Exam01Soln
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Exam01Soln

Course Number: MAE 320, Fall 2008

College/University: WVU

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Name MAE 320 Thermodynamics Exam 1 February 16, 2009 Each problem is worth the points indicated. Closed book and closed notes. Cell phone must be turned off. No group work. You must be completed by 8:55 AM with the exam and appendix tables turned in. Clearly state any assumptions or unknowns. 1. For the following problems, place the correct answer next to the problem number in the space provided. No other credit...

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320 Name MAE Thermodynamics Exam 1 February 16, 2009 Each problem is worth the points indicated. Closed book and closed notes. Cell phone must be turned off. No group work. You must be completed by 8:55 AM with the exam and appendix tables turned in. Clearly state any assumptions or unknowns. 1. For the following problems, place the correct answer next to the problem number in the space provided. No other credit will be given. Each question is worth 5 points each; Problem 1 is worth a total of 40 points. 1a. B The density and the specific volume of a simple compressible system are known. The number of additional intensive, independent properties needed to fix the state of this system is (A) 0 (B) 1 "The state of a simple compressible substance is fixed by two intensive, (C) 2 independent properties. Specific volume and density are dependent, so (D) 3 they count as one property. Therefore, we need one more property." (E) 4 1b. D A pressure gage connected to a tank reads 55 kPa at a location where the atmospheric pressure is 72.1 cmHg. The density of mercury is 13,600 kg/m3. The absolute pressure in the tank is Pabs = Pamb + Pgage = gh + Pgage (A) 41 kPa (B) 56 kPa kg m 1kN 1kPa (C) 82 kPa = 13600 3 9.81 2 ( 0.721m ) + 55kPa 2 2 m s (D) 151 kPa 1000kg m / s 1kN / m (E) 196 kPa = 151kPa 1c. D A 1.5-kW electric resistance heater in a room is turned on and kept on for 20 min. The amount of energy transferred to the room by the heater is (A) 1.5 kJ & W = Wt (B) 60 kJ 60s 1kJ / s (C) 750 kJ = (1.5kW )( 20 min ) 1 min kW (D) 1800 kJ (E) 3600 kJ = 1800kJ 1d. C The compressibility chart (Figure A-15) can be used to determine (A) the enthalpy of an ideal gas. (B) the quality of a saturated mixture. (C) the deviation from ideal gas behavior. (D) All of the above. (E) None of the above. Name 1e. D Why does one use the van der Waals (or Beattie-Bridgeman and Benedict-WebbRuebin) equation? (A) They provides for a more accurate representation of real gas behavior. (B) They allow for a closed form solution of property data for computer computation. (C) Most gasses have constants in these equations readily available. (D) All of the above. (E) None of the above. 1f. A A rigid tank contains 0.8 kg of steam at 1.0 MPa and 300 C. The volume of the tank is most closely to (A) 0.206 m3 v = V / m V = mv (B) 0.324 m3 From Table A - 6 v = 0.25799 m3 / kg 3 (C) 4.85 m V = ( 0.8kg ) ( 0.25799m3 / kg ) = 0.2064m3 (D) 36.9 m3 3 (E) 310 m 1g. State the first law of thermodynamics in sentence format. Do not give a formula and do not state the first law equation formula in sentence format. The first law of thermodynamics states that energy can be neither created nor destroyed during a process; it can only change forms. See page 70. 1h. Explain why the "" is used for changes in work and heat transfer but the "" or "d" is used for changes in internal energy, kinetic energy, and potential energy. The is used to represent that work and heat are path functions and are inexact differentials while properties such as U, KE, and PE are point functions and the changes on rely on the difference between the initial and final states. See page 63. Name 2. (a) Locate the state P=1400 kPa, u=2698.9 kJ/kg for water on a P-v diagram, sketch (not to scale) with respect to the saturated liquid and saturated vapor lines. (b) Label the temperature curve through the state. Make sure your temperature curve extends from the compressed liquid region into the superheat region. (c) Place the value of the specific volume for the state on the specific volume axis and the value of the pressure for the state on the pressure axis. (d) Place the value of the saturation pressure for this state (temperature) on the P-axis. (20 points) P 3976.2 kPa 1400 kPa T = 250 C v 0.16356 m3/kg Name 3. Water is pumped from a 100 m deep well to a 50 m high storage tank. Determine the required power, in kW, to pump 700 liters per minute. (20 points) The power required to pump the water is the product of the mass flow rate and the specific potential energy. & & & W = m * pe = mgz & = Vgz The selection of the density can be assumed to be 1000 kg/m3, the gravity is 9.81 m/s2 and the change in elevation is 50-(-100)=150 m. kg Liters m & W = 1000 3 700 9.81 2 (150m ) m min s kg - m / s 2 Liters 1N 1m3 1 min 2 2 m min kg - m / s 1000Liters 60s Nm 1kN = 171675 s 1000N kN 1kW = 17.1675 s 1kN / s = 17.17 kW = 1030050000 Name 4. Consider a wind turbine with a 10 m diameter rotor that is rotating at 25 rpm under steady wind at an average velocity of 20 km/h. The turbine has an efficiency of 33%. If the density of air is 1.20 kg/m3, determine (a) the power produced, in kW, and (b) the revenue generated by the wind turbine per year if the electric power produced is sold to the utility at $0.05/kWh. The definition of efficiency for a turbine is = or & W & Wshaft out & E = mech ,fluid & Wuseful & Winput useful & = Winput where & mV 2 & & & Winput = E mech ,fluid = mke = 2 and & m = AV So that AV & Wuseful = 2 d 2 A= 4 2 3 kg (10 ) 2 km 1h 1000m m 20 1 .2 3 h 3600s km m 4 = 0.33 2 2 3 kg m m N = 2666.47 3 3 2 m s kg - m / s 3 = 2666.47 N - m 1J s N-m J W = 2666.47 sJ/s kW = 2666.47W 1000W = 2.666kW The money gained by selling this electric is & Money = W * t * Unit Cost days 24hours = ( 2.66647kW ) 365 ( $0.05 / kWh ) year day = $1170 / year Name MAE 320 Thermodynamics Exam 1 February 16, 2009 Each problem is worth the points indicated. Closed book and closed notes. Cell phone must be turned off. No group work. You must be completed by 8:55 AM with the exam and appendix tables turned in. Clearly state any assumptions or unknowns. 1. For the following problems, place the correct answer next to the problem number in the space provided. No other credit will be given. Each question is worth 5 points each; Problem 1 is worth a total of 40 points. 1a. A The density and the internal energy of a simple compressible system are known. The number of additional intensive, independent properties needed to fix the state of this system is (A) 0 (B) 1 "The state of a simple compressible substance is fixed by two intensive, (C) 2 independent Internal properties. energy and density are independent, so they (D) 3 count as two properties. Therefore, we do not need any more properties" (E) 4 1b. E A pressure gage connected to a tank reads 100 kPa at a location where the atmospheric pressure is 72.1 cmHg. The density of mercury is 13,600 kg/m3. The absolute pressure in the tank is Pabs = Pamb + Pgage = gh + Pgage (A) 41 kPa (B) 56 kPa kg m 1kN 1kPa (C) 82 kPa = 13600 3 9.81 2 ( 0.721m ) + 100kPa 2 2 m s 1000kg m / s 1kN / m (D) 151 kPa (E) 196 kPa = 196kPa 1c. E A 1.5-kW electric resistance heater in a room is turned on and kept on for 40 min. The amount of energy transferred to the room by the heater is & (A) 1.5 kJ W = Wt (B) 60 kJ 60s 1kJ / s (C) 750 kJ = (1.5kW )( 40 min ) 1 min kW (D) 1800 kJ (E) 3600 kJ = 3600kJ 1d. C The compressibility chart (Figure A-15) can be used to determine (A) the enthalpy of an ideal gas. (B) the quality of a saturated mixture. (C) the deviation from ideal gas behavior. (D) All of the above. (E) None of the above. Name 1e. D Why does one use the van der Waals (or Beattie-Bridgeman and Benedict-WebbRuebin) equation? (A) They provides for a more accurate representation of real gas behavior. (B) They allow for a closed form solution of property data for computer computation. (C) Most gasses have constants in these equations readily available. (D) All of the above. (E) None of the above. 1f. B A rigid tank contains 5.5 kg of steam at 4.0 MPa and 300 C. The volume of the tank is most closely to (A) 0.206 m3 v = V / m V = mv (B) 0.324 m3 From Table A - 6 v = 0.05887 m3 / kg 3 (C) 4.85 m V = ( 5.5kg ) ( 0.05887m 3 / kg ) = 0.324m3 (D) 36.9 m3 3 (E) 310 m 1g. State the first law of thermodynamics in sentence format. Do not give a formula and do not state the first law equation formula in sentence format. The first law of thermodynamics states that energy can be neither created nor destroyed during a process; it can only change forms. See page 70. 1h. Explain why the "" is used for changes in work and heat transfer but the "" or "d" is used for changes in internal energy, kinetic energy, and potential energy. The is used to represent that work and heat are path functions and are inexact differentials while properties such as U, KE, and PE are point functions and the changes on rely on the difference between the initial and final states. See page 63. Name 2. (a) Locate the state P=1600 kPa and u=258.47 kJ/kg for R134a on a P-v diagram, sketch (not to scale) with respect to the saturated liquid and saturated vapor lines. (b) Label the temperature curve through the state. Make sure your temperature curve extends from the compressed liquid region into the superheat region. (c) Place the value of the specific volume for the state on the specific volume axis and the value of the pressure for the state on the pressure axis. (d) Place the value of the saturation pressure for this state (temperature) on the P-axis. (20 points) P 1600 kPa T = 57.88 C v 0.012123 m3/kg Name 3. Water is pumped from a 10 m deep well to a 50 m high storage tank. Determine the required power, in kW, to pump 70 liters per minute. (20 points) The power required to pump the water is the product of the mass flow rate and the specific potential energy. & & & W = m * pe = mgz & = Vgz The selection of the density can be assumed to be 1000 kg/m3, the gravity is 9.81 m/s2 and the change in elevation is 50-(-100)=150 m. kg Liters m & W = 1000 3 70 9.81 2 ( 60m ) m s min kg - m / s 2 Liters 1N 1m3 1 min 2 2 m min kg - m / s 1000Liters 60s Nm 1kN = 686.7 s 1000N kN 1kW = 0.6867 s 1kN / s = 0.687 kW = 41202000 Name 4. Consider a wind turbine with a 20 m diameter rotor that is rotating at 25 rpm under steady wind at an average velocity of 10 km/h. The turbine has an efficiency of 25%. If the density of air is 1.20 kg/m3, determine (a) the power produced, in kW, and (b) the revenue generated by the wind turbine per year if the electric power produced is sold to the utility at $0.06/kWh. The definition of efficiency for a turbine is = or & W & Wshaft out & E = mech ,fluid & Wuseful & Winput useful & = Winput where & mV 2 & & & Winput = E mech ,fluid = mke = 2 and & m = AV So that AV & Wuseful = 2 d 2 A= 4 2 3 kg ( 20 ) 2 km 1h 1000m m 10 1.2 3 h 3600s km m 4 = 0.25 2 2 3 kg m m N = 1010.0 3 3 2 m s kg - m / s 3 = 1010.0 N - m 1J s N-m J W = 1010.0 sJ/s kW = 1010.0W 1000W = 1.01kW The money gained by selling this electric is & Money = W * t * Unit Cost days 24hours = (1.010kW ) 365 ( $0.06 / kWh ) year day = $531 / year Name Useful and Useless Equations; Manometer: P = g h Barometer: P = g h Pressure Pgage = Pabs - Patm and Pvac = Patm - Pabs & & & & Mass Flow m = and = AV ( Volumetric Flow Rate) dT & 4 4 & & Qcond = - A k t , Q conv = h A ( Ts - Tf ) , Q rad = A ( Ts - Tsurr ) dx dE system & & Energy Balance: E in - E out = E system or Ein - Eout = U + KE + PE and E in - E out = dt Or ( Qin - Qout ) + (Win - Wout ) + ( Emass , in - Emass , out ) = U + KE + PE r V2 & & & Where Q = Q t , W = W t , and E = E t and PE = mgZ , KE = m 2 Modes of Heat Transfer: General Heat Equation Electrical Work We = Efficiency: 2 1, along path 2 Q = Q12 , and General Work Equation sh 1, along path 2 W = Fds = W12 1 2 V I dt = VI t , Shaft Work W 1 th = Desired Result , combustion Required Input 1 2 = 2 nT , Spring Work Wspring = k ( x2 - x12 ) 2 & & W W Q = out , generator = electrical output , motor = mechanical output & & HV Wmechanical input Welectrical input y= Quality: x = mass saturated vapor mass total Pv RT Y = y f + x( y g - y f ) m , where y is a property (v, u, h) = y f + x y fg where yfg = yg - yf Ideal Gas: Pv = RT Compressibility Factor: Z = 2 27R 2Tcr RT vactual a and b = cr , v R = van der Waals: P + 2 ( v - b ) = R T , where a = RTcr / Pcr 64Pcr 8Pcr v R T c A a b Beattie-Bridgeman: P = u2 1 - v + B ) - 2 , where A = A 0 1- and B = B0 1- 3 ( v vT v v v R T C 0 1 bR u T - a a c ( - / v2 ) Benedict-Webb-Rubin: P = u + B0 R u T - A 0 - 2 2 + + 6 + 3 2 1 + 2 e v T v v3 v vT v
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