This is an **unformatted preview**. Sign up to view the full document.

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 0 Given t = 1.8t 32 Find t () 40 Ans. 1.5 By definition: P= F A F = mass g Note: Pressures are in gauge pressure. P 3000bar D 4mm A 4 D 2 A 12.566 mm 2 F PA g 9.807 m s 2 mass F g mass 384.4 kg Ans. 1.6 By definition: P 3000atm P= D F A 0.17in F = mass g A 4 D 2 A 0.023 in 2 F PA g 32.174 ft sec 2 mass F g mass 1000.7 lbm Ans. 1.7 Pabs = gh Patm 13.535 gm cm 3 g 9.832 m s 2 h 56.38cm Patm 101.78kPa Pabs gh Patm Pabs 176.808 kPa Ans. 1.8 13.535 gm cm 3 g 32.243 ft s 2 h 25.62in Patm 29.86in_Hg Pabs gh Patm Pabs 27.22 psia Ans. 1 1.10 Assume the following: 13.5 gm cm 3 g 9.8 m s 2 P 400bar h P g h 302.3 m Ans. 1.11 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 0.40kg g 9.81 m s 2 x 1.08cm F mass g F 3.924 N Ks F x 4 10 Ks 3 363.333 N m On Mars: x 0.40cm FMars mass FMars Kx FMars mK gMars gMars 0.01 mK kg Ans. 1.12 Given: d P= dz g and: = MP RT Substituting: d P= dz MP g RT PDenver Separating variables and integrating: Psea 1 dP = P zDenver Mg dz RT 0 After integrating: ln PDenver Psea = Mg zDenver RT Mg zDenver RT Taking the exponential of both sides and rearranging: Psea 1atm PDenver = Psea e gm mol M 29 g 9.8 m s 2 2 R 82.06 cm atm mol K 3 T ( 10 273.15) K zDenver 1 mi Mg zDenver RT Mg 0.194 PDenver Psea e RT zDenver PDenver 0.823 atm Ans. PDenver 1.13 The same proportionality applies as in Pb. 1.11. 0.834 bar Ans. gearth 32.186 ft s 2 gmoon 5.32 ft s 2 lmoon 18.76 learth lmoon gearth gmoon learth 113.498 M wmoon learth lbm M gmoon M 113.498 lbm 18.767 lbf Ans. wmoon Ans. 1.14 costbulb hr 5.00dollars 10 day 1000hr costelec hr 0.1dollars 70W 10 day kW hr costbulb 18.262 dollars yr costelec 25.567 dollars yr costtotal costbulb costelec costtotal 43.829 dollars yr Ans. 1.15 D 1.25ft mass 250lbm g 32.169 ft s 2 3 Patm 30.12in_Hg A 4 D 2 A 1.227 ft 2 (a) F Patm A mass g F 2.8642 10 lbf 3 Ans. (b) Pabs F A Pabs 16.208 psia Ans. (c) l 1.7ft Work F l Work 4.8691 3 10 ft lbf Ans. PE mass g l PE 424.9 ft lbf Ans. 1.16 D 0.47m mass 150kg g 9.813 m s 2 Patm 101.57kPa A 4 D 2 A 0.173 m 2 (a) F (b) Pabs Patm A F A mass g F Pabs 1.909 10 N 4 Ans. 110.054 kPa Ans. (c) l 0.83m Work F l Work 15.848 kJ Ans. EP mass g l EP 1.222 kJ Ans. 1.18 mass 1250kg u 40 m s EK 1 2 mass u 2 EK 1000 kJ Ans. Work EK Work 1000 kJ Ans. 1.19 Wdot = mass g h 0.91 0.92 time Wdot 200W g 9.8 m s 2 h 50m 4 mdot Wdot g h 0.91 0.92 25.00 ton mdot 0.488 kg s Ans. 1.22 a) cost_coal MJ 29 kg 2.00 gal cost_coal 0.95 GJ 1 cost_gasoline 37 GJ m 3 cost_gasoline 14.28 GJ 1 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 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. A Function being fit: f ( A B C) T e B T C First derivative of the function with respect to parameter A d f ( A B C) T dA exp A B T C First derivative of the function with respect to parameter B d f ( A B C) T dB 1 T C exp A B T C First derivative of the function with respect to parameter C d f ( A B C) T dC 18.5 9.5 0.2 11.8 t 23.1 32.7 44.4 52.1 63.3 75.5 B ( T C) 2 exp A B T C 3.18 5.48 9.45 16.9 Psat 28.2 41.9 66.6 89.5 129 187 6 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 a1 T a1 T a2 a2 Guess values of parameters 15 guess 3000 50 F ( a) T 1 exp a0 T a2 a1 T a2 2 a1 T a2 a1 T a2 exp a0 Apply the genfit function A B C Compare fit with data. 200 A genfit ( Psat guess F) T 13.421 2.29 10 3 B C Ans. 69.053 150 Psat f ( A B C) T 100 50 0 240 260 280 300 T 320 340 360 To find the normal boiling point, find the value of T for which Psat = 1 atm. 7 Psat 1atm Tnb A B Psat ln kPa C K Tnb 329.154 K Tnb 1.25 a) t1 C2 273.15K 56.004 degC dollars gal dollars gal Ans. 1970 C1 ( 1 i) t2 t2 t1 2000 C1 C2 0.35 i 5% 1.513 The increase in price of gasoline over this period kept pace with the rate of inflation. b) t1 1970 Given C2 C1 t2 = (1 2000 i) t2 t1 C1 i 16000 dollars yr i C2 5.511 % 80000 dollars yr Find ( i) 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 Chapter 2 - Section A - Mathcad Solutions 2.1 (a) Mwt 35 kg g 9.8 m s 2 z 5m Work Mwt g z Work 1.715 kJ Ans. (b) Utotal Work Utotal 1.715 kJ Ans. (c) By Eqs. (2.14) and (2.21): dU 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 Utotal Q 1.715 kJ Ans. (e) 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 2.4 The electric power supplied to the motor must equal the work done by the motor plus the heat generated by the motor. i 9.7amp E 110V Wdotmech 1.25hp Wdotelect i E Wdotelect 1.067 10 W 3 Qdot Wdotelect Wdotmech Qdot 134.875 W Ans. 2.5 Eq. (2.3): U = Q t W Step 1 to 2: Ut12 200J W12 6000J Q12 Ut12 W12 Q12 5.8 10 J 3 Ans. Step 3 to 4: Q34 800J W34 300J Ut34 Q34 W34 t Ut34 500 J t Ans. 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 U values for all of the steps must sum to zero. t Ut41 4700J Ut23 Ut12 Ut34 Ut41 Ut23 Step 2 to 3: 4000 J Ans. Ut23 4 10 J 3 Q23 3800J W23 Ut23 Q23 W23 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 W41 W12341 W12 W23 W34 W41 4.5 10 J 3 Ans. Step 4 to 1: Ut41 4700J W41 4.5 10 J 3 Q41 Ut41 W41 Q41 200 J Ans. Note: Q12341 = W12341 2.11 The enthalpy change of the water = work done. M 20 kg CP 4.18 kJ kg degC t 10 degC Wdot 0.25 kW M CP t Wdot 0.929 hr Ans. 2.12 Q 7.5 kJ U 12 kJ W U Q W U 12 kJ 19.5 kJ 12 kJ Ans. Ans. Q U Q 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 40 kg mt 5 kg Cc 0.50 kJ kg degC Ct 0.5 kJ kg degC Cw 4.18 kJ kg degC tc 500 degC t1 25 degC t2 30 degC (guess) Given mc Cc t2 tc = mw Cw mt Ct t2 t1 t2 Find t2 11 t2 27.78 degC Ans. 2.15 mass 1 kg CV 4.18 kJ kg K (a) T 1K Ut mass CV T Ut 4.18 kJ Ans. (b) g 9.8 m s 2 EP Ut z EP mass g z 426.531 m Ans. (c) EK Ut u EK 1 mass 2 u 91.433 m s Ans. 2.17 z 50m 1000 kg m 3 u 5 m s 2 D mdot 2m uA A mdot 4 D 2 A 4 kg 3.142 m 1.571 10 s 3 Wdot mdot g z Wdot 7.697 10 kW Ans. 2.18 (a) U1 kJ 762.0 kg P1 1002.7 kPa V1 cm 1.128 gm 3 H1 U1 P1 V 1 H1 763.131 kJ kg Ans. (b) U2 kJ 2784.4 kg P2 1500 kPa V2 cm 169.7 gm 3 H2 U2 P2 V 2 U U2 U1 H H2 H1 U 2022.4 kJ kg Ans. H 2275.8 kJ kg Ans. 12 2.22 D1 2.5cm u1 2 m s D2 5cm (a) For an incompressible fluid, =constant. By a mass balance, mdot = constant = u 1A1 = u2A2 u2 (b) u1 1 2 D1 D2 u2 2 2 u2 0.5 m s Ans. EK 1 2 u1 2 EK 1.875 J kg Ans. 2.23 Energy balance: mdot3 H3 mdot1 H1 mdot2 H2 = Qdot Mass balance: mdot3 mdot1 mdot2 = 0 Therefore: mdot1 H3 mdot Cp T3 H1 T1 mdot2 H3 H2 = Qdot T2 = Qdot or mdot2 CP T3 T3 CP mdot1 mdot2 = Qdot mdot1 CP T1 mdot2 CP T2 mdot1 1.0 kg s T1 25degC mdot2 0.8 kg s T2 75degC Qdot 30 kJ s CP 4.18 kJ kg K T3 Qdot mdot1 CP T1 mdot1 mdot2 CP T2 mdot2 CP T3 43.235 degC Ans. 2.25By Eq. (2.32a): H By continuity, incompressibility u2 = u1 u = 0 2 A1 2 H = CP T A2 CP 4.18 kJ kg degC 13 2 u = u1 2 A1 A2 2 1 2 u = u1 2 D1 D2 4 1 SI units: u1 2 14 m s D1 2.5 cm D2 3.8 cm T D2 u1 2 CP 7.5cm 1 D1 D2 4 T 0.019 degC Ans. T u1 2 2 CP 1 D1 D2 4 T 0.023 degC Ans. Maximum T change occurrs for infinite D2: D2 cm T u1 2 2 CP 1 D1 D2 4 T 0.023 degC Ans. 2.26 T1 300K T2 520K u1 50 10 kmol hr m s u2 3.5 m s molwt 29 kg kmol Wsdot 98.8kW ndot CP 7 R 2 H CP T2 T1 H 6.402 10 3 kJ kmol By Eq. (2.30): Qdot H u2 2 2 u1 2 2 molwt ndot 2 Wsdot Qdot 9.904 kW Ans. 2.27By Eq. (2.32b): H= u 2 gc also V2 V1 = 2 T 2 P1 T 1 P2 2 By continunity, constant area u2 = u1 V2 V1 14 u2 = u1 T 2 P1 T 1 P2 u = u2 u1 2 2 u = u1 2 T 2 P1 T 1 P2 2 1 H = CP T = ft s 7 R T2 2 T1 T1 P1 100 psi P2 20 psi u1 20 579.67 rankine R 3.407 ft lbf mol rankine molwt 28 gm mol T2 578 rankine (guess) Given T2 7 2 R T2 T1 = T2 u1 2 2 T 2 P1 T 1 P2 2 1 molwt Find T2 578.9 rankine Ans. ( 119.15 degF) 2.28 u1 3 m s u2 200 m s H1 2 334.9 2 kJ kg H2 2726.5 kJ kg By Eq. (2.32a): Q H2 H1 u2 u1 2 Q 2411.6 kJ kg Ans. 2.29 u1 30 u2 m s m 500 s H1 3112.5 kJ kg H2 2945.7 kJ kg (guess) By Eq. (2.32a): Given H2 H1 = u1 2 u2 2 2 u2 Find u2 u2 578.36 m s Ans. 3 D1 5 cm V1 388.61 cm gm 15 V2 667.75 cm 3 gm Continuity: D2 D1 u1 V2 u2 V1 D2 1.493 cm Ans. 2.30 (a) t1 30 degC t2 250 degC n 3 mol CV 20.8 J mol degC By Eq. (2.19): Q n CV t2 t1 Q 13.728 kJ Ans. Take into account the heat capacity of the vessel; then mv 100 kg cv 0.5 kJ kg degC Q mv cv n CV t2 t1 Q 11014 kJ Ans. (b) t1 200 degC t2 40 degC n 4 mol CP 29.1 joule mol degC Q n CP t2 t1 By Eq. (2.23): Q 18.62 kJ Ans. 2.31 (a) t1 70 degF t2 350 degF n 3 mol CV 5 BTU mol degF By Eq. (2.19): Q n CV t2 t1 Q 4200 BTU Ans. Take account of the heat capacity of the vessel: mv 200 lbm cv 0.12 BTU lbm degF Q mv cv n CV t2 t1 Q 10920 BTU Ans. (b) t1 400 degF t2 150 degF n 4 mol 16 CP 7 BTU mol degF By Eq. (2.23): Q n CP t2 t1 Q 7000 BTU Ans. 2.33 H1 1322.6 BTU lbm H2 1148.6 BTU lbm u1 10 ft s V1 3.058 ft 3 lbm 2 V2 78.14 ft 3 lbm 4 lb D1 3 in D2 10 in mdot 4 D 1 u1 V1 mdot 3.463 10 sec u2 mdot 4 V2 u2 2 22.997 D2 ft sec 2 Eq. (2.32a): Ws Wdot Ws mdot H2 H1 Wdot u2 2 u1 2 Ws Ans. 173.99 BTU lb 39.52 hp 2.34 H1 307 BTU lbm H2 330 BTU lbm u1 20 ft s molwt 44 gm mol V1 9.25 ft 3 lbm V2 0.28 ft 3 lbm D1 4 in D2 1 in mdot 4 D1 u1 V1 2 mdot 679.263 lb hr u2 mdot 4 V2 u2 2 9.686 D2 ft sec Ws 5360 BTU lbmol Eq. (2.32a): Q H2 H1 u2 2 u1 2 17 2 Ws molwt Q 98.82 BTU lbm Qdot mdot Q Qdot 67128 BTU hr Ans. 2.36 T1 300 K P 1 bar n 1 kg 28.9 gm mol 3 n 34.602 mol V1 3 bar cm T1 83.14 mol K P V1 cm 24942 mol V2 W= n V1 P dV = n P V 1 V2 = n P V1 3 V1 Whence W n P 2 V1 W 172.61 kJ Ans. Given: T2 = T1 V2 V1 = T1 3 Whence T2 H 17.4 3 T1 kJ mol CP 29 joule mol K H CP T2 T1 Ans. Q n H Q 602.08 kJ Ans. U Q n W U 12.41 kJ mol 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 J mol K T1 293.15 K T2 333.15 K P1 1000 kPa P2 100 kPa (a) Cool at const V1 to P2 (b) Heat at const P2 to T2 CP 7 R 2 CV 5 R 2 Ta2 T1 P2 P1 Ta2 29.315 K 18 Tb T2 Ta2 Tb 303.835 K Ta Ta2 T1 Ta 263.835 K Hb Ua V1 Ha C P Tb CV Ta R T1 P1 Ua Hb Ua 2.437 8.841 5.484 10 3 J mol 3 J 10 mol V2 R T2 P2 10 V1 V 1 P2 10 3 3 m mol Ha V2 3 J 0.028 m 3 mol P1 7.677 mol Ub U Hb Ua Ha kg m 2 5 2 5 3 P2 V 2 Ub Hb V1 U 0.831 kJ Ub 6.315 Ans. 10 3 J mol mol kJ mol Ans. H H 1.164 2.39 996 9.0 10 1 4 kg ms D 0.0001 Note: D = /D in this solution D cm u 1 m 5 s 5 22133 Re D u Re 55333 110667 276667 19 0.00635 fF 0.3305 ln 0.27 D 7 Re 0.9 2 fF 0.00517 0.00452 0.0039 0.313 mdot u 4 D 2 mdot 1.956 kg 1.565 s 9.778 Ans. 0.632 P L 2 fF D u 2 P L kPa 11.254 m 0.206 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 0.573 Wdot 1.009 10 kW 3 Wdot kW Cost 799924 dollars Ans. 20 Chapter 3 - Section A - Mathcad Solutions 3.1 = 1 d dT = 1 d dP T P At constant T, the 2nd equation can be written: d = dP ln ( ) 1.01 ln 2 1 = P 44.1810 6 bar 1 2 = 1.01 1 P P 225.2 bar 3 P2 = 226.2 bar Ans. 3.4 b 2700 bar c cm 0.125 gm V2 P1 1 bar P2 500 bar Since Work = V1 P2 P dV a bit of algebra leads to Work c P1 P P b dP Work 0.516 J gm Ans. Alternatively, formal integration leads to Work c P2 P1 b ln P2 P1 b b Work 0.516 J gm Ans. 3.5 = a bP a 3.9 10 6 atm 1 b 0.1 10 9 atm 2 P1 1 atm P2 3000 atm V 1 ft 3 (assume const.) Combine Eqs. (1.3) and (3.3) for const. T: P2 Work V P1 ( a b P)P dP 21 Work 16.65 atm ft 3 Ans. 3.6 1.2 10 3 degC 3 1 CP 0.84 kJ kg degC M 5 kg V1 1 m 1590 kg P 1 bar t1 0 degC t2 20 degC With beta independent of T and with P=constant, dV = V Vtotal dT M V V2 Vtotal V1 exp 7.638 t2 10 5 t1 m 3 V V2 V1 Ans. Work P Vtotal (Const. P) Work 7.638 joule Ans. Q M CP t2 t1 Q 84 kJ Ans. Htotal Q Htotal 84 kJ Ans. Utotal Q Work Utotal 83.99 kJ 7 R 2 Ans. 5 2 3.8 P1 8 bar P2 1 bar T1 600 K CP CV R (a) Constant V: W= 0 and U = Q = CV T T2 T1 P2 P1 T T2 T1 T 525 K U CV T Q and U 10.91 kJ mol Ans. H CP T H 15.28 kJ mol Ans. (b) Constant T: U= H= 0 Work and 10.37 Q= W kJ mol Work R T1 ln P2 P1 Q and Ans. (c) Adiabatic: Q= 0 22 and U = W = CV T 1 CP CV U T2 T1 P2 P1 T2 331.227 K T T2 T1 CV T H CP T W and U 5.586 kJ mol Ans. H 7.821 kJ mol Ans. 3.9 P4 2bar CP 7 R 2 CV 5 R 2 P1 10bar T1 600K V1 R T1 P1 R CP V1 4.988 10 3 3 m mol Step 41: Adiabatic T4 T1 P4 P1 T4 378.831 K 3 J U41 CV T1 T4 U41 4.597 10 H41 CP T1 T4 H41 6.436 10 mol 3 J mol Q41 J 0 mol Q41 J 0 mol W41 U41 W41 4.597 10 3 3 J mol P2 3bar T2 600K V2 R T2 P2 V2 m 0.017 mol Step 12: Isothermal U12 0 J mol U12 0 J mol H12 0 J mol H12 0 J mol 23 Q12 W12 R T1 ln Q12 P2 P1 Q12 6.006 10 3 J mol 3 J W12 6.006 10 mol P3 2bar V3 V2 T3 P3 V 3 R T3 400 K 3 J Step 23: Isochoric U23 CV T3 T2 H23 CP T3 T2 Q23 CV T3 T2 W23 0 J mol mol 3 J H23 5.82 10 mol 3 J Q23 4.157 10 mol J W23 0 mol U23 4.157 10 P4 2 bar T4 378.831 K V4 CV T4 R T4 P4 T3 V4 U34 m 0.016 mol 439.997 J mol J 3 Step 34: Isobaric U34 H34 CP T4 T3 H34 615.996 mol Q34 CP T4 T3 Q34 615.996 W34 R T4 T3 W34 175.999 J mol J mol 3.10 For all parts of this problem: T2 = T1 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 12 bar 24 V1 12 m 3 V2 1m 3 (a) Work = n R T ln P2 Work P1 Work P1 V1 ln P2 P1 Ans. 2982 kJ (b) Step 1: adiabatic compression to P2 1 5 3 Vi V1 P1 P2 (intermediate V) Vi 2.702 m 3 W1 P2 V i P1 V 1 1 W1 3063 kJ Step 2: cool at const P2 to V2 W2 P2 V 2 Vi W2 2042 kJ Work W1 W2 Work 5106 kJ Ans. (c) Step 1: adiabatic compression to V2 Pi P1 V1 V2 (intermediate P) Pi 62.898 bar W1 Pi V 2 P1 V 1 1 W1 7635 kJ Step 2: No work. Work W1 Work 7635 kJ Ans. (d) Step 1: heat at const V1 to P2 W1 = 0 Step 2: cool at const P2 to V2 W2 P2 V 2 V1 Work W2 Work 13200 kJ Ans. (e) Step 1: cool at const P1 to V2 W1 P1 V 2 V1 25 W1 1100 kJ Step 2: heat at const V2 to P2 W2 = 0 Work W1 Work 1100 kJ Ans. 3.17(a) No work is done; no heat is transferred. U = t T= 0 T2 = T1 = 100 degC Not reversible (b) The gas is returned to its initial state by isothermal compression. Work = n R T ln 3 V1 V2 V2 4 3 but n R T = P2 V 2 V1 4m m 3 P2 6 bar Work P2 V2 ln V1 V2 P2 500 kPa Work 878.9 kJ Ans. 3.18 (a) P1 100 kPa T1 303.15 K CP 7 2 R CV 5 R 2 CP CV 1 Adiabatic compression from point 1 to point 2: Q12 U12 0 kJ mol CV T2 T1 U12 = W12 = CV T12 H12 CP T2 T1 T2 W12 T1 P2 P1 U12 U12 3.679 kJ mol H12 5.15 kJ mol W12 3.679 kJ mol Ans. Cool at P2 from point 2 to point 3: T3 T1 H23 CP T3 T2 Q23 H23 U23 CV T3 T2 W23 U23 Q23 26 H23 5.15 kJ mol U23 3.679 kJ mol Ans. Q23 5.15 kJ mol W23 1.471 kJ mol Ans. Isothermal expansion from point 3 to point 1: U31 = Q31 H31 = 0 W31 P3 P2 W31 R T3 ln P1 P3 W31 4.056 kJ mol Q31 4.056 kJ mol Ans. FOR THE CYCLE: U= H= 0 Q Q12 Q23 Q31 Work Work W12 1.094 W23 kJ mol W31 Q 1.094 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 W12 0.8 W12 4.598 kJ mol Q12 U12 W12 Q12 0.92 kJ mol Step 23: W23 W23 0.8 W23 1.839 kJ mol Q23 U23 W23 Q23 5.518 kJ mol Step 31: W31 W31 0.8 W31 3.245 kJ mol Q31 W31 Q31 27 3.245 kJ mol FOR THE CYCLE: Q Q12 Q23 Q31 Work W12 W23 W31 Q 3.192 kJ mol Work 3.192 kJ mol 3.19Here, V represents total volume. P1 1000 kPa V1 1m 3 V2 5 V1 T1 600 K CP 21 joule mol K CV CP R CP CV (a) Isothermal: T2 T1 Work = n R T1 ln T2 600 K V1 V2 200 kPa P2 P1 V1 V2 P2 Work Ans. Work P1 V1 ln V1 V2 V1 V2 1609 kJ Ans. (b) Adiabatic: P2 P1 T2 T1 P2 V 2 P1 V 1 T2 208.96 K P2 69.65 kPa Ans. Work P2 V 2 P1 V 1 1 Work 994.4 kJ Ans, (c) Restrained adiabatic: Work = U = Pext V Pext 100 kPa Work Pext V2 V1 Work 400 kJ Ans. n P1 V 1 R T1 U = n CV T T2 Work n CV T1 T2 442.71 K Ans. P2 P1 V 1 T2 V 2 T1 28 P2 147.57 kPa Ans. 3.20 T1 423.15 K P1 8bar P3 3 bar CP 7 2 R CV 5 2 R T2 T1 T3 323.15 K Step 12: H12 0 kJ mol U12 0 kJ mol If r= V1 V2 = V1 V3 Then r T 1 P3 T 3 P1 W12 R T1 ln r () W12 2.502 kJ mol W23 0 Q12 kJ mol W12 Q12 2.502 kJ mol Step 23: U23 CV T3 T2 Q23 U23 H23 CP T3 T2 Q23 2.079 kJ mol U23 2.079 kJ mol H23 2.91 kJ mol Process: Work W12 W23 Work 2.502 kJ mol Ans. Q Q12 Q23 Q 0.424 kJ mol Ans. H H12 H23 H 2.91 kJ mol Ans. U U12 U23 U 2.079 kJ mol Ans. 29 3.21 By Eq. (2.32a), unit-mass basis: molwt 28 gm mol H 1 2 u = 0 2 But H = CP T Whence T= u2 2 u1 2 2 CP CP R 7 2 molwt u1 2.5 m s u2 2 2 50 m s t1 150 degC t2 t1 u2 u1 2 CP t2 148.8 degC Ans. 3.22 CP P1 7 R 2 1 bar CV P3 5 R 2 10 bar T1 303.15 K T3 403.15 K U CV T3 T1 H CP T3 T1 U 2.079 kJ mol Ans. H 2.91 kJ mol Ans. Each part consists of two steps, 12 & 23. (a) T2 T3 P2 P1 T2 T1 Work W23 W23 R T2 ln P3 P2 Work 6.762 kJ mol Ans. Q U Work Q 4.684 kJ mol Ans. 30 (b) P2 P1 T2 T3 U12 CV T2 T1 H12 CP T2 T1 Q12 H12 W12 U12 Q12 W12 W23 0.831 kJ mol W23 R T2 ln P3 P2 W23 7.718 kJ mol Work W12 Work 6.886 kJ mol Ans. Q U Work Q 4.808 kJ mol Ans. (c) T2 H23 U23 T1 CP T3 CV T3 P2 T2 T2 P3 W12 Q23 R T1 ln H23 P2 P1 W23 U23 Q23 Work W12 W23 Work 4.972 kJ mol Ans. Q U Work Q 2.894 kJ mol Ans. For the second set of heat-capacity values, answers are (kJ/mol): U = 1.247 U = 2.079 (a) Work = 6.762 Q = 5.515 (b) Work = 6.886 Q = 5.639 (c) Work = 4.972 Q = 3.725 31 3.23 T1 303.15 K T2 T1 T3 393.15 K P1 1 bar P3 12 bar CP 7 R 2 CV 5 R 2 For the process: U CV T3 T1 H CP T3 T1 U 1.871 kJ mol H 2.619 kJ mol Ans. Step 12: P2 kJ mol P3 T1 T3 W12 W12 R T1 ln P2 P1 W12 5.608 Q12 Q12 5.608 kJ mol Step 23: For the process: W23 Work 0 kJ mol W23 Q23 U W12 Q Q12 Q23 Work 5.608 kJ mol Q 3.737 kJ mol Ans. 3.24 W12 = 0 Work = W23 = P2 V3 V 2 = R T3 T2 But T3 = T1 So... Work = R T2 Therefore T1 Also W = R T1 ln P P1 350 K ln T2 T1 P = P1 T1 T2 T1 800 K P1 4 bar P P1 exp T2 T1 T1 P 2.279 bar Ans. 32 3.25 VA 256 cm 3 Define: P P1 = r r 0.0639 Assume ideal gas; let V represent total volume: P1 V B = P 2 V A VB From this one finds: P P1 = VA VA VB VB VA ( 1) r r VB 3750.3 cm 3 Ans. 3.26 T1 300 K P1 1 atm CP 7 R 2 CV CP R CP CV The process occurring in section B is a reversible, adiabatic compression. Let P ( )= P2 final nA = nB TA ( )= TA final TB ( )= TB final Since the total volume is constant, nA R TA 2 nA R T1 = P2 P1 TB or TA TB 2 T1 = P2 P1 1 (1) (a) P2 TA 1.25 atm P2 P1 TB TB T1 P2 P1 UA UB (2) 2 T1 Q = nA Define q= Q nA q CV TA TB 2 T1 (3) TB 319.75 K TA 430.25 K q 3.118 kJ mol Ans. 33 (b) Combine Eqs. (1) & (2) to eliminate the ratio of pressures: TA 425 K (guess) 1 TB 300 K Given TB = T1 TA TB 2 T1 TB Find TB TB 319.02 K 1.24 atm Ans. Ans. P2 q P1 TA TB 2 T1 TB 2 T1 (1) P2 CV TA q 2.993 kJ mol Ans. (c) TB 325 K 1 By Eq. (2), P2 P1 TB T1 P2 P1 P2 1.323 atm Ans. TA 2 T1 TB (1) TA 469 K Ans. q CV TA TB 2 T1 q 4.032 kJ mol Ans. (d) Eliminate TA TB from Eqs. (1) & (3): q 3 kJ mol P2 1 q P1 2 T1 C V P1 P2 1.241 atm Ans. TB T1 P2 P1 P2 P1 (2) TB 319.06 K Ans. TA 2 T1 TB (1) 34 TA 425.28 K Ans. 3.30 B 242.5 cm 3 C 25200 cm 6 T 2 373.15 K mol mol P1 1 bar P2 55 bar B' B RT B' 7.817 10 3 1 bar C' C 2 B 2 R T 2 C' 3.492 10 5 1 bar 2 (a) Solve virial eqn. for initial V. Guess: V1 RT P1 Given P1 V 1 = 1 RT B V1 C V1 2 V1 Find V1 V1 cm 30780 mol 3 Solve virial eqn. for final V. Guess: V2 RT P2 Given P2 V 2 RT = 1 B V2 C 2 V2 V2 Find V2 V2 cm 241.33 mol 3 Eliminate P from Eq. (1.3) by the virial equation: V2 Work RT V1 1 B V C V 2 1 dV V Work 12.62 kJ mol Ans. (b) Eliminate dV from Eq. (1.3) by the virial equation in P: P2 dV = R T P 1 2 C' dP W RT P1 1 P C' P dP W 35 12.596 kJ mol Ans. 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 Tr T Tc Tr 1.056 Pc 50.4 bar P 12 bar Pr P Pc Pr 0.238 0.087 (a) (guess) B 140 cm 3 mol C 7200 cm 6 2 V mol RT P V 2066 cm 3 mol Given PV = 1 RT B V C V 2 V Find ( ) V V 0.422 Tr 1.6 cm 1919 mol 3 Z PV RT Z 0.929 Ans. (b) B0 0.083 B0 0.304 B1 0.139 0.172 Tr 4.2 B1 2.262 10 3 Z 1 B0 B1 Pr Tr Z 0.932 V ZRT P V cm Ans. 1924 mol 3 (c) For Redlich/Kwong EOS: 1 0.5 0 0.08664 0.42748 Table 3.1 ( ) Tr Tr Table 3.1 q Tr Tr Tr Eq. (3.54) T r Pr Pr Tr Eq. (3.53) 36 Calculate Z Guess: Z 0.9 Given Eq. (3.52) Z= 1 T r Pr q Tr T r Pr Z Z T r Pr T r Pr Z T r Pr Z Find Z) ( Z 0.928 V Z RT P V 1916.5 cm 3 mol Ans. (d) For SRK EOS: 1 0 0.08664 1 2 0.42748 Table 3.1 Tr q Tr 1 Tr 0.480 1.574 0.176 2 1 Tr 2 Table 3.1 Eq. (3.54) Guess: Tr Z T r Pr 0.9 Pr Tr Eq. (3.53) Calculate Z Given Eq. (3.52) Z= 1 T r Pr q Tr T r Pr Z Z T r Pr T r Pr Z T r Pr Z Find Z) ( Z 0.928 V Z RT P V 1918 cm 3 mol Ans. (e) For Peng/Robinson EOS: 1 2 1 2 0.07779 0.45724 1 2 Table 3.1 Tr 1 0.37464 Tr Tr 1.54226 0.26992 2 1 Tr Pr Tr 2 Table 3.1 q Tr Eq. (3.54) 37 T r Pr Eq. (3.53) Calculate Z Guess: Z 0.9 Given Eq. (3.52) Z= 1 T r Pr q Tr T r Pr Z Z T r Pr T r Pr Z T r Pr Z Find ( Z) Z 0.92 V ZRT P V 1900.6 cm 3 mol Ans. 3.33 Tc 305.3 K T 323.15 K Tr T Tc Tr 1.058 Pc 48.72 bar P 15 bar Pr P Pc Pr 0.308 0.100 (a) (guess) B cm 156.7 mol PV = 1 RT 3 C C V 2 9650 cm 6 2 V mol RT P V cm 1791 mol 3 Given B V V Find ( V) V cm 1625 mol 3 Z PV RT Z 0.907 Ans. (b) B0 0.083 0.422 Tr 1.6 B0 0.302 B1 0.139 0.172 Tr 4.2 B1 3.517 10 3 Z 1 B0 B1 Pr Tr Z 0.912 V ZRT P V cm Ans. 1634 mol 3 (c) For Redlich/Kwong EOS: 1 0 0.08664 38 0.42748 Table 3.1 ( ) Tr Tr 0.5 Table 3.1 q Tr Tr Tr Eq. (3.54) T r Pr Pr Tr Eq. (3.53) Calculate Z Guess: Z 0.9 Given Eq. (3.52) Z= 1 T r Pr q Tr T r Pr Z Z T r Pr T r Pr Z T r Pr Z Find Z) ( Z 0.906 V Z RT P V cm 1622.7 mol 3 Ans. (d) For SRK EOS: 1 0 0.08664 1 2 0.42748 Table 3.1 Tr q Tr 1 Tr 0.480 1.574 0.176 2 1 Tr 2 Table 3.1 Eq. (3.54) Tr T r Pr Pr Tr Eq. (3.53) Calculate Z Guess: Z 0.9 Given Eq. (3.52) Z= 1 T r Pr q Tr T r Pr Z Z T r Pr T r Pr Z T r Pr Z Find Z) ( Z 0.907 V Z RT P V 1624.8 cm 3 mol Ans. (e) For Peng/Robinson EOS: 1 2 1 2 39 0.07779 0.45724 Table 3.1 1 2 Tr 1 0.37464 Tr Tr 1.54226 0.26992 2 1 Tr Pr Tr 2 Table 3.1 q Tr Eq. (3.54) T r Pr Eq. (3.53) Calculate Z Guess: Z 0.9 Given Eq. (3.52) Z= 1 T r Pr q Tr T r Pr Z Z T r Pr T r Pr Z T r Pr Z Find ( Z) Z 0.896 V ZRT P V cm 1605.5 mol 3 Ans. 3.34 Tc Pc 318.7 K T 348.15 K Tr Pr T Tc P Pc Tr Pr 1.092 37.6 bar P 15 bar 0.399 0.286 (guess) (a) 3 B 194 cm mol C 15300 cm 6 2 V mol RT P V 1930 cm 3 mol Given PV = 1 RT B V C V 2 V Find ( V) V cm 1722 mol 3 Z PV RT Z 0.893 Ans. (b) B0 0.083 0.422 Tr 1.6 B0 0.283 40 B1 0.139 0.172 Tr 4.2 B1 0.02 3 Z 1 B0 B1 Pr Tr Z 0.899 V Z RT P V 1734 cm mol Ans. (c) For Redlich/Kwong EOS: 1 0 0.08664 0.42748 Table 3.1 ( ) Tr Tr 0.5 Table 3.1 q Tr Tr Tr Eq. (3.54) T r Pr Pr Tr Eq. (3.53) Guess: Z 0.9 Calculate Z Given Eq. (3.52) Z= 1 T r Pr q Tr T r Pr Z Z T r Pr T r Pr Z T r Pr Z Find Z) ( Z 0.888 V Z RT P V cm 1714.1 mol 3 Ans. (d) For SRK EOS: 1 0 0.08664 1 2 0.42748 Table 3.1 Table 3.1 Tr q Tr 1 Tr 0.480 1.574 0.176 2 1 Tr 2 Eq. (3.54) Tr T r Pr Pr Tr Eq. (3.53) 41 Calculate Z Guess: Eq. (3.52) Given Z 0.9 Z= 1 T r Pr q Tr T r Pr Z Z T r Pr T r Pr Z T r Pr Z Find ( Z) Z 0.895 V ZRT P V 1726.9 cm 3 mol Ans. (e) For Peng/Robinson EOS: 1 2 1 2 0.07779 0.45724 1 2 Table 3.1 Tr 1 0.37464 Tr Tr 1.54226 0.26992 2 1 Tr Pr Tr 2 Table 3.1 q Tr Eq. (3.54) Guess: T r Pr Z 0.9 Eq. (3.53) Calculate Z Given Eq. (3.52) Z= 1 T r Pr q Tr T r Pr Z Z T r Pr T r Pr Z T r Pr Z Find ( Z) Z 0.882 V ZRT P V cm 1701.5 mol 3 Ans. 3.35 T 523.15 K P 3 1800 kPa (a) B cm 152.5 mol C 5800 cm 6 2 V mol RT (guess) P Given PV = 1 RT B V C V 2 V Find ( V) Z PV RT V 2250 42 cm 3 mol Z 0.931 Ans. (b) Tc 647.1 K Pc 220.55 bar 0.345 Tr T Tc Pr P Pc B0 0.083 0.422 Tr 1.6 Tr 0.808 Pr 0.082 B0 0.51 B1 0.139 0.172 Tr 4.2 B1 0.281 Z 1 B0 3 B1 Pr Tr V Z RT P Z 0.939 V cm 2268 mol Ans. 3 (c) Table F.2: molwt gm 18.015 mol V cm molwt 124.99 gm cm 2252 mol 3 or V 6 2 Ans. 3.37 B cm 53.4 mol 3 C 2620 cm D 5000 cm 9 3 n mol mol mol T 273.15 K Given PV RT = 1 B V 10 C V 2 D V 3 fP V) ( Find V) ( i 0 10 Pi 10 20 i bar Vi RT Pi (guess) Zi fPi Vi Pi RT Eq. (3.12) Z1i 1 B Pi RT Eq. (3.38) Z2i 1 2 1 4 B Pi RT Eq. (3.39) 43 110 -10 20 40 60 80 Zi 1 0.953 0.906 0.861 Z1i 1 0.953 0.906 0.859 0.812 0.765 0.718 0.671 0.624 0.577 0.53 Z2i 1 0.951 0.895 0.83 0.749 0.622 0.5+0.179i 0.5+0.281i 0.5+0.355i 0.5+0.416i 0.5+0.469i Pi 100 120 140 160 180 200 bar 0.819 0.784 0.757 0.74 0.733 0.735 0.743 Note that values of Z from Eq. (3.39) are not physically meaningful for pressures above 100 bar. 1 0.9 Zi Z1 i Z2 i 0.8 0.7 0.6 0.5 0 50 100 Pi bar 1 150 200 44 3.38 (a) Propane: Tc 369.8 K Pc 42.48 bar 0.152 T 313.15 K P 13.71 bar Tr T Tc Tr 0.847 Pr P Pc Pr 0.323 For Redlich/Kwong EOS: 1 0 0.08664 0.42748 Table 3.1 ( ) Tr Tr 0.5 Table 3.1 Eq. (3.53) q Tr Tr Tr Eq. (3.54) T r Pr Pr Tr Calculate Z for liquid by Eq. (3.56) Guess: Z 0.01 Given Z= T r Pr Z T r Pr Z T r Pr 1 q Tr T r Pr T r Pr Z Z Find Z) Z ( 0.057 V Z RT P V 108.1 cm 3 mol Ans. Calculate Z for vapor by Eq. (3.52) Guess: Z 0.9 Given Z= 1 T r Pr q Tr T r Pr Z Z Z T r Pr T r Pr Z Find Z) ( Z 0.789 V Z RT P V cm 1499.2 mol 3 Ans. 45 Rackett equation for saturated liquid: cm 3 Tr T Tc Tr 0.847 Vc 200.0 mol Zc 0.276 V V c Zc 1 Tr 0.2857 V 94.17 cm 3 mol Ans. For saturated vapor, use Pitzer correlation: 0.422 Tr 1.6 B0 0.083 B0 0.468 B1 0.139 0.172 Tr 4.2 B1 0.207 V RT P R B0 B1 Tc Pc V 1.538 10 3 3 cm mol Ans. 46 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 (c) 122.7 (d) 133.6 (e) 148.9 (f) 158.3 (g) 170.4 (h) 187.1 (i) 153.2 (j) 164.2 (k) 179.1 (l) 201.4 (m) (n) (o) (p) (q) (r) (s) (t) 61.7 64.1 66.9 70.3 64.4 67.4 70.8 74.8 1174.7 920.3 717.0 1516.2 1216.1 971.1 768.8 1330.3 1057.9 835.3 645.8 1252.5 1006.9 814.5 661.2 1318.7 1046.6 835.6 669.5 98.1 102.8 109.0 125.4 130.7 137.4 146.4 133.9 140.3 148.6 160.6 53.5 55.1 57.0 59.1 54.6 56.3 58.3 60.6 1228.7 990.4 805.0 1577.0 1296.8 1074.0 896.0 1405.7 1154.3 955.4 795.8 1276.9 1038.5 853.4 707.8 1319.0 1057.2 856.4 700.5 47 3.39 (a) Propane Tc 369.8 K Pc 42.48 bar 0.152 T ( 40 273.15)K T 313.15 K P 13.71 bar Tr T Tc Tr 0.847 Pr P Pc Pr 0.323 From Table 3.1 for SRK: 1 0 0.08664 0.42748 1 2 Tr q Tr 1 Tr Tr 0.480 1.574 0.176 2 1 Tr 2 Eq. (3.54) T r Pr Z Pr Tr 0.01 Eq. (3.53) Calculate Z for liquid by Eq. (3.56) Guess: Given Z= T r Pr Z T r Pr Z T r Pr 1 q Tr T r Pr T r Pr 3 Z Z Find Z) ( Z 0.055 V Z RT P V cm 104.7 mol Ans. Calculate Z for vapor by Eq. (3.52) Guess: Z 0.9 Given Z= 1 T r Pr q Tr T r Pr Z Z T r Pr T r Pr Z T r Pr 3 Z Find Z) ( Z 0.78 V Z RT P V 1480.7 cm mol Ans. 48 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 (c) 118.2 (d) 128.5 (e) 142.1 (f) 150.7 (g) 161.8 (h) 177.1 (i) 146.7 (j) 156.9 (k) 170.7 (l) 191.3 (m) (n) (o) (p) (q) (r) (s) (t) 61.2 63.5 66.3 69.5 61.4 63.9 66.9 70.5 1157.8 904.9 703.3 1487.1 1189.9 947.8 747.8 1305.3 1035.2 815.1 628.5 1248.9 1003.2 810.7 657.4 1296.8 1026.3 817.0 652.5 98.1 102.8 109.0 125.4 130.7 137.4 146.4 133.9 140.3 148.6 160.6 53.5 55.1 57.0 59.1 54.6 56.3 58.3 60.6 1228.7 990.4 805.0 1577.0 1296.8 1074.0 896.0 1405.7 1154.3 955.4 795.8 1276.9 1038.5 853.4 707.8 1319.0 1057.2 856.4 700.5 49 3.40 (a) Propane Tc 369.8 K Pc 42.48 bar 0.152 T ( 40 273.15)K T 313.15 K P 13.71 bar Tr T Tc Tr 0.847 Pr P Pc Pr 0.323 From Table 3.1 for PR: 1 2 Tr 1 2 1 0.37464 1 2 1.54226 0.26992 2 1 Tr 2 0.07779 0.45724 q Tr Tr Tr Eq. (3.54) T r Pr Z Pr Tr 0.01 Eq. (3.53) Calculate Z for liquid by Eq. (3.56) Guess: Given Z= T r Pr Z T r Pr Z T r Pr 1 q Tr T r Pr T r Pr 3 Z Z Find Z) ( Z 0.049 V Z RT P V cm 92.2 mol Ans. Calculate Z for vapor by Eq. (3.52) Guess: Z 0.6 Given Z= 1 T r Pr q Tr T r Pr Z Z T r Pr T r Pr Z T r Pr Z Find Z) ( Z 0.766 V Z RT P V 1454.5 cm 3 mol Ans. 50 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 1131.8 879.2 678.1 1453.5 1156.3 915.0 715.8 1271.9 1002.3 782.8 597.3 1233.0 987.3 794.8 641.6 1280.2 1009.7 800.5 636.1 98.1 102.8 109.0 125.4 130.7 137.4 146.4 133.9 140.3 148.6 160.6 53.5 55.1 57.0 59.1 54.6 56.3 58.3 60.6 1228.7 990.4 805.0 1577.0 1296.8 1074.0 896.0 1405.7 1154.3 955.4 795.8 1276.9 1038.5 853.4 707.8 1319.0 1057.2 856.4 700.5 51 (c) 104.4 (d) 113.7 (e) 125.2 (f) 132.9 (g) 143.0 (h) 157.1 (i) 129.4 (j) 138.6 (k) 151.2 (l) 170.2 (m) (n) (o) (p) (q) (r) (s) (t) 54.0 56.0 58.4 61.4 54.1 56.3 58.9 62.2 3.41 (a) For ethylene, molwt 28.054 gm Tc mol 282.3 K Pc 50.40 bar 0.087 T 328.15 K P 35 bar Tr T Tc Pr P Pc Tr 1.162 Pr 0.694 From Tables E.1 & E.2: Z0 0.838 Z1 0.033 Z Z0 Z1 Z 0.841 n 18 kg molwt Vtotal ZnRT P Vtotal 0.421 m 3 Ans. (b) T 323.15 K P 115 bar Vtotal 0.25 m 3 Tr T Tc Tr 1.145 0.482 Pr Z1 P Pc 0.126 Pr 2.282 From Tables E.3 & E.4: Z0 Z Z0 Z1 Z 0.493 n P Vtotal ZRT n 2171 mol mass n molwt mass 60.898 kg Ans. 3.42 Assume validity of Eq. (3.38). P1 1bar T1 300K V1 cm 23000 mol 3 Z1 P1 V 1 R T1 Z1 0.922 B R T1 Z1 P1 1 B 1.942 10 3 3 cm mol With this B, recalculate at P2 P2 5bar 3 3 cm Z2 1 B P2 R T1 Z2 0.611 V2 52 R T1 Z2 P2 V2 3.046 10 mol Ans. 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 1.6 B0 0.146 B1 0.139 0.172 Tr 4.2 B1 0.104 V RT P B0 B1 R Tc Pc V cm 989 mol 3 Ans. 3 For an ideal gas: V RT P Tc 3 V cm 1044 mol Pc 3.44 T 320 K 0.152 P 16 bar 369.8 K 42.48 bar Vc cm 200 mol Zc 0.276 molwt 44.097 gm mol Tr T Tc 1 Tr Tr 0.2857 0.865 Pr P Pc Pr 0.377 3 Vliq V c Zc Vliq 96.769 cm mol Vtank 0.35 m 3 mliq 0.8 Vtank Vliq molwt mliq 127.594 kg Ans. B0 0.083 0.422 Tr 1.6 B0 0.449 B1 0.139 0.172 Tr 4.2 B1 0.177 53 Vvap RT P B0 B1 R Tc Pc Vvap 1.318 10 3 3 cm mol mvap 0.2 Vtank Vvap molwt mvap 2.341 kg Ans. 3.45 T 298.15 K Tc 425.1 K Tr T Tc P Tr 0.701 P 2.43 bar Pc 37.96 bar 3 Pr Pc Pr 0.064 0.200 Vvap 16 m molwt 58.123 gm mol B0 0.083 0.422 Tr 1.6 B0 0.661 B1 0.139 0.172 Tr 4.2 B1 0.624 3 3 cm V RT P B0 B1 R Tc Pc V 9.469 10 mol mvap Vvap V molwt mvap 98.213 kg Ans. 3.46 (a) T 333.15 K Tc 305.3 K Tr T Tc Tr 1.091 P 14000 kPa Pc 48.72 bar Pr 3 P Pc Pr 2.874 0.100 Vtotal 0.15 m molwt 30.07 gm mol From tables E.3 & E.4: Z0 0.463 54 Z1 0.037 Z Z0 Z1 Z 0.459 V Z RT P V 90.87 cm 3 mol methane Vtotal V molwt methane 49.64 kg Ans. (b) V Vtotal 40 kg P 20000 kPa P V = Z R T = Z R Tr Tc or Tr = Z where PV R Tc 29.548 mol kg Whence Tr = 0.889 Z at Pr P Pc Pr 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 1.283 T and Tr Tc Z 0.693 Whence T 391.7 K or 118.5 degC Ans. 3.47 Vtotal 0.15 m 3 T 298.15 K Tc 282.3 K Pc 50.40 bar P V = P r Pc V = Z R T 0.087 molwt 28.054 gm mol V Vtotal 40 kg molwt Pr = Z or where RT Pc V 4.675 Whence Pr = 4.675 Z at Tr T Tc Tr 1.056 55 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 1.582 and Z 0.338 P Pc Pr P 79.73 bar Ans. 3.48 mwater 15 kg Vtotal 0.4 m 3 V Vtotal mwater V 26.667 cm 3 gm Interpolate in Table F.2 at 400 degC to find: P = 9920 kPa Ans. 3.49 T1 298.15 K Tc 305.3 K Tr1 Pr1 T1 Tc P1 Pc Tr1 Pr1 0.977 P1 Vtotal 2200 kPa 0.35 m 3 Pc 48.72 bar 0.100 0.452 From Tables E.1 & E.2: Z0 .8105 Z1 0.0479 Z Z0 Z1 Z 0.806 V1 Z R T1 P1 V1 cm 908 mol 3 T2 493.15 K Tr2 T2 Tc Tr2 1.615 Assume Eq. (3.38) applies at the final state. B0 0.083 0.422 Tr2 1.6 B0 0.113 B1 0.139 0.172 Tr2 4.2 B1 0.116 P2 V1 B0 R T2 B1 R Tc Pc 56 P2 42.68 bar Ans. 3.50 T 303.15 K Tc 3 304.2 K Tr T Tc Tr 0.997 Vtotal 0.5 m Pc 73.83 bar 0.224 molwt 44.01 gm mol B0 0.083 0.422 Tr 1.6 B0 0.341 B1 0.139 0.172 Tr 4.2 B1 0.036 3 3 cm V Vtotal 10 kg molwt RT V B0 B1 R Tc Pc V 2.2 10 mol P P 10.863 bar Ans. 3.51 Basis: 1 mole of LIQUID nitrogen Tn 77.3 K Tc 126.2 K Tr Tn Tc Tr 0.613 P 1 atm Pc 34.0 bar Pr P Pc Pr 3 0.03 0.038 molwt 28.014 gm mol Vliq 34.7 cm B0 0.083 0.422 Tr 1.6 B0 0.842 B1 0.139 0.172 Tr 4.2 B1 1.209 Z 1 B0 B1 Pr Tr Z 0.957 57 nvapor P Vliq Z R Tn nvapor 5.718 10 3 mol Final conditions: ntotal 1 mol nvapor V 2 Vliq ntotal V 69.005 cm 3 mol T 298.15 K Tr T Tc Tr 2.363 Pig RT V Pig 359.2 bar Use Redlich/Kwong at so high a P. 0.08664 0.42748 2 2 ( ) Tr Tr .5 Tr 0.651 a Tr R Tc Pc Eq. (3.42) b R Tc Pc Eq. (3.43) a 3 3 bar cm 0.901 m 2 mol b cm 26.737 mol 450.1 bar 3 P RT V b a V ( b) V Eq. (3.44) P Ans. 3.52 For isobutane: Tc 408.1 K Pc 36.48 bar V1 1.824 cm 3 gm T1 300 K P1 4 bar T2 415 K P2 75 bar Tr1 T1 Tc Pr1 P1 Pc Tr2 T2 Tc Pr2 P2 Pc Tr1 0.735 Pr1 0.11 Tr2 1.017 Pr2 2.056 58 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 Z RT with Z from Eq. (3.57) and Tables E.3 and E.4. Thus Vc 262.7 cm 3 mol 0.181 Z0 r2 0.3356 Z1 r2 0.0756 Z Z0 Z1 Z 0.322 r1 r2 P2 V c Z R T2 1.774 Eq. (3.75): V2 V1 V2 cm 2.519 gm 3 Ans. 3.53 For n-pentane: Tc 469.7 K Pc 33.7 bar 1 0.63 gm cm 3 T1 291.15 K P1 Pr1 1 bar P1 Pc T2 Tr2 413.15 K T2 Tc P2 Pr2 120 bar P2 Pc Tr1 Tr1 T1 Tc 0.62 Pr1 r1 0.03 Tr2 r2 0.88 Pr2 3.561 From Fig. (3.16): 2.69 r2 2.27 By Eq. (3.75), 2 1 r1 2 0.532 gm cm 3 Ans. 3.54 For ethanol: Tc 513.9 K T 453.15 K Tr T Tc Tr 0.882 Pc 61.48 bar P 200 bar Pr P Pc Pr 3.253 Vc 167 cm 3 mol molwt 59 46.069 gm mol From Fig. 3.16: r 2.28 r Vc molwt = r c= r Vc 0.629 gm cm 3 Ans. 3.55 For ammonia: Tc 405.7 K T 293.15 K Tr T Tc Tr 0.723 Pc 112.8 bar cm 72.5 mol 3 P 857 kPa Pr P Pc 0.253 Pr 0.076 Vc Zc 0.242 0.2857 Eq. (3.72): Vliquid 0.422 Tr 1.6 V c Zc 1 Tr Vliquid cm 27.11 mol 3 B0 0.083 B0 0.627 B1 0.139 0.172 Tr 4.2 B1 0.534 Vvapor RT P B0 B1 R Tc Pc Vvapor cm 2616 mol 3 3 V Vvapor Vliquid V cm 2589 mol Ans. 60 Alternatively, use Tables E.1 & E.2 to get the vapor volume: Z0 0.929 Z1 0.071 Z Z0 Z1 Z 3 0.911 Vvapor ZRT P Vvapor 2591 cm mol 3 V Vvapor Vliquid V 2564 cm mol Ans. 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: R ft atm 0.7302 lbmol rankine 3 T 519.67 rankine P 1 atm V 1400 ft 3 n PV RT T Tc n 3.689 lbmol For methane at 3000 psi and 60 degF: Tc 190.6 1.8 rankine T 519.67 rankine Tr Tr 1.515 Pc 45.99 bar P 3000 psi Pr P Pc Pr 4.498 0.012 From Tables E.3 & E.4: Z0 0.819 ZnRT P Z1 0.234 Z 3 Z0 Z1 Ans. Z 0.822 Vtank Vtank 5.636 ft 61 3.59 T 25K P 3.213bar Calculate the effective critical parameters for hydrogen by equations (3.58) and (3.56) Tc 1 43.6 21.8K 2.016T K Tc 30.435 K Pc 1 20.5 44.2K 2.016T bar Pc 10.922 bar 0 Pr P Pc Pr 0.294 Tr T Tc Tr 0.821 3 Initial guess of volume: V RT P V cm 646.903 mol Use the generalized Pitzer correlation B0 Z 1 0.083 0.422 Tr 1.6 B0 0.495 B1 0.139 0.172 Tr 4.2 B1 0.254 B0 B1 Pr Tr Z 0.823 Ans. Experimental: Z = 0.7757 For Redlich/Kwong EOS: 1 0.5 0 0.08664 0.42748 Table 3.1 Tr Tr Table 3.1 q Tr Tr Tr Eq. (3.54) T r Pr Pr Tr Eq. (3.53) 62 Calculate Z Guess: Z 0.9 Given Eq. (3.52) Z= 1 T r Pr q Tr T r Pr Z Z Z T r Pr T r Pr Z Find Z) ( Z 0.791 Ans. Experimental: Z = 0.7757 3.61For methane: 0.012 Tc 190.6K Pc 45.99bar T 288.706 K At standard condition: Pitzer correlations: T ( 60 32) 5 9 273.15 K P 1atm Tr T Tc Tr 1.515 Pr B1 P Pc 0.139 0.172 Tr 4.2 Pr B1 0.022 B0 Z0 0.083 0.422 Tr Pr Tr 1.6 B0 0.134 0.109 1 B0 Z0 0.998 Z1 B1 Pr Tr Z1 0.00158 Z Z0 Z1 Z 0.998 V1 Z RT P V1 m 0.024 mol P 300psi 3 (a) At actual condition: Pitzer correlations: T T ( 50 32) 5 9 273.15 K 283.15 K Tr T Tc 1.6 Tr 1.486 Pr P Pc Pr 0.45 B0 0.083 0.422 Tr B0 0.141 B1 0.139 0.172 Tr 4.2 B1 0.106 63 Z0 1 B0 Pr Tr Z0 0.957 Z1 B1 Pr Tr Z1 0.0322 3 Z Z0 Z1 3 Z 0.958 V2 ZRT P V2 0.00109 3 m mol q1 150 10 6 ft day q2 q1 V2 V1 3 kmol q2 6.915 10 6 ft day Ans. (b) n1 q1 V1 n1 7.485 10 hr Ans. (c) D 22.624in q2 A A 4 D 2 A 0.259 m 2 u u 8.738 m s Ans. 64 3.62 0.012 0.087 0.1 0.140 0.152 0.181 0.187 0.19 0.191 0.194 0.196 0.2 0.205 0.21 0.21 0.212 0.218 0.23 0.235 0.252 0.262 0.28 0.297 0.301 0.302 0.303 0.31 0.322 0.326 ZC 0.286 0.281 0.279 0.289 0.276 0.282 0.271 0.267 0.277 0.275 0.273 0.274 0.273 0.273 0.271 0.272 0.275 0.272 0.269 0.27 0.264 0.265 0.256 0.266 0.266 0.263 0.263 0.26 0.261 65 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. m slope ZC ( 0.091) b intercept ZC ( 0.291) 2 r corr ZC ( 0.878) r 0.771 0.29 ZC m 0.28 b 0.27 0.26 0.25 0 0.1 0.2 0.3 0.4 The equation of the line is: Zc = 0.291 0.091 Ans. 3.65 Cp 7 R 2 Cv 5 2 R Cp Cv 1.4 T1 298.15K P1 1bar P2 5bar T3 T1 P3 5bar Step 1->2 Adiabatic compression 1 T2 U12 T1 P2 P1 Cv T2 T1 T2 472.216 K U12 3.618 H12 Cp T2 T1 H12 5.065 kJ mol kJ Ans. mol Ans. Q12 0 kJ mol Q12 W12 0 kJ mol kJ mol Ans. Ans. W12 U12 3.618 Step 2->3 Isobaric cooling U23 Cv T3 T2 U23 3.618 kJ mol kJ Ans. H23 Cp T3 T2 H23 5.065 Q23 H23 Q23 5.065 mol kJ Ans. W23 R T3 T2 W23 1.447 mol kJ Ans. mol Ans. Step 3->1 Isothermal expansion U31 Cv T1 T3 U31 0 H31 Cp T1 T3 66 H31 0 kJ mol kJ Ans. Ans. mol Q31 W31 R T3 ln Q31 P1 P3 Q31 3.99 W31 kJ mol kJ 3.99 mol Ans. Ans. For the cycle Qcycle Q12 Q23 Q31 Qcycle 1.076 kJ Ans. mol Wcycle W12 W23 W31 Wcycle 1.076 kJ Ans. mol Now assume that each step is irreversible with efficiency: 80% Step 1->2 Adiabatic compression W12 W12 W12 4.522 kJ mol Ans. Q12 U12 W12 Q12 0.904 kJ mol Ans. Step 2->3 Isobaric cooling W23 W23 W23 1.809 kJ mol Ans. Q23 U23 W23 Q23 5.427 kJ mol Ans. Step 3->1 Isothermal expansion W31 W31 W31 3.192 Q31 U31 W31 Q31 3.192 kJ mol kJ Ans. mol Ans. For the cycle Qcycle Q12 Q23 Q31 Qcycle Wcycle W12 W23 W31 Wcycle kJ Ans. mol kJ 3.14 mol Ans. 3.14 67 3.67 a) PV data are taken from Table F.2 at pressures above 1atm. 125 150 175 P 200 225 250 275 300 Z PVM RT kPa 2109.7 1757.0 1505.1 V 1316.2 1169.2 1051.6 955.45 875.29 cm 3 gm T ( 300 273.15) K gm mol M 18.01 1 VM 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 A Zi i 1 Xi i B intercept ( Y) X 6 5 cm 10 2 B cm 128.42 mol 3 Ans. slope ( Y) X mol cm 3 A 1.567 5 mol 3 mol X 0 10 5 mol 3 8 10 cm cm Below is a plot of the data along with the linear fit and the extrapolation to the y-intercept. 68 115 120 (Z-1)/p 125 130 0 2 10 (Z-1)/p Linear fit 5 4 10 p 5 6 10 5 8 10 5 b) Repeat part a) for T = 350 C PV data are taken from Table F.2 at pressures above 1atm. 125 150 175 P 200 225 250 275 300 Z PVM RT kPa 2295.6 1912.2 1638.3 V 1432.8 1273.1 1145.2 1040.7 953.52 cm gm 3 T ( 350 273.15)K gm mol M 18.01 1 VM 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 i 69 1 Xi i B intercept ( X Y) B 105.899 cm 3 mol Ans. A slope ( X Y) A 5 mol 3 1.784 5 mol 3 6 5 cm 10 2 mol X 0 mol cm 3 10 8 10 cm cm Below is a plot of the data along with the linear fit and the extrapolation to the y-intercept. 90 95 (Z-1)/p 100 105 110 0 2 10 (Z-1)/p Linear fit 5 4 10 p 5 6 10 5 8 10 5 c) Repeat part a) for T = 400 C PV data are taken from Table F.2 at pressures above 1atm. 125 150 175 P 200 225 250 275 300 kPa 2481.2 2066.9 1771.1 V 1549.2 1376.6 1238.5 1125.5 1031.4 70 cm gm 3 T ( 400 273.15)K M 18.01 gm mol Z PV M RT 1 VM 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 i 1 Xi i B intercept ( X Y) 6 5 cm 10 2 B 89.902 cm 3 mol Ans. A slope ( X Y) A 2.044 5 mol 3 mol X 0 mol cm 3 10 5 mol 3 8 10 cm cm Below is a plot of the data along with the linear fit and the extrapolation to the y-intercept. 70 75 (Z-1)/p 80 85 90 0 2 10 (Z-1)/p Linear fit 5 4 10 p 5 6 10 5 8 10 5 71 3.70 Create a plot of (Z 1) Z Tr Pr vs Pr Z Tr Data from Appendix E at Tr = 1 0.01 0.05 0.10 Pr 0.20 0.40 0.60 0.80 Pr Z Tr 0.9967 0.9832 0.9659 Z 0.9300 0.8509 0.7574 0.6355 Tr 1 X Y (Z 1) Z Tr Pr Create a linear fit of Y vs X Slope Intercept Rsquare slope ( X Y) intercept ( X Y) corr ( X Y) 0.28 Slope Intercept Rsquare 0.033 0.332 0.9965 Y Slope X Intercept 0.3 0.32 0.34 0 0.2 0.4 0.6 X 0.8 1 1.2 1.4 The second virial coefficient (Bhat) is the value when X -> 0 Bhat Intercept B0 0.083 0.422 Tr 72 Bhat 1.6 0.332 0.339 Ans. Ans. By Eqns. (3.65) and (3.66) These values differ by 2%. B0 3.71 Use the SRK equation to calculate Z Tc 150.9 K T ( 30 T 273.15)Kr T Tc P Tr 2.009 Pc 48.98 bar P 300 bar Pr Pc Pr 6.125 0.0 1 0 0.08664 1 0.42748 Table 3.1 2 Tr q Tr Calculate Z 1 Tr Tr 0.480 1.574 0.176 2 1 Tr 2 Table 3.1 Eq. (3.54) Guess: T r Pr Z 0.9 Pr Tr Eq. (3.53) Given Eq. (3.52) Z= 1 T r Pr q Tr T r Pr Z Z T r Pr T r Pr Z T r Pr Z Find Z) ( Z 1.025 V Z RT P V cm Ans. 86.1 mol 3 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 Vt 0.4 1800 cm 3 cm 5 mol 33.0 mol 3 Vt 555 cm 3 V Vt n 73 V 111 cm 3 mol Use SRK equation to calculate pressure. Tc 308.3 K T ( 125 273.15) K Tr T Tc Tr 1.291 Pc 61.39 bar 0.0 1 0 0.08664 1 2 0.42748 Table 3.1 Tr q Tr 1 Tr Tr 0.480 1.574 0.176 2 1 Tr 2 Table 3.1 Eq. (3.54) a Tr Pc R Tc 2 2 Eq. (3.45) b R Tc Pc Eq. (3.46) 3 a 3 3 bar cm 3.995 m 2 mol b cm 36.175 mol P RT V b a V (V b) T ( 10 273.15)K P 197.8 bar Ans. 3.73 mass 35000kg 0.152 Tc 369.8K Pc 3 42.48bar M n 44.097 7.937 gm mol 5 Zc 0.276 Vc 200.0 cm mol n mass M 10 mol a) Estimate the volume of gas using the truncated virial equation Tr T Tc Tr 0.766 P 1atm Pr P Pc B0 0.083 0.422 Tr 1.6 Eq. (3-65) B1 0.139 0.172 Tr 4.2 Eq. (3-66) B0 0.564 74 B1 0.389 Z 1 B0 B1 Pr Tr Z 0.981 7 3 3 Vt ZnRT P Vt 3 2.379 6 10 m m This would require a very large tank. If the D tank were spherical the diameter would be: 0.2857 Vt D 32.565 m b) Calculate the molar volume of the liquid with the Rackett equation(3.72) Vliq P V c Zc 6.294atm 1 Tr Vliq Pr B1 Pr Tr Z 85.444 0.15 cm 3 mol P Pc Pr Z 1 B0 ZRT P Vtank 90% 0.878 3 3 cm Vvap Guess: Given Vvap 90% Vliq n 10% Vtank Vvap 3.24 10 mol Vtank Vliq = n Vtank Vtank Find Vtank 75.133 m 6 3 This would require a small tank. If the tank D were spherical, the diameter would be: 3 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 Chapter 4 - Section A - Mathcad Solutions 4.1 (a) T0 473.15 K T 1373.15 K n 10 mol For SO2: A 5.699 B 0.801 10 3 C 0.0 D 1.015 10 5 H R ICPH T0 T A B C D H 47.007 kJ mol Q n H Q (b) T0 470.073 kJ Ans. 523.15 K T 1473.15 K n 3 12 mol For propane:A 1.213 B 28.785 10 C 8.824 10 6 D 0 H R ICPH T0 T A B C 0.0 H 161.834 kJ mol Q n H Q 1.942 10 kJ Ans. 3 4.2 (a) T0 473.15 K n 10 mol Q 3 800 kJ For ethylene: A 1.424 B 14.394 10 K C 4.392 10 K 2 6 2 (guess) Given Q= nR A T0 1 B 2 T0 2 T 2 1 C 3 T0 3 3 1 Find 2.905 T0 T 1374.5 K Ans. (b) T0 533.15 K n 15 mol Q 3 2500 kJ For 1-butene: A 1.967 B 31.630 10 K 76 C 9.873 10 K 2 6 3 (guess) Given Q= nR Find A T0 1 B 2 T0 2 2.652 2 1 T0 C 3 T0 3 3 1 T T 1413.8 K Ans. 6 (c) T0 500 degF n 40 lbmol Q 10 BTU Values converted to SI units T0 533.15K n 1.814 10 mol 4 Q 3 1.055 10 kJ 6 For ethylene: A 1.424 B 14.394 10 K 2 C 4.392 10 K 2 6 2 (guess) Given Q= nR Find A T0 1 B 2 T0 2 T 1 C 3 T0 3 3 1 2.256 T0 T 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 1 atm T0 122 degF V 250 ft 3 T 3 932 degF Convert given values to SI units V 7.079 m T ( T 32degF) 273.15K T0 T0 32degF 273.15K T 773.15 K T0 323.15 K n PV R T0 n 266.985 mol For air: A 3.355 B 0.575 10 3 C 0.0 D 0.016 10 5 H R ICPH T0 T A B C D 77 H 13.707 kJ mol Q n H Q 4.4 molwt 3.469 3 10 BTU Ans. 100.1 gm mol T0 323.15 K 4 T 1153.15 K n 10000 kg molwt n 9.99 10 mol For CaCO3: A 12.572 B 2.637 10 3 C 0.0 D 3.120 10 5 H R ICPH T0 T A B C D H 9.441 10 4 J mol Q n H Q 9.4315 10 kJ 6 Ans. 4.7 Let step 12 represent the initial reversible adiabatic expansion, and step 23 the final constant-volume heating. T1 P2 298.15 K 101.3 kPa T3 P3 298.15 K 104.0 kPa P1 T2 121.3 kPa T3 P2 P3 CP 30 J mol K (guess) R CP T2 290.41 K Given T2 = T1 P2 P1 CP Find CP CP 56.95 J Ans. mol K 4.9a) Acetone: Tc 508.2K Pc 47.01bar Tn 329.4K Hn 29.10 kJ mol Trn Tn Tc Trn 0.648 Use Eq. (4.12) to calculate H at Tn ( Hncalc) 1.092 ln Hncalc R Tn Pc bar Trn 1.013 0.930 Hncalc 30.108 kJ mol Ans. 78 To compare with the value listed in Table B.2, calculate the % error. %error Hncalc Hn Hn %error 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 % 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% 79 b) 469.7 Tc 507.6 562.2 560.4 K 33.70 Pc 30.25 48.98 43.50 366.3 273.15 K bar 25.79 Hn 28.85 kJ 30.72 mol 29.97 72.150 J 36.0 Tn 68.7 80.0 80.7 H25 366.1 433.3 gm 392.5 M 86.177 gm 78.114 mol 82.145 Tr1 Tn Tc 0.658 0.673 0.628 0.631 Tr2 ( 25 273.15) K Tc H2 H25 M 26.429 H2 31.549 H1 Hn Tr1 kJ 33.847 mol 32.242 1 1 Tr2 Tr1 0.38 H2calc H1 Eq. (4.13) %error H2calc H2 0.072 H2 26.448 H2calc 31.533 kJ Ans. 33.571 mol 26.429 H2 31.549 kJ 33.847 mol %error 0.052 0.814 1.781 % 32.816 32.242 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 (a) T 459.67 5 V 1.934 5 0 0.012 i 1 5 18.787 21.162 Data: P 23.767 26.617 29.726 t 5 10 15 xi ti 1 459.67 yi ln Pi slope slope ( x y) slope 4952 dPdT ( P) 3 T 2 slopedPdT 0.545 H T V dPdT 5.4039 H 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 H = 85.817 ( 90.111) ( 85.834) (b) (c) H = 81.034 ( 81.136) (d) H = 76.007 ( 75.902) (e) H = 69.863 ( 69.969) 4.11 119.377 M 32.042 153.822 gm mol 536.4 Tc 512.6 K Pc 556.4 Hexp is the given value at the normal boiling point. 81 54.72 80.97 bar Tn 45.60 334.3 337.9 K 349.8 Tn Tc H is the value at 0 degC. Tr1 273.15K Tc Tr2 270.9 H 1189.5 217.8 J gm 246.9 Hexp 1099.5 194.2 J gm 0.509 Tr1 0.533 0.491 1 1 Tr2 Tr1 0.38 0.623 Tr2 0.659 0.629 (a) By Eq. (4.13) Hn H PCE Hn Hexp Hexp 100% This is the % error 245 Hn J 1055.2 gm 193.2 0.77 PCE 4.03 % 0.52 (b) By Eq. (4.12): Hn R Tn M 1.092 ln Pc bar Tr2 1.013 0.930 PCE Hn Hexp Hexp 100% 247.7 Hn 1195.3 192.3 J gm 0.34 PCE 8.72 0.96 % 4.12 Acetone 0.307 Vc Tr Tc 3 508.2K 329.4K 0.648 Pc P Pr 47.01bar 1atm P Pc Zc Hn Pr 0.233 29.1 kJ mol cm 209 mol Tn Tc Tn Tr 0.022 82 Generalized Correlations to estimate volumes Vapor Volume B0 0.083 0.422 Tr B1 0.139 1.6 B0 0.762 Eq. (3.65) 0.172 Tr 4.2 B1 Pr Tr 0.924 Eq. (3.66) Z 1 B0 Pr Tr B1 Z 0.965 (Pg. 102) 3 4 cm V Z R Tn P 2 V 2.609 10 mol Liquid Volume 1 Tr 7 Vsat V c Zc Eq. (3.72) Vsat H= T V cm 70.917 mol d Psat dT 3 Combining the Clapyeron equation (4.11) B T C A with Antoine's Equation Psat = e gives H= T V ( T 3 4 cm B C) 2 A B ( C) T e V A V Vsat V B 2.602 2756.22 10 mol C 228.060 14.3145 83 A B Tn 273.15K Hcalc Tn V Tn K B 273.15K 2 C e C K kPa K Hcalc 29.662 kJ Ans. mol %error Hcalc Hn Hn %error 1.9 % 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% 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 348.15 K P 100 kPa (guess) 5622.7 K T Given ln P = 48.157543 kPa P 5622.7 K T 2 4.70504 ln T K 0.029 bar K 3 P Find ( ) dPdT P 4.70504 T dPdT P 87.396 kPa H joule 31600 mol Vliq cm 96.49 mol Clapeyron equation: dPdT = H T V Vliq V = vapor molar volume. V Vliq H T dPdT cm 1369.5 mol 3 Eq. (3.39) B PV V RT 1 B Ans. 4.14 (a) Methanol: Tc 512.6K Pc 80.97bar Tn 3 337.9K AL 13.431 BL 51.28 10 CL 131.13 10 6 CPL ( ) T AL BL K BV 85 T CL K 2 T 2 R AV 2.211 12.216 10 3 CV 3.450 10 6 CPV ( ) T AV BV K T CV K 2 T 2 R P 3bar Tsat 368.0K T1 300K T2 500K Estimate Hv using Riedel equation (4.12) and Watson correction (4.13) Trn Tn Tc Trn 0.659 Trsat Tsat Tc Trsat 0.718 1.092 ln Hn Pc bar Trn 1.013 0.930 R Tn Hn 38.301 kJ mol kJ mol Hv Hn Tsat 1 1 Trsat Trn 0.38 Hv T2 35.645 H T1 CPL ( ) T T d Hv T sat CPV ( ) T T d H 49.38 kJ mol Ans. n 100 kmol hr Q n H Q 1.372 10 kW 3 (b) Benzene: Hv = 28.273 kJ mol H = 55.296 kJ mol Q = 1.536 10 kW 3 (c) Toluene Hv = 30.625 kJ mol H = 65.586 kJ mol Q = 1.822 10 kW 3 4.15 Benzene Tc 562.2K Pc 48.98bar Tn 353.2K T1sat 451.7K T2sat 358.7K Cp 162 J mol K 86 Estimate Hv using Riedel equation (4.12) and Watson correction (4.13) Trn Tn Tc Trn 0.628 Tr2sat T2sat Tc Tr2sat 0.638 1.092 ln Hn Pc bar Trn 1.013 0.930 R Tn Hn 30.588 kJ mol Hv Hn 1 1 Tr2sat Trn 0.38 Hv 30.28 kJ mol Assume the throttling process is adiabatic and isenthalpic. Guess vapor fraction (x): x 0.5 Given Cp T1sat T2sat = x Hv x Find ( x) x 0.498 Ans. 4.16 (a) For acetylene: Tc 308.3 K Pc 61.39 bar Tn 189.4 K T 298.15 K Trn Tn Tc Trn 0.614 Tr T Tc Tr 0.967 ln Hn R Tn 1.092 1 1 Tr Trn J mol Pc bar 0.930 0.38 1.013 Trn Hn 16.91 kJ mol kJ mol Hv Hn Hv 6.638 Hf 227480 H298 Hf Hv H298 220.8 kJ mol Ans. 87 (b) For 1,3-butadiene: H298 = 88.5 kJ mol (c) For ethylbenzene: H298 = 12.3 kJ mol (d) For n-hexane: H298 = 198.6 kJ mol (e) For styrene: H298 = 103.9 kJ mol 4.17 1st law: dQ = dU dW = CV dT P dV (A) Ideal gas: PV= RT and P dV V dP = R dT Whence Since V dP = R dT P V = const then P dV P V 1 (B) dV = V dP from which V dP = P dV Combines with (B) to yield: P dV = R dT 1 Combines with (A) to give: dQ = CV dT R dT 1 or dQ = CP dT R dT R dT 1 which reduces to dQ = CP dT R dT 1 (C) R dT or dQ = CP R 1 Since CP is linear in T, the mean heat capacity is the value of CP at the arithmetic mean temperature. Thus Tam 675 88 CPm R 3.85 0.57 10 3 Tam 950 K T1 400 K 1.55 Integrate (C): T2 R T2 Q CPm R 1 T1 Q 6477.5 J mol Ans. P1 1 bar P2 P1 T2 T1 1 P2 11.45 bar Ans. 4.18 For the combustion of methanol: CH3OH(g) + (3/2)O2(g) = CO2(g) + 2H2O(g) H298 H298 393509 676485 2 ( 241818) ( 200660) For 6 MeOH: H298 = 4 058 910 J Ans. 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 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 2 n 2 11.286 i 1 4 A 3.639 5.457 B 3.470 3.280 0.506 1.045 1.450 0.593 10 K 3 0.227 1.157 D 0.121 0.040 10 K 5 2 A i ni Ai B i ni Bi D i ni D i A 54.872 B 0.012 T 1 K D 1.621 10 K 5 2 For the products, HP = R T0 CP R dT T0 298.15K The integral is given by Eq. (4.7). Moreover, by an energy balance, H298 HP = 0 2 (guess) Given H298 = R A T0 8.497 1 B 2 T0 2 T 2 1 D T0 1 Find T T0 2533.5 K 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 2 nn = 14.107 2 T = 2198.6 K Ans. (c) nO = 1.5 2 nn = 16.929 2 T = 1950.9 K T = 1609.2 K Ans. Ans. (d) nO = 3.0 2 nn = 22.571 2 90 (e) 50% xs air preheated to 500 degC. For this process, Hair H298 HP = 0 773.15) Hair = MCPH ( 298.15 For one mole of air: MCPH 773.15 298.15 3.355 0.575 10 For 4.5/0.21 = 21.429 moles of air: Hair = n R MCPH T Hair Hair 3 0.0 0.016 10 5 = 3.65606 21.429 8.314 3.65606 ( 298.15 309399 J mol H298 0.506 B 1.045 1.450 0.593 773.15) J mol The energy balance here gives: 1.5 n 2 2 16.929 Hair HP = 0 3.639 A 5.457 3.470 3.280 0.227 10 K 3 D 1.157 0.121 0.040 10 K 5 2 A i ni Ai 78.84 B i ni Bi 1 0.016 K D i ni D i 1.735 10 K 5 2 A B D 2 (guess) Given H298 Hair = R A T0 1 B 2 T0 2 2 1 D T0 Find 7.656 91 1 T T0 K T 2282.5 K K Ans. 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 (b) -905,468 J (c) -71,660 J (d) -61,980 J (e) -367,582 J (f) -2,732,016 J (g) -105,140 J (h) -38,292 J (i) 164,647 J (j) -48,969 J (k) -149,728 J (l) -1,036,036 J (m) 207,436 J (n) 180,500 J (o) 178,321 J (p) -132,439 J (q) -44,370 J (r) -68,910 J (s) -492,640 J (t) 109,780 J (u) 235,030 J (v) -132,038 J (w) -1,807,968 J (x) 42,720 J (y) 117,440 J (z) 175,305 J 92 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. Ho298 T/ K A (a) (b ) (f ) (i ) (j ) (l ) (m) (n ) (o ) (r ) (t ) (u ) (v ) (w) (x ) (y ) 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 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 4.23 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 4.24 q 150 10 6 ft 3 day T ( 60 5 32) K 9 273.15K T 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: HfCH4 74520 J mol HfO2 0 J mol HfCO2 393509 J mol HfH2Oliq 285830 J mol Hc HfCO2 2 HfH2Oliq HfCH4 2 HfO2 HigherHeatingValue Hc Hc 8.906 10 5 J mol Assuming methane is an ideal gas at standard conditions: n q P RT n 1.793 10 5dollar GJ 8 mol day 5 dollar n HigherHeatingValue 4.25 7.985 10 day Ans. Calculate methane standard heat of combustion with water as liquid product Standard Heats of Formation: CH4 + 2O2 --> CO2 +2H2O HfCH4 HfCO2 HcCH4 HcCH4 74520 J mol J mol HfO2 0 J mol 285830 J mol 393509 HfCO2 890649 HfH2Oliq HfCH4 2 HfH2Oliq J mol 94 2 HfO2 Calculate ethane standard heat of combustion with water as liquid product: Standard Heats of Formation:C2H6 + 7/2O2 --> 2CO2 +3H2O HfC2H6 83820 J mol HcC2H6 2 HfCO2 3 HfH2Oliq HfC2H6 7 2 HfO2 HcC2H6 1560688 J mol Calculate propane standard heat of combustion with water as liquid product Standard Heats of Formation:C3H8 + 5O2 --> 3CO2 +4H2O HfC3H8 104680 J mol HcC3H8 3 HfCO2 4 HfH2Oliq HfC3H8 5 HfO2 HcC3H8 2219.167 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 c) 0.85 HcCH4 0.05 HcC2H6 0.07 HcC2H6 0.03 HcC3H8 0.03 HcC3H8 946.194 kJ mol kJ mol 932.875 Gas b) has the highest standard heat of combustion. 4.26 2H2 + O2 = 2H2O(l) Ans. Hf1 2 ( 285830) J C + O2 = CO2(g) Hf2 393509 J N2(g)+2H2O(l)+CO2(g)=(NH2)2CO(s)+3/2O2 H 631660 J . N2(g)+2H2(g)+C(s)+1/2O2(g)=(NH2)2CO(s) H298 Hf1 Hf2 H 95 H298 333509 J Ans. 4.28 On the basis of 1 mole of C10H18 (molar mass = 162.27) Q 43960 162.27 J Q 7.133 10 J 6 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 H U= H ( )= PV H ngas H R T ngas ( 10 14.5)mol 6 298.15 K Q R T ngas 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 Moles oxygen Moles nitrogen n1 n2 n3 96 1 2 1.3 2.6 79 21 n2 n3 2.6 9.781 Total moles of dry gases entering n n1 n2 n3 n 13.381 At 30 degC the vapor pressure of water is 4.241 kPa. Moles of water vapor entering: n4 4.241 13.381 101.325 4.241 n4 (1) (2) (3) (4) 0.585 Leaving: CO2 -- 1 mol H2O -- 2.585 mol O2 -- 2.6 - 2 = 0.6 mol N2 -- 9.781 mol By an energy balance on the furnace: Q= H= H298 HP For evaluation of HP we number species as above. 5.457 1.045 B 1.450 0.506 0.593 10 3 1 n 2.585 0.6 9.781 i 1 4 R 1.157 D 0.121 0.227 0.040 10 5 A 3.470 3.639 3.280 8.314 J mol K A i ni Ai B i ni Bi 3 D i ni Di A 48.692 B 10.896983 10 C 0 D 5.892 10 4 The TOTAL value for MCPH of the product stream: HP R M CPH ( 303.15K 1773.15K A B C D) ( 1773.15 303.15)K HP 732.013 kJ mol From Example 4.7: H298 802625 J mol Q HP H298 Q = 70 612 J 97 Ans. 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 n 1.578 2.585 1.578 Moles water condensing: 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 n H50 Ans. 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 ENERGY BALANCE: H= HR H298 HP = 0 REACTANTS: 1=NH3; 2=O2; 3=N2 4 n 6.5 24.45 i 1 3 A 3.578 3.639 3.280 B 3.020 0.506 10 0.593 3 0.186 D 0.227 10 0.040 5 A i ni Ai B i ni Bi D i ni Di A 118.161 B 0.02987 C 0.0 D 1.242 10 5 TOTAL mean heat capacity of reactant stream: HR R MCPH ( 348.15K 298.15K A B C D) ( 298.15K 348.15K) HR 52.635 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 2.5 n 3.2 4.8 24.45 i 1 5 A 3.578 3.639 3.387 3.470 3.280 B 3.020 0.506 0.629 1.450 0.593 10 K 3 0.186 0.227 D 0.014 0.121 0.040 10 K 5 2 A i ni Ai B i ni Bi D i ni Di A 119.65 B 1 0.027 K D 8.873 10 K 4 2 By the energy balance and Eq. (4.7), we can write: (guess) 2 T0 298.15K 99 Given H298 H R = R A T0 1 B 2 T0 2 2 1 D T0 Find 1 3.283 T T0 T 978.9 K Ans. 4.31 C2H4(g) + H2O(g) = C2H5OH(l) BASIS: 1 mole ethanol produced Energy balance: n 1mol H= Q= HR H298 H298 [ 277690 ( 52510 241818) ] J mol H298 8.838 10 4 J mol Reactant stream consists of 1 mole each of C2H4 and H2O. i 1 2 n 1.424 3.470 1 1 B 14.394 1.450 10 3 A C 4.392 0.0 10 6 D 0.0 0.121 10 5 A i ni Ai B i ni Bi C i ni Ci D i ni Di A 4.894 B 0.01584 C 4.392 10 6 D 1.21 10 4 HR R M CP ( H 298.15K 593.15K A B C D)( 298.15K 593.15K) HR 2.727 10 4 J mol Q HR H298 1mol Q 115653 J Ans. 100 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 H298 4.003 10 4 J mol Q= HR H298 HP REACTANTS: 1=CH4; 2=H2O 1.702 3.470 9.081 1.450 i 1 2 2.164 0.0 n 0.2 0.4 10 6 A B 10 3 C D 0.0 0.121 10 5 A i ni Ai B i ni Bi C i ni Ci D i ni Di A 1.728 B 2.396 10 3 C 4.328 10 7 D 4.84 10 3 HR RI CPH ( 773.15K 298.15K A B C D) HR 1.145 10 4 J mol PRODUCTS: 1=CO2; 2=CO; 3=H2O; 4=H2 0.0275 n 0.1725 0.1725 0.6275 A 5.457 3.376 3.470 3.249 B 1.045 0.557 1.450 0.422 10 3 1.157 D 0.031 0.121 0.083 10 5 101 i 1 4 A i ni Ai B i ni Bi 4 D i ni Di A 3.37 B 6.397 10 C 0.0 D 3.579 10 3 HP HP RI CPH ( 298.15K 1123.15K A B C D) 4 J 2.63 10 mol Q HR H298 HP mol Q 54881 J Ans. 4.33 CH4 + 2O2 = CO2 + 2H2O(g) C2H6 + 3.5O2 = 2CO2 + 3H2O(g) H298a 802625 H298b 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 Q 8 10 5 J mol Energy balance: Q = H = H298 HP PRODUCTS: 1=CO2; 2=H2O; 3=O2; 4=N2 HP = Q H298 1.157 1.25 n 2.25 1.9 16.082 i 1 4 A 5.457 3.470 3.639 3.280 B 1.045 1.450 0.506 0.593 10 K 3 D 0.121 0.227 0.040 10 K 5 2 A i ni Ai B i ni Bi D i ni Di 4 2 A 74.292 B 0.015 102 1 K C 0.0 D 9.62 10 K By the energy balance and Eq. (4.7), we can write: T0 Given 303.15K 2 (guess) Q H298 = R A T0 1 B 2 T0 2 2 1 Find D T0 1.788 T T0 1 Ans. T 542.2 K 4.34 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 0.129 A 5.699 3.639 8.060 B 0.801 0.506 10 1.056 3 1.015 D 0.227 10 2.028 D i 5 i 1 3 A i ni Ai B i ni Bi 2.58 10 103 ni Di D 1.16 10 4 A 0.06985 B 7 C 0 Hnet R M CPH ( 298.15K 773.15K A B C D)( 773.15K 298.15K) Hnet 77.617 J mol H773 H298 Hnet H773 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: Energy balance: moles CO = moles H2O = 0.2 moles CO2 = moles H2 = 0.3 Q= H= HR H298 HP H298 0.3 [ 393509 ( 110525 J 214818) ] mol 0.557 1.450 H298 2.045 10 4 J mol Reactants: 1: CO 2: H2O n i 0.5 0.5 1 2 A 3.376 3.470 B 10 3 D 0.031 0.121 10 5 A i ni Ai B i ni Bi D i ni Di A 3.423 B 1.004 10 3 C 0 D 4.5 10 3 HR R M CPH ( 298.15K 398.15K A B C D)( 298.15K 398.15K) HR 3.168 10 3 J mol Products: 1: CO 2: H2O 3: CO2 4: H2 3.376 A 3.470 5.457 3.249 104 0.2 n 0.2 0.3 0.3 0.557 B 1.450 1.045 0.422 10 3 0.031 D 0.121 1.157 0.083 10 5 i 1 4 A i ni Ai 3.981 B i ni Bi 8.415 10 4 D i ni Di 3.042 10 4 A B C 0 D HP HP Q R MCPH ( 298.15K 698.15K A B C D) ( 698.15K 1.415 10 HR 4 J 298.15K) mol HP mol Q 9470 J Ans. H298 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 Since air is 21 mol % O2, 15.124 x 100.175 0.21 = 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: BTU 209.13 lbm lbm 6 Q 0.3 19000 Q 1.192 10 BTU Reaction upon which net heating value is based: 106 OIL + (21.024)O2 = (14.8)CO2 + (12.448 + 0.116)H2O(g) + (0.149)N2 H298a 19000 209.13 BTU H298a 3.973 10 BTU 6 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 393509) 0.42993 BTU Guess: y 50 (y)H2O(l) = (y)H2O(g) H298c ( y) 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 3 11.8 n y) ( 5.913 79.278 15.797 y A r 1.986 T 1.045 0.557 B 400 459.67 1.8 T 477.594 5.457 3.376 3.639 3.280 3.470 1.157 0.031 3 5 0.506 10 0.593 1.450 D 0.227 10 0.040 0.121 i 1 5 A ( y) i n y) i Ai B ( y) ( i n y) i Bi D ( y) ( i n y) i Di ( D ( y) T0 2 T T0 1.602 CP ( y) r A ( y) B ( y) T0 2 1 107 Given CP ()( y 400 77)BTU = Q H298 () y y Find y () y Whence 49.782 (lbmol H2O evaporated) y 18.015 209.13 4.288 (lb H2O evap. per lb oil burned) Ans. 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 HP H298 ( 135100 2 227480) 0.242 J 2 H298 5.169 10 J 3 Products: 0.242 n 0.379 0.379 i 1 3 A i 4.736 A 3.280 6.132 ni Ai B 1.359 0.593 10 1.952 3 0.725 D 0.040 1.299 10 5 B i ni Bi D i ni Di A 4.7133 B 1.2934 10 3 C 0 D 6.526 10 4 HP R M CPH ( 298.15K 873.15K A B C D)( 873.15K 298.15K)mol HP 2.495 10 J 4 HP 2.495 10 J 4 Q H298 HP Q 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 For this reaction, H298 [ ( 241818) 2 4 ( 92307) ] J mol H298 1.144 10 5 J mol Evaluate H823 by Eq. (4.21) with T0 298.15K T 823.15K 1: H2O 2: Cl2 3: HCl 4=O2 2 n 2 4 1 i 1 4 A 3.470 4.442 3.156 3.639 B 1.45 0.089 0.623 0.506 10 3 0.121 D 0.344 0.151 0.227 10 5 A i ni Ai B i ni Bi D i ni Di 8.23 10 4 A H823 H298 0.439 MCPH T0 T B A 8 10 B 5 C 0 D C D R T T0 H823 117592 J mol Heat transferred per mol of entering gas mixture: Q H823 4 0.45 mol Q 13229 J Ans. 4.39 CO2 + C = 2CO 2C + O2 = 2CO Eq. (4.21) applies to each reaction: H298a 172459 H298b J (a) mol J (b) 221050 mol For (a): 2 n 1 1 A 3.376 1.771 5.457 109 0.557 B 0.771 10 1.045 3 0.031 D 0.867 10 1.157 5 i 1 3 A i ni Ai B i ni Bi 4 D i ni Di A H1148a 0.476 B 7.02 10 C 0 D 1.962 10 5 H298a R M CP 298.15K 1148.15K H A B C D ( 1148.15K 298.15K) H1148a 1.696 10 5 J mol For (b): 2 n 1 2 i 1 3 A 3.376 3.639 1.771 B 0.557 0.506 10 0.771 3 0.031 D 0.227 10 0.867 5 A i ni Ai B B i 4 ni Bi C 0 D i ni Di 1.899 10 5 A H1148b 0.429 9.34 10 D H298b R M CP 298.15K 1148.15K H 5 J A B C D ( 1148.15K 298.15K) H1148b 2.249 10 mol The combined heats of reaction must be zero: nCO 2 H1148a nO 2 H1148b = 0 2 nCO Define: r= nO r H1148b H1148a r 1.327 2 110 For 100 mol flue gas and x mol air, moles are: Flue gas Air Feed mix CO2 CO O2 N2 12.8 3.7 5.4 78.1 0 0 0.21x 0.79x 12.8 3.7 5.4 + 0.21x 78.1 + 0.79x Whence in the feed mix: r= 12.8 5.4 0.21 x x 12.5 5.4 r mol 0.21 x 19.155 mol Flue gas to air ratio = 100 19.155 5.221 Ans. Product composition: nCO 3.7 2 ( 12.8 5.4 0.21 19.155) nCO nN 2 48.145 93.232 nN 2 78.1 0.79 19.155 nCO nCO nN 2 Mole % CO = 100 34.054 Ans. Mole % N2 = 100 34.054 65.946 4.40 CH4 + 2O2 = CO2 + 2H2O(g) CH4 + (3/2)O2 = CO + 2H2O(g) H298a 802625 J mol H298b 519641 J mol 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 mol O2 2.538 79 21 9.548 mol N2 111 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 0.282 H298b H298 6.747 10 5 J mol Moles H2O formed by reaction = 0.94 2.0 1.88 Moles O2 consumed by reaction = 2 0.658 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 0.282 n 1.880 0.799 9.608 i 1 5 A i 5.457 3.376 A 3.470 3.639 3.280 B 1.045 0.557 1.450 10 0.506 0.593 3 1.157 0.031 D 0.121 0.227 0.040 10 5 ni Ai 45.4881 B B i ni Bi 3 D i ni Di 3.396 10 298.15K) 4 A HP HP 9.6725 10 C 0 D R M CPH ( 298.15K 483.15K A B C D)( 483.15K 7.541 10 4 J mol Hrx H298 HP Hrx 599.252 kJ mol kg 34.0 sec Energy balance: HH2O mdotH2O Hrx ndotfuel = 0 mdotH2O 112 From Table C.1: HH2O ( 398.0 104.8) kJ kg ndotfuel HH2O mdotH2O Hrx ndotfuel 16.635 mol sec Volumetric flow rate of fuel, assuming ideal gas: V ndotfuel R 298.15 K 101325 Pa V m 0.407 sec 3 Ans. 4.41 C4H8(g) = C4H6(g) + H2(g) 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 1 n 1 1 2.734 A 3.249 1.967 i 1 3 B T0 298.15 K T 798.15 K by Eq. (4.21): Evaluate H798 1: C4H6 2: H2 3: C4H8 26.786 0.422 31.630 10 3 8.882 C 0.0 9.873 10 6 0.0 D 0.083 10 0.0 5 A i ni Ai B i ni Bi 3 C i ni Ci 7 D i ni Di 3 A 4.016 B 4.422 10 C 9.91 10 D 8.3 10 H798 H298 MCPH 298.15K 798.15K 5 J A B C D R T T0 H798 1.179 10 mol Q 0.33 mol H798 Q 38896 J 113 Ans. 4.42Assume Ideal Gas and P = 1 atm P 1atm R 7.88 10 3 BTU mol K a) T0 ( 70 459.67) rankine T T0 20rankine Q 12 BTU sec T0 294.261 K 3 T 305.372 K ICPH T0 T 3.355 0.575 10 0 0.016 10 5 38.995 K ndot Q R ICPH T0 T 3.355 0.575 10 3 0 3 0.016 10 5 ndot 39.051 3 mol s Vdot ndot R T0 P Vdot m 0.943 s T T0 Vdot ft 33.298 sec Q 12 kJ s Ans. b) T0 R ( 24 273.15) K 3 13K 8.314 10 kJ mol K 3 ICPH T0 T 3.355 0.575 10 0 0.016 10 5 45.659 K ndot Q R ICPH T0 T 3.355 0.575 10 3 0 0.016 10 5 ndot 31.611 3 mol s Vdot ndot R T0 P Vdot m 0.7707 s Ans. 4.43Assume Ideal Gas and P = 1 atm P 1atm a) T0 ( 94 459.67) rankine 3 T ( 68 459.67) rankine R 1.61 10 atm ft mol rankine 3 Vdot ft 50 sec 3 ndot P Vdot R T0 114 ndot 56.097 mol s T0 307.594 K 3 T 0 293.15 K 0.016 10 5 ICPH T0 T 3.355 0.575 10 R 7.88 10 3 BTU 50.7 K mol K 3 Q R ICPH T0 T 3.355 0.575 10 0 0.016 10 5 ndot 22.4121 BTU Ans. sec Q b) T0 ( 35 273.15)K 3 5 atm m T ( 25 273.15)K R 8.205 10 m 1.5 sec 3 mol K ndot 3 Vdot P Vdot R T0 0 0.016 10 5 ndot 59.325 mol s ICPH T0 T 3.355 0.575 10 R Q 35.119 K 8.314 10 3 kJ mol K 3 R ICPH T0 T 3.355 0.575 10 0 0.016 10 5 ndot 17.3216 kJ Ans. s Q 4.44 First calculate the standard heat of combustion of propane C3H8 + 5O2 = 3CO2(g) + 4H2O (g) H298 H298 3 393509 J mol 4 241818 J mol 104680 J mol 2.043 10 dollars gal 6 J mol 80% Cost 2.20 115 Estimate the density of propane using the Rackett equation Tc T 369.8K ( 25 273.15) K 1 Tr Zc 0.276 Tr T Tc Vc Tr cm 200.0 mol 0.806 3 0.2857 Vsat V c Zc Vsat cm 89.373 mol 3 Heating_cost Vsat Cost H298 Heating_cost Heating_cost 0.032 dollars MJ dollars 10 BTU 6 Ans. 33.528 4.45 T0 ( 25 273.15) K Q Q T ( 500 273.15) K 3 a) Acetylene R ICPH T0 T 6.132 1.952 10 2.612 10 4 J 0 1.299 10 5 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 4.46 T0 Q ( 25 30000 273.15)K J mol T ( 500 273.15)K a) Acetylene T Given Q = R ICPH T0 T 6.132 1.952 10 Find ( T) T 835.369 K T 3 0 1.299 10 5 273.15K 562.2 degC The calculations are repeated and the answers are in the following table: 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( 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 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 ( 250 y 273.15) K 0.5 Q 11500 J mol a) Guess mole fraction of methane: Given y ICPH T0 T 1.702 9.081 10 (1 y 3 2.164 10 3 6 0 R 6 = Q y) ICPH T0 T 1.131 19.225 10 Find ( y) y 0.637 Ans. 117 5.561 10 0 R b) T0 ( 100 273.15) K T ( 400 y 273.15)K 0.5 Q 54000 J mol Guess mole fraction of benzene Given y ICPH T0 T ( 1 y c) T0 0.206 39.064 10 3 13.301 10 3 6 0 R 6 = Q y)ICPH T0 T Find y () ( 150 3.876 63.249 10 Ans. T ( 250 y 20.928 10 0 R y 0.245 273.15) K 273.15)K 0.5 Q 17500 J mol Guess mole fraction of toluene Given y ICPH T0 T 0.290 47.052 10 ( 1 y 3 15.716 10 3 6 0 R 6 = Q y)ICPH T0 T 1.124 55.380 10 Find y () y 0.512 Ans. 18.476 10 0 R 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. I e m e a e Pi h nt r di t nc Se ton I ci TH1 Pi h a End nc t TH1 Se ton I ci I Se ton I ci THi T THi Se ton I ci I TC1 TCi TH2 TC1 TCi TC2 TH2 T TC2 118 To solve the problem, apply an energy balance around each section of the exchanger. T H1 Section I balance: mdotC HC1 HCi = ndotH THi CP dT T Hi Section II balance: mdotC HCi HC2 = ndotH TH2 CP dT If the pinch is intermediate, then THi = TCi + T. If the pinch is at the end, then TH2 = TC2 + T. a) TH1 1000degC TC1 100degC TCi 100degC TC2 25degC T 10degC HC1 2676.0 kJ kg HCi 419.1 3 kJ kg HC2 104.8 kJ kg 5 For air from Table C.1:A 3.355 B 0.575 10 1 C 0 D 0.016 10 Assume as a basis ndot = 1 mol/s. ndotH kmol s Assume pinch at end: TH2 TC2 T Guess: mdotC 1 kg s THi 110degC Given mdotC HC1 mdotC HCi 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 kg s mdotC THi mdotC ndotH Find mdotC THi THi 170.261 degC mdotC 11.255 0.011 kg mol Ans. THi TCi 70.261 degC TH2 119 TC2 10 degC 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 500degC TC1 100degC TCi 100degC TC2 25degC T 10degC HC1 2676.0 kJ kg HCi 419.1 kJ kg HC2 104.8 kJ kg Assume as a basis ndot = 1 mol/s. ndotH 1 kmol s Assume pinch is intermediate: THi TCi T Guess: mdotC 1 kg s TH2 110degC Given mdotC HC1 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 3 kg mdotC HCi mdotC TH2 48.695 degC mdotC 5.03 kg s mdotC ndotH 5.03 10 mol Ans. THi TCi 10 degC 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 H0f3 393.509 kJ mol H0f4 285.830 kJ mol M3 44 gm mol H0r 6 H0f3 6 H0f4 H0f1 6 H0f2 H0r 2801.634 kJ mol Ans. b) energy_per_kg 150 kJ kg mass_person 57kg mass_glucose mass_person energy_per_kg H0r M1 mass_glucose 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 2.216 10 kg 8 Ans. 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 H0f1 74.520 kJ mol H0f2 83.820 kJ mol H0f3 0 kJ mol H0f4 393.509 kJ mol H0f5 241.818 kJ mol a) H0c 1.05 H0f4 2 H0f5 0.85 H0f1 0.10 H0f2 1.05 H0f3 H0c 825.096 kJ mol 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 0.5 2.05mol 121 n3 1.025 mol Excess O2 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 ( 25 273.15) K T2 ( 600 273.15) K A n3 3.639 n4 6.311 n5 3.470 n6 3.280 3 mol B n3 0.506 n4 0.805 n5 1.450 mol n6 0.593 10 C n3 0 n4 0 n5 0 mol n6 0 10 6 D n3 ( 0.227) n4 ( 0.906) n5 0.121 mol n6 0.040 10 5 Q H0c ICPH T1 T2 A B C D R Q 529.889 kJ mol Ans. 122 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 = 1 TC TH TC 323.15 K TH 798.15 K 148.78 kJ s QH 250 kJ s Work QH 1 TC TH or Work Work QH 148.78 kW which is the power. Ans. Work By Eq. (5.1), QC QC 101.22 kJ s Ans. 5.3 (a) Let symbols Q and Work represent rates in kJ/s TH 750 K TC 300 K Work 95000 kW By Eq. (5.8): Work QH 1 TC TH QH 0.6 Work 5 1.583 10 kW Ans. But = Whence QH QC (b) QH QH QC Work QC 6.333 4 10 kW Ans. 0.35 Work QH Work QH QC 5 2.714 10 kW Ans. 1.764 5 10 kW Ans. 5.4 (a) TC 303.15 K TH 623.15 K Carnot 1 TC TH 0.55 123 Carnot 0.282 Ans. (b) 0.35 Carnot 0.55 Carnot 0.636 By Eq. (5.8), TH TC 1 Carnot TH 833.66 K Ans. 5.7 Let the symbols represent rates where appropriate. Calculate mass rate of LNG evaporation: V 9000 m s 3 P 1.0133 bar T 298.15 K molwt 17 gm mol mLNG PV molwt RT mLNG 6254 kg s Maximum power is generated by a Carnot engine, for which Work QC = QH QC QC = QH QC 1= TH TC 1 TH 303.15 K TC QC 113.7 K 6 QC Work 512 kJ mLNG kg TH TC 3.202 10 kW 6 QC 1 Work 5.336 10 kW Ans. QH QC Work QH 8.538 10 kW 6 Ans. 5.8 Take the heat capacity of water to be constant at the valueCP 4.184 (a) T1 273.15 K T2 373.15 K Q CP T2 kJ kg K T1 Q 418.4 kJ kg K kJ kg SH2O Sres CP ln Q T2 T2 T1 SH2O 1.305 Sres 124 1.121 kJ kg K Ans. Stotal SH2O Sres Stotal 0.184 kJ kgK Ans. (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. Sres Q 2 1 323.15 K 1 373.15 K Sres 1.208 kJ kgK Stotal Sres SH2O Stotal 0.097 kJ kgK Ans. (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. 3 5.9 P1 1 bar T1 500 K V 0.06m n P1 V R T1 n 1.443 mol CV 5 2 R Q 15000 J (a) Const.-V heating; U= Q W = Q = n CV T2 T1 T2 T1 Q n CV T2 1 10 K 3 By Eq. (5.18), P2 P1 S = n CP ln T2 T1 T2 T1 R ln P2 P1 But = Whence S n CV ln T2 T1 S 20.794 J K Ans. (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 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: 7 R 2 SA SA CP ln 8.726 463.15 343.15 J mol K SB CP ln 473.15 593.15 SB 6.577 J mol K Ans. (b) For both cases: Stotal SA SB Stotal 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 CP ln 8.726 463.15 343.15 J mol K SB CP ln 353.15 473.15 SB 8.512 J mol K Ans. Stotal SA SB Stotal 0.214 J mol K Ans. 5.16 By Eq. (5.8), dW dQ = 1 T T dW = dQ T dQ T Since dQ/T = dS, dW = dQ T dS Integration gives the required result. T1 600 K T2 400 K T 300 K Q CP T2 T1 Q 5.82 10 3 J mol 126 S CP ln Q T2 T1 T S S 11.799 J mol K Work Work 2280 J mol Ans. Q Q Work Q 3540 J mol Ans. Sreservoir Q T Sreservoir 11.8 J mol K Ans. S Sreservoir 0 J mol K Process is reversible. 5.17 TH1 600 K TC1 300 K TH2 300 K TH1 TC2 TC1 250 K For the Carnot engine, use Eq. (5.8): W QH1 = 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 300K P1 1.2bar T2 450K P2 6bar Cp 7 R 2 H C p T2 T1 R ln 3 J H P2 P1 4.365 10 3 J mol Ans. S Cp ln T2 T1 S 1.582 J mol K Ans. (b) H = 5.82 10 mol S = 1.484 J mol K 127 (c) H = 3.118 10 3 J mol 3 J S = 4.953 J mol K (d) H = 3.741 10 mol 3 J S = 2.618 J mol K (e) H = 6.651 10 mol S = 3.607 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 T3 V 3 1 = T4 V 4 1 For constant-volume step 4 to 1: For isobaric step 2 to 3: V1 = V4 P2 T2 = P3 T3 Solving these 4 equations for T4 yields: T4 = T1 7 R 2 T2 T3 Cp Cv 5 R 2 Cp Cv 1.4 T1 ( 200 273.15) K T2 T4 ( 1000 273.15) K T3 ( 1700 273.15) K T4 T1 T2 T3 1 873.759 K Eq. (A) p. 306 1 T4 T3 T1 T2 0.591 Ans. 128 5.21 CV CP R P1 2 bar P2 7 bar T1 298.15 K CP CV 1.4 With the reversible work given by Eq. (3.34), we get for the actual W: 1 Work 1.35 R T1 1 P2 1 Work 3.6 10 3 J P1 U = C V T2 T1 Whence mol But Q = 0, and W = T2 T1 Work CV T2 471.374 K S CP ln T2 T1 R ln P2 P1 S 2.914 J mol K Ans. 5.25 P 4 T 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 T dP = dT P (A) PV= RT P dV V dP = R dT P dV = R dT V dP = R dT RT dP P 129 In combination with (A) this becomes P dV = R dT Moreover, R dT = 2 R dT P3 = P 1 T1 T3 = P1 T1 T2 V1 W31 = V3 P dV = 2 R T 1 U31 W31 = CV T1 2R T1 T3 = 2 R T1 T3 2 R T1 R T1 T2 Q31 = T3 T2 Q31 = CV T3 = C P W23 Q31 = Wnet Qin = W12 W31 CP 7 R 2 T1 P1 700 K T2 P3 W12 2.91 350 K T1 T2 3 J 1.5 bar P1 10 W12 R T2 T1 mol 3 J W23 W31 Q31 R T2 ln P3 P1 T2 T1 T2 W23 W31 Q31 2.017 10 mol 3 J 2 R T1 CP W12 5.82 10 mol R W23 Q31 1.309 10 4 J mol W31 0.068 Ans. 130 5.26 T 403.15 K P1 2.5 bar P2 6.5 bar Tres J mol K 298.15 K By Eq. (5.18), S R ln P2 P1 S 7.944 Ans. With the reversible work given by Eq. (3.27), we get for the actual W: Work 1.3 R T ln P2 P1 (Isothermal compresion) Work 4.163 10 3 J mol Q Work Q here is with respect to the system. So for the heat reservoir, we have Sres Stotal Q Tres S Sres Sres Stotal 13.96 J mol K J mol K Ans. 6.02 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 3 0.0 1.015 10 5 S 536.1 J K 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 3 8.824 10 6 0.0 S 2018.7 J K Ans. 131 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 S S 10 mol n R ICPS 473.15K 1374.5K 1.424 14.394 10 900.86 J K Ans. 3 4.392 10 6 0.0 (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 S S 15 mol n R ICPS 533.15K 1413.8K 1.967 31.630 10 2657.5 J K Ans. 3 9.873 10 6 0.0 (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 S S 18140 mol n R ICPS 533.15K 1202.9K 1.424 14.394 10 1.2436 10 6 J 3 4.392 10 6 0.0 K 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 T1 T2 298.15 K 248.15 K 348.15 K T0 ( 1 Temperature of entering air Temperature of chilled air Temperature of warm air x)CP T2 T0 = 0 x C P T1 x 0.3 (guess) 132 Given x 1 x = T2 T1 T0 T0 x Find x () 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 R P0 5 bar P T2 T0 1 bar P P0 Stotal x CP ln T1 T0 ( 1 x)CP ln R ln Stotal 12.97 J mol K Ans. Since this is positive, there is no violation of the second law. 5.30 T1 523.15 K T2 353.15 K P1 3 bar P2 1 bar Tres CV Sres 303.15 K Work 1800 J mol CP Q Q 7 2 R CP R Q Q= Sres U Work 5.718 J mol K C V T2 T1 3 J Work Tres 1.733 10 mol S CP ln T2 T1 R ln P2 P1 S 2.301 J mol K Stotal S Sres Stotal 3.42 J PROCESS IS POSSIBLE. mol K 133 5.33 For the process of cooling the brine: CP 3.5 kJ kg K T 40 K mdot 20 kg sec t 0.27 T1 ( 273.15 25) K T1 298.15 K T2 ( 273.15 15) K T2 258.15 K T ( 273.15 30) K T 303.15 K H CP T H 140 kJ kg kJ kg K S CP ln T2 T1 Wdotideal Wdot S 0.504 Eq. (5.26): By Eq. (5.28): mdot H T S Wdotideal Wdot 256.938 kW Wdotideal t 951.6 kW Ans. 5.34 E 110 volt i 9.7 amp T 300 K Wdotmech 1.25 hp Wdotelect iE Wdotelect 1.067 10 W 3 At steady state: Qdot Wdotelect Wdotmech = d t U = 0 dt Qdot T SdotG = d t S = 0 dt Qdot Wdotelect Wdotmech Qdot 134.875 W SdotG Qdot T SdotG 0.45 W K Ans. 134 5.35 25 ohm i 2 10 amp T 300 K Wdotelect i Wdotelect 2.5 10 W 3 At steady state: Qdot Wdotelect = d t U = 0 dt Qdot Wdotelect Qdot T SdotG = d t S = 0 dt SdotG Qdot T Qdot 2.5 10 watt 3 SdotG 8.333 watt K Ans. 5.38 mdot 10 kmol hr T1 ( 25 273.15) K P1 10bar P2 1.2bar Cp 7 R 2 Cv Cp R Cp Cv 7 5 (a) Assuming an isenthalpic process: T2 T2 T1 T2 298.15 K Ans. S (b) = R T1 Cp 1 R T dT ln P2 P1 P2 P1 Eq. (5.14) S T2 7 R ln T1 2 R ln S 17.628 J mol K Ans. (c) SdotG mdot S SdotG 48.966 W K Ans. 3 J (d) T ( 20 273.15) K Wlost T S Wlost 5.168 10 mol Ans. 5.39(a) T1 500K P1 6bar T2 371K P2 1.2bar Cp 7 R 2 T 300K Basis: 1 mol n 1mol H n C p T2 T1 Ws 135 H Ws 3753.8 J Ans. S n Cp ln T2 T1 Wideal R ln P2 P1 S 4.698 J K Ans. Eq. (5.27) Eq. (5.30) H T S Wideal 5163 J Wlost SG Wideal Wlost T Wideal 5163J 2953.9J 4193.7J Ws Wlost SG Wlost 1409.3J 493J 1130J 1409.3 J 4.698 SG 4.698 1.643 3.767 Ans. Ans. Eq. (5.39) Ws (a) (b) (c) J K 3753.8J 2460.9J 3063.7J J K J K J K (d) 3853.5J 4952.4J 1098.8J 3.663 J K J (e) 3055.4J 4119.2J 1063.8J 3.546 K 5.41 P1 2500kPa P2 S SdotG 150kPa 0.023 kJ mol K T 300K mdot 20 S SdotG R ln P2 P1 mol sec mdot S T SdotG W 0.468 kJ sec K Ans. Ans. TH TC Wdotlost 5.42 QH 1kJ Wdotlost TH TC 140.344 kW ( 250 ( 25 0.45kJ 273.15)K 273.15)K 523.15 K 298.15 K W actual QH actual 0.45 136 max 1 TC TH max 0.43 Since actual> max, the process is impossible. 5.43 QH 150 kJ Q1 50 kJ Q2 100 kJ TH 550 K T1 350 K T2 250 K T 300 K (a) SG QH TH Q1 T1 Q2 T2 SG 0.27 kJ K Ans. (b) Wlost T SG Wlost 81.039 kJ Ans. 5.44 Wdot 750 MW TH ( 315 273.15)K TC TC ( 20 273.15)K TH (a) max 588.15 K 0.502 Ans. 293.15 K 1 TC TH max QdotH Wdot max QdotC QdotC QdotH Wdot (minimum value) 745.297 MW (b) 0.6 max QdotH Wdot Wdot QdotH 10 MW 3 2.492 10 W 9 QdotC QdotH QdotC 1.742 3 (actual value) River temperature rise: Vdot m 165 s 1 gm cm 3 Cp 1 cal gm K T QdotC Vdot Cp T 2.522 K Ans. 137 5.46 T1 P1 ( 20 5bar 273.15)K T2 P2 ( 27 1atm 273.15) K T3 ( 22 273.15)K First check the First Law using Eqn. (2.33) neglect changes in kinetic and potential energy. H 6 7 R ICPH T1 T2 3.355 0.575 10 1 7 3 0 0.016 10 5 ICPH T1 T3 3.355 0.575 10 10 4 kJ 3 0 0.016 10 5 R H 8.797 mol H is essentially zero so the first law is satisfied. Calculate the rate of entropy generation using Eqn. (5.23) SG 6 7 R ICPS T1 T2 3.355 0.575 10 1 7 3 0 3 0.016 10 5 R ICPS T1 T3 3.355 0.575 10 kJ mol K Since SG 0 0.016 10 5 R ln P2 P1 SG 0.013 0, this process is possible. 5.47 a) Vdot P ft 100000 hr 1atm 3 T1 T ( 70 ( 70 459.67)rankine T2 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 3 0 0.016 10 3 5 5 R ICPS T1 T2 3.355 0.575 10 0 0.016 10 Wideal 1.776 hp 138 b) Vdot P 3000 1atm m 3 hr T1 T ( 25 ( 25 273.15) K 273.15) K T2 (8 273.15) K Assume air is an Ideal Gas R T1 Calculate ideal work using Eqn. (5.26) ndot P Vdot ndot 34.064 mol s Wideal ndot R ICPH T1 T2 3.355 0.575 10 T 3 0 0.016 10 3 5 5 R ICPS T1 T2 3.355 0.575 10 0 0.016 10 Wideal 5.48 T1 1.952 kW 459.67) rankine T2 ( 300 459.67) rankine ( 2000 Cp ( ) T T ( 70 3.83 0.000306 T rankine R Hv Tsteam 970 BTU lbm M 29 gm mol 459.67) rankine ( 212 459.67) rankine a) First apply an energy balance on the boiler to get the ratio of steam flow rate to gas flow rate.: T2 ndotgas T1 Cp ( ) T T d T2 mdotsteam Hv = 0 Cp ( ) T T d mdotndot T1 Hv mdotndot 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 Calculate entropy generation per lbmol of gas: SdotG ndotgas Ssteam = mdotsteam ndotgas Hv Tsteam T2 Ssteam Sgas BTU lb rankine 3 kg Ssteam 1.444 Sgas T1 C p ( T) T dT Sgas Sgas SdotG 9.969 10 BTU mol lb rankine SdotG mdotndot Ssteam 11.756 BTU lbmol rankine Calculate lost work by Eq. (5.34) Wlost SdotG T Hv Tsteam Wlost 6227 BTU Ans. lbmol 1.444 BTU lb rankine BTU lb b) Hsteam Hv Ssteam Ssteam Wideal Hsteam T Ssteam Wideal 205.071 Calculate lbs of steam generated per lbmol of gas cooled. T2 C p ( T ) dT mn T1 Hv mn 15.043 lb lbmol Use ratio to calculate ideal work of steam per lbmol of gas Wideal mn 3.085 T2 10 3 BTU lbmol Ans. c) Hgas T1 Cp ( T) dT Hgas T Sgas Wideal 9.312 10 3 BTU Wideal lbmol Ans. 140 5.49 T1 ( 1100 273.15) K T2 ( 150 273.15) K Cp ( ) T T ( 25 3.83 0.000551 T R K Hv Tsteam 2256.9 ( 100 kJ kg M 29 gm mol 273.15) K 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 ndotgas T1 Cp ( ) T T d T2 mdotsteam Hv = 0 Cp ( ) T T d mdotndot T1 Hv mdotndot 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 Hv Tsteam T2 Ssteam Sgas 3 Ssteam 6.048 10 J kg K Sgas T1 Cp ( ) T dT T Sgas Sgas SdotG 41.835 J mol K SdotG mdotndot Ssteam 49.708 J mol K Ans. Calculate lost work by Eq. (5.34) Wlost SdotG T Wlost 14.8 kJ mol 141 b) Hsteam Hv Ssteam Hv Tsteam Ssteam 6.048 10 kJ kg 3 J kg K Wideal Hsteam T Ssteam Wideal 453.618 Calculate lbs of steam generated per lbmol of gas cooled. T2 Cp ( T) dT mn T1 Hv mn 15.135 gm mol Use ratio to calculate ideal work of steam per lbmol of gas Wideal mn 6.866 T2 kJ mol Ans. c) Hgas T1 Cp ( T) dT Hgas T Sgas Wideal 21.686 kJ mol T 4.392 10 6 Wideal Ans. 5.50 T1 a) ( 830 273.15)K T2 ( 35 273.15)K 3 ( 25 0 273.15)K Sethylene Sethylene Qethylene Qethylene Wlost T R ICPS T1 T2 1.424 14.394 10 0.09 kJ mol K R ICPH T1 T2 1.424 14.394 10 60.563 kJ mol Qethylene 3 4.392 10 6 0 Sethylene 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 Sethylene QC T = 0 Solving for QC gives: QC T Sethylene QC 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 Chapter 6 - Section A - Mathcad Solutions 6.7 At constant temperature Eqs. (6.25) and (6.26) can be written: dS = V dP and dH = 1 T V dP For an estimate, assume properties independent of pressure. T 270 K P1 3 3 m 381 kPa P2 1200 kPa 3 V 1.551 10 kg 2.095 10 K 1 S V P2 P1 H 1 T V P2 P1 S 2.661 J kg K Tc Ans. H 551.7 J kg Ans. 6.8 Isobutane: 408.1 K Zc 0.282 CP 2.78 P1 4000 kPa J gm K 3 P2 2000 kPa molwt gm 58.123 mol Vc cm 262.7 mol 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 360 K 361 Tr T Tc 0.88 Tr 0.882 0.885 (The elements are denoted by subscripts 1, 2, & 3 2 7 131.604 V 132.138 132.683 V V c Zc 1 Tr cm mol 3 Assume that changes in T and V are negligible during throtling. Then Eq. (6.8) is integrated to yield: 144 H= T S V P but H= 0 Then at 360 K, S V 1 P2 T1 P1 S 0.733 J mol K Ans. We use the additional values of T and V to estimate the volume expansivity: V V3 V1 V 1.079 cm 3 mol T T3 T1 T 2K 1 V1 V T 4.098835 10 3 K 1 Assuming properties independent of pressure, Eq. (6.29) may be integrated to give S = CP T T V P P P2 P1 P 2 10 kPa 3 Whence T T1 CP S V1 P molwt T 0.768 K Ans. 6.9 T 298.15 K 250 10 6 P1 K 1 1 bar 45 10 6 P2 1500 bar bar 1 V1 cm 1003 kg 3 By Eq. (3.5), V1 2 V2 V2 Vave V1 exp 970.287 cm P2 3 P1 V2 cm 937.574 kg 3 Vave kg By Eqs. (6.28) & (6.29), H Vave 1 T P2 P1 U H P2 V 2 P1 V 1 H 134.6 kJ kg Ans. U 5.93 kJ kg Ans. S Vave P2 P1 Q T S Work U Q S 0.03636 kJ kg K Ans. Q 10.84 145 kJ Ans. Work kg 4.91 kJ kg Ans. 6.10 For a constant-volume change, by Eq. (3.5), T2 T1 5 P2 1 P1 = 0 T1 298.15 K 5 1 T2 P1 323.15 K 1 bar 36.2 10 T2 K 4.42 10 bar P2 6.14 --- 6.16 300 175 575 500 325 175 575 T 650 300 400 150 575 375 475 K T1 P1 P2 205.75 bar Ans. for Parts (a) through (n): 61.39 48.98 48.98 37.96 73.83 34.99 45.60 .187 .000 .210 .200 .224 .048 .193 bar Vectors containing T, P, Tc, Pc, and 40 75 30 50 60 60 35 P 50 35 70 50 15 25 75 bar Tc 308.3 150.9 562.2 425.1 304.2 132.9 556.4 K Pc 553.6 282.3 373.5 126.2 568.7 369.8 365.6 40.73 50.40 89.63 34.00 24.90 42.48 46.65 .210 .087 .094 .038 .400 .152 .140 Tr T Tc Pr P Pc 146 6.14 Redlich/Kwong equation: Pr Tr 0.08664 0.42748 Eq. (3.53) q Tr z 1 q z z z 1.5 Eq. (3.54) Guess: Given Z i z= 1 q Find z () Eq. (3.52) 1 14 Ii ln Z Z i qi i qi i Eq. (6.65b) HRi SRi R Ti Z i qi i qi 1 i 1.5 qi Ii Eq. (6.67) The derivative in these 0.5 qi Ii Eq. (6.68) equations equals -0.5 SRi R ln Z Z i qi 0.695 0.605 0.772 0.685 0.729 0.75 0.709 0.706 0.771 0.744 0.663 0.766 0.775 0.75 HRi -2.302103 -2.068103 -3.319103 -4.503103 -2.3103 -1.362103 -4.316103 -5.381103 -1.764103 -2.659103 -1.488103 -3.39103 -2.122103 -3.623103 J mol -5.461 -8.767 -4.026 -6.542 -5.024 -5.648 -5.346 -5.978 -4.12 -4.698 -7.257 -4.115 -3.939 -5.523 J mol K Ans. 147 6.15 Soave/Redlich/Kwong equation: 0.08664 0.5 2 0.42748 Pr Tr c 0.480 1.574 0.176 2 1 c 1 Tr Eq. (3.53) q Eq. (3.54) Tr Guess: Given z z= 1 1 q z z z Eq. (3.52) Z q Find z () The derivative in the following equations equals: ci i 1 14 Ii ln Z Z Tri i i qi i qi 0.5 Tri i 0.5 i Eq. (6.65b) HRi R Ti Z i qi 1 ci 1 qi Ii 0.5 Eq. (6.67) SRi R ln Z i qi i ci Tri i qi Ii Eq. (6.68) Z i qi 0.691 0.606 0.774 0.722 0.741 0.768 0.715 0.741 0.774 0.749 0.673 0.769 0.776 0.787 HRi -2.595103 -2.099103 -3.751103 -4.821103 -2.585103 -1.406103 -4.816103 -5.806103 -1.857103 -2.807103 -1.527103 -4.244103 -2.323103 -3.776103 SRi J mol -6.412 -8.947 -4.795 -7.408 -5.974 -6.02 -6.246 -6.849 -4.451 -5.098 -7.581 -5.618 -4.482 -6.103 J mol K Ans. 148 6.16 Peng/Robinson equation: 1 2 1 2 0.07779 0.5 0.45724 2 c Pr Tr 0.37464 1.54226 0.26992 2 1 Guess: c 1 z Tr Eq. (3.53) q Eq. (3.54) Tr 1 Given z= 1 q z z z Eq. (3.52) Z q 0.5 Find ( z) The derivative in the following equations equals: ci i 1 14 Ii 1 2 i qi Tri i ln 2 1 Z Z ci i qi i qi i i Eq. (6.65b) HRi R Ti Z Tri i 0.5 1 qi Ii 0.5 Eq. (6.67) SRi Z i qi 0.667 0.572 0.754 0.691 0.716 0.732 0.69 0.71 0.752 0.725 0.64 0.748 0.756 0.753 R ln Z HRi i qi i ci Tri i qi Ii Eq. (6.68) SRi -2.655103 -2.146103 -3.861103 -4.985103 -2.665103 -1.468103 -4.95103 -6.014103 -1.917103 -2.896103 -1.573103 -4.357103 -2.39103 -3.947103 J mol -6.41 -8.846 -4.804 -7.422 -5.993 -6.016 -6.256 -6.872 -4.452 -5.099 -7.539 -5.631 -4.484 -6.126 J mol K Ans. 149 Lee/Kesler Correlation --- By linear interpolation in Tables E.1--E.12: h0 equals ( ) HR RTc ( ) SR R 0 0 h1 equals ( ) HR RTc ( ) SR R 1 1 h equals HR RTc SR R s0 equals s1 equals s equals .686 .590 .774 .675 .725 .744 Z0 .705 .699 .770 .742 .651 .767 .776 .746 Z1 .093 .155 .024 .118 .008 .165 .019 .102 .001 .007 .144 .034 .032 .154 h0 .950 1.709 .705 1.319 .993 1.265 .962 1.200 .770 .875 1.466 .723 .701 1.216 h1 1.003 .471 .591 .437 .635 .184 .751 .444 .550 .598 .405 .631 .604 .211 Z Z0 Z1 Eq. (3.57) h h0 h1 (6.85) HR ( Tc R) h 150 .711 1.110 .497 .829 .631 .710 s0 .674 .750 .517 .587 .917 .511 .491 .688 .961 .492 .549 .443 .590 .276 s1 .700 .441 .509 .555 .429 .589 .563 .287 HRi -0.891 -1.11 -0.612 -0.918 -0.763 -0.723 -0.809 -0.843 -0.561 -0.639 -0.933 -0.747 -0.577 -0.728 -2.916103 -2.144103 -3.875103 -4.971103 -2.871103 -1.407103 -5.121103 -5.952103 -1.92103 -2.892103 -1.554103 -4.612103 -2.438103 -3.786103 s s0 s1 SR ( s R) Eq. (6.86) Zi 0.669 0.59 0.769 0.699 0.727 0.752 0.701 0.72 0.77 0.743 0.656 0.753 0.771 0.768 hi -1.138 -1.709 -0.829 -1.406 -1.135 -1.274 -1.107 -1.293 -0.818 -0.931 -1.481 -0.975 -0.793 -1.246 si SRi J mol -7.405 -9.229 -5.091 -7.629 -6.345 -6.013 -6.727 -7.005 -4.667 -5.314 -7.759 -6.207 -4.794 -6.054 J mol K Ans. 151 6.17 t 50 273.15 K The pressure is the vapor pressure given by the Antoine equation: T 323.15 K t T P () t d dt exp 13.8858 2788.51 t 220.79 P 36.166 kPa P ( ) 36.166 50 P () 1.375 t dPdt 1.375 kPa 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: 0.210 Tc 562.2 K Pc 48.98 bar Zc 0.271 Vc 259 cm 3 mol Tr T Tc Tr 0.575 Pr P Pc Pr 0.007 By Eqs. (3.65), (3.66), (3.61), & (3.63) B0 0.083 0.422 Tr 1.6 B0 0.941 B1 0.139 0.172 Tr 4.2 B1 3 4 cm 1.621 Vvap RT P 1 B0 B1 Pr Tr 1 Tr 2/7 Vvap 7.306 10 mol cm 3 By Eq. (3.72), Vliq V c Zc Vliq 93.151 mol Solve Eq. (6.72) for the latent heat and divide by T to get the entropy change of vaporization: S dPdt Vvap Vliq S 100.34 J mol K Ans. (b) Here for the entropy change of vaporization: S RT dPdt P 152 S 102.14 J mol K Ans. 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 x Hfusion = 0 CP 4.226 J gm K t 6K Hfusion 333.4 joule gm x CP t Hfusion x 0.076 Ans. The entropy change for the two steps is: T2 273.15 K T1 x Hfusion T2 ( 273.15 6) K 3 S CP ln T2 T1 S 1.034709 10 J Ans. gm K The freezing process itself is irreversible, because it does not occur at the equilibrium temperature of 0 degC. 6.21 Data, Table F.4: H1 1156.3 BTU lbm H2 1533.4 BTU lbm S1 1.7320 BTU S2 lbm rankine 1.9977 BTU lbm rankine H H2 H1 S S2 S1 H 377.1 BTU lbm S 0.266 BTU Ans. lbm rankine For steam as an ideal gas, apply Eqs. (4.9) and (5.18). [t in degF] T1 ( 227.96 459.67)rankine T2 ( 1000 459.67)rankine P1 20 psi P2 50 psi T1 382.017 K T2 810.928 K 153 molwt 18 lb lbmol 3 5 H H R MCPH T1 T2 3.470 1.450 10 molwt 372.536 BTU lbm Ans. 0.0 0.121 10 T2 T1 R MCPS T1 T2 3.470 1.450 10 S BTU lbm rankine 3 0.0 0.121 10 5 ln T2 T1 ln P2 P1 molwt S 0.259 Ans. 6.22 Data, Table F.2 at 8000 kPa: Vliq cm 1.384 gm 3 Hliq 3 1317.1 J gm J gm Sliq 3.2076 J gm K J gm K Vvap cm 23.525 gm 6 Hvap 2759.9 Svap 0.15 10 3 cm 2 Vvap 6 5.7471 mliq mliq Htotal Stotal 0.15 10 3 cm 2 Vliq 54.191 kg mliq Hliq mliq Sliq mvap mvap 3.188 kg Htotal Stotal mvap Hvap mvap Svap 80173.5 kJ 192.145 kJ K Ans. Ans. 154 6.23 Data, Table F.2 at 1000 kPa: Vliq 1.127 cm 3 gm Hliq 3 762.605 J gm Sliq 2.1382 J gm K Vvap 194.29 cm gm Hvap 2776.2 J gm Svap 6.5828 J gm K Let x = fraction of mass that is vapor (quality) x 0.5 (Guess) Given x Vvap ( 1 x)Vliq = 70 30 x Find x () x 0.013 H ( 1 x)Hliq x Hvap S S ( 1 x)Sliq J gm K x Svap Ans. H 789.495 J gm 2.198 6.24 Data, Table F.3 at 350 degF: Vliq ft 0.01799 lbm 3 Vvap ft 3.342 lbm 3 Hliq 321.76 BTU lbm Hvap 1192.3 BTU lbm mliq mvap = 3 lbm mvap Vvap = 50 mliq Vliq mliq 50 mliq Vliq = 3 lbm Vvap mliq 1 3 lbm 50 Vliq Vvap mliq 2.364 lb mvap 3 lbm mliq mvap 0.636 lb Htotal mliq Hliq mvap Hvap Htotal 1519.1 BTU Ans. 155 6.25 V 1 cm 0.025 gm 3 Data, Table F.1 at 230 degC: 3 Vliq 1.209 cm gm Hliq 3 990.3 J gm Sliq 2.6102 J gm K Vvap 71.45 cm gm Hvap 2802.0 J gm Svap 6.2107 J gm K V= ( 1 x)Vliq x Vvap x V Vvap Vliq Vliq H ( 1 x)Hliq x Hvap S ( 1 x)Sliq x Svap x 0.552 H 1991 J gm S 4.599 J gm K Ans. 6.26 Vtotal = mtotal Vliq mvap Vlv Table F.1, 150 degC: Vtotal 0.15 m 3 Vvap 3 cm 392.4 gm 3 Table F.1, 30 degC: Vliq cm 1.004 gm Vlv cm 32930 gm 3 mtotal Vtotal Vvap mvap Vtotal mtotal Vliq Vlv mtotal 0.382 kg mvap 4.543 10 3 kg mliq mtotal mvap Vtot.liq mliq Vliq mliq 377.72 gm Vtot.liq 379.23 cm 3 Ans. 156 6.27 Table F.2, 1100 kPa: Hliq 781.124 J gm Hvap 2779.7 Interpolate @101.325 kPa & 105 degC: H2 2686.1 J gm J gm Const.-H throttling: H2 = Hliq x Hvap Hliq x H2 Hvap Hliq Hliq x 0.953 Ans. 6.28 Data, Table F.2 at 2100 kPa and 260 degC, by interpolation: H1 2923.5 J gm J S1 6.5640 J gm K molwt 18.015 gm mol H2 2923.5 gm 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 S1 S 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 BTU lbm S1 1.5703 BTU lbm rankine H2 1257.7 BTU lbm 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 For steam as an ideal gas, there would be no temperature change and the entropy change would be given by: P1 300 psi P2 20 psi molwt 18 lb lbmol R ln S P2 P1 S 0.299 BTU lbm rankine Ans. molwt 6.30 Data, Table F.2 at 500 kPa and 300 degC S1 7.4614 J gm K The final state is at this entropy and a pressure of 50 kPa. This is a state of wet steam, for which Sliq 1.0912 J gm K Svap 7.5947 J gm K Hliq 340.564 J gm Hvap 2646.9 J gm S2 = S1 = Sliq x Svap Sliq x S1 Svap Sliq Sliq J gm x 0.98 H2 Hliq x Hvap Hliq H2 2599.6 Ans. 6.31 Vapor pressures of water from Table F.1: At 25 degC: Psat 3.166 kPa P 101.33 kPa xwater Psat P xwater 0.031 Ans. At 50 degC: Psat 12.34 kPa xwater Psat P xwater 0.122 Ans. 158 6.32 Process occurs at constant total volume: Vtotal ( 0.014 0.021)m 3 Data, Table F.1 at 100 degC: Uliq 419.0 J gm Uvap 3 2506.5 J gm Vliq 1.044 cm 3 3 gm Vvap 1673.0 cm 3 gm mliq 0.021 m Vliq mvap 0.014 m Vvap 4 mass mliq mvap x mvap mass x 4.158 10 (initial quality) This state is first reached as saturated liquid at 349.83 degC V2 Vtotal mass V2 cm 1.739 gm 3 For this state, P = 16,500.1 kPa, and U2 Q 1641.7 J gm U1 Uliq x Uvap Uliq U1 419.868 J gm U2 U1 3 Q 1221.8 J gm Ans. 6.33 Vtotal 0.25 m Data, Table F.2, sat. vapor at 1500 kPa: V1 cm 131.66 gm 3 U1 2592.4 J gm mass Vtotal V1 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: x 0.75 V2 = V1 = Vliq x Vvap Vliq Whence x= V1 Vvap Vliq Vliq (A) Find P for which (A) yields the value x = 0.75 for wet steam 159 Since the liquid volume is much smaller than the vapor volume, we make a preliminary calculation to estimate: Vvap V1 x Vvap 175.547 cm 3 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 782.41 J gm Uvap 2584.9 J gm U2 Uliq x Uvap Uliq U2 2134.3 J gm Q mass U2 U1 Q 869.9 kJ Ans. 6.34 Table F.2,101.325 kPa: Vliq cm 1.044 gm 3 Vvap cm 1673.0 gm 0.02 m Vliq mvap mtotal 3 3 Uliq mvap J 418.959 gm 1.98 m Vvap Vliq Uliq 3 Uvap mtotal J 2506.5 gm mliq mvap cm 3 mliq x V1 U1 x Vvap x Uvap Vliq Uliq V1 U1 98.326 gm J gm x 0.058 540.421 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 U1 Q 41860.5 kJ Ans. 160 6.35 Data, Table F.2 at 800 kPa and 350 degC: V1 354.34 cm 3 gm U1 2878.9 J gm Vtotal 0.4 m 3 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 2638.7 J gm Q Vtotal V1 U2 U1 Q 271.15 kJ Ans. 6.36 Data, Table F.2 at 800 kPa and 200 degC: U1 2629.9 J gm S1 6.8148 J gm K mass 1 kg (a) Isothermal expansion to 150 kPa and 200 degC U2 Q 2656.3 J gm S1 S2 Q 7.6439 J gm K Ans. T 473.15 K mass T S2 392.29 kJ Also: Work mass U2 U1 Q Work 365.89 kJ (b) Constant-entropy expansion to 150 kPa. The final state is wet steam: Sliq 1.4336 J gm K Svap 7.2234 J gm K Uliq 444.224 J gm Uvap 2513.4 J gm x S1 Svap Sliq Sliq x 0.929 3 J U2 Uliq x Uvap Uliq U2 2.367 10 gm W mass U2 U1 W 262.527 kJ Ans. 161 6.37 Data, Table F.2 at 2000 kPa: x 0.94 Hvap 2797.2 J gm Hliq 908.589 J gm H1 Hliq x Hvap Hliq H1 2.684 10 3 J gm mass 1 kg For superheated vapor at 2000 kPa and 575 degC, by interpolation: H2 3633.4 J gm Q mass H2 H1 Q 949.52 kJ Ans. 6.38 First step: Q12 = 0 W12 = U2 U1 Second step: For process: W23 = 0 Q = U3 U2 Q23 = U3 W = U2 U2 U1 Table F.2, 2700 kPa: Uliq 977.968 J gm Uvap 2601.8 J gm Sliq 2.5924 J gm K Svap 6.2244 J gm K 3 J x1 0.9 U1 Uliq x1 Uvap Uliq U1 2.439 10 gm S1 Sliq x1 Svap Sliq S1 5.861 2 3 m 10 2 s K Table F.2, 400 kPa: Sliq 1.7764 J gm K Svap 6.8943 J gm K Uliq 604.237 J gm 3 Uvap 2552.7 J gm Vliq cm 1.084 gm Vvap 462.22 cm 3 gm 162 Since step 1 is isentropic, S2 = S1 = Sliq x2 Svap Sliq x2 S1 Svap Sliq Sliq 3 J x2 0.798 U2 Uliq x2 Uvap Uliq U2 2.159 10 gm 3 V2 Vliq x2 Vvap Vliq V2 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 2560.7 J gm Whence Q U3 U2 Work U2 U1 Q 401.317 J gm Ans. Work 280.034 J gm Ans. 6.39 Table F.2, 400 kPa & 175 degC: Table F.1,sat. vapor, 175 degC U1 2605.8 J gm J gm S1 7.0548 J gm K U2 2578.8 S2 6.6221 J gm K mass 4 kg T ( 175 273.15)K Q mass T S2 S1 W mass U2 U1 Q Q 775.66 kJ Ans. W 667.66 kJ Ans. 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 J gm K 163 H H2 H1 H 600.1 J gm Ans. S S2 S1 S 0.115 J gm K Ans. (b) T1 ( 450 273.15)K T2 ( 140 273.15)K T1 723.15 K T2 413.15 K P1 3000 kPa P2 235 kPa Eqs. (6.95) & (6.96) for an ideal gas: Hig gm mol 3 5 R ICPH T1 T2 3.470 1.450 10 0.0 0.121 10 molwt 18 molwt R ICPS T1 T2 3.470 1.450 10 Sig Hig (c) Tc molwt 3 0.0 0.121 10 5 ln P2 P1 620.6 647.1 K J gm Pc 220.55 bar Sig 0.0605 0.345 J gm K Ans. Tr1 T1 Tc Pr1 P1 Pc Tr2 T2 Tc Pr2 P2 Pc Tr1 1.11752 Pr1 0.13602 Tr2 0.63846 Pr2 0.01066 The generalized virial-coefficient correlation is suitable here H H HRB Tr1 Pr1 R Tc HRB Tr2 Pr2 molwt J Ans. 593.95 gm Hig S Sig R SRB Tr2 Pr2 SRB Tr1 Pr1 molwt S 0.078 J gm K Ans. 164 6.41 Data, Table F.2 superheated steam at 550 kPa and 200 degC: V1 385.19 cm 3 gm U1 2640.6 J gm S1 7.0108 J gm K 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 U2 U1 Step 2--3: Isentropic expansion to initial T. Q12 322.5 J gm Q23 = 0 S3 = S 2 S3 7.5782 J gm K Step 3--1: Constant-T compression to initial P. T 473.15 K Q31 T S1 S3 Q31 268.465 J gm For the cycle, the internal energy change = 0. Wcycle = Qcycle = Q12 Q31 = Wcycle Q12 1 Q31 Q12 0.1675 Ans. 6.42 Table F.4, sat.vapor, 300(psi): T1 ( 417.35 459.67) rankine H1 1202.9 BTU lbm T1 877.02 rankine S1 1.5105 BTU lbm rankine Superheated steam at 300(psi) & 900 degF H2 1473.6 BTU lbm S2 1.7591 BTU lbm rankine S3 S2 Q12 H2 H1 Q31 T 1 S1 S3 Q31 218.027 BTU lbm 165 For the cycle, the internal energy change = 0. Wcycle = Qcycle = Q12 Q31 = Wcycle Q12 Whence 1 Q31 Q12 0.1946 Ans. 6.43 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 0.95 S2 = Sliq 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 J gm K Svap 7.0520 J gm K S2 Sliq x Svap Sliq S2 6.7798 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 t1 559.16 (degC) H1 3598.0 kJ kg Superheat: t ( 559.16 212.37)K t 346.79 K Wdot Ans. (b) mdot 5 kg sec Wdot mdot H2 H1 4760 kW Ans. 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 3205.4 kJ kg 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: Sliq 0.6493 kJ kg K Svap 8.1511 kJ kg K Hliq 191.832 kJ kg Hvap 2584.8 kJ kg x2 S2 Svap Sliq Sliq x2 0.879 H' Hliq x2 Hvap Hliq H' 2.294 10 3 kJ H2 H1 kg H' H1 0.681 Ans. 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 Sliq 1.0261 kJ kg K Svap 7.6709 kJ kg K Hliq 317.16 kJ kg Hvap 2636.9 kJ kg x2 S2 Svap Sliq Sliq x2 0.95 H' Hliq x2 Hvap Hliq H2 H1 H' 2.522 10 3 kJ kg H' H1 0.78 Ans. 6.47 Table F.2 at 1600 kPa and 225 degC: P 1600 kPa V cm 132.85 gm 3 H 2856.3 J gm S 6.5503 J gm K Table F.2 (ideal-gas values, 1 kPa and 225 degC) Hig 2928.7 J gm Sig 10.0681 J gm K P0 VR 1 kPa T R molwt P T ( 225 273.15)K T 498.15 K V The enthalpy of an ideal gas is independent of pressure, but the entropy DOES depend on P: HR H Hig cm 3 Sig R molwt 72.4 J gm ln P P0 SR S Sig Sig VR 10.96 gm HR SR 0.11 J Ans. gm K Reduced conditions: 0.345 Tc 647.1 K Pc 220.55 bar Tr T Tc Tr 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 4.2 B1 0.377 168 By Eqs. (3.61) + (3.62) & (3.63) along with Eq. (6.40) Z 1 B0 B1 Pr Tr Z 0.935 VR RT ( Z 1) P molwt HR R Tc molwt HRB Tr Pr 3 SR R molwt SRB Tr Pr VR 9.33 cm gm HR 53.4 J gm SR 0.077 J gm K Ans. 6.48 P 1000 kPa T ( 179.88 273.15) K T 453.03 K (Table F.2) molwt 3 18.015 gm mol 3 Vl cm 1.127 gm Vv cm 194.29 gm Vlv Hlv Vv Vl Hl Sl 762.605 J gm Hv Sv 2776.2 J gm J gm K Hv Hl 2.1382 J gm K 3 6.5828 Slv Sv Sl Vlv cm 193.163 gm Hlv 2.014 10 3 J gm Slv 4.445 J gm K (a) Gl Hl T Sl G l 206.06 J gm Gv Hv T Sv G v 206.01 J gm (b) Slv 4.445 J gm K r Hlv T r 3 4.445 J gm K (c) VR Vv T R molwt P 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 Hig 2841.1 J gm Sig 9.8834 J gm K P0 1 kPa The enthalpy of an ideal gas is independent of pressure; the entropy DOES depend on P: HR SR Hv Sv Hig Sig Sig Sig HR P R ln P0 molwt 64.9 J gm Sig 3.188 J gm K Ans. SR 0.1126 J gm K Ans. (d) Assume ln P vs. 1/T linear and fit three data pts @ 975, 1000, & 1050 kPa. 975 Data: 178.79 t 179.88 182.02 (degC) pp 1000 kPa 1050 i 1 3 xi dPdT ti 1 yi 273.15 P T 2 ln ppi kPa Slope slope ( y) Slope x 4717 Slope K dPdT 22.984 kPa K Slv Vlv dPdT Slv 4.44 J gm K Ans. Reduced conditions: 0.345 Tc 647.1 K Pc 220.55 bar Tr T Tc Tr 0.7001 Pr P Pc Pr 0.0453 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 0.943 VR RT ( 1) Z P molwt 170 HR R Tc molwt HRB Tr Pr SR R SRB Tr Pr molwt VR 11.93 cm 3 gm HR 43.18 J gm SR 0.069 J gm K Ans. 6.49 T ( 358.43 459.67) rankine T 818.1 rankine P 150 psi (Table F.4) molwt 3 18.015 gm mol Vl ft 0.0181 lbm Vv ft 3.014 lbm 3 Vlv Vv Vl Hl 330.65 BTU lbm Hv 1194.1 BTU lbm Hlv Slv Hv Hl Sl 0.5141 BTU lbm rankine 3 Sv 1.5695 BTU lbm rankine Sv Sl Vlv ft 2.996 lbm Hlv 863.45 BTU lbm (a) Gl Hl T Sl Gv Hv T Sv Gl 89.94 BTU lbm Gv 89.91 BTU lbm (b) Slv 1.055 BTU lbm rankine r Hlv T r 1.055 BTU lbm rankine (c) VR Vv T R molwt P VR 0.235 ft 3 lbm Ans. 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 The enthalpy of an ideal gas is independent of pressure; the entropy DOES depend on P: HR Hv Hig HR Sig 28.5 BTU lbm Ans. Sig P R ln P0 molwt Sv Sig Sig 0.552 BTU lbm rankine SR SR 0.0274 BTU lbm rankine Ans. (d) Assume ln P vs. 1/T linear and fit threedata points (@ 145, 150, & 155 psia) 145 Data: 355.77 t 358.43 361.02 (degF) pp 150 psi 155 i 1 3 xi ti 1 459.67 yi ln ppi psi Slope Slope slope ( y) x 8.501 10 3 dPdT P T 2 Slope rankine dPdT 1.905 psi rankine Slv Vlv dPdT Slv 1.056 BTU Ans. lbm rankine Reduced conditions: 0.345 Tc 647.1 K Pc 220.55 bar Tr T Tc Tr 0.7024 Pr P Pc 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 4.2 B1 0.62 172 By Eqs. (3.61) + (3.62) & (3.63) along with Eq. (6.40) Z 1 B0 B1 Pr Tr Z 0.942 VR RT P molwt (Z 1) HR R Tc molwt HRB Tr Pr 3 SR R SRB Tr Pr molwt VR 0.1894 ft lbm HR 19.024 BTU lbm SR 0.0168 BTU lbm rankine Ans. 6.50 For propane: Tc 369.8 K Pc 42.48 bar 0.152 T ( 195 273.15) K T 468.15 K P 135 bar P0 Pr 3.178 1 bar Tr T Tc Tr 1.266 Pr P Pc Use the Lee/Kesler correlation; by interpolation, Z0 0.6141 Z1 0.1636 Z 3 Z0 Z1 Z 0.639 V ZRT P V cm 184.2 mol Ans. HR0 2.496 R Tc HR1 0.586 R Tc HR0 7.674 10 3 J mol HR1 1.802 10 3 J mol SR0 1.463 R SR1 0.717 R SR0 12.163 J mol K SR1 5.961 J mol K HR HR0 HR1 3 J SR SR0 SR1 HR 7.948 10 mol SR 3 13.069 J mol K 6 H R ICPH 308.15K T 1.213 28.785 10 173 8.824 10 0.0 HR S R ICPS 308.15K T 1.213 28.785 10 J mol 3 8.824 10 J mol K 6 0.0 ln P P0 SR H 6734.9 Ans. S 15.9 Ans. 6.51 For propane: Tc 369.8 K Pc 42.48 bar 0.152 T ( 70 273.15)K T 343.15 K P0 101.33 kPa P 1500 kPa Tr T Tc Tr 0.92793 Pr P Pc Pr 0.35311 Assume propane an ideal gas at the initial conditions. Use generalized virial correlation at final conditions. H R Tc HRB Tr Pr H 1431.3 J mol J mol K Ans. S R SRB Tr Pr ln P P0 S 25.287 Ans. 6.52 For propane: 0.152 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: 3 P 1 bar A 6.72219 B 1.33236 C 2.13868 D 1.38551 Given () T 1 T Tc 1.5 Guess: 3 T 6 200 K P = Pc exp T A () B T () T C () T () T D () T 1 T Find T) ( 230.703 K 174 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) Pc exp 230.703 K A ( T) B ( T) 1.5 C ( T) kPa K ( T) 3 D ( T) 6 1 d P ( T) dT Pr P Pc B0 T 4.428 dPdT T Tc 0.172 Tr 4.2 4.428124 kPa K P 1 bar Pr 0.024 Tr Tr 0.624 B0 0.083 0.422 Tr 1.6 0.815 B1 0.139 B1 2 7 1.109 Vvap RT P 1 B0 3 4 cm B1 Pr Tr Vliq V c Zc 1 Tr Vvap 1.847 10 mol Vliq dPdT Vliq Hlv 75.546 cm 3 mol 10 4 J Hlv T Vvap 1.879 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 0 R Tc and 175 r1 = H R 1 R Tc T1 370 K P1 200 bar Tr T1 Tc Tr 1.001 Pr P1 Pc Pr 4.708 By interpolation, find: r0 3.773 r1 3.568 By Eq. (6.85) H1 R Tc r0 r1 H1 1.327 10 4 J 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 8.824 10 6 0.0 H2 1.048 10 4 J mol For Step (3) the enthalpy change is given by Eq. (6.87), for which Tr 230.703 K Tc Tr 0.6239 Pr 1 bar Pc Pr 0.0235 H3 H3 R Tc HRB Tr Pr 232.729 J mol For Step (4), H4 = x Hlv For the process, H1 H2 H3 x Hlv = 0 x H1 H2 Hlv H3 x 0.136 Ans. 6.53 For 1,3-butadiene: 0.190 Tc 425.2 K Pc 42.77 bar Zc 0.267 Vc cm 220.4 mol 3 Tn 268.7 K T 380 K P 1919.4 kPa T0 273.15 K P0 101.33 kPa Tr T Tc Tr 0.894 Pr P Pc Pr 0.449 176 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 Vvap HR0 HR0 0.7442 ZRT P 0.689 R Tc 2.436 10 Z1 0.1366 Vvap HR1 HR1 Z Z0 cm 3 Z1 Ans. Z 0.718 1182.2 mol 0.892 R Tc 3.153 10 3 J 3 J mol mol SR0 SR0 HR HR Hvap 0.540 R 4.49 HR0 3.035 SR1 SR1 SR SR 0.888 R 7.383 SR0 5.892 3 J mol K HR1 10 3 J J mol K SR1 mol J mol K 8.882 10 6 6 R ICPH T0 T 2.734 26.786 10 3 0.0 HR Svap R ICPS T0 T 2.734 26.786 10 Hvap 6315.9 J mol 8.882 10 0.0 ln P P0 SR Ans. Svap 1.624 J mol K Ans. For saturated vapor, by Eqs. (3.63) & (4.12) 2 Vliq V c Zc 1 Tr 7 Vliq cm 109.89 mol 3 Ans. 177 1.092 ln Hn R Tn Pc bar Tn Tc 1.013 Hn 22449 0.930 J mol By Eq. (4.13) H Hn 1 1 Tr Tn Tc 0.38 H 14003 J mol Hliq Hvap H Hliq 7687.4 J mol Ans. Sliq Svap H T Sliq 38.475 J mol K Ans. 6.54 For n-butane: 0.200 Tc 425.1 K Pc 37.96 bar Zc P 0.274 1435 kPa Vc T0 cm 255 mol 3 Tn P0 272.7 K 101.33 kPa T Tr 370 K T Tc 273.15 K Tr 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 0.7692 Z1 0.1372 Z Z0 Z1 3 Z 0.742 V ZRT P V cm 1590.1 mol Ans. HR0 0.607 R Tc HR1 0.831 R Tc HR0 2.145 10 3 J mol 178 HR1 2.937 10 3 J mol SR0 SR0 HR HR Hvap 0.485 R 4.032 HR0 2.733 SR1 J SR1 SR SR 3 0.835 R 6.942 SR0 5.421 mol K HR1 10 3 J J mol K SR1 mol J mol K 6 R ICPH T0 T 1.935 36.915 10 3 11.402 10 6 0.0 HR Svap R ICPS T0 T 1.935 36.915 10 J mol 11.402 10 0.0 ln P P0 SR Hvap 7427.4 Ans. Svap 4.197 J mol K Ans. For saturated vapor, by Eqs. (3.72) & (4.12) 1 Tr 2/7 Vliq V c Zc Vliq cm 123.86 mol 3 Ans. 1.092 ln Hn R Tn Pc bar Tn Tc 1.013 Hn 22514 0.930 J mol By Eq. (4.13) H Hn 1 1 Tr Tn Tc 0.38 H 15295.2 J mol Hliq Sliq Hvap Svap H H T Hliq Sliq 7867.8 J mol J mol K Ans. 37.141 Ans. 179 6.55 Under the stated conditions the worst possible cycling of demand can be represented as follows: 10, kg/ 000 hr Dem and ( hr kg/ ) 2/ hr 3 1/ hr 3 1 hr 6, 000 tm e i 4, kg/ 000 hr netst age or ofst eam netdepl i eton ofst eam This situation is also represented by the equation: 4000 where 10000 1 = 6000 = time of storage liquid 2 Solution gives hr 3 The steam stored during this leg is: mprime mprime 6000 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 m2 = Hprime Hf2 Vf2 H1 Vtank P2 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 P1 Hfg2 Vfg2 m2 m1 = Hprime H1 Eq. (A) However M1 v1 = m2 V2 = Vtank and therefore = V1 V2 180 Making this substitution and rearranging we get Hprime Hf2 Vf2 Hfg2 Vfg2 V2 P2 P1 Hfg2 Vfg2 = Hprime V1 H1 In this equation we can determine from the given information everything except Hprime and Vprime. These quantities are expressed by H1 = Hf1 x1 Hfg1 and V1 = Vf1 x1 Vfg1 Therefore our equation becomes (with Hprime = Hg2) Hg2 Hf2 V2 Vf2 Hfg2 Vfg2 P2 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 therefore x2 Vf2 19 = 1 0.05V2 = x2 Vg2 or x2 = Vf2 19Vg2 Vf2 x2 Vf2 x2 Vg2 Then V2 = Vg2 0.05 Vf2 19Vg2 Vf2 = 20 19 Vf2 1 Vg2 Eq. (C) Now we need property values: Initial state in accumulator is wet steam at 700 kPa. We find from the steam tables Hf1 697.061 kJ Hg1 kg 2762.0 kJ kg Hfg1 Hg1 Hf1 Hfg1 2064.939 kJ kg P1 700kPa 181 Vf1 1.108 cm 3 gm Vg1 272.68 cm 3 gm Vfg1 Vg1 Vf1 Vfg1 271.572 cm 3 gm Final state in accumulator is wet steam at 1000 kPa. From the steam tables P2 1000kPa Hf2 762.605 kJ kg 3 Hg2 2776.2 kJ kg Hfg2 3 Hg2 Hf2 Hfg2 2013.595 kJ kg 3 Vf2 1.127 cm gm Vg2 194.29 cm gm Vfg2 Vg2 Vf2 Vfg2 193.163 cm gm Solve Eq. (C) for V2 V2 Vg2 0.05 Vf2 19Vg2 Vf2 V2 1.18595 10 3 3m kg Next solve Eq. (B) for x1 Given Guess: x1 0.1 Hg2 Hf2 V2 Vf2 Hfg2 Vfg2 P2 P1 Hfg2 Vfg2 = Hg2 Hf1 Vf1 x1 Hfg1 x1 Vfg1 x1 Find x1 x1 4.279 10 4 Thus V1 Vf1 x1 Vfg1 V1 1.22419 cm 3 Eq. (A) gives V1 m2 = V2 m1 gm and 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 m2 m2 m1 = 2667lb 3.873 182 m1 m2 Find m1 m2 10 kg 10 kg 4 Finally, find the tank volume Vtank m2 V2 Vtank 45.9 m 3 Ans. Note that just to store 1333.3 kg of saturated vapor at 1000 kPa would require a volume of: 1333.3kg Vg2 259 m 3 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 = U1 U2 Hprime Hg2 = Hprime Hprime Uf1 Uf2 2.776 = Hprime Hprime 3 kJ Hf1 Hf2 Hprime Given m2 m1 Hprime 10 kg = Hprime Hprime Hf1 Hf2 m1 m2 m1 = 2667lb m2 Find m1 m2 m2 V 3.837 m2 Vf2 0.95 10 kg 3 4 V 45.5 m 0.140 Ans. Tc P 6.56 Propylene: T 365.6 K 38 bar Pc P0 46.65 bar 1 bar 400.15 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: 183 Tr T Tc Tr 1.095 Pr P Pc Pr 0.815 Step (1): Use the Lee/Kesler correlation, interpolate. H0 0.863 R Tc H1 0.534 R Tc HR H0 H1 H0 2.623 10 3 J mol H1 1.623 10 3 J mol HR 2.85 10 3 J mol S0 0.565 R S1 0.496 R SR S0 S1 S0 4.697 J mol K S1 4.124 J mol K SR 5.275 J mol K Step (2): For the heat capacity of propylene, A 1.637 B 22.706 10 K 3 C 6.915 10 K 2 6 Solve energy balance for final T. See Eq. (4.7). 1 (guess) Given HR = R Find AT 1 B 2 T 2 0.908 2 1 Tf 3 C 3 T 3 3 1 T Tf 6 363.27 K Ans. Sig Sig R ICPS T Tf 1.637 22.706 10 22.774 J mol K 6.915 10 0.0 ln P0 P S SR Sig S 28.048 J mol K Ans. 184 6.57 Propane: 0.152 Tc 369.8 K Pc 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 1.144 Pr P Pc 3 J Pr 0.518 Step (1): Use the generalized virial correlation HR R Tc HRB Tr Pr HR 1.366 10 mol SR R SRB Tr Pr SR 2.284 J mol K 6 Step (2): For the heat capacity of propane, A 1.213 B 28.785 10 K 3 C 8.824 10 K 2 Solve energy balance for final T. See Eq. (4.7). 1 (guess) Given HR = R Find AT 1 B 2 T 2 2 1 Tf 3 C 3 T 3 3 1 0.967 T Tf 408.91 K Ans. Sig Sig R ICPS T Tf 1.213 28.785 10 22.415 J mol K 8.824 10 6 0.0 ln P0 P S SR Sig S 24.699 185 J mol K Ans. 6.58 For propane: Tc 369.8 K Pc 42.48 bar 0.152 T ( 100 273.15)K T 373.15 K P0 1 bar P 10 bar Tr T Tc Tr 1.009 Pr P Pc Pr 0.235 Assume ideal gas at initial conditions. Use virial correlation at final conditions. H R Tc HRB Tr Pr H 801.9 J mol J mol K Ans. S R SRB Tr Pr ln P P0 S 20.639 Ans. 6.59 H2S: 0.094 Tc 373.5 K Pc 89.63 bar T1 400 K P1 5 bar T2 600 K P2 25 bar Tr1 T1 Tc Pr1 P1 Pc Tr2 T2 Tc Pr2 P2 Pc Tr1 1.071 Pr1 0.056 Tr2 1.606 Pr2 0.279 Use generalized virial-coefficient correlation for both sets of conditions. Eqs. (6.91) & (6.92) are written H R ICPH T1 T2 3.931 1.490 10 0.0 0.232 10 R Tc HRB Tr2 Pr2 HRB Tr1 Pr1 3 5 S R ICPS T1 T2 3.931 1.490 10 R SRB Tr2 Pr2 3 0.0 0.232 10 5 ln P2 P1 SRB Tr1 Pr1 H 7407.3 J mol S 1.828 J mol K Ans. 186 6.60 Carbon dioxide: 0.224 Tc 304.2 K Pc 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: Tr T Tc Tr 1.046 Pr P Pc Pr 0.217 Step (1): Use the generalized virial correlation HR R Tc HRB Tr Pr HR 587.999 J mol SR R SRB Tr Pr SR 1.313 J mol K Step (2): For the heat capacity of carbon dioxide, A 5.457 B 1.045 10 K 3 D 1.157 10 K 5 2 Solve energy balance for final T. See Eq. (4.7). Given 1 (guess) HR = R A T Find 1 B 2 T 2 0.951 2 1 Tf D T T 3 1 Tf 302.71 K 5 Ans. Sig R ICPS T Tf 5.457 1.045 10 J mol K 0.0 1.157 10 ln P0 P Sig 21.047 S SR Sig S 22.36 J mol K Ans. 187 6.61 T0 523.15 K P0 3800 kPa P 120 kPa S 0 J mol K For the heat capacity of ethylene: A 1.424 B 14.394 10 K 3 C 4.392 10 2 6 K (a) For the entropy change of an ideal gas, combine Eqs. (5.14) & (5.15) with D = 0: 0.4 (guess) Given S = R A ln Find B T0 0.589 C T0 2 1 2 1 ln P P0 Tf T0 3 Tf 308.19 K 6 Ans. Hig R ICPH T0 Tf 1.424 14.394 10 4.392 10 0.0 Hig Ws 1.185 10 4 J mol Ws 11852 J mol Hig Ans. (b) Ethylene: 0.087 Tc 282.3 K Pc 50.40 bar Tr0 T0 Tc Tr0 1.85317 Pr0 P0 Pc Pr0 0.75397 At final conditions as calculated in (a) Tr T Tc Tr 1.12699 Pr P Pc Pr 0.02381 Use virial-coefficient correlation. The entropy change is now given by Eq. (6.92): 0.5 (guess) Given 188 S = R A ln SRB B T0 T0 Tc C T0 2 1 2 1 ln P P0 Pr SRB Tr0 Pr0 Find T T0 T 303.11 K Ans. Tr T Tc Tr 1.074 The work is given by Eq. (6.91): Hig R ICPH T0 T 1.424 14.394 10 3 4.392 10 6 0.0 Hig 1.208 10 4 J mol Ws Hig R Tc HRB Tr Pr HRB Tr0 Pr0 Ws 11567 J mol Ans. 6.62 T0 S 493.15 K 0 J mol K P0 30 bar P 2.6 bar For the heat capacity of ethane: A 1.131 B K (a) For the entropy change of an ideal gas, combine Eqs. (5.14) & (5.15) with D = 0: (guess) Given 0.4 19.225 10 K 3 C 5.561 10 2 6 S = R A ln Find B T0 C T0 2 1 2 1 ln P P0 0.745 T T0 3 T 367.59 K 6 Ans. Hig R ICPH T0 T 1.131 19.225 10 189 5.561 10 0.0 Hig 8.735 10 3 J mol Ws Hig Ws 8735 J mol Ans. (b) Ethane: 0.100 Tc 305.3 K Pc 48.72 bar Tr0 T0 Tc Tr0 1.6153 Pr0 P0 Pc Pr0 0.61576 At final conditions as calculated in (a) Tr ( ) T T Tc Tr ( ) 1.20404 T Pr P Pc Pr 0.05337 Use virial-coefficient correlation. The entropy change is now given by Eq. (6.83): 0.5 (guess) Given S = R A ln SRB B T0 T0 Tc C T0 2 1 2 1 ln P P0 Pr SRB Tr0 Pr0 Tr Find T T T0 T 362.73 K Ans. Tc Tr 1.188 The work is given by Eq. (6.91): Hig R ICPH T0 T 1.131 19.225 10 3 5.561 10 6 0.0 Hig 9.034 10 3 J mol Ws Hig R Tc HRB Tr Pr HRB Tr0 Pr0 Ws 8476 J mol Ans. 190 6.63 n-Butane: 0.200 Tc 425.1 K Pc 37.96 bar T0 323.15 K P0 1 bar P 7.8 bar S 0 J mol K For the heat capacity of n-butane: A 1.935 B 36.915 10 K 3 C 11.402 10 K 2 6 Tr0 T0 Tc Tr0 0.76017 Pr0 P0 Pc P Pr0 0.02634 Pr Pc HRB0 Pr 0.205 0.05679 HRB Tr0 Pr0 = 0.05679 The entropy change is given by Eq. (6.92) combined with Eq. (5.15) with D = 0: (guess) 0.4 Given S = R A ln SRB T0 Tc B T0 Pr C T0 2 1 2 1 ln P P0 SRB Tr0 Pr0 Find 1.18 T T0 T 381.43 K Ans. Tr T Tc Tr 0.89726 The work is given by Eq. (6.91): Hig R ICPH T0 T 1.935 36.915 10 3 11.402 10 6 0.0 Hig 6.551 10 3 J mol Ws Hig R Tc HRB Tr Pr HRB Tr0 Pr0 Ws 5680 J mol Ans. 191 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 3344.6 kJ kg S1 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 H2 H1 T S2 S1 Wideal 1224.3 kJ kg Ans. 6.65 Sat. liquid at 325 K (51.85 degC), Table F.1: Hliq kJ 217.0 kg Sliq kJ 0.7274 kg K Vliq cm 1.013 gm 3 Psat P1 12.87 kPa 8000 kPa For the compressed liquid at 325 K and 8000 kPa, apply Eqs. (6.28) and (6.29) with T 325 K 460 10 6 K 1 H1 Hliq Vliq 1 T P1 Psat H1 223.881 kJ kg S1 Sliq Vliq P1 Psat S1 0.724 kJ kg K For sat. vapor at 8000 kPa, from Table F.2: H2 2759.9 kJ kg S2 5.7471 kJ kg K T 300 K Heat added in boiler: Q H2 H1 Q 2536 kJ kg Maximum work from steam, by Eq. (5.27): Wideal H1 H2 T S1 S2 192 Wideal 1029 kJ kg Work as a fraction of heat added: Frac Wideal Q Frac 0.4058 Ans. The heat not converted to work ends up in the surroundings. SdotG.surr Q Wideal T 10 kg sec SdotG.surr 50.234 kW K SdotG.system S1 S2 10 kg sec SdotG.system 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 Qdot 2536 SdotG kg sec kg kW Qdot 50.234 K T kJ 10 T ( 600 273.15)K T Ans. 873.15 K SdotG 21.19 kW K By Eq. (5.34) T 300 K Wdotlost T SdotG Wdotlost 6356.9 kW Ans. 6.67 For sat. liquid water at 20 degC, Table F.1: H1 83.86 kJ kg S1 0.2963 kJ kg K For sat. liquid water at 0 degC, Table F.1: H0 0.04 kJ kg S0 0.0000 kJ kg K For ice at at 0 degC: H2 H0 333.4 kJ kg S2 S0 333.4 kJ 273.15 kg K 193 H2 333.44 kJ kg S2 1.221 kJ kg K T 293.15 K mdot 0.5 kg sec t 0.32 By Eqs. (5.26) and (5.28): Wdotideal Wdot mdot H2 H1 T S2 S1 Wdot Wdotideal 13.686 kW Wdotideal t 42.77 kW Ans. 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 S2 2676.0 kJ kg S1 Q' 7.3554 kJ kg K kJ kg H2 T 0.0 kJ kg 0.0 kJ kg K 2000 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 T Sapparatus.reservoir Happaratus.reservoir = H2 H1 Q' Sapparatus.reservoir = S2 S1 Q' T' Wideal = 0.0 kJ kg T' 450 K Given (Guess) kJ 0 = H2 H1 kg Find ( ) T' Q' T S2 T' S1 Q' T' 409.79 K Ans. T' (136.64 degC) 194 6.69 From Table F.4 at 200(psi): H1 1222.6 BTU lbm S1 1.5737 BTU lbm rankine (at 420 degF) (Sat. liq. and vapor) Hliq BTU 355.51 lbm Hvap 1198.3 BTU lbm Sliq 0.5438 BTU lbm rankine Svap 1.5454 BTU lbm rankine x 0.96 H2 Hliq x Hvap 3 BTU Hliq S2 Sliq x Svap Sliq H2 1.165 10 lbm S2 1.505 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 0.5 H1 H Hvap 0.5 H2 H 1193.6 BTU lbm Ans. (wet steam) x Hliq Hliq x 0.994 S Sliq x Svap Sliq S 1.54 BTU lbm rankine By Eq. (5.22) on the basis of 1 pound mass of exit steam, SG S 0.5 S1 0.5 S2 SG 2.895 10 4 BTU lbm rankine Ans. 6.70 From Table F.3 at 430 degF (sat. liq. and vapor): Vliq ft 0.01909 lbm 3 Vvap ft 1.3496 lbm 3 Vtank 80 ft 3 Uliq 406.70 BTU lbm Uvap 1118.0 BTU lbm mliq 4180 lbm VOLliq mliq Vliq 195 VOLliq 79.796 ft 3 VOLvap mvap U1 Vtank VOLvap Vvap VOLliq VOLvap mvap 0.204 ft 3 0.151 lbm 406.726 BTU lbm mliq Uliq mliq mvap Uvap mvap U1 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 0 H dm = 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 mvap Have m = 0 Have 1203.5 m1 BTU lbm mass mliq m2 ( mass) Property values below are for sat. liq. and vap. at 420 degF ft 0.01894 lbm 395.81 BTU lbm Vtank m2 ( mass) Uliq 3 Vliq Vvap ft 1.4997 lbm 1117.4 BTU lbm 3 Uliq Uvap V2 ( mass) U2 ( mass) mass x mass) ( Uliq V2 ( mass) Vliq Vvap Vliq x mass) Uvap ( 50 lbm (Guess) 196 Given mass = m1 U1 U2 ( mass) Have U2 ( mass) mass Find ( mass) mass 55.36 lbm Ans. 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 J gm K V1 64.721 cm 3 gm Vtank 50 m 3 S2 = S1 = 6.7093 J gm K By interpolation in Table F.2 at this entropy and 3500 kPa: V2 cm 78.726 gm 3 t2 = 362.46 C Ans. m1 Vtank V1 m2 Vtank V2 m m1 m2 m 137.43 kg Ans. 6.72 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 Similarly, Whence mt 1000 kg mt Vt = Vliq mt Q = Hliq mt y Hlv y Vlv = 0 H mt Required data from Table F.1 are: H 209.3 kJ kg Vliq 1.251 cm 3 At 50 degC: At 250 degC: Hliq kJ 1085.8 kg kJ 1714.7 kg gm 3 Hlv y Q Vlv cm 48.79 gm Vliq mt Vlv mt Hliq H y 25.641 kg Q 832534 kJ Ans. kJ kg y Hlv 6.73 Given: C 0.43 kJ kg K Vtank T1 0.5 m 295 K 3 Hin mtank 120.8 30 kg Data for saturated nitrogen vapor: 80 85 90 T 95 100 105 110 K 1.396 2.287 3.600 P 5.398 bar 7.775 10.83 14.67 V 0.1640 0.1017 0.06628 0.04487 0.03126 0.02223 0.01598 m kg 3 198 78.9 82.3 85.0 H 86.8 87.7 87.4 85.6 By Eq. (2.29) multiplied through by dt, d nt Ut H dm = dQ kJ kg 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 mvap Tvap Vvap Hvap Uvap 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 Also, Hin mvap = Q = mtank C Tvap mvap = Vtank Vvap T1 (A) (B) Calculate internal-energy values for saturated vapor nitrogen at the given values of T: 56.006 U ( H P V) 59.041 61.139 U 62.579 63.395 63.325 62.157 kJ kg Fit tabulated data with cubic spline: Us lspline ( U) T interp ( T U t) Us (guess) Vs lspline ( V) T interp ( T V t) Vs Uvap () t Tvap Vvap () t 100 K Combining Eqs. (A) & (B) gives: 199 Given Uvap Tvap Tvap mvap Hin = mtank C T1 Tvap Vvap Tvap Vtank Find Tvap Vtank Vvap Tvap Tvap mvap 97.924 K 13.821 kg Ans. 6.74 The result of Part (a) of Pb. 3.15 applies, with m replacing n: m2 U2 Whence Also U2 = Uliq.2 V2 = Vliq.2 H m1 U1 H = Q= 0 m2 H x2 Ulv.2 x2 Vlv.2 V2 = Vtank m2 U2 = m1 H U1 Eliminating x2 from these equations gives Vtank m2 H Uliq.2 m2 Vliq.2 Vlv.2 Ulv.2 = m1 H U1 which is later solved for m2 Vtank 50 m 3 m1 16000 kg V1 V1 Vtank m1 3.125 10 3 3m kg Data from Table F.1 Vliq.1 cm 1.003 gm 3 @ 25 degC: Vlv.1 cm 43400 gm 3 Uliq.1 104.8 kJ kg Ulv.1 2305.1 kJ kg 200 x1 x1 V1 Vliq.1 Vlv.1 U1 5 Uliq.1 104.913 x1 Ulv.1 kJ kg 4.889 10 U1 Data from Table F.2 @ 800 kPa: Vliq.2 1.115 cm 3 Uliq.2 3 720.043 gm 1.115) cm Ulv.2 Ulv.2 H kJ kg kJ kg Vlv.2 Vlv.2 ( 240.26 m 0.239 kg 3 gm ( 2575.3 720.043) 3 kJ 1.855 kJ kg 10 kg Data from Table F.2 @ 1500 kPa: 2789.9 m1 H m2 H U1 Vtank Ulv.2 Vlv.2 Ulv.2 Vlv.2 m2 2.086 10 kg 4 Uliq.2 Vliq.2 msteam m2 m1 msteam 4.855 10 kg 3 Ans. 6.75 The result of Part (a) of Pb. 3.15 applies, with Whence U2 = H n1 = Q = 0 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 1 100 P2 200 300 400 384.09 384.82 t2 385.57 386.31 387.08 303316 3032.17 V2 1515.61 1010.08 757.34 3 cm 3 gm i 1 5 Vtank 1.75 m massi Vtank V2 i 5.77 mass 10 3 T rises very slowly as P increases 3 2 massi 0.577 1.155 1.733 2.311 kg 1 0 0 200 P2 i 400 6.76 Vtank Vliq Hliq x1 2m 3 3 Data from Table F.2 @ 3000 kPa: Vvap Hvap V1 cm 1.216 gm 1008.4 0.1 cm 66.626 gm 2802.3 Vliq 3 kJ kg kJ kg Vliq m1 Vtank V1 x1 Vvap 3 3m V1 7.757 10 kg m1 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 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 2. Exit of Hliq 0.6 m1 kg of liquid from the tank: 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 Vliq 0.6 m1 Vliq = 0 y= 0.6 m1 Vliq Vvap Vliq Q= 0.6 m1 Vliq Hvap Vvap Vliq Hliq 0.6 m1 Hliq 0.6 m1 = H mtank But H = Hliq and mtank and therefore the last two terms of the energy equation cancel: Q 0.6 m1 Vliq Hvap Vvap Vliq Hliq Q 5159 kJ Ans. 6.77 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 Also mdot1 = mdot3 mdot2 mdot3 5 kg sec Whence mdot2 mdot3 H1 H1 H2 H3 mdot1 mdot3 mdot2 mdot2 0.484 kg sec Ans. mdot1 4.516 kg sec Ans. 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: Vliq cm 1.012 gm 3 Hliq 209.3 kJ kg Sliq 0.7035 kJ kg K Psat 12.34 kPa T 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: V ( 1.015 1.010) 3 3 cm cm 3 gm T 10 K P 3100 kPa 4 1 V 5 10 1 Vliq V T gm 4.941 10 K Apply Eqs. (6.28) & (6.29) at constant T: H1 Hliq Vliq 1 T P Psat H1 211.926 S1 Sliq Vliq P Psat 204 S1 0.702 kJ kg kJ kg K By Eq. (2.30), neglecting kinetic and potential energies and setting the heat and work terms equal to zero: H3 mdot3 Also H1 mdot1 H2 mdot2 = 0 mdot2 = mdot3 mdot1 Whence mdot1 mdot3 H3 H2 H1 H2 mdot1 1.89 kg sec Ans. mdot2 mdot3 mdot1 mdot2 13.11 kg sec For adiabatic conditions, Eq. (5.22) becomes SdotG S3 mdot3 S1 mdot1 S2 mdot2 SdotG 1.973 kJ sec K 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 Hliq H3 mdot3 H1 mdot1 H2 mdot2 = 0 Also mdot3 = mdot2 mdot1 mdot2 mdot1 H1 H3 H2 H3 mdot2 3.241 kg sec Ans. For adiabatic conditions, Eq. (5.22) becomes 205 S2 Sliq mdot3 mdot2 mdot1 SdotG S3 mdot3 S1 mdot1 S2 mdot2 SdotG 3.508 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 12 bar P2 2 bar n1 1 mol n2 2.5 mol CP 7 R 2 CP 29.099 J mol K 1st law: T 600 K (guess) T1 n2 CP T T2 = 0 J Given n1 CP T T Find ( ) T T 542.857 K Ans. 2nd law: P 5 bar (guess) = 0 J K Given n1 CP ln T P R ln T1 P1 T P n2 CP ln R ln T2 P2 P 4.319 bar P Find ( ) P Ans. 6.81 molwt 28.014 lb lbmol CP R 7 2 molwt CP 0.248 BTU lbm rankine Ms = steam rate in lbm/sec Mn = nitrogen rate in lbm/sec Mn 40 lbm sec 206 (1) = sat. liq. water @ 212 degF entering (2) = exit steam at 1 atm and 300 degF (3) = nitrogen in at 750 degF (4) = nitrogen out at 325 degF BTU lbm BTU lbm T3 T4 0.3121 BTU lbm rankine BTU lbm rankine 1209.67 rankine 784.67 rankine H1 H2 180.17 S1 S2 (Table F.3) 1192.6 1.8158 (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 Given Ms 3 lbm sec Ms H2 (guess) Q = 60 Mn CP T4 Ms BTU Ms lbm Ms H1 T3 = 60 lbm sec BTU lbm Find Ms 3.933 Ans. Eq. (5.22) here becomes SdotG = Ms S2 S1 Mn S4 S3 Q T S4 T S3 = CP ln T4 T3 Q 60 BTU Ms lbm Q 235.967 BTU sec 529.67 rankine SdotG SdotG Ms S2 2.064 S1 Mn CP ln T4 T3 Q T BTU sec rankine Ans. 207 6.82 molwt 28.014 gm mol CP R 2 molwt 7 CP 1.039 J gm K Ms = steam rate in kg/sec Mn= nitrogen rate in kg/sec Mn 20 kg sec (1) = sat. liq. water @ 101.33 kPa entering (2) = exit steam at 101.33 kPa and 150 degC (3) = nitrogen in @ 400 degC T3 673.15 K (4) = nitrogen out at 170 degC T4 443.15 K H1 419.064 kJ kg S1 1.3069 kJ kg K (Table F.2) H2 2776.2 kJ kg S2 7.6075 kJ kg K (Table F.2) 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 kg sec (guess) Q = 80 kJ Ms kg Given Ms H2 H1 Mn CP T4 T3 = 80 kJ Ms kg Ms Find Ms Ms 1.961 kg sec Ans. Eq. (5.22) here becomes SdotG = Ms S2 S1 Mn S4 S3 T Q T 298.15 K Q T S4 S3 = CP ln Ms S2 4.194 T4 T3 S1 Mn CP ln Ans. Q 80 kJ kg Ms SdotG SdotG T4 T3 kJ sec K 208 6.86 Methane = 1; propane = 2 T 1 363.15 K P 2 5500 kPa y1 0.5 y2 1 y1 0.012 0.152 Zc1 0.286 Zc2 0.276 Tc1 190.6 K Tc2 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 y2 Tc2 Ppc y1 Pc1 y2 Pc2 Tpc 280.2 K Ppc 44.235 bar Tpr T Tpc Tpr 1.296 Ppr P Ppc Ppr 1.243 By interpolation in Tables E.3 and E.4: Z0 0.8010 Z1 0.1100 y1 1 y2 2 0.082 Z Z0 Z1 Z 0.81 For the molar mass of the mixture, we have: gm molwt molwt y1 16.043 y2 44.097 mol 30.07 gm mol V ZRT P molwt V cm 14.788 gm 3 mdot 3 4 cm 1.4 kg sec u 30 m sec 2 Vdot V mdot Vdot 2.07 10 sec A Vdot u A 6.901 cm D 4A D 2.964 cm Ans. 209 6.87 Vectors containing T, P, Tc, and Pc for the calculation of Tr and Pr: 500 400 450 600 620 T 250 150 500 450 400 1.176 1.315 0.815 0.971 425.2 304.2 552.0 617.7 617.2 Tc 190.6 154.6 469.7 430.8 374.2 0.468 2.709 0.759 0.948 20 200 60 20 20 P 90 20 10 35 15 Pc 42.77 73.83 79.00 21.10 36.06 45.99 50.43 33.70 78.84 40.60 Tr T Tc Pr P Pc Tr 1.005 1.312 0.97 1.065 1.045 1.069 Pr 0.555 1.957 0.397 0.297 0.444 0.369 Parts (a), (g), (h), (i), and (j) --- By virial equation: 500 150 T 500 K P 450 400 T Tc 20 20 10 bar Tc 35 15 P Pc 425.2 154.6 469.7 K Pc 430.8 374.2 42.77 50.43 33.70 bar 78.84 40.6 .190 .022 .252 .245 .327 Tr Pr 210 1.176 0.97 Tr 1.065 1.045 1.069 0.422 Tr 1.6 0.468 0.397 Pr 0.297 0.444 0.369 0.172 Tr 4.2 B0 0.073 Eq. (3.65) B1 0.139 Eq. (3.66) DB0 0.675 Tr 2.6 Eq. (6.89) DB1 0.722 Tr 5.2 Eq. (6.90) 0.253 0.37 B0 0.309 0.321 0.306 B1 0.052 0.056 6.718 4.217 9.009 10 10 10 3 3 3 0.443 0.73 DB0 0.574 0.603 0.568 DB1 0.311 0.845 0.522 0.576 0.51 Combine Eqs. (3.61) + (3.62), (3.63), and (6.40) and the definitions of Tr and Pr to get: VR R Tc B0 Pc B1 HR R Tc Pr B0 Tr DB0 ( B1 Tr DB1) Eq. (6.87) SR R Pr DB0 DB1 Eq. (6.88) 211 200.647 94.593 VR 355.907 146.1 232.454 cm 3 1.377 10 559.501 HR 1.226 10 1.746 10 1.251 10 3 1.952 2.469 J mol J mol K 3 3 3 SR 1.74 2.745 2.256 mol Parts (b), (c), (d), (e), and (f) --- By Lee/Kesler correlation: By linear interpolation in Tables E.1--E.12: DEFINE: h0 equals ( ) HR RTc ( ) SR R 0 0 ( ) HR h1 equals RTc s1 equals ( ) SR R 2.008 4.445 h0 3.049 0.671 1.486 1 1 h equals HR RTc SR R s0 equals .663 .124 Z0 .278 .783 .707 1.137 4.381 s0 2.675 0.473 0.824 400 450 T 600 K 620 250 P s1 Z1 s equals 0.208 .050 .088 .036 0.138 0.405 5.274 2.910 0.557 0.289 200 60 20 20 90 bar 0.233 5.121 h1 2.970 0.596 0.169 304.2 552.0 Tc 617.7 K 617.2 190.6 .224 .111 .492 .303 .012 212 Z s HR Z0 s0 Z1 Eq. (3.57) s1 (6.86) h h0 h1 Eq. (6.85) ( h Tc R) SR ( s R) 0.71 0.118 Z 0.235 0.772 0.709 HR 5.21 2.301 10 3 10.207 4 4 10 41.291 J mol 2.316 10 4.37 10 SR 34.143 5.336 6.88 J mol K 3 3 2.358 10 48.289 VR T (Z R P 549.691 1) VR 1.909 10 587.396 67.284 3 cm mol 3 And. 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) 650 300 600 T 350 400 200 450 250 K 60 100 100 P 75 150 75 80 100 bar 562.2 304.2 304.2 Tc1 305.3 373.5 190.6 190.6 126.2 K Tc2 553.6 132.9 568.7 282.3 190.6 126.2 469.7 154.6 K 213 48.98 73.83 73.83 Pc1 48.72 89.63 45.99 45.99 34.00 bar 40.73 34.99 24.90 Pc2 50.40 45.99 34.00 33.70 50.43 .5 Tc2) Ppc P Ppc 44.855 54.41 49.365 K bar .210 .224 .224 1 .100 .094 .012 .012 .038 .5 Pc2) .5 1 2 .210 .048 .400 .087 .012 .038 .252 .022 .5 2 Tpc ( Tc1 .5 T Tpc ( Pc1 .5 Tpr Ppr 557.9 218.55 436.45 Tpc 293.8 282.05 158.4 330.15 140.4 1.165 1.373 1.375 Tpr 1.191 1.418 1.263 1.363 1.781 Ppr Ppc 0.21 0.136 0.312 bar 49.56 67.81 39.995 39.845 42.215 1.338 1.838 2.026 1.513 2.212 1.875 2.008 2.369 214 0.094 0.053 0.025 0.132 0.03 Lee/Kesler Correlation --- By linear interpolation in Tables E.1--E.12: .6543 .7706 .7527 Z0 .6434 .7744 .6631 .7436 .9168 .890 .658 .729 s0 .944 .704 .965 .750 .361 Z1 .1219 .1749 .1929 .1501 .1990 .1853 .1933 .1839 h0 1.395 1.217 1.346 1.510 1.340 1.623 1.372 0.820 h1 .461 .116 .097 .400 .049 .254 .110 0.172 .466 .235 .242 s1 .430 .224 .348 .250 .095 0 h0 equals ( HR) RTpc ( SR) R h1 equals ( HR) RTpc ( SR) R 1 h equals HR RTpc SR R 0 1 s0 equals s1 equals s equals Z Z0 Z1 Eq. (3.57) h h0 h1 Eq. (6.85) s s0 s1 Eq. (6.86) 215 HR ( hTpc R) 0.68 0.794 0.813 SR ( R) s 6919.583 2239.984 4993.974 HR 3779.762 J 8.213 5.736 6.689 SR 8.183 J Z 0.657 0.785 0.668 0.769 0.922 3148.341 mol 2145.752 3805.813 951.151 5.952 mol K Ans. 8.095 6.51 3.025 6.95 Tc 647.1K Pc 220.55bar At Tr = 0.7: T 0.7 Tc T 452.97 K Find Psat in the Saturated Steam Tables at T = 452.97 K T1 451.15K P1 957.36kPa T2 453.15K P2 1002.7kPa Psat P2 T2 Psat Pc P1 ( T T1 T1) P1 Psat 998.619 kPa Psat 9.986 bar Psatr Psatr 0.045 1 log Psatr 0.344 Ans. This is very close to the value reported in Table B.1 ( = 0.345). 6.96 Tc 374.2K Pc 40.60bar At Tr = 0.7: T 0.7 Tc T 471.492 rankine T T 459.67rankine T 11.822 degF Find Psat in Table 9.1 at T = 11.822 F T1 10degF P1 26.617psi T2 15degF P2 29.726psi 216 Psat P2 T2 P1 (T T1 T1) P1 Psat 27.75 psi Psat 1.913 bar Psatr Psat Pc Psatr 0.047 1 log Psatr 0.327 Ans. This is exactly the same as the value reported in Table B.1. 6.101 For benzene a) 0.210 Tc 562.2K Pc 48.98bar Zc 0.271 Tn 353.2K Trn Tn Tc Trn 0.628 Psatrn 1 atm Pc Psatrn 0.021 lnPr0 ( Tr) 5.92714 6.09648 Tr 15.6875 Tr 1.28862 ln ( Tr) 0.169347 Tr 6 6 Eqn. (6.79) lnPr1 ( Tr) 15.2518 13.4721 ln ( Tr) 0.43577 Tr Eqn. (6.80) ln Psatrn lnPr0 Trn lnPr1 Trn Eqn. (6.81). 0.207 lnPsatr ( Tr) lnPr0 ( Tr) lnPr1 ( Tr) 2 7 Eqn. (6.78) Zsatliq Psatrn Trn 0.083 Zc 1 1 T rn Eqn. (3.73) Zsatliq 0.00334 B0 0.422 Trn 1.6 Eqn. (3.65) Z0 1 B0 Psatrn Trn Eqn. (3.64) B0 0.805 Z0 0.974 B1 0.139 0.172 Trn 4.2 Eqn. (3.66) Z1 B1 Psatrn Trn Equation following Eqn. (3.64) B1 1.073 217 Z1 0.035 Zsatvap Z0 Z1 Eqn. (3.57) Zsatvap 0.966 Zlv Zsatvap Zsatliq 2 Zlv 0.963 Hhatlv d dTrn lnPsatr Trn Trn Zlv Hhatlv 6.59 Hlv R Tc Hhatlv Hlv 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: 0.224 Tc Tt 304.2K 216.55K Pc Pt 73.83bar 5.170bar At the triple point: a) At Tr = 0.7 T 0.7Tc T 212.94 K Ttr Tt Tc Ttr 0.712 Ptr Pt Pc Ptr 6 0.07 lnPr0 ( ) Tr 5.92714 6.09648 Tr 1.28862 ln ( ) 0.169347 Tr Tr 6 Eqn. (6.79) lnPr1 ( ) Tr 15.2518 15.6875 Tr 13.4721 ln ( ) 0.43577 Tr Tr Eqn. (6.80) ln Ptr lnPr0 Ttr lnPr1 Ttr Eqn. (6.81). 218 0.224 Ans. This is exactly the same value as given in Table B.1 b) Psatr 1atm Pc Psatr 0.014 Guess: Trn 0.7 Given ln Psatr = lnPr0 Trn lnPr1 Trn Trn Find Trn Trn 0.609 Tn Trn Tc Tn 185.3 K Ans. This seems reasonable; a Trn of about 0.6 is common for triatomic species. 219 Chapter 7 - Section A - Mathcad Solutions 7.1 u2 325 m sec R 8.314 J mol K molwt 28.9 gm CP mol 7 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 u2 2 2 = 0 But H = CP T Whence T u2 2 2 CP T 52.45 K Ans. 7.4 From Table F.2 at 800 kPa and 280 degC: H1 3014.9 kJ kg 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 V2 cm 531.21 gm 3 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 u2 2 2 = 0 Whence u2 2 H2 H1 u2 565.2 m sec 2 Ans. By Eq. (2.27), A2 mdot V2 u2 A2 7.05 cm Ans. 7.5 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 H1 3014.9 kJ kg S1 7.1595 kJ kg K S2 = S 1 Interpolations in Table F.2 at several pressures and at the given entropy yield the following values: 400 425 P 450 kPa 475 500 kg sec 2855.2 2868.2 H2 2880.7 2892.5 2903.9 kJ kg 531.21 507.12 V2 485.45 465.69 447.72 mdot V2 u2 cm 3 gm mdot 0.75 565.2 541.7 u2 2 H2 H1 A2 7.05 7.022 A2 7.028 cm 7.059 7.127 2 u2 m 518.1 sec 494.8 471.2 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 s pmin Given pmin 1 5 cspline P A2 400 kPa pi Pi A( ) P (guess) 2 a2 i A2 i interp s p a2 P cm d A pmin = 0 kPa dpmin 431.78 kPa Ans. pmin A pmin Find pmin 2 7.021 cm Ans. 221 Show spline fit graphically: 7.13 p 400 kPa 401 kPa 500 kPa 7.11 A2 cm i 7.09 7.07 2 A( ) p cm 2 7.05 7.03 7.01 400 420 440 Pi 460 480 500 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. 800 775 P 750 kPa 725 700 H2 2956.0 2948.5 2940.8 2932.8 2924.9 V2 u2 kJ kg 294.81 302.12 V2 309.82 317.97 326.69 cm gm 3 u2 2 H2 H1 A2 = mdot 222 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 V2 u2 5.553 5.552 5.557 5.577 cm sec kg 2 At the throat, mdot A2 u2 3 V2 3 A2 mdot 6 cm 2 1.081 kg sec Ans. 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 1.4098 S1 Svap Sliq Sliq ft sec kJ kg K Svap 7.2479 kJ kg K x x 0.966 7.10 u1 230 u2 2000 ft sec From Table F.4 at 130(psi) and 420 degF: H1 1233.6 Btu lbm S1 1.6310 u1 2 Btu lbm rankine u2 2 By Eq. (2.32a), H H2 2 1154.8 Btu lbm Btu lbm Btu lbm rankine H 78.8 Btu lbm H2 H1 H From Table F.4 at 35(psi), we see that the final state is wet steam: Hliq Sliq 228.03 Btu lbm Btu lbm rankine Hvap Svap 1167.1 0.3809 1.6872 223 x H2 Hvap Hliq Hliq x 0.987 (quality) S2 Sliq x Svap Sliq S2 1.67 BTU lbm rankine SdotG S2 S1 SdotG 0.039 Btu lbm rankine Ans. 7.11 u2 580 m T2 sec ( 273.15 15)K molwt 28.9 gm CP mol R 2 molwt 7 By Eq. (2.32a), H= u1 2 u2 2 2 = u2 2 2 But u2 2 H = CP T T Whence T 2 CP 167.05 K Ans. Initial t = 15 + 167.05 = 182.05 degC Ans. 7.12 Values from the steam tables for saturated-liquid water: At 15 degC: V cm 1.001 gm 3 T 288.15 K Enthalpy difference for saturated liquid for a temperature change from 14 to 15 degC: H ( 67.13 58.75) 4 J gm t 2K Cp H t 1.5 10 K P 4 atm Cp 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 T V 1 Cp T P 1 joule 9.86923 cm3 atm T 0.093 K The entropy change for this process is given by Eq. (7.26): S Cp ln T T T V P S 293.15 K J gm or 1.408 10 3 J gm K Apply Eq. (5.36) with Q=0: Wlost T S Wlost T 0.413 Wlost 0.413 kJ kg Ans. 7.13--7.15 350 T1 350 250 400 304.2 Tc 282.3 126.2 369.8 K K P2 1.2bar 80 P1 60 60 20 bar 73.83 Pc 50.40 34.00 42.48 bar .224 .087 .038 .152 5.457 A 1.424 3.280 1.213 0.0 C 4.392 0.0 8.824 10 K 2 6 1.045 B 14.394 .593 28.785 1.157 D 0.0 0.040 0.0 10 K 5 2 10 K 3 225 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 1.084 Pr P1 Pc Pr 1.19 1.765 0.471 0.08664 Tr T1 Tc Tr 1.24 1.981 1.082 7.13 Redlich/Kwong equation: 0.42748 Pr Tr Guess: Eq. (3.53) q Tr z 1 1.5 Eq. (3.54) Given Z i z= 1 Find z () q z z z Eq. (3.52) q 1 4 Ii ln Z Z i qi i qi i Eq. (6.65b) HRi SRi R T1i R ln Z Z i qi i qi 1 i 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 302 232 385 K 226 Z i qi 0.721 0.773 0.956 0.862 2.681 HR 2.253 kJ 0.521 mol 5.177 SR 4.346 1.59 2.33 J mol K 1.396 T2 T1 Cp R A B T1 2 1 C 2 T1 3 2 1 D T1 2 T2 HR Cp T1 S Cp ln 31.545 T2 T1 R ln P2 P1 SR 279.971 T2 302.026 232.062 384.941 7.14 Soave/Redlich/Kwong equation: 2 K Ans. S 29.947 J 31.953 mol K Ans. 22.163 0.08664 0.42748 0.5 2 c 0.480 Pr Tr 1.574 0.176 1 c 1 Tr Eq. (3.53) q Eq. (3.54) Tr Guess: z 1 Given z= 1 q z z z i Eq. (3.52) Z q Find ( z) i 1 4 Ii ln Z Z i qi i qi Eq. (6.65b) HRi R T1i Z i qi 1 ci Tri i 0.5 1 qi Ii Eq. (6.67) 227 SRi R ln Z i qi i ci Tri i 0.5 qi Ii Eq. (6.68) 0.5 The derivative in these equations equals: Now iterate for T2: ci Tri i 273 Guesses T2 300 232 384 K Z i qi 0.75 0.79 0.975 0.866 2.936 HR 2.356 kJ 0.526 mol 6.126 SR 4.769 J 1.789 mol K 1.523 2.679 T2 T1 Cp R A B T1 2 1 C 2 T1 3 2 1 D T1 2 272.757 T2 HR Cp T1 T2 299.741 231.873 383.554 K Ans. 31.565 S Cp ln T2 T1 R ln P2 P1 SR S 30.028 J 32.128 mol K 22.18 Ans. 228 7.15 Peng/Robinson equation: 1 2 1 2 0.07779 0.45724 2 c 0.37464 1.54226 0.26992 2 1 c 1 Tr 0.5 Pr Tr Eq. (3.53) q Eq. (3.54) Tr Guess: Given Z i z z= 1 1 q z z z Eq. (3.52) q 1 4 Find z () Ii 1 2 2 ln Z Z i qi i qi 0.5 i i Eq. (6.65b) HRi R T1i Z i qi 1 ci Tri i 1 qi Ii 0.5 Eq. (6.67) SRi R ln Z i qi i ci Tri i qi Ii Eq. (6.68) The derivative in these equations equals: Now iterate for T2: 270 Guesses T2 297 229 383 K ci Tri i 0.5 229 Z i qi 0.722 0.76 0.95 0.85 3.041 HR 2.459 0.6 1.581 kJ mol 6.152 SR 4.784 J 1.847 mol K 2.689 T2 T1 Cp R A B T1 2 1 C 2 T1 3 2 1 D T1 2 269.735 T2 HR Cp T1 T2 297.366 229.32 382.911 K Ans. 31.2 S Cp ln T2 T1 R ln P2 P1 SR S 29.694 J 31.865 mol K Ans. 22.04 7.18 Wdot H1 3500 kW 3462.9 kJ kg Data from Table F.2: H2 2609.9 kJ kg S1 7.3439 kJ kg K By Eq. (7.13), mdot Wdot H2 H1 mdot 4.103 kg sec Ans. For isentropic expansion, exhaust is wet steam: Sliq 0.8321 S2 Svap Sliq Sliq kJ kg K Svap 7.9094 kJ kg K S2 S1 x x 0.92 (quality) 230 Hliq 251.453 kJ kg Hvap 2609.9 kJ kg 3 kJ H'2 Hliq x Hvap Hliq H'2 2.421 10 kg H2 H'2 H1 H1 0.819 Ans. 7.19 The following vectors contain values for Parts (a) through (g). For intake conditions: 3274.3 kJ kg 6.5597 kJ kg K 3509.8 kJ kg 6.8143 kJ kg K 0.80 0.77 0.82 0.75 0.75 0.80 0.75 kJ 3634.5 kg H1 kJ 6.9813 kg K S1 kJ 3161.2 kg 6.4536 kJ kg K kJ 2801.4 kg kJ 6.4941 kg K 1444.7 Btu lbm 1.6000 Btu lbm rankine 1389.6 Btu lbm 1.5677 Btu lbm rankine 231 For discharge conditions: 0.9441 kJ kg K 7.7695 kJ kg K 0.8321 kJ kg K 7.9094 kJ kg K 0.6493 kJ kg K 8.1511 kJ kg K Sliq 1.0912 kJ kg K Svap 7.5947 kJ kg K S'2 = S1 1.5301 kJ kg K 7.1268 kJ kg K 0.1750 Btu lbm rankine 1.9200 Btu lbm rankine 0.2200 Btu lbm rankine kJ kg 1.8625 Btu lbm rankine 80 kg sec 289.302 2625.4 kJ kg 251.453 kJ kg 2609.9 kJ kg 90 kg sec 191.832 kJ kg 2584.8 kJ kg 70 kg sec Hliq 340.564 kJ kg Hvap 2646.0 kJ kg mdot 65 kg sec 504.701 kJ kg 2706.3 kJ kg 50 kg sec 94.03 Btu lbm 1116.1 Btu lbm 150 lbm sec 120.99 Btu lbm 1127.3 Btu lbm 100 lbm sec 232 x'2 S1 Svap Sliq Sliq H'2 Hliq x'2 Hvap Hliq H H2 Hvap H'2 Hliq H1 H2 H1 H Wdot H mdot x2 Hliq S2 Sliq x2 Svap Sliq H2 H2 H2 H2 H2 H2 H2 1 2 3 4 5 6 7 S2 2423.9 2535.9 kJ 2467.8 kg 2471.4 2543.4 1 2 3 4 5 7.1808 7.6873 7.7842 7.1022 6.7127 kJ kg K S2 S2 S2 S2 S2 S2 Ans. 1031.9 Btu 1057.4 lbm 6 7 1.7762 Btu 1.7484 lbm rankine 68030 87653 81672 Wdot 44836 kW 12900 65333 35048 Wdot 91230 117544 109523 60126 17299 87613 46999 hp Ans. 233 7.20 T 423.15 K P0 8.5 bar P 1 bar For isentropic expansion, S 0 J mol K For the heat capacity of nitrogen: A 3.280 B 0.593 10 K 3 D 0.040 10 K 5 2 For the entropy change of an ideal gas, combine Eqs. (5.14) & (5.15) with C = 0. Substitute: 0.5 (guess) Given S = R A ln B T D T 2 1 1 ln 2 T P P0 Find T0 T0 762.42 K Ans. Thus the initial temperature is 489.27 degC 7.21 T1 1223.15 K P1 10 bar P2 1.5 bar CP 32 J mol K 0.77 Eqs. (7.18) and (7.19) derived for isentropic compression apply equally well for isentropic expansion. They combine to give: R CP W's C P T1 P2 P1 1 W's 15231 J mol Ws W's H Ws 234 Ws 11728 J mol Ans. Eq. (7.21) also applies to expansion: T2 T1 H CP T2 856.64 K Ans. 7.22 Isobutane: Tc 408.1 K Pc 36.48 bar 0.181 T0 523.15 K P0 5000 kPa P 500 kPa S 0 J mol K For the heat capacity of isobutane: A T0 Tc 1.677 B 37.853 10 K 1.282 3 C P0 Pc P 11.945 10 K 2 6 Tr0 Tr0 Pr0 Pr0 1.3706 Pr Pc Pr 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 C T0 2 1 2 1 ln P P0 Pr SRB Tr0 Pr0 Find T T0 T 445.71 K Tr T Tc Tr 1.092 235 The enthalpy change is given by Eq. (6.91): Hig Hig H' R ICPH T0 T 1.677 37.853 10 11.078 kJ mol 3 11.945 10 6 0.0 Hig 8331.4 R Tc HRB Tr Pr J mol HRB Tr0 Pr0 H' The actual enthalpy change from Eq. (7.16): 0.8 Wdot ndot ndot H 700 mol sec Wdot H H' 4665.6 kW H Ans. 6665.1 J mol The actual final temperature is now found from Eq. (6.91) combined with Eq (4.7), written: 0.7 Given (guess) H = R A T0 1 Tc HRB Find B 2 T0 2 T0 Pr Tc 0.875 T 2 1 C 3 T0 3 3 1 HRB Tr0 Pr0 T0 T 457.8 K Ans. 7.23 From Table F.2 @ 1700 kPa & 225 degC: H1 2851.0 kJ kg S1 x2 6.5138 kJ kg K Sliq 0.6493 kJ kg K At 10 kPa: 0.95 236 Hliq 191.832 kJ kg Hvap 2584.8 kJ kg Svap 8.1511 kJ kg K mdot 0.5 kg sec Wdot 180 kW H2 Hliq x2 Hvap Hliq H H2 H1 H2 2.465 10 3 kJ kg H 385.848 kJ kg (a) Qdot mdot H Wdot Qdot 12.92 kJ sec Ans. (b) For isentropic expansion to 10 kPa, producing wet steam: x'2 S1 Svap Sliq Sliq H'2 Hliq x'2 Hvap Hliq x'2 0.782 H'2 2.063 10 3 kJ kg Ans. Wdot' mdot H'2 H1 Wdot' 394.2 kW 7.24 T0 673.15 K P0 8 bar P 1 bar For isentropic expansion, For the heat capacity of carbon dioxide: S 0 J mol K 5 2 A 5.457 B 1.045 10 K 3 D 1.157 10 K 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 D T0 2 1 2 T' T0 1 ln P P0 Find 0.693 237 T' 466.46 K H' R ICPH T0 T' 5.457 1.045 10 9.768 kJ mol Work H 3 0.0 1.157 10 5 H' 0.75 H Work H' 7.326 kJ mol Work 7.326 kJ mol Ans. For the enthalpy change of an ideal gas, combine Eqs. (4.2) and (4.7) with C = 0: Given H = R A T0 1 B 2 T0 2 0.772 2 1 D T0 T0 1 Find T T 519.9 K Ans. Thus the final temperature is 246.75 degC 7.25 Vectors containing data for Parts (a) through (e): 500 450 T1 525 475 550 H [ ( Cp T2 P1 6 5 10 7 4 T1) ] R Cp 371 376 T2 458 372 403 P2 1.2 2.0 3.0 1.5 1.2 Cp 3.5 4.0 5.5 R 4.5 2.5 Ideal gases with constant heat capacities HS Cp T1 P2 P1 1 Eq. (7.22) Applies to expanders as well as to compressors 238 0.7 0.803 H HS 0.649 0.748 0.699 7.26 Cp 7 2 R ndot 175 mol sec T1 550K P1 6bar P2 1.2bar Guesses: Given 0.75 Wdot 600kW Wdot = Wdot 0.065 .08 ln Wdot kW ndot Cp T1 P2 P1 R Cp 1 Find ( Wdot) Wdot kW Wdot 594.716 kW Ans. 0.065 0.08 ln 0.576 Ans. For an expander operating with an ideal gas with constant Cp, one can show that: R Cp T2 T1 1 P2 P1 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 10 3 J K sec Ans. 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) A second relation follows from Eq. (7.16), written: [x is quality] H = Hvap - 3207.1 = ( HS) = (0.75)[ (x)(Hvap) + (1-x)(Hliq) - 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 Sl 503.7 1.5276 Hv Sv 2706.0 7.1293 The two equations for x are: xH Hv 801.7 .75 Hl .75 ( Hv Hl) xS 6.7093 Sl Sv Sl 0.924 xS 0.925 The trial values given produce: xH 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 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: cm 3 V 1001 kg Cp 4.190 kJ kg degC H kJ kg T T Eqs. (7.16) and (7.24) combine to give: Ws H (7.14) Ws 0.223 V ( P2 P1) Eq. (7.25) with =0 is solved for T: H V ( P2 Cp P1) 0.044 degC Ans. 7.30 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): 753.15 673.15 T0 773.15 K 723.15 755.37 200 150 ndot 175 100 0.5 453.59 241 6 bar 5 bar P0 7 bar 8 bar 95 psi 0.80 0.75 mol sec 1 bar 1 bar P 1 bar 2 bar 15 psi J mol K S i 0 1 5 0.78 0.85 0.80 For the heat capacity of nitrogen: A 0.5 Given 3.280 (guess) B 0.593 10 K 3 D 0.040 10 K 5 2 S = R A ln B T0 D T0 2 2 1 2 1 ln P P0 Tau T0 P0 P Find 460.67 431.36 i Tau T0 P0 Pi i i Ti T0 i i T 453.48 K 494.54 455.14 H'i R ICPH T0 Ti 3.280 0.593 10 i 3 0.0 0.040 10 5 8879.2 7279.8 H' 9714.4 6941.7 9112.1 J mol 7103.4 5459.8 H H' J mol H 7577.2 5900.5 7289.7 0.5 Given (guess) H = R A T0 Tau T0 H 1 Find B 2 T0 2 2 1 D T0 i 1 i Tau T0 Hi Ti T0 i i 242 520.2 492.62 T 525.14 K 529.34 516.28 Ans. 1421 819 Wdot ndot H Wdot 1326 kW Ans. 590 1653 7.31 Property values and data from Example 7.6: H1 3391.6 kJ kg S1 6.6858 kJ kg K mdot 59.02 kg sec H2 2436.0 kJ kg S2 7.6846 kJ kg K Wdot 56400 kW T 300 K By Eq. (5.26) Wdotideal t mdot H2 H1 T t S2 0.761 S1 Ans. Wdotideal 74084 kW Wdot Wdotideal The process is adiabatic; Eq. (5.33) becomes: SdotG mdot S2 S1 SdotG 58.949 kW K Ans. Wdotlost T SdotG Wdotlost 17685 kW Ans. 7.32 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 kJ kg K 243 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 kJ kg K kJ kg K Sliq 0.8932 Slv 6.9391 0.72 For isentropic expansion of steam in the turbine: S'3 S'3 H23 H23 x3 x3 S2 6.519 kJ kg K H'3 437.996 H3 Hliq Hlv 0.883 x'3 x'3 H2 kJ kg S'3 Sliq Slv H'3 H'3 Hliq 2.174 x'3 Hlv 10 3 kJ 0.811 H3 H3 S3 S3 kg H2 2.345 H23 10 3 kJ kg Sliq 7.023 x3 Slv kJ kg K mol sec 197.96)K For the exhaust gases: T1 T1 ndot 125 T2 T2 ( 273.15 673.15 K 18 400)K ( 273.15 471.11 K molwt gm mol 3 Hgas Sgas R MCPH T1 T2 3.34 1.12 10 R MCPS T1 T2 3.34 1.12 10 0.0 0.0 T2 T2 T1 T1 3 0.0 0.0 ln 244 Hgas 6.687 10 3 kJ kmol Sgas 11.791 kJ kmol K Energy balance on boiler: mdot ndot Hgas H2 H1 mdot 0.30971 kg sec (a) Wdot mdot H3 H2 Wdot 135.65 kW Ans. (b) By Eq. (5.25): T 293.15 K Wdotideal ndot Hgas mdot H3 H1 T ndot Sgas mdot S3 314.302 kW t S1 t Wdotideal Wdot Wdotideal 0.4316 Ans. (c) For both the boiler and the turbine, Eq. (5.33) applies with Q = 0. For the boiler: SdotG Boiler: ndot Sgas mdot S2 S1 SdotG 0.4534 kW K Ans. For the turbine: SdotG mdot S3 S2 Turbine: SdotG 0.156 kW K Ans. (d) Wdotlost.boiler 0.4534 kW K T Wdotlost.boiler 132.914 kW Wdotlost.turbine 0.1560 kW T K Wdotlost.turbine Fractionboiler 45.731 kW Fractionboiler Wdotlost.boiler Wdotideal 0.4229 Ans. 245 Fractionturbine Note that: Wdotlost.turbine Wdotideal t Fractionturbine Fractionturbine 1 0.1455 Ans. Fractionboiler 7.34 From Table F.2 for sat. vap. at 125 kPa: H1 2685.2 kJ kg S1 7.2847 kJ kg K kJ kg K For isentropic expansion, S'2 = S1 = 7.2847 Interpolation in Table F.2 at 700 kPa for the enthalpy of steam with this entropy gives H'2 H2 3051.3 kJ kg 0.78 H kJ kg H'2 H1 H 469.359 kJ kg H1 H H2 3154.6 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 Ans. mdot H Wdot 1173.4 kW Ans. 246 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): 298.15 353.15 T0 303.15 373.15 299.82 338.71 100 100 ndot 150 50 0.5 453.59 0.5 453.59 mol sec K 101.33 kPa 375 kPa P0 100 kPa 500 kPa 14.7 psi 55 psi 375 kPa 1000 kPa P 500 kPa 1300 kPa 55 psi 135 psi 0.75 0.70 0.80 0.75 0.75 0.70 i S 0 J mol K 1 6 For the heat capacity of air: A 0.5 Given 3.355 (guess) B 0.575 10 K 3 D 0.016 10 K 5 2 S = R A ln B T0 D T0 2 2 1 2 1 ln P P0 Tau T0 P0 P Find 247 i Tau T0 P0 Pi i i 431.06 464.5 Ti T0 i i 476.19 T 486.87 434.74 435.71 K H'i R ICPH T0 Ti 3.355 0.575 10 i 3 0.0 0.016 10 5 3925.2 3314.6 5133.2 H' J 3397.5 mol 3986.4 2876.6 5233.6 4735.1 H H' H 6416.5 4530 5315.2 4109.4 J mol 1.5 (guess) Given Tau T0 H = R A T0 H Find 1 B 2 T0 2 i 2 1 D T0 Hi 1 Tau T0 i Ti T0 i i Wdot ndot H 248 474.68 511.58 T 518.66 524.3 479.01 476.79 K 702 635 Wdot 1291 304 1617 1250 hp 523 474 Wdot 962 227 1205 932 kW Ans. 7.36 Ammonia: Tc 405.7 K Pc 112.8 bar 0.253 T0 294.15 K P0 200 kPa P 1000 kPa S 0 J mol K For the heat capacity of ammonia: A 3.578 B 3.020 10 K 0.725 3 D P0 Pc P 0.186 10 K 5 2 Tr0 T0 Tc Tr0 Pr0 Pr0 0.0177 Pr Pc Pr 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 Tc D T0 2 1 2 1 ln P P0 SRB Pr SRB Tr0 Pr0 Find 1.437 T T0 T 422.818 K Tr 249 T Tc Tr 1.042 Hig Hig H' H' R ICPH T0 T 3.578 3.020 10 4.826 Hig 4652 3 0.0 0.186 10 5 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: 1.4 Given (guess) H = R A T0 1 Tc HRB B 2 T0 2 T0 Pr Tc 1.521 2 1 D T0 1 HRB Tr0 Pr0 Find T Tr T0 T Tc T Tr 447.47 K 1.103 Ans. S R A ln B T0 D 2 2 T0 SRB Tr0 Pr0 1 1 ln P P0 SRB Tr Pr S 2.347 J mol K Ans. 250 7.37 Propylene: Tc 365.6 K Pc 46.65 bar 0.140 T0 303.15 K P0 11.5 bar P 18 bar S 0 J mol K For the heat capacity of propylene: A 1.637 B 22.706 10 K 3 C 6.915 10 K 2 6 Tr0 T0 Tc Tr0 0.8292 Pr0 P0 Pc P Pr0 Pr 0.2465 0.386 Pr Use generalized second-virial correlation: Pc The entropy change is given by Eq. (6.92) combined with Eq. (5.15) with D = 0: 1.1 (guess) Given S = R A ln SRB B T0 T0 Tc C T0 2 1 2 1 ln P P0 Pr SRB Tr0 Pr0 Find 1.069 T T0 T 324.128 K Tr T Tc 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 3 6.915 10 6 0.0 Hig 1.409 10 3 J mol H' Hig R Tc HRB Tr Pr 251 HRB Tr0 Pr0 H' 964.1 J mol The actual enthalpy change from Eq. (7.17): 0.80 H H' H 1205.2 J mol ndot 1000 mol sec Wdot ndot H Wdot 1205.2 kW Ans. The actual final temperature is now found from Eq. (6.91) combined with Eq (4.7), written: 1.1 (guess) Given H = R A T0 1 Tc HRB Find 7.38 Methane: B 2 T0 2 T0 Pr Tc T 2 1 C 3 T0 3 3 1 HRB Tr0 Pr0 1.079 Tc 190.6 K T0 Pc T 327.15 K Ans. 0.012 45.99 bar T0 308.15 K P0 3500 kPa P 5500 kPa S 0 J mol K For the heat capacity of methane: A 1.702 B 9.081 10 K 3 C 2.164 10 K 2 6 Tr0 T0 Tc Tr0 1.6167 Pr0 P0 Pc P Pr0 0.761 Pr Pc Pr 1.196 252 Use generalized second-virial correlation: The entropy change is given by Eq. (6.92) combined with Eq. (5.15) with D = 0: 1.1 Given (guess) S = R A ln SRB B T0 T0 Tc C T0 2 1 2 1 ln P P0 Pr SRB Tr0 Pr0 Find 1.114 T Tr T0 T Tc T Tr 343.379 K 1.802 The enthalpy change for the final T is given by Eq. (6.91), with HRB for this T: Hig Hig H' H' R ICPH T0 T 1.702 9.081 10 1.298 10 Hig 1158.8 3 J 3 2.164 10 6 0.0 mol HRB Tr0 Pr0 R Tc HRB Tr Pr J mol The actual enthalpy change from Eq. (7.17): 0.78 mol sec H H' H 1485.6 J mol Ans. ndot 1500 Wdot ndot H Wdot 2228.4 kW The actual final temperature is now found from Eq. (6.91) combined with Eq (4.7), written: 1.1 (guess) 253 Given H = R A T0 1 Tc HRB Find B 2 T0 2 T0 Pr Tc T 2 1 C 3 T0 3 3 1 HRB Tr0 Pr0 1.14 T0 T 351.18 K Ans. 7.39 From the data and results of Example 7.9, T1 293.15 K T2 428.65 K P1 140 kPa P2 560 kPa Work 5288.3 J T mol 293.15 K 3 6 H H R ICPH T1 T2 1.702 9.081 10 5288.2 J mol 2.164 10 0.0 S R ICPS T1 T2 1.702 9.081 10 J mol K 3 2.164 10 6 0.0 ln P2 P1 S 3.201 Since the process is adiabatic: SG S SG 3.2012 J mol K Ans. Wideal H T S Wideal 4349.8 J mol Ans. Wlost T S Wlost 938.4 J mol Ans. Wideal t Work t 0.823 Ans. 254 7.42 P1 1atm T1 ( 35 273.15) K T1 308.15 K P2 50atm T2 ( 200 273.15) K T2 473.15 K 0.65 Vdot 0.5 m 3 sec Cp 3.5 R V R T1 P1 ndot Vdot V ndot 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: 1 N r= P2 P1 (where r is the pressure ratio in each stage and N is the number of stages.) ( T2 T1) T1 Eq. (7.23) may be solved for T2prime: T'2 T'2 415.4 K R1 N Cp Eq. (7.18) written for a single stage is: T'2 = T1 P2 P1 Put in logarithmic form and solve for N: N R Cp ln ln P2 P1 T'2 T1 N 3.743 (a) Although any number of stages greater than this would serve, design for 4 stages. 1 N (b) Calculate r for 4 stages: N 4 r P2 P1 r 2.659 Power requirement per stage follows from Eq. (7.22). In kW/stage: R Cp Wdotr ndot Cp T1 r 1 Wdotr 255 87.944 kW Ans. (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 mdotw 7.44 Qdotr Hw mdotw 1.052 kg sec Ans. (in each interchanger) 300 290 T1 295 K 300 305 464 547 T2 455 K 505 496 P2 P1 2.0 1.5 1.2 bar 1.1 1.5 6 5 6 bar 8 7 Cp 3.5 2.5 4.5 R 5.5 4.0 H [ ( Cp T2 T1) ] R Cp Ideal gases with constant heat capacities HS Cp T1 P2 P1 1 256 (7.22) 3.219 3.729 HS 4.745 5.959 4.765 kJ mol 0.675 HS H 0.698 0.793 0.636 0.75 Ans. 7.47 The following vectors contain values for Parts (a) through (e). Intake conditions first: 298.15 363.15 T1 333.15 K 294.26 366.48 2000 kPa 5000 kPa P2 5000 kPa 20 atm 1500 psi 100 kPa 200 kPa P1 20 kPa 1 atm 15 psi 0.75 0.70 0.75 0.70 0.75 mdot 20 kg 30 kg 15 kg 50 lb 80 lb 1 sec 257.2 696.2 523.1 217.3 714.3 10 K 6 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 1.002 1.038 By Eq. (7.24) cm gm 3 4.15 4.20 CP 4.20 4.185 4.20 HS V P2 P1 H HS kJ kg K 257 1.906 4.973 HS 5.065 1.929 10.628 kJ kg 2.541 7.104 H 6.753 2.756 14.17 0.188 kJ kg By Eq. (7.25) T H V 1 T1 CP P2 P1 0.807 T 0.612 K 0.227 1.506 50.82 213.12 Wdot H mdot Wdot 101.29 kW Wdot 62.5 514.21 298.338 363.957 T2 T1 T T2 333.762 K 294.487 367.986 t2 t2 T2 K 1 2 3 68.15 285.8 135.84 hp 83.81 689.56 Ans. 25.19 90.81 60.61 degC 273.15 t2 t2 t2 t2 T2 K 1.8 459.67 4 5 70.41 202.7 degF t2 258 7.48 Results from Example 7.10: H 11.57 kJ kg W 11.57 kJ kg S 0.0090 kJ kg K T 300 K Wideal H T S Wideal t W Wideal 8.87 kJ kg Ans. t 0.767 Ans. Since the process is adiabatic. SG S SG 9 10 3 kJ kg K Ans. Wlost T S Wlost 2.7 kJ kg Ans. 7.53 T1 T3 ( 25 ( 200 273.15) K 273.15) K P1 P3 1.2bar 5bar P2 5bar Cpv 105 J mol K Hlv 30.72 kJ mol 0.7 Estimate the specific molar volume of liquid benzene using the Rackett equation (3.72). From Table B.1 for benzene: Tc 562.2K Zc 0.271 Vc 259 cm 3 mol From Table B.2 for benzene: Tn ( 80.0 2 7 273.15) K Trn Tn Tc Assume Vliq = Vsat: V V c Zc 1 T rn Eq. (3.72) V cm 96.802 mol 3 Calculate pump power Ws V P2 P1 Ws 0.053 kJ mol Ans. 259 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: 13.7819 B 2726.81 C 217.572 Tsat A B ln P2 kPa C degC Tsat 142.77 degC Tsat 273.15K Tsat 415.9 K Tsat Estimate the heat of vaporization at Tsat using Watson's method From Table B.2 Hlv At 80 C: 30.72 kJ mol Tsat Tc Tr1 ( 80 273.15) K Tc Tr1 0.38 0.628 Tr2 Tr2 0.74 Hlv2 Hlv 1 1 Tr2 Tr1 Eq. (4.13) Hlv2 26.822 kJ mol Calculate the heat exchanger heat duty. Q R ICPH T2 Tsat 0.747 67.96 10 Hlv2 Cpv T3 Tsat 51.1 kJ mol Ans. 3 37.78 10 6 0 Q 260 7.54 T1 ( 25 273.15)K P1 1.2bar P2 1.2bar T3 ( 200 273.15)K P3 5bar Cpv 105 J mol K 0.75 Calculate the compressor inlet temperature. Combining equations (7.17), (7.21) and (7.22) yields: T2 1 T3 R Cpv T2 408.06 K 1 P3 P2 1 T2 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: 13.7819 B 2726.81 C 217.572 Tsat A B ln P1 kPa C degC Tsat 85.595 degC Tsat Tsat 273.15K Tsat 358.7 K 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 Tr1 ( 80 273.15) K Tc Tr1 0.38 0.628 Tr2 Tsat Tc Tr2 0.638 Hlv2 Q Hlv 1 1 Tr2 Tr1 Eq. (4.13) 3 Hlv2 30.405 6 kJ mol R ICPH T1 Tsat 0.747 67.96 10 Hlv2 Cpv T2 Tsat 37.78 10 0 Q 44.393 kJ mol Ans. 7.57 ndot 100 kmol hr P1 1.2bar T1 300K P2 6bar Cp 50.6 J mol K 0.70 Assume the compressor is adaiabatic. R Cp T2 Wdots Wdote P2 P1 T1 (Pg. 77) T2 Wdots Wdote 390.812 K ndot Cp T2 Wdots T1 127.641 kW 182.345 kW C_compressor 3040dollars Wdots kW 0.952 C_compressor 307452 dollars Ans. C_motor 380dollars Wdote kW 0.855 C_motor 32572 dollars Ans. 262 7.59 T1 375K P1 18bar P2 1.2bar For ethylene: 0.087 Tc 282.3K Pc 50.40bar Tr1 T1 Tc Tr1 1.328 Pr1 P1 Pc Pr1 0.357 Pr2 P2 Pc Pr2 6 0.024 A 1.424 B 14.394 10 3 C 4.392 10 D 0 a) For throttling process, assume the process is adiabatic. Find T2 such that H = 0. H = Cpmig T2 T1 HR2 HR1 Eq. (6-93) 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 = 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 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 T2 T1 R SRB Tr1 Pr1 R ln P2 P1 Eq. (6-94) S 22.128 J mol K Ans. 263 b) For expansion process. 70% First find T2 for isentropic expansion. Solve Eq. (6-94) with S = 0. Guess: Given T2 T1 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 Tc Tr2 Eq. (6-94) 0.779 Now calculate the isentropic enthalpy change, HS. HR2 HRB Tr2 Pr2 R Tc HS R MCPH T1 T2 A B C D T2 T1 HRB Tr2 Pr2 R Tc HRB Tr1 Pr1 6.423 10 3 J R Tc HS mol Calculate actual enthalpy change using the expander efficiency. H HS H 4.496 10 3 J 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 Ans. 264 Now recalculate S at calculated T2 S R MCPS T1 T2 A B C D ln R SRB Tr2 Pr2 T2 T1 R SRB Tr1 Pr1 R ln P2 P1 Eq. (6-94) S 7.77 J mol K 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 Cpgas 200 J mol K 150 J mol K Hlv 35000 J mol Light oil: Exit stream: b) T2 T3 25degC Cpoil 200degC 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 Solving for D/F gives: DF Cpgas T3 Hlv T1 T2 Cpoil T3 DF 0.643 Ans. c) 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 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 S2 6.9636 At point 4: Table F.1, H4 209.3 At point 1: H1 H4 At point 3: Table F.1, Hliq H4 Hlv 2382.9 x3 0.96 H3 Hliq x3 Hlv H3 2496.9 Sliq For isentropic expansion, 0.7035 S'3 S2 Slv 7.3241 x'3 S'3 Sliq Slv x'3 0.855 H'3 Hliq x'3 Hlv H'3 2246 H3 turbine H2 H2 turbine H'3 0.805 Ans. Ws H3 H2 QH H2 H1 Ws 1.035 10 3 QH 3.322 10 3 Ws cycle QH cycle 0.311 Ans. 266 8.2 mdot 1.0 (kg/s) 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 0.741 x4 S1 Sliq Svap Sliq 3 x4 0.241 (c) The rate of heat addition, Step 1--2: Qdot12 mdot ( H2 H1) Qdot12 1.931 10 (kJ/s) (d) The rate of heat rejection, Step 3--4: H3 Hliq x3 ( Hvap Hliq) H4 Hliq x4 ( Hvap Hliq) H3 1.919 10 3 H4 699.083 Qdot34 mdot ( H4 H3) Qdot34 1.22 10 3 (kJ/s) (e) Wdot12 0 Wdot34 0 Wdot23 mdot ( H3 H2) Wdot23 873.222 Wdot41 mdot ( H1 H4) Wdot41 0.368 161.617 (f) Wdot23 Wdot41 Qdot12 Note that the first law is satisfied: Q Qdot12 Qdot34 W Wdot23 Wdot41 Q W 0 267 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 kJ kg 6.9013 kJ kg K 3529.6 kJ kg 6.9485 kJ kg K 3635.4 H2 kJ kg S2 6.9875 kJ kg K kJ 3475.6 kg 6.9145 kJ kg K 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 kJ kg 2584.8 kJ kg 251.453 kJ kg 2609.9 kJ kg 191.832 Hliq kJ kg Hvap 2584.8 kJ kg kJ 419.064 kg 2676.0 kJ kg 180.17 BTU lbm 1150.5 BTU lbm 69.73 BTU lbm 1105.8 BTU lbm 268 0.6493 kJ kg K 8.1511 kJ kg K 0.8321 kJ kg K 7.9094 kJ kg K 0.6493 Sliq kJ kg K 8.1511 Svap kJ kg K 1.3069 kJ kg K 7.3554 kJ kg K 0.3121 BTU lbm rankine 1.7568 BTU lbm rankine 0.1326 BTU lbm rankine 1.9781 BTU lbm rankine cm 1.010 gm 3 cm 1.017 gm 3 0.80 0.75 0.80 turbine pump 0.75 0.75 0.80 0.75 0.75 0.75 1.010 Vliq cm gm 3 cm 1.044 gm 3 0.78 0.78 0.80 0.0167 ft 3 lbm 3 ft 0.0161 lbm 269 80 100 Wdot 70 50 50 80 10 kW 3 10000 kPa 7000 kPa P1 8500 kPa 6500 kPa 950 psi 1125 psi P4 10 kPa 20 kPa 10 kPa 101.33 kPa 14.7 psi 1 psi Wpump Vliq P1 pump P4 H4 Hliq H1 H4 Wpump S'3 = S2 x'3 S2 Svap Sliq Sliq H'3 Hliq x'3 Hvap Hliq H3 H2 turbine H'3 H2 Wturbine H3 H2 mdot Wdot Wturbine Wpump QdotH QdotC H2 QdotH H1 mdot Wdot Answers follow: mdot1 mdot2 mdot3 mdot4 QdotH 70.43 108.64 62.13 67.29 kg sec 1 2 3 4 240705 355111 kJ 213277 sec 205061 QdotH QdotH QdotH QdotH QdotH mdot5 mdot6 145.733 lbm 153.598 sec 5 6 192801 BTU 228033 sec 270 QdotC QdotC 1 2 3 4 0.332 160705 255111 kJ 0.282 Wdot QdotH 0.328 0.244 0.246 0.333 QdotC QdotC 143277 sec 155061 QdotC QdotC 8.4 5 6 145410 BTU 152208 sec Subscripts refer to Fig. 8.3. Saturated liquid at 50 kPa (point 4) V4 cm 1.030 gm 3 H4 340.564 kJ kg P4 P1 3300 kPa 50 kPa Saturated liquid and vapor at 50 kPa: Hliq H4 kJ kg K Hvap 2646.0 kJ kg kJ kg K P1 Wpump 3.348 kJ kg Sliq 1.0912 Svap Wpump H1 7.5947 By Eq. (7.24), V 4 P4 kJ kg H1 H4 Wpump 343.911 The following vectors give values for temperatures of 450, 550, and 650 degC: 3340.6 H2 3565.3 3792.9 kJ kg S2 7.0373 7.3282 7.5891 kJ kg K 271 S'3 H'3 QH S2 Hliq H2 0.914 x'3 x'3 Hvap H1 S'3 Svap Sliq Sliq H'3 H2 Wpump Hliq Wturbine Wturbine QH 0.297 0.314 0.332 Ans. x'3 0.959 0.999 8.5 Subscripts refer to Fig. 8.3. Saturated liquid at 30 kPa (point 4) V4 cm 1.022 gm 3 H4 289.302 kJ kg P1 30 kPa Saturated liquid and vapor at 30 kPa: Hliq H4 kJ kg K Hvap 2625.4 kJ kg 7.7695 kJ kg K 5000 P4 7500 10000 kPa Sliq 0.9441 Svap By Eq. (7.24), Wpump V 4 P4 P1 kJ kg 294.381 H1 H4 Wpump H1 296.936 299.491 The following vectors give values for pressures of 5000, 7500, and 10000 kPa at 600 degC 3664.5 H2 3643.7 3622.7 272 7.2578 kJ kg S2 7.0526 6.9013 kJ kg K S'3 H'3 QH S2 Hliq H2 0.925 x'3 x'3 Hvap H1 S'3 Svap Sliq Sliq H'3 H2 Wpump Hliq Wturbine Wturbine QH 0.359 0.375 0.386 Ans. x'3 0.895 0.873 8.6 From Table F.2 at 7000 kPa and 640 degC: H1 3766.4 kJ kg S1 7.2200 kJ kg K S'2 S1 For sat. liq. and sat. vap. at 20 kPa: Hliq Sliq 251.453 kJ kg Hvap Svap 2609.9 kJ kg kJ kg K 0.8321 kJ kg K 7.9094 The following enthalpies are interpolated in Table F.2 at four values for intermediate pressure P2: 725 P2 750 775 800 kPa 3023.9 H'2 3032.5 3040.9 3049.0 W12 H'2 H1 3187.3 H2 kJ 3200.5 kg kJ kg 0.78 579.15 W12 H2 H1 7.4939 W12 572.442 kJ 565.89 kg 559.572 3194 S2 7.4898 7.4851 7.4797 kJ kg K 3206.8 273 where the entropy values are by interpolation in Table F.2 at P2. x'3 S2 Svap Sliq Sliq H'3 W12 H'3 Hliq x'3 Hvap Hliq W23 W H2 W 20.817 7.811 5.073 17.723 kJ kg W23 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 765.16 kPa (P2) Also needed are values of H2 and S2 at this pressure. Again we do linear interpolations: linterp P2 H2 765.16 kPa linterp P2 S2 765.16 kPa 3197.9 kJ kg kJ kg K H2 S2 3197.9 kJ kg kJ kg K 7.4869 7.4869 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 H'3 H'3 H2 H1 kJ kg Hliq x'3 x'3 W23 W23 S2 Svap 0.94 Sliq Sliq 568.5 Hliq 2.469 x'3 Hvap 10 3 kJ H'3 568.46 H2 kJ kg kg 274 Work W12 W23 Work 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 Sliq Sliq H'3 Hliq x'3 Hvap Hliq x'3 0.903 H'3 2.38 10 3 kJ kg W' H'3 H1 W' 1386.2 kJ kg Whence the overall efficiency is: Work overall W' overall 0.8202 Ans. 275 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 H3 2770.6 H2 kJ kg H3 0.78 2.918 10 3 kJ WI kg WI H'3 521.586 H2 kJ kg WI Isentropic expansion to 20 kPa: S'4 Hliq Sliq S2 251.453 kJ kg kJ Exhaust is wet: for sat. liq. & vap.: Hvap Svap 2609.9 kJ kg kJ kg K 0.8321 kg K 7.9094 276 x'4 x'4 H4 S'4 Svap 0.876 H2 Sliq Sliq H'4 H'4 Hliq 2.317 x'4 Hvap 10 3 kJ Hliq kg H'4 H2 3 H4 2.564 10 3 kJ kg H5 Hliq V5 V 5 P6 1.017 P5 cm gm H6 H6 P5 20 kPa P6 4500 kPa Wpump Wpump H5 Wpump kJ kg 5.841 kJ kg 257.294 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: Hsat.liq kJ 558.5 kg Psat 294.26 kPa Vsat.liq 1.073 cm 3 gm Also by approximation, the definition of the volume expansivity yields: 1 Vsat.liq 9.32 10 1.083 20 4 1 1.063 cm gm K 3 P1 P6 K 277 By Eq. (7.25), H1 Hsat.liq Vsat.liq 1 T1 P1 Psat H1 561.305 kJ kg By an energy balance on the feedwater heater: mass H1 H3 H6 H7 kg mass 0.13028 kg Ans. Work in 2nd section of turbine: WII ( kg 1 mass) H4 H3 WII 307.567 kJ Wnet WI Wpump 1 kg WII Wnet 823.3 kJ QH H2 H1 1 kg QH 2878 kJ Wnet QH 0.2861 Ans. 8.8 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 BTU lbm 0.78 WI H'3 H2 H3 H2 WI H3 1242.2 BTU lbm WI 219.024 BTU lbm S3 1.7431 BTU lbm rankine 278 Isentropic expansion to 1(psia): S'4 S2 Exhaust is wet: for sat. liq. & vap.: Hliq 69.73 BTU lbm Hvap 1105.8 BTU lbm Sliq 0.1326 BTU lbm rankine Svap 1.9781 BTU lbm rankine x'4 S'4 Svap Sliq Sliq H'4 Hliq x'4 Hvap Hliq x'4 0.831 H'4 931.204 BTU lbm H4 H2 H'4 H2 H4 1047.8 BTU lbm x4 H4 Hvap Hliq Hliq S4 Sliq x4 Svap Sliq x4 P5 0.944 S4 1.8748 BTU lbm rankine V5 ft 0.0161 lbm 3 1 psi H5 Hliq Wpump V 5 P6 P5 Wpump 2.489 BTU lbm P6 650 psi H6 H5 Wpump H6 72.219 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 459.67 rankine t1 270.01 279 At this temperature, 270.01 degF, interpolation in Table F.3 gives: Psat Hsat.liq 41.87 psi 238.96 BTU lbm Vsat.liq 0.1717 ft 3 lbm BTU lbm rankine Ssat.liq 0.3960 The definition of the volume expansivity yields: 1 Vsat.liq 4.95 10 0.01726 20 5 0.01709 ft 3 lbm rankine P1 P6 1 rankine BTU lbm BTU lbm rankine By Eq. (7.25) and (7.26), H1 S1 Hsat.liq Ssat.liq Vsat.liq 1 Vsat.liq P1 T1 Psat P1 Psat H1 S1 257.6 0.397 By an energy balance on the feedwater heater: mass H1 H3 H6 H7 lbm mass 0.18687 lbm Ans. Work in 2nd section of turbine: WII Wnet QH 1 lbm WI H2 Wnet QH mass H4 H3 WII WII Wnet QH 158.051 BTU 374.586 BTU 1.204 10 BTU 3 Wpump 1 lbm H1 1 lbm 0.3112 Ans. 280 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 H'3 H3 S2 3142.6 H2 By double interpolation in Table F.2, kJ kg H3 H10 0.80 3.244 10 3 kJ WI WI H'3 407.6 H2 kJ kg WI kg From Table F.1: 829.9 kJ kg 281 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 Sliq kJ 251.453 kg 0.8321 kJ kg K Hvap Svap 2609.9 kJ kg kJ kg K Vliq 1.017 cm 3 gm 7.9094 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 H6 60.09 Hliq V 6 P2 P6 Tsat V6 tsat Vliq 273.15 K P6 20 kPa Wpump Wpump [Eq. (7.24)] 8.238 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 20 10 4 1 1.012 cm 3 gm K CP 272.0 10 4.18 230.2 kJ kg K K CP kJ kg K Solving Eq. (7.25) for delta T gives: T67 H67 Vliq 1 CP Tsat P2 P6 T67 0.678 K t7 tsat T67 K t9 190 2 282 t7 t7 t8 t9 5 t7 60.768 t8 130.38 From Table F.1: H8 547.9 kJ kg H7 Hliq H67 t9 125.38 T9 273.15 t9 K H7 259.691 kJ kg 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 Vsat.9 cm 1.065 gm 3 Psat.9 234.9 kPa Hsat.1 V9 kJ 807.5 kg ( 1.075 Vsat.1 3 cm 1.142 gm 3 Psat.1 ( 1.156 1255.1 kPa 3 cm 1.056) gm V1 cm 1.128) gm T 20 K 1 9 V9 T 1 1 Vsat.1 V1 T Vsat.9 9 8.92 10 4 1 K 1 1.226 10 3 1 K H9 Hsat.9 Vsat.9 1 9 T9 P2 Psat.9 H9 530.9 kJ kg T1 ( 273.15 190)K T1 1 T1 463.15 K H1 Hsat.1 Vsat.1 1 P2 Psat.1 H1 810.089 kJ kg Now we can make an energy balance on feedwater heater I to find the mass of steam condensed: mI H1 H3 H9 H10 kg 283 mI 0.11563 kg Ans. 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 kg Then H4 H4 H2 2.941 H'4 10 3 kJ H2 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 Then S2 Svap 0.889 Sliq Sliq H'5 H'5 Hliq 2.349 x'5 Hvap 10 3 kJ Hliq kg H5 2609.4 kJ kg H5 H2 H'5 H2 The work of the turbine is: Wturbine WI 1 kg 1 kg mI H4 H3 1 kg mI mII H5 H4 936.2 kJ QH H2 H1 1 kg QH 2.842 10 kJ 3 Wturbine Wturbine Wpump 1 kg QH 0.3265 Ans. 284 8.10 Isobutane: Tc 408.1 K Pc 36.48 bar 0.181 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 533.15 K P0 4800 kPa P 450 kPa S 0 J mol K For the heat capacity of isobutane: A 1.677 B 37.853 10 K 3 C 11.945 10 K 2 6 Tr0 T0 Tc Tr0 1.3064 Pr0 Pr P0 Pc P Pc Pr0 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 C T0 2 1 2 1 ln P P0 Pr SRB Tr0 Pr0 Find 0.852 T T0 T 454.49 K Tr T Tc Tr 1.114 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 3 11.945 10 6 0.0 Hig 1.141 10 4 J mol 285 Hturbine Hturbine Hig 8850.6 R Tc HRB Tr Pr 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 Avp 450 kPa 14.57100 Bvp Cvp 2606.775 Cvp 274.068 tsat Avp Bvp VP ln kPa 3 tsat 34 Tsat Tsat tsat 273.15 K 307.15 K Vc cm 262.7 mol Zc 2 0.282 Trsat Tsat Tc 3 Trsat 0.753 Vliq Wpump V c Zc 1 T rsat 7 Vliq P Wpump cm 112.362 mol 488.8 Vliq P0 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 Hig Ha Ha 1.756 Hig 18082 10 4 J mol HRB Tr Pr R Tc HRB Trsat Pr J mol For the condensation process, we estimate the latent heat by Eqs. (4.12) and (4.13): Tn 261.4 K Pc bar Trn 0.38 Trn 1.013 Tn Tc Trn 0.641 R Tn 1.092 ln Hn 0.930 Hn 2.118 10 4 J mol Hb Qdotout Hn 1 1 Trsat Trn Ha Hb Hb Qdotout 18378 J mol mdot Qdotin Qdotout Wturbine 4360 kW Wpump mdot Qdotin 1000 kW Qdotin 0.187 Ans. 5360 kW 8.11 Isobutane: Tc 408.1 K Pc 36.48 bar 0.181 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 J mol K P0 3400 kPa P 450 kPa molwt 58.123 gm mol 287 For the heat capacity of isobutane: A 1.677 T0 Tc B 37.853 10 K 1.0124 3 C 11.945 10 K 2 6 Tr0 Tr0 Pr0 Pr P0 Pc P Pc Pr0 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: 0.8 Given (guess) S = R A ln SRB T0 Tc B T0 Pr C T0 2 1 2 1 ln P P0 SRLK0 Find 0.809 T Tr T0 T Tc T 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 Hig Hturbine Hturbine R ICPH T0 T 1.677 37.853 10 9.3 10 3 J 3 11.945 10 6 0.0 mol R Tc HRB Tr Pr J mol 288 Hig 4852.6 HRLK0 Wturbine Hturbine 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: Vliq 112.36 cm 3 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 5834 kW Wpump Ans. 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 Hig 307.15K Trsat Tsat Tc 3 Trsat 11.945 10 0.753 6 R ICPH T Tsat 1.677 37.853 10 2.817 kJ mol R Tc HRB Trsat Pr J mol 0.0 Ha Ha Hig 2975 HRB Tr Pr 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. For the heater/boiler: Qdotin Wdot Qdotout Qdotin 33387 kW Ans. Ans. Wdot Qdotin 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 0.8 mdot W'turbine W'pump W'pump Wdot 414.3 J mol Ans. 4475 kW 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 28805 kW Wturbine 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 W'pump Wpump mdot 0.134 Ans. Qdotin 33280 kW Ans. Wdot Qdotin 290 8.13 Refer to Fig. 8.10. CP 7 R 2 PC 1 bar TC 293.15 K PD 5 bar 1.4 By Eq. (3.30c): 1 PC V C = PD V D 1 VC VD = PD PC or r 1 PD PC r 3.157 Ans. Eq. (3.30b): TD TC PD PC TA QDA CP QDA 1500 J mol QDA = CP TA TD R TC TD TA 515.845 K re = VB VA = VC VA = PC R TA PA PA PD re T C PA T A PC re 8.14 2.841 Ans. 3 Ratio 5 7 9 Eq. (8.12) now becomes: 1 Ratio = PB PA 1.35 0.248 0.341 0.396 0.434 Ans. 1 1 Ratio 291 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) R CP 2 WAB = CP TA PB PA 1 = CP TA cr 7 1 WCD = CP TC PD PC R CP 2 1 = CP TC er 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 2 7 er 0.5 (guess) 2 7 Given TC er 1 = TA cr 1 er er Find er) ( 0.552 292 By Eq. (7.18): TD = TC PD PC 2 7 R CP This may be written: TD TC er 1 By Eq. (7.11) uE 2 uD = 2 2 PD V D 1 1 PE PD (A) We note the following: er = PD PC cr = PB PA = PC PE cr er = PD PE The following substitutions are made in (A): uD = 0 1 = 2 R = 7 CP PD V D = R T D molwt 29 PE PD gm mol = 1 cr er Then 2 7 uE 2 R 7 TD 1 2 molwt 1 cr er uE 843.4 m sec Ans. PE 1 bar PD cr er PE PD 3.589 bar Ans. 8.17 TA 305 K PA 1.05bar PB 7.5bar 0.8 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 3 0.0 0.016 10 5 R Cpmair 29.921 J mol K 293 Compressor: R Cpmair Wsair TB Cpmair TA Wsair Cpmair PB PA 1 Wsair 8.292 10 3 J mol TA TB 582.126 K Combustion: 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 HR HR 1000K N 57.638 (This is the final value after iteration) 582.03)K 4.217 R ( 298.15 300)K Cpmair N ( 298.15 4.896 10 5 J mol The product stream contains: 1 mol CO2, 2mol H2O, 0.79N mol N2, and (0.21N-2) mol O2 1 n 2 .79 N .21 N i 1 4 5.457 A 2 3.470 3.280 3.639 B 1.045 1.450 0.593 0.506 10 3 1.157 D 0.121 0.040 0.227 10 5 294 ni i 58.638 A i ni Ai B i ni Bi D i ni D i 5 A CpmP HP 198.517 B 0.036 D 1.387 5 10 R MCPH 298.15K 1000.K 198.517 0.0361 0.0 CpmP TC 298.15K HP 1.292 J mol 1.3872 10 10 6 J mol From Ex. 4.7: H298 HP 136.223 802625 HR H298 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 1.3872 10 5 R J mol K For 58.638 moles of combustion product: R Cpm Ws 58.638 Cpm TC PD PC 1 Ws 1.214 10 6 J mol TD TC Ws Cpm TD 343.123 K 295 (Final result of iteration.) Ans. Wsnet Ws Wsair N Wsnet 7.364 10 5 J mol Ans. (J per mole of methane) Parts (b) and (c) are solved in exactly the same way, with the following results: (b) TC 1200 N 37.48 Wsnet 7.365 10 5 TD 5 343.123 (c) TC 1500 N 24.07 Wsnet 5.7519 10 TD 598.94 8.18 tm 0.35 me 0.95 line_losses 20% Cost_fuel 4.00 dollars GJ Cost_electricity tm Cost_fuel 1 me ( line_losses) Cost_electricity 0.05 cents kW hr Ans. This is about 1/2 to 1/3 of the typical cost charged to residential customers. 8.19 TC 111.4K TH 300K Hnlv HE 8.206 Carnot 1 TC TH Carnot kJ mol HE 0.629 0.6 Carnot 0.377 Assume as a basis: W 1kJ QH QC Hnlv W HE QH 2.651 kJ QC QH 1 HE QC 1.651 kJ W 0.201 mol kJ Ans. 296 8.20 TH ( 27 273.15)K TC (6 273.15)K a) Carnot 1 TC TH Carnot 0.07 Ans. b) actual Carnot 0.6 2 3 actual 0.028 Ans. 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 Chapter 9 - Section A - Mathcad Solutions 9.2 TH ( 20 273.15) K TH 293.15 K TC ( 20 273.15) K TC 253.15 K QdotC Carnot 125000 kJ day TC TH TC (9.3) 0.6 Carnot 3.797 Wdot QdotC (9.2) Wdot 0.381 kW Cost 0.08 kW hr Wdot Cost 267.183 dollars yr Ans. 9.4 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 0.21868 P1 P2 138.83 138.83 State 2, Sat. Vapor at TH: H2 State 3, Wet Vapor at TC: Hliq 116.166 S2 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 S2 Sliq Svap Sliq x3 0.971 x4 S1 Sliq Svap Sliq x4 0.302 (c) Heat addition, Step 4--3: H3 Hliq x3 ( Hvap Hliq) H4 Hliq x4 ( Hvap Hliq) H3 101.888 H4 42.118 Q43 ( H3 H4) 298 Q43 59.77 (Btu/lbm) (d) Heat rejection, Step 2--1: Q21 ( H1 H2) Q21 71.223 (Btu/lbm) (e) W21 0 W43 0 W32 ( H2 H3) W32 14.278 W14 ( H4 H1) W14 2.825 (f) Q43 W14 W32 5.219 Note that the first law is satisfied: Q Q21 Q43 W W32 W14 Q W 0 9.7 TC 298.15 K TH 523.15 K (Engine) (Refrigerator) T'C 273.15 K T'H 298.15 K 1 TC TH By Eq. (5.8): Carnot Carnot 0.43 By Eq. (9.3): T'C Carnot T'H T'C Carnot 10.926 By definition: = Wengine QH = Q'C Q'C Wrefrig 35 kJ sec But Wengine = Wrefrig QH Q'C Carnot Carnot Whence QH 7.448 kJ sec Ans. Given that: 0.6 Carnot 0.6 Carnot 6.556 QH Q'C QH 20.689 kJ sec Ans. 299 9.8 (a) QC 4 kJ sec W 1.5 kW QC W 2.667 Ans. (b) QH QC W QH 5.5 kJ sec Ans. (c) = TC TH TC TH ( 40 273.15)K TH 313.15 K TC TH 1 TC 227.75 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. 489.67 479.67 T2 469.67 rankine 459.67 449.67 0.79 0.78 0.77 0.76 0.75 QdotC 600 500 400 300 200 Btu sec 107.320 105.907 H2 104.471 103.015 101.542 Btu lbm 0.22244 0.22325 S2 0.22418 0.22525 0.22647 H4 37.978 Btu lbm Btu lbm rankine T4 539.67 rankine From Table 9.1 for sat. liquid S'3 = S2 (isentropic compression) 300 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 H'3 116.5 117.2 117.9 Btu lbm H23 H'3 H2 H3 H1 H2 H4 H23 24.084 30.098 kJ kg 273.711 276.438 H3 279.336 283.026 286.918 8.653 kJ kg H1 88.337 kJ kg H23 36.337 43.414 50.732 mdot QdotC H2 H1 7.361 mdot 6.016 4.613 3.146 lbm sec Ans. 689.6 595.2 Btu sec QdotH mdot H4 H3 QdotH 494 386.1 268.6 94.5 100.5 Ans. Wdot mdot H23 Wdot 99.2 90.8 72.4 kW Ans. 301 6.697 QdotC Wdot 5.25 4.256 3.485 2.914 Ans. TC T2 TC TH T4 9.793 7.995 Carnot TH TC Carnot 6.71 5.746 4.996 Ans. 9.10 Subscripts in the following refer to Fig. 9.1. All property values come from Tables F.1 and F.2. T2 ( 4 273.15)K 1200 kJ kg T4 H2 ( 34 273.15)K kJ kg S2 0.76 9.0526 kJ kg K QdotC H4 kJ sec 2508.9 142.4 S'2 = S2 (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 kJ kg H1 H4 402.368 302 H3 mdot H2 H23 QdotC H3 mdot 2.911 10 3 kJ kg H2 H1 H3 0.507 kg Ans. sec kJ Ans. sec Ans. QdotH Wdot mdot H4 mdot H23 QdotH Wdot 1404 204 kW Ans. QdotC Wdot T2 Carnot 5.881 T4 T2 Carnot 9.238 Ans. 9.11 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 Sliq 0.01733 Btu lbm rankine Svap 0.22714 Btu lbm rankine For sat. liquid at Point 4 (80 degF): H4 37.978 Btu lbm S4 0.07892 Btu lbm rankine (a) Isenthalpic expansion: H1 H4 QdotC 5 Btu sec mdot QdotC H2 H1 mdot 0.0796 lbm sec Ans. (b) Isentropic expansion: S1 S4 x1 S1 Svap Sliq Sliq H1 Hliq x1 Hvap Hliq H1 34.892 BTU lbm mdot QdotC H2 H1 mdot 0.0759 lbm sec Ans. (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 117.5 Btu lbm S2A 0.262 Btu lbm rankine mdot QdotC H2A H4 mdot 0.0629 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 H2 a 304 H3 3.5896 Ans. In the second case H1 has its last calculated value [Part (b)]: H2 b H1 H2 b H3 3.7659 Ans. In Part (c), compression is at constant entropy of 0.262 to the final pressure. Again from Fig. G.2: H3 138 Btu lbm Wdot Wdot H3 1.289 H2A mdot BTU sec (Last calculated value of mdot) QdotC c Wdot c 3.8791 Ans. 9.12 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 Btu lbm S2A 0.2435 Btu lbm rankine Btu lbm R For sat. liquid at Point 4 (80 degF): H4 37.978 S4 0.07892 Energy balance, heat exchanger: H1 H4 H2A H2 H1 27.885 BTU lbm mdot 25.634 lbm sec QdotC 2000 Btu sec mdot QdotC H2 H1 305 For compression at constant entropy of 0.2435 to the final pressure of 101.37(psia), by Fig. G.2: H'3 127 Btu lbm 0.75 Hcomp H'3 H2A Wdot mdot Hcomp 25.634 lbm sec Wdot 396.66 kW Hcomp Ans. 14.667 Btu lbm mdot 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 QdotC H2 H4 H'3 116 Btu lbm Hcomp Hcomp H'3 H2 Wdot mdot Hcomp lbm sec 13.457 Btu lbm mdot 29.443 Wdot 418.032 kW Ans. 9.13 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 Btu lbm 113.3 H'3 116.5 119.3 Btu lbm H1 H4 306 (a) By Eq. (9.4): H2 H'3 H1 H2 8.294 5.528 4.014 H2 Since H = H3 H2 Ans. (b) H H'3 0.75 Eq. (9.4) now becomes H2 H1 H 6.221 4.146 3.011 Ans. 9.14 WINTER TH 293.15 Wdot 1.5 TC QdotH = 0.75 TH TH TC Wdot = QdotH TH TC Given 250 (Guess) TH TC Wdot = TH 0.75 TH TC TC TC Find TC 268.94 K Ans. Minimum t = -4.21 degC 307 SUMMER TC QdotC 298.15 0.75 TH TC TH TC Wdot = TC QdotC TH 300 (Guess) Given Wdot 0.75 TH TC = TH TC TC TH Find TH TH 322.57 K 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 H9 284.7 kJ kg H15 1186.7 1056.4 kJ kg By Eq. (9.8): z H4 H9 H15 H15 z 0.17 0.351 Ans. 9.17 Advertized combination unit: TH ( 150 459.67)rankine TC ( 30 459.67)rankine TH 609.67 rankine TC 489.67 rankine QC 50000 Btu hr WCarnot QC 308 TH TC TC WCarnot 12253 Btu hr WI 1.5 WCarnot WI 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 QC QH 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 TH TC TC WCarnot 9190 Btu hr Work 1.5 WCarnot Work 13785 Btu hr The total power required is WII QH Work WII 82165 Btu hr NO CONTEST 9.18 TC 210 T'H 260 T'C 255 TH 305 By Eq. (9.3): TC TH TC I 0.65 TC T'H TC II 0.65 T'C TH T'C WCarnot = QC WI = QC I WII = QC II Define r as the ratio of the actual work, WI + WII, to the r Carnot work: 9.19 1 I 1 II r 1.477 Ans. This problem is just a reworking of Example 9.3 with different values of x. It could be useful as a group project. 309 9.22 TH 290K TC 250K Carnot Ws 0.40kW TC Carnot TH TC 6.25 3 2 65% Carnot 4.063 Ans. QC 9.23 Ws QC 1.625 -3 Q 10 kgm sec H 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 ( 10 273.15) K T4 ( 30 273.15) K T2 T1 Calculate the high and low operating pressures using the given vapor pressure equation Guess: PL 1bar PH 4104.67 T1 K 2bar PL Given ln PL bar = 45.327 5.146 ln T1 K 615.0 bar T1 K 2 PL Find PL PL 6.196 bar PH Given ln PH bar = 45.327 4104.67 T4 K 5.146 ln T4 K 615.0 bar T4 K 2 PH Find PH PH 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 ndottoluene R ICPH T1 T2 15.133 6.79 10 3 16.35 10 6 0 QdotC 177.536 kW 310 Since the throttling process is adiabatic: H4 = H1 But: Hliq4 = Hliq1 x1 Hlv1 so: Hliq4 T4 Hliq1 = x1 Hlv and: Hliq4 Hliq1 = Vliq P4 P1 T1 Cpliq ( T) dT Estimate Vliq using the Rackett Eqn. 0.253 Tc 405.7K Pc 3 112.80bar Zc 0.242 Vc cm 72.5 mol Tn 239.7K Hlvn 23.34 kJ mol Tr ( 20 273.15)K Tc 2 Tr 0.723 3 Vliq V c Zc 1 Tr 7 Vliq cm 27.112 mol Estimate Hlv at 10C using Watson correlation T1 Tn Tr1 Trn 0.591 Trn Tc Tc Tr1 kJ mol 0.698 Hlv Hliq41 Hlvn Vliq PH 1 1 PL Tr1 Trn 0.38 Hlv 20.798 R ICPH T1 T4 22.626 100.75 10 3 192.71 10 6 0 Hliq41 1.621 kJ mol x1 Hliq41 Hlv x1 0.078 For the evaporator H12 = H2 H1 = H1vap H1liq x1 Hlv = 1 x1 Hlv H12 ndot 1 x1 Hlv H12 ndot 311 19.177 kJ mol Ans. QdotC H12 9.258 mol sec Chapter 10 - Section A - Mathcad Solutions 10.1 Benzene: Toluene: A1 := 13.7819 A2 := 13.9320 A1 B1 T degC +C1 B1 := 2726.81 B2 := 3056.96 A2 B2 T degC C1 := 217.572 C2 := 217.625 Psat1 ( T) := e kPa Psat2 ( T) := e Guess: +C2 kPa (a) Given: x1 := 0.33 Given T := 100 degC y1 := 0.5 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 y1 = 0.545 Ans. P = 109.303 kPa Ans. T := 100 degC Guess: 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 Ans. P = 92.156 kPa Ans. P := 120 kPa Guess: y1 := 0.5 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. (d) Given: y1 := 0.33 Given P := 120 kPa Guess: x1 := 0.33 T := 100 degC x1 Psat1 ( T) + ( 1 x1) Psat2 ( T) = P 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 Psat1 ( T) + ( 1 x1) Psat2 ( T) = P x1 Psat1 ( T) = y1 P x1 := 0.33 y1 := 0.5 x1 := Find ( x1 , y1) y1 (f) z1 := 0.33 Guess: Given x1 = 0.282 Ans. y1 = 0.484 Ans. x1 = 0.282 L := 0.5 z1 = L x1 + V y1 L+V = 1 y1 = 0.484 V := 0.5 L := Find ( L , V) V (g) Vapor Fraction: Liquid Fraction: V = 0.238 L = 0.762 Ans. 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 10.2 Pressures in kPa; temperatures in degC (a) Antoine coefficients: Benzene=1; Ethylbenzene=2 A1 := 13.7819 A2 := 13.9726 Psat1 ( T) := exp A1 B1 := 2726.81 B2 := 3259.93 C1 := 217.572 C2 := 212.300 B2 Psat2 ( T) := exp A2 T + C2 P-x-y diagram: T := 90 y1 ( x1) := x1 Psat1 ( T) P ( x1 ) P ( x1) := x1 Psat1 ( T) + ( 1 x1) Psat2 ( T) T-x-y diagram: P' := 90 B1 T + C1 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) ) + ( 1 x1) Psat2 ( T ( x1) ) x1 Psat1 ( T ( x1) ) x1 := 0 , 0.05 .. 1.0 150 140 130 P ( x1 ) P ( x1 ) 50 100 120 T ( x1 ) 110 100 T ( x1 ) 90 80 70 0 0 x1 , y1 ( x1 ) 0.5 1 60 0 x1 , y'1 ( x1 ) 0.5 1 314 (b) Antoine coefficients: 1-Chlorobutane=1; Chlorobenzene=2 A1 := 13.7965 A2 := 13.8635 Psat1 ( T) := exp A1 B1 := 2723.73 B2 := 3174.78 B1 T + C1 C1 := 218.265 C2 := 211.700 Psat2 ( T) := exp A2 B2 T + C2 P-x-y diagram: T := 90 y1 ( x1) := x1 Psat1 ( T) P ( x1 ) P ( x1) := x1 Psat1 ( T) + ( 1 x1) Psat2 ( T) T-x-y diagram: 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) ) + ( 1 x1) Psat2 ( T ( x1) ) x1 Psat1 ( T ( x1) ) x1 := 0 , 0.05 .. 1.0 160 130 122.5 115 113.33 P ( x1 ) P ( x1 ) 66.67 107.5 T ( x1 ) 100 T ( x1 ) 92.5 85 77.5 20 0 0.5 1 70 0 0.5 1 x1 , y1 ( x1 ) x1 , y'1 ( x1 ) 315 10.3 Pressures in kPa; temperatures in degC (a) Antoine coefficinets: n-Pentane=1; n-Heptane=2 A1 := 13.7667 A2 := 13.8622 Psat1 ( T) := exp A1 B1 := 2451.88 B2 := 2911.26 C1 := 232.014 C2 := 216.432 Psat2 ( T) := exp A2 B1 T + C1 B2 T + C2 T := 55 P := Psat1 ( T) + Psat2 ( T) 2 x1 Psat1 ( T) P 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 := y1 = 0.89 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 z1 := x1 , x1 + 0.01 .. y1 V is obviously linear in z1: 1 x1 y1 V ( z1) := z1 x1 y1 x1 V ( z1 ) 0.5 0 0.45 0.5 0.55 0.6 0.65 z1 0.7 0.75 0.8 0.85 316 (b) At fixed T and z1, calculate x1, y1 and P as functions of fraction vapor (V). z1 := 0.5 Guess: Given x := 0.5 y := 0.5 p := Psat1 ( T) + Psat2 ( T) 2 Three equations relate x1, y1, & P for given V: 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 P ( V) := f ( V) 3 V := 0 , 0.1 .. 1.0 1 Plot P, x1 and y1 vs. vapor fraction (V) 150 100 P ( V) 50 x1 ( V) y1 ( V) 0.5 0 0 0.5 V 1 0 0 0.5 V 1 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. Benzene: A1 := 13.7819 B1 := 2726.81 B2 := 3259.93 A2 10.7 C1 := 217.572 C2 := 212.300 B2 T degC +C2 Ethylbenzene A2 := 13.9726 A1 B1 T degC +C1 Psat1 ( T) := e kPa 317 Psat2 ( T) := e kPa (a) Given: x1 := 0.35 Given y1 := 0.70 Guess: T := 116 degC P := 132 kPa x1 Psat1 ( T) + ( 1 x1) Psat2 ( T) = P x1 Psat1 ( T) = y1 P T := Find ( T , P) P T = 134.1 degC Ans. 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 (c) T = 91.44 deg_C P = 118.72 kPa P = 66.38 kPa P = 36.02 kPa (d) T = 72.43 deg_C 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 2726.81 B := 3056.96 3259.93 217.572 C := 217.625 212.300 zi := 1 n i := 1 .. n Bi T degC +Ci T := 110 degC P := 90 kPa Psat ( i , T) := e kPa ki := Psat ( i , T) P Guess: V := 0.5 318 Given i=1 n 1 + V ( ki 1) V = 0.836 zi ki = 1 Eq. (10.17) V := Find ( V) Ans. yi := 1 + V ( ki 1) yi P Psat ( i , T) zi ki Eq. (10.16) 0.371 y = 0.339 0.29 0.142 x = 0.306 0.552 0.188 x = 0.334 0.478 0.238 x = 0.345 0.417 0.293 x = 0.342 0.366 Ans. xi := Ans. (b) T = 110 deg_C P = 100 kPa (c) T = 110 deg_C P = 110 kPa V = 0.575 0.441 y = 0.333 0.226 0.508 y = 0.312 0.18 0.572 y = 0.284 0.144 V = 0.352 (d) T = 110 deg_C P = 120 kPa V = 0.146 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 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) z1 := 0.75 z2 := 1 z1 Ai Bi T degC +Ci i := 1 .. n T := ( 340 273.15) degC P := 115 kPa Psat ( i , T) := e Guess: Given kPa ki := Psat ( i , T) P V := 0.5 i=1 n 1 + V ( ki 1) V = 0.656 yi := zi ki = 1 Ans. Eq. (10.17) V := Find ( V) Eq. (10.16) 1 + V ( ki 1) yi P Psat ( i , T) y1 V z1 y1 = 0.678 y1 = 0.320 y1 = 0.682 zi ki y1 = 0.805 Ans. xi := x1 = 0.644 Ans. r := (b) (c) (d) x1 = 0.285 x1 = 0.183 x1 = 0.340 r = 0.705 V = 0.547 V = 0.487 V = 0.469 Ans. r = 0.741 r = 0.624 r = 0.639 320 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 x1 + x2 = 1 Solve for x1 and y1: x1 := P Psat2 H1 Psat2 3 y2 P = x2 Psat2 P = H1 x1 + ( 1 x1) Psat2 P = y1 P + y2 P y1 := H1 x1 P Ans. x1 = 4.502 10 y1 = 0.9 10.16 Pressures in kPa Psat1 := 32.27 Psat2 := 73.14 A := 0.67 2 ( x1 , x2) := exp A x1 z1 := 0.65 1 ( x1 , x2) := exp A x2 ( 2 ) ( 2 ) P ( x1 , x2) := x1 1 ( x1 , x2) Psat1 + x2 2 ( x1 , x2) Psat2 (a) BUBL P calculation: Pbubl := P ( x1 , x2) DEW P calculation: Guess: Given x1 := 0.5 x1 := z1 Pbubl = 56.745 y1 := z1 P' := x2 := 1 x1 Ans. y2 := 1 y1 Psat1 + Psat2 2 y1 P' = x1 1 ( x1 , 1 x1) Psat1 P' = x1 1 ( x1 , 1 x1) Psat1 ... + ( 1 x1) 2 ( x1 , 1 x1) Psat2 x1 := Find ( x1 , P') Pdew 321 Pdew = 43.864 Ans. 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) := P ( x1 , 1 x1) x1 := 0.75 x2 := 1 x1 x1 1 ( x1 , 1 x1) Psat1 The fraction vapor, by material balance is: V := y1 ( x1) x1 z1 x1 V = 0.379 P ( x1 , x2) = 51.892 Ans. (c) See Example 10.3(e). 12.0 := 1 ( 0 , 1) Psat1 Psat2 12.1 := Psat1 2 ( 1 , 0) Psat2 12.0 = 0.862 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 A := 0.95 2 ( x1 , x2) := exp A x1 1 ( x1 , x2) := exp A x2 ( 2 ) ( 2 ) P ( x1 , x2) := x1 1 ( x1 , x2) Psat1 + x2 2 ( x1 , x2) Psat2 y1 ( x1) := x1 1 ( x1 , 1 x1) Psat1 P ( x1 , 1 x1) x1 := 0.05 Pbubl = 47.971 y1 ( x1) = 0.196 (b) DEW P calculation: Guess: x1 := 0.1 y1 := 0.05 P' := x2 := 1 x1 Ans. y2 := 1 y1 Psat1 + Psat2 2 (a) BUBL P calculation: Pbubl := P ( x1 , x2) 322 Given y1 P' = x1 1 ( x1 , 1 x1) Psat1 P' = x1 1 ( x1 , 1 x1) Psat1 ... + ( 1 x1) 2 ( x1 , 1 x1) Psat2 x1 := Find ( x1 , P') Pdew (c) Azeotrope Calculation: Guess: Given y1 = x1 := 0.8 Pdew = 42.191 x1 = 0.0104 Ans. y1 := x1 P := x1 0 Psat1 + Psat2 2 x1 1 x1 = y1 x1 1 ( x1 , 1 x1) Psat1 P P = x1 1 ( x1 , 1 x1) Psat1 + ( 1 x1) 2 ( x1 , 1 x1) Psat2 xaz1 yaz := Find ( x , y , P) 1 1 1 Paz 10.18 Psat1 := 75.20 kPa At the azeotrope: Therefore ln1 = A x2 2 xaz1 0.857 yaz = 0.857 1 Paz 81.366 Ans. Psat2 := 31.66 kPa y1 = x1 2 1 = Psat1 Psat2 2 and x1 := 0.294 ln i = P Psati x2 := 1 x1 ln2 = A x1 ln 2 1 = A x1 x2 ( 2 2 ) Psat1 Psat2 2 2 Whence For A := x1 := 0.6 A = 2.0998 x2 x1 x2 := 1 x1 323 1 := exp A x2 y1 := ( 2 ) 2 := exp A x1 P = 90.104 kPa Psat1 := 1.24 x1 := 0.65 ( 2 ) P := x1 1 Psat1 + x2 2 Psat2 y1 = 0.701 Psat2 := 0.89 x2 := 1 x1 2 := exp A x1 y1 := Ans. x1 1 Psat1 P 10.19 Pressures in bars: A := 1.8 1 := exp A x2 ( 2 ) P = 1.671 ( 2 ) P := x1 1 Psat1 + x2 2 Psat2 y1 = 0.6013 By a material balance, V= (c) Guess: z1 x1 y1 x1 For 0V 1 x1 1 Psat1 P Answer to Part (b) 0.6013 z1 0.65 Ans. (a) Azeotrope calculation: x1 := 0.6 y1 := x1 P := Psat1 + Psat2 2 2 1 ( x1) := exp A ( 1 x1) 2 ( x1) := exp A x1 ( 2 ) Given P = x1 1 ( x1) Psat1 + ( 1 x1) 2 ( x1) Psat2 x1 0 x1 1 x1 = y1 y1 = x1 1 ( x1) Psat1 P x1 y1 := Find ( x1 , y1 , P) P x1 0.592 y1 = 0.592 P 1.673 Ans. 324 10.20 Antoine coefficients: Acetone(1): Methanol(2): A1 := 14.3145 A2 := 16.5785 P in kPa; T in degC B1 := 2756.22 B2 := 3638.27 C1 := 228.060 C2 := 239.500 P1sat ( T) := exp A1 B1 T + C1 P2sat ( T) := exp A2 B2 T + C2 A := 0.64 x1 := 0.175 z1 := 0.25 p := 100 (kPa) 1 ( x1 , x2) := exp A x2 ( 2 ) 2 ( x1 , x2) := exp A x1 ( 2 ) P ( x1 , T) := x1 1 ( x1 , 1 x1) P1sat ( T) ... + ( 1 x1) 2 ( x1 , 1 x1) P2sat ( T) y1 ( x1 , T) := Guesses: Given F= L+V z1 F = x1 L + y1 ( x1 , T) V p = P ( x1 , T) x1 1 ( x1 , 1 x1) P1sat ( T) P ( x1 , T ) F := 1 T := 100 V := 0.5 L := 0.5 L V := Find ( L , V , T) T T = 59.531 (degC) L 0.431 V = 0.569 T 59.531 y1 ( x1 , T) = 0.307 Ans. 325 10.22 x1 := 0.002 A1 := 10.08 A1 y1 := 0.95 B1 := 2572.0 B1 Guess: T := 300 K B2 := 6254.0 A2 B2 A2 := 11.63 Psat1 ( T) := e x2 := 1 x1 Given T K bar y2 := 1 y1 = x2 2 y1 x1 1 y2 Psat2 ( T) := e 1 := e 0.93 x2 2 T K bar 2 := e 0.93 x1 2 Psat1 ( T) Psat2 ( T) T := Find ( T) T = 376.453 K Ans. P := x1 1 Psat1 ( T) y1 P = 0.137 bar Ans. 326 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) 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 Component methane ethylene ethane xi 0.100 0.500 0.400 b) DEW P T=-60 F (-51.11 C) 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 Component methane ethylene ethane yi 0.500 0.250 0.250 c) BUBL T P=250 psia (17.24 bar) 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 P=250 psia (17.24 bar) 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 Component methane ethylene ethane d) DEW T Component methane ethylene ethane yi 0.430 0.360 0.210 327 Problem 10.26 a) BUBL P T=60 C (140 F) 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 Component ethane propane isobutane isopentane xi 0.10 0.20 0.30 0.40 b) DEW P T=60 C (140 F) 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 Component ethane propane isobutane isopentane yi 0.48 0.25 0.15 0.12 c) BUBL T P=15 bar (217.56 psia) 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 Component ethane propane isobutane isopentane xi 0.14 0.13 0.25 0.48 d) DEW T P=15 bar (217.56 psia) 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 Component ethane propane isobutane isopentane yi 0.42 0.30 0.15 0.13 328 Problem 10.27 FLASH T=80 F (14.81 C) 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 Component methane ethane propane n-butane 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 Problem 10.28 First calculate equilibrium composition T=95 C (203 F) 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 Component n-butane n-hexane xi 0.25 0.75 Now calculate liquid fraction from mole balances z1= x1= y1= L= 0.5 0.25 0.633 0.347 ANSWER Problem 10.29 FLASH P = 2.00 atm (29.39 psia) T = 200 F (93.3 C) Fraction condensed L= 0.73 ANSWER xi=yi/Ki 0.191 0.455 0.354 SUM = 1.000 Component n-pentane n-hexane n-heptane 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 = 1.000 329 Problem 10.30 FLASH T=40 C (104 F) 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 Component ethane propane n-butane 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 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 Problem 10.31 FLASH T=70 F (21.11 C) 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 Component ethane propane i-butane n-butane 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 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 1.000 330 Problem 10.32 FLASH T=-15 C (5 F) 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 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 Target: y1=0.8 Component methane ethane propane n-butane zi 0.30 0.10 0.30 0.30 L= 0.8145 xi=yi/Ki 0.162 0.103 0.352 0.364 SUM = 0.982 Component methane ethane propane n-butane zi 0.30 0.10 0.30 0.30 L= 0.6850 xi=yi/Ki 0.073 0.088 0.376 0.424 SUM = 0.960 Component methane ethane propane n-butane zi 0.30 0.10 0.30 0.30 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 Problem 10.33 First calculate vapor composition and temperature on top tray BUBL T: P=20 psia 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 Component n-butane n-pentane xi 0.50 0.50 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 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 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 10.35 a) The equation from NIST is: Mi = ki yi P Eq. (1) The equation for Henry's Law is:i Hi = yi P x Eq. (2) Solving to eliminate P gives: Hi By definition: Mi = ni ns Ms Eq. (3) ki xi where M is the molar mass and the subscript s refers to the solvent. = Mi Dividing by the toal number of moles gives: Mi = xi xs Ms Eq. (4) Combining Eqs. (3) and (4) gives: Hi = 1 xs Ms ki If xi is small, then x s is approximately equal to 1 and: Hi = b) For water as solvent: Ms 18.015 For CO2 in H2O: ki 0.034 gm mol mol 1 Ms ki Eq. (5) kg bar Hi By Eq. (5): Hi 1 Ms ki 1633 bar Ans. The value is Table 10.1 is 1670 bar. The values agree within about 2%. 10.36 Acetone: 14.3145 2756.22 T degC 228.060 Psat1 ( ) T e 14.8950 kPa 3413.10 T degC 250.523 Acetonitrile Psat2 ( ) T e kPa a) Find BUBL P and DEW P values T 50degC x1 0.5 y1 0.5 333 BUBLP x1 Psat1 ( T) 1 x1 Psat2 ( T) BUBLP 0.573 atm Ans. DEWP 1 y1 Psat1 ( T) 1 y1 DEWP 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 x1 0.5 y1 0.5 Guess: T 50degC Given x1 Psat1 ( T) 1 x1 Psat2 ( T) = P BUBLT Find ( T) x1 Psat1 ( T) = y1 P Find x1 T BUBLT 1 46.316 degC Ans. y1 P Given x1 DEWT x1 Psat2 ( T) = 1 51.238 degC Ans. DEWT 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 2726.81 T degC 217.572 Benzene: Psat1 ( T) e 13.9320 kPa 3056.96 T degC 217.625 Toluene: Psat2 ( T) e kPa a) Calculate the equilibrium composition of the liquid and vapor at the flash T and P T 90degC P 75kPa Guess: x1 0.5 y1 0.5 334 Given x1 Psat1 ( )= y1 P T Find x1 y1 1 x1 Psat2 ( )= 1 T y1 0.458 y1 P x1 y1 x1 0.252 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 0.1604 y1 0.2919 x2 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 x1 Psat2 ( )= 1 T Find y2 y3 y1 y3 P y1 y2 y3 = 1 y2 y3 y2 0.608 y3 0.1 Ans. Conclusion: An air leak is consistent with the measured compositions. 10.38 yO21 0.0387 yN21 0.7288 yCO21 0.0775 yH2O1 0.1550 ndot 10 kmol hr T1 16.3872 100degC 3885.70 T degC T2 25degC P 1atm 230.170 PsatH2O ( ) T e kPa 335 Calculate the mole fraction of water in the exit gas if the exit gas is saturated with water. yH2O2 PsatH2O T2 P yH2O2 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. Guess: ndotvap yO22 Given ndot 2 0.0387 ndotliq yN22 ndotvap ndot 2 0.7288 yCO22 0.0775 ndot = ndotliq Overall balance O2 balance N2 balance CO2 balance Summation equation ndot yO21 = ndotvap yO22 ndot yN21 = ndotvap yN22 ndot yCO21 = ndotvap yCO22 yO22 yN22 yCO22 yH2O2 = 1 ndotliq ndotvap yO22 yN22 yCO22 ndotliq yO22 Find ndotliq ndotvap yO22 yN22 yCO22 1.276 0.044 kmol hr yN22 ndotvap 0.835 336 8.724 kmol hr 0.089 yH2O2 0.031 yCO22 Apply an energy balance around the cooler to calculate heat transfer rate. HlvH2O Qdot 40.66 kJ mol T1 T1 273.15K T2 3 T2 273.15K 5 5 5 ndotvap yO22 R ICPH T1 T2 3.639 0.506 10 0 3 3 3 0.227 10 ndotvap yN22 R ICPH T1 T2 3.280 0.539 10 ndotvap yCO22 R ICPH T1 T2 5.457 1.045 10 ndotvap yH2O2 R ICPH T1 T2 3.470 1.450 10 HlvH2O ndotliq Qdot 19.895 kW Ans. 0 0.040 10 0 1.157 10 5 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 xC4 xC5 P 18psia Ans. 0.05 0.85 0.10 KC3 KC4 KC5 3.9 0.925 0.23 The vapor mole fractions must sum to 1. xC3 KC3 xC4 KC4 xC5 KC5 1.004 337 10.40 H2S + 3/2 O2 -> H2O + SO2 By a stoichiometric balance, calculate the following total molar flow rates Feed: Products ndotH2S 10 kmol hr ndotO2 3 ndotH2S 2 ndotSO2 ndotH2S ndotH2O ndotH2S 16.3872 3885.70 T degC Exit conditions: 230.170 P 1atm T2 70degC PsatH2O ( ) T e kPa a) Calculate the mole fraction of H2O and SO2 in the exiting vapor stream assuming vapor is saturated with H2O yH2Ovap ySO2 1 PsatH2O T2 P yH2Ovap yH2Ovap 0.308 Ans. ySO2 0.692 Ans. b) Calculate the vapor stream molar flow rate using balance on SO 2 ndotvap ndotSO2 ySO2 ndotvap 14.461 kmol hr Ans. Calculate the liquid H2O flow rate using balance on H2O ndotH2Ovap ndotvap yH2Ovap ndotH2Ovap 4.461 kmol hr ndotH2Oliq ndotH2O ndotH2Ovap ndotH2Oliq 5.539 kmol Ans. hr 338 10.41 NCL 0.01 kg kg MH2O 18.01 gm mol Mair 29 gm mol a) YH2O NCL Mair MH2O YH2O 0.0161 yH2O YH2O 1 YH2O yH2O 0.0158 Ans. b) P 1atm ppH2O 16.3872 yH2O P ppH2O 1.606 kPa Ans. 3885.70 T degC 230.170 c) PsatH2O ( ) T e kPa Guess: T Find ( ) T 20degC Given Tdp 14.004 degC yH2O P = PsatH2O ( ) Tdp T Tdp Tdp 32degF Tdp 57.207 degF Ans. 10.42 ndot1 50 kmol hr 16.3872 Tdp1 20degC Tdp2 10degC P 1atm 3885.70 T degC 230.170 MH2O 18.01 PsatH2O ( ) T e kPa gm mol y1 PsatH2O Tdp1 P y1 0.023 y2 PsatH2O Tdp2 P y2 0.012 By a mole balances on the process Guess: ndot2liq ndot1 ndot2vap ndot1 339 Given ndot1 y1 = ndot2vap y2 ndot2liq H2O balance ndot1 = ndot2vap ndot2liq Overall balance ndot2liq ndot2vap ndot2vap Find ndot2liq ndot2vap 49.441 kmol hr ndot2liq 0.559 kmol hr mdot2liq ndot2liq MH2O mdot2liq 10.074 kg hr Ans. 10.43Benzene: Cyclohexane: A1 A2 13.7819 13.6568 B1 B2 kPa 2726.81 2723.44 C1 C2 217.572 220.618 Psat1 ( T) exp A1 B1 T degC C1 Psat2 ( T) exp A2 B2 T degC kPa C2 Guess: T 66degC Given Psat1 ( T) = Psat2 ( T) T Find ( T) The Bancroft point for this system is: Psat1 ( T) 39.591 kPa T 52.321 degC Ans. Com ponent1 Benzene 2- t Bu anol Acet ti oni le r Com ponent2 Cyclohexane W at er Et hanol T (C) 52.3 8 .7 7 65.8 P ( Pa) k 39.6 64 .2 60 .6 340 Chapter 11 - Section A - Mathcad Solutions 11.1 For an ideal gas mole fraction = volume fraction CO2 (1): x1 0.7 V1 0.7m 3 N2 (2): x2 0.3 V2 0.3m 3 i 1 2 P 1bar T ( 25 273.15) K P n i Vi RT n 40.342 mol S nR i xi ln xi S 204.885 J K Ans. 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 nAr 4 mol 2.5 mol TN2 TAr [75 ( ( 130 273.15)K] 273.15)K PN2 PAr 30 bar 20 bar TN2 348.15 K TAr 403.15 K i 1 2 ntotal nN2 nAr x1 nN2 ntotal x2 nAr ntotal x1 0.615 x2 0.385 CvAr 3 R 2 CvN2 5 R 2 CpAr CvAr R CpN2 CvN2 R Find T after mixing by energy balance: T TN2 2 TAr (guess) Given nN2 CvN2 T TN2 = nAr CvAr TAr 341 T T Find T) ( T 273.15 K 90 degC Find P after mixing: P PN2 2 PAr (guess) Given nN2 nAr R T P = nN2 R TN2 PN2 nAr R TAr 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 nN2 CpN2 ln T TN2 T TAr R ln P PN2 P PAr SN2 11.806 J K J K SAr nAr CpAr ln R ln SAr 9.547 Smix ntotal R i xi ln xi J K Smix 36.006 J K S SN2 SAr Smix S 38.27 Ans. 11.3 mdotN2 molwtN2 2 kg sec gm mol mdotH2 molwtH2 0.5 kg sec gm mol i 1 2 28.014 2.016 molarflowN2 molarflowtotal mdotN2 molwtN2 molarflowN2 molarflowH2 mdotH2 molwtH2 319.409 mol sec molarflowH2 molarflowtotal 342 y1 molarflowN2 molarflowtotal y1 0.224 y2 molarflowH2 molarflowtotal y2 0.776 S R molarflowtotal i yi ln yi S 1411 J secK Ans. 11.4 T1 448.15 K T2 308.15 K P1 3 3 bar P2 6 1 bar For methane: MCPHm MCPH T1 T2 1.702 9.081 10 2.164 10 0.0 MCPSm For ethane: MCPS T1 T2 1.702 9.081 10 3 2.164 10 6 0.0 MCPHe MCPH T1 T2 1.131 19.225 10 3 5.561 10 6 0.0 MCPSe MCPS T1 T2 1.131 19.225 10 3 5.561 10 6 0.0 MCPHmix 0.5 MCPHm 0.5 MCPHe MCPHmix MCPSmix 6.21 6.161 MCPSmix H 0.5 MCPSm 0.5 MCPSe T1 R ln P2 P1 R MCPHmix T2 H 7228 J mol S R MCPSmix ln T2 T1 R 2 0.5 ln ( ) 0.5 The last term is the entropy change of UNmixing J T 300 K S 15.813 mol K Wideal H T S Wideal 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 S R y1 ln y1 y2 ln y2 S 4.273 J mol K 3 J By Eq. (5.27): Wideal T S Wideal 1.282 10 mol By Eq. (5.28): Work Wideal t Work 25638 J mol Ans. 11.16 0 10 20 40 60 P 80 100 200 300 400 500 bar 1.000 0.985 0.970 0.942 0.913 Z 0.885 0.869 0.765 0.762 0.824 0.910 Fi Zi Pi 1 ln 1 0 1 1 end rows ( P) i 2 end Fi is a well behaved function; use the trapezoidal rule to integrate Eq. (11.35) numerically. Ai i Fi Fi 1 Pi 2 Pi 1 ln i ln i 1 i Pi Ai exp ln i fi Generalized correlation for fugacity coefficient: For CO2: Tc 304.2 K Pc 73.83 bar 0.224 T P G ( P) ( 150 273.15) K Tr T Tc Tr 1.391 exp Pc Tr B0 Tr B1 Tr 344 fG ( P) G ( P) P Pi bar Calculate values: 10 20 40 60 80 100 200 300 400 500 fi i 0.993 0.978 0.949 0.922 0.896 0.872 0.77 0.698 0.656 0.636 bar 9.925 19.555 37.973 55.332 71.676 87.167 153.964 209.299 262.377 317.96 400 fi bar 300 200 0.8 i G Pi f G Pi 0.6 bar 100 0 0.4 0 200 Pi 400 600 0 200 Pi bar 400 600 bar Agreement looks good up to about 200 bar (Pr=2.7 @ Tr=1.39) 11.17 For SO2: Tc T 430.8 K 600 K 1.393 Pc P Pr 78.84 bar 300 bar P Pc Pr 0.245 Tr T Tc Tr 3.805 For the given conditions, we see from Fig. 3.14 that the Lee/Kesler correlation is appropriate. 345 Data from Tables E.15 & E.16 and by Eq. (11.67): 0 0.672 1 1.354 0 1 0.724 f P GRRT ln f 217.14 bar GRRT 0.323 Ans. 11.18 Isobutylene: Tc 417.9 K Pc 40.00 bar 0.194 a) At 280 degC and 20 bar: T ( 280 273.15)K P 20 bar Tr ( ) T T Tc Tr ( ) 1.3236 T Pr ( ) P P Pc Pr ( ) 0.5 P At these conditions use the generalized virial-coeffieicnt correlation. f PHIB Tr ( )Pr ( ) T P P T ( 280 f 18.76 bar 273.15)K Ans. P 100 bar b) At 280 degC and 100 bar: Tr ( ) 1.3236 T Pr ( ) 2.5 P At these conditions use the Lee/Kesler correlation, Tables E.15 & E.16 and Eq. (11.67). 0 0.7025 1 1.2335 0 1 f P 0.732 f 73.169 bar Ans. 11.19 The following vectors contain data for Parts (a) and (b): (a) = Cyclopentane; (b) = 1-butene Tc 511.8 420.0 0.273 0.277 383.15 393.15 K Pc 45.02 40.43 258 239.3 275 34 346 bar cm mol 3 0.196 0.191 Zc Vc Tn 322.4 266.9 5.267 25.83 K T K P bar Psat bar Tr T Tc Tr 0.7486 0.9361 Psatr Psat Pc Psatr 0.117 0.6389 Calculate the fugacity coefficient at the vapor pressure by Eq. (11.68): (a) PHIB Tr Psatr 1 1 1 0.900 (b) PHIB Tr Psatr 2 2 2 0.76 Eq. (3.72), the Rackett equation: Tr T Tc Tr 0.749 0.936 Eq. (11.44): 2 Vsat V c Zc 1 Tr 7 Vsat 107.546 cm3 133.299 mol f PHIB Tr Psatr 11.78 20.29 Psat exp Vsat ( Psat) P RT f bar Ans. 11.21 Table F.1, 150 degC: Psat 476.00 kPa molwt 18 gm mol Vsat 1.091 cm molwt gm 3 3 T ( 150 273.15)K P 150 bar Vsat cm 19.638 mol T 423.15 K Equation Eq. (11.44) with satPsat = fsat r exp Vsat P RT Psat r 1.084 r= f fsat = 1.084 Ans. 347 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 3121.2 273.15) K 459.67) rankine J gm 6.2915 S1 J gm K H1 Btu 1389.6 lbm 1.5677 Btu lbm rankine T1 Table F.2: (a) 300 kPa & 400 degC; (b) 50(psia) & 800 degF: T2 3275.2 H2 J gm S2 8.0338 J gm K Btu 1431.7 lbm 1.9227 Btu lbm rankine Eq. (A) on page 399 may be recast for this problem as: r exp molwt R H2 T1 H1 S2 S1 f2 f1 r 0.0377 0.0542 Ans. (a) r= f2 f1 = 0.0377 (b) r= = 0.0542 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 420.0 Pc 33.70 40.0 40.43 313.0 Vc 238.9 239.3 348 0.252 bar 0.194 0.191 0.270 Zc 0.275 0.277 cm 3 309.2 Tn 266.3 266.9 K mol 200 P 300 bar 150 0.6583 Tr 0.6372 0.6355 Psat 1.01325 1.01325 bar 1.01325 0.0301 Pr Psat Pc Pr 0.0253 0.0251 Tr Tn Tc Calculate the fugacity coefficient at the nbp by Eq. (11.68): (a) (b) PHIB Tr Pr 1 1 PHIB Tr Pr 2 1 0.9572 2 2 0.9618 (c) PHIB Tr Pr 3 3 3 0.9620 1 Tr 0.2857 Eq. (3.72): Vsat V c Zc Eq. (11.44): f PHIB Tr Pr 2.445 f 3.326 bar 1.801 Psat exp Vsat ( Psat) P R Tn Ans. 11.24 (a) Chloroform: Tc 536.4 K Pc 3 54.72 bar 0.222 Zc 0.293 Vc cm 239.0 mol Tn 334.3 K Psat 22.27 bar T 473.15 K Tr T Tc Tr 0.882 2 Trn Tn Tc Trn 0.623 3 Eq. (3.72): Vsat V c Zc 1 Trn 7 Vsat cm 94.41 mol 349 Calculate fugacity coefficients by Eqs. (11.68): P Pc if P Psat Pr ( P ) Tr Pr ( P ) ( P) exp B0 Tr B1 Tr f ( P) ( P) P ( Psat) Psat exp Vsat ( P Psat) RT ( P) P if P Psat ( P) ( Psat) Psat P exp Vsat ( P RT Psat) 0 bar 0.5 bar 40 bar 40 Psat bar ( P) Psat bar f ( P) bar 30 0.8 P bar 20 0.6 10 0 0 20 40 0.4 0 20 P bar 40 P P bar bar (b) Isobutane Zc T Tc Vc Tr 408.1 K 262.7 T Tc Vsat Pc 3 36.48 bar 261.4 K Tn Tc Vsat 0.181 Psat 5.28 bar 0.282 cm mol Tr Tn 0.767 2 313.15 K Trn Trn 0.641 3 Eq. (3.72): V c Zc 1 Trn 7 cm 102.107 mol 350 Calculate fugacity coefficients by Eq. (11.68): Pr ( ) P P Pc if P Psat ( )P P () P exp Pr ( ) P Tr B0 Tr B1 Tr fP) ( ( )Psat exp Psat Vsat ( Psat) P RT () P P if P Psat () ( ) P Psat Psat P exp Vsat ( P RT Psat) 0 bar 0.5 bar 10 bar 10 Psat bar 0.8 Psat bar fP) ( bar P bar 5 () P 0.6 0 0 5 10 0.4 0 5 P bar 10 P P bar bar 11.25 Ethylene = species 1; Propylene = species 2 Tc 282.3 365.6 K Pc 50.40 46.65 bar cm mol 3 w 0.087 0.140 Zc 0.281 0.289 Vc P i 131.0 188.4 y1 T n 423.15 K 2 30 bar 1 n 351 0.35 y2 k 1 1 n y1 j 1 n By Eqs. (11.70) through (11.74) wi i j wj 2 1 3 Tc i j Tci Tc j Zc Zci 2 Zc j i j Vc Vci Vc j 1 3 3 Zc Pc i j i j i j 2 T R Tc i j Vc i j Tr i j Tc Tr 1.499 1.317 1.317 1.157 50.345 48.189 48.189 46.627 K Zc bar i j Vc 157.966 cm3 157.966 188.4 mol 131 0.087 0.114 0.114 0.14 Tc 282.3 321.261 Pc 321.261 365.6 0.281 0.285 0.285 0.289 By Eqs. (3.65) and (3.66): B0i j B0 B0 Tr 0.138 0.189 i j B1i j B1 Tr i j 0.189 0.251 i j B1i j B1 0.108 0.085 0.085 0.046 59.892 99.181 99.181 cm3 159.43 mol Bi j R Tc i j B0i j Pc i j B By Eq. (11.64): i j 2 Bi j Bi i P RT Bj j 20.96 cm3 mol 20.96 0 0 hatk exp Bk k 1 2 i j yi y j 2 i k i j fhatk hatk yk P hat 0.957 0.875 352 fhat 10.053 17.059 bar Ans. For an ideal solution , id = pure species Pr P k Pck Pr 0.595 0.643 Pr idk exp Tr k B0k k k k B1k k k k fhatid k idk yk P id 0.95 0.873 fhatid 9.978 17.022 bar Ans. Alternatively, Pr P i j Pr idk exp k k k k Pc i j Tr B0k k k k B1k k id 0.95 0.873 11.27 Methane = species 1 Ethane = species 2 Propane = species 3 0.21 y 0.43 0.36 T 373.15 K 0.012 P 35 bar 0.286 w 0.100 0.152 45.99 Zc 0.279 0.276 98.6 cm mol 3 190.6 Tc 305.3 K 369.8 Pc 48.72 bar 42.48 Vc 145.5 200.0 n 3 i 1 n j 1 n k 1 n By Eqs. (11.70) through (11.74) wi i j wj 2 1 3 Tc i j Tci Tc j Zc Zci 2 Zc j i j Vc Vci Vc j 1 3 3 Zc Pc i j i j i j 2 R Tc i j Vc i j 353 1.958 1.547 1.406 Tr T i j Tc Tr 1.547 1.222 1.111 1.406 1.111 1.009 i j 98.6 Vc 120.533 120.533 143.378 145.5 171.308 200 cm 3 mol 143.378 171.308 45.964 47.005 43.259 Pc 47.005 48.672 45.253 bar 43.259 45.253 42.428 190.6 Tc 241.226 241.226 265.488 305.3 336.006 K 369.8 0.012 0.056 0.082 0.056 0.1 0.126 0.082 0.126 0.152 0.286 0.282 0.281 Zc 0.282 0.279 0.278 0.281 0.278 0.276 265.488 336.006 By Eqs. (3.65) and (3.66): B0i j B0 Tr B1i j i j B1 Tr i j Bi j R Tc i j B0i j Pc i j i j B1i j By Eq. (11.64): i j 0 30.442 107.809 0 23.482 0 2 Bi j Bi i Bj j 30.442 cm 3 mol 107.809 23.482 hatk exp P RT Bk k 1 2 i j yi y j 2 i k i j fhatk hatk yk P hat 1.019 0.881 0.775 354 7.491 fhat 13.254 bar 9.764 Ans. For an ideal solution , id = pure species 0.761 Prk P Pck Pr 0.718 0.824 0.977 fhatid k idk exp Prk Tr k k B0k k k k B1k k 7.182 fhatid 13.251 bar 9.569 Ans. idk yk P id 0.88 0.759 11.28 Given: (a) GE RT = 2.6 x1 1.8 x2 x1 x2 Substitute x2 = 1 - x1: .8 x1 1.8 x1 1 x1 = 1.8 x1 x1 2 GE = RT 0.8 x1 3 Apply Eqs. (11.15) & (11.16) for M = GE/RT: ln 1 = GE RT d 1 x1 GE RT dx1 ln 2 = GE RT d x1 GE RT dx1 d GE RT dx1 = 1.8 2 x1 2.4 x1 2 2 ln 1 = 1.8 ln 2 = 2 x1 2 x1 1.4 x1 1.6 x1 3 3 1.6 x1 Ans. (b) Apply Eq. (11.100): GE = x1 RT 1.8 1 x1 2 x1 x1 2 1.4 x1 2 3 1.6 x1 3 1.6 x1 This reduces to the initial condition: 355 (c) Divide Gibbs/Duhem eqn. (11.100) by dx1: d ln 1 dx1 x1 x2 d ln 2 dx1 = 0 Differentiate answers to Part (a): d ln 1 = 2 dx1 x1 2 2.8 x1 4.8 x1 2 d ln 2 = 2 x1 dx1 3 4.8 x1 2 d ln 1 = 2 x1 dx1 d ln 1 = 1 dx1 2.8 x1 4.8 x1 x2 x1 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 When x1 = 0, we see from the 3rd eq. of Part (c) that (e) DEFINE: g = GE/RT g x1 ln 1 x1 ln 2 x1 ln 1 () 0 d ln 1 dx1 d ln 2 dx1 = 0 Q.E.D. = 0 Q.E.D. 1.8 x1 1.8 x1 1.8 2 x1 2 0.8 x1 3 2 2 x1 1.6 x1 1.4 x1 3 1.6 x1 3 ln 2 () 1 2.6 x1 0 0.1 1.0 356 0 g x1 ln 1 x1 ln 2 x1 1 0 ln 1 ( ) 2 ln 2 ( ) 1 3 0 H H1bar H2bar 0.2 0.4 0.6 0.8 x1 0.02715 11.32 87.5 265.6 417.4 534.5 531.7 421.1 347.1 VE 321.7 276.4 252.9 190.7 178.1 138.4 98.4 37.6 10.0 357 0.09329 0.17490 0.32760 0.40244 0.56689 0.63128 x1 0.66233 0.69984 0.72792 0.77514 0.79243 0.82954 0.86835 0.93287 0.98233 253 n rows x1 i 1 n x1 0 0.01 1 (a) Guess: x1 1 F x1 x1 2 a x1 x1 x1 a b c 3000 b 3000 c a 250 3.448 10 3 3 1 linfitx1 VE F b c Ans. 3.202 10 244.615 3 x1 1 600 VEi x1 ( 1 x1) a b x1 c ( x1) 2 400 200 0 0 0.2 0.4 i 0.6 0.8 x1 x1 By definition of the excess properties V = x1 x2 a E b x1 c x1 3 (c 2 d 3 E V = 4 c x1 dx1 Vbar1 E b) x1 2 2 (b 2 a) x1 a = x2 = x1 2 a 2 b x1 3 c x1 Vbar2 E 2 a b 2 (b c) x1 3 c x1 2 (b) To find the maximum, set dVE/dx1 = 0 and solve for x1. Then use x1 to find VEmax. Guess: Given 4 c ( x1) x1 3 x1 0.5 3 (c b) ( x1) x1 2 2 (b a) x1 a= 0 Find ( x1) 0.353 Ans. 358 VEmax x1 ( 1 x1) a b x1 2 c x1 2 VEmax 3c( ) x1 2 536.294 Ans. (c) VEbar1 ( ) x1 ( 1 x1) a 2 2 b x1 VEbar2 ( ) x1 x1 0 0.01 1 4000 ( ) a x1 b 2( b c)x1 3c( ) x1 2 2000 VEbar 1 ( ) x1 x1 VEbar 2 ( ) 0 2000 0 0.2 0.4 x1 x1 0.6 0.8 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 466 809 cm mol 3 P n 2 bar 2 y1 i 0.5 1 n y2 j 1 1 n y1 276 466 By Eq. (11.61): B i 359 yi y j Bi j j B 504.25 cm 3 mol Use a spline fit of B as a function of T to find derivatives: 331 b11 276 235 cm 3 980 b22 809 684 cm 3 558 b12 466 399 cm 3 mol mol mol 50 t 75 100 vs11 lspline ( t b11) B11 ( T) 273.15 K 323.15 t 348.15 K 373.15 3 interp ( vs11 t b11 T) B11 ( T) cm 276 mol vs22 lspline ( t b22) B22 ( T) interp ( vs22 t b22 T) B22 ( T) cm 809 mol 3 vs12 lspline ( t b12) B12 ( T) interp ( vs12 t b12 T) B12 ( T) cm 466 mol 3 dBdT d d B12 ( T) B11 ( T) dT dT d dT B12 ( T) dBdT 1.92 3.18 d B22 ( T) dT cm 3.18 5.92 mol K 3 3 Differentiate Eq. (11.61): dBdT i j yi y j dBdTi dBdT j cm 3.55 mol K By Eq. (3.38): Z 1 BP RT Z 0.965 V ZR T P HRRT R T By Eq. (6.55): HRRT By Eq. (6.56): P R B T dBdT HRRT SRR 0.12 HR SRR P dBdT R 0.085 SR SRR R V 13968 cm 3 mol HR 348.037 360 J mol SR 0.71 J mol K Ans. 11.34 Propane = 1; n-Pentane = 2 T B ( 75 273.15)K 466 809 cm 3 P 2 bar n y1 2 0.5 y2 1 y1 276 466 i 1 n mol j 1 n ij 2 Bi j Bii B j j By Eqs. (11.63a) and (11.63b): hat1 ( ) y1 exp P B1 1 RT ( 1 2 y1) 1 2 2 hat2 ( ) y1 y1 exp P B2 2 RT y1 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 y1 361 0.6 0.8 0.0426 11.36 23.3 45.7 66.5 86.6 118.2 144.6 176.6 HE 195.7 204.2 191.7 174.1 141.0 116.8 85.6 43.5 22.6 500 a b c 0 0.0817 0.1177 0.1510 0.2107 0.2624 0.3472 x1 0.4158 0.5163 0.6156 0.6810 0.7621 0.8181 0.8650 0.9276 0.9624 n x1 rows x1 i 1 n 0 0.01 1 (a) Guess: x1 1 F x1 x1 2 3 a x1 x1 x1 b 100 c 0.01 a 539.653 1.011 10 913.122 3 1 1 linfit x1 HE F b c Ans. x1 HEi x1 ( 1 x1) a b x1 c ( x1) 2 100 200 300 0 0.2 0.4 i 0.6 0.8 x1 x1 362 By definition of the excess properties H = x1 x2 a d dx1 E E b x1 3 c x1 3( c 2 H = 4 c x1 E b) x1 2 2( b 2 a)x1 a Hbar1 = x2 = x1 2 a 2 b x1 3 c x1 Hbar2 E 2 a b 2( b 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: Given x1 2 x1 0.5 3 HE ( ) x1 3( c 0.512 b x1 x1 ( 1 2 x1) a a)x1 b x1 a= 0 c x1 4c( ) x1 Find ( ) x1 x1 ( 1 b)( ) x1 Ans. c x1 2 2( b x1 x1) a HEmin (c) HEmin 204.401 Ans. HEbar1 ( ) x1 HE ( ) ( x1 1 d x1) HE ( ) x1 dx1 HEbar2 ( ) x1 x1 0 0.01 1 500 0 500 HE ( ) x1 x1 d HE ( ) x1 dx1 HEbar 1 ( ) x1 x1 HEbar 2 ( ) 1000 0 0.2 0.4 x1 363 0.6 0.8 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) (1) = Acetone (2) = 1,3-butadiene y1 0.28 y2 Tc 1 y1 508.2 425.2 T K Zc ( 60 273.15) K Vc P 209 170 kPa cm mol 3 w n 2 0.307 0.190 0.233 0.267 220.4 i 1 n wi wj 2 j 1 n ki j 0 0.307 0.2485 0.082 Eq. (11.70) i j 0.2485 0.082 508.2 0.19 0.126 0.126 0.152 464.851 425.2 0 K Eq. (11.71) Tci j Tc Tc i j 1 ki j Tc 464.851 369.8 Zc Eq. (11.73) Zci j i Zc 2 1 0.233 0.25 j Zc 1 3 0.25 0.267 0.276 209 0 214.65 3 Vc Eq. (11.74) Vci j 3 i Vc 3 j 2 Vc 214.65 220.4 200 0 cm mol 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 Tri j T Tci j 0.656 0.717 0.717 0.784 B0i j B0 Tri j Pri j P Pci j 0.036 0.038 Pr 0.038 0.04 0.824 0 Tr Eq. (3.65) 0.74636 B0 0.6361 0.16178 Eq. (3.66) B1i j B1 Tri j 0.6361 0.5405 0.27382 0.16178 0.27382 0.33295 0.874 B1 0.558 0.098 0.558 0.098 0.34 0.028 0.028 0.027 Eq. (11.69a) + (11.69b) Bi j R Tci j B0i j Pci j i j B1i j B n 910.278 665.188 n 665.188 cm3 499.527 mol cm 3 Eq. (11.61) B i 1 j 1 yi y j Bi j B 598.524 mol Eq. (3.38) Z V 1 BP RT Z V 0.675 Tri j 2.6 0.963 1.5694 10 3 4 cm RT Z P mol Ans. 0.722 Tri j 5.2 Eq. (6.89) dB0dTri j Eq. (6.90) dB1dTri j 365 Differentiating Eq. (11.61) and using Eq. (11.69a) + (11.69b) n n dBdT i 1 j 1 yi y j R dB0dTri j Pci j i j dB1dTri j Eq. (6.55) HR PT B T dBdT HR 344.051 Eq. (6.56) SR P dBdT SR 0.727 J mol J Ans. Eq. (6.54) GR BP 3 GR 101.7 mol K J Ans. mol Ans. (b) cm V = 15694 mol SR = 1.006 HR = 450.322 GR = 125.1 J mol J mol K 3 J mol (c) V = 24255 cm HR = 175.666 mol J mol J SR = 0.41 mol K GR = 53.3 J mol (d) V = 80972 cm 3 mol HR = 36.48 J mol SR = 0.097 J mol K 3 GR = 8.1 J mol (e) cm V = 56991 mol HR = 277.96 J mol SR = 0.647 J mol K GR = 85.2 J mol 366 Data for Problems 11.38 - 11.40 325 200 575 T 350 300 525 225 200 P 15 100 40 35 50 10 25 75 1.054 1.325 1.023 Tc 308.3 150.9 562.2 304.2 282.3 507.6 190.6 126.2 Pc 61.39 48.98 48.98 73.83 50.40 30.25 45.99 34.00 .187 .000 .210 .224 .087 .301 .012 .038 0.244 2.042 0.817 Pr P Pc Tr T Tc Tr 1.151 1.063 1.034 1.18 1.585 Pr 0.474 0.992 0.331 0.544 2.206 11.38 Redlich/Kwong Equation: 0.02 0.133 0.069 Pr Tr Eq. (3.53) 0.08664 0.42748 4.559 3.234 4.77 0.036 0.081 0.028 0.04 0.121 q Tr 1.5 Eq. (3.54) q 3.998 4.504 4.691 3.847 2.473 Guess: z 1 367 Given z= 1 q z z z Eq. (3.52) Z q Find z () i 1 8 Ii i qi ln Z Z i qi i qi i qi i Eq. (6.65) i exp Z i Pi i qi 0.925 0.722 0.668 0.887 0.639 0.891 0.881 0.859 1 ln Z i qi Ii Eq. (11.37) fi Z i 0.93 0.744 0.749 0.896 0.73 0.9 0.89 0.85 fi 13.944 74.352 29.952 31.362 36.504 8.998 22.254 63.743 11.39 Soave/Redlich/Kwong Equation 2 0.08664 0.5 0.42748 2 c 0.480 1.574 0.176 0.02 0.133 0.069 1 c 1 Tr 4.49 3.202 4.737 q Eq. (3.54) q Pr Tr Eq. (3.53) 0.036 0.081 0.028 0.04 0.121 3.79 4.468 4.62 3.827 2.304 Tr Guess: z 1 368 Given z= 1 q z z z Eq. (3.52) Z q Find z () i 1 8 Ii ln Z Z ln Z i qi i qi i qi i Eq. (6.65) i exp Z i Pi i qi 0.927 0.729 0.673 0.896 0.646 0.893 0.882 0.881 i qi 1 i qi Ii Eq. (11.37) fi Z i 0.931 0.748 0.751 0.903 0.733 0.902 0.891 0.869 fi 13.965 74.753 30.05 31.618 36.66 9.018 22.274 65.155 11.40 Peng/Robinson Equation 1 2 1 2 2 0.07779 0.45724 2 c 0.37464 1.54226 0.26992 0.018 0.12 0.062 1 c 1 Tr 0.5 5.383 3.946 5.658 Pr Tr Eq.(3.53) 0.032 0.073 0.025 0.036 0.108 q Tr Eq.(3.54) q 4.598 5.359 5.527 4.646 2.924 369 Guess: z 1 q z z z Eq. (3.52) Z q Find ( z) Given z = 1 i 1 8 Ii 1 2 2 ln Z Z i qi i qi i i Eq. (6.65) i exp Z i Pi i qi 1 i qi 0.918 0.69 0.647 0.882 0.617 0.881 0.865 0.845 ln Z i qi i i qi Ii Eq. (11.37) fi fi Z 0.923 0.711 0.73 0.89 0.709 0.891 0.876 0.832 13.842 71.113 29.197 31.142 35.465 8.91 21.895 62.363 BY GENERALIZED CORRELATIONS Parts (a), (d), (f), and (g) --- Virial equation: 325 T 350 525 225 T Tc Tc 308.3 304.2 507.6 190.6 P 15 35 10 25 Pc 61.39 73.83 30.25 45.99 .187 .224 .301 .012 Tr Pr P Pc Evaluation of : B0 B0 ( Tr) Eq. (3.65) B1 B1 ( Tr) Eq. (3.66) 370 DB0 0.675 Tr 2.6 Eq. (6.89) DB1 0.722 Tr 5.2 Eq. (6.90) 0.932 Pr B0 exp Tr B1 Eq. (11.60) 0.904 0.903 0.895 (a) (d) (f) (g) Parts (b), (c), (e), and (h) --- Lee/Kesler correlation: Interpolate in Tables E.13 - E.16: .7454 0 .7517 .7316 .8554 1 1.1842 0.9634 0.9883 1.2071 0.000 0.210 0.087 0.038 0.745 0 1 Eq. (11.67): 0.746 0.731 0.862 (b) (c) (e) (h) 11.43 ndot1 x1 2 kmol hr ndot2 x1 4 kmol hr x2 ndot3 1 x1 ndot1 x2 ndot2 0.667 ndot1 ndot3 0.333 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 St 8.82 W K Ans. 371 11.44 For air entering the process: xO21 0.21 xN21 0.79 For the enhanced air leaving the process: xO22 0.5 xN22 0.5 ndot2 50 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 O 2 xN21 ndotair = xN22 ndot2 ndotair ndotO2 Find ndotair ndotO2 Mole balance on N2 ndotair 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 R xO21 ln xO21 xN21 ln xN21 Entropy change of mixing S23 R xO22 ln xO22 xN22 ln xN22 Total rate of entropy generation: SdotG ndotair S12 ndot2 S23 SdotG 152.919 W K Ans. 372 10 11.50 T 544.0 273.15K 932.1 J mol 30 K 50 GE 513.0 494.2 HE 893.4 845.9 J mol Assume Cp is constant. Then HE is of the form: Find a and c using the given HE and T values. a c HE = c aT slope ( HE) T intercept ( HE) T a c 2.155 1.544 J mol K 10 3 J mol T bT c GE is of the form: GE = a T ln T K Rearrange to find b using estimated a and c values along with GE and T data. GE B a T ln T T K T c 13.543 B 13.559 13.545 J mol K Use averaged b value 3 Bi b i 1 3 b 13.549 J mol K Now calculate HE, GE and T*SE at 25 C using a, b and c values. HE ( ) T aT c J Ans. mol HE [25 ( 273.15) ] 901.242 K GE ( ) T a T ln T K T bT ( c GE [25 273.15) ] 522.394 K J Ans. mol J Ans. mol TSE ( ) T HE ( ) GE ( ) T T TSE [25 ( 273.15) ] 378.848 K 373 Chapter 12 - Section A - Mathcad Solutions 12.1 Methanol(1)/Water(2)-- VLE data: 39.223 42.984 48.852 52.784 P 56.652 60.614 63.998 67.924 70.229 72.832 kPa T 333.15 K 0.1686 0.2167 0.3039 0.3681 x1 0.4461 0.5282 0.6044 0.6804 0.7255 0.7776 0.5714 0.6268 0.6943 0.7345 y1 0.7742 0.8085 0.8383 0.8733 0.8922 0.9141 n 10 i 1 n Number of data points: Calculate x2 and y2: n x2 rows ( ) P 1 x1 y2 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 y2 P 2 x1 Psat1 x2 Psat2 GERT x1 ln 1 x2 ln 2 374 i 1 2 3 4 5 6 7 8 9 10 1 i 1.572 1.47 1.32 1.246 1.163 1.097 1.05 1.031 1.021 1.012 2 i 1.013 1.026 1.075 1.112 1.157 1.233 1.311 1.35 1.382 1.41 ln 1 i 0.452 0.385 0.278 0.22 0.151 0.093 0.049 0.031 0.021 0.012 ln 2 i 0.013 0.026 0.073 0.106 0.146 0.209 0.271 0.3 0.324 0.343 GERTi 0.087 0.104 0.135 0.148 0.148 0.148 0.136 0.117 0.104 0.086 0.5 0.4 ln 1 i ln 2 i GERT i 0.3 0.2 0.1 0 0 0.2 0.4 0.6 0.8 x1 i (a) Fit GE/RT data to Margules eqn. by linear least squares: VXi x1 i VYi GERTi x1 x2 i i intercept ( VX VY) 0.683 A12 Ans. Slope Slope A12 A12 slope ( VX VY) 0.208 Intercept 0.683 Intercept Intercept A21 A21 Slope 0.475 375 The following equations give CALCULATED values: 1 ( x2) x1 2 ( x2) x1 j 1 101 exp x2 exp x1 2 2 A12 A21 2 A21 2 A12 A12 x1 A21 x2 X1 j .01 j .01 X2 j 1 X1 j pcalc j X1 j 1 X1 X2 Psat1 j j j X2 j 2 X1 X2 Psat2 j j X1 Y1calc 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 60 50 j Pi kPa pcalc kPa 40 j pcalc kPa 30 20 10 0 0.2 0.4 i i j 0.6 j 0.8 x1 y1 X1 Y1calc P-x data P-y data P-x calculated P-y calculated 376 Pcalc i x1 1 x1 x2 Psat1 i i i x1 1 x1 x2 Psat1 i i i Pcalc i x2 2 x1 x2 Psat2 i i i y1calc i RMS deviation in P: Pi RMS i Pcalc n 2 i RMS 0.399 kPa (b) Fit GE/RT data to van Laar eqn. by linear least squares: VXi Slope Slope a12 a12 x1 i VYi x1 x2 i i GERTi intercept ( VX VY) 1.418 slope ( VX VY) 0.641 1 Intercept 0.705 Intercept Intercept a21 a21 a12 x1 a21 x2 2 1 ( Slope Intercept) 0.485 2 Ans. 1 ( x1 x2) exp a12 1 2 ( x1 x2) exp a21 1 X1 X2 a21 x2 a12 x1 j 1 101 j j .01 j 1 X1 .00999 j (To avoid singularities) 377 pcalc j X1 j 1 X1 X2 Psat1 j j X2 j 2 X1 X2 Psat2 j j Pcalc i x1 1 x1 x2 Psat1 i i i x2 2 x1 x2 Psat2 i i i X1 Y1calc j j 1 X1 X2 Psat1 j j pcalc j y1calc i x1 1 x1 x2 Psat1 i i i Pcalc i P-x,y Diagram: van Laar eqn. fit to GE/RT data. 90 80 Pi kPa 70 60 50 j Pi kPa pcalc kPa 40 j pcalc kPa 30 20 10 0 0.2 i 0.4 i j 0.6 j 0.8 1 x1 y1 X1 Y1calc P-x data P-y data P-x calculated P-y calculated RMS deviation in P: Pi RMS i Pcalc n 2 i RMS 0.454 kPa 378 (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 12 0.5 GERTi 21 1.0 x2 2 12 21 i x1 ln x1 i i i i 12 21 x2 ln x2 i i x1 12 21 Minimize SSE 12 12 21 0.476 1.026 Ans. 21 12 exp x2 1 ( x2) x1 21 12 x1 x2 x1 x2 12 x1 21 x2 12 exp 2 ( x2) x1 x1 21 12 x1 x2 x2 x2 21 x1 21 x1 j 1 101 X1 X1 j .01 j .01 X2 X2 j 1 X1 j pcalc Pcalc j j 1 X1 X2 Psat1 j j j 2 X1 X2 Psat2 j j i x1 1 x1 x2 Psat1 i i i X1 x2 2 x1 x2 Psat2 i i i x1 1 x1 x2 Psat1 i i i Pcalc i Y1calc j 1 X1 X2 Psat1 j j j pcalc j y1calc i 379 P-x,y diagram: Wilson eqn. fit to GE/RT data. 90 80 Pi kPa 70 60 50 j Pi kPa pcalc kPa 40 j pcalc kPa 30 20 10 0 0.2 i 0.4 i j 0.6 j 0.8 1 x1 y1 X1 Y1calc P-x data P-y data P-x calculated P-y calculated RMS deviation in P: Pi RMS i Pcalc n 2 i RMS 0.48 kPa (d) BARKER'S METHOD by non-linear least squares. Margules equation. Guesses for parameters: answers to Part (a). 1 x1 x2 A 12 A 21 2 x1 x2 A 12 A 21 exp ( ) A12 x2 exp ( ) A21 x1 2 2 2 A21 2 A12 A12 x1 A21 x2 380 Minimize the sum of the squared errors using the Mathcad Minimize function. Guesses: SSE A12 A21 i A12 0.5 Pi x1 A21 1.0 2 x1 x2 A12 A21 Psat1 i 1 i i i x2 2 x1i x2i A 12 A 21 Psat2 A12 A21 A12 Minimize SSE A12 A21 0.758 0.435 Ans. A21 pcalc j X1 j j 1 X 1 j X 2 j A 12 A 21 Psat1 2 X 1 j X 2 j A 12 A 21 Psat2 1 X 1 j X 2 j A 12 A 21 Psat1 X2 X1 Y1calc j j pcalc j Pcalc i x1 i i 1 x1i x2i A 12 A 21 Psat1 2 x1i x2i A 12 A 21 Psat2 1 x1i x2i A 12 A 21 Psat1 x2 x1 y1calc i i Pcalc i RMS deviation in P: Pi RMS i Pcalc n 2 i RMS 0.167 kPa 381 P-x-y diagram, Margules eqn. by Barker's method 90 Pi kPa 80 70 60 50 j Pi kPa pcalc kPa 40 j pcalc kPa 30 20 10 0 0.2 i 0.4 i j 0.6 j 0.8 1 x1 y1 X1 Y1calc P-x data P-y data P-x calculated P-y calculated Residuals in P and y1 1 Pi Pcalc kPa i 0.5 y1 y1calc 100 i i 0 0.5 0 0.2 0.4 0.6 0.8 x1 Pressure residuals y1 residuals i 382 (e) BARKER'S METHOD by non-linear least squares. van Laar equation. Guesses for parameters: answers to Part (b). j 1 101 X1 j .01 j .00999 a12 x1 a21 x2 2 X2 j 1 X1 j 1 x1 x2 a12 a21 exp a12 1 2 x1 x2 a12 a21 exp a21 1 a21 x2 a12 x1 2 Minimize the sum of the squared errors using the Mathcad Minimize function. Guesses: SSE a12 a21 i a12 0.5 Pi x1 a21 1.0 2 x1 x2 a12 a21 Psat1 i 1 i i i x2 2 x1i x2i a12 a21 Psat2 a12 a21 Minimize SSE a12 a21 a12 a21 0.83 0.468 Ans. pcalc j X1 j j 1 X 1 j X 2 j a12 a21 Psat1 2 X 1 j X 2 j a12 a21 Psat2 1 X 1 j X 2 j a12 a21 Psat1 X2 X1 Y1calc Pcalc j j pcalc x1 i i j i 1 x1i x2i a12 a21 Psat1 2 x1i x2i a12 a21 Psat2 1 x1i x2i a12 a21 Psat1 x2 x1 y1calc i i Pcalc i 383 RMS deviation in P: Pi RMS i Pcalc n 2 i RMS 0.286 kPa P-x,y diagram, van Laar Equation by Barker's Method 90 80 70 Pi kPa 60 Pi kPa pcalc kPa 50 j pcalc kPa j 40 30 20 10 0 0.2 i 0.4 i j 0.6 j 0.8 1 x1 y1 X1 Y1calc P-x data P-y data P-x calculated P-y calculated 384 Residuals in P and y1. 1 Pi Pcalc kPa i 0.5 y1 y1calc 100 i i 0 0.5 0 0.2 0.4 0.6 0.8 x1 Pressure residuals y1 residuals i (f) BARKER'S METHOD by non-linear least squares. Wilson equation. Guesses for parameters: answers to Part (c). j 1 101 1 x1 x2 12 21 X1 j .01 j exp .01 x2 x2 12 12 X2 j 1 X1 j ln x1 x2 x1 21 12 x2 x1 21 2 x1 x2 12 21 exp ln x2 x1 x1 x1 x2 21 12 12 21 x2 x1 21 Minimize the sum of the squared errors using the Mathcad Minimize function. Guesses: 12 0.5 21 385 1.0 SSE 12 21 i Pi x1 x1 x2 i 1 i i i 12 12 21 Psat1 21 Psat2 2 x2 2 x1i x2i 12 21 Minimize SSE 12 12 21 0.348 1.198 Ans. 21 pcalc j X1 j j 1 X1 j X2 j 2 X1 j X2 j 12 12 21 Psat1 21 Psat2 X2 X1 Y1calc j j 1 X1 j X2 j 12 j 21 Psat1 pcalc Pcalc i x1 i i 1 x1i x2i 2 x1i x2i 12 12 21 Psat1 21 Psat2 x2 x1 y1calc i i 1 x1i x2i 12 i 21 Psat1 Pcalc RMS deviation in P: 2 i Pi RMS i Pcalc n RMS 0.305kPa 386 P-x,y diagram, Wilson Equation by Barker's Method 90 Pi kPa 80 70 60 50 j Pi kPa pcalc kPa 40 j pcalc kPa 30 20 10 0 0.2 i 0.4 i j 0.6 j 0.8 1 x1 y1 X1 Y1calc P-x data P-y data P-x calculated P-y calculated Residuals in P and y1. 1 Pi Pcalc i kPa 0.5 y1 y1calc 100 i i 0 0.5 0 0.2 0.4 0.6 0.8 x1 Pressure residuals y1 residuals i 387 12.3 Acetone(1)/Methanol(2)-- VLE data: 72.278 75.279 77.524 78.951 82.528 86.762 90.088 93.206 95.017 P 96.365 97.646 98.462 99.811 99.950 100.278 100.467 100.999 101.059 99.877 99.799 kPa T 328.15 K 0.0287 0.0570 0.0858 0.1046 0.1452 0.2173 0.2787 0.3579 0.4050 x1 0.4480 0.5052 0.5432 0.6332 0.6605 0.6945 0.7327 0.7752 0.7922 0.9080 0.9448 0.0647 0.1295 0.1848 0.2190 0.2694 0.3633 0.4184 0.4779 0.5135 y1 0.5512 0.5844 0.6174 0.6772 0.6926 0.7124 0.7383 0.7729 0.7876 0.8959 0.9336 n y2 20 1 y1 i 1 n Number of data points: Calculate x2 and y2: n x2 rowsP) ( 1 x1 Vapor Pressures from equilibrium data: Psat1 96.885 kPa Psat2 68.728 kPa 388 Calculate EXPERIMENTAL values of activity coefficients and excess Gibbs energy. y1 P 1 y2 P 2 x1 Psat1 1 1 2 3 4 5 6 7 8 9 x2 Psat2 2 i 1.013 1.011 1.006 1.002 1.026 1.027 1.057 1.103 1.13 1.14 1.193 1.2 1.278 1.317 1.374 1.431 1.485 1.503 1.644 1.747 GERT x1 ln 1 x2 ln 2 i i 1.682 1.765 1.723 1.706 1.58 1.497 1.396 1.285 1.243 1.224 1.166 1.155 1.102 1.082 1.062 1.045 1.039 1.037 1.017 1.018 ln 1 i 0.52 0.568 0.544 0.534 0.458 0.404 0.334 0.25 0.218 0.202 0.153 0.144 0.097 0.079 0.06 0.044 0.039 0.036 0.017 0.018 ln 2 i 0.013 0.011 5.81510-3 1.97510-3 0.026 0.027 0.055 0.098 0.123 0.131 0.177 0.182 0.245 0.275 0.317 0.358 0.395 0.407 0.497 0.558 GERTi 0.027 0.043 0.052 0.058 0.089 0.108 0.133 0.152 0.161 0.163 0.165 0.162 0.151 0.145 0.139 0.128 0.119 0.113 0.061 0.048 10 11 12 13 14 15 16 17 18 19 20 389 0.6 ln 1 i ln 2 i GERTi 0.4 0.2 0 0 0.2 0.4 x1 i 0.6 0.8 (a) Fit GE/RT data to Margules eqn. by linear least squares: VXi Slope Slope A12 A12 x1 i VYi GERTi x1 x2 i i intercept ( VX VY) 0.708 A12 Ans. slope ( VX VY) 0.018 Intercept 0.708 Intercept Intercept A21 A21 Slope 0.69 The following equations give CALCULATED values: 1 ( x2) x1 exp x2 2 A12 2 A21 A12 x1 2 ( x2) x1 j 1 101 X1 exp x1 2 A21 X1 j 2 A12 .01 j .01 X2 A21 x2 X2 j 1 X1 j pcalc j j 1 X1 X2 Psat1 j j j j 2 X1 X2 Psat2 j j X1 Y1calc j 1 X1 X2 j j Psat1 pcalc j 390 P-x,y Diagram: Margules eqn. fit to GE/RT data. 105 100 Pi kPa 95 90 85 j Pi kPa pcalc kPa 80 j pcalc kPa 75 70 65 0 0.2 0.4 i i j 0.6 j 0.8 x1 y1 X1 Y1calc P-x data P-y data P-x calculated P-y calculated Pcalc i x1 1 x1 x2 Psat1 i i i x2 2 x1 x2 Psat2 i i i y1calc i x1 1 x1 x2 Psat1 i i i Pcalc i RMS deviation in P: Pi RMS i Pcalc n 2 i RMS 0.851 kPa 391 (b) Fit GE/RT data to van Laar eqn. by linear least squares: x1 x2 VXi Slope Slope a12 a12 x1 i VYi i i GERTi intercept ( VX VY) 1.442 1 ( Slope 0.686 2 slope ( VX VY) 0.015 1 Intercept 0.693 Intercept Intercept a21 a21 a12 x1 a21 x2 2 Intercept) Ans. 1 ( x2) x1 exp a12 1 2 ( x2) x1 exp a21 1 X1 X2 a21 x2 a12 x1 j 1 101 j j .01 j 1 X1 .00999 j (To avoid singularities) pcalc Pcalc j X1 j 1 X1 X2 j j Psat1 X2 j 2 X1 X2 j j Psat2 i x1 1 x1 x2 Psat1 i i i x2 2 x1 x2 Psat2 i i i X1 Y1calc j j 1 X1 X2 Psat1 j j pcalc j y1calc i x1 1 x1 x2 Psat1 i i i Pcalc i 392 P-x,y Diagram: van Laar eqn. fit to GE/RT data. 105 100 Pi kPa 95 90 85 j Pi kPa pcalc kPa 80 j pcalc kPa 75 70 65 0 0.2 i 0.4 i j 0.6 j 0.8 1 x1 y1 X1 Y1calc P-x data P-y data P-x calculated P-y calculated RMS deviation in P: Pi RMS i Pcalc n 2 i RMS 0.701 kPa (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: 12 0.5 21 1.0 SSE 12 21 i GERTi x1 ln x1 i i x2 ln x2 i i x2 2 i i 12 21 x1 393 12 21 Minimize SSE 12 12 21 0.71 0.681 Ans. 21 12 exp x2 1 ( x2) x1 21 12 x1 x2 x1 x2 12 x1 21 x2 12 exp 2 ( x2) x1 x1 21 12 x1 x2 x2 x2 21 x1 21 x1 j 1 101 X1 j .01 j .01 X2 j 1 X1 j pcalc j X1 j 1 X1 X2 Psat1 j j X2 j 2 X1 X2 Psat2 j j Pcalc i x1 1 x1 x2 Psat1 i i i x2 2 x1 x2 Psat2 i i i X1 Y1calc j 1 X1 X2 Psat1 j j j pcalc j y1calc i x1 1 x1 x2 Psat1 i i i Pcalc i 394 P-x,y diagram: Wilson eqn. fit to GE/RT data. 105 100 Pi kPa 95 90 85 j Pi kPa pcalc kPa 80 j pcalc kPa 75 70 65 0 0.2 i 0.4 i j 0.6 j 0.8 1 x1 y1 X1 Y1calc P-x data P-y data P-x calculated P-y calculated RMS deviation in P: Pi RMS i Pcalc n 2 i RMS 0.361 kPa (d) BARKER'S METHOD by non-linear least squares. Margules equation. Guesses for parameters: answers to Part (a). 1 x1 x2 A 12 A 21 2 x1 x2 A 12 A 21 exp ( x2) exp ( x1) 2 2 A12 A21 2 A21 2 A12 A12 x1 A21 x2 395 Minimize the sum of the squared errors using the Mathcad Minimize function. Guesses: SSE A12 A21 i A12 0.5 Pi x1 A21 1.0 2 x1 x2 A12 A21 Psat1 i 1 i i i x2 2 x1i x2i A 12 A 21 Psat2 A12 A21 A12 Minimize SSE A12 A21 0.644 0.672 Ans. A21 pcalc j X1 j j 1 X 1 j X 2 j A 12 A 21 Psat1 2 X 1 j X 2 j A 12 A 21 Psat2 1 X 1 j X 2 j A 12 A 21 Psat1 X2 X1 Y1calc j j pcalc j Pcalc i x1 i i 1 x1i x2i A 12 A 21 Psat1 2 x1i x2i A 12 A 21 Psat2 1 x1i x2i A 12 A 21 Psat1 x2 x1 y1calc i i Pcalc i RMS deviation in P: Pi RMS i Pcalc n 2 i RMS 0.365 kPa 396 P-x-y diagram, Margules eqn. by Barker's method 105 Pi kPa 100 95 90 85 j Pi kPa pcalc kPa 80 j pcalc kPa 75 70 65 0 0.2 i 0.4 i j 0.6 j 0.8 1 x1 y1 X1 Y1calc P-x data P-y data P-x calculated P-y calculated Residuals in P and y1 2 Pi Pcalc kPa i 1 y1 y1calc 100 i i 0 1 0 0.2 0.4 x1 i 0.6 0.8 Pressure residuals y1 residuals 397 (e) BARKER'S METHOD by non-linear least squares. van Laar equation. Guesses for parameters: answers to Part (b). j 1 101 X1 j .01 j .00999 a12 x1 a21 x2 2 X2 j 1 X1 j 1 x1 x2 a12 a21 exp a12 1 2 x1 x2 a12 a21 exp a21 1 a21 x2 a12 x1 2 Minimize the sum of the squared errors using the Mathcad Minimize function. Guesses: SSE a12 a21 i a12 0.5 Pi x1 a21 1.0 2 x1 x2 a12 a21 Psat1 i 1 i i i x2 2 x1i x2i a12 a21 Psat2 a12 a21 Minimize SSE a12 a21 a12 a21 0.644 0.672 Ans. pcalc j X1 j j 1 X 1 j X 2 j a12 a21 Psat1 2 X 1 j X 2 j a12 a21 Psat2 1 X 1 j X 2 j a12 a21 Psat1 X2 X1 Y1calc Pcalc j j pcalc x1 i i j i 1 x1i x2i a12 a21 Psat1 2 x1i x2i a12 a21 Psat2 1 x1i x2i a12 a21 Psat1 x2 x1 y1calc i i Pcalc i 398 RMS deviation in P: Pi RMS i Pcalc n 2 i RMS 0.364 kPa P-x,y diagram, van Laar Equation by Barker's Method 105 100 95 Pi kPa 90 Pi kPa pcalc kPa 85 j pcalc kPa j 80 75 70 65 0 0.2 i 0.4 i j 0.6 j 0.8 1 x1 y1 X1 Y1calc P-x data P-y data P-x calculated P-y calculated 399 Residuals in P and y1. 1.5 1 Pi Pcalc kPa i 0.5 0 0.5 1 y1 y1calc 100 i i 0 0.2 0.4 x1 i 0.6 0.8 Pressure residuals y1 residuals (f) BARKER'S METHOD by non-linear least squares. Wilson equation. Guesses for parameters: answers to Part (c). 1 101 X1 12 21 j j .01 j .01 x2 x2 12 12 X2 j 1 X1 j 1 x1 x2 exp ln x1 x2 x1 21 12 x2 x1 21 2 x1 x2 12 21 exp ln x2 x1 x1 x1 x2 21 12 12 21 x2 x1 21 Minimize the sum of the squared errors using the Mathcad Minimize function. Guesses: 12 0.5 21 400 1.0 SSE 12 21 i Pi x1 x1 x2 i 1 i i i 12 12 21 Psat1 21 Psat2 2 x2 2 x1i x2i 12 21 Minimize SSE 12 12 21 0.732 0.663 Ans. 21 pcalc j X1 j j 1 X1 j X2 j 2 X1 j X2 j 12 12 21 Psat1 21 Psat2 X2 X1 Y1calc j j 1 X1 j X2 j 12 j 21 Psat1 pcalc Pcalc i x1 i i 1 x1i x2i 2 x1i x2i 12 12 21 Psat1 21 Psat2 x2 x1 y1calc i i 1 x1i x2i 12 i 21 Psat1 Pcalc RMS deviation in P: Pi RMS i Pcalc n 2 i RMS 0.35 kPa 401 P-x,y diagram, Wilson Equation by Barker's Method 105 Pi kPa 100 95 90 85 j Pi kPa pcalc kPa 80 j pcalc kPa 75 70 65 0 0.2 i 0.4 i j 0.6 j 0.8 1 x1 y1 X1 Y1calc P-x data P-y data P-x calculated P-y calculated Residuals in P and y1. 2 Pi Pcalc kPa i 1 y1 y1calc 100 i i 0 1 0 0.2 0.4 x1 i 0.6 0.8 Pressure residuals y1 residuals 402 12.6 Methyl t-butyl ether(1)/Dichloromethane--VLE data: T 83.402 82.202 80.481 76.719 72.442 68.005 P 65.096 59.651 56.833 53.689 51.620 50.455 49.926 49.720 x2 1 x1 kPa 308.15 K 0.0141 0.0253 0.0416 0.0804 0.1314 0.1975 y1 0.2457 0.3686 0.4564 0.5882 0.7176 0.8238 0.9002 0.9502 0.0330 0.0579 0.0924 0.1665 0.2482 0.3322 x1 0.3880 0.5036 0.5749 0.6736 0.7676 0.8476 0.9093 0.9529 y2 Psat2 1 y1 85.265 kPa Psat1 49.624 kPa Calculate EXPERIMENTAL values of activity coefficients and excess Gibbs energy. y1 P 1 y2 P 2 x1 Psat1 x2 Psat2 GERT x1 ln 1 x2 ln 2 GERTx1x2 GERT x1 x2 n rows ( ) P n 14 i 1 n 403 (a) Fit GE/RT data to Margules eqn. by nonlinear least squares. Minimize sum of the squared errors using the Mathcad Minimize function. Guesses: A12 0.3 A21 0.5 A21 x1 i C A12 x2 i 0.2 C x1 x2 i i SSE A12 A21 C i GERTi x1 x2 i 2 i A12 A21 C Minimize SSE A12 A21 C A12 A21 C 0.336 0.535 0.195 Ans. (b) Plot data and fit GeRTx1x2 ( x2) x1 GeRT ( x2) x1 ln 1 ( x2) x1 A21 x1 A12 x2 C x1 x2 GeRTx1x2 ( x2)x1 x2 x1 x2 2 A12 2 A21 A12 C x1 3 C x1 2 ln 2 ( x2) x1 j 1 101 x1 2 A21 2 A12 X1 j A21 C x2 3 C x2 X2 j 2 .01 j .01 1 X1 j GERTx1x2 i GeRTx1x2 X1 X2 j j 0 0.1 0.2 ln 1 i ln 1 X1 X2 j j 0.3 0.4 ln 2 i ln 2 X1 X2 j j 0.5 0.6 0 0.2 i 0.4 j i 0.6 j i 0.8 j x1 X1 x1 X1 x1 X1 404 (c) Plot Pxy diagram with fit and data 1 ( x1 x2) exp ln 1 ( x1 x2) 2 ( x1 x2) exp ln 2 ( x1 x2) j Pcalc j X1 1 X1 X2 Psat1 j j j X2 j 2 X1 X2 Psat2 j j X1 y1calc j 1 X1 X2 Psat1 j j Pcalc j P-x,y Diagram from Margules Equation fit to GE/RT data. 90 Pi kPa 80 Pi kPa 70 j Pcalc kPa 60 j Pcalc kPa 50 40 0 0.2 0.4 i i j 0.6 j 0.8 x1 y1 X1 y1calc P-x data P-y data P-x calculated P-y calculated (d) Consistency Test: GERTi 1i 2 i GeRT x1 x2 i i GERTi ln 1 2i ln 1 x1 x2 i i 2 x1 x2 i i ln 405 0.004 0 GERTi 0 ln 1 2i 0.025 0.004 0 0.5 1 0.05 0 0.5 1 x1 i x1 i Calculate mean absolute deviation of residuals 4 mean GERT 9.391 10 mean ln 1 2 0.021 (e) Barker's Method by non-linear least squares: Margules Equation 1 x1 x2 A 12 A 21 C exp ( ) A12 x2 2 2 A21 2 A12 C x1 3 C x1 2 x1 x2 A 12 A 21 C exp ( ) A21 x1 2 2 A12 2 A21 C x2 3 C x2 Minimize sum of the squared errors using the Mathcad Minimize function. Guesses: A12 0.3 Pi i A21 x1 0.5 C 0.2 2 SSE A12 A21 C x1 x2 A12 A21 C Psat1 i 1 i i i x2 2 x1i x2i A 12 A 21 C Psat2 A12 A21 C Minimize SSE A12 A21 C A12 A21 C 0.364 0.521 0.23 Ans. 406 Plot P-x,y diagram for Margules Equation with parameters from Barker's Method. Pcalc j X1 j j 1 X 1 j X 2 j A 12 A 21 C Psat1 2 X 1 j X 2 j A 12 A 21 C Psat2 X2 X1 y1calc j j 1 X 1 j X 2 j A 12 A 21 C Psat1 Pcalc 90 j Pi kPa 80 Pi kPa 70 j Pcalc kPa 60 j Pcalc kPa 50 40 0 0.2 0.4 i i j 0.6 j 0.8 x1 y1 X1 y1calc P-x data P-y data P-x calculated P-y calculated Pcalc i x1 i i 1 x1i x2i A 12 A 21 C Psat1 2 x1i x2i A 12 A 21 C Psat2 x2 x1 y1calc i i 1 x1i x2i A 12 A 21 C Psat1 Pcalc i 407 Plot of P and y1 residuals. 0.8 0.6 Pi Pcalc i kPa 0.4 y1 y1calc 100 0.2 i i 0 0.2 0 0.5 1 x1 i Pressure residuals y1 residuals RMS deviations in P: Pi RMS i Pcalc n 2 i RMS 0.068 kPa 408 12.8 (a) Data: 0.0523 0.1299 0.2233 0.2764 0.3482 0.4187 x1 0.5001 0.5637 0.6469 0.7832 0.8576 0.9388 0.9813 1 1.202 1.307 1.295 1.228 1.234 1.180 1.129 1.120 1.076 1.032 1.016 1.001 1.003 i 1 n n 13 2 1.002 1.004 1.006 1.024 1.022 1.049 1.092 1.102 1.170 1.298 1.393 1.600 1.404 n rows x1 x1 ln 1 i i x2 i 1 x1 i GERTi x2 ln 2 i i (b) Fit GE/RT data to Margules eqn. by linear least-squares procedure: Xi Slope Slope A12 A12 x1 i Yi slope ( Y) X GERTi x1 x2 i i intercept ( Y) X 0.286 A12 Ans. Intercept Intercept A21 A21 0.247 Intercept 0.286 exp x2 exp x1 2 2 Slope 0.534 1 ( x2) x1 2 ( x2) x1 A12 A21 2 A21 2 A12 A12 x1 A21 x2 GeRT ( x2) x1 x1 ln 1 ( x2) x1 x2 ln 2 ( x2) x1 409 Plot of data and correlation: 0.5 GERT i GeRT x1 x2 i i ln 1 i ln 1 x1 x2 i i ln 2 i ln 2 x1 x2 i i 0.4 0.3 0.2 0.1 0 0 0.2 0.4 x1 i 0.6 0.8 (c) Calculate and plot residuals for consistency test: GERTi GeRT x1 x2 i i GERTi 1i 2 i ln 1 2i ln 1 x1 x2 i i 2 x1 x2 i i ln 410 GERTi 3.31410-3 -2.26410-3 -3.1410-3 -2.99810-3 -2.87410-3 -2.2210-3 -2.17410-3 -1.55310-3 -8.74210-4 2.94410-4 5.96210-5 9.02510-5 4.23610-4 ln 1 2i 0.098 -9.15310-5 -0.021 0.026 -0.019 5.93410-3 0.028 -9.5910-3 9.13910-3 -5.61710-4 -0.011 0.028 -0.168 0.1 0.05 ln 1 2i 0 0 0.5 x1 i 1 Calculate mean absolute deviation of residuals: mean GERT 1.615 10 3 mean ln 1 2 0.03 Based on the graph and mean absolute deviations, the data show a high degree of consistency 12.9 Acetonitrile(1)/Benzene(2)-- VLE data T 318.15 K 0.1056 0.1818 0.2783 0.3607 0.4274 31.957 33.553 35.285 36.457 36.996 P 37.068 36.978 36.778 35.792 34.372 32.331 30.038 kPa 0.0455 0.0940 0.1829 0.2909 0.3980 x1 0.5069 0.5458 0.5946 0.7206 0.8145 0.8972 0.9573 411 y1 0.4885 0.5098 0.5375 0.6157 0.6913 0.7869 0.8916 x2 Psat1 1 x1 27.778 kPa y2 1 y1 Psat2 29.819 kPa Calculate EXPERIMENTAL values of activity coefficients and excess Gibbs energy. y1 P 1 y2 P 2 x1 Psat1 x2 Psat2 GERT x1 ln 1 x2 ln 2 GERTx1x2 GERT x1 x2 n rows ( ) P n 12 i 1 n (a) Fit GE/RT data to Margules eqn. by nonlinear least squares. Minimize sum of the squared errors using the Mathcad Minimize function. Guesses: A12 0.3 A21 0.5 C 0.2 SSE A12 A21 C i GERTi A21 x1 i A12 x2 i C x1 x2 i i x1 x2 i i 2 A12 A21 C Minimize SSE A12 A21 C A12 A21 C 1.128 1.155 0.53 Ans. (b) Plot data and fit GeRTx1x2 ( x2) x1 GeRT ( x2) x1 ln 1 ( x2) x1 A21 x1 A12 x2 C x1 x2 GeRTx1x2 ( x2)x1 x2 x1 x2 2 A12 2 A21 A12 C x1 3 C x1 2 ln 2 ( x2) x1 j 1 101 x1 2 A21 2 A12 X1 j A21 C x2 3 C x2 X2 j 2 .01 j .01 1 X1 j 412 1.2 GERTx1x2 i GeRTx1x2 X1 X2 j j 1 0.8 0.6 ln 1 i ln 1 X1 X2 j j ln 2 i ln 2 X1 X2 j j 0.4 0.2 0 0 0.2 i 0.4 j i 0.6 j i 0.8 j x1 X1 x1 X1 x1 X1 (c) Plot Pxy diagram with fit and data 1 ( x1 x2) exp ln 1 ( x1 x2) 2 ( x1 x2) exp ln 2 ( x1 x2) j Pcalc j X1 1 X1 X2 j j Psat1 X2 j 2 X1 X2 j j Psat2 X1 y1calc j j 1 X1 X2 Psat1 j j Pcalc j 413 P-x,y Diagram from Margules Equation fit to GE/RT data. 38 Pi kPa 36 Pi kPa 34 Pcalc kPa 32 j 30 Pcalc kPa j 28 26 0 0.2 0.4 i i j 0.6 j 0.8 x1 y1 X1 y1calc P-x data P-y data P-x calculated P-y calculated (d) Consistency Test: GERTi 1i 2 i GeRT x1 x2 i i GERTi ln 1 2i ln 1 x1 x2 i i 2 x1 x2 i i ln 0.004 0 GERTi 0 ln 1 2i 0.025 0.004 0 0.5 1 0.05 0 0.5 1 x1 i x1 i 414 Calculate mean absolute deviation of residuals 4 mean GERT 6.237 10 mean ln 1 2 0.025 (e) Barker's Method by non-linear least squares: Margules Equation 1 x1 x2 A 12 A 21 C exp ( ) A12 x2 2 2 A21 2 A12 C x1 3 C x1 2 x1 x2 A 12 A 21 C exp ( ) A21 x1 2 2 A12 2 A21 C x2 3 C x2 Minimize sum of the squared errors using the Mathcad Minimize function. Guesses: A12 0.3 Pi i A21 x1 0.5 C 0.2 2 SSE A12 A21 C x1 x2 A12 A21 C Psat1 i 1 i i i x2 2 x1i x2i A 12 A 21 C Psat2 A12 A21 C Minimize SSE A12 A21 C A12 A21 C 1.114 1.098 0.387 Ans. Plot P-x,y diagram for Margules Equation with parameters from Barker's Method. Pcalc j X1 j j 1 X 1 j X 2 j A 12 A 21 C Psat1 2 X 1 j X 2 j A 12 A 21 C Psat2 1 X 1 j X 2 j A 12 A 21 C Psat1 X2 X1 y1calc j j Pcalc j 415 38 Pi kPa 36 Pi kPa 34 Pcalc kPa 32 j Pcalc kPa 30 j 28 26 0 0.2 0.4 i i j 0.6 j 0.8 x1 y1 X1 y1calc P-x data P-y data P-x calculated P-y calculated Pcalc i x1 i i 1 x1i x2i A 12 A 21 C Psat1 2 x1i x2i A 12 A 21 C Psat2 x2 x1 y1calc i i 1 x1i x2i A 12 A 21 C Psat1 Pcalc i 416 Plot of P and y1 residuals. 0.6 0.4 Pi Pcalc i kPa 0.2 0 0.2 0.4 y1 y1calc 100 i i 0 0.5 1 x1 i Pressure residuals y1 residuals RMS deviations in P: Pi RMS i Pcalc n 2 i RMS 0.04 kPa 417 12.12 It is impractical to provide solutions for all of the systems listed in the table on Page 474 we present as an example only the solution for the system 1-propanol(1)/water(2). Solutions for the other systems can be obtained by rerunning the following Mathcad program with the appropriate parameter values substituted for those given. The file WILSON.mcd reproduces the table of Wilson parameters on Page 474 and includes the necessary Antoine coefficients. Antoine coefficients: 1-Propanol: Water: A1 A2 16.1154 16.3872 B1 B2 3483.67 K 3885.70 K C1 C2 205.807 K 230.170 K Psat1 ( T) exp A1 (T B1 273.15 K) C1 kPa Psat2 ( ) T exp A2 (T B2 273.15 K) C2 kPa Parameters for the Wilson equation: V1 cm 75.14 mol 775.48 3 V2 cm 18.07 mol 1351.90 3 a12 cal mol a21 cal mol 12 ( ) T a12 V2 exp RT V1 12 ( ) T x1 21 ( ) T a21 V1 exp RT V2 21 ( ) T x1 21 ( ) T exp x2 1 ( x2 T) x1 x2 12 ( ) x2 T x1 x2 12 ( ) T exp 2 ( x2 T) x1 x1 12 ( ) T x1 x2 12 ( ) x2 T x2 x1 418 21 ( ) T x1 21 ( ) T 21 ( ) T P-x,y diagram at T ( 60 273.15) K Guess: P 70 kPa Given P = x1 1 ( x1 1 x1 T) Psat1 ( T) ( 1 x1) 2 ( x1 1 x1 T) Psat2 ( T) Peq x1) ( Find P) ( yeq x1) ( x1 1 ( x1 1 x1 T) Psat1 ( T) Peq x1) ( x 0 0.05 1.0 x 0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5 0.55 0.6 0.65 0.7 0.75 0.8 0.85 0.9 0.95 1 yeq x) ( 0 0.315 0.363 0.383 0.395 0.404 0.413 0.421 0.431 0.441 0.453 0.466 0.483 0.502 0.526 0.556 0.594 0.646 0.718 0.825 1 Peq x) ( kPa 20.007 28.324 30.009 30.639 30.97 31.182 31.331 31.435 31.496 31.51 31.467 31.353 31.148 30.827 30.355 29.686 28.759 27.491 25.769 23.437 20.275 419 P,x,y Diagram at T 32 333.15 K 30 28 Peq x) ( kPa Peq x) ( kPa 26 24 22 20 0 0.2 0.4 0.6 0.8 x yeq x) ( 12.13 It is impractical to provide solutions for all of the systems listed in the table on Page 474; we present as an example only the solution for the system 1-propanol(1)/water(2). Solutions for the other systems can be obtained by rerunning the following Mathcad program with the appropriate parameter values substituted for those given. The file WILSON.mcd reproduces the table of Wilson parameters on Page 474 and includes the necessary Antoine coefficients. Antoine coefficients: 1-Propanol: A1 Water: Psat1 ( ) T 16.1154 16.3872 B1 B2 3483.67 K 3885.70 K kPa C1 C2 205.807 K 230.170 K A2 exp A1 ( T B1 273.15 K) C1 B2 273.15 K) C2 420 Psat2 ( ) T exp A2 ( T kPa Parameters for the Wilson equation: V1 75.14 cm 3 mol cal mol V2 18.07 cm 3 mol cal mol a12 775.48 a21 1351.90 12 ( T) a12 V2 exp RT V1 exp x2 12 ( T) x1 21 ( T) a21 V1 exp RT V2 21 ( T) 1 ( x1 x2 T) x2 12 ( T) x1 x2 x2 12 ( T) x1 21 ( T) exp 2 ( x1 x2 T) x1 12 ( T) x1 x2 12 ( T) x2 x1 21 ( T) x2 x1 21 ( T) 21 ( T) T-x,y diagram at P Guess: Given 101.33 kPa 273.15) K T ( 90 P = x1 1 ( x1 1 x1 T) Psat1 ( T) ( 1 x1) 2 ( x1 1 x1 T) Psat2 ( T) Find T) ( x1 1 ( x1 1 x1 Teq ( x1) ) Psat1 ( Teq ( x1) ) P Teq ( x1) yeq ( x1) x 0 0.05 1.0 421 x 0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5 0.55 0.6 0.65 0.7 0.75 0.8 0.85 0.9 0.95 1 yeq x () 0 0.304 0.358 0.381 0.395 0.407 0.418 0.429 0.44 0.453 0.468 0.484 0.504 0.527 0.555 0.589 0.631 0.686 0.759 0.858 1 Teq x () K 373.149 364.159 362.476 361.836 361.49 361.264 361.101 360.985 360.911 360.881 360.904 360.99 361.154 361.418 361.809 362.364 363.136 364.195 365.644 367.626 370.349 T,x,y Diagram at P 375 101.33 kPa Teq( ) 370 x K Teq( ) x K 365 360 0 0.2 0.4 0.6 0.8 1 x yeq x) ( 422 12.14 It is impractical to provide solutions for all of the systems listed in the table on Page 474; we present as an example only the solution for the system 1-propanol(1)/water(2). Solutions for the other systems can be obtained by rerunning the following Mathcad program with the appropriate parameter values substituted for those given. The file NRTL.mcd reproduces the table of NRTL parameters on Page 474 and includes the necessary Antoine coefficients. Antoine coefficients: 1-Propanol: A1 Water: Psat1 ( T) 16.1154 16.3872 B1 B2 B1 273.15 K) 3483.67 K 3885.70 K kPa C1 C2 205.807 K 230.170 K A2 exp A1 (T C1 Psat2 ( T) exp A2 (T B2 273.15 K) C2 kPa Parameters for the NRTL equation: b12 500.40 b12 RT exp 12 ( T) cal mol b21 1636.57 cal mol b21 RT exp 2 0.5081 12 ( T) G12 ( T) 21 ( T) G21 ( T) 21 ( T) 1 ( x1 x2 T) exp x2 2 21 ( T) G21 ( T) x1 x2 G21 ( T) G12 ( T) 12 ( T) x1 G12 ( T) ) 2 ( x2 2 ( x1 x2 T) exp x1 2 12 ( T) G12 ( T) 2 x2 x1 G12 ( T) G21 ( T) 21 ( T) x2 G21 ( T) ) 423 ( x1 2 P-x,y diagram at T ( 60 273.15)K Guess: P 70 kPa Given P = x1 1 ( 1 x1 T)Psat1 ( ) x1 T ( x1) 2 ( 1 x1 T)Psat2 ( ) 1 x1 T Peq x1) ( Find P) ( yeq x1) ( x1 1 ( 1 x1 x1 T)Psat1 ( ) T Peq x1) ( x 0 0.05 1.0 x 0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5 0.55 0.6 0.65 0.7 0.75 0.8 0.85 0.9 0.95 1 yeq x () 0 0.33 0.373 0.382 0.386 0.39 0.395 0.404 0.414 0.427 0.442 0.459 0.479 0.503 0.531 0.564 0.606 0.659 0.732 0.836 1 Peq x () kPa 20.007 28.892 30.48 30.783 30.876 30.959 31.048 31.127 31.172 31.163 31.085 30.922 30.657 30.271 29.74 29.03 28.095 26.868 25.256 23.124 20.275 424 P,x,y Diagram at T 35 333.15 K Peq ( x) kPa 30 Peq ( x) kPa 25 20 0 0.2 0.4 x yeq ( x) 0.6 0.8 12.15 It is impractical to provide solutions for all of the systems listed in the table on Page 474; we present as an example only the solution for the system 1-propanol(1)/water(2). Solutions for the other systems can be obtained by rerunning the following Mathcad program with the appropriate parameter values substituted for those given. The file NRTL.mcd reproduces the table of NRTL parameters on Page 474 and includes the necessary Antoine coefficients. Antoine coefficients: 1-Propanol: A1 16.1154 Water: Psat1 ( T) B1 B2 3483.67 K 3885.70 K kPa C1 C2 205.807 K 230.170 K A2 exp A1 16.3872 (T B1 273.15 K) C1 Psat2 ( T) exp A2 (T B2 273.15 K) C2 kPa Parameters for the NRTL equation: b12 500.40 cal mol b21 1636.57 425 cal mol 0.5081 12 ( ) T b12 RT 21 ( ) T b21 RT G12 ( ) T exp 12 ( ) T G21 ( ) T exp 2 21 ( ) T 1 ( x2 T) x1 exp x2 2 G21 ( ) T 21 ( ) T x1 x2 G21 ( ) T G12 ( ) 12 ( ) T T ( x2 x1 G12 ( ) T) 2 2 ( x2 T) x1 exp x1 2 G12 ( ) T 12 ( ) T x2 x1 G12 ( ) T G21 ( ) 21 ( ) T T ( x1 x2 G21 ( ) T) 2 2 T-x,y diagram at P 101.33 kPa Guess: T ( 90 273.15)K Given P = x1 1 ( 1 x1 T)Psat1 ( ) x1 T ( x1) 2 ( 1 x1 T)Psat2 ( ) 1 x1 T Teq x1) ( Find T) ( yeq x1) ( x1 1 ( 1 x1 x1 Teq x1) Psat1 ( ( ) ( ) Teq x1 ) P 426 x 0 0.05 1.0 x 0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5 0.55 0.6 0.65 0.7 0.75 0.8 0.85 0.9 0.95 1 yeq ( x) 0 0.32 0.377 0.394 0.402 0.408 0.415 0.424 0.434 0.447 0.462 0.48 0.5 0.524 0.552 0.586 0.629 0.682 0.754 0.853 1 Teq ( x) K 373.149 363.606 361.745 361.253 361.066 360.946 360.843 360.757 360.697 360.676 360.709 360.807 360.985 361.262 361.66 362.215 362.974 364.012 365.442 367.449 370.349 T,x,y Diagram at P 375 101.33 kPa Teq ( x) 370 K Teq ( x) K 365 360 0 0.2 0.4 x yeq ( x) 427 0.6 0.8 1 12.16 It is impractical to provide solutions for all of the systems listed in the table on Page 474; we present as an example only the solution for the system 1-propanol(1)/water(2). Solutions for the other systems can be obtained by rerunning the following Mathcad program with the appropriate parameter values substituted for those given. The file WILSON.mcd reproduces the table of Wilson parameters on Page 474 and includes the necessary Antoine coefficients. Antoine coefficients: 1-Propanol: Water: Psat1 ( ) T A1 A2 exp A1 16.1154 16.3872 B1 B2 3483.67 K 3885.70 K kPa C1 C2 205.807 K 230.170 K ( T B1 273.15 K) C1 Psat2 ( ) T exp A2 ( T B2 273.15 K) C2 kPa Parameters for the Wilson equation: V1 cm 75.14 mol 775.48 3 V2 cm 18.07 mol 1351.90 3 a12 cal mol a21 cal mol 12 ( ) T a12 V2 exp RT V1 21 ( ) T a21 V1 exp RT V2 exp x2 1 ( x2 T) x1 12 ( ) T x1 x2 12 ( ) x2 T x1 x2 12 ( ) T 21 ( ) T x1 21 ( ) T exp 2 ( x2 T) x1 x1 12 ( ) T x1 x2 12 ( ) x2 T x2 x1 428 21 ( ) T x1 21 ( ) T 21 ( ) T (a) BUBL P: T ( 60 273.15) K x1 0.3 x2 1 x1 Guess: P 101.33 kPa y1 0.4 y2 1 y1 Given y1 P = x1 1 ( x1 x2 T) Psat1 ( T) y1 y2 P = x2 2 ( x1 x2 T) Psat2 ( T) y2 = 1 Pbubl y1 y2 Pbubl 31.33 kPa Find ( P y1 y2) y1 0.413 y2 0.587 Ans. (b) DEW P: T ( 60 273.15) K y1 0.3 y2 1 y1 Guess: Given P 101.33 kPa x1 0.1 x1 x2 x2 = 1 1 x1 y1 P = x1 1 ( x1 x2 T) Psat1 ( T) y2 P = x2 2 ( x1 x2 T) Psat2 ( T) Pdew x1 x2 Pdew 27.79 kPa Find ( P x1 x2) x1 0.042 x2 0.958 Ans. (c) P,T-flash Calculation P Pdew 2 Pbubl T ( 60 273.15) K z1 0.3 Guess: V 0.5 x1 y1 0.1 0.1 x2 y2 1 1 y1 x1 Given y1 = x1 1 ( x1 x2 T) Psat1 ( T) P x1 x2 = 1 y2 = x2 2 ( x1 x2 T) Psat2 ( T) P 429 y1 y2 = 1 x1 ( 1 V) y1 V = z1 Eq. (10.15) x1 x2 y1 y2 V x1 0.08 Find x1 x2 y1 y2 V) ( x2 0.92 y1 0.351 y2 0.649 V 0.813 (d) Azeotrope Calculation Test for azeotrope at: T ( 60 273.15)K 1 ( 1 T) 21.296 0 2 ( 0 T) 4.683 1 120 1 ( 1 T)Psat1 ( ) 0 T Psat2 ( ) T 120 21.581 121 Psat1 ( ) T 2 ( 0 T)Psat2 ( ) 1 T 121 0.216 Since one of these values is >1 and the other is <1, an azeotrope exists. See Ex. 10.3(e) Guess: P 101.33 kPa x1 y1 0.3 0.3 x2 y2 1 1 y1 x1 Given y1 P = x1 1 ( x2 T)Psat1 ( ) x1 T y2 P = x2 2 ( x2 T)Psat2 ( ) x1 T x1 x2 = 1 y1 y2 = 1 x1 = y1 x1 x2 y1 y2 Paz Paz 31.511 kPa Find x1 x2 y1 y2 P) ( x1 0.4386 430 y1 0.4386 Ans. 12.17 It is impractical to provide solutions for all of the systems listed in the table on Page 474; we present as an example only the solution for the system 1-propanol(1)/water(2). Solutions for the other systems can be obtained by rerunning the following Mathcad program with the appropriate parameter values substituted for those given. The file NRTL.mcd reproduces the table of NRTL parameters on Page 474 and includes the necessary Antoine coefficients. Antoine coefficients: 1-Propanol: A1 Water: Psat1 ( T) 16.1154 16.3872 B1 B2 B1 273.15 K) 3483.67 K 3885.70 K kPa C1 C2 205.807 K 230.170 K A2 exp A1 (T C1 Psat2 ( T) exp A2 (T B2 273.15 K) C2 kPa Parameters for the NRTL equation: b12 500.40 b12 RT exp 12 ( T) cal mol b21 1636.57 cal mol b21 RT exp 2 0.5081 12 ( T) 21 ( T) G21 ( T) G12 ( T) 21 ( T) 1 ( x1 x2 T) exp x2 2 21 ( T) G21 ( T) x1 x2 G21 ( T) G12 ( T) 12 ( T) x1 G12 ( T) ) 2 ( x2 2 ( x1 x2 T) exp x1 2 12 ( T) G12 ( T) 2 x2 x1 G12 ( T) G21 ( T) 21 ( T) x2 G21 ( T) ) 431 ( x1 2 (a) BUBL P: T ( 60 273.15)K x1 0.3 x2 1 x1 Guess: P 101.33 kPa y1 0.4 y2 1 y1 Given y1 P = x1 1 ( x2 T)Psat1 ( ) x1 T y1 y2 P = x2 2 ( x2 T)Psat2 ( ) x1 T y2 = 1 Pbubl y1 y2 Pbubl 31.05 kPa Find ( y1 y2) P y1 0.395 y2 0.605 Ans. (b) DEW P: T ( 60 273.15)K y1 0.3 y2 1 y1 Guess: P 101.33 kPa x1 0.1 x2 1 x1 Given y1 P = x1 1 ( x2 T)Psat1 ( ) x1 T x1 y2 P = x2 2 ( x2 T)Psat2 ( ) x1 T x2 = 1 Pdew x1 x2 Pdew 27.81 kPa Find ( x1 x2) P x1 0.037 x2 0.963 Ans. (c) P,T-flash Calculation P Pdew 2 Pbubl T ( 60 273.15)K z1 0.3 Guess: V 0.5 x1 y1 0.1 0.1 x2 y2 1 1 y1 x1 Given y1 = x1 1 ( x2 T)Psat1 ( ) x1 T P x1 x2 = 1 y2 = x2 2 ( x2 T)Psat2 ( ) x1 T P y1 y2 = 1 x1 ( 1 V) y1 V = z1 Eq. (10.15) 432 x1 x2 y1 y2 V x1 0.06 Find ( x1 x2 y1 y2 V) x2 0.94 y1 0.345 y2 0.655 V 0.843 (d) Azeotrope Calculation Test for azeotrope at: T ( 60 273.15) K 1 ( 0 1 T) 19.863 2 ( 1 0 T) 4.307 120 1 ( 0 1 T) Psat1 ( T) Psat2 ( T) 120 20.129 121 Psat1 ( T) 2 ( 1 0 T) Psat2 ( T) 121 0.235 Since one of these values is >1 and the other is <1, an azeotrope exists. See Ex. 10.3(e). Guess: P 101.33 kPa x1 y1 0.3 0.3 x2 y2 1 1 x1 x1 Given y1 P = x1 1 ( x1 x2 T) Psat1 ( T) y2 P = x2 2 ( x1 x2 T) Psat2 ( T) x1 x2 = 1 y1 y2 = 1 x1 = y1 x1 x2 y1 y2 Paz Paz 31.18 kPa Find ( x1 x2 y1 y2 P) x1 0.4187 433 y1 0.4187 Ans. 12.18 It is impractical to provide solutions for all of the systems listed in the table on Page 474; we present as an example only the solution for the system 1-propanol(1)/water(2). Solutions for the other systems can be obtained by rerunning the following Mathcad program with the appropriate parameter values substituted for those given. The file WILSON.mcd reproduces the table of Wilson parameters on Page 474 and includes the necessary Antoine coefficients. Antoine coefficients: 1-Propanol: Water: Psat1 ( ) T A1 A2 exp A1 16.1154 16.3872 B1 B2 3483.67 K 3885.70 K kPa C1 C2 205.807 K 230.170 K ( T B1 273.15 K) C1 Psat2 ( ) T exp A2 ( T B2 273.15 K) C2 kPa Parameters for the Wilson equation: V1 cm 75.14 mol 775.48 3 V2 cm 18.07 mol 1351.90 3 a12 cal mol a21 cal mol 12 ( ) T a12 V2 exp RT V1 21 ( ) T a21 V1 exp RT V2 21 ( ) T x1 21 ( ) T exp x2 1 ( x2 T) x1 12 ( ) T x1 x2 12 ( ) x2 T x1 x2 12 ( ) T exp 2 ( x2 T) x1 x1 12 ( ) T x1 x2 12 ( ) x2 T x2 x1 21 ( ) T 21 ( ) T x1 21 ( ) T 434 (a) BUBL T: P 101.33 kPa x1 0.3 x2 1 x1 Guess: T ( 60 273.15) K y1 0.3 y2 1 y1 Given y1 P = x1 1 ( x1 x2 T) Psat1 ( T) y1 y2 P = x2 2 ( x1 x2 T) Psat2 ( T) y2 = 1 Tbubl y1 y2 Tbubl 361.1 K Find ( T y1 y2) y1 0.418 y2 0.582 Ans. (b) DEW T: P 101.33 kPa y1 0.3 y2 1 x1 Guess: T ( 60 273.15) K x1 0.1 x1 x2 x2 = 1 1 y1 Given y1 P = x1 1 ( x1 x2 T) Psat1 ( T) y2 P = x2 2 ( x1 x2 T) Psat2 ( T) Tdew x1 x2 Tdew 364.28 K Find ( T x1 x2) x1 0.048 x2 0.952 Ans. (c) P,T-flash Calculation T Tdew 2 Tbubl P 101.33 kPa z1 0.3 Guess: V 0.5 x1 y1 0.1 0.1 x2 y2 1 1 y1 x1 Given y1 = x1 1 ( x1 x2 T) Psat1 ( T) P x1 x2 = 1 y2 = x2 2 ( x1 x2 T) Psat2 ( T) P y1 y2 = 1 x1 ( 1 V) y1 V = z1 Eq. (10.15) 435 x1 x2 y1 y2 V x1 0.09 Find x1 x2 y1 y2 V) ( x2 0.91 y1 0.35 y2 0.65 V 0.807 (d) Azeotrope Calculation Test for azeotrope at: P 101.33 kPa C1 273.15 K Tb1 A1 B1 P ln kPa B2 A2 P ln kPa Tb1 370.349 K Tb2 C2 273.15 K Tb2 373.149 K 1 ( 1 Tb2) 16.459 0 2 ( 0 Tb1) 3.779 1 120 1 ( 1 T)Psat1 ( ) 0 Tb2 P 120 121 19.506 121 P 2 ( 0 T)Psat2 ( ) 1 Tb1 0.281 Since one of these values is >1 and the other is <1, an azeotrope exists. See Ex. 10.3(e). Guesses: T ( 60 273.15)K x1 0.4 x2 1 y1 y1 x2 = 1 y2 = 1 x1 = y1 0.4 y2 1 x1 Given y1 P = x1 1 ( x2 T)Psat1 ( ) x1 x1 T y2 P = x2 2 ( x2 T)Psat2 ( ) y1 x1 T 436 x1 x2 y1 y2 Taz Taz 360.881 K x1 0.4546 y1 0.4546 Ans. Find ( x1 x2 y1 y2 T) 12.19 It is impractical to provide solutions for all of the systems listed in the table on page 474; we present as an example only the solution for the system 1-propanol(1)/water(2). Solutions for the other systems can be obtained by rerunning the following Mathcad program with the appropriate parameter values substituted for those given. The file NRTL.mcd reproduces the table of NRTL parameters on Page 474 and includes the necessary Antoine coefficients. Antoine coefficients: 1-Propanol: A1 Water: Psat1 ( T) 16.1154 16.3872 B1 B2 B1 273.15 K) 3483.67 K 3885.70 K kPa C1 C2 205.807 K 230.170 K A2 exp A1 (T C1 Psat2 ( T) exp A2 (T B2 273.15 K) C2 kPa Parameters for the NRTL equation: b12 12 ( T) G12 ( T) 500.40 b12 RT exp cal mol b21 1636.57 cal mol b21 RT exp 0.5081 21 ( T) 12 ( T) G21 ( T) 21 ( T) 437 1 ( x2 T) x1 exp x2 2 21 ( ) T G21 ( ) T x1 x2 G21 ( ) T G12 ( ) 12 ( ) T T x1 G12 ( ) T) 2 2 ( x2 2 ( x2 T) x1 exp x1 2 12 ( ) T G12 ( ) T x2 x1 G12 ( ) T G21 ( ) 21 ( ) T T x2 G21 ( ) T) 2 2 ( x1 (a) BUBL T: P 101.33 kPa x1 0.3 x2 1 x1 Guess: T ( 60 273.15)K y1 0.3 y2 1 y1 Given y1 P = x1 1 ( x2 T)Psat1 ( ) x1 T y1 y2 P = x2 2 ( x2 T)Psat2 ( ) x1 T y2 = 1 Tbubl y1 y2 Tbubl 360.84 K P Find ( y1 y2) T y1 0.415 y2 y1 0.585 0.3 Ans. (b) DEW T: 101.33 kPa y2 1 x1 Guess: T ( 90 273.15)K x1 0.05 x2 1 y1 Given y1 P = x1 1 ( x2 T)Psat1 ( ) x1 T y2 P = x2 2 ( x2 T)Psat2 ( ) x1 T x1 x2 = 1 Tdew x1 x2 Tdew 364.27 K Find ( x1 x2) T x1 0.042 x2 0.958 Ans. 438 (c) P,T-flash Calculation T Tdew 2 Tbubl P 101.33 kPa z1 0.3 Guess: V 0.5 x1 y1 0.1 0.1 x2 y2 1 1 y1 x1 Given y1 = x1 1 ( x1 x2 T) Psat1 ( T) P x1 x2 = 1 y2 = x2 2 ( x1 x2 T) Psat2 ( T) P y1 y2 = 1 x1 ( 1 x1 x2 y1 y2 V x1 0.069 V) y1 V = z1 Eq. (10.15) Find ( x1 x2 y1 y2 V) x2 0.931 y1 0.352 y2 0.648 V 0.816 (d) Azeotrope Calculation Test for azeotrope at: P 101.33 kPa Tb1 A1 B1 P ln kPa B2 A2 P ln kPa 14.699 C1 273.15 K Tb1 370.349 K Tb2 C2 273.15 K Tb2 373.149 K 1 ( 0 1 Tb2) 2 ( 1 0 Tb1) 4.05 439 120 1 ( 1 T)Psat1 ( ) 0 Tb2 P 120 17.578 121 P 2 ( 0 T)Psat2 ( ) 1 Tb1 121 0.27 Since one of these values is >1 and the other is <1, an azeotrope exists. See Ex. 10.3(e). Guesses: T ( 90 273.15)K x1 0.4 x2 1 y1 y1 0.4 y2 1 x1 Given y1 P = x1 1 ( x2 T)Psat1 ( ) x1 x1 T x2 = 1 y2 P = x2 2 ( x2 T)Psat2 ( ) y1 x1 T x1 x2 y1 y2 Taz Taz 360.676 K x1 0.4461 y1 Find x1 x2 y1 y2 T) ( y2 = 1 x1 = y1 0.4461 Ans. 12.20 Molar volumes & Antoine coefficients: 74.05 V 40.73 18.07 A 14.3145 16.5785 16.3872 B 2756.22 3638.27 3885.70 kPa Ci 0 228.060 C 239.500 230.170 Psat ( T) i exp Ai Bi T K T ( 65 273.15) K 273.15 161.88 291.27 0 107.38 0 cal mol Wilson parameters: a 583.11 1448.01 469.55 440 ( i j T) Vj Vi exp ai j RT i 1 3 j 1 3 p 1 3 (a) BUBL P calculation: No iteration required. x1 0.3 x2 0.4 x3 1 x1 x2 ( i x T) exp 1 ln j xj xp p j ( i j T) ( p i T) ( p j T) xj Pbubl i xi ( i x T) Psat ( i T) yi xi ( i x T) Psat ( i T) Pbubl 0.527 y 0.367 0.106 (b) DEW P calculation: Pbubl 117.1 kPa Ans. y1 Guess: Given 0.3 0.05 y2 x2 0.4 0.2 y3 x3 1 1 y1 x1 y2 x2 P Pbubl x1 P y1 = x1 ( 1 x T) Psat ( 1 T) P y3 = x3 ( 3 x T) Psat ( 3 T) x1 x2 x3 Pdew 441 P y2 = x2 ( 2 x T) Psat ( 2 T) xi = 1 i Find x1 x2 x3 P 0.035 x 0.19 0.775 (c) P,T-flash calculation: z1 Guess: Given P y1 = x1 ( x T)Psat1 T) 1 ( P y2 = x2 ( x T)Psat2 T) 2 ( P y3 = x3 ( x T)Psat3 T) 3 ( xi = 1 i i Pdew 69.14 kPa Ans. P z3 Pdew 2 1 Pbubl T 338.15 K 0.3 V z2 0.5 0.4 z1 z2 Use x from DEW P and y from BUBL P as initial guess. x1 ( 1 x2 ( 1 x3 ( 1 V) y1 V = z1 V) y2 V = z2 V) y3 V = z3 yi = 1 x1 x2 x3 y1 y2 y3 V 0.109 x 0.345 0.546 Find x1 x2 x3 y1 y2 y3 V 0.391 y 0.426 0.183 442 V 0.677 Ans. 12.21 Molar volumes & Antoine coefficients: Antoine coefficients: 74.05 V 40.73 18.07 T ( 65 273.15)K A 14.3145 16.5785 16.3872 Psat ( i T) B 2756.22 3638.27 3885.70 exp Ai Bi T K 228.060 C 239.500 230.170 kPa Ci 273.15 NRTL parameters: 0 0.3084 0.3084 0.5343 0 0.2994 0 0 b 222.64 184.70 631.05 0 253.88 0 cal mol 0.5343 0.2994 1197.41 845.21 bi j RT i 1 3 j 1 3 i j Gi j l 1 3 k 1 3 (a) BUBL P calculation: No iteration required. x1 0.3 x2 0.4 x3 1 x1 x2 exp i j i j j i G j i xj ( i x T) exp j Gl i xl l xk k j G k j x j Gi j j l k i j l Gl j xl G l j xl Pbubl i xi ( i x T) Psat ( i T) 0.525 yi xi ( i x T) Psat ( i T) Pbubl y 0.37 0.105 Pbubl 115.3 kPa Ans. 443 (b) DEW P calculation: y1 0.3 y2 0.4 y3 1 y1 y2 Guess: x1 0.05 x2 0.2 x3 1 x1 x2 P Pbubl Given 2 ( P y1 = x1 ( x T)Psat1 T) P y2 = x2 ( x T)Psat2 T) 1 ( P y3 = x3 ( x T)Psat3 T) 3 ( xi = 1 i x1 x2 x3 Pdew 0.038 x 0.192 0.77 (c) P,T-flash calculation: Find x1 x2 x3 P Pdew 68.9 kPa Ans. P Pdew 2 Pbubl T 338.15 K z1 0.3 z2 0.4 z3 1 z1 z2 Guess: V 0.5 Use x from DEW P and y from BUBL P as initial guess. 1 ( Given P y1 = x1 ( x T)Psat1 T) x1 ( 1 V) y1 V = z1 P y2 = x2 ( x T)Psat2 T) 2 ( x2 ( 1 V) y2 V = z2 P y3 = x3 ( x T)Psat3 T) 3 ( x3 ( 1 V) y3 V = z3 xi = 1 i yi = 1 i 444 x1 x2 x3 y1 y2 y3 V 0.118 x 0.347 0.534 Find x1 x2 x3 y1 y2 y3 V 0.391 y 0.426 0.183 V 0.667 Ans. 12.22 Molar volumes & Antoine coefficients: 74.05 V 40.73 18.07 A 14.3145 16.5785 16.3872 B 2756.22 3638.27 3885.70 kPa Ci 228.060 C 239.500 230.170 Psat ( i T) exp Ai Bi T K P 101.33kPa 273.15 0 Wilson parameters: a 583.11 161.88 291.27 0 107.38 0 cal mol 1448.01 469.55 ( i j T) Vj Vi exp ai j RT i 1 3 j 1 3 p 1 3 (a) x1 BUBL T calculation: 0.3 x2 0.4 x3 445 1 x1 x2 ( x T) i exp 1 ln j xj xp p j ( j T) i ( i T) p ( j T) p xj Guess: Given T 300K y1 0.3 y2 0.3 y3 1 y1 y2 P y1 = x1 ( x T)Psat1 T) 1 ( P y3 = x3 ( x T)Psat3 T) 3 ( y1 y2 y3 Tbubl Find y1 y2 y3 T P y2 = x2 ( x T)Psat2 T) 2 ( P= i xi ( x T)Psati T) i ( 0.536 y 0.361 0.102 (b) DEW T calculation: y1 Guess: Given P y1 = x1 ( x T)Psat1 T) P y2 = x2 ( x T)Psat2 T) 1 ( 2 ( P y3 = x3 ( x T)Psat3 T) 3 ( i Tbubl 334.08K Ans. 0.3 0.05 y2 x2 0.4 0.2 y3 x3 1 1 y1 x1 y2 x2 T Tbubl x1 xi = 1 446 x1 x2 x3 Tdew Find x1 x2 x3 T 0.043 x 0.204 0.753 (c) P,T-flash calculation: z1 Guess: Tdew 347.4 K Ans. T z3 Tdew 2 1 Tbubl T 340.75 K 0.3 V z2 0.5 0.2 z1 z2 Use x from DEW P and y from BUBL P as initial guess. x1 ( 1 x2 ( 1 x3 ( 1 Given P y1 = x1 ( 1 x T) Psat ( 1 T) P y2 = x2 ( 2 x T) Psat ( 2 T) P y3 = x3 ( 3 x T) Psat ( 3 T) xi = 1 i i V) y1 V = z1 y2 V = z2 y3 V = z3 V) V) yi = 1 x1 x2 x3 y1 y2 y3 V 447 Find x1 x2 x3 y1 y2 y3 V 0.125 x 0.17 0.705 0.536 y 0.241 0.223 V 0.426 Ans. 12.23 Molar volumes & Antoine coefficients: Antoine coefficients: 74.05 V 40.73 18.07 P 101.33kPa A 14.3145 16.5785 16.3872 Psati T) ( B 2756.22 3638.27 3885.70 exp Ai Bi T K 228.060 C 239.500 230.170 kPa Ci 273.15 NRTL parameters: 0 0.3084 0.3084 0.5343 0 0.2994 0 0 b 222.64 184.70 631.05 0 253.88 0 bi j RT cal mol 0.5343 0.2994 1197.41 845.21 l exp i k 1 3 1 3 j 1 3 1 3 ( j T) i i j G ( j T) i ( j T) i (a) BUBL T calculation: x1 0.3 x2 0.4 x3 1 x1 x2 ( i T)G ( i T)x j j j ( x T) i exp j G ( i T)xl l l xk ( j T)G ( j T) k k x j G ( j T) i j l ( j T) i k G ( j T)xl l l G ( j T)xl l 448 Guess: T 300K y1 0.3 y2 0.3 y3 1 y1 y2 Given P y1 = x1 ( 1 x T) Psat ( 1 T) P y2 = x2 ( 2 x T) Psat ( 2 T) P y3 = x3 ( 3 x T) Psat ( 3 T) P= i xi ( i x T) Psat ( i T) y1 y2 y3 Tbubl 0.533 y 0.365 0.102 (b) DEW T calculation: Tbubl 334.6 K Find y1 y2 y3 T Ans. y1 Guess: 0.3 0.05 y2 0.4 y3 1 y1 y2 x1 x2 0.2 x3 1 x1 x2 T Tbubl Given P y1 = x1 ( 1 x T) Psat ( 1 T) P y2 = x2 ( 2 x T) Psat ( 2 T) P y3 = x3 ( 3 x T) Psat ( 3 T) xi = 1 i x1 x2 x3 Tdew 0.046 x 0.205 0.749 449 Find x1 x2 x3 T Tdew 347.5 K Ans. (c) P,T-flash calculation: z1 Guess: T z3 Tdew 2 1 Tbubl T 341.011 K 0.3 V z2 0.5 0.2 z1 z2 Use x from DEW P and y from BUBL P as initial guess. x1 ( 1 x2 ( 1 x3 ( 1 Given P y1 = x1 ( x T)Psat1 T) 1 ( P y2 = x2 ( x T)Psat2 T) 2 ( P y3 = x3 ( x T)Psat3 T) 3 ( xi = 1 i i V) y1 V = z1 V) y2 V = z2 V) y3 V = z3 yi = 1 x1 x2 x3 y1 y2 y3 V 0.133 x 0.173 0.694 Find x1 x2 x3 y1 y2 y3 V 0.537 y 0.238 0.225 V 0.414 Ans. 450 12.26 x1 0.4 x2 1 x1 V1 110 cm 3 mol V2 90 cm 3 mol VE x1 x2 x1 x2 45 x1 25 x2 cm 3 mol VE x1 x2 7.92 cm 3 mol By Eq. (12.27): V x1 x2 VE x1 x2 x1 V1 x2 V2 V x1 x2 105.92 cm 3 mol By Eqs. (11.15) & (11.16): Vbar1 V x 1 x2 d x2 V x1 x2 dx1 Vbar1 cm 190.28 mol 49.68 cm mol 3 3 Ans. Vbar2 Vbar2 V x1 x2 x1 d dx1 V x1 x2 Check by Eq. (11.11): V x1 Vbar1 3 x2 Vbar2 V 3 105.92 cm mol 3 OK 12.27 V1 cm 58.63 mol V2 3 cm 118.46 mol moles1 750 cm V1 moles2 1500 cm V2 3 moles moles1 moles2 moles 25.455 mol x1 moles1 moles x1 0.503 cm mol 3 x2 1 x1 cm 0.256 mol 3 VE x1 x2 1.026 0.220 x1 x2 VE By Eq. (12.27), V VE x1 V1 451 x2 V 2 V 88.136 cm 3 mol Vtotal V moles Vtotal 2243 cm 3 Ans. For an ideal solution, Eq. (11.81) applies: Vtotal x1 V 1 x2 V2 moles Vtotal 2250 cm 3 Ans. 12.28 LiCl.2H2O ---> Li + 1/2 Cl2 + 2 H2 + O2 (1) Li + 1/2 Cl2 + 10 H2O ---> LiCl(10 H2O) (2) 2(H2 + 1/2 O2 ---> H2O) (3) -------------------------------------------------------------------LiCl.2H2O + 8 H2O(l) ---> LiCl(10 H2O) H1 ( 1012650)J (Table C.4) H2 H3 441579 J 2 ( 285830 J) (Pg. 457) (Table C.4) H H1 H2 H3 H 589 J (On the basis of 1 mol of solute) Since there are 11 moles of solution per mole of solute, the result on the basis of 1 mol of solution is H 11 53.55 J Ans. 12.29 2(HCl + 2.25 H2O -----> HCl(2.25 H2O)) (1) HCl(4.5 H2O) -----> HCl + 4.5 H2O (2) ---------------------------------------------HCl(4.5 H2O) + HCl -----> 2 HCl(2.25 H2O) H1 2 ( 50.6 kJ) (Fig. 12.14 @ n=2.25) H2 62 kJ (Fig. 12.14 @ n=4.5 with sign change) H H1 H2 H 39.2 kJ Ans. 452 12.30 Calculate moles of LiCl and H2O in original solution: nLiCl nLiCl 0.1 125 42.39 kmol nH2O nH2O n'LiCl nH2O nLiCl 0.9 125 18.015 kmol 3 0.295 kmol 6.245 10 mol 20 kmol 42.39 21.18 Moles of LiCl added: n'LiCl 0.472 kmol Mole ratio, original solution: Mole ratio, final solution: nH2O nLiCl n'LiCl 8.15 nLiCl n'LiCl 0.7667 kmol 0.2949(LiCl(21.18 H2O) ---> LiCl + 21.18 H2O) (1) 0.7667(LiCl + 8.15 H2O ---> LiCl(8.15 H2O)) (2) --------------------------------------------------------------------------------------0.2949 LiCL(21.18 H2O) + 0.4718 LiCl ---> 0.7667 LiCl(8.145 H2O) H1 nLiCl 35 kJ mol (Fig. 12.14, n=21.18) H2 nLiCl H1 n'LiCl Q 32 kJ mol (Fig. 12.14, n=8.15) Q H2 14213 kJ Ans. 12.31 Basis: 1 mole of 20% LiCl solution entering the process. Assume 3 steps in the process: 1. Heat M1 moles of water from 10 C to 25 C 2. Unmix 1 mole (0.8 moles water + 0.2 moles LiCl) of 20 % LiCl solution 3. Mix (M1 + 0.8) moles of water and 0.2 moles of LiCl 453 Step 1: From Steam Tables H1 H1 104.8 1.132 kJ kg 41.99 kJ kg 18.015 kg kmol kJ mol Step 2: From Fig. 12.14 with n = 4 moles H2O/mole solute: H2 25.5 kJ mol Step 3: Guess M1 and find H3 solution from Figure 12.14. Calculate H for process. Continue to guess M1 until H =0 for adiabatic process. M1 1.3 mol n3 n3 H H x 0.8 mol 10.5 M1 0.2 mol H3 33.16 kJ mol M1 H1 0.061 kJ 0.2 mol M1 1 mol 0.2 mol H2 0.2 mol H3 Close enough x 0.087 Ans. 12.32 H2O @ 5 C -----> H2O @ 25 C (1) LiCl(3 H2O) -----> LiCl + 3 H2O (2) LiCl + 4 H2O -----> LiCl(4 H2O) (3) -------------------------------------------------------------------------H2O @ 5 C + LiCl(3 H2O) -----> LiCl(4 H2O) H1 H2 H3 H 104.8 kJ kg kJ mol 21.01 kJ gm 18.015 kg mol H1 1.509 kJ mol 20.756 From p. 457 ( H LiCl(s) - H LiCl in 3 mol H2O) From Figure 12.14 H3 0.2 mol 454 25.5 H1 kJ mol H2 H 646.905 J Ans. 12.33 (a) LiCl + 4 H2O -----> LiCl(4H2O) H 25.5 kJ From Figure 12.14 mol 0.2 mol H 5.1 kJ Ans. (b) LiCl(3 H2O) -----> LiCl + 3 H2O (1) LiCl + 4 H2O -----> LiCl(4 H2O) (2) ----------------------------------------------------LiCl(3 H2O) + H2O -----> LiCl(4 H2O) H1 20.756 kJ mol From p. 457 ( H LiCl(s) - H LiCl in 3 mol H2O) H2 25.5 kJ From Figure 12.14 mol H 0.2 mol H1 H2 H 0.949 kJ Ans. (c) LiCl*H2O -----> Li +1/2 Cl2 + H2 + 1/2 O2 (1) H2 + 1/2 O2 -----> H2O (2) Li + 1/2 Cl2 -----> LiCl (3) LiCl + 4 H2O -----> LiCl(4 H2O) (4) ---------------------------------------------------------------------LiCl*H2O + 3 H2O -----> LiCl(4 H2O) H1 H2 H3 H4 H (d) 712.58 kJ mol kJ mol kJ mol From p. 457 for LiCl.H2O From Table C.4 Hf H2O(l) From p. 457 for LiCl 285.83 408.61 25.5 0.2 mol kJ mol H1 From Figure 12.14 H2 H3 H4 H 1.472 kJ Ans. LiCl + 4 H2O -----> LiCl(4 H2O) (1) 4/9 (LiCl(9 H2O) -----> LiCl + 9 H2O) (2) --------------------------------------------------------------5/9 LiCl + 4/9 LiCl(9 H2O) -----> LiCl(4 H2O) 455 H1 H2 H (e) 25.5 4 9 kJ mol kJ mol H1 From Figure 12.14 ( ) 32.4 From Figure 12.14 0.2 mol H2 H 2.22 kJ Ans. 5/6 (LiCl(3 H2O) -----> LiCl + 3 H2O) (1) 1/6 (LiCl(9 H2O) -----> LiCl + 9 H2O) (2) LiCl + 4 H2O -----> LiCl(4 H2O) (3) -----------------------------------------------------------------------5/6 LiCl(3 H2O) + 1/6 LiCl(9 H2O) -----> LiCl(4 H2O) H1 H2 H3 kJ 5 ( 20.756) mol 6 kJ 1 ( ) 32.4 mol 6 25.5 kJ mol H1 From p. 457 ( H LiCl(s) - H LiCl in 3 mol H2O) From Figure 12.14 From Figure 12.14 H (f) 0.2 mol H2 H3 H 0.561 kJ Ans. 5/8 (LiCl*H2O -----> Li +1/2 Cl2 + H2 + 1/2 O2) (1) 5/8 (H2 + 1/2 O2 -----> H2O) (2) 3/8 (LiCl(9 H2O) -----> LiCl + 9 H2O) (3) 5/8 (Li + 1/2 Cl2 -----> LiCl (4) LiCl + 4 H2O -----> LiCl(4 H2O) (5) ---------------------------------------------------------------------------------------5/8 LiCl*H2O + 3/8 LiCl(9 H2O) -----> LiCl(4 H2O) kJ 5 ( 712.58) mol 8 5 8 H1 H2 From p. 457 for LiCl.H2O From Table C.4 Hf H2O(l) ( 285.83) kJ mol 456 H3 kJ 3 ( 32.4) mol 8 From Figure 12.14 H4 5 8 ( 408.61) kJ mol From p. 457 for LiCl H5 25.5 kJ mol From Figure 12.14 H 0.2 mol H1 H2 H3 H4 H5 H 0.403 kJ Ans. 12.34 BASIS: 1 second, during which the following are mixed: (1) 12 kg hydrated (6 H2O) copper nitrate (2) 15 kg H2O n1 12 kmol 295.61 sec 0.041 kmol sec n2 15 kmol 18.015 sec n1 n2 6 n1 0.833 n2 kmol sec 26.51 Mole ratio, final solution: n1 6(H2 + 1/2 O2 ---> H2O(l)) Cu + N2 + 3 O2 ---> Cu(NO3)2 (1) (2) Cu(NO3)2.6H2O ---> Cu + N2 + 6 O2 + 6 H2 (3) Cu(NO3)2 + 20.51 H2O ---> Cu(NO3)2(20.51 H2O) (4) -----------------------------------------------------------------------------------------------Cu(NO3)2.6H2O + 14.51 H2O(l) ---> Cu(NO3)2(20.51 H2O) H1 6 ( 285.83 kJ) (Table C.4) H2 302.9 kJ H3 ( 2110.8 kJ) H4 47.84 kJ H H1 H2 H3 H4 H 45.08 kJ 457 This value is for 1 mol of the hydrated copper nitrate. On the basis of 1 second, kJ H Ans. Q 1830 Q n 1 sec mol 12.35 LiCl.3H2O ---> Li + 1/2 Cl2 + 3H2 + 3/2 O2 (1) 3(H2 + 1/2 O2 ---> H2O(l)) (2) 2(Li + 1/2 Cl2 + 5 H2O ---> LiCl(5H2O)) (3) LiCl(7H2O) ---> Li + 1/2 Cl2 + 7 H2O (4) ------------------------------------------------------------------------------LiCl(7H2O) + LiCl.3H2O ---> 2 LiCl(5H2O) H1 1311.3 kJ H2 3 ( 285.83 kJ) (Table C.4) H3 2 ( 436.805 kJ) H4 ( 439.288 kJ) (Pg. 457) H H1 H2 H3 H4 Ans. H 19.488 kJ Q H Q 19.488 kJ 12.36 Li + 1/2 Cl2 + (n+2)H2O ---> LiCl(n+2 H2O) (1) 2(H2 + 1/2 O2 ---> H2O) (2) LiCl.2H2O ---> Li + 1/2 Cl2 + 2H2 + O2 (3) -------------------------------------------------------------------------------------LiCl.2H2O + n H2O ---> LiCl(n+2 H2O) H2 2 ( 285.83 kJ) H3 1012.65 kJ (Table C.4) Since the process is isothermal, H= H1 H2 H3 Since it is also adiabatic, H= 0 Therefore, H1 H2 H3 H1 440.99 kJ Interpolation in the table on pg. 457 shows that the LiCl is dissolved in 8.878 mol H2O. xLiCl 1 9.878 xLiCl 0.1012 Ans. 458 12.37 Data: 10 15 20 25 n 50 100 300 500 1000 862.74 867.85 870.06 871.07 Hf 872.91 kJ 873.82 874.79 875.13 875.54 Hf HfCaCl2 Htilde 795.8 kJ Ca + Cl2 + n H2O ---> CaCl2(n H2O) CaCl2(s) ---> Ca + Cl2 -------------------------------------------CaCl2(s) + n H2O ---> CaCl2(n H2O) From Table C.4: i 1 rows ( n) 65 HfCaCl2 70 Hf i HfCaCl2 kJ 75 80 10 100 ni 459 1 10 3 12.38 CaCl2 ---> Ca + Cl2 (1) 2(Ca + Cl2 + 12.5 H2O ---> CaCl2(12.5 H2O) (2) CaCl2(25 H2O) ---> Ca + Cl2 + 25 H2O (3) -----------------------------------------------------------------------------------CaCl2(25 H2O) + CaCl2 ---> 2 CaCl2(12.5 H2O) H1 795.8 kJ (Table C.4) H2 2 ( 865.295 kJ) H3 871.07 kJ H H1 H2 H3 Q H Q 63.72 kJ Ans. 12.39 The process may be considered in two steps: Mix at 25 degC, then heat/cool solution to the final temperature. The two steps together are adiabatic and the overall enthalpy change is 0. Calculate moles H2O needed to form solution: 85 18.015 n 15 110.986 n 34.911 Moles of H2O per mol CaCl2 in final solution. Moles of water added per mole of CaCl2.6H2O: n 6 28.911 Basis: 1 mol of Cacl2.6H2O dissolved CaCl2.6H2O(s) ---> Ca + Cl2 + 6 H2 + 3 O2 (1) Ca + Cl2 + 34.991 H2O --->CaCl2(34.911 H2O) (2) 6(H2 + 1/2 O2 ---> H2O) (3) --------------------------------------------------------------------------------------CaCl2.6H2O + 28.911 H2O ---> CaCl2(34.911 H2O) H1 2607.9 kJ H3 6 ( 285.83 kJ) (Table C.4) H2 871.8 kJ (Pb. 12.37) H298 H1 H2 H3 for reaction at 25 degC H298 21.12 kJ msoln ( 110.986 34.911 18.015)gm msoln 739.908 gm 460 CP 3.28 kJ kg degC H298 CP T = 0 T H298 msoln CP T 8.702 degC T 25 degC T T 16.298 degC Ans. 12.43 m1 150 lb (H2SO4) m2 350 lb (25% soln.) H1 8 BTU lbm H2 23 BTU lbm (Fig. 12.17) 100 % m1 m1 25 % m2 m2 47.5 % (Final soln.) m3 m1 m2 H3 90 BTU lbm (Fig. 12.17) Q m3 H3 m1 H1 m2 H2 Q 38150 BTU Ans. 12.44 Enthalpies from Fig. 12.17. x1 0.5 x2 1 x1 H 69 BTU lbm (50 % soln) H1 20 BTU lbm (pure H2SO4) H2 108 BTU lbm (pure H2O) HE H x1 H1 x2 H2 HE 133 BTU Ans. lbm 12.45 (a) m1 400 lbm (35% soln. at 130 degF) m2 175 lbm (10% soln. at 200 degF) H1 100 BTU lbm H2 152 BTU lbm (Fig. 12.19) 35 % m1 m1 10 % m2 m2 27.39 % (Final soln) m3 m1 m2 H3 41 BTU lbm (Fig. 12.19) 461 Q m3 H3 m1 H1 m2 H2 Q 43025 BTU Ans. (b) Adiabatic process, Q = 0. H3 m1 H1 m2 H2 m3 H3 115.826 BTU lbm From Fig. 12.19 the final soln. with this enthalpy has a temperature of about 165 degF. 12.46 m1 25 lbm sec (feed rate) x1 0.2 H1 24 BTU lbm (Fig. 12.17 at 20% & 80 degF) H2 55 BTU lbm (Fig. 12.17 at 70% and 217 degF) [Slight extrapolation] x2 0.7 H3 1157.7 BTU lbm (Table F.4, 1.5(psia) & 217 degF] m2 x1 m1 x2 m2 H2 m2 7.143 lbm sec m3 m1 m2 m3 17.857 lbm sec Q m3 H3 m1 H1 Q 20880 BTU sec Ans. 12.47 Mix m1 lbm NaOH with m2 lbm 10% soln. @ 68 degF. BASIS: m2 1 lbm x3 0.35 x2 0.1 m1 1 lbm (guess) m3 m1 m2 Given m1 m2 = m3 m1 m1 x2 m2 = x3 m3 0.385 lbm m1 m3 Find m1 m3 m3 1.385 lbm 462 From Example 12.8 and Fig. 12.19 H1 478.7 BTU lbm m2 H2 H2 43 BTU lbm BTU lbm H3 m1 H1 m3 H3 164 From Fig. 12.19 at 35% and this enthalpy, we find the temperature to be about 205 degF. 12.48 First react 1 mol SO3(l) with 1 mol H2O(l) to form 1 mol H2SO4(l): SO3(l) + H2O(l) ---> H2SO4(l) With data from Table C.4: H298 [ 813989 ( 441040 285830) ]J H298 mH2SO4 0.5 8.712 10 J 4 Mix 1 mol or 98.08 gm H2SO4(l) with m gm H2O to form a 50% solution. mH2SO4 mH2O 98.08 gm msoln mH2SO4 msoln Data from Fig. 12.17: HH2SO4 HH2O Hsoln Hmix Hmix Q 0 BTU lbm [pure acid @ 77 degF (25 degC)] 45 BTU lbm BTU lbm [pure water @ 77 degF (25 degC)] 70 [50% soln. @ 140 degF (40 deg C)] mH2SO4 HH2SO4 mH2O HH2O msoln Hsoln 18.145 kg H298 msoln BTU lbm Q 283 BTU lbm Ans. Hmix 463 12.49 m1 140 lbm x1 0.15 m2 230 lbm x2 0.8 H1 65 BTU lb (Fig. 12.17 at 160 degF) H2 102 BTU lb (Fig. 12.17 at 100 degF) m3 m1 m2 x3 m1 x1 m2 x2 m3 x3 55.4 % Q 20000 BTU H3 Q m1 H1 m3 m2 H2 H3 92.9 BTU lbm From Fig. 12.17 find temperature about 118 degF 12.50 Initial solution (1) at 60 degF; Fig. 12.17: m1 1500 lbm x1 0.40 H1 98 BTU lbm Saturated steam at 1(atm); Table F.4: m3 m2 m1 m2 H2 1150.5 BTU lbm x3 m2 x1 m1 m1 m2 H3 m2 m1 H1 m2 H2 m3 m2 m2 125 lbm x3 m2 36.9 % H3 m2 2 BTU lbm The question now is whether this result is in agreement with the value read from Fig. 12.17 at 36.9% and 180 degF. It is close, but we make a second calculation: m2 120 lbm x3 m2 37% H3 m2 5.5 BTU lbm This is about as good a result as we can get. 464 12.51 Initial solution (1) at 80 degF; Fig. 12.17: m1 1 lbm x1 0.45 H1 95 BTU lbm Saturated steam at 40(psia); Table F.4: m3 m2 m1 m2 H2 1169.8 BTU lbm x3 m2 x1 m1 m1 m2 H3 m2 m1 H1 m2 H2 m3 m2 m2 0.05 lbm x3 m2 42.9 % H3 m2 34.8 BTU lbm The question now is whether this result is in agreement with the value read from Fig. 12.17 at 36.9% and 180 degF. It is close, but we make a second calculation: m2 0.048 lbm x3 m2 42.9 % H3 m2 37.1 BTU lbm This is about as good a result as we can get. 12.52 Initial solution (1) at 80 degF; Fig. 12.19: m1 1 lbm x1 0.40 H1 77 BTU lbm Saturated steam at 35(psia); Table F.4: H2 1161.1 BTU lbm x3 0.38 m2 x1 m1 x3 m1 m3 m1 m2 m3 1.053 lbm m2 0.053 lbm H3 m1 H1 m2 H2 m3 H3 131.2 BTU lbm We see from Fig. 12.19 that for this enthalpy at 38% the temperature is about 155 degF. 465 12.53 Read values for H, H1, & H2 from Fig. 12.17 at 100 degF: H 56 BTU lbm H1 8 BTU lbm H2 68 BTU lbm x1 0.35 x2 1 x1 H H x1 H 1 x2 H2 H 103 BTU lbm Ans. 12.54 BASIS: 1(lbm) of soln. Read values for H1 & H2 from Fig. 12.17 at 80 degF: H1 Q= 4 BTU lbm x1 H 1 H2 48 BTU lbm x1 0.4 x2 1 x1 H= H x2 H2 = 0 H x1 H 1 x2 H2 H 30.4 BTU lbm From Fig. 12.17, for a 40% soln. to have this enthalpy the temperature is well above 200 degF, probably about 250 degF. 12.55 Initial solution: x1 2 98.08 2 98.08 15 18.015 x1 0.421 Final solution: x2 3 98.08 3 98.08 14 18.015 x2 0.538 Data from Fig. 12.17 at 100 degF: HH2O 68 BTU lbm HH2SO4 9 BTU lbm H1 75 BTU lbm H2 101 BTU lbm 466 Unmix the initial solution: Hunmix Hunmix x1 HH2SO4 118.185 BTU lbm 1 x1 HH2O H1 React 1 mol SO3(g) with 1 mol H2O(l) to form 1 mol H2SO4(l). We neglect the effect of Ton the heat of reaction, taking the value at 100 degF equal to the value at 77 degF (25 degC) J mol J mol HfH2O HfSO3 Hrx 1.324 10 5 J HfSO3 HfH2SO4 Hrx 395720 HfH2O 285830 J mol 813989 HfH2SO4 mol Finally, mix the constituents to form the final solution: Hmix Q H2 x2 HH2SO4 1 x2 HH2O Hmix 137.231 BTU lbm Hunmix ( 98.08 2 1 lbmol Hrx Hmix ( 98.08 3 15 18.015)lb 14 18.015)lb Q 76809 BTU Ans. 12.56 Read values for H(x1=0.65), H1, & H2 from Fig. 12.17 at 77 degF: H 125 BTU lbm x2 H1 1 0 BTU lbm H2 H 45 BTU lbm x1 H 1 BTU lbm x1 0.65 x1 H x2 H2 Ans. H 140.8 467 From the intercepts of a tangent line drawn to the 77 degF curve of Fig. 12.17 at 65%, find the approximate values: Hbar1 136 BTU lbm Hbar2 103 BTU lbm Ans. 12.57 Graphical solution: If the mixing is adiabatic and water is added to bring the temperature to 140 degF, then the point on the H-x diagram of Fig. 12.17 representing the final solution is the intersection of the 140-degF isotherm with a straight line between points representing the 75 wt % solution at 140 degF and pure water at 40 degF. This intersection gives x3, the wt % of the final solution at 140 degF: x3 42 % m1 1 lb By a mass balance: x3 = 0.75 m1 m1 m2 m2 0.75 m1 x3 m1 m2 0.786 lbm Ans. 12.58 (a) m1 x1 25 lbm 0 m2 x2 40 lbm 1 m3 75 lbm x3 0.25 Enthalpy data from Fig. 12.17 at 120 degF: H1 88 BTU lbm H2 14 BTU lbm H3 7 BTU lbm m4 m1 m2 m3 m4 140 lbm x4 x1 m1 x2 m2 m4 x3 m3 x4 0.42 H4 63 BTU lbm (Fig. 12.17) Q m4 H4 m1 H1 m2 H2 m3 H3 Q 11055 BTU Ans. 468 (b) First step: m1 40 lb x1 1 H1 14 BTU lbm m2 75 lb x2 0.25 H2 7 BTU lbm m3 m1 m2 x3 x1 m1 x2 m2 m3 H3 Q m1 H1 m3 m2 H2 x3 0.511 H3 95.8 BTU lbm From Fig. 12.17 at this enthalpy and wt % the temperature is about 100 degF. 12.59 BASIS: 1 mol NaOH neutralized. For following reaction; data from Table C.4: NaOH(s) + HCl(g) ---> NaCl(s) + H2O(l) H298 [ 411153 285830 5 ( 425609 92307) ]J H298 1.791 10 J NaOH(s) + HCl(g) ---> NaCl(s) + H2O(l) (1) NaOH(inf H2O) ---> NaOH(s) + inf H2O (2) HCl(9 H2O) ---> HCl(g) + 9 H2O(l) (3) NaCl(s) + inf H2O ---> NaCl(inf H2O) (4) ---------------------------------------------------------------------------------------NaOH(inf H2O) + HCl(9 H2O) ---> NaCl(inf H2O) H1 H298 H2 44.50 kJ H3 68.50 kJ H4 3.88 kJ H H1 H2 H3 H4 Q H Q 62187 J Ans. 469 12.60 First, find heat of solution of 1 mole of NaOH in 9 moles of H2O at 25 degC (77 degF). Weight % of 10 mol-% NaOH soln: x1 HH2O Hsoln 1 40.00 1 40.00 9 18.015 45 BTU lbm BTU lbm BTU lbm x1 19.789 % (Table F.3, sat. liq. at 77 degF) 35 (Fig. 12.19 at x1 and 77 degF) HNaOH 478.7 [Ex. 12.8 (p. 468 at 68 degF] Correct NaOH enthalpy to 77 degF with heat capacity at 72.5 degF (295.65 K); Table C.2: T 295.65 K molwt 3 40.00 gm mol Cp R molwt 0.121 16.316 10 K T Cp 0.245 BTU lbm rankine BTU lbm HNaOH HNaOH Cp ( 77 68)rankine HNaOH 480.91 H H Hsoln 0.224 x1 HNaOH kJ gm 1 x1 HH2O This is for 1 gm of SOLUTION. However, for 1 mol of NaOH, it becomes: H H x1 molwt H 45.259 kJ mol 470 Now, on the BASIS of 1 mol of HCl neutralized: NaOH(s) + HCl(g) ---> NaCl(s) + H2O(l) (1) HCl(inf H2O) ---> HCl(g) + inf H2O (2) NaOH(9 H2O) ---> NaOH(s) + 9 H2O (3) NaCl + inf H2O ---> NaCl(inf H2O) (4) --------------------------------------------------------------------------------------HCl(inf H2O) + NaOH(9 H2O) ---> NaCl(inf H2O) H1 H2 H3 H4 H 179067 J 74.5 kJ 45.259 kJ 3.88 kJ H1 H2 (Pb. 12.59) (Fig. 12.14 with sign change) (See above; note sign change) (given) H3 H3 Q H Q 14049 J Ans. 12.61 Note: The derivation of the equations in part a) can be found in Section B of this manual. 0.1 0.2 0.3 0.4 0.5 x1 0.6 0.7 0.8 0.85 0.9 0.95 HE 73.27 144.21 208.64 262.83 302.84 323.31 320.98 279.58 237.25 178.87 100.71 kJ kg HE x1 x2 x2 1 x1 H 471 In order to take the necessary derivatives of H, we will fit the data to a HE 3 2 third order polynomial of the form H = = a bx.1 c x1 d x1 . x1 x2 Use the Mathcad regress function to find the parameters a, b, c and d. w w n a b c d 2 3 H w w n regress x1 kJ kg 3 3 3 735.28 824.518 195.199 914.579 3 a b c d kJ kg H x1 a b x1 c x1 d x1 Using the equations given in the problem statement and taking the derivatives of the polynomial analytically: HEbar1 x1 1 x1 2 H x1 x1 b 2 c x1 3 d x1 2 kJ kg kJ kg HEbar2 x1 x1 2 H x1 1 x1 b 2 c x1 3 d x1 2 472 0 500 1000 (kJ/kg) 1500 2000 2500 0 0.2 H/x1x2 HEbar1 HEbar2 0.4 x1 0.6 0.8 12.62 Note: This problem uses data from problem 12.61 0.1 0.2 0.3 0.4 0.5 x1 0.6 0.7 0.8 0.85 0.9 0.95 HE 73.27 144.21 208.64 262.83 302.84 323.31 320.98 279.58 237.25 178.87 100.71 kJ kg x2 1 x1 H HE x1 x2 473 Fit a third order polynomial of the form HE x1 x2 = a bx.1 c x1 2 d x1 3 . Use the Mathcad regress function to find the parameters a, b, c and d. w w n a b c d regress x1 H w w n kJ kg 3 3 3 735.28 824.518 195.199 914.579 3 a b c d By the equations given in problem 12.61 H x1 a H b x1 c x1 2 d x1 3 kJ kg H x1 Hbar1 x1 x1 x1 1 1 x1 2 2 x1 x1 x1 b 2 c x1 3 d x1 2 H kJ kg Hbar2 x1 x1 H x1 1 x1 b 2 c x1 3 d x1 2 kJ kg At time , let: x1 = mass fraftion of H2SO4 in tank m = total mass of 90% H2SO4 added up to time H = enthalpy of H2SO4 solution in tank at 25 C H2 = enthalpy of pure H2O at 25 C H1 = enthalpy of pure H2SO4 at 25 C H3 = enthalpy of 90% H2SO4 at 25 C Material and energy balances are then written as: x1 ( 4000 Q= m) = 0.9m m) H Solving for m: m x1 ( 4000kg)x1 0.9 x1 Eq. (A) Ht = ( 4000 4000H2 m H3 474 x1 H1 x2 H2 and since T is constant at 25 C, we set H1 = H2 = 0 at this T, making H = H. The energy balance then becomes: Eq. (B) Q= ( 4000 m) H m H3 Since H = H Applying these equations to the overall process, for which: 6hr x1 0.5 H3 H Define quantities as a function of x 1 H( ) 0.9 H3 H 178.737 kJ kg kJ kg H( ) 0.5 303.265 Q x1 m x1 Qtx1 4000kg ( 4000kg) 1 x 0.9 x1 m x1 H x1 m x1 H3 m ( ) 5000 kg 0.5 6 4000kg m x1 H m x1 H3 Qt0.5) ( 1.836 10 kJ Since the heat transfer rate q is constant: q Qtx1 Q x1 q and x1 Eq. (C) The following is probably the most elegant solution to this problem, and it leads to the direct calculation of the required rates, dm r= d When 90% acid is added to the tank it undergoes an enthalpy change equal to: 0.9Hbar1+0.1Hbar2-H3, where Hbar1 and Hbar2 are the partial enthalpies of H2SO4 and H2O in the solution of mass fraction x1 existing in the tank at the instant of addition. This enthalpy change equals the heat required per kg of 90% acid to keep the temperature at 25 C. Thus, r x1 q 0.9 Hbar1 x1 0.1 Hbar2 x1 475 H3 Plot the rate as a function of time 1200 x1 0 0.01 0.5 1100 1000 r x1 kg hr 900 800 700 600 0 1 2 3 x1 hr 4 5 6 12.64 mdot1 20000 lb hr x1 0.8 T1 T2 T3 120degF H1 H2 H3 92 BTU lb Enthalpies from Fig. 12.17 x2 x3 0.0 40degF 7 BTU lb 70 BTU lb 0.5 140degF a) Use mass balances to find feed rate of cold water and product rate. Guess: Given mdot2 mdot1 mdot1 mdot2 = mdot3 mdot3 2mdot1 Total balance H2SO4 balance mdot1 x1 mdot2 x2 = mdot3 x3 476 mdot2 mdot3 Find mdot2 mdot3 mdot2 12000 lb mdot3 hr 32000 lb Ans. hr b) Apply an energy balance on the mixer Qdot mdot3 H3 mdot1 H1 mdot2 H2 Qdot 484000 BTU hr Since Qdot is negative, heat is removed from the mixer. c) For an adiabatic process, Qdot is zero. Solve the energy balance to find H3 H3 mdot1 H1 mdot2 H2 H3 54.875 BTU lb mdot3 From Fig. 12.17, this corresponds to a temperature of about 165 F 12.65 Let L = total moles of liquid at any point in time and Vdot = rate at which liquid boils and leaves the system as vapor. dL = Vdot dt d L x1 An unsteady state species balance on water yields: = y1 Vdot dt An unsteady state mole balance yields: Expanding the derivative gives: L dx1 dt dx1 dt dx1 dt dx1 dt dx1 x1 dL = Vdot y1 dt Substituting -Vdot for dL/dt: L x1 ( Vdot)= y1 Vdot Rearranging this equation gives: L = x1 = y1 = dL L y1 Vdot x1 dL dt Substituting -dL/dt for Vdot: L Eliminating dt and rearranging: y1 477 x1 At low concentrations y1 and x1 can be related by: y1 = Psat1 inf1 P x1 = K1 x1 dx1 K1 1 x1 where: dL L K1 = inf1 Psat1 P Substituting gives: = Integrating this equation yields: ln Lf L0 = 1 K1 1 ln x1f x10 where L0 and x10 are the initial conditions of the system For this problem the following values apply: L0 T 1mol x10 P 600 10 6 x1f 50 10 6 130degC 1atm inf1 5.8 Psat1 exp 16.3872 3885.70 T degC K1 kPa 230.170 Psat1 270.071 kPa K1 Psat1 inf1 P 15.459 Lf norg0 norg0 L0 exp L0 1 1 K1 x10 1 ln x1f x10 Lf norgf norgf 0.842 mole Lf 1 x1f 0.999 mole norg0 norgf 0.842 mole %lossorg norg0 %lossorg 15.744 % Ans. The water can be removed but almost 16% of the organic liquid will be removed with the water. 478 12.69 1 - Acetone 2- Methanol T ( 50 273.15) K For Wilson equation a12 161.88 cal mol a21 cal 583.11 mol 0.708 V1 74.05 V1 cm 3 mol a21 RT V2 40.73 cm 3 mol V2 12 V1 exp a12 RT ln inf1 12 21 V2 21 exp 21 0.733 ln 12 1 ln inf1 0.613 Ans. From p. 445 ln inf2 ln 21 1 12 ln inf2 0.603 Ans. From Fig. 12.9(b) For NRTL equation ln inf1 = 0.62 ln inf2 = 0.61 b12 12 184.70 cal mol b21 12 222.64 cal mol 0.3048 b12 RT 0.288 b21 21 RT 21 0.347 G12 exp 12 G12 21 0.916 12 exp G21 12 exp 21 G21 0.9 From p. 446 ln inf1 ln inf1 0.611 ln inf2 12 21 exp 21 ln inf2 0.600 Both estimates are in close agreement with the values from Fig. 12.9 (b) 479 12.71 Psat1 183.4kPa Psat2 96.7kPa x1 0.253 y1 0.456 P 139.1kPa Check whether or not the system is ideal using Raoult's Law (RL) PRL x1 Psat1 1 1 x1 Psat2 2 PRL 118.635 kPa Since PRL<P, and are not equal to 1. Therefore, we need a model for GE/RT. A two parameter model will work. From Margules Equation: GE = x1 x2 A21 x1 RT 2 A12 x2 Eq. (12.10a) Eq. (12.10b) ln 1 = x2 ln 2 = x1 Find 1 A12 A21 2 A21 2 A12 A12 x1 A21 x2 2 and 2 at x1=0.253 from the given data. 1 y1 P 1 x1 Psat1 1 1.367 1 2 y1 P x1 Psat2 2 1 1.048 Use the values of find A12 and A21. and 2 at x1=0.253 and Eqs. (12.10a) and (12.10b) to Guess: A12 0.5 A21 0.5 Given ln 1 = 1 ln 2 = x1 2 x1 2 A12 2 A21 A21 A12 x1 1 x1 Eq. (12.10a) Eq. (12.10b) A21 2 A12 A12 A12 A21 1 x1 Find A12 A21 0.644 A21 0.478 exp 1 2 x1 2 A12 2 A21 A21 A12 x1 1 x1 2 x1 exp x1 A21 2 A12 480 a) x1 0.5 y1 x1 1 x1 Psat1 P y1 0.743 Ans. P x1 1 x1 Psat1 1 x1 2 x1 Psat2 P 160.148 kPa Ans. b) 1inf exp A12 1inf Psat1 1inf 1.904 2inf exp A21 2inf 1.614 120 Since Psat2 12 120 3.612 121 Psat1 2inf Psat2 121 1.175 remains above a value of 1, an azeotrope is unlikely based on the assumption that the model of GE/RT is reliable. 12.72 P 108.6kPa x1 0.389 T ( 35 273.15) K Psat1 120.2kPa Psat2 73.9kPa Check whether or not the system is ideal using Raoult's Law (RL) PRL x1 Psat1 1 1 x1 Psat2 2 PRL 91.911 kPa Since PRL < P, and are not equal to 1. Therefore, we need a model for GE/RT. A one parameter model will work. Assume a model of the form: GE = A x1 x2 RT 2 1 = exp A x2 2 = exp A x.1 2 Since we have no y1 value, we must use the following equation to find A: P = x1 1 Psat1 x2 2 Psat2 481 Use the data to find the value of A Guess: A 1 Given P = x1 exp A 1 A x1 2 Psat1 1 x1 exp A x1 2 Psat2 A Find A) ( 0.677 1 x1 exp A 1 x1 1 x1 Psat1 P x1 2 2 x1 exp A x1 Ans. P 2 a) y1 b) P c) 1inf 120 y1 1 1inf 0.554 2 x1 Psat2 x1 1 x1 Psat1 exp ( A) 1inf Psat1 x1 110.228 kPa 2inf Ans. 1.968 1.968 2inf exp ( A) Psat1 2inf Psat2 Psat2 120 3.201 121 121 0.826 Since 12 ranges from less than 1 to greater than 1 an azeotrope is likely based on the assumption that our model is reliable. 482 Chapter 13 - Section A - Mathcad Solutions Note: For the following problems the variable kelvin is used for the SI unit of absolute temperature so as not to conflict with the variable K used for the equilibrium constant 13.4 H2(g) + CO2(g) = H2O(g) + CO(g) = i i= 1 1 1 1= 0 1 2 n0 = 1 1= 2 By Eq. (13.5). yH = yCO = 2 2 yH2O = yCO = 2 By Eq. (A) and with data from Example 13.13 at 1000 K: T 1000 kelvin G 1 2 ( 395790) RT 2 1 2 ln J mol 1 2 ( 192420 2 200240) J mol 2 2 ln 2 Guess: Given d d e e 0.5 e = 0 G J mol e Find e e 0.45308 0.3 0.31 0.6 2.082 2.084 G 10 5 2.086 2.088 0.2 0.3 0.4 0.5 0.6 483 13.5 (a) H2(g) + CO2(g) = H2O(g) + CO(g) = i i= 1 1 1 1= 0 1 2 n0 = 1 1= 2 By Eq. (13.5). yH = yCO = 2 2 yH2O = yCO = 2 By Eq. (A) and with data from Example 13.13 at 1100 K: T 1100 kelvin G 1 2 ( 395960) RT 2 1 2 ln J mol 1 2 ( 187000 2 209110) J mol 2 2 ln 2 Guess: Given d d e e 0.5 e = 0 G J mol e Find e e 0.502 Ans. 0.35 0.36 0.65 2.102 2.103 2.104 G 10 5 2.105 2.106 2.107 0.3 0.35 0.4 0.45 0.5 0.55 0.6 0.65 484 (b) = H2(g) + CO2(g) = H2O(g) + CO(g) i= i 1 1 1 1= 0 1 2 n0 = 1 1= 2 By Eq. (13.5), yH = yCO = 2 2 yH2O = yCO = 2 By Eq. (A) and with data from Example 13.13 at 1200 K: T 1200 kelvin G 1 2 ( 396020) J RT 2 1 2 ln mol 1 2 ( 181380 2 217830) J mol 2 2 ln 2 Guess: Given d d e e 0.1 e = 0 G J mol e Find e e 0.53988 Ans. 0.4 0.41 0.7 2.121 2.122 2.123 G 10 5 2.124 2.125 2.126 2.127 0.35 0.4 0.45 0.5 0.55 0.6 0.65 0.7 485 (c) = H2(g) + CO2(g) = H2O(g) + CO(g) i= i 1 1 1 1= 0 1 2 n0 = 1 1= 2 By Eq, (13.5), yH = yCO = 2 2 yH2O = yCO = 2 By Eq. (A) and with data from Example 13.13 at 1300 K: T 1300 kelvin G 1 2 ( 396080) RT 2 1 2 ln J mol 1 2 ( 175720 2 226530) J mol 2 2 ln 2 Guess: d d e e 0.6 J mol Given G e = 0 e Find e e 0.57088 Ans. 0.4 0.41 0.7 2.14 2.142 G 10 5 2.144 2.146 2.148 0.35 0.4 0.45 0.5 0.55 0.6 0.65 0.7 486 13.6 H2(g) + CO2(g) = H2O(g) + CO(g) = i i= 1 1 1 1= 0 1 2 n0 = 1 1= 2 By Eq, (13.5), yH = yCO = 2 2 yH2O = yCO = 2 With data from Example 13.13, the following vectors represent values for Parts (a) through (d): 1000 T 1100 1200 1300 kelvin 3130 G 150 3190 6170 J mol Combining Eqs. (13.5), (13.11a), and (13.28) gives 2 2 1 2 2 1 2 = 1 2 = K = exp G RT 0.4531 exp G RT 0.5021 1 Ans. 0.5399 0.5709 13.11 4HCl(g) + O2(g) = 2H2O(g) + 2Cl(g) = 1 H298 n0 = 6 114408 J mol T 773.15 kelvin J mol T0 298.15 kelvin G298 75948 487 The following vectors represent the species of the reaction in the order in which they appear: 4 1 2 2 end rowsA) ( i Ai i 3.156 A 3.639 3.470 4.442 B 0.623 0.506 1.450 0.089 10 3 0.151 D 0.227 0.121 0.344 10 5 i 1 end i Bi i A B D i i Di A 0.439 B 8 10 5 C 0 D 8.23 10 4 G H298 T H298 G298 T0 R IDCPH T0 T A B C D R T IDCPS T0 T A B C D G 1.267 10 4 J mol K 7.18041 K exp G RT By Eq. (13.5) 1 6 yHCl = yH2O = 5 6 2 6 4 yO2 = yCl2 = 2 6 Apply Eq. (13.28); 0.5 4 (guess) Find Given 5 6 2 5 4 6 1 4 = 2K 0.793 yHCl yO2 1 6 yH2O 2 6 yCl2 2 6 yHCl 0.3508 yO2 0.0397 yH2O 488 0.3048 yCl2 0.3048 Ans. 13.12 N2(g) + C2H2(g) = 2HCN(g) = 0 n0 = 2 This is the reaction of Pb. 4.21(x). From the answers for Pbs. 4.21(x), 4.22(x), and 13.7(x), find the following values: H298 42720 J mol G298 39430 J mol A 0.060 B 0.173 10 3 C 0 D 0.191 10 5 T 923.15 kelvin T0 298.15 kelvin G H298 H298 G298 T0 R IDCPH T0 T A B C D R T IDCPS T0 T A B C D T G 3.242 10 4 J mol K exp G RT K 0.01464 By Eq. (13.5), yN2 = 1 2 yC2H4 = 1 2 yHCN = 2e 2 = By Eq. (13.28), 0.5 (guess) Given 1 2 2 1 2 = K Find 0.057 yN2 yC2H4 1 2 yHCN yN2 0.4715 yC2H4 0.4715 yHCN 0.057 Ans. Given the assumption of ideal gases, P has no effect on the equilibrium composition. 489 13.13 CH3CHO(g) + H2(g) = C2H5OH(g) = 1 n0 = 2.5 This is the reaction of Pb. 4.21(r). From the answers for Pbs. 4.21(r), 4.22(r), and 13.7(r), find the following values: H298 68910 J mol G298 39630 J mol 6 A 1.424 B 1.601 10 3 C 0.156 10 D 0.083 10 5 T 623.15 kelvin T0 298.15 kelvin G H298 H298 G298 T0 R IDCPH T0 T A B C D R T IDCPS T0 T A B C D T G 6.787 10 3 J mol 1 2.5 K exp G RT 1.5 2.5 K 3.7064 By Eq. (13.5), yCH3CHO = yH2 = yC2H5OH = 2.5 By Eq. (13.28), Given 0.5 2.5 (guess) Find 1 1.5 = 3K 0.818 yCH3CHO 1 2.5 yH2 1.5 2.5 yC2H5OH 2.5 yCH3CHO 0.108 yH2 0.4053 yC2H5OH 0.4867 Ans. If the pressure is reduced to 1 bar, Given 2.5 1 1.5 = 1K Find 0.633 yCH3CHO 1 2.5 yH2 1.5 2.5 yC2H5OH 2.5 yCH3CHO 0.1968 yH2 0.4645 490 yC2H5OH 0.3387 Ans. 13.14 C6H5CH:CH2(g) + H2(g) = C6H5.C2H5(g) = 1 n0 = 2.5 This is the REVERSE reaction of Pb. 4.21(y). From the answers for Pbs. 4.21(y), 4.22(y), and 13.7(y) WITH OPPOSITE SIGNS, find the following values: H298 117440 J mol G298 3 83010 J mol 6 A 4.175 B 4.766 10 C 1.814 10 D 0.083 10 5 T 923.15 kelvin T0 298.15 kelvin G H298 T H298 G298 T0 R IDCPH T0 T A B C D R T IDCPS T0 T A B C D G 2.398 10 3 J mol K exp G RT 1 2.5 K 1.36672 By Eq. (13.5), yC6H5CHCH2 = yH2 = 1.5 2.5 yC6H5C2H5 = 2.5 By Eq. (13.28), 0.5 (guess) Given 2.5 1 1.5 = 1.0133 K Find 0.418 yC6H5CHCH2 1 2.5 yH2 1.5 2.5 yC6H5C2H5 2.5 yC6H5CHCH2 0.2794 yH2 0.5196 yC6H5C2H5 0.201 Ans. 491 13.15 Basis: 1 mole of gas entering, containing 0.15 mol SO2, 0.20 mol O2, and 0.65 mol N2. SO2 + 0.5O2 = SO3 = 0.5 n0 = 1 By Eq. (13.5), ySO2 = 0.15 1 0.5 yO2 = 0.20 1 0.5 0.5 ySO3 = 1 0.5 From data in Table C.4, H298 98890 J mol G298 70866 J mol The following vectors represent the species of the reaction in the order in which they appear: 1 0.5 1 end rowsA) ( i Ai i 5.699 A 3.639 8.060 B 0.801 0.506 10 1.056 3 1.015 D 0.227 10 2.028 5 i B i 1 end i Bi A D i i Di A 0.5415 B 2 10 6 C 0 D 8.995 10 4 T 753.15 kelvin T0 298.15 kelvin G H298 H298 G298 T0 R IDCPH T0 T A B C D R T IDCPS T0 T A B C D T G 2.804 10 4 J mol K exp 0.1 G RT (guess) K 88.03675 By Eq. (13.28), Given 1 0.15 0.5 0.2 0.5 0.5 0.5 492 = K Find 0.1455 By Eq. (13.4), nSO3 = = 0.1455 By Eq. (4.18), H753 H298 R IDCPH T0 T A B C D H753 98353 J mol Q H753 Q 14314 J Ans. mol 13.16 C3H8(g) = C2H4(g) + CH4(g) = 1 Basis: 1 mole C3H8 feed. By Eq. (13.4) nC3H8 = 1 Fractional conversion of C3H8 = n0 nC3H8 n0 = 1 1 1 = By Eq. (13.5), yC3H8 = 1 1 yC2H4 = 1 yCH4 = 1 From data in Table C.4, H298 82670 J mol G298 42290 J mol The following vectors represent the species of the reaction in the order in which they appear: 1 1 1 end rowsA) ( i Ai i 1.213 A 1.424 1.702 B 28.785 14.394 10 9.081 3 8.824 C 4.392 10 2.164 6 i 1 end i Bi i A B C i i Ci A 1.913 B 5.31 10 3 C 2.268 10 6 D 0 (a) T 625 kelvin T0 298.15 kelvin 493 G H298 H298 G298 T0 R IDCPH T0 T A B C D R T IDCPS T0 T A B C D T G 2187.9 J mol K exp G RT (guess) K 1.52356 By Eq. (13.28), 2 0.5 Given 1 1 = K Find 0.777 This value of epsilon IS the fractional conversion. Ans. 2 (b) 0.85 K 1 4972.3 1 J mol K 2.604 G R T ln ( K) G Ans. The problem now is to find the T which generates this value. It is not difficult to find T by trial. This leads to the value: T = 646.8 K Ans. 13.17 C2H6(g) = H2(g) + C2H4(g) = 1 Basis: 1 mole entering C2H6 + 0.5 mol H2O. n0 = 1.5 By Eq. (13.5), yC2H6 = 1 1.5 yH = 1.5 yC2H4 = 1.5 From data in Table C.4, H298 136330 J mol G298 100315 J mol The following vectors represent the species of the reaction in the order in which they appear: 494 1 1 1 A 1.131 3.249 1.424 5.561 C 0.0 4.392 10 6 19.225 B 0.422 14.394 0.0 D 0.083 10 0.0 5 10 3 end rowsA) ( i Ai i i 1 end i Bi A B i C i i Ci D i i Di A 3.542 B 4.409 10 3 C 1.169 10 6 D 8.3 10 3 T 1100 kelvin T0 298.15 kelvin G H298 T H298 G298 T0 R IDCPH T0 T A B C D R T IDCPS T0 T A B C D G 5.429 10 3 J mol 0.5 2 K exp G RT K 1.81048 By Eq. (13.28), (guess) Given 1.5 1 = K Find 0.83505 By Eq. (13.4), nC2H6 = 1 nH2 = nC2H4 = n= 1 yC2H6 1 1 yH2 1 yC2H4 1 yC2H6 0.0899 yC2H4 0.4551 yH2 0.4551 Ans. 495 13.18 C2H5CH:CH2(g) = CH2:CHCH:CH2(g) + H2(g) (1) (2) (3) = 1 Number the species as shown. Basis is 1 mol species 1 + x mol steam. n0 = 1 x y1 = 1 1 x y2 = y3 = 1 x By Eq. (13.5), = 0.10 From data in Table C.4, H298 109780 J mol G298 79455 J mol The following vectors represent the species of the reaction in the order in which they appear: 1 1 1 A 1.967 2.734 3.249 9.873 C 8.882 10 0.0 6 31.630 B 26.786 10 0.422 3 0.0 D 0.0 0.083 10 5 end A rows ( A) i Ai i i B i 1 end i Bi C i i Ci D i i Di A T G 4.016 950 kelvin H298 B 4.422 10 T0 3 C 9.91 10 7 D 8.3 10 3 298.15 kelvin H298 G298 T0 R IDCPH T0 T A B C D R T IDCPS T0 T A B C D T 496 G 4.896 10 3 J mol K exp G RT x K 0.53802 By Eq. (13.28), ( )( ) 1 0.1 0.1 1 = K Since 0.10 1 x = K K 0.10 0.843 x 0.10 1 x 6.5894 (a) y1 1 1 x yH2O 1 0.2 y1 y1 0.0186 yH2O ysteam 0.7814 0.8682 Ans. Ans. (b) ysteam 6.5894 7.5894 13.19 C4H10(g) = CH2:CHCH:CH2(g) + 2H2(g) (1) (2) (3) = 2 Number the species as shown. Basis is 1 mol species 1 + x mol steam entering. n0 = 1 x By Eq. (13.5), y1 = 1 1 x 2 y2 = 1 x 2 = 0.12 From data in Table C.4, y3 = 2 y2 = 0.24 H298 235030 J mol G298 166365 J mol The following vectors represent the species of the reaction in the order in which they appear: 1 1 2 A 1.935 2.734 3.249 B 36.915 26.786 10 0.422 3 497 11.402 C 8.882 0.0 end rows ( A) i Ai i 0.0 10 6 D 0.0 0.083 10 5 i 1 end i Bi i A B C i i Ci D i i Di A 7.297 B 9.285 10 3 C 2.52 10 6 D 1.66 10 4 T 925 kelvin T0 298.15 kelvin G H298 T H298 G298 T0 R IDCPH T0 T A B C D R T IDCPS T0 T A B C D G 9.242 10 3 J mol K exp G RT 2 K 1 x 2 0.30066 By Eq. (13.28), ( 0.12) ( 0.24) 1 = K Because 0.12 1 x 2 = K K ( 0.24) 2 x 0.12 1 2 x 4.3151 0.839 (a) y1 1 1 x 2 yH2O 1 0.36 y1 y1 0.023 yH2O 0.617 Ans. (b) ysteam 4.3151 5.3151 ysteam 0.812 Ans. 498 13.20 1/2N2(g) + 3/2H2(g) = NH3(g) = 1 Basis: 1/2 mol N2, 3/2 mol H2 feed n0 = 2 This is the reaction of Pb. 4.21(a) with all stoichiometric coefficients divided by two. From the answers to Pbs. 4.21(a), 4.22(a), and 13.7(a) ALL DIVIDED BY 2, find the following values: H298 46110 J mol G298 16450 3 J mol A 2.9355 B 2.0905 10 C 0 D 0.3305 10 5 (a) T 300 kelvin T0 298.15 kelvin G H298 T H298 G298 T0 R IDCPH T0 T A B C D R T IDCPS T0 T A B C D G P 1 1.627 10 4 J mol 1 K exp G RT K 679.57 P0 From Pb. 13.9 for ideal gases: 1 yNH3 1 1.299 K P P0 0.5 0.9664 2 yNH3 0.9349 Ans. (b) For yNH3 = 0.5 by the preceding equation 2 3 Solving the next-to-last equation for K with P = P0 gives: 1 K 1 2 1 1.299 K 6.1586 499 Find by trial the value of T for which this is correct. It turns out to be T = 399.5 kelvin Ans. (c) For P = 100, the preceding equation becomes 1 K 1 2 1 129.9 K 0.06159 Another solution by trial for T yields T = 577.6 kelvin Ans. (d) Eq. (13.27) applies, and requires fugacity coefficients, which can be evaluated by the generalized second-virial correlation. Since iteration will be necessary, we assume a starting T of 583 K for which: T 583kelvin P 100bar For NH3(1): Tc1 405.7kelvin Pc1 112.8bar Pr1 P Pc1 1 0.253 0.887 Tr1 For N2(2): T Tc1 Tr1 1.437 Pc2 Pr1 2 Tc2 126.2kelvin 34.0bar 0.038 Tr2 583 126.2 Tr2 4.62 Pr2 100 34.0 Pr2 2.941 For H2(3), estimate critical constants using Eqns. (3.58) and (3.59) Tc3 1 43.6 21.8 2.016 T kelvin kelvin Tc3 42.806 K Tr3 T Tc3 Tr3 13.62 Pc3 1 20.5 44.2 T 2.016 kelvin bar Pc3 19.757 bar Pr3 P Pc3 Pr3 5.061 3 0 500 Therefore, i 1 3 1 2 3 PHIB Tr1 Pr1 PHIB Tr2 Pr2 PHIB Tr3 Pr3 0.924 1.034 1.029 1 i 0.5 1.5 i i 1.184 The expression used for K in Part (c) now becomes: 1 K 1 2 1 K 129.9 1.184 0.07292 Another solution by trial for T yields T = 568.6 K Ans. Of course, the INITIAL assumption made for T was not so close to the final T as is shown here, and several trials were in fact made, but not shown here. The trials are made by simply changing numbers in the given expressions, without reproducing them. 13.21 CO(g) + 2H2(g) = CH3OH(g) Basis: 1 mol CO, 2 mol H2 feed From the data of Table C.4, = 2 n0 = 3 H298 90135 J mol G298 24791 J mol This is the reaction of Ex. 4.6, Pg. 142 from which: A (a) T 7.663 B 10.815 10 T0 3 C 3.45 10 6 D 0.135 10 5 300 kelvin 298.15 kelvin 501 G H298 T H298 G298 T0 R IDCPH T0 T A B C D R T IDCPS T0 T A B C D G P 1 2.439 10 4 J mol 1 K exp G RT K 1.762 10 4 P0 By Eq. (13.5), with the species numbered in the order in which they appear in the reaction, y1 = 1 3 2 y2 = 2 3 2 2 y3 = 3 2 By Eq. (13.28), 0.8 (guess) Given 3 4 1 2 2 3 = P P0 2 K Find 0.9752 y3 (b) 3 y3 2 0.5 3 y3 2 y3 1 y3 0.9291 Ans. By the preceding equation 0.75 Solution of the equilibrium equation for K gives K 3 4 1 2 2 3 K 27 Find by trial the value of T for which this is correct. It turns out to be: T = 364.47 kelvin Ans. 502 (c) For P = 100 bar, the preceding equation becomes K 3 4 1 2 2 100 3 2 K 2.7 10 3 Another solution by trial for T yields T = 516.48 kelvin Ans. (d) Eq. (13.27) applies, and requires fugacity coefficients. Since iteration will be necessary, assume a starting T of 528 K, for which: T 528kelvin P 100bar For CO(1): Tc1 132.9kelvin Pc1 34.99bar 1 0.048 Tr1 T Tc1 Tc3 Tr1 3.973 Pr1 P Pc1 Pr1 2.858 For CH3OH(3): 512.6kelvin Pc3 80.97bar 3 0.564 Tr T Tc3 Tr 1.03 Pr P Pc3 Pr 1.235 By Eq. (11.67) and data from Tables E.15 & E.16. 3 0.6206 0.9763 3 3 0.612 For H2(2), the reduced temperature is so large that it may be assumed ideal: Therefore: i 1 3 1 PHIB Tr1 Pr1 1.0 0.612 1 2 1 1.032 1 0.612 i i i 503 0.5933 The expression used for K in Part (c) now becomes: 3 4 1 2 2 K 100 3 2 0.593 K 1.6011 10 3 Another solution by trial for T yields: T = 528.7 kelvin Ans. 13.22 CaCO3(s) = CaO(s) + CO2(g) Each species exists PURE as an individual phase, for which the activity is f/f0. For the two species existing as solid phases, f and f0 are for practical purposes the same, and the activity is unity. If the pure CO2 is assumed an ideal gas at 1(atm), then for CO2 the activity is f/f0 = P/P0 = P (in bar). As a result, Eq. (13.10) becomes K = P = 1.0133, and we must find the T for which K has this value. From the data of Table C.4, H298 178321 J mol G298 130401 J mol The following vectors represent the species of the reaction in the order in which they appear: 1 1 1 A 12.572 6.104 5.457 B 2.637 0.443 10 1.045 i Ai 3 3.120 D 1.047 10 1.157 i Bi 5 i 1 3 A i B i D i i Di A T 1.011 1151.83 kelvin B 1.149 10 T0 3 C 0 D 9.16 10 4 298.15 kelvin G H298 T H298 G298 T0 R IDCPH T0 T A B C D R T IDCPS T0 T A B C D 504 G 126.324 J mol K exp G RT K 1.0133 Thus T = 1151.83 kelvin Ans. Although a number of trials were required to reach this result, only the final trial is shown. A handbook value for this temperature is 1171 K. 13.23 NH4Cl(s) = NH3(g) + HCl(g) The NH4Cl exists PURE as a solid phase, for which the activity is f/f0. Since f and f0 are for practical purposes the same, the activity is unity. If the equimolar mixture of NH3 and HCl is assumed an ideal gas mixture at 1.5 bar, then with f0 = 1 bar the activity of each gas species is its partial pressure, (0.5)(1.5) = 0.75. As a result, Eq. (13.10) becomes K = (0.75)(0.75) = 0.5625 , and we must find the T for which K has this value. From the given data and the data of Table C.4, H298 176013 J mol G298 91121 J mol The following vectors represent the species of the reaction in the order in which they appear: 1 1 1 A 5.939 3.578 3.156 B 16.105 3.020 0.623 i Ai 0.0 10 3 D 0.186 10 0.151 5 i 1 3 A i B i i Bi D i i Di 3 A T 0.795 623.97 kelvin B 0.012462 T0 298.15 kelvin C 0 D 3.5 10 G H298 T H298 G298 T0 R IDCPH T0 T A B C D R T IDCPS T0 T A B C D 505 G Thus 2.986 10 3 J mol K exp G RT K 0.5624 T = 623.97 K Ans. Although a number of trials were required to reach this result, only the final trial is shown. 13.25 NO(g) + (1/2)O2(g) = NO2(g) = 0.5 yNO2 yNO yO2 0.5 = yNO2 yNO ( ) 0.21 0.5 = K T 298.15 kelvin From the data of Table C.4, G298 K 1.493 (guesses) 35240 J mol K yNO exp 10 G298 RT 12 10 6 yNO2 0.5 10 6 Given yNO yNO2 yNO2 = ( ) 0.21 K yNO yNO2 yNO yNO = 5 10 7.307 10 6 Find yNO yNO2 12 This is about 7 10 6 ppm (a negligible concentration) Ans. 13.26 C2H4(g) + (1/2)O2(g) = <(CH2)2>O(g) = 0.5 See Example 13.9, Pg. 508-510 From Table C.4, H298 105140 J mol G298 81470 J mol Basis: 1 mol C2H4 entering reactor. Moles O2 entering: nO2 1.25 0.5 Moles N2 entering: nN2 nO2 79 21 506 n0 1 nO2 nN2 n0 3.976 Index the product species with the numbers: 1 = ethylene 2 = oxygen 3 = ethylene oxide 4 = nitrogen The numbers of moles in the product stream are given by Eq. (13.5). For the product stream, data from Table C.1: Guess: 1 n nO2 0.5 0.8 1.424 A 3.639 0.385 3.280 0.0 10 6 2 14.394 B 0.506 23.463 0.593 10 kelvin 3 nN2 4.392 C 0.0 9.296 0.0 D 1 10 kelvin 5 2 0.227 0.0 0.040 0.5 1 0 kelvin i 1 4 A i n i Ai B i n i Bi C i n i Ci D i n i Di y n n0 0.5 K i y i i K 15.947 The energy balance for the adiabatic reactor is: H298 HP = 0 For the second term, we combine Eqs. (4.3) & (4.7). The three equations together provide the energy balance. 507 For the equilibrium state, apply a combination of Eqs. (13.11a) & (13.18).The reaction considered here is that of Pb. 4.21(g), for which the following values are given in Pb. 4.23(g): A D Guess: 3.629 0.114 10 kelvin 3 A T0 1 3 3 5 2 B T0 8.816 10 3 kelvin C 4.904 10 6 2 kelvin 298.15 kelvin idcph B 2 T0 2 2 1 C 3 T0 1 D T0 1 idcps A ln B T0 C T0 D 2 1 2 2 1 T0 idcph Given H298 = R A T0 1 3 3 130.182 kelvin idcps 0.417 B 2 T0 2 2 1 C 3 H298 T0 1 D T0 H298 idcps 1 K = exp G298 1 T0 R T0 R T0 0.88244 3.18374 idcph Find 508 0.0333 y 0.88244) ( 0.052 0.2496 0.6651 T T0 T 949.23 kelvin Ans. Ans. 13.27 CH4(g) = C(s) + 2H2(g) = 1 (gases only) The carbon exists PURE as an individual phase, for which the activity is unity. Thus we leave it out of consideration. From the data of Table C.4, H298 74520 J mol G298 50460 J mol The following vectors represent the species of the reaction in the order in which they appear: 1 1 2 A 1.702 1.771 3.249 2.164 6 9.081 B 0.771 10 0.422 3 0.0 10 i 1 3 C 0.0 0.0 D 0.867 10 0.083 5 A i i Ai B i i Bi C i i Ci D i i Di A T 6.567 B 7.466 10 T0 3 C 2.164 10 6 D 7.01 10 4 923.15 kelvin 298.15 kelvin 509 G H298 T H298 G298 T0 R IDCPH T0 T A B C D R T IDCPS T0 T A B C D G 1.109 10 4 J mol n0 = 1 K exp G RT 1 1 2 K 4.2392 By Eq. (13.5), yCH4 = yH2 = 2 2 2 1 (a) By Eq. (13.28), 2 1 1 = 4 1 = K K 4 K 0.7173 (fraction decomposed) yCH4 1 1 yH2 2 1 yCH4 0.1646 yH2 (b) For a feed of 1 mol CH4 and 1 mol N2, 0.8354 n0 = 2 Ans. By Eq. (13.28), .8 (guess) Given 2 2 2 1 = K Find 0.7893 (fraction decomposed) yCH4 1 2 yH2 2 2 yN2 1 yCH4 yH2 yH2 0.5659 yCH4 0.0756 yN2 0.3585 Ans. 510 13.28 1/2N2(g) + 1/2O2(g) = NO(g) = 0 (1) This is the reaction of Pb. 4.21(n) with all stoichiometric coefficients divided by two. From the answers to Pbs. 4.21(n), 4.22(n), and 13.7(n) ALL DIVIDED BY 2, find the following values: H298 90250 J mol G298 86550 3 J mol 5 A 0.0725 B 0.0795 10 C 0 D 0.1075 10 T 2000 kelvin T0 298.15 kelvin G H298 T H298 G298 T0 R IDCPH T0 T A B C D R T IDCPS T0 T A B C D 4 J G 6.501 10 mol K1 exp G RT = 0.5 K1 0.02004 (2) 1/2N2(g) + O2(g) = NO2(g) From the data of Table C.4, H298 33180 J mol G298 51310 J mol The following vectors represent the species of the reaction in the order in which they appear: 0.5 1 1 i 1 3 A 3.280 3.639 4.982 B 0.593 0.506 10 1.195 3 0.040 D 0.227 10 0.792 i Bi 5 A i i Ai B i D i i Di A 0.297 B 3.925 10 4 C 0 D 5.85 10 4 T 2000 kelvin T0 298.15 kelvin 511 G H298 T H298 G298 T0 R IDCPH T0 T A B C D R T IDCPS T0 T A B C D 10 5 J G 1.592 mol K2 exp G RT K2 6.9373 10 5 With the assumption of ideal gases, we apply Eq. (13.28): (1) yN2 yNO 0.5 yO2 0.5 = yNO ( ) 0.7 0.5 ( ) 0.05 0.5 = K1 yNO (2) K1 ( ) 0.7 P0 1 0.5 ( ) 0.05 200 0.5 yNO 3.74962 10 3 Ans. P yNO2 yN2 0.5 = yO2 yNO2 ( ) 0.7 0.5 = ( ) 0.05 0.5 P P0 0.5 K2 yNO2 P P0 0.5 K2 ( ) 0.7 ( ) 0.05 yNO2 4.104 10 5 Ans. 13.29 2H2S(g) + SO2(g) = 3S(s) + 2H2O(g) The sulfur exists PURE as a solid phase, for which the activity is f/f0. Since f and f0 are for practical purposes the same, the activity is unity, and it is omitted from the equilibrium equation. Thus for the gases only, = 1 From the given data and the data of Table C.4, H298 145546 J mol G298 89830 J mol 512 The following vectors represent the species of the reaction in the order in which they appear: 2 1 3 2 i 1 4 A 3.931 5.699 4.114 3.470 B 1.490 0.801 1.728 1.450 10 3 0.232 D 1.015 0.783 0.121 10 5 A i i Ai B i i Bi D i i Di A 5.721 B 6.065 10 3 C 0 D 6.28 10 4 T 723.15 kelvin T0 298.15 kelvin G H298 T H298 G298 T0 R IDCPH T0 T A B C D R T IDCPS T0 T A B C D 4 J G 1.538 10 mol K exp G RT (basis) K 12.9169 By Eq. (13.5), gases only: n0 = 3 yH2S = 2 3 2 ySO2 = 1 3 yH2O = 2 3 By Eq. (13.28), 0.5 2 (guess) Given 2 2 2 3 2 = 8K 1 Find 0.767 Percent conversion of reactants = PC PC = ni0 ni0 ni 100 = i ni0 100 [By Eq. (13.4)] 513 Since the reactants are present in the stoichiometric proportions, for each reactant, ni0 = i Whence PC 100 PC 76.667 Ans. 13.30 N2O4(g) = 2NO2(g) (a) (b) = 1 Data from Tables C.4 and C.1 provide the following values: H298 57200 J mol G298 5080 J mol T0 298.15 kelvin T 350 kelvin 3 A 1.696 B 0.133 10 C 0 D 1.203 10 5 G H298 T H298 G298 T0 R IDCPH T0 T A B C D R T IDCPS T0 T A B C D G 3.968 10 3 J mol K exp G RT K 3.911 Basis: 1 mol species (a) initially. Then ya = 1 1 yb = 2 1 2 1 2 1 K 4P K = P P0 1 K (a) P 5 P0 1 0.4044 ya 1 1 ya 0.4241 Ans. (b) P 1 P0 1 K 4P K 0.7031 514 By Eq. (4.18), at 350 K: H H298 R IDCPH T0 T A B C D H 56984 J mol This is Q per mol of reaction, which is 0.7031 0.4044 0.299 Whence Q H Q 17021 J mol Ans. 13.31 By Eq. (13.32), 2 2 K= xB B xA A 1 = 2 xA xA A B ln a = 0.1 xB ln b = 0.1 xA 1 = Whence 2 2 K= 1 xA xA exp 0.1 xA xA xA 2 exp 0.1 xB exp 0.1 xA xB K= 1 xA xA exp 0.1 2 xA J mol 1 K = exp G RT G 1000 T 298.15 kelvin xA .5 (guess) Given xA 1 xA xA exp 0.1 2 xA Ans. 1 = exp G RT xA Find xA 0.3955 For an ideal solution, the exponential term is unity: Given 1 xA xA = exp G RT xA Find xA xA 0.4005 This result is high by 0.0050. Ans. 515 13.32 H2O(g) + CO(g) = H2(g) + CO2(g) = 0 From the the data of Table C.4, H298 41166 J mol G298 28618 J mol T0 298.15 kelvin T 800 kelvin 3 A 1.860 B 0.540 10 C 0 D 1.164 10 5 G H298 T H298 G298 T0 R IDCPH T0 T A B C D R T IDCPS T0 T A B C D 3 J G 9.668 10 mol = 0 K exp G RT K 4.27837 (a) No. Since , at low pressures P has no effect (b) No. K decreases with increasing T. (The standard heat of reaction is negative.). (c) Basis: 1 mol CO, 1 mol H2, w mol H2O feed. From the problem statement, nCO nCO nH2 nCO2 = 0.02 By Eq. (13.4), nCO = 1 nH2 = 1 NCO2 = 1 1 1 = 1 2 = 0.02 0.96 1.02 0.941 Let z = w/2 = moles H2O/mole "Water gas". By Eq. (13.5), yH2O = 2z w = 2 2z 2 w yCO = 1 2 2z yH2 = 1 2 2z 516 yCO2 = 2 2z By Eq. (13.28) z 2 (guess) Given 1 2z 1 = K z Find z () z 4.1 Ans. (d) 2CO(g) = CO2(g) + C(s) = 1 (gases) Data from Tables C.4 and C.1: H298 172459 J mol G298 120021 J mol D 1.962 10 5 A G 0.476 H298 B 0.702 10 3 C 0 T H298 G298 T0 R IDCPH T0 T A B C D R T IDCPS T0 T A B C D G 3.074 10 4 J mol K exp G RT K 101.7 By Eq. (13.28), gases only, with P = P0 = 1 bar yCO2 yCO 2 = K = 101.7 for the reaction AT EQUILIBRIUM. If the ACTUAL value of this ratio is GREATER than this value, the reaction tries to shift left to reduce the ratio. But if no carbon is present, no reaction is possible, and certainly no carbon is formed. The actual value of the ratio in the equilibrium mixture of Part (c) is yCO2 2 2z yCO 1 2 2z yCO2 0.092 yCO 5.767 10 3 RATIO yCO2 yCO 2 RATIO 2.775 10 3 No carbon can deposit from the equilibrium mixture. 517 13.33 CO(g) + 2H2(g) = CH3OH(g) = 2 (1) This is the reaction of Pb. 13.21, where the following parameter values are given: H298 90135 J mol G298 24791 J mol T 550 kelvin T0 298.15 kelvin A 7.663 B 10.815 10 3 C 3.45 10 6 D 0.135 10 5 G H298 H298 G298 T0 R IDCPH T0 T A B C D R T IDCPS T0 T A B C D T G 3.339 10 4 J mol K1 exp G RT = 0 K1 6.749 (2) 10 4 H2(g) + CO2(g) = CO(g) + H2O(g) From the the data of Table C.4, H298 41166 J mol G298 28618 J mol T 550 kelvin T0 298.15 kelvin The following vectors represent the species of the reaction in the order in which they appear: 1 1 1 1 i 1 4 A 3.249 5.457 3.376 3.470 B 0.422 1.045 0.557 1.450 i Ai 0.083 10 3 D 1.157 0.031 0.121 10 5 A i B i i Bi D i i Di A 1.86 B 5.4 10 4 518 C 0 D 1.164 10 5 G H298 T H298 G298 T0 R IDCPH T0 T A B C D R T IDCPS T0 T A B C D 4 J G 1.856 10 mol K2 exp G RT K2 0.01726 Basis: 1 mole of feed gas containing 0.75 mol H2, 0.15 mol CO, 0.05 mol CO2, and 0.05 mol N2. Stoichiometric numbers, i.j i= j 1 2 H2 CO CO2 CH3OH H2O _______________________________________________ -2 -1 -1 1 0 -1 1 0 0 1 By Eq. (13.7) 0.75 yH2 = 1 2 1 2 1 2 2 0.15 yCO = 1 1 2 2 1 1 0.05 yCO2 = 1 2 1 yCH3OH = 1 2 1 yH2O = 2 1 2 1 P 100 P0 1 1 By Eq. (13.40), 0.1 2 0.1 (guesses) Given 1 1 2 1 2 2 = 1 2 0.75 2 1 0.15 1 2 0.15 P P0 2 K1 2 2 2 2 0.75 2 1 0.05 = K2 1 2 Find 1 2 519 1 0.1186 2 8.8812 10 3 0.75 yH2 1 2 1 2 1 2 2 0.15 yCO 1 1 2 2 1 1 0.05 yCO2 1 2 1 yCH3OH 1 2 1 yH2O 2 1 2 1 yN2 1 yH2 yCO yCO2 yCH3OH yH2O yH2 0.6606 yCO 0.0528 yCO2 0.0539 Ans. yCH3OH 0.1555 yH2O 0.0116 yN2 0.0655 13.34 CH4(g) + H2O(g) = CO(g) + 3H2(g) From the the data of Table C.4, = 2 (1) H298 205813 J mol G298 141863 J mol The following vectors represent the species of the reaction in the order in which they appear: 1 1 1 3 2.164 C 0.0 0.0 0.0 10 6 1.702 A 3.470 3.376 3.249 0.0 D 0.121 0.031 0.083 10 5 9.081 B 1.450 0.557 0.422 10 3 i 1 4 A i i Ai B i i Bi C i i Ci D i i Di A 7.951 B 8.708 10 3 C 520 2.164 10 6 D 9.7 10 3 T G 1300 kelvin H298 T0 298.15 kelvin T H298 G298 T0 R IDCPH T0 T A B C D R T IDCPS T0 T A B C D 10 5 J G 1.031 mol K1 exp G RT K1 13845 = 0 (2) This is the reaction of Pb. 13.32, where parameter values are given: H298 A G H2O(g) + CO(g) = H2(g) + CO2(g) 41166 J mol 0.540 10 G298 3 28618 J mol D 1.164 10 5 1.860 H298 B C 0.0 T H298 G298 T0 R IDCPH T0 T A B C D R T IDCPS T0 T A B C D 10 3 J G 5.892 mol K2 exp G RT K2 0.5798 (a) No. Primary reaction (1) shifts left with increasing P. (b) No. Primary reaction (1) shifts left with increasing T. (c) The value of K1 is so large compared with the value of K2 that for all practical purposes reaction (1) may be considered to go to completion. With a feed equimolar in CH4 and H2O, no H2O then remains for reaction (2). In this event the ratio, moles H2/moles CO is very nearly equal to 3.0. 521 (d) With H2O present in an amount greater than the stoichiometric ratio, reaction (2) becomes important. However, reaction (1) for all practical purposes still goes to completion, and may be considered to provide the feed for reaction (2). On the basis of 1 mol CH4 and 2 mol H2O initially, what is left as feed for reaction (2) is: 1 mol H2O, 1 mol CO, and 3 mol H2; n0 = 5. Thus, for reaction (2) at equilibrium by Eq. (13.5): yCO = yH2O = 1 5 yCO2 = 5 yH2 = 3 5 By Eq. (13.28), 0.5 (guess) Given 3 1 2 = K2 Find 0.1375 Ratio = yH2 yCO Ratio 3 1 Ratio 3.638 Ans. (e) One practical way is to add CO2 to the feed. Some H2 then reacts with the CO2 by reaction (2) to form additional CO and to lower the H2/CO ratio. (f) 2CO(g) = CO2(g) + C(s) = 1 (gases) This reaction is considered in the preceding problem, Part (d), from which we get the necessary parameter values: H298 172459 J mol G298 120021 J mol For T = 1300 K, T 1300 kelvin T0 298.15 kelvin A 0.476 B 0.702 10 3 C 0.0 D 1.962 10 5 G H298 T H298 G298 T0 R IDCPH T0 T A B C D R T IDCPS T0 T A B C D 522 G 5.673 10 4 J mol K exp G RT K 5.255685 10 3 As explained in Problem 13.32(d), the question of carbon deposition depends on: RATIO = yCO2 yCO 2 When for ACTUAL compositions the value of this ratio is greater than the equilibrium value as given by K, there can be no carbon deposition. Thus in Part (c), where the CO2 mole fraction approaches zero, there is danger of carbon deposition. However, in Part (d) there can be no carbon deposition, because Ratio > K: Ratio 1 5 2 Ratio 0.924 13.37 Formation reactions: C + 2H2 = CH4 H2 + (1/2)O2 = H2O C + (1/2)O2 = CO C + O2 = CO2 Elimination first of C and then of O2 leads to a pair of reactions: CH4 + H2O = CO + 3H2 CO + H2O = CO2 + H2 (1) (2) There are alternative equivalent pairs, but for these: Stoichiometric numbers, i= j 1 2 -1 0 -1 -1 1 -1 0 1 523 i.j CH4 H2O CO CO2 H2 j _________________________________________________ 3 1 2 0 For initial amounts: 2 mol CH4 and 3 mol H2O, n0 = 5, and by Eq. (13.7): 2 yCH4 = yCO2 = 1 3 yH2O = yH2 = 1 2 5 2 1 2 5 2 1 2 yCO = 1 2 5 2 1 3 1 5 5 2 1 2 1 By Eq. (13.40), with P = P0 = 1 bar yCO yH2 = k1 yCH4 yH2O 3 yCO2 yH2 yCO yH2O = k2 From the data given in Example 13.14, G1 27540 J mol G2 3130 J mol T 1000 kelvin K1 exp G1 RT K2 K2 exp 1.457 G2 RT K1 1 27.453 1.5 2 1 2 (guesses) 3 1 1 2 1 2 Given 2 1 3 2 1 3 2 5 2 1 2 = K1 2 3 1 1 2 3 = K2 1 2 Find 1 2 1 1.8304 2 0.3211 2 yCH4 5 1 3 yH2O 5 524 1 2 2 1 2 1 yCO 1 2 5 2 1 yCO2 2 5 2 1 yH2 3 1 2 5 2 1 yCH4 0.0196 yH2O 0.098 yCO 0.1743 yCO2 0.0371 yH2 0.6711 These results are in agreement with those of Example 13.14. 13.39Phase-equilibrium equations: Ethylene oxide(1): p1 = y1 P = 415x1 P 101.33 kPa x1 = x2 = y1 P 415kPa y2 P Psat2 Water(2): x2 Psat2 = y2 P Psat2 3.166 kPa (steam tables) Ethylene glycol(3): Psat3 = 0.0 y3 = 0.0 Therefore, y2 = 1 y1 and x3 = 1 x2 x3 For the specified standard states: (CH2)2O(g) + H2O(l) = CH2OH.CH2OH(l) By Eq. (13.40) and the stated assumptions, k= y1 3 x3 P P0 = 2 x2 x3 y1 x2 T 298.15kelvin Data from Table C.4: G298 k 6.018 72941 10 12 J mol Ans. k exp G298 RT 525 So large a value of k requires either y1 or x2 to approach zero. If y1 approaches zero, y2 approaches unity, and the phase-equilibrium expression for water(2) makes x2 = 32, which is impossible. Thus x2 must approach zero, and the phase-equilibrium equation requires y2 also to approach zero. This means that for all practical purposes the reaction goes to completion. For initial amounts of 3 moles of ethylene oxide and 1 mole of water, the water present is entirely reacted along with 1 mole of the ethylene oxide. Conversion of the oxide is therefore 33.3 %. 1 13.41 a) Stoichiometric coefficients: 1 Initial numbers of n0 moles j 50 50 kmol hr 0 0 1 0 1 0 1 1 Number of components: i 1 4 Number of reactions: n0 n0i i 1 2 100 kmol hr vj i i j v yA 1 1 0.05 n0 Given values: Guess: yB 0.10 yC 0.4 yD 0.4 1 1 kmol hr 2 1 kmol hr Given yA = n01 n0 1 1 2 2 yB = n02 n0 1 1 2 Eqn. (13.7) yC = n03 n0 1 1 2 2 yD = n04 n0 1 2 2 yC yD Find yC yD 1 2 1 2 1 44.737 kmol hr 2 2.632 kmol hr 526 (i) nA n01 1 2 nA 2.632 nB n02 1 nB 2 5.263 kmol hr kmol nC n03 1 nC 42.105 hr kmol nD n04 2 nD 2.632 hr kmol n nA nB nC nD n 52.632 hr kmol hr Ans. (ii) yC 0.8 yD 0.05 Ans. 40 1 b) Stoichiometric coefficients: 1 2 0 1 Number of reactions: 1 1 0 Initial numbers of n0 moles 40 kmol hr 0 0 j 1 2 Number of components: i 1 4 vj i i j v 1 2 n0 n0i i n0 80 kmol hr Given values: yC 0.52 yD 0.04 Guess: yA 0.4 yB 0.4 1 1 kmol hr 2 1 kmol hr Given yA = n01 n0 1 1 2 2 2 1 yB = n02 n0 1 1 2 2 2 2 2 Eqn. (13.7) yC = n03 n0 1 2 2 yD = n04 n0 1 2 2 527 yA yB 1 2 Find yA yB 1 2 1 26 kmol hr 2 2 kmol hr yA 0.24 yB 0.2 nA n01 1 2 nA 12 nB n02 1 2 2 nB 10 kmol hr kmol nC n03 1 nC 26 hr kmol Ans. nD n04 2 nD 2 hr kmol hr 1 c) Stoichiometric coefficients: 1 1 0 1 Number of reactions: 100 1 1 0 Initial numbers of n0 moles 0 0 0 j kmol hr Number of components: i 1 4 1 2 vj i i j v 1 1 n0 n0i i n0 100 kmol hr Given values: yC 0.3 yD 0.1 Guess: yA 0.4 yB 0.4 1 1 kmol hr 2 1 kmol hr Given yA = n01 n0 1 1 2 2 0 yB = n02 n0 1 1 2 2 Eqn. (13.7) yC = n03 n0 1 1 2 yD = n04 n0 1 2 2 528 yA yB 1 2 Find yA yB 1 2 1 37.5 kmol hr 2 12.5 kmol hr yA 0.4 yB 0.2 nA n01 1 2 nA 50 nB n02 1 2 nB 25 kmol hr kmol nC n03 1 nC 37.5 hr kmol Ans. nD n04 2 nD 12.5 hr kmol hr 1 1 d) Stoichiometric coefficients: 1 1 0 1 1 Number of reactions: 40 1 0 0 Initial numbers of n0 moles 60 0 0 0 kmol hr Number of components: i 1 5 j 1 2 kmol hr vj i i j v 1 0 0.25 n0 i n0i n0 100 Given values: yC yD 0.20 Guess: yA 0.2 yB 0.4 yE 0.1 1 1 kmol hr 2 1 kmol hr 529 Given yA = n01 n0 1 1 2 yB = n02 n0 1 1 2 Eqn. (13.7) yC = n03 n0 1 1 yD = n04 n0 2 1 yE = n05 n0 2 1 yA yB yE 1 2 Find yA yB yE 1 2 1 20 kmol hr 2 16 kmol hr (i) (ii) nA n01 1 2 nA nB n02 1 2 nB nC nD n03 n04 1 nC nD kmol hr kmol 24 hr kmol 20 hr 4 16 kmol hr kmol yA 0.05 yB 0.3 Ans. yC yD 0.25 0.2 2 nE n05 2 nE 16 hr yE 0.2 13.45 C2H4(g) + H2O(g) -> C2H5OH(g) T0 298.15kelvin P0 1bar T 400kelvin P 2bar 1 = C2H4(g) H0f1 52500 J mol G0f1 68460 J mol 2 = H2O(g) H0f2 241818 J mol J G0f2 228572 3 = C2H5OH(g) H0f3 235100 mol G0f3 168490 J mol J mol 530 H0 H0f1 H0f2 H0f3 H0 45.782 G0 G0f1 G0f2 G0f3 G0 8.378 kJ mol kJ mol A ( 1.424) ( 3.470) ( 3.518) A 3 1.376 B [( 14.394) ( 1.450) ( 20.001) ]10 B 4.157 10 3 C [ ( 4.392) () ( 6.002) 0 ]10 6 C 1.61 10 6 D [ () ( 0 0.121) () 0 ]10 5 D K0 1.21 10 29.366 4 a) K0 exp G0 R T0 H0 1 R T0 Eqn. (13.21) K298 K298 Ans. b) K1 exp T0 T Eqn. (13.22) K1 9.07 10 3 K2 exp 1 IDCPH T0 T A B C D T IDCPS T0 T A B C D K0 K1 K2 Eqn. (13.20) K2 0.989 Eqn. (13.23) K400 K400 0.263 Ans. c) Assume as a basis there is initially 1 mol of C2H4 and 1 mol of H2O 1 y1 = 2 e e 1 y2 = 2 e e y3 = e 2 e Assuming ideal gas behavior y3 y1 y2 = K P P0 e Substituting results in the following expression: 2 e 1 2 e 1 e 2 = K400 e e P P0 531 Solve for e using a Mathcad solve block. P P0 e Guess: e 0.5 e Given 2 e 1 2 1 e 1 e 2 e e = K400 e e Find e e 0.191 1 y2 y2 y1 y1 e e 2 0.447 2 0.447 y3 y3 e 2 0.105 e Ans. d) Since a decrease in pressure will cause a shift on the reaction to the left and the mole fraction of ethanol will decrease. 13.46 H2(g) + O2(g) -> H2O2(g) H0fH2O2 S0H2 136.1064 kJ mol T 298.15kelvin P 1bar 130.680 J mol kelvin J mol kelvin S0O2 S0O2 205.152 J mol kelvin S0H2O2 232.95 S0fH2O2 G0f S0H2 H0fH2O2 S0H2O2 S0fH2O2 G0f 102.882 kJ mol J mol kelvin Ans. T S0fH2O2 105.432 532 13.48 C3H8(g) -> C3H6(g) + H2(g) (I) C3H8(g) -> C2H4(g) + CH4(g) (II) T0 298.15kelvin P0 1bar T 750kelvin P 1.2bar 1 = C3H8 (g) H0f1 104680 J mol G0f1 24290 J mol 2 = C3H6 (g) H0f2 19710 J mol G0f2 62205 J mol 3 = H2 (g) H0f3 0 J mol G0f3 0 J mol 4 = C2H4 (g) H0f4 52510 J mol G0f4 68460 J mol 5 = CH4 (g) H0f5 74520 J mol G0f5 50460 J mol Calculate equilibrium constant for reaction I: H0I H0f1 H0f2 H0f3 H0I 124.39 G0I G0f1 G0f2 G0f3 G0I 86.495 kJ mol kJ mol AI ( 1.213) ( 1.637) ( 3.249) AI 3 3.673 BI [( 28.785) ( 22.706) ( 0.422) ]10 BI 5.657 10 3 CI [ ( 8.824) ( 6.915) () 0 ]10 6 CI 1.909 10 6 DI [ () () ( 0 0 0.083) ]10 5 DI KI0 KI1 KI2 8.3 0 10 3 KI0 KI1 KI2 exp G0I R T0 H0I 1 R T0 1 Eqn. (13.21) exp T0 T Eqn. (13.22) 1.348 10 13 exp IDCPH T0 T AI BI CI DI T IDCPS T0 T AI BI CI DI 533 1.714 Eqn. (13.23) KI KI0 KI1 KI2 Eqn. (13.20) KI 0.016 Calculate equilibrium constant for reaction II: H0II H0f1 H0f4 H0f5 H0II 82.67 kJ mol kJ G0II G0f1 G0f4 G0f5 G0II 42.29 mol AII ( 1.213) ( 1.424) ( 1.702) AII 3 1.913 BII [ ( 28.785) ( 14.394) ( 9.081) ] 10 BII 6 5.31 10 3 CII [ ( 8.824) ( 4.392) ( 2.164) ] 10 CII 2.268 10 6 DII [ ( 0) ( 0) ( 0) ] 10 5 DII KII0 KII1 0 3.897 5.322 10 10 8 8 KII0 KII1 KII2 exp G0II R T0 H0II 1 R T0 Eqn. (13.21) exp T0 T Eqn. (13.22) exp 1 IDCPH T0 T AII BII CII DII T IDCPS T0 T AII BII CII DII KII2 1.028 Eqn. (13.23) KII KII0 KII1 KII2 Eqn. (13.20) KII 21.328 Assume an ideal gas and 1 mol of C3H8 initially. 1 y1 = 1 I I II II y2 = I 1 y3 = II I I 1 I II y4 = II 1 y5 = II II Eqn. (13.7) II I 1 I The equilibrium relationships are: y2 y3 P0 = KI P y1 y4 y5 y1 = KII P0 P Eqn. (13.28) 534 Substitution yields the following equations: I I 1 I II 1 I I II II I II 1 1 = KI P0 P II II 1 I II 1 I I II II I II 1 1 = KII P0 P Use a Mathcad solve block to solve these two equations for I and II. Note that the equations have been rearranged to facilitate the numerical solution. Guess: 0.5 0.5 I II Given I I II 1 1 = KI II P0 P P0 P 1 1 1 1 I I I I II II II II I II I II 1 I I II 1 = KII II I Find I II II I 0.026 II 0.948 1 y1 1 I I II II II y2 I 1 y3 II II I I 1 I II y4 y1 1 y5 II I 1 I II 0.01298 y2 0.0132 y3 0.0132 y4 535 0.4803 y5 0.4803 A summary of the values for the other temperatures is given in the table below. T= y1 y2 y3 y4 y5 7 K 50 0.0130 0.0132 0.0132 0.48 03 0.48 03 1000 K 0.00047 0.034 0.034 0.46 58 0.46 58 1250 K 0.00006 0.059 3 0.059 3 0.4407 0.4407 13.49 n-C4H10(g) -> iso-C4H10(g) T0 298.15kelvin P0 1bar T 425kelvin P 15bar 1 = n-C4H10(g) H0f1 125790 2 = iso-C4H10(g) H0f2 134180 J mol J G0f1 16570 mol G0f2 20760 J mol J mol H0 H0f1 H0f2 H0 8.39 kJ mol kJ G0 G0f1 G0f2 G0 4.19 mol A 0.258 A B ( 1.935) ( 1.677) [( 36.915) ( 37.853) ]10 3 B 9.38 10 4 C [11.402) ( 11.945) ( ]10 6 C 5.43 10 7 D [ () () 0 0 ]10 5 D K0 0 5.421 Ans. a) K0 exp G0 R T0 H0 1 R T0 Eqn. (13.21) b) K1 exp T0 T Eqn. (13.22) K1 0.364 K2 exp 1 IDCPH T0 T A B C D T IDCPS T0 T A B C D 536 K2 1 Eqn. (13.23) Ke K0 K1 K2 Eqn. (13.20) Ke 1.974 Ans. Assume as a basis there is initially 1 mol of n-C4H10(g) y1 = 1 e y2 = e a) Assuming ideal gas behavior y2 y1 = Ke e Substitution results in the following expression: 1 = Ke e e Solving for Ke yields the following analytical expression for 1 e 1 Ke e e 0.336 e y1 1 y1 0.664 y2 y2 0.336 Ans. b) Assume the gas is an ideal solution. In this case Eqn. (13.27) applies. yi i = i P P0 K Eqn. (13.27) Substituting for yi yields: 1 e e 1 2 = K e= 2 2 This can be solved analytically for e to get: Ke 1 Calculate i for each pure component using the PHIB function. For n-C4H10: 1 0.200 Tc1 425.1kelvin Pc1 37.96bar Tr1 1 T Tc1 Tr1 1 1 Pr1 1 P Pc1 Pr1 0.395 PHIB Tr1 Pr1 0.872 For iso-C4H10: 2 0.181 Tc2 408.1kelvin Pc2 36.48bar Tr2 T Tc2 Tr2 1.041 Pr2 537 P Pc2 Pr2 0.411 2 PHIB Tr2 Pr2 e 2 2 0.884 e Solving for yields: 2 e 2 Ke 1 0.339 y1 1 e y1 0.661 y2 e y2 0.339 Ans. The values of y1 and y2 calculated in parts a) and b) differ by less than 1%. Therefore, the effects of vapor-phase nonidealities is here minimal. 538 Chapter 14 - Section A - Mathcad Solutions 14.1 A12 := 0.59 A21 := 1.42 2 T := ( 55 + 273.15) K Margules equations: 1 ( x1) := exp ( 1 x1) A12 + 2 ( A21 A12) x1 2 ( x1) := exp x1 A21 + 2 ( A12 A21) ( 1 x1) 2 Psat1 := 82.37 kPa (a) Psat2 := 37.31 kPa BUBL P calculations based on Eq. (10.5): Pbubl ( x1) := x1 1 ( x1) Psat1 + ( 1 x1) 2 ( x1) Psat2 y1 ( x1) := x1 := 0.25 x1 := 0.50 x1 := 0.75 x1 1 ( x1) Psat1 Pbubl ( x1) Pbubl ( x1) = 64.533 kPa Pbubl ( x1) = 80.357 kPa Pbubl ( x1) = 85.701 kPa y1 ( x1) = 0.562 y1 ( x1) = 0.731 y1 ( x1) = 0.808 (b) BUBL P calculations with virial coefficients: B11 := 963 cm 3 mol B22 := 1523 cm 3 mol B12 := 52 cm 3 mol 12 := 2 B12 B11 B22 B11 ( P Psat1) + P y22 12 1 ( P , T , y1 , y2) := exp R T B22 ( P Psat2) + P y12 12 2 ( P , T , y1 , y2) := exp R T 539 Guess: x1 := 0.25 P := Psat1 + Psat2 2 Given y1 := 0.5 y2 := 1 y1 y1 1 ( P , T , y1 , y2) P = x1 1 ( x1) Psat1 y2 2 ( P , T , y1 , y2) P = ( 1 x1) 2 ( x1) Psat2 y2 = 1 y1 y1 y2 := Find ( y1 , y2 , P) P x1 := 0.50 Given y1 0.558 y2 = 0.442 P 63.757 kPa y1 1 ( P , T , y1 , y2) P = x1 1 ( x1) Psat1 y2 2 ( P , T , y1 , y2) P = ( 1 x1) 2 ( x1) Psat2 y2 = 1 y1 y1 y2 := Find ( y1 , y2 , P) P x1 := 0.75 Given y1 0.733 y2 = 0.267 P 79.621 kPa y1 1 ( P , T , y1 , y2) P = x1 1 ( x1) Psat1 y2 2 ( P , T , y1 , y2) P = ( 1 x1) 2 ( x1) Psat2 y2 = 1 y1 y1 y2 := Find ( y1 , y2 , P) P y1 0.812 y2 = 0.188 P 85.14 kPa 540 14.3 T := 200 K H1 := 200 bar P := 30 bar B := 105 cm 3 y1 := 0.95 mol Assume Henry's law applies to methane(1) in the liquid phase, and that the Lewis/Randall rule applies to the methane in the vapor: fhat1 = H1 x1 By Eq. (11.36): l fhat1 = y1 1 P 1 := exp B P R T 1 = 0.827 v Equate the liquid- and vapor-phase fugacities and solve for x1: x1 := y1 1 P H1 x1 = 0.118 Ans. 14.4 Pressures in kPa Data: 0.000 0.0895 0.1981 0.3193 x1 := 0.4232 0.5119 0.6096 0.7135 12.30 15.51 18.61 21.63 P := 24.01 25.92 27.96 30.12 x2 := 1 x1 0.000 0.2716 0.4565 0.5934 y1 := 0.6815 0.7440 0.8050 0.8639 Psat2 := P1 i := 2 .. rows ( P) (a) It follows immediately from Eq. (12.10a) that: ln 1 = A12 Combining this with Eq. (12.10a) yields the required expression 541 (b) (c) Henry's constant will be found as part of the solution to Part (c) BARKER'S METHOD by non-linear least squares. Margules equation. The most satisfactory procedure for reduction of this set of data is to find the value of Henry's constant by regression along with the Margules parameters. 1 ( x1 , x2 , A12 , A21) := exp ( x2) A12 + 2 ( A21 A12) x1 2 2 2 ( x1 , x2 , A12 , A21) := exp ( x1) A21 + 2 ( A12 A21) x2 Guesses: H1 := 50 A21 := 0.2 A12 := 0.4 Mininize the sums of the squared errors by setting sums of derivatives equal to zero. Given 0= i 2 d H1 Pi x1 1 x1 , x2 , A12 , A21 ... i i i exp ( A12) dA12 + x2i 2 x1i , x2i , A12 , A21 Psat2 ( ( ( ( ) ) ) ) 0= i 2 d H1 Pi x1 1 x1 , x2 , A12 , A21 ... i i i exp ( A12) dA21 + x2i 2 x1i , x2i , A12 , A21 Psat2 2 d H1 Pi x1 1 x1 , x2 , A12 , A21 ... i i i exp ( A12) dH1 + x2i 2 x1i , x2i , A12 , A21 Psat2 0= i ( ( ) ) A12 A21 := Find ( A12 , A21 , H1) H 1 A12 0.348 A21 = 0.178 H 1 51.337 Ans. 542 (d) 1 ( x1 , x2) := exp x22 A12 + 2 ( A21 A12) x1 2 ( x1 , x2) := exp x1 A21 + 2 ( A12 A21) x2 2 Pcalc := x1 1 x1 , x2 i i i i ( ) exp( A12) + x2i 2 ( x1i , x2i) Psat2 H1 y1calc := i H1 x1 1 x1 , x2 i i i exp ( A 12) Pcalc i ( ) 0.2 0 PiPcalc i (y1iy1calci) 100 0.2 0.4 0.6 0 0.2 0.4 x1 i 0.6 0.8 Pressure residuals y1 residuals Fit GE/RT data to Margules eqn. by least squares: i := 2 .. rows ( P) Given 0= y2 := 1 y1 y1 Pi i d x1 ln H1 dA12 i x1i exp ( A ) 12 y P + x ln 2i i 2i x2 Psat2 i 543 i ... A21 x1 ... x1 x2 i i i + A12 x2i 2 0= i y1 Pi i d x1 ln H1 dA21 i x1i exp ( A ) 12 y P + x ln 2i i 2i x2 Psat2 i y1 Pi i d x1 ln H1 dH1 i x1 i exp ( A ) 12 y2 Pi + x ln i 2i x2 Psat2 i ... A21 x1 ... x1 x2 i i i + A12 x2i 2 2 0= i ... A21 x1 ... x1 x2 i i i + A12 x2i A12 A21 := Find ( A12 , A21 , H1) H 1 2 2 A12 0.375 A21 = 0.148 H 1 53.078 Ans. 1 ( x1 , x2) := exp x2 A12 + 2 ( A21 A12) x1 2 ( x1 , x2) := exp x1 A21 + 2 ( A12 A21) x2 H1 Pcalc := x1 1 x1 , x2 + x2 2 x1 , x2 Psat2 i i i i exp ( A 12) i i i ( ) ( ) y1calc := i H1 x1 1 x1 , x2 i i i exp ( A 12) Pcalc i ( ) 544 0 0.2 PiPcalc i (y1iy1calci) 100 0.4 0.6 0.8 0 0.2 0.4 x1 i 0.6 0.8 Pressure residuals y1 residuals 14.5 Pressures in kPa Data: 0.3193 0.4232 0.5119 0.6096 x1 := 0.7135 0.7934 0.9102 1.000 x2 := 1 x1 21.63 24.01 25.92 27.96 P := 30.12 31.75 34.15 36.09 0.5934 0.6815 0.7440 0.8050 y1 := 0.8639 0.9048 0.9590 1.000 Psat1 := P8 i := 1 .. 7 (a) It follows immediately from Eq. (12.10a) that: ln 2 = A21 Combining this with Eq. (12.10a) yields the required expression. (b) Henry's constant will be found as part of the solution to Part (c). 545 (c) BARKER'S METHOD by non-linear least squares. Margules equation. The most satisfactory procedure for reduction of this set of data is to find the value of Henry's constant by regression along with the Margules parameters. 1 ( x1 , x2 , A12 , A21) := exp ( x2) A12 + 2 ( A21 A12) x1 2 2 2 ( x1 , x2 , A12 , A21) := exp ( x1) A21 + 2 ( A12 A21) x2 Guesses: H2 := 14 A21 := 0.148 A12 := 0.375 Mininize the sums of the squared errors by setting sums of derivatives equal to zero. Given 0 = i d 2 dA Pi x1i 1 x1i , x2i , A12 , A21 Psat1 ... 12 H2 + x2i 2 x1i , x2i , A12 , A21 exp ( A21) ( ) ( ( ) ) 0= dA21 Pi x1 1 ( x1 , x2 , A12 , A21) Psat1 ... i d H2 + x2 2 x1 , x2 , A12 , A21 i i i exp ( A21) i i i 2 0= i d 2 dH Pi x1i 1 x1i , x2i , A12 , A21 Psat1 ... 2 H2 + x2 2 x1 , x2 , A12 , A21 i i i exp ( A21) ( ) ( ) A12 A21 := Find ( A12 , A21 , H2) H 2 A12 0.469 A21 = 0.279 H 2 14.87 Ans. (d) 1 ( x1 , x2) := exp x22 A12 + 2 ( A21 A12) x1 2 ( x1 , x2) := exp x1 A21 + 2 ( A12 A21) x2 2 546 H2 Pcalc := x1 1 x1 , x2 Psat1 + x2 2 x1 , x2 i i i i i i i exp ( A 21) x1 1 x1 , x2 Psat1 i i i y1calc := i Pcalc i ( ) ( ) ( ) The plot of residuals below shows that the procedure used (Barker's method with regression for H2) is not in this case very satisfactory, no doubt because the data do not extend close enough to x1 = 0. 1 0 PiPcalc i 1 2 3 4 (y1iy1calci) 100 0.2 0.4 0.6 x1 i 0.8 Pressure residuals y1 residuals Fit GE/RT data to Margules eqn. by least squares: i := 1 .. 7 Given 0= y2 := 1 y1 i y1 Pi d x1 ln i ... dA12 i x1i Psat1 y2 Pi i + x ln 2i H2 x2 i exp ( A ) 21 A21 x1 ... x1 x2 i i i + A12 x2i 2 547 0= i y1 Pi d x1 ln i ... dA21 i x1i Psat1 y2 Pi i + x ln 2i H2 x2i exp ( A ) 21 A21 x1 ... x1 x2 i i i + A12 x2i 2 2 0= i y1 Pi d x1 ln i ... dH2 i x1i Psat1 y2 Pi i + x ln 2i H2 x2 i exp ( A ) 21 A21 x1 ... x1 x2 i i i + A12 x2i A12 A21 := Find ( A12 , A21 , H2) H 2 2 2 A12 0.37 A21 = 0.204 H 2 15.065 Ans. 1 ( x1 , x2) := exp x2 A12 + 2 ( A21 A12) x1 2 ( x1 , x2) := exp x1 A21 + 2 ( A12 A21) x2 H2 Pcalc := x1 1 x1 , x2 Psat1 + x2 2 x1 , x2 i i i i i i i exp ( A 21) y1calc := i ( ) ( ) x1 1 x1 , x2 Psat1 i i i ( ) Pcalc i 548 0 PiPcalc 0.2 i (y1iy1calci) 100 0.4 0.6 0.3 0.4 0.5 0.6 x1 i 0.7 0.8 0.9 1 Pressure residuals y1 residuals This result is considerably improved over that obtained with Barker's method. 14.6 Pressures in kPa Data: 15.79 17.51 18.15 19.30 19.89 P := 21.37 24.95 29.82 34.80 42.10 0.0 0.0932 0.1248 0.1757 0.2000 x1 := 0.2626 0.3615 0.4750 0.5555 0.6718 0.0 0.1794 0.2383 0.3302 0.3691 y1 := 0.4628 0.6184 0.7552 0.8378 0.9137 i := 2 .. rows ( P) (a) x2 := 1 x1 Psat2 := P1 It follows immediately from Eq. (12.10a) that: ln 1 = A12 Combining this with Eq. (12.10a) yields the required expression 549 (b) Henry's constant will be found as part of the solution to Part (c) (c) BARKER'S METHOD by non-linear least squares. Margules equation. The most satisfactory procedure for reduction of this set of data is to find the value of Henry's constant by regression along with the Margules parameters. 1 ( x1 , x2 , A12 , A21) := exp ( x2) A12 + 2 ( A21 A12) x1 2 2 2 ( x1 , x2 , A12 , A21) := exp ( x1) A21 + 2 ( A12 A21) x2 Guesses: H1 := 35 A21 := 1.27 A12 := 0.70 Mininize the sums of the squared errors by setting sums of derivatives equal to zero. Given 0= i 2 d H1 Pi x1 1 x1 , x2 , A12 , A21 ... i i i exp ( A12) dA12 + x2i 2 x1i , x2i , A12 , A21 Psat2 ( ( ) ) 0= i 2 d H1 Pi x1 1 x1 , x2 , A12 , A21 ... i i exp ( A12) dA21 i + x2i 2 x1i , x2i , A12 , A21 Psat2 ( ( ) ) 0= i 2 d H1 Pi x1 1 x1 , x2 , A12 , A21 ... i i i exp ( A12) dH1 + x2i 2 x1i , x2i , A12 , A21 Psat2 ( ( ) ) A12 A21 := Find ( A12 , A21 , H1) H 1 550 A12 0.731 A21 = 1.187 H 1 32.065 Ans. (d) 1 ( x1 , x2) := exp x2 A12 + 2 ( A21 A12) x1 2 2 ( x1 , x2) := exp x1 A21 + 2 ( A12 A21) x2 2 Pcalc := x1 1 x1 , x2 i i i i ( ) exp( A12) + x2i 2 ( x1i , x2i) Psat2 H1 y1calc := i H1 x1 1 x1 , x2 i i i exp ( A 12) Pcalc i ( ) 0.5 0 PiPcalc i 0.5 1 1.5 2 (y1iy1calci) 100 0 0.1 0.2 0.3 x1 i 0.4 0.5 0.6 0.7 Pressure residuals y1 residuals Fit GE/RT data to Margules eqn. by least squares: i := 2 .. rows ( P) Given 0= y2 := 1 y1 i y1 Pi i d x1 ln H1 dA12 i x1 i exp ( A ) 12 y2 Pi + x ln i 2i x2 Psat2 i 551 ... A21 x1 ... x1 x2 i i i + A12 x2i 2 0= i y1 Pi i d x1 ln H1 dA21 i x1 i exp ( A ) 12 y2 Pi + x ln i 2i x2 Psat2 i y1 Pi i d x1 ln H1 dH1 i x1i exp ( A ) 12 y P + x ln 2i i 2i x2 Psat2 i ... A21 x1 ... x1 x2 i i i + A12 x2i 2 2 0= i ... A21 x1 ... x1 x2 i i i + A12 x2i A12 A21 := Find ( A12 , A21 , H1) H 1 2 2 A12 0.707 A21 = 1.192 H 1 33.356 Ans. 1 ( x1 , x2) := exp x2 A12 + 2 ( A21 A12) x1 2 ( x1 , x2) := exp x1 A21 + 2 ( A12 A21) x2 H1 Pcalc := x1 1 x1 , x2 + x2 2 x1 , x2 Psat2 i i i i exp ( A 12) i i i x1 1 x1 , x2 y1calc := i i i i ( ) ( ) ( ) exp( A12) H1 i Pcalc 552 0 0.5 PiPcalc i 1 1.5 2 2.5 (y1iy1calci) 100 0 0.1 0.2 0.3 x1 i 0.4 0.5 0.6 0.7 Pressure residuals y1 residuals 14.7 Pressures in kPa Data: 0.1757 0.2000 0.2626 0.3615 0.4750 x1 := 0.5555 0.6718 0.8780 0.9398 1.0000 x2 := 1 x1 19.30 19.89 21.37 24.95 29.82 P := 34.80 42.10 60.38 65.39 69.36 0.3302 0.3691 0.4628 0.6184 0.7552 y1 := 0.8378 0.9137 0.9860 0.9945 1.0000 Psat1 := P10 i := 1 .. 9 (a) It follows immediately from Eq. (12.10a) that: ln 2 = A21 Combining this with Eq. (12.10a) yields the required expression. (b) Henry's constant will be found as part of the solution to Part (c). 553 (c) BARKER'S METHOD by non-linear least squares. Margules equation. The most satisfactory procedure for reduction of this set of data is to find the value of Henry's constant by regression along with the Margules parameters. 1 ( x1 , x2 , A12 , A21) := exp ( x2) A12 + 2 ( A21 A12) x1 2 2 2 ( x1 , x2 , A12 , A21) := exp ( x1) A21 + 2 ( A12 A21) x2 Guesses: H2 := 4 A21 := 1.37 A12 := 0.68 Mininize the sums of the squared errors by setting sums of derivatives equal to zero. Given 0= dA12 Pi x1 1 ( x1 , x2 , A12 , A21) Psat1 ... i d H2 + x2 2 x1 , x2 , A12 , A21 i i i exp ( A21) i i i ( ) 2 0= i d 2 dA Pi x1i 1 x1i , x2i , A12 , A21 Psat1 ... 21 H2 + x2i 2 x1i , x2i , A12 , A21 exp ( A21) ( ) ( ) 0= dH2 Pi x1 1 ( x1 , x2 , A12 , A21) Psat1 ... i d H2 + x2 2 x1 , x2 , A12 , A21 i i i exp ( A21) i i i ( ) 2 A12 A21 := Find ( A12 , A21 , H2) H 2 A12 0.679 A21 = 1.367 H 2 3.969 Ans. 554 (d) 1 ( x1 , x2) := exp x22 A12 + 2 ( A21 A12) x1 2 ( x1 , x2) := exp x1 A21 + 2 ( A12 A21) x2 2 H2 Pcalc := x1 1 x1 , x2 Psat1 + x2 2 x1 , x2 i i i i i i i exp ( A 21) y1calc := i ( ) ( ) x1 1 x1 , x2 Psat1 i i i ( ) Pcalc 1 i PiPcalc 0 i (y1iy1calci) 100 1 2 0 0.2 0.4 x1 i 0.6 0.8 Pressure residuals y1 residuals Fit GE/RT data to Margules eqn. by least squares: i := 1 .. 9 Given 0= y2 := 1 y1 i y1 Pi d x1 ln i ... dA12 i x1i Psat1 y2 Pi i + x ln 2i H2 x2 i exp ( A ) 21 A21 x1 ... x1 x2 i i i + A12 x2i 2 555 0= i y1 Pi d x1 ln i ... dA21 i x1i Psat1 y2 Pi i + x ln 2i H2 x2 i exp ( A ) 21 y1 Pi d x1 ln i ... dH2 i x1i Psat1 y2 Pi i + x ln 2i H2 x2i exp ( A ) 21 A21 x1 ... x1 x2 i i i + A12 x2i A21 x1 ... x1 x2 i i i + A12 x2i 2 2 0= i A12 A21 := Find ( A12 , A21 , H2) H 2 2 2 A12 0.845 A21 = 1.229 H 2 4.703 Ans. 1 ( x1 , x2) := exp x2 A12 + 2 ( A21 A12) x1 2 ( x1 , x2) := exp x1 A21 + 2 ( A12 A21) x2 H2 Pcalc := x1 1 x1 , x2 Psat1 + x2 2 x1 , x2 i i i i i i i exp ( A 21) y1calc := i ( ) ( ) x1 1 x1 , x2 Psat1 i i i ( ) Pcalc i 556 1 PiPcalc 0 i (y1iy1calci) 100 1 2 0 0.2 0.4 x1 i 0.6 0.8 Pressure residuals y1 residuals 14.8 (a) Data from Table 12.1 15.51 18.61 21.63 24.01 P := 25.92 kPa x1 := 27.96 30.12 31.75 34.15 n := rows ( P) Psat1 := 36.09kPa 0.0895 0.1981 0.3193 0.4232 0.5119 y1 := 0.6096 0.7135 0.7934 0.9102 n=9 0.2716 0.4565 0.5934 0.6815 0.7440 1 := 0.8050 0.8639 0.9048 0.9590 i 1.304 1.188 1.114 1.071 1.044 2 := 1.023 1.010 1.003 0.997 i i 1.009 1.026 1.050 1.078 1.105 1.135 1.163 1.189 1.268 i i := 1 .. n x2 := 1 x1 y2 := 1 y1 Psat2 := 12.30kPa T := ( 50 + 273.15)K Data reduction with the Margules equation and Eq. (10.5): 1 := i y1 Pi i x1 Psat1 i 2 := i y2 Pi i x2 Psat2 i 557 i := 1 .. n GERTi := x1 ln 1 + x2 ln 2 i i i i A12 := 0.1 A21 := 0.3 ( ) ( ( ) ) Guess: f ( A12 , A21) := i=1 n 2 GERTi A21 x1i + A12 x2i x1i x2i A12 := Minimize ( f , A12 , A21) A21 A12 = 0.374 A21 = 0.197 Ans. RMS Error: RMS := RMS = 1.033 10 3 i=1 n 2 GERTi A21 x1i + A12 x2i x1i x2i ( ) n x1 := 0 , 0.01 .. 1 0.1 GERT i A21 x1+A12 ( 1x1) x1 ( 1x1) 0.05 0 0 0.2 0.4 i 0.6 0.8 x1 , x1 Data reduction with the Margules equation and Eq. (14.1): cm B11 := 1840 mol 3 cm B22 := 1800 mol 3 cm B12 := 1150 mol 3 12 := 2 B12 B11 B22 B11 ( Pi Psat1) + Pi y2 2 12 i 1 := exp i R T 558 ( ) 1 := i y1 1 Pi i i x1 Psat1 i B22 ( Pi Psat2) + Pi y1 2 12 i 2 := exp i R T i := 1 .. n GERTi := x1 ln 1 + x2 ln 2 i i i i A12 := 0.1 A21 := 0.3 ( ) 2 := i y2 2 Pi i i x2 Psat2 i ( ) ( ( ) ) Guess: f ( A12 , A21) := i=1 n 2 GERTi A21 x1i + A12 x2i x1i x2i A12 := Minimize ( f , A12 , A21) A21 n A12 = 0.379 A21 = 0.216 Ans. RMS Error: RMS := 4 i=1 2 GERTi A21 x1i + A12 x2i x1i x2i ( ) n RMS = 9.187 10 x1 := 0 , 0.01 .. 1 0.1 GERT i A21 x1+A12 ( 1x1) x1 ( 1x1) 0.05 0 0 0.5 x1 , x1 i 1 The RMS error with Eqn. (14.1) is about 11% lower than the RMS error with Eqn. (10.5). Note: The following problem was solved with the temperature (T) set at the normal boiling point. To solve for another temperature, simply change T to the approriate value. 559 14.9 (a) Acetylene: T := Tn Tc := 308.3K Tr := T Tc Pc := 61.39bar Tr = 0.614 Tn := 189.4K For Redlich/Kwong EOS: := 1 := 0 1 := 0.08664 := 0.42748 ( Tr ) R Tc 2 2 Table 3.1 ( Tr) := Tr q ( Tr) := 2 Table 3.1 a ( Tr) := ( Tr , Pr) := Guess: Eq. (3.45) Pc Pr Tr zv := 0.9 zv ( Tr , Pr) Eq. (3.53) ( Tr ) Tr Eq. (3.54) Define Z for the vapor (Zv) Given Eq. (3.52) zv = 1 + ( Tr , Pr) q ( Tr) ( Tr , Pr) Zv ( Tr , Pr) := Find ( zv) Define Z for the liquid (Zl) Given Eq. (3.56) ( zv + ( Tr , Pr) ) ( zv + ( Tr , Pr) ) zl := 0.01 Guess: zl = ( Tr , Pr) + zl + ( Tr , Pr) zl + ( Tr , Pr) ( )( ) 1 + ( Tr , Pr) zl q ( T r ) ( T r , Pr ) To find liquid root, restrict search for zl to values less than 0.2, zl < 0.2 Zl ( Tr , Pr) := Find ( zl) Define I for liquid (Il) and vapor (Iv) Il ( Tr , Pr) := 1 1 ln Zl ( Tr , Pr) + ( Tr , Pr) Zl ( Tr , Pr) + ( Tr , Pr) Zv ( Tr , Pr) + ( Tr , Pr) Zv ( Tr , Pr) + ( Tr , Pr) 560 Iv ( Tr , Pr) := ln Eq. (6.65b) lnl ( Tr , Pr) := Zl ( Tr , Pr) 1 ln Zl ( Tr , Pr) ( Tr , Pr) q ( Tr) Il ( Tr , Pr) Eq. (11.37) lnv ( Tr , Pr) := Zv ( Tr , Pr) 1 ln Zv ( Tr , Pr) ( Tr , Pr) q ( Tr) Iv ( Tr , Pr) Guess Psat: Psatr := Given 1bar Pc ( ) ( ) lnl ( Tr , Psatr) = lnv ( Tr , Psatr) Zl ( Tr , Psatr) = 4.742 10 Psat = 1.6 bar Ans. 3 Psatr := Find ( Psatr) Zv ( Tr , Psatr) = 0.965 Psatr = 0.026 Psat := Psatr Pc The following table lists answers for all parts. Literature values are interpolated from tables in Perry's Chemical Engineers' Handbook, 6th ed. The last column shows the percent difference between calculated and literature values at 0.85Tc. These range from 0.1 to 27%. For the normal boiling point (Tn), Psat should be 1.013 bar. Tabulated results for Psat do not agree well with this value. Differences range from 3 to > 100%. Tn (K) Psat (bar) 0.85 Tc (K) Psat (bar) Psat (bar) % Difference @ Tn @ 0.85 Tc Lit. Values 189.4 1.60 262.1 20.27 19.78 2.5% Acetylene 87.3 0.68 128.3 20.23 18.70 8.2% Argon 353.2 1.60 477.9 16.028 15.52 3.2% Benzene 272.7 1.52 361.3 14.35 12.07 18.9% n-Butane 0.92 113.0 15.2 12.91 17.7% Carbon Monoxide 81.7 447.3 2.44 525.0 6.633 5.21 27.3% n-Decane 169.4 1.03 240.0 17.71 17.69 0.1% Ethylene 371.6 2.06 459.2 7.691 7.59 1.3% n-Heptane 111.4 0.71 162.0 19.39 17.33 11.9% Methane 77.3 0.86 107.3 14.67 12.57 16.7% Nitrogen 14.10 (a) Acetylene: := 0.187 T := Tn Tc := 308.3K Pc := 61.39bar Tn := 189.4K Tr := T Tc Note: For solution at 0.85T c, set T := 0.85Tc. For SRK EOS: := 1 := 0 := 0.08664 561 Tr = 0.614 := 0.42748 Table 3.1 1 2 2 ( Tr , ) := 1 + ( 0.480 + 1.574 0.176 ) 1 Tr 2 Table 3.1 Tr , R Tc a ( Tr) := Pc q ( Tr) := Tr , Tr ( ) 2 2 Eq. (3.45) ( Tr , Pr) := Pr Tr ( ) Eq. (3.54) Eq. (3.53) Define Z for the vapor (Zv) Given Eq. (3.52) Guess: zv := 0.9 zv ( Tr , Pr) zv = 1 + ( Tr , Pr) q ( Tr) ( Tr , Pr) Zv ( Tr , Pr) := Find ( zv) Define Z for the liquid (Zl) Given Eq. (3.56) ( zv + ( Tr , Pr) ) ( zv + ( Tr , Pr) ) Guess: zl := 0.01 zl = ( Tr , Pr) + zl + ( Tr , Pr) zl + ( Tr , Pr) ( )( ) 1 + ( Tr , Pr) zl q ( T r ) ( T r , Pr ) To find liquid root, restrict search for zl to values less than 0.2, zl < 0.2 Zl ( Tr , Pr) := Find ( zl) Define I for liquid (Il) and vapor (Iv) Il ( Tr , Pr) := 1 1 ln Zl ( Tr , Pr) + ( Tr , Pr) Zl ( Tr , Pr) + ( Tr , Pr) Zv ( Tr , Pr) + ( Tr , Pr) Zv ( Tr , Pr) + ( Tr , Pr) Eq. (6.65b) Iv ( Tr , Pr) := ln 562 lnl ( Tr , Pr) := Zl ( Tr , Pr) 1 ln Zl ( Tr , Pr) ( Tr , Pr) q ( Tr) Il ( Tr , Pr) Eq. (11.37) lnv ( Tr , Pr) := Zv ( Tr , Pr) 1 ln Zv ( Tr , Pr) ( Tr , Pr) q ( Tr) Iv ( Tr , Pr) Guess Psat: Psatr := Given 2bar Pc ( ) ( ) lnl ( Tr , Psatr) = lnv ( Tr , Psatr) Zl ( Tr , Psatr) = 3.108 10 Psat = 1.073 bar Ans. 3 Psatr := Find ( Psatr) Zv ( Tr , Psatr) = 0.975 Psatr = 0.017 Psat := Psatr Pc The following table lists answers for all parts. Literature values are interpolated from tables in Perry's Chemical Engineers' Handbook, 6th ed. The last column shows the percent difference between calculated and literature values at 0.85Tc. These range from less than 0.1 to 2.5%. For the normal boiling point (Tn), Psat should be 1.013 bar. Tabulated results for Psat agree well with this value. Differences range from near 0 to 6%. Tn (K) Psat (bar) 0.85 Tc (K) Psat (bar) Psat (bar) % Difference @ Tn @ 0.85 Tc Lit. Values 189.4 1.073 262.1 20.016 19.78 1.2% 87.3 0.976 128.3 18.79 18.70 0.5% 353.2 1.007 477.9 15.658 15.52 0.9% 272.7 1.008 361.3 12.239 12.07 1.4% 81.7 1.019 113.0 12.871 12.91 -0.3% 447.3 1.014 525.0 5.324 5.21 2.1% 169.4 1.004 240.0 17.918 17.69 1.3% 371.6 1.011 459.2 7.779 7.59 2.5% 111.4 0.959 162.0 17.46 17.33 0.8% 77.3 0.992 107.3 12.617 12.57 0.3% Acetylene Argon Benzene n-Butane Carbon Monoxide n-Decane Ethylene n-Heptane Methane Nitrogen 14.10 (b) Acetylene: := 0.187 T := Tn Tc := 308.3K Pc := 61.39bar Tn := 189.4K Tr := T Tc Note: For solution at 0.85T c, set T := 0.85Tc. For PR EOS: := 1 + 2 := 1 2 := 0.07779 563 Tr = 0.614 := 0.45724 Table 3.1 1 2 2 Table 3.1 ( Tr , ) := 1 + ( 0.37464 + 1.54226 0.26992 ) 1 Tr 2 a ( Tr) := q ( Tr) := Tr , R Tc Pc ( ) 2 2 Eq. (3.45) ( Tr , Pr) := Pr Tr Tr , Tr ( ) Eq. (3.54) Eq. (3.53) Define Z for the vapor (Zv) Given Eq. (3.52) Guess: zv := 0.9 zv ( Tr , Pr) zv = 1 + ( Tr , Pr) q ( Tr) ( Tr , Pr) Zv ( Tr , Pr) := Find ( zv) Define Z for the liquid (Zl) Given Eq. (3.56) ( zv + ( Tr , Pr) ) ( zv + ( Tr , Pr) ) Guess: zl := 0.01 zl = ( Tr , Pr) + zl + ( Tr , Pr) zl + ( Tr , Pr) ( )( ) 1 + ( Tr , Pr) zl q ( T r ) ( T r , Pr ) To find liquid root, restrict search for zl to values less than 0.2,zl < 0.2 Zl ( Tr , Pr) := Find ( zl) Define I for liquid (Il) and vapor (Iv) Il ( Tr , Pr) := 1 1 ln Zl ( Tr , Pr) + ( Tr , Pr) Zl ( Tr , Pr) + ( Tr , Pr) Zv ( Tr , Pr) + ( Tr , Pr) Zv ( Tr , Pr) + ( Tr , Pr) Eq. (6.65b) Iv ( Tr , Pr) := ln 564 lnl ( Tr , Pr) := Zl ( Tr , Pr) 1 ln Zl ( Tr , Pr) ( Tr , Pr) q ( Tr) Il ( Tr , Pr) Eq. (11.37) lnv ( Tr , Pr) := Zv ( Tr , Pr) 1 ln Zv ( Tr , Pr) ( Tr , Pr) q ( Tr) Iv ( Tr , Pr) Guess Psat: Given Psatr = 0.018 Psat := Psatr Pc Psatr := 2bar Pc ( ) ( ) lnl ( Tr , Psatr) = lnv ( Tr , Psatr) Zl ( Tr , Psatr) = 2.795 10 Psat = 1.09 bar Ans. 3 Psatr := Find ( Psatr) Zv ( Tr , Psatr) = 0.974 The following table lists answers for all parts. Literature values are interpolated from tables in Perry's Chemical Engineers' Handbook, 6th ed. The last column shows the percent difference between calculated and literature values at 0.85Tc. These range from less than 0.1 to 1.2%. For the normal boiling point (Tn), Psat should be 1.013 bar. Tabulated results for Psat agree well with this value. Differences range from near 0 to 7.6%. Tn (K) Psat (bar) 0.85 Tc (K) Psat (bar) Psat (bar) % Difference @ Tn @ 0.85 Tc Lit. Values 189.4 1.090 262.1 19.768 19.78 -0.1% 87.3 1.015 128.3 18.676 18.70 -0.1% 353.2 1.019 477.9 15.457 15.52 -0.4% 272.7 1.016 361.3 12.084 12.07 0.1% 81.7 1.041 113.0 12.764 12.91 -1.2% 447.3 1.016 525.0 5.259 5.21 0.9% 169.4 1.028 240.0 17.744 17.69 0.3% 371.6 1.012 459.2 7.671 7.59 1.1% 111.4 0.994 162.0 17.342 17.33 0.1% 77.3 1.016 107.3 12.517 12.57 -0.4% Acetylene Argon Benzene n-Butane Carbon Monoxide n-Decane Ethylene n-Heptane Methane Nitrogen 14.12 (a) van der Waals Eqn. := 0 ( Tr) Tr := Tr := 0.7 1 8 := Pr Tr 27 64 ( Tr) := 1 := 0 q ( Tr) := ( Tr , Pr) := zv := 0.9 (guess) zv ( Tr , Pr) ( zv) 2 Given zv = 1 + ( Tr , Pr) q ( Tr) ( Tr , Pr) Eq. (3.52) 565 Zv ( Tr , Pr) := Find ( zv) zl := .01 Given (guess) zl = ( Tr , Pr) + ( zl) 2 1 + ( Tr , Pr) zl q ( Tr) ( Tr , Pr) Eq. (3.56) zl < 0.2 Zl ( Tr , Pr) := Find ( zl) Iv ( Tr , Pr) := ( Tr , Pr) Zv ( Tr , Pr) Il ( Tr , Pr) := ( Tr , Pr) Zl ( Tr , Pr) Case II, pg. 218. By Eq. (11.39): lnv ( Tr , Pr) := Zv ( Tr , Pr) 1 ln ( Zv ( Tr , Pr) ( Tr , Pr) ) q ( Tr) Iv ( Tr , Pr) lnl ( Tr , Pr) := Zl ( Tr , Pr) 1 ln ( Zl ( Tr , Pr) ( Tr , Pr) ) q ( Tr) Il ( Tr , Pr) Psatr := .1 Given lnl ( Tr , Psatr) lnv ( Tr , Psatr) = 0 Zl ( Tr , Psatr) = 0.05 Psatr := Find ( Psatr) Psatr = 0.2 Zv ( Tr , Psatr) = 0.839 lnl ( Tr , Psatr) = 0.148 := 1 log ( Psatr) (b) lnv ( Tr , Psatr) = 0.148 ( Tr , Psatr) = 0.036 = 0.302 Ans. Redlich/Kwong Eqn.Tr := 0.7 := 0 .5 := 1 ( Tr) := Tr q ( Tr) := := 0.08664 := 0.42748 ( Tr) Tr ( Tr , Pr) := Pr Tr Guess: zv := 0.9 Given zv = 1 + ( Tr , Pr) q ( Tr) ( Tr , Pr) Zv ( Tr , Pr) := Find ( zv) Guess: zl := .01 zv ( zv + ( Tr , Pr) ) zv ( Tr , Pr) Eq. (3.52) 566 Given zl < 0.2 zl = ( Tr , Pr) + zl ( zl + ( Tr , Pr) ) Zl ( Tr , Pr) := Find ( zl) 1 + ( Tr , Pr) zl q ( Tr) ( Tr , Pr) Eq. (3.55) Iv ( Tr , Pr) := ln Zv ( Tr , Pr) + ( Tr , Pr) Il ( Tr , Pr) := ln Zl ( Tr , Pr) + ( Tr , Pr) Zv ( Tr , Pr) Zl ( Tr , Pr) By Eq. (11.39): lnv ( Tr , Pr) := Zv ( Tr , Pr) 1 ln ( Zv ( Tr , Pr) ( Tr , Pr) ) q ( Tr) Iv ( Tr , Pr) lnl ( Tr , Pr) := Zl ( Tr , Pr) 1 ln ( Zl ( Tr , Pr) ( Tr , Pr) ) q ( Tr) Il ( Tr , Pr) Psatr := .1 Given lnl ( Tr , Psatr) = lnv ( Tr , Psatr) Zl ( Tr , Psatr) = 0.015 Psatr := Find ( Psatr) Psatr = 0.087 ( Tr , Psatr) = 0.011 Zv ( Tr , Psatr) = 0.913 lnv ( Tr , Psatr) = 0.083 lnl ( Tr , Psatr) = 0.083 := 1 log ( Psatr) = 0.058 Ans. 14.15 (a) x1 := 0.1 Guess: x2 := 1 x1 A12 := 2 2 x1 := 0.9 A21 := 2 x2 := 1 x1 1 ( A21 , A12) := exp x2 A12 + 2 ( A21 A12) x1 1( A21 , A12) := exp x2 A12 + 2 ( A21 A12) x1 2 2 ( A21 , A12) := exp x1 A21 + 2 ( A12 A21) x2 2 2( A21 , A12) := exp x1 A21 + 2 ( A12 A21) x2 2 567 Given x1 1 ( A21 , A12) = x1 1( A21 , A12) x2 2 ( A21 , A12) = x2 2( A21 , A12) A12 := Find ( A12 , A21) A21 (b) x1 := 0.2 Guess: A21 = 2.747 A12 = 2.747 Ans. x2 := 1 x1 A12 := 2 2 x1 := 0.9 A21 := 2 x2 := 1 x1 1 ( A21 , A12) := exp x2 A12 + 2 ( A21 A12) x1 1( A21 , A12) := exp x2 A12 + 2 ( A21 A12) x1 2 2 ( A21 , A12) := exp x1 A21 + 2 ( A12 A21) x2 2 2( A21 , A12) := exp x1 A21 + 2 ( A12 A21) x2 2 Given x1 1 ( A21 , A12) = x1 1( A21 , A12) x2 2 ( A21 , A12) = x2 2( A21 , A12) A12 := Find ( A12 , A21) A21 (c) x1 := 0.1 Guess: A12 = 2.148 A21 = 2.781 Ans. x2 := 1 x1 A12 := 2 2 x1 := 0.8 A21 := 2 x2 := 1 x1 1 ( A21 , A12) := exp x2 A12 + 2 ( A21 A12) x1 1( A21 , A12) := exp x2 A12 + 2 ( A21 A12) x1 2 2 ( A21 , A12) := exp x1 A21 + 2 ( A12 A21) x2 2 2( A21 , A12) := exp x1 A21 + 2 ( A12 A21) x2 2 568 Given x1 1 ( A21 , A12) = x1 1( A21 , A12) x2 2 ( A21 , A12) = x2 2( A21 , A12) A12 := Find ( A12 , A21) A21 14.16 (a) x1 := 0.1 Guess: Given A12 = 2.781 A21 = 2.148 Ans. x2 := 1 x1 a12 := 2 x1 := 0.9 a21 := 2 x2 := 1 x1 2 2 a12 x1 a12 x1 exp a12 1 + x1 = exp a12 1 + x1 a21 x2 a21 x2 2 2 a21 x2 a21 x2 exp a21 1 + x2 = exp a21 1 + x2 a12 x1 a12 x1 a12 := Find ( a12 , a21) a21 (b) x1 := 0.2 Guess: Given a12 = 2.747 a21 = 2.747 Ans. x2 := 1 x1 a12 := 2 x1 := 0.9 a21 := 2 x2 := 1 x1 2 2 a12 x1 a12 x1 exp a12 1 + x1 = exp a12 1 + x1 a21 x2 a21 x2 2 2 a21 x2 a21 x2 exp a21 1 + x2 = exp a21 1 + x2 a12 x1 a12 x1 a12 := Find ( a12 , a21) a21 a12 = 2.199 a21 = 2.81 Ans. 569 (c) x1 := 0.1 Guess: x2 := 1 x1 a12 := 2 x1 := 0.8 a21 := 2 x2 := 1 x1 Given 2 2 a12 x1 a12 x1 exp a12 1 + x1 = exp a12 1 + x1 a21 x2 a21 x2 2 2 a21 x2 a21 x2 exp a21 1 + x2 = exp a21 1 + x2 a12 x1 a12 x1 a12 := Find ( a12 , a21) a21 a12 = 2.81 a21 = 2.199 Ans. 14.18 (a) a := 975 T := 250 .. 450 2.1 b := 18.4 A ( T) := c := 3 a + b c ln ( T) T A ( T) 2 1.9 250 300 350 T 400 450 Parameter A = 2 at two temperatures. The lower one is an UCST, because A decreases to 2 as T increases. The higher one is a LCST, because A decreases to 2 as T decreases. Guess: Given x0 x := 0.25 A ( T) ( 1 2 x) = ln x 0.5 1 x x Eq. (E), Ex. 14.5 x2 ( T) := 1 x1 ( T) x1 ( T) := Find ( x) 570 UCST := 300 (guess) Given A ( UCST) = 2 UCST := Find ( UCST) UCST = 272.93 LCST := 400 (guess) Given A ( LCST) = 2 LCST := Find ( LCST) LCST = 391.21 Plot phase diagram as a function of T T1 := 225 , 225.1 .. UCST T2 := LCST .. 450 500 T1 T1 T2 T2 400 300 200 0.2 0.3 0.4 0.5 0.6 0.7 0.8 x1 ( T1 ) , x2 ( T1 ) , x1 ( T2 ) , x2 ( T2 ) (b) a := 540 T := 250 .. 450 2.5 b := 17.1 A ( T) := c := 3 a + b c ln ( T) T A ( T) 2 1.5 250 300 350 T 400 450 Parameter A = 2 at a single temperature. It is a LCST, because A decreases to 2 as T decreases. 571 Guess: Given x0 x := 0.25 A ( T) ( 1 2 x) = ln x 0.5 1 x x Eq. (E), Ex. 14.5 x1 ( T) := Find ( x) LCST := 350 (guess) Given A ( LCST) = 2 LCST := Find ( LCST) LCST = 346 Plot phase diagram as a function of T 450 T := LCST .. 450 T T 400 350 300 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 x1 ( T ) , 1x1 ( T) (c) a := 1500 T := 250 .. 450 b := 19.9 A ( T) := c := 3 a + b c ln ( T) T 3 2.5 A ( T) 2 1.5 250 300 350 T 400 450 Parameter A = 2 at a single temperature. It is an UCST, because A decreases to 2 as T increases. 572 Guess: x := 0.25 A ( T) ( 1 2 x) = ln x 0.5 1 x x Eq. (E), Ex. 14.5 Given x0 x1 ( T) := Find ( x) UCST := 350 (guess) Given A ( UCST) = 2 UCST := Find ( UCST) UCST = 339.66 Plot phase diagram as a function of T 350 T := UCST .. 250 T T 300 250 0 0.2 0.4 0.6 0.8 x1 ( T ) , 1x1 ( T) 14.20 Guess: Given x1 := 0.5 x1 := 0.5 Write Eq. (14.74) for species 1: 2 2 x1 exp 0.4 ( 1 x1) = x1 exp 0.8 ( 1 x1) x1 1 x1 + x1 1 x1 = 1 (Material balance) x1 := Find ( x1 , x1) x1 x1 = 0.371 x1 = 0.291 Ans. 573 14.22 Temperatures in kelvins; pressures in kPa. P1sat ( T) := exp 19.1478 5363.7 T 2048.97 T water P := 1600 SF6 P2sat ( T) := exp 14.6511 Find 3-phase equilibrium temperature and vapor-phase composition (pp. 594-5 of text): Guess: Given T := 300 P = P1sat ( T) + P2sat ( T) Tstar := Find ( T) Tstar = 281.68 P1sat ( Tstar) 6 y1star 10 = 695 P Find saturation temperatures of pure species 2: y1star := Guess: Given T := 300 P2sat ( T) = P T2 := Find ( T) T2 = 281.71 P2sat ( T) P P1sat ( T) TI := Tstar , Tstar + 0.01 .. Tstar + 6 y1I ( T) := P Because of the very large difference in scales appropriate to regions I and II [Fig. 14.21(a)], the txy diagram is presented on the following page in two parts, showing regions I and II separately. TII := Tstar , Tstar + 0.0001 .. T2 y1II ( T) := 1 281.7 TII Tstar 281.69 281.68 0 100 200 300 6 400 500 6 600 700 y1II ( TII) 10 , y1II ( TII) 10 574 288 286 TI Tstar 284 282 280 650 700 750 800 850 6 900 950 6 1000 1050 y1I ( TI) 10 , y1I ( TI) 10 14.24 Temperatures in deg. C; pressures in kPa P1sat ( T) := exp 13.9320 3056.96 T + 217.625 Toluene P := 101.33 Water 3885.70 P2sat ( T) := exp 16.3872 T + 230.170 Find the three-phase equilibrium T and y: Guess: Given y1star := T := 25 P = P1sat ( T) + P2sat ( T) P1sat ( Tstar) P Tstar := Find ( T) y1star = 0.444 Tstar = 84.3 For z1 < y1*, first liquid is pure species 2. y1 := 0.2 Given Guess: Tdew := Tstar Tdew := Find ( Tdew) Tdew = 93.855 Ans. y1 = 1 P2sat ( Tdew) P For z1 > y1*, first liquid is pure species 1. 575 y1 := 0.7 Given Guess: y1 = Tdew := Tstar Tdew := Find ( Tdew) Tdew = 98.494 Ans. P1sat ( Tdew) P In both cases the bubblepoint temperature is T*, and the mole fraction of the last vapor is y1*. 14.25 Temperatures in deg. C; pressures in kPa. P1sat ( T) := exp 13.8622 2910.26 T + 216.432 3885.70 T + 230.170 n-heptane P := 101.33 water P2sat ( T) := exp 16.3872 Find the three-phase equilibrium T and y: Guess: Given y1star := T := 50 P = P1sat ( T) + P2sat ( T) P1sat ( Tstar) P Tstar := Find ( T) y1star = 0.548 Tstar = 79.15 Since 0.35<y1*, first liquid is pure species 2. y1 ( T) := 1 P2sat ( T) P Tdew := Tstar Tdew := Find ( Tdew) Tdew = 88.34 Find temperature of initial condensation at y1=0.35: y10 := 0.35 Given Guess: y1 ( Tdew) = y10 Define the path of vapor mole fraction above and below the dew point. y1path ( T) := if ( T > Tdew , y10 , y1 ( T) ) T := 100 , 99.9 .. Tstar Path of mole fraction heptane in residual vapor as temperature is decreased. No vapor exists below Tstar. 576 100 95 90 T 85 80 75 Tstar Tdew 0.3 0.35 0.4 0.45 0.5 0.55 y1path ( T ) 14.26 Pressures in kPa. P1sat := 75 P2sat := 110 A := 2.25 2 1 ( x1) := exp A ( 1 x1) 2 ( x1) := exp A x1 ( 2 ) Find the solubility limits: Guess: Given x1 = 0.224 x1 := 0.1 A ( 1 2 x1) = ln 1 x1 x1 x1 := Find ( x1) x1 = 0.776 x1 := 1 x1 Find the conditions for VLLE: Guess: Pstar := P1sat y1star := 0.5 Given Pstar = x1 1 ( x1) P1sat + ( 1 x1) 2 ( x1) P2sat y1star Pstar = x1 1 ( x1) P1sat Pstar := Find ( Pstar , y1star) y1star Pstar = 160.699 y1star = 0.405 Calculate VLE in two-phase region. Modified Raoult's law; vapor an ideal gas. Guess: x1 := 0.1 P := 50 577 Given P = x1 1 ( x1) P1sat + ( 1 x1) 2 ( x1) P2sat y1 ( x1) := x1 1 ( x1) P1sat P ( x1) P ( x1) := Find ( P) Plot the phase diagram. Define liquid equilibrium line: PL ( x1) := if ( P ( x1) < Pstar , P ( x1) , Pstar) Define vapor equilibrium line: PV ( x1) := if ( P ( x1) < Pstar , P ( x1) , Pstar) Define pressures for liquid phases above Pstar: Pliq := Pstar .. Pstar + 10 x1 := 0 , 0.01 .. 1 200 175 PL ( x1) 150 PV ( x1) Pliq Pliq 125 100 75 50 0 0.2 0.4 0.6 0.8 1 Pstar x1 , y1 ( x1) , x1 , x1 x1 := 0 , 0.05 .. 0.2 x1 = 0 0.05 0.1 0.15 0.2 PL ( x1) = 110 133.66 147.658 155.523 159.598 y1 ( x1) = 0 0.214 0.314 0.368 0.397 578 x1 := 1 , 0.95 .. 0.8 x1 = 1 0.95 0.9 0.85 0.8 PL ( x1) = 75 113.556 137.096 150.907 158.506 y1 ( x1) = 1 0.631 0.504 0.444 0.414 x1 = 0.224 x1 = 0.776 y1star = 0.405 14.27 Temperatures in deg. C; pressures in kPa. Water: P1sat ( T) := exp 16.3872 3885.70 T + 230.170 2451.88 T + 232.014 2910.26 T + 216.432 z3 := 1 z1 z2 n-Pentane: P2sat ( T) := exp 13.7667 P3sat ( T) := exp 13.8622 z1 := 0.45 z2 := 0.30 n-Heptane: P := 101.33 (a) Calculate dew point T and liquid composition assuming the hydrocarbon layer forms first: Tdew1 := 100 x2 := z2 x3 := 1 x2 Guess: Given P = x2 P2sat ( Tdew1) + x3 P3sat ( Tdew1) z3 P = x3 P3sat ( Tdew1) x2 + x3 = 1 x2 x3 := Find ( x2 , x3 , Tdew1) Tdew1 Tdew1 = 66.602 x3 = 0.706 579 x2 = 0.294 Calculate dew point temperature assuming the water layer forms first: x1 := 1 Given Guess: Tdew2 := 100 Tdew2 := Find ( Tdew2) Tdew2 = 79.021 Since Tdew2 > Tdew1, the water layer forms first (b) Calculate the temperature at which the second layer forms: Guess: Tdew3 := 100 y1 := z1 Given x2 := z2 y2 := z2 x3 := 1 x2 y3 := z3 x1 P1sat ( Tdew2) = z1 P P = P1sat ( Tdew3) + x2 P2sat ( Tdew3) + x3 P3sat ( Tdew3) y1 P = P1sat ( Tdew3) y2 z2 = y3 z3 y1 + y2 + y3 = 1 x2 + x3 = 1 y2 P = x2 P2sat ( Tdew3) y1 y2 y3 := Find ( y1 , y2 , y3 , Tdew3 , x2 , x3) Tdew3 x2 x3 y1 = 0.288 Tdew3 = 68.437 (c) y2 = 0.388 x2 = 0.1446 y3 = 0.324 x3 = 0.8554 Calculate the bubble point given the total molar composition of the two phases x2 := z2 z2 + z3 x3 := z3 z2 + z3 Tbubble := Tdew3 x2 = 0.545 x3 = 0.455 580 Given P = P1sat ( Tbubble) + x2 P2sat ( Tbubble) + x3 P3sat ( Tbubble) Tbubble := Find ( Tbubble) P1sat ( Tbubble) P x2 P2sat ( Tbubble) y2 := P x3 P3sat ( Tbubble) y3 := P y1 := Tbubble = 48.113 y1 = 0.111 y2 = 0.81 y3 = 0.078 14.28 Temperatures in deg. C; pressures in kPa. Water: P1sat ( T) := exp 16.3872 3885.70 T + 230.170 2451.88 T + 232.014 2910.26 T + 216.432 z3 := 1 z1 z2 n-Pentane: P2sat ( T) := exp 13.7667 P3sat ( T) := exp 13.8622 z1 := 0.32 z2 := 0.45 n-Heptane: P := 101.33 (a) Guess: Given Calculate dew point T and liquid composition assuming the hydrocarbon layer forms first: Tdew1 := 70 x2 := z2 x3 := 1 x2 P = x2 P2sat ( Tdew1) + x3 P3sat ( Tdew1) z3 P = x3 P3sat ( Tdew1) x2 + x3 = 1 x2 x3 := Find ( x2 , x3 , Tdew1) Tdew1 Tdew1 = 65.122 x3 = 0.686 581 x2 = 0.314 Calculate dew point temperature assuming the water layer forms first: x1 := 1 Given Guess: Tdew2 := 70 Tdew2 := Find ( Tdew2) Tdew2 = 70.854 Since Tdew1>Tdew2, a hydrocarbon layer forms first (b) Calculate the temperature at which the second layer forms: Guess: Tdew3 := 100 y1 := z1 Given x2 := z2 y2 := z2 x3 := 1 x2 y3 := z3 x1 P1sat ( Tdew2) = z1 P P = P1sat ( Tdew3) + x2 P2sat ( Tdew3) + x3 P3sat ( Tdew3) y1 P = P1sat ( Tdew3) y2 z2 = y3 z3 y1 + y2 + y3 = 1 x2 + x3 = 1 y2 P = x2 P2sat ( Tdew3) y1 y2 y3 := Find ( y1 , y2 , y3 , Tdew3 , x2 , x3) Tdew3 x2 x3 y1 = 0.24 Tdew3 = 64.298 (c) y2 = 0.503 x2 = 0.2099 y3 = 0.257 x3 = 0.7901 Calculate the bubble point given the total molar composition of the two phases x2 := z2 z2 + z3 x3 := z3 z2 + z3 Tbubble := Tdew3 x2 = 0.662 582 x3 = 0.338 Given P = P1sat ( Tbubble) + x2 P2sat ( Tbubble) + x3 P3sat ( Tbubble) Tbubble = 43.939 y1 = 0.09 y2 = 0.861 y3 = 0.049 Tbubble := Find ( Tbubble) P1sat ( Tbubble) P x2 P2sat ( Tbubble) y2 := P x3 P3sat ( Tbubble) y3 := P y1 := 14.32 := 0.302 0.224 Tc := 748.4 K 304.2 Pc := 40.51 bar 73.83 P := 10bar , 20bar .. 300bar T := 353.15K Use SRK EOS From Table 3.1, p. 98 of text: := 1 := 0 := 0.08664 := 0.42748 T Tr := Tc := 1 + 0.480 + 1.574 0.176 1 Tr 2 ( )( 0.5 ) 2 Eq. (14.32) 2 2 R Tc Eq. (14.31) a := Pc R Tc b := Pc 6.842 kg m5 a= 0.325 s2 mol2 2 ( P) := z2 := 1 b2 P R T Eq. (14.33) 1.331 10 4 m3 b= 2.968 10 5 mol q2 := a2 b2 R T Eq. (14.34) (guess) 583 Given z2 = 1 + 2 ( P) q2 2 ( P) Z2 ( P) := Find ( z2) I2 ( P) := ln ( z2 2 ( P) z2 + 2 ( P) z2 + 2 ( P) )( ) Eq. (14.36) Z 2 ( P) + 2 ( P) Z 2 ( P) Eq. (6.65b) For simplicity, let 1 represent the infinite-dilution value of the fugacity coefficient of species 1 in solution. Eq. (14.103): 1 ( P) := exp l12 := 0.088 b1 ( Z2 ( P) 1) ln ( Z2 ( P) 2 ( P) ) ... b2 0.5 b1 a1 + q2 2 ( 1 l12) I2 ( P) b2 a2 V1 := 124.5 cm 3 Psat1 := 0.0102bar mol Eqs. (14.98) and (14.99), with sat1 = 1 and (P - Psat1) = P, combine to give: y1 ( P) := Psat1 P 1 ( P) exp P V 1 R T 0.1 0.01 y1 ( P) 1 .10 1 .10 3 4 0 50 100 150 P bar 200 250 300 584 14.33 := 0.302 0.038 Tc := 748.4 K 126.2 Pc := 40.51 bar 34.00 P := 10bar , 20bar .. 300bar T := 308.15K (K) Use SRK EOS From Table 3.1, p. 98 of text: := 1 := 0 := 0.08664 := 0.42748 T Tr := Tc 2 0.5 := 1 + 0.480 + 1.574 0.176 1 Tr ( )( ) 2 2 2 R Tc a := Pc Eq. (14.31) R Tc b := Pc Eq. (14.32) 7.298 kg m5 a= 0.067 s2 mol2 2 ( P) := z2 := 1 Given z2 = 1 + 2 ( P) q2 2 ( P) Z2 ( P) := Find ( z2) I2 ( P) := ln b2 P R T Eq. (14.33) 1.331 10 4 m3 b= 2.674 10 5 mol q2 := a2 b2 R T Eq. (14.34) (guess) ( z2 2 ( P) z2 + 2 ( P) z2 + 2 ( P) )( ) Eq. (14.36) Z 2 ( P) + 2 ( P) Z 2 ( P) Eq. (6.65b) For simplicity, let 1 represent the infinite-dilution value of the fugacity coefficient of species 1 in solution. 585 l12 := 0.0 Eq. (14.103): 1 ( P) := exp b1 ( Z2 ( P) 1) ln ( Z2 ( P) 2 ( P) ) ... b2 0.5 b1 a1 + q2 2 ( 1 l12) I2 ( P) b2 a2 4 Psat1 := 2.9 10 bar V1 := 125 cm 3 mol Eqs. (14.98) and (14.99), with sat1 = 1 and (P - Psat1) = P, combine to give: y1 ( P) := Psat1 P 1 ( P) exp P V 1 R T 10 y1 ( P) 10 5 1 0 50 100 150 P bar 200 250 300 Note: y axis is log scale. 586 14.45 A labeled diagram of the process is given below. The feed stream is taken as the phase and the solvent stream is taken as the phase. F nF xF1 = 0.99 xF2 = 0.01 S nS xS3 = 1.0 Feed Mixer/ Settler Solvent R nR x1 x2 = 0.001 E nE x2 x3 Define the values given in the problem statement. Assume as a basis a feed rate nF = 1 mol/s. nF := 1 mol s xF1 := 0.99 x2 := 0.001 xF2 := 0.01 x1 := 1 x2 xS3 := 1 Apply mole balances around the process as well as an equilibrium relationship From p. 585 A12 := 1.5 A23 := 0.8 2 2 ( x2) := exp A12 ( 1 x2) 2 2 ( x2) := exp A23 ( 1 x2) Material Balances nS + nF = nE + nR nS = x3 nE xF1 nF = x1 nR (Total) (Species 3) (Species 1) Substituting the species balances into the total balance yields xF1 1 nS + nF = nS + nF x3 x1 Solving for the ratio of solvent to feed (nS/nF) gives nS nF = x1 xF1 x3 1 x3 x1 587 We need x3. Assume exiting streams are at equilibrium. Here, the only distributing species is 2. Then x2 2 = x2 2 Substituting for 2 and 2 2 2 x2 exp A12 1 x2 = x2 exp A23 1 x2 ( ) ( ) Solve for x2 using Mathcad Solve Block Guess: Given 2 2 x2 exp A12 1 x2 = x2 exp A23 1 x2 x2 := 0.5 ( ) ( ) x2 := Find x2 ( ) x2 = 0.00979 x3 := 1 x2 x3 = 0.9902 From above, the equation for the ratio nS/nF is: nSnF := x1 xF1 x3 1 x3 x1 Ans. Ans. a) nSnF = 0.9112 b) x2 = 0.00979 c) "Good chemistry" here means that species 2 and 3 "like" each other, as evidenced by the negative GE23. "Bad chemistry" would be reflected in a positive GE23, with values less than (essential) but perhaps near to GE12. 14.46 1 - n-hexane 2 - water Since this is a dilute system in both phases, Eqns. (C) and (D) from Example 14.4 on p. 584 can be used to find 1 and 2. x1 := 520 10 6 x2 := 1 x1 x2 := 2 10 6 x1 := 1 x2 588 1 := 2 := x1 x1 1 x1 1 x1 1 = 1.923 10 2 = 4.997 10 3 Ans. 5 Ans. 14.50 1 - butanenitrile Psat1 := 0.07287bar 2- benzene Psat2 := 0.29871bar cm 3 V1 := 90 V2 := 92 3 cm 3 mol cm 3 mol cm 3 B1 , 1 := 7993 T := 318.15K i := 1 .. 2 mol B2 , 2 := 1247 cm mol B1 , 2 := 2089 x1 := 0.4819 x2 := 1 x1 mol B2 , 1 := B1 , 2 y1 := 0.1813 y2 := 1 y1 P := 0.20941bar k := 1 .. 2 j := 1 .. 2 yi P xi Psati Term A is calculated using the given data. term_Ai := Term B is calculated using Eqns. (14.4) and (14.5) j , i := 2 B j , i B j , j Bi , i hati := exp P 1 Bi , i + 2 R T sati := exp y jyk( 2 j , i j , k) j k Bi , i Psati R T term_Bi := hati sati Term C is calculated using Eqn. (11.44) fsati := sati Psati term_A = Vi ( P Psati) fi := sati Psati exp R T term_B = 589 term_Ci := fsati fi 1.081 1.108 0.986 1.006 term_C = 1 1 Ans. 14.51 a) Equivalent to d2(G/RT)/dx12 = 0, use d2(GE/RT)/dx12 = -1/x1x2 For GE/RT = Ax1x2 = A(x1-x12) d(GE/RT)/dx1 = A(1-2x1) d2(GE/RT)/dx12 = -2A Thus, -2A = -1/x1x2 or 2Ax1x2 = 1. Substituting for x2: x1-x12 = 1/(2A) or x12-x1+1/(2A) = 0. 1+ The solution to this equation yields two roots: x1 = 1 and The two roots are symmetrical around x1 = 1/2 Note that for: A<2: No real roots A = 2: One root, x1 = 1/3 (consolute point) A>2: Two real roots, x1 > 0 and x1 <1 b) Plot the spinodal curve along with the solubility curve 540K T + 21.1 3 ln T K Both curves are symmetrical around x1 = 1/2. Create functions to represent the left and right halves of the curves. From Fig. 14.15: A ( T) := From above, the equations for the spinodal curves are: xspr1 ( T) := xr := 0.7 1 1 A ( T) 2 + 2 2 A ( T) xl := 0.3 xspl1 ( T) := 1 1 A ( T) 2 2 2 A ( T) x1 = 1 2 1 2 2 A 2 A From Eq. (E) in Example 14.5, the solubility curves are solved using a Solve Block: 590 Given A ( T) ( 1 2xr) = ln A ( T) ( 1 2xl) = ln 1 xr xr xr > 0.5 xr1 ( T) := Find ( xr) Given 1 xl xl xl < 0.5 xl1 ( T) := Find ( xl) Find the temperature of the upper consolute point. T := 300K Given A ( T) = 2 Tu := Find ( T) Tu = 345.998 K T := 250K .. 346K 360 340 320 300 280 260 240 0.1 0.2 xl1 xr1 xspl1 xpr1 0.3 0.4 0.5 0.6 0.7 0.8 14.54 The solution is presented for one of the systems given. The solutions for the other systems follow in the same manner. f) 1- Carbon tetrachloride 1 := 0.193 A1 := 14.0572 Psat1 ( T) := exp A1 Tc1 := 556.4K B1 := 2914.23 Pc1 := 45.60bar C1 := 232.148 kPa T 273.15 + C 1 K B1 591 2 - n-heptane 2 := 0.350 A2 := 13.8622 Psat2 ( T) := exp A2 Tc2 := 540.2K B2 := 2910.26 Pc2 := 27.40bar C2 := 216.432 kPa T 273.15 + C 2 K B2 T := ( 100 + 273.15)K Tr1 := Tr2 := T Tc1 T Tc2 Tr1 = 0.671 Tr2 = 0.691 Psat1r := Psat2r := Psat1 ( T) Pc1 Psat2 ( T) Pc2 Psat1r = 0.043 Psat2r = 0.039 Using Wilson's equation 12 := 1.5410 21 := 0.5197 1 ( x1) := exp ln x1 + ( 1 x1) 12 ... 12 21 + 1 x ( 1) x + 1 x 1 x + x ( 1) 1 21 1) 12 1 ( 2 ( x1) := exp ln ( 1 x1) + x1 21 ... 12 21 + x ( 1) x + 1 x 1 x + x ( 1) 1 21 1) 12 1 ( For part i, use the modified Raoult's Law. Define the pressure and vapor mole fraction y1 as functions of the liquid mole fraction, x 1. Pi ( x1) := x1 1 ( x1) Psat1 ( T) + ( 1 x1) 2 ( x1) Psat2 ( T) yi1 ( x1) := x1 1 ( x1) Psat1 ( T) Pi ( x1) Modified Raoult's Law: Eqn. (10.5) 592 For part ii, assume the vapor phase is an ideal solution. Use Eqn. (11.68) and the PHIB function to calculate hat and sat. sat1 := PHIB Tr1 , Psat1r , 1 ( ( ) ) sat1 = 0.946 1 ( P) := hat1 ( P) sat1 P hat1 ( P) := PHIB Tr1 , , 1 P c1 sat2 := PHIB Tr2 , Psat2r , 2 sat2 = 0.95 2 ( P) := hat2 ( P) sat2 P hat2 ( P) := PHIB Tr2 , , 2 P c2 Solve Eqn. (14.1) for y1 and P given x1. Guess: Given y1 1 ( P) P = x1 1 ( x1) Psat1 ( T) Eqn. (14.1) y1 := 0.5 P := 1bar ( 1 y1) 2 (P) P = ( 1 x1) 2 ( x1) Psat2 (T) fii ( x1) := Find ( P , y1) fii is a vector containing the values of P and y 1. Extract the pressure, P and vapor mole fraction, y1 as functions of the liquid mole fraction. Pii ( x1) := fii ( x1) 0 yii1 ( x1) := fii ( x1) 1 Plot the results in Mathcad x1 := 0 , 0.1 .. 1.0 593 2 1.9 1.8 Pi ( x1 ) bar Pi ( x1 ) bar Pii ( x1 ) bar Pii ( x1 ) bar 1.5 1.4 1.3 1.2 1.1 1 1.6 1.7 0 0.2 P-x Raoult's P-y Raoult's P-x Gamma/Phi P-y Gamma/Phi x1 , yi1 ( x1 ) , x1 , yii1 ( x1 ) 0.4 0.6 0.8 594 Chapter 15 - Section A - Mathcad Solutions 15.1 Initial state: Liquid water at 70 degF. H1 38.05 BTU lbm S1 0.0745 BTU lbm rankine (Table F.3) Final state: Ice at 32 degF. H2 T (a) ( 0.02 ( 70 143.3) BTU lbm S2 0.0 143.3 491.67 BTU lbm rankine 459.67)rankine Point A: sat. vapor at 32 degF. Point C: sat. liquid at 70 degF. P = 85.79(psia). Point D: Mix of sat. liq. & sat. vapor at 32 degF with the enthalpy of Point C. Point B: Superheated vapor at 85.79(psia) and the entropy of Point A. Data for Points A, C, & D from Table 9.1. Data for Point B from Fig. G.2. 595 Wideal H2 H1 T S2 S1 Wideal 12.466 BTU lbm mdot 1 lbm sec Wdotideal mdot Wideal Wdotideal 13.15 kW Ans. (b) For the Carnot heat pump, heat equal to the enthalpy change of the water is extracted from a cold reservoir at 32 degF, with heat rejection to the surroundings at 70 degF. TC 491.67 rankine TH TH TC T QC H2 Work H1 QC BTU lbm 181.37 BTU lbm Work QC TC 14.018 Wdot t mdot Work Wdot t 14.79 kW Ans. Wdotideal Wdot 0.889 Ans. The only irreversibility is the transfer of heat from the water as it cools from 70 to 32 degF to the cold reservoir of the Carnot heat pump at 70 degF. (c) Conventional refrigeration cycle under ideal conditions of operation: Isentropic compression, infinite flow rate of cooling water, & minimum temp. difference for heat transfer = 0. For sat. liquid and vapor at 32 degF, by interpolation in the table: HA 107.60 BTU lbm SA 0.2223 BTU lbm rankine For sat. liquid at 70 degF: HC 34.58 BTU lbm HD HC For superheated vapor at 85.79(psia) and S = 0.2223: HB 114 BTU lbm 596 Refrigerent circulation rate: H2 mdot H1 1 lbm sec mdot 2.484 lbm sec Ans. Ans. HA mdot HB Wdotideal HD HA Wdot t Wdot t 16.77 kW 0.784 Wdot The irreversibilities are in the throttling process and in heat transfer in both the condenser and evaporator, where there are finite temperature differences. (d) Practical cycle. 0.75 Point A: Sat. vapor at 24 degF. Point B: Superheated vapor at 134.75(psia). Point D: Mix of sat. liq. and sat. vapor at 24 degF with H of point C, Point C: Sat. Liquid at 98 degF. (Note that minimum temp. diff. is not at end of condenser, but it is not practical to base design on 8-degF temp. diff. at pinch. See sketch.) For sat. liquid and vapor at 24 degF: Hliq 19.58 BTU lbm BTU lbm rankine Hvap 106.48 BTU lbm BTU lbm rankine HA Hvap Sliq 0.0433 Svap 0.2229 SA Svap 597 For sat. liquid at 98 degF, P=134.75(psia): HC 44.24 BTU lbm SC 0.0902 BTU lbm rankine For isentropic compression, the entropy of Point B is 0.2229 at P=134.75(psia). From Fig. G.2, H'B 118 BTU lbm BTU lbm BTU lbm rankine xD Svap Sliq HB HA H'B HA HB 121.84 The entropy at this H is read from Fig. G.2 at P=134.75(psia) HD HC xD HD Hvap 0.094 SB 0.228 Hliq Hliq xD 0.284 SD Sliq SD BTU lbm rankine Refrigerent circulation rate: lbm sec mdot 2.914 lbm sec Ans. H2 mdot H1 1 HA mdot HB Wdotideal HD HA Wdot Wdot 47.22 kW t Wdot T SA t 0.279 ( 70 Ans. 459.67)rankine THERMODYNAMIC ANALYSIS Wdotlost.compressor Qdotcondenser mdot T SB HB SC 598 mdot HC mdot T Wdotlost.condenser SB Qdotcondenser Wdotlost.throttle mdot T SD SC Wdotlost.evaporator T mdot SA SD lbm S 2 S1 1 sec 27.85% 17.59% 30.02% 14.02% 10.52% Wdotideal Wdotlost.compressor Wdotlost.condenser Wdotlost.throttle Wdotlost.evaporator 13.152 kW 8.305 kW 14.178 kW 6.621 kW 4.968 kW The percent values above express each quantity as a percentage of the actual work, to which the quantities sum. 15.2 Assume ideal gases. Data from Table C.4 H298 S298 282984 J H298 G298 G298 S298 257190 J J K 298.15 K 86.513 BASIS: 1 mol CO and 1/2 mol O2 entering with accompanying N2=(1/2)(79/21)=1.881 mol nCO 1 mol nair 2.381 mol nCO2 1 mol nN2 1.881 mol 599 (a) Isothermal process at 298.15 K: Since the enthalpy change of mixing for ideal gases is zero, the overall enthalpy change for the process is H y1 H298 nN2 nair For unmixing the air, define y1 0.79 y2 1 y1 By Eq. (12.35) with no minus sign: Sunmixing Sunmixing nair R y1 ln y1 10.174 J K y2 ln y2 For mixing the products of reaction, define y1 nCO2 nN2 nCO2 nCO2 y1 0.347 y2 y2 ln y2 1 y1 15.465 J K Smixing nN2 R y1 ln y1 Smixing S T Sunmixing 300 K S298 Wideal Smixing H 600 S S Wideal 81.223 J K Ans. T 259 kJ (b) Adiabatic combustion: Heat-capacity data for the product gases from Table C.1: A nCO2 5.457 mol B nCO2 1.045 mol D nCO2 1.157 mol T nN2 3.280 A 3 11.627 3 nN2 0.593 10 B 2.16 10 nN2 0.040 10 5 D 1.082 10 5 For the products, HP = R T0 CP R dT T0 298.15 K The integral is given by Eq. (4.7). Moreover, by an energy balance, H298 Guess Given HP = 0 2 A 11.627 B 2.160 10 K 2 3 . D 1.082 10 K 5 2 H298 = R mol A T0 Find 8.796 1 B 2 T0 2 T 601 1 T D T0 1 T0 2622.603 K For the cooling process from this temperature to the final temperature of 298.15 K, the entropy change is calculated by ICPS 2622.6 298.15 11.627 2.160 10 ICPS H H t 3 0.0 1.082 10 5 = 29.701 246.934 J K 29.701 H298 2.83 10 J 5 S R mol ICPS H T S S Wideal.cooling Wideal.cooling t 208904 J Ans. Ans. Wideal.cooling Wideal 0.8078 The surroundings increase in entropy in the amount: Q H298 Wideal.cooling S Q T S 246.93 J K The irreversibility is in the combustion reaction. Ans. 15.3 For the sat. steam at 2700 kPa, Table F.2: H1 2801.7 kJ kg S1 6.2244 kJ kg K For the sat. steam at 275 kPa, Table F.2: H2 2720.7 kJ kg S2 7.0201 602 kJ kg K For sat. liquid and vapor at 1000 kPa, Table F.2: Hliq Hvap 762.6 kJ kg kJ kg Sliq Svap 2.1382 kJ kg K kJ kg K Tsat 453.03K 2776.2 6.5828 (a) Assume no heat losses, no shaft work, and negligible changes in kinetic and potential energy. Then by Eqs. (2.30) and (5.22) for a completely reversible process: H fs ( mdot)= 0 S fs ( mdot)= 0 We can also write a material balance, a quantity requirement, and relation between H3 and S3 which assumes wet steam at point 3. The five equations (in 5 unknowns) are as follows: Guesses: mdot1 H3 Given H3 mdot3 S3 mdot3 0.1 H1 2 kg s mdot2 S3 mdot1 Sliq mdot3 H3 Hliq mdot1 mdot2 H2 Tsat H1 mdot1 S1 mdot1 H2 mdot2 = 0 S2 mdot2 = 0 H3 kJ s kJ sK Hliq mdot3 = 300 kJ s mdot3 = mdot1 S3 = Sliq mdot1 mdot2 mdot3 H3 S3 mdot2 Hliq H3 Tsat Find mdot1 mdot2 mdot3 H3 S3 603 mdot1 H3 0.086 kg s 3 kJ mdot2 S3 0.064 kg s mdot3 Ans. 0.15 kg s 2.767 10 kg 6.563 kJ kg K Steam at Point 3 is indeed wet. (b) Turbine: Constant-S expansion of steam from Point 1 to 1000 kPa results in wet steam of quality x'turb x'turb turb S1 Svap 0.919 0.78 Sliq Sliq H'turb H'turb Hturb Hturb Hliq 2.614 H1 2.655 x'turb Hvap 10 3 kJ Hliq kg H1 turb H'turb 10 3 kJ kg xturb xturb Hturb Hvap 0.94 Hliq Hliq Sturb Sturb Sliq 6.316 xturb Svap kJ kg K Sliq Compressor: Constant-S compression of steam from Point 2 to 1000 kPa results in superheated steam. Interpolation in Table F.2 yields H'comp 2993.5 kJ kg comp 0.75 Hcomp H2 H'comp comp H2 Hcomp kJ kg K 3084.4 kJ kg By interpolation: Scomp 7.1803 604 The energy balance, mass balance, and quantity requirement equations of Part (a) are still valid. In addition, The work output of the turbine equals the work input of the compressor. Thus we have 4 equations (in 4 unknowns): kg kg Guesses: mdot1 0.086 mdot2 0.064 s s mdot3 Given Hcomp H3 mdot3 0.15 kg s H3 2770. kJ kg H2 mdot2 = H1 mdot1 mdot2 Hturb H1 mdot1 kJ s Hliq mdot3 = 300 kJ s H2 mdot2 = 0 H3 mdot3 = mdot1 mdot1 mdot2 mdot3 H3 kg s kg s Find mdot1 mdot2 mdot3 H3 mdot1 mdot3 0.10608 mdot2 H3 0.04274 kg s Ans. 0.14882 2.77844 3 kJ 10 kg Steam at Point 3 is slightly superheated. By interpolation, S3 6.5876 kJ kg K 300K (assumed) THERMODYNAMIC ANALYSIS T By Eq. (5.25), with the enthalpy term equal to zero: Wdotideal T mdot3 S3 mdot1 S1 605 mdot2 S2 Wdotideal Wdotlost.turb 6.014 kW T mdot1 Sturb S1 S2 Wdotlost.comp T mdot2 Scomp Wdotlost.mixing Wdotlost.turb Wdotlost.comp Wdotlost.mixing T mdot3 S3 mdot1 Sturb mdot2 Scomp 2.9034 kW 2.054 kW 1.0561 kW 48.2815% 34.1565% 17.5620% The percent values above express each quantity as a percentage of the absolute value of the ideal work, to which the quantities sum. 15.4 Some property values with reference to Fig. 9.1 are given in Example 9.1. Others come from Table 9.1 or Fig. G.2. For sat. liquid and vapor at the evaporator temperature of 0 degF: Hliq 12.090 BTU lbm BTU lbm rankine Hvap 103.015 BTU lbm Svap 0.22525 Sliq 0.02744 BTU lbm rankine For sat. liquid at the condenser outlet temperature of 80 degF: H4 37.978 BTU lbm S2 Hliq Hliq S4 0.07892 BTU lbm rankine H2 x1 x1 Hvap H1 Hvap 0.285 Svap H1 S1 S1 606 H4 Sliq 0.084 x1 Svap BTU lbm rankine Sliq From Example 9.1(b) for the compression step: H 17.48 BTU lbm H3 H2 H H3 120.5 BTU lbm From Fig. G.2 at H3 and P = 101.37(psia): S3 0.231 BTU lbm rankine lbm hr 10 4 BTU mdot 1845.1 Wdot mdot H Wdot 3.225 hr The purpose of the condenser is to transfer heat to the surroundings. Thus the heat transferred in the condenser is Q in the sense of Chapter 15; i.e., it is heat transfer to the SURROUNDINGS, taken here to be at a temperature of 70 degF. Internal heat transfer (within the system) is not Q. The heat transferred in the evaporator comes from a space maintained at 10 degF, which is part of the system, and is treated as an internal heat reservoir. The ideal work of the process is that of a Carnot engine operating between the temperature of the refrigerated space and the temperature of the surroundings. T QdotC ( 70 459.67)rankine 120000 BTU hr TH TC TH TC T ( 10 459.67)rankine 4 BTU Wdotideal Wdotlost.comp Qdot H4 QdotC TC Wdotideal S2 Qdot 1.533 10 hr T mdot S3 H3 mdot T mdot S4 T mdot S1 1.523 10 5 BTU hr Wdotlost.cond Wdotlost.throttle S3 S4 607 Qdot T mdot S2 S1 H1 H2 mdot T TC The final term accounts for the entropy change of the refrigerated space (an internal heat reservoir). Wdotideal Wdotlost.comp Wdotlost.cond Wdotlost.throttle Wdotlost.evap Wdotlost.evap 15329.9 BTU hr 47.53% 5619.4 BTU hr BTU hr BTU hr BTU hr 17.42% 3625.2 11.24% 4730.2 14.67% 2947.6 9.14% The percent values above express each quantity as a percentage of the actual work, to which they sum: Wdot 32252.3 BTU hr 15.5 The discussion at the top of the second page of the solution to the preceding problem applies equally here. In each case, T ( 70 459.67) rankine TH T The following vectors refer to Parts (a)-(e): 40 30 tC 20 10 0 608 600 500 QdotC 400 300 200 BTU sec TC tC 459.67 rankine Wdotideal QdotC TH TC TC For sat. liquid and vapor at the evaporator temperature, Table 9.1: 21.486 18.318 Hliq 15.187 12.090 9.026 0.04715 0.04065 Sliq 0.03408 0.02744 0.02073 BTU Svap lbm rankine BTU lbm 107.320 105.907 Hvap 104.471 103.015 101.542 BTU lbm H2 Hvap 0.22244 0.22325 0.22418 0.22525 0.22647 BTU lbm rankine S2 Svap For sat. liquid at the condenser temperature: H4 37.978 BTU lbm S4 0.07892 BTU lbm rankine H1 H4 x1 H1 Hvap Hliq Hliq S1 Sliq x1 Svap Sliq From the results of Pb. 9.9, we find: 117.7 118.9 H3 120.1 121.7 123.4 BTU lbm From these values we must find the corresponding entropies from Fig. G.2. They are read at the vapor pressure for 80 degF of 101.37 kPa. The flow rates come from Problem 9.9: 609 0.227 0.229 S3 0.231 0.234 0.237 BTU lbm rankine 8.653 7.361 mdot 6.016 4.613 3.146 T mdot S3 H4 H3 mdot S3 S4 S1 mdot Qdot lbm sec Wdotlost.comp Qdot S2 Wdotlost.cond Wdotlost.throttle Wdotlost.evap T mdot S4 T mdot S1 T mdot S2 T H1 H2 TC The final term accounts for the entropy change of the refrigerated space (an internal heat reservoir). Wdot mdot H3 36.024 40.844 H2 20.9 22.419 Wdotlost.comp 21.732 21.379 17.547 BTU sec Wdotideal BTU 41.695 sec 38.325 30.457 610 11.149 10.52 Wdotlost.cond 9.444 7.292 5.322 12.991 11.268 Wdotlost.evap 9.406 7.369 5.122 BTU sec BTU sec 8.754 10.589 Wdotlost.throttle 11.744 11.826 10.322 BTU sec 89.818 95.641 Wdot 94.024 86.194 68.765 BTU sec In each case the ideal work and the lost work terms sum to give the actual work, and each term may be expressed as a percentage of the actual work. 15.6 The discussion at the top of the second page of the solution to Problem 15.4 applies equally here. T TC ( 70 ( 30 459.67)rankine 459.67)rankine TH TC TH QdotC T 2000 BTU sec Wdotideal QdotC TC Wdotideal 163.375 BTU sec For sat. liquid and vapor at the evaporator temperature, Table 9.1: Hliq 18.318 BTU lbm BTU lbm Sliq 0.04065 BTU lbm rankine BTU lbm rankine Hvap H2 105.907 Hvap Svap S2 0.22325 Svap 611 For sat. liquid at the condenser temperature: H4 37.978 BTU lbm S4 0.07892 BTU lbm rankine From Problem 9.12, H2A 116. BTU lbm BTU lbm S2A 0.2435 BTU lbm rankine H3 H2A 14.667 H3 130.67 BTU lbm From Fig. G.2 at this enthalpy and 33.11(psia): S3 0.2475 BTU lbm rankine Energy balance on heat exchanger: H1 x1 x1 H4 H1 Hvap 0.109 H2A Hliq Hliq H2 H1 S1 S1 27.885 BTU lbm Sliq Sliq 0.061 x1 Svap BTU lbm rankine Upstream from the throttle (Point 4A) the state is subcooled liquid with the enthalpy: H4A H1 The entropy at this point is essentially that of sat. liquid with this enthalpy; by interpolation in Table 9.1: S4A 0.05986 BTU lbm rankine lbm sec From Problem 9.12: mdot 25.634 612 Wdotlost.comp Qdot H4 T mdot S3 H3 mdot T mdot S4 T mdot S1 S2A Wdotlost.cond Wdotlost.throttle S3 S4A Qdot Wdotlost.evap T mdot S2 S1 H1 H2 T mdot TC The final term accounts for the entropy change of the refrigerated space (an internal heat reservoir). Wdotlost.exchanger Wdot mdot H3 Wdotideal Wdotlost.comp Wdotlost.cond Wdotlost.throttle Wdotlost.evap Wdotlost.exchanger BTU sec T mdot S2A H2A 163.38 BTU sec S2 S4A S4 43.45% 54.31 BTU sec BTU sec 14.45% 87.08 23.16% 9.98 BTU sec BTU sec BTU sec 2.65% 45.07 11.99% 16.16 4.30% Wdot 375.97 The figures on the right are percentages of the actual work, to which the terms sum. 613 15.7 Compression to a pressure at which condensation in coils occurs at 110 degC. Table F.1 gives this sat. pressure as 143.27 kPa comp 0.75 419.1 kJ kg kJ kg H1 H2 S1 S2 1.3069 kJ kg K kJ kg K (sat. liquid) 2676.0 7.3554 (sat. vapor) For isentropic compression to 143.27 kPa, we find by double interpolation in Table F.2: H'3 2737.0 kJ kg H3 H2 H'3 H2 H3 2757.3 kJ kg comp By more double interpolation in Table F.2 at 143.27 kPa, S3 7.4048 kJ kg K By an energy balance, assuming the slurry passes through unchanged, H4 H1 H3 H2 H4 500.4 kJ kg 614 This enthalpy is a bit larger than that of sat. liquid at 110 degC; find quality and then the entropy: Hliq Slv 461.3 kJ kg kJ kg K Hlv x4 2230.0 H4 kJ kg Sliq x4 1.4185 kJ kg K 5.8203 Hliq Hlv 0.018 kg sec S4 T Sliq x4 Slv S4 1.5206 kJ kg K mdot 0.5 300 K Wdotideal Wdotlost.evap mdot H4 mdot T mdot T H2 H1 S4 S3 T S3 S2 S4 S2 S1 S1 Wdotlost.comp Wdot mdot H3 Wdotideal Wdotlost.evap Wdotlost.comp Wdot 8.606 kW 24.651 kW 7.41 kW 21.16% 60.62% 18.22% 40.667 kW The figures on the right are percentages of the actual work, to which the terms sum. 15.8 A thermodynamic analysis requires an exact definition of the overall process considered, and in this case we must therefore specify the source of the heat transferred to the boiler. Since steam leaves the boiler at 900 degF, the heat source may be considered a heat reservoir at some higher temperature. We assume in the following that this temperature is 950 degF. The assumption of a different temperature would provide a variation in the solution. 615 The ideal work of the process in this case is given by a Carnot engine operating between this temperature and that of the surroundings, here specified to be 80 degF. We take as a basis 1 lbm of H2O passing through the boiler. Required property values come from Pb. 8.8. TH ( 459.67 950)rankine TC ( 459.67 80)rankine T TC Subscripts below correspond to points on figure of Pb. 8.7. H1 H2 H3 H4 H5 H7 QH H2 257.6 1461.2 1242.2 1047.8 69.7 250.2 BTU lbm S1 S2 S3 S4 S5 S7 Wideal 0.3970 1.6671 1.7431 1.8748 0.1326 0.4112 BTU lbm rankine H1 1 lbm QH 1 TC TH For purposes of thermodynamic analysis, we consider the following 4 parts of the process: The boiler/heat reservoir combination The turbine The condenser and throttle valve The pump and feedwater heater Wlost.boiler.reservoir m 0.18688 lbm T S2 S1 1 lbm QH TH (From Pb. 8.8) Wlost.turbine T m S3 S2 1 lbm m S4 S2 The purpose of the condenser is to transfer heat to the surroundings. The amount of heat is Q 1 lbm H5 1 lbm m H4 m H7 Q 829.045 BTU 616 Wlost.cond.valve T 1 lbm S5 1 lbm m S4 m S7 Q Wlost.pump.heater T 1 lbm S1 S5 m S7 S3 The absolute value of the actual work comes from Pb. 8.8: Wabs.value = 374.61 BTU Wlost.boiler.reservoir Wlost.turbine Wlost.cond.valve Wlost.pump.heater Wideal 742.82 BTU 50.43% 30.24% 13.30% 4.90% 1.13% 224.66 BTU 98.81 BTU 36.44 BTU 8.36 BTU (absolute value) The numbers on the right are percentages of the absolute value of the ideal work, to which they sum. 15.9 Refer to Figure 9.7, page 330 The analysis presented here is for the liquefaction section to the right of the dashed line. Enthalpy and entropy values are those given in Ex. 9.3 plus additional values from the reference cited on page 331 at conditions given in Ex. 9.3. Property values: H4 H5 H7 H9 H10 1140.0 kJ kg kJ kg S4 S5 S7 S9 S10 9.359 kJ kg K kJ kg K kJ kg K kJ kg K kJ kg K 1009.7 8.894 719.8 kJ kg kJ kg kJ kg 7.544 285.4 4.928 796.9 9.521 617 H14 1042.1 kJ kg S14 11.015 kJ kg K H15 1188.9 kJ kg S15 11.589 kJ kg K H6 H5 S6 S5 H11 H5 S11 S5 H12 H10 S12 S10 H13 H10 S13 S10 T 295K The basis for all calculations is 1 kg of methane entering at point 4. All work quantities are in kJ. Results given in Ex. 9.3 on this basis are: Fraction of entering methane that is liquefied: Fraction of entering methane passing through the expander: On this basis also Eq. (5.26) for Ideal Work, Eq. (5.33) for Entropy Generation,and Eq. (5.34) for Lost Work can be written: z 0.113 x ( m) H fs 0.25 T ( m) SG = S fs Wideal = Q Wlost = T SG T ______________________________________________________________ ( m) S fs Wideal Wideal H15 ( 1 489.001 z) H9 z kJ kg H4 T S15 ( 1 z) S9 z S4 Wout H12 H11 x Wout kJ kg (a) Heat Exchanger I: SG.a SG.a 0.044 kJ kg K S5 S4 S15 S14 ( 1 z) kJ kg Wlost.a T SG.a Wlost.a 13.021 (b) Heat Exchanger II: SG.b SG.b 0.313 kJ kg K S7 S6 ( 1 x) Wlost.b S14 S13 ( 1 kJ kg z) Wlost.b T SG.b 92.24 618 (c) Expander: SG.c S12 S11 x SG.c 0.157 kJ kg K Wlost.c T SG.c Wlost.c 46.241 kJ kg (d) Throttle: SG.d 0.964 SG.d kJ kg K S9 z Wlost.d S10 ( 1 T SG.d z x) S7 ( 1 Wlost.d x) 284.304 kJ kg Entropy-generation analysis: kJ/kg-K Percent of S_Ga 0.044 2.98% S_Gb 0.313 21.18% S_Gc 0.157 10.62% S_Gd 0.964 1.478 65.22% 100.00% Work analysis, Eq. (15.3): kJ/kg Percent of 10.88% Wout Wlost.a 53.20 13.02 2.66% Wlost.b 92.24 18.86% Wlost.c 46.24 9.46% Wlost.d 284.30 58.14% 489.00 Note that: = Wideal 100.00% 619 Chapter 16 - Section A - Mathcad Solutions 16.10(Planck's constant) h 6.626 10 34 Js (Boltzmann's constant) 23 J k 1.381 10 K (Avagodro's number) NA 6.023 10 mol 23 1 P 1bar T 298.15K V RT P V 0.025 m 3 mol 39.948 a) For Argon: gm mol M NA 3 2 Sig R ln 2 MkT h 2 Ve NA 5 2 Sig 154.84 J mol K J mol K Ans. 83.800 b) For Krypton: M gm mol NIST value: 154.84 NA 3 Sig R ln 2 MkT h 2 2 Ve NA 5 2 Sig 164.08 J mol K Ans. NIST value: 164.05 J mol K 131.30 c) For Xenon M gm mol NA 3 Sig R ln 2 MkT h 2 2 Ve NA 5 2 Sig 164.08 J mol K J mol K Ans. NIST value: 169.68 620 Chapter 1 - Section B - Non-Numerical Solutions 1.1 This system of units is the English-system equivalent of SI. Thus, gc = 1(lbm )(ft)(poundal)1 (s)2 1.2 (a) Power is power, electrical included. Thus, kgm2 Nm energy [=] [=] s3 s time (b) Electric current is by denition the time rate of transfer of electrical charge. Thus Power [=] Charge [=] (electric current)(time) [=] As (c) Since power is given by the product of current and electric potential, then kgm2 power [=] current As3 (d) Since (by Ohms Law) current is electric potential divided by resistance, Electric potential [=] kgm2 electric potential [=] 3 current A2 s (e) Since electric potential is electric charge divided by electric capacitance, Resistance [=] A2 s charge [=] Capacitance [=] electric potential kgm2 4 1.3 The following are general: ln x = ln 10 log10 x P sat /kPa = P sat /torr 100 kPa 750.061 torr (A) (B) (C) t/ C = T /K 273.15 By Eqs. (B) and (A), ln P sat /kPa = ln 10 log10 P sat /torr + ln 100 750.061 The given equation for log10 P sat /torr is: log10 P sat /torr = a b t/ C + c Combining these last two equations with Eq. (C) gives: ln P sat /kPa = ln 10 a b T /K 273.15 + c + ln 100 750.061 = 2.3026 a b T /K 273.15 + c 2.0150 Comparing this equation with the given equation for ln P sat /kPa shows that: A = 2.3026 a 2.0150 B = 2.3026 b 621 C = c 273.15 1.9 Reasons result from the fact that a spherical container has the minimum surface area for a given interior volume. Therefore: (a) A minimum quantity of metal is required for tank construction. (b) The tensile stress within the tank wall is everywhere uniform, with no sites of stress concentration. Moreover, the maximum stress within the tank wall is kept to a minimum. (c) The surface area that must be insulated against heat transfer by solar radiation is minimized. 1.17 Kinetic energy as given by Eq. (1.5) has units of massvelocity2 . Its fundamental units are therefore: E K [=] kgm2 s2 [=] Nm [=] J Potential energy as given by Eq. (1.7) has units of masslengthacceleration. Its fundamental units are therefore: E P [=] kgmms2 [=] Nm [=] J 1.20 See Table A.1, p. 678, of text. 1(atm) 1 bar = 1/0.986923 = 1.01325 bar 1(Btu) 1 kJ = 1/0.947831 = 1.05504 kJ 1(hp) 0.75 kW = 1/1.34102 = 0.745701 kW 1(in) 2.5 cm = 2.54 cm exactly, by denition (see p. 651 of text) 1(lbm ) 0.5 kg = 0.45359237 kg exactly, by denition (see p. 651 of text) 1(mile) 1.6 km = 5280/3280.84 = 1.60934 km 1(quart) 1 liter = 1000/(264.172 4) = 0.94635 liter (1 liter 1000 cm3 ) 1(yard) 1 m = (0.0254)(36) = 0.9144 m exactly, by denition of the (in) and the (yard) An additional item could be: 1(mile)(hr)1 0.5 m s1 = (5280/3.28084)(1/3600) = 0.44704 m s1 1.21 One procedure here, which gives results that are internally consistent, though not exact, is to assume: 1 Year [=] 1 Yr [=] 364 Days This makes 1 Year equivalent to exactly 52 7-Day Weeks. Then the average Month contains 30 1 Days 3 and 4 1 Weeks. With this understanding, 3 1 Year [=] 1 Yr [=] 364 Days [=] (364)(24)(3600) = 31,449,600 Seconds Whence, 1 Sc [=] 31.4496 Second 1 Mn [=] 314.496 Second 1 Hr [=] 3144.96 Second 1 Dy [=] 31449.6 Second 1 Wk [=] 314496. Second 1 Mo [=] 3144960 Second 1 Second [=] 0.031797 Sc 1 Minute [=] 60 Second [=] 0.19078 Mn 1 Hour [=] 3600 Second [=] 1.14469 Hr 1 Day [=] (24)(3600) Second [=] 2.74725 Dy 1 Week [=] (7)(24)(3600) Second [=] 1.92308 Wk 1 Month [=] (4 1 )(7)(24)(3600) Second[=] 0.83333 Mo 3 The nal item is obviously also the ratio 10/12. 622 Chapter 2 - Section B - Non-Numerical Solutions 2.3 Equation (2.2) is here written: U t + E P + E K = Q + W (a) In this equation W does not include work done by the force of gravity on the system. This is accounted for by the E K term. Thus, W = 0. (b) Since the elevation of the egg decreases, sign(E P ) is (). (c) The egg is at rest both in its initial and nal states; whence E K = 0. (d) Assuming the egg does not get scrambled, its internal energy does not change; thus U t = 0. (e) The given equation, with U t = E K = W = 0, shows that sign(Q) is (). A detailed examination of the process indicates that the kinetic energy of the egg just before it strikes the surface appears instantly as internal energy of the egg, thus raising its temperature. Heat transfer to the surroundings then returns the internal energy of the egg to its initial value. 2.6 If the refrigerator is entirely contained within the kitchen, then the electrical energy entering the refrigerator must inevitably appear in the kitchen. The only mechanism is by heat transfer (from the condenser of the refrigerator, usually located behind the unit or in its walls). This raises, rather than lowers, the temperature of the kitchen. The only way to make the refrigerator double as an air conditioner is to place the condenser of the refrigerator outside the kitchen (outdoors). 2.7 According to the phase rule [Eq. (2.7)], F = 2 + N . According to the laboratory report a pure material (N = 1) is in 4-phase ( = 4) equilibrium. If this is true, then F = 2 4 + 1 = 1. This is not possible; the claim is invalid. 2.8 The phase rule [Eq. (2.7)] yields: F = 2 + N = 2 2 + 2 = 2. Specication of T and P xes the intensive state, and thus the phase compositions, of the system. Since the liquid phase is pure species 1, addition of species 2 to the system increases its amount in the vapor phase. If the composition of the vapor phase is to be unchanged, some of species 1 must evaporate from the liquid phase, thus decreasing the moles of liquid present. 2.9 The phase rule [Eq. (2.7)] yields: F = 2 + N = 2 2 + 3 = 3. With only T and P xed, one degree of freedom remains. Thus changes in the phase compositions are possible for the given T and P. If ethanol is added in a quantity that allows T and P to be restored to their initial values, the ethanol distributes itself between the phases so as to form new equilibrium phase compostions and altered amounts of the vapor and liquid phases. Nothing remains the same except T and P. 2.10 (a) Since F = 3, xing T and P leaves a single additional phase-rule variable to be chosen. (b) Adding or removing liquid having the composition of the liquid phase or adding or removing vapor having the composition of the vapor phase does not change the phase compositions, and does not alter the intensive state of the system. However, such additions or removals do alter the overall composition of the system, except for the unusual case where the two phase compositions are the same. The overall composition, depending on the relative amounts of the two phases, can range from the composition of the liquid phase to that of the vapor phase. 2.14 If the uid density is constant, then the compression becomes a constant-V process for which the work is zero. Since the cylinder is insulated, we presume that no heat is transferred. Equation (2.10) then shows that U = 0 for the compression process. 623 2.16 Electrical and mechanical irreversibilities cause an increase in the internal energy of the motor, manifested by an elevated temperature of the motor. The temperature of the motor rises until a dynamic equilibrium is established such that heat transfer from the motor to the srroundings exactly compensates for the irreversibilities. Insulating the motor does nothing to decrease the irreversibilities in the motor and merely causes the temperature of the motor to rise until heat-transfer equilibrium is reestablished with the surroundings. The motor temperature could rise to a level high enough to cause damage. 2.19 Let symbols without subscripts refer to the solid and symbols with subscript w refer to the water. Heat transfer from the solid to the water is manifested by changes in internal energy. Since energy is t conserved, U t = Uw . If total heat capacity of the solid is C t (= mC) and total heat capacity of t the water is Cw (= m w Cw ), then: t C t (T T0 ) = Cw (Tw Tw0 ) or Tw = Tw0 Ct (T T0 ) t Cw (A) This equation relates instantaneous values of Tw and T . It can be written in the alternative form: t t T C t T0 C t = Tw0 Cw Tw Cw or t t Tw0 Cw + T0 C t = Tw Cw + T C t (B) The heat-transfer rate from the solid to the water is given as Q = K (Tw T ). [This equation implies that the solid is the system.] It may also be written: Ct dT = K (Tw T ) d (C) In combination with Eq. (A) this becomes: Ct Ct dT = K Tw0 t (T T0 ) T Cw d or dT =K d T T0 Tw0 T t t Cw C = T K 1 1 + t t Cw C +K T0 Tw0 + t t Cw C Dene: K 1 1 + t Cw Ct K T0 Tw0 + t Cw Ct where both and are constants. The preceding equation may now be written: dT = T d Rearrangement yields: 1 d( T ) dT = d = T T Integration from T0 to T and from 0 to gives: T 1 ln T0 = 624 which may be written: T = exp( ) T0 When solved for T and rearranged, this becomes: T = + T0 exp( ) where by the denitions of and , Tw C t + T0 C t = 0 tw Cw + C t When = 0, the preceding equation reduces to T = T0 , as it should. When = , it reduces to T = /. Another form of the equation for / is found when the numerator on the right is replaced by Eq. (B): t Tw Cw + T C t = t Cw + C t By inspection, T = / when Tw = T , the expected result. 2.20 The general equation applicable here is Eq. (2.30): H + 1 u 2 + zg m 2 fs = Q + Ws (a) Write this equation for the single stream owing within the pipe, neglect potential- and kineticenergy changes, and set the work term equal to zero. This yields: ( H )m = Q (b) The equation is here written for the two streams (I and II) owing in the two pipes, again neglecting any potential- and kinetic-energy changes. There is no work, and the the heat transfer is internal, between the two streams, making Q = 0. Thus, ( H )I m I + ( H )II m II = 0 (c) For a pump operating on a single liquid stream, the assumption of negligible potential- and kineticenergy changes is reasonable, as is the assumption of negligible heat transfer to the surroundings. Whence, ( H )m = W (d) For a properly designed gas compressor the result is the same as in Part (c). (e) For a properly designed turbine the result is the same as in Part (c). (f ) The purpose of a throttle is to reduce the pressure on a owing stream. One usually assumes adiabatic operation with negligible potential- and kinetic-energy changes. Since there is no work, the equation is: H =0 (g) The sole purpose of the nozzle is to produce a stream of high velocity. The kinetic-energy change must therefore be taken into account. However, one usually assumes negligible potential-energy change. Then, for a single stream, adiabatic operation, and no work: H + 1 u2 m = 0 2 The usual case is for a negligible inlet velocity. The equation then reduces to: H + 1 u2 = 0 2 2 625 2.21 We reformulate the denition of Reynolds number, with mass owrate m replacing velocity u: m = u A = u 2 D 4 Solution for u gives: u= 4 m D2 Whence, Re 4 m 4 m D u D = = 2 D D (a) Clearly, an increase in m results in an increase in Re. (b) Clearly, an increase in D results in a decrease in Re. 2.24 With the tank as control volume, Eqs. (2.25) and (2.29) become: dm +m =0 dt and d(mU ) +Hm =0 dt Expanding the derivative in the second equation, and eliminating m by the rst equation yields: m dm dm dU =0 H +U dt dt dt Multiply by dt and rearrange: dm dU = m H U Substitution of H for H requires the assumption of uniform (though not constant) conditions throughout the tank. This requires the absence of any pressure or temperature gradients in the gas in the tank. 2.32 From the given equation: P= RT V b V2 V1 By Eq. (1.3), W = P dV = V2 V1 RT d(V b) V b Whence, W = RT ln V1 b V2 b 2.35 Recall: Whence, d(P V ) = P d V + V d P d W = V d P d(P V ) and and d W = P d V W = V dP (P V ) By Eq. (2.4), d Q = dU d W and dU = d H P d V V d P the preceding equation becomes d Q = d H V d P By Eq. (2.11), U = H P V With d W = P d V Whence, Q= H V dP 626 . . 2.38 (a) By Eq. (2.24a), m = u A With m, A, and all constant, u must also be constant. With q = u A, q is also constant. . . . (b) Because mass is conserved, m must be constant. But n = M/m may change, because M may change. At the very least, depends on T and P. Hence u and q can both change. 2.40 In accord with the phase rule, the system has 2 degrees of freedom. Once T and P are specied, the intensive state of the system is xed. Provided the two phases are still present, their compositions cannot change. 2.41 In accord with the phase rule, the system has 6 degrees of freedom. Once T and P are specied, 4 remain. One can add liquid with the liquid-phase composition or vapor with the vapor-phase composition or both. In other words, simply change the quantities of the phases. . 2.43 Let n represent the moles of air leaving the home. By an energy balance, . dn dU d(nU ) . . +U =n H +n Q=nH+ dt dt dt But a material balance yields Then dn . n = dt . dU dn +n Q = (H U ) dt dt or . dU dn +n Q = P V dt dt 2.44 (a) By Eq. (2.32a): By Eq. (2.24a): H2 H1 + 1 (u 2 u 2 ) = 0 1 2 2 . . 4 m m = u= D2 A Then u2 2 u2 1 = 4 2 . m2 2 1 1 4 4 D1 D2 and given H2 H1 = 1 (P2 P1 ) 1 1 (P2 P1 ) + 2 4 2 . m2 2 2 4 4 D1 D2 4 4 D1 D2 =0 1/2 . Solve for m: . m = 2(P1 P2 ) 4 4 4 D1 D2 4 4 D1 D2 . (b) Proceed as in part (a) with an extra term, Here solution for m yields: . m = 2 (P1 P2 ) 2 C(T2 T1 ) 4 2 4 4 D1 D2 4 4 D1 D2 1/2 Because the quantity in the smaller square brackets is smaller than the leading term of the preceding result, the effect is to decrease the mass owrate. 627 Chapter 3 - Section B - Non-Numerical Solutions 3.2 Differentiate Eq. (3.2) with respect to P and Eq. (3.3) with respect to T : P = T 1 V2 V P T V T + P 1 V 2V P T = + 2V P T T = P 1 V2 V T P V P T 1 V 2V T P = 2V P T Addition of these two equations leads immediately to the given equation. One could of course start with Eq. (3.4) and apply the condition for an exact differential, but this topic is not covered until Chapter 6. 3.3 The Tait equation is given as: V = V0 1 AP B+P where V0 , A, and B are constants. Application of Eq. (3.3), the denition of , requires the derivative of this equation: V P T = V0 AV0 AP A = + 2 B+P (B + P) B+P 1 + P B+P Multiplication by 1/V in accord with Eq. (3.3), followed by substitution for V0 /V by the Tait equation leads to: AB = (B + P)[B + (1 A)P] 3.7 (a) For constant T , Eq. (3.4) becomes: dV = dP V Integration from the initial state (P1 , V1 ) to an intermediate state (P, V ) for constant gives: ln V = (P P1 ) V1 Whence, V = V1 exp[ (P P1 )] = V1 exp( P) exp( P1 ) If the given equation applies to the process, it must be valid for the initial state; then, A(T ) = V1 exp( P1 ), and V = A(T ) exp( P) (b) Differentiate the preceding equation: Therefore, W = = V2 V1 d V = A(T ) exp( P)d P P2 P1 P d V = A(T ) P exp( P)d P A(T ) [( P1 + 1) exp( P1 ) ( P2 + 1) exp( P2 )] 628 With V1 = A(T ) exp( P1 ) and V2 = A(T ) exp( P2 ), this becomes: W = 1 [( P1 + 1)V1 ( P2 + 1)V2 ] or W = P1 V1 P2 V2 + V1 V2 3.11 Differentiate Eq. (3.35c) with respect to T : T 1 P [(1)/]1 dT dP =T + P (1)/ dz dz 1 dT P (1)/ d P =0 + P (1)/ dz dz P Algebraic reduction and substitution for d P/dz by the given equation yields: T P dT 1 =0 (Mg) + dz For an ideal gas T/P = 1/R. This substitution reduces the preceding equation to: Mg dT = R dz 1 3.12 Example 2.13 shows that U2 = H . If the gas is ideal, H = U + P V = U + RT For constant C V , Whence, U2 U = C V (T2 T ) and and U2 U = RT C V (T2 T ) = RT C P CV R T2 T = = CV CV T When C P /C V is set equal to , this reduces to: T2 = T This result indicates that the nal temperature is independent of the amount of gas admitted to the tank, a result strongly conditioned by the assumption of no heat transfer between gas and tank. 3.13 Isobaric case ( = 0). Here, Eqs. (3.36) and (3.37) reduce to: W = RT1 (1 1) and Q= RT1 (1 1) 1 Both are indeterminate. The easiest resolution is to write Eq. (3.36) and (3.37) in the alternative but equivalent forms: W = RT1 1 T2 1 T1 and Q= ( )RT1 ( 1)( 1) T2 1 T1 from which we nd immediately for = 0 that: W = R(T2 T1 ) and Q= R (T2 T1 ) = C P (T2 T1 ) 1 629 Isothermal case ( = 1). Equations (3.36) and (3.37) are both indeterminate of form 0/0. Application of lH pitals rule yields the appropriate results: o W = RT1 ln P2 P1 and Q = RT1 ln P2 P1 Note that if y P2 P1 (1)/ then 1 dy = 2 d P2 P1 (1)/ ln P2 P1 Adiabatic case ( = ). In this case simple substitution yields: W = RT1 1 P2 P1 ( 1)/ 1 and Q=0 Isochoric case ( = ). Here, simple substitution yields: W =0 and Q= RT1 1 RT1 P2 1 = 1 P1 T2 1 = C V (T2 T1 ) T1 3.14 What is needed here is an equation relating the heat transfer to the quantity of air admitted to the tank and to its temperature change. For an ideal gas in a tank of total volume V t at temperature T , n1 = P1 V t RT and n2 = P2 V t RT The quantity of air admitted to the tank is therefore: V t (P2 P1 ) RT The appropriate energy balance is given by Eq. (2.29), which here becomes: n = (A) d(nU )tank n H = Q dt where the prime ( ) identies the entrance stream of constant properties. Multiplying by dt and integrating over the time of the process yields: n 2 U2 n 1 U1 n H = Q With n = n 2 n 1 , n 2 (U2 H ) n 1 (U1 H ) = Q Because U2 = H2 RT and U1 = H1 RT , this becomes: n 2 (H2 H RT ) n 1 (U1 H RT ) = Q or n 2 [C P (T T ) RT ] n 1 [C P (T T ) RT ] = Q Because n = n 2 n 1 , this reduces to: Q = n [C P (T T ) RT ] Given: V t = 100, 000 cm3 T = 298.15 K T = 318.15 K P1 = 101.33 kPa P2 = 1500 kPa 630 By Eq. (A) with R = 8, 314 cm3 kPa mol1 K1 , n = (100, 000)(1500 101.33) = 56.425 mol (8, 314)(298.15) With R = 8.314 J mol1 K1 and C P = (7/2)R, the energy equation gives: Q = (56.425)(8.314) 7 (298.15 318.15) 298.15 = 172, 705.6 J 2 or Q = 172.71 kJ 3.15 (a) The appropriate energy balance is given by Eq. (2.29), here written: d(nU )tank n H = Q dt where the prime ( ) identies the entrance stream of constant properties. Multiplying by dt and integrating over the time of the process yields: n 2 U2 n 1 U1 n H = Q Since n = n 2 n 1 , rearrangement gives: n 2 (U2 H ) n 1 (U1 H ) = Q (b) If the gas is ideal, H = U + P V = U + RT Whence for an ideal gas with constant heat capacities, U2 H = U2 U RT = C V (T2 T ) RT Substitute R = C P C V : Similarly, and U2 H = C V T2 C V T C P T + C V T = C V T2 C P T U1 H = C V T1 C P T n 2 (C V T2 C P T ) n 1 (C V T1 C P T ) = Q Note also: n2 = P2 Vtank RT2 n1 = P1 Vtank RT1 (c) If n 1 = 0, n 2 (C V T2 C P T ) = Q (d) If in addition Q = 0, C V T2 = C P T and T2 = CP CV T Whence, T2 = T (e) 1. Apply the result of Part (d), with = 1.4 and T = 298.15 K: T2 = (1.4)(298.15) = 417.41 K 631 Then, with R = 83.14 bar cm3 mol1 K1 : n2 = (3)(4 106 ) P2 Vtank = 345.8 mol = (83.14)(417.41) RT2 2. Heat transfer between gas and tank is: Q = m tank C(T2 T ) where C is the specic heat of the tank. The equation of Part (c) now becomes: n 2 (C V T2 C P T ) = m tank C(T2 T ) Moreover n2 = P2 Vtank RT2 These two equations combine to give: P2 Vtank (C V T2 C P T ) = m tank C(T2 T ) RT2 With C P = (7/2)R and C V = C P R = (7/2)R R = (5/2)R, this equation becomes: R P2 Vtank (5T2 7T ) = m tank C(T2 T ) 2 RT2 Note: R in the denominator has the units of P V ; R in the numerator has energy units. Given values in the appropriate units are: m tank = 400 kg C = 460 J mol1 kg1 T = 298.15 K P2 = 3 bar Appropriate values for R are therefore: Vtank = 4 106 cm3 R(denominator) = 83.14 bar cm3 mol1 K1 Numerically, R(numerator) = 8.314 J mol1 K1 8.314 (3)(4 106 ) = (400)(460)(T2 298.15) [(5)(T2 ) (7)(298.15)] 2 (83.14)(T2 ) Solution for T2 is by trial, by an iteration scheme, or by the solve routine of a software package. The result is T2 = 304.217 K. Then, n2 = (3)(4 106 ) P2 Vtank = 474.45 mol = (83.14)(304.217) RT2 3.16 The assumption made in solving this problem is that the gas is ideal with constant heat capacities. The appropriate energy balance is given by Eq. (2.29), here written: d(nU )tank +Hn =Q dt Multiplied by dt it becomes: d(nU ) + H dn = d Q 632 where n and U refer to the contents of the tank, and H and n refer to the exit stream. Since the stream bled from the tank is merely throttled, H = H , where H is the enthalpy of the contents of the tank. By material balance, dn = dn. Thus, n dU + U dn H dn = Q Also, Thus, or dU = C V dT or n dU (H U )dn = d Q d Q = mC dT H U = P V = RT nC V dT RT dn = mC dT where m is the mass of the tank, and C is its specic heat. R d(nC V + mC) R d(nC V ) R dT = dn = = C V nC V + mC C V nC V + mC nC V + mC T Integration yields: ln T2 T1 = n 2 C V + mC R ln n 1 C V + mC CV or T2 = T1 n 2 C V + mC n 1 C V + mC R/C V In addition, n1 = P1 Vtank RT1 and n2 = P2 Vtank RT2 These equations may be solved for T2 and n 2 . If mC >>> nC V , then T2 = T1 . If mC = 0, then we recover the isentropic expansion formulas. 3.27 For an ideal gas, Whence, U = CV T U= P V = RT CV R (P V ) = R T (P V ) But 1 CV CV = = 1 C P CV R Therefore : U= 1 (P V ) 1 3.28 Since Z = P V /RT the given equation can be written: V = RT + B RT P Differentiate at constant T : dV = RT dP P2 V2 The isothermal work is then: W = V1 P d V = RT P2 P1 1 dP P Whence, W = RT ln P2 P1 Compared with Eq. (3.27) 3.29 Solve the given equation of state for V : V = RT +b RT P 633 Whence, V P = T RT P2 By denition [Eq. (3.3)]: 1 V V P T Substitution for both V and the derivative yields: = P2 RT RT +b RT P Solve the given equation of state for P: P= RT V b+ Differentiate: P T = V R V b+ RT + RT d dT T V b+ RT 2 By the equation of state, the quantity in parentheses is RT /P; substitution leads to: P T V P + = T P RT 2 d dT T 3.31 When multiplied by V /RT , Eq. (3.42) becomes: Z= a(T )V /RT V a(T )V /RT V 2 = V b V + ( + )bV + b2 V b (V + b)(V + b) Substitute V = 1/: Z= 1 a(T ) 1 RT 1 + ( + )b + (b)2 1 b Expressed in series form, the rst term on the right becomes: 1 = 1 + b + (b)2 + 1 b The nal fraction of the second term becomes: 1 = 1 ( + )b + [( + )2 ](b)2 + 1 + ( + )b + (b)2 Combining the last three equations gives, after reduction: Z =1+ b ( + )a(T )b 2 a(T ) + + b2 + RT RT Equation (3.12) may be written: Comparison shows: Z = 1 + B + C 2 + B =b a(T ) RT and C = b2 + ( + )ba(T ) RT 634 For the Redlich/Kwong equation, the second equation becomes: C = b2 + a(T ) ba(T ) =b b+ RT RT Values for a(T ) and b are found from Eqs. (3.45) and (3.46), with numerical values from Table 3.1: b= 0.08664RTc Pc 0.42748RTc a(T ) = Tr1.5 Pc RT The numerical comparison is an open-ended problem, the scope of which must be decided by the instructor. 3.36 Differentiate Eq. (3.11): Z P = B + 2C P + 3D P 2 + T Whence, Z P =B T,P=0 Equation (3.12) with V = 1/: Differentiate: Z = 1 + B + C 2 + D 3 + Z = B + 2C + 3D 2 + T Whence, Z =B T,=0 3.56 The compressibility factor is related to the measured quantities by: Z= M PV t PV t = m RT n RT (A) By Eq. (3.39), B = (Z 1)V = (Z 1)M V t m (B) (a) By Eq. (A), dT dm dV t dP dM dZ + t + = T m V P M Z (C) Thus Max |% Z | |% M| + |% P| + |% V t | + |% m| + |% T | Assuming approximately equal error in the ve variables, a 1% maximum error in Z requires errors in the variables of <0.2%. (b) By Eq. (B), dm dM dV t Z dZ dB + t + = m M V Z 1 Z B By Eq. (C), Z dB = Z 1 B dT dP T P + 2Z 1 Z 1 dm dM dV t + t m M V Therefore 635 Max |% B| Z Z 1 |% P| + |% T | + 2Z 1 Z 1 |% V t | + |% M| + |% m| For Z 0.9 and for approximately equal error in the ve variables, a 1% maximum error in B requires errors in the variables of less than about 0.02%. This is because the divisor Z 1 0.1. In the limit as Z 1, the error in B approaches innity. 3.57 The Redlich/Kwong equation has the following equivalent forms, where a and b are constants: Z= a V 3/2 (V + b) RT V b P= a RT 1/2 V b T V (V + b) From these by differentiation, Z V = T a(V b)2 b RT 3/2 (V + b)2 RT 3/2 (V b)2 (V + b)2 (A) P V = T a(2V + b)(V b)2 RT 3/2 V 2 (V + b)2 T 1/2 V 2 (V b)2 (V + b)2 (B) In addition, we have the mathematical relation: Z P = T ( Z / V )T ( P/ V )T (C) Combining these three equations gives Z P = T aV 2 (V b)2 b RT 3/2 V 2 (V + b)2 a RT (2V + b)(V b)2 R 2 T 5/2 V 2 (V + b)2 (D) For P 0, V , and Eq. (D) becomes: P0 lim Z P = T b a/RT 3/2 RT For P , V b, and Eq. (D) becomes: P lim Z P = T b RT 3.60 (a) Differentiation of Eq. (3.11) gives: Z P = B + 2C P + 3D P 2 + T whence P0 lim Z P =B T If the limiting value of the derivative is zero, then B = 0, and B = B RT = 0 (b) For simple uids, = 0, and Eqs. (3.52) and (3.53) combine to give B 0 = B Pc /RTc . If B = 0, then by Eq. (3.65), 0.422 B 0 = 0.083 1.6 = 0 Tr 636 and Tr = 0.422 0.083 (1/1.6) = 2.763 3.63 Linear isochores require that ( P/ T )V = Constant. (a) By Eq. (3.4) applied to a constant-V process: P T R V P T = V P T = = V (b) For an ideal gas P V = RT , and V (c) Because a and b are constants, differentiation of Eq. (3.42) yields: R V b In each case the quantities on the right are constant, and so therefore is the derivative. 3.64 (a) Ideal gas: Low P, or low , or large V and/or high T . See Fig. 3.15 for quantitative guidance. (b) Two-term virial equation: Low to modest P. See Fig. 3.14 for guidance. (c) Cubic EOS: Gases at (in principle) any conditions. (d) Lee/Kesler correlation: Same as (c), but often more accurate. Note that corresponding states correlations are strictly valid for non-polar uids. (e) Incompressible liquids: Liquids at normal T s and Ps. Inappropriate where changes in V are required. (f ) Rackett equation: Saturated liquids; a corresponding states application. (g) Constant , liquids: Useful where changes in V are required. For absolute values of V , a reference volume is required. (h) Lydersen correlation for liquids: a corresponding-states method applicable to liquids at extreme conditions. 3.66 Write Eq. (3.12) with 1/ substituted everywhere for V . Subtract 1 from each side of the equation and divide by . Take the limit as 0. 3.68 Follow the procedure laid out on p. 93 with respect to the van der Waals equation to obtain from Eq. (3.42) the following three more-general equations: 1 + (1 ) 2 = 3Z c 2 = 3Z c ( + ) ( 2 + 1) + ( + 1) + 3 = Zc where by denition [see Eqs. (3.45) and (3.46)]: b Pc RTc and ac Pc R 2 Tc2 For a given EOS, and are xed, and the above set represents 3 equations in 3 unknowns, , , and Z c . Thus, for a given EOS the value of Z c is preordained, unrelated to experimental values of Z c . 637 PROPRIETARY MATERIAL. 2005 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. (a, b) For the Redlich/Kwong and Soave/Redlich/Kwong equations, = 0 and = 1. Substitution of these values into the 3-equation set allows their solution to yield: Zc = 1 3 = 0.086640 = 0.427480 (c) For the Peng/Robinson equation, = 1 2 and = 1 + 2. As for the Soave and SRK equations the 3-equation set can be solved (with considerably greater difculty) to yield: Z c = 0.30740 3.69 Equation (3.12): Eliminate : = 0.077796 where = 0.457236 = P/Z RT Z = 1 + B + C 2 + . . . Z =1+ C P2 BP + 2 2 2 + ... Z R T Z RT Z =1+ 2 P2 C P2 B Pc Pr Pr + C Pr + . . . + 2 c2 2 r 2 + . . . = 1 + B Z 2 Tr2 Z Tr R Tc Z Tr RTc Z Tr Rearrange: (Z 1)Z Tr Pr + . . . = B +C Z Tr Pr B = lim (Z 1)Z Tr /Pr Pr 0 3.74 In a cylinder lled with 1 mole of an ideal gas, the molecules have kinetic energy only, and for a given T and P occupy a volume V ig . (a) For 1 mole of a gas with molecules having kinetic energy and purely attractive interactions at the same T and P, the intermolecular separations are smaller, and V < V ig . In this case Z < 1. (b) For 1 mole of a gas with molecules having kinetic energy and purely repulsive interactions at the same T and P, the intermolecular separations are larger, and V > V ig . In this case Z > 1. (c) If attractive and repulsive interactions are both present, they tend to cancel each other. If in balance, then the average separation is the same as for an ideal gas, and V = V ig . In this case Z = 1. 3.75 van der Waals EOS: P= a RT 2 V b V Z= Set V = 1/: Z= a b a 1 =1+ RT 1 b RT 1 b a V V b V RT whence Z rep = b 1 b Z attr = a RT 638 3.76 Write each modication in Z -form, (a) Z= The required behavior is: (b) Z= a V RT V b V lim Z = 1 a RT V lim Z = 1 The required behavior is: (c) Z= a V 2 RT (V b) V lim Z = a RT V lim Z = 1 The required behavior is: (d) Z =1 a 1 V b V RT V lim Z = 0 lim Z = 1 a a =1 RT V RT V Although lim Z = 1 as required, the equation makes Z linear in ; i.e., a 2-term virial EOS in . Such an equation is quite inappropriate at higher densities. V 3.77 Refer to Pb. 2.43, where the general equation was developed; . dU dn +n Q = P V dt dt For an ideal gas, n= PV t RT and dn = dt PV t RT 2 dT dt Note that P V t /R = const. Also for an ideal gas, dU = C V dT whence dT dU = CV dt dt . PV t Q = RT RT 2 P V t dT dT PV t dT = CP CV + RT dt dt RT dt t2 t1 Integration yields: ln R T2 = CP PV t T1 . Q dt 3.78 By Eq. (3.4), dV = dT d P V where and are average values Integrate: ln D2 Vt V2 = ln 2t = ln 2 = ln 2 V1 V1 D1 D1 + D D1 2 = ln 1 + D D1 2 = (T2 T1 ) (P2 P1 ) ln(1.0035)2 = 250 106 (40 10) 45 106 (P2 6) Solution for P2 yields: P2 = 17.4 bar 639 Chapter 4 - Section B - Non-Numerical Solutions 4.5 For consistency with the problem statement, we rewrite Eq. (4.8) as: CP = A + C B T1 ( + 1) + T12 ( 2 + + 1) 3 2 where T2 /T1 . Dene C Pam as the value of C P evaluated at the arithmetic mean temperature Tam . Then: 2 C Pam = A + BTam + C Tam where Tam T1 ( + 1) T1 + T1 T2 + T1 = = 2 2 2 and 2 Tam = T12 2 ( + 2 + 1) 4 Whence, C B T1 ( + 1) + T12 ( 2 + 2 + 1) 4 2 Dene as the difference between the two heat capacities: C Pam = A + C P C Pam = C T12 2 + + 1 2 + 2 + 1 4 3 This readily reduces to: C T12 ( 1)2 12 Making the substitution = T2 /T1 yields the required answer. = 4.6 For consistency with the problem statement, we rewrite Eq. (4.8) as CP = A + D B T1 ( + 1) + 2 T12 where T2 /T1 . Dene C Pam as the value of C P evaluated at the arithmetic mean temperature Tam . Then: D C Pam = A + BTam + 2 Tam As in the preceding problem, Tam = T1 ( + 1) 2 and 2 Tam = T12 2 ( + 2 + 1) 4 Whence, C Pam = A + 4D B T1 ( + 1) + 2 2 2 T1 ( + 2 + 1) Dene as the difference between the two heat capacities: C P C Pam = D T12 4 1 2 + 2 + 1 This readily reduces to: = D T12 1 +1 2 Making the substitution = T2 /T1 yields the required answer. 640 4.8 Except for the noble gases [Fig. (4.1)], C P increases with increasing T . Therefore, the estimate is likely to be low. 4.27 (a) When the water formed as the result of combustion is condensed to a liquid product, the resulting latent-heat release adds to the heat given off as a result of the combustion reaction, thus yielding a higher heating value than the lower heating value obtained when the water is not condensed. (b) Combustion of methane(g) with H2 O(g) as product (LHV): C(s) + O2 (g) CO2 (g) 2H2 (g) + O2 (g) 2H2 O(g) CH4 (g) C(s) + 2H2 (g) H298 = 393,509 H298 = (2)(241,818) H298 = 74,520 CH4 (g) + 2O2 (g) CO2 (g) + 2H2 O(g) H298 = 802,625 J (LHV) Combustion of methane(g) with H2 O(l) as product (HHV): CH4 (g) + 2O2 (g) CO2 (g) + 2H2 O(g) 2H2 O(g) 2H2 O(l) H298 = 802,625 H298 = (2)(44,012) CH4 (g) + 2O2 (g) CO2 (g) + 2H2 O(l) H298 = 890,649 J (HHV) (c) Combustion of n-decane(l) with H2 O(g) as product (LHV): 10 C(s) + 10 O2 (g) 10 CO2 (g) 11 H2 (g) + 5 1 O2 (g) 11 H2 O(g) 2 H298 = (10)(393,509) H298 = (11)(241,818) H298 = 249,700 C10 H22 (l) 10 C(s) + 11 H2 (g) C10 H22 (l) + 15 1 O2 (g) 10 CO2 (g) + 11 H2 O(g) 2 H298 = 6,345,388 J (LHV) Combustion of n-decane(l) with H2 O(l) as product (HHV): C10 H22 (l) + 15 1 O2 (g) 10 CO2 (g) + 11 H2 O(g) 2 H298 = 6,345,388 H298 = (11)(44,012) 11 H2 O(g) 11 H2 O(l) C10 H22 (l) + 15 1 O2 (g) 10 CO2 (g) + 11 H2 O(l) 2 H298 = 6,829,520 J (HHV) 4.49 Saturated because the large H lv overwhelms the sensible heat associated with superheat. H lv . Water because it is cheap, available, non-toxic, and has a large The lower energy content is a result of the decrease in H lv with increasing T , and hence P. However, higher pressures allow higher temperature levels. 641 Chapter 5 - Section B - Non-Numerical Solutions 5.1 Shown to the right is a P V diagram with two adiabatic lines 1 2 and 2 3, assumed to intersect at point 2. A cycle is formed by an isothermal line from 3 1. An engine traversing this cycle would produce work. For the cycle U = 0, and therefore by the rst law, Q + W = 0. Since W is negative, Q must be positive, indicating that heat is absorbed by the system. The net result is therefore a complete conversion of heat taken in by a cyclic process into work, in violation of Statement 1a of the second law (Pg. 160). The assumption of intersecting adiabatic lines is therefore false. Q = U t + E K + E P 5.5 The energy balance for the over-all process is written: Assuming the egg is not scrambled in the process, its internal-energy change after it returns to its initial temperature is zero. So too is its change in kinetic energy. The potential-energy change, however, is negative, and by the preceding equation, so is Q. Thus heat is transferred to the surroundings. The total entropy change of the process is: t Stotal = S t + Ssurr Just as U t for the egg is zero, so is S t . Therefore, t Stotal = Ssurr = Q Q surr = T T Since Q is negative, Stotal is positive, and the process is irreversible. 5.6 By Eq. (5.8) the thermal efciency of a Carnot engine is: T = 1 TC H Differentiate: TC TH = 1 TH and TH TC = TC 1 TC = 2 TH TH TH Since TC /TH is less unity, the efciency changes more rapidly with TC than with TH . So in theory it is more effective to decrease TC . In practice, however, TC is xed by the environment, and is not subject to control. The practical way to increase is to increase TH . Of course, there are limits to this too. 5.11 For an ideal gas with constant heat capacities, and for the changes T1 T2 and P1 P2 , Eq. (5.14) can be rewritten as: P2 T2 R ln S = C P ln P1 T1 (a) If P2 = P1 , S P = C P ln T2 T1 If V2 = V1 , T2 P2 = T1 P1 Whence, SV = C P ln T2 T1 R ln T2 T1 = C V ln T2 T1 642 Since C P > C V , this demonstrates that (b) If T2 = T1 , ST = R ln P2 P1 SP > SV . If V2 = V1 , P2 T2 = P1 T1 Whence, SV = C P ln P2 P1 R ln P2 P1 = C V ln P2 P1 This demonstrates that the signs for ST and SV are opposite. 5.12 Start with the equation just preceding Eq. (5.14) on p. 170: ig dP C dT C dT dS d ln P = P = P P R T R T R ig For an ideal gas P V = RT , and ln P + ln V = ln R + ln T . Therefore, dT dV dP = + T V P ig or ig dV dT dP = V T P Whence, ig dV dT C dT dS = + = P V T R T R ig dT CP + d ln V 1 T R Because (C P /R) 1 = C V /R, this reduces to: C dT dS + d ln V = V R T R ig Integration yields: S = R T T0 V C V dT + ln V0 R T ig ********************** As an additional part of the problem, one could ask for the following proof, valid for constant heat capacities. Return to the original equation and substitute dT /T = d P/P + d V /V : C dV C dP dP C dV C dP dS + P = V + P = P R V R P P R V R P R ig ig ig ig Integration yields: V C C P S + P ln = V ln V0 R P0 R R ig ig 5.13 As indicated in the problem statement the basic differential equations are: d W d Q H d QC = 0 (A) TH d QH = TC d QC (B) where Q C and Q H refer to the reservoirs. 643 t t (a) With d Q H = C H dTH and d Q C = CC dTC , Eq. (B) becomes: t TH C H dTH = t TC CC dTC or C t dTH dTC = H t C C TH TC Whence, d ln TC = d ln TH where t CH t CC Integration from TH0 and TC0 to TH and TC yields: TC = TC0 TH TH0 or TC = TC0 TH TH0 t t (b) With d Q H = C H dTH and d Q C = CC dTC , Eq. (A) becomes: t t d W = C H dTH + CC dTC Integration yields: t t W = C H (TH TH0 ) + CC (TC TC0 ) Eliminate TC by the boxed equation of Part (a) and rearrange slightly: W = t C H TH0 TH t 1 + CC TC0 TH0 TH TH0 1 (c) For innite time, TH = TC T , and the boxed equation of Part (a) becomes: T = TC0 T TH0 +1 = TC0 TH0 T From which: T = (TC0 )1/( +1) T (TH0 ) /( +1) = TC0 (TH0 ) and T = (TC0 )1/( TH0 +1) (TH0 ) /( +1)1 Because /( + 1) 1 = 1/( T = TH0 + 1), then: TC0 TH0 1/( +1) and T TH0 = TC0 TH0 /( +1) Because TH = T , substitution of these quantities in the boxed equation of Part (b) yields: W = t C H TH0 TC0 TH0 1/( +1) 1 + t CC TC0 TC0 TH0 /( +1) 1 5.14 As indicated in the problem statement the basic differential equations are: d W d Q H d QC = 0 (A) TH d QH = TC d QC (B) where Q C and Q H refer to the reservoirs. 644 t (a) With d Q C = CC dTC , Eq. (B) becomes: TH d QH = t TC CC dTC or t d Q H = CC TH dTC TC Substitute for d Q H and d Q C in Eq. (A): t d W = CC TH Integrate from TC0 to TC : t W = CC TH ln dTC t + CC dTC TC TC t + CC (TC TC0 ) TC0 or t W = CC TH ln TC0 + TC TC0 TC (b) For innite time, TC = TH , and the boxed equation above becomes: t W = CC TH ln TC0 + TH TC0 TH . 5.15 Write Eqs. (5.8) and (5.1) in rate form and combine to eliminate | Q H |: . . . . |W | TC |W | = |W | + | Q| = 1r or . . =1 1r TH |W | + | Q C | . With | Q C | = k A(TC )4 = k A(r TH )4 , this becomes: where r TC TH . |W | . r 1 1 = |W | 1r 1r = k Ar 4 (TH )4 or A= . |W | k(TH )4 1 (1 r )r 3 Differentiate, noting that the quantity in square brackets is constant: . . |W | 1 3 |W | dA = + = k(TH )4 (1 r )r 4 (1 r )2r 3 k(TH )4 dr 4r 3 (1 r )2r 4 Equating this equation to zero, leads immediately to: 4r = 3 or r = 0.75 5.20 Because W = 0, Eq. (2.3) here becomes: Q= U t = mC V T A necessary condition for T to be zero when Q is non-zero is that m = . This is the reason that natural bodies (air and water) that serve as heat reservoirs must be massive (oceans) or continually renewed (rivers). 5.22 An appropriate energy balance here is: Q= Ht = 0 m 1 T1 + m 2 T2 m1 + m2 Applied to the process described, with T as the nal temperature, this becomes: m 1 C P (T T1 ) + m 2 C P (T T2 ) = 0 whence T = (1) If m 1 = m 2 , 645 T = (T1 + T2 )/2 The total entropy change as a result of temperature changes of the two masses of water: T T + m 2 C P ln T2 T1 Equations (1) and (2) represent the general case. If m 1 = m 2 = m, S t = m 1 C P ln (2) S t = mC P ln T2 T1 T2 or S t = 2mC P ln T T1 T2 Because T = (T1 + T2 )/2 > T1 T2 , S t is positive. 5.23 Isentropic processes are not necessarily reversible and adiabatic. The term isentropic denotes a process for which the system does not change in entropy. There are two causes for entropy changes in a system: The process may be internally irreversible, causing the entropy to increase; heat may be transferred between system amd surroundings, causing the entropy of the system to increase or decrease. For processes that are internally irreversible, it is possible for heat to be transferred out of the system in an amount such that the entropy changes from the two causes exactly compensate each other. One can imagine irreversible processes for which the state of the system is the same at the end as at the beginning of the process. The process is then necessarily isentropic, but neither reversible nor adiabatic. More generally, the system conditions may change in such a way that entropy changes resulting from temperature and pressure changes compensate each other. Such a process is isentropic, but not necessarily reversible. Expansion of gas in a piston/cylinder arrangement is a case in point. It may be reversible and adiabatic, and hence isentropic. But the same change of state may be irreversible with heat transfer to the surroundings. The process is still isentropic, but neither reversible nor adiabatic. An isentropic process must be either reversible and adiabatic or irreversible and non-adiabatic. 5.24 By denition, C P dT T0 T T T0 By inspection, one sees that for both T > T0 and T0 > T the numerators and denominators of the above fractions have the same sign. Thus, for both cases C P H is positive. CP H = T T0 C P dT = T0 T Similarly, CP S dT dT T0 T CP T T = = ln(T0 /T ) ln(T /T0 ) T T0 CP By inspection, one sees that for both T > T0 and T0 > T the numerators and denominators of the above fractions have the same sign. Thus, for both cases C P S is positive. When T = T0 , both the numerators and denominators of the above fractions become zero, and the fractions are indeterminate. Application of lH pitals rule leads to the result: C P H = C P S = C P . o 5.31 The process involves three heat reservoirs: the house, a heat sink; the furnace, a heat source; and the surroundings, a heat source. Notation is as follows: |Q| |Q F | |Q | Heat transfer to the house at temperature T Heat transfer from the furnace at TF Heat transfer from the surroundings at T The rst and second laws provide the two equations: |Q| = |Q F | + |Q | and |Q | |Q| |Q F | =0 T TF T 646 Combine these equations to eliminate |Q |, and solve for |Q F |: |Q F | = |Q| T T TF T TF T With T = 295 K TF = 810 K T = 265 K and |Q| = 1000 kJ The result is: |Q F | = 151.14 kJ Shown to the right is a scheme designed to accomplish this result. A Carnot heat engine operates with the furnace as heat source and the house as heat sink. The work produced by the engine drives a Carnot refrigerator (reverse Carnot engine) which extracts heat from the surroundings and discharges heat to the house. Thus the heat rejected by the Carnot engine (|Q 1 |) and by the Carnot refrigerator (|Q 2 |) together provide the heat |Q| for the house. The energy balances for the engine and refrigerator are: |W |engine = |Q F | |Q 1 | Equation (5.7) may be applied to both the engine and the refrigerator: |W |refrig = |Q 2 | |Q | T |Q | TF |Q F | = = T |Q 2 | T |Q 1 | Combine the two pairs of equations: |W |engine = |Q 1 | TF T TF 1 = |Q 1 | T T |W |refrig = |Q 2 | 1 T T = |Q 2 | T T T Since these two quantities are equal, |Q 1 | T T TF T = |Q 2 | T T or |Q 2 | = |Q 1 | Because the total heat transferred to the house is |Q| = |Q 1 | + |Q 2 |, |Q| = |Q 1 | + |Q 1 | TF T T T But |Q 1 | = |Q F | T TF TF T TF T = |Q 1 | 1 + T T T T = |Q 1 | whence |Q| = |Q F | T TF TF T T T TF T T T Solution for |Q F | yields the same equation obtained more easily by direct application of the two laws of thermodynamics to the overall result of the process. 5.32 The process involves three heat reservoirs: the house, a heat source; the tank, a heat source; and the surroundings, a heat sink. Notation is as follows: 647 |Q| |Q | |Q | Heat transfer from the tank at temperature T Heat transfer from the house at T Heat transfer to the surroundings at T The rst and second laws provide the two equations: |Q| + |Q | = |Q | and |Q | |Q | |Q| =0 T T T Combine these equations to eliminate |Q |, and solve for |Q|: |Q| = |Q | T T T T T T With T = 448.15 K T = 297.15 K T = 306.15 K and |Q | = 1500 kJ The result is: |Q| = 143.38 kJ Shown to the right is a scheme designed to accomplish this result. A Carnot heat engine operates with the tank as heat source and the surroundings as heat sink. The work produced by the engine drives a Carnot refrigerator (reverse Carnot engine) which extracts heat |Q | from the house and discharges heat to the surroundings. The energy balances for the engine and refrigerator are: |W |engine = |Q| |Q 1 | Equation (5.7) may be applied to both the engine and the refrigerator: |W |refrig = |Q 2 | |Q | Combine the two pairs of equations: |W |engine = |Q| 1 T T T |Q 1 | = T |Q| T |Q 2 | = T |Q | = |Q| T T T |W |refrig = |Q | T T = |Q | T t T Since these two quantities are equal, |Q| T T T T = |Q | T T or |Q| = |Q | T T T T T T 5.36 For a closed system the rst term of Eq. (5.21) is zero, and it becomes: . . Qj d(m S)cv = SG 0 + T, j dt j 648 . where Q j is here redened to refer to the system rather than to the surroundings. Nevertheless, the sect ond term accounts for the entropy changes of the surroundings, and can be written simply as d Ssurr /dt: t . d(m S)cv d Ssurr = SG 0 dt dt or T t . d Scv d Ssurr = SG 0 dt dt Multiplication by dt and integration over nite time yields: t Scv + t Ssurr 0 or Stotal 0 5.37 The general equation applicable here is Eq. (5.22): . (S m)fs j . . Qj = SG 0 T, j (a) For a single stream owing within the pipe and with a single heat source in the surroundings, this becomes: . . Q . = SG 0 ( S)m T (b) The equation is here written for two streams (I and II) owing in two pipes. Heat transfer is . internal, between the two streams, making Q = 0. Thus, . . . ( S)I m I + ( S)II m II = SG 0 (c) For a pump operatiing on a single stream and with the assumption of negligible heat transfer to the surroundings: . . ( S)m = SG 0 (d) For an adiabatic gas compressor the result is the same as for Part (c). (e) For an adiabatic turbine the result is the same as for Part (c). (f ) For an adiabatic throttle valve the result is the same as for Part (c). (g) For an adiabatic nozzle the result is the same as for Part (c). 5.40 The gure on the left below indicates the direct, irreversible transfer of heat |Q| from a reservoir at T1 to a reservoir at T2 . The gure on the right depicts a completely reversible process to accomplish the same changes in the heat reservoirs at T1 and T2 . 649 The entropy generation for the direct heat-transfer process is: SG = |Q| 1 1 T1 T2 = |Q| T1 T2 T1 T2 For the completely reversible process the net work produced is Wideal : |W1 | = |Q| T1 T T1 and |W2 | = |Q| T2 T T2 Wideal = |W1 | |W2 | = T |Q| T1 T2 T1 T2 This is the work that is lost, Wlost , in the direct, irreversible transfer of heat |Q|. Therefore, Wlost = T |Q| T1 T2 = T SG T1 T2 Note that a Carnot engine operating between T1 and T2 would not give the correct Wideal or Wlost , because the heat it transfers to the reservoir at T2 is not Q. 5.45 Equation (5.14) can be written for both the reversible and irreversible processes: Sirrev = Tirrev T0 CP ig P dT ln P T Srev = Tirrev Trev ig Trev T0 CP ig P dT ln P T By difference, with Srev = 0: Sirrev = CP dT T Since Sirrev must be greater than zero, Tirrev must be greater than Trev . 650 Chapter 6 - Section B - Non-Numerical Solutions 6.1 By Eq. (6.8), H S =T P and isobars have positive slope Differentiate the preceding equation: 2 H S2 = P T S P Combine with Eq. (6.17): 2 H S2 = P T CP and isobars have positive curvature. 6.2 (a) Application of Eq. (6.12) to Eq. (6.20) yields: C P P = T {V T ( V / T ) P } T P or C P P = T V T T P 2V T 2 P V T P Whence, C P P = T T 2V T 2 P For an ideal gas: V T = P R P and 2V T 2 =0 P (b) Equations (6.21) and (6.33) are both general expressions for d S, and for a given change of state both must give the same value of d S. They may therefore be equated to yield: (C P C V ) dT = T P T dV + V V T dP P Restrict to constant P: C P = CV + T P T V V T P By Eqs. (3.2) and (6.34): V T = V P and P T = V Combine with the boxed equation: C P C V = T V 6.3 By the denition of H , U = H P V . Differentiate: U T = P H T P P V T or P U T = CP P P V T P 651 Substitute for the nal derivative by Eq. (3.2), the denition of : U T = CP PV P Divide Eq. (6.32) by dT and restrict to constant P. The immediate result is: U T = CV + T P P T P V V T P Solve for the two derivatives by Eqs. (6.34) and (3.2); substitution gives: U T = CV + P (T P)V 6.4 (a) In general, dU = C V dT + T P T P dV V (6.32) By the equation of state, P= RT V b whence P T = V P R = T V b Substituting this derivative into Eq. (6.32) yields dU = C V dT , indicating that U = f (T ) only. (b) From the denition of H , From the equation of state, d H = dU + d(P V ) d(P V ) = R dT + b d P Combining these two equations and the denition of part (a) gives: d H = C V dT + R dT + b d P = (C V + R)dT + b d P Then, H T = CV + R P By denition, this derivative is C P . Therefore C P = C V + R. Given that C V is constant, then so is C P and so is C P /C V . (c) For a mechanically reversible adiabatic process, dU = d W . Whence, by the equation of state, C V dT = P d V = d(V b) RT d V = RT V b V b or R dT d ln(V b) = CV T But from part (b), R/C V = (C P C V )/C V = 1. Then d ln T = ( 1)d ln(V b) From which: or d ln T + d ln(V b) 1 = 0 T (V b) 1 = const. Substitution for T by the equation of state gives P(V b)(V b) 1 = const. R or P(V b) = const. 652 6.5 It follows immediately from Eq. (6.10) that: V = G P and T S= G T P Differentation of the given equation of state yields: V = RT P and S= d (T ) R ln P dT Once V and S (as well as G) are known, we can apply the equations: H = G +TS These become: H = (T ) T d (T ) dT and U = H P V = H RT and U = (T ) T d (T ) RT dT By Eqs. (2.16) and (2.20), CP = H T and P CV = U T V Because is a function of temperature only, these become: C P = T d2 dT 2 and C V = T d2 R = CP R dT 2 The equation for V gives the ideal-gas value. The equations for H and U show these properties to be functions of T only, which conforms to ideal-gas behavior. The equation for S shows its relation to P to be that of an ideal gas. The equations for C P and C V show these properties to be functions of T only, which conforms to ideal-gas behavior, as does the result, C P = C V + R. We conclude that the given equation of state is consistent with the model of ideal-gas behavior. 6.6 It follows immediately from Eq. (6.10) that: V = G P and T S= G T P Differentation of the given equation of state yields: V =K and S= d F(T ) dT Once V and S (as well as G) are known, we can apply the equations: H = G +TS These become: H = F(T ) + K P T d F(T ) dT and U = H PV = H PK and U = F(T ) T d F(T ) dT By Eqs. (2.16) and (2.20), CP = H T and P CV = U T V 653 Because F is a function of temperature only, these become: C P = T d2 F dT 2 and C V = T d2 F = CP dT 2 The equation for V shows it to be constant, independent of both T and P. This is the denition of an incompressible uid. H is seen to be a function of both T and P, whereas U , S, C P , and C V are functions of T only. We also have the result that C P = C V . All of this is consistent with the model of an incompressible uid, as discussed in Ex. 6.2. 6.11 Results for this problem are given in the text on page 217 by Eqs. (6.61), (6.62) and (6.63) for G R , H R , and S R respectively. 6.12 Parameter values for the van der Waals equation are given by the rst line of Table 3.1, page 98. At the bottom of page 215, it is shown that I = /Z . Equation (6.66b) therefore becomes: q GR = Z 1 ln(Z ) Z RT For given T and P, Z is found by solution of Eq. (3.52) for a vapor phase or Eq. (3.56) for a liquid phase with = = 0. Equations (3.53) and (3.54) for the van der Waals equation are: = Pr 8Tr and q= 27 8Tr With appropriate substitutions, Eqs. (6.67) and (6.68) become: q HR = Z 1 Z RT and SR = ln(Z ) R 6.13 This equation does not fall within the compass of the generic cubic, Eq. (3.42); so we start anew. First, multiply the given equation of state by V /RT : a V PV exp = V RT V b RT Substitute: Z PV RT V = 1 a q b RT Then, Z= 1 exp(qb) 1 b With the denition, b, this becomes: Z= 1 exp(q ) 1 (A) Because = P/Z RT , = bP Z RT Given T and P, these two equations may be solved iteratively for Z and . Because b is a constant, Eqs. (6.58) and (6.59) may be rewritten as: 654 GR = RT 0 (Z 1) 0 d + Z 1 ln Z (B) HR = RT Z T d + Z 1 (C) In these equations, Z is given by Eq. (A), from which is also obtained: ln Z = ln(1 ) q and Z T = q exp(q ) T (1 ) The integrals in Eqs. (B) and (C) must be evaluated through the exponential integral, E(x), a special function whose values are tabulated in handbooks and are also found from such software packages as MAPLE R . The necessary equations, as found from MAPLE R , are: 0 (Z 1) d = exp(q){E[q(1 )] E(q)} E(q ) ln(q ) where is Eulers constant, equal to 0.57721566. . . . and T 0 Z T s = q exp(q){E[q(1 )] E(q)} Once values for G R /RT and H R /RT are known, values for S R /R come from Eq. (6.47). The difculties of integration here are one reason that cubic equations have found greater favor. 6.18 Assume the validity for purposes of interpolation of Eq. (6.75), and write it for T2 , T , and T1 : ln P2sat = A B T2 (A) ln P sat = A B T (B) ln P1sat = A B T1 (C) Subtract (C) from (A): ln P2sat =B P1sat 1 1 T2 T1 =B (T2 T1 ) T1 T2 Subtract (C) from (B): ln P sat =B P1sat 1 1 T T1 =B (T T1 ) T1 T The ratio of these two equations, upon rearrangement, yields the required result. 6.19 Write Eq. (6.75) in log10 form: log P sat = A B T (A) Apply at the critical point: log Pc = A B Tc (B) 655 By difference, log Pr sat = B 1 1 T Tc =B Tr 1 T (C) If P sat is in (atm), then application of (A) at the normal boiling point yields: log 1 = A B Tn or A= B Tn With Tn /Tc , Eq. (B) can now be written: log Pc = B 1 1 Tc Tn =B Tc Tn Tn Tc =B 1 Tn Whence, B= Equation (C) then becomes: log Pr sat = Tn 1 Tn 1 log Pc Tr 1 log Pc = T 1 Tr 1 log Pc Tr Apply at Tr = 0.7: log(Pr sat )Tr =0.7 = 3 7 1 log Pc By Eq. (3.48), Whence, = 1.0 log(Pr sat )Tr =0.7 = 3 7 1 log Pc 1 6.83 The slopes of isobars and isochores on a T S diagram are given by Eqs. (6.17) and (6.30): T S = P T CP and T S = V T CV Both slopes are necessarily positive. With C P > C V , isochores are steeper. An expression for the curvature of isobars results from differentiation of the rst equation above: 2T S2 = P 1 CP T S P T C2 P C P S = P T T 2 2 CP CP C P T P T S = P T T 1 2 CP CP C P T P With C P = a + bT , C P T =b P and 1 T CP C P T =1 P a bT = a + bT a + bT Because this quantity is positive, so then is the curvature of an isobar. 6.84 Division of Eq. (6.8) by d S and restriction to constant T yields: H S =T +V T P S By Eq. (6.25), T P S = T 1 V Therefore, H S =T T 1 1 = (T 1) 656 Also, 2 H S2 = T 1 2 S = T 1 2 P T P S = T 1 2 P T 1 V Whence, 2 H S2 = T 1 3V P T By Eqs. (3.2) and (3.38): = 1 V V T and P V = RT +B P Whence, V T = P dB R + dT P and = 1 V dB R + dT P Differentiation of the second preceding equation yields: P = T R V P2 dB R + dT P 1 V2 V P = T 1 R (V ) 2 V V P2 V P T From the equation of state, V P = T RT P2 Whence, P = T R RT R (T 1) = + 2 2 V P2 V P VP Clearly, the signs of quantity (T 1) and the derivative on the left are the same. The sign is determined from the relation of and V to B and d B/dT : T V T 1 = dB R + dT P dB dB RT B T +T dT 1 = dT 1= P RT RT +B +B P P In this equation d B/dT is positive and B is negative. Because RT /P is greater than |B|, the quantity T 1 is positive. This makes the derivative in the rst boxed equation positive, and the second derivative in the second boxed equation negative. 6.85 Since a reduced temperature of Tr = 2.7 is well above normal temperatures for most gases, we expect on the basis of Fig. 3.10 that B is () and that d B/dT is (+). Moreover, d 2 B/dT 2 is (). By Eqs. (6.54) and (6.56), By Eqs. (3.38) and (6.40), GR = BP V R = B, and S R = P(d B/dT ) Whence, both G R and S R are (). From the denition of G R , H R = G R + T S R , and H R is (). and V R is (). Combine the equations above for G R , S R , and H R : HR = P B T dB dT Whence, HR T =P P dB d2 B dB T 2 dT dT dT = P T d2 B dT 2 R Therefore, C P HR T is (+). P (See Fig. 6.5.) 657 6.89 By Eq. (3.5) at constant T : P = V 1 P1 ln V1 (A) (a) Work d W = P d V = 1 1 V 1 P1 d V = ln V d V P1 + ln V1 d V ln V1 V2 V1 W = 1 ln V d V P1 + 1 ln V1 (V2 V1 ) W = 1 1 [(V2 ln V2 V2 ) (V1 ln V1 V1 )] P1 (V2 V1 ) (V2 ln V1 V1 ln V1 ) = V2 1 + V1 V2 P1 (V2 V1 ) V2 ln V1 By Eq. (3.5), ln V2 = (P2 P1 ) V1 whence W = P1 V1 P2 V2 V2 V1 (b) Entropy By Eq. (A), By Eq. (6.29), P = d S = V d P ln V1 ln V P1 and d P = 1 d ln V dS = V d ln V = d V and S= (V2 V1 ) (c) Enthalpy Substitute for d P: By Eq. (6.28), d H = (1 T )V d P d H = (1 T )V 1 T 1 dV d ln V = H= 1 T (V1 V2 ) These equations are so simple that little is gained through use of an average V . For the conditions given in Pb. 6.9, calculations give: W = 4.855 kJ kg1 S = 0.036348 kJ kg1 K1 M P H = 134.55 kJ kg1 6.90 The given equation will be true if and only if dP = 0 T The two circumstances for which this condition holds are when ( M/ P)T = 0 or when d P = 0. The former is a property feature and the latter is a process feature. 6.91 Neither C P ig T H ig H ig H ig ig = CP + = T P P V P T P V nor ( T / P)V is in general zero for an ideal gas. T P V H ig P = S H ig P + T H ig T P T P S = CP ig T P S 658 T P = S T S ig P S ig P = T T CP T ig S ig P T H ig P ig =T S S ig P Neither T nor ( S / P)T is in general zero for an ideal gas. The difculty here is that the expression independent of pressure is imprecise. 6.92 For S = S(P, V ): dS = S P dP + V S V dV P By the chain rule for partial derivatives, dS = S T V T P dP + V S T P T V dV P With Eqs. (6.30) and (6.17), this becomes: dS = CV T T P dP + V CP T T V dV P 6.93 By Eq. (6.31), P=T P T V U V T (a) For an ideal gas, P= RT V and P T = V R V Therefore RT RT = V V U V and T U V =0 T (b) For a van der Waals gas, P= a RT 2 V b V and P T = V R V b Therefore RT a RT 2 = V b V b V U V and T U V = T a V2 (c) Similarly, for a Redlich/Kwong uid nd: U V = T T 1/2 V (V (3/2)A + b) where A = a(Tc ) Tc 1 2 6.94 (a) The derivatives of G with respect to T and P follow from Eq, (6.10): S = G T and P V = G P T Combining the denition of Z with the second of these gives: Z P PV = RT RT G P T 659 Combining Eqs. (2.11) and (3.63) and solving for U gives U = G + T S P V . Replacing S and V by their derivatives gives: U =GT G T P P G P T Developing an equation for C V is much less direct. First differentiate the above equation for U with respect to T and then with respect to P: The two resulting equations are: U T = T P 2G T 2 2 P P 2G T P U P = T T G T P P 2G P2 T From the denition of C V and an equation relating partial derivatives: CV U T = V U T + P U P T P T V Combining the three equations yields: C V = T 2G T 2 P P 2G T P T 2G T P +P 2G P2 T P T V Evaluate ( P/ T )V through use of the chain rule: P T = V P V T V T = P ( V / T ) P ( V / P)T The two derivatives of the nal term come from differentiation of V = (G/ P)T : V T = P 2G P T and V P = T 2G P2 T Then P T = V ( 2 G/ T ) P ( 2 G/ P 2 )T and C V = T 2G T 2 P P 2G T P + T 2G T P +P 2G P2 T ( 2 G/ P T ) ( 2 G/ P 2 )T Some algebra transforms this equation into a more compact form: C V = T 2G T 2 +T P ( 2 G/ T P)2 ( 2 G/ P 2 )T (b) The solution here is analogous to that of part (a), but starting with the derivatives inherent in Eq. (6.9). 6.97 Equation (6.74) is exact: H lv d ln P sat = R Z lv d(1/T ) The right side is approximately constant owing to the qualitatively similar behaviior of Z lv . Both decrease monotonically as T increases, becoming zero at the critical point. H lv and 660 6.98 By the Clapeyron equation: If the ratio S sl to H sl S sl d P sat = = T V sl V sl dT sl V is assumed approximately constant, then P sat = A + BT If the ratio H sl to V sl is assumed approximately constant, then P sat = A + B ln T 6.99 By Eq, (6.73) and its analog for sv equilibrium: sat d Psv dT = t Pt Htsv Pt Htsv RTt2 RTt2 Z tsv sat d Plv dT = t Pt Htlv Pt Htlv RTt2 RTt2 Z tlv sat d Psv dT t sat d Plv dT t Pt RTt2 Htsv Htlv Because Htsv Htlv = Htsl is positive, then so is the left side of the preceding equation. 6.100 By Eq. (6.72): H lv d P sat = T V lv dT But V lv = RT P sat Z lv whence H lv d ln P sat = RT 2 Z lv dT lv (6.73) H 1 H lv Tc H lv d ln Pr sat = 2 = = 2 Z lv lv 2 Z lv Tr RTc Tr Z RT dTr 6.102 Convert c to reduced conditions: c d ln P sat d ln T = T =Tc d ln Prsat d ln Tr = Tr Tr =1 d ln Prsat dTr = Tr =1 d ln Prsat dTr Tr =1 From the Lee/Kesler equation, nd that d ln Prsat dTr = 5.8239 + 4.8300 Tr =1 Thus, c (L/K) = 5.82 for = 0, and increases with increasing molecular complexity as quantied by . 661 Chapter 7 - Section B - Non-Numerical Solutions 7.2 (a) Apply the general equation given in the footnote on page 266 to the particular derivative of interest here: S T T = S P P T P S The two partial derivatives on the right are found from Eqs. (6.17) and (6.16); thus, T P = S T CP V T P For gases, this derivative is positive. It applies to reversible adiabatic expansions and compressions in turbines and compressors. (b) Application of the same general relation (page 266) yields: T V = U T U V U V T The two partial derivatives on the right are found from Eqs. (2.16) and (6.31); thus, T V = U 1 CV PT P T V For gases, this may be positive or negative, depending on conditions. Note that it is zero for an ideal gas. It applies directly to the Joule expansion, an adiabatic expansion of gas conned in a portion of a container to ll the entire container. 7.3 The equation giving the thermodynamic sound speed appears in the middle of page 257. As written, it implicitly requires that V represent specic volume. This is easily conrmed by a dimensional analysis. If V is to be molar volume, then the right side must be divided by molar mass: c2 = V2 M P V (A) S Applying the equation given in the footnote on page 266 to the derivative yields: P V = S P S V S V P This can also be written: P V = S P T V T S V S T P T V = P T S V S T P P T V T V P Division of Eq. (6.17) by Eq. (6.30) shows that the rst product in square brackets on the far right is the ratio C P /C V . Reference again to the equation of the footnote on page 266 shows that the second product in square brackets on the far right is ( P/ V )T , which is given by Eq. (3.3). Therefore, P V = S CP CV P V = T CP CV 1 V 662 Substitute into Eq. (A): c2 = V CP MC V or c= V CP MC V (a) For an ideal gas, V = RT /P and = 1/P. Therefore, cig = RT C P M CV (b) For an incompressible liquid, V is constant, and = 0, leading to the result: c = . This of course leads to the conclusion that the sound speed in liquids is much greater than in gases. 7.6 As P2 decreases from an initial value of P2 = P1 , both u 2 and m steadily increase until the critical pressure ratio is reached. At this value of P2 , u 2 equals the speed of sound in the gas, and further reduction in P2 does not affect u 2 or m. 7.7 The mass-ow rate m is of course constant throughout the nozzle from entrance to exit. The velocity u rises monotonically from nozzle entrance (P/P1 = 1) to nozzle exit as P and P/P1 decrease. The area ratio decreases from A/A1 = 1 at the nozzle entrance to a minimum value at the throat and thereafter increases to the nozzle exit. 7.8 Substitution of Eq. (7.12) into (7.11), with u 1 = 0 gives: u2 throat = 2 2 P1 V1 1 +1 1 = P1 V1 2 +1 where V1 is specic volume in m3 kg1 and P1 is in Pa. The units of u 2 throat are then: N m3 kg1 = N m kg1 = kg m s2 m kg1 = m2 s2 m2 With respect to the nal term in the preceding equation, note that P1 V1 has the units of energy per unit mass. Because 1 N m = 1 J, equivalent units are Jkg1 . Moreover, P1 V1 = RT1 /M; whence Pa m3 kg1 = u2 throat = RT1 M 2 +1 With R in units of J(kg mol)1 K1 , RT1 /M has units of Jkg1 or m2 s2 . 663 7.16 It is shown at the end of Ex. 7.5 that the Joule/Thomson inversion curve is the locus of states for which ( Z / T ) P = 0. We apply the following general equation of differential calculus: x y = z x y + w x w y w y z Z T = P Z T + Z T T P Whence, Z T = Z T P Z T T P Because P = Z RT , = P Z RT and T = P P R 1 Z +T (Z T )2 Z T P Setting ( Z / T ) P = 0 in each of the two preceding equations reduces them to: Z T = Z T T and P T = P P = 2 T Z RT Combining these two equations yields: T Z T = Z T (a) Equation (3.42) with van der Waals parameters becomes: P= a RT 2 V b V Multiply through by V /RT , substitute Z = P V /RT , V = 1/, and rearrange: Z= a 1 RT 1 b In accord with Eq. (3.51), dene q a/b RT . In addition, dene b. Then, Z= 1 q 1 (A) Differentiate: Z T = Z T = dq dT By Eq. (3.54) with (Tr ) = 1 for the van der Waals equation, q = dq = dT / Tr . Whence, 1 Tr2 dTr = dT 1 Tr2 Tc = q 1 = T T Tr Then, Z T = ( ) q T = q T In addition, Z =b T Z = T b qb (1 )2 664 Substitute for the two partial derivatives in the boxed equation: T b q qb = (1 )2 T or q = Whence, 1 =1 2q q (1 )2 (B) By Eq. (3.46), Pc = RTc /b. Moreover, P = Z RT . Division of the second equation by the rst gives Pr = ZbT / Tc . Whence Pr = Z Tr (C) These equations allow construction of a Tr vs. Pr inversion curve as in Fig. 7.2. For a given value of Tr , calculate q. Equation (B) then gives , Eq. (A) gives Z , and Eq. (C) gives Pr . (b) Proceed exactly as in Part (a), with exactly the same denitions. This leads to a new Eq. (A): Z= By Eq. (3.54) with (Tr ) = Tr0.5 for the Redlich/Kwong equation, q = q 1 1+ 1 (A) / Tr1.5 . This leads to: 1.5 q dq = T dT and Z T = 1.5 q T (1 + ) Moreover, Z T = bq b 2 (1 + )2 (1 ) 2 Substitution of the two derivatives into the boxed equation leads to a new Eq. (B): q= 1+ 1 As in Part (a), for a given Tr , calculate q, and solve Eq. (B) for , by trial or a by a computer routine. As before, Eq. (A) then gives Z , and Eq. (C) of Part (a) gives Pr . 7.17 (a) Equal to. (b) Less than. (c) Less than. (d) Equal to. (e) Equal to. 1 2.5 + 1.5 (B) 7.28 When a saturated liquid is expanded in a turbine some of the liquid vaporizes. A turbine properly designed for expansion of liquids cannot handle the much larger volumes resulting from the formation of vapor. For example, if saturated liquid at 5 bar expands isentropically to 1 bar, the fraction of the original liquid that vaporizes is found as follows: l v v l S2 = S2 + x2 (S2 S2 ) = S1 or v x2 = Were the expansion irreversible, the fraction of liquid vaporized would be even greater. 7.33 Apply Eq. (2.29) to this non-steady-state process, with n replacing m, with the tank as control volume, and with a single inlet stream. Since the process is adiabatic and the only work is shaft work, this equation may be multiplied by dt to give: d(nU )tank H dn = d Ws 665 l 1.8604 1.3027 S1 S2 = 0.0921 = l v 7.3598 1.3027 S2 S2 Because the inlet stream has constant properties, integration from beginning to end of the process yields: Ws = n 2 U 2 n 1 U 1 n H where the subscripted quantities refer to the contents of the tank and n and H refer to the inlet stream. Substitute n = n 2 n 1 and H = U + P V = U + RT : Ws = n 2U2 n 1U1 (n 2 n 1 )(U + RT ) = n 2 (U2 U RT ) n 1 (U1 U RT ) With U = C V T for an ideal gas with constant heat capacities, this becomes: Ws = n 2 [C V (T2 T ) RT ] n 1 [C V (T1 T ) RT ] However, T = T1 , and therefore: Ws = n 2 [C V (T2 T1 ) RT1 ] + n 1 RT1 By Eq. (3.30b), T2 = P2 P1 ( 1)/ ) Moreover, n1 = P1 Vtank RT1 and n2 = P2 Vtank RT2 With = 1.4, T2 = 573.47 K. Then, with R = 8.314 m3 kPa kmol1 K1 , n1 = (101.33)(20) = 0.8176 kmol (8.314)(298.15) and n2 = (1000)(20) = 4.1948 kmol (8.314)(573.47) Substitution of numerical values into the boxed equation, with R = 8.314 kJ kmol1 K1 , gives: Ws = 15, 633 kJ 7.40 Combine Eqs. (7.13) and (7.17): Ws = n H =n ( H )S By Eq. (6.8), Assume now that RT1 /P1 . Then ( H )S = V dP = V P P is small enough that V , an average value, can be approximated by V1 = ( H )S = RT1 P1 P and RT1 Ws = n P1 P Equation (7.22) is the usual equation for isentropic compression of an ideal gas with constant heat capacities. For irreversible compression it can be rewritten: nC P T1 Ws = P2 P1 R/C P R/C P 1 For P sufciently small, the quantity in square brackets becomes: P2 P1 R/C P 1= 1+ P P1 1 1+ R P C P P1 1 The boxed equation is immediately recovered from this result. 666 7.41 The equation immediately preceding Eq. (7.22) page 276 gives T2 = T1 . With this substitution, Eq. (7.23) becomes: 1 T1 T1 = T1 1 + T2 = T1 + The entropy generation SG is simply S for the compression process, for which Eq. (5.14) may be rewritten: P2 R/C P C P T2 C P P2 C P T2 S ln ln = ln ln = P1 R T1 R P1 T1 R R Combine the two preceding equations: 1 CP S ln 1 + = R R CP ln ln = R 1+ 1 Whence, + 1 CP SG ln = R R 7.43 The relevant fact here is that C P increases with increasing molecular complexity. Isentropic compression work on a mole basis is given by Eq. (7.22), which can be written: Ws = C P T1 ( 1) where P2 P1 R/C P This equation is a proper basis, because compressor efciency and owrate n are xed. With all other variables constant, differentiation yields: d d Ws = T1 ( 1) + C P dC P dC P From the denition of , ln = P2 R ln P1 CP whence P2 R 1 d d ln ln = = 2 P1 dC P dC P CP Then, P2 R d ln = 2 P1 dC P CP and R P2 d Ws ln = T1 1 P1 CP dC P = T1 ( 1 ln ) When = 1, the derivative is zero; for > 1, the derivative is negative (try some values). Thus, the work of compression decreases as C P increases and as the molecular complexity of the gas increases. 7.45 The appropriate energy balance can be written: W = H Q. Since Q is negative (heat transfer is out of the system), the work of non-adiabatic compression is greater than for adiabatic compression. Note that in order to have the same change in state of the air, i.e., the same H , the irreversibilities of operation would have to be quite different for the two cases. 7.46 There is in fact no cause for concern, as adiabatic compression sends the steam further into the superheat region. 667 7.49 (a) This result follows immediately from the last equation on page 267 of the text. (b) This result follows immediately from the middle equation on page 267 of the text. (c) This result follows immediately from Eq. (6.19) on page 267 of the text. T Z Z but by (a), this is zero. = (d) T P V P V P (e) Rearrange the given equation: V ( P/ T )V V = = P ( P/ V )T T T P T = V V T P For the nal equality see footnote on p. 266. This result is the equation of (c). 7.50 From the result of Pb. 7.3: c = V CP 1 M CV where = 1 V V P T With V = RT +B P then V P = T RT P2 Also, let = CP CV Then c = PV BP = 1+ = (RT + B P) RT MRT MRT RT M c= B RT + RT M RT P M A value for B at temperature T may be extracted from a linear t of c vs. P. 7.51 (a) On the basis of Eq. (6.8), write: HS = ig V ig d P = RT dP P (const S) HS = V dP = Z RT dP P (const S) HS HS ig = Z RT d P (const S) P Z RT d P (const S) P By extension, and with equal turbine efciencies, . W H = . ig = Z H ig W 7.52 By Eq. (7.16), H = ( H ) S For C P = constant, T2 T1 = [(T2 ) S T1 ] P2 P1 R/C P For an ideal gas with constant C P , (T2 ) S is related to T1 by (see p. 77): (T2 ) S = T1 Combine the last two equations, and solve for T2 : T2 = T1 1 + P2 P1 R/C P 1 668 From which = T2 1 T1 P2 P1 R/C P Note that < 1 1 Results: For T2 = 318 K, = 1.123; For T2 = 348 K, = 1.004; For T2 = 398 K, = 0.805. Only T2 = 398 K is possible. 7.55 The proposal of Pb. 7.53, i.e., pumping of liquid followed by vaporization. The reason is that pumping a liquid is much less expensive than vapor compression. 7.56 What is required here is the lowest saturated steam temperature that satises the T constraint. Data from Tables F.2 and B.2 lead to the following: Benzene/4.5 bar; n-Decane/17 bar; Ethylene glycol/33 bar; o-Xylene/9 bar 669 Chapter 8 - Section B - Non-Numerical Solutions 8.12 (a) Because Eq. (8.7) for the efciency Diesel includes the expansion ratio, re VB /V A , we relate this quantity to the compression ratio, r VC /VD , and the Diesel cutoff ratio, rc V A /VD . Since VC = VB , re = VC /V A . Whence, VA VC /VD r = rc = = VD VC /V A re or rc 1 = r re Equation (8.7) can therefore be written: Diesel 1 =1 (rc /r ) (1/r ) rc /r 1/r 1 (1/r ) =1 1/r 1 rc 1 rc 1 or Diesel = 1 1 r rc 1 (rc 1) (b) We wish to show that: rc 1 >1 (rc 1) or more simply xa 1 >1 a(x 1) Taylors theorem with remainder, taken to the 1st derivative, is written: g = g(1) + g (1) (x 1) + R where, R g [1 + (x 1)] (x 1)2 2! (0 < < 1) Then, x a = 1 + a (x 1) + 1 a (a 1) [1 + (x 1)]a2 (x 1)2 2 Note that the nal term is R. For a > 1 and x > 1, R > 0. Therefore: x a > 1 + a (x 1) x a 1 > a (x 1) and rc 1 >1 (rc 1) (c) If = 1.4 and r = 8, then by Eq. (8.6): Otto = 1 1 8 0.4 and 0.4 Otto = 0.5647 rc = 2 Desiel = 1 1 8 21.4 1 1.4(2 1) and Diesel = 0.4904 rc = 3 Desiel = 1 1 8 0.4 31.4 1 1.4(3 1) and Diesel = 0.4317 670 8.15 See the gure below. In the regenerative heat exchanger, the air temperature is raised in step B B , while the air temperature decreases in step D D . Heat addition (replacing combustion) is in step B C. By denition, W AB WC D Q BC where, W AB = (H B H A ) = C P (TB TA ) WC D = (H D HC ) = C P (TD TC ) Q B C = C P (TC TB ) = C P (TC TD ) Whence, = TB T A TA TB + TC TD =1 TC TD TC TD By Eq. (3.30b), TB = T A PB PA ( 1)/ and TD = TC PD PC ( 1)/ = TC PA PB ( 1)/ TA Then, =1 PB PA ( 1)/ 1 TC 1 PA PB ( 1)/ Multiplication of numerator and denominator by (PB /PA )( 1)/ gives: =1 TA TC PB PA ( 1)/ 671 8.21 We give rst a general treatment of paths on a P T diagram for an ideal gas with constant heat capacities undergoing reversible polytropic processes. Equation (3.35c), p. 78, may be rewritten as P = K T /(1) ln P = ln K + ln T 1 dT dP = 1 T P P dP = 1T dT (A) Sign of d P/dT is that of 1, i.e., + = 0 d P/dT = 0 = 1 d P/dT = Constant P Constant T Special cases By Eq. (A), d2 P = 2 1 dT P 1 dP 2 T T dT = 1 1T P P T 1T P d2 P = ( 1)2 T 2 dT 2 (B) Sign of d 2 P/dT 2 is that of , i.e., + For a constant-V process, P varies with T in accord with the ideal-gas law: P = RT /V or P = K T With respect to the initial equation, P = K T /(1) , this requires = . Moreover, d P/dT = K and d 2 P/dT 2 = 0. Thus a constant-V process is represented on a P T diagram as part of a straight line passing through the origin. The slope K is determined by the initial P T coordinates. For a reversible adiabatic process (an isentropic process), = . In this case Eqs. (A) and (B) become: P dP = 1T dT P d2 P = ( 1)2 T 2 dT 2 We note here that /( 1) and /( 1)2 are both > 1. Thus in relation to a constant-V process the isentropic process is represented by a line of greater slope and greater curvature for the same T and P. Lines characteristic of the various processes are shown on the following diagram. = =1 P = =0 0 0 T The required sketches appear on the following page. (Courtesy of Prof. Mark T. Swihart, State University of New York at Buffalo.) 672 P P 0 0 T Figure 1: The Carnot cycle 0 0 T Figure 2: The Otto cycle P P 0 0 T Figure 3: The Diesel cycle 0 0 T Figure 4: The Brayton cycle 8.23 This is a challenging and open-ended problem for which we offer no solution. Problem 8.21 may offer some insight. 673 Chapter 9 - Section B - Non-Numerical Solutions 9.1 Since the object of doing work |W | on a heat pump is to transfer heat |Q H | to a heat sink, then: What you get = |Q H | What you pay for = |W | Whence |Q H | |W | For a Carnot heat pump, = TH |Q H | = TH TC |Q H | |Q C | 9.3 Because the temperature of the nite cold reservoir (contents of the refrigerator) is a variable, use differential forms of Carnots equations, Eqs. (5.7) and (5.8): TH d QH = TC d QC and dW = 1 TC TH d QH In these equations Q C and Q H refer to the reservoirs. With d Q H = C t dTC , the rst of Carnots equations becomes: dTC d Q H = C t TH TC Combine this equation with the second of Carnots equations: d W = C t TH dTC + C t dTC TC Integration from TC = TH to TC = TC yields: W = C t TH ln TC + C t (TC TH ) TH or W = C t TH ln TC TH 1 + TH TC 9.5 Differentiation of Eq. (9.3) yields: TC = TH TH TC 1 = + 2 (TH TC )2 (TH TC ) TH TC and TH = TC TC (TH TC )2 Because TH > TC , the more effective procedure is to increase TC . For a real refrigeration system, increasing TC is hardly an option if refrigeration is required at a particular value of TC . Decreasing TH is no more realistic, because for all practical purposes, TH is xed by environmental conditions, and not subject to control. 9.6 For a Carnot refrigerator, is given by Eq. (9.3). Write this equation for the two cases: = TC TH TC and = TC T H TC Because the directions of heat transfer require that TH > T H and TC < TC , a comparison shows that < and therefore that is the more conservative value. 674 9.20 On average, the coefcient of performance will increase, thus providing savings on electric casts. On the other hand, installation casts would be higher. The proposed arrangement would result in cooling of the kitchen, as the refrigerator would act as an air conditioner. This would be detrimental in the winter, but benecial in the summer, at least in temperate climates. 9.21 = 0.6 Carnot = 0.6 TC TH TC If < 1, then TC < TH /1.6. For TH = 300 K, then TC < 187.5 K, which is most unlikely. 675 Chapter 10 - Section B - Non-Numerical Solutions 10.5 For a binary system, the next equation following Eq. (10.2) shows that P is linear in x1 . Thus no maximum or minimum can exist in this relation. Since such an extremum is required for the existence of an azeotrope, no azeotrope is possible. 10.6 (a) Because benzene and toluene are chemically similar and the pressure is only 1(atm), this system can be modeled by Raoults law to a good approximation. (b) Although n-hexane and n-heptane are chemically similar, a pressure of 25 bar is too high for modeling this system by Raoults law. (c) At 200 K, hydrogen is supercritical, and modeling the hydrogen/propane system at this temperature by Raoults law is out of the question, because no value of P sat for hydrogen is known. (d) Because isooctane and n-octane are chemically similar and at a temperature (373.15 K) close to their normal boiling points, this system can be modeled by Raoults law to a good approximation. (e) Water and n-decane are much too dissimilar to be modeled by Raoults law, and are in fact only slightly soluble in one another at 300 K. 10.12 For a total volume V t of an ideal gas, P V t = n RT . Multiply both sides by yi , the mole fraction of species i in the mixture: yi P V t = n i RT or pi V t = mi RT Mi where m i is the mass of species i, Mi is its molar mass, and pi is its partial pressure, dened as pi yi P. Solve for m i : Mi pi V t mi = RT Applied to moist air, considered a binary mixture of air and water vapor, this gives: m H2 O = M H 2 O p H2 O V t RT and m air = Mair pair V t RT (a) By denition, h m H2 O m air or h= MH2 O pH2 O Mair pair Since the partial pressures must sum to the total pressure, pair = P pH2 O ; whence, h= p H2 O MH2 O Mair P pH2 O (b) If air is in equilibrium with liquid water, then the partial pressure of water vapor in the air equals the vapor pressure of the water, and the preceding equation becomes: h sat = PHsat MH2 O 2O Mair P PHsat 2O 676 (c) Percentage humidity and relative humidity are dened as follows: h pc pH2 O P PHsat h 2O (100) sat = sat PH2 O P pH2 O h and h rel p H2 O (100) PHsat 2O Combining these two denitions to eliminate pH2 O gives: h pc = h rel P PHsat (h rel /100) 2O P PHsat 2O 10.14 Because the vapor space above the liquid phase is nearly pure gas, Eq. (10.4) becomes P = xi Hi . For the same mole fraction of gas dissolved in the liquid phase, P is then proportional to Hi . Values given in Table 10.1 indicate that were air used rather than CO2 , P would be about 44 times greater, much too high a pressure to be practical. 10.15 Because Henrys constant for helium is very high, very little of this gas dissolves in the blood streams of divers at approximately atmospheric pressure. 10.21 By Eq. (10.5) and the given equations for ln 1 and ln 2 , 2 y1 P = x1 exp(Ax2 )P1sat and 2 y2 P = x2 exp(Ax1 )P2sat These equations sum to give: 2 2 P = x1 exp(Ax2 )P1sat + x2 exp(Ax1 )P2sat Dividing the equation for y1 P by the preceding equation yields: y1 = 2 x1 exp(Ax2 )P1sat 2 2 x1 exp(Ax2 )P1sat + x2 exp(Ax1 )P2sat For x1 = x2 this equation obviously reduces to: P= P1sat P1sat + P2sat 10.23 A little reection should convince anyone that there is no other way that BOTH the liquid-phase and vapor-phase mole fractions can sum to unity. 10.24 By the denition of a K -value, y1 = K 1 x1 and y2 = K 2 x2 . Moreover, y1 + y2 = 1. These equations combine to yield: K 1 x1 + K 2 x2 = 1 Solve for x1 : or K 1 x1 + K 2 (1 x1 ) = 1 x1 = 1 K2 K1 K2 Substitute for x1 in the equation y1 = K 1 x1 : y1 = K 1 (1 K 2 ) K1 K2 677 Note that when two phases exist both x1 and y1 are independent of z 1 . By a material balance on the basis of 1 mole of feed, x1 L + y1 V = z 1 or x1 (1 V) + y1 V = z 1 Substitute for both x1 and y1 by the equations derived above: K 1 (1 K 2 ) 1 K2 V = z1 (1 V) + K1 K2 K1 K2 Solve this equation for V: V= z 1 (K 1 K 2 ) (1 K 2 ) (K 1 1)(1 K 2 ) Note that the relative amounts of liquid and vapor phases do depend on z 1 . 10.35 Molality Mi = xi ni = x s Ms ms where subscript s denotes the solvent and Ms is the molar mass of the solvent. The given equation may therefore be written: xi = ki yi P x s Ms or xi 1 x s Ms k i = yi P Comparison with Eq. (10.4) shows that Hi = 1 x s Ms k i or for xi 0 Hi = 1 Ms k i For water, Ms = 18.015 g mol1 Thus, or 0.018015 kg mol1 . Hi = 1 = 1633 bar (0.018015)(0.034) This is in comparison with the value of 1670 bar in Table 10.1. 678 Chapter 11 - Section B - Non-Numerical Solutions 11.6 Apply Eq. (11.7): Ti (nT ) n i =T P,T,n j n n i =T T,P,n j Pi (n P) n i =P P,T,n j n n i =P T,P,n j 11.7 (a) Let m be the mass of the solution, and dene the partial molar mass by: m i m n i T,P,n j Let Mk be the molar mass of species k. Then m= m n i nk Mk = ni Mi + n j M j k j ( j = i) and = T,P,n j (n i Mi ) n i = Mi T,P,n j Whence, m i = Mi (b) Dene a partial specic property as: Mi M t m i = T,P,m j M t n i T,P,m j n i m i T,P,m j If Mi is the molar mass of species i, ni = mi Mi and n i m i = T,P,m j 1 Mi Because constant m j implies constant n j , the initial equation may be written: Mi Mi = Mi 11.8 By Eqs. (10.15) and (10.16), dV V1 = V + x2 d x1 and dV V2 = V x1 d x1 Because V = 1 then 1 d dV = 2 d x1 d x1 whence 1 x2 d 1 = V1 = 2 d x1 1 x2 d d x1 = d 1 x2 2 d x1 1 x1 d 1 = V2 = + 2 d x1 1+ x1 d d x1 = d 1 + x1 2 d x1 With 2 = a0 + a1 x 1 + a2 x 1 and d = a1 + 2a2 x1 d x1 these become: 1 2 V1 = 2 [a0 a1 + 2(a1 a2 )x1 + 3a2 x1 ] and 1 2 V2 = 2 (a0 + 2a1 x1 + 3a2 x1 ) 679 11.9 For application of Eq. (11.7) all mole fractions must be eliminated from the given equation by the relation xi = n i /n: n1n2n3 C n M = n 1 M1 + n 2 M2 + n 3 M3 + n2 For M1 , (n M) n 1 T,P,n 2 ,n 3 = M1 + n 2 n 3 C 2n 1 1 3 2 n n n n 1 T,P,n 2 ,n 3 Because n = n 1 + n 2 + n 3 , n n 1 =1 T,P,n 2 ,n 3 Whence, n1 n2n3 C M1 = M1 + 2 1 2 n n and M1 = M1 + x2 x3 [1 2x1 ]C Similarly, M2 = M2 + x1 x3 [1 2x2 ]C and M3 = M3 + x1 x2 [1 2x3 ]C One can readily show that application of Eq. (11.11) regenerates the original equation for M. The innite dilution values are given by: Mi = Mi + x j xk C ( j, k = i) Here x j and xk are mole fractions on an i-free basis. 11.10 With the given equation and the Daltons-law requirement that P = P= RT V i pi , then: yi Z i i i For the mixture, P = Z RT /V . These two equations combine to give Z = 11.11 The general principle is simple enough: yi Z i . Given equations that represent partial properties Mi , MiR , or MiE as functions of composition, one may combine them by the summability relation to yield a mixture property. Application of the dening (or equivalent) equations for partial properties then regenerates the given equations if and only if the given equations obey the Gibbs/Duhen equation. 11.12 (a) Multiply Eq. (A) of Ex. 11.4 by n (= n 1 + n 2 ) and eliminate x1 by x1 = n 1 /(n 1 + n 2 ): n H = 600(n 1 + n 2 ) 180 n 1 20 n3 1 (n 1 + n 2 )2 Form the partial derivative of n H with respect to n 1 at constant n 2 : H1 = 600 180 20 n3 n2 2n 3 3n 2 1 1 1 1 + 40 = 420 60 (n 1 + n 2 )3 (n 1 + n 2 )2 (n 1 + n 2 )2 (n 1 + n 2 )3 3 2 H1 = 420 60 x1 + 40 x1 Whence, Form the partial derivative of n H with respect to n 2 at constant n 1 : H2 = 600 + 20 2 n3 1 (n 1 + n 2 )3 or 3 H2 = 600 + 40 x1 680 (b) In accord with Eq. (11.11), 2 3 3 H = x1 (420 60 x1 + 40 x1 ) + (1 x2 )(600 + 40 x1 ) Whence, 3 H = 600 180 x1 20 x1 (c) Write Eq. (11.14) for a binary system and divide by d x1 : x1 d H2 d H1 =0 + x2 d x1 d x1 Differentiate the the boxed equations of part (a): d H1 2 = 120 x1 + 120 x1 = 120 x1 x2 d x1 and d H2 2 = 120 x1 d x1 Multiply each derivative by the appropriate mole fraction and add: 2 2 120 x1 x2 + 120x1 x2 = 0 (d) Substitute x1 = 1 and x2 = 0 in the rst derivative expression of part (c) and substitute x1 = 0 in the second derivative expression of part (c). The results are: d H1 d x1 = x1 =1 d H2 d x1 =0 x1 =0 (e) 11.13 (a) Substitute x2 = 1 x1 in the given equation for V and reduce: 3 2 V = 70 + 58 x1 x1 7 x1 Apply Eqs. (11.15) and (11.16) to nd expressions for V1 and V2 . First, dV 2 = 58 2 x1 21 x1 d x1 Then, 3 2 V1 = 128 2 x1 20 x1 + 14 x1 and 3 2 V2 = 70 + x1 + 14 x1 681 (b) In accord with Eq. (11.11), 2 3 2 3 V = x1 (128 2 x1 20 x1 + 14 x1 ) + (1 x1 )(70 + x1 + 14 x1 ) Whence, 3 2 V = 70 + 58 x1 x1 7 x1 which is the rst equation developed in part (a). (c) Write Eq. (11.14) for a binary system and divide by d x1 : x1 d V2 d V1 =0 + x2 d x1 d x1 Differentiate the the boxed equations of part (a): d V1 2 = 2 40 x1 + 42 x1 d x1 and d V2 2 = 2 x1 + 42 x1 d x1 Multiply each derivative by the appropriate mole fraction and add: 2 2 x1 (2 40 x1 + 42 x1 ) + (1 x1 )(2 x1 + 42 x1 ) = 0 The validity of this equation is readily conrmed. (d) Substitute x1 = 1 in the rst derivative expression of part (c) and substitute x1 = 0 in the second derivative expression of part (c). The results are: d V1 d x1 = x1 =1 d V2 d x1 =0 x1 =0 (e) 11.14 By Eqs. (11.15) and (11.16): dH H1 = H + x2 d x1 and dH H2 = H x1 d x1 682 Given that: Then, after simplication, H = x1 (a1 + b1 x1 ) + x2 (a2 + b2 x2 ) dH = a1 + 2b1 x1 (a2 + 2b2 x2 ) d x1 Combining these equations gives after reduction: H1 = a1 + b1 x1 + x2 (x1 b1 x2 b2 ) and H2 = a2 + b2 x2 x1 (x1 b1 x2 b2 ) These clearly are not the same as the suggested expressions, which are therefore not correct. Note that application of the summability equation to the derived partial-property expressions reproduces the original equation for H . Note further that differentiation of these same expressions yields results that satisfy the Gibbs/Duhem equation, Eq. (11.14), written: x1 d H2 d H1 =0 + x2 d x1 d x1 The suggested expresions do not obey this equation, further evidence that they cannot be valid. 11.15 Apply the following general equation of differential calculus: x y = z x y + w x w y w y z (n M) n i = T,P,n j (n M) n i + T,V,n j (n M) V T,n V n i T,P,n j Whence, M Mi = Mi + n V T,n V n i or T,P,n j M Mi = Mi n V T,n V n i T,P,n j By denition, (nV ) Vi n i =n T,P,n j V n i +V T,P,n j or n V n i = Vi V T,P,n j Therefore, M Mi = Mi + (V Vi ) V T,x 11.20 Equation (11.59) demonstrates that ln i is a partial property with respect to G R /RT . Thus ln i = G i /RT . The partial-property analogs of Eqs. (11.57) and (11.58) are: ln i P = T,x Vi R RT and ln i T = P,x HiR RT 2 The summability and Gibbs/Duhem equations take on the following forms: GR = RT xi ln i i and i xi d ln i = 0 (const T, P) 683 11.26 For a pressure low enough that Z and ln are given approximately by Eqs. (3.38) and (11.36): Z =1+ BP RT and ln = BP RT then: ln Z 1 11.28 (a) Because Eq. (11.96) shows that ln i is a partial property with respect to G E/RT , Eqs. (11.15) and (11.16) may be written for M G E/RT : ln 1 = d(G E/RT ) GE + x2 d x1 RT ln 2 = d(G E/RT ) GE x1 d x1 RT Substitute x2 = 1 x1 in the given equaiton for G E/RT and reduce: GE 2 3 = 1.8 x1 + x1 + 0.8 x1 RT whence d(G E/RT ) 2 = 1.8 + 2 x1 + 2.4 x1 d x1 Then, 3 2 ln 1 = 1.8 + 2 x1 + 1.4 x1 1.6 x1 and 3 2 ln 2 = x1 1.6 x1 (b) In accord with Eq. (11.11), GE 2 3 2 3 = x1 ln 1 + x2 ln 2 = x1 (1.8 + 2 x1 + 1.4 x1 1.6 x1 ) + (1 x1 )(x1 1.6 x1 ) RT Whence, GE 3 2 = 1.8 x1 + x1 + 0.8 x1 RT which is the rst equation developed in part (a). (c) Write Eq. (11.14) for a binary system with Mi = ln i and divide by d x1 : x1 d ln 2 d ln 1 =0 + x2 d x1 d x1 Differentiate the the boxed equations of part (a): d ln 1 2 = 2 + 2.8 x1 4.8 x1 d x1 and d ln 2 2 = 2 x1 4.8 x1 d x1 Multiply each derivative by the appropriate mole fraction and add: 2 2 x1 (2 + 2.8 x1 4.8 x1 ) + (1 x1 )(2 x1 4.8 x1 ) = 0 The validity of this equation is readily conrmed. (d) Substitute x1 = 1 in the rst derivative expression of part (c) and substitute x1 = 0 in the second derivative expression of part (c). The results are: d ln 1 d x1 = x1 =1 d ln 2 d x1 =0 x1 =0 684 (e) 11.29 Combine denitions of the activity coefcient and the fugacity coefcients: i fi /xi P f i /P or i = i i Note: See Eq. (14.54). E 11.30 For C P = const., the following equations are readily developed from those given in the last column of Table 11.1 (page 415): E H E = CP T and SE = G E T P,x E = CP T T Working equations are then: E S1 = E H1E G 1 T1 and E E E S2 = S1 + C P T T E H2E = H1E + C P T and E E G 2 = H2E T2 S2 For T1 = 298.15, T2 = 328.15, T = 313.15 and given in the following table: T = 30, results for all parts of the problem are I. E II. For C P = 0 E G1 H1E E S1 E CP E S2 H2E E G2 E S2 H2E E G2 (a) (b) (c) (d) (e) (f) (g) 622 1095 407 632 1445 734 759 1920 1595 984 208 605 416 1465 4.354 1.677 1.935 2.817 2.817 3.857 2.368 4.2 3.3 2.7 23.0 11.0 11.0 8.0 3.951 1.993 1.677 0.614 1.764 2.803 1.602 1794 1694 903 482 935 86 1225 497.4 1039.9 352.8 683.5 1513.7 833.9 699.5 4.354 1.677 1.935 2.817 2.817 3.857 2.368 1920 1595 984 208 605 416 1465 491.4 1044.7 348.9 716.5 1529.5 849.7 688.0 685 11.31 (a) Multiply the given equation by n (= n 1 + n 2 ), and convert remaining mole fractions to ratios of mole numbers: n2n3 n1n3 n1n2 nG E + A23 + A13 = A12 n n n RT Differentiation with respect to n 1 in accord with Eq. (11.96) yields [(n/n 1 )n 2 ,n 3 = 1]: ln 1 = A12 n 2 1 n1 n n2 + A13 n 3 1 n1 n n2 A23 n2n3 n2 = A12 x2 (1 x1 ) + A13 x3 (1 x1 ) A23 x2 x3 Similarly, ln 2 = A12 x1 (1 x2 ) A13 x1 x3 + A23 x3 (1 x2 ) ln 3 = A12 x1 x2 + A13 x1 (1 x3 ) + A23 x2 (1 x3 ) (b) Each ln i is multiplied by xi , and the terms are summed. Consider the rst terms on the right of each expression for ln i . Multiplying each of these terms by the appropriate xi and adding gives: 2 2 A12 (x1 x2 x1 x2 + x2 x1 x2 x1 x1 x2 x3 ) = A12 x1 x2 (1 x1 + 1 x2 x3 ) = A12 x1 x2 [2 (x1 + x2 + x3 )] = A12 x1 x2 An analogous result is obtained for the second and third terms on the right, and adding them yields the given equation for G E/RT . (c) For innite dilution of species 1, For pure species 1, For innite dilution of species 2, For innite dilution of species 3, x1 = 0: x1 = 1: x2 = 0: x3 = 0: ln 1 (x1 = 0) = A12 x2 + A13 x3 A23 x2 x3 ln 1 (x1 = 1) = 0 2 ln 1 (x2 = 0) = A13 x3 2 ln 1 (x3 = 0) = A12 x2 11.35 By Eq. (11.87), written with M G and with x replaced by y: G E = B P y1 B11 P y2 B22 P or GE = GR Equations (11.33) and (11.36) together give G iR = Bii P. Then for a binary mixture: G E = P(B y1 B11 y2 B22 ) i yi G iR Combine this equation with the last equation on Pg. 402: G E = 12 P y1 y2 From the last column of Table 11.1 (page 415): S E = G E T P,x Because 12 is a function of T only: SE = d12 P y1 y2 dT By the denition of G E , H E = G E + T S E ; whence, H E = 12 T d12 dT P y1 y2 E Again from the last column of Table 11.1: C P = HE T P,x This equation and the preceding one lead directly to: E C P = T d 2 12 P y1 y2 dT 2 686 11.41 From Eq. (11.95): (G E /RT ) T = P H E RT 2 or (G E /T ) T = P H E T2 To an excellent approximation, write: (G E /T ) T P H E (G E /T ) 2 Tmean T From the given data: 0.271 785/323 805/298 (G E /T ) = 0.01084 = = 25 323 298 T and 1060 H E = 0.01082 = 2 3132 Tmean The data are evidently thermodynamically consistent. 11.42 By Eq. (11.14), the Gibbs/Duhem equation, x1 d M2 d M1 =0 + x2 d x1 d x1 Given that M1 = M1 + Ax2 and M2 = M2 + Ax1 then d M1 = A d x1 and d M2 =A d x1 Then d M2 d M1 = x1 A + x2 A = A(x2 x1 ) = 0 + x2 d x1 d x1 The given expressions cannot be correct. x1 2 2 M E = Ax1 x2 11.45 (a) For nd 2 E M1 = Ax1 x2 (2 3x1 ) and 2 E M2 = Ax1 x2 (2 3x2 ) Note that at both x1 = 0 (x2 = 1) and x1 = 1 (x2 = 0), In particular, E E ( M1 ) = ( M2 ) = 0 E E M1 = M2 = 0 E E Although M E has the same sign over the whole composition range, both M1 and M2 change sign, which is unusual behavior. Find also that E E d M2 d M1 = 2Ax1 (1 6x1 x2 ) = 2Ax2 (1 6x1 x2 ) and d x1 d x1 The two slopes are thus of opposite sign, as required; they also change sign, which is unusual. E E d M2 d M1 =0 = 2A and For x1 = 0 d x1 d x1 For x1 = 1 E d M1 =0 d x1 and E d M2 = 2A d x1 (b) For M E = A sin( x1 ) nd: E M1 = A sin( x1 ) + A x2 cos( x1 ) and E M2 = A sin( x1 ) A x1 cos( x1 ) E E d M2 d M1 = A 2 x1 sin( x1 ) = A 2 x2 sin( x1 ) and d x1 d x1 The two slopes are thus of opposite sign, as required. But note the following, which is unusual: For x1 = 0 and x1 = 1 E d M1 =0 d x1 and E d M2 =0 d x1 PLOTS OF THE FUNCTIONS ARE SHOWN ON THE FOLLOWING PAGE. 687 Pb. 11.45 (a) A MEi 10 A . xi i 2. 0 .. 100 1 xi MEbar1i 3. 1 .00001 .01 . i xi 2 xi 2 xi . 2 A . xi . 1 xi 2 3 . xi MEbar2i 2 1.5 MEi 1 A . xi . xi . 1 MEbar1 i 0.5 0 0.5 MEbar2i 0 0.2 0.4 xi 0.6 0.8 1 Pb. 11.45 (b) MEi A . sin p .x i p (pi prints as bf p) .x .x i i MEbar1i MEbar2i 40 30 MEi A . sin A . sin A.p . 1 xi . cos p p .x i p A . p . xi . cos .x i 20 MEbar1 i 10 0 10 MEbar2i 0 0.2 0.4 xi 0.6 0.8 1 687A 11.46 By Eq. (11.7), (n M) Mi = n i = M +n T,P,n j M n i T,P,n j At constant T and P, dM = k M xk d xk T,P,x j Divide by dn i with restriction to constant n j ( j = i): M n i = T,P,n j k M xk T,P,x j xk n i nj With nk xk = n xk n i = nj n k n2 (k = i) M n i = T,P,n j 1 n xk k=i M xk 1 ni n n2 (k = i) T,P,x j M 1 + (1 xi ) xi n T,P,x j = 1 n M xi T,P,x j 1 n xk k M xk T,P,x j Mi = M + M xi T,P,x j k xk M xk T,P,x j For species 1 of a binary mixture (all derivatives at constant T and P): M1 = M + M x1 x2 x1 M x1 x2 x2 M x2 x1 = M + x2 M x1 x2 M x2 x1 Because x1 + x2 = 1, the partial derivatives in this equation are physically unrealistic; however, they do have mathematical signicance. Because M = M(x1 , x2 ), we can quite properly write: dM = M x1 x2 d x1 + M x2 x1 d x2 Division by d x1 yields: dM = d x1 M x1 + x2 M x2 x1 d x2 = d x1 M x1 x2 M x2 x1 wherein the physical constraint on the mole fractions is recognized. Therefore dM M1 = M + x 2 d x1 The expression for M2 is found similarly. 688 11.47 (a) Apply Eq. (11.7) to species 1: (n M E ) E M1 = n 1 n2 Multiply the given equation by n and eliminate the mole fractions in favor of mole numbers: n M E = An 1 n 2 1 1 + n 1 + Bn 2 n 2 + Bn 1 E M1 = An 2 1 1 + n 1 + Bn 2 n 2 + Bn 1 + n1 B 1 (n 1 + Bn 2 )2 (n 2 + Bn 1 )2 Conversion back to mole fractions yields: E M1 = Ax2 1 1 + x2 + Bx1 x1 + Bx2 x1 B 1 + 2 (x2 + Bx1 )2 (x1 + Bx2 ) The rst term in the rst parentheses is combined with the rst term in the second parentheses and the second terms are similarly combined: E M1 = Ax2 x1 1 1 x1 + Bx2 x1 + Bx2 + Bx1 1 1 x2 + Bx1 x2 + Bx1 Reduction yields: 2 E M1 = Ax2 1 B + (x1 + Bx2 )2 (x2 + Bx1 )2 Similarly, 2 E M2 = Ax1 B 1 + 2 (x2 + Bx1 )2 (x1 + Bx2 ) (b) The excess partial properties should obey the Gibbs/Duhem equation, Eq. (11.14), when written for excess properties in a binary system at constant T and P: x1 E E d M2 d M1 =0 + x2 d x1 d x1 If the answers to part (a) are mathematically correct, this is inevitable, because they were derived from a proper expression for M E . Furthermore, for each partial property MiE , its value and derivative with respect to xi become zero at xi = 1. 1 1 E E +1 ( M 2 ) = A 1 + (c) ( M 1 ) = A B B 11.48 By Eqs. (11.15) and (11.16), written for excess properties, nd: E d2 M E d M1 = x2 2 d x1 d x1 E d2 M E d M2 = x1 2 d x1 d x1 E At x1 = 1, d M1 /d x1 = 0, and by continuity can only increase or decrease for x1 < 1. Therefore the 2 E 1 /d x1 is the same as the sign of d 2 M E /d x1 . Similarly, at x1 = 0, d M2 /d x1 = 0, and by E sign of d M 2 E 2 E the same argument the sign of d M2 /d x1 is of opposite sign as the sign of d M /d x1 . 689 11.49 The claim is not in general valid. 1 V V T V id = P i xi Vi id = i 1 xi Vi xi i Vi T = P 1 i xi Vi xi Vi i i The claim is valid only if all the Vi are equal. 690 Chapter 12 - Section B - Non-Numerical Solutions 12.2 Equation (12.1) may be written: yi P = xi i Pi sat . Summing for i = 1, 2 gives: P = x1 1 P1sat + x2 2 P2sat . Differentiate at constant T : d2 d1 dP 2 + 1 + P2sat x2 = P1sat x1 d x1 d x1 d x1 Apply this equation to the limiting conditions: For x1 = 0 : x2 = 1 1 = 1 2 = 1 d2 =0 d x1 d1 =0 d x1 For x1 = 1 : x2 = 0 1 = 1 2 = 2 Then, dP d x1 x1 =0 = P1sat 1 P2sat or dP d x1 x1 =0 + P2sat = P1sat 1 dP d x1 x1 =1 = P1sat P2sat 2 or dP d x1 x1 =1 P1sat = P2sat 2 Since both Pi sat and i are always positive denite, it follows that: dP d x1 x1 =0 P2sat and dP d x1 x1 =1 P1sat 12.4 By Eqs. (12.15), Therefore, ln 2 ln 1 = Ax2 and 2 ln 2 = Ax1 1 2 2 = A(x2 x1 ) = A(x2 x1 ) = A(1 2x1 ) 2 By Eq. (12.1), y1 x2 P2sat 1 = = y2 x1 P1sat 2 y1 /x1 y2 /x2 P2sat P1sat = 12 r Whence, ln(12 r ) = A(1 2x1 ) az x1 If an azeotrope exists, 12 = 1 at 0 1. At this value of x1 , az ln r = A(1 2x1 ) The quantity A(1 2x1 ) is linear in x1 , and there are two possible relationships, depending on the sign of A. An azeotrope exhists whenever |A| | ln r |. NO azeotrope can exist when |A| < | ln r |. 12.5 Perhaps the easiest way to proceed here is to note that an extremum in ln 1 is accompanied by the opposite extremum in ln 2 . Thus the difference ln 1 ln 2 is also an extremum, and Eq. (12.8) becomes useful: d(G E/RT 1 = ln 1 ln 2 = ln d x1 2 Thus, given an expression for G E/RT = g(x1 ), we locate an extremum through: d ln(1 /2 ) d 2 (G E/RT ) =0 = 2 d x1 d x1 691 For the van Laar equation, write Eq. (12.16), omitting the primes ( ): x1 x2 GE = A12 A21 A RT where A A12 x1 + A21 x2 Moreover, dA = A12 A21 d x1 and d 2A =0 2 d x1 Then, d(G E/RT ) = A12 A21 d x1 x1 x2 d A x2 x1 2 A d x1 A x2 x1 2x1 x2 d A dA x1 x2 d 2A x2 x1 d A 2 d 2 (G E/RT ) + 2 = A12 A21 2 2 3 dx 2 A2 A d x1 A d x1 d x1 A A d x1 1 = A12 A21 2x1 x2 2(x2 x1 ) d A 2 + 2 A3 d x1 A A dA d x1 2 = dA dA 2A12 A21 + x1 x2 A2 (x2 x1 )A 3 d x1 d x1 A 2 = 2A12 A21 A3 A + x2 dA d x1 x1 dA A d x1 This equation has a zero value if either A12 or A21 is zero. However, this makes G E/RT everywhere zero, and no extremum is possible. If either quantity in parentheses is zero, substitution for A and d A/d x1 reduces the expression to A12 = 0 or A21 = 0, again making G E/RT everywhere zero. We conclude that no values of the parameters exist that provide for an extremum in ln(1 /2 ). The Margules equation is given by Eq. (12.9b), here written: GE = Ax1 x2 RT where A = A21 x1 + A12 x2 dA = A21 A12 d x1 d 2A =0 2 d x1 Then, dA d(G E/RT ) = A(x2 x1 ) + x1 x2 d x1 d x1 d 2A dA dA d 2 (G E/RT ) + x1 x2 2 + (x2 x1 ) = 2A + (x2 x1 ) 2 d x1 d x1 d x1 d x1 = 2A + 2(x2 x1 ) dA dA A = 2 (x1 x2 ) d x1 d x1 This equation has a zero value when the quantity in square brackets is zero. Then: (x2 x1 ) dA A = (x2 x1 )(A21 A12 ) A21 x1 A12 x2 = A21 x2 + A12 x1 2(A21 x1 + A12 x2 ) = 0 d x1 Substituting x2 = 1 x1 and solving for x1 yields: x1 = A21 2A12 3(A21 A12 ) or x1 = (r 2) 3(r 1) r A21 A12 692 When r = 2, x1 = 0, and the extrema in ln 1 and ln 2 occur at the left edge of a diagram such as those of Fig. 12.9. For values of r > 2, the extrema shift to the right, reaching a limiting value for r = at x1 = 1/3. For positive values of the parameters, in all of these cases A21 > A12 , and the intercepts of the ln 2 curves at x1 = 1 are larger than the intercepts of the ln 1 curves at x1 = 0. When r = 1/2, x1 = 1, and the extrema in ln 1 and ln 2 occur at the right edge of a diagram such as those of Fig. 12.9. For values of r < 1/2, the extrema shift to the left, reaching a limiting value for r = 0 at x1 = 2/3. For positive values of the parameters, in all of these cases A21 < A12 , and the intercepts of the ln 1 curves at x1 = 0 are larger than the intercepts of the ln 2 curves at x1 = 1. No extrema exist for values of r between 1/2 and 2. 12.7 Equations (11.15) and (11.16) here become: ln 1 = d(G E/RT ) GE + x2 d x1 RT and ln 2 = d(G E/RT ) GE x1 d x1 RT (a) For simplicity of notation, omit the primes that appear on the parameters in Eqs. (12.16) and (12.17), and write Eq. (12.16) as: x1 x2 GE = A12 A21 D RT where D A12 x1 + A21 x2 Then, x1 x2 x2 x1 d(G E/RT ) 2 (A12 A21 ) = A12 A21 D D d x1 and ln 1 = A12 A21 x1 x2 + x2 D x1 x2 x2 x1 2 (A12 A21 ) D D = 2 x1 x2 A12 A21 2 (A12 A21 ) x1 x2 + x2 x1 x2 D D = 2 2 A12 A21 x2 A12 A21 x2 (A21 x2 + A21 x1 ) (D A12 x1 + A21 x1 ) = D2 D2 = 2 A12 A2 x2 21 = A12 D2 A21 x2 D 2 = A12 D A21 x2 2 = A12 A12 x1 + A21 x2 A21 x2 2 ln 1 = A12 1 + A12 x1 A21 x2 2 The equation for ln 2 is derived in analogous fashion. (b) With the understanding that T and P are constant, ln 1 = (nG E/RT ) n 1 n2 and Eq. (12.16) may be written: A12 A21 n 1 n 2 nG E = nD RT where 693 n D = A12 n 1 + A21 n 2 Differentiation in accord with the rst equation gives: ln 1 = A12 A21 n 2 n1 1 n D (n D)2 (n D) n 1 n2 ln 1 = A12 x1 A12 A21 x2 n1 A12 A21 n 2 1 A12 = 1 D D nD nD 2 A12 A2 x2 A12 A21 x2 A12 A21 x2 21 A21 x2 = (D A12 x1 ) = D2 D2 D2 = The remainder of the derivation is the same as in Part (a). 12.10 This behavior requires positive deviations from Raoults law over part of the composition range and negative deviations over the remainder. Thus a plot of G E vs. x1 starts and ends with G E = 0 at x1 = 0 and x1 = 1 and shows positive values over part of the composition range and negative values over the remainder, with an intermediate crossing of the x1 axis. Because these deviations are usually quite small, the vapor pressures P1sat and P2sat must not be too different, otherwise the dewpoint and bubblepoint curves cannot exhibit extrema. 12.11 Assume the Margules equation, Eq. (12.9b), applies: GE = x1 x2 (A21 x1 + A12 x2 ) RT and 1 GE (equimolar) = (A12 + A21 ) 8 RT But [see page 438, just below Eq. (12.10b)]: A12 = ln 1 A21 = ln 2 1 GE (equimolar) = (ln 1 + ln 2 ) 8 RT or 1 GE (equimolar) = ln(1 2 ) 8 RT 12.24 (a) By Eq. (12.6): GE = x1 ln 1 + x2 ln 2 RT 2 2 = x1 x2 (0.273 + 0.096 x1 ) + x2 x1 (0.273 0.096 x2 ) = x1 x2 (0.273 x2 + 0.096 x1 x2 + 0.273 x1 0.096 x1 x2 ) = x1 x2 (0.273)(x1 + x2 ) GE = 0.273 x1 x2 RT (b) The preceding equation is of the form from which Eqs. (12.15) are derived. From these, 2 ln 1 = 0.273 x2 and 2 ln 2 = 0.273 x1 (c) The equations of part (b) are not the reported expressions, which therefore cannot be correct. See Problem 11.11. 12.25 Write Eq. (11.100) for a binary system, and divide through by d x1 : x1 d ln 2 d ln 1 =0 + x2 d x1 d x1 whence x1 d ln 1 x1 d ln 1 d ln 2 = = x2 d x2 x2 d x1 d x1 694 Integrate, recalling that ln 2 = 1 for x1 = 0: ln 2 = ln(1) + 2 (a) For ln 1 = Ax2 , x1 0 x1 d ln 1 d x1 = x2 d x2 x1 0 x1 d ln 1 d x1 x2 d x2 d ln 1 = 2Ax2 d x2 Whence ln 2 = 2A x1 0 x1 d x1 or 2 ln 2 = Ax1 By Eq. (12.6), 2 (b) For ln 1 = x2 (A + Bx2 ), GE = Ax1 x2 RT d ln 1 2 2 = 2x2 (A + Bx2 ) + x2 B = 2Ax2 + 3Bx2 = 2Ax2 + 3Bx2 (1 x1 ) d x2 Whence 2 ln 2 = Ax1 + ln 2 = 2A 3B 2 3 x Bx1 2 1 x1 0 x1 d x1 + 3B x1 0 x1 d x1 3B x1 0 2 x1 d x1 or 2 ln 2 = x1 A + B 3B 2 Bx1 = x1 A + (1 + 2x2 ) 2 2 Apply Eq. (12.6): 3B GE 2 2 Bx1 ) = x1 x2 (A + Bx2 ) + x2 x1 (A + 2 RT Algebraic reduction can lead to various forms of this equation; e.g., B GE = x1 x2 A + (1 + x2 ) 2 RT 2 2 (c) For ln 1 = x2 (A + Bx2 + C x2 ), d ln 1 2 2 2 3 = 2x2 (A + Bx2 + C x2 ) + x2 (B + 2C x2 ) = 2Ax2 + 3Bx2 + 4C x2 d x2 = 2Ax2 + 3Bx2 (1 x1 ) + 4C x2 (1 x1 )2 Whence or ln 2 = 2A x1 0 x1 d x1 + 3B x1 0 x1 0 x1 (1 x1 )d x1 + 4C x1 0 x1 0 x1 (1 x1 )2 d x1 x1 0 3 x1 d x1 ln 2 = (2A + 3B + 4C) ln 2 = x1 d x1 (3B + 8C) 2 x1 2 x1 d x1 + 4C 2A + 3B + 4C 2 3B + 8C 3 3 4 x1 + C x1 2 ln 2 = x1 A + 8C 3B + 2C B + 3 2 2 x1 + C x1 695 or 2 ln 2 = x1 A + C B 2 (1 + 2x2 ) + (1 + 2x2 + 3x2 ) 3 2 The result of application of Eq. (12.6) reduces to equations of various forms; e.g.: C B GE 2 = x1 x2 A + (1 + x2 ) + (1 + x2 + x2 ) 3 2 RT 12.40 (a) As shown on page 458, x1 = 1 1+n and H= H x1 H= H (1 + n) Eliminating 1 + n gives: (A) Differentiation yields: H d x1 1 d H d H = 2 = x1 d n dn x1 d n H 1 d H 2 x1 d x1 x1 d x1 dn where 1 d x1 2 = x1 = (1 + n)2 dn Whence, d H = dn H x1 dHE d H = H E x1 d x1 d x1 Comparison with Eq. (11.16) written with M H E , dH H2E = H E x1 d x1 E shows that d H = H2E dn (b) By geometry, with reference to the following gure, d H = dn HI n Combining this with the result of Part (a) gives: H2E = HI n From which, Substitute: H= I = H n H2E HE H = x1 x1 and n= x2 x1 696 Whence, I = H E x2 H2E x2 HE H2E = x1 x1 x1 However, by the summability equation, H E x2 H2E = x1 H1E Then, I = H1E 12.41 Combine the given equation with Eq. (A) of the preceding problem: H = x2 (A21 x1 + A12 x2 ) With x2 = 1 x1 and x1 = 1/(1 + n) (page 458): x2 = n 1+n The preceding equations combine to give: H= n 1+n A12 n A21 + 1+n 1+n (a) It follows immediately from the preceding equation that: n0 lim H =0 (b) Because n/(1 + n) 1 for n , it follows that: n lim H = A12 (c) Analogous to Eq. (12.10b), page 438, we write: Eliminate the mole fractions in favor of n: H2E = 1 1+n 2 2 H2E = x1 [A21 + 2(A12 A21 )x2 ] A21 + 2(A12 A21 ) n 1+n In the limit as n 0, this reduces to A21 . From the result of Part (a) of the preceding problem, it follows that n0 lim d H = A21 dn 12.42 By Eq. (12.29) with M H , H=H i xi Hi . Differentiate: H t = P,x H t P,x i xi Hi t P,x With H t CP, P,x this becomes t t0 H t = CP P,x i xi C Pi = t t0 CP Therefore, H H0 d( H ) = C P dt H= H0 + C P dt 697 12.61 (a) From the denition of M: Differentiate: M E = x1 x2 M (A) dM dME = M(x2 x1 ) + x1 x2 d x1 d x1 (B) Substitution of Eqs. ( A) & (B) into Eqs. (11.15) & (11.16), written for excess properties, yields the required result. (b) The requested plots are found in Section A. 12.63 In this application the microscopic state of a particle is its species identity, i.e., 1, 2, 3, . . . . By assumption, this label is the only thing distinguishing one particle from another. For mixing, t t t S t = Smixed Sunmixed = Smixed i Sit where the total emtropies are given by Eq. (5.42). Thus, for an unmixed species i, and for the mixed system of particles, Sit = k ln i = k ln Ni ! =0 Ni ! t Smixed = k ln N! N1 ! N2 ! N3 ! Combining the last three equations gives: S t = k ln N! N1 ! N2 ! N3 ! From which: 1 N! 1 St St S = (ln N ! ln = = = N N1 ! N2 ! N3 ! N kN R(N /N A ) R ln Ni !) i ln N ! N ln N N and ln Ni ! Ni ln Ni Ni 1 S (N ln N N N R Ni ln Ni + i i Ni ) = 1 (N ln N N xi N ln xi N ) i = 1 (N ln N N xi N ln xi i i xi N ln N ) = i xi ln x1 12.66 Isobaric data reduction is complicated by the fact that both composition and temperature vary from point to point, whereas for isothermal data composition is the only signicant variable. (The effect of pressure on liquid-phase properties is assumed negligible.) Because the activity coefcients are strong functions of both liquid composition and T , which are correlated, it is quite impossible without additional information to separate the effect of composition from that of T . Moreover, the Pi sat values depend strongly on T , and one must have accurate vapor-pressure data over a temperature range. 12.67 (a) Written for G E , Eqs. (11.15) and (11.16) become: dG E G 1 = G E + x2 d x1 E and dG E G 2 = G E x1 d x1 E Divide through by RT ; dene G GE ; RT note by Eq. (11.91) that G iE = ln i RT Then ln 1 = G + x2 dG d x1 and ln 2 = G x1 dG d x1 Given: GE = A1/k x! x2 RT with 698 A x 1 Ak + x 2 Ak 21 12 Whence: G = x1 x2 A1/k and d A1/k dG + A1/k (x2 x1 ) = x1 x2 d x1 d x1 1 A1/k k dA 1 d A1/k (A21 Ak ) = = A(1/k)1 12 k A d x1 k d x1 and A1/k k dG (A21 Ak )+A1/k (x2 x1 ) = x1 x2 12 kA d x1 Finally, 2 ln 1 = x2 A1/k (Ak Ak )x1 21 12 +1 kA Similarly, 2 ln 2 = x1 A1/k 1 (Ak Ak )x2 21 12 kA (b) Appropriate substitition in the preceding equations of x1 = 1 and x1 = 0 yields: ln 1 = A1/k = (Ak )1/k = A12 12 (c) Let g ln 2 = A1/k = (Ak )1/k = A21 21 If k = 1, If k = 1, GE = A1/k = (x1 Ak + x2 Ak )1/k 12 21 x1 x2 RT g = x1 A21 + x2 A12 (Margules equation) A21 A12 (van Laar equation) x1 A12 + x2 A21 For k = 0, , +, indeterminate forms appear, most easily resolved by working with the logarithm: 1 ln g = ln(x1 Ak + x2 Ak )1/k = ln x1 Ak + x2 Ak 21 12 21 12 k g = (x1 A1 + x2 A1 )1 = 21 12 Apply lH pitals rule to the nal term: o d ln x1 Ak + x2 Ak x1 Ak ln A21 + x2 Ak ln A12 21 12 21 12 = k dk x1 A21 + x2 Ak 12 (A) Consider the limits of the quantity on the right as k approaches several limiting values. For k 0, x1 x2 ln g x1 ln A21 + x2 ln A12 = ln A21 + ln A12 and x1 x2 g = A21 A12 For k , Assume A12 /A21 > 1, and rewrite the right member of Eq. (A) as x1 ln A21 + x2 (A12 /A21 )k ln A12 x1 + x2 (A12 /A21 )k For k , Whence For k +, Whence k lim (A12 /A21 )k 0 g = A21 and k lim ln g = ln A21 except at x1 = 0 where g = A12 and k k lim (A12 /A21 )k g = A12 lim ln g = ln A12 except at x1 = 1 where g = A21 If A12 /A21 < 1 rewrite Eq. (A) to display A21 /A12 . 699 12.68 Assume that Eq. (12.1) is the appropriate equilibrium relation, written as xe e Pesat = xe e Pesat = ye P e EtOH Because P is low, we have assumed ideal gases, and for small xe let e e . For volume fraction e in the vapor, the ideal-gas assumption provides ev ye , and for the liquid phase, with xe small l e = xe Vel xe Vel xe Vel Vb xb Vb xe Vel + xb Vb b blood Then Vb l sat P ev P Ve e e e Ve P volume % EtOH in blood Vb e Pesat volume % EtOH in gas 12.70 By Eq. (11.95), HE = T RT E (G E /RT ) T P,x G = x1 ln(x1 + x2 RT 12 ) x2 ln(x2 + x1 21 ) (12.18) (G E /RT ) T E x d 21 d 12 x2 x1 dT dT = x2 + x1 21 x1 + x2 12 x1 x2 d 12 H dT = x1 x2 T x1 + x2 RT ij 12 d 21 dT + x2 + x1 21 = ai j Vj exp RT Vi (i = j) (12.24) Vj d ij = Vi dT exp ai j RT 12 a12 ai j = RT 2 ij ai j RT 2 H E = x1 x2 x1 + x2 + 12 21 a21 x2 + x1 21 E Because C P = d H E /dT , differentiate the preceding expression and reduce to get: E x2 21 (a21 /RT )2 x1 12 (a12 /RT )2 CP + = x1 x2 (x2 + x1 21 )2 (x1 + x2 12 )2 R Because 12 and 21 E must always be positive numbers, C P must always be positive. 700 Chapter 13 - Section B - Non-Numerical Solutions 13.1 (a) = i 4NH3 (g) + 5O2 (g) 4NO(g) + 6H2 O(g) i = 4 5 + 4 + 6 = 1 n0 = i0 =2+5=7 By Eq. (13.5), yNH3 = 2 4 7+ yO2 = 5 5 7+ yNO = 4 7+ yH2 O = 6 7+ (b) = 2H2 S(g) + 3O2 (g) 2H2 O(g) + 2SO2 (g) i = 2 3 + 2 + 2 = 1 i n0 = i0 =3+5=8 By Eq. (13.5), yH2 S = 3 2 8 yO2 = 5 3 8 yH2 O = 2 8 ySO2 = 2 8 (c) = i 6NO2 (g) + 8NH3 (g) 7N2 (g) + 12H2 O(g) i = 6 8 + 7 + 12 = 5 n0 = i0 =3+4+1=8 By Eq. (13.5), yNO2 = 3 6 8 + 5 yNH3 = 4 8 8 + 5 yN2 = 1 + 7 8 + 5 yH2 O = 12 8 + 5 13.2 C2 H4 (g) + 1 O2 (g) (CH2 )2 O(g) 2 (1) (2) C2 H4 (g) + 3O2 (g) 2CO2 (g) + 2H2 O(g) The stoichiometric numbers i, j are as follows: i= C2 H4 O2 (CH2 )2 O CO2 H2 O j j 1 1 1 2 1 0 0 1 2 2 1 3 0 2 2 0 n0 = i0 =2+3=5 By Eq. (13.7), yC2 H4 = 2 1 2 5 1 1 2 yO2 = 3 1 1 32 2 5 1 2 1 y (CH2 )2 O = 1 5 1 1 2 yCO2 = 22 5 1 1 2 yH2 O = 22 5 1 1 2 701 13.3 CO2 (g) + 3H2 (g) CH3 OH(g) + H2 O(g) CO2 (g) + H2 (g) CO(g) + H2 O(g) The stoichiometric numbers i, j are as follows: (1) (2) i= CO2 H2 CH3 OH CO H2 O j j 1 2 1 1 3 1 1 0 0 1 1 1 2 0 n0 = i0 =2+5+1=8 By Eq. (13.7), yCO2 = 2 1 2 8 21 yH2 = 5 31 2 8 21 yCH3 OH = 1 8 21 yCO = 1 + 2 8 21 yH2 O = 1 + 2 8 21 13.7 The equation for G , appearing just above Eq. (13.18) is: H0 G = T ( H0 T0 G ) + R 0 T T0 CP dT RT R T T0 C P dT R T To calculate values of G , one combines this equation with Eqs. (4.19) and (13.19), and evaluates parameters. In each case the value of H0 = H298 is tabulated in the solution to Pb. 4.21. In addition, the values of A, B, C, and D are given in the solutions to Pb. 4.22. The required values of G = G in J mol1 are: 0 298 (a) 32,900; (f ) 2,919,124; (i) 113,245; (n) 173,100; (r) 39,630; (t) 79,455; (u) 166,365; (x) 39,430; (y) 83,010 13.8 The relation of K y to P and K is given by Eq. (13.28), which may be concisely written: Ky = P P K (a) Differentiate this equation with respect to T and combine with Eq. (13.14): Ky T = P P P Ky H d ln K Ky d K dK = = Ky = RT 2 dT K dT dT Substitute into the given equation for (e / T ) P : e T = P K y de RT 2 d K y H (b) The derivative of K y with respect to P is: Ky P = T P P 1 1 K = K P P P P P 1 K y 1 = P P 702 Substitute into the given equation for (e / P)T : e P = T K y de () P d Ky (c) With K y i (yi )i , ln K y = i i ln yi . Differentiation then yields: 1 d Ky = K y de i i dyi yi de (A) Because yi = n i /n, 1 n i dn 1 dn i dyi = 2 = n n de n de de dn dn i yi de de But Whence, n i = n i0 + i e dn i = i de and n = n 0 + e and dn = de Therefore, i yi dyi = n 0 + e de Substitution into Eq. (A) gives 1 d Ky K y de = i i yi i yi n 0 + e m i=1 = 1 n 0 + e m i i2 i yi = 1 n 0 + e i2 i yi k k=1 In this equation, both K y and n 0 + e (= n) are positive. It remains to show that the summation term is positive. If m = 2, this term becomes 2 (y2 1 y1 2 )2 2 1 1 (1 + 2 ) + 2 2 (1 + 2 ) = y1 y2 y2 y1 where the expression on the right is obtained by straight-forward algebraic manipulation. One can proceed by induction to nd the general result, which is m i=1 i2 i yi m m m k k k=1 = i (yk i yi k )2 yi yk (i < k) All quantities in the sum are of course positive. 13.9 1 N (g) 2 2 + 3 H2 (g) NH3 (g) 2 For the given reaction, = 1, and for the given amounts of reactants, n 0 = 2. By Eq. (13.5), yN2 = 1 (1 2 e ) 2 e yH2 = 3 (1 2 e ) 2 e yNH3 = e 2 e By Eq. (13.28), yNH3 1/2 3/2 yN2 yH2 = [ 1 (1 2 P e (2 e ) =K 3 P e )]1/2 [ 2 (1 e )]3/2 703 Whence, e (2 e ) = (1 e )2 1 2 1/2 3 2 3/2 K P P = 1.299K P P This may be written: where, r e 2 2 r e + (r 1) = 0 r 1 + 1.299K P P The roots of the quadratic are: e = 1 1 = 1 r 1/2 r 1/2 Because e < 1, e = 1 r 1/2 , e = 1 1 + 1.299K P P 1/2 13.10 The reactions are written: Mary: 2NH3 + 3NO 3H2 O + 5 N2 2 (A) (B) Paul: Peter: 4NH3 + 6NO 6H2 O + 5N2 3H2 O + 5 N2 2NH3 + 3NO 2 (C) Each applied Eqs. (13.11b) and (13.25), here written: ln K = G /RT For reaction (A), G = 3 G fH A and K = (P ) i ( fi )i 2O 2 G fNH 3 G fNO 3 For Marys reaction = 1 , and: 2 K A = (P ) 1 2 ff3H 2O 5/2 ffN 2 ff2NH ff3NO 3 and ln K A = G A RT For Pauls reaction = 1, and K B = (P ) 1 ff6H 2O ff5N 2 ff4NH ff6NO 3 and ln K B = 2 G A RT For Peters reaction = 1 , and: 2 K C = (P ) 1 2 ff2NH ff3NO 3 ff3H 2O 5/2 ffN 2 and ln K C = G A RT In each case the two equations are combined: Mary: (P ) 1 2 ff3H 2O 5/2 ffN 2 ff2NH ff3NO 3 = exp G A RT 704 Paul: (P ) 1 ff6H 2O ff5N 2 ff4NH ff6NO 3 G A = exp RT 2 Taking the square root yields Marys equation. Peter: (P ) 2 1 ff2NH ff3NO 3 ff3H 2O 5/2 ffN 2 = exp G A RT 1 Taking the reciprocal yields Marys equation. 13.24 Formation reactions: 1 N 2 2 + 3 H2 NH3 2 (1) 1 N 2 2 + 1 O2 NO 2 (2) 1 N 2 2 + O2 NO2 (3) H 2 + 1 O2 H 2 O 2 (4) Combine Eq. (3) with Eq. (1) and with Eq. (2) to eliminate N2 : NO2 + 3 H2 NH3 + O2 2 (5) NO2 1 O2 + NO 2 (6) The set now comprises Eqs. (4), (5), and (6); combine Eq. (4) with Eq. (5) to eliminate H2 : (7) NO2 + 3 H2 O NH3 + 1 3 O2 4 2 Equations (6) and (7) represent a set of independent reactions for which r = 2. Other equivalent sets of two reactions may be obtained by different combination procedures. By the phase rule, F = 2 + N r s = 21+520 F =4 13.35 (a) Equation (13.28) here becomes: yB = yA P P 0 K =K Whence, yB = K (T ) 1 yB (b) The preceding equation indicates that the equilibrium composition depends on temperature only. However, application of the phase rule, Eq. (13.36), yields: F =2+211=2 This result means in general for single-reaction equilibrium between two species A and B that two degrees of freedom exist, and that pressure as well as temperature must be specied to x the equilibrium state of the system. However, here, the specication that the gases are ideal removes the pressure dependence, which in the general case appears through the i s. 13.36 For the isomerization reaction in the gas phase at low pressure, assume ideal gases. Equation (13.28) then becomes: yB = yA P P 0 K =K whence 1 yA = K (T ) yA 705 Assume that vapor/liquid phase equilibrium can be represented by Raoults law, because of the low pressure and the similarity of the species: xA PAsat (T ) = yA P (a) Application of Eq. (13.36) yields: and (1 xA )PBsat (T ) = (1 yA )P F = 2 + N r = 22+21 = 1 (b) Given T , the reaction-equilibriuum equation allows solution for yA . The two phase-equilibrium equations can then be solved for xA and P. The equilibrium state therefore depends solely on T . 13.38 (a) For low pressure and a temperature of 500 K, the system is assumed to be a mixture of ideal gases, for which Eq. (13.28) is appropriate. Therefore, yMX = yOX P P 0 KI = KI yPX = yOX P P 0 K II = K II yEB = yOX P P 0 K III = K III (b) These equation equations lead to the following set: yMX = K I yOX (1) yPX = K II yOX (2) yEB = K III yOX (3) The mole fractions must sum to unity, and therefore: yOX + K I yOX + K II yOX + K III yOX = yOX (1 + K I + K II + K III ) = 1 yOX = 1 1 + K I + K II + K III (4) (c) With the assumption that combine to give: C P = 0 and therefore that K 2 = 1, Eqs. (13.20), (13.21), and (13.22) K = K 0 K 1 = exp G 298 exp RT0 H298 1 H298 T0 1 T RT0 Whence, K = exp 298.15 500 (8.314)(298.15) G 298 The data provided lead to the following property changes of reaction and equilibrium constants at 500 K: Reaction H298 G 298 K I II III 1,750 1,040 10,920 3,300 1,000 8,690 2.8470 1.2637 0.1778 706 (d) Substitution of numerical values into Eqs. (1), (2), (3), and (4) yields the following values for the mole fractions: yOX = 0.1891 yMX = 0.5383 yPX = 0.2390 yEB = 0.0336 13.40 For the given owrates, n A0 = 10 and n B0 = 15, nA nB nC nD n with n A0 the limiting reactant without (II) = n A0 I II = n B0 I = I II = II = n 0 I II Use given values of YC and SC/D to nd I and II : YC = I II n A0 and SC/D = I II II Solve for I and II : I = SC/D + 1 n A0 YC = SC/D 2+1 2 10 0.40 = 6 II = 10 0.40 n A0 YC =2 = 2 SC/D nA nB nC nD n = 10 6 2 = 15 6 =62 =2 = 17 =2 =9 =4 =2 yA yB yC yD = 2/17 = 9/17 = 4/17 = 2/17 =1 = 0.1176 = 0.5295 = 0.2353 = 0.1176 13.42 A compound with large positive G f has a disposition to decompose into its constituent elements. Moreover, large positive G f often implies large positive H . Thus, if any decomposition product f is a gas, high pressures can be generated in a closed system owing to temperature increases resulting from exothermic decomposition. 13.44 By Eq. (13.12), G i i G i G P and from Eq. (6.10), (G i / P)T = Vi = T i i G i P = T i i Vi For the ideal-gas standard state, G P Vi = RT /P . Therefore = T i i RT P = RT P and G (P2 ) G (P1 ) = RT ln P2 P1 13.47 (a) For isomers at low pressure Raoults law should apply: P = x A PAsat + x B PBsat = PBsat + x A (PAsat PBsat ) For the given reaction with an ideal solution in the liquid phase, Eq. (13.33) becomes: Kl = 1 xA xB = xA xA from which xA = 1 Kl + 1 707 The preceding equation now becomes, P = 1 Kl 1 +1 PBsat + Kl 1 +1 PAsat P= Kl Kl + 1 PBsat + Kl 1 +1 PAsat (A) P = PBsat For K l = 0 (b) Given Raoults law: P = PAsat For K l = 1 = x A + x B = yA P P =P sat + y B PBsat PA yB yA sat + PBsat PA P= y A /PAsat PAsat PBsat PAsat PBsat 1 sat sat sat = sat = PA + y A (PBsat PAsat ) y A PB + y B PA + y B /PB For the given reaction with ideal gases in the vapor phase, Eq. (13.28) becomes: yB = Kv yA whence yA = Kv 1 +1 Elimination of y A from the preceding equation and reduction gives: P= (K v + 1)PAsat PBsat K v PAsat + PBsat (B) P = PBsat For K v = 0 P = PAsat For K v = (c) Equations (A) and (B) must yield the same P. Therefore Kl Kl + 1 PBsat + 1 l +1 K PAsat = (K v + 1)PAsat PBsat K v PAsat + PBsat Some algebra reduces this to: P sat Kv = Bsat PA Kl (d) As mentioned already, the species (isomers) are chemically similar, and the low pressure favors ideal-gas behavior. (e) F = N + 2 r = 2 + 2 2 1 = 1 Thus xing T should sufce. 708 Chapter 14 - Section B - Non-Numerical Solutions 14.2 Start with the equation immediately following Eq. (14.49), which can be modied slightly to read: ln i = ln Z (nG R/RT ) (n Z ) +1 +n n i n i n i where the partial derivatives written here and in the following development without subscripts are understood to be at constant T , n/ (or /n), and n j . Equation (6.61) after multiplication by n can be written: 2 3 nG R n ln Z + n 2 (nC) = 2n(n B) n 2 n RT Differentiate: 3 (nG R/RT ) (n B + n Bi ) + =2 2 n n n i 2 (2n 2 C + n 2 Ci ) n ln Z ln Z n i or ln Z 3 (nG R/RT ) ln Z = 2(B + Bi ) + 2 (2C + Ci ) n n i 2 n i By denition, Bi (n B) n i and T,n j Ci (nC) n i T,n j The equation of state, Eq. (3.40), can be written: Z = 1 + B + C 2 or n Z = n + n(n B) 2 + n 2 (nC) n n 2 Differentiate: (n Z ) (n B + n Bi ) + =1+ n n n i (2n 2 C + n 2 Ci ) or (n Z ) = 1 + (B + Bi ) + 2 (2C + Ci ) n i When combined with the two underlined equations, the initial equation reduces to: ln i = 1 + (B + Bi ) + 1 2 (2C + Ci ) 2 The two mixing rules are: 2 2 B = y1 B11 + 2y1 y2 B12 + y2 B22 3 2 2 3 C = y1 C111 + 3y1 y2 C112 + 3y1 y2 C122 + y2 C222 Application of the denitions of Bi and Ci to these mixing rules yields: 2 2 B1 = y1 (2 y1 )B11 + 2y2 B12 y2 B22 2 2 2 3 C1 = y1 (3 2y1 )C111 + 6y1 y2 C112 + 3y2 (1 2y1 )C122 2y2 C222 2 2 B2 = y1 B11 + 2y1 B12 + y2 (2 y2 )B22 3 2 2 2 C2 = 2y1 C111 + 3y1 (1 2y2 )C112 + 6y1 y2 C122 + 2y2 (3 2y2 )C222 709 In combination with the mixing rules, these give: B + B1 = 2(y1 B11 + y2 B12 ) 2 2 2C + C1 = 3(y1 C111 + 2y1 y2 C112 + y2 C122 ) B + B2 = 2(y2 B22 + y1 B12 ) 2 2 2C + C2 = 3(y2 C222 + 2y1 y2 C122 + y1 C112 ) In combination with the boxed equation these expressions along with Eq. (3.40) allow calculation of ln 1 and ln 2 . 14.11 For the case described, Eqs. (14.1) and (14.2) combine to give: yi P = xi Pi sat isat i If the vapor phase is assumed an ideal solution, i = i , and yi P = xi Pi sat isat i When Eq. (3.38) is valid, the fugacity coefcient of pure species i is given by Eq. (11.36): ln i = Bii P RT and isat = Bii Pi sat RT Therefore, ln Bii (Pi sat P) Bii P Bii Pi sat isat = = ln isat ln i = RT RT RT i For small values of the nal term, this becomes approximately: Bii (Pi sat P) isat =1+ RT i Whence, yi P = xi Pi sat 1 + Bii (Pi sat P) RT or yi P xi Pi sat = xi Pi sat Bii (Pi sat P) RT Write this equation for species 1 and 2 of a binary mixture, and sum. This yields on the left the difference between the actual pressure and the pressure given by Raoults law: P P(RL) = x1 B11 P1sat (P1sat P) + x2 B22 P2sat (P2sat P) RT Because deviations from Raoults law are presumably small, P on the right side may be replaced by its Raoults-law value. For the two terms, P1sat P = P1sat x1 P1sat x2 P2sat = P1sat (1 x2 )P1sat x2 P2sat = x2 (P1sat P2sat ) P2sat P = P2sat x1 P1sat x2 P2sat = P2sat x1 P1sat (1 x1 )P2sat = x1 (P2sat P1sat ) Combine the three preceding equations: P P(RL) = x1 x2 B11 (P1sat P2sat )P1sat x1 x2 B22 (P1sat P2sat )P2sat RT = x1 x2 (P1sat P2sat ) (B11 P1sat B22 P2sat ) RT 710 Rearrangement yields the following: P P(RL) = x1 x2 (P1sat P2sat )2 RT B11 P1sat B22 P2sat P1sat P2sat = (B11 B22 )P2sat x1 x2 (P1sat P2sat )2 B11 + P1sat P2sat RT = B22 x1 x2 (P1sat P2sat )2 (B11 ) 1 + 1 B11 RT P2sat P1sat P2sat Clearly, when B22 = B11 , the term in square brackets equals 1, and the pressure deviation from the Raoults-law value has the sign of B11 ; this is normally negative. When the virial coefcients are not equal, a reasonable assumption is that species 2, taken here as the heavier species (the one with the smaller vapor pressure) has the more negative second virial coefcient. This has the effect of making the quantity in parentheses negative and the quantity in square brackets < 1. However, if this latter quantity remains positive (the most likely case), the sign of B11 still determines the sign of the deviations. 14.13 By Eq. (11.90), the denition of i , Whence, ln i = ln fi ln xi ln f i 1 1 d fi 1 d ln fi d ln i = = i d xi xi xi d xi d xi f Combination of this expression with Eq. (14.71) yields: 1 d fi >0 fi d xi Because fi 0, d fi >0 d xi (const T, P) By Eq. (11.46), the denition of fi , RT d fi d ln fi di = = RT d xi d xi fi d xi Combination with Eq. (14.72) yields: di >0 d xi (const T, P) 14.14 Stability requires that G < 0 (see Pg. 575). The limiting case obtains when event Eq. (12.30) becomes: G E = RT xi ln xi i G = 0, in which For an equimolar solution xi = 1/N where N is the number of species. Therefore, G E (max) = RT i 1 1 = RT ln N N i 1 ln N = RT ln N N For the special case of a binary solution, N = 2, and G E (max) = RT ln 2 711 14.17 According to Pb. 11.35, G E = 12 P y1 y2 or 12 P GE y1 y2 = RT RT This equation has the form: GE = Ax1 x2 RT for which it is shown in Examples 14.5 and 14.6 that phase-splitting occurs for A > 2. Thus, the formation of two immiscible vapor phases requires: 12 P/RT > 2. Suppose T = 300 K and P = 5 bar. The preceding condition then requires: 12 > 9977 cm3 mol1 for vapor-phase immiscibility. Such large positive values for 12 are unknown for real mixtures. (Examples of gas/gas equilibria are known, but at conditions outside the range of applicability of the two-term virial EOS.) 14.19 Consider a quadratic mixture, described by: GE = Ax1 x2 RT It is shown in Example 14.5 that phase splitting occurs for such a mixture if A > 2; the value of A = 2 corresponds to a consolute point, at x1 = x2 = 0.5. Thus, for a quadratic mixture, phase-splitting obtains if: 1 1 G E > 2 RT = 0.5RT 2 2 This is a model-dependent result. Many liquid mixtures are known which are stable as single phases, even though G E > 0.5RT for equimolar composition. 14.21 Comparison of the Wilson equation, Eq. (12.18) with the modied Wilson equation shows that (G E/RT )m = C(G E/RT ), where subscript m distinguishes the modied Wilson equation from the original Wilson equation. To simplify, dene g (G E/RT ); then gm = Cg ngm = Cng (ng) (ngm ) =C n 1 n 1 ln(1 )m = C ln 1 where the nal equality follows from Eq. (11.96). Addition and subtraction of ln x1 on the left side of this equation and of C ln x1 on the right side yields: ln(x1 1 )m ln x1 = C ln(x1 1 ) C ln x1 or Differentiate: ln(x1 1 )m = C ln(x1 1 ) (C 1) ln x1 d ln(x1 1 ) C 1 d ln(x1 1 )m =C x1 d x1 d x1 As shown in Example 14.7, the derivative on the right side of this equation is always positive. However, for C sufciently greater than unity, the contribution of the second term on the right can make d ln(x1 1 )M <0 d x1 over part of the composition range, thus violating the stability condition of Eq. (14.71) and implying the formation of two liquid phases. 14.23 (a) Refer to the stability requirement of Eq. (14.70). For instability, i.e., for the formation of two liquid phases, 1 d 2 (G E /RT ) < 2 x1 x2 d x1 712 over part of the composition range. The second derivative of G E must be sufciently negative so as to satisfy this condition for some range of x1 . Negative curvature is the norm for mixtures for which G E is positive; see, e.g., the sketches of G E vs. x1 for systems (a), (b), (d), (e), and (f ) in Fig. 11.4. Such systems are candidates for liquid/liquid phase splitting, although it does not in fact occur for the cases shown. Rather large values of G E are usually required. (b) Nothing in principle precludes phase-splitting in mixtures for which G E < 0; one merely requires that the curvature be sufciently negative over part of the composition range. However, positive curvature is the norm for such mixtures. We know of no examples of liquid/liquid phasesplitting in systems exhibiting negative deviations from ideal-solution behavior. 14.29 The analogy is Raoults law, Eq. (10.1), applied at constant P (see Fig. 10.12): yi P = xi Pi sat If the vapor phase in VLE is ideal and the liquid molar volumes are negligible (assumptions inherent in Raoults law), then the Clausius/Clapeyron equation applies (see Ex. 6.5): Hilv d ln Pi sat = RT 2 dT Integration from the boiling temperature Tbi at pressure P (where Pi sat = P) to the actual temperature T (where Pi sat = Pi sat ) gives: T Hilv P sat dT ln i = 2 P Tbi RT Combination with Eq. (10.1) yields: yi = xi exp T Tbi Hilv dT RT 2 which is an analog of the Case I SLE equations. 14.30 Consider binary (two-species) equilibrium between two phases of the same kind. Equation (14.74) applies: xi i = xi i (i = 1, 2) If phase is pure species 1 and phase is pure species 2, then x1 = 1 = 1 and x2 = 2 = 1. Hence, x1 1 = x1 1 = 1 and x2 2 = x2 2 = 1 The reasoning applies generally to (degenerate) N -phase equilibrium involving N mutually immiscible species. Whence the cited result for solids. 14.31 The rules of thumb are based on Case II binary SLE behavior. For concreteness, let the solid be pure species 1 and the solvent be liquid species 2. Then Eqs. (14.93) and (14.92a) apply: x1 = 1 = exp sl H1 RTm 1 T Tm 1 T (a) Differentiate: sl H1 d x1 = 1 RT 2 dT Thus d x1 /dT is necessarily positive: the solid solubility x1 increases with increasing T . (b) Equation (14.92a) contains no information about species 2. Thus, to the extent that Eqs. (14.93) and (14.92a) are valid, the solid solubility x1 is independent of the identity of species 2. 713 (c) Denote the two solid phases by subscripts A and B. Then, by Eqs. (14.93) and (14.92a), the solubilities x A and x B are related by: xA = exp xB H sl (Tm B Tm A ) RTm A Tm B sl HB where by assumption, sl HA = H sl Accordingly, x A /x B > 1 if and only if TA < TB , thus validating the rule of thumb. (d) Identify the solid species as in Part (c). Then x A and x B are related by: sl sl ( H B H A )(Tm T ) xA = exp RTm T xB where by assumption, Tm A = Tm B Tm sl HA < sl H B , in Notice that Tm > T (see Fig. 14.21b). Then x A /x B > 1 if and only if accord with the rule of thumb. 14.34 The shape of the solubility curve is characterized in part by the behavior of the derivative dyi /d P (constant T ). A general expression is found from Eq. (14.98), y1 = P1sat P/F1 , where the enhancement factor F1 depends (at constant T ) on P and y1 . Thus, P sat P sat dy1 = 1 2 F1 + 1 P P dP F1 P + y1 F1 y1 P dy1 dP = y1 + y1 P ln F1 P + y1 ln F1 y1 P dy1 dP Whence, dy1 = dP y1 ln F1 P y1 1 P 1 y1 ln F1 y1 (A) P This is a general result. An expression for F1 is given by Eq. (14.99): F1 sat V s (P P1sat ) 1 exp 1 RT 1 From this, after some reduction: ln F1 P = y1 ln 1 P + y1 V1s RT and ln F1 y1 = P ln 1 y1 P Whence, by Eq. (A), dy1 = dP ln 1 y1 P y1 1 + y1 ln 1 y1 1 V1s + P RT P (B) 714 This too is a general result. If the two-term virial equation in pressure applies, then ln 1 is given by Eq. (11.63a), from which: ln 1 P = y1 1 2 (B11 + y2 12 ) RT and ln 1 y1 = P 2y2 12 P RT Whence, by Eq. (B), dy1 = dP y1 2 1 V1s B11 y2 12 P RT 2y1 y2 12 P 1 RT The denominator of this equation is positive at any pressure level for which Eq. (3.38) is likely to be valid. Hence, the sign of dy1 /d P is determined by the sign of the group in parentheses. For very low pressures the 1/P term dominates and dy1 /d P is negative. For very high pressures, 1/P is small, and dy1 /d P can be positive. If this is the case, then dy1 /d P is zero for some intermediate pressure, and the solubility y1 exhibits a minimum with respect to pressure. Qualitatively, these features are consistent with the behavior illustrated by Fig. 14.23. However, the two-term virial equation is only valid for low to moderate pressures, and is unable to mimic the change in curvature and attening of the y1 vs. P curve observed for high pressures for the naphthalene/CO2 system. 14.35 (a) Rewrite the UNILAN equation: m ln(c + Pes ) ln(c + Pes ) (A) 2s As s 0, this expression becomes indeterminate. Application of lH pitals rule gives: o n= s0 lim n = lim m s0 2 Pes Pes + c + Pes c + Pes = m 2 P P + c+P c+P or lims0 n = mP c+P which is the Langmuir isotherm. (b) Henrys constant, by denition: k lim dn dP P0 Differentiate Eq. (A): m dn = 2s dP es es c + Pes c + Pes Whence, k= m 2s es es c c = m cs es es 2 or k= m sinh s cs (c) All derivatives of n with respect to P are well-behaved in the zero-pressure limit: lim P0 m dn sinh s = cs dP 715 m d 2n = 2 sinh 2s P0 d P 2 c s 2m d 3n = 3 sinh 3s lim 3 P0 d P c s lim Etc. Numerical studies show that the UNILAN equation, although providing excellent overall correlation of adsorption data at low-to-moderate surface coverage, tends to underestimate Henrys constant. 14.36 Start with Eq. (14.109), written as: ln(P/n) = ln k + n 0 (z 1) dn +z1 n With z = 1 + Bn + Cn + , this becomes: 3 ln(P/n) = ln k + 2Bn + Cn 2 + 2 Thus a plot of ln(P/n) vs. n produces ln k as the intercept and 2B as the limiting slope (for n 0). Alternatively, a polynomial curve t of ln(P/n) in n yields ln k and 2B as the rst two coefcients. 2 14.37 For species i in a real-gas mixture, Eqs. (11.46) and (11.52) give: i = At constant temperature, g g g i (T ) + RT ln yi i P di = RT d ln yi i P With di = di , Eq. (14.105) then becomes: a d + d ln P + xi d ln yi i = 0 RT i (const T ) For pure-gas adsorption, this simplies to: a d = d ln P + d ln (const T ) (A) RT which is the real-gas analog of Eq. (14.107). On the left side of Eq. (A), introduce the adsorbate compressibility factor z through z a/RT = A/n RT : dn a d = dz + z n RT where n is moles adsorbed. On the right side of Eq. (A), make the substitution: (B) dP (C) P which follows from Eq. (11.35). Combination of Eqs. ( A), (B), and (C) gives on rearrangement (see Sec. 14.8): dP dn n dz + (Z 1) d ln = (1 z) P n P which yields on integration and rearrangement: d ln = (Z 1) n = k P exp P 0 (Z 1) dP exp P n 0 (1 z) dn +1z n This equation is the real-gas analog of Eq. (14.109). 716 14.39 & 14.40 Start with Eq. (14.109). With z = (1 bm)1 , one obtains the isotherm: n = k P(1 bn) exp bn 1 bn (A) For bn sufciently small, exp bn 1 bn 1 bn 1 bn Whence, by Eq. (A), n k P(1 2bn) or n kP 1 + 2bk P which is the Langmuir isotherm. With z = 1 + n, the adsorption isotherm is: n = k P exp(2n) from which, for n sufciently small, the Langmuir isotherm is again recovered. 14.41 By Eq. (14.107) with a = A/n, dP Ad =n P RT The denition of and its derivative are: A RT and d = Ad RT Whence, d = n dP P (A) By Eq. (14.128), the Raoults law analogy, xi = yi P/Pi . Summation for given P yields: yi xi = P Pi i i (B) By general differentiation, d i xi = P d i yi + Pi i yi dP Pi (C) The equation, i x i = 1, is an approximation that becomes increasingly accurate as the solution procedure converges. Thus, by rearrangement of Eq. (B), yi = Pi i xi i P = 1 P With P xed, Eq. (C) can now be written in the simple but approximate form: d i xi = dP P Equation (A) then becomes: d = n d i xi or = n i xi i where we have replaced differentials by deviations. The deviation in value must be unity. Therefore, yi 1 xi = P Pi i i xi is known, since the true 717 By Eq. (14.132), n= i 1 (xi /n i ) Combine the three preceding equations: P = i i yi 1 Pi (xi /n i ) When xi = yi P/Pi , the Raoults law analogy, is substituted the required equation is reproduced: P = i P i yi 1 Pi yi Pi n i 14.42 Multiply the given equation for G E/RT by n and convert all mole fractions to mole numbers: n2n3 n1n3 n1n2 nG E + A23 + A13 = A12 n n n RT Apply Eq. (11.96) for i = 1: ln 1 = A12 n 2 n2n3 1 n1 1 n1 A23 2 + A13 n 3 n n n2 n n2 = A12 x2 (1 x1 ) + A13 x3 (1 x1 ) A23 x2 x3 Introduce solute-free mole fractions: x2 x2 = x2 1 x1 x2 + x3 and x3 = x3 1 x1 Whence, For x1 0, ln 1 = A12 x2 (1 x1 )2 + A13 x3 (1 x1 )2 A23 x2 x3 (1 x1 )2 ln 1 = A12 x2 + A13 x3 A23 x2 x3 Apply this equation to the special case of species 1 innitely dilute in pure solvent 2. In this case, x2 = 1, x3 = 0, and ln 1,2 = A12 Also ln 1,3 = A13 Whence, ln 1 = x2 ln 1,2 + x3 ln 1,3 A23 x2 x3 In logarithmic form the equation immediately following Eq. (14.24) on page 552 may be applied to the several innite-dilution cases: ln H1 = ln f 1 + ln 1 Whence, or ln H1,2 = ln f 1 + ln 1,2 ln H1,3 = ln f 1 + ln 1,3 ln H1 ln f 1 = x2 (ln H1,2 ln f 1 ) + x3 (ln H1,3 ln f 1 ) A23 x2 x3 ln H1 = x2 ln H1,2 + x3 ln H1,3 A23 x2 x3 718 14.43 For the situation described, Figure 14.12 would have two regions like the one shown from to , probably one on either side of the minimum in curve II. 14.44 By Eq. (14.136) with V2 = V2 : V2 = ln(x2 2 ) RT Represent ln 2 by a Taylor series: ln 2 = ln 2 |x1 =0 + d ln 2 d x1 x1 =0 x1 + 1 d 2 ln 2 2 2 d x1 x1 =0 2 x1 + But at x1 = 0 (x2 = 1), both ln 2 and its rst derivative are zero. Therefore, ln 2 = 1 2 d 2 ln 2 2 d x1 x1 =0 2 x1 + Also, ln x2 = ln(1 x1 ) = x1 2 x4 x3 x1 1 1 4 3 2 Therefore, ln(x2 2 ) = + ln x2 + ln 2 = x1 1 2 1 1 2 d 2 ln 2 2 d x1 x1 =0 2 x1 and 1 1 V2 1 =1+ 2 2 x1 RT d 2 ln 2 2 d x1 x1 + x1 =0 Comparison with the given equation shows that: B= 1 1 1 2 2 d 2 ln 2 2 d x1 x1 =0 14.47 Equation (11.95) applies: E (G E/RT ) T = P,x HE RT 2 For the partially miscible system G /RT is necessarily large, and if it is to decrease with increasing T , the derivative must be negative. This requires that H E be positive. 14.48 (a) In accord with Eqs. (14.1) and (14.2), yi i P = xi i Pi sat isat Ki i Pi sat isat yi = P xi i 12 1 P1sat 1sat 2 K1 sat = 2 P2sat 1 2 K2 (b) 12 (x1 = 0) = P1sat 1 (P sat ) 1 P1sat 1 (P1sat ) 2 (P2sat ) = 1 sat 1 sat sat sat P2 P2sat 1 (P2 ) 1 (P2 ) 2 (P2 ) 12 (x1 = 1) = (P sat ) P sat 1 (P1sat ) 2 (P1sat ) P1sat = 1 sat 2 1 sat sat sat 2 (P2sat ) 2 P2 1 (P1 ) 2 (P2 ) 2 P2 The nal fractions represent corrections to modied Raoults law for vapor nonidealities. 719 (c) If the vapor phase is an ideal solution of gases, then i = i for all compositions. 14.49 Equation (11.98) applies: ln i T = P,x HiE RT 2 Assume that H E and HiE are functions of composition only. Then integration from Tk to T gives: ln HE i (x, T ) = i R i (x, Tk ) T Tk HE dT = i R T2 1 1 Tk T = HiE RT T 1 Tk i (x, T ) = i (x, Tk ) exp HiE RT T 1 Tk 14.52 (a) From Table 11.1, p. 415, nd: G E T = S E = 0 P,x and G E is independent of T . Therefore FR (x) GE = RT RT (b) By Eq. (11.95), (G E /RT ) T = P,x HE =0 RT 2 GE = FA (x) RT (c) For solutions exhibiting LLE, G E /RT is generally positive and large. Thus and are positive for LLE. For symmetrical behavior, the magic number is A = 2: A<2 homogeneous; A=2 consolute point; A>2 LLE With respect to Eq. (A), increasing T makes G E /RT smaller. thus, the consolute point is an upper consolute point. Its value follows from: =2 RTU TU = 2R The shape of the solubility curve is as shown on Fig. 14.15. 14.53 Why? Because they are both nontoxic, relatively inexpensive, and readily available. For CO2 , its Tc is near room temperature, making it a suitable solvent for temperature-sensitive materials. It is considereably more expensive than water, which is probably the cheapest possible solvent. However, both Tc and Pc for water are high, which increases heating and pumping costs. 720 Chapter 16 - Section B - Non-Numerical Solutions 16.1 The potential is displayed as follows. Note that K is used in place of k as a parameter to avoid confusion with Boltzmanns constant. Combination of the potential with Eq. (16.10) yields on piecewise integration the following expression for B: 2 B = N A d 3 1 + (K 3 1) 1 e/kT (l 3 K 3 ) e /kT 1 3 From this expression, 1 dB (K 3 1) e/kT + (l 3 K 3 ) e = kT 2 dT /kT according to which d B/dT = 0 for T and also for an intermediate temperature Tm : Tm = k ln + K3 1 l3 K 3 That Tm corresponds to a maximum is readily shown by examination of the second derivative d 2 B/dT 2 . 16.2 The table is shown below. Here, contributions to U (long range) are found from Eq. (16.3) [for U (el)], Eq. (16.4) [for U (ind)], and Eq. (16.5) [for U (disp)]. Note the following: 1. As also seen in Table 16.2, the magnitude of the dispersion interaction in all cases is substantial. 2. U (el), hence f (el), is identically zero unless both species in a molecular pair have non-zero permanent dipole moments. 3. As seen for several of the examples, the fractional contribution of induction forces can be substantial for unlike molecular pairs. Roughly: f (ind) is larger, the greater the difference in polarity of the interacting species. 721 Molecular Pair C6 /1078 J m6 f (el) f (ind) f (disp) f (el)/ f (disp) CH4 /C7 H16 CH4 /CHCl3 CH4 /(CH3 )2 CO CH4 /CH3 CN C7 H16 /CHCl3 C7 H16 /(CH3 )2 CO C7 H16 /CH3 CN CHCl3 /(CH3 )2 CO CHCl3 /CH3 CN (CH3 )2 CO/CH3 CN 49.8 34.3 24.9 22.1 161.9 119.1 106.1 95.0 98.3 270.3 0 0 0 0 0 0 0 0.143 0.263 0.806 0 0.008 0.088 0.188 0.008 0.096 0.205 0.087 0.151 0.052 1.000 0.992 0.912 0.812 0.992 0.904 0.795 0.770 0.586 0.142 0 0 0 0 0 0 0 0.186 0.450 5.680 16.3 Water (H2 O), a highly polar hydrogen donor and acceptor, is the common species for all four systems; in all four cases, it experiences strong attractive interactions with the second species. Here, interactions between unlike molecular pairs are stronger than interactions between pairs of molecules of the same kind, and therefore H is negative. (See the discussion of signs for H E in Sec. 16.7.) 16.4 Of the eight potential combinations of signs, two are forbidden by Eq. (16.25). Suppose that H E is negative and S E is positive. Then, by Eq. (16.25), G E must be negative: the sign combination G E , H E , and S E is outlawed. Similar reasoning shows that the combination G E , H E , and S E is inconsistent with Eq. (16.25). All other combinations are possible in principle. 16.5 In Series A, hydrogen bonding occurs between the donor hydrogens of CH2 Cl2 and the electron-rich benzene molecule. In series B, a charge-transfer complex occurs between acetone and the aromatic benzene molecule. Neither cyclohexane nor n-hexane offers the opportunity for these special solvation interactions. Hence the mixtures containing benzene have more negative (smaller positive) values of H E than those containing cyclohexane and n-hexane. (See Secs. 16.5 and 16.6.) 16.6 (a) Acetone/cyclohexane is an NA/NP system; one expects G E , H E , and S E . (b) Acetone/dichloromethane is a solvating NA/NA mixture. Here, without question, one will see G E , H E , and S E . (c) Aniline/cyclohexane is an AS/NP mixture. Here, we expect either Region I or Region II behavior: G E and H E , with S E or . [At 323 K (50 C), experiment shows that S E is for this system.] (d) Benzene/carbon disulde is an NP/NP system. We therefore expect G E , H E , and S E . (e) Benzene/n-hexane is NP/NP. Hence, G E , H E , and S E . (f ) Chloroform/1,4-dioxane is a solvating NA/NA mixture. Hence, G E (g) Chloroform/n-hexane is NA/NP. Hence, G , H , and S . (h) Ethanol/n-nonane is an AS/NP mixture, and ethanol is a very strong associator. Hence, we expect Region II behavior: G E , H E , and S E . 16.7 By denition, i j 2 Bi j 1 2 E E E , HE , and S E . Bii + B j j At normal temperature levels, intermolecular attractions prevail, and the second virial coefcients are negative. (See Sec. 16.2 for a discussion of the connection between intermolecular forces and the second virial coefcient.) If interactions between unlike molecular pairs are weaker than interactions between pairs of molecules of the same kind, |Bi j | < 1 |Bii + B j j | 2 722 and hence (since each B is negative) i j > 0. If unlike interactions are stronger than like interactions, |Bi j | > 1 |Bii + B j j | 2 Hence i j < 0. For identical interactions of all molecular pairs, Bi j = Bii = B j j , and i j = 0 The rationalizations of signs for H E of binary liquid mixtures presented in Sec. 16.7 apply approximately to the signs of 12 for binary gas mixtures. Thus, positive 12 is the norm for NP/NP, NA/NP, and AS/NP mixtures, whereas 12 is usually negative for NA/NA mixtures comprising solvating species. One expects 12 to be essentially zero for ideal solutions of real gases, e.g., for binary gas mixtures of the isomeric xylenes. 16.8 The magnitude of Henrys constant Hi is reected through Henrys law in the solubility of solute i in a liquid solvent: The smaller Hi , the larger the solubility [see Eq. (10.4)]. Hence, molecular factors that inuence solubility also inuence Hi . In the present case, the triple bond in acetylene and the double bond in ethylene act as proton acceptors for hydrogen-bond formation with the donor H in water, the triple bond being the stronger acceptor. No hydrogen bonds form between ethane and water. Because hydrogen-bond formation between unlike species promotes solubility through smaller values of G E and i than would otherwise obtain, the values of Hi are in the observed order. 16.9 By Eq. (6.70), H = T S . For the same temperaature and pressure, less structure or order means larger S. Consequently, S sl , S lv , and S sv are all positive, and so therefore are H sl , H lv , and H sv . 16.11 At the normal boiling point: Therefore H R,l = H R,v H lv H v H l = (H v H ig ) (H l H ig ) = H R,v H R,l H lv At 1(atm), H R,v should be negligible relative to H lv . Then H R,l H lv . Because the normal boiling point is a representative T for typical liquid behavior, and because H R reects intermolecular forces, H lv has the stated feature. H lv (H2 O) is much larger than H lv (CH4 ) because of the strong hydrogen bonding in liquid water. R,l R,l 16.12 By denition, write C lP = C P +C P , where C P is the residual heat capacity for the liquid phase. ig R,l Also by denition, C P = ( H R,l / T ) P . By assumption (modest pressure levels) C P C v . P ig Thus, R,l C lP C v + P H R,l T P For liquids, H is highly negative, becoming less so as T increases, owing to diminution of interR,l molecular forces (see, e.g., Fig. 6.5 or Tables E.5 and E.6). Thus C P is positive, and C lP > C v . P 16.13 The ideal-gas equation may be written: Vt = N RT n RT = NA P P RT Vt = NA P N The quantity V t /N is the average volume available to a particle, and the average length available is about: RT 1/3 V t 1/3 = NA P N 83.14 cm3 bar mol1 K1 300 K V t 1/3 = 34.6 1010 m or 34.6 A = N 6.023 1023 mol1 1 bar 106 cm3 m3 For argon, this is about 10 diameters. See comments on p. 649 with respect to separations at which attractions become negligible. 1/3 723 ... View Full Document

This **preview** has intentionally **blurred** sections. Sign up to view the full version.