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Unformatted text preview: 101 CHAPTER 10
The correspondence between the new problem set and the previous 4th edition chapter 8 problem set. New 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 Old new new 1 2 3 5 6 7 9 15 16 10 11 49 47 13 14 38 52 8 New 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 Old new new 18 new 19 20 21 new new 23 24 new new 25 26 27 32 28 33 34 New 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 Old 35 41 42 46 54 56 39 55 53 new 43 51 17 29 30 31 36 40 44 57 The problems that are labeled advanced starts at number 53. The English unit problems are: New 61 62 63 64 65 66 67 68 69 70 Old new 58 60 63 61 80 62 new 65 66 New 71 72 73 74 75 76 77 78 79 80 Old 67 new 68 new 69 70 72 new 79 84 New 81 82 83 84 85 86 87 Old 75 83 77 64 71 73 78 102 10.1 Calculate the reversible work and irreversibility for the process described in Problem 5.18, assuming that the heat transfer is with the surroundings at 20C. C.V.: A + B. This is a control mass. Continuity equation: Energy: m2  (mA1 + mB1) = 0 ; m2u2  mA1uA1  mB1uB1 = 1Q2  1W2 PB = PB1 = const. System: if VB 0 piston floats if VB = 0 then P2 < PB1 and v = VA/mtot see PV 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 State B1: Table B.1.2 sup. vapor vB1 = 1.0315 m3/kg, uB1 = 2965.5 kJ/kg mB1 = VB1/vB1 = 0.9695 kg => At (T2 , PB1) m2 = mTOT = 1.56 kg 2 V2 v2 = 0.7163 > v a = VA/mtot = 0.641 so VB2 > 0 so now state 2: P2 = PB1 = 300 kPa, T2 = 200 C => u 2 = 2650.7 kJ/kg and V2 = m2 v2 = 1.560.7163 = 1.117 m3 (we could also have checked Ta at: 300 kPa, 0.641 m3/kg => T = 155 C)
ac 1W2 = 1Q 2 PBdVB = PB1(V2  V1) = PB1(V2  V1) = 264.82 kJ tot
B = m2u2  mA1uA1  mB1uB1 + 1W2 = 484.7 kJ From the results above we have : sA1 = 7.3593, sB1 = 8.0329, s2 = 7.3115 kJ/kg K
rev 1W2 = To(S2  S1)  (U2  U1) + 1Q2(1  To/TH) = 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 ac 1I 2 = 1W2  1W2 = 6.6  (264.82) = 271.4 kJ 103 10.2 Calculate the reversible work and irreversibility for the process described in Problem 5.65, assuming that the heat transfer is with the surroundings at 20C. P 2 1 Linear spring gives
1W2 = PdV = 2(P1 + P2)(V2  V1) 1 v = m(u2  u1) + 1W2 Equation of state: PV = mRT 1Q 2 State 1: V1 = mRT1/P1 = 2 x 0.18892 x 673.15 /500 = 0.5087 m3 State 2: V2 = mRT2/P2 = 2 x 0.18892 x 313.15 /300 = 0.3944 m3
1W2 = 2(500 + 300)(0.3944  0.5087) = 45.72 kJ 1 From Figure 5.10: Cp(Tavg) = 45/44 = 1.023 Cv = 0.83 = C p  R For comparison the value from Table A.5 at 300 K is Cv = 0.653 kJ/kg K
1Q 2 rev 1W2 = = mCv(T2  T1) + 1W2 = 2 x 0.83(40  400)  45.72 = 643.3 kJ To(S2  S1)  (U2  U1) + 1Q2(1  To/TH)
ac = Tom(s2  s1)+ 1W2  1Q2To/To = Tom[CP ln(T2 / T1)  R ln(P2 / P1)] + 1W2  1Q2 = 293.15 x 2 [ 1.023 ln(313/673)  0.1889 ln(300/500)]  45.72 + 643.3 = 402.6  45.72 + 643.3 = 195.0 kJ
1I 2 rev ac ac = 1W2  1W2 = 195.0  (45.72) = 240.7 kJ 10.3 The compressor in a refrigerator takes refrigerant R134a in at 100 kPa, 20C and compresses it to 1 MPa, 40C. With the room at 20C find the minimum compressor work. Solution: C.V. Compressor out to ambient. Minimum work in is the reversible work. SSSF, 1 inlet and 2 exit energy Eq.: wc = h1  h2 + qrev Entropy Eq.: s2 = s1 + dq/T + sgen = s1 + qrev/T0 + 0 => qrev = T0(s2 s1) wc min = h1  h2 + T0(s2 s1) = 387.22  420.25 + 293.15 (1.7148  1.7665) = 48.19 kJ/kg 104 10.4 Calculate the reversible work out of the twostage turbine shown in Problem 6.41, assuming the ambient is at 25C. Compare this to the actual work which was found to be 18.08 MW. C.V. Turbine. SSSF, 1 inlet and 2 exits. Use Eq. 10.12 for each flow stream with q = 0 for adiabatic turbine. Table B.1: h1 = 3373.7, h2 = 2755.9, h3 = 251.4 + 0.9 x 2358.3 = 2373.9 s1 = 6.5966, s2 = 6.8382, s3 = 0.8319 + 0.9 x 7.0766 = 7.2009 . . . . . . . Wrev = T0(m2s2 + m3s3  m1s1)  (m2h2 + m3h3  m1h1) = 298.15(5 6.8382 + 15 7.2009  20 6.5966) (5 2755.9 + 15 2373.9  20 3373.7) . . = 21.14 MW = Wac + Qrev = 3062.7 kW + 18084 kW 10.5 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 T A as shown in Fig. P10.5. . 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 T A = 20C, TC = 5C, TF = 10C, and QF = 3 kW. C.V. Refrigerator (heat pump), SSSF, no external flows except heat transfer. . . . . Energy Eq.: QF + Qc + W = QA (amount rejected to ambient) Reversible gives minimum work in as from Eq. 10.1 or 10.9 on rate form. TA . TA . . 293.15 293.15 W = QF 1  + Qc 1  = 31  + 31  263.15 278.15 TF TC = 0.504 kW 10.6 (negative so work goes in) An air compressor takes air in at the state of the surroundings 100 kPa, 300 K. The air exits at 400 kPa, 200C at the rate of 2 kg/s. Determine the minimum compressor work input. C.V. Compressor, SSSF, minimum work in is reversible work. 1 = 0 at ambient conditions s0  s2 = sT0  sT2  R ln(P0/P2) = 6.86926  7.3303  0.287 ln(100/400) = 0.06317 2 = h2  h0 + T0(s0  s2) = 475.79  300.473 + 300(s0  s2) 2 = 156.365 kJ/kg . . . WREV = m(21) = 312.73 kW = Wc 105 10.7 A supply of steam at 100 kPa, 150C is needed in a hospital for cleaning purposes at a rate of 15 kg/s. A supply of steam at 150 kPa, 250C is available from a boiler and tap water at 100 kPa, 15C is also available. The two sources are then mixed in an SSSF mixing chamber to generate the desired state as output. Determine the rate of irreversibility of the mixing process.
1 3 2 B.1.1: h1 = 62.99 s1 = 0.2245 B.1.3: h2 = 2972.7 s2 = 7.8437 B.1.3: h3 = 2776.4 s3 = 7.6133 C.V. Mixing chamber, SSSF . . . Cont.: m1 + m2 = m3 . . . Energy: m1h1 + m2h2 = m3h3 . . . . Entropy: m1s1 + m2s2 + Sgen = m3s3 . . 2776.4  62.99 m2/m3 = (h3  h1)/(h2  h1) = = 0.9325 2972.7  62.99 . . m2 = 13.988 kg/s, m1 = 1.012 kg/s . . . . . I = T0Sgen = T0(m3s3  m1s1  m2s2) = 298.15 (15 7.6133  1.012 0.2245  13.988 7.8437) = 1269 kW 10.8 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.
1 3 2 C.V. Mixing chamber . . . . Cont.: m1 + m2 = m3 = 2m1 . . . Energy: m1h1 + m1h2 = 2m1h3 Properties from Table A.7 h3 = (h1 + h2)/2 = (300.473 + 1635.8)/2 = 968.14 kJ/kg sT3 = 8.9867 . . . . Entropy Eq.: m1s1 + m1s2 + Sgen = 2m1s3 . . Sgen/2m1 = s3  (s1 + s2)/2 = 8.9867  (6.86926  8.61208)/2 = 1.246 kJ/kg K 106 10.9 A steam turbine receives steam at 6 MPa, 800C. 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 20C, 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 + 0.95152373.14 = 2483.9 wT,s = 4132.7  2483.9 = 1648.79 kJ/kg C.V. Actual turbine wT,ac = wT,s = 1483.91 kJ/kg = hi  he,ac  qloss he,ac = hi  qloss  wT,ac = 4132.7  49.7  1483.91 = 2599.1 Actual exit state: P,h sat. vap., se,ac = 8.0085 kJ kg C.V. Reversible process, work from Eq.10.12 qR = T0(se,ac  si) = 293.15 (8.0085  7.6566) = 103.15 wR = hi  he,ac + qR = 4132.7  2599.1 + 103.16 = 1636.8 kJ/kg 10.10 A 2kg piece of iron is heated from room temperature 25C to 400C by a heat source at 600C. What is the irreversibility in the process? C.V. Iron out to 600C source, which is a control mass. Energy Eq.: mFe(u2  u1) = 1Q2  1W2 =>
1W2 Process: Constant pressure = PmFe(v2  v1) 1Q2 = mFe(h2  h1) = mFeC(T2  T1)
1Q2 = 2 0.42 (400  25) = 315 kJ mFe(s2  s1) = 1Q2/Tres + 1S2 gen
1S2 gen = mFe(s2  s1)  1Q2/Tres = mFeC ln (T2/T1)  1Q2/Tres = 2 0.42 ln 673.15 315 = 0.3233 kJ/K 298.15 873.15 1I2 = To (1S2 gen ) = 298.15 0.3233 = 96.4 kJ 107 10.11 A 2kg/s flow of steam at 1 MPa, 700C should be brought to 500C by spraying in liquid water at 1 MPa, 20C in an SSSF setup. Find the rate of irreversibility, assuming that surroundings are at 20C. C.V. Mixing chamber, SSSF. State 1 is superheated vapor in, state 2 is compressed liquid in, and state 3 is flow out. No work or heat transfer. Cont.: Energy: Entropy: m3 = m1 + m2 m3h3 = m1h1 + m2h2 m3s3 = m1s1 + m2s2 + Sgen 2 1 3 Table B.1.3: h1 = 3923.1, s1 = 8.2731, h3 = 3478.5, s3 = 7.7622, For state 2 interpolate between, saturated liquid 20C table B.1.1 and, compressed liquid 5 MPa, 20C 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 108 10.12 Fresh water can be produced from saltwater by evaporation and subsequent condensation. An example is shown in Fig. P10.12 where 150kg/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 20C 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 1 2 3 T [C] 30 25 25 . . . C.V. Valve. m1 = mex = m2 + m3 Energy Eq.: h1 = 125.77 , h1 = he ; 4 5 23 6 7 17 8 20 Entropy Eq.: s1 + sgen = se s1 = 0.4369 , P2 = Psat 2 = T3) = 3.169 kPa (T he = h1 xe = (125.77  104.87)/2442.3 = 0.008558 se = 0.3673 + 0.008558 8.1905 = 0.4374 sgen = se  s1 = 0.4374  0.4369 = 0.000494 . . I = mT0sgen = 150 293.15 0.000494 = 21.72 kW C.V. Condenser. h2 = 2547.2 , s2 = 8.558 , h5 = 96.5 , s5 = 0.3392 . . m2 = (1  xe)m1 = 148.716 kg/s h7 = 71.37 , s7 = 0.2535 , h8 = 83.96 , s8 = 0.2966 . . . . m2h2 + m7h7 = m2h5 + m7h8 . . 2547.2  96.5 kg m7 = m2 (h2  h5)/(h8  h7) = 148.716 = 28948 83.96  71.37 s . . . . . 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 109 10.13 An air compressor receives atmospheric air at T0 = 17C, 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 ; Table A.7: he,s = 617.51 wc,in,s = 617.51  290.58 = 326.93 Actual: wc,in,ac = wc,in,s/c = 371.51 ; wc,in,ac + hi = he,ac + qloss => h e,ac = 290.58 + 371.51  32.693 = 629.4 => 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. 10.14 A piston/cylinder has forces on the piston so it keeps constant pressure. It contains 2 kg of ammonia at 1 MPa, 40C and is now heated to 100C by a reversible heat engine that receives heat from a 200C 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 se,s = si Pr,e,s = Pr,i (Pe/Pi) = 0.9917 14 = 13.884 qloss = 32.693 NH 3 QL
H.E. = mam(s2  s1)Tres W Process: P = const. 1W2 = P(v2  v1)mam WH.E. = mam(s2  s1)Tres  mam(h2  h1) QH
200 C Table B.2.2: h1 = 1508.5, s1 = 5.1778, h2 = 1664.3, s2 = 5.6342 WH.E. = 2 [(5.6342  5.1778)473.15  (1664.3  1508.5)] = 120.3 kJ 1010 10.15 Air flows through a constant pressure heating device, shown in Fig. P10.15. 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. C.V. Total out to T0 s1 + qrev/T0 = s2 0 h1 + qrev  wrev = h2 0 qrev = T0(s2  s1) 0 (same as Eq. 10.12) h2  h1 = T0(s2  s1)  wrev Constant Cp gives: Cp(T2  T1) = T0Cp ln (T2/T ) + 200 1 T2  T0ln(T2/T1) = T1 + 200/Cp T1 = 300 , Cp = 1.004 , T0 = 300 T2  300 ln (T2/300) = 300 + (200/1.004) = 499.3 At 600 K LHS = 392 (too low) At 800 K LHS = 505.75 Linear interpolation gives T2 = 790 K (LHS = 499.5 OK) 10.16 Air enters the turbocharger compressor (see Fig. P10.16), of an automotive engine at 100 kPa, 30C, and exits at 170 kPa. The air is cooled by 50C 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 compressioncooling process. a) Compressor. First ideal which is reversible adiabatic, constant s: T2S = T1
k1 k (P )
1 P2 = 303.2 (170) 100 0.286 = 352.9 K wS = CP0(T1  T2S) = 1.004(303.2  352.9) = 49.9 w = wS/S = 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 = 46.3C b) Irreversibility from Eq.10.13 with rev. work from Eq.10.12, (q = 0 at TH) 319.4 170 kJ s3  s1 = 1.004 ln  0.287 ln = 0.1001 kg K 303.2 100 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 1011 10.17 A car airconditioning unit has a 0.5kg aluminum storage cylinder that is sealed with a valve and it contains 2 L of refrigerant R134a at 500 kPa and both are at room temperature 20C. It is now installed in a car sitting outside where the whole system cools down to ambient temperature at 10C. What is the irreversibility of this process? C.V. Aluminum and R134a 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 Table B.5.2: v1 = 0.04226, u1 = 411.65  5000.04226 = 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 + 0.4216185.7 = 264.9, s2 = 0.9507 + 0.42160.7812 = 1.2801
1Q2 1S2 gen 1I2 = 0.5 (27) + 0.0473(264.9  390.5) =  19.44 kJ 19.44 = 0.003815 kJ/K 263.15 = 0.5 (0.09716) + 0.0473(1.2801  1.7342) + = T0 ( 1S2 gen ) = 263.15 0.003815 = 1.0 kJ 10.18 A steady combustion of natural gas yields 0.15 kg/s of products (having approximately the same properties as air) at 1100C, 100 kPa. The products are passed through a heat exchanger and exit at 550C. 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 20C? PROD.e PROD.i Heat T i 1100 o o Exchanger 550 o ( C) 1100 C C . e QH . 550 Heat WNET Engine . S QL . . QH = mPRODCP0(TL  Te)PROD = 0.15 1.004(1100  550) = 82.83 . . . SL =  Sie = mPRODCP0 ln(Ti/Te)PROD = 0.15 1.004 ln(1373.15/823.15) = 0.077 kW/K . . QL = TL SL = 293.15 0.077 = 22.57 kW . . . WNET = QH  QL = 82.83  22.57 = 60.26 kW
T L= 20 C
o TL 20 s 1012 10.19 A counterflowing heat exchanger cools air at 600 K, 400 kPa to 320 K using a supply of water at 20C, 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).
2 1 W 4 AIR C.V. Total mah1 + mH2Oh3 = mah2 + mH2Oh4 + W mas1 + mH2Os3 = mas2 + mH2Os4 H2O (sgen = 0) 3 Table A.7: h1 = 607.316, B.1.1: sT1 = 7.57638 Table A.7: h2 = 320.576, sT2 = 6.93413, 4: P4 = P3, s4 Table B.1.2: h3 = 83.96, s3 = 0.2966 s4 = (ma/mH2O)(s1  s2) + s3 = (1/0.1)(7.57638  6.93413) + 0.2966 = 6.7191 x4 = (6.71911.530)/5.597 = 0.9271, T4 = 120.20C h4 = 504.68 + 0.9271 2201.96 = 2546.1 kJ/kg, W = ma(h1  h2) + mH2O(h3  h4) = 1(607.32  320.58) + 0.1(83.96  2546.1) = 40.53 kW 10.20 Water as saturated liquid at 200 kPa goes through a constant pressure heat exchanger as shown in Fig. P10.20. The heat input is supplied from a reversible heat pump extracting heat from the surroundings at 17C. 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 + Q/T0 = m1s2 . . Substitute Q0 into energy equation and divide by m1 h1  T0s1 + win = h2  T0s2 LHS = 504.7  290.15 1.5301 + 4060/2 = 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 600 C: At 700 C: h2  T0s2 = 3703.96  290.15 8.7769 = 1157.34 kJ/kg h2  T0s2 = 3927.66  290.15 9.0194 = 1310.68 kJ/kg Linear interpolation T2 = 667C = (h2  T0s2)  (h1  T0s1) = win = 1200 kJ/kg = 1260.7  504.7 + 290.15 1.5301 1200 kJ/kg 1013 10.21 Calculate the irreversibility for the process described in Problem 6.63, assuming that heat transfer is with the surroundings at 17C. I = To Sgen so apply 2nd law out to To = 17 oC m2 s2  m1s1 = misi + 1Q2 / To + 1S2gen ToSgen = To ( m2 s2  m1s1  mi si )  1Q2 mi = 3.082 kg m2 =3.982 kg ToSgen = 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 m1 = 0.90 kg  (  819.2 kJ) = 290.15 (0.0956  1.4705) + 819.2 = 420.3 kJ 10.22 The hightemperature heat source for a cyclic heat engine is a SSSF heat exchanger where R134a enters at 80C, saturated vapor, and exits at 80C, saturated liquid at a flow rate of 5 kg/s. Heat is rejected from the heat engine to a SSSF heat exchanger where air enters at 150 kPa and ambient temperature 20C, and exits at 125 kPa, 70C. 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. R134a Heat Exchanger, mR134a = 5 kg/s Inlet: T1 = 80oC, sat. vap. x1 = 1.0, Table B.5.1 h1 = hg = 429.189 kJ/kg, s1 = sg = 1.6862 kJ/kgK Exit: T2 = 80oC, sat. liq. x2 = 0.0 h2 = hf = 322.794 kJ/kg, s2 = sf = 1.3849 kJ/kgK C.V. Air Heat Exchanger, C p = 1.004 kJ/kgK, R = 0.287 kJ/kgK Inlet: T3 = 20oC, P3 = 150 kPa Exit: T4 = 70oC, P4 = 125 kPa . . . . 2nd Law: I = To Snet = mR134a (s2  s1) + mair (s4  s3) s4  s3 = Cpln(T4/T3)  Rln(P4/P3) = 0.2103 kJ/kgK . . . mair = [I  mR134a (s2  s1)]/(s4  s3) = 10.0 kg/s . . . . . 1st Law for each line: Q + mhin = mhex + W; W = 0 . . . . R134a: 1Q2 = mR134a(h2  h1) = 532 kW; QH = 1Q2 . . . . . Air: 3Q4 = mair (h4  h3) = mair (T4  T3) = 501.8 kW; QL = 3Q4 . . . . . Wnet = QH  QL = 30.2 kW; th = Wnet / QH = 0.057, or 5.7% 1014 10.23 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 500C Find the change in availability of the control mass for each of the three cases. a) = Wel = 10 kJ b) = Wspring = 10 kJ 298.15 c) = [1  (T0/TH)] Qout = 1 10 = 6.14 kJ 773.15 10.24 Calculate the availability of the water at the initial and final states of Problem 8.32, and the irreversibility of the process. 1: u1 = 83.94 2: u2 = 3124.3 0: uo = 104.86 , s1= 0.2966 s2= 7.7621 so= 0.3673
ac 1W2 = ac 1Q 2 203 kJ v1= 0.001 = 3243.4 v2= 0.354 TH = 873.15 K vo= 0.001003 , = (u  Tos)  (uo  Toso) + Po( v  vo) 1 = (83.94  298.150.2966)  (104.86  298.150.3673) + 100 (0.00...
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 Spring '08
 Lewis
 Thermodynamics, Energy, Heat, Heat engine, kJ/kg, 20°C

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