quiz-2-solution - MECH 310 Thermodynamics I American...

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Unformatted text preview: MECH 310 Thermodynamics I American University of Beirut, Fall 2007 Quiz 2 January 11, 2007 Handout # Quiz 2 • This is a 120 minutes exam. • You allowed to bring in 3 cheat sheets in addition to the thermodynamic tables. • You are advised to read the whole exam before you start. • Make sure you state all the assumptions you make and that you clearly identify any control mass or control volume you utilize in your analysis. • Good luck! Problem 1[30 points] - Hydroelectric Power In the hydroelectric power plant sketched in Figure 1, water flows to the turbine from a large reservoir at a rate of 105 kg/s. The difference in elevation between the water reservoir and the turbine exit plane is 180 m. The turbine is well insulated and its efficiency is 97%. The water velocity at the exit plane of the turbine is 6 m/s. The water reservoir temperature is T0 = 280 K. (a) What is the water temperature at the turbine exit plane? (b) How much entropy is generated by irreversibility in the turbine? (c) If the difference in elevation between the turbine exit plane and the discharge pond is 3 m, what is the water temperature at the discharge pond? How much entropy is generated by the irreversbility in the discharge? water reservoir work out turbine exit plane discharge pond Figure 1: Schematic for problem 1. 1 Solution (a) Applying the first law for a control volume consisting of the water reservoir and the falling water until the inlet to the turbine (CV1 in Figure 2) and noting that the flow is steady, no heat or work interactions are taking place, the velocity of water in the reservoir is negligible, and that the flow is isothermal, we get 2 (Vin )turbine 2 = gH ⇒ (Vin )turbine = 2 2gH = 60 m/s where g is gravity. CV1 water reservoir work out CV2 turbine exit plane discharge pond CV3 Figure 2: Control volumes for problem 1. Next we apply the first law for a control volume consisting of the turbine (CV2 in Figure 2). Noting that the turbine is thermally insulated, the flow is steady, and changes in potential energy across the turbine are negligible, we get 0 = −(W → )turbine + m (hin − hout ) + ˙ m ˙ 2 2 (Vin )turbine − (Vout )turbine 2 The efficiency of the turbine is the ratio of the actual work produced to the maximum work that can be produced. The maximum work corresponds to the isentropic process in the turbine and to zero outlet velocity, η= 2 2 W→ 2m (hin − hout ) + m ((Vin )turbine − (Vout )turbine ) ˙ ˙ = 2 (W → )max 2m (hin − hout,s ) + m ((Vin )turbine ) ˙ ˙ Now using the relation dh = T ds + v dp and noting that the turbine inlet and outlet pressure are atmospheric, then for an isentropic process hin = hout,s . Also note that hin − hout = cp (Tin − Tout ), then η= 2 2 W→ 2cp (Tin − Tout ) + (Vin − Vout ) = 2 (W → )max Vin 2 Which leads to Tout = Tin + 1 2 2 (1 − η )Vin − Vout = 288.6 K 2cp (b) Applying 2nd law for the turbine control volume and noting that the turbine is thermally insulated, then for an isobaric process of an incompressible liquid Tout ˙ Sirr = m(sout − sin ) = mcp ln ˙ ˙ = 12663 kJ/s K Tin (c) Assuming that the discharge process (CV3 in Figure 2) is adiabatic (this claim is not entirely true since some heat is transferred from the water to the surrounding), the first lay yields for a steady process , 0 = m(hin − hout ) + ˙ m ˙ 2 2 Vin − Vout + mg (zin − zout ) ˙ 2 with Vout = 0 and hin − hout = cp (Tin − Tout ), we get 2 Vin 0 = cp (Tin − Tout ) + + g (zin − zout ) 2 V2 g ⇒ Tout = Tin + in + (zin − zout ) = 288.6 + 11.46 = 300 K 2cp cp The second law applied for the control volume consisting of the turbine discharge and the discharge pond assuming adiabatic process leads to Tout ˙ = 16310 kJ/s K Sirr = m(sout − sin ) = mcp ln ˙ ˙ Tin 3 Problem 2 [20 points] - Homer Simpson: “Lisa... In this house, we obey the laws of thermodynamics” US Patent No.: US 6,962,052 B2 (2005) by “inventor” Haim Goldenblum. . W→ . Q← Figure 3: Schematic for problem 2. ˙ (a) Does the proposed device obey the first law of thermodynamics? If heat Q enters the heat exchanger, how much work per unit time does the device produce? (b) Does the device obey the second law of thermodynamics? Why? 4 Solution (a) Applying first law for the cyclic device, we get W → = Q← So the device obeys the first law of thermodynamics. (b) The device violated the Kelvin-Planck statement of the second law since it consists of a cyclic device producing work while exchanging heat with a single thermodynamic reservoir. 5 Problem 3 [25 points] 1- Air is contained in the insulated cylinder shown in the figure below. At this point Air is containedis atthe insulatedºC and theshown in volume is 15 L. The pistonpoint the air the air in 140 kPa, 25 cylinder cylinder the figure below. At this crossis at 140 sectional ◦ C and0.045cylinder the spring is linear with springcross-sectional area is kPa, 25 area is the m2, and volume is 15 L. The piston constant 35 kN/m. 0.045 m2 ,Tand the is opened,linear with spring constant 35 kN/m. Theinto theis opened, he valve spring is and air from the line at 700 kPa, 25 ºC flows valve and air from the line at 700essure 25 ◦ Ces 700 kPa, the cylinder until the pressure reaches cylinder until the pr kPa, reach flows into and the valve is closed. Find the final emp the valve 700 kPa, tand erature is closed. Find the final temperature. F made of a rock bed for problem 3. 2- A thermal storage is igure 4: Schematic having a volume of 2 m3, density 2750 3 kg/m and specific heat of 0.89 kJ/kg K. The bed is heated by solar energy until its temperature reaches 400 K. If a reversible heat engine is connected between the rock bed and the environment thermal reservoir at temperature To. a) Find the amount of work produced as the temperature of the rock bed decreases to To. b) If the environmental thermal reservoir is replaced by a rigid vessel containing 10 kg of air that is initially at To, what would be the amount of work produced? 6 Solution Applying the first law for the control volume consisting of the air inside the cylinder and noting that the process is unsteady and the cylinder is insulated, then m2 u2 − m1 u1 = −W → + min hin (1) where kinetic and potential energy effects are neglected. Now conservation of mass requires min = m2 − m1 Assuming the process to be quasi-static, the work done by the system must balance the work done on the environment plus the work done in compressing the spring W→ = Noting that ∆z = z − z1 = volume at initial state V2 V1 V1 −V , Apiston p dV = V2 V1 p1 − kspring (∆z ) Apiston dV where z1 and V1 correspond height of piston and 1 kspring (V2 − V1 )2 2 A2 piston W → = p1 (V2 − V1 ) + Equation (1) becomes m2 u2 − m1 u1 = p1 (V2 − V1 ) + kspring (V2 − V1 )2 + (m2 − m1 )hin 2A2 piston Using the ideal gas law pV = mRT and noting u = cv T , h = cp T , cv kspring cp (p2 V2 − p1 V1 ) = p1 (V2 − V1 ) + (V2 − V1 )2 + 2 R 2Apiston R p2 V2 p1 V1 − Tin T2 T1 (2) The unknowns in the above equation are the volume and temperature at the final state. The additional equation needed is naturally the second law of thermodynamics applied for the control volume between initial and final states. Noting that the cylinder is insulated and assuming that the process is irreversible, we get m2 s2 − m1 s1 = min sin = (m2 − m1 )sin ⇒ m2 (s2 − sin ) = m1 (s1 − sin ) p2 T1 p1 T2 ⇒ m2 cp ln − R ln = m1 cp ln − R ln Tin pin Tin pin p2 V2 T2 p2 p1 V1 T1 p1 cp ln = cp ln ⇒ − R ln − R ln RT2 Tin pin RT1 Tin pin (3) So basically we solve for T2 by substituting expression for V2 obtained from equation (3) in equation (2) to get an equation in which the only unknown is T2 . 7 2- A thermal storage is made of a rock bed having a volume of 2 m3, density 2750 Problem a4 [25 points] of 0.89 kJ/kg K. The bed is heated by solar energy until kg/m3 nd specific heat its temperature reaches 400 K. If a reversible heat engine is connected between 3 A thermal storage is madeironment thermahaving airvolume erature 3 , o. the rock bed and the env of a rock bed l reservo at temp of 2 m T density 2750 kg/m and specific heat of 0.89 kJ/kg K. The bed is heated by solar energy until its temperature reaches a400 ind thea reversible heat engine is connected ature of the rock bed bed and the ) F K. If amount of work produced as the temper between the rock environmentecreases to To. d thermal reservoir at temperature To . a) FindbtheIf the environmental thermal as the temperature of the rock bed decreases to To . ) amount of work produced reservoir is replaced by a rigid vessel b) If the environmental thermal reservoir is replaced what wouldvessel containing 10 kg of containing 10 kg of air that is initially at To, by a rigid be the amount of work produced? air that is initially at To , what would be the amount of work produced? Figure 5: Schematic for problem 4. 8 Solution (a) First law for the cyclic engine gives ∆E = 0 = Q← − Q→ − W → H L Second law for the reversible cyclic engine gives ← → ∆S = 0 = SH − SL = δ Q ← Q→ H −L T To where the temperature of the rock bed T decreases from TH =400 K to To . Now applying the first law for the rock bed over an infinitesimal process from T to T − dT , we get δQ← = −mrb crb dT = −(ρV c)rb dT H Substituting in the entropy balance equation, we get To TH (4) ⇒ Q→ L Also from equation (4), −(ρV c)rb dT Q→ − L =0 T To TH = (ρV c)rb To ln To Q← = (ρV c)rb (TH − To ) H Finally the work output is W → = Q← − Q→ = (ρV c)rb (TH − To ) − To ln H L TH To (b) In this case, not only the temperature of the rock bed is decreasing as it is giving heat to the cyclic engine, but also the temperature of the water reservoir is increasing as it is receiving heat from the cyclic engine. The process will start from the initial state and continue until both reservoirs reach the same final temperature Tf . First law for the rock bed, we have δQ← = −mrb crb dT H ⇒ Q← = mrb crb (TH − Tf ) H First law for the water reservoir δQ→ = mw cw dT L → ⇒ QL = mw cw (Tf − To ) 9 Entropy balance for an infinitesimal reversible process of the cyclic engine yields, Tf δQ→ δQ← L H − =0 T T TH To TH Tf ⇒ (mc)rb ln = (mc)w ln Tf To Tf C C ⇒ Tf = To w TH rb 1 Cw +Crb where Cw = mw cw and Crb = mrb crb . Substituting Tf in Q← and Q→ , we get H L C C W → = Q← − Q→ = Crb TH − To w TH rb L H 1 Cw +Crb − Cw C C To w TH rb 1 Cw +Crb − To 10 ...
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