This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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 (0.001002  0.001003) = 0.159 kJ/kg 2 = (3124.3  298.157.7621)  (104.86  298.150.3673) + 100 (0.35411  0.001003) = 850 kJ/kg 1I2 = m(1  2) + [1(T0/TH)]1Q2  1W2 + Po( V2  V1) = 849.84 + (1 298.15 ) 3243.4  203 + 100 (0.3541  0.001) 873.15 so OK]
ac ac = 849.84 + 2135.9  203 + 35.31 = 1118. kJ [(Sgen = 3.75 kJ/K ToSgen = 1118 kJ 10.25 A steady stream of R22 at ambient temperature, 10C, and at 750 kPa enters a solar collector. The stream exits at 80C, 700 kPa. Calculate the change in availability of the R22 between these two states.
R22 i Solar Collector e Inlet (T,P) Table B.4.1 (liquid) 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 ie = e  i = (he  hi)  T0(se  si) = (305.912  56.463)  283.2(1.0761  0.2173) = 6.237 kJ/kg 1015 10.26 Nitrogen flows in a pipe with velocity 300 m/s at 500 kPa, 300C. What is its availability with respect to an ambient at 100 kPa, 20C? = h1  h0 + (1/2)V12  T0(s1  s0) = Cp(T1  T0) + (1/2)V12  T0(Cp ln(T1/T0)  R ln(P1/P0)) = 1.0426(300  20) + (300 2/2000) 573.15 500  293.151.042 ln  0.2968ln = 271.96 kJ/kg 293.15 100 10.27 A 10kg iron disk brake on a car is initially at 10C. Suddenly the brake pad hangs up, increasing the brake temperature by friction to 110C 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%. All the friction work is turned into internal energy of the disk brake. m(u2  u1) = 1Q2  1W2 1Q2 = mFeCFe(T2  T1)
1Q2 = 10 0.45 (110  10) = 450 kJ Neglect the work to the surroundings at P 0 = m(u2  u1)  T0m(s2  s1) 383.15 m(s2s1) = mC ln(T2/T1) = 10 0.45 ln = 1.361 kJ/K 283.15 = 450  283.15 1.361 = 64.63 kJ Wengine = thQgas = 1Q2 = Friction work Qgas = 1Q2/th = 450/0.35 = 1285.7 kJ 1016 10.28 A 1 kg block of copper at 350C is quenched in a 10 kg oil bath initially at ambient temperature of 20C. Calculate the final uniform temperature (no heat transfer to/from ambient) and the change of availability of the system (copper and oil). C.V. Copper and oil. Cco = 0.42 Coil = 1.8 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 (2  1) = u2  u1  To(s2  s1) = mC [(T2  T1)  Toln (T2 / T1) ] mcv(2  1)cv + moil (2  1)oil = = 0.42 [(322.5)  293.15 ln 300.65 300.65 ] + 10 1.8 [7.5  293.15 ln ] 623.15 293.15 =  45.713 + 1.698 =  44.0 kJ 10.29 Calculate the availability of the system (aluminum plus gas) at the initial and final states of Problem 8.74, and also the process irreversibility. State 1: T1 = 200 oC, v1 = V1/ m = 0.05 / 1.1186 = 0.0447 v2 = v1 (2 / 1.5) (298.15 / 473.15) = 0.03756 The metal does not change volume, so the combined is using Eq.10.22 as State 2: 1 = mgasgas + mAlAl = mgas[u1uoTo(s1  so)]cv + mgasPo(v1vo) + mAl[u1uo To(s1so)]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 1 = 1.1186 [ 0.653(20025)  298.15 (0.842 ln 473.15 2000  0.18892 ln ) 298.15 100 473.15 ] 298.15 + 100 (0.0447  0.5633 ) ] + 4 0.90 [ 200 25  298.15 ln = 128.88 + 134.3 = 263.2 kJ 2 = 1.1186 [ 0.653(25  25)  298.15 (0.842 ln 298.15 1500  0.18892 ln ) 298.15 100 298.15 ] 298.15 + 100 (0.03756  0.5633 ) ] + 4 0.9 [ 25 25  298.15 ln = 111.82 + 0 = 111.82 kJ 1I2 = 1  2 + [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 ToSgen = 164.58 so OK ] 1017 10.30 Consider the springtime melting of ice in the mountains, which gives cold water running in a river at 2C while the air temperature is 20C. What is the availability of the water (SSSF) relative to the temperature of the ambient? = h1  h0  T0(s1  s0) flow availability from Eq.10.19 Approximate both states as saturated liquid = 8.392  83.96  293.15(0.03044  0.2966) = 2.457 kJ/kg Why is it positive? As the water is brought to 20C it can be heated with qL from a heat engine using qH from atmosphere TH = T0 thus giving out work. 10.31 Refrigerant R12 at 30C, 0.75 MPa enters a SSSF device and exits at 30C, 100 kPa. Assume the process is isothermal and reversible. Find the change in availability of the refrigerant. hi = 64.539, si = 0.2397, he = 209.866, se = 0.8482 = he  hi  T0(se  si) = 209.866  64.539  298.15(0.8482  0.2397) = 36.1 kJ/kg 10.32 A geothermal source provides 10 kg/s of hot water at 500 kPa, 150C 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. SSSF with no work or heat transfer. . . . Cont. Eq.: m1 = m2 + m3 ; Vap. . . . 2 1 Energy Eq.: m1h1 = m2h2 + m3h3 Liq. . . . . 3 Entropy Eq.: m1s1 + Sgen = m2s2 + m3s3 B.1.1: B.1.2: ho = 104.87, so = 0.3673, h1 = 632.18, s1= 1.8417 h2 = 2706.63, s2= 7.1271, h3 = 504.68, s3= 1.530 . . h h h1 = xh2 + (1  x) h3 => x = m2/m1 = 1 3 = 0.0579 h2  h3 . . . . m2 = xm1 = 0.579 kg/s m3 = (1x)m1 = 9.421 kg/s . Sgen = 0.579 7.1271 + 9.421 1.53  10 1.8417 = 0.124 kW/K 1 = 632.18  104.87  298.15 (1.8417  0.3673) = 87.72 2 = 2706.63  104.87  298.15 (7.1271  0.3673) = 586.33 3 = 504.68  104.87  298.15 (1.53  0.3673) = 53.15 . . . m1 1 = 877.2 kW m22 = 339.5 kW m33 = 500.7 kW . . . . I = m1 1  m2 2  m33 = 37 kW Flow availability Eq.10.22: = (h  Tos)  (ho  Toso) = h  ho  To(s  so) 1018 10.33 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? C.V. Heat exchanger. SSSF, no work term. qout = h1  h2 = 1635.8  503.36 = 1132.4 kJ/kg T = (1  o ) qout = 1  2 TH
Heat exch 1 QH
H.E. . 2 . W 1  2 = h1  h2  To ( s1  s2 ) = 1132.4  298.15 (8.612  7.3869) = 1132.4  356.3 = 767.1 kJ/kg T 1  o = (1  2 ) / qout = 767.1 / 1132.4 = 0.6774 TH To = 0.3226 => TH = 924 K TH 10.34 A wooden bucket (2 kg) with 10 kg hot liquid water, both at 85C, is lowered 400 m down into a mineshaft. What is the availability of the bucket and water with respect to the surface ambient at 20C? 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.10.22 1  0 = mwood[u1  u0  T0(s1 s0)] + mH2O[u1 u0 T0(s1 s0)] + mtotg(z1 z0) = 2[1.26(85  20)  293.15 1.26 ln 273.15 + 85 ] + 10[ 355.82  83.94 293.15 . QL  293(1.1342  0.2966)] + 12 9.807 (400) /1000 = 15.85 + 263.38  47.07 = 232.2 kJ 1019 10.35 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, 300C, and the other at 1 MPa, 50C. The partition ruptures, and the nitrogen comes to a uniform state at 100C. Assuming the surroundings are at 25C find the actual heat transfer and the irreversibility in the process. m2 = mA + mB = PA1VA RTA1 + PB1VB RTB1 = 20000.1 10000.1 + 0.2968573.15 0.2968323.15 = 1.176 + 1.043 = 2.219 kg P2 = m2RT2/Vtot = 2.219 0.2968 373.15/0.2 = 1228.8 kPa Energy Eq.:
1Q2 mA(u2  u1)A + mB(u2  u1)B = 1Q2  0 / = mACv(T2  T1)A + mBCv(T2  T1)B = 1.176 0.745 (100  300) + 1.043 0.745 (100  50) = 136.4 kJ mA(s2  s1)A + mB(s2  s1)B = 1Q2/Tsur + 1Ss gen 373.15 1228.8 (s2  s1)A = 1.042 ln  0.2968 ln = 0.09356 573.15 2000 373.15 1228.8 (s2  s1)B = 1.042 ln  0.2968 ln = 0.0887 323.15 1000
1S2,gen = 1.176 (0.09356) + 1.043 0.0887 + 136.4/298.15 = 0.4396 kJ/K I = (T0)(1S2,gen) = 131.08 kJ 10.36 An air compressor is used to charge an initially empty 200L tank with air up to 5 MPa. The air inlet to the compressor is at 100 kPa, 17C and the compressor isentropic efficiency is 80%. Find the total compressor work and the change in availability of the air. C.V. Tank + compressor (constant inlet conditions) USUF, no heat transfer. Continuity: m2  m1 = min ( m1 = 0 ) Energy: m2u2 = minhin  1W2 Entropy: m2s2 = minsin + 1S2 gen 1w2,s = hin  u2,s = 346.77 Reversible compressor: 1S2 GEN = 0 s2 = sin Pr2 = Prin(P2/Pin) = 49.495 T2,s = 855, u2,s = 637.2 Actual compressor: 1w2,AC = 1w2,s/c = 433.46 u2,AC = hin  1w2,AC = 723.89 T2,AC = 958 K v2 = RT2/P2 = 0.055 State 2 u, P m2 = V2/v2 = 3.636 1W2 = m2(1w2,AC) = 1576 kJ m2(2  1) = 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 1020 10.37 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 secondlaw efficiency of the compressor process? 800 kPa T2s = 300(8)0.286 = 543.8 K Ws = 1.004(543.8  300) = 244.6 kJ/kg 2 100 kPa 2s Ws 244.6 w = = = 287.8 K 0.85 s 300 K 1 w 287.8 = 300 + = 586.8 K T 2 = T1 + s CP0 1.004 s2  s1 = 1.004 ln(586.8/300)  0.287 ln 8 = 0.07645 2  1 = (h2  h1)  T0(s2  s1) = 287.8  300(0.07645) = 264.9 kJ/kg 2nd law efficiency: 2nd Law = (2  1)/(w) = 264.9/287.8 = 0.92 10.38 Steam enters a turbine at 25 MPa, 550C and exits at 5 MPa, 325C at a flow rate of 70 kg/s. Determine the total power output of the turbine, its isentropic efficiency and the second law efficiency. hi = 3335.6, si = 6.1765, he = 2996.5, se = 6.3289 Actual turbine: wT,ac = hi  he = 339.1 kJ/kg Isentropic turbine: se,s = si he,s = 2906.6 wT,s = hi  he,s = 429 kJ/kg Rev. turbine: wrev = wT,ac + T0(se  si) = 339.1 + 45.44 = 384.54 kJ/kg T = wT,ac/wT,s = 339.1/429 = 0.79 II = wT,ac/wrev = 339.1/384.54 = 0.88 10.39 A compressor is used to bring saturated water vapor at 1 MPa up to 17.5 MPa, where the actual exit temperature is 650C. Find the irreversibility and the secondlaw efficiency. Inlet state: Table B.1.2 hi = 2778.1, si = 6.5864 Actual compressor Table B.1.3 Energy Eq. Actual compressor: h e,ac = 3693.9 , se,ac = 6.7356 wc,ac = he,ac  hi = 915.8 kJ/kg From Eq.10.14: i = T0(se,ac  si) = 298.15 (6.7356  6.5864) = 44.48 kJ/kg From Eq.10.13: wrev = i + w c,ac = 915.8 + 44.48 = 871.32 II = wrev/wc,ac = 871.32/915.8 = 0.951 1021 10.40 A flow of steam at 10 MPa, 550C goes through a twostage 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 T1 2 T2 3 CV: T1, h1 = 3500.9 kJ/kg, s1 = 6.7561 kJ/kg K Isentropic s2s = s1 h2s = 3017.9 wT1,s = h1  h2s = 483 kJ/kg Actual T1: wT1,ac = T1 wT1,s = 410.55 = h 1  h2ac s2ac = 6.8782 h2ac = h1  wT1,ac = 3090.35, CV: T2, s3s = s2ac = 6.8782 x3s = (6.87821.091)/6.5029 = 0.8899, h3s = 340.47 + 0.8899 2305.4 = 2392.2 kJ/kg wT2,s = h2ac  h3s = 698.15 wT2,ac = T2 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(s2acs1) = 298.15(6.8782  6.7561) = 36.4 kJ/kg II = wT1,ac/wT1 = 0.918
R R wT1 = wT1,ac + i = 447 kJ/kg,
R Actual T2: iT2,ac = T0(s3acs2ac) = 298.15(7.1736  6.8782) = 88.07 kJ/kg wT2 = wT2,ac + iT2,ac = 681.5, II = wT2,ac/wT2 = 0.871 10.41 Consider the twostage turbine in the previous problem as a single turbine from inlet to final actual exit and find its secondlaw efficiency. From solution to Problem 10.40 we have wT,ac = wT1,ac + wT2,ac = h1  h3ac = 1004 kJ/kg The actual compared to the reversible turbine has, Eq.10.14, i = qR = T0(s3  si) = qR + qR = 124.47 kJ/kg T1 T2 wR = wT,ac + i = 1128.5 kJ/kg II = wT,ac/wR = 0.89 1022 10.42 The simple steam power plant shown in Problem 6.39 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. h6 = 2393.2, s6 = 7.5501, h5 = 3404.3, s5 = 6.8953 6 = h6  h0  T0(s6  s0) = 2393.2  104.89  298.15(6.8953  0.3674) = 146.79 kJ/kg w rev = 5  6 = h5  h6  T0(s5  s6) = 1206.3 kJ/kg. wac = h5  h6 = 1011.1 ; II = wac/wrev = 0.838 10.43 A compressor takes in saturated vapor R134a at 20C and delivers it at 30C, 0.4 MPa. Assuming that the compression is adiabatic, find the isentropic efficiency and the second law efficiency. Table B.5 hi = 386.08, si = 1.7395, he,ac = 423.22, se,ac = 1.7895 Ideal exit: Pe, se,s = si he,s = 408.51 Isentropic compressor wc,s = he,s  hi = 22.43 Actual compressor wc,ac = he,ac  hi = 37.14 Reversible between inlet and actual exit wc,rev = hi  he,ac  T0(si  se,ac) = 37.14  298.15(1.7395  1.7895) = 22.23 TH,c = (wc,s/wc,ac) = (22.43/37.14) = 0.604 II = (wc,rev/wc,ac) = (22.23/37.14) = 0.599 10.44 Steam is supplied in a line at 3 MPa, 700C. 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 , s2 > s1 = 7.7571 , h2, P2 s2 = 7.9425 1  2 = h1h2 T0(s1s2) = 0 298.15(7.75717.9425) = 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 = wT,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 II = 835.2/917.63 = 0.91 1023 10.45 The condenser in a refrigerator receives R134a at 700 kPa, 50C and it exits as saturated liquid at 25C. The flowrate is 0.1 kg/s and the condenser has air flowing in at ambient 15C and leaving at 35C. Find the minimum flow rate of air and the heat exchanger secondlaw efficiency.
3 2 4 AIR 1 R134a C.V. Total heat exchanger. m1h1 + mah3 = m1h2 + mah4 ma = m1 h1  h2 h4  h3 = 0.1 436.89  234.59 = 1.007 kg/s 1.004(35  15) 1  2 = h1  h2  T0(s1  s2) = 436.89  234.59  288.15(1.7919  1.1201) = 8.7208 4  3 = h4  h3  T0(s4  s3) = 1.004(35  15)  288.15 1.004 ln II = ma(4  3)/m1(1  2) = 308.15 = +0.666 288.15 1.007(0.666) = 0.77 0.1(8.7208) 10.46 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 700C brings the steam to 600C. Find the secondlaw efficiency for this process. Note that no formula is given for this particular case so determine a reasonable expression for it. 1: Table B.1.2 P 1 , x1 v1 = 0.001093 + 0.50.3738 = 0.188, h1 = 640.21 + 0.52108.47 = 1694.5 , s1 = 1.8606 + 0.54.9606 = 4.341 2: P2 = P1,T2 v2 = 0.8041, h2 = 3701.7 , s2 = 8.3521 m(u2  u1) = 1Q2  1W2 = 1Q2  P(V2  V1)
1Q2 1W2 = m(u2  u1) + Pm(v2  v1) = m(h2  h1) = 2007.2 kJ = Pm(v2  v1) = 308.05 kJ P0m(v2  v1) = 61.61 kJ 1W2 to atm = Useful work out = 1W2  1W2 to atm = 246.44 298.15 reservoir = (1  T0/Tres)1Q2 = 1 2007.2 = 1392.2 973.15 II = Wnet/ = 0.177 1024 10.47 Air flows into a heat engine at ambient conditions 100 kPa, 300 K, as shown in Fig. P10.47. 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 = 300.473 + 1200  300  822.202 = 378.27 kJ/kg TH = 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 = 3007.88514  6.86925  0.287 ln + 300 100 750 (300/1500) 1200 = 184.764 kJ/kg wrev = hi  he + q1500  q750 + q0 = wac + q0 = 563.03 kJ/kg II = wac/wrev = 378.27/563.03 = 0.672 10.48 Consider the highpressure closed feedwater heater in the nuclear power plant described in Problem 6.42. Determine its secondlaw efficiency. For this case with no work the second law efficiency is from Eq. 10.25: II = m16(18  16)/m17(17  15) Properties (taken from computer software): h15 = 585 s15 = 1.728 h16 = 565 s16 = 1.6603 h17 = 2593 s17 = 6.1918 h18 = 688 s18 = 1.954 18  16 = h18  h16  T0(s18  s16) = 35.433 17  15 = h17  h15  T0(s17  s15) = 677.12 II = (75.6 35.433)/(4.662 677.12) = 0.85 1025 10.49 Consider a gasoline engine for a car as an SSSF device where air and fuel enters at the surrounding conditions 25C, 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 secondlaw efficiencies. 1 AIR 2 C.V. Total out to reservoirs mah1 + QH = mah2 + W + Qout mas1 + QH/TH + Sgen = mass + Qout/T0 h1 = 298.61 sT1 = 6.8631 h2 = 1046.22 sT1 = 8.1349 W Qout QH T0 TH wac = W/ma = h1  h2 + qH  qout = 298.6  1046.22 + 2200  750 = 702.4 kJ/kg TH = w/qH = 702.4/2200 = 0.319 sgen = s2  s1 + qout qH 750 2200 = 8.1349  6.8631 + = 2.565 kJ/kg K 298.15 1800 T0 TH sgen = 0 and qR from T0, no qout 0 itot = (T0)sgen = 764.8 kJ/kg For reversible case have 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 II = wac/wrev = 0.479 10.50 Air enters a steadyflow 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 Inlet: Ti = 1600 K, Table A.7: hi = 1757.3 kJ/kg, si = 8.1349 kJ/kg K 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: i  e = w/2ndLaw = 711.1/0.85 = 836.6 kJ/kg i  e = (hi  he)  To(si  se) = 836.6 kJ/kg hi  he = w = 711.1 kJ/kg, assume To = 25oC si  se = 0.4209 kJ/kgK si  se = se  si  R ln(Pi/Pe) = 0.4209 Pi = 3003 kPa
o o o o => Pe/Pi = 30.03; 1026 10.51 Consider the two turbines in Problem 9.72. What is the secondlaw efficiency for the combined system? From the solution to Problem 9.72 we have s1 = 6.2828, s2 = 6.7240, s5 = 6.5233, mT1/mtot = 0.6848 mT1/mtot = 0.3152 => wac = Wnet/mtot = 301.775 kJ/kg C.V. Total system out to T0 mT1/mtoth1 + mT2/mtoth2 + qrev = h5 + wrev wqc + qrev = wrev 0 0 mT1/mtots1 + mT2/mtots2 + qrev = s5 0 qrev = 298.15[6.5233  0.6848 6.2825  0.3152 6.724] = 30.242 kJ/kg 0 wrev = wac + qrev = 301.775 + 30.242 = 332 kJ/kg 0 II = wac/wrev = 301.775/332 = 0.909 10.52 Air in a piston/cylinder arrangement is at 110 kPa, 25C, 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 25C 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 1027 Advanced Problems 10.53 Refrigerant22 is flowing in a pipeline at 10C, 600 kPa, with a velocity of 200 m/s, at a steady flowrate of 0.1 kg/s. It is desired to decelerate the fluid and increase its pressure by installing a diffuser in the line (a diffuser is basically the opposite of a nozzle in this respect). The R22 exits the diffuser at 30C, with a velocity of 100 m/s. It may be assumed that the diffuser process is SSSF, polytropic, and internally reversible. Determine the diffuser exit pressure and the rate of irreversibility for the process. C.V. Diffuser out to T0, Int. Rev. flow sgen R22 = 0 / m1 = m2 , h1 + (1/2)V12 + q = h2 + (1/2)V22 s1 + dq/T + sgen = s2 = s1 + q/T0 + sgen Rev. Polytropic Process:
n w12 = 0 =  v dP + (V12  V22)/2 /
1
2 2 LHS = v dP = n1(P2v2  P1v1) = 2(V1  V2) = 2(2002  1002)/1000 = 15 kJ/kg P1v1n = P2v2n n = ln(P2/P1)/ln(v1/v2) Inlet state: v1 = 0.04018, s1 = 0.9295, h1 = 255.1 Exit state: T2,V2, ? only P2 unknown. Trial and error on P2 to give LHS = 15: P2 = 1 MPa v2 = 0.0246,
1000 0.04018 n = ln 600 / ln 0.0246 = 1.0412 1 v2 = v(P2), n = n(P2) LHS = (1.0412/0.0412)(1000 0.0246  600 0.04018) = 12.43 LHS too small so P2 > 1.0 MPa Assume sat. vapor @ 30C, P2 = 1.1919 MPa v2 = 0.01974, LHS =
1.1919 0.04018 n = ln 0.6 /ln0.01974 = 0.9657 0.9657 (1191.9 0.01974  600 0.04018) = 16.33 > 15 0.0343 P2 = 1.1 MPa v2 = 0.021867, n = 0.9963, LHS = 14.596 Interpolate to get the final state: P2 = 1.121 MPa, v2 = 0.021353, n = 0.9887, LHS = 15.01 OK s2 = 0.8967 h2 = 260.56 Find q from energy equation (Assume T0 = T1 = 10C) q = h2  h1 + 2(V22  V12) = 260.56  255.1  15 = 9.54 kJ/kg . I = m[T0(s2  s1)  q ]= 0.1[283.15(0.8967  0.9295) + 9.54] = 0.0253 kW
1 1028 10.54 A piston/cylinder contains ammonia at 20C, quality 80%, and a volume of 10 L. A force is now applied to the piston so it compresses the ammonia in an adiabatic process to a volume of 5 L, where the piston is locked. Now heat transfer with the ambient takes place so the ammonia reaches the temperature of the ambient at 20C. Find the work and heat transfer. If it is done in a reversible process, how much work and heat transfer would be involved? P Process 1 2 3 1: v1 = 0.49926 m3/kg, s1 = 4.5658 kJ/kg K u1 = 1152.25  94.9 = 1057.4 kJ/kg m = V1/v1 = 0.010/0.49926 = 0.02 kg V V2 V1 Assume reversible adiabatic compression m(u2  u1) =  1W2 ; m(s2  s1) = 0, v2 = v1V2/V1 = 0.2496 For 2C, v2 s = 4.431 2: s2 = s1, v1 Trial and error on T2. For 0C, v2 s = 4.6925, T2 = 1C, x = 0.832, u2 = 1125 kJ/kg
1W2 = m(u2  u1) =  1.345 kJ, (1Q2 = 0) 3: T3, v3 = v2 P3 = 537 kPa, u3 = 1351.6, s3 = 5.382
2Q3 = m(u3  u2) = 0.02(1351.6  1125) = 4.532 kJ Qrev = mT0(s3  s1) = 0.02293.15(5.382  4.5658) = 4.785 kJ Wrev = Qrev  m(u3  u1) = 4.785  0.02(1351.61057.4) = 1.1 kJ 1029 10.55 Consider the irreversible process in Problem 8.34. 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 = 47.3476 + 26.7428 = 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 1030 10.56 Water in a piston/cylinder is at 100 kPa, 34C, shown in Fig. P10.56. The cylinder has stops mounted so Vmin = 0.01 m3 and Vmax = 0.5 m3. The piston is loaded with a mass and outside P0, so a pressure inside of 5 MPa will float it. Heat of 15000 kJ from a 400C source is added. Find the total change in availability of the water and the total irreversibility.
P 1a 1 2 1b v CV water plus cyl. wall out to reservoir m2 = m1 = m, m(u2 u1) = 1Q2  1W2 m(s2  s1) = 1Q2/Tres + 1S2 gen System eq: States must be on the 3 lines Here Peq = 5 MPa, so since P1 < Peq V1 = Vmin = 0.01 m3 1: v1 = 0.0010056 m3/kg, u1 = 142.49 kJ/kg, s1 = 0.4916 kJ/kg K m = V1/v1 = 9.9443 kg Heat added gives q = Q/m = 1508.4 kJ/kg so check if we go from 1 past 1a (Peq,v1) and past 1b (Peq,vmax) 1a: T = 40C, u1a = 166.95 1q1a = u1a  u1 = 24.5 kJ/kg 1b: T = 337.5C, u1b= 2780.94
1q1b = as 1w1b = Peq(v1b  v1) = 246.4 h1b  u1  Peqv1 = 2884.8 The state is seen to fall between 1a and 1b so P2 = Peq.
1w2 = Peq(v2  v1), u2  u1 = 1q2  1w2 u2 + Peqv2 = h2 = 1q2 + u1 + Peqv1 = 1655.9 < hg(P2) => T2 = 264C x2 = (1655.9  1154.21)/1640.12 = 0.3059, v2 = 0.001286 + 0.3059 0.03815 = 0.01295, s2 = 2.9201 + 0.3059 3.0532 = 3.8542 The nonflow availability change of the water (Eq.10.22) m(2  1) = m[u2  u1  T0(s2  s1) + P0(v2  v1)] = 9.944[1591.15  142.49  298.15(3.8542  0.4916) + 100(0.01295  0.0010056)] = 9.944 447.3 = 4448 kJ I = T(1S2 gen) = T [m(s2  s1)  1Q2/Tres ]= 298.15 [ 9.944(3.8542  0.4916)  15000/673.15] = 3326 kJ (Piston and atm. have an increase in availability) 1031 10.57 A rock bed consists of 6000 kg granite and is at 70C. A small house with lumped mass of 12000 kg wood and 1000 kg iron is at 15C. They are now brought to a uniform final temperature with no external heat transfer. a. For a reversible process, find the final temperature and the work done in the process. b. If they are connected thermally by circulating water between the rock bed and the house, find the final temperature and the irreversibility of the process, assuming that surroundings are at 15C. a) For a reversible process a heat engine is installed between the rockbed and the house. Take C.V. Total mrock(u2  u1) + mwood(u2  u1) + mFe(u2  u1) = 1W2 mrock(s2  s1) + mwood(s2  s1) + mFe(s2  s1) = 0 / (mC)rockln T2 T1 + (mC)woodln T2 T1 + (mC)Feln T2 T1 =0 / 6000 0.89 ln (T2/343.15) + 12000 1.26 ln (T2/288.15) 0 + 1000 0.46 ln (T2/288.15) = / => T2 = 301.3 K + (12000 1.26 + 460)(301.3  288.15) 1W2 = 18602 kJ b) No work, no Q irreversible process. (mC)rock(T2  70) + (mCwood + mCFe)(T2  15) = 0 / T2 = 29.0C = 302.2 K Sgen = mi(s2  s1)i = 5340 ln
1I2 1W2 = 6000 0.89(301.3  343.15) 302.2 302.2 + 15580 ln = 63.13 kJ/K 343.15 288.15 = (T0)1S2,gen = 288.15 63.13 = 18191 kJ 1032 10.58 Consider the heat engine in Problem 10.47. The exit temperature was given as 800 K, but what are the theoretical limits for this temperature? Find the lowest and the highest, assuming that the heat transfers are as given. For an exit temperature that is the average of the highest and lowest possible, find the firstand secondlaw efficiencies for the heat engine. The lowest exhaust temperature will occur when the maxumum 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 Te,min = 446 K The maximum exhaust temperature occurs with no work out hi + qH = qm + he he = 300.473 + 1200  300 = 1200.5 Te,max = 1134 K Te = (1/2)(Te,min + Te,max) = 790 K see solution for 10.47 to follow calculations he = 811.23, s = 7.8713 Te TH = wac/qH = 0.32 wac = 300.47 + 1200  300  811.23 = 389.24, q0 = 300[7.8713  6.86926  0.287 ln wrev = wac + q0 = 569.85, 100 300 1200 + ] = 180.61 kJ/kg 100 750 1500 II = wac/wrev = 389.24/569.85 = 0.683 1033 10.59 Air in a piston/cylinder arrangement, shown in Fig. P10.59, 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 to raise the piston. 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 20C.
P 2 Pstop P1 V1 m2 = m1 = m = P1V1/RT1 = 200 0.5/0.287 300 = 1.161 kg m(u2  u1) = 1Q2  1W2
W 1 2 Vstop 1 m(s2  s1) = dQ/T + 1S2 gen Process: P = P0 + (VV0) if V Vstop Information: Pstop = P0 + (VstopV0) Eq. of state Tstop = T1PstopVstop/P1V1 = 1200 < T2 So the piston will hit the stops => V 2 = Vstop
1W2 = 2(P 1 1Q2 = 1 + Pstop)(Vstop V1) = 2(200 + 400)(1  0.5) = 150 kJ 1 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 T RES
1S2 gen tot = 1I2 m(s2  s2)  1Q2/TRES = 1.034 kJ/K = T0(1S2 gen) = 293.15 1.034 = 303 kJ 1034 10.60 Consider two rigid containers each of volume 1 m 3 containing air at 100 kPa, 400 K. An internally reversible Carnot heat pump is then thermally connected between them so it heats one up and cools the other down. In order to transfer heat at a reasonable rate, the temperature difference between the working substance inside the heat pump and the air in the containers is set to 20C. The process stops when the air in the coldest tank reaches 300 K. Find the final temperature of the air that is heated up, the work input to the heat pump, and the overall secondlaw efficiency. A H.P. QA QB W
H.P. B The high and the low temperatures in the heat pump are TA+20 and TB20, respectively. Since TA and TB change during the process, the coefficient of performance changes, and so it must be integrated. dUA = mAduA = mACvdTA = dQA dU = mBduB = mBcvdTB = dQB B Carnot heat pump: dQA dQB = TA + 20 TB  20 = mACvdTA dTA = mBCvdTB dTB (dTA)/(TA + 20) =  (dTB)/(TB  20) dTA TA2 + 20 dTB TB2  20 = ln =  = ln TA1 + 20 TB  20 TB1  20 TA + 20 TA1 = TB1 = 400 K, TB2 = 300 K ln TA2 + 20 420 = ln 300  20 TA2 = 550 K 380 mA = mB = P1V1/RT1 = (100 1)/(0.287 400) = 0.871 kg WH.P. = QA  QB = mCv(TA2  TA1) + mCv(TB2  TB1) = 0.871 0.717 (550  400 + 300  400) = 31.2 kJ Second law efficiency is total increase in air availability over the work input. = mCv(T2  T1) + P0(V  V1)  T0(s2  s1) 2 A = QA  T0m(s2  s1)A B = QB  T0m(s2  s1)B A + B = WH.P.  mT0[ (s2  s1)A + (s2  s1)B] To find entropies we need the pressures (P2/P1) = T2/T1 for both A and B s2  s1 = Cpln(T2/T1)  R ln(T2/T1) = Cvln(T2/T1) tot = WH.P.  mT0Cv[ln(T2/T1)A + ln(T2/T1)B] 550 300 = 31.2  0.871 298.15 0.7175 ln + ln = 25.47 400 400 II = tot/WH.P. = 25.47/31.2 = 0.816 1035 ENGLISH UNIT PROBLEMS
10.61ECalculate the reversible work and irreversibility for the process described in Problem 5.122, assuming that the heat transfer is with the surroundings at 68 F.
P 2 v 1 Linear spring gives 1 = PdV = (P1 + P2)(V2  V1) 2 1Q 2 = m(u2  u1) + 1W2 Equation of state: PV = mRT
1W2 State 1: V1 = mRT1/P1 = State 2: V2 = mRT2/P2 =
1W2 1 4 35.1 (750 + 460) = 16.85 ft3 70 144 4 35.1 (75 + 460) = 11.59 ft3 45 144 = 2(70 + 45)(11.59 16.85) x144/778 = 55.98 Btu From Table C.7 Cp(Tavg) = [(69270)/(1200537)]/M = 10.45/44.01 = 0.2347 Btu/lbm R CV = Cp R = 0.2375 35.10/778 = 0.1924
1Q 2 rev 1W2 = = mCV(T2  T1) + 1W2 = 4x 0.1924(75  750)  55.98 = 575.46 Btu To(S2  S1)  (U2  U1) + 1Q2(1  To/TH) = Tom(s2  s1)+ 1W2  1Q2To/To
ac ac = Tom[CP ln(T2 / T1)  R ln(P2 / P1)] + 1W2  1Q2 = 527.7 x 4 [ 0.2347 ln(535/1210)  0.0451 ln(45/70)]  55.98 + 575.46 = 362.24  55.98 + 575.46 = 157.2 Btu
1I 2 rev ac = 1W2  1W2 = 157.2  (55.98) = 213.2 Btu 10.62EThe compressor in a refrigerator takes refrigerant R134a in at 15 lbf/in.2, 0 F and compresses it to 125 lbf/in.2, 100 F. With the room at 70 F find the reversible heat transfer and the minimum compressor work. 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) qrev = 529.67(0.41262  0.42288) = 5.43 Btu/lbm wc min = h1  h2 + T0(s2 s1) = 167.193  181.059  5.43 = 19.3 Btu/lbm 1036 10.63EA supply of steam at 14.7 lbf/in.2, 320 F is needed in a hospital for cleaning purposes at a rate of 30 lbm/s. A supply of steam at 20 lbf/in. 2, 500 F is available from a boiler and tap water at 14.7 lbf/in.2, 60 F 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 C.8.1 h1 = 28.08, s1 = 0.05555 C.8.2 h2 = 1286.8, s2 = 1.8919 C.8.2 h3 = 1202.1 , s3 = 1.828 C.V. Mixing chamber . . . m1 + m2 = m3 . . . m1h1 + m2h2 = m3h3 . . . . m1s1 + m2s2 + Sgen = m3s3 . . 1202.1  28.08 m2/m3 = (h3  h1)/(h2  h1) = = 0.9327 1286.8  28.08 . . m2 = 27.981 lbm/s , m1 = 2.019 lbm/s . . . . . I = T0Sgen = T0(m3s3  m1s1  m2s2) = 536.67 (30 1.828  2.019 0.05555  27.981 1.8919) = 961 Btu/s 10.64EA 4lbm piece of iron is heated from room temperature 77 F to 750 F by a heat source at 1100 F. What is the irreversibility in the process? C.V. Iron out to 1100 F 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 = 4 0.107 (750  77) = 288.04 Btu mFe(s2  s1) = 1Q2/Tres + 1S2 gen
1S2 gen = mFe(s2  s1)  1Q2/Tres = mFeC ln (T2/T1)  1Q2/Tres = 4 0.107 ln 1209.67 288.04 = 0.163 Btu/R 536.67 1559.67 1I2 = To (1S2 gen ) = 536.67 0.163 = 87.57 Btu 1037 10.65EFresh water can be produced from saltwater by evaporation and subsequent condensation. An example is shown in Fig. P10.12 where 300lbm/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 68 F, 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 T [F] 86 2 77 3 77 4 5 74 6 7 63 8 68 . . . C.V. Valve: m1 = mex = m2 + m3, Energy Eq.: h1 = he ; Entropy Eq.: si + sgen = se h1 = 54.08 si = 0.1043 P2 = Psat(T2 = T3) = 0.4641 he = h1 xe = (54.08  45.08)/1050.0 = 0.008571 se = 0.08769 + 0.008571 1.9565 = 0.1045 sgen = se  si = 0.1045  0.1043 = 0.0002 . . I = mT0sgen = 300 528 0.0002 = 31.68 Btu/s C.V. Condenser. h2 = 1095.1 , s2 = 2.044 , h5 = 42.09 , s5 = 0.0821 h7 = 31.08 , s7 = 0.0613 , h8 = 36.09 , s8 = 0.0708 . . . . Energy Eq.: m2h2 + m7h7 = m2h5 + m7h8 . . 1095.1  42.09 lbm m7 = m2 (h2  h5)/(h8  h7) = 297.44 = 62516 36.09  31.08 s . . . . . Entropy Eq.: m2s2 + m7s7 + Sgen = m2s5 + m7s8 . . . . I = T0Sgen = T0[m2(s5  s2) + m7(s8  s7)] = 528[297.44(0.0821  2.044) + 62516(0.0708  0.0613)] = 528 10.354 = 5467 Btu/s . . m2 = (1  xe)mi = 297.44 lbm/s 1038 10.66EAir flows through a constant pressure heating device as shown in Fig. P10.15. It is heated up in a reversible process with a work input of 85 Btu/lbm air flowing. The device exchanges heat with the ambient at 540 R. The air enters at 540 R, 60 lbf/in.2. Assuming constant specific heat develop an expression for the exit temperature and solve for it. C.V. Total out to T0 Entropy Eq.: Energy Eq.: s1 + qrev/T0 = s2 0 h1 + qrev  wrev = h2 0 qrev = T0(s2  s1) 0 h2  h1 = T0(s2  s1)  wrev Constant Cp gives: Cp(T2  T1) = T0Cp ln (T2/T ) + 85 1 T2  T0ln(T2/T1) = T1 + 85/Cp T1 = 540 Cp = 0.24 T0 = 540 T2  540 ln (T2/540) = 540 + (85/0.24) = 894.17 At 1400 R LHS = 885.56 (too low) T2 = 1414 R At 1420 R (LHS = 894.19 OK) LHS = 897.9 Interpolate to get 10.67EAir enters the turbocharger compressor of an automotive engine at 14.7 lbf/in2, 90 F, and exits at 25 lbf/in2, as shown in Fig. P10.16. The air is cooled by 90 F 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. 2 Cooler Comp. 1 Air 3 . Qc T3 = T2  90 F, P2 = 25 lbf/in2 s COMP = 0.75, P1 = 14.7 lbf/in2 T1 = 90 F = 550 R T2s = T1 . Wc (P )
1 P2 k1 k = 550 25 (14.7) 0.286 = 640.2 R ws = CP0(T2s  T1) = 0.24(640.2  550) = 21.65 w =  ws/s = 21.65/0.75 = 28.87 = CP0(T2  T1) = 0.24(T2  550) => T2 = 670.3 R T3 = 670.3  90 = 580.3 R b) Irreversibility from Eq.10.13 with rev. work from Eq.10.12, (q = 0 at TH) s3  s1 = 0.24 ln 580.3 53.34 25 ln = 0.0235 Btu/lbm R 550 778 14.7 i = T(s3  s1)  (h3  h1)  w = T(s3  s1)  CP(T3  T1)  CP(T1  T2) = 550(0.0235)  0.24(90) = +8.7 Btu/lbm 1039 10.68ECalculate the irreversibility for the process described in Problem 6.101, assuming that the heat transfer is with the surroundings at 61 F. I = To Sgen so apply 2nd law out to To = 61 F m2 s2  m1s1 = misi + 1Q2 / To + 1S2gen ToSgen = To ( m2 s2  m1s1  mi si )  1Q2 m1 = 2.104 lbm mi = 7.152 lbm Use from table C.4: Cp = 0.24 m2 =9.256 lbm R = 53.34 / 778 = 0.06856 ToSgen = I = To [ m1 (s2  s1) + mi (s2  si)]  1Q2 = 520.7[2.104(Cp ln 630 60 630 60  R ln ) +7.152(Cpln  R ln )] 519.7 45 1100 75  (  868.9) = 520.7 (0.05569  0.8473) + 868.9 = 456.7 Btu 10.69EA control mass gives out 1000 Btu of energy in the form of a. Electrical work from a battery b. Mechanical work from a spring c. Heat transfer at 700 F Find the change in availability of the control mass for each of the three cases. a) = Wel = 1000 Btu b) = Wspring = 1000 Btu T0 537 c) = 1  = Qout = 1 1000 = 537 Btu 1160 TH 10.70EA steady stream of R22 at ambient temperature, 50 F, and at 110 lbf/in. 2 enters a solar collector. The stream exits at 180 F, 100 lbf/in.2. Calculate the change in availability of the R22 between these two states.
R22 i Solar Collector e 50 F 110 lbf/in2 hi = 24.275 si = 0.0519 180 F 100 lbf/in2 he = 132.29 se = 0.2586 ie = e  i = (he  hi)  T0(se  si) = (132.29  24.275) 510(0.2586  0.0519) = 2.6 Btu/lbm 1040 10.71EA 20lbm iron disk brake on a car is at 50 F. Suddenly the brake pad hangs up, increasing the brake temperature by friction to 230 F 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%. All the friction work is turned into internal energy of the disk brake. m(u2  u1) = 1Q2  1W2 1Q2 = mFeCFe(T2  T1)
1Q2 = 20 0.107 (230  50) = 385.2 Btu Neglect the work to the surroundings at P 0
690 m(s2  s1) = mC ln(T2 / T1) = 20 0.107 ln 510 = 0.6469 = m(u2  u1)  T0m(s2  s1) = 385.2  510 0.6469 = 55.28 Btu Wengine = thQgas = 1Q2 = Friction work Qgas = 1Q2/th = 385.2/0.35 = 1100 Btu 10.72ECalculate the availability of the system (aluminum plus gas) at the initial and final states of Problem 8.113, and also the irreversibility. State 1: T1 = 400 F v1 = 2/2.862 = 0.6988 P1 = 300 psi Ideal gas v2 = v1(300 / 220)(537 / 860) = 0.595 ; vo = 8.904 = RTo / Po The metal does not change volume so the terms as Eq.10.22 are added 1 = mgasgas + mAlAl 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 860 35.1 300 1 = 2.862[0.156(40077)  537(0.201 ln ln ) 537 778 14.7 144 860 + 14.7 (0.6988  8.904 )( ) ] + 8 0.21 [ 400 77  537 ln ] 778 537 = 143.96 + 117.78 = 261.74 Btu 220 537 35.1 ln ) 2 = 2.862[ 0.156(77  77)  537 (0.201 ln 14.7 537 778 144 537 + 14.7(0.595  8.904 )( )] + 8 0.21 [ 77 77  537 ln ] 778 537 = 122.91 + 0 = 122.91 Btu 1I2 = 1  2 + (1  To/TH )1Q2  1W2AC + Pom( V2  V1) = 261.74  122.91 + 0  (14.29) + 14.7 2.862 = 152.3 Btu [(Sgen = 0.2837 Btu/R ToSgen = 152.3 so OK ]
144 778 ( 0.595  0.6988) 1041 10.73EConsider the springtime melting of ice in the mountains, which gives cold water running in a river at 34 F while the air temperature is 68 F. What is the availability of the water (SSSF) relative to the temperature of the ambient? = h1  h0  T0(s1  s0) flow availability from Eq.10.19 Approximate both states as saturated liquid = 1.9973  36.088  527.67 (0.00405  0.07081) = 1.136 Btu/lbm Why is it positive? As the water is brought to 68 F it can be heated with qL from a heat engine using qH from atmosphere TH = T0 thus giving out work. 10.74EA geothermal source provides 20 lbm/s of hot water at 80 lbf/in. 2, 300 F flowing into a flash evaporator that separates vapor and liquid at 30 lbf/in.2. Find the three fluxes of availability (inlet and two outlets) and the irreversibility rate. C.V. Flash evaporator chamber. SSSF with no work or heat transfer. . . . Cont. Eq.: m1 = m2 + m3 ; Vap. . . . 2 1 Energy Eq.: m1h1 = m2h2 + m3h3 Liq. . . . . 3 Entropy Eq.: m1s1 + Sgen = m2s2 + m3s3 C.8.1: so = 0.08769, h1 = 269.73, s1= 0.4372 C.8.1: h2 = 1164.3, s2= 1.6997, h3 = 218.9, s3= 0.36815 . . h h h1 = xh2 + (1  x) h3 => x = m2/m1 = 1 3 = 0.05376 h2  h3 . . . . m3 = (1x)m1 = 18.925 lbm/s m2 = xm1 = 1.075 lbm/s . Sgen = 0.579 7.1271 + 9.421 1.53  10 1.8417 = 0.124 kW/K Flow availability Eq.10.22: = (h  Tos)  (ho  Toso) = h  ho  To(s  so) 1 = 269.73  45.08  537 (0.4372  0.08769) = 36.963 2 = 1164.3  45.08  537 (1.6997  0.08769) = 253.57 3 = 218.9  45.08  537 (0.36815  0.08769) = 23.21 . . . m1 1 = 739.3 Btu/s m22 = 272.6 Btu/s m33 = 439.3 Btu/s . . . . I = m1 1  m2 2  m33 = 27.4 Btu/s ho = 45.08, 1042 10.75EA wood bucket (4 lbm) with 20 lbm hot liquid water, both at 180 F, is lowered 1300 ft down into a mineshaft. What is the availability of the bucket and water with respect to the surface ambient at 70 F? v1 v0 for both wood and water 1  0 = mwood[u1  u0 T0(s1  s0)] + mH2O[u1  u0 T0(s1  s0)] + mtotg(z1 z0) = 4[0.3(180  70)  0.3 530 ln 640 ] + 20[147.76  38.09 530  530(0.263  0.074)] + 24 (g/gc) (1300 / 778) = 12.05 + 199.3  40.1 = 171.25 Btu 10.76EAn air compressor is used to charge an initially empty 7ft3 tank with air up to 750 lbf/in.2. The air inlet to the compressor is at 14.7 lbf/in.2, 60 F and the compressor isentropic efficiency is 80%. Find the total compressor work and the change in energy of the air. C.V. Tank + compressor (constant inlet conditions) Continuity: Entropy: m2  0 = min Energy: m2u2 = minhin  1W2 m2s2 = minsin + 1S2 GEN To use isentropic efficiency we must calc. ideal device Reversible compressor: 1S2 GEN = 0 s2 = sin Pr2 = Prin(P2/Pin) = 0.9767 (750/14.7) = 49.832 T2,s = 1541 R u2,s = 274.49 Btu/lbm 1w2,s = hin  u2,s = 124.38  274.49 = 150.11 Actual compressor: 1w2,AC = 1w2,s/c = 187.64 Btu/lbm u2,AC = hin  1w2,AC = 312 T2,AC = 1729 R Final state 2 u, P v2 = RT2/P2 = 0.854 ft3/lbm m2 = V2/v2 = 8.2 lbm 1W2 = m2(1W2,AC) = 1539 Btu m2(2  1) = m2 [u2  u1 + P0(v2  v1)  T0(s2  s1) ] = 8.2 312  88.733 + 14.7(0.854  13.103)  1.63074 53.34 750 ln 778 14.7 [ 144  520 1.9311 778 ( )] = 8.2 173.94 = 1426.3 Btu 1043 10.77EA compressor is used to bring saturated water vapor at 150 lbf/in. 2 up to 2500 lbf/in.2, where the actual exit temperature is 1200 F. Find the irreversibility and the second law efficiency. Inlet state: Table C.8.1 hi = 1194.9 Btu/lbm, si = 1.5704 Btu/lbm R Actual compressor C.8.2: he = 1587.7 Btu/lbm, se = 1.6101 Btu/lbm R Energy Eq. actual compressor: wc,ac = he  hi = 392.8 Btu/lbm Eq.10.14: i = T0(se  si) = 536.67 (1.6101  1.5704) = 21.3 Btu/lbm Eq.10.13: wrev = i + w c,ac = 21.3 + 1194.9  1587.7 = 371.5 Btu/lbm II = wrev/(wc,ac) = 371.5 / 392.8 = 0.946 10.78EThe simple steam power plant in Problem 6.91, shown in Fig P6.39 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. Turbine from 6.91 C.8: h6 = 1028.7 ho = 45.08 s6= 1.8053 so= 0.08769 h5 = 1455.6 s5= 1.6408 6 = h6  ho  To ( s6  so ) = 61.26 Btu/lbm wrev = 5  6 = h5  h6  To( s5  s6 ) = 515.2 Btu/lbm wAC = h5  h6 = 426.9 II = wac / wrev = 426.9 / 515.2 = 0.829 10.79ESteam is supplied in a line at 450 lbf/in.2, 1200 F. A turbine with an isentropic efficiency of 85% is connected to the line by a valve and it exhausts to the atmosphere at 14.7 lbf/in.2. If the steam is throttled down to 300 lbf/in.2 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. C.V. Valve: h2 = h1 = 1630.8, s2 > s1, h2, P2 s2 = 1.86348 Ideal turbine: Actual turbine: s3 = s2 h3s = 1230.88 wT,s = h2h3s = 399.92 wT,ac = TwT,s = 339.93 Btu/lbm 2  1 = h2  h1  T0(s2  s1) = 536.67(1.86348  1.8192) = 23.76 Btu/lbm h3ac = h2  wTac = 1290.87 s3ac = 1.9299 wrev = 2  3 = 1630.8  1290.87  536.67(1.86348  1.9299) = 375.58 II = wac/wrev = 339.93/375.58 = 0.905 1044 10.80EA piston/cylinder arrangement has a load on the piston so it maintains constant pressure. It contains 1 lbm of steam at 80 lbf/in.2, 50% quality. Heat from a reservoir at 1300 F brings the steam to 1000 F. Find the secondlaw efficiency for this process. Note that no formula is given for this particular case, so determine a reasonable expression for it. 1: P1, x1 v1 = 2.7458 h1 = 732.905 s1 = 1.0374 2: P2 = P1,T2 v2 = 10.831 h2 = 1532.6 s2 = 1.9453 m(u2  u1) = 1Q2  1W2 = 1Q2  P(V2  V1)
1Q2 1W2 = m(u2  u1) + Pm(v2  v1) = m(h2  h1) = 799.7 Btu = Pm(v2  v1) = 119.72 Btu P0m(v2  v1) = 22 Btu 1W2 to atm = Useful work out = 1W2  1W2 to atm = 119.72  22 = 97.72 Btu T0 536.67 reservoir = 1 1Q2 = 1 799.7 = 556 Btu 1759.67 Tres nII = Wnet/ = 0.176 10.81EAir flows into a heat engine at ambient conditions 14.7 lbf/in.2, 540 R, as shown in Fig. P10.47. Energy is supplied as 540 Btu per lbm air from a 2700 R source and in some part of the process a heat transfer loss of 135 Btu per lbm air happens at 1350 R. The air leaves the engine at 14.7 lbf/in.2, 1440 R. Find the first and the secondlaw efficiencies. C.V. Engine out to reservoirs hi + qH = qL + he + w wac = 129.18 + 540  135  353.483 = 180.7 TH = w/qH = 180.7/540 = 0.335 For second law efficiency also a q to/from ambient si + (qH/TH) + (q0/T0) = (qloss/Tm) + se q0 = T0[se  si + (qloss/Tm)  (qH/TH)] 135 540 = 5401.88243  1.6398 + = 77.02 1350 2700 wrev = hi  he + qH  qloss + q0 = wac + q0 = 257.7 II = wac/wrev = 180.7/257.7 = 0.70 1045 10.82EConsider a gasoline engine for a car as an SSSF device where air and fuel enters at the surrounding conditions 77 F, 14.7 lbf/in.2 and leaves the engine exhaust manifold at 1800 R, 14.7 lbf/in.2 as products assumed to be air. The engine cooling system removes 320 Btu/lbm air through the engine to the ambient. For the analysis take the fuel as air where the extra energy of 950 Btu/lbm of air released in the combustion process, is added as heat transfer from a 3240 R reservoir. Find the work out of the engine, the irreversibility per poundmass of air, and the first and secondlaw efficiencies. 2 C.V. Total out to reservoirs 1 mah1 + QH = mah2 + W + Qout AIR W mas1 + QH/TH + Sgen = mass + Qout/T0 h1 = 128.381 sT1 = 1.63831 Q Q out T0 H TH h2 = 449.794 sT1 = 1.94209 wac = W/ma = h1  h2 + qH  qout = 128.38  449.794 + 950  320 = 308.6 Btu/lbm TH = w/qH = 308.6/950 = 0.325 itot = (T0)sgen = T0(s2  s1) + qout qHT0/TH
536.67 = 536.67(1.94209  1.63831) + 320  950 3240 = 325.67 Btu/lbm For reversible case have sgen = 0 and qR from T0, no qout 0 qR = T0(s2  s1)  (T0/TH)qH = itot  qout = 5.67 Btu/lbm 0,in wrev = h1  h2 + qH + qR = wac + itot = 634.3 Btu/lbm 0,in II = wac/wrev = 0.486 10.83EConsider the two turbines in Problem 9.119, shown in Fig. P9.72. What is the secondlaw efficiency of the combined system? From the solution to 9.119 we have s1 = 1.499 s2 = 1.582 s5 = 1.5651 mT1/mtot = 0.525 mT1/mtot = 0.475 wac = Wnet/mtot = 146.317 Btu/lbm C.V. Total system out to T0 mT1/mtoth1 + mT2/mtoth2 + qrev = h5 + wrev wqc + qrev = wrev 0 0 mT1/mtots1 + mT2/mtots2 + qrev = s5 0 qrev = 536.67[1.5651  0.525 1.499  0.475 1.582] = 14.316 Btu/lbm 0 wrev = wac + qrev = 146.317 + 14.316 = 160.63 0 II = wac/wrev = 146.317/160.63 = 0.911 1046 10.84E(Adv.) Refrigerant22 is flowing in a pipeline at 40 F, 80 lbf/in.2, with a velocity of 650 ft/s, at a steady flowrate of 0.2 lbm/s. It is desired to decelerate the fluid and increase its pressure by installing a diffuser in the line (a diffuser is basically the opposite of a nozzle in this respect). The R22 exits the diffuser at 80 F, with a velocity of 160 ft/s. It may be assumed that the diffuser process is SSSF, polytropic, and internally reversible. Determine the diffuser exit pressure and the rate of irreversibility for the process. Note: The exit velocity should be 320 ft/s at 80 F C.V. Diffuser out to T0, Int. Rev. flow sgen R22 = 0 / m1 = m2 , h1 + (1/2)V12 + q = h2 + (1/2)V22 s1 + dq/T + sgen = s2 = s1 + q/T0 + sgen Rev. Process: w12 = 0 =  v dP + (V12  V22)/2 / Polytropic process: LHS = v dP = (P v  P1v1) = (V12  V22)/2 n1 2 2 = (1/2)(6502  3202)/32.174 = 4974.5 ftlbf P1v1n = P2v2n n = ln(P2/P1)/ln(v1/v2) Inlet state: v1 = 0.68782, s1 = 0.2211, h1 = 108.347 Exit state: T2,V2, ? only P2 unknown. P2 = 125 psi v2 = 0.46219 n = ln(125/80)/ln(0.68782/0.46219) = 1.12257 LHS = 0.12257 (125 0.462187  80 0.68782) 144 = 3624 P2 = 150 psi v2 = 0.370548, n = 1.01627, LHS = 5006 Interpolate to get P2 = 149.4 psi. Take 150 psi as OK. P2 = 150 psi v2 = 0.370548, h2 = 111.556 s2 = 0.21484 Find q from energy equation (Assume T0 = T1 = 40 F) q = h2  h1 + (V22  V12)/2 = 111.556  108.347  (4974.5/778) = 3.185 Btu/lbm i = /m = T0(s2  s1)  q = 499.67(0.21484  0.2211) + 3.185 = 0.057 Btu/lbm I 1.12257 n v2 = v(P2), n = n(P2) Trial and error on P 2 to give LHS = 4974.5 ftlbf: 1047 10.85E(Adv.) Water in a piston/cylinder is at 14.7 lbf/in.2, 90 F, as shown in Fig. P10.56. The cylinder has stops mounted so that Vmin = 0.36 ft3 and Vmax = 18 ft3. The piston is loaded with a mass and outside P0, so a pressure inside of 700 lbf/in.2 will float it. Heat of 14000 Btu from a 750 F source is added. Find the total change in availability of the water and the total irreversibility.
P 1a 1 2 1b v CV water plus cyl. wall out to reservoir m2 = m1 = m, m(u2 u1) = 1Q2  1W2 m(s2  s1) = 1Q2/Tres + 1S2 gen System eq: States must be on the 3 lines Here Peq = 700 lbf/in2, since P1 < Peq => V1 = Vmin = 0.36 ft3 1: v1 = 0.016099, u1 = 58.07, m = V1/v1 = 22.362 lbm Heat added gives q = Q/m = 626.062 Btu/lbm so check if we go from 1 past 1a (Peq,v1) and past 1b (Peq,vmax) 1a: T 100 F, u1a 67.8 1b: T 600 F, u1b= 1180 =>
1q1a = u1a  u1 9.73 h1b  u1  Peqv1 = 1221.8 1q1b = The state is seen to fall between 1a and 1b so P2 = Peq.
1w2 = Peq(v2  v1), u2  u1 = 1q2  1w2 u2 + Peqv2 = h2 = 1q2 + u1 + Peqv1 = 686.22 < hg(P2) x2 = 0.3132, v2 = 0.3043, s2 = 0.9044 T2 = 467 F,
1S2 gen V2 = 6.805 = m(s2  s1)  1Q2/Tres = 22.362(0.9044  0.11165)  14000/1210 = 6.157 Btu/R 1I2 = T0(1S2,gen) = 3304 Btu The nonflow availability change of the water, Eq.10.22 2  1 = u2  u1 + P0(v2  v1)  T0(s2  s1) = 646.794  58.07 + 14.7(0.3043  0.016099)  536.67(0.9044  0.11165) = 164.06 Btu/lbm m = 3669 Btu 144 778 1048 10.86EA rock bed consists of 12000 lbm granite and is at 160 F. A small house with lumped mass of 24000 lbm wood and 2000 lbm iron is at 60 F. They are now brought to a uniform final temperature with no external heat transfer. a. For a reversible process, find the final temperature and the work done in the process. b. If they are connected thermally by circulating water between the rock bed and the house, find the final temperature and the irreversibility of the process assuming that surroundings are at 60 F. a) For a reversible process a heat engine is installed between the rockbed and the house. Take C.V. Total mrock(u2  u1) + mwood(u2  u1) + mFe(u2  u1) = 1W2 mrock(s2  s1) + mwood(s2  s1) + mFe(s2  s1) = 0 / (mC)rockln T2 T1 + (mC)woodln T2 T1 + (mC)Feln T2 T1 =0 / 12000 0.243 ln (T2/619.67) + 24000 0.42 ln (T2/519.67) 0 + 2000 0.107 ln (T2/519.67) = / T2 = 540.3 R 1W2 = 12000 0.243(540.3  619.67) + (24000 0.42 + 314)(540.3  519.67) 1W2 = 19078 Btu b) No work, no Q irreversible process. (mC)rock(T2  160) + (mCwood + mCFe)(T2  60) = 0 / T2 = 82 F = 541.7 R Sgen = mi(s2  s1)i = 2916 ln
1I2 541.7 541.7 + 10294 ln = 35.26 Btu/R 619.67 519.67 = T0Sgen = 519.67 35.26 = 18324 Btu 1049 10.87EAir in a piston/cylinder arrangement, shown in Fig. P10.59, is at 30 lbf/in.2, 540 R with a volume of 20 ft3. If the piston is at the stops the volume is 40 ft3 and a pressure of 60 lbf/in.2 is required. The air is then heated from the initial state to 2700 R by a 3400 R reservoir. Find the total irreversibility in the process assuming surroundings are at 70 F.
P 2 Pstop P1 V1 m2 = m1 = m = P1V1/RT1 = 30 20 144/(53.34 540) = 3.0 lbm m(u2  u1) = 1Q2  1W2
W 1 2 Vstop 1 m(s2  s1) = dQ/T + 1S2 gen Process: P = P0 + (VV0) if V Vstop Information: Pstop = P0 + (VstopV0) Eq. of state Tstop = T1PstopVstop/P1V1 = 2160 < T2 So the piston hits the stops => V 2 = Vstop => P2 = 2.5 P1
1W2 = 2(P 1 1Q2 = 1 + Pstop)(Vstop V1) = 2(30 + 60)(40  20) = 166.6 Btu 1 m(u2  u1) + 1W2 = 3(518.165  92.16) + 166.6 = 1444.6 Btu
o o s2  s1 = sT2  sT1 R ln(P2/P1) = 2.0561  1.6398  53.34 ln (2.5)/778 = 0.3535 Btu/lbm R Take control volume as total out to reservoir at T RES
1S2 gen tot = 1I2 m(s2  s2)  1Q2/TRES = 0.6356 Btu/R = T0(1S2 gen) = 530 0.6356 = 337 Btu 111 CHAPTER 11
The correspondence between the new problem set and the previous 4th edition chapter 9 problem set. New 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 94 95 96 97 98 99 Old New 2 New 3 4a 4b New 5 New 6 7 8 1 13 New 14 15 mod 16 17 New 18 New New 19 mod 20 mod 22 New 23 mod 24 a,b 24 c,d mod New New New 75 76 77 78 New 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 59 60 61 62 100 101 102 103 104 105 Old 25 New 26 new 27 New 28 New 30 31 mod 32 New New 33 34 35 36 37 38 New 39 40 41 42 43 44 mod 44 mod 45 46 New 47 48 49 79 80 81 82 83 85 New 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 93 106 107 108 109 Old 50 51 52 53 New 54 55 56 57 mod 58 mod New New 59 mod 60 61 mod New New New 62 63 New 64 65 66 67 68 New 69 New 70 71 87 88 89 91 112 The problems that are labeled advanced are: New 110 111 112 Old 9 21 New New 113 114 115 Old 84 86 90 New 116 Old 92 The English unit problems are: New 117 118 119 120 121 122 123 124 125 126 127 128 129 Old New 94 95 a 95 b 93 99 100 98 101 mod 102 mod 103 New 105 New 130 131 132 133 134 135 136 137 138 139 140 141 142 Old 106 107 108 109 110 111 112 113 114 115 mod 116 mod 117 mod New New 143 144 145 146 147 148 149 150 151 152 153 154 155 Old New 118 119 New 121 122 123 124 125 127 128 130 96 113 11.1 A steam power plant as shown in Fig. 11.3 operating in a Rankine cycle has saturated vapor at 3.5 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. C.V. Pump Rev adiabatic wP = h2  h1 ; s2 = s1 since incompressible it is easier to find work as w p = v dP = v1 (P2  P1) = 0.00101 (3500  10) = 3.525 kJ/kg => h 2 = h1  wp = 191.81 + 3.525 = 195.33 C.V. Boiler : qh = h3  h2 = 2803.43  195.33 = 2608.1 kJ/kg C.V. Turbine : wt = h3  h4 ; s4 = s3 s4 = s3 = 6.1252 = 0.6492 + x4 (7.501) => x4 = 0.73 => h 4 = 191.81 + 0.73 (2392.82) = 1938.57 wt = 2803.43  1938.57 = 864.9 kJ/kg C.V. Condenser : qL = h4  h1 = 1938.57  191.81 = 1746.8 kJ/kg cycle = wnet / qH = (wt + wp) / qH = (864.9  3.5) / 2608.1 = 0.33 11.2 Consider a solarenergypowered ideal Rankine cycle that uses water as the working fluid. Saturated vapor leaves the solar collector at 175C, and the condenser pressure is 10 kPa. Determine the thermal efficiency of this cycle. H2O ideal Rankine cycle T3 = 175C P3 = PG 175C = 892 kPa CV: turbine s4 = s3 = 6.6256 = 0.6493 + x4 7.5009 x4 = 0.797 h4 = 191.83 + 0.797 2392.8 = 2098.3 wT = h3  h4 = 2773.6  2098.3 = 675.3 kJ/kg wP = v1(P2  P1) = 0.00101(892  10) = 0.89 wNET = wT + wP = 675.3  0.89 = 674.4 kJ/kg h2 = h1  wP = 191.83 + 0.89 = 192.72 qH = h3  h2 = 2773.6  192.72 = 2580.9 kJ/kg TH = wNET/qH = 674.4/2580.9 = 0.261
2 1 4 s 3 T 114 11.3 A utility runs a Rankine cycle with a water boiler at 3.5 MPa and the cycle has the highest and lowest temperatures of 450C and 45C respectively. Find the plant efficiency and the efficiency of a Carnot cycle with the same temperatures. Solution: 1: 45oC , x = 0 => h1 = 188.42 , v1 = 0.00101 , Psat = 9.6 kPa 3: 3.5 MPa , 450oC => h 3 = 3337.2 , s3 = 7.0051 wP = h2  h1 ; s2 = s1 since incompressible it is easier to find work as wp = v dP = v1 (P2  P1) = 0.00101 (3500  9.6) = 3.525 => h 2 = h1  wp = 188.42 + 3.525 = 191.95 C.V. Boiler : qh = h3  h2 = 3337.2  191.95 = 3145.3 C.V. Turbine : wt = h3  h4 ; s4 = s3 s4 = s3 = 7.0051 = 0.6386 + x4 (7.5261) wt = 3337.2  2214.2 = 1123 kJ/kg C.V. Condenser : qL = h4  h1 = 2214.2  188.42 = 2025.78 kJ/kg cycle = wnet / qH = (wt + wp) / qH = (1123  3.5) / 3145.3 = 0.356 273.15 + 45 = 0.56 carnot = 1  TL / TH = 1 273.15 + 450 11.4 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. C.V. Turbine wT = h3  h4; s4 = s3 x4 = 0.8459 => h 4 = 188.42 + 0.8459 (2394.77) = 2214.2 => C.V. Pump Rev adiabatic 4: 15 kPa, x4 = 0.95 => s4 = 7.6458 , h4 = 2480.4 3: s3 = s4, P3 h3 = 4036.7, T3 = 758C wT = h3  h4 = 4036.7  2480.4 = 1556.3 . . m = WT/wT = 7.5 1000/1556.3 = 4.82 kg/s 115 11.5 A supply of geothermal hot water is to be used as the energy source in an ideal Rankine cycle, with R134a as the cycle working fluid. Saturated vapor R134a leaves the boiler at a temperature of 85C, and the condenser temperature is 40C. Calculate the thermal efficiency of this cycle. CV: Pump (use R134a Table B.5)
2 T wP = h2  h1 = vdP v1(P2P1)
1 D 2 1 3 85 C 40 C s
o o = 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 => h4 = 256.54 + 0.9346 163.28 = 409.14 Energy Eq.: 4 x 4 = 0.9346 wT = h3  h4 = 428.1  409.14 = 18.96 kJ/kg wNET = wT + wP = 18.96  1.67 = 17.29 kJ/kg TH = wNET/qH = 17.29/169.89 = 0.102 11.6 Do Problem 11.5 with R22 as the working fluid. Same Ts diagram and analysis as in problem 11.5 with R22 T CV: Pump (use R22 Table B.4)
2 wP = h2  h1 = vdP v1(P2P1)
1 D 2 1 3 85 C 40 C s
o o = 0.000884(4037  1534) = 2.21 kJ/kg h2 = h1  wP = 94.27 + 2.21 = 96.48 CV: Boiler: CV: Turbine s4 = s3 = 0.7918 = 0.3417 + x4 0.5329, => h4 = 94.27 + 0.8446 166.88 = 235.22 wT = h3  h4 = 253.69  235.22 = 18.47 kJ/kg 4 qH = h3  h2 = 253.69  96.48 = 157.21 kJ/kg x4 = 0.8446 TH = wNET/qH = (18.47  2.21)/157.21 = 0.1034 116 11.7 Do Problem 11.5 with ammonia as the working fluid. Solution:
T CV: Pump (use Ammonia Table B.2) 2 w = h  h = vdP = v (P P ) P 2 1 1 1 2 1 D 2 1 3 85 C 40 C s
o o = 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 3.5083 h4 = 371.43 + 0.8643 1098.8 = 1321.13 Energy Eq.: wT = h3  h4 = 1447.8  1321.13 = 126.67 kJ/kg wNET = wT + wP = 126.67  5.27 = 121.4 kJ/kg TH = wNET/qH = 121.4/1071.1 = 0.113 4 => x4 = 0.8643 117 11.8 Consider the ammonia Rankinecycle power plant shown in Fig. P11.8, a plant that was designed to operate in a location where the ocean water temperature is 25C near the surface and 5C at some greater depth. 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? a) Turbine s2S = s1 = 5.0863 = 0.8779 + x2S 4.3269 x2S = 0.9726 h2S = 227.08 + 0.9726 1225.09 = 1418.6 wST = h1  h2S = 1460.29  1418.6 = 41.69 wT = S wST = 0.80 41.69 = 33.35 kJ/kg . . WT = mwT = 1000 33.35 = 33 350 kW Pump: wSP v3(P4  P3) = 0.0016(857  615) = 0.387 wP = wSP/S = 0.387/0.80 = 0.484 kJ/kg . . WP = mwP = 1000(0.484) = 484 kW b) h2 = h1  wT = 1460.29  33.35 = 1426.94 . Qto low T H2O = 1000(1426.94  227.08) = 1.1999106 kW 1.1999106 . mlow T H2O = = 142 840 kg/s 29.38  20.98 h4 = h3  wP = 227.08 + 0.48 = 227.56 . Qfrom high T H2O = 1000(1460.29  227.56) = 1.2327106 kW 1.2327106 . mlow T H2O = = 147 280 kg/s 104.87  96.50 c) . . 33 350  484 = 0.027 TH = WNET/QH = 1.2327106
T 1 4 4s 3 2s 2 20 C o 10 C s o 118 11.9 Do Problem 11.8 with R134a as the working fluid in the Rankine cycle. Solution: a) Turbine s2S = s1 = 1.7183 = 1.0485 + x2S 0.6733 x2S = 0.9948 h2S = 213.58 + 0.9948 190.65 = 403.24 wST = h1  h2S = 409.84  403.24 = 6.6 kJ/kg wT = S wST = 0.80 6.6 = 5.28 kJ/kg . . WT = mwT = 5280 kW Pump: wSP v3(P4  P3) = 0.000794(572.8  415.8) = 0.125 kJ/kg . . WP = mwP = 156 kW wP = wSP/S = 0.156 => b) h2 = h1  wT = 409.84  5.28 = 404.56 . Qto low T H2O = 1000(404.56  213.58) = 190980 kW . mlow T H2O = 190980 = 22736 kg/s 29.38  20.98
4 4s 3 2s 2 T 1 20 C o 10 C s o h4 = h3  wP = 213.58 + 0.156 = 213.74 . Qfrom high T H2O = 1000(409.84  213.74) = 196100 kW . mlow T H2O = 196100 = 23429 kg/s 104.87  96.50 . . 5280  156 = 0.026 c) TH = WNET/QH = 196100 119 11.10 Consider the boiler in Problem 11.5 where the geothermal hot water brings the R134a to saturated vapor. Assume a counter flowing heat exchanger arrangement. The geothermal water temperature should be equal to or greater than the R134a 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 95C, what is the maximum power output of this cycle for R134a as the working fluid? (hint: split the heat exchanger C.V. into two so the pinch point with T = 0, T = 85C 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 23 into two parts as shown in the diagram.
3 sat. vap liquid
R134a 2 D LIQUID HEATER R134a 85o C BOILER
. QAB
A sat liq . QBC at 85 oC
B liquid H2O out C liquid H2O
95 oC liq H2O at 85 oC Write the enrgy equation for the first section AB and D3: . . 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  h C) hC = 323.1, TC = 77.2C > T2 OK CV Pump: CV: Turbine s4 = s3 = 1.6782 = 1.1909 + x4 0.5214 => h4 = 256.54 + 0.9346 163.28 = 409.14 Energy Eq.: wT = h3  h4 = 428.1  409.14 = 18.96 kJ/kg x 4 = 0.9346 wP = v1(P2P1) = 0.000873(2926.2  1017.0) = 1.67 kJ/kg Cycle: wNET = wT + wP = 18.96  1.67 = 17.29 kJ/kg . . WNET = mR134AwNET = 0.8813 17.29 = 15.24 kW 1110 11.11 Do the previous problem with R22 as the working fluid. A flow with 2 kg/s of water is available at 95 oC 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.
3 sat. vap liquid R22 2 D LIQUID HEATER BOILER
. QAB
A R22 sat liq . QBC at 85 oC
B 85o C liquid H2O out C liquid H2O
95 oC liq H2O at 85 oC . . QAB = mH2O(hA  hB) = 2(397.94  355.88) = 84.12 kW . . = mR22(253.69  165.09) mR22 = 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 TC = 77.2 oC OK . From Problem 11.6 then: Wnet = 0.949 (18.47  2.21) = 15.43 kW 11.12 The power plant in Problem 11.1 is modified to have a superheater section following the boiler so the steam leaves the super heater at 3.5 MPa, 400C. Find the specific work and heat transfer in each of the ideal components and the cycle efficiency. C.V. Tubine: Energy: wT,s = h3  h4s; Entropy: s4s = s3 = 6.8404 x4s = (6.8404  0.6492)/7.501 = 0.8254 ; h4s = 191.81 + 0.8254 2392.82 = 2166.8 kJ/kg wT,s = 3222.24  2166.8 = 1055.4 kJ/kg C.V. Pump: wP = v dP = v1(P2  P1) = 0.00101(3500  10) = 3.525 kJ/kg h2 = h1  wP = 191.81 + 3.525 = 195.34 C.V. Condenser: qC = h4  h1 = 2166.8  191.81 = 1975 C.V. Boiler: qH = h3  h2 = 3222.24  195.34 = 3027 CYCLE = wNET/qH = (1055.4  3.53)/3027 = 0.3475 1111 11.13 A steam power plant has a steam generator exit at 4 MPa, 500C and a condenser exit temperature of 45C. Assume all components are ideal and find the cycle efficiency and the specific work and heat transfer in the components.
T 3 From the Rankine cycle we have the states: 1: 45C x = 0 , v1 = 0.00101 , h1 = 188.45 / 2: 4 MPa
2 1 4 s 3: 4 MPa, 500C , h3 = 3445.3 , s3 = 7.0901 4: Psat(45C) = 9.593 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 qL,out = h4  h1 = 2241.3  188.42 = 2052.9 kJ/kg C.V. Condenser: TH = wnet/qH = (wT + wP)/qH = (1204  4.03)/3252.8 = 0.369 11.14 Consider an ideal Rankine cycle using water with a highpressure 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 hightemperature energy source is the hot exhaust gas from a gasturbine engine. Calculate the thermal efficiency of the cycle if the state entering the turbine is 25 MPa, 500C, and the condenser pressure is 5 kPa. What is the steam quality at the turbine exit? s4 = s3 = 5.9592 = 0.4764 + x4 7.9187 x4 = 0.6924 Very low for a turbine exhaust s1 = 0.4764 , h1 = 137.82 h4 = 1816 , h3 = 3162.4 s2 = s1 => h2 = 162.8 wNET = h3  h4  (h2  h1) = 1321.4 qH = h3  h2 = 2999.6, = wNET/qH = 0.44
3 25 MPa o 500 C 5 kPa 1 4 s T 2 1112 11.15 Steam enters the turbine of a power plant at 5 MPa and 400C, 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 Ts = 85 % State 1 : T1 = 400oC , P1 = 5 MPa , superheated h1 = 3195.6 kJ/kg , s1 = 6.6458 kJ/kgK State 3 : P3 = P2 = 10 kPa , sat liq , x3 = 0 T3 = 45.8oC , h3 = hf = 191.8 kJ/kg , v3 = vf = 0.00101 m3/kg C.V Turbine : 1st Law: qT + h1 = h2 + wT ; qT = 0 wT = h1  h2 , Assume Turbine is isentropic s2s = s1 = 6.6458 kJ/kgK , s2s = sf + x2s sfg , solve for x2s = 0.7994 h2s = hf + x2shfg = 1091.0 kJ/kg wTs = h1  h2s = 1091 kJ/kg , wT = TswTs = 927.3 kJ/kg . . WT m= = 21.568 kg/s , h2 = h1  wT = 2268.3 kJ/kg wT C.V. Condenser: 1st Law : qc + h2 = h3 + wc ; wc = 0 . . qc = h3  h2 = 2076.5 kJ/kg , Qc = m qc = 44786 kW wps =  v dP = v3(P4  P3) = 5.04 kJ/kg 1st Law : qp + h3 = h4 + wp ; qp = 0 h4 = h3  wp = 196.8 kJ/kg C.V Boiler : 1st Law : qB + h4 = h1 + wB ; wB = 0 qB = h1  h4 = 2998.8 kJ/kg (or qB = wT + wB  q c) wnet = wT + wP = 922.3 kJ/kg th = wnet / qB = 0.307 Carnot cycle : TH = T1 = 400oC , TL = T3 = 45.8oC th = T H  TL TH = 0.526 C.V. Pump: Assume Isentropic 1113 11.16 Consider an ideal steam reheat cycle where steam enters the highpressure turbine at 3.5 MPa, 400C, and then expands to 0.8 MPa. It is then reheated to 400C and expands to 10 kPa in the lowpressure turbine. Calculate the cycle thermal efficiency and the moisture content of the steam leaving the lowpressure turbine. P3 = 3.5 MPa, T3 = T5 = 400 oC P4 = P5 = 0.8 MPa, P6 = 10 kPa From solution 11.12: wP = 3.525 kJ/kg, h2 = 195.34
2 1 T 3.5 MPa o 3 5 400 C 4 6
s 10 kPa h3 = 3222.2, s3 = 6.8404 qH1 = h3h2 = 3222.3195.35 = 3026.9 s4 = s3 => h4 = 2851.6 ; h5 = 3267.1 qH = qH1 + h5  h4 = 3026.9 +3267.1  2851.6 = 3442.4 kJ/kg s6 = s5 = 7.5715 => x6 = (7.57150.6492)/7.501 = 0.9229 ; h6 = 2400 wT,tot = h3 h4 + h5  h6 = 3222.2  2851.6 + 3267.1  2400 = 1237.8 kJ/kg CYCLE = (1237.8  3.525)/3442.4 = 0.3586 11.17 The reheat pressure effect the operating variables and thus turbine performance. Repeat Problem 11.16 twice, using 0.6 and 1.0 MPa for the reheat pressure. P3 = 3.5 MPa, T3 = T5 = 400 oC P4 = P5 = 0.6 MPa, P6 = 10 kPa From solution 11.12: wP = 3.525 kJ/kg, h3 = 3222.2, h2 = 195.34
2 1 T 3.5 MPa o 3 5 400 C 4 6
s 10 kPa s3 = 6.8404 qH1 = h3h2 = 3222.2195.34 = 3026.9 wT,tot = h3 h4 + h5  h6, qH = qH1 + h5  h4, s4 = s3, s6 = s5 For P4=P5 = 1 MPa: s5 = 7.465, state 4 is superheated vapor For P4=P5 = 0.6 MPa: s5 = 7.7078, state 4 is superheated vapor x6 0.9087 0.9410 P4=P5 h4 1 2900.7 0.6 2793.2 h5 3263.9 3270.3 h6 2366 2443.5 wT 1219.5 1255.9 qH 3390.1 3504.0 CYCLE 0.3587 0.3574 Notice the very small changes in efficiency. 1114 11.18 The effect of a number of reheat stages on the ideal steam reheat cycle is to be studied. Repeat Problem 11.16 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. P4 = P5= 1.2 MPa, P6 = P7 = 0.2 MPa From solution to 11.12, wP = 3.525 kJ/kg, h2 = 195.34 3: h3 = 3222.2, s3 = 6.8404 4: P4, s4 = s3 sup. vap. h4 = 2940.9 5: h5 = 3260.7 , s5 = 7.3773 6: P6, s6 = s5 sup. vap. h6 = 2811.2 7: h7 = 3276.5, s7 = 8.2217 8: P8, s8 = s7 sup. vap. h8 = 2607.9 T 3.5 MPa o 3 5 7 400 C 4 2 1 6 8
s 10 kPa wT = (h3  h4) + (h5  h6) + (h7  h8) = 281.4 + 449.5 + 668.6 = 1399.5 kJ/kg qH = (h3  h2) + (h5  h4) + (h7  h6) = 3027 + 319.8 + 465.3 = 3812.1 kJ/kg TH = (wT + wP)/qH = (1399.5  3.53)/3812.1 = 0.366 11.19 A closed feedwater heater in a regenerative steam power cycle heats 20 kg/s of water from 100C, 20 MPa to 250C, 20 MPa. The extraction steam from the turbine enters the heater at 4 MPa, 275C, and leaves as saturated liquid. What is the required mass flow rate of the extraction steam? 3 2 1 From table B.1 B.1.4: 100C, 20 MPa B.1.4: 250C, 20 MPa h1 = 434.06 h2 = 1086.75 h3 = 2886.2 h4 = 1087.31 4 B.1.3: 4 MPa, 275C B.1.2: 4 MPa, sat. liq. C.V. Feedwater Heater . . . . Energy Eq.: m1h1 + m3h3 = m1h2 + m3h4 . . m3 = m1(h1h2)/(h4h3) = 7.257 kg/s 1115 11.20 An open feedwater heater in a regenerative steam power cycle receives 20 kg/s of water at 100C, 2 MPa. The extraction steam from the turbine enters the heater at 2 MPa, 275C, and all the feedwater leaves as saturated liquid. What is the required mass flow rate of the extraction steam? Solution: C.V Feedwater heater . . . m1 + m2 = m3 . . . . . m1h1 + m2h2 = m3h3 = (m1 + m2) h3 2 3 1 Table B.1.1 and Table B.1.4 at 100 C interpolate to 2 MPa: h1 = 420.4 Table B.1.3: h2 = 2963, Table B.1.2: h3 = 908.8 . . h3  h1 908.8  420.4 m2 = m1 = 20 = 4.755 kg/s 2963  908.8 h2  h3 Remark: h1 was interpolated between sat liq and compressed liquid at 5 MPa both at 100oC. The saturated liquid value 419.02 could have been used with very small error. 11.21 A power plant with one closed feedwater heater has a condenser temperature of 45C, a maximum pressure of 5 MPa, and boiler exit temperature of 900C. 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 200C. Find the fraction of extraction steam flow and the two specific pump work inputs. From turbine s1 = 0.6387, h1 = 188.45 v1 = 0.00101 s4 = 2.1387, h4 = 762.81 T6 => h 6 = 853.9 6 7 Pump 2 C.V. Turbine: s3 = sIN = 7.9593, T3 = 573.8, 4 h3 = 3640.58 5 2 3 Pump 1 1 From condenser C.V. P1: wP1 = v1(P2  P1) = 5.04 h2 = h1  wP1 = 193.49 C.V. P2: wP2 = v4(P7  P4) = 4.508 h7 = h4  wP2 = 767.31 . . . . . . . C.V. Tot: m6 = m1 + m3 , m6/m1 = 1 + x, m3/m1 = x (1 + x)h6 = xh3  xwP2  wP1 + h1 = xh3  xwP2 + h2 . . x = (h6 h2)/(h3 + wP2  h6) = 0.2364, m3/m6 = 0.1912 1116 11.22 A power plant with one open feedwater heater has a condenser temperature of 45C, a maximum pressure of 5 MPa, and boiler exit temperature of 900C. 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: State out of boiler 5 h5 = 4378.82 , s5 = 7.9593 C.V. Turbine: s7 = s6 = s5 P6 , s6 => h6 = 3640.6, T6 = 574oC C.V Pump P1 wP1 = h2  h1 = v1(P2  P1) = 0.00101(1000  9.6) = 1.0 kJ/kg => h 2 = h1  wP1 = 188.42 + 1.0 = 189.42 kJ/kg . . C.V. Feedwater heater: Call m6 / mtot = x Energy Eq.: x= C.V Pump P2 wP2 = h4  h3 = v3(P4  P3) = 0.001127(5000  1000) = 4.5 kJ/kg (1  x) h2 + x h6 = 1 h3 = 762.79  189.42 = 0.1661 3640.6  189.42 From turbine To boiler 4 Pump 2 3 6 Pump 1 1 2 From condenser h3  h2 h6  h2 1117 11.23 A steam power plant operates with a boiler output of 20 kg/s steam at 2 MPa, 600C. The condenser operates at 50C dumping energy to a river that has an average temperature of 20C. 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 5C how much water should be pumped from the river to the heat exchanger (condenser)? Solution: The setup is as shown in Fig. 11.10, 1: 50oC sat liq. v1 = 0.001012 , h1 = 209.31 2: 600 kPa 3: 600 kPa 5: P , T sat liq. h5 = 3690.1 s 2 = s1 h3 = hf = 670.54 s5 = 7.7023 Condenser: To river 7 Ex turbine From 1 river To pump 1 6: 600 kPa s6 = s5 => h 6 = 3270.0 CV P1 wP1 = v1(P2  P1) = 0.001012 (600  12.35) = 0.595 h2 = h1  wP1 = 209.9 C.V FWH x h6 + (1 x) h2 = h3 x= CV Turbine s 7 = s6 = s5 => x 7 = 0.9493 , h7 = 2471.17 kJ/kg h3  h2 h6  h2 = 670.54  209.9 = 0.1505 3270.0  209.9 qL = h7  h1 = 2471.17  209.31 = 2261.86 kJ/kg . . QL = (1  x) mqL = 0.85 20 2261.86 = 38429 kW . . = mH2O hH2O = m (20.93) = 38429 kW . m = 1836 kg/s CV Condenser + Heat Exchanger . QL . . 0 = mH2O (s7  s1) (1  x) + Sg TL . 38429 Sg =  20 0.85 (7.7023  0.7037) = 12.184 kW/K 293.15 1118 11.24 Consider an ideal steam regenerative cycle in which steam enters the turbine at 3.5 MPa, 400C, 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: C.V. Turbine 2nd Law s7 = s6 = s5 = 6.8404 kJ/kg K P6 , s6 => h6 = 2851 (sup vap) C.V Pump P1 wP1 = h2  h1 = v1(P2  P1) = 0.00101(800  10) = 0.798 kJ/kg => h 2 = h1  wP1 = 191.81 + 0.798 = 192.61 . . Call m6 / mtot = x (1  x) h2 + x h6 = 1 h3 x= C.V Pump P2 wP2 = h4  h3 = v3(P4  P3) = 0.001115(3500  800) = 3.01 h4 = h3  wP2 = 721.1 + 3.01 = 724.1 CV Boiler qH = h5  w4 = 3222.2 + 724.1 = 2498.1 CV Turbine . . WT / m5 = h5  h6 + (1  x) (h6  h7) s7 = s6 = s5 = 6.8404 => x7 = 6.8404  0.6492 7.501 h3  h2 h6  h2 = 721.1  192.61 = 0.1988 2851  192.61 x7 = 0.8254 => h7 = 191.81 + x7 2392.82 = 2166.8 wT = 3222.2  2851 + (1  0.1988) ( 2851  2166.8) = 919.38 wnet = wT + (1  x) wP1 + wP2 = 919.38 + (1  0.1968)(0.798)  3.01 = 915.7 kJ/kg cycle = wnet / qH = 915.7 / 2498.1 = 0.366 1119 11.25 Repeat Problem 11.24, 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 3.5 MPa. Condensate from the feedwater heater is drained through a trap to the condenser. Solution: C.V. Turbine, 2nd law: s4 = s5 = s6 = 6.8404 kJ/kg K h4 = 3222.24 , h5 = 2851 => x 6 = (6.8404  0.6492)/7.501 = 0.8254 h6 = 191.81 + x6 2392.82 = 2166.8
2 3 BOILER
4 TURBINE. FW HTR Trap 7 P
1 5 6 COND. Assume feedwater heater exit at the T of the condensing steam C.V Pump wP = h2  h1 = v1(P2  P1) = 0.00101(3500  10) = 3.525 h2 = h1  wP = 191.81 + 3.525 = 195.3 T3 = Tsat (P5) = 170.43 , h3 = hf = h7 = 721.1 C.V FWH . . m5 / m3 = x , x= h3  h2 h5  hf 800 = Energy Eq.: h2 + x h5 = h3 + h7 x 721.1  195.3 = 0.2469 2851  721.1 wT = h4  h5 + (1  x)(h5  h6) = 3222.2  2851 + 0.7531 ( 2851  2166.8) = 886.51 wnet = wT + wP = 886.51  3.53 = 883 qH = h4  h3 = 3222.24  721.1 = 2501.1 cycle = wnet / qH = 883 / 2501.1 = 0.353 1120 11.26 A steam power plant has high and low pressures of 25 MPa and 10 kPa, and one open feedwater heater operating at 1 MPa with the exit as saturated liquid. The maximum temperature is 800C 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 Ts 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: h5 = 4047.1, s5 = 6.9345, h1 = 191.83, v1 = 0.00101 h3 = 762.8 v3 = 0.001127 Pump P1: wP1 = v1(P2  P1) = 0.00101 990 = 1 kJ/kg h2 = h1  wP1 = 192.83 kJ/kg Turbine 56: s6 = s5 h6 = 2948 kJ/kg wT56 = h5  h6 = 1099.1 kJ/kg . . . . . Feedwater Heater (mTOT = m5): xm5h6 + (1  x)m5h2 = m5h3 x= h3  h2 h6  h2 = 762.8  192.83 = 0.2069 2948  192.83 To get state 7 into condenser consider turbine. s7 = s6 = s5 x7 = (6.9345  0.6493)/7.5009 = 0.83793 h7 = 191.81 + 0.83793 2392.82 = 2196.8 kJ/kg Find specific turbine work to get total flow rate . . . . . WT = mTOTh5  xmTOTh6  (1  x)mTOTh7 = mTOT 1694.9 . mTOT = 5000/1694.9 = 2.95 kg/s . . QL = mTOT (1x) (h7h1) = 2.95 0.7931(2196.8191.83) = 4691 kW 1121 11.27 Do Problem 11.26 with a closed feedwater heater instead of an open and a drip pump to add the extraction flow to the feed water line at 25 MPa. Assume the temperature is 175C after the drip pump flow is added to the line. One main pump brings the water to 25 MPa from the condenser. Solution: v1 = 0.00101 , h1 = 191.83 T4 = 175oC ; h4 = 741.16 h6 = 4047.08 , s6 = 6.9345 h6a = hf 1MPa = 762.79 , v6a = 0.001127 C.V Pump 1 wP1 = h2  h1 = v1(P2  P1) = 0.00101(25000  10) = 25.24 kJ/kg => h 2 = h1  wP1 = 191.83 + 25.24 = 217.07 C.V Pump 2 wP2 = h6b  h6a = v6a(P6b  P6a) = 0.001127(25000  1000) = 27.05 kJ/kg Turbine section 1: C.V FWH + P2 . . m6 / m4 = x x h6 + (1  x) h2 + x (wP2)= h4 x= Turbine: h4  h 2 h6  h2  wP2 = 741.16  217.07 = 0.19 2948  217.07 + 27.05 s 6 = s5 P6 = 1 MPa => h6 = 2948 kJ/kg From turbine 6 4 6b Pump 2 6a 3 From condenser 1 2 Pump 1 s7 = s6 = s5 & P7 = 10 kPa 6.9345  0.6493 = 0.83793 7.5009 => x 7 = . . WT = m5 [ h5  h6 + (1  x) (h6  h7) ] . . = m5 [ 4047.08  2948 + (2948  2196.8)] = m5 1707.6 . . . WT = 5000 kW = m5 1707.6 => m5 = 2.928 kg/s . . QL = m5(1  x) (h7  h1) = 2.928 0.81 (2196.8  191.83) = 4755 kW 1122 11.28 Consider an ideal combined reheat and regenerative cycle in which steam enters the highpressure turbine at 3.5 MPa, 400C, 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 400C 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. 5: h5 = 3222.24, 7: h7 = 3267.07, 3: h3 = hf = 721.1 C.V. T1 s 5 = s6 => h6 = 2851
4 s5 = 6.8404 s7 = 7.5715
5 7 T1 1x x HTR
3 6 T2
8 wT1 = h5  h6 = 3222.2  2851 = 371.2 C.V. Pump 1 wP1 = h2  h1 = v1(P2  P1) = 0.00101(800  10) = 0.798 => h 2 = h1  wP1 = 191.81 + 0.798 = 192.61 C.V. FWH x h6 + (1  x) h2 = h3 721.1  192.61 x= = = 0.1988 h6  h2 2851  192.61 C.V. Pump 2 h3  h2
P 1x
2 COND. P
1 T 400 oC 4 5 7 6 2 3 1 8 10 kPa s wP2 = h4  h3 = v3(P4  P3) = 0.001115(3500  800) = 3.01 => h 4 = h3  wP2 = 721.1 + 3.01 = 724.11 qH = h5  h4 + (1  x)(h7  h6 ) = 2498.13 + 333.35 = 2831.5 C.V. Turbine 2 s 7 = s8 => x8 = (7.5715  0.6492)/7.501 = 0.92285 h8 = hf + x8 hfg = 191.81 + 0.92285 2392.82 = 2400.0 wT2 = h7  h8 = 3267.07  2400.02 = 867.05 wnet = wT1 + (1  x) wT2 + (1  x) wP1 + wP2 = 371.2 + 694.7  0.64  3.01 = 1062.3 kJ/kg cycle = wnet / qH = 1062.25 / 2831.5 = 0.375 1123 11.29 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 highpressure turbine at 8 MPa, 550C, 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 550C. The reheated steam is then expanded in the lowpressure turbine to 10 kPa. Determine the steam flow rate to the highpressure turbine and the power required to drive each of the pumps. 7 a) T 5 5 550 oC 7 HI P LOW P T1 T2 4 6 6 10 kPa 8 2 3 6a 1 8 COND. HTR s 4
2 P 3 P 1 b) wP12 = 0.00101(600  10) = 0.6 kJ/kg h2 = h1  wP12 = 191.8 + 0.6 = 192.4 wP34 = 0.00101(8000  600) = 8.1 kJ/kg h4 = h3  wP34 = 670.6 + 8.1 = 678.7 ; s6 = s5 = 6.8778 T6 = 182.32 oC h5 = 3521.0, h6 = 2810.0, h7 = 3591.9, s8 = s7 = 8.1348 = 0.6493 + x8 7.5009 x8 = 0.9979 h8 = 191.83 + 0.9979 2392.8 = 2579.7 CV: heater Cont: m6a + m2 = m3 = 1 kg, 1st law: m6a = CV: turbine wT = (h5  h6) + (1  m6a)(h7  h8) = 3521  2810 + 0.8173(3591.9  2579.7) = 1538.2 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 m6ah6 + m2h2 = m3h3 670.6  192.4 = 0.1827, m2 = m7 = 1  m6a = 0.8173 2810.0  192.4 1124 11.30 The low pressure turbine in a reheat and regenerative cycle receives 10 kg/s steam at 600 kPa, 550C. The turbine exhausts to a condenser operating at 10 kPa. The condenser cooling water temperature is restricted to a maximum of 10C increase so what is the needed flow rate of the cooling water? The steam velocity in the turbinecondenser connecting pipe is restricted to a maximum of 100 m/s, what is the diameter of the connecting pipe? Solution: 7 a) T 5 550 oC 5 7 HI P LOW P T1 T2 4 6 6 10 kPa 8 2 3 6a T COND. 1 8 HTR s 4 2 1 V8 = 100 m/s, TCOOL H2O = 10 oC 3 P P h7 = 3591.83, s7 = 8.13465 CV T2 => s8 = s7 = 8.13465 = 0.6492 + x8 7.501 => v8 = vf + x8 vfg = 0.00101 + x8 14.67254 = 14.643 h8 = hf + x8 hfg = 191.81 + x8 2392.82 = 2579.7 CV CONDENSER: . . QL = m(h8 h1) = 10(2579.7  191.81) = 23879 kW hH2O = h30  h20 = 125.77  83.94 = 41.83 (=CP liq(T) = 4.184 10) . . mH2O = QL/ hH2O = 23879 / 41.83 = 571 kg/s Turbine  condenser connecting pipe: . . m = AV = AV/v => A = mv/V = 10 14.643/100 = 1.464 m2 D = (4A/)1/2 = 1.37 m x8 = 0.99793 1125 11.31 A steam power plant has a high pressure of 5 MPa and maintains 50C in the condenser. The boiler exit temperature is 600C. All the components are ideal except the turbine which has an actual exit state of saturated vapor at 50C. Find the cycle efficiency with the actual turbine and the turbine isentropic efficiency. Simple Rankine cycle. Boiler exit: h3 = 3666.5 , s3 = 7.2588 Ideal Turbine: 4s: 50C, s = s3 => x = (7.2588  0.7037)/7.3725 = 0.88913, h4s = 209.31 + 0.88913 2382.75 = 2327.88 => Condenser exit: Actual turbine: wTs = h3  h4s = 1338.62 h1 = 209.31 , Actual turbine exit: h4ac = hg = 2592.1 wTac = h3  h4ac = 1074.4 T = wTac / wTs = 0.803: Isentropic Efficiency Pump: wP = v1( P2  P1) = 0.001012(500012.35) = 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 cycle = (wTac + wP) / qH = 0.31: Cycle Efficiency 11.32 A steam power cycle has a high pressure of 3.5 MPa and a condenser exit temperature of 45C. The turbine efficiency is 85%, and other cycle components are ideal. If the boiler superheats to 800C, find the cycle thermal efficiency. Basic Rankine cycle as shown in Figure 11.3 in the main text. C.V. Turb.: wT = h3  h4, s4 = s3 + sT,GEN Ideal: s4 = s3 = 7.9134 => h4S = 2503.2, wT,S = 1640.44 Actual: wT,AC = wT,S = 1394.38 C.V. Pump: C.V. Boiler: wP = v dP v1(P2  P1) = 3.52 kJ/kg qH = h3  h2 = h3  h1 + wP = 3951.7 = (wT,AC + wP)/qH = (1394.4  3.5)/3951.7 = 0.352 1126 11.33 A steam power plant operates with with a high pressure of 5 MPa and has a boiler exit temperature of of 600C receiving heat from a 700C source. The ambient at 20C provides cooling for the condenser so it can maintain 45C 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. Take CV around each component (all SSSF) in standard Rankine Cycle. 1: v = 0.00101; h = 188.42, s = 0.6386 (saturated liquid at 45C). 3: h = 3666.5, s = 7.2588 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 ws = h3  h4s = 3666.5  2295 = 1371.5, 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(50009.59) = 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 1127 11.34 Repeat Problem 11.26 assuming the turbine has an isentropic efficiency of 85%. The physical components and the Ts 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: h5 = 4047.1, s5 = 6.9345, h3 = 762.8 , v3 = 0.001127 Pump P1: wP1 = v1(P2  P1) = 0.00101 990 = 1 kJ/kg h2 = h1  wP1 = 192.83 kJ/kg Turbine 56: s6 = s5 h6 = 2948 kJ/kg wT56,s = h5  h6 = 1099.1 kJ/kg wT56,AC = 1099.1 0.85 = 934.24 wT56,AC = h5  h6AC h6AC = h5  wT56,AC = 4047.1  934.24 = 3112.86 . . . . . Feedwater Heater (mTOT = m5): xm5h6AC + (1  x)m5h2 = m5h3 x= h3  h2 h6  h2 = 762.8  192.83 = 0.1952 3112.86  192.83 h1 = 191.83, v1 = 0.00101 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 = 0.83793 , h7s = 2196.8 wT57,s = h5  h7s = 4047.1  2196.8 = 1850.3 wT57,AC = wT57,sT = 1572.74 = h5  h7AC => h 7AC = 2474.4 Find specific turbine work to get total flow rate . WT . 5000 mTOT = = = 3.453 kg/s xwT56 + (1x)wT57 0.1952934.24 + 0.80481572.74 . . QL = mTOT(1  x)(h7  h1) = 3.453 0.8048(2474.4  191.83) = 6343 kW 1128 11.35 Steam leaves a power plant steam generator at 3.5 MPa, 400C, and enters the turbine at 3.4 MPa, 375C. The isentropic turbine efficiency is 88%, and the turbine exhaust pressure is 10 kPa. Condensate leaves the condenser and enters the pump at 35C, 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, 30C. Calculate the 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 25C.
2 1 T ST. GEN.
6 TURBINE. = 0.88
5 4 3 COND. s1 = 6.8405 , 5 5S 6 4 3.5 MPa 3.4 MPa 1 400 oC 375 oC 2 10 kPa 3S 3 s P 1: h1 = 3222.3, 2: h2 = 3165.7, 3s: s3S = s2 x3S = 0.8157, h3S = 2143.6 wT,S = h2  h3S = 3165.7  2143.6 = 1022.1 wT,AC = wT,S = 899.4 kJ/kg, 3ac: s2 = 6.7675 h3 = h2  wT,AC = 2266.3 wP,S = vf(P5  P4) = 0.001006(3700  10) = 3.7 kJ/kg wP,AC = wP,S/P = 4.6 kJ/kg qH = h1  h6 = 3222.3  129.0 = 3093.3 kJ/kg = wNET/qH = (899.4  4.6)/3093.3 = 0.289 C.V. Line from 1 to 2: Energy Eq.: w = 0, / 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 1129 11.36 For the steam power plant described in Problem 11.1, 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. CV Pump, Rev & Adiabatic: wPs = h2s  h1 = v1(P2  P1) = 0.00101(3500  10) = 3.525 ; wPac = wPs / P = 3.525/0.8 = 4.406 = h2a  h1 h2a = wPac + h1 = 4.406 + 191.81 = 196.2 kJ/kg CV Boiler: CV Turbine: qH = h3  h2a = 2803.43  196.2 = 2607.2 kJ/kg wTs = 2803.43  1938.57 = 864.9 kJ/kg wTac = wTs T = 735.2 = h3  h4a h4a = h3  wTac = 2803.43  735.2 = 2068.2 CV Condensor: qL = h4a  h1 = 2068.2  191.81 = 1876.4 cycle = (wTac + wPac) / qH = (735.2  4.41) / 2607.2 = 0.28 This compares to 0.33 for the ideal case. 11.37 A small steam power plant has a boiler exit of 3 MPa, 400C 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. CV Turbine (Ideal): s4s = s3 = 6.9211, x4s = (6.9211  1.091)/6.5029 = 0.8965 h4s = 2407.35, h3 = 3230.8 CV Turbine (Actual): wTac = T wTs = 658.76 = h3  h4ac, => h 4ac = 2572 . . 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 => h 2 = h1 wP = 340.47 + 3.04 = 343.51 kJ/kg C.V. Boiler: qH = h3  h2 = 3230.8  343.51 = 2887.3 kJ/kg cycle = (wTac + wP) / qH = (658.76  3.04) / 2887.3 = 0.227 => wTs = h3  h4s = 823.45 s2s = s1 1130 11.38 In a particular reheatcycle power plant, steam enters the highpressure turbine at 5 MPa, 450C and expands to 0.5 MPa, after which it is reheated to 450C. The steam is then expanded through the lowpressure turbine to 7.5 kPa. Liquid water leaves the condenser at 30C, 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 T 5 MPa 0.5 MPa o 3 450 C 5 7.5 kPa 4S 4 6S 6 s
2 3 HP LP TURBINE.
6 2 2S 1 4 ST1= ST2 0.87 = P = SP 0.82
=> x 4S = 0.999 COND.
1 a) s4S = s3 = 6.8185 = 1.8606 + x4S 4.9606 h4S = 640.21 + 0.999 2108.5 = 2746.6 wT1,S = h3  h4S = 3316.1  2746.6 = 569.5 kJ/kg wT1 = T1,S wT1,S = 0.87 569.5 = 495.5 kJ/kg h4ac = 3316.1  495.5 = 2820.6 s6S = s5 = 7.9406 = 0.5764 + x6S 7.675 wT2,S = h5  h6S = 3377.9  2477.3 = 900.6 wT2 = 0.87 900.6 = 783.5 wT1+T2 = 783.5 + 495.5 = 1279 kJ/kg . . m = WT/wT1+T2 = 10000/1279 = 7.82 kg/s b) wP,S = (0.001004)(5000  7.5) = 5.01 kJ/kg wP = wSP/SP = 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 TH = wN/qH = 1272.9/3743.2 = 0.34 x6S = 0.9595 h6S = 168.79 + 0.9595 2406 = 2477.3 kJ/kg 1131 11.39 A supercritical steam power plant has a high pressure of 30 MPa and an exit condenser temperature of 50C. The maximum temperature in the boiler is 1000C 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?
3 2b 2 1a 1 1b 30 MPa 1000 C 1 MPa 3b 3a 50 C 4s 4ac s Basically a Rankine Cycle 1: 50 C, 12.35 kPa, h = 209.31, s = 0.7037 2: 30 MPa 3: 30 MPa, 1000 C, h = 4554.7, s = 7.2867 4AC: 50C, x = 1, h = 2592.1 T a) C.V. Turbine Ideal: s4S = s3 x4S = 0.8929, h4S = 2336.8 => wT,S = h3  h4S = 2217.86 Actual: wT,AC = h3h4AC = 1962.6, = wT,AC/wT,S = 0.885 b) P2 1b 2b m tot 3b m1 1b: Sat liq. 179.91C, h = 762.81 3a: 1 MPa, s = s3 > h3a = 3149.09, T3a = 345.96 > wT1s = 1405.6 3b: 1 MPa, wT1ac = wT1s = 1243.96 wT1ac = h3h3b => h 3b = 3310.74 1a: wP1 = v1(P1aP1) 1 h1a = h1  wP1 = 210.31 1a P1 1 . . . . 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 = h3xh3b  (1x)h4AC = 1834.7 => mTOT = 13.63 kg/s d) C.V. No FWH, Pump Ideal: wP = h2S  h1, s2S = s1 Steam table h2S = 240.1, T2S = 51.2C 1 FWH, CV: P2. s2b = s1b = 2.1386 => T 2b = 183.9C 1132 11.40 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 600C in an external gasfired superheater. The steam enters the turbine, which has one (opentype) 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 6 6 600o C
SUP. HT. TURBINE.
7 8 0.4 MPa 4 2 3 1 5 7s 7 7.5 kPa Q
5 REACT. 4 HTR.
3 COND.
2 1 8s 8 P P
wP12 = 0.001008(400  7.5) = 0.4 kJ/kg h2 = h1  wP12 = 168.8 + 0.4 = 169.2 s . WNET = 1 MW , ST = 0.87 wP34 = 0.001084(5000  400) = 5.0 kJ/kg h4 = h3  wP34 = 604.7 + 5.0 = 609.7 s7S = s6 = 7.2589, P7=0.4 MPa => T7S= 221.2 oC, h7S = 2904.5 h6  h7 = ST(h6  h7S) 3666.5  h7 = 0.87(3666.5  2904.5) = 662.9 h7 = 3003.6 s8S = s6 = 7.2589 = 0.5764 + x8S 7.6750 h8S = 168.8 + 0.8707 2406.0 = 2263.7 h6  h8 = ST(h6  h8S) 3666.5  h8 = 0.87(3666.5  2263.7) = 1220.4 h8 = 2446.1 CV: heater cont: m2 + m7 = m3 = 1.0 kg, Energy Eq.: m2h2 + m7h7 = m3h3 x8S = 0.8707 m7 = (604.7169.2)/(3003.6169.2) = 0.1536 CV: turbine wT = (h6  h7) + (1  m7)(h7  h8) = 3666.53003.6 + 0.8464(3003.62446.1) = 1134.8 kJ/kg 1133 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 11.41 A cogenerating steam power plant, as in Fig. 11.17, operates with a boiler output of 25 kg/s steam at 7 MPa, 500C. 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. Pump 1: Inlet state is saturated liquid: h1 = 168.79, v1 = 0.001008 wP1 =  v dP = v1 ( P2  P1) = 0.001008(70007.5) = 7.05 kJ/kg  wP1 = h2  h1 => h2 = h1  wP1 = 175.84, T5 = 42C Turbine: h5 = 3410.3, s5 = 6.7975 P6, s6 = s5 => x6 = 0.9952, h6 = 2738.6 P7, s7 = s5 => x7 = 0.8106, h7 = 2119.0 . . . WT = m5 ( h5  h6) + 0.80m5 ( h6  h7) = 25 (3410.3  2738.6) + 20 (2738.6  2119) = 16792.5 + 12392 = 29.185 MW . . Qproc = m6(h6  h8) = 5(2738.6  417.46) = 11.606 MW 1134 11.42 A 10 kg/s steady supply of saturatedvapor 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 20C, 100 kPa, is pumped to a pressure of 5 MPa and then fed to a steam generator. It may be assumed that the isentropic efficiency of the pump is 75%, and that of the turbine is 85%. What is the steam temperature exiting the steam generator? 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? State 3: P3 = 5 MPa State 4: P4 = 500 kPa, sat. vap. > x4 = 1.0, T4 = 151.9C h4 = hg = 2748.7 kJ/kg, s4 = sg = 6.8212 kJ/kgK State 1: T1 = 20C, P1 = 100 kPa h1 = hf = 83.94 kJ/kg, v1 = vf = 0.001002 m3/kg State 2: P2 = 5 MPa (a) C.V.: Turbine, Assume reversible & adiabatic, s3 = s4s s3 = s4s = sf + x4s sfg = 1.8606 + x4s4.9606 h4s = hf + x4s hfg = 640.2 + x4s2108.5 ts = = 0.85 wts h3  h4s Trial and Error to find T3 @ P3 = 5 MPa Assume T3 = 400C > s3 = 6.6458 kJ/kgK, h3 = 3195.6 kJ/kg s4s = s3 > x4s = 0.9646, h3  h 4 h4s = hf + x4s hfg = 2674.1 kJ/kg wt = h 3  h4 = 0.857 ts; T3 = 400C h3  h4s (b) With Cogeneration; C.V. Pump wPs =  vdP = v1( P2 P1) = 4.91 kJ/kg wPw = wPs / Ps = 6.55 kJ/kg
st 1 Law: q + h1 = h2 + w; q = 0; h2 = h1  wPw = 90.5 kJ/kg C.V. Steam Generator: qw = h3  h2 = 3105.1 kJ/kg Without Cogeneration; C.V. Pump: wPs = v1( P4 P1) = 0.4 kJ/kg wPw/o = wPs / Ps = 0.53 kJ/kg; h2 = h1  wPw/o = 84.5 kJ/kg C.V. Steam Generator: qw/o = h4 h2 = 2664.2 kJ/kg . Additional Heat Transfer: qw  qw/o = 440.9 kJ/kg; Qextra = 4409 kW Difference in Net Power: w diff = (wt + wPw)  wPw/o, wt = h3  h4 = 446.9 kJ/kg . wdiff = 440.9 kJ/kg, Wdiff = 4409 kW 1135 11.43 An industrial application has the following steam requirement: one 10kg/s stream at a pressure of 0.5 MPa and one 5kg/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, 500C 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 lowpressure boiler without cogeneration. Assume that for each, 20C liquid water is pumped to the required pressure and fed to a boiler.
BOILER.
2 3 10 MPa, 500 C . W HPT o . W P P
1 . QH 20 C
o HP TURB. = 0.85 s 4 H 2O IN {1.4 MPa 5 kg/s = 0.85 s
LP TURB.
5 STEAM . W LPT MPa {0.5kg/s 10 STEAM a) highpressure turbine s4S = s3 = 6.5966 T4S = 219.9 oC, h4S = 2852.6 wS HPT = h3  h4S = 3373.7  2852.6 = 521.1 wHPT = SwS HPT = 0.85 521.1 = 442.9 h4 = h3  w = 3373.7442.9 = 2930.8 T4 = 251.6C, s4 = 6.7533 lowpressure turbine s5S = s4 = 6.7533 = 1.8607 + x5S 4.9606, x5S = 0.9863 h5S = 640.23 + 0.9863 2108.5 = 2719.8 wS LPT = h4  h5S = 2930.8  2719.8 = 211.0 wLPT = SwS LPT = 0.85 211.0 = 179.4 kJ/kg 1136 h5 = h4  w = 2930.8  179.4 = 2751.4 > hG OK . WTURB = 15 442.9 + 10 179.4 = 8438 kW . b) wP = 15[0.001002(10000  2.3)] = 150.3 kW h2 = h1  wP = 83.96 + 10.02 = 94.0 . . QH = m1(h3  h2) = 15(3373.7  94.0) = 49196 kW 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 20 oC water.
1 2 3 5 kg/s o 20 C P . Q 56 5 kg/s SAT. VAP. AT 1.4 MPa 10 kg/s o 20 C 4 5 6 P . W P2 10 kg/s SAT. VAP. AT 0.5 MPa h2 = h1  wP = 83.96  0.001002(1400  2.3) = 85.4 . . Q3 = m1(h3  h2) = 5(2790.0  85.4) = 13523 kW 2 . WP1 = 5 14.0 = 7 kW h5 = h4  wP2 = 83.96 + 0.001002(500  2.3) = 84.5 . . Q = m4(h6  h5) = 10(2748.7  84.5) = 26642 kW 5 6 . WP2 = 10 0.5 = 5 kW . Total QH = 13523 + 26642 = 40165 kW 1137 11.44 In a cogenerating steam power plant the turbine receives steam from a highpressure steam drum and a lowpressure steam drum as shown in Fig. P11.44. The condenser is made as two closed heat exchangers used to heat water running in a separate loop for district heating. The hightemperature heater adds 30 MW and the lowtemperature 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. Assume a reversible turbine . . . . . m1h1 + m2h2 = m3h3 + m4h4 + WT . . . m1s1 + m2s2 = mTOTsmix h1 = 3445.9 s1 = 6.9108
3 4 2 1 WT Turbine . h2 = 2855.4 s2 = 7.0592 . mTOT = 27 kg/s sMIX = 6.9383 s3 = sMIX h3 = 2632.4 x3 = 0.966 s4 = sMIX h4 = 2413.5 x4 = 0.899 . WT = 22 3445.9 + 5 2855.4  13 2632.4  14 2413.5 = 22077 kW = 22 MW . . District heating line QTOT = m(h95  h60) = 60935 kW OK, this matches close enough . . . . C.V. Both heaters: m3h3 + m4h4  QTOT = mTOThEX 13 2632.4  14 2413.5  60935 = 7075.2 = 27 hEX hEX = 262 hf TEX = 62.5C 1138 11.45 A boiler delivers steam at 10 MPa, 550C to a twostage 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, 90C to the feedwater line. The remainder of the steam continues through the lowpressure 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? h5 = 3500.9, s5 = 6.7567 4s: s6S = s5 h6S = 2932.1 wT1,S = h5  h6S = 568.8 wT1,AC = 483.5 h6AC = h5  wT1,AC = 3017.4 6ac: P6, h6AC s6AC = 6.9129 7s: s7S = s6AC h7S = 2189.9 wT2,S = h6AC  h7S = 827.5 3
P2 5 T1 Boiler 6 7 4 8 Process heat 5 MW 2 P1 1 C T2 wT2,AC = 622 = h 6AC  h7AC h7AC = 2355.4 8: h8 = 377.6 qPROC = h6AC  h8 = 2639.8 . . . 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 1139 11.46 Consider an ideal airstandard Brayton cycle in which the air into the compressor is at 100 kPa, 20C, and the pressure ratio across the compressor is 12:1. The maximum temperature in the cycle is 1100C, 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.
P 2 s 1 P 3 s s 2 P 4 v
k1 P2 k a) T2 = T1 = 293.2(12)0.286 = 596.8 K P1 T 3 P 4 P s s 1 P1 = 100 kPa T1 = 20 oC P2 P1 = 12 T3 = 1100 oC . m = 10 kg/s wC = 1w2 = CP0(T2  T1) = 1.004(596.8  293.2) = 304.8 kJ/kg
k1 P4 k 1 0.286 T4 = T3 = 1373.2 = 674.7 K 12 P3 wT = CP0(T3  T4) = 1.004(1373.2  674.7) = 701.3 kJ/kg . . . . WC = mwC = 3048 kW, WT = mwT = 7013 kW qH = CP0(T3  T2) = 1.004(1373.2  596.8) = 779.5 kJ/kg TH = wNET/qH = (701.3  304.8)/779.5 = 0.509 b) v4 = RT4/P4 = v2 = RT2/P2 = mep = wNET v4  v2 0.287 674.7 = 1.9364 m3/kg 100 0.287 596.8 = 0.1427 1200 396.5 = 221 kPa 1.9364  0.1427 = Too low for a reciprocating machine (compared with corresponding values for Otto and Diesel cycles.) 1140 11.47 Repeat Problem 11.46, but assume variable specific heat for the air, table A.7. a) From A.7: s 2 = s1 h1 = 293.6, Pr1 = 1.0286 Pr2 = Pr1 (P2/P1) = 1.0286 12 = 12.343 T2 = 590 K, h2 = 597.3 kJ/kg wC = 1w2 = h2  h1 = 597.3  293.6 = 303.7 kJ/kg From A.7: h3 = 1483.3, Pr3 = 333.59 s4 = s3 Pr4 = Pr3 (P4/P3) = 333.59 (1/12) = 27.8 T4 = 735 K, h4 = 751.2 kJ/kg wT = h3  h4 = 1483.2  751.2 = 732 kJ/kg . . . . WC = mwC = 3037 kW, WT = mwT = 7320 kW qH = h3  h2 = 1483.2  597.3 = 885.9 kJ/kg wNET = 732  303.7 = 428.3 kJ/kg TH = wNET/qH = 428.3/885.9 = 0.483 b) v4 = (0.287 735)/100 = 2.1095 m3/kg v2 = (0.287 590)/1200 = 0.1411 mep = 428.3/(2.1095  0.1411) = 217.6 kPa 11.48 An ideal regenerator is incorporated into the ideal airstandard Brayton cycle of Problem 11.46. Find the thermal efficiency of the cycle with this modification.
T 3 s x 2 s 1 s y 4 Problem 11.46 + ideal regen. From 11.46: wT = 701.3, wC = 304.8 kJ/kg wNET = 396.5 kJ/kg Ideal regen.: TX = T4 = 674.7 K qH = h3  hX = 1.004(1373.2  674.7) = 701.3 kJ/kg = wT TH = wNET/qH = 396.5/701.3 = 0.565 1141 11.49 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. Combustion: h3 = h2 + qH; 2w3 = 0 and Tmax = T3 = 1200 K h2 = h3  qH = 1277.8  670 = 607.8 T2 600 K; Pr2 = 13.0923 ; T1 = 300 K; Pr1 = 1.1146 Ideal Compression: P2 / P1 = Pr2 / Pr1 = 11.75 Ideal Expansion: Pr4 = Pr3 / (P3 / P4) = 191.174 / 11.75 = 16.27 linear interpolation T4 636 K, h4 = 645.7 wT = h3  h4 = 1277.8  645.7 = 632.1 wC = h2  h1 = 607.8  300.47 = 307.3 wnet = wT + wC = 632.1  307.3 = 324.8 = wnet / qH = 324.8 / 670 = 0.485 11.50 A large stationary Brayton cycle gasturbine 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?
T T1 = 300 K, P2/P1 = 14, T3 = 1600 K a) Assume const CP0: T2 = T1(P2/P1
k1 )k 3 s 2 4 s2 = s1 = 300(14)
0.286 = 638.1 K s 1 s wC = w12 = h2  h1 = CP0(T2  T1) = 1.004 (638.1  300) = 339.5 kJ/kg Also, s4 = s3 T4 = T3(P4/P3
k1 )k = 1600 (1/14)0.286 = 752.2 K wT = w34 = 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 = 100000/511.7 = 195.4 kg/s . . WT = mwT = 195.4 851.2 = 166.32 MW wC/wT = 339.5/851.2 = 0.399 b) qH = CP0(T3  T2) = 1.004 (1600  638.1) = 965.7 kJ/kg TH = wNET/qH = 511.7/965.7 = 0.530 1142 11.51 Repeat Problem 11.50, but assume that the compressor has an isentropic efficiency of 85% and the turbine an isentropic efficiency of 88%. Same as problem 11.50 , except SC = 0.85 & ST = 0.88 P1 = 100 kPa, T1 = 300 K, P2 = 1400 kPa, T3 = 1600 K a) From solution 11.50: T2S = 638.1 K, wSC = 339.5 kJ/kg wC = wSC/SC = 339.5/0.85 = 399.4 kJ/kg = CP0(T2T1) T2 = T1  wc/CP0 = 300 + 399.4 = 697.8 K 1.004 From solution 11.50: T4S = 752.2 K, wST = 851.2 kJ/kg wT = ST wST = 0.88 851.2 = 749.1 kJ/kg = C P0(T3T4) T4 = T3  wT/CP0 = 1600 749.1 = 853.9 K 1.004 wNET = 749.1  399.4 = 349.7 kJ/kg . . m = WNET/wNET = 100000/349.7 = 286.0 kg/s . . wT = mwT = 286.0749.1 = 214.2 MW wC/wT = 399.4/749.1 = 0.533 b) qH = CP0(T3  T2) = 1.004(1600  697.8) = 905.8 kJ/kg TH = wN/qH = 349.7/905.8 = 0.386 11.52 Repeat Problem 11.51, but include a regenerator with 75% efficiency in the cycle.
T x' 2s 2 x 4s 1 s 3 4 Same as 11.51, but with a regenerator hX  h2 TX  T2 TX  697.8 = = REG = 0.75 = T4  T2 853.9  697.8 ' hX  h2 TX = 814.9 K a) Turbine and compressor work not affected by regenerator. b) qH = CP0(T3  TX) = 1.004(1600  814.9) = 788.2 kJ/kg TH = wNET/qH = 349.7/788.2 = 0.444 1143 11.53 A gas turbine with air as the working fluid has two ideal turbine sections, as shown in Fig. P11.53, 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, Tr2 = 445 h2 = 446.74, wC = 446.74  290.43 = 156.3 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 1646.74  1046.22 = 0.50 1646.74  446.74 h3  h2 . . . WT1 = WC,in wT1 = wC = 156.3 = h 3  h4 x= = 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 h3  hMIX 1144 11.54 The gasturbine cycle shown in Fig. P11.54 is used as an automotive engine. In the first turbine, the gas expands to pressure P , just low enough for this turbine 5 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 P , the 5 net specific work output of the engine, and the mass flow rate through the engine. Find also the air temperature entering the burner T , and the thermal efficiency of 3 the engine.
k1 P2 k a) 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 wT1 = wC = 201.6 = C P0(T4  T5) = 1.004(1600  T5) T5 = 1399.2 K
k1 T5 k 1399.23.5 s5 = s4 P5 = P4 = 600 = 375 kPa 1600 T4 k1 P6 k 1000.286 b) s6 = s5 T6 =T5 = 1399.2 = 958.8 K 375 P5 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 TH = wNET/qH = 442.2/643.8 = 0.687 1145 11.55 Repeat Problem 11.54, but assume that the compressor has an efficiency of 82%, that both turbines have efficiencies of 87%, and that the regenerator efficiency is 70%.
k1 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/SC = 201.6/0.82 = 245.8 kJ/kg = wT1 = CP0(T4  T5) = 1.004(1600  T5) T5 = 1355.2 K wST1 = wT1/ST1 = 245.8/0.87 = 282.5 = CP0(T4  T5S) = 1.004(1600  T5S) s5S = s4 P5 = P4(T5S/T4 b) P6 = 100 kPa, s6S = s5
k1 P6 k 100 0.286 T6S = T5 = 1355.2 = 985.2K 304.9 P5 k k1 ) T5S = 1318.6 K = 600( 1318.6 3.5 ) = 304.9 kPa 1600 wST2 = CP0(T5T6S) = 1.004(1355.2 985.2) = 371.5 kJ/kg wT2 = ST2 wST2 = 0.87 371.5 = 323.2 kJ/kg 323.2 = CP0(T5T6) = 1.004(1355.2 T6) T6 = 1033.3K . . m = WNET/wNET = 150/323.2 = 0.464 kg/s c) wC = 245.8 = C P0(T2  T1) = 1.004(T2 300) T2 = 544.8 K REG = h3  h2 h6  h2 = T 3  T2 = = 0.7 T6  T2 1033.3  544.8 T3  544.8 T3 = 886.8 K qH = CP0(T4  T3) = 1.004(1600 886.8) = 716 kJ/kg TH = wNET/qH = 323.2/716 = 0.451 1146 11.56 Repeat the questions in Problem 11.54 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. 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 P3 = 0.98 600 = 588 kPa, P4 = 0.98 588 = 576.2 kPa s5 = s4 P5 = P4(T5S/T4
k k1 ) k1 = 576.2( 1399.2 3.5 ) = 360.4 kPa 1600 b) P6 = 100/0.98 = 102 kPa, s6S = s5 P6 k 102 0.286 T6 = T5 = 1399.2 = 975.2K 292.8 P5 wST2 = CP0(T5T6) = 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 TH = wNET/qH = 425.7/627.3 = 0.678
k1 1147 11.57 Consider an ideal gasturbine 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 20C, and the temperature entering each turbine is 1100C. An ideal regenerator is also incorporated into the cycle. Determine the compressor work, the turbine work, and the thermal efficiency of the cycle.
10 REG CC
5 6 9 I.C.
1 2 4 COMP COMP TURB
7 TURB
8 CC P2/P1 = P4/P3 = P6/P7 = P8/P9 = 8.0 P1 = 100 kPa T1 = T3 = 20 oC, T6 = T8 = 1100 oC Assume const. specific heat s2 = s1 and s4 = s3 P2 k T4 = T2 = T1 = 293.2(8)0.286 = 531.4 K P 1
k1 T 6 5 7 4 3 2 1 8 9 10 s Total wC = 2 (w12) = 2CP0(T2  T1) = 2 1.004(531.4  293.2) = 478.1 kJ/kg P7 10.286 Also s6 = s7 and s8 = s9: T7 = T9 = T6 k = 1373.2 = 757.6 K P 8 6 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 TH = wNET/qH = 757.4/1235.5 = 0.613
k1 1148 11.58 Repeat Problem 11.57, but assume that each compressor stage and each turbine stage has an isentropic efficiency of 85%. Also assume that the regenerator has an efficiency of 70%. T T1 = T3 = 20 oC, T6 = T8 = 1100 oC 8 6 P2/P1 = P4/P3 = P6/P7 = P8P = 8.0 9 From solution 11.57: T4S = T2S = 531.4 K, wSC = 478.1 kJ/kg T7S = T9S = 757.6 K, wST = 1235.5 kJ/kg wC = wSC/SC = 478.1/0.85 = 562.5 kJ/kg w12 = w34 = 562.5/2 = 281.3 kJ/kg wT = ST wST = 0.85 1235.5 = 1050.2 kJ/kg w67 = w89 = 1050.2/2 = 525.1 kJ/kg T2 = T4 = T1 + (w12 P0) = 293.2 + /C /C T7 = T9 = T6  (+w67 P0) = 1373.2 REG = h5  h4 h9  h4 = T 5  T4 T9  T 4 = 281.3 = 573.5 K 1.004 525.1 = 849.9 K 1.004 = 0.7 T5 = 767 K
5 4 4S 3 2S 1 7S 2 7 9S 9 s T5  573.5 849.9  573.5 qH = CP0(T6  T5) + CP0(T8  T7) = 1.004(1373.2  767.0) + 1.004(1373.2  849.9) = 1133.5 kJ/kg wNET = wT + wC = 1050.2  562.5 = 487.7 TH = wNET/qH = 487.7/1133.5 = 0.430 1149 11.59 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 secondlaw efficiency of the turbine and the cycle thermal efficiency. State 1: P1 = 100 kPa, T1 = 300 K P2 = 5 P1 = 500 kPa;
st 1 Law: q + hi = he + w; q = 0 State 7: P7 = Po = 100 kPa State 3: T3 = 330 K; State 6: T6 = 1500 K, P6 = P4 P4 = 5 P3 = 2500 kPa => wc1 = h1  h2 = CP (T1  T2) wc1 = wc1 s/ = 214.6 Ideal compression T2s = T1 (P2/P1)(k1)/k = 475.4 K wc1 s = CP (T1  T2s) = 176.0 kJ/kg, T2 = T1  wc1/CP = 513.9 K T4s = T3 (P4/P3)(k1)/k = 475.4 K wc2 s = CP (T3  T4s) = 193.6 kJ/kg; T4 = T3  wc2 / CP = 565.2 K Ideal Turbine (reversible and adiabatic) T7s = T6(P7/P6)(k1)/k = 597.4 K => w Ts = CP (T6  T7s) = 905.8 kJ/kg
st 1 Law Turbine: wc2 = 236.1 kJ/kg q + h6 = h7 + w; q=0 wT = h6  h7 = CP (T6  T7) = Ts wTs = 0.86 905.8 = 779.0 kJ/kg T7 = T6  wT/ CP = 1500  779/1.004 = 723.7 K s6  s7 = CP ln T6 T7  R ln P6 P7 = 0.1925 kJ/kg K 6  7 = (h6  h7)  To(s6  s7) = 779.0  298.15(0.1925) = 836.8 kJ/kg 2nd Law = d) wT 67 = 779.0 / 836.8 = 0.931 wnet = wT + wc1 + wc2 = 328.3 kJ/kg w=0 th = qH / wnet ; st 1 Law Combustor: q + hi = he + w; qc = h6  h5 = CP (T6  T5) Regenerator: reg = T 5  T4 T7  T4 = 0.8 > T5 = 692.1 K qH = qc = 810.7 kJ/kg; th = 0.405 1150 11.60 A twostage air compressor has an intercooler between the two stages as shown in Fig. P11.60. 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, see Problem 9.130, 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. Optimal intercooler pressure P2 = 1: h1 = 290.43, Pr1 = 0.9899 C.V.: C1 wC1 = h2  h1, s2 = s1 Pr2 = Pr1(P2/P1) = 3.9596, T2 = 430, h2 = 431.95 wC1 = 431.95  290.43 = 141.5 kJ/kg C.V.Cooler: T3 = T1 h3 = h1 qOUT = h2  h3 = h2  h1 = wC1 = 141.5 kJ/kg C.V.: C2 T3 = T1, s4 = s3 Pr4 = Pr3(P4/P3) = Pr1(P2/P1) = 3.9596, T4 = T2 Thus we get wC2 = wC1 = 141.5 kJ/kg 11.61 Repeat Problem 11.60 when the intercooler brings the air to T = 320 K. The 3 corrected formula for the optimal pressure is P =[ P P (T /T )n/(n1)]1/2 see
2 1 4 3 1 100 1600 = 400 kPa Problem 9.131, where n is the exponent in the assumed polytropic process. 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 1151 11.62 Consider an ideal airstandard Ericsson cycle that has an ideal regenerator as shown in Fig. P11.62. 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. T P P2 = P3 = 1 MPa
2 T 1 P 3 3
T T 4 T1 = T2 = 300 K P1 = 100 kPa q = 4q1 (ideal reg.) P
P 4 v P 2 T 1 s 2 3 qH = 3q4 & wT = qH rp = P2/P1 = 10 TH = CARNOT TH. = 1  TL/TH = 0.7 T3 = T4 = TH = 1000 K P2 1000 = 198.25 qL = wC = v dP = RT1ln = 0.287 300 ln 100 P1 wT = qH = v dP = RT3ln(P4/P3) = 660.8 kJ/kg 11.63 An airstandard Ericsson cycle has an ideal regenerator. Heat is supplied at 1000C and heat is rejected at 20C. 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. See the cycle diagrams in solution to 11.62: T3 = T4 = 1273.15 K, P1 = 70 kPa, T1 = T2 = 293.15 K, qH = 600 kJ/kg Ideal regenerator: 2q3 = 4q1, wT = qH = 600 kJ/kg TH = CARNOT = 1  293.15/1273.15 = 0.7697 wNET = THqH = 0.7697 600 = 461.82 qL = wC = 600  461.82 = 138.2 1152 11.64 Consider an ideal airstandard cycle for a gasturbine, 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.
T 2 BURN 3 3 s4 s 5 P = 90 kPa P 2 COMP 1 TURB 5 4 NOZ
1 s C.V. Comp.: s2 = s1 Pr2 = Pr1 (P2/P1) = 0.9899 14 = 13.8586 h2 = 617.2, T2 = 609 K wC = h2  h1 = 617.2  290.43 = 326.8 kJ/kg C.V. Turb.: wT = h3  h4 = wC h4 = h3 + wC = 1635.8  326.8 = 1309 Pr4 = 209.1, T4 = 1227 P4 = Pr4 (P3/Pr3) = 209.1(1260/483.155) = 545.3 kPa C.V. Nozzle: s2 = s1 Pr5 = Pr4 (P5/P4) = 209.1(90/545.3) = 34.51 => T5 = 778 K, h5 = 798.1 kJ/kg V5 = 2 1000 510.9 = 1010.8 m/s (1/2)V2 = h4  h5 = 510.9 5 1153 11.65 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. C.V. Turb.: hi = 1336.7, Pri = 226, se = si Pre = Pri (Pe/Pi) = 226(250/1500) = 37.667 Te = 796, he = 817.9, wT,s = 1336.7  817.9 = 518.8 wT,AC = wT,s T = 441 = he,AC  hi he,AC = 895.7 Te,AC = 866 K, Pre,AC = 52.21 C.V. Nozzle: (1/2)V2 = hi  he; se = si e Pre = Pri (Pe/Pi) = 52.21(100/250) = 20.884 Te,s = 681 K, he,s = 693.1 kJ/kg
2 (1/2)Ve,s = hi  he,s = 895.7  693.1 = 202.6 kJ/kg 2 2 (1/2)Ve,AC = (1/2)Ve,s NOZ = 192.47 kJ/kg Ve,AC = 2 1000 192.47 = 620 m/s 1154 11.66 Repeat Problem 11.64, but assume that the isentropic compressor efficiency is 87%, the isentropic turbine efficiency is 89%, and the isentropic nozzle efficiency is 96%. Same as 11.64 except SC = 0.87, ST = 0.89, SN = 0.96 Assume const. specific heat P1 = P5 = 90 kPa, T1 = 290 K P2/P1 = 14, T3 = 1500 K From solution 11.64: T2S = 609 K, wSC = 326.8 kJ/kg wC = wSC/SC = 326.8/0.87 = 375.6 kJ/kg wT = wC = 375.6 = h 3  h4 h4 = 1635.8  375.6 = 1260.2, T4 = 1185 K wST = wT/ST = 375.6/0.89 = 422.0 kJ/kg = 1635.8  h4s h4s = 1213.8, T4s = 1145, Pr4s = 158.06 s4S= s3 P4 = P3 Pr4s/Pr3 = 1260158.06/483.155 = 412 kPa s5S = s4 Pr5 = Pr4 P5/P4 = 181.690/412 = 39.67 T5S = 807, h5S = 829.8,
2 V5S/2000 = h4  h5S = 1260.2  829.8 = 430.4 2 => V2/2000 = SN V5s/2000 = 0.96 430.4 = 413.2 kJ/kg 5 T 3 4 5 2S 1 2 5S s V5 = 2000*413.2 = 909 m/s 1155 11.67 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. . . . C.V. Shaft: WT = m(h3  h4) = WC . . h3  h4 = WC / m = 4000/9 = 444.4 h4 = 1757.3 444.4 = 1135.8 wTa = wC = 444.4 wTs = wTa / = 522.82 = h 3  h4s T4s 1163 K h4s = 1234.5 Pr = 168.28, P4 = (Pr4 / Pr3) P3 = (168.28/ 634.967)2000 = 530 kPa 4a: 530 kPa, h = 1312.85, T 1230 K, Pr4 = 211.92 5: 100 kPa Pr5s = Pr4(100/530) = 40 h5s = 830.95
2 0.5V5s = h4a  h5s = 481.9 kJ/kg
2 2 0.5V5a = (0.5V5s) = 457.808
2 V5a= 957 m/s T5a 830 K h5a = h4  0.5V5a = 1312.85 457.808 = 855 1156 11.68 A jet aircraft is flying at an altitude of 4900 m, where the ambient pressure is approximately 55 kPa and the ambient temperature is 18C. 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.
3 4 5 2 1 X T Ambient TX = 18oC = 255.2 K, PX = 55 kPa = P5 also VX = 280 m/s Assume that the air at this state is reversibly decelerated to zero velocity and then enters the comp. at 1.
s P amb P2/P1 = 14 & T3 = 1450 K (280)2 = 294.3 K 2 1000 1.0035 T 1 = TX + VX 2 1000
k 2 = 255.2 + T1k1 294.33.5 = 90.5 kPa P1 = PX = 55 255.2 TX Then, T2 = T1 (P2/P1
k1 )k = 294.3(14)0.286 = 626.0 K T4 = 1118.3 K wC = w12 = CP0(T2T1) = 1.004(1450  T4) P3 = P2 = 14 90.5 = 1267 kPa P4 =
k k1 P3 (T4/T3) k1 )k = 1267(1118.3/1450)3.5 = 510 kPa = 1118.3(55/510)
0.286 T5 = T4 (P5/P4 V2 5 2 1000 = 591.5 K = CP0(T4  T5) = 1.004(1118.3  591.5) = 528.7 kJ/kg V5 = 1028 m/s 1157 11.69 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. Compression 1 to 2: s2 = s1 vr2 = vr1/8 = 179.49/8 = 22.436, T2 = 673 K, u2 = 491.5, Pr2 = 20 P2 = Pr2(P1/Pr1) = 20(95/1.1146) = 1705 kPa Compression 2 to 3: u3 = u2 + qH = 491.5 + 1300 = 1791.5 T3 = 2118 K P3 = P2 (T3/T2) = 1705(2118/673) = 5366 kPa 11.70 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, u2 = 496.94 P2 = P1(Pr2/Pr1) = 90(20.784/0.995) = 1880 kPa
1 2 w = u1  u2 = 207.19  496.94 = 289.75 kJ/kg 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 P4 = P3(Pr4/Pr3) = 4976(57.773/1051) = 273.5 kPa
3 4 w = u3  u4 = 1486.33  665.8 = 820.5 kJ/kg wNET = 3w4 + 1w2 = 820.5  289.75 = 530.8 kJ/kg = wNET/qH = 530.8/989.39 = 0.536 1158 11.71 To approximate an actual sparkignition engine consider an airstandard 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, 10C. 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.
P T s 4 v 1 2 s 1 v s v v 3 s 4 3 v 2 s qH = 1800 kJ v1/v2 = 7 P1 = 90 kPa T1 = 10 oC a) P2 = P1(v1/v2) = 90(7)1.4 = 1372 kPa T2 = T1(P2/P1)(v2/v1)= 283.15 15.245 1 = 616.6 K 7 k T3 = T2 + qH/CV0 = 616.6 + 1800/0.717 = 3127 K P3 = P2T3/T2= 1372 3127 / 616.6 = 6958 kPa b) c) TH = 1  T1/T2 = 1  283.15/616.5 = 0.541 wNET = TH qH = 0.544 1800 = 979.2 kJ/kg v1 = RT1/P1 = (0.287 283.2)/90 = 0.9029 v2 = (1/7) v1 = 0.1290 m3/kg mep = wNET v1v2 = 979.2 = 1265 kPa 0.9029  0.129 1159 11.72 Repeat Problem 11.71, 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). Table A.7 is used with interpolation. a) T1 = 283.15 K, u1 = 202.3, vr1 = 207.94 s2 = s1 vr2 = vr1(v2/v1) = 207.94(1/7) = 29.705 T2 = 606.7 K, u2 = 440.2 => u3 = 440.2 + 1800 = 2240.2 => w12 = u2  u1 = 237.9, vr3 = 0.3402 T3 = 2575.5 K , P3 = 90 7 2575.5 / 283.15 = 5730 kPa b) vr4 = vr3 7 = 2.3814
3w 4 => T 4 = 1437 K; u4 = 1147 = u3  u4 = 2240.2  1147 = 1093.2 wnet = 1093.2  237.9 = 855.3 kJ/kg TH = wnet / qH = 855.3 / 1800 = 0.475 (c) mep = 855.3 / (0.9029  0.129) = 1105 kPa 11.73 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. Standard Otto Cycle T3 = 2050 K u2 = u3  qH T2 = T3  qH / Cvo = 2050  1000 / 0.717 = 655.3 K P2 = P1(T2 / T1)k/(k1) = 90(655.3/290) 3.5 = 1561 kPa CR = v1 / v2 = (T2 / T1)1/(k1) = (655.3 / 290) 2.5 = 7.67 1w2 = u2  u1 = Cvo( T2  T1) = 0.717(655.3  290) = 262 kJ / kg P3 = P2T3 / T2 = 1561 2050 / 655.3 = 4883 kPa 1160 11.74 Answer the same three questions for the previous problem, but use variable heat capacities (use table A.7). T3 = 2050 K u3 = 1725.714 u2 = u3  qH = 1725.714  1000 = 725.714 T2 = 960.56 vr1 = 195.361 => vr2 = 8.216 v1 / v2 = vr1 / vr2 = 195.361 / 8.216 = 23.78 1w2 = u2  u1 = 725.714  207.19 = 518.5 kJ/kg P3=P2T3 / T2 = P1( T3 / T1)( v1 / v3) = 90(2050 / 290)23.78 = 15129 kPa 11.75 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 11.71 using these values. same as 11.71 except v1/v2 = 10 & qH = 1700 kJ a) P2 = P1(v1/v2) = 90(10)1.4 = 2260.7 kPa T2 = T1(v1/v2)
k1 k = 283.15(10)0.4 = 711.2 K T3 = T2 + qH / Cvo = 711.2 + 1700 / 0.7165 = 3084 K P3 = P2(T3 / T2) = 2260.73084 / 711.2 = 9803 kPa TH = 1  T1/T2 = 1  283.15/711.2 = 0.602 wnet = TH qH = 0.6 1700 = 1023.4 kJ/kg v1 = RT1/P1 = 0.287283.15/90= 0.9029, v2 = v1/10= 0.0903 mep = wNET/(v1  v2)= 1023.4 / (0.9029  0.0903) = 1255 kPa 1161 11.76 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 11.71 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 11.71 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 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
1 2 w = Pdv = 11.4(T2  T1) = 0.4 (606.6 283.15)= 239.3 R 0.287 3 4 w = Pdv = R(T4  T3)/(1  1.5) = 0.287(1181.93127)/0.5 = 1116.5 wNET = 1116.5  239.3 = 877.2 CYCLE = wNET/qH = 877.2/1800 = 0.487 mep = wNET/(v1  v2) = 877.2/(0.9029  0.129) = 1133 kPa Note a smaller wNET, CYCLE, mep compared to an ideal cycle. 1162 11.77 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 twothirds of the total qH occurs at constant volume and the last onethird 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, 20C, 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 4 s 2 s 1 5 v 2 s 1 v s 5 v T 3 3 4 s P1 = 90 kPa, T1 = 20oC rV = v1/v2 = 7 a) q23 = (2/3) 2100 = 1400 kJ/kg; q34 = 2100/3 = 700 b) P2 = P1(v1/v2)k = 90(9)1.4 = 1951 kPa T2 = T1(v1/v2)k1 = 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 P4 T1 7350.8 293.15 = = = = 7.131 90 3357 v4 v4 P1 T4 T5 = T4(v4/v5)k1 = 3357(1/7.131)0.4 = 1530 K qL = CV0(T5T1) = 0.717(1530  293.15) = 886.2 kJ/kg TH = 1  qL/qH = 1  886.2/2100 = 0.578 Std. Otto Cycle: TH = 1  (9)0.4 = 0.585, small difference v1 1163 11.78 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. P2 = 95(20) 1.4 = 6297.5 kPa T2 = T1(v1 / v2)k1 = 290200.4 = 961 K P3 = P2(T3 / T2) = 6297.5 (1800 / 961) = 11796 kPa 1w2 = u2  u1 Cvo( T2  T1) = 0.717(961  290) = 481.1 kJ / kg T4 = T3( v3 / v4)0.4 = 1800(1 / 20) 0.4 = 543 K
3w 4 = u3  u4 Cvo(T3  T2) = 0.717(1800  543) = 901.3 kJ / kg wnet = 3w4 + 1w2 = 901.3  481.1 = 420.2 kJ / kg = wnet / qH = 1 (1 / 20) 0.4 = 0.698 ( = 420.2 / 0.717(1800  961)) 11.79 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. Power from mean effective pressure. mep = wnet / (vmax  vmin) > wnet = mep(vmax  vmin) V = Bore2 0.25 S = 0.12 0.25 0.11 = 0.000864 m3 W = mep(Vmax  Vmin) = 1400 0.25 0.12 0.11 = 1.2096 Work per Cylinders per Power Stroke . W = W n RPM 0.5 (cycles / min)(min / 60 s)(kJ / cycle) = 1.2096 6 2000 0.5 (1/60) = 121 kW = 162 hp 1164 11.80 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. P2 = P3 = 7000 kPa => v1 / v2 = (P2/P11/ k = (7000 / 200)0.7143 = 12.67 ) T2 = T1(P2 / P1)(k1) / k = 300(7000 / 200) 0.2857 = 828.4 v3 / v2 = T3 / T2 = 1500 / 828.4 = 1.81 v4 / v3 = v1 / v3 = (v1 / v2)( v2 / v3) = 12.67 / 1.81 = 7 T4 = T3(v3 / v4)k1 = (1500 / 7) 0.4 = 688.7 qL = Cvo(T4  T1) = 0.717(688.7  300) = 278.5 qH = h3  h2 Cpo(T3  T2) = 1.004(1500  828.4) = 674 = 1  qL / qH = 1 278.5 / 674 = 0.587 wnet = qnet = 674  278.5 = 395.5 vmax = v1 = R T1 / P1 = 0.287300 / 200 = 0.4305 m3 / kg vmin = vmax / (v1 / v2) = 0.4305 / 12.67 = 0.034 m3 / kg mep = wnet / (vmax  vmin) = 395.5 / (0.4305  0.034) = 997 kPa 11.81 Consider an ideal airstandard 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%? Diesel cycle: T2 = T1(v1/v2) P1 = 95 kPa,
k1 T1 = 290 K, v1/v2 = 20, TH = 0.6 = 290(20)0.4 = 961.2 K v2 = 0.876 / 20 = 0.0438 v1 = 0.287 290/95 = 0.876 = v4, v3 = v2(T3/T2) = 0.043883(T3/961.2) = 0.0000456 T3 T3 = T4 (v4/v3)
k1 0.876 =( )0.4 T4 = 0.019345 T1.4 3 0.0000456 T3 T 4  T1 k(T3  T2) =10.019345 T1.4  290 3 1.4(T3  961.2) TH = 0.60 = 1  0.019345 T1.4  0.56 T3 + 248.272 = 0 3 3050: LHS = +1.06 T3 = 3040 K 3040: LHS = 0.036, 1165 11.82 Consider an ideal Stirlingcycle engine in which the state at the beginning of the isothermal compression process is 100 kPa, 25C, the compression ratio is 6, and the maximum temperature in the cycle is 1100C. Calculate the maximum cycle pressure and the thermal efficiency of the cycle with and without regenerators.
P 3 3 v 2 T T 4 v 1 v 2 T 1 s v v T T 4 Ideal Stirling cycle T1 = T2 = 25 oC P1 = 100 kPa v1/v2 = 6 T3 = T4 = 1100 oC a) T1 = T2 P2 = P1(v1/v2) = 100 6 = 600 kPa V2 = V3 P3 = P2T3/T2 = 6001373.2/298.2 = 2763 kPa b) w34 = q34 = RT3 ln(v4/v3) = 0.2871373.2ln6 = 706.1 kJ/kg q23 = CV0(T3  T2) = 0.7165(1100  25) = 770.2 kJ/kg w12 = q12 = RT1 ln(v1/v2) = 0.287 298.2 ln(6) = 153.3 kJ/kg wNET = 706.1  153.3 = 552.8 kJ/kg NO REGEN = 552.8 = 0.374, 706.1+770.2 WITH REGEN = 552.8 = 0.783 706.1 11.83 An airstandard Stirling cycle uses helium as the working fluid. The isothermal compression brings helium from 100 kPa, 37C 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 , Cvo = 3.1156 Compression/expansion: v4 / v3 = v1 / v2 = P2 / P1 = 600 / 100 = 6 1 > 2 1w2 = q12 = P dv = R T1ln(v1 / v2) = RT1ln (P /P1) 2 = 2.077 310 ln6 = 1153.7 kJ/kg 2 > 3 : 2w3 = 0; 3 > 4: q23 = Cvo(T3  T2) = 3.1156(1200  310) = 2773 kJ/kg v4 w4 = q34 = R T3ln = 2.0771200 ln6 = 4465.8 kJ/kg 3 v3 q41 = Cvo(T4  T1) = 2773 kJ/kg 4 > 1 4w1 = 0; cycle = (1w2+3w4)/(q23+q34) = (1153.7+4465.8) /(2773 +4465.8) = 0.458 1166 11.84 Consider an ideal airstandard Stirling cycle with an ideal regenerator. The minimum pressure and temperature in the cycle are 100 kPa, 25C, the compression ratio is 10, and the maximum temperature in the cycle is 1000C. 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 solution 11.82, with P1 = 100 kPa, T1 = T2 = 25oC,
1 2 v1/v2 = 10, o T3 = T4 = 1000 C From 12 at const T: w = 1q2 = T1(s2  s1) = RT1ln(v1/v2) = 0.287 298.2 ln(10) = 197.1 kJ/kg From 23 at const V:
2 3 w =0 / q23 = CV0(T3  T2) = 0.7165(1000  25) = 698.6 From 34 at const T; = +RT3 ln From 41 at const V;
3 4 w = 3q4 = T3(s4  s3) = 0.287 1237.2 ln(10) = 841.4 kJ/kg w =0 / v4 v3 4 1 q41 = CV0(T1  T4) = 0.7165(25  1000) = 698.6 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, NOTE: TH = wNET qH = 644.3 = 0.766 841.4 qH = q34 = RT3 ln(10), qL = q12 = RT1 ln(10) TH = (qH  qL)/qH = (T3  T1)/T = (975/1273.2) = 0.766 = Carnot efficiency 3 1167 11.85 The airstandard Carnot cycle was not shown in the text; show the Ts diagram for this cycle. In an airstandard 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. = 0.6 = 1  TH/TL TH = TL/0.4 = 700 K P1 = 100 kPa
1 T H T L T 2 3 P2 = P1(TH/TL)k1 = 2.47 MPa [or P2 = P1(Pr2/Pr1) = 2.645 MPa] 1 4 s 11.86 Air in a piston/cylinder goes through a Carnot cycle in which T = 26.8C and the L total cycle efficiency is = 2/3. Find T , the specific work and volume ratio in H the adiabatic expansion for constant C , C . Repeat the calculation for variable p v heat capacities. Carnot cycle: Same Ts diagram as in previous problem. = 1  TL/TH = 2/3 TH = 3 TL = 3 300 = 900 K Adiabatic expansion 3 to 4:
3 4 Pvk = constant R (T  T ) = u3  u4 1k 4 3 w = (P4v4  P3v3/(1  k) = = Cv(T3  T4) = 0.7165(900  300) = 429.9 kJ/kg v4/v3 = (T3/T4)1/(k  1) = 32.5 = 15.6 For variable Cp, Cv we get, TH = 3 TL = 900 K
3 4 w = u3  u4 = 674.824  214.364 = 460.5 kJ/kg v4/v3 = vr4/vr3 = 179.49/9.9169 = 18.1 1168 11.87 Consider an ideal refrigeration cycle that has a condenser temperature of 45C and an evaporator temperature of 15C. Determine the coefficient of performance of this refrigerator for the working fluids R12 and R22.
T Ideal Ref. Cycle Tcond = 45 oC = T3 Tevap = 15 oC 2 3 1 s 4 h1, kJ/kg s 2 = s1 P2, MPa T2, oC h2, kJ/kg h3=h4, kJ/kg wC = h2h1 qL = h1h4 =qL/wC R12 180.97 0.7051 1.0843 54.7 212.63 79.71 31.66 101.26 3.198 R22 244.13 0.9505 1.729 74.4 289.26 100.98 45.13 143.15 3.172 11.88 The environmentally safe refrigerant R134a is one of the replacements for R12 in refrigeration systems. Repeat Problem 11.87 using R134a and compare the result with that for R12. T h1 = 389.2, s2 = s1 = 1.7354 h3 = 264.11, P3 = P2 = 1.16 MPa At 1 MPa, T2 = 45.9, h2 = 426.8 At 1.2 MPa,T2 = 53.3, h2 = 430.7 T2 = 51.8, h2 = 429.9 wC = h2  h1 = 429.9  389.2 = 40.7 kJ/kg qL = h1  h4 = h1  h3 = 389.2  264.11 = 125.1 kJ/kg = qL/(wC) = 125.1/40.7 = 3.07
45 C 15 C
o o 2 3 4 1 s 1169 11.89 A refrigerator using R22 is powered by a small natural gas fired heat engine with a thermal efficiency of 25%. The R22 condenses at 40C and it evaporates at 20C 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? Evaporator: Inlet State is sat. liqvap with h4 = h3 =94.272 kJ / kg The exit state is sat. vap. with h = 242.055 kJ / kg qL = h1  h4 = h1  h3 = 147.78 kJ / kg Compressor: Inlet State 1 and Exit State 2 about 1.6 MPa wC = h2  h1 ; 2: T2 70C s2 = s1 = 0.9593 kJ / kg K h2 = 287.17 kJ / kg wC = h2  h1 = 45.11 kJ / kg Condenser: Brings it to sat liq at state 3 qH = h2  h3 = 287.17  94.272 = 192.9 kJ / kg Overall Refrigerator: = qL / wC = 147.78 / 45.11 = 3.276 Heat Engine: . . . . WHE = HEQ1 = WC = QL / . . QL / Q1 = = 0.253.276 = 0.819 11.90 A refrigerator with R12 as the working fluid has a minimum temperature of 10C and a maximum pressure of 1 MPa. Assume an ideal refrigeration cycle as in Fig. 11.32. Find the specific heat transfer from the cold space and that to the hot space, and the coefficient of performance. h1 = 183.058, s1 = 0.7014, h3 = 76.155 s2 = s1 & P2 h2 210.16 qL = h1  h4 = h1  h3 = 183.058  76.155 = 106.9 kJ/kg qH = h2  h3 = 210.16  76.155 = 134 kJ/kg b = qL/(wc) = qL/(qH  qL) = 3.945 1170 11.91 A refrigerator in a meat warehouse must keep a low temperature of 15C and the outside temperature is 20C. It uses R12 as the refrigerant which must remove 5 kW from the cold space. Find the flow rate of the R12 needed assuming a standard vapor compression refrigeration cycle with a condenser at 20C. Basic refrigeration cycle: T1 = T4 = 15C, T3 = 20C Table B.3: h4 = h3 = 54.27 ; h1 = hg = 180.974 . . . QL = mR12 qL = mR12(h1  h4) qL = 180.974  54.874 = 126.1 kJ/kg . mR12 = 5.0 / 126.1 = 0.03965 kg / s 11.92 A refrigerator with R12 as the working fluid has a minimum temperature of 10C and a maximum pressure of 1 MPa. The actual adiabatic compressor exit temperature is 60C. 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. State 1: Inlet to compressor, sat. vap. 10C, h1 = 183.058, s1 = 0.7014 State 2: Actual compressor exit, h2AC = 217.81 State 3: Exit condenser, sat. liq. 1MPa, h3 = 76.155 State 4: Exit valve, h4 = h3 C.V. Evaporator: qL = h1  h4 = h1  h3 = 106.9 kJ/kg C.V. Ideal Compressor: wC,S = h2,S  h1, s2,S = s1 State 2s: T2,S = 49.66, h2,S = 209.9, wC,S = 26.842 C.V. Actual Compressor: wC = h2,AC  h1 = 34.752 kJ/kg = qL/wC = 3.076, C = wC,S/wC = 0.7724 C.V. Condenser: qH = h2,AC  h3 = 141.66 kJ/kg 1171 11.93 Consider an ideal heat pump that has a condenser temperature of 50C and an evaporator temperature of 0C. Determine the coefficient of performance of this heat pump for the working fluids R12, R22, and ammonia.
T Ideal Heat Pump: Tcond = 50 oC = T3 Tevap = 0 oC 2 3 1 s 4 R12 h1, kJ/kg s 2 = s1 P2, MPa T2, oC h2, kJ/kg h3=h4, kJ/kg wC = h2h1 qH = h2h3 =qH/wC 187.53 0.6965 1.2193 56.7 211.95 84.94 24.42 127.01 5.201 R22 249.95 0.9269 1.9423 72.2 284.25 107.85 34.3 176.4 5.143 NH3 1442.32 5.3313 2.0333 115.6 1672.84 421.58 230.52 1251.26 5.428 11.94 The air conditioner in a car uses R134a and the compressor power input is 1.5 kW bringing the R134a from 201.7 kPa to 1200 kPa by compression. The cold space is a heat exchanger that cools atmospheric air from the outside down to 10C and blows it into the car. What is the mass flow rate of the R134a and what is the low temperature heat transfer rate. How much is the mass flow rate of air at 10C? Standard Refrigeration Cycle Table B.5 h1 = 392.285; s1 = 1.7319 ; h4 = h3 = 266 C.V. Compressor (assume ideal) . . m1 = m2 wC = h2  h1; s2 = s1 + sgen P2, s = s1 > h2 = 429.5 > wC = 37.2 . . . 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 . . . mairhair = QL mairCpT . . mair = QL / (CpT) = 5.21 / (1.00420) = 0.26 kg / s 1172 11.95 A refrigerator using R134a is located in a 20C room. Consider the cycle to be ideal, except that the compressor is neither adiabatic nor reversible. Saturated vapor at 20C enters the compressor, and the R134a exits the compressor at 50C. The condenser temperature is 40C. 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. Table B.5: P2 = P3 = Psat 40C = 1017 kPa, s2 1.7472 , h2 430.87 ; h4 = h3 = 256.54 kJ/kg s1 = 1.7395 , h1 = 386.08 = qL / wC > wC = qL / = (h1 h4) / = (386.08  256.54) / 2.3 = 56.32 . . WC = m wC = 11.26 kW C.V. Compressor h1 + wC + q = h 2 > qin = h2  h1  wC = 430.87  386.08  56.32 = 11.53 s1 + dQ/T + sgen = s2 sgen = s2  s1  q / To = 1.7472  1.7395 + (11.53 / 293.15) = 0.047 . . Sgen = m sgen = 0.2 0.047 = 0.0094 kW / K 11.96 A small heat pump unit is used to heat water for a hotwater supply. Assume that the unit uses R22 and operates on the ideal refrigeration cycle. The evaporator temperature is 15C and the condenser temperature is 60C. 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 60C. T Ideal R22 heat pump o o T1 = 15 C, T3 = 60 C 2 h1 = 255.02, s2 = s1 = 0.9062 P2 = P3 = 2.427 MPa, h3 = 122.18 T2 = 78.4 oC, h2 = 282.86 wC = h2  h1 = 27.84 qH = h2  h3 = 160.68 To heat 0.1 kg/s of water from 15 oC to 60 oC, . . 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
4 1 s 3 i.e. a heat loss 1173 11.97 The refrigerant R22 is used as the working fluid in a conventional heat pump cycle. Saturated vapor enters the compressor of this unit at 10C; its exit temperature from the compressor is measured and found to be 85C. If the isentropic efficiency of the compressor is estimated to be 70%, what is the coefficient of performance of the heat pump? T R22 heat pump: 2 o TEVAP = 10 C, S COMP = 0.70
2S T2 = 85 C Isentropic compressor: s2S = s1 = 0.9129 but P2 unknown. Trial & error. Assume P2 = 2.11 MPa o 3 4 1 s At P2,s2S: T2S = 72.1oC, h2S = 281.59, At P2,T2 : h2 = 293.78 calculate S COMP = (h2S  h1)/(h2  h1) = 281.59  253.42 = 0.698 0.70 293.28  253.42 OK P2 = 2.11 MPa = P3 T3 = 53.7 oC, h3 = 112.99, wC = h2  h1 = 40.36 qH = h2  h3 = 180.79, = qH/wC = 4.48 1174 11.98 In an actual refrigeration cycle using R12 as the working fluid, the refrigerant flow rate is 0.05 kg/s. Vapor enters the compressor at 150 kPa, 10C, and leaves at 1.2 MPa, 75C. The power input to the compressor is measured and found be 2.4 kW. The refrigerant enters the expansion valve at 1.15 MPa, 40C, and leaves the evaporator at 175 kPa, 15C. Determine the entropy generation in the compression process, the refrigeration capacity and the coefficient of performance for this cycle. Actual ref. cycle 2 T P1 = 0.15 MPa, P2 = 1.2 MPa 1: comp. inlet T1 = 10 oC, T2 = 75 oC
3 4 5 1 s P3 = 1.15 MPa, P5 = 0.175 MPa T3 = 40 oC, T5 = 15 oC 5: evap. exit . . WCOMP = 2.4 kW in, m = 0.05 kg/s Table B.3 h1 = 184.619, s1 = 0.7318, h2 = 226.543, s2 = 0.7404 s1 + dq/T + sgen = s2 CV Compressor: h1 + qCOMP = h2 + wCOMP ; wCOMP = 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 b) qL = h5  h4 = 181.024  74.527 = 106.5 kJ/kg . . QL = mqL = 0.05 106.5 = 5.325 kW = qL/wCOMP = 106.5/48.0 = 2.219 c) 11.99 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 50C, and saturated vapor leaves the evaporator at 10C. 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.
T NH3 absorption cycle:
o sat. vapor at 50 oC exits the generator 50 C o sat. vapor at 10 C exits the evaporator qH = qGEN = 7000 kJ/kg NH3 out of gen. 10 oC GEN. EXIT EVAP EXIT s 1 2 qL = h2  h1 = hG 10oC  hF 50oC = 1452.2  421.6 = 1030.6 kJ/kg qL/qH = 1030.6/7000 = 0.147 1175 11.100 The performance of an ammonia absorption cycle refrigerator is to be compared with that of a similar vaporcompression system. Consider an absorption system having an evaporator temperature of 10C and a condenser temperature of 50C. The generator temperature in this system is 150C. In this cycle 0.42 kJ is transferred to the ammonia in the evaporator for each kilojoule transferred from the hightemperature source to the ammonia solution in the generator. To make the comparison, assume that a reservoir is available at 150C, and that heat is transferred from this reservoir to a reversible engine that rejects heat to the surroundings at 25C. This work is then used to drive an ideal vaporcompression system with ammonia as the refrigerant. Compare the amount of refrigeration that can be achieved per kilojoule from the hightemperature source with the 0.42 kJ that can be achieved in the absorption system.
T H = 50 C T ' = 150 C H Q ' = 1 kJ H REV. H.E.
' QL o o T 3 3 1 s 2 QH COND. 2 COMP. WC 4 EVAP. 1 QL T L = 10 C
o T1 = 10 C
4 o T L'= 25 C o h1 = 1430.8 , s1 = 5.4673 h4 = h3 = 421.48 For the rev. heat engine: 298.2 TH = 1  TL/TH = 1 = 0.295 423.2 WC = TH Q = 0.295 kJ H For the NH3 refrig. cycle: s2 = s1 = 5.4673 P2 = P3 = 2033 kPa , => T2 135C Use 2000 kPa Table h2 1724 wC = h2  h1 = 1724  1430.8 = 293.2 qL = h1  h4 = 1430.8  421.48 = 1009.3 = qL/wC = 1009.3 / 293.2 = 3.44 QL = wC = 3.44 0.295 = 1.015 kJ based on assumption of ideal heat engine & refrig. cycle. 1176 11.101 A heat exchanger is incorporated into an ideal airstandard refrigeration cycle, as shown in Fig. P11.101. 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.
q 6 1 COMP
L T
3 2 2 qH 3 4 EXP 5 4 6 5 1 s T1 = T3 = 15 oC = 288.2 K, P1 = 100 kPa, P2 = 1.4 MPa T4 = T6 = 50 oC = 223.2 K, s2 = s1 T2 = T1(P2/P1
k1 )k = 288.2(1400/100)0.286 = 613 K wC = w12 = CP0(T2  T1) = 1.0035(613  288.2) = 326 kJ/kg s5 = s4 T5= T4(P5/P4) k = 223.2(100/1400)
k1 0.286 = 104.9 K wE = CP0(T4  T5) = 1.0035(223.2  104.9) = 118.7 kJ/kg wNET = 118.7  326.0 = 207.3 kJ/kg qL = CP0(T6  T5) = 1.0035(223.2  104.9) = 118.7 kJ/kg = qL/wNET = 118.7/207.3 = 0.573 11.102 Repeat Problem 11.101, but assume an isentropic efficiency of 75% for both the compressor and the expander. 2 From solution 11.101 : T2S = 613 K, wSC = 326 kJ/kg 2S T T5S = 104.9 K, wSE = 118.7 wC = wSC / SC = 326/0.75 = 434.6 kJ/kg wE = SE wSE = 0.75 118.7 = 89.0 kJ/kg = CP0(T4T5) = 1.004(223.2 T5) T5 = 134.5 K wNET = 89.0  434.6 = 345.6 kJ/kg qL = CP0(T6  T5) = 1.004(223.2  134.5) = 89.0 kJ/kg = qL/(wNET) = 89.0/345.6 = 0.258
3 4 6 5S 5 1 s 1177 11.103 Repeat Problems 11.101 and 11.102, but assume that helium is the cycle working fluid instead of air. Discuss the significance of the results. a) Problem 11.101 , except for Helium 14000.40 100 0.40 T2 = 288.2 = 828.2 K, T5 = 223.2 = 77.7 K 100 1400 wC = w12 = 5.1926(828.2  288.2) = 2804.1 kJ/kg wE = 5.1926(223.2  77.7) = 755.5 kJ/kg wNET = 755.5  2804.1 = 2048.6 kJ/kg qL = 5.1926(223.2  77.7) = 755.5 kJ/kg = 755.5/2048.6 = 0.369 b) Problem 11.102 , except for Helium From part a): T2S = 828.2 K, wSC = 2804.1 T5S = 77.7 K, wSE = 755.5 wC = wSC/SC = 2804.1/0.75 = 3738.8 kJ/kg wE = SE wSE = 0.75 755.5 = 566.6 kJ/kg wNET = 566.6  3738.8 = 3172.2 kJ/kg wE = 566.6 = 5.1926(223.2  T5) T5 = 114.1 K qL = 5.1926(223.2  114.1) = 566.6 kJ/kg = 566.6/3172.2 = 0.179 1178 11.104 A binary system power plant uses mercury for the hightemperature cycle and water for the lowtemperature cycle, as shown in Fig. 11.39. The temperatures and pressures are shown in the corresponding Ts diagram. The maximum temperature in the steam cycle is where the steam leaves the superheater at point 4 where it is 500C. 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 0.04 1.60 Tg, C 309 562 hf, kJ/kg hg, kJ/kg sf kJ/kgK 42.21 75.37 335.64 364.04 0.1034 0.1498 sg, kJ/kgK 0.6073 0.4954 a) For the mercury cycle: 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 = b) 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 TH = wNET/qH = 2277.8/4306.8 = 0.529 1179 11.105 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 500C. The available hightemperature source is the exhaust of 175 kg/s air at 600C from a gas turbine. If the boiler operates as a counterflowing heat exchanger where the temperature difference at the pinch point is 20C, find the maximum water mass flow rate possible and the air exit temperature. T wP = h2  h1 = v1(P2  P1) = 0.00101(3000  10) = 3.02 h2 = h1  wP = 191.83 + 3.02 = 194.85 h3 = 3456.5, hair,in = 903.16 State 2a: T2a = TSAT = 233.9 C h2a = 1008.42 hair(T2a + 20) = 531.28 Air temperature should be 253.9C at the point where the water is at state 2a. C.V. Section 2a3, ia . . mH O(h3  h2a) = mair(hi  ha)
2 3 2a 2 1 s e a i HEAT EXCH 2 2a 3 . 903.16  531.28 mH O = 175 = 26.584 kg/s 3456.5  1008.42 2 . . Take C.V. Total: mH O(h3  h2) = mair(hi  he) 2 . . he = hi  mH O(h3  h2)/mair
2 = 903.6  26.584(3456.5  194.85)/175 = 408.13 Te = 406.7 K = 133.6 C, Te > T2 = 46.5 C OK. 11.106 A simple Rankine cycle with R22 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 R22 boiler. Diesel inlet conditions are 100 kPa, 20C, the compression ratio is 20, and the maximum temperature in the cycle is 2800C. Saturated vapor R22 leaves the bottoming cycle boiler at 110C, and the condenser temperature is 30C. 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 200C in the R22 boiler. 1180
T v 2 s 1 v 3 s 4 DIESEL AIRSTD CYCLE s 6 5 8 s 7 T IDEAL R12 RANKINE BOTTOMING CYCLE P1 = 100 kPa, T1 = 20 oC, T7 = 110 oC, x7 = 1.0 . v1/v2 = 20, T3 = 2800oC, T5 = T8 = 30oC, WDIESEL = 1.0 MW a) T2 = T1(v1/v2)k1 = 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 v3k1 0.133 070.4 = 1469.6 K T4 = T3 = 3073.2 0.8415 v4 qL = 0.717(293.2  1469.6) = 843.5 wNET = 2109.8  843.5 = 1266.3 kJ/kg . . mAIR = WNET/wNET = 1000/1266.3 = 0.79 kg/s b) s8 = s7 = 0.60758 = 0.2399 + x8 0.4454, x8 = 0.8255 h8 = 64.59 + 0.8255 135.03 = 176.1 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 qH = h7  h6 = 198.0  67.1 = 130.9 kJ/kg . QH available from Diesel exhaust cooled to 200 oC: . QH = 0.79 0.717(1469.6  473.2) = 564 kW . . mR12 = QH/qH = 564/130.9 = 4.309 kg/s . WR12 = 4.309(21.9  2.5) = 83.6 kW 1181 11.107 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 lowtemperature 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 = T 3m , TLN = 75 K = T 4N = T1N TLm = T4m = T1m = T3N  10, Trial and error on T3N or TLm. For each cycle we have, wC = h2  h1, s2 = s1, qH = h2  h3, qL = h1  h4 = h1  h3 Nitrogen: T4 = T1 = 75 K h1 = 74.867 s1 = 5.4609 N2 a) b) c) T3 120 115 110 h3 17.605 34.308 48.446 P2 2.5125 1.9388 1.4672 h2 202.96 188.35 173.88 wc 128.1 113.5 99.0 qH 220.57 222.66 222.33 qL 92.47 109.18 123.31 Methane: T3 = 180 K h3 = 0.5 P2 = 3.28655 MPa CH4 a) b) c) T4 110 105 100 h1 221 212.2 202.9 s1 9.548 9.691 9.851 h2 540.3 581.1 629.7 wc 319.3 368.9 426.8 qH 540.8 581.6 630.2 qL 221.5 212.7 203.4 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) . . . . = QL 75 K/Wtot in = qLn/[wcn (mm/mn)wcm] Case a) b) c) . . mm/mn 0.996 1.047 1.093 . . wcn+(mm/mn)wcm 446.06 499.65 565.49 0.207 0.219 0.218 A maximum coeff. of performance is between case b) and c). 1182 11.108 A cascade system is composed of two ideal refrigeration cycles, as shown in Fig. 11.41. The hightemperature cycle uses R22. Saturated liquid leaves the condenser at 40C, and saturated vapor leaves the heat exchanger at 20C. The lowtemperature cycle uses a different refrigerant, R23 (Fig. G.3 or the software). Saturated vapor leaves the evaporator at 80C, and saturated liquid leaves the heat exchanger at 10C. Calculate the ratio of the mass flow rates through the two cycles and the coefficient of performance of the system.
2' COND C R22 3' T3 = 40o C ' x '= 0.0
3 T1 = 20 C 1' ' x1 = 1.0 '
2 o 4' 3 o T3 = 10 C x = 0.0 3 T1 = 80o C 1 x1 = 1.0
T ' 2 ' 3 C R23 EVAP
T 2 3 1 s 4 ' 4 ' 1 s 4 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 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 s 1.76 1.76 h  h 242.1  94.3 1 4 . . a) m/m = = = 0.672 405  185 h2  h3 . . b) QL/m = h1  h4 = 330  185 = 145 . . . .  WTOT/m = (h2  h1) + (m/m)(h  h ) 2 1 = (405  330) + (1/0.672)(289  242.1) = 144.8 . = QL/WTOT = 145/144.8 = 1.0 1183 11.109 Consider an ideal dualloop heatpowered refrigeration cycle using R12 as the working fluid, as shown in Fig. P11.109. Saturated vapor at 105C leaves the boiler and expands in the turbine to the condenser pressure. Saturated vapor at  15C 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 45C 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 Q /Q L H.
T 1 TURB. 6 BOIL. 5 P 7 COND. 3 COMP. 2 E V A P . 4 . QL 6 5 3 4 7 1 s 2 T3 = 45 C sat. liq. T1 = 15 oC sat. vap. o P5 = P6 = 3.6509 MPa; T6 = 105 oC sat. vap.; P2 = P3 = P7 = 1.0843 MPa h1 = 180.846; h3 = h4 = 76.647; h6 = 206.099 s7 = s6 = 0.6319 = 0.2875 + x7 0.3931; x7 = 0.8761 h7 = 79.647 + 0.8761 125.074 = 189.226 s2 = s1 = 0.7046, P2 => T2 = 54.7 oC, h2 = 212.401 a) CV: turbine + compressor . . . . . . . . Cont: m1 = m2, m6 = m7 ; 1st law: m1h1 + m6h6 = m2h2 + m7h7 . . m6/m1 = (212.401  180.846)/(206.099  189.226) = 1.870 b) CV: pump wP = v3(P5  P3) = 0.000811(3651  1084) = 2.082 kJ/kg h5 = h3  wP = 81.729 . . CV: evaporator QL = m1(h1  h4) . . CV: boiler QH = m6(h6  h5) . . QL m1(h1  h4) 180.846  79.647 = = 0.435 = . = . QH m6(h6  h5) 1.87(206.099  81.729) 1184 11.110 Find the availability of the water at all four states in the Rankine cycle described in Problem 11.12. Assume that the hightemperature source is 500C and the lowtemperature reservoir is at 25C. 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? Reference State: 100 kPa, 25C, s0 = 0.3674, h0 = 104.89 1 = h1  h0  T0(s1  s0) = 191.83  104.89  298.15(0.6493  0.3674) = 2.89 2 = 195.35  104.89  298.15(0.6493  0.3674) = 1 + 3.525 = 6.42 3 = 3222.3  104.89  298.15(6.8405  0.3674) = 1187.5 4 = 3  wT,s = 131.96 H = (1  T0/TH)qH = 0.6144 3027 = 1859.7 L = (1  T0/T0)qC = 0 / II = wNET/H = (1055.5  3.53)/1859.7 = 0.5657 Notice TH > T3, TL < T4 = T1 so cycle is externally irreversible. Both q H and qC over finite T. 1185 11.111 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, 600C, 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 0.00101(20000  10) = 20.2 kJ/kg h2 = h1 + wP = 191.8 + 20.2 = 212.0 s4 = s3 = 6.5048 = 0.6493 + x 4 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 TH = wN qH = 1457.7 = 0.438 3325.6
1 ST. GEN. TURBINE. 4 COND.
2 P 1 T 20 MPa o 3 600 C 2 1 10 kPa 4 s b) one feedwater heater wP12 = 0.00101(1000  10) = 1.0 kJ/kg h2 = h1 + wP12 = 191.8 + 1.0 = 192.8 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 5 ST. GEN.
6 TURBINE.
7 HTR.
3 COND.
1 2 P
4 P 1186 x6 = 0.9816 h6 = 762.8 + 0.9816 2015.3 = 2741.1 CV: heater const: m3 = m6 + m2 = 1.0 kg 1st law: m6h6 + m2h2 = m3h3 m6 = 762.8  192.8 = 0.2237 2741.1  192.8
4 2 3 1 20 MPa 5 600 C 1 MPa 6 7 s 10 kPa
o T 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 TH = 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
6 7 ST. GEN. TURBINE.
8 9 10 HP HTR
5 LP HTR
1 3 COND. P
4 P
2 P 1187 wP56 = 0.001217(20000  3000) = 20.7 kJ/kg h6 = h5 + wP56 = 1008.4 + 20.7 = 1029.1 s8 = s7 = 6.5048 T8 = 293.2 oC at P8 = 3 MPa h8 = 2974.8 s9 = s8 = 6.5048 = 1.5301 + x9 5.5970
T
o 600 C 6 4 5 2 3 1 80 MPa 7 3 MPa 8 0.2 MPa 10 kPa 9 10 s x9 = 0.8888 => h9 = 504.7 + 0.888 2201.9 = 2461.8 CV: high pressure heater cont: m8 = cont: m9 = m5 = m4 + m8 = 1.0 kg ; 1008.4  507.7 = 0.2030 2974.8  507.7 m9 + m2 = m3 = m4 ; 1st law: m5h5 = m4h4 + m8h8 m4 = 0.7970 1st law: m9h9 + m2h2 = m3h3 CV: low pressure heater 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 TH = wN/qH = 1224.8/2508.5 = 0.488 1188 11.112 Find the availability of the water at all the states in the steam power plant described in Problem 11.36. Assume the heat source in the boiler is at 600C and the lowtemperature reservoir is at 25C. Give the second law efficiency of all the components. From solution to 11.1 and 11.36 : States 1: 3: 4a: 1  2a  3  4a  1 hf = 191.81, h3 = 2803.43, h4a = 2068.2, sf = 0.6492 s3 = 6.1252 s4a = 6.5313, x4a = 0.7842 2a: h2a = 196.2, s2a 0.6628 = h  ho  To(s  so) ho = 104.89, so = 0.3674 1= 191.81  104.89  298.15(0.6492  0.3674) = 2.90 2a = 196.2  104.89  298.15(0.6628  0.3674) = 3.24 3 = 2803.43  104.89  298.15(6.1252  0.3674) = 981.85 4a = 2068.2  104.89  298.15(6.5313  0.3674) = 125.54 II Pump = (2a  1) / wp ac = (3.24  2.9) / 4.406 = 0.077 II Boiler = (3  2a) / [(1 To/TH) qH] = (981.85  3.24) / [0.6582607.2] = 0.57 II Turbine = wT ac / (3  4a) = 735.2 / (981.85  125.54) = 0.859 II Cond = amb / (4a  1) = 0 1189 11.113 The power plant shown in Fig. 11.40 combines a gasturbine cycle and a steamturbine cycle. The following data are known for the gasturbine cycle. Air enters the compressor at 100 kPa, 25C, 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 1250C, the exhaust pressure is 100 kPa, and the isentropic turbine efficiency is 87%; the cycle exhaust temperature from the heat exchanger is 200C. The following data are known for the steamturbine 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 500C and the isentropic turbine efficiency is 87%. Determine a. The mass flow rate of air in the gasturbine 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 /P1 = 14 = 0.87
SC . HEAT
2 3 T3 = 1250o C
. COMP AIR
1 5 GAS TURB W NET CT P4 = 100 kPa
4 = 0.87
ST P 1= 100 kPa T1 = 25o C T5 = 200 oC HEAT EXCH
7 6 T7 = 500 C
. o P6 = P7 = 12.5 MPa H 2O STEAM TURB
8 WST ST 0.87 = = 0.85
SP 9 P COND P8 = P9 = 10 kPa
h1 = 298.66, h5 = 475.84 a) From Air Tables, A.7: Pr1 = 1.0913, 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/SC = 335.82/0.87 = 386 = h1  h2 At T3 = 1523.2 K: Pr3 = 515.493, h3 = 1663.91 h2 = 684.66 1190 . . 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 wST = h3  h4S = 1663.91  812.68 = 851.23 wT = ST wST = 0.87 851.23 = 740.57 = h 3  h4 => h4 = 923.34 kJ/kg Steam cycle: wSP 0.00101(12500  10) = 12.615 wP =  wSP/SP = 12.615/0.85 = 14.84 h6 = h9  wP = 191.83 + 14.84 = 206.67 At 12.5 MPa, 500 oC: h7 = 3341.7, s7 = 6.4617 . mH h4  h5 923.34  475.84 . = mAIR = 61.27 = 8.746 kg/s O 3341.7  206.67 h7  h6 2 c) s8S = s7 = 6.4617 = 0.6492 + x8S 7.501, x8S = 0.7749 h8S = 191.81 + 0.7749 2392.8 = 2046.0 wST = h7  h8S = 3341.7  2046.0 = 1295.7 wT = ST 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 . . TH = WNET/QH = 31.455/60 = 0.524 1191 11.114 For Problem 11.105, 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 11.105 : . mH2O = 26.584 kg/s, h2 = 194.85, s2 = 0.6587 h3 = 3456.5, s3 = 7.2338, s = 7.9820, s = 7.1762 Ti Te hi = 903.16, he = 408.13 3  2 = h3  h2  T0(s3  s2) = 1301.28 kJ/kg i  e = hi  he  T0(s  s ) = 254.78 kJ/kg Ti Te . (3  2)mH2O 1301.28 26.584 . II = = = 0.776 (i  e)mair 254.78 175 1192 11.115 One means of improving the performance of a refrigeration system that operates over a wide temperature range is to use a twostage compressor. Consider an ideal refrigeration system of this type that uses R12 as the working fluid, as shown in Fig. P11.115. Saturated liquid leaves the condenser at 40C and is throttled to 20C. The liquid and vapor at this temperature are separated, and the liquid is throttled to the evaporator temperature, 70C. Vapor leaving the evaporator is compressed to the saturation pressure corresponding to 20C, 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, 40C. 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 twostage compressor unit studied in this problem.
ROOM . +QH COND 1 T 5
o 5 COMP. ST.2 4 SAT.LIQ. o 40 C 40 C 20 C
o o 1 6 2 7 4 3 8 s 9 SAT.VAP. o 20 C FLASH MIX.CHAM
3 9 COMP. ST.1 EVAP . 2 70 C CHAMBER 6 SAT.LIQ. o 20 C R12 ref. with 2stage compression 7 8 QL
COLD SPACE SAT.VAP. o 70 C 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 1193 QL = m3(h8  h7) = 0.647(155.536  17.8) qL = 89.1 kJ/kgm1 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 const. press. 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 CV: 2nd stage compressor P4 = 0.1509 MPa = P9 = P3 P5 = Psat 40oC = 0.9607 MPa, s5 = s4 T5 = 70oC, h5 = 225.8 CV: condenser 1st law: qH = h1  h5 = 74.527  225.8 = 151.27 2 stage = qL/(qH  qL) = 89.1/(151.27  89.1) = 1.433 b) 1 stage compression h3 = h4 = 74.527 h1 = 155.536 qL = h1  h4 = 81.0 kJ/kg s1 = s2 = 0.7744 P2 = 0.9607 MPa o T2 = 80.9 C, h2 = 234.0 40o C 70 C
o T 2 3 4 1 s qH = h2  h3 = 234.0  74.527 = 159.47 kJ/kg 1 stage = qL/(qH  qL) = 81.0/(159.47  81.0) = 1.032 1194 11.116 A jet ejector, a device with no moving parts, functions as the equivalent of a coupled turbinecompressor unit (see Problems 9.82 and 9.90). Thus, the turbinecompressor in the dualloop 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 airconditioning 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 . L H a. Assume an ideal cycle. b. Assume an ejector efficiency of 20% (see Problem 9.90).
VAP o 150 C BOIL. . QH
11 o 2 T 2 JET EJECT.
3
1' 3 11
1 4 2' COND. 30 C HP P.
10 9 o 4 VAP o 10 C 9 5,10 8 7 6 1 s 5 6 FLASH CH.
7 T1 = T7 = 10 C T2 = 150 oC T4 = 30 oC T9 = 20 oC Assume T5 = T10 o LIQ o 10 C LP P. 20 C CHILL . QL
8 (from mixing streams 4 & 9). P3 = P4 = P5 = P8 = P9 = P10= PG 30 oC = 4.246 kPa
o P11 = P2 = PG 150oC = 475.8 kPa, P1 = P6 = P7 = PG 10 C = 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 s = s1 & s = s2 (1' & 2' at P3 = P4) 1 2 . . then, m1(h  h1) = m2(h2  h ) 1 2 From s = s2 = 0.4369 + x 8.0164; x = 0.7985 2 2 2 h = 125.79 + 0.7985 2430.5 = 2066.5 2 1195 From s = s1 = 8.9008 T = 112 C, h = 2710.4 1 1 1 2746.5  2066.5 . . m1/m2 = = 3.5677 2710.4  2519.8 Also h4 = 125.79, h7 = 42.01, h9 = 83.96 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 LP pump: wLP P = 0.0010(4.246  1.2276) = 0.003 h8 = h7  wLP P = 42.01 + 0.003 = 42.01 . . . Chiller: QL = m7(h9h8) = 202.627(83.96  42.01) = 8500 kW (for m2 = 1) HP pump: wHP P = 0.001002(475.8  4.246) = 0.47 h11 = h10  wHP P = 84.88 + 0.47 = 85.35 . . 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 . Then, QL = 39.762(83.96  42.01) = 1668 kW h11 = 85.69 + 0.47 = 86.16 . QH = 1(2746.5  86.16) = 2660.3 kW . . & QL/QH = 1668/2660.3 = 0.627 1196 English Unit Problems
11.117E A steam power plant, as shown in Fig. 11.3, operating in a Rankine cycle has saturated vapor at 600 lbf/in.2 leaving the boiler. The turbine exhausts to the condenser operating at 2 lbf/in.2. Find the specific work and heat transfer in each of the ideal components and the cycle efficiency. 1: h1 = 93.73, v1 = 0.01623, 3: h3 = hg = 1204.06, s3 = sg = 1.4464 144 = 1.8 Btu/lbm C.V. Pump: wP = v dP = v1(P2  P1) = 0.01623(600  2) 778 h2 = h1  wP = 93.73 + 1.8 = 95.81 Btu/lbm C.V. Boiler: qH = h3  h2 = 1204.06  95.53 = 1108.53 Btu/lbm C.V. Tubine: wT = h3  h4, s4 = s3 s4 = s3 = 1.4464 = 0.1744 + x4 1.7461 h4 = 93.73 + 0.7285 1022.2 = 838.42 wT = 1204.06  838.42 = 365.63 Btu/lbm CYCLE = (wT + wP)/qH = (365.63  1.8)/1108.53 = 0.33 C.V. Condenser: qL = h4  h1 = 838.42  93.73 = 744.69 Btu/lbm => x 4 = 0.7285, 11.118E Consider a solarenergypowered ideal Rankine cycle that uses water as the working fluid. Saturated vapor leaves the solar collector at 350 F, and the condenser pressure is 1 lbf/in. 2. Determine the thermal efficiency of this cycle.
T H2O ideal Rankine cycle CV: turbine s4 = s3 = 1.5793 = 0.13266 + x4 1.8453 x4 = 0.784 h4 = 69.74 + 0.784 1036.0 = 881.9 wT = h3  h4 = 1193.1  881.9 = 311.2 Btu/lbm wP = vdP v1(P2  P1) = 0.016136(67  1) 144 / 778 = 0.2 wNET = wT + wP = 311.2  0.2 = 311 h2 = h1  wP = 69.7 + 0.2 = 69.9 Btu/lbm qH = h3  h2 = 1193.1  69.9 = 1123.2 Btu/lbm TH = wNET/qH = 311/1123.2 = 0.277
3 2 1 4 s 1197 11.119E A supply of geothermal hot water is to be used as the energy source in an ideal Rankine cycle, with R134a as the cycle working fluid. Saturated vapor R134a leaves the boiler at a temperature of 180 F, and the condenser temperature is 100 F. Calculate the thermal efficiency of this cycle.
T a) From the R134a tables, h1 = 108.86 v1 = 0.01387 P1 = 138.93 P2 = P3 = 400.4 h3 = 184.36 s3 = 0.402 CV. Pump: wP = v1(P2  P1) = 0.01387(400.4138.93)
2 1 D 3 180 F 100 F s 4 144 = 0.671 Btu/lbm = h2  h1 778 h2 = h1  wP = 108.86 + 0.671 = 109.53 CV: Turbine s4 = s3 x4 = (0.402  0.2819)/0.1272 = 0.9442 h4 = 176.08, wT = h3  h4 = 8.276 btu/lbm CV: Boiler qH = h3  h2 = 184.36  109.53 = 74.83 Btu/lbm TH = (wT + w12)/qH = (8.276  0.671)/74.83 = 0.102 11.120E Do Problem 11.119 with R22 as the working fluid. Same diagram as in problem 11.119, now from the R22 tables, h1 = 39.267, v1 = 0.01404, P1 = 210.6, P2 = P3 = 554.8, h3 = 110.07, s3 = 0.1913 CV: Pump 144 = 0.894 Btu/lbm wP = v1(P2P1) = 0.01404 (554.8210.6) 778 h2 = h1  wP = 39.267 + 0.894 = 40.16 CV: Turbine s4 = s3 x4 = (0.1913  0.07942)/0.13014 = 0.9442 h4 = 101.885, wT = h3  h4 = 8.185 Btu/lbm CV: Boiler qH = h3  h2 = 110.07  40.16 = 69.91 Btu/lbm TH = (wT + w12)/qH = (8.185  0.894)/157.21 = 0.104 1198 11.121E The power plant in Problem 11.117 is modified to have a superheater section following the boiler so the steam leaves the super heater at 600 lbf/in.2, 700 F. Find the specific work and heat transfer in each of the ideal components and the cycle efficiency. h3 = 1350.6, s3 = 1.5871, h1 = 94.01, v1 = 0.01623 C.V. Pump: wP = v dP = v2(P2  P1) = 0.01623(600  2)(144 / 778) = 1.8 Btu/lbm h2 = h1  wP = 95.81 Btu/lbm C.V. Boiler: qH = h3  h2 = 1350.6  95.81 = 1254.79 Btu/lbm C.V. Tubine: wT = h3  h4, s4 = s3 x4 = 0.8093, h4 = 921.23 wT = 1350.6  921.23 = 429.37 Btu/lbm CYCLE = (wT + wP)/qH = (429.37  1.8)/1254.79 = 0.341 C.V. Condenser: qL = h4  h1 = 921.23  94.01 = 827.22 Btu/lbm 11.122E Consider a simple ideal Rankine cycle using water 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 hightemperature energy source is the hot exhaust gas from a gasturbine engine. Calculate the thermal efficiency of the cycle if the state entering the turbine is 3500 lbf/in.2, 1100 F, and the condenser pressure is 1 lbf/in.2. What is the steam quality at the turbine exit? s4 = s3 = 1.5193 = 0.1326 + x41.8453 x4 = 0.7515 very low for a turbine exhaust h4 = 848.3, h1 = 69.73, h3 = 1496 s2 = s1 = 0.1326 => h2 = 80.13 qH = h3  h2 = 1415.9 Btu/lbm wnet = h3  h4  (h2  h1) = 637.3 Btu/lbm = wnet/qH = 637.3/1415.9 = 0.45
2 3500 lbf/in 3 1100 F 2 1 lbf/in T 2 1 4 s 1199 11.123E Consider an ideal steam reheat cycle in which the steam enters the highpressure turbine at 600 lbf/in.2, 700 F, and then expands to 120 lbf/in.2. It is then reheated to 700 F and expands to 2 lbf/in.2 in the lowpressure turbine. Calculate the thermal efficiency of the cycle and the moisture content of the steam leaving the lowpressure turbine. Basic cycle as in 11.121 plus reheat. From solution 11.121: wP = 1.8 Btu/lbm, h2 = 95.81 h3 = 1350.6, s3 = 1.5871 qH1 = h3h2 = 1350.695.81 = 1254.79 CV T1: wT1 = h3  h4; s4 = s3 => x4 = 0.9986, h4 = 1190 Btu/lbm CV T2: wT2 = h5  h6; s6 = s5 = 1.7825, h5 = 1378.2 => x 6 = 0.9213, h6 = 1035.7, wT2 = 1378.2  1035.7 = 342.5 wT,tot = wT1 + wT2 = 160.6 + 342.5 = 503.1 Btu/lbm qH = qH1 + h5  h4 = 1254.79 + 1378.2  1190 = 1443 Btu/lbm CYCLE = wT,tot/qH = 503.1/1443 = 0.349
T 3 5 2 1 4 6 s 11.124E A closed feedwater heater in a regenerative steam power cycle heats 40 lbm/s of water from 200 F, 2000 lbf/in.2 to 450 F, 2000 lbf/in.2. The extraction steam from the turbine enters the heater at 500 lbf/in.2, 550 F and leaves as saturated liquid. What is the required mass flow rate of the extraction steam? 3 2 1 From the steam tables: h1 = 172.6 h2 = 431.14 all h3 = 1266.6 h4 = 449.5 Btu/lbm 4 C.V. Feedwater Heater . . m3 = m1(h1  h2)/(h4  h3) = 40 lbm 172.6  431.14 = 12.656 s 449.5  1266.6 11100 11.125E Consider an ideal steam regenerative cycle in which steam enters the turbine at 600 lbf/in.2, 700 F, and exhausts to the condenser at 2 lbf/in.2. Steam is extracted from the turbine at 120 lbf/in. 2 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 poundmass of steam. From solution 11.121: h5 = 1350.6, s5 = 1.5871 h7 = 921.23, h1 = 94.01 2nd law: s6 = s5 => x6 = 0.999 5
ST. GEN. TURBINE. 6 4
P FW HTR COND. 7 h6 = 1190 Btu/lbm Also: h3 = 312.66 wP12 = 0.01623(120  2)144/778 = 0.35 wP34 = 0.0179(600  120)144/778 = 1.59 h2 = 94.36, h4 = 314.25 2 P 3 1 T 600 psi 5 4 2 1
CV: feedwater heater cont : m3 = m6 + m2 = 1 lbm, 1st law: m6h6 + m2h2 = m3h3 120 psi 3 6 7 2 psi s m6(1190) + (1  m6)(94.36) = 312.66 m6 = 0.1992, m2 = 0.8008 CV: turbine wT = (h5  h6) + (1  m6)(h6  h7) = (1350.6  1190) + 0.8008(1190  921.2) = 375.9 Btu/lbm CV: pumps wP = m2wP12 + m3wP34 = 0.8008 0.35 + 1 1.59 = 1.87 Btu/lbm wNET = wT  wP = 375.9  1.87 = 374 Btu/lbm CV: steam generator qH = h5  h4 = 1350.6  314.25 = 1036.35 Btu/lbm TH = wNET/qH = 374/1036.35 = 0.361 11101 11.126E Consider an ideal combined reheat and regenerative cycle in which steam enters the highpressure turbine at 500 lbf/in.2, 700 F, and is extracted to an open feedwater heater at 120 lbf/in.2 with exit as saturated liquid. The remainder of the steam is reheated to 700 F at this pressure, 120 lbf/in.2, and is fed to the lowpressure turbine. The condenser pressure is 2 lbf/in.2. Calculate the thermal efficiency of the cycle and the net work per poundmass of steam.
7 5 5: h5 = 1356.66, 7: h7 = 1378.17, s5 = 1.6112 s7 = 1.7825
T1 1x
6 3: h3 = hf = 312.59, v3 = 0.01788 C.V. T1 s 5 = s6 => h6 = 1209.76
4 T2
8 x HTR
3 wT1 = h5  h6 = 1356.66  1209.76 = 146.9 Btu/lbm C.V. Pump 1 wP1 = h2  h1 = v1(P2  P1) = 0.01623(120  2) = 0.354 1x
2 COND. P
1 P => h 2 = h1  wP1 = 93.73 + 0.354 = 94.08 Btu/lbm C.V. FWH x h6 + (1  x) h2 = h3 312.59  94.08 x= = = 0.1958 1209.76  94.08 h6  h2 C.V. Pump 2 h3  h2 T 700 F 4 5 7 6 2 3 1 8 2 psi s wP2 = h4  h3 = v3(P4  P3) = 0.01788(500  120)(144/778) = 1.26 Btu/lbm => h 4 = h3  wP2 = 312.59 + 1.26 = 313.85 Btu/lbm qH = h5  h4 + (1  x)(h7  h6 ) = 1042.81 + 135.43 = 1178.2 Btu/lbm C.V. Turbine 2 s 7 = s8 => x8 = (1.7825  0.1744)/1.746 = 0.921 h8 = hf + x8 hfg = 93.73 + 0.921 1022.2 = 1035.2 wT2 = h7  h8 = 1378.17  1035.2 = 342.97 wnet = wT1 + (1  x) wT2 + (1  x) wP1 + wP2 = 146.9 + 275.8  0.285  1.26 = 421.15 kJ/kg cycle = wnet / qH = 421.15 / 1178.2 = 0.357 11102 11.127E A steam power cycle has a high pressure of 500 lbf/in. 2 and a condenser exit temperature of 110 F. The turbine efficiency is 85%, and other cycle components are ideal. If the boiler superheats to 1400 F, find the cycle thermal efficiency. C.V. Turb.: wT = h3  h4, s4 = s3 + sT,GEN Ideal: s4S = s3 = 1.8704 => x4S = 0.9519, h4S = 1059.7 Btu/lbm => wT,S = 1741  1059.7 = 681.3 Btu/lbm Actual: wT,AC = wT,S = 579.1 Btu/lbm C.V. Pump: wP = vdP v1(P2  P1) = 0.016166(5001.28)144/778 = 1.49 Btu/lbm C.V. Boiler: qH = h3  h2 = h3  h1 + wP = 1661.5 Btu/lbm qL = h4ac  h1 = 1161.9  78.01 = 1083.9 C.V. Condenser: = (wT,AC  wP)/qH = (579.1  1.49)/1661.5 = 0.348 11.128E The steam power cycle in Problem 11.117 has an isentropic efficiency of the turbine of 85% and that for the pump it is 80%. Find the cycle efficiency and the specific work and heat transfer in the components. States numbered as in fig 9.3 of text. CV Pump: wP,S = v1(P2P1) = 0.01623(600  2)144/778 = 1.8 Btu/lbm wP,AC = 1.8/0.8 = 2.245 Btu/lbm h2 = h1  wP,AC = 93.73 + 2.245 = 95.975 CV Turbine: wT,S = h3  h4s , s4 = s3 = 1.4464 Btu/lbm R wT,S = 1204.06  838.42 = 365.64 Btu/lbm wT,AC = h3  h4AC = 365.64 0.85 = 310.78 h4AC = 893.3 CV Boiler: qH = h3  h2 = 1204.06  95.975 = 1108.1 Btu/lbm qL = h4AC  h1 = 893.3  93.73 = 799.5 Btu/lbm CYCLE = (wT + wP)/qH = (310.78  2.245)/1108.1 = 0.278 Compared to (365.641.8)/1108.5 = 0.33 in the ideal case. 11103 11.129E Steam leaves a power plant steam generator at 500 lbf/in.2, 650 F, and enters the turbine at 490 lbf/in.2, 625 F. The isentropic turbine efficiency is 88%, and the turbine exhaust pressure is 1.7 lbf/in.2. Condensate leaves the condenser and enters the pump at 110 F, 1.7 lbf/in.2. The isentropic pump efficiency is 80%, and the discharge pressure is 520 lbf/in.2. The feedwater enters the steam generator at 510 lbf/in.2, 100 F. Calculate the thermal efficiency of the cycle and the entropy generation of the flow in the line between the steam generator exit and the turbine inlet, assuming an ambient temperature of 77 F.
2 1 T 500 lbf/in 1 2 ST. GEN. TURBINE. sT =
6 5 2 650 F 625 F 1.7 lbf/in
2 0.88 3 COND. P
4 5 5s 6 4 3s 3 s ST = 0.88, SP = 0.80 h1 = 1328.0, h2 = 1314.0 => x 3S = 0.79745 s3S = s2 = 1.5752 = 0.16483 + x3S1.7686 h3S = 88.1 + 0.797 451025.4 = 905.8 wST = h2  h3S = 1314.0  905.8 = 408.2 wT = STwST = 0.88408.2 = 359.2 Btu/lbm h3 = h2  wT = 1314.0  359.2 = 954.8 wSP = 0.016166(5201.7) 144 = 1.55 Btu/lbm 778 wp = wSP/SP = 1.55/0.80 = 1.94 Btu/lbm qH = h1  h6 = 1328.0  68.1 = 1259.9 Btu/lbm TH = wNET/qH = (359.2  1.94)/1259.9 = 0.284 C.V. Line from 1 to 2: Energy Eq.: w = 0, / q = h2  h1 = 1314  1328 =  14 Btu/lbm => Entropy Eq.: s1 + sgen + q/T0 = s2 sgen = s2  s1 q/T0 = 1.5752  1.586  (14/536.7) = 0.0153 Btu/lbm R 11104 11.130E 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 700 lbf/in.2 exits this heat exchanger and is then superheated to 1100 F in an external gasfired superheater. The steam enters the turbine, which has one (opentype) feedwater extraction at 60 lbf/in.2. The isentropic turbine efficiency is 87%, and the condenser pressure is 1 lbf/in.2. Determine the heat transfer in the reactor and in the superheater to produce a net power output of 1000 Btu/s.
700 lbf/in 2
6 SUP. HT. T 6 1100 F 60 lbf/in 2 TURBINE.
7 8 Q
5 REACT. 4 4 2 3 1 5 7s 7 HTR.
3 1 lbf/in 2 COND.
2 1 8s 8 P P s . WNET = 1000 Btu/s, ST = 0.87 wP12 = 0.016136(60  1)144/778 = 0.18 Btu/lbm h2 = h1  wP12 = 69.73 + 0.18 = 69.91 Btu/lbm wP34 = 0.017378(700  60)144/778 = 2.06 Btu/lbm h4 = h3  wP34 = 262.24 + 2.06 = 264.3 Btu/lbm s7S=s6 = 1.7682, P7 => T7S = 500.8 F, h7S = 1283.4 h7 = h6  ST(h6  h7S) = 1625.8  0.87(1625.8  1283.4) = 1327.9 s8S = s6 = 1.7682 = 0.13264 + x8S 1.8453 => h8S = 69.73 + 0.8863 1036 = 987.9 Btu/lbm h8 = h6  ST(h6  h8S) = 1625.8  0.87(1625.8  987.9) = 1070.8 CV: heater: cont: m2 + m7 = m3 = 1.0 lbm, 1st law: m2h2 + m7h7 = m3h3 m7 = (262.2469.91) / (1327.969.91) = 0.1529 CV: turbine: wT = (h6  h7) + (1  m7)(h7  h8) = 1625.81327.9 + 0.8471(1327.91070.8) = 515.7 Btu/lbm CV pumps: wP = m1wP12 + m3wP34 = (0.84710.18 + 12.06) = 2.2 Btu/lbm . wNET = 515.7  2.2 = 513.5 Btu/lbm => m = 1000/513.5 = 1.947 lbm/s . . CV: reactor QREACT = m(h5h4) = 1.947(1202  264.3) = 1825.7 Btu/s . . CV: superheater QSUP = m(h6  h5) = 1.947(1625.8  1202) = 825 Btu/s x 8S = 0.8863 11105 11.131E A boiler delivers steam at 1500 lbf/in.2, 1000 F to a twostage turbine as shown in Fig. 11.17. After the first stage, 25% of the steam is extracted at 200 lbf/in.2 for a process application and returned at 150 lbf/in. 2, 190 F to the feedwater line. The remainder of the steam continues through the lowpressure turbine stage, which exhausts to the condenser at 2 lbf/in. 2. One pump brings the feedwater to 150 lbf/in.2 and a second pump brings it to 1500 lbf/in.2. 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 5000 Btu/s of power, how much power can then be cogenerated by the turbine? 3: h3 = 1490.3, s3 = 1.6001 4s: s4S = s3 h4S = 1246.6 wT1,S = h3  h4S = 243.7 wT1,AC = 207.15 Btu/lbm h4AC = h3  wT1,AC = 1283.16 4ac: P4, h4AC s4AC = 1.6384 5s: s5S = s4AC h5S = 951.3 wT2,S = h4AC  h5S = 331.9 wT2,AC = 265.5 = h 4AC  h5AC h5AC = 1017.7 Btu/lbm 7: h7 = 158.02 ; qPROC = h4AC  h7 = 1125.1 Btu/lbm . . . m4 = Q/qPROC = 5000/1125.1 = 4.444 lbm/s = 0.25 mTOT . . . . . mTOT = m3 = 17.776 lbm/s, m5 = m3  m4 = 13.332 lbm/s . . . . WT = m3h3  m4h4AC  m5h5AC = 7221 Btu/s
2 P2 7 1 Proc. 5000 Btu/s 6 P1 C B 4 5 3 T1 T2 11106 11.132E A large stationary Brayton cycle gasturbine power plant delivers a power output of 100000 hp to an electric generator. The minimum temperature in the cycle is 540 R, and the maximum temperature is 2900 R. The minimum pressure in the cycle is 1 atm, and the compressor pressure ratio is 14 to 1. Calculate the power output of the turbine, the fraction of the turbine output required to drive the compressor and the thermal efficiency of the cycle? Brayton: . wNET = 100 000 hp P1 = 1 atm, T1 = 540 R P2/P1 = 14, T3 = 2900 R a) Assume const CP0: s 2 = s1 T2 = T 1
k1 k T 3 s 2 s 1 s 4 (P )
1 P2 = 540(14)0.286 = 1148.6 R wC = w12 = h2  h1 = CP0(T2T1) = 0.24(1148.6540) = 146.1 Btu/lbm As s4 = s3 T4 = T3 P4
3 k1 k (P ) = 2900( 1 0.286 ) = 1363.3 R 14 wT = w34 = h3  h4 = CP0(T3T4) = 0.24(29001363.3) = 368.8 Btu/lbm wNET = wT + wC = 368.8  146.1 = 222.7 Btu/lbm . . m = WNET/wNET = 100 0002544/222.7 = 1 142 344 lbm/h . . WT = mwT = 165 600 hp, wC/wT = 0.396 b) qH = CP0(T3T2) = 0.24(29001148.6) = 420.3 Btu/lbm TH = wNET/qH = 222.7/420.3 = 0.530 11107 11.133E An ideal regenerator is incorporated into the ideal airstandard Brayton cycle of Problem 11.132. Calculate the cycle thermal efficiency with this modification.
T 3 s x 2 s 1 s y 4 Problem 9.108 + ideal regen., where : wT = 368.8, wC = 146.1 Btu/lbm wNET = 222.7 Btu/lbm Ideal regen.: TX = T4 = 1363.3 R qH = h3  hX = 0.24(2900  1363.3) = 368.8 Btu/lbm = w T TH = wNET/qH = 222.7/368.8 = 0.604 11.134E Consider an ideal gasturbine 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 14 lbf/in.2, the temperature entering each compressor is 70 F, and the temperature entering each turbine is 2000 F. An ideal regenerator is also incorporated into the cycle. Determine the compressor work, the turbine work, and the thermal efficiency of the cycle.
10 REG CC
5 6 9 I.C.
1 2 4 COMP COMP TURB
7 TURB
8 CC P2/P1 = P4/P3 = P6/P7 = P8/P9 = 8.0 P1 = 14 lbf/in2 T1 = T3 = 70 F, T6 = T8 = 2000 F Assume const. specific heat s2 = s1 and s4 = s3 T4 = T2 = T1(P2/P1
k1 )k T 6 5 7 4 3 2 1 8 9 10 s = 529.67(8)0.2857 = 959.4 R Total compressorwork wC = 2 (w12) = 2CP0(T2  T1) = 2 0.24(959.4  529.67) = 206.3 Btu/lbm 11108 Also s6 = s7 and s8 = s9 P7 k 10.2857 T7 = T9 = T6 = 2459.67 = 1357.9 R P 8 6 Total turbinework wT = 2 w67 = 2CP0(T6  T7) = 2 0.24(2459.67  1357.9) = 528.85 Btu/lbm wNET = 528.85  206.3 = 322.55 Btu/lbm Ideal regenerator: T5 = T9, T10 = T4 qH = (h6  h5) + (h8  h7) = 2CP0(T6  T5) = 2 0.24(2459.67  1357.9) = wT = 528.85 Btu/lbm TH = wNET/qH = 322.55/528.85 = 0.61
k1 11.135E Repeat Problem 11.134, but assume that each compressor stage and each turbine stage has an isentropic efficiency of 85%. Also assume that the regenerator has an efficiency of 70%. T4S = T2S = 959.4 R, wCS = 206.3 T7S = T9S = 1357.9 R, wTS = 528.85 wC = wSC/SC = 242.7 Btu/lbm
4S 3 5 4 7S 7 9S T 6 8 9 w12 = w34 = 242.7/2 = 121.35 /C T2 = T4 = T1 + (w12 P0) = 529.67 + 121.35/0.24 = 1035.3 R wT = T wTS = 449.5 2S 2 1 s T7 = T9 = T6  (+w67 P0) = 2459.67  449.5/20.24 = 1523 R /C REG = h5  h4 h9  h4 = T 5  T4 = = 0.7 T9  T4 1523  1035.3 T5  1035.3 T5 = 1376.7 R qH = CP0(T6  T5) + CP0(T8  T7) = 0.24(2459.67  1376.7) + 0.24(2459.67  1523) = 484.7 Btu/lbm wNET = wT + wC = 449.5  242.7 = 206.8 Btu/lbm TH = wNET/qH = 206.8/484.7 = 0.427 11109 11.136E An airstandard Ericsson cycle has an ideal regenerator as shown in Fig. P11.62. Heat is supplied at 1800 F and heat is rejected at 68 F. Pressure at the beginning of the isothermal compression process is 10 lbf/in.2. The heat added is 275 Btu/lbm. Find the compressor work, the turbine work, and the cycle efficiency. T P T4 = T3 = 1800 F
2 T 1 P 3 3
T T 4 T1 = T2 = 68 F = 527.7R P1 = 10 lbf/in2 q = 4q1 (ideal reg.) P
P 4 v P 2 T 1 s 2 3 qH = 3q4 & wT = qH = 275 Btu/lbm TH = CARNOT TH. = 1  TL/TH = 1  527.7/2349.7 = 0.775 wnet = TH qH = 0.775 275 = 213.13 Btu/lbm qL = wC = 275  213.13 = 61.88 Btu/lbm 11.137E The turbine in a jet engine receives air at 2200 R, 220 lbf/in.2. It exhausts to a nozzle at 35 lbf/in.2, which in turn exhausts to the atmosphere at 14.7 lbf/in.2. 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. C.V. Turb.: hi = 560.588, Pri = 206.092, se = si Pre = Pri (Pe/Pi) = 206.092(35/220) = 32.787 Te = 1381, he = 338.13, wT,s = 560.588  338.13 = 222.46 wT,AC = wT,s T = 189.09 = he,AC  hi he,AC = 371.5 Te,AC = 1508.4 R, Pre,AC = 45.917 C.V. Nozzle: (1/2)V2 = hi  he; se = si e Pre = Pri (Pe/Pi) = 45.917(14.7/35) = 19.285 Te,s = 1199.6 R, he,s = 291.3 Btu/lbm
2 (1/2)Ve,s = hi  he,s = 371.5  291.3 = 80.2 Btu/lbm 2 2 (1/2)Ve,AC = (1/2)Ve,s NOZ = 76.19 Btu/lbm Ve,AC = 2 25037 76.19 = 1953 ft/s 11110 11.138E Air flows into a gasoline engine at 14 lbf/in.2, 540 R. The air is then compressed with a volumetric compression ratio of 8 1. In the combustion process 560 Btu/lbm of energy is released as the fuel burns. Find the temperature and pressure after combustion. Compression 1 to 2: s2 = s1 vr2 = vr1/8 = 179.49/8 = 22.436 T2 = 1212 R, u2 = 211.31 Btu/lbm, Pr2 = 20 P2 = Pr2(P1/Pr1) = 20(14/1.1146) = 251.2 lbf/in2 Compression 2 to 3: u3 = u2 + qH = 211.31 + 560 = 771.3, T3 = 3817 R P3 = P2 (T3/T2) = 251.2(3817/1212) = 791 lbf/in2 11.139E To approximate an actual sparkignition engine consider an airstandard Otto cycle that has a heat addition of 800 Btu/lbm of air, a compression ratio of 7, and a pressure and temperature at the beginning of the compression process of 13 lbf/in.2, 50 F. Assuming constant specific heat, with the value from Table C.4, determine the maximum pressure and temperature of the cycle, the thermal efficiency of the cycle and the mean effective pressure.
P T s 4 v 1 2 s 1 v s v v 3 s 4 3 v 2 s qH = 800 Btu v1/v2 = 7 P1 = 13 lbf/in2, T1 = 50 F v1 = RT1/P1 = 53.34510/13144 = 14.532 ft3/lbm v2 = v1/7 = 2.076 ft3/lbm a) P2 = P1(v1/v2) = 13(7)1.4 = 198.2 lbf/in2 T2 = T1(v1/v2)
k1 k = 510(7)0.4 = 1110.7 R T3 = T2 + qH/CV0 = 1110.7 + 800/0.171 = 5789 R P3 = P2T3/T2= 198.2 5789/1110.7 = 1033 lbf/in2 b) c) TH = 1  (T1/T2) = 1  510/1110.7 = 0.541 wNET = TH qH = 0.541 800 = 432.8 Btu mep = wNET v1v2 = 432.8778 = 188 lbf/in2 (14.5322.076)144 11111 11.140E 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 downwards motion in the expansion stroke. Therefore consider a cycle identical to the Otto cycle, except that the first twothirds of the total qH occurs at constant volume and the last onethird occurs at constant pressure. Assume the total qH is 700 Btu/lbm, that the state at the beginning of the compression process is 13 lbf/in.2, 68 F, 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 4 s 2 s 1 5 v 2 s 1 v s 5 v T 3 3 4 s P1 = 13, T1 = 527.67 R rV = v1/v2 = 7 2 Btu q23 = 700 = 466.7 3 lbm 1 Btu q34 = 700 = 233.3 3 lbm P2 = P1(v1/v2) = 13(9)1.4 = 281.8 lbf/in2 T2 = T1(v1/v2)
k1 k = 527.67(9)0.4 = 1270.7 R T3 = T2 + q23/CV0 = 1270.7 + 466.7/0.171 = 4000 R P3 = P2(T3/T2) = 281.8 4000/1270.7 = 887.1 lbf/in2 = P4 T4 = T3 + q34/CP0 = 4000 + 233.3/0.24 = 4972 R v5 v4 = v1 v4 = (P4/P1) (T1/T4) =
k1 88.1 527.67 = 7.242 13 4972
0.4 T5 = T4(v4/v5) = 4972(1/7.242) = 2252 R qL = CV0(T5T1) = 0.171(2252  527.67) = 294.9 Btu/lbm TH = 1  qL/qH = 1  294.9/700 = 0.579 Std Otto cycle: TH = 1  (9)0.4 = 0.585 11112 11.141E 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 11.139 but assume 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 11.139 : T3 = 5789 R, P3 = 1033 lbf/in2 v3 = RT3/P3 = v2 = 2.076 ft3/lbm, v4 = v1 = 14.532 except for process 3 to 4: Pv1.5 = constant. P4 = P3(v3/v4)1.5 = 1033(1/7)1.5 = 55.78 lbf/in2 T4 = T3(v3/v4)0.5 = 5789(1/7)0.5 = 2188 R
1w2 = P dv = R(T2  T1)/(1  1.4) = 53.34(1110.7  510)/(0.4778) = 102.96 Btu/lbm = P dv = R(T4  T3)/(1  1.5) = 53.34(2188  5789)/(0.5778) = 493.8 Btu/lbm 3w4 wNET = 493.8  102.96 = 390.84 Btu/lbm CYCLE = wNET/qH = 390.84/700 = 0.488 mep = wNET/(v1v2) = 390.84778/(14.532  2.076) = 169.5 lbf/in2 Notice a smaller wNET, CYCLE, mep compared to ideal cycle. 11.142E A diesel engine has a bore of 4 in., a stroke of 4.3 in. 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 200 lbf/in. 2. With a total of 6 cylinders find the engine power in Btu/s and horsepower, hp. Power from mean effective pressure. mep = wnet / (vmax  vmin) > wnet = mep(vmax  vmin) V = Bore2 0.25 S = 42 0.25 4.3 = 54.035 in3 W = mep(Vmax  Vmin) = 20054.035/(12778) = 1.1575 Btu Work per Cylinders per Power Stroke . W = WnRPM0.5 (cycles / min)(min / 60 s)(Btu / cycle) = 1.1575620000.5(1/60) = 115.75 Btu/s = 115.75 3600/2544.43 hp = 164 hp 11113 11.143E At the beginning of compression in a diesel cycle T = 540 R, P = 30 2 and the state after combustion (heat addition) is 2600 R and 1000 lbf/in. lbf/in.2. Find the compression ratio, the thermal efficiency and the mean effective pressure. P2 = P3 = 1000 lbf/in2 T2 = T1(P2/P1)(k1)/k = 540(1000/30) 0.2857 = 1470.6 v1/v2 = (P2/P1)1/k = (1000/30)0.7143 = 12.24 v3/v2 = T3/T2 = 2600/1470.6 = 1.768 v4/v3 = v1/v3 = (v1/v2)(v2/v3) = 12.24/1.768 = 6.923 T4 = T3(v3/v4)k1 = 2600*6.9230.4 = 1199 qL = CV(T4  T1) = 0.171(1199540) = 112.7 qH = h3  h2 = CP (T3  T2) = 0.24(2600  1470.6) = 271.1 = 1  qL/qH = 1  112.7 / 271.1 = 0.5843 wnet = qnet = 271.1  112.7 = 158.4 vmax = v1 = RT1/P1 = 53.34 540/(30 144) = 6.6675 vmin = vmax(v1/v2) = 6.6675 / 12.24 = 0.545 mep = [158.4/(6.6675  0.545)] (778/144) = 139.8 lbf/in2 11114 11.144E Consider an ideal airstandard diesel cycle where the state before the compression process is 14 lbf/in.2, 63 F and the compression ratio is 20. Find the maximum temperature(by iteration) in the cycle to have a thermal efficiency of 60%. Diesel cycle: P1= 14, T1 = 522.67 R, v1/v2 = 20, TH = 0.60 T2 = T1(v1/v2) v1 =
k1 = 522.67(20)0.4 = 1732.4 R 53.34522.67 13.829 = 13.829, v2 = = 0.6915 20 14144 v3 = v2T3/T2 = 0.6915T3/1732.4 = 0.000399 T3 T3 T4 = 13.829 (v ) =(0.000399 T )
k1 3 3 v4 0.4 T4 = 0.0153 T3
1.4 1.4 TH = 0.60 = 1 T4T1 k(T3T2)
1.4 =1 0.0153 T3 522.67 1.4(T31732.4) 0.0153 T3  0.56 T3 + 447.5 = 0 5500R: LHS = 5.04, 5450R: LHS = 0.5 Linear interpolation, T3 = 5455 R 5100R: LHS = 35.54, 11115 11.145E Consider an ideal Stirlingcycle engine in which the pressure and temperature at the beginning of the isothermal compression process are 14.7 lbf/in.2, 80 F, the compression ratio is 6, and the maximum temperature in the cycle is 2000 F. Calculate the maximum pressure in the cycle and the thermal efficiency of the cycle with and without regenerators.
P 3 T T 3
v 2 T 4 v 1 v T 4 Ideal Stirling cycle T1 = T2 = 80 F P1 = 14.7 lbf/in2 v1 v2 v 2 T 1 v =6 s T3 = T4 = 2000 F T1 = T2 P2 = P1 v1/v2 = 14.76 = 88.2 V2 = V3 P3 = P2 T3/T2 = 88.2 w34 = q34 = RT3 ln (v4/v3) = (53.34/778) 2460 ln 6 = 302.2 Btu/lbm q23 = CV0(T3T2) = 0.171(2460540) = 328.3 Btu/lbm w12 = q12 = RT1 ln v1 v2 =53.34 540 ln 6 = 66.3 Btu/lbm 778 2460 = 401.8 lbf/in2 540 wNET = 302.2  66.3 = 235.9 Btu/lbm NO REGEN = 235.9 = 0.374, 302.2+328.3 235.9 = 0.781 302.2 WITH REGEN = 11116 11.146E An ideal airstandard Stirling cycle uses helium as working fluid. The 2 2 isothermal compression brings the helium from 15 lbf/in. , 70 F to 90 lbf/in. . The exspansion takes place at 2100 R and there is no regenerator. Find the work and heat transfer in all four processes per lbm helium and the cycle efficiency. Substance helium C.4: R = 386 ftlbf/lbmR CV = 0.753 v4/v3 = v1/v2 = P2/P1 = 90/15 = 6 1 > 2: 2 > 3: 3 > 4: 4 > 1: 1w2 = 1q2 = P dV = 386 530 ln(6)/778 = 471.15 Btu/lbm
2w 3 3w 4 4w 1 = 0; 2q3 = CP (T3  T2) = 0.753(2100  530) = 1182.2 = 3q4 = RT3ln(v4/v3) = 386 2100 ln(6)/778 = 1866.8 = 0;
4q 1 = CP (T4  T1) = 1182.2 Cycle = wnet/ qH = (471.15 + 1866.0) / (1182.2 + 1866.8) = 0.458 11.147E The airstandard Carnot cycle was not shown in the text; show the Ts diagram for this cycle. In an airstandard Carnot cycle the low temperature is 500 R and the efficiency is 60%. If the pressure before compression and after heat rejection is 14.7 lbf/in.2, find the high temperature and the pressure just before heat addition. = 0.6 = 1  TH/TL TH = TL/0.4 = 500/0.4 = 1250 R P2 = P1(TH/TL
1 k1 ) T H T L T 2 3 = 14.7( 1250 3.5 ) 500 1 4 s = 363.2 lbf/in 2 [or P2 = P1(Pr2/Pr1) = 14.7 22.48 / 0.8515 = 388 lbf/in2] 11117 11.148E Air in a piston/cylinder goes through a Carnot cycle in which TL = 80.3 F and the total cycle efficiency is = 2/3. Find TH, the specific work and volume ratio in the adiabatic expansion for constant Cp, Cv. Repeat the calculation for variable heat capacities. Carnot cycle: Same process diagram as in previous problem. = 1  TL/TH = 2/3 TH = 3 TL = 3 540 = 1620 R Adiabatic expansion 3 to 4:
3 4 Pvk = constant w = (P4v4  P3v3)/(1  k) = [R/(1k)](T4  T3) = u3  u4 = Cv(T3  T4) = 0.171(1620  540) = 184.68 Btu/lbm v4/v3 = (T3/T4)1/(k  1) = 32.5 = 15.6 For variable Cp, Cv we get, TH = 3 TL = 1620 R
3 4 w = u3  u4 = 290.13  92.16 = 197.97 Btu/lbm v4/v3 = vr4/vr3 = 179.49/9.9289 = 18.1 11.149E Consider an ideal refrigeration cycle that has a condenser temperature of 110 F and an evaporator temperature of 5 F. Determine the coefficient of performance of this refrigerator for the working fluids R12 and R22.
T Ideal Ref. Cycle Tcond = 110 F = T3 Tevap = 5 F Use Table C.10 for R22 Use computer table for R12 h1, Btu/lbm s 2 = s1 P2, lbf/in2 T2 , F h2, Btu/lbm h3=h4, Btu/lbm wC = h2h1 qL = h1h4 =qL/(wC) R12 77.803 0.16843 151.11 127.29 91.107 33.531 13.3 44.27 3.33 2 3 1 s 4 R22 104.954 0.22705 241.04 161.87 123.904 42.446 18.95 62.51 3.30 11118 11.150E The environmentally safe refrigerant R134a is one of the replacements for R12 in refrigeration systems. Repeat Problem 11.149 using R134a and compare the result with that for R12.
T Ideal Ref. Cycle Tcond = 110 F Tevap = 5 F h1 = 167.325 Btu/lbm s2 = s1 = 0.4145 Btu/lbm R P2 = 161.124 lbf/in2, T2 = 122.2 F, h3=h4 = 112.455 Btu/lbm 2 3 1 s 4 h2 = 184.44 Btu/lbm wC = h2h1 = 17.115 Btu/lbm ; qL = h1h4 = 54.87 Btu/lbm =qL/(wC) = 3.206 11.151E Consider an ideal heat pump that has a condenser temperature of 120 F and an evaporator temperature of 30 F. Determine the coefficient of performance of this heat pump for the working fluids R12, R22, and ammonia.
T Ideal Heat Pump Tcond = 120 F Tevap = 30 F Use Table C.9 for NH3 Use Table C.10 for R22 Use computer table for R12 R12 h1, Btu/lbm 80.42 s 2 = s1 P2, lbf/in2 T2 , F h2, Btu/lbm h3=h4, Btu/lbm wC = h2h1 qH = h2h3 =qH/(wC) 0.1665 172.3 132.2 91.0 36.011 10.58 54.995 5.198 2 3 1 s 4 R22 107.28 0.2218 274.6 160.4 122.17 45.71 14.89 76.46 5.135 NH3 619.58 1.2769 286.5 239.4 719.5 178.83 99.92 540.67 5.411 11119 11.152E The refrigerant R22 is used as the working fluid in a conventional heat pump cycle. Saturated vapor enters the compressor of this unit at 50 F; its exit temperature from the compressor is measured and found to be 185 F. If the isentropic efficiency of the compressor is estimated to be 70%, what is the coefficient of performance of the heat pump?
T R22 heat pump: T2 = 185 F TEVAP = 50 F, S COMP = 0.70 Isentropic compressor: s2S = s1 = 0.2180 but P2 unknown. Trial & error. Assume P2 = 307 lbf/in2 At P2,s2S: T2S = 162 F, h2S = 121.07, At P2,T2: h2 = 126.24 calculate S COMP= h2S  h1 h2  h1 = 121.07  108.95 = 0.701 0.70 126.24  108.95
3 2 2S 4 1 s OK P2 = 307 lbf/in2 = P3 T3 = 128.8 F, h3 = 48.66 wC = h2  h1 = 17.29, = qH/(wC) = 4.49 qH = h2  h3 = 77.58, 11.153E 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 120 F, and saturated vapor leaves the evaporator at 50 F. If 3000 Btu of heat is required in the generator (solar collector) per poundmass of ammonia vapor generated, determine the overall performance of this system. NH3 absorption cycle: sat. vapor at 120 F exits the generator. Sat. vapor at 50 F exits the evaporator qH = qGEN = 3000 Btu/lbm NH3 out of generator. qL = h2  h1 = hG 50 F  hF 120 F = 624.28  178.79 = 445.49 Btu/lbm qL/qH = 445.49/3000 = 0.1485
T GEN. EXIT EVAP EXIT s 120F 50 F 1 2 11120 11.154E Consider an ideal dualloop heatpowered refrigeration cycle using R12 as the working fluid, as shown in Fig. P11.109. Saturated vapor at 220 F leaves the boiler and expands in the turbine to the condenser pressure. Saturated vapor at 0 F 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 110 F 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.
T 1 TURB. 6 BOIL. 5 P 7 COND. 3 COMP. 2 E V A P . 4 . QL 6 5 3 4 7 1 s 2 T F 1 2 3 4 5 6 7 0 110 0 220 110 P lbf/in 23.849 151.11 151.11 23.849 524.43 524.43 151.11
2 h Btu/lbm 77.271 s Btu/lbm R 168.88 168.88 Computer tables for properties. P2=P3=PSAT at 110 F P5=P6=PSAT at 220 F s2=s1=0.168 88 h2=91.277 Pump work: wP = h5h3 v5(P5P3) 33.531 33.531 89.036 0.067 45 0.067 45 0.151 49 0.151 49 wP = 0.0129(524.4  151.1) 144 = 0.894 778 h5 = 33.531 + 0.894 = 34.425 Btu/lbm (1x7) = 0.162 79  0.151 49 0.011 30 = = 0.1187 0.095 34 0.095 34 h7 = 87.844  0.1187(54.313) = 81.397 Btu/lbm 11121 CV: turbine + compressor . . . . Continuity Eq.: m1 = m2, m6 = m7 . . . . Energy Eq.: m1h1 + m6h6 = m2h2 + m7h7 89.03681.397 7.639 . . m1/m6 = = = 0.545, 91.27777.271 14.006 CV: pump: . . m6/m1 = 1.833 wP = v3(P5P3), h5 = h3  wP . . . . CV evaporator: QL = m1(h1h4), CV boiler: QH = m6(h6h5) . m1(h1h4) . . 77.27133.531 = = 0.436 = QL/QH = . m6(h6h5) 1.833(89.03634.425) 11.155E (Adv.) Find the availability of the water at all four states in the Rankine cycle described in Problem 11.121. Assume the hightemperature source is 900 F and the lowtemperature reservoir is at 65 F. Determine the flow of availability in or out of the reservoirs per poundmass of steam flowing in the cycle. What is the overall cycle second law efficiency? Ref. state 14.7 lbf/in2, 77F, h0 = 45.08 Btu/lbm, From solution to 11.121: s1 = 0.17497, h1 = 94.01, s2 = 0.175 = s1, h2 = 95.81, s4 = s3 = 1.5871 Btu/lbm R h4 = 921.23 s0 = 0.08774 Btu/lbm R h3 = 1350.6, = h  h0  T0(s  s0) 1 = 94.01  45.08  536.67(0.17497  0.08774) = 2.116 2 = 3.92, 3 = 500.86, 4 = 71.49 Btu/lbm H = (1  T0/TH)qH = 0.6054 1254.79 = 759.65 Btu/lbm L = (1  T0/T0)qC = 0 / II = wNET/H = (429.37  1.8)/759.65 = 0.563 Notice TH > T3, TL < T4 = T1, so cycle is externally irreversible. Both q and H qC over finite T. 121 CHAPTER 12
The correspondence between the new problem set and the previous 4th edition chapter 11 problem set. New 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 Old New New New 1 New 3 new new new new new new 5 new 4 new new new 6 7 new new 12 8a mod 8b mod 10 new 11 13 new New 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 59 60 Old 15 new new 16 17 new new 20 new new 24 new 26 new 23 32 40 37 46 28 new new 39 new new new 35 new 27 21 New 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 Old new 22 25 38 50 new 43 44 47 48 49 52 41 9 14 33 mod 34 mod 36 42 51 The problems that are labeled advanced start at number 74. 122 The English unit problems are: New 81 82 83 84 85 86 87 88 89 Old new new new new new 77 78 79 82 New 90 91 92 93 94 95 96 97 98 99 Old 80 84 86 85 mod 87 91 93 94 88 97 New 100 101 102 103 Old 96 99 83 98 123 12.1 A gas mixture at 20C, 125 kPa is 50% N2, 30% H2O and 20% O2 on a mole basis. Find the mass fractions, the mixture gas constant and the volume for 5 kg of mixture. From Eq. 12.3: ci = yi Mi/ yjMj MMIX = yjMj = 0.528.013 + 0.318.015 + 0.231.999 = 14.0065 + 5.4045 + 6.3998 = 25.811 cN2 = 14.0065 / 25.811 = 0.5427, cH2O = 5.4045 / 25.811 = 0.2094 cO2 = 6.3998 / 25.811 = 0.2479, sums to 1 OK  RMIX = R/MMIX = 8.3145 / 25.811 = 0.3221 kJ/kg K V = mRMIX T/P = 50.3221393.15/125 = 5.065 m3 12.2 A 100 m3 storage tank with fuel gases is at 20C, 100 kPa containing a mixture of acetylene C2H2, propane C3H8 and butane C4H10. A test shows the partial pressure of the C2H2 is 15 kPa and that of C3H8 is 65 kPa. How much mass is there of each component? Assume ideal gases, then the ratio of partial to total pressure is the mole fraction, y = P/Ptot yC2H2 = 15/100 = 0.15, yC3H8 = 65/100 = 0.65, yC4H10 = 20/100 = 0.20  ntot = PV/RT = 100 100 / (8.31451 293.15) = 4.1027 kmoles mC2H2 = (nM)C2H2 = yC2H2ntotMC2H2 = 0.154.102726.038 = 16.024 kg mC3H8 = (nM)C3H8 = yC3H8ntotMC3H8 = 0.654.102744.097 = 117.597 kg mC4H10 = (nM)C4H10 = yC4H10ntotMC4H10 = 0.204.102758.124 = 47.693 kg 12.3 A mixture of 60% N2, 30% Ar and 10% O2 on a mole basis is in a cylinder at 250kPa, 310 K and volume 0.5 m3. Find the mass fractions and the mass of argon. From Eq. 12.3: ci = yi Mi/ yjMj MMIX = yjMj = 0.628.013 + 0.339.948 + 0.131.999 = 31.992 cN2 = (0.628.013) / 31.992 = 0.5254, cAr = (0.339.948) / 31.992 = 0.3746 cO2 = (0.131.999) / 31.992 = 0.1, sums to 1 OK  RMIX = R/MMIX = 8.3145 / 31.992 = 0.260 kJ/kg K mMIX = PV/(RMIX T) = 2500.5 / 0.26310 = 1.551 kg mAr = cAr mMIX = 0.37461.551 = 0.581 kg 124 12.4 A carbureted internal combustion engine is converted to run on methane gas (natural gas). The airfuel ratio in the cylinder is to be 20 to 1 on a mass basis. How many moles of oxygen per mole of methane are there in the cylinder? The mass ratio mAIR/mCH = 20, so relate mass and mole 4 nAIR nCH4 = n = m/M (m ) M m
AIR CH4 CH4/MAIR = 20 16.04/28.97 = 11.0735 nO2 nCH4 = 0.2111.0735 = 2.325 mole O2/mole CH4 12.5 Weighing of masses gives a mixture at 60C, 225 kPa with 0.5 kg O2, 1.5 kg N2 and 0.5 kg CH4. Find the partial pressures of each component, the mixture specific volume (mass basis), mixture molecular weight and the total volume. From Eq. 12.4: yi = (mi /Mi) / mj/Mj ntot = mj/Mj = (0.5/31.999) + (1.5/28.013) + (0.5/16.04) = 0.015625 + 0.053546 + 0.031172 = 0.100343 yO2 = 0.015625/0.100343 = 0.1557, yCH4 = 0.031172/0.100343 = 0.3107 PO2 = yO2 Ptot = 0.1557225 = 35 kPa, PN2 = yN2 Ptot = 0.5336225 = 120 kPa PCH4 = yCH4 Ptot = 0.3107225 = 70 kPa  Vtot = ntot RT/P = 0.100343 8.31451 333.15 / 225 = 1.235 m3 v = Vtot/mtot = 1.235 / (0.5 + 1.5 + 0.5) = 0.494 m3/kg MMIX = yjMj = mtot/ntot = 2.5/0.100343 = 24.914 yN2 = 0.053546/0.100343 = 0.5336, 125 12.6 At a certain point in a coal gasification process, a sample of the gas is taken and stored in a 1L cylinder. An analysis of the mixture yields the following results: Component H2 CO CO2 N2 Percent by volume 25 40 15 20 Determine the mass fractions and total mass in the cylinder at 100 kPa, 20C. How much heat transfer must be transferred to heat the sample at constant volume from the initial state to 100C? Volume fractions same as mole fractions so From Eq. 12.3: ci = yi Mi/ yjMj yi kmoli/kmol H2 CO CO2 N2 0.25 0.40 0.15 0.20 Mi kgi/kmoli 2.016 28.01 44.01 28.013 = kgi/kg = kgi/kmol = 0.504/23.9121 = 11.204/23.9121 = 6.6015/23.9121 = 5.6026/23.9121 MMIX = 23.9121  RMIX = R/MMIX = 8.3145/23.9121 = 0.34771 kJ/kg/K m = PV/RT = 100103/0.34771 293.15 = 9.81104 kg CV0 MIX = ciCV0 i = 0.021110.0849 + 0.46850.7445 + 0.27610.6529 + 0.23430.7448 = 0.9164 kJ/kg K Q 12 = U12 = mCV0(T2T1) = 9.81104 0.9164(10020) = 0.0719 kJ 12.7 A pipe, cross sectional area 0.1 m2, carries a flow of 75% O2 and 25% N2 by mole with a velocity of 25 m/s at 200 kPa, 290 K. To install and operate a mass flow meter it is necessary to know the mixture density and the gas constant. What are they? What mass flow rate should the meter then show? From Eq. 12.3: ci = yi Mi/ yjMj MMIX = yjMj = 0.7531.999 + 0.2528.013 = 31.0025  RMIX = R/MMIX = 8.3145 / 31.0025 = 0.2682 kJ/kg K v = RMIX T/P = 0.2682 290 / 200 = 0.38889 m3/kg = 1/v = 2.5714 kg/m3 . m = AV = 2.57140.125 = 6.429 kg/s = ci = kgi/kg = 0.0211 = 0.4685 = 0.2761 = 0.2343 126 12.8 A pipe flows 0.05 kmole a second mixture with mole fractions of 40% CO2 and 60% N2 at 400 kPa, 300 K. Heating tape is wrapped around a section of pipe with insulation added and 2 kW electrical power is heating the pipe flow. Find the mixture exit temperature. C.V. Pipe heating section. Assume no heat loss to the outside, ideal gases. . .  . . Energy Eq.: Q = m(he  hi) = n(he  hi) = nCP mix(Te  Ti)   CP mix = yi Ci = 0.4 37.056 + 0.6 29.189 = 32.336 kJ/kmole . . Te = Ti + Q / nCP mix = 300 + 2/(0.05 32.336) = 301.2 K 12.9 A rigid insulated vessel contains 0.4 kmol of oxygen at 200 kPa, 280 K separated by a membrane from 0.6 kmol carbon dioxide at 400 kPa, 360 K. The membrane is removed and the mixture comes to a uniform state. Find the final temperature and pressure of the mixture. C.V. Total vessel. Control mass with two different initial states. Mass: n = nO2 + nCO2 = 0.4 + 0.6 = 1.0 kmole Process: V = constant (rigid) => W = 0, insulated => Q = 0   Energy: U2  U1 = 0  0 = n O2 CV O2(T2  T1 O2) + nCO2CV CO2(T2  T1 CO2) Initial state from ideal gas Table A.5   CV O2 = 0.661831.999 = 21.177, CV CO2 = 0.652944.01 = 28.73 kJ/kmole  O2 : VO2 = nRT1/P = 0.4 8.3145 280/200 = 4.656 m3,  CO2 : VCO2 = nRT1/P = 0.6 8.3145 360/400 = 4.49 m3     nO2 CV O2 T2 + nCO2CV CO2 T2 = nO2 CV O2 T1 O2 + nCO2CV CO2 T1 CO2 (8.4708 + 17.238 ) T2 = 2371.8 + 6205.68 = 8577.48 kJ => T2 = 333.6 K  P2 = nRT2/V = 1 8.3145 333.6 / 9.146 = 303.3 kPa 12.10 An insulated gas turbine receives a mixture of 10% CO 2, 10% H2O and 80% N2 on a mole basis at 1000 K, 500 kPa. The volume flow rate is 2 m 3/s and its exhaust is at 700 K, 100 kPa. Find the power output in kW using constant specific heat from A.5 at 300 K. C.V. Turbine, SSSF, 1 inlet, 1 exit flow with an ideal gas mixture, q = 0. . .  . . Energy Eq.: WT = m(hi  he) = n(hi  he) = nCP mix(Ti  Te) . .   PV = nRT => n = PV / RT = 5002/(8.31451000) = 0.1203 kmole/s   CP mix = yi Ci = 0.1 44.01 0.842 + 0.1 18.015 1.872 + 0.8 28.013 1.042 = 30.43 kJ/kmol K . WT = 0.1203 30.43 (1000  700) = 1098 kW 127 12.11 Solve Problem 12.10 using the values of enthalpy from Table A.8. C.V. Turbine, SSSF, 1 inlet, 1 exit flow with an ideal gas mixture, q = 0. . .  . .  Energy Eq.: WT = m(hi  he) = n(hi  he) = n [ yj (hi  he)j ] . .   PV = nRT => n = PV / RT = 5002/(8.31451000) = 0.1203 kmole/s . WT = 0.1203[0.1(33397  17754) + 0.1(26000  14190) + 0.8(21463  11937)] = 1247 kW 12.12 Consider Problem 12.10 and find the value for the mixture heat capacity, mole basis and the mixture ratio of specific heats, kmix, both estimated at 850 K from values (differences) of h in Table A.8. With these values make an estimate for the reversible adiabatic exit temperature of the turbine at 100 kPa.  We will find the individual heat capacities by : CP i = (h900 h800)/(900  800)   CP CO2 = (28030  22806)/100 = 52.24; CP H2O = (21937  18002)/100 = 39.35  CP N2 = (18223  15046)/100 = 31.77   CP mix = yi CP i = 0.1 52.24 + 0.1 39.35 + 0.8 31.77 = 34.575 kJ/kmol      CV mix = CP mix  R = 34.575  8.3145 = 26.26, k = CP mix/CV mix = 1.3166 Rev. ad. turbine Te = => s = constant. Assume const. avg. heat capacities:
0.3166 k1 Ti(Pe/Pi) k = 1000 (100/500)1.3166 = 1000 (0.2)0.2405 = 679 K 12.13 The gas mixture from Problem 12.6 is compressed in a reversible adiabatic process from the initial state in the sample cylinder to a volume of 0.2 L. Determine the final temperature of the mixture and the work done during the process. CP0 MIX = CV0 MIX + RMIX = 0.9164 + 0.34771 = 1.2641 k = CP0/CV0 = 1.2641/0.9164 = 1.379 T2 = T 1 (V ) V
2 1 k1 = 293.15 1 (0.2) 0.3805 = 539.5 K W12 =  U12 = mCV0(T2T1) =  9.81104 0.9164(539.5293.15) =  0.22 kJ 128 12.14 Three SSSF flows are mixed in an adiabatic chamber at 150 kPa. Flow one is 2 kg/s of O2 at 340 K, flow two is 4 kg/s of N2 at 280 K and flow three is 3 kg/s of CO2 at 310 K. All flows are at 150 kPa the same as the total exit pressure. Find the exit temperature and the rate of entropy generation in the process. C.V. Mixing chamber, no heat transfer, no work. . . . . Continuity: m1 + m2 + m3 = m4 . . . . Energy: m1h1 + m2h2 + m3h3 = m4h4 . . . . . Entropy: m1s1 + m2s2 + m3s3 + Sgen = m4s4 1 O2 N2 CO2 2 3 4 mix Assume ideal gases and since T is close to 300 K use heat capacity from A.5 . . . m1CP O2(T1  T4) + m2CP N2(T2  T4) + m3CP CO2(T3  T4)= 0 2 0.922 340 + 4 1.042 280 + 3 0.842 310 = (2 0.922 + 4 1.042 + 3 0.842 ) T4 => 2577.06 = 8.538 T 4 => T4 = 301.83 K State 4 is a mixture so the component exit pressure is the partial pressure. For each component se  si = CP ln(Te / Ti)  R ln(Pe / Pi) and the pressure ratio is Pe / Pi = y P4 / Pi = y for each. m 2 4 3 = + + = 0.0625 + 0.1428 + 0.06817 = 0.2735 M 32 28.013 44.01 n= 0.0625 0.1428 0.06817 = 0.2285, yN2 = = 0.5222, yCO2 = = 0.2493 yO2 = 0.2735 0.2735 0.2735 . . . . Sgen = m1(s4  s1) + m2(s4  s2) + m3(s4  s3) = 2 [ 0.922 ln(301.83/340)  0.2598 ln(0.2285)] + 4 [ 1.042 ln(301.83/280)  0.2968 ln(0.5222)] + 3 [ 0.842 ln(301.83/310)  0.1889 ln(0.2493)] = 0.5475 + 1.084 + 0.2399 = 1.871 kW/K 129 12.15 Carbon dioxide gas at 320 K is mixed with nitrogen at 280 K in a SSSF insulated mixing chamber. Both flows are at 100 kPa and the mole ratio of carbon dioxide to nitrogen is 2 1. Find the exit temperature and the total entropy generation per mole of the exit mixture. . . CV mixing chamber, SSSF. The inlet ratio is nCO = 2 nN and assume no 2 2 external heat transfer, no work involved. . . . . Continuity: nCO + 2nN = nex = 3nN ; 2 2 2 . . Energy Eq.: nN (hN + 2hCO ) = 3nN hmix ex 2 2 2 2  Take 300K as reference and write h = h300 + CPmix(T300). CP N (Ti N 300) + 2CP CO (Ti CO 300) = 3CP mix(Tmix ex300) 2 2 2 2 CP mix = yiCP i = (29.178 + 2 37.05)/3 = 34.42 3CP mixTmix ex = CP N2Ti N2 + 2CP CO2Ti CO2 = 31882 Tmix ex = 308.7 K; Pex N = Ptot/3; Pex CO = 2Ptot/3 2 2 . . ...  .  Sgen = nexsex(ns)iCO (ns)iN = nN (sesi)N + 2nN (sesi)CO 2 2 2 2 2 2 Tex  Tex . .  Sgen/3nN2 = [CPN2ln  Rln yN2 + 2CPCO2ln  2 Rln yCO2]/3 TiN2 TiCO2 = [2.8485 + 9.1343  2.6607+6.742]/3 = 5.35 kJ/kmol mix K 12.16 A piston/cylinder contains 0.5 kg argon and 0.5 kg hydrogen at 300 K, 100 kPa. The mixture is compressed in an adiabatic process to 400 kPa by an external force on the piston. Find the final temperature, the work and the heat transfer in the process. C.V. Mixture in cylinder. Control mass with adiabatic process: 1Q2 = 0 Cont.Eq.: m 2 = m1 = m ; Energy Eq.: m(u2  u1) =  1W2 assume reversible. m(s2  s1) = dQ/T + 1S2 gen = 0 + 0 Entropy Eq.: Rmix = ciRi = 0.5 0.2081 + 0.5 4.1243 = 2.1662 CP mix = ciCPi = 0.5 0.52 + 0.5 14.209 = 7.3645 CV = CP  R = 7.36452.1662 = 5.1983, k = CP / CV = 1.417 s2 = s1 => T2 = T1(P2/P1)(k1)/k = 300 (400/100)0.2943 = 451 K
1W2 = m(u1  u2) = mCV (T1  T2) = 15.1983(300  451) = 784.9 kJ 1210 12.17 Natural gas as a mixture of 75% methane and 25% ethane by volume is flowing to a compressor at 17C, 100 kPa. The reversible adiabatic compressor brings the flow to 250 kPa. Find the exit temperature and the needed work per kg flow. C.V. Compressor. SSSF, adiabatic q = 0, reversible sgen = 0 Energy Eq.: w = hex  hin ; Entropy Eq.: se = si si + sgen = si = se Process: reversible => sgen = 0 => Assume ideal gas mixture and constant heat capacity, so we need k and CP CP = yi CPi = 0.75 2.254 16.043 + 0.25 1.766 30.07 = 40.3966 Mmix = 0.75 16.043 + 0.25 30.07 = 19.55  Cp = CP/ Mmix = 2.066 , R = R/ Mmix = 8.31451 / 19.55 = 0.4253 Cv = Cp  R = 1.6407 , k = Cp/ Cv = 1.259 Te = Ti (Pe/ Pi)(k1)/k = 290 (250/100) 0.2057 = 350.2 K wc in = CP (Te Ti) = 2.066 (350.2 290) = 124.3 kJ/kg 12.18 Take Problem 12.15 with inlet temperature of 1400 K for the carbon dioxide and 300 K for the nitrogen. First estimate the exit temperature with the specific heats from Table A.5 and use this to start iterations using A.8 to find the exit temperature. . . CV mixing chamber, SSSF. The inlet ratio is nCO = 2 nN and assume no 2 2 external heat transfer, no work involved. CP CO2 = 44.01 0.842 = 37.06 CP N2 = 28.013 1.042 = 29.189 . . . . Cont. Eq.: 0 = nin  nex; Energy Eq.: 0 = nin hin  nex hex . . 0 = 2nN2 CP CO2(Tin Tex) CO2 + nN2 CP N2 (Tin Tex) N2 0 = 2 37.06 (1400Tex) + 29.189 (300Tex) 0 = 103768 + 8756.7 103.309 Tex _ Tex = 1089 K . . . From Table A.8: nin hin = nN2 [2 55895 + 1 54] = nN2 111844 . . . @ 1000K : nex hex = nN2 [2 33397 + 21463] = nN2 88257 . . . @ 1100K : nex hex = nN2 [2 38885 + 24760] = nN2 102530 . . . @ 1200K : nex hex = nN2 [2 44473 + 28109] = nN2 117055 Now linear interpolation between 1100 K and 1200 K Tex = 1100 + 100 111844102530 = 1164 K 117055102530 1211 12.19 A mixture of 60% helium and 40% nitrogen by volume enters a turbine at 1 MPa, 800 K at a rate of 2 kg/s. The adiabatic turbine has an exit pressure of 100 kPa and an isentropic efficiency of 85%. Find the turbine work. Assume ideal gas mixture and take CV as turbine. wT s = hi  hes, ses = si, Tes = Ti(Pe/Pi)(k1)/k CP mix = 0.6 5.1926 4.003 + 0.4 1.0416 28.013 = 24.143  (k1)/k = R/CP mix = 8.31451/24.143 = 0.3444 Mmix= 0.64.003 + 0.428.013 = 13.607, CP=CP/Mmix= 1.774 Tes = 800(100/1000)0.3444 = 362 K, wTs = CP(TiTes) = 777 . wT ac = wTs = 660.5 kJ/kg; W = mwTs = 1321 kW 12.20 A mixture of 50% carbon dioxide and 50% water by mass is brought from 1500 K, 1 MPa to 500 K, 200 kPa in a polytropic process through a SSSF device. Find the necessary heat transfer and work involved using values from Table A.5. Process Pvn = constant leading to n ln(v2/v1) = ln(P1/P2); v = RT/P n = ln(1000/200)/ln(500 1000/200 1500) =3.1507 Rmix = ciRi = 0.5 0.18892 + 0.5 0.46152 = 0.32522 CP mix = ciCPi = 0.5 0.8418 + 0.5 1.8723 = 1.357 n nR w = vdP = (PevePivi)= (T T ) = 476.4 kJ/kg n1 n1 e i q = hehi + w = CP(TeTi) + w = 880.6 kJ/kg 12.21 Solve Problem 12.20 using specific heats CP = h/T, from Table A.8 at 1000 K. CP CO2 = 3888528030 = 1.223 (1100900)44.01 3019021937 CP H2O = = 2.2906 (1100900)18.015 CP mix = ciCPi = 0.5 1.223 + 0.5 2.2906 = 1.7618 Rmix = ciRi = 0.5 0.18892 + 0.5 0.46152 = 0.32522 Process Pvn = C => n = ln(P1/P2) / ln(v2/v1) = 3.1507 n nR w = vdP = (P v P v )= (T T ) = 476.4 kJ/kg n1 e e i i n1 e i q = hehi + w = 1.7618(5001500) + 476.4 = 1285.4 kJ/kg 1212 12.22 A 50/50 (by mole) gas mixture of methane CH 4 and ethylene C2H4 is contained in a cylinder/piston at the initial state 480 kPa, 330 K, 1.05 m3. The piston is now moved, compressing the mixture in a reversible, polytropic process to the final state 260 K, 0.03 m3. Calculate the final pressure, the polytropic exponent, the work and heat transfer and net entropy change for the process. Ideal gas mixture: CH4, C2H4, 50% each by mol => yCH4 = yC2H4 = 0.5 State 1:  n = P1V1/RT1 = 480 1.05 / (8.31451 330) = 0.18369 kmol Cv mix = yi Mi Cvi = 31.477 kJ/kmolK V1 T2 = 13236 kPa State 2: T2 = 260 K, V2 = 0.03 m3, Ideal gas => P2 = P1 V T
2 1 Process: PVn = constant, V1n1 T2 V1 = , ln = (n1) ln => n = 0.933 T1 V2 T1 V2 1 => 1W2 = P dv = 1n(P 2V2 P1V1) = 1595.7 kJ  Energy Eq.: 1Q2 = n(u2 u1) + 1W2 = n Cv mix (T2 T1) + 1W2 T2 = 0.18369 31.477(260330) 1595.7 = 2000.4 kJ  s2 s1 = Cv mix ln (T2 / T1) + R ln (V2 / V1) = 37.065 kJ/kmolK 2nd Law:  s Snet = n(s2  1)  1Q2 /To with T0 = 260 K Snet = 0.18369 (37.065)  (2000.4)/260 = 0.8854 kJ/K 12.23 A mixture of 2 kg oxygen and 2 kg of argon is in an insulated piston cylinder arrangement at 100 kPa, 300 K. The piston now compresses the mixture to half its initial volume. Find the final pressure, temperature and the piston work. C.V. Mixture. Control mass, boundary work and no Q, assume reversible. Energy Eq.: u2  u1 = 1q2  1w2 =  1w2 ; Entropy Eq.: s2s1 = 0 + 0 = 0 Process: constant s => Pvk = constant, v2 = v1/2, Assume ideal gases (T1 >> TC ) and use kmix and Cv mix for properties. Rmix = ciRi = 0.5 0.25983 + 0.5 0.20813 = 0.234 kJ/kg mix CPmix = ciCPi = 0.5 0.9216 + 0.5 0.5203 = 0.721 kJ/kg mix Cvmix = CPmix Rmix = 0.487, kmix = CPmix/Cvmix = 1.4805 P2 = P1(v1/v2)k = P1(2)k = 100(2)1.4805 = 279 kPa T2 = T1(v1/v2)k1 = T1(2)k1 = 300(2)0.4805 = 418.6 K
1W2 = mtot (u1u2) = mtot Cv(T1T2) = 40.487 (300  418.6) = 231 kJ 1213 12.24 Two insulated tanks A and B are connected by a valve. Tank A has a volume of 1 m3 and initially contains argon at 300 kPa, 10C. Tank B has a volume of 2 m3 and initially contains ethane at 200 kPa, 50C. The valve is opened and remains open until the resulting gas mixture comes to a uniform state. Determine the final pressure and temperature. C.V. Tanks A + B. Control mass no W, no Q. Energy Eq.: U2U1 = 0 = nArCV0(T2TA1) + nC H CVO(T2TB1) 2 6  nAr = PA1VA/RTA1 = (300 1) / (8.31451 283.15) = 0.1274 kmol  nC2H6 = PB1 B/RTB1 = (200 2) / (8.31451 323.15) = 0.1489 kmol V Continuity Eq.: Energy Eq.: n2 = nAr + nC
2H6 = 0.2763 kmol 0.1274 39.948 0.3122 (T2  283.2) + 0.1489 30.07 1.4897 (T2  323.2) = 0 Solving, T2 = 315.5 K  P2 = n2RT2/(VA+VB) = 0.27638.3145315.5/3 = 242 kPa 1214 12.25 Reconsider the Problem 12.24, but let the tanks have a small amount of heat transfer so the final mixture is at 400 K. Find the final pressure, the heat transfer and the entropy change for the process. C.V. Both tanks. Control mass with mixing and heating of two ideal gases.  nAr = PA1VA/RTA1 = 3001 = 0.1274 kmol 8.3145283.2 2002 = 0.1489 kmol 8.3145323.2
2H6  nC2H6 = PB1 B/RTB1 = V Continuity Eq.: Energy Eq.: n2 = nAr + nC = 0.2763 kmol U2U1 = nArCV0(T2TA1) + nC H CVO(T2TB1) = 1Q2 2 6  P2 = n2RT2/(VA+VB) = 0.27638.3145400 / 3 = 306.3 kPa
1Q 2 = 0.127439.9480.3122(400  283.15) + 0.148930.071.4897(400  323.15) = 698.3 kJ SSYS = nArSAr + nC2H6SC2H6 SSURR = 1Q2/TSURR; yAr = 0.1274/0.2763 = 0.4611 T2  yArP2 SAr = CP Ar ln  R ln TA1 PA1 = 39.9480.5203 ln 0.4611306.3 400  8.3145 ln 283.15 300 = 13.445 kJ/kmol K SC T2  yC2H6P2 = CC2H6 ln  R ln 2H6 TB1 PB1 = 30.071.7662 ln 0.5389306.3 400  8.3145 ln 323.15 200 = 12.9270 kJ/kmol K Assume the surroundings are at 400 K (it heats the gas) SNET = nArSAr + nC2H6SC2H6 + SSURR = 0.127413.445 + 0.148912.9270  698.3/400 = 1.892 kJ/K 1215 12.26 A piston/cylinder contains helium at 110 kPa at ambient temperature 20C, and initial volume of 20 L as shown in Fig. P12.26. The stops are mounted to give a maximum volume of 25 L and the nitrogen line conditions are 300 kPa, 30C. The valve is now opened which allows nitrogen to flow in and mix with the helium. The valve is closed when the pressure inside reaches 200 kPa, at which point the temperature inside is 40C. Is this process consistent with the second law of thermodynamics? P1 = 110 kPa, T1 = 20 oC, V1 = 20 L, Vmax = 25 L = V2 P2 = 200 kPa, T2 = 40 oC, Pi = 300 kPa, Ti = 30 oC Const. P to stops, then const. V = Vmax Qcv = U2  U1 + Wcv  nihi, Wcv = P1(V2  V1) = n2h2  n1h1  nihi  (P2  P1)V2 = nA(hA2  hAi) + nB(hB2  hB1)  (P2  P1)V2 nB = n1 = P1V1/RT1 = 1100.02/8.3145293.2 = 0.0009 kmol n2 = nA + nB = P2V2/RT2 = 2000.025/8.3145313.2 = 0.00192 kmol, nA = n2  nB = 0.00102 kmol Qcv = 0.0010228.0131.0416(4030) + 0.00094.0035.1926(4020)  (200  110) 0.025 = 0.298 + 0.374  2.25 =  1.578 kJ Sgen = n2s2  n1s1  nisi  Qcv/T0 = nA(sA2  sAi) + nB(sB2  sB1)  Qcv/T0 yA2 = 0.00102/0.00192 = 0.5313, yB2 = 0.4687 313.2  0.5313*200 sA2  sAi = 29.178 ln  R ln = 9.5763 303.2 300 313.2  0.4687*200  R ln = 2.7015 sB2  sB1 = 20.786 ln 293.2 110 Sgen = 0.001029.5763 + 0.00092.7015 + 1.578/293.2 = 0.0176 kJ/K > 0 Satisfies 2nd law. 1216 12.27 Repeat Problem 12.17 for an isentropic compressor efficiency of 82%. C.V. Compressor. SSSF, adiabatic q = 0, reversible sgen = 0 Energy Eq.: w = hex  hin ; Entropy Eq.: si + sgen = si = se se = si Process: For ideal compressor reversible => sgen = 0 => Assume ideal gas mixture and constant heat capacity, so we need k and CP CP = yi CPi = 0.75 2.254 16.043 + 0.25 1.766 30.07 = 40.3966 Mmix = 0.75 16.043 + 0.25 30.07 = 19.55  Cp = CP/ Mmix = 2.066 , R = R/ Mmix = 8.31451 / 19.55 = 0.4253 Cv = Cp  R = 1.6407 , k = Cp/ Cv = 1.259 Te = Ti (Pe/ Pi)(k1)/k = 290 (250/100) 0.2057 = 350.2 K wc in = CP (Te Ti) = 2.066 (350.2 290) = 124.3 kJ/kg wc actual = wc in/ = 124.3/0.82 = 151.6 kJ/kg = Cp (Te actual  Ti) => Te actual = T + w c actual/CP = 290 + 151.6 / 2.066 = 363.4 K 12.28 A spherical balloon has an initial diameter of 1 m and contains argon gas at 200 kPa, 40C. The balloon is connected by a valve to a 500L rigid tank containing carbon dioxide at 100 kPa, 100C. The valve is opened, and eventually the balloon and tank reach a uniform state in which the pressure is 185 kPa. The balloon pressure is directly proportional to its diameter. Take the balloon and tank as a control volume, and calculate the final temperature and the heat transfer for the process. PA1VA1 2000.5236 A = = 1.606 kg VA1 = 13 = 0.5236, mA1 = RTA1 6 0.208 13313.2 mB1 = PB1VB1/RTB1 = 1000.50/0.18892373.2 = 0.709 kg B P2VA2 = PA1VA1 VA2 = VA1 2: Uniform ideal gas mix. :
1/3 1/3 CO 2 (PP ) = 0.5236(185) 200
2 3 A1 3 = 0.4144 m3 P2(VA2+VB) = (mARA+mBRB)T2 T2 = 185(0.4144+0.50) / (1.6060.20813 + 0.7090.18892) = 361.3 K W12 = P2VA2PA1VA1 1(1/3) = 1850.41442000.5236 = 21.0 kJ (4/3) Q = mACV0A(T2TA1) + mBCV0B(T2TB1) + W12 = 1.6060.3122(361.3313.2) + 0.7090.6529(361.3373.2)  21.0 = 18.6  21.0 = 2.4 kJ 1217 12.29 A large SSSF air separation plant takes in ambient air (79% N2, 21% O2 by volume) at 100 kPa, 20C, at a rate of 1 kmol/s. It discharges a stream of pure O2 gas at 200 kPa, 100C, and a stream of pure N 2 gas at 100 kPa, 20C. The plant operates on an electrical power input of 2000 kW. Calculate the net rate of entropy change for the process. Air 79 % N2 21 % O2 P1 = 100 kPa T1 = 20 oC . n1 = 1 kmol/s
1 2 pure O 2 pure N 2 3 P2 = 200 kPa T2 = 100 oC P3 = 100 kPa T3 = 20 oC . WIN = 2000 kW . . .  . . . . Sgen = nisi  QCV/T0 = (n2 2 + n3 3  n1 1)  QCV/T0 s s s . . . . . . QCV = nhi + WCV = nO2CP0 O2(T2T1) + nN2CP0 N2(T3T1) + WCV = 0.21 32 0.9216 (10020) + 0  2000 = 1505 kW nisi = 0.21 29.49 ln 293.2  8.3145 ln 21 . [ 373.2 200 + 0.79 [0  8.3145 ln (100/79)]= 3.9894 kW/K . Sgen = 1505/293.2  3.9894 = 1.1436 kW/K 1218 12.30 An insulated vertical cylinder is fitted with a frictionless constant loaded piston of cross sectional area 0.1 m 2 and the initial cylinder height of 1.0 m. The cylinder contains methane gas at 300 K, 150 kPa, and also inside is a 5L capsule containing neon gas at 300 K, 500 kPa. The capsule now breaks, and the two gases mix together in a constant pressure process. What is the final temperature, final cylinder height and the net entropy change for the process. Ap = 0.1 m2, h = 1.0 m => Vtot = Va1 + Vb1 = 0.1 m3 Methane: M = 16.04 kg/kmol, Cp = 2.2537 kJ/kgK, R = 0.51835 kJ/kgK Neon: M = 20.183 kg/kmol, Cp = 1.0299 kJ/kgK, R = 0.41195 kJ/kgK State 1: Methane, Tal = 300 K, Pa1 = 150 kPa, Va1 = Vtot Vb1 = 0.095 m3 Neon, Tbl = 300 K, Pb1 = 500 kPa, Vb1 = 5 L ma = mb = PalVal RaTal PblVbl RbTbl = 0.0916 kg, na = = 0.0202 kg, nb = ma Ma mb Mb = 0.00571 kmol = 0.001 kmol State 2: Mix, P2mix = Pa1 = 150 kPa Energy Eq: 1Q2 = ma(ua2ua1) + mb( ub2  ub1) + 1W2;
1W2 1Q2 =0 = P dV = P2(V2V1) tot; P2V2 = mtotRmixT2 = (maRa + mbRb)T2 Assume Constant Specific Heat 0 = maCpa(T2Ta1) + mbCpb 2Tb1 + (maRa + mbRb)T2  P2V1 tot (T Solving for T2 = 293.3 K V2 = (maRa + mbRb)T2 P2 = 0.1087 m3, h2 = V2 Ap = 1.087 m
1Q 2 Entropy Eq.: na na+nb Snet = ma(sa2 sa1) + mb(sb2sb1) = 0.851, yb = 1  ya = 0.149 T0 ; 1Q2 =0 ya = T2 yaP2 = 0.02503 kJ/kgK sa2  sa1 = Cpa ln  Raln Ta1 Pa1 T2 ybP2  Rb ln = 1.2535 kJ/kgK sb2  sb1 = Cpb ln Tb1 Pb1 Snet = 0.0276 kJ/K 1219 12.31 The only known sources of helium are the atmosphere (mole fraction approximately 5 106) and natural gas. A large unit is being constructed to separate 100 m3/s of natural gas, assumed to be 0.001 He mole fraction and 0.999 CH 4. The gas enters the unit at 150 kPa, 10C. Pure helium exits at 100 kPa, 20C, and pure methane exits at 150 kPa, 30C. Any heat transfer is with the surroundings at 20C. Is an electrical power input of 3000 kW sufficient to drive this unit? P2 = 100 kPa 1 2 He 0.999 CH4 T = 20 oC 0.001 He at 150 kPa, 10 oC . V1 = 100 m3/s
2 CH 4 . WCV = 3000 kW 3 P3 = 140 kPa T3 = 30 oC . . n1 = P1V1/RT1 = 150100/(8.3145283.2) = 6.37 kmol/s . . . => n2 = 0.001; n1 = 0.006 37; n3 = 6.3636 CP CH4= 16.042.2537 = 36.149 CP He= 4.0035.1926 = 20.786, . . . . .  . . . QCV = n2h2 + n3h3  n1h1 + WCV = n2CP0 He(T2T1) + n3CP0 CH (T3T1) + WCV 4 = 0.0063720.786(2010)+6.363636.149(3010)+(3000) = +1600 kW . . . . . s s s Sgen = n2 2 + n3 3  n1 1  QCV/T0 = 0.00637 20.786 ln [ 293.2 100  8.3145 ln 283.2 0.001150 ]  1600/293.2 + 6.3636 36.149 ln = +13.498 kW/K > 0 [ 303.2 140  8.3145 ln 283.2 0.999150 12.32 A steady flow of 0.01 kmol/s of 50% carbon dioxide and 50% water at 1200K and 200 kPa is used in a heat exchanger where 300 kW is extracted from the flow. Find the flow exit temperature and the rate of change of entropy using Table A.8. C.V. Heat exchanger, SSSF, 1 inlet, 1 exit, no work. Continuity Eq.: yCO2 = yH2O = 0.5 .  . . . . Energy Eq.: Q = m(he  hi) = n(he  hi) => he = hi + Q/n Inlet state: Table A.8 hi = 0.5 44473 + 0.5 34506 = 39489.5 kJ/kmol . . Exit state: he = hi + Q/n = 39489.5  300/0.01 = 9489.5 kJ/kmol Trial and error for T with h values from Table A.8 @500 K he = 0.5(8305 + 6922) = 7613.5 @600 K he = 0.5(12906 + 10499) = 11702.5 Interpolate to have the right h: T = 545.9 K 1220 12.33 An insulated rigid 2 m3 tank A contains CO2 gas at 200C, 1MPa. An uninsulated rigid 1 m3 tank B contains ethane, C2H6, gas at 200 kPa, room temperature 20C. The two are connected by a oneway check valve that will allow gas from A to B, but not from B to A. The valve is opened and gas flows from A to B until the pressure in B reaches 500 kPa and the valve is closed. The mixture in B is kept at room temperature due to heat transfer. Find the total number of moles and the ethane mole fraction at the final state in B. Find the final temperature and pressure in tank A and the heat transfer to/from tank B. Tank A: VA = 2 m3, state A1 : CO2, TA1 = 200C = 473.2 K, PA1 = 1 MPa C v0 CO2 = 0.653 44.01 = 28.74, C P0 CO2 = 0.842 44.01 = 37.06 Tank B: VB = 1 m3, state B1: C2H6, TB1 = 20C = 293.2 K, PB1 = 200 kPa Slow Flow A to B to PB2 = 500 kPa and assume TB2 = TB1 = T0 PB1VB = nB1RTB1 , yC2H6 B2 = nB1 = PB1VB RTB1 nB1 nB2 = PB1 PB2 = PB2VB = nB2 mix RTB2 200 = 0.400 500 = 200 1 = 0.08204 kmol R 293.2 = 500 1 = 0.2051 kmol R 293.2 nB2 mix = PB2VB RTB2 nCO2 B2 = 0.2051 0.08201 = 0.12306 kmol 10002 = 0.50833 kmol ; nA2 = nA1  nCO2 B2 = 0.38527 kmol RTA1 R473.2 =0=n u n u +n h C.V. A: All CO Energy Eq.: Q nA1 = =
2 CV A A2 A2 A1 A1 ave ave PA1VA 0 = nA2 C v0 CO2TA2  nA1 C v0 CO2TA1 + nave C P0 CO2 ( TA1 + TA2)/2 0 = 28.74( 0.38527 TA2  0.50833473.2) + 0.12306 37.06(473.2 + TA2)/2 => PA2 = C.V. B: QCV B + nBi h Bi ave = (nu) B2  (nu) B1 = (nu)CO2 B2 + (nu)C2H6 B2  (nu)C2H6 B1 QCV B = 0.12306 28.74 293.2 + 0 0.12306 37.06 (473.2 + 436.9)/2 = 1037.8 kJ TA2 = 436.9 K = 0.38527 R 436.9 = 700 kPa 2 nA2 RTA2 VA 1221 12.34 A tank has two sides initially separated by a diaphragm. Side A contains 1 kg of water and side B contains 1.2 kg of air, both at 20C, 100 kPa. The diaphragm is now broken and the whole tank is heated to 600C by a 700C reservoir. Find the final total pressure, heat transfer and total entropy generation. C.V. Total tank out to reservoir. Energy Eq.: U2U1 = ma(u2u1)a + mv(u2u1)v = 1Q2 V2 = VA + VB = mvvv1 + mava1 = 0.001 + 1.009 = 1.01 m3 Entropy Eq.: S2S1 = ma(s2s1)a + mv(s2s1)v = 1Q2/Tres + Sgen Volume: vv2 = V2/mv = 1.01, T2 => P2v = 400 kPa va2 = V2/ma = 0.8417, T2 => P2a = mRT2/V2 = 297.7 kPa P2tot = P2v + P2a = 697.7 kPa Water table B.1: u1 = 83.95, u2 = 3300, s1 = 0.2966, s2 = 8.4558 Air table A.7: u1 = 293, u2 = 652.3, sT1 = 2.492, sT2 = 3.628 From energy equation we have
1Q2 = 1(3300  83.95) + 1.2(652.3  293) = 3647.2 kJ From the entropy equation we have Sgen = 1(8.4558  0.2966) + 1.2[3.628  2.492  0.287ln(301.6/100)]  3647.2 / 973.2 = 5.4 kJ/K 1222 12.35 A 0.2 m3 insulated, rigid vessel is divided into two equal parts A and B by an insulated partition, as shown in Fig. P12.35. The partition will support a pressure difference of 400 kPa before breaking. Side A contains methane and side B contains carbon dioxide. Both sides are initially at 1 MPa, 30C. A valve on side B is opened, and carbon dioxide flows out. The carbon dioxide that remains in B is assumed to undergo a reversible adiabatic expansion while there is flow out. Eventually the partition breaks, and the valve is closed. Calculate the net entropy change for the process that begins when the valve is closed.
A CH 4 B CO 2 PMAX = 400 kPa, VA1 = VB1 = 0.1 m3 PA1 = PB1 = 1 MPa TA1 = TB1 = 30 oC = 303.2 K CO2 inside B: sB2 = sB1 to PB2 = 600 kPa (PA2 = 1000 kPa) 600 1.289 = 270.4 K For CO2, k = 1.289 => TB2 = 303.2 1000 nB2 = PB2VB2/RTB2 = 6000.1/8.3145270.4 = 0.026 688
0.289 ( ) nA2 = nA1 = 10000.1/8.3145303.2 = 0.039 668 kmol Q23 = 0 = n 3u3  ni2ui2 + 0 = n A2CV0 A(T3TA2) + nB2CV0 B(T3TB2) = 0 i 0.039 66816.041.736(T3303.2) + 0.026 68844.010.653(T3270.4) = 0 Solve T3 = 289.8 K P3 = 0.0663568.3145289.8/0.2 = 799.4 kPa PA3 = 0.5978799.4 = 477.9 kPa , PB3 = P3  PA3 = 321.5 kPa    = 16.042.254 ln(289.8)  8.3145 ln 477.9 = 4.505 kJ/kmol K sA3 sA2 303.2 1000    = 44.010.842 ln(289.8)  8.3145 ln 321.5 = 7.7546 kJ/kmol K sB3 sB2 270.4 600 SNET = 0.0396684.505 + 0.0266887.7546 = +0.3857 kJ/K 1223 12.36 Consider 100 m3 of atmospheric air which is an airwater vapor mixture at 100 kPa, 15C, and 40% relative humidity. Find the mass of water and the humidity ratio. What is the dew point of the mixture? Airvap P = 100 kPa, T= 15 oC, = 40% Pg = Psat15 = 1.705 kPa mv = P vV RvT = => Pv = Pg = 0.41.705 = 0.682 kPa 0.682100 = 0.513 kg 0.461288.15 Pa = Ptot Pv1 = 100 0.682 = 99.32 ma = w1 = P aV RaT mv ma = = 99.32100 = 120.1 kg 0.287288.15 0.513 = 0.0043 120.1 T = 1.4 oC Tdew is T when Pv = Pg = 0.682 kPa; 12.37 The products of combustion are flowing through a SSSF heat exchanger with 12% CO2, 13% H2O and 75% N2 on a volume basis at the rate 0.1 kg/s and 100 kPa. What is the dewpoint temperature? If the mixture is cooled 10C below the dewpoint temperature, how long will it take to collect 10 kg of liquid water? yH
2O = 0.13; PH2O = 0.13100 = 13 kPa, TDEW = 50.95 oC Cool to 40.95 oC PG = 7.805 kPa yH nH
2O = 7.805/100 = nH2O(v)/(nH2O(v) + 0.87) = 0.07365 per kmol mix in 2O(v) nLIQ = 0.13  0.07365 = 0.05635 MMIX IN = 0.1244.01 + 0.1318.015 + 0.7528.013 = 28.63 kg/kmol . . nMIX IN = mTOTAL/MMIX IN = 0.1/28.63 = 0.003493 kmol/s . nLIQ COND = 0.003 4930.05635 = 0.000 197 kmol/s . or mLIQ COND = 0.000 197 18.015 = 0.003 55 kg/s For 10 kg, takes ~ 47 minutes 1224 12.38 A new highefficiency home heating system includes an airtoair heat exchanger which uses energy from outgoing stale air to heat the fresh incoming air. If the outside ambient temperature is 10C and the relative humidity is 30%, how much water will have to be added to the incoming air, if it flows in at the rate of 1 m3/s and must eventually be conditioned to 20C and 40% relative humidity? Outside ambient air: PV1 = 1PG1 = 0.300.2602 = 0.078 kPa Assuming P1 = P2 = 100 kPa, => PA1 = 100  0.078 = 99.922 kPa . PA1V1 99.9221 0.078 . mA = = = 1.3228 kg/s w1 = 0.622 = 0.00049 99.922 RAT1 0.287263.2 Conditioned to : T2 = 20 oC, 2 = 0.40 w2 = 0.622 0.9356 = 0.00587 99.064 PV2 = 2PG2 = 0.402.339 = 0.9356 => . . mLIQ IN = mA(w2w1) = 1.3228(0.00587  0.00049) = 0.00712 kg/s = 25.6 kg/h 12.39 Consider 100 m3 of atmospheric air at 100 kPa, 25C, and 80% relative humidity. Assume this is brought into a basement room where it cools to 15C, 100 kPa. How much liquid water will condense out? Pg = Psat25 = 3.169 kPa => Pv = Pg = 0.8 3.169 = 2.535 kPa P vV 2.535100 mv1 = = = 1.844 kg RvT 0.461298.15 Pg15C = 1.705 < Pv1 => State 2 is saturated 2 = 100% , Pv2 = Pg2 = 1.705 P vV 1.705100 mv2 = = = 1.2835 kg RvT 0.461288.15 mliq = mv1  mv2 = 1.844 1.2835 = 0.56 kg 12.40 A flow of 2 kg/s completely dry air at T1, 100 kPa is cooled down to 10C by spraying liquid water at 10C, 100 kPa into it so it becomes saturated moist air at 10C. The process is SSSF with no external heat transfer or work. Find the exit moist air humidity ratio and the flow rate of liquid water. Find also the dry air inlet temperature T1. 2: saturated Pv = Pg = 1.2276 kPa and hfg (10C)= 2477.7 kJ/kg w2 = 0.622 1.2276/ (100  1.2276) = 0.00773 C.V. Box . . . . . ma + mliq = m a(1 + w2) => mliq = w2 ma = 0.0155kg/s . . . ma ha1  mliq hf = ma (ha2 + w2 hg2) ha1  ha2 = Cpa (T1 T2) = w2 hg2 w2 hf = w2 hfg = 19.15 => T1 = 29.1C 1225 12.41 A piston/cylinder has 100 kg of saturated moist air at 100 kPa, 5C. If it is heated to 45C in an isobaric process, find 1Q2 and the final relative humidity. If it is compressed from the initial state to 200 kPa in an isothermal process, find the mass of water condensing. Energy Eq.: m(u2u1) = Q12  W12, ~ Initial state 1: 1 = 100%, h1 = 39 kJ/kg dry air, w1 = 0.054 ma = mtot/(1+ w1) = 99.463 kg, mv1 = w1ma = 0.537 kg Case a: P = constant => W12 = mP(v2v1) => ~ ~ Q12 = m(u2u1) + W12 = m(h2h1) = ma(h2  h1) ~ State 2: w2 = w1, T2 => h2 = 79 , 2 = 9.75% Q12 = 99.463 (79  39) = 3979 kJ. Case b: T = constant & 2 = 100% => Pv = Pg = 0.8721 kPa w2 = 0.622 Pv2/Pa2 = 0.622 Pg2/(Ptot2Pg2) = 0.002697 mv2 = w2 ma = 0.268 kg, mliq = mv1  mv2 = 0.269 kg 12.42 A flow moist air at 100 kPa, 40C, 40% relative humidity is cooled to 15C in a constant pressure SSSF device. Find the humidity ratio of the inlet and the exit flow, and the heat transfer in the device per kg dry air. . . . C.V. Cooler. mv1 = mliq + mv2 Psychrometric chart: State 2: T < Tdew = 23C => 2 = 100% . . . . ~ ~ mv2/ma = 2 = 0.0107 , h2 = 62 mv1/ma = 1 = 0.018, h1 = 106; . . ~ mliq/ma = 1 2 = 0.0073 , hf = 62.98 . . ~ . . ~ m q = m h  m h  m h =>
a out a 1 liq f a 2 ~ ~ qout = h1  (1 2) hf  h2 = 106 0.0073 62.98 62 = 43.54 kJ/kgdry air Tables: Pg1 = 7.384 kPa Pv1 = 2.954 1 = 0.0189 Pv2 = 1.705 kPa = Pg2 => 2 = 0.0108 hv1 = 2574.3 , hv2 = 2528.9 , hf = 62.98 qout = CP (T1  T2) + 1hv1  2 hv2  (1 2) hf = 0.717(40  15) + 0.0189 2574.3  0.0108 2528.9  0.0073 62.98 = 38.8 kJ/kg dry air 1226 12.43 Ambient moist air enters a steadyflow airconditioning unit at 102 kPa, 30C, with a 60% relative humidity. The volume flow rate entering the unit is 100 L/s. The moist air leaves the unit at 95 kPa, 15C, with a relative humidity of 100%. Liquid condensate also leaves the unit at 15C. Determine the rate of heat transfer for this process. State 1: PV1 = 1PG1 = 0.604.246 = 2.5476 w1 = 0.622 2.5476/(102  2.5476) = 0.01593 . PA1V1 99.450.1 . = = 0.1143 kg/s mA = RAT1 0.287303.2 PV2 = PG2 = 1.705, w2 = 0.622 1.705/(95  1.705) = 0.01137 . . . . . . Energy Eq.: QCV + mAhA1 + mV1hV1 = mAhA2 + mV2hA2 + m3hL3 . . QCV/mA = CP0A(T2T1) + w2hV2  w1hV1 + (w1w2)hL3 = 1.004(1530) + 0.011372528.9  0.015932556.2 + 0.0045663.0 = 26.732 kJ/kg air . Q CV = 0.1143(26.73) = 3.055 kW 12.44 A steady supply of 1.0 m3/s air at 25C, 100 kPa, 50% relative humidity is needed to heat a building in the winter. The outdoor ambient is at 10C, 100 kPa, 50% relative humidity. What are the required liquid water input and heat transfer rates for this purpose? Air: R = 0.287 kJ/kg K, Cp = 1.004 kJ/kgK State 1: T1 = 10C, 1 = 50%, P1 = 100 kPa Pv1= 1Pg1 = 0.6138 kPa, Pg1= 1.2276 kPa, Pa1 = P1 Pv1 = 99.39 kPa => 1 = 0.622 Pv1/Pa1 = 0.003841 . State 2: T2 = 25C, P2 = 100 kPa, 2 = 50%, V a2 = 1 m3/s Pg2 = 3.169 kPa, Pv2= 2Pg2 = 1.5845 kPa, P = P2  Pv2 = 98.415 kPa, 2 = 0.622 Pv2/Pa2 = 0.010014 . a2 . m a2 = Pa2 V a2/RTa2 = 98.415 1/(0.287 298.15) = 1.15 kg/s Steam tables B.1.1: hv1 = 2519.7 kJ/kg, hv2 = 2547.2 kJ/kg State 3: Assume: Liq. Water at T3 = 25C, hf3 = 104.9 kJ/kg . . . . . Conservation of Mass: ma1 = ma2, mf3 = mv2  mv1 . . mf3= ma2(2  1) = 1.15 0.006173 = 0.0071 kg/s . . . . . . 1stLaw: Q + ma1ha1 + mv1hv1 + mf3hf3 = ma2ha2 +mv2hv2 . . mf3 . Q . = Cp(T2 T1) + 2hv2  1hv1  . hf3 => Q = 34.76 kW ma ma 1227 12.45 Consider a 500L rigid tank containing an airwater vapor mixture at 100 kPa, 35C, with a 70% relative humidity. The system is cooled until the water just begins to condense. Determine the final temperature in the tank and the heat transfer for the process. Pv1 = PG1 = 0.75.628 = 3.9396 kPa Since mv = const & V = const & also Pv = PG2: PG2 = Pv1 T2/T1 = 3.9396 T2/308.2 = 0.01278 T2 Assume T2 = 30oC: Assume T2 = 25oC: 0.01278303.2 = 3.875 = 4.246 = PG 30C / 0.01278298.2 = 3.811 = 3.169 = PG 25C / interpolating T2 = 28.2 oC w2 = w1 = 0.622 3.9396 = 0.025 51 (1003.9369) ma = Pa1V/RaT1 = (1003.94)0.5/(0.287308.2) = 0.543 kg 1st law: Q12 = U2U1 = ma(ua2ua1) + mv(uv2uv1) = 0.717(28.2  35) + 0.02551 (2414.2  2423.4) = 5.11 kJ/kg Q12 = 0.543(5.11) = 2.77 kJ 12.46 Air in a piston/cylinder is at 35C, 100 kPa and a relative humidity of 80%. It is now compressed to a pressure of 500 kPa in a constant temperature process. Find the final relative and specific humidity and the volume ratio V2/V1. Check to see if the second state is saturated or not. First assume no water is condensed 2: w2 = 0.622 Pv2/(P2Pv2) w2 = w1 => Pv2 = 22.568 > Pg = 5.628 kPa Conclusion is state 2 is saturated 2 = 100%, w2 = 0.622 Pg/(P2Pg) = 0.00699 To get the volume ratio, write the ideal gas law for the vapor phases V2 = Va2 + Vv2 + Vf2 = (maRa + mv2Rv)T/P2 + mliqvf V1 = Va1 + Vv1 = (maRa + mv1Rv)T/P1 Take the ratio and devide through with maRaT/P2 to get V2/V1 = (P1/P2)[1 + 0.622w2 + (w1w2)P2vf/RaT] / [1+0.622 w1] = 0.1973 The liquid contribution is nearly zero (=0.000126) in the numerator. 1: w1 = 0.029 1228 12.47 A 300L rigid vessel initially contains moist air at 150 kPa, 40C, with a relative humidity of 10%. A supply line connected to this vessel by a valve carries steam at 600 kPa, 200C. The valve is opened, and steam flows into the vessel until the relative humidity of the resultant moist air mixture is 90%. Then the valve is closed. Sufficient heat is transferred from the vessel so the temperature remains at 40C during the process. Determine the heat transfer for the process, the mass of steam entering the vessel, and the final pressure inside the vessel.
i H 2O Pv1 = 1PG1 = 0.17.384 = 0.7384 kPa Pv2 = 0.97.384 = 6.6456 kPa Pa2 = Pa1 = 150  0.738 = 149.262 kPa
AIR + H 2O w1 = 0.622 w2 = 0.622 0.7384 = 0.003 08 149.262 6.6456 = 0.0277 149.262 ma = 149.2620.3/(0.287313.2) = 0.5 kg P2 = 149.262 + 6.6456 = 155.9 kPa mvi = 0.5(0.0277  0.00308) = 0.0123 kg uv1 = uv2 uG at 40 oC and ua2 = ua1 QCV = ma(ua2  ua1) + mv2uv2  mv1uv1  mvihi = mvi(uG at T  hi) = 0.0123(2430.1  2850.1) = 5.15 kJ 1229 12.48 A combination air cooler and dehumidification unit receives outside ambient air at 35C, 100 kPa, 90% relative humidity. The moist air is first cooled to a low temperature T2 to condense the proper amount of water, assume all the liquid leaves at T2. The moist air is then heated and leaves the unit at 20C, 100 kPa, relative humidity 30% with volume flow rate of 0.01 m3/s. Find the temperature T2, the mass of liquid per kilogram of dry air and the overall heat transfer rate. 1 COOL 2'
. Q C LIQ OUT . QH 2 MIX IN HEAT MIX OUT CV a) Pv1 = 1PG1 = 0.9 5.628 = 5.0652 kPa w1 = 0.622 5.0652 = 0.033 19 1005.0652 Pv3 = 3PG3 = 0.3 2.339 = 0.7017 kPa 0.7017 w2 = w3 = 0.622 = 0.0044 1000.7017 . . mLIQ 2/ma = w1  w2 = 0.033 19  0.0044 = 0.028 79 kg/kg air PG2 = Pv3 = 0.7017 kPa T2 = 1.7 oC b) For a C.V. around the entire unit . . . QCV = QH + QC Net heat transfer, 1st law: . . . . QCV/ma = (ha3ha1) + w3hv3  w1hv1 + mL2 hL2/ma = 1.004(2035) + 0.00442538.1  0.033 192565.3 + 0.028 797.28 = 88.82 kJ/kg air . Pa3V3 (1000.7017)0.01 . ma = = = 0.0118 kg/s RaT3 0.287293.2 . QCV = 0.0118(88.82) = 1.05 kW 1230 12.49 A rigid container, 10 m3 in volume, contains moist air at 45C, 100 kPa, = 40%. The container is now cooled to 5C. Neglect the volume of any liquid that might be present and find the final mass of water vapor, final total pressure and the heat transfer. CV container. m 2 = m1 ; m2u2  m1u1 = 1Q2 Tdew = 27.7C 1: 45C, = 40% => w1 = 0.0236 , Final state T2 < Tdew so condensation, 2 = 100% Pv1 = 0.4 Pg = 0.4*9.593 = 3.837 kPa, Pa1 = Ptot  Pv1 = 97.51 kPa ma = Pa1V/RT1 = 10.679 kg, mv1 = w1 ma = 0.252 kg Pv2 = Pg2 = 0.8721 kPa, Pa2 = Pa1T2/T1 = 85.25 kPa P2 = Pa2 + Pv2 = 86.12 kPa mv2 = Pv2V/RvT2 = 0.06794 kg mf2 = mv1  mv2 = 0.184 kg The heat transfer from the energy equation becomes
1Q 2 (=V/vg= 0.06797 steam table) = ma(u2u1)a + mv2ug2 + mf2uf2  mv1ug1 = ma Cv(T2T1) + mv2 2382.3 + mf2 20.97  mv1 2436.8 = 306.06 + 161.853 + 3.858  614.07 =  754.4 kJ 12.50 A saturated airwater vapor mixture at 20 oC, 100 kPa, is contained in a 5m3 closed tank in equilibrium with 1 kg of liquid water. The tank is heated to 80oC. Is there any liquid water in the final state? Find the heat transfer for the process. AIR
+ a) Since VLIQ = mLIQvF 0.001 m3, VGAS V Q12 1 = 1.00 Pv1 = PG1 = 2.339 kPa w1 = 0.622 2.339 /(100  2.339) = 0.0149 = 97.6614.999 = 5.802 kg 0.287293.2 Pa2 = 97.661 => mv1 = w1ma = 0.086 kg VAP LIQ Pa1V RaT1 ma = At state 2: 353.2 4.999 = 117.623 kPa 293.2 5 wMAX 2 = 0.622 47.39 / 117.623 = 0.2506 But w2 ACTUAL = 0.086 + 1.0 = 0.1872 < wMAX 2 5.802 No liquid at 2 Q12 = ma (ua2ua1) + mv2 uv2  mv1uv1  mliq 1uliq 1 = 5.802 0.717(80  20) + 1.086 2482.2  0.086 2402.9  1 84.0 = 249.6 + 2695.7  206.65  84 = 2655 kJ 1231 12.51 An airwater vapor mixture enters a steady flow heater humidifier unit at state 1: 10C, 10% relative humidity, at the rate of 1 m3/s. A second airvapor stream enters the unit at state 2: 20C, 20% relative humidity, at the rate of 2 m3/s. Liquid water enters at state 3: 10C, at the rate of 400 kg per hour. A single airvapor flow exits the unit at state 4: 40C. Calculate the relative humidity of the exit flow and the rate of heat transfer to the unit. Assume: P = 100 kPa . State 1: T1 = 10C, 1 = 10%, Va1 = 1 m3/s Pv1= 1Pg1 = 0.1228 kPa, Pa1 = P  Pv1 = 99.877 kPa . Pv1 Pa1Va1 . 1 = 0.622 = 0.000765, ma1 = = 1.2288 kg/s Pa1 RTa1 . . mv1 = 1m a1 = 0.00094 kg/s, hv1 = hg1 = 2519.7 kJ/kg . State 2: T2 = 20C, 2 = 20%, Va2 = 2 m3/s Pg1= 1.2276 kPa, Pv2= Pg2 = 0.4677 kPa, Pa2 = P  Pv2 = 99.532 kPa . Pv2 Pa2 V a2 . 2 = 0.622 = 0.002923, m a2 = = 2.3656 kg/s Pa2 RTa2 . . mv2 = 2ma2 = 0.00691 kg/s, hv2 = hg2 = 2538.1 kJ/kg . State 3: Liquid. T3 = 10C, mf3 = 400 kg/hr = 0.1111 kg/s, hf3 = 42 kJ/kg Pg2 = 2.3385 kPa, State 4: T4 = 40C Continuity Eq. air: . . . m a4 = m a2 + m a1 = 3.5944, . . . . mv4 = mv1 + mv2 + mf3 = 0.11896 kg/s Pv4 = 5.052 kPa Continuity Eq. water: . m v4 Pv4 4 = . = 0.0331 = 0.622 m a4 PPv4 Pg4 = 7.384 kPa, 4 = Pv4 Pg4 = 0.684, hv4 = hg4 = 2574.3 kJ/kg . . . . . . . . 1stLaw: Q + ma1ha1 + mv1hv1 + ma2ha2 +mv2hv2 + mf3hf3 = ma4ha4 +mv4hv4 . Q = 1.004(3.5944 40  1.2288 10  2.3656 20) + 0.11896 2574.3  0.00094 2519.7  0.00691 2538.1  0.1111 42.0 = 366 kW 1232 12.52 In a hot and dry climate, air enters an airconditioner unit at 100 kPa, 40C, and 5% relative humidity, at the steady rate of 1.0 m3/s. Liquid water at 20C is sprayed into the air in the AC unit at the rate 20 kg/hour, and heat is rejected from the unit at at the rate 20 kW. The exit pressure is 100 kPa. What are the exit temperature and relative humidity? . State 1: T1 = 40C, P1 = 100 kPa, 1 = 5%, Va1 = 1 m3/s Pg1= 7.3837 kPa, Pv1= 1Pg1 = 0.369 kPa, Pa1 = P Pv1 = 99.63 kPa . Pv1 Pa1Va1 . 1 = 0.622 = 0.0023, ma1 = = 1.108 kg/s, hv1 = 2574.3 kJ/kg Pa1 RTa1 . State 2 : Liq. Water. 20C, mf2 = 20 kg/hr = 0.00556 kg/s, hf2 = 83.9 kJ/kg . . . . . Conservation of Mass: ma1 = ma3, mv1 + m12 = mv3 . . 3 = (mf2 / ma1) + 1 = ( 0.00556/1.108 ) + 0.0023 = 0.0073 State 3 : P3 = 100 kPa and Pv3 = P33/(0.622 + 3) = 1.16 kPa . . . . . . . 1stLaw: Q + ma1ha1 + mv1hv1 + mf2hf2 = ma3ha3 +mv3hv3; Q =  20 kW . . . (ha3ha1) + 3hv3 = 1hv1 + (mf2hf2  Q )/ma1 = 24.39; Unknowns: ha3, hv3 Trial and Error for T3; T3 = 301.6 K, Pg3 = 3.901 kPa , 3 = Pv3 Pg3 = 0.297 12.53 A waterfilled reactor of 1 m3 is at 20 MPa, 360C and located inside an insulated containment room of 100 m3 that contains air at 100 kPa and 25C. Due to a failure the reactor ruptures and the water fills the containment room. Find the final pressure. CV Total container. mv(u2u1) + ma(u2u1) = 1Q2  1W2 = 0 Initial water: Initial air: v1 = 0.0018226, u1 = 1702.8, mv = V/v = 548.67 kg ma = PV/RT = 10099/0.287298.2 = 115.7 kg Substitute into energy equation 548.67 (u2  1702.8) + 115.70.717 (T2  25) = 0 u2 + 0.1511 T2 = 1706.6 Trial and error 2phase T = 150 LHS = 1546 & v2 = V2/mv = 0.18226 m3/kg (T guess, v2 => x2 => u 2 => LHS) T = 160 LHS = 1820.2 T = 155 LHS = 1678.1 => T = 156 C LHS = 1705.7 OK x2 = 0.5372, Psat = 557.5 kPa Pa2 = Pa1V1T2/V2T1 = 100 99 429.15 / (100298.15) = 142.5 kPa => P2 = Pa2 + Psat = 700 kPa. 1233 12.54 Use the psychrometric chart to find the missing property of: , , Twet, Tdry a. Tdry = 25C, =80% c.Tdry = 20C, and = 0.008 a. 25C, = 80% b. 15C, = 100% c. 20C, = 0.008 d. 25C, Twet = 23C b. Tdry =15C, =100% d. Tdry = 25C, Twet = 23C => = 0.016; Twet = 22.3C => = 0.0106; Twet = 15C => = 57%; Twet = 14.4C => = 0.017; = 86% 12.55 Use the psychrometric chart to find the missing property of: , , Twet, Tdry a. = 50%, = 0.012 c. = 0.008 and Twet =17C a. = 50%, = 0.012 b. Twet = 15C, =60% b. Twet =15C, = 60%. d. Tdry = 10C, = 0.006 => Tdry = 23.5C, Twet = 20.6 C => Tdry = 20.2C, = 0.0086 c. = 0.008, Twet =17C => Tdry = 27.2C, = 37% d. Tdry= 10C, = 0.006 => = 80%, Twet = 8.2C 12.56 For each of the states in Problem 12.55 find the dew point temperature. The dew point is the state with the same humidity ratio (abs humidity ) and completely saturated = 100%. From psychrometric chart: a. b. Tdew = 16.8C Tdew = 12C c. Tdew = 10.9C d. Tdew = 6.5C Pv = Pg = Ptot /[ + 0.622] = Psat(Tdew) in B.1.1 Finding the solution from the tables is done for cases a,c and d as Eq.12.21 solve: For case b use energy Eq. 12.23 to find 1 first from Tad sat = Twet. 1234 12.57 One means of airconditioning hot summer air is by evaporative cooling, which is a process similar to the SSSF adiabatic saturation process. Consider outdoor ambient air at 35C, 100 kPa, 30% relative humidity. What is the maximum amount of cooling that can be achieved by such a technique? What disadvantage is there to this approach? Solve the problem using a first law analysis and repeat it using the psychrometric chart, Fig. F.5.
Ambient Air 1 3 Liquid Cooled 2 Air P1 = P2 = 100 kPa T1 = 35 oC, 1 = 30% Pv1 = 1Pg1 = 0.305.628 = 1.6884 1 = 0.6221.6884/98.31 = 0.01068 For adiabatic saturation (Max. cooling is for 2 = 1), 1st law, Eq.12.23 1 (hv1  hf2) = Cp(T2  T1) + 2 hfg2 2 = 1 & 2 = 0.622 PG2/(P2  PG2) Only one unknown: T2 . Trial and error on energy equation: CpT2 + 2 hfg2 + 1 hf2 = CpT1 + 1hv1 = 1.004 35 + 0.01068 2565.3 = 62.537 T2 = 20 oC: PG2 = 2.339, hf2 = 83.94 , hfg2 = 2454.12 => 2 = 0.622 2.339/ 97.661 = 0.0149 LHS = 1.004 20 + 0.0149 2454.1 + 0.01068 83.94 = 57.543 T2 = 25 oC: PG2 = 3.169, hf2 = 104.87 , hfg2 = 2442.3 => 2 = 0.622 3.169/ 96.831 = 0.02036 LHS = 1.004 25 + 0.02036 2442.3 + 0.01068 104.87 = 75.945 linear interpolation: b) T2 = 21.4 oC chart F.5 : Ad. sat. T WetBulbTemperature 21.5 oC 1235 12.58 Use the formulas and the steam tables to find the missing property of: , , and Tdry, total pressure is 100 kPa; repeat the answers using the psychrometric chart a. = 50%, = 0.010 b. Twet =15C, = 50% c. Tdry = 25C, Twet = 21C a. From Eq.12.21 Pv = P /(0.622 + ) = 100 0.01/0.632 = 1.582 kPa From Eq.12.18 Pg = Pv/ = 1.582/0.5 = 3.165 kPa => T = 25 C b. Assume Twet is adiabatic saturation T and use energy Eq.12.23 At 15C: Pg = 1.705 => = 0.622 1.705/(100  1.705) = 0.01079 LHS = 1 (hv1  hf2) + CpT1 = RHS = CpT2 + 2 hfg2 RHS = 1.00415 + 0.01079 2465.93 = 41.667 kJ/kg 1 = 0.622 Pg/(100  Pg) where Pg is at T1. => Trial and error. T = 21.4C, 1 = 0.008 LHS25C = 49.98, LHS20C = 38.3 c. At 21C:: Pg = 2.505 => 2 = 0.622 2.505/(100  2.505) = 0.016 hf2 = 88.126 and hfg2 = 2451.76 kJ/kg, hv1 = 2547.17 From Eq.12.23: 1 = [Cp(T2T1) + 2 hfg2 ]/(hv1  hf2) = 0.0143 Pv = P /(0.622 + ) = 2.247, Using the psychrometric chart F.5: a: Tdry = 25.3 C b. Tdry = 21.6C, = 0.008 c: = 0.0141, = 7172% = 2.247/3.169 = 0.71 12.59 Compare the weather two places where it is cloudy and breezy. At beach A it is 20C, 103.5 kPa, relative humidity 90% and beach B has 25C, 99 kPa, relative humidity 20%. Suppose you just took a swim and came out of the water. Where would you feel more comfortable and why? Your skin being wet and air is flowing over it will feel Twet. With the small difference in pressure from 100 kPa use the psychrometric chart. A: 20C, = 90% B: 25C, = 20% => Twet = 18.7C => Twet = 12.3C At beach A it is comfortable, at B it feels chilly. 1236 12.60 Ambient air at 100 kPa, 30C, 40% relative humidity goes through a constant pressure heat exchanger in a SSSF process. In one case it is heated to 45C and in another case it is cooled until it reaches saturation. For both cases find the exit relative humidity and the amount of heat transfer per kilogram dry air. . . . . CV heat exchanger: mAi = mAe, mvi = mve, we = wi ~ (ha + whv)i + q = (h a + whv)e = he, Using the psychrometric chart: i: ~ Case I) e: Te = 45 oC, we = wi => he = 92, e = 17%, q = 9276 = 16 kJ/kg dry air ~ Case II) e: w e = wi, e = 100% => he = 61, Te = 14.5oC q = 6176 = 15 kJ/kg dry air 12.61 A flow, 0.2 kg/s dry air, of moist air at 40C, 50% relative humidity flows from the outside state 1 down into a basement where it cools to 16C, state 2. Then it flows up to the living room where it is heated to 25C, state 3. Find the dew point for state 1, any amount of liquid that may appear, the heat transfer that takes place in the basement and the relative humidity in the living room at state 3. Solve using psychrometric chart: a) Tdew = 27.2 (w = w1, = 100%) ~ w1 = 0.0232, h1 = 118.2 b) T2 < Tdew so we have 2 = 100% liquid water appear in the basement. ~ => w2 = 0.0114 h2 = 64.4 and from steam tbl. hf = 67.17 . . mliq = mair (w1w2) = 0.2(0.02320.0114) = 0.00236 kg/s . ~ . . ~ c) Energy equation: mair h1 = mliq hf + mair h2 + Qout Qout = 0.2[118.2  64.4  0.011867.17] = 10.6 kW d) w3 = w2 = 0.0114 & 25C => 3 = 58%. If you solve by the formula's and the tables the numbers are: Pg40 = 7.384 ; Pv1 = 0.57.384 = 3.692 kPa w1 = 0.622 3.692 / (100  3.692) = 0.02384 Pv1 = Pg (Tdew) => Tdew 1 = 27.5 C 2: = 100%, Pv2 = Pg2 = 1.832 kPa, w2 = 0.6221.832/98.168 = 0.0116 . . mliq = mair (w1w2) = 0.20.01223 = 0.00245 kg/s 3: w3 = w2 => Pv3 = Pv2 = 1.832 & Pg3 = 3.169 3 = Pv/Pg = 1.832/3.169 = 57.8% ~ ~ q = he  hi ~ wi = 0.0104, hi = 76 1237 12.62 A flow of air at 5C, = 90%, is brought into a house, where it is conditioned to 25C, 60% relative humidity. This is done in a SSSF process with a combined heaterevaporator where any liquid water is at 10C. Find any flow of liquid, and the necessary heat transfer, both per kilogram dry air flowing. Find the dew point for the final mixture. CV heater and evaporator. Use psychrometric chart. ~ Inlet: w1 = 0.0048, h1 = 37.5, hf = 42.01 ~ Exit: w2 = 0.0118, h2 = 75, Tdew = 16.5C From these numbers we see that water and heat must be added. Continuity eq. and energy equation give . . mLIQ IN/mA = w2  w1 = 0.007 kg/kg dry air ~ ~ q = h2  h1  (w2w1)hf = 37.3 kJ/kg dry air 12.63 Atmospheric air at 35C, relative humidity of 10%, is too warm and also too dry. An air conditioner should deliver air at 21C and 50% relative humidity in the amount of 3600 m3 per hour. Sketch a setup to accomplish this, find any amount of liquid (at 20C) that is needed or discarded and any heat transfer. CV air conditioner. Check from psychrometric chart. ~ Inlet: w1 = 0.00305, hmix,1 = 62.8, hf,20 = 83.96 ~ w2 = 0.0076, hmix,2 = 60.2 Water must be added. Continuity and energy equations . . . . . ~ . . ~ mA(1+w1) + mliq = mA(1+w2); mAh1mix + mliqhf + QCV = mAh2mix . . mA = PaV/RT = (101.325  1.25)1/0.287294.15 = 1.1845 kg/s . . mliq = mA(w2  w1) = 0.00539 kg/s = 19.4 kg/h . QCV = 1.1845(60.2  62.8  0.0045583.96) =  3.53 kW Exit:
Liquid water Cooler Exit Inlet 1238 12.64 In a car's defrost/defog system atmospheric air, 21C, relative humidity 80%, is taken in and cooled such that liquid water drips out. The now dryer air is heated to 41C and then blown onto the windshield, where it should have a maximum of 10% relative humidity to remove water from the windshield. Find the dew point of the atmospheric air, specific humidity of air onto the windshield, the lowest temperature and the specific heat transfer in the cooler. Solve using the psychrometric chart Air inlet: T = 21C, = 80% Air exit: T = 41C, = 10% => w1 = 0.0124, Tdew = 17.3C => w3 = 0.0044, Tdew = 1.9C To remove enough water we must cool to the exit T dew, followed by heating to Tex. The enthalpies are ~ ~ h1 = 72, h2 = 32.5, hf(1.9C) = 8 CV cooler: . . mliq /mair = w1w3 = 0.0124  0.0044 = 0.008 kg liq/kg air . ~ ~ q = Q/mair = h2 + (w1  w3) hf  h1 = 32.5 + 0.0088  72 = 39.4 kJ/kg dry air If the steam and air tables are used the numbers are 1: Pg1 = 2.505, Pv1 = 2.004 => w1 = 0.01259 ~ hg1 = 2539.9, ha1 = 294.3 => h1 = 326.3 3: Pg3 = 7.826, Pv3 = 0.783 => w3 = 0.00486 2: wg3 = w3 => T2 = T3dew = 3.3C, hf2 = 13.77 ~ hg2 = 2507.4, ha2 = 276.56 => h2 = 288.75 . . mliq /mair = 0.00773, q = 288.75 + 0.00773 13.77  326.3 = 37.45 kJ/kg air 12.65 Two moist air streams with 85% relative humidity, both flowing at a rate of 0.1 kg/s of dry air are mixed in a SSSF setup. One inlet flowstream is at 32.5C and the other at 16C. Find the exit relative humidity. ~ ~ 2: w2 = 0.0094, h2 = 60 1: w1 = 0.0266, h1 = 120 CV mixing chamber. Continuity Eq. water: Energy Eq.: . . . mair w1 + mair w2 = 2mair wex; . ~ . ~ . ~ mair h1 + mair h2 = 2mair hex ~ ~ wex = (w1 + w2)/2 = 0.018 ; ~ ex = (h1 + h2)/2 = 90 h ~ exit: wex, hex => Tex = 24.5C, = 94% 1239 12.66 A flow of moist air at 21C, 60% relative humidity should be produced from mixing of two different moist air flows. Flow 1 is at 10C, relative humidity 80% and flow 2 is at 32C and has Twet = 27C. The mixing chamber can be followed by a heater or a cooler. No liquid water is added and P = 100 kPa. Find the two controls one is the ratio of the two mass flow rates ma1/ma2 and the other is the heat transfer in the heater/cooler per kg dry air. ~ ~ w1 = 0.006 , h 1 = 45 2: w2 = 0.0208 , h 2 = 105 ~ 4: w4 = 0.0091 , h 4 = 64 , Tdew 4 = 12.5C C.V : Total Setup . . . . ma1 w1 + ma2 w2 = (ma1 + ma2) w4 . . ~ . ~ . . ~ ma1 h 1 + ma2 h 2 + Qa1 = (ma1 + ma2) h 4 w4 w2 . . x = ma1/ma2 => x w1 + w2 = (1+x) w4 => x = = 3.773 w w 1:
1 4 . . . ~ ~ ~ ~ q = Qa1/(ma1 + ma2) = h 4  [x/(1+x)] h 1  [1/(1+x)] h2 = 64 0.7905 45  0.2095 105 = 6.43 kJ/kgdry air 12.67 Consider two states of atmospheric air. (1) 35C, Twet = 18C and (2) 26.5C, = 60%. Suggest a system of devices that will allow air in a SSSF process to change from (1) to (2) and from (2) to (1). Heaters, coolers (de)humidifiers, liquid traps etc. are available and any liquid/solid flowing is assumed to be at the lowest temperature seen in the process. Find the specific and relative humidity for state 1, dew point for state 2 and the heat transfer per kilogram dry air in each component in the systems. Use the psychrometric chart ~ ~ 1: w1= 0.006, h1 = 70.5, 1 = 18%, Tdew = 6.5C, hdew = 42 ~ ~ 2: w2= 0.013, h2 = 79.4, 2 = 60%, Tdew = 18C, hdew = 71 Since w2 > w1 water must be added in process I to II and removed in the process II to I. Water can only be removed by cooling below dew point temperature so I to II: Adiab. sat I to Dew,II, then heater from Dew,II to II II to I: Cool to Dew,I then heat Dew,I to I The first one can be done because Tdew II = Tad sat I ~ ~ I to II: q = hII  hdewII = 79.4  71 = 8.4 kJ/kg air ~ ~ II to I: qcool = hII  hdewI  (w2w1)hf(at TdewI) = 79.4  0.007 27.29 = 37.2 kJ/kg air ~ ~ qheat = hI  hdewI = 70.5  42 = 28.5 kJ/kg air 1240 12.68 An insulated tank has an air inlet, 1 = 0.0084, and an outlet, T2 = 22C, 2 = 90% both at 100 kPa. A third line sprays 0.25 kg/s of water at 80C, 100 kPa. For a SSSF operation find the outlet specific humidity, the mass flow rate of air needed and the required air inlet temperature, T 1. Take CV tank in SSSF. Continuity and energy equations are: . . . . . ~ . ~ Continuity: m3 + ma w1 = ma w2; Energy Eq.: m3hf + ma h1 = ma h2 All state properties are known except T1. From the psychrometric chart we get ~ State 2: w2 = 0.015, h2 = 79.5 State 3: hf = 334.91 (steam tbl) . . ma = m3/(w2w1) = 0.25/(0.0150.0084) = 37.88 kg/s ~ ~ h1 = h2  (w2w1)hf = 79.5  0.0066 334.91 = 77.3 ~ Chart (w1, h1) => T1 = 36.5C 12.69 You have just washed your hair and now blow dry it in a room with 23C, = 60%, (1). The dryer, 500 W, heats the air to 49C, (2), blows it through your hair where the air becomes saturated (3), and then flows on to hit a window where it cools to 15C (4). Find the relative humidity at state 2, the heat transfer per kilogram of dry air in the dryer, the air flow rate, and the amount of water condensed on the window, if any. The blowdryer heats the air at constant specific humidity to 2 and it then goes through an adiabatic saturation process to state 3, finally cooling to 4. ~ 1: 23C, 60% rel hum => w1 = 0.0104, h1 = 69 ~ 2: w2 = w1, T2 => 2 = 15%, h2 = 95 CV. 1 to 2: w 2 = w1; CV. 2 to 3: ~ ~ q = h2  h1 = 95  69 = 26 kJ/kg dry air . ma = Q/q = 0.5/26 = 0.01923 kg/s . . . ~ . . ~ w3  w2 = mliq /ma ; ma h2 + mliq hf = ma h3 3: = 100% => T3 = Twet,2 = 24.8C, w3 = 0.0198 4: = 100%, T4 => w4 = 0.01065 . . mliq = (w3w4)ma = (0.01980.01065)0.01923 = 0.176 g/s If the steam tables and formula's are used then we get hg1 = 2543.5, hg2 = 2590.3, Pg1 = 2.837, Pv1 = 1.7022, Pg2 = 11.8, w1 = 0.01077, w2 = w1 , Pv2 = Pv1 2 = Pv2/Pg2 = 14.4%, hf3 = 114, Trial and error for adiabatic saturation temperature. T3 = 25C, w3 = 0.02, Pv4 =Pg4 = 1.705 kPa, w4 = 0.6221.705/(1001.705) = 0.0108 1241 12.70 A watercooling tower for a power plant cools 45C liquid water by evaporation. The tower receives air at 19.5C, = 30%, 100 kPa that is blown through/over the water such that it leaves the tower at 25C, = 70%. The remaining liquid water flows back to the condenser at 30C having given off 1 MW. Find the mass flow rate of air, and the amount of water that evaporates. CV Total cooling tower, SSSF. . . Water in air: win + mevap/ma = wex . ~ . . ~ . . Energy: ma hin + m1 h45 = ma hex + (m1  mevap) h30 ~ Inlet: 19.5C, 30% rel hum => win = 0.0041, hin = 50 ~ Exit : 25C, 70% rel hum => wex = 0.0138, hex = 80 Take the two water flow difference to mean the 1 MW . . . . Q = m1 h45  (m1  mevap) h30 = 1 MW . ~ ~ . . ma(hexhin) = ma(8050) = 1000 kW => ma = 33.33 kg/s . . mevap = (wex  win) ma = 0.0097 33.33 = 0.323 kg/s The needed makeup water flow could be added to give a slightly different meaning to the 1 MW. 12.71 An indoor pool evaporates 1.512 kg/h of water, which is removed by a dehumidifier to maintain 21C, = 70% in the room. The dehumidifier, shown in Fig. P12.71, is a refrigeration cycle in which air flowing over the evaporator cools such that liquid water drops out, and the air continues flowing over the condenser. For an air flow rate of 0.1 kg/s the unit requires 1.4 kW input to a motor driving a fan and the compressor and it has a coefficient of performance, = QL/ Wc = 2.0. Find the state of the air as it returns to the room and the compressor work input. The unit must remove 1.512 kg/h liquid to keep steady state in the room. As water condenses out state 2 is saturated. ~ 1: 21 C, 70% => w1 = 0.0108, h1 = 68.5 . . . . CV 1 to 2: mliq = ma(w1  w2) => w2 = w1  mliq /ma ~ ~ qL = h1  h2  (w1  w2) hf2 w2 = 0.0108  1.512/36000.1 = 0.0066 ~ 2: w2, 100% => T2 = 8C, h2 =45, hf2 = 33.6 qL = 68.5  45  0.004233.6 = 23.36 kJ/kg dry air . . ~ ~ CV Total system : h3 = h1 + Wel/ma  (w1w2) hf = 68.5 + 14  0.14 = 82.36 kJ/kg dry air ~ 3: w3 = w2, h3 => T3 = 46C, 3 = 1112% . . Wc = ma qL/ = 1.165 kW 1242 12.72 To refresh air in a room, a counterflow heat exchanger, see Fig. P12.72, is mounted in the wall, drawing in outside air at 0.5C, 80% relative humidity and pushing out room air, 40C, 50% relative humidity. Assume an exchange of 3 kg/min dry air in a SSSF device, and also that the room air exits the heat exchanger to the atmosphere at 23C. Find the net amount of water removed from the room, any liquid flow in the heat exchanger and (T, ) for the fresh air entering the room. ~ State 1: w1 = 0.0232, h1 = 119.2, Tdew,1 = 27C The room air is cooled to 23C < Tdew1 so liquid will form in the exit flow channel and state 2 is saturated. ~ 2: 23C, = 100% => w2 = 0.0178, h2 = 88, hf2 = 96.52 ~ 3: 0.5C, = 80% => w3 = 0.0032, h3 = 29.2 CV 1 to 2: . . mliq,2 = ma (w1  w2) = 3 (0.0232  0.0178) = 0.0162 kg/min . . . CV room: mv,out = ma (w1  w4) = ma (w1  w3) CV Heat exchanger: ~ ~ ~ ~ h4 = h3 + h1  h2  (w1w2) hf2 = 29.2 + 119.2  88  0.005496.52 = 59.9 kJ/kg dry air ~ 4: w4 = w3, h4 => T4 = 32.5C, = 12% = 3(0.02320.0032) = 0.06 kg/min ~ ~ ~ ~ ma(h4  h3) = ma(h1  h2)  mliqhf2 1243 12.73 Steam power plants often utilize large cooling towers to cool the condenser cooling water so it can be recirculated; see Fig. P12.73. The process is essentially evaporative adiabatic cooling, in which part of the water is lost and must therefore be replenished. Consider the setup shown in Fig. P12.73, in which 1000 kg/s of warm water at 32C from the condenser enters the top of the cooling tower and the cooled water leaves the bottom at 20C. The moist ambient air enters the bottom at 100 kPa, dry bulb temperature of 18C and a wet bulb temperature of 10C. The moist air leaves the tower at 95 kPa, 30C, and relative humidity of 85%. Determine the required mass flow rate of dry air, and the fraction of the incoming water that evaporates and is lost. Air + vap. P4 = 95 kPa 4 1 T1 = 32 oC T4 = 30 oC . 4 = 0.85 m1 = 1000 kg/s LIQ H 2 O P2 = 100 kPa T2 = 18 oC WBT2 = 10 oC
2 3 2 T3 = 20 oC T2 = 10 oC, w2 = 0.622 2 = 1.0 1.2276 = 0.00773 1001.2276 Air + vap.
2' 18 C LIQ IN w2 = (ha2ha2) + w2hFG2 hv2  hF 2 = o 1.0035(1018) + 0.007732477.7 = 0.00446 2534.4  42.0 3.609 Pv4 = 0.85 4.246 = 3.609, w4 = 0.622 = 0.02456 953.609 . . . . . . . m1 + mv2 = m3 + mv4 Cons. mass: ma2 = ma4 = ma, . . . . . or m3 = m1 + ma(w2w4) and set r = m1/ma . . . . . . 1st law: m1h1 + maha2 + mv2hv2 = m3h3 + maha4 + mv4hv4 or r h1 + (ha2ha4) + w2hv2 = (r + w2  w4)h3 + w4hv4 r(h1h3) = ha4  ha2 + w4hv4  w2hv2  (w4w2)h3 r(134.1583.96) = 1.004(3018) + 0.024 562556.3  0.004 462534.4  0.020183.96 . . . r = m1/ma = 1.232 ma = 811.7 kg/s . . . m3 = m1 + ma(w2w4) = 1000  811.70.0201 = 983.7 . . m/m1 = 0.0163 1244 12.74 A semipermeable membrane is used for the partial removal of oxygen from air that is blown through a grain elevator storage facility. Ambient air (79% nitrogen, 21% oxygen on a mole basis) is compressed to an appropriate pressure, cooled to ambient temperature 25C, and then fed through a bundle of hollow polymer fibers that selectively absorb oxygen, so the mixture leaving at 120 kPa, 25C, contains only 5% oxygen. The absorbed oxygen is bled off through the fiber walls at 40 kPa, 25C, to a vacuum pump. Assume the process to be reversible and adiabatic and determine the minimum inlet air pressure to the fiber bundle. A = N2 ;
0.79 A + 0.21 B P1 = ? 1 3 0.1684 B B = O2
2 0.79 A + 0.0416 B (0.79/0.95) MIX 5 % O2 P2 = 120 kPa P3 = 40 kPa All T = 25 oC Let sA1 = sB1 = 0 at T = 25 oC & P1 sMIX 1 = 0 + 0  y A1R ln yA1  yB1R ln yB1 sMIX 2 = 0 + 0  R ln (P2/P1)  yA2R ln yA2  yB2R ln yB2 Pure B:  3 = 0  R ln (P3/P1) s . . . For n  = n  + n s s s
1 1 2 2 3 3 R 0.8316 ln (P2/P1)  0.79 ln 0.95  0.0416 ln 0.05  0.1684 ln (P3/P1) + 0.79 ln 0.79 + 0.21 ln 0.21 = 0 0.8316 ln (P2/P1) + 0.1684 ln (P3/P1) = 0.3488 + 4.6025  ln P1 = 0.3488 P1 min = 141 kPa For P1 > P1 min . S > 0 [ 1245 12.75 A 100L insulated tank contains N2 gas at 200 kPa and ambient temperature 25 C. The tank is connected by a valve to a supply line flowing CO2 at 1.2 MPa, 90 C. A mixture of 50% N2, 50% CO2 by mole should be obtained by opening the valve and allowing CO2 flow in to an appropriate pressure is reached and close the valve. What is the pressure? The tank eventually cools to ambient temperature. Find the net entropy change for the overall process. CO 2 i V = 100 L, P1 = 200 kPa, T1 = T0 = 25 oC Pi = 1.2 MPa, Ti = 90 oC At state 2: yN2 = yCO2 = 0.50 n2 CO2 = n2 N2 = n1 N2 = P1V/RT1 N = 200*0.1/8.3145*298.2 = 0.00807 kmol n2 = 0.01614 kmol 1st law: nihi = n2u2  n1u1 , for const specific heats niCPoiTi = (niCVoi+n1CVo1)T2  n1CVo1T1 But ni = n1 CPoiTi = CVoiT2 + CVo1(T2T1) 44.010.842363.2 = 44.010.653 T2 + 28.0130.745(T2298.2) T2 = 396.7 K P2 = n2RT2/V = 0.016148.3145396.7/0.1 = 532 kPa Cool to T3 = T0 = 298.2 K P3 = P2 T3/T2 = 532 298.2/396.7 = 400 kPa Q23 = n2CVo2(T3T2) = 0.016 14(0.528.0130.745 + 0.544.010.653)(298.2396.7) = 39.4 kJ s s s SNET = n3 3  n1 1  ni i  Q23/T0 = n [(s )   ] + n [(s )  s ]  Q /T s
i CO2 3 i 1 N2 3 1 23 0 2 = 0.00807(44.010.8418 ln 0.5400 298.2  8.3145 ln ) 363.2 1200 0.5400 39.4 )298.2 200 + 0.00807(  8.3145 ln = +0.0613 + 0 + 0.1321 = +0.1934 kJ/K 1246 12.76 A cylinder/piston loaded with a linear spring contains saturated moist air at 120 kPa, 0.1 m3 volume and also 0.01 kg of liquid water, all at ambient temperature 20C. The piston area is 0.2 m2, and the spring constant is 20 kN/m. This cylinder is attached by a valve to a line flowing dry air at 800 kPa, 80C. The valve is opened, and air flows into the cylinder until the pressure reaches 200 kPa, at which point the temperature is 40C. Determine the relative humidity at the final state, the mass of air entering the cylinder and the work done during the process. P1 = 120 kPa, T1 = 20 oC = T0, V1 = 0.1 m3, mLIQ 1 = 0.01 kg, AP = 0.2 m2, ks = 20 kN/m Pi = 800 kPa P2 = 200 kPa Ti = 80 oC T2 = 40 oC
2 A+V i
DRY AIR P2 = P1 + (ks/Ap)(V2V1) 200 = 120 + (20/0.22)(V20.1) V2 = 0.26 m3 (or = w1 = 0.6222.339/117.66 = 0.012 36) 2.3390.1 = 0.001 73 ( = w1mA1 ) 0.461 52293.2 1 = 1.0 mv1 = Pv1V1 RvT1 Assume no liquid at state 2 mv2 = mv1 + mL1 = 0.01173 kg Pv2 = a) 2 = mvRvT2 0.0011 730.461 52313.2 = = 6.521 kPa V2 0.26 6.521 = 0.883 7.384 PA1V1 RAT1 = 117.660.1 = 0.1398 0.287293.2 b) mA1 = mA2 = 193.4790.26 = 0.5596 0.287313.2 mAi = mA2  mA1 = 0.4198 kg 1 1 c) WCV = PdV = (P1+P2)(V2V1) = (120+200)(0.260.1) = 25.6 kJ 2 2 1247 12.77 Consider the previous problem and additionally determine the heat transfer. Show that the process does not violate the second law. a) QCV = mA2hA2  mA1hA1  mAihAi + mv2hv2 mv1hv1  mL1hL1  P2V2 + P1V1 + WCV = 0.55961.004313.2  0.13981.004293.2  0.41981.004353.2 + 0.0011732574.3  0.001732538.4 0.01 83.9  2000.26 + 1200.1 + 25.6 = 3.48 kJ b) SCV = S2  S1 = mA2sA2  mA1sA1 + mvsv2  mv1sv1  mL1sL1 SSURR =  QCV/T0  mAisAi SNET = mA2sA2  mA1sA1  mAisAi + mvsv2 mv1sv1  mL1sL1  QCV/T0 = mA2(sA2sAi) + mA1(sAisA1) + mvsv2 mv1sv1  mL1sL1  QCV/T0 sA2  sAi = 1.004 ln sAi  sA1 = 1.004 ln 313.2 193.479  0.287 ln = +0.2866 353.2 800 313.2 800  0.287 ln = 0.3633 293.2 117.66 sv2 = sG2  Rv ln 2 = 8.2569  0.46152 ln 0.883 = 8.3143 sv1 = sG1  Rv ln 1 = 8.6671 0 = 8.6671 sL1 = sF1 = 0.2966 SNET = 0.5596 (+0.2868) + 0.1398 (0.3633) + 0.011738.3143 0.001738.6671 0.010.2966 + 3.48/293.2 = +0.201 kJ/K 1248 12.78 The airconditioning by evaporative cooling in Problem 12.57 is modified by adding a dehumidification process before the water spray cooling process. This dehumidification is achieved as shown in Fig. P12.78 by using a desiccant material, which absorbs water on one side of a rotating drum heat exchanger. The desiccant is regenerated by heating on the other side of the drum to drive the water out. The pressure is 100 kPa everywhere and other properties are on the diagram. Calculate the relative humidity of the cool air supplied to the room at state 4, and the heat transfer per unit mass of air that needs to be supplied to the heater unit. States as noted on Fig. P12.78, text page 466. At state 1, 35 oC: PV1 = 1PG1 = 0.305.628 = 1.6884 w1 = 0.6221.6884/98.31 = 0.010 68 At T3 = 25 oC: w3 = w2 = w1/2 = 0.00534 Evaporative cooling process to state 4, where T4 = 20oC As in Eq. 12.23: w3(hV3hL4) = CP0A(T4T3) + w4hFG4 0.005 34(2547.283.9) = 1.004(2025) + w42454.2 w4 = 0.0074 = 0.622PV4/(100PV4) PV4 = 1.176 kPa, 4 = 1.176/2.339 = 0.503 At T5 = 25 oC, w5 = w4 = 0.0074 Evaporative cooling process to state 6, where T6 = 20oC w5(hV5hL6) = CP0A(T6T5) + w6hFG6 0.0074(2547.283.9) = 1.004(2025) + w62454.2 For adiabatic heat exchanger, . . . . . mA2 = mA3 = mA6 = mA7 = mA, => w6 = 0.009 47 Also w2 = w3, w6 = w7 hA2 + w2hV2 + hA6 + w6hV6 = hA3 + w3hV3 + hA7 + w7hV7 or CP0AT7 + w6(hV7hV6) = CP0A(T2+T6T3) + w2(h hV3) V2 1.004 T7 + 0.009 47(hV7  2538.1) = 1.004(60 + 20  25) + 0.005 34(2609.62547.2) = 55.526 By trial and error, T7 = 54.7 oC, hV7 = 2600.3 For the heater 78, w8 = w7, . . Q/mA = CP0A(T8T7) + w7(h hV7) V8 = 1.004(8054.7) + 0.009 47(2643.72600.3) = 25.8 kJ/kg dry air 1249 12.79 A vertical cylinder is fitted with a piston held in place by a pin, as shown in Fig. P12.79. The initial volume is 200 L and the cylinder contains moist air at 100 kPa, 25C, with wetbulb temperature of 15C. The pin is removed, and at the same time a valve on the bottom of the cylinder is opened, allowing the mixture to flow out. A cylinder pressure of 150 kPa is required to balance the piston. The valve is closed when the cylinder volume reaches 100 L, at which point the temperature is that of the surroundings, 15C. a. Is there any liquid water in the cylinder at the final state? b. Calculate the heat transfer to the cylinder during the process. c. Take a control volume around the cylinder, calculate the entropy change of the control volume and that of the surroundings. From Fig. F.5, at T1 = 25 oC & WBT1 = 15 oC a) w1 = 0.006 67 for saturation at state 2, where P2 = PEXT = 150 kPa, T2 = 15 oC; Max PV2 = PG2 = 1.705 kPa wMAX 2 = 0.6221.705/(1501.705) = 0.00715 > w 1 not saturated at 2, no liquid b) w1 = 0.006 67 = 0.622 mA1 = mA2 = PV1 100PV1 PV1 = 1.061 kPa (1001.061)0.2 = 0.2312 kg, mV1 = w1mA1 = 0.00154 kg 0.287298.2 (1501.592)0.1 = 0.1794 kg, mV2 = w2mA2 = 0.0012 kg 0.287288.2 WCV = PEXTdV = PEXT(V2V1) = 150(0.10.2) = 15 kJ QCV = mA2uA2  mA1uA1 + mV2uV2  mV1uV1 + (mAEhAE AVE + mVEhVE AVE) + WCV = 0.17940.717288.2  0.23120.717298.2 + 0.00122396.0  0.001542409.8 + 0.05181.004(298.2 +288.2)/2 + 0.00034(2547.2+2528.9)/2  15.0 = 12.1 kJ c) SNET = mA2(sA2 sAE AVE) + mA1(sAE AVE  sA1) + mV2(sV2sVE AVE) + mV1(sVE AVE  sV1)  QCV/T0 1250 Assume that the mixture exiting is throttled across the value to 100 kPa and then discharged. Therefore, since the composition is constant. PAE = PA1 = 98.94 kPa, PVE = PV1 = 1.061 kPa Also, PA2 = 148.41 kPa & PV2 = 1.592 kPa Using TAE TVE 293.2 K & T0 = 15 oC = 288.2 K SNET = 0.1794( 1.004 ln 288.2 148.425  0.287 ln ) 293.2 98.94 293.2 0) 298.2 288.2 1.575  0.461 52 ln ) 293.2 1.05 293.2  0 )  (12.1)/288.2 298.2 + 0.2312( 1.004 ln + 0.0012( 1.8723 ln + 0.001 54( 1.8723 ln = +0.0138 kJ/K 1251 12.80 Ambient air is at a condition of 100 kPa, 35C, 50% relative humidity. A steady stream of air at 100 kPa, 23C, 70% relative humidity, is to be produced by first cooling one stream to an appropriate temperature to condense out the proper amount of water and then mix this stream adiabatically with the second one at ambient conditions. What is the ratio of the two flow rates? To what temperature must the first stream be cooled? P1 = P2 = 100 kPa 1
5 2 COOL . Q COOL 3 4 . Q MIX= 0 MIX T1 = T2 = 35 oC 1 = 2 = 0.50, 4 = 1.0 P5 = 100, T5 = 23 oC 5 = 0.70 2.814 = 0.0180 1002.814 LIQ H 2 O Pv1 = Pv2 = 0.55.628 = 2.814 kPa => w1 = w2 = 0.622 Pv5 = 0.72.837 = 1.9859 kPa => w5 = 0.622 C.V.: Mixing chamber: cons. mass: Energy Eq.: Call the mass flow ratio 1.9859 = 0.0126 1001.9859 r = ma2/ma1 w1 + r w4 =(1+ r)w5 ha1 + w1hv1 + rha4 + rw4hv4 = (1+r)h a5 + (1+r)w5hv5 0.018 + rw4 = (1+r) 0.0126 or 0.0180.0126 = , r= 0.0126w4 ma1 ma2 with w4 = 0.622 PG4 100PG4 1.004308.2 + 0.0182565.3 + r1.004T4 + r w4hv4 = (1+r)1.004 296.2 + (1+r)0.01262543.6 or r 1.004T4 + w4hG4  329.3 + 26.2 = 0 PG4 = 0.8721, hG4 = 2510.5 [ Assume T4 = 5 oC w4 = 0.6220.8721/(1000.8721) = 0.0055 r = ma2/ma1 = 0.0180.0126 = 0.7606 0.01260.0055 OK 0.7606[1.004278.2 + 0.00552510.5  329.6] + 26.2 = 1.42 0 1252 ENGLISH UNIT PROBLEMS
12.81EA gas mixture at 70 F, 18 lbf/in.2 is 50% N2, 30% H2O and 20% O2 on a mole basis. Find the mass fractions, the mixture gas constant and the volume for 10 lbm of mixture. From Eq. 12.3: ci = yi Mi/ yjMj MMIX = yjMj = 0.5 28.013 + 0.3 18.015 + 0.2 31.999 = 14.0065 + 5.4045 + 6.3998 = 25.811 cN2 = 14.0065 / 25.811 = 0.5427, cH2O = 5.4045 / 25.811 = 0.2094 cO2 = 6.3998 / 25.811 = 0.2479, sums to 1 OK  RMIX = R/MMIX = 1545.36 / 25.811 = 59.87 lbf ft/lbm R V = mRMIX T/P = 10 59.87 710 / 18 144 = 164 ft3 12.82EWeighing of masses gives a mixture at 80 F, 35 lbf/in.2 with 1 lbm O2, 3 lbm N2 and 1 lbm CH4. Find the partial pressures of each component, the mixture specific volume (mass basis), mixture molecular weight and the total volume. From Eq. 12.4: yi = (mi /Mi) / mj/Mj ntot = mj/Mj = (1/31.999) + (3/28.013) + (1/16.04) = 0.031251 + 0.107093 + 0.062344 = 0.200688 yO2 = 0.031251/0.200688 = 0.1557, yCH4 = 0.062344/0.200688 = 0.3107 PO2 = yO2 Ptot = 0.1557 35 = 5.45 lbf/in.2, PN2 = yN2 Ptot = 0.5336 35 = 18.676 lbf/in.2, PCH4 = yCH4 Ptot = 0.3107 35 = 10.875 lbf/in.2  Vtot = ntot RT/P = 0.200688 1545 539.7 / (35 144) = 33.2 ft3 v = Vtot/mtot = 33.2 / (1 + 3 + 1) = 6.64 ft3/lbm MMIX = yjMj = mtot/ntot = 5/0.200688 = 24.914 yN2 = 0.107093/0.200688 = 0.5336, 1253 12.83EA pipe flows 0.15 lbmol a second mixture with molefractions of 40% CO2 and 60% N2 at 60 lbf/in.2, 540 R. Heating tape is wrapped around a section of pipe with insulation added and 2 Btu/s electrical power is heating the pipe flow. Find the mixture exit temperature. C.V. Pipe heating section. Assume no heat loss to the outside, ideal gases. . .  . . Energy Eq.: Q = m(he  hi) = n(he  hi) = nCP mix(Te  Ti)   CP mix = yi Ci = 0.4 8.846 + 0.6 6.975 = 7.723 Btu/lbmole . . Te = Ti + Q / nCP mix = 540 + 2/(0.15 7.723) = 541.7 R 12.84EAn insulated gas turbine receives a mixture of 10% CO 2, 10% H2O and 80% N2 on a mole basis at 1800 R, 75 lbf/in.2. The volume flow rate is 70 ft3/s and its exhaust is at 1300 R, 15 lbf/in.2. Find the power output in Btu/s using constant specific heat from C.4 at 540 R. C.V. Turbine, SSSF, 1 inlet, 1 exit flow with an ideal gas mixture, q = 0. . .  . . Energy Eq.: WT = m(hi  he) = n(hi  he) = nCP mix(Ti  Te) . .   PV = nRT => n = PV / RT = 75 144 70 / (1545.4 1800) = 0.272 lbmole/s   CP mix = yi Ci = 0.1 44.01 0.201 + 0.1 18.015 0.447 + 0.8 28.013 0.249 = 7.27 Btu/lbmol R . WT = 0.272 72.7 (1800  1300) = 988.7 Btu/s 12.85ESolve Problem 12.84 using the values of enthalpy from Table C.7 C.V. Turbine, SSSF, 1 inlet, 1 exit flow with an ideal gas mixture, q = 0. . .  . Energy Eq.: WT = m(hi  he) = n(hi  he) . .   PV = nRT => n = PV / RT = 75 144 70 / (1545.4 1800) = 0.272 lbmole/s . WT = 0.272 [0.1(14358  8121) + 0.1(11178  6468.5) + 0.8(9227  5431)] = 1123.7 Btu/s 1254 12.86ECarbon dioxide gas at 580 R is mixed with nitrogen at 500 R in a SSSF insulated mixing chamber. Both flows are at 14.7 lbf/in.2 and the mole ratio of carbon dioxide to nitrogen is 2 1. Find the exit temperature and the total entropy generation per mole of the exit mixture. . . CV mixing chamber, SSSF. The inlet ratio is nCO2 = 2 nN2 and assume no external heat transfer, no work involved. . . . . . . nCO + 2nN = nex = 3nN ; nN (hN + 2hCO ) = 3nN hmix ex
2 2 2 2 2 2 2  Take 540R as reference and write h = h540 + CPmix(T540). CP N (Ti N 540) + 2CP CO (Ti CO 540) = 3CP mix(Tmix ex 540)
2 2 2 2 CP mix = yiCP i = (29.178 + 237.05)/3 = 8.2718 3CP mixTmix ex = CP N Ti N + 2CP CO Ti CO = 13837
2 2 2 2 Tmix ex = 557.6 R; Pex N = Ptot/3; Pex CO = 2Ptot/3 . . ...  .  Sgen = nexsex(ns)iCO  (ns)iN = nN (se  si)N + 2nN (se  si)CO
2 2 2 2 2 2 2 2 Tex Tex . . Sgen/nN = CPN ln  Rln yN + 2CPCO ln  2 Rln yCO 2 2 TiN 2 2 TiCO 2
2 2 = 0.7575 + 2.1817  0.7038 + 1.6104 = 3.846 Btu/lbmol N2 R 12.87EA mixture of 60% helium and 40% nitrogen by volume enters a turbine at 150 lbf/in.2, 1500 R at a rate of 4 lbm/s. The adiabatic turbine has an exit pressure of 15 lbf/in.2 and an isentropic efficiency of 85%. Find the turbine work. Assume ideal gas mixture and take CV as turbine. wT s = hihes, ses = si, Tes = Ti(Pe/Pi)(k1)/k CP mix = 0.6 1.25 4.003 + 0.4 0.248 28.013 = 5.7811  (k1)/k = R/CP mix = 1545/(5.7811778) = 0.3435 Mmix= 0.64.003 + 0.428.013 = 13.607, CP=CP/Mmix= 0.4249 Tes= 1500(15/150)0.3435 = 680 R, wTs = CP(TiTes) = 348.4 . . wT ac = wTs = 296.1 Btu/lbm; W = mwTs = 1184 Btu/s 1255 12.88EA mixture of 50% carbon dioxide and 50% water by mass is brought from 2800 R, 150 lbf/in.2 to 900 R, 30 lbf/in.2 in a polytropic process through a SSSF device. Find the necessary heat transfer and work involved using values from C.4. Process Pvn = constant leading to n ln(v2/v1) = ln(P1/P2); v = RT/P n = ln(150/30)/ln(900 150/30 2800) =3.3922 Rmix = ciRi = (0.5 35.1 + 0.5 85.76)/778 = 0.07767 CP mix = ciCPi = 0.5 0.203 + 0.5 0.445 = 0.324 n nR Btu w = vdP =  (PevePivi) = (T T ) = 209.3 n1 n1 e i lbm q = hehi + w = CP(TeTi) + w = 406.3 Btu/lbm 12.89EA mixture of 4 lbm oxygen and 4 lbm of argon is in an insulated piston cylinder arrangement at 14.7 lbf/in.2, 540 R. The piston now compresses the mixture to half its initial volume. Find the final pressure, temperature and the piston work. Since T1 >> TC assume ideal gases. u2u1 = 1q2  1w2 =  1w2 ; Pvk = constant, v2 = v1/2 P2 = P1(v1/v2)k = P1(2)k; T2 = T1(v1/v2)k1 = T1(2)k1 Find kmix to get P2,T2 and Cv mix for u2u1 Rmix = ciRi = (0.5 48.28 + 0.5 38.68)/778 = 0.055887 CPmix = ciCPi = 0.5 0.219 + 0.5 0.1253 = 0.17215 Cvmix = CPmix Rmix = 0.11626, kmix = CPmix/Cvmix = 1.4807 P2 = 14.7(2)1.4805= 41.03 lbf/in2, T2 = 540*20.4805= 753.5 R
1 2 1 s2s1 = 0 w = u1u2= Cv(T1T2)= 0.11626(540753.5) = 24.82 Btu/lbm W2 = mtot 1w2 = 8 (24.82) = 198.6 Btu 1256 12.90ETwo insulated tanks A and B are connected by a valve. Tank A has a volume of 30 ft3 and initially contains argon at 50 lbf/in.2, 50 F. Tank B has a volume of 60 ft3 and initially contains ethane at 30 lbf/in.2, 120 F. The valve is opened and remains open until the resulting gas mixture comes to a uniform state. Find the final pressure and temperature and the entropy change for the process. a) 1st law: U2U1 = 0 = nArCV0(T2TA1) + nC H CVO(T2TB1)
2 6 5014430 nAr = PA1VA/RTA1 = = 0.2743 lbmol 1545509.7 nC 3014460 = PB1 B/RTB1 = V = 0.2894 lbmol H 1545579.7 2 6
H 2 6 n2 = nAr + nC = 0.5637 lbmol 0.274339.9480.0756(T2509.7) + 0.289430.070.361(T2509.7) = 0 Solving, T2 = 565.1 R 0.56371545565.1 = 38 lbf/in2 P2 = n2RT2/(VA+VB) = 90144 b) SSURR = 0 SNET = SSYS = nArSAr + nC H SC
2 6 H 2 6 yAr = 0.2743/0.5637 = 0.4866 T2  yArP2 SAr = CP Ar ln  R ln TA1 PA1 = 39.9480.1253 ln 565.1 1545 0.486638 ln = 2.4919 Btu/lbmol R 509.7 778 50 T2  yC2H6P2  R ln SC2H6 = CC2H6 ln TB1 PB1 = 30.070.427 ln 565.1 1545 0.513438 ln 579.7 778 30 = 0.5270 Btu/lbmol R SNET = 0.27432.4919 + 0.28940.5270 = 0.836 Btu/R 1257 12.91EA large SSSF air separation plant takes in ambient air (79% N2, 21% O2 by volume) at 14.7 lbf/in.2, 70 F, at a rate of 2 lb mol/s. It discharges a stream of pure O2 gas at 30 lbf/in.2, 200 F, and a stream of pure N2 gas at 14.7 lbf/in.2, 70 F. The plant operates on an electrical power input of 2000 kW. Calculate the net rate of entropy change for the process. Air 79 % N2 21 % O2 P1 = 14.7
3 1 2 pure O 2 pure N 2 P2 = 30 T2 = 200 F P3 = 14.7 . T1 = 70 F T3 = 70 F WIN = 2000 kW . n1 = 2 lbmol/s . . QCV . QCV . dSNET . . =+ nisi = + (n2 2 + n3 3  n1 1) s s s dt T0 i T0 . . . . . . QCV = nhi + WCV = nO CP0 O (T2T1) + nN CP0 N (T3T1) + WCV 2 2 2 2 = 0.212 [320.213(20070)] + 0  20003412/3600 = +382.6  1895.6 = 1513 Btu/s 660 1545 30 . nisi = 0.212 320.219 ln ln 530 778 0.2114.7 [ + 0.792 0  [ 1545 14.7 ln 778 0.7914.7 = 1.9906 Btu/R s dSNET dt =+ 1513  1.9906 = 0.864 Btu/R s 530 1258 12.92EA tank has two sides initially separated by a diaphragm. Side A contains 2 lbm of water and side B contains 2.4 lbm of air, both at 68 F, 14.7 lbf/in.2. The diaphragm is now broken and the whole tank is heated to 1100 F by a 1300 F reservoir. Find the final total pressure, heat transfer, and total entropy generation. U2U1 = ma(u2u1)a + mv(u2u1)v = 1Q2 S2S1 = ma(s2s1)a + mv(s2s1)v = 1Q2/T + Sgen V2 = VA + VB = mvvv1 + mava1 = 0.0321 + 31.911 = 31.944 ft3 vv2 = V2/mv = 15.9718, T2 => P2v = 58.7 lbf/in2 va2 = V2/ma = 13.3098, T2 => P2a = mRT2/V2 = 43.415 lbf/in2 P2tot = P2v + P2a = 102 lbf/in2 Water: u1 = 36.08, u2 = 1414.3, s1 = 0.0708, s2 = 2.011 Air: u1 = 90.05, u2 = 278.23, sT1 = 1.6342, sT2 = 1.9036
1 2 Q = 2(1414.3  36.08) + 2.4(278.23  90.05) = 3208 Btu Sgen = 2(2.0110.0708) + 2.4[1.9036  1.6342  (53.34/778)ln(43.415/14.7)]  3208/1760 = 3.8804 + 0.4684  1.823 = 2.526 Btu/R 12.93EConsider a volume of 2000 ft3 that contains an airwater vapor mixture at 14.7 lbf/in.2, 60 F, and 40% relative humidity. Find the mass of water and the humidity ratio. What is the dew point of the mixture? Airvap P = 14.7 lbf/in.2, T= 60 F, = 40% Pg = Psat60 = 0.256 lbf/in.2 Pv = Pg = 0.4 0.256 = 0.1024 lbf/in.2 mv1 = P vV RvT = 0.1024 144 2000 = 0.661 lbm 85.76 520 Pa = Ptot Pv1 = 14.7 0.1024 = 14.598 ma = w1 = P aV RaT mv ma = = 14.598 144 2000 = 151.576 lbm 53.34 520 0.661 = 0.00436 151.576 T = 35.5 F Tdew is T when Pg(Tdew) = 0.1024; 1259 12.94EConsider a 10ft3 rigid tank containing an airwater vapor mixture at 14.7 lbf/in.2, 90 F, with a 70% relative humidity. The system is cooled until the water just begins to condense. Determine the final temperature in the tank and the heat transfer for the process. Pv1 = PG1 = 0.7 0.6988 = 0.489 lbf/in2 Since mv = const & V = const & also Pv = PG2: PG2 = Pv1T2/T1 = 0.489T2/549.7 For T2 = 80 F: For T2 = 70 F: 0.489539.7/549.7 = 0.4801 = 0.5073 ( = PG at 80 F ) / 0.489529.7/549.7 = 0.4712 = 0.3632 ( = PG at 70 F ) / interpolating T2 = 78.0 F w2 = w1 = 0.622 ma = Pa1V RaT1 = 0.489 = 0.0214 (14.70.489) 14.21114410 = 0.698 lbm 53.34549.7 1st law: Q12 = U2U1 = ma(ua2ua1) + mv(uv2uv1) = 0.698[0.171(78  90) + 0.0214(1036.3  1040.2)] = 0.698(2.135 Btu/lbm air) = 1.49 Btu 12.95EAir in a piston/cylinder is at 95 F, 15 lbf/in.2 and a relative humidity of 80%. It is now compressed to a pressure of 75 lbf/in.2 in a constant temperature process. Find the final relative and specific humidity and the volume ratio V2/V1. Check if the second state is saturated or not. First assume no water is condensed 1: Pv1= 1PG1 = 0.66, w1 = 0.6220.66/14.34 = 0.0286 2: w2 = 0.622 Pv2/(P2Pv2) = w1 => Pv2 = 3.297 > Pg = 0.825 lbf/in2 Conclusion is state 2 is saturated 2 = 100%, w2 = 0.622 Pg/(P2Pg) = 0.00692 To get the volume ratio, write the ideal gas law for the vapor phases V2 = Va2 + Vv2 + Vf2 = (maRa + mv2Rv)T/P2 + mliqvf V1 = Va1 + Vv1 = (maRa + mv1Rv)T/P1 Take the ratio and devide through with maRaT/P2 to get V2/V1 = (P1/P2)[1 + 0.622w2 + (w1w2)P2vf/RaT]/(1+0.622 w1) = 0.1974 The liquid contribution is nearly zero (=0.000127) in the numerator. 1260 12.96EA 10ft3 rigid vessel initially contains moist air at 20 lbf/in.2, 100 F, with a relative humidity of 10%. A supply line connected to this vessel by a valve carries steam at 100 lbf/in.2, 400 F. The valve is opened, and steam flows into the vessel until the relative humidity of the resultant moist air mixture is 90%. Then the valve is closed. Sufficient heat is transferred from the vessel so the temperature remains at 100 F during the process. Determine the heat transfer for the process, the mass of steam entering the vessel, and the final pressure inside the vessel. i Airvap mix: P1 = 20 lbf/in2, T1 = 560 R
H 2O 1 = 0.10, T2 = 560 R, 2 = 0.90 Pv1 = 1PG1 = 0.10.9503 = 0.095 lbf/in2
AIR + H 2O Pv2 = 0.90.9503 = 0.8553 lbf/in2 Pa2 = Pa1 = P1  Pv1 = 20  0.095 = 19.905 w1 = 0.6220.095/19.905 = 0.002 96 w2 = 0.6220.8553/19.905 = 0.026 64 w= mv ma mvi = ma(w2w1), ma = 19.90514410 = 0.96 lbm 53.34560 P2 = 19.905 + 0.855 = 20.76 lbf/in2 mvi = 0.96(0.02664  0.00296) = 0.0227 lbm CV: vessel QCV = ma(ua2ua1) + mv2uv2  mv1uv1  mvihi uv uG at T uv1 = uv2 = uG at 100 F, ua2 = ua1 QCV = mvi(uG at T  hi) = 0.0227(1043.51227.5) = 4.18 Btu 1261 12.97EA waterfilled reactor of 50 ft3 is at 2000 lbf/in.2, 550 F and located inside an insulated containment room of 5000 ft3 that has air at 1 atm. and 77 F. Due to a failure the reactor ruptures and the water fills the containment room. Find the final pressure. CV Total container. Initial air: Energy: mv(u2u1) + ma(u2u1) = 1Q2  1W2 = 0 mv = V/v = 2335.7 lbm Initial water: v1 = 0.021407, u1 = 539.24, Substitute into energy equation 2335.7 (u2539.24) + 366.040.171 (T277) = 0 u2 + 0.0268 T2 = 541.3 T = 300 LHS = 550.789 & v2 = V2/mv = 2.1407 ft3/lbm Trial and error 2phase (T guess, v2 => x2 => u 2 => LHS) T = 290 LHS = 506.05 T2 = 298 F, x2 = 0.3198, Psat = 65 lbf/in2, LHS=541.5 OK Pa2 = Pa1V1T2/V2T1 = 14.74950757.7/5000536.67 = 20.55 lbf/in2 => P2 = Pa2 + Psat = 85.55 lbf/in2 12.98EAtmospheric air at 95 F, relative humidity of 10%, is too warm and also too dry. An air conditioner should deliver air at 70 F and 50% relative humidity in the amount of 3600 ft3 per hour. Sketch a setup to accomplish this, find any amount of liquid (at 68 F) that is needed or discarded and any heat transfer. CV air conditioner. Check from psychrometric chart, inlet 1, exit 2. ~ ~ In: w1 = 0.0034, hmix,1 = 27, hf,68 = 36.08, Ex: w2 = 0.0078, hmix,2 = 25.4 Water must be added. Continuity and energy equations . . . . . . . mA(1 + w1) + mliq = mA(1 + w2) & mAh1mix + mliqhf + QCV = mAh2mix . . mtot = PVtot /RT = 14.73600144/53.34529.67 = 270 lbm/h . . mA = mtot/(1+w2) = 267.91 lbm/h . . mliq = mA(w2  w1) = 267.91(0.0078  0.0034) = 1.179 lbm/h . QCV = 267.91(25.4270.004436.08) =  471.2 Btu/h If the properties are calculated from the tables then Pg1 = 0.8246, hg1 = 1102.9, Pv1 = 1 Pg1 = 0.08246, Pg2 = 0.36324, hg2 = 1092 Pv2 = 2 Pg2 = 0.1816 ma = PV/RT = 14.74950144/53.34536.67 = 366.04 lbm w1 = 0.622Pv1/(Ptot Pv1) = 0.0035, w2 = 0.00778 ~ ~ h2h1 = 0.24(7095) + 0.007781092  0.00351102.9 = 1.364 . mliq = 267.91(0.00778  0.0035) = 1.147 lbm/h . QCV = 267.91(1.364  0.0042836.08) =  406.8 Btu/h 1262 12.99ETwo moist air streams with 85% relative humidity, both flowing at a rate of 0.2 lbm/s of dry air are mixed in a SSSF setup. One inlet flowstream is at 90 F and the other at 61 F. Find the exit relative humidity. CV mixing chamber. . . . ma w1 + ma w2 = 2ma wex; . . . ma h1 + ma h2 = 2ma hex 1: w1 = 0.0262, h1 = 50.4 2: w2 = 0.0097, h2 = 25.2 wex = (w1 + w2)/2 = 0.018 ; hex = (h1 + h2)/2 = 37.8 Exit: wex, hex => Tex = 75 F, = 95% 12.100E An indoor pool evaporates 3 lbm/h of water, which is removed by a dehumidifier to maintain 70 F, = 70% in the room. The dehumidifier is a refrigeration cycle in which air flowing over the evaporator cools such that liquid water drops out, and the air continues flowing over the condenser, as shown in Fig. P12.71. For an air flow rate of 0.2 lbm/s the unit requires 1.2 Btu/s input to a motor driving a fan and the compressor and it has a coefficient of performance, = Q_L /W_c = 2.0. Find the state of the air after the evaporator, T2, 2, 2 and the heat rejected. Find the state of the air as it returns to the room and the compressor work input. The unit must remove 3 lbm/h liquid to keep steady state in the room. As water condenses out state 2 is saturated. 1: 70 F, 70% => w1 = 0.011, h1 = 28.8 . . . . CV 1 to 2: mliq = ma(w1  w2) => w2 = w1  mliq /ma qL = h1  h2  (w1  w2) hf2 w2 = 0.011  3/(3600 0.2) = 0.0068 ~ 2: w2, 100% => T2 = 47 F, h2 = 18.8, hf2 = 15.04 qL = 28.8  18.8  0.004215.04 = 9.94 Btu/lbm dry air . . ~ ~ CV Total system : h3 = h1 + Wel/ma  (w1w2) hf = 28.8 + 1.2/0.2  0.0632 = 34.74 Btu/lbm dry air 3: w3 = w2, h3 => T3 = 112 F, 3 = 12% . . Wc = ma qL/ = 1 Btu/s 1263 12.101E To refresh air in a room, a counterflow heat exchanger is mounted in the wall, as shown in Fig. P12.72. It draws in outside air at 33 F, 80% relative humidity and draws room air, 104 F, 50% relative humidity, out. Assume an exchange of 6 lbm/min dry air in a SSSF device, and also that the room air exits the heat exchanger to the atmosphere at 72 F. Find the net amount of water removed from room, any liquid flow in the heat exchanger and (T, ) for the fresh air entering the room. State 1: w1 = 0.0236, h1 = 51, Tdew,1 = 81.5 F The room air is cooled to 72 F < Tdew1 so liquid will form in the exit flow channel and state 2 is saturated. 2: 72 F, = 100% => w2 = 0.017, h2 = 36, hf2 = 40.09 3: 33 F, = 80% => w3 = 0.0031, h3 = 11.3 . . CV 1 to 2: mliq,2 = ma (w1  w2) = 6 (0.0236  0.017) = 0.04 lbm/min . . . CV room: mv,out = ma (w1  w4) = ma (w1  w3) = 6(0.02360.0031) = 0.123 lbm/min . ~ ~ . ~ ~ . CV Heat exchanger: ma(h4  h3) = ma(h1  h2)  mliq hf2 h4 = h3 + h1  h2  (w1w2) hf2 = 11.3 + 51  36  0.0066 40.09 = 26 Btu/lbm dry air 4: w4 = w3, h4 => T4 = 94 F, = 9% 1264 12.102E A 4ft3 insulated tank contains nitrogen gas at 30 lbf/in.2 and ambient temperature 77 F. The tank is connected by a valve to a supply line flowing carbon dioxide at 180 lbf/in.2, 190 F. A mixture of 50 mole percent nitrogen and 50 mole percent carbon dioxide is to be obtained by opening the valve and allowing flow into the tank until an appropriate pressure is reached and the valve is closed. What is the pressure? The tank eventually cools to ambient temperature. Calculate the net entropy change for the overall process.
CO 2 i V = 4 ft3, P1 = 30 lbf/in2, T1 = T0 = 77 F Pi = 180 lbf/in.2, Ti = 190 F n2 CO At state 2: yN = yCO = 0.50 2 2 = n2 N = n1 N = P1V/RT1
2 2 N 2 2 = 304144/(1545536.67) = 0.02084 lbmol n2 = 0.04168 lbmol 1st law: nihi = n2u2  n1u1 , for const specific heats niCPoiTi = (niCVoi+n1CVo1)T2  n1CVo1T1 But ni = n1 CPoiTi = CVoiT2 + CVo1(T2T1) 44.010.201649.67 = 44.010.156 T2 + 28.0130.178(T2536.67) T2 = 710.9 R P2 = n2RT2/V = 0.041681545710.9/4144 = 79.48 lbf/in2 Cool to T3 = T0 = 77 F = 536.67 R P3 = P2 T3/T2 = 79.48 536.67/710.9 = 60 lbf/in2 Q23 = n2CVo2(T3T2) = 0.04168(0.528.0130.178 + 0.544.010.156)(536.67710.9) =  43.0 Btu SNET = n3 3  n1 1  ni i  Q23/T0 s s s = ni[(sCO )3   i] + n1[(sN )3  s1]  Q23/T0 s
2 2 = 0.02084(44.010.201 ln + 0.02084(  1.98589 ln 0.560 536.67  1.98589 ln ) 649.67 180 0.560 43.0 ) 30 536.67 = +0.03893 + 0 + 0.0801 = +0.119 Btu/R 1265 12.103E Ambient air is at a condition of 14.7 lbf/in.2, 95 F, 50% relative humidity. A steady stream of air at 14.7 lbf/in.2, 73 F, 70% relative humidity, is to be produced by first cooling one stream to an appropriate temperature to condense out the proper amount of water and then mix this stream adiabatically with the second one at ambient conditions. What is the ratio of the two flow rates? To what temperature must the first stream be cooled? 1 P1 = P2 = P5 = 14.7 lbf/in2 5
2 COOL . Q COOL 3 4 . Q MIX= 0 MIX T1 = T2 = 95 F 1 = 2 = 0.50, 4 = 1.0 T5 = 73 F, 5 = 0.70 LIQ H 2 O Pv1 = Pv2 = 0.50.8246 = 0.4123 w1 = w2 = 0.622 0.4123 = 0.0179 14.70.4123 => w5 = 0.622 0.2845 = 0.0123 14.70.2845 Pv5 = 0.70.4064 = 0.2845 MIX: Call the mass flow ratio r = ma2/ma1 Conservation of water mass: Energy Eq.: ha1 + w1hv1 + rha4 + rw4hv4 = (1+r)h a5 + (1+r)w5hv5 or r= 0.0179 + rw4 = (1+r) 0.0123 w4 = 0.622 PG4 14.7PG4 w1 + r w4 =(1+ r)w5 0.01790.0123 , with 0.0123w4 0.24555 + 0.01791107.2 + r0.24T4 + rw4hv4 = (1+r)0.24533 + (1+r)0.01231093.3 or r 0.24T4 + w4hG4  141.4 + 11.66 = 0 PG4 = 0.121 66, hG4 = 1078.9 [ Assume T4 = 40 F w4 = 0.622 ma2 ma1 = 0.121 66 = 0.0052 14.70.121 66 0.01790.0123 = 0.7887 0.01230.0052 OK 0.7887[0.24500 + 0.00521078.9  141.4] + 11.66 = 0.29 0 => T4 = 40 F 131 CHAPTER 13
The correspondence between the new problem set and the previous 4th edition chapter 10 problem set: New 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 Old 1 10 2 3 4 5 6 7 9 12 13 14 15 16 17 18 19 21 22 23 24 New 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 Old 25 new 26 mod 27 28 new 30 31 mod new 35 36 37 39 new 40 new 41 38 45 44 47 New 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 Old 48 new 49 new new 51 53 55 56 61 63 new 65 66 new 67 68 new new The problems that are labeled advanced are: New 62 63 64 65 66 Old 8 29 20 46 50 New 67 68 69 70 71 Old 52 70 mod 57 58 62 New 72 73 74 Old new new new 132 The Englishunit problems are: New 75 76 77 78 79 80 81 82 Old 74 75 76 78 79 80 82 83 New 83 84 85 86 87 88 89 90 Old 84 81 86 mod 87 88 89 90 91 New 91 92 93 94 95 96 97 Old 92 93 96 97 98 99 101 mod indicates a modification from the previous problem that changes the solution but otherwise is the same type problem. 133 The following table gives the values for the compressibility, enthalpy departure and the entropy departure along the saturated liquidvapor 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 0.94 0.92 0.90 0.88 0.86 0.84 0.82 0.80 0.78 0.76 0.74 0.72 0.70 0.68 0.66 0.64 0.60 0.58 0.54 0.52 0.78 0.69 0.61 0.53 0.46 0.40 0.35 0.30 0.25 0.21 0.18 0.15 0.12 0.10 0.08 0.06 0.05 0.03 0.02 0.01 0.0007 0.14 0.12 0.10 0.09 0.08 0.07 0.06 0.05 0.04 0.035 0.03 0.025 0.02 0.017 0.014 0.01 0.009 0.005 0.004 0.002 0.0014 0.54 0.59 0.64 0.67 0.70 0.73 0.76 0.79 0.81 0.83 0.85 0.87 0.88 0.90 0.91 0.92 0.94 0.95 0.96 0.98 0.98 3.65 3.81 3.95 4.07 4.17 4.26 4.35 4.43 4.51 4.58 4.65 4.72 4.79 4.85 4.92 4.98 5.04 5.16 5.22 5.34 5.41 1.39 1.19 1.03 0.90 0.78 0.69 0.60 0.52 0.46 0.40 0.34 0.29 0.25 0.21 0.18 0.15 0.12 0.08 0.06 0.03 0.02 3.45 3.74 4.00 4.25 4.49 4.73 4.97 5.22 5.46 5.72 5.98 6.26 6.54 6.83 7.14 7.47 7.81 8.56 8.97 9.87 10.38 1.10 0.94 0.82 0.72 0.64 0.57 0.50 0.45 0.39 0.35 0.31 0.27 0.23 0.20 0.17 0.15 0.12 0.08 0.07 0.04 0.03 134 13.1 A special application requires R12 at 140C. It is known that the triplepoint temperature is 157C. Find the pressure and specific volume of the saturated vapor at the required condition. The lowest temperature in Table B.3 for R12 is 90oC, so it must be extended to 140oC using the Clapeyron Eq. 13.7 integrated as in example 13.1 Table B.3: R= at T1 = 90oC = 183.2 K, P1 = 2.8 kPa. 8.3145 = 0.068 76 120.914 P hfg (T  T1) 189.748 (133.2  183.2) ln = = = 5.6543 P1 R T T 0.068 76 133.2 183.2
1 P = 2.8 exp(5.6543) = 0.0098 kPa 135 13.2 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 2 P PP 3 s P P PP
1 2 T
4 3 4 TT v sfg at T qH = TsFG; vfg at T qL = (TT)sFG ; wNET = qH  qL = TsFG Problem similar to development in section 13.1 for shaft work, here boundary movement work, w = Pdv 3 2 4 1 4 wNET = P(v2v1) + Pdv + (PP)(v4v3) + Pdv Approximating, P Pdv (P  ) (v v ); 3 2 2
2 3 Pdv 1 (P  2 ) (v1v4) P & collecting terms, v2+v3 v +v wNET P[( )  ( 12 4)] 2 (the smaller the P, the better the approximation) sFG P v +v v +v T ( 22 3)  ( 12 4) v3 v2 = vG , v4 v1 = vF In the limit as T 0: lim P dPSAT sFG & T0 = = dT vFG T 136 13.3 Ice (solid water) at 3C, 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, hF = 0.01 kJ/kg, Tabel B.1.5: Clapeyron P vI = 0.001 0908, hI = 333.4 kJ/kg dPIF dT dPIF dT = hFhI (vFvI)T = 333.4 = 13 442 0.0000908273.16 T = 13 442(3  0.01) = 40 460 kPa P = Ptp + P = 40 461 kPa 13.4 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, hFG T(vGvF) PG = 70(0.525 015) (76.117.41) = 215.7 kJ/kg (207.8) 7565 sFG = hFG/T = 3.081 kJ/kg K (2.97) Comparison not very close because PG not linear function of T. Using 71 K & 69 K, hFG = 70(0.525 015) (44.5633.24) = 208.0 7169 (1938.81084.2) = 134.82 kJ/kg (134.17) 115105 At 110 K, hFG 110(0.014 342) sFG = 134.82 = 1.226 kJ/kg K (1.22) 110 137 13.5 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 30 MPa H 2O
T.P. T dPIF dT = hIF TvIF const 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 dT = 333.41 = 13 442 kPa/K 273.16(0.000 090 8) (30 0000.6) = 2.2 oC (13 442) at P = 30 MPa, T 0.01 + 13.6 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. Helium at 4.22 K: P1 = 0.1013 MPa, hFG = 83.3 kJ/kmol dPSAT dT = hFG TvFG hFGPSAT RT2 ln P2 P1 = hFG 1 1 [T  T ] R 1 2 For T2 = 1.0 K: 101.3 For T2 = 0.5 K: ln P2 101.3 ln P2 = 83.3 1 1 [4.22  1.0] 8.3145 => P2 = 0.048 kPa = 48 Pa = 83.3 1 1 [4.22  0.5] 8.3145 P2 = 2.1601106 kPa = 2.1601103 Pa 138 13.7 A certain refrigerant vapor enters an SSSF constant pressure condenser at 150 kPa, 70C, 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 ln (150) = 8.15  1000/T2 => T2 = 318.5 K = 45.3oC = T3 q13 = h3 h1 = (h3 h2) + (h2 h1) =  hFG T3 + CP0(T2 T1) dPG dT = hFG TvFG , RT vFG vG = , PG dPG dT = PG d ln PG dT = hFG RT2 PG d ln PG dT = +1000/T2 = hFG/RT2 hFG = 1000 R = 1000 8.3145/100 = 83.15 q13 = 83.15 + 0.7(45.3  70) = 100.44 kJ/kg . QCOND = 1.5(100.44) = 150.6 kW 13.8 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 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 90C: Psat = 38.1 kPa, hIG = 574.5 kJ/kg. Estimate the pressure in the wall cavity at 100 K. 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 P2 P1 = hIG R dPSUB dT = hIG TvIG hIGPSUB RT2 1 1 574.5 1 1 [183.2  100] = 0.188 92 [183.2  100] = 13.81 P2 = 3.83105 kPa = 3.83102 Pa or P2 = P1 1.005106 139 13.9 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). TNBP = normal boiling pt T. TNMP = normal melting pt T. TTP = triple point T. 1) TTP TNMP
TTP P TP P Solid 101.3 kPa PTP Liquid Vap. T TP T
NMP T T NBP 2) h FG (1/P ) dP 2 dT SAT SAT RT
TNMP 0.1013 MPa Since hFG const hFG NBP we ca perform the integral over temperature 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 PSUB PTP h hIG 1 IG = (1/PSUB) dPSUB 2 dT [T  1 ] R TP T RT
P TP TTP P SUB T ln or PSUB = fn(T) 1310 13.10 Derive expressions for ( T/v)u and for (h/s)v that do not contain the properties h, u, or s. P P  T( )v T u u T (v )u =  (v)T (T)v = C v / (see Eqs. 13.33 and 13.34) (Eq.13.20) h P T As dh = Tds + vdP => ( )v = T + v( )v = T  v( )s s s v P T( )v T s s T (v )s =  (v)T (T)v =  C v But / (Eq.13.22) (s )v = T + vT (T)v C
v h P 13.11 Derive expressions for ( h/v)T and for (h/T )v that do not contain the properties h, u, or s. Find (h)T and (h )v v T
and use Eq.13.22 =T dh = Tds + vdP (h)T = T(s )T + v(P)T v v v (P)v + v(P)T T v Also for the second first derivative use Eq.13.28 s (h )v = T(T)v + v(P)v = Cv + v(P)v T T T 13.12 Develop an expression for the variation in temperature with pressure in a constant entropy process, (T/P)s, that only includes the properties PvT and the specific heat, Cp. (P)s = s T (P)T (T)P
s s v ) T P T v == ( ) (CP/T) CP T P ( {(P)T = (T)P, Maxwell relation Eq. 13.23 and the other is Eq.13.27} v 1311 13.13 Determine the volume expansivity, P, and the isothermal compressibility, T, for water at 20C, 5 MPa and at 300C, and 15 MPa using the steam tables. Water at 20oC, 5 MPa (compressed liquid) P = v 1 v (T)P 1(T)P v v Estimate by finite difference. Using values at 0oC, 20oC and 40oC, 1 0.001 0056  0.000 9977 = 0.000 1976/oC 0.000 9995 40  0 v 1 v (P)T  1(P)T v v P T =  Using values at saturation, 5 MPa and 10 MPa, 1 0.000 9972  0.001 0022 = 0.000 50 MPa1 0.000 9995 10  0.0023 T  Water at 300oC, 15 MPa (compressed liquid) 1 0.001 4724  0.001 3084 = 0.002 977/oC 0.001 377 320  280 1 0.001 3596  0.001 3972 = 0.002 731 MPa1 0.001 377 20  10 P T  13.14 Sound waves propagate through a media as pressure waves that cause the media to go through isentropic compression and expansion processes. The speed of sound c is defined by c 2 = (P/ )s and it can be related to the adiabatic compressibility, which for liquid ethanol at 20C is 940 m2/N. Find the speed of sound at this temperature. c2 = (P)s = v (P)s = v
2 1 1 v v P s = ( ) 1 s From Table A.4 for ethanol, c= = 783 kg/m3
1/2 (940101 12 ) 783 = 1166 m/s 1312 13.15 Consider the speed of sound as defined in Problem 13.14. Calculate the speed of sound for liquid water at 20C, 2.5 MPa and for water vapor at 200C, 300 kPa using the steam tables. From problem 13.14 : c2 = (P)s = v (P)s v
2 Liquid water at 20oC, 2.5 MPa, assume (P)s (P) v v
c2 =  T Using saturated liquid at 20oC and compressed liquid at 20oC, 5 MPa, 50.0023 (0.001 002+0.000 9995) (0.000 99950.001 002) = 2.00210 2
2 6 => c = 1415 m/s Superheated vapor water at 200oC, 300 kPa v = 0.7163, s = 7.3115 At P = 200 kPa & s = 7.3115: T = 157oC, v = 0.9766 At P = 400 kPa & s = 7.3115: T = 233.8oC, v = 0.5754 c2 = (0.7163)2 => 0.4000.200 (0.57540.9766) = 0.255810 c = 506 m/s
6 13.16 Find the speed of sound for air at 20C, 100 kPa using the definition in Problem 13.14 and relations for polytropic processes in ideal gases. From problem 13.14 : c2 = (P)s = v (P)s v
2 For ideal gas and isentropic process, Pvk = const P = Cvk P = kCvk1 = kPv1 v c2 = v2(kPv1) = kPv = kRT c = kRT = 1.40.287293.151000 = 343.2 m/s 1313 13.17 A cylinder fitted with a piston contains liquid methanol at 20C, 100 kPa and volume 10 L. The piston is moved, compressing the methanol to 20 MPa at constant temperature. Calculate the work required for this process. The isothermal compressibility of liquid methanol at 20C is 1220 (m)2/N. v w12 = Pdv = P P T dPT =  v T PdPT
2 2 ( )
2 1 1 For v constant & T constant the integral can be evaluated w12 = v T 2 (P 2 2  P1 ) For liquid methanol, from Table A.4: = 787 m3/kg V1 = 10 L, W12 = m = 0.01787 = 7.87 kg 0.011220 (20)2  (0.1)2 2 [ = 2440 J = 2.44 kJ 13.18 A piston/cylinder contains 5 kg of butane gas at 500 K, 5 MPa. The butane expands in a reversible polytropic process with polytropic exponent, n = 1.05, until the final pressure is 3 MPa. Determine the final temperature and the work done during the process. C4H10 m = 5 kg T1 = 500 K P1 = 5 MPa Rev. polytropic process (n=1.05), P2 = 3 MPa Tr1 = 500 5 = 1.176, Pr1 = = 1.316 425.2 3.8 From Fig. D.1: Z1 = 0.68 V1 =
n P1V1 50.680.14304500 = 0.0486 m3 5000 =
n P2V2 => = V2 = 0.0486 1 1.05 (5/3) = 0.0791 m3 Z2Tr2 = P2V2 mRTC 30000.0791 = 0.7803 50.143 04425.2 at Pr2 = 3/3.8 = 0.789 Trial & error: Tr2 = 1.062, Z2 = 0.735
2 => T2 = 451.6 K W12 = PdV = 1 P2V2  P1V1 1n = 30000.0791  50000.0486 = 114 kJ 1  1.05 1314 13.19 Show that the two expressions for the JouleThomson coefficient J given by Eq. 13.54 are valid. j = ( Also j = T h )h =  (P)T P v= /(h )P = [T(v )P  v]/C T T
(T)P =
= P v RT2 Z ( ) CPP T P P ZRT , P ZR RT Z + ( ) P P T P [ZRT + RT (Z)P  v]/C P P T
2 13.20 A 200L rigid tank contains propane at 9 MPa, 280C. The propane is then allowed to cool to 50C as heat is transferred with the surroundings. Determine the quality at the final state and the mass of liquid in the tank, using the generalized compressibility chart, Fig. D.1. Propane C3H8: V = 0.2 m3, P1 = 9 MPa, T1 = 280oC = 553.2 K cool to T2 = 50 oC = 323.2 K From Table A.2: Pr1 = TC = 369.8 K, PC = 4.25 MPa 9 553.2 = 2.118, Tr1 = = 1.496 From Fig. D.1: Z1 = 0.825 4.25 369.8 Z1RT1 0.8250.188 55553.2 v2 = v1 = = = 0.00956 P1 9 000 PG2 = 0.45 4250 = 1912 kPa From Fig. D.1 at Tr2 = 0.874, vG2 = 0.71 0.188 55 323.2/1912 = 0.02263 vF2 = 0.075 0.188 55 323.2/1912 = 0.00239 0.00956 = 0.002 39 + x 2(0.02263  0.00239) mLIQ 2 = (10.354)0.2/0.00956 = 13.51 kg => x 2 = 0.354 1315 13.21 A rigid tank contains 5 kg of ethylene at 3 MPa, 30C. It is cooled until the ethylene reaches the saturated vapor curve. What is the final temperature? C2 H4 T 1 2 v
Pr2 = Pr1 Trial & error: Tr2 ZG2 0.866 0.72 Z2Tr2 Z1Tr1 V = const m = 5 kg P1 = 3 MPa T1 = 30 oC = 303.2 K cool to x2 = 1.0 Pr1 = 3 = 0.595, 5.04 Tr1 = 303.2 = 1.074 282.4 Fig. D.1: = 0.595 ZG2Tr2 Z1 = 0.82 = 0.6756 ZG2Tr2 0.821.074 Pr2 0.42 Pr2 CALC 0.421 ~ OK => T2 = 244.6 K 13.22 Two uninsulated tanks of equal volume are connected by a valve. One tank contains a gas at a moderate pressure P 1, and the other tank is evacuated. The valve is opened and remains open for a long time. Is the final pressure P2 greater than, equal to, or less than P1/2? VA = VB V2 = 2V1, T2 = T1 = T P2 V1 Z2 mRT 1 Z2 = = 2 Z1 P1 V2 Z1 mRT If T < TB, Z2 > Z1 1 > P1 2 P2 P1 < 1 2 P2
Z 1 1.0 2 2 P2 T < TB 1 P1 P T > TB A GAS B EVAC. If T > TB, Z2 < Z1 1316 13.23 Show that van der Waals equation can be written as a cubic equation in the compressibility factor involving the reduced pressure and reduced temperature as 27 Pr2 27 Pr Z Z3 ( + 1 ) Z2 + =0 8Tr 64 T2 512 Tr 3 r Pr van der Waals equation, Eq.13.55:
2 2 27 R TC a= 64 PC P= RT a vb v2 RTC 8PC b= v2(vb) multiply equation by P Get: v3  (b + RT 2 a ab )v +( )v =0 P P P Multiply by Get: Pv P3 3 3 and substitute Z = RT R T bP aP abP2 Z3 ( + 1) Z2 + ( 2 2) Z ( 3 3) = 0 RT RT R T 27 Pr 27 Pr2 Z Z3 ( + 1 ) Z2 + =0 8Tr 64 T2 512 Tr 3 r Pr Substitute for a and b, get: Where Pr = P , Pc Tr = T Tc 1317 13.24 Develop expressions for isothermal changes in enthalpy and in entropy for both van der Waals equation and RedlichKwong equation of state. RT a van der Waals equation of state: P=  vb v2 R (P)v = vb T RT (u)T = T(P)v  P = vb  RT + va vb v T
2 2 2 P a 1 1  P]dv = 2dv = a(  ) (u2u1)T = [T v1 v2 T v v ( ) 1 1 1 1 (h2h1)T = (u2u1)T + P2v2  P1v1 = P2v2  P1v1 + a(  ) v1 v2
2 v2b P R dv = dv = R ln (s2s1)T = T v vb v1b 2 ( ) ( ) 1 1 RedlichKwong equation of state: R a (P)v = vb + 2v(v+b)T T
2 P= RT a vb v(v+b)T1/2 3/2 v2+b v2 3a 3a (u2u1)T = dv = ln[( )(v +b)] v2 2v(v+b)T1/2 2bT1/2 1
1 (h2h1)T = P2v2  P1v1 2 3a ln 2bT1/2 [(vv+b)(vv+b)]
2 2 2 1 R a/2 (s2s1)T = + vb v(v+b)T3/2 dv [ 1 = R ln a (v b) 2bT ln[(vv+b)(vv+b)] v b
2 1 2 1 3/2 2 1 1318 13.25 Determine the reduced Boyle temperature as predicted by an equation of state (the experimentally observed value for most substances is about 2.5), using the van der Waals equation and the RedlichKwong equation. Note: It is helpful to use Eqs. 13.47 and 13.48 in addition to Eq. 13.46 The Boyle temp. is that T at which lim Z ( ) =0 P0 P T lim Z1 1 lim lim Z But P0( )T = P0 = (v  RT) P0 P0 RT P P van der Waals: multiply by vb = & or P= RT a  vb v2 vb , get P vRT a(1b/v) =b P Pv only at TBoyle RT a(vb) or P Pv2 a(10) lim Z RT P0( )T = b  =0 RT P TBoyle = a 27 = T = 3.375 TC Rb 8 C P= RT a  vb v(v+b)T1/2 RedlichKwong: as in the first part, get RT a(1b/v) v=b P Pv(1+b/v)T1/2 a(10) lim Z & RT P0( )T = b  =0 P Pv(1+0)T1/2
5/2 only at TBoyle or 3/2 TBoyle = 2 PC a 0.427 48 R TC = Rb RPC 0.08 664 R TC TBoyle = ( 0.427 48 2/3 ) TC = 2.9 TC 0.086 64 1319 13.26 Consider a straight line connecting the point P = 0, Z = 1 to the critical point P = PC, Z = ZC on a Z versus P compressibility diagram. This straight line will be tangent to one particular isotherm at low pressure. The experimentally determined value is about 0.8 TC. Determine what value of reduced temperature is predicted by an equation of state, using the van der Waals equation and the RedlichKwong equation. See also note for Problem 13.25. ZC  1 PC  0
1.0 Z ZC 0 C.P. PC P slope = lim Z ( ) for T = T P0 P T From solution 13.25 But also equals 1 lim lim Z (P)T = P0 Z1 = RT lim (v  RT) P0 P0 P0 P VDW: using solution 13.25: 1Z or Z 1 1 a lim Z (P)T = C = RT[b  RT] P0 PC ( P C)(RT)2 + bRT  a = 0
C 2 27 R TC a= , 64 PC 2 3 Substituting ZC = , 8
2 b= RTC 8PC Tr = 0.727 40 Tr + 8 Tr  27 = 0 solving, RedlichKwong: using solution 13.25, ZC1 1 a lim Z (P)T = P = RT[b  RT 3/2] or P0 C Substitute get 1 ZC = , 3 a = 0.42748 R2TC PC
5/2 ( P C)R2T 5/2 + bRT 3/2  a = 0
C 1Z , b = 0.08664 RTC PC 2 5/2 3/2 Tr + 0.086 64 Tr  0.427 48 = 0 3 solving, Tr = 0.787 1320 13.27 Determine the 2nd virial coefficient B(T) using the van der Waals equation and the RedlichKwong equation of state. Find also its value at the critical point (the  experimentally observed value is about 0.34 RTC/PC).  RT  lim where Eq.13.47: = v From Eq.13.51: B(T) =  P0 P   b RT a v van der Waals: P =   2 which we can multiply by , get P v b v     RT a(v b) b=   RT = b  a(1b/v ) v  2 or v  P P Pv Pv Taking the limit for P > 0 then we get :  RTC 1 27 TC  B(T) = b  a/RT = (  ) PC 8 64 T where a,b are from Eq.13.59. At T = TC then we have   RTC 19 RTC B(TC) = (  ) = 0.297 64 PC PC For Redlich Kwong the result becomes   a(1 b/v )   RT = b  v   P Pv (1 + b/v ) T1/2 a B(T) = b   3/2 RT => Now substitute Eqs.13.61 and 13.62 for a and b,  RTC TC3/2 B(T) = 0.08664  0.42748 T PC [ and evaluated at TC it becomes   RTC RTC B(TC) = 0.08664  0.42748 = 0.341 PC PC [ 1321 13.28 One early attempt to improve on the van der Waals equation of state was an expression of the form RT a P= vb v2T Solve for the constants a, b, and v using the same procedure as for the van der c Waals equation. From the equation of state take the first two derivatives of P with v: P RT (v )T =  (vb)2 + v2a and 3 T 2RT (v2)T =  (vb)3  v6a 4 T 2P Since both these derivatives are zero at the critical point: RT 2a 2RT 6a and =0 2+ 3 =0 (vb) v T (vb)3 v4T RTC a  2 Also, PC = vCb v T C C solving these three equations: vC = 3b,
2 27 R TC a= , 64 PC 3 b= RTC 8PC 13.29 Use the equation of state from the previous problem and determine the Boyle temperature. RT a P=  2 vb v T vb RT a(1b/v) Multiplying by gives: v  b =  P P PvT Using solution from 13.25 for TBoyle: a (v  RT)= b  a(10) = b  RT2 = 0 at TBoyle P0 P RTT lim or TBoyle = a = Rb
2 27 R TC 1 8PC = 64 PC R RTC 3 27 T 8 C 1322 13.30 Calculate the difference in internal energy of the idealgas value and the realgas value for carbon dioxide at the state 20C, 1 MPa, as determined using the virial equation of state, including second virial coefficient terms. For carbon dioxide we have: B = 0.128 m3/kmol, T(dB/dT) = 0.266 m3/kmol, both at 20C. virial eq.: v uu =  * P= RT BRT + 2 ; v v (T) v = v + BR + RT (dB) v2 v2 dT P R [ v RT2 dB P (T) v  P dv =  v2 (dT ) dv =  RT T (dB) v dT [ [ Solution of virial equation (quadratic formula):   RT 8.3145293.15  1 RT  v= 1 + 1 + 4BP/RT where: = = 2.43737 2 P 1000 P [  1 v = 2.43737 2 [1 + 1 + 4(0.128)/2.43737 = 2.3018 m /kmol
3 Using the minussign root of the quadratic formula results in a compressibility factor < 0.5, which is not consistent with such a truncated equation of state. uu* = 8.3145 293.15 0.266 =  281.7 kJ/kmol 2.3018 [ 13.31 Refrigerant123, dichlorotrifluoroethane, which is currently under development as a potential replacement for environmentally hazardous refrigerants, undergoes an isothermal SSSF process in which the R123 enters a heat exchanger as saturated liquid at 40C and exits at 100 kPa. Calculate the heat transfer per kilogram of R123, using the generalized charts, Fig. D.2 R123: M = 152.93, TC = 456.9 K, PC = 3.67 MPa
1 2 Heat Exchanger SSSF
Tr1 = Tr2 = 313.2/456.9 = 0.685, From Fig. D.2: Pr1 = 0.084, T1 = T2 = 40 oC, x1 = 0 P2 = 100 kPa From D.1: P1 = 0.0843670 = 308 kPa Pr2 = 0.1/3.67 = 0.027 (h*  h)1/RTC = 4.9 s1 + dq/T + sgen = s2, sgen > 0 P2 < P1 with no work done, so process is irreversibel. Energy Eq.: q + h1 = h2, Entropy Eq.: (h*h)2/RTC = 0.056 From Fig. D.2: h2  h1 = 8.3145 456.9 [0.056 + 0 + 4.901]/152.93= 120.4 kJ/kg 1323 13.32 Calculate the heat transfer during the process described in Problem 13.18. From solution 13.18, V1 = 0.0486 m3, V2 = 0.0791 m3, Tr1 = 1.176, Pr1 = 1.316 ; From Fig. D.2:
* * W12 = 114 kJ (h* h)2 = 0.90 RTC Tr2 = 1.062, Pr2 = 0.789, T2 = 451.6 K (h* h)1 = 1.30 RTC , h2  h1 = 1.7164(451.6  500) = 83.1 kJ/kg h2  h1 = 83.1 + 8.3145425.2 (0.90 + 1.30) = 58.8 kJ/kg 58.124 U2  U1 = m(h2  h1)  P2V2 + P1V1 = 5(58.8)  30000.0791 + 50000.0486 = 288.3 kJ Q12 = U2  U1 + W12 = 174.3 kJ 13.33 Saturated vapor R22 at 30C is throttled to 200 kPa in an SSSF process. Calculate the exit temperature assuming no changes in the kinetic energy, using the generalized charts, Fig. D.2 and the R22 tables, Table B.4 R22 throttling process 1st law: h2h1 = (h2h2) + (h2h1) + (h1h1) = 0
* * * * a) Generalized Chart, Fig. D.2, R = 8.31451/86.469 = 0.096156 303.2 * = 0.821 => (h1h1) = 0.096156 369.3 (0.53) = 18.82 Tr1 = 369.3 For CP0, use h values from Table B.4 at low pressure. CP0 278.115  271.594) / (30  20) = 0.6521 kJ/kg K Substituting: (h2h2) + 0.6521(T230) + 18.82 = 0 at Pr2 = 200/4970 = 0.040 Assume T2 = 5.0 oC => Tr2 =278.2/369.3 = 0.753 (h2h2) = RT 0.07 = 0.096156 369.3 (0.07) = 2.49 Substituting : T2 = 5.0 oC b) R22 tables, B.4: at T1 = 30 oC, x1 = 1.0 => h1 = 259.12 2.49 + 0.6521(5.030) + 18.82 = 0.03 0
* * h2 = h1 = 259.12 , P2 = 0.2 MPa => T2 = 4.7 oC 1324 13.34 250L tank contains propane at 30C, 90% quality. The tank is heated to 300C. Calculate the heat transfer during the process. V = 250 L = 0.25 m3 T1 = 30 oC = 303.2 K, x1 = 0.90 Heat to T2 = 300 oC = 573.2 K 1 v M = 44.094, TC = 369.8 K, PC = 4.25 MPa R = 0.188 55, CP0 = 1.6794 C3 H8 T 2 Tr1 = 0.82 Fig. D.1: Z1 = (1 x1) Zf1 + x1 Zg1 = 0.1 0.05 + 0.9 0.785 = 0.711 Fig D.2: Pr
SAT h1h1 RTc
SAT * = 0.1 4.43 + 0.9 0.52 = 0.911 = 1.275 MPa = 0.30 P1 m= 12750.25 = 7.842 kg 0.7110.188 55303.2 7.842Z20.188 55573.2 0.254250 = Z2 1.254 Pr2 = at Tr2 = 1.55 Trial and error on Pr2 Pr2 = 0.743 => P2 = 3.158 MPa, Z2 = 0.94 , (h* h)2 = 0.35 RTC (h2h1) = 1.6794(30030)
* * * * = 453.4 kJ/kg (h1h1) = 0.9110.188 55369.8 = 63.5 kJ/kg (h2h2) = 0.350.188 55369.8 = 24.4 kJ/kg Q12 = m(h2h1)  (P2P1)V = 7.842(24.4+453.4+63.5)  (31581275)0.25 = +3862  471 = 3391 kJ 1325 13.35 The new refrigerant fluid R123 (see Table A.2) is used in a refrigeration system that operates in the ideal refrigeration cycle, except the compressor is neither reversible nor adiabatic. Saturated vapor at 26.5C enters the compressor and superheated vapor exits at 65C. Heat is rejected from the compressor as 1 kW and the R123 flow rate is 0.1 kg/s. Saturated liquid exits the condenser at 37.5C. Specific heat for R123 is Cp =0.6 kJ/kg. Find the coefficient of performance. R123: Tc = 456.9 K, Pc = 3.67 MPa, M = 152.93 kg/kmol, R = 0.05438 kJ/kg K State 1: T1 = 26.5oC = 246.7 K, sat vap., x1 = 1.0 Tr1 = 0.54, Fig D.1, Pr1 = 0.01, P1 = Pr1Pc = 37 kPa h1h1 = 0.03 RTC = 0.8 kJ/kg State 2: T2 = 65oC = 338.2 K State 3: T = 37.5oC = 310.7 K, sat. liq., x = 0
3 3 Fig. D.2, * Tr3 = 0.68, Fig. D.1: Pr3 = 0.08, P3 = Pr3Pc = 294 kPa P2 = P3 = 294 kPa, Pr2 = 0.080, Tr2 = 0.74, Fig. D.2:
* * h2h2 = 0.25 RTC = 6.2 kJ/kg h3h3 = 4.92 RTC = 122.2 kJ/kg State 4: T4 = T1 = 246.7 K, h4 = h3 1st Law Evaporator: qL + h4 = h1 + w; w = 0, qL = h1  h3 = (h1  h1)
* * * * * * h4 = h3 + (h1  h3) + (h3  h3) h1  h3 = Cp(T1  T3) = 38.4 kJ/kg, qL = 0.8 38.4 + 122.2 = 83.0 kJ/kg . . 1st Law Compressor: q + h1 = h2 + wc; Q = 1.0 kW, m = 0.1 kg/s wc = h1  h2 + q; h1  h2 = (h1  h1) h1  h2 = CP (T1  T2) = 54.9 kJ/kg, wc = 0.8 54.9 + 6.2 10.0 = 59.5 kJ/kg = qL/wc = 83.0/59.5 = 1.395
* * * + (h1  h2) + (h2  h2) * * * 1326 13.36 A cylinder contains ethylene, C2H4, at 1.536 MPa, 13C. It is now compressed in a reversible isobaric (constant P) process to saturated liquid. Find the specific work and heat transfer. Ethylene C2H4 P1 = 1.536 MPa = P2 , T1 = 13oC = 260.2 K State 2: saturated liquid, x2 = 0.0 Tr1 = 260.2 1.536 = 0.921 Pr1 = Pr2 = = 0.305 282.4 5.04
* From Figs. D.1, D.2: Z1 = 0.85 , (h1h1)/RTc = 0.40 Z1RT1 0.850.29637260.2 v1 = = = 0.042675 P1 1536 (h1h1) = 0.296 37282.40.40 = 33.5 From Figs. D.1, D.2: T2 = 0.824282.4 = 232.7 K Z2 = 0.05 , (h2h2)/RTc = 4.42 v2 =
* * * * Z2RT2 0.050.29637232.7 = = 0.002245 P2 1536 (h2h2) = 0.296 37282.44.42 = 369.9 (h2h1) = CP0(T2T1) = 1.5482(232.7260.2) = 42.6 w12 = Pdv = P(v2v1) = 1536(0.002 2450.042 675) = 62.1 kJ/kg q12 = (u2u1) + w12 = (h2h1) = 369.9  42.6 + 33.5 = 379 kJ/kg
* 1327 13.37 A piston/cylinder initially contains propane at T = 7C, quality 50%, and volume 10L. A valve connecting the cylinder to a line flowing nitrogen gas at T = 20C, P = 1 MPa is opened and nitrogen flows in. When the valve is closed the cylinder contains a gas mixture of 50% nitrogen, 50% propane on a mole basis at T = 20C, P = 500 kPa. What is the cylinder volume at the final state and how much heat transfer took place? State 1: Propane, T1 = 7oC, x1 = 0.5, V1 = 10 L Tcp = 369.8 K, Pcp = 4.25 kPa, Cpp = 1.679 kJ/kgK, M = 44.097 kg/kmol Fig. D.1: Tr1 = 0.72, Pr1 = 0.12, P1 = Pr1Pc = 510 kPa Fig. D.1: Zf1 = 0.020, Zg1 = 0.88, Z1 = (1  x1)Zf + x1Zg = 0.45  n1 = P1V1/(Z1RT1) = 510 0.01/(0.45 8.3145 266.2) = 0.00512 kmol *   * *  h 1 = h 1o + CPP(T1  To) + (h 1  h 1 ) ; h 1o = 0, *   *   (h 1h 1)f /RTc = 4.79, (h 1h 1)g /RTc = 0.25 *  *  *  h 1  h 1 = (1  x1) (h 1  h 1)f + x1 (h 1  h 1)g = 7748 kJ/kmol  h 1 = 0 + 1.679 44.094(7  20)  7748 = 9747 kJ/kmol Inlet: Nitrogen, Ti = 20oC, Pi = 1.0 MPa, Tcn = 126.2 K, Pcn = 3.39 MPa, Cpn = 1.042 kJ/kgK, M = 28.013 kg/kmol *  Tri = 2.323, Pri = 0.295, h i h i = 0.06 8.3145 126.2 = 62.96 kJ/kmol  *   * * h i = h io + CPn (Ti  To) + (h i  h i ) ; h io = 0, Ti  To = 0 State 2: 50% Propane, 50% Nitrogen by mol, T2 = 20oC, P2 = 500 kPa Tcmix = yiTci = 248 K, Pcmix = yiPci = 3.82 MPa *   Tr2 = 1.182, Pr2 = 0.131, Z2 = 0.97, (h 2h 2)/RTc = 0.06  *   * * h 2 = h 2o + CPmix(T2  To) + (h 2  h 2 ) ;; h 2o = 0, T2  To = 0  a) ni = n1 => n 2 = n1 + ni = 0.1024, V2 = n2Z2RT2/P2 = 0.0484 m3  b) 1st Law: Qcv + nihi = n2u2  n21u21 + Wcv; u = h  Pv Wcv = (P1 + P2)(V2  V1)/2 = 19.88 kJ Qcv = n2h2  n1h1  nihi  P2V2 + P1V1 + Wcv   h i = 62.96 kJ/kmol, h 2 = 123.7 kJ/kmol, Qcv = 50.03 kJ 1328 13.38 An ordinary lighter is nearly full of liquid propane with a small amount of vapor, the volume is 5 cm3 and temperature is 23C. The propane is now discharged slowly such that heat transfer keeps the propane and valve flow at 23C. Find the initial pressure and mass of propane and the total heat transfer to empty the lighter. Propane C3H8 T1 = 23oC = 296.2 K = constant, x1 = 0.0 V1 = 5 cm3 = 5106 m3, From Figs. D.1 and D.2, Tr1 = 296.2/369.8 = 0.804 Z1 = 0.04 P1 = PG T1 = 0.254.25 = 1.063 MPa, (h1h1) = 0.188 55369.84.51 = 314.5 m1 =
* 10635106 = = 0.00238 kg Z1RT1 0.040.188 55296.2 Therefore, m2 much smaller than m1 ( 9.0 106 kg) P1V1 State 2: Assume vapor at 100 kPa, 23oC QCV = m2u2  m1u1 + mehe = m2h2  m1h1  (P2P1)V + (m1m2)he = m2(h2he) + m1(heh1)  (P2P1)V (he  h1) = 0 + 0 + 314.5 QCV = 0 + 0.00238(314.5)  (1001063)5106 = 0.753 kJ 1329 13.39 An uninsulated piston/cylinder contains propene, C3H6, at ambient temperature, 19C, with a quality of 50% and a volume of 10 L. The propene now expands very slowly until the pressure in the cylinder drops to 460 kPa. Calculate the mass of propene, the work and heat transfer for this process. Propene C3H6: From Fig. D.1: Pr1 = Pr sat = 0.25, From D.1: m= P1V1 Z1RT1 = T1 = 19oC = 292.2 K, x1 = 0.50, V1 = 10 L Tr1 = 292.2/364.9 = 0.80, P1 = 0.25 4.6 = 1.15 MPa Z1 = 0.5 0.04 + 0.5 0.805 = 0.4225 11500.010 = 0.471 kg 0.42250.197 58292.2 Assume reversible and isothermal process (slow, no friction, not insulated) Q12 = m(u2u1) + W12
2 2 W12 = PdV (cannot integrate); 1 Q12 = TdS = Tm(s2s1) 1 From Figs. D.2 and D.3: h1h1 = 0.19758 364.9(0.5 4.51 + 0.5 0.46) = 179.2 (s1s1) = 0.197 58 (0.5 5.46 + 0.5 0.39) = 0.5779 (h2h1) = 0 and
* * * * (s2s1) = 0  0.197 58 ln * * 460 = + 0.1829 1161 At Tr2 = 0.80, Pr2 = 0.10, from D.1, D.2 and D.3, Z2 = 0.93 (h2h2) = 0.197 58 364.9 0.16 = 11.5 (s2s2) = 0.197 58 0.13 = 0.0257 (s2s1) = 0.0257 + 0.1829 + 0.5779 = 0.7351 q12 = 292.2 0.7351 = 214.8 kJ/kg (h2h1) = 11.5 + 0 + 179.2 = 167.7 u2u1 = (h2h1) + RT(Z1Z2) = 167.7 + 0.197 58292.2(0.42250.93) = 138.4 w12 = q12  (u2u1) = 214.8  138.4 = 76.4 kJ/kg W12 = m w12 = 36.0 kJ => Q12 = m q12 = 101.2 kJ
* * 1330 13.40 A 200L rigid tank contains propane at 400 K, 3.5 MPa. A valve is opened, and propane flows out until half the initial mass has escaped, at which point the valve is closed. During this process the mass remaining inside the tank expands according to the relation Pv1.4 = constant. Calculate the heat transfer to the tank during the process. C3H8: V = 200 L, T1 = 400 K, P1 = 3.5 MPa Flow out to m2 = m1/2 ; Pv1.4 = const inside Tr1 = v1 = 400 3.5 = 1.082, Pr1 = = 0.824 Fig D.1: Z1 = 0.74 369.8 4.25 0.740.188 55400 = 0.01594, v2 = 2v1 = 0.03188 3500 0.2 1 m1 = = 12.55 kg, m2 = m1 = 6.275 kg, 0.015 94 2 P 2 = P1 (v ) v
2 1 1.4 = 3500 = 1326 kPa 21.4 Trial & error: saturated with 1.326 Pr2 = = 0.312 T2 = 0.826369.8 = 305.5 K & 4.25 P2v2 = Z2RT2 Z2 = 13260.03188 = 0.734 0.188 55305.5 Z2 = ZF2 + x2(ZG2  ZF2) = 0.734 = 0.05 + x2(0.780.05) (h1h1) = 0.188 55369.8(0.9) = 62.8 (h2h1) = 1.6794(305.5400) = 158.7 (h2h2) = (h2hF2)  x2hFG2 = 0.188 55369.8 4.41  0.937(4.410.55) = 55.3 1st law: QCV = m2h2  m1h1 + (P1P )V + mehe AVE 2 Let h1 = 0 then
* * * * * * * * * => x2 = 0.937 [ h1 = 0 + (h1h1) = 62.8
* * h2 = h1 + (h2h1) + (h2h2) = 0  158.7 55.3 = 214.0 he AVE = (h1+h2)/2 = 138.4 QCV = 6.275(214.0)  12.55(62.8) + (35001326)0.2 + 6.275(138.4) = 981.4 kJ 1331 13.41 A newly developed compound is being considered for use as the working fluid in a small Rankinecycle power plant driven by a supply of waste heat. Assume the cycle is ideal, with saturated vapor at 200C entering the turbine and saturated liquid at 20C exiting the condenser. The only properties known for this compound are molecular weight of 80 kg/kmol, ideal gas heat capacity CPO= 0.80 kJ/kg K and TC= 500 K, PC= 5 MPa. Calculate the work input, per kilogram, to the pump and the cycle thermal efficiency.
1 Turbine . QH Ht. Exch Cond . WT T1 = 200oC = 473.2 K, x1 = 1.0 T3 = 20oC = 293.2 K, x3 = 0.0 Properties known: M = 80, CPO = 0.8 kJ/kg K
2 TC = 500 K, PC = 5.0 MPa Tr1 = 473.2 293.2 = 0.946 , Tr3 = = 0.586 500 500 P 4 . W P 3 From Fig. D.1, Pr1 = 0.72, P1 = 0.725 = 3.6 MPa = P4 Pr3 = 0.023, P3 = 0.115 MPa = P2 , ZF3 = 0.004 vF3 = ZF3RT3 P3
4 = 0.0048.3145293.2 = 0.00106 11580 wP =  vdP vF3(P4 P3) = 0.00106(3600115) = 3.7 kJ/kg
3 qH + h4 = h1 , but h3 = h4 + wP => From Fig. D.2: (h1h1) = 0.103 935001.25 = 64.9 (h3h3) = 0.103 935005.2 = 270.2
* * * q H = (h1h3) + wP (h1h3) = CP0(T1T3) = 0.80(20020) = 144.0 (h1h3) = 64.9 + 144.0 + 270.2 = 349.3 * 1332 qH = 349.3 + (3.7) = 345.6 kJ/kg Turbine, (s2  s1) = 0 = (s2  s2)+(s2  s1) + (s1  s1) From Fig. D.3, (s1s1) = 0.103930.99 = 0.1029 (s2s1) = 0.80 ln Substituting, s2s2 = +0.1029  0.0250 = 0.0779 = (s2sF2)  x2sFG2 0.0779 = 0.103 938.85  x20.103 93(8.850.06) (h2h2) = (h2hF2)  x2hFG2 From Fig. D.2, hFG2 = 0.10393500(5.20.07) = 266.6 (h2h2) = 270.2 0.922 266.6 = 25.0 wT = (h1h2) = 64.9 + 144.0 + 25.0 = 104.1 kJ/kg TH = wNET qH = 104.13.7 = 0.29 345.6
* * * * * * * * * * * * 293.2 115  0.103 93 ln = 0.0250 473.2 3600 => x 2 = 0.922 1333 13.42 A geothermal power plant on the Raft river uses isobutane as the working fluid. The fluid enters the reversible adiabatic turbine, as shown in Fig. P13.42, at 160C, 5.475 MPa and the condenser exit condition is saturated liquid at 33C. Isobutane has the properties Tc= 408.14 K, Pc= 3.65 MPa, CP0= 1.664 kJ/kg K and ratio of specific heats k = 1.094 with a molecular weight as 58.124. Find the specific turbine work and the specific pump work. Turbine inlet: T1 = 160oC , P1 = 5.475 MPa Condenser exit: T3 = 33oC , x3 = 0.0, Tr3 = 306.2 / 408.1 = 0.75 From Fig. D.1: Pr3 = 0.16, Z3 = 0.03 From Fig. D.2 & D.3: (h1h1) = 0.143 05408.12.84 = 165.8 (s1s1) = 0.143 052.15 = 0.3076 (s2s1) = 1.664 ln
* * * * * * => P2 = P3 = 0.16 3.65 = 0.584 MPa Pr1 = 5.475 / 3.65 = 1.50 Tr1 = 433.2 / 408.1 = 1.061, 306.2 0.584  0.143 05 ln = 0.2572 433.2 5.475 (s2s2) = (s2sF2)  x2sFG2 = 0.143 056.12  x20.143 05(6.120.29) = 0.8755  x20.8340 (s2s1) = 0 = 0.8755 + x20.8340  0.2572 + 0.3076 (h2h1) = CP0(T2T1) = 1.664(306.2  433.2) = 211.3 From Fig. D.2:, (h2h2) = (h2hF2)  x2hFG2 = 0.143 05408.1[4.690.99(4.690.32)] = 273.8  0.99 255.1 = 21.3 Turbine: Pump: wT = (h1h2) = 165.8 + 211.3 + 21.3 = 66.8 kJ/kg vF3 = ZF3RT3 P3 = 0.030.143 05306.2 = 0.00225 584
* * * * => x2 = 0.99 wP =  v dP vF3(P4 P3) = 0.00225 (5475584) = 11.0 kJ/kg 1334 13.43 Carbon dioxide collected from a fermentation process at 5C, 100 kPa should be brought to 243 K, 4 MPa in an SSSF process. Find the minimum amount of work required and the heat transfer. What devices are needed to accomplish this change of state? 278.2 100 Tri = = 0.915 Pri = = 0.0136 304.1 7380 From D.2 and D.3 : (h*h) /RTC = 0.02, (s*s)ri/R = 0.01
ri Tre = 243 4 = 0.80, Pre = = 0.542 304.1 7.38 (h*h) /RTC = 4.5
re
* * * From D.2 and D.3:
* (s*s)re/R = 4.74 (hihe) =  (hi hi) + (hi he) + (hehe) =  0.188 92304.10.01 + 0.8418(278.2243) + 0.188 92304.14.5
* * * * = 287.6 kJ/kg (sise) =  (si si) + (si se) + (sese) =  0.188 920.01 + 0.8418 ln(278.2/243)  0.188 92 ln(0.1/4) + 0.188 924.74 = 1.7044 kJ/kg K wrev = (hihe) T0(sise) = 287.6  278.2(1.7044) = 186.6 kJ/kg qrev = (hehi) + wrev = 287.6 186.6 = 474.2 kJ/kg 1335 13.44 An insulated cylinder fitted with a frictionless piston contains saturatedvapor carbon dioxide at 0oC, at which point the cylinder volume is 20 L. The external force on the piston is now slowly decreased, allowing the carbon dioxide to expand until the temperature reaches  30 oC. Calculate the work done by the CO2during this process. CO2: Tc = 304.1 K, Pc = 7.38 MPa, Cp = 0.842 kJ/kgK, R = 0.18892 kJ/kg K State 1: T1 = 0oC, sat. vap., x1 = 1.0, V1 = 20 L Tr1 = 0.9, P1 = Pr1Pc = 0.53 7380 = 3911 kPa, Z1 = Zg = 0.67
* (h1  h1)g P 1V 1 * = 0.9 RTC, (s1  s1)g/R = 0.72, m = = 2.262 kg Z1RT1 State 2: T2 = 30oC Tr2 = 0.8, P2 = Pr2Pc = 0.25 7380 = 1845 kPa 2nd Law: Snet =m(s2  s1)  1Q2/T ;
* * * * 1Q 2 = 0, Snet = 0 s2  s1 = (s2  s2) s2  s1 = CP ln
* * * + (s2  s1) + (s1  s1) = 0
 R ln P2 P1
* T2 T1 = 0.044 kJ/kgK, s1  s1 = 0.136 kJ/kgK
* * s2  s2 = 0.180 kJ/kg K,
* * (s2  s2)f = 5.46 R, (s2  s2)g = 0.39 R
* (s2  s2) = (1x2)(s2  s2)f + x2 (s2  s2)g
1st Law:
1Q 2 x2 = 0.889 u = h  Pv = m(u2  u1) + 1W2 ; 1Q2 = 0,
* * * * Z2 = (1  x2)Zf + x2Zg = 0.111 0.04 + 0.889 0.81 = 0.725; (h2  h1) = (h2  h2)
* * + (h2  h1) + (h1  h1)
* h2  h1 = Cp(T2  T1) = 25.3 kJ/kg, (h1  h1) = 51.7 kJ/kg (h2  h2)f
* * = 4.51 RTC ,
* (h2  h2)g = 0.46 RTC
* * (h2  h2) = (1  x2)(h2  h2)f + x2 (h2  h2)g =
h2  h1 = 52.2 25.3 + 51.7 = 25.8 kJ/kg 52.2 kJ/kg u2  u1 = (h2  h1)  Z2RT2 + Z1RT1 = 25.8 0.725 0.18892 243.2 + 0.67 0.18892 273.2 = 24.5 kJ/kg 1W2 = 55.4 kJ 1336 13.45 An evacuated 100L rigid tank is connected to a line flowing R142b gas, chlorodifluoroethane, at 2 MPa, 100C. The valve is opened, allowing the gas to flow into the tank for a period of time and then it is closed. Eventually, the tank cools to ambient temperature, 20C, at which point it contains 50% liquid, 50% vapor, by volume. Calculate the quality at the final state and the heat transfer for the process. The idealgas specific heat of R142b is C = 0.787 kJ/kg K. Rigid tank V = 100 L, m1 = 0 Line: R142b CH3CClF2 M = 100.495, TC = 410.3 K, PC = 4.25 MPa, CP0 = 0.787 kJ/kg K  R = R/M = 8.31451 / 100.495 = 0.082 73 kJ/kg K Line Pi = 2 MPa, Ti = 100 oC, Flow in to T2 = T0 = 20oC VLIQ 2 = VVAP 2 = 50 L Continuity: mi = m2 ; Energy: From D.2 at i: Pri = 2 / 4.25 = 0.471, (hi hi) = 0.082 73410.30.72 = 24.4 (h2hi ) = CP0(T2Ti) = 0.787(20100) = 63.0 From D.2: Tr2 = 293.2 = 0.715 => P2 = 0.1154250 = 489 kPa 410.3
* * * QCV + mihi = m2u2 = m2h2  P2V Tri = 373.15 / 410.3 = 0.91 sat. liq.: ZF = 0.02, (h*hF) = RTC4.85 = 164.6 sat. vap.: ZG = 0.88, (h*hG) = RTC0.25 = 8.5 mLIQ 2 = mVAP 2 = P2VLIQ 2 ZFRT2 ZGRT2
* = 4890.050 = 50.4 kg 0.020.082 73293.2 m2 = 51.55 kg P2VVAP 2 = 1.15 kg, x2 = mVAP 2/m2 = 0.0223 (h2h2) = (1x2)(h2hF2) + x2(h2hG2) = 0.9777 164.6 + 0.0223 8.5 = 161.1 QCV = m2(h2hi)  P2V = 51.55(161.163.0+24.4)  4890.10 = 10 343 kJ
* * 1337 13.46 A cylinder fitted with a movable piston contains propane, initially at 67oC and 50 % quality, at which point the volume is 2 L. The piston has a crosssectional area of 0.2 m2. The external force on the piston is now gradually reduced to a final value of 85 kN, during which process the propane expands to ambient temperature, 4oC. Any heat transfer to the propane during this process comes from a constanttemperature reservoir at 67oC, while any heat transfer from the propane goes to the ambient. It is claimed that the propane does 30 kJ of work during the process. Does this violate the second law? +Q from Tres = 67oC C3 H8 Fext Q to Environment To = 4oC Fext 2 = 85 kN Propane: Tc= 369.8 K, Pc = 4.25 MPa, R = 0.18855 kJ/kg K, Cp = 1.679 kJ/kg K State 1: T1 = 67oC = 340.2 K, x1 = 0.5, V1 = 2.0 L Tr1 = 0.92, Fig D.1, Pr1 = 0.61, P1 = Pr1Pc = 2.592 MPa Zf1 = 0.10, Zg1 = 0.64, Z1 = (1  x1)Zf1 + x1Zg1 = 0.37 P 1V 1 * * m= = 0.218 kg, (h1  h1)f = 3.95 RTc , (h1  h1)g = 1.03 RTc Z1RT1 (s1  s1)f = 4.0 R , (s1  s1)g = 0.82 R
State 2: T2 = 4oC = 277.2 K, Fext 2 = 85 kN Tr2 = 0.75, P2 = Pr2 Pc = 0.165 4250 = 701 kPa P2 = Fext 2/Ap = 425 kPa, Pr2 = 0.10, Z2 = 0.92,
* sat sat * * P2 < P2 State 2 is a vapor V2 = mZ2RT2/P2 = 0.0247 m3 s2  s2 = 0.16 R = 0.0302 kJ/kg K
1W2 * sat h2  h2 = 0.18 RTc =12.6 kJ/kg, 1st Law: 1Q2 = m(u2  u1) + 1W2;
1Q 2 * * * = 30 kJ, u = h  Pv = m(h2  h1)  P2V2 + P1V1 + 1W2 (h2  h1) = (h2  h2)
* + (h2  h1) + (h1  h1)
* * * (h1  h1) = (1  x1)(h1  h1)f + x1 (h1  h1)g =
h2  h1 = Cp(T2  T1) = 105.8 kJ/kg
1Q 2 * * 173.6 kJ/kg = 0.218 (12.6  105.8 + 173.6)  4250.0247 + 25920.002 + 30 = 36.7 kJ 1338 2nd Law: 1Q 2 Snet = m(s2  s1)  ; Tres = 67oC = 340.2 K T * s2  s1 = (s2  s2)
* * + (s2  s1) + (s1  s1)
* * * * * s1  s1 = (1  x1)(s1  s1)f + x1 (s1  s1)g = 0.4544 kJ/kgK s2  s1 = Cpln
* T2 T1  R ln P2 P1 = 0.0030 kJ/kg K Snet = 0.218 (0.03020.0030+0.4544) 36.7/340.2 = 0.0161 kJ/K; Snet < 0 Process is Impossible 13.47 Consider the following equation of state, expressed in terms of reduced pressure and temperature: Pr (1  T62 ) Z=1+ 14 Tr r What does this equation predict for enthalpy departure from the ideal gas value at the state Pr = 0.4, Tr = 0.9 ? What does this equation predict for the reduced Boyle temperature? Pr Pv a) Z= =1+ (1  T62) RT 14 Tr r 6Tc2 RT RTc v= + (1  T2 ) ; P 14Pc RTc 18RTc v vT = T p 14Pc 14PcT3 RTc v 18 h  h* = [v  T ] dP = (1  T 2) Pr = 0.606 RTc 14 T p r 0 Z 1 b) = (1  T62) 14PcTr P T r
P
3 v R 12RTc = + T p P 14PcT3 3 =>
6 = 2.45 Lim Z P0 P T = 0 at Tboyle (1  T62) = 0
r Tr = 1339 13.48 Saturated liquid ethane at 2.44 MPa enters (SSSF) a heat exchanger and is brought to 611 K at constant pressure, after which it enters a reversible adiabatic turbine where it expands to 100 kPa. Find the heat transfer in the heat exchanger, the turbine exit temperature and turbine work. From D.2, Pr1 = 2.44/4.88 = 0.50 , Tr1 = 0.89, T1 = 0.89305.4 = 271.8 K (h1h1) = 0.2765305.44.12 = 347.9 (h2h1) = 1.766 (611  271.8) = 599.0 Pr2 = 0.50 , Tr2 = 611/305.4 = 2.00 From D.2: (h2h2) = RTc 0.14 = 0.2765305.40.14 = 11.8
* * * * q = (h2h1) = 11.8 + 599.0 + 347.9 = 935.1 kJ/kg From D.3, (s2s2) = 0.27650.05 = 0.0138 611 Assume T3 = 368 K , Tr3 = 1.205 at Pr3 = 0.020 (s3s2) = 0.8954 + 0.8833 = 0.0121 From D.3, (s3s3) = 0.27650.01 = 0.0028 (s3s2) = 0.0028  0.0121 + 0.0138 0 Therefore, T3 = 368 K From D.2, (h3h3) = 0.2765305.40.01 = 0.8 w = (h2h3) = 11.8 + 1.766 (611  368) + 0.8 = 418.1 kJ/kg
* * * * * (s3s2) = 1.766 ln * * T3  0.2765 ln 100 2440 1340 13.49 A flow of oxygen at 230 K, 5 MPa is throttled to 100 kPa in an SSSF process. Find the exit temperature and the entropy generation. Tri = 230 5 = 1.488, Pri = = 0.992, 154.6 5.04
* Pre = 0.1 = 0.02 5.04 From D.2: First law: (hi hi) = 0.2598154.60.50 = 20.1 (hehi) = 0 =  (h ehe) + (hehi ) + (hi hi) (hehi ) = 0.9216(208  230) = 20.3
* * * * * * Assume Te = 208 K , Tre = 1.345: From D.2: Check first law From D.3, (si si) = 0.25980.25 = 0.0649 (sesi ) = 0.9216 ln
* * * * (hehe) = 0.2598154.60.01 = 0.4 (hehi) = 0.4 20.3 + 20.1 0 OK
* => Te = 208 K and (sese) = 0.25980.01 = 0.0026 208 0.1  0.2598 ln = 0.9238 230 5 (sesi) = 0.0026 + 0.9238 + 0.0649 = 0.9861 kJ/kg K 13.50 A cylinder contains ethylene, C2H4, at 1.536 MPa, 13C. It is now compressed isothermally in a reversible process to 5.12 MPa. Find the specific work and heat transfer. Ethylene C2H4 P1 = 1.536 MPa , T2 = T1 = 13oC = 260.2 K Tr2 = Tr1 = 260.2 / 282.4 = 0.921 , Pr1 = 1.536 / 5.04 = 0.305 From D.1, D.2 and D.3:
* Z1 = 0.85 (s1s1) = 0.29640.30 = 0.0889
* * (h1h1) = 0.2964282.40.40 = 33.5 and From D.1, D.2 and D.3:
* Z2 = 0.17 , Pr2 = 5.12/5.04 = 1.016 (comp. liquid) (s2s2) = 0.29643.6 = 1.067 5.12 = 0.3568 1.536 (h2h2) = 0.2964282.44.0 = 334.8 and Ideal gas: (h2h1) = 0
* * and (s2s1) = 0  0.2964 ln * * q12 = T(s2s1) = 260.2(1.067  0.3568 + 0.0889) = 347.3 kJ/kg (h2h1) = 334.8 + 0 + 33.5 = 301.3 (u2u1) = (h2h1)  RT(Z2Z1) = 301.3  0.2964260.2(0.170.85) = 248.9 w12 = q12  (u2u1) = 347.3 + 248.9 = 98.4 kJ/kg 1341 13.51 A control mass of 10 kg butane gas initially at 80C, 500 kPa, is compressed in a reversible isothermal process to onefifth of its initial volume. What is the heat transfer in the process? Butane C4H10: m = 10 kg, T1 = 80 oC, P1 = 500 kPa Compressed, reversible T = const, to V2 = V1/5 Tr1 = 353.2 500 = 0.831, Pr1 = = 0.132 425.2 3800 Z1 = 0.92, (s1 s1) = 0.1430.16 = 0.0230
* From D.1 and D.3: Z1RT1 0.920.143353.2 v1 = = = 0.09296 m3/kg P1 500 v2 = v1/5 = 0.01859 m3/kg At Tr2 = Tr1 = 0.831 From D.1: PG = 0.3253800 = 1235 kPa sat. liq.: ZF = 0.05, sat. vap.: ZG = 0.775, Therefore vF = vG = 0.050.143353.2 = 0.00205 1235 0.7750.143353.2 = 0.0317 1235 (s*sF) = R5.08 = 0.7266 (s*sG) = R0.475 = 0.0680 Since vF < v2 < vG x2 = (v2vF)/(vGvF) = 0.5578 (s2s2) = (1x2)(s2sF2) + x2(s2sG2) = 0.44220.7266 + 0.55780.0680 = 0.3592 (s2s1) = CP0 ln (T2/T1)  R ln (P2/P1) = 0  0.143 ln (1235/500) = 0.1293 (s2s1) = 0.3592  0.1293 + 0.0230 = 0.4655 Q12 = Tm(s2s1) = 353.210(0.4655) = 1644 kJ
* * * * * 1342 13.52 An uninsulated compressor delivers ethylene, C2H4, to a pipe, D = 10 cm, at 10.24 MPa, 94C and velocity 30 m/s. The ethylene enters the compressor at 6.4 MPa, 20.5C and the work input required is 300 kJ/kg. Find the mass flow rate, the total heat transfer and entropy generation, assuming the surroundings are at 25C. 293.7 6.4 = 1.040 , Pri = = 1.270 282.4 5.04 From D.2 and D.3, Tri =
* (hi hi) = 0.296 37282.42.65 = 221.8 (si si) = 0.296 372.08 = 0.6164 Tre = 367.2 10.24 = 1.30 , Pre = = 2.032 => From D.1: 282.4 5.04 ZeRTe 0.690.296 37367.2 ve = = = 0.0073 m3/kg Pe 10 240 => Ze = 0.69
* 2 Ae = De = 0.007 85 m2 4 From D.2 and D.3,
* . AeVe 0.007 8530 m= = = 32.26 kg/s ve 0.0073 (hehe) = 0.296 37282.41.6 = 133.9 (sese) = 0.296 370.90 = 0.2667 (hehi ) = 1.5482(367.2293.7) = 113.8 (sesi ) = 1.5482 ln
* * * * * 367.2 10.24  0.296 37 ln = 0.2065 293.7 6.4 (hehi) = 133.9 + 113.8 + 221.8 = 201.7 kJ/kg (sesi) = 0.2667 + 0.2065 + 0.6164 = 0.5562 kJ/kg K First law: 302  300 = 97.9 kJ/kg q = (hehi) + KEe + w = 201.7 + 21000 . . Qcv = mq = 32.26(97.9) = 3158 kW . Qcv . . 3158 Sgen =  + m(se  si) = + + 32.26(0.5562) = 28.53 kW/K 298.2 T0 1343 13.53 A distributor of bottled propane, C3H8, needs to bring propane from 350 K, 100 kPa to saturated liquid at 290 K in an SSSF process. If this should be accomplished in a reversible .setup given the surroundings at 300 K, find the ratio of the volume flow . rates Vin/Vout, the heat transfer and the work involved in the process. From Table A.2: Tri = From D.1, D.2 and D.3, Zi = 0.99 (hi hi) = 0.1886369.80.03 = 2.1 (si si) = 0.18860.02 = 0.0038 290 = 0.784, 369.8 From D.1, D.2 and D.3, Tre = Pre = 0.22 , Pe = 0.224.25 = 0.935 MPa (hehe) = 0.1886369.84.57 = 318.6 (sese) = 0.18865.66 = 1.0672 (hehi ) = 1.679(290  350) = 100.8 (sesi )
* * * * * * * * 350 = 0.946 , 369.8 Pri = 0.1 = 0.024 4.25 and Ze = 0.036 = 1.679 ln 290 0.935  0.1886 ln = 0.7373 350 0.1 (hehi) = 318.6  100.8 + 2.1 = 417.3 (sesi) = 1.0672  0.7373 + 0.0038 = 1.8007 . ZiTi/Pi Vin 0.99 350 0.935 . = = = 310.3 0.1 Vout ZeTe/Pe 0.036 290 wrev = (hihe) T0(sise) = 417.3  300(1.8007) = 122.9 kJ/kg qrev = (hehi) + wrev = 417.3 122.9 = 540.2 kJ/kg 1344 13.54 Saturatedliquid ethane at T1 = 14C is throttled into a SSSF mixing chamber at the rate of 0.25 kmol/s. Argon gas at T2 = 25C, P2 = 800 kPa, enters the chamber at the rate of 0.75 kmol/s. Heat is transferred to the chamber from a heat source at a constant temperature of 150 oC at a rate such that a gas mixture exits the chamber at T3 = 120oC, P3 = 800 kPa. Find the rate of heat transfer and the rate of entropy generation. . Argon, Ta2 = 25oC, P2 = 800 kPa, n2 = 0.75 kmol/s Tca = 150 K, Pca = 4.87 MPa, Ma = 39.948 kg/kmol, Cpa = 0.52 kJ/kg K ha3  ha2 = MaCpa(T3  Ta2) = 1973.4 kJ/kmol . Inlet: Ethane, Tb1 = 14oC, sat. liq., xb1 = 0, n1 = 0.25 kmol/s Tcb = 305.4 K, Pcb = 4.88 MPa, Mb = 30.07 kg/kmol, Cpb = 1.766 kJ/kgK Tr1 = 0.94, Pb1 = Pr1Pcb = 0.69 4880 = 3367 kPa     s s h b1  h b1 = 3.81 RTcb = 9674.5 kJ/kmol,  b1   b1 = 3.74 R = 31.1   h b3  h b1 = MbCpb(T3  Tb1) = 5629.6 kJ/kmol Exit: Mix, T3 = 120oC, P3 = 800 kPa consider this an ideal gas mixture. . . . . . . SSSF Energy Eq.: n1hb1 + n2ha2 +Q = n3h3 = n1hb3 + n2ha3 . . . Q = n1(hb3  hb1) + n2(ha3  ha2) = 0.25 (5629.6 + 9674.5) + 0.75(1973.4) = 5306 kW . . . . Entropy Eq.: Sgen = n1(sb3   b1) + n2(sa3   a2)  Q/TH ; s s . . . . ya = n2/ntot = 0.75; yb = n1/ntot = 0.25 T yP    = M C ln 3   ln a 3 = 8.14 kJ/kmolK sa3 sa2 R a pa T P
a2 a2 TH = 150oC T y P    = M C ln 3   ln b 3 +    = sb3 sb1 R sb1 sb1 b pb T Pb1 b1 = 40.172 + 31.1 = 71.27 kJ/kmol K . Sgen = 0.25 71.27 + 0.75 8.14  5306 / 423 = 11.38 kW/K 1345 13.55 One kilogram per second water enters a solar collector at 40C and exits at 190C, as shown in Fig. P13.55. The hot water is sprayed into a directcontact heat exchanger (no mixing of the two fluids) used to boil the liquid butane. Pure saturatedvapor butane exits at the top at 80C and is fed to the turbine. If the butane condenser temperature is 30C and the turbine and pump isentropic efficiencies are each 80%, determine the net power output of the cycle. H2O cycle: solar energy input raises 1 kg/s of liquid H2O from 40oC to 190oC. Therefore, corresponding heat input to the butane in the heat exchanger is . . QH = m(hF 190 ChF 40 C)H2O = 1(807.62167.57) = 640.05 kW
1 Turbine . QH Ht. Exch Cond P 4 . W P 3 . WT C4H10 cycle T1 = 80 oC, x1 = 1.0 ; T3 = 30 oC, x3 = 0.0 ST = SP = 0.80
2 Tr1 = 353.2 = 0.831 425.2 From D.1, D.2 and D.3: P1 = 0.3253800 = 1235 kPa (h1h1) = 0.143 04425.20.56 = 34.1 (s1s1) = 0.143 040.475 = 0.0680
* * Tr3 = 303.2 = 0.713 425.2 From D.1, D.2 and D.3: P3 = 0.1133800 = 429 kPa sat. liq.: (h*hF) = RTC4.81 = 292.5 ; sat. vap.: (h*hG) = RTC0.235 = 14.3 ; /R), s1 is larger than sG at T3. To demonstrate, (s1sG3) = 1.7164 ln
* * (s*sF) = R6.64 = 0.950 (s*sG) = R0.22 = 0.031 Because of the combination of properties of C 4H10 (particularly the large CP0 353.2 1235  0.143 04 ln = 0.1107 303.2 429 (s1sG3) = 0.0680 + 0.1107 + 0.031 = +0.0737 kJ/kg K 1346 T 1 3 2s 2 s From D.2 and D.3: so that T2S will be > T3, as shown in the Ts diagram. A number of other heavy hydrocarbons also exhibit this behavior. Assume T2S = 315 K, Tr2S = 0.741 (h2Sh2S) = RTC0.21 = 12.8 (s1s2S) = 1.7164 ln
* * * and (s2Ss2S) = R0.19 = 0.027 * 353.2 1235  0.143 04 ln = +0.0453 315 429 (s1s2S) = 0.0680 + 0.0453 + 0.027 0 T2S = 315 K (h1h2S) = 1.7164(353.2315) = 65.6 wST = h1h2S = 34.1 + 65.6 + 12.8= 44.3 kJ/kg wT = SwST = 0.8044.3 = 35.4 kJ/kg At state 3, v3 = 0.0190.143 04303.2 = 0.001 92 m3/kg 429
* * wSP v3(P4P3) = 0.001 92(1235429) = 1.55 kJ/kg wP = wSP SP = 1.55 = 1.94 kJ/kg 0.8 wNET = wT + wP = 35.4  1.94= 33.46 kJ/kg For the heat exchanger, . . QH = 640.05 = mC4H10(h1h4) But h1h4 = h1h3+wP
* * * * h1h3 = (h1h1) + (h1h3) + (h3h3) = 34.1 + 1.716(80  30) + 292.5 = 344.2 kJ/kg Therefore, 640.05 . mC4H10 = = 1.87 kg/s 344.21.94 . . WNET = mC4H10wNET = 1.87 33.46 = 62.57 kW 1347 13.56 A line with a steady supply of octane, C8H18, is at 400C, 3 MPa. What is your best estimate for the availability in an SSSF setup where changes in potential and kinetic energies may be neglected? Availability of Octane at 3 = 1.205, 2.49 From D.2 and D.3, Pri =
* Ti = 400 oC, Pi = 3 MPa 673.2 = 1.184 568.8
* Tri = (h1h1) = 0.072 79568.81.13 = 46.8 ; 100 kPa 298.2 0.1 = 0.524 , Pr0 = = 0.040 568.8 2.49 From D.2 and D.3, Tr0 =
* (s1s1) = 0.072 790.69 = 0.05 Exit state in equilibrium with the surroundings, assume T0 = 298.2 K, P0 = (h0h0) = RTC5.4 = 223.6
* * and (s0s0) = R10.37 = 0.755 * (hi h0) = 1.7113(673.2298.2) = 641.7 (si s0) = 1.7113 ln
* * 673.2 3  0.072 79 ln = 1.1459 298.2 0.1 (hih0) = 46.8 + 641.7 + 223.6 = 818.5 (sis0) = 0.05 + 1.1459 + 0.755 = 1.8509 i = wrev = (hih0)  T0(sis0) = 818.5  298.2(1.8509) = 266.6 kJ/kg 1348 13.57 A piston/cylinder contains ethane gas, initially at 500 kPa, 100 L, and at ambient temperature, 0C. The piston is now moved, compressing the ethane until it is at 20C, with a quality of 50%. The work required is 25% more than would have been required for a reversible polytropic process between the same initial and final states. Calculate the heat transfer and the net entropy change for the process. Ethane: Tc = 305.4 K, Pc = 4.88 MPa, R = 0.2765 kJ/kgK, Cp = 1.766 kJ/kg K State 1: Tr1 = 0.895, Pr1 = 0.102 Z1 = 0.95 v1 = Z1RT1/P1 = 0.1435 m3/kg, m1 = V1/v1 = 0.697 kg (h1  h1) = 0.13RTc = 11.0 kJ/kg, (s1  s1) = 0.09 R = 0.025 kJ/kg K
State 2: T2 = 20oC, x2 = 0.5, 1W2 = 1.25Wrev Tr2 = 0.96, Pr2 = 0.78, P2 = Pr2Pc = 3806 kPa Zf2 = 0.14, Zg2 = 0.54, Z2 = (1  x2)Zf + x2Zg = 0.34 * * (h2  h2) = (1  x2) 3.65 RTc + x2 (1.39 RTc) = 212.8 kJ/kg (s2  s2) = (1  x2) 3.45 R + x2 1.10 R = 0.629 kJ/kg K
v2 = Z2RT2/P2 = 0.0072 m3/kg, P1V1 = P2V2 , Wrev = P dV =
n n * * V2 = mv2 = 0.005 m3 n = 0.6783 ln P2 P1 = n ln V1 V2 = 96.3 kJ, 1W2 = 1.25Wrev = 120.4 kJ 1n a) 1st Law: 1Q2 = m(u2  u1) + 1W2; u = h  Pv u2  u1 = (h2  h1)  (P2v2  P1v1) h2  h1 = = (h2  h2)
* * * P 2V 2  P 1V 1 + (h2  h1) + (h1  h1) * * * (h2  h1) = Cp(T2  T1) = 35.3 kJ/kg,
h2  h1 = 212.8 +35.3 + 11.0 = 166.5 kJ/kg, u2  u1 = 122.2 kJ/kg 1Q2 = 0.697(122.2)  120.4 = 205.6 kJ b) 2nd Law: Snet = m(s2  s1)  1Q2 /To; To = 0oC s2  s1 = (s2  s2)
* * * + (s2  s1) + (s1  s1) * * * (s2  s1) = Cp ln(T2 / T1)  R ln(P2 / P1) = 0.436 kJ/kg K,
205.6 Snet = 0.697(0.629  0.436 + 0.025) + = 0.028 kJ/K 273.2 1349 13.58 The environmentally safe refrigerant R152a (see Problem 13.65) is to be evaluated as the working fluid for a heat pump system that will heat winter households in two different climates. In the colder climate the cycle evaporator temperature is 20C, and the more moderate climate the evaporator temperature is 0C. In both climates the cycle condenser temperature is 30C. For this study assume all processes are ideal. Determine the cycle coefficient of performance for the two climates. R152a difluoroethane. From 13.65: CP0 = 0.996 Ideal Heat Pump TH = 30 oC From A.2: M = 66.05, R = 0.125 88, T C = 386.4 K, PC = 4.52 MPa CASE II: T1 = 0 oC 303.2 = 0.785 386.4 Pr3 = Pr2 = 0.22 => Tr3 =
1 v CASE I: T1 = 20 oC
T 3 4 2 P3 = P2 = 994 kPa Sat.liq.: h3  h3 = 4.56RTC = 221.8 * CASE I)
* T1 = 20 oC = 253.2 K, Tr1 = 0.655, Pr1 = 0.058 P1 = 262 kPa and s1  s1 = 0.14R = 0.0176
* * h1  h1 = 0.14RTC = 6.8
* Assume T2 = 307 K, Tr2 = 0.795 given Pr2 = 0.22 From D.2, D.3: s2  s2 = 0.34R = 0.0428 ; s2  s1 = 0.996 ln
* * h2  h2 = 0.40RTc = 19.5 307 994  0.125 88 ln = 0.0241 253.2 262 s2  s1 = 0.0428 + 0.0241 + 0.0176 = 0.001 0 OK h2  h1 = 19.5 + 0.996(307253.2) + 6.8 = 40.9 h2  h3 = 19.5 + 0.996(307303.2) + 221.8 = 206.1 = qH wIN = h2  h3 h2  h1 = 206.1 = 5.04 40.9 1350 case II) T1 = 0 oC = 273.2 K, Tr1 = 0.707 => Pr1 = 0.106, P1 = 479 kPa h1  h1 = 0.22RTC = 10.7
* * and s1  s1 = 0.21R = 0.0264
* * Assume T2 = 305 K, Tr2 = 0.789 s2  s2 = 0.35R = 0.0441 s2  s1 = 0.996 ln
* * and h2  h2 = 0.38RTC = 18.5 305.0 994  0.125 88 ln = 0.0178 273.2 479 s2  s1 = 0.0441 + 0.0178 + 0.0264 = 0.0001 0 OK h2  h1 = 18.5 + 0.996(305.0273.2) + 10.7 = 23.9 h2  h3 = 18.5 + 0.996(305.0303.2) + 221.8 = 205.1 = h2  h3 h2  h1 = 205.1 = 8.58 23.9 13.59 Repeat the calculation for the coefficient of performance of the heat pump in the two climates as described in Problem 13.58 using R12 as the working fluid. Compare the two results. Problem the same as 13.58 , exept working fluid is R12 For R12: At T3 = 30 oC: CASE I) T1 = 20 oC, P3 = P2 = 0.745 MPa, h3 = 64.6 h1 = 178.7, s1 = 0.7087 T2 = 39.8 oC At s2 = s1 & P2 h2 = 206.8 = CASE II: h2  h3 h2  h1 = 206.8  64.6 142.2 = = 5.06 206.8  178.7 28.1 h1 = 187.5, s1 = 0.6965 T1 = 0oC, At s2 = s1 & P2 = T = 34.6oC , h = 203.0 203.064.6 138.4 = = 8.93 203.0187.5 15.5 1351 13.60 One kmol/s of saturated liquid methane, CH4, at 1 MPa and 2 kmol/s of ethane, C2H6, at 250C, 1 MPa are fed to a mixing chamber with the resultant mixture exiting at 50C, 1 MPa. Assume that Kay's rule applies to the mixture and determine the heat transfer in the process. Control volume the mixing chamber, inlet CH4 is 1, inlet C2H6 is 2 and the exit state is 3. Energy equation is . . . . QCV = n3 h3  n1 h1  n2 h2 Select the ideal gas reference temperature to be T3 and use the generalized charts for all three states. Pr1 = Prsat = 1/4.60 = 0.2174 => Trsat = 0.783, T1 = 0.783 190.4 = 149.1 K, h1 = 4.57 Pr2 = 1/4.88 = 0.205, Tr2 = 523/305.4 = 1.713 h1 = C1(T1  T3)  h1 RTc = 36.15(149.1  323.2)  4.57 8.3145 190.4 = 13528 kJ/kmol h2 = C2(T2  T3)  h2 RTc = 53.11(250  50)  0.08 8.3145 305.4 = 10419 kJ/kmol Tcmix = (1 190.4 + 2 305.4)/3 = 267.1 K Pcmix = (1 4.60 + 2 4.88)/3 = 4.79 MPa Tr3 = 323.2/267.1 = 1.21 , Pr3 = 1/4.79 = 0.21 h3 = 0  0.15 267.1 8.3145 =  333 kJ/kmol . QCV = 3(333)  1(13528)  2(10419) =  8309 kW 13.61 Consider the following reference state conditions: the entropy of real saturated liquid methane at 100C is to be taken as 100 kJ/kmol K, and the entropy of hypothetical ideal gas ethane at 100C is to be taken as 200 kJ/kmol K. Calculate the entropy per kmol of a real gas mixture of 50% methane, 50% ethane (mole basis) at 20C, 4 MPa, in terms of the specified reference state values, and assuming Kay's rule for the real mixture behavior. CH : T = 100 oC, s = 100 kJ/kmol K
4 0 LIQ 0 * s C2H6: T0 = 100 oC, P0 = 1 MPa,  0 = 200 kJ/kmol K Also for CH4: TC = 190.4 K, PC = 4.60 MPa For a 50% mixture: Tcmix = 0.5 190.4 + 0.5 305.4 = 247.9 K Pcmix = 0.5 4.60 + 0.5 4.88 = 4.74 MPa 1352 IG MIX at T0(=100 oC), P0(=1 MPa): CH4: Tr0 = 0.91 ,
G PG = 0.57 4.60 = 2.622 MPa
G  * s s0 CH4 =  LIQ 0 P + (s*sLIQ)at P  R ln (P0/PG) = 100 + 4.018.3145  8.3145 ln (1/2.622) = 141.36 * s0 MIX = 0.5141.36 + 0.5200  8.3145(0.5 ln 0.5 + 0.5 ln 0.5) = 176.44 CP0 MIX = 0.516.042.254 + 0.530.071.766 = 44.629 293.2 4 *  8.3145 ln = 188.41 kJ/kmol K sTP MIX = 176.44 + 44.629 ln 173.2 1 For the mixture at T,P: Tr = 1.183, Pr = 0.844 * sTP MIX   TP MIX = 0.43638.3145 = 3.63 kJ/kmol K s Therefore, sTP MIX = 188.41  3.63 = 184.78 kJ/kmol K An alternative is to form the ideal gas mixture at T,P instead of at T0,P0 : T P  * s sTP CH4 =  LIQ 0 + (s*sLIQ) + CP0 CH4 ln  R ln T0 PG
P ,T G 0 at P , T G 0 = 100 + 33.34 + 16.042.254 ln 293.2 4  8.3145 ln 173.2 2.6 = 100 + 33.34 + 19.03  3.53 = 148.84 293.2 4 * sTP C2H6 = 200 + 30.071.766 ln  8.3145 ln 173.2 1 = 200 + 27.96  11.53 = 216.43 * sTP MIX = 0.5148.84 + 0.5216.43  8.3145(0.5 ln 0.5 + 0.5 ln 0.5) = 188.41 sTP MIX = 188.41  3.63 = 184.78 kJ/kmol K 1353 Advanced Problems
13.62 An experiment is conducted at 100C inside a rigid sealed tank containing liquid R22 with a small amount of vapor at the top. When the experiment is done the container and the R22 warms up to room temperature of 20C. What is the pressure inside the tank during the experiment? If the pressure at room temperature should not exceed 1 MPa, what is the maximum percent of liquid by volume that can be used during the experiment? T R22 tables Go to 70 oC a) For hFG const & vFG vG RT/PG hFG 1 1 ln PG0 R T0 T1 PG1
2 20 C
o o [ 1 100 C v extrapolating from 70 oC (Table B.4.1) to TAVE = 85 oC, hFG 256.5 Also R = 8.3145 = 0.096 15 86.469 For T0 = 203.2 K & T1 = 173.2 K ln 256.5 1 1 (20.5)= 0.096 15 [203.2  173.2] PG1 PG1 = 2.107 kPa b) Extrapolating vF from 70 oC to T1 = 100 oC vF1 0.000 634 Also vG1 RT1/PG1 = 0.096 15173.2 = 7.9037 2.107 Since v1 = v2 vF2 = 0.000 824 0.000 824 = 0.000 634 + x 17.9031 => x 1 = 2.404105 VLIQ 1 m = (1x1)vF1 = 0.000 634, VVAP 1 m = x1vG1 = 0.000 190 % LIQ, by vol. = 0.000 634 100 = 76.9 % 0.000 824 1354 13.63 Determine the lowpressure JouleThomson inversion temperature from the condition in Eq. 13.54 as predicted by an equation of state, using the van der Waals equation and the RedlichKwong equation. v T( ) P  v T T j = ( )n = , But CP P v (T)P = R (vb) (T)v (v )T
P P a) van der Waal VDW: v (T)P =  [ (vb)2 + v3]
v2 and then rearranging, v2 RT 2a Substitute into j eq'n & multiply by 1 CP j = [ 1 b RT( )(1b/v)+ 2a 1b/v 1 2 2a ) v RT( 1b/v lim
P0 j only at TINV Then or = 1 2a [b + RT]= 0 CP TINV = 2a 27 = T = 6.75 TC Rb 4 C
1 b) RK: P P R 2a = (T)v vb + v(v+b)T3/2 RT a(2v+b) (v )T = (vb)2 + v2(v+b)2 T1/2 Substituting as in part a)
2a bRT (1b/v)2 (1+b/v)2 T1/2 RT a(2+b/v) (1b/v)2 v(1+b/v)2 T1/2 1 j = then lim 1 CP [ P0 j = 1 a [b  5 RT3/2]= 0 only at TINV CP 2 5a 2/3 ) = 5.3 TC 2Rb or TINV = ( 1355 13.64 Suppose the following information is available for a given pure substance: vx Known: T PSAT = fn(T) 5 4
6 3 1 2 s vapor P = fn(T,v) values of vF, PC, & TC vapor CV at vx Outline the procedure that should be followed to develop a table of thermodynamic properties comparable to Tables 1,2 and 3 of the steam tables. 1) from vapor pressure curve, get P SAT at each T. 2) vF values are known. 3) reference state: T0 & sat. liquid. Set 4) from eq'n of state, find v1 = vG at T0 h0 = 0, s0 = 0, u0 = 0  P0v0 then vFG(T0) = vG(T0)  vF(T0) 5) from clapeyron eq'n, using slope of vapor pressure curve & vFG, find hFG(T0) then h1 = hG(T0) = 0 + h FG(T0) 6) u1 = uG(T0) = h1  P1v1; uFG(T0) = u1  u0 hFG(T0) 7) sFG(T0) = ; s1 = sG(T0) = 0 + sFG(T0) T0 8) find P2 from eq'n of state ( T2 = T0, v2 = vx )
2 P 9) u2  u1 = [T( )v  P]dvT T
1 evaluate from eq'n of state then h2 = u2 + P2v2
2 P 10) s2  s1 = ( )v dvT, find s2. T
1 3 11) u3  u2 = Cvx dT, find u3. 2 12) find P3 from eq'n of state ( T3, v3 = vx ) 13) h3 = u3 + P3v3 1356 3 14) s3  s2 = (Cvx/T) dT, find s3 2 15) at any point 4 (T4=T3) given by desired P4, find v4 from eq'n of state. 16) find P5 = PSAT at T5 = T3 from vap. pressure. 17) find v5 = vG(T5) from eq'n of state. 18) find u5, h5, s5 as in steps 9 & 10. 19) find hFG(T5) from Clapeyron eq'n as in step 5. 20) h6 = hF(T5) = h5  hFG(T5) 21) sFG(T5) = hFG(T5) , s6 = sF(T5) = sF  sFG(T5) T5 22) u6 = h6  P6v6, uFG(T5) = u5  u6 23) pick a different T (instead of T3) and repeat steps 11 to 22. For T T5 steps 16 to 22 are eliminated. 1357 13.65 The refrigerant R152a, difluoroethane, is tested by the following procedure. A 10L evacuated tank is connected to a line flowing saturatedvapor R152a at 40C. The valve is then opened, and the fluid flows in rapidly, so that the process is essentially adiabatic. The valve is to be closed when the pressure reaches a certain value P2, and the tank will then be disconnected from the line. After a period of time, the temperature inside the tank will return to ambient temperature, 25C, through heat transfer with the surroundings. At this time, the pressure inside the tank must be 500 kPa. What is the pressure P2 at which the valve should be closed during the filling process? The ideal gas specific heat of R152a is CP0 = 0.996 kJ/kg K. M = 66.05, TC = 386.4 K, PC = 4.52 MPa, T3 = T0 = 25oC, P3 = 500 kPa, R = R/M = 8.3145/66.05 = 0.12588 Tr3 = 298.2/386.4 = 0.772, Pr3 = 500/4520 = 0.111 (h*h)3 = 0.19 RTC R152a CHF2CH3 : A.2: From D.1 and D.2 at 3: Z3 = 0.92, m3 = m2 = mi = P 3V Z3RT3 = 5000.010 = 0.145 kg 0.920.125 88298.2 hi = u2 = h2  Z2RT2
* Filling process: or
* Energy Eq.:
* (h2h2) + CP0(T2Ti) + (hi hi)  P2V/m2 = 0 From D.2 with Tri = 313.2/386.4 = 0.811, (hi hi) = 0.125 88386.40.49 = 23.8 ; Assume P2 = 575 kPa, Pr2 = 0.127 Now assume T2 = 339 K, Tr2 = 0.877 => From D.1: Z2 = 0.93 Z2T2 0.93339 Z3T3 0.92298.2 = = 0.5483 = = 0.5487 P2 P3 575 500 Pi = 0.2764520 = 1248 kPa T2 = 339 K is the correct T2 for the assumed P2 of 575 kPa. Now check the 1st law to see if 575 kPa is the correct P 2. From D.2, h2h2 = 0.125 88386.40.17 = 8.3 5750.010 = +1.5 0 0.1456 1st law sum = +4.2)
* Substituting into 1st law, 8.3 + 0.996(339313.2) + 23.8 P2 = 575 kPa (Note: for P2 = 580 kPa, T2 = 342 K, 1358 13.66 An insulated cylinder has a piston loaded with a linear spring (spring constant of 600 kN/m) and held by a pin, as shown in Fig. P13.66. The cylinder crosssectional area is 0.2 m2, the initial volume is 0.1 m3, and it contains carbon dioxide at 2.5 MPa, 0C. The piston mass and outside atmosphere add a force per unit area of 250 kPa, and the spring force would be zero at a cylinder volume of 0.05 m3. Now the pin is pulled out; what is the final pressure inside the cylinder? AP = 0.2 m3, VFs = 0 = 0.05 m3 , ks = 600 kN/m P0 P0 = 250 kPa, P1 = 2.5 MPa, V1 = 0.1 m3, T1 = 0oC R = 0.1889, C = 0.842 ks Any point: PEXT = P0 + 2 (VV0) Ap 600 Pin pulled: PEXT = 250 + 2 (0.10.05) = 1000 kPa 0.2 But P1 = 2500 kPa Irr. process, expansion to 1000 < P2 < 2500 600 (V 0.05) = 15 000 V2  500 0.22 2 CO2 Q = 0 At 2: P2 = 250 +
2 1 W12 = PEXTdV = (PEXT 1 + PEXT 2)(V2V1) 2
1 Q12 = 0 = m(h2h1)  P2V2 + P1V1 + W12 h2h1 = (h2h2) + CP0(T2T1) + (h1h1) Tr1 = 273.2 2.5 = 0.898, Pr1 = = 0.339 , 304.1 7.38 => m= From Fig. D.1, Z1 = 0.82
* * Also P1V1 = mZ1RT1 From D.2,
* 25000.1 = 5.91 kg 0.820.188 92273.2 (h1h1) = 0.490.1889304.1 = 28.2 => V2 = 1655+500 = 0.1437 m3 15 000 Assume P2 = 1655 kPa Z2Tr2 = P2V2 mRTC = 16550.1437 = 0.7004 5.910.188 92304.1 Z2 = 0.850 , ZTr = 0.7004 ~ OK At Tr2 = 0.824 and Pr2 = 0.224 => T2 = 250.6 K 1359 (h2h2) = 0.37 0.1889 304.1 = 21.3 (h2h1) = 0.842(250.6  273.2) = 19.0 W12 = 0.5 (1000 + 1655) (0.14370.10) = 58.0 kJ Q12 = 5.91(21.319.0+28.2)  16550.1437 + 250 + 58.0 = 1.3 0 OK (Note: for P2 = 1660 kPa, Q12 = +2.7) P2 = 1655 kPa
* * * 13.67 A 10 m3 storage tank contains methane at low temperature. The pressure inside is 700 kPa, and the tank contains 25% liquid and 75% vapor, on a volume basis. The tank warms very slowly because heat is transferred from the ambient. a. What is the temperature of the methane when the pressure reaches 10 MPa? b. Calculate the heat transferred in the process, using the generalized charts. c. Repeat parts (a) and (b), using the methane tables, Table B.7. Discuss the differences in the results. CH4: V = 10 m3, P1 = 700 kPa VLIQ 1 = 2.5 m3, VVAP 1 = 7.5 m3 a) Pr1 = 0.70 10 = 0.152, Pr2 = = 2.174 4.60 4.60 From D.1: ZF1 = 0.025, ZG1 = 0.87 & T1 = 0.74 190.4 = 140.9 K vF1 = vG1 = 0.0250.518 35140.9 = 0.00261 700 0.870.518 35140.9 = 0.0908 700 2.5 7.5 mLIQ 1 = = 957.9 kg, mVAP 1 = = 82.6 kg 0.00261 0.0908 Total m = 1040.3 kg v2 = v1 = Z20.518 35190.4Tr2 V 10 = = 0.00961 = m 1040.5 10 000 or Z2Tr2 = 0.9737 at Pr2 = 2.174 By trial and error 1360 Tr2 = 1.334 & Z2 = 0.73, T2 = 1.334190.4 = 254.0 K b) 1st law: Q12 = m(u2u1) = m(h2h1)  V(P2P1) Using D.2 & x1 =
* * 82.6 = 0.0794 1040.5 (h1h1) = (h1hF1)  x1hFG1 = 0.518 35190.4 4.720.0794(4.720.29) (h2h1) = 2.2537(254.0140.9) = 254.9 (h2h2) = 0.518 35190.4(1.47) = 145.1 (h2h1) = 145.1 + 254.9 + 431.1 = 540.9 kJ/kg Q12 = 1040.5(540.9)  10(10 000700) = 469 806 kJ c) Using Table B.7 for CH4 T1 = TSAT 1 = 141.7 K, vF1 = 0.002 675, uF1 = 178.47
* * * [ = 431.1 vG1 = 0.090 45 , uG1 = 199.84 mLIQ 1 = 2.5 7.5 = 934.6, mVAP 1 = = 82.9 0.002 675 0.090 45 m = 1017.5 kg and v2 = 10 = 0.009 828 m3/kg 1017.5 Total mass T2 = 259.1 K At v2 & P2 = 10 MPa u2 = 296.11 Q12 = m(u2u1) = 1017.5296.11  934.6(178.47)  82.9(199.84) = 451 523 kJ 1361 13.68 Calculate the difference in entropy of the idealgas value and the realgas value for carbon dioxide at the state 20C, 1 MPa, as determined using the virial equation of state. Use numerical values given in Problem 13.30. CO2 at T = 20oC, P = 1 MPa
RT/P* RT/P* P * sP*  sP = T v dv ; ID Gas, ( ) R * P sP*  sP = dv = R ln * v P
v(P) RT/P* v(P) Therefore, at P: sP  sP = R ln * P P + dv P* T v ( ) v(P) virial: P= RT BRT + 2 and v v (P)v = R + BR + RT(dB) v v T v dT
2 2 Integrating, P RT dB  sP = R ln * + R ln * + R B + T dT P P v RT dB 1 = R ln + B+T Pv dT v Using values for CO2 from solution 13.30,
* sP [ ( ( )]( ( )) ] ( [ 1 P* v RT ) 1  *   = 8.3145 ln 2.437 37 + 0.128 + 0.266 sP sP 2.3018 2.3018 = 0.9743 kJ/kmol K [ ) 13.69 Carbon dioxide gas enters a turbine at 5 MPa, 100C, and exits at 1 MPa. If the isentropic efficiency of the turbine is 75%, determine the exit temperature and the secondlaw efficiency. CO2 turbine: S = w/wS = 0.75 inlet: T1 = 100oC, P1 = 5 MPa, exhaust: P2 = 1 MPa a) Pr1 =
* 5 373.2 1 = 0.678, Tr1 = = 1.227, Pr2 = = 0.136 7.38 304.1 7.38 From D.2 and D.3, (h1h1) = 0.188 92304.10.52 = 29.9 (s1s1) = 0.188 920.30 = 0.0567
* 1362 Assume T2S = 253 K, Tr2S = 0.832 From D.2 and D.3:
* (h2Sh2S) = RTC0.20 = 11.5 * (s2Ss2S) = R0.17 = 0.0321 (s2Ss1) = 0.8418 ln
* * 253 1  0.188 92 ln = 0.0232 373.2 5 (s2Ss1) = 0.0321  0.0232 + 0.0567 0 T2S = 253 K (h2Sh1) = 0.8418(253373.2) = 101.2 wS = (h1h2S) = 29.9 + 101.2 + 11.5 = 82.8 kJ/kg w = SwS = 0.7582.8 = 62.1 kJ/kg = (h 1h1) + (h1h2) + (h2h2) Assume T2 = 275 K, Tr2 = 0.904 (h1h2) = 0.8418(373.2275) = 82.7 From D.2 and D.3, (h2h2) = RTC0.17 = 9.8 ; Substituting, w = 29.9 + 82.7 + 9.8 = 62.7 62.1 T2 = 275 K b) (s2s1) = 0.8418 ln
* * * * * * * * * * * * (s2s2) = R0.13 = 0.0245 275 1  0.188 92 ln = +0.0470 373.2 5 (s2s1) = 0.0245 + 0.0470 + 0.0567 = +0.0792 Assuming T0 = 25 oC, (12) = (h1  h2)  T0(s1  s2) = 62.1 + 298.2(0.0792) = 85.7 kJ/kg 2nd Law = w 62.1 = = 0.725 12 85.7 1363 13.70 A 4 m3 uninsulated storage tank, initially evacuated, is connected to a line flowing ethane gas at 10 MPa, 100C. The valve is opened, and ethane flows into the tank for a period of time, after which the valve is closed. Eventually, the whole system cools to ambient temperature, 0C, at which time the it contains onefourth liquid and threefourths vapor, by volume. For the overall process, calculate the heat transfer from the tank and the net change of entropy. Rigid tank V = 4 m3, m1 = 0 Line: C2H6 at Pi = 10 MPa, Ti = 100 oC Flow in, then cool to M = 30.07, Pri =
* T2 = T0 = 0 oC, VLIQ 2 = 1 m3 & VVAP 2 = 3 m3 R = 0.2765, CP0 = 1.766 Tri = 373.2 = 1.225 305.4
* 10 = 2.049, 4.88 From D.2 and D.3, (hi hi) = 0.2765305.42.0 = 168.9 and Tr2 = 273.2 = 0.895 305.4 (si si) = 0.27651.22 = 0.3373 From D.1, D.2 and D.3, P2 = PG = 0.514880 = 2489 kPa sat. liq.: ZF = 0.087 ; (h*hF) = RTC4.09 = 345.4 ; (s*sF) = R4.3 = 1.189 sat. vap. : ZG = 0.68 ; (h*hG) = RTC0.87 = 73.5 ; (s*sG) = R0.70 = 0.193 mLIQ 2 = mVAP 2 = 24891 = 378.7 kg 0.0870.2765273.2 24893 = 145.4 kg 0.680.2765273.2 => x2 = 145.4 = 0.277 524.1
* * * * m2 = 524.1 kg 1st law: QCV = m2u2  mihi = m2(h2hi)  P2V = m2[(h2h2) + (h2hi ) + (hi hi)] P2V (h2hi ) = 1.7662(0100) = 176.6 (h2h2) = (1x2)(h2hF2) + x2(h2hG2) = 0.723 345.4 + 0.277 73.5 = 270.1 QCV = 524.1[270.1  176.6 + 168.9] 2489 4 = 155 551 kJ
* * * * * 1364 SNET = m2(s2si)  QCV/T0 (s2si) = (s2s2) + (s2si ) + (si si) (s2si ) = 1.7662 ln
* * * * * * * * 273.2 2.489  0.2765 ln = 0.1664 373.2 10
* (s2s2) = (1x2)(s2sF2) + x2(s2sG2) = 0.723 1.189 + 0.277 0.193 = 0.9131 (s2si) = 0.9131  0.1664 + 0.3373 = 0.7422 SNET = 524.1(0.7422) 155 551 = 180.4 kJ/K 273.2 13.71 The environmentally safe refrigerant R142b (see Problem 13.45) is to be evaluated as the working fluid in a portable, closedcycle power plant, as shown in Fig. P13.71. The aircooled condenser temperature is fixed at 50C, and the maximum cycle temperature is fixed at 180C, because of concerns about thermal stability. The isentropic efficiency of the expansion engine is estimated to be 80%, and the minimum allowable quality of the fluid exiting the expansion engine is 90%. Calculate the heat transfer from the condenser, assuming saturated liquid at the exit. Determine the maximum cycle pressure, based on the specifications listed T 1 R142b CH3CClF2 M = 100.495, TC = 410.3 K, PC = 4.25 MPa
4 3 2S2 CP0 = 0.787 kJ/kg K R = 0.082 73 kJ/kg K
s T1 = 180 oC, T2 = T3 = 50 oC, x2 = 0.90, S EXP ENG = 0.80 a) Tr2 = 323.2 = 0.788 410.3 From D.1, D.2 and D.3: P2 = P3 = 0.234250 = 978 kPa sat. liq.: (h*hF) = RTC4.55 = 154.4 ; sat. vap.: (h*hG) = RTC0.42 = 14.3 ; (s*sF) = R5.62 = 0.4649 (s*sG) = R0.37 = 0.0306 Condenser, 1st law qCOND = h3  h2 = hF  (hF + x2hFG) = x2hFG = 0.90(154.4  14.3) = 126.1 kJ/kg 1365 b) expansion engine, efficiency S EXP ENGINE = w12 w12S s2S  s1 = 0 s2  s1 > 0 1st law & 2nd law, ideal engine: 1st law & 2nd law, real engine:
* * * * w12S = h1  h2S, w12 = h1  h2, (h1h2) = (h1h2S) = 0.787(18050) = +102.3
* * (s1s2) = * * (s1s2S) P1 453.2 = 0.787 ln  0.08273 ln 323.2 978 P1 978
* = 0.266 05  0.08273 ln
* * * (h2h2) = (1x2)(h2hF2) + x2S(h2hG2) = 0.1 154.4 + 0.9 14.3 = 28.3 (h2Sh2S) = (1x2S)(h2ShF2) + x2S(h2ShG2) = 154.4  x2S140.1 (s2Ss2S) = (1x2S)(s2SsF2) + x2S(s2SsG2) = 0.4649  x2S0.4343 Substituting,
* (s1s1) * * * * * + 0.266 05  0.082 73 ln P1 978 + 0.4649  x2S 0.4343 = 0 (Eq.1) Also (h1h1) + 102.3 + 28.3
* (h1h1) * + 102.3 + 154.4  x2S140.1 = 0.80 (Eq.2) P1(kPa) 453.2 , Tr1 = = 1.105 Both Pr1 = 410.3 4250 Trial & Error solution: Assume P1 = 9300 kPa, Pr1 = 2.188 From D.2 and D.3: (h1h1) = RTC3.05 = 103.5 and Substituting into eq. 1, 0.1737 + 0.730 75  0.1863  x2S0.4343 = 0 => x2S = 0.8537
* * (s1s1) = R2.10 = 0.1737 Substituting into eq. 2, 27.1 103.4 + 130.5 = = 0.806 0.8 103.4 + 256.7  0.8537 140.1 33.6 P1 = 9300 kPa 1366 13.72 The refrigerant fluid R21 (see Table A.2) is to be used as the working fluid in a solar energy powered Rankinecycle type power plant. Saturated liquid R21 enters the pump, state 1, at 25oC, and saturated vapor enters the turbine, state 3, at 88oC. For R21 : CP0 = 0.582 kJ/kg K and find the boiler heat transfer q23 and the thermal efficiency of the cycle. R21: Tc = 451.6 K, Pc = 5.18 MPa, M = 102.925 kg/kmol, R = 0.08079 kJ/kgK State 1: T1 = 25oC, x1 = 0 From D.1 and D.2: Tr1 = 0.66, Pr1 = 0.06, P1 = Pr1Pc = 311 kPa Z1 = 0.01 , (h1  h1)f = 4.98 RTc , * v1 = Z1RT1/P1 = 0.000775 m3/kg State 3: T3 = 88oC, x3 = 1.0 Tr3 = 0.8, Pr3 = 0.25, P3 = Pr3Pc = 1295 kPa (h3  h3)g = 0.46 RTc , (s3  s3)g = 0.39 R
=> qH = h3  h2 a) 1st Law Boiler: qH + h2 = h3 + w; w = 0 1st Law Pump: q + h1 = h2 + wp; q = 0 wp =  v dP = v1(P2  P1) = 0.000775(1295  311) = 0.76 kJ/kg h3  h1 = (h3  h3)
* * * * * + (h3  h1) + (h1  h1) * * * h3  h1 = CP (T3  T1) = 0.582(88  25) = 36.7 kJ/kg, h3  h1 = 36.7 + RTc( 0.46 + 4.98) = 201.6 kJ/kg h2 = h1  wp; qH = h3  h1 + wp = 201.6  0.76 = 200.8 kJ/kg b) th = wnet/qH; wnet = wt + wp 1st Law Turbine: q + h3 = h4 + wt; q = 0 wt = h3  h4 = (h3  h3)
* * * * + (h3  h4) + (h4  h4)
T4 = T1
* * * * h3  h4 = Cp(T3  T4) = 36.7 kJ/kg;
* h4  h4 = (1  x4)(h4  h4)f + x4 (h4  h4)g; x4 = ? 2nd Law Turbine:
* Snet = m(s3  s4)  Q/T; Assume Isentropic s3  s4 = (s3  s3)
* * + (s3  s4) + (s4  s4) = 0
4 * * * (s3  s4) = CP ln T T3  R ln P3 P4 = 0.0033 kJ/kg K 1367 s4  s4 = 0.0348 kJ/kgK, s4  s4 = (1  x4)(s4  s4)f + x4 (s4  s4)g = 7.47 R  x4(7.47  0.15) R Tr4 = Tr1 = 0.66,
* * * * * (s4  s4)f = 7.47 R, (s4  s4)g = 0.15 R
=> x 4 = 0.962 (h4  h4)g = 0.15 RTc wnet = 30.7 kJ/kg
* * * 0.0348 = 7.47 R  x4(7.47  0.15) R (h4  h4)f = 4.98 RTc , h4  h4 = 11.5,
* wt = h3  h4 = 31.5 kJ/kg; th = wnet/qH = 0.153 13.73 A cylinder/piston contains a gas mixture, 50% CO2 and 50% C2H6 (mole basis) at 700 kPa, 35C, at which point the cylinder volume is 5 L. The mixture is now compressed to 5.5 MPa in a reversible isothermal process. Calculate the heat transfer and work for the process, using the following model for the gas mixture: a. Ideal gas mixture. b. Kay's rule and the generalized charts. c. The van der Waals equation of state. a) Ideal gas mixture U2  U1 = mCv mix(T2  T1) = 0 Q12 = W12 = P dV = P1V1 ln(V2/V1) =  P1V1 ln(P2/P1) =  700 0.005 ln(5500/700) = 7.71 kJ b) Kay's rule Tcmix = 0.5 304.1 + 0.5 305.4 = 304.75 K Pcmix = 0.5 7.38 + 0.5 4.88 = 6.13 MPa Tr1 = 308.15/304.75 = 1.011, P r1 = 0.7/6.13 = 0.1142 Z1 = 0.96, h1 = 0.12, s1 = 0.08 n = P1V1/Z1R T1 = 700*0.005 = 0.00142 kmol 0.962*8.3145*308.15 Tr2 = Tr1 , Pr2 = 5.5/6.13 = 0.897, Z2 = 0.58, h2 = 1.35, s2 = 1.0   h2  h1 = (h2  h1)  R Tc(h2  h1) = 0  8.3145 304.75(1.35  0.12) =  3117   u2  u1 = h2  h1 + RT(Z1  Z2) =  3117 1368 + 8.3145 308.15(0.96  0.58) = 2143 kJ/kmol  s Q12 = nT(s2   1)T = 0.00142 308.15 8.3145[ 0  ln(5.5/0.7) 1.0 W12 = Q12 + 0.08 ] =  10.85 kJ   n(u  u ) = 10.85  0.00142(2143) =  7.81 kJ
2 1 c) van der waal's equation For CO2 : b = R Tc/8Pc = 8.3145 304.1/8 7380 = 0.04282 a = 27 Pc b2 = 27 7380 0.042822 = 365.45 For C2H6 : b = R Tc/8Pc = 8.3145 305.4/8 4880 = 0.06504 a = 27 Pc b2 = 27 4880 0.065042 = 557.41 amix = (0.5 365.45 + 0.5 557.41)2 = 456.384 bmix = 0.5 0.04282 + 0.5 0.06504 = 0.05393 8.3145*308.2 456.384   2  700 = 0 v1  0.05393 v1 By trial and error: v = 3.5329 m3/kmol
1 8.3145*308.2 456.384   2  5500 = 0 v2  0.05393 v2 By trial and error: v2 = 0.2815 m3/kmol n = V1/v1 = 0.005/3.5329 = 0.00142    ) = n R T ln v2  b Q12 = nT(s2 s1 T v b
1 = 0.00142 8.3145 308.2 ln 0.2815  0.05392 =  9.93 kJ 3.5329  0.05392 U2U1 = 0.00142 456.39(3.53291  0.28151) = 2.12 kJ Q12 = U2U1 + W12 => W12 = 9.93 (2.12) = 7.81 kJ 1369 13.74 A gas mixture of a known composition is frequently required for different purposes, e.g., in the calibration of gas analyzers. It is desired to prepare a gas mixture of 80% ethylene and 20% carbon dioxide (mole basis) at 10 MPa, 25C in an uninsulated, rigid 50L tank. The tank is initially to contain CO2 at 25C and some pressure P1. The valve to a line flowing C2H4 at 25C, 10 MPa, is now opened slightly, and remains open until the tank reaches 10 MPa, at which point the temperature can be assumed to be 25C. Assume that the gas mixture so prepared can be represented by Kay's rule and the generalized charts. Given the desired final state, what is the initial pressure of the carbon dioxide, P1? Determine the heat transfer and the net entropy change for the process of charging ethylene into the tank. Pi =10 MPa A A = C2H4, B = CO2 T1 = 25 oC P2 = 10 MPa, T2 = 25 oC yA2 = 0.8, yB2 = 0.2 a) Mixture at 2 : PC2 = 0.8 5.04 + 0.2 7.38 = 5.508 MPa TC2 = 0.8 282.4 + 0.2 304.1 = 286.7 K Tr2 = 298.15/286.7 = 1.040; Pr2 = 10/5.508 = 1.816 D.1 : n2 = Z2 = 0.32
Ti = 25 C V=0.05 m 3
o B P 2V 10 0000.05 = 0.6302 kmol  = Z2RT2 0.328.3145298.2 nA2 = ni = 0.8 n2 = 0.5042 kmol C2H4 nB2 = n1 = 0.2 n2 = 0.1260 kmol CO2 298.2 = 0.981 304.1 n1ZB1RT1 0.126 ZB1 8.3145298.2 Pr1 = = = 0.8466 ZB1 PCBV 73800.05 Tr1 = By trial & error: Pr1 = 0.618 & ZB1 = 0.73 P1 = 0.618 7.38 = 4.56 MPa b) 1st law: QCV + nihi = n2u2  n1u1 = n2h2  n1h1  (P2P1)V  *  *  * or QCV = n2(h2h2)  n1(h1h1)  ni(hihi )  (P2P1)V ` 1370 * * * (since Ti = T1 = T2, hi = h1 = h2) * (h1h1) = 0.838.3145304.1 = 2099 kJ/kmol * (h2h2) = 3.408.3145286.7 = 8105 kJ/kmol Tri = 298.2 10 = 1.056, Pri = = 1.984 282.4 5.04 * (hi hi) = 3.358.3145282.4 = 7866 kJ/kmol QCV = 0.6302(8105)  0.126(2099)  0.5042(7866)  (10 0004560)0.05 = 1149 kJ SCV = n2 2  n1 1 , SSURR =  QCV/T0  ni i s s s SNET = n2 2  n1 1  QCV/T0  ni i s s s
* * Let  A0 =  B0 = 0 at T0 = 25 oC, P0 = 0.1 MPa s s Then  MIX 0 = 8.3145 (0.8 ln 0.8 + 0.2 ln 0.2) = 4.161 kJ/kmol K s  =  * + (s* s* ) + (s s* ) s1 sB0  P1 T1  P0 T0 B  1  P1 T1 B = 0 + (08.3145 ln 4.56 )  0.60 8.3145 = 36.75 kJ/kmol K 0.1 *  =  * + (s* s* ) + (s s* ) si sA0  Pi Ti  P0 T0 A  i  Pi Ti A = 0 + (08.3145 ln 10 )  2.448.3145 = 58.58 kJ/kmol K 0.1  = * * *  * s2 sMIX 0 + (sP2 T2sP0 T0)MIX + (s2sP2 T2)MIX = 4.161 + (08.3145 ln 10 )  2.5518.3145 = 55.34 kJ/kmol K 0.1 SNET = 0.6302(55.33)  0.126(36.75)  0.5042(58.58) + 1149/298.2 = +3.15 kJ/K 1371 English Unit Problems
13.75EA special application requires R22 at 150 F. It is known that the triplepoint temperature is less than 150 F. Find the pressure and specific volume of the saturated vapor at the required condition. The lowest temperature in Table C.10 for R22 is 100 F, so it must be extended to 150 F using the Clapeyron eqn. At T1 = 100 F = 359.7 R , P1 = 2.398 lbf/in.2 and R= 1.9859 = 0.022 97 Btu/lbm R 86.469 P hfg (TT1) 107.9 (309.7359.7) ln = = = 2.1084 P1 R T T 0.022 97 309.7 359.7 1 P = 0.2912 lbf/in.2 vg = RT 0.022 97 778 309.7 = = 132 ft3/lbm Pg 0.2912 144 13.76EIce (solid water) at 27 F, 1 atm is compressed isothermally until it becomes liquid. Find the required pressure. Water, triple point T = 32.02 F = 491.69 R, P = 0.088 67 lbf/in.2 vF = 0.016 022 vI = 0.017 473 hF = 0.00 dPIF dT P = hFhI (vFvI)T dT = hI = 143.34 143.34778.2 = 1085.8 0.001 451491.69144 dPIF T = 1085.8(2732.02) = 5450.7 lbf/in.2 P = Ptp + P = 5451 lbf/in.2 1372 13.77E Using thermodynamic data for water from Tables C.8.1 and C.8.3, estimate the freezing temperature of liquid water at a pressure of 5000 lbf/in.2.
P H 2O
T.P. T dPIF dT = hIF TvIF constant At the triple point, vIF = vF  vI = 0.016 022  0.017 473 = 0.001 451 ft3/lbm hIF = hF  hI = 0.0  (143.34) = 143.34 Btu/lbm dPIF dT = 143.34 778.2 = 1085.8 lbf/in.2 R 491.69(0.001 451) 144 (50000.09) = 27.4 F (1085.8) at P = 5000 lbf/in2, T 32.02 + 13.78E Determine the volume expansivity, P, and the isothermal compressibility, T , for water at 50 F, 500 lbf/in.2 and at 500 F, 1500 lbf/in.2 using the steam tables. Water at 50 F, 500 lbf/in.2 (compressed liquid) P = v 1 v (T)P 1(T)P v v 1 0.016 106  0.015 994 = 0.000 103 F1 0.015 998 100  32 v 1 v (P)T  1(P)T v v Using values at 32 F, 50 F and 100 F P T =  Using values at saturation, 500 and 1000 lbf/in.2 1 0.015 971  0.016 024 T = 0.000 0033 in.2/lbf 0.015 998 1000  0.178 Water at 500 F, 1500 lbf/in.2 (compressed liquid) 1 0.021 579  0.019 264 P = 0.001 143 F1 0.020 245 550  450 T 1 0.020 139  0.020 357 = 0.000 0108 in.2/lbf 0.020 245 2000  1000 1373 13.79E Sound waves propagate through a media as pressure waves that causes the media to go through isentropic compression and expansion processes. The speed of sound c is defined by c2 = (P/)s and it can be related to the adiabatic compressibility, which for liquid ethanol at 70 F is 6.4 in.2/lbf. Find the speed of sound at this temperature. c2 = (P)s = v (P)s = v
2 1 1 v v P s = ( ) 1 s From Table C.3 for ethanol, c= 32.174144 (6.410 48.9)
6 1/2 = 48.9 lbm/ft3 = 3848 ft/s 13.80E Consider the speed of sound as defined in Problem 13.79. Calculate the speed of sound for liquid water at 50 F, 250 lbf/in.2 and for water vapor at 400 F, 80 lbf/in.2 using the steam tables. From problem 13.79 : c2 = (P)s = v (P)s v
2 T Liquid water at 50 F, 250 lbf/in.2 Assume (P)s (P) v v
2 Using saturated liquid at 50 F and compressed liquid at 50 F, 500 lbf/in.2, c2 =  (0.016024+0.015998) ((5000.18)14432.174) = 22.83210 2 0.0159980.016024 6 c = 4778 ft/s Superheated vapor water at 400 F, 80 lbf/in.2 v = 6.217, s = 1.6790 At P = 60 lbf/in.2 & s = 1.6790: T = 343.8 F, v = 7.7471 At P = 100 lbf/in.2 & s = 1.6790: T = 446.2 F, v = 5.2394 c2 = (6.217)2 c = 1690 ft/s ((10060)14432.174) = 2.85610 5.23947.7471 6 1374 13.81E A cylinder fitted with a piston contains liquid methanol at 70 F, 15 lbf/in.2 and volume 1 ft3. The piston is moved, compressing the methanol to 3000 lbf/in.2 at constant temperature. Calculate the work required for this process. The isothermal compressibility of liquid methanol at 70 F is 8.3 103 in.2/lbf. v w12 = Pdv = P P T dPT =  v T PdPT
2 2 ( ) 1 1 For v const & T const. V1 = 1.0 ft3, W12 = => w 12 =  v T (P 2 2 2  P1 2 ) For liquid methanol, from Table C.3 : = 49.1 lbm/ft3 m = 1.0 49.1 = 49.1 lbm 1.00.0083 (3000)2  (15)2 144 = 37 349 ft lbf = 48.0 Btu 2 [ 13.82E A piston/cylinder contains 10 lbm of butane gas at 900 R, 750 lbf/in.2. The butane expands in a reversible polytropic process with polytropic exponent, n = 1.05, until the final pressure is 450 lbf/in.2. Determine the final temperature and the work done during the process. C4H10 , m = 10 lbm , T1 = 900 R , P1 = 750 lbf/in.2 Rev. polytropic process (n=1.05), P2 = 450 lbf/in.2 Tr1 = 900 750 = 1.176, Pr1 = = 1.361 765.4 551 V1 =
n P1V1 => From Fig. D.1 : Z1 = 0.67 10 0.67 26.58 900 750 144 =
n P2V2 = 1.484 ft3
1 1.05 750 V2 = 1.484 450 = ( ) = 2.414 ft3 = 0.7688 Z2Tr2 = P2V2 mRTC 450 144 2.414 10 26.58 765.4 at Pr2 = 450/551 = 0.817 Trial & error:
2 Tr2 = 1.068, P2V2  P1V1 1n = Z2 = 0.72 => T2 = 817.4 R W12 = PdV = 1 750 (450 2.4141.05 1.484 ) 144 = 98.8 Btu 778 1 1375 13.83E A 7ft3 rigid tank contains propane at 1300 lbf/in.2, 540 F. The propane is then allowed to cool to 120 F as heat is transferred with the surroundings. Determine the quality at the final state and the mass of liquid in the tank, using the generalized compressibility charts. Propane C3H8: V = 7.0 ft3, P1 = 1300 lbf/in.2, T1 = 540 F = 1000 R cool to T2 = 120 F = 580 R From Table C.1 : TC = 665.6 R, PC = 616 lbf/in.2 1300 1000 = 2.110, Tr1 = = 1.502 616 665.6 From D.1: Z1 = 0.83 Pr1 = Z1RT1 0.83 35.04 1000 v2 = v1 = = P1 1300 144 0.715 35.04 580 265 144 0.075 35.04 580 265 144 = 0.1554 ft3/lbm From D.1 at Tr2 = 0.871, saturated => PG2 = 0.43 616 = 265 lbf/in.2 vG2 = vF2 = = 0.3808 = 0.0399 => x2 = 0.3388 0.1554 = 0.0399 + x 2(0.37810.0399) mLIQ 2 = (1  0.3388) 7.0 = 29.8 lbm 0.1554 1376 13.84E A rigid tank contains 5 lbm of ethylene at 450 lbf/in.2, 90 F. It is cooled until the ethylene reaches the saturated vapor curve. What is the final temperature? C2H4 m = 5 lbm
C 2H 4 P1 = 450 lbf/in2, T1 = 90 F = 249.7 R
1 2 v T Pr1 = 450 = 0.616, 731 Tr1 = 549.7 = 1.082 508.3 Fig. D.1 Z1 = 0.82 Pr2 = Pr1 Z2Tr2 Z1Tr1 = 0.616 ZG2Tr2 0.82 1.082 Pr2 CALC 0.432 = 0.6943 ZG2Tr2 Trial & error: Tr2 ZG2 0.871 0.715 Pr2 0.43 ~ OK => T2 = 442.7 R 13.85E Calculate the difference in internal energy of the idealgas value and the realgas value for carbon dioxide at the state 70 F, 150 lbf/in. 2, as determined using the virial equation of state. For carbon dioxide at 70 F, B = 2.036 ft3/lb mol , T(dB/dT) = 4.236 ft3/lb mol Solution: virial:
* CO2 at 70 F, 150 lbf/in2 P= RT BRT + 2 ; v v (P)v = R + BR + RT(dB) v v T v dT
2 2 RT2 dB P RT2 dB T  P dv = uu =  v v2 dT dv =  v dT T v v [( ) ] ( )
3 B = 2.036 ft3/lbmol T (dB)= 4.236 ft /lbmol dT  = 1 RT 1 + 1+4BP/RT v 2 P RT 1545529.7 But = = 37.8883 P 150144 [ v = 0.537.8883 1 + [ 1+4(2.036)/37.8883 = 35.7294 ft /lbmol
3 1.9859529.7  4.236 = 123.9 Btu/lbmol uu* = 35.7294 1377 13.86E Calculate the heat transfer during the process described in Problem 13.81. From solution 13.82, V1 = 1.473 ft3, Tr1 = 1.176, From D.1:
* * V2 = 2.396 ft3, W12 = 98.8 Btu Pr2 = 0.817, T2 = 817.4 R = 0.95 Pr1 = 1.361, Tr2 = 1.068, (h h) RT
C * = 1.36, 1 (h h) RT
C * 2 h2  h1 = 0.415 (817.4  900) = 34.3 Btu/lbm h2  h1 = 34.3 + 26.58765.4 (0.95 + 1.36) = 23.6 Btu/lbm 778 U2U1 = m(h2h1)  P2V2 + P1V1 = 10(23.6)  4501442.414 7501441.484 + = 231.1 Btu 778 778 Q12 = U2U1 + W12 = 132.3 Btu 1378 13.87E Saturated vapor R22 at 90 F is throttled to 30 lbf/in.2 in a SSSF process. Calculate the exit temperature assuming no changes in the kinetic energy, using the generalized charts, Fig. D.2 and repeat using the R22 tables, Table C.10. R22 throttling process
2 T1 = 90 F, x1 = 1.00, P2 = 30 lbf/in. Energy Eq.: h2h1 = (h2h2) + (h2h1) + (h1h1) = 0 Tr1 = 549.7 = 0.827 664.7 * * * * Generalized charts, From D.2:
* 1.9859664.7 (0.55) = 8.40 86.469 To get CP0, use h values from Table C.10 at low pressure. (h1h1) = CP0 121.867118.724 = 0.1572 10080 (h2h2) + 0.1572 (T2  30) + 8.40 = 0
* Substituting into energy Eq.: at Pr2 = 30 = 0.042 721 Assume T2 = 43.4 F = 503.1 R => Tr2 = (h2h2) = 1.9859664.7 (0.07) = 1.07 86.469 Substituting,
* 503.4 = 0.757 664.7 1.07 + 0.1572(43.4  90) + 8.40 = 0.005 0 T2 = 43.4 F b) R22 tables, C.10: T1 = 90 F, x1 = 1.0 => h 1 = 111.62 h2 = h1 = 111.62 , P2 = 30 lbf/in.2 => T2 = 42.1 F 1379 13.88E A 10ft3 tank contains propane at 90 F, 90% quality. The tank is heated to 600 F. Calculate the heat transfer during the process.
C 3H 8 T 2 1 v V = 10 ft3 T1 = 90 F = 549.7 R, x1 = 0.90 Heat to T2 = 600 F = 1059.7 R M = 44.094, TC = 665.6 R PC = 616 lbf/in.2 R = 35.04, CP0 = 0.407 Tr1 = 0.826 Figs. D.1 and D.2 Z1 = 0.1 0.053 + 0.9 0.78 = 0.707, h1h1 RTc Pr
SAT * = 0.1 4.4 + 0.9 0.55 = 0.935 = 0.31 ; P1
SAT = 0.31 616 = 191 lbf/in.2 = 20.2 lbm Z2 1.183 m= 191 144 10 0.707 35.04 549.7 616 144 10 Pr2 = 20.2 Z 2 35.04 1059.7 = at Tr2 = 1.592 Pr2 = 0.79 Trial & error: Z2 = 0.94 * * * * P2 = 490 lbf/in.2 h2h2 RTc
* = 0.36 (h2h1) = 0.407(60090) = 207.6 Btu/lbm (h1h1) = 0.935 35.04 665.9/ 778 = 28.0 (h2h2) = 0.36 35.04 665.9/ 778 = 10.8 Q12 = m(h2h1)  (P2P1)V = 20.2 (10.8 + 207.6 + 28.0)  (490  191) = +4541  553 = 3988 Btu 144 60 778 1380 13.89E A newly developed compound is being considered for use as the working fluid in a small Rankinecycle power plant driven by a supply of waste heat. Assume the cycle is ideal, with saturated vapor at 400 F entering the turbine and saturated liquid at 70 F exiting the condenser. The only properties known for this compound are molecular weight of 80 lbm/lbmol, ideal gas heat capacity Cpo = 0.20 Btu/lbm R and Tc = 900 R, Pc = 750 lbf/in.2. Calculate the work input, per lbm, to the pump and the cycle thermal efficiency.
1 Turbine . QH Ht. Exch Cond P 4 . W P 3 . WT T1 = 400 F = 860 R x1 = 1.0 T3 = 70 F = 530 R
2 x3 = 0.0 Properties known: M = 80, CPO = 0.2 Btu/lbm R TC = 900 R, PC = 750 lbf/in.2 Tr1 = 860 530 = 0.956, Tr3 = = 0.589 900 900 From D.1: Pr1 = 0.76, P1 = 0.76 750 = 570 lbf/in.2 = P4 Pr3 = 0.025, P3 = 19 lbf/in.2 = P2, ZF3 = 0.0045 vF3 = ZF3RT3/P3 =
4 0.0045 1545 530 = 0.0168 19 144 80 144 = 1.71 Btu/lbm 778 wP =  vdP vF3(P4 P3) = 0.0168 (570  19) 3 qH + h4 = h1 , but h3 = h4 + wP => q H = (h1h3) + wP From D.2, (h1h1) = (1.9859/80) 900 1.34 = 30.0 (h3h3) = (1.9859/80) 900 5.2 = 116.1 (h1h3) = CP0(T1T3) = 0.2(40070) = 66.0 (h1h3) = 30.0 + 66.0 + 116.1 = 152.1 qH = 152.1 + (1.71) = 150.4 Btu/lbm
* * * * 1381 Turbine, (s2s1) = 0 = (s2s2)+(s2s1)+(s1s1) From D.3, (s1s1) = (1.9859/80) 1.06 = 0.0263 (s2s1) = 0.20 ln Substituting, (s2s2) =  0.0124 + 0.0263 = 0.0139 = (s2sF2)  x2sFG2 0.0139 = 0.024 82 8.77  x2 0.024 82 (8.77  0.075) => x 2 = 0.9444 (h2h2) = (h2hF2)  x2hFG2 From D.2, hfg2 = 0.024 82 900 (5.2  0.07) = 114.6 (h2h2) = 116.1  0.9444 114.6 = 7.9 wT = (h1h2) = 30.0 + 66.0 + 7.9 = 43.9 Btu/lbm TH = wNET qH = 43.9  1.7 = 0.281 150.4
* * * * * * * * * * * * 530 19  0.024 82 ln = 0.0124 860 570 1382 13.90E A 7ft3 rigid tank contains propane at 730 R, 500 lbf/in.2. A valve is opened, and propane flows out until half the initial mass has escaped, at which point the valve is closed. During this process the mass remaining inside the tank expands according to the relation Pv1.4 = constant. Calculate the heat transfer to the tank during the process. C3H8: V = 7.0 ft3, T1 = 400 K, P1 = 500 lbf/in.2 Flow out to m2 = m1/2; Tr1 = v1 = Pv1.4 = const inside => From D.1: Z1 = 0.76 730 500 = 1.097, Pr1 = = 0.812 665.6 616 0.76 35.04 730 = 0.270, 500 144 m1 = 7.0/0.270 = 25.92 lbm, P 2 = P1 v2 = 2v1 = 0.54 m2 = m1/2 = 12.96 lbm, (v )
2 v1 1.4 = 500 = 189.5 lbf/in.2 21.4 Trial & error: saturated with 189.5 Pr2 = = 0.308 T2 = 0.825 665.6 = 549.4 R & 616 P2v2 = Z2RT2 Z2 = 189.5 144 0.54 = 0.764 35.04 549.4 Z2 = ZF2 + x2(ZG2  ZF2) 0.764 = 0.052 + x 2(0.78  0.052) =>
* * x2 = 0.978 (h1h1) = 35.04 665.6 0.85 / 778 = 25.5 (h2h1) = 0.407 (549.4  730) = 73.4 (h2h2) = (h2hF2)  x2hFG2 = 18.7 1st law:
* * * * 35.04 665.6 778 [4.41  0.978(4.41  0.54)] = QCV = m2h2  m1h1 + (P1P )V + mehe AVE 2 h1 = 0 + (h1h1) = 25.5
* * Let h1 = 0 then:
* * * h2 = h1 + (h2h1) + (h2h2) = 0  73.4  18.7 = 92.1 he AVE = (h1 + h2)/2 = 58.8 QCV = 12.96 (92.1)  25.92(25.5) + (500189.5)7.0 = 892 Btu 144 + 12.96(58.8) 778 1383 13.91E A geothermal power plant on the Raft river uses isobutane as the working fluid as shown in Fig. P13.42. The fluid enters the reversible adiabatic turbine at 320 F, 805 lbf/in.2 and the condenser exit condition is saturated liquid at 91 F. Isobutane has the properties Tc = 734.65 R, Pc = 537 lbf/in.2, Cpo = 0.3974 Btu/lbm R and ratio of specific heats k = 1.094 with a molecular weight as 58.124. Find the specific turbine work and the specific pump work. R = 26.58 ft lbf/lbm R = 0.034 166 Btu/lbm R Turbine inlet: T1 = 320 F , P1 = 805 lbf/in.2 Condenser exit: T3 = 91 F , x3 = 0.0 ; From D.1 : Pr3 = 0.165, Z3 = 0.0275 P2 = P3 = 0.165 537 = 88.6 lbf/in.2 Tr1 = 779.7 / 734.7 = 1.061, From D.2 and D.3, (h1h1) = 0.034 166 734.7 2.85 = 71.5 (s1s1) = 0.034 166 2.15 = 0.0735 (s2s1) = 0.3974 ln
* * * * * * Tr3 = 550.7 / 734.7 = 0.75 Pr1 = 805 / 537 = 1.499 550.7 88.6  0.034 166 ln = 0.0628 779.7 805 (s2s2) = (s2sF2)  x2sFG2 = 0.034 166 6.12  x2 0.034 166(6.12  0.29) = 0.2090  x2 0.1992 (s2s1) = 0 = 0.2090 + x2 0.19920.0628 + 0.0735 (h2h1) = CP0(T2T1) = 0.3974(550.7779.7) = 91.0 From D.2, (h2h2) = (h2hF2)  x2hFG2 = 0.034 166734.7[4.69  0.9955(4.69  0.32)] = 117.7  0.9955 109.7 = 8.5 Turbine: Pump wT = (h1h2) = 71.5 + 91.0 + 8.5 = 28.0 Btu/lbm vF3 =
4 * * * * => x2 = 0.9955 ZF3RT3 P3 = 0.0275 26.58 550.7 = 0.031 55 88.6 144 144 wP =  vdP vF3(P4 P3) = 0.031 55(80588.6) = 4.2 Btu/lbm 778
3 1384 13.92E Carbon dioxide collected from a fermentation process at 40 F, 15 lbf/in.2 should be brought to 438 R, 590 lbf/in.2 in a SSSF process. Find the minimum amount of work required and the heat transfer. What devices are needed to accomplish this change of state? 35.1 R= = 0.045 12 Btu/lbm R 778 Tri = 500 = 0.913 , 547.4 Pri = 15 = 0.014 1070 s*s ( R ) = 0.01 R From D.2 and D.3: Tre = 438 = 0.80, 547.4
* h*h (RT ) = 0.02, C Pre = 590 = 0.551 1070 sese = 4.70 R
* From D.2 and D.3:
* hehe = 4.50 RTc,
* * * (hihe) =  (hi hi) + (hi he) + (hehe) =  0.045 12 547.4 0.02 + 0.203(500  438) + 0.045 12 547.4 4.50 = 123.2 Btu/lbm (sise) =  (si si) + (si se) + (sese) =  0.045 120.01 + 0.203 ln = 0.4042 Btu/lbm R wrev = (hihe) T0(sise) = 123.2  500(0.4042) = 78.4 Btu/lbm qrev = (hehi) + wrev = 123.2  78.9 = 202.1 Btu/lbm 15 500  0.045 12 ln + 0.045 124.70 590 438
* * * * 1385 13.93E A control mass of 10 lbm butane gas initially at 180 F, 75 lbf/in.2, is compressed in a reversible isothermal process to onefifth of its initial volume. What is the heat transfer in the process? Butane C4H10: m = 10 lbm, T1 = 180 F, P1 = 75 lbf/in.2 Compressed, reversible T = const, to V2 = V1/5 Tr1 = 640 75 = 0.836, Pr1 = = 0.136 => From D.1 and D.3: Z 1 = 0.92 765.4 551 (s1s1) = 0.16 * 26.58 = 0.0055 778 Z1RT1 0.92 26.58 640 v1 = = = 1.449 ft3/lbm P1 75 144 v2 = v1/5 = 0.2898 ft3/lbm At Tr2 = Tr1 = 0.836 From D.1 and D.3: sat. liq.: sat. vap.: Therefore vF = vG = 0.058 26.58 640 = 0.0366 187 144 0.77 26.58 640 = 0.4864 187 144 x2 = v2vF vGvF = 0.563 PG = 0.34 551 =187 lbf/in.2 (s*sF) = R 5.02 = 0.1715 (s*sG) = R 0.49 = 0.0167 ZF = 0.058 ; ZG = 0.765 ; Since vF < v2 < vG twophase (s2s2) = (1x2)(s2sF2) + x2(s2sG2)
* * * = 0.437 0.1715 + 0.563 0.0167 = 0.0843
* * (s2s1) = CP0 ln T2 T1  R ln P2 P1 =0 26.58 187 ln = 0.0312 778 75 (s2s1) = 0.0843  0.0312 + 0.0055 = 0.110 Btu/lbm R Q12 = Tm(s2s1) = 64010(0.110) = 704 Btu 1386 13.94E A cylinder contains ethylene, C2H4, at 222.6 lbf/in.2, 8 F. It is now compressed isothermally in a reversible process to 742 lbf/in.2. Find the specific work and heat transfer. Ethylene C2H4 P1 = 222.6 lbf/in.2, T2 = T1 = 8 F = 467.7 R State 2: P2 = 742 lbf/in.2 R = 55.07 ft lbf/lbm R = 0.070 78 Btu/lbm R 467.7 222.6 Tr2 = Tr1 = = 0.920 Pr1 = = 0.305 508.3 731 From D.1, D.2 and D.3:
* Z1 = 0.85 , ( ) h*h RTC 1 = 0.40 (h1h1) = 0.070 78 508.3 0.40 = 14.4 (s1s1) = 0.070 78 0.30 = 0.0212 742 = 1.015 (comp. liquid) Pr2 = 731 From D.1, D.2 and D.3: Z2 = 0.17 (h2h2) = 0.070 78 508.3 4.0 = 143.9 (s2s2) = 0.070 78 3.6 = 0.2548 (h2h1) = 0 (s2s1) = 0  0.070 78 ln
* * * * * * * 742 = 0.0852 222.6 q12 = T(s2s1) = 467.7(0.2548  0.0852 + 0.0212) = 149.1 Btu/lbm (h2h1) = 143.9 + 0 + 14.4 = 129.5 (u2u1) = (h2h1)  RT(Z2Z1) = 129.5  0.070 78 467.7 (0.17  0.85) = 107.0 w12 = q12  (u2u1) = 149.1 + 107.0 = 42.1 Btu/lbm 1387 13.95E A cylinder contains ethylene, C 2H4, at 222.6 lbf/in.2, 8 F. It is now compressed in a reversible isobaric (constant P) process to saturated liquid. Find the specific work and heat transfer. Ethylene C2H4 P1 = 222.6 lbf/in.2 = P2, T1 = 8 F = 467.7 R State 2: saturated liquid, x2 = 0.0 R = 55.07 ft lbf/lbm R = 0.070 78 Btu/lbm R 467.7 222.6 = 0.920 Pr1 = Pr2 = = 0.305 Tr1 = 508.3 731 From D.1 and D.2: Z1RT1 P1 Z1 = 0.85 , ( ) h*h RTC 1 = 0.40 v1 =
* = 0.85 55.07 467.7 = 0.683 222.6 144 (h1h1) = 0.070 78 508.3 0.40 = 14.4 From D.1 and D.2: Z2 = 0.05 , v2 =
* * T2 = 0.822 508.3 = 417.8 R
2 ( )
= h*h RTC = 4.42 Z2RT2 P2 0.05 55.07 417.8 = 0.035 89 222.6 144 (h2h2) = 0.070 78 508.3 4.42 = 159.0 (h2h1) = CP0(T2T1) = 0.411(417.8  467.7) = 20.5 144 w12 = Pdv = P(v2v1) = 222.6(0.035 89  0.683) = 26.7 Btu/lbm 778
1 2 * q12 = (u2  u1) + w12 = (h2h1) = 159.0  20.5 + 14.4 = 165.1 Btu/lbm 1388 13.96E A distributor of bottled propane, C3H8, needs to bring propane from 630 R, 14.7 lbf/in.2 to saturated liquid at 520 R in a SSSF process. If this should be accomplished in a reversible setup given the surroundings at 540 R, find the ratio of the volume flow rates V.in / V.out, the heat transfer and the work involved in the process. R = 35.04/778 = 0.045 04 Btu/lbm R 630 14.7 Tri = = 0.947 Pri = = 0.024 665.6 616 From D.1, D.2 and D.3 : Zi = 0.99 (hi hi) = 0.045 03 665.6 0.03 = 0.9 (si si) = 0.045 04 0.02 = 0.0009 Tre = 520/665.6 = 0.781, From D.1, D.2 and D.3 : Pre = = 0.21 , Pe = 0.21 616 = 129 lbf/in.2 Ze = 0.035 (hehe) = 0.045 04 665.6 4.58 = 137.3 (sese) = 0.045 04 5.72 = 0.2576 (hehi ) = 0.407 (520  630) = 44.8 (sesi ) = 0.407 ln
* * * * * * * * 520 132  0.045 04 ln = 0.1770 630 14.7 (hehi) = 137.3  44.8 + 0.9 = 181.2 (sesi) = 0.2576  0.1759 + 0.0009 = 0.4326 . ZiTi/Pi Vin 0.99 630 129 . = = = 300.7 Vout ZeTe/Pe 0.035 520 14.7 wrev = (hihe) T0(sise) = 181.2  540(0.4326) = 52.4 Btu/lbm qrev = (hehi) + wrev = 181.2  52.4 = 233.6 Btu/lbm 1389 13.97E A line with a steady supply of octane, C8H18, is at 750 F, 440 lbf/in.2. What is your best estimate for the availability in an SSSF setup where changes in potential and kinetic energies may be neglected? Availability of Octane at Ti = 750 F, Pi = 440 lbf/in.2 R = 13.53 ft lbf/lbm R = 0.017 39 Btu/lbm R 440 1209.7 Pri = = 1.219, Tri = = 1.182 361 1023.8 From D.2 and D.3: (h1h1) = 0.017 39 1023.8 1.15 = 20.5 (s1s1) = 0.017 39 0.71 = 0.0123 Exit state in equilibrium with the surroundings Assume T0 = 77 F, P0 = 14.7 lbf/in.2 536.7 14.7 = 0.524 , Pr0 = = 0.041 1023.8 361 From D.2 and D.3: Tr0 =
* * * (h0h0) = RTC 5.41 = 96.3
* * and (s0s0) = R 10.38 = 0.1805 * (hi h0) = 0.409(1209.7  536.7) = 275.3 (si s0) = 0.409 ln
* * 1209.7 440  0.017 39 ln = 0.2733 536.7 14.7 (hih0) = 20.5 + 275.3 + 96.3 = 351.1 (sis0) = 0.0123 + 0.2733 + 0.1805 = 0.4415 i = wrev = (hih0)  T0(sis0) = 351.1  536.7 (0.4415) = 114.1 Btu/lbm 141 CHAPTER 14
The correspondence between the new problem set and the previous 4th edition chapter 12 problem set. New 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 Old new 3 new new 1 2 6 4 5 10 7 8 9 11 12 13 new 15 16 18 19 new 20 21 22 23 24 25 26 27 New 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 59 60 Old 28 new 29 30 31 32 33 34 36 37 39 40 mod 42 17 43 44 46 mod 49 50 47 48 52 new 56 57 new 58 59 67 65 New 61 62 63 64 65 66 67 68 69 70 71 72 73 74 Old new 66 69 70 71 14 38 41 54 60 61 62 68 new The problems that are labeled advanced start at number 66. 142 The English unit problems are: New Old New 80 81 82 83 84 85 86 87 88 89 Old 78 79 80 81 82 83 84 85 86 87 New 90 91 92 93 94 95 96 97 98 Old 88 89 90 92 mod 96 97 98 100 101 75 76 77 78 79 72 73 74 76 77 143 14.1 Calculate the theoretical airfuel ratio on a mass and mole basis for the combustion of ethanol, C2H5OH. Reaction Eq.: C2H5OH + O2(O2 + 3.76N2) => aCO 2+ bH2O + cN2 Balance C: 2 = a Balance H: 6 = 2b => b = 3 => O2 = 3 Balance O: 1 + 2O2 = 2a + b = 4 + 3 = 7 (air/fuel)mol = O2(1 + 3.76)/1 = 3 4.76 = 14.28 (air/fuel)mass = (O2MO2 + N2 MN2)/MFuel = (331.999 + 11.2828.013)/46.069 = 8.943 14.2 In a combustion process with decane, C10H22, and air, the dry product mole fractions are 86.9% N2, 1.163% O2, 10.975% CO2 and 0.954% CO. Find the equivalence ratio and the percent theoretical air of the reactants. x C10H22 + O (O2 + 3.76 N2) a H2O + b CO2 + c CO + d N2 + e O2
2 Stoichiometric combustion: = 1, c = 0, e = 0,
2 C balance: b = 10x H balance: a = 22x/2 = 11x, O balance: 2 O = a + 2b = 11x + 20x = 31x O = 15.5x, N = 58.28x
2 2 => (A/F)s = (O O 3.76 = 86.9
2 2 + N2 )/x = 73.78
2 Actual combustion: C balance: d = 86.9 O = 23.11 10x = 10.975 + 0.954 = 11.929
2 => x = 1.1929 (A/F)ac = O 4.76/1.1929 = 92.22 = (A/F)ac / (A/F)s = 92.22 / 73.78 = 1.25 or 125% 14.3 A certain fuel oil has the composition C10H22. If this fuel is burned with 150% theoretical air, what is the composition of the products of combustion? C10H22 + O (O2 + 3.76 N2) a H2O + b CO2 + c N2 + d O2
2 Stoichiometric combustion: = 1, d = 0, H balance: a = 22/2 = 11, => O2 = 15.5 O2 balance:
2 C balance: b = 10 O balance: 2 O = a + 2b = 11 + 20 = 31 = 1.5 => O2 = 1.515.5 = 23.25 b = 10, N balance: c = 23.253.76 = 87.42 Actual case: H balance: a = 11, C balance: d = 23.25  10  11/2 = 7.75 (excess oxygen) 144 14.4 Natural gas B from Table 14.2 is burned with 20% excess air. Determine the composition of the products. The reaction equation (stoich. complete comb.) with the fuel composition is: 60.1 CH4 + 14.8 C2H6 + 13.4 C3H8 + 4.2 C4H10 + 7.5 N2 + O2 (O2 + 3.76 N2) C balance: H balance: O balance: N2 balance: a H2O + b CO2 + c N2 60.1 + 214.8 + 313.4 + 44.2 = b = 146.7 460.1 + 614.8 + 813.4 + 104.2 = 2a = 478.4 => a = 239.2 2 O2 = a + 2b = 239.2 + 2146.7 7.5 + 3.76 O2 = c = 1008.8 so now more O2 and N2 c = 7.5 + 3.76319.56 = 1209 => O2 = 266.3 20% excess air: O2 = 1.2266.3 = 319.56 Extra oxygen: d = 319.56  266.3 = 53.26, Products: 14.5 239.2 H2O + 146.7 CO2 + 1209 N2 + 53.26 O2 A Pennsylvania coal contains 74.2% C, 5.1% H, 6.7% O, (dry basis, mass percent) plus ash and small percentages of N and S. This coal is fed into a gasifier along with oxygen and steam, as shown in Fig. P14.5. The exiting product gas composition is measured on a mole basis to: 39.9% CO, 30.8% H2, 11.4% CO2, 16.4% H2O plus small percentages of CH4, N2, and H2S. How many kilograms of coal are required to produce 100 kmol of product gas? How much oxygen and steam are required? Number of kmol per 100 kg coal: C : n = 74.2/12.01 = 6.178 O2: n = 6.7/31.999 = 0.209 (6.178 C + 2.53 H2 + 0.209 O2)x + y H2O + z O2 in and 39.9 CO + 30.8 H2 + 11.4 CO2 + 16.4 H2O out in 100 kmol of mix out C balance: H2 balance: 6.178 x = 39.9 + 11.4 x = 8.304 y = 26.191 H2: n = 5.1/2.016 = 2.530 2.538.304 + y = 30.8 + 16.4 O2 balance: 0.209 8.304 + 26.191 39.9 16.4 +z= + 11.4 + z = 24.719 2 2 2 Therefore, for 100 kmol of mixture out require: 830.4 kg of coal 26.191 kmol of steam 24.719 kmol of oxygen 145 14.6 Repeat Problem 14.5 for a certain Utah coal that contains, according to the coal analysis, 68.2% C, 4.8% H, 15.7% O on a mass basis. The exiting product gas contains 30.9% CO, 26.7% H2, 15.9% CO2 and 25.7% H2O on a mole basis. Number of kmol per 100 kg coal: C : 68.2/12.01 = 5.679 O2: 15.7/32.00 = 0.491 (5.679 C + 2.381 H2 + 0.491 O2)x + y H2O + z O2 in 30.9 CO + 26.7 H2 + 15.9 CO2 + 25.7 H2O out in 100 kmol of mix out C: H 2: O 2: 5.679x = 30.9 + 15.9 x = 8.241 y = 32.778 2.381 8.241 + y = 26.7 + 25.7 0.491 8.241 + H2: 4.8/2.016 = 2.381 32.778 30.9 25.7 +z= + 15.9 + 2 2 2 z = 23.765 Therefore, for 100 kmol of mixture out, require: 824.1 kg of coal 32.778 kmol of steam 23.765 kmol of oxygen 14.7 A sample of pine bark has the following ultimate analysis on a dry basis, percent by mass: 5.6% H, 53.4% C, 0.1% S, 0.1% N, 37.9% O and 2.9% ash. This bark will be used as a fuel by burning it with 100% theoretical air in a furnace. Determine the airfuel ratio on a mass basis. Converting the Bark Analysis from a mass basis: Substance c/M = kmol / 100 kg coal Product oxygen required S 0.1/32 0.003 SO2 0.003 H2 5.6/2 2.80 H 2O 1.40 C 53.4/12 4.45 CO2 4.45 O2 37.9/32 1.184 N2 0.1/28 0.004 Combustion requires: 0.003 + 1.40 + 4.45 = 5.853 kmol O2 there is in the bark 1.184 kmol O2 so the net from air is 4.669 kmol O2 AF = (4.669 + 4.669 3.76) 28.97 kg air = 6.44 100 kg bark 146 14.8 Liquid propane is burned with dry air. A volumetric analysis of the products of combustion yields the following volume percent composition on a dry basis: 8.6% CO2, 0.6% CO, 7.2% O2 and 83.6% N2. Determine the percent of theoretical air used in this combustion process. a C3H8 + b O2 + c N2 8.6 CO2 + 0.6 CO + d H2O + 7.2 O2 + 83.6 N2 C balance: H2 balance: N2 balance: O2 balance: 3a = 8.6 + 0.6 = 9.2 a = 3.067 4a = d d = 12.267 c = 83.6 b = 8.6 + 0.6 12.267 + + 7.2 = 22.234 2 2 AirFuel ratio = Theoretical: 22.234 + 83.6 = 34.51 3.067 C3H8 + 5 O2 + 18.8 N2 3 CO2 + 4 H2O + 18.8 N2 theo. AF ratio = % theoretical air = 5 + 18.8 = 23.8 1 34.51 100 % = 145 % 23.8 14.9 A fuel, CxHy, is burned with dry air and the product composition is measured on a dry basis to be: 9.6% CO2, 7.3% O2 and 83.1% N2. Find the fuel composition (x/y) and the percent theoretical air used. FuCxHy + O O2 + 3.76O N2 9.6CO2 + 7.3O2 + 83.1N2 + H OH2O
2 2 2 N2 balance: 3.76O = 83.1 O = 22.101
2 2 O2 balance: O = 9.6 + 7.3 + 2 H
2 1 2O H 2O = 10.402 H balance: Fu y = 2 H C balance: Fu x = 9.6 2O = 20.804 Fuel composition ratio = x/y = 9.6/20.804 = 0.461 Theoretical air = O AC 2 O stoich 9.6 + 1 29.804 2 4 = 22.101 = 1.493 147 14.10 Many coals from the western United States have a high moisture content. Consider the following sample of Wyoming coal, for which the ultimate analysis on an asreceived basis is, by mass: Component Moisture H C S N O Ash % mass 28.9 3.5 48.6 0.5 0.7 12.0 5.8 This coal is burned in the steam generator of a large power plant with 150% theoretical air. Determine the airfuel ratio on a mass basis. Converting from mass analysis: Substance c/M = kmol / 100 kg coal Product oxygen required S 0.5/32 0.0156 SO2 0.0156 H2 3.5/2 1.75 H 2O 0.875 C 4.86/12 4.05 CO2 4.05 O2 12/32 0.375 N2 0.7/28 0.025 Combustion requires then oxygen as: 0.0156 + 0.875 + 4.05 = 4.9406 The coal does include 0.375 O2 so only 4.5656 O2 from air/100 kg coal AF = 1.5 (4.5656 + 4.5656 3.76) 28.97/100 = 9.444 kg air/kg coal 14.11 Pentane is burned with 120% theoretical air in a constant pressure process at 100 kPa. The products are cooled to ambient temperature, 20C. How much mass of water is condensed per kilogram of fuel? Repeat the answer, assuming that the air used in the combustion has a relative humidity of 90%. C5H12 + 1.2 8 (O2 + 3.76 N2) 5 CO2 + 6 H2O + 0.96 O2 + 36.1 N2 Products cooled to 20oC, 100 kPa, so for H2O at 20C: Pg = 2.339 kPa yH2O MAX = Pg/P = Therefore, nH2O MAX 2.339 = => n H2O MAX = 1.007 < n H2O 100 nH2O MAX + 42.06 nH2O LIQ = 6  1.007 = 4.993 nH2O VAP = 1.007, mH2O LIQ = 4.993 18.015 = 1.247 kg/kg fuel 72.151 => w1 = 0.622 2.105 = 0.013375 97.895 Pv1 = 0.9 2.339 = 2.105 kPa nH2O IN = 0.013375 28.97 (9.6 + 36.1) = 0.983 kmol 18.015 nH2O OUT = 0.983 + 6 = 6.983 => nH2O LIQ = 6.983  1.007 = 5.976 kmol nH2O LIQ = 5.976 18.015 = 1.492 kg/kg fuel 72.151 148 14.12 The coal gasifier in an integrated gasification combined cycle (IGCC) power plant produces a gas mixture with the following volumetric percent composition: Product % vol. CH4 0.3 H 2 29.6 CO 41.0 CO2 10.0 N 2 0.8 H 2O 17.0 H 2S 1.1 NH3 0.2 This gas is cooled to 40C, 3 MPa, and the H2S and NH are removed in 3 water scrubbers. Assuming that the resulting mixture, which is sent to the combustors, is saturated with water, determine the mixture composition and the theoretical airfuel ratio in the combustors. CH4 H2 CO CO2 N2 n 0.3 yH2O = nV 29.6 where 41.0 10.0 0.8 81.7 , nV+81.7 nV = number of moles of water vapor Cool to 40C PG = 7.384, P = 3000 kPa nV 7.384 = yH2O MAX = 3000 nV+81.7 a) Mixture composition: CH4 H2 0.3 kmol 29.6 CO 41.0 nV = 0.2016 CO2 10.0 N2 0.8 H2O(v) 0.2016 81.9016 kmol (from 100 kmol of the original gas mixture) 0.3 CH4 + 0.6 O2 0.3 CO2 + 0.6 H2O 29.6 H2 + 14.8 O2 29.6 H2O 41 CO + 20.5 O2 41 CO2 Number of moles of O2 = 0.6 + 14.8 + 20.5 = 35.9 Number of moles of air = 35.9 + 3.76 35.9 (N2) Air Fuel ratio = 28.97(35.9 + 3.76(35.9)) 0.3(16) + 29.6(2) + 41(28) + 10(44) + 0.8(28) + 0.2016(18) = 2.95 kg air/kg fuel 149 14.13 The hot exhaust gas from an internal combustion engine is analyzed and found to have the following percent composition on a volumetric basis at the engine exhaust manifold. 10% CO2, 2% CO, 13% H2O, 3% O2 and 72% N2. This gas is fed to an exhaust gas reactor and mixed with a certain amount of air to eliminate the carbon monoxide, as shown in Fig. P14.13. It has been determined that a mole fraction of 10% oxygen in the mixture at state 3 will ensure that no CO remains. What must the ratio of flows be entering the reactor? Exhaust gas at state 1: CO2 10 %, H2O 13%, CO 2%, O2 3%, N2 72% Exhaust gas at state 3: CO = 0 %, O2 = 10 %
2 Air 1 Exh. gas Reactor 3 gas out 0.02 CO + x O2 + 3.76x N2 0.02 CO2 + (x0.01) O2 + 3.76x N2 At 3: CO2 = 0.10 + 0.02 = 0.12, O2 = (x0.01) + 0.03 + x + 0.02 H2O = 0.13 N2 = 0.72 + 3.76x or nTOT = 0.12 + 0.13 + x + 0.02 + 0.72 + 3.76x = 0.99 + 4.76x yO2 = 0.16 = or x+0.02 0.99+4.76x = x = 0.151 air 2 Exh. Gas 1 4.76x kmol air = 0.718 1 kmol Exh. gas 14.14 Methanol, CH3OH, is burned with 200% theoretical air in an engine and the products are brought to 100 kPa, 30C. How much water is condensed per kilogram of fuel? CH3OH + O {O2 + 3.76 N2} CO2 + 2 H2O + 3.76 O N 2
2 2 Stoichiometric O 2S = 1.5 O 2 AC =3 Actual products: CO2 + 2 H2O + 1.5 O2 +11.28 N2 Psat(30C) = 4.246 kPa yH2O = 0.04246 = H2O 1 + H2O + 1.5 + 11.28 H2O = 0.611 H2O cond = 2  0.611 = 1.389 MFu = 32.042 MH2O 1.389 18 kg H2O = = 0.781 kg fuel MFu 32.042 1410 14.15 The output gas mixture of a certain airblown coal gasifier has the composition of producer gas as listed in Table 14.2. Consider the combustion of this gas with 120% theoretical air at 100 kPa pressure. Determine the dew point of the products and find how many kilograms of water will be condensed per kilogram of fuel if the products are cooled 10C below the dewpoint temperature. {3 CH4 + 14 H2 + 50.9 N2 + 0.6 O2 + 27 CO + 4.5 CO2} + 31.1 O2 + 116.9 N2 34.5 CO2 + 20 H2O + 5.2 O2 + 167.8 N2 Products: yH2O = yH2O MAX = PG/100 = PG = 8.79 kPa At T = 33.2C, P G = 5.13 kPa yH2O = mH2O LIQ = nH2O 5.13 = 100 nH2O+34.5+5.2+167.8 nH2O = 11.22 = 0.0639 kg/kg fuel 20 34.5 + 20 + 5.2 + 167.8 TDEW PT = 43.2C 8.78(18) 3(16) +14(2) +50.9(28) +0.6(32) +27(28) +4.5(44) 14.16 Pentene, C5H10 is burned with pure oxygen in an SSSF process. The products at one point are brought to 700 K and used in a heat exchanger, where they are cooled to 25C. Find the specific heat transfer in the heat exchanger. C5H10 + O O2 5 CO2 + 5 H2O O = 7.5
2 2 . . 5 nF hCO + 5 nF hH
2 2 . . + Q = 5 nF h CO O f 2 . + (5  x) nF h H liq Find x: yH = Pg(25) Ptot = 0.0313 = 2O . + (x) nF h H vap 2O 2O max x x = 0.1614 5+x Out of the 5 H2O only 0.1614 still vapor. . Q     = 5 hCO2,700 + (5x)(hf liq  hf vap  h700) + x(hf vap  hf vap  h700) nF = 5(17761) + 4.84(4401114184)  0.16(14184) = 372738 kJ/kmol Fu 1411 14.17 Butane gas and 200% theoretical air, both at 25C, enter a SSSF combustor. The products of combustion exits at 1000 K. Calculate the heat transfer from the combustor per kmol of butane burned. C4H10 + (1/) O2(O2 + 3.76 N2) a CO2 + b H2O + c N2 + d O2 First we need to find the stoichiometric air ( = 1, d = 0 ) C balance: 4 = a, H balance: 10 = 2b => b=5 O balance: 2O2 = 2a + b = 8 + 5 = 13 => O2 = 6.5 Now we can do the actual air: (1/) = 2 => O2 = 2 6.5 = 13 N balance: c = 3.76 O2 = 48.88, O balance: d = 13  6.5 = 6.5 q = HR  HP = HR  HP  HP = M(HRP)  HP Table 14.3: HRP = 45714, The rest of the values are from Table A.8     h CO2 = 33397, h N2 = 21463, h O2 = 22703, h H2O = 26000 kJ/kmol q = 58.124(45714)  (4 33397 + 5 26000 + 48.88 21463 + 6.5 22703) = 2 657 080  1 460 269 = 1 196 811 kJ/kmol butane 14.18 Liquid pentane is burned with dry air and the products are measured on a dry basis as: 10.1% CO2, 0.2% CO, 5.9% O2 remainder N2. Find the enthalpy of formation for the fuel and the actual equivalence ratio. FuC5H12 + O O2 + 3.76 O N2 2 2 o o o o x H2O + 10.1 CO2 + 0.2 CO + 5.9 O2 + 83.8 N2 Balance of C: 5 Fu = 10.1 + 0.2 Fu = 2.06 Balance of H: 12 Fu = 2 x x = 6 Fu = 12.36 Balance of O: 2 O = x + 20.2 + 0.2 + 2 5.9 O = 22.28
2 2 Balance of N: 2 3.76 O = 83.8 2 O = 22.287 OK
2 2 O for 1 kmol fuel = 10.816
2 = 1, C5H12 + 8 O2 + 8 3.76 N2 6 H2O + 5 CO2 + 30.08 N2    HRP = HP  HR = 6 hf H2O + 5 hf CO2  hf fuel 14.3: HRP = 44983 72.151  hf fuel = 172 998 kJ/kmol = O2 stoich/O2 AC = 8/10.816 = 0.74 1412 14.19 A rigid vessel initially contains 2 kmol of carbon and 2 kmol of oxygen at 25C, 200 kPa. Combustion occurs, and the resulting products consist of 1 kmol of carbon dioxide, 1 kmol of carbon monoxide, and excess oxygen at a temperature of 1000 K. Determine the final pressure in the vessel and the heat transfer from the vessel during the process. 2 C + 2 O2 1 CO2 + 1 CO + 2 O2 Process P 2 = P1 H1 = 0 H2 = 1(393522 + 33397) + 1(110527 + 21686) + (1/2)(0 + 22703) = 437615 kJ Q = (U2  U1) = (H2  H1)  n2RT2 + n1RT1 1 2 = (437615  0)  8.3145(2.5 1000  2 298.2) = 453442 kJ 14.20 In a test of rocket propellant performance, liquid hydrazine (N2H4) at 100 kPa, 25C, and oxygen gas at 100 kPa, 25C, are fed to a combustion chamber in the ratio of 0.5 kg O2/kg N2H4. The heat transfer from the chamber to the surroundings is estimated to be 100 kJ/kg N2H4. Determine the temperature of the products exiting the chamber. Assume that only H2O, H2, and N2 are present. The enthalpy of formation of liquid hydrazine is +50417 kJ/kmol.
1
1 V = constant, n2T2 n1T1 = 200 C: solid, n1(GAS) = 2, n2(GAS) = 2.5 2.5 1000 = 838.4 kPa 2 298.2 Liq. N2H4: 100 kPa, 25 C Gas O2: 100 kPa, 25oC
2 o Comb. Chamber 3 Products . . Q/mN2H4 = 100 kJ/kg . . . . mO2/mN2H4 = 0.5 = 32nO2/32nN2H4 Energy Eq.:
1 and QCV = HP  HR = 100 32.045 = 3205 kJ/kmol fuel 1 N2H4 + 2 O2 H2O + H2 + N2 HR = 1(50417) + 2(0) = 50417 kJ HP = 241826 + hH2O + hH2 + hN2 HP = hH2O + hH2 + hN2 = 241826 + 50417 3205 = 289038 Table A.8 : 2800 K HP = 282141 Interpolate to get 3000 K HP = 307988
1 TP = 2854 K 1413 14.21 Repeat the previous problem, but assume that saturatedliquid oxygen at 90 K is used instead of 25C oxygen gas in the combustion process. Use the generalized charts to determine the properties of liquid oxygen. Problem same as 14.20, except oxygen enters at 2 as saturated liquid at 90 K. . . . . . . mO2/mN2H4 = 0.5 = 32nO2/32nN2H4 and Q/mN2H4 = 100 kJ/kg Energy Eq.: QCV = HP  HR = 100 32.045 = 3205 kJ/kmol fuel 1 N2H4 + 2 O2 H2O + H2 + N2 ~ hf = 5.2
1 Reaction equation: At 90 K, Tr2 = 90/154.6 = 0.582 * Figure D.2, (h  h) = 8.3145 154.6 5.2 = 6684 kJ/kmol hAT 2 = 6684 + 0.922 32(90  298.15) = 12825 kJ/kmol HR = 50417 + 2(0  12825) = 44005 kJ, 1st law: hP = hH
2O 2 2 1 HP = 241826 + hH + hN = Qcv + HR  H = 282626 P From Table A.8, HP 2800K = 282141, HP 3000K = 307988 Therefore, T P = 2804 K 14.22 The combustion of heptane C7H16 takes place in a SSSF burner where fuel and air are added as gases at P , T . The mixture has 125% theoretical air and the o o products are going through a heat exchanger where they are cooled to 600 K.Find the heat transfer from the heat exchanger per kmol of heptane burned. The reaction equation for stiochiometric ratio is: C7H16 + vO2 (O2 + 3.76 N2) => 7CO2 + 8 H2O + vO2 3.76 N2 So the balance (C and H was done in equation) of oxygen gives vO2 = 7 + 4 = 11, and actual one is 111.25 = 13.75. Now the actual reaction equation is: C7H16 + 13.75 O2 + 51.7 N2 => 7CO2 + 8 H2O + 51.7 N2+ 2.75 O2 To find the heat transfer take control volume as combustion chamber and heat exchanger Q = HPo + HP  HRo = HRPo + HP Q = 100.205(44922) + 7(12906) + 8(10499) + 51.7(8894) + 2.75(9245) HR + Q = HP => = 4 501 409 + 90342 + 83922 + 459 819.8 + 25423.75 =  3 841 831 kJ/kmol fuel 1414 14.23 Ethene, C2H4, and propane, C3H8, in a 1 1 mole ratio as gases are burned with 120% theoretical air in a gas turbine. Fuel is added at 25C, 1 MPa and the air comes from the atmosphere, 25C, 100 kPa through a compressor to 1 MPa and mixed with the fuel. The turbine work is such that the exit temperature is 800 K with an exit pressure of 100 kPa. Find the mixture temperature before combustion, and also the work, assuming an adiabatic turbine. =1 C2H4 + C3H8 + 8 O2 + 30.08 N2 5 CO2 + 6 H2O + 30.08 N2 = 1.2 C2H4 + C3H8 + 9.6 O2 + 36.096 N2 5 CO2 + 6 H2O + 1.6 O2 + 36.096 N2 45.696 kmol air per 2 kmol fuel C.V. Compressor (air flow) Energy Eq.:
2 1 wc = h2  h1 Entropy Eq.: s2 = s1 Pr = Pr P2/P1 = 13.573 T2 air = 570.8 K wc = 576.44  298.34 = 278.1 kJ/kg = 8056.6 kJ/kmol air C.V. Mixing Chanber (no change in composition) . . . nairhair in + nFu1h1 in + nFu2h2 in = (SAME)exit CP F1 + CP F2Texit  T0 = 45.696 CP airT2 air  Texit C2H4: CP F1 = 43.43, C3H8: CP F2 = 74.06, CP air = 29.07 45.696CP airT2 + CP F1 + CP F2T0 Texit = = 548.7 K CP F1 + CP F2 + 45.696 CP air Dew Point Products: yH PH
2O 2 2O = 5 + 6 + 1.6 + 36.096 = 0.1232 6 = yH OPtot = 123.2 kPa Tdew = 105.5C = Hin  Hout = HR  HP 800
2 C.V. Turb. + combustor + mixer + compressor (no Q) wnet (800K out so no liquid H2O)
2 2 2   = hf C2H4 + hf C3H8  5 hCO  6 hH O  1.6 hO  36.096 hN = 2 576 541 2 kmol Fu wT= wnet + wcomp = 2 944 695
kJ 2 kmol Fu kJ 1415 14.24 One alternative to using petroleum or natural gas as fuels is ethanol (C 2H5OH), which is commonly produced from grain by fermentation. Consider a combustion process in which liquid ethanol is burned with 120% theoretical air in an SSSF process. The reactants enter the combustion chamber at 25C, and the products exit at 60C, 100 kPa. Calculate the heat transfer per kilomole of ethanol, using the enthalpy of formation of ethanol gas plus the generalized charts. C2H5OH + 1.2 3 (O2 + 3.76 N2) 2CO2 + 3H2O + 0.6O2 + 13.54N2 0 Fuel: hf = 235000 kJ/kmol for IG yH2O MIX = Products at 60C, 100 kPa nV MAX 19.94 = => n V MAX = 4.0 > 3 No liq. 100 nV MAX+2+0.6+13.54 Tr = 298.2 / 513.9 = 0.58 hf = 5.23 from Figure D.2 * (h  h)LIQ = 8.3145 513.9 5.23 = 22347 kJ/kmol HR = 1(235000  22347) + 0 + 0 = 257 347 kJ HP = 2(393522 + 1327) + 3(241826 + 1178) + 0.6(0+1032) + 13.54(0+1020) = 1 491 904 kJ QCV = HP  HR = 1 234 557 kJ 14.25 Another alternative to using petroleum or natural gas as fuels is methanol, CH3OH, which can be produced from coal. Both methanol and ethanol have been used in automotive engines. Repeat the previous problem using liquid methanol as the fuel instead of ethanol. CH3OH + 1.2 1.5 (O2 + 3.76 N2) 1 CO2 + 2 H2O + 0.3 O2 + 6.77 N2 React at 25 oC, Prod at 60 oC = 333.2 K, 100 kPa yH2O MAX = nV MAX 19.94 = => n V MAX = 2.0 > 2 No liq. 100 nV MAX+1+0.3+6.77 o CH3OH: hf = 201 300 kJ, TC = 512.6 K ~ Tr = 298.15 / 512.6 = 0.582 hf = 5.22 Figure D.2  (h*h)LIQ = 8.3145 512.6 5.22 = 22 248 HR = 1 hLIQ = 201 300  22 248 = 223 548 kJ HP = 1(393 522 + 1327) + 2(241 826 + 1178) + 0.3(1032) + 6.77(1020) = 866 276 kJ Q = HP  HR = 642 728 kJ 1416 14.26 Another alternative fuel to be seriously considered is hydrogen. It can be produced from water by various techniques that are under extensive study. Its biggest problem at the present time is cost, storage, and safety. Repeat Problem 14.24 using hydrogen gas as the fuel instead of ethanol. H2 + 1.2 0.5 O2 + 1.2 3.76 0.5 N2 1 H2O + 0.1 O2 + 2.256 N2 Products at 60C, 100 kPa yH2O MAX = nV MAX 19.94 = 100 nV MAX+0.1+2.256 Solving, nV MAX = 0.587 < 1 => nV = 0.587, nLIQ = 0.413 HR = 0 + 0 + 0 = 0 HP = 0.413(285830 + 18.015(251.1  104)) + 0.587(241826 + 1178) + 0.1(0 + 1032) + 2.256(0 + 1020) = 255 816 kJ Q CV = HP  HR = 255 816 kJ 14.27 Hydrogen peroxide, H2O2, enters a gas generator at 25C, 500 kPa at the rate of 0.1 kg/s and is decomposed to steam and oxygen exiting at 800 K, 500 kPa. The resulting mixture is expanded through a turbine to atmospheric pressure, 100 kPa, as shown in Fig. P14.27. Determine the power output of the turbine, and the heat transfer rate in the gas generator. The enthalpy of formation of liquid H2O2 is  187 583 kJ/kmol. . mH2O2 1 0.1 . H2O2 H2O + O2 nH2O2 = = = 0.00294 kmol/s 2 34.015 M . . nMIX = nH2O2 1.5 = 0.00441 kmol/s
2 1 CP0 MIX = 1.872 18.015 + 0.922 31.999 = 32.317 3 3 CV0 MIX = 32.317  8.3145 = 24.0 => kMIX = 32.317/24.0 = 1.3464 CV: turbine. Assume reversible s3 = s2 100 0.2573 T3 = T2( ) = 800( ) = 528.8 K 500 P2 w = CP0(T2  T3) = 32.317(800  528.8) = 8765 kJ/kmol . WCV = 0.00441 8765 = 38.66 kW CV: Gas Generator . H1 = 0.00294(187 583 + 0) = 551.49 . H2 = 0.00294(241 826 + 18002) + 0.00147(0 + 15836) = 634.76 . . . QCV = H2  H1 = 634.76 + 551.49 = 83.27 kW P3
k1 k 1417 14.28 In a new highefficiency furnace, natural gas, assumed to be 90% methane and 10% ethane (by volume) and 110% theoretical air each enter at 25C, 100 kPa, and the products (assumed to be 100% gaseous) exit the furnace at 40C, 100 kPa. What is the heat transfer for this process? Compare this to an older furnace where the products exit at 250C, 100 kPa. 0.90CH4 + 0.10C2 H6 Furnace 25 C 110% Air
o Prod. 40 oC 100 kPa 0.9 CH4 + 0.1 C2H6 + 1.1 2.15 O2 + 3.76 2.365 N2 1.1 CO2 + 2.1 H2O + 0.215 O2 + 8.892 N2 HR = 0.9(74873) + 0.1(84740) = 75860 kJ HP = 1.1(393 522 + 562) + 2.1(241 826 + 504) + 0.215(441) + 8.892(437) = 935052 kJ assuming all gas QCV = HP  HR = 859 192 kJ/kmol fuel b) TP = 250 oC HP = 1.1(393 522 + 9346) + 2.1(241 826 + 7740) + 0.215(6808) + 8.892(6597) = 854 050 kJ QCV = HP  HR = 778 190 kJ/kmol fuel 14.29 Repeat the previous problem, but take into account the actual phase behavior of the products exiting the furnace. Same as 14.28, except check products for saturation at 40 oC, 100 kPa yV MAX = nV MAX 7.384 = => 100 nV MAX+10.207 nLIQ = 2.1  0.814 = 1.286 Solving, nV MAX = 0.814 nV = 0.814, HLIQ = 1.286[285 830 + 18.015(167.6  104.9)] = 366125 kJ HGAS = 1.1(393 522 + 562) + 0.814(241826 + 504 ) + 0.215(441) + 8.892(437) = 624711 kJ QCV = HP  HR = (366 125  624 711) + 75860 = 914 976 kJ/kmol fuel b) no liquid, answer the same as 14.28. 1418 14.30 Methane, CH4, is burned in an SSSF process with two different oxidizers: Case A: Pure oxygen, O2 and case B: A mixture of O2 + x Ar. The reactants are supplied at T0, P0 and the products should for both cases be at 1800 K. Find the required equivalence ratio in case (A) and the amount of Argon, x, for a stoichiometric ratio in case (B). a) Stoichiometric has = 2, actual has: CH4 + O2 CO2 + 2H2O + (  2)O2 HR = HP + HP 1800 => HP 1800=  HRP = 50010 16.04 = 802 160 h = 79432, hH O= 62693, hO = 51674 CO
2 2 2 HP 1800= 101470 + 51674 = 802160 b) = 13.56, = 6.78 CH4 + 2 O2 + 2x Ar CO2 + 2H2O + 2x Ar HP 1800= 79432 + 2 62693 + 2x 0.5203 39.948(1800  298) = 204818 + x62438 802160 = 204818 + x 62438 x = 9.567 14.31 Butane gas at 25C is mixed with 150% theoretical air at 600 K and is burned in an adiabatic SSSF combustor. What is the temperature of the products exiting the combustor? 25 C GAS 150% Air 600 K o C H 4 10 Adiab. Comb. QCV = 0 Prod. at Tp C4H10 + 1.56.5 (O2 + 3.76 N2 ) 4 CO2 + 5 H2O + 3.25 O2 + 36.66 N2 Reactants: hAIR = 9.75(9245) + 36.66(8894) = 416193 kJ ; o hC4H10 = hf IG = 126 200 kJ => H R = +289 993 kJ * * * * HP = 4(393522 + hCO2) + 5(241826 + hH2O) + 3.25 hO2 + 36.66 hN2 Energy Eq.: HP  HR = 0 * * * * 4 hCO2 + 5 hH2O + 3.25 hO2 + 36.66 hN2 = 3 073 211 Trial and Error: LHS2000 K = 2 980 000, LHS2200 K = 3 369 866 Linear interpolation to match RHS => TP = 2048 K 1419 14.32 In a rocket, hydrogen is burned with air, both reactants supplied as gases at Po, To. The combustion is adiabatic and the mixture is stoichiometeric (100% theoretical air). Find the products dew point and the adiabatic flame temperature (~2500 K). The reaction equation is: H2 + vO2 (O2 + 3.76 N2) => H2O + 3.76 vO2 N2 The balance of hydrogen is done, now for oxygen we need v O2 = 0.5 and thus we have 1.88 for nitrogen. yv = 1/(1+1.88) = 0.3472 => Pv = 101.325 0.3472 = 35.18 kPa = Pg Table B.1.2: Tdew = 72.6 C. HR = HP => 0 = 241826 + hwater + 1.88 hnitrogen Find now from table A.13 the two enthalpy terms At 2400 K : HP = 93741 + 1.88 70640 = 226544 At 2600 K : HP = 104520 + 1.88 77963 = 251090.4 Then interpolate to hit 241826 to give T = 2524.5 K 14.33 Liquid butane at 25C is mixed with 150% theoretical air at 600 K and is burned in an adiabatic SSSF combustor. Use the generalized charts for the liquid fuel and find the temperature of the products exiting the combustor. o TC = 425.2 K 25 C LIQ. C4 H 10 Prod. Adiab. TR = 0.701 150% Air Comb. at Tp 600 K QCV = 0 o hC4H10 = hf IG + (hLIQ  h*) see Fig. D.2 = 126200 + (4.85 8.3145 425.2) = 143 346 kJ C4H10 + 1.5 6.5 O2 + 3.76 9.75 N2 4 CO2 + 5 H2O + 3.25 O2 + 36.66 N2 hAIR = 9.75(9245) + 36.66(8894) = 416 193 kJ HR = 416 193 143 346 = +272 847 kJ * * * * HP = 4(393522 + hCO2) + 5(241826 + hH2O) + 3.25 hO2 + 36.66 hN2 Energy Eq.: HP  HR = 0 * * * * 4 hCO2 + 5 hH2O + 3.25 hO2 + 36.66 hN2 = 3 056 065 Trial and Error: LHS2000 K = 2 980 000, LHS2200 K = 3 369 866 Linear interpolation to match RHS => TP = 2039 K 1420 14.34 A stoichiometric mixture of benzene, C6H6, and air is mixed from the reactants flowing at 25C, 100 kPa. Find the adiabatic flame temperature. What is the error if constant specific heat at T0 for the products from Table A.5 are used? C6H6 + O O2 + 3.76 O N2 6CO2 + 3H2O + 3.76 O N 2
2 2 2 O = 6 + 3/2 = 7.5 N = 28.2
2 2 HP=HP + HP = HR = HR HP = HRP = 40576 78.114 = 3169554 HP 2600K= 3280600, HP 2400K= 2968000 TAD= 2529 K iCPi = 6 37.048 + 3 33.73 + 28.2 29.178 = 1146.3 T = HP/iCPi = 2765 TAD= 3063 K, 21% high 14.35 Liquid nbutane at T0, is sprayed into a gas turbine with primary air flowing at 1.0 MPa, 400 K in a stoichiometric ratio. After complete combustion, the products are at the adiabatic flame temperature, which is too high, so secondary air at 1.0 MPa, 400 K is added, with the resulting mixture being at 1400 K. Show that Tad > 1400 K and find the ratio of secondary to primary air flow. C.V. Combustion Chamber. C4H10 + 6.5 O2 + 6.5 3.76 N2 5 H2O + 4 CO2 + 24.44 N2 Primary Air Fuel Combustion Chamber Secondary TAD Air To turbine 1400 K Mixing Energy Eq.: Hair + Hfuel = HR = HP HP + HP = HR + HR HP = HR + HR  HP HP = 45344 58.124 + 6.5(3.76 2971 + 3027) = 2727861 HP 1400 = 5.43491 + 4.55895 + 24.44 34936 = 1294871 < HP Try TAD > 1400: HP = 2658263 @2400 K, C.V. Mixing Ch. HP = 2940312 @2600 K Air Second: O2 sO2 + 3.76 N2 P + O2 second Hair = HP 1400 + O2 second Hair 1400 O2 second= HP  HP 1400 Hair 1400  Hair 400 = 1432990 = 9.3 168317  14198 ratio = O2 sec/O2 prim = 9.3/6.5 = 1.43 1421 14.36 Consider the gas mixture fed to the combustors in the integrated gasification combined cycle power plant, as described in Problem 14.12. If the adiabatic flame temperature should be limited to 1500 K, what percent theoretical air should be used in the combustors? Product % vol. CH4 0.3 H 2 29.6 CO 41.0 CO2 10.0 N 2 0.8 H 2O 17.0 H 2S 1.1 NH3 0.2 Mixture may be saturated with water so the gases are ( H 2S and NH3 out) CH4 H2 CO CO2 N2 n 0.3 29.6 41.0 10.0 0.8 81.7 yV MAX = 7.384/3000 = nV/(nV + 81.7) Solving, nV = 0.2 kmol, rest condensed {0.3 CH4 + 29.6 H2 + 41.0 CO + 10.0 CO2 + 0.8 N2
+ 0.2 H2O + 35.9x O2 + 3.76 35.9x N2} 51.3 CO2 + 30.4 H2O + 35.9(x  1) O2 + (135.0x + 0.8) N2 For the fuel gas mixture, nCP0 MIX = 0.3 16.04 2.2537 + 29.6 2.016 14.2091 + 41.0 28.01 1.0413 + 10.0 44.01 0.8418 + 0.8 28.013 1.0416 + 0.2 18.015 1.8723 = 2455.157 0 nhf MIX = 0.3(74873) + 29.6(0) + 41.0(110527) + 10.0(393522) + 0.8(0) + 0.2(241826) = 8537654 kJ At 40C, for the fuel mixture: HMIX = 8537654 + 2455.157(40  25) = 8500827 kJ Assume air enters at 25C: hAIR = 0 Products at 1500 K: HP = 51.3(393522 + 61705) + 30.4(241826 + 48149) + 35.9(x  1)(0 + 40600) + (135x + 0.8)(0 + 38405) = 24336806 + 6642215x 1st law: HP = HR = HMIX x= +24336809  8500827 = 2.384 6642215 or 238 % theo. air 1422 14.37 Acetylene gas at 25C, 100 kPa is fed to the head of a cutting torch. Calculate the adiabatic flame temperature if the acetylene is burned with a. 100% theoretical air at 25C. b. 100% theoretical oxygen at 25C. a) C2H4 + 2.5 O2 + 2.5 3.76 N2 2 CO2 + 1 H2O + 9.4 N2 o HR = hf C2H2 = +226 731 kJ * * * HP = 2(393 522 + hCO2) + 1(241 826 + hH2O) + 9.4 hN2 QCV = HP  HR = 0 * * * 2 hCO2 + 1 hH2O + 9.4 hN2 = 1 255 601 kJ LHS3000 = 1 303 775 Trial and Error A.8: LHS2800 = 1 198 369, Linear interpolation: TPROD = 2909 K b) C2H2 + 2.5 O2 2 CO2 + H2O HR = +226 731 kJ ; * * HP = 2(393 522 + hCO2) + 1(241 826 + hH2O) * * 2 hCO2 + 1 hH2O = 1 255 601 2 343 782 + 302 295 = 989 859 2 337 288 + 296 243 = 970 819 Difference, extrapolating At 6000 K (limit of A.8) At 5900 K or 19040/100 K change TPROD 6000 + 265 742/190.40 7400 K 14.38 Ethene, C2H4, burns with 150% theoretical air in an SSSF constantpressure process with reactants entering at P0, T0. Find the adiabatic flame temperature. C2H4 + 3(O2 + 3.76N2) 2CO2 + 2H2O + 11.28N2 C2H4 + 4.5(O2 + 3.76N2) 2CO2 + 2H2O + 1.5 O2 + 16.92N2 yH
2O = 2/(2 + 2 + 1.5 + 16.92) = 0.0892 => Pv = 9.041 TDEW = 43.8C HP = HP + 2hCO + 2hH 2  HR = hf Fu 2O + 1.5hO + 16.92hN
2 2 HP + HP = HR
kJ HP = HRP = 28.054 47158 = 1322970.5 kmol Fu Initial guess based on N2 from A.8 HP(2000) = 1366982, T1 = 2100 K => TAD 1950 K HP(1900) = 1278398 1423 14.39 Solid carbon is burned with stoichiometric air in an SSSF process. The reactants at T0, P0 are heated in a preheater to T2 = 500 K as shown in Fig. P14.39, with the energy given by the product gases before flowing to a second heat exchanger, which they leave at T0. Find the temperature of the products T4, and the heat transfer per kmol of fuel (4 to 5) in the second heat exchanger. Control volume: Total minus last heat exchanger. C + O2 + 3.76N2 CO2 + 3.76N2 Energy Eq.: HR = HR = HP = HP + HP = hf CO + hCO + 3.76hN 3 3 2 2 2 hf CO = 393 522, HP
2 3 2400 =381 436, HP 3 2500 = 401 339 T3 = Tad.flame = 2461 K Control volume: Total. Then energy equation:  HR + Q = HP kJ    Q = HRP = hf CO2  0 = 393 522 kmol fuel 1424 14.40 A study is to be made using liquid ammonia as the fuel in a gasturbine engine. Consider the compression and combustion processes of this engine. a. Air enters the compressor at 100 kPa, 25C, and is compressed to 1600 kPa, where the isentropic compressor efficiency is 87%. Determine the exit temperature and the work input per kilomole. b. Two kilomoles of liquid ammonia at 25C and x times theoretical air from the compressor enter the combustion chamber. What is x if the adiabatic flame temperature is to be fixed at 1600 K? Air P1 = 100 kPa T1 = 25 oC
1 COMP. W
k1 k 2 P2 = 1600 kPa S COMP = 0.87 a) ideal compressor process (adiabatic reversible): s2S = s1 T2S = T1( ) P1 P2 = 298.2( 1600 0.286 ) = 659 K 100 wS = CP0(T2S  T1) = 1.004(659  298.2) = 362.2 Real process: w = wS/S = 362.2/0.87 = 416.3 kJ/kg T2 = T1  w/CP0 = 298.2 + 416.3/1.004 = 713 K Also w = 416.3 28.97 = 12060 kJ/kmol b) 2 liq NH3, 25 C Air 100 kPa, 25 oC
o COMP W COMB. CHAMBER Q = 0 Prod. PP = 1600 kPa TP = 1600 K 2 NH3 + 1.5x O2 + 5.64x N2 3 H2O + 1.5(x  1) O2 + (5.64x + 1) N2 Using Tables A.16 and A.2, hNH3 = 45720 + 17.031(298.36  1530.04) = 66697 kJ/kmol HR = 2(66697) + 0 = 133394 kJ W = 12060 7.14x = 86108 x kJ HP = 3(241826 + 52907) + 1.5(x  1)(0 + 44267) + (5.64x + 1)(0 + 41904) = 302739x  591254 1st law: HR = HP + W = 133394 = 302739 x  591254  86108 x => x = 2.11 1425 14.41 A closed, insulated container is charged with a stoichiometric ratio of oxygen and hydrogen at 25C and 150 kPa. After combustion, liquid water at 25C is sprayed in such that the final temperature is 1200 K. What is the final pressure? H2 + 2O2 H2O P: 1 H2O + xiH2O 3U2  U1 = xihi = xihf liq = 1 + xiHP  HR  1 + xiRTP + 2RTR 1 HR = , HP = 241826 + 34506 = 207320, hf liq= 285830 Substitute xi( 285830 + 207320 + 8.3145 1200) =
3 207320  8.3145 1200  2298.15= 213579 xi = 3.116 P1V1 = nRRT1, P2V1 = npRTp P2 = P11 + xiTP 3 2T1 = 150 4.116 1200
3 2 = 1657 kPa 298.15 14.42 Wet biomass waste from a foodprocessing plant is fed to a catalytic reactor, where in an SSSF process it is converted into a lowenergy fuel gas suitable for firing the processing plant boilers. The fuel gas has a composition of 50% methane, 45% carbon dioxide, and 5% hydrogen on a volumetric basis. Determine the lower heating value of this fuel gas mixture per unit volume. For 1 kmol fuel gas, 0.5 CH4 + 0.45 CO2 + 0.05 H2 + 1.025 O2 (0.5 + 0.45) CO2 + 1.05 H2O The lower heating value is with water vapor in the products. Since the 0.45 CO2 cancels, hRP = 0.5(393522) + 1.05(241826)  0.5(74873)  0.05(0) = 413242 kJ/kmol fuel gas With n 100 = P/RT = = 0.04033 kmol/m3 V 8.3145 298.2 LHV = +413242 0.04033 = 16 666 kJ/m3 1426 14.43 Determine the lower heating value of the gas generated from coal as described in Problem 14.12. Do not include the components removed by the water scrubbers. The gas from problem 14.12 is saturated with water vapor. Lower heating value LHV has water as vapor. LHV = HRP = HP  HR Only CH4, H2 and CO contributes. From 14.12 the gas mixture after the scrubbers has i = 81.9 of composition: 0.3CH4 + 29.6H2 + 41CO + 10CO2 + 0.8N2 + 0.2016H2O    LHV = [0.3HRPCH4 + 29.6HRPH2 + 41HRPCO]/81.9 = [0.3(50010 16.043) + 29.6(241826) + 41(393522 + 110527)]/81.9 = 232009 kmol gas
kJ 14.44 Propylbenzene, C9H12, is listed in Table 14.3, but not in table A.9. No molecular weight is listed in the book. Find the molecular weight, the enthalpy of formation for the liquid fuel and the enthalpy of evaporation. ^ M = 9 12.011 + 12 1.0079 = 120.194 Hfg = 384 kJ/kg C9H12 + 12 O2 9 CO2 + 6 H2O       HRP = ihfi  hfFu hfFu = ihfi  hRP P P    ^ hfFu = 9hfCO2 + 6hfH2O g + M(41219)liq Fu H2O vap = 38383.5 kJ/kmol 14.45 Determine the higher heating value of the sample Wyoming coal as specified in Problem 14.10. 0 Note: For SO2 , hf = 296842 kJ/kmol From solution 14.10 , for 100 kg of coal there are 4.05 kmol CO2, 1.75 kmol H2O, 0.0156 kmol SO2 in the products. Therefore, hRP0 = 4.05(393522) + 1.75(285830) + 0.0156(296842) = 2098597 kJ/100 kg coal So that HHV = +20986 kJ/kg coal 1427 14.46 Consider natural gas A and natural gas D, both of which are listed in Table 14.2. Calculate the enthalpy of combustion of each gas at 25C, assuming that the products include vapor water. Repeat the answer for liquid water in the products. Natural Gas A 0.939 CH4 + 0.036 C2H6 + 0.012 C3H8 + 0.013 C4H10 + 2.1485 O2 + 3.76 2.1485 N2 1.099CO2 + 2.099 H2O + 8.0784 N2 HR = 0.939(74878) + 0.036(84740) + 0.012(103900) + 0.013(126200) = 76244 kJ a) vapor H2O HP = 1.099(393522) + 2.099(241826) = 940074 hRP = HP  HR = 863830 kJ/kmol b) Liq. H2O HP = 1.099(393522) + 2.099(285830) = 1032438 hRP = 956194 kJ/kmol Natural Gas D: 0.543 CH4 + 0.163 C2H6 + 0.162 C3H8 + 0.074 C4H10 + 0.058 N2 + O2 + N2 1.651 CO2 + 2.593 H2O + N2 HR = 0.543(74873) + 0.163(84740) + 0.162(130900) + 0.074(126200) = 80639 kJ a) vapor H2O HP = 1.651(393522) + 2.593(241826) = 1276760 kJ hRP = 1196121 kJ/kmol b) Liq. H2O HP = 1.651(393522) + 2.593(285830) = 1390862 kJ hRP = 1310223 kJ/kmol 1428 14.47 Blast furnace gas in a steel mill is available at 250C to be burned for the generation of steam. The composition of this gas is, on a volumetric basis, Component CH4 H2 CO CO2 N2 H2O Percent by volume 0.1 2.4 23.3 14.4 56.4 3.4 Find the lower heating value (kJ/m3) of this gas at 250C and ambient pressure. Of the six components in the gas mixture, only the first 3 contribute to the heating value. These are, per kmol of mixture: 0.024 H2, For these components, 0.024 H2 + 0.001 CH4 + 0.233 CO + 0.1305 O2 0.026 H2O + 0.234 CO2 The remainder need not be included in the calculation, as the contributions to reactants and products cancel. For the lower HV(water as vapor) at 250C hRP = 0.026(241826+7742) + 0.234(393522+9348)  0.024(0+6558)  0.001(74873+2.25416.04(25025))  0.233(110527+6625)  v0 = R To/Po = 8.3145 523.2/100 = 43.5015 m3/kmol LHV = +72573 / 43.5015 = 1668 kJ/m3 14.48 The enthalpy of formation of magnesium oxide, MgO(s), is 601827 kJ/kmol at 25C. The melting point of magnesium oxide is approximately 3000 K, and the increase in enthalpy between 298 and 3000 K is 128449 kJ/kmol. The enthalpy of sublimation at 3000 K is estimated at 418000 kJ/kmol, and the specific heat of magnesium oxide vapor above 3000 K is estimated at 37.24 kJ/kmol K. a. Determine the enthalpy of combustion per kilogram of magnesium. b. Estimate the adiabatic flame temperature when magnesium is burned with theoretical oxygen. a) 1 Mg + O2 MgO(s) 2 hCOMB = hCOMB/M = h/M = 601827/24.32 = 24746 kJ/kg f (MgO = vapor phase) 0.1305(0+6810) = 72 573 kmol fuel
kJ 0.001 CH4, 0.233 CO b) assume TR = 25C and also that T P > 3000 K, 1st law: QCV = HP  HR = 0, but HR = 0 HP = h + (h3000  h298)SOL + hSUB + CP VAP(TP  3000) f = 601827 + 128449 + 418000 + 37.24(TP  3000) = 0 Solving, TP = 4487 K 1429 14.49 A rigid container is charged with butene, C4H8, and air in a stoichiometric ratio at P0, T0. The charge burns in a short time with no heat transfer to state 2. The products then cool with time to 1200 K, state 3. Find the final pressure, P3, the total heat transfer, 1Q3, and the temperature immediately after combustion, T2. The reaction equation is, having used C and H atom balances: C4H8 + O O2 + 3.76 N2 4 CO2 + 4 H2O + 3.76 O N 2 2 2 Counting now the oxygen atoms we get O = 6.
2 C.V. analysis gives: U2  U1 = Q  W = Q = H2  H1  P2V2 + P1V1 = H2  H1  Rn2T2  n1T1 ^ H2  H1 = HP 1200  HR = HP  HR + HP = M HRP + HP = 2542590 + 950055 = 1592535 ^ Where M = 56.108 and n1= 1 + 6 4.76 = 29.56, n2 = 4 + 4 + 6 3.76 = 30.56, able A.8 at 1200 K: hCO = 44473,
2 hH O=34506,
2 hN =28109.
2 kJ Now solving for the heat transfer: Q = 1592535  8.3145(30.56 1200  29.56 298.15) = 1824164 kmol fuel To get the pressure, assume ideal gases: n2RT2 n2T2 P2 = = P1 = 421.6 kPa V2 n1T1 Before heat transfer takes place we have constant U so: U1  U1 = 0 = H1  H1  n2RT1 + n1RT1
a a a Now split the enthalpy H1 = HP + HPT1 and arrange things with the a a unknowns on LHS and knowns on RHS: HP n2RT = HR  HP  n1RT1= 2 542 590  73278 = 2 469 312 Trial and error leads to: LHS (3000 K) = 3 209 254  30.56 8.31451 3000 = 2 446 980 LHS (3000 K) = 3 471 331  30.56 8.31451 3200 = 2 658 238 linear interpolation T = 3021 K 1430 14.50 In an engine a mixture of liquid octane and ethanol, mole ratio 9 1, and stoichiometric air are taken in at T0, P0. In the engine the enthalpy of combustion is used so that 30% goes out as work, 30% goes out as heat loss and the rest goes out the exhaust. Find the work and heat transfer per kilogram of fuel mixture and also the exhaust temperature. 0.9 C8H18 + 0.1 C2H5OH + 11.55 O2 + 43.428 N2 8.4 H2O + 7.4 CO2 + 43.428 N2 For 0.9 octane + 0.1 ethanol kJ  HRP mix= 0.9HRP C8H18 + 0.1HRP C2H5OH = 4690690.3 kmol ^ ^ ^ Mmix = 0.9 Moct + 0.1 Malc = 107.414 Energy:   hin + qin = hex + ex = hex + hex + ex  ex + hex  qin = HRP mix    hex  hin = HRP mix kJ kJ  ex= qin = 0.3HRP= 1407207 kmol = 13101 kg Fu kJ  hprod = hex = 0.4HRP= 1 876 276 kmol Fu hprod = 8.4 hH 2O + 7.4hCO + 43.428 hN
2 2 satisfied for T = 1216 K 14.51 Consider the same situation as in the previous problem. Find the dew point temperature of the products. If the products in the exhaust are cooled to 10C, find the mass of water condensed per kilogram of fuel mixture. Reaction equation with 0.9 octane and 0.1 ethanol is 0.9 C8H18 + 0.1 C2H5OH + 11.55 O2 + 43.428 N2 8.4 H2O + 7.4 CO2 + 43.428 N2 yH
2 = O 8.4 = 0.1418 8.4 + 7.4 + 43.428
2 PH 2O = yH OPtot = 14.3 kPa
2O Tdew= 52.9 C yH
2 2O 10 C PH = 1.2276 = 0.012276 =
kmol x x + 7.4 + 43.428 x = 0.6317 H O= 7.77 kmol Fu mix  H O 18.015 mH
2 = O cond 2 107.414 = 1.303 kmol Fu mix kmol 1431 14.52 Calculate the irreversibility for the process described in Problem 14.19. From solution 14.19, Reactants: 2 C(s) + 2 O2 at 25C, 200 kPa 1 Product mixture: 1 CO2 + 1 CO + O2 at 1000 K, 838.4 kPa 2 & 1Q2 = 453442 kJ Reactants: SR = 2(5.740) + 2(205.148  8.31451 ln Products: ni CO2 CO O2 1.0 1.0 0.5 yi 0.40 0.40 0.20  si 269.299 234.538 243.579 Rln yiP P0 259.238 224.477 239.281 Si 200 ) = 410.250 kJ/K 100 10.061 10.061 4.298 SP = 1.0(259.238) + 1.0(224.477) + 0.5(239.281) = 603.355 kJ/K I = T0(SP  SR)  1Q2 = 298.15(603.355  410.250)  (453442) = +511 016 kJ 1432 14.53 Pentane gas at 25C, 150 kPa enters an insulated SSSF combustion chamber. Sufficient excess air to hold the combustion products temperature to 1800 K enters separately at 500 K, 150 kPa. Calculate the percent theoretical air required and the irreversibility of the process per kmol of pentane burned. C5H12 + 8XO2 + 30.08XN2 5CO2 + 6H2O + 8(X1)O2 + 30.08XN2 1st Law: Qcv + HR = HP + Wcv; Wcv = 0 , Qcv = 0 => HR = HP o  Reactants: C5H12 : h f from A.9 and h 500 for O2 and N2 from A.8 o   HR =(h f )C5H12 + 8X h O2 + 30.08X h N2 = 146 500 + 8X 6086 + 30.08 X 5911 = 226 491 X  146 500 o  o    HP = 5(h f + h )CO2 + 6(h f + h )H2O + 8(X1) h O2 + 30.08 X h N2 = 5(393522 + 79432) + 6(241826 + 62693) + 8(X1) 51674 + 30.08 X 48979 = 1 886 680 X  3 058 640 Energy Eq. solve for X; => X = 1.754
o b) Reactants: Pi = 150 kPa, Po = 100 kPa,  f s ni yi o / o s f s 500 348.945 220.693 206.74  yiPi  R ln Po 3.371 9.604 1.411 Si kJ kmol K C5H12 O2 N2 1 8X 30.08 X 1 0.21 0.79 352.316 211.089 208.151 SR = niSi = 14410.249 kJ/K Products: Pe = 150 kPa, Po = 100 kPa ni yi o s 1800 302.969 259.452 264.797 248.304  yiPe  R ln Po 18.550 17.027 16.988 1.045 Si kJ kmol K CO2 H2O O2 N2 5 6 8(X1) 30.08X 0.0716 0.086 0.0864 0.756 321.519 276.479 281.785 247.259 SP = niSi = 17732.073 kJ/K; I = To(SP  SR) = 298.15(17 732.07  14 410.25) = 990 MJ 1433 14.54 Consider the combustion of methanol, CH3OH, with 25% excess air. The combustion products are passed through a heat exchanger and exit at 200 kPa, 40C. Calculate the absolute entropy of the products exiting the heat exchanger per kilomole of methanol burned, using appropriate reference states as needed. CH3OH + 1.25 1.5 (O2 + 3.76 N2) CO2 + 2 H2O + 0.38 O2 + 7.05 N2 Products exit at 40 oC, 200 kPa, check for saturation: nV MAX 7.384 yV MAX = = = 200 P nV MAX + 1 + 0.38 + 7.05 nV = nV MAX = 0.323 Gas mixture: ni CO2 H 2O O2 N2 1.0 0.323 0.38 7.05 yi 0.1143 0.0369 0.0434 0.8054
i i PG nLIQ = 1.677  si 215.633 190.485 206.592 193.039  yiP Rln P0 +12.270 +21.671 +20.322 3.964 Si 227.903 212.156 226.914 189.075 SGAS MIX = n S = 1715.64 kJ/K sLIQ = 69.950 + 18.015(0.5725  0.3674) = 73.645 SLIQ = 1.677 73.645 = 123.50 kJ/K SPROD = 1715.64 + 123.50 = 1839.14 kJ/K 1434 14.55 The turbine in Problem 14.23 is adiabatic. Is it reversible, irreversible, or impossible? Inlet to the turbine is the exit from the mixing of air and fuel at 1 MPa. From solution to 14.23, we have: CP C H = 43.43, CP C H = 74.06, Tturbine,in = 548.7 K
2 2 3 8 C2H4 + C3H8 + 9.6 O2 + 36.096 N2 5 CO2 +6 H2O + 1.6 O2 + 36.096 N2 dQ Sex  Sin = + Sgen = Sgen T Inlet: 1 MPa, 548.7 K ni C2H4 C3H8 O2 N2 1 1 9.6 36.096 yi 0.02097 0.02097 0.2013 0.7568  SFu = Si + CP Fu lnT/T0  si  yiP Rln P0 245.82 315.09 223.497 209.388 12.989 12.989 5.816 16.828 Si 258.809 328.079 217.681 192.56 Sin = 258.809 + 328.079 + 9.6 217.681 + 36.096 192.56 = 9627.3 ni CO2 H 2O O2 N2 Sex 5 6 1.6 36.096 yi 0.1027 0.1232 0.0329 0.7413  si 257.496 223.826 235.92 221.016  yiP Rln P0 18.925 17.409 28.399 2.489 Si 276.421 241.235 264.319 223.505 = 5 276.421 + 6 241.235 1.6 264.319 + 36.096 223.505 = 11320 2kmol Fu K
kJ Sgen = Sex  Sin = 1693 2kmol Fu K > 0 kJ Possible, but one should check the state after combustion to account for generation by combustion alone and then the turbine expansion separately. 1435 14.56 Saturated liquid butane enters an insulated constant pressure combustion chamber at 25C, and x times theoretical oxygen gas enters at the same P and T. The combustion products exit at 3400 K. With complete combustion find x. What is the pressure at the chamber exit? and what is the irreversibility of the process? Butane: T1 = To = 25oC, sat liq., x1 = 0, Tc = 425.2 K, Pc = 3.8 MPa Fig. D.1: Tr1 = 0.7, Pr1 = 0.1, P1 = Pr1Pc = 380 kPa Figs. D.2 and D.3:  *    *  h 1  h 1 f = 4.85 RTc , (s 1  s 1)f = 6.8 R Oxygen: T2 = To = 25oC, X  Theoretical O2 Products: T3 = 3400 K, Assumes complete combustion C4H10 + 6.5XO2 4CO2 + 5H2O + 6.5(X1)O2 1st Law: Q cv + HR = HP + Wcv; Qcv = 0, Wcv = 0 o  HR = n(h f + h )C4H10 = 1(126200 + 17146) = 143346 kJ o  n(h f + h )CO2 = 4(393522 + 177836) = 862744 kJ o  n(h f + h )H2O = 5(241826 + 149073) = 463765 kJ o  n(h f + h )O2 = 6.5(X1)(0 + 114101) = (X1)741656.5 kJ o  Products: CO2 H2O O2 HP = ni (h f + h )i = 741656.5X  2068165.5
X = 2.594 Pe = Pi = 380 kPa HP = HR solve for X; Assume that the exit pressure equals the inlet pressure:  sC P1 o   P1 * s = [ f  R ln  (s 1   1)f] ;  O = [ o  R ln ] s s s 4H10 2 Po Po + [205.48  11.10] 6.5 2.594 = 3516.45 kJ/K SR = SC4H10 + SO2 = [306.647  11.10  56.539] Products: ni yi o si 341.988 293.550 289.499  yiPe  R ln Po 2.016 0.158 5.904 Si kJ kmol K CO2 H2O O2 4 5 10.368 0.2065 0.2582 0.5353 344.004 293.708 283.595 SP = niSi = 5784.87 kJ/K; I = To(SP  SR) = 298.15 (5784.87  3516.45) = 676329 kJ 1436 14.57 An inventor claims to have built a device that will take 0.001 kg/s of water from the faucet at 10C, 100 kPa, and produce separate streams of hydrogen and oxygen gas, each at 400 K, 175 kPa. It is stated that this device operates in a 25C room on 10kW electrical power input. How do you evaluate this claim? Liq H2O 10oC, 100 kPa 0.001 kg/s H 2O H 2 + 2 O 2 Hi  He = [285830 + 18.015(42.01  104.89)]  2961  2 (3027) = 291437 kJ (Si  Se) = [69.950 + 18.015(0.151  0.3674)]  (139.219  8.3145 ln 1.75)  2 (213.873  8.3145 ln 1.75) = 173.124 WREV = (Hi  He)  T0(Si  Se) = 291437  298.15(173.124) = 239820 kJ . WREV = (0.001/18.015)(239820) = 13.31 kW . . . I = WREV  WCV = 13.31  (10) < 0 Impossible 14.58 Two kilomoles of ammonia are burned in an SSSF process with x kmol of oxygen. The products, consisting of H2O, N2, and the excess O2, exit at 200C, 7 MPa. a. Calculate x if half the water in the products is condensed. b. Calculate the absolute entropy of the products at the exit conditions. 2NH3 + xO2 3H2O + N2 + (x  1.5)O2 Products at 200 oC, 7 MPa with a) yH2O VAP = PG/P = nH2O LIQ = nH2O VAP = 1.5 => x = 5.757 Si 181.785 187.503 185.671
1 1 1 H2 gas O2 gas . WCV = 10 kW each at 400 K 175 kPa T0 = 25 oC 1.5538 1.5 = 7 1.5 + 1 + x  1.5 yi 0.222 0.630 0.148  si 204.595 218.985 205.110 b) SPROD = SGAS MIX + SH2O LIQ Gas mixture: H 2O O2 N2 ni 1.5 4.257 1.0 Rln(yiP/P0) 22.810 31.482 19.439 SGAS MIX = 1.5(181.785) + 4.257(187.503) + 1.0(185.67) = 1256.55 kJ/K SH2O LIQ = 1.5[69.950 + 18.015(2.3223  0.3674)] = 157.75 kJ/K SPROD = 1256.55 + 157.75 = 1414.3 kJ/K 1437 14.59 Consider the SSSF combustion of propane at 25C with air at 400 K. The products exit the combustion chamber at 1200 K. It may be assumed that the combustion efficiency is 90%, and that 95% of the carbon in the propane burns to form carbon dioxide; the remaining 5% forms carbon monoxide. Determine the ideal fuelair ratio and the heat transfer from the combustion chamber. Ideal combustion process, assumed adiabatic, excess air to keep 1200 K out. C3H8 + 5x O2 + 18.8x N2 3 CO2 + 4 H2O + 5(x  1) O2 + 18.8x N2 HR = 103900 +5x(0 + 3027) + 18.8x(0 + 2971) = 103900 + 70990x HP = 3(393522 + 44473) + 4(241826 + 34506) + 5(x  1)(0 + 29761) + 18.8x(0 + 28109) = 2025232 + 677254x 1st law: HP  HR = 0 Solving, x = 3.169 FAIDEAL = 1/(23.8 3.169) = 0.01326 b) FAACTUAL = 0.01326/0.90 = 0.01473 C3H8 + 14.26 O2 + 53.62 N2 2.85 CO2 + 0.15 CO + 4 H2O + 9.335 O2 + 53.62 N2 HR = 103900 + 14.26(0 + 3027) + 53.62(0 + 2971) = +98570 kJ HP = 2.85(393522 + 44473) + 0.15(110527 + 28427) + 4(241826 + 34506) + 9.335(0 + 29761) + 53.62(0 + 28109) = 51361 kJ QCV = HP  HR = 149931 kJ 1438 14.60 Graphite, C, at P0, T0 is burned with air coming in at P0, 500 K in a ratio so the products exit at P0, 1200 K. Find the equivalence ratio, the percent theoretical air, and the total irreversibility. C + b(O2 + 3.76 N2) CO2 + (  1) O2 + 3.76 N2 HP = HR HP 1200  HR = HR  HP 44473 + (  1) 29761 + 3.7628109  ( 6086 + 3.765911) = 0  (393522) = 3.536 Sgen = SP  SR = s
PR 2 2  R ln( y) R: yO = 0.21, yN = 0.79
2 P: yO = 0.1507, yN = 0.79, yCO = 0.0593
2 2 SP = 279.39 + 2.536 250.011 + 13.295 234.227 = 4027.5 SR = 5.74 + 3.536(220.693 + 3.76 206.74) = 3534.8 For the pressure correction the term with the nitrogen drops out (same y). R  ln(y) = 8.3145(2.8235 + 1.8927  1.5606) = 26.236
PR kJ Sgen = 4027.5  3534.8 + 26.236 = 518.94 I = T0 Sgen = 154721 kmol C 1439 14.61 A gasoline engine is converted to run on propane. Assume the propane enters the engine at 25C, at the rate 40 kg/h. Only 90% theoretical air enters at 25C such that 90% of the C burns to form CO2, and 10% of the C burns to form CO. The combustion products also include H2O, H2 and N2, exit the exhaust at 1000 K. Heat loss from the engine (primarily to the cooling water) is 120 kW. What is the power output of the engine? What is the thermal efficiency? . Propane: T = 25oC, m = 40 kg/hr, M = 44.094 kg/kmol
1 Air: T2 = 25oC, 90% theoretical Air produces 90% CO 2, 10% CO Products: T3 = 1000 K, CO2, CO, H2O, H2, N2 C3H8 + 4.5O2 + 16.92N2 2.7 CO2 + 0.3CO + 3.3H2O + 0.7H2 + 16.92N2 . . nC3H8 = m/(M3600) = 0.000252 kmol/s . . . 1st Law: Q + H = H + W ; Q = 120 kW
R P o HR = nC3H8 h f = 103 900 kJ Products: CO2 CO H 2O H2 N2 o  nCO2(h f + h ) = 2.7(393522 + 33397) = 972337.5 kJ o  nCO(h f + h ) = 0.3(110527 + 21686) = 26652 kJ o  nH2O(h f + h ) = 3.3(241826 + 26000) = 712226 kJ o  nH2(h f + h ) = 0.7(0 + 20663) = 14464.1 kJ o  nN2(h f + h ) = 16.92(0 + 21463) = 363154 kJ o  HP = ni (h f + h )i = 1 333 598 kJ . . . W = Q + n(HR  HP) = 189.9 kW C3H8: Table 14.3 HRPo = 50343 kJ/kg . . HHV = nC3H8 M(HRPo) = 559.4 kW . . th = W/HHV = 0.339 1440 14.62 A small aircooled gasoline engine is tested, and the output is found to be 1.0 kW. The temperature of the products is measured to 600 K. The products are analyzed on a dry volumetric basis, with the result: 11.4% CO2, 2.9% CO, 1.6% O2 and 84.1% N2. The fuel may be considered to be liquid octane. The fuel and air enter the engine at 25C, and the flow rate of fuel to the engine is 1.5 104 kg/s. Determine the rate of heat transfer from the engine and its thermal efficiency. a C8H18 + b O2 + 3.76b N2 11.4 CO2 + 2.9 CO + c H2O + 1.6 O2 + 84.1 N2 b= 84.1 1 = 22.37, a = (11.4 + 2.9) = 1.788 3.76 8 c = 9a = 16.088 C8H18 + 12.5 O2 + 47.1 N2 6.38 CO2 + 1.62 CO + 9 H2O + 0.89 O2 + 47.1 N2 0 HR = hf C8H18 = 250105 kJ HP = 6.38(393522 + 15788) + 1.62(110527 + 10781) + 9(241826 + 12700) + 0.89(0 + 11187) + 47.1(0 + 10712) = 4119174 kJ H P  HR = 4119174  (250105) = 3869069 kJ . . HP  HR = (0.00015/114.23)(3869069) = 5.081 kW . QCV = 5.081 + 1.0 = 4.081 kW . QH = 0.00015(47893) = 7.184 kW . . TH = WNET/QH = 1.0/7.184 = 0.139 14.63 A gasoline engine uses liquid octane and air, both supplied at P , T , in a 0 0 stoichiometric ratio. The products (complete combustion) flow out of the exhaust valve at 1100 K. Assume that the heat loss carried away by the cooling water, at 100C, is equal to the work output. Find the efficiency of the engine expressed as (work/lower heating value) and the second law efficiency. C8H18 + O O2 + 3.76 N2 8 CO2 + 9 H2O + 47 N2 2 2 O = 16 + 9 O = 12.5
2 2 LHV = 44425 kg fuel LHV = 5.07476106 kmol fuel HP 1100 = 8 38885 + 9 30190 + 47 24760 = 1746510 C.V. Total engine kJ kJ 1441 Hin = Hex + W + Qloss = Hex + 2 W 2 W = Hin  Hex= HR  H= HRP + HR HP 1100 = 5.07476106 + 0  1746510 = 3328250 W = 1.664106 kmol fuel 1.664106 W th = = = 0.328 LHV 5.07476106 Find entropies in and out:  inlet: SFu = 360.575 1  SO = 205.148  8.3145 ln = 218.12 4.76 2 3.76  = 193.57 SN = 191.609  8.3145 ln 4.76 2  Sin = 360.575 + 12.5 218.12 + 47 193.57 = 12185 8  exit: SCO = 275.528  8.3145 ln = 292.82 64 2 9 SH O = 236.732  8.3145 ln = 253.04 64 2 47  SN = 231.314  8.1345 ln = 233.88 64 2  Sex = 8 292.82 + 9 253.04 + 47 233.88 = 15612 Assume the same Q loss out to 100C reservoir in the reversible case and compute Q0 :
rev Sin + Q0 /T0 = Sex + Qloss/Tres rev kJ Q0 rev = T0( Sex  Sin) + Qloss T0/Tres = 298.15(15612  12185) + 1.664106 298.15/373.15 = 2.351106 kmol fuel
kJ Hin + Q0 = Hex + Wrev + Qloss Wrev = Hin  Hex  Qloss + Q0 = Wac + Q0 = 4.015106 kmol fuel II = Wac/ Wrev =1.664106/4.015106 = 0.414
rev rev kJ rev 1442 14.64 In Example 14.16, a basic hydrogenoxygen fuel cell reaction was analyzed at 25C, 100 kPa. Repeat this calculation, assuming that the fuel cell operates on air at 25C, 100 kPa, instead of on pure oxygen at this state. Anode: 2 H2 4 e + 4 H+ Cathode: 4 H+ + 4 e 1 O2 + 2 H2O Overall: 2 H2 + 1 O2 2 H2O Example 14.16: G25C = 474283 kJ (for pure O2) For PO2 = 0.21 0.1: SO = 205.148 8.3145 ln0.21 = 218.124
2 S = 2(69.950) 2(130.678) 1(218.124) = 339.58 kJ/K G25C = 571660 298.15(339.58) = 470414 kJ E = +470414/(96487 4) = 1.219 V 1443 14.65 Consider a methaneoxygen fuel cell in which the reaction at the anode is CH4 + 2H2O CO2 + 8e + 8H+ The electrons produced by the reaction flow through the external load, and the positive ions migrate through the electrolyte to the cathode, where the reaction is 8 e + 8 H+ + 2 O2 4 H2O a. Calculate the reversible work and the reversible EMS for the fuel cell operating at 25C, 100 kPa. b. Repeat part (a), but assume that the fuel cell operates at 600 K instead of at room temperature. CH4 + 2H2O CO2 + 8e + 8H+ and 8e + 8H+ + 2CO2 4H2O Overall CH4 + 2O2 CO2 + 2H2O a) 25 oC assume all liquid H2O and all comp. at 100 kPa H25 C = 393522 + 2(285830) (74873) 0 = 890 309 kJ S25 C = 213.795 + 2(69.950) 186.251 2(205.148) =  242.852 kJ/K G25 C = 890309 298.15(242.852) =  817 903 kJ Wrev = G0 = +817903 kJ G0 +817903 0 = = 1.06 V E = 96485 8 96485 8 b) 600 K assume all gas H2O and all comp. at 100 kPa H600 K = 1(393522 + 12906) + 2(241826 + 10499)  2(0 + 9245)  1[74873 + 16.04 2.2537(600  298.2)] = 797797 kJ S600 K = 1(243.284) + 2(213.051)  1(186.251 + 16.04 2.2537 ln = 4.9606 kJ/K G600 K = H600 K  TS600 K = 797797  600(4.9606) = 800773 kJ Wrev = +800773 kJ E0= +800773 = 1.037 V 96485 8
0 0 0 0 0 0 0 0 600 )  2(226.450) 298.2 1444 Advanced Problems 14.66 A gas mixture of 50% ethane and 50% propane by volume enters a combustion chamber at 350 K, 10 MPa. Determine the enthalpy per kilomole of this mixture relative to the thermochemical base of enthalpy using Kay's rule. * hMIX O = 0.5(84740) + 0.5(103900) = 94320 kJ/kmol CP0 MIX = 0.5 30.07 1.7662 + 0.5 44.097 1.67 = 63.583 * * h350  h298 = 63.583(350  298.2) = 3294 kJ/kmol Kay's rule: TC MIX = 0.5 305.4 + 0.5 369.8 = 337.6 K PC MIX = 0.5 4.88 + 0.5 4.25 = 4.565 MPa Tr = 350/337.6 = 1.037, P r = 10/4.565 = 2.19
* From Fig. D.2: h  h = 8.3145 337.6 3.53 = 9909 kJ/kmol hMIX 350K,10MPa = 94320 + 3294  9909 = 100935 kJ/kmol 1445 14.67 A mixture of 80% ethane and 20% methane on a mole basis is throttled from 10 MPa, 65C, to 100 kPa and is fed to a combustion chamber where it undergoes complete combustion with air, which enters at 100 kPa, 600 K. The amount of air is such that the products of combustion exit at 100 kPa, 1200 K. Assume that the combustion process is adiabatic and that all components behave as ideal gases except the fuel mixture, which behaves according to the generalized charts, with Kay's rule for the pseudocritical constants. Determine the percentage of theoretical air used in the process and the dewpoint temperature of the products. Reaction equation: 0 Fuel mix: hf FUEL = 0.2(74873) + 0.8(84740) = 82767 kJ/kmol CP0 FUEL = 0.2 2.2537 16.04 + 0.8 1.7662 30.07 = 49.718 * hFUEL = 49.718(65  25) = 1989 kJ/kmol TCA = 305.4 K, TCB = 190.4 K Tc mix=282.4 K PCA = 4.88, PCB=4.60 Pc mix= 4.824 MPa Tr = 338.2/282.4 = 1.198, P r = 10/4.824 = 2.073  (h*  h)FUEL IN = 8.31451 282.4 2.18 = 5119
kJ hFUEL IN = 82767 + 1989  5119 = 85897 kmol 1st law: 1.8(393522 + 44473) + 2.8(241826 + 34506) + 3.2(x  1)(29761) + (12.03x)(28109) + 85897  (3.2x)(9245)  (12.03x)(8894) = 0 a) x = 4.104 or 410.4 % b) nP = 1.8 + 2.8 + 3.2(4.104  1) + 12.03 4.104 = 63.904 yH2O = 2.8/63.904 = PV/100 ; PV = 4.38 kPa, T = 30.5C 1446 14.68 Gaseous propane mixes with air, both supplied at 500 K, 0.1 MPa. The mixture goes into a combustion chamber and products of combustion exit at 1300 K, 0.1 MPa. The products analyzed on a dry basis are 11.42% CO2, 0.79% CO, 2.68% O2, and 85.11% N2 on a volume basis. Find the equivalence ratio and the heat transfer per kmol of fuel. C3H8 + O2 + 3.76 N2 CO2 + H2O + 3.76 N2 = 3, = 4, = + ( A/F) S = 4.76 = 23.8 hP = hP + jh(1300 K) q = hP  hR = hP  hR + jh(1300 K) = H + jh(1300 K) =5 2 = ( A/F) /( A/F) S = 1.1123 % Theoretical Air = 111.23 % jh(1300 K) = 983230.7 kmol Fu
q = 1182480 kmol Fu
kJ kJ 1447 14.69 A closed rigid container is charged with propene, C H , and 150% theoretical air 3 6 at 100 kPa, 298 K. The mixture is ignited and burns with complete combustion. Heat is transferred to a reservoir at 500 K so the final temperature of the products is 700 K. Find the final pressure, the heat transfer per kmole fuel and the total entropy generated per kmol fuel in the process. C3H6 + O O2 + 3.76 N2 3 CO2 + 3 H2O + x N2 2 Oxygen, O2 , balance: Actual Combustion: 2 O = 6 + 3 = 9
2 O = 4.5
2 = 1.5 O 2 ac = 1.5 4.5 = 6.75 C3H6 + 6.75 O2 + 25.38 N2 3 CO2 + 3 H2O + 25.38 N2 + 2.25 O2 P 2 = P1 npT2 nRT1 = 100 33.63 700 = 238.3 kPa 33.13 298.15
kJ HP 700 = 317754 + 314190 + 25.3811937 + 2.2512499 = 426916 kmol fuel U2  U1 = 1Q2  0 = H2  H1  n2RT2 + n1RT1 1 2 Q = HRP + HP 700  nPRT2 + n1RT1 = 45780 42.081 + 426916  33.63 8.3145 700
kJ + 33.13 8.3145 298.15 = 1.613106 kmol fuel  Reactants: ni yi si Rln(yiP/P0) C3H8 O2 N2 1.0 6.75 0.0302 0.2037 267.066 205.143 191.609 29.104 13.228 2.216
kJ Si 296.17 218.376 189.393 25.38 0.7661 S1= 296.17 + 6.75 218.376 + 25.38 189.393 = 6577 kmol fuel K  Products: ni yi si Rln(yiP/P0) Si CO2 H 2O O2 N2 3 3 2.25 25.38 0.0892 0.0892 0.0669 0.7547 250.752 218.739 231.465 216.865 +12.875 +12.875 +15.266  4.88 263.627 231.614 246.731 211.985 S2 = 3(263.627 + 231.614) + 2.25 246.731 + 25.38 211.985 = 7421 kJ/kmol fuel K 1.613106 kJ S2 gen= S2  S1  1Q2/Tres = 7421  6577 + = 4070 kmol fuel K 1 500 1448 14.70 Consider one cylinder of a sparkignition, internalcombustion engine. Before the compression stroke, the cylinder is filled with a mixture of air and methane. Assume that 110% theoretical air has been used, that the state before compression is 100 kPa, 25C. The compression ratio of the engine is 9 to 1. a. Determine the pressure and temperature after compression, assuming a reversible adiabatic process. b. Assume that complete combustion takes place while the piston is at top dead center (at minimum volume) in an adiabatic process. Determine the temperature and pressure after combustion, and the increase in entropy during the combustion process. c. What is the irreversibility for this process? 1 CH4 + 1.1 2 O2 + 3.76 2.2 N2 1 CO2 + 2 H2O + 0.2 O2 + 2 H2O P1 = 100 kPa, T1 = 298.2 K, V2/V1 = 1/8, Rev. Ad. s2 = s1 Assume T2 ~ 650 K TAVE ~ 475 K Table A.6: CP0 CH4 = 44.887, CP0 O2 = 30.890, CP0 N2 = 29.415 CP0 MIX = (1 44.887 + 2.2 30.890 + 8.27 29.415)/11.47 = 31.047  CV0 MIX = CP0  R = 22.732, k = CP0/CV0 = 1.366 a) T2 = T1(V1/V2)k1 = 298.2 (9)0.366 = 666.4 K (avg OK) P2 = P1(V1/V2)k = 100 (9)1.366 = 2011 kPa b) comb. 23 const. vol., Q = 0 Q = 0 = (H3  H2)  R(n3T3  n2T2) 2 3 0 H2 = 1 hf CH4 + n2 CP0 MIX (T2  T1) H2 = 74873 + 11.47 31.047(666.4 298.2) = +56246 kJ * * * * H3 = 1(393522 + hCO2) + 2(241826 + hH2O) + 0.2 hO2 + 8.27 hN2 Substituting, * * * * 1 hCO2 + 2 hH2O + 0.2 hO2 + 8.27 hN2  95.366 T3  869868 = 0 Trial & error: T3 = 2907 K 1 147072 + 2 121377 + 0.2 94315 + 8.27 89274  95.366 2907  869868 0 OK P 3 = P2 n3T3 n2T2 = P2 T3 T2 = 2011 2907 = 8772 kPa 666.4 1449 c) state 1 REAC CH4 O2 N2 ni 1 2.2 8.27 11.47 S2 = S1 = niSi = 2295.17 kJ/K state 3 PROD CO2 H 2O O2 N2 yi 0.0872 0.1918 0.7210  si 186.251 205.148 191.609  R ln(yiP/P0) +20.283 +13.730 +2.720 Si 206.534 218.878 194.329 ni 1 2 0.2 8.27 11.47 yi 0.0872 0.1744 0.0174 0.7210  si 332.213 284.753 283.213 265.726  R ln(yiP/P0) 16.916 22.680 3.516 34.480 Si 315.297 262.073 279.697 231.246 S3 = niSi = 2807.79 kJ/K I = T0(S3  S2) = 298.2(2807.79  2295.17) = 152860 kJ 1450 14.71 Consider the combustion process described in Problem 14.67. a. Calculate the absolute entropy of the fuel mixture before it is throttled into the combustion chamber. b. Calculate the irreversibility for the overall process. From solution 14.67 , fuel mixture 0.8 C2H6 + 0.2 CH4 at 65C, 10 MPa CP0 FUEL = 49.718 kJ/kmol K. Using Kay's rule: Tr1 = 1.198, Pr1 = 2.073 and x = 410.4 % theoretical air or 13.13 O2 + 49.36 N2 in at 600 K, 100 kPa out at 100 kPa, 1200 K and 1.8CO2 + 2.8H2O + 9.93O2 + 49.36N2 a)  0 FUEL = 0.2(186.251) + 0.8(229.597) s
*  8.3145(0.2 ln 0.2 + 0.8 ln 0.8) = 225.088 sTP = 49.718 ln From Fig. D.3:
* 338.2 10  8.3145 ln = 32.031 298.2 0.1 * (s s)FUEL = 1.37 8.3145 = 11.391 sFUEL = 225.088  32.031  11.391 = 181.66 kJ/kmol K b) Air at 600 K, 100 kPa ni O2 N2 13.13 49.36 yi 0.21 0.79  si 226.45 212.177 Rln(yiP/P0) +12.976 +1.96 Si 239.426 214.137 SAIR = niSi = 13713.47 kJ/K SR = 181.66 + 13713.47 = 13895.1 kJ/K Products at 1200 K, 100 kPa PROD ni yi CO2 H 2O O2 N2 1.8 2.8 9.93 49.36 0.0282 0.0438 0.1554 0.7726 o si 279.390 240.485 250.011 234.227 Rln(yiP/P0) +29.669 +26.008 +15.479 +2.145 Si 309.059 266.493 265.490 236.372 SP = niSi = 15606.1 kJ/K I = T0(SP  SR)  QCV = 298.15(15606.1  13895.1) + 0 = 510132 kJ 1451 14.72 Liquid acetylene, C2H2, is stored in a highpressure storage tank at ambient temperature, 25C. The liquid is fed to an insulated combustor/steam boiler at the steady rate of 1 kg/s, along with 140% theoretical oxygen, O2, which enters at 500 K, as shown in Fig. P14.72. The combustion products exit the unit at 500 kPa, 350 K. Liquid water enters the boiler at 10C, at the rate of 15 kg/s, and superheated steam exits at 200 kPa. a.Calculate the absolute entropy, per kmol, of liquid acetylene at the storage tank state. b. Determine the phase(s) of the combustion products exiting the combustor boiler unit, and the amount of each, if more than one. c. Determine the temperature of the steam at the boiler exit.  a) C2H2: SIG 25C = 200.958 TR = 298.2/308.3 = 0.967 => From Fig. D.1:
1 PR = 0.82
1 P1 = 0.82 6.14 = 5.03 MPa, Sliq T
1 P1  ( S*  S) 1 = 3.33R = 27.687
1 T1 = ST 0 P0  + T  R lnP1/P + ( S  S*) P = 140.695 kmol K kJ b) 1 C2H2 + 1.4 2.5 O2 2 CO2 + 1 H2O + 1 O2 H1 = 226731 + (3.56 R 308.3) = 217605 kJ H2 = 3.5(0 + 6086) = 21301 kJ Products T3 = 350 K = 76.8C PG = 41.8 kPa yV max = PG P = nV max 41.8 = 0.0836 = nV max = 0.2737 = n V gas mix 500 nV max + 2 + 1 nliq = 1  0.2737 = 0.7263 Gas Mix = 2 CO2 + 0.2737 H2O + 1 O2 c) Hliq = 0.7263(285830 + 18.015(321.5  104.9)) = 204764 kJ
3 Hgas mix = 2(393522 + 2036) + 0.2737(241826 + 1756) + 1541
3 = 847138 kJ H3 = Hliq + Hgas mix = 204 764 847 138 = 1 051 902 kJ
3 3 H3  H1  H2 = 1 290 808 kJ . . . . or H3  H1  H2 = 1290808/26.038 = 49574 kW = mH Oh4  h5 2 h5 = 42.01 + 49574 = 3346.9 T5 = 433.4 C 15 1452 14.73 Natural gas (approximate it as methane) at a ratio of 0.3 kg/s is burned with 250% theoretical air in a combustor at 1 MPa where the reactants are supplied at T0. Steam at 1 MPa, 450C at a rate of 2.5 kg/s is added to the products before they enter an adiabatic turbine with an exhaust pressure of 150 kPa. Determine the turbine inlet temperature and the turbine work assuming the turbine is reversible. CH4 + O O2 + 3.76 N2 CO2 + 2 H2O + 7.52 N2 2 2 O = 2 + 2 O = 2
2 2 => Actual O = 2 2.5 = 5
2 CH4 + 5 O2 + 18.8 N2 CO2 + 2 H2O + 3 O2 + 18.8 N2 C.V. combustor and mixing chamber HR + nH OhH O in = HP ex
2 2 . . nH2O mH2OMFu 2.5 16.043 kmol steam nH O = . = . = = 7.421 kmol fuel 2 nFu mFuMH2O 0.3 18.015 Energy equation becomes nH Ohex  hinH O + hCO + 2hH O + 3hO + 18.8hN ex 2 2 2 2 2 2 = HRP = 50010 16.043 = 802 310 hex  hinH O = hH O ex  15072.5, so then: 2 2
kJ hCO2 + 9.421hH2O + 3hO2 + 18.8hN2ex = 914163 kmol fuel Trial and error on Tex Tex = 1000 K LHS = 749956 ; Tex = 1200 K LHS = 987286 => Tex = 1100 K LHS = 867429 Tex 1139 K = Tin turbine If air then Tex turbine 700 K and Tavg 920 K. Find CP mix between 900 and 1000K. From Table A.8: 53.67 + 9.421(40.63) + 3(34.62) + 18.8(32.4) CP mix = niCPi/ ni = 32.221 = 35.673 kJ/kmol K CV mix = CP mix  R = 27.3587, kmix= 1.304 Tex turbine = 1139 (150 / 1000)0.2331= 732 K H732 = 19370.6 + 9.421(15410) + 3(13567) + 18.8(12932) = 448371 wT = Hin  Hex = Hin  Hex = 914163  448371 = 465792 kmol fuel . . ^ WT = nFuwT = mFuwT/MFu = (0.3 465792)16.043 = 8710 kW
kJ 1453 14.74 Liquid hexane enters a combustion chamber at 31C, 200 kPa, at the rate 1 kmol/s 200% theoretical air enters separately at 500 K, 200 kPa, and the combustion products exit at 1000 K, 200 kPa. The specific heat of ideal gas hexane is C = p 143 kJ/kmol K. Hexane: Tc = 507.5 K, Pc = 3010 kPa Tr1 = 0.6, Fig. D.1: Prg = 0.028, Pg1 = Pr1Pc = 84.47 kPa *    * s Figs D.2 and D.3: (h 1  h 1) f = 5.16 RTc , (s 1   1) f = 8.56 R Air: T2 = 500 K, P2 = 200 kPa, 200% theoretical air Products: T3 = 1000 K, P3 = 200 kPa  o *  * * *  a) h C6H14 = h f  (h 1  h 1)f + (h 1  h 0) + (h 0  h 0) *  * *  o h 0  h 0 = 0 , h 1  h 0 = CP (T1  To) = 858 kJ/kmol, hf = 167300 kJ/kmol *   h 1  h 1 = 5.16 8.3145 507.5 = 21773 kJ/kmol, h C6H14 = 188215 kJ/kmol T1  P1 o *   s s s s C6H14 =  T + CP ln  R ln + ( 1   1) To Po o T P o  +  ln 1   ln 1 = 387.979 + 2.85 5.763 = 385.066 kJ/kmolK R s T CP To Po o  *   = 8.568.3145 = 71.172 kJ/kmolK,  s1 s1 s C6H14 = 313.894 kJ/kmolK b) C6H14 + 19O2 + 71.44N2 6CO2 + 7H2O + 9.5O2 + 71.44N2 T3 Tc prod = yiTci = 179.3 K, Tr3 = = 5.58 Ideal Gas Tc prod c) 1st Law: Q + HR = HP + W; W=0 => Q = HP  H R    HR = (h )C6H14 + 19h O2 + 71.44 h N2 = 188 215 + 19 6086 + 71.44 5911 = 349701 kJ/kmol fuel o  o  o  o  HP = 6(h f + h )CO2 = 7(h f + h )H2O + 9.5 (h f + h )O2 + 71.44(h f + h )N2 CO2 H2O O2  (o + h ) = (393522 + 33397) = 360125 kJ/kmol hf  (o + h ) = (241826 + 26000) = 215826 kJ/kmol hf  (o + h ) = (0 + 22703) = 22703 kJ/kmol hf 1454  (o + h ) = (0 + 21463) = 21463 kJ/kmol hf . HP = 1922537 kJ; Q = 2272238 kW . . . d) I = To n (SP  SR)  Q; To = 25oC N2  o  yO2P2  yN2P2 o SR = ( )C6H14 + 19 ( 500  Rln s s )O2 + 71.44(s 500  Rln ) Po Po N2 ()C6H14 = 313.894 kJ/kmol K, (500)O2 = 220.693 kJ/kmol K s s
o (500)N2 = 206.740 kJ/kmol K, s
o . SR = 19141.9 kW/K ni CO2 H2O O2 N2 6 7 9.5 yi yO2 = 0.21, yN2 = 0.79 Products: o si 269.299 232.739 243.579  yiPe  R ln Po 17.105 15.829 13.291 Si (kJ/kmolK) 286.404 248.568 256.87 224.684 0.0639 0.0745 0.1011 71.44 0.7605 228.171 3.487 SP = ni i = 21950.1 kJ/K; s . . . I = To n (SP  SR)  Q = 3109628 kW 1455 English unit problems
14.75E Pentane is burned with 120% theoretical air in a constant pressure process at 14.7 lbf/in2. The products are cooled to ambient temperature, 70 F. How much mass of water is condensed per poundmass of fuel? Repeat the answer, assuming that the air used in the combustion has a relative humidity of 90%. C5H12 + 1.2 8 (O2 + 3.76 N2) 5 CO2 + 6 H2O + 0.96 O2 + 36.1 N2 Products cooled to 70 F, 14.7 lbf/in2 a) for H2O at 70 F: PG = 0.3632 lbf/in2 nH2O MAX 0.3632 yH2O MAX = = = 14.7 P nH2O MAX + 42.06 Solving, nH2O MAX = 1.066 < n H2O Therefore, nH2O VAP = 1.066, nH2O LIQ = 6  1.066 = 4.934 mH2O LIQ = 4.934 18.015 = 1.232 lbm/lbm fuel 72.151 PG b) Pv1 = 0.9 0.3632 = 0.3269 lbf/in2 w1 = 0.622 0.3269 = 0.014 147 14.373 28.97 (9.6 + 36.1) = 1.040 lbmol 18.015 nH2O IN = 0.014147 nH2O OUT = 1.04 + 6 = 7.04 nH2O LIQ = 7.04  1.066 = 5.974 lb mol nH2O LIQ = 5.974 18.015 = 1.492 lbm/lbm fuel 72.151 1456 14.76E The output gas mixture of a certain airblown coal gasifier has the composition of producer gas as listed in Table 14.2. Consider the combustion of this gas with 2 120% theoretical air at 14.7 lbf/in. pressure. Find the dew point of the products and the mass of water condensed per poundmass of fuel if the products are cooled 20 F below the dew point temperature? a) {3 CH4 + 14 H2 + 50.9 N2 + 0.6 O2 + 27 CO + 4.5 CO2} + 31.1 O2 + 116.9 N2 34.5 CO2 + 20 H2O + 5.2 O2 + 167.8 N2 Products: yH2O = yH2O MAX = PG 14.7 = 20 34.5 + 20 + 5.2 + 167.8 PG = 1.2923 lbf/in2 TDEW PT = 110.4 F b) At T = 90.4 F, PG = 0.7089 lbf/in2 nH2O 0.7089 yH2O = = 14.7 nH2O + 34.5 + 5.2 + 167.8 nH2O LIQ = 20  10.51 = 9.49 lb mol mH2O LIQ = 9.49(18) 3(16)+14(2)+50.9(28)+0.6(32)+27(28)+4.5(44) => n H2O = 10.51 = 0.069 lbm/lbm fuel 14.77E Pentene, C5H10 is burned with pure oxygen in an SSSF process. The products at one point are brought to 1300 R and used in a heat exchanger, where they are cooled to 77 F. Find the specific heat transfer in the heat exchanger. C5H10 + O O2 5 CO2 + 5 H2O, stoichiometric: O = 7.5
2 2 Heat exchanger in at 1300 R, out at 77 F, so some water will condense. 5 H2O (5  x)H2Oliq + x H2Ovap yH
2Omax = Pg 77 Ptot = 0.464 x = 0.03158 = x = 0.163 14.696 5+x Q q= . = 5hex  hinCO + 5hex  hinH O  (5 x)hfg H O 2 2 vap nfuel 2 = 164340 lb mol fuel
Btu 1457 14.78E A rigid vessel initially contains 2 pound mole of carbon and 2 pound mole of oxygen at 77 F, 30 lbf/in. 2. Combustion occurs, and the resulting products consist of 1 pound mole of carbon dioxide, 1 pound mole of carbon monoxide, and excess oxygen at a temperature of 1800 R. Determine the final pressure in the vessel and the heat transfer from the vessel during the process. 1 2 C + 2 O2 1 CO2 + 1 CO + O2 2 V = constant, C: solid, n1(GAS) = 2, n2(GAS) = 2.5 P 2 = P1 H1 = 0 H2 = 1(169184 + 14358) + 1(47518 + 9323) + 2(0 + 9761) = 188141 Btu
1 2 n2T2 n1T1 = 30 2.5 1800 lbf = 125.8 2 in 2 536.7 1 Q = (U2U1) = (H2H1)  n2RT2 + n1RT1 = (188141  0)  1.98589(2.5 1800  2 536.67) = 194945 Btu 14.79E In a test of rocket propellant performance, liquid hydrazine (N2H4) at 14.7 lbf/in.2, 77 F, and oxygen gas at 14.7 lbf/in.2, 77 F, are fed to a combustion chamber in the ratio of 0.5 lbm O2/lbm N2H4. The heat transfer from the chamber to the surroundings is estimated to be 45 Btu/lbm N2H4. Determine the temperature of the products exiting the chamber. Assume that only H2O, H2, and N2 are present. The enthalpy of formation of liquid hydrazine is +21647 Btu/lb mole.
1 N2H4 O2
2
1 Comb. Chamber 3 Products N2H4 + 2 O2 H2O + H2 + N2 nO /nFu (mO 32)/(mFu 32) = 0.5 ;
2 2 . QCV/mFu = 45 QCV = 45 32.045 = 1442 lb mol fu . C.V. combustion chamber: nFuh1 + nO h2 + QCV = ntoth3 2 Btu  or  H1 + H2 + QCV = HP 3 3 => HR + QCV = HP + HP 3 Btu HP = HR  HP + QCV = 21647 + 103966  1442 = 124171 lb mol fuel Trial and error on T3: T3 = 5000 HP = 120071, T3 = 5133 R T3 = 5200 HP = 126224 1458 14.80E Repeat the previous problem, but assume that saturatedliquid oxygen at 170 R is used instead of 77 F oxygen gas in the combustion process. Use the generalized charts to determine the properties of liquid oxygen. Problem the same as 14.79, except oxygen enters at 2 as saturated liquid at 170 R. At 170 R, Tr2 = 170  = 0.61 h f = 5.1 278.6 From Fig. D.2: * (h  h) = 1.98589 278.6 5.1 = 2822 Btu/lbmol HP = HR + HR  HP + QCV = 21647  0.5(2822)
3 + 0.5(0.219)(170  536.67)(32) + 103966  1442 = 121475 With HP
3 5000 R = 120071, HP 3 5200 R = 126224 T3 = 5045 R 1459 14.81E Ethene, C2H4, and propane, C3H 8 , in a 1 1 mole ratio as gases are burned with 120% theoretical air in a gas turbine. Fuel is added at 77 F, 150 lbf/in.2 and the air comes from the atmosphere, 77 F, 15 lbf/in.2 through a compressor to 150 lbf/in.2 and mixed with the fuel. The turbine work is such that the exit temperature is 1500 R with an exit pressure of 14.7 lbf/in. 2. Find the mixture temperature before combustion, and also the work, assuming an adiabatic turbine. C2H4 + C3H8 + O2(O2+3.76N2) 5CO2 + 6H2O + N2N2 = 1 O2 = 8 = 1.2 O2 = 9.6 so we have 45.696 lbmol air per 2 lbmol fuel C2H4 + C3H8 + 9.6(O2 + 3.76N2) 5CO2 + 6H2O + 1.6 O2 + 36.096N2 C.V. Compressor (air flow) wc,in = h2  h1 s2 = s1 Pr1 = 1.0907 Pr2 = Pr1P2/P1 = 10.907 T2 air = 1027.3 R wc,in = 247.81  128.38 = 119.53 Btu/lbm = 3462.4 Btu/lbmol air C.V. Mixing chamber nairhair in + nfu1hfu1 + nfu2hfu2 = (same)exit (CPF1 + CPF2)(Tex  T0) = 45.696 CP air(T2 air  Tex) CPF1 = 11.53, CPF2 = 17.95, CP air = 6.953 45.696CP airT2 + (CPF1 + CPF2)T0 = 985.6 R = Tin combust Tex = CPF1 + CPF2 + 45.696CP air Turbine work: take C.V. total and subtract compressor work. Wtotal = Hin  Hout = HR  HP,1500 = h F1 + h F2  5hCO2  6hH2O  36.096hN2  1.6hO2 f f = 22557 + (44669)  5(10557  169184)  6(8306  103966)  36.096 6925  1.6 7297.5 = 1083342 Btu/2 lbmol Fuel wT = wtot + wc,in = 1083342 + 3462.4 45.696 = 1 241 560 Btu/2 lbmol fuel 1460 14.82E One alternative to using petroleum or natural gas as fuels is ethanol (C2H5OH), which is commonly produced from grain by fermentation. Consider a combustion process in which liquid ethanol is burned with 120% theoretical air in an SSSF process. The reactants enter the combustion chamber at 77 F, and the products exit at 140 F, 14.7 lbf/in.2. Calculate the heat transfer per pound mole of ethanol, using the enthalpy of formation of ethanol gas plus the generalized tables or charts. C2H5OH + 1.2 3(O2 + 3.76N2) 2CO2 + 3H2O + 0.6O2 + 13.54N2 Products at 140 F, 14.7 lbf/in2, yH2O = 2.892/14.7 = nv/(2 + 0.6 + nv + 13.54) nv = 3.953 > 3 no condensation h = 101032 Btu/lbmol as gas f Tr = 536.67/925 = 0.58 D.2: h/RTc = 5.23 HR = 101032  5.23 1.98589 925 + 0 + 0 = 110639 Btu/ lbmol fuel HP = 2(169184 + 570) + 3(47518 + 506.5) + 0.6(443.7) + 13.54(438.4) = 472060 Btu/ lbmol fuel QCV = HP  HR = 361421 Btu/lbmol fuel 1461 14.83E Hydrogen peroxide, H2O2, enters a gas generator at 77 F, 75 lbf/in. 2 at the rate of 0.2 lbm/s and is decomposed to steam and oxygen exiting at 1500 R, 75 lbf/in.2. The resulting mixture is expanded through a turbine to atmospheric pressure, 14.7 lbf/in.2, as shown in Fig. P14.27. Determine the power output of the turbine, and the heat transfer rate in the gas generator. The enthalpy of formation of liquid H2 O2 is 80541 Btu/lb mol. H2O2 H 2 O + 2 O2 1 nFu = mFu/MFu = 0.2/34.015 = 0.00588 lbmol/s nex,mix = 1.5 nFu = 0.00882 lbmol/s
2 1 Cp mix = 3 0.445 18.015 + 3 0.219 31.999 = 7.6804 Cv mix = Cp mix  1.98588 = 5.6945 ; kmix = Cp mix/Cv mix = 1.3487 Reversible turbine T3 = T2 (P3/P2)(k1)/k = 1500 (14.7/75)0.2585 = 984.3 R  w = Cp(T2  T3) = 7.6804(1500  984.3) = 3960.8 Btu/lbmol WCV = nmix w = 0.00882 3960.8 = 34.9 Btu/s C.V. Gas generator QCV = H2  H1 = 0.00588 (103966 + 8306) + 0.00294(7297.5)  0.00588(80541) = 67.45 Btu/s 1462 14.84E In a new highefficiency furnace, natural gas, assumed to be 90% methane and 10% ethane (by volume) and 110% theoretical air each enter at 77 F, 14.7 2 lbf/in. , and the products (assumed to be 100% gaseous) exit the furnace at 100 F, 2 14.7 lbf/in. . What is the heat transfer for this process? Compare this to an older 2 furnace where the products exit at 450 F, 14.7 lbf/in. .
0.90CH 4+ 0.10C H 6 2 Furnace 77 F 110% Air Prod. 100 F 2 14.7 lbf/in HR = 0.9(32190) + 0.1(36432) = 32614 Btu 0.9 CH4 + 0.1 C2H6 + 1.1 2.15 O2 + 3.76 2.365 N2 1.1 CO2 + 2.1 H2O + 0.215 O2 + 8.892 N2 a) TP = 100 F HP = 1.1(169184 + 206) + 2.1(103966 + 185) + 0.215(162) + 8.892(160) = 402360 Btu, assuming all gas QCV = HP  HR = 369746 Btu/lb mol fuel b) TP = 450 F HP = 1.1(169184 + 3674) + 2.1(103966 + 3057) + 0.215(2688) + 8.892(2610) = 370184 Btu QCV = HP  HR = 337570 Btu/lb mol fuel 14.85E Repeat the previous problem, but take into account the actual phase behavior of the products exiting the furnace. Same as 14.84 , except possible condensation. a) 100 F, 14.7 lbf/in2 yv max = 0.9503 / 14.7 = nv max/[nv max + 10.207] nv max = 0.705 nv = 0.705; nliq = 2.1  0.705 = 1.395 Hliq = 1.395[122885 + 18.015(68.05  45.09)] = 170847 Hgas = 1.1(169184 + 206) + 0.705(103966 + 185) + 0.215(162) + 8.892(160) = 257584 HP = Hliq + Hgas = 428431 QCV = HP  HR = 395817 Btu/lbmol fuel b) Older furnace has no condensation, so same as in 14.84. 1463 14.86E Methane, CH , is burned in an SSSF process with two different oxidizers: A. Pure oxygen, 4 2 and B a mixture of O2 + x Ar. The reactants are supplied at T , O 0 P and the products in are at 3200 R both cases. Find the required equivalence 0 in case A and the amount of Argon, x, for a stoichiometric ratio in case B. ratio a) CH4 + O2 CO2 + 2H2O + (  2)O2 s = 2 for stoichiometric mixture. HP 3200 = HR = HP + HP 3200 hCO2 = 33579, hH2O = 26479, hO2 = 21860 HP = HR  H = HRP = (50010/2.326) 16.04 = 344867 P = 33579 + 2 26479 + (  2)21860 = 42817 + 21860 = 13.8175 = /s = 6.91 b) CH4 + 2O2 + 2xAr CO2 + 2H2O + 2xAr HP = 33579 + 2 26479 + 2 0.1253 39.948 (3200  536.67) = 86537 + x 26662.5 = 344867 x = 9.689 14.87E Butane gas at 77 F is mixed with 150% theoretical air at 1000 R and is burned in an adiabatic SSSF combustor. What is the temperature of the products exiting the combustor? C4H10 + 1.5 6.5(O2 + 3.76N2) 4CO2 + 5H2O + 3.25 O2 + 36.66N2 HR = HR + Hair,in HP = HP + 4hCO + 5hH
2 2O + 3.25hO + 36.66hN
2 2 HP = HR HP = HR  H + Hair,in P
45714 HP = HRP + Hair,in = 2.326 58.124 + 9.75 3366 + 36.66 3251 = 1294339 Btu/lbmol fuel = 4hCO2 + 5hH2O + 3.25hO2 + 36.66hN2 at Tad HP,3600R = 1281185 HP,3800R = 1374068 Tad = 3628 R 1464 14.88E Liquid nbutane at T , is sprayed into a gas turbine with primary air flowing at 2 150 lbf/in. , 700 R 0 a stoichiometric ratio. After complete combustion, the in products are at the adiabatic flame temperature, which is too high so secondary 2 air at 150 lbf/in. , 700 R is added, with the resulting mixture being at 2500 R. Show that T > 2500 R and find the ratio of secondary to primary air flow. ad C4H10 + 6.5(O2 + 3.76N2) 4CO2 + 5H2O + 24.44N2
Fuel Primary air Tad COMBUSTOR MIXING 2500 R 2nd.air C.V. Combustor HR = Hair + HFu = HP = HP + HP = HR + HR HP = HR  H + HR = HRP + HR P = 45344 58.124/2.326 + 6.5(1158 + 3.76 1138) = 1168433 Btu/lbmol fuel HP,2500R = 4 23755 + 5 18478 + 24.44 14855 = 550466 HP > HP,2500R Tad > 2500 R (If iteration Tad 4400 R) C.V. Mixing chamber HP + O2 2ndHair,700 = HP,2500R + O2 2ndHair,2500R O2 2nd = HP  HP,2500 Hair,2500  Hair,700 = 1168433  550466 = 9.344 71571  5437 Ratio = O2 2nd/O2 Prim. = 9.344/6.5 = 1.44 1465 2 14.89E Acetylene gas at 77 F, 14.7 lbf/in. is fed to the head of a cutting torch. Calculate the adiabatic flame temperature if the acetylene is burned with 100% theoretical air at 77 F. Repeat the answer for 100% theoretical oxygen at 77 F. a) C2H4 + 2.5 O2 + 2.5 3.76 N2 2 CO2 + 1 H2O + 9.4 N2 o HR = hf C2H2 = 97477 Btu * * * HP = 2(169184 + hCO2) + 1(103966 + hH2O) + 9.4 hN2 QCV = HP  HR = 0 * * * 2 hCO2 + 1 hH2O + 9.4 hN2 = 539811 Btu Trial and Error: TPROD = 5236 R 2 147196 + 121488 + 9.4 89348 = 1255751 OK b) C2H2 + 2.5 O2 2 CO2 + H2O HR = 97477 Btu * * HP = 2(169184 + hCO2) + 1(103966 + hH2O) * * 2 hCO2 + 1 hH2O = 539811 At 10000 R (limit of C.7): 2 135426 + 118440 = 389292 At 9500 R: 2 127734 + 111289 = 366757 or 4507/100 R change, TPROD 10000 + Difference, extrapolating 150519 13340 R 45.07 14.90E Ethene, C2H 4 , burns with 150% theoretical air in an SSSF constantpressure process with reactants entering at P , T . Find the adiabatic flame temperature. 0 0 C2H4 + 1.5 3(O2 + 3.76N2) 2CO2 + 2H2O + 1.5O2 + 16.92N2 HP = HR = HP + HP = HR HP = HR  H = HRP = 28.054 47158/2.326 = 568775 P HP = 2hCO2 + 2hH2O + 1.5hO2 + 16.92hN2 Trial and error on Tad... HP,3400R = 545437 Tad = 3510 R HP,3600R = 587736 1466 14.91E Solid carbon is burned with stoichiometric air in an SSSF process, as shown in Fig. P14.39. The reactants at T , P are heated in a preheater to T = 900 R with 0 2 the energy given by the products 0before flowing to a second heat exchanger, which they leave at T . Find the temperature of the products T , and the heat 0 transfer per lb mol of fuel (4 to 5) in the second heat exchanger. 4 a) Following the flow we have: Inlet T1, after preheater T2, after mixing and combustion chamber T3, after preheater T4, after last heat exchanger T5 = T1. b) Products out of preheater T 4. Control volume: Total minus last heat exchanger. C + O2 + 3.76N2 CO2 + 3.76N2 Energy Eq.: HR = HR = HP = HP + HP = hf CO + hCO + 3.76hN 3 3 2 2 2 h CO = 169184, HP f
2 3 4400 = 167764, HP 3 4600 = 177277 T3 = Tad.flame = 4430 R c) Control volume total. Then energy equation: HR + q = HP Btu  q = HRP = hf CO2  0 = 169184 lbmol fuel 14.92E A closed, insulated container is charged with a stoichiometric ratio of oxygen and 2 hydrogen at 77 F and 20 lbf/in. . After combustion, liquid water at 77 F is sprayed in such that the final temperature is 2100 R. What is the final pressure? H2 + 2O2 H2O ; 1 P: 1 H2O + xiH2O 3 U2  U1 = xihi = xihf liq = 1 + xiHP  HR  1 + xiRTP + 2RTR HR = , HP = 103966 + 14218.5 = 89747.5, h liq= 122885 f Substitute xi(122885 + 89747.5 + 1.98588 2100) = 89747.5  1.98588(2100  2 536.67)= 92319.2 xi = 3.187 P1V1 = nRRT1, P2V2 = npRTp P2 = P1(1 + xi)TP 1.5 T1 = 20(4.187)(2100) = 218.5 lbf/in2 1.5(536.67)
3 1467 14.93E Blast furnace gas in a steel mill is available at 500 F to be burned for the generation of steam. The composition of this gas is, on a volumetric basis, Component CH4 H2 CO CO2 N2 H2O Percent by volume 0.1 2.4 23.3 14.4 56.4 3.4 3 Find the lower heating value (Btu/ft ) of this gas at 500 F and P . 0 Of the six components in the gas mixture, only the first 3 contribute to the heating value. These are, per lb mol of mixture: 0.024 H2, 0.001 CH4, 0.233 CO For these components, 0.024 H2 + 0.001 CH4 + 0.233 CO + 0.1305 O2 0.026 H2 + 0.234 CO2 The remainder need not be included in the calculation, as the contributions to reactants and products cancel. For the lower HV(water vapor) at 500 F hRP = 0.026(103966 + 3488) + 0.234(169184 + 4229)  0.024(0 + 2101)  0.001(32190 + 0.538 16.04(50077))  0.233(47518 + 2981)  0.1305(0 + 3069) = 31257 Btu/lb mol fuel RT0 1545 536.7  = v0 = = 391.47 ft3/lb mol P0 14.7 144 LHV = 31 680 / 391.47 = 79.85 Btu/ft3 1468 14.94E Two pound moles of ammonia are burned in an SSSF process with x lb mol of oxygen. The products, consisting of H2O, N2, and the excess O2, exit at 400 F, 2 1000 lbf/in. . a. Calculate x if half the water in the products is condensed. b. Calculate the absolute entropy of the products at the exit conditions. 2NH3 + xO2 3H2O + N2 + (x  1.5)O2 Products at 400 F, 1000 lbf/in2 with nH2O LIQ= nH2O VAP= 1.5 a) yH2O VAP = x = 5.070 b) SPROD = SGAS MIX + SH2O LIQ Gas mixture: ni H 2O O2 N2 1.5 3.57 1.0 yi 0.2471 0.5881 0.1648  si 48.939 52.366 49.049 yiP P0 43.335 45.040 44.249 Si PG P = 247.1 1.5 = 1000 1.5 + 1 + x  1.5 Rln 5.604 7.326 4.800 SGAS MIX = 1.5(43.335) + 3.57(45.040) + 1.0(44.249) = 270.04 Btu/R SH2O LIQ = 1.5[16.707 + 18.015(0.5647  0.0877)] = 37.95 Btu/R SPROD = 270.04 + 37.95 = 307.99 Btu/R 1469 14.95E Graphite, C, at P , T is burned with air coming in at P , 900 R in a ratio so the 0 0 products exit at P0 , 2200 R. Find the equivalence ratio, the percent theoretical air 0 and the total irreversibility. C + (O2 + 3.76N2) CO2 + (  1)O2 + 3.76N2 HP = HR HP,2200  HR = HR  H = HRP P 19659 + (  1)13136 + 3.76 12407  (2616 + 3.76 2541) = 0  (169184) 6523 + 47616.2 = 169184 = 3.416 or 342 % theoretical air Equivalence ratio = 1/ = 0.293 S gen = sP  sR = R: yO = 0.21
2 (s  Rln y )  (s  Rln y )
i i i i i i P R yN = 0.79
2 P: yO = 0.1507
2 yN = 0.79
2 yCO = 0.0593
2 Sgen = 66.952 + 5.610 + 2.416(59.844 + 3.758) + 3.76 3.416(56.066 + 0.468)  1.371  3.416(52.686 + 3.099 + 3.76(49.353 + 0.468)) = 120.5 Btu/lbmol C R I = T0Sgen = 64677 Btu/lbmol C 1470 14.96E A small aircooled gasoline engine is tested, and the output is found to be 2.0 hp. The temperature of the products is measured and found to be 730 F. The products are analyzed on a dry volumetric basis, with the following result 11.4% CO 2, 2.9% CO, 1.6% O2 and 84.1% N2 . The fuel may be considered to be liquid octane. The fuel and air enter the engine at 77 F, and the flow rate of fuel to the engine is 1.8 lbm/h. Determine the rate of heat transfer from the engine and its thermal efficiency. a C8H18 + b O2 + 3.76b N2 11.4 CO2 + 2.9 CO + c H2O + 1.6 O2 + 84.1 N2 b = 84.1/3.76 = 22.37, a = (1/8)(11.4 + 2.9) = 1.788, c = 9a = 16.088 C8H18 + 12.5 O2 + 47.1 N2 6.38 CO2 + 1.62 CO + 9 H2O + 0.89 O2 + 47.1 N2  a) HR = hf C8H18 = 107526 Btu HP = 6.38(169184 + 6807) + 1.62(47518 + 4647) + 9(103966 + 5475) + 0.89(0+4822) + 47.1(0 + 4617) = 1770092 Btu HPHR = 1770092  (107526) = 1662566 Btu . . 1.8 HPHR = (1662566) = 26198 Btu/h 114.23 . QCV = 26198 + 2.0(2544) = 21110 Btu/h . b) QH = 1.8 20590 = 37062 . WNET = 2.0 2544 = 5088 ; TH = 5088 = 0.137 37062 1471 14.97E A gasoline engine uses liquid octane and air, both supplied at P , T , in a stoichiometric ratio. The products (complete combustion) flow out of 0 exhaust the 0 valve at 2000 R. Assume that the heat loss carried away by the cooling water, at 200 F, is equal to the work output. Find the efficiency of the engine expressed as (work/lower heating value) and the second law efficiency. C8H18 + 12.5(O2 + 3.76N2) 8CO2 + 9H2O + 47N2 LHV = 44425 114.232/2.326 = 2181753 Btu/lbmol fuel HP,2000 = 8 16982 + 9 13183 + 47 10804 = 643470 C.V. Total engine: Hin = Hex + W + Qloss = Hex + 2W W = (Hin  Hex)/2 = (HR  HP)/2 = (H  HP,2000)/2 RP = (2181753  643470)/2 = 769142 Btu/lbmol fuel TH = W/LHV = 769142/2181753 = 0.353 For 2nd law efficiency we must find reversible work Sin =  fuel + 12.5(sO + 3.76sN ) s
2 2 = 86.122 + 12.5 [48.973  1.98589 ln (1/4.76)] + 47[45.739  1.98589 ln (3.76/4.76)] = 2908.8 Btu/lbmol fuel R Sex = 8sCO + 9sH
2 2O + 47sN = 8[65.677  1.98589 ln (8/64)]
2 + 9[56.619  1.98589 ln (9/64)] + 47[55.302  1.98589 ln (47/64)] = 3731.1 Btu/lbmol fuel R Assume the same Qloss out to 200 F = 659.67 R reservoir and compute Qrev: 0 Sin + Qrev/T0 = Sex + Qloss/Tres 0 Qrev = T0(Sex  Sin) + QlossT0/Tres 0 = 536.67(3731.1  2908.8) + 769142*536.67/659.67 = 1067034 Btu/lbmol fuel Wrev = Hin  Hex  Qloss + Qrev = Wac + Qrev 0 0 = 769142 + 1067034 = 1836176 II = Wac/Wrev = 769142/1836176 = 0.419 1472 14.98EIn Example 14.16, a basic hydrogenoxygen fuel cell reaction was analyzed at 25C, 100 kPa. Repeat this calculation, assuming that the fuel cell operates on air at 77 F, 14.7 lbf/in.2, instead of on pure oxygen at this state. Anode: 2 H2 4 e + 4 H+ Cathode: 4 H+ + 4 e 1 O2 + 2 H2O Overall: 2 H2 + 1 O2 2 H2O Example 14.16 : G25C = 474283 kJ/kmol Or G77 F = 203904 Btu/lbmol PO = yO2 P = 0.21 14.7 = 3.087 lbf/in2
2  = 48.973 1.98589 ln 0.21 = 52.072 sO
2 S = 2(16.707) 2(31.186) 1(52.072) =  81.03 Btu/R H = 2(122885) 2(0) 1(0) = 245770 Btu G77 F = 245770 536.67(81.03) = 202 284 Btu E = G/N0e ne = 202284 2.326/(96485 4) = 1.219 V 151 CHAPTER 15
The correspondence between the new problem set and the previous 4th edition chapter 13 problem set. New 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 Old 1 mod 2 mod new 11 new 12 new 13 14 new 16 new 17 new 15 new 18 20 new 21 New 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 Old 22 19 23 24 25 27 28 new 29 30 new new 31 32 33 35 34 new 36 37 mod New 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 Old 38 39 40 41 42 43 45 47 49 mod 50 51 52 new new new 44 48 The problems that are labeled advanced start at number 53. The English unit problems are: New 58 59 60 61 62 63 Old 53 new 58 59 60 61 New 64 65 66 67 68 69 Old 62 63 64 new new 65 New 70 71 72 73 74 75 Old 67 68 69 70 71 72 152 15.1 Carbon dioxide at 15 MPa is injected into the top of a 5km deep well in connection with an enhanced oilrecovery process. The fluid column standing in the well is at a uniform temperature of 40C. What is the pressure at the bottom of the well assuming ideal gas behavior? (Z1Z2) = 5000 m, P1 = 15 MPa T = 40 oC = const Equilibrium at constant T wREV = 0 = g + PE = RT ln (P2/P1) + g(Z2Z1) = 0 ln (P2/P1) = 9.8075000 = 0.8287 10000.188 92313.2 Z1 1 CO 2 Z2 2 P2 = 15 exp(0.8287) = 34.36 MPa 15.2 Consider a 2kmdeep gas well containing a gas mixture of methane and ethane at a uniform temperature of 30C. The pressure at the top of the well is 14 MPa, and the composition on a mole basis is 90% methane, 10% ethane. Determine the pressure and composition at the bottom of the well, assuming an ideal gas mixture. Z1 (Z1Z2) = 2000 m, Let A = CH4, B = C2H6 1 T = 30 oC = const Z2 From section 15.1, for A to be at equilibrium between 1 and 2: WREV = 0 = n A(GA1GA2) + nAMAg(Z1Z2) Similarly, for B: WREV = 0 = n B(GB1GB2) + nBMBg(Z1Z2) Using eq. 15.10 for A: RT ln (PA2/PA1) = MAg(Z1Z2) with a similar expression for B. Now, ideal gas mixture, Substituting: ln yA2P2 yA1P1 = MAg(Z1Z2) RT and ln PA1 = yA1P, etc. = MBg(Z1Z2) RT
GAS MIX A+B 2 P1 = 14 MPa, yA1 = 0.90, yB1 = 0.10 yB2P2 yB1P1 ln (yA2P2) = ln(0.914) + 16.049.807(2000) = 2.6585 10008.3145303.2 => y A2P2 = 14.2748 ln (yB2P2) = ln(0.114) + => 30.079.807(2000) = 0.570 43 10008.3145303.2 y B2P2 = (1yA2)P2 = 1.76903 yA2 = 0.8897 Solving: P2 = 16.044 MPa & 153 15.3 Using the same assumptions as those in developing Eq. d in Example 15.1: Develop an expression for pressure at the bottom of a deep column of liquid in terms of the isothermal compressibility, T. For liquid water at 20C, T=0.0005 [1/MPa]. Use the result of the first question to estimate the pressure in the Pacific ocean at the depth of 3 km. d gT = v (1 TP) dPT d gT + g dz = 0 v(1 TP) dPT =  g dz v (1 TP) dPT + g dz = 0 and integrate P (1 P) dP = + g +Hdz T T P 0 v 0 P (1 => 1 g P  P0  T [P2  P02] = H 2 v 1 1 g P) = P0  T P02 + H 2 T 2 v v = vf 20C = 0.001002; H = 3000 m , g = 9.80665 m/s2; T = 0.0005 1/MPa P (1  2 0.0005P) = 0.101  2 0.0005 0.1012 + [9.80665 3000/0.001002] 106 = 29.462 MPa, P = 29.682 MPa which is close to P
1 1 15.4 Calculate the equilibrium constant for the reaction O2 <=> 2O at temperatures of 298 K and 6000 K. Reaction At 25 oC (298.15 K): 0 0 H0 = 2hf O  1hf O2 = 2(249 170)  1(0) = 498 340 kJ/kmol 0 0 S0 = 2sO  1sO2 = 2(161.059)  1(205.148) = 116.97 kJ/kmol K G0 = H0  TS0 = 498 340  298.15116.97 = 463 465 kJ/kmol G0 463 465 = 186.961 ln K =   = RT 8.3145298.15 At 6000 K: H0 = 2(249 170+121 264)  (0+224 210) = 516 658 kJ/kmol S0 = 2(224.597) 1(313.457) = 135.737 kJ/kmol K G0 = 516 658  6000135.737 = 297 764 kJ/kmol ln K = +297 764 = +5.969 8.31456000 O2 <=> 2O 154 15.5 Calculate the equilibrium constant for the reaction H2 <=> 2H at a temperature of 2000 K, using properties from Table A.8. Compare the result with the value listed in Table A.10. hH = 52942
2 sH = 188.419
2 o hf = 0 o h f = 217999 hH = 35375 sH = 154.279 G0 = H  TS = HRHS  HLHS T (S0RHS  S0LHS) = 2 (35375+217999) 52943 2000(2154.279182.419) = 213528 ln K = G0/RT = 213528 / (8.31452000) = 12.8407 Table A.10 ln K = 12.841 OK Plot to scale the values of lnK versus 1/T for the reaction 2CO2 <=> 2CO + O 2. Write an equation for lnK as a function of temperature. 2 CO2 2 CO + 1 O2 T(K) 2000 2400 2800 3200 3600 104 5.000 4.167 3.571 3.125 2.778 1 T ln K 13.266 7.715 3.781 0.853 1.408
8 4 0 4 8 12 0 1 2 3 4 5
4 1 10 x _ T 15.6 T(K) 4000 4500 5000 5500 6000 104 1 T 2.500 2.222 2.000 1.818 1.667
almost ln K 3.204 4.985 6.397 7.542 8.488 For the range below ~ 5000 K, ln K A + B/T Using values at 2000 K & 5000 K A = 19.5056 B = 65 543 K linear 155 15.7 Calculate the equilibrium constant for the reaction:2CO <=> 2CO + O at 3000 2 2 K using values from Table A.8 and compare the result to Table A.10. hCO = 93504 hCO = 152853
2 0 hf CO = 110527 0 hf CO = 393522
2 sCO = 273.607 sCO = 334.17
2 hO = 98013
2 sO = 284.466
2 2 2 2 2 G0 = H  TS = 2 HCO + HO 2 HCO  T (2sCO + sO  2sCO ) = 2 (93504 110527) + 98013 + 0 2(152853393522) 3000(2273.607 + 284.466  2334.17) = 55285 ln K = G0/RT = 55285/ (8.314513000) = 2.2164 Table A.10 ln K = 2.217 OK 15.8 Pure oxygen is heated from 25C to 3200 K in an SSSF process at a constant pressure of 200 kPa. Find the exit composition and the heat transfer. The only reaction will be the dissociation of the oxygen O2 2O ; K(3200) = 0.046467 Look at initially 1 mol Oxygen and shift reaction with x nO2 = 1  x; nO = 2x; ntot = 1 + x; yi = ni/ntot y2 P O 4x2 1 + x 8x2 21 K= ( ) = 2= y0 Po (1 + x)2 1  x 1  x2
2 K/8 x = 0.07599; y0 = 0.859; y0 = 0.141 1 + K/8 2 q = n0 exh0 ex + n0exhOex  h0 in = (1 + x)(y 0 h0 + y0hO)  0 x2 =
2 2 2 2 2 h0 = 106022;
2 hO = 249170 + 60767 q = 145015 kJ/kmol O2 q = q/32 = 4532 kJ/kg (=3316.5 if no reaction) 156 15.9 Pure oxygen is heated from 25C, 100 kPa to 3200 K in a constant volume container. Find the final pressure, composition, and the heat transfer. As oxygen is heated it dissociates O2 2O ln Keq = 3.069 C. V. Heater: U2  U1 = 1Q2 = H2  H1  P2v + P1v  Per mole O2: 1q2 = h2  h1 + R(T1  (n2/n1)T2) Shift x in reaction 1 to have final composition: (1  x)O 2 + 2xO n1 = 1 yO
22 n2 = 1  x + 2x = 1 + x = (1  x)/(1 + x) ; yO2 = 2x/(1 + x) P2/Po = (1 + x)T2/T1 Ideal gas and V2 = V1 P2 = P1n2T2/n1T1 Keq = e 3.069 y2 P2 O 2x 2 1 + x 1 + x T2 = ( )=( ) ( )( )( ) 1+x 1x 1 y02 Po T1 4x2 T1 3.069 = e = 0.00433 x=0.0324 1  x T2 (nO )2 = 0.9676, (nO)2 = 0.0648, n2 = 1.0324 2 q = 0.9676(106022) + 0.0648(249170 + 60767)  1 2 + 8.3145(298.15  1.0324(3200)) = 97681 kJ/kmolO2 yO 2=0.9676/1.0324 = 0.937; yO2=0.0648/1.0324 = 0.0628 2 15.10 Nitrogen gas, N2 , is heated to 4000 K, 10 kPa. What fraction of the N2 is dissociated to N at this state? N2 <=> 2 N Initial Change Equil. yN2 =
2 @ T = 4000 K, lnK = 12.671 K = 3.14x106 1 x 1x 0 2x 2x ntot = 1  x + 2x = 1 + x 2x 1x , yN = 1+x 1+x 3.14x10 6 = 4x2 10 1  x2 100 => x = 0.000886 yN = 0.0018 yN P 21 K= ; => yN2 Po nN2 = 0.99911, nN = 0.00177, yN2 = 0.9982, 157 15.11 Hydrogen gas is heated from room temperature to 4000 K, 500 kPa, at which state the diatomic species has partially dissociated to the monatomic form. Determine the equilibrium composition at this state. H2 2H x +2x (2x)2 P K= (P0)21 (1x)(1+x) Equil. nH2 = 1  x nH = 0 + 2x n =1+ x at 4000 K: ln K = 0.934 => K = 2.545 x = 0.3360 2.545 x2 = 0.127 25 = 4(500/100) 1x2 Solving, nH2 = 0.664, nH = 0.672, ntot = 1.336 yH2 = 0.497, yH = 0.503 15.12 Consider the chemical equilibrium involving H2O, H2, CO, and CO2, and no other substances. Show that the equilibrium constant at any temperature can be found using values from Table A.10 only. The reaction between the species is (1) From A.10 we find: (2) 2 H2O 2 H2 + 1 O2, Form (1) from (2) and (3) as: (1) = From A.10 1 1 (3)  (2) 2 2 ln K => ln K = (ln K3  ln K2)/2 (3) 2 CO2 2 CO + 1 O2 1 CO2 + 1 H2 1 CO + 1 H2O at any T is listed for reactions (2) and (3) 158 15.13 One kilomole Ar and one kilomole O2 is heated up at a constant pressure of 100 kPa to 3200 K, where it comes to equilibrium. Find the final mole fractions for Ar, O2, and O. The only equilibrium reaction listed in the book is dissociation of O2. So assuming that we find in Table A.10: Ar + O2 Ar + (1  x)O2 + 2x O The atom balance already shown in above equation can also be done as Species Start Total The total number of moles is 1 Ar 1 0 1 O2 0 x 1x 2x 2x so O ln(K) = 3.072 Change ntot = 1 + 1x + 2x = 2 + x yAr = 1/(2 + x); yO = 1  x/(2 + x); yO = 2x/(2 + x) 2 and the definition of the equilibrium constant (Ptot = Po) becomes K=e
3.072 4x2 = 0.04633 = = y02 (2 + x)(1  x) y2 O The equation to solve becomes from the last expression (K + 4)x2 + Kx  2K = 0 If that is solved we get x = 0.0057 0.1514 = 0.1457; yO = 0.1358; y02 = 0.3981; x must be positive yAr = 0.4661 159 15.14 A piston/cylinder contains 0.1 kmol hydrogen and 0.1 kmol Ar gas at 25C, 200 kPa. It is heated up in a constant pressure process so the mole fraction of atomic hydrogen is 10%. Find the final temperature and the heat transfer needed. When gas is heated up H2 splits partly into H as H2 2H Component Initial Shift Final H2 0.1 x and the gas is diluted with Ar Ar 0.1 0 H 0 2x 2x Total = 0.2 + x x = 0.010526 0.1x 0.1 yH = 0.1 = 2x/(0.2+x) 2x = 0.02+0.1x ntot = 0.21053 yH2 = 0.425 = [(0.1x)/(0.2+x)]; Do the equilibrium constant: K(T) = yH yH2
2 yAr = 1 rest = 0.475 P 0.01 (P0)21 = (0.425) (200) = 0.047059 100 ln (K) = 3.056 so from Table A.10 interpolate to get T = 3110 K To do the energy eq., we look up the enthalpies in Table A.8 at 3110K hH2 = 92829.1; hH = 217999 + 58447.4 = 276445.4 (= hf + h) hAr = CP(3110298.15) = 20.7863 (3110298.13) = 58447.9 (same as h for H) Now get the total number of moles to get nH = 0.021053; nH2 = ntot 1x = 0.08947; 2+x nAr = 0.1 Since pressure is constant W = PV and Q becomes differences in h Q = nh = 0.08947 92829.1 0 + 0.021053 276446.4 0 + 0.1 58447.9 = 19970 kJ 1510 15.15 Air (assumed to be 79% nitrogen and 21% oxygen) is heated in an SSSF process at a constant pressure of 100 kPa, and some NO is formed. At what temperature will the mole fraction of N.O be 0.001? 0.79 N2 + 0.21 O2 heated at 100 kPa, forms NO N2 + O2 2 NO x x +2x nN2 = 0.79  x nO2 = 0.21  x nNO = 0 ntot = 1.0 At exit, yNO = 0.001 = 2x 1.0 x = 0.0005 + 2x nN2 = 0.7895, nO2 = 0.2095 K= yNO yN2yO2
2 P 10 (P0)0 = 0.78950.2095 = 6.046106 T = 1444 K 6 or ln K = 12.016 From Table A.10, 1511 15.16 Saturated liquid butane enters an insulated constant pressure combustion chamber at 25C, and x times theoretical oxygen gas enters at the same pressure and temperature. The combustion products exit at 3400 K. Assuming that the products are a chemical equilibrium gas mixture that includes CO, what is x? Butane: T1 = 25oC, sat. liq., x1 = 0, Tc = 425.2 K, Pc = 3.8 Mpa Tr1 = 0.7, Figs. D.1and D.2, Pr1 = 0.10, P1 = Pr1Pc = 380 kPa  *  h 1  h 1f = 4.85 RTc Oxygen: T = 25oC, X * theoretical air
2 Products: T3 = 3400 K C4H10 + 6.5XO2 => ?CO 2 + 5H2O + ?O2 2CO2 <=> 2CO + O2 Initial Change Equil. 4 2a 42a 0 2a 2a 6.5(X1) a 6.5(X1) + a nCO = 4  2a, nCO = 2a, nO = 6.5(X1) + a, nH2O = 5, ntot = 2.5 + a + 6.5X 2 2 2a 4  2a 6.5(x  1) + a yCO = , yCO2 = , yO2 = 2.5 + a +6.5X 2.5 + a +6.5X 2.5 + a +6.5X yCOyO P12+12 a 2 6.5X  6.5 + a P1 K= 2 2 = 2a 6.5X  2.5 + a Po Po y CO
2 2 @ T3 = 3400 K, ln(K) = 0.346, K = 1.4134 Equation 1. a 2 6.5X  6.5 + a 1.4134 = (3.76) 2a 6.5X  2.5 + a Need a second equation: 1st Law: Qcv + HR = HP + Wcv; Qcv = 0, Wcv = 0 o  HR = (h f + h )C H = (126200  17146) = 143346 kJ
4 10 Products @ 3400 K: o  o  o  o  HP = n(h f + h )CO2 + n(h f + h )CO + n(h f + h )O2 + n(h f + h )H2O = (4  2a)(393522 + 177836) + 2a(110527 + 108440) + [6.5(X  1) + a](0 + 114101) + 5(241826 + 149073) = 463765 kJ/kmol H P = HR => 1924820 = 541299a + 741656.5 X a 0.87, X 1.96 Equation 2. Two equations and two unknowns, solve for X and a. 1512 15.17 The combustion products from burning pentane, C5H12, with pure oxygen in a stoichiometric ratio exists at 2400 K, 100 kPa. Consider the dissociation of only CO2 and find the equilibrium mole fraction of CO. C5H12 + 8 O2 5 CO2 + 6 H2O At 2400K, ln K = 7.715 K = 4.461x104 Initial Change Equil. 2 CO2 2 CO + 1 O2 5 2z 52z 0 +2z 2z 0 +z z Assuming P = Po = 0.1 MPa, and ntot = 5 + z + 6 = 11 + z y2 yO CO K= y2 CO
2 2 P 2z 2 z ( 0) = (1) = 4.461x104 ; P 5  2z 11 + z z = 0.291 yCO = 0.0515 Trial & Error (compute LHS for various values of z): nCO = 4.418;
2 nCO = 0.582; nO = 0.291
2 => 15.18 Find the equilibrium constant for the reaction 2NO + O 2NO from the 2 2 elementary reactions in Table A.10 to answer which of the nitrogen oxides, NO or NO , is the more stable at ambient conditions? What about at 2000 K? 2 2 NO + O2 2 NO2 (1) But N2 + O2 2 NO (2) N2 + 2 O2 2 NO2 (3) Reaction 1 = Reaction 3  Reaction 2 G1 = G3  G2
0 0 0 => ln K1 = ln K3  ln K2 At 25 oC, from Table A.10: or ln K1 = 41.355  (69.868) = +28.513 K1 = 2.4161012 an extremely large number, which means reaction 1 tends to go very strongly from left to right. At 2000 K: ln K1 = 19.136  (7.825) = 11.311 or K1 = 1.224105 meaning that reaction 1 tends to go quite strongly from right to left. 1513 15.19 Methane in equilibrium with carbon and hydrogen as: CH4 C + 2H2. has lnK = 0.3362 at 800 K. For a mixture at 100 kPa, 800 K find the equilibrium mole fractions of all components (CH 4, C, and H2 neglect hydrogen dissociation).Redo the molefractions for a mixture state of 200 kPa, 800 K. CH4 C + 2H2 ln K =  0.3362 nCH = 1x; nC = x; nH = 2x; ntot = 1+2x 4 2 y2 yC
H2 yCH 4 P 1+21 = K = e 0.3362 P o x 1+2x 4x (1+2x) 100 = 1x 100 1+2x 2 2 2 x3 0.71448 = = = 0.17862 2 4 (1+2x) (1x) Solve by trial and error or successive substitutions x = 0.5 x = 0.7 Interpolate LHS = 0.0625 LHS = 0.198495 x = 0.6814 x = 0.6 x 0.67 LHS = 0.11157 LHS = 0.16645 LHS = 0.17787 x = 0.682 x yC = = 0.2885 1+2x 1x yCH4 = = 0.1345 1+2x 2x yH2 = = 0.577 1+2x P 200 P = 100 = 2 o x = 0.4 x = 0.44 x3 0.71448 = = 0.044655 2 (1+2x) (1x) 4 22 x = 0.45 x = 0.446 LHS = 0.045895 LHS = 0.03292 LHS = 0.04304 1x = 0.293; yCH4 = 1+2x x 2x yC = = 0.236;yH = = 0.471 2 1+2x 1+2x 1514 15.20 A mixture of 1 kmol carbon dioxide, 2 kmol carbon monoxide, and 2 kmol oxygen, at 25C, 150 kPa, is heated in a constant pressure SSSF process to 3000 K. Assuming that only these same substances are present in the exiting chemical equilibrium mixture, determine the composition of that mixture. initial mix: Equil. mix: CO2, CO, O2 at Const. 1 CO2, 2 CO, T = 3000 K,
Pressure Q 2 O2 Reaction initial change equil. 2 CO2 1 2x (12x)
1 P = 150 kPa 2 CO 2 +2x (2+2x) + O2 2 +x (2+x) From A.10 at 3000 K: K = 0.1090 For each n > 0 1 < x < +2
2 yCOyO2 P 1 1+x 2 K= 2 (P0) = 4(12x) (2+x)(150) 5+x 100 y
CO2 or 1+x (12x) (2+x)= 0.018 167, 5+x
2 Trial & error: x = 0.521 nCO2 = 2.042 nCO = 0.958 y = 0.4559 nO2 = 1.479 CO2 = 0.2139 y nTOT = 4.479 CO yO2 = 0.3302 15.21 Repeat the previous problem for an initial mixture that also includes 2 kmol of nitrogen, which does not dissociate during the process. Same as Prob. 15.20 except also 2 kmol of inert N2 Equil.: nCO2 = (12x), nCO = (2+2x), n O2 = (2+x), n N2 = 2 yCOyO2 P 1 1+x 2 (P0) = 4(12x) (2+x)(150) K= 2 7+x 100 y
CO2 2 or ( 1+x 2 2+x ) ( )= 0.018167 Trial & error: x = 0.464 12x 7+x nO2 = 1.536 nN2 = 2.0 nCO2 = 1.928 nCO = 1.072 yCO2 = 0.295 y = 0.164 nTOT = 6.536 CO yO2 = 0.235 yN2 = 0.306 1515 15.22 Complete combustion of hydrogen and pure oxygen in a stoichiometric ratio at P , T to form water would result in a computed adiabatic flame temperature of 0 0 4990 K for an SSSF setup. a. How should the adiabatic flame temperature be found if the equilibrium reaction 2H + O H O is considered? Disregard all other possible reactions 2 2 2 (dissociations) and show the final equation(s) to be solved. b. Find the equilibrium composition at 3800 K, again disregarding all other reactions. c. Which other reactions should be considered and which components will be present in the final mixture? a) 2H2 + O2 2H2O HP = HR = Ho + HP = Ho = P R Species Initial Shift Final y2 H Keq = ( ) y2 yO P0 H2 2
2 2O H2 2 2x 22x O2 1 x 1x H 2O 2x 2x P 1 , ntot = 22x + 1x + 2x = 3x (1) + hH2O) = Hp = (22x)hH + (1x)hO + 2x(ho fH2O
2 4x2 (3x)2 3x x2(3x) = = Keq(T) Keq = (3x)2 (22x)2 1x (1x)3 ho = 241826; hH (T), hO (T), hH O(T) fH2O
2 2 2 (2) Trial and Error (solve for x,T) using Eqs. (1) and (2). yO = 0.15;
2 yH = 0.29; yH
2 2O = 0.56] b) At 3800 K Keq = e1.906 (Reaction is times 1 of table) x2(3x)(1x)3 = e1.906 = 6.726 x 0.5306 yH
2 = O 2x 1x 22x = 0.43; yO = = 0.19; yH = = 0.38 3x 3x 2 3x 2 H2 2 H O2 2 O 2 H2O H2 + 2 OH c) Other possible reactions from table A.10 1516 15.23 Gasification of char (primarily carbon) with steam following coal pyrolysis yields a gas mixture of 1 kmol CO and 1 kmol H2. We wish to upgrade the hydrogen content of this syngas fuel mixture, so it is fed to an appropriate catalytic reactor along with 1 kmol of H 2O. Exiting the reactor is a chemical equilibrium gas mixture of CO, H 2, H2O, and CO2 at 600 K, 500 kPa. Determine the equilibrium composition. Note: see Problem 15.12.
1 CO + 1 H 2 1 H 2O Chem. Equil. Mix CO, H2, H2O, CO2 600 K 500 kPa (1) 1 CO + 1 H2O 1 CO2 + 1 H2 x x +x +x (3) 2 CO2 2 CO + 1 O2 (2) 2 H2O 2 H2 + 1 O2 (1) = 1 1 (2)  (3) 2 2 1 [85.79(92.49)]= +3.35, 2 nH2O = 1x, K1 = 28.503 n H2 = 1 + x ln K1 = Equilibrium: nCO = 1x, nCO2 = 0 + x, n = 3,
28.503 = yCO2yH2 P yCO2yH2 0 K= ( ) =y y yCOyH2O P0 CO H2O x(1+x) x = 0.7794 (1x)2 n y CO 0.2206 0.0735 H2O 0.2206 0.0735 CO2 H2 0.7794 1.7794 0.2598 0.5932 % 7.35 7.35 26.0 59.3 1517 15.24 A gas mixture of 1 kmol carbon monoxide, 1 kmol nitrogen, and 1 kmol oxygen at 25C, 150 kPa, is heated in a constant pressure SSSF process. The exit mixture can be assumed to be in chemical equilibrium with CO 2, CO, O2, and N2 present. The mole fraction of CO2 at this point is 0.176. Calculate the heat transfer for the process. initial mix: Equil. mix: Const. 1 CO, 1 O2, CO2, CO, O2, N2
Pressure Q 1 N2 yCO2 = 0.176 P = 150 kPa + O2 1 x (1x) also, N2 1 0 1 reaction initial change equil. 2 CO2 0 +2x 2x 2x 3x 2 CO 1 2x (12x) yCO2 = 0.176 = x = 0.242 65 nCO2 = 0.4853 nO2 = 0.7574 yCO2 = 0.176 yO2 = 0.2747 nCO = 0.5147 nN2 = 1 yCO = 0.1867 yCOyO2 P 1 0.186720.2747 150 K= 2 (P0) = 0.1762 (100) = 0.4635 y
CO2 2 From A.10, TPROD = 3213 K From A.9, HR = 110 527 kJ HP = 0.4853(393 522 + 166 134) + 0.5147(110 527 + 101 447) + 0.7574(0 + 106 545) + 1(0 + 100 617) = +66 284 kJ QCV = HP  HR = 66 284  (110 527) = +176 811 kJ 1518 15.25 A rigid container initially contains 2 kmol of carbon monoxide and 2 kmol of oxygen at 25C, 100 kPa. The content is then heated to 3000 K at which point an equilibrium mixture of CO2, CO, and O2 exists. Disregard other possible species and determine the final pressure, the equilibrium composition and the heat transfer for the process. 2 CO + 2 O2 2 CO2 + O2 Initial 2 Shift 2x Final 22x yCO = 22x ; 4x 2 2x+x 2x yO =
2 Species: CO O2 CO2 0 2x 2x : ntot = 22x + 2x + 2x = 4x 2x ; 4x yCO =
2 2x 4x + 2xhCO  2hoCO  2hoO f f
2 2 U2  U1 = 1Q2 = H2  H1  P2v + P1v = (2  2x)hCO 2 + (2  x)hO
22 2  R(4  x)T2 + 4RT1 Keq = e2.217 =
2 2 1 2 (P ) y0 yCO o 2 y2 CO P = 4x2 4x 4T1 4(1x)2 2x (4x)T2 x 2 1 1 T2 2.217 ( ) = e = 23.092 1x 2x 4 T1 x = 0.8382; yCO = 0.102; yO = 0.368;
2 yCO = 0.53
2 P2 = P1(4x)T2/4T1 = 100(3.1618)(3000/4)(298.15) = 795.4 kPa
1 2 Q = 0.3236(110527 + 93504) + 1.1618(98013) + 1.6764(393522 + 152853)  2(110527)  2() + 8.3145(4(298.15)  3000(3.1618)) = 142991 kJ 1519 15.26 One approach to using hydrocarbon fuels in a fuel cell is to "reform" the hydrocarbon to obtain hydrogen, which is then fed to the fuel cell. As a part of the analysis of such a procedure, consider the reforming section a. Determine the equilibrium constant for this reaction at a temperature of 800 K. b. One kilomole each of methane and water are fed to a catalytic reformer. A mixture of CH4, H 2O, H2, and CO exits in chemical equilibrium at 800 K, 100 kPa; determine the equilibrium composition of this mixture. For CH4, use CP0 at ave. temp., 550 K. Table A.6, CP0 = 49.316 kJ/kmol K 0 0 a) h800 K = hf + CP0T = 74 873 + 49.316(800298.2) = 50 126 kJ/kmol 800 0 = 234.918 kJ/kmol K s800 K = 186.251 + 49.316 ln 298.2 For
0 CH4 + H2O 3H2 + CO H800 K = 3(0+14 681) + 1(110 527+15 174)  1(50 126)  1(241 826+18 002) = +222 640 kJ S800 K = 3(159.554) + 1(227.277)  1(234.918)  1(223.826) = +247.195 kJ/K G0 = H0  TS0 = 222 640  800(247.195) = +24 884 kJ G0 24 884 ln K =   = = 3.7411 => K = 0.0237 RT 8.3144800 b) CH4 + 2O H + CO 3 H2 initial change equil. nTOTAL = 2 + 2x K=
2 P 2 (3x)3x (P0) = (1x)(1x)(2+2x)2(100) 100 yCH4yH2O 0 1 x (1x) 1 x (1x) 0 +3x 3x 0 +x x yH2yCO 3 or or x x (1x) (1+x)
2 2 = 40.0237 = 0.003 51 271 Solving, x = 0.2365 x2 = 1x2 0.003 51 = 0.059 25 nCH4 = 0.7635 nH2 nCO nH2O = 0.7635 nTOT y y = 0.7095 y = 0.2365 y = 2.473 CH4 = 0.3087 = 0.3087 = 0.0956 H2O H2 CO = 0.2870 1520 15.27 In a test of a gasturbine combustor, saturatedliquid methane at 115 K is to be burned with excess air to hold the adiabatic flame temperature to 1600 K. It is assumed that the products consist of a mixture of CO , H O, N , O , and NO in 2 2 2 2 chemical equilibrium. Determine the percent excess air used in the combustion, and the percentage of NO in the products. CH4 + 2x O2 + 7.52x N2 1 CO2 + 2 H2O + (2x2) O2 + 7.52x N2 Then init ch. equil. N2 + O2 = 2 NO 7.52x 2x2 0 a a +2a (7.52xa) (2x2a) 2a nTOT = 1 + 9.52x 1600 K: ln K = 10.55, K = 2.628105 yNO P 0 4a2 5 2.62810 K = ( ) = y y = (7.52xa)(2x2a) yN2yO2 P0 N2 O2 From A.9 and B.7, HR = 1[74 873 + 16.043(274.7624.1)]+ 0 + 0 = 89 292 kJ (Air assumed 25 oC) HP = 1(393 522 + 67 569) + 2(241 826 + 52 907) + (7.52xa)(41 904) + (2x2a)(44 267) + 2a(90 291 + 43 319) = 792 325 + 403 652 x + 181 049 a Assume a ~ 0, then from Subst. 2.628105 a2 = , (13.098a)(1.483a) 4 Use this a in 1st law x= 703 042  181 0490.0113 = 1.7366 403 652
5 2 2 Also CO2 H2O 1 0 1 2 0 2 yNO HP  HR = 0 x = 1.7417 get a 0.0113 2.62810 a2 = , a = 0.0112 (13.059a)(1.4732a) 4 % excess air = 73.7 % % NO = 20.0112100 = 0.128 % 1+9.521.7366 x = 1.7366 1521 15.28 The van't Hoff equation H d ln K =  2 dTPo RT relates the chemical equilibrium constant K to the enthalpy of reaction Ho. From the value of K in Table A.10 for the dissociation of hydrogen at 2000 K and the value of Ho calculated from Table A.8 at 2000 K use van't Hoff equation to predict the constant at 2400 K. H2 2H H = 2 (35375+217999) 52942 = 453806 lnK2000 = 12.841; Assume H is constant and integrate the Van't Hoff equation 2000 H 1 1  2 lnK2400  lnK2000 = (H/RT )dT =   ( ) R T2400 T2000 2400 lnK2400 = lnK2000 + H ( 1 T2400 T2000 1 )/R
o = 12.841 + 453806 ( = 8.293 65 ) / 8.31451 = 12.841 + 4.548 12000 Table A.10 lists 8.280 (H not exactly constant) 1522 15.29 Catalytic gas generators are frequently used to decompose a liquid, providing a desired gas mixture (spacecraft control systems, fuel cell gas supply, and so forth). Consider feeding pure liquid hydrazine, N H , to a gas generator, from 2 4 which exits a gas mixture of N , H , and NH in chemical equilibrium at 100C, 2 2 3 350 kPa. Calculate the mole fractions of the species in the equilibrium mixture. Initially, 2 N2H4 1 N2 + 1 H2 + 2 NH3 then, initial change equil. yNH3 yN2yH2
3 2 N2 1 x (1x) P (P0)
2 + 3 H2 1 3x (13x) 2 NH3 2 +2x (2+2x) nTOTAL = (42x) K= = (2+2x)2(42x)2 350 2 ( ) (1x)(13x)3 100 CP0 = 17.032.130 = 36.276 At 100 oC = 373.2 K, for NH3 use A.5 0 hNH3 = 45 720 + 36.276(373.2298.2) = 42 999 kJ/kmol  0 = 192.572 + 36.276 ln 373.2 = 200.71 kJ/kmol K sNH3 298.2 Using A.8, H100 C = 2(42 999)  1(0+2188)  3(0+2179) = 94 723 kJ S100 C = 2(200.711)  1(198.155)  3(137.196) = 208.321 kJ/K G100 C = H0  TS0 = 94 723  373.2(208.321) = 16 978 kJ G0 +16 978 ln K =   = = 5.4716 RT 8.3145373.2 Therefore, => K = 237.84
0 0 0 [ (1+x)(2x) (13x) 2 237.843.52 1 = = 182.096 (1x)(13x) 16 By trial and error, x = 0.226 y = 0.2181 nN2 = 0.774 nNH3 = 2.452 N2 y = 0.0908 nH2 = 0.322 nTOT = 3.518 H2 y NH3 = 0.6911 1523 15.30 Acetylene gas at 25C is burned with 140% theoretical air, which enters the burner at 25C, 100 kPa, 80% relative humidity. The combustion products form a mixture of CO2, H 2O, N2, O2, and NO in chemical equilibrium at 2200 K, 100 kPa. This mixture is then cooled to 1000 K very rapidly, so that the composition does not change. Determine the mole fraction of NO in the products and the heat transfer for the overall process. C2H2 + 3.5 O2 + 13.16 N2 + water 2 CO2 + 1 H2O + 1 O2 + 13.16 N2 + water water: PV = 0.83.169 = 2.535 kPa nV = nA PV/PA = (3.5+13.16) 2.535/97.465 = 0.433 So, total H2O in products is 1.433. a) reaction: N2 + O2 <> 2 NO change : x x +2x K = 0.001 074 nH2O = 1.433, nO2 = 1x, n TOT = 17.593 nCO2 = 2, at 2200 K, from A.10: Equil. products: nN2 = 13.16x, nNO = 0+2x, (2x)2 = 0.001 074 => x = 0.0576 K= (1x)(13.16x) yNO = 20.0576 = 0.006 55 17.593 b) Final products (same composition) at 1000 K HR = 1(226 731) + 0.433(241 826) = 122 020 kJ HP = 2(393 522 + 33 397) + 1.433(241 826+26 000) + 0.9424(0+22 703) + 13.1024(0+21 463) + 0.1152(90 291+22 229) = 713 954 kJ QCV = HP  HR = 835 974 kJ 1524 15.31 The equilibrium reaction as: CH4 C + 2H2. has ln K = 0.3362 at 800 K and lnK = 4.607 at 600 K. By noting the relation of K to temperature show how you would interpolate ln K in (1/T) to find K at 700 K and compare that to a linear interpolation. A.10: ln K =  0.3362 at 800K ln K = 4.607 at 600K 1 1 800 1 700 800 700 lnK700 = lnK800 + (4.607+0.3362) = 0.3362 + (4.2708) 1 1 800 1 600 800 600 = 2.1665 Linear interpolation: lnK700 = lnK600 + = 4.607 + 700  600 (lnK800  lnK600) 800 600 1 (0.3362 + 4.607) = 2.4716 2 o 15.32 Use the information in Problem 15.31 to estimate the enthalpy of reaction, H , at 700 K using Van't Hoff equation with finite differences for the derivatives.  2 dlnK = [H/RT ]dT or solve for H  2 dlnK  2 lnK H = RT = RT dT T = 8.31451 700 2 0.3362 + 4.607 = 86998 kJ/kmol 800  600 [Remark: compare this to A.8 values + A.5, A.9, H = HC + 2HH  HCH = 0.61 12 (700298) + 2 11730
2 4 2.254 16.04 (700298)  (74873) = 86739 ] 1525 15.33 A step in the production of a synthetic liquid fuel from organic waste matter is the following conversion process: 1 kmol of ethylene gas (converted from the waste) at 25C, 5 MPa, and 2 kmol of steam at 300C, 5 MPa, enter a catalytic reactor. An ideal gas mixture of ethanol, ethylene, and water in chemical equilibrium leaves the reactor at 700 K, 5 MPa. Determine the composition of the mixture and the heat transfer for the reactor. IG chem. equil. o 1 C 2H 4 mixture 25 C, 5 MPa C2H5OH, C2H4, H2O 2 H 2O 700 K, 5 MPa 300 oC, 5 MPa A.6 at ~ 500 K: 1 C2H4 + 1 H2O 1 C2H5OH init 1 2 0 CP0 C2H4 = 62.3 ch. equil.
0 x (1x) x (2x) +x x a) H700 K = 1(235 000 + 115(700298.2))  1(+52 467 + 62.3(700298.2))  1(241 826 + 14 190) = 38 656 kJ S700 K = 1(282.444 + 115 ln = 110.655 kJ/K G700 K = H0  TS0 = +38 803 kJ G0 ln K =  = 6.667 RT => K = 0.001 272 = yC2H5OH yC2H4yH2O
0 0 700 700 )  1(219.330 + 62.3 ln )  1(218.739) 298.2 298.2 (PP )
0 1 x (1x)(3x)= 0.0012725.0 = 0.0636 2x 0.1 By trial and error: x = 0.0404 => C 2H5OH: n = 0.0404, y = 0.01371 C2H4: n = 0.9596, y = 0.3242, b) Reactants: C2H4: H2O: n = 1.9596 , y = 0.6621 Tr = 298.2/282.4 = 1.056, P r = 5/5.04 = 0.992  A.15: (h*h) = 1.308.3145282.4 = 3062 kJ HC2H4 = 1(+52 467  3062) = +49 405 kJ H2O, LIQ Ref. + St. Table: HH2O = 2(285830 + 18.015(2924.5104.9)) = 470 070 kJ HPROD = 0.0404(235 000 + 115(700298.2)) + 0.9596(+52 467 + 62.3(700298.2)) + 1.9596(241 826 + 14 190) = 379 335 kJ QCV = HP  HR = +41 330 kJ 1526 15.34 Methane at 25C, 100 kPa, is burned with 200% theoretical oxygen at 400 K, 100 kPa, in an adiabatic SSSF process, and the products of combustion exit at 100 kPa. Assume that the only significant dissociation reaction in the products is that of carbon dioxide going to carbon monoxide and oxygen. Determine the equilibrium composition of the products and also their temperature at the combustor exit. Combustion: CH4 + 4O2 CO2 + 2H2O + 2O2 2 CO + O2 0 +2x 2x
2 Dissociation: 2 CO2 initial change equil. 1 2x 12x ,H2O 2 0 2 inert 2 +x 2+x nTOT = 5+x yCOyO2 P x 2 P (P0)= (0.5x) (2+x)(P0) Equil. Eq'n: K = 2 5+x y
CO2 or x K (0.5x) (2+x)= (P/P0) 5+x
2 (1) 1st law: HP  HR = 0 (12x)(393 522 + hCO2) + 2x(110 527 + hCO) + 2(241 826 + hH2O) + (2+x)hO2  1(74 873)  4(3027) = 0 or (12x)hCO2 + 2xhCO + 2hH2O + (2+x)hO2 + 565 990x  814 409 = 0 Assume TP = 3256 K. From A.10: K = 0.6053 Solving (1) by trial & error, x = 0.2712 Substituting x and the h values from A.8 (at 3256 K) into (2) 0.4576168 821 + 0.5424103 054 + 2140 914 + 2.2712108 278 + 565 9900.2712  814 409 0 OK TP = 3256 K & x = 0.2712 nCO2 = 0.4576, nCO = 0.5424, nH2O = 2.0, nO2 = 2.2712 1527 15.35 Calculate the irreversibility for the adiabatic combustion process described in the previous problem. From solution of Prob. 15.34 , it is found that the product mixture consists of 0.4576 CO2, 0.5424 CO, 2.0 H2O & 2.2712 O2 at 3256 K, 100 kPa. The reactants include 1 CH4 at 25 oC, 100 kPa and 4 O2 at 400 K, 100 kPa. Reactants: SR = 1(186.251) + 4(213.873) = 1041.74 kJ/K Products: ni CO2 CO H 2O O2 0.4576 0.5424 2.0 2.2712 yi 0.0868 0.1029 0.3794 0.4309 0 si 339.278 276.660 291.099 287.749  yiP R ln P0 +20.322 +18.907 +8.058 +7.000 * Si 359.600 295.567 299.157 294.749 SP = 0.4576(359.600) + 0.5424(295.567) + 2.0(299.157) + 2.2712(294.749) = 1592.62 kJ/K I = T0(SPSR)  QCV = 298.15(1592.62  1041.74)  0 = 164 245 kJ 15.36 In rich (too much fuel) combustion the excess fuel may be broken down to give H2 and CO may form. In the products at 1200 K, 200 kPa the reaction called the water gas reaction may take place: CO2 + H2 H2O + CO Find the equilibrium constant for this reaction from the elementary reactions. (1) (2) (1) = 1 CO2 + 1 H2 1 CO + 1 H2O 2 H 2O 2 H 2 + 1 O 2, 1 1 (3)  (2) 2 2 => (3) 2 CO2 2 CO + 1 O2 ln K = (ln K3  ln K2)/2 From A.10 at 1200 K for reactions (2) and (3) ln K2 =  36.363, ln K3 =  35.736 => ln K = 0.3135 => K = 1.3682 1528 15.37 An important step in the manufacture of chemical fertilizer is the production of ammonia, according to the reaction: N 2 + 3H2 2NH3 a. Calculate the equilibrium constant for this reaction at 150C. b. For an initial composition of 25% nitrogen, 75% hydrogen, on a mole basis, calculate the equilibrium composition at 150C, 5 MPa. 1N2 + 3H2 <=> 2NH3 at 150 oC o hNH3 150 C = 45 720 + 2.1317.031(15025) = 41 186 423.2 o sNH3 150 C = 192.572 + 2.1317.031 ln = 205.272 298.2 H150 C = 2(41 186)  1(0+3649)  3(0+3636) = 96 929 kJ S150 C = 2(205.272)  1(201.829)  3(140.860) = 213.865 kJ/K G150 C = 96 929  423.2(213.865) = 6421 kJ/kmol ln K = +6421 = 1.8248, K = 6.202 8.3144423.2
0 0 o b) nNH3 = 2x, nN2 = 1x, nH2 = 33x K= yNH3 yN2yH2 or or
3 2 ( ) P P0 2 (2x)222(2x)2 P = 33(1x)4 P0
2 ( ) 2 x (1x) (2x) 1x 2 = 27 5 2 6.202 ( ) = 26165 16 0.1 x (1x)(2x) = 161.755 1x NH3 N2 H2 n 1.843 0.0785 0.2355 y 0.8544 0.0364 0.1092 Trial & Error: x = 0.9215 1529 15.38 A space heating unit in Alaska uses propane combustion is the heat supply. Liquid propane comes from an outside tank at 44C and the air supply is also taken in from the outside at 44C. The airflow regulator is misadjusted, such that only 90% of the theoretical air enters the combustion chamber resulting in incomplete combustion. The products exit at 1000 K as a chemical equilibrium gas mixture including only CO2, CO, H2O, H2, and N2. Find the composition of the products. Propane: Liquid, T1 = 44oC = 229.2 K Air: T2 = 44oC = 229.2 K, 90% Theoretical Air Products: T3 = 1000 K, CO2, CO, H2O, H2, N2 Theoretical Air: C3H8 + 5O2 + 18.8N2 => 3CO2 + 4H2O + 18.8N2 90% Theoretical Air: C3H8 + 4.5O2 + 16.92N2 => aCO2 + bCO + cH2O + dH2 + 16.92N2 Carbon: Oxygen: Reaction: Initial: Change: Equil: a+b=3 2a + b + c = 9 Where: 2 a 3 CO + H2O b x bx c x cx CO2 a x a+x + H2 d x d+x Hydrogen: c + d = 4 Chose an Initial guess such as: a = 2, b = 1, c = 4, d = 0 Note: A different initial choice of constants will produce a different value for x, but will result in the same number of moles for each product. nCO2 = 2 + x, n CO = 1  x, nH2O = 4  x, nH2 = x, nN2 = 16.92 The reaction can be broken down into two known reactions to find K (1) (2) 2CO2 2CO + O2 @ 1000 K ln(K1) = 47.052 2H2O 2H2 + O2 @ 1000 K ln(K2) = 46.321 For the overall reaction: lnK = (ln(K2)  ln(K1))/2 = 0.3655; K = 1.4412 K= yCO2yH2 P 1+111 yCO2yH2 (2 + x)x = = 1.4412 = P yCOyH2O o yCOyH2O (1  4)(4  x) => x = 0.6462 nCO = 0.3538 nH2 = 0.6462 nN2 = 16.92 nCO2 = 2.6462 nH2O = 3.3538 1530 15.39 One kilomole of carbon dioxide, CO2 , and 1 kmol of hydrogen, H2 at room temperature, 200 kPa is heated to 1200 K at 200 kPa. Use the water gas reaction, see problem 15.36, to determine the mole fraction of CO. Neglect dissociations of H2 and O2. 1 CO2 + 1 H2 1 CO + 1 H2O Initial Shift Total 1 x 1x 1 x 1x 0 +x x 0 +x x; ntot = 2 yH2O = yCO = x/2, yH2 = yCO2 = (1x)/2 From solution to problem 15.36, K = 1.3682 (x/2)(x/2) x2 =K= 1x 1x (1x)2 ( )( ) 2 2 x = 1.1697 / 2.1697 = 0.5391 yH2O = yCO = x/2 = 0.27, yH2 = yCO2 = (1x)/2 = 0.23 => x = 1.1697 1x 1531 15.40 Consider the production of a synthetic fuel (methanol) from coal. A gas mixture of 50% CO and 50% H leaves a coal gasifier at 500 K, 1 MPa, and enters a 2 catalytic converter. A gas mixture of methanol, CO and H2 in chemical equilibrium with the reaction: CO + 2H 2 <=> CH3OH leaves the converter at the same temperature and pressure, where it is known that ln K = 5.119. a. Calculate the equilibrium composition of the mixture leaving the converter. b. Would it be more desirable to operate the converter at ambient pressure? 1 CO 1 H2 Reaction: initial change equil. a) K=
CONVERTER Equil. Mix + CH3OH, CO, H2 500 K, 1 MPa CH3OH 0 +x x => x(1x) K P 2 = ( ) (12x)2 4 P0 CO 1 x (1x) 2 H2 1 2x (12x) yCH3OH P 2 x 22x 2 P 2 2 ( 0) = (1x)(12x) ( 0) P y y P
CO H2 ln K = 5.119, K = 0.005 98 => x = 0.1045 n H2 = 1  2x = 0.791 yH2 = 0.4417 x(1x) 0.005 98 1 2 = (0.1) = 0.1495 4 (12x)2 nCH3OH = x = 0.1045, yCH3OH n CO = 1x = 0.8955, yCO = 0.5000, = 0.0583, b) For P = 0.1 MPa x(1x) 0.005 98 0.1 2 = (0.1) = 0.001 495 4 (12x)2 x is much smaller (~ 0.0015) not good 1532 15.41 Consider the following coal gasifier proposed for supplying a syngas fuel to a gas turbine power plant. Fifty kilograms per second of dry coal (represented as 48 kg C plus 2 kg H) enter the gasifier, along with 4.76 kmol/s of air and 2 kmol/s of steam. The output stream from this unit is a gas mixture containing H2, CO, N2, CH4, and CO2 in chemical equilibrium at 900 K, 1 MPa. a. Set up the reaction and equilibrium equation(s) for this system, and calculate the appropriate equilibrium constant(s). b. Determine the composition of the gas mixture leaving the gasifier. a) Entering the gasifier: 4 C + 1 H2 + 1 O2 + 3.76 N2 + 2 H2O Since the chem. equil. outlet mixture contains no C, O2 or H2O, we must first consider "preliminary" reaction (or reactions) to eliminate those substances in terms of substances that are assumed to be present at equilibrium. One possibility is 4 C + 1 O2 + 2 H2O 4 CO + 2 H2 such that the "initial" composition for the equilibrium reaction is 4 CO + 3 H2 + 3.76 N2 (or convert equal amounts of CO and H2 to half of CH4 and CO2  also present at equilibrium. The final answer will be the same.) reaction initial change equil. 2 CO 4 2x (42x) nTOT = 10.76  2x For CH4 at 600 K (formula in Table A.6), CP0 = 52.22 At 900 K 0 hCH4 = 74 873 + 52.22(900  298.2) = 43 446 kJ/kmol  0 = 186.251 + 52.22 ln (900 / 298.2) = 243.936 kJ/kmol K sCH4 (The integratedequation values are 43 656 and 240.259) H900 K = 1(43 447) + 1(393 522 + 28 030)  2(110 527 + 18 397)  2(0 + 17 657) = 259 993 kJ S900 K = 1(243.935) + 1(263.646)  2(231.074)  2(163.060) = 280.687 kJ/K G900 K = 259 993  900(280.687) = 7375 kJ
0 0 0 + 2 H2 3 2x (32x) CH4 0 +x x + CO2 0 +x x also N2 3.76 0 3.76 1533 +7375 = 0.9856, 8.3145900 ln K = K = 2.679 b) K= yCH4yCO2 P 2 xx(10.762x)2 P 2 (P0) = (42x)2(32x)2 (P0) 2 2 y y
CO H2 or x(10.762x) P 1 = 0 K= 2.679 = 16.368 (42x)(32x) P 0.1 x = 1.2781 nN2 = 3.76 yN2 = 0.458 nH2 = 0.444, nCH4 = nCO2 = 1.278, yH2 = 0.054, yCH4 = yCO2 = 0.156, By trial & error, nCO = 1.444, yCO = 0.176, 1534 15.42 Ethane is burned with 150% theoretical air in a gas turbine combustor. The products exiting consist of a mixture of CO , H O, O , N , and NO in chemical 2 2 2 2 equilibrium at 1800 K, 1 MPa. Determine the mole fraction of NO in the products. Is it reasonable to ignore CO in the products? Combustion: C2H6 + 5.25 O2 + 19.74 N2 2 CO2 + 3 H2O + 1.75 O2 + 19.74 N2 Products at 1800 K, 1 MPa Equilibrium mixture: CO2, H2O, O2, N2, NO N2 initial change equil. 19.74 x 19.74x + O2 1.75 x 1.75x 2 NO 0 +2x 2x Equil. comp. nCO2 = 2, nO2 = 1.75x, nNO = 2x , nH2O = 3, nN2 = 19.74x P 0 4x2 ( ) = (19.74x)(1.75x) K = 1.19210 = yN2yO2 P0
4 yNO 2 Solving, x = 0.031 75 yNO = b) initial 20.031 75 = 0.0024 26.49 2 CO2 2 change 2a equil. 22a
2 2 CO + 0 +2a 2a O2 0 +2x 2x yCOyO2 P 1 2a 2 1 8 K = 4.19410 = 2 (P0) =(22a) (1.75x+a)0.1 26.49+a y
CO2 This equation should be solved simultaneously with the equation solved in part a) (modified to include the unknown a). Since x was found to be small and also a will be very small, the two are practically independent. Therefore, use the value x = 0.031 75 in the equation above, and solve for a. a (1a) (1.750.031 75+a) = (0.1)4.194108 26.49+a 1.0
2 Solving, a = 0.000 254 or yCO = 1.92105 negligible for most applications. 1535 15.43 One kilomole of liquid oxygen, O2, at 93 K, and x kmol of gaseous hydrogen, H , 2 at 25C, are fed to an SSSF combustion chamber. x is greater than 2, such that there is excess hydrogen for the combustion process. There is a heat loss from the chamber of 1000 kJ per kmol of reactants. Products exit the chamber at chemical equilibrium at 3800 K, 400 kPa, and are assumed to include only H2O, H2, and O. a. Determine the equilibrium composition of the products and also x, the amount of H2 entering the combustion chamber. b. Should another substance(s) have been included in part (a) as being present in the products? Justify your answer. x H2 + 1 O2 2 H2O + (x2) H2 (1) 1 H2O 1 H2 + 1 O shift a +a x2+a +a a and a > 0 a < 2 and ntot = x + a ln K2 = 1.906 ln K3 = 0.017 => K1 = 0.3823 Equil 2a (2) 2 H2O 2 H2 + 1 O2 (3) 1 O2 2 O ln K1 = 0.5( ln K2 + ln K3 ) = 0.9615 Equil.: K1 (x2+a)a 0.3823 = 0.95575 1 = (2a)(x+a) = 4 (P/Po) 1st law: Q + HR = HP , Q = (1+x)(1000) kJ * Table A.8: hIG = 5980 kJ/kmol [or = 0.922 32(93  298.2) =  6054 kJ/kmol ] Fig. D.2: Tr = 93/154.6 = 0.601, hf = 5.16R154.6 = 6633 * HR = x(0 + 0) + 1(0 + hIG + hf) = 1(5980  6633) =  12613 kJ HP = (2a)(241 826 + 171 981) + (x2+a)(0 + 119077) + a(249170 + 73424) = 119077 x + 511516 a  377844 = Q + HR = 1000  1000 x  12613 Rearrange eq. to: x + 4.2599 a = 3.03331 (1.03331 + 5.2599 a) a = 0.095575 (2a)(3.033313.2599 a) x = 2.1898 yO = a = 0.083 x+a ln K = 0.984, ln K = 0.017 Substitute it into the equilibrium eq.: Solve a = 0.198, LHS = 0.09547, yH2O = 2a = 0.755, x+a yH2 = x2+a = 0.162, x+a Other substances and reactions: H2 <=> 2 H, 2 H2O <=> H2 + 2 OH, O2 <=> 2 O, : : ln K = 0.201, All are significant as K's are of order 1. 1536 15.44 Butane is burned with 200% theoretical air, and the products of combustion, an equilibrium mixture containing only CO , H O, O , N , NO, and NO , exit from 2 2 2 2 2 the combustion chamber at 1400 K, 2 MPa. Determine the equilibrium composition at this state. Combustion: C4H10 + 13 O2 + 48.9 N2 4 CO2 + 5 H2O + 6.5 O2 + 48.9 N2 Dissociation: 1) N2 + O2 2 NO a +2a 2) change N2 + 2O2 2 NO2 b 2b +2b change a At equilibrium: nN2 = 48.9ab nH2O = 5 nCO2 = 4 nO2 = 6.5a2b nNO = 2a nNO2 = 2b nTOT = 64.4b At 1400 K, from A.10: K1 = K1 = 3.761106, K2 = 9.0261010 K2 = (2b)2(64.4b) P 1 (P0) 2 (6.5a2b) (48.9ab) (2a)2 ; (48.9ab)(6.5a2b) As K1 and K2 are both very small, with K2 << K1, the unknowns a & b will both be very small, with b << a. From the equilibrium eq.s, for a first trial a~ 1 2 1 K148.96.5 ~ 0.0173 ;b ~ 6.5 2 2 48.9 K 2 ~ 0.000 38 0.1 64.4 Then by trial & error, 3.761106 a2 = = 0.940 25106 (48.9ab)(6.5a2b) 4 b (64.4b) = (6.5a2b)2(48.9ab) nCO2 = 4 , nH2O = 5 ,
2 9.0261010( 4 2 ) 0.1 = 45.131010 Solving, a = 0.017 27, b = 0.000 379 nN2 = 48.882 , nO2 = 6.482 , yCO2 = 0.062 11 , yH2O = 0.077 64 , nNO = 0.034 54 , yNO = 0.000 55 , nNO2 = 0.000 76 yNO2 = 0.000 01 yN2 = 0.759 04 , yO2 = 0.100 65 1537 15.45 A mixture of 1 kmol water and 1 kmol oxygen at 400 K is heated to 3000 K, 200 kPa, in an SSSF process. Determine the equilibrium composition at the outlet of the heat exchanger, assuming that the mixture consists of H2O, H2, O2, and OH. Reactions and equilibrium eq'ns the same as in example 15.8 (but different initial composition). At equil.: nH2O = 12a2b, nH2 = 2a+b, nO2 = 1+a nOH = 2b, K1 = 0.002 062, 2a+b 1+a P ( ); 12a2b 2+a+b P0 nTOT = 2+a+b K2 = 0.002 893
2 P 2a+b 2b (12a2b) (P0) 2+a+b Since T = 3000 K is the same, the two equilibrium constants are the same: The two equilibrium equations are K1 = K2 = which must be solved simultaneously for a & b. If solving manually, it simplifies the solution to divide the first by the second, which leaves a quadratic equation in a & b  can solve for one in terms of the other using the quadratic formula (with the root that gives all positive moles). This reduces the problem to solving one equation in one unknown, by trial & error. Solving => b = 0.116, a = 0.038 => nH2O = 0.844, nH2 = 0.0398, nO2 = 0.962, nOH = 0.232, nTOT = 2.0778 yH2O = 0.4062, yH2 = 0.0191, yO2 = 0.4630, yOH = 0.1117 1538 15.46 One kilomole of air (assumed to be 78% nitrogen, 21% oxygen, and 1% argon) at room temperature is heated to 4000 K, 200 kPa. Find the equilibrium composition at this state, assuming that only N2, O2, NO, O, and Ar are present. 1 kmol air (0.78 N2, 0.21 O2, 0.01 Ar) heated to 4000 K, 200 kPa. Equil.: 1) N2 + O2 2 NO +2a nN2 = 0.78a nO2 = 0.21ab nAr = 0.01 2) O2 2 O nO = 2b nNO = 2a ntot = 1+b
0 4a2 (200) K1 = 0.0895 = (0.78a)(0.21ab) 100 change a a change b +2b 4b2 K2 = 2.221 = (200) (1+b)(0.21ab) 100 Divide 1st eq'n by 2nd and solve for a as function(b), using K1 P X = ( 0)= 0.0806 K2 P Get Xb2 a= [1+ 2(1+b) Also K2 b2 = = 0.277 63 (1+b)(0.21ab) 4(P/P0) Assume Therefore, nN2 = 0.7504 nO = 0.2560 yN2 = 0.6652 yO = 0.2269 nO2 = 0.0524 nNO = 0.0592 yO2 = 0.0465 yNO = 0.0525 nAr = 0.01 yAr = 0.0089 b = 0.1280 From (1), get a = 0.0296 Then, check a & b in (2) OK (2) 1+ 40.78(1+b) Xb2 (1) 1539 15.47 Acetylene gas and x times theoretical air (x > 1) at room temperature and 500 kPa are burned at constant pressure in an adiabatic SSSF process. The flame temperature is 2600 K, and the combustion products are assumed to consist of N 2, O2, CO2, H2O, CO, and NO. Determine the value of x. Combustion: C2H2 + 2.5x O2 + 9.4x N2 2 CO2 + H2O + 2.5(x1)O2 + 9.4xN2 Eq. products 2600 K, 500 kPa: N2, O2, CO2, H2O, CO & NO 2 Reactions: 1) 2 CO2 2 CO + O2 +2a +a nN2 = 9.4xb , 2) N2 + O2 2 NO b +2b nNO = 2b nCO = 2a , change 2a Equil. Comp.: change b nH2O = 1 , nO2 = 2.5x  2.5 + a  b , nCO2 = 2  2a , nTOT = 11.9x + 0.5 + a K2 = 4.913103 At 2600 K, from A.10: K1 = 3.721103, K1 3.721103 a 2 2.5x2.5+ab = =( ) ( ) 1a 11.9x+0.5+a 5 (P/P0) (2b)2 K2 = 4.91310 = (9.4b)(2.5x2.5+ab)
3 Also, from the 1st law: HP  HR = 0 where HR = 1(+226 731) + 0 + 0 = +226 731 kJ HP = (9.4xb)(0+77 963) + (2.5x2.5+ab)(0+82 225) + (22a)(393 522+128 074) + 1(241 826+104 520) + 2a(110 527+78 679) + 2b(90 291+80 034) Substituting, EQ1: 988 415x + 549 425a + 180 462b  1 100 496 = 0 which results in a set of 3 equations in the 3 unknowns x, a, b. Assume x = 1.07 Then EQ2: 3.721103 a 2 0.175+ab =( ) ( ) 1a 13.233+a 5 4.913103 b2 EQ3: = 4 (10.058b)(0.175+a+b) Solving, a = 0.1595, b = 0.0585 Then checking in EQ1, 988 4151.07 + 549 4250.1595 + 180 4620.0585  1 100 496 0 Therefore, x = 1.07 1540 15.48 One kilomole of water vapor at 100 kPa, 400 K, is heated to 3000 K in a constant pressure SSSF process. Determine the final composition, assuming that H 2O, H2, H, O2, and OH are present at equilibrium. Reactions: 1) 2 H 2O 2 H 2 + O2 +2a H2 2 H +2c nH = nTOT = 2c 1+a+b+c 2) 2 H2O H2 + 2 OH 2b +b +2b change 2a 3) +a change change c At equilibrium (3000 K, 100 kPa) nH2O = 12a2b nO2 = a nH2 = K1 K2 (P/P0) K3 2a+bc nOH = 2b 2.062103 2a+bc 2 = =( )( a ) 0 12a2b 1+a+b+c 1 (P/P ) =
2 2.893103 2a+bc 2b =( )(12a2b) 1+a+b+c 1 2.496102 (2a)2 = = 1 (2a+bc)(1+a+b+c) (P/P0) These three equations must be solved simultaneously for a, b & c: a = 0.0622, and b = 0.0570, c = 0.0327 nH2O = 0.7616 yH2O = 0.6611 nH2 = 0.1487 yH2 = 0.1291 nO2 = 0.0622 yO2 = 0.0540 nOH = 0.1140 yOH = 0.0990 nH = 0.0654 yH = 0.0568 1541 15.49 Operation of an MHD converter requires an electrically conducting gas. It is proposed to use helium gas "seeded" with 1.0 mole percent cesium, as shown in Fig. P15.49. The cesium is partly ionized (Cs Cs` + e) by heating the mixture to 1800 K, 1 MPa, in a nuclear reactor to provide free electrons. No helium is ionized in this process, so that the mixture entering the converter consists of He, Cs, Cs`, and e. Determine the mole fraction of electrons in the mixture at 1800 K, where ln K = 1.402 for the cesium ionization reaction described. Reaction: CS CS + e, init change 0.01 x 0 +x x 0 +x x
+ Also He 0.99 0 0.99 ; total: 1 + x
0 ln K = 1.402 => K = 4.0633 Equil (0.01x) K= yeyCs+ P x 0 = 0.01x yCs P ( )( x )(1+x)(PP ) or x x (0.01x)(1+x)= 4.0633 / (1/0.1) = 0.40633 x = 0.009767 ye = x/(1+x) = 0.00967 Quadratic equation: 15.50 One kilomole of argon gas at room temperature is heated to 20000 K, 100 kPa. Assume that the plasma in this condition consists of an equilibrium mixture of Ar, Ar`, Ar``, and e according to the simultaneous reactions 1) Ar Ar+ + e2) Ar+ Ar++ + e The ionization equilibrium constants for these reactions at 20000 K have been calculated from spectroscopic data as ln K1 = 3.11 and ln K2 = 4.92. Determine the equilibrium composition of the plasma. 1) Ar Ar+ + ech. a +a +a 2) Ar+ Ar++ + ech. b +b +b Equil. Comp.: nAr = 1a, nAr+ = ab, nAr++ = b, ne = a+b, nTOT = 1+a+b K1 = K2 = yAr+ ye P (ab)(a+b) 0 = (1a)(1+a+b) (1) = 22.421 yAr P ( ) yAr++ye P b(a+b) 0 = (ab)(1+a+b) (1) = 0.0073 yAr+ P ( ) By trial & error: a = 0.978 57, b = 0.014 13 nAr = 0.02143, nAr+ = 0.96444, nAr++ = 0.01413, ne = a0.9927 yAr = 0.0107, yAr+ = 0.484, yAr++ = 0.0071, ye = 0.4982 1542 15.51 Plot to scale the equilibrium composition of nitrogen at 10 kPa over the temperature range 5000 K to 15000 K, assuming that N 2 , N, N` , and e  are present. For the ionization reaction N N` + e, the ionization equilibrium constant K has been calculated from spectroscopic data as T [K] 10000 12 000 14 000 16 000 100K 6.26 102 1.51 15.1 92 1) N2 2N 2) N N+ + ene = b change a +2a change b +b +b Equil. Comp.: nN2 = 1a, nN = 2ab, nN+ = b, EQ1: K1 = yN yN2
2 (2ab) (PP )= (1a)(1+a+b)(PP )
0 0 EQ2: yN+ ye P b2 P K2 = 0 = (2ab)(1+a+b) yN P P0 ( ) ( )
4a2 10 ( ) (1a2) 100 For T < 10 000 K: b ~ 0 so neglect EQ2: K1 = 118 260 T To extrapolate K1 above 6000 K: ln K1 16.845  (from values at 5000 K & 6000 K) T(K) 5000 6000 7000 8000 10000 K1 0.0011 0.0570 0.9519 7.866 151.26 a 0.0524 0.3532 0.8391 0.9755 0.9987 yN 0.0996 0.5220 0.9125 0.9876 0.9993 yN2 0.9004 0.4780 0.0875 0.0124 0.0007 For T > 10 000 K: T(K) 10 000 12 000 14 000 16 000 K2 a 1.0 => b 0.1577 0.7244 1.5512 1.8994 b2 10 b2 K2 = ( ) = (4b2) 0.1 (2b)(2+b) 100 yN 0.8538 0.4862 0.1264 0.0258 yN2 0.0731 0.2659 0.4368 0.4871 6.26104 1.51102 0.151 0.92 Note that b 0 is not a very good approximation in the vicinity of 10 000 K. In this region, it would be better to solve the original set simultaneously for a & b. The answer would be approximately the same. 1543 15.52 Hydrides are rare earth metals, M, that have the ability to react with hydrogen to form a different substance MH with a release of energy. The hydrogen can then x be released, the reaction reversed, by heat addition to the MH . In this reaction x only the hydrogen is a gas so the formula developed for the chemical equilibrium is inappropriate. Show that the proper expression to be used instead of Eq. 15.34 is ln (PH2/Po) = G0/RT when the reaction is scaled to 1 kmol of H2. 1 M + x H2 <=> MH x 2 At equilibrium GP = GR , assume g of the solid is a function of T only. 0 0 0 gMHx = hMHx  TsMHx = gMHx , 0 0 0 gM = hM  TsM = gM 0 0 0 gH2 = hH2  TsH2 + RT ln(PH2/Po) = gH2 + RT ln(PH2/Po) GP = GR: 1 0 1 0 gMHx = gM + x gH2 = gM + x[gH2 + RT ln(PH2/Po)] 2 2 0 0 0 0 0 G0 = gMHx  gM  x gH2/2 = gMHx  gM Scale to 1 mole of hydrogen ~ 0 0 G0 = (gMHx  gM)/(x/2) = RT ln(PH2/Po) which is the desired result. 1544 Advanced Problems
15.53 Repeat Problem 15.1 using the generalized charts, instead of ideal gas behavior. (Z1Z2) = 5000 m, P1 = 15 MPa
Z1 1 CO 2 Z2 2 T = 40 oC = const 313.2 15 Tr = = 1.03, Pr1 = = 2.033 304.1 7.38 Equilibrium: wREV = 0 = g + PE 9.8075000 = 49.04 kJ/kg 1000 g2  g1 = h2  h1 T(s2  s1) = g(Z1  Z2) = From Figures D.2 and D.3, h1  h1 = RTc 3.54 = 203.4 kJ/kg ; s1  s1 = R 2.61 = 0.4931 kJ/kg K h2  h 1 = 0 ;
* * * * s2  s1 = 0  R ln(P2 /P1) = 0.18892 ln( P2 /15) * * Trial and error. Assume P2 = 55 MPa (Pr2 = 55/7.38 = 7.45) h2  h2 = RTc 3.60 = 206.8 kJ/kg ; s2  s2 = R 2.14 = 0.4043 kJ/kg K g = 206.8 + 0 + 203.4  313.2[0.4043  0.18892 ln(55/15) + 0.4931] = 45.7 Too low so assume P2 = 60 MPa (Pr2 = 60/7.38 = 8.13) h2  h2 = RTc 3.57 = 205.1 kJ/kg ; s2  s2 = R 2.11 = 0.3986 kJ/kg K g = 205.1 + 0 + 203.4  313.2[0.3986  0.18892 ln(60/15) + 0.4931] = 50.7 Make linear interpolation P2 = 58 MPa
* * * * 15.54 Derive the van't Hoff equation given in problem 15.28, using Eqs.15.12 and 15.15. Note: the d(g/T) at constant P for each component can be expressed using the relations in Eqs. 13.18 and 13.19. 0 0 0 0 Eq. 15.12: G0 = vC gC + vD gD  vA gA  vB gB Eq. 15.15: lnK = G0/RT Eq. 13.19: G0 =  T S0 1 dlnK d G0 1 dG0 G0 dG0 0 =(  )=+ =  [G T ] dT dT RT RT dT RT2 RT2 dT 1 dg 0 0 =  2 [G + T S ] used Eq.13.19 =s dT RT 1 =  2 0 RT 1545 15.55 A coal gasifier produces a mixture of 1 CO and 2H2 that is then fed to a catalytic converter to produce methane. A chemicalequilibrium gas mixture containing CH4, CO, H2, and H2O exits the reactor at 600 K, 600 kPa. Determine the mole fraction of methane in the mixture. CO Initial Change Equil. 1 x 1x + 3H2 CH4 + H2O 2 3x 23x 0 x x 0 x x n = (1  x) + (2  3x) + x + x = 3  2x yCH4yH2O P (1+113) = 2 x2 P 2 K= = P 3 (1x)(23x)3 Po yCOyH2 o  lnK =  Go/R; Go = Ho  TSo o  o  HP = nCH4 [h f + CP(T  To)] + nH2O (h f + h ) = [74873 + 2.25416.04(600  298.15)] + (241826 + 10499) = 295290 o  HR = nCO (h f + h ) = 75188 kJ 600 = HP  HR = 295290  (75188) = 220102 kJ
o  o s (s T)CH4 =  To + CPln(T/To) = 186.251 + 2.25416.04 ln(600/298.2) = 211.549 o  + nH2 (f + h ) = 1(110527 + 8942) + 3(0 + 8799) h
o o (s T )H2O = 213.051 kJ/kmolK; SP = 424.6 kJ/K o o (s T)CO = 218.321 kJ/kmolK, (s T)H2 = 151.078 kJ/kmolK
o o o o o S600 = SP  SR = (n T)CH4 + (n T )H2O  (ns T)CO  (ns T)H2 s s = (211.549 + 213.051)  (218.321 + 3 151.078) = 246.955 kJ/K Go = Ho  TSo = 220 102  600(246.955) = 71929 kJ, lnK = (71915)/(8.31451600) = 14.418 Solve for x, x = 0.6667, ntot = 1.6667, => K = 1.827106 yCH4 = 0.4 1546 15.56 Dry air is heated from 25C to 4000 K in a 100kPa constantpressure process. List the possible reactions that may take place and determine the equilibrium composition. Find the required heat transfer. Air assumed to be 21% oxygen and 79% nitrogen by volume. From the elementary reactions we have at 4000 K (A.10) (1) O2 <=> 2 O (2) N2 <=> 2 N K1 = 2.221 = y O/yO2 K2 = 3.141 106 = yN/yN2 K3 = 0.08955 = yNO/yN2 yO2
2 2 2 (3) N2 + O2 <=> 2 NO Call the shifts a,b,c respectively so we get nO2 = 0.21ac, nO = 2a, nN2 = 0.79bc, nN = 2b, nNO = 2c, ntot = 1+a+b From which the molefractions are formed and substituted into the three equilibrium equations. The result is K1 = 2.221 = y O/yO2 = 4a2/[(1+a+b)(0.21ac)] K2 = 3.141 106 = yN/yN2 = 4b2/[(1+a+b)(0.79bc)] K3 = 0.08955 = y NO/yN2 yO2 = 4c2/[(0.79bc)(0.21ac)] which gives 3 eqs. for the unknowns (a,b,c). Trial and error assume b = c = 0 solve for a from K 1 then for c from K3 and finally given the (a,c) solve for b from K2. The order chosen according to expected magnitude K1>K3>K2 a = 0.15, b = 0.000832, c = 0.0244 => nO2 = 0.0356, nO = 0.3, nN2 = 0.765, nN = 0.00167, nNO = 0.049 Q = Hex  Hin = nO2hO2 + nN2hN2 + nO(hfO + hO) + nN(hfN + hN) + nNO(hfNO + hNO)  0 = 0.0356 138705 + 0.765 130027 + 0.3(249170 + 77675) + 0.00167(472680 + 77532) + 0.049(90291 + 132671) = 214 306 kJ/kmol air [If no reac. Q = nO2hO2 + nN2hN2 = 131 849 kJ/kmol air]
2 2 2 1547 15.57 Methane is burned with theoretical oxygen in an SSSF process, and the products exit the combustion chamber at 3200 K, 700 kPa. Calculate the equilibrium composition at this state, assuming that only CO , CO, H O, H , O , and OH are 2 2 2 2 present. Combustion: Dissociation reactions: 1) 2 H 2O 2 H 2 + O2 +2a +a 2 CO2 2 CO + O2 +2c +c nO2 = nOH = a+c 2b nCO2 = nCO = nTOT = Products at 3200 K, 700 kPa K1 = 0.007 328 = ( K2 = 0.012 265 = ( K3 = 0.426 135 = ( 2a+b 2 a+c ) (3+a+b+c)(700) 22a2b 100
2 2b 2a+b ) (3+a+b+c)(700) 22a2b 100 CH4 + 2 O2 CO2 + 2 H2O 2) 2 H2O H2 + 2 OH +b +2b change 2a 3) change 2b change 2c At equilibrium: NH2O = 22a2b NH2 = 2a+b 12c 2c 3+a+b+c 2c 2 ) ( a+c )(700) 12c 3+a+b+c 100 These 3 equations must be solved simultaneously for a, b, & c. If solving by hand divide the first equation by the second, and solve for c = fn(a,b). This reduces the solution to 2 equations in 2 unknowns. Solving, a = 0.024, b = 0.1455, c = 0.236 Substance: n y H 2O 1.661 0.4877 H2 0.1935 0.0568 O2 0.260 0.0764 OH 0.291 0.0855 CO2 0.528 0.1550 CO 0.472 0.1386 1548 ENGLISH UNIT PROBLEMS
15.58ECarbon dioxide at 2200 lbf/in.2 is injected into the top of a 3mi deep well in connection with an enhanced oil recovery process. The fluid column standing in the well is at a uniform temperature of 100 F. What is the pressure at the bottom of the well assuming ideal gas behavior? (Z1Z2) = 3 miles = 15 840 ft P1 = 2200 lbf/in2, T = 100 F = const Equilibrium and ideal gas beahvior wREV = 0 = g + PE = RT ln (P2/P1) + g(Z2Z1) = 0 32.215 840 = 0.8063 32.235.1559.7 Z1 1 CO 2 Z2 2 ln (P2/P1) = P2 = 2200 exp(0.8063) = 4927 lbf/in2 15.59E Calculate the equilibrium constant for the reaction O2 <=> 2O at temperatures of 537 R and 10 000 R. Find the change in Gibbs function at the two T's from Table C.7: H0 = 2(249 170+121 264)  (0+224 210) = 516 658 kJ/kmol S0 = 2(224.597) 1(313.457) = 135.737 kJ/kmol K G0 = 516 658  6000135.737 = 297 764 kJ/kmol ln K = 537 R: +297 764 = +5.969 8.31456000 0 0 H0 = 2hf O  1hf O2 = 2 107 124 = 214 248 ; 0 0 S0 = 2sO  1sO2 = 2 38.442  48.973 = 27.911 G0 = H0  TS0 = 214 248  537 27.911 = 199 260 G0 +199 260  =  185.85 ln K =   = RT 1.98589537 0 0 10 000 R: H0 = 2hf O  1hf O2 = 2 (107 124 + 47897)  87997 = 222 045; 0 0 S0 = 2sO  1sO2 = 2 53.210  74.034 = 32.386 G0 = H0  TS0 = 222 045  10 000 32.386 = 101 815 ln K =  G0/RT = 101 815 / (1.98589 10 000) = +5.127 1549 15.60E Pure oxygen is heated from 77 F to 5300 F in an SSSF process at a constant pressure of 30 lbf/in.2. Find the exit composition and the heat transfer. The only reaction will be the dissociation of the oxygen O2 2O ; K(5300 F) = K(3200 K) = 0.046467 Look at initially 1 mol Oxygen and shift the above reaction with x nO2 = 1  x; nO = 2x; ntot = 1 + x; yi = ni/ntot y2 P O 4x2 1 + x 8x2 K= ( )21 = 2= y02 Po (1 + x)2 1  x 1  x2 K/8 x = 0.07599; y02 = 0.859; y0 = 0.141 1 + K/8 q = n02exh02ex + n0exhOex  h02in = (1 + x)(y 02h02 + y0hO)  0 h02 = 45 581; hO = 107124 + 26125 = 133 249 q = 1.076(0.859 45581 + 0.141 133249) = 62345 Btu/lbmol O2 q = q/32 = 1948 Btu/lbm (=1424 if no dissociation) x2 = 15.61E Pure oxygen is heated from 77 F, 14.7 lbf/in.2 to 5300 F in a constant volume container. Find the final pressure, composition, and the heat transfer As oxygen is heated it dissociates O2 2O C. V. Heater: Per mole O2: n1 = 1 ln Keq = 3.069 U2  U1 = 1Q2 = H2  H1  P2v + P1v  1q2 = h2  h1 + R(T1  (n2/n1)T2) (1  x)O2 + 2xO n2 = 1  x + 2x = 1 + x yO = 2x/(1 + x) Shift x in reaction final composition: yO2 = (1  x)/(1 + x) ; Ideal gas and V2 = V1 P2 = P1n2T2/n1T1 P2/Po = (1 + x)T2/T1 Keq = e 3.069 y2 P2 O 2x 2 1 + x 1 + x T2 = ( )=( ) ( )( )( ) 1+x 1x 1 y02 Po T1 4x2 T1 3.069 = e = 0.00433 x=0.0324 1  x T2 (nO2)2 = 0.9676, (nO)2 = 0.0648, n2 = 1.0324 1q2 = 0.9676(45581) + 0.0648(107124 + 26125)  + 1.98589(536.67  1.0324 5760) = 41996 Btu/lbmol O2 yO2=0.9676 / 1.0324 = 0.937; yO =0.0648/1.0324 = 0.0628 1550 15.62E Air (assumed to be 79% nitrogen and 21% oxygen) is heated in an SSSF process at a constant pressure of 14.7 lbf/in.2, and some NO is formed. At what temperature will the mole fraction of NO be 0.001? 0.79N2 + 0.21O2 N2 + O2 2 NO x x +2x heated at 14.7 lbf/in2, forms NO At exit, yNO = 0.001 nN2 = 0.79  x nO2 = 0.21  x nNO = 0 n = 1.0 yNO = 0.001 = K= yNO yN2yO2
2 + 2x 2x 1.0 x = 0.0005
6 nN2 = 0.7895, or nO2 = 0.2095 ln K = 12.016 P 10 (P0)0 = 0.78950.2095 = 6.046106 T = 1444 K = 2600 R From Table A.10, 15.63E The combustion products from burning pentane, C5H12, with pure oxygen in a stoichiometric ratio exists at 4400 R. Consider the dissociation of only CO2 and find the equilibrium mole fraction of CO. C5H12 + 8 O2 5 CO2 + 6 H2O At 4400 R, ln K = 7.226 K = 7.272x104 Assuming P = Po = 0.1 MPa, y2 yO CO K= y2 CO2
2 2 CO2 2 CO + 1 O2 Initial 5 Change Equil. 52z 0 2z 2z 0 +2z z +z P 2z 2 z ( 0) = ( )( )(1); T & E : z = 0.2673 5  2z 5 + z P nCO = 0.5346; nO = 0.2673
2 nCO = 4.4654;
2 yCO = 0.1015 1551 15.64E A gas mixture of 1 pound mol carbon monoxide, 1 pound mol nitrogen, and 1 pound mol oxygen at 77 F, 20 lbf/in.2, is heated in a constant pressure SSSF process. The exit mixture can be assumed to be in chemical equilibrium with CO2, CO, O2, and N2 present. The mole fraction of CO2 at this point is 0.176. Calculate the heat transfer for the process. initial mix: 1 CO, 1 O2, 1 N2
Q Const. Pressure Equil. mix: CO2, CO, O2, N2 yCO2 = 0.176 P = 20 lbf/in2 + O2 1 x (1x) also, N2 1 0 1 reaction initial change equil. 2 CO2 0 +2x 2x 2x 3x 2 CO 1 2x (12x) yCO2 = 0.176 = x = 0.242 65 y = 0.176 nCO2 = 0.4853 nO2 = 0.757 35 CO2 yCO = 0.186 67 nCO = 0.5147 nN2 = 1 y = 0.274 67 O2 yCOyO2 P 1 0.186 6720.274 67 20 (P0) = (14.504)= 0.426 07 K= 2 0.1762 y
CO2 2 Since Table A.10 corresponds to a pressure P0 of 100 kPa, which is 14.504 lbf/in2. Then, from A.10, TPROD = 3200 K = 5760 R HR = 47 518 Btu HP = 0.4853(169 184+71 075) + 0.5147(47 518+43 406) + 0.757 35(0+45 581) + 1(0+43 050) = +27 842 Btu QCV = HP  HR = 27 842  (47 518) = +75 360 Btu 1552 15.65EIn a test of a gasturbine combustor, saturatedliquid methane at 210 R is to be burned with excess air to hold the adiabatic flame temperature to 2880 R. It is assumed that the products consist of a mixture of CO2, H2O, N2, O2, and NO in chemical equilibrium. Determine the percent excess air used in the combustion, and the percentage of NO in the products. CH4 + 2x O2 + 7.52x N2 1 CO2 + 2 H2O + (2x2) O2 + 7.52x N2 Then init ch. equil. N2 + O2 2 NO 0 +2a 2a Also CO2 H2O 1 0 1 2 0 2 7.52x 2x2 a a (7.52xa) (2x2a) nTOT = 1 + 9.52x 2880 R: ln K = 10.55, yNO
2 K = 2.628105
2 yNO P 4a2 2.628105 K = (P0)0 = y y = (7.52xa)(2x2a) yN2yO2 N2 O2 HR = 1[32 190 + (18544300)]+ 0 + 0 (assume 77 F) = 38 344 Btu HP = 1(169 184 + 29 049) + 2(103 966 + 22 746) + (7.52xa)(18 015) + (2x2a)(19 031) + 2a(38 818 + 18 624) = 340 639 + 173 535 x + 77 838 a Assume Subst. 2.628105 a2 = , (13.1a)(1.484a) 4 Use this a in 1st law x= 302 295  77 8380.01125 = 1.737 173 535 2.62810 a2 = , a = 0.0112 (13.062a)(1.474a) 4 x = 1.737 % NO = % excess air = 73.7 % 20.0112100 = 0.128 % 1+9.521.737
5 a ~ 0, then from HP  HR = 0 x = 1.742 get a 0.01125 1553 15.66E Acetylene gas at 77 F is burned with 140% theoretical air, which enters the burner at 77 F, 14.7 lbf/in.2, 80% relative humidity. The combustion products form a mixture of CO2, H2O, N2, O2, and NO in chemical equilibrium at 3500 F, 14.7 lbf/in.2. This mixture is then cooled to 1340 F very rapidly, so that the composition does not change. Determine the mole fraction of NO in the products and the heat transfer for the overall process. C2H2 + 3.5 O2 + 13.16 N2 + water 2 CO2 + 1 H2O + 1 O2 + 13.16 N2 + water Water: PV = 0.80.46 = 0.368 lbf/in2 PV 0.368 nV = nA = (3.5+13.16) = 0.428 14.332 PA So, total H2O in products is : 1 + nV = 1.428. a) reaction: change : N2 + O2 2 NO x x +2x K = 0.001 074 at 3500F = 3960 R (=2200 K), from A.10: Equilibrium products: nCO2 = 2, nH2O = 1.428, nO2 = 1x, nNO = 0+2x, nN2 = 13.16x, K= n TOT = 17.588 (2x)2 = 0.001 074 (1x)(13.16x) x = 0.0576 20.0576 = 0.006 55 17.588 By trial and error, yNO = b) Final products (same composition) at 1340 F = 1800 R HR = 1(97 476) + 0.428(103 966) = 52 979 Btu HP = 2(169 184+14 358) + 1.428(103 966+11 178) + 0.9424(0+9761) + 13.1024(0+9227) + 0.1152(38 818+9557) = 306 486 Btu QCV = HP  HR = 359 465 Btu 1554 15.67E The equilibrium reaction with methane as CH4 C + 2H2 has ln K = 0.3362 at 1440 R and ln K = 4.607 at 1080 R. By noting the relation of K to temperature, show how you would interpolate lnK in (1/T) to find K at 1260 R and compare that to a linear interpolation. ln K = 0.3362 at 1440 R ln K = 4.607 at 1080 R 1 1 1260 1440 lnK1260 = lnK1440 + (4.607+0.3362) 1 1 1080 1440 1440 1 1260 = 0.3362 + (4.2708) = 2.1665 1440 1 1080 Linear interpolation: lnK1260 = lnK1080 + = 4.607 + 1260  1080 (lnK1440  lnK1080) 1440  1080 1 (0.3362 + 4.607) = 2.4716 2 o 15.68E Use the information in problem 15.67 to estimate the enthalpy of reaction, H , at 1260 R using the van't Hoff equation (see problem 15.28) with finite differences for the derivatives.  2 dlnK = [H/RT ]dT or solve for H  2 dlnK  2 lnK H = RT = RT dT T = 1.98589 1260 2 0.3362 + 4.607 = 37403 Btu/lb mol 1440  1080 [Remark: compare this to C.7 values + C.4, C.12, H = HC + 2HH  HCH = 0.146 12 (1260537) + 2 5044
2 4 0.538 16.043 (1260537) (32190) = 37304 ] 1555 15.69E An important step in the manufacture of chemical fertilizer is the production of ammonia, according to the reaction: N 2 + 3H2 2NH3 a. Calculate the equilibrium constant for this reaction at 300 F. b. For an initial composition of 25% nitrogen, 75% hydrogen, on a mole basis, calculate the equilibrium composition at 300 F, 750 lbf/in.2. 1N2 + 3H2 <=> 2NH3 at 300 F o a) hNH3 300 F = 19 656 + 0.50917.031(30077) = 17723 760 o sNH3 300 F = 45.969 + 0.50917.031 ln = 48.980 537 H300 F = 2(17723)  1(0 + 1557)  3(0 + 1552) = 41 659 Btu S300 F = 2(48.98)  1(48.164)  3(33.60) = 51.0 Btu/R G300 F = 41 659  760(51.0) = 2899 Btu ln K = b) +2899 = 1.9208, 1.98589760 nN2 = 1x, K = 6.826
0 0 o nNH3 = 2x, K= yNH3 yN2yH2
2 3 2 nH2 = 33x (PP )
0 2 = (2x)222(2x)2 P 33(1x)4 P0 ( ) 2 or or x (1x) (2x) 1x 2 = 27 750 2 6.826 ( ) = 29985 16 14.7 x (1x)(2x)= 173.16 1x NH3 N2 H2 n 1.848 0.0758 0.2273 y 0.8591 0.0352 0.1057 Trial & Error: x = 0.9242 1556 15.70E Ethane is burned with 150% theoretical air in a gas turbine combustor. The products exiting consist of a mixture of CO2, H2O, O2, N2, and NO in chemical equilibrium at 2800 F, 150 lbf/in.2. Determine the mole fraction of NO in the products. Is it reasonable to ignore CO in the products? Combustion: C2H6 + 5.25 O2 + 19.74 N2 2 CO2 + 3 H2O + 1.75 O2 + 19.74 N2 Products at 2800 F, 150 lbf/in2 a) Equilibrium mixture: CO2, H2O, O2, N2, NO N2 initial change equil. 19.74 x 19.74x + O2 1.75 x 1.75x 2 NO 0 +2x 2x Equil. comp. nCO2 = 2, nH2O = 3, nO2 = 1.75x, nN2 = 19.74x, nNO = 2x P 0 4x2 ( ) = (19.74x)(1.75x) K = 1.28310 = yN2yO2 P0
4 yNO 2 Solving, x = 0.032 95 20.032 95 = 0.002 49 26.49 2 CO 0 +2a 2a
2 yNO = b) 2 CO2 initial change equil. 2 2a 22a + O2 0 +2x 2x yCOyO2 P 1 2a 2 150 8 K = 5.25910 = 2 (P0) = (22a) (1.75x+a)(14.504) 26.49+a y
CO2 Since Table A.10 corresponds to a pressure P0 of 100 kPa, which is 14.504 lbf/in2. This equation should be solved simultaneously with the equation solved in part a) (modified to include the unknown a). Since x was found to be small and also a will be very small, the two are practically independent. Therefore, use the value x = 0.032 95 in the equation above, and solve for a. a (1a) (1.750.032 95+a)=(14.504)5.259108 26.49+a 150
2 Solving, a = 0.000 28 or yCO = 2.1105 negligible for most applications. 1557 15.71E One pound mole of air (assumed to be 78% nitrogen, 21% oxygen, and 1% argon) at room temperature is heated to 7200 R, 30 lbf/in.2. Find the equilibrium composition at this state, assuming that only N2, O2, NO, O, and Ar are present. 1 lbmol air (0.78 N2, 0.21 O2, 0.01 Ar) heated to 7200 R, 30 lbf/in2. 1) N2 + O2 2 NO a +2a 2) change O2 2 O b +2b change a nN2 = nO2 = 0.78a 0.21ab Equil.: nAr = nO = 0.01 2b nNO = n = 2a 1+b 4a2 30 0 (14.504) K1 = 0.0895 = (0.78a)(0.21ab) 4b2 30 K2 = 2.221 = (14.504) (1+b)(0.21ab) Divide 1st eq'n by 2nd and solve for a as function(b), using K1 P X= ( ) = 0.083 35 K2 P0 Get Xb2 a= [1 + 2(1+b) Also 1+ 40.78(1+b) Xb2 (1) (2) K2 b2 = = 0.268 44 (1+b)(0.21ab) 4(P/P0) Assume b = 0.1269 Then, check a & b in (2) OK Therefore, Subst. n y N2 0.7501 0.6656 O2 0.0532 0.0472 Ar 0.01 0.0089 O 0.2538 0.2252 From (1), get a = 0.0299 NO 0.0598 0.0531 1558 15.72E Dry air is heated from 77 F to 7200 R in a 14.7 lbf/in.2 constantpressure process. List the possible reactions that may take place and determine the equilibrium composition. Find the required heat transfer. Air assumed to be 21% oxygen and 79% nitrogen by volume. From the elementary reactions at 4000 K = 7200 R (A.10): (1) O2 <=> 2 O (2) N2 <=> 2 N K1 = 2.221 = y O/yO2 K2 = 3.141 106 = yN/yN2 K3 = 0.08955 = yNO/yN2 yO2 nN2 = 0.79bc, nN = 2b,
2 2 2 (3) N2 + O2 <=> 2 NO nO2 = 0.21ac, nNO = 2c, Call the shifts a,b,c respectively so we get nO = 2a, ntot = 1+a+b From which the molefractions are formed and substituted into the three equilibrium equations. The result is corrected for 1 atm = 14.7 lbf/in2 = 101.325 kPa versus the tables 100 kPa K1 = 2.1511 = y O/yO2 = 4a2/[(1+a+b)(0.21ac)] K2 = 3.042106 = yN/yN2 = 4b2/[(1+a+b)(0.79bc)] K3 = 0.08955 = y NO/yN2 yO2 = 4c2/[(0.79bc)(0.21ac)] which give 3 eqs. for the unknowns (a,b,c). Trial and error assume b = c = 0 solve for a from K 1 then for c from K3 and finally given the (a,c) solve for b from K2. The order chosen according to expected magnitude K1>K3>K2 a = 0.15, b = 0.000832, c = 0.0244 => nO2 = 0.0356, nO = 0.3, nN2 = 0.765, nN = 0.00167, nNO = 0.049 Q = Hex  Hin = nO2hO2 + nN2hN2 + nO(hfO + hO) + nN(hfN + hN) + nNO(hfNO + hNO)  0 = 0.035659632 + 0.76555902 + 0.3(107124 + 33394) + 0.00167(203216 + 33333) + 0.049(38818 + 57038) = 92 135 Btu/lbmol air [If no reaction: Q = nO2hO2 + nN2hN2 = 56 685 Btu/lbmol air]
2 2 2 1559 15.73E Acetylene gas and x times theoretical air (x > 1) at room temperature and 75 lbf/in.2 are burned at constant pressure in an adiabatic SSSF process. The flame temperature is 4600 R, and the combustion products are assumed to consist of N2, O2, CO2, H2O, CO, and NO. Determine the value of x. Combustion: C2H2 + 2.5x O2 + 9.4x N2 2 CO2 + H2O + 2.5(x1)O2 + 9.4x N2 Eq. products 4600 R, 75 lbf/in2: N2, O2, CO2, H2O, CO, NO 2 Reactions: 1) 2 CO2 2 CO + O2 +2a +a 2) change N2 + O2 2 NO b b +2b change 2a Equil. Comp.: 9.4xb nN2 = nO2 = nCO = nTOT = nCO2 = nH2O = nNO = 22a 1 2b 2.5x2.5+ab 2a 11.9x+0.5+a At 4600 R, from A.17: K1 = 2.359103, K2 = 4.249103 2.359103 a 2 2.5x2.5+ab = 4.622104 =( ) ( ) 0 = 1a 11.9x+0.5+a 5.103 (P/P ) (2b)2 K2 = 4.24910 = (9.4b)(2.5x2.5+ab)
3 K1 Also, from the 1st law: HP  HR = 0 where HR = 1(+97 477) + 0 + 0 = +97 497 Btu HP = (9.4xb)(0+32 817) + (2.5x2.5+ab)(0+34 605) + (22a)(169 184+53 885) + 1(103 966+43 899) + 2a(47 518+33 122) + 2b(38 818+31 161) Substituting, 394 992 x + 236 411 a + 72 536 b  377 178 = 97477 which results in a set of 3 equations in the 3 unknowns x,a,b. Trial and error solution from the last eq. and the ones for K1 and K2. The result is x = 1.12 , a = 0.1182, b = 0.05963 1560 15.74E One pound mole of water vapor at 14.7 lbf/in. 2, 720 R, is heated to 5400 R in a constant pressure SSSF process. Determine the final composition, assuming that H2O, H2, H, O2, and OH are present at equilibrium. Reactions: 1) 2 H 2O 2 H 2 + O2 +2a +a H2 2 H 2) change 2 H2O H2+ 2 OH 2b +b +2b change 2a 3) change c +2c At equilibrium (5400 R, 14.7 lbf/in2) nH2O = 12a2b nOH = nH2 = nO2 = K1 K2 K3 2a+bc a nH = nTOT = 2b 2c 1+a+b+c 2.062103 2a+bc 2 a = =( ) (1+a+b+c) 12a2b 1.03 (P/P0)
2 2.893103 2a+bc 2b =( )(12a2b) 0 = 1+a+b+c 1.03 (P/P ) 2.496102 (2a)2 = = 1.03 (2a+bc)(1+a+b+c) (P/P0) These three equations must be solved simultaneously for a, b & c: and a = 0.0622, b = 0.0570, c = 0.0327 nH2O = 0.7616 yH2O = 0.6611 nH2 = 0.1487 yH2 = 0.1291 nO2 = 0.0622 yO2 = 0.0540 1561 15.75E Methane is burned with theoretical oxygen in an SSSF process, and the products exit the combustion chamber at 5300 F, 100 lbf/in.2. Calculate the equilibrium composition at this state, assuming that only CO2, CO, H2O, H2, O2, and OH are present. Combustion: 1) CH4 + 2 O2 CO2 + 2 H2O At equilibrium: nH2O = nH2 = nO2 = nOH = nCO2 = nCO = nTOT = Products at 5300F, 100 lbf/in2 K1 = 0.007 328 = ( K2 = 0.012 265 = ( K3 = 0.426 135 = ( 2a+b 2 ) ( a+c )( 100 ) 22a2b 3+a+b+c 14.504
2 2b 2a+b 100 ) (3+a+b+c)(14.504) 22a2b Dissociation reactions: 2 H 2O 2 H 2 + O2 +2a +a 2 H2O H2 + 2 OH +b +2b 2 CO2 2 CO + O2 +2c +c 22a2b 2a+b a+c 2b 12c 2c 3+a+b+c change 2a 2) change 2b 3) change 2c 2c 2 a+c 100 ) (3+a+b+c)(14.504) 12c These 3 equations must be solved simultaneously for a, b, & c. If solving by hand divide the first equation by the second, and solve for c = fn(a,b). This reduces the solution to 2 equations in 2 unknowns. Solving, a = 0.0245, b = 0.1460, c = 0.2365 Substance: n y H2O 1.659 0.4871 H2 0.195 0.0573 O2 0.260 0.0763 OH 0.292 0.0857 CO2 0.527 0.1547 CO 0.473 0.1389 161 CHAPTER 16
Notice that most of the solutions are done using the computer tables, which includes the steam tables, air table, compressible flow table and the normal shock table. This significantly reduces the amount of time it will take to solve a problem, so this should be considered in problem assignments and exams. Changes of problems from the 4th edition Chapter 14 to the new Chapter 16 are: New 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 New Old 1 new 2 new new 3 new new 5 6 7 new new new new 8 9 mod skip a) 10 11 12 Old New 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 New Old 13 14 new 15 16 17 18 19 20 21 22 23 24 25 26 new 27 28 29 new Old New 41 42 43 44E 45E 46E 47E 48E 49E 50E 51E 52E 53E 54E 55E 56E 57E Old 30 31 4 39 38 new new 40 new new new 43 44 45 46 47 new New Old 162 16.1 Steam leaves a nozzle with a pressure of 500 kPa, a temperature of 350C, and a velocity of 250 m/s. What is the isentropic stagnation pressure and temperature? h0 = h1 + V2/2 = 3167.7 + 2502/2000 = 3198.4 kJ/kg 1 s0 = s1 = 7.6329 kJ/kg K Computer software: (ho, so) To = 365 C, Po = 556 kPa 16.2 An object from space enters the earth's upper atmosphere at 5 kPa, 100 K with a relative velocity of 2000 m/s or more . Estimate the object`s surface temperature. ho1  h1 = V2/2 = 20002/2000 = 2000 kJ/kg 1 ho1 = h1 + 2000 = 100 + 2000 = 2100 kJ/kg => T = 1875 K The value for h from ideal gas table A.7 was estimated since the lowest T in the table is 200 K. 16.3 The products of combustion of a jet engine leave the engine with a velocity relative to the plane of 400 m/s, a temperature of 480C, and a pressure of 75 kPa. Assuming that k = 1.32, Cp = 1.15 kJ/kg K for the products, determine the stagnation pressure and temperature of the products relative to the airplane. ho1  h1 = V2/2 = 4002/2000 = 80 kJ/kg 1 To1  T1 = (ho1  h1)/Cp = 80/1.15 = 69.6 K To1 = 480 + 273.15 + 69.6 = 823 K Po1 = P1(To1/T1)k/(k1) = 75(823/753.15)4.125 = 108 kPa 16.4 A meteorite melts and burn up at temperatures of 3000 K. If it hits air at 5 kPa, 50 K how high a velocity should it have to experience such a temperature? Assume we have a stagnation T = 3000 K h1 + V2/2 = hstagn. 1 Use table A.7, hstagn. = 3525.36, h1 = 50 (remember convert to J/kg = m2/s2) V2/2 = 3525.36 50 = 3475.4 kJ/kg 1 V1 = 2 3475.4 1000 = 2636 m/s 163 16.5 I drive down the highway at 110 km/h on a day with 25C, 101.3 kPa. I put my hand , cross sectional area 0.01 m2, flat out the window. What is the force on my hand and what temperature do I feel? The air stagnates on the hand surface : Use constant heat capacity Tstagn. = T1 + V2/2 1 0.5 1102 (1000/3600)2 = 25 + = 25.465C Cp 1004 h1 + V2/2 = hstagn. 1 Assume a reversible adiabatic compression Pstagn. = P1 (Tstagn./T1)k/(k1) = 101.3 (298.615/298.15)3.5 = 101.85 kPa 16.6 Air leaves a compressor in a pipe with a stagnation temperature and pressure of 150C, 300 kPa, and a velocity of 125 m/s. The pipe has a crosssectional area of 0.02 m2. Determine the static temperature and pressure and the mass flow rate. ho1  h1 = V2/2 = 1252/2000 = 7.8125 kJ/kg 1 To1  T1 = (ho1  h1)/Cp = 7.8125/1.004 = 7.8 K T1 = To1  T = 150  7.8 = 142.2 C = 415.4 K P1 = Po1(T1/To1)k/(k1) = 300(415.4/423.15)3.5 = 281 kPa AV P1AV1 281.2(0.02)(125) . m = AV = = = = 5.9 kg/s v 0.287(415.4) RT1 16.7 A stagnation pressure of 108 kPa is measured for an air flow where the pressure is 100 kPa and 20C in the approach flow. What is the incomming velocity? Assume a reversible adiabatic compression To1 = T1 (Po1/P1)(k1)/k = 293.15 ( 108 0.2857 ) = 299.667 K 100 V2/2 = ho1  h1 = Cp (To1  T1) = 6.543 1 V1 = 2 6.543 1000 = 114.4 m/s 164 16.8 A jet engine receives a flow of 150 m/s air at 75 kPa, 5C across an area of 0.6 m 2 with an exit flow at 450 m/s, 75 kPa, 600 K. Find the mass flow rate and thrust. . m = AV; ideal gas = P/RT . m = (P/RT)AV = ( 75 ) 0.6 150 = 0.9395 0.6 150 0.287 278.15 = 84.555 kg/s . Fnet = m (VexVin) = 84.555 (450150) = 25367 N 16.9 A water cannon sprays 1 kg/s liquid water at a velocity of 100 m/s horizontally out from a nozzle. It is driven by a pump that receives the water from a tank at 15C, 100 kPa. Neglect elevation differences and the kinetic energy of the water flow in the pump and hose to the nozzle. Find the nozzle exit area, the required pressure out of the pump and the horizontal force needed to hold the cannon. 0.001001 . . m = AV = AV/v A= mv/V = 1 = 1.0 105 m2 100 . . . 2 _ Wp = mwp = mv(Pex  Pin) = mVex/2
2 Pex = Pin + Vex/2v = 100 + 1002/2 1000 0.001 = 150 kPa . F = mVex = 1 100 = 100 N 16.10 An irrigation pump takes water from a lake and discharges it through a nozzle as shown in Fig. P16.10. At the pump exit the pressure is 700 kPa, and the temperature is 20C. The nozzle is located 10 m above the pump and the atmospheric pressure is 100 kPa. Assuming reversible flow through the system determine the velocity of the water leaving the nozzle. Assume we can neglect kinetic energy in the pipe in and out of the pump. Incompressible flow so Bernoulli's equation applies (V1 V2 V3 0) v(P3  P2)+ (V2  V2)/2 + g(Z3  Z2) = 0 3 2 P 3 = P2 g(Z3  Z2) v = 700 9.807(10) = 602 kPa 1000(0.001002) V2/2 = v(P3  P4) 4 V4 = 2v(P3  P4) = 2 0.001002 502.1 1000 = 31.72 m/s 165 16.11 A water turbine using nozzles is located at the bottom of Hoover Dam 175 m below the surface of Lake Mead. The water enters the nozzles at a stagnation pressure corresponding to the column of water above it minus 20% due to friction. The temperature is 15C and the water leaves at standard atmospheric pressure. If the flow through the nozzle is reversible and adiabatic, determine the velocity and kinetic energy per kilogram of water leaving the nozzle. P = gZ = gZ 9.807 175 = = 1714.5 kPa v 0.001001 1000 Pac = 0.8 P = 1371.6 kPa
2 vP = Vex/2 Vex = 2vP Vex = 2 0.001001 1000 1371.6 = 62.4 m/s 2 Vex/2 = vP = 1.373 kJ/kg 16.12 A water tower on a farm holds 1 m3 liquid water at 20C, 100 kPa in a tank on top of a 5 m tall tower. A pipe leads to the ground level with a tap that can open a 1.5 cm diameter hole. Neglect friction and pipe losses and estimate the time it will take to empty the tank for water. Bernoulli Equation: Pe = Pi; Vi = 0; Ze = 0; Zi = H V2/2 = gZi e Ve = 2gZ = 2 9.807 5 = 9.9 m/s . m = AVe = AVe/v = m/t m = V/v; A = D2/4 = 0.0152 / 4 = 1.77 104 m2 t = t = mv/AVe = V/AVe 1 = 571.6 sec = 9.53 min 1.77 10 4 9.9 166 16.13 Find the speed of sound for air at 100 kPa at the two temperatures 0C and 30C. Repeat the answer for carbon dioxide and argon gases. From eq. 16.28 we have c0 = kRT = c30 = 1.4 0.287 273.15 1000 = 331 m/s 1.4 0.287 303.15 1000 = 349 m/s For Carbon Dioxide: R = 0.1889, k = 1.289 c0 = c30 = 1.289 0.1889 273.15 1000 = 257.9 m/s 1.289 0.1889 303.15 1000 = 271.7 m/s For Argon: R = 0.2081, k = 1.667 c0 = c30 = 1.667 0.2081 273.15 1000 = 307.8 m/s 1.667 0.2081 303.15 1000 = 324.3 m/s 16.14 If the sound of thunder is heard 5 seconds after seing the lightning and the weather is 20C how far away is the lightning taking place? The sound travels with the speed of sound in air (ideal gas). L = c t = kRT t = 1.4 0.287 293.15 1000 5 = 1716 m 16.15 Estimate the speed of sound for steam directly from Eq. 16.28 and the steam tables for a state of 6 MPa, 400C. Use table values at 5 and 7 MPa at the same entropy as the wanted state. Eq. 16.28 is then done by finite difference. Find also the answer for the speed of sound assuming steam is an ideal gas. c2 = ( P P )s = ()s State 6 MPa, 400C s = 6.5407 7 MPa, s v = 0.04205; = 1/v = 23.777 5 MPa, s v = 0.05467; = 1/v = 18.2909 c2 = 7000  5000 = 364.56 1000 c = 603.8 m/s 23.777  18.2909 14190  10499 = 36.91; 100 Cv = Cp  R = 36.91  8.31451 = 28.595 Cp = k = Cp/Cv = 1.2908; c = kRT = R = 0.4615 1.2908 0.4615 673.15 1000 = 633.2 m/s 167 16.16 A convergentdivergent nozzle has a throat diameter of 0.05 m and an exit diameter of 0.1 m. The inlet stagnation state is 500 kPa, 500 K. Find the back pressure that will lead to the maximum possible flow rate and the mass flow rate for three different gases as: air; hydrogen or carbon dioxide. There is a maximum possible flow when M=1 at the throat, T* = 2 2 k 2 1 To; P* = Po ( )k1; * = o ( )k1 k+1 k+1 k+1 . m = *A*V = *A*c = P*A* k/RT* A* = D2/4 = 0.001963 m2 * 2.209 0.154 3.318 . m 1.944 0.515 2.273 k a) b) c) 1.400 1.409 1.289 T* 416.7 415.1 436.9 P* 264.1 263.4 273.9 c 448.2 1704.5 348.9 AE/A* = (DE/D*)2 = 4. There are 2 possible solutions corresponding to points c and d in Fig. 16.13 and Fig. 16.17. For these we have ME a) b) c) 0.1466 0.1464 0.1483 PE/Po 0.985 0.985 0.986 ME 2.940 2.956 2.757 PE/Po 0.0298 0.0293 0.0367 PB = PE 0.985 500 = 492.5 kPa all cases point c PB = PE = a) 14.9 b) 14.65 c) 18.35 kPa, point d 168 16.17 Air is expanded in a nozzle from 2 MPa, 600 K, to 200 kPa. The mass flow rate through the nozzle is 5 kg/s. Assume the flow is reversible and adiabatic and determine the throat and exit areas for the nozzle.
Mach # 2 k P * = Po k1 k+1
Velocity = 2 0.5283 = 1.056 MPa T* = To 2/(k+1) = 600 0.8333 = 500 K v* = RT*/P* = 0.287 500/1056 Area Density 0.2 MPa 2.0 MPa P = 0.1359 m3/kg kRT* = 1.4 1000 0.287 500 = 448.2 m/s . A* = mv*/c* = 5 0.1359/448.2 = 0.00152 m2 c* = P2/Po = 200/2000 = 0.1 M* = 1.701 = V2/c* 2 V2 = 1.701 448.2 = 762.4 m/s T2 = 500 0.5176 = 258.8 K v2 = RT2/P2 = 0.287 258.8/200 = 0.3714 m3/kg . A2 = mv2/V2 = 5 0.3714/762.4 = 0.002435 m2 16.18 Consider the nozzle of Problem 16.17 and determine what back pressure will cause a normal shock to stand in the exit plane of the nozzle. This is case g in Fig. 16.17. What is the mass flow rate under these conditions? PE/Po = 200/2000 = 0.1 ; ME = 2.1591 = Mx My = 0.5529 ; Py/Px = 5.275 Py = 5.275 200 = 1055 kPa . m = 5 kg/s same as in Problem 16.9 169 16.19 At what Mach number will the normal shock occur in the nozzle of Problem 16.17 if the back pressure is 1.4 MPa? (trial and error on Mx) Relate the inlet and exit conditions to the shock conditions with reversible flow before and after the shock. It becomes trial and error. Assume Mx = 1.8 My = 0.6165 ; Poy/Pox = 0.8127 AE/A* = A2/A* = 0.002435/0.001516 = 1.6062 x Ax/A*x = 1.439 ; Ax/A*y = 1.1694 AE/A* = (AE/A*)(Ax/A*)/(Ax/A*) = y x y x ME = 0.5189 ; PE/Poy = 0.8323 PE = (PE/Poy)(Poy/Pox)Pox = 0.8323 0.8127 2000 = 1353 kPa So select the mach number a little less Mx = 1.7 My = 0.64055 ; Poy/Pox = 0.85573 Ax/A*x = 1.3376 ; Ax/A* = 1.1446 y AE/A* = (AE/A*)(Ax/A*)/(Ax/A*) = y x y x ME = 0.482 ; PE/Poy = 0.853 PE = (PE/Poy)(Poy/Pox)Pox = 0.853 0.85573 2000 = 1459.9 kPa Now interpolate between the two Mx = 1.756 My = 0.6266 ; Poy/Pox = 0.832 Ax/A*x = 1.3926 ; Ax/A* = 1.1586 y AE/A* = 1.6062 1.1586/1.3926 = 1.3363 y ME = 0.5 ; PE/Poy = 0.843 PE = 0.843 0.832 2000 = 1402.7 kPa OK 1.6062(1.1446) = 1.3744 1.3376 1.6062(1.1694) = 1.3053 1.439 1610 16.20 Consider the nozzle of Problem 16.17. What back pressure will be required to cause subsonic flow throughout the entire nozzle with M = 1 at the throat? AE/A* = 0.002435/0.001516 = 1.6062 ME = 0.395 ; PE/Po = 0.898 PE = 0.898 2000 = 1796 kPa 16.21 Determine the mass flow rate through the nozzle of Problem 16.17 for a back pressure of 1.9 MPa. PE/Pox = 1.9/2.0 = 0.95 ME = 0.268 TE = (T/To)To = 0.9854 600 = 591.2 K cE = kRTE = 1.4 1000 0.287 591.2 = 487.4 m/s VE = MEcE = 0.268 487.4 = 130.6 m/s vE = RT/P = 0.287 591.2/1900 = 0.0893 m3/kg . m = AEVE/vE = 0.002435 130.6/0.0893 = 3.561 kg/s 16.22 At what Mach number will the normal shock occur in the nozzle of Problem 16.16 flowing with air if the back pressure is halfway between the pressures at c and d in Fig. 16.17? First find the two pressure that will give exit at c and d. See solution to 16.8 a) AE/A* = (DE/D*) = 4 2 PE = 492.5 (c) 16.9 (d) PE = (492.5 + 14.9)/2 = 253.7 kPa Assume ME = 2.4 My = 0.5231 ; Poy/Pox = 0.54015 Ax/A* = 2.4031 ; Ax/A* = 1.298 x y AE/A* = (AE/A*)(Ax/A*) / (Ax/A*) = 4 1.298/2.4031 = 2.1605 y x y x ME = 0.2807 ; PE/Poy = 0.94675 PE = (PE/Poy)(Poy/Pox)Pox = 0.94675 0.54015 500 = 255.7 kPa Repeat if Mx = 2.5 PE = 233.8 kPa Interpolate to match the desired pressure => Mx = 2.41 1611 16.23 A convergent nozzle has minimum area of 0.1 m2 and receives air at 175 kPa, 1000 K flowing with 100 m/s. What is the back pressure that will produce the maximum flow rate and find that flow rate? P* 2 k =( )k1 = 0.528 Po k+1 Find Po: h0 = h1 + V2/2 = 1046.22 + 1002/2000 = 1051.22 1 T0 = Ti + 4.4 = 1004.4 from table A.7 P0 = Pi (T0/Ti)k/(k1) = 175 (1004.4/1000)3.5 = 177.71 P* = 0.528 Po = 0.528 177.71 = 93.83 kPa T* = 0.8333 To = 836.97 K * = P* 93.83 = = 0.3906 RT* 0.287 836.97 Critical Pressure Ratio V = c = kRT* = 1.4 1000 0.287 836.97 = 579.9 m/s . m = AV = 0.3906 0.1 579.9 = 22.65 kg/s 16.24 A convergentdivergent nozzle has a throat area of 100 mm 2 and an exit area of 175 mm2. The inlet flow is helium at a total pressure of 1 MPa, stagnation temperature of 375 K. What is the back pressure that will give sonic condition at the throat, but subsonic everywhere else? AE/A* = 175/100 = 1.75 ; kHe = 1.667 ME = 0.348 ; PE/PO = 0.906 PE = 0.906 1000 = 906 kPa 1612 16.25 A nozzle is designed assuming reversible adiabatic flow with an exit Mach number of 2.6 while flowing air with a stagnation pressure and temperature of 2 MPa and 150C, respectively. The mass flow rate is 5 kg/s, and k may be assumed to be 1.40 and constant. a. Determine the exit pressure, temperature, and area, and the throat area. b. Suppose that the back pressure at the nozzle exit is raised to 1.4 MPa, and that the flow remains isentropic except for a normal shock wave. Determine the exit Mach number and temperature, and the mass flow rate through the nozzle. (a) From Table A.11: ME = 2.6 PE = 2.0 0.05012 = 0.1002 MPa T* = 423.15 0.8333 = 352.7 K P* = 2.0 0.5283 = 1.057 MPa c* = 1.4 1000 0.287 352.7 = 376.5 m/s v* = 0.287 352.7/1057 = 0.0958 m3/kg A* = 5 0.0958/376.5 = 1.272 103m2 AE = 1.272 103 2.896 = 3.68 103m2 TE = 423.15 0.42517 = 179.9 K Assume Mx = 2 then My = 0.57735, Poy/Pox = 0.72088, AE/A* = 2.896 x Ax/A* = 1.6875, Ax/A* = 1.2225, x y AE/A* = 2.896 1.2225/1.6875 = 2.098 y ME = 0.293, PE/Poy = 0.94171 PE = 0.94171 0.72088 2.0 = 1.357 MPa, OK TE = 0.98298 423.15 = 416 K, . m = 5 kg/s 1613 16.26 A jet plane travels through the air with a speed of 1000 km/h at an altitude of 6 km, where the pressure is 40 kPa and the temperature is 12C. Consider the inlet diffuser of the engine where air leaves with a velocity of 100 m/s. Determine the pressure and temperature leaving the diffuser, and the ratio of inlet to exit area of the diffuser, assuming the flow to be reversible and adiabatic. V = 1000 km/h = 277.8 m/s, h1 = 261.48, Pr1 = 0.6862 ho1 = 261.48 + 277.8 2/2000 = 300.07 To1 = 299.7 K, Pro1 = 1.1107 Po1 = 40 1.1107/0.6862 = 64.74 kPa h2 = 300.07  1002/2000 = 295.07 T2 = 294.7 K, Pr2 = 1.0462 v 1 = 0.287 261.15/40 = 1.874 P2 = 64.74 1.0462/1.1107 = 61 kPa v2 = 0.287 294.7/61 = 1.386 A1/A2 = (1.874/1.386)(100/277.8) = 0.487 16.27 A 1m3 insulated tank contains air at 1 MPa, 560 K. The tank is now discharged through a small convergent nozzle to the atmosphere at 100 kPa. The nozzle has an exit area of 2 105 m2. a. Find the initial mass flow rate out of the tank. b. Find the mass flow rate when half the mass has been discharged. c. Find the mass of air in the tank and the mass flow rate out of the tank when the nozzle flow changes to become subsonic. a. The back pressure ratio: PB/Po1 = 100/1000 = 0.1 < (P*/Po)crit = 0.5283 so the initial flow is choked with the maximum possible flow rate. ME = 1 ; PE = 0.5283 1000 = 528.3 kPa TE = T* = 0.8333 560 = 466.7 K VE = c = kRT* = 1.4 1000 0.287 466.7 = 433 m/s vE = RT*/PE = 0.287 466.7/528.3 = 0.2535 m3/kg 1614 . m1 = AVE/vE = 2 105 433/0.2535 = 0.0342 kg/s b. The initial mass is m1 = P1V/RT1 = 1000 1/0.287 560 = 6.222 kg with a mass at state 2 as m2 = m1/2 = 3.111 kg. Assume an adiabatic reversible expansion of the mass that remains in the tank. P2 = P1(v1/v2)k = 100 0.51.4 = 378.9 kPa T2 = T1(v1/v2)k1 = 560 0.50.4 = 424 K The pressure ratio is still less than critical and the flow thus choked. PB/Po2 = 100/378.9 = 0.264 < (P*/Po)crit ME = 1 ; PE = 0.5283 378.9 = 200.2 kPa TE = T* = 0.8333 424 = 353.7 K VE = c = kRT* = 1.4 1000 0.287 353.7 = 377 m/s
5 210 (377)(200.2) . = 0.0149 kg/s m2 = AVEPE/RTE = 0.287(353.7) c. The flow changes to subsonic when the pressure ratio reaches critical. PB/Po3 = 0.5283 Po3 = 189.3 kPa v1/v3 = (Po3/P1)1/k = (189.3/1000)0.7143 = 0.3046 m3 = m1v1/v3 = 1.895 kg T3 = T1(v1/v3)k1 = 560 0.30460.4 = 348 K PE = PB = 100 kPa ; ME = 1 TE = 0.8333 348 = 290 K ; VE =
5 kRTE = 341.4 m/s 210 (341.4)(100) . = 0.0082 kg/s m3 = AVEPE/RTE = 0.287(290) 1615 16.28 A 1m3 uninsulated tank contains air at 1 MPa, 560 K. The tank is now discharged through a small convergent nozzle to the atmosphere at 100 kPa while heat transfer from some source keeps the air temperature in the tank at 560 K. The nozzle has an exit area of 2 105 m2. a. Find the initial mass flow rate out of the tank. b. Find the mass flow rate when half the mass has been discharged. c. Find the mass of air in the tank and the mass flow rate out of the tank when the nozzle flow changes to become subsonic. a. b. Same solution as in 16.27 a) From solution 16.27 b) we have m2 = m1/2 = 3.111 kg P2 = P1/2 = 500 kPa ; T2 = T1 ; PB/P2 = 100/500 = 0.2 < (P*/Po)crit The flow is choked and the velocity is: VE = c = kRT* = 433 m/s from 16.27 a) PE = 0.5283 500 = 264.2 kPa ; ME = 1 210 (433)(264.2) . = 0.01708 kg/s m2 = AVEPE/RTE = 0.287(466.7) c. Flow changes to subsonic when the pressure ratio reaches critical. PB/Po = 0.5283 ; P3 = 189.3 kPa m3 = m1P3/P1 = 1.178 kg ; T3 = T1 VE = 433 m/s 210 (433)(189.3) . = 0.01224 kg/s m3 = AVEPE/RTE = 0.287(466.7)
5 5 16.29 The products of combustion enter a convergent nozzle of a jet engine at a total pressure of 125 kPa, and a total temperature of 650C. The atmospheric pressure is 45 kPa and the flow is adiabatic with a rate of 25 kg/s. Determine the exit area of the nozzle. The critical pressure: Pcrit = P2 = 125 0.5283 = 66 kPa > Pamb The flow is then choked. V 2 = c2 = T2 = 923.15 0.8333 = 769.3 K 1.4 1000 0.287 769.3 = 556 m/s v2 = 0.287 769.3/66 = 3.3453 A2 = 25 3.3453/556 = 0.1504 m2 1616 16.30 Air is expanded in a nozzle from 700 kPa, 200C, to 150 kPa in a nozzle having an efficiency of 90%. The mass flow rate is 4 kg/s. Determine the exit area of the nozzle, the exit velocity, and the increase of entropy per kilogram of air. Compare these results with those of a reversible adiabatic nozzle. T2s = T1(P2/P1)(k1)/k = 473.2 (150/700)0.286 = 304.6 K V2s2 = 2 1000 1.0035(473.2  304.6) = 338400 V22 = 0.9 338400 V2 = 552 m/s h2 + V2/2 = h1 T2 = T1  V2/2Cp 2 2 T2 = 473.2  5522/(2 1000 1.004) = 321.4 K ; v2 = 0.287 321.4/150 = 0.6149 m3/kg A2 = 4 0.6149/552 = 0.00446 m2 = 4460 mm2 321.4 150 s2  s1 = 1.0035 ln  0.287 ln = 0.0539 kJ/kg K 473.2 700 16.31 Repeat Problem 16.26 assuming a diffuser efficiency of 80%. Same as problem 16.26, except D = 0.80. We thus have from 16.26 h3  h1 ho1  h1 = h3  261.48 300.07  261.48 = 0.8
h 01 02 3 2 1 s h3 = 292.35, Pr3 = 1.0129 Po2 = P3 = 40 1.0129/0.6862 = 59.04 kPa Pro2 = Pro1 = 1.1107 h2 = 300.07  1002/2000 = 295.07 T2 = 294.7 K, Pr2 = 1.0462 P2 = 59.04 1.0462/1.1107 = 55.6 kPa v2 = 0.287 294.7/55.6 = 1.521 A1/A2 = (1.874/1.521)(100/277.8) = 0.444 1617 16.32 Consider the diffuser of a supersonic aircraft flying at M = 1.4 at such an altitude that the temperature is 20C, and the atmospheric pressure is 50 kPa. Consider two possible ways in which the diffuser might operate, and for each case calculate the throat area required for a flow of 50 kg/s. a. The diffuser operates as reversible adiabatic with subsonic exit velocity. b. A normal shock stands at the entrance to the diffuser. Except for the normal shock the flow is reversible and adiabatic, and the exit velocity is subsonic. This is shown in Fig. P16.32. a. Assume a convergentdivergent diffuser with M = 1 at the throat. Relate the inlet state to the sonic state P1/Po = 0.31424 ; P*/Po1 = 0.5283 P* = c* = 0.5283 0.8333 50 = 84 kPa ; T* = ; 253.2 = 293.7 K 0.31424 0.71839 kRT* = 1.4 1000 0.287 293.7 = 343.5 m/s v* = RT*/P* = 0.287 293.7/84 = 1.0035 . A* = mv*/c* = 50 1.0035/343.5 = 0.1461 m2 b. Across the shock we have My = 0.7397 ; Py = 50 2.12 = 106 kPa ; Ty = 253.2 1.2547 = 317.7 K P* = T* = 0.5283 106 = 80.6 kPa 0.6952 0.8333 317.7 = 293.7 K, c* = 343.5 m/s 0.9011 v* = 0.287 293.7/80.6 = 1.0458 A* = 50 1.0458/343.5 = 0.1522 m2 1618 16.33 Air enters a diffuser with a velocity of 200 m/s, a static pressure of 70 kPa, and a temperature of 6C. The velocity leaving the diffuser is 60 m/s and the static pressure at the diffuser exit is 80 kPa. Determine the static temperature at the diffuser exit and the diffuser efficiency. Compare the stagnation pressures at the inlet and the exit. To1 = T1 + V2/2Cp = 267.15 + 2002/(2000 1.004) = 287.1 K 1 To2 = To1 T2 = To2  V2/2Cp = 287.1  602/(2000 1.004) = 285.3 K 2 To1  T1 T1 To2  T2 T2
ex k1 Po1  P1 = k P1 Po1  P1 = 18.25 Po1 = 88.3 kPa Po2  P2 = 1.77 Po2 = 81.8 kPa k  1 Po2  P2 = k P2 Ts = T1 (Po2/P1)k1/k = 267.15 1.0454 = 279.3 K D = Ts  T1 To1  T1
ex = 279.3  267.15 = 0.608 287.1  267.15 16.34 Steam at a pressure of 1 MPa and temperature of 400C expands in a nozzle to a pressure of 200 kPa. The nozzle efficiency is 90% and the mass flow rate is 10 kg/s. Determine the nozzle exit area and the exit velocity. First do the ideal reversible adiabatic nozzle s2s= s1= 7.4651 T2s = 190.4C ; h2s = 2851, h1= 3263.9 Now the actual nozzle can be calculated h1  h2ac = D(h1  h2s) = 0.9(3263.9  2851) = 371.6 h2ac = 2892.3 , T2 = 210.9C, v2 = 1.1062 V2 = 2000(3263.9  2892.3) = 862 m/s . A2 = mv2/V2 = 10 1.1062/862 = 0.01283 m2 1619 16.35 Steam at 800 kPa, 350C flows through a convergentdivergent nozzle that has a throat area of 350 mm2. The pressure at the exit plane is 150 kPa and the exit velocity is 800 m/s. The flow from the nozzle entrance to the throat is reversible and adiabatic. Determine the exit area of the nozzle, the overall nozzle efficiency, and the entropy generation in the process. ho1 = 3161.7, so1 = 7.4089 P*/Po1 = (2/(k+1))k/(k1) = 0.54099 P* = 432.7 kPa At *: (P*,s* = so1) h* = 2999.3, v* = 0.5687 h = V2/2 V* = 2000(3161.72999.3) = 569.9 m/s . m = AV*/v* = 350 106 569.9/0.5687 = 0.3507 kg/s he = ho1  V2/2 = 3161.7  8002/2 1000 = 2841.7 e Exit: Pe, he: ve = 1.395, se =7.576 . Ae = mve/Ve = 0.3507 1.395/800 = 6.115 104 m2 sgen = se  so1 = 7.576  7.4089 = 0.167 kJ/kg K 16.36 Air at 150 kPa, 290 K expands to the atmosphere at 100 kPa through a convergent nozzle with exit area of 0.01 m2. Assume an ideal nozzle. What is the percent error in mass flow rate if the flow is assumed incompressible? Pe k1 Te = Ti ( ) k = 258.28 K Pi V2/2 = hi  he = Cp (Ti  Te) = 1.004 (290  258.28) = 31.83 e Ve = 252.3 m/s; ve = RTe Pe = 0.287 258.28 = 0.7412 100 0.01 252.3 . m = AVe / ve = = 3.4 kg/s 0.7413 Incompressible Flow: vi = RT/P = 0.287 290/150 = 0.55487 m3/kg V2/2 = v P = vi (Pi  Pe) = 0.55487 (150  100) = 27.74 kJ/kg e . => Ve = 235 m/s => m = AVe / vi = 0.01 235 / 0.55487 = 4.23 kg/s . mincompressible 4.23 . = = 1.25 about 25% overestimation. 3.4 mcompressible 1620 16.37 A sharpedged orifice is used to measure the flow of air in a pipe. The pipe diameter is 100 mm and the diameter of the orifice is 25 mm. Upstream of the orifice, the absolute pressure is 150 kPa and the temperature is 35C. The pressure drop across the orifice is 15 kPa, and the coefficient of discharge is 0.62. Determine the mass flow rate in the pipeline. k1P 0.4 15 T = Ti = 308.15 = 8.8 1.4 150 k Pi vi = RTi/Pi = 0.5896 Pe = 135 kPa, Te = 299.35, ve = 0.6364 . . mi = me Vi = Ve(De/Di)2 vi/ve = 0.0579 hi  he = V2(1  0.05792)/2 = Cp(Ti  Te) e Vs =
e 2 1000 1.0035 8.8/(1  0.0579)2 = 133.1 m/s . m = 0.62 (/4) (0.025)(2133.1/0.6364) = 0.06365 kg/s 16.38 A critical nozzle is used for the accurate measurement of the flow rate of air. Exhaust from a car engine is diluted with air so its temperature is 50C at a total pressure of 100 kPa. It flows through the nozzle with throat area of 700 mm2 by suction from a blower. Find the needed suction pressure that will lead to critical flow in the nozzle, the mass flow rate and the blower work, assuming the blower exit is at atmospheric pressure 100 kPa. P* = 0.5283 Po = 52.83 kPa, T* = 0.8333 To = 269.3 K v* = RT*/P* = 0.287 269.3/52.83 = 1.463 c* = kRT* = 1.4 1000 0.287 269.3 = 328.9 m/s . m = Ac*/v* = 700 106 328.9/1.463 = 0.157 kg/s 1621 16.39 A convergent nozzle is used to measure the flow of air to an engine. The atmosphere is at 100 kPa, 25C. The nozzle used has a minimum area of 2000 mm2 and the coefficient of discharge is 0.95. A pressure difference across the nozzle is measured to 2.5 kPa. Find the mass flow rate assuming incompressible flow. Also find the mass flow rate assuming compressible adiabatic flow. Assume Vi 0, vi = RTi/Pi = 0.287 298.15/100 = 0.8557 Ve,s2/2 = hi  he,s = vi(Pi  Pe) = 2.1393 kJ/kg Ve,s = 2 1000 2.1393 = 65.41 m/s . ms = AVe,s/vi = 2000 106 65.41/0.8557 = 0.153 kg/s . . ma = CDms = 0.1454 kg/s Te,s = Ti (Pe/Pi)(k1)/k = 298.15(97.5/100)0.2857 = 296 K h = CpT = 1.0035 2.15 = 2.1575 = Ve,s2/2 Ve,s = 2 1000 2.1575 = 65.69 m/s ve,s = 0.287 296/97.5 = 0.8713 . ms = AVe,s/ve,s = 2000 106 65.69/0.8713 = 0.1508 kg/s . . ma = CDms = 0.1433 kg/s 1622 16.40 A convergent nozzle with exit diameter of 2 cm has an air inlet flow of 20C, 101 kPa (stagnation conditions). The nozzle has an isentropic efficiency of 95% and the pressure drop is measured to 50 cm water column. Find the mass flow rate assuming compressible adiabatic flow. Repeat calculation for incompressible flow. Convert P to kPa: P = 50 cm H2O = 0.5 9.8064 = 4.903 kPa T0 = 20C = 293.15 K Assume inlet Vi = 0 P0 = 101 kPa Pe = P0  P = 101  4.903 = 96.097 kPa Pe k1 96.097 0.2857 Te = T0 ( ) k = 293.15 ( ) = 289.01 101 P0 V2/2 = hi  he = Cp (Ti  Te) = 1.004 (293.15  289.01) e = 4.1545 kJ/kg = 4254.5 J/kg Ve 2 /2 = Ve2s/2 = 0.95 4154.5 = 3946.78 ac Te ac = Ti e ac = Pe RTp Ve 2 /2 ac Cp = = 293.15 3.9468 = 289.2 1.0035 => Ve = 91.15 m/s Ve ac = 88.85 m/s 96.097 = 1.158 kg/m3 0.287 289.2 . m = AV = 1.158 0.022 88.85 = 0.0323 kg/s 4 16.41 Steam at 600 kPa, 300C is fed to a set of convergent nozzles in a steam turbine. The total nozzle exit area is 0.005 m 2 and they have a discharge coefficient of 0.94. The mass flow rate should be estimated from the measurement of the pressure drop across the nozzles, which is measured to be 200 kPa. Determine the mass flow rate. se,s = si = 7.3724 kJ/kg K ; ve,s = 0.5932, hi = 3061.6 Ve,s = 2 1000(3061.6  2961) = 448.55 m/s (Pe, se,s) he,s = 2961 kJ/kg . m = 0.005 448.55/0.5932 = 3.781 kg/s 1623 16.42 The coefficient of discharge of a sharpedged orifice is determined at one set of conditions by use of an accurately calibrated gasometer. The orifice has a diameter of 20 mm and the pipe diameter is 50 mm. The absolute upstream pressure is 200 kPa and the pressure drop across the orifice is 82 mm of mercury. The temperature of the air entering the orifice is 25C and the mass flow rate measured with the gasometer is 2.4 kg/min. What is the coefficient of discharge of the orifice at these conditions? P = 82 101.325/760 = 10.93 kPa k1 0.4 T = Ti P/Pi = 298.15 10.93/200 = 4.66 1.4 k vi = RTi/Pi = 0.4278, ve = RTe/Pe = 0.4455 Vi = VeAevi/Aive = 0.1536 Ve (V2  V2)/2 = V2(1  0.15362)/2 = hi  he = CpT e i e Ve = 2 1000 1.004 4.66/(1  0.15362) = 97.9 m/s . m = AeVe/ve = 0.022 97.9/0.4455 = 0.069 kg/s 4 CD = 2.4/60 0.069 = 0.58 1624 16.43 (Adv.) Atmospheric air is at 20C, 100 kPa with zero velocity. An adiabatic reversible compressor takes atmospheric air in through a pipe with crosssectional area of 0.1 m2 at a rate of 1 kg/s. It is compressed up to a measured stagnation pressure of 500 kPa and leaves through a pipe with crosssectional area of 0.01 m2. What is the required compressor work and the air velocity, static pressure and temperature in the exit pipeline? C.V. compressor out to standing air and exit to stagnation point. . . . _ m ho1 + Wc = m(h + V2/2)ex = mho,ex . . o mso1 = msex Pr,o,ex = Pr,o1 (Pst,ex/Po1) = 1.028(500/100) = 5.14 To,ex = 463, ho,ex = 465.38, ho1 = 209.45 . _ Wc = m(ho,ex  ho1) = 1(465.38  209.45) = 255.9 kW Pex = Po,ex(Tex/To,ex)k/(k1)
2 Tex = To,ex  Vex/2Cp . m = 1 kg/s = (AV)ex = PexAVex/RTex . Now select 1 unknown amongst Pex, Tex, Vex and write the continuity eq. m and solve the nonlinear equation. Say, use Tex then Vex = 2Cp(To,ex  Tex) 2Cp(To,ex  Tex)/RTex o . m = 1 kg/s = Pex(Tex/To,ex)k/k1A solve for Tex/To,ex (close to 1) Tex = 462.6 K Vex= 28.3 m/s, Pex = 498.6 kPa 1625 English Unit Problems
16.44E Steam leaves a nozzle with a velocity of 800 ft/s. The stagnation pressure is 100 lbf/in2, and the stagnation temperature is 500 F. What is the static pressure and temperature? h1 = ho1 V2/2gc= 1 8002 Btu 1279.1 778 = 1266.3 lbm 2 32.174 s1 = s0 = 1.7085 Btu/lbm R (h, s) Computer table P1 = 88 lbf/in.2, T = 466 F 16.45E Air leaves the compressor of a jet engine at a temperature of 300 F, a pressure of 45 lbf/in2, and a velocity of 400 ft/s. Determine the isentropic stagnation temperature and pressure. ho1  h1 = V2/2gc = 4002/2 32.174 778 = 3.2 Btu/lbm 1 To1  T  1 = (ho1  h1)/Cp = 3.2/0.24 = 13.3 To1 = T + T = 300 + 13.3 = 313.3 F = 773 R Po1 = P1 To1/T1k1 = 45(773/759.67)3.5 = 47.82 lbf/in2 k 16.46E A meteorite melts and burn up at temperatures of 5500 R. If it hits air at 0.75 lbf/in.2, 90 R how high a velocity should it have to reach such temperature? Assume we have a stagnation T = 5500 R h1 + V2/2 = hstagn. 1 Extrapolating from table C.6, hstagn. = 1546.5, h1 = 21.4 V2/2 = 1546.5 21.4 = 1525.1 Btu/lbm 1 V1 = 2 32.174 778 1525.1 = 8738 ft/s 1626 16.47E A jet engine receives a flow of 500 ft/s air at 10 lbf/in.2, 40 F inlet area of 7 ft2 with an exit at 1500 ft/s, 10 lbf/in.2, 1100 R. Find the mass flow rate and thrust. . m = AV; ideal gas = P/RT . m = (P/RT)AV = 10 144 7 500 = 189.1 lbm/s 53.34 499.7 . Fnet = m (Vex  Vin) = 189.1 (1500  500) / 32.174 = 5877 lbf 16.48E A water turbine using nozzles is located at the bottom of Hoover Dam 575 ft below the surface of Lake Mead. The water enters the nozzles at a stagnation pressure corresponding to the column of water above it minus 20% due to friction. The temperature is 60 F and the water leaves at standard atmospheric pressure. If the flow through the nozzle is reversible and adiabatic, determine the velocity and kinetic energy per kilogram of water leaving the nozzle. P = gZ = g(Z/v)/gc = 575/(0.016035 144) = 249 lbf/in.2 gc and Bernoulli vP = Vex2/2 Pac = 0.8P = 199.2 lbf/in.2 Vex = 2v P = 2g Z = 2 32.174 575 = 192.4 ft/s 2 Vex/2 = vP = gZ/gc = 575/778 = 0.739 Btu/lbm 16.49E Find the speed of sound for air at 15 lbf/in.2, at the two temperatures of 32 F and 90 F. Repeat the answer for carbon dioxide and argon gases. From eq. 16.28 we have c32 = c90 = kRT = 1.4 32.174 53.34 491.7 = 1087 ft/s 1.4 32.174 53.34 549.7 = 1149 ft/s For Carbon Dioxide: R = 35.1, k = 1.289 c32 = c90 = 1.289 32.174 35.1 491.7 = 846 ft/s 1.289 32.174 35.1 549.7 = 894.5 ft/s For Argon: R = 38.68, k = 1.667 c32 = c90 = 1.667 32.174 38.68 491.7 = 1010 ft/s 1.667 32.174 38.68 549.7 = 1068 ft/s 1627 16.50E Air is expanded in a nozzle from 300 lbf/in.2, 1100 R to 30 lbf/in. 2. The mass flow rate through the nozzle is 10 lbm/s. Assume the flow is reversible and adiabatic and determine the throat and exit areas for the nozzle. 2 k P * = Po k1 k+1
Velocity Area 300 psia P Density 30 psia Mach # = 300 0.5283 = 158.5 lbf/in.2. T* = To 2/(k+1) = 1100 0.8333 = 916.6 R v* = RT*/P* = 53.34 916.6/(158.5 144) = 2.1421 ft3/lbm kRT* = 1.4 32.174 53.34 916.6 = 1484 ft/s . A* = mv*/c* = 10 2.1421/1484 = 0.0144 ft2 c* = P2/Po = 30/300 = 0.1 Table A.11 M* = 1.701 = V2/c* 2 V2 = 1.701 1484 = 2524 ft/s T2 = 916.6 0.5176 = 474.4 R v2 = RT2/P2 = 53.34 474.4/(30 144) = 5.8579 ft3/lbm . A2 = mv2/V2 = 10 5.8579 / 2524 = 0.0232 ft2 1628 16.51E A convergent nozzle has a minimum area of 1 ft2 and receives air at 25 lbf/in.2, 1800 R flowing with 330 ft/s. What is the back pressure that will produce the maximum flow rate and find that flow rate? P* 2 k =( )k1 = 0.528 Po k+1 Find Po: Critical Pressure Ratio from table C.6 Cp = (463.445  449.794)/50 = 0.273 T0 = Ti + V2/2Cp h0 = h1 + V2/2 1 T0 = 1800 + 3302/2 = 1807.97 => T * = 0.8333 To = 1506.6 R 32.174 778 0.273 P0 = Pi (T0/Ti)k/(k1) = 25 (1807.97/1800)3.5 = 25.39 P* = 0.528 Po = 0.528 25.39 = 13.406 lbf/in2 13.406 144 P* = = = 0.024 RT* 53.34 1506.6
* V = c = kRT* = 1.4 53.34 1506.6 32.174 = 1902.6 ft/s . m = AV = 0.024 1 1902.6 = 45.66 lbm/s 16.52E A jet plane travels through the air with a speed of 600 mi/h at an altitude of 20000 ft, where the pressure is 5.75 lbf/in. 2 and the temperature is 25 F. Consider the diffuser of the engine where air leaves at with a velocity of 300 ft/s. Determine the pressure and temperature leaving the diffuser, and the ratio of inlet to exit area of the diffuser, assuming the flow to be reversible and adiabatic. V = 600 mi/h = 880 ft/s v1 = 53.34 484.67/(5.75 144) = 31.223 , h1 = 115.91, Pr1 = 0.7637 ho1 = 115.91 + 880 2/(2 32.174 778) = 131.38 Btu/lbm Pro1 = 1.1822 Po1 = 5.75 1.1822 = 8.9 lbf/in.2 0.7637 h2 = 131.38  3002/(2 32.174 778) = 129.58 T2 = 542 R, Pr2 = 1.1267 => P2 = 8.9 1.1267/1.1822 = 8.48 lbf/in.2 v2 = 53.34 542/(8.482 144) = 23.67 ft3/lbm A1/A2 = (31.223/23.67)(300/880) = 0.45 1629 16.53E The products of combustion enter a nozzle of a jet engine at a total pressure of 18 lbf/in.2, and a total temperature of 1200 F. The atmospheric pressure is 6.75 lbf/in.2. The nozzle is convergent, and the mass flow rate is 50 lbm/s. Assume the flow is adiabatic. Determine the exit area of the nozzle. Pcrit = P2 = 18 0.5283 = 9.5 lbf/in.2 > Pamb The flow is then choked. T2 = 1660 0.8333 = 1382 R V 2 = c2 = 1.4 32.174 53.34 1382 = 1822 ft/s v2 = 53.34 1382/9.5 144 = 53.9 A2 = 50 53.9/1822 = 1.479 ft2 16.54E Repeat Problem 16.52 assuming a diffuser efficiency of 80%. From solution to 16.52 h1 = 115.91, ho1 = 131.38, Pr1 = 0.7637, v1 = 31.223 h2 = 129.58, Pr2 = 1.1267, Pro1 = 1.1822 D = (h3  h1)/(ho1  h1) = 0.8 h3 = 128.29, Pr3 = 1.088 Po2=P3 = 5.75 1.088/0.7637 = 8.192 lbf/in.2 P2 = Po2Pr2/Pro1 = 8.192 1.1267/1.1822 = 7.807 lbf/in.2 T2 = 542 R v2 = 53.34 542 144 = 25.716 ft3/lbm 7.807 A1/A2 = v1V2/v2V1 = 31.223 300/(25.716 880) = 0.414 1630 16.55E A 50ft3 uninsulated tank contains air at 150 lbf/in.2, 1000 R. The tank is now discharged through a small convergent nozzle to the atmosphere at 14.7 lbf/in.2 while heat transfer from some source keeps the air temperature in the tank at 1000 R. The nozzle has an exit area of 2 104 ft2. a. Find the initial mass flow rate out of the tank. b. Find the mass flow rate when half the mass has been discharged. c. Find the mass of air in the tank and the mass flow rate out of the tank when the nozzle flow changes to become subsonic. PB/Po = 14.7/150 = 0.098 < (P*/Po)crit = 0.5283 a. The flow is choked, max possible flow rate ME =1 ; PE = 0.5283 150 = 79.245 lbf/in.2 TE = T* = 0.8333 1000 = 833.3 R VE = c = kRT* = 1.4 53.34 833.3 32.174 = 1415 ft/s vE = RT*/PE = 53.34 833.3/(79.245 144) = 3.895 ft3/lbm . Mass flow rate is : m1 = AVE/vE = 2 104 1415/3.895 = 0.0727 lbm/s b. m1 = P1V/RT1 = 150 50 144/53.34 1000 = 20.247 lbm m2 = m1/2 = 10.124 lbm, P2=P1/2 = 75 lbf/in.2 ; T = T1 2 PB/P2= 14.7/75 = 0.196 < (P*/Po)crit The flow is choked and the velocity is the same as in 1) PE = 0.5283 75 = 39.623 lbf/in.2 ; ME =1 2 104 1415 39.623 144 . = 0.0303 lbm/s m2 = AVEPE/RTE = 53.34 1000 c. Flow changes to subsonic when the pressure ratio reaches critical. PB/Po = 0.5283 P3 = 27.825 lbf/in.2 m3 = m1P3/P1 = 3.756 lbm ; T3 = T1 VE = 1415 ft/s 2 104 1415 27.825 144 . = 0.02125 lbm/s m3 = AVEPE/RTE = 53.34 1000 1631 16.56E Air enters a diffuser with a velocity of 600 ft/s, a static pressure of 10 lbf/in.2, and a temperature of 20 F. The velocity leaving the diffuser is 200 ft/s and the static pressure at the diffuser exit is 11.7 lbf/in. 2 . Determine the static temperature at the diffuser exit and the diffuser efficiency. Compare the stagnation pressures at the inlet and the exit. To1 = T1 + To2 = To1 V2 1 2gcCp = 480 + 6002/(2 32.174 778 0.24) = 510 R T2 = To2  V2/2Cp = 510  2002/(2 32.174 0.24 778) = 506.7 R 2 To2  T2 T2 = k1 Po2  P2 Po2  P2 = 0.267 Po2 = 11.97 lbf/in.2 k P2 Tex,s = T1 (Po2/P1)(k1)/k = 480 1.0528 = 505.3 R D = Tex,s  T1 To1  T1 = 505.3  480 = 0.844 51  480 1632 16.57E A convergent nozzle with exit diameter of 1 in. has an air inlet flow of 68 F, 14.7 lbf/in.2 (stagnation conditions). The nozzle has an isentropic efficiency of 95% and the pressure drop is measured to 20 in. water column. Find the mass flow rate assuming compressible adiabatic flow. Repeat calculation for incompressible flow. Convert P to lbf/in2 P = 20 in H2O = 20 0.03613 = 0.7226 lbf/in2 T0 = 68 F = 527.7 R Assume inlet Vi = 0 P0 = 14.7 lbf/in2 Pe = P0  P = 14.7  0.7226 = 13.977 lbf/in2 Pe k1 13.977 0.2857 Te = T0 ( ) k = 527.7 ( ) = 520.15 14.7 P0 V2/2 = hi  he = Cp (Ti  Te) = 0.24 (527.7  520.15) = 1.812 Btu/lbm e Ve 2 /2 = V2/2 = 0.95 1.812 = 1.7214 Btu/lbm ac e Ve ac = Te ac = Ti e ac = Pe RTe ac 2 32.174 1.7214 778 = 293.6 ft/s Ve 2 /2 ac Cp = 1.7214 = 520.53 0.24 = 527.7  13.977 144 = 0.07249 lbm/ft3 53.34 520.53 1 . m = AV = 0.07249 ( )2 293.6 = 0.116 lbm/s 4 12 Incompressible: i = P0 RT0 = 14.7 144 = 0.0752 lbm/ft3 53.34 527.7 V2/2 = vi (Pi  Pe) = e P 0.7226 144 = = 1.7785 Btu/lbm i 0.0752 778 Ve 2 /2 = V2/2 = 0.95 1.7785 = 1.6896 Btu/lbm ac e Ve ac = 2 32.174 1.6896 778 = 290.84 ft/s . 1 m = AV = 0.0752 ( )2 290.84 = 0.119 lbm/s 4 12 ...
View
Full
Document
This note was uploaded on 04/18/2008 for the course MTLE 4100 taught by Professor Lewis during the Spring '08 term at Rensselaer Polytechnic Institute.
 Spring '08
 Lewis

Click to edit the document details