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

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

  1. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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