2 = P2V/RT2 = 5000 × 0.25/(0.287 × T2) = 4355.4/T2 Substitute into energy equation u2 + 0.00306 T2 = 260.32 Now trial and error on T2 T2 = 360 => LHS = 258.63 (low); T2 = 370 => LHS = 265.88 (high) Interpolation T2 = 362.3 K (LHS = 260.3 OK) m2 = 4355.4/362.3 = 12.022 kg ; P3 = m2RT3/V = 4140 kPa Now use the energy equation from the beginning to the final state 1Q3 = m2u3 - m1u1 - minhin = (12.022 - 0.29) 214.36 - 11.732 × 260.32 = -539.2 kJ Entropy equation from state 1 to state 3 with change in s from Eq.8.28 Sgen = m2s3 - m1s1 - minsin - 1Q3/T = m2(s3 -sin) - m1(s1 - sin) - 1Q3/T = 12.022[6.8693 - 6.7256 - R ln(4140/6000)] - 0.29[6.8693 - 6.7256 - R ln(100/6000)] + 539.2/300 = 4.423 kJ/K P T line 2 3 T2 6 MPa line 300 260 1 v 2 3 v=C 5 MPa 100 kPa 1 s Problem could have been solved with constant specific heats from A.5 in which case we would get the energy explicit in T2 (no iterations). Sonntag, Borgnakke and van Wylen 9.130 A horizontal, insulated cylinder has a frictionless piston held against stops by an external force of 500 kN. The piston cross-sectional area is 0.5 m2, and the initial volume is 0.25 m3. Argon gas in the cylinder is at 200 kPa, 100°C. A valve is now opened to a line flowing argon at 1.2 MPa, 200°C, and gas flows in until the cylinder pressure just balances the external force, at which point the valve is closed. Use constant heat capacity to verify that the final temperature is 645 K and find the total entropy generation. Solution: The process has inlet flow, no work (volume constant) and no heat transfer. Continuity Eq.6.15: m2 − m1 = mi Energy Eq.6.16: m2 u2 − m1u1 = mi hi m1= P1V1/RT1 = 200 ×0.25/(0.2081 ×373.15) = 0.644 kg 500 P2 = 0.5 = 1000 kPa ⇒ Force balance: P2A = F For argon use constant heat capacities so the energy equation is: m2 CVo T2 – m1 CVo T1 = (m2 – m1 ) CPo T in We know P2 so only 1 unknown for state 2. Use ideal gas law to write m2T2 = P2V1/R and m1 T1 = P1V1/R and divide the energy equation with CVo to solve for the change in mass (P2 V1 – P1V1)/R = (m2 – m1 ) (CPo/CVo ) T in (m2 – m1 ) = (P2 – P1)V1/(R k T in ) = (1000 - 200)×0.25/(0.2081×1.667×473.15) = 1.219 kg m2 = 1.219 + 0.644 = 1.863 kg. T2 = P2V1/(m2R) = 1000×0.25/(1.863×0.2081) = 645 K Entropy Eq.9.12: OK m2s2 - m1s1 = misi + 0 + 1S2 gen 1S2 gen = m1(s2 - s1) + (m2 - m1)(s2 - si) T2 P2 = m1 Cp lnT - R ln P [ 1 T P + (m2 - m1)[Cp ln T2i - R ln P2i ] 1 645 1000 = 0.644[ 0.52 ln 373.15 - 0.2081 ln 200 ] 645 1000 + 1.219[ 0.52 ln 473.15 - 0.2081 ln 1200] = - 0.03242 + 0.24265 = 0.21 kJ/K Sonntag, Borgnakke and van Wylen 9.131 A rigid 1.0 m3 tank contains water initially at 120°C, with 50 % liquid and 50% vapor, by volume. A pressure-relief valve on the top of the tank is set to 1.0 MPa (the tank pressure cannot exceed 1.0 MPa - water will be discharged instead). Heat is now transferred to the tank from a 200°C heat source until the tank contains saturated vapor at 1.0 MPa. Calculate the heat transfer to the tank and show that this process does not violate the second law. Solution: C.V. Tank and walls out to the source. Neglect storage in walls. There is flow out and no boundary or shaft work. m2 − m1 = − me Continuity Eq.6.15: Energy Eq.6.16: m2 u2 − m1u1 = - mehe + 1Q2 Entropy Eq.9.12: m2s2 − m1s1 = − mese + ∫ dQ/T + 1S2 gen State 1: T1 = 120oC, Table B.1.1 vf = 0.00106 m3/kg, mliq = 0.5V1/vf = 471.7 kg vg = 0.8919 m3/kg, mg = 0.5V1/vg = 0.56 kg, m1 = 472.26 kg, x1 = mg/m1 = 0.001186 u1 = uf + x1ufg = 503.5 + 0.001186×2025.8 = 505.88 kJ/kg, s1 = sf + x1sfg = 1.5275 + 0.001186×5.602 = 1.5341 kJ/kg-K State 2: P2 = 1.0 MPa, sat. vap. x2 = 1.0, V2 = 1m3 v2 = vg = 0.19444 m3/kg, m2 = V2/v2 = 5.14 kg u2 = ug = 2583.6 kJ/kg, s2 = sg = 6.5864 kJ/kg-K Exit: Pe = 1.0 MPa, sat. vap. xe = 1.0, se = sg = 6.5864 kJ/kg, he = hg = 2778.1 kJ/kg, me = m1 - m2 = 467.12 kg From the energy equation we get 1Q2 = m2 u2 − m1u1 + mehe = 1 072 080 kJ From the entropy Eq.9.24 (with 9.25 and 9.26) we get 1Q2 S2 gen = m2s2 − m1s1 + mese − T ; TH = 200oC = 473 K 1 H 1S2 gen = ∆Snet = 120.4 kJ/K > 0 Process Satisfies 2nd Law Sonntag, Borgnakke and van Wylen 9.132 A certain industrial process requires a steady 0.5 kg/s of air at 200 m/s, at the condition of 150 kPa, 300 K. This air is to be the exhaust from a specially designed turbine whose inlet pressure is 400 kPa. The turbine process may be assumed to be reversible and polytropic, with polytropic exponent n = 1.20. a) What is the turbine inlet temperature? b) What are the power output and heat transfer rate for the turbine? c) Calculate the rate of net entropy increase, if the heat transfer comes from a source at a temperature 100°C higher than the turbine inlet temperature. Solution: C.V. Turbine, this has heat transfer, PVn = Constant, n = 1.2 Exit: Te = 300K, Pe = 150 kPa, Ve = 200 m/s a) Process polytropic Eq.8.37: n-1 Te / Ti = (Pe/Pi) n => Ti = 353.3 K . . . . mi(h + V2/2)in + Q = mex(h + V2/2)ex + WT b) 1st Law Eq.6.12: Reversible shaft work in a polytropic process, Eq.9.14 and Eq.9.19: n 2 2 2 2 wT = −∫ v dP + ( Vi − Ve )/2 = − n-1(Peve - Pivi) + ( Vi − Ve )/2 n 2 = − n-1R(Te -Ti) − Ve /2 = 71.8 kJ/kg . . WT = mwT = 35.9 kW Assume constant specific heat in the energy equation . . . 2 Q = m[CP (Te -Ti) + Ve /2 ] + WT = 19.2 kW c) 2nd Law Eq.9.7 or 9.23 with change in entropy from Eq.8.25: . . . dSnet/dt = Sgen = m(se -si) - QH/TH, TH = Ti + 100 = 453.3 K se - si = Cpln(Te / Ti) - R ln(Pe / Pi) = 0.1174 kJ/kg K dSnet/dt = 0.5×0.1174 - 19.2/453.3 = 0.0163 kW/K P T i n = k = 1.4 i n=1 e n = 1.2 v n=1 e n = 1.2 s Sonntag, Borgnakke and van Wylen 9.133 Assume both the compressor and the nozzle in Problem 9.37 have an isentropic efficiency of 90% the rest being unchanged. Find the actual compressor work and its exit temperature and find the actual nozzle exit velocity. 1 T 2 3 P2 = P3 C.V. Ideal compressor, inlet: 1 exit: 2 3 -W 145 P1 5 Energy Eq.6.13: Entropy Eq.9.8: Adiabatic : q = 0. Reversible: sgen = 0 s h1 + 0 = wC + h2; s1 + 0/T + 0 = s2 - wCs = h2 - h1 , Properties use air Table A.5: s2 = s1 kJ kJ CPo = 1.004 kg K, R = 0.287 kg K, k = 1.4, Process gives constant s (isentropic) which with constant CPo gives Eq.8.32 => ⇒ k-1 T2 = T1( P2/P1) k = 290 (400/100) 0.2857 = 430.9 K −wCs = CPo(T2 – T1) = 1.004 (430.9 – 290) = 141.46 kJ/kg The ideal nozzle then expands back down to state 1 (constant s). The actual compressor discharges at state 3 however, so we have: wC = wCs/ηC = -157.18 ⇒ T3 = T1 - wC/Cp = 446.6 K Nozzle receives air at 3 and exhausts at 5. We must do the ideal (exit at 4) first. s4 = s3 ⇒ Eq.8.32: k-1 T4 = T3 (P4/P3) k = 300.5 K 12 12 2 Vs = Cp(T3 - T4) = 146.68 ⇒ 2 Vac = 132 kJ/kg ⇒ Vac = 513.8 m/s If we need it, the actual nozzle exit (5) can be found: T5 = T3 - V2 /2Cp = 315 K ac Sonntag, Borgnakke and van Wylen Problems solved with Pr and vr functions 9.28 A compressor receives air at 290 K, 100 kPa and a shaft work of 5.5 kW from a gasoline engine. It should deliver a mass flow rate of 0.01 kg/s air to a pipeline. Find the maximum possible exit pressure of the compressor. Solution: . C.V. Compressor, Steady single inlet and exit flows. Adiabatic: Q = 0. Continuity Eq.6.11: Energy Eq.6.12: Entropy Eq.9.8: . . . mi = me = m, . . . mhi = mhe + WC, . . . msi + Sgen = mse . ( Reversible Sgen = 0 ) . . .. Wc = mwc => -wc = -W/m = 5.5/0.01 = 550 kJ/kg Use Table A.7, hi = 290.43 kJ/kg, Pr i = 0.9899 he = hi + (-wc) = 290.43 + 550 = 840.43 kJ/kg A.7 => Te = 816.5 K, Pr e = 41.717 Pe = Pi (Pr e/Pr i) = 100 × (41.717/0.9899) = 4214 kPa P i T e ∆ h = 550 kJ/kg e i i v s e -WC Sonntag, Borgnakke and van Wylen 9.32 Do the previous problem using the air tables in A.7 The exit nozzle in a jet engine receives air at 1200 K, 150 kPa with neglible kinetic energy. The exit pressure is 80 kPa and the process is reversible and adiabatic. Use constant heat capacity at 300 K to find the exit velocity. Solution: C.V. Nozzle, Steady single inlet and exit flow, no work or heat transfer. 2 Energy Eq.6.13: hi = he + Ve /2 ( Zi = Ze ) Entropy Eq.9.8: se = si + ∫ dq/T + sgen = si + 0 + 0 Process: sgen = 0 as used above leads to se = si hi = 1277.8 kJ/kg, Pr i = 191.17 q = 0, Inlet state: The constant s is done using the Pr function from A.7.2 Pr e = Pr i (Pe / Pi) = 191.17 (80/150) = 101.957 Interpolate in A.7 => 101.957 – 91.651 Te = 1000 + 50 111.35 – 91.651 = 1026.16 K he = 1046.2 + 0.5232 × (1103.5 – 1046.2) = 1076.2 kJ/kg 2 From the energy equation we have Ve /2 = hi - he , so then Ve = 2 (hi - he) = P T i 2(1277.8 - 1076.2) × 1000 = 635 m/s Hi P e e v s Low P Low V i Hi V Sonntag, Borgnakke and van Wylen 9.34 Air enters a turbine at 800 kPa, 1200 K, and expands in a reversible adiabatic process to 100 kPa. Calculate the exit temperature and the work output per kilogram of air, using a. The ideal gas tables, Table A.7 b. Constant specific heat, value at 300 K from table A.5 Solution: air i C.V. Air turbine. Adiabatic: q = 0, reversible: sgen = 0 . W Turbine Energy Eq.6.13: Entropy Eq.9.8: e wT = hi − he , s e = si hi = 1277.8 kJ/kg, Pr i = 191.17 The constant s process is done using the Pr function from A.7.2 a) Table A.7: 100 ⇒ Pr e = Pr i (Pe / Pi) = 191.17 800 = 23.896 ⇒ Te = 705.7 K, he = 719.7 kJ/kg w = hi - he = 1277.8 – 719.7 = 558.1 kJ/kg Interpolate in A.7.1 b) Table A.5: CPo = 1.004 kJ/kg K, R = 0.287 kJ/kg K, k = 1.4, then from Eq.8.32 Te = Ti (Pe/Pi) k-1 k 1000.286 = 1200 800 = 662.1 K w = CPo(Ti - Te) = 1.004(1200 - 662.1) = 539.8 kJ/kg Sonntag, Borgnakke and van Wylen 9.69 An old abandoned saltmine, 100 000 m3 in volume, contains air at 290 K, 100 kPa. The mine is used for energy storage so the local power plant pumps it up to 2.1 MPa using outside air at 290 K, 100 kPa. Assume the pump is ideal and the process is adiabatic. Find the final mass and temperature of the air and the required pump work. Solution: C.V. The mine volume and the pump Continuity Eq.6.15: m2 - m1 = min Energy Eq.6.16: m2u2 - m1u1 = 1Q2 - 1W2 + minhin Entropy Eq.9.12: m2s2 - m1s1 = ⌠dQ/T + 1S2 gen + minsin ⌡ Process: Adiabatic 1Q2 = 0 , Process ideal 1S2 gen = 0 , s1 = sin ⇒ m2s2 = m1s1 + minsin = (m1 + min)s1 = m2s1 ⇒ s2 = s1 2100 Constant s ⇒ Pr2 = Pr i (P2 / Pi) = 0.9899 100 = 20.7879 ⇒ T2 = 680 K , u2 = 496.94 kJ/kg A.7.2 m1 = P1V1/RT1 = 100×105/(0.287 × 290) = 1.20149 × 105 kg m2 = P2V2/RT2 = 100 × 21×105/(0.287 × 680) = 10.760 × 105 kg ⇒ min = 9.5585×105 kg 1W2 = minhin + m1u1 - m2u2 = min(290.43) + m1(207.19) - m2(496.94) = -2.322 × 108 kJ P s=C 2 T T2 400 290 1, i v 2 100 kPa 1, i s Sonntag, Borgnakke and van Wylen 9.89 Calculate the air temperature and pressure at the stagnation point right in front of a meteorite entering the atmosphere (-50 °C, 50 kPa) with a velocity of 2000 m/s. Do this assuming air is incompressible at the given state and repeat for air being a compressible substance going through an adiabatic compression. Solution: 121 2 Kinetic energy: 2 V = 2 (2000) /1000 = 2000 kJ/kg vatm = RT/P = 0.287 × 223/50 =1.28 m3/kg Ideal gas: a) incompressible Energy Eq.6.13: 1 ∆h = 2 V2 = 2000 kJ/kg If A.5 ∆T = ∆h/Cp = 1992 K unreasonable, too high for that Cp Use A.7: 1 hst = ho + 2 V2 = 223.22 + 2000 = 2223.3 kJ/kg Tst = 1977 K Bernoulli (incompressible) Eq.9.17: 1 ∆P = Pst - Po = 2 V2/v = 2000/1.28 = 1562.5 kPa Pst = 1562.5 + 50 = 1612.5 kPa b) compressible Tst = 1977 K the same energy equation. From A.7.2: Stagnation point Pr st = 1580.3; Free Pr o = 0.39809 Pr st 1580.3 Pst = Po × P = 50 × 0.39809 ro = 198 485 kPa Notice that this is highly compressible, v is not constant. Sonntag, Borgnakke and van Wylen 9.127 Supercharging of an engine is used to increase the inlet air density so that more fuel can be added, the result of which is an increased power output. Assume that ambient air, 100 kPa and 27°C, enters the supercharger at a rate of 250 L/s. The supercharger (compressor) has an isentropic efficiency of 75%, and uses 20 kW of power input. Assume that the ideal and actual compressor have the same exit pressure. Find the ideal specific work and verify that the exit pressure is 175 kPa. Find the percent increase in air density entering the engine due to the supercharger and the entropy generation. ex . -Wc C.V.: Air in compressor (steady flow) in . . . . Energy: mhin - W = mhex Assume: Q = 0 ⇐ . . . . Cont: min = mex = m = V/vin = 0.29 kg/s . . . Entropy: msin + Sgen = msex Inlet state: vin = RTin/Pin = 0.8614 m3/kg, Pr in = 1.1167 . . ηc = wC s/wC ac => -WS = -WAC × ηc = 15 kW .. -wC s = -WS/m = 51.724 kJ/kg, Table A.7: -wC ac = 68.966 kJ/kg hex s = hin - wC s = 300.62 + 51.724 = 352.3 kJ/kg ⇒ Tex s = 351.5 K, Pr ex = 1.949 Pex = Pin × Pr ex/Pr in = 100 × 1.949 / 1.1167 = 174.5 kPa The actual exit state is hex ac = hin - wC ac = 369.6 kJ/kg ⇒ Tex ac = 368.6 K vex = RTex/Pex = 0.606 m3/kg ρex/ρin = vin/vex = 0.8614/0.606 = 1.42 or 42 % increase sgen = sex - sin = 7.0767 - 6.8693 - 0.287 ln(174/100)] = 0.0484 kJ/kg K SOLUTION MANUAL SI UNIT PROBLEMS CHAPTER 10 SONNTAG • BORGNAKKE • VAN WYLEN FUNDAMENTALS of Thermodynamics Sixth Edition Sonntag, Borgnakke and van Wylen CONTENT SUBSECTION PROB NO. Correspondence table Concept-Study Guide Problems Available energy, reversible work Irreversibility Availability (exergy) Device Second Law Efficiency Exergy Balance Equation Review Problems Problems re-solved using Pr and vr functions from table A.7.2: 31, 61 English Unit Problems 1-20 21-35 36-50 51-68 69-83 84-94 95-106 107-145 Sonntag, Borgnakke and van Wylen Correspondence List 6th edition CHAPTER 10 Sonntag/Borgnakke/Wylen The correspondence between the new problem set and the previous 5th edition chapter 10 problem set. Study guide problems 10.1-10.20 are all new. New 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 Old new 23 new 3 new 4 5 new 6 9 13 15 14 57a new new new 2 7 new 8 12 21 10 16 11 17 22 35 57b New 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 Old 25 30 32 new 33 new 26 18 new 31 36 20 new 24 27 28 29 34 37 43 new 45 38 39 40 42 new 44 47 50 New 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 Old new new new new 19 new new new new new new new new new new 1 46 48 49 new 52 55 58 59 new new Sonntag, Borgnakke and van Wylen The English unit problems are: The correspondence between the new English unit problem set and the previous 5th edition chapter 10 problem set with the current set of SI problems. New 107 108 109 110 111 112 113 114 115 116 117 118 119 5th new new new new new new 69 62 new 66 86a new 63 SI 12 14 15 16 18 20 22 24 23 32 34 37 39 New 120 121 122 123 124 125 126 127 128 129 130 131 132 5th 65 68 64 67 86b 70 73 74 76 new 71 72 75 SI 42 43 44 45 50 51 52 53 61 63 65 67 68 New 133 134 135 136 137 138 139 140 141 142 143 144 145 5th new new 77mod 78 79mod 81 new new 61 80 82 new 87 SI 82 71 74 76 78 79 87 89 96 97 99 104 Sonntag, Borgnakke and van Wylen Concept-Study Guide Problems 10.1 Can I have any energy transfer as heat transfer that is 100% available? By definition the possible amount of work that can be obtained equals the exergy (availability). The maximum is limited to that out of a reversible heat engine, if constant T then that is the Carnot heat engine To )Q T So we get a maximum for an infinite high temperature T, where we approach an efficiency of one. In practice you do not have such a source (the closest would be solar radiation) and secondly no material could contain matter at very high T so a cycle process can proceed (the closest would be a plasma suspended by a magnetic field as in a tokamak). W = (1 − 10.2 Is energy transfer as work 100% available? Yes. By definition work is 100% exergy or availability. 10.3 We cannot create nor destroy energy, but how about available energy? Yes. Every process that is irreversible to some degree destroys exergy. This destruction is directly proportional to the entropy generation. 10.4 Energy can be stored as internal energy, potential energy or kinetic energy. Are those energy forms all 100% available? The internal energy is only partly available, a process like an expansion can give out work or if it cools by heat transfer out it is a Q out that is only partly available as work. Potential energy like from gravitation, mgH, or a compressed spring or a charged battery are forms that are close to 100% available with only small losses present. Kinetic energy like in a fly-wheel or motion of a mass can be transferred to work out with losses depending on the mechanical system. Sonntag, Borgnakke and van Wylen 10.5 All the energy in the ocean is that available? No. Since the ocean is at the ambient T (it is the ambient) it is not possible to extract any work from it. You can extract wave energy (wind generated kinetic energy) or run turbines from the tide flow of water (moon generated kinetic energy). However, since the ocean temperature is not uniform there are a few locations where cold and warmer water flows close to each other like at different depths. In that case a heat engine can operate due to the temperature difference. 10.6 Does a reversible process change the availability if there is no work involved? Yes. There can be heat transfer involved and that has an availability associated with it, which then equals the change of availability of the substance. 10.7 Is the reversible work between two states the same as ideal work for the device? No. It depends on the definition of ideal work. The ideal device does not necessarily have the same exit state as the actual device. An ideal turbine is approximated as a reversible adiabatic device so the ideal work is the isentropic work. The reversible work is between the inlet state and the actual exit state that do not necessarily have the same entropy. 10.8 When is the reversible work the same as the isentropic work? That happens when the inlet and exit states (or beginning and end states) have the same entropy. 10.9 If I heat some cold liquid water to To, do I increase its availability? No. You decrease its availability by bringing it closer to To, where it has zero availability, if we neglect pressure effects. Any substance at a T different from ambient (higher or lower) has a positive availability since you can run a heat engine using the two temperatures as the hot and cold reservoir, respectively. For a T lower than the ambient it means that the ambient is the hot side of the heat engine. Sonntag, Borgnakke and van Wylen 10.10 Are reversible work and availability (exergy) connected? Yes. They are very similar. Reversible work is usually defined as the reversible work that can be obtained between two states, inlet-exit or beginning to end. Availability is a property of a given state and defined as the reversible work that can be obtained by changing the state of the substance from the given state to the dead state (ambient). 10.11 Consider availability (exergy) associated with a flow. The total exergy is based on the thermodynamic state, the kinetic and potential energies. Can they all be negative? No. By virtue of its definition kinetic energy can only be positive. The potential energy is measured from a reference elevation (standard sea level or a local elevation) so it can be negative. The thermodynamic state can only have a positive exergy the smallest it can be is zero if it is the ambient dead state. Sonntag, Borgnakke and van Wylen 10.12 A flow of air at 1000 kPa, 300 K is throttled to 500 kPa. What is the irreversibility? What is the drop in flow availability? A throttle process is constant enthalpy if we neglect kinetic energies. Process: h e = hi so ideal gas o => Te = Ti o Entropy Eq.: se - si = sgen = sTe - sTi – R ln Pe Pe = 0 – R ln Pi Pi sgen = - 0.287 ln (500 / 1000) = 0.2 kJ/kg K Eq.10.11: i = To sgen = 298 0.2 = 59.6 kJ/kg The drop in availability is exergy destruction, which is the irreversibility ∆ψ = i = 59.6 kJ/kg P 1000 e i Phigh cb Plow 500 T i Pi i e v Pe e s Sonntag, Borgnakke and van Wylen 10.13 A steam turbine inlet is at 1200 kPa, 500oC. The actual exit is at 300 kPa with an actual work of 407 kJ/kg. What is its second law efficiency? The second law efficiency is the actual work out measured relative to the reversible work out, Eq. 10.29. Steam turbine To = 25°C = 298.15 K Inlet state: Table B.1.3 hi = 3476.28 kJ/kg; Actual turbine energy Eq.: si = 7.6758 kJ/kg K he = hi - wac = 3476.28 – 407 = 3069.28 kJ/kg Te = 300oC; Actual exit state: Table B.1.3 se = 7.7022 kJ/kg K From Eq.10.9, wrev = (hi - Tosi) – (he - Tose) = (hi - he) + To(se - si) = (3476.28 – 3069.28) + 298.15(7.7022 – 7.6758) = 407 + 7.87 = 414.9 kJ/kg ηII = wac/wrev = 407 / 414.9 = 0.98 P 1200 300 T i 500 e 300 v i Pi Pe e s Sonntag, Borgnakke and van Wylen 10.14 A heat exchanger increases the availability of 3 kg/s water by 1650 kJ/kg using 10 kg/s air coming in at 1400 K and leaving with 600 kJ/kg less availability. What are the irreversibility and the second law efficiency? C.V. Heat exchanger, steady flow 1 inlet and 1 exit for air and water each. The two flows exchange energy with no heat transfer to/from the outside. 4 1 air 2 3 water The irreversibility is the destruction of exergy (availability) so .. . . I = Φdestruction = Φin - Φout = 10 × 600 – 3 × 1650 = 1050 kW The second law efficiency, Eq.10.32 . . 3 × 1650 ηII = Φout / Φin = = 0.825 10 × 600 Sonntag, Borgnakke and van Wylen 10.15 A heat engine receives 1 kW heat transfer at 1000 K and gives out 600 W as work with the rest as heat transfer to the ambient. What are the fluxes of exergy in and out? To . . 298.15 Exergy flux in: ΦH = 1 – QH = 1 – 1 kW = 0.702 kW TH 1000 To . . Exergy flux out: ΦL = 1 – QL = 0 TL ( TL = To ) . . The other exergy flux out is the power Φout = W = 0.6 kW 1000 K QH = 1 kW HE W = 600 W QL Tamb Sonntag, Borgnakke and van Wylen 10.16 A heat engine receives 1 kW heat transfer at 1000 K and gives out 600 W as work with the rest as heat transfer to the ambient. Find its first and second law efficiencies. First law efficiency is based on the energies .. 0.6 ηI = W/QH = = 0.6 1 The second law efficiency is based on work out versus availability in To . . 298.15 Exergy flux in: ΦH = 1 – QH = 1 – 1 kW = 0.702 kW TH 1000 . W 0.6 ηII = . = = 0.855 0.702 ΦH Notice the exergy flux in is equal to the Carnot heat engine power output given 1 kW at 1000 K and rejecting energy to the ambient. 1000 K QH = 1 kW HE W = 600 W QL Tamb 10.17 Is the exergy equation independent of the energy and entropy equations? No. The exergy equation is derived from the other balance equations by defining the exergy from the state properties and the reference dead state. Sonntag, Borgnakke and van Wylen 10.18 A heat pump has a coefficient of performance of 2 using a power input of 2 kW. Its low temperature is To and the high temperature is 80oC, with an ambient at To. Find the fluxes of exergy associated with the energy fluxes in and out. First let us do the energies in and out . QH . . COP = β = . => QH = β W = 2 × 2 kW = 4 kW W Energy Eq.: . . . QL = QH – W = 4 – 2 = 2 kW To . . ( TL = To ) Exergy flux in: ΦL = 1 – QL = 0 TL . . Exergy flux in: ΦW = W = 2 kW To . . 298.15 Exergy flux out: ΦH = 1 – QH = 1 – 4 kW = 0.623 kW TH 353.15 Remark: The process then destroys (2 – 0.623) kW of exergy. o 80 C QH HP W = 2 kW QL To Sonntag, Borgnakke and van Wylen 10.19 Use the exergy balance equation to find the efficiency of a steady state Carnot heat engine operating between two fixed temperature reservoirs? The exergy balance equation, Eq.10.36, for this case looks like To . To . . 0 = 1 – QH - 1 – QL - W + 0 + 0 – 0 – 0 TH TL Steady state (LHS = 0 and dV/dt = 0, no mass flow terms, Carnot cycle so reversible and the destruction is then zero. From the energy equation we have . . . 0 = QH - QL - W which we can subtract from the exergy balance equation to get To . To . QH + Q 0= – TH TL L Solve for one heat transfer in terms of the other TL . . Q QL = TH H The work from the energy equation is TL . . . . W = QH - QL = QH [ 1 TH from which we can read the Carnot cycle efficiency as we found in Chapter 7. Sonntag, Borgnakke and van Wylen 10.20 Find the second law efficiency of the heat pump in problem 10.18. The second law efficiency is a ratio of exergies namely what we want out divided by what we have to put in. Exergy from first term on RHS Eq. 10.36 To . . ΦH = 1 – QH; TH . . QH = β W = 2 × 2 kW = 4 kW . . ΦH To QH 298.15 4 = 0.31 ηII = . = 1 – . = 1 – TH 353.15 2 W W o 80 C QH HP W = 2 kW QL To Sonntag, Borgnakke and van Wylen Available Energy, Reversible work 10.21 Find the availability of 100 kW delivered at 500 K when the ambient is 300 K. Solution: The availability of an amount of heat transfer equals the possible work that can be extracted. This is the work out of a Carnot heat engine with heat transfer to the ambient as the other reservoir. The result is from Chapter 7 as also shown in Eq. 10.1 and Eq. 10.36 To . . . 300 Φ = Wrev HE = (1 – ) Q = (1 – ) 100 kW = 40 kW T 500 Sonntag, Borgnakke and van Wylen 10.22 A control mass gives out 10 kJ of energy in the form of a. Electrical work from a battery b. Mechanical work from a spring c. Heat transfer at 500°C Find the change in availability of the control mass for each of the three cases. Solution: a) Work is availability b) Work is availability ∆Φ = −Wel = -10 kJ ∆Φ = −Wspring = -10 kJ c) Give the heat transfer to a Carnot heat engine and W is availability T0 ∆Φ = −[1 − Qout = −1 - 298.15 10 = −6.14 kJ 773.15 TH Sonntag, Borgnakke and van Wylen 10.23 A heat engine receives 5 kW at 800 K and 10 kW at 1000 K rejecting energy by heat transfer at 600 K. Assume it is reversible and find the power output. How much power could be produced if it could reject energy at To = 298 K? Solution: C.V. The heat engine, this is in steady state. . . . . Energy Eq.: 0 = Q1 + Q2 – QL – W . . . Q1 Q2 QL Entropy Eq.: 0 = + – +0 T1 T2 TL Q1 Q2 HE . Now solve for QL from the entropy equation TL . TL . . 600 600 Q1 + Q2 = ×5+ × 10 = 9.75 kW QL = T1 T2 800 1000 Substitue into the enrgy equation and solve for the work term . . . . W = Q1 + Q2 – QL = 5 + 10 – 9.75 = 5.25 kW For a low temperature of 298 K we can get . 298 . QL2 = Q = 4.843 kW 600 L . . . . W = Q1 + Q2 – QL = 5 + 10 – 4.843 = 10.16 kW Remark: Notice the large increase in the power output. W QL Sonntag, Borgnakke and van Wylen 10.24 The compressor in a refrigerator takes refrigerant R-134a in at 100 kPa, −20°C and compresses it to 1 MPa, 40°C. With the room at 20°C find the minimum compressor work. Solution: 1 C.V. Compressor out to ambient. Minimum work in is the reversible work. Steady flow, 1 inlet and 2 exit Energy Eq.: 2 -WC wc = h1 - h2 + qrev ⌠ Entropy Eq.: s2 = s1 + ⌡dq/T + sgen = s1 + qrev/To + 0 => qrev = To(s2 - s1) wc min = h1 - h2 + To(s2 - s1) = 387.22 - 420.25 + 293.15 × (1.7148 - 1.7665) = -48.19 kJ/kg Sonntag, Borgnakke and van Wylen 10.25 Find the specific reversible work for a steam turbine with inlet 4 MPa, 500°C and an actual exit state of 100 kPa, x = 1.0 with a 25°C ambient. Solution: Steam turbine To = 25°C = 298.15 K Inlet state: Table B.1.3 hi = 3445.2 kJ/kg; si = 7.090 kJ/kg K Exit state: Table B.1.2 he = 2675.5 kJ/kg; se = 7.3593 kJ/kg K From Eq.9.39, wrev = (hi - Tosi) – (he - Tose) = (hi - he) + To(se - si) = (3445.2 – 2675.5) + 298.2(7.3593 – 7.0900) = 769.7 + 80.3 = 850.0 kJ/kg P T i i i e e e WT v s Sonntag, Borgnakke and van Wylen 10.26 Calculate the reversible work out of the two-stage turbine shown in Problem 6.82, assuming the ambient is at 25°C. Compare this to the actual work which was found to be 18.08 MW. C.V. Turbine. Steady flow, 1 inlet and 2 exits. Use Eq. 10.12 for each flow stream with q = 0 for adiabatic turbine. 1 Supply state 1: 20 kg/s at 10 MPa, 500°C 2 Process steam 2: 5 kg/s, 0.5 MPa, 155°C, Exit state 3: 20 kPa, x = 0.9 WT Table B.1.3: h1 = 3373.7, h2 = 2755.9 kJ/kg, 3 s1 = 6.5966, Table B.1.2: s2 = 6.8382 kJ/kg K HP h3 = 251.4 + 0.9 × 2358.3 = 2373.9 kJ/kg, s3 = 0.8319 + 0.9 × 7.0766 = 7.2009 kJ/kg K . rev . . . . . . W = (m1h1 - m2h2 - m3h3) - T0(m1s1 - m2s2 - m3s3) = 20 × 3373.7 − 5 × 2755.9 − 15 × 2373.9 − 298.15 (20 × 6.5966 - 5 × 6.8382 + 15 × 7.2009) . . = 21.14 MW = Wac + Qrev = 18084 kW + 3062.7 kW LP Sonntag, Borgnakke and van Wylen 10.27 A household refrigerator has a freezer at TF and a cold space at TC from which energy is removed and rejected to the ambient at TA as shown in Fig. P10.27. . Assume that the rate of heat transfer from the cold space, QC, is the same as from . the freezer, QF, find an expression for the minimum power into the heat pump. . Evaluate this power when TA = 20°C, TC = 5°C, TF = −10°C, and QF = 3 kW. Solution: C.V. Refrigerator (heat pump), Steady, no external flows except heat transfer. . . . . Energy Eq.: QF + Qc + W = QA (amount rejected to ambient) QC QF REF W QA Reversible gives minimum work in as from Eq. 10.1 or 10.9 on rate form. TA . TA . . 293.15 293.15 + 3 1 − W = QF 1 − + Qc 1 − = 3 1 − TF TC 263.15 278.15 = -0.504 kW (negative so work goes in) Sonntag, Borgnakke and van Wylen 10.28 Find the specific reversible work for a R-134a compressor with inlet state of –20°C, 100 kPa and an exit state of 600 kPa, 50°C. Use a 25°C ambient temperature. Solution: This is a steady state flow device for which the reversible work is given by Eq.10.9. The compressor is also assumed to be adiabatic so q = 0 wrev = To(se - si) - (he - hi) Table B.5.2: hi = 387.22 kJ/kg; si = 1.7665 kJ/kg K he = 438.59 kJ/kg; se = 1.8084 kJ/kg K wrev = 298.15 (1.8084 - 1.7665) - (438.59 - 387.22) = -38.878 kJ/kg P T e e e i i v i WC in s Sonntag, Borgnakke and van Wylen 10.29 An air compressor takes air in at the state of the surroundings 100 kPa, 300 K. The air exits at 400 kPa, 200°C at the rate of 2 kg/s. Determine the minimum compressor work input. C.V. Compressor, Steady flow, minimum work in is reversible work. ψ1 = 0 at ambient conditions ° ° s0 - s2 = sT0 - sT2 - R ln(P0/P2) = 6.86926 - 7.3303 - 0.287 ln(100/400) = -0.06317 kJ/kg K ψ2 = h2 - h0 + T0(s0 - s2) = 475.79 - 300.473 + 300 (-0.06317) = 156.365 kJ/kg . . REV . = m(ψ2 - ψ1) = 312.73 kW = Wc -W Sonntag, Borgnakke and van Wylen 10.30 A steam turbine receives steam at 6 MPa, 800°C. It has a heat loss of 49.7 kJ/kg and an isentropic efficiency of 90%. For an exit pressure of 15 kPa and surroundings at 20°C, find the actual work and the reversible work between the inlet and the exit. C.V. Reversible adiabatic turbine (isentropic) wT = hi - he,s ; se,s = si = 7.6566 kJ/kg K, hi = 4132.7 kJ/kg xe,s = (7.6566 - 0.7548)/7.2536 = 0.9515, he,s = 225.91 + 0.9515×2373.14 = 2483.9 kJ/kg wT,s = 4132.7 - 2483.9 = 1648.79 kJ/kg C.V. Actual turbine wT,ac = ηwT,s = 1483.91 kJ/kg = hi - he,ac - qloss ⇒ he,ac = hi - qloss - wT,ac = 4132.7 - 49.7 - 1483.91 = 2599.1 kJ/kg Actual exit state: P,h ⇒ sat. vap., se,ac = 8.0085 kJ/kg K C.V. Reversible process, work from Eq.10.12 qR = T0(se,ac - si) = 293.15 × (8.0085 - 7.6566) = 103.15 kJ kg wR = hi - he,ac + qR = 4132.7 - 2599.1 + 103.16 = 1636.8 kJ/kg Sonntag, Borgnakke and van Wylen 10.31 An air compressor receives atmospheric air at T0 = 17°C, 100 kPa, and compresses it up to 1400 kPa. The compressor has an isentropic efficiency of 88% and it loses energy by heat transfer to the atmosphere as 10% of the isentropic work. Find the actual exit temperature and the reversible work. C.V. Compressor Isentropic: wc,in,s = he,s - hi ; se,s = si From table A.7.1 and entropy equation we get o o sTe s = sTi + R ln (Pe/Pi) = 6.83521 + 0.287 ln(14) = 7.59262 Back interpolate in Table A.7: ⇒ he,s = 617.23 kJ/kg wc,in,s = 617.23 - 290.43 = 326.8 kJ/kg Actual: wc,in,ac = wc,in,s/ηc = 371.36 ; qloss = 32.68 kJ/kg wc,in,ac + hi = he,ac + qloss => he,ac = 290.43 + 371.36 - 32.68 = 629.1 kJ/kg => Te,ac = 621 K Reversible: wrev = hi - he,ac + T0(se,ac - si) = 290.43 - 629.1 + 290.15 × (7.6120 - 6.8357) = -338.67 + 225.38 = -113.3 kJ/kg Since qloss is also to the atmosphere it is the net q exchanged with the ambient that explains the change in s. Sonntag, Borgnakke and van Wylen 10.32 Air flows through a constant pressure heating device, shown in Fig. P10.32. It is heated up in a reversible process with a work input of 200 kJ/kg air flowing. The device exchanges heat with the ambient at 300 K. The air enters at 300 K, 400 kPa. Assuming constant specific heat develop an expression for the exit temperature and solve for it by iterations. C.V. Total out to T0 Energy Eq.: h1 + qrev - wrev = h2 0 Entropy Eq.: s1 + qrev/T0 = s2 0 ⇒ qrev = T0(s2 - s1) 0 h2 - h1 = T0(s2 - s1) - wrev (same as Eq. 10.12) Constant Cp gives: Cp(T2 - T1) = T0Cp ln (T2/T1) + 200 The energy equation becomes T2 200 T2 - T0 ln( T ) = T1 + C 1 p T1 = 300 K, Cp = 1.004 kJ/kg K, T0 = 300 K T2 - 300 ln ( T2 200 ) = 300 + = 499.3 K 300 1.004 Now trial and error on T2 At 600 K LHS = 392 (too low) At 800 K LHS = 505.75 Linear interpolation gives T2 = 790 K (LHS = 499.5 OK) Sonntag, Borgnakke and van Wylen 10.33 A piston/cylinder has forces on the piston so it keeps constant pressure. It contains 2 kg of ammonia at 1 MPa, 40°C and is now heated to 100°C by a reversible heat engine that receives heat from a 200°C source. Find the work out of the heat engine. C.V. Ammonia plus heat engine Energy: mam(u2 - u1) = 1Q2,200 - WH.E. - 1W2,pist Entropy: mam(s2 - s1) = 1Q2/Tres + 0 => 1Q2 NH 3 = mam(s2 - s1)Tres QL Process: P = const. ⇒ 1W2 = P(v2 - v1)mam Substitute the piston work term and heat transfer into the energy equation W HE cb QH WH.E. = mam(s2 - s1)Tres - mam(h2 - h1) o 200 C Table B.2.2: h1 = 1508.5 kJ/kg, s1 = 5.1778 kJ/kg K, h2 = 1664.3 kJ/kg, s2 = 5.6342 kJ/kg K WH.E. = 2 × [(5.6342 - 5.1778)473.15 - (1664.3 - 1508.5)] = 120.3 kJ Sonntag, Borgnakke and van Wylen 10.34 A rock bed consists of 6000 kg granite and is at 70°C. A small house with lumped mass of 12000 kg wood and 1000 kg iron is at 15°C. They are now brought to a uniform final temperature with no external heat transfer by connecting the house and rock bed through some heat engines. If the process is reversible, find the final temperature and the work done in the process. Solution: Take C.V. Total (rockbed and heat engine) Energy Eq.: mrock(u2 - u1) + mwood(u2 - u1) + mFe(u2 - u1) = -1W2 Entropy Eq.: / mrock(s2 - s1) + mwood(s2 - s1) + mFe(s2 - s1) = 0 (mC)rockln T2 T2 T2 + (mC)woodln + (mC)Feln = 0 / T1 T1 T1 6000 × 0.89 ln (T2/343.15) + 12000 × 1.26 ln (T2/288.15) + 1000 × 0.46 ln (T2/288.15) = 0 / => T2 = 301.3 K Now from the energy equation -1W2 = 6000 × 0.89(301.3 - 343.15) + (12000 × 1.26 + 460)(301.3 - 288.15) ⇒ 1W2 = 18 602 kJ W HE cb Q H QL H O U S E Sonntag, Borgnakke and van Wylen 10.35 An air flow of 5 kg/min at 1500 K, 125 kPa goes through a constant pressure heat exchanger, giving energy to a heat engine shown in Figure P10.35. The air exits at 500 K and the ambient is at 298 K, 100 kPa. Find the rate of heat transfer delivered to the engine and the power the engine can produce. Solution: C.V. Heat exchanger . . Continuity eq.: m1 = m2 ; . . . Energy Eq.6.12: m1h1 = m1h2 + QH 2 1 Table A.7.1: h1 = 1635.8 kJ/kg, QH h2 = 503.36 kJ/kg, s1 = 8.61209 kJ/kg K s2 = 7.38692 kJ/kg K W HE QL Ambient . . 5 kg kJ (1635.8 – 503.36) = 94.37 kW QH = m(h1 – h2) = 60 s kg C.V. Total system for which we will write the second law. . . . . Entropy Equation 9.8: m s1 + Sgen = m s2 + QL/To . Process: Assume reversible Sgen = 0, and P = C for air . . 5 kg kJ (8.61209 – 7.38692) QL = To m (s1 – s2) = 298 K 60 s kg K = 30.425 kW Energy equation for the heat engine gives the work as . . . W = QH - QL = 94.37 – 30.425 = 63.9 kW Sonntag, Borgnakke and van Wylen Irreversibility 10.36 Calculate the irreversibility for the condenser in Problem 9.53 assuming an ambient temperature at 17°C. Solution: C.V. Condenser. Steady state with no shaft work term. . . . Energy Equation 6.12: m hi + Q = mhe . . . . Entropy Equation 9.8: m si + Q/T + Sgen = m se Properties are from Table B.1.2 hi = 225.91 + 0.9 × 2373.14 = 2361.74 kJ/kg , he= 225.91 kJ/kg si = 0.7548 + 0.9 × 7.2536 = 7.283 kJ/kg K, se = 0.7548 kJ/kg K From the energy equation . . . Qout = –Q = m (hi – he) = 5(2361.74 – 225.91) = 10679 kW From the entropy equation . . . Sgen = m (se – si) + Qout/T = 5(0.7548 – 7.283) + 10679/(273 + 17) = –35.376 + 36.824 = 1.448 kW/K . From Eq.10.11 times m, . . I = To Sgen = 290 × 1.448 = 419.9 kW Sonntag, Borgnakke and van Wylen 10.37 A constant pressure piston/cylinder contains 2 kg of water at 5 MPa and 100oC. Heat is added from a reservoir at 700oC to the water until it reaches 700oC. We want to find the total irreversibility in the process. Solution: C.V. Piston cylinder out to the reservoir (incl. the walls). Energy Eq.: m(u2 - u1) = 1Q2 - 1W2 Entropy Eq.: m(s2 - s1) = 1Q2/Tres + 1S2 gen State 1: State 2: H2O h1 = 422.71 kJ/kg, s1 = 1.303 kJ/kg K h2 = 3900.13 kJ/kg, s2 = 7.5122 kJ/kg K Process: P = C => 1W2 = P(V2 – V1) 700 C From the energy equation we get 1Q2 = m(u2 - u1) + 1W2 = m(h2 - h1) = 2(3900.13 – 422.71) = 6954.8 kJ From the entropy equation we get 6954.8 kJ 1Q2 1S2 gen = m(s2 - s1) – T = 2(7.5122 – 1.303) - 273 + 700 = 5.2717 K res Now the irreversibility is from Eq. 10.19 kJ 1I2 = m 1i2 = To 1S2 gen = 298.15 K × 5.2717 K = 1572 kJ Sonntag, Borgnakke and van Wylen 10.38 Calculate the reversible work and irreversibility for the process described in Problem 5.97, assuming that the heat transfer is with the surroundings at 20°C. Solution: P 2 1 Linear spring gives 1 ⌡ 1W2 = ⌠PdV = 2(P1 + P2)(V2 - V1) 1Q2 = m(u2 - u1) + 1W2 v Equation of state: PV = mRT State 1: V1 = mRT1/P1 = 2 × 0.1889 × 673.15 /500 = 0.5087 m3 State 2: V2 = mRT2/P2 = 2 × 0.1889 × 313.15 /300 = 0.3944 m3 1 1W2 = 2(500 + 300)(0.3944 - 0.5087) = -45.72 kJ From Figure 5.11: Cp(Tavg) = 5.25 R = 0.99 ⇒ Cv = 0.803 = Cp - R For comparison the value from Table A.5 at 300 K is Cv = 0.653 kJ/kg K 1Q2 = mCv(T2 - T1) + 1W2 = 2 × 0.803(40 - 400) - 45.72 = -623.9 kJ rev 1W2 = To(S2 - S1) - (U2 - U1) + 1Q2 (1 - To/TH) ac = Tom(s2 - s1)+ 1W2 - 1Q2 To/To ac = Tom[CP ln(T2 / T1) − R ln(P2 / P1)] + 1W2 - 1Q2 = 293.15 × 2 [ 0.99 ln(313/673) - 0.1889 ln(300/500)] - 45.72 + 623.9 = -387.8 - 45.72 + 623.9 = 190.4 kJ rev ac 1I2 = 1W2 - 1W2 = 190.4 - (-45.72) = 236.1 kJ Sonntag, Borgnakke and van Wylen 10.39 A supply of steam at 100 kPa, 150°C is needed in a hospital for cleaning purposes at a rate of 15 kg/s. A supply of steam at 150 kPa, 250°C is available from a boiler and tap water at 100 kPa, 15°C is also available. The two sources are then mixed in a mixing chamber to generate the desired state as output. Determine the rate of irreversibility of the mixing process. C.V. Mixing chamber, Steady flow . . . Continuity Eq.: m 1 + m2 = m3 . . . Energy Eq.: m1h1 + m2h2 = m3h3 . . . . Entropy Eq.: m1s1 + m2s2 + Sgen = m3s3 Table properties B.1.1: h1 = 62.99 kJ/kg, s1 = 0.2245 kJ/kg K B.1.3: h2 = 2972.7 kJ/kg, s2 = 7.8437 kJ/kg K B.1.3: h3 = 2776.4 kJ/kg, s3 = 7.6133 kJ/kg K From the energy equation we get .. 2776.4 - 62.99 = 0.9325 m2/m3 = (h3 - h1)/(h2 - h1) = 2972.7 - 62.99 . . m2 = 13.988 kg/s, m1 = 1.012 kg/s From the entropy equation we get . . . . . I = T0Sgen = T0(m3s3 - m1s1 - m2s2) = 298.15 × (15 × 7.6133 - 1.012 × 0.2245 - 13.988 × 7.8437) = 1269 kW T cb 1 MIXING 2 CHAMBER 3 1 2 3 s Sonntag, Borgnakke and van Wylen 10.40 The throttle process in Example 6.5 is an irreversible process. Find the reversible work and irreversibility assuming an ambient temperature at 25°C. Solution: C.V. Throttle. Steady state, adiabatic q = 0 and no shaft work w = 0. Inlet state: B.2.1 hi = 346.8 kJ/kg; si = 1.2792 kJ/kg K Energy Eq.6.13: he = h i Exit state: P = 291 kPa, he = hi B.2.1 which is two-phase se = sf + xsfg = 0.5408 + 0.1638 × 4.9265 = 1.3478 kJ/kg K The reversible work is the difference in availability also equal to the expression in Eq.10.9 or 10.36 and 10.37 wrev = ψi - ψe = (hi - Tosi) – (he - Tose) = (hi - he) + To(se - si) = 0 + 298.15 (1.2792 - 1.3478) = 20.45 kJ/kg i = wrev - w = 20.45 - 0 = 20.45 kJ/kg i e P T P i i e h=C T v e h=C v Sonntag, Borgnakke and van Wylen 10.41 Two flows of air both at 200 kPa of equal flow rates mix in an insulated mixing chamber. One flow is at 1500 K and the other is at 300 K. Find the irreversibility in the process per kilogram of air flowing out. C.V. Mixing chamber . . . . Continuity Eq..: m 1 + m2 = m3 = 2 m1 . . . Energy Eq.: m1h1 + m1h2 = 2 m1h3 . . . . Entropy Eq.: m1s1 + m1s2 + Sgen = 2 m1s3 Properties from Table A.7 h3 = (h1 + h2)/2 = (300.473 + 1635.8)/2 = 968.14 kJ/kg ° ⇒ sT3 = 8.0474 kJ/kg K From the entropy equation . . Sgen/2m1 = s3 − (s1 + s2)/2 = 8.0474 - (6.86926 + 8.61208)/2 = 0.30673 kJ/kg K .. . . i = I/2m1 = T Sgen/2m1 = 298.15 × 0.30673 = 91.45 kJ/kg T 1 MIXING 2 CHAMBER 2 3 200 kPa 3 1 s Sonntag, Borgnakke and van Wylen 10.42 Fresh water can be produced from saltwater by evaporation and subsequent condensation. An example is shown in Fig. P10.42, where 150-kg/s saltwater, state 1, comes from the condenser in a large power plant. The water is throttled to the saturated pressure in the flash evaporator and the vapor, state 2, is then condensed by cooling with sea water. As the evaporation takes place below atmospheric pressure, pumps must bring the liquid water flows back up to P0. Assume that the saltwater has the same properties as pure water, the ambient is at 20°C and that there are no external heat transfers. With the states as shown in the table below find the irreversibility in the throttling valve and in the condenser. State T [°C] 1 30 2 25 h [kJ/kg] s [kJ/kg K] 125.77 0.4369 3 25 4 -- 2547.2 8.558 5 23 6 -- 96.5 0.3392 7 17 71.37 83.96 0.2535 0.2966 C.V. Valve. P2 = Psat(T2 = T3) = 3.169 kPa . . . Continuity Eq.: m1 = mex = m2 + m3 Energy Eq.: h 1 = he ; Entropy Eq.: s1 + sgen = se he = h1 ⇒ xe = (125.77 - 104.87)/2442.3 = 0.008558 ⇒ se = 0.3673 + 0.008558 × 8.1905 = 0.4374 kJ/kg K . . m2 = (1 - xe)m1 = 148.716 kg/s sgen = se - s1 = 0.4374 - 0.4369 = 0.000494 kJ/kg K .. I = mT0sgen = 150 × 293.15 × 0.000494 = 21.72 kW C.V. Condenser. Energy Eq.: . . . . m2h2 + m7h7 = m2h5 + m7h8 ⇒ . . 2547.2 - 96.5 kg = 28 948 m7 = m2 × (h2 - h5)/(h8 - h7) = 148.716 × 83.96 - 71.37 s . . . . . Entropy Eq.: m2s2 + m7s7 + Sgen = m2s5 + m7s8 . . . . I = T0Sgen = T0 [m2(s5 - s2) + m7(s8 - s7)] = 293.15[148.716(0.3392 - 8.558) + 28948(0.2966 - 0.2535)] = 293.15 × 25.392 = 7444 kW 8 20 Sonntag, Borgnakke and van Wylen 10.43 Calculate the irreversibility for the process described in Problem 6.133, assuming that heat transfer is with the surroundings at 17°C. Solution: C.V. Cylinder volume out to To = 17 oC. Continuity Eq.6.15: m2 - m1 = min Energy Eq.6.16: m2u2 - m1u1 = minhline + 1Q2 - 1W2 Entropy Eq.9.12: m2 s2 - m1s1 = misi + 1Q2 / To + 1S2 gen Process: P1 is constant to stops, then constant V to state 2 at P2 P1V 300 × 0.25 State 1: P1, T1 m1 = = = 0.90 kg RT1 0.287 × 290.2 State 2: Open to P2 = 400 kPa, T2 = 350 K 400 × 1 = 3.982 kg 0.287 × 350 mi = 3.982 - 0.90 = 3.082 kg m2 = AIR Only work while constant P 1W2 = P1(V2 - V1) = 300(1 - 0.25) = 225 kJ Energy eq.: 1Q2 = m2u2 - m1u1 + 1W2 - mihi = 3.982 × 0.717 × 350 - 0.90 × 0.717 × 290.2 + 225 - 3.082 × 1.004 × 600 = -819.2 kJ Entropy eq. gives To 1S2 gen = I = To [ m1 (s2 - s1) + mi (s2 - si)] - 1Q2 = 290.15[0.9(Cp ln 350 400 350 400 - R ln ) + 3.082(Cpln - R ln )] 290 300 600 500 - ( - 819.2 kJ) = 290.15 (0.0956 - 1.4705) + 819.2 = 420.3 kJ Sonntag, Borgnakke and van Wylen 10.44 A 2-kg piece of iron is heated from room temperature 25°C to 400°C by a heat source at 600°C. What is the irreversibility in the process? Solution: C.V. Iron out to 600°C source, which is a control mass. Energy Eq.: mFe(u2 - u1) = 1Q2 - 1W2 Entropy Eq.: mFe(s2 - s1) = 1Q2/Tres + 1S2 gen Process: Constant pressure => 1W2 = PmFe(v2 - v1) ⇒ 1Q2 = mFe(h2 - h1) = mFeC(T2 - T1) = 2 × 0.42 × (400 - 25) = 315 kJ 1S2 gen = mFe(s2 - s1) - 1Q2/Tres = mFeC ln (T2/T1) - 1Q2/Tres = 2 × 0.42 × ln 1I2 673.15 315 = 0.3233 kJ/K 298.15 873.15 = To (1S2 gen ) = 298.15 × 0.3233 = 96.4 kJ Fe A real flame may be more than 600°C, but a little away from it where the gas has mixed with some air it may be 600°C. Sonntag, Borgnakke and van Wylen 10.45 Air enters the turbocharger compressor (see Fig. P10.45), of an automotive engine at 100 kPa, 30°C, and exits at 170 kPa. The air is cooled by 50°C in an intercooler before entering the engine. The isentropic efficiency of the compressor is 75%. Determine the temperature of the air entering the engine and the irreversibility of the compression-cooling process. Solution: a) Compressor. First ideal which is reversible adiabatic, constant s: P2 P1 k-1 k () T2S = T1 (170) 100 0.286 = 303.2 = 352.9 K wS = CP0(T1 - T2S) = 1.004(303.2 - 352.9) = -49.9 kJ/kg Now the actual compressor w = wS/ηS = -49.9/0.75 = -66.5 kJ/kg = CP(T1 - T2) ⇒ T2 = 369.5 K T3(to engine) = T2 - ∆TINTERCOOLER = 369.5 - 50 = 319.5 K = 46.3°C b) Irreversibility from Eq.10.13 with rev. work from Eq.10.12, (q = 0 at TH) kJ 319.4 170 s3 - s1 = 1.004 ln - 0.287 ln = -0.1001 303.2 100 kg K i = T(s3 - s1) - (h3 - h1) - w = T(s3 - s1) - CP(T3 - T1) - CP(T1 - T2) = 303.2(-0.1001) - 1.004(-50) = +19.8 kJ/kg Cooler -Q C 3 2 -WC 1 Compressor cb Engine Exhaust Sonntag, Borgnakke and van Wylen 10.46 A 2-kg/s flow of steam at 1 MPa, 700°C should be brought to 500°C by spraying in liquid water at 1 MPa, 20°C in an steady flow. Find the rate of irreversibility, assuming that surroundings are at 20°C. C.V. Mixing chamber, Steady flow. State 1 is superheated vapor in, state 2 is compressed liquid in, and state 3 is flow out. No work or heat transfer. • Continuity Eq.6.9: • Energy Eq.6.10: 2 • • • 1 • • 3 m3h3 = m1h1 + m2h2 • Entropy Eq.9.7: Table B.1.3: • m 3 = m1 + m2 • m3s3 = m1s1 + m2s2 + Sgen h1 = 3923.1 kJ/kg, s1 = 8.2731 kJ/kg K, h3 = 3478.5 kJ/kg, s3 = 7.7622 kJ/kg K, For state 2 interpolate between, saturated liquid 20°C table B.1.1 and, compressed liquid 5 MPa, 20°C from Table B.1.4: h2 = 84.9, s2 = 0.2964 • • x = m2/m1 = (h3 - h1)/(h2 - h3) = 0.13101 • ⇒ m2 = 2 × 0.131 = 0.262 kg/s ; • • • • • m3 = 2 + 0.262 = 2.262 kg/s Sgen = m3s3 - m1s1 - m2s2 = 0.9342 kW/K .. . . • I = Wrev - Wac = Wrev = ToSgen = 293.15 × 0.9342 = 273.9 kW Sonntag, Borgnakke and van Wylen 10.47 A car air-conditioning unit has a 0.5-kg aluminum storage cylinder that is sealed with a valve and it contains 2 L of refrigerant R-134a at 500 kPa and both are at room temperature 20°C. It is now installed in a car sitting outside where the whole system cools down to ambient temperature at −10°C. What is the irreversibility of this process? C.V. Aluminum and R-134a Energy Eq.: mAl(u2 - u1)Al + mR(u2 - u1)R = 1Q2 - 1W2 (1W2 = 0) Entropy Eq.: mAL(s2 - s1)Al + mR(s2 - s1)R = 1Q2/T0 + 1S2 gen (u2 - u1)Al = Cv,Al(T2 - T1) = 0.9(-10 - 20) = - 27 kJ/kg (s2 - s1)Al = Cp,Al ln(T2/T1) = 0.9 ln(263.15/293.15) = -0.09716 kJ/kg K Table B.5.2: v1 = 0.04226 m3/kg, u1 = 390.5 kJ/kg, s1 = 1.7342 kJ/kg K, v2 = v1 = 0.04226 & T2 => mR134a = V/v1 = 0.0473 kg x2 = (0.04226 - 0.000755)/0.09845 = 0.4216 u2 = 186.57 + 0.4216×185.7 = 264.9 kJ/kg, s2 = 0.9507 + 0.4216×0.7812 = 1.2801 kJ/kg K 1Q2 = 0.5 × (-27) + 0.0473(264.9 - 390.5) = - 19.44 kJ 1S2 gen = 0.5 (-0.09716) + 0.0473(1.2801 - 1.7342) + 1I2 = T0 ( 1S2 gen ) = 263.15 × 0.003815 = 1.0 kJ 19.44 = 0.003815 kJ/K 263.15 Sonntag, Borgnakke and van Wylen 10.48 The high-temperature heat source for a cyclic heat engine is a steady flow heat exchanger where R-134a enters at 80°C, saturated vapor, and exits at 80°C, saturated liquid at a flow rate of 5 kg/s. Heat is rejected from the heat engine to a steady flow heat exchanger where air enters at 150 kPa and ambient temperature 20°C, and exits at 125 kPa, 70°C. The rate of irreversibility for the overall process is 175 kW. Calculate the mass flow rate of the air and the thermal efficiency of the heat engine. C.V. R-134a Heat Exchanger, . mR134a = 5 kg/s, Table B.5.1 Inlet: T1 = 80oC, sat. vapor x1 = 1.0, QH h1 = hg = 429.189 kJ/kg, W HE s1 = sg = 1.6862 kJ/kg-K QL Exit: T2 = 80oC, sat. liquid x2 = 0.0 h2 = hf = 322.794 kJ/kg, 2 1 3 4 s2 = sf = 1.3849 kJ/kg-K C.V. Air Heat Exchanger, Cp = 1.004 kJ/kg-K, R = 0.287 kJ/kg-K Inlet: T3 = 20oC, P3 = 150 kPa Exit: T4 = 70oC, P4 = 125 kPa T4 P4 s4 - s3 = Cp ln ( ) – R ln( ) = 0.2103 kJ/kg-K T3 P3 . 2nd Law for the total system as control volume (since we know I ): . . . . I = To Snet = mR134a (s2 - s1) + mair(s4 - s3) .. . mair = [I - mR134a (s2 - s1)]/(s4 - s3) = 10.0 kg/s . . . . . 1st Law for each line: Q + mhin = mhex + W; W = 0 . . . R-134a: 1Q2 = -QH = mR134a(h2 - h1) = -532 kW . . . . Air: QL = 3Q4 = mair(h4 - h3) = mair Cp(T4 - T3) = 501.8 kW Control volume heat engine . . . Wnet = QH - QL = 532 – 501.8 = 30.2 kW; . . or 5.7% ηth = Wnet / QH = 0.057, Sonntag, Borgnakke and van Wylen 10.49 A rigid container with volume 200 L is divided into two equal volumes by a partition. Both sides contains nitrogen, one side is at 2 MPa, 300°C, and the other at 1 MPa, 50°C. The partition ruptures, and the nitrogen comes to a uniform state at 100°C. Assuming the surroundings are at 25°C find the actual heat transfer and the irreversibility in the process. Solution: C.V. Total container Continuity Eq.: m 2 – mA – mB = 0 Energy Eq.: mA(u2 - u1)A + mB(u2 - u1)B = 1Q2 - 1W2 Entropy Eq.: mA(s2 - s1)A + mB(s2 - s1)B = 1Q2/Tsur + 1Ss gen Process: V=C => 1W2 =0 From the initial state we get the mass as PA1VA PB1VB + m2 = mA + mB = RTA1 RTB1 = 1000×0.1 2000×0.1 + = 1.176 + 1.043 = 2.219 kg 0.2968×573.15 0.2968×323.15 P2 = m2RT2/Vtot = 2.219 × 0.2968 × 373.15/0.2 = 1228.8 kPa From the energy equation we get the heat transfer as the change in U 1Q2 = mACv(T2 - T1)A + mBCv(T2 - T1)B = 1.176 × 0.745 × (100 - 300) + 1.043 × 0.745 × (100 - 50) = -136.4 kJ The entropy changes are found from Eq.8.25 373.15 1228.8 (s2 - s1)A = 1.042 × ln - 0.2968 × ln = -0.09356 kJ/kg K 573.15 2000 (s2 - s1)B = 1.042 × ln 373.15 1228.8 - 0.2968 × ln = 0.0887 kJ/kg K 323.15 1000 The entropy generation follows from the entropy equation 1S2,gen = 1.176× (-0.09356) + 1.043× 0.0887 + 136.4/298.15 = 0.4396 kJ/K Now the irreversibility comes from Eq. 10.19 1I2 = T0 × 1S2,gen = 131.08 kJ Sonntag, Borgnakke and van Wylen 10.50 A rock bed consists of 6000 kg granite and is at 70°C. A small house with lumped mass of 12000 kg wood and 1000 kg iron is at 15°C. They are now brought to a uniform final temperature by circulating water between the rock bed and the house. Find the final temperature and the irreversibility of the process, assuming an ambient at 15°C. C.V. Total Rockbed and house. No work, no Q irreversible process. Energy Eq.: (mC)rock(T2 - 70) + (mCwood + mCFe)(T2 - 15) = 0 / T2 = 29.0°C = 302.2 K Entropy Eq.: S2 – S1 = ∑mi(s2 - s1)i = 0 + Sgen Sgen = ∑mi(s2 - s1)i = 5340 ln 1I2 302.2 302.2 + 15580 ln = 63.13 kJ/K 343.15 288.15 = (T0)1S2,gen = 288.15 × 63.13 = 18191 kJ cb Q H O U S E Sonntag, Borgnakke and van Wylen Availability (exergy) 10.51 A steady stream of R-22 at ambient temperature, 10°C, and at 750 kPa enters a solar collector. The stream exits at 80°C, 700 kPa. Calculate the change in availability of the R-22 between these two states. Solution: SOLAR COLLECTOR inlet Inlet (T,P) Table B.4.1 (liquid): exit hi = 56.46 kJ/kg, si = 0.2173 kJ/kg K Exit (T,P) Table B.4.2 (sup. vap.): he = 305.91 kJ/kg, se = 1.0761 kJ/kg K From Eq.10.24 or 10.37 ∆ψie = ψe - ψi = (he - hi) - T0(se - si) = (305.912 - 56.463) - 283.2(1.0761 - 0.2173) = 6.237 kJ/kg Sonntag, Borgnakke and van Wylen 10.52 Consider the springtime melting of ice in the mountains, which gives cold water running in a river at 2°C while the air temperature is 20°C. What is the availability of the water relative to the temperature of the ambient? Solution: ψ = h1 - h0 - T0(s1 - s0) flow availability from Eq.10.24 Approximate both states as saturated liquid from Table B.1.1 ψ = 8.392 - 83.96 - 293.15(0.03044 - 0.2966) = 2.457 kJ/kg Why is it positive? As the water is brought to 20°C it can be heated with qL from a heat engine using qH from atmosphere TH = T0 thus giving out work. Sonntag, Borgnakke and van Wylen 10.53 A geothermal source provides 10 kg/s of hot water at 500 kPa, 150°C flowing into a flash evaporator that separates vapor and liquid at 200 kPa. Find the three fluxes of availability (inlet and two outlets) and the irreversibility rate. C.V. Flash evaporator chamber. Steady flow with no work or heat transfer. Cont. Eq.: Energy Eq.: Entropy Eq.: B.1.1: B.1.2: . . . m1 = m2 + m3 ; . . . m1h1 = m2h2 + m3h3 . . . . m1s1 + Sgen = m2s2 + m3s3 ho = 104.87, h2 = 2706.63, 1 Vap. 2 Liq. 3 so = 0.3673, h1 = 632.18, s1 = 1.8417 s2 = 7.1271, h3 = 504.68, s3 = 1.530 .. h -h h1 = xh2 + (1 - x) h3 => x = m2/m1 = 1 3 = 0.0579 h -h . . . 2 3. m2 = xm1 = 0.579 kg/s m3 = (1-x)m1 = 9.421 kg/s . Sgen = 0.579 × 7.1271 + 9.421 × 1.53 - 10 × 1.8417 = 0.124 kW/K Flow availability Eq.10.22: ψ = (h - Tos) - (ho - Toso) = h - ho - To(s - so) ψ1 = 632.18 - 104.87 - 298.15 (1.8417 - 0.3673) = 87.72 kJ/kg ψ2 = 2706.63 - 104.87 - 298.15 (7.1271 - 0.3673) = 586.33 kJ/kg ψ3 = 504.68 - 104.87 - 298.15 (1.53 - 0.3673) = 53.15 kJ/kg . . . m1 ψ1 = 877.2 kW m2ψ2 = 339.5 kW m3ψ3 = 500.7 kW . . .. I = m1 ψ1 - m2 ψ2 - m3ψ3 = 37 kW Sonntag, Borgnakke and van Wylen 10.54 Find the availability at all 4 states in the power plant of Problem 9.42 with an ambient at 298 K. Solution: Flow availability from Eq.10.24 neglecting kinetic and potential energy is: ψ = h - h0 - T0(s - s0) so we need (h,s) for all four states. QH 1 WT 4 WP, in 3 2 . QL C.V. Turbine. Entropy Eq.9.8: Table B.1.2 P1 = P4 = 20 MPa, T1 = 700 °C h1 = 3809.1 kJ/kg, s1 = 6.7993 kJ/kg K P2 = P3 = 20 kPa, T3 = 40 °C State 3: (P, T) Comp. liquid, take sat. liquid Table B.1.1 h3 = 167.5 kJ/kg, v3 = 0.001008 m3/kg s2 = s1 = 6.7993 kJ/kg K s2 = 0.8319 + x2 × 7.0766 => x2 = 0.8433 h2 = 251.4 + 0.8433× 2358.33 = 2240.1 kJ/kg wT = h1 - h2 = 3809.1 - 2240.1 = 1569 kJ/kg CV. Pump, property relation in Eq.9.13 gives work from Eq.9.18 as wP = - v3( P4 - P3) = -0.001008(20000 – 20) = -20.1 kJ/kg h4 = h3 - wP = 167.5 + 20.1 = 187.6 kJ/kg Flow availability from Eq.10.24 and notice that since turbine work and pump work are reversible they represent also change in avalability. ψ1 = h1 - h0 - T0(s1 - s0) = 3809.1 – 104.87 - 298 (6.7993 – 0.3673) = 1787.5 kJ/kg ψ2 = h2 - h0 - T0(s2 - s0) = ψ1 - wT = 1787.5 - 1569 = 218.5 kJ/kg ψ3 = h3 - h0 - T0(s3 - s0) = 167.5 - 104.87 - 298(0.5724 - 0.3673) = 1.51 kJ/kg ψ4 = h4 - h0 - T0(s4 - s0) = ψ3 - wP = 1.51 + 20.1 = 21.61 kJ/kg Sonntag, Borgnakke and van Wylen 10.55 Air flows at 1500 K, 100 kPa through a constant pressure heat exchanger giving energy to a heat engine and comes out at 500 K. What is the constant temperature the same heat transfer should be delivered at to provide the same availability? Solution: C.V. Heat exchanger . . Continuity eq.: m1 = m2 ; . . . 2 1 Energy Eq.6.12: m1h1 = m1h2 + QH QH W HE Table A.7.1: h1 = 1635.8 kJ/kg, h2 = 503.36 kJ/kg, s1 = 8.61209 kJ/kg K s2 = 7.38692 kJ/kg K QL Ambient qout = h1 - h2 = 1635.8 - 503.36 = 1132.4 kJ/kg Availability from heat transfer at T: Eq.10.37: ∆ψ = (1 - To )q = ψ1 - ψ2 TH out ψ1 - ψ2 = h1 - h2 - To ( s1 - s2 ) = 1132.4 - 298.15 (8.6121 - 7.38692) = 1132.4 - 356.3 = 767.1 kJ/kg 1- To = (ψ1 - ψ2 ) / qout = 767.1 / 1132.4 = 0.6774 TH To = 0.3226 => TH TH = 924 K Sonntag, Borgnakke and van Wylen 10.56 Calculate the change in availability (kW) of the two flows in Problem 9.61. Solution: The two flows in the heat exchanger exchanges energy and thus also exergy (availability). Fist find state 4 Air A.7: h1 = 1046.22, h2 = 401.3 kJ/kg, o 4 2 1 air 3 water o sT1 = 8.1349, sT2 = 7.1593 kJ/kg K Water B.1.1: h3 = 83.94 kJ/kg, s3 = 0.2966 kJ/kg K . . Energy Eq.6.10: mAIR∆hAIR = mH2O∆hH2O . . h4 - h3 = (mAIR/mH2O)(h1 - h2) = (2/0.5)644.92 = 2579.68 kJ/kg h4 = h3 + 2579.68 = 2663.62 < hg at 200 kPa T4 = Tsat = 120.23°C, x4 = (2663.62 – 504.68)/2201.96 = 0.9805, s4 = 1.53 + x4 5.597 = 7.01786 kJ/kg K We consider each flow separately and for each flow availability is Eq.10.24, include mass flow rate as in Eq.10.36, use To = 20 C For the air flow: . . m1(ψ1 - ψ2 ) = m1 [ h1 - h2 - To ( s1 - s2 ) ] 125 = 2 [ 1046.22 - 401.3 - 293.2(8.1349 - 7.1593 - 0.287 ln )] 100 = 2 (644.92 - 267.22 ) = 755.4 kW For the water flow: . . m3(ψ4 - ψ3 ) = m3 [ h4 - h3 - To ( s4 - s3 ) ] = 0.5 [ 2663.62 - 83.94 - 293.2(7.01786 - 0.2966)] = 0.5[ 2579.68 - 1970.7 ] = 304.7 kW Sonntag, Borgnakke and van Wylen 10.57 Nitrogen flows in a pipe with velocity 300 m/s at 500 kPa, 300°C. What is its availability with respect to an ambient at 100 kPa, 20°C? Solution: From the availability or exergy in Eq.10.24 2 ψ = h1 - h0 + (1/2)V1 - T0(s1 - s0) T1 P1 2 = Cp(T1 - T0) + (1/2)V1 - T0[Cp ln( ) - R ln( ) ] T0 P0 = 1.042(300 - 20) + 3002 573.15 500 - 293.151.042 ln - 0.2968 ln100 2000 293.15 = 272 kJ/kg Notice that the high velocity does give a significant contribution. Sonntag, Borgnakke and van Wylen 10.58 A steady combustion of natural gas yields 0.15 kg/s of products (having approximately the same properties as air) at 1100°C, 100 kPa. The products are passed through a heat exchanger and exit at 550°C. What is the maximum theoretical power output from a cyclic heat engine operating on the heat rejected from the combustion products, assuming that the ambient temperature is 20°C? Solution: C.V. Heat exchanger . . Continuity eq.: mi = me ; . . . Energy Eq.6.12: mihi = mihe + QH . . QH = miCP0(Ti - Te) = 0.15 × 1.004(1100 - 550) = 82.83 kW We do not know the H.E efficiency, high T not constant. C.V. Total heat exchanger plus heat engine, reversible process. . . . Entropy Eq.: misi + 0 = mise + QL/TL Ti . . . QL = TL mi (si – se) = TL miCP0 ln ( ) T e = 293.15 × 0.15 × 1.004 ln ( 1373.15 ) = 22.57 kW 823.15 her we used Eq.8.25 for the change in s of the air. Energy Eq. heat engine: . . . WNET = QH - QL = 82.83 - 22.57 = 60.26 kW T QH e W HE w TL e i i qL s QL Ambient Sonntag, Borgnakke and van Wylen 10.59 Find the change in availability from inlet to exit of the condenser in Problem 9.42. Solution: Condenser of Prob. 9.42 has inlet equal to turbine exit. State 2: P2 = 20 kPa; s2 = s1 = 6.7993 kJ/kg K => x2 = (6.7993 – 0.8319)/7.0766 = 0.8433 h2 = 2240.1 kJ/kg State 3: P2 = P3; T3 = 40°C; Compressed liquid assume sat.liq. same T Table B.1.1 h3 = 167.5 kJ/kg; s3 = 0.5724 kJ/kg K From Eq.10.24 or 10.37 ψ3 - ψ2 = (h3 - Tos3) – (h2 - Tos2) = (h3 - h2) – To(s3 - s2) = (167.5 – 2240.1) – 298.2(0.5724 − 6.7993 ) = −2072.6 + 1856.9 = -215.7 kJ/kg Sonntag, Borgnakke and van Wylen 10.60 Refrigerant R-12 at 30°C, 0.75 MPa enters a steady flow device and exits at 30°C, 100 kPa. Assume the process is isothermal and reversible. Find the change in availability of the refrigerant. Solution: Table B.3.1: Table B.3.2: hi = 64.59 kJ/kg, si = 0.2399 kJ/kg K, compr. liquid. he = 210.02 kJ/kg, se = 0.8488 kJ/kg K, sup. vapor From Eq. 10.24 or 10.37 ∆ψ = he - hi - T0(se - si) = 210.02 - 64.59 - 298.15(0.8488 - 0.2399) = -36.1 kJ/kg T P 745 kPa 750 100 i 30 e i e v Remark: Why did the availability drop? The exit state is much closer to the ambient dead state, so it lost its ability to expand and do work. s Sonntag, Borgnakke and van Wylen 10.61 An air compressor is used to charge an initially empty 200-L tank with air up to 5 MPa. The air inlet to the compressor is at 100 kPa, 17°C and the compressor isentropic efficiency is 80%. Find the total compressor work and the change in availability of the air. C.V. Tank + compressor Transient process with constant inlet conditions, no heat transfer. Continuity: m2 - m1 = min ( m1 = 0 ) Energy: m2u2 = minhin - 1W2 Entropy: m2s2 = minsin + 1S2 gen Reversible compressor: 1S2 GEN =0 ⇒ s2 = sin State 1: v1 = RT1/P1 = 0.8323 m3/kg, State inlet, Table A.7.1: hin = 290.43 kJ/kg, Eq.8.28: o o sT2 = sTin + R ln ( Table A.7.1 ⇒ o sTin = 6.83521 kJ/kg K P2 5000 ) = 6.83521 + 0.287 ln ( ) = 7.95796 Pin 100 T2,s = 854.6 K, u2,s = 637.25 kJ/kg ⇒ 1w2,s = hin - u2,s = 290.43 – 637.25 = -346.82 kJ/kg Actual compressor: 1w2,AC = 1w2,s/ηc = -433.53 kJ/kg u2,AC = hin - 1w2,AC = 290.43 –(-433.53) = 723.96 kJ/kg o ⇒ T2,AC = 958.5 K, State 2 u, P sT2 ac = 8.08655 kJ/kg K v2 = RT2/P2 = 0.05502 m3/kg so m2 = V2/v2 = 3.635 kg ⇒ 1W2 = m2 (1w2,AC) = -1575.9 kJ m2(φ2 - φ1) = m2[u2 - u1 + P0(v2 - v1) - T0(s2 - s1)] = 3.635 [723.96 - 207.19 + 100(0.05502 - 0.8323) - 290[8.08655 6.83521 - 0.287 ln(5000/100)] = 1460.4 kJ − 1W2 cb Sonntag, Borgnakke and van Wylen 10.62 Water as saturated liquid at 200 kPa goes through a constant pressure heat exchanger as shown in Fig. P10.62. The heat input is supplied from a reversible heat pump extracting heat from the surroundings at 17°C. The water flow rate is 2 kg/min and the whole process is reversible, that is, there is no overall net entropy change. If the heat pump receives 40 kW of work find the water exit state and the increase in availability of the water. C.V. Heat exchanger + heat pump. . . . . . . . . . m1 = m2 = 2 kg/min, m1h1 + Q0 + Win = m1h2, m1s1 + Q0/T0 = m1s2 . . Substitute Q0 into energy equation and divide by m1 h1 - T0s1 + win = h2 - T0s2 LHS = 504.7 - 290.15 × 1.5301 + 40×60/2 = 1260.7 kJ/kg State 2: P2 , h2 - T0s2 = 1260.7 kJ/kg At sat. vap. hg - T0sg = 638.8 so state 2 is superheated vapor at 200 kPa. At 600oC: h2 - T0s2 = 3703.96 - 290.15 × 8.7769 = 1157.34 kJ/kg At 700oC: h2 - T0s2 = 3927.66 - 290.15 × 9.0194 = 1310.68 kJ/kg Linear interpolation ⇒ T2 = 667°C ∆ψ = (h2 - T0s2) - (h1 - T0s1) = win = 1200 kJ/kg = 1260.7 - 504.7 + 290.15 × 1.5301 ≈ 1200 kJ/kg T Q1 1 W HP w To 2 1 2 qo s Q0 Ambient Sonntag, Borgnakke and van Wylen 10.63 An electric stove has one heating element at 300oC getting 500 W of electric power. It transfers 90% of the power to 1 kg water in a kettle initially at 20oC, 100 kPa, the rest 10% leaks to the room air. The water at a uniform T is brought to the boiling point. At the start of the process what is the rate of availability transfer by: a) electrical input b) from heating element and c) into the water at Twater. a) Work is availability . . Φ = W = 500 W b) Heat transfer at 300oC is only partly availability To . . 293.15 Φ = 1 – Q = 1 – 500 = 244 W TH 273.15 + 300 c) Water receives heat transfer at 20oC as 90% of 500 W To . . 293.15 Q = 1 – 450 = 0 W Φ = 1 – Twater 273.15 + 20 500 W at 300oC Sonntag, Borgnakke and van Wylen 10.64 Calculate the availability of the water at the initial and final states of Problem 8.70, and the irreversibility of the process. State properties s1 = 0.2966 kJ/kg K, , v1 = 0.001 m3/kg 1: u1 = 83.94 kJ/kg, 2: u2 = 3124.3 kJ/kg, s2 = 7.7621 kJ/kg K, v2 = 0.354 m3/kg 0: uo = 104.86 kJ/kg, ac so = 0.3673 kJ/kg K, vo = 0.001003 m3/kg ac Process transfers: 1W2 = 203 kJ, 1Q2 = 3243.4 kJ, TH = 873.15 K φ = (u - Tos) - (uo - Toso) + Po( v - vo) φ1 = (83.94 - 298.15×0.2966) - (104.86 - 298.15×0.3673) + 100 (0.001002 - 0.001003) = 0.159 kJ/kg φ2 = (3124.3 - 298.15×7.7621) - (104.86 - 298.15×0.3673) + 100 (0.35411 - 0.001003) = 850 kJ/kg ac ac 1I2 = m(φ1 - φ2) + [1 - (T0/TH)]1Q2 - 1W2 + Po( V2 - V1) = -849.84 + (1 - 298.15 ) 3243.4 - 203 + 100 (0.3541 - 0.001) 873.15 = -849.84 + 2135.9 - 203 + 35.31 = 1118. kJ [(Sgen = 3.75 kJ/K ToSgen = 1118 kJ so OK] Sonntag, Borgnakke and van Wylen 10.65 A 10-kg iron disk brake on a car is initially at 10°C. Suddenly the brake pad hangs up, increasing the brake temperature by friction to 110°C while the car maintains constant speed. Find the change in availability of the disk and the energy depletion of the car’s gas tank due to this process alone. Assume that the engine has a thermal efficiency of 35%. Solution: All the friction work is turned into internal energy of the disk brake. Energy eq.: m(u2 - u1) = 1Q2 - 1W2 ⇒ 1Q2 = mFeCFe(T2 - T1) 1Q2 = 10 × 0.45 × (110 - 10) = 450 kJ Neglect the work to the surroundings at P0, so change in availability is from Eq.10.27 ∆φ = m(u2 - u1) - T0m(s2 - s1) Change in s for a solid, Eq.8.20 383.15 m(s2-s1) = mC ln(T2/T1) = 10 × 0.45 × ln = 1.361 kJ/K 283.15 ∆φ = 450 - 283.15 × 1.361 = 64.63 kJ Wengine = ηthQgas = 1Q2 = Friction work Qgas = 1Q2/ηth = 450/0.35 = 1285.7 kJ Sonntag, Borgnakke and van Wylen 10.66 A 1 kg block of copper at 350°C is quenched in a 10 kg oil bath initially at ambient temperature of 20°C. Calculate the final uniform temperature (no heat transfer to/from ambient) and the change of availability of the system (copper and oil). Solution: C.V. Copper and oil. Cco = 0.42 kJ/kg K, Coil = 1.8 kJ/kg K m2u2 - m1u1 = 1Q2 - 1W2 = 0 = mcoCco(T2 - T1)co + (mC)oil(T2 - T1)oil 1 × 0.42 ( T2 - 350) + 10 × 1.8 (T2 - 20) = 0 18.42 T2 = 507 => T = 27.5°C = 300.65 K For each mass copper and oil, we neglect work term (v = C) so Eq.10.22 is (φ2 - φ1) = u2 - u1 - To(s2 - s1) = mC [(T2 - T1) - Toln (T2 / T1) ] mcv(φ2 - φ1)cv + moil (φ2 - φ1)oil = = 0.42 × [(-322.5) - 293.15 ln 300.65 300.65 + 10 × 1.8 [7.5 - 293.15 ln 293.15 623.15 = - 45.713 + 1.698 = - 44.0 kJ Cu Oil Sonntag, Borgnakke and van Wylen 10.67 Calculate the availability of the system (aluminum plus gas) at the initial and final states of Problem 8.137, and also the process irreversibility. State 1: T1 = 200 oC, v1 = V1/ m = 0.05 / 1.1186 = 0.0447 m3/kg State 2: v2 = v1 × (2 / 1.5) × (298.15 / 473.15) = 0.03756 m3/kg The metal does not change volume, so the combined is using Eq.10.22 as φ1 = mgasφgas + mAlφAl = mgas[u1-uo-To(s1 - so)]cv + mgasPo(v1-vo) + mAl[u1-uo -To(s1-so)]Al T P = mgasCv (T1 - To) - mgasTo [Cp ln 1 - R ln 1 ] + mgasPo (v1 - vo) To Po + mAl [C (T1 - To) - ToC ln (T1/To) ]Al φ1 = 1.1186 [ 0.653(200-25) - 298.15 (0.842 ln 473.15 2000 - 0.18892 ln ) 298.15 100 + 100 (0.0447 - 0.5633 ) ] + 4 × 0.90 [ 200 -25 - 298.15 ln 473.15 298.15 = 128.88 + 134.3 = 263.2 kJ 298.15 1500 φ2 = 1.1186 [ 0.653(25 - 25) - 298.15 (0.842 ln - 0.18892 ln ) 298.15 100 + 100 (0.03756 - 0.5633 ) ] + 4 × 0.9 [ 25 -25 - 298.15 ln 298.15 298.15 = 111.82 + 0 = 111.82 kJ The irreversibility is as in Eq.10.28 1I2 = φ1 - φ2 + [1 - (T0/TH)] 1Q2 - 1W2AC + Pom( V2 - V1) = 263.2 - 111.82 + 0 - (-14) + 100 × 1.1186 (0.03756 - 0.0447) = 164.58 kJ [(Sgen = 0.552 Al ToSgen = 164.58 Q CO 2 Tamb so OK ] Sonntag, Borgnakke and van Wylen 10.68 A wooden bucket (2 kg) with 10 kg hot liquid water, both at 85°C, is lowered 400 m down into a mineshaft. What is the availability of the bucket and water with respect to the surface ambient at 20°C? C.V. Bucket and water. Both thermal availability and potential energy terms. v1 ≈ v0 for both wood and water so work to atm. is zero. Use constant heat capacity table A.3 for wood and table B.1.1 (sat. liq.) for water. From Eq.10.27 φ1 - φ0 = mwood[u1 - u0 - T0(s1- s0)] + mH2O[u1- u0- T0(s1- s0)] + mtotg(z1- z0) 273.15 + 85 = 2[1.26(85 - 20) - 293.15× 1.26 ln + 10[ 355.82 - 83.94 293.15 - 293(1.1342 - 0.2966)] + 12 × 9.807 × (-400) /1000 = 15.85 + 263.38 - 47.07 = 232.2 kJ Sonntag, Borgnakke and van Wylen Device Second-Law Efficiency 10.69 Air enters a compressor at ambient conditions, 100 kPa, 300 K, and exits at 800 kPa. If the isentropic compressor efficiency is 85%, what is the second-law efficiency of the compressor process? Solution: T 800 kPa 2 2s 1 Ideal (isentropic, Eq.8.32) T2s = 300(8)0.286 = 543.8 K -ws = 1.004(543.8 - 300) = 244.6 kJ/kg -ws 244.6 -w = = = 287.8 kJ/kg K 0.85 ηs -w 287.8 T2 = T1 + = 300 + = 586.8 K CP0 1.004 100 kPa 300 K s Eq.8.25: s2 - s1 = 1.004 ln(586.8/300) - 0.287 ln 8 = 0.07645 Availability, Eq.10.24 ψ2 - ψ1 = (h2 - h1) - T0(s2 - s1) = 287.8 - 300(0.07645) = 264.9 kJ/kg 2nd law efficiency, Eq.10.29 or 10.30 (but for a compressor): η2nd Law = ψ2 - 1 264.9 = = 0.92 -w 287.8 Sonntag, Borgnakke and van Wylen 10.70 A compressor takes in saturated vapor R-134a at −20°C and delivers it at 30°C, 0.4 MPa. Assuming that the compression is adiabatic, find the isentropic efficiency and the second law efficiency. Solution: Table B.5 Inlet: Actual exit: hi = 386.08 kJ/kg, si = 1.7395 kJ/kg K, he,ac = 423.22 kJ/kg, se,ac = 1.7895 kJ/kg K Ideal exit: Pe, se,s = si ⇒ he,s = 408.51 kJ/kg Isentropic compressor wc,s = he,s - hi = 22.43 kJ/kg Actual compressor wc,ac = he,ac - hi = 37.14 kJ/kg Reversible between inlet and actual exit Eq.10.9 -wc,rev = hi - he,ac - T0(si - se,ac) = -37.14 - 298.15(1.7395 - 1.7895) = -22.23 Eq.9.27: ηs = (wc,s/wc,ac) = (22.43/37.14) = 0.604 Second law efficiency for compressor, Eq.10.32 (modified) ηII = (wc,rev/wc,ac) = (22.23/37.14) = 0.599 Sonntag, Borgnakke and van Wylen 10.71 A steam turbine has inlet at 4 MPa, 500°C and actual exit of 100 kPa, x = 1.0. Find its first law (isentropic) and its second law efficiencies. Solution: C.V. Steam turbine Energy Eq.6.13: w = h i - he Entropy Eq.9.8: se = si + sgen Inlet state: Table B.1.3 hi = 3445.2 kJ/kg; si = 7.0900 kJ/kg K Exit (actual) state: Table B.1.2 he = 2675.5; Actual turbine energy equation w = hi - he = 769.7 kJ/kg Ideal turbine reversible process so sgen = 0 se = 7.3593 kJ/kg K giving ses = si = 70900 = 1.3025 + xes × 6.0568 xes = 0.9555, hes = 417.4 + 0.9555 × 2258.0 = 2575.0 kJ/kg The energy equation for the ideal gives ws = hi - hes = 870.2 kJ/kg The first law efficiency is the ratio of the two work terms ηs = w/ws = 0.885 The reversible work for the actual turbine states is, Eq.10.9 wrev = (hi - he) + To(se - si) = 769.7 + 298.2(7.3593 – 7.0900) = 769.7 + 80.3 = 850.0 kJ/kg Second law efficiency Eq.10.29 η2nd Law = w/wrev = 769.7/850.0 = 0.906 Sonntag, Borgnakke and van Wylen 10.72 The condenser in a refrigerator receives R-134a at 700 kPa, 50°C and it exits as saturated liquid at 25°C. The flowrate is 0.1 kg/s and the condenser has air flowing in at ambient 15°C and leaving at 35°C. Find the minimum flow rate of air and the heat exchanger second-law efficiency. 4 3 AIR 2 1 • • C.V. Total heat exchanger. Energy Eq.6.10 R-134a ⇒ ma = m1 × • • • • m1h1 + mah3 = m1h2 + mah4 h1 - h2 436.89 - 234.59 = 0.1 × 1.004(35 - 15) = 1.007 kg/s h4 - h3 Availability from Eq.10.24 ψ1 - ψ2 = h1 - h2 - T0(s1 - s2) = 436.89 - 234.59 - 288.15(1.7919 - 1.1201) = 8.7208 kJ/kg ψ4 - ψ3 = h4 - h3 - T0(s4 - s3) = 1.004(35 - 15) - 288.15 × 1.004 × ln 308.15 = +0.666 kJ/kg 288.15 Efficiency from Eq.10.30 • • ηII = ma(ψ4 - ψ3)/m1(ψ1 - ψ2) = 1.007(0.666) = 0.77 0.1(8.7208) Sonntag, Borgnakke and van Wylen 10.73 Steam enters a turbine at 25 MPa, 550°C and exits at 5 MPa, 325°C at a flow rate of 70 kg/s. Determine the total power output of the turbine, its isentropic efficiency and the second law efficiency. Solution: hi = 3335.6 kJ/kg, si = 6.1765 kJ/kg K, he = 2996.5 kJ/kg, se = 6.3289 kJ/kg K Actual turbine: wT,ac = hi - he = 339.1 kJ/kg Isentropic turbine: se,s = si ⇒ he,s = 2906.6 kJ/kg wT,s = hi - he,s = 429 kJ/kg Rev. turbine: wrev = wT,ac + T0(se - si) = 339.1 + 45.44 = 384.54 kJ/kg Eq.9.27: ηT = wT,ac/wT,s = 339.1/429 = 0.79 Eq.10.29: ηII = wT,ac/wrev = 339.1/384.54 = 0.88 Sonntag, Borgnakke and van Wylen 10.74 A compressor is used to bring saturated water vapor at 1 MPa up to 17.5 MPa, where the actual exit temperature is 650°C. Find the irreversibility and the second-law efficiency. Solution: Inlet state: Table B.1.2 hi = 2778.1 kJ/kg, si = 6.5864 kJ/kg K Actual compressor Table B.1.3: he,ac = 3693.9 kJ/kg, se,ac = 6.7356 kJ/kg K Energy Eq. Actual compressor: -wc,ac = he,ac - hi = 915.8 kJ/kg From Eq.10.11: i = T0(se,ac - si) = 298.15 (6.7356 - 6.5864) = 44.48 kJ/kg From Eq.10.10: wrev = i + wc,ac = -915.8 + 44.48 = -871.32 kJ/kg ηII = -wrev/wc,ac = 871.32/915.8 = 0.951 Sonntag, Borgnakke and van Wylen 10.75 A flow of steam at 10 MPa, 550°C goes through a two-stage turbine. The pressure between the stages is 2 MPa and the second stage has an exit at 50 kPa. Assume both stages have an isentropic efficiency of 85%. Find the second law efficiencies for both stages of the turbine. 1 2 T1 3 T2 Actual T1: CV: T1, h1 = 3500.9 kJ/kg, s1 = 6.7561 kJ/kg K Isentropic s2s = s1 ⇒ h2s = 3017.9 kJ/kg wT1,s = h1 - h2s = 483 kJ/kg wT1,ac = ηT1 wT1,s = 410.55 = h1 - h2ac h2ac = h1 - wT1,ac = 3090.35, s2ac = 6.8782 CV: T2, s3s = s2ac = 6.8782 ⇒ x3s = (6.8782-1.091)/6.5029 = 0.8899, h3s = 340.47 + 0.8899 × 2305.4 = 2392.2 kJ/kg wT2,s = h2ac - h3s = 698.15 ⇒ wT2,ac = ηT2 wT2,s = 593.4 kJ/kg ⇒ h3ac = 2496.9, x3ac = (2496.9 - 340.47)/2305.4 =0.9354, s3ac = 1.091 + 0.9354 × 6.5029 = 7.1736 kJ/kg K Actual T1: R iT1,ac = T0(s2ac-s1) = 298.15(6.8782 - 6.7561) = 36.4 kJ/kg ⇒ wT1 = wT1,ac + i = 447 kJ/kg, R ηII = wT1,ac/wT1 = 0.918 Actual T2: iT2,ac = T0(s3ac-s2ac) = 298.15(7.1736 - 6.8782) = 88.07 kJ/kg R ⇒ wT2 = wT2,ac + iT2,ac = 681.5, R ηII = wT2,ac/wT2 = 0.871 Sonntag, Borgnakke and van Wylen 10.76 The simple steam power plant shown in Problem 6.99 has a turbine with given inlet and exit states. Find the availability at the turbine exit, state 6. Find the second law efficiency for the turbine, neglecting kinetic energy at state 5. Solution: interpolation or software: h5 = 3404.3 kJ/kg, s5 = 6.8953 kJ/kg K Table B.1.2: x6 = 0.92 so h6 = 2393.2 kJ/kg, s6 = 7.5501 kJ/kg K Flow availability (exergy) from Eq.10.24 ψ6 = h6 - h0 - T0(s6 - s0) = 2393.2 - 104.89 - 298.15(6.8953 - 0.3674) = 146.79 kJ/kg In the absence of heat transfer the work is form Eq.10.9 or 10.39 wrev = ψ5 - ψ6 = h5 - h6 - T0(s5 - s6) = 1206.3 kJ/kg ηII = wac/wrev = 0.838 wac = h5 - h6 = 1011.1 kJ/kg; T P 5 5 6 v 6 s Sonntag, Borgnakke and van Wylen 10.77 A steam turbine inlet is at 1200 kPa, 500oC. The actual exit is at 200 kPa, 300oC. What are the isentropic efficiency and its second law efficiency? Solution: C.V. Turbine actual, steady state and adiabatic. Inlet state: Table B.1.3: hi = 3476.28 kJ/kg, si = 7.6758 kJ/kg K Exit state: Table B.1.3: he = 3071.79 kJ/kg, se = 7.8926 kJ/kg K Energy Eq.: wTac = hi - he = 3476.28 – 3071.79 = 404.49 kJ/kg C.V. Turbine isentropic, steady state, reversible and adiabatic. Isentropic exit state: 200 kPa, s = si => hes = 2954.7 kJ/kg wT s = hi - hes = 3476.28 – 2954.7 = 521.58 kJ/kg Energy eq.: ηI = wTac/wT s = 404.49 = 0.776 521.58 Reversible work for actual turbine is from Eq.10.9 or 10.39 rev wT = ψi - ψe = hi - he - T0(si - se) = wTac - T0(si - se) = 404.49 – 298.15(7.6758 – 7.8926) = 469.13 kJ/kg Then the second law efficiency is in Eq.10.29 rev 404.49 ηII = wTac/wT = = 0.862 469.13 T P i i e ac es e ac v s Sonntag, Borgnakke and van Wylen 10.78 Steam is supplied in a line at 3 MPa, 700°C. A turbine with an isentropic efficiency of 85% is connected to the line by a valve and it exhausts to the atmosphere at 100 kPa. If the steam is throttled down to 2 MPa before entering the turbine find the actual turbine specific work. Find the change in availability through the valve and the second law efficiency of the turbine. Take C.V. as valve and a C.V. as the turbine. Valve: h2 = h1 = 3911.7 kJ/kg, s2 > s1 = 7.7571 kJ/kg K, h2, P2 ⇒ s2 = 7.9425 kJ/kg K ψ1 - ψ2 = h1−h2 −T0(s1-s2) = 0 -298.15(7.7571-7.9425) = 55.3 kJ/kg So some potential work is lost in the throttling process. Ideal turbine: s3 = s2 ⇒ h3s = 2929.13 wT,s = 982.57 kJ/kg wT,ac = h2 - h3ac = ηwT,s = 835.2 kJ/kg h3ac = 3911.7 - 835.2 = 3076.5 ⇒ s3ac = 8.219 kJ/kg K wrev = h2 - h3ac - T0(s2 - s3ac) = 835.2 - 298.15(7.9425 - 8.219) = 917.63 kJ/kg ⇒ ηII = 835.2/917.63 = 0.91 Sonntag, Borgnakke and van Wylen 10.79 Air flows into a heat engine at ambient conditions 100 kPa, 300 K, as shown in Fig. P10.79. Energy is supplied as 1200 kJ per kg air from a 1500 K source and in some part of the process a heat transfer loss of 300 kJ/kg air happens at 750 K. The air leaves the engine at 100 kPa, 800 K. Find the first and the second law efficiencies. C.V. Engine out to reservoirs hi + q1500 = q750 + he + w wac = 300.47 + 1200 - 300 - 822.20 = 378.27 kJ/kg ηTH = w/q1500 = 0.3152 For second law efficiency also a q to/from ambient si + (q1500/TH) + (q0/T0) = (q750/Tm) + se q0 = T0(se - si) + (T0/Tm)q750 - (T0/TH)q1500 100 300 300 = 3007.88514 - 6.86925 - 0.287 ln + 100 750 -(300/1500) 1200 = 184.764 kJ/kg wrev = hi - he + q1500 - q750 + q0 = wac + q0 = 563.03 kJ/kg ηII = wac/wrev = 378.27/563.03 = 0.672 Sonntag, Borgnakke and van Wylen 10.80 Air enters a steady-flow turbine at 1600 K and exhausts to the atmosphere at 1000 K. The second law efficiency is 85%. What is the turbine inlet pressure? C.V.: Turbine, exits to atmosphere so assume Pe = 100 kPa o Inlet: Ti = 1600 K, Table A.7: hi = 1757.3 kJ/kg, si = 8.1349 kJ/kg K o Exit: Te = 1000 K, he = 1046.2 kJ/kg, se = 8.6905 kJ/kg K 1st Law: q + hi = he + w; q = 0 => w = (hi - he) = 711.1 kJ/kg 2nd Law: ψi - ψe = w/η2ndLaw = 711.1/0.85 = 836.6 kJ/kg ψi - ψe = (hi - he) - To(si - se) = 836.6 kJ/kg si - se = 0.4209 kJ/kg-K hi - he = w = 711.1 kJ/kg, assume To = 25oC o o si - se = se - si - R ln(Pi/Pe) = 0.4209 kJ/kg K Pi = 3003 kPa => Pe/Pi = 30.03; Sonntag, Borgnakke and van Wylen 10.81 Calculate the second law efficiency of the counter flowing heat exchanger in Problem 9.61 with an ambient at 20°C. Solution: C.V. Heat exchanger, steady flow 1 inlet and 1 exit for air and water each. The two flows exchange energy with no heat transfer to/from the outside. 4 2 1 air 3 water Heat exchanger Prob 9.61 with To = 20°C solve first for state 4. . . Energy Eq.6.10: mAIR∆hAIR = mH2O∆hH2O From A.7: h1 - h2 = 1046.22 – 401.3 = 644.92 kJ/kg From B.1.2 h3 = 83.94 kJ/kg; s3 = 0.2966 kJ/kg K . . h4 - h3 = (mAIR/mH2O)(h1 - h2) = (2/0.5)644.92 = 2579.68 kJ/kg h4 = h3 + 2579.68 = 2663.62 < hg at 200 kPa T4 = Tsat = 120.23°C, x4 = (2663.62 – 504.68)/2201.96 = 0.9805, s4 = 1.53 + x4 5.597 = 7.01786 kJ/kg K We need the change in availability for each flow from Eq.10.24 (ψ1 - ψ2) = (h1 - h2) + To(s2 - s1) = (1046.2 – 401.3) + 293.2(7.1593 – 8.1349 – 0.287 ln(100/125) = 644.9 + 293.2(-0.91156) = 377.6 kJ/kg (ψ4 - ψ3) = (h4 - h3) + To(s4 - s3) = (2663.6 – 83.9) – 293.2(7.0179 – 0.2966) = 2579.9 – 1970.7 = 609.0 Efficiency from Eq.10.30 . . η2nd Law = [mw(ψ4 - ψ3)]/[mA(ψ1 - ψ2)] = (0.5 × 609.0)/(2 × 377.6) = 0.403 Sonntag, Borgnakke and van Wylen 10.82 Calculate the second law efficiency of the coflowing heat exchanger in Problem 9.62 with an ambient at 17°C. 4 Solution: C.V. Heat exchanger, steady 2 flows in and two flows out. 1 2 3 First solve for the exit temperature in Problem 9.62 C.V. Heat exchanger, steady 2 flows in and two flows out. . . . . Energy Eq.6.10: mO2h1 + mN2h3 = mO2h2 + mN2h4 Same exit tempearture so T4 = T2 with values from Table A.5 . . . . mO2CP O2T1 + mN2CP N2T3 = (mO2CP O2 + mN2CP N2)T2 T2 = 0.25 × 0.922× 290 + 0.6 × 1.042 × 500 379.45 = 0.8557 0.25 × 0.922 + 0.6 × 1.042 = 443.4 K The second law efficiency for a heat exchanger is the ratio of the availability gain by one fluid divided by the availability drop in the other fluid. We thus have to find the change of availability in both flows. For each flow availability is Eq.10.24 include mass flow rate as in Eq.10.36 For the oxygen flow: . . mO2(ψ2 - ψ1 ) = mO2 [ h2 - h1 - To ( s2 - s1 ) ] . = mO2 [ CP(T2 - T1) - To [ CP ln(T2 / T1) − R ln(P2 / P1) ] . = mO2CP [ T2 - T1 - Toln(T2 / T1) ] = 0.25 × 0.922 [ 443.4 - 290 - 290 ln(443.4/290) ] = 6.977 kW For the nitrogen flow . . mN2(ψ3 - ψ4 ) = mN2CP [ T3 - T4 - Toln(T3 / T4) ] = 0.6 × 1.042 [ 500 - 443.4 - 290 ln(500/443.4) ] = 13.6 kW From Eq.10.30 . mO2(ψ1 - ψ2) 6.977 = = 0.513 η2nd Law = . mN2(ψ3 - ψ4) 13.6 Sonntag, Borgnakke and van Wylen 10.83 A heat exchanger brings 10 kg/s water from 100oC to 500oC at 2000 kPa using air coming in at 1400 K and leaving at 460 K. What is the second law efficiency? Solution: C.V. Heat exchanger, steady flow 1 inlet and 1 exit for air and water each. The two flows exchange energy with no heat transfer to/from the outside. We need to find the air mass flow rate. Energy Eq.: 2 4 3 air 1 water . . mH2O(h2 - h1) = mair(h3 - h4) h2 - h1 . . 3467.55 - 420.45 = 10 = 28.939 kg/s mair = mH2O h3 - h4 1515.27 - 462.34 Availability increase of the water flow . . mH2O(ψ2 - ψ1) = mH2O[h2 - h1 - To(s2 - s1)] = 10 [ 3467.55 – 420.45 – 298.15(7.4316 – 1.3053)] = 10 [ 3047.1 – 1826.56 ] = 12 205 kW Availability decrease of the air flow . . mair(ψ3 - ψ4) = mair[h3 - h4 – To(s3 - s4)] = 28.939 [1515.27 – 462.34 – 298.15(8.52891 – 7.30142)] = 19 880 kW . mH2O(ψ2 - ψ1) 12 205 η2nd Law = . = = 0.614 mair(ψ3 - ψ4) 19 880 Sonntag, Borgnakke and van Wylen Exergy Balance Equation 10.84 Find the specific flow exergy in and out of the steam turbine in Example 9.1 assuming an ambient at 293 K. Use the exergy balance equation to find the reversible specific work. Does this calculation of specific work depend on To? Solution: The specific flow exergy is from Eq. 10.37 12 ψi = hi + Vi – Tosi – (ho – Toso) 2 Reference state: ho = 83.94 kJ/kg, so = 0.2966 kJ/kg K, ho – Toso = -2.9638 kJ/kg The properties are listed in Example 9.1 so the specific flow exergies are ψi = 3051.2 + 1.25 – 293 × 7.1228 – (-2.9638) = 968.43 kJ/kg ψe = 2655.0 + 20 – 293 × 7.1228 – (-2.9638) = 590.98 kJ/kg The reversible work is from Eq.10.39, with q = 0 and sgen = 0, so w = ψi – ψe = 968.43 – 590.98 = 377.45 kJ/kg The offset To terms drop out as we take the difference and also (si = se) ψi – ψe = hi – he – To(si – se) = hi – he Notice since the turbine is reversible we get the same as in Example 9.1 Sonntag, Borgnakke and van Wylen 10.85 A counterflowing heat exchanger cools air at 600 K, 400 kPa to 320 K using a supply of water at 20°C, 200 kPa. The water flow rate is 0.1 kg/s and the air flow rate is 1 kg/s. Assume this can be done in a reversible process by the use of heat engines and neglect kinetic energy changes. Find the water exit temperature and the power out of the heat engine(s). air 1 HE W HE W HE W QL QL QL 3 2 QH QH QH 4 water C.V. Total Energy eq.: • • • • mah1 + mH2Oh3 = mah2 + mH2Oh4 + W Entropy Eq,: • • • • mas1 + mH2Os3 = mas2 + mH2Os4 • Table A.7: h1 = 607.316 kJ/kg, (sgen = 0) s° = 7.57638 kJ/kg K T1 Table A.7: h2 = 320.576 kJ/kg, s° = 6.93413 kJ/kg K, T2 Table B.1.1: h3 = 83.96 kJ/kg, s3 = 0.2966 kJ/kg K From the entropy equation we first find state 4 • • s4 = (ma/mH2O)(s1 - s2) + s3 = (1/0.1)(7.57638 - 6.93413) + 0.2966 = 6.7191 4: P4 = P3, s4 ⇒ Table B.1.2: x4 = (6.7191-1.530)/5.597 = 0.9271, h4 = 504.68 + 0.9271 × 2201.96 = 2546.1 kJ/kg, T4 = 120.20°C From the energy equation • • • W = ma(h1 - h2) + mH2O(h3 - h4) = 1(607.32 - 320.58) + 0.1(83.96 - 2546.1) = 40.53 kW Sonntag, Borgnakke and van Wylen 10.86 Evaluate the steady state exergy fluxes due to a heat transfer of 250 W through a wall with 600 K on one side and 400 K on the other side. What is the exergy destruction in the wall. Solution: . Exergy flux due to a Q term Eq.10.36: To . . Φ Q = (1 – )Q T To . . 298 ) Q = (1 – ) 250 = 125.8 W Φ 1 = (1 – T1 600 To . . 298 Φ 2 = (1 – ) Q = (1 – ) 250 = 63.8 W T2 400 1 250 W Steady state state so no .storage and Eq.10.36 is . . 0 = Φ1 - Φ2 - Φdestr. . . . Φdestr. = Φ1 - Φ2 = 125.8 – 63.8 = 62 W 600 K 2 400 K Sonntag, Borgnakke and van Wylen 10.87 A heat engine operating with an evironment at 298 K produces 5 kW of power output with a first law efficiency of 50%. It has a second law efficiency of 80% and TL = 310 K. Find all the energy and exergy transfers in and out. Solution: From the definition of the first law efficiency . . 5 QH = W / η = = 10 kW . . . 0.5 Energy Eq.: QL = QH - W = 10 – 5 = 5 kW . . ΦW = W = 5 kW .. From the definition of the second law efficiency η = W/ΦH, this requires that we assume the availability delivered at. 310 K is lost and not counted otherwise the .. efficiency should be η = W/(ΦH - ΦL). To . . 5 Φ H = (1 – ) QH = = 6.25 kW TH 0.8 To . . 298 Φ L = (1 – ) Q = (1 – ) 5 = 0.194 kW TL L 310 . Notice from the ΦH form we could find the single characteristic TH as To . (1 – ) = 6.25 kW / QH = 0.625 => TH = 795 K TH Sonntag, Borgnakke and van Wylen 10.88 Consider the condenser in Problem 9.42. Find the specific energy and exergy that are given out, assuming an ambient at 20oC. Find also the specific exergy destruction in the process. Solution: Condenser from state 2 to state 3 QH 1 4 WT WP, in 2 P2 = P3 = 20 kPa T3 = 40 °C State 1: (P, T) Table B.1.3 h1 = 3809.1 kJ/kg, s1 = 6.7993 kJ/kg K . QL 3 C.V. Turbine. Entropy Eq.9.8: Table B.1.2 s2 = s1 = 6.7993 kJ/kg K s2 = 0.8319 + x2 × 7.0766 => x2 = 0.8433 h2 = 251.4 + 0.8433× 2358.33 = 2240.1 kJ/kg State 3: (P, T) Compressed liquid, take sat. liq. Table B.1.1 h3 = 167.54 kJ/kg, s3 = 0.5724 kJ/kg K C.V. Condenser Energy Eq.: qL = h2 – h3 = 2240.1 – 167.54 = 2072.56 kJ/kg Exergy Eq.: ∆ψ = ψ2 – ψ3 = h2 – h3 –To(s2 – s3) = 2072.56 – 293.15(6.7993 – 0.5724) = 247.1 kJ/kg going out Since all the exergy that goes out ends up at the ambient where it has zero exergy, the destruction equals the outgoing exergy. ψdestr = ∆ψ = 247.1 kJ/kg Notice the condenser gives out a large amount of energy byt little exergy. Sonntag, Borgnakke and van Wylen 10.89 The condenser in a power plant cools 10 kg/s water at 10 kPa, quality 90% so it comes out as saturated liquid at 10 kPa. The cooling is done by ocean-water coming in at ambient 15oC and returned to the ocean at 20oC. Find the transfer out of the water and the transfer into the ocean-water of both energy and exergy (4 terms). Solution: C.V. Water line. No work but heat transfer out. . . Energy Eq.: Qout = m (h1 – h2) = 10(2345.35 – 191.81) = 21 535 kW C.V. Ocean water line. No work but heat transfer in equals water heattransfer out Energy Eq.: q = h4 - h3 = 83.94 – 62.98 = 20.96 kJ/kg . . mocean = Qout /q = 21 535 / 20.96 = 1027.4 kg/s Exergy out of the water follows Eq.10.37 . . . . Φout = m(ψ1 - mψ2 ) = m [ h1 - h2 - To ( s1 - s2) ] = 10 [ 2345.35 – 191.81 – 288.15(7.4001 – 0.6492)] = 2082.3 kW Exergy into the ocean water . . . Φocean = mocean(ψ4 - ψ3) = mocean [ h4 - h3 – To(s4 - s3)] = 1027.4 [ 20.96 – 288.15(0.2966 – 0.2245)] = 189.4 kW Notice there is a large amount of energy exchanged but very little exergy. Sonntag, Borgnakke and van Wylen 10.90 Use the exergy equation to analyze the compressor in Example 6.10 to find its second law efficiency assuming an ambient at 20oC. C.V. The R-134a compressor. Steady flow. We need to find the reversible work and compare that to the actual work. Exergy eq.: 10.36: . . . rev 0 = m(ψ1 - mψ2 ) + (-Wcomp) + 0 . . rev -Wcomp = m [ h2 - h1 - To ( s2 - s1 ) ] . . ac = -Wcomp - mTo ( s2 - s1 ) = 5 kW – 0.1 kg/s × 293.15 K × (1.7768 – 1.7665) = 4.7 kW . rev . ac 4.7 ηII = -Wcomp / -Wcomp = = 0.94 5 For a real device this is a little high. kJ kg K Sonntag, Borgnakke and van Wylen 10.91 Consider the car engine in Example 7.1 and assume the fuel energy is delivered at a constant 1500 K. The 70% of the energy that is lost is 40% exhaust flow at 900 K and the remainder 30% heat transfer to the walls at 450 K goes on to the coolant fluid at 370 K, finally ending up in atmospheric air at ambient 20oC. Find all the energy and exergy flows for this heat engine. Find also the exergy destruction and where that is done. . . From the example in the text we get: QL = 0.7 QH = 233 kW This is separated into two fluxes: . . QL1 = 0.4 QH = 133 kW @900 K . . QL2 = 0.3 QH = 100 kW @450 K . = QL3 = 100 kW @370 K . = QL4 = 100 kW @293 K Gases 1500 K Steel Glycol 450 K 370 K Air flow 293 K Radiator Assume all the fuel energy is delivered at 1500 K then that has an exergy of To . . 293 ΦQH = (1 – ) QH = (1 – ) 333 = 267.9 kW TH 1500 Sonntag, Borgnakke and van Wylen 10.92 Estimate some reasonable temperatures to use and find all the fluxes of exergy in the refrigerator given in Example 7.2 We will assume the following temperatures: Ambient: T = 20oC usually it is the kitchen air. Low T: T = 5oC (refrigerator) T= -10oC (freezer) . . ΦW = W = 150 W To . Tamb . . Φ H = (1 – ) QH = (1 – )Q =0 TH Tamb H To . . 293 Φ L = (1 – ) QL = (1 – ) 250 = -13.5 W TL 278 I.e. the flux goes into the cold space! Why? As you cool it T < To and you increase its availability (exergy), it is further away from the ambient. Sonntag, Borgnakke and van Wylen 10.93 Use the exergy equation to evaluate the exergy destruction for Problem 10.44. A 2-kg piece of iron is heated from room temperature 25°C to 400°C by a heat source at 600°C. What is the irreversibility in the process? Solution: C.V. Iron out to 600°C source, which is a control mass. To Exergy Eq.10.42: Φ2 - Φ1 = (1 – ) Q - W + Po(V2 – V1) - 1Φ2 destr. TH 1 2 1 2 To evaluate it we need the heat transfer and the change in exergy Eq.10.43 Φ2 - Φ1 = mFe(u2 - u1) + Po(V2 – V1) - mFeTo(s2 - s1) Energy Eq.5.11: mFe(u2 - u1) = 1Q2 - 1W2 Process: Constant pressure => 1W2 = PmFe(v2 - v1) ⇒ 1Q2 = mFe(h2 - h1) = mFeC(T2 - T1) = 2 × 0.42 × (400 - 25) = 315 kJ 1Φ2 destr. To ) Q − 1W2 − mFe(u2 - u1) + mFeTo(s2 - s1) TH 1 2 To = (1 – ) Q − 1Q2 + mFeTo(s2 - s1) TH 1 2 673 298 ) 315 – 315 + 2 × 0.42 × 298 ln = 96.4 kJ = (1 298 873 = (1 – Notice the destruction is equal to 1I2 = To Sgen Sonntag, Borgnakke and van Wylen 10.94 Use the exergy balance equation to solve for the work in Problem 10.33. A piston/cylinder has forces on the piston so it keeps constant pressure. It contains 2 kg of ammonia at 1 MPa, 40°C and is now heated to 100°C by a reversible heat engine that receives heat from a 200°C source. Find the work out of the heat engine. Solution: To evaluate it we need the change in exergy Eq.10.43 Φ2 - Φ1 = mam(u2 - u1) + Po(V2 – V1) - mamTo(s2 - s1) The work in Eq.10.44 ( W = WH.E. + 1W2,pist) is from the exergy Eq.10.42 W = Po(V2 – V1) + (1 – = (1 – To ) Q – (Φ2 – Φ1) – 0 TH 1 2 To ) Q – mam(u2 – u1) + mamTo(s2 - s1) TH 1 2 Now we must evaluate the three terms on the RHS and the work 1W2,pist. State 1: u1 = 1369.8 kJ/kg, v1 = 0.13868 m3/kg, s1 = 5.1778 kJ/kg K State 2: u2 = 1490.5 kJ/kg, v2 = 0.17389 m3/kg, s2 = 5.6342 kJ/kg K 1W2,pist = mamP(v2 - v1) = 2 × 1000 (0.17389 - 0.13868) = 70.42 kJ C.V. Heat engine and ammnia (otherwise we involve another Q) Entropy: mam(s2 - s1) = 1Q2/TH + 0 => 1Q2 NH 3 QL = TH mam(s2 - s1) = 473.15 × 2 (5.6342 – 5.1778) = 431.89 kJ Substitute this heat transfer into the work term W HE cb QH o 200 C W = (1 - 298.15 ) 431.89 – 2(1490.5–1369.8) + 2×298.15(5.6342–5.1778) 473.15 = 159.74 – 241.4 + 272.15 = 190.49 kJ WH.E. = W - 1W2,pist = 190.49 – 70.42 = 120.0 kJ Sonntag, Borgnakke and van Wylen Review Problems 10.95 A small air gun has 1 cm3 air at 250 kPa, 27oC. The piston is a bullet of mass 20 g. What is the potential highest velocity with which the bullet can leave? Solution: The availability of the air can give the bullet kinetic energy expressed in the exergy balance Eq.10.42 (no heat transfer and reversible), Φ2 - Φ1 = m(u2 - u1) + Po(V2 – V1) - mTo(s2 - s1) = -1W2 + Po(V2 – V1) Ideal gas so: m = PV/RT = 250 × 1 × 10-6 = 2.9 × 10-6 kg 0.287 × 300 The second state with the lowest exergy to give maximum velocity is the dead state and we take To = 20oC. Now solve for the work term 1W2 = -m(u2 - u1) + mTo(s2 - s1) T2 P2 = mCv(T1 – T2) + mTo [ Cp ln( ) – R ln( ) ] T1 P1 = 2.9 × 10-6 [ 0.717(27 – 20) + 293.15 (1.004 ln 1 293 100 – 0.287 ln )] 300 250 2 = 0.0002180 kJ = 0.218 J = 2 mbulletVex 2 Vex = 2 × 0.218/0.020 = 4.67 m/s Comment: Notice that an isentropic expansion from 250 kPa to 100 kPa will give the final air temperature as 230.9 K but less work out. The above process is not adiabatic but Q is transferred from ambient at To. Sonntag, Borgnakke and van Wylen 10.96 Calculate the reversible work and irreversibility for the process described in Problem 5.134, assuming that the heat transfer is with the surroundings at 20°C. C.V.: A + B. This is a control mass. Continuity equation: m2 - (mA1 + mB1) = 0 ; Energy: m2u2 - mA1uA1 - mB1uB1 = 1Q2 - 1W2 System: if VB ≥ 0 piston floats ⇒ PB = PB1 = const. if VB = 0 then P2 < PB1 and v = VA/mtot see P-V diagram P State A1: Table B.1.1, x = 1 a vA1 = 1.694 m3/kg, uA1 = 2506.1 kJ/kg PB1 mA1 = VA/vA1 = 0.5903 kg 2 State B1: Table B.1.2 sup. vapor vB1 = 1.0315 m3/kg, uB1 = 2965.5 kJ/kg V2 mB1 = VB1/vB1 = 0.9695 kg => m2 = mTOT = 1.56 kg At (T2 , PB1) v2 = 0.7163 > va = VA/mtot = 0.641 so VB2 > 0 so now state 2: P2 = PB1 = 300 kPa, T2 = 200 °C => u2 = 2650.7 kJ/kg and V2 = m2 v2 = 1.56 × 0.7163 = 1.117 m3 (we could also have checked Ta at: 300 kPa, 0.641 m3/kg => T = 155 °C) ac ⌠ 1W2 = ⌡PBdVB = PB1(V2 - V1)B = PB1(V2 - V1)tot = -264.82 kJ 1Q2 = m2u2 - mA1uA1 - mB1uB1 + 1W2 = -484.7 kJ From the results above we have : sA1 = 7.3593 kJ/kg K, sB1 = 8.0329 kJ/kg K, s2 = 7.3115 kJ/kg K rev 1W2 = To(S2 - S1) - (U2 - U1) + 1Q2(1 - To/TH) ac = To(m2s2 - mA1sA1 - mB1sB1) + 1W2 - 1Q2To/TH = 293.15 (1.5598 × 7.3115 - 0.5903 × 7.3593 - 0.9695 × 8.0329) + (-264.82) - (-484.7) × 293.15 / 293.15 = -213.3 - 264.82 + 484.7 = 6.6 kJ rev ac 1I2 = 1W2 - 1W2 = 6.6 - (-264.82) = 271.4 kJ Sonntag, Borgnakke and van Wylen 10.97 A piston/cylinder arrangement has a load on the piston so it maintains constant pressure. It contains 1 kg of steam at 500 kPa, 50% quality. Heat from a reservoir at 700°C brings the steam to 600°C. Find the second-law efficiency for this process. Note that no formula is given for this particular case so determine a reasonable expression for it. Solution: 1: Table B.1.2 P1, x1 ⇒ v1 = 0.001093 + 0.5×0.3738 = 0.188 m3/kg, h1 = 640.21 + 0.5×2108.47 = 1694.5 kJ/kg, s1 = 1.8606 + 0.5×4.9606 = 4.341 kJ/kg K 2: P2 = P1,T2 ⇒ v2 = 0.8041, h2 = 3701.7 kJ/kg, s2 = 8.3521 kJ/kg K Energy Eq.: 1Q2 1W2 m(u2 - u1) = 1Q2 - 1W2 = 1Q2 - P(V2 - V1) = m(u2 - u1) + Pm(v2 - v1) = m(h2 - h1) = 2007.2 kJ = Pm(v2 - v1) = 308.05 kJ 1W2 to atm = P0m(v2 - v1) = 61.61 kJ Useful work out = 1W2 - 1W2 to atm = 246.44 kJ 298.15 ∆φreservoir = (1 - T0/Tres)1Q2 = 1 2007.2 = 1392.2 kJ 973.15 ηII = Wnet/∆φ = 0.177 Sonntag, Borgnakke and van Wylen 10.98 Consider the high-pressure closed feedwater heater in the nuclear power plant described in Problem 6.102. Determine its second-law efficiency. For this case with no work the second law efficiency is from Eq. 10.25: • • ηII = m16(ψ18 - ψ16)/m17(ψ17 - ψ15) Properties (taken from computer software): h [kJ/kg] h15 = 585 h16 = 565 h17 = 2593 h18 = 688 s [kJ/kgK] s15 = 1.728 s16 = 1.6603 s17 = 6.1918 s18 = 1.954 The change in specific flow availability becomes ψ18 - ψ16 = h18 - h16 - T0(s18 - s16) = 35.433 kJ/kg ψ17 - ψ15 = h17 - h15 - T0(s17 - s15) = 677.12 kJ/kg ηII = (75.6 × 35.433)/(4.662 × 677.12) = 0.85 Sonntag, Borgnakke and van Wylen 10.99 Consider a gasoline engine for a car as a steady device where air and fuel enters at the surrounding conditions 25°C, 100 kPa and leaves the engine exhaust manifold at 1000 K, 100 kPa as products assumed to be air. The engine cooling system removes 750 kJ/kg air through the engine to the ambient. For the analysis take the fuel as air where the extra energy of 2200 kJ/kg of air released in the combustion process, is added as heat transfer from a 1800 K reservoir. Find the work out of the engine, the irreversibility per kilogram of air, and the first- and second-law efficiencies. C.V. Total out to reservoirs • • • • • Energy Eq.: mah1 + QH = mah2 + W + Qout Entropy Eq.: • • • • Air intake filter Fuel line Radiator 1 W Shaft power Qo 2 Exhaust flow • mas1 + QH/TH + Sgen = mass + Qout/T0 To Coolant flow Burning of the fuel releases • QH at TH. From the air Table A.7 kJ/kg kJ/kg K ° = 6.8631 h1 = 298.61 sT1 h2 = 1046.22 s° = 8.1349 T1 • • wac = W/ma = h1 - h2 + qH - qout = 298.6 - 1046.22 + 2200 - 750 = 702.4 kJ/kg ηTH = w/qH = 702.4/2200 = 0.319 sgen = s2 - s1 + qout qH 750 2200 = 8.1349 - 6.8631 + = 2.565 kJ/kg K T0 TH 298.15 1800 itot = (T0)sgen = 764.8 kJ/kg For reversible case have sgen = 0 and qR from T0, no qout 0 qR = T0(s2 - s1) - (T0/TH)qH = 14.78 kJ/kg 0,in wrev = h1 - h2 + qH + qR = wac + itot = 1467.2 kJ/kg 0,in ηII = wac/wrev = 0.479 Sonntag, Borgnakke and van Wylen 10.100 Consider the nozzle in Problem 9.112. What is the second law efficiency for the nozzle? A nozzle in a high pressure liquid water sprayer has an area of 0.5 cm2. It receives water at 250 kPa, 20°C and the exit pressure is 100 kPa. Neglect the inlet kinetic energy and assume a nozzle isentropic efficiency of 85%. Find the ideal nozzle exit velocity and the actual nozzle mass flow rate. Solution: C.V. Nozzle. Liquid water is incompressible v ≈ constant, no work, no heat transfer => Bernoulli Eq.9.17 12 V – 0 = v(Pi - Pe) = 0.001002 ( 250 – 100) = 0.1503 kJ/kg 2 ex 2 × 0.1503 × 1000 J/kg = 17.34 m s Vex = -1 This was the ideal nozzle now we can do the actual nozzle, Eq. 9.30 12 12 Vex ac = η Vex = 0.85 × 0.1503 = 0.12776 kJ/kg 2 2 Vex ac = 2 × 0.12776 × 1000 J/kg = 15.99 m s -1 The second law efficiency is the actual nozzle compare to a reversible process between the inlet and actual exit states. However here there is no work so the actual exit state then must have the reversible possible kinetic energy. Energy actual nozzle: 12 hi + 0 = he + 2Vex ac same Z, no q and no w. The reversible process has zero change in exergies from Eq.10.36 as 0 = 0 – 0 + 0 + ψi - ψe – 0 12 ψi = ψe = hi + 0 – To si = he + 2Vex rev – Tose 12 12 V = hi - he + To (se - si) = 2Vex ac + To sgen 2 ex rev We can not get properties for these states accurately enough by interpolation to carry out the calculations. With the computer program we can get: Inlet: hi = 84.173 kJ/kg, si = 0.29652 kJ/kg K 12 he s = 84.023 kJ/kg, Te s = 19.998°C, 2Vex = 0.15 kJ/kg 12 Exit,ac: V = 0.1275 kJ/kg, he = 84.173 – 0.1275 = 84.0455 kJ/kg 2 ex ac (P, h) => se = 0.29659 kJ/kg K, T = 20.003°C Exit,s: 12 12 V =V + To sgen = 0.1275 + 293.15(0.29659 – 0.29652) 2 ex rev 2 ex ac = 0.148 kJ/kg ηII = 0.1275/0.148 = 0.86 Sonntag, Borgnakke and van Wylen 10.101 Air in a piston/cylinder arrangement is at 110 kPa, 25°C, with a volume of 50 L. It goes through a reversible polytropic process to a final state of 700 kPa, 500 K, and exchanges heat with the ambient at 25°C through a reversible device. Find the total work (including the external device) and the heat transfer from the ambient. C.V. Total out to ambient ma(u2 - u1) = 1Q2 - 1W2,tot , ma(s2 - s1) = 1Q2/T0 ma = 110 × 0.05/0.287 × 298.15 = 0.0643 kg 1Q2 = T0ma(s2 - s1) = 298.15 × 0.0643[7.3869 - 6.8631 - 0.287 ln (700/110)] = -0.14 kJ 1W2,tot = 1Q2 - ma(u2 - u1) = -0.14 - 0.0643 × (359.844 - 213.037) = -9.58 kJ Sonntag, Borgnakke and van Wylen 10.102 Consider the irreversible process in Problem 8.128. Assume that the process could be done reversibly by adding heat engines/pumps between tanks A and B and the cylinder. The total system is insulated, so there is no heat transfer to or from the ambient. Find the final state, the work given out to the piston and the total work to or from the heat engines/pumps. C.V. Water mA + mB + heat engines. No Qexternal, only 1W2,cyl + WHE m2 = mA1 + mB1 = 6 kg, m2u2 - mA1uA1 - mB1uB1 = -1W2,cyl - WHE // m2s2 - mA1sA1 - mB1sB1 = 0 + 0 vA1 = 0.06283 uA1 = 3448.5 sA1 = 7.3476 VA = 0.2513 m3 vB1 = 0.09053 uB1 = 2843.7 sB1 = 6.7428 VB = 0.1811 m3 m2s2 = 4×7.3476 + 2×6.7428 = 42.876 ⇒ s2 = 7.146 kJ/kg K If P2 < Plift = 1.4 MPa then V2’ = VA + VB = 0.4324 m3 , v2’ = 0.07207 m3/kg (Plift , s2) ⇒ v2 = 0.20135 ⇒ V2 = 1.208 m3 > V2’ OK ⇒ P2 = Plift = 1.4 MPa u2 = 2874.2 kJ/kg 1W2,cyl = Plift(V2 - VA - VB) = 1400×(1.208 - 0.4324) = 1085.84 kJ WHE = mA1uA1 + mB1uB1 - m2u2 - 1W2,cyl = 4 × 3447.8 + 2 × 2843.7 - 6 × 2874.2 - 1085.84 = 1147.6 kJ Sonntag, Borgnakke and van Wylen 10.103 Consider the heat engine in Problem 10.79. The exit temperature was given as 800 K, but what are the theoretical limits for this temperature? Find the lowest and the highest, assuming the heat transfers are as given. For each case give the first and second law efficiency. The lowest exhaust temperature will occur when the maximum amount of work is delivered which is a reversible process. Assume no other heat transfers then 2nd law: si + qH/TH + 0 = se + qm/Tm / se - si = qH/TH - qm/Tm = s° - s° - R ln(Pe/Pi) Te Ti s° = s° + R ln(Pe/Pi) + qH/TH - qm/Tm Te Ti = 6.86926 + 0.287 ln(100/100) + 1200/1500 - 300/750 = 7.26926 kJ/kg K Table A.7.1 ⇒ Te,min = 446 K, he = 447.9 kJ/kg hi + q1500 = q750 + he + w wrev = hi + q1500 - q750 - he = 300.47 + 1200 - 300 - 447.9 = 752.57 kJ/kg w rev 752.57 = = 0.627 ηI = ηTH = q1500 1200 The second law efficiency measures the work relative to the source of availability and not q1500. So ηII = wrev 752.57 752.57 = = = 0.784 (1- To/TH)q1500 (1 - 300/1500)1200 960 The maximum exhaust temperature occurs with no work out hi + qH = qm + he ⇒ he = 300.473 + 1200 - 300 = 1200.5 kJ/kg Table A.7.1 ⇒ Te,max = 1134 K Now : wac = 0 so ηI = ηII = 0 Sonntag, Borgnakke and van Wylen 10.104 Air in a piston/cylinder arrangement, shown in Fig. P10.104, is at 200 kPa, 300 K with a volume of 0.5 m3. If the piston is at the stops, the volume is 1 m3 and a pressure of 400 kPa is required. The air is then heated from the initial state to 1500 K by a 1900 K reservoir. Find the total irreversibility in the process assuming surroundings are at 20°C. Solution: Energy Eq.: m(u2 - u1) = 1Q2 - 1W2 Entropy Eq,: m(s2 - s1) = ⌠ dQ/T + 1S2 gen ⌡ Process: Information: P = P0 + α(V-V0) if V ≤ Vstop Pstop = P0 + α(Vstop-V0) Eq. of state ⇒ Tstop = T1PstopVstop/P1V1 = 1200 < T2 So the piston will hit the stops => V2 = Vstop P2 = (T2/Tstop) Pstop = (1500/1200) 400 = 500 kPa = 2.5 P1 State 1: P1V1 m2 = m1 = RT1 200 × 0.5 = 0.287 × 300 = 1.161 kg P 2 1a 1 v v1 1 1W2 = 2(P1 1Q2 = vstop Air Q Tres 1 + Pstop)(Vstop- V1) = 2(200 + 400)(1 - 0.5) = 150 kJ M(u2 - u1) + 1W2 = 1.161(1205.25 - 214.36) + 150 = 1301 kJ o o s2 - s1 = sT2 - sT1 -R ln(P2/P1) = 8.6121 - 6.8693 - 0.287 ln 2.5 = 1.48 kJ/kg K Take control volume as total out to reservoir at TRES 1S2 gen tot = 1I2 m(s2 - s2) - 1Q2/TRES = 1.034 kJ/K = T0(1S2 gen) = 293.15 × 1.034 = 303 kJ Sonntag, Borgnakke and van Wylen 10.105 A jet of air at 200 m/s flows at 25oC, 100 kPa towards a wall where the jet flow stagnates and leaves at very low velocity. Consider the process to be adiabatic and reversible. Use the exergy equation and the second law to find the stagnation temperature and pressure. Solution: C.V. From free flow to stagnation.point. .Reversible adiabatic steady flow. . Exergy Eq.10.36: 0 = mψi - mψe - Φdestr. . . . . . . Entropy Eq.: 0 = msi - mse + ∫ mdq/T + msgen = msi - mse + 0 + 0 . Process: Reversible Φdestr. = 0, sgen = 0, adiabatic q = 0 12 From exergy Eq.: ψe - ψi = 0 = he – Tose – hi + Tosi – Vi 2 so entropy terms drop out From entropy Eq.: s e = si , 12 12 Exergy eq. now leads to: he = hi + Vi => Te = Ti + Vi /Cp 2 2 2 J/kg 1 200 Te = 25 + = 44.92oC 2 1004 J/kg K Eq.8.32: Pe = Pi ( Te/Te ) State i is the free stream state. State e is the stagnation state. k k-1 273 + 44.921.4 / 0.4 = 100 = 125.4 kPa 273 +25 i e cb Sonntag, Borgnakke and van Wylen 10.106 Consider the light bulb in Problem 8.123. What are the fluxes of exergy at the various locations mentioned? What are the exergy destruction in the filament, the entire bulb including the glass and the entire room including the bulb? The light does not affect the gas or the glass in the bulb but it gets absorbed on the room walls. A small halogen light bulb receives an electrical power of 50 W. The small filament is at 1000 K and gives out 20% of the power as light and the rest as heat transfer to the gas, which is at 500 K; the glass is at 400 K. All the power is absorbed by the room walls at 25oC. Find the rate of generation of entropy in the filament, in the total bulb including glass and the total room including bulb. Solution: . g leads Wel = 50 W a . s QRAD = 10 W . glass QCOND = 40 W We will assume steady state and no storage in the bulb, air or room walls. C.V. Filament steady-state . . . Energy Eq.5.31: dEc.v./dt = 0 = Wel – QRAD – QCOND . . QRAD QCOND . Entropy Eq.8.43: dSc.v./dt = 0 = – – + Sgen TFILA TFILA . . . . 50 = 0.05 W/K Sgen = (QRAD + QCOND)/TFILA = Wel/TFILA = 1000 C.V. Bulb including glass . . QRAD leaves at 1000 K QCOND leaves at 400 K . . Sgen = ∫ dQ/T = -(-10/1000) – (-40/400) = 0.11 W/K C.V. Total room. All energy leaves at 25°C . . . Eq.5.31: dEc.v./dt = 0 = Wel – QRAD – QCOND . QTOT . Eq.8.43: dSc.v./dt = 0 = – + Sgen TWALL . QTOT . Sgen = = 50/(25+273) = 0.168 W/K TWALL Sonntag, Borgnakke and van Wylen Problems Solved Using Pr and vr Functions 10.31 An air compressor receives atmospheric air at T0 = 17°C, 100 kPa, and compresses it up to 1400 kPa. The compressor has an isentropic efficiency of 88% and it loses energy by heat transfer to the atmosphere as 10% of the isentropic work. Find the actual exit temperature and the reversible work. C.V. Compressor Isentropic: wc,in,s = he,s - hi ; se,s = si Table A.7: Pr,e,s = Pr,i × (Pe/Pi) = 0.9917 × 14 = 13.884 ⇒ he,s = 617.51 kJ/kg wc,in,s = 617.51 - 290.58 = 326.93 kJ/kg Actual: wc,in,ac = wc,in,s/ηc = 371.51 ; qloss = 32.693 kJ/kg wc,in,ac + hi = he,ac + qloss => he,ac = 290.58 + 371.51 - 32.693 = 629.4 kJ/kg => Te,ac = 621 K Reversible: wrev = hi - he,ac + T0(se,ac - si) = 290.58 - 629.4 + 290.15 × (7.6121 - 6.8357) = -338.82 + 225.42 = -113.4 kJ/kg Since qloss is also to the atmosphere it is not included as it will not be reversible. Sonntag, Borgnakke and van Wylen 10.61 An air compressor is used to charge an initially empty 200-L tank with air up to 5 MPa. The air inlet to the compressor is at 100 kPa, 17°C and the compressor isentropic efficiency is 80%. Find the total compressor work and the change in availability of the air. Solution: C.V. Tank + compressor Transient process with constant inlet conditions, no heat transfer. Continuity: m2 - m1 = min ( m1 = 0 ) Energy: m2u2 = minhin - 1W2 Entropy: m2s2 = minsin + 1S2 gen Reversible compressor: 1S2 GEN = 0 ⇒ s2 = sin State 1: v1 = RT1/P1 = 0.8323 m3/kg, State inlet, Table A.7.1: hin = 290.43 kJ/kg, Table A.7.2 Prin = 0.9899 o sTin = 6.8352 kJ/kg K used for constant s process Table A.7.2 ⇒ Pr2 = Prin(P2/Pin) = 0.9899 × (5000/100) = 49.495 ⇒ T2,s = 855 K, u2,s = 637.2 kJ/kg ⇒ 1w2,s = hin - u2,s = 290.43 – 637.2 = -346.77 kJ/kg Actual compressor: 1w2,AC = 1w2,s/ηc = -433.46 kJ/kg u2,AC = hin - 1w2,AC = 290.43 – (-433.46) = 723.89 kJ/kg Backinterpolate in Table A.7.1 o ⇒ T2,AC = 958 K, sT2,AC = 8.0867 kJ/kg K ⇒ v2 = RT2/P2 = 0.055 m3/kg State 2 u, P m2 = V2/v2 = 3.636 kg ⇒ 1W2 = m2(1w2,AC) = -1576 kJ m2(φ2 - φ1) = m2[u2 - u1 + P0(v2 - v1) - T0(s2 - s1)] = 3.636 [723.89 - 207.19 + 100(0.055 - 0.8323) - 290[8.0867 6.8352 - 0.287 ln(5000/100)] = 1460.3 kJ Here we used Eq.8.28 for the change in entropy. SOLUTION MANUAL SI UNIT PROBLEMS CHAPTER 11 SONNTAG • BORGNAKKE • VAN WYLEN FUNDAMENTALS of Thermodynamics Sixth Edition CONTENT SUBSECTION PROB NO. Correspondence table Concept-Study guide problems 1-20 Rankine cycles, power plants Simple cycles 21-35 Reheat cycles 36-40 Open feedwater heaters 41-47 Closed feedwater heaters 48-52 Nonideal cycles 53-62 Cogeneration 63-67 Brayton cycles, gas turbines 68-73 Regenerators, Intercoolers, nonideal cycles 74-84 Ericsson Cycles 85-86 Jet engine cycles 87-92 Otto cycles 93-105 Diesel cycles 106-112 Stirling and Carnot cycles 113-118 Refrigeration cycles 119-133 Ammonia absorption cycles 134-135 Air-standard refrigeration cycles 136-139 Combined cycles 140-145 Availability or Exergy Concepts 146-151 Review Problems 152-166 Problems re-solved with the Pr, vr functions from A.7.2: 79, 81, 93, 94, 100, 103, 110, 118 CORRESPONDANCE TABLE The correspondence between the new problem set and the 5th edition chapter 11 problem set. Problems 11.1-20 are all new New 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 5th 1 mod 2 3 mod new 4 5 6 7 10 11 8 mod new 12 mod 13 14 mod new 16 mod new 17 mod 18 mod 20 22 new 23 24 mod 40 mod 26 mod 19 21 25 mod New 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 5th 27 mod new 15 31 32 mod 33 36 mod 37 34 35 38 39 41 42 mod 44 45 new 46 47 49 50 new new 48 54 52 mod 60 new 53 51 New 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 5th 61 57 59 mod 56 62 63 64 new 65 67 68 new 69 mod 70 mod 71 new new new 73 74 75 new 72 76 77 new 79 78 80 new For many of the cycle problems we recommend that the students be allowed to use the software for properties to reduce the time spent on interpolations. New 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 5th new 81 82 83 84 85 86 a 86 b 90 87 88 89 91 92 93 94 mod 95 new 96 new New 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 5th new new 98 99 100 new 101 103 mod 102 104 105 106 108 mod 109 107 new 110 new 112 new New 151 152 153 154 155 156 157 158 159 160 161 162 163 164 165 166 5th 114 new 9 mod 29 15 28 mod 40 43 55 58 mod 59 70 111 113 115 116 The correspondence between the new English unit problem set and the previous 5th edition chapter 11 problem set and the current SI problems. New 167 168 169 170 171 172 173 174 175 176 177 178 179 180 181 182 183 5th 117 mod 118 mod new 119 120 new 121 mod 122 mod 123 mod 125 mod 124 127 mod 128 mod 129 131 mod 132 new SI 21 22 24 26 27 32 33 35 37 45 48 55 57 60 66 71 72 New 184 185 186 187 188 189 190 191 192 193 194 195 196 197 198 199 200 5th new 133 136 137 138 139 new new 141 140 142 143 144 145 146 147 148a SI 73 74 86 89 93 95 97 98 104 105 107 109 112 113 114 116 117 New 201 202 203 204 205 206 207 208 209 210 211 212 213 214 215 216 217 5th 148b new 149 150 151 new new new 155 new new 154 126 130 134 135 153 SI 118 120 121 125 130 137 146 147 148 150 144 157 160 160 134 Concept-Study Guide Problems 11.1 Is a steam power plant running in a Carnot cycle? Name the four processes. No. It runs in a Rankine cycle. 1-2: 2-3: 3-4: 4-1: An isentropic compression (constant s) An isobaric heating (constant P) An isentropic expansion (constant s) An isobaric cooling, heat rejection (constant P) Pump Boiler Turbine Condenser 11.2 Consider a Rankine cycle without superheat. How many single properties are needed to determine the cycle? Repeat the answer for a cycle with superheat. a. No superheat. Two single properties. High pressure (or temperature) and low pressure (or temperature). This assumes the condenser output is saturated liquid and the boiler output is saturated vapor. Physically the high pressure is determined by the pump and the low temperature is determined by the cooling medium. b. Superheat. Three single properties. High pressure and temperature and low pressure (or temperature). This assumes the condenser output is saturated liquid. Physically the high pressure is determined by the pump and the high temperature by the heat transfer from the hot source. The low temperature is determined by the cooling medium. 11.3 Which component determines the high pressure in a Rankine cycle? What determines the low pressure? The high pressure in the Rankine cycle is determined by the pump. The low pressure is determined as the saturation pressure for the temperature you can cool to in the condenser. 11.4 Mention two benefits of a reheat cycle. The reheat raises the average temperature at which you add heat. The reheat process brings the states at the lower pressure further out in the superheated vapor region and thus raises the quality (if two-phase) in the last turbine section. 11.5 What is the difference between an open and a closed feedwater heater? The open feedwater heater mixes the two flows at the extraction pressure and thus requires two feedwater pumps. The closed feedwater heater does not mix the flows but let them exchange energy (it is a two fluid heat exchanger). The flows do not have to be at the same pressure. The condensing source flow is dumped into the next lower pressure feedwater heater or the condenser or it is pumped up to line pressure by a drip pump and added to the feedwater line. 11.6 Can the energy removed in a power plant condenser be useful? Yes. In some applications it can be used for heating buildings locally or as district heating. Other uses could be to heat green houses or as general process steam in a food process or paper mill. These applications are all based on economics and scale. The condenser then has to operate at a higher temperature than it otherwise would. 11.7 In a cogenerating power plant, what is cogenerated? The electricity is cogenerated. The main product is a steam supply. 11.8 Why is the back work ratio in the Brayton cycle much higher than in the Rankine cycle? Recall the expression for shaft work in a steady flow device w = − ⌠ v dP ⌡ The specific volume in the compressor is not so much smaller than the specific volume in the turbine of the Brayton cycle as it is in the pump (liquid) compared to turbine (superheated vapor) in the Rankine cycle. 11.9 The Brayton cycle has the same 4 processes as the Rankine cycle, but the T-s and P-v diagrams look very different; why is that? The Brayton cycle have all processes in the superheated vapor (close to ideal gas) region. The Rankine cycle crosses in over the two-phase region. 11.10 Is it always possible to add a regenerator to the Brayton cycle? What happens when the pressure ratio is increased? No. When the pressure ratio is high, the temperature after compression is higher than the temperature after expansion. The exhaust flow can then not heat the flow into the combustor. 11.11 Why would you use an intercooler between compressor stages? The cooler provides two effects. It reduces the specific volume and thus reduces the work in the following compressor stage. It also reduces the temperature into the combustor and thus lowers the peak temperature. This makes the control of the combustion process easier (no autoignition or uncontrollable flame spread), it reduces the formation of NOx that takes place at high temperatures and lowers the cooling requirements for the chamber walls. 11.12 The jet engine does not produce shaft work; how is power produced? The turbine produces just enough shaft work to drive the compressor and it makes a little electric power for the aircraft. The power is produced as thrust of the engine. In order to exhaust the gases at high speed they must be accelerated so the high pressure in the turbine exit provides that force (high P relative to ambient). The high P into the turbine is made by the compressor, that pushes the flow backwards, and thus has a net resulting force forwards on the blades transmitted to the shaft and the aircraft. The outer housing also has a higher pressure inside that gives a net component in the forward direction. 11.13 How is the compression in the Otto cycle different from the Brayton cycle? The compression in an Otto cycle is a volume reduction dictated by the piston motion. The physical handles are the volumes V1 and V2. The compression in a Brayton cycle is the compressor pushing on the flow so it determines the pressure. The physical control is the pressure P2. 11.14 Does the inlet state (P1, T1) have any influence on the Otto cycle efficiency? How about the power produced by a real car engine? Very little. The efficiency for the ideal cycle only depends on compression ratio when we assume cold air properties. The u’s are slightly non-linear in T so there will be a small effect. In a real engine there are several effects. The inlet state determines the density and thus the total mass in the chamber. The more mass the more energy is released when the fuel burns, the peak P and T will also change which affects the heat transfer loss to the walls and the formation of Nox (sensitive to T). The combustion process may become uncontrollable if T is too high (knocking). Some increase in P1 like that done by a turbo-charger or super-charger increases the power output and if high, it must be followed by an intercooler to reduce T1. If P1 is too high the losses starts to be more than the gain so there is an optimum level. 11.15 How many parameters do you need to know to completely describe the Otto cycle? How about the Diesel cycle? Otto cycle. State 1 (2 parameters) and the compression ratio CR and the energy release per unit mass in the combustion, a total of 4 parameters. With that information you can draw the diagrams in Figure 11.28. Another way of looking at it is four states (8 properties) minus the four process equations (s2 = s1, v3 = v2, s4 = s3 and v4 = v1) gives 4 unknowns. Diesel cycle. Same as for the Otto cycle namely 4 parameters. The only difference is that one constant v process is changed to a constant P process. 11.16 The exhaust and inlet flow processes are not included in the Otto or Diesel cycles. How do these necessary processes affect the cycle performance? Due to the pressure loss in the intake system and the dynamic flow process we will not have as much mass in the cylinder nor as high a P as in a reversible process. The exhaust flow requires a slightly higher pressure to push the flow out through the catalytic converter and the muffler (higher back pressure) and the pressure loss in the valve so again there is a loss relative to a reversible process. Both of these processes subtracts a pumping work from the net work out of the engine and a lower charge mass gives less power (not necessarily lower efficiency) than other wise could be obtained. 11.17 A refrigerator in my 20oC kitchen uses R-12 and I want to make ice cubes at –5o C. What is the minimum high P and the maximum low P it can use? Since the R-12 must give heat transfer out to the kitchen air at 20oC, it must at least be that hot at state 3. From Table B.3.1: P3 = P2 = Psat = 567 kPa is minimum high P. Since the R-12 must absorb heat transfer at the freezers –5oC, it must at least be that cold at state 4. From Table B.3.1: P1 = P4 = Psat = 261 kPa is maximum low P. 11.18 How many parameters are needed to completely determine a standard vapor compression refrigeration cycle? Two parameters: The high pressure and the low pressure. This assumes the exit of the condenser is saturated liquid and the exit of the evaporator is saturated vapor. 11.19 Why would one consider a combined cycle system for a power plant? For a heat pump or refrigerator? Dual cycle or combined cycle systems have the advantage of a smaller difference between the high and low ranges for P and T. The heat can be added at several different temperatures reducing the difference between the energy source T and the working substance T. The working substance vapor pressure at the desired T can be reduced from a high value by adding a topping cycle with a different substance or have a higher low pressure at very low temperatures. 11.20 Since any heat transfer is driven by a temperature difference, how does that affect all the real cycles relative to the ideal cycles? . Heat transfers are given as Q = CA ∆T so to have a reasonable rate the area and the temperature difference must be large. The working substance then must have a different temperature than the ambient it exchanges energy with. This gives a smaller temperature difference for a heat engine with a lower efficiency as a result. The refrigerator or heat pump must have the working substance with a higher temperature difference than the reservoirs and thus a lower coefficient of performance (COP). The smaller CA is the larger ∆T must be for a certain magnitude of the heat transfer rate. This can be a design problem, think about the front end air intake grill for a modern car which is very small compared to a car 20 years ago. Simple Rankine cycles 11.21 A steam power plant as shown in Fig. 11.3 operating in a Rankine cycle has saturated vapor at 3.0 MPa leaving the boiler. The turbine exhausts to the condenser operating at 10 kPa. Find the specific work and heat transfer in each of the ideal components and the cycle efficiency. Solution: C.V. Pump Reversible and adiabatic. s 2 = s1 Energy: wp = h2 - h1 ; Entropy: since incompressible it is easier to find work (positive in) as wp = ∫ v dP = v1 (P2 - P1) = 0.00101 (3000 - 10) = 3.02 kJ/kg => h2 = h1 + wp = 191.81 + 3.02 = 194.83 kJ/kg C.V. Boiler : qH = h3 - h2 = 2804.14 - 194.83 = 2609.3 kJ/kg C.V. Turbine : wT = h3 - h4 ; s4 = s3 s4 = s3 = 6.1869 = 0.6492 + x4 (7.501) => x4 = 0.7383 => h4 = 191.81 + 0.7383 (2392.82) = 1958.34 kJ/kg wT = 2804.14 - 1958.34 = 845.8 kJ/kg C.V. Condenser : qL = h4 - h1 = 1958.34 - 191.81 = 1766.5 kJ/kg ηcycle = wnet / qH = (wT + wp) / qH = (845.8 - 3.0) / 2609.3 = 0.323 Boiler Turbine T 3 3 QB 2 WP 1 2 WT 1 4 Condenser Q 4 s 11.22 Consider a solar-energy-powered ideal Rankine cycle that uses water as the working fluid. Saturated vapor leaves the solar collector at 175°C, and the condenser pressure is 10 kPa. Determine the thermal efficiency of this cycle. Solution: C.V. H2O ideal Rankine cycle State 3: ⇒ T3 = 175°C P3 = PG 175°C = 892 kPa, s3 = 6.6256 CV Turbine adiabatic and reversible so second law gives s4 = s3 = 6.6256 = 0.6493 + x4 × 7.5009 => x4 = 0.797 h4 = 191.83 + 0.797 × 2392.8 = 2098.3 kJ/kg The energy equation gives wT = h3 - h4 = 2773.6 - 2098.3 = 675.3 kJ/kg C.V. pump and incompressible liquid gives work into pump wP = v1(P2 - P1) = 0.00101(892 - 10) = 0.89 kJ/kg h2 = h1 + wP = 191.83 + 0.89 = 192.72 kJ/kg C.V. boiler gives the heat transfer from the energy equation as qH = h3 - h2 = 2773.6 - 192.72 = 2580.9 kJ/kg The cycle net work and efficiency are found as wNET = wT - wP = 675.3 - 0.89 = 674.4 kJ/kg ηTH = wNET/qH = 674.4/2580.9 = 0.261 Q RAD T Turbine Solar collector 3 3 2 2 WT WP 1 4 1 Condenser 4 s Q 11.23 A utility runs a Rankine cycle with a water boiler at 3.0 MPa and the cycle has the highest and lowest temperatures of 450°C and 45°C respectively. Find the plant efficiency and the efficiency of a Carnot cycle with the same temperatures. Solution: The states properties from Tables B.1.1 and B.1.3 1: 45oC , x = 0 => h1 = 188.42 , v1 = 0.00101 , Psat = 9.6 kPa 3: 3.0 MPa , 450oC => h3 = 3344 , s3 = 7.0833 C.V. Pump Reversible and adiabatic. Energy: wp = h2 - h1 ; Entropy: s 2 = s1 since incompressible it is easier to find work (positive in) as wp = ∫ v dP = v1 (P2 - P1) = 0.00101 (3000 - 9.6) = 3.02 kJ/kg => h2 = h1 + wp = 188.42 + 3.02 = 191.44 kJ/kg C.V. Boiler : qH = h3 - h2 = 3344 - 191 = 3152.56 kJ/kg C.V. Turbine : wT = h3 - h4 ; s4 = s3 s4 = s3 = 7.0833 = 0.6386 + x4 (7.5261) => x4 = 0.8563 => h4 = 188.42 + 0.8563 (2394.77) = 2239.06 kJ/kg wT = 3344 – 2239.06 = 1105 kJ/kg C.V. Condenser : qL = h4 - h1 = 2239.06 - 188.42 = 2050.64 kJ/kg ηcycle = wnet / qH = (wT + wp) / qH = (1105 - 3.02) / 3152.56 = 0.349 ηcarnot = 1 - TL / TH = 1 Boiler 273.15 + 45 = 0.56 273.15 + 450 Turbine T 3 3 QB 2 WP 1 2 WT 1 4 Condenser Q 4 s 11.24 A Rankine cycle uses ammonia as the working substance and powered by solar energy. It heats the ammonia to 140oC at 5000 kPa in the boiler/superheater. The condenser is water cooled and the exit kept at 25oC. Find (T, P and x if applicable) for all four states in the cycle. Solution: Based on the standard Rankine cycle and Table B.2 and Table A.4 for Cp. State 1: Saturated liquid. P1 = Psat = 1003 kPa, x1 = 0 State 2: P2 = 5000 kPa, consider C.V. pump Energy: h2 - h1 = wp = v1 (P2 - P1) = 0.001658 (5000 – 1003) = 6.627 kJ/kg T2 = T1 + (h2 - h1)/Cp = 25 + 6.627/4.84 = 26.4oC State 3: Table B.2.2 140oC at 5000 kPa, s = 4.9068 kJ/kg K State 4: P4 = P1 = 1003 kPa. Consider the turbine for which s4 = s3. s3 < sg = 5.0293 kJ/kg K at 25oC x4 = (s3 – sf)/sfg = (4.9068 – 1.121)/3.9083 = 0.96866 P T 3 3 2 2 1 4 v 1 4 s 11.25 A steam power plant operating in an ideal Rankine cycle has a high pressure of 5 MPa and a low pressure of 15 kPa. The turbine exhaust state should have a quality of at least 95% and the turbine power generated should be 7.5 MW. Find the necessary boiler exit temperature and the total mass flow rate. Solution: C.V. Turbine assume adiabatic and reversible. Energy: wT = h3 - h4; Entropy: s 4 = s3 Since the exit state is given we can relate that to the inlet state from entropy. 4: 15 kPa, x4 = 0.95 => s4 = 7.6458 kJ/kg K, h4 = 2480.4 kJ/kg 3: s3 = s4, P3 ⇒ h3 = 4036.7 kJ/kg, T3 = 758°C wT = h3 - h4 = 4036.7 - 2480.4 = 1556.3 kJ/kg . . m = WT/wT = 7.5 × 1000/1556.3 = 4.82 kg/s P T 3 2 1 3 4 v 2 1 4 s 11.26 A supply of geothermal hot water is to be used as the energy source in an ideal Rankine cycle, with R-134a as the cycle working fluid. Saturated vapor R-134a leaves the boiler at a temperature of 85°C, and the condenser temperature is 40°C. Calculate the thermal efficiency of this cycle. Solution: CV: Pump (use R-134a Table B.5) 2 ⌠ wP = h2 - h1 = ⌡ vdP ≈ v1(P2-P1) 1 = 0.000873(2926.2 - 1017.0) = 1.67 kJ/kg h2 = h1 + wP = 256.54 + 1.67 = 258.21 kJ/kg CV: Boiler qH = h3 - h2 = 428.10 - 258.21 = 169.89 kJ/kg CV: Turbine s4 = s3 = 1.6782 = 1.1909 + x4 × 0.5214 => x4 = 0.9346 h4 = 256.54 + 0.9346 × 163.28 = 409.14 kJ/kg Energy Eq.: wT = h3 - h4 = 428.1 - 409.14 = 18.96 kJ/kg wNET = wT - wP = 18.96 - 1.67 = 17.29 kJ/kg ηTH = wNET/qH = 17.29/169.89 = 0.102 3 QH WT T 2 WP, in 1 4 . QL 3 2 1 4 s 11.27 Do Problem 11.26 with R-22 as the working fluid. A supply of geothermal hot water is to be used as the energy source in an ideal Rankine cycle, with R-134a as the cycle working fluid. Saturated vapor R-134a leaves the boiler at a temperature of 85°C, and the condenser temperature is 40°C. Calculate the thermal efficiency of this cycle. Solution: CV: Pump (use R-22 Table B.4) 2 wP = h2 - h1 = ⌠ vdP ≈ v1(P2-P1) = 0.000884(4037 - 1534) = 2.21 kJ/kg ⌡ 1 h2 = h1 + wP = 94.27 + 2.21 = 96.48 kJ/kg CV: Boiler: qH = h3 - h2 = 253.69 - 96.48 = 157.21 kJ/kg CV: Turbine s4 = s3 = 0.7918 = 0.3417 + x4 × 0.5329, => x4 = 0.8446 h4 = 94.27 + 0.8446 × 166.88 = 235.22 wT = h3 - h4 = 253.69 - 235.22 = 18.47 kJ/kg ηTH = wNET/qH = (18.47 - 2.21)/157.21 = 0.1034 3 QH WT T 2 WP, in 1 4 . QL 3 2 1 4 s 11.28 Do Problem 11.26 with ammonia as the working fluid. A supply of geothermal hot water is to be used as the energy source in an ideal Rankine cycle, with R-134a as the cycle working fluid. Saturated vapor R-134a leaves the boiler at a temperature of 85°C, and the condenser temperature is 40°C. Calculate the thermal efficiency of this cycle. Solution: CV: Pump (use Ammonia Table B.2) wP = h2 - h1 = ⌠2 vdP = v1(P2-P1) ⌡1 = 0.001725(4608.6 - 1554.9) = 5.27 kJ/kg h2 = h1 + wP = 371.43 + 5.27 = 376.7 kJ/kg CV: Boiler qH = h3 - h2 = 1447.8 - 376.7 = 1071.1 kJ/kg CV: Turbine s4 = s3 = 4.3901 = 1.3574 + x4 × 3.5088 => x4 = 0.8643 h4 = 371.43 + 0.8643 × 1098.8 = 1321.13 kJ/kg Energy Eq.: wT = h3 - h4 = 1447.8 - 1321.13 = 126.67 kJ/kg wNET = wT - wP = 126.67 - 5.27 = 121.4 kJ/kg ηTH = wNET/qH = 121.4/1071.1 = 0.113 3 QH WT T 2 WP, in 1 4 . QL 3 2 1 4 s 11.29 Consider the boiler in Problem 11.26 where the geothermal hot water brings the R-134a to saturated vapor. Assume a counter flowing heat exchanger arrangement. The geothermal water temperature should be equal to or greater than the R-134a temperature at any location inside the heat exchanger. The point with the smallest temperature difference between the source and the working fluid is called the pinch point. If 2 kg/s of geothermal water is available at 95°C, what is the maximum power output of this cycle for R-134a as the working fluid? (hint: split the heat exchanger C.V. into two so the pinch point with ∆T = 0, T = 85°C appears). 2 kg/s of water is available at 95 oC for the boiler. The restrictive factor is the boiling temperature of 85° C. Therefore, break the process up from 2-3 into two parts as shown in the diagram. liquid 2 LIQUID HEATER R-134a liquid H2O out R-134a BOILER sat liq . -QBC at 85 oC C 3 sat. vap D 85o C . -QAB B liq H2O at 85 oC A liquid H2O 95 oC Write the energy equation for the first section A-B and D-3: . . -QAB = mH2O(hA - hB) = 2(397.94 - 355.88) = 84.12 kW . . = mR134A(428.1 - 332.65) ⇒ mR134A = 0.8813 kg/s To be sure that the boiling temp. is the restrictive factor, calculate TC from the energy equation for the remaining section: . -QAC = 0.8813(332.65 - 258.21) = 65.60 kW = 2(355.88 - hC) ⇒ hC = 323.1 kJ/kg, TC = 77.2°C > T2 OK CV Pump: CV: Turbine: wP = v1(P2-P1) = 0.000873(2926.2 - 1017.0) = 1.67 kJ/kg s4 = s3 = 1.6782 = 1.1909 + x4 × 0.5214 => x4 = 0.9346 h4 = 256.54 + 0.9346 × 163.28 = 409.14 kJ/kg Energy Eq.: wT = h3 - h4 = 428.1 - 409.14 = 18.96 kJ/kg Cycle: wNET = wT - wP = 18.96 - 1.67 = 17.29 kJ/kg . . WNET = mR134AwNET = 0.8813 × 17.29 = 15.24 kW 11.30 Do the previous problem with R-22 as the working fluid. A flow with 2 kg/s of water is available at 95oC for the boiler. The restrictive factor is the boiling temperature of 85oC. Therefore, break the process up from 23 into two parts as shown in the diagram. 2 liquid LIQUID HEATER R-22 3 sat. vap D sat liq . -QBC at 85 oC C liquid H2O out R-22 BOILER 85o C . -QAB B A liquid H2O 95 oC liq H2O at 85 oC . . -QAB = mH2O(hA - hB) = 2(397.94 - 355.88) = 84.12 kW . . = mR-22(253.69 - 165.09) ⇒ mR-22 = 0.949 kg/s To verify that TD = T3 is the restrictive factor, find TC. . -QAC = 0.949(165.09 - 96.48) = 65.11 = 2.0(355.88 - hC) hC = 323.32 kJ/kg ⇒ TC = 77.2oC OK State 1: 40oC, 1533.5 kPa, v1 = 0.000884 m3/kg CV Pump: wP = v1(P2 -P1) = 0.000884(4036.8 - 1533.5) = 2.21 kJ/kg CV: Turbine s4 = s3 = 0.7918 = 0.3417 + x4 × 0.5329 => x4 = 0.8446 h4 = 94.27 + 0.8446 × 166.88 = 235.22 kJ/kg Energy Eq.: Cycle: wT = h3 - h4 = 253.69 - 235.22 = 18.47 kJ/kg wNET = wT - wP = 18.47 - 2.21 = 16.26 kJ/kg . . WNET = mR22wNET = 0.949 × 16.26 = 15.43 kW 11.31 Consider the ammonia Rankine-cycle power plant shown in Fig. P11.31. The plant was designed to operate in a location where the ocean water temperature is 25°C near the surface and 5°C at some greater depth. The mass flow rate of the working fluid is 1000 kg/s. a. Determine the turbine power output and the pump power input for the cycle. b. Determine the mass flow rate of water through each heat exchanger. c. What is the thermal efficiency of this power plant? Solution: a) C.V. Turbine. Assume reversible and adiabatic. s2 = s1 = 5.0863 = 0.8779 + x2 × 4.3269 => x2 = 0.9726 h2 = 227.08 + 0.9726 × 1225.09 = 1418.6 kJ/kg wT = h1 - h2 = 1460.29 - 1418.6 = 41.69 kJ/kg . . WT = mwT = 1000 × 41.69 = 41 690 kW Pump: wP ≈ v3(P4 - P3) = 0.0016(857 - 615) = 0.387 kJ/kg . . WP = mwP = 1000 × 0.387 = 387 kW b) Consider to condenser heat transfer to the low T water . Qto low T H2O = 1000(1418.6 - 227.08) = 1.1915×106 kW . 1.1915×106 mlow T H2O = = 141 850 kg/s 29.38 - 20.98 h4 = h3 + wP = 227.08 + 0.39 = 227.47 kJ/kg Now consider the boiler heat transfer from the high T water . Qfrom high T H2O = 1000(1460.29 - 227.47) = 1.2328×106 kW c) . 1.2328×106 mhigh T H2O = = 147 290 kg/s 104.87 - 96.50 . . 41 690 - 387 ηTH = WNET/QH = = 0.033 1.2328×106 T 1 QH WT 4 WP, in 3 2. QL 1 4 3 2 s 11.32 A smaller power plant produces 25 kg/s steam at 3 MPa, 600oC in the boiler. It cools the condenser with ocean water coming in at 12oC and returned at 15oC so the condenser exit is at 45oC. Find the net power output and the required mass flow rate of ocean water. Solution: The states properties from Tables B.1.1 and B.1.3 1: 45oC, x = 0: h1 = 188.42 kJ/kg, v1 = 0.00101 m3/kg, Psat = 9.59 kPa 3: 3.0 MPa, 600oC: h3 = 3682.34 kJ/kg, s3 = 7.5084 kJ/kg K C.V. Pump Reversible and adiabatic. Energy: wp = h2 - h1 ; Entropy: s 2 = s1 since incompressible it is easier to find work (positive in) as wp = ∫ v dP = v1 (P2 - P1) = 0.00101 (3000 - 9.6) = 3.02 kJ/kg C.V. Turbine : wT = h3 - h4 ; s4 = s3 s4 = s3 = 7.5084 = 0.6386 + x4 (7.5261) => x4 = 0.9128 => h4 = 188.42 + 0.9128 (2394.77) = 2374.4 kJ/kg wT = 3682.34 – 2374.4 = 1307.94 kJ/kg . . WNET = m(wT – wp) = 25 (1307.94 – 3.02) = 32.6 MW C.V. Condenser : qL = h4 - h1 = 2374.4 - 188.42 = 2186 kJ/kg . . . QL = mqL = 25 × 2186 = 54.65 MW = mocean Cp ∆T . . mocean = QL / Cp ∆T = 54 650 / (4.18 × 3) = 4358 kg/s Boiler Turbine T 3 3 QB 2 WP 1 2 WT 4 Condenser 1 Q 4 s 11.33 The power plant in Problem 11.21 is modified to have a super heater section following the boiler so the steam leaves the super heater at 3.0 MPa, 400°C. Find the specific work and heat transfer in each of the ideal components and the cycle efficiency. Solution: C.V. Tubine: Energy: wT,s = h3 - h4; Entropy: ⇒ x4 = s4 = s3 = 6.9211 kJ/kg K s4 - sf 6.9211 - 0.6492 = = 0.83614 ; sfg 7.501 h4 = 191.81 + 0.83614 × 2392.82 = 2192.5 kJ/kg wT,s = 3230.82 - 2192.5 = 1038.3 kJ/kg ⌠ C.V. Pump: wP = ⌡v dP = v1(P2 - P1) = 0.00101(3000 - 10) = 3.02 kJ/kg ⇒ h2 = h1 + wP = 191.81 + 3.02 = 194.83 kJ/kg C.V. Condenser: qC = h4 - h1 = 2192.5 - 191.81 = 2000.7 kJ/kg C.V. Boiler: qH = h3 - h2 = 3230.82 – 194.83 = 3036 kJ/kg ηCYCLE = wNET/qH = 1038.3 – 3.02 = 0.341 3036 P T 3 3 2 2 1 4 v 1 4 s 11.34 A steam power plant has a steam generator exit at 4 MPa, 500°C and a condenser exit temperature of 45°C. Assume all components are ideal and find the cycle efficiency and the specific work and heat transfer in the components. Solution: From the Rankine cycle we have the states: 1: 45°C x = 0 , v1 = 0.00101 m3/kg, h1 = 188.45 kJ/kg / 3: 4 MPa, 500°C , h3 = 3445.3 kJ/kg, s3 = 7.0901 kJ/kg K C.V. Turbine: s4 = s3 ⇒ x4 = (7.0901 - 0.6386)/7.5261 = 0.8572, h4 = 188.42 + 0.8572 × 2394.77 = 2241.3 wT = h3 - h4 = 3445.3 - 2241.3 = 1204 kJ/kg C.V. Pump: wP = v1(P2 - P1) = 0.00101(4000 - 9.6) = 4.03 kJ/kg wP = h2 - h1 ⇒ h2 = 188.42 + 4.03 = 192.45 kJ/kg C.V. Boiler: qH = h3 - h2 = 3445.3 - 192.45 = 3252.8 kJ/kg C.V. Condenser: qL,out = h4 - h1 = 2241.3 - 188.42 = 2052.9 kJ/kg ηTH = wnet/qH = (wT + wP)/qH = (1204 - 4.03)/3252.8 = 0.369 Boiler Turbine T 3 3 QB 2 WP 1 2 WT 4 Condenser 1 Q 4 s 11.35 Consider an ideal Rankine cycle using water with a high-pressure side of the cycle at a supercritical pressure. Such a cycle has a potential advantage of minimizing local temperature differences between the fluids in the steam generator, such as the instance in which the high-temperature energy source is the hot exhaust gas from a gas-turbine engine. Calculate the thermal efficiency of the cycle if the state entering the turbine is 30 MPa, 550°C, and the condenser pressure is 5 kPa. What is the steam quality at the turbine exit? Solution: For the efficiency we need the net work and steam generator heat transfer. C.V. Pump. For this high exit pressure we use Table B.1.4 State 1: s1 = 0.4764 kJ/kg K, h1 = 137.82 kJ/kg Entropy Eq.: s2 = s1 => h2 = 168.36 kJ/kg wp = h2 - h1 = 30.54 kJ/kg C.V. Turbine. Assume reversible and adiabatic. Entropy Eq.: s4 = s3 = 6.0342 = 0.4764 + x4 × 7.9187 x4 = 0.70186 Very low for a turbine exhaust h4 = 137.79 + x4 × 2423.66 = 1838.86 , h3 = 3275.36 kJ/kg wT = h3 - h4 = 1436.5 kJ/kg Steam generator: qH = h3 - h2 = 3107 kJ/kg wNET = wT − wp = 1436.5 – 30.54 = 1406 kJ/kg η = wNET/qH = 1406 / 3107 = 0.45 P 2 T 3 30 MPa 3 2 1 4 v 5 kPa 1 4 s Reheat Cycles 11.36 A smaller power plant produces steam at 3 MPa, 600oC in the boiler. It keeps the condenser at 45oC by transfer of 10 MW out as heat transfer. The first turbine section expands to 500 kPa and then flow is reheated followed by the expansion in the low pressure turbine. Find the reheat temperature so the turbine output is saturated vapor. For this reheat find the total turbine power output and the boiler heat transfer. Boiler Turbine 3 cb T QH 4 4 2 WP WT 5 2 1 6 Condenser 1 3 MPa 5 3 9.59 kPa 6 QL s The states properties from Tables B.1.1 and B.1.3 1: 45oC, x = 0: h1 = 188.42 kJ/kg, v1 = 0.00101 m3/kg, Psat = 9.59 kPa 3: 3.0 MPa, 600oC: h3 = 3682.34 kJ/kg, s3 = 7.5084 kJ/kg K 6: 45oC, x = 1: h6 = 2583.19 kJ/kg, s6 = 8.1647 kJ/kg K C.V. Pump Reversible and adiabatic. Energy: wp = h2 - h1 ; Entropy: s 2 = s1 since incompressible it is easier to find work (positive in) as wp = ∫ v dP = v1 (P2 - P1) = 0.00101 (3000 - 9.59) = 3.02 kJ/kg h2 = h1 + wp = 188.42 + 3.02 = 191.44 kJ/kg C.V. HP Turbine section Entropy Eq.: => h4 = 3093.26 kJ/kg; T4 = 314oC s 4 = s3 C.V. LP Turbine section Entropy Eq.: s6 = s5 = 8.1647 kJ/kg K => State 5: 500 kPa, s5 C.V. Condenser. => state 5 h5 = 3547.55 kJ/kg, T5 = 529oC Energy Eq.: qL = h6 – h1 = hfg = 2394.77 kJ/kg . . m = QL / qL = 10 000 / 2394.77 = 4.176 kg/s Both turbine sections . . . WT,tot = mwT,tot = m(h3 - h4 + h5 - h6) = 4.176 (3682.34 - 3093.26 +3547.55 – 2583.19) = 6487 kW Both boiler sections . . QH = m(h3 - h2 + h5 - h4) = 4.176 (3682.34 – 191.44 + 3547.55 - 3093.26) = 16 475 kW 11.37 Consider an ideal steam reheat cycle where steam enters the high-pressure turbine at 3.0 MPa, 400°C, and then expands to 0.8 MPa. It is then reheated to 400°C and expands to 10 kPa in the low-pressure turbine. Calculate the cycle thermal efficiency and the moisture content of the steam leaving the low-pressure turbine. Solution: C.V. Pump reversible, adiabatic and assume incompressible flow wP = v1(P2 - P1) = 0.00101(3000 - 10) = 3.02 kJ/kg, h2 = 191.81 + 3.02 = 194.83 kJ/kg Boiler Turbine 3 cb 3 MPa T 3 QH 4 2 WP WT 5 1 4 2 1 6 Condenser 5 10 kPa 6 QL s C.V. HP Turbine section P3 = 3 MPa, T3 = 400oC => h3 = 3230.82 kJ/kg, s3 = 6.9211 kJ/kg K s4 = s3 => h4 = 2891.6 kJ/kg; C.V. LP Turbine section State 5: 400oC, 0.8 MPa => h5 = 3267.1 kJ/kg, s5 = 7.5715 kJ/kg K Entropy Eq.: s6 = s5 = 7.5715 kJ/kg K => x6 = two-phase state s6 - sf 7.5715 - 0.6492 = = 0.92285 = 0.923 sfg 7.501 h6 = 191.81 + 0.92285 × 2392.82 = 2400 kJ/kg wT,tot = h3 - h4 + h5 - h6 = 3230.82 - 2891.6+3267.1 - 2400 = 1237.8 kJ/kg qH1 = h3 - h2 = 3230.82 - 194.83 = 3036 kJ/kg qH = qH1 + h5 - h4 = 3036 + 3267.1 - 2891.6 = 3411.5 kJ/kg ηCYCLE = (1237.8 - 3.02)/3411.5 = 0.362 11.38 A smaller power plant produces 25 kg/s steam at 3 MPa, 600oC in the boiler. It cools the condenser with ocean water so the condenser exit is at 45oC. There is a reheat done at 500 kPa up to 400oC and then expansion in the low pressure turbine. Find the net power output and the total heat transfer in the boiler. Solution: The states properties from Tables B.1.1 and B.1.3 1: 45oC, x = 0: h1 = 188.42 kJ/kg, v1 = 0.00101 m3/kg, Psat = 9.59 kPa 3: 3.0 MPa, 600oC: 5: 500 kPa, 400oC: h3 = 3682.34 kJ/kg, h5 = 3271.83 kJ/kg, s3 = 7.5084 kJ/kg K s5 = 7.7937 kJ/kg K C.V. Pump Reversible and adiabatic. Incompressible flow so Energy: wp = h2 - h1 = v1(P2 - P1) = 0.00101 (3000 - 9.6) = 3.02 kJ/kg C.V. LP Turbine section Entropy Eq.: s6 = s5 = 7.7937 kJ/kg K => x6 = (s6 - sf)/sfg = two-phase state 7.7937 - 0.6386 = 0.9507 7.5261 h6 = 188.42 + 0.9507 × 2394.77 = 2465.1 kJ/kg Both turbine sections wT,tot = h3 - h4 + h5 - h6 = 3682.34 - 3093.26 + 3271.83 – 2465.1 = 1395.81 kJ/kg . . . . Wnet = WT - Wp = m(wT,tot – wp) = 25 (1395.81 – 3.02) = 34 820 kW Both boiler sections . . QH = m(h3 - h2 + h5 - h4) = 25 (3682.34 – 191.44 + 3271.83 - 3093.26) = 91 737 kW Boiler Turbine 3 cb QH T WP 1 3 MPa 5 4 4 2 3 WT 5 2 1 6 Condenser QL 9.59 kPa 6 s 11.39 The reheat pressure effect the operating variables and thus turbine performance. Repeat Problem 11.37 twice, using 0.6 and 1.0 MPa for the reheat pressure. Solution Boiler Turbine 3 cb 3 MPa T 3 QH 4 2 WP WT 5 1 4 Condenser 10 kPa 2 1 6 5 6 QL s C.V. Pump reversible, adiabatic and assume incompressible flow wP = v1(P2 - P1) = 0.00101(3000 - 10) = 3.02 kJ/kg, h2 = h1 + wP = 191.81 + 3.02 = 194.83 kJ/kg State 3: 3 MPa, 400oC => h3 = 3230.82 kJ/kg, s3 = 6.9211 kJ/kg K Low T boiler section: qH1 = h3 - h2 = 3230.82 - 194.83 = 3035.99 kJ/kg State 4: P4, s4 = s3 For P4 = 1 MPa: h4 = 2940.85 kJ/kg state 4 is sup. vapor State 5: 400oC, P5 = P4 => h5 = 3263.9 kJ/kg, s5 = 7.465 kJ/kg K, For P4 = 0.6 MPa: h4 = 2793.2 kJ/kg state 4 is sup. vapor State 5: 400oC, P5 = P4 State 6: 10 kPa, s 6 = s5 => h5 = 3270.3 kJ/kg, s5 = 7.7078 kJ/kg K, => x6 = (s6 - sf)/sfg Total turbine work: wT,tot = h3- h4 + h5 - h6 Total boiler H.Tr.: qH = qH1 + h5 - h4 Cycle efficiency: ηCYCLE = (wT,tot – wP)/qH x6 P4=P5 1 0.9087 0.6 0.9410 h6 2366 2443.5 wT 1187.9 1228.0 Notice the very small changes in efficiency. qH 3359.0 3437.7 ηCYCLE 0.3527 0.3563 11.40 The effect of a number of reheat stages on the ideal steam reheat cycle is to be studied. Repeat Problem 11.37 using two reheat stages, one stage at 1.2 MPa and the second at 0.2 MPa, instead of the single reheat stage at 0.8 MPa. C.V. Pump reversible, adiabatic and assume incompressible flow, work in wP = v1(P2 - P1) = 0.00101(3000 - 10) = 3.02 kJ/kg, h2 = h1 + wP = 191.81 + 3.02 = 194.83 kJ/kg P4 = P5 = 1.2 MPa, P6 = P7 = 0.2 MPa 3 MPa T 357 3: h3 = 3230.82 kJ/kg, s3 = 6.9211 kJ/kg K 4: P4, s4 = s3 ⇒ sup. vap. h4 = 2985.3 5: h5 = 3260.7 kJ/kg, s5 = 7.3773 kJ/kg K 6: P6, s6 = s5 ⇒ sup. vapor 4 2 1 o 400 C 10 kPa 6 h6 = 2811.2 kJ/kg 8 s 7: h7 = 3276.5 kJ/kg, s7 = 8.2217 kJ/kg K 8: P8, s8 = s7 ⇒ sup. vapor h8 = 2607.9 kJ/kg Total turbine work, same flow rate through all sections wT = (h3 - h4) + (h5 - h6) + (h7 - h8) = 245.5 + 449.5 + 668.6 = 1363.6 kJ/kg Total heat transfer in boiler, same flow rate through all sections qH = (h3 - h2) + (h5 - h4) + (h7 - h6) = 3036 + 319.8 + 465.3 = 3821.1 kJ/kg Cycle efficiency: ηTH = wT - wP 1363.6 - 3.02 = = 0.356 qH 3821.1 Open Feedwater Heaters 11.41 An open feedwater heater in a regenerative steam power cycle receives 20 kg/s of water at 100°C, 2 MPa. The extraction steam from the turbine enters the heater at 2 MPa, 275°C, and all the feedwater leaves as saturated liquid. What is the required mass flow rate of the extraction steam? Solution: From turbine 6 The complete diagram is as in Feedwater to P2 Feedwater Figure 11.8 in main text. from P1 heater 2 3 C.V Feedwater heater Continuity Eq.: Energy Eq.: . . . m 2 + m6 = m3 . . . . . m2h2 + m6h6 = m3h3 = (m2 + m6) h3 Table B.1.4: h2 = 420.45 kJ/kg, Table B.1.2: h3 = 908.77 kJ/kg Table B.1.3: h6 = 2963 kJ/kg, this is interpolated With the values substituted into the energy equation we get . . h3 - h2 908.77 - 420.45 m6 = m2 = 20 × = 4.754 kg/s 2963 - 908.77 h6 - h3 Remark: For lower pressures at state 2 where Table B.1.4 may not have an entry the corresponding saturated liquid at same T from Table B.1.1 is used. 11.42 A power plant with one open feedwater heater has a condenser temperature of 45°C, a maximum pressure of 5 MPa, and boiler exit temperature of 900°C. Extraction steam at 1 MPa to the feedwater heater is mixed with the feedwater line so the exit is saturated liquid into the second pump. Find the fraction of extraction steam flow and the two specific pump work inputs. Solution: From turbine 6 Pump 1 1 To boiler The complete diagram is as in Figure 11.8 in the main text. From 4 3 FWH condenser 2 Pump 2 State out of boiler 5: h5 = 4378.82 kJ/kg, s5 = 7.9593 kJ/kg K C.V. Turbine reversible, adiabatic: s7 = s6 = s5 State 6: P6 , s6 => h6 = 3640.6 kJ/kg, T6 = 574oC C.V Pump P1 wP1 = h2 - h1 = v1(P2 - P1) = 0.00101(1000 - 9.6) = 1.0 kJ/kg => h2 = h1 + wP1 = 188.42 + 1.0 = 189.42 kJ/kg C.V. Feedwater heater: Energy Eq.: x= Call . . m6 / mtot = x (the extraction fraction) (1 - x) h2 + x h6 = 1 h3 h3 - h2 762.79 - 189.42 = = 0.1661 h6 - h2 3640.6 - 189.42 C.V Pump P2 wP2 = h4 - h3 = v3(P4 - P3) = 0.001127(5000 - 1000) = 4.5 kJ/kg 11.43 A Rankine cycle operating with ammonia is heated by some low temperature source so the highest T is 120oC at a pressure of 5000 kPa. Its low pressure is 1003 kPa and it operates with one open feedwater heater at 2033 kPa. The total flow rate is 5 kg/s. Find the extraction flow rate to the feedwater heater assuming its outlet state is saturated liquid at 2033 kPa. Find the total power to the two pumps. 5 MPa T 5 5 TURBINE HP LP STEAM GEN 4 6 7 FWH 3 4 COND. 2 P2 2.03 MPa 6 23 1 1 MPa 7 s cb P1 1 State 1: x1 = 0, h1 = 298.25 kJ/kg, v1 = 0.001658 m3/kg State 3: x3 = 0, h3 = 421.48 kJ/kg, v3 = 0.001777 m3/kg State 5: h5 = 421.48 kJ/kg, s5 = 4.7306 kJ/kg K State 6: s6 = s5 => x6 = (s6 – sf)/sfg = 0.99052, h6 = 1461.53 kJ/kg C.V Pump P1 wP1 = h2 - h1 = v1(P2 - P1) = 0.001658(2033 - 1003) = 1.708 kJ/kg => h2 = h1 + wP1 = 298.25 + 1.708 = 299.96 kJ/kg . . C.V. Feedwater heater: Call m6 / mtot = x (the extraction fraction) Energy Eq.: (1 - x) h2 + x h6 = 1 h3 h3 - h2 762.79 - 189.42 = = 0.1046 h6 - h2 3640.6 - 189.42 . . mextr = x mtot = 0.1046 × 5 = 0.523 kg/s . . m1 = (1-x) mtot = (1 – 0.1046) 5 = 4.477 kg/s x= C.V Pump P2 wP2 = h4 - h3 = v3(P4 - P3) = 0.001777(5000 - 2033) = 5.272 kJ/kg Total pump work . . . Wp = m1wP1 + mtot wP2 = 4.477 × 1.708 + 5 × 5.272 = 34 kW 11.44 A steam power plant operates with a boiler output of 20 kg/s steam at 2 MPa, 600°C. The condenser operates at 50°C dumping energy to a river that has an average temperature of 20°C. There is one open feedwater heater with extraction from the turbine at 600 kPa and its exit is saturated liquid. Find the mass flow rate of the extraction flow. If the river water should not be heated more than 5°C how much water should be pumped from the river to the heat exchanger (condenser)? Solution: The setup is as shown in Fig. 11.10. Condenser 1: 50oC sat liq. v1 = 0.001012 m3/kg, h1 = 209.31 kJ/kg 2: 600 kPa s2 = s 1 3: 600 kPa, sat liq. 7 Ex turbine To river h3 = hf = 670.54 kJ/kg river 5: (P, T) h5 = 3690.1 kJ/kg, To pump 1 s5 = 7.7023 kJ/kg K 6: 600 kPa, s6 = s5 From 1 => h6 = 3270.0 kJ/kg CV P1 wP1 = v1(P2 - P1) = 0.001012 (600 - 12.35) = 0.595 kJ/kg h2 = h1 + wP1 = 209.9 kJ/kg C.V FWH x h6 + (1 -x) h2 = h3 x= h3 - h2 670.54 - 209.9 = = 0.1505 h6 - h2 3270.0 - 209.9 . . m6 = x m5 = 0.1505 × 20 = 3 kg/s CV Turbine: s7 = s6 = s5 => x7 = 0.9493 , h7 = 2471.17 kJ/kg CV Condenser qL = h7 - h1 = 2471.17 - 209.31 = 2261.86 kJ/kg The heat transfer out of the water from 7 to 1 goes into the river water . . QL = (1 - x) mqL = 0.85 × 20 × 2261.86 = 38 429 kW . . . = mH2O ∆hH2O = mH2O (hf25 - hf20) = m (20.93) . m = 38 429 / 20.93 = 1836 kg/s 11.45 Consider an ideal steam regenerative cycle in which steam enters the turbine at 3.0 MPa, 400°C, and exhausts to the condenser at 10 kPa. Steam is extracted from the turbine at 0.8 MPa for an open feedwater heater. The feedwater leaves the heater as saturated liquid. The appropriate pumps are used for the water leaving the condenser and the feedwater heater. Calculate the thermal efficiency of the cycle and the net work per kilogram of steam. Solution: This is a standard Rankine cycle with an open FWH as shown in Fig.11.10 C.V Pump P1 wP1 = h2 - h1 = v1(P2 - P1) = 0.00101(800 - 10) = 0.798 kJ/kg => h2 = h1 + wP1 = 191.81 + 0.798 = 192.61 kJ/kg . . Call m6 / mtot = x (the extraction fraction) C.V. FWH (1 - x) h2 + x h6 = 1 h3 x= h3 - h2 721.1 - 192.61 = = 0.1958 h6 - h2 2891.6 - 192.61 C.V Pump P2 wP2 = h4 - h3 = v3(P4 - P3) = 0.001115(3000 - 800) = 2.45 kJ/kg h4 = h3 + wP2 = 721.1 + 2.45 = 723.55 kJ/kg CV Boiler: qH = h5 - h4 = 3230.82 - 723.55 = 2507.3 kJ/kg CV Turbine 2nd Law s7 = s6 = s5 = 6.9211 kJ/kg K P6 , s6 => h6 = 2891.6 kJ/kg (superheated vapor) s7 = s6 = s5 = 6.9211 => => x7 = 6.9211 - 0.6492 = 0.83614 7.501 h7 = 191.81 + x7 2392.82 = 2192.55 kJ/kg Turbine has full flow in HP section and fraction 1-x in LP section . . WT / m5 = h5 - h6 + (1 - x) (h6 - h7) wT = 3230.82 – 2891.6 + (1 - 0.1988) ( 2891.6 – 2192.55) = 899.3 P2 has the full flow and P1 has the fraction 1-x of the flow wnet = wT - (1 - x) wP1 - wP2 = 899.3 - (1 - 0.1988)0.798 – 2.45 = 896.2 kJ/kg ηcycle = wnet / qH = 896.2 / 2507.3 = 0.357 11.46 In one type of nuclear power plant, heat is transferred in the nuclear reactor to liquid sodium. The liquid sodium is then pumped through a heat exchanger where heat is transferred to boiling water. Saturated vapor steam at 5 MPa exits this heat exchanger and is then superheated to 600°C in an external gas-fired superheater. The steam enters the turbine, which has one (open-type) feedwater extraction at 0.4 MPa. The isentropic turbine efficiency is 87%, and the condenser pressure is 7.5 kPa. Determine the heat transfer in the reactor and in the superheater to produce a net power output of 1 MW. Solution: The complete cycle diagram is similar to Figure 11.8 except the boiler is sparated into a section heated by the reactor and a super heater section. T 6 SUPER HEATER Q 0.4 MPa TURBINE 8 REACTOR P2 COND. 2 P1 7 23 1 FWH 3 4 5 4 7 5 5 MPa 6 7.5 kPa 8 s 1 CV. Pump P1 wP1 = 0.001008(400 - 7.5) = 0.4 kJ/kg ; h2 = h1 + wP1 = 168.8 + 0.4 = 169.2 kJ/kg CV. Pump P2 wP2 = 0.001084(5000 - 400) = 5.0 kJ/kg h4 = h3 + wP2 = 604.7 + 5.0 = 609.7 kJ/kg C.V. Turbine (to get exit state properties) s7 = s6 = 7.2589, P7 = 0.4 MPa => T7 = 221.2oC, h7 = 2904.5 kJ/kg s8 = s6 = 7.2589 = 0.5764 + x8 × 7.6750 x8 = 0.8707 h8 = 168.8 + 0.8707 × 2406.0 = 2263.7 kJ/kg CV: Feedwater heater FWH (to get the extraction fraction x7) . . . . Divide the equations with the total mass flow rate m3 = m4 = m5 = m6 Continuity: x2 + x7 = x3 = 1.0 , Energy Eq.: x2h2 + x7h7 = h3 x7 = (604.7-169.2)/(2904.5-169.2) = 0.1592 CV: Turbine (to get the total specific work) Full flow from 6 to 7 and the fraction (1 - x7) from 7 to 8. wT = (h6 - h7) + (1 - x7)(h7 - h8) = 3666.5-2904.5 + 0.8408(2904.5-2263.7) = 1300.8 kJ/kg CV: Pumps (P1 has x1 = 1 - x7, P2 has the full flow x3 = 1) wP = x1wP1 + x3wP2 = 0.8408 × 0.4 + 1 × 5.0 = 5.3 kJ/kg . wNET = 1300.8 - 5.3 = 1295.5 => m = 1000/1295.5 = 0.772 kg/s CV: Reactor (this has the full flow) . . QREACT = m(h5 - h4) = 0.772(2794.3 - 609.7) = 1686 kW CV: Superheater (this has the full flow) . . QSUP = m(h6 - h5) = 0.772 (3666.5 - 2794.3) = 673 kW 11.47 A steam power plant has high and low pressures of 20 MPa and 10 kPa, and one open feedwater heater operating at 1 MPa with the exit as saturated liquid. The maximum temperature is 800°C and the turbine has a total power output of 5 MW. Find the fraction of the flow for extraction to the feedwater and the total condenser heat transfer rate. The physical components and the T-s diagram is as shown in Fig. 11.10 in the main text for one open feedwater heater. The same state numbering is used. From the Steam Tables: State 5: (P, T) h5 = 4069.8 kJ/kg, s5 = 7.0544 kJ/kg K, State 1: (P, x = 0) h1 = 191.81 kJ/kg, v1 = 0.00101 m3/kg State 3: (P, x = 0) h3 = 762.8 kJ/kg, v3 = 0.001127 m3/kg Pump P1: wP1 = v1(P2 - P1) = 0.00101 × 990 = 1 kJ/kg h2 = h1 + wP1 = 192.81 kJ/kg Turbine 5-6: s6 = s5 ⇒ h6 = 3013.7 kJ/kg wT56 = h5 - h6 = 4069.8 – 3013.7 = 1056.1 kJ/kg . . Feedwater Heater (mTOT = m5): ⇒ x= . . . xm5h6 + (1 - x)m5h2 = m5h3 h3 - h2 762.8 - 192.81 = = 0.2021 h6 - h2 3013.7 - 192.81 To get state 7 into condenser consider turbine. s7 = s6 = s5 ⇒ x7 = (7.0544 - 0.6493)/7.5009 = 0.85391 h7 = 191.81 + 0.85391 × 2392.82 = 2235.1 kJ/kg Find specific turbine work to get total flow rate . . . . WT = mTOTh5 - xmTOTh6 - (1 - x)mTOTh7 = . . = mTOT × (h5 - xh6 - (1 - x)h7) = mTOT × 1677.3 . mTOT = 5000/1677.3 = 2.98 kg/s . . QL = mTOT (1-x) (h7-h1) = 2.98 × 0.7979(2235.1 - 191.81) = 4858 kW Closed Feedwater Heaters 11.48 A closed feedwater heater in a regenerative steam power cycle heats 20 kg/s of water from 100°C, 20 MPa to 250°C, 20 MPa. The extraction steam from the turbine enters the heater at 4 MPa, 275°C, and leaves as saturated liquid. What is the required mass flow rate of the extraction steam? Solution: The schematic is from Figure 11.11 has the feedwater from the pump coming at state 2 being heated by the extraction flow coming from the turbine state 6 so the feedwater leaves as saturated liquid state 4 and the extraction flow leaves as condensate state 6a. 6 From table B.1 h kJ/kg h4 = 1086.75 h6 = 2886.2 B.1.2: 4 MPa, sat. liq. 6a h2 = 434.06 B.1.3: 4 MPa, 275°C 2 B.1.4: 100°C, 20 MPa B.1.4: 250°C, 20 MPa 4 h6a = 1087.31 C.V. Feedwater Heater Energy Eq.: . . . . m2h2 + m6h6 = m2h4 + m6h6a Since all four state are known we can solve for the extraction flow rate . h2 - h4 . = 7.257 kg/s m6 = m2 h6a - h6 11.49 A power plant with one closed feedwater heater has a condenser temperature of 45°C, a maximum pressure of 5 MPa, and boiler exit temperature of 900°C. Extraction steam at 1 MPa to the feedwater heater condenses and is pumped up to the 5 MPa feedwater line where all the water goes to the boiler at 200°C. Find the fraction of extraction steam flow and the two specific pump work inputs. Solution: s1 = 0.6387 kJ/kg K, h1 = 188.45 kJ/kg v1 = 0.00101 m3/kg, s4 = 2.1387 kJ/kg K, h4 = 762.81 kJ/kg From turbine 6 3 Pump 1 5 2 1 From condenser 7 T6 => h6 = 853.9 kJ/kg Pump 2 4 C.V. Turbine: Reversible, adiabatic so constant s from inlet to extraction point s3 = sIN = 7.9593 kJ/kg K => T3 = 573.8, h3 = 3640.6 kJ/kg C.V. P1: wP1 = v1(P2 - P1) = 5.04 kJ/kg ⇒ h2 = h1 + wP1 = 193.49 kJ/kg C.V. P2: wP2 = v4(P7 - P4) = 4.508 kJ/kg ⇒ h7 = h4 + wP2 = 767.31 kJ/kg C.V. Total FWH and pumps: The extraction fraction is: Continuity Eq.: Energy: x= .. x = m3/m6 . . . m 6 = m1 + m3 , 1 = (1-x) + x (1 - x)(h1 + wP1) + x(h3 +wP2) = h6 h6 - h2 853.9 - 193.49 = = 0.1913 h3 + wP2 - h2 3640.6 + 4.508 - 193.49 .. m3/m6 = x = 0.1913 11.50 Repeat Problem 11.45, but assume a closed instead of an open feedwater heater. A single pump is used to pump the water leaving the condenser up to the boiler pressure of 3.0 MPa. Condensate from the feedwater heater is drained through a trap to the condenser. Solution: C.V. Turbine, 2nd law: s4 = s5 = s6 = 6.9211 kJ/kg K BOILER 4 3 h4 = 3230.82 , h5 = 2891.6 TURBINE FW HTR 5 => x6 = (6.9211 - 0.6492)/7.501 Trap = 0.83614 h6 = 191.81 + x6 2392.82 2 =2192.55 kJ/kg 6 7 1 COND. P Assume feedwater heater exit at the T of the condensing steam C.V Pump wP = h2 - h1 = v1(P2 - P1) = 0.00101(3000 - 10) = 3.02 kJ/kg h2 = h1 + wP = 191.81 + 3.02 = 194.83 kJ/kg T3 = Tsat (P5) = 170.43°C, h3 = hf = h7 = 721.1 kJ/kg C.V FWH . . m5 / m 3 = x , x= Energy Eq.: h 2 + x h5 = h 3 + h 7 x h3 - h2 721.1 - 194.83 = = 0.2425 h5 - hf 800 2891.6 - 721.1 Turbine work with full flow from 4 to 5 fraction 1-x flows from 5 to 6 wT = h4 - h5 + (1 - x)(h5 - h6) = 3230.82 – 2891.6 + 0.7575 (2891.6 - 2192.55) = 868.75 kJ/kg wnet = wT - wP = 868.75 - 3.02 = 865.7 kJ/kg qH = h4 - h3 = 3230.82 - 721.1 = 2509.7 kJ/kg ηcycle = wnet / qH = 865.7 / 2509.7 = 0.345 11.51 Do Problem 11.47 with a closed feedwater heater instead of an open and a drip pump to add the extraction flow to the feed water line at 20 MPa. Assume the temperature is 175°C after the drip pump flow is added to the line. One main pump brings the water to 20 MPa from the condenser. Solution: From v1 = 0.00101 m3/kg, From turbine 6 1 condenser h1 = 191.81 kJ/kg 3 4 T4 = 175oC; h4 = 751.66 kJ/kg Pump 1 2 h6a = hf 1MPa = 762.79 kJ/kg, 6b 3 v6a = 0.001127 m /kg 6a Pump 2 Turbine section 1: s6 = s5 = 7.0544 kJ/kg K P6 = 1 MPa => h6 = 3013.7 kJ/kg C.V Pump 1 wP1 = h2 - h1 = v1(P2 - P1) = 0.00101(20 000 - 10) = 20.19 kJ/kg => h2 = h1 + wP1 = 191.81 + 20.19 = 212.0 kJ/kg C.V Pump 2 wP2 = h6b - h6a = v6a(P6b - P6a) = 0.001127(20 000 - 1000) = 21.41 kJ/kg . . C.V FWH + P2 select the extraction fraction to be x = m6 / m4 x h6 + (1 - x) h2 + x (wP2) = h4 x= Turbine: h4 - h 2 751.66 - 212.0 = = 0.191 h6 - h2 - wP2 3013.7 - 212.0 + 21.41 s7 = s6 = s5 & P7 = 10 kPa => x 7 = 7.0544 - 0.6493 = 0.85391 7.5009 h7 = 191.81 + 0.85391 × 2392.82 = 2235.1 kJ/kg wT = [ h5 - h6 + (1 - x) (h6 - h7) ] = [ 4069.8 – 3013.7 + 0.809 (3013.7 - 2235.1)] = 1686 kJ/kg . . . . WT = 5000 kW = m5 × wT = m5 × 1686 kJ/kg => m5 = 2.966 kg/s . . QL = m5(1 - x) (h7 - h1) = 2.966 × 0.809 (2235.1 - 191.81) = 4903 kW 11.52 Assume the powerplant in Problem 11.43 has one closed feedwater heater instead of the open FWH. The extraction flow out of the FWH is saturated liquid at 2033 kPa being dumped into the condenser and the feedwater is heated to 50oC. Find the extraction flow rate and the total turbine power output. 5 MPa T 5 5 TURBINE HP LP STEAM GEN 6 7 6a FWH 3,4 COND. 2 6a 3,4 2 2.03 MPa 6 1 1 MPa 7 s cb P1 1 State 1: x1 = 0, h1 = 298.25 kJ/kg, v1 = 0.001658 m3/kg State 3: h3 = hf + (P3–Psat)vf = 421.48 + (5000–2033)0.001777 = 426.75 kJ/kg State 5: h5 = 421.48 kJ/kg, s5 = 4.7306 kJ/kg K State 6: s6 = s5 => State 6a: x6a = 0 => State 7: s7 = s5 => x6 = (s6 – sf)/sfg = 0.99052, h6 = 1461.53 kJ/kg h6a = 421.48 kJ/kg x7 = (s7 – sf)/sfg = 0.9236, h7 = 1374.43 kJ/kg C.V Pump P1 wP1 = h2 - h1 = v1(P2 - P1) = 0.001658(5000 - 1003) = 6.627 kJ/kg => h2 = h1 + wP1 = 298.25 + 6.627 = 304.88 kJ/kg C.V. Feedwater heater: Energy Eq.: Call . . m6 / mtot = x (the extraction fraction) h2 + x h6 = 1 h3 + x h6a h3 - h 2 426.75 - 304.88 = = 0.1172 h6 - h6a 1461.53 - 421.48 . . mextr = x mtot = 0.1172 × 5 = 0.586 kg/s x= Total turbine work . . . WT = mtot(h5 – h6) + (1 – x)mtot (h6 – h) = 5(1586.3 – 1461.53) + (5 – 0.586)(1461.53 – 1374.43) = 1008 kW Nonideal Cycles 11.53 Steam enters the turbine of a power plant at 5 MPa and 400°C, and exhausts to the condenser at 10 kPa. The turbine produces a power output of 20 000 kW with an isentropic efficiency of 85%. What is the mass flow rate of steam around the cycle and the rate of heat rejection in the condenser? Find the thermal efficiency of the power plant and how does this compare with a Carnot cycle. . Solution: WT = 20 000 kW and ηTs = 85 % State 3: State 1: h3 = 3195.6 kJ/kg , s3 = 6.6458 kJ/kgK P1 = P4 = 10 kPa , sat liq , x1 = 0 T1 = 45.8oC , h1 = hf = 191.8 kJ/kg , v1 = vf = 0.00101 m3/kg C.V Turbine : 1st Law: qT + h3 = h4 + wT ; qT = 0 wT = h3 - h4 , Assume Turbine is isentropic s4s = s3 = 6.6458 kJ/kgK , s4s = sf + x4s sfg , solve for x4s = 0.7994 h4s = hf + x4shfg = 1091.0 kJ/kg wTs = h3 - h4s = 1091 kJ/kg , wT = ηTswTs = 927.3 kJ/kg . . WT m= = 21.568 kg/s , wT C.V. Condenser: 1st Law : h4 = h3 - wT = 2268.3 kJ/kg h4 = h1 + qc + wc ; qc = h4 - h1 = 2076.5 kJ/kg , wc = 0 . . Qc = m qc = 44 786 kW C.V. Pump: Assume adiabatic, reversible and incompressible flow wps = ∫ v dP = v1(P2 - P1) = 5.04 kJ/kg 1st Law : C.V Boiler : 1st Law : h2 = h1 + wp = 196.8 kJ/kg qB + h2 = h3 + wB ; wB = 0 qB = h3 - h2 = 2998.8 kJ/kg wnet = wT - wP = 922.3 kJ/kg ηth = wnet / qB = 0.307 Carnot cycle : TH = T3 = 400oC , TL = T1 = 45.8oC ηth = TH - TL = 0.526 TH 11.54 A steam power plant has a high pressure of 5 MPa and maintains 50°C in the condenser. The boiler exit temperature is 600°C. All the components are ideal except the turbine which has an actual exit state of saturated vapor at 50°C. Find the cycle efficiency with the actual turbine and the turbine isentropic efficiency. Solution: A standard Rankine cycle with an actual non-ideal turbine. Boiler exit: h3 = 3666.5 kJ/kg, s3 = 7.2588 kJ/kg K 4s: 50°C, s = s3 => x = (7.2588 - 0.7037)/7.3725 = 0.88913, Ideal Turbine: h4s = 209.31 + 0.88913 × 2382.75 = 2327.88 kJ/kg => wTs = h3 - h4s = 1338.62 kJ/kg Condenser exit: h1 = 209.31 , Actual turbine: wTac = h3 - h4ac = 1074.4 kJ/kg Actual turbine exit: h4ac = hg = 2592.1 ηT = wTac / wTs = 0.803: Isentropic Efficiency Pump: wP = v1( P2 - P1) = 0.001012(5000-12.35) = 5.05 kJ/kg h2 = h1 + wP = 209.31 + 5.05 = 214.36 kJ/kg qH = h3 - h2 = 3666.5 - 214.36 = 3452.14 kJ/kg ηcycle = (wTac - wP) / qH = 0.31: Cycle Efficiency 11.55 A steam power cycle has a high pressure of 3.0 MPa and a condenser exit temperature of 45°C. The turbine efficiency is 85%, and other cycle components are ideal. If the boiler superheats to 800°C, find the cycle thermal efficiency. Solution: Basic Rankine cycle as shown in Figure 11.3 in the main text. C.V. Turbine: wT = h3 - h4, s4 = s3 + sT,GEN Ideal Table B.1.3: s4 = s3 = 7.9862 kJ/kg K => x4s = (7.9862 – 0.6386)/7.5261 = 0.9763 h4s = hf + x hfg = 188.42 + 0.9763 × 2394.77 = 2526.4 kJ/kg wTs = h3 - h4s = 4146 – 2526.4 = 1619.6 kJ/kg Actual: wT,AC = η × wT,S = 0.85 × 1619.6 = 1376.66 kJ/kg wP = ∫ v dP ≈ v1(P2 - P1) = 0.00101 (3000 - 9.6) = 3.02 kJ/kg C.V. Pump: h2 = h1 + wP = 188.42 + 3.02 = 191.44 kJ/kg C.V. Boiler: qH = h3 - h2 = 4146 – 191.44 = 3954.6 kJ/kg η = (wT,AC - wP)/qH = (1376.66 – 3.02)/3954.6 = 0.347 P T 3 3 2 2 4ac 1 4s v 4ac 1 4s s 11.56 A steam power plant operates with with a high pressure of 5 MPa and has a boiler exit temperature of of 600°C receiving heat from a 700°C source. The ambient at 20°C provides cooling for the condenser so it can maintain 45°C inside. All the components are ideal except for the turbine which has an exit state with a quality of 97%. Find the work and heat transfer in all components per kg water and the turbine isentropic efficiency. Find the rate of entropy generation per kg water in the boiler/heat source setup. Solution: Take CV around each component steady state in standard Rankine Cycle. 1: v = 0.00101; h = 188.42, s = 0.6386 (saturated liquid at 45°C). 3: h = 3666.5 kJ/kg, s = 7.2588 kJ/kg K superheated vapor 4ac: h = 188.42 + 0.97 × 2394.8 = 2511.4 kJ/kg CV Turbine: no heat transfer q = 0 wac = h3 - h4ac = 3666.5 - 2511.4 = 1155.1 kJ/kg Ideal turbine: s4 = s3 = 7.2588 => x4s = 0.88, h4s = 2295 kJ/kg ws = h3 - h4s = 3666.5 - 2295 = 1371.5 kJ/kg, Eff = wac / ws = 1155.1 / 1371.5 = 0.842 CV Condenser: no shaft work w = 0 qout = h4ac - h1 = 2511.4 - 188.42 = 2323 kJ/kg CV Pump: no heat transfer, q = 0 incompressible flow so v = constant w = v(P2- P1) = 0.00101(5000-9.59) = 5.04 kJ/kg CV Boiler: no shaft work, w = 0 qH = h3 - h2 = h3 - h1 - wP = 3666.5 - 188.42 -5.04 = 3473 kJ/kg s2 + (qH/ TH) + sGen = s3 and s2 = s1 (from pump analysis) sgen = 7.2588 - 0.6386 - 3473/(700+273) = 3.05 kJ/kg K 11.57 For the steam power plant described in Problem 11.21, assume the isentropic efficiencies of the turbine and pump are 85% and 80%, respectively. Find the component specific work and heat transfers and the cycle efficiency. Solution: This is a standard Rankine cycle with actual non-ideal turbine and pump. CV Pump, Rev & Adiabatic: wPs = h2s - h1 = v1(P2 - P1) = 0.00101(3000 - 10) = 3.02 kJ/kg; s2s = s1 wPac = wPs / ηP = 3.02/0.8 = 3.775 kJ/kg = h2a - h1 h2a = wPac + h1 = 3.775 + 191.81 = 195.58 kJ/kg CV Boiler: qH = h3 - h2a = 2804.14 – 195.58 = 2608.56 kJ/kg C.V. Turbine : wT = h3 - h4 ; s4 = s3 s4 = s3 = 6.1869 = 0.6492 + x4 (7.501) => x4 = 0.7383 => h4 = 191.81 + 0.7383 (2392.82) = 1958.34 kJ/kg wTs = 2804.14 - 1958.34 = 845.8 kJ/kg wTac = wTs × ηT = 718.9 = h3 - h4a h4a = h3 - wTac = 2804.14 - 718.9 = 2085.24 kJ/kg CV Condenser: qL = h4a - h1 = 2085.24 - 191.81 = 1893.4 kJ/kg ηcycle = (wTac - wPac) / qH = (718.9 – 3.78) / 2608.56 = 0.274 This compares to 0.32 for the ideal case. Boiler T Turbine 3 3 2 QB 2 WP 1 WT 1 4s 4ac 4 Condenser Q s state 2s and 2ac nearly the same 11.58 A small steam power plant has a boiler exit of 3 MPa, 400°C while it maintains 50 kPa in the condenser. All the components are ideal except the turbine which has an isentropic efficiency of 80% and it should deliver a shaft power of 9.0 MW to an electric generator. Find the specific turbine work , the needed flow rate of steam and the cycle efficiency. Solution: This is a standard Rankine cycle with an actual non-ideal turbine. CV Turbine (Ideal): s4s = s3 = 6.9211 kJ/kg K, x4s = (6.9211 - 1.091)/6.5029 = 0.8965 h4s = 2407.35 kJ/kg, h3 = 3230.8 kJ/kg => wTs = h3 - h4s = 823.45 kJ/kg CV Turbine (Actual): wTac = ηT × wTs = 658.76 = h3 - h4ac, => h4ac = 2572 kJ/kg . . m = W / wTac = 9000/658.76 = 13.66 kg/s C.V. Pump: wP = h2 - h1 = v1(P2 - P1) = 0.00103 (3000 - 50) = 3.04 kJ/kg => h2 = h1 + wP = 340.47 + 3.04 = 343.51 kJ/kg C.V. Boiler: qH = h3 - h2 = 3230.8 - 343.51 = 2887.3 kJ/kg ηcycle = (wTac - wP) / qH = (658.76 - 3.04) / 2887.3 = 0.227 11.59 Repeat Problem 11.47 assuming the turbine has an isentropic efficiency of 85%. The physical components and the T-s diagram is as shown in Fig. 11.10 in the main text for one open feedwater heater. The same state numbering is used. From the Steam Tables: State 5: (P, T) h5 = 4069.8 kJ/kg, s5 = 7.0544 kJ/kg K, State 1: (P, x = 0) h1 = 191.81 kJ/kg, v1 = 0.00101 m3/kg State 3: (P, x = 0) h3 = 762.8 kJ/kg, v3 = 0.001127 m3/kg Pump P1: wP1 = v1(P2 - P1) = 0.00101 × 990 = 1 kJ/kg h2 = h1 + wP1 = 192.81 kJ/kg Turbine 5-6: s6 = s5 ⇒ h6 = 3013.7 kJ/kg wT56,s = h5 - h6 = 4069.8 – 3013.7 = 1056.1 kJ/kg ⇒ wT56,AC = 1056.1 × 0.85 = 897.69 kJ/kg wT56,AC = h5 - h6AC ⇒ h6AC = h5 - wT56,AC = 4069.8 - 897.69 = 3172.11 kJ/kg . . . . . Feedwater Heater (mTOT = m5): xm5h6AC + (1 - x)m5h2 = m5h3 ⇒ x= h3 - h2 762.8 - 192.81 = = 0.1913 h6 - h2 3172.11 - 192.81 To get the turbine work apply the efficiency to the whole turbine. (i.e. the first section should be slightly different). s7s = s6s = s5 ⇒ x7s = (7.0544 – 0.6493)/7.5009 = 0.85391, h7s = 191.81 + 0.85391 × 2392.82 = 2235.1 kJ/kg wT57,s = h5 - h7s = 4069.8 - 2235.1 = 1834.7 kJ/kg wT57,AC = wT57,sηT = 1559.5 = h5 - h7AC => h7AC = 2510.3 kJ/kg Find specific turbine work to get total flow rate . WT . 5000 mTOT = = = 3.489 kg/s xwT56 + (1-x)wT57 0.1913×897.69 + 0.8087×1559.5 . . QL = mTOT(1 - x)(h7 - h1) = 3.489 × 0.8087(2510.3 - 191.81) = 6542 kW 11.60 Steam leaves a power plant steam generator at 3.5 MPa, 400°C, and enters the turbine at 3.4 MPa, 375°C. The isentropic turbine efficiency is 88%, and the turbine exhaust pressure is 10 kPa. Condensate leaves the condenser and enters the pump at 35°C, 10 kPa. The isentropic pump efficiency is 80%, and the discharge pressure is 3.7 MPa. The feedwater enters the steam generator at 3.6 MPa, 30°C. Calculate the thermal efficiency of the cycle and the entropy generation for the process in the line between the steam generator exit and the turbine inlet, assuming an ambient temperature of 25°C. 2 T 1 TURBINE. ST. GEN. 6 η = 0.88 5 4 5 5s 64 3 COND. 3.5 MPa 3.4 MPa o 1 400 C o 375 C 2 10 kPa 3s 3 s P 1: h1 = 3222.3 kJ/kg, s1 = 6.8405 kJ/kg K, 2: s2 = 6.7675 kJ/kg K h2 = 3165.7 kJ/kg, 3s: s3S = s2 ⇒ x3S = 0.8157, h3S = 2143.6 kJ/kg wT,S = h2 - h3S = 3165.7 - 2143.6 = 1022.1 kJ/kg wT,AC = ηwT,S = 899.4 kJ/kg, 3ac: h3 = h2 - wT,AC = 2266.3 kJ/kg -wP,S = vf(P5 - P4) = 0.001006(3700 - 10) = 3.7 kJ/kg -wP,AC = -wP,S/ηP = 4.6 kJ/kg qH = h1 - h6 = 3222.3 - 129.0 = 3093.3 kJ/kg η = wNET/qH = (899.4 - 4.6)/3093.3 = 0.289 C.V. Line from 1 to 2: w = 0, / Energy Eq.: q = h2 - h1 = 3165.7 - 3222.3 = - 56.6 kJ/kg Entropy Eq.: s1 + sgen + q/T0 = s2 => sgen = s2 - s1 -q/T0 = 6.7675 - 6.8405 - (-56.6/298.15) = 0.117 kJ/kg K 11.61 In a particular reheat-cycle power plant, steam enters the high-pressure turbine at 5 MPa, 450°C and expands to 0.5 MPa, after which it is reheated to 450°C. The steam is then expanded through the low-pressure turbine to 7.5 kPa. Liquid water leaves the condenser at 30°C, is pumped to 5 MPa, and then returned to the steam generator. Each turbine is adiabatic with an isentropic efficiency of 87% and the pump efficiency is 82%. If the total power output of the turbines is 10 MW, determine the mass flow rate of steam, the pump power input and the thermal efficiency of the power plant. 5 5 MPa T 3 2 2S 1 3 0.5 MPa o 450 C 5 7.5 kPa LP TURBINE. HP 4S 4 4 ηST1= ηST2 0.87 = 6S 6 2 s P 6 COND. = η SP 0.82 a) s4S = s3 = 6.8185 = 1.8606 + x4S × 4.9606 => 1 x4S = 0.999 h4S = 640.21 + 0.999 × 2108.5 = 2746.6 kJ/kg wT1,S = h3 - h4S = 3316.1 - 2746.6 = 569.5 kJ/kg wT1 = ηT1,S × wT1,S = 0.87 × 569.5 = 495.5 kJ/kg h4ac = 3316.1 - 495.5 = 2820.6 kJ/kg s6S = s5 = 7.9406 = 0.5764 + x6S × 7.675 ⇒ x6S = 0.9595 h6S = 168.79 + 0.9595 × 2406 = 2477.3 kJ/kg wT2,S = h5 - h6S = 3377.9 - 2477.3 = 900.6 kJ/kg wT2 = 0.87 × 900.6 = 783.5 kJ/kg . . m = WT/(wT1 + wT2) = 10000/(783.5 + 495.5) = 7.82 kg/s b) -wP,S = (0.001004)(5000 - 7.5) = 5.01 kJ/kg -wP = -wSP/ηSP = 5.01/0.82 = 6.11 kJ/kg . . WP = wPm = -7.82 × 6.11 = -47.8 kW c) qH = (h3 - h2) + (h5 - h4) = 3316.1 - 130.2 + 3377.9 - 2820.6 = 3743.2 kJ/kg wN = 1279.0 - 6.11 = 1272.9 kJ/kg ηTH = wN/qH = 1272.9/3743.2 = 0.34 11.62 A supercritical steam power plant has a high pressure of 30 MPa and an exit condenser temperature of 50°C. The maximum temperature in the boiler is 1000°C and the turbine exhaust is saturated vapor There is one open feedwater heater receiving extraction from the turbine at 1MPa, and its exit is saturated liquid flowing to pump 2. The isentropic efficiency for the first section and the overall turbine are both 88.5%. Find the ratio of the extraction mass flow to total flow into turbine. What is the boiler inlet temperature with and without the feedwater heater? Basically a Rankine Cycle 1: 50°C, 12.35 kPa, h = 209.31 kJ/kg, s = 0.7037 kJ/kg K 2: 30 MPa 3: 30 MPa, 1000 °C, h = 4554.7 kJ/kg, s = 7.2867 kJ/kg K 4AC: 50°C, x = 1, h = 2592.1 kJ/kg 3 T 2b 2 1a 1b 1 30 MPa 1000 C 1 MPa 3b 3a 50 C 4s 4ac s a) C.V. Turbine Ideal: s4S = s3 ⇒ x4S = 0.8929, h4S = 2336.8 kJ/kg => wT,S = h3 - h4S = 2217.86 kJ/kg Actual: wT,AC = h3 - h4AC = 1962.6 kJ/kg, η = wT,AC/wT,S = 0.885 b) P2 1b 2b m tot 3b m1 1a P1 1 1b: Sat liq. 179.91°C, h = 762.81 kJ/kg 3a: 1 MPa, s = s3 -> h3a = 3149.09 kJ/kg, T3a = 345.96 -> wT1s = 1405.6 kJ/kg 3b: 1 MPa, wT1ac = ηwT1s = 1243.96 kJ/kg wT1ac = h3-h3b => h3b = 3310.74 kJ/kg 1a: wP1 = v1(P1a-P1) ≈ 1 kJ/kg h1a = h1 + wP1 = 210.31 kJ/kg . . . . C.V. Feedwater Heater: mTOTh1b = m1h3b + (mTOT - m1)h1a .. ⇒ m1/mTOT = x = (h1b - h1a)/(h3b - h1a) = 0.178 . . . . c) C.V. Turbine: (mTOT)3 = (m1)3b + (mTOT - m1)4AC . . . . . _ WT = mTOTh3 - m1h3b - (mTOT - m1)h4AC = 25 MW = mTOTwT . wT = h3-xh3b - (1-x)h4AC = 1834.7 kJ/kg => mTOT = 13.63 kg/s d) C.V. No FWH, Pump Ideal: wP = h2S - h1, s2S = s1 Steam table ⇒ h2S = 240.1 kJ/kg, T2S = 51.2°C 1 FWH, CV: P2. s2b = s1b = 2.1386 kJ/kg K => T2b = 183.9°C Cogeneration 11.63 A cogenerating steam power plant, as in Fig. 11.13, operates with a boiler output of 25 kg/s steam at 7 MPa, 500°C. The condenser operates at 7.5 kPa and the process heat is extracted as 5 kg/s from the turbine at 500 kPa, state 6 and after use is returned as saturated liquid at 100 kPa, state 8. Assume all components are ideal and find the temperature after pump 1, the total turbine output and the total process heat transfer. Solution: Pump 1: Inlet state is saturated liquid: h1 = 168.79 kJ/kg, v1 = 0.001008 m3/kg wP1 = ∫ v dP = v1 ( P2 - P1) = 0.001008( 100 - 7.5) = 0.093 kJ/kg wP1 = h2 - h1 => h2 = h1 + wP1 = 168.88 kJ/kg, T2 = 40.3°C Turbine: h5 = 3410.3 kJ/kg, s5 = 6.7974 kJ/kg K P6, s6 = s5 => x6 = 0.9952, h6 = 2738.6 kJ/kg P7, s7 = s5 => x7 = 0.8106, h7 = 2119.0 kJ/kg From the continuity equation we have the full flow from 5 to 6 and the remainder after the extraction flow is taken out flows from 6 to 7. . . . WT = m5 ( h5 - h6) + 0.80m5 ( h6 - h7) = 25 (3410.3 - 2738.6) + 20 (2738.6 - 2119) = 16 792.5 + 12 392 = 29.185 MW . . Qproc = m6(h6 - h8) = 5(2738.6 - 417.46) = 11.606 MW Steam generator T Turbine 5 QH 4 P2 5 6 Thermal process WP2 8 2 3 Mixer WT 6 7 4 3 2 Condenser P1 1 QL WP1 8 1 7 s 11.64 A 10 kg/s steady supply of saturated-vapor steam at 500 kPa is required for drying a wood pulp slurry in a paper mill. It is decided to supply this steam by cogeneration, that is, the steam supply will be the exhaust from a steam turbine. Water at 20°C, 100 kPa, is pumped to a pressure of 5 MPa and then fed to a steam generator with an exit at 400°C. What is the additional heat transfer rate to the steam generator beyond what would have been required to produce only the desired steam supply? What is the difference in net power? Solution: Desired exit State 4: P4 = 500 kPa, sat. vap. => x4 = 1.0, T4 = 151.9°C h4 = hg = 2748.7 kJ/kg, s4 = sg = 6.8212 kJ/kg-K Inlet State: 20°C, 100 kPa h1 = hf = 83.94 kJ/kg, v1 = vf = 0.001002 m3/kg Without Cogeneration; The water is pumped up to 500 kPa and then heated in the steam generator to the desired exit T. C.V. Pump: wPw/o = v1( P4- P1) = 0.4 kJ/kg h2 = h1 + wPw/o = 84.3 kJ/kg C.V. Steam Generator: qw/o = h4 - h2 = 2664.4 kJ/kg With Cogeneration; The water is pumped to 5 MPa, heated in the steam generator to 400°C and then flows through the turbine with desired exit state. C.V. Pump: wPw = ∫ vdP = v1( P2- P1) = 4.91 kJ/kg h2 = h1 + wPw = 88.85 kJ/kg C.V. Steam Generator: Exit 400°C, 5 MPa => h3 = 3195.64 kJ/kg qw = h3 - h2 = 3195.64 - 88.85 = 3106.8 kJ/kg C.V.: Turbine, Inlet and exit states given wt = h3 - h4 = 3195.64 - 2748.7 = 446.94 kJ/kg Comparison Additional Heat Transfer: qw - qw/o = 3106.8 - 2664.4 = 442.4 kJ/kg . . Qextra = m(qw - qw/o) = 4424 kW Difference in Net Power: wdiff = (wt - wPw) + wPw/o, wdiff = 446.94 - 4.91 + 0.4 = 442.4 kJ/kg . . Wdiff = mwdiff = 4424 kW By adding the extra heat transfer at the higher pressure and a turbine all the extra heat transfer can come out as work (it appears as a 100% efficiency) 11.65 In a cogenerating steam power plant the turbine receives steam from a highpressure steam drum and a low-pressure steam drum as shown in Fig. P11.65. The condenser is made as two closed heat exchangers used to heat water running in a separate loop for district heating. The high-temperature heater adds 30 MW and the low-temperature heaters adds 31 MW to the district heating water flow. Find the power cogenerated by the turbine and the temperature in the return line to the deaerator. Solution: Inlet states from Table B.1.3 h1 = 3445.9 kJ/kg, s1 = 6.9108 kJ/kg K h2 = 2855.4 kJ/kg, s2 = 7.0592 kJ/kg K 2 . WT 1 Turbine . . . mTOT = m1 + m2 = 27 kg/s Assume a reversible turbine and the two flows can mix without s generation. 3 4 Energy Eq.6.10: . . . . . m1h1 + m2h2 = m3h3 + m4h4 + WT Entropy Eq.9.7: . . . m1s1 + m2s2 = mTOTsmix ⇒ sMIX = 6.9383 kJ/kg K State 3: s3 = sMIX ⇒ h3 = 2632.4 kJ/kg, x3 = 0.966 State 4: s4 = sMIX ⇒ h4 = 2413.5 kJ/kg, x4 = 0.899 . WT = 22 × 3445.9 + 5 × 2855.4 - 13 × 2632.4 - 14 × 2413.5 = 22 077 kW = 22 MW . . District heating line QTOT = m(h95 - h60) = 60 935 kW OK, this matches close enough . . . . C.V. Both heaters: m3h3 + m4h4 - QTOT = mTOThEX 13 × 2632.4 - 14 × 2413.5 – 60 935 = 7075.2 = 27 × hEX hEX = 262 ≈ hf ⇒ TEX = 62.5°C 11.66 A boiler delivers steam at 10 MPa, 550°C to a two-stage turbine as shown in Fig. 11.17. After the first stage, 25% of the steam is extracted at 1.4 MPa for a process application and returned at 1 MPa, 90°C to the feedwater line. The remainder of the steam continues through the low-pressure turbine stage, which exhausts to the condenser at 10 kPa. One pump brings the feedwater to 1 MPa and a second pump brings it to 10 MPa. Assume the first and second stages in the steam turbine have isentropic efficiencies of 85% and 80% and that both pumps are ideal. If the process application requires 5 MW of power, how much power can then be cogenerated by the turbine? Solution: 5: h5 = 3500.9, s5 = 6.7567 kJ/kg K 5 First ideal turbine T1 T1 6s: s6S = s5 ⇒ h6S = 2932.1 kJ/kg 6 wT1,S = h5 - h6S = 568.8 kJ/kg 7 Now the actual turbine T1 4 ⇒ wT1,AC = 483.5 kJ/kg h6AC = h5 - wT1,AC = 3017.4 6ac: P6, h6AC ⇒ s6AC = 6.9129 kJ/kg K T2 Boiler P2 8 3 Process heat 5 MW 2 P1 1 C First ideal turbine T2 (it follows the actual T1) State 7s: s7S = s6AC ⇒ h7S = 2189.9 kJ/kg wT2,S = h6AC - h7S = 827.5 kJ/kg wT2,AC = ηwT2,S = 622 = h6AC - h7AC ⇒ h7AC = 2355.4 kJ/kg Now do the process heat requirement 8: h8 = 377.6 kJ/kg, qPROC = h6AC - h8 = 2639.8 kJ/kg . . . m6 = Q/qPROC = 5000/2639.8 = 1.894 kg/s = 0.25 mTOT . . . . . ⇒ mTOT = m5 = 7.576 kg/s, m7 = m5 - m6 = 5.682 kg/s . . . . WT = m5h5 - m6h6AC - m7h7AC = 7424 kW 11.67 A smaller power plant produces 25 kg/s steam at 3 MPa, 600 C, in the boiler. It cools the condenser to an exit of 45C and the cycle is shown in Fig. P11.67. There is an extraction done at 500 kPa to an open feedwater heater, and in addition a steam supply of 5 kg/s is taken out and not returned. The missing 5 kg/s water is added to the feedwater heater from a 20C, 500 kPa source. Find the needed extraction flow rate to cover both the feedwater heater and the steam supply. Find the total turbine power output. Solution: The states properties from Tables B.1.1 and B.1.3 1: 45oC, x = 0: h1 = 188.42 kJ/kg, v1 = 0.00101 m3/kg, Psat = 9.59 kPa 5: 3.0 MPa, 600oC: 3: 500 kPa, x = 0: h5 = 3682.34 kJ/kg, s5 = 7.5084 kJ/kg K h3 = 640.21 kJ/kg 8: h8 = 84.41 kJ/kg 6: 500 kPa, s6 = s5 from HP turbine, h6 = 3093.26 kJ/kg C.V. Pump 1. Reversible and adiabatic. Incompressible so v = constant Energy: wp1 = h2 - h1 = ∫ v dP = v1(P2 - P1) = 0.00101 (500 - 9.6) = 0.495 kJ/kg h2 = h1 + wp1 = 188.42 + 0.495 = 188.915 kJ/kg C.V. Turbine sections Entropy Eq.: s7 = s5 = 7.5084 kJ/kg K => two-phase state s7 = 7.5084 = 0.6386 + x7 × 7.5261 ⇒ x7 = 0.9128 h7 = 188.42 + 0.9128 × 2394.77 = 2374.4 kJ/kg .. C.V. Feedwater heater, including the make-up water flow, x = m6/m5. . . . . . . Energy eq.: m8h8 + (m5 - m6)h2 + (m6 - m8)h6 = m5h3 . Divide by m5 and solve for x .. h3 - h2 + (h6 - h8) m8/ m5 640.21 - 188.915 + (3093.26 - 84.41)5/25 = x= h6 - h2 3093.26 - 188.915 = 0.3626 . . m6 = x m5 = 0.3626 × 25 = 9.065 kg/s C.V. Turbine energy equation . . . . WT = m5h5 - m6h6 - m7h7 = 25 × 3682.34 – 9.065 × 3093.26 – 16.935 × 2374.4 = 26 182 kW Brayton Cycles, Gas Turbines 11.68 Consider an ideal air-standard Brayton cycle in which the air into the compressor is at 100 kPa, 20°C, and the pressure ratio across the compressor is 12:1. The maximum temperature in the cycle is 1100°C, and the air flow rate is 10 kg/s. Assume constant specific heat for the air, value from Table A.5. Determine the compressor work, the turbine work, and the thermal efficiency of the cycle. Solution: Compression ratio 3 T P P2 = 12 P1 23 P Max temperature 4 2 T3 = 1100oC s s P = 100 kPa . m = 10 kg/s 1 s v 4 1 The compression is reversible and adiabatic so constant s. From Eq.8.32 k-1 P2 k T2 = T1 = 293.2(12)0.286 = 596.8 K P1 Energy equation with compressor work in wC = -1w2 = CP0(T2 - T1) = 1.004(596.8 - 293.2) = 304.8 kJ/kg The expansion is reversible and adiabatic so constant s. From Eq.8.32 k-1 P4 k 1 0.286 T4 = T3 = 1373.2 = 674.7 K 12 P3 Energy equation with turbine work out wT = CP0(T3 - T4) = 1.004(1373.2 - 674.7) = 701.3 kJ/kg Scale the work with the mass flow rate . . . . WC = mwC = 3048 kW, WT = mwT = 7013 kW Energy added by the combustion process qH = CP0(T3 - T2) = 1.004(1373.2 - 596.8) = 779.5 kJ/kg ηTH = wNET/qH = (701.3 - 304.8)/779.5 = 0.509 11.69 Repeat Problem 11.68, but assume variable specific heat for the air, table A.7. Consider an ideal air-standard Brayton cycle in which the air into the compressor is at 100 kPa, 20°C, and the pressure ratio across the compressor is 12:1. The o maximum temperature in the cycle is 1100 C, and the air flow rate is 10 kg/s. Assume constant specific heat for the air, value from Table A.5. Determine the compressor work, the turbine work, and the thermal efficiency of the cycle. Solution: From A.7: h1 = 293.6 kJ/kg, o sT1 = 6.84597 kJ/kg K The compression is reversible and adiabatic so constant s. From Eq.8.28 o o s2 = s1 ⇒ sT2 = sT1 + Rln(P2/P1) = 6.84597 + 0.287ln12 = 7.55914 ⇒ T2 = 590 K, h2 = 597.2 kJ/kg Energy equation with compressor work in wC = -1w2 = h2 - h1 = 597.2 - 293.6 = 303.6 kJ/kg The expansion is reversible and adiabatic so constant s. From Eq.8.28 From A.7: h3 = 1483.1, o o sT3 = 8.50554 o s4 = s3 ⇒ sT4 = sT3 + Rln(P4/P3) = 8.50554 + 0.287ln(1/12) = 7.79237 ⇒ T4 = 734.8 K, h4 = 751.1 kJ/kg Energy equation with turbine work out wT = h3 - h4 = 1483.1 - 751.1 = 732 kJ/kg Scale the work with the mass flow rate . . ⇒ WC = mwC = 3036 kW, . . WT = mwT = 7320 kW Energy added by the combustion process qH = h3 - h2 = 1483.1 - 597.2 = 885.9 kJ/kg wNET = wT - wC = 732 - 303.6 = 428.4 kJ/kg ηTH = wNET/qH = 428.4/885.9 = 0.484 11.70 A Brayton cycle inlet is at 300 K, 100 kPa and the combustion adds 670 kJ/kg. The maximum temperature is 1200 K due to material considerations. What is the maximum allowed compression ratio? For this calculate the net work and cycle efficiency assuming variable specific heat for the air, table A.7. Solution: Combustion: h3 = h2 + qH; 2w3 = 0 and Tmax = T3 = 1200 K h2 = h3 - qH = 1277.8 - 670 = 607.8 kJ/kg From Table A.7.1 o o T2 ≈ 600 K; sT2 = 7.57638 ; T1 = 300 K; sT1 = 6.86926 kJ/kg K Reversible adiabatic compression leads to constant s, from Eq.8.28: o o P2 / P1 = exp[ (sT2 - sT1)/R ] = exp(2.4638) = 11.75 Reversible adiabatic expansion leads to constant s, from Eq.8.28 o o sT4 = sT3 + R ln(P4 / P3) = 8.34596 + 0.287 ln(1 / 11.75) = 7.6388 kJ/kgK From Table A.7.1 by linear interpolation T4 ≈ 636.6 K, h4 = 645.97 kJ/kg wT = h3 - h4 = 1277.8 - 645.97 = 631.8 kJ/kg wC = h2 - h1 = 607.8 - 300.47 = 307.3 kJ/kg wnet = wT - wC = 631.8 - 307.3 = 324.5 kJ/kg η = wnet / qH = 324.5 / 670 = 0.484 11.71 A large stationary Brayton cycle gas-turbine power plant delivers a power output of 100 MW to an electric generator. The minimum temperature in the cycle is 300 K, and the maximum temperature is 1600 K. The minimum pressure in the cycle is 100 kPa, and the compressor pressure ratio is 14 to 1. Calculate the power output of the turbine. What fraction of the turbine output is required to drive the compressor? What is the thermal efficiency of the cycle? Solution: P 2 s 2 = s1 ⇒ T2 = T1(P2/P1) k-1 k 4 1 Solve using constant CP0 Compression in compressor: 3 T Brayton cycle so this means: Minimum T: T1 = 300 K Maximum T: T3 = 1600 K Pressure ratio: P2/P1 = 14 P = 100 kPa s Implemented in Eq.8.32 = 300(14)0.286 = 638.1 K wC = h2 - h1 = CP0(T2 - T1) = 1.004 (638.1 - 300) = 339.5 kJ/kg Expansion in turbine: s 4 = s3 ⇒ T4 = T3(P4/P3) k-1 k Implemented in Eq.8.32 = 1600 (1/14)0.286 = 752.2 K wT = h3 − h4 = CP0(T3 − T4) = 1.004 (1600 − 752.2) = 851.2 kJ/kg wNET = 851.2 - 339.5 = 511.7 kJ/kg Do the overall net and cycle efficiency . . m = WNET/wNET = 100000/511.7 = 195.4 kg/s . . WT = mwT = 195.4 × 851.2 = 166.32 MW wC/wT = 339.5/851.2 = 0.399 Energy input is from the combustor qH = CP0(T3 - T2) = 1.004 (1600 - 638.1) = 965.7 kJ/kg ηTH = wNET/qH = 511.7/965.7 = 0.530 11.72 o A Brayton cycle produces 14 MW with an inlet state of 17 C, 100 kPa, and a compression ratio of 16:1. The heat added in the combustion is 960 kJ/kg. What are the highest temperature and the mass flow rate of air, assuming cold air properties? Solution: Efficiency is from Eq.11.8 . Wnet wnet -0.4/1.4 -(k-1)/k η= = = 1 - rp = 1 - 16 = 0.547 . qH QH from the required power we can find the needed heat transfer . . 14 000 QH = Wnet / η = = 25 594 kW 0.547 . . m = QH / qH = 25 594 kW/ 960 kJ/kg = 26.66 kg/s Temperature after compression is (k-1)/k 0.4/1.4 T2 = T1 rp = 290 × 16 = 640.35 K The highest temperature is after combustion 960 T3 = T2 + qH/Cp = 640.35 + = 1596.5 K 1.004 11.73 Do the previous problem with properties from table A.7.1 instead of cold air properties. Solution: With the variable specific heat we must go through the processes one by one to get net work and the highest temperature T3. From A.7.1: o h1 = 290.43 kJ/kg, sT1 = 6.83521 kJ/kg K The compression is reversible and adiabatic so constant s. From Eq.8.28 o o s2 = s1 ⇒ sT2 = sT1 + Rln(P2/P1) = 6.83521 + 0.287 ln16 = 7.63094 ⇒ T2 = 631.9 K, h2 = 641 kJ/kg Energy equation with compressor work in wC = -1w2 = h2 - h1 = 641 - 290.43 = 350.57 kJ/kg Energy Eq. combustor: h3 = h2 + qH = 641 + 960 = 1601 kJ/kg o State 3: (P, h): T3 = 1471 K, sT3 = 8.58811 kJ/kg K The expansion is reversible and adiabatic so constant s. From Eq.8.28 o o s4 = s3 ⇒ sT4 = sT3 + Rln(P4/P3) = 8.58811 + 0.287ln(1/16) = 7.79238 ⇒ T4 = 734.8 K, h4 = 751.11 kJ/kg Energy equation with turbine work out wT = h3 - h4 = 1601 - 751.11 = 849.89 kJ/kg Now the net work is wnet = wT - wC = 849.89 – 350.57 = 499.32 kJ/kg The total required power requires a mass flow rate as . . Wnet 14 000 kW m= = = 28.04 kg/s wnet 499.32 kJ/kg Regenerators, Intercoolers, and Nonideal Cycles 11.74 An ideal regenerator is incorporated into the ideal air-standard Brayton cycle of Problem 11.68. Find the thermal efficiency of the cycle with this modification. Consider an ideal air-standard Brayton cycle in which the air into the compressor is at 100 kPa, 20°C, and the pressure ratio across the compressor is 12:1. The maximum temperature in the cycle is 1100°C, and the air flow rate is 10 kg/s. Assume constant specific heat for the air, value from Table A.5. Determine the compressor work, the turbine work, and the thermal efficiency of the cycle. Solution: Compression ratio 3 T P P2 = 12 P1 23 Max temperature x 4 2 T3 = 1100oC s s y . P = 100 kPa s m = 10 kg/s 1 v 4 1 The compression is reversible and adiabatic so constant s. From Eq.8.32 k-1 P2 k T2 = T1 = 293.2(12)0.286 = 596.8 K P1 Energy equation with compressor work in wC = h2 - h1 = CP0(T2 - T1) = 1.004(596.8 - 293.2) = 304.8 kJ/kg The expansion is reversible and adiabatic so constant s. From Eq.8.32 k-1 P4 k 1 0.286 T4 = T3 = 1373.2 = 674.7 K 12 P3 Energy equation with turbine work out wT = CP0(T3 - T4) = 1.004(1373.2 - 674.7) = 701.3 kJ/kg Ideal regenerator: TX = T4 = 674.7 K qH = h3 - hX = 1.004(1373.2 - 674.7) = 701.3 kJ/kg = wT ηTH = wNET/qH = (701.3 - 304.8)/701.3 = 0.565 11.75 The gas-turbine cycle shown in Fig. P11.75 is used as an automotive engine. In the first turbine, the gas expands to pressure P5, just low enough for this turbine to drive the compressor. The gas is then expanded through the second turbine connected to the drive wheels. The data for the engine are shown in the figure and assume that all processes are ideal. Determine the intermediate pressure P5, the net specific work output of the engine, and the mass flow rate through the engine. Find also the air temperature entering the burner T3, and the thermal efficiency of the engine. a) Consider the compressor k-1 P2 k s2 = s1 ⇒ T2 = T1 = 300(6)0.286 = 500.8 K P1 -wC = -w12 = CP0(T2 - T1) = 1.004(500.8 - 300) = 201.6 kJ/kg Consider then the first turbine work wT1 = -wC = 201.6 = CP0(T4 - T5) = 1.004(1600 - T5) ⇒ T5 = 1399.2 K T5 s5 = s4 ⇒ P5 = P4 T4 k-1 k 1399.23.5 = 600 = 375 kPa 1600 k-1 b) P6 k 1000.286 s6 = s5 ⇒ T6 =T5 = 1399.2 = 958.8 K 375 P5 The second turbine gives the net work out wT2 = CP0(T5 - T6) = 1.004(1399.2 - 958.8) = 442.2 kJ/kg . . m = WNET/wT2 = 150/442.2 = 0.339 kg/s c) Ideal regenerator ⇒ T3 = T6 = 958.8 K qH = CP0(T4 - T3) = 1.004(1600 - 958.8) = 643.8 kJ/kg ηTH = wNET/qH = 442.2/643.8 = 0.687 11.76 Repeat Problem 11.71, but include a regenerator with 75% efficiency in the cycle. A large stationary Brayton cycle gas-turbine power plant delivers a power output of 100 MW to an electric generator. The minimum temperature in the cycle is 300 K, and the maximum temperature is 1600 K. The minimum pressure in the cycle is 100 kPa, and the compressor pressure ratio is 14 to 1. Calculate the power output of the turbine. What fraction of the turbine output is required to drive the compressor? What is the thermal efficiency of the cycle? Solution: Both compressor and turbine are reversible and adiabatic so constant s, Eq.8.32 relates then T to P assuming constant heat capacity. Compressor: ⇒ T2 = T1(P2/P1) k-1 k = 300(14)0.286 = 638.1 K wC = h2 - h1 = CP0(T2 - T1) = 1.004 (638.1 - 300) = 339.5 kJ/kg Turbine s4 = s3 ⇒ T4 = T3(P4/P3) k-1 k = 1600 (1/14)0.286 = 752.2 K wT = h3 − h4 = CP0(T3 − T4) = 1.004 (1600 − 752.2) = 851.2 kJ/kg wNET = 851.2 - 339.5 = 511.7 kJ/kg . . m = WNET/wNET = 100 000/511.7 = 195.4 kg/s . . WT = mwT = 195.4 × 851.2 = 166.32 MW wC/wT = 339.5/851.2 = 0.399 x' x 2 1 For the regenerator 3 T ηREG = 0.75 = 4 P = 100 kPa hX - h2 TX - T2 TX - 638.1 = = hX' - h2 T4 - T2 752.2 - 638.1 ⇒ TX = 723.7 K s Turbine and compressor work not affected by regenerator. Combustor needs to add less energy with the regenerator as qH = CP0(T3 - TX) = 1.004(1600 – 723.7) = 879.8 kJ/kg ηTH = wNET/qH = 511.7/879.8 = 0.582 11.77 A two-stage air compressor has an intercooler between the two stages as shown in Fig. P11.77. The inlet state is 100 kPa, 290 K, and the final exit pressure is 1.6 MPa. Assume that the constant pressure intercooler cools the air to the inlet temperature, T3 = T1. It can be shown that the optimal pressure, P2 = (P1P4)1/2, for minimum total compressor work. Find the specific compressor works and the intercooler heat transfer for the optimal P2. Solution: Optimal intercooler pressure P2 = 100 × 1600 = 400 kPa o 1: h1 = 290.43, sT1 = 6.83521 C.V. C1: wC1 = h2 - h1, s2 = s1 o leading to Eq.8.28 o ⇒ sT2 = sT1 + R ln(P2/P1) = 6.83521 + 0.287 ln 4 = 7.2331 ⇒ T2 = 430.3 K, h2 = 432.05 kJ/kg wC1 = 432.05 - 290.43 = 141.6 kJ/kg C.V. Cooler: T3 = T1 ⇒ h3 = h1 qOUT = h2 - h3 = h2 - h1 = wC1 = 141.6 kJ/kg C.V. C2: T3 = T1, s4 = s3 o o o and since sT3 = sT1 , P4/P3 = P2/P1 o o ⇒ sT4 = sT3 + R ln(P4/P3) = sT2 , so we have T4 = T2 Thus we get wC2 = wC1 = 141.6 kJ/kg P T 4 1600 kPa 4 3 2 400 kPa 2 100 kPa 1 v 3 1 s 11.78 A two-stage compressor in a gas turbine brings atmospheric air at 100 kPa, 17oC to 500 kPa, then cools it in an intercooler to 27oC at constant P. The second stage brings the air to 1000 kPa. Assume both stages are adiabatic and reversible. Find the combined specific work to the compressor stages. Compare that to the specific work for the case of no intercooler (i.e. one compressor from 100 to 1000 kPa). Solution: C.V. Stage 1: 1 => 2 Reversible and adiabatic gives constant s which from Eq.8.32 gives: T2 = T1 (P2/P1)(k-1)/k = 290 (500/100) 0.2857 = 459.3 K wc1in = CP( T2 - T1) = 1.004(459.3 –290) = 187.0 kJ/kg C.V. Stage 2: 3 => 4 Reversible and adiabatic gives constant s which from Eq.8.32 gives: T4 = T3 (P4/P3)(k-1)/k = 300 (1000/500) 0.2857 = 365.7 K wc2in = CP( T4 - T3) = 1.004(365.7 – 300) = 65.96 kJ/kg wtot = wc1 + wc2 = 187 + 65.96 = 253 kJ/kg The intercooler reduces the work for stage 2 as T is lower and so is specific volume. C.V. One compressor 1 => 5 Reversible and adiabatic gives constant s which from Eq.8.32 gives: T5 = T1 (P5/P1)(k-1)/k = 290 (1000/100) 0.2857 = 559.88 K win = CP( T5 - T1) = 1.004(559.88 –290) = 271 kJ/kg P T 4 5 1000 kPa 2 500 kPa 5 4 3 2 100 kPa 1 3 v 1 s The reduction in work due to the intercooler is shaded in the P-v diagram. 11.79 A gas turbine with air as the working fluid has two ideal turbine sections, as shown in Fig. P11.79, the first of which drives the ideal compressor, with the second producing the power output. The compressor input is at 290 K, 100 kPa, and the exit is at 450 kPa. A fraction of flow, x, bypasses the burner and the rest (1 − x) goes through the burner where 1200 kJ/kg is added by combustion. The two flows then mix before entering the first turbine and continue through the second turbine, with exhaust at 100 kPa. If the mixing should result in a temperature of 1000 K into the first turbine find the fraction x. Find the required pressure and temperature into the second turbine and its specific power output. C.V.Comp.: -wC = h2 - h1; s2 = s1 Reversible and adiabatic gives constant s which from Eq.8.32 gives: T2 = T1 (P2/P1)(k-1)/k = 290 (450/100) 0.2857 = 445.7 K h2 = 447.75 kJ/kg, -wC = 447.75 - 290.43 = 157.3 kJ/kg C.V.Burner: h3 = h2 + qH = 447.75 + 1200 = 1647.75 kJ/kg ⇒ T3 = 1510 K C.V.Mixing chamber: (1 - x)h3 + xh2 = hMIX = 1046.22 kJ/kg x= h3 - hMIX h3 - h2 = 1647.75 - 1046.22 = 0.5013 1647.75 - 447.75 . . . WT1 = WC,in ⇒ wT1 = -wC = 157.3 = h3 - h4 h4 = 1046.22 - 157.3 = 888.9 kJ/kg ⇒ T4 = 860 K P4 = PMIX(T4/TMIX)k/(k-1) = 450 × (860/1000)3.5 = 265 kPa s4 = s5 ⇒ T5 = T4 (P5/P4)(k-1)/k = 860 (100/265)0.2857 = 651 K h5 = 661.2 kJ/kg wT2 = h4 - h5 = 888.9 - 661.2 = 227.7 kJ/kg 11.80 Repeat Problem 11.71, but assume that the compressor has an isentropic efficiency of 85% and the turbine an isentropic efficiency of 88%. Solution: Brayton cycle so this means: Minimum T: T1 = 300 K Maximum T: T3 = 1600 K Pressure ratio: P2/P1 = 14 Solve using constant CP0 Ideal compressor: 3 P 2s 2 1 s 2 = s1 ⇒ T2s = T1(P2/P1) T k-1 k 4 4s P = 100 kPa s Implemented in Eq.8.32 = 300(14)0.286 = 638.1 K wCs = h2 - h1 = CP0(T2 - T1) = 1.004 (638.1 - 300) = 339.5 kJ/kg Actual compressor ⇒ wC = wSC/ηSC = 339.5/0.85 = 399.4 kJ/kg = CP0(T2-T1) ⇒ T2 = T1 + wc/CP0 = 300 + 399.4/1.004 = 697.8 K Ideal turbine: s 4 = s3 ⇒ T4s = T3(P4/P3) Implemented in Eq.8.32 k-1 k = 1600 (1/14)0.286 = 752.2 K wTs = h3 − h4 = CP0(T3 − T4) = 1.004 (1600 − 752.2) = 851.2 kJ/kg Actual turbine ⇒ wT = ηST wST = 0.88 × 851.2 = 749.1 kJ/kg = CP0(T3-T4) ⇒ T4 = T3 - wT/CP0 = 1600 - 749.1/1.004 = 853.9 K Do the overall net and cycle efficiency wNET = 749.1 - 399.4 = 349.7 kJ/kg . . m = WNET/wNET = 100000/349.7 = 286.0 kg/s . . WT = mwT = 286.0×749.1 = 214.2 MW wC/wT = 399.4/749.1 = 0.533 Energy input is from the combustor qH = CP0(T3 - T2) = 1.004(1600 - 697.8) = 905.8 kJ/kg ηTH = wNET/qH = 349.7/905.8 = 0.386 11.81 Repeat Problem 11.77 when the intercooler brings the air to T3 = 320 K. The corrected formula for the optimal pressure is P = [ P P (T /T )n/(n-1)]1/2 see 2 14 3 1 Problem 9.184, where n is the exponent in the assumed polytropic process. Solution: The polytropic process has n = k (isentropic) so n/(n - 1) = 1.4/0.4 = 3.5 P2 = 400 (320/290)3.5 = 475.2 kPa C.V. C1: s2 = s1 ⇒ T2 = T1 (P2/P1) k-1 k = 290 (475.2/100)0.2857 = 452.67 K -wC1 = h2 - h1 = Cp(T2 − T1) = 1.004(452.67 – 290) = 163.3 kJ/kg C.V. Cooler: qOUT = h2 - h3 = 1.004(452.67 – 320) = 133.2 kJ/kg C.V. C2: s4 = s3 ⇒ T4 = T3 (P4/P3) k-1 k = 320 (1600/475.2)0.2857 = 452.67 K -wC2 = h4 - h3 = Cp(T2 − T1) = 1.004(452.67 – 320) = 133.2 kJ/kg 11.82 Consider an ideal gas-turbine cycle with two stages of compression and two stages of expansion. The pressure ratio across each compressor stage and each turbine stage is 8 to 1. The pressure at the entrance to the first compressor is 100 kPa, the temperature entering each compressor is 20°C, and the temperature entering each turbine is 1100°C. An ideal regenerator is also incorporated into the cycle. Determine the compressor work, the turbine work, and the thermal efficiency of the cycle. Solution: 10 REG I.C. 9 CC 1 5 2 COMP 6 4 COMP TURB TURB 7 8 CC T P2/P1 = P4/P3 = P6/P7 = P8/P9 = 8.0 P1 = 100 kPa T1 = T3 = 20oC, T6 = T8 = 1100oC Assume constant specific heat s2 = s1 and s4 = s3 ⇒ 6 5 4 8 7 2 9 10 k-1 P2 k s 3 1 T4 = T2 = T1 = 293.2(8)0.286 = 531.4 K P1 Total wC = 2 × w12 = 2CP0(T2 - T1) = 2 × 1.004(531.4 - 293.2) = 478.1 kJ/kg k-1 P7 k 10.286 Also s6 = s7 and s8 = s9: ⇒ T7 = T9 = T6 = 1373.2 = 757.6 K 8 P6 Total wT = 2 × w67 = 2CP0(T6 - T7) = 2 × 1.004(1373.2 - 756.7) = 1235.5 kJ/kg wNET = 1235.5 - 478.1 = 757.4 kJ/kg Ideal regenerator: T5 = T9, T10 = T4 ⇒ qH = (h6 - h5) + (h8 - h7) = 2CP0(T6 - T5) = 2 × 1.004(1373.2 - 757.6) = 1235.5 kJ/kg ηTH = wNET/qH = 757.4/1235.5 = 0.613 11.83 A gas turbine cycle has two stages of compression, with an intercooler between the stages. Air enters the first stage at 100 kPa, 300 K. The pressure ratio across each compressor stage is 5 to 1, and each stage has an isentropic efficiency of 82%. Air exits the intercooler at 330 K. Calculate the temperature at the exit of each compressor stage and the total specific work required. Solution: State 1: P1 = 100 kPa, T1 = 300 K State 3: T3 = 330 K P2 = 5 P1 = 500 kPa; Energy Eq.: wc1 + h1 = h2 => Ideal C1 constant s, Eq.8.32: P4 = 5 P3 = 2500 kPa wc1 = h2 - h1 = CP(T2 - T1) T2s = T1 (P2/P1)(k-1)/k = 475.4 K wc1 s = CP(T2s - T1) = 176.0 kJ/kg, Actual Eq.9.28: wc1 = wc1 s/η = 176/0.82 = 214.6 kJ/kg T2 = T1 + wc1/CP = 513.7 K Ideal C2 constant s, Eq.8.32: T4s = T3 (P4/P3)(k-1)/k = 552.6 K wc2 s = CP(T4s - T3 ) = 193.4 kJ/kg; Actual Eq.9.28: wc2 = wc2 s/η = 235.9 kJ/kg T4 = T3 + wc2 / CP = 565 K Total work in: w = wc1 + wc2 = 214.6 + 235.9 = 450.5 kJ/kg P T 4s 4ac 4s 4ac 2s 2ac 3 1 3 v 2s 2ac 1 s 11.84 Repeat the questions in Problem 11.75 when we assume that friction causes pressure drops in the burner and on both sides of the regenerator. In each case, the pressure drop is estimated to be 2% of the inlet pressure to that component of the system, so P3 = 588 kPa, P4 = 0.98 P3 and P6 = 102 kPa. Solution: k-1 P2 k a) From solution 11.75: T2 = T1 = 300(6)0.286 = 500.8 K P1 -wC = -w12 = CP0(T2 - T1) = 1.004(500.8 - 300) = 201.6 kJ/kg P3 = 0.98 × 600 = 588 kPa, P4 = 0.98 × 588 = 576.2 kPa k s5 = s4 ⇒ P5 = P4(T5S/T4)k-1 = 576.2( 1399.2 3.5 ) = 360.4 kPa 1600 b) P6 = 100/0.98 = 102 kPa, s6S = s5 k-1 P6 k 102 0.286 T6 = T5 = 1399.2 = 975.2 K 292.8 P5 wST2 = CP0(T5-T6) = 1.004(1399.2 - 975.2) = 425.7 kJ/kg . . m = WNET/wNET = 150/425.7 = 0.352 kg/s c) T3 = T6 = 975.2 K qH = CP0(T4 - T3) = 1.004 (1600 - 975.2) = 627.3 kJ/kg ηTH = wNET/qH = 425.7/627.3 = 0.678 Ericsson Cycles 11.85 Consider an ideal air-standard Ericsson cycle that has an ideal regenerator as shown in Fig. P11.85. The high pressure is 1 MPa and the cycle efficiency is 70%. Heat is rejected in the cycle at a temperature of 300 K, and the cycle pressure at the beginning of the isothermal compression process is 100 kPa. Determine the high temperature, the compressor work, and the turbine work per kilogram of air. P T P 2 3 3 T4 T T P 1 P ⇒ qH = 3q4 & wT = qH rp = P2/P1 = 10 P 4 2 T v P2 = P3 = 1 MPa T1 = T2 = 300 K P1 = 100 kPa q = -4q1 (ideal reg.) 23 1 s ηTH = ηCARNOT TH. = 1 - TL/TH = 0.7 ⇒ T3 = T4 = TH = 1000 K P2 1000 qL = -wC = ⌠v dP = RT1ln = 0.287 × 300 × ln = 198.25 ⌡ 100 P1 wT = qH = -⌠v dP = -RT3ln(P4/P3) = 660.8 kJ/kg ⌡ 11.86 An air-standard Ericsson cycle has an ideal regenerator. Heat is supplied at 1000°C and heat is rejected at 20°C. Pressure at the beginning of the isothermal compression process is 70 kPa. The heat added is 600 kJ/kg. Find the compressor work, the turbine work, and the cycle efficiency. Solution: Identify the states Heat supplied at high temperature Heat rejected at low temperature Beginning of the compression: q = -4q1 Ideal regenerator: 23 ⇒ T3 = T4 = 1000°C = 1273.15 K T1 = T2 = 20°C = 293.15 K P1 = 70 kPa ⇒ qH = 3q4 = 600 kJ/kg wT = qH = 600 kJ/kg ηTH = ηCARNOT = 1 - 293.15 = 0.7697 1273.15 wNET = ηTHqH = 0.7697 × 600 = 461.82 kJ/kg qL = -wC = 600 - 461.82 = 138.2 kJ/kg P T P 2 3 3 T4 T T P 1 P P 4 2 T v 1 s Jet Engine Cycles 11.87 Consider an ideal air-standard cycle for a gas-turbine, jet propulsion unit, such as that shown in Fig. 11.27. The pressure and temperature entering the compressor are 90 kPa, 290 K. The pressure ratio across the compressor is 14 to 1, and the turbine inlet temperature is 1500 K. When the air leaves the turbine, it enters the nozzle and expands to 90 kPa. Determine the pressure at the nozzle inlet and the velocity of the air leaving the nozzle. Solution: 2 3 3 T BURNER 4 5 P COMPR. TURBINE 1 4 2 1 NOZ C.V. Compressor: Reversible and adiabatic o P = 90 kPa 5 s 2 = s1 s From Eq.8.28 o ⇒ sT2 = sT1 + R ln(P2/P1) = 6.83521 + 0.287 ln 14 = 7.59262 kJ/kg K From A.7 h2 = 617.2 kJ/kg, T2 = 609.4 K wC = h2 - h1 = 617.2 - 290.43 = 326.8 kJ/kg C.V. Turbine: wT = h3 - h4 = wC and s4 = s3 ⇒ h4 = h3 - wC = 1635.8 - 326.8 = 1309 o ⇒ sT4 = 8.37142 kJ/kg K, T4 = 1227 K o o P4 = P3 exp[(sT4 - sT3)/R] = 1260 exp[ (8.37142 - 8.61208)/0.287 ] = 1260 exp(-0.83854) = 544.8 kPa C.V. Nozzle: s5 = s4 = s3 so from Eq.8.28 o o ⇒ sT5 = sT3 + R ln(P5/P3) = 8.61208 + 0.287 ln (1/14) = 7.85467 kJ/kgK => From A.7 T5 = 778 K, h5 = 798.2 kJ/kg Now the energy equation (1/2)V2 = h4 - h5 = 510.8 ⇒ 5 V5 = 2 × 1000 × 510.8 = 1011 m/s 11.88 The turbine section in a jet engine receives gas (assume air) at 1200 K, 800 kPa with an ambient atmosphere at 80 kPa. The turbine is followed by a nozzle open to the atmosphere and all the turbine work drives a compressor receiving air at 85 kPa, 270 K with the same flow rate. Find the turbine exit pressure so the nozzle has an exit velocity of 800 m/s. To what pressure can the compressor bring the incomming air? Solution: C.V. Reversible and adiabatic turbine and nozzle. This gives constant s, from Eq.8.32 we can relate the T’s and P’s State 1: 1200 K, 800 kPa State 3: 80 kPa; s3 = s1 Eq.8.32: T3 = T1 (P3/P1)(k-1)/k = 1200(80/800) 0.2857 = 621.56 K Energy: h1 + 0 = h3 + (1/2)V2 + wT = h2 + wT 3 wT = h1 - h3 - (1/2)V2 ≅ CP(T1 - T3) - (1/2)V2 3 3 = 1.004(1200 – 621.56) – (1/2) × 8002/1000 = 580.75 – 320 = 260.75 kJ/kg C.V. Nozzle alone to establish state 2. h2 = h3 + (1/2)V2 3 T2 = T3 + (1/2)V2/CP = 621.56 + 320/1.004 = 940.29 K 3 P2 = P1 + (T2/T1)k/(k-1) = 800 × (940.29/1200)3.5 = 340.7 kPa C.V. Compressor wc = he - hi = wT = 260.75 kJ/kg Te = Ti + wc/ CP = 270 + 260.75/1.004 = 529.71 K Reversible adiabatic compressor, constant s gives relation in Eq.8.32 Pe = Pi × (Te/Ti)k/(k-1) = 85 × (529.71/270)3.5 = 899 kPa T 1 1 TURBINE NOZZLE 2 3 2 3 s 11.89 The turbine in a jet engine receives air at 1250 K, 1.5 MPa. It exhausts to a nozzle at 250 kPa, which in turn exhausts to the atmosphere at 100 kPa. The isentropic efficiency of the turbine is 85% and the nozzle efficiency is 95%. Find the nozzle inlet temperature and the nozzle exit velocity. Assume negligible kinetic energy out of the turbine. Solution: o C.V. Turbine: hi = 1336.7, sTi = 8.3940, ses = si o then from Eq.8.28 o ⇒ sTes = sTi + R ln(Pe/Pi) = 8.3940 + 0.287 ln (250/1500) = 7.8798 kJ/kg K Table A.7.1 Tes = 796 K, hes = 817.9 kJ/kg, Energy Eq.: wT,s = hi - hes = 1336.7 - 817.9 = 518.8 kJ/kg wT,AC = wT,s × ηT = 441 kJ/kg = he,AC - hi Eq.9.27: o ⇒ he,AC = 895.7 ⇒ Te,AC = 866 K, sTe = 7.9730 kJ/kg K o C.V. Nozzle: hi = 895.7 kJ/kg, sTi = 7.9730 KJ/kgK, ses = si then from Eq.8.28 o o ⇒ sTes = sTi + R ln(Pe/Pi) = 7.9730 + 0.287 ln (100/250) = 7.7100 kJ/kgK Table A.7.1 ⇒ Te,s = 681 K, he,s = 693.1 kJ/kg 2 Energy Eq.: (1/2)Ve,s = hi - he,s = 895.7 - 693.1 = 202.6 kJ/kg Eq.9.30: Ve,AC = 2 2 (1/2)Ve,AC = (1/2)Ve,s × ηNOZ = 192.47 kJ/kg 2 × 1000 × 192.47 = 620 m/s 11.90 Consider an air standard jet engine cycle operating in a 280K, 100 kPa environment. The compressor requires a shaft power input of 4000 kW. Air enters the turbine state 3 at 1600 K, 2 MPa, at the rate of 9 kg/s, and the isentropic efficiency of the turbine is 85%. Determine the pressure and temperature entering the nozzle at state 4. If the nozzle efficiency is 95%, determine the temperature and velocity exiting the nozzle at state 5. Solution: . . . C.V. Shaft: WT = m(h3 - h4) = WC . . CV Turbine: h3 - h4 = WC / m = 4000/9 = 444.4 kJ/kg h4 = 1757.3 – 444.4 = 1312.9 kJ/kg Work back to the ideal turbine conditions wTa = wC = 444.4 ⇒ wTs = wTa / η = 522.82 = h3 - h4s h4s = 1234.5 ⇒ Eq.9.27: T4s ≈ 1163 K, sT4s = 8.3091 kJ/kg K o o o s4s - s3 = 0 = sT4s - sT3 - R ln(P4/P3 ) 0 = 8.3091 - 8.6905 - 0.287 ln(P4/2000) => P4 = 530 kPa o State 4 from A.7.1: h4 = 1312.9, T4 = 1229.8 K, sT4 = 8.3746 kJ/kg K First consider the reversible adiabatic (isentropic) nozzle so from Eq.8.28 o o s5s - s4 = 0 = sT5s - sT4 - R ln(P5/P4 ) o sT5s = 8.3746 + 0.287 ln(100/530) = 7.8960 kJ/kg K Table A.7.1: T5s = 808.1 K, h5s = 831.0 kJ/kg 2 ⇒ 0.5V5s = h4 - h5s = 1312.9 - 831.0 = 481.9 kJ/kg Now consider the actual nozzle 2 2 0.5V5a = η(0.5V5s) = 457.81 ⇒ V5a= 957 m/s Eq.9.30: 2 h5a = h4 - 0.5V5a = 1312.9 – 457.81 = 855.1 kJ/kg ⇒ T5a ≈ 830 K 11.91 A jet aircraft is flying at an altitude of 4900 m, where the ambient pressure is approximately 55 kPa and the ambient temperature is −18°C. The velocity of the aircraft is 280 m/s, the pressure ratio across the compressor is 14:1 and the cycle maximum temperature is 1450 K. Assume the inlet flow goes through a diffuser to zero relative velocity at state 1. Find the temperature and pressure at state 1 and the velocity (relative to the aircraft) of the air leaving the engine at 55 kPa. Solution: T Ambient TX = -18oC = 255.2 K, PX = 55 kPa = P5 also VX = 280 m/s 3 4 5 2 1 x P = 55 kPa Assume that the air at this state is reversibly decelerated to zero velocity and then enters the compressor at 1. P2/P1 = 14 & T3 = 1450 K s C.V. Diffuser section. 2 EnergyEq.: VX (280)2 T1 = TX + = 255.2 + = 294.3 K 2 × 1000 2 × 1000 × 1.0035 k T1 k-1 294.33.5 Eq.8.32: P1 = PX = 55 = 90.5 kPa 255.2 TX C.V. Compressor, isentropic so use Eq.8.32 and then energy equation k-1 )k T2 = T1 (P2/P1 = 294.3(14)0.286 = 626.0 K wC = -1w2 = CP0(T2-T1) = 1.004(1450 - T4) ⇒ T4 = 1118.3 K Pressure ratio: P3 = P2 = 14 × 90.5 = 1267 kPa C.V. Turbine, isentropic so use Eq.8.32 k P4 = P3 (T4/T3)k-1 = 1267(1118.3/1450)3.5 = 510 kPa C.V. Nozzle, isentropic so use Eq.8.32 and energy equation T5 = T4 (P5/P4) k-1 k = 1118.3(55/510) 0.286 = 591.5 K V2 5 = CP0(T4 - T5) = 1.004(1118.3 - 591.5) = 528.7 kJ/kg 2 × 1000 ⇒ V5 = 1028 m/s 11.92 An afterburner in a jet engine adds fuel after the turbine thus raising the pressure and temperature due to the energy of combustion. Assume a standard condition of 800 K, 250 kPa after the turbine into the nozzle that exhausts at 95 kPa. Assume the afterburner adds 450 kJ/kg to that state with a rise in pressure for same specific volume, and neglect any upstream effects on the turbine. Find the nozzle exit velocity before and after the afterburner is turned on. Solution: Before afterburner is on: 1: 800 K; 250 kPa and 2: 95 kPa After afterburner is on: 3: v = v1 and 4: 95 kPa T 1 2 3 v1 3 4 1 2 P = 95 kPa 4 s Assume reversible adiabatic nozzle flow, then constant s from Eq.8.32 T2 = T1 (P2/P1)(k-1)/k = 800 × (95/250) Energy Eq.: V2 = 0.2857 = 606.8 K (1/2)V2 = CP(T1 - T2) 2 2 CP(T1 - T2) = 2 × 1004(800 - 606.8) = 622.8 m/s Add the qAB at assumed constant volume then energy equation gives T3 = T1 + qAB/Cv = 800 + 450/0.717 = 1427.6 K v3 = v1 => P3 = P1( T3/T1) = 250 × 1427.6/800 = 446.1 kPa Reversible adiabatic expansion, again from Eq.8.32 T4 = T3 (P4/P3)(k-1)/k = 1427.6 × (95/446.1) V2 = 2 CP(T3 - T4) = 0.2857 = 917.7 K 2 × 1004(1427.6 - 917.7) = 1012 m/s Otto Cycles 11.93 Air flows into a gasoline engine at 95 kPa, 300 K. The air is then compressed with a volumetric compression ratio of 8:1. In the combustion process 1300 kJ/kg of energy is released as the fuel burns. Find the temperature and pressure after combustion using cold air properties. Solution: Solve the problem with constant heat capacity. Compression 1 to 2: s2 = s1 ⇒ From Eq.8.33 and Eq.8.34 T2 = T1 (v1/v2) k-1 0.4 = 300 × 8 k 1.4 P2 = P1×(v1/v2) = 95 × 8 = 689.2 K = 1746 kPa Combustion 2 to 3 at constant volume: u3 = u2 + qH T3 = T2 + qH/Cv = 689.2 + 1300/0.717 = 2502 K P3 = P2 × (T3/T2) = 1746 (2502 / 689.2) = 6338 kPa P T 3 3 v 2 s 4 1v 2 1 4 s 11.94 A gasoline engine has a volumetric compression ratio of 9. The state before compression is 290 K, 90 kPa, and the peak cycle temperature is 1800 K. Find the pressure after expansion, the cycle net work and the cycle efficiency using properties from Table A.5. Compression 1 to 2: s2 = s1 ⇒ From Eq.8.33 and Eq.8.34 T2 = T1 (v1/v2) k-1 0.4 = 290 × 9 k 1.4 P2 = P1× (v1/v2) = 90 × 9 = 698.4 K = 1950.7 kPa Combustion 2 to 3 at constant volume: v 3 = v2 qH = u3 – u2 = Cv(T3 – T2) = 0.717 (1800 – 698.4) = 789.85 kJ/kg P3 = P2 × (T3/T2) = 1950.7 (1800 / 698.4) = 5027.6 kPa s4 = s3 ⇒ From Eq.8.33 and Eq.8.34 Expansion 3 to 4: T4 = T3 (v3/v4) k-1 = 1800 × (1/9) 0.4 = 747.4 K P4 = P3(T4/T3)(v3/v4) = 5027.6 (747.4/1800) (1/9) = 232 kPa Find now the net work 1w2 = u1 - u2 = Cv(T1 - T2) = 0.717(290 – 698.4) = -292.8 kJ/kg 3w4 = u3 - u4 = Cv(T3 - T4) = 0.717(1800 – 747.4) = 754.7 kJ/kg Net work and overall efficiency wNET = 3w4 + 1w2 = 754.7 - 292.8 = 461.9 kJ/kg η = wNET/qH = 461.9/789.85 = 0.585 Comment: We could have found η from Eq.11.18 and then wNET = ηqH. P T 3 3 v 2 s 4 1v 2 1 4 s 11.95 To approximate an actual spark-ignition engine consider an air-standard Otto cycle that has a heat addition of 1800 kJ/kg of air, a compression ratio of 7, and a pressure and temperature at the beginning of the compression process of 90 kPa, 10°C. Assuming constant specific heat, with the value from Table A.5, determine the maximum pressure and temperature of the cycle, the thermal efficiency of the cycle and the mean effective pressure. Solution: P T 3 3 4 2 4 1v 2 1 v s Compression: Reversible and adiabatic so constant s from Eq.8.33-34 k P2 = P1(v1/v2) = 90(7)1.4 = 1372 kPa T2 = T1(v1/v2) k-1 = 283.2 × (7)0.4 = 616.6 K Combustion: constant volume T3 = T2 + qH/CV0 = 616.6 + 1800/0.717 = 3127 K P3 = P2T3/T2= 1372 × 3127 / 616.6 = 6958 kPa Efficiency and net work ηTH = 1 - T1/T2 = 1 - 283.2/616.5 = 0.541 wnet = ηTH × qH = 0.541 × 1800 = 973.8 kJ/kg Displacement and Pmeff v1 = RT1/P1 = (0.287 × 283.2)/90 = 0.9029 m3/kg v2 = (1/7) v1 = 0.1290 m3/kg Pmeff = wNET 973.8 = = 1258 kPa v1-v2 0.9029 - 0.129 11.96 A gasoline engine has a volumetric compression ratio of 8 and before compression has air at 280 K, 85 kPa. The combustion generates a peak pressure of 6500 kPa. Find the peak temperature, the energy added by the combustion process and the exhaust temperature. Solution: Solve the problem with cold air properties. Compression. Isentropic so we use Eqs.8.33-8.34 k P2 = P1(v1/v2) = 85(8)1.4 = 1562 kPa T2 = T1(v1/v2) k-1 = 280(8)0.4 = 643.3 K Combustion. Constant volume T3 = T2 (P3/P2) = 643.3 × 6500/1562 = 2677 K qH = u3 - u2 ≈ Cv(T3 - T2) = 0.717 (2677 – 643.3) = 1458 kJ/kg Exhaust. Isentropic expansion so from Eq.8.33 T4 = T3/80.4 = 2677/2.2974 = 1165 K P T 3 3 2 4 4 1v 2 1 v s 11.97 A gasoline engine has a volumetric compression ratio of 10 and before compression has air at 290 K, 85 kPa in the cylinder. The combustion peak pressure is 6000 kPa. Assume cold air properties. What is the highest temperature in the cycle? Find the temperature at the beginning of the exhaust (heat rejection) and the overall cycle efficiency. Solution: Compression. Isentropic so we use Eqs.8.33-8.34 k P2 = P1(v1/v2) = 85 (10)1.4 = 2135.1 kPa T2 = T1(v1/v2) k-1 = 290 (10)0.4 = 728.45 K Combustion. Constant volume T3 = T2 (P3/P2) = 728.45 × 6000/2135.1 = 2047 K Exhaust. Isentropic expansion so from Eq.8.33 T4 = T3 / (v1/v2) k-1 = T3 / 100.4 = 2047 / 2.5119 = 814.9 K Overall cycle efficiency is from Eq.11.18, rv = v1/v2 1-k η = 1 − rv = 1 − 10 -0.4 = 0.602 Comment: No actual gasoline engine has an efficiency that high, maybe 35%. 11.98 A for stroke gasoline engine has a compression ratio of 10:1 with 4 cylinders of total displacement 2.3 L. the inlet state is 280 K, 70 kPa and the engine is running at 2100 RPM with the fuel adding 1800 kJ/kg in the combustion process. What is the net work in the cycle and how much power is produced? solution: Overall cycle efficiency is from Eq.11.18, rv = v1/v2 1-k ηTH = 1 − rv = 1 − 10 -0.4 = 0.602 wnet = ηTH × qH = 0.602 × 1800 = 1083.6 kJ/kg We also need specific volume to evaluate Eqs.11.15 to 11.17 v1 = RT1 / P1 = 0.287 × 280 / 70 = 1.148 m3/kg Pmeff = wnet wnet 1083.6 = 1 = 1.148 × 0.9 = 1048.8 kPa v1 – v2 v (1 – ) 1 rv Now we can find the power from Eq.11.17 . RPM 1 2100 1 W = Pmeff Vdispl = 1048.8 × 0.0023 × × = 42.2 kW 60 2 60 2 11.99 A gasoline engine takes air in at 290 K, 90 kPa and then compresses it. The combustion adds 1000 kJ/kg to the air after which the temperature is 2050 K. Use the cold air properties (i.e. constant heat capacities at 300 K) and find the compression ratio, the compression specific work and the highest pressure in the cycle. Solution: Standard Otto Cycle Combustion process: T3 = 2050 K; u2 = u3 - qH T2 = T3 - qH / Cvo = 2050 - 1000 / 0.717 = 655.3 K Compression process P2 = P1(T2 / T1)k/(k-1) = 90(655.3/290) 3.5 = 1561 kPa CR = v1 / v2 = (T2 / T1)1/(k-1) = (655.3 / 290) 2.5 = 7.67 -1w2 = u2 - u1 = Cvo( T2 - T1) = 0.717(655.3 - 290) = 262 kJ / kg Highest pressure is after the combustion P3 = P2T3 / T2 = 1561 × 2050 / 655.3 = 4883 kPa P T 3 3 4 2 4 1v 2 1 v s 11.100 Answer the same three questions for the previous problem, but use variable heat capacities (use table A.7). A gasoline engine takes air in at 290 K, 90 kPa and then compresses it. The combustion adds 1000 kJ/kg to the air after which the temperature is 2050 K. Use the cold air properties (i.e. constant heat capacities at 300 K) and find the compression ratio, the compression specific work and the highest pressure in the cycle. Solution: Standard Otto cycle, solve using Table A.7.1 Combustion process: T3 = 2050 K ; u3 = 1725.7 kJ/kg u2 = u3 - qH = 1725.7 - 1000 = 725.7 kJ/kg o ⇒ T2 = 960.5 K ; sT2 = 8.0889 kJ/kg K Compression 1 to 2: s2 = s1 ⇒ From Eq.8.28 o o o o 0 = sT2 - sT1 - R ln(P2/P1) = sT2 - sT1 - R ln(Τ2v1/T1v2) = 8.0889 - 6.8352 - 0.287 ln(960.5/290) - 0.287 ln(v1/v2) Solving => v1 / v2 = 23.78 Comment: This is much too high for an actual Otto cycle. -1w2 = u2 - u1 = 725.7 - 207.2 = 518.5 kJ/kg Highest pressure is after combustion P3 = P2T3 / T2 = P1(T3 / T1)(v1 / v3) = 90 × (2050 / 290) × 23.78 = 15 129 kPa P T 3 3 4 2 4 1v 2 1 v s 11.101 When methanol produced from coal is considered as an alternative fuel to gasoline for automotive engines, it is recognized that the engine can be designed with a higher compression ratio, say 10 instead of 7, but that the energy release with combustion for a stoichiometric mixture with air is slightly smaller, about 1700 kJ/kg. Repeat Problem 11.95 using these values. Solution: P T 3 3 4 2 4 1v 2 1 v s Compression: Reversible and adiabatic so constant s from Eq.8.33-34 k P2 = P1(v1/v2) = 90(10)1.4 = 2260.7 kPa T2 = T1(v1/v2) k-1 = 283.15(10)0.4 = 711.2 K Combustion: constant volume T3 = T2 + qH / Cvo = 711.2 + 1700 / 0.717 = 3082 K P3 = P2(T3 / T2) = 2260.7×3082 / 711.2 = 9797 kPa Efficiency, net work, displacement and Pmeff ηTH = 1 - T1/T2 = 1 - 283.15/711.2 = 0.602 wnet = ηTH × qH = 0.6 × 1700 = 1023.4 kJ/kg v1 = RT1/P1 = 0.287×283.15/90 = 0.9029 m3/kg, v2 = v1/10 = 0.0903 m3/kg Pmeff = wnet = 1023.4 / (0.9029 - 0.0903) = 1255 kPa v1 – v2 11.102 A gasoline engine receives air at 10 C, 100 kPa, having a compression ratio of 9:1 by volume. The heat addition by combustion gives the highest temperature as 2500 K. use cold air properties to find the highest cycle pressure, the specific energy added by combustion, and the mean effective pressure. Solution: P T 3 3 4 2 4 1v 2 1 v s Compression: Reversible and adiabatic so constant s from Eq.8.33-34 k P2 = P1(v1/v2) = 100 (9)1.4 = 2167.4 kPa T2 = T1(v1/v2) k-1 = 283.15 (9)0.4 = 681.89 K Combustion: constant volume P3 = P2(T3 / T2) = 2167.4 × 2500 / 681.89 = 7946.3 kPa qH = u3 – u2 = Cvo(T3 - T2) = 0.717 (2500 – 681.89) = 1303.6 kJ/kg Efficiency, net work, displacement and Pmeff ηTH = 1 - T1/T2 = 1 - 283.15/681.89 = 0.5847 wnet = ηTH × qH = 0.5847 × 1303.6 = 762.29 kJ/kg v1 = RT1/P1 = 0.287 × 283.15 / 100 = 0.81264 m3/kg, v2 = v1/10 = 0.081264 m3/kg Pmeff = wnet 762.29 = = 1055 kPa v1 – v2 0.81264 - 0.081264 11.103 Repeat Problem 11.95, but assume variable specific heat. The ideal gas air tables, Table A.7, are recommended for this calculation (and the specific heat from Fig. 5.10 at high temperature). Solution: Table A.7 is used with interpolation. T1 = 283.2 K, o u1 = 202.3 kJ/kg, sT1 = 6.8113 kJ/kg K Compression 1 to 2: s2 = s1 ⇒ From Eq.8.28 o o o o 0 = sT2 - sT1 - R ln(P2/P1) = sT2 - sT1 - R ln(Τ2v1/T1v2) o o sT2 - R ln(Τ2/T1) = sT1 + R ln(v1/v2) = 6.8113 + 0.287 ln 7 = 7.3698 This becomes trial and error so estimate first at 600 K and use A.7.1. LHS600 = 7.5764 - 0.287 ln(600/283.2) = 7.3609 (too low) LHS620 = 7.6109 - 0.287 ln(620/283.2) = 7.3860 (too high) Interpolate to get: T2 = 607.1 K, u2 = 440.5 kJ/kg => -1w2 = u2 - u1 = 238.2 kJ/kg, u3 = 440.5 + 1800 = 2240.5 => T3 = 2575.8 K , o sT3 = 9.2859 kJ/kgK P3 = 90 × 7 × 2575.8 / 283.2 = 5730 kPa Expansion 3 to 4: o s 4 = s3 ⇒ From Eq.8.28 as before o sT4 - R ln(Τ4/T3) = sT3 + R ln(v3/v4) = 9.2859 + 0.287 ln(1/7) = 8.7274 This becomes trial and error so estimate first at 1400 K and use A.7.1. LHS1400 = 8.5289 - 0.287 ln(1400/2575.8) = 8.7039 (too low) LHS1450 = 8.5711 - 0.287 ln(1450/2575.8) = 8.7360 (too high) Interpolation ⇒ T4 = 1436.6 K, u4 = 1146.9 kJ/kg 3w4 = u3 - u4 = 2240.5 - 1146.9 = 1093.6 kJ/kg Net work, efficiency and mep wnet = 3w4 + 1w2 = 1093.6 - 238.2 = 855.4 kJ/kg ηTH = wnet / qH = 855.4 / 1800 = 0.475 v1 = RT1/P1 = (0.287 × 283.2)/90 = 0.9029 m3/kg v2 = (1/7) v1 = 0.1290 m3/kg Pmeff = wnet = 855.4 / (0.9029 - 0.129) = 1105 kPa v1 – v2 11.104 It is found experimentally that the power stroke expansion in an internal combustion engine can be approximated with a polytropic process with a value of the polytropic exponent n somewhat larger than the specific heat ratio k. Repeat Problem 11.95 but assume that the expansion process is reversible and polytropic (instead of the isentropic expansion in the Otto cycle) with n equal to 1.50. See solution to 11.95 except for process 3 to 4. T3 = 3127 K, P3 = 6.958 MPa v3 = RT3/P3 = v2 = 0.129 m3/kg, v4 = v1 = 0.9029 m3/kg Process: Pv1.5 = constant. P4 = P3(v3/v4)1.5 = 6958 (1/7)1.5 = 375.7 kPa T4 = T3(v3/v4)0.5 = 3127(1/7)0.5 = 1181.9 K R 0.287 w = ⌠Pdv = 1-1.4(T2 - T1) = -0.4 (606.6 -283.15)= -239.3 kJ/kg ⌡ 12 w = ⌠Pdv = R(T4 - T3)/(1 - 1.5) ⌡ 34 = -0.287(1181.9-3127)/0.5 = 1116.5 kJ/kg wNET = 1116.5 - 239.3 = 877.2 kJ/kg ηCYCLE = wNET/qH = 877.2/1800 = 0.487 Pmeff = wnet = 877.2/(0.9029 - 0.129) = 1133 kPa v1 – v2 Note a smaller wNET, ηCYCLE, Pmeff compared to an ideal cycle. 11.105 In the Otto cycle all the heat transfer qH occurs at constant volume. It is more realistic to assume that part of qH occurs after the piston has started its downward motion in the expansion stroke. Therefore, consider a cycle identical to the Otto cycle, except that the first two-thirds of the total qH occurs at constant volume and the last one-third occurs at constant pressure. Assume that the total qH is 2100 kJ/kg, that the state at the beginning of the compression process is 90 kPa, 20°C, and that the compression ratio is 9. Calculate the maximum pressure and temperature and the thermal efficiency of this cycle. Compare the results with those of a conventional Otto cycle having the same given variables. P 3 T 4 3 s s v 5 2 2 s 1 s 1 5 a) q23 = (2/3) × 2100 = 1400 kJ/kg; v s v b) P1 = 90 kPa, T1 = 20oC rV = v1/v2 = 7 4 q34 = 2100/3 = 700 kJ/kg P2 = P1(v1/v2)k = 90(9)1.4 = 1951 kPa T2 = T1(v1/v2)k-1 = 293.15(9)0.4 = 706 K T3 = T2 + q23/CV0 = 706 + 1400/0.717 = 2660 K P3 = P2T3/T2 = 1951(2660/706) = 7350.8 kPa = P4 T4 = T3 + q34/CP0 = 2660 + 700/1.004 = 3357 K v5 v4 = v1 v4 = P4 T1 7350.8 293.15 ×= × = 7.131 90 P1 T4 3357 T5 = T4(v4/v5)k-1 = 3357(1/7.131)0.4 = 1530 K qL = CV0(T5-T1) = 0.717(1530 - 293.15) = 886.2 kJ/kg ηTH = 1 - qL/qH = 1 - 886.2/2100 = 0.578 Std. Otto Cycle: ηTH = 1 - (9)-0.4 = 0.585, small difference Diesel Cycles 11.106 A diesel engine has a state before compression of 95 kPa, 290 K, and a peak pressure of 6000 kPa, a maximum temperature of 2400 K. Find the volumetric compression ratio and the thermal efficiency. Solution: Standard Diesel cycle and we will use cold air properties. Compression process (isentropic) from Eqs.8.32-8.34: k (P2/P1) = (v1/v2) = CR1.4 CR = v1/v2 = (P2/P1) T2 = T1(P2/P1) k-1/k 1/k = (6000/95) 1/1.4 = 290 × (6000/95) = 19.32 0.2857 = 947.9 K Combustion and expansion volumes v3 = v2 × T3/T2 = v1 T3/(T2 × CR) ; v4 = v1 Expansion process, isentropic from Eq.8.32 T4 = T3 (v3/v4) k-1 = T3 [T3/ (CR × T2)] k-1 0.4 = 2400 × [ 2400/(19.32 × 947.9) ] = 1064.6 K Efficiency from Eq.11.7 1 T4 - T1 1 1064.6 – 290 η=1– =1– = 0.619 k T3 - T2 1.4 2400 – 947.9 P T 2 3 s 3 4 P s 4 1v 2 1 v s 11.107 A diesel engine has a bore of 0.1 m, a stroke of 0.11 m and a compression ratio of 19:1 running at 2000 RPM (revolutions per minute). Each cycle takes two revolutions and has a mean effective pressure of 1400 kPa. With a total of 6 cylinders find the engine power in kW and horsepower, hp. Solution: Work from mean effective pressure, Eq.11.15. wnet Pmeff = => wnet = Pmeff (vmax - vmin) vmax – vmin The displacement is ∆V = πBore2 × 0.25 × S = π × 0.12 × 0.25 × 0.11 = 0.000864 m3 Work per cylinder per power stroke, Eq.11.16 W = Pmeff(Vmax - Vmin) = 1400 × 0.000864 kPa m3 = 1.2096 kJ/cycle Only every second revolution has a power stroke so we can find the power, see also Eq.11.17 . W = W × Ncyl × RPM × 0.5 (cycles / min)×(min / 60 s)×(kJ / cycle) = 1.2096 × 6 × 2000 × 0.5 × (1/60) = 121 kW = 162 hp The conversion factor from kW to hp is from Table A.1 under power. 11.108 A diesel engine has a compression ratio of 20:1 with an inlet of 95 kPa, 290 K, state 1, with volume 0.5 L. The maximum cycle temperature is 1800 K. Find the maximum pressure, the net specific work and the thermal efficiency. Solution: Compression process (isentropic) from Eqs.8.33-34 T2 = T1(v1 / v2)k-1 = 290 × 200.4 = 961 K P2 = 95×(20) 1.4 = 6297.5 kPa ; v2 = v1/20 = RT1/(20 P1) = 0.043805 -1w2 = u2 - u1 ≈ Cvo( T2 - T1) = 0.717 (961 - 290) = 481.1 kJ/kg Combustion at constant P which is the maximum presssure v3 = v2 T3 /T2 = 0.043805 × 1800/961 = 0.08205 P3 = P2 = 6298 kPa ; 2w3 = P (v3 - v2) = 6298 × (0.08215 - 0.043805) = 241.5 kJ/kg 2q3 = u3 - u2 + 2w3 = h3 - h2 = Cpo(T3 - T2) = 1.004(1800 - 961) = 842.4 Expansion process (isentropic) from Eq.8.33 T4 = T3( v3 / v4)0.4 = 1800 (0.08205 / 0.8761) 0.4 = 698 K 3w4 = u3 - u4 ≈ Cvo(T3 - T4) = 0.717 (1800 - 698) = 790.1 kJ/kg Cycle net work and efficiency wnet = 2w3 + 3w4 + 1w2 = 241.5 + 790.1 - 481.1 = 550.5 kJ/kg η = wnet / qH = 550.5/ 842.4 = 0.653 P T 2 s s 3 P 3 2 4 1v 1 4 v s 11.109 At the beginning of compression in a diesel cycle T = 300 K, P = 200 kPa and after combustion (heat addition) is complete T = 1500 K and P = 7.0 MPa. Find the compression ratio, the thermal efficiency and the mean effective pressure. Solution: Standard Diesel cycle. See P-v and T-s diagrams for state numbers. Compression process (isentropic) from Eqs.8.33-8.34 P2 = P3 = 7000 kPa => v1 / v2 = (P2/P1)1/ k = (7000 / 200)0.7143 = 12.67 T2 = T1(P2 / P1)(k-1) / k = 300(7000 / 200) 0.2857= 828.4 K Expansion process (isentropic) first get the volume ratios v3 / v2 = T3 / T2 = 1500 / 828.4 = 1.81 v4 / v3 = v1 / v3 = (v1 / v2)( v2 / v3) = 12.67 / 1.81 = 7 The exhaust temperature follows from Eq.8.33 T4 = T3(v3 / v4)k-1 = (1500 / 7) 0.4 = 688.7 K qL = Cvo(T4 - T1) = 0.717(688.7 - 300) = 278.5 kJ/kg qH = h3 - h2 ≈ Cpo(T3 - T2) = 1.004(1500 - 828.4) = 674 kJ/kg Overall performance η = 1 - qL / qH = 1- 278.5 / 674 = 0.587 wnet = qnet = qH - qL = 674 - 278.5 = 395.5 kJ/kg vmax = v1 = R T1 / P1 = 0.287×300 / 200 = 0.4305 m3/kg vmin = vmax / (v1 / v2) = 0.4305 / 12.67 = 0.034 m3/kg Pmeff = wnet = 395.5 / (0.4305 - 0.034) = 997 kPa vmax – vmin P T 2 s s 3 P 3 2 4 1v 1 4 v s Remark: This is a too low compression ratio for a practical diesel cycle. 11.110 Do problem 11.106, but use the properties from A.7 and not the cold air properties. A diesel engine has a state before compression of 95 kPa, 290 K, and a peak pressure of 6000 kPa, a maximum temperature of 2400 K. Find the volumetric compression ratio and the thermal efficiency. Solution: Compression: s2 = s1 => from Eq.8.28 ° ° sT2 = sT1 + R ln(P2 / P1) = 6.8352 + 0.287 ln(6000/95) = 8.025 kJ/kg K A.7.1 => T2 = 907.6 K; h2 = 941.16; ° h3 = 2755.8 kJ/kg; sT3 = 9.19586 kJ/kg K qH = h3 - h2 = 2755.8 – 941.16 = 1814.2 kJ/kg CR = v1/v2 = (T1/T2)(P2/P1) = (290/907.6) × (6000/ 95) = 20.18 Expansion process ° ° ° sT4 = sT3 + R ln(P4 / P3) = sT3 + R ln(T4 / T3) + R ln(v3/v4) v3/v4 = v3/v1= (v2/v1) × (T3/T2) = (T3/T2) (1/CR) = (2400/907.6) (1/20.18) = 0.13104 ° ° sT4 - R ln(T4 / T3) = sT3 + R ln(v3/v4) = 9.1958 + 0.287 ln 0.13104 = 8.61254 ° Trial and error on T4 since it appears both in sT4 and the ln function T4 =1300 LHS = 8.4405 – 0.287 ln (1300/2400) = 8.616 T4 = 1250 LHS = 8.3940 – 0.287 ln (1250/2400) = 8.5812 Now Linear interpolation T4 = 1295 K, u4 = 1018.26 kJ/kg qL = u4 - u1 = 1018.26 – 207.19 = 811.08 kJ/kg η = 1 – (qL/ qH) = 1 – (811.08/1814.2) = 0.553 P T 2 3 s 3 4 P s 4 1v 2 1 v s 11.111 A diesel engine has air before compression at 280 K, 85 kPa. The highest temperature is 2200 K and the highest pressure is 6 MPa. Find the volumetric compression ratio and the mean effective pressure using cold air properties at 300 K. Solution: k k Compression (P2/P1) = (v1/v2) = CR CR = v1/v2 = (P2/P1) T2 = T1(P2/P1) k-1/k 1/k = (6000/85) 1/1.4 = 280 × (6000/85) = 20.92 0.2857 = 944.8 K Combustion. Highest temperature is after combustion. qH = h3 - h2 = CP(T3 –T 2) = 1.004(2200 – 944.8) = 1260.2 kJ/kg Expansion T4 = T3 (v3/v4) k-1 = T3 [ T3/ (CR × T2) ] k-1 0.4 = 2200 × (2200/20.92 × 944.8) = 914.2 K qL = u4 - u1 = CV( T4 - T1) = 0.717(914.2 – 280) = 454.7 kJ/kg v1 = RT1/P1 = 0.287 × 280/85 = 0.9454 m3/kg Displacement and mep from net work v1 - v2 = v1- v1/CR = v1[1 – (1/CR)] = 0.9002 m3/kg Pmeff = wnet/(v1 – v2) = (qH - qL)/( v1 - v2) = (1260.2 – 454.7)/0.9002 = 894.8 kPa P T 2 3 s 3 4 P s 4 1v 2 1 v s 11.112 Consider an ideal air-standard diesel cycle in which the state before the compression process is 95 kPa, 290 K, and the compression ratio is 20. Find the maximum temperature (by iteration) in the cycle to have a thermal efficiency of 60%? Solution: Diesel cycle: P1 = 95 kPa, T1 = 290 K, v1/v2 = 20, ηTH = 0.6 Since the efficiency depends on T3 and T4, which are connected through the expansion process in a nonlinear manner we have an iterative problem. T2 = T1(v1/v2) k-1 = 290(20)0.4 = 961.2 K v1 = 0.287 × 290/95 = 0.876 m3/kg = v4, v2 = v1/CR = 0.876 / 20 = 0.0438 m3/kg v3 = v2(T3/T2) = 0.0438 (T3/961.2) = 0.0000456 T3 T3 = T4 (v4/v3) k-1 =( 0.876 )0.4 ⇒ T4 = 0.019345 T1.4 3 0.0000456 T3 Now substitute this into the formula for the efficiency ηTH = 0.60 = 1 - T4 - T1 k(T3 - T2) =1- 0.019345 × T1.4 - 290 3 1.4(T3 - 961.2) ⇒ 0.019345 × T1.4 - 0.56 × T3 + 248.272 = 0 3 Trial and error on this non-linear equation in T3 3050 K: LHS = +1.06 Linear interpolation T3 = 3040 K 3040 K: LHS = -0.036, Stirling-cycle engine 11.113 Consider an ideal Stirling-cycle engine in which the state at the beginning of the isothermal compression process is 100 kPa, 25°C, the compression ratio is 6, and the maximum temperature in the cycle is 1100°C. Calculate the maximum cycle pressure and the thermal efficiency of the cycle with and without regenerators. P T 3 3 T v 2 4 v v T T4 v T3 = T4 = 1100 oC 2 1 T Ideal Stirling cycle T1 = T2 = 25 oC P1 = 100 kPa CR = v1/v2 = 6 1 s v Isothermal compression (heat goes out) T1 = T2 ⇒ P2 = P1(v1/v2) = 100 × 6 = 600 kPa w = 1q2 = -RT1 ln(v1/v2) = -0.287× 298.2 ln(6) = -153.3 kJ/kg 12 Constant volume heat addition V2 = V3 ⇒ P3 = P2T3/T2 = 600×1373.2/298.2 = 2763 kPa q23 = u3 – u2 = Cv o(T3 - T2) = 0.717 (1100 - 25) = 770.8 kJ/kg Isothermal expansion (heat comes in) w34 = q34 = RT3 ln(v4/v3) = 0.287 × 1373.2 × ln6 = 706.1 kJ/kg wnet = 706.1 - 153.3 = 552.8 kJ/kg Efficiency without regenerator, (q23 and q34 are coming in from source) ηNO REGEN = wnet 552.8 = = 0.374, q23 + q34 770.8 + 706.1 Efficiency with regenerator, (Now only q34 is coming in from source) ηWITH REGEN = wnet 552.8 = = 0.783 q34 706.1 11.114 An air-standard Stirling cycle uses helium as the working fluid. The isothermal compression brings helium from 100 kPa, 37°C to 600 kPa. The expansion takes place at 1200 K and there is no regenerator. Find the work and heat transfer in all of the 4 processes per kg helium and the thermal cycle efficiency. Helium table A.5: R = 2.077 kJ/kg K, Cvo = 3.1156 kJ/kg K Compression/expansion: v4 / v3 = v1 / v2 = P2 / P1 = 600 / 100 = 6 1 -> 2 -1w2 = -q12 = ∫ P dv = R T1ln(v1 / v2) = RT1ln (P2 /P1) 2 -> 3 : 3 -> 4: = 2.077 × 310 × ln 6 = 1153.7 kJ/kg 2w3 = 0; q23 = Cvo(T3 - T2) = 3.1156(1200 - 310) = 2773 kJ/kg v4 3w4 = q34 = R T3lnv = 2.077×1200 ln 6 = 4465.8 kJ/kg 3 4 -> 1 4w1 = 0; ηcycle = q41 = Cvo(T4 - T1) = -2773 kJ/kg 1w2 + 3w4 q23 + q34 = -1153.7 + 4465.8 = 0.458 2773 + 4465.8 11.115 Consider an ideal air-standard Stirling cycle with an ideal regenerator. The minimum pressure and temperature in the cycle are 100 kPa, 25°C, the compression ratio is 10, and the maximum temperature in the cycle is 1000°C. Analyze each of the four processes in this cycle for work and heat transfer, and determine the overall performance of the engine. Ideal Stirling cycle diagram as in Fig. 11.31, with P1 = 100 kPa, T1 = T2 = 25oC, v1/v2 = 10, From 1-2 at const T: T3 = T4 = 1000oC w = 1q2 = T1(s2 - s1) 12 = -RT1ln(v1/v2) = -0.287 × 298.2 × ln(10) = -197.1 kJ/kg From 2-3 at const V: w =0 / 23 q23 = CV0(T3 - T2) = 0.717 (1000 - 25) = 699 kJ/kg From 3-4 at const T; = +RT3 × ln w = 3q4 = T3(s4 - s3) 34 v4 v3 From 4-1 at const V; = 0.287 × 1237.2 × ln(10) = 841.4 kJ/kg w =0 / 41 q41 = CV0(T1 - T4) = 0.717 (25 - 1000) = -699 kJ/kg wNET = -197.1 + 0 + 841.4 + 0 = 644.3 kJ/kg Since q23 is supplied by -q41 (regenerator) qH = q34 = 841.4 kJ/kg, ηTH = NOTE: qH = q34 = RT3 × ln(10), wNET qH = 644.3 = 0.766 841.4 qL = -1q2 = RT1 × ln(10) qH - qL T3 - T1 975 ηTH = = = = 0.766 = Carnot efficiency qH T3 1273.2 11.116 The air-standard Carnot cycle was not shown in the text; show the T–s diagram for this cycle. In an air-standard Carnot cycle the low temperature is 280 K and the efficiency is 60%. If the pressure before compression and after heat rejection is 100 kPa, find the high temperature and the pressure just before heat addition. Solution: Carnot cycle efficiency from Eq.7.5 η = 0.6 = 1 - TH/TL ⇒ TH = TL/0.4 = 700 K Just before heat addition is state 2 and after heat rejection is state 1 so P1 = 100 kPa and the isentropic compression is from Eq.8.32 P2 = P1(TH/TL 1 k-1 ) = 2.47 MPa P T 2 T s TH 3 s 1 T 4 TL v 2 1 3 4 s 11.117 Air in a piston/cylinder goes through a Carnot cycle in which TL = 26.8°C and the total cycle efficiency is η = 2/3. Find TH, the specific work and volume ratio in the adiabatic expansion for constant CP, Cv. Solution: Carnot cycle efficiency Eq.7.5: η = 1 - TL/TH = 2/3 ⇒ TH = 3 × TL = 3 × 300 = 900 K Pvk = constant, work from Eq.8.38 (n = k) R 3w4 = (P4v4 - P3v3)/(1 - k) = 1 - k(T4 - T3) = u3 - u4 Adiabatic expansion 3 to 4: = Cv(T3 - T4) = 0.717(900 - 300) = 429.9 kJ/kg v4/v3 = (T3/T4)1/(k - 1) = 32.5 = 15.6 P T 2 T s TH 3 s 1 T 4 TL v 2 1 3 4 s 11.118 Do the previous problem 11.117 using values from Table A.7.1. Air in a piston/cylinder goes through a Carnot cycle in which TL = 26.8°C and the total cycle efficiency is η = 2/3. Find TH, the specific work and volume ratio in the adiabatic expansion. Solution: Carnot cycle efficiency Eq.7.5: η = 1 - TL/TH = 2/3 ⇒ TH = 3 × TL = 3 × 300 = 900 K From A.7.1: ° u3 = 674.82 kJ/kg, sT3 = 8.0158 kJ/kg K ° u4 = 214.36 kJ/kg, sT4 = 6.8693 kJ/kg K Energy equation with q = 0 3w4 = u3 - u4 = 674.82 - 214.36 = 460.5 kJ/kg Entropy equation, constant s ° ° ° sT4 = sT3 + R ln(P4 / P3) = sT3 + R ln(T4 / T3) + R ln(v3/v4) => 6.8693 = 8.0158 + 0.287 ln(300/900) + 0.287 ln(v3/v4) => v4/v3 = 18.1 P T 2 T s TH 3 s 1 T 4 TL v 2 1 3 4 s Refrigeration cycles 11.119 A refrigerator with R-12 as the working fluid has a minimum temperature of −10°C and a maximum pressure of 1 MPa. Assume an ideal refrigeration cycle as in Fig. 11.24. Find the specific heat transfer from the cold space and that to the hot space, and the coefficient of performance. Solution: Exit evaporator sat. vapor −10°C from B.3.1: h1 = 183.19, s1 = 0.7019 kJ/kgK Exit condenser sat. liquid 1 MPa from B.3.1: h3 = 76.22 kJ/kg Compressor: s2 = s1 & P2 from B.3.2 ⇒ h2 ≈ 210.1 kJ/kg Evaporator: qL = h1 - h4 = h1 - h3 = 183.19 - 76.22 = 107 kJ/kg Condenser: qH = h2 - h3 = 210.1 - 76.22 = 133.9 kJ/kg COP: β = qL/wc = qL/(qH - qL) = 3.98 T Ideal refrigeration cycle Pcond = P3= P2 = 1 MPa Tevap = -10oC = T1 Properties from Table B.3 2 3 4 1 s 11.120 Consider an ideal refrigeration cycle that has a condenser temperature of 45°C and an evaporator temperature of −15°C. Determine the coefficient of performance of this refrigerator for the working fluids R-12 and R-22. Solution: T Ideal refrigeration cycle Tcond = 45oC = T3 Tevap = -15oC = T1 2 3 4 1 s Compressor Exp. valve Evaporator Property for: h1, kJ/kg s2 = s1, kJ/kg K P2, MPa T2, oC h2, kJ/kg wC = h2 - h1 h3 = h4, kJ/kg qL = h1 - h4 β = qL/wC R-12, B.3 180.97 0.7051 1.0843 54.7 R-22, B.4 244.13 0.9505 1.729 74.4 212.63 31.66 79.71 101.26 3.198 289.26 45.13 100.98 143.15 3.172 The value of h2 is taken from the computer program as it otherwise will be a double interpolation due to the value of P2. 11.121 The environmentally safe refrigerant R-134a is one of the replacements for R-12 in refrigeration systems. Repeat Problem 11.120 using R-134a and compare the result with that for R-12. Consider an ideal refrigeration cycle that has a condenser temperature of 45°C and an evaporator temperature of −15°C. Determine the coefficient of performance of this refrigerator for the working fluids R-12 and R-22. Solution: T Ideal refrigeration cycle Tcond = 45oC = T3 2 3 Tevap = -15oC = T1 4 1 s Compressor Exp. valve Evaporator Property for: h1, kJ/kg s2 = s1, kJ/kg K P2, MPa T2, oC h2, kJ/kg wC = h2 - h1 h3 = h4, kJ/kg qL = h1 - h4 β = qL/wC R-12, B.3 180.97 0.7051 1.0843 54.7 R-134a, B.5 389.2 1.7354 1.16 51.8* 212.63 31.66 79.71 101.26 3.198 429.9* 40.7 264.11 125.1 3.07 * To get state 2 an interpolation is needed: At 1 MPa, s = 1.7354 : T = 45.9 and h = 426.8 kJ/kg At 1.2 MPa, s = 1.7354 : T = 53.3 and h = 430.7 kJ/kg make a linear interpolation to get properties at 1.16 MPa 11.122 A refrigerator using R-22 is powered by a small natural gas fired heat engine with a thermal efficiency of 25%, as shown in Fig.P11.122. The R-22 condenses at 40°C and it evaporates at −20°C and the cycle is standard. Find the two specific heat transfers in the refrigeration cycle. What is the overall coefficient of performance as QL/Q1? Solution: Evaporator: Inlet State is saturated liq-vap with h4 = h3 =94.27 kJ/kg The exit state is saturated vapor with h1 = 242.06 kJ/kg qL = h1 - h4 = h1 - h3 = 147.79 kJ/kg Compressor: Inlet State 1 and Exit State 2 about 1.6 MPa wC = h2 - h1 ; s2 = s1 = 0.9593 kJ/kgK 2: T2 ≈ 70°C h2 = 287.2 kJ/kg wC = h2 - h1 = 45.14 kJ/kg Condenser: Brings it to saturated liquid at state 3 qH = h2 - h3 = 287.2 - 94.27 = 192.9 kJ/kg Overall Refrigerator: β = qL / wC = 147.79 / 45.14 = 3.274 Heat Engine: . . . . WHE = ηHEQ1 = WC = QL / β . . QL / Q1 = ηβ = 0.25 × 3.274 = 0.819 T Ideal refrigeration cycle Tcond = 40oC = T3 2 3 Tevap = -20oC = T1 Properties from Table B.4 4 1 s 11.123 A refrigerator in a meat warehouse must keep a low temperature of -15°C and the outside temperature is 20°C. It uses R-12 as the refrigerant which must remove 5 kW from the cold space. Find the flow rate of the R-12 needed assuming a standard vapor compression refrigeration cycle with a condenser at 20°C. Solution: Basic refrigeration cycle: Table B.3: T1 = T4 = -15°C, T3 = 20°C h4 = h3 = 54.87 kJ/kg; h1 = hg = 180.97 kJ/kg . . . QL = mR-12 × qL = mR-12(h1 - h4) qL = 180.97 - 54.87 = 126.1 kJ/kg . mR-12 = 5.0 / 126.1 = 0.03965 kg/s T Ideal refrigeration cycle Tcond = 20oC 2 3 Tevap = -15oC = T1 Properties from Table B.3 4 1 s 11.124 A refrigerator with R-12 as the working fluid has a minimum temperature of −10°C and a maximum pressure of 1 MPa. The actual adiabatic compressor exit temperature is 60°C. Assume no pressure loss in the heat exchangers. Find the specific heat transfer from the cold space and that to the hot space, the coefficient of performance and the isentropic efficiency of the compressor. Solution: State 1: Inlet to compressor, sat. vapor -10°C, h1 = 183.19 kJ/kg, s1 = 0.7019 kJ/kg K State 2: Actual compressor exit, h2AC = 217.97 kJ/kg State 3: Exit condenser, sat. liquid 1MPa, h3 = 76.22 kJ/kg State 4: Exit valve, h4 = h3 C.V. Evaporator: qL = h1 - h4 = h1 - h3 = 107 kJ/kg C.V. Ideal Compressor: wC,S = h2,S - h1, s2,S = s1 State 2s: 1 MPa, s = 0.7019 kJ/kg K; T2,S = 49.66°C, h2,S = 210.1 kJ/kg wC,S = h2,S - h1 = 26.91 kJ/kg C.V. Actual Compressor: wC = h2,AC - h1 = 34.78 kJ/kg β= qL = 3.076, ηC = wC,S/wC = 0.774 wC C.V. Condenser: qH = h2,AC - h3 = 141.75 kJ/kg Ideal refrigeration cycle with actual compressor Pcond = P3= P2 = 1 MPa T2 = 60oC Tevap = -10oC = T1 Properties from Table B.3 T 2s 2ac 3 4 1 s 11.125 Consider an ideal heat pump that has a condenser temperature of 50°C and an evaporator temperature of 0°C. Determine the coefficient of performance of this heat pump for the working fluids R-12, R-22, and ammonia. Solution: T Ideal heat pump Tcond = 50oC = T3 2 3 Tevap = 0oC = T1 4 1 s C.V. Compressor Exp. valve Condenser Property for: From Table: h1, kJ/kg s2 = s1, kJ/kgK P2, MPa T2, oC h2, kJ/kg wC = h2 - h1 h3 = h4, kJ/kg qH = h2 - h 3 β′ =qH/wC R-12 B.3 187.53 0.6965 1.2193 56.7 R-22 B.4 249.95 0.9269 1.9423 72.2 NH3 B.2 1442.32 5.3313 2.0333 115.6 211.95 24.42 84.94 127.01 5.201 284.25 34.3 107.85 176.4 5.143 1672.84 230.52 421.58 1251.26 5.428 11.126 The air conditioner in a car uses R-134a and the compressor power input is 1.5 kW bringing the R-134a from 201.7 kPa to 1200 kPa by compression. The cold space is a heat exchanger that cools atmospheric air from the outside 30°C down to 10°C and blows it into the car. What is the mass flow rate of the R-134a and what is the low temperature heat transfer rate. How much is the mass flow rate of air at 10°C? Standard Refrigeration Cycle Table B.5: h1 = 392.28 kJ/kg; s1 = 1.7319 kJ/kg K; h4 = h3 = 266 C.V. Compressor (assume ideal) . . m1 = m2 wC = h2 - h1; s2 = s1 + sgen P2, s = s1 => h2 = 429.5 kJ/kg => wC = 37.2 kJ/kg . . . m wC = WC => m = 1.5 / 37.2 = 0.0403 kg/s C.V. Evaporator . . QL = m(h1 - h4) = 0.0405(392.28 - 266) = 5.21 kW C.V. Air Cooler . . . mair∆hair = QL ≈ mairCp∆T . . mair = QL / (Cp∆T) = 5.21 / (1.004×20) = 0.26 kg / s T 2 Ideal refrigeration cycle Pcond = 1200 kPa = P3 Pevap = 201.7 kPa = P1 3 4 1 s 11.127 A refrigerator using R-134a is located in a 20°C room. Consider the cycle to be ideal, except that the compressor is neither adiabatic nor reversible. Saturated vapor at -20°C enters the compressor, and the R-134a exits the compressor at 50°C. The condenser temperature is 40°C. The mass flow rate of refrigerant around the cycle is 0.2 kg/s, and the coefficient of performance is measured and found to be 2.3. Find the power input to the compressor and the rate of entropy generation in the compressor process. Solution: h4 = h3 = 256.54 kJ/kg Table B.5: P2 = P3 = Psat 40C = 1017 kPa, s2 ≈ 1.7472 kJ/kg K, h2 ≈ 430.87 kJ/kg; s1 = 1.7395 kJ/kg K, h1 = 386.08 kJ/kg β = qL / wC -> wC = qL / β = (h1- h4) / β = (386.08 - 256.54) / 2.3 = 56.32 . . WC = m wC = 11.26 kW C.V. Compressor h1 + wC + q = h2 -> qin = h2 - h1 - wC = 430.87 - 386.08 - 56.32 = -11.53 kJ/kg i.e. a heat loss s1 + ∫ dQ/T + sgen = s2 sgen = s2 - s1 - q / To = 1.7472 - 1.7395 + (11.53 / 293.15) = 0.047 kJ/kg K . . Sgen = m sgen = 0.2 × 0.047 = 0.0094 kW / K Ideal refrigeration cycle with actual compressor Tcond = 40oC T 2s 2ac 3 T2 = 50oC Tevap = -20oC = T1 Properties from Table B.5 4 1 s 11.128 A refrigerator has a steady flow of R-22 as saturated vapor at –20°C into the adiabatic compressor that brings it to 1000 kPa. After the compressor, the temperature is measured to be 60°C. Find the actual compressor work and the actual cycle coefficient of performance. Solution: Table B.4.1: h1 = 242.06 kJ/kg, s1 = 0.9593 kJ/kg K P2 = P3 = 1000 kPa, h4 = h3 = hf = 72.86 kJ/kg h2 ac = 286.97 kJ/kg C.V. Compressor (actual) Energy Eq.: wC ac = h2 ac - h1 = 286.97 – 242.06 = 44.91 kJ/kg C.V. Evaporator Energy Eq.: qL = h1- h4 = h1- h3 = 242.06 – 72.86 = 169.2 kJ/kg β= qL 169.2 = = 3.77 wC ac 44.91 Ideal refrigeration cycle with actual compressor Tcond = 23.4oC = Tsat 1000 kPa T 2s 2ac 3 T2 = 60oC Tevap = -20oC = T1 Properties from Table B.4 4 1 s 11.129 A small heat pump unit is used to heat water for a hot-water supply. Assume that the unit uses R-22 and operates on the ideal refrigeration cycle. The evaporator temperature is 15°C and the condenser temperature is 60°C. If the amount of hot water needed is 0.1 kg/s, determine the amount of energy saved by using the heat pump instead of directly heating the water from 15 to 60°C. Solution: Ideal R-22 heat pump T 2 o o T1 = 15 C, T3 = 60 C From Table B.4.1 h1 = 255.02 kJ/kg, s2 = s1 = 0.9062 kJ/kg K P2 = P3 = 2.427 MPa, h3 = 122.18 kJ/kg Entropy compressor: s2 = s1 => 3 4 1 T2 = 78.4oC, h2 = 282.86 kJ/kg Energy eq. compressor: wC = h2 - h1 = 27.84 kJ/kg Energy condenser: qH = h2 - h3 = 160.68 kJ/kg To heat 0.1 kg/s of water from 15oC to 60oC, . . QH2O = m(∆h) = 0.1(251.11 - 62.98) = 18.81 kW Using the heat pump . . WIN = QH2O(wC/qH) = 18.81(27.84/160.68) = 3.26 kW a saving of 15.55 kW s 11.130 The refrigerant R-22 is used as the working fluid in a conventional heat pump cycle. Saturated vapor enters the compressor of this unit at 10°C; its exit temperature from the compressor is measured and found to be 85°C. If the compressor exit is at 2 MPa what is the compressor isentropic efficiency and the cycle COP? Solution: R-22 heat pump: Table B.4 State 1: TEVAP = 10oC, x = 1 h1 = 253.42 kJ/kg, s1 = 0.9129 kJ/kg K State 2: T2, P2: C.V. Compressor Energy Eq.: 4 2 1 s wC ac = h2 - h1 = 295.17 – 253.42 = 41.75 kJ/kg η= T2S = 69oC, h2S = 280.2 kJ/kg wC s h2S - h1 280.2 - 253.42 = = = 0.6414 wC ac h2 - h1 295.17 - 253.42 C.V. Condenser Energy Eq.: qH = h2 - h3 = 295.17 – 109.6 = 185.57 kJ/kg COP Heat pump: 2s 3 h2 = 295.17 kJ/kg State 2s: 2 MPa , s2S = s1 = 0.9129 kJ/kg Efficiency: T β= qH 185.57 = = 4.44 wC ac 41.75 11.131 A refrigerator in a laboratory uses R-22 as the working substance. The high pressure is 1200 kPa, the low pressure is 201 kPa, and the compressor is reversible. It should remove 500 W from a specimen currently at –20°C (not equal to T in the cycle) that is inside the refrigerated space. Find the cycle COP and the electrical power required. Solution: State 1: 201 kPa, x = 1, Table B.4.1: h1 = 239.92 kJ/kg, s1 = 0.9685 kJ/kg K State 3: 1200 kPa, x = 0, Table B.4.1: h3 = 81.57 kJ/kg C.V. Compressor Energy Eq.: wC = h2 - h1 Entropy Eq.: s2 = s1 + sgen = s1 State 2: 1.2 MPa , s2 = s1 = 0.9685 kJ/kg, T2 ≈ 60oC, h2 = 285.21 kJ/kg wC = h2 - h1 = 285.21 – 239.92 = 45.29 kJ/kg Energy Eq. evaporator: COP Refrigerator: Power: qL = h1 – h4 = h1 – h3 = 239.92 – 81.57 = 158.35 kJ/kg β= qL 158.35 = = 3.5 wC 45.29 . . WIN = QL / β = 500 W/ 3.5 = 142.9 W 11.132 Consider the previous problem and find the two rates of entropy generation in the process and where they occur. Solution: From the basic cycle we know that entropy is generated in the valve as the throttle process is irreversible. State 1: 201 kPa, x = 1, Table B.4.1: h1 = 239.92 kJ/kg, s1 = 0.9685 kJ/kg K State 3: 1200 kPa, x = 0, Table B.4.1: h3 = 81.57 kJ/kg, s3 = 0.30142 kJ/kg K Energy Eq. evaporator: qL = h1 – h4 = h1 – h3 = 239.92 – 81.57 = 158.35 kJ/kg Mass flow rate: . . m = QL / qL = 0.5 / 158.35 = 0.00316 kg/s C.V. Valve Energy Eq.: h4 = h3 = 81.57 kJ/kg x4 = => x4 = (h4 – hf)/hfg 81.57 - 16.19 = 0.29223 223.73 s4 = sf + x4 sfg = 0.067 + x4 × 0.9015 = 0.33045 kJ/kg K Entropy Eq.: sgen = s4 - s3 = 0.33045 – 0.30142 = 0.02903 kJ/kg K . . Sgen valve = msgen = 0.00316 × 0.02903 = 0.0917 W/K There is also entropy generation in the heat transfer process from the specimen at –20°C to the refrigerant T = -25°C = Tsat (201 kPa). . . Sgen inside = QL [ 1 Tspecimen – 1 1 1 = 500 (248 – 253) = 0.04 W/K TL 11.133 In an actual refrigeration cycle using R-12 as the working fluid, the refrigerant flow rate is 0.05 kg/s. Vapor enters the compressor at 150 kPa, −10°C, and leaves at 1.2 MPa, 75°C. The power input to the compressor is measured and found be 2.4 kW. The refrigerant enters the expansion valve at 1.15 MPa, 40°C, and leaves the evaporator at 175 kPa, −15°C. Determine the entropy generation in the compression process, the refrigeration capacity and the coefficient of performance for this cycle. Solution: Actual refrigeration cycle 1: compressor inlet T1 = -10oC, P1 = 150 kPa 2: compressor exit T2 = 75oC, P2 = 1.2 MPa 3: Expansion valve inlet T3 = 40oC P3 = 1.15 MPa 5: evaporator exit Table B.3 2 T 3 5 4 1 T5 = -15oC, P5 = 175 kPa h1 = 184.619, s1 = 0.7318, h2 = 226.543, s2 = 0.7404 CV Compressor: h1 + qCOMP + wCOMP = h2 ; s1 + ∫ dq/T + sgen = s2 . . wCOMP = WCOMP/m = 2.4/0.05 = 48.0 kJ/kg qCOMP = h2 - wCOMP - h1 = 226.5 - 48.0 - 184.6 = -6.1 kJ/kg sgen = s2 - s1 - q / To = 0.7404 - 0.7318 + 6.1/298.15 = 0.029 kJ / kg K C.V. Evaporator qL = h5 - h4 = 181.024 - 74.527 = 106.5 kJ/kg . . ⇒ QL = mqL = 0.05 × 106.5 = 5.325 kW COP: β = qL/wCOMP = 106.5/48.0 = 2.219 s Ammonia absorption cycles 11.134 Consider a small ammonia absorption refrigeration cycle that is powered by solar energy and is to be used as an air conditioner. Saturated vapor ammonia leaves the generator at 50°C, and saturated vapor leaves the evaporator at 10°C. If 7000 kJ of heat is required in the generator (solar collector) per kilogram of ammonia vapor generated, determine the overall performance of this system. Solution; T NH3 absorption cycle: sat. vapor at 50oC exits the generator sat. vapor at 10oC exits the evaporator qH = qGEN = 7000 kJ/kg NH3 out of gen. Exit generator 1 Evaporator exit 2 C.V. Evaporator qL = h2 - h1 = hg 10oC - hf 50oC = 1452.2 - 421.6 = 1030.6 kJ/kg COP ⇒ qL/qH = 1030.6/7000 = 0.147 s 11.135 The performance of an ammonia absorption cycle refrigerator is to be compared with that of a similar vapor-compression system. Consider an absorption system having an evaporator temperature of −10°C and a condenser temperature of 50°C. The generator temperature in this system is 150°C. In this cycle 0.42 kJ is transferred to the ammonia in the evaporator for each kilojoule transferred from the high-temperature source to the ammonia solution in the generator. To make the comparison, assume that a reservoir is available at 150°C, and that heat is transferred from this reservoir to a reversible engine that rejects heat to the surroundings at 25°C. This work is then used to drive an ideal vapor-compression system with ammonia as the refrigerant. Compare the amount of refrigeration that can be achieved per kilojoule from the high-temperature source with the 0.42 kJ that can be achieved in the absorption system. Solution: o T T H = 50 C o T ' = 150 C H QH Q' = 1 kJ H 2 3 3 CONDENSER 2 W C REV. H.E. 4 1 s COMP. ' QL EVAPORATOR T ' = 25oC L QL 1 4 T1 = -10 oC h1 = 1430.8 , s1 = 5.4673 h4 = h3 = 421.48 o TL = -10 C For the rev. heat engine: 298.2 ′′ ηTH = 1 - TL/TH = 1 = 0.295 423.2 ′ ⇒ WC = ηTH QH = 0.295 kJ For the NH3 refrig. cycle: s2 = s1 = 5.4673 P2 = P3 = 2033 kPa , => Use 2000 kPa Table T2 ≈ 135°C h2 ≈ 1724 wC = h2 - h1 = 1724 - 1430.8 = 293.2 kJ/kg qL = h1 - h4 = 1430.8 - 421.48 = 1009.3 kJ/kg β = qL/wC = 1009.3 / 293.2 = 3.44 ⇒ QL = βwC = 3.44 × 0.295 = 1.015 kJ based on assumption of ideal heat engine & refrigeration cycle. Air standard refrigeration cycles 11.136 The formula for the coefficient of performance when we use cold air properties is not given in the text. Derive the expression for COP as function of the compression ratio similar to how the Brayton cycle efficiency was found. Definition of COP: β= qL qL 1 = = wnet qH - qL qH -1 qL From the refrigeration cycle we get the ratio of the heat transfers as qH Cp(T2 - T3) T2(1 - T3/T2) = = qL Cp(T1 - T4) T1(1 - T4/T1) The pressure ratios are the same and we have isentropic compression/expansion P2 P3 T2k/(k-1) T3k/(k-1) == = P1 P4 T1 T4 so now we get T2 T3 T4 T3 = or = T1 T4 T1 T2 The heat transfer ratio simplifies to qH T2 = qL T1 and so the COP reduces to β= 1 1 = T2 P2(k-1)/k -1 -1 T1 P1 11.137 A heat exchanger is incorporated into an ideal air-standard refrigeration cycle, as shown in Fig. P11.137. It may be assumed that both the compression and the expansion are reversible adiabatic processes in this ideal case. Determine the coefficient of performance for the cycle. Solution: T qL 2 6 3 1 2 qH 5 3 6 4 COMP 1 4 5 EXP s Standard air refrigeration cycle with T1 = T3 = 15 oC = 288.2 K, P1 = 100 kPa, P2 = 1.4 MPa T4 = T6 = -50 oC = 223.2 K We will solve the problem with cold air properties. Compressor, isentropic s2 = s1 so from Eq.8.32 ⇒ T2 = T1(P2/P1) k-1 k = 288.2(1400/100)0.286 = 613 K wC = -w12 = CP0(T2 - T1) = 1.004(613 - 288.2) = 326 kJ/kg Expansion in expander (turbine) s5 = s4 ⇒ T5 = T4(P5/P4) k-1 k = 223.2(100/1400) 0.286 = 104.9 K wE = CP0(T4 - T5) = 1.004(223.2 - 104.9) = 118.7 kJ/kg Net cycle work wNET = wE - wC = 118.7 - 326.0 = -207.3 kJ/kg qL = CP0(T6 - T5) = wE = 118.7 kJ/kg Overall cycle performance, COP β = qL/wNET = 118.7 / 207.3 = 0.573 11.138 Repeat Problems 11.137, but assume that helium is the cycle working fluid instead of air. Discuss the significance of the results. A heat exchanger is incorporated into an ideal air-standard refrigeration cycle, as shown in Fig. P11.137. It may be assumed that both the compression and the expansion are reversible adiabatic processes in this ideal case. Determine the coefficient of performance for the cycle. Solution: T q 2 L 6 3 1 2 qH 5 3 6 4 COMP 1 4 EXP 5 s Standard air refrigeration cycle with helium and states as T1 = T3 = 15 oC = 288.2 K, P1 = 100 kPa, P2 = 1.4 MPa T4 = T6 = -50 oC = 223.2 K Compressor, isentropic s2 = s1 so from Eq.8.32 k-1 k 14000.40 = 288.2 = 828.2 K 100 wC = -w12 = CP0(T2 - T1) = 5.193(828.2 - 288.2) = 2804.1 kJ/kg ⇒ T2 = T1(P2/P1) Expansion in expander (turbine) k-1 k 100 0.40 = 223.2 = 77.7 K 1400 wE = CP0(T4 - T5) = 15.193(223.2 - 77.7) = 755.5 kJ/kg s5 = s4 ⇒ T5 = T4(P5/P4) Net cycle work wNET = 755.5 - 2804.1 = -2048.6 kJ/kg qL = CP0(T6 - T5) = 5.193(223.2 - 77.7) = 755.5 kJ/kg Overall cycle performance, COP β = qL/wNET = 755.5/2048.6 = 0.369 Notice that the low temperature is lower and work terms higher than with air. It is due to the higher heat capacity CP0 and ratio of specific heats ( k = 1 2/3). The expense is a lower COP requiring more work input per kJ cooling. 11.139 Repeat Problem 11.137, but assume an isentropic efficiency of 75% for both the compressor and the expander. Standard air refrigeration cycle with T1 = T3 = 15 oC = 288.2 K, P1 = 100 kPa, P2 = 1.4 MPa T4 = T6 = -50 oC = 223.2 K We will solve the problem with cold air properties. Ideal compressor, isentropic s2S = s1 so from Eq.8.32 ⇒ T2S = T1(P2/P1) k-1 k = 288.2(1400/100)0.286 = 613 K wSC = -w12 = CP0(T2S - T1) = 1.004(613 - 288.2) = 326 kJ/kg The actual compressor wC = wSC / ηSC = 326/0.75 = 434.6 kJ/kg Expansion in ideal expander (turbine) s5 = s4 ⇒ T5S = T4(P5/P4) k-1 k = 223.2(100/1400) 0.286 = 104.9 K wE = CP0(T4 - T5) = 1.004(223.2 - 104.9) = 118.7 kJ/kg 2 The actual expander (turbine) wE = ηSE × wSE = 0.75 × 118.7 = 89.0 kJ/kg = CP0(T4-T5) = 1.004(223.2 - T5) T ⇒ T5 = 134.5 K 2S 3 1 4 6 5S 5 wNET = 89.0 - 434.6 = -345.6 kJ/kg s qL = CP0(T6 - T5) = 1.004(223.2 - 134.5) = 89.0 kJ/kg β = qL/(-wNET) = 89.0/345.6 = 0.258 Combined Cycles 11.140 A binary system power plant uses mercury for the high-temperature cycle and water for the low-temperature cycle, as shown in Fig. 11.39. The temperatures and pressures are shown in the corresponding T–s diagram. The maximum temperature in the steam cycle is where the steam leaves the superheater at point 4 where it is 500°C. Determine the ratio of the mass flow rate of mercury to the mass flow rate of water in the heat exchanger that condenses mercury and boils the water and the thermal efficiency of this ideal cycle. The following saturation properties for mercury are known P, MPa Tg, °C hf, kJ/kg hg, kJ/kg sf kJ/kgK 0.04 309 42.21 335.64 0.1034 1.60 562 75.37 364.04 0.1498 Solution: For the mercury cycle: sg, kJ/kgK 0.6073 0.4954 sd = sc = 0.4954 = 0.1034 + xd × 0.5039, xd = 0.7779 hb = ha - wP HG ≈ ha ( since vF is very small) qH = hc - ha = 364.04 - 42.21 = 321.83 kJ/kg qL = hd - ha = 270.48 - 42.21 = 228.27 kJ/kg For the steam cycle: s5 = s4 = 7.0097 = 0.6493 + x5 × 7.5009, x5 = 0.8480 h5 = 191.83 + 0.848 × 2392.8 = 2220.8 wP ≈ v1(P2 - P1) = 0.00101(4688 - 10) = 4.7 kJ/kg h2 = h1 + wP = 191.8 + 4.7 = 196.5 qH (from Hg) = h3 - h2 = 2769.9 - 196.5 = 2600.4 qH (ext. source) = h4 - h3 = 3437.4 - 2796.9 = 640.5 CV: Hg condenser - H2O boiler: mHg/mH2O = 1st law: mHg(hd - ha) = mH2O(h3 - h2) 2796.9 - 196.5 = 11.392 270.48 - 42.21 qH TOTAL = (mHg/mH2O)(hc - hb) + (h4 - h3) (for 1 kg H2O) = 11.392 × 321.83 + 640.5 = 4306.8 kJ All qL is from the H2O condenser: qL = h5 - h1 = 2220.8 - 191.8 = 2029.0 kJ wNET = qH - qL = 4306.8 - 2029.0 = 2277.8 kJ ηTH = wNET/qH = 2277.8/4306.8 = 0.529 11.141 A Rankine steam power plant should operate with a high pressure of 3 MPa, a low pressure of 10 kPa, and the boiler exit temperature should be 500°C. The available high-temperature source is the exhaust of 175 kg/s air at 600°C from a gas turbine. If the boiler operates as a counterflowing heat exchanger where the temperature difference at the pinch point is 20°C, find the maximum water mass flow rate possible and the air exit temperature. Solution: T C.V. Pump wP = h2 - h1 = v1(P2 - P1) = 0.00101(3000 - 10) = 3.02 kJ/kg h2 = h1 + wP = 191.83 + 3.02 = 194.85 kJ/kg Heat exchanger water states State 2a: T2a = TSAT = 233.9 °C h2a = 1008.42 kJ/kg State 3: h3 = 3456.5 kJ/kg 3 2a 2 1 s e Heat exchanger air states inlet: hair,in = 903.16 kJ/kg State 2a: hair(T2a + 20) = 531.28 kJ/kg a i HEAT EXCH 2 3 2a Air temperature should be 253.9°C at the point where the water is at state 2a. C.V. Section 2a-3, i-a . . mH2O(h3 - h2a) = mair(hi - ha) . 903.16 - 531.28 mH2O = 175 = 26.584 kg/s 3456.5 - 1008.42 . . Take C.V. Total: mH2O(h3 - h2) = mair(hi - he) . . ⇒ he = hi - mH2O(h3 - h2)/mair = 903.6 - 26.584(3456.5 - 194.85)/175 = 408.13 kJ/kg ⇒ Te = 406.7 K = 133.6 °C, Te > T2 = 46.5 °C OK. 11.142 A simple Rankine cycle with R-22 as the working fluid is to be used as a bottoming cycle for an electrical generating facility driven by the exhaust gas from a Diesel engine as the high temperature energy source in the R-22 boiler. Diesel inlet conditions are 100 kPa, 20°C, the compression ratio is 20, and the maximum temperature in the cycle is 2800°C. Saturated vapor R-22 leaves the bottoming cycle boiler at 110°C, and the condenser temperature is 30°C. The power output of the Diesel engine is 1 MW. Assuming ideal cycles throughout, determine a. The flow rate required in the diesel engine. b. The power output of the bottoming cycle, assuming that the diesel exhaust is cooled to 200°C in the R-22 boiler. T 2 1 T AIR-STD DIESEL CYCLE 3 P 4 7 6 5 v s IDEAL R-12 RANKINE BOTTOMING CYCLE 8 s Diesel cycle information given means: Inlet state: P1 = 100 kPa, T1 = 20 oC, Compression ratio: v1/v2 = 20, High temperature: T3 = 2800oC, . Power output: WDIESEL = 1.0 MW Rankine cycle information given means: Boiler exit state: T7 = 110 oC, x7 = 1.0 Condenser temperature: a) T5 = T8 = 30oC Consider the Diesel cycle T2 = T1(v1/v2)k-1 = 293.2(20)0.4 = 971.8 K P2 = P1(v1/v2)k = 100(20)1.4 = 6629 kPa qH = CP0(T3 - T2) = 1.004(3073.2 - 971.8) = 2109.8 kJ/kg v1 = 0.287 × 293.2 0.8415 = 0.8415, v2 = = 0.04208 20 100 v3 = v2(T3/T2) = 0.04208(3073.2/971.8) = 0.13307 v3k-1 0.133 070.4 T4 = T3 = 3073.2 = 1469.6 K 0.8415 v4 qL = 0.717(293.2 - 1469.6) = -843.5 kJ/kg wNET = 2109.8 - 843.5 = 1266.3 kJ/kg . . mAIR = WNET/wNET = 1000/1266.3 = 0.79 kg/s b) Consider the Rankine cycle s8 = s7 = 0.60758 = 0.2399 + x8 × 0.4454, x8 = 0.8255 h8 = 64.59 + 0.8255 × 135.03 = 176.1 kJ/kg wT = h7 - h8 = 198.0 - 176.1 = 21.9 kJ/kg -wP = v5(P6 - P5) = 0.000774(3978.5 - 744.9) = 2.50 h6 = h5 - wP = 64.6 + 2.5 = 67.1 kJ/kg qH = h7 - h6 = 198.0 - 67.1 = 130.9 kJ/kg Connecting the two cycles. . QH available from Diesel exhaust cooled to 200 oC: . QH = 0.79 × 0.717(1469.6 - 473.2) = 564 kW . . ⇒ mR-12 = QH/qH = 564/130.9 = 4.309 kg/s . WR-12 = 4.309(21.9 - 2.5) = 83.6 kW Comment: The heat exchange process between the two cycles is not realistic. The exhaust must be expanded down to 100 kPa from state 4 and then flow at constant P through a heat exchanger. 11.143 A cascade system is composed of two ideal refrigeration cycles, as shown in Fig. 11.41. The high-temperature cycle uses R-22. Saturated liquid leaves the condenser at 40°C, and saturated vapor leaves the heat exchanger at −20°C. The low-temperature cycle uses a different refrigerant, R-23. Saturated vapor leaves the evaporator at −80°C, h = 330 kJ/kg, and saturated liquid leaves the heat exchanger at −10°C, h = 185 kJ/kg. R-23 out of the compressor has h = 405 kJ/kg. Calculate the ratio of the mass flow rates through the two cycles and the coefficient of performance of the system. T3 = 40oC ' 3' COND 2' sat. liquid C R-22 sat. vapor o -20 C 1' 4' o 2 1 3 sat. liquid R-23 C o T 1 = -80 C sat. vapor 3 T = -10 C 4 EVAP T T 2' 2 3' 3 4' 1' 4 1 s 1′ 2′ 3′ 4′ T,oC -20 71 40 -20 P 0.245 1.534 1.534 h 242.1 289.0 94.3 94.3 s s 0.9593 0.9593 1 2 3 4 T,oC -80 50 -10 -80 P 0.12 1.90 1.90 0.12 h 330 405 185 185 h1 - h4 242.1 - 94.3 ′′ .. m/m′ = = = 0.672 405 - 185 h2 - h3 qL = h1 - h4 = 330 - 185 = 145 kJ/kg . . .. - WTOT/m = (h2 - h1) + (m′/m)(h2 - h1) ′′ = (405 - 330) + (1/0.672)(289 - 242.1) = 144.8 kJ/kg . β = QL/-WTOT = 145/144.8 = 1.0 s 1.76 1.76 11.144 Consider an ideal dual-loop heat-powered refrigeration cycle using R-12 as the working fluid, as shown in Fig. P11.87. Saturated vapor at 105°C leaves the boiler and expands in the turbine to the condenser pressure. Saturated vapor at −15°C leaves the evaporator and is compressed to the condenser pressure. The ratio of the flows through the two loops is such that the turbine produces just enough power to drive the compressor. The two exiting streams mix together and enter the condenser. Saturated liquid leaving the condenser at 45°C is then separated into two streams in the necessary proportions. Determine the ratio of mass flow rate through the power loop to that through the refrigeration loop. Find also the . . performance of the cycle, in terms of the ratio QL /QH. Solution: T 1 TURB. COMP. 7 2 6 BOIL. E V A P . COND. 5 . QL 6 5 2 3 7 4 1 4 s 3 T1 = -15 oC sat. vap. P Table B.3.1 T6 = 105oC sat. vapor Table B.3.1 T3 = 45oC sat. liquid => => P5 = P6 = 3.6509 MPa P2 = P3 = P7 = 1.0843 MPa h1 = 180.97; h3 = h4 = 79.71; h6 = 206.57 C.V. Turbine s7 = s6 = 0.6325 = 0.2877 + x7 × 0.3934; x7 = 0.8765 h7 = 79.71 + 0.8765 × 125.16 = 189.41 C.V. Compressor (computer tables are used for this due to value of P) s2 = s1 = 0.7051, P2 => T2 = 54.7oC, h2 = 212.6 kJ/kg CV: turbine + compressor Continuity Eq.: . . . . m1 = m2, m6 = m7 ; Energy Eq.: . . . . m1h1 + m6h6 = m2h2 + m7h7 .. m6/m1 = (212.6 - 180.97)/(206.57 - 189.41) = 1.843 CV: pump wP = v3(P5 - P3) = 0.000811(3651 - 1084) = 2.082 kJ/kg h5 = h3 + wP = 81.79 kJ/kg . . CV: evaporator ⇒ QL = m1(h1 - h4) CV: boiler . . ⇒ QH = m6(h6 - h5) . . QL m1(h1 - h4) 180.97 - 79.71 β= =. = = 0.44 . 1.843(206.57 - 81.79) QH m6(h6 - h5) 11.145 For a cryogenic experiment heat should be removed from a space at 75 K to a reservoir at 180 K. A heat pump is designed to use nitrogen and methane in a cascade arrangement (see Fig. 11.41), where the high temperature of the nitrogen condensation is at 10 K higher than the low-temperature evaporation of the methane. The two other phase changes take place at the listed reservoir temperatures. Find the saturation temperatures in the heat exchanger between the two cycles that gives the best coefficient of performance for the overall system. The nitrogen cycle is the bottom cycle and the methane cycle is the top cycle. Both std. refrigeration cycles. THm = 180 K = T3m , TLN = 75 K = T4N = T1N TLm = T4m = T1m = T3N - 10, Trial and error on T3N or TLm. For each cycle we have, -wC = h2 - h1, s2 = s1, Nitrogen: T4 = T1 = 75 K N2 T3 h3 a) 120 -17.605 b) 115 -34.308 c) 110 -48.446 -qH = h2 - h3, qL = h1 - h4 = h1 - h3 ⇒ h1 = 74.867 kJ/kg, s1 = 5.4609 kJ/kg K P2 h2 -wc -qH 2.5125 202.96 128.1 220.57 1.9388 188.35 113.5 222.66 1.4672 173.88 99.0 222.33 Methane: T3 = 180 K ⇒ h3 = -0.5 kJ/kg, P2 = 3.28655 MPa CH4 T4 h1 s1 h2 -wc -qH a) 110 221 9.548 540.3 319.3 540.8 b) 105 212.2 9.691 581.1 368.9 581.6 c) 100 202.9 9.851 629.7 426.8 630.2 The heat exchanger that connects the cycles transfers a Q . . . . .. QHn = qHn mn = QLm = qLm mm => mm/mn = qHn/qLm The overall unit then has . . . . . QL 75 K = mn qLn ; Wtot in = - (mnwcn + mmwcm) . . .. β = QL 75 K/Wtot in = qLn/[-wcn -(mm/mn)wcm] .. .. β mm/mn wcn+(mm/mn)wcm a) 0.996 446.06 0.207 b) 1.047 499.65 0.219 c) 1.093 565.49 0.218 A maximum coeff. of performance is between case b) and c). Case qL 92.47 109.18 123.31 qL 221.5 212.7 203.4 Availabilty or Exergy Concepts 11.146 Find the flows and fluxes of exergy in the condenser of Problem 11.32. Use those to determine the second law efficiency. For this case we select To = 12°C = 285 K, the ocean water temperature. 1 The states properties from Tables B.1.1 and B.1.3 1: 45oC, x = 0: h1 = 188.42 kJ/kg, 6 cb 3: 3.0 MPa, 600oC: s3 = 7.5084 kJ/kg K C.V. Turbine : wT = h3 - h4 ; s4 = s3 s4 = s3 = 7.5084 = 0.6386 + x4 (7.5261) => 4 5 x4 = 0.9128 => h4 = 188.42 + 0.9128 (2394.77) = 2374.4 kJ/kg C.V. Condenser : qL = h4 - h1 = 2374.4 - 188.42 = 2186 kJ/kg . . . QL = mqL = 25 × 2186 = 54.65 MW = mocean Cp ∆T . . mocean = QL / Cp ∆T = 54 650 / (4.18 × 3) = 4358 kg/s The net drop in exergy of the water is . . Φwater = mwater [h4 – h1 – To(s4 – s1)] = 25 [ 2374.4 – 188.4 – 285 (7.5084 – 0.6386)] = 54 650 – 48 947 = 5703 kW The net gain in exergy of the ocean water is . . Φocean = mocean[h6 – h5 – To(s6 – s5)] T6 . = mocean[Cp(T6 – T5) – ToCp ln( ) ] T5 273 + 15 = 4358 [ 4.18(15 – 12) – 285 × 4.18 ln 273 + 12 = 54 650 – 54 364 = 286 kW The second law efficiency is . . 286 ηII = Φocean / Φwater = = 0.05 5703 In reality all the exergy in the ocean water is destroyed as the 15°C water mixes with the ocean water at 12°C after it flows back out into the ocean and the efficiency does not have any significance. Notice the small rate of exergy relative to the large rates of energy being transferred. 11.147 Find the availability of the water at all four states in the Rankine cycle described in Problem 11.33. Assume that the high-temperature source is 500°C and the lowtemperature reservoir is at 25°C. Determine the flow of availability in or out of the reservoirs per kilogram of steam flowing in the cycle. What is the overall cycle second law efficiency? Solution: Reference State: 100 kPa, 25°C, so = 0.3674 kJ/kg K, ho = 104.89 kJ/kg ψ1 = h1 - ho - To(s1 - so) = 191.83 - 104.89 - 298.15(0.6493 - 0.3674) = 2.89 kJ/kg ψ2 = 195.35 - 104.89 - 298.15(0.6493 - 0.3674) = ψ1 + 3.525 = 6.42 kJ/kg ψ3 = 3222.3 - 104.89 - 298.15(6.8405 - 0.3674) = 1187.5 kJ/kg ψ4 = ψ3 - wT,s = 131.96 kJ/kg ∆ψH = (1 - To/TH)qH = 0.6144 × 3027 = 1859.7 kJ/kg ∆ψL = (1 - To/To)qC = 0 kJ/kg ηII = wNET/∆ψH = (1055.5 - 3.53)/1859.7 = 0.5657 Notice— TH > T3, TL < T4 = T1 so cycle is externally irreversible. Both qH and qC over finite ∆T. 11.148 Find the flows of exergy into and out of the feedwater heater in Problem 11.43. State 1: x1 = 0, h1 = 298.25 kJ/kg, v1 = 0.001658 m3/kg State 3: x3 = 0, h3 = 421.48 kJ/kg, v3 = 0.001777 m3/kg State 5: h5 = 421.48 kJ/kg, s5 = 4.7306 kJ/kg K State 6: s6 = s5 => x6 = (s6 – sf)/sfg = 0.99052, h6 = 1461.53 kJ/kg C.V Pump P1 wP1 = h2 - h1 = v1(P2 - P1) = 0.001658(2033 - 1003) = 1.708 kJ/kg => h2 = h1 + wP1 = 298.25 + 1.708 = 299.96 kJ/kg C.V. Feedwater heater: Energy Eq.: Call . . m6 / mtot = x (the extraction fraction) (1 - x) h2 + x h6 = 1 h3 x= 3 h3 - h2 762.79 - 189.42 = = 0.1046 h6 - h2 3640.6 - 189.42 . . mextr = x mtot = 0.1046 × 5 = 0.523 kg/s . . m2 = (1-x) mtot = (1 – 0.1046) 5 = 4.477 kg/s FWH 6 x 1-x 2 Reference State: 100 kPa, 20°C, so = 6.2826 kJ/kg K, ho = 1516.1 kJ/kg ψ2 = h2 - ho - To(s2 - so) = 299.96 - 1516.1 - 293.15(1.121 - 6.2826) = 296.21 kJ/kg ψ6 = 1461.53 - 1516.1 - 293.15(4.7306 - 6.2826) = 400.17 kJ/kg ψ3 = 421.48 - 1516.1 - 293.15(1.5121 - 6.2826) = 303.14 kJ/kg The rate of exergy flow is then . . Φ2 = m2ψ2 = 4.477 × 296.21 = 1326 kW . . Φ6 = m6ψ6 = 0.523 × 400.17 = 209.3 kW . . Φ3 = m3ψ3 = 5.0 × 303.14 = 1516 kW The mixing is destroying 1326 + 209 – 1516 = 19 kW of exergy 11.149 Find the availability of the water at all the states in the steam power plant described in Problem 11.57. Assume the heat source in the boiler is at 600°C and the low-temperature reservoir is at 25°C. Give the second law efficiency of all the components. From solution to 11.21 and 11.57 : States h [kJ/kg] s [kJ/kg K] 0 104.89 0.3674 1 sat liq. 191.81 0.6492 2a 195.58 0.6529 3 2804.14 6.1869 4a (x = 0.7913) 2085.24 6.5847 The entropy for state 2a was done using the compressed liquid entry at 2MPa at the given h. You could interpolate in the compressed liquid tables to get at 3 MPa or use the computer tables to be more accurate. Definition of flow exergy: ψ = h - ho - To(s - so) ψ1= 191.81 - 104.89 - 298.15(0.6492 - 0.3674) = 2.90 kJ/kg ψ2a = 195.58 - 104.89 - 298.15(0.6529 - 0.3674) = 5.57 kJ/kg ψ3 = 2804.14 - 104.89 - 298.15(6.1869 - 0.3674) = 964.17 kJ/kg ψ4a = 2085.24 - 104.89 - 298.15(6.5847 - 0.3674) = 126.66 kJ/kg ηII Pump = (ψ2a - ψ1) / wp ac = (5.57 - 2.9) / 3.775 = 0.707 ηII Boiler = (ψ3 - ψ2a) / [(1- To/TH) qH] = (964.17 - 3.18) / [0.658×2608.6] = 0.56 ηII Turbine = wT ac / (ψ3 - ψ4a) = 718.9 / (964.17 - 126.66) = 0.858 ηII Cond = ∆ψamb / (ψ4a - ψ1) = 0 Remark: Due to the interpolation the efficiency for the pump is not quite correct. It should have a second law efficiency greater than the isentropic efficiency. 11.150 Consider the Brayton cycle in Problem 11.72. Find all the flows and fluxes of exergy and find the overall cycle second-law efficiency. Assume the heat transfers are internally reversible processes, and we then neglect any external irreversibility. Solution: Efficiency is from Eq.11.8 . Wnet wnet -0.4/1.4 -(k-1)/k η= = = 1 - rp = 1 - 16 = 0.547 . qH QH from the required power we can find the needed heat transfer . . 14 000 = 25 594 kW QH = Wnet / η = 0.547 . . m = QH / qH = 25 594 kW/ 960 kJ/kg = 26.66 kg/s Temperature after compression is (k-1)/k 0.4/1.4 T2 = T1 rp = 290 × 16 = 640.35 K The highest temperature is after combustion 960 T3 = T2 + qH/Cp = 640.35 + = 1596.5 K 1.004 For the exit flow I need the exhaust temperature k-1 − k T4 = T3 rp . . ηII = WNET/ΦH = 1596.5 × 16−0.2857 = 723 K since the low T exergy flow out is lost The high T exergy input from combustion is . . . ΦH = m(ψ3 - ψ2) = m[h3 – h2 – T(s3 – s2)] = 26.66 [960 – 298 × 1.004 ln ( 1596.5 )] = 18 303 kW 640.35 . . ηII = WNET/ΦH = 14 000 / 18 303 = 0.765 . . . Φflow in = m(ψ4 - ψo) = m[h4 – ho – T(s4 – so)] = 26.66 [ 1.004(17 – 25) – 298 × 1.004 ln ( 290 ) ] = 2.0 kW 298 . . . Φflow out = m(ψ1 - ψo) = m[h1 – ho – T(s1 – so)] = 26.66 [ 1.004(723 – 298) – 298 × 1.004 ln ( 723 ) ] = 4302 kW 298 11.151 For Problem 11.141, determine the change of availability of the water flow and that of the air flow. Use these to determine a second law efficiency for the boiler heat exchanger. From solution to 11.141 : . mH2O = 26.584 kg/s, h2 = 194.85 kJ/kg, s2 = 0.6587 kJ/kg K h3 = 3456.5 kJ/kg, s3 = 7.2338, s° = 7.9820, s° = 7.1762 kJ/kg K Ti Te hi = 903.16 kJ/kg, he = 408.13 kJ/kg ψ3 - ψ2 = h3 - h2 - T0(s3 - s2) = 1301.28 kJ/kg ψi - ψe = hi - he - T0(s° - s° ) = 254.78 kJ/kg Ti Te ηII = . (ψ3 - ψ2)mH2O . (ψi - ψe)mair = 1301.28 × 26.584 = 0.776 254.78 × 175 Review Problems 11.152 A simple steam power plant is said to have the four states as listed: 1: (20oC, 100 kPa), 2: (25oC, 1 MPa), 3: (1000oC, 1 MPa), 4: (250oC, 100 kPa) with an energy source at 1100oC and it rejects energy to a 0oC ambient. Is this cycle possible? Are any of the devices impossible? Solution: The cycle should be like Figure 11.3 for an ideal or Fig.11.9 for an actual pump and turbine in the cycle. We look the properties up in Table B.1: State 1: h1 = 83.94 , s1 = 0.2966 State 2: h2 = 104.87, s2 = 0.3673 State 3: h3 = 4637.6 , s3 = 8.9119 State 4: h4 = 2974.3, s4 = 8.0332 We may check the overall cycle performance Boiler: qH = h3 - h2 = 4637.6 - 104.87 = 4532.7 kJ/kg Condenser: qL = h4 - h1 = 2974.3 - 83.94 = 2890.4 kJ/kg ηcycle = qnet / qH = (qH − qL) / qH = 1642.3 / 4532.7 = 0.362 ηcarnot = 1 - TL / TH = 1 - 273.15 = 0.80 > ηcycle OK 273.15 + 1100 Check the second law for the individual devices: C.V. Boiler plus wall to reservoir qH 4532.7 sgen = s3 - s2 = 8.9119 - 0.3673 = 5.24 kJ/kg K > 0 OK Tres 1373 C.V. Condenser plus wall to reservoir qL 2890.4 sgen = s1 - s4 + = 0.2966 - 8.0332 + = 2.845 kJ/kg K > 0 OK Tres 273 C.V. Pump: wp = h2 - h1 = 20.93 kJ/kg ; sgen = s2 - s1 = 0.3673 - 0.2966 = 0.0707 kJ/kg K > 0 OK C.V. Turbine: wT = h3 - h4 = 4637.6 - 2974.3 = 1663.3 kJ/kg sgen = s4 - s3 = 8.0332 - 8.9119 = - 0.8787 kJ/kg K sgen < 0 NOT POSSIBLE T 3 QH 3 WT 2 2 WP, in 1 4. QL 4 1 s 11.153 Do Problem 11.31 with R-134a as the working fluid in the Rankine cycle. Consider the ammonia Rankine-cycle power plant shown in Fig. P11.31, a plant that was designed to operate in a location where the ocean water temperature is 25°C near the surface and 5°C at some greater depth. The mass flow rate of the working fluid is 1000 kg/s. a. Determine the turbine power output and the pump power input for the cycle. b. Determine the mass flow rate of water through each heat exchanger. c. What is the thermal efficiency of this power plant? Solution: a) Turbine s2 = s1 = 1.7183 = 1.0485 + x2 × 0.6733 => x2 = 0.9948 h2 = 213.58 + 0.9948 × 190.65 = 403.24 kJ/kg wT = h1 - h2 = 409.84 - 403.24 = 6.6 kJ/kg . . WT = mwT = 6600 kW Pump: wP ≈ v3(P4 - P3) = 0.000794(572.8 - 415.8) = 0.125 kJ/kg . . WP = mwP = 125 kW wP = wP /ηS = 0.125 => b) Consider the condenser heat transfer to the low T water . Qto low T H2O = 1000(403.24 - 213.58) = 189 660 kW . mlow T H2O = 189660 = 22 579 kg/s 29.38 - 20.98 h4 = h3 - wP = 213.58 + 0.125 = 213.71 kJ/kg Now consider the boiler heat transfer from the high T water . Qfrom high T H2O = 1000(409.84 - 213.71) = 196 130 kW . mhigh T H2O = c) 196130 = 23 432 kg/s 104.87 - 96.50 . . 6600 - 125 ηTH = WNET/QH = = 0.033 196130 T 1 QH WT 4 WP, in 3 2. QL 1 4 3 2 s 11.154 An ideal steam power plant is designed to operate on the combined reheat and regenerative cycle and to produce a net power output of 10 MW. Steam enters the high-pressure turbine at 8 MPa, 550°C, and is expanded to 0.6 MPa, at which pressure some of the steam is fed to an open feedwater heater, and the remainder is reheated to 550°C. The reheated steam is then expanded in the low-pressure turbine to 10 kPa. Determine the steam flow rate to the high-pressure turbine and the power required to drive each of the pumps. a) 7 T 5 o 550 C 5 7 HI P LOW P T1 T2 4 6 6 10 kPa 8 3 2 6a 1 8 COND. HTR s 4 2 3 P P 1 b) -wP12 = 0.00101(600 - 10) = 0.6 kJ/kg h2 = h1 - wP12 = 191.8 + 0.6 = 192.4 kJ/kg -wP34 = 0.00101(8000 - 600) = 8.1 kJ/kg h4 = h3 - wP34 = 670.6 + 8.1 = 678.7 ; h5 = 3521.0 kJ/kg, s6 = s5 = 6.8778 ⇒ T6 = 182.32 oC h6 = 2810.0 kJ/kg, h7 = 3591.9, s8 = s7 = 8.1348 = 0.6493 + x8 × 7.5009 ⇒ x8 = 0.9979 h8 = 191.83 + 0.9979 × 2392.8 = 2579.7 kJ/kg CV: heater Cont: m6a + m2 = m3 = 1 kg, 1st law: m6a = m6ah6 + m2h2 = m3h3 670.6 - 192.4 = 0.1827, m2 = m7 = 1 - m6a = 0.8173 2810.0 - 192.4 CV: turbine wT = (h5 - h6) + (1 - m6a)(h7 - h8) = 3521 - 2810 + 0.8173(3591.9 - 2579.7) = 1538.2 kJ/kg CV: pumps wP = m2wP12 + m4wP34 = 0.8214×(-0.6) + 1×(-8.1) = -8.6 kJ/kg wN = 1538.2 - 8.6 = 1529.6 kJ/kg (m5) . . m5 = WN/wN = 10000/1529.6 = 6.53 kg/s 11.155 Steam enters the turbine of a power plant at 5 MPa and 400°C, and exhausts to the condenser at 10 kPa. The turbine produces a power output of 20 000 kW with an isentropic efficiency of 85%. What is the mass flow rate of steam around the cycle and the rate of heat rejection in the condenser? Find the thermal efficiency of the power plant and how does this compare with a Carnot cycle. . Solution: WT = 20 000 kW and ηTs = 85 % State 3: State 1: h3 = 3195.6 kJ/kg , s3 = 6.6458 kJ/kgK P1 = P4 = 10 kPa , sat liq , x1 = 0 T1 = 45.8oC , h1 = hf = 191.8 kJ/kg , v1 = vf = 0.00101 m3/kg C.V Turbine : 1st Law: qT + h3 = h4 + wT ; qT = 0 wT = h3 - h4 , Assume Turbine is isentropic s4s = s3 = 6.6458 kJ/kgK , s4s = sf + x4s sfg , solve for x4s = 0.7994 h4s = hf + x4shfg = 1091.0 kJ/kg wTs = h3 - h4s = 1091 kJ/kg , wT = ηTswTs = 927.3 kJ/kg . . WT m= = 21.568 kg/s , wT C.V. Condenser: 1st Law : h4 = h3 - wT = 2268.3 kJ/kg h4 = h1 + qc + wc ; qc = h4 - h1 = 2076.5 kJ/kg , wc = 0 . . Qc = m qc = 44786 kW C.V. Pump: Assume adiabatic, reversible and incompressible flow wps = ∫ v dP = v1(P2 - P1) = 5.04 kJ/kg 1st Law : C.V Boiler : 1st Law : h2 = h1 + wp = 196.8 kJ/kg qB + h2 = h3 + wB ; wB = 0 qB = h3 - h2 = 2998.8 kJ/kg wnet = wT - wP = 922.3 kJ/kg ηth = wnet / qB = 0.307 Carnot cycle : TH = T3 = 400oC , TL = T1 = 45.8oC ηth = TH - TL = 0.526 TH 11.156 Consider an ideal combined reheat and regenerative cycle in which steam enters the high-pressure turbine at 3.0 MPa, 400°C, and is extracted to an open feedwater heater at 0.8 MPa with exit as saturated liquid. The remainder of the steam is reheated to 400°C at this pressure, 0.8 MPa, and is fed to the lowpressure turbine. The condenser pressure is 10 kPa. Calculate the thermal efficiency of the cycle and the net work per kilogram of steam. Solution: In this setup the flow is separated into fractions x and 1-x after coming out of T1. The two flows are recombined in the FWH. C.V. T1 s6 = s5 = 6.9211 kJ/kg K => h6 = 2891.6 kJ/kg wT1 = h5 - h6 = 3230.82 – 2891.6 = 339.22 kJ/kg C.V. Pump 1: wP1 = h2 - h1 = v1(P2 - P1) = 0.00101(800 - 10) = 0.798 kJ/kg => h2 = h1 + wP1 = 191.81 + 0.798 = 192.61 kJ/kg 7 T o 400 C 5 5 7 T2 T1 4 3 2 1 6 8 x 8 s 4 P2 C.V. FWH, 6 1-x 10 kPa 1-x FWH 3 COND. 2 P1 h3 = hf = 721.1 Energy equation per unit mass flow exit at 3: h3 - h2 721.1 - 192.61 x h6 + (1 - x) h2 = h3 => x = = = 0.1958 h6 - h2 2891.6 - 192.61 C.V. Pump 2 wP2 = h4 - h3 = v3(P4 - P3) = 0.001115(3000 - 800) = 2.45 kJ/kg => h4 = h3 + wP2 = 721.1 + 2.45 = 723.55 kJ/kg C.V. Boiler/steam generator including reheater. Total flow from 4 to 5 only fraction 1-x from 6 to 7 qH = h5 - h4 + (1 - x)(h7 - h6 ) = 2507.3 + 301.95 = 2809.3 kJ/kg 1 C.V. Turbine 2 s8 = s7 = 7.5715 kJ/kg K => x8 = (7.5715 - 0.6492)/7.501 = 0.92285 h8 = hf + x8 hfg = 191.81 + 0.92285 × 2392.82 = 2400.0 kJ/kg wT2 = h7 - h8 = 3267.07 - 2400.02 = 867.05 kJ/kg Sum the work terms to get net work. Total flow through T1 only fraction 1-x through T2 and P1 and after FWH we have the total flow through P2. wnet = wT1 + (1 - x) wT2 - (1 - x) wP1 - wP2 = 339.2 + 697.3 - 0.64 – 2.45 = 1033.41 kJ/kg ηcycle = wnet / qH = 1033.41 / 2809.3 = 0.368 11.157 In one type of nuclear power plant, heat is transferred in the nuclear reactor to liquid sodium. The liquid sodium is then pumped through a heat exchanger where heat is transferred to boiling water. Saturated vapor steam at 5 MPa exits this heat exchanger and is then superheated to 600°C in an external gas-fired superheater. The steam enters the turbine, which has one (open-type) feedwater extraction at 0.4 MPa. The isentropic turbine efficiency is 87%, and the condenser pressure is 7.5 kPa. Determine the heat transfer in the reactor and in the superheater to produce a net power output of 1 MW. 5 MPa T o 6 6 600 C SUP. HT. TURBINE. 0.4 MPa Q 7 5 8 REACT. 23 1 HTR. 3 4 P 5 7s 4 COND. 2 7 7.5 kPa 8s 8 s . WNET = 1 MW , ηST = 0.87 1 P -wP12 = 0.001008(400 - 7.5) = 0.4 kJ/kg h2 = h1 - wP12 = 168.8 + 0.4 = 169.2 kJ/kg -wP34 = 0.001084(5000 - 400) = 5.0 kJ/kg h4 = h3 - wP34 = 604.7 + 5.0 = 609.7 kJ/kg s7S = s6 = 7.2589, P7=0.4 MPa => T7S = 221.2 oC, h7S = 2904.5 kJ/kg h6 - h7 = ηST(h6 - h7S) ⇒ 3666.5 - h7 = 0.87(3666.5 - 2904.5) = 662.9 ⇒ h7 = 3003.6 kJ/kg s8S = s6 = 7.2589 = 0.5764 + x8S × 7.6750 ; x8S = 0.8707 h8S = 168.8 + 0.8707 × 2406.0 = 2263.7 kJ/kg h6 - h8 = ηST(h6 - h8S) ⇒ 3666.5 - h8 = 0.87(3666.5 - 2263.7) = 1220.4 ⇒ h8 = 2446.1 kJ/kg CV: heater cont: m2 + m7 = m3 = 1.0 kg, Energy Eq.: m7 = (604.7-169.2)/(3003.6-169.2) = 0.1536 m2h2 + m7h7 = m3h3 CV: turbine wT = (h6 - h7) + (1 - m7)(h7 - h8) = 3666.5-3003.6 + 0.8464(3003.6-2446.1) = 1134.8 kJ/kg CV: pumps wP = m1wP12 + m3wP34 = 0.8464(-0.4) + 1(-5.0) = -5.3 kJ/kg . wNET = 1134.8 - 5.3 = 1129.5 => m = 1000/1129.5 = 0.885 kg/s CV: reactor . . QREACT = m(h5 - h4) = 0.885(2794.3 - 609.7) = 1933 kW CV: superheater . QSUP = 0.885(h6 - h5) = 0.885(3666.5 - 2794.3) = 746 kW 11.158 An industrial application has the following steam requirement: one 10-kg/s stream at a pressure of 0.5 MPa and one 5-kg/s stream at 1.4 MPa (both saturated or slightly superheated vapor). It is obtained by cogeneration, whereby a highpressure boiler supplies steam at 10 MPa, 500°C to a turbine. The required amount is withdrawn at 1.4 MPa, and the remainder is expanded in the lowpressure end of the turbine to 0.5 MPa providing the second required steam flow. Assuming both turbine sections have an isentropic efficiency of 85%, determine the following. a. The power output of the turbine and the heat transfer rate in the boiler. b. Compute the rates needed were the steam generated in a low-pressure boiler without cogeneration. Assume that for each, 20°C liquid water is pumped to the required pressure and fed to a boiler. Solution: o 10 MPa, 500 C BOILER 2 3 . . W P 1 . QH P o 20 C H2O IN W HPT HP TURB. ηs = 0.85 4 1.4 MPa STEAM 5 kg/s ηs = 0.85 . LP TURB. 0.5 MPa 5 W PT L STEAM 10 kg/s a) With cogeneration high-pressure turbine, first the ideal then the actual. s4S = s3 = 6.5966 kJ/kg K ⇒ T4S = 219.9 oC, h4S = 2852.6 kJ/kg wS HPT = h3 - h4S = 3373.7 - 2852.6 = 521.1 kJ/kg actual turbine from Eq.9.27 wHPT = ηSwS HPT = 0.85 × 521.1 = 442.9 kJ/kg h4 = h3 - w = 3373.7-442.9 = 2930.8 kJ/kg ⇒ T4 = 251.6°C, s4 = 6.7533 kJ/kg K low-pressure turbine first the ideal then the actual s5S = s4 = 6.7533 = 1.8607 + x5S × 4.9606, x5S = 0.9863 h5S = 640.23 + 0.9863 × 2108.5 = 2719.8 kJ/kg wS LPT = h4 - h5S = 2930.8 - 2719.8 = 211.0 kJ/kg actual turbine from Eq.9.27 wLPT = ηSwS LPT = 0.85 × 211.0 = 179.4 kJ/kg h5 = h4 - w = 2930.8 - 179.4 = 2751.4 > hG OK . WTURB = 15 × 442.9 + 10 × 179.4 = 8438 kW . WP = 15[0.001002(10000 - 2.3)] = 150.3 kW h2 = h1 + wP = 83.96 + 10.02 = 94.0 kJ/kg . . QH = m1(h3 - h2) = 15(3373.7 - 94.0) = 49196 kW b) Without cogeneration This is to be compared to the amount of heat required to supply 5 kg/s of 1.4 MPa sat. vap. plus 10 kg/s of 0.5 MPa sat. vap. from 20oC water. WP1 5 kg/s 1 2Q 3 2 3 Sat. vapor 1.4 MPa o 20 C WP2 10 kg/s o 20 C 4 5Q 6 5 Sat. vapor 6 0.5 MPa Pump 1 and boiler 1 wP = 0.001002(1400 - 2.3) = 14.0 kJ/kg, h2 = h1 + wP = 83.96 + 14.0 = 85.4 kJ/kg . . Q3 = m1(h3 - h2) = 5(2790.0 - 85.4) = 13 523 kW 2 . WP1 = 5 × 14.0 = 7 kW Pump 2 and boiler 2 h5 = h4 + wP2 = 83.96 + 0.001002(500 - 2.3) = 84.5 kJ/kg . . Q6 = m4(h6 - h5) = 10(2748.7 - 84.5) = 26 642 kW 5 . WP2 = 10 × 0.5 = 5 kW . Total QH = 13523 + 26642 = 40 165 kW 11.159 Repeat Problem 11.75, but assume that the compressor has an efficiency of 82%, that both turbines have efficiencies of 87%, and that the regenerator efficiency is 70%. k-1 P2 k a) From solution 11.54: T2 = T1 = 300(6)0.286 = 500.8 K P1 -wC = -w12 = CP0(T2 - T1) = 1.004(500.8 - 300) = 201.6 kJ/kg -wC = -wSC/ηSC = 201.6/0.82 = 245.8 kJ/kg = wT1 ⇒ = CP0(T4 - T5) = 1.004(1600 - T5) T5 = 1355.2 K wST1 = wT1/ηST1 = 245.8/0.87 = 282.5 kJ/kg = CP0(T4 - T5S) = 1.004(1600 - T5S) k s5S = s4 ⇒ P5 = P4(T5S/T4)k-1 = 600( ⇒ T5S = 1318.6 K 1318.6 3.5 ) = 304.9 kPa 1600 b) P6 = 100 kPa, s6S = s5 k-1 P6 k 100 0.286 T6S = T5 = 1355.2 = 985.2K 304.9 P5 wST2 = CP0(T5-T6S) = 1.004(1355.2- 985.2) = 371.5 kJ/kg wT2 = ηST2 × wST2 = 0.87 × 371.5 = 323.2 kJ/kg 323.2 = CP0(T5-T6) = 1.004(1355.2 -T6) ⇒ T6 = 1033.3K . . m = WNET/wNET = 150/323.2 = 0.464 kg/s c) wC = 245.8 = CP0(T2 - T1) = 1.004(T2 – 300) ⇒ T2 = 544.8 K ηREG = h3 - h2 h6 - h2 = T3 - T2 T6 - T2 = T3 - 544.8 1033.3 - 544.8 = 0.7 ⇒ T3 = 886.8 K qH = CP0(T4 - T3) = 1.004(1600 – 886.8) = 716 kJ/kg ηTH = wNET/qH = 323.2/716 = 0.451 11.160 Consider a gas turbine cycle with two stages of compression and two stages of expansion. The pressure ratio across each compressor stage and each turbine stage is 8 to 1. The pressure at the entrance of the first compressor is 100 kPa, the temperature entering each compressor is 20oC, and the temperature entering each turbine is 1100oC. A regenerator is also incorporated into the cycle and it has an efficiency of 70%. Determine the compressor work, the turbine work, and the thermal efficiency of the cycle. See Fig.11.23 for the configuration. P2/P1 = P4/P3 = P6/P7 = P8/P9 = 8.0 T 6 P1 = 100 kPa 5 T1 = T3 = 20 oC, T6 = T8 = 1100 oC 4 Assume constant specific heat s2 = s1 and s4 = s3 8 7 2 9 10 3 1 s k-1 P2 k T4 = T2 = T1 = 293.15(8)0.286 = 531 K P1 Total -wC = 2 × (-w12) = 2CP0(T2 - T1) = 2 × 1.004(531 - 293.15) = 477.6 kJ/kg k-1 P7 k 10.286 Also s6 = s7 and s8 = s9: ⇒ T7 = T9 = T6 = 1373.15 = 758 K 8 P6 Total wT = 2 × w67 = 2CP0(T6 - T7) = 2 × 1.004(1373.15 - 758) = 1235.2 kJ/kg wNET = 1235.2 - 477.6 = 757.6 kJ/kg Ideal regenerator: T5 = T9, T10 = T4 so the actual one has ηREG = h5 - h4 h9 - h4 = T5 - T4 T9 - T4 = T5 - 531 758 - 531 = 0.7 ⇒ T5 = 689.9 K ⇒ qH = (h6 - h5) + (h8 - h7) = CP0(T6 - T5) + CP0(T8 - T7) = 1.004(1373.15 – 689.9) + 1.004 (1373.15 – 758) = 1303.6 kJ/kg ηTH = wNET/qH = 757.6/1303.6 = 0.581 11.161 A gas turbine cycle has two stages of compression, with an intercooler between the stages. Air enters the first stage at 100 kPa, 300 K. The pressure ratio across each compressor stage is 5 to 1, and each stage has an isentropic efficiency of 82%. Air exits the intercooler at 330 K. The maximum cycle temperature is 1500 K, and the cycle has a single turbine stage with an isentropic efficiency of 86%. The cycle also includes a regenerator with an efficiency of 80%. Calculate the temperature at the exit of each compressor stage, the second-law efficiency of the turbine and the cycle thermal efficiency. State 1: P1 = 100 kPa, T1 = 300 K State 7: P7 = Po = 100 kPa State 3: T3 = 330 K; State 6: T6 = 1500 K, P6 = P4 P2 = 5 P1 = 500 kPa; P4 = 5 P3 = 2500 kPa Ideal compression T2s = T1 (P2/P1)(k-1)/k = 475.4 K 1st Law: q + hi = he + w; q = 0 => wc1 = h1 - h2 = CP(T1 - T2) wc1 s = CP(T1 - T2s) = -176.0 kJ/kg, T2 = T1 - wc1/CP = 513.9 K wc1 = wc1 s/ η = -214.6 T4s = T3 (P4/P3)(k-1)/k = 475.4 K wc2 s = CP(T3 - T4s) = -193.6 kJ/kg; wc2 = -236.1 kJ/kg T4 = T3 - wc2 / CP = 565.2 K Ideal Turbine (reversible and adiabatic) T7s = T6(P7/P6)(k-1)/k = 597.4 K => wTs = CP(T6 - T7s) = 905.8 kJ/kg 1st Law Turbine: q + h 6 = h7 + w; q=0 wT = h6 - h7 = CP(T6 - T7) = ηTs wTs = 0.86 × 905.8 = 779.0 kJ/kg T7 = T6 - wT/ CP = 1500 - 779/1.004 = 723.7 K T6 P6 s6 - s7 = CP ln - R ln = -0.1925 kJ/kg K T7 P7 ψ6 - ψ7 = (h6 - h7) - To(s6 - s7) = 779.0 - 298.15(-0.1925) = 836.8 kJ/kg wT η2nd Law = = 779.0 / 836.8 = 0.931 ψ6-ψ7 d) ηth = qH / wnet ; wnet = wT + wc1 + wc2 = 328.3 kJ/kg st 1 Law Combustor: q + hi = he + w; w = 0 qc = h6 - h5 = CP(T6 - T5) T5 - T4 Regenerator: ηreg = = 0.8 -> T5 = 692.1 K T7 - T4 qH = qc = 810.7 kJ/kg; ηth = 0.405 11.162 A gasoline engine has a volumetric compression ratio of 9. The state before compression is 290 K, 90 kPa, and the peak cycle temperature is 1800 K. Find the pressure after expansion, the cycle net work and the cycle efficiency using properties from Table A.7. Use table A.7 and interpolation. Compression 1 to 2: s2 = s1 ⇒ From Eq.8.28 o o o o 0 = sT2 - sT1 - R ln(P2/P1) = sT2 - sT1 - R ln(Τ2v1/T1v2) o o sT2 - R ln(Τ2/T1) = sT1 + R ln(v1/v2) = 6.83521 + 0.287 ln 9 = 7.4658 This becomes trial and error so estimate first at 680 K and use A.7.1. LHS680 = 7.7090 - 0.287 ln(680/290) = 7.4644 (too low) LHS700 = 7.7401 - 0.287 ln(700/290) = 7.4872 (too high) Interpolate to get: T2 = 681.23 K, u2 = 497.9 kJ/kg P2 = P1 (Τ2/T1) (v1/v2) = 90 (681.23 / 290) × 9 = 1902.7 kPa 1w2 = u1 - u2 = 207.2 - 497.9 = -290.7 kJ/kg Combustion 2 to 3: constant volume v3 = v2 qH = u3 - u2 = 1486.3 - 497.9 = 988.4 kJ/kg P3 = P2(T3/T2) = 1902.7 (1800/681.2) = 5028 kPa Expansion 3 to 4: o s 4 = s3 ⇒ From Eq.8.28 as before o sT4 - R ln(Τ4/T3) = sT3 + R ln(v3/v4) = 8.8352 + 0.287 ln(1/9) = 8.2046 This becomes trial and error so estimate first at 850 K and use A.7.1. LHS850 = 7.7090 - 0.287 ln(850/1800) = 8.1674 (too low) LHS900 = 7.7401 - 0.287 ln(900/1800) = 8.2147 (too high) Interpolation ⇒ T4 = 889.3 K, u4 = 666 kJ/kg P4 = P3(T4/T3)(v3/v4) = 5028 (889.3/1800) (1/9) = 276 kPa 3w4 = u3 - u4 = 1486.3 - 666.0 = 820.3 kJ/kg Net work and overall efficiency wNET = 3w4 + 1w2 = 820.3 - 290.7 = 529.6 kJ/kg η = wNET/qH = 529.6/988.4 = 0.536 11.163 The effect of a number of open feedwater heaters on the thermal efficiency of an ideal cycle is to be studied. Steam leaves the steam generator at 20 MPa, 600°C, and the cycle has a condenser pressure of 10 kPa. Determine the thermal efficiency for each of the following cases. A: No feedwater heater. B: One feedwater heater operating at 1 MPa. C: Two feedwater heaters, one operating at 3 MPa and the other at 0.2 MPa. a) no feed water heater 2 3 ⌠ wP = ⌡ vdP 1 ≈ 0.00101(20000 - 10) = 20.2 kJ/kg h2 = h1 + wP = 191.8 + 20.2 = 212.0 s4 = s3 = 6.5048 ST. GEN. TURBINE. 4 COND. = 0.6493 + x4 × 7.5009 x4 = 0.78064 h4 = 191.83 + 0.780 64 × 2392.8 = 2059.7 wT = h3 - h4 = 3537.6 - 2059.7 = 1477.9 kJ/kg wN = wT - wP = 1477.9 - 20.2 = 1457.7 qH = h3 - h2 = 3537.6 - 212.0 = 3325.6 2 1 P 20 MPa T o 3 600 C 10 kPa 2 4 1 s ηTH = wN qH = 1457.7 = 0.438 3325.6 b) one feedwater heater wP12 = 0.00101(1000 - 10) = 1.0 kJ/kg h2 = h1 + wP12 = 191.8 + 1.0 = 192.8 5 ST. GEN. TURBINE. 6 wP34 = 0.001127 (20000- 1000) = 21.4 kJ/kg h4 = h3 + wP34 = 762.8 + 21.4 = 784.2 s6 = s5 = 6.5048 = 2.1387 + x6 × 4.4478 7 HTR. 3 COND. 1 P 4 2 P x6 = 0.9816 20 MPa T h6 = 762.8 + 0.9816 × 2015.3 = 2741.1 CV: heater const: m3 = m6 + m2 = 1.0 kg 1st law: m6h6 + m2h2 = m3h3 762.8 - 192.8 m6 = = 0.2237 2741.1 - 192.8 o 600 C 5 1 MPa 4 6 23 1 10 kPa 7 s m2 = 0.7763, h7 = 2059.7 ( = h4 of part a) ) CV: turbine wT = (h5 - h6) + m2(h6 - h7) = (3537.6 - 2741.1) + 0.7763(2741.1 - 2059.7) = 1325.5 kJ/kg CV: pumps wP = m1wP12 + m3wP34 = 0.7763(1.0) + 1(21.4) = 22.2 kJ/kg wN = 1325.5 - 22.2 = 1303.3 kJ/kg CV: steam generator qH = h5 - h4 = 3537.6 - 784.2 = 2753.4 kJ/kg ηTH = wN/qH = 1303.3/2753.4 = 0.473 c) two feedwater heaters wP12 = 0.00101 × (200 - 10) = 0.2 kJ/kg h2 = wP12 + h1 = 191.8 + 0.2 = 192.0 wP34 = 0.001061 × (3000 - 200) = 3.0 kJ/kg h4 = h3 + wP34 = 504.7 + 3.0 = 507.7 7 TURBINE. ST. GEN. 8 9 HP HTR LP HTR 5 COND. 1 3 P 6 10 P 4 P 2 wP56 = 0.001217(20000 - 3000) = 20.7 kJ/kg h6 = h5 + wP56 = 1008.4 + 20.7 = 1029.1 s8 = s7 = 6.5048 T = 293.2 oC 8 at P8 = 3 MPa h = 2974.8 8 s9 = s8 = 6.5048 = 1.5301 + x9 × 5.5970 T 80 MPa o 600 C 7 3 MPa 6 45 23 1 8 0.2 MPa 10 kPa 9 10 s x9 = 0.8888 => h9 = 504.7 + 0.888 × 2201.9 = 2461.8 kJ/kg CV: high pressure heater cont: m5 = m4 + m8 = 1.0 kg ; m8 = 1008.4 - 507.7 = 0.2030 2974.8 - 507.7 CV: low pressure heater cont: m9 + m2 = m3 = m4 ; m9 = 1st law: m5h5 = m4h4 + m8h8 m4 = 0.7970 1st law: m9h9 + m2h2 = m3h3 0.7970(504.7 - 192.0) = 0.1098 2461.8 - 192.0 m2 = 0.7970 - 0.1098 = 0.6872 CV: turbine wT = (h7 - h8) + (1 - m8)(h8 - h9) + (1 - m8 - m9)(h9 - h10) = (3537.6 - 2974.8) + 0.797(2974.8 - 2461.8) + 0.6872(2461.8 - 2059.7) = 1248.0 kJ/kg CV: pumps wP = m1wP12 + m3wP34 + m5wP56 = 0.6872(0.2) + 0.797(3.0) + 1(20.7) = 23.2 kJ/kg wN = 1248.0 - 23.2 = 1224.8 kJ/kg CV: steam generator qH = h7 - h6 = 3537.6 - 1029.1 = 2508.5 kJ/kg ηTH = wN/qH = 1224.8/2508.5 = 0.488 11.164 The power plant shown in Fig. 11.40 combines a gas-turbine cycle and a steamturbine cycle. The following data are known for the gas-turbine cycle. Air enters the compressor at 100 kPa, 25°C, the compressor pressure ratio is 14, and the isentropic compressor efficiency is 87%; the heater input rate is 60 MW; the turbine inlet temperature is 1250°C, the exhaust pressure is 100 kPa, and the isentropic turbine efficiency is 87%; the cycle exhaust temperature from the heat exchanger is 200°C. The following data are known for the steam-turbine cycle. The pump inlet state is saturated liquid at 10 kPa, the pump exit pressure is 12.5 MPa, and the isentropic pump efficiency is 85%; turbine inlet temperature is 500°C and the isentropic turbine efficiency is 87%. Determine a. The mass flow rate of air in the gas-turbine cycle. b. The mass flow rate of water in the steam cycle. c. The overall thermal efficiency of the combined cycle. . QH = 60 MW P2 /P = 14 1 HEAT 2 3 T3 = 1250o C η = 0.87 SC . GAS TURB COMP W NET CT P4 = 100 kPa AIR 5 1 P 1= 100 kPa T = 25o C 1 4 ST T5 = 200 oC HEAT EXCH o 7 6 SP = 0.85 T7 = 500 C P6 = P7 = 12.5 MPa . 9 P WST STEAM TURB H 2O η η = 0.87 ηST 0.87 = 8 COND a) From Air Tables, A.7: Pr1 = 1.0913, h1 = 298.66, P8 = P9= 10 kPa h5 = 475.84 kJ/kg s2 = s1 Pr2S = Pr1(P2/P1) = 1.0913 × 14 = 15.2782 T2S = 629 K, h2S = 634.48 wSC = h1 - h2S = 298.66 - 634.48 = -335.82 kJ/kg wC = wSC/ηSC = -335.82/0.87 = -386 = h1 - h2 ⇒ h2 = 684.66 kJ/kg At T3 = 1523.2 K: Pr3 = 515.493, h3 = 1663.91 kJ/kg . . mAIR = QH/(h3 - h2) = b) 60 000 = 61.27 kg/s 1663.91 - 684.66 Pr4S = Pr3(P4/P3) = 515.493(1/14) = 36.8209 => T4S = 791 K, h4S = 812.68 kJ/kg wST = h3 - h4S = 1663.91 - 812.68 = 851.23 kJ/kg wT = ηST × wST = 0.87 × 851.23 = 740.57 = h3 - h4 => h4 = 923.34 kJ/kg Steam cycle: -wSP ≈ 0.00101(12500 - 10) = 12.615 kJ/kg -wP = - wSP/ηSP = 12.615/0.85 = 14.84 kJ/kg h6 = h9 - wP = 191.83 + 14.84 = 206.67 kJ/kg At 12.5 MPa, 500 oC: h7 = 3341.7 kJ/kg, s7 = 6.4617 kJ/kg K h4 - h5 . 923.34 - 475.84 = mAIR = 61.27 = 8.746 kg/s O h7 - h6 3341.7 - 206.67 2 . mH c) s8S = s7 = 6.4617 = 0.6492 + x8S × 7.501, x8S = 0.7749 h8S = 191.81 + 0.7749 × 2392.8 = 2046.0 kJ/kg wST = h7 - h8S = 3341.7 - 2046.0 = 1295.7 kJ/kg wT = ηST × wST = 0.87 × 1295.7 = 1127.3 kJ/kg . . . WNET = m(wT+wC)AIR +m(wT+wP)H O 2 = 61.27(740.57 - 386.0) + 8.746(1127.3 - 14.84) = 21725 + 9730 = 31455 kW = 31.455 MW . . ηTH = WNET/QH = 31.455/60 = 0.524 11.165 One means of improving the performance of a refrigeration system that operates over a wide temperature range is to use a two-stage compressor. Consider an ideal refrigeration system of this type that uses R-12 as the working fluid, as shown in Fig. P11.165. Saturated liquid leaves the condenser at 40°C and is throttled to −20°C. The liquid and vapor at this temperature are separated, and the liquid is throttled to the evaporator temperature, −70°C. Vapor leaving the evaporator is compressed to the saturation pressure corresponding to −20°C, after which it is mixed with the vapor leaving the flash chamber. It may be assumed that both the flash chamber and the mixing chamber are well insulated to prevent heat transfer from the ambient. Vapor leaving the mixing chamber is compressed in the second stage of the compressor to the saturation pressure corresponding to the condenser temperature, 40°C. Determine a. The coefficient of performance of the system. b. The coefficient of performance of a simple ideal refrigeration cycle operating over the same condenser and evaporator ranges as those of the two-stage compressor unit studied in this problem. ROOM . T +Q H 5 COND 5 1 o SAT.LIQ. o 40 C 40 C o -20 C COMP. ST.2 o SAT.VAP. o -20 C 4 MIX.CHAM 3 -70 C 2 2 49 3 7 -20 C R-12 refrigerator with 2-stage compression 7 EVAP 8 s 6 SAT.LIQ. o COMP. ST.1 . -Q L SAT.VAP. o -70 C 6 FLASH CHAMBER 9 8 1 COLD SPACE CV: expansion valve, upper loop h2 = h1 = 74.527 = 17.8 + x2 × 160.81; x2 = 0.353 m3 = x2m2 = x2m1 = 0.353 kg ( for m1=1 kg) m6 = m1 - m3 = 0.647 kg CV: expansion valve, lower loop h7 = h6 = 17.8 = -26.1 + x7 × 181.64, x7 = 0.242 QL = m3(h8 - h7) = 0.647(155.536 - 17.8) qL = 89.1 kJ/kg-m1 CV: 1st stage compressor s8 = s9 = 0.7744, P9 = PSAT -20 oC = 0.1509 MPa ⇒ T9 = 9 oC, h9 = 196.3 kJ/kg CV: mixing chamber (assume constant pressure mixing) 1st law: m6h9 + m3h3 = m1h4 or h4 = 0.647 × 196.3 + 0.353 × 178.61 = 190.06 kJ/kg h4, P4 = 0.1509 MPa ⇒ T4 = -1.0 oC, s4 = 0.7515 kJ/kg K CV: 2nd stage compressor P4 = 0.1509 MPa = P9 = P3 P5 = Psat 40oC = 0.9607 MPa, s5 = s4 ⇒ T5 = 70oC, h5 = 225.8 kJ/kg CV: condenser 1st law: -qH = h1 - h5 = 74.527 - 225.8 = -151.27 kJ/kg β2 stage = qL/(qH - qL) = 89.1/(151.27 - 89.1) = 1.433 b) 1 stage compression h3 = h4 = 74.53 kJ/kg h1 = 155.54 kJ/kg qL = h1 - h4 = 81.0 kJ/kg s1 = s2 = 0.7744 P2 = 0.9607 MPa T 2 40o C 3 o -70 C ⇒ T2 = 80.9 oC, h2 = 234.0 4 qH = h2 - h3 = 234.0 - 74.53 = 159.47 kJ/kg β1 stage = qL/(qH - qL) = 81.0/(159.47 - 81.0) = 1.032 1 s 11.166 A jet ejector, a device with no moving parts, functions as the equivalent of a coupled turbine-compressor unit (see Problems 9.82 and 9.90). Thus, the turbinecompressor in the dual-loop cycle of Fig. P11.109 could be replaced by a jet ejector. The primary stream of the jet ejector enters from the boiler, the secondary stream enters from the evaporator, and the discharge flows to the condenser. Alternatively, a jet ejector may be used with water as the working fluid. The purpose of the device is to chill water, usually for an air-conditioning system. In this application the physical setup is as shown in Fig. P11.116. Using the data given on the diagram, evaluate the performance of this cycle in terms of the ratio Q /Q . LH a. Assume an ideal cycle. b. Assume an ejector efficiency of 20% (see Problem 9.90). T 2 VAP o 150 C 2 1' JET 3 BOIL. 11 4 EJECT. . 1 2' QH 9 5,10 3 VAP 8 76 1 o 10 C 11 COND. o 30 C HP P. 10 o 4 5 FLASH CH. 6 9 o LIQ o 10 C 7 20 C CHILL . QL 8 LP P. T1 = T7 = 10 C T2 = 150 oC T4 = 30 oC T9 = 20 oC Assume T5 = T10 (from mixing streams 4 & 9). P3 = P4 = P5 = P8 = P9 = P10 = PG 30 oC = 4.246 kPa P11 = P2 = PG 150oC = 475.8 kPa, P1 = P6 = P7 = PG 10oC = 1.2276 kPa . . . . . . . . Cont: m1 + m9 = m5 + m10, m5 = m6 = m7 + m1 . . . . . . . . m7 = m8 = m9, m10 = m11 = m2, m3 = m4 . . . a) m1 + m2 = m3; ideal jet ejector s1 = s1 & s2 = s2 (1' & 2' at P3 = P4) ′ ′ . . then, m1(h1 - h1) = m2(h2 - h2) ′ ′ s From s2 = s2 = 0.4369 + x2 × 8.0164; x2 = 0.7985 ′ ′ ′ h2 = 125.79 + 0.7985 × 2430.5 = 2066.5 kJ/kg ′ From s1 = s1 = 8.9008 ⇒ T1 = 112 °C, h1 = 2710.4 kJ/kg ′ ′ ′ .. 2746.5 - 2066.5 = 3.5677 ⇒ m1/m2 = 2710.4 - 2519.8 Also h4 = 125.79 kJ/kg, h7 = 42.01 kJ/kg, h9 = 83.96 kJ/kg Mixing of streams 4 & 9 ⇒ 5 & 10: . . . . . . (m1 + m2)h4 + m7h9 = (m7 + m1 + m2)h5 = 10 Flash chamber (since h6 = h5) : .. . . (m7+m1)h5 = 10 = m1h1 + m7h1 . ⇒ using the primary stream m2 = 1 kg/s: . . 4.5677 × 125.79 + m7 × 83.96 = (m7 + 4.5677)h5 . . & (m7 + 3.5677)h5 = 3.5677 × 2519.8 + m7 × 42.01 . Solving, m7 = 202.627 & h5 = 84.88 kJ/kg LP pump: -wLP P = 0.0010(4.246 - 1.2276) = 0.003 kJ/kg h8 = h7 - wLP P = 42.01 + 0.003 = 42.01 kJ/kg . . . Chiller: QL = m7(h9-h8) = 202.627(83.96 - 42.01) = 8500 kW (for m2 = 1) HP pump: -wHP P = 0.001002(475.8 - 4.246) = 0.47 kJ/kg h11 = h10 - wHP P = 84.88 + 0.47 = 85.35 kJ/kg Boiler: . . Q11 = m11(h2 - h11) = 1(2746.5 - 85.35) = 2661.1 kW .. ⇒ QL/QH = 8500/2661.1 = 3.194 .. .. b) Jet eject. eff. = (m1/m2)ACT/(m1/m2)IDEAL = 0.20 .. ⇒ (m1/m2)ACT = 0.2 × 3.5677 = 0.7135 . using m2 = 1 kg/s: . . 1.7135 × 125.79 + m7 × 83.96 = (m7 + 1.7135)h5 . . & (m7 + 0.7135)h5 = 0.7135 × 2519.8 + m7 × 42.01 Solving, . m7 = 39.762 & h5 = h10 = 85.69 kJ/kg . Then, QL = 39.762(83.96 - 42.01) = 1668 kW h11 = 85.69 + 0.47 = 86.16 kJ/kg . QH = 1(2746.5 - 86.16) = 2660.3 kW .. & QL/QH = 1668/2660.3 = 0.627 Problems solved using Table A.7.2 11.79 A gas turbine with air as the working fluid has two ideal turbine sections, as shown in Fig. P11.79, the first of which drives the ideal compressor, with the second producing the power output. The compressor input is at 290 K, 100 kPa, and the exit is at 450 kPa. A fraction of flow, x, bypasses the burner and the rest (1 − x) goes through the burner where 1200 kJ/kg is added by combustion. The two flows then mix before entering the first turbine and continue through the second turbine, with exhaust at 100 kPa. If the mixing should result in a temperature of 1000 K into the first turbine find the fraction x. Find the required pressure and temperature into the second turbine and its specific power output. C.V.Comp.: -wC = h2 - h1; s2 = s1 Pr2 = Pr1(P2/P1) = 0.9899(450/100) = 4.4545, T2 = 445 K h2 = 446.74, -wC = 446.74 - 290.43 = 156.3 kJ/kg C.V.Burner: h3 = h2 + qH = 446.74 + 1200 = 1646.74 kJ/kg ⇒ T3 = 1509 K C.V.Mixing chamber: (1 - x)h3 + xh2 = hMIX = 1046.22 kJ/kg x= h3 - hMIX h3 - h2 = 1646.74 - 1046.22 = 0.50 1646.74 - 446.74 . . . WT1 = WC,in ⇒ wT1 = -wC = 156.3 = h3 - h4 h4 = 1046.22 - 156.3 = 889.9 ⇒ T4 = 861 K P4 = (Pr4/PrMIX)PMIX = (51/91.65) × 450 = 250.4 kPa s4 = s5 ⇒ Pr5 = Pr4(P5/P4) = 51(100/250.4) = 20.367 h5 = 688.2 T5 = 676 K wT2 = h4 - h5 = 889.9 - 688.2 = 201.7 kJ/kg 11.81 Repeat Problem 11.77 when the intercooler brings the air to T3 = 320 K. The corrected formula for the optimal pressure is P = [ P P (T /T )n/(n-1)]1/2 see 2 14 3 1 Problem 9.184, where n is the exponent in the assumed polytropic process. Solution: The polytropic process has n = k (isentropic) so n/(n - 1) = 1.4/0.4 = 3.5 P2 = 400 (320/290)3.5 = 475.2 kPa C.V. C1: s2 = s1 ⇒ Pr2 = Pr1(P2/P1) = 0.9899(475.2/100) = 4.704 ⇒ T2 = 452 K, h2 = 453.75 -wC1 = h2 - h1 = 453.75 - 290.43 = 163.3 kJ/kg C.V. Cooler: qOUT = h2 - h3 = 453.75 - 320.576 = 133.2 kJ/kg C.V. C2: s4 = s3 ⇒ Pr4 = Pr3(P4/P3) = 1.3972(1600/475.2) = 4.704 ⇒ T4 = T2 = 452 K, h4 = 453.75 -wC2 = h4 - h3 = 453.75 - 320.576 = 133.2 kJ/kg 11.93 Air flows into a gasoline engine at 95 kPa, 300 K. The air is then compressed with a volumetric compression ratio of 8:1. In the combustion process 1300 kJ/kg of energy is released as the fuel burns. Find the temperature and pressure after combustion using cold air properties. Solution: Solve the problem with variable heat capacity, use A.7.1 and A.7.2. P T 3 3 v 2 4 1v s 4 2 1 s Compression 1 to 2: s2 = s1 ⇒ From A.7.2 vr2 = vr1 8 T2 = 673 K, P2 = P1 × Pr2 Pr1 = 179.49 = 22.436, 8 u2 = 491.5 kJ/kg, = 20 × Pr2 = 20 95 = 1705 kPa 1.1146 Compression 2 to 3: u3 = u2 + qH = 491.5 + 1300 = 1791.5 kJ/kg T3 = 2118 K P3 = P2 × (T3/T2) = 1705 × 2118 = 5366 kPa 673 11.94 A gasoline engine has a volumetric compression ratio of 9. The state before compression is 290 K, 90 kPa, and the peak cycle temperature is 1800 K. Find the pressure after expansion, the cycle net work and the cycle efficiency using properties from Table A.7. Use table A.7 and interpolation. Compression 1 to 2: s2 = s1 ⇒ vr2 = vr1(v2/v1) vr2 = 196.37/9 = 21.819 ⇒ T2 ≅ 680 K, Pr2 ≅ 20.784, P2 = P1(Pr2/Pr1) = 90 (20.784 / 0.995) = 1880 kPa w = u1 - u2 = 207.19 - 496.94 = -289.75 kJ/kg 12 Combustion 2 to 3: qH = u3 - u2 = 1486.33 - 496.94 = 989.39 kJ/kg P3 = P2(T3/T2) = 1880 (1800 / 680) = 4976 kPa Expansion 3 to 4: s4 = s3 ⇒ vr4 = vr3 × 9 = 1.143 × 9 = 10.278 ⇒ T4 = 889 K, Pr4 = 57.773, u4 = 665.8 kJ/kg P4 = P3(Pr4/Pr3) = 4976 (57.773 / 1051) = 273.5 kPa w = u3 - u4 = 1486.33 - 665.8 = 820.5 kJ/kg 34 wNET = 3w4 + 1w2 = 820.5 - 289.75 = 530.8 kJ/kg η = wNET/qH = 530.8/989.39 = 0.536 u2 = 496.94 11.100 Answer the same three questions for the previous problem, but use variable heat capacities (use table A.7). A gasoline engine takes air in at 290 K, 90 kPa and then compresses it. The combustion adds 1000 kJ/kg to the air after which the temperature is 2050 K. Use the cold air properties (i.e. constant heat capacities at 300 K) and find the compression ratio, the compression specific work and the highest pressure in the cycle. Solution: Standard Otto cycle, solve using Table A.7.1 and Table A.7.2 Combustion process: T3 = 2050 K ; u3 = 1725.7 kJ/kg u2 = u3 - qH = 1725.7 - 1000 = 725.7 kJ/kg ⇒ T2 = 960.5 K ; vr2 = 8.2166 Compression 1 to 2: s2 = s1 ⇒ From the vr function v1/v2 = vr1/vr2 = 195.36/8.2166 = 23.78 Comment: This is much too high for an actual Otto cycle. -1w2 = u2 - u1 = 725.7 - 207.2 = 518.5 kJ/kg Highest pressure is after combustion P3 = P2T3 / T2 = P1(T3 / T1)(v1 / v3) = 90 × (2050 / 290) × 23.78 = 15 129 kPa P T 3 3 4 2 4 1v 2 1 v s 11.103 Repeat Problem 11.95, but assume variable specific heat. The ideal gas air tables, Table A.7, are recommended for this calculation (and the specific heat from Fig. 5.10 at high temperature). Solution: Table A.7 is used with interpolation. T1 = 283.2 K, u1 = 202.3 kJ/kg, vr1 = 210.44 Compression 1 to 2: s2 = s1 ⇒ From definition of the vr function vr2 = vr1 (v2/v1) = 210.4 (1/7) = 30.063 Interpolate to get: T2 = 603.9 K, u2 = 438.1 kJ/kg => -1w2 = u2 - u1 = 235.8 kJ/kg, u3 = 438.1 + 1800 = 2238.1 => T3 = 2573.4 K , vr3 = 0.34118 P3 = 90 × 7 × 2573.4 / 283.2 = 5725 kPa Expansion 3 to 4: s 4 = s3 ⇒ From the vr function as before vr4 = vr3 (v4/v3) = 0.34118 (7) = 2.3883 Interpolation ⇒ T4 = 1435.4 K, u4 = 1145.8 kJ/kg 3w4 = u3 - u4 = 2238.1 - 1145.8 = 1092.3 kJ/kg Net work, efficiency and mep wnet = 3w4 + 1w2 = 1092.3 - 235.8 = 856.5 kJ/kg ηTH = wnet / qH = 856.5 / 1800 = 0.476 v1 = RT1/P1 = (0.287 × 283.2)/90 = 0.9029 m3/kg v2 = (1/7) v1 = 0.1290 m3/kg Pmeff = wnet = 856.5 / (0.9029 - 0.129) = 1107 kPa v1 – v2 11.110 Do problem 11.106, but use the properties from A.7 and not the cold air properties. A diesel engine has a state before compression of 95 kPa, 290 K, and a peak pressure of 6000 kPa, a maximum temperature of 2400 K. Find the volumetric compression ratio and the thermal efficiency. Solution: Compression: s2 = s1 => From definition of the Pr function Pr2 = Pr1 (P2/P1) = 0.9899 (6000/95) = 62.52 A.7.1 => T2 = 907 K; h2 = 941.0 kJ/kg; h3 = 2755.8 ; vr3 = 0.43338 qH = h3 - h2 = 2755.8 – 941.0 = 1814.8 kJ/kg CR = v1/v2 = (T1/T2)(P2/P1) = (290/907) × (6000/ 95) = 20.19 Expansion process vr4 = vr3 (v4 / v3) = vr3 (v1 / v3) = vr3 (v1 / v2) × (T2/T3) = vr3 CR × (T2/T3) = 0.43338 × 20.19 × (907/2400) = 3.30675 Linear interpolation T4 = 1294.8 K, u4 = 1018.1 kJ/kg qL = u4 - u1 = 1018.1 – 207.2 = 810.9 kJ/kg η = 1 – (qL/ qH) = 1 – (810.9/1814.8) = 0.553 P T 2 3 s 3 4 P s 4 1v 2 1 v s 11.118 Do the previous problem 11.117 using values from Table A.7.1. and A.7.2 Air in a piston/cylinder goes through a Carnot cycle in which TL = 26.8°C and the total cycle efficiency is η = 2/3. Find TH, the specific work and volume ratio in the adiabatic expansion. Solution: Carnot cycle efficiency Eq.7.5: η = 1 - TL/TH = 2/3 ⇒ TH = 3 × TL = 3 × 300 = 900 K From A.7.1: u3 = 674.82 kJ/kg, vr3 = 9.9169 u4 = 214.36 kJ/kg, vr4 = 179.49 Energy equation with q = 0 3w4 = u3 - u4 = 674.82 - 214.36 = 460.5 kJ/kg Entropy equation, constant s expressed with the vr function v4/v3 = vr4/vr3 = 179.49 / 9.9169 = 18.1 P T 2 T s TH 3 s 1 T 4 TL v 2 1 3 4 s CHAPTER 13 SOLUTION MANUAL SONNTAG • BORGNAKKE • VAN WYLEN FUNDAMENTALS of Thermodynamics Sixth Edition Sonntag, Borgnakke and van Wylen th Fundamentals of Thermodynamics 6 Edition Sonntag, Borgnakke and van Wylen CONTENT CHAPTER 13 SUBSECTION PROB NO. Correspondence table Study guide problems Clapeyron equation Property Relations, Maxwell, and those for Enthalpy, internal Energy and Entropy Volume Expansivity and Compressibility Equations of State Generalized Charts Mixtures Review problems 34-40 41-52 53-68 69-94 95-106 107-119 English unit problems 120-145 1-20 21-33 Sonntag, Borgnakke and van Wylen CHAPTER 13 th 6 ed. CORRESPONDANCE TABLE The new problem set relative to the problems in the fifth edition. New 5th 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 1 3 new 2 4 new new new 6 7 5 9 8 new 11 new 10 12 new new 13 new 16 new 17 new 14 new 15 New 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 5th new new new new new 22 25 24a 24b 47b 29 23 27a 27b 28 30 68 new new 20 21 31 38 new new 36 43 47a new 49 New 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 5th 50 33 34 35 39 42 56 44 45 48 51 52 53 58a 58b new new new 54 new new new 60 37 61 73a,b 73a,c 26 40 41 New 110 111 112 113 114 115 116 117 118 119 5th 46 55 57 62 65 69 70 67 74a 74b Sonntag, Borgnakke and van Wylen The English-unit problems are: New 120 121 122 123 124 125 126 127 128 129 5th 75 76 new 77 78 81 79 80 new 85 SI 21mod 22 27 31 41 45 47 49 51 65 New 130 131 132 133 134 135 136 137 138 139 5th 83 84 82 86 95 92 87 88 94 91 SI 69 70 73 74 75 76 81 82 80 85 New 140 141 142 143 144 145 5th 97 93 96 new 90 89 SI 86 90 92 95 108 109 mod indicates a modification from the previous problem that changes the solution but otherwise is the same type problem. Sonntag, Borgnakke and van Wylen The following table gives the values for the compressibility, enthalpy departure and the entropy departure along the saturated liquid-vapor boundary. These are used for all the problems using generalized charts as the figures are very difficult to read accurately (consistently) along the saturated liquid line. It is suggested that the instructor hands out copies of this page or let the students use the computer for homework solutions. Tr Pr Zf Zg d(h/RT)f d(h/RT)g d(s/R)f d(s/R)g 0.96 0.78 0.14 0.54 3.65 1.39 3.45 1.10 0.94 0.69 0.12 0.59 3.81 1.19 3.74 0.94 0.92 0.61 0.10 0.64 3.95 1.03 4.00 0.82 0.90 0.53 0.09 0.67 4.07 0.90 4.25 0.72 0.88 0.46 0.08 0.70 4.17 0.78 4.49 0.64 0.86 0.40 0.07 0.73 4.26 0.69 4.73 0.57 0.84 0.35 0.06 0.76 4.35 0.60 4.97 0.50 0.82 0.30 0.05 0.79 4.43 0.52 5.22 0.45 0.80 0.25 0.04 0.81 4.51 0.46 5.46 0.39 0.78 0.21 0.035 0.83 4.58 0.40 5.72 0.35 0.76 0.18 0.03 0.85 4.65 0.34 5.98 0.31 0.74 0.15 0.025 0.87 4.72 0.29 6.26 0.27 0.72 0.12 0.02 0.88 4.79 0.25 6.54 0.23 0.70 0.10 0.017 0.90 4.85 0.21 6.83 0.20 0.68 0.08 0.014 0.91 4.92 0.18 7.14 0.17 0.66 0.06 0.01 0.92 4.98 0.15 7.47 0.15 0.64 0.05 0.009 0.94 5.04 0.12 7.81 0.12 0.60 0.03 0.005 0.95 5.16 0.08 8.56 0.08 0.58 0.02 0.004 0.96 5.22 0.06 8.97 0.07 0.54 0.01 0.002 0.98 5.34 0.03 9.87 0.04 0.52 0.0007 0.0014 0.98 5.41 0.02 10.38 0.03 Sonntag, Borgnakke and van Wylen Concept-Study Guide Problems 13.1 Mention two uses of the Clapeyron equation. If you have experimental information about saturation properties down to a certain temperature Clapeyron equation will allow you to make an intelligent curve extrapolation of the saturated pressure versus temperature function Psat(T) for lower temperatures. From Clapeyrons equation we can calculate a heat of evaporation, heat of sublimation or heat of fusion based on measurable properties P, T and v. 13.2 The slope dP/dT of the vaporization line is finite as you approach the critical point, yet hfg and vfg both approach zero. How can that be? The slope is dP/dT = hfg / Tvfg Recall the math problem what is the limit of f(x)/g(x) when x goes towards a point where both functions f and g goes towards zero. A finite limit for the ratio is obtained if both first derivatives are different from zero so we have dP/dT → [dhfg /dT] / d(Tvfg)/dT as T → Tc 13.3 In view of Clapeyron’s equation and Fig. 3.7, is there something special about ice I versus the other forms of ice? Yes. The slope of the phase boundary dP/dT is negative for ice I to liquid whereas it is positive for all the other ice to liquid interphases. This also means that these other forms of ice are all heavier than liquid water. The pressure must be more than 200 MPa = 2000 atm so even the deepest ocean cannot reach that pressure (recall about 1 atm per 10 meters down). 13.4 If we take a derivative as (∂P/∂T)v in the two-phase region, see Figs. 3.18 and 3.19, does it matter what v is? How about T? In the two-phase region, P is a function only of T, and not dependent on v. Sonntag, Borgnakke and van Wylen 13.5 Sketch on a P-T diagram how a constant v line behaves in the compressed liquid region, the two-phase L-V region and the superheated vapor region? P v < vc Cr.P. S L V vsmall > vc vmedium vlarge T P v > vc L C.P. vmedium S V T v v large Sonntag, Borgnakke and van Wylen 13.6 If I raise the pressure in an isentropic process, does h go up or down? Is that independent upon the phase? Tds = 0 = dh – vdP , so h increases as P increases, for any phase. The magnitude is proportional to v (i.e. large for vapor and small for liquid and solid phases) 13.7 If I raise the pressure in an isothermal process does h go up or down for a liquid or solid? What do you need to know if it is a gas phase? Eq. 13.25: ∂h ∂v (∂P)T = v – T (∂T)P = v[1 - Tα P] Liquid or solid, α P is very small, h increases with P ; For a gas, we need to know the equation of state. 13.8 The equation of state in Example 13.3 was used as explicit in v. Is it explicit in P? Yes, the equation can be written explicitly in P. P = RT / [v + C/T3] 13.9 Over what range of states are the various coefficients in Section 13.5 most useful? For solids or liquids, where the coefficients are essentially constant over a wide range of P’s and T’s. 13.10 For a liquid or a solid is v more sensitive to T or P? How about an ideal gas? For a liquid or solid, v is more sensitive to T than P. For an ideal gas, v = RT/P , varies directly with T, inversely with P. Sonntag, Borgnakke and van Wylen 13.11 If I raise the pressure in a solid at constant T, does s go up or down? In Example 13.4, it is found that change in s with P at constant T is negatively related to volume expansivity (a positive value for a solid), dsT = - v α P dPT , so raising P decreases s. 13.12 Most equations of state are developed to cover which range of states? Most equations of state are developed to cover the gaseous phase, from low to moderate densities. Many cover high-density regions as well, including the compressed liquid region. 13.13 Is an equation of state valid in the two-phase regions? No. In a two-phase region, P depends only on T. There is a discontinuity at each phase boundary. 13.14 As P → 0, the specific volume v → ∞. For P → ∞, does v → 0? At very low P, the substance will be essentially an ideal gas, Pv = RT, so that v becomes very large. However at very high P, the substance eventually must become a solid, which cannot be compressed to a volume approaching zero. 13.15 Must an equation of state satisfy the two conditions in Eqs. 13.50 and 13.51? It has been observed from experimental measurements that substances do behave in that manner. If an equation of state is to be accurate in the near-critical region, it would have to satisfy these two conditions. If the equation is simple it may be overly restrictive to inpose these as it may lead to larger inaccuracies in other regions. Sonntag, Borgnakke and van Wylen 13.16 At which states are the departure terms for h and s small? What is Z there? Departure terms for h and s are small at very low pressure or at very high temperature. In both cases, Z is close to 1. 13.17 What is the benefit of the generalized charts? Which properties must be known besides the charts themselves? The generalized charts allow for the approximate calculations of enthalpy and entropy changes (and P,v,T behavior), for processes in cases where specific data or equation of state are not known. They also allow for approximate phase boundary determinations. It is necessary to know the critical pressure and temperature, as well as ideal-gas specific heat. 13.18 What does it imply if the compressibility factor is larger than 1? Compressibility factor greater than one results from domination of intermolecular forces of repulsion (short range) over forces of attraction (long range) – either high temperature or very high density. This implies that the density is lower than what is predicted by the ideal gas law, the ideal gas law assumes the molecules (atoms) can be pressed closer together. 13.19 The departure functions for h and s as defined are always positive. What does that imply for the real substance h and s values relative to ideal gas values? Real-substance h and s are less than the corresponding ideal-gas values. 13.20 What is the benefit of Kay’s rule versus a mixture equation of state? Kay’s rule for a mixture is not nearly as accurate as an equation of state for the mixture, but it is very simple to use. Sonntag, Borgnakke and van Wylen Clapeyron Equation 13.21 A special application requires R-12 at −140°C. It is known that the triple-point temperature is −157°C. Find the pressure and specific volume of the saturated vapor at the required condition. The lowest temperature in Table B.3 for R-12 is -90oC, so it must be extended to -140oC using the Clapeyron Eq. 13.7 integrated as in example 13.1 Table B.3: at T1 = -90oC = 183.2 K, P1 = 2.8 kPa. 8.3145 = 0.068 76 kJ/kg K 120.914 P hfg (T - T1) 189.748 (133.2 - 183.2) ln = = = -5.6543 P1 R T × T1 0.068 76 133.2 × 183.2 R= P = 2.8 exp(-5.6543) = 0.0098 kPa Sonntag, Borgnakke and van Wylen 13.22 Ice (solid water) at −3°C, 100 kPa, is compressed isothermally until it becomes liquid. Find the required pressure. Water, triple point T = 0.01oC , P = 0.6113 kPa Table B.1.1: vf = 0.001 m3/kg, hf = 0.01 kJ/kg, Tabel B.1.5: Clapeyron vi = 0.001 0908 m3/kg, hi = -333.4 kJ/kg hf - hi dPif 333.4 = = = -13 442 kPa/K dT (vf - vi)T -0.0000908 × 273.16 ∆P ≈ dPif dT ∆T = -13 442(-3 - 0.01) = 40 460 kPa P = Ptp + ∆P = 40 461 kPa Sonntag, Borgnakke and van Wylen 13.23 An approximation for the saturation pressure can be ln Psat = A – B/T, where A and B are constants. Which phase transition is that suitable for, and what kind of property variations are assumed? Clapeyron Equation expressed for the three phase transitions are shown in Eqs. 13.5-13.7. The last two leads to a natural log function if integrated and ideal gas for the vapor is assumed. dPsat hevap = Psat dT RT2 where hevap is either hfg or hig. Separate the variables and integrate -1 Psat dPsat = hevap R-1 T-2 dT ln Psat = A – B/T ; B = hevap R-1 if we also assume hevap is constant and A is an integration constant. The function then applies to the liquid-vapor and the solid-vapor interphases with different values of A and B. As hevap is not excactly constant over a wide interval in T means that the equation cannot be used for the total domain. Sonntag, Borgnakke and van Wylen 13.24 In a Carnot heat engine, the heat addition changes the working fluid from saturated liquid to saturated vapor at T, P. The heat rejection process occurs at lower temperature and pressure (T − ∆T), (P − ∆P). The cycle takes place in a piston cylinder arrangement where the work is boundary work. Apply both the first and second law with simple approximations for the integral equal to work. Then show that the relation between ∆P and ∆T results in the Clapeyron equation in the limit ∆T → dT. T T T −∆T 1 P P 2 2 P T P-∆P 3 4 4 T-∆T 3 s sfg at T qH = Tsfg; 1 P-∆P v vfg at T qL = (T-∆T)sfg ; wnet = qH - qL = ∆Tsfg Problem similar to development in section 13.1 for shaft work, here boundary movement work, w = ⌠ Pdv ⌡ 3 4 ⌠ ⌠ wNET = P(v2-v1) + ⌡ Pdv + (P - ∆P)(v4 - v3) + ⌡ Pdv 2 1 Approximating, 3 4 ∆P ⌠ Pdv ≈ (P - ) (v3 - v2); ⌡ 2 ⌠ Pdv ≈ ⌡ 2 wNET ≈ ∆P[( Collecting terms: 1 v2+v3 v +v ) - ( 12 4)] 2 (the smaller the ∆P, the better the approximation) sfg ∆P ⇒ ≈1 1 ∆T (v + v ) − (v + v ) 2 In the limit as 2 3 ∆T → 0: 2 1 4 v3 → v2 = vg , lim ∆P dPsat sfg = & ∆T→0 = ∆T dT vfg v4 → v1 = vf (P - ∆2P) (v1 - v4) Sonntag, Borgnakke and van Wylen 13.25 Calculate the values hfg and sfg for nitrogen at 70 K and at 110 K from the Clapeyron equation, using the necessary pressure and specific volume values from Table B.6.1. dPg hfg sfg Clapeyron equation Eq.13.7: = = dT Tvfg vfg For N2 at 70 K, using values for Pg from Table B.6 at 75 K and 65 K, and also vfg at 70 K, ∆Pg 76.1-17.41 = 70(0.525 015) = 215.7 kJ/kg (207.8) 75-65 ∆Τ sfg = hfg/T = 3.081 kJ/kg K (2.97) hfg ≈ T(vg-vf) ( ) Comparison not very close because Pg not linear function of T. Using 71 K & 69 K from the software, (44.56-33.24) = 208.0 kJ/kg 71-69 1938.8-1084.2 ≈ 110(0.014 342)( ) = 134.82 kJ/kg (134.17) 115-105 hfg = 70(0.525 015) At 110 K, hfg sfg = 134.82 = 1.226 kJ/kg K (1.22) 110 Sonntag, Borgnakke and van Wylen 13.26 Ammonia at –70oC is used in a special application at a quality of 50%. Assume the only table available is B.2 that goes down to –50oC. To size a tank to hold 0.5 kg with x = 0.5, give your best estimate for the saturated pressure and the tank volume. To size the tank we need the volume and thus the specific volume. If we do not have the table values for vf and vg we must estimate those at the lower T. We therefore use Clapeyron equation to extrapolate from –50oC to –70oC to get the saturation pressure and thus vg assuming ideal gas for the vapor. The values for vf and hfg do not change significantly so we estimate Between -50oC and –70oC: vf = 0.001375 m3/kg, hfg = 1430 kJ/kg The integration of Eq.13.7 is the same as in Example 13.1 so we get P2 hfg T2 - T1 1430 -70 + 50 = ( )= = -1.2923 P1 R T2T1 0.4882 203.15 × 223.15 P2 = P1 exp(-1.2923) = 40.9 exp(-1.2923) = 11.2 kPa 0.4882 × 203.15 vg = RT2/P2 = = 8.855 m3/kg 11.2 ln v2 = (1-x) vf + x vg = 0.5 × 0.001375 + 0.5 × 8.855 = 4.428 m3/kg V2 = mv2 = 2.214 m3 P A straight line extrapolation will give a negative pressure. T -70 -50 Sonntag, Borgnakke and van Wylen 13.27 The saturation pressure can be approximated as ln Psat = A – B/T, where A and B are constants. Use the steam tables and determine A and B from properties at 25o C only. Use the equation to predict the saturation pressure at 30oC and compare to table value. dPsat = Psat (-B)(-T-2) dT so we notice from Eq.13.7 and Table values from B.1.1 and A.5 that hfg 2442.3 B= = = 5292 K R 0.4615 Now the constant A comes from the saturation pressure as 5292 A = ln Psat + B/T = ln 3.169 + = 18.9032 273.15 + 25 Use the equation to predict the saturation pressure at 30C as 5292 ln Psat = A – B/T = 18.9032 = 1.4462 273.15 + 30 Psat = 4.2469 kPa compare this with the table value of Psat = 4.246 kPa and we have a very close approximation. ln Psat = A – B/T ⇒ Sonntag, Borgnakke and van Wylen 13.28 Using the properties of water at the triple point, develop an equation for the saturation pressure along the fusion line as a function of temperature. Solution: The fusion line is shown in Fig. 3.4 as the S-L interphase. From Eq.13.5 we have dPfusion hif = dT Tvif Assume hif and vif are constant over a range of T’s. We do not have any simple models for these as function of T other than curve fitting. Then we can integrate the above equation from the triple point (T1, P1) to get the pressure P(T) as P – P1 = hif T ln T1 vif Now take the properties at the triple point from B.1.1 and B.1.5 P1 = 0.6113 kPa, T1 = 273.16 K vif = vf – vi = 0.001 – 0.0010908 = - 9.08 × 10−5 m3/kg hif = hf – hi = 0.0 – (-333.4) = 333.4 kJ/kg The function that approximates the pressure becomes P = 0.6113 – 3.672 × 106 ln T T1 [kPa] Sonntag, Borgnakke and van Wylen 13.29 Helium boils at 4.22 K at atmospheric pressure, 101.3 kPa, with hfg = 83.3 kJ/kmol. By pumping a vacuum over liquid helium, the pressure can be lowered, and it may then boil at a lower temperature. Estimate the necessary pressure to produce a boiling temperature of 1 K and one of 0.5 K. Solution: Helium at 4.22 K: P1 = 0.1013 MPa, dPSAT dT = hFG TvFG ≈ hFGPSAT RT2 ⇒ ln hFG = 83.3 kJ/kmol P2 P1 = hFG 1 1 [T − T ] R 1 2 For T2 = 1.0 K: ln P2 101.3 For T2 = 0.5 K: ln P2 101.3 = 83.3 [ 1 − 1] 8.3145 4.22 1.0 = 83.3 1 1 [4.22 − 0.5] 8.3145 => P2 = 0.048 kPa = 48 Pa P2 = 2.1601×10-6 kPa = 2.1601 × 10-3 Pa Sonntag, Borgnakke and van Wylen 13.30 A certain refrigerant vapor enters a steady flow constant pressure condenser at 150 kPa, 70°C, at a rate of 1.5 kg/s, and it exits as saturated liquid. Calculate the rate of heat transfer from the condenser. It may be assumed that the vapor is an ideal gas, and also that at saturation, vf << vg. The following quantities are known for this refrigerant: ln Pg = 8.15 - 1000/T ; CP = 0.7 kJ/kg K with pressure in kPa and temperature in K. The molecular weight is 100. Refrigerant: State 1 T1 = 70oC P1 = 150 kPa State 2 P2 = 150 kPa x2 = 1.0 State 3 P3 = 150 kPa x3 = 0.0 Get the saturation temperature at the given pressure ln (150) = 8.15 - 1000/T2 => T2 = 318.5 K = 45.3oC = T3 q = h3 - h1 = (h3 - h2) + (h2 - h1) = - hfg T3 + CP0(T2 - T1) 13 dPg hfg = , dT Tvfg vfg ≈ vg = RT , Pg dPg d ln Pg hfg = Pg = P dT dT RT2 g d ln Pg = +1000/T2 = hfg/RT2 dT hfg = 1000 × R = 1000 × 8.3145/100 = 83.15 kJ/kg q = -83.15 + 0.7(45.3 - 70) = -100.44 kJ/kg . QCOND = 1.5(-100.44) = -150.6 kW 13 T P 1 3 2 3 1 v 2 s Sonntag, Borgnakke and van Wylen 13.31 Using thermodynamic data for water from Tables B.1.1 and B.1.5, estimate the freezing temperature of liquid water at a pressure of 30 MPa. P H2O dPif hif = ≈ const dT Tvif 30 MPa T.P. T At the triple point, vif = vf - vi = 0.001 000 - 0.001 090 8 = -0.000 090 8 m3/kg hif = hf - hi = 0.01 - (-333.40) = 333.41 kJ/kg dPif 333.41 = = -13 442 kPa/K dT 273.16(-0.000 090 8) ⇒ at P = 30 MPa, (30 000-0.6) T ≈ 0.01 + = -2.2 oC (-13 442) Sonntag, Borgnakke and van Wylen 13.32 Small solid particles formed in combustion should be investigated. We would like to know the sublimation pressure as a function of temperature. The only information available is T, hFG for boiling at 101.3 kPa and T, hIF for melting at 101.3 kPa. Develop a procedure that will allow a determination of the sublimation pressure, Psat(T). P TNBP = normal boiling pt T. Solid TNMP = normal melting pt T. 101.3 kPa PTP TTP = triple point T. Vap. T TTP TNMPT NBP 1) TTP ≈ TNMP TTP PTP 2) Liquid ⌠ hFG (1/PSAT) dPSAT ≈ 2 dT ⌠ ⌡ ⌡ RT 0.1013 MPa TNMP Since hFG ≈ const ≈ hFG NBP the integral over temperature becomes ln PTP 0.1013 ≈ hFG NBP R [T 1 NBP - 1 TTP → get PTP 3) hIG at TP = hG - hI = (hG - hF) + (hF - hI) ≈ hFG NBP + hIF NMP Assume hIG ≈ const. again we can evaluate the integral ln PSUB PTP PSUB T hIG 1 ⌠ hIG 1 = ⌡ (1/PSUB) dPSUB ≈ 2 dT ≈ [T − T] ⌠ R TP ⌡ RT PTP or PSUB = fn(T) TTP Sonntag, Borgnakke and van Wylen 13.33 A container has a double wall where the wall cavity is filled with carbon dioxide at room temperature and pressure. When the container is filled with a cryogenic liquid at 100 K the carbon dioxide will freeze so that the wall cavity has a mixture of solid and vapor carbon dioxide at the sublimation pressure. Assume that we do not have data for CO2 at 100 K, but it is known that at −90°C: Psat = 38.1 kPa, hIG = 574.5 kJ/kg. Estimate the pressure in the wall cavity at 100 K. Solution: For CO2 space: at T1 = -90 oC = 183.2 K , P1 = 38.1 kPa, hIG = 574.5 kJ/kg For T2 = TcO2 = 100 K: Clapeyron ln dPSUB hIG hIGPSUB = ≈ dT TvIG RT2 P2 hIG 1 1 574.5 1 1 = [183.2 − 100] = 0.188 92 [183.2 − 100] = -13.81 P1 R or P2 = P1 × 1.005×10-6 ⇒ P2 = 3.83×10-5 kPa = 3.83×10-2 Pa Sonntag, Borgnakke and van Wylen Property Relations 13.34 Use Gibbs relation du = Tds – Pdv and one of Maxwell’s relations to find an expression for (∂u/∂P)T that only has properties P, v and T involved. What is the value of that partial derivative if you have an ideal gas? du = Tds – Pdv divide this by dP so we get ∂u ∂s ∂v ∂v ∂v = T – P = –T – P ∂PT ∂PT ∂PT ∂TP ∂PT where we have used Maxwell Eq.13.23. Now for an ideal gas we get RT Ideal gas: Pv = RT ⇒ v = P then the derivatives are R ∂v ∂v and =P = –RTP–2 ∂TP ∂PT and the derivative of u is R ∂u ∂v ∂v = –T – P = –T P – P( –RTP–2) = 0 ∂PT ∂TP ∂PT This confirm that u is not sensitive to P and only a function of T. Sonntag, Borgnakke and van Wylen 13.35 Start from Gibbs relation dh = Tds + vdP and use one of Maxwell’s equation to get (∂h/∂v)T in terms of properties P, v and T. Then use Eq.13.24 to also find an expression for (∂h/∂T)v. Find ∂h (∂h)T and (∂T)v ∂v dh = Tds + vdP ⇒ and use Eq.13.22 ∂s (∂h)T = T (∂v)T + v(∂P)T ∂v ∂v =T ∂P (∂T)v + v(∂P)T ∂v Also for the second first derivative use Eq.13.28 ∂h s ∂P ∂P (∂T)v = T(∂∂T)v + v(∂T)v = Cv + v(∂T)v Sonntag, Borgnakke and van Wylen 13.36 From Eqs. 13.23 and 13.24 and the knowledge that Cp > Cv what can you conclude about the slopes of constant v and constant P curves in a T-s diagram? Notice that we are looking at functions T(s, P or v given). Solution: The functions and their slopes are: ∂T Constant v: T(s) at that v with slope ∂s v ∂T Constant P: T(s) at that P with slope ∂s P Slopes of these functions are now evaluated using Eq.13.23 and Eq.13.24 as ∂T ∂s -1 T = = C ∂s P ∂TP p ∂T ∂s -1 T = = C ∂s v ∂Tv v Since we know Cp > Cv then it follows that T/Cv > T/Cp and therefore ∂T ∂T > ∂s P ∂s v which means that constant v-lines are steeper than constant P lines in a T-s diagram. Sonntag, Borgnakke and van Wylen 13.37 Derive expressions for (∂T/∂v)u and for (∂h/∂s)v that do not contain the properties h, u, or s. Use Eq. 13.30 with du = 0. ∂P P - T( )v ∂T ∂T ∂u ∂u (see Eqs. 13.33 and 13.34) ( ∂v )u = - (∂v)T (∂T)v = C v / As dh = Tds + vdP => ( ∂h )v = T + v(∂P)v = T - v(∂T)s ∂s ∂s ∂v But ∂s ∂s (∂T)s = - (∂v)T/(∂T)v = ∂v ⇒ v ∂P (∂h)v = T + CT (∂T)v ∂s v ∂P ) ∂T v Cv (Eq.13.20) T( (Eq.13.22) Sonntag, Borgnakke and van Wylen 13.38 Develop an expression for the variation in temperature with pressure in a constant entropy process, (∂T/∂P)s, that only includes the properties P–v–T and the specific heat, Cp. Follow the development for Eq.13.32. ∂s ∂v (∂P)T -(∂T)P T ∂v (∂T)s = - ∂s = - (C /T) = C (∂T)P ∂P P P (∂T)P ∂s ∂v {(∂P)T = -(∂T)P, Maxwell relation Eq. 13.23 and the other is Eq.13.27} Sonntag, Borgnakke and van Wylen 13.39 Use Eq. 13.34 to get an expression for the derivative (∂T/∂v)s. What is the general shape of a constant s process curve in a T-v diagram? For an ideal gas can you say a little more about the shape? Equation 13.34 says ∂P dT + ( )v dv T ∂T so then in a constant s process we have ds = 0 and we find T ∂P (∂T)s = − C (∂T)v ∂v v ds = Cv As T is higher the slope is steeper (but negative) unless the last term (∂P/∂T)v counteracts. If we have an ideal gas this last term can be determined ∂P P = RT/v ⇒ (∂T)v = R v T P (∂T)s = − C R = − C ∂v vv v and we see the slope is steeper for higher P and a little lower for higher T as Cv is an increasing function of T. Sonntag, Borgnakke and van Wylen 13.40 Evaluate the isothermal changes in the internal energy, the enthalpy and the entropy for an ideal gas. Confirm the results in Chapters 5 and 8. We need to evaluate duT, dhT and dsT for an ideal gas: P = RT/v. From Eq.13.31 we get duT = [ T ∂P (∂T)v – P ] dvT = [ T ( R ) – P ] dvT = [ P – P] dvT = 0 v From Eq.13.27 we get using v = RT/P dhT = [ v – T ( ∂v ) ] dPT = [ v – T ( R ) ] dPT = [ v – v ] dPT = 0 P ∂T P These two equations confirms the statements in chapter 5 that u and h are functions of T only for an ideal gas. From eq.13.32 or Eq.13.34 we get dsT = – ( =– ∂v ) dP = ∂T P T ∂P (∂T)v dvT R R dP = dv PT vT so the change in s can be integrated to find s2 – s1 = –R ln P2 v2 = R ln P1 v1 when T2 = T1 Sonntag, Borgnakke and van Wylen Volume Expansivity and Compressibility 13.41 Determine the volume expansivity, αP, and the isothermal compressibility, βT, for water at 20°C, 5 MPa and at 300°C, and 15 MPa using the steam tables. Water at 20oC, 5 MPa (compressed liquid) 1 ∂v 1 ∆v αP = ( )P ≈ ( )P Estimate by finite difference. v ∂T v ∆T Using values at 0oC, 20oC and 40oC, αP ≈ 1 0.001 0056 - 0.000 9977 = 0.000 1976 oC-1 0.000 9995 40 - 0 βT = - ∆v 1 ∂v (∂P)T ≈ - 1(∆P)T v v Using values at saturation, 5 MPa and 10 MPa, βT ≈ - 1 0.000 9972 - 0.001 0022 = 0.000 50 MPa-1 0.000 9995 10 - 0.0023 Water at 300oC, 15 MPa (compressed liquid) 1 0.001 4724 - 0.001 3084 = 0.002 977 oC-1 0.001 377 320 - 280 1 0.001 3596 - 0.001 3972 βT ≈ = 0.002 731 MPa-1 0.001 377 20 - 10 αP ≈ Sonntag, Borgnakke and van Wylen 13.42 What are the volume expansivity αp, the isothermal compressibility βT, and the adiabatic compressibility βs for an ideal gas? The volume expansivity from Eq.13.37 and ideal gas v = RT/P gives 1R 1 1 ∂v αp = ( )P = ( ) = vP T v ∂T The isothermal compressibility from Eq.13.38 and ideal gas gives 1 ∂v 1 1 βT = − ( )T = − ( − RT P−2 ) = v ∂P v P The adiabatic compressibility βs from Eq.13.40 and ideal gas 1 ∂v β s = − ( )s v ∂P From Eq.13.32 we get for constant s (ds = 0) T ∂v v T (∂T)s = C