Solucionario Introducción a la Termodinamica en Ingeniería Química: Smith, Van Ness & Abbott

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Solution Manual-Chemical Engineering Thermodynamics

Solution Manual-Chemical Engineering Thermodynamics

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  • 1. Chapter 1 - Section A - Mathcad Solutions 1.4 The equation that relates deg F to deg C is: t(F) = 1.8 t(C) + 32. Solve this equation by setting t(F) = t(C). Guess solution: t t = 1.8t Given 0 Find t () 32 Ans. 40 P= 1.5 By definition: F A F = mass g A P 3000bar D 4mm F PA g 9.807 m D 4 F mass 2 g s P= 1.6 By definition: F A A 3000atm D 0.17in F PA g 32.174 ft 4 D mass 2 sec gh 13.535 1.8 gm 3 101.78kPa 13.535 g 9.832 mass 2 384.4 kg Ans. 2 F g A mass 2 0.023 in 1000.7 lbm Ans. gm 3 29.86in_Hg m h 2 56.38cm s Pabs g gh 32.243 Patm ft Pabs h 2 176.808 kPa Ans. 25.62in s cm Patm 12.566 mm Patm cm Patm A F = mass g P 1.7 Pabs = 2 Note: Pressures are in gauge pressure. Pabs gh 1 Patm Pabs 27.22 psia Ans.
  • 2. 1.10 Assume the following: 13.5 gm g 3 m 9.8 2 s cm P 1.11 P h 400bar h g Ans. 302.3 m The force on a spring is described by: F = Ks x where Ks is the spring constant. First calculate K based on the earth measurement then gMars based on spring measurement on Mars. On Earth: F = mass g = K x mass g 0.40kg 9.81 m x 2 1.08cm s F F mass g F x Ks 3.924 N Ks 363.333 N m On Mars: x 0.40cm gMars 1.12 Given: FMars gMars FMars mass d P= dz g 0.01 FMars Kx and: mK = 4 MP RT Substituting: Separating variables and integrating: 1 dP = P Psea 1atm ln M 29 2 mK PDenver Psea = PDenver = Psea e gm mol d P= dz zDenver Mg zDenver RT Mg zDenver RT g 9.8 MP g RT Mg dz RT 0 Psea Taking the exponential of both sides and rearranging: 3 Ans. kg PDenver After integrating: 10 m 2 s
  • 3. 3 R 82.06 cm atm T mol K Mg zDenver RT Mg RT ( 10 zDenver 273.15) K 0.194 zDenver Psea e PDenver 0.823 atm Ans. PDenver PDenver 1 mi 0.834 bar Ans. 1.13 The same proportionality applies as in Pb. 1.11. gearth ft 32.186 gmoon 2 5.32 M lmoon 1.14 gearth gmoon costbulb Ans. 18.767 lbf Ans. hr 0.1dollars 70W 10 day kW hr hr 5.00dollars 10 day 1000hr costelec dollars yr costelec 25.567 dollars yr costtotal 43.829 dollars yr Ans. 18.262 costtotal costbulb D 18.76 113.498 113.498 lbm wmoon M gmoon costbulb 1.15 learth M learth lbm wmoon lmoon 2 s s learth ft 1.25ft costelec mass 250lbm g 32.169 ft 2 s 3
  • 4. Patm (a) F 30.12in_Hg Patm A l Work 1.7ft D D 2 3 10 lbf Work PE mass g l mass 1.227 ft 2 Ans. Ans. 16.208 psia F l 0.47m A 2.8642 Pabs PE 1.16 4 F mass g F A (b) Pabs (c) A 3 10 ft lbf Ans. 4.8691 Ans. 424.9 ft lbf g 150kg 9.813 m 2 s Patm (a) F Patm A l 0.83m EP 1.18 mass EK 2 A 4 110.054 kPa Work EP mass g l u 1250kg 1 2 mass u 2 40 EK m s Work EK Wdot = 200W Ans. 1000 kJ mass g h 0.91 0.92 time Wdot Ans. 1000 kJ g 9.8 m 2 s 4 h 2 0.173 m Ans. 10 N F l Work 1.19 1.909 Pabs Work D 4 F mass g F A (b) Pabs (c) A 101.57kPa 50m Ans. 15.848 kJ Ans. 1.222 kJ Ans.
  • 5. Wdot mdot 1.22 a) cost_coal mdot g h 0.91 0.92 0.488 25.00 ton cost_coal MJ 29 kg cost_gasoline 0.95 GJ kg s 1 2.00 gal 37 GJ cost_gasoline Ans. 14.28 GJ 1 3 m cost_electricity 0.1000 kW hr cost_electricity 27.778 GJ 1 b) The electrical energy can directly be converted to other forms of energy whereas the coal and gasoline would typically need to be converted to heat and then into some other form of energy before being useful. The obvious advantage of coal is that it is cheap if it is used as a heat source. Otherwise it is messy to handle and bulky for tranport and storage. Gasoline is an important transportation fuel. It is more convenient to transport and store than coal. It can be used to generate electricity by burning it but the efficiency is limited. However, fuel cells are currently being developed which will allow for the conversion of gasoline to electricity by chemical means, a more efficient process. Electricity has the most uses though it is expensive. It is easy to transport but expensive to store. As a transportation fuel it is clean but batteries to store it on-board have limited capacity and are heavy. 5
  • 6. 1.24 Use the Matcad genfit function to fit the data to Antoine's equation. The genfit function requires the first derivatives of the function with respect to the parameters being fitted. B T C A Function being fit: f ( A B C) T e First derivative of the function with respect to parameter A d f ( A B C) T dA B exp A T C First derivative of the function with respect to parameter B d f ( A B C) T dB 1 T C B exp A T C First derivative of the function with respect to parameter C d f ( A B C) T dC B ( T 2 exp A C) 18.5 3.18 9.5 5.48 0.2 11.8 t 9.45 16.9 23.1 32.7 Psat 28.2 41.9 44.4 66.6 52.1 89.5 63.3 129 75.5 187 6 B T C
  • 7. T t 273.15 lnPsat ln ( ) Psat Array of functions used by Mathcad. In this case, a0 = A, a1 = B and a2 = C. exp a0 exp a0 F ( a) T a1 T a2 a1 T 1 exp a0 T a2 a1 T a2 2 Guess values of parameters 15 a2 guess a1 T exp a0 3000 50 a2 a1 T a2 Apply the genfit function A 13.421 A B B genfit ( Psat guess F) T C 2.29 C 3 10 Ans. 69.053 Compare fit with data. 200 150 Psat f ( A B C) T 100 50 0 240 260 280 300 320 340 360 T To find the normal boiling point, find the value of T for which Psat = 1 atm. 7
  • 8. Psat 1atm B Tnb A Tnb 1.25 a) t1 1970 C2 C1 ( 1 t2 i) 273.15K 2000 dollars gal C1 0.35 1.513 329.154 K Ans. 56.004 degC C2 t2 t1 Tnb C K Psat ln kPa i 5% dollars gal The increase in price of gasoline over this period kept pace with the rate of inflation. b) t1 1970 Given t2 C2 C1 = (1 2000 i) C1 t2 t1 i 16000 dollars yr Find ( i) i C2 80000 dollars yr 5.511 % The salary of a Ph. D. engineer over this period increased at a rate of 5.5%, slightly higher than the rate of inflation. c) This is an open-ended problem. The strategy depends on age of the child, and on such unpredictable items as possible financial aid, monies earned by the child, and length of time spent in earning a degree. 8
  • 9. Chapter 2 - Section A - Mathcad Solutions 2.1 (a) Mwt g 35 kg 9.8 m z 2 5m s Work (b) Work Mwt g z Utotal Utotal Work (c) By Eqs. (2.14) and (2.21): dU 1.715 kJ Ans. 1.715 kJ Ans. d ( )= CP dT PV Since P is constant, this can be written: MH2O CP dT = MH2O dU MH2O P dV Take Cp and V constant and integrate: MH2O CP t2 t1 = Utotal kJ MH2O 30 kg CP 4.18 t1 20 degC kg degC t2 t1 Utotal MH2O CP t2 20.014 degC Ans. (d) For the restoration process, the change in internal energy is equal but of opposite sign to that of the initial process. Thus Q (e) Utotal Q 1.715 kJ Ans. In all cases the total internal energy change of the universe is zero. 2.2 Similar to Pb. 2.1 with mass of water = 30 kg. Answers are: (a) W = 1.715 kJ (b) Internal energy change of the water = 1.429 kJ (c) Final temp. = 20.014 deg C (d) Q = -1.715 kJ 9
  • 10. 2.4 The electric power supplied to the motor must equal the work done by the motor plus the heat generated by the motor. E i 9.7amp Wdotelect Qdot 2.5 i E Wdotelect t Eq. (2.3): U = Q Step 1 to 2: Ut12 Wdotelect 1.067 3 134.875 W Ans. W W12 200J Ut12 Q34 Q12 800J Ut34 1.25hp 10 W Qdot Wdotmech Q12 Step 3 to 4: Wdotmech 110V W12 6000J W34 Q34 Ut34 W34 3 5.8 Ans. 10 J 300J Ans. 500 J t t Step 1 to 2 to 3 to 4 to 1: Since U is a state function, U for a series of steps that leads back to the initial state must be zero. Therefore, the sum of the t U values for all of the steps must sum to zero. Ut41 4700J Ut23 Step 2 to 3: Ut23 W23 Ut34 Ut41 Ans. 4000 J Ut23 Ut12 4 Ut23 3 Q23 Q23 3800J W23 10 J 200 J Ans. For a series of steps, the total work done is the sum of the work done for each step. W12341 1400J 10
  • 11. W41 W12341 Step 4 to 1: W12 W23 Ut41 Ut41 Ans. 10 J 3 4.5 Q41 W41 3 4.5 W41 4700J Q41 Note: W41 W34 10 J 200 J Ans. Q12341 = W12341 2.11 The enthalpy change of the water = work done. M Wdot 2.12 Q CP 20 kg 4.18 kJ kg degC t 10 degC M CP t 0.25 kW 0.929 hr Wdot Ans. U Q 12 kJ W U 19.5 kJ Ans. Q 12 kJ U W U 7.5 kJ Q 12 kJ Ans. 2.13Subscripts: c, casting; w, water; t, tank. Then mc Uc mw Uw mt Ut = 0 Let C represent specific heat, C = CP = CV Then by Eq. (2.18) mc Cc tc mw Cw tw mt Ct tt = 0 mc 2 kg mw Cc 0.50 kJ kg degC Ct mt 40 kg tc 500 degC Given t1 mc Cc t2 t2 0.5 5 kg kJ kg degC Cw 25 degC t2 30 degC tc = mw Cw mt Ct t2 Find t2 11 t2 4.18 kJ kg degC (guess) t1 27.78 degC Ans.
  • 12. 2.15 mass (a) (b) T g CV 1 kg Ut 1K 9.8 m Ut mass CV T EP 2 kJ kg K 4.18 Ans. 4.18 kJ Ut s z (c) 2.17 EP EK z z mass g 426.531 m Ans. EK u Ut 1 mass 2 1000 50m u kg 5 A 3 A 2m mdot Wdot uA mdot mdot g z Wdot 4 D 2 (b) 2 4 kg 1.571 10 7.697 s 3 10 kW U1 P1 1002.7 kPa U1 H1 763.131 U2 kJ 2784.4 kg H2 U2 P1 V 1 kJ kg kJ kg cm 1.128 gm V1 Ans. 3 P2 U P2 V 2 2022.4 Ans. 3 kJ 762.0 kg U Ans. 3.142 m H1 2.18 (a) m s m s u m D 91.433 Ans. 1500 kPa U2 U1 H 12 cm 169.7 gm V2 H 2275.8 kJ kg H2 H1 Ans.
  • 13. 2.22 D1 (a) u1 2.5cm m s D2 5cm For an incompressible fluid, =constant. By a mass balance, mdot = constant = u 1A1 = u2A2 u2 (b) D1 u1 u2 D2 1 EK 2 2 2 u2 1 2 2.23 Energy balance: Mass balance: u1 2 mdot3 H3 mdot1 1.0 Qdot 30 Qdot H1 T1 CP 4.18 s mdot1 mdot2 H2 = Qdot H2 = Qdot mdot2 CP T3 25degC kJ Ans. kg mdot2 H3 T1 kg s J mdot2 = 0 mdot2 = Qdot mdot1 CP T1 Ans. 1.875 mdot1 mdot Cp T3 T3 CP mdot1 m s mdot1 H1 mdot1 H3 or 0.5 EK mdot3 Therefore: T3 2 T2 = Qdot mdot1 CP T1 mdot2 CP T2 mdot2 0.8 kg s T2 75degC kJ kg K mdot2 CP T2 mdot2 CP T3 43.235 degC 2 2.25By Eq. (2.32a): By continuity, incompressibility H u = 0 2 A1 u2 = u1 13 A2 H = CP T CP 4.18 kJ kg degC Ans.
  • 14. 2 u = u1 u1 u1 2 2 CP D2 2 u = u1 1 m s 14 D1 D1 2 1 A2 SI units: T 2 A1 2 4 1 D2 D1 D2 2.5 cm 3.8 cm 4 T 0.019 degC Ans. T D2 0.023 degC Ans. 7.5cm u1 T 2 2 CP D1 1 4 D2 Maximum T change occurrs for infinite D2: D2 cm u1 T 2.26 T1 2 CP 300K Wsdot H 2 D1 1 T D2 T2 u1 520K ndot 98.8kW CP T2 4 50 H T1 10 0.023 degC m s u2 kmol hr 3.5 10 molwt 29 kg kmol 7 R 2 CP 6.402 m s Ans. 3 kJ kmol By Eq. (2.30): Qdot H 2 u2 2 2 u1 2 molwt ndot Wsdot Qdot 2 2.27By Eq. (2.32b): By continunity, constant area H= u2 = u1 u 2 gc V2 u2 = u1 V1 14 also T 2 P1 T 1 P2 Ans. 9.904 kW V2 V1 = 2 T 2 P1 T 1 P2 2 u = u2 u1 2
  • 15. 2 u = u1 P1 R T2 T2 P2 molwt mol rankine 7 2 u1 20 psi ft lbf 28 20 7 R T2 2 ft s T1 T1 579.67 rankine H2 2726.5 kJ kg kJ kg Ans. gm mol (guess) 578 rankine Given H = CP T = 1 T 1 P2 100 psi 3.407 2 T 2 P1 2 2 R T2 T1 = T2 Find T2 u1 2 2 T 2 P1 T 1 P2 1 molwt Ans. 578.9 rankine ( 119.15 degF) 2.28 u1 3 m s u2 200 m s H1 2 By Eq. (2.32a): 2.29 u1 u2 m s m 500 s 30 By Eq. (2.32a): Q H2 H1 3112.5 u1 u2 H1 334.9 kJ kg 2 Q 2 kJ H2 kg 2411.6 2945.7 (guess) 2 H2 Given u2 578.36 m s H1 = u2 u1 2 5 cm V1 388.61 u2 2 Ans. 3 3 D1 kJ kg cm gm 15 V2 667.75 cm gm Find u2
  • 16. Continuity: 2.30 (a) t1 D2 t2 30 degC CV u1 V2 u2 V1 D1 D2 Ans. 1.493 cm n 250 degC 3 mol J mol degC 20.8 By Eq. (2.19): Q n CV t2 Q t1 Ans. 13.728 kJ Take into account the heat capacity of the vessel; then cv mv Q (b) 100 kg mv cv t1 200 degC CP 29.1 n CV CV Q Q Ans. 11014 kJ n 40 degC 4 mol joule mol degC Q n CP t2 t2 70 degF 5 kJ kg degC t1 t2 By Eq. (2.23): 2.31 (a) t1 t2 0.5 BTU mol degF n CV t2 Q t1 18.62 kJ n 350 degF 3 mol By Eq. (2.19): Q t1 4200 BTU Ans. Take account of the heat capacity of the vessel: mv Q mv cv cv 200 lbm (b) t1 400 degF n CV t2 0.12 BTU lbm degF Q t1 t2 150 degF 16 10920 BTU n Ans. 4 mol Ans.
  • 17. CP 7 BTU mol degF Q H1 2.33 n CP t2 1322.6 V1 Q t1 BTU BTU lbm 1148.6 3 V2 lbm 78.14 ft 4 mdot mdot V2 10 22.997 u2 Wdot 2.34 H1 V1 H2 307 9.25 BTU lbm ft H2 330 V2 u1 lbm ft 0.28 173.99 BTU lb Ans. 39.52 hp BTU 3 lbm Ws 2 Wdot Ws mdot u1 u2 H1 20 ft s molwt 44 3 D1 lbm D2 4 in 1 in 2 mdot u2 4 D1 u1 mdot V1 V2 mdot u2 679.263 9.686 2 4 D2 2 Eq. (2.32a): Q H2 H1 10 in sec 2 D2 Eq. (2.32a): Ws D2 ft sec 2 4 3 in 4 lb 3.463 V1 mdot 10 D1 lbm 2 u2 ft s u1 3 2 D 1 u1 Ans. 7000 BTU H2 lbm ft 3.058 By Eq. (2.23): ft sec u1 u2 2 17 lb hr 2 Ws Ws molwt 5360 Q BTU lbmol 98.82 BTU lbm gm mol
  • 18. Qdot 2.36 T1 Qdot mdot Q P 300 K 67128 BTU hr Ans. 1 kg n 1 bar 28.9 V1 n gm 34.602 mol mol 3 3 bar cm T1 83.14 mol K P cm 24942 mol V1 V2 P dV = n P V 1 W= n V2 = n P V1 3 V1 V1 Whence W Given: V2 V1 joule 29 Q n H U mol K Q = T1 3 Whence T1 Q W U n CP T2 602.08 kJ 12.41 kJ mol Ans. 172.61 kJ H T2 = T1 CP W n P 2 V1 T2 H 3 T1 17.4 kJ mol Ans. Ans. Ans. 2.37 Work exactly like Ex. 2.10: 2 steps, (a) & (b). A value is required for PV/T, namely R. R 8.314 T1 (a) Cool at const V1 to P2 (b) Heat at const P2 to T2 Ta2 T1 P2 P1 Ta2 293.15 K T2 333.15 K P1 J mol K 1000 kPa P2 100 kPa 7 R 2 CP 29.315 K 18 CV 5 R 2
  • 19. Tb T2 Hb C P Tb Hb Ua CV Ta Ua R T1 V1 P1 Ta2 Tb V1 303.835 K 2.437 8.841 Ta 10 5.484 10 mol Ha Ua V 1 P2 P1 Ha Ub Hb P2 V 2 V1 Ub U Ub U 0.831 H 2.39 Ua Ha Hb H 1.164 996 kg 9.0 10 3 D 5 2 kJ mol ms 1 cm u 5 1 m 5 s 5 22133 Re D u Re Ta 263.835 K mol 3 J mol V2 3 R T2 V2 P2 7.677 6.315 0.028 m mol 3 J 10 mol 3 J 10 mol Ans. mol 4 kg m 2 kJ T1 3 J 3 3 m 10 Ta2 55333 110667 276667 19 Ans. D 0.0001 Note: D = /D in this solution
  • 20. 0.00635 fF 0.3305 ln 0.27 D 7 Re 0.9 2 0.00517 fF 0.00452 0.0039 0.313 mdot u 4 D 2 mdot 1.956 kg 1.565 s Ans. 9.778 0.632 P L 2 fF D 2 P L u 0.206 kPa 11.254 m Ans. 3.88 2.42 mdot 4.5 kg s H1 761.1 kJ kg H2 536.9 kJ kg Assume that the compressor is adiabatic (Qdot = 0). Neglect changes in KE and PE. Wdot Cost mdot H2 15200 H1 Wdot Wdot 1.009 3 10 kW 0.573 Cost kW 20 799924 dollars Ans.
  • 21. Chapter 3 - Section A - Mathcad Solutions 3.1 1 d = 1 = dT d dP P T At constant T, the 2nd equation can be written: d = 2 ln dP 6 44.1810 P = 1 bar 2 = 1.01 1 ln ( ) 1.01 P P Ans. P2 = 226.2 bar 225.2 bar 3 3.4 b cm 0.125 gm c 2700 bar P1 P2 1 bar 500 bar V2 Since Work = a bit of algebra leads to P dV V1 P2 Work c P P b dP Work 0.516 Work 0.516 P1 J Ans. gm Alternatively, formal integration leads to Work 3.5 = a P1 c P2 P1 a bP P2 1 atm b ln P2 b P1 b 3.9 10 6 atm 1 b V 3000 atm 0.1 10 1 ft 3 9 J gm Ans. atm 2 (assume const.) Combine Eqs. (1.3) and (3.3) for const. T: P2 Work ( a V Work b P)P dP P1 21 16.65 atm ft 3 Ans. 1
  • 22. 3.6 1.2 10 V1 3 degC 1 CP 0.84 kJ kg degC M 5 kg t2 20 degC 3 m 1 P 1590 kg t1 1 bar 0 degC With beta independent of T and with P=constant, dV = V V2 dT Vtotal Vtotal M V Work V1 exp P Vtotal M CP t2 P2 (a) Constant V: T2 T1 U H R T1 ln (c) Adiabatic: T1 P1 Ans. 84 kJ 84 kJ Ans. 83.99 kJ Ans. T T1 U 10.91 15.28 H= 0 Q= 0 22 kJ mol kJ mol Work and CV U = Q = CV T T2 and 7 R 2 CP 600 K and Q Ans. Utotal U= P2 7.638 joule Work H CP T (b) Constant T: Work Q 1 bar V1 Ans. Q Q and CV T m t1 T P1 3 V2 Htotal W= 0 P2 10 Q Utotal 8 bar 5 V Work 7.638 Htotal P1 t1 (Const. P) Q 3.8 t2 525 K Ans. Ans. and 10.37 Q= W kJ mol Ans. U = W = CV T 5 2 R
  • 23. 1 CP T2 CV P2 T1 U and U 5.586 3.9 P4 2bar CP H kJ 7 R 2 10bar T1 Step 41: Adiabatic T 331.227 K Ans. mol T4 R T1 P1 V1 P4 T1 kJ mol Ans. 5 R 2 CV 600K 7.821 T2 CP T H CV T W P1 T2 P1 V1 4.988 T4 3 3 m 10 378.831 K mol R CP P1 3 J U41 CV T1 T4 U41 4.597 10 H41 CP T1 T4 H41 6.436 10 Q41 3bar T2 Step 12: Isothermal Q41 U41 W41 4.597 mol J 0 mol W41 P2 J 0 mol mol 3 J V2 600K U12 0 J mol H12 0 J mol 23 R T2 P2 3 J 10 3 V2 m 0.017 mol U12 0 J mol H12 0 J mol mol T1
  • 24. Q12 W12 P3 2bar V3 Q12 Q12 P1 W12 P3 V 3 T3 V2 Step 23: Isochoric P2 R T1 ln T3 R U23 CV T3 H23 CP T3 T2 Q23 2 bar T4 CV T3 W23 P4 0 T2 R T4 V4 3 J 10 mol 3 J 6.006 10 400 K 3 J 4.157 10 P4 3 V4 m 0.016 mol U34 CV T4 T3 U34 439.997 H34 CP T4 T3 H34 615.996 Q34 CP T4 T3 Q34 W34 R T4 T3 W34 3.10 For all parts of this problem: T2 = T1 mol mol 3 J H23 5.82 10 mol 3 J Q23 4.157 10 mol J W23 0 mol J mol 378.831 K Step 34: Isobaric U23 T2 6.006 615.996 175.999 J mol J mol J mol J mol and Also Q = Work and all that remains is U= H= 0 to calculate Work. Symbol V is used for total volume in this problem. P1 1 bar P2 V1 12 bar 24 3 12 m V2 3 1m
  • 25. (a) P2 Work = n R T ln Work P1 Work P1 V1 ln P2 P1 Ans. 2982 kJ (b) Step 1: adiabatic compression to P2 1 5 3 Vi P2 V i W1 V1 P1 P2 (intermediate V) P1 V 1 Vi W1 3063 kJ W2 1 3 2.702 m 2042 kJ Step 2: cool at const P2 to V2 W2 P2 V 2 Work W1 Vi W2 Work 5106 kJ Ans. (c) Step 1: adiabatic compression to V2 Pi W1 P1 V1 (intermediate P) V2 Pi V 2 1 Step 2: No work. Work (d) Step 1: heat at const V1 to P2 62.898 bar W1 P1 V 1 Pi 7635 kJ W1 Work 7635 kJ Ans. W2 Work 13200 kJ Ans. W1 = 0 Step 2: cool at const P2 to V2 W2 P2 V 2 V1 Work (e) Step 1: cool at const P1 to V2 W1 P1 V 2 W1 V1 25 1100 kJ
  • 26. Step 2: heat at const V2 to P2 Work 3.17(a) W2 = 0 Work W1 No work is done; no heat is transferred. t U = (b) T= 0 Not reversible T2 = T1 = 100 degC The gas is returned to its initial state by isothermal compression. Work = n R T ln 3 V1 3.18 (a) P1 P2 V2 ln 7 2 but V2 4 3 3 P2 V1 6 bar 878.9 kJ Ans. Work V2 T1 500 kPa 303.15 K CP 5 R 2 CV R n R T = P2 V 2 m P2 100 kPa CP V1 V2 4m Work CV Adiabatic compression from point 1 to point 2: Q12 0 kJ mol 1 U12 = W12 = CV T12 U12 CV T2 U12 3.679 T1 kJ mol T2 H12 CP T2 W12 H12 5.15 T1 kJ W12 mol T1 3.679 P2 P1 U12 kJ mol Cool at P2 from point 2 to point 3: T3 Ans. 1100 kJ H23 T1 U23 CV T3 CP T3 W23 T2 26 Q23 T2 U23 Q23 H23 Ans.
  • 27. H23 5.15 Q23 5.15 kJ U23 mol kJ W23 mol kJ mol 3.679 Ans. kJ 1.471 Ans. mol Isothermal expansion from point 3 to point 1: U31 = H31 = 0 Q31 4.056 W31 P2 P1 R T3 ln P3 W31 W31 P3 kJ mol FOR THE CYCLE: Q Q Q12 1.094 Q23 Q31 U= 4.056 kJ mol Ans. H= 0 Work kJ mol W12 Work Q31 1.094 W23 W31 kJ mol (b) If each step that is 80% efficient accomplishes the same change of state, all property values are unchanged, and the delta H and delta U values are the same as in part (a). However, the Q and W values change. Step 12: W12 Q12 Step 23: W23 Q23 Step 31: W31 Q31 W12 W12 W23 0.8 U23 4.598 Q12 0.92 kJ mol W23 0.8 U12 kJ W12 1.839 kJ mol mol W31 0.8 W31 Q23 5.518 kJ mol W31 W23 3.245 kJ mol Q31 27 3.245 kJ mol
  • 28. FOR THE CYCLE: Q Q12 Q Q23 3.192 Work kJ mol W12 Work Q31 3.192 W23 W31 kJ mol 3.19Here, V represents total volume. P1 CP 21 CV joule mol K CP T2 208.96 K P1 V 1 n P1 V 1 T2 Work n CV P2 P1 V2 Ans. T1 P2 V 2 P1 V 1 69.65 kPa Ans. Work Work = Work Pext V2 994.4 kJ Ans, U = Pext V V1 Work U = n CV T R T1 V1 Ans. 1609 kJ V2 P2 P1 200 kPa T2 1 100 kPa V2 V1 P1 (c) Restrained adiabatic: Pext P2 Work P2 P2 V 2 V1 P2 600 K T2 Work CV V2 (b) Adiabatic: 600 K CP V1 P1 V1 ln T1 5 V1 R Work = n R T1 ln T1 Work V2 1m (a) Isothermal: T2 3 V1 1000 kPa T2 V 1 T2 V 2 T1 28 442.71 K Ans. P2 T1 147.57 kPa Ans. 400 kJ Ans.
  • 29. 3.20 T1 CP P1 423.15 K 7 2 Step 12: If V1 V2 = 2.502 Step 23: V1 V3 kJ mol 0 Process: 2.079 U12 mol mol R T1 ln r () Q12 2.502 CV T3 CP T3 T2 2.079 W23 kJ mol H23 Work Q Q23 2.502 0.424 2.91 kJ mol kJ mol kJ mol H H12 H23 H 2.91 U U12 U23 U 2.079 29 kJ mol T2 H23 W12 Q12 kJ U23 U23 Q 323.15 K W12 W12 U23 Work 0 T 3 P1 kJ mol kJ mol T3 T 1 P3 r Q12 W23 T1 kJ Then Q23 Q23 0 3 bar T2 R 2 H12 r= W12 5 CV R P3 8bar kJ mol kJ mol Ans. Ans. Ans. Ans.
  • 30. 3.21 By Eq. (2.32a), unit-mass basis: But CP u1 2.5 t2 3.22 28 Whence H = CP T R 7 2 molwt molwt u2 1 2 H u2 T= m s u2 t1 gm mol 2 2 u1 2 CP 50 m s t1 u1 t2 2 CP 7 R 2 CV 5 R 2 P1 1 bar P3 148.8 degC Ans. 10 bar CV T3 U 2.079 T1 mol CP T3 H Ans. T3 303.15 K H T1 kJ 2.91 T1 kJ mol Each part consists of two steps, 12 & 23. (a) T2 W23 T3 R T2 ln P2 P3 P2 P1 150 degC 2 2 CP U 2 u = 0 T2 T1 Work W23 Work 6.762 Q U Q 4.684 30 kJ mol Ans. Work kJ mol Ans. Ans. 403.15 K
  • 31. (b) P2 T2 P1 H12 CP T2 W12 U12 W23 R T2 ln Work W12 Q (c) U T2 U12 T3 CV T2 Q12 Q12 P3 H12 W12 T1 0.831 W23 P2 Work W23 Q Work P2 T1 P3 T1 W12 kJ mol 7.718 6.886 4.808 kJ mol kJ mol kJ mol CP T3 T2 Q23 CV T3 T2 W23 U23 Work 4.972 P1 H23 U23 Ans. P2 R T1 ln H23 Ans. Work Q W12 U W23 Q Work 2.894 Q23 kJ mol kJ mol For the second set of heat-capacity values, answers are (kJ/mol): U = 2.079 U = 1.247 (a) Work = 6.762 Q = 5.515 (b) Work = 6.886 Q = 5.639 (c) Work = 4.972 Q = 3.725 31 Ans. Ans.
  • 32. 3.23 T1 303.15 K T2 T1 T3 393.15 K P1 1 bar P3 12 bar CP 7 R 2 For the process: U U Step 12: W12 CV T3 1.871 P2 5.608 kJ mol mol For the process: Work W12 Q 3.24 Work 5.608 W12 = 0 Work = W23 = P2 V3 But T3 = T1 Also W = R T1 ln ln P T2 T1 P = P1 T1 P1 exp T2 T2 R T1 ln 5.608 P P1 Ans. P2 U P1 kJ mol W23 kJ Q mol 3.737 V 2 = R T3 kJ mol Ans. T2 Work = R T2 So... mol Q23 kJ mol W23 Q23 kJ Q12 W12 Step 23: Q12 2.619 T1 W12 T3 0 CP T3 H kJ T1 P3 Q12 H T1 5 R 2 CV T1 Therefore T1 T1 32 T1 800 K P 350 K 2.279 bar P1 Ans. 4 bar
  • 33. 3.25 3 VA P Define: 256 cm P1 r 0.0639 VB 3750.3 cm CV = r CP Assume ideal gas; let V represent total volume: P1 V B = P 2 V A P P1 3.26 T1 = VA VA VA ( 1) r VB VB 300 K From this one finds: VB P1 r 1 atm CP 7 R 2 3 Ans. CP R CV The process occurring in section B is a reversible, adiabatic compression. Let TA ( )= TA final P ( )= P2 final TB ( )= TB final Since the total volume is constant, nA = nB nA R TA 2 nA R T1 = P2 P1 TB TA TB 2 T1 = P2 P1 or (1) 1 (a) P2 TA 2 T1 TB 1.25 atm P2 P1 Q = nA TB Define q= TB TA 319.75 K T1 Q nA q 430.25 K 33 P2 (2) P1 UA CV TA q UB TB 2 T1 3.118 kJ mol (3) Ans.
  • 34. (b) Combine Eqs. (1) & (2) to eliminate the ratio of pressures: (guess) 425 K TB 300 K TB Find TB TB TA 319.02 K Ans. P2 1.24 atm Ans. kJ mol Ans. 1 TB = T1 Given P2 P1 q (c) TA P1 TA 2 T1 P2 P2 (1) TB P1 kJ mol 2.993 1 T1 CV TA 3 q 2 T1 By Eq. (2), TB Eliminate q (1) TB TB P2 (d) 2 T1 TB 325 K q TB 2 T1 CV TA TB TA Ans. TA 469 K Ans. q 2 T1 TA 1.323 atm TB q P1 2 T1 C V 4.032 kJ mol Ans. from Eqs. (1) & (3): P2 1.241 atm Ans. (2) P2 TB 319.06 K Ans. (1) TA 425.28 K Ans. P1 1 TB T1 TA 2 T1 P2 P1 P2 P1 TB 34
  • 35. 6 3 3.30 B P1 B' cm 242.5 C 25200 mol mol P2 7.817 T B C 10 3 1 bar B 2 373.15 K 2 55 bar B' 1 bar RT C' cm R T 2 C' 2 3.492 10 5 1 2 bar (a) Solve virial eqn. for initial V. RT P1 Guess: V1 Given P1 V 1 = 1 RT B V1 3 C V1 2 V1 V2 Find V1 cm 30780 mol cm 241.33 mol V1 Solve virial eqn. for final V. Guess: RT P2 V2 P2 V 2 Given RT = 1 B V2 3 C V2 2 V2 Find V2 Eliminate P from Eq. (1.3) by the virial equation: V2 Work RT 1 B V C V 2 1 dV V Work 12.62 kJ mol V1 (b) Eliminate dV from Eq. (1.3) by the virial equation in P: P2 1 dV = R T P C' dP W 1 P RT 2 P1 W 35 12.596 kJ mol Ans. C' P dP Ans.
  • 36. Note: The answers to (a) & (b) differ because the relations between the two sets of parameters are exact only for infinite series. 3.32 Tc 282.3 K T 298.15 K Pc 50.4 bar P 12 bar T Tc Tr Pr Tr Pr P Pc 6 3 B Given cm 140 C mol PV = 1 RT 7200 cm mol B V V (b) B0 V Find ( ) V Tr V Tr Z 1 (c) B0 cm 1919 mol B1 B1 4.2 Tr Z PV RT Z B0 1.6 Pr 3 V 2066 cm mol 2 0.172 0.139 RT P C 0.422 0.083 V 2 3 B1 0.238 (guess) 0.087 (a) 1.056 Ans. Z 0.929 V cm Ans. 1924 mol 0.304 2.262 0.932 10 V 3 3 ZRT P For Redlich/Kwong EOS: 0 1 ( ) Tr T r Pr Tr 0.5 Pr Tr Table 3.1 Eq. (3.53) 36 Table 3.1 0.42748 0.08664 q Tr Tr Tr Eq. (3.54)
  • 37. Calculate Z Guess: Given T r Pr 0.9 Eq. (3.52) Z= 1 Z Z Z Find Z) ( (d) q Tr Z T r Pr Z T r Pr Z T r Pr 3 Z RT V 0.928 T r Pr V P 1916.5 cm mol Ans. For SRK EOS: 2 1 Tr 1 0.480 Tr q Tr (e) 1 Tr Guess: Z Table 3.1 2 Pr T r Pr Tr T r Pr q Tr Z Find Z) ( 0.9 Z T r Pr Z V 0.928 T r Pr T r Pr Z T r Pr 3 Z RT V P 1918 1 2 0.45724 0.07779 2 1 q Tr cm mol Ans. For Peng/Robinson EOS: 1 Tr Eq. (3.53) Tr Eq. (3.52) Given Z 0.176 Eq. (3.54) Calculate Z Z= 1 1.574 2 Table 3.1 0.42748 0.08664 0 1 1 0.37464 Tr 1.54226 0.26992 Eq. (3.54) Tr 37 2 T r Pr 1 Tr Pr Tr 2 Table 3.1 2 Table 3.1 Eq. (3.53)
  • 38. Calculate Z Guess: 0.9 Eq. (3.52) Given Z= 1 T r Pr Z Z Find ( Z) q Tr Z Z T r Pr Z T r Pr 3 V 1900.6 cm mol 305.3 K Pc T 323.15 K Tr T Tc Tr P 15 bar Pr P Pc Pr 0.308 (guess) 0.100 6 3 (a) cm 156.7 mol B Given PV = 1 RT B V C 9650 cm mol V (b) B0 V Find ( V) Tr B1 V Tr (c) B0 cm 1625 mol B1 B1 4.2 Tr Z PV Z B0 1.6 Pr 3 V cm 1791 mol 2 0.172 0.139 RT P C 0.422 0.083 V 2 3 1 Ans. 1.058 48.72 bar 3.33 Tc Z Z T r Pr ZRT P V 0.92 T r Pr Ans. Z RT 0.907 V cm Ans. 1634 mol 0.302 3.517 0.912 V 10 3 ZRT P 3 For Redlich/Kwong EOS: 1 0 0.08664 38 0.42748 Table 3.1
  • 39. ( ) Tr 0.5 Tr Table 3.1 Pr T r Pr Tr q Tr Eq. (3.54) Tr Eq. (3.53) Tr Calculate Z Guess: Given T r Pr 0.9 Eq. (3.52) Z= 1 Z Z Z Find Z) ( (d) q Tr Z T r Pr Z Z T r Pr cm 1622.7 mol V P Ans. For SRK EOS: 0.42748 0.08664 0 1 Tr 1 0.480 Tr q Tr 0.176 Eq. (3.54) Guess: 2 1 Tr T r Pr Tr Calculate Z Z Table 3.1 2 Pr Eq. (3.53) Tr 0.9 Eq. (3.52) Given Z= 1 Z 1.574 Table 3.1 2 1 (e) T r Pr 3 Z RT V 0.906 T r Pr T r Pr Find Z) ( q Tr Z T r Pr Z Z V 0.907 T r Pr Z RT P T r Pr Z T r Pr 3 V 1624.8 cm mol Ans. For Peng/Robinson EOS: 1 2 1 0.07779 2 39 0.45724 Table 3.1
  • 40. 1 Tr 1 0.37464 Tr q Tr 1.54226 0.26992 Eq. (3.54) 1 Guess: Z Tr Table 3.1 2 Pr T r Pr Tr Calculate Z 2 2 Eq. (3.53) Tr 0.9 Eq. (3.52) Given Z= 1 T r Pr q Tr Z Z T r Pr Z T r Pr Z T r Pr 3 ZRT V 0.896 T r Pr cm 1605.5 mol V Z Find ( Z) 3.34 Tc 318.7 K T 348.15 K Tr T Tc Tr Pc 37.6 bar P 15 bar Pr P Pc Pr Ans. 1.092 0.399 P 0.286 (guess) 6 3 (a) B Given 194 cm C mol PV = 1 RT 15300 cm mol B V V cm 1722 mol Find ( V) (b) B0 0.083 V 1930 mol 2 3 V 3 cm C V V 2 RT P 0.422 Tr 1.6 B0 40 Z 0.283 PV RT Z 0.893 Ans.
  • 41. B1 0.172 0.139 Tr Z 1 B0 (c) B1 4.2 Pr B1 Z Tr 0.02 V 0.899 3 Z RT V P 1734 cm mol Ans. For Redlich/Kwong EOS: ( ) Tr Tr 0.5 Table 3.1 Pr T r Pr Table 3.1 0.42748 0.08664 0 1 Tr q Tr Eq. (3.54) Tr Eq. (3.53) Tr Calculate Z Guess: Given T r Pr 0.9 Eq. (3.52) Z= 1 Z Z Find Z) ( (d) q Tr Z Z T r Pr Z T r Pr Z T r Pr 3 Z RT V 0.888 T r Pr cm 1714.1 mol V P Ans. For SRK EOS: 0.42748 0.08664 0 1 1 Tr q Tr 1 0.480 Tr 1.574 0.176 Eq. (3.54) Tr 41 2 1 T r Pr Tr Table 3.1 2 Table 3.1 2 Pr Tr Eq. (3.53)
  • 42. Calculate Z Guess: Eq. (3.52) Given Z= 1 T r Pr Z (e) q Tr Z Find ( Z) Z 0.9 Z T r Pr Z Z T r Pr T r Pr 3 ZRT P V 0.895 T r Pr V 1726.9 Ans. mol For Peng/Robinson EOS: 1 1 2 0.45724 0.07779 2 1 Tr 1 0.37464 Tr q Tr T r Pr Guess: Z T r Pr q Tr Z Find ( Z) Tr 2 Table 3.1 2 Pr Eq. (3.53) Tr 0.9 523.15 K Z T r Pr Z ZRT P V 0.882 P T r Pr cm 152.5 mol B Given Z PV RT T r Pr Z T r Pr 3 cm 1701.5 mol V Ans. 1800 kPa 6 3 (a) 1 Table 3.1 Eq. (3.52) Z= 1 T 2 0.26992 Tr Given Z 1.54226 Eq. (3.54) Calculate Z 3.35 cm C mol C B V PV = 1 RT V 2 2 V RT (guess) P V 5800 cm Find ( V) 3 V 2250 42 cm mol Z 0.931 Ans.
  • 43. (b) Tc Tr Tr B1 647.1 K Pc T Tc Pr 0.139 0.172 Tr (c) Table F.2: B0 Pc 4.2 0.422 0.083 Tr B0 0.082 B1 0.51 Z 0.281 1.6 1 B0 B1 Pr Tr 3 Z RT P V P Pr 0.808 0.345 220.55 bar Z molwt V V cm molwt 124.99 gm V 0.939 cm 2268 mol cm 2252 mol Ans. 3 gm 18.015 mol 3 or cm 53.4 mol B T Zi Z1i C 2620 cm mol 2 D 5000 cm mol 3 n mol 273.15 K PV Given i 9 6 3 3.37 Ans. RT 0 10 Pi fPi Vi Pi RT 1 B Pi RT = 1 10 B V 10 C V 2 D V fP V) ( 3 RT Pi Vi 20 i bar Find V) ( (guess) Eq. (3.12) Eq. (3.38) 43 Z2i 1 2 1 4 B Pi RT Eq. (3.39)
  • 44. 1·10 -10 Z2i Z1i Zi 20 1 1 1 40 0.953 0.953 0.951 60 0.906 0.906 0.895 0.861 0.859 0.83 0.749 80 Pi 100 bar 0.819 0.812 120 0.784 0.765 0.622 140 0.757 0.718 0.5+0.179i 160 0.74 0.671 0.5+0.281i 180 0.733 0.624 0.5+0.355i 200 0.735 0.577 0.5+0.416i 0.743 0.53 0.5+0.469i Note that values of Z from Eq. (3.39) are not physically meaningful for pressures above 100 bar. 1 0.9 Zi 0.8 Z1 i Z2 i 0.7 0.6 0.5 0 50 100 Pi bar 44 150 1 200
  • 45. 3.38 (a) Propane: Tc Tc 313.15 K P Tr T Pc T Tr 369.8 K 0.847 Pr 0.152 42.48 bar 13.71 bar P Pc Pr 0.323 For Redlich/Kwong EOS: ( ) Tr Tr 0.5 Table 3.1 Pr T r Pr Table 3.1 0.42748 0.08664 0 1 Tr q Tr Eq. (3.54) Tr Eq. (3.53) Tr Calculate Z for liquid by Eq. (3.56) Guess: Z 0.01 Given Z= Z T r Pr Z Find Z) Z ( T r Pr 0.057 Z Calculate Z for vapor by Eq. (3.52) Z T r Pr q Tr T r Pr 3 Z RT P V 1 T r Pr V Guess: 108.1 cm Ans. mol Z 0.9 V cm 1499.2 mol Given Z= 1 Z T r Pr Find Z) ( q Tr Z T r Pr Z Z Z V 0.789 45 T r Pr T r Pr Z RT P 3 Ans.
  • 46. Rackett equation for saturated liquid: T Tc Tr Tr 0.847 3 Vc V 200.0 V c Zc cm Zc mol 1 Tr 0.276 3 0.2857 V 94.17 cm Ans. mol For saturated vapor, use Pitzer correlation: B0 0.083 0.422 Tr B1 0.139 V 0.468 B1 0.207 0.172 Tr RT P B0 1.6 4.2 R B0 B1 Tc V Pc 46 1.538 3 3 cm 10 mol Ans.
  • 47. Parts (b) through (t) are worked exactly the same way. All results are summarized as follows. Volume units are cu.cm./mole. R/K, Liq. R/K, Vap. Rackett Pitzer (a) 108.1 1499.2 94.2 1537.8 (b) 114.5 1174.7 98.1 1228.7 (c) 122.7 920.3 102.8 990.4 (d) 133.6 717.0 109.0 805.0 (e) 148.9 1516.2 125.4 1577.0 (f) 158.3 1216.1 130.7 1296.8 (g) 170.4 971.1 137.4 1074.0 (h) 187.1 768.8 146.4 896.0 (i) 153.2 1330.3 133.9 1405.7 (j) 164.2 1057.9 140.3 1154.3 (k) 179.1 835.3 148.6 955.4 (l) 201.4 645.8 160.6 795.8 (m) 61.7 1252.5 53.5 1276.9 (n) 64.1 1006.9 55.1 1038.5 (o) 66.9 814.5 57.0 853.4 (p) 70.3 661.2 59.1 707.8 (q) 64.4 1318.7 54.6 1319.0 (r) 67.4 1046.6 56.3 1057.2 (s) 70.8 835.6 58.3 856.4 (t) 74.8 669.5 60.6 700.5 47
  • 48. 3.39 (a) Propane T Tc 273.15)K T Tr ( 40 T Tc Tr Pc 369.8 K Pr 0.847 0.152 42.48 bar P P Pc 13.71 bar Pr 313.15 K 0.323 From Table 3.1 for SRK: 0.42748 0.08664 0 1 2 1 Tr 1 0.480 Tr q Tr 1.574 0.176 Eq. (3.54) 2 1 2 Tr Pr T r Pr Tr Calculate Z for liquid by Eq. (3.56) Guess: Eq. (3.53) Tr Z 0.01 Given Z= Z T r Pr Find Z) ( Z Z T r Pr 0.055 Z Calculate Z for vapor by Eq. (3.52) Z T r Pr q Tr T r Pr 3 Z RT P V 1 T r Pr cm 104.7 mol V Guess: Z Ans. 0.9 Given Z= 1 Z T r Pr Find Z) ( q Tr Z 0.78 Z T r Pr Z V 48 T r Pr Z T r Pr Z RT P T r Pr 3 V 1480.7 cm mol Ans.
  • 49. Parts (b) through (t) are worked exactly the same way. All results are summarized as follows. Volume units are cu.cm./mole. SRK, Liq. SRK, Vap. Rackett Pitzer (a) 104.7 1480.7 94.2 1537.8 (b) 110.6 1157.8 98.1 1228.7 (c) 118.2 904.9 102.8 990.4 (d) 128.5 703.3 109.0 805.0 (e) 142.1 1487.1 125.4 1577.0 (f) 150.7 1189.9 130.7 1296.8 (g) 161.8 947.8 137.4 1074.0 (h) 177.1 747.8 146.4 896.0 (i) 146.7 1305.3 133.9 1405.7 (j) 156.9 1035.2 140.3 1154.3 (k) 170.7 815.1 148.6 955.4 (l) 191.3 628.5 160.6 795.8 (m) 61.2 1248.9 53.5 1276.9 (n) 63.5 1003.2 55.1 1038.5 (o) 66.3 810.7 57.0 853.4 (p) 69.5 657.4 59.1 707.8 (q) 61.4 1296.8 54.6 1319.0 (r) 63.9 1026.3 56.3 1057.2 (s) 66.9 817.0 58.3 856.4 (t) 70.5 652.5 60.6 700.5 49
  • 50. 3.40 (a) Propane T ( 40 Tr Tc T Tc Pc 369.8 K T 273.15)K Tr P P Pc 13.71 bar Pr 313.15 K Pr 0.847 0.152 42.48 bar 0.323 From Table 3.1 for PR: 2 1 Tr 1 1 0.37464 1 2 0.26992 2 Eq. (3.54) 1 2 Pr T r Pr Tr Tr 0.45724 0.07779 2 Tr q Tr 1.54226 Calculate Z for liquid by Eq. (3.56) Guess: Eq. (3.53) Tr Z 0.01 Given Z= Z T r Pr Find Z) ( Z Z T r Pr 0.049 Z Calculate Z for vapor by Eq. (3.52) Z T r Pr q Tr T r Pr 3 Z RT P V 1 T r Pr cm 92.2 mol V Guess: Z Ans. 0.6 Given Z= 1 Z T r Pr Find Z) ( q Tr Z 0.766 Z T r Pr Z V 50 T r Pr Z T r Pr Z RT P T r Pr 3 V 1454.5 cm mol Ans.
  • 51. Parts (b) through (t) are worked exactly the same way. All results are summarized as follows. Volume units are cu.cm./mole. PR, Liq. PR, Vap. Rackett Pitzer (a) 92.2 1454.5 94.2 1537.8 (b) 97.6 1131.8 98.1 1228.7 (c) 104.4 879.2 102.8 990.4 (d) 113.7 678.1 109.0 805.0 (e) 125.2 1453.5 125.4 1577.0 (f) 132.9 1156.3 130.7 1296.8 (g) 143.0 915.0 137.4 1074.0 (h) 157.1 715.8 146.4 896.0 (i) 129.4 1271.9 133.9 1405.7 (j) 138.6 1002.3 140.3 1154.3 (k) 151.2 782.8 148.6 955.4 (l) 170.2 597.3 160.6 795.8 (m) 54.0 1233.0 (n) 56.0 987.3 55.1 1038.5 (o) 58.4 794.8 57.0 853.4 (p) 61.4 641.6 59.1 707.8 (q) 54.1 1280.2 54.6 1319.0 (r) 56.3 1009.7 56.3 1057.2 (s) 58.9 800.5 58.3 856.4 (t) 62.2 636.1 60.6 700.5 53.5 1276.9 51
  • 52. 3.41 (a) For ethylene, molwt 28.054 gm Tc mol 282.3 K Pc 50.40 bar 0.087 T Tr Tc 328.15 K P 35 bar P Pc Pr T Tr 1.162 Pr 0.694 From Tables E.1 & E.2: Z Z0 Z1 Z Z0 Z1 0.838 0.033 0.841 Vtotal ZnRT P Vtotal 0.421 m P 115 bar Vtotal 0.25 m Tr 1.145 Pr P Pc From Tables E.3 & E.4: Z0 0.482 Z1 0.126 18 kg n molwt (b) T 323.15 K T Tc Tr Z Z0 Z Z1 mass Pr ZRT mass n molwt Ans. 3 P Vtotal n 0.493 3 n 2.282 2171 mol Ans. 60.898 kg 3.42 Assume validity of Eq. (3.38). 3 P1 1bar Z1 P1 V 1 R T1 T1 Z1 300K 0.922 cm 23000 mol V1 R T1 Z1 P1 B With this B, recalculate at P2 Z2 1 B P2 R T1 Z2 0.611 P2 V2 52 R T1 Z2 P2 B 1 3 3 cm 1.942 10 mol 5bar V2 3.046 10 3 3 cm mol Ans.
  • 53. 3.43 T 753.15 K Tc 513.9 K Tr T Tc Tr 1.466 P 6000 kPa Pc 61.48 bar Pr P Pc Pr 0.976 0.645 B0 0.083 0.422 Tr B1 0.172 0.139 Tr RT P V B0 For an ideal gas: 3.44 T 320 K 0.104 3 V cm 989 mol V Pc cm 1044 mol Ans. 3 RT P V P 0.146 B1 4.2 Tc B1 R B0 1.6 Tc 369.8 K Pc Zc 16 bar 42.48 bar 0.276 molwt 3 Vc 0.152 Tr Vliq Vtank cm 200 mol T Tc Tr V c Zc 1 Tr 3 B0 0.083 Tr B1 0.139 1.6 0.172 Tr 4.2 Pr gm mol 0.377 3 cm Vliq 0.8 Vtank Vliq molwt 0.422 P Pc 0.2857 mliq 0.35 m Pr 0.865 44.097 B0 0.449 B1 0.177 53 96.769 mliq 127.594 kg mol Ans.
  • 54. RT P Vvap B0 B1 R Tc 3 3 cm Vvap 1.318 mvap Pc 10 2.341 kg mol 0.2 Vtank mvap Vvap Ans. molwt 3.45 T B0 Tc 425.1 K 2.43 bar Pc 37.96 bar Pr Vvap 0.083 0.422 Tr B1 0.139 RT P 1.6 0.172 Tr V T Tr 0.200 P 298.15 K 4.2 B0 3 gm mol 0.624 Tc B1 R V Pc Vvap mvap 0.064 0.661 B1 58.123 0.701 Pr Pc molwt 16 m B0 Tr Tc P mvap V molwt 9.469 98.213 kg 3 3 cm 10 mol Ans. P 333.15 K Tc 305.3 K Tr T Tc Tr 1.091 14000 kPa Pc 48.72 bar Pr P Pc Pr 2.874 0.100 3.46 (a) T Vtotal From tables E.3 & E.4: Z0 3 molwt 0.15 m 0.463 54 Z1 0.037 30.07 gm mol
  • 55. Z Z0 Vtotal methane (b) Z Z1 methane V molwt Vtotal V P 40 kg or Tr = Whence V 0.459 Z Tr = PV 29.548 R Tc at 90.87 mol P V = Z R T = Z R Tr Tc 20000 kPa 0.889 Z V P cm Ans. 49.64 kg where 3 Z RT Pr P Pc Pr mol kg 4.105 This equation giving Tr as a function of Z and Eq. (3.57) in conjunction with Tables E.3 & E.4 are two relations in the same variables which must be satisfied at the given reduced pressure. The intersection of these two relations can be found by one means or another to occur at about: Tr Whence T V or T or 3 0.15 m Vtotal T 298.15 K molwt 0.087 50.40 bar P V = P r Pc V = Z R T 40 kg molwt Whence 0.693 118.5 degC Ans. Pc 282.3 K Pr = Z Tr Tc 391.7 K 3.47 Vtotal Tc and 1.283 Z RT Pc V where Pr = 4.675 Z at Tr 55 T Tc 4.675 Tr 1.056 28.054 gm mol
  • 56. This equation giving Pr as a function of Z and Eq. (3.57) in conjunction with Tables E.3 & E.4 are two relations in the same variables which must be satisfied at the given reduced temperature. The intersection of these two relations can be found by one means or another to occur at about: Pr and 1.582 3.48 mwater Z P Pc Pr 3 Vtotal 15 kg P 0.338 V 0.4 m Interpolate in Table F.2 at 400 degC to find: 298.15 K Tc 305.3 K Tr1 P1 2200 kPa Pc 48.72 bar Pr1 3 T2 Z0 mwater T1 Tc P1 Pc 3 V 26.667 Z Z1 Z1 0.422 Tr2 B1 0.139 T2 Tc Z R T1 P1 Tr2 0.806 B0 1.6 0.172 Tr2 Tr1 0.977 Pr1 0.452 1.615 3 V1 B1 4.2 0.113 0.116 R T2 P2 V1 B0 gm 0.0479 V1 .8105 Tr2 493.15 K 0.083 cm Ans. Assume Eq. (3.38) applies at the final state. B0 Ans. 0.100 0.35 m From Tables E.1 & E.2: Z0 Z Vtotal P = 9920 kPa 3.49 T1 Vtotal 79.73 bar B1 R P2 Tc Pc 56 42.68 bar Ans. cm 908 mol
  • 57. 3.50 T B0 0.5 m 0.083 0.139 73.83 bar 0.422 Tr B1 304.2 K Pc 3 Vtotal B0 4.2 V 10 kg molwt 0.224 0.997 molwt 44.01 0.036 Vtotal V Tr Tc 0.341 B1 1.6 0.172 Tr T Tr Tc 303.15 K 3 3 cm 2.2 10 mol RT P V B0 B1 R P Tc 10.863 bar Ans. Pc 3.51 Basis: 1 mole of LIQUID nitrogen P B0 Tc 126.2 K Tr 1 atm Pc 34.0 bar Pr 0.038 Tn molwt 77.3 K 0.083 0.139 1 B0 4.2 B1 Pr Tr 0.842 B1 0.172 Tr Z mol B0 1.6 1.209 Z 0.957 57 Tr P Pc Vliq 0.613 Pr Tc gm 0.422 Tr B1 28.014 Tn 0.03 3 34.7 cm gm mol
  • 58. P Vliq nvapor nvapor Z R Tn 5.718 10 3 mol Final conditions: ntotal T 1 mol V nvapor RT Pig Pig V V T Tc 69.005 Tr ntotal Tr 298.15 K 3 2 Vliq cm 2.363 mol 359.2 bar Use Redlich/Kwong at so high a P. 2 2 a Tr R Tc Pc a 3 3 bar cm 0.901 m 2 b Eq. (3.42) RT V b Tr .5 R Tc Tr 0.651 Eq. (3.43) Pc 3 b mol P ( ) Tr 0.42748 0.08664 a V ( b) V Eq. (3.44) cm 26.737 mol P 450.1 bar Ans. 3 3.52 For isobutane: T1 Tr1 Tr1 300 K T1 Tc 0.735 Tc 408.1 K Pc 36.48 bar V1 1.824 P1 4 bar T2 415 K P2 cm 75 bar Pr1 Pr1 P1 Tr2 Pc Tr2 0.11 58 T2 Tc 1.017 Pr2 Pr2 P2 Pc 2.056 gm
  • 59. From Fig. (3.17): r1 2.45 The final T > Tc, and Fig. 3.16 probably should not be used. One can easily show that r= P Vc with Z from Eq. (3.57) and Tables E.3 and E.4. Thus Z RT 3 Vc Z 262.7 Z0 cm Z0 0.181 mol Z Z1 Eq. (3.75): V1 r2 Tc 3 r1 cm 2.519 gm V2 Pc 469.7 K 1.774 r2 Z R T2 r2 3.53 For n-pentane: 0.0756 P2 V c 0.322 V2 Z1 0.3356 33.7 bar Ans. 0.63 1 gm 3 cm T1 Tr1 Tr1 P1 291.15 K T1 Pr1 Tc T2 1 bar P1 Tr2 Pc T2 0.03 Tr2 2.69 r2 r2 2 2 1 0.532 r1 3.54 For ethanol: Tc Pc Pc 3.561 gm 3 Ans. cm 513.9 K T 453.15 K Tr T Tc Tr 0.882 61.48 bar P 200 bar Pr P Pc Pr 3.253 3 Vc P2 2.27 From Fig. (3.16): By Eq. (3.75), Pr2 0.88 r1 120 bar Pr2 Tc Pr1 0.62 P2 413.15 K 167 cm mol molwt 59 46.069 gm mol
  • 60. From Fig. 3.16: r 2.28 = r r c= 0.629 Vc r Vc gm Ans. 3 cm molwt 3.55 For ammonia: Tc 405.7 K T 293.15 K Tr T Tc Tr 0.723 Pc 112.8 bar P 857 kPa Pr P Pc Pr 0.076 Vc cm 72.5 mol Zc 0.242 3 Eq. (3.72): B0 Vliquid 0.083 0.139 Vvapor cm 27.11 mol Vliquid B0 4.2 B0 B1 R 0.627 B1 1.6 0.172 Tr RT P 3 0.2857 0.422 Tr B1 V c Zc 1 Tr 0.253 0.534 Tc Pc 3 Vvapor cm 2616 mol 3 V Vvapor Vliquid V 60 cm 2589 mol Ans.
  • 61. Alternatively, use Tables E.1 & E.2 to get the vapor volume: Z0 Z1 0.929 Vvapor Z 0.071 Z0 Z Z1 0.911 3 ZRT P Vvapor cm 2591 mol 3 V Vvapor V Vliquid cm 2564 Ans. mol 3.5810 gal. of gasoline is equivalent to 1400 cu ft. of methane at 60 degF and 1 atm. Assume at these conditions that methane is an ideal gas: 3 R ft atm 0.7302 lbmol rankine V 1400 ft T P 1 atm n 3 519.67 rankine PV RT n 3.689 lbmol For methane at 3000 psi and 60 degF: Tc 190.6 1.8 rankine T 519.67 rankine Tr T Tc Tr 1.515 Pc 45.99 bar P 3000 psi Pr P Pc Pr 4.498 Z 0.822 0.012 From Tables E.3 & E.4: Z0 0.819 Vtank ZnRT P Z1 Z 0.234 Vtank 61 5.636 ft Z0 3 Z1 Ans.
  • 62. 3.59 T P 25K 3.213bar Calculate the effective critical parameters for hydrogen by equations (3.58) and (3.56) 43.6 Tc 1 K 1 30.435 K Pc 10.922 bar 2.016T 20.5 Pc Tc bar 21.8K 44.2K 2.016T 0 P Pc Pr Pr Initial guess of volume: T Tr 0.294 Tr cm 646.903 mol 3 RT P V 0.821 V Tc Use the generalized Pitzer correlation B0 0.083 0.422 Tr Z 1 B0 1.6 B0 B1 0.495 B1 0.139 0.172 Tr Pr Tr Z 0.823 Ans. 4.2 B1 0.254 Experimental: Z = 0.7757 For Redlich/Kwong EOS: 0 1 Tr T r Pr Tr 0.5 Pr Tr Table 3.1 Eq. (3.53) 62 Table 3.1 0.42748 0.08664 q Tr Tr Tr Eq. (3.54)
  • 63. Calculate Z Guess: Given T r Pr 0.9 Eq. (3.52) Z= 1 Z Z q Tr Z Find Z) ( 3.61For methane: Z Z Tc 0.012 T T r Pr T r Pr Ans. 0.791 At standard condition: Z T r Pr Experimental: Z = 0.7757 Pc 190.6K ( 60 32) 5 9 45.99bar 273.15 K T P Tr T Tc B0 0.083 Z0 Z 1 Tr 0.422 Tr Pr B0 Z0 1.6 Pitzer correlations: T Tc Tr 0.083 0.422 Tr B1 0.139 1.6 0.172 Tr 4.2 0.134 B1 0.139 0.172 T T Tr Z RT P ( 50 32) 5 9 0.109 Tr 0.00158 3 V1 273.15 K m 0.024 mol P 300psi Pr 0.45 283.15 K 1.486 B0 B1 B1 V1 0.998 B1 Pr Z1 0.998 4.2 0.022 Z1 P Pc Tr Z Z1 (a) At actual condition: B0 B0 Z0 Tr Pr 1.515 1atm Pr Pitzer correlations: 288.706 K 0.141 0.106 63 Pr P Pc
  • 64. Z0 Z 1 B0 Z0 Pr Tr Z1 6 ft q1 150 10 (b) n1 (c) D u Z0 0.957 Z Z1 0.958 q1 n1 7.485 q2 V1 q1 V1 22.624in q2 A A u 4 D 8.738 2 m s 64 3 kmol 10 hr A Ans. 0.0322 3 V2 P V2 q2 day Z1 Tr ZRT V2 3 Pr B1 6.915 Ans. 2 0.259 m 0.00109 10 6 ft m mol 3 day Ans.
  • 65. 3.62 0.012 0.286 0.087 0.281 0.1 0.279 0.140 0.289 0.152 0.276 0.181 0.282 0.187 0.271 0.19 0.267 0.191 0.277 0.194 0.275 0.196 0.273 0.2 0.274 0.205 0.273 0.21 0.273 0.21 ZC Use the first 29 components in Table B.1 sorted so that values are in ascending order. This is required for the Mathcad slope and intercept functions. 0.271 m slope b intercept r corr 0.212 0.272 0.218 0.275 0.23 0.272 0.235 0.269 0.252 0.264 0.28 0.265 ZC 0.297 0.256 m 0.301 0.266 0.302 0.266 0.303 0.263 0.31 0.263 0.322 0.26 0.326 0.261 ( 0.091) 0.27 0.262 ZC ZC ZC ( 0.291) 2 ( 0.878) r 0.771 0.29 0.28 b 0.27 0.26 0.25 0 0.1 0.2 The equation of the line is: Zc = 0.291 0.091 65 0.3 0.4 Ans.
  • 66. 5 7 R 2 Cv 298.15K P1 P2 T1 T3 5bar P3 Cp 1bar 5bar 3.65 Cp T1 2 R 1.4 Cv Step 1->2 Adiabatic compression 1 T2 T1 P2 T2 P1 472.216 K U12 Cv T2 T1 U12 3.618 H12 Cp T2 T1 H12 5.065 Q12 W12 0 kJ mol kJ mol kJ mol kJ mol Q12 0 U12 W12 3.618 Ans. Ans. Ans. kJ mol Ans. Step 2->3 Isobaric cooling U23 Cv T3 T2 U23 3.618 H23 Cp T3 T2 H23 5.065 Q23 W23 R T3 Q23 H23 W23 T2 5.065 1.447 kJ mol kJ mol kJ mol kJ mol Ans. Ans. Ans. Ans. Step 3->1 Isothermal expansion U31 Cv T1 T3 U31 0 H31 Cp T1 T3 H31 0 66 kJ mol kJ mol Ans. Ans.
  • 67. Q31 R T3 ln W31 P1 Q31 Q31 P3 W31 kJ mol kJ 3.99 mol 3.99 Ans. Ans. For the cycle Qcycle Q12 Wcycle Q23 W12 Qcycle Q31 W23 Wcycle W31 1.076 1.076 kJ Ans. mol kJ Ans. mol Now assume that each step is irreversible with efficiency: 80% Step 1->2 Adiabatic compression W12 Q12 W12 U12 W12 Q12 W12 4.522 kJ mol kJ mol 0.904 Ans. Ans. Step 2->3 Isobaric cooling W23 Q23 W23 U23 W23 1.809 kJ mol kJ mol Q23 5.427 W31 W23 3.192 Ans. Ans. Step 3->1 Isothermal expansion W31 Q31 W31 U31 Q31 W31 3.192 kJ mol kJ mol Ans. Ans. For the cycle Qcycle Q12 Wcycle W12 Q23 W23 Qcycle Q31 Wcycle W31 67 kJ Ans. mol kJ 3.14 mol Ans. 3.14
  • 68. 3.67 a) PV data are taken from Table F.2 at pressures above 1atm. 125 150 1757.0 175 P 2109.7 1505.1 200 1316.2 cm 1169.2 gm V kPa 225 250 300 Z 955.45 T ( 300 273.15) K M 1051.6 275 3 18.01 gm mol 875.29 1 VM PVM RT i 0 7 If a linear equation is fit to the points then the value of B is the y-intercept. Use the Mathcad intercept function to find the y-intercept and hence, the value of B Yi Zi 1 3 Xi B intercept ( Y) X A i slope ( Y) X Ans. 6 5 cm 10 2 i A cm 128.42 mol 1.567 B mol X 0 mol 3 cm 10 5 mol 3 cm 8 10 5 mol 3 cm Below is a plot of the data along with the linear fit and the extrapolation to the y-intercept. 68
  • 69. 115 (Z-1)/p 120 125 130 0 2 10 5 4 10 5 6 10 p 5 8 10 5 (Z-1)/p Linear fit b) Repeat part a) for T = 350 C PV data are taken from Table F.2 at pressures above 1atm. 125 150 1912.2 175 P 2295.6 1638.3 200 1432.8 225 V kPa 1273.1 250 300 Z 1040.7 T ( 350 273.15)K M 1145.2 275 3 cm gm 18.01 gm mol 953.52 1 VM PVM RT i 0 7 If a linear equation is fit to the points then the value of B is the y-intercept. Use the Mathcad intercept function to find the y-intercept and hence, the value of B Yi Zi 1 3 Xi i B intercept ( X Y) i 69 B 105.899 cm mol Ans.
  • 70. A A slope ( X Y) 6 5 cm 10 2 1.784 mol X 0 mol 3 10 cm 5 mol 3 8 10 5 mol 3 cm cm Below is a plot of the data along with the linear fit and the extrapolation to the y-intercept. 90 (Z-1)/p 95 100 105 110 0 2 10 5 4 10 5 6 10 p 5 8 10 5 (Z-1)/p Linear fit c) Repeat part a) for T = 400 C PV data are taken from Table F.2 at pressures above 1atm. 125 150 2066.9 175 P 2481.2 1771.1 200 1549.2 225 kPa V 1376.6 250 1238.5 275 1125.5 300 1031.4 70 3 cm gm T ( 400 273.15)K M 18.01 gm mol
  • 71. 1 VM PV M RT Z i 0 7 If a linear equation is fit to the points then the value of B is the y-intercept. Use the Mathcad intercept function to find the y-intercept and hence, the value of B Yi Zi 1 3 Xi B i intercept ( X Y) B i A A slope ( X Y) 89.902 cm mol Ans. 6 5 cm 10 2 2.044 mol X 0 mol 3 cm 10 5 mol 3 8 10 5 mol 3 cm cm Below is a plot of the data along with the linear fit and the extrapolation to the y-intercept. 70 (Z-1)/p 75 80 85 90 0 2 10 5 4 10 p (Z-1)/p Linear fit 71 5 6 10 5 8 10 5
  • 72. 3.70 Create a plot of (Z 1) Z Tr vs Pr Pr Z Tr Data from Appendix E at Tr = 1 0.01 0.9967 0.05 0.9832 0.10 0.9659 Z 0.20 Pr 0.9300 0.40 X 0.7574 0.80 1 0.8509 0.60 Tr 0.6355 Pr (Z Y Z Tr 1) Z Tr Pr Create a linear fit of Y vs X Slope slope ( X Y) Intercept Rsquare Slope intercept ( X Y) 0.033 Intercept corr ( X Y) Rsquare 0.332 0.9965 0.28 0.3 Y Slope X Intercept 0.32 0.34 0 0.2 0.4 0.6 0.8 1 1.2 1.4 X The second virial coefficient (Bhat) is the value when X -> 0 Bhat Intercept By Eqns. (3.65) and (3.66) Bhat B0 0.083 0.422 Tr These values differ by 2%. 72 1.6 B0 0.332 0.339 Ans. Ans.
  • 73. 3.71 Use the SRK equation to calculate Z T T 273.15)Kr Tc 150.9 K T ( 30 Pc 48.98 bar P 300 bar Tr Pr Pc 2.009 Pr Tc P 6.125 0.0 0.42748 Table 3.1 0.08664 0 1 2 1 Tr 1 0.480 Tr q Tr 0.176 Eq. (3.54) 1 Tr Table 3.1 2 Pr T r Pr Tr Calculate Z Guess: Z Eq. (3.53) Tr 0.9 Eq. (3.52) Given Z= 1 Z 1.574 2 T r Pr q Tr Z Find Z) ( Z T r Pr Z V 1.025 T r Pr T r Pr Z T r Pr 3 Z RT P cm Ans. 86.1 mol V This volume is within 2.5% of the ideal gas value. 3.72 After the reaction is complete, there will be 5 moles of C 2H2 and 5 moles of Ca(OH)2. First calculate the volume available for the gas. n 5mol V Vt 3 0.4 1800 cm 3 cm 5 mol 33.0 mol Vt n Vt 3 555 cm 3 V 73 111 cm mol
  • 74. Use SRK equation to calculate pressure. T Tc 308.3 K Pc ( 125 61.39 bar T Tc Tr 273.15) K Tr 0.0 0.42748 Table 3.1 0.08664 0 1 2 1 Tr 1 1.574 0.176 2 2 2 Pc Eq. (3.45) 3 3 bar cm 3.995 m 2 RT V b 3.73 mass b Tc 0.152 ( 10 Vc Pc 369.8K 200.0 197.8 bar Ans. 273.15)K 3 0.276 cm 36.175 mol P b) T 35000kg Eq. (3.46) Pc 3 a V (V R Tc b mol Zc Tr Eq. (3.54) R Tc Tr P 1 Table 3.1 2 Tr a a 0.480 Tr q Tr 1.291 cm mol n M 42.48bar mass M n 44.097 7.937 gm mol 5 10 mol a) Estimate the volume of gas using the truncated virial equation Tr B0 T Tr Tc 0.083 0.422 Tr B0 1.6 0.766 P Eq. (3-65) B1 Pr 1atm 0.139 0.172 Tr B1 0.564 74 0.389 4.2 P Pc Eq. (3-66)
  • 75. Z 1 B0 B1 Pr Z Tr ZnRT Vt 0.981 Vt P 3 This would require a very large tank. If the D tank were spherical the diameter would be: 7 3 2.379 6 Vt 3 10 m m D 32.565 m b) Calculate the molar volume of the liquid with the Rackett equation(3.72) Vliq V c Zc P 1 0.2857 Vvap 3 Vliq 6.294atm Z 1 Tr P Pc Pr B0 B1 85.444 Pr 0.878 Pr Tr ZRT P Vvap Guess: Vtank Given 90% mol 0.15 Z cm 3 3 cm 3.24 10 mol 90% Vliq n Vtank 10% Vtank = n Vvap Vtank Find Vtank Vtank Vliq 75.133 m This would require a small tank. If the tank D were spherical, the diameter would be: 3 3 6 Vtank D 5.235 m Although the tank is smaller, it would need to accomodate a pressure of 6.294 atm (92.5 psi). Also, refrigeration would be required to liquify the gaseous propane stream. 75
  • 76. Chapter 4 - Section A - Mathcad Solutions 4.1 (a) T0 T 473.15 K For SO2: A B 5.699 n 1373.15 K 0.801 10 H 47.007 C 5 D 0.0 1.015 10 R ICPH T0 T A B C D H 3 10 mol kJ T 523.15 K For propane:A 28.785 10 H 161.834 3 Ans. 12 mol C 6 8.824 10 D 0 R ICPH T0 T A B C 0.0 H 470.073 kJ n 1473.15 K B 1.213 n H Q (b) T0 Q mol kJ mol n 473.15 K For ethylene: A 2 (guess) Q= nR A T0 3 10 kJ Ans. 800 kJ 3 14.394 10 K B B 2 T0 2 1 n A 1.967 2 T 2.905 533.15 K For 1-butene: 1.424 1.942 6 4.392 10 C K 2 Given Find (b) T0 Q 10 mol n H Q 4.2 (a) T0 Q C 3 T0 3 1 T T0 15 mol Q 31.630 10 K B 76 3 1 Ans. 1374.5 K 2500 kJ 3 C 9.873 10 K 2 6
  • 77. (guess) 3 Q= nR Given A T0 2 n 500 degF C 3 T0 3 1 T 2.652 Find (c) T0 B 2 T0 2 1 3 T T0 1 Ans. 1413.8 K 6 Q 40 lbmol 10 BTU Values converted to SI units T0 n 533.15K For ethylene: A Q= nR Q 10 mol B 14.394 10 K 1.424 2 (guess) 4 1.814 6 1.055 10 kJ 3 4.392 10 C K Given A T0 1 Find B 2 T0 2 2 T 2.256 C 3 T0 3 1 T T0 3 6 2 1 1202.8 K Ans. T = 1705.4degF 4.3 Assume air at the given conditions an ideal gas. Basis of calculation is 1 second. P T0 1 atm ( T T PV R T0 T0 773.15 K n For air: H 32degF) 273.15K T0 n A T 250 ft V Convert given values to SI units T 3 V 122 degF 932 degF 7.079 m 3 T0 32degF 273.15K 323.15 K 266.985 mol 3.355 B 0.575 10 R ICPH T0 T A B C D 77 3 C 0.0 D 0.016 10 5
  • 78. kJ 13.707 4.4 molwt Q gm mol 100.1 T0 10000 kg molwt n n For CaCO3: A 323.15 K B 10 mol 2.637 10 H 9.441 3 C D 0.0 3.120 10 5 R ICPH T0 T A B C D H 1153.15 K 3 10 BTU Ans. 4 9.99 12.572 3.469 T mol n H Q H 4 J 10 Q mol n H 6 Q Ans. 9.4315 10 kJ 4.7 Let step 12 represent the initial reversible adiabatic expansion, and step 23 the final constant-volume heating. T1 298.15 K T3 298.15 K P1 121.3 kPa P2 101.3 kPa P3 104.0 kPa T2 T3 CP 30 T2 290.41 K Given J mol K T2 = T1 4.9a) Acetone: Tc Hn (guess) 29.10 P2 P2 P3 R CP CP P1 Pc 508.2K kJ 47.01bar Tn Trn mol CP Find CP Tc 56.95 Tn 329.4K Trn J Ans. mol K 0.648 Use Eq. (4.12) to calculate H at Tn ( Hncalc) 1.092 ln Hncalc R Tn Pc 1.013 bar 0.930 Trn 78 Hncalc 30.108 kJ mol Ans.
  • 79. To compare with the value listed in Table B.2, calculate the % error. %error Hncalc Hn %error Hn 3.464 % Values for other components in Table B.2 are given below. Except for acetic acid, acetonitrile. methanol and nitromethane, agreement is within 5% of the reported value. Acetone Acetic Acid Acetonitrile Benz ene iso-Butane n-Butane 1-Butanol Carbon tetrachloride Chlorobenz ene Chloroform Cyclohexane Cyclopentane n-Decane Dichloromethane Diethyl ether Ethanol Ethylbenz ene Ethylene glycol n-Heptane n-Hexane M ethanol M ethyl acetate M ethyl ethyl k etone Nitromethane n-Nonane iso-Octane n-Octane n-Pentane Phenol 1-Propanol 2-Propanol Toluene W ater o-Xylene m-Xylene p-Xylene Hn (k mol) J/ 30.1 40.1 33.0 30.6 21.1 22.5 41.7 29.6 35.5 29.6 29.7 27.2 40.1 27.8 26.6 40.2 35.8 51.5 32.0 29.0 38 .3 30.6 32.0 36.3 37.2 30.7 34.8 25.9 46.6 41.1 39.8 33.4 42.0 36.9 36.5 36.3 79 % error 3.4% 69.4% 9.3% -0.5% -0.7% 0.3% -3.6% -0.8 % 0.8 % 1.1% -0.9% -0.2% 3.6% -1.0% 0.3% 4.3% 0.7% 1.5% 0.7% 0.5% 8 .7% 1.1% 2.3% 6.7% 0.8 % -0.2% 1.2% 0.3% 1.0% -0.9% -0.1% 0.8 % 3.3% 1.9% 2.3% 1.6%
  • 80. b) 33.70 469.7 Tc 507.6 562.2 30.25 Pc K 48.98 36.0 68.7 H25 273.15 K 80.0 366.1 72.150 J M 433.3 gm 392.5 Tn Tr2 Tc ( 25 273.15) K Tc H2 H25 M 0.673 H2 0.628 H1 31.533 Hn 31.549 kJ 33.847 mol 32.242 1 Tr2 1 0.38 Tr1 Eq. (4.13) %error 26.448 H2calc H1 26.429 0.631 H2calc 86.177 gm 78.114 mol 82.145 0.658 Tr1 kJ 30.72 mol 29.97 366.3 80.7 Tr1 28.85 Hn bar 43.50 560.4 Tn 25.79 kJ Ans. 33.571 mol 32.816 31.549 kJ 33.847 mol 32.242 H2 H2 0.072 26.429 H2 H2calc %error 0.052 0.814 % 1.781 The values calculated with Eq. (4.13) are within 2% of the handbook values. 4.10 The ln P vs. 1/T relation over a short range is very nearly linear. Our procedure is therefore to take 5 points, including the point at the temperature of interest and two points on either side, and to do a linear least-squares fit, from which the required derivative in Eq. (4.11) can be found. Temperatures are in rankines, pressures in psia, volumes in cu ft/lbm, and enthalpies in Btu/lbm. The molar mass M of tetrafluoroethane is 102.04. The factor 5.4039 converts energy units from (psia)(cu ft) to Btu. 80
  • 81. (a) T 459.67 V 5 1.934 18.787 P 0 23.767 t 5 26.617 dPdT T H 2 ti 1 459.67 yi ln Pi 15 slope ( x y) slope ( P) 3 xi 10 29.726 slope 1 5 5 21.162 Data: i 0.012 4952 slopedPdT T V dPdT 5.4039 H 0.545 90.078 Ans. The remaining parts of the problem are worked in exactly the same way. All answers are as follows, with the Table 9.1 value in ( ): (a) H = 90.078 ( 90.111) (b) H = 85.817 ( 85.834) (c) H = 81.034 ( 81.136) (d) H = 76.007 ( 75.902) (e) H = 69.863 ( 69.969) 119.377 4.11 M 32.042 153.822 H is the value at 0 degC. 536.4 mol Tc 334.3 512.6 K Pc 80.97 bar Tn 337.9 K 556.4 gm 54.72 45.60 349.8 Hexp is the given value at the normal boiling point. 81 Tr1 273.15K Tc Tr2 Tn Tc
  • 82. 270.9 H 1189.5 0.509 246.9 J gm Hexp 217.8 J gm 1099.5 Hn Hn Hexp Hexp Hn PCE 0.38 Tr1 0.52 Hexp 1195.3 1.092 ln R Tn Hn Hexp 4.03 % M Pc 1.013 bar 0.930 Tr2 100% 0.34 247.7 Hn 1 H Tr2 0.77 J 1055.2 gm 193.2 (b) By Eq. (4.12): PCE 1 0.629 This is the % error 100% 245 Hn 0.659 Tr2 0.491 194.2 (a) By Eq. (4.13) PCE 0.533 Tr1 0.623 J gm PCE 8.72 % 0.96 192.3 4.12 Acetone 0.307 Tc 508.2K Pc Tn 329.4K P 1atm Pr P Pc 47.01bar Zc 0.233 3 Vc Tr cm 209 mol Tn Tc Tr 0.648 82 Hn 29.1 Pr kJ mol 0.022
  • 83. Generalized Correlations to estimate volumes Vapor Volume B0 0.422 0.083 Tr B1 Tr 4.2 Pr Z 1 V B1 Pr Z R Tn P Tr Tr 0.762 Eq. (3.65) B1 0.172 0.139 B0 B0 1.6 0.924 Eq. (3.66) Z V (Pg. 102) 0.965 3 4 cm 2.609 10 mol Liquid Volume 2 Vsat V c Zc 1 Tr 3 7 Eq. (3.72) Vsat Combining the Clapyeron equation (4.11) gives A Vsat 14.3145 V B 2.602 A B H= T V V 2 e C) 3 4 cm 10 d Psat dT T C Psat = e ( T V H= T V B A with Antoine's Equation cm 70.917 mol mol 2756.22 C 83 228.060 B ( C) T
  • 84. B A Tn 273.15K Hcalc B Tn V Tn C K e 2 273.15K kPa K C K Hcalc 29.662 kJ Ans. mol %error Hcalc Hn %error 1.9 % Hn The table below shows the values for other components in Table B.2. Values agree within 5% except for acetic acid. Acetone Acetic Acid Acetonitrile Benz ene isoButane nButane 1Butanol Carbon tetrachloride Chlorobenz ene Chloroform Cyclohexane Cyclopentane nDecane Dichloromethane Diethyl ether Ethanol Ethylbenz ene Ethylene glycol nHeptane nHexane M ethanol M ethyl acetate M ethyl ethyl k etone Nitromethane nNonane isoOctane nOctane nPentane Phenol 1- panol Pro Hn ( J/ k mol) 29 .7 37.6 31.3 30.8 21.2 22.4 43.5 29 .9 35.3 29 .3 29 .9 27.4 39 .6 28 .1 26 .8 39 .6 35.7 53.2 31.9 29 .0 36 .5 30.4 31.7 34.9 37.2 30.8 34.6 25.9 45.9 41.9 84 % error 1.9 % 58 .7% 3.5% 0.2% 0.7% 0.0% 0.6 % 0.3% 0.3% 0.1% 0.1% 0.4% 2.2% 0.2% 0.9 % 2.8 % 0.5% 4.9 % 0.4% 0.4% 3.6 % 0.2% 1.3% 2.6 % 0.7% 0.1% 0.6 % 0.2% - % 0.6 1.1%
  • 85. p 2-Propanol Toluene W ater o-Xylene m-Xylene p-Xylene 40.5 33.3 41.5 36.7 36.2 35.9 1.7% 0.5% 2.0% 1.2% 1.4% 0.8% 4.13 Let P represent the vapor pressure. T ln Given P P 348.15 K P = 48.157543 kPa Find ( ) dPdT P P 5622.7 K T P Clapeyron equation: B dPdT = Pc AL 13.431 BL 2.211 0.029 bar K 3 cm 96.49 mol Vliq H T dPdT 1 512.6K AV dPdT 3 4.14 (a) Methanol: Tc AL 4.70504 T T K 4.70504 ln H T V Vliq Vliq PV V RT CPL ( ) T 5622.7 K T joule 31600 mol V = vapor molar volume. V Eq. (3.39) 2 H 87.396 kPa (guess) 100 kPa BL K BV 85 cm 1369.5 mol B Tn 80.97bar 51.28 10 T CL K 2 T 2 3 Ans. 337.9K CL 131.13 10 CV 3.450 10 6 R 12.216 10 3 6
  • 86. CPV ( ) T P BV AV CV T K K Tsat 3bar 2 T 2 R T1 368.0K T2 300K 500K Estimate Hv using Riedel equation (4.12) and Watson correction (4.13) Tn Trn Trn Tc 1.092 ln Hn Pc Hv Trsat 1 Trn kJ mol 35.645 kJ mol T2 CPL ( ) T T d 100 38.301 Hv Hv CPV ( ) T T d T1 n 0.718 0.38 Tsat H Trsat Hn Tc R Tn Trn 1 Hn Tsat 1.013 bar 0.930 Trsat 0.659 kmol hr H 49.38 T sat Q n H Q 1.372 3 10 kW kJ mol Ans. (b) Benzene: Hv = 28.273 kJ mol H = 55.296 kJ mol Q = 1.536 10 kW (c) Toluene Hv = 30.625 kJ mol H = 65.586 kJ mol Q = 1.822 10 kW 4.15 Benzene Tc T1sat 562.2K 451.7K Pc 48.98bar T2sat 86 358.7K 3 3 Tn 353.2K Cp 162 J mol K
  • 87. Estimate Hv using Riedel equation (4.12) and Watson correction (4.13) Trn Tn Trn Tc Hn Hv Hn 30.588 30.28 R Tn Trn Tr2sat 1 0.638 Hn Tc 1.013 bar 0.930 1 Tr2sat Hv Pc 1.092 ln T2sat Tr2sat 0.628 0.38 Trn kJ mol kJ mol Assume the throttling process is adiabatic and isenthalpic. Guess vapor fraction (x): x Given Cp T1sat 0.5 4.16 (a) For acetylene: Tc Tn ln Hn R Tn 1.092 1 Tr 1 Hn Hf 227480 0.498 Tn 61.39 bar Tr 0.614 Pc bar J mol T Tc Tr 1.013 Trn 16.91 Hv 0.930 Hn 6.638 0.38 Trn Hv x Find ( x) Ans. 189.4 K 298.15 K Trn Tc Pc 308.3 K T Trn x T2sat = x Hv H298 Hf 87 Hv 0.967 kJ mol kJ mol H298 220.8 kJ mol Ans.
  • 88. kJ mol (b) For 1,3-butadiene: H298 = 88.5 (c) For ethylbenzene: H298 = 12.3 (d) For n-hexane: H298 = 198.6 (e) For styrene: H298 = 103.9 4.17 1st law: dQ = dU Ideal gas: Since mol kJ mol and V dP = R dT V dP = P P (B) 1 V R dT which reduces to or dQ = P dV = R 1 R dT 1 dQ = CV dT R dT 1 R dT 1 dQ = CP dT CP dV = V dP dV Combines with (A) to give: dQ = CP dT V dP = R dT P dV Combines with (B) to yield: or (A) P dV P dV then P V = const from which kJ dW = CV dT PV= RT Whence kJ mol R dT 1 R dT (C) Since CP is linear in T, the mean heat capacity is the value of CP at the arithmetic mean temperature. Thus Tam 88 675
  • 89. CPm R 3.85 0.57 10 Integrate (C): P1 4.18 R 950 K P2 P1 T1 400 K 1.55 J Q 6477.5 P2 11.45 bar H298 = 4 058 910 J R T2 1 1 bar Tam T2 CPm Q 3 Ans. Ans. T1 mol 1 T2 T1 Ans. For the combustion of methanol: CH3OH(g) + (3/2)O2(g) = CO2(g) + 2H2O(g) H298 393509 H298 676485 For 6 MeOH: 2 ( 241818) ( 200660) For the combustion of 1-hexene: C6H12(g) + 9O2(g) = 6CO2(g) + 6H2O(g) H298 H298 6 ( 393509) 6 ( 241818) ( 41950) 3770012 H298 = 3 770 012 J Ans. Comparison is on the basis of equal numbers of C atoms. 4.19 C2H4 + 3O2 = 2CO2 + 2H2O(g) H298 [ ( 241818) 2 ( 393509) 52510] 2 J mol Parts (a) - (d) can be worked exactly as Example 4.7. However, with Mathcad capable of doing the iteration, it is simpler to proceed differently. 89
  • 90. Index the product species with the numbers: 1 = oxygen 2 = carbon dioxide 3 = water (g) 4 = nitrogen (a) For the product species, no excess air: 0 3.639 0.506 2 5.457 1.045 n A 10 B 2 3.470 3.280 3 1.157 10 K 0.593 ni Ai B A 1 4 A 0.121 0.040 D ni Bi ni D i i i i B 54.872 5 2 D K 1.450 11.286 i 0.227 0.012 1 K 5 2 D 1.621 10 K T0 298.15K T For the products, CP HP = R dT R T0 The integral is given by Eq. (4.7). Moreover, by an energy balance, H298 HP = 0 (guess) 2 H298 = R A T0 Given 8.497 Find 1 T T0 B 2 T0 2 2 T 1 D T0 2533.5 K 1 Ans. Parts (b), (c), and (d) are worked the same way, the only change being in the numbers of moles of products. (b) nO = 0.75 nn = 14.107 T = 2198.6 K Ans. (c) nO = 1.5 nn = 16.929 T = 1950.9 K Ans. (d) nO = 3.0 nn = 22.571 T = 1609.2 K Ans. 2 2 2 2 2 2 90
  • 91. (e) 50% xs air preheated to 500 degC. For this process, Hair H298 HP = 0 Hair = MCPH ( 298.15 773.15) For one mole of air: 3 MCPH 773.15 298.15 3.355 0.575 10 0.0 0.016 10 5 = 3.65606 For 4.5/0.21 = 21.429 moles of air: Hair = n R MCPH T Hair Hair 21.429 8.314 3.65606 ( 298.15 309399 J mol J mol The energy balance here gives: H298 1.5 n 773.15) 3.639 5.457 HP = 0 0.227 0.506 2 Hair 1.045 A 2 B 3.470 1.450 3.280 16.929 A 3 ni Ai B B 5 10 K 0.121 2 0.040 ni Bi D i 78.84 1.157 D 0.593 i A 10 K ni D i i 1 0.016 K D 5 2 1.735 10 K 2 (guess) H298 Find Hair = R A T0 1 D T0 Given B 2 T0 2 2 1 1 7.656 T 91 T0 K T 2282.5 K K Ans.
  • 92. 4.20 n-C5H12 + 8O2 = 5CO2 + 6H2O(l) By Eq. (4.15) with data from Table C.4: H298 5 ( 393509) 6 ( 285830) ( 146760) H298 = 3 535 765 J 4.21 Ans. The following answers are found by application of Eq. (4.15) with data from Table C.4. (a) -92,220 J (n) 180,500 J (b) -905,468 J (o) 178,321 J (c) -71,660 J (p) -132,439 J (d) -61,980 J (q) -44,370 J (e) -367,582 J (r) -68,910 J (f) -2,732,016 J (s) -492,640 J (g) -105,140 J (t) 109,780 J (h) -38,292 J (u) 235,030 J (i) 164,647 J (v) -132,038 J (j) -48,969 J (w) -1,807,968 J (k) -149,728 J (x) 42,720 J (l) -1,036,036 J (y) 117,440 J (m) 207,436 J (z) 175,305 J 92
  • 93. 4.22 The solution to each of these problems is exactly like that shown in Example 4.6. In each case the value of Ho298 is calculated in Problem 4.21. Results are given in the following table. In the first column the letter in ( ) indicates the part of problem 4.21 appropriate to the value. T/ K (a) (b ) (f ) (i ) (j ) (l ) (m) (n ) (o ) (r ) (t ) (u ) (v ) (w) (x ) (y ) 4.23 A 873.15 773.15 923.15 973.15 583.15 683.15 850.00 1350.00 1073.15 723.15 733.15 750.00 900.00 673.15 648.15 1083.15 -5.871 1.861 6.048 9.811 -9.523 -0.441 4.575 -0.145 -1.011 -1.424 4.016 7.297 2.418 2.586 0.060 4.175 103 B 4.181 -3.394 -9.779 -9.248 11.355 0.004 -2.323 0.159 -1.149 1.601 -4.422 -9.285 -3.647 -4.189 0.173 -4.766 106 C 0.000 0.000 0.000 2.106 -3.450 0.000 0.000 0.000 0.000 0.156 0.991 2.520 0.991 0.000 0.000 1.814 10-5 D -0.661 2.661 7.972 -1.067 1.029 -0.643 -0.776 0.215 0.916 -0.083 0.083 0.166 0.235 1.586 -0.191 0.083 IDCPH/ J -17,575 4,729 15,635 25,229 -10,949 -2,416 13,467 345 -9,743 -2,127 7,424 12,172 3,534 4,184 125 12,188 Ho298 J HoT/ -109,795 -900,739 -2,716,381 189,876 -59,918 -1,038,452 220,903 180,845 168,578 -71,037 117,204 247,202 -128,504 -1,803,784 42,845 129,628 This is a simple application of a combination of Eqs. (4.18) & (4.19) with evaluated parameters. In each case the value of Ho298 is calculated in Pb. 4.21. The values of A, B, C and D are given for all cases except for Parts (e), (g), (h), (k), and (z) in the preceding table. Those missing are as follows: Part No. A 103 B 106 C 10-5 D (e) -7.425 20.778 0.000 3.737 (g) -3.629 8.816 -4.904 0.114 (h) -9.987 20.061 -9.296 1.178 (k) 1.704 -3.997 1.573 0.234 (z) -3.858 -1.042 0.180 0.919 93
  • 94. 4.24 q 6 ft 150 10 3 T day ( 60 5 32) K 9 T 273.15K 288.71 K P 1atm The higher heating value is the negative of the heat of combustion with water as liquid product. Calculate methane standard heat of combustion with water as liquid product: CH4 + 2O2 --> CO2 +2H2O Standard Heats of Formation: J mol HfCH4 74520 HfCO2 393509 Hc HfCO2 HfO2 J mol J mol HfH2Oliq 2 HfH2Oliq HigherHeatingValue 0 285830 HfCH4 J mol 2 HfO2 5 J Hc Hc 8.906 10 mol Assuming methane is an ideal gas at standard conditions: n q P RT n HigherHeatingValue 4.25 8 mol n 1.793 10 5dollar GJ day 5 dollar 7.985 10 day Ans. Calculate methane standard heat of combustion with water as liquid product Standard Heats of Formation: CH4 + 2O2 --> CO2 +2H2O J mol HfCH4 74520 HfCO2 393509 HcCH4 HfCO2 HcCH4 890649 J mol HfO2 0 J mol HfH2Oliq 2 HfH2Oliq J mol 94 HfCH4 285830 J mol 2 HfO2
  • 95. Calculate ethane standard heat of combustion with water as liquid product: Standard Heats of Formation:C2H6 + 7/2O2 --> 2CO2 +3H2O J mol HfC2H6 83820 HcC2H6 2 HfCO2 HcC2H6 1560688 3 HfH2Oliq HfC2H6 7 2 HfO2 J mol Calculate propane standard heat of combustion with water as liquid product Standard Heats of Formation:C3H8 + 5O2 --> 3CO2 +4H2O J mol HfC3H8 104680 HcC3H8 3 HfCO2 HcC3H8 2219.167 4 HfH2Oliq HfC3H8 5 HfO2 kJ mol Calculate the standard heat of combustion for the mixtures a) 0.95 HcCH4 0.02 HcC2H6 0.02 HcC3H8 921.714 kJ mol b) 0.90 HcCH4 0.05 HcC2H6 0.03 HcC3H8 946.194 kJ mol c) 0.85 HcCH4 0.07 HcC2H6 0.03 HcC3H8 932.875 kJ mol Gas b) has the highest standard heat of combustion. 4.26 2H2 + O2 = 2H2O(l) Hf1 C + O2 = CO2(g) Hf2 N2(g)+2H2O(l)+CO2(g)=(NH2)2CO(s)+3/2O2 H Ans. 2 ( 285830) J 393509 J 631660 J . N2(g)+2H2(g)+C(s)+1/2O2(g)=(NH2)2CO(s) H298 Hf1 Hf2 H298 H 95 333509 J Ans.
  • 96. 4.28 On the basis of 1 mole of C10H18 (molar mass = 162.27) Q 43960 162.27 J 6 Q 7.133 10 J This value is for the constant-volume reaction: C10H18(l) + 14.5O2(g) = 10CO2(g) + 9H2O(l) Assuming ideal gases and with symbols representing total properties, Q= T U= H ( )= PV 298.15 K H Q H R T ngas ngas R T ngas ( 10 14.5)mol 6 H 7.145 10 J This value is for the constant-V reaction, whereas the STANDARD reaction is at const. P.However, for ideal gases H = f(T), and for liquids H is a very weak function of P. We therefore take the above value as the standard value, and for the specified reaction: C10H18(l) + 14.5O2(g) = 10CO2(g) + 9H2O(l) H 9H2O(l) = 9H2O(g) Hvap 9 44012 J ___________________________________________________ C10H18(l) + 14.5O2(g) = 10CO2(g) + 9H2O(g) H298 4.29 H Hvap H298 6748436 J Ans. FURNACE: Basis is 1 mole of methane burned with 30% excess air. CH4 + 2O2 = CO2 + 2H2O(g) Entering: Moles methane n1 1 Moles oxygen n2 2 1.3 Moles nitrogen n3 2.6 96 79 21 n2 2.6 n3 9.781
  • 97. Total moles of dry gases entering n n1 n2 n n3 13.381 At 30 degC the vapor pressure of water is 4.241 kPa. Moles of water vapor entering: n4 Leaving: 4.241 13.381 101.325 4.241 n4 CO2 -- 1 mol H2O -- 2.585 mol O2 -- 2.6 - 2 = 0.6 mol N2 -- 9.781 mol 0.585 (1) (2) (3) (4) By an energy balance on the furnace: Q= H= H298 For evaluation of 1 A 0.6 9.781 i HP we number species as above. 5.457 2.585 n HP 3.470 3.639 1.045 1.450 B R 8.314 B 48.692 0.121 D 0.227 0.040 D ni Bi i i 10.896983 10 3 ni Di C D 0 4 5.892 10 The TOTAL value for MCPH of the product stream: HP R M CPH ( 303.15K 1773.15K A B C D) ( 1773.15 HP 732.013 kJ mol From Example 4.7: Q HP H298 5 10 J mol K i A 3 0.593 ni Ai B A 10 0.506 3.280 1 4 1.157 H298 802625 Q = 70 612 J 97 J mol Ans. 303.15)K
  • 98. HEAT EXCHANGER: Flue gases cool from 1500 degC to 50 degC. The partial pressure of the water in the flue gases leaving the furnace (in kPa) is pp n2 n1 n2 n3 n4 101.325 pp 18.754 The vapor pressure of water at 50 degC (exit of heat exchanger) is 12.34 kPa, and water must condense to lower its partial pressure to this value. Moles of dry flue gases: n n1 n3 n4 n 11.381 Moles of water vapor leaving the heat exchanger: n2 12.34 n 101.325 12.34 n2 Moles water condensing: 1.578 n 2.585 1.578 Latent heat of water at 50 degC in J/mol: H50 2382.918.015 J mol Sensible heat of cooling the flue gases to 50 degC with all the water as vapor (we assumed condensation at 50 degC): Q R MCPH ( 323.15 K 1773.15 K A B C D)( 323.15 1773.15) K Q = 766 677 J 4.30 4NH3(g) + 5O2(g) = 4NO(g) + 6H2O(g) BASIS: 4 moles ammonia entering reactor Moles O2 entering = (5)(1.3) = 6.5 Moles N2 entering = (6.5)(79/21) = 24.45 Moles NH3 reacting = moles NO formed = (4)(0.8) = 3.2 Moles O2 reacting = (5)(0.8) = 4.0 Moles water formed = (6)(0.8) = 4.8 98 n H50 Ans.
  • 99. ENERGY BALANCE: H= HR H298 HP = 0 REACTANTS: 1=NH3; 2=O2; 3=N2 4 n 3.578 6.5 A 3.639 24.45 i 3.020 B 0.506 10 3.280 A 1 3 118.161 ni Ai B 3 5 D 0.227 10 0.593 B 0.040 D ni Bi ni Di i i i A 0.186 C 0.02987 5 D 0.0 1.242 10 TOTAL mean heat capacity of reactant stream: HR HR R MCPH ( 348.15K 298.15K A B C D) ( 298.15K 52.635 348.15K) kJ mol The result of Pb. 4.21(b) is used to get J H298 0.8 ( 905468) mol PRODUCTS: =NH3; 2=O2; 3=NO; 4=H2O; 5=N2 1 0.8 3.020 2.5 n 3.578 3.639 0.506 3.2 A 3.387 B 0.629 0.186 10 K 0.227 3 D 0.014 4.8 1.450 3.280 0.593 2 0.121 24.45 i 3.470 5 10 K 0.040 1 5 A ni Ai B 119.65 D B 1 0.027 K By the energy balance and Eq. (4.7), we can write: (guess) 2 T0 298.15K 99 ni Di i i i A ni Bi D 4 8.873 10 K 2
  • 100. H298 B 2 T0 2 1 2 1 1 T 3.283 Find 4.31 H R = R A T0 D T0 Given T T0 Ans. 978.9 K C2H4(g) + H2O(g) = C2H5OH(l) n BASIS: 1 mole ethanol produced H= Q= Energy balance: H298 [ 277690 HR ( 52510 1mol H298 241818) ] J mol 4 J H298 8.838 10 mol Reactant stream consists of 1 mole each of C2H4 and H2O. 1 2 n i 1.424 A 3.470 1 1 14.394 B 1.450 ni Ai B A B HR Q 4.392 C 10 0.0 C 6 D C 0.01584 0.0 ni Di i 4.392 10 6 D R M CP ( H 298.15K 593.15K A B C D)( 298.15K 4 1.21 10 593.15K) 4 J 2.727 10 HR mol Q H298 1mol 100 5 10 0.121 D ni Ci i i 4.894 HR 3 ni Bi i A 10 115653 J Ans.
  • 101. 4.32 One way to proceed is as in Example 4.8 with the alternative pair of reactions: CH4 + H2O = CO + 3H2 H298a 205813 CH4 + 2H2O = CO2 + 4H2 H298b 164647 BASIS: 1 mole of product gases containing 0.0275 mol CO2; 0.1725 mol CO; & H2O 0.6275 mol H2 Entering gas, by carbon & oxygen balances: 0.0275 + 0.1725 = 0.2000 mol CH4 0.1725 + 0.1725 + 2(0.0275) = 0.4000 mol H2O J H298 0.1725 H298a 0.0275 H298b mol The energy balance is written Q= HR H298 1.702 3.470 9.081 B 1.450 ni Ai B A HR HR 10 i 3 B 2.164 C 0.0 0.4 10 6 D 2.396 10 3 C 0.0 5 10 0.121 D ni Ci ni Di i i i 1.728 mol 0.2 n 1 2 C ni Bi i A 4.003 10 HP REACTANTS: 1=CH4; 2=H2O A 4 J H298 4.328 10 7 D 3 4.84 10 RI CPH ( 773.15K 298.15K A B C D) 4 J 1.145 10 mol PRODUCTS: 1=CO2; 2=CO; 3=H2O; 4=H2 0.0275 n 0.1725 0.1725 0.6275 5.457 A 3.376 3.470 1.045 B 3.249 0.557 1.450 0.422 101 1.157 10 3 D 0.031 0.121 0.083 5 10
  • 102. i A 1 4 B ni Ai HP B 3.37 ni Di i i i A D ni Bi 6.397 10 4 C 3 D 0.0 3.579 10 RI CPH ( 298.15K 1123.15K A B C D) 4 J 2.63 10 mol HP Q HR H298 Q HP mol 4.33 CH4 + 2O2 = CO2 + 2H2O(g) Ans. 54881 J H298a H298b C2H6 + 3.5O2 = 2CO2 + 3H2O(g) 802625 1428652 BASIS: 1 mole fuel (0.75 mol CH4; 0.25 mol C2H6) burned completely with 80% xs. air. O2 in = 1.8[(0.75)(2) + (0.25)(3.5)] = 4.275 mol N2 in = 4.275(79/21) = 16.082 mol Product gases: CO2 = 0.75 + 2(0.25) = 1.25 mol H2O = 2(0.75) + 3(0.25) = 2.25 mol O2 = (0.8/1.8)(4.275) = 1.9 mol N2 = 16.082 mol H298 0.75 H298a 0.25 H298b J mol Energy balance: Q = H = H298 HP PRODUCTS: 1=CO2; 2=H2O; 3=O2; 4=N2 1.25 n 5.457 3.470 1.450 A 1.9 1 4 B 3.639 16.082 i 8 10 HP = Q 1.045 2.25 5 Q 3.280 74.292 10 K 3 D B 102 0.227 D ni Bi 0.015 0.121 5 10 K 0.040 ni Di i i i H298 1.157 0.593 ni Ai B A A 0.506 J mol 1 K C 0.0 D 4 2 9.62 10 K 2
  • 103. By the energy balance and Eq. (4.7), we can write: T0 303.15K Q 2 (guess) H298 = R A T0 2 1 1 1.788 4.34 B 2 T0 2 1 D T0 Given T T0 T Find Ans. 542.2 K BASIS: 1 mole of entering gases containing 0.15 mol SO2; 0.20 mol O2; 0.65 mol N2 SO2 + 0.5O2 = SO3 Conversion = 86% SO2 reacted = SO3 formed = (0.15)(0.86) = 0.129 mol O2 reacted = (0.5)(0.129) = 0.0645 mol Energy balance: H773 = HR H298 HP Since HR and HP cancel for the gas that passes through the converter unreacted, we need consider only those species that react or are formed. Moreover, the reactants and products experience the same temperature change, and can therefore be considered together. We simply take the number of moles of reactants as being negative. The energy balance is then written: H773 = H298 Hnet H298 [ 395720 ( 296830) ]0.129 J mol 1: SO2; 2: O2; 3: SO3 0.129 n 0.0645 5.699 A 0.129 i 1 3 3.639 B 8.060 A A 0.06985 0.506 10 3 5 D 0.227 10 2.028 1.056 ni Ai B i 1.015 0.801 ni Bi D ni Di i B 2.58 10 103 i 7 C 0 D 4 1.16 10
  • 104. Hnet R M CPH ( 298.15K 773.15K A B C D)( 773.15K Hnet 77.617 H773 298.15K) J mol H298 H773 Hnet 12679 J mol Ans. 4.35 CO(g) + H2O(g) = CO2(g) + H2(g) BASIS: 1 mole of feed consisting of 0.5 mol CO and 0.5 mol H2O. Moles CO reacted = moles H2O reacted = moles CO2 formed = moles H2 formed = (0.6)(0.5) = 0.3 Product stream: moles CO = moles H2O = 0.2 moles CO2 = moles H2 = 0.3 Energy balance: Q= H298 0.3 [ 393509 H= HR H298 J 214818) ] mol ( 110525 HP 4 J H298 2.045 10 mol Reactants: 1: CO 2: H2O 0.5 n i 0.5 3.376 A 3.470 1.450 ni Ai B A 1 2 0.557 B HR HR 0.031 D B ni Di i i 3.423 5 10 0.121 D 1.004 10 3 0 D C R M CPH ( 298.15K 398.15K A B C D)( 298.15K 3 4.5 10 398.15K) 3 J 3.168 10 Products: 0.2 0.3 0.3 mol 1: CO 2: H2O 3: CO2 4: H2 0.2 n 3 ni Bi i A 10 3.376 A 3.470 5.457 0.557 B 3.249 1.450 1.045 0.422 104 0.031 10 3 D 0.121 1.157 0.083 5 10
  • 105. i 1 4 A ni Ai B i A ni Bi D i 3.981 B 8.415 10 i 4 C 4 0 D HP R MCPH ( 298.15K 698.15K A B C D) ( 698.15K HP 1.415 10 Q ni Di 3.042 10 298.15K) 4 J HR mol H298 HP mol Q 9470 J Ans. 4.36 BASIS: 100 lbmol DRY flue gases containing 3.00 lbmol CO2 and 11.80 lbmol CO x lbmol O2 and 100-(14.8-x)= 85.2-x lbmol N2. The oil therefore contains 14.80 lbmol carbon;a carbon balance gives the mass of oil burned: 14.8 12.011 lbm 0.85 209.133 lbm The oil also contains H2O: 209.133 0.01 lbmol 0.116 lbmol 18.015 Also H2O is formed by combustion of H2 in the oil in the amount 209.133 0.12 lbmol 2.016 12.448 lbmol Find amount of air entering by N2 & O2 balances. N2 entering in oil: 209.133 0.02 lbmol 28.013 0.149 lbmol lbmol N2 entering in the air=(85.2-x)-0.149 =85.051-x lbmol O2 in flue gas entering with dry air = 3.00 + 11.8/2 + x + 12.448/2 = 15.124 + x lbmol (CO2) (CO) (O2) (H2O from combustion) Total dry air = N2 in air + O2 in air = 85.051 - x + 15.124 + x = 100.175 lbmol 105
  • 106. Since air is 21 mol % O2, 0.21 = 15.124 x 100.175 x ( 0.21 100.175 15.124)lbmol x 5.913 lbmol O2 in air = 15.124 + x = 21.037 lbmols N2 in air = 85.051 - x = 79.138 lbmoles N2 in flue gas = 79.138 + 0.149 = 79.287 lbmols [CHECK: Total dry flue gas = 3.00 + 11.80 + 5.913 + 79.287 = 100.00 lbmol] Humidity of entering air, sat. at 77 degF in lbmol H2O/lbmol dry air, P(sat)=0.4594(psia) 0.4594 14.696 0.4594 0.03227 lbmol H2O entering in air: 0.03227 100.175 lbmol 3.233 lbmol If y = lbmol H2O evaporated in the drier, then lbmol H2O in flue gas = 0.116+12.448+3.233+y = 15.797 + y Entering the process are oil, moist air, and the wet material to be dried, all at 77 degF. The "products" at 400 degF consist of: 3.00 lbmol CO2 11.80 lbmol CO 5.913 lbmol O2 79.287 lbmol N2 (15.797 + y) lbmol H2O(g) Energy balance: Q= H= H298 HP where Q = 30% of net heating value of the oil: Q 0.3 19000 BTU 209.13 lbm lbm Reaction upon which net heating value is based: 106 Q 6 1.192 10 BTU
  • 107. OIL + (21.024)O2 = (14.8)CO2 + (12.448 + 0.116)H2O(g) + (0.149)N2 H298a 19000 209.13 BTU 6 H298a 3.973 10 BTU To get the "reaction" in the drier, we add to this the following: (11.8)CO2 = (11.8)CO + (5.9)O2 H298b 11.8 ( 110525 (y)H2O(l) = (y)H2O(g) H298c ( y) 393509) 0.42993 BTU Guess: y 50 44012 0.42993 y BTU [The factor 0.42993 converts from joules on the basis of moles to Btu on the basis of lbmol.] Addition of these three reactions gives the "reaction" in the drier, except for some O2, N2, and H2O that pass through unchanged. Addition of the corresponding delta H values gives the standard heat of reaction at 298 K: H298 ( y) H298a H298b H298c ( y) For the product stream we need MCPH: 1: CO2 2: CO 3:O2 4: N2 5: H2O T0 298.15 r 1.986 400 T 459.67 1.8 T 477.594 3 1.045 1.157 11.8 n y) ( 5.457 3.376 0.557 0.031 5.913 A i B 3 5 D 0.227 10 0.593 0.040 3.470 y 1 5 A ( y) 1.450 0.121 n y) i Ai B ( y) ( i T T0 0.506 10 3.280 79.278 15.797 3.639 1.602 n y) i Bi D ( y) ( n y) i Di ( i CP ( y) 107 r A ( y) i B ( y) T0 2 1 D ( y) T0 2
  • 108. CP ()( y 400 Given y y H298 () y Find y () (lbmol H2O evaporated) 49.782 y 18.015 209.13 Whence 77)BTU = Q (lb H2O evap. per lb oil burned) Ans. 4.288 4.37 BASIS: One mole of product gas containing 0.242 mol HCN, and (1-0.242)/2 = 0.379 mol each of N2 and C2H2. The energy balance is Q= H= H298 H298 HP ( 135100 2 227480) 0.242 J 2 3 H298 5.169 10 J Products: 0.242 n 4.736 0.379 A 3.280 0.379 ni Ai 0.725 3 0.593 10 D 0.040 1.952 B 4.7133 B D ni Bi 1.2934 10 ni Di i 3 4 0 D C HP R M CPH ( 298.15K 873.15K A B C D)( 873.15K HP 6.526 10 2.495 10 J Q 298.15K)mol 4 HP H298 Q HP 4 2.495 10 J 30124 J Ans. 4.38 BASIS: 1 mole gas entering reactor, containing 0.6 mol HCl, 0.36 mol O2, and 0.04 mol N2. HCl reacted = (0.6)(0.75) = 0.45 mol 4HCl(g) + O2(g) = 2H2O(g) + 2Cl2(g) 108 5 10 1.299 i i A B 6.132 1 3 A i 1.359
  • 109. For this reaction, H298 [ ( 241818) 2 Evaluate T0 4 ( 92307) ] 5 J H298 1.144 10 mol by Eq. (4.21) with H823 T 298.15K J mol 823.15K 1: H2O 2: Cl2 3: HCl 4=O2 2 3.470 2 n 4.442 A 4 0.089 B 3.156 1 i 1.45 0.623 3.639 A 1 4 ni Ai 10 3 D 0.344 0.151 0.506 B D ni Di i i 0.439 H823 H298 MCPH T0 T H823 117592 B 5 8 10 A B 5 10 0.227 ni Bi i A 0.121 C 0 D R T 4 D T0 C 8.23 10 J mol Heat transferred per mol of entering gas mixture: Q H823 4 Q 0.45 mol 4.39 CO2 + C = 2CO Ans. J (a) mol J (b) 221050 mol H298a 2C + O2 = 2CO 13229 J 172459 H298b Eq. (4.21) applies to each reaction: For (a): 2 n 1 1 3.376 A 1.771 0.557 B 0.771 10 5.457 1.045 109 0.031 3 D 5 0.867 10 1.157
  • 110. i A 1 3 B ni Ai H1148a B 0.476 i 4 7.02 10 H298a R M CP 298.15K 1148.15K H H1148a ni Di i i A D ni Bi A C B C D 0 5 1.962 10 D ( 1148.15K 298.15K) 5 J 1.696 10 mol For (b): 2 n 3.376 1 A 3.639 2 i 1 3 0.557 B H1148b ni Ai 5 D 0.227 10 B 0.867 ni Bi D ni Di i 0.429 B 9.34 10 H298b R M CP 298.15K 1148.15K H H1148b 3 0.771 i A 0.506 10 1.771 A 0.031 i 4 C A B 0 C D D ( 1148.15K 5 J 2.249 10 mol The combined heats of reaction must be zero: nCO 2 H1148a nO H1148b = 0 2 nCO Define: r= 2 nO H1148b r H1148a 2 110 r 5 1.899 10 1.327 298.15K)
  • 111. For 100 mol flue gas and x mol air, moles are: Flue gas Air Feed mix CO2 12.8 0 12.8 CO 3.7 0 3.7 O2 5.4 0.21x 5.4 + 0.21x N2 78.1 0.79x 78.1 + 0.79x Whence in the feed mix: r= 12.5 5.4 r mol 0.21 x 12.8 5.4 0.21 x x 19.155 mol 100 19.155 Flue gas to air ratio = Ans. 5.221 Product composition: nCO 3.7 nN 78.1 2 2 ( 12.8 0.21 19.155) 0.79 19.155 nCO Mole % N2 = 100 48.145 93.232 2 nCO Mole % CO = nCO nN 5.4 nN 100 34.054 Ans. 2 34.054 65.946 H298a 802625 J mol H298b 4.40 CH4 + 2O2 = CO2 + 2H2O(g) 519641 J mol CH4 + (3/2)O2 = CO + 2H2O(g) BASIS: 1 mole of fuel gas consisting of 0.94 mol CH4 and 0.06 mol N2 Air entering contains: 1.35 2 0.94 2.538 79 21 2.538 9.548 mol O2 mol N2 111
  • 112. Moles CO2 formed by reaction = 0.94 0.7 0.658 Moles CO formed by reaction = 0.94 0.3 0.282 H298 0.658 H298a 5 J H298 0.282 H298b Moles H2O formed by reaction = 0.94 2.0 Moles O2 consumed by reaction = 2 0.658 6.747 10 mol 1.88 3 0.282 2 1.739 Product gases contain the following numbers of moles: (1) (2) (3) (4) (5) CO2: 0.658 CO: 0.282 H2O: 1.880 O2: 2.538 - 1.739 = 0.799 N2: 9.548 + 0.060 = 9.608 0.658 1.045 1.157 0.282 n 5.457 3.376 0.557 0.031 1.880 A 3.470 B 1.450 10 0.799 3.280 D 0.593 0.121 0.506 9.608 i 3.639 3 1 5 A ni Ai B A 0.227 0.040 ni Bi i D B 9.6725 10 i 3 C R M CPH ( 298.15K 483.15K A B C D)( 483.15K HP 7.541 10 HH2O mdotH2O 4 0 D HP Energy balance: ni Di i 45.4881 5 10 3.396 10 298.15K) 4 J mol Hrx H298 Hrx ndotfuel = 0 112 HP Hrx kJ mol kg 34.0 sec 599.252 mdotH2O
  • 113. From Table C.1: HH2O ( 398.0 104.8) kJ kg HH2O mdotH2O ndotfuel ndotfuel Hrx 16.635 mol sec Volumetric flow rate of fuel, assuming ideal gas: ndotfuel R 298.15 K V 3 m 0.407 sec V 101325 Pa 4.41 C4H8(g) = C4H6(g) + H2(g) Ans. H298 109780 J mol BASIS: 1 mole C4H8 entering, of which 33% reacts. The unreacted C4H8 and the diluent H2O pass throught the reactor unchanged, and need not be included in the energy balance. Thus T T0 1 n by Eq. (4.21): Evaluate H798 1: C4H6 2: H2 3: C4H8 1 2.734 A 3.249 26.786 B 0.422 1.967 i 298.15 K 798.15 K 1 8.882 10 3 C 0.0 0.0 31.630 6 10 D 9.873 5 0.083 10 0.0 1 3 ni Ai B A 4.016 B H298 H798 1.179 10 Q 3 C 9.91 10 MCPH 298.15K 798.15K 5 J 0.33 mol H798 mol Q 38896 J 113 Ans. ni Di i i 4.422 10 H798 D ni Ci i i A C ni Bi A B 7 D C 3 8.3 10 D R T T0
  • 114. 4.42Assume Ideal Gas and P = 1 atm P R 1atm a) T0 ( 70 T0 3 BTU 7.88 10 mol K T 294.261 K ICPH T0 T 3.355 0.575 10 3 T0 T 459.67) rankine BTU sec 305.372 K 5 0 0.016 10 R ICPH T0 T 3.355 0.575 10 Vdot b) T0 38.995 K ( 24 3 5 0 0.016 10 T T0 ndot 39.051 Vdot ft 33.298 sec Q 13K 12 kJ s 3 5 0 0.016 10 45.659 K Q R ICPH T0 T 3.355 0.575 10 3 0 5 0.016 10 Vdot 4.43Assume Ideal Gas and P = 1 atm ( 94 459.67) rankine 1.61 10 3 Vdot 31.611 P T ( 68 m 0.7707 s 1atm 459.67) rankine 3 atm ft mol rankine 3 ft 50 sec ndot mol s 3 ndot R T0 P a) T0 Ans. kJ mol K ndot Vdot mol s 3 m 0.943 s Vdot 273.15) K 8.314 10 3 3 ndot R T0 P ICPH T0 T 3.355 0.575 10 R 12 Q ndot R Q 20rankine ndot P Vdot R T0 114 ndot 56.097 mol s Ans.
  • 115. T0 307.594 K T ICPH T0 T 3.355 0.575 10 3 293.15 K 5 0 0.016 10 50.7 K 3 BTU R 7.88 10 Q R ICPH T0 T 3.355 0.575 10 mol K 3 0 5 0.016 10 ndot Q b) T0 R ( 35 273.15)K T 273.15)K mol K 3 m 1.5 sec 3 P Vdot R T0 ndot ICPH T0 T 3.355 0.575 10 3 0 ndot 5 0.016 10 59.325 8.314 10 Q R ICPH T0 T 3.355 0.575 10 35.119 K 3 0 5 0.016 10 Q ndot 17.3216 4.44 First calculate the standard heat of combustion of propane C3H8 + 5O2 = 3CO2(g) + 4H2O (g) H298 Cost 3 393509 J mol 4 241818 6 J 2.043 10 2.20 mol s kJ mol K R H298 BTU Ans. sec 3 5 atm m 8.205 10 Vdot ( 25 22.4121 dollars gal mol 80% 115 J mol 104680 J mol kJ Ans. s
  • 116. Estimate the density of propane using the Rackett equation 3 Tc T Vsat 369.8K ( 25 Zc 273.15) K V c Zc Heating_cost 1 Tr 0.276 Tr Vc T Tc Tr cm 200.0 mol 0.806 3 0.2857 Vsat Vsat Cost cm 89.373 mol Heating_cost H298 Heating_cost 0.032 dollars MJ 33.528 Ans. dollars 6 10 BTU 4.45 T0 ( 25 a) Acetylene 273.15) K T ( 500 273.15) K Q R ICPH T0 T 6.132 1.952 10 Q 2.612 10 3 0 5 1.299 10 4 J mol The calculations are repeated and the answers are in the following table: J/mol a) Acetylene 26, 120 b) Ammonia 20, 200 c) nbutane 71, 964 d) Carbon diox ide 21, 779 e) Carbon monox ide 14, 457 f) Ethane 3 420 8, g) Hydrogen 13866 , h) Hydrogen chloride 14, 040 i) M ethane 233 ,18 j Nitric ox ) ide 14, 0 73 k) Nitrogen 14, 276 l) Nitrogen diox ide 20, 846 m) Nitrous ox ide 22, 019 n) Ox ygen 15, 052 o) Propylene 46, 147 116
  • 117. 4.46 T0 ( 25 Q 273.15)K 30000 a) Acetylene T ( 500 273.15)K J mol Given Q = R ICPH T0 T 6.132 1.952 10 T Find ( T) T 835.369 K T 3 273.15K 0 5 1.299 10 562.2 degC The calculations are repeated and the answers are in the following table: a) Acetylene b) Ammonia c) n-butane d) Carbon dioxide e) Carbon monoxide f) Ethane g) Hydrogen h) Hydrogen chloride i) Methane j) Nitric oxide k) Nitrogen l) Nitrogen dioxide m) Nitrous oxide n) Oxygen o) Propylene 4.47 T0 ( 25 273.15)K T( K) 835.4 964.0 534.4 932.9 1248.0 690.2 1298.4 1277.0 877.3 1230.2 1259.7 959.4 927.2 1209.9 636.3 T T( C) 562.3 690.9 261.3 659.8 974.9 417.1 1025.3 1003.9 604.2 957.1 986.6 686.3 654.1 936.8 363.2 ( 250 a) Guess mole fraction of methane: y 273.15) K Q 11500 0.5 Given y ICPH T0 T 1.702 9.081 10 (1 y 3 2.164 10 y) ICPH T0 T 1.131 19.225 10 Find ( y) y 0.637 Ans. 117 3 6 = Q 0 R 5.561 10 6 0 R J mol
  • 118. b) T0 ( 100 273.15) K T ( 400 Guess mole fraction of benzene 273.15)K y Q 54000 J mol 0.5 Given y ICPH T0 T ( 1 3 0.206 39.064 10 y)ICPH T0 T y Find y () y c) T0 ( 150 3 3.876 63.249 10 273.15) K 6 13.301 10 = Q 0 R 6 20.928 10 0 R Ans. 0.245 T ( 250 Guess mole fraction of toluene 273.15)K y Q 17500 J mol 0.5 Given y ICPH T0 T 0.290 47.052 10 ( 1 y 3 15.716 10 y)ICPH T0 T 1.124 55.380 10 Find y () y 0.512 3 6 = Q 0 R 18.476 10 6 0 R Ans. 4.48 Temperature profiles for the air and water are shown in the figures below. There are two possible situations. In the first case the minimum temperature difference, or "pinch" point occurs at an intermediate location in the exchanger. In the second case, the pinch occurs at one end of the exchanger. There is no way to know a priori which case applies. Pi h a End nc t I e m e a e Pi h nt r di t nc TH1 Se ton I ci I Se ton I ci TH1 Se ton I ci THi Se ton I ci I THi T TH2 TC1 TC1 TCi TCi TC2 118 TH2 T TC2
  • 119. To solve the problem, apply an energy balance around each section of the exchanger. T H1 Section I balance: mdotC HC1 HCi = ndotH CP dT THi T Hi Section II balance: HC2 = ndotH mdotC HCi CP dT TH2 If the pinch is intermediate, then THi = TCi + T. If the pinch is at the end, then TH2 = TC2 + T. a) TH1 T 10degC TC1 2676.0 TCi For air from Table C.1:A kJ kg 3.355 B 100degC TC2 25degC HCi 100degC HC1 1000degC 419.1 kJ kg HC2 104.8 0.575 10 3 C 0 D kJ kg 5 0.016 10 kmol s Assume as a basis ndot = 1 mol/s. ndotH Assume pinch at end: TH2 TC2 THi 110degC Guess: mdotC 1 kg s 1 T Given mdotC HC1 mdotC HCi mdotC THi mdotC ndotH THi TCi HCi = ndotH R ICPH THi TH1 A B C D Energy balances on Section I and HC2 = ndotH R ICPH TH2 THi A B C D II 170.261 degC mdotC Find mdotC THi THi 0.011 kg mol 70.261 degC Ans. TH2 119 TC2 10 degC 11.255 kg s
  • 120. Since the intermediate temperature difference, THi - TCi is greater than the temperature difference at the end point, TH2 - TC2, the assumption of a pinch at the end is correct. b) TH1 TC1 T 10degC 2676.0 TCi kJ kg 100degC TC2 25degC HCi 100degC HC1 500degC 419.1 kJ kg HC2 104.8 1 kmol s THi TCi T TH2 110degC Assume as a basis ndot = 1 mol/s. ndotH Assume pinch is intermediate: Guess: mdotC 1 kg s kJ kg Given mdotC HC1 mdotC HCi mdotC TH2 mdotC ndotH THi TCi HCi = ndotH R ICPH THi TH1 A B C D Energy balances on Section I and HC2 = ndotH R ICPH TH2 THi A B C D II Find mdotC TH2 TH2 5.03 10 3 kg 10 degC mol 48.695 degC mdotC 5.03 kg s Ans. TH2 TC2 23.695 degC Since the intermediate temperature difference, THi - TCi is less than the temperature difference at the end point, TH2 - TC2, the assumption of an intermediate pinch is correct. 4.50a) C6H12O6(s) + 6 O2(g)= 6 CO2(g) + 6 H2O(l) 1 = C6H12O6 , 2 = O2 , 3 = CO2 , 4 = H2O H0f1 1274.4 kJ mol H0f2 120 0 kJ mol M1 180 gm mol
  • 121. H0f3 H0r 393.509 6 H0f3 kJ mol 6 H0f4 b) energy_per_kg mass_glucose H0f4 150 kJ kg H0f1 285.830 mass_person energy_per_kg H0r M3 H0r 6 H0f2 mass_person kJ mol 44 2801.634 gm mol kJ mol Ans. 57kg mass_glucose M1 0.549 kg Ans. c) 6 moles of CO2 are produced for every mole of glucose consumed. Use molecular mass to get ratio of mass CO2 produced per mass of glucose. 6 275 10 mass_glucose 6 M3 M1 8 Ans. 2.216 10 kg 4.51 Assume as a basis, 1 mole of fuel. 0.85 (CH4(g) + 2 O2(g) = CO2(g) + 2 H2O(g)) 0.10(C2H6 (g) + 3.5 O2(g) = 2 CO2(g) + 3 H2O(g)) -----------------------------------------------------------------0.85 CH4(g) + 0.10 C2H6(g) + 2.05 O2(g) = 1.05 CO2(g) + 2 H2O(g) 1 = CH4, 2 = C2H6, 3 = O2, 4 = CO2, 5 = H2O 6 = N2 kJ mol H0f1 74.520 H0f4 393.509 a) H0c H0c 825.096 kJ mol 2 H0f5 83.820 H0f5 kJ mol 1.05 H0f4 kJ mol H0f2 241.818 0.85 H0f1 H0f3 0 kJ mol kJ mol 0.10 H0f2 1.05 H0f3 Ans. b) For complete combustion of 1 mole of fuel and 50% excess air, the exit gas will contain the following numbers of moles: n3 n3 0.5 2.05mol 121 1.025 mol Excess O2
  • 122. n4 1.05mol n5 2mol n6 0.05mol 79 1.5 2.05mol 21 n6 11.618 mol Total N2 Air and fuel enter at 25 C and combustion products leave at 600 C. T1 A B C D Q ( 25 T2 273.15) K n3 3.639 n4 6.311 ( 600 n5 3.470 273.15) K n6 3.280 mol n3 0.506 n4 0.805 n5 1.450 n6 0.593 10 3 mol n3 0 n4 0 n5 0 n6 0 10 6 mol n3 ( 0.227) n4 ( 0.906) n5 0.121 5 n6 0.040 10 mol H0c ICPH T1 T2 A B C D R 122 Q 529.889 kJ mol Ans.
  • 123. Chapter 5 - Section A - Mathcad Solutions 5.2 Let the symbols Q and Work represent rates in kJ/s. Then by Eq. (5.8) Work = QH TC TC = 1 TH TH 323.15 K Work TC QH 1 250 kJ s kJ s Work or 148.78 Work TH By Eq. (5.1), QH 798.15 K 148.78 kW which is the power. Ans. QC QH QC Work 101.22 kJ s Ans. 5.3 (a) Let symbols Q and Work represent rates in kJ/s TH TC 750 K By Eq. (5.8): But TC 1 Work = QH Work 300 K 0.6 TH Whence 95000 kW QH Work QH QC (b) QH QH 0.35 QC 5.4 (a) TC 303.15 K Carnot 1 TC TH QC Work QH TH Work 6.333 QH Work 5 1.583 10 kW Ans. 5 2.714 10 kW Ans. QC 1.764 4 10 kW Ans. 5 10 kW Ans. 623.15 K 0.55 123 Carnot 0.282 Ans.
  • 124. (b) 0.35 Carnot By Eq. (5.8), Carnot 0.55 TC TH 1 0.636 833.66 K Ans. TH Carnot 5.7 Let the symbols represent rates where appropriate. Calculate mass rate of LNG evaporation: 3 V 9000 molwt m P s 17 gm mLNG mol T 1.0133 bar PV molwt RT 298.15 K mLNG 6254 kg s Maximum power is generated by a Carnot engine, for which Work QH = QC QC TH 512 Work QH QC TH TC QH 1= QC TH TC 1 1 113.7 K QC kJ mLNG kg QC = TC 303.15 K QC 5.8 QC 3.202 Work QH Work 6 10 kW 6 10 kW 8.538 Ans. 6 5.336 Ans. 10 kW Take the heat capacity of water to be constant at the valueCP (a) T1 273.15 K SH2O Sres CP ln Q T2 T2 T2 T1 373.15 K Q CP T2 SH2O 1.305 1.121 kJ kg K Q kJ kg K Sres T1 124 Ans. 4.184 418.4 kJ kg K kJ kg
  • 125. Stotal SH2O Sres Stotal 0.184 kJ Ans. kgK (b) The entropy change of the water is the same as in (a), and the total heat transfer is the same, but divided into two halves. Q Sres 2 Stotal 1 373.15 K 1 323.15 K Sres Sres Stotal SH2O 0.097 1.208 kJ kJ kgK Ans. kgK (c) The reversible heating of the water requires an infinite number of heat reservoirs covering the range of temperatures from 273.15 to 373.15 K, each one exchanging an infinitesimal quantity of heat with the water and raising its temperature by a differential increment. 5.9 P1 T1 P1 V n R T1 (a) Const.-V heating; T2 T1 Q n CV But P2 P1 = 1.443 mol U= Q T2 By Eq. (5.18), 1 T1 Whence 3 0.06m 5 CV 2 R W = Q = n CV T2 Q 15000 J T1 3 10 K S = n CP ln T2 V 500 K n 1 bar S T2 T1 R ln n CV ln T2 T1 P2 P1 S 20.794 (b) The entropy change of the gas is the same as in (a). The entropy change of the surroundings is zero. Whence Stotal = 10.794 J K Ans. The stirring process is irreversible. 125 J K Ans.
  • 126. 5.10 (a) The temperature drop of the second stream (B) in either case is the same as the temperature rise of the first stream CP (A), i.e., 120 degC. The exit temperature of the second stream is therefore 200 degC. In both cases we therefore have: SA SA 8.726 463.15 CP ln SB 343.15 J mol K 473.15 593.15 CP ln SB 7 R 2 6.577 J Ans. mol K (b) For both cases: Stotal SA Stotal SB 2.149 J mol K Ans. (c) In this case the final temperature of steam B is 80 degC, i.e., there is a 10-degC driving force for heat transfer throughout the exchanger. Now SA SA 8.726 463.15 343.15 CP ln Stotal SB J mol K SA SB SB dQ Since dQ/T = dS, 8.512 Stotal dW 5.16 By Eq. (5.8), Ans. J mol K dW = dQ T dW = dQ J mol K 0.214 T = 1 353.15 473.15 CP ln Ans. T dQ T T dS Integration gives the required result. T1 Q CP T2 T2 600 K T1 Q 5.82 T 400 K 126 3 J 10 mol 300 K
  • 127. S CP ln Work Q Q T2 T1 T Q 5.17 TH1 11.799 Work S Q Work Q Sreservoir S S 2280 3540 Sreservoir T J mol K Sreservoir 0 600 K TC1 J mol K J mol Ans. J mol 11.8 Ans. J Ans. mol K Process is reversible. 300 K TH2 For the Carnot engine, use Eq. (5.8): W QH1 TC2 300 K = TH1 250 K TC1 TH1 The Carnot refrigerator is a reverse Carnot engine. W TH2 TC2 = Combine Eqs. (5.8) & (5.7) to get: TC2 QC2 Equate the two work quantities and solve for the required ratio of the heat quantities: TC2 TH1 TC1 Ans. r r 2.5 TH1 TH2 TC2 5.18 (a) T1 H S (b) C p T2 Cp ln T2 T1 T2 1.2bar T1 R ln P2 mol 4.365 S P1 3 J H = 5.82 10 450K H P1 300K 1.582 S = 1.484 127 P2 6bar 3 J 10 mol J mol K J mol K Ans. Ans. Cp 7 R 2
  • 128. 3 J (c) H = 3.118 10 (d) H = 3.741 10 (e) H = 6.651 10 S = 4.953 mol 3 J S = 2.618 mol 3 J J mol K J mol K S = 3.607 mol J mol K 5.19This cycle is the same as is shown in Fig. 8.10 on p. 305. The equivalent states are A=3, B=4, C=1, and D=2. The efficiency is given by Eq. (A) on p. 305. Temperature T4 is not given and must be calaculated. The following equations are used to derive and expression for T4. For adiabatic steps 1 to 2 and 3 to 4: T1 V 1 1 = T2 V 2 1 1 T3 V 3 For constant-volume step 4 to 1: P2 T2 = P3 T3 Solving these 4 equations for T4 yields: T4 = T1 7 R 2 T1 ( 200 T4 T1 T3 Eq. (A) p. 306 1 ( 1000 T4 T2 T3 1.4 Cv T2 273.15) K T2 Cp 5 R 2 Cv 1 V1 = V4 For isobaric step 2 to 3: Cp = T4 V 4 T3 ( 1700 873.759 K 1 T4 T1 T3 T2 128 273.15) K 0.591 Ans. 273.15) K
  • 129. 5.21 CV CP P1 R P2 2 bar CP 7 bar T1 298.15 K 1.4 CV With the reversible work given by Eq. (3.34), we get for the actual W: 1 Work 1.35 P2 R T1 1 3 J Work 3.6 10 P1 1 But Q = 0, and W = mol Whence S 5.25 P T2 CP ln T1 T 4 R ln T1 Work P2 S 2.914 P1 T2 T1 T2 U = C V T2 471.374 K J mol K CV Ans. 800 Step 1-2: Volume decreases at constant P. Heat flows out of the system. Work is done on the system. W12 = P V2 V1 = R T2 T1 Step 2-3: Isothermal compression. Work is done on the system. Heat flows out of the system. W23 = R T2 ln P3 P2 = R T2 ln P3 P1 Step 3-1: Expansion process that produces work. Heat flows into the system. Since the PT product is constant, P dT T dP = 0 PV= RT P dV = R dT T dP = dT P P dV V dP = R dT (A) V dP = R dT RT dP P 129
  • 130. In combination with (A) this becomes P dV = R dT R dT = 2 R dT Moreover, P3 = P 1 T1 T3 = P1 T1 T2 V1 P dV = 2 R T 1 W31 = T3 = 2 R T1 T2 V3 Q31 = U31 Q31 = CV = CP W31 = CV T1 2R Wnet W12 = Qin T1 T3 2 R T1 T3 = C P W23 R T1 T3 T2 W31 Q31 7 R 2 W23 W31 Q31 R T2 R T2 ln W12 T2 350 K 1.5 bar P3 P1 T1 W23 T2 3 J T1 T2 2.017 10 W31 T2 2.91 W23 P1 R T1 W12 P3 2 R T1 CP 700 K P1 W12 T1 5.82 10 Q31 1.309 10 W31 0.068 Q31 130 10 mol 3 J mol 3 J mol 4 J mol Ans.
  • 131. 5.26 T 403.15 K P1 By Eq. (5.18), S P2 2.5 bar R ln P2 P1 S 7.944 Tres 6.5 bar 298.15 K J mol K Ans. With the reversible work given by Eq. (3.27), we get for the actual W: Work Q 1.3 R T ln Work P2 P1 (Isothermal compresion) Work 4.163 10 3 J mol Q here is with respect to the system. So for the heat reservoir, we have Sres Q Sres Tres Stotal S Stotal Sres 13.96 6.02 J Ans. mol K J mol K Ans. 5.27 (a) By Eq. (5.14) with P = const. and Eq. (5.15), we get for the entropy change of 10 moles n 10 mol S n R ICPS 473.15K 1373.15K 5.699 0.640 10 S 536.1 J K 3 0.0 5 1.015 10 Ans. (b) By Eq. (5.14) with P = const. and Eq. (5.15), we get for the entropy change of 12 moles n 12 mol S n R ICPS 523.15K 1473.15K 1.213 28.785 10 S 2018.7 J K Ans. 131 3 8.824 10 6 0.0
  • 132. 5.28 (a) The final temperature for this process was found in Pb. 4.2a to be 1374.5 K. The entropy change for 10 moles is then found as follows n 10 mol S n R ICPS 473.15K 1374.5K 1.424 14.394 10 S 900.86 J K 3 4.392 10 6 0.0 Ans. (b) The final temperature for this process was found in Pb. 4.2b to be 1413.8 K. The entropy change for 15 moles is then found as follows: n 15 mol S n R ICPS 533.15K 1413.8K 1.967 31.630 10 S 2657.5 J 3 9.873 10 6 0.0 Ans. K (c) The final temperature for this process was found in Pb. 4.2c to be 1202.9 K. The entropy change for 18.14 kg moles is then found as follows n 18140 mol S S n R ICPS 533.15K 1202.9K 1.424 14.394 10 6 J 1.2436 10 K 3 4.392 10 6 0.0 Ans. 5.29 The relative amounts of the two streams are determined by an energy balance. Since Q = W = 0, the enthalpy changes of the two streams must cancel. Take a basis of 1 mole of air entering, and let x = moles of chilled air. Then 1 - x = the moles of warm air. T0 298.15 K Temperature of entering air T1 248.15 K Temperature of chilled air T2 348.15 K Temperature of warm air x C P T1 x 0.3 T0 ( 1 x)CP T2 T0 = 0 (guess) 132
  • 133. x Given 1 x = T2 T0 T1 T0 x x Find x () 0.5 Thus x = 0.5, and the process produces equal amounts of chilled and warmed air. The only remaining question is whether the process violates the second law. On the basis of 1 mole of entering air, the total entropy change is as follows. CP 7 2 P0 R Stotal x CP ln Stotal 12.97 P 5 bar T1 ( 1 T0 J mol K 1 bar T2 x)CP ln R ln T0 P P0 Ans. Since this is positive, there is no violation of the second law. 5.30 T1 Tres 303.15 K CV CP R Q= Sres Tres CP ln Stotal S T2 T1 R ln Sres P1 353.15 K Work Q Sres S T2 523.15 K 1800 U J mol Work 5.718 Q J mol K P2 CP Q S P1 Stotal 133 3.42 3 bar 7 2 P2 1 bar R C V T2 T1 Work 3 J 1.733 10 2.301 mol J mol K J PROCESS IS POSSIBLE. mol K
  • 134. 5.33 For the process of cooling the brine: kJ T mdot 40 K CP 3.5 T1 ( 273.15 25) K T1 ( 273.15 15) K T2 ( 273.15 30) K T 0.27 258.15 K T kg sec 298.15 K T2 20 303.15 K kg K H S CP ln kJ kg H T2 T1 Eq. (5.26): Wdotideal By Eq. (5.28): 140 S CP T 0.504 mdot H kJ kg K T S Wdotideal Wdotideal Wdot t Wdot 256.938 kW 951.6 kW Ans. t 5.34 E i 110 volt Wdotmech 1.25 hp Wdotelect At steady state: Qdot Wdotelect Qdot T SdotG = Qdot SdotG Wdotelect T 9.7 amp 300 K Wdotelect iE Wdotmech = 1.067 d t U = 0 dt d t S = 0 dt Qdot Qdot T 134 134.875 W SdotG Wdotmech 0.45 W K Ans. 3 10 W
  • 135. 5.35 i 25 ohm 2 Wdotelect T 10 amp Wdotelect i At steady state: 300 K 3 2.5 10 W d t U = 0 dt SdotG 8.333 kmol hr T1 ( 25 Cp R T2 Cp 1 R T dT ln watt K P2 10bar P2 1.2bar 7 5 Cv T2 Qdot T Ans. Cp (a) Assuming an isenthalpic process: S (b) = R Wdotelect SdotG P1 273.15) K Cv 10 Qdot d t S = 0 dt 10 watt 7 R 2 Cp SdotG = 3 2.5 5.38 mdot Wdotelect = Qdot T Qdot Qdot T2 T1 298.15 K Ans. Eq. (5.14) P1 T1 T2 7 R ln T1 2 S (c) SdotG (d) T 5.39(a) T1 T H R ln mdot S ( 20 273.15) K P2 P1 SdotG 48.966 Wlost T 500K P1 300K Basis: 1 mol n C p T2 T1 S 6bar T2 n Ws 135 17.628 W K S 371K J mol K Ans. Ans. Wlost P2 5.168 3 J 10 1.2bar mol Cp Ans. 7 R 2 1mol H Ws 3753.8 J Ans.
  • 136. S n Cp ln T2 T1 P2 R ln S P1 Eq. (5.27) Wideal H Eq. (5.30) Wlost Wideal Eq. (5.39) SG T S Ws SG T K Ans. 1409.3 J Wlost Wlost J 5163 J Ans. J K Ans. Wideal Wideal Ws 4.698 4.698 Wlost SG (a) 5163J 1409.3J 4.698 J K (b) 2460.9J 2953.9J 493J 1.643 J K (c) 3063.7J 4193.7J 1130J 3.767 J K (d) 3853.5J 4952.4J 1098.8J 3.663 (e) 5.41 3753.8J 3055.4J 4119.2J 1063.8J 3.546 P1 S R ln SdotG QH P2 S P1 mdot S Wdotlost 5.42 P2 2500kPa 1kJ W 0.023 SdotG T SdotG W QH actual K mdot 20 mol sec mol K kJ sec K Ans. 140.344 kW Ans. TH ( 250 TC actual 300K K J kJ 0.468 Wdotlost 0.45kJ T 150kPa J ( 25 0.45 136 273.15)K 273.15)K TH 523.15 K TC 298.15 K
  • 137. TC 1 max actual> max, Since 5.43 QH the process is impossible. 50 kJ Q2 100 kJ T1 350 K T2 250 K QH Q1 Q2 TH T1 T2 SG 0.27 550 K (a) SG (b) Wlost 5.44 0.43 Q1 150 kJ TH max TH Wlost T SG (a) TH 750 MW 1 max Ans. Ans. 81.039 kJ TC ( 20 TC 273.15)K 273.15)K 293.15 K Ans. 0.502 max TH 300 K kJ K 588.15 K TC Wdot QdotH ( 315 TH Wdot T QdotC QdotH Wdot QdotC 745.297 MW max (b) 0.6 QdotC QdotH Wdot QdotH max Wdot QdotC 1.742 (minimum value) QdotH 3 Cp 1 cal gm K T m 165 s QdotC Vdot 137 Cp 1 9 10 W (actual value) 10 MW 3 River temperature rise: Vdot 2.492 gm 3 cm T 2.522 K Ans.
  • 138. 5.46 T1 ( 20 P1 273.15)K T2 P2 5bar ( 27 273.15) K 1atm T3 ( 22 273.15)K First check the First Law using Eqn. (2.33) neglect changes in kinetic and potential energy. 6 H 7 1 7 H 3 R ICPH T1 T2 3.355 0.575 10 ICPH T1 T3 3.355 0.575 10 8.797 10 4 kJ mol 3 5 0 0 0.016 10 0.016 10 5 R H is essentially zero so the first law is satisfied. Calculate the rate of entropy generation using Eqn. (5.23) 6 SG R ICPS T1 T2 3.355 0.575 10 7 1 7 SG P R ICPS T1 T3 3.355 0.575 10 0.013 5.47 a) Vdot 3 kJ mol K Since SG 5 0 3 0.016 10 5 0 0.016 10 R ln P2 P1 0, this process is possible. 3 ft 100000 hr T1 459.67)rankine T2 T 1atm ( 70 ( 70 459.67)rankine ( 20 459.67)rankine Assume air is an Ideal Gas ndot P Vdot R T1 ndot 258.555 lbmol hr Calculate ideal work using Eqn. (5.26) Wideal ndot R ICPH T1 T2 3.355 0.575 10 T Wideal 3 0 R ICPS T1 T2 3.355 0.575 10 1.776 hp 138 5 0.016 10 3 0 5 0.016 10
  • 139. 3 b) Vdot P 3000 m T1 1atm ( 25 273.15) K T hr ( 25 273.15) K T2 (8 273.15) K 0 0.016 10 Assume air is an Ideal Gas P Vdot ndot ndot 34.064 mol s R T1 Calculate ideal work using Eqn. (5.26) Wideal ndot R ICPH T1 T2 3.355 0.575 10 T Wideal 5.48 T1 R ICPS T1 T2 3.355 0.575 10 3 5 5 0 0.016 10 1.952 kW ( 2000 Cp ( ) T T 3 459.67) rankine 3.83 ( 70 0.000306 T2 T rankine ( 300 Hv R 459.67) rankine Tsteam 970 459.67) rankine BTU M lbm ( 212 29 gm mol 459.67) rankine a) First apply an energy balance on the boiler to get the ratio of steam flow rate to gas flow rate.: T2 ndotgas Cp ( ) T T d mdotsteam Hv = 0 T1 T2 Cp ( ) T T d mdotndot T1 mdotndot Hv 15.043 lb lbmol Calculate the rate of entropy generation in the boiler. This is the sum of the entropy generation of the steam and the gas. SdotG = SdotGsteam SdotGgas 139
  • 140. Calculate entropy generation per lbmol of gas: SdotG ndotgas mdotsteam = Ssteam ndotgas Sgas Hv Ssteam Ssteam Tsteam T2 Sgas C p ( T) dT 1.444 Sgas 9.969 SdotG 11.756 Wlost T 6227 BTU lb rankine 10 3 kg mol lb rankine T1 SdotG mdotndot Ssteam Sgas BTU BTU lbmol rankine Calculate lost work by Eq. (5.34) Wlost SdotG T b) Hsteam Hv Wideal Hsteam Hv Ssteam T Ssteam Tsteam Ssteam BTU Ans. lbmol Wideal 1.444 BTU lb rankine 205.071 BTU lb Calculate lbs of steam generated per lbmol of gas cooled. T2 C p ( T ) dT mn T1 mn Hv 15.043 lb lbmol Use ratio to calculate ideal work of steam per lbmol of gas Wideal mn 3.085 10 3 BTU lbmol Ans. T2 c) Cp ( T) dT Hgas T1 Wideal Hgas T Sgas Wideal 140 9.312 3 BTU 10 lbmol Ans.
  • 141. 5.49 T1 ( 1100 Cp ( ) T T 273.15) K 3.83 ( 25 T2 0.000551 T R K ( 150 Hv 273.15) K 273.15) K 2256.9 Tsteam ( 100 kJ kg M 29 gm mol 273.15) K a) First apply an energy balance on the boiler to get the ratio of steam flow rate to gas flow rate.: T2 Cp ( ) T T d ndotgas mdotsteam Hv = 0 T1 T2 Cp ( ) T T d T1 mdotndot mdotndot Hv 15.135 gm mol Calculate the rate of entropy generation in the boiler. This is the sum of the entropy generation of the steam and the gas. SdotG = SdotGsteam SdotGgas Calculate entropy generation per lbmol of gas: SdotG ndotgas Ssteam = mdotsteam ndotgas Ssteam Sgas Hv Ssteam Tsteam T2 Sgas Cp ( ) T dT T Sgas 6.048 41.835 T1 SdotG mdotndot Ssteam Sgas SdotG 49.708 Wlost 14.8 3 10 J kg K J mol K J mol K Calculate lost work by Eq. (5.34) Wlost SdotG T 141 kJ mol Ans.
  • 142. b) Hsteam Hv Wideal Hsteam Hv Ssteam T Ssteam Tsteam Ssteam 3 6.048 Wideal 10 453.618 J kg K kJ kg Calculate lbs of steam generated per lbmol of gas cooled. T2 Cp ( T) dT T1 mn mn Hv 15.135 gm mol Use ratio to calculate ideal work of steam per lbmol of gas Wideal mn 6.866 kJ mol Ans. T2 c) Cp ( T) dT Hgas T1 Wideal 5.50 T1 a) Hgas ( 830 Sethylene Sethylene Qethylene T Sgas 273.15)K T2 ( 35 21.686 0.09 kJ mol 273.15)K R ICPS T1 T2 1.424 14.394 10 3 Ans. T 4.392 10 ( 25 6 273.15)K 0 kJ mol K R ICPH T1 T2 1.424 14.394 10 3 4.392 10 6 0 kJ mol Qethylene 60.563 Wlost Sethylene T Wideal Qethylene Wlost 33.803 kJ mol Now place a heat engine between the ethylene and the surroundings. This would constitute a reversible process, therefore, the total entropy generated must be zero. calculate the heat released to the surroundings for Stotal = 0. 142
  • 143. Sethylene QC T = 0 Solving for QC gives: QC QC T Sethylene 26.76 kJ mol Now apply an energy balance around the heat engine to find the work produced. Note that the heat gained by the heat engine is the heat lost by the ethylene. QH Qethylene WHE QH QC WHE 33.803 kJ mol The lost work is exactly equal to the work that could be produced by the heat engine 143
  • 144. Chapter 6 - Section A - Mathcad Solutions 6.7 At constant temperature Eqs. (6.25) and (6.26) can be written: dS = and V dP dH = 1 T V dP For an estimate, assume properties independent of pressure. T V S S 6.8 P1 270 K 1.551 10 3 3 m 2.661 P1 P2 1200 kPa 3 K H Ans. Tc 551.7 Zc 408.1 K 1 1 H P1 J kg K Isobutane: 2.095 10 kg V P2 P2 381 kPa T V P2 J kg Ans. 0.282 CP 2.78 Vc cm 262.7 mol 4000 kPa molwt 2000 kPa P1 gm 58.123 mol J gm K 3 Eq. (3.63) for volume of a saturated liquid may be used for the volume of a compressed liquid if the effect of pressure on liquid volume is neglected. 359 T 0.88 360 K Tr 361 T Tc Tr 0.882 0.885 (The elements are denoted by subscripts 1, 2, & 3 2 V V c Zc 1 Tr 131.604 7 V 132.138 3 cm mol 132.683 Assume that changes in T and V are negligible during throtling. Then Eq. (6.8) is integrated to yield: 144
  • 145. H= T S V 1 P2 T1 S but V P Then at 360 K, H= 0 P1 S 0.733 J Ans. mol K We use the additional values of T and V to estimate the volume expansivity: 3 V V3 V V1 1 V V1 T 1.079 cm T mol 4.098835 3 10 T3 T T1 2K 1 K Assuming properties independent of pressure, Eq. (6.29) may be integrated to give S = CP Whence 6.9 T T T S T1 T 6 P P1 V1 P P1 K P2 T molwt CP 298.15 K 250 10 P V P 1 1 bar P2 6 45 10 bar 3 2 10 kPa Ans. 0.768 K 1500 bar 3 1 cm 1003 kg V1 3 By Eq. (3.5), Vave V2 V1 Vave 2 Vave 1 H 134.6 S S T kJ kg P2 970.287 cm P1 V2 U kJ kg K Ans. Q Q H U P1 P1 By Eqs. (6.28) & (6.29), kg Ans. Vave P2 0.03636 P2 3 V2 H V1 exp cm 937.574 kg 5.93 T S 10.84 145 P2 V 2 kJ kg P1 V 1 Ans. Work kJ Ans. Work kg 4.91 U kJ kg Q Ans.
  • 146. 6.10 For a constant-volume change, by Eq. (3.5), T2 T1 36.2 10 T2 P2 6.14 --- 6.16 P2 5 K T1 P1 = 0 T1 1 298.15 K 4.42 10 P1 P2 5 205.75 bar Vectors containing T, P, Tc, Pc, and T2 bar P1 1 323.15 K 1 bar Ans. for Parts (a) through (n): 300 40 308.3 61.39 .187 175 75 150.9 48.98 .000 575 30 562.2 48.98 .210 500 50 425.1 37.96 .200 325 60 304.2 73.83 .224 175 60 132.9 34.99 .048 575 T 35 K P 556.4 bar Tc 45.60 K Pc .193 bar 650 50 553.6 40.73 .210 300 35 282.3 50.40 .087 400 70 373.5 89.63 .094 150 50 126.2 34.00 .038 575 15 568.7 24.90 .400 375 25 369.8 42.48 .152 475 75 365.6 46.65 .140 Tr T Tc Pr P Pc 146
  • 147. 6.14 Redlich/Kwong equation: Pr Tr Eq. (3.53) q z Given z= 1 i q z q Eq. (3.52) z z Find z () Ii HRi R Ti SRi R ln Z i qi Eq. (3.54) 1.5 1 1 14 Z 0.42748 Tr Guess: Z 0.08664 Z ln Z i qi Z i Eq. (6.65b) i qi 1 i qi 1.5 qi Ii Eq. (6.67) The derivative in these i i qi 0.5 qi Ii Eq. (6.68) equations equals -0.5 HRi SRi -2.302·103 J 0.695 -5.461 0.605 -2.068·103 mol -8.767 0.772 -3.319·103 -4.026 0.685 -4.503·103 -6.542 0.729 -2.3·103 -5.024 0.75 -1.362·103 -5.648 0.709 -4.316·103 -5.346 0.706 -5.381·103 -5.978 0.771 -1.764·103 -4.12 0.744 -2.659·103 -4.698 0.663 -1.488·103 -7.257 0.766 -3.39·103 -4.115 0.775 -2.122·103 -3.939 0.75 -3.623·103 -5.523 147 J mol K Ans.
  • 148. 6.15 Soave/Redlich/Kwong equation: 0.08664 1 0.42748 2 0.5 c 1 z z= 1 0.480 1.574 0.176 z q Eq. (3.52) Tr Z z z The derivative in the following equations equals: ci q Tri Find z () 0.5 i i 1 14 HRi R Ti Z Ii i qi ln 1 Z i qi Z Eq. (6.65b) i qi 0.5 Tri ci i 1 qi Ii Eq. (6.67) i SRi R ln Z i qi i ci Tri 0.5 qi Ii Eq. (6.68) i Z i qi HRi SRi J mol 0.691 -2.595·103 0.606 -2.099·103 0.774 -3.751·103 0.722 -7.408 0.741 -2.585·103 -5.974 0.768 -1.406·103 -6.02 0.715 -4.816·103 -6.246 0.741 -5.806·103 -6.849 0.774 -1.857·103 -4.451 0.749 -2.807·103 -5.098 0.673 -1.527·103 -7.581 0.769 -4.244·103 -5.618 0.776 -2.323·103 -4.482 0.787 -3.776·103 -6.103 J mol K -4.795 -4.821·103 148 -6.412 -8.947 2 Eq. (3.54) Eq. (3.53) q 1 Given Pr Tr Tr Guess: c Ans.
  • 149. 6.16 Peng/Robinson equation: Guess: Given c 2 0.5 c 1 0.37464 Pr Tr Tr z 1 2 0.45724 0.07779 1 1 2 1.54226 0.26992 Eq. (3.53) q Eq. (3.54) Tr 1 z= 1 z q z Eq. (3.52) Z z The derivative in the following equations equals: ci Tri q Find ( z) 0.5 i i 1 14 HRi 1 Ii R Ti Z 2 i qi Z 2 1 i qi i Z ln i qi i ci R ln Z i qi i ci Eq. (6.65b) 0.5 Tri 1 qi Ii i SRi Eq. (6.67) 0.5 Tri Eq. (6.68) qi Ii i Z i qi 2 HRi SRi 0.667 -2.655·103 J -6.41 0.572 -2.146·103 mol -8.846 0.754 -3.861·103 -4.804 0.691 -4.985·103 -7.422 0.716 -2.665·103 -5.993 0.732 -1.468·103 -6.016 0.69 -4.95·103 -6.256 0.71 -6.014·103 -6.872 0.752 -1.917·103 -4.452 0.725 -2.896·103 -5.099 0.64 -1.573·103 -7.539 0.748 -4.357·103 -5.631 0.756 -2.39·103 -4.484 0.753 -3.947·103 -6.126 149 J mol K Ans.
  • 150. Lee/Kesler Correlation --- By linear interpolation in Tables E.1--E.12: 0 h0 equals ( ) HR RTc 1 h1 equals 0 s0 equals ( ) SR R .686 ( ) HR RTc ( ) SR SR R s equals R .950 .471 .705 .024 1.003 1.709 .155 .774 RTc 1 s1 equals .093 .590 HR h equals .591 .675 1.319 .437 .725 .008 .993 .635 .744 Z0 .118 .165 1.265 .184 .705 .699 .770 Z1 .019 .962 h0 .102 1.200 .001 h1 .751 .444 .770 .550 .742 .007 .875 .598 .651 .144 1.466 .405 .767 .034 .723 .631 .776 .032 .701 .604 1.216 .211 .746 Z Z0 .154 Z1 Eq. (3.57) h 150 h0 h1 (6.85) HR ( Tc R) h
  • 151. .711 1.110 .492 .497 .549 .829 .443 .631 .590 .710 s0 .961 .276 .674 .750 .700 s1 .441 .517 .287 0.669 -1.138 SRi HRi si hi Zi Eq. (6.86) .563 .688 ( s R) .589 .491 SR .429 .511 s1 .555 .917 s0 .509 .587 s -0.891 -2.916·103 J mol -7.405 0.59 -1.709 -1.11 -2.144·103 0.769 -0.829 -0.612 -3.875·103 -5.091 0.699 -1.406 -0.918 -4.971·103 -7.629 0.727 -1.135 -0.763 -2.871·103 -6.345 0.752 -1.274 -0.723 -1.407·103 -6.013 0.701 -1.107 -0.809 -5.121·103 -6.727 0.72 -1.293 -0.843 -5.952·103 -7.005 0.77 -0.818 -0.561 -1.92·103 -4.667 0.743 -0.931 -0.639 -2.892·103 -5.314 0.656 -1.481 -0.933 -1.554·103 -7.759 0.753 -0.975 -0.747 -4.612·103 -6.207 0.771 -0.793 -0.577 -2.438·103 -4.794 0.768 -1.246 -0.728 -3.786·103 -6.054 J mol K 151 -9.229 Ans.
  • 152. 6.17 T t 50 273.15 K The pressure is the vapor pressure given by the Antoine equation: T P () t d t 323.15 K 2788.51 t 220.79 exp 13.8858 P ( ) 36.166 50 P P () 1.375 t dPdt 36.166 kPa 1.375 kPa dt K (a) The entropy change of vaporization is equal to the latent heat divided by the temperature. For the Clapeyron equation, Eq. (6.69), we need the volume change of vaporization. For this we estimate the liquid volume by Eq. (3.63) and the vapor volume by the generalized virial correlation. For benzene: Tc T Tc 3 Vc 259 cm mol Pc 562.2 K Tr 0.210 Tr Zc 48.98 bar P Pc Pr 0.575 Pr 0.271 0.007 By Eqs. (3.65), (3.66), (3.61), & (3.63) B0 0.422 0.083 Tr Vvap RT P 1.6 1 By Eq. (3.72), B0 B1 0.941 0.139 0.172 Tr B0 Vliq B1 Pr Vvap Tr V c Zc 1 Tr B1 4.2 1.621 3 4 cm 7.306 10 mol 3 2/7 Vliq 93.151 cm mol Solve Eq. (6.72) for the latent heat and divide by T to get the entropy change of vaporization: S dPdt Vvap S Vliq 100.34 J mol K Ans. 102.14 J mol K Ans. (b) Here for the entropy change of vaporization: S RT dPdt P S 152
  • 153. 6.20 The process may be assumed to occur adiabatically and at constant pressure. It is therefore isenthalpic, and may for calculational purposes be considered to occur in two steps: (1) Heating of the water from -6 degC to the final equilibrium temperature of 0 degC. (2) Freezing of a fraction x of the water at the equilibrium T. Enthalpy changes for these two steps sum to zero: CP t CP x Hfusion = 0 Hfusion 333.4 joule gm J gm K 4.226 t CP t x x Hfusion 6K 0.076 Ans. The entropy change for the two steps is: T2 S T1 273.15 K CP ln ( 273.15 x Hfusion T2 T2 T1 6) K S 1.034709 10 3 J Ans. gm K The freezing process itself is irreversible, because it does not occur at the equilibrium temperature of 0 degC. H1 1156.3 BTU lbm H2 1.7320 BTU S2 lbm rankine 1.9977 BTU 1533.4 S1 6.21 Data, Table F.4: H H2 H1 S S2 H 377.1 BTU lbm S 0.266 lbm BTU lbm rankine S1 BTU Ans. lbm rankine For steam as an ideal gas, apply Eqs. (4.9) and (5.18). [t in degF] T1 ( 227.96 P1 20 psi T1 382.017 K T2 459.67)rankine P2 50 psi T2 810.928 K 153 ( 1000 459.67)rankine
  • 154. molwt 18 lb lbmol R MCPH T1 T2 3.470 1.450 10 molwt H H 372.536 BTU lbm 3 5 0.0 0.121 10 T2 T1 Ans. R MCPS T1 T2 3.470 1.450 10 S 3 5 0.0 0.121 10 ln T2 T1 ln P2 P1 molwt S 0.259 BTU lbm rankine Ans. 6.22 Data, Table F.2 at 8000 kPa: 3 Vliq cm 1.384 gm Hliq 1317.1 J gm 3 Vvap cm 23.525 gm Hvap 2759.9 6 mliq mliq mvap 54.191 kg mliq Hliq Stotal mliq Sliq J gm Svap J gm K 5.7471 J gm K 6 0.15 10 3 cm 2 Vliq Htotal 3.2076 Sliq mvap mvap Hvap 0.15 10 3 cm 2 Vvap 3.188 kg Htotal 154 Ans. Stotal mvap Svap 80173.5 kJ 192.145 kJ K Ans.
  • 155. 6.23 Data, Table F.2 at 1000 kPa: 3 Vliq 1.127 Hliq gm 762.605 J gm Sliq Hvap cm 2776.2 J gm Svap 3 Vvap 194.29 cm gm Let x = fraction of mass that is vapor (quality) x Vvap 70 30 x 2.1382 6.5828 0.5 ( 1 H ( 1 x)Hliq H 789.495 J gm ( 1 2.198 J gm K (Guess) 0.013 S x Hvap gm K Find x () S x)Vliq = x x Given J x)Sliq x Svap J gm K Ans. 6.24 Data, Table F.3 at 350 degF: 3 3 Vliq Hliq 321.76 Vvap mliq 3 lbm 1 Htotal lbm 1192.3 BTU lbm mvap = 3 lbm mvap Vvap = 50 mliq Vliq mliq mliq mvap BTU ft 3.342 lbm Hvap ft 0.01799 lbm mliq 50 Vliq 50 mliq Vliq = 3 lbm Vvap 2.364 lb Vvap 3 lbm mvap mliq mliq Hliq mvap Hvap Htotal 155 0.636 lb 1519.1 BTU Ans.
  • 156. 3 6.25 1 cm 0.025 gm V Data, Table F.1 at 230 degC: 3 Vliq 1.209 cm Hliq gm 990.3 J gm 3 Vvap 71.45 cm Hvap gm J gm 2802.0 x Vvap x H ( 1 x)Hliq x Hvap S x 6.26 x)Vliq 0.552 Vtotal = mtotal Vliq Vtotal 3 0.15 m mtotal ( 1 Vliq x)Sliq S 4.599 x Svap J gm K 0.382 kg mliq mtotal mvap mliq 377.72 gm Ans. 3 Table F.1, 150 degC: Vvap Vtotal Vlv mtotal Vliq Vlv mvap 4.543 3 10 Vtot.liq Vtot.liq 379.23 cm kg mliq Vliq 156 cm 392.4 gm 3 cm 1.004 gm mvap Vvap J gm K mvap Vlv Vliq Vtotal J gm K Vliq Vvap J gm 6.2107 3 Table F.1, 30 degC: mtotal 1991 2.6102 Svap V V= ( 1 H Sliq 3 Ans. cm 32930 gm
  • 157. 6.27 Table F.2, 1100 kPa: Hliq 781.124 J gm Hvap Interpolate @101.325 kPa & 105 degC: Const.-H throttling: H2 = Hliq H2 x 6.28 Hvap H2 x Hvap 2779.7 2686.1 J gm J gm Hliq Hliq x Hliq Ans. 0.953 Data, Table F.2 at 2100 kPa and 260 degC, by interpolation: H1 H2 J 2923.5 2923.5 gm J gm S1 6.5640 J gm K molwt 18.015 gm mol Final state is at this enthalpy and a pressure of 125 kPa. By interpolation at these conditions, the final temperature is 224.80 degC and S2 7.8316 J gm K S S2 S S1 1.268 J gm K Ans. For steam as an ideal gas, there would be no temperature change and the entropy change would be given by: P1 2100 kPa P2 125 kPa S P2 R ln P1 molwt S 1.302 J gm K Ans. 6.29 Data, Table F.4 at 300(psia) and 500 degF: H1 1257.7 H2 1257.7 BTU lbm BTU lbm S1 1.5703 BTU lbm rankine Final state is at this enthalpy and a pressure of 20(psia). By interpolation at these conditions, the final temperature is 438.87 degF and S2 1.8606 BTU lbm rankine S 157 S2 S1 S 0.29 BTU lbm rankine
  • 158. For steam as an ideal gas, there would be no temperature change and the entropy change would be given by: P1 P2 300 psi 6.30 lb lbmol P1 S molwt 0.299 BTU lbm rankine Ans. Data, Table F.2 at 500 kPa and 300 degC S1 7.4614 J gm K 1.0912 Hliq 340.564 S2 = S1 = Sliq H2 Hliq The final state is at this entropy and a pressure of 50 kPa. This is a state of wet steam, for which J gm K Sliq 6.31 18 P2 R ln S molwt 20 psi Svap Hvap J gm x Svap x Hvap 7.5947 J gm K 2646.9 Sliq x H2 Hliq J gm S1 Svap Sliq Sliq x 0.98 J gm Ans. xwater 0.031 Ans. xwater 0.122 Ans. 2599.6 Vapor pressures of water from Table F.1: At 25 degC: Psat P xwater 101.33 kPa At 50 degC: Psat xwater 3.166 kPa Psat P 12.34 kPa Psat P 158
  • 159. 6.32 Process occurs at constant total volume: Vtotal 3 ( 0.014 0.021)m Data, Table F.1 at 100 degC: Uliq 419.0 J gm 1.044 cm 1673.0 mvap 0.014 m Vvap Vliq mvap x mass V2 4 4.158 10 3 Vtotal V2 mass mliq gm 3 3 x mass gm cm Vvap gm 0.021 m mliq J 2506.5 3 3 Vliq Uvap cm 1.739 gm mvap (initial quality) This state is first reached as saturated liquid at 349.83 degC For this state, P = 16,500.1 kPa, and U2 Q 1641.7 U2 J gm Uliq Q U1 U1 1221.8 x Uvap J gm Uliq U1 419.868 J gm Ans. 3 6.33 Vtotal 0.25 m Data, Table F.2, sat. vapor at 1500 kPa: 3 V1 cm 131.66 gm U1 2592.4 J gm Of this total mass, 25% condenses making the quality 0.75 Since the total volume and mass don't change, we have for the final state: V2 = V1 = Vliq x= V1 Vvap Vliq Vliq x Vvap (A) Vtotal mass V1 x Whence Vliq Find P for which (A) yields the value x = 0.75 for wet steam 159 0.75
  • 160. Since the liquid volume is much smaller than the vapor volume, we make a preliminary calculation to estimate: Vvap V1 3 Vvap x 175.547 cm gm This value occurs at a pressure a bit above 1100 kPa. Evaluate x at 1100 and 1150 kPa by (A). Interpolate on x to find P = 1114.5 kPa and Uliq U2 Q 782.41 Uliq J gm x Uvap mass U2 Uvap 2584.9 J gm U2 2134.3 Q Uliq 869.9 kJ U1 J gm Ans. 3 3 Vliq Uvap 6.34 Table F.2,101.325 kPa: Uliq J 418.959 gm Vvap J 2506.5 gm cm 1673.0 gm mliq cm 1.044 gm 0.02 m Vliq 3 3 mvap 1.98 m Vvap mtotal mliq mvap x mvap mtotal 3 V1 Vliq x Vvap Vliq U1 Uliq x Uvap Uliq cm V1 98.326 U1 540.421 x gm 0.058 J gm Since the total volume and the total mass do not change during the process, the initial and final specific volumes are the same. The final state is therefore the state for which the specific volume of saturated vapor is 98.326 cu cm/gm. By interpolation in Table F.1, we find t = 213.0 degC and U2 2598.4 J gm Q mtotal U2 160 U1 Q 41860.5 kJ Ans.
  • 161. 6.35 Data, Table F.2 at 800 kPa and 350 degC: 3 V1 cm 354.34 U1 gm 2878.9 J gm Vtotal 3 0.4 m The final state at 200 degC has the same specific volume as the initial state, and this occurs for superheated steam at a pressure between 575 and 600 kPa. By interpolation, we find P = 596.4 kPa and U2 J gm 2638.7 Q Vtotal V1 U2 Q U1 271.15 kJ Ans. 6.36 Data, Table F.2 at 800 kPa and 200 degC: U1 J gm 2629.9 S1 6.8148 J gm K mass 1 kg (a) Isothermal expansion to 150 kPa and 200 degC U2 J gm 2656.3 Q mass T S2 Also: J gm K S2 Q S1 Work 7.6439 392.29 kJ mass U2 U1 T 473.15 K Ans. Work Q 365.89 kJ (b) Constant-entropy expansion to 150 kPa. The final state is wet steam: J gm K Svap 7.2234 J gm K J gm Uvap 2513.4 J gm Sliq 1.4336 Uliq 444.224 x S1 Svap Sliq x Sliq U2 Uliq x Uvap W mass U2 U2 Uliq W U1 161 0.929 2.367 3 J 10 262.527 kJ gm Ans.
  • 162. 6.37 Data, Table F.2 at 2000 kPa: x H1 Hvap 0.94 Hliq x Hvap 2797.2 H1 Hliq J Hliq gm 2.684 10 908.589 3 J mass gm J gm 1 kg For superheated vapor at 2000 kPa and 575 degC, by interpolation: H2 3633.4 J gm 6.38 First step: Q mass H2 Q H1 Ans. 949.52 kJ Q12 = 0 W12 = U2 U1 Second step: W23 = 0 Q23 = U3 U2 For process: Q = U3 Table F.2, 2700 kPa: Uliq 977.968 Sliq 2.5924 x1 0.9 U1 S1 Uliq Sliq W = U2 U2 U1 J gm Uvap 2601.8 J gm J gm K Svap 6.2244 J gm K x1 Uvap x1 Svap Uliq U1 Sliq S1 3 J 2.439 10 5.861 2 3 m 10 2 gm s K Table F.2, 400 kPa: J gm K Svap 6.8943 J gm K J gm Uvap 2552.7 J gm Sliq 1.7764 Uliq 604.237 Vliq cm 1.084 gm 3 3 162 Vvap 462.22 cm gm
  • 163. Since step 1 is isentropic, S2 = S1 = Sliq U2 Uliq x2 Svap x2 Uvap S1 x2 Sliq Svap x2 Sliq 2.159 U2 Uliq Sliq 10 0.798 3 J gm 3 V2 Vliq x2 Vvap V2 Vliq 369.135 cm gm V3 = V2 and the final state is sat. vapor with this specific volume. Interpolate to find that this V occurs at T = 509.23 degC and U3 J gm Q 2560.7 401.317 Whence J gm Q Ans. U3 Work U2 Work 280.034 U2 J gm Ans. U1 2605.8 J gm S1 7.0548 Table F.1,sat. vapor, 175 degC U2 2578.8 J gm S2 6.6221 mass T 6.39 Table F.2, 400 kPa & 175 degC: 4 kg Q mass T S2 Q S1 775.66 kJ Ans. ( 175 273.15)K W mass U2 W U1 667.66 kJ 6.40 (a)Table F.2, 3000 kPa and 450 degC: H1 3344.6 J gm S1 7.0854 J gm K Table F.2, interpolate 235 kPa and 140 degC: H2 2744.5 J gm S2 7.2003 163 J gm K Ans. U1 Q J gm K J gm K
  • 164. H H2 S S2 (b) T1 H H1 S1 ( 450 S J gm Ans. J gm K Ans. 600.1 0.115 T2 273.15)K ( 140 273.15)K T1 723.15 K T2 413.15 K P1 3000 kPa P2 235 kPa gm mol 3 5 R ICPH T1 T2 3.470 1.450 10 0.0 0.121 10 Eqs. (6.95) & (6.96) for an ideal gas: Hig molwt molwt 3 R ICPS T1 T2 3.470 1.450 10 Sig Tr1 Tr1 5 ln 0.0 0.121 10 620.6 J gm 647.1 K T1 Sig Pc Pr1 H S S P1 Tc 1.11752 Pr1 Tr2 Pc 0.13602 J gm K Tr2 T2 Tc 0.63846 HRB Tr1 Pr1 R Tc HRB Tr2 Pr2 molwt J Ans. 593.95 gm Hig Sig 0.078 R SRB Tr2 Pr2 J gm K SRB Tr1 Pr1 molwt Ans. 164 Ans. 0.345 220.55 bar P1 0.0605 Pr2 Pr2 The generalized virial-coefficient correlation is suitable here H P2 molwt Hig (c) Tc 18 P2 Pc 0.01066
  • 165. 6.41 Data, Table F.2 superheated steam at 550 kPa and 200 degC: 3 V1 385.19 cm U1 gm 2640.6 J gm S1 J gm K 7.0108 Step 1--2: Const.-V heating to 800 kPa. At the initial specific volume and this P, interpolation gives t = 401.74 degC, and U2 2963.1 J gm S2 7.5782 J gm K Q12 S3 S3 = S 2 Q23 = 0 7.5782 U1 Q12 Step 2--3: Isentropic expansion to initial T. U2 322.5 J gm J gm K Step 3--1: Constant-T compression to initial P. T 473.15 K Q31 T S1 Q31 S3 268.465 J gm For the cycle, the internal energy change = 0. Wcycle = Qcycle = Q12 1 Q31 Q31 = 0.1675 Q12 Wcycle Q12 Ans. 6.42 Table F.4, sat.vapor, 300(psi): T1 T1 459.67) rankine H1 ( 417.35 1202.9 877.02 rankine S1 1.5105 BTU lbm BTU lbm rankine Superheated steam at 300(psi) & 900 degF H2 Q12 1473.6 H2 BTU lbm H1 S2 Q31 1.7591 BTU lbm rankine T 1 S1 165 S3 S3 Q31 S2 218.027 BTU lbm
  • 166. For the cycle, the internal energy change = 0. Wcycle = Qcycle = Q12 1 6.43 Q31 Q31 Q12 Whence Ans. 0.1946 Q12 Wcycle = Data, Table F.2, superheated steam at 4000 kPa and 400 degC: S1 6.7733 J gm K For both parts of the problem: S2 S1 (a)So we are looking for the pressure at which saturated vapor has the given entropy. This occurs at a pressure just below 575 kPa. By interpolation, P2 = 572.83 kPa Ans. (b)For the wet vapor the entropy is given by x S2 = Sliq 0.95 x Svap Sliq So we must find the presure for which this equation is satisfied. This occurs at a pressure just above 250 kPa. At 250 kPa: Sliq 1.6071 S2 Sliq S2 6.7798 J gm K x Svap Svap 7.0520 J gm K Sliq J gm K Slightly > 6.7733 By interpolation P2 = 250.16 kPa Ans. 6.44 (a) Table F.2 at the final conditions of saturated vapor at 50 kPa: S2 7.5947 kJ kg K H2 2646.0 kJ kg S1 S2 Find the temperature of superheated vapor at 2000 kPa with this entropy. It occurs between 550 and 600 degC. By interpolation 166
  • 167. t1 (degC) 559.16 Superheat: (b) mdot 5 t kg sec H1 ( 559.16 t 212.37)K Wdot kJ kg 3598.0 mdot H2 H1 Ans. 346.79 K 4760 kW Ans. Wdot 6.45 Table F.2 for superheated vapor at the initial conditions, 1350 kPa and 375 degC, and for the final condition of sat. vapor at 10 kPa: H1 kJ kg 3205.4 S1 7.2410 kJ kg K H2 2584.8 kJ kg If the turbine were to operate isentropically, the final entropy would be S2 S1 Table F.2 for sat. liquid and vapor at 10 kPa: kJ kg K Svap 8.1511 kJ kg K kJ kg Hvap 2584.8 kJ kg Sliq 0.6493 Hliq 191.832 x2 S2 Svap Sliq Sliq x2 H' Hliq 0.879 x2 Hvap H' 2.294 H2 H1 3 kJ 10 kg Ans. 0.681 H' H1 Hliq 6.46 Table F.2 for superheated vapor at the initial conditions, 1300 kPa and 400 degC, and for the final condition of 40 kPa and 100 degC: H1 3259.7 kJ kg S1 7.3404 kJ kg K H2 2683.8 kJ kg If the turbine were to operate isentropically, the final entropy would be S2 S1 Table F.2 for sat. liquid and vapor at 40 kPa: 167
  • 168. Sliq 1.0261 kJ kg K Svap 7.6709 Hliq 317.16 kJ kg Hvap 2636.9 S2 x2 Sliq Svap H2 x2 Sliq kJ kg K kJ kg H' Hliq 0.95 H' 2.522 H1 6.47 Table F.2 at 1600 kPa and 225 degC: H 3 kJ kg P 1600 kPa S 6.5503 J gm K P0 1 kPa 498.15 K VR V 3 cm 132.85 gm 10 Hliq Ans. 0.78 H' H1 V x2 Hvap J gm 2856.3 J gm K Table F.2 (ideal-gas values, 1 kPa and 225 degC) Hig T J gm 2928.7 ( 225 Sig 10.0681 T 273.15)K T R molwt P The enthalpy of an ideal gas is independent of pressure, but the entropy DOES depend on P: HR H Sig Hig R molwt 3 VR 10.96 cm gm HR Reduced conditions: Tr T Tc 72.4 ln P P0 SR J gm S SR Sig 0.11 Sig J Ans. gm K 0.345 Tr Tc 647.1 K Pc 220.55 bar 0.76982 Pr P Pc Pr 0.072546 The generalized virial-coefficient correlation is suitable here B0 0.083 0.422 Tr 1.6 B0 0.558 B1 0.139 0.172 Tr 168 4.2 B1 0.377
  • 169. By Eqs. (3.61) + (3.62) & (3.63) along with Eq. (6.40) Z 1 B0 B1 R Tc HR Pr Z Tr molwt R SR HRB Tr Pr molwt 3 VR 6.48 9.33 P cm HR gm 53.4 T 1000 kPa (Table F.2) SR 273.15) K 18.015 J gm Hv 2776.2 J gm K Sv 6.5828 762.605 Sl 2.1382 (b) (c) Slv VR T Sl G l 4.445 Vv 453.03 K J gm K T R molwt P Hlv 206.06 gm r 10 Gv 3 J Hv Hl Sv Slv gm Hv Vl Slv J gm K Vv Hlv gm 2.014 J Vlv J 3 cm 193.163 gm Hl Ans. gm K gm mol Vv Hl J T cm 194.29 gm Vl (a) Gl 0.077 3 3 cm 1.127 gm Vlv SRB Tr Pr J gm ( 179.88 molwt RT ( Z 1) P molwt VR 0.935 4.445 T Sv G v Hlv r T Sl J gm K 206.01 4.445 J gm K 3 VR cm 14.785 gm Ans. For enthalpy and entropy, assume that steam at 179.88 degC and 1 kPa is an ideal gas. By interpolation in Table F.2 at 1 kPa: 169 J gm
  • 170. Hig 2841.1 J gm Sig 9.8834 J P0 gm K 1 kPa The enthalpy of an ideal gas is independent of pressure; the entropy DOES depend on P: HR Hv SR Sv Sig P R ln P0 molwt Sig Hig HR Sig 64.9 J gm Sig 3.188 Ans. SR 0.1126 J gm K J gm K Ans. (d) Assume ln P vs. 1/T linear and fit three data pts @ 975, 1000, & 1050 kPa. 975 Data: pp 178.79 1000 kPa t 1050 xi 1 yi 273.15 ti P dPdT T Slv 2 ln ppi Slope kPa dPdT Slope K 1 3 slope ( y) Slope x 22.984 4717 kPa K J gm K Ans. Slv 4.44 0.345 Reduced conditions: T Tc i 182.02 Tc 647.1 K Pc 220.55 bar 0.7001 Pr P Pc Pr 0.0453 Vlv dPdT Tr (degC) 179.88 Tr The generalized virial-coefficient correlation is suitable here B0 0.083 0.422 Tr 1.6 B0 0.664 B1 0.139 0.172 Tr 4.2 B1 0.63 By Eqs. (3.61) + (3.62) & (3.63) along with Eq. (6.40) Z 1 B0 B1 Pr Tr Z 170 0.943 VR RT ( 1) Z P molwt
  • 171. R Tc HR molwt R SRB Tr Pr molwt SR HRB Tr Pr 3 VR cm 11.93 6.49 T HR gm ( 358.43 43.18 J SR gm T 459.67) rankine (Table F.4) Vv BTU lbm Hv 1194.1 BTU lbm rankine Sv 1.5695 330.65 Sl 0.5141 Gl (b) (c) Slv VR Gv Hv Gv BTU lbm rankine T R molwt P r T VR Hl Sv Sl T Sv 89.91 Hlv r Hv BTU lbm 863.45 BTU lbm 1.055 Vv Hlv Vl Slv lbm rankine Vv Hlv BTU T Sl 89.94 150 psi Vlv BTU lbm 3 ft 2.996 lbm Hl P gm mol ft 3.014 lbm Hl (a) Gl 18.015 Ans. gm K 3 3 Vl J 818.1 rankine molwt ft 0.0181 lbm Vlv 0.069 0.235 ft BTU lbm 1.055 BTU lbm rankine 3 Ans. lbm For enthalpy and entropy, assume that steam at 358.43 degF and 1 psi is an ideal gas. By interpolation in Table F.4 at 1 psi: Hig 1222.6 BTU lbm Sig 2.1492 171 BTU lbm rankine P0 1 psi
  • 172. The enthalpy of an ideal gas is independent of pressure; the entropy DOES depend on P: HR Hv P R ln P0 molwt Sig SR HR Hig Sv Sig 28.5 Sig 0.0274 Ans. BTU lbm rankine 0.552 SR Sig BTU lbm BTU lbm rankine Ans. (d) Assume ln P vs. 1/T linear and fit threedata points (@ 145, 150, & 155 psia) 145 Data: pp 355.77 150 psi t 155 xi ti 1 459.67 (degF) 358.43 P T Slv 2 yi ln ppi Slope psi slope ( y) x Slv Vlv dPdT T Tc Tr 0.345 Tc 1.905 1.056 10 psi rankine BTU Ans. lbm rankine Pc 647.1 K Pr 0.7024 3 8.501 dPdT Slope rankine Reduced conditions: Tr 1 3 361.02 Slope dPdT i P Pc 220.55 bar Pr 0.0469 The generalized virial-coefficient correlation is suitable here B0 0.083 0.422 Tr 1.6 B0 0.66 B1 0.139 0.172 Tr 172 4.2 B1 0.62
  • 173. By Eqs. (3.61) + (3.62) & (3.63) along with Eq. (6.40) Z 1 HR VR 6.50 B0 R Tc ft Z Tr HR lbm 19.024 For propane: Tc T ( 195 T Tr T Tc 273.15) K Tr BTU lbm SR (Z BTU lbm rankine 1) P0 135 bar P Pc Pr Ans. 0.152 42.48 bar P 468.15 K 1.266 0.0168 Pc 369.8 K P molwt R SRB Tr Pr molwt SR 3 RT VR 0.942 HRB Tr Pr molwt 0.1894 Pr B1 Pr 1 bar 3.178 Use the Lee/Kesler correlation; by interpolation, Z0 V Z Z1 ZRT P 0.1636 V 0.6141 cm 184.2 mol 0.586 R Tc HR1 1.802 SR1 0.717 R J mol K SR1 5.961 HR1 SR HR0 7.674 SR0 1.463 R SR0 12.163 HR H 0.639 Ans. HR1 2.496 R Tc HR0 Z Z1 3 HR0 HR Z0 3 J 10 mol 3 J 7.948 10 SR0 SR mol R ICPH 308.15K T 1.213 28.785 10 173 13.069 3 10 3 J mol J mol K SR1 J mol K 8.824 10 6 0.0 HR
  • 174. S R ICPS 308.15K T 1.213 28.785 10 H 6734.9 J mol Ans. 6.51 For propane: T ( 70 Tr T Tc S Tr 6 8.824 10 15.9 J mol K 0.0 ln P P0 SR Ans. 369.8 K Pc 42.48 bar 0.152 343.15 K P0 101.33 kPa P 1500 kPa Tc T 273.15)K 3 P Pc Pr 0.92793 Pr 0.35311 Assume propane an ideal gas at the initial conditions. Use generalized virial correlation at final conditions. H S H R SRB Tr Pr P P0 ln 6.52 For propane: 1431.3 J mol Ans. S R Tc HRB Tr Pr 25.287 J mol K Ans. 0.152 3 cm Vc 200.0 Zc 0.276 Pc 42.48 bar Tc 369.8 K mol If the final state is a two-phase mixture, it must exist at its saturation temperature at 1 bar. This temperature is found from the vapor pressure equation: P A 1 bar D 1.38551 Given P = Pc exp T () T A () B T Find T) ( B 6.72219 T Tc 1 () T 1.5 C Guess: C 1 T 1.33236 () T 230.703 K 174 () T 3 D T () T 6 2.13868 200 K
  • 175. The latent heat of vaporization at the final conditions will be needed for an energy balance. It is found by the Clapeyron equation. We proceed exactly as in Pb. 6.17. P ( T) A ( T) Pc exp P d P ( T) dT 230.703 K 1 bar Tr Vvap Vvap Hlv 3 ( T) D kPa K Pr 0.815 B1 6 ( T) dPdT 0.024 B1 0.139 Pr Vliq Tr 4.428124 T Tr Tc 0.172 Tr B0 4.2 V c Zc 3 4 cm 10 T Vvap B0 1.6 1 1.847 C 4.428 Pc 0.422 RT P 1.5 ( T) P Pr 0.083 ( T) 1 T B0 B Tr kPa K 0.624 B1 1.109 2 1 Tr 7 3 Vliq mol Vliq dPdT Hlv 75.546 1.879 cm mol 4 J 10 mol ENERGY BALANCE: For the throttling process there is no enthalpy change. The calculational path from the initial state to the final is made up of the following steps: (1) Transform the initial gas into an ideal gas at the initial T & P. (2) Carry out the temperature and pressure changes to the final T & P in the ideal-gas state. (3) Transform the ideal gas into a real gas at the final T & P. (4) Partially condense the gas at the final T & P. The sum of the enthalpy changes for these steps is set equal to zero, and the resulting equation is solved for the fraction of the stream that is liquid. For Step (1), use the generalized correlation of Tables E.7 & E.8, and let r0 = H R R Tc 0 and r1 = 175 H R R Tc 1
  • 176. T1 Tr P1 370 K T1 Tr Tc By Eq. (6.85) r0 H1 P1 Pr 1.001 By interpolation, find: 200 bar r1 3.773 R Tc r0 Pr Pc 3.568 H1 r1 4.708 4 J 1.327 10 mol For Step (2) the enthalpy change is given by Eq. (6.95), for which H2 R ICPH T1 T 1.213 28.785 10 3 6 8.824 10 0.0 4 J H2 1.048 10 mol For Step (3) the enthalpy change is given by Eq. (6.87), for which Tr H3 230.703 K Tc Tr R Tc HRB Tr Pr H3 232.729 H1 Pr For Step (4), 0.0235 H4 = x Hlv J mol For the process, x 1 bar Pc Pr 0.6239 H2 H1 H3 x H2 0.136 H3 x Hlv = 0 Ans. Hlv 6.53 For 1,3-butadiene: Tc 425.2 K Vc 0.190 cm 220.4 mol Tn 268.7 K 101.33 kPa 3 Pc T Tr 42.77 bar 380 K T Tc Zc 0.267 P 1919.4 kPa T0 273.15 K P0 Tr 0.894 Pr P Pc Pr 176 0.449
  • 177. Use Lee/Kesler correlation. HOWEVER, the values for a saturated vapor lie on the very edge of the vapor region, and some adjacent numbers are for the liquid phase. These must NOT be used for interpolation. Rather, EXTRAPOLATIONS must be made from the vapor side. There may be some choice in how this is done, but the following values are as good as any: Z0 0.7442 Z1 0.1366 Z Z0 Z1 0.718 3 Vvap ZRT P Vvap HR0 0.689 R Tc HR1 0.892 R Tc HR0 2.436 HR1 3.153 SR0 0.540 R SR1 0.888 R SR0 4.49 SR1 7.383 HR HR Hvap Svap Z 10 3 J mol J mol K HR0 3.035 HR1 SR 3 J 10 SR0 SR mol 6315.9 J mol Ans. 3 Svap Ans. mol 3 J 10 mol J mol K SR1 5.892 R ICPH T0 T 2.734 26.786 10 R ICPS T0 T 2.734 26.786 10 Hvap 1182.2 cm J mol K 3 6 8.882 10 8.882 10 1.624 6 0.0 J mol K 0.0 ln HR P P0 SR Ans. For saturated vapor, by Eqs. (3.63) & (4.12) 2 Vliq V c Zc 1 Tr 3 7 Vliq 177 cm 109.89 mol Ans.
  • 178. 1.092 ln Hn R Tn Pc Hn Tn 0.930 By Eq. (4.13) 1.013 bar Tc H Hn 1 Tr Hvap Sliq Svap 0.38 H Tn 1 Hliq 14003 J mol Tc H Hliq 7687.4 H T Sliq 38.475 6.54 For n-butane: J mol 22449 J Ans. mol J mol K Ans. Tc 425.1 K Vc 0.200 cm 255 mol Tn 272.7 K 101.33 kPa 3 Pc T Tr Zc 37.96 bar 370 K 0.274 P T0 273.15 K P0 Tr T Tc 1435 kPa 0.87 Pr P Pc Pr 0.378 Use Lee/Kesler correlation. HOWEVER, the values for a saturated vapor lie on the very edge of the vapor region, and some adjacent numbers are for the liquid phase. These must NOT be used for interpolation. Rather, EXTRAPOLATIONS must be made from the vapor side. There may be some choice in how this is done, but the following values are as good as any: Z0 V Z 0.1372 ZRT P Z0 V Z1 0.7692 Z Z1 cm 1590.1 mol 3 HR0 2.145 HR1 10 3 J mol 178 0.831 R Tc HR1 0.607 R Tc HR0 Ans. 2.937 3 J 10 mol 0.742
  • 179. SR0 0.485 R SR0 4.032 HR HR Hvap Svap SR1 SR1 J mol K HR0 2.733 HR1 mol R ICPH T0 T 1.935 36.915 10 7427.4 J mol SR0 SR 3 R ICPS T0 T 1.935 36.915 10 Hvap 6.942 SR 3 J 10 0.835 R Ans. SR1 5.421 3 J mol K 6 11.402 10 11.402 10 Svap J mol K 6 4.197 0.0 0.0 HR P ln SR P0 J mol K Ans. For saturated vapor, by Eqs. (3.72) & (4.12) Vliq V c Zc 1 Tr 3 2/7 Vliq 1.092 ln Hn R Tn Pc By Eq. (4.13) H Hvap Sliq Svap J mol Hn 22514 H Tn 15295.2 Tc 1 Hn 1 Hliq Ans. 1.013 bar 0.930 cm 123.86 mol Tr 0.38 Tn Tc H Hliq 7867.8 J mol Ans. H T Sliq 37.141 J mol K Ans. 179 J mol
  • 180. 6.55 Under the stated conditions the worst possible cycling of demand can be represented as follows: 10, kg/ 000 hr 1/ hr 3 2/ hr 3 Dem and ( hr kg/ ) 1 hr tm e i 6, 000 4, kg/ 000 hr netst age or ofst eam netdepl i eton ofst eam This situation is also represented by the equation: 4000 = 6000 10000 1 = time of storage liquid 2 Solution gives hr 3 The steam stored during this leg is: mprime where 6000 mprime kg hr 4000 kg hr 1333.3 kg We consider this storage leg, and for this process of steam addition to a tank the equation developed in Problem 6-74 is applicable: m1 Hprime H1 Vtank P2 m2 = Hprime Hf2 Vf2 P1 Hfg2 Vfg2 Hfg2 Vfg2 We can replace Vtank by m2V2, and rearrange to get m2 m1 Hprime Hf2 Vf2 Hfg2 Vfg2 V 2 P2 However M1 v1 = m2 V2 = Vtank P1 Hfg2 and therefore 180 = Hprime Vfg2 m2 m1 = V1 V2 H1 Eq. (A)
  • 181. Making this substitution and rearranging we get Hprime Hf2 Vf2 Hfg2 Vfg2 P2 V2 Hfg2 P1 Vfg2 Hprime = H1 V1 In this equation we can determine from the given information everything except Hprime and Vprime. These quantities are expressed by H1 = Hf1 and x1 Hfg1 V1 = Vf1 x1 Vfg1 Therefore our equation becomes (with Hprime = Hg2) Hg2 Hf2 Hfg2 Vf2 Vfg2 P2 V2 P1 Hfg2 Vfg2 = Hg2 Hf1 Vf1 x1 Hfg1 Eq. (B) x1 Vfg1 In this equation only x1 is unknown and we can solve for it as follows. First we need V2: From the given information we can write: 0.95V2 = 1 x2 Vf2 therefore 19 = Then V2 = 1 0.05V2 = x2 Vg2 x2 Vf2 x2 Vg2 Vf2 Vg2 0.05 or 19Vg2 Vf2 Vf2 x2 = = 19Vg2 20 19 Vf2 Vf2 Eq. (C) 1 Vg2 Now we need property values: Initial state in accumulator is wet steam at 700 kPa. P1 700kPa We find from the steam tables Hf1 697.061 kJ Hg1 kg 2762.0 kJ kg 181 Hfg1 Hg1 Hf1 Hfg1 2064.939 kJ kg
  • 182. Vf1 1.108 3 3 3 cm Vg1 gm 272.68 cm Vfg1 gm Vf1 Vfg1 Vg1 Final state in accumulator is wet steam at 1000 kPa. From the steam tables Hf2 762.605 kJ kg Hg2 2776.2 1.127 Hfg2 Hf2 Hfg2 Hg2 cm Vg2 gm 194.29 gm 1000kPa 2013.595 kJ kg 3 3 3 Vf2 kJ kg P2 271.572 cm cm Vfg2 gm Vf2 Vfg2 Vg2 193.163 cm gm Solve Eq. (C) for V2 Vf2 Vg2 V2 19Vg2 0.05 V2 Vf2 Next solve Eq. (B) for x1 1.18595 Guess: x1 10 3 3m kg 0.1 Given Hg2 Hf2 Hfg2 Vf2 Vfg2 P2 V2 x1 x1 Find x1 4.279 Hfg2 P1 10 Vfg2 = Hg2 Hf1 Vf1 x1 Hfg1 x1 Vfg1 4 3 Thus V1 Vf1 V1 x1 Vfg1 V1 m2 = V2 m1 Eq. (A) gives 1.22419 and cm gm mprime = m2 m1 = 2667kg Solve for m1 and m2 using a Mathcad Solve Block: mprime Guess: m1 m2 m1 2 Given m1 m2 m1 3.752 = 4 V1 V2 10 kg m2 m1 = 2667lb m2 3.873 182 4 10 kg m1 m2 Find m1 m2
  • 183. Finally, find the tank volume Vtank m2 V2 3 Vtank 45.9 m Ans. Note that just to store 1333.3 kg of saturated vapor at 1000 kPa would require a volume of: 3 1333.3kg Vg2 259 m One can work this problem very simply and almost correctly by ignoring the vapor present. By first equation of problem 3-15 m2 m1 Hprime = Hprime Hprime U1 U2 Hg2 = Hprime Uf1 Hprime Uf2 Hprime 2.776 = Hprime Hf1 Hprime Hf2 3 kJ 10 kg Given m2 m1 = Hprime Hprime m1 Hf1 Hf2 m2 3.837 V m2 m1 = 2667lb m2 Find m1 m2 4 10 kg m2 Vf2 0.95 V 6.56 Propylene: 3 45.5 m Ans. 0.140 T Tc 400.15 K P 365.6 K 38 bar Pc 46.65 bar P0 1 bar The throttling process, occurring at constant enthalpy, may be split into two steps: (1) Transform into an ideal gas at the initial conditions, evaluating property changes from a generalized correlation. (2) Change T and P in the ideal-gas state to the final conditions, evaluating property changes by equations for an ideal gas. Property changes for the two steps sum to the property change for the process. For the initial conditions: 183
  • 184. T Tc Tr Tr P Pc Pr 1.095 Pr 0.815 Step (1): Use the Lee/Kesler correlation, interpolate. H0 H0 2.623 10 S0 4.697 mol J mol K H1 1.623 0.496 R S1 3 J 4.124 HR 0.534 R Tc S1 0.565 R S0 H1 0.863 R Tc 3 J 10 mol HR SR J mol K H0 H1 3 J 2.85 S0 SR 10 mol S1 5.275 J mol K Step (2): For the heat capacity of propylene, A B 1.637 22.706 10 K 3 6.915 10 C K 6 2 Solve energy balance for final T. See Eq. (4.7). (guess) 1 HR = R AT Given 1 B 2 T 2 2 Tf 0.908 Find C 3 T 3 1 3 Tf T Sig R ICPS T Tf 1.637 22.706 10 Sig 22.774 SR Sig S 184 6.915 10 6 363.27 K Ans. J mol K S 3 1 28.048 J mol K 0.0 ln Ans. P0 P
  • 185. 6.57 Propane: Tc 0.152 Pc 369.8 K 42.48 bar P0 1 bar P 22 bar T 423 K The throttling process, occurring at constant enthalpy, may be split into two steps: (1) Transform into an ideal gas at the initial conditions, evaluating property changes from a generalized correlation. (2) Change T and P in the ideal-gas state to the final conditions, evaluating property changes by equations for an ideal gas. Property changes for the two steps sum to the property change for the process. For the initial conditions: Tr T Tc Tr P Pc Pr 1.144 Pr 0.518 Step (1): Use the generalized virial correlation 3 J HR R Tc HRB Tr Pr HR 1.366 10 SR R SRB Tr Pr SR 2.284 J mol K mol Step (2): For the heat capacity of propane, A B 1.213 3 28.785 10 K 8.824 10 C K 6 2 Solve energy balance for final T. See Eq. (4.7). 1 (guess) HR = R AT Given 1 B 2 T 0.967 Find 2 2 C 3 T 3 1 Tf Sig R ICPS T Tf 1.213 28.785 10 Sig 22.415 J mol K S SR Sig S 24.699 185 J mol K 1 Tf T 3 3 8.824 10 Ans. 6 Ans. 408.91 K 0.0 ln P0 P
  • 186. 6.58 For propane: T ( 100 T Tc Pc 42.48 bar T 273.15)K Tr 0.152 369.8 K Tc P0 1 bar P 10 bar Pr P Pc Pr 0.235 373.15 K Tr 1.009 Assume ideal gas at initial conditions. Use virial correlation at final conditions. H S R SRB Tr Pr 6.59 H2S: T1 Tr1 Tr1 J mol Ans. J mol K Ans. H ln P P0 801.9 S R Tc HRB Tr Pr 20.639 0.094 P1 400 K T1 Tc 373.5 K Pc 89.63 bar 5 bar T2 600 K P2 25 bar P1 Pr1 Tc Pr1 1.071 T2 Tr2 Pc Tc Tr2 0.056 Pr2 Pr2 1.606 P2 Pc 0.279 Use generalized virial-coefficient correlation for both sets of conditions. Eqs. (6.91) & (6.92) are written 3 5 H R ICPH T1 T2 3.931 1.490 10 0.0 0.232 10 R Tc HRB Tr2 Pr2 HRB Tr1 Pr1 S R ICPS T1 T2 3.931 1.490 10 R SRB Tr2 Pr2 H 7407.3 J mol 3 0.0 5 0.232 10 ln P2 P1 SRB Tr1 Pr1 S 186 1.828 J mol K Ans.
  • 187. 6.60 Carbon dioxide: Tc 0.224 Pc 304.2 K 73.83 bar P0 101.33 kPa P 1600 kPa T 318.15 K Throttling process, constant enthalpy, may be split into two steps: (1) Transform to ideal gas at initial conditions, generalized correlation for property changes. (2) Change T and P of ideal gas to final T & P. Property changes by equations for an ideal gas. Assume ideal gas at final T & P. Sum property changes for the process. For the initial T & P: T Tc Tr Tr P Pr 1.046 Pr Pc 0.217 Step (1): Use the generalized virial correlation R Tc HRB Tr Pr SR HR R SRB Tr Pr 587.999 SR HR 1.313 J mol J mol K Step (2): For the heat capacity of carbon dioxide, A 1.045 10 K B 5.457 3 5 D 1.157 10 K 2 Solve energy balance for final T. See Eq. (4.7). Given 1 (guess) HR = R A T 1 2 1 R ICPS T Tf 5.457 1.045 10 Sig 21.047 SR Sig T Tf T 3 0.0 S 22.36 J mol K 187 302.71 K 5 1.157 10 J mol K S 1 D Tf 0.951 Find Sig B 2 T 2 Ans. ln P0 P Ans.
  • 188. 6.61 T0 P0 523.15 K S J mol K 0 A P 3800 kPa 120 kPa For the heat capacity of ethylene: 14.394 10 B 1.424 3 C K 4.392 10 6 2 K (a) For the entropy change of an ideal gas, combine Eqs. (5.14) & (5.15) with D = 0: (guess) 0.4 Given S = R A ln Hig Tf 1.185 Ws 10 3 308.19 K 4.392 10 6 Ans. 0.0 mol Ws (b) Ethylene: 11852 Tc 0.087 Tr0 Tc Tf T0 ln 4 J Hig T0 P P0 1 2 R ICPH T0 Tf 1.424 14.394 10 Hig Tr0 C T0 0.589 Find 1 2 B T0 1.85317 Pr0 J mol Ans. 282.3 K P0 Pc Pc 50.40 bar Pr0 0.75397 At final conditions as calculated in (a) Tr T Tc Tr 1.12699 Pr P Pc Use virial-coefficient correlation. The entropy change is now given by Eq. (6.92): 0.5 (guess) Given 188 Pr 0.02381
  • 189. S = R A ln SRB C T0 Pr Tc Tc 2 T T0 Tr T 1 ln P P0 SRB Tr0 Pr0 T Find Tr T0 1 2 B T0 303.11 K Ans. 1.074 The work is given by Eq. (6.91): Hig R ICPH T0 T 1.424 14.394 10 Hig Ws Hig Ws 6.62 T0 S A 1.208 11567 0.0 HRB Tr0 Pr0 Ans. P0 P 30 bar 2.6 bar For the heat capacity of ethane: B 1.131 6 mol J mol J mol K 4.392 10 4 J R Tc HRB Tr Pr 493.15 K 0 10 3 19.225 10 K 3 C 5.561 10 6 2 K (a) For the entropy change of an ideal gas, combine Eqs. (5.14) & (5.15) with D = 0: (guess) Given 0.4 S = R A ln Find Hig B T0 1 2 C T0 2 T 0.745 R ICPH T0 T 1.131 19.225 10 189 T0 3 P P0 1 ln T 367.59 K 5.561 10 6 0.0 Ans.
  • 190. Hig 8.735 Ws 3 J 10 mol Ws Hig (b) Ethane: J mol Tr0 Tc Ans. Tc 0.100 T0 Tr0 8735 Pc 48.72 bar Pr0 P0 Pr0 1.6153 Pc 305.3 K 0.61576 At final conditions as calculated in (a) T Tc Tr ( ) T P Pc Pr Tr ( ) 1.20404 T Pr 0.05337 Use virial-coefficient correlation. The entropy change is now given by Eq. (6.83): (guess) 0.5 S = R A ln B T0 SRB Tr Given T0 Tc Pr 2 T T0 Tr Tc ln 1 P P0 SRB Tr0 Pr0 T Find T 1 2 C T0 362.73 K 1.188 The work is given by Eq. (6.91): Hig Hig R ICPH T0 T 1.131 19.225 10 9.034 Ws Hig Ws 8476 10 5.561 10 3 J mol R Tc HRB Tr Pr J mol 3 Ans. 190 HRB Tr0 Pr0 6 0.0 Ans.
  • 191. 6.63 n-Butane: T0 S A 425.1 K Pc P0 1 bar 37.96 bar P 7.8 bar For the heat capacity of n-butane: 36.915 10 K B 1.935 T0 Tr0 323.15 K J mol K 0 Tc 0.200 Tr0 Tc 3 C K Pr0 Pc P Pr HRB Tr0 Pr0 2 P0 Pr0 0.76017 6 11.402 10 Pr Pc = 0.05679 0.02634 0.205 HRB0 0.05679 The entropy change is given by Eq. (6.92) combined with Eq. (5.15) with D = 0: (guess) 0.4 S = R A ln Given B T0 SRB T0 Tc Find T Tc Tr Pr Tr 1 2 ln T T0 381.43 K 0.89726 The work is given by Eq. (6.91): Hig R ICPH T0 T 1.935 36.915 10 Hig 6.551 Ws Hig Ws 5680 3 11.402 10 3 J 10 mol R Tc HRB Tr Pr J mol P P0 SRB Tr0 Pr0 T 1.18 1 2 C T0 HRB Tr0 Pr0 Ans. 191 6 0.0 Ans.
  • 192. 6.64 The maximum work results when the 1 kg of steam is reduced in a completely reversible process to the conditions of the surroundings, where it is liquid at 300 K (26.85 degC). This is the ideal work. From Table F.2 for the initial state of superheated steam: H1 kJ 3344.6 S1 kg 7.0854 kJ kg K From Table F.1, the state of sat. liquid at 300 K is essentially correct: H2 112.5 kJ kg S2 0.3928 kJ kg K T 300 K By Eq. (5.27), Wideal 6.65 H2 H1 T S2 Wideal S1 1224.3 kJ kg Ans. Sat. liquid at 325 K (51.85 degC), Table F.1: Hliq kJ 217.0 kg Psat 12.87 kPa P1 Sliq 325 K H1 Hliq S1 Sliq cm 1.013 gm Vliq For the compressed liquid at 325 K and 8000 kPa, apply Eqs. (6.28) and (6.29) with 8000 kPa T 3 kJ 0.7274 kg K 460 10 Vliq 1 T Vliq P1 P1 6 K 1 H1 Psat 223.881 S1 Psat 0.724 kJ kg kJ kg K For sat. vapor at 8000 kPa, from Table F.2: H2 2759.9 kJ kg S2 Q Heat added in boiler: 5.7471 H2 kJ kg K H1 T Q 300 K 2536 kJ kg Maximum work from steam, by Eq. (5.27): Wideal H1 H2 T S1 S2 192 Wideal 1029 kJ kg
  • 193. Work as a fraction of heat added: Frac Wideal Frac Q Ans. 0.4058 The heat not converted to work ends up in the surroundings. Q SdotG.surr Wideal T SdotG.system S1 kg 10 SdotG.surr sec S2 10 kg 50.234 SdotG.system sec kW K 50.234 kW K Obviously the TOTAL rate of entropy generation is zero. This is because the ideal work is for a completely reversible process. 6.66 Treat the furnace as a heat reservoir, for which kg sec kg kW Qdot 50.234 K T Qdot 2536 SdotG kJ T 10 ( 600 SdotG 273.15)K 21.19 kW K T 873.15 K Ans. By Eq. (5.34) T 6.67 Wdotlost 300 K Wdotlost T SdotG For sat. liquid water at 20 degC, Table F.1: H1 kJ kg 83.86 S1 0.2963 kJ kg K For sat. liquid water at 0 degC, Table F.1: H0 0.04 kJ S0 kg 0.0000 kJ kg K For ice at at 0 degC: H2 H0 333.4 kJ kg S2 S0 193 333.4 kJ 273.15 kg K 6356.9 kW Ans.
  • 194. kJ H2 333.44 T S2 293.15 K kg 1.221 mdot 0.5 kJ kg K kg 0.32 t sec By Eqs. (5.26) and (5.28): Wdotideal mdot H2 H1 T S2 Wdotideal Wdot S1 Wdotideal Wdot 42.77 kW 13.686 kW Ans. t 6.68 This is a variation on Example 5.6., pp. 175-177, where all property values are given. We approach it here from the point of view that if the process is completely reversible then the ideal work is zero. We use the notation of Example 5.6: H1 2676.0 S2 0.0 kJ kg kJ kg K S1 Q' 2000 kJ kg K 7.3554 kJ kg H2 T kJ kg 0.0 273.15 K The system consists of two parts: the apparatus and the heat reservoir at elevated temperature, and in the equation for ideal work, terms must be included for each part. Wideal = Happaratus.reservoir Happaratus.reservoir = H2 Sapparatus.reservoir = S2 T' 450 K Given H1 S1 (Guess) kJ 0 = H2 H1 kg T' T Sapparatus.reservoir Q' Wideal = 0.0 Q' T' Q' T S2 S1 T' Find ( ) T' Q' T' 409.79 K (136.64 degC) 194 kJ kg Ans.
  • 195. 6.69 From Table F.4 at 200(psi): H1 BTU 1222.6 S1 lbm Hliq BTU 355.51 lbm Sliq 0.5438 Hliq H2 1.165 10 BTU x Hvap (at 420 degF) lbm rankine Hvap BTU lbm rankine H2 1.5737 1198.3 Svap 1.5454 (Sat. liq. and vapor) BTU lbm BTU x lbm rankine S2 3 BTU lbm Sliq x Svap S2 Hliq 1.505 0.96 Sliq BTU lbm rankine Neglecting kinetic- and potential-energy changes, on the basis of 1 pound mass of steam after mixing, Eq. (2.30) yields for the exit stream: H x S 0.5 H1 H Hvap Sliq H Hliq x Svap Sliq 1.54 BTU lbm 0.994 S Hliq 1193.6 x 0.5 H2 (wet steam) Ans. BTU lbm rankine By Eq. (5.22) on the basis of 1 pound mass of exit steam, SG 6.70 S 0.5 S1 SG 0.5 S2 2.895 10 4 BTU lbm rankine Ans. From Table F.3 at 430 degF (sat. liq. and vapor): 3 3 Vliq ft 0.01909 lbm Vvap ft 1.3496 lbm Uliq 406.70 BTU lbm Uvap 1118.0 VOLliq BTU lbm VOLliq mliq Vliq 195 Vtank mliq 79.796 ft 3 80 ft 3 4180 lbm
  • 196. VOLvap Vtank VOLliq VOLvap VOLvap mvap mvap Vvap mliq Uliq U1 mvap Uvap mliq U1 mvap 0.204 ft 3 0.151 lbm 406.726 BTU lbm By Eq. (2.29) multiplied through by dt, we can write, d mt Ut H dm = 0 (Subscript t denotes the contents of the tank. H and m refer to the exit stream.) m Integration gives: m2 U2 m1 U1 H dm = 0 0 From Table F.3 we see that the enthalpy of saturated vapor changes from 1203.9 to 1203.1(Btu/lb) as the temperature drops from 430 to 420 degF. This change is so small that use of an average value for H of 1203.5(Btu/lb) is fully justified. Then m2 U2 m1 m1 U1 mliq Have m = 0 Have mvap 1203.5 m2 ( mass) m1 Property values below are for sat. liq. and vap. at 420 degF 3 3 Vliq ft 0.01894 lbm Vvap ft 1.4997 lbm Uliq 395.81 BTU lbm Uvap 1117.4 V2 ( mass) U2 ( mass) mass Vtank m2 ( mass) Uliq x mass) ( x mass) Uvap ( 50 lbm (Guess) 196 Uliq BTU lbm V2 ( mass) Vliq Vvap Vliq BTU lbm mass
  • 197. m1 U1 U2 ( mass) Have U2 ( mass) Given mass = mass Find ( mass) mass Ans. 55.36 lbm 6.71 The steam remaining in the tank is assumed to have expanded isentropically. Data from Table F.2 at 4500 kPa and 400 degC: S1 6.7093 3 J gm K S2 = S1 = 6.7093 J gm K V1 64.721 cm Vtank gm 3 50 m By interpolation in Table F.2 at this entropy and 3500 kPa: 3 cm 78.726 gm V2 Vtank m1 6.72 m2 V1 Ans. t2 = 362.46 C Vtank V2 m m1 m2 m 137.43 kg Ans. This problem is similar to Example 6.8, where it is shown that Q= mt Ht H mt Here, the symbols with subscript t refer to the contents of the tank, whereas H refers to the entering stream. We illustrate here development of a simple expression for the first term on the right. The1500 kg of liquid initially in the tank is unchanged during the process. Similarly, the vapor initially in the tank that does NOT condense is unchanged. The only two enthalpy changes within the tank result from: 1. Addition of 1000 kg of liquid water. This contributes an enthalpy change of Hliq mt 2. Condensation of y kg of sat. vapor to sat. liq. This contributes an enthalpy change of y Hliq Thus Hvap = y Hlv mt Ht = Hliq mt y Hlv 197
  • 198. Similarly, mt Vt = Vliq mt Whence mt Q = Hliq mt H At 250 degC: 209.3 Hliq Vliq mt 6.73 y Vlv mt Hliq H Given: C kJ kg K 3 Vliq cm 1.251 gm 3 kJ 1714.7 kg cm 48.79 gm Vlv 25.641 kg y Hlv Vtank 0.43 kJ kg kJ 1085.8 kg Hlv Q H mt Required data from Table F.1 are: 1000 kg At 50 degC: y y Hlv y Vlv = 0 T1 Q Ans. 832534 kJ 3 0.5 m 295 K kJ kg Hin 120.8 mtank 30 kg Data for saturated nitrogen vapor: 80 0.1640 85 2.287 0.1017 90 T 1.396 3.600 0.06628 95 K P 5.398 bar V 0.04487 100 7.775 0.03126 105 10.83 0.02223 110 14.67 0.01598 198 3 m kg
  • 199. 78.9 At the point when liquid nitrogen starts to accumulate in the tank, it is filled with saturated vapor nitrogen at the final temperature and having properties 82.3 85.0 H kJ kg 86.8 87.7 mvap Tvap Vvap Hvap Uvap 87.4 85.6 By Eq. (2.29) multiplied through by dt, d nt Ut H dm = dQ Subscript t denotes the contents of the tank; H and m refer to the inlet stream. Since the tank is initially evacuated, integration gives mvap Uvap Hin mvap = Q = mtank C Tvap Also, mvap = (A) T1 Vtank (B) Vvap Calculate internal-energy values for saturated vapor nitrogen at the given values of T: 56.006 U ( H 59.041 P V) 61.139 U 62.579 kJ kg 63.395 63.325 62.157 Fit tabulated data with cubic spline: Us lspline ( U) T Uvap () t Tvap Vs interp ( T U t) Us 100 K lspline ( V) T Vvap () t (guess) Combining Eqs. (A) & (B) gives: 199 interp ( T V t) Vs
  • 200. Given Uvap Tvap Tvap mvap 6.74 Hin = mtank C T1 Tvap Vvap Tvap Vtank Find Tvap Tvap 97.924 K mvap 13.821 kg Vtank Vvap Tvap Ans. The result of Part (a) of Pb. 3.15 applies, with m replacing n: m2 U2 H m1 U1 H = Q= 0 Whence Also m2 H U2 = m1 H U2 = Uliq.2 x2 Ulv.2 V2 = Vliq.2 x2 Vlv.2 U1 V2 = Vtank m2 Eliminating x2 from these equations gives Vtank m2 H m2 Uliq.2 Vliq.2 Vlv.2 Ulv.2 = m1 H U1 which is later solved for m2 Vtank 3 50 m m1 16000 kg V1 V1 Data from Table F.1 @ 25 degC: 3 Vliq.1 cm 1.003 gm Uliq.1 104.8 kJ kg 3 Vlv.1 cm 43400 gm Ulv.1 2305.1 200 kJ kg Vtank m1 3.125 10 3 3m kg
  • 201. V1 x1 x1 Vliq.1 U1 4.889 10 5 Uliq.1 x1 Ulv.1 U1 Vlv.1 104.913 kJ kg Data from Table F.2 @ 800 kPa: Vliq.2 1.115 720.043 kJ kg Ulv.2 ( 2575.3 Ulv.2 3 cm 1.855 Uliq.2 gm 3 Vlv.2 ( 240.26 Vlv.2 1.115) cm m 0.239 kg gm 720.043) 3 Data from Table F.2 @ 1500 kPa: m1 H U1 Vtank m2 H msteam 6.75 Uliq.2 m2 Vliq.2 m1 H 2789.9 3 kJ 10 kg kJ kg Ulv.2 Vlv.2 m2 Ulv.2 2.086 4 10 kg Vlv.2 msteam 4.855 3 10 kg The result of Part (a) of Pb. 3.15 applies, with Whence kJ kg Ans. n1 = Q = 0 U2 = H From Table F.2 at 400 kPa and 240 degC H = 2943.9 kJ kg Interpolation in Table F.2 will produce values of t and V for a given P where U = 2943.9 kJ/kg. 201
  • 202. 1 384.09 303316 100 384.82 3032.17 1515.61 V2 385.57 t2 200 P2 300 387.08 gm 1010.08 400 i 386.31 3 cm 757.34 1 5 Vtank 3 1.75 m Vtank massi V2 i 5.77 10 3 T rises very slowly as P increases 0.577 mass 1.155 3 kg 1.733 2 massi 2.311 1 0 0 200 P2 6.76 3 Vtank 400 i Data from Table F.2 @ 3000 kPa: 2m 3 3 Vliq cm 1.216 gm Vvap cm 66.626 gm Hliq 1008.4 kJ kg Hvap 2802.3 x1 0.1 V1 V1 Vliq kJ kg x1 Vvap 7.757 10 3 3m kg Vliq m1 m1 Vtank V1 257.832 kg The process is the same as that of Example 6.8, except that the stream flows out rather than in. The energy balance is the same, except for a sign: Q= mt Ht H mtank 202
  • 203. where subscript t denotes conditions in the tank, and H is the enthalpy of the stream flowing out of the tank. The only changes affecting the enthalpy of the contents of the tank are: 1. Evaporation of y kg of sat. liq.: y Hvap Hliq 2. Exit of of liquid from the tank: 0.6 m1 kg 0.6 m1 Hliq Thus mt Ht = y Hvap Hliq 0.6 m1 Hliq Similarly, since the volume of the tank is constant, we can write, mt Vt = y Vvap Whence Q= But y= Vliq 0.6 m1 Vliq = 0 0.6 m1 Vliq Vvap Vliq 0.6 m1 Vliq Hvap Vvap Vliq Hliq and H = Hliq 0.6 m1 Hliq 0.6 m1 = H mtank mtank and therefore the last two terms of the energy equation cancel: Q 6.77 0.6 m1 Vliq Hvap Vvap Vliq Hliq Q 5159 kJ Ans. Data from Table F.1 for sat. liq.: H1 100.6 kJ kg (24 degC) H3 355.9 kJ kg (85 degC) Data from Table F.2 for sat. vapor @ 400 kPa: H2 2737.6 kJ kg By Eq. (2.30), neglecting kinetic and potential energies and setting the heat and work terms equal to zero: H3 mdot3 H1 mdot1 H2 mdot2 = 0 203
  • 204. Also mdot1 = mdot3 Whence mdot3 H1 mdot2 mdot1 mdot3 mdot2 0.484 H1 kg mdot3 mdot2 5 kg sec Ans. sec H3 H2 mdot2 kg sec Ans. mdot1 4.516 6.78 Data from Table F.2 for sat. vapor @ 2900 kPa: H3 2802.2 kJ kg S3 6.1969 kJ kg K mdot3 15 kg sec Table F.2, superheated vap., 3000 kPa, 375 degC: H2 3175.6 kJ kg S2 6.8385 kJ kg K Table F.1, sat. liq. @ 50 degC: 3 Vliq cm 1.012 gm Hliq Psat 12.34 kPa T 209.3 kJ kg Sliq 0.7035 kJ kg K 323.15 K Find changes in H and S caused by pressure increase from 12.34 to 3100 kPa. First estimate the volume expansivity from sat. liq, data at 45 and 55 degC: 3 V V ( 1.015 5 10 1.010) cm T gm 3 3 cm P 10 K 1 Vliq gm V T 3100 kPa 4.941 Apply Eqs. (6.28) & (6.29) at constant T: H1 Hliq S1 Sliq Vliq 1 Vliq P T P kJ kg kJ Psat 204 H1 211.926 S1 Psat 0.702 kg K 10 4 K 1
  • 205. By Eq. (2.30), neglecting kinetic and potential energies and setting the heat and work terms equal to zero: H3 mdot3 H1 mdot1 H2 mdot2 = 0 Also mdot2 = mdot3 Whence mdot1 mdot2 mdot3 mdot1 mdot3 H3 H2 H1 H2 kg sec mdot1 mdot2 mdot1 1.89 13.11 Ans. kg sec For adiabatic conditions, Eq. (5.22) becomes SdotG S3 mdot3 SdotG 1.973 S1 mdot1 kJ sec K S2 mdot2 Ans. The mixing of two streams at different temperatures is irreversible. 6.79 Table F.2, superheated vap. @ 700 kPa, 200 degC: H3 2844.2 kJ kg S3 6.8859 kJ kg K Table F.2, superheated vap. @ 700 kPa, 280 degC: H1 3017.7 kJ kg S1 7.2250 kJ kg K mdot1 50 kg sec Table F.1, sat. liq. @ 40 degC: Hliq 167.5 kJ kg Sliq 0.5721 kJ kg K By Eq. (2.30), neglecting kinetic and potential energies and setting the heat and work terms equal to zero: H2 Also mdot2 H3 mdot3 Hliq mdot3 = mdot2 mdot1 H1 H3 H1 mdot1 H2 mdot2 = 0 mdot1 H3 mdot2 H2 For adiabatic conditions, Eq. (5.22) becomes 205 3.241 kg sec Ans.
  • 206. S2 mdot3 Sliq SdotG S3 mdot3 SdotG 3.508 mdot2 S1 mdot1 mdot1 S2 mdot2 kJ sec K Ans. The mixing of two streams at different temperatures is irreversible. 6.80 Basis: 1 mol air at 12 bar and 900 K (1) + 2.5 mol air at 2 bar and 400 K (2) = 3.5 mol air at T and P. T1 900 K T2 400 K P1 n1 1 mol n2 2.5 mol CP 600 K 2 bar CP 29.099 (guess) 1st law: T Given n1 CP T T T1 n2 CP T P Given T 5 bar T P R ln T1 P1 T P n2 CP ln R ln T2 P2 P Find ( ) P molwt 28.014 lb lbmol CP Ms 542.857 K 4.319 bar R 7 2 molwt = nitrogen rate in lbm/sec = 0 Ans. J K Ans. CP 0.248 BTU lbm rankine = steam rate in lbm/sec Mn J mol K (guess) n1 CP ln P 7 R 2 T2 = 0 J Find ( ) T 2nd law: 6.81 P2 12 bar Mn 206 40 lbm sec
  • 207. (1) = sat. liq. water @ 212 degF entering (2) = exit steam at 1 atm and 300 degF (3) = nitrogen in at 750 degF T3 1209.67 rankine (4) = nitrogen out at 325 degF T4 784.67 rankine H1 180.17 H2 1192.6 BTU lbm S1 lbm BTU lbm rankine (Table F.3) S2 BTU 0.3121 1.8158 BTU lbm rankine (Table F.4) Eq. (2.30) applies with negligible kinetic and potential energies and with the work term equal to zero and with the heat transfer rate given by Ms 3 Given Ms lbm (guess) sec Ms H2 H1 Q = 60 Mn CP T4 Find Ms Ms T3 = 60 3.933 lbm sec BTU Ms lbm BTU lbm Ms Ans. Eq. (5.22) here becomes SdotG = Ms S2 S4 S3 = CP ln T S1 Mn S4 T4 Q T3 Q T S3 60 BTU Ms lbm 529.67 rankine SdotG Ms S2 SdotG 2.064 S1 Mn CP ln BTU sec rankine T4 T3 Ans. 207 Q T Q 235.967 BTU sec
  • 208. 6.82 molwt 28.014 gm CP mol R 2 molwt 7 CP 1.039 J gm K Ms = steam rate in kg/sec Mn= nitrogen rate in kg/sec kg sec Mn 20 (3) = nitrogen in @ 400 degC T3 673.15 K (4) = nitrogen out at 170 degC T4 443.15 K (1) = sat. liq. water @ 101.33 kPa entering (2) = exit steam at 101.33 kPa and 150 degC H1 419.064 H2 2776.2 kJ kg S1 S2 kJ kg 1.3069 7.6075 kJ kg K (Table F.2) kJ (Table F.2) kg K By Eq. (2.30), neglecting kinetic and potential energies and setting the work term to zero and with the heat transfer rate given by Ms 1 Given Ms kg sec (guess) Ms H2 H1 Ms Find Ms Q = 80 Mn CP T4 1.961 T3 = 80 kg kJ Ms kg kJ Ms kg Ans. sec Eq. (5.22) here becomes SdotG = Ms S2 S4 S3 = CP ln SdotG Ms S2 SdotG 4.194 S1 Mn S4 T4 T T3 S1 kJ sec K Q T S3 Mn CP ln Ans. 208 T4 T3 298.15 K Q T Q 80 kJ kg Ms
  • 209. 6.86 Methane = 1; propane = 2 T 363.15 K 1 Tc1 0.012 2 190.6 K y2 Tc2 0.5 1 y1 5500 kPa P y1 0.152 Zc1 0.286 Zc2 0.276 369.8 K Pc1 45.99 bar Pc2 42.48 bar The elevated pressure here requires use of either an equation of state or the Lee/Kesler correlation with pseudocritical parameters. We choose the latter. Tpc y1 Tc1 Tpc Ppc Tpr T Tpr Tpc 1.296 44.235 bar Ppr 280.2 K y1 Pc1 Ppc y2 Tc2 y2 Pc2 P Ppc Ppr 1.243 By interpolation in Tables E.3 and E.4: Z0 Z1 0.8010 y1 1 y2 2 0.1100 Z 0.082 Z0 For the molar mass of the mixture, we have: gm molwt molwt y1 16.043 y2 44.097 mol V Vdot D ZRT P molwt V mdot 4A 30.07 3 V Vdot D cm 14.788 gm mdot 3 4 cm 2.07 10 2.964 cm 209 sec Ans. Z Z1 1.4 A Vdot u 0.81 gm mol kg sec u A 30 m sec 2 6.901 cm
  • 210. 6.87 Vectors containing T, P, Tc, and Pc for the calculation of Tr and Pr: 500 425.2 20 42.77 400 304.2 200 73.83 450 552.0 60 79.00 600 617.7 20 21.10 620 T 617.2 Tc 20 P 36.06 Pc 250 190.6 90 45.99 150 154.6 20 469.7 10 33.70 450 430.8 35 78.84 400 374.2 15 Pr P Pc 40.60 1.176 0.468 1.315 2.709 0.815 0.759 0.971 Tr T Tc 50.43 500 Tr 0.948 1.005 Pr 0.555 1.312 1.957 0.97 0.397 1.065 0.297 1.045 0.444 1.069 0.369 Parts (a), (g), (h), (i), and (j) --- By virial equation: 500 425.2 42.77 .190 150 T 20 20 154.6 50.43 .022 500 K P 10 bar Tc 469.7 K Pc 33.70 bar .252 450 35 430.8 78.84 .245 400 15 374.2 40.6 .327 Tr T Tc Pr P Pc 210
  • 211. 1.176 0.97 Tr 0.468 0.397 1.065 Pr 0.297 1.045 0.444 1.069 0.369 B0 0.073 0.422 Eq. (3.65) B1 1.6 0.172 0.139 4.2 Tr DB0 0.675 2.6 Tr Eq. (6.89) DB1 Tr 0.052 0.37 B1 0.321 6.718 4.217 0.306 5.2 Eq. (6.90) 0.443 9.009 10 10 10 3 DB0 3 0.311 0.73 0.056 0.309 0.722 Tr 0.253 B0 Eq. (3.66) 0.845 0.574 DB1 0.522 0.603 0.568 3 0.576 0.51 Combine Eqs. (3.61) + (3.62), (3.63), and (6.40) and the definitions of Tr and Pr to get: Tc B0 Pc B1 VR R HR R Tc Pr B0 Tr DB0 SR R Pr DB0 DB1 ( B1 Tr DB1) Eq. (6.88) 211 Eq. (6.87)
  • 212. 3 1.377 10 200.647 94.593 VR 355.907 146.1 1.952 559.501 3 cm 2.469 3 HR 1.226 10 mol 3 J SR mol 2.745 1.746 10 2.256 3 232.454 J mol K 1.74 1.251 10 Parts (b), (c), (d), (e), and (f) --- By Lee/Kesler correlation: By linear interpolation in Tables E.1--E.12: 0 DEFINE: h0 equals 1 ( ) HR ( ) HR h1 equals RTc RTc 0 s0 equals h equals HR RTc s equals SR R 1 ( ) SR R s1 equals ( ) SR R .663 2.008 0.233 .124 Z0 0.208 .050 4.445 5.121 .278 Z1 .088 h0 3.049 h1 2.970 .783 .036 0.671 0.596 .707 0.138 1.486 0.169 1.137 4.381 s0 0.405 5.274 2.675 s1 2.910 0.473 0.557 0.824 0.289 400 304.2 .224 450 T 200 60 552.0 .111 617.7 K .492 600 K P 20 Tc bar 620 20 617.2 .303 250 90 190.6 .012 212
  • 213. Z s Z1 Eq. (3.57) Z0 s0 HR s1 h0 ( h Tc R) SR Eq. (6.85) ( s R) 3 5.21 10 2.301 0.118 0.235 h1 (6.86) 0.71 Z h 10.207 4 10 4 HR 2.316 10 0.772 4.37 41.291 J mol SR J mol K 5.336 3 10 0.709 34.143 6.88 3 2.358 10 48.289 VR T (Z R P 549.691 1) VR 3 3 1.909 10 cm mol And. 587.396 67.284 The Lee/Kesler tables indicate that the state in Part (c) is liquid. 6.88 Vectors containing T, P, Tc1, Tc2, Pc1, Pc2, 1, and 2 for Parts (a) through (h) 650 562.2 553.6 300 100 304.2 132.9 600 T 60 100 304.2 568.7 350 400 K P 75 150 bar Tc1 305.3 373.5 K Tc2 282.3 190.6 200 75 190.6 126.2 450 80 190.6 469.7 250 100 126.2 154.6 213 K
  • 214. 48.98 .210 .210 73.83 34.99 .224 .048 73.83 Pc1 40.73 24.90 .224 .400 48.72 bar 89.63 50.40 Pc2 45.99 1 bar .100 .087 2 .094 .012 45.99 34.00 .012 .038 45.99 33.70 .012 .252 34.00 50.43 .038 .022 Tpc ( Tc1 .5 Tpr T Tpc .5 Tc2) Ppc ( Pc1 .5 .5 Pc2) .5 P Ppc Ppr 557.9 0.21 218.55 54.41 0.136 436.45 Tpc 44.855 49.365 0.312 293.8 282.05 K Ppc 49.56 67.81 bar 0.094 0.053 158.4 39.995 0.025 330.15 39.845 0.132 140.4 42.215 0.03 1.165 1.373 1.838 1.375 Tpr 1.338 2.026 1.191 1.418 Ppr 1.513 2.212 1.263 1.875 1.363 2.008 1.781 2.369 214 1 .5 2
  • 215. Lee/Kesler Correlation --- By linear interpolation in Tables E.1--E.12: .6543 1.395 .461 .7706 .1749 1.217 .116 .7527 Z0 .1219 .1929 1.346 .097 .6434 Z1 .7744 .1501 1.510 h0 .1990 1.340 h1 .400 .049 .6631 .1853 1.623 .254 .7436 .1933 1.372 .110 .9168 .1839 0.820 0.172 .890 .466 .658 .235 .729 .242 .944 s0 .430 s1 .704 .224 .965 .348 .750 .250 .361 .095 0 h0 equals s0 equals Z Z0 s s0 ( HR) RTpc ( SR) R 1 h1 equals 0 s1 equals Z1 Eq. (3.57) s1 h Eq. (6.86) 215 ( HR) RTpc ( SR) h equals s equals SR R 1 R h0 HR RTpc h1 Eq. (6.85)
  • 216. HR ( hTpc R) SR ( R) s 0.68 8.213 0.794 2239.984 5.736 0.813 Z 6919.583 4993.974 6.689 0.657 3779.762 HR 0.785 J SR 3148.341 mol 8.183 5.952 mol K Ans. 0.668 2145.752 8.095 0.769 3805.813 6.51 0.922 6.95 Tc J 951.151 3.025 Pc 647.1K At Tr = 0.7: T 220.55bar T 0.7 Tc 452.97 K Find Psat in the Saturated Steam Tables at T = 452.97 K T1 P1 451.15K P2 T2 Psat P1 ( T T1 Psat Pc Psatr T2 957.36kPa Psat T1) P1 Psatr 453.15K 998.619 kPa 1 0.045 P2 Psat 1002.7kPa 9.986 bar 0.344 log Psatr Ans. This is very close to the value reported in Table B.1 ( = 0.345). 6.96 Tc At Tr = 0.7: T Pc 374.2K T T T 471.492 rankine T 0.7 Tc 459.67rankine 40.60bar 11.822 degF Find Psat in Table 9.1 at T = 11.822 F T1 10degF P1 T2 26.617psi 216 15degF P2 29.726psi
  • 217. P2 T2 Psat P1 (T T1 Psat Pc Psatr T1) Psatr P1 Psat Psat 27.75 psi 1 0.047 1.913 bar Ans. 0.327 log Psatr This is exactly the same as the value reported in Table B.1. 6.101 For benzene a) Tc Tn Trn Tc 562.2K Pc Trn 0.210 0.628 Psatrn lnPr0 ( Tr) 5.92714 lnPr1 ( Tr) 15.2518 6.09648 Tr 15.6875 Tr lnPr0 ( Tr) atm Pc Psatrn 353.2K 0.021 6 0.169347 Tr Eqn. (6.79) 13.4721 ln ( Tr) 0.43577 Tr 6 Eqn. (6.80) Eqn. (6.81). lnPr1 Trn lnPsatr ( Tr) 1 Tn 0.271 1.28862 ln ( Tr) lnPr0 Trn ln Psatrn Zc 48.98bar lnPr1 ( Tr) 0.207 Eqn. (6.78) 2 Zsatliq B0 Psatrn Trn 0.083 Zc 0.422 B1 Eqn. (3.73) Eqn. (3.65) 1.6 0.805 0.139 0.172 Trn B1 7 Zsatliq 4.2 Eqn. (3.66) Z0 1 Z0 Trn B0 1 1 T rn Z1 B1 217 Psatrn Eqn. (3.64) 0.974 Z1 1.073 B0 0.00334 Trn Psatrn Trn 0.035 Equation following Eqn. (3.64)
  • 218. Zsatvap Z0 Zlv Zsatvap d Hhatlv Hlv Eqn. (3.57) Z1 dTrn Zsatvap 0.966 Zlv Zsatliq lnPsatr Trn Trn 2 0.963 Hhatlv Zlv Hlv R Tc Hhatlv 6.59 30.802 kJ mol Ans. This compares well with the value in Table B.2 of 30.19 kJ/mol The results for the other species are given in the table below. EstimatedValue (kJ/mol) Table B.2 (kJ/mol) 30.80 30.72 Benz ene 21.39 21.30 isoButane 29.81 29.82 Carbon tetrachlorid e 30.03 29.97 Cy clohex ane 39.97 38.75 nDecane 29.27 28.85 n- ane Hex 34.70 34.41 nOctane 33.72 33.18 Toluene 37.23 36.24 o- lene Xy 6.103 For CO2: Tc a) At Tr = 0.7 Ttr Tt Ttr Tc Pc 73.83bar 216.55K Pt 5.170bar T At the triple point: 304.2K Tt 0.224 0.7Tc T Ptr 0.712 212.94 K Pt Ptr Pc 0.07 lnPr0 ( ) Tr 5.92714 6.09648 Tr 1.28862 ln ( ) 0.169347 Tr Tr Eqn. (6.79) lnPr1 ( ) Tr 15.2518 15.6875 Tr 13.4721 ln ( ) 0.43577 Tr Tr 6 Eqn. (6.80) ln Ptr lnPr0 Ttr lnPr1 Ttr 6 Eqn. (6.81). 218 0.224 Ans.
  • 219. This is exactly the same value as given in Table B.1 b) Psatr 1atm Pc 0.014 ln Psatr = lnPr0 Trn Given Trn Psatr 0.609 Tn Trn Tc Guess: Trn lnPr1 Trn 0.7 Trn Tn Find Trn 185.3 K Ans. This seems reasonable; a Trn of about 0.6 is common for triatomic species. 219
  • 220. Chapter 7 - Section A - Mathcad Solutions 7.1 u2 325 m R sec 8.314 J mol K molwt 28.9 7 gm CP mol R 2 molwt With the heat, work, and potential-energy terms set equal to zero and with the initial velocity equal to zero, Eq. (2.32a) reduces to H 2 u2 Whence 7.4 But = 0 2 T u2 H = CP T 2 T 2 CP 52.45 K Ans. From Table F.2 at 800 kPa and 280 degC: H1 kJ kg 3014.9 S1 7.1595 kJ kg K Interpolation in Table F.2 at P = 525 kPa and S = 7.1595 kJ/(kg*K) yields: H2 kJ 2855.2 kg 3 V2 cm 531.21 gm mdot 0.75 kg sec With the heat, work, and potential-energy terms set equal to zero and with the initial velocity equal to zero, Eq. (2.32a) reduces to: H 2 u2 2 Whence = 0 u2 2 H2 H1 m sec u2 By Eq. (2.27), 7.5 A2 mdot V2 u2 565.2 A2 7.05 cm 2 Ans. Ans. The calculations of the preceding problem may be carried out for a series of exit pressures until a minimum cross-sectional area is found. The corresponding pressure is the minimum obtainable in the converging nozzle. Initial property values are as in the preceding problem. 220
  • 221. H1 3014.9 kJ S1 kg 7.1595 kJ S2 = S 1 kg K Interpolations in Table F.2 at several pressures and at the given entropy yield the following values: 400 425 P 531.21 2855.2 2868.2 450 kPa 507.12 kJ kg 2880.7 H2 485.45 V2 475 465.69 500 mdot 2892.5 2903.9 447.72 0.75 kg sec u2 2 H2 565.2 7.022 m 518.1 sec 494.8 u2 u2 7.05 541.7 mdot V2 A2 H1 A2 2 7.028 cm 7.059 7.127 471.2 Fit the P vs. A2 data with cubic spline and find the minimum P at the point where the first derivative of the spline is zero. i 1 5 s cspline P A2 pmin pi 400 kPa Pi a2 A( ) P i A2 i interp s p a2 P (guess) 2 Given pmin cm d A pmin = 0 kPa dpmin 431.78 kPa Ans. pmin A pmin 221 Find pmin 2 7.021 cm Ans. 3 cm gm
  • 222. Show spline fit graphically: p 400 kPa 401 kPa 500 kPa 7.13 7.11 A2 7.09 i 2 cm 7.07 A( ) p 2 cm 7.05 7.03 7.01 400 420 440 460 Pi 480 500 p kPa kPa 7.9 From Table F.2 at 1400 kPa and 325 degC: H1 3096.5 kJ kg S1 7.0499 kJ kg K S2 S1 Interpolate in Table F.2 at a series of downstream pressures and at S = 7.0499 kJ/(kg*K) to find the minimum cross-sectional area. 800 294.81 775 P 2956.0 2948.5 302.12 750 kPa H2 2940.8 kJ kg V2 309.82 725 317.97 700 u2 2932.8 2924.9 326.69 2 H2 H1 A2 = 222 V2 u2 mdot 3 cm gm
  • 223. Since mdot is constant, the quotient V2/u2 is a measure of the area. Its minimum value occurs very close to the value at vector index i = 3. 5.561 5.553 V2 5.552 u2 2 cm sec kg 5.557 5.577 At the throat, A2 A2 u2 3 V2 mdot 2 6 cm mdot 1.081 3 kg Ans. sec At the nozzle exit, P = 140 kPa and S = S1, the initial value. From Table F.2 we see that steam at these conditions is wet. By interpolation, Sliq x 7.10 u1 kJ kg K 1.4098 S1 Svap 230 Sliq Svap x Sliq ft sec 7.2479 kJ kg K 0.966 u2 2000 ft sec From Table F.4 at 130(psi) and 420 degF: H1 Btu lbm 1233.6 S1 2 By Eq. (2.32a), H2 H1 1.6310 H H H2 Btu lbm rankine u2 u1 2 H 2 1154.8 78.8 Btu lbm From Table F.4 at 35(psi), we see that the final state is wet steam: Hliq 228.03 Sliq 0.3809 Btu lbm Btu lbm rankine Hvap 1167.1 Svap 1.6872 223 Btu lbm Btu lbm rankine Btu lbm
  • 224. H2 x Hvap S2 u2 x Svap S2 580 x Sliq ( 273.15 By Eq. (2.32a), But T 1.67 BTU lbm rankine 0.039 15)K molwt H= u1 28.9 u2 2 gm CP mol Ans. R 2 molwt 7 2 2 2 Btu lbm rankine = u2 2 Whence H = CP T u2 (quality) SdotG S1 m T2 sec 0.987 S2 Hliq Sliq SdotG 7.11 Hliq 2 T 2 CP Ans. 167.05 K Initial t = 15 + 167.05 = 182.05 degC Ans. 7.12 Values from the steam tables for saturated-liquid water: 3 At 15 degC: V cm 1.001 gm T 288.15 K Enthalpy difference for saturated liquid for a temperature change from 14 to 15 degC: H ( 67.13 1.5 10 K 58.75) J gm t 2K 4 P 4 atm Cp Cp H t 4.19 J gm K Apply Eq. (7.25) to the constant-enthalpy throttling process. Assumes very small temperature change and property values independent of P. 224
  • 225. T V 1 joule 1 P T T 9.86923 cm3 atm Cp 0.093 K The entropy change for this process is given by Eq. (7.26): S T Cp ln T T Apply Eq. (5.36) with Q=0: Wlost T S Wlost 7.13--7.15 P2 350 T1 350 250 T J gm Wlost 80 P1 K 60 60 bar 20 73.83 282.3 126.2 Pc K 369.8 .224 50.40 .087 bar 34.00 .038 42.48 .152 5.457 1.045 1.424 14.394 B 3.280 1.213 10 4.392 0.0 3 K .593 28.785 0.0 C or 1.2bar 304.2 A 1.408 10 1.157 6 10 K 2 3 J gm K 293.15 K 0.413 400 Tc S V P D 8.824 0.0 0.040 0.0 225 5 10 K 2 0.413 kJ kg Ans.
  • 226. As in Example 7.4, Eq. (6.93) is applied to this constant-enthalpy process. If the final state at 1.2 bar is assumed an ideal gas, then Eq. (A) of Example 7.4 (pg. 265) applies. Its use requires expressions for HR and Cp at the initial conditions. 1.151 T1 Tr 1.24 Tr Tc 1.084 P1 Pc Pr 1.981 Pr 1.082 7.13 0.42748 0.08664 Eq. (3.53) Tr 1.765 0.471 Redlich/Kwong equation: Pr 1.19 q Eq. (3.54) 1.5 Tr Guess: z Given z= 1 Z q Find z () i 1 4 1 q Ii HRi R T1i Z SRi R ln Z i qi i qi z Eq. (3.52) z z ln 1 i Z i qi Z i Eq. (6.65b) i qi 1.5 qi Ii Eq. (6.67) The derivative in these 0.5 qi Ii Eq. (6.68) equations equals -0.5 The simplest procedure here is to iterate by guessing T2, and then calculating it. 280 Guesses T2 302 232 K 385 226
  • 227. Z i qi 2.681 5.177 0.721 0.773 2.253 kJ 0.521 mol HR 0.956 0.862 Cp HR Cp 2.33 B T1 2 R A C 2 T1 3 1 T1 S Cp ln 302.026 K 232.062 0.480 Guess: P2 P1 SR Ans. 0.42748 1 c 1 0.5 2 Tr Eq. (3.54) q Tr 1 Given z= 1 i Ii HRi R ln 0.08664 2 0.176 Eq. (3.53) z 1 4 T2 T1 2 22.163 1.574 Pr Tr T1 J 31.953 mol K Soave/Redlich/Kwong equation: c D 1 29.947 S Ans. 384.941 7.14 2 31.545 279.971 T2 J mol K 1.59 1.396 T2 T1 T2 4.346 SR R T1i Z ln z q Z i qi Eq. (3.52) Z z z i qi Z i q Eq. (6.65b) i qi 1 ci Tri i 227 0.5 1 qi Ii Eq. (6.67) Find ( z)
  • 228. SRi R ln Z i qi Tri ci i 0.5 Eq. (6.68) qi Ii i The derivative in these equations equals: Tri ci 0.5 i Now iterate for T2: 273 Guesses 300 T2 232 K 384 Z i qi 2.936 0.75 0.79 HR 0.975 0.866 6.126 2.356 kJ 0.526 mol 4.769 J 1.789 mol K SR 1.523 T2 T1 Cp R A 2.679 B T1 2 1 C 2 T1 3 2 1 D T1 2 272.757 T2 HR Cp T1 299.741 T2 231.873 K Ans. 383.554 31.565 S Cp ln T2 T1 R ln P2 P1 SR S 30.028 32.128 mol K 22.18 228 J Ans.
  • 229. 7.15 Peng/Robinson equation: 1 c 2 1 0.37464 1.54226 Pr Tr z q i 1 4 HRi 2 1 c 1 Tr z q Eq. (3.52) z z Find z () 1 Ii R T1i Z 2 ln Z i qi i Z i qi i 2 i qi 1 ci Tri Eq. (6.65b) 0.5 1 qi Ii R ln Z i qi i ci Tri Eq. (6.67) 0.5 Eq. (6.68) qi Ii i The derivative in these equations equals: ci Tri i Now iterate for T2: 270 Guesses T2 297 229 K 383 229 0.5 0.5 Tr Eq. (3.54) q i SRi 0.45724 1 z= 1 Z 0.07779 0.26992 Eq. (3.53) Guess: Given 2 2
  • 230. Z i qi 3.041 6.152 0.722 0.76 HR 0.95 0.85 2.459 0.6 kJ mol 4.784 J 1.847 mol K SR 1.581 T2 T1 Cp R A 2.689 B T1 2 C 2 T1 3 1 2 D 1 T1 2 269.735 HR Cp T2 T1 297.366 T2 Ans. K 229.32 382.911 31.2 S T2 T1 Cp ln R ln P2 P1 SR 29.694 J 31.865 mol K S Ans. 22.04 7.18 Wdot H1 3500 kW 3462.9 kJ kg Data from Table F.2: 2609.9 H2 kJ kg S1 4.103 kg sec Ans. 7.3439 kJ kg K S2 S1 By Eq. (7.13), Wdot H2 H1 mdot mdot For isentropic expansion, exhaust is wet steam: Sliq x 0.8321 S2 Svap kJ kg K Sliq Sliq Svap x 230 7.9094 0.92 kJ kg K (quality)
  • 231. Hliq H'2 251.453 Hliq x Hvap H2 Hvap Hliq H'2 H1 2609.9 2.421 kJ kg 3 kJ 10 kg H1 H'2 7.19 kJ kg 0.819 Ans. The following vectors contain values for Parts (a) through (g). For intake conditions: 3274.3 kJ kg 6.5597 3509.8 kJ kg 6.8143 kJ 3634.5 kg H1 kJ 3161.2 kg 1389.6 Btu lbm Btu lbm kg K kJ kg K kJ 6.9813 kg K 6.4536 S1 kJ 2801.4 kg 1444.7 kJ kJ kg K kJ 6.4941 kg K 1.6000 1.5677 231 Btu lbm rankine Btu lbm rankine 0.80 0.77 0.82 0.75 0.75 0.80 0.75
  • 232. For discharge conditions: 0.9441 0.8321 0.6493 Sliq 1.0912 1.5301 kJ 7.7695 kg K kJ kg K 7.9094 kJ 8.1511 kg K kJ kg K kJ kg K 7.1268 Btu 0.1750 1.9200 lbm rankine Btu lbm rankine 0.2200 7.5947 Svap 1.8625 kJ kg K kJ kg K kJ kg K kJ kg K S'2 = S1 kJ kg K Btu lbm rankine Btu lbm rankine 289.302 2625.4 kJ kg 80 kg sec 251.453 kJ kg 2609.9 kJ kg 90 kg sec 191.832 Hliq kJ kg kJ kg 2584.8 kJ kg 70 kg sec 340.564 kJ kg 2646.0 kJ kg 65 kg sec 504.701 kJ kg 2706.3 kJ kg 50 kg sec Btu lbm 1116.1 Btu lbm 150 Btu lbm 1127.3 Btu lbm 100 94.03 120.99 Hvap 232 mdot lbm sec lbm sec
  • 233. x'2 S1 Svap H x2 Sliq H'2 H2 Hvap H2 H2 H2 H2 H2 H2 H2 1 H'2 Sliq H1 Hliq Hliq Hliq H2 H1 S2 Sliq Wdot x2 Svap Sliq 3 S2 kJ 2467.8 kg 2471.4 2 7.6873 7.7842 S2 4 2543.4 kJ kg K 7.1022 3 S2 Ans. 6.7127 5 S2 1031.9 Btu 1057.4 lbm 6 S2 7 1.7762 Btu 1.7484 lbm rankine 68030 91230 87653 117544 81672 Wdot H mdot 7.1808 S2 5 7 H 1 2423.9 2535.9 6 Hliq S2 2 4 x'2 Hvap 109523 44836 kW Wdot 60126 12900 17299 65333 87613 35048 46999 233 hp Ans.
  • 234. 7.20 T P0 423.15 K P 8.5 bar 1 bar For isentropic expansion, S J 0 mol K For the heat capacity of nitrogen: A B 3.280 3 0.593 10 K 5 D 0.040 10 K 2 For the entropy change of an ideal gas, combine Eqs. (5.14) & (5.15) with C = 0. Substitute: (guess) 0.5 Given S = R A ln B 1 D T 1 T 2 2 T T0 Find ln P P0 T0 Ans. 762.42 K Thus the initial temperature is 489.27 degC 7.21 T1 CP 32 P1 1223.15 K J mol K P2 10 bar 1.5 bar 0.77 Eqs. (7.18) and (7.19) derived for isentropic compression apply equally well for isentropic expansion. They combine to give: W's C P T1 Ws W's P2 R CP W's 1 P1 H Ws 234 15231 J mol Ws 11728 J mol Ans.
  • 235. Eq. (7.21) also applies to expansion: T2 7.22 H T1 T2 CP Isobutane: Tc 408.1 K Pc T0 P0 5000 kPa P 523.15 K S 0 J mol K A 0.181 36.48 bar 500 kPa For the heat capacity of isobutane: 37.853 10 K B 1.677 T0 Tr0 Ans. 856.64 K Tr0 Tc 3 C K Pr0 1.282 11.945 10 Pr P0 2 Pr0 Pc P Pr Pc 6 1.3706 0.137 The entropy change is given by Eq. (6.92) combined with Eq. (5.15) with D = 0: 0.5 (guess) Given S = R A ln SRB B T0 T0 Tc Find T Tr Pr 1 2 C T0 2 SRB Tr0 Pr0 T0 T Tc T Tr 235 445.71 K 1.092 1 ln P P0
  • 236. The enthalpy change is given by Eq. (6.91): Hig R ICPH T0 T 1.677 37.853 10 Hig 11.078 H' Hig H' 8331.4 3 6 11.945 10 0.0 kJ mol R Tc HRB Tr Pr HRB Tr0 Pr0 J mol The actual enthalpy change from Eq. (7.16): 0.8 Wdot ndot 700 mol sec ndot H H Wdot H' H 6665.1 J mol Ans. 4665.6 kW The actual final temperature is now found from Eq. (6.91) combined with Eq (4.7), written: 0.7 (guess) Given H = R A T0 1 Tc HRB Find 7.23 B 2 T0 2 T0 Pr Tc 0.875 2 C 1 3 3 3 T0 1 HRB Tr0 Pr0 T T0 T 457.8 K Ans. From Table F.2 @ 1700 kPa & 225 degC: H1 2851.0 At 10 kPa: kJ kg S1 6.5138 x2 0.95 236 kJ kg K Sliq 0.6493 kJ kg K
  • 237. Hliq kJ 191.832 mdot 0.5 Hvap kg kg 2584.8 Wdot sec Svap 8.1511 H H1 2.465 10 (a) Qdot mdot H (b) For isentropic expansion to 10 kPa, producing wet steam: S1 Svap x'2 Hliq H2 Hliq H2 3 kJ H Wdot Sliq H'2 T0 mdot H'2 Hliq H'2 Sliq 2.063 385.848 Qdot kg 0.782 Wdot' 7.24 kJ kg K 180 kW H2 x'2 x2 Hvap kJ kg 12.92 x'2 Hvap P0 673.15 K 8 bar P For isentropic expansion, For the heat capacity of carbon dioxide: A 5.457 B Ans. Hliq kg Ans. 394.2 kW 1 bar S 0 J mol K 3 1.045 10 K kJ sec 3 kJ 10 Wdot' H1 kJ kg 5 D 1.157 10 K 2 For the entropy change of an ideal gas, combine Eqs. (5.14) & (5.15) with C = 0: (guess) 0.5 Given S = R A ln B T0 T0 Find 1 D 2 2 T' 0.693 237 1 T0 T' ln P P0 466.46 K
  • 238. H' H' 9.768 0.0 1.157 10 5 kJ mol 0.75 H 3 R ICPH T0 T' 5.457 1.045 10 Work Work H' H 7.326 Work 7.326 kJ mol Ans. kJ mol For the enthalpy change of an ideal gas, combine Eqs. (4.2) and (4.7) with C = 0: Given H = R A T0 B 2 T0 2 1 Find 2 0.772 1 T D T0 T0 1 T Ans. 519.9 K Thus the final temperature is 246.75 degC 7.25 Vectors containing data for Parts (a) through (e): 500 371 1.2 3.5 450 T1 6 5 376 2.0 4.0 525 P1 10 T2 458 P2 3.0 Cp 5.5 R 475 HS 372 1.5 4.5 550 H 7 4 403 1.2 2.5 [ ( Cp T2 Cp T1 Ideal gases with constant heat capacities T1) ] P2 P1 R Cp Eq. (7.22) Applies to expanders as well as to compressors 1 238
  • 239. 0.7 0.803 H 0.649 HS 0.748 0.699 7.26 Cp 7 2 R ndot Guesses: mol 175 T1 sec 0.75 Wdot 550K P1 6bar P2 1.2bar 600kW Given Wdot = 0.065 Wdot .08 ln Wdot kW Find ( Wdot) 0.065 ndot Cp T1 Wdot P2 P1 1 Ans. 594.716 kW Wdot kW 0.08 ln R Cp 0.576 Ans. For an expander operating with an ideal gas with constant Cp, one can show that: T2 P2 P1 T1 1 R Cp 1 T2 433.213 K By Eq. (5.14): S R T2 Cp ln T1 R ln P2 P1 S 6.435 J mol K By Eq. (5.37), for adiabatic operation : SdotG ndot S SdotG 1.126 239 3 10 J K sec Ans.
  • 240. 7.27 Properties of superheated steam at 4500 kPa and 400 C from Table F.2, p. 742. H1 3207.1 S1 6.7093 If the exhaust steam (Point 2, Fig. 7.4) is "dry," i.e., saturated vapor, then isentropicexpansion to the same pressure (Point 2', Fig. 7.4) must produce "wet" steam, withentropy: S2 = S1 = 6.7093 = (x)(Svap) + (1-x)(Sliq) [x is quality] A second relation follows from Eq. (7.16), written: HS) = (0.75)[ (x)(Hvap) + (1-x)(Hliq) - 3207.1] H = Hvap - 3207.1 = ( Each of these equations may be solved for x. Given a final temperature and the corresponding vapor pressure, values for Svap, Sliq, Hvap, and Hliq are found from the table for saturated steam, and substitution into the equations for x produces two values. The required pressure is the one for which the two values of x agree. This is clearly a trial process. For a final trial temperature of 120 degC, the following values of H and S for saturated liquid and saturated vapor are found in the steam table: Hl 503.7 Hv 2706.0 Sl 1.5276 Sv 7.1293 The two equations for x are: xH Hv 801.7 .75 Hl .75 ( Hv Hl) xS The trial values given produce: xH 6.7093 Sl Sv Sl 0.924 xS 0.925 These are sufficiently close, and we conclude that: t=120 degC; P=198.54 kPa If were 0.8, the pressure would be higher, because a smaller pressure drop would be required to produce the same work and H. 240
  • 241. 7.29 P1 5 atm P2 1 atm T1 15 degC 0.55 Data in Table F.1 for saturated liquid water at 15 degC give: 3 V 1001 cm kg Cp 4.190 kJ kg degC Eqs. (7.16) and (7.24) combine to give: Ws H (7.14) Ws 0.223 Eq. (7.25) with =0 is solved for T: H P1) kJ kg T T 7.30 V ( P2 H V ( P2 P1) Cp Ans. 0.044 degC Assume nitrogen an ideal gas. First find the temperature after isentropic expansion from a combination of Eqs. (5.14) & (5.15) with C = 0. Then find the work (enthalpy change) of isentropic expansion by a combination of Eqs. (4.2) and (4.7) with C = 0. The actual work (enthalpy change) is found from Eq. (7.20). From this value, the actual temperature is found by a second application of the preceding equation, this time solving it for the temperature. The following vectors contain values for Parts (a) through (e): 753.15 1 bar 673.15 T0 6 bar 5 bar 1 bar P0 773.15 K 7 bar P 1 bar 723.15 8 bar 2 bar 755.37 95 psi 15 psi 200 150 ndot 0.80 0.75 175 mol sec 0.78 100 0.85 0.5 453.59 0.80 241 S i 0 1 5 J mol K
  • 242. For the heat capacity of nitrogen: A 3.280 3 0.593 10 B 5 D 0.040 10 K 2 K (guess) 0.5 Given S = R A ln B T0 1 D 2 2 2 T0 Tau T0 P0 P Find ln 1 P P0 Tau T0 P0 Pi i i i 460.67 431.36 Ti T0 i T i 453.48 K 494.54 455.14 H'i 3 R ICPH T0 Ti 3.280 0.593 10 5 0.0 0.040 10 i 8879.2 7279.8 H' 7103.4 5459.8 9714.4 J mol H H H' 7577.2 6941.7 0.5 5900.5 9112.1 J mol 7289.7 (guess) Given H = R A T0 Tau T0 H 1 B 2 T0 2 2 Find i 242 1 1 D T0 Tau T0 i Hi Ti T0 i i
  • 243. 520.2 492.62 T 1421 819 525.14 K Ans. Wdot ndot H 1326 kW Ans. Wdot 529.34 516.28 7.31 590 1653 Property values and data from Example 7.6: kg H1 3391.6 kJ kg S1 6.6858 kJ kg K mdot H2 2436.0 kJ kg S2 7.6846 kJ kg K Wdot 56400 kW T 300 K Wdotideal 74084 kW Wdotideal t 59.02 sec By Eq. (5.26) mdot H2 H1 Wdot Wdotideal T t S2 S1 Ans. 0.761 The process is adiabatic; Eq. (5.33) becomes: SdotG mdot S2 Wdotlost 7.32 SdotG S1 Wdotlost T SdotG 58.949 kW K 17685 kW Ans. Ans. For sat. vapor steam at 1200 kPa, Table F.2: H2 2782.7 kJ kg S2 6.5194 kJ kg K The saturation temperature is 187.96 degC. The exit temperature of the exhaust gas is therefore 197.96 degC, and the temperature CHANGE of the exhaust gas is -202.04 K. For the water at 20 degC from Table F.1, H1 83.86 kJ kg S1 0.2963 243 kJ kg K
  • 244. The turbine exhaust will be wet vapor steam. For sat. liquid and sat. vapor at the turbine exhaust pressure of 25 kPa, the best property values are found from Table F.1 by interpolation between 64 and 65 degC: kJ kJ Hliq 272.0 Hlv 2346.3 kg kg Sliq 0.8932 kJ kg K Slv 6.9391 kJ kg K 0.72 For isentropic expansion of steam in the turbine: S'3 S2 S'3 6.519 x'3 kJ kg K x'3 S'3 Sliq H'3 0.811 Hliq H'3 Slv 2.174 H23 H'3 H2 H3 H2 H23 H23 437.996 kJ kg H3 2.345 10 S3 Sliq x3 Slv S3 7.023 kJ kg K For the exhaust gases: ndot 125 mol sec T1 ( 273.15 T2 ( 273.15 T1 673.15 K T2 471.11 K H3 x3 x3 Hliq Hlv 0.883 molwt 18 400)K 3 kJ kg 197.96)K gm mol Hgas R MCPH T1 T2 3.34 1.12 10 Sgas R MCPS T1 T2 3.34 1.12 10 244 3 3 0.0 0.0 T2 0.0 0.0 ln T2 T1 T1 x'3 Hlv 3 kJ 10 kg
  • 245. Hgas 6.687 3 kJ 10 Sgas kmol 11.791 kJ kmol K Energy balance on boiler: mdot ndot Hgas H2 H1 (a) Wdot mdot mdot H3 T 314.302 kW S1 Wdot Wdotideal t sec 293.15 K ndot Hgas mdot H3 H1 T ndot Sgas mdot S3 Wdotideal kg 135.65 kW Ans. Wdot H2 (b) By Eq. (5.25): Wdotideal 0.30971 t Ans. 0.4316 (c) For both the boiler and the turbine, Eq. (5.33) applies with Q = 0. For the boiler: SdotG ndot Sgas Boiler: mdot S2 SdotG S1 0.4534 kW K For the turbine: SdotG mdot S3 Turbine: SdotG 0.156 (d) Wdotlost.boiler 0.4534 Wdotlost.turbine Fractionboiler kW 0.1560 K T kW T K Wdotlost.boiler Wdotideal 245 kW K Ans. S2 Ans. Wdotlost.boiler 132.914 kW Wdotlost.turbine 45.731 kW Fractionboiler 0.4229 Ans.
  • 246. Wdotlost.turbine Fractionturbine Note that: 7.34 t 0.1455 Ans. Fractionturbine Wdotideal Fractionboiler Fractionturbine 1 From Table F.2 for sat. vap. at 125 kPa: H1 kJ kg 2685.2 S1 7.2847 kJ kg K For isentropic expansion, S'2 = S1 = 7.2847 kJ kg K Interpolation in Table F.2 at 700 kPa for the enthalpy of steam with this entropy gives H'2 3051.3 H2 H1 kJ kg H 0.78 H2 H 3154.6 H'2 kJ kg H1 H 469.359 kJ kg Ans. Interpolation in Table F.2 at 700 kPa for the entropy of steam with this enthalpy gives S2 mdot 2.5 kg sec Wdot 7.4586 kJ kg K mdot H 246 Ans. Wdot 1173.4 kW Ans.
  • 247. 7.35 Assume air an ideal gas. First find the temperature after isentropic compression from a combination of Eqs. (5.14) & (5.15) with C = 0. Then find the work (enthalpy change) of isentropic compression by a combination of Eqs. (4.2) and (4.7) with C = 0. The actual work (enthalpy change) is found from Eq. (7.20). From this value, the actual temperature is found by a second application of the preceding equation, this time solving it for the temperature. The following vectors contain values for Parts (a) through (f): 298.15 101.33 kPa 375 kPa 353.15 375 kPa 1000 kPa 303.15 T0 373.15 100 kPa P0 K 500 kPa P 500 kPa 1300 kPa 299.82 14.7 psi 55 psi 338.71 55 psi 135 psi 100 0.75 100 0.70 150 ndot 0.80 mol sec 50 S 0.75 0.5 453.59 1 6 0.75 0.5 453.59 i J mol K 0 0.70 For the heat capacity of air: A 3.355 0.5 0.575 10 K 3 5 2 (guess) B D 0.016 10 K Given S = R A ln B T0 2 2 T0 Tau T0 P0 P 1 D Find 1 2 i 247 ln P P0 Tau T0 P0 Pi i i
  • 248. 431.06 464.5 Ti T0 i 476.19 i T K 486.87 434.74 435.71 H'i 3 R ICPH T0 Ti 3.355 0.575 10 5 0.0 0.016 10 i 3925.2 3314.6 5133.2 J 3397.5 mol H' 3986.4 2876.6 5233.6 4735.1 H' H 6416.5 H 4530 J mol 5315.2 4109.4 1.5 (guess) Given H = R A T0 Tau T0 H Wdot 1 B 2 T0 2 ndot H Find i 248 2 1 Tau T0 i D T0 Hi 1 Ti T0 i i
  • 249. 474.68 702 523 511.58 635 474 518.66 T 524.3 1291 Wdot K 304 962 Wdot hp 479.01 1617 1205 476.79 7.36 1250 932 Ammonia: 0 Tc 405.7 K Pc 294.15 K P0 200 kPa J mol K A P 1000 kPa For the heat capacity of ammonia: 3.020 10 K B 3.578 T0 Tr0 0.253 112.8 bar T0 S Ans. kW 227 Tr0 Tc 3 0.186 10 K P0 Pr0 0.725 5 D Pr0 Pc P Pr Pr Pc 2 0.0177 0.089 Use generalized second-virial correlation: The entropy change is given by Eq. (6.92) combined with Eq. (5.15); C = 0: 1.4 (guess) Given S = R A ln B T0 T0 SRB Find T0 Tc 1 D 2 ln 1 2 Pr T P0 SRB Tr0 Pr0 1.437 P Tr 249 T0 T Tc T Tr 422.818 K 1.042
  • 250. Hig Hig 4.826 H' Hig H' 4652 3 R ICPH T0 T 3.578 3.020 10 0.0 5 0.186 10 kJ mol R Tc HRB Tr Pr HRB Tr0 Pr0 J mol The actual enthalpy change from Eq. (7.17): 0.82 H' H H 5673.2 J mol The actual final temperature is now found from Eq. (6.91) combined with Eq (4.7), written: (guess) 1.4 Given 1 Tc HRB Find B 2 T0 2 T0 Pr Tc 2 1.521 H = R A T0 T R A ln B T0 SRB Tr Pr S 2.347 J mol K T0 T Tc D 2 2 T0 SRB Tr0 Pr0 Ans. 250 1 HRB Tr0 Pr0 Tr S D T0 1 1 T 447.47 K Tr 1.103 1 ln P P0 Ans.
  • 251. 7.37 Propylene: 0 A 365.6 K Pc 303.15 K P0 11.5 bar J 22.706 10 B 1.637 3 K P0 Pr0 0.8292 6.915 10 C K Tr0 Tc P 18 bar For the heat capacity of propylene: mol K T0 Tr0 0.140 46.65 bar T0 S Tc Pr 2 Pr0 Pc P Pr Pc 6 0.2465 0.386 Use generalized second-virial correlation: The entropy change is given by Eq. (6.92) combined with Eq. (5.15) with D = 0: (guess) 1.1 Given S = R A ln B T0 T0 1 2 C T0 2 Tc Find Pr T P P0 SRB Tr0 Pr0 1.069 SRB ln 1 T0 T T Tc Tr 324.128 K Tr 0.887 The enthalpy change for the final T is given by Eq. (6.91), with HRB for this T: Hig R ICPH T0 T 1.637 22.706 10 Hig 1.409 H' Hig 3 6.915 10 3 J 10 mol R Tc HRB Tr Pr 251 HRB Tr0 Pr0 6 0.0
  • 252. H' J mol 964.1 The actual enthalpy change from Eq. (7.17): ndot H' H 0.80 1000 mol Wdot H Wdot ndot H J mol 1205.2 Ans. 1205.2 kW sec The actual final temperature is now found from Eq. (6.91) combined with Eq (4.7), written: (guess) 1.1 Given H = R A T0 B 2 T0 2 T0 Pr Tc 1 Tc HRB 7.38 Methane: C 3 T0 3 1 T T0 0 Tr0 Tc 190.6 K Pc T0 Tc Ans. 308.15 K P0 0.012 45.99 bar 3500 kPa J mol K A 1 327.15 K T0 S 3 HRB Tr0 Pr0 T 1.079 Find 2 P 5500 kPa For the heat capacity of methane: B 1.702 Tr0 9.081 10 K 1.6167 3 C K Pr0 Pr 252 2.164 10 P0 Pc P Pc 6 2 Pr0 Pr 0.761 1.196
  • 253. Use generalized second-virial correlation: The entropy change is given by Eq. (6.92) combined with Eq. (5.15) with D = 0: (guess) 1.1 Given S = R A ln B T0 T0 1 2 C T0 2 Tc Find Pr T P P0 SRB Tr0 Pr0 1.114 SRB ln 1 T0 T T Tc Tr 343.379 K Tr 1.802 The enthalpy change for the final T is given by Eq. (6.91), with HRB for this T: Hig Hig 1.298 10 H' Hig H' 1158.8 3 R ICPH T0 T 1.702 9.081 10 2.164 10 6 0.0 3 J mol R Tc HRB Tr Pr HRB Tr0 Pr0 J mol The actual enthalpy change from Eq. (7.17): 0.78 ndot 1500 H mol sec Wdot H' ndot H H Wdot 1485.6 J mol 2228.4 kW Ans. The actual final temperature is now found from Eq. (6.91) combined with Eq (4.7), written: 1.1 (guess) 253
  • 254. Given H = R A T0 B 2 T0 2 T0 Pr Tc 1 Tc HRB 7.39 C 3 T0 3 1 3 1 HRB Tr0 Pr0 T 1.14 Find 2 T T0 Ans. 351.18 K From the data and results of Example 7.9, T1 Work T2 293.15 K 5288.3 P1 428.65 K J T mol R ICPH T1 T2 1.702 9.081 10 H 5288.2 S 3 R ICPS T1 T2 1.702 9.081 10 3.201 560 kPa 293.15 K H S P2 140 kPa 6 2.164 10 0.0 J mol 3 6 2.164 10 0.0 ln P2 P1 J mol K Since the process is adiabatic: SG Wideal Wlost H T SG S T S S Wideal Wlost Wideal t Work 254 3.2012 t J mol K 4349.8 938.4 0.823 J mol J mol Ans. Ans. Ans. Ans.
  • 255. 7.42 P1 1atm T1 ( 35 P2 50atm T2 ( 200 T1 473.15 K Cp 273.15) K 308.15 K T2 273.15) K 3.5 R 3 0.65 V Vdot R T1 P1 0.5 ndot m sec Vdot ndot V 19.775 mol sec With compression from the same initial conditions (P1,T1) to the same final conditions (P2,T2) in each stage, the same efficiency in each stage, and the same power delivered to each stage, the applicable equations are: P2 P1 r= 1 N (where r is the pressure ratio in each stage and N is the number of stages.) Eq. (7.23) may be solved for T2prime: T'2 T'2 N R Cp P2 P1 ln ln T1) T1 Eq. (7.18) written for a single stage is: 415.4 K T'2 = T1 ( T2 R1 N Cp Put in logarithmic form and solve for N: P2 P1 N T'2 (a) Although any number of stages greater than this would serve, design for 4 stages. 3.743 T1 (b) Calculate r for 4 stages: N 4 r P2 P1 1 N r 2.659 Power requirement per stage follows from Eq. (7.22). In kW/stage: Wdotr ndot Cp T1 r R Cp 1 Wdotr 255 87.944 kW Ans.
  • 256. (c) Because the gas (ideal) leaving the intercooler and the gas entering the compressor are at the same temperature (308.15 K), there is no enthalpy change for the compressor/interchanger system, and the first law yields: Qdotr Wdotr Qdotr 87.944 kW Ans. Heat duty = 87.94 kW/interchanger (d) Energy balance on each interchanger (subscript w denotes water): With data for saturated liquid water from the steam tables: kJ kJ Hw ( 188.4 104.8) Hw 83.6 kg kg Qdotr mdotw mdotw Hw 1.052 kg sec Ans. (in each interchanger) 7.44 300 290 T1 2.0 1.5 295 K P1 1.2 bar 300 305 1.5 464 6 3.5 547 T2 1.1 5 2.5 455 K P2 6 bar Cp 4.5 R 505 HS 5.5 496 H 8 7 4.0 [ ( Cp T2 Cp T1 Ideal gases with constant heat capacities T1) ] P2 P1 R Cp (7.22) 1 256
  • 257. 3.219 0.675 3.729 HS kJ mol 4.745 0.698 HS H 5.959 0.636 4.765 7.47 Ans. 0.793 0.75 The following vectors contain values for Parts (a) through (e). Intake conditions first: 298.15 20 kg 363.15 T1 100 kPa 200 kPa 30 kg P1 333.15 K 20 kPa mdot 15 kg 294.26 1 atm 50 lb 366.48 15 psi 80 lb 2000 kPa 0.75 257.2 5000 kPa 0.70 696.2 5000 kPa 0.75 523.1 20 atm 0.70 0.75 10 K 6 217.3 1500 psi P2 1 sec 714.3 From the steam tables for sat.liq. water at the initial temperature (heat capacity calculated from enthalpy values): 1.003 1.036 V 1.017 4.15 4.20 3 cm gm 4.20 CP 1.002 4.185 1.038 kJ kg K 4.20 By Eq. (7.24) HS V P2 257 P1 H HS
  • 258. 1.906 4.973 HS 2.541 7.104 kJ 5.065 H kg 6.753 1.929 kg 2.756 10.628 kJ 14.17 0.188 By Eq. (7.25) T H V 1 T1 P2 0.807 P1 T CP 0.612 K 0.227 1.506 50.82 213.12 Wdot H mdot Wdot 68.15 285.8 101.29 kW Wdot 135.84 hp 62.5 83.81 514.21 689.56 298.338 363.957 T2 T1 T T2 333.762 K 294.487 367.986 t2 t2 T2 K 273.15 t2 t2 t2 T2 K t2 1.8 459.67 t2 1 25.19 2 90.81 degC 60.61 3 4 5 258 70.41 202.7 degF Ans.
  • 259. 7.48 Results from Example 7.10: H kJ 11.57 W kg 11.57 Wideal 300 K H T Wideal T 8.87 kJ kg S kJ kg 0.0090 kJ kg K Wideal S t Ans. t W 0.767 Ans. Since the process is adiabatic. SG S Wlost 7.53 SG T T1 ( 25 T3 S ( 200 Cpv 105 9 3 kJ 10 Wlost 2.7 kJ kg P1 273.15) K J mol K Ans. 1.2bar P3 273.15) K Ans. kg K 5bar Hlv 30.72 P2 5bar kJ mol 0.7 Estimate the specific molar volume of liquid benzene using the Rackett equation (3.72). 3 From Table B.1 for benzene: Tc 562.2K From Table B.2 for benzene: Tn ( 80.0 Zc 0.271 Vc 273.15) K Trn 259 cm mol Tn Tc 2 Assume Vliq = Vsat: V V c Zc 1 T rn 3 7 Eq. (3.72) Calculate pump power Ws V P2 P1 Ws 259 0.053 kJ mol Ans. V cm 96.802 mol
  • 260. Assume that no temperature change occurs during the liquid compression. Therefore: T2 T1 Estimate the saturation temperature at P = 5 bar using the Antoine Equation and values from Table B.2 For benzene from A Table B.2: B Tsat A ln 13.7819 B C degC Tsat P2 kPa Tsat 2726.81 C 217.572 Tsat 415.9 K 142.77 degC Tsat 273.15K Estimate the heat of vaporization at Tsat using Watson's method From Table B.2 Hlv At 80 C: ( 80 Tr1 Hlv2 273.15) K Tc Hlv 1 Tr2 1 Tr1 30.72 kJ mol Tr1 0.628 Tr2 Tsat Tr2 Tc 0.38 Eq. (4.13) Hlv2 26.822 Calculate the heat exchanger heat duty. Q R ICPH T2 Tsat 0.747 67.96 10 Hlv2 Cpv T3 Tsat Q 51.1 kJ mol Ans. 260 3 37.78 10 6 0 0.74 kJ mol
  • 261. 7.54 T1 ( 25 T3 ( 200 Cpv 273.15)K 273.15)K 105 P1 P3 P2 1.2bar 1.2bar 5bar J mol K 0.75 Calculate the compressor inlet temperature. Combining equations (7.17), (7.21) and (7.22) yields: T3 T2 1 P3 1 T2 1 P2 408.06 K T2 R Cpv 273.15K 134.91 degC Calculate the compressor power Ws Cpv T3 T2 Ws 6.834 kJ mol Ans. Calculate the heat exchanger duty. Note that the exchanger outlet temperature, T2, is equal to the compressor inlet temperature. The benzene enters the exchanger as a subcooled liquid. In the exchanger the liquid is first heated to the saturation temperature at P1, vaporized and finally the vapor is superheated to temperature T 2. Estimate the saturation temperature at P = 1.2 bar using the Antoine Equation and values from Table B.2 For benzene from A Table B.2: B Tsat A ln P1 B 13.7819 C degC Tsat kPa Tsat 2726.81 C 217.572 Tsat 358.7 K 85.595 degC Tsat 273.15K Estimate the heat of vaporization at Tsat using Watson's method From Table B.2 At 25 C: Hlv 30.72 kJ mol 261 From Table B.1 for benzene: Tc 562.2K
  • 262. ( 80 Tr1 273.15) K Tr1 Tc Hlv2 Hlv 1 Tr2 1 Tr1 Eq. (4.13) Q 44.393 3 R ICPH T1 Tsat 0.747 67.96 10 Hlv2 Cpv T2 Tsat 7.57 ndot Cp 100 50.6 mol Tr2 Tc 0.638 0.38 Q kJ Tsat Tr2 0.628 Hlv2 37.78 10 30.405 6 kJ mol 0 Ans. kmol hr P1 T1 1.2bar J mol K P2 300K 6bar 0.70 Assume the compressor is adaiabatic. T2 P2 P1 Wdots Wdote T1 (Pg. 77) ndot Cp T2 T2 T1 Wdots 127.641 kW Wdote 3040dollars 380dollars 390.812 K Wdots Wdots C_compressor C_motor R Cp 182.345 kW 0.952 C_compressor kW Wdote 307452 dollars Ans. 0.855 C_motor kW 262 32572 dollars Ans.
  • 263. 7.59 T1 18bar For ethylene: T1 Tr1 Tr1 Tc P2 1.2bar 0.087 P1 375K Tc 282.3K P2 Pr2 A B 1.424 14.394 10 3 Pc C 4.392 10 6 0.357 Pr2 Pc 50.40bar Pr1 P1 Pr1 1.328 Pc 0.024 D 0 a) For throttling process, assume the process is adiabatic. Find T2 such that H = 0. H = Cpmig T2 T1 HR2 Eq. (6-93) HR1 Use the MCPH function to calculate the mean heat capacity and the HRB function for the residual enthalpy. Guess: T2 T1 Given 0 J mol T2 = MCPH T1 T2 A B C D R T2 T1 T2 Pr2 R Tc HRB R Tc HRB Tr1 Pr1 Tc Find T2 T2 365.474 K Ans. Tr2 T2 Tc Tr2 1.295 Calculate change in entropy using Eq. (6-94) along with MCPS function for the mean heat capacity and SRB function for the residual entropy. S R MCPS T1 T2 A B C D ln R SRB Tr2 Pr2 S 22.128 J mol K T2 T1 R SRB Tr1 Pr1 Ans. 263 R ln P2 P1 Eq. (6-94)
  • 264. b) For expansion process. 70% First find T2 for isentropic expansion. Solve Eq. (6-94) with S = 0. Guess: T2 T1 Given 0 T2 P2 J R ln = R MCPS T1 T2 A B C D ln T1 mol K P1 T2 Pr2 SRB R R SRB Tr1 Pr1 Tc T2 Find T2 T2 219.793 K Tr2 T2 Eq. (6-94) Tr2 Tc Now calculate the isentropic enthalpy change, HS. HR2 HS HS HRB Tr2 Pr2 R Tc R MCPH T1 T2 A B C D T2 T1 HRB Tr2 Pr2 R Tc HRB Tr1 Pr1 6.423 R Tc 3 J 10 mol Calculate actual enthalpy change using the expander efficiency. H HS H 4.496 3 J 10 mol Find T2 such that H matches the value above. Given HS = MCPH T1 T2 A B C D R T2 T1 T2 Pr2 R Tc HRB R Tc HRB Tr1 Pr1 Tc T2 Find T2 T2 268.536 K 264 Ans. 0.779
  • 265. Now recalculate S at calculated T2 S R MCPS T1 T2 A B C D ln T1 R SRB Tr1 Pr1 R SRB Tr2 Pr2 S 7.77 J mol K T2 R ln P2 Eq. (6-94) P1 Ans. Calculate power produced by expander kJ Ans. P H P 3.147 mol The advantage of the expander is that power can be produced in the expander which can be used in the plant. The disadvantages are the extra capital and operating cost of the expander and the low temperature of the gas leaving the expander compared to the gas leaving the throttle valve. 7.60 Hydrocarbon gas: T1 500degC Light oil: 25degC Cpoil Exit stream: b) T2 T3 Cpgas 150 J mol K J mol K J mol 200degC 200 Hlv 35000 Assume that the oil vaporizes at 25 C. For an adiabatic column, the overall energy balance is as follows. F Cpgas T3 T1 D Hlv Coilp T3 T2 = 0 DF 0.643 Solving for D/F gives: DF c) Cpgas T3 Hlv T1 Cpoil T3 T2 Ans. Using liquid oil to quench the gas stream requires a smaller oil flow rate. This is because a significant portion of the energy lost by the gas is used to vaporize the oil. 265
  • 266. Chapter 8 - Section A - Mathcad Solutions 8.1 With reference to Fig. 8.1, SI units, At point 2: Table F.2, H2 3531.5 At point 4: Table F.1, H4 209.3 At point 1: H1 H4 At point 3: Table F.1, Hliq H4 x3 H3 Hliq 0.96 Sliq For isentropic expansion, x'3 H'3 S'3 Slv x'3 Hlv H3 turbine Ws Ws H3 Hlv x3 Hlv H2 S'3 S2 0.855 H'3 2246 H2 H'3 1.035 3 10 H2 QH 3.322 Ws cycle QH cycle 266 Ans. 0.805 turbine QH H2 H3 Slv 0.7035 x'3 Sliq Hliq S2 H1 10 0.311 3 Ans. 6.9636 2382.9 2496.9 7.3241
  • 267. 8.2 1.0 (kg/s) mdot The following property values are found by linear interpolation in Table F.1: State 1, Sat. Liquid at TH: H1 860.7 S1 2.3482 P1 3.533 State 2, Sat. Vapor at TH: H2 2792.0 S2 6.4139 P2 3.533 State 3, Wet Vapor at TC: Hliq 112.5 Hvap 2550.6 P3 1616.0 State 4, Wet Vapor at TC: Sliq 0.3929 Svap 8.5200 P4 1616.0 (a) The pressures in kPa appear above. (b) Steps 2--3 and 4--1 (Fig. 8.2) are isentropic, for which S3=S2 and S1=S4. Thus by Eq. 6.82): x3 S2 Sliq Svap Sliq x3 S1 Sliq Svap Sliq x4 0.741 x4 0.241 (c) The rate of heat addition, Step 1--2: Qdot12 mdot ( H2 Qdot12 H1) 1.931 3 10 (kJ/s) (d) The rate of heat rejection, Step 3--4: Hliq x3 ( Hvap Hliq) H3 1.919 H4 10 Qdot34 3 mdot ( H4 (e) Wdot12 699.083 1.22 873.222 mdot ( H3 H2) Wdot23 Wdot41 mdot ( H1 H4) Wdot41 Wdot23 Wdot41 Qdot12 161.617 0.368 Note that the first law is satisfied: Q Qdot12 Q W Qdot34 W Wdot23 0 267 Wdot41 3 10 0 Wdot23 (f) x4 ( Hvap Hliq) Qdot34 H3) Wdot34 0 Hliq H4 H3 (kJ/s)
  • 268. 8.3 The following vectors contain values for Parts (a) through (f). Enthalpies and entropies for superheated vapor, Tables F.2 and F.4 @ P2 and T2 (see Fig. 8.4): 3622.7 6.9013 kJ kg K 3529.6 kJ kg 6.9485 kJ kg K 3635.4 H2 kJ kg kJ kg 6.9875 kJ kg K 6.9145 kJ kg K S2 kJ 3475.6 kg 1507.0 BTU lbm 1.6595 BTU lbm rankine 1558.8 BTU lbm 1.6759 BTU lbm rankine Sat. liq. and sat. vap. values from Tables F.2 and F.4 @ P3 = P4: 191.832 2584.8 kJ kg 251.453 kJ kg 2609.9 kJ kg 191.832 Hliq kJ kg kJ kg 2584.8 kJ kg 2676.0 kJ kg Hvap kJ 419.064 kg 180.17 BTU lbm 1150.5 BTU lbm 69.73 BTU lbm 1105.8 BTU lbm 268
  • 269. 0.6493 kJ kg K 8.1511 kJ kg K 0.8321 kJ kg K 7.9094 kJ kg K kJ 0.6493 kg K Sliq 0.1326 Svap kJ 1.3069 0.3121 8.1511 7.3554 kg K BTU lbm rankine 1.7568 BTU 1.9781 lbm rankine kJ kg K kJ kg K BTU lbm rankine BTU lbm rankine 3 cm 1.010 gm 3 cm 1.017 gm 0.80 1.010 cm gm Vliq 3 cm 1.044 gm 0.0167 ft 0.75 0.75 3 0.75 0.80 turbine 0.78 0.80 pump 0.75 0.78 0.80 3 lbm 3 ft 0.0161 lbm 269 0.75 0.75
  • 270. 80 10000 kPa 10 kPa 100 7000 kPa 20 kPa 70 Wdot 50 3 8500 kPa P1 10 kW 10 kPa P4 6500 kPa 101.33 kPa 50 950 psi 14.7 psi 80 1125 psi 1 psi Vliq P1 Wpump P4 H4 pump S'3 = S2 H3 H2 x'3 S2 Sliq Svap turbine H'3 Sliq H2 Hliq H1 H'3 Hliq H4 Wpump x'3 Hvap Hliq H2 QdotH Wdot Wturbine Wpump H3 H2 H1 mdot QdotC mdot Wturbine QdotH Wdot Answers follow: QdotH mdot1 mdot2 108.64 mdot3 62.13 mdot4 1 70.43 67.29 QdotH kg sec 2 QdotH 3 QdotH 240705 355111 kJ 213277 sec 205061 4 mdot5 mdot6 153.598 QdotH 145.733 lbm 5 sec QdotH 6 270 192801 BTU 228033 sec
  • 271. 0.332 QdotC 1 160705 0.282 QdotC 255111 kJ Wdot 0.328 QdotC 143277 sec QdotH 0.244 2 3 155061 QdotC 0.246 4 0.333 QdotC 145410 BTU 5 QdotC 152208 6 8.4 sec Subscripts refer to Fig. 8.3. Saturated liquid at 50 kPa (point 4) 3 V4 cm 1.030 gm H4 P4 3300 kPa P1 kJ kg 340.564 50 kPa Saturated liquid and vapor at 50 kPa: Hliq H4 Sliq 1.0912 Hvap kJ kg K 2646.0 Svap 7.5947 By Eq. (7.24), Wpump H1 H1 H4 Wpump kJ kg kJ kg K V 4 P4 343.911 P1 Wpump 3.348 kJ kg kJ kg The following vectors give values for temperatures of 450, 550, and 650 degC: 3340.6 H2 3565.3 7.0373 kJ kg S2 3792.9 7.3282 7.5891 271 kJ kg K
  • 272. S'3 S2 H'3 Hliq QH H2 S'3 x'3 x'3 Hvap Hliq Sliq Svap Wturbine Sliq H'3 Wturbine H1 H2 Wpump QH 0.914 8.5 0.959 0.314 0.999 x'3 0.297 0.332 Ans. Subscripts refer to Fig. 8.3. Saturated liquid at 30 kPa (point 4) 3 V4 cm 1.022 gm H4 289.302 kJ kg P1 30 kPa Saturated liquid and vapor at 30 kPa: Hliq H4 Hvap 5000 kJ kg 2625.4 P4 7500 10000 Sliq 0.9441 kJ kg K By Eq. (7.24), Svap Wpump 7.7695 V 4 P4 kJ kg K P1 294.381 H1 H4 Wpump H1 296.936 kJ kg 299.491 The following vectors give values for pressures of 5000, 7500, and 10000 kPa at 600 degC 3664.5 H2 3643.7 7.2578 kJ kg S2 3622.7 7.0526 6.9013 272 kJ kg K kPa
  • 273. S'3 S2 H'3 Hliq QH H2 S'3 x'3 x'3 Hvap Hliq Sliq Svap Sliq Wturbine H'3 H2 Wturbine H1 Wpump QH 0.925 8.6 0.895 0.375 0.873 x'3 0.359 0.386 Ans. From Table F.2 at 7000 kPa and 640 degC: H1 kJ 3766.4 S1 kg 7.2200 kJ kg K S'2 S1 For sat. liq. and sat. vap. at 20 kPa: kJ kg Hvap 2609.9 kJ kg kJ kg K Svap 7.9094 kJ kg K Hliq 251.453 Sliq 0.8321 The following enthalpies are interpolated in Table F.2 at four values for intermediate pressure P2: 3023.9 725 P2 750 775 H'2 kPa W12 H'2 579.15 W12 572.442 kJ 565.89 kg 559.572 3040.9 kJ kg 3049.0 800 0.78 3032.5 H1 H2 3187.3 H2 3194 kJ 3200.5 kg 3206.8 273 H1 W12 7.4939 S2 7.4898 7.4851 7.4797 kJ kg K
  • 274. where the entropy values are by interpolation in Table F.2 at P2. x'3 S2 Svap W23 W Sliq H'3 Sliq H'3 W12 Hliq H2 x'3 Hvap Hliq 20.817 W23 W 7.811 kJ kg 5.073 17.723 The work difference is essentially linear in P2, and we interpolate linearly to find the value of P2 for which the work difference is zero: linterp W P2 0.0 kJ kg (P2) 765.16 kPa Also needed are values of H2 and S2 at this pressure. Again we do linear interpolations: linterp P2 H2 765.16 kPa 3197.9 kJ kg H2 3197.9 kJ kg linterp P2 S2 765.16 kPa 7.4869 kJ kg K S2 7.4869 kJ kg K We can now find the temperature at this state by interplation in Table F.2. This gives an intermediate steam temperature t2 of 366.6 degC. The work calculations must be repeated for THIS case: W12 W12 H2 H1 568.5 H'3 Hliq H'3 2.469 x'3 kJ x'3 kg x'3 Hvap 10 Hliq W23 3 kJ W23 kg 274 S2 Svap Sliq Sliq 0.94 H'3 568.46 H2 kJ kg
  • 275. Work W12 Work W23 1137 kJ kg For a single isentropic expansion from the initial pressure to the final pressure, which yields a wet exhaust: x'3 S1 Svap x'3 H'3 H'3 Hliq x'3 Hvap H'3 Sliq 0.903 W' Sliq 2.38 10 W' H1 Hliq 3 kJ 1386.2 kg kJ kg Whence the overall efficiency is: Work overall overall W' 275 0.8202 Ans.
  • 276. 8.7 From Table F.2 for steam at 4500 kPa and 500 degC: H2 3439.3 kJ kg S2 7.0311 kJ kg K S'3 S2 By interpolation at 350 kPa and this entropy, H'3 2770.6 H3 H2 kJ 0.78 kg WI H3 WI 2.918 10 3 kJ kg Isentropic expansion to 20 kPa: S'4 Exhaust is wet: for sat. liq. & vap.: S2 Hliq 251.453 Sliq 0.8321 kJ kg kJ kg K Hvap 2609.9 Svap 7.9094 276 kJ kg kJ kg K H'3 H2 WI 521.586 kJ kg
  • 277. x'4 S'4 Sliq Svap x'4 H2 H'4 2.317 10 H4 H2 Hliq H'4 0.876 H4 H'4 Sliq x'4 Hvap 2.564 10 Hliq 3 kJ kg 3 kJ kg 3 H5 Hliq V5 V 5 P6 Wpump Wpump 5.841 cm 1.017 P5 20 kPa H6 H5 Wpump H6 257.294 gm P5 kJ kg P6 4500 kPa kJ kg For sat. liq. at 350 kPa (Table F.2): H7 584.270 kJ kg t7 138.87 (degC) We need the enthalpy of compressed liquid at point 1, where the pressure is 4500 kPa and the temperature is: t1 138.87 6 T1 t1 273.15 K t1 132.87 At this temperature, 132.87 degC, interpolation in Table F.1 gives: 3 kJ 558.5 kg Hsat.liq Psat 294.26 kPa Vsat.liq 1.073 Also by approximation, the definition of the volume expansivity yields: 1 1.083 20 Vsat.liq 9.32 10 1.063 3 cm gm K 4 1 K 277 P1 P6 cm gm
  • 278. By Eq. (7.25), H1 Hsat.liq Vsat.liq 1 T1 P1 H1 Psat 561.305 kJ kg By an energy balance on the feedwater heater: mass H1 H6 H3 H7 mass kg Ans. 0.13028 kg Work in 2nd section of turbine: WII ( kg 1 Wnet mass) H4 WI Wpump 1 kg QH QH 8.8 H2 WII WII 2878 kJ 307.567 kJ Wnet H3 823.3 kJ H1 1 kg Wnet Ans. 0.2861 QH Refer to figure in preceding problem. Although entropy values are not needed for most points in the process, they are recorded here for future use in Problem 15.8. From Table F.4 for steam at 650(psia) & 900 degF: H2 1461.2 BTU lbm S2 1.6671 BTU lbm rankine S'3 S2 By interpolation at 50(psia) and this entropy, H'3 1180.4 H3 H2 S3 1.7431 BTU lbm WI WI 0.78 H3 1242.2 BTU lbm rankine 278 BTU lbm H'3 WI 219.024 H2 BTU lbm
  • 279. S'4 Isentropic expansion to 1(psia): S2 Exhaust is wet: for sat. liq. & vap.: 69.73 Sliq 0.1326 x'4 S'4 Hvap BTU lbm rankine Sliq 1.9781 BTU lbm rankine H2 H'4 H4 Hvap Hliq 0.944 1047.8 S4 Hliq 931.204 H4 H2 Hliq Sliq S4 H4 H'4 Sliq 0.831 x4 BTU lbm H'4 Svap x'4 x4 1105.8 Svap BTU lbm Hliq x'4 Hvap 1.8748 Hliq BTU lbm BTU lbm x4 Svap Sliq BTU lbm rankine 3 P5 Wpump P6 H5 1 psi V 5 P6 P5 Wpump H6 650 psi V5 Hliq H5 2.489 ft 0.0161 lbm H6 72.219 BTU lbm Wpump BTU lbm For sat. liq. at 50(psia) (Table F.4): H7 250.21 BTU lbm t7 281.01 S7 0.4112 BTU lbm rankine We need the enthalpy of compressed liquid at point 1, where the pressure is 650(psia) and the temperature is t1 281.01 11 T1 t1 279 459.67 rankine t1 270.01
  • 280. At this temperature, 270.01 degF, interpolation in Table F.3 gives: Psat 41.87 psi Hsat.liq Vsat.liq 238.96 0.1717 Ssat.liq 0.3960 ft 3 lbm BTU lbm BTU lbm rankine The definition of the volume expansivity yields: 0.01726 1 ft 20 Vsat.liq 4.95 0.01709 10 5 3 P1 P6 H1 257.6 BTU lbm S1 lbm rankine 0.397 BTU lbm rankine 1 rankine By Eq. (7.25) and (7.26), H1 Hsat.liq S1 Vsat.liq 1 Ssat.liq Vsat.liq T1 P1 P1 Psat Psat By an energy balance on the feedwater heater: mass H1 H6 H3 H7 lbm mass 0.18687 lbm Ans. WII 158.051 BTU Work in 2nd section of turbine: WII 1 lbm Wnet WI QH H2 Wnet QH mass H4 H3 Wpump 1 lbm WII Wnet H1 1 lbm QH 0.3112 280 Ans. 374.586 BTU 1.204 3 10 BTU
  • 281. 8.9 Steam at 6500 kPa & 600 degC (point 2) Table F.2: H2 3652.1 kJ kg S2 7.1258 kJ kg K P2 6500 kPa At point 3 the pressure must be such that the steam has a condensation temperature in feedwater heater I of 195 degC, 5 deg higher than the temperature of the feed water to the boiler at point 1. Its saturation pressure, corresponding to 195 degC, from Table F.1, is 1399.0 kPa. The steam at point 3 is superheated vapor at this pressure, and if expansion from P2 to P3 is isentropic, S'3 S2 H'3 3142.6 H3 H2 By double interpolation in Table F.2, kJ kg WI From Table F.1: 0.80 H3 H10 3.244 WI 829.9 281 3 kJ 10 kJ kg kg H'3 WI 407.6 H2 kJ kg
  • 282. Similar calculations are required for feedwater heater II. At the exhaust conditions of 20 kPa, the properties of sat. liq. and sat. vap. are: Hliq kJ 251.453 kg Sliq 0.8321 Hvap Svap kJ kg K 2609.9 7.9094 3 kJ Vliq kg 1.017 cm gm kJ kg K If we find t7, then t8 is the mid-temperature between t7 and t1(190 degC), and that fixes the pressure of stream 4 so that its saturation temperature is 5 degC higher. At point 6, we have saturated liquid at 20 kPa, and its properties from Table F.2 are: tsat 60.09 Tsat H6 Hliq V6 Wpump Wpump V 6 P2 8.238 P6 tsat 273.15 K Vliq P6 20 kPa [Eq. (7.24)] kJ kg H67 Wpump We apply Eq. (7.25) for the calculation of the temperature change from point 6 to point 7. For this we need values of the heat capacity and volume expansivity of water at about 60 degC. They can be estimated from data in Table F.1: 1 Vliq 5.408 1.023 1.012 20 10 3 cm CP gm K 4 1 CP K 272.0 kJ kg K 230.2 10 4.18 kJ kg K Solving Eq. (7.25) for delta T gives: H67 T67 t7 Vliq 1 Tsat P2 P6 190 t7 T67 CP tsat T67 K t9 2 282 t7 t8 0.678 K t9 5
  • 283. t7 60.768 t8 130.38 H7 Hliq H67 H7 259.691 kJ kg H8 From Table F.1: t9 T9 125.38 547.9 kJ kg 273.15 t9 K At points 9 and 1, the streams are compressed liquid (P=6500 kPa), and we find the effect of pressure on the liquid by Eq. (7.25). Values by interpolation in Table F.1 at saturation temperatures t9 and t1: Hsat.9 kJ 526.6 kg Hsat.1 kJ 807.5 kg 3 Vsat.9 cm 1.065 gm Vsat.1 cm 1.142 gm Psat.9 234.9 kPa Psat.1 1255.1 kPa 3 3 3 V9 T cm 1.056) gm ( 1.075 V1 1 20 K 9 9 H9 Hsat.9 Vsat.9 1 T1 ( 273.15 Hsat.1 Vsat.1 1 1 T 1 V1 Vsat.1 T 4 1 9 T9 10 1 T1 P2 1 K P2 Psat.9 Psat.1 1.226 H9 530.9 T1 8.92 190)K H1 ( 1.156 V9 Vsat.9 cm 1.128) gm 10 463.15 K H1 810.089 kJ kg kJ kg Now we can make an energy balance on feedwater heater I to find the mass of steam condensed: mI H1 H9 H3 H10 mI kg 283 0.11563 kg Ans. 3 1 K
  • 284. The temperature at point 8, t8 = 130.38 (see above) is the saturation temperture in feedwater heater II. The saturation pressure by interpolation in Table F.1 is 273.28 kPa. Isentropic expansion of steam from the initial conditions to this pressure results in a slightly superheated vapor, for which by double interpolation in Table F.2: H'4 2763.2 kJ H4 H2 H4 Then kg H'4 2.941 H2 3 kJ 10 kg We can now make an energy balance on feedwater heater II to find the mass of steam condensed: mII H9 H7 1 kg H4 mI H10 H8 H8 mII 0.09971 kg Ans. The final stage of expansion in the turbine is to 20 kPa, where the exhaust is wet. For isentropic expansion, x'5 x'5 S2 Svap Sliq H'5 0.889 H5 Then H2 Hliq H'5 Sliq 2.349 H'5 x'5 Hvap Hliq 3 kJ 10 kg H2 H5 2609.4 kJ kg The work of the turbine is: Wturbine Wturbine WI 1 kg 1 kg mI H4 H3 1 kg mI mII H5 H4 936.2 kJ Wturbine QH H2 Wpump 1 kg H1 1 kg 0.3265 QH 284 QH Ans. 3 2.842 10 kJ
  • 285. 8.10 Tc Isobutane: Pc 408.1 K 0.181 36.48 bar For isentropic expansion in the turbine, let the initial state be represented by symbols with subscript zero and the final state by symbols with no subscript. Then T0 P0 533.15 K S 0 J mol K A 37.853 10 K B Tr0 Tc 450 kPa For the heat capacity of isobutane: 1.677 T0 Tr0 P 4800 kPa 3 11.945 10 C K Pr0 1.3064 Pr 6 2 P0 Pr0 Pc P Pc Pr 1.3158 0.123 Use generalized second-virial correlation: The entropy change is given by Eq. (6.92) combined with Eq. (5.15) with D = 0: 0.8 (guess) Given S = R A ln SRB B T0 T0 Tc Pr 1 2 C T0 2 ln P P0 SRB Tr0 Pr0 T 0.852 Find 1 Tr T T0 T Tc Tr 454.49 K 1.114 The enthalpy change for this final temperature is given by Eq. (6.91), with HRB at the above T: Hig Hig R ICPH T0 T 1.677 37.853 10 1.141 10 4 J mol 285 3 11.945 10 6 0.0
  • 286. Hturbine Hig R Tc HRB Tr Pr Hturbine 8850.6 J mol HRB Tr0 Pr0 Wturbine Hturbine The work of the pump is given by Eq. (7.24), and for this we need an estimate of the molar volume of isobutane as a saturated liquid at 450 kPa. This is given by Eq. (3.72). The saturation temperature at 450 kPa is given by the Antoine equation solved for t degC: VP 450 kPa Avp 14.57100 Bvp Bvp tsat Avp Cvp VP ln kPa 2606.775 tsat Cvp 34 Tsat tsat Tsat 3 Vc 274.068 cm 262.7 mol Zc 0.282 Trsat 273.15 K 307.15 K Tsat Trsat Tc 0.753 2 Vliq Wpump V c Zc 1 T rsat Vliq P0 3 7 Vliq P cm 112.362 mol Wpump 488.8 J mol The flow rate of isobutane can now be found: mdot 1000 kW Wturbine Wpump mdot 119.59 mol sec Ans. The enthalpy change of the isobutane in the cooler/condenser is calculated in two steps: a. Cooling of the vapor from 454.48 to 307.15 K b. Condensation of the vapor at 307.15 K Enthalpy change of cooling: HRB at the initial state has already been calculated. For saturated vapor at 307.15 K: Hig R ICPH T Tsat 1.677 37.853 10 286 3 11.945 10 6 0.0
  • 287. Hig 4 J 1.756 Ha Hig Ha 18082 10 mol R Tc HRB Trsat Pr HRB Tr Pr J mol For the condensation process, we estimate the latent heat by Eqs. (4.12) and (4.13): Tn 261.4 K R Tn 1.092 ln Hn 0.930 Hb Hn 1 Trsat 1 Pc bar mdot Qdotin Wturbine 0.641 1.013 Hn Trn 4 J 2.118 10 0.38 Hb Ha 18378 mol J mol Hb Wpump mdot 4360 kW 8.11 Isobutane: Tc Trn Tc Trn Qdotout Qdotout Tn Trn Qdotout 1000 kW Qdotin Qdotin 408.1 K 5360 kW 0.187 Pc 36.48 bar 0.181 Ans. For isentropic expansion in the turbine, let the initial (inlet) state be represented by symbols with subscript zero and the final (exit) state by symbols with no subscript. Then T0 S 413.15 K 0 P0 3400 kPa J mol K 287 P 450 kPa molwt 58.123 gm mol
  • 288. For the heat capacity of isobutane: A 1.677 B T0 Tr0 37.853 10 11.945 10 C K Tr0 Tc 3 K 1.0124 2 P0 Pr0 Pr0 Pc P Pc Pr 6 Pr 0.932 0.123 Use Lee/Kesler correlation for turbine-inlet state, designating values by HRLK and SRLK: HRLK0 1.530 SRLK0 1.160 The entropy change is given by Eq. (6.92) combined with Eq. (5.15) with D = 0: (guess) 0.8 Given S = R A ln B T0 T0 SRB Pr Tc Find 1 2 C T0 1 2 ln P P0 SRLK0 0.809 T T0 T T Tc Tr Tr 334.08 K 0.819 The enthalpy change for this final temperature is given by Eq. (6.91), with HRB at the above T: Hig R ICPH T0 T 1.677 37.853 10 Hig 9.3 3 11.945 10 6 0.0 3 J 10 mol Hturbine Hig R Tc HRB Tr Pr Hturbine 4852.6 J mol HRLK0 Wturbine 288 Hturbine
  • 289. The work of the pump is given by Eq. (7.24), and the required value for the molar volume of saturated-liquid isobutane at 450 kPa (34 degC) is the value calculated in Problem 8.10: 3 Vliq 112.36 cm mol Wpump Vliq P0 P Wpump 331.462 J mol For the cycle the net power OUTPUT is: mdot kg 75 molwt sec Wdot Wdot mdot Wturbine Wpump Ans. 5834 kW The enthalpy change of the isobutane in the cooler/condenser is calculated in two steps: a. Cooling of the vapor from 334.07 to 307.15 K b. Condensation of the vapor at 307.15 K Enthalpy change of cooling: HRB at the initial state has already been calculated. For saturated vapor at 307.15 K it was found in Problem 8.10 as: Tsat Hig 307.15K Tsat Trsat Trsat Tc R ICPH T Tsat 1.677 37.853 10 Hig 2.817 Hig R Tc HRB Trsat Pr Ha 2975 11.945 10 6 0.0 kJ mol Ha 3 0.753 HRB Tr Pr J mol For the condensation process, the enthalpy change was found in Problem 8.10: Hb 18378 J mol Qdotout Qdotout 289 mdot Ha Hb 27553 kW Ans.
  • 290. For the heater/boiler: Qdotin Wdot Qdotout Qdotin Wdot Qdotin 33387 kW Ans. Ans. 0.175 We now recalculate results for a cycle for which the turbine and pump each have an efficiency of 0.8. The work of the turbine is 80% of the value calculated above, i.e., W'turbine 0.8 Wturbine W'turbine 3882 J mol The work of the pump is: W'pump Wdot Wpump W'pump 0.8 mdot W'turbine W'pump Wdot 414.3 J mol 4475 kW Ans. The decrease in the work output of the turbine shows up as an increase in the heat transferred out of the cooler condenser. Thus Qdotout Qdotout Qdotout Wturbine 28805 kW W'turbine mdot Ans. The increase in pump work shows up as a decrease in the heat added in the heater/boiler. Thus Qdotin Qdotin Wdot Qdotin W'pump Wpump mdot 0.134 290 Ans. Qdotin 33280 kW Ans.
  • 291. CP PC 1 bar By Eq. (3.30c): 7 R 2 TC 8.13 Refer to Fig. 8.10. 293.15 K PD PC V C = PD V D 1 VC VD 1 PD = 1.4 5 bar or PD r PC r PC Ans. 3.157 1 Eq. (3.30b): TD QDA = CP TA TC PD QDA PC QDA TA TD CP TD 1500 TA J mol 515.845 K R TC re = VB VA = VC VA = PC R TA PA PD T C PA re T A PC PA re 8.14 2.841 Ans. 3 5 Ratio Ratio = 7 PB PA 1.35 9 Eq. (8.12) now becomes: 0.248 1 1 0.341 1 Ratio 0.396 0.434 291 Ans.
  • 292. 8.16 Figure shows the air-standard turbojet power plant on a PV diagram. 7 TA 303.15 K TC 1373.15 K R CP 2 By Eq. (7.22) WAB = CP TA WCD = CP TC PB R CP 1 = CP TA cr PA PD 2 R CP 1 2 1 = CP TC er PC 7 7 1 where cr is the compression ratio and er is the expansion ratio. Since the two work terms are equal but of opposite signs, cr 6.5 er 0.5 (guess) 2 Given TC er 7 2 1 = TA cr 7 er Find er) ( er 292 1 0.552
  • 293. By Eq. (7.18): PD TD = TC R CP PC 2 This may be written: TD TC er 7 1 By Eq. (7.11) uE 2 2 2 uD = PD V D 1 PE 1 (A) PD We note the following: er = PD cr = PC PB PA = PC cr er = PE PD PE The following substitutions are made in (A): 1 uD = 0 = 2 R = 7 CP Then molwt uE 2 PE 8.17 TA R 7 TD 1 2 molwt PD 1 bar 305 K PA PE PD V D = R T D 1 cr er 29 2 7 cr er PE PB 1.05bar PD = 1 cr er gm mol uE 843.4 m sec Ans. PD 3.589 bar Ans. 0.8 7.5bar Assume air to be an ideal gas with mean heat capacity (final temperature by iteration): Cpmair MCPH 298.15K 582K 3.355 0.575 10 Cpmair 29.921 J mol K 293 3 0.0 0.016 10 5 R
  • 294. Compressor: TB PB Cpmair TA Wsair R Cpmair PA Wsair TA TB Cpmair Combustion: Wsair 1 8.292 10 3 J mol 582.126 K CH4 + 2O2 = CO2 + 2H2O Basis: Complete combustion of 1 mol CH4. Reactants are N mol of air and 1mol CH4. Because the combustion is adiabatic, the basic equation is: HR H298 HP = 0 For H_R, the mean heat capacities for air and methane are required. The value for air is given above. For methane the temperature change is very small; use the value given in Table C.1 for 298 K: 4.217*R. The solution process requires iteration for N. Assume a value for N until the above energy balance is satisfied. (a) TC 1000K HR 57.638 (This is the final value after iteration) N Cpmair N ( 298.15 HR 4.896 582.03)K 4.217 R ( 298.15 300)K 5 J 10 mol The product stream contains: 1 mol CO2, 2mol H2O, 0.79N mol N2, and (0.21N-2) mol O2 1 5.457 2 n A .79 N .21 N i 2 3.470 3.280 1.045 B 3.639 1.450 0.593 0.506 1 4 294 1.157 10 3 D 0.121 0.040 0.227 5 10
  • 295. ni 58.638 A ni Ai i i A CpmP HP ni Bi D i 198.517 B CpmP TC 298.15K HP H298 802625 H298 HP 136.223 ni D i i 0.036 D MCPH 298.15K 1000.K 198.517 0.0361 0.0 From Ex. 4.7: HR B 1.292 1.387 10 5 1.3872 10 5 R 6 J 10 mol J mol J (This result is sufficiently close to zero.) mol Thus, N = 57.638 moles of air per mole of methane fuel. Ans. Assume expansion of the combustion products in the turbine is to 1(atm), i.e., to 1.0133 bar: PD 1.0133bar PC 7.5bar The pertinent equations are analogous to those for the compressor. The mean heat capacity is that of the combustion gases, and depends on the temperature of the exhaust gases from the turbine, which must therefore be found by iteration. For an initial calculation use the mean heat capacity already determined. This calculation yields an exhaust temperature of about 390 K. Thus iteration starts with this value. Parameters A, B, and D have the final values determined above. Cpm Cpm MCPH 1000K 343.12K 198.517 0.0361 0.0 1.849 10 3 5 1.3872 10 R J For 58.638 moles of combustion product: mol K R Ws TD 58.638 Cpm TC PD Cpm 1 PC TC Ws Cpm TD 343.123 K 295 Ws 6 J 1.214 10 mol (Final result of iteration.) Ans.
  • 296. Wsnet Ws Wsair N Wsnet 5 J 7.364 10 Ans. mol (J per mole of methane) Parts (b) and (c) are solved in exactly the same way, with the following results: 5 (b) TC N 37.48 Wsnet 7.365 10 (c) 8.18 1200 1500 N 24.07 Wsnet 5.7519 10 me 0.95 line_losses TC 0.35 tm TD TD 5 343.123 598.94 20% Cost_fuel 4.00 dollars GJ Cost_fuel Cost_electricity tm Cost_electricity 0.05 1 me ( cents kW hr line_losses) Ans. This is about 1/2 to 1/3 of the typical cost charged to residential customers. 8.19 TC TH 111.4K 1 Carnot TC Carnot TH Assume as a basis: QH W W QH Hnlv 300K 0.629 QC 2.651 kJ QC W 0.201 mol kJ 0.6 kJ mol Carnot HE 0.377 1kJ HE Hnlv HE 8.206 Ans. 296 QH 1 HE QC 1.651 kJ
  • 297. 8.20 TH ( 27 a) Carnot b) actual TC 273.15)K 1 (6 TC Carnot TH Carnot 0.6 273.15)K 0.07 Ans. actual 0.028 Ans. 2 3 c) The thermal efficiency is low and high fluid rates are required to generate reasonable power. This argues for working fluids that are relatively inexpensive. Candidates that provide reasonable pressures at the required temperature levels include ammonia, n-butane, and propane. 297
  • 298. Chapter 9 - Section A - Mathcad Solutions 9.2 TH ( 20 TC ( 20 QdotC 0.08 Cost 9.4 TC QdotC Wdot kW hr 253.15 K kJ day TC TH 293.15 K TC 273.15) K 125000 Carnot TH 273.15) K (9.3) 0.6 (9.2) Wdot Wdot Cost 3.797 Carnot 0.381 kW 267.183 dollars yr Ans. Basis: 1 lbm of tetrafluoroethane The following property values are found from Table 9.1: State 1, Sat. Liquid at TH: H1 44.943 S1 0.09142 P1 138.83 State 2, Sat. Vapor at TH: H2 116.166 S2 0.21868 P2 138.83 State 3, Wet Vapor at TC: Hliq 15.187 Hvap 104.471 P3 26.617 State 4, Wet Vapor at TC: Sliq 0.03408 Svap 0.22418 P4 26.617 (a) The pressures in (psia) appear above. (b) Steps 3--2 and 1--4 (Fig. 8.2) are isentropic, for which S3=S2 and S1=S4. Thus by Eq. 6.82): x3 0.971 x4 S1 Sliq Svap Sliq x3 ( Hvap Hliq) H4 Hliq H4 42.118 Q43 59.77 S2 Sliq Svap Sliq x3 x4 0.302 (c) Heat addition, Step 4--3: H3 Hliq H3 101.888 Q43 ( H3 H4) 298 x4 ( Hvap Hliq) (Btu/lbm)
  • 299. (d) Heat rejection, Step 2--1: W21 ( H2 W14 ( H4 0 H3) W32 14.278 H1) W14 2.825 H2) Q43 W14 W32 (f) (Btu/lbm) 71.223 0 W32 (e) ( H1 Q21 W43 Q21 5.219 Note that the first law is satisfied: Q 9.7 Q21 W Q43 W32 Q W14 W TC 298.15 K TH 523.15 K (Engine) T'C 273.15 K T'H 298.15 K 0 (Refrigerator) By Eq. (5.8): Carnot By Eq. (9.3): Carnot By definition: = But TH T'H T'C Wengine 0.43 Carnot 10.926 = QH Given that: 0.6 QH Wrefrig kJ sec 35 QH 7.448 Carnot 0.6 Carnot 20.689 Q'C Q'C Q'C QH Carnot Q'C Carnot T'C Wengine = Wrefrig Whence QH TC 1 kJ Ans. sec 299 Carnot kJ sec Ans. 6.556
  • 300. 9.8 (a) QC kJ sec 4 W QC (c) QC TH TC ( 40 TC 1 5.5 TH TC TH kJ QH W TC = Ans. 2.667 W (b) QH 1.5 kW 227.75 K Ans. sec 273.15)K TH 313.15 K Ans. or -45.4 degC 9.9 The following vectors contain data for parts (a) through (e). Subscripts refer to Fig. 9.1. Values of H2 and S2 for saturated vapor come from Table 9.1. 489.67 600 479.67 0.78 500 469.67 rankine 0.77 459.67 0.76 300 449.67 T2 0.79 0.75 200 107.320 H2 0.22325 104.471 103.015 Btu lbm S2 Btu lbm rankine 0.22647 539.67 rankine S'3 = S2 0.22418 0.22525 101.542 T4 400 0.22244 105.907 QdotC H4 37.978 Btu lbm (isentropic compression) 300 From Table 9.1 for sat. liquid Btu sec
  • 301. The saturation pressure at Point 4 from Table 9.1 is 101.37(psia). For isentropic compression, from Point 2 to Point 3', we must read values for the enthalpy at Point 3' from Fig. G.2 at this pressure and at the entropy values S2. This cannot be done with much accuracy. The most satisfactory procedure is probably to read an enthalpy at S=0.22 (H=114) and at S=0.24 (H=126) and interpolate linearly for intermediate values of H. This leads to the following values (rounded to 1 decimal): 115.5 116.0 Btu lbm H23 H'3 H2 117.2 117.9 H3 H2 H1 116.5 H'3 H23 H4 88.337 kJ kg H23 273.711 30.098 H1 24.084 276.438 36.337 kJ kg H3 279.336 43.414 283.026 50.732 kJ kg 286.918 8.653 mdot 7.361 QdotC H2 mdot H1 6.016 lbm sec Ans. 4.613 3.146 689.6 595.2 QdotH mdot H4 H3 QdotH Btu sec 494 Ans. 386.1 268.6 94.5 100.5 Wdot mdot H23 Wdot 99.2 90.8 72.4 301 kW Ans.
  • 302. 6.697 QdotC 5.25 Wdot 4.256 Ans. 3.485 2.914 TC T2 TH T4 9.793 7.995 TC Carnot TH Carnot TC Ans. 6.71 5.746 4.996 9.10 Subscripts in the following refer to Fig. 9.1. All property values come from Tables F.1 and F.2. T2 ( 4 QdotC H4 273.15)K 1200 142.4 kJ sec kJ kg T4 ( 34 H2 2508.9 S'2 = S2 273.15)K kJ kg 0.76 S2 9.0526 kJ kg K (isentropic compression) The saturation pressure at Point 4 from Table F.1 is 5.318 kPa. We must find in Table F.2 the enthalpy (Point 3') at this pressure and at the entropy S2. This requires double interpolation. The pressure lies between entries for pressures of 1 and 10 kPa, and linear interpolation with P is unsatisfactory. Steam is here very nearly an ideal gas, for which the entropy is linear in the logarithm of P, and interpolation must be in accord with this relation. The enthalpy, on the other hand, changes very little with P and can be interpolated linearly. Linear interpolation with temperture is satisfactory in either case. The result of interpolation is H'3 2814.7 kJ kg H23 H23 H'3 H2 402.368 302 kJ kg H1 H4
  • 303. H3 H2 H23 H3 QdotC mdot H2 mdot H1 QdotH mdot H4 Wdot mdot H23 QdotC H3 0.507 9.11 T2 kg Ans. sec kJ Ans. sec 204 kW Ans. T2 T4 kg 1404 QdotH Wdot 10 Ans. 5.881 Wdot Carnot 3 kJ 2.911 Carnot 9.238 Ans. Parts (a) & (b): subscripts refer to Fig. 9.1 At the conditions of Point 2 [t = -15 degF and P = 14.667(psia)] for sat. liquid and sat. vapor from Table 9.1: Hliq 7.505 Btu lbm Hvap 303 100.799 Btu lbm H2 Hvap
  • 304. Sliq Btu lbm rankine 0.01733 Svap 0.22714 Btu lbm rankine For sat. liquid at Point 4 (80 degF): H4 37.978 Btu S4 lbm (a) Isenthalpic expansion: QdotC Btu sec 5 H1 0.07892 H2 mdot H1 (b) Isentropic expansion: S1 x1 Sliq Svap H1 Sliq QdotC mdot H2 S1 Hliq x1 Hvap mdot H1 lbm rankine H4 QdotC mdot Btu 0.0759 0.0796 sec Ans. S4 H1 Hliq lbm lbm 34.892 BTU lbm Ans. sec (c) The sat. vapor from the evaporator is superheated in the heat exchanger to 70 degF at a pressure of 14.667(psia). Property values for this state are read (with considerable uncertainty) from Fig. G.2: H2A mdot 117.5 Btu lbm H4 0.262 mdot QdotC H2A Btu S2A 0.0629 lbm rankine lbm sec Ans. (d) For isentropic compression of the sat. vapor at Point 2, S3 Svap and from Fig. G.2 at this entropy and P=101.37(psia) H3 118.3 Btu lbm Eq. (9.4) may now be applied to the two cases: In the first case H1 has the value of H4: H2 a H4 H3 H2 a 304 3.5896 Ans.
  • 305. In the second case H1 has its last calculated value [Part (b)]: H2 H1 H3 b H2 Ans. 3.7659 b In Part (c), compression is at constant entropy of 0.262 to the final pressure. Again from Fig. G.2: H3 138 Btu H3 Wdot lbm Wdot 1.289 QdotC c 9.12 c Wdot H2A mdot (Last calculated value of mdot) BTU sec Ans. 3.8791 Subscripts: see figure of the preceding problem. At the conditions of Point 2 [sat. vapor, t = 20 degF and P = 33.110(psia)] from Table 9.1: H2 105.907 Btu lbm S2 0.22325 Btu lbm rankine At Point 2A we have a superheated vapor at the same pressure and at 70 degF. From Fig. G.2: H2A 116 Btu lbm S2A Btu lbm rankine 0.2435 For sat. liquid at Point 4 (80 degF): H4 37.978 Btu lbm S4 0.07892 Btu lbm R Energy balance, heat exchanger: H1 QdotC H4 H2A 2000 H2 Btu sec H1 mdot 305 27.885 BTU lbm QdotC H2 H1 mdot 25.634 lbm sec
  • 306. For compression at constant entropy of 0.2435 to the final pressure of 101.37(psia), by Fig. G.2: H'3 Wdot mdot 127 Btu lbm 0.75 mdot Hcomp 25.634 lbm sec Hcomp Wdot H'3 Hcomp 396.66 kW H2A 14.667 Btu lbm Ans. If the heat exchanger is omitted, then H1 = H4. Points 2A & 2 coincide, and compression is at a constant entropy of 0.22325 to P = 101.37(psia). mdot Wdot mdot 9.13 QdotC H2 H4 H'3 116 Btu lbm mdot Hcomp 29.443 lbm sec Hcomp Wdot H'3 Hcomp 418.032 kW H2 13.457 Btu lbm Ans. Subscripts refer to Fig. 9.1. At Point 2 [sat. vapor @ 10 degF] from Table 9.1: H2 104.471 Btu lbm S2 0.22418 Btu lbm R S'3 S2 H values for sat. liquid at Point 4 come from Table 9.1 and H values for Point 3` come from Fig. G.2. The vectors following give values for condensation temperatures of 60, 80, & 100 degF at pressures of 72.087, 101.37, & 138.83(psia) respectively. 31.239 H4 37.978 44.943 113.3 Btu lbm H'3 116.5 119.3 306 Btu lbm H1 H4
  • 307. (a) By Eq. (9.4): 8.294 H2 H'3 (b) H1 H2 H Ans. 5.528 4.014 H'3 H2 0.75 Since H = H3 H2 Eq. (9.4) now becomes H2 6.221 H1 9.14 Ans. 4.146 H 3.011 WINTER TH Wdot 293.15 1.5 QdotH = 0.75 TH TC TH TC Wdot = QdotH TH TC 250 (Guess) Given TH TC Wdot = TH 0.75 TH TC TC Find TC TC 268.94 K Ans. Minimum t = -4.21 degC 307
  • 308. SUMMER TC 298.15 QdotC 0.75 TH TC TH TC Wdot = TC QdotC TH (Guess) 300 Given Wdot 0.75 TH TC TH 322.57 K TH TC TC Find TH TH = Ans. Maximum t = 49.42 degC 9.15 and 9.16 Data in the following vectors for Pbs. 9.15 and 9.16 come from Perry's Handbook, 7th ed. H4 1033.5 785.3 kJ kg By Eq. (9.8): 9.17 H9 z 284.7 kJ kg H4 H15 H9 1186.7 H15 H15 z 1056.4 0.17 kJ kg Ans. 0.351 Advertized combination unit: TH ( 150 TH 50000 Btu hr WCarnot ( 30 TC 609.67 rankine QC TC 459.67)rankine 489.67 rankine QC 308 TH TC TC 459.67)rankine WCarnot 12253 Btu hr
  • 309. WI WI 1.5 WCarnot 18380 Btu hr This is the TOTAL power requirement for the advertized combination unit. The amount of heat rejected at the higher temperature of 150 degF is QH WI QH QC 68380 Btu hr For the conventional water heater, this amount of energy must be supplied by resistance heating, which requires power in this amount. For the conventional cooling unit, TH ( 120 459.67) rankine WCarnot QC Work TH TC WCarnot TC Work 1.5 WCarnot 9190 Btu hr Btu hr 13785 The total power required is WII 9.18 QH TC 210 WII Work T'H 82165 T'C 260 Btu hr NO CONTEST TH 255 305 By Eq. (9.3): TC TH WCarnot = TC QC I 0.65 TC T'H WI = II TC QC WII = 0.65 9.19 TH T'C QC II I Define r as the ratio of the actual work, WI + WII, to the r Carnot work: T'C 1 1 I r 1.477 Ans. II This problem is just a reworking of Example 9.3 with different values of x. It could be useful as a group project. 309
  • 310. 9.22 TH TC 290K Ws 250K TC Carnot TH 4.063 65% Carnot 6.25 Carnot TC 0.40kW Ans. QC 9.23 QC Ws 3 -3 Q 10 kgm sec H 1.625 2 Ws QC QH 2.025 kW Follow the notation from Fig. 9.1 With air at 20 C and the specification of a minimum approach T = 10 C: T1 ( 10 T4 273.15) K ( 30 T2 273.15) K T1 Calculate the high and low operating pressures using the given vapor pressure equation Guess: PL PH 1bar 2bar PL PL Given ln 4104.67 = 45.327 bar T1 5.146 ln T1 615.0 K T1 K PL PL Find PL bar 2 K 6.196 bar PH PH Given ln 4104.67 = 45.327 bar T4 5.146 ln T4 615.0 K T4 K PH Find PH PH bar 2 K 11.703 bar Calculate the heat load ndottoluene 50 kmol hr T1 ( 100 273.15) K T2 ( 20 273.15) K Using values from Table C.3 QdotC QdotC ndottoluene R ICPH T1 T2 15.133 6.79 10 177.536 kW 310 3 16.35 10 6 0
  • 311. Since the throttling process is adiabatic: But: Hliq4 = Hliq1 H4 = H1 x1 Hlv1 so: Hliq4 Hliq1 = x1 Hlv T4 and: Hliq4 Hliq1 = Vliq P4 P1 Cpliq ( T) dT T1 Estimate Vliq using the Rackett Eqn. Pc 112.80bar Vc cm 72.5 mol Tn 239.7K 273.15)K Tc Tr 0.723 Tc 0.253 405.7K 3 Zc 0.242 ( 20 Tr Hlvn 23.34 kJ mol 2 Vliq 1 Tr V c Zc 3 7 cm 27.112 mol Vliq Estimate Hlv at 10C using Watson correlation T1 Tn Tr1 Trn 0.591 Trn Tc Tc Hlv Hliq41 Hlvn Vliq PH Hliq41 1.621 1 Tr1 1 0.698 0.38 Trn PL Tr1 Hlv 20.798 R ICPH T1 T4 22.626 kJ mol 100.75 10 Hliq41 x1 kJ mol x1 Hlv 3 192.71 10 0.078 For the evaporator H12 = H2 H1 = H1vap H12 x1 ndot 1 QdotC H12 Hlv H1liq x1 Hlv = 1 H12 ndot 311 19.177 9.258 x1 Hlv kJ mol mol sec Ans. 6 0
  • 312. Chapter 10 - Section A - Mathcad Solutions 10.1 Benzene: A1 := 13.7819 B1 := 2726.81 C1 := 217.572 Toluene: A2 := 13.9320 B2 := 3056.96 C2 := 217.625 B1 A1 − Psat1 ( T) := e T degC (a) Given: x1 := 0.33 Given +C1 A2 − ⋅ kPa Psat2 ( T) := e Guess: T := 100⋅ degC B2 T degC y1 := 0.5 +C2 ⋅ kPa P := 100⋅ kPa x1⋅ Psat1 ( T) + ( 1 − x1) ⋅ Psat2 ( T) = P x1⋅ Psat1 ( T) = y1⋅ P ⎛ y1 ⎞ := Find ( y1 , P) ⎜ ⎝P⎠ (b) Given: y1 := 0.33 Given Ans. y1 = 0.545 P = 109.303 kPa Guess: T := 100⋅ degC Ans. x1 := 0.33 P := 100⋅ kPa x1⋅ Psat1 ( T) + ( 1 − x1) ⋅ Psat2 ( T) = P x1⋅ Psat1 ( T) = y1⋅ P ⎛ x1 ⎞ := Find ( x1 , P) ⎜ ⎝P⎠ (c) Given: x1 := 0.33 Given x1 = 0.169 P := 120⋅ kPa Ans. Guess: P = 92.156 kPa y1 := 0.5 Ans. T := 100⋅ degC x1⋅ Psat1 ( T) + ( 1 − x1) ⋅ Psat2 ( T) = P x1⋅ Psat1 ( T) = y1⋅ P ⎛ y1 ⎞ := Find ( y1 , T) ⎜ T⎠ ⎝ y1 = 0.542 312 Ans. T = 103.307 degC Ans.
  • 313. (d) Given: y1 := 0.33 P := 120⋅ kPa Guess: x1 := 0.33 T := 100⋅ degC x1⋅ Psat1 ( T) + ( 1 − x1) ⋅ Psat2 ( T) = P Given x1⋅ Psat1 ( T) = y1⋅ P ⎛ x1 ⎞ := Find ( x1 , T) ⎜ T⎠ ⎝ (e) Given: Given x1 = 0.173 Ans. T = 109.131 degC Ans. T := 105⋅ degC P := 120⋅ kPa Guess: x1 := 0.33 y1 := 0.5 x1⋅ Psat1 ( T) + ( 1 − x1) ⋅ Psat2 ( T) = P x1⋅ Psat1 ( T) = y1⋅ P ⎛ x1 ⎞ := Find ( x1 , y1) ⎜ y1 ⎠ ⎝ x1 = 0.282 z1 := 0.33 x1 = 0.282 y1 = 0.484 Guess: L := 0.5 V := 0.5 Given z1 = L⋅ x1 + V⋅ y1 (f) Ans. y1 = 0.484 Ans. L+V = 1 ⎛L ⎞ := Find ( L , V) ⎜ ⎝V ⎠ (g) Vapor Fraction: V = 0.238 Ans. Liquid Fraction: L = 0.762 Ans. Benzene and toluene are both non-polar and similar in shape and size. Therefore one would expect little chemical interaction between the components. The temperature is high enough and pressure low enough to expect ideal behavior. 313
  • 314. 10.2 Pressures in kPa; temperatures in degC (a) Antoine coefficients: Benzene=1; Ethylbenzene=2 A1 := 13.7819 B1 := 2726.81 C1 := 217.572 A2 := 13.9726 B2 := 3259.93 C2 := 212.300 ⎛ Psat1 ( T) := exp ⎜ A1 − B1 ⎞ T + C1 ⎠ ⎝ B2 ⎞ ⎛ Psat2 ( T) := exp ⎜ A2 − T + C2 ⎠ ⎝ P-x-y diagram: T := 90 P ( x1) := x1⋅ Psat1 ( T) + ( 1 − x1) ⋅ Psat2 ( T) T-x-y diagram: y1 ( x1) := x1⋅ Psat1 ( T) P ( x1 ) P' := 90 Guess t for root function: t := 90 T ( x1) := root ⎡x1⋅ Psat1 ( t) + ( 1 − x1) ⋅ Psat2 ( t) − P' , t⎤ ⎣ ⎦ y'1 ( x1) := x1⋅ Psat1 ( T ( x1) ) x1⋅ Psat1 ( T ( x1) ) + ( 1 − x1) ⋅ Psat2 ( T ( x1) ) x1 := 0 , 0.05 .. 1.0 150 140 130 P ( x1 ) 120 100 T ( x1 ) 110 100 T ( x1 ) 90 P ( x1 ) 50 80 70 0 0 0.5 x1 , y1 ( x1 ) 60 1 314 0 0.5 x1 , y'1 ( x1 ) 1
  • 315. (b) Antoine coefficients: 1-Chlorobutane=1; Chlorobenzene=2 A1 := 13.7965 B1 := 2723.73 C1 := 218.265 A2 := 13.8635 B2 := 3174.78 C2 := 211.700 ⎛ Psat1 ( T) := exp ⎜ A1 − ⎝ P-x-y diagram: ⎛ B1 ⎞ T + C1 ⎠ ⎝ T := 90 P ( x1) := x1⋅ Psat1 ( T) + ( 1 − x1) ⋅ Psat2 ( T) T-x-y diagram: B2 ⎞ T + C2 ⎠ Psat2 ( T) := exp ⎜ A2 − y1 ( x1) := x1⋅ Psat1 ( T) P ( x1 ) P' := 90 Guess t for root function: t := 90 T ( x1) := root ⎡x1⋅ Psat1 ( t) + ( 1 − x1) ⋅ Psat2 ( t) − P' , t⎤ ⎣ ⎦ y'1 ( x1) := x1⋅ Psat1 ( T ( x1) ) x1⋅ Psat1 ( T ( x1) ) + ( 1 − x1) ⋅ Psat2 ( T ( x1) ) x1 := 0 , 0.05 .. 1.0 130 160 122.5 115 113.33 107.5 T ( x1 ) 100 T ( x1 ) P ( x1 ) P ( x1 ) 92.5 66.67 85 77.5 20 0 0.5 x1 , y1 ( x1 ) 70 1 315 0 0.5 x1 , y'1 ( x1 ) 1
  • 316. 10.3 Pressures in kPa; temperatures in degC (a) Antoine coefficinets: n-Pentane=1; n-Heptane=2 A1 := 13.7667 B1 := 2451.88 C1 := 232.014 A2 := 13.8622 B2 := 2911.26 C2 := 216.432 ⎛ B1 ⎞ T + C1 ⎠ Psat1 ( T) := exp ⎜ A1 − ⎝ ⎛ Psat2 ( T) := exp ⎜ A2 − ⎝ ⎛ Psat1 ( T) + Psat2 ( T) ⎞ 2 ⎝ ⎠ P := ⎜ T := 55 B2 ⎞ T + C2 ⎠ P = 104.349 Since for Raoult's law P is linear in x, at the specified P, x1 must be 0.5: x1 := 0.5 y1 := x1⋅ Psat1 ( T) y1 = 0.89 P For a given pressure, z1 ranges from the liquid composition at the bubble point to the vapor composition at the dew point. Material balance: z1 = x1⋅ ( 1 − V) + y1⋅ V V ( z1) := z1 := x1 , x1 + 0.01 .. y1 z1 − x1 y1 − x1 V is obviously linear in z1: 1 x1 y1 V ( z1 ) 0.5 0 0.45 0.5 0.55 0.6 0.65 0.7 z1 316 0.75 0.8 0.85
  • 317. (b) At fixed T and z1, calculate x1, y1 and P as functions of fraction vapor (V). z1 := 0.5 Guess: x := 0.5 ⎛ Psat1 ( T) + Psat2 ( T) ⎞ 2 ⎝ ⎠ p := ⎜ y := 0.5 Three equations relate x1, y1, & P for given V: Given p = x⋅ Psat1 ( T) + ( 1 − x) ⋅ Psat2 ( T) y⋅ p = x⋅ Psat1 ( T) z1 = ( 1 − V) ⋅ x + V⋅ y f ( V) := Find ( x , y , p) x1 ( V) := f ( V) 1 y1 ( V) := f ( V) 2 Plot P, x1 and y1 vs. vapor fraction (V) P ( V) := f ( V) 3 V := 0 , 0.1 .. 1.0 150 1 100 x1 ( V) P ( V) 50 0 0.5 y1 ( V) 0 0.5 0 1 0 0.5 1 V V 10.4 Each part of this problem is exactly like Problem 10.3, and is worked in exactly the same way. All that is involved is a change of numbers. In fact, the Mathcad solution for Problem 10.3 can be converted into the solution for any part of this problem simply by changing one number, the temperature. 10.7 Benzene: A1 := 13.7819 B1 := 2726.81 C1 := 217.572 Ethylbenzene A2 := 13.9726 B2 := 3259.93 C2 := 212.300 A1 − Psat1 ( T) := e B1 T degC +C1 A2 − ⋅ kPa Psat2 ( T) := e 317 B2 T degC +C2 ⋅ kPa
  • 318. y1 := 0.70 Guess: T := 116⋅ degC P := 132⋅ kPa (a) Given: x1 := 0.35 Given x1⋅ Psat1 ( T) + ( 1 − x1) ⋅ Psat2 ( T) = P x1⋅ Psat1 ( T) = y1⋅ P ⎛T ⎞ ⎜ := Find ( T , P) ⎝P ⎠ Ans. T = 134.1 degC P = 207.46 kPa Ans. For parts (b), (c) and (d) use the same structure. Set the defined variables and change the variables in the Find statement at the end of the solve block. (b) T = 111.88⋅ deg_C P = 118.72⋅ kPa (c) T = 91.44⋅ deg_C P = 66.38⋅ kPa (d) T = 72.43⋅ deg_C P = 36.02⋅ kPa To calculate the relative amounts of liquid and vapor phases, one must know the composition of the feed. 10.8 To increase the relative amount of benzene in the vapor phase, the temperature and pressure of the process must be lowered. For parts (c) and (d), the process must be operated under vacuum conditions. The temperatures are well within the bounds of typical steam and cooling water temperatures. 10.9 ⎛ 13.7819 ⎞ (1) = benzene ⎜ A := ⎜ 13.9320 ⎟ (2) = toluene (3) = ethylbenzene ⎜ 13.9726 ⎠ ⎝ (a) n := rows ( A) Ai− Psat ( i , T) := e i := 1 .. n ⎛ 2726.81 ⎞ ⎜ B := ⎜ 3056.96 ⎟ ⎜ 3259.93 ⎠ ⎝ T := 110⋅ degC ⎛ 217.572 ⎞ ⎜ C := ⎜ 217.625 ⎟ ⎜ 212.300 ⎠ ⎝ P := 90⋅ kPa zi := 1 n Bi T degC +Ci ⋅ kPa ki := 318 Psat ( i , T) P Guess: V := 0.5
  • 319. n zi⋅ ki ∑ 1 + V⋅ ( ki − 1) V := Find ( V) V = 0.836 Given i=1 yi := xi := zi⋅ ki 1 + V⋅ ( ki − 1) = 1 Ans. Psat ( i , T) T = 110⋅ deg_C Ans. V = 0.575 ⎛ 0.441 ⎞ ⎜ y = ⎜ 0.333 ⎟ ⎜ 0.226 ⎠ ⎝ V = 0.352 ⎛ 0.238 ⎞ ⎜ x = ⎜ 0.345 ⎟ ⎜ 0.417 ⎠ ⎝ ⎛ 0.508 ⎞ ⎜ y = ⎜ 0.312 ⎟ ⎜ 0.18 ⎠ ⎝ V = 0.146 ⎛ 0.293 ⎞ ⎜ x = ⎜ 0.342 ⎟ ⎜ 0.366 ⎠ ⎝ ⎛ 0.572 ⎞ ⎜ y = ⎜ 0.284 ⎟ ⎜ 0.144 ⎠ ⎝ P = 110⋅ kPa (d) T = 110⋅ deg_C Ans. ⎛ 0.188 ⎞ ⎜ x = ⎜ 0.334 ⎟ ⎜ 0.478 ⎠ ⎝ P = 100⋅ kPa (c) ⎛ 0.371 ⎞ ⎜ y = ⎜ 0.339 ⎟ ⎜ 0.29 ⎠ ⎝ ⎛ 0.142 ⎞ ⎜ x = ⎜ 0.306 ⎟ ⎜ 0.552 ⎠ ⎝ Eq. (10.16) yi⋅ P (b) T = 110⋅ deg_C Eq. (10.17) P = 120⋅ kPa 10.10 As the pressure increases, the fraction of vapor phase formed (V) decreases, the mole fraction of benzene in both phases increases and the the mole fraction of ethylbenzene in both phases decreases. 319
  • 320. ⎛ 14.3145 ⎞ ⎛ 2756.22 ⎞ ⎛ 228.060 ⎞ 10.11 (a) (1) = acetone A := ⎜ B := ⎜ C := ⎜ (2) = acetonitrile ⎝ 14.8950 ⎠ ⎝ 3413.10 ⎠ ⎝ 250.523 ⎠ n := rows ( A) i := 1 .. n z1 := 0.75 T := ( 340 − 273.15) ⋅ degC P := 115⋅ kPa z2 := 1 − z1 Bi Ai− degC Psat ( i , T) := e Guess: T +Ci ⋅ kPa ki := Psat ( i , T) P V := 0.5 n zi⋅ ki ∑ Given i=1 1 + V⋅ ( ki − 1) V := Find ( V) = 1 Ans. V = 0.656 Eq. (10.16) yi := xi := r := Eq. (10.17) zi⋅ ki yi⋅ P Psat ( i , T) y1⋅ V z1 y1 = 0.805 Ans. x1 = 0.644 Ans. r = 0.705 1 + V⋅ ( ki − 1) Ans. (b) x1 = 0.285 y1 = 0.678 V = 0.547 r = 0.741 (c) x1 = 0.183 y1 = 0.320 V = 0.487 r = 0.624 (d) x1 = 0.340 y1 = 0.682 V = 0.469 r = 0.639 320
  • 321. 10.13 H1 := 200⋅ bar Psat2 := 0.10⋅ bar P := 1⋅ bar Assume at 1 bar that the vapor is an ideal gas. The vapor-phase fugacities are then equal to the partial presures. Assume the Lewis/Randall rule applies to concentrated species 2 and that Henry's law applies to dilute species 1. Then: y1⋅ P = H1⋅ x1 y2⋅ P = x2⋅ Psat2 P = y1⋅ P + y2⋅ P P = H1⋅ x1 + ( 1 − x1) ⋅ Psat2 x1 + x2 = 1 Solve for x1 and y1: x1 := P − Psat2 y1 := H1 − Psat2 −3 x1 = 4.502 × 10 H1⋅ x1 P Ans. y1 = 0.9 10.16 Pressures in kPa Psat1 := 32.27 Psat2 := 73.14 ( γ 1 ( x1 , x2) := exp A⋅ x2 2 ) A := 0.67 z1 := 0.65 ( γ 2 ( x1 , x2) := exp A⋅ x1 2 ) P ( x1 , x2) := x1⋅ γ 1 ( x1 , x2) ⋅ Psat1 + x2⋅ γ 2 ( x1 , x2) ⋅ Psat2 (a) BUBL P calculation: x1 := z1 x2 := 1 − x1 Pbubl := P ( x1 , x2) Pbubl = 56.745 DEW P calculation: y1 := z1 Guess: P' := Given x1 := 0.5 Ans. y2 := 1 − y1 Psat1 + Psat2 2 y1⋅ P' = x1⋅ γ 1 ( x1 , 1 − x1) ⋅ Psat1 P' = x1⋅ γ 1 ( x1 , 1 − x1) ⋅ Psat1 ... + ( 1 − x1) ⋅ γ 2 ( x1 , 1 − x1) ⋅ Psat2 ⎛ x1 ⎞ := Find ( x1 , P') ⎜ Pdew ⎠ ⎝ Pdew = 43.864 321 Ans.
  • 322. The pressure range for two phases is from the dewpoint to the bubblepoint: From 43.864 to 56.745 kPa (b) BUBL P calculation: y1 ( x1) := x1 := 0.75 x2 := 1 − x1 x1⋅ γ 1 ( x1 , 1 − x1) ⋅ Psat1 P ( x1 , 1 − x1) The fraction vapor, by material balance is: V := z1 − x1 V = 0.379 y1 ( x1) − x1 P ( x1 , x2) = 51.892 Ans. (c) See Example 10.3(e). γ 1 ( 0 , 1) ⋅ Psat1 α 12.0 := α 12.1 := Psat2 α 12.0 = 0.862 Psat1 γ 2 ( 1 , 0) ⋅ Psat2 α 12.1 = 0.226 Since alpha does not pass through 1.0 for 0<x1<1, there is no azeotrope. 10.17 Psat1 := 79.8 Psat2 := 40.5 ( γ 1 ( x1 , x2) := exp A⋅ x2 2 ) A := 0.95 ( γ 2 ( x1 , x2) := exp A⋅ x1 P ( x1 , x2) := x1⋅ γ 1 ( x1 , x2) ⋅ Psat1 + x2⋅ γ 2 ( x1 , x2) ⋅ Psat2 y1 ( x1) := x1⋅ γ 1 ( x1 , 1 − x1) ⋅ Psat1 P ( x1 , 1 − x1) (a) BUBL P calculation: Pbubl := P ( x1 , x2) x1 := 0.05 x2 := 1 − x1 Pbubl = 47.971 Ans. y1 ( x1) = 0.196 (b) DEW P calculation: Guess: y1 := 0.05 x1 := 0.1 P' := 322 y2 := 1 − y1 Psat1 + Psat2 2 2 )
  • 323. y1⋅ P' = x1⋅ γ 1 ( x1 , 1 − x1) ⋅ Psat1 Given P' = x1⋅ γ 1 ( x1 , 1 − x1) ⋅ Psat1 ... + ( 1 − x1) ⋅ γ 2 ( x1 , 1 − x1) ⋅ Psat2 ⎛ x1 ⎞ := Find ( x1 , P') ⎜ Pdew ⎠ ⎝ Pdew = 42.191 Ans. x1 = 0.0104 (c) Azeotrope Calculation: Guess: x1 := 0.8 y1 := x1 x1⋅ γ 1 ( x1 , 1 − x1) ⋅ Psat1 Given y1 = P P := Psat1 + Psat2 x1 ≥ 0 2 x1 ≤ 1 x1 = y1 P = x1⋅ γ 1 ( x1 , 1 − x1) ⋅ Psat1 + ( 1 − x1) ⋅ γ 2 ( x1 , 1 − x1) ⋅ Psat2 ⎛ xaz1 ⎞ ⎜ ⎜ yaz ⎟ := Find ( x , y , P) 1 1 ⎜ 1⎟ ⎜ Paz ⎠ ⎝ 10.18 Psat1 := 75.20⋅ kPa y1 = x1 γ2 γ1 2 = Psat1 and γi = lnγ2 = A⋅ x1 ⎛ Psat1 ⎞ ⎛ γ2⎞ ln ⎜ ⎝ γ1⎠ A := ⎝ Psat2 ⎠ 2 A = 2.0998 2 x2 − x1 For x1 := 0.6 x2 := 1 − x1 323 x2 := 1 − x1 ( 2 2 = A⋅ x1 − x2 ln ⎜ Whence P Psati x1 := 0.294 Psat2 2 lnγ1 = A⋅ x2 Ans. Psat2 := 31.66⋅ kPa At the azeotrope: Therefore ⎛ xaz1 ⎞ ⎜ ⎛ 0.857 ⎞ ⎜ ⎜ yaz ⎟ = ⎜ 0.857 ⎟ ⎜ 1⎟ ⎜ ⎜ Paz ⎝ 81.366 ⎠ ⎠ ⎝ )
  • 324. ( γ 1 := exp A⋅ x2 ) ( 2 x1⋅ γ 1⋅ Psat1 y1 := ) 2 γ 2 := exp A⋅ x1 P = 90.104 kPa P 10.19 Pressures in bars: P := x1⋅ γ 1⋅ Psat1 + x2⋅ γ 2⋅ Psat2 Ans. y1 = 0.701 Psat1 := 1.24 x1 := 0.65 A := 1.8 ( γ 1 := exp A⋅ x2 Psat2 := 0.89 x2 := 1 − x1 ) ( 2 γ 2 := exp A⋅ x1 P := x1⋅ γ 1⋅ Psat1 + x2⋅ γ 2⋅ Psat2 y1 = 0.6013 y1 := ) 2 x1⋅ γ 1⋅ Psat1 P Answer to Part (b) P = 1.671 By a material balance, V= z1 − x1 For y1 − x1 (c) 0≤V ≤1 Ans. (a) Azeotrope calculation: Guess: x1 := 0.6 y1 := x1 2 γ 1 ( x1) := exp ⎡ A⋅ ( 1 − x1) ⎤ ⎣ ⎦ Given y1 = 0.6013 ≤ z1 ≤ 0.65 P := Psat1 + Psat2 2 ( γ 2 ( x1) := exp A⋅ x1 ) 2 P = x1⋅ γ 1 ( x1) ⋅ Psat1 + ( 1 − x1) ⋅ γ 2 ( x1) ⋅ Psat2 x1⋅ γ 1 ( x1) ⋅ Psat1 P x1 ≥ 0 ⎛ x1 ⎞ ⎜ ⎜ y1 ⎟ := Find ( x1 , y1 , P) ⎜ ⎝P⎠ x1 ≤ 1 x1 = y1 ⎛ x1 ⎞ ⎛ 0.592 ⎞ ⎜ ⎜ ⎜ y1 ⎟ = ⎜ 0.592 ⎟ ⎜ ⎜ ⎝ P ⎠ ⎝ 1.673 ⎠ 324 Ans.
  • 325. 10.20 Antoine coefficients: P in kPa; T in degC Acetone(1): A1 := 14.3145 B1 := 2756.22 C1 := 228.060 Methanol(2): A2 := 16.5785 B2 := 3638.27 C2 := 239.500 ⎛ P1sat ( T) := exp ⎜ A1 − ⎝ A := 0.64 x1 := 0.175 ( ⎛ B1 ⎞ T + C1 ⎠ γ 1 ( x1 , x2) := exp A⋅ x2 2 ⎝ z1 := 0.25 ) p := 100 (kPa) ( γ 2 ( x1 , x2) := exp A⋅ x1 P ( x1 , T) := x1⋅ γ 1 ( x1 , 1 − x1) ⋅ P1sat ( T) ... + ( 1 − x1) ⋅ γ 2 ( x1 , 1 − x1) ⋅ P2sat ( T) y1 ( x1 , T) := Guesses: x1⋅ γ 1 ( x1 , 1 − x1) ⋅ P1sat ( T) P ( x1 , T ) V := 0.5 L := 0.5 F := 1 T := 100 Given F= L+V z1⋅ F = x1⋅ L + y1 ( x1 , T) ⋅ V ⎛L ⎞ ⎜ ⎜ V ⎟ := Find ( L , V , T) ⎜T ⎝ ⎠ T = 59.531 (degC) B2 ⎞ T + C2 ⎠ P2sat ( T) := exp ⎜ A2 − p = P ( x1 , T) ⎛ L ⎞ ⎛ 0.431 ⎞ ⎜ ⎜ ⎜ V ⎟ = ⎜ 0.569 ⎟ ⎜T ⎜ ⎝ ⎠ ⎝ 59.531 ⎠ y1 ( x1 , T) = 0.307 Ans. 325 2 )
  • 326. 10.22 x1 := 0.002 y1 := 0.95 B1 := 2572.0 A1 := 10.08 A1 − Psat1 ( T) := e x2 := 1 − x1 Given P := Guess: A2 := 11.63 B1 Psat2 ( T) := e 0.93 ⋅ x2 y2 := 1 − y1 Psat2 ( T) = x1⋅ γ 1⋅ Psat1 ( T) B2 := 6254.0 A2 − ⎛T⎞ ⎜ ⎝ K ⎠ ⋅ bar Psat1 ( T) y1 T := 300⋅ K 2 γ 1 := e x2⋅ γ 2⋅ y1 x1⋅ γ 1⋅ y2 T := Find ( T) P = 0.137 bar 326 Ans. B2 ⎛T⎞ ⎜ ⎝ K ⎠ ⋅ bar 0.93 ⋅ x1 2 γ 2 := e T = 376.453 K Ans.
  • 327. Problems 10.25 to 10.34 have been solved using MS-EXCEL 2000 We give the resulting spreadsheets. Problem 10.25 a) BUBL P T=-60 F (-51.11 C) Component methane ethylene ethane xi 0.100 0.500 0.400 b) DEW P T=-60 F (-51.11 C) Component methane ethylene ethane yi 0.500 0.250 0.250 c) BUBL T P=250 psia (17.24 bar) Component methane ethylene ethane T=-50 F T=-60 F T=-57 F (-49.44 C) ANSWER xi Ki yi=Ki*xi Ki yi=Ki*xi Ki yi=Ki*xi 0.120 4.900 0.588 4.600 0.552 4.700 0.564 0.400 0.680 0.272 0.570 0.228 0.615 0.246 0.480 0.450 0.216 0.380 0.182 0.405 0.194 SUM = 1.076 SUM = 0.962 SUM = 1.004 close enough d) DEW T P=250 psia (17.24 bar) Component methane ethylene ethane yi 0.430 0.360 0.210 P=200 psia P=250 psia P=215 psia (14.824 bar) ANSWER Ki yi=Ki*xi Ki yi=Ki*xi Ki yi=Ki*xi 5.600 0.560 4.600 0.460 5.150 0.515 0.700 0.350 0.575 0.288 0.650 0.325 0.445 0.178 0.380 0.152 0.420 0.168 SUM = 1.088 SUM = 0.900 SUM = 1.008 close enough P=190 psia P=200 psia (13.79 bar) ANSWER Ki xi=yi/Ki Ki xi=yi/Ki 5.900 0.085 5.600 0.089 0.730 0.342 0.700 0.357 0.460 0.543 0.445 0.562 SUM = 0.971 SUM = 1.008 close enough T=-40 F T = -50 F T = -45 F (-27.33 C) ANSWER Ki xi=yi/Ki Ki xi=yi/Ki Ki xi=yi/Ki 5.200 0.083 4.900 0.088 5.050 0.085 0.800 0.450 0.680 0.529 0.740 0.486 0.520 0.404 0.450 0.467 0.485 0.433 SUM = 0.937 SUM = 1.084 SUM = 1.005 close enough 327
  • 328. Problem 10.26 a) BUBL P Component ethane propane isobutane isopentane b) DEW P Component ethane propane isobutane isopentane c) BUBL T Component ethane propane isobutane isopentane d) DEW T Component ethane propane isobutane isopentane T=60 C (140 F) xi 0.10 0.20 0.30 0.40 P=200 psia P=50 psia P=80 psia (5.516 bar) ANSWER Ki yi=Ki*xi Ki yi=Ki*xi Ki yi=Ki*xi 2.015 0.202 6.800 0.680 4.950 0.495 0.620 0.124 2.050 0.410 1.475 0.295 0.255 0.077 0.780 0.234 0.560 0.168 0.071 0.028 0.205 0.082 0.12 0.048 SUM = 0.430 SUM = 1.406 SUM = 1.006 close enough T=60 C (140 F) yi 0.48 0.25 0.15 0.12 P=80 psia P=50 psia P=52 psia (3.585 bar) ANSWER Ki xi=yi/Ki Ki xi=yi/Ki Ki xi=yi/Ki 4.950 0.097 6.800 0.071 6.600 0.073 1.475 0.169 2.050 0.122 2.000 0.125 0.560 0.268 0.780 0.192 0.760 0.197 0.12 1.000 0.205 0.585 0.195 0.615 SUM = 1.534 SUM = 0.970 SUM = 1.010 close enough P=15 bar (217.56 psia) xi 0.14 0.13 0.25 0.48 T=220 F T=150 F T=145 F (62.78 C) ANSWER Ki yi=Ki*xi Ki yi=Ki*xi Ki yi=Ki*xi 5.350 0.749 3.800 0.532 3.700 0.518 2.500 0.325 1.525 0.198 1.475 0.192 1.475 0.369 0.760 0.190 0.720 0.180 0.57 0.274 0.27 0.130 0.25 0.120 SUM = 1.716 SUM = 1.050 SUM = 1.010 close enough P=15 bar (217.56 psia) yi 0.42 0.30 0.15 0.13 T=150 F Ki xi=yi/Ki 3.800 0.111 1.525 0.197 0.760 0.197 0.27 0.481 SUM = 0.986 T=145 F T=148 F (64.44 C) ANSWER Ki xi=yi/Ki Ki xi=yi/Ki 3.700 0.114 3.800 0.111 1.475 0.203 1.500 0.200 0.720 0.208 0.740 0.203 0.25 0.520 0.26 0.500 SUM = 1.045 SUM = 1.013 close enough 328
  • 329. Problem 10.27 FLASH Component methane ethane propane n-butane T=80 F (14.81 C) zi 0.50 0.10 0.20 0.20 V= 0.855 Ki yi 10.000 0.575 2.075 0.108 0.680 0.187 0.21 0.129 SUM = 1.000 P=250 psia (17.24 bar) Fraction condensed L= 0.145 ANSWER xi=yi/Ki 0.058 0.052 0.275 0.616 SUM = 1.001 Problem 10.28 First calculate equilibrium composition T=95 C (203 F) Component n-butane n-hexane xi 0.25 0.75 P=80 psia P=65 psia P=69 psia (4.83 bar) ANSWER Ki yi=Ki*xi Ki yi=Ki*xi Ki yi=Ki*xi 2.25 0.5625 2.7 0.675 2.6 0.633 0.45 0.3375 0.51 0.3825 0.49 0.3675 SUM = 0.9000 SUM = 1.0575 SUM = 1.0005 Close enough Now calculate liquid fraction from mole balances ANSWER z1= x1= y1= L= 0.5 0.25 0.633 0.347 Problem 10.29 FLASH Component n-pentane n-hexane n-heptane P = 2.00 atm (29.39 psia) T = 200 F (93.3 C) zi 0.25 0.45 0.30 V= 0.266 Ki yi 2.150 0.412 0.960 0.437 0.430 0.152 SUM = 1.000 Fraction condensed L= 0.73 ANSWER xi=yi/Ki 0.191 0.455 0.354 SUM = 1.000 329
  • 330. Problem 10.30 FLASH Component ethane propane n-butane T=40 C (104 F) V= 0.60 P=110 psia zi Ki yi 0.15 5.400 0.223 0.35 1.900 0.432 0.50 0.610 0.398 SUM = 1.053 Fraction condensed L= 0.40 P=100 psia xi=yi/Ki Ki yi 0.041 4.900 0.220 0.227 1.700 0.419 0.653 0.540 0.373 0.921 SUM = 1.012 ANSWER P=120 psia (8.274 bar) xi=yi/Ki Ki yi xi=yi/Ki 0.045 4.660 0.219 0.047 0.246 1.620 0.413 0.255 0.691 0.525 0.367 0.699 0.982 SUM = 0.999 1.001 Fraction condensed L= 0.80 P=40 psia xi=yi/Ki Ki yi 0.004 9.300 0.035 0.039 3.000 0.107 0.508 1.150 0.558 0.472 0.810 0.370 1.023 SUM = 1.071 ANSWER P=44 psia (3.034 bar) xi=yi/Ki Ki yi xi=yi/Ki 0.004 8.500 0.034 0.004 0.036 2.700 0.101 0.037 0.485 1.060 0.524 0.494 0.457 0.740 0.343 0.464 0.982 SUM = 1.002 1.000 Problem 10.31 FLASH Component ethane propane i-butane n-butane T=70 F (21.11 C) zi 0.01 0.05 0.50 0.44 V= 0.20 P=50 psia Ki yi 7.400 0.032 2.400 0.094 0.925 0.470 0.660 0.312 SUM = 0.907 330
  • 331. Problem 10.32 FLASH Component methane ethane propane n-butane Component methane ethane propane n-butane Component methane ethane propane n-butane T=-15 C (5 F) zi 0.30 0.10 0.30 0.30 zi 0.30 0.10 0.30 0.30 zi 0.30 0.10 0.30 0.30 Target: y1=0.8 P=300 psia V= 0.1855 Ki yi 5.600 0.906 0.820 0.085 0.200 0.070 0.047 0.017 SUM = 1.079 L= 0.8145 xi=yi/Ki 0.162 0.103 0.352 0.364 SUM = 0.982 P=150 psia V= 0.3150 Ki yi 10.900 0.794 1.420 0.125 0.360 0.135 0.074 0.031 SUM = 1.086 L= 0.6850 xi=yi/Ki 0.073 0.088 0.376 0.424 SUM = 0.960 P=270 psia (18.616 bar) V= 0.2535 L= 0.7465 ANSWER Ki yi xi=yi/Ki 6.200 0.802 0.129 0.900 0.092 0.103 0.230 0.086 0.373 0.0495 0.020 0.395 SUM = 1.000 SUM = 1.000 331
  • 332. Problem 10.33 First calculate vapor composition and temperature on top tray BUBL T: P=20 psia Component n-butane n-pentane xi 0.50 0.50 T=70 F T=60 F T=69 F (20.56 C) ANSWER Ki yi=Ki*xi Ki yi=Ki*xi Ki yi=Ki*xi 1.575 0.788 1.350 0.675 1.550 0.775 0.450 0.225 0.360 0.180 0.440 0.220 SUM = 1.013 SUM = 0.855 SUM = 0.995 close enough Using calculated vapor composition from top tray, calculate composition out of condenser FLASH P=20 psia (1.379 bar) V= 0.50 Component n-butane n-pentane L= 0.50 T=70 F zi Ki yi 0.78 1.575 0.948 0.22 0.450 0.137 SUM = 1.085 xi=yi/Ki 0.602 0.303 0.905 T=60 F (15.56 C) Ki yi 1.350 0.890 0.360 0.116 SUM = 1.007 ANSWER xi=yi/Ki 0.660 0.324 0.983 Problem 10.34 FLASH T=40 C (104 F) V= 0.60 Component methane n-butane L= 0.40 P=350 psia zi Ki yi 0.50 7.900 0.768 0.50 0.235 0.217 SUM = 0.986 P=250 psia xi=yi/Ki Ki yi 0.097 11.000 0.786 0.924 0.290 0.253 1.021 SUM = 1.038 332 ANSWER P=325 psia (7.929 bar) xi=yi/Ki Ki yi xi=yi/Ki 0.071 8.400 0.772 0.092 0.871 0.245 0.224 0.914 0.943 SUM = 0.996 1.006 close enough
  • 333. Eq. (1) 10.35 a) The equation from NIST is: Mi = ki yi P The equation for Henry's Law is:i Hi = yi P x Solving to eliminate P gives: By definition: Mi Hi ni = ns Ms Eq. (2) Mi Eq. (3) ki xi where M is the molar mass and the subscript s refers to the solvent. = Dividing by the toal number of moles gives: Mi = Combining Eqs. (3) and (4) gives: Hi = xi 1 xs Ms ki If xi is small, then x s is approximately equal to 1 and: Hi = b) For water as solvent: Ms For CO2 in H2O: By Eq. (5): Hi ki Eq. (4) xs Ms gm mol mol 1 Ms ki Eq. (5) 18.015 0.034 kg bar 1 Ms ki Hi 1633 bar Ans. The value is Table 10.1 is 1670 bar. The values agree within about 2%. 10.36 14.3145 Acetone: Psat1 ( ) T Psat2 ( ) T T degC e 14.8950 Acetonitrile 2756.22 T a) Find BUBL P and DEW P values T 50degC x1 0.5 y1 333 kPa 3413.10 degC e 228.060 0.5 250.523 kPa
  • 334. BUBLP x1 Psat1 ( T) 1 x1 Psat2 ( T) BUBLP DEWP 1 DEWP 1 y1 Psat1 ( T) y1 0.573 atm Ans. 0.478 atm Ans. Psat2 ( T) At T = 50 C two phases will form between P = 0.478 atm and 0.573 atm b) Find BUBL T and DEW T values P 0.5atm Given y1 0.5 x1 Psat1 ( T) BUBLT Given x1 1 BUBLT 1 x1 Psat1 ( T) = y1 P DEWT T 50degC x1 Psat2 ( T) = P Find ( T) x1 Guess: 0.5 Find x1 T DEWT Ans. 46.316 degC x1 Psat2 ( T) = 1 51.238 degC y1 P Ans. At P = 0.5 atm, two phases will form between T = 46.3 C and 51.2 C 10.37 Calculate x and y at T = 90 C and P = 75 kPa 13.7819 Benzene: Psat1 ( T) Psat2 ( T) T degC e 13.9320 Toluene: 2726.81 kPa 3056.96 T degC e 217.572 217.625 kPa a) Calculate the equilibrium composition of the liquid and vapor at the flash T and P T 90degC P 75kPa Guess: 334 x1 0.5 y1 0.5
  • 335. 1 x1 Psat1 ( )= y1 P T Given x1 Psat2 ( )= 1 T y1 P x1 x1 Find x1 y1 y1 y1 0.252 0.458 The equilibrium compositions do not agree with the measured values. b) Assume that the measured values are correct. Since air will not dissolve in the liquid to any significant extent, the mole fractions of toluene in the liquid can be calculated. x1 y1 0.1604 x2 0.2919 1 x1 x2 0.8396 Now calculate the composition of the vapor. y3 represents the mole fraction of air in the vapor. Guess: y2 0.5 y3 1 y2 y1 Given 1 y3 y1 y3 P Find y2 y3 y2 0.1 y1 y2 y3 = 1 0.608 y3 y2 x1 Psat2 ( )= 1 T Ans. Conclusion: An air leak is consistent with the measured compositions. 10.38 yO21 ndot yN21 0.0387 10 kmol T1 hr 16.3872 PsatH2O ( ) T e yCO21 0.7288 T2 100degC 3885.70 T degC 230.170 kPa 335 0.0775 yH2O1 25degC P 1atm 0.1550
  • 336. Calculate the mole fraction of water in the exit gas if the exit gas is saturated with water. PsatH2O T2 yH2O2 yH2O2 P 0.0315 This is less than the mole fraction of water in the feed. Therefore, some of the water will condense. Assume that two streams leave the process: a liquid water stream at rate ndotliq and a vapor stream at rate ndotvap. Apply mole balances around the cooler to calculate the exit composition of the vapor phase. ndotvap Given ndot 2 yO22 Guess: 0.0387 ndot = ndotliq ndot 2 ndotliq yN22 0.7288 yCO22 ndotvap 0.0775 Overall balance ndot yO21 = ndotvap yO22 O2 balance ndot yN21 = ndotvap yN22 N2 balance ndot yCO21 = ndotvap yCO22 CO2 balance yO22 yN22 yCO22 yH2O2 = 1 Summation equation ndotliq ndotvap yO22 Find ndotliq ndotvap yO22 yN22 yCO22 yN22 yCO22 ndotliq yO22 1.276 0.044 kmol hr yN22 ndotvap 0.835 8.724 yCO22 336 kmol hr 0.089 yH2O2 0.031
  • 337. Apply an energy balance around the cooler to calculate heat transfer rate. HlvH2O Qdot 40.66 kJ T1 mol T1 273.15K T2 ndotvap yO22 R ICPH T1 T2 3.639 0.506 10 3 ndotvap yN22 R ICPH T1 T2 3.280 0.539 10 3 ndotvap yH2O2 R ICPH T1 T2 3.470 1.450 10 HlvH2O ndotliq 0.227 10 5 0 0.040 10 3 3 0 0 0.121 10 10.39 Assume the liquid is stored at the bubble point at T = 40 F Taking values from Fig 10.14 at pressure: xC3 0.05 KC3 3.9 xC4 0.85 KC4 0.925 xC5 0.10 KC5 0.23 The vapor mole fractions must sum to 1. xC3 KC3 xC4 KC4 xC5 KC5 1.004 337 P 18psia 5 1.157 10 Ans. 19.895 kW 273.15K 5 0 ndotvap yCO22 R ICPH T1 T2 5.457 1.045 10 Qdot T2 Ans. 5
  • 338. 10.40 H2S + 3/2 O2 -> H2O + SO2 By a stoichiometric balance, calculate the following total molar flow rates kmol Feed: ndotH2S 10 Products ndotSO2 ndotH2S hr 3 ndotH2S 2 ndotO2 ndotH2O ndotH2S Exit conditions: P T2 1atm 3885.70 16.3872 70degC PsatH2O ( ) T T degC e 230.170 kPa a) Calculate the mole fraction of H2O and SO2 in the exiting vapor stream assuming vapor is saturated with H2O PsatH2O T2 yH2Ovap ySO2 P 1 yH2Ovap yH2Ovap ySO2 Ans. 0.308 Ans. 0.692 b) Calculate the vapor stream molar flow rate using balance on SO 2 ndotvap ndotSO2 ndotvap ySO2 14.461 kmol hr Ans. Calculate the liquid H2O flow rate using balance on H2O ndotH2Ovap ndotH2Oliq ndotH2Ovap ndotvap yH2Ovap ndotH2O ndotH2Ovap 338 ndotH2Oliq 4.461 5.539 kmol hr kmol Ans. hr
  • 339. 10.41 NCL a) 0.01 YH2O yH2O b) P kg Mair MH2O YH2O 1 YH2O PsatH2O ( ) T e 10.42 ndot1 Tdp kmol hr Tdp1 T degC e PsatH2O Tdp1 P 1.606 kPa Ans. ppH2O yH2O P 230.170 Guess: kPa Tdp 32degF 20degC Tdp2 Tdp T 20degC Find ( ) T 57.207 degF Ans. P 10degC y1 MH2O 230.170 18.01 kPa y2 0.023 By a mole balances on the process Guess: ndot2liq mol 1atm 3885.70 16.3872 y1 T 14.004 degC PsatH2O ( ) T gm Ans. yH2O P = PsatH2O ( ) Tdp T 50 29 3885.70 degC Given Tdp 0.0158 ppH2O 1atm Mair 0.0161 yH2O 16.3872 c) 18.01 YH2O kg NCL gm mol MH2O ndot1 ndot2vap ndot1 339 PsatH2O Tdp2 P y2 gm mol 0.012
  • 340. ndot1 y1 = ndot2vap y2 Given ndot1 = ndot2vap ndot2liq kmol hr ndot2liq ndot2vap 49.441 mdot2liq ndot2liq MH2O 10.43Benzene: mdot2liq Given kmol 10.074 hr kg hr Ans. 13.7819 B1 2726.81 C1 217.572 A2 13.6568 B2 2723.44 C2 220.618 exp A1 B1 T degC Guess: T 0.559 A1 Cyclohexane: Psat2 ( T) Overall balance ndot2liq Find ndot2liq ndot2vap ndot2vap Psat1 ( T) H2O balance ndot2liq exp A2 kPa C1 B2 T degC kPa C2 66degC Psat1 ( T) = Psat2 ( T) T Find ( T) The Bancroft point for this system is: Psat1 ( T) 39.591 kPa Com ponent1 Benzene 2- t Bu anol Acet ti oni le r Com ponent2 Cyclohexane W at er Et hanol T 52.321 degC T (C) 52.3 8 .7 7 65.8 340 P ( Pa) k 39.6 64 .2 60 .6 Ans.
  • 341. Chapter 11 - Section A - Mathcad Solutions 11.1 For an ideal gas mole fraction = volume fraction 3 CO2 (1): x1 0.7 V1 0.7m N2 (2): x2 0.3 V2 0.3m i P 1bar T 1 2 P 3 ( 25 273.15) K Vi i n n RT S nR 40.342 mol S xi ln xi 204.885 J K Ans. i 11.2 For a closed, adiabatic, fixed-volume system, U =0. Also, for an ideal gas, U = Cv T. First calculate the equilibrium T and P. nN2 4 mol TN2 [75 ( 273.15)K] PN2 30 bar nAr 2.5 mol TAr ( 130 273.15)K PAr 20 bar TN2 348.15 K TAr 403.15 K ntotal nN2 x1 nAr x1 CvAr CpAr CvAr CvN2 3 R 2 R CvN2 nN2 ntotal 0.615 x2 1 2 nAr ntotal x2 0.385 T T 5 R 2 CpN2 i R Find T after mixing by energy balance: T TN2 Given TAr 2 nN2 CvN2 T (guess) TN2 = nAr CvAr TAr 341 Find T) (
  • 342. T 273.15 K 90 degC Find P after mixing: PN2 P PAr (guess) 2 Given nN2 nAr R T P = nN2 R TN2 nAr R TAr PN2 PAr P Find ( P) P 24.38 bar Calculate entropy change by two-step path: 1) Bring individual stream to mixture T and P. 2) Then mix streams at mixture T and P. SN2 SAr nAr CpAr ln T TN2 nN2 CpN2 ln Smix ntotal T TAr R R ln SN2 P PAr xi ln xi J K SAr P PN2 11.806 9.547 J K Smix R ln 36.006 J K i S 11.3 SN2 mdotN2 molwtN2 SAr 2 Smix kg sec molarflowtotal 38.27 mdotH2 28.014 molarflowN2 S gm mol mdotN2 molwtN2 molarflowN2 molwtH2 J K 0.5 Ans. kg sec 2.016 molarflowH2 gm mol i mdotH2 molwtH2 molarflowH2 molarflowtotal 342 1 2 319.409 mol sec
  • 343. y1 molarflowN2 y1 molarflowtotal S R molarflowtotal molarflowH2 y2 0.224 y2 molarflowtotal S yi ln yi J secK 1411 0.776 Ans. i 11.4 T1 T2 448.15 K P1 308.15 K P2 3 bar 1 bar For methane: MCPHm MCPSm 3 MCPH T1 T2 1.702 9.081 10 MCPS T1 T2 1.702 9.081 10 6 2.164 10 3 6 2.164 10 0.0 0.0 For ethane: MCPHe MCPH T1 T2 1.131 19.225 10 MCPSe MCPS T1 T2 1.131 19.225 10 3 6 5.561 10 3 6 5.561 10 0.0 0.0 MCPHmix 0.5 MCPHm 0.5 MCPHe MCPHmix 6.21 MCPSmix 0.5 MCPSm 0.5 MCPSe MCPSmix 6.161 H R MCPHmix T2 S R MCPSmix ln H T1 T2 T1 R ln P2 7228 J mol R 2 0.5 ln ( ) 0.5 P1 The last term is the entropy change of UNmixing J T 300 K S 15.813 mol K Wideal H T Wideal S 2484 J mol Ans. 11.5 Basis: 1 mole entering air. y1 0.21 y2 0.79 t 0.05 T 300 K Assume ideal gases; then H= 0 The entropy change of mixing for ideal gases is given by the equation following Eq. (11.26). For UNmixing of a binary mixture it becomes: 343
  • 344. S R y1 ln y1 S y2 ln y2 By Eq. (5.27): Wideal By Eq. (5.28): Work T 4.273 J mol K Wideal Wideal 25638 t 0 40 mol Ans. 0.942 60 P 0.970 J mol 0.985 20 10 1.000 10 11.16 3 J 1.282 Work S 0.913 80 Z bar ln 1 0 end 1 rows ( P) 1 0.885 100 0.869 200 0.765 300 0.762 400 0.824 500 0.910 i 2 end Zi Fi 1 Pi Fi is a well behaved function; use the trapezoidal rule to integrate Eq. (11.35) numerically. Fi Ai i Fi 1 Pi 2 Pi 1 ln i fi exp ln i ln i 1 Ai i Pi Generalized correlation for fugacity coefficient: For CO2: Tc Pc 304.2 K T ( 150 273.15) K Tr T Tc 0.224 73.83 bar Tr P G ( P) exp Pc Tr B0 Tr B1 Tr 344 fG ( P) G ( P) P 1.391
  • 345. Pi fi bar bar i 10 0.993 9.925 20 0.978 19.555 40 0.949 37.973 60 0.922 55.332 80 0.896 71.676 100 0.872 87.167 200 0.77 153.964 300 0.698 209.299 400 0.656 262.377 500 0.636 317.96 Calculate values: 400 300 fi 0.8 i bar G Pi f G Pi 200 0.6 0.4 bar 0 200 400 100 0 600 0 200 400 Pi Pi bar 600 bar Agreement looks good up to about 200 bar (Pr=2.7 @ Tr=1.39) 11.17 For SO2: Tc 430.8 K Pc T Tr T Tc 600 K P Tr 1.393 Pr 78.84 bar 0.245 300 bar P Pc Pr For the given conditions, we see from Fig. 3.14 that the Lee/Kesler correlation is appropriate. 345 3.805
  • 346. Data from Tables E.15 & E.16 and by Eq. (11.67): 0 0.672 1.354 1 f GRRT P f 417.9 K a) At 280 degC and 20 bar: Tr ( ) T T Tc T ln GRRT 217.14 bar Tc 11.18 Isobutylene: 0.724 0 1 Pc ( 280 0.194 40.00 bar P 273.15)K Pr ( ) P Tr ( ) 1.3236 T Ans. 0.323 P Pc 20 bar Pr ( ) 0.5 P At these conditions use the generalized virial-coeffieicnt correlation. f PHIB Tr ( )Pr ( ) T P f P b) At 280 degC and 100 bar: T ( 280 Ans. 18.76 bar P 273.15)K 100 bar Pr ( ) 2.5 P Tr ( ) 1.3236 T At these conditions use the Lee/Kesler correlation, Tables E.15 & E.16 and Eq. (11.67). 1 0.7025 0 1 f 0.732 0 f 1.2335 P Ans. 73.169 bar 11.19 The following vectors contain data for Parts (a) and (b): (a) = Cyclopentane; (b) = 1-butene Tc Zc T 511.8 420.0 K 0.273 Vc 0.277 383.15 393.15 Pc K P 45.02 40.43 239.3 34 346 0.191 3 258 275 0.196 bar bar cm mol Tn Psat 322.4 266.9 5.267 25.83 K bar
  • 347. T Tr 0.7486 Tr Tc Psat Pc Psatr 0.9361 Psatr 0.117 0.6389 Calculate the fugacity coefficient at the vapor pressure by Eq. (11.68): (a) PHIB Tr Psatr (b) PHIB Tr Psatr 1 2 1 2 0.900 2 1 0.76 Eq. (3.72), the Rackett equation: T Tc Tr 0.749 Tr 0.936 Eq. (11.44): 2 Vsat f f 11.21 V c Zc 1 Tr PHIB Tr Psatr 11.78 20.29 107.546 cm3 133.299 mol 7 Vsat Psat exp Vsat ( Psat) P RT Ans. bar Table F.1, 150 degC: Psat molwt 476.00 kPa 18 3 cm molwt gm Vsat 1.091 Vsat cm 19.638 mol T ( 150 T P 273.15)K 150 bar = 1.084 Ans. 423.15 K 3 Equation Eq. (11.44) with r exp Vsat P RT Psat satPsat = fsat r 1.084 347 r= f fsat gm mol
  • 348. 11.22 The following vectors contain data for Parts (a) and (b): molwt 18 gm mol Table F.2: (a) 9000 kPa & 400 degC; (b) 1000(psia) & 800 degF: T1 ( 400 ( 800 459.67) rankine 3121.2 H1 273.15) K J gm 6.2915 S1 Btu 1389.6 lbm 1.5677 J gm K Btu lbm rankine Table F.2: (a) 300 kPa & 400 degC; (b) 50(psia) & 800 degF: T2 3275.2 H2 J gm 8.0338 S2 Btu 1431.7 lbm 1.9227 J gm K Btu lbm rankine Eq. (A) on page 399 may be recast for this problem as: r (a) exp r= molwt R f2 f1 H2 H1 S2 T1 = 0.0377 (b) r= 0.0377 r S1 f2 0.0542 = 0.0542 f1 Ans. 11.23 The following vectors contain data for Parts (a), (b), and (c): (a) = n-pentane (b) = Isobutylene (c) = 1-Butene: 469.7 Tc 417.9 K 33.70 Pc 40.0 420.0 Zc 313.0 0.194 bar 40.43 0.270 0.252 0.275 0.277 Vc 238.9 239.3 348 0.191 309.2 3 cm mol Tn 266.3 266.9 K T1
  • 349. 200 P 1.01325 300 bar Psat 1.01325 bar 150 Tr 1.01325 0.6583 Tn 0.6372 Tr Tc 0.0301 Psat Pc Pr 0.6355 0.0253 Pr 0.0251 Calculate the fugacity coefficient at the nbp by Eq. (11.68): (a) (b) (c) PHIB Tr Pr 1 1 PHIB Tr Pr 1 0.9572 2 0.9618 PHIB Tr Pr 3 3 3 0.9620 2 Eq. (3.72): Eq. (11.44): 2 Vsat f V c Zc 1 Tr PHIB Tr Pr 0.2857 Psat exp Vsat ( Psat) P R Tn 2.445 f Ans. 3.326 bar 1.801 11.24 (a) Chloroform: Tc Pc 334.3 K 0.882 Trn 0.222 54.72 bar Tn 536.4 K 3 Zc T 0.293 473.15 K Eq. (3.72): Vc cm 239.0 mol Tr T Tc Vsat Tr 2 V c Zc 1 Trn 349 Psat Tn Tc Trn 22.27 bar 0.623 3 7 Vsat cm 94.41 mol
  • 350. Calculate fugacity coefficients by Eqs. (11.68): Pr ( P ) P Pc f ( P) if P Psat ( P) P ( P) if P Psat ( P) P ( P) exp Pr ( P ) B0 Tr Tr ( Psat) Psat exp Psat ( Psat) P B1 Tr Vsat ( P Psat) RT Vsat ( P exp Psat) RT 0 bar 0.5 bar 40 bar 40 f ( P) Psat bar 30 Psat bar 0.8 bar P 20 ( P) 0.6 bar 10 0 0 20 0.4 40 0 20 P P bar bar (b) Isobutane Tc 40 P bar 408.1 K Pc 36.48 bar Tn 261.4 K 0.767 Trn 0.181 3 Zc T 0.282 313.15 K Eq. (3.72): Vc Tr 262.7 T Tc Vsat cm mol Tr 2 V c Zc 1 Trn 350 Psat Tn Tc Trn 5.28 bar 0.641 3 7 Vsat cm 102.107 mol
  • 351. Calculate fugacity coefficients by Eq. (11.68): Pr ( ) P P Pc fP) ( if P Psat ( )P P () P if P Psat () ( ) P Psat P () P exp Pr ( ) P P exp B1 Tr Vsat ( Psat) P RT ( )Psat exp Psat Psat B0 Tr Tr Vsat ( P Psat) RT 0 bar 0.5 bar 10 bar 10 Psat bar fP) ( Psat bar 0.8 bar 5 P () P bar 0.6 0 0 5 0.4 10 0 P P bar bar 5 10 P bar 11.25 Ethylene = species 1; Propylene = species 2 282.3 Tc 365.6 Pc K 0.281 Zc Vc 0.289 T 423.15 K P n 2 i 30 bar 1 n 50.40 46.65 131.0 188.4 y1 0.140 3 cm mol 0.35 j 1 n 351 0.087 w bar y2 1 k 1 n y1
  • 352. By Eqs. (11.70) through (11.74) wi wj i j Tc i j 2 1 3 Vci Vc Vc j i j Tr Tci Tc j Zc i j 1.499 1.317 Tr Tc 157.966 cm3 157.966 188.4 mol 131 0.087 0.114 1.317 1.157 50.345 48.189 Pc 48.189 46.627 282.3 321.261 321.261 Tc 0.114 0.14 2 R Tc i j Vc i j i j Vc Zc j i j Pc T i j i j 1 3 3 2 Zci Zc 365.6 bar 0.281 0.285 K Zc 0.285 0.289 By Eqs. (3.65) and (3.66): B0i j B0 Bi j B0 Tr B1i j i j 0.138 0.189 0.189 0.251 B1 R Tc i j B0i j Pc i j B1i j B i j B1 Tr i j 0.108 0.085 0.085 0.046 99.181 cm3 159.43 mol 59.892 99.181 By Eq. (11.64): i j 2 Bi j hatk exp Bi i P RT 20.96 cm3 mol 20.96 0 0 Bj j Bk k 1 2 yi y j 2 i k i fhatk hatk yk P hat 0.957 0.875 352 i j j fhat 10.053 17.059 bar Ans.
  • 353. For an ideal solution , id = pure species Pr P k Pr Pck fhatid idk 0.643 exp 0.95 id idk yk P k Pr 0.595 Tr k B0k k k k 9.978 fhatid 0.873 k k B1k k Ans. bar 17.022 Alternatively, Pr Pr P i j idk Pc exp i j k k Tr B0k k T 0.873 35 bar 0.012 w 0.43 0.286 0.100 Zc 0.279 0.152 0.276 45.99 0.36 98.6 190.6 305.3 K Pc 48.72 bar 369.8 n P 373.15 K 0.21 Tc 0.95 id k k 11.27 Methane = species 1 Ethane = species 2 Propane = species 3 y k k B1k k Vc 145.5 42.48 i 3 j 1 n 200.0 k 1 n 1 n By Eqs. (11.70) through (11.74) wi i j Vc i j wj Tc i j 2 Vci 1 3 Vc j Tci Tc j Zc Zci i j 1 3 3 Pc i j 2 Zc j 2 Zc R Tc i j Vc i j i j 353 3 cm mol
  • 354. 1.958 1.547 1.406 Tr T i j 1.547 1.222 1.111 Tr Tc i j 1.406 1.111 1.009 98.6 Vc 120.533 143.378 120.533 145.5 171.308 143.378 171.308 3 cm mol 200 45.964 47.005 43.259 47.005 48.672 45.253 bar 0.056 43.259 45.253 42.428 Pc 0.012 0.056 0.082 0.082 0.126 0.152 190.6 Tc 305.3 265.488 336.006 0.126 0.286 0.282 0.281 241.226 265.488 241.226 0.1 Zc 336.006 K 0.282 0.279 0.278 0.281 0.278 0.276 369.8 By Eqs. (3.65) and (3.66): B0i j Bi j B0 Tr B1i j B1 Tr i j i j R Tc i j B0i j Pc i j B1i j i j By Eq. (11.64): i j 2 Bi j 0 Bi i Bj j 30.442 107.809 30.442 0 23.482 107.809 23.482 hatk exp P RT Bk k 1 2 yi y j 2 i k i fhatk hatk yk P 0.881 0.775 354 mol 0 i j j 1.019 hat 3 cm 7.491 fhat 13.254 bar 9.764 Ans.
  • 355. For an ideal solution , id = pure species 0.761 P Pck Prk Pr idk 0.718 exp 0.824 Prk Tr idk yk P k id 7.182 fhatid 0.88 GE RT 2.6 x1 = 13.251 bar 9.569 0.759 11.28 Given: k k B1k k k k 0.977 fhatid B0k k 1.8 x2 x1 x2 Substitute x2 = 1 - x1: (a) GE = RT .8 x1 1.8 x1 1 2 x1 = 1.8 x1 x1 3 0.8 x1 Apply Eqs. (11.15) & (11.16) for M = GE/RT: ln 1 = d GE RT dx1 GE RT ln 2 = 1 = 1.8 ln 1 = 1.8 2 x1 GE RT d x1 dx1 2 x1 2 x1 ln 2 = 2.4 x1 1.4 x1 2 2 1.6 x1 3 Ans. 3 1.6 x1 (b) Apply Eq. (11.100): GE = x1 RT 1.8 1 2 x1 x1 x1 1.4 x1 2 2 1.6 x1 1.6 x1 3 This reduces to the initial condition: 355 3 GE RT d x1 GE RT dx1 Ans.
  • 356. (c) Divide Gibbs/Duhem eqn. (11.100) by dx1: x1 d ln 1 dx1 x2 d ln 2 = 0 dx1 Differentiate answers to Part (a): d ln 1 = 2 dx1 2.8 x1 x1 d ln 1 = 2 x1 dx1 x2 d ln 1 = 1 dx1 4.8 x1 4.8 x1 2 3 2 4.8 x1 2.8 x1 x1 d ln 2 = 2 x1 dx1 2 2 x1 4.8 x1 2 These two equations sum to zero in agreement with the Gibbs/Duhem equation. (d) When x1 = 1, we see from the 2nd eq. of Part (c) that d ln 1 When x1 = 0, we see from the 3rd eq. of Part (c) that d ln 2 dx1 dx1 = 0 Q.E.D. = 0 Q.E.D. (e) DEFINE: g = GE/RT g x1 1.8 x1 ln 1 x1 1.8 ln 2 x1 x1 ln 1 () 0 1.8 2 2 3 x1 0.8 x1 2 x1 1.4 x1 2 1.6 x1 3 1.6 x1 3 ln 2 () 1 356 2.6 x1 0 0.1 1.0
  • 357. 0 g x1 1 ln 1 x1 ln 2 x1 0 ln 1 ( ) 2 ln 2 ( ) 1 3 0 0.2 0.4 0.6 0.8 x1 H H1bar H2bar 0.02715 0.09329 417.4 0.32760 534.5 0.40244 531.7 0.56689 421.1 0.63128 x1 265.6 0.17490 11.32 87.5 347.1 0.66233 0.69984 253 VE 321.7 276.4 0.72792 252.9 0.77514 190.7 0.79243 178.1 0.82954 138.4 0.86835 98.4 0.93287 37.6 0.98233 10.0 357 n x1 rows x1 i 1 n 0 0.01 1
  • 358. (a) Guess: a x1 1 2 x1 F x1 3000 x1 1 3000 c a x1 3 x1 1 b a b 3.448 3 10 b 3.202 10 c linfitx1 VE F c x1 250 3 Ans. 244.615 600 400 VEi x1 ( 1 x1) a b x1 c ( x1) 2 200 0 0 0.2 0.4 0.6 0.8 x1 x1 i By definition of the excess properties E V = x1 x2 a b x1 d 3 E V = 4 c x1 dx1 Vbar1 Vbar2 E E = x2 = x1 2 2 c x1 3 (c b) x1 a a 2 2 b x1 b 2 3 c x1 2 (b c) x1 2 (b a) x1 a 2 3 c x1 2 (b) To find the maximum, set dVE/dx1 = 0 and solve for x1. Then use x1 to find VEmax. Guess: x1 0.5 Given 4 c ( x1) x1 3 3 (c Find ( x1) b) ( x1) x1 2 2 (b 0.353 a) x1 Ans. 358 a= 0
  • 359. VEmax x1 ( 1 x1) a (c) VEbar1 ( ) x1 ( 1 x1 c x1 ( ) a x1 b VEmax 2 b x1 2 x1) a 2 VEbar2 ( ) x1 2 b x1 Ans. 3c( ) x1 2( b 536.294 2 c)x1 2 3c( ) x1 0 0.01 1 4000 2000 VEbar 1 ( ) x1 x1 VEbar 2 ( ) 0 2000 0 0.2 0.4 0.6 0.8 x1 x1 Discussion: a) Partial property for species i goes to zero WITH ZERO SLOPE as xi -> 1. b) Interior extrema come in pairs: VEbar min for species 1 occurs at the same x1 as VEbar max for species 2, and both occur at an inflection point on the VE vs. x1 plot. c) At the point where the VEbar lines cross, the VE plot shows a maximum. 11.33 Propane = 1; n-Pentane = 2 T B ( 75 273.15)K 276 466 466 809 By Eq. (11.61): P 2 bar y1 n 2 i 0.5 y2 1 1 n j 1 n y1 3 cm mol B yi y j Bi j i 359 j 3 B 504.25 cm mol
  • 360. Use a spline fit of B as a function of T to find derivatives: 331 b11 980 3 cm 276 b22 mol 809 b12 mol 466 684 235 50 t 558 3 cm 3 cm mol 399 323.15 75 t 273.15 K 348.15 K 100 373.15 3 lspline ( t b11) B11 ( T) vs11 interp ( vs11 t b11 T) B11 ( T) cm 276 mol B22 ( T) cm 809 mol B12 ( T) cm 466 mol 1.92 3.18 3 3 lspline ( t b22) B22 ( T) vs22 interp ( vs22 t b22 T) 3 lspline ( t b12) B12 ( T) vs12 dBdT interp ( vs12 t b12 T) d d B12 ( T) B11 ( T) dT dT d B12 ( T) dT dBdT d B22 ( T) dT cm 3.18 5.92 mol K 3 Differentiate Eq. (11.61): dBdT yi y j dBdTi dBdT j i By Eq. (3.38): Z 1 By Eq. (6.55): HRRT By Eq. (6.56): BP RT P R 3 V 13968 cm mol HR 348.037 360 ZR T P V 0.965 dBdT HRRT P dBdT R SRR j Z B T cm 3.55 mol K 0.12 SRR 0.085 J mol SR HR SR 0.71 HRRT R T SRR R J mol K Ans.
  • 361. 11.34 Propane = 1; n-Pentane = 2 T ( 75 273.15)K P 2 bar y1 3 276 ij cm n 466 B 466 2 809 mol 0.5 y2 j 1 n i 1 n 2 Bi j Bii B j j By Eqs. (11.63a) and (11.63b): hat1 ( ) y1 exp P B1 1 RT hat2 ( ) y1 exp P B2 2 RT y1 ( 1 2 y1 2 y1) 1 2 1 2 0 0.1 1.0 1 0.99 0.98 hat1 ( ) y1 0.97 hat2 ( ) y1 0.96 0.95 0.94 0 0.2 0.4 0.6 y1 361 0.8 1 y1
  • 362. 0.0426 0.0817 45.7 0.1177 66.5 0.1510 86.6 0.2107 118.2 0.2624 144.6 0.3472 11.36 23.3 176.6 0.4158 x1 195.7 HE 0.5163 0.6156 85.6 0.9276 43.5 0.9624 22.6 (a) Guess: a x1 1 2 x1 3 x1 0 0.01 1 116.8 0.8650 1 n 141.0 0.8181 i 174.1 0.7621 rows x1 x1 191.7 0.6810 F x1 n 204.2 x1 1 x1 1 x1 500 b 100 c 0.01 a b linfit x1 HE F c 539.653 b 1.011 10 c a 913.122 0 100 HEi x1 ( 1 x1) a b x1 c ( x1) 2 200 300 0 0.2 0.4 0.6 x1 x1 i 362 0.8 3 Ans.
  • 363. By definition of the excess properties E H = x1 x2 a d b x1 E dx1 H = 4 c x1 Hbar1 Hbar2 E E 2 = x2 3( c b) x1 a 2 b x1 a 2 = x1 3 2 c x1 b 2 2( b 3 c x1 2( b a)x1 a 2 c)x1 3 c x1 2 (b) To find the minimum, set dHE/dx1 = 0 and solve for x1. Then use x1 to find HEmin. Guess: x1 HE ( ) x1 3 Given x1 0.5 4c( ) x1 Find ( ) x1 HEmin x1 x1 ( 1 x1) a 3( c x1 ( 1 2 b)( ) x1 x1) a 2( b a)x1 x1 2 b x1 c x1 HEmin HEbar1 ( ) x1 HE ( ) ( x1 1 a= 0 HE ( ) x1 x1 204.401 d x1) HE ( ) x1 dx1 d HE ( ) x1 dx1 0 0.01 1 500 HEbar 1 ( ) x1 x1 HEbar 2 ( ) 0 500 1000 0 2 c x1 Ans. 0.512 HEbar2 ( ) x1 (c) b x1 0.2 0.4 0.6 x1 363 0.8 Ans.
  • 364. Discussion: a) Partial property for species i goes to zero WITH ZERO SLOPE as xi -> 1. b) Interior extrema come in pairs: HEbar min for species 1 occurs at the same x1 as HEbar max for species 2, and both occur at an inflection point on the H E vs. x1 plot. c) At the point where the HEbar lines cross, the HE plot shows a minimum. 11.37 (a) y1 (1) = Acetone 0.28 0.307 w n 0.190 2 (2) = 1,3-butadiene y2 T 1 508.2 Tc i y1 j ki j 170 kPa 209 Vc 0.267 1 n P 273.15) K 0.233 K Zc 425.2 1 n ( 60 220.4 3 cm mol 0 0.307 0.2485 0.082 Eq. (11.70) wi wj 0.2485 2 0.19 0.082 i j 0.126 0.126 0.152 508.2 Tc Tc i Zc Eq. (11.73) Zci j j Zc i 1 ki j Zc 0 K 3 Vc i 2 209 3 j 0.25 0.267 0.276 1 3 1 Eq. (11.74) Vci j 425.2 0.233 0.25 j 2 Vc Tc 464.851 369.8 Eq. (11.71) Tci j 464.851 Vc 0 214.65 214.65 220.4 200 3 cm mol 0 47.104 45.013 Eq. (11.72) Pci j Zci j R Tci j Vci j Pc 45.013 42.826 bar 42.48 0 Note: the calculated pure species Pc values in the matrix above do not agree exactly with the values in Table B.1 due to round-off error in the calculations. 364
  • 365. Tri j Tr T Tci j 0.036 0.038 0.656 0.717 Pr 0.717 0.784 Eq. (3.65) P Pci j Pri j 0.038 0.04 0.824 B0i j 0 B0 Tri j 0.74636 B1i j 0.5405 0.27382 0.16178 Eq. (3.66) 0.16178 0.6361 B0 0.6361 0.27382 0.33295 B1 Tri j 0.874 0.558 B1 0.558 0.098 0.34 0.098 Eq. (11.69a) + (11.69b) 0.028 910.278 665.188 n 1 V 1 j RT Z P Eq. (6.89) dB0dTri j 665.188 cm3 499.527 mol 3 yi y j Bi j BP RT Z i j B1i j n B i Eq. (3.38) 0.027 R Tci j B0i j Pci j Bi j B Eq. (11.61) 0.028 V Tri j 2.6 365 598.524 mol 1 Z 0.675 B cm 0.963 1.5694 Eq. (6.90) 3 4 cm 10 mol dB1dTri j Ans. 0.722 Tri j 5.2
  • 366. Differentiating Eq. (11.61) and using Eq. (11.69a) + (11.69b) n n dBdT yi y j i 1 j 1 Eq. (6.55) HR Eq. (6.56) SR Eq. (6.54) GR PT B T R dB0dTri j Pci j dBdT i j dB1dTri j 0.727 GR BP 344.051 SR P dBdT HR 101.7 mol K J mol 3 (b) cm V = 15694 mol SR = 1.006 HR = 450.322 J mol K GR = 125.1 3 (c) V = 24255 cm mol GR = 53.3 V = 80972 cm HR = 36.48 mol SR = 0.097 J mol K GR = 8.1 J mol J mol 3 (e) cm V = 56991 mol SR = 0.647 HR = 277.96 J GR = 85.2 mol K 366 J mol J mol 3 (d) J mol J mol HR = 175.666 J SR = 0.41 mol K J mol J J mol J mol Ans. Ans. Ans.
  • 367. Data for Problems 11.38 - 11.40 325 15 308.3 61.39 .187 200 100 150.9 48.98 .000 575 40 562.2 48.98 .210 73.83 .224 50.40 .087 350 T 300 P 35 304.2 Tc 50 Pc 282.3 525 10 507.6 30.25 .301 225 25 190.6 45.99 .012 200 75 126.2 34.00 .038 1.054 1.325 T Tc Tr 2.042 1.023 Tr 0.244 0.817 1.151 Pr 1.063 P Pc Pr 0.474 0.992 1.034 0.331 1.18 0.544 1.585 2.206 11.38 Redlich/Kwong Equation: 0.08664 0.42748 0.02 0.133 Eq. (3.53) 3.234 0.069 Pr Tr 4.559 4.77 0.036 0.081 q 1.5 Tr Eq. (3.54) q 3.998 4.504 0.028 0.04 z 3.847 0.121 Guess: 4.691 2.473 1 367
  • 368. Given i z= 1 Ii 1 8 exp Z i fi z q Z ln i qi Eq. (3.52) Z z z i qi Z 1 ln Z i q Find z () Eq. (6.65) i qi i qi i qi Ii Eq. (11.37) i Pi Z i qi fi i 0.925 0.93 13.944 0.722 0.744 74.352 0.668 0.749 29.952 0.887 0.896 31.362 0.639 0.73 36.504 0.891 0.9 8.998 0.881 0.89 22.254 0.859 0.85 63.743 11.39 Soave/Redlich/Kwong Equation c 0.480 1.574 0.176 0.08664 2 1 c 1 0.42748 0.5 2 Tr 0.02 0.133 Eq. (3.53) 3.202 0.069 Pr Tr 4.49 4.737 0.036 0.081 q Tr Eq. (3.54) q 3.79 4.468 0.028 0.04 z 3.827 0.121 Guess: 4.62 2.304 1 368
  • 369. Given i z= 1 Ii 1 8 exp Z i fi Z z q i qi Eq. (3.52) z z ln Z i qi i i qi q Find z () Eq. (6.65) i qi ln Z 1 Z Z i qi Ii Eq. (11.37) i Pi i qi fi i 0.927 0.931 13.965 0.729 0.748 74.753 0.673 0.751 30.05 0.896 0.903 31.618 0.646 0.733 36.66 0.893 0.902 9.018 0.882 0.891 22.274 0.881 0.869 65.155 11.40 Peng/Robinson Equation 1 c 2 0.37464 1 1.54226 2 0.07779 0.26992 0.45724 2 1 c 1 0.5 Tr 0.018 0.12 Eq.(3.53) 3.946 0.062 Pr Tr 5.383 5.658 0.032 0.073 q Tr Eq.(3.54) q 4.598 5.359 0.025 5.527 0.036 4.646 0.108 2.924 369 2
  • 370. Guess: z 1 Given z = 1 i exp Z fi z i qi i Pi Z 2 1 Eq. (3.52) Z z 1 Ii 1 8 i z q ln 2 ln Z Z i qi i Z i qi q Find ( z) i i qi i qi i Eq. (6.65) qi Ii Eq. (11.37) fi i 0.918 0.923 13.842 0.69 0.711 71.113 0.647 0.73 29.197 0.882 0.89 31.142 0.617 0.709 35.465 0.881 0.891 8.91 0.865 0.876 21.895 0.845 0.832 62.363 BY GENERALIZED CORRELATIONS Parts (a), (d), (f), and (g) --- Virial equation: 325 T 350 525 308.3 Tc 225 Tr T Tc 304.2 507.6 15 P 190.6 Pr 35 10 25 61.39 73.83 .224 30.25 .301 45.99 Pc .187 .012 P Pc Evaluation of : B0 B0 ( Tr) Eq. (3.65) 370 B1 B1 ( Tr) Eq. (3.66)
  • 371. 0.675 DB0 Eq. (6.89) 2.6 0.722 DB1 5.2 Tr Eq. (6.90) Tr 0.932 Pr B0 exp Tr Eq. (11.60) B1 (a) (d) (f) (g) 0.904 0.903 0.895 Parts (b), (c), (e), and (h) --- Lee/Kesler correlation: Interpolate in Tables E.13 - E.16: .7454 0 1.1842 0.9634 0.9883 1 .7316 .8554 0.210 0.087 1.2071 .7517 0.000 0.038 0.745 0 1 (b) (c) (e) (h) 0.746 Eq. (11.67): 0.731 0.862 11.43 ndot1 x1 2 kmol hr ndot1 ndot3 ndot2 x1 4 kmol hr 0.333 ndot3 x2 1 x1 ndot1 ndot2 x2 0.667 a) Assume an ideal solution since n-octane and iso-octane are non-polar and very similar in chemical structure. For an ideal solution, there is no heat of mixing therefore the heat transfer rate is zero. b) St R x1 ln x1 x2 ln x2 ndot3 371 St 8.82 W K Ans.
  • 372. 11.44 For air entering the process: 0.21 xN21 0.79 For the enhanced air leaving the process: xO22 0.5 xN22 0.5 ndot2 50 xO21 mol sec a) Apply mole balances to find rate of air and O2 fed to process Guess: ndotair 40 mol sec ndotO2 10 mol sec Given xO21 ndotair ndotO2 = xO22 ndot2 Mole balance on N2 xN21 ndotair = xN22 ndot2 ndotair ndotO2 ndotair Mole balance on O 2 Find ndotair ndotO2 31.646 mol sec Ans. ndotO2 18.354 mol sec Ans. b) Assume ideal gas behavior. For an ideal gas there is no heat of mixing, therefore, the heat transfer rate is zero. c) To calculate the entropy change, treat the process in two steps: 1. Demix the air to O2 and N2 2. Mix the N2 and combined O2 to produce the enhanced air Entropy change of demixing S12 Entropy change of mixing S23 R xO21 ln xO21 R xO22 ln xO22 Total rate of entropy generation: SdotG SdotG 372 ndotair S12 152.919 W K xN21 ln xN21 xN22 ln xN22 ndot2 S23 Ans.
  • 373. 10 11.50 T 932.1 544.0 30 K GE 273.15K J mol 513.0 50 HE 893.4 J mol 845.9 494.2 Assume Cp is constant. Then HE is of the form: HE = c aT Find a and c using the given HE and T values. a slope ( HE) T a c intercept ( HE) T c 2.155 mol K 3 J 1.544 GE is of the form: GE = a T ln J 10 mol T K T bT c Rearrange to find b using estimated a and c values along with GE and T data. GE a T ln B T K T 13.543 c B T 13.559 J mol K 13.545 Use averaged b value 3 Bi b i 1 b 3 13.549 J mol K Now calculate HE, GE and T*SE at 25 C using a, b and c values. HE ( ) T aT HE [25 ( c T K GE ( ) T a T ln TSE ( ) T HE ( ) GE ( ) T T T bT 273.15) ] 901.242 K J Ans. mol ( c GE [25 273.15) ] 522.394 K J Ans. mol TSE [25 ( 373 273.15) ] 378.848 K J Ans. mol
  • 374. Chapter 12 - Section A - Mathcad Solutions 12.1 Methanol(1)/Water(2)-- VLE data: T 333.15 K 39.223 0.5714 42.984 0.2167 0.6268 48.852 0.3039 0.6943 52.784 P 0.1686 0.3681 0.7345 56.652 60.614 0.4461 x1 kPa 0.7742 y1 0.5282 0.8085 63.998 0.6044 0.8383 67.924 0.6804 0.8733 70.229 0.7255 0.8922 72.832 0.7776 0.9141 Number of data points: n Calculate x2 and y2: x2 rows ( ) P 1 n 10 y2 x1 i 1 n 1 y1 Vapor Pressures from equilibrium data: Psat1 84.562 kPa Psat2 19.953 kPa Calculate EXPERIMENTAL values of activity coefficients and excess Gibbs energy. y1 P 1 x1 Psat1 y2 P 2 GERT x2 Psat2 374 x1 ln 1 x2 ln 2
  • 375. 1 i ln 2 i ln 1 i 2 i i GERTi 1 1.572 1.013 0.452 0.013 0.087 2 1.47 1.026 0.385 0.026 0.104 3 1.32 1.075 0.278 0.073 0.135 4 1.246 1.112 0.22 0.106 0.148 5 1.163 1.157 0.151 0.146 0.148 6 1.097 1.233 0.093 0.209 0.148 7 1.05 1.311 0.049 0.271 0.136 8 1.031 1.35 0.031 0.3 0.117 9 1.021 1.382 0.021 0.324 0.104 10 1.012 1.41 0.012 0.343 0.086 0.2 0.4 0.5 0.4 ln 1 i 0.3 ln 2 i GERT i 0.2 0.1 0 0 0.6 x1 0.8 i (a) Fit GE/RT data to Margules eqn. by linear least squares: VXi Slope Slope x1 i slope ( VX VY) 0.208 VYi GERTi x1 x2 i i Intercept intercept ( VX VY) Intercept 0.683 A12 Intercept A21 Slope A12 0.683 A21 0.475 375 A12 Ans.
  • 376. The following equations give CALCULATED values: 2 1 ( x2) x1 2 ( x2) x1 j exp x2 exp x1 2 X1 j j X1 Y1calc 2 A21 A12 x1 A21 2 A12 A21 x2 X1 1 101 pcalc A12 j .01 j 1 X1 X2 Psat1 j j j j X2 .01 X2 j 2 X1 X2 Psat2 j j j 1 X1 X2 Psat1 j j pcalc j P-x,y Diagram: Margules eqn. fit to GE/RT data. 90 80 Pi kPa 70 Pi 60 kPa pcalc 50 j kPa pcalc kPa 40 j 30 20 10 0 0.2 0.4 0.6 x1 y1 X1 Y1calc i P-x data P-y data P-x calculated P-y calculated 376 i j 0.8 j 1 X1 j
  • 377. Pcalc x1 1 x1 x2 Psat1 i i i i y1calc x2 2 x1 x2 Psat2 i i i x1 1 x1 x2 Psat1 i i i Pcalc i i RMS deviation in P: Pi RMS 2 Pcalc i RMS n 0.399 kPa i (b) Fit GE/RT data to van Laar eqn. by linear least squares: VXi x1 VYi i x1 x2 i i GERTi Slope slope ( VX VY) Intercept intercept ( VX VY) Slope 0.641 Intercept 1.418 a12 a12 1 Intercept 0.705 1 ( x1 x2) 2 ( x1 x2) j a21 1 101 a21 a12 x1 exp a12 1 X2 j j 0.485 Ans. 2 a21 x2 exp a21 1 X1 1 ( Slope Intercept) a21 x2 a12 x1 .01 j 1 2 X1 .00999 j 377 (To avoid singularities)
  • 378. pcalc Pcalc X1 j 1 X1 X2 Psat1 j j j x1 1 x1 x2 Psat1 i i i i X1 Y1calc j j X2 j 2 X1 X2 Psat2 j j x2 2 x1 x2 Psat2 i i i 1 X1 X2 Psat1 j j pcalc y1calc x1 1 x1 x2 Psat1 i i i Pcalc i i j P-x,y Diagram: van Laar eqn. fit to GE/RT data. 90 80 Pi 70 kPa Pi 60 kPa pcalc 50 j kPa pcalc kPa 40 j 30 20 10 0 0.2 0.4 0.6 x1 y1 X1 Y1calc i i j 0.8 j P-x data P-y data P-x calculated P-y calculated RMS deviation in P: Pi RMS Pcalc n 2 i RMS i 378 0.454 kPa 1
  • 379. (c) Fit GE/RT data to Wilson eqn. by non-linear least squares. Minimize the sum of the squared errors using the Mathcad Minimize function. Guesses: SSE 0.5 12 12 GERTi 21 1.0 21 x1 ln x1 i i 2 x2 i 12 i x2 ln x2 i i x1 Minimize SSE 12 12 x1 exp Pcalc X1 j x1 X1 j X1 Y1calc j x1 21 12 21 x2 x2 12 x1 x1 21 21 .01 j .01 1 X1 X2 Psat1 j j X2 j x1 1 x1 x2 Psat1 i i i i x2 x2 1 101 x2 12 12 x1 2 ( x2) x1 pcalc x2 x1 Ans. 21 12 1 ( x2) x1 1.026 21 21 exp x2 0.476 21 12 j 21 i j X2 j 1 X1 j 2 X1 X2 Psat2 j j x2 2 x1 x2 Psat2 i i i 1 X1 X2 Psat1 j j j pcalc y1calc i x1 1 x1 x2 Psat1 i i i Pcalc i j 379
  • 380. P-x,y diagram: Wilson eqn. fit to GE/RT data. 90 80 Pi kPa 70 Pi 60 kPa pcalc 50 j kPa pcalc 40 j 30 kPa 20 10 0 0.2 0.4 0.6 x1 y1 X1 Y1calc i i j 0.8 j P-x data P-y data P-x calculated P-y calculated RMS deviation in P: Pi RMS Pcalc n 2 i RMS 0.48 kPa i (d) BARKER'S METHOD by non-linear least squares. Margules equation. Guesses for parameters: answers to Part (a). 2 1 x1 x2 A 12 A 21 exp ( ) A12 x2 2 x1 x2 A 12 A 21 exp ( ) A21 x1 2 380 2 A21 A12 x1 2 A12 A21 x2 1
  • 381. Minimize the sum of the squared errors using the Mathcad Minimize function. Guesses: A12 SSE A12 A21 0.5 A21 Pi x1 i x2 X1 j X2 X1 Y1calc A21 2 X 1 j X 2 j A 12 A 21 Psat2 j 1 X 1 j X 2 j A 12 A 21 Psat1 pcalc j 1 x1i x2i A 12 A 21 Psat1 i x2 2 x1i x2i A 12 A 21 Psat2 x1 1 x1i x2i A 12 A 21 Psat1 i y1calc 0.435 j j x1 i 0.758 1 X 1 j X 2 j A 12 A 21 Psat1 j i i Pcalc i RMS deviation in P: Pi RMS Pcalc n 2 i 2 2 x1i x2i A 12 A 21 Psat2 i Minimize SSE A12 A21 A21 Pcalc x1 x2 A12 A21 Psat1 i 1 i i A12 A12 pcalc 1.0 RMS i 381 0.167 kPa Ans.
  • 382. P-x-y diagram, Margules eqn. by Barker's method 90 80 Pi kPa 70 Pi 60 kPa pcalc 50 j kPa pcalc 40 j 30 kPa 20 10 0 0.2 0.4 0.6 x1 y1 X1 Y1calc i i j 0.8 1 j P-x data P-y data P-x calculated P-y calculated Residuals in P and y1 1 Pi Pcalc 0.5 i kPa y1 y1calc 100 i i 0 0.5 0 0.2 0.4 x1 Pressure residuals y1 residuals 382 i 0.6 0.8
  • 383. (e) BARKER'S METHOD by non-linear least squares. van Laar equation. Guesses for parameters: answers to Part (b). j X1 1 101 1 x1 x2 a12 a21 .01 j j .00999 exp a12 1 2 x1 x2 a12 a21 exp a21 1 X2 j 1 X1 j 2 a12 x1 a21 x2 2 a21 x2 a12 x1 Minimize the sum of the squared errors using the Mathcad Minimize function. Guesses: a12 SSE a12 a21 0.5 a21 Pi x1 i a12 X1 j x2 i X1 Y1calc j a12 x1 i 2 X 1 j X 2 j a12 a21 Psat2 j 1 X 1 j X 2 j a12 a21 Psat1 j 1 x1i x2i a12 a21 Psat1 i x2 2 x1i x2i a12 a21 Psat2 x1 1 x1i x2i a12 a21 Psat1 i y1calc i i Pcalc i 383 0.83 a21 j pcalc 2 2 x1i x2i a12 a21 Psat2 1 X 1 j X 2 j a12 a21 Psat1 j X2 Pcalc x1 x2 a12 a21 Psat1 i 1 i i Minimize SSE a12 a21 a21 pcalc 1.0 0.468 Ans.
  • 384. RMS deviation in P: Pi RMS Pcalc 2 i RMS n 0.286 kPa i P-x,y diagram, van Laar Equation by Barker's Method 90 80 70 Pi kPa 60 Pi kPa pcalc 50 j kPa pcalc j 40 kPa 30 20 10 0 0.2 0.4 0.6 x1 y1 X1 Y1calc i i P-x data P-y data P-x calculated P-y calculated 384 j 0.8 j 1
  • 385. Residuals in P and y1. 1 Pi Pcalc 0.5 i kPa y1 y1calc 100 i i 0 0.5 0 0.2 0.4 x1 0.6 0.8 i Pressure residuals y1 residuals (f) BARKER'S METHOD by non-linear least squares. Wilson equation. Guesses for parameters: answers to Part (c). j 1 101 1 x1 x2 X1 12 21 .01 j j exp ln x1 x2 2 x1 x2 12 21 exp .01 x2 x1 x2 x1 X1 j 21 12 x2 x1 21 21 21 12 x1 1 j 12 12 ln x2 x1 X2 x2 12 x2 x1 21 Minimize the sum of the squared errors using the Mathcad Minimize function. Guesses: 12 0.5 21 385 1.0
  • 386. SSE 12 Pi 21 x1 i x1 x2 i 1 i i x2 2 x1i x2i i 12 X1 j 1 X1 j X2 j j X2 X1 Y1calc 2 X1 j X2 j 12 21 Psat2 j 1 X1 j X2 j 12 21 Psat1 pcalc x1 1 x1i x2i i j 21 Psat1 12 x2 2 x1i x2i 12 21 Psat2 x1 1 x1i x2i 12 21 Psat1 i y1calc 1.198 21 Psat1 12 j j i 0.348 21 i i Pcalc i RMS deviation in P: Pi RMS Pcalc 2 i RMS n i 386 2 21 Psat2 12 Minimize SSE 21 Pcalc 12 21 12 pcalc 21 Psat1 12 0.305kPa Ans.
  • 387. P-x,y diagram, Wilson Equation by Barker's Method 90 80 Pi kPa 70 Pi 60 kPa pcalc 50 j kPa pcalc 40 j 30 kPa 20 10 0 0.2 0.4 0.6 x1 y1 X1 Y1calc i i j 0.8 1 j P-x data P-y data P-x calculated P-y calculated Residuals in P and y1. 1 Pi Pcalc i 0.5 kPa y1 y1calc 100 i i 0 0.5 0 0.2 0.4 x1 Pressure residuals y1 residuals 387 i 0.6 0.8
  • 388. 12.3 Acetone(1)/Methanol(2)-- VLE data: T 328.15 K 72.278 0.0647 75.279 0.0570 0.1295 77.524 0.0858 0.1848 78.951 0.1046 0.2190 82.528 0.1452 0.2694 86.762 0.2173 0.3633 90.088 0.2787 0.4184 93.206 0.3579 0.4779 95.017 P 0.0287 0.4050 0.5135 96.365 97.646 kPa 0.4480 x1 y1 0.5052 0.5512 0.5844 98.462 0.5432 0.6174 99.811 0.6332 0.6772 99.950 0.6605 0.6926 100.278 0.6945 0.7124 100.467 0.7327 0.7383 100.999 0.7752 0.7729 101.059 0.7922 0.7876 99.877 0.9080 0.8959 99.799 0.9448 0.9336 Number of data points: n Calculate x2 and y2: x2 rowsP) ( 1 x1 n y2 Vapor Pressures from equilibrium data: Psat1 96.885 kPa Psat2 388 68.728 kPa 20 i 1 n 1 y1
  • 389. Calculate EXPERIMENTAL values of activity coefficients and excess Gibbs energy. y1 P 1 x1 Psat1 ln 1 i i i GERT x2 Psat2 2 1 i y2 P 2 x1 ln 1 ln 2 i x2 ln 2 GERTi 0.013 0.027 0.568 0.011 0.043 0.544 5.815·10-3 0.052 1.002 0.534 1.975·10-3 0.058 1.58 1.026 0.458 0.026 0.089 6 1.497 1.027 0.404 0.027 0.108 7 1.396 1.057 0.334 0.055 0.133 8 1.285 1.103 0.25 0.098 0.152 9 1.243 1.13 0.218 0.123 0.161 10 1.224 1.14 0.202 0.131 0.163 11 1.166 1.193 0.153 0.177 0.165 12 1.155 1.2 0.144 0.182 0.162 13 1.102 1.278 0.097 0.245 0.151 14 1.082 1.317 0.079 0.275 0.145 15 1.062 1.374 0.06 0.317 0.139 16 1.045 1.431 0.044 0.358 0.128 17 1.039 1.485 0.039 0.395 0.119 18 1.037 1.503 0.036 0.407 0.113 19 1.017 1.644 0.017 0.497 0.061 20 1.018 1.747 0.018 0.558 0.048 1 1.682 1.013 0.52 2 1.765 1.011 3 1.723 1.006 4 1.706 5 389
  • 390. 0.6 0.