Che 211- che thermosolutions-7th edition

CHE 211- Che ThermoSolutions-7th edition
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Unformatted text preview: Chapter 1 - Section A - Mathcad Solutions 1.4 The equation that relates deg F to deg C is: t(F) = 1.8 t(C) + 32. Solve this equation by setting t(F) = t(C). Guess solution: t t = 1.8t Given 0 Find t () 32 Ans. 40 P= 1.5 By definition: F A F = mass g A P 3000bar D 4mm F PA g 9.807 m D 4 F mass 2 g s P= 1.6 By definition: F A A 3000atm D 0.17in F PA g 32.174 ft 4 D mass 2 sec gh 13.535 1.8 gm 3 101.78kPa 13.535 g 9.832 mass 2 384.4 kg Ans. 2 F g A mass 2 0.023 in 1000.7 lbm Ans. gm 3 29.86in_Hg m h 2 56.38cm s Pabs g gh 32.243 Patm ft Pabs h 2 176.808 kPa Ans. 25.62in s cm Patm 12.566 mm Patm cm Patm A F = mass g P 1.7 Pabs = 2 Note: Pressures are in gauge pressure. Pabs gh 1 Patm Pabs 27.22 psia Ans. 1.10 Assume the following: 13.5 gm g 3 m 9.8 2 s cm P 1.11 P h 400bar h g Ans. 302.3 m The force on a spring is described by: F = Ks x where Ks is the spring constant. First calculate K based on the earth measurement then gMars based on spring measurement on Mars. On Earth: F = mass g = K x mass g 0.40kg 9.81 m x 2 1.08cm s F F mass g F x Ks 3.924 N Ks 363.333 N m On Mars: x 0.40cm gMars 1.12 Given: FMars gMars FMars mass d P= dz g 0.01 FMars Kx and: mK = 4 MP RT Substituting: Separating variables and integrating: 1 dP = P Psea 1atm ln M 29 2 mK PDenver Psea = PDenver = Psea e gm mol d P= dz zDenver Mg zDenver RT Mg zDenver RT g 9.8 MP g RT Mg dz RT 0 Psea Taking the exponential of both sides and rearranging: 3 Ans. kg PDenver After integrating: 10 m 2 s 3 R 82.06 cm atm T mol K Mg zDenver RT Mg RT ( 10 zDenver 273.15) K 0.194 zDenver Psea e PDenver 0.823 atm Ans. PDenver PDenver 1 mi 0.834 bar Ans. 1.13 The same proportionality applies as in Pb. 1.11. gearth ft 32.186 gmoon 2 5.32 M lmoon 1.14 gearth gmoon costbulb Ans. 18.767 lbf Ans. hr 0.1dollars 70W 10 day kW hr hr 5.00dollars 10 day 1000hr costelec dollars yr costelec 25.567 dollars yr costtotal 43.829 dollars yr Ans. 18.262 costtotal costbulb D 18.76 113.498 113.498 lbm wmoon M gmoon costbulb 1.15 learth M learth lbm wmoon lmoon 2 s s learth ft 1.25ft costelec mass 250lbm g 32.169 ft 2 s 3 Patm (a) F 30.12in_Hg Patm A l Work 1.7ft D D 2 3 10 lbf Work PE mass g l mass 1.227 ft 2 Ans. Ans. 16.208 psia Fl 0.47m A 2.8642 Pabs PE 1.16 4 F mass g F A (b) Pabs (c) A 3 10 ft lbf Ans. 4.8691 Ans. 424.9 ft lbf g 150kg 9.813 m 2 s Patm (a) F Patm A l 0.83m EP 1.18 mass EK 2 A 4 110.054 kPa Work EP mass g l u 1250kg 1 2 mass u 2 40 EK m s Work EK Wdot = 200W Ans. 1000 kJ mass g h 0.91 0.92 time Wdot Ans. 1000 kJ g 9.8 m 2 s 4 h 2 0.173 m Ans. 10 N Fl Work 1.19 1.909 Pabs Work D 4 F mass g F A (b) Pabs (c) A 101.57kPa 50m Ans. 15.848 kJ Ans. 1.222 kJ Ans. Wdot mdot 1.22 a) cost_coal mdot g h 0.91 0.92 0.488 25.00 ton cost_coal MJ 29 kg cost_gasoline 0.95 GJ kg s 1 2.00 gal 37 GJ cost_gasoline Ans. 14.28 GJ 1 3 m cost_electricity 0.1000 kW hr cost_electricity 27.778 GJ 1 b) The electrical energy can directly be converted to other forms of energy whereas the coal and gasoline would typically need to be converted to heat and then into some other form of energy before being useful. The obvious advantage of coal is that it is cheap if it is used as a heat source. Otherwise it is messy to handle and bulky for tranport and storage. Gasoline is an important transportation fuel. It is more convenient to transport and store than coal. It can be used to generate electricity by burning it but the efficiency is limited. However, fuel cells are currently being developed which will allow for the conversion of gasoline to electricity by chemical means, a more efficient process. Electricity has the most uses though it is expensive. It is easy to transport but expensive to store. As a transportation fuel it is clean but batteries to store it on-board have limited capacity and are heavy. 5 1.24 Use the Matcad genfit function to fit the data to Antoine's equation. The genfit function requires the first derivatives of the function with respect to the parameters being fitted. B TC A Function being fit: f ( A B C) T e First derivative of the function with respect to parameter A d f ( A B C) T dA B exp A T C First derivative of the function with respect to parameter B d f ( A B C) T dB 1 T C B exp A T C First derivative of the function with respect to parameter C d f ( A B C) T dC B ( T 2 exp A C) 18.5 3.18 9.5 5.48 0.2 11.8 t 9.45 16.9 23.1 32.7 Psat 28.2 41.9 44.4 66.6 52.1 89.5 63.3 129 75.5 187 6 B T C T t 273.15 lnPsat ln ( ) Psat Array of functions used by Mathcad. In this case, a0 = A, a1 = B and a2 = C. exp a0 exp a0 F ( a) T a1 T a2 a1 T 1 exp a0 T a2 a1 T a2 2 Guess values of parameters 15 a2 guess a1 T exp a0 3000 50 a2 a1 T a2 Apply the genfit function A 13.421 A B B genfit ( Psat guess F) T C 2.29 C 3 10 Ans. 69.053 Compare fit with data. 200 150 Psat f ( A B C) T 100 50 0 240 260 280 300 320 340 360 T To find the normal boiling point, find the value of T for which Psat = 1 atm. 7 Psat 1atm B Tnb A Tnb 1.25 a) t1 1970 C2 C1 ( 1 t2 i) 273.15K 2000 dollars gal C1 0.35 1.513 329.154 K Ans. 56.004 degC C2 t2 t1 Tnb CK Psat ln kPa i 5% dollars gal The increase in price of gasoline over this period kept pace with the rate of inflation. b) t1 1970 Given t2 C2 C1 = (1 2000 i) C1 t2 t1 i 16000 dollars yr Find ( i) i C2 80000 dollars yr 5.511 % The salary of a Ph. D. engineer over this period increased at a rate of 5.5%, slightly higher than the rate of inflation. c) This is an open-ended problem. The strategy depends on age of the child, and on such unpredictable items as possible financial aid, monies earned by the child, and length of time spent in earning a degree. 8 Chapter 2 - Section A - Mathcad Solutions 2.1 (a) Mwt g 35 kg 9.8 m z 2 5m s Work (b) Work Mwt g z Utotal Utotal Work (c) By Eqs. (2.14) and (2.21): dU 1.715 kJ Ans. 1.715 kJ Ans. d ( )= CP dT PV Since P is constant, this can be written: MH2O CP dT = MH2O dU MH2O P dV Take Cp and V constant and integrate: MH2O CP t2 t1 = Utotal kJ MH2O 30 kg CP 4.18 t1 20 degC kg degC t2 t1 Utotal MH2O CP t2 20.014 degC Ans. (d) For the restoration process, the change in internal energy is equal but of opposite sign to that of the initial process. Thus Q (e) Utotal Q 1.715 kJ Ans. Ans. In all cases the total internal energy change of the universe is zero. 2.2 Similar to Pb. 2.1 with mass of water = 30 kg. Answers are: (a) W = 1.715 kJ (b) Internal energy change of the water = 1.429 kJ (c) Final temp. = 20.014 deg C (d) Q = -1.715 kJ 9 2.4 The electric power supplied to the motor must equal the work done by the motor plus the heat generated by the motor. E i 9.7amp Wdotelect Qdot 2.5 i E Wdotelect t Eq. (2.3): U =Q Step 1 to 2: Ut12 Wdotelect 1.067 3 134.875 W Ans. W W12 200J Ut12 Q34 Q12 800J Ut34 1.25hp 10 W Qdot Wdotmech Q12 Step 3 to 4: Wdotmech 110V W12 6000J W34 Q34 Ut34 W34 3 5.8 Ans. 10 J 300J Ans. 500 J t t Step 1 to 2 to 3 to 4 to 1: Since U is a state function, U for a series of steps that leads back to the initial state must be zero. Therefore, the sum of the t U values for all of the steps must sum to zero. Ut41 4700J Ut23 Step 2 to 3: Ut23 W23 Ut34 Ut41 Ans. 4000 J Ut23 Ut12 4 Ut23 3 Q23 Q23 3800J W23 10 J 200 J Ans. For a series of steps, the total work done is the sum of the work done for each step. W12341 1400J 10 W41 W12341 Step 4 to 1: W12 W23 Ut41 Ut41 Ans. 10 J 3 4.5 Q41 W41 3 4.5 W41 4700J Q41 Note: W41 W34 10 J 200 J Ans. Q12341 = W12341 2.11 The enthalpy change of the water = work done. M Wdot 2.12 Q CP 20 kg 4.18 kJ kg degC t 10 degC M CP t 0.25 kW 0.929 hr Wdot Ans. U Q 12 kJ W U 19.5 kJ Ans. Q 12 kJ U W U 7.5 kJ Q 12 kJ Ans. 2.13Subscripts: c, casting; w, water; t, tank. Then mc Uc mw Uw mt Ut = 0 Let C represent specific heat, C = CP = CV Then by Eq. (2.18) mc Cc tc mw Cw tw mt Ct tt = 0 mc 2 kg mw Cc 0.50 kJ kg degC Ct mt 40 kg tc 500 degC Given t1 mc Cc t2 t2 0.5 5 kg kJ kg degC Cw 25 degC t2 30 degC tc = mw Cw mt Ct t2 Find t2 11 t2 4.18 kJ kg degC (guess) t1 27.78 degC Ans. 2.15 mass (a) (b) T g CV 1 kg Ut 1K 9.8 m Ut mass CV T EP 2 kJ kg K 4.18 Ans. 4.18 kJ Ut s z (c) 2.17 EP EK z z mass g 426.531 m Ans. EK u Ut 1 mass 2 1000 50m u kg 5 A 3 A 2m mdot Wdot uA mdot mdot g z Wdot 4 D 2 (b) 2 4 kg 1.571 10 7.697 s 3 10 kW U1 P1 1002.7 kPa U1 H1 763.131 U2 kJ 2784.4 kg H2 U2 P1 V 1 kJ kg kJ kg cm 1.128 gm V1 Ans. 3 P2 U P2 V 2 2022.4 Ans. 3 kJ 762.0 kg U Ans. 3.142 m H1 2.18 (a) m s m s u m D 91.433 Ans. 1500 kPa U2 U1 H 12 cm 169.7 gm V2 H 2275.8 kJ kg H2 H1 Ans. 2.22 D1 (a) u1 2.5cm m s D2 5cm For an incompressible fluid, =constant. By a mass balance, mdot = constant = u 1A1 = u2A2 u2 (b) D1 u1 u2 D2 1 EK 2 2 2 u2 1 2 2.23 Energy balance: Mass balance: u1 2 mdot3 H3 mdot1 1.0 Qdot 30 Qdot H1 T1 CP 4.18 s mdot1 mdot2 H2 = Qdot H2 = Qdot mdot2 CP T3 25degC kJ Ans. kg mdot2 H3 T1 kg s J mdot2 = 0 mdot2 = Qdot mdot1 CP T1 Ans. 1.875 mdot1 mdot Cp T3 T3 CP mdot1 m s mdot1 H1 mdot1 H3 or 0.5 EK mdot3 Therefore: T3 2 T2 = Qdot mdot1 CP T1 mdot2 CP T2 mdot2 0.8 kg s T2 75degC kJ kg K mdot2 CP T2 mdot2 CP T3 43.235 degC 2 2.25By Eq. (2.32a): By continuity, incompressibility H u =0 2 A1 u2 = u1 13 A2 H = CP T CP 4.18 kJ kg degC Ans. 2 u = u1 u1 u1 2 2 CP D2 2 u = u1 1 m s 14 D1 D1 2 1 A2 SI units: T 2 A1 2 4 1 D2 D1 D2 2.5 cm 3.8 cm 4 T 0.019 degC Ans. T D2 0.023 degC Ans. 7.5cm u1 T 2 2 CP D1 1 4 D2 Maximum T change occurrs for infinite D2: D2 cm u1 T 2.26 T1 2 CP 300K Wsdot H 2 D1 1 T D2 T2 u1 520K ndot 98.8kW CP T2 4 50 H T1 10 0.023 degC m s u2 kmol hr 3.5 10 molwt 29 kg kmol 7 R 2 CP 6.402 m s Ans. 3 kJ kmol By Eq. (2.30): Qdot H 2 u2 2 2 u1 2 molwt ndot Wsdot Qdot 2 2.27By Eq. (2.32b): By continunity, constant area H= u2 = u1 u 2 gc V2 u2 = u1 V1 14 also T 2 P1 T 1 P2 Ans. 9.904 kW V2 V1 = 2 T 2 P1 T 1 P2 2 u = u2 u1 2 2 u = u1 P1 R T2 T2 P2 molwt mol rankine 7 2 u1 20 psi ft lbf 28 20 7 R T2 2 ft s T1 T1 579.67 rankine H2 2726.5 kJ kg kJ kg Ans. gm mol (guess) 578 rankine Given H = CP T = 1 T 1 P2 100 psi 3.407 2 T 2 P1 2 2 R T2 T1 = T2 Find T2 u1 2 2 T 2 P1 T 1 P2 1 molwt Ans. 578.9 rankine ( 119.15 degF) 2.28 u1 3 m s u2 200 m s H1 2 By Eq. (2.32a): 2.29 u1 u2 m s m 500 s 30 By Eq. (2.32a): Q H2 H1 3112.5 u1 u2 H1 334.9 kJ kg 2 Q 2 kJ H2 kg 2411.6 2945.7 (guess) 2 H2 Given u2 578.36 m s H1 = u2 u1 2 5 cm V1 388.61 u2 2 Ans. 3 3 D1 kJ kg cm gm 15 V2 667.75 cm gm Find u2 Continuity: 2.30 (a) t1 D2 t2 30 degC CV u1 V2 u2 V1 D1 D2 Ans. 1.493 cm n 250 degC 3 mol J mol degC 20.8 By Eq. (2.19): Q n CV t2 Q t1 Ans. 13.728 kJ Take into account the heat capacity of the vessel; then cv mv Q (b) 100 kg mv cv t1 200 degC CP 29.1 n CV CV Q Q Ans. 11014 kJ n 40 degC 4 mol joule mol degC Q n CP t2 t2 70 degF 5 kJ kg degC t1 t2 By Eq. (2.23): 2.31 (a) t1 t2 0.5 BTU mol degF n CV t2 Q t1 18.62 kJ n 350 degF 3 mol By Eq. (2.19): Q t1 4200 BTU Ans. Take account of the heat capacity of the vessel: mv Q mv cv cv 200 lbm (b) t1 400 degF n CV t2 0.12 BTU lbm degF Q t1 t2 150 degF 16 10920 BTU n Ans. 4 mol Ans. CP 7 BTU mol degF Q H1 2.33 n CP t2 1322.6 V1 Q t1 BTU BTU lbm 1148.6 3 V2 lbm 78.14 ft 4 mdot mdot V2 10 22.997 u2 Wdot 2.34 H1 V1 H2 307 9.25 BTU lbm ft H2 330 V2 u1 lbm ft 0.28 173.99 BTU lb Ans. 39.52 hp BTU 3 lbm Ws 2 Wdot Ws mdot u1 u2 H1 20 ft s molwt 44 3 D1 lbm D2 4 in 1 in 2 mdot u2 4 D1 u1 mdot V1 V2 mdot u2 679.263 9.686 2 4 D2 2 Eq. (2.32a): Q H2 H1 10 in sec 2 D2 Eq. (2.32a): Ws D2 ft sec 2 4 3 in 4 lb 3.463 V1 mdot 10 D1 lbm 2 u2 ft s u1 3 2 D 1 u1 Ans. 7000 BTU H2 lbm ft 3.058 By Eq. (2.23): ft sec u1 u2 2 17 lb hr 2 Ws Ws molwt 5360 Q BTU lbmol 98.82 BTU lbm gm mol Qdot 2.36 T1 Qdot mdot Q P 300 K 67128 BTU hr Ans. 1 kg n 1 bar 28.9 V1 n gm 34.602 mol mol 3 3 bar cm T1 83.14 mol K P cm 24942 mol V1 V2 P dV = n P V 1 W= n V2 = n P V1 3 V1 V1 Whence W Given: V2 V1 joule 29 Q nH U mol K Q = T1 3 Whence T1 Q W U n CP T2 602.08 kJ 12.41 kJ mol Ans. 172.61 kJ H T2 = T1 CP W n P 2 V1 T2 H 3 T1 17.4 kJ mol Ans. Ans. Ans. 2.37 Work exactly like Ex. 2.10: 2 steps, (a) & (b). A value is required for PV/T, namely R. R 8.314 T1 (a) Cool at const V1 to P2 (b) Heat at const P2 to T2 Ta2 T1 P2 P1 Ta2 293.15 K T2 333.15 K P1 J mol K 1000 kPa P2 100 kPa 7 R 2 CP 29.315 K 18 CV 5 R 2 Tb T2 Hb C P Tb Hb Ua CV Ta Ua R T1 V1 P1 Ta2 Tb V1 303.835 K 2.437 8.841 Ta 10 5.484 10 mol Ha Ua V 1 P2 P1 Ha Ub Hb P2 V 2 V1 Ub U Ub U 0.831 H 2.39 Ua Ha Hb H 1.164 996 kg 9.0 10 3 D 5 2 kJ mol ms 1 cm u 5 1m 5s 5 22133 Re D u Re Ta 263.835 K mol 3J mol V2 3 R T2 V2 P2 7.677 6.315 0.028 m mol 3J 10 mol 3J 10 mol Ans. mol 4 kg m 2 kJ T1 3J 3 3m 10 Ta2 55333 110667 276667 19 Ans. D 0.0001 Note: D = /D in this solution 0.00635 fF 0.3305 ln 0.27 D 7 Re 0.9 2 0.00517 fF 0.00452 0.0039 0.313 mdot u 4 D 2 mdot 1.956 kg 1.565 s Ans. 9.778 0.632 PL 2 fF D 2 PL u 0.206 kPa 11.254 m Ans. 3.88 2.42 mdot 4.5 kg s H1 761.1 kJ kg H2 536.9 kJ kg Assume that the compressor is adiabatic (Qdot = 0). Neglect changes in KE and PE. Wdot Cost mdot H2 15200 H1 Wdot Wdot 1.009 3 10 kW 0.573 Cost kW 20 799924 dollars Ans. Chapter 3 - Section A - Mathcad Solutions 3.1 1d = 1 = dT d dP P T At constant T, the 2nd equation can be written: d = 2 ln dP 6 44.1810 P = 1 bar 2 = 1.01 1 ln ( ) 1.01 P P Ans. P2 = 226.2 bar 225.2 bar 3 3.4 b cm 0.125 gm c 2700 bar P1 P2 1 bar 500 bar V2 Since Work = a bit of algebra leads to P dV V1 P2 Work c P P b dP Work 0.516 Work 0.516 P1 J Ans. gm Alternatively, formal integration leads to Work 3.5 =a P1 c P2 P1 a bP P2 1 atm b ln P2 b P1 b 3.9 10 6 atm 1 b V 3000 atm 0.1 10 1 ft 3 9 J gm Ans. atm 2 (assume const.) Combine Eqs. (1.3) and (3.3) for const. T: P2 Work ( a V Work b P)P dP P1 21 16.65 atm ft 3 Ans. 1 3.6 1.2 10 V1 3 degC 1 CP 0.84 kJ kg degC M 5 kg t2 20 degC 3 m 1 P 1590 kg t1 1 bar 0 degC With beta independent of T and with P=constant, dV = V V2 dT Vtotal Vtotal MV Work V1 exp P Vtotal M CP t2 P2 (a) Constant V: T2 T1 U H R T1 ln (c) Adiabatic: T1 P1 Ans. 84 kJ 84 kJ Ans. 83.99 kJ Ans. T T1 U 10.91 15.28 H=0 Q=0 22 kJ mol kJ mol Work and CV U = Q = CV T T2 and 7 R 2 CP 600 K and Q Ans. Utotal U= P2 7.638 joule Work H CP T (b) Constant T: Work Q 1 bar V1 Ans. Q Q and CV T m t1 T P1 3 V2 Htotal W= 0 P2 10 Q Utotal 8 bar 5 V Work 7.638 Htotal P1 t1 (Const. P) Q 3.8 t2 525 K Ans. Ans. and 10.37 Q=W kJ mol Ans. U = W = CV T 5 2 R 1 CP T2 CV P2 T1 U and U 5.586 3.9 P4 2bar CP H kJ 7 R 2 10bar T1 Step 41: Adiabatic T 331.227 K Ans. mol T4 R T1 P1 V1 P4 T1 kJ mol Ans. 5 R 2 CV 600K 7.821 T2 CP T H CV T W P1 T2 P1 V1 4.988 T4 3 3m 10 378.831 K mol R CP P1 3J U41 CV T1 T4 U41 4.597 10 H41 CP T1 T4 H41 6.436 10 Q41 3bar T2 Step 12: Isothermal Q41 U41 W41 4.597 mol J 0 mol W41 P2 J 0 mol mol 3J V2 600K U12 0 J mol H12 0 J mol 23 R T2 P2 3J 10 3 V2 m 0.017 mol U12 0 J mol H12 0 J mol mol T1 Q12 W12 P3 2bar V3 Q12 Q12 P1 W12 P3 V 3 T3 V2 Step 23: Isochoric P2 R T1 ln T3 R U23 CV T3 H23 CP T3 T2 Q23 2 bar T4 CV T3 W23 P4 0 T2 R T4 V4 3J 10 mol 3J 6.006 10 400 K 3J 4.157 10 P4 3 V4 m 0.016 mol U34 CV T4 T3 U34 439.997 H34 CP T4 T3 H34 615.996 Q34 CP T4 T3 Q34 W34 R T4 T3 W34 3.10 For all parts of this problem: T2 = T1 mol mol 3J H23 5.82 10 mol 3J Q23 4.157 10 mol J W23 0 mol J mol 378.831 K Step 34: Isobaric U23 T2 6.006 615.996 175.999 J mol J mol J mol J mol and Also Q = Work and all that remains is U= H=0 to calculate Work. Symbol V is used for total volume in this problem. P1 1 bar P2 V1 12 bar 24 3 12 m V2 3 1m (a) P2 Work = n R T ln Work P1 Work P 1 V 1 ln P2 P1 Ans. 2982 kJ (b) Step 1: adiabatic compression to P2 1 5 3 Vi P2 V i W1 V1 P1 P2 (intermediate V) P1 V 1 Vi W1 3063 kJ W2 1 3 2.702 m 2042 kJ Step 2: cool at const P2 to V2 W2 P2 V 2 Work W1 Vi W2 Work 5106 kJ Ans. (c) Step 1: adiabatic compression to V2 Pi W1 P1 V1 (intermediate P) V2 Pi V 2 1 Step 2: No work. Work (d) Step 1: heat at const V1 to P2 62.898 bar W1 P1 V 1 Pi 7635 kJ W1 Work 7635 kJ Ans. W2 Work 13200 kJ Ans. W1 = 0 Step 2: cool at const P2 to V2 W2 P2 V 2 V1 Work (e) Step 1: cool at const P1 to V2 W1 P1 V 2 W1 V1 25 1100 kJ Step 2: heat at const V2 to P2 Work 3.17(a) W2 = 0 Work W1 No work is done; no heat is transferred. t U= (b) T=0 Not reversible T2 = T1 = 100 degC The gas is returned to its initial state by isothermal compression. Work = n R T ln 3 V1 3.18 (a) P1 P2 V2 ln 7 2 but V2 4 3 3 P2 V1 6 bar 878.9 kJ Ans. Work V2 T1 500 kPa 303.15 K CP 5 R 2 CV R n R T = P2 V 2 m P2 100 kPa CP V1 V2 4m Work CV Adiabatic compression from point 1 to point 2: Q12 0 kJ mol 1 U12 = W12 = CV T12 U12 CV T2 U12 3.679 T1 kJ mol T2 H12 CP T2 W12 H12 5.15 T1 kJ W12 mol T1 3.679 P2 P1 U12 kJ mol Cool at P2 from point 2 to point 3: T3 Ans. 1100 kJ H23 T1 U23 CV T3 CP T3 W23 T2 26 Q23 T2 U23 Q23 H23 Ans. H23 5.15 Q23 5.15 kJ U23 mol kJ W23 mol kJ mol 3.679 Ans. kJ 1.471 Ans. mol Isothermal expansion from point 3 to point 1: U31 = H31 = 0 Q31 4.056 W31 P2 P1 R T3 ln P3 W31 W31 P3 kJ mol FOR THE CYCLE: Q Q Q12 1.094 Q23 Q31 U= 4.056 kJ mol Ans. H=0 Work kJ mol W12 Work Q31 1.094 W23 W31 kJ mol (b) If each step that is 80% efficient accomplishes the same change of state, all property values are unchanged, and the delta H and delta U values are the same as in part (a). However, the Q and W values change. Step 12: W12 Q12 Step 23: W23 Q23 Step 31: W31 Q31 W12 W12 W23 0.8 U23 4.598 Q12 0.92 kJ mol W23 0.8 U12 kJ W12 1.839 kJ mol mol W31 0.8 W31 Q23 5.518 kJ mol W31 W23 3.245 kJ mol Q31 27 3.245 kJ mol FOR THE CYCLE: Q Q12 Q Q23 3.192 Work kJ mol W12 Work Q31 3.192 W23 W31 kJ mol 3.19Here, V represents total volume. P1 CP 21 CV joule mol K CP T2 208.96 K P1 V 1 n P1 V 1 T2 Work n CV P2 P1 V2 Ans. T1 P2 V 2 P1 V 1 69.65 kPa Ans. Work Work = Work Pext V2 994.4 kJ Ans, U = Pext V V1 Work U = n CV T R T1 V1 Ans. 1609 kJ V2 P2 P1 200 kPa T2 1 100 kPa V2 V1 P1 (c) Restrained adiabatic: Pext P2 Work P2 P2 V 2 V1 P2 600 K T2 Work CV V2 (b) Adiabatic: 600 K CP V1 P 1 V 1 ln T1 5 V1 R Work = n R T1 ln T1 Work V2 1m (a) Isothermal: T2 3 V1 1000 kPa T2 V 1 T2 V 2 T1 28 442.71 K Ans. P2 T1 147.57 kPa Ans. 400 kJ Ans. 3.20 T1 CP P1 423.15 K 7 2 Step 12: If V1 V2 = 2.502 Step 23: V1 V3 kJ mol 0 Process: 2.079 U12 mol mol R T1 ln r () Q12 2.502 CV T3 CP T3 T2 2.079 W23 kJ mol H23 Work Q Q23 2.502 0.424 2.91 kJ mol kJ mol kJ mol H H12 H23 H 2.91 U U12 U23 U 2.079 29 kJ mol T2 H23 W12 Q12 kJ U23 U23 Q 323.15 K W12 W12 U23 Work 0 T 3 P1 kJ mol kJ mol T3 T 1 P3 r Q12 W23 T1 kJ Then Q23 Q23 0 3 bar T2 R 2 H12 r= W12 5 CV R P3 8bar kJ mol kJ mol Ans. Ans. Ans. Ans. 3.21 By Eq. (2.32a), unit-mass basis: But CP u1 2.5 t2 3.22 28 Whence H = CP T R 7 2 molwt molwt u2 1 2 H u2 T= m s u2 t1 gm mol 2 2 u1 2 CP 50 m s t1 u1 t2 2 CP 7 R 2 CV 5 R 2 P1 1 bar P3 148.8 degC Ans. 10 bar CV T3 U 2.079 T1 mol CP T3 H Ans. T3 303.15 K H T1 kJ 2.91 T1 kJ mol Each part consists of two steps, 12 & 23. (a) T2 W23 T3 R T 2 ln P2 P3 P2 P1 150 degC 2 2 CP U 2 u =0 T2 T1 Work W23 Work 6.762 Q U Q 4.684 30 kJ mol Ans. Work kJ mol Ans. Ans. 403.15 K (b) P2 T2 P1 H12 CP T2 W12 U12 W23 R T 2 ln Work W12 Q (c) U T2 U12 T3 CV T2 Q12 Q12 P3 H12 W12 T1 0.831 W23 P2 Work W23 Q Work P2 T1 P3 T1 W12 kJ mol 7.718 6.886 4.808 kJ mol kJ mol kJ mol CP T3 T2 Q23 CV T3 T2 W23 U23 Work 4.972 P1 H23 U23 Ans. P2 R T 1 ln H23 Ans. Work Q W12 U W23 Q Work 2.894 Q23 kJ mol kJ mol For the second set of heat-capacity values, answers are (kJ/mol): U = 2.079 U = 1.247 (a) Work = 6.762 Q = 5.515 (b) Work = 6.886 Q = 5.639 (c) Work = 4.972 Q = 3.725 31 Ans. Ans. 3.23 T1 303.15 K T2 T1 T3 393.15 K P1 1 bar P3 12 bar CP 7 R 2 For the process: U U Step 12: W12 CV T3 1.871 P2 5.608 kJ mol mol For the process: Work W12 Q 3.24 Work 5.608 W12 = 0 Work = W23 = P2 V3 But T3 = T1 Also W = R T 1 ln ln P T2 T1 P = P1 T1 P1 exp T2 T2 R T1 ln 5.608 P P1 Ans. P2 U P1 kJ mol W23 kJ Q mol 3.737 V 2 = R T3 kJ mol Ans. T2 Work = R T2 So... mol Q23 kJ mol W23 Q23 kJ Q12 W12 Step 23: Q12 2.619 T1 W12 T3 0 CP T3 H kJ T1 P3 Q12 H T1 5 R 2 CV T1 Therefore T1 T1 32 T1 800 K P 350 K 2.279 bar P1 Ans. 4 bar 3.25 3 VA P Define: 256 cm P1 r 0.0639 VB 3750.3 cm CV =r CP Assume ideal gas; let V represent total volume: P1 V B = P 2 V A P P1 3.26 T1 = VA VA VA ( 1) r VB VB 300 K From this one finds: VB P1 r 1 atm CP 7 R 2 3 Ans. CP R CV The process occurring in section B is a reversible, adiabatic compression. Let TA ( )= TA final P ( )= P2 final TB ( )= TB final Since the total volume is constant, nA = nB nA R TA 2 nA R T1 = P2 P1 TB TA TB 2 T1 = P2 P1 or (1) 1 (a) P2 TA 2 T1 TB 1.25 atm P2 P1 Q = nA TB Define q= TB TA 319.75 K T1 Q nA q 430.25 K 33 P2 (2) P1 UA CV TA q UB TB 2 T1 3.118 kJ mol (3) Ans. (b) Combine Eqs. (1) & (2) to eliminate the ratio of pressures: (guess) 425 K TB 300 K TB Find TB TB TA 319.02 K Ans. P2 1.24 atm Ans. kJ mol Ans. 1 TB = T1 Given P2 P1 q (c) TA P1 TA 2 T1 P2 P2 (1) TB P1 kJ mol 2.993 1 T1 CV TA 3 q 2 T1 By Eq. (2), TB Eliminate q (1) TB TB P2 (d) 2 T1 TB 325 K q TB 2 T1 CV TA TB TA Ans. TA 469 K Ans. q 2 T1 TA 1.323 atm TB q P1 2 T1 C V 4.032 kJ mol Ans. from Eqs. (1) & (3): P2 1.241 atm Ans. (2) P2 TB 319.06 K Ans. (1) TA 425.28 K Ans. P1 1 TB T1 TA 2 T1 P2 P1 P2 P1 TB 34 6 3 3.30 B P1 B' cm 242.5 C 25200 mol mol P2 7.817 T B C 10 31 bar B 2 373.15 K 2 55 bar B' 1 bar RT C' cm RT 2 C' 2 3.492 10 51 2 bar (a) Solve virial eqn. for initial V. RT P1 Guess: V1 Given P1 V 1 =1 RT B V1 3 C V1 2 V1 V2 Find V1 cm 30780 mol cm 241.33 mol V1 Solve virial eqn. for final V. Guess: RT P2 V2 P2 V 2 Given RT =1 B V2 3 C V2 2 V2 Find V2 Eliminate P from Eq. (1.3) by the virial equation: V2 Work RT 1 B V C V 2 1 dV V Work 12.62 kJ mol V1 (b) Eliminate dV from Eq. (1.3) by the virial equation in P: P2 1 dV = R T P C' dP W 1 P RT 2 P1 W 35 12.596 kJ mol Ans. C' P dP Ans. Note: The answers to (a) & (b) differ because the relations between the two sets of parameters are exact only for infinite series. 3.32 Tc 282.3 K T 298.15 K Pc 50.4 bar P 12 bar T Tc Tr Pr Tr Pr P Pc 6 3 B Given cm 140 C mol PV =1 RT 7200 cm mol B V V (b) B0 V Find ( ) V Tr V Tr Z 1 (c) B0 cm 1919 mol B1 B1 4.2 Tr Z PV RT Z B0 1.6 Pr 3 V 2066 cm mol 2 0.172 0.139 RT P C 0.422 0.083 V 2 3 B1 0.238 (guess) 0.087 (a) 1.056 Ans. Z 0.929 V cm Ans. 1924 mol 0.304 2.262 0.932 10 V 3 3 ZRT P For Redlich/Kwong EOS: 0 1 ( r) T T r Pr Tr 0.5 Pr Tr Table 3.1 Eq. (3.53) 36 Table 3.1 0.42748 0.08664 q Tr Tr Tr Eq. (3.54) Calculate Z Guess: Given T r Pr 0.9 Eq. (3.52) Z=1 Z Z Z Find Z) ( (d) q Tr Z T r Pr Z T r Pr Z T r Pr 3 Z RT V 0.928 T r Pr V P 1916.5 cm mol Ans. For SRK EOS: 2 1 Tr 1 0.480 Tr q Tr (e) 1 Tr Guess: Z Table 3.1 2 Pr T r Pr Tr T r Pr q Tr Z Find Z) ( 0.9 Z T r Pr Z V 0.928 T r Pr T r Pr Z T r Pr 3 Z RT V P 1918 1 2 0.45724 0.07779 2 1 q Tr cm mol Ans. For Peng/Robinson EOS: 1 Tr Eq. (3.53) Tr Eq. (3.52) Given Z 0.176 Eq. (3.54) Calculate Z Z=1 1.574 2 Table 3.1 0.42748 0.08664 0 1 1 0.37464 Tr 1.54226 0.26992 Eq. (3.54) Tr 37 2 T r Pr 1 Tr Pr Tr 2 Table 3.1 2 Table 3.1 Eq. (3.53) Calculate Z Guess: 0.9 Eq. (3.52) Given Z=1 T r Pr Z Z Find ( Z) q Tr Z Z T r Pr Z T r Pr 3 V 1900.6 cm mol 305.3 K Pc T 323.15 K Tr T Tc Tr P 15 bar Pr P Pc Pr 0.308 (guess) 0.100 6 3 (a) cm 156.7 mol B Given PV =1 RT B V C 9650 cm mol V (b) B0 V Find ( V) Tr B1 V Tr (c) B0 cm 1625 mol B1 B1 4.2 Tr Z PV Z B0 1.6 Pr 3 V cm 1791 mol 2 0.172 0.139 RT P C 0.422 0.083 V 2 3 1 Ans. 1.058 48.72 bar 3.33 Tc Z Z T r Pr ZRT P V 0.92 T r Pr Ans. Z RT 0.907 V cm Ans. 1634 mol 0.302 3.517 0.912 V 10 3 ZRT P 3 For Redlich/Kwong EOS: 1 0 0.08664 38 0.42748 Table 3.1 ( r) T 0.5 Tr Table 3.1 Pr T r Pr Tr q Tr Eq. (3.54) Tr Eq. (3.53) Tr Calculate Z Guess: Given T r Pr 0.9 Eq. (3.52) Z=1 Z Z Z Find Z) ( (d) q Tr Z T r Pr Z Z T r Pr cm 1622.7 mol V P Ans. For SRK EOS: 0.42748 0.08664 0 1 Tr 1 0.480 Tr q Tr 0.176 Eq. (3.54) Guess: 2 1 Tr T r Pr Tr Calculate Z Z Table 3.1 2 Pr Eq. (3.53) Tr 0.9 Eq. (3.52) Given Z=1 Z 1.574 Table 3.1 2 1 (e) T r Pr 3 Z RT V 0.906 T r Pr T r Pr Find Z) ( q Tr Z T r Pr Z Z V 0.907 T r Pr Z RT P T r Pr Z T r Pr 3 V 1624.8 cm mol Ans. For Peng/Robinson EOS: 1 2 1 0.07779 2 39 0.45724 Table 3.1 1 Tr 1 0.37464 Tr q Tr 1.54226 0.26992 Eq. (3.54) 1 Guess: Z Tr Table 3.1 2 Pr T r Pr Tr Calculate Z 2 2 Eq. (3.53) Tr 0.9 Eq. (3.52) Given Z=1 T r Pr q Tr Z Z T r Pr Z T r Pr Z T r Pr 3 ZRT V 0.896 T r Pr cm 1605.5 mol V Z Find ( Z) 3.34 Tc 318.7 K T 348.15 K Tr T Tc Tr Pc 37.6 bar P 15 bar Pr P Pc Pr Ans. 1.092 0.399 P 0.286 (guess) 6 3 (a) B Given 194 cm C mol PV =1 RT 15300 cm mol B V V cm 1722 mol Find ( V) (b) B0 0.083 V 1930 mol 2 3 V 3 cm C V V 2 RT P 0.422 Tr 1.6 B0 40 Z 0.283 PV RT Z 0.893 Ans. B1 0.172 0.139 Tr Z 1 B0 (c) B1 4.2 Pr B1 Z Tr 0.02 V 0.899 3 Z RT V P 1734 cm mol Ans. For Redlich/Kwong EOS: ( r) T Tr 0.5 Table 3.1 Pr T r Pr Table 3.1 0.42748 0.08664 0 1 Tr q Tr Eq. (3.54) Tr Eq. (3.53) Tr Calculate Z Guess: Given T r Pr 0.9 Eq. (3.52) Z=1 Z Z Find Z) ( (d) q Tr Z Z T r Pr Z T r Pr Z T r Pr 3 Z RT V 0.888 T r Pr cm 1714.1 mol V P Ans. For SRK EOS: 0.42748 0.08664 0 1 1 Tr q Tr 1 0.480 Tr 1.574 0.176 Eq. (3.54) Tr 41 2 1 T r Pr Tr Table 3.1 2 Table 3.1 2 Pr Tr Eq. (3.53) Calculate Z Guess: Eq. (3.52) Given Z=1 T r Pr Z (e) q Tr Z Find ( Z) Z 0.9 Z T r Pr Z Z T r Pr T r Pr 3 ZRT P V 0.895 T r Pr V 1726.9 Ans. mol For Peng/Robinson EOS: 1 1 2 0.45724 0.07779 2 1 Tr 1 0.37464 Tr q Tr T r Pr Guess: Z T r Pr q Tr Z Find ( Z) Tr 2 Table 3.1 2 Pr Eq. (3.53) Tr 0.9 523.15 K Z T r Pr Z ZRT P V 0.882 P T r Pr cm 152.5 mol B Given Z PV RT T r Pr Z T r Pr 3 cm 1701.5 mol V Ans. 1800 kPa 6 3 (a) 1 Table 3.1 Eq. (3.52) Z=1 T 2 0.26992 Tr Given Z 1.54226 Eq. (3.54) Calculate Z 3.35 cm C mol C B V PV =1 RT V 2 2 V RT (guess) P V 5800 cm Find ( V) 3 V 2250 42 cm mol Z 0.931 Ans. (b) Tc Tr Tr B1 647.1 K Pc T Tc Pr 0.139 0.172 Tr (c) Table F.2: B0 Pc 4.2 0.422 0.083 Tr B0 0.082 B1 0.51 Z 0.281 1.6 1 B0 B1 Pr Tr 3 Z RT P V P Pr 0.808 0.345 220.55 bar Z molwt V V cm molwt 124.99 gm V 0.939 cm 2268 mol cm 2252 mol Ans. 3 gm 18.015 mol 3 or cm 53.4 mol B T Zi Z1i C 2620 cm mol 2 D 5000 cm mol 3 n mol 273.15 K PV Given i 9 6 3 3.37 Ans. RT 0 10 Pi fPi Vi Pi RT 1 B Pi RT =1 10 B V 10 C V 2 D V fP V) ( 3 RT Pi Vi 20 i bar Find V) ( (guess) Eq. (3.12) Eq. (3.38) 43 Z2i 1 2 1 4 B Pi RT Eq. (3.39) 1·10 -10 Z2i Z1i Zi 20 1 1 1 40 0.953 0.953 0.951 60 0.906 0.906 0.895 0.861 0.859 0.83 0.749 80 Pi 100 bar 0.819 0.812 120 0.784 0.765 0.622 140 0.757 0.718 0.5+0.179i 160 0.74 0.671 0.5+0.281i 180 0.733 0.624 0.5+0.355i 200 0.735 0.577 0.5+0.416i 0.743 0.53 0.5+0.469i Note that values of Z from Eq. (3.39) are not physically meaningful for pressures above 100 bar. 1 0.9 Zi 0.8 Z1 i Z2 i 0.7 0.6 0.5 0 50 100 Pi bar 44 150 1 200 3.38 (a) Propane: Tc Tc 313.15 K P Tr T Pc T Tr 369.8 K 0.847 Pr 0.152 42.48 bar 13.71 bar P Pc Pr 0.323 For Redlich/Kwong EOS: ( r) T Tr 0.5 Table 3.1 Pr T r Pr Table 3.1 0.42748 0.08664 0 1 Tr q Tr Eq. (3.54) Tr Eq. (3.53) Tr Calculate Z for liquid by Eq. (3.56) Guess: Z 0.01 Given Z= Z T r Pr Z Find Z) Z ( T r Pr 0.057 Z Calculate Z for vapor by Eq. (3.52) Z T r Pr q Tr T r Pr 3 Z RT P V 1 T r Pr V Guess: 108.1 cm Ans. mol Z 0.9 V cm 1499.2 mol Given Z=1 Z T r Pr Find Z) ( q Tr Z T r Pr Z ZZ V 0.789 45 T r Pr T r Pr Z RT P 3 Ans. Rackett equation for saturated liquid: T Tc Tr Tr 0.847 3 Vc V 200.0 V c Zc cm Zc mol 1 Tr 0.276 3 0.2857 V 94.17 cm Ans. mol For saturated vapor, use Pitzer correlation: B0 0.083 0.422 Tr B1 0.139 V 0.468 B1 0.207 0.172 Tr RT P B0 1.6 4.2 R B0 B1 Tc V Pc 46 1.538 3 3 cm 10 mol Ans. Parts (b) through (t) are worked exactly the same way. All results are summarized as follows. Volume units are cu.cm./mole. R/K, Liq. R/K, Vap. Rackett Pitzer (a) 108.1 1499.2 94.2 1537.8 (b) 114.5 1174.7 98.1 1228.7 (c) 122.7 920.3 102.8 990.4 (d) 133.6 717.0 109.0 805.0 (e) 148.9 1516.2 125.4 1577.0 (f) 158.3 1216.1 130.7 1296.8 (g) 170.4 971.1 137.4 1074.0 (h) 187.1 768.8 146.4 896.0 (i) 153.2 1330.3 133.9 1405.7 (j) 164.2 1057.9 140.3 1154.3 (k) 179.1 835.3 148.6 955.4 (l) 201.4 645.8 160.6 795.8 (m) 61.7 1252.5 53.5 1276.9 (n) 64.1 1006.9 55.1 1038.5 (o) 66.9 814.5 57.0 853.4 (p) 70.3 661.2 59.1 707.8 (q) 64.4 1318.7 54.6 1319.0 (r) 67.4 1046.6 56.3 1057.2 (s) 70.8 835.6 58.3 856.4 (t) 74.8 669.5 60.6 700.5 47 3.39 (a) Propane T Tc 273.15)K T Tr ( 40 T Tc Tr Pc 369.8 K Pr 0.847 0.152 42.48 bar P P Pc 13.71 bar Pr 313.15 K 0.323 From Table 3.1 for SRK: 0.42748 0.08664 0 1 2 1 Tr 1 0.480 Tr q Tr 1.574 0.176 Eq. (3.54) 2 1 2 Tr Pr T r Pr Tr Calculate Z for liquid by Eq. (3.56) Guess: Eq. (3.53) Tr Z 0.01 Given Z= Z T r Pr Find Z) ( Z Z T r Pr 0.055 Z Calculate Z for vapor by Eq. (3.52) Z T r Pr q Tr T r Pr 3 Z RT P V 1 T r Pr cm 104.7 mol V Guess: Z Ans. 0.9 Given Z=1 Z T r Pr Find Z) ( q Tr Z 0.78 Z T r Pr Z V 48 T r Pr Z T r Pr Z RT P T r Pr 3 V 1480.7 cm mol Ans. Parts (b) through (t) are worked exactly the same way. All results are summarized as follows. Volume units are cu.cm./mole. SRK, Liq. SRK, Vap. Rackett Pitzer (a) 104.7 1480.7 94.2 1537.8 (b) 110.6 1157.8 98.1 1228.7 (c) 118.2 904.9 102.8 990.4 (d) 128.5 703.3 109.0 805.0 (e) 142.1 1487.1 125.4 1577.0 (f) 150.7 1189.9 130.7 1296.8 (g) 161.8 947.8 137.4 1074.0 (h) 177.1 747.8 146.4 896.0 (i) 146.7 1305.3 133.9 1405.7 (j) 156.9 1035.2 140.3 1154.3 (k) 170.7 815.1 148.6 955.4 (l) 191.3 628.5 160.6 795.8 (m) 61.2 1248.9 53.5 1276.9 (n) 63.5 1003.2 55.1 1038.5 (o) 66.3 810.7 57.0 853.4 (p) 69.5 657.4 59.1 707.8 (q) 61.4 1296.8 54.6 1319.0 (r) 63.9 1026.3 56.3 1057.2 (s) 66.9 817.0 58.3 856.4 (t) 70.5 652.5 60.6 700.5 49 3.40 (a) Propane T ( 40 Tr Tc T Tc Pc 369.8 K T 273.15)K Tr P P Pc 13.71 bar Pr 313.15 K Pr 0.847 0.152 42.48 bar 0.323 From Table 3.1 for PR: 2 1 Tr 1 1 0.37464 1 2 0.26992 2 Eq. (3.54) 1 2 Pr T r Pr Tr Tr 0.45724 0.07779 2 Tr q Tr 1.54226 Calculate Z for liquid by Eq. (3.56) Guess: Eq. (3.53) Tr Z 0.01 Given Z= Z T r Pr Find Z) ( Z Z T r Pr 0.049 Z Calculate Z for vapor by Eq. (3.52) Z T r Pr q Tr T r Pr 3 Z RT P V 1 T r Pr cm 92.2 mol V Guess: Z Ans. 0.6 Given Z=1 Z T r Pr Find Z) ( q Tr Z 0.766 Z T r Pr Z V 50 T r Pr Z T r Pr Z RT P T r Pr 3 V 1454.5 cm mol Ans. Parts (b) through (t) are worked exactly the same way. All results are summarized as follows. Volume units are cu.cm./mole. PR, Liq. PR, Vap. Rackett Pitzer (a) 92.2 1454.5 94.2 1537.8 (b) 97.6 1131.8 98.1 1228.7 (c) 104.4 879.2 102.8 990.4 (d) 113.7 678.1 109.0 805.0 (e) 125.2 1453.5 125.4 1577.0 (f) 132.9 1156.3 130.7 1296.8 (g) 143.0 915.0 137.4 1074.0 (h) 157.1 715.8 146.4 896.0 (i) 129.4 1271.9 133.9 1405.7 (j) 138.6 1002.3 140.3 1154.3 (k) 151.2 782.8 148.6 955.4 (l) 170.2 597.3 160.6 795.8 (m) 54.0 1233.0 (n) 56.0 987.3 55.1 1038.5 (o) 58.4 794.8 57.0 853.4 (p) 61.4 641.6 59.1 707.8 (q) 54.1 1280.2 54.6 1319.0 (r) 56.3 1009.7 56.3 1057.2 (s) 58.9 800.5 58.3 856.4 (t) 62.2 636.1 60.6 700.5 53.5 1276.9 51 3.41 (a) For ethylene, molwt 28.054 gm Tc mol 282.3 K Pc 50.40 bar 0.087 T Tr Tc 328.15 K P 35 bar P Pc Pr T Tr 1.162 Pr 0.694 From Tables E.1 & E.2: Z Z0 Z1 Z Z0 Z1 0.838 0.033 0.841 Vtotal ZnRT P Vtotal 0.421 m P 115 bar Vtotal 0.25 m Tr 1.145 Pr P Pc From Tables E.3 & E.4: Z0 0.482 Z1 0.126 18 kg n molwt (b) T 323.15 K T Tc Tr Z Z0 Z Z1 mass Pr ZRT mass n molwt Ans. 3 P Vtotal n 0.493 3 n 2.282 2171 mol Ans. 60.898 kg 3.42 Assume validity of Eq. (3.38). 3 P1 1bar Z1 P1 V 1 R T1 T1 Z1 300K 0.922 cm 23000 mol V1 R T1 Z1 P1 B With this B, recalculate at P2 Z2 1 B P2 R T1 Z2 0.611 P2 V2 52 R T1 Z2 P2 B 1 3 3 cm 1.942 10 mol 5bar V2 3.046 10 3 3 cm mol Ans. 3.43 T 753.15 K Tc 513.9 K Tr T Tc Tr 1.466 P 6000 kPa Pc 61.48 bar Pr P Pc Pr 0.976 0.645 B0 0.083 0.422 Tr B1 0.172 0.139 Tr RT P V B0 For an ideal gas: 3.44 T 320 K 0.104 3 V cm 989 mol V Pc cm 1044 mol Ans. 3 RT P V P 0.146 B1 4.2 Tc B1 R B0 1.6 Tc 369.8 K Pc Zc 16 bar 42.48 bar 0.276 molwt 3 Vc 0.152 Tr Vliq Vtank cm 200 mol T Tc Tr V c Zc 1 Tr 3 B0 0.083 Tr B1 0.139 1.6 0.172 Tr 4.2 Pr gm mol 0.377 3 cm Vliq 0.8 Vtank Vliq molwt 0.422 P Pc 0.2857 mliq 0.35 m Pr 0.865 44.097 B0 0.449 B1 0.177 53 96.769 mliq 127.594 kg mol Ans. RT P Vvap B0 B1 R Tc 3 3 cm Vvap 1.318 mvap Pc 10 2.341 kg mol 0.2 Vtank mvap Vvap Ans. molwt 3.45 T B0 Tc 425.1 K 2.43 bar Pc 37.96 bar Pr Vvap 0.083 0.422 Tr B1 0.139 RT P 1.6 0.172 Tr V T Tr 0.200 P 298.15 K 4.2 B0 3 gm mol 0.624 Tc B1 R V Pc Vvap mvap 0.064 0.661 B1 58.123 0.701 Pr Pc molwt 16 m B0 Tr Tc P mvap V molwt 9.469 98.213 kg 3 3 cm 10 mol Ans. P 333.15 K Tc 305.3 K Tr T Tc Tr 1.091 14000 kPa Pc 48.72 bar Pr P Pc Pr 2.874 0.100 3.46 (a) T Vtotal From tables E.3 & E.4: Z0 3 molwt 0.15 m 0.463 54 Z1 0.037 30.07 gm mol Z Z0 Vtotal methane (b) Z Z1 methane V molwt Vtotal V P 40 kg or Tr = Whence V 0.459 Z Tr = PV 29.548 R Tc at 90.87 mol P V = Z R T = Z R Tr Tc 20000 kPa 0.889 Z V P cm Ans. 49.64 kg where 3 Z RT Pr P Pc Pr mol kg 4.105 This equation giving Tr as a function of Z and Eq. (3.57) in conjunction with Tables E.3 & E.4 are two relations in the same variables which must be satisfied at the given reduced pressure. The intersection of these two relations can be found by one means or another to occur at about: Tr Whence T V or T or 3 0.15 m Vtotal T 298.15 K molwt 0.087 50.40 bar P V = P r Pc V = Z R T 40 kg molwt Whence 0.693 118.5 degC Ans. Pc 282.3 K Pr = Z Tr Tc 391.7 K 3.47 Vtotal Tc and 1.283 Z RT Pc V where Pr = 4.675 Z at Tr 55 T Tc 4.675 Tr 1.056 28.054 gm mol This equation giving Pr as a function of Z and Eq. (3.57) in conjunction with Tables E.3 & E.4 are two relations in the same variables which must be satisfied at the given reduced temperature. The intersection of these two relations can be found by one means or another to occur at about: Pr and 1.582 3.48 mwater Z P Pc Pr 3 Vtotal 15 kg P 0.338 V 0.4 m Interpolate in Table F.2 at 400 degC to find: 298.15 K Tc 305.3 K Tr1 P1 2200 kPa Pc 48.72 bar Pr1 3 T2 Z0 mwater T1 Tc P1 Pc 3 V 26.667 Z Z1 Z1 0.422 Tr2 B1 0.139 T2 Tc Z R T1 P1 Tr2 0.806 B0 1.6 0.172 Tr2 Tr1 0.977 Pr1 0.452 1.615 3 V1 B1 4.2 0.113 0.116 R T2 P2 V1 B0 gm 0.0479 V1 .8105 Tr2 493.15 K 0.083 cm Ans. Assume Eq. (3.38) applies at the final state. B0 Ans. 0.100 0.35 m From Tables E.1 & E.2: Z0 Z Vtotal P = 9920 kPa 3.49 T1 Vtotal 79.73 bar B1 R P2 Tc Pc 56 42.68 bar Ans. cm 908 mol 3.50 T B0 0.5 m 0.083 0.139 73.83 bar 0.422 Tr B1 304.2 K Pc 3 Vtotal B0 4.2 V 10 kg molwt 0.224 0.997 molwt 44.01 0.036 Vtotal V Tr Tc 0.341 B1 1.6 0.172 Tr T Tr Tc 303.15 K 3 3 cm 2.2 10 mol RT P V B0 B1 R P Tc 10.863 bar Ans. Pc 3.51 Basis: 1 mole of LIQUID nitrogen P B0 Tc 126.2 K Tr 1 atm Pc 34.0 bar Pr 0.038 Tn molwt 77.3 K 0.083 0.139 1 B0 4.2 B1 Pr Tr 0.842 B1 0.172 Tr Z mol B0 1.6 1.209 Z 0.957 57 Tr P Pc Vliq 0.613 Pr Tc gm 0.422 Tr B1 28.014 Tn 0.03 3 34.7 cm gm mol P Vliq nvapor nvapor Z R Tn 5.718 10 3 mol Final conditions: ntotal T 1 mol V nvapor RT Pig Pig V V T Tc 69.005 Tr ntotal Tr 298.15 K 3 2 Vliq cm 2.363 mol 359.2 bar Use Redlich/Kwong at so high a P. 2 2 a Tr R Tc Pc a 3 3 bar cm 0.901 m 2 b Eq. (3.42) RT Vb Tr .5 R Tc Tr 0.651 Eq. (3.43) Pc 3 b mol P ( r) T 0.42748 0.08664 a V ( b) V Eq. (3.44) cm 26.737 mol P 450.1 bar Ans. 3 3.52 For isobutane: T1 Tr1 Tr1 300 K T1 Tc 0.735 Tc 408.1 K Pc 36.48 bar V1 1.824 P1 4 bar T2 415 K P2 cm 75 bar Pr1 Pr1 P1 Tr2 Pc Tr2 0.11 58 T2 Tc 1.017 Pr2 Pr2 P2 Pc 2.056 gm From Fig. (3.17): r1 2.45 The final T > Tc, and Fig. 3.16 probably should not be used. One can easily show that r= P Vc with Z from Eq. (3.57) and Tables E.3 and E.4. Thus Z RT 3 Vc Z 262.7 Z0 cm Z0 0.181 mol Z Z1 Eq. (3.75): V1 r2 Tc 3 r1 cm 2.519 gm V2 Pc 469.7 K 1.774 r2 Z R T2 r2 3.53 For n-pentane: 0.0756 P2 V c 0.322 V2 Z1 0.3356 33.7 bar Ans. 0.63 1 gm 3 cm T1 Tr1 Tr1 P1 291.15 K T1 Pr1 Tc T2 1 bar P1 Tr2 Pc T2 0.03 Tr2 2.69 r2 r2 2 2 1 0.532 r1 3.54 For ethanol: Tc Pc Pc 3.561 gm 3 Ans. cm 513.9 K T 453.15 K Tr T Tc Tr 0.882 61.48 bar P 200 bar Pr P Pc Pr 3.253 3 Vc P2 2.27 From Fig. (3.16): By Eq. (3.75), Pr2 0.88 r1 120 bar Pr2 Tc Pr1 0.62 P2 413.15 K 167 cm mol molwt 59 46.069 gm mol From Fig. 3.16: r 2.28 = r r c= 0.629 Vc r Vc gm Ans. 3 cm molwt 3.55 For ammonia: Tc 405.7 K T 293.15 K Tr T Tc Tr 0.723 Pc 112.8 bar P 857 kPa Pr P Pc Pr 0.076 Vc cm 72.5 mol Zc 0.242 3 Eq. (3.72): B0 Vliquid 0.083 0.139 Vvapor cm 27.11 mol Vliquid B0 4.2 B0 B1 R 0.627 B1 1.6 0.172 Tr RT P 3 0.2857 0.422 Tr B1 V c Zc 1 Tr 0.253 0.534 Tc Pc 3 Vvapor cm 2616 mol 3 V Vvapor Vliquid V 60 cm 2589 mol Ans. Alternatively, use Tables E.1 & E.2 to get the vapor volume: Z0 Z1 0.929 Vvapor Z 0.071 Z0 Z Z1 0.911 3 ZRT P Vvapor cm 2591 mol 3 V Vvapor V Vliquid cm 2564 Ans. mol 3.5810 gal. of gasoline is equivalent to 1400 cu ft. of methane at 60 degF and 1 atm. Assume at these conditions that methane is an ideal gas: 3 R ft atm 0.7302 lbmol rankine V 1400 ft T P 1 atm n 3 519.67 rankine PV RT n 3.689 lbmol For methane at 3000 psi and 60 degF: Tc 190.6 1.8 rankine T 519.67 rankine Tr T Tc Tr 1.515 Pc 45.99 bar P 3000 psi Pr P Pc Pr 4.498 Z 0.822 0.012 From Tables E.3 & E.4: Z0 0.819 Vtank ZnRT P Z1 Z 0.234 Vtank 61 5.636 ft Z0 3 Z1 Ans. 3.59 T P 25K 3.213bar Calculate the effective critical parameters for hydrogen by equations (3.58) and (3.56) 43.6 Tc 1 K 1 30.435 K Pc 10.922 bar 2.016T 20.5 Pc Tc bar 21.8K 44.2K 2.016T 0 P Pc Pr Pr Initial guess of volume: T Tr 0.294 Tr cm 646.903 mol 3 RT P V 0.821 V Tc Use the generalized Pitzer correlation B0 0.083 0.422 Tr Z 1 B0 1.6 B0 B1 0.495 B1 0.139 0.172 Tr Pr Tr Z 0.823 Ans. 4.2 B1 0.254 Experimental: Z = 0.7757 For Redlich/Kwong EOS: 0 1 Tr T r Pr Tr 0.5 Pr Tr Table 3.1 Eq. (3.53) 62 Table 3.1 0.42748 0.08664 q Tr Tr Tr Eq. (3.54) Calculate Z Guess: Given T r Pr 0.9 Eq. (3.52) Z=1 Z Z q Tr Z Find Z) ( 3.61For methane: ZZ Tc 0.012 T T r Pr T r Pr Ans. 0.791 At standard condition: Z T r Pr Experimental: Z = 0.7757 Pc 190.6K ( 60 32) 5 9 45.99bar 273.15 K T P Tr T Tc B0 0.083 Z0 Z 1 Tr 0.422 Tr Pr B0 Z0 1.6 Pitzer correlations: T Tc Tr 0.083 0.422 Tr B1 0.139 1.6 0.172 Tr 4.2 0.134 B1 0.139 0.172 T T Tr Z RT P ( 50 32) 5 9 0.109 Tr 0.00158 3 V1 273.15 K m 0.024 mol P 300psi Pr 0.45 283.15 K 1.486 B0 B1 B1 V1 0.998 B1 Pr Z1 0.998 4.2 0.022 Z1 P Pc Tr Z Z1 (a) At actual condition: B0 B0 Z0 Tr Pr 1.515 1atm Pr Pitzer correlations: 288.706 K 0.141 0.106 63 Pr P Pc Z0 Z 1 B0 Z0 Pr Tr Z1 6 ft q1 150 10 (b) n1 (c) D u Z0 0.957 Z Z1 0.958 q1 n1 7.485 q2 V1 q1 V1 22.624in q2 A A u 4 D 8.738 2 m s 64 3 kmol 10 hr A Ans. 0.0322 3 V2 P V2 q2 day Z1 Tr ZRT V2 3 Pr B1 6.915 Ans. 2 0.259 m 0.00109 10 6 ft m mol 3 day Ans. 3.62 0.012 0.286 0.087 0.281 0.1 0.279 0.140 0.289 0.152 0.276 0.181 0.282 0.187 0.271 0.19 0.267 0.191 0.277 0.194 0.275 0.196 0.273 0.2 0.274 0.205 0.273 0.21 0.273 0.21 ZC Use the first 29 components in Table B.1 sorted so that values are in ascending order. This is required for the Mathcad slope and intercept functions. 0.271 m slope b intercept r corr 0.212 0.272 0.218 0.275 0.23 0.272 0.235 0.269 0.252 0.264 0.28 0.265 ZC 0.297 0.256 m 0.301 0.266 0.302 0.266 0.303 0.263 0.31 0.263 0.322 0.26 0.326 0.261 ( 0.091) 0.27 0.262 ZC ZC ZC ( .291) 0 2 ( 0.878) r 0.771 0.29 0.28 b 0.27 0.26 0.25 0 0.1 0.2 The equation of the line is: Zc = 0.291 0.091 65 0.3 0.4 Ans. 5 7 R 2 Cv 298.15K P1 P2 T1 T3 5bar P3 Cp 1bar 5bar 3.65 Cp T1 2 R 1.4 Cv Step 1->2 Adiabatic compression 1 T2 T1 P2 T2 P1 472.216 K U12 Cv T2 T1 U12 3.618 H12 Cp T2 T1 H12 5.065 Q12 W12 0 kJ mol kJ mol kJ mol kJ mol Q12 0 U12 W12 3.618 Ans. Ans. Ans. kJ mol Ans. Step 2->3 Isobaric cooling U23 Cv T3 T2 U23 3.618 H23 Cp T3 T2 H23 5.065 Q23 W23 R T3 Q23 H23 W23 T2 5.065 1.447 kJ mol kJ mol kJ mol kJ mol Ans. Ans. Ans. Ans. Step 3->1 Isothermal expansion U31 Cv T1 T3 U31 0 H31 Cp T1 T3 H31 0 66 kJ mol kJ mol Ans. Ans. Q31 R T 3 ln W31 P1 Q31 Q31 P3 W31 kJ mol kJ 3.99 mol 3.99 Ans. Ans. For the cycle Qcycle Q12 Wcycle Q23 W12 Qcycle Q31 W23 Wcycle W31 1.076 1.076 kJ Ans. mol kJ Ans. mol Now assume that each step is irreversible with efficiency: 80% Step 1->2 Adiabatic compression W12 Q12 W12 U12 W12 Q12 W12 4.522 kJ mol kJ mol 0.904 Ans. Ans. Step 2->3 Isobaric cooling W23 Q23 W23 U23 W23 1.809 kJ mol kJ mol Q23 5.427 W31 W23 3.192 Ans. Ans. Step 3->1 Isothermal expansion W31 Q31 W31 U31 Q31 W31 3.192 kJ mol kJ mol Ans. Ans. For the cycle Qcycle Q12 Wcycle W12 Q23 W23 Qcycle Q31 Wcycle W31 67 kJ Ans. mol kJ 3.14 mol Ans. 3.14 3.67 a) PV data are taken from Table F.2 at pressures above 1atm. 125 150 1757.0 175 P 2109.7 1505.1 200 1316.2 cm 1169.2 gm V kPa 225 250 300 Z 955.45 T ( 300 273.15) K M 1051.6 275 3 18.01 gm mol 875.29 1 VM PVM RT i 07 If a linear equation is fit to the points then the value of B is the y-intercept. Use the Mathcad intercept function to find the y-intercept and hence, the value of B Yi Zi 1 3 Xi B intercept ( Y) X A i slope ( Y) X Ans. 6 5 cm 10 2 i A cm 128.42 mol 1.567 B mol X 0 mol 3 cm 10 5 mol 3 cm 8 10 5 mol 3 cm Below is a plot of the data along with the linear fit and the extrapolation to the y-intercept. 68 115 (Z-1)/p 120 125 130 0 2 10 5 4 10 5 6 10 p 5 8 10 5 (Z-1)/p Linear fit b) Repeat part a) for T = 350 C PV data are taken from Table F.2 at pressures above 1atm. 125 150 1912.2 175 P 2295.6 1638.3 200 1432.8 225 V kPa 1273.1 250 300 Z 1040.7 T ( 350 273.15)K M 1145.2 275 3 cm gm 18.01 gm mol 953.52 1 VM PVM RT i 07 If a linear equation is fit to the points then the value of B is the y-intercept. Use the Mathcad intercept function to find the y-intercept and hence, the value of B Yi Zi 1 3 Xi i B intercept ( X Y) i 69 B 105.899 cm mol Ans. A A slope ( X Y) 6 5 cm 10 2 1.784 mol X 0 mol 3 10 cm 5 mol 3 8 10 5 mol 3 cm cm Below is a plot of the data along with the linear fit and the extrapolation to the y-intercept. 90 (Z-1)/p 95 100 105 110 0 2 10 5 4 10 5 6 10 p 5 8 10 5 (Z-1)/p Linear fit c) Repeat part a) for T = 400 C PV data are taken from Table F.2 at pressures above 1atm. 125 150 2066.9 175 P 2481.2 1771.1 200 1549.2 225 kPa V 1376.6 250 1238.5 275 1125.5 300 1031.4 70 3 cm gm T ( 400 273.15)K M 18.01 gm mol 1 VM PV M RT Z i 07 If a linear equation is fit to the points then the value of B is the y-intercept. Use the Mathcad intercept function to find the y-intercept and hence, the value of B Yi Zi 1 3 Xi B i intercept ( X Y) B i A A slope ( X Y) 89.902 cm mol Ans. 6 5 cm 10 2 2.044 mol X 0 mol 3 cm 10 5 mol 3 8 10 5 mol 3 cm cm Below is a plot of the data along with the linear fit and the extrapolation to the y-intercept. 70 (Z-1)/p 75 80 85 90 0 2 10 5 4 10 p (Z-1)/p Linear fit 71 5 6 10 5 8 10 5 3.70 Create a plot of (Z 1) Z Tr vs Pr Pr Z Tr Data from Appendix E at Tr = 1 0.01 0.9967 0.05 0.9832 0.10 0.9659 Z 0.20 Pr 0.9300 0.40 X 0.7574 0.80 1 0.8509 0.60 Tr 0.6355 Pr (Z Y Z Tr 1) Z Tr Pr Create a linear fit of Y vs X Slope slope ( X Y) Intercept Rsquare Slope intercept ( X Y) 0.033 Intercept corr ( X Y) Rsquare 0.332 0.9965 0.28 0.3 Y Slope X Intercept 0.32 0.34 0 0.2 0.4 0.6 0.8 1 1.2 1.4 X The second virial coefficient (Bhat) is the value when X -> 0 Bhat Intercept By Eqns. (3.65) and (3.66) Bhat B0 0.083 0.422 Tr These values differ by 2%. 72 1.6 B0 0.332 0.339 Ans. Ans. 3.71 Use the SRK equation to calculate Z T T 273.15)Kr Tc 150.9 K T ( 30 Pc 48.98 bar P 300 bar Tr Pr Pc 2.009 Pr Tc P 6.125 0.0 0.42748 Table 3.1 0.08664 0 1 2 1 Tr 1 0.480 Tr q Tr 0.176 Eq. (3.54) 1 Tr Table 3.1 2 Pr T r Pr Tr Calculate Z Guess: Z Eq. (3.53) Tr 0.9 Eq. (3.52) Given Z=1 Z 1.574 2 T r Pr q Tr Z Find Z) ( Z T r Pr Z V 1.025 T r Pr T r Pr Z T r Pr 3 Z RT P cm Ans. 86.1 mol V This volume is within 2.5% of the ideal gas value. 3.72 After the reaction is complete, there will be 5 moles of C 2H2 and 5 moles of Ca(OH)2. First calculate the volume available for the gas. n 5mol V Vt 3 0.4 1800 cm 3 cm 5 mol 33.0 mol Vt n Vt 3 555 cm 3 V 73 111 cm mol Use SRK equation to calculate pressure. T Tc 308.3 K Pc ( 125 61.39 bar T Tc Tr 273.15) K Tr 0.0 0.42748 Table 3.1 0.08664 0 1 2 1 Tr 1 1.574 0.176 2 2 2 Pc Eq. (3.45) 3 3 bar cm 3.995 m 2 RT Vb 3.73 mass b Tc 0.152 ( 10 Vc Pc 369.8K 200.0 197.8 bar Ans. 273.15)K 3 0.276 cm 36.175 mol P b) T 35000kg Eq. (3.46) Pc 3 a V (V R Tc b mol Zc Tr Eq. (3.54) R Tc Tr P 1 Table 3.1 2 Tr a a 0.480 Tr q Tr 1.291 cm mol n M 42.48bar mass M n 44.097 7.937 gm mol 5 10 mol a) Estimate the volume of gas using the truncated virial equation Tr B0 T Tr Tc 0.083 0.422 Tr B0 1.6 0.766 P Eq. (3-65) B1 Pr 1atm 0.139 0.172 Tr B1 0.564 74 0.389 4.2 P Pc Eq. (3-66) Z 1 B0 B1 Pr Z Tr ZnRT Vt 0.981 Vt P 3 This would require a very large tank. If the D tank were spherical the diameter would be: 73 2.379 6 Vt 3 10 m m D 32.565 m b) Calculate the molar volume of the liquid with the Rackett equation(3.72) Vliq V c Zc P 1 0.2857 Vvap 3 Vliq 6.294atm Z 1 Tr P Pc Pr B0 B1 85.444 Pr 0.878 Pr Tr ZRT P Vvap Guess: Vtank Given 90% mol 0.15 Z cm 3 3 cm 3.24 10 mol 90% Vliq n Vtank 10% Vtank =n Vvap Vtank Find Vtank Vtank Vliq 75.133 m This would require a small tank. If the tank D were spherical, the diameter would be: 3 3 6 Vtank D 5.235 m Although the tank is smaller, it would need to accomodate a pressure of 6.294 atm (92.5 psi). Also, refrigeration would be required to liquify the gaseous propane stream. 75 Chapter 4 - Section A - Mathcad Solutions 4.1 (a) T0 T 473.15 K For SO2: A B 5.699 n 1373.15 K 0.801 10 H 47.007 C 5 D 0.0 1.015 10 R ICPH T0 T A B C D H 3 10 mol kJ T 523.15 K For propane:A 28.785 10 H 161.834 3 Ans. 12 mol C 6 8.824 10 D 0 R ICPH T0 T A B C 0.0 H 470.073 kJ n 1473.15 K B 1.213 nH Q (b) T0 Q mol kJ mol n 473.15 K For ethylene: A 2 (guess) Q= nR A T0 3 10 kJ Ans. 800 kJ 3 14.394 10 K B B 2 T0 2 1 n A 1.967 2 T 2.905 533.15 K For 1-butene: 1.424 1.942 6 4.392 10 C K 2 Given Find (b) T0 Q 10 mol nH Q 4.2 (a) T0 Q C 3 T0 3 1 T T0 15 mol Q 31.630 10 K B 76 3 1 Ans. 1374.5 K 2500 kJ 3 C 9.873 10 K 2 6 (guess) 3 Q= nR Given A T0 2 n 500 degF C 3 T0 3 1 T 2.652 Find (c) T0 B 2 T0 2 1 3 T T0 1 Ans. 1413.8 K 6 Q 40 lbmol 10 BTU Values converted to SI units T0 n 533.15K For ethylene: A Q= nR Q 10 mol B 14.394 10 K 1.424 2 (guess) 4 1.814 6 1.055 10 kJ 3 4.392 10 C K Given A T0 1 Find B 2 T0 2 2 T 2.256 C 3 T0 3 1 T T0 3 6 2 1 1202.8 K Ans. T = 1705.4degF 4.3 Assume air at the given conditions an ideal gas. Basis of calculation is 1 second. P T0 1 atm ( T T PV R T0 T0 773.15 K n For air: H 32degF) 273.15K T0 n A T 250 ft V Convert given values to SI units T 3 V 122 degF 932 degF 7.079 m 3 T0 32degF 273.15K 323.15 K 266.985 mol 3.355 B 0.575 10 R ICPH T0 T A B C D 77 3 C 0.0 D 0.016 10 5 kJ 13.707 4.4 molwt Q gm mol 100.1 T0 10000 kg molwt n n For CaCO3: A 323.15 K B 10 mol 2.637 10 H 9.441 3 C D 0.0 3.120 10 5 R ICPH T0 T A B C D H 1153.15 K 3 10 BTU Ans. 4 9.99 12.572 3.469 T mol nH Q H 4J 10 Q mol nH 6 Q Ans. 9.4315 10 kJ 4.7 Let step 12 represent the initial reversible adiabatic expansion, and step 23 the final constant-volume heating. T1 298.15 K T3 298.15 K P1 121.3 kPa P2 101.3 kPa P3 104.0 kPa T2 T3 CP 30 T2 290.41 K Given J mol K T2 = T1 4.9a) Acetone: Tc Hn (guess) 29.10 P2 P2 P3 R CP CP P1 Pc 508.2K kJ 47.01bar Tn Trn mol CP Find CP Tc 56.95 Tn 329.4K Trn J Ans. mol K 0.648 Use Eq. (4.12) to calculate H at Tn ( Hncalc) 1.092 ln Hncalc R Tn Pc 1.013 bar 0.930 Trn 78 Hncalc 30.108 kJ mol Ans. To compare with the value listed in Table B.2, calculate the % error. %error Hncalc Hn %error Hn 3.464 % Values for other components in Table B.2 are given below. Except for acetic acid, acetonitrile. methanol and nitromethane, agreement is within 5% of the reported value. Acetone Acetic Acid Acetonitrile Benz ene iso-Butane n-Butane 1-Butanol Carbon tetrachloride Chlorobenz ene Chloroform Cyclohexane Cyclopentane n-Decane Dichloromethane Diethyl ether Ethanol Ethylbenz ene Ethylene glycol n-Heptane n-Hexane M ethanol M ethyl acetate M ethyl ethyl k etone Nitromethane n-Nonane iso-Octane n-Octane n-Pentane Phenol 1-Propanol 2-Propanol Toluene W ater o-Xylene m-Xylene p-Xylene Hn (k mol) J/ 30.1 40.1 33.0 30.6 21.1 22.5 41.7 29.6 35.5 29.6 29.7 27.2 40.1 27.8 26.6 40.2 35.8 51.5 32.0 29.0 38 .3 30.6 32.0 36.3 37.2 30.7 34.8 25.9 46.6 41.1 39.8 33.4 42.0 36.9 36.5 36.3 79 % error 3.4% 69.4% 9.3% -0.5% -0.7% 0.3% -3.6% -0.8 % 0.8 % 1.1% -0.9% -0.2% 3.6% -1.0% 0.3% 4.3% 0.7% 1.5% 0.7% 0.5% 8 .7% 1.1% 2.3% 6.7% 0.8 % -0.2% 1.2% 0.3% 1.0% -0.9% -0.1% 0.8 % 3.3% 1.9% 2.3% 1.6% b) 33.70 469.7 Tc 507.6 562.2 30.25 Pc K 48.98 36.0 68.7 H25 273.15 K 80.0 366.1 72.150 J M 433.3 gm 392.5 Tn Tr2 Tc ( 25 273.15) K Tc H2 H25 M 0.673 H2 0.628 H1 31.533 Hn 31.549 kJ 33.847 mol 32.242 1 Tr2 1 0.38 Tr1 Eq. (4.13) %error 26.448 H2calc H1 26.429 0.631 H2calc 86.177 gm 78.114 mol 82.145 0.658 Tr1 kJ 30.72 mol 29.97 366.3 80.7 Tr1 28.85 Hn bar 43.50 560.4 Tn 25.79 kJ Ans. 33.571 mol 32.816 31.549 kJ 33.847 mol 32.242 H2 H2 0.072 26.429 H2 H2calc %error 0.052 0.814 % 1.781 The values calculated with Eq. (4.13) are within 2% of the handbook values. 4.10 The ln P vs. 1/T relation over a short range is very nearly linear. Our procedure is therefore to take 5 points, including the point at the temperature of interest and two points on either side, and to do a linear least-squares fit, from which the required derivative in Eq. (4.11) can be found. Temperatures are in rankines, pressures in psia, volumes in cu ft/lbm, and enthalpies in Btu/lbm. The molar mass M of tetrafluoroethane is 102.04. The factor 5.4039 converts energy units from (psia)(cu ft) to Btu. 80 (a) T 459.67 V 5 1.934 18.787 P 0 23.767 t 5 26.617 dPdT T H 2 ti 1 459.67 yi ln Pi 15 slope ( x y) slope ( P) 3 xi 10 29.726 slope 15 5 21.162 Data: i 0.012 4952 slopedPdT T V dPdT 5.4039 H 0.545 90.078 Ans. The remaining parts of the problem are worked in exactly the same way. All answers are as follows, with the Table 9.1 value in ( ): (a) H = 90.078 ( 90.111) (b) H = 85.817 ( 85.834) (c) H = 81.034 ( 81.136) (d) H = 76.007 ( 75.902) (e) H = 69.863 ( 69.969) 119.377 4.11 M 32.042 153.822 H is the value at 0 degC. 536.4 mol Tc 334.3 512.6 K Pc 80.97 bar Tn 337.9 K 556.4 gm 54.72 45.60 349.8 Hexp is the given value at the normal boiling point. 81 Tr1 273.15K Tc Tr2 Tn Tc 270.9 H 1189.5 0.509 246.9 J gm Hexp 217.8 J gm 1099.5 Hn Hn Hexp Hexp Hn PCE 0.38 Tr1 0.52 Hexp 1195.3 1.092 ln R Tn Hn Hexp 4.03 % M Pc 1.013 bar 0.930 Tr2 100% 0.34 247.7 Hn 1 H Tr2 0.77 J 1055.2 gm 193.2 (b) By Eq. (4.12): PCE 1 0.629 This is the % error 100% 245 Hn 0.659 Tr2 0.491 194.2 (a) By Eq. (4.13) PCE 0.533 Tr1 0.623 J gm PCE 8.72 % 0.96 192.3 4.12 Acetone 0.307 Tc 508.2K Pc Tn 329.4K P 1atm Pr P Pc 47.01bar Zc 0.233 3 Vc Tr cm 209 mol Tn Tc Tr 0.648 82 Hn 29.1 Pr kJ mol 0.022 Generalized Correlations to estimate volumes Vapor Volume B0 0.422 0.083 Tr B1 Tr 4.2 Pr Z 1 V B1 Pr Z R Tn P Tr Tr 0.762 Eq. (3.65) B1 0.172 0.139 B0 B0 1.6 0.924 Eq. (3.66) Z V (Pg. 102) 0.965 3 4 cm 2.609 10 mol Liquid Volume 2 Vsat V c Zc 1 Tr 3 7 Eq. (3.72) Vsat Combining the Clapyeron equation (4.11) gives A Vsat 14.3145 V B 2.602 A B H=T V V 2 e C) 3 4 cm 10 d Psat dT TC Psat = e ( T V H=T V B A with Antoine's Equation cm 70.917 mol mol 2756.22 C 83 228.060 B ( C) T B A Tn 273.15K Hcalc B Tn V Tn C K e 2 273.15K kPa K C K Hcalc 29.662 kJ Ans. mol %error Hcalc Hn %error 1.9 % Hn The table below shows the values for other components in Table B.2. Values agree within 5% except for acetic acid. Acetone Acetic Acid Acetonitrile Benz ene isoButane nButane 1Butanol Carbon tetrachloride Chlorobenz ene Chloroform Cyclohexane Cyclopentane nDecane Dichloromethane Diethyl ether Ethanol Ethylbenz ene Ethylene glycol nHeptane nHexane M ethanol M ethyl acetate M ethyl ethyl k etone Nitromethane nNonane isoOctane nOctane nPentane Phenol 1- panol Pro Hn ( J/ k mol) 29 .7 37.6 31.3 30.8 21.2 22.4 43.5 29 .9 35.3 29 .3 29 .9 27.4 39 .6 28 .1 26 .8 39 .6 35.7 53.2 31.9 29 .0 36 .5 30.4 31.7 34.9 37.2 30.8 34.6 25.9 45.9 41.9 84 % error 1.9 % 58 .7% 3.5% 0.2% 0.7% 0.0% 0.6 % 0.3% 0.3% 0.1% 0.1% 0.4% 2.2% 0.2% 0.9 % 2.8 % 0.5% 4.9 % 0.4% 0.4% 3.6 % 0.2% 1.3% 2.6 % 0.7% 0.1% 0.6 % 0.2% -% 0.6 1.1% p 2-Propanol Toluene W ater o-Xylene m-Xylene p-Xylene 40.5 33.3 41.5 36.7 36.2 35.9 1.7% 0.5% 2.0% 1.2% 1.4% 0.8% 4.13 Let P represent the vapor pressure. T ln Given P P 348.15 K P = 48.157543 kPa Find ( ) dPdT P P 5622.7 K T P Clapeyron equation: B dPdT = Pc AL 13.431 BL 2.211 0.029 bar K 3 cm 96.49 mol Vliq H T dPdT 1 512.6K AV dPdT 3 4.14 (a) Methanol: Tc AL 4.70504 T T K 4.70504 ln H T V Vliq Vliq PV V RT CPL ( ) T 5622.7 K T joule 31600 mol V = vapor molar volume. V Eq. (3.39) 2 H 87.396 kPa (guess) 100 kPa BL K BV 85 cm 1369.5 mol B Tn 80.97bar 51.28 10 T CL K 2 T 2 3 Ans. 337.9K CL 131.13 10 CV 3.450 10 6 R 12.216 10 3 6 CPV ( ) T P BV AV CV T K K Tsat 3bar 2 T 2 R T1 368.0K T2 300K 500K Estimate Hv using Riedel equation (4.12) and Watson correction (4.13) Tn Trn Trn Tc 1.092 ln Hn Pc Hv Trsat 1 Trn kJ mol 35.645 kJ mol T2 CPL ( ) T Td 100 38.301 Hv Hv CPV ( ) T Td T1 n 0.718 0.38 Tsat H Trsat Hn Tc R Tn Trn 1 Hn Tsat 1.013 bar 0.930 Trsat 0.659 kmol hr H 49.38 T sat Q nH Q 1.372 3 10 kW kJ mol Ans. (b) Benzene: Hv = 28.273 kJ mol H = 55.296 kJ mol Q = 1.536 10 kW (c) Toluene Hv = 30.625 kJ mol H = 65.586 kJ mol Q = 1.822 10 kW 4.15 Benzene Tc T1sat 562.2K 451.7K Pc 48.98bar T2sat 86 358.7K 3 3 Tn 353.2K Cp 162 J mol K Estimate Hv using Riedel equation (4.12) and Watson correction (4.13) Trn Tn Trn Tc Hn Hv Hn 30.588 30.28 R Tn Trn Tr2sat 1 0.638 Hn Tc 1.013 bar 0.930 1 Tr2sat Hv Pc 1.092 ln T2sat Tr2sat 0.628 0.38 Trn kJ mol kJ mol Assume the throttling process is adiabatic and isenthalpic. Guess vapor fraction (x): x Given Cp T1sat 0.5 4.16 (a) For acetylene: Tc Tn ln Hn R Tn 1.092 1 Tr 1 Hn Hf 227480 0.498 Tn 61.39 bar Tr 0.614 Pc bar J mol T Tc Tr 1.013 Trn 16.91 Hv 0.930 Hn 6.638 0.38 Trn Hv x Find ( x) Ans. 189.4 K 298.15 K Trn Tc Pc 308.3 K T Trn x T2sat = x Hv H298 Hf 87 Hv 0.967 kJ mol kJ mol H298 220.8 kJ mol Ans. kJ mol (b) For 1,3-butadiene: H298 = 88.5 (c) For ethylbenzene: H298 = 12.3 (d) For n-hexane: H298 = 198.6 (e) For styrene: H298 = 103.9 4.17 1st law: dQ = dU Ideal gas: Since mol kJ mol and V dP = R dT V dP = P P (B) 1 V R dT which reduces to or dQ = P dV = R 1 R dT 1 dQ = CV dT R dT 1 R dT 1 dQ = CP dT CP dV = V dP dV Combines with (A) to give: dQ = CP dT V dP = R dT P dV Combines with (B) to yield: or (A) P dV P dV then P V = const from which kJ dW = CV dT PV= RT Whence kJ mol R dT 1 R dT (C) Since CP is linear in T, the mean heat capacity is the value of CP at the arithmetic mean temperature. Thus Tam 88 675 CPm R 3.85 0.57 10 Integrate (C): P1 4.18 R 950 K P2 P1 T1 400 K 1.55 J Q 6477.5 P2 11.45 bar H298 = 4 058 910 J R T2 1 1 bar Tam T2 CPm Q 3 Ans. Ans. T1 mol 1 T2 T1 Ans. For the combustion of methanol: CH3OH(g) + (3/2)O2(g) = CO2(g) + 2H2O(g) H298 393509 H298 676485 For 6 MeOH: 2 ( 241818) ( 200660) For the combustion of 1-hexene: C6H12(g) + 9O2(g) = 6CO2(g) + 6H2O(g) H298 H298 6 ( 393509) 6 ( 241818) ( 41950) 3770012 H298 = 3 770 012 J Ans. Comparison is on the basis of equal numbers of C atoms. 4.19 C2H4 + 3O2 = 2CO2 + 2H2O(g) H298 [ ( 241818) 2 ( 393509) 52510] 2 J mol Parts (a) - (d) can be worked exactly as Example 4.7. However, with Mathcad capable of doing the iteration, it is simpler to proceed differently. 89 Index the product species with the numbers: 1 = oxygen 2 = carbon dioxide 3 = water (g) 4 = nitrogen (a) For the product species, no excess air: 0 3.639 0.506 2 5.457 1.045 n A 10 B 2 3.470 3.280 3 1.157 10 K 0.593 ni Ai B A 14 A 0.121 0.040 D ni Bi ni D i i i i B 54.872 52 D K 1.450 11.286 i 0.227 0.012 1 K 52 D 1.621 10 K T0 298.15K T For the products, CP HP = R dT R T0 The integral is given by Eq. (4.7). Moreover, by an energy balance, H298 HP = 0 (guess) 2 H298 = R A T0 Given 8.497 Find 1 T T0 B 2 T0 2 2 T 1 D T0 2533.5 K 1 Ans. Parts (b), (c), and (d) are worked the same way, the only change being in the numbers of moles of products. (b) nO = 0.75 nn = 14.107 T = 2198.6 K Ans. (c) nO = 1.5 nn = 16.929 T = 1950.9 K Ans. (d) nO = 3.0 nn = 22.571 T = 1609.2 K Ans. 2 2 2 2 2 2 90 (e) 50% xs air preheated to 500 degC. For this process, Hair H298 HP = 0 Hair = MCPH ( 298.15 773.15) For one mole of air: 3 MCPH 773.15 298.15 3.355 0.575 10 0.0 0.016 10 5 = 3.65606 For 4.5/0.21 = 21.429 moles of air: Hair = n R MCPH T Hair Hair 21.429 8.314 3.65606 ( 98.15 2 309399 J mol J mol The energy balance here gives: H298 1.5 n 773.15) 3.639 5.457 HP = 0 0.227 0.506 2 Hair 1.045 A 2 B 3.470 1.450 3.280 16.929 A 3 ni Ai B B 5 10 K 0.121 2 0.040 ni Bi D i 78.84 1.157 D 0.593 i A 10 K ni D i i 1 0.016 K D 52 1.735 10 K 2 (guess) H298 Find Hair = R A T0 1 D T0 Given B 2 T0 2 2 1 1 7.656 T 91 T0 K T 2282.5 K K Ans. 4.20 n-C5H12 + 8O2 = 5CO2 + 6H2O(l) By Eq. (4.15) with data from Table C.4: H298 5 ( 393509) 6 ( 285830) ( 146760) H298 = 3 535 765 J 4.21 Ans. The following answers are found by application of Eq. (4.15) with data from Table C.4. (a) -92,220 J (n) 180,500 J (b) -905,468 J (o) 178,321 J (c) -71,660 J (p) -132,439 J (d) -61,980 J (q) -44,370 J (e) -367,582 J (r) -68,910 J (f) -2,732,016 J (s) -492,640 J (g) -105,140 J (t) 109,780 J (h) -38,292 J (u) 235,030 J (i) 164,647 J (v) -132,038 J (j) -48,969 J (w) -1,807,968 J (k) -149,728 J (x) 42,720 J (l) -1,036,036 J (y) 117,440 J (m) 207,436 J (z) 175,305 J 92 4.22 The solution to each of these problems is exactly like that shown in Example 4.6. In each case the value of Ho298 is calculated in Problem 4.21. Results are given in the following table. In the first column the letter in ( ) indicates the part of problem 4.21 appropriate to the value. T/ K (a) (b ) (f ) (i ) (j ) (l ) (m) (n ) (o ) (r ) (t ) (u ) (v ) (w) (x ) (y ) 4.23 A 873.15 773.15 923.15 973.15 583.15 683.15 850.00 1350.00 1073.15 723.15 733.15 750.00 900.00 673.15 648.15 1083.15 -5.871 1.861 6.048 9.811 -9.523 -0.441 4.575 -0.145 -1.011 -1.424 4.016 7.297 2.418 2.586 0.060 4.175 103 B 4.181 -3.394 -9.779 -9.248 11.355 0.004 -2.323 0.159 -1.149 1.601 -4.422 -9.285 -3.647 -4.189 0.173 -4.766 106 C 0.000 0.000 0.000 2.106 -3.450 0.000 0.000 0.000 0.000 0.156 0.991 2.520 0.991 0.000 0.000 1.814 10-5 D -0.661 2.661 7.972 -1.067 1.029 -0.643 -0.776 0.215 0.916 -0.083 0.083 0.166 0.235 1.586 -0.191 0.083 IDCPH/ J -17,575 4,729 15,635 25,229 -10,949 -2,416 13,467 345 -9,743 -2,127 7,424 12,172 3,534 4,184 125 12,188 Ho298 J HoT/ -109,795 -900,739 -2,716,381 189,876 -59,918 -1,038,452 220,903 180,845 168,578 -71,037 117,204 247,202 -128,504 -1,803,784 42,845 129,628 This is a simple application of a combination of Eqs. (4.18) & (4.19) with evaluated parameters. In each case the value of Ho298 is calculated in Pb. 4.21. The values of A, B, C and D are given for all cases except for Parts (e), (g), (h), (k), and (z) in the preceding table. Those missing are as follows: Part No. A 103 B 106 C 10-5 D (e) -7.425 20.778 0.000 3.737 (g) -3.629 8.816 -4.904 0.114 (h) -9.987 20.061 -9.296 1.178 (k) 1.704 -3.997 1.573 0.234 (z) -3.858 -1.042 0.180 0.919 93 4.24 q 6 ft 150 10 3 T day ( 60 5 32) K 9 T 273.15K 288.71 K P 1atm The higher heating value is the negative of the heat of combustion with water as liquid product. Calculate methane standard heat of combustion with water as liquid product: CH4 + 2O2 --> CO2 +2H2O Standard Heats of Formation: J mol HfCH4 74520 HfCO2 393509 Hc HfCO2 HfO2 J mol J mol HfH2Oliq 2 HfH2Oliq HigherHeatingValue 0 285830 HfCH4 J mol 2 HfO2 5J Hc Hc 8.906 10 mol Assuming methane is an ideal gas at standard conditions: n q P RT n HigherHeatingValue 4.25 8 mol n 1.793 10 5dollar GJ day 5 dollar 7.985 10 day Ans. Calculate methane standard heat of combustion with water as liquid product Standard Heats of Formation: CH4 + 2O2 --> CO2 +2H2O J mol HfCH4 74520 HfCO2 393509 HcCH4 HfCO2 HcCH4 890649 J mol HfO2 0 J mol HfH2Oliq 2 HfH2Oliq J mol 94 HfCH4 285830 J mol 2 HfO2 Calculate ethane standard heat of combustion with water as liquid product: Standard Heats of Formation:C2H6 + 7/2O2 --> 2CO2 +3H2O J mol HfC2H6 83820 HcC2H6 2 HfCO2 HcC2H6 1560688 3 HfH2Oliq HfC2H6 7 2 HfO2 J mol Calculate propane standard heat of combustion with water as liquid product Standard Heats of Formation:C3H8 + 5O2 --> 3CO2 +4H2O J mol HfC3H8 104680 HcC3H8 3 HfCO2 HcC3H8 2219.167 4 HfH2Oliq HfC3H8 5 HfO2 kJ mol Calculate the standard heat of combustion for the mixtures a) 0.95 HcCH4 0.02 HcC2H6 0.02 HcC3H8 921.714 kJ mol b) 0.90 HcCH4 0.05 HcC2H6 0.03 HcC3H8 946.194 kJ mol c) 0.85 HcCH4 0.07 HcC2H6 0.03 HcC3H8 932.875 kJ mol Gas b) has the highest standard heat of combustion. 4.26 2H2 + O2 = 2H2O(l) Hf1 C + O2 = CO2(g) Hf2 N2(g)+2H2O(l)+CO2(g)=(NH2)2CO(s)+3/2O2 H Ans. 2 ( 285830) J 393509 J 631660 J . N2(g)+2H2(g)+C(s)+1/2O2(g)=(NH2)2CO(s) H298 Hf1 Hf2 H298 H 95 333509 J Ans. 4.28 On the basis of 1 mole of C10H18 (molar mass = 162.27) Q 43960 162.27 J 6 Q 7.133 10 J This value is for the constant-volume reaction: C10H18(l) + 14.5O2(g) = 10CO2(g) + 9H2O(l) Assuming ideal gases and with symbols representing total properties, Q= T U= H ( V)= P 298.15 K H Q H R T ngas ngas R T ngas ( 10 14.5)mol 6 H 7.145 10 J This value is for the constant-V reaction, whereas the STANDARD reaction is at const. P.However, for ideal gases H = f(T), and for liquids H is a very weak function of P. We therefore take the above value as the standard value, and for the specified reaction: C10H18(l) + 14.5O2(g) = 10CO2(g) + 9H2O(l) H 9H2O(l) = 9H2O(g) Hvap 9 44012 J ___________________________________________________ C10H18(l) + 14.5O2(g) = 10CO2(g) + 9H2O(g) H298 4.29 H Hvap H298 6748436 J Ans. FURNACE: Basis is 1 mole of methane burned with 30% excess air. CH4 + 2O2 = CO2 + 2H2O(g) Entering: Moles methane n1 1 Moles oxygen n2 2 1.3 Moles nitrogen n3 2.6 96 79 21 n2 2.6 n3 9.781 Total moles of dry gases entering n n1 n2 n n3 13.381 At 30 degC the vapor pressure of water is 4.241 kPa. Moles of water vapor entering: n4 Leaving: 4.241 13.381 101.325 4.241 n4 CO2 -- 1 mol H2O -- 2.585 mol O2 -- 2.6 - 2 = 0.6 mol N2 -- 9.781 mol 0.585 (1) (2) (3) (4) By an energy balance on the furnace: Q= H= H298 For evaluation of 1 A 0.6 9.781 i HP we number species as above. 5.457 2.585 n HP 3.470 3.639 1.045 1.450 B R 8.314 B 48.692 0.121 D 0.227 0.040 D ni Bi i i 10.896983 10 3 ni Di C D 0 4 5.892 10 The TOTAL value for MCPH of the product stream: HP R M CPH ( 303.15K 1773.15K A B C D) ( 1773.15 HP 732.013 kJ mol From Example 4.7: Q HP H298 5 10 J mol K i A 3 0.593 ni Ai B A 10 0.506 3.280 14 1.157 H298 802625 Q = 70 612 J 97 J mol Ans. 303.15)K HEAT EXCHANGER: Flue gases cool from 1500 degC to 50 degC. The partial pressure of the water in the flue gases leaving the furnace (in kPa) is pp n2 n1 n2 n3 n4 101.325 pp 18.754 The vapor pressure of water at 50 degC (exit of heat exchanger) is 12.34 kPa, and water must condense to lower its partial pressure to this value. Moles of dry flue gases: n n1 n3 n4 n 11.381 Moles of water vapor leaving the heat exchanger: n2 12.34 n 101.325 12.34 n2 Moles water condensing: 1.578 n 2.585 1.578 Latent heat of water at 50 degC in J/mol: H50 2382.918.015 J mol Sensible heat of cooling the flue gases to 50 degC with all the water as vapor (we assumed condensation at 50 degC): Q R MCPH ( 323.15 K 1773.15 K A B C D)( 23.15 3 1773.15) K Q = 766 677 J 4.30 4NH3(g) + 5O2(g) = 4NO(g) + 6H2O(g) BASIS: 4 moles ammonia entering reactor Moles O2 entering = (5)(1.3) = 6.5 Moles N2 entering = (6.5)(79/21) = 24.45 Moles NH3 reacting = moles NO formed = (4)(0.8) = 3.2 Moles O2 reacting = (5)(0.8) = 4.0 Moles water formed = (6)(0.8) = 4.8 98 n H50 Ans. ENERGY BALANCE: H= HR H298 HP = 0 REACTANTS: 1=NH3; 2=O2; 3=N2 4 n 3.578 6.5 A 3.639 24.45 i 3.020 B 0.506 10 3.280 A 13 118.161 ni Ai B 3 5 D 0.227 10 0.593 B 0.040 D ni Bi ni Di i i i A 0.186 C 0.02987 5 D 0.0 1.242 10 TOTAL mean heat capacity of reactant stream: HR HR R MCPH ( 348.15K 298.15K A B C D) ( 298.15K 52.635 348.15K) kJ mol The result of Pb. 4.21(b) is used to get J H298 0.8 ( 905468) mol PRODUCTS: =NH3; 2=O2; 3=NO; 4=H2O; 5=N2 1 0.8 3.020 2.5 n 3.578 3.639 0.506 3.2 A 3.387 B 0.629 0.186 10 K 0.227 3 D 0.014 4.8 1.450 3.280 0.593 2 0.121 24.45 i 3.470 5 10 K 0.040 15 A ni Ai B 119.65 D B 1 0.027 K By the energy balance and Eq. (4.7), we can write: (guess) 2 T0 298.15K 99 ni Di i i i A ni Bi D 4 8.873 10 K 2 H298 B 2 T0 2 1 2 1 1 T 3.283 Find 4.31 H R = R A T0 D T0 Given T T0 Ans. 978.9 K C2H4(g) + H2O(g) = C2H5OH(l) n BASIS: 1 mole ethanol produced H=Q= Energy balance: H298 [ 277690 HR ( 2510 5 1mol H298 241818) J mol 4J H298 8.838 10 mol Reactant stream consists of 1 mole each of C2H4 and H2O. 12n i 1.424 A 3.470 1 1 14.394 B 1.450 ni Ai B A B HR Q 4.392 C 10 0.0 C 6 D C 0.01584 0.0 ni Di i 4.392 10 6 D R M CP ( H 298.15K 593.15K A B C D)( 98.15K 2 4 1.21 10 593.15K) 4J 2.727 10 HR mol Q H298 1mol 100 5 10 0.121 D ni Ci i i 4.894 HR 3 ni Bi i A 10 115653 J Ans. 4.32 One way to proceed is as in Example 4.8 with the alternative pair of reactions: CH4 + H2O = CO + 3H2 H298a 205813 CH4 + 2H2O = CO2 + 4H2 H298b 164647 BASIS: 1 mole of product gases containing 0.0275 mol CO2; 0.1725 mol CO; & H2O 0.6275 mol H2 Entering gas, by carbon & oxygen balances: 0.0275 + 0.1725 = 0.2000 mol CH4 0.1725 + 0.1725 + 2(0.0275) = 0.4000 mol H2O J H298 0.1725 H298a 0.0275 H298b mol The energy balance is written Q= HR H298 1.702 3.470 9.081 B 1.450 ni Ai B A HR HR 10 i 3 B 2.164 C 0.0 0.4 10 6 D 2.396 10 3 C 0.0 5 10 0.121 D ni Ci ni Di i i i 1.728 mol 0.2 n 12 C ni Bi i A 4.003 10 HP REACTANTS: 1=CH4; 2=H2O A 4J H298 4.328 10 7 D 3 4.84 10 RI CPH ( 773.15K 298.15K A B C D) 4J 1.145 10 mol PRODUCTS: 1=CO2; 2=CO; 3=H2O; 4=H2 0.0275 n 0.1725 0.1725 0.6275 5.457 A 3.376 3.470 1.045 B 3.249 0.557 1.450 0.422 101 1.157 10 3 D 0.031 0.121 0.083 5 10 i A 14 B ni Ai HP B 3.37 ni Di i i i A D ni Bi 6.397 10 4 C 3 D 0.0 3.579 10 RI CPH ( 298.15K 1123.15K A B C D) 4J 2.63 10 mol HP Q HR H298 Q HP mol 4.33 CH4 + 2O2 = CO2 + 2H2O(g) Ans. 54881 J H298a H298b C2H6 + 3.5O2 = 2CO2 + 3H2O(g) 802625 1428652 BASIS: 1 mole fuel (0.75 mol CH4; 0.25 mol C2H6) burned completely with 80% xs. air. O2 in = 1.8[(0.75)(2) + (0.25)(3.5)] = 4.275 mol N2 in = 4.275(79/21) = 16.082 mol Product gases: CO2 = 0.75 + 2(0.25) = 1.25 mol H2O = 2(0.75) + 3(0.25) = 2.25 mol O2 = (0.8/1.8)(4.275) = 1.9 mol N2 = 16.082 mol H298 0.75 H298a 0.25 H298b J mol Energy balance: Q = H = H298 HP PRODUCTS: 1=CO2; 2=H2O; 3=O2; 4=N2 1.25 n 5.457 3.470 1.450 A 1.9 14 B 3.639 16.082 i 8 10 HP = Q 1.045 2.25 5 Q 3.280 74.292 10 K 3 D B 102 0.227 D ni Bi 0.015 0.121 5 10 K 0.040 ni Di i i i H298 1.157 0.593 ni Ai B A A 0.506 J mol 1 K C 0.0 D 42 9.62 10 K 2 By the energy balance and Eq. (4.7), we can write: T0 303.15K Q 2 (guess) H298 = R A T0 2 1 1 1.788 4.34 B 2 T0 2 1 D T0 Given T T0 T Find Ans. 542.2 K BASIS: 1 mole of entering gases containing 0.15 mol SO2; 0.20 mol O2; 0.65 mol N2 SO2 + 0.5O2 = SO3 Conversion = 86% SO2 reacted = SO3 formed = (0.15)(0.86) = 0.129 mol O2 reacted = (0.5)(0.129) = 0.0645 mol Energy balance: H773 = HR H298 HP Since HR and HP cancel for the gas that passes through the converter unreacted, we need consider only those species that react or are formed. Moreover, the reactants and products experience the same temperature change, and can therefore be considered together. We simply take the number of moles of reactants as being negative. The energy balance is then written: H773 = H298 Hnet H298 [ 395720 ( 296830) ]0.129 J mol 1: SO2; 2: O2; 3: SO3 0.129 n 0.0645 5.699 A 0.129 i 13 3.639 B 8.060 A A 0.06985 0.506 10 3 5 D 0.227 10 2.028 1.056 ni Ai B i 1.015 0.801 ni Bi D ni Di i B 2.58 10 103 i 7 C 0 D 4 1.16 10 Hnet R M CPH ( 298.15K 773.15K A B C D)( 73.15K 7 Hnet 77.617 H773 298.15K) J mol H298 H773 Hnet 12679 J mol Ans. 4.35 CO(g) + H2O(g) = CO2(g) + H2(g) BASIS: 1 mole of feed consisting of 0.5 mol CO and 0.5 mol H2O. Moles CO reacted = moles H2O reacted = moles CO2 formed = moles H2 formed = (0.6)(0.5) = 0.3 Product stream: moles CO = moles H2O = 0.2 moles CO2 = moles H2 = 0.3 Energy balance: Q= H298 0.3 [ 393509 H= HR H298 J 214818) mol ( 110525 HP 4J H298 2.045 10 mol Reactants: 1: CO 2: H2O 0.5 n i 0.5 3.376 A 3.470 1.450 ni Ai B A 12 0.557 B HR HR 0.031 D B ni Di i i 3.423 5 10 0.121 D 1.004 10 3 0D C R M CPH ( 298.15K 398.15K A B C D)( 98.15K 2 3 4.5 10 398.15K) 3J 3.168 10 Products: 0.2 0.3 0.3 mol 1: CO 2: H2O 3: CO2 4: H2 0.2 n 3 ni Bi i A 10 3.376 A 3.470 5.457 0.557 B 3.249 1.450 1.045 0.422 104 0.031 10 3 D 0.121 1.157 0.083 5 10 i 14A ni Ai B i A ni Bi D i 3.981 B 8.415 10 i 4 C 4 0D HP R MCPH ( 298.15K 698.15K A B C D) ( 698.15K HP 1.415 10 Q ni Di 3.042 10 298.15K) 4J HR mol H298 HP mol Q 9470 J Ans. 4.36 BASIS: 100 lbmol DRY flue gases containing 3.00 lbmol CO2 and 11.80 lbmol CO x lbmol O2 and 100-(14.8-x)= 85.2-x lbmol N2. The oil therefore contains 14.80 lbmol carbon;a carbon balance gives the mass of oil burned: 14.8 12.011 lbm 0.85 209.133 lbm The oil also contains H2O: 209.133 0.01 lbmol 0.116 lbmol 18.015 Also H2O is formed by combustion of H2 in the oil in the amount 209.133 0.12 lbmol 2.016 12.448 lbmol Find amount of air entering by N2 & O2 balances. N2 entering in oil: 209.133 0.02 lbmol 28.013 0.149 lbmol lbmol N2 entering in the air=(85.2-x)-0.149 =85.051-x lbmol O2 in flue gas entering with dry air = 3.00 + 11.8/2 + x + 12.448/2 = 15.124 + x lbmol (CO2) (CO) (O2) (H2O from combustion) Total dry air = N2 in air + O2 in air = 85.051 - x + 15.124 + x = 100.175 lbmol 105 Since air is 21 mol % O2, 0.21 = 15.124 x 100.175 x ( 0.21 100.175 15.124)lbmol x 5.913 lbmol O2 in air = 15.124 + x = 21.037 lbmols N2 in air = 85.051 - x = 79.138 lbmoles N2 in flue gas = 79.138 + 0.149 = 79.287 lbmols [CHECK: Total dry flue gas = 3.00 + 11.80 + 5.913 + 79.287 = 100.00 lbmol] Humidity of entering air, sat. at 77 degF in lbmol H2O/lbmol dry air, P(sat)=0.4594(psia) 0.4594 14.696 0.4594 0.03227 lbmol H2O entering in air: 0.03227 100.175 lbmol 3.233 lbmol If y = lbmol H2O evaporated in the drier, then lbmol H2O in flue gas = 0.116+12.448+3.233+y = 15.797 + y Entering the process are oil, moist air, and the wet material to be dried, all at 77 degF. The "products" at 400 degF consist of: 3.00 lbmol CO2 11.80 lbmol CO 5.913 lbmol O2 79.287 lbmol N2 (15.797 + y) lbmol H2O(g) Energy balance: Q= H= H298 HP where Q = 30% of net heating value of the oil: Q 0.3 19000 BTU 209.13 lbm lbm Reaction upon which net heating value is based: 106 Q 6 1.192 10 BTU OIL + (21.024)O2 = (14.8)CO2 + (12.448 + 0.116)H2O(g) + (0.149)N2 H298a 19000 209.13 BTU 6 H298a 3.973 10 BTU To get the "reaction" in the drier, we add to this the following: (11.8)CO2 = (11.8)CO + (5.9)O2 H298b 11.8 ( 110525 (y)H2O(l) = (y)H2O(g) H298c ( y) 393509) 0.42993 BTU Guess: y 50 44012 0.42993 y BTU [The factor 0.42993 converts from joules on the basis of moles to Btu on the basis of lbmol.] Addition of these three reactions gives the "reaction" in the drier, except for some O2, N2, and H2O that pass through unchanged. Addition of the corresponding delta H values gives the standard heat of reaction at 298 K: H298 ( y) H298a H298b H298c ( y) For the product stream we need MCPH: 1: CO2 2: CO 3:O2 4: N2 5: H2O T0 298.15 r 1.986 400 T 459.67 1.8 T 477.594 3 1.045 1.157 11.8 n y) ( 5.457 3.376 0.557 0.031 5.913 A i B 3 5 D 0.227 10 0.593 0.040 3.470 y 1 5 A ( y) 1.450 0.121 n y) i Ai B ( y) ( i T T0 0.506 10 3.280 79.278 15.797 3.639 1.602 n y) i Bi D ( y) ( n y) i Di ( i CP ( y) 107 r A ( y) i B ( y) T0 2 1 D ( y) T0 2 CP ()( y 400 Given y y H298 () y Find y () (lbmol H2O evaporated) 49.782 y 18.015 209.13 Whence 77)BTU = Q (lb H2O evap. per lb oil burned) Ans. 4.288 4.37 BASIS: One mole of product gas containing 0.242 mol HCN, and (1-0.242)/2 = 0.379 mol each of N2 and C2H2. The energy balance is Q= H= H298 H298 HP ( 135100 2 227480) 0.242 J 2 3 H298 5.169 10 J Products: 0.242 n 4.736 0.379 A 3.280 0.379 ni Ai 0.725 3 0.593 10 D 0.040 1.952 B 4.7133 B D ni Bi 1.2934 10 ni Di i 3 4 0D C HP R M CPH ( 298.15K 873.15K A B C D)( 73.15K 8 HP 6.526 10 2.495 10 J Q 298.15K)mol 4 HP H298 Q HP 4 2.495 10 J 30124 J Ans. 4.38 BASIS: 1 mole gas entering reactor, containing 0.6 mol HCl, 0.36 mol O2, and 0.04 mol N2. HCl reacted = (0.6)(0.75) = 0.45 mol 4HCl(g) + O2(g) = 2H2O(g) + 2Cl2(g) 108 5 10 1.299 i i A B 6.132 1 3A i 1.359 For this reaction, H298 [ ( 241818) 2 Evaluate T0 4 ( 92307) ] 5J H298 1.144 10 mol by Eq. (4.21) with H823 T 298.15K J mol 823.15K 1: H2O 2: Cl2 3: HCl 4=O2 2 3.470 2 n 4.442 A 4 0.089 B 3.156 1 i 1.45 0.623 3.639 A 14 ni Ai 10 3 D 0.344 0.151 0.506 B D ni Di i i 0.439 H823 H298 MCPH T0 T H823 117592 B 5 8 10 A B 5 10 0.227 ni Bi i A 0.121 C 0 DRT 4 D T0 C 8.23 10 J mol Heat transferred per mol of entering gas mixture: Q H823 4 Q 0.45 mol 4.39 CO2 + C = 2CO Ans. J (a) mol J (b) 221050 mol H298a 2C + O2 = 2CO 13229 J 172459 H298b Eq. (4.21) applies to each reaction: For (a): 2 n 1 1 3.376 A 1.771 0.557 B 0.771 10 5.457 1.045 109 0.031 3 D 5 0.867 10 1.157 i A 13 B ni Ai H1148a B 0.476 i 4 7.02 10 H298a R M CP 298.15K 1148.15K H H1148a ni Di i i A D ni Bi A C B C D 0 5 1.962 10 D ( 148.15K 1 298.15K) 5J 1.696 10 mol For (b): 2 n 3.376 1 A 3.639 2 i 13 0.557 B H1148b ni Ai 5 D 0.227 10 B 0.867 ni Bi D ni Di i 0.429 B 9.34 10 H298b R M CP 298.15K 1148.15K H H1148b 3 0.771 i A 0.506 10 1.771 A 0.031 i 4 C A B 0 C D D ( 148.15K 1 5J 2.249 10 mol The combined heats of reaction must be zero: nCO 2 H1148a nO H1148b = 0 2 nCO Define: r= 2 nO H1148b r H1148a 2 110 r 5 1.899 10 1.327 298.15K) For 100 mol flue gas and x mol air, moles are: Flue gas Air Feed mix CO2 12.8 0 12.8 CO 3.7 0 3.7 O2 5.4 0.21x 5.4 + 0.21x N2 78.1 0.79x 78.1 + 0.79x Whence in the feed mix: r= 12.5 5.4 r mol 0.21 x 12.8 5.4 0.21 x x 19.155 mol 100 19.155 Flue gas to air ratio = Ans. 5.221 Product composition: nCO 3.7 nN 78.1 2 2 ( 12.8 0.21 19.155) 0.79 19.155 nCO Mole % N2 = 100 48.145 93.232 2 nCO Mole % CO = nCO nN 5.4 nN 100 34.054 Ans. 2 34.054 65.946 H298a 802625 J mol H298b 4.40 CH4 + 2O2 = CO2 + 2H2O(g) 519641 J mol CH4 + (3/2)O2 = CO + 2H2O(g) BASIS: 1 mole of fuel gas consisting of 0.94 mol CH4 and 0.06 mol N2 Air entering contains: 1.35 2 0.94 2.538 79 21 2.538 9.548 mol O2 mol N2 111 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 5J H298 0.282 H298b Moles H2O formed by reaction = 0.94 2.0 Moles O2 consumed by reaction = 2 0.658 6.747 10 mol 1.88 3 0.282 2 1.739 Product gases contain the following numbers of moles: (1) (2) (3) (4) (5) CO2: 0.658 CO: 0.282 H2O: 1.880 O2: 2.538 - 1.739 = 0.799 N2: 9.548 + 0.060 = 9.608 0.658 1.045 1.157 0.282 n 5.457 3.376 0.557 0.031 1.880 A 3.470 B 1.450 10 0.799 3.280 D 0.593 0.121 0.506 9.608 i 3.639 3 1 5A ni Ai B A 0.227 0.040 ni Bi i D B 9.6725 10 i 3 C R M CPH ( 298.15K 483.15K A B C D)( 83.15K 4 HP 7.541 10 HH2O mdotH2O 4 0D HP Energy balance: ni Di i 45.4881 5 10 3.396 10 298.15K) 4J mol Hrx H298 Hrx ndotfuel = 0 112 HP Hrx kJ mol kg 34.0 sec 599.252 mdotH2O From Table C.1: HH2O ( 398.0 104.8) kJ kg HH2O mdotH2O ndotfuel ndotfuel Hrx 16.635 mol sec Volumetric flow rate of fuel, assuming ideal gas: ndotfuel R 298.15 K V 3 m 0.407 sec V 101325 Pa 4.41 C4H8(g) = C4H6(g) + H2(g) Ans. H298 109780 J mol BASIS: 1 mole C4H8 entering, of which 33% reacts. The unreacted C4H8 and the diluent H2O pass throught the reactor unchanged, and need not be included in the energy balance. Thus T T0 1 n by Eq. (4.21): Evaluate H798 1: C4H6 2: H2 3: C4H8 1 2.734 A 3.249 26.786 B 0.422 1.967 i 298.15 K 798.15 K 1 8.882 10 3 C 0.0 0.0 31.630 6 10 D 9.873 5 0.083 10 0.0 13 ni Ai B A 4.016 B H298 H798 1.179 10 Q 3 C 9.91 10 MCPH 298.15K 798.15K 5J 0.33 mol H798 mol Q 38896 J 113 Ans. ni Di i i 4.422 10 H798 D ni Ci i i A C ni Bi A B 7 D C 3 8.3 10 DRT T0 4.42Assume Ideal Gas and P = 1 atm P R 1atm a) T0 ( 70 T0 3 BTU 7.88 10 mol K T 294.261 K ICPH T0 T 3.355 0.575 10 3 T0 T 459.67) rankine BTU sec 305.372 K 5 0 0.016 10 R ICPH T0 T 3.355 0.575 10 Vdot b) T0 38.995 K ( 24 3 5 0 0.016 10 T T0 ndot 39.051 Vdot ft 33.298 sec Q 13K 12 kJ s 3 5 0 0.016 10 45.659 K Q R ICPH T0 T 3.355 0.575 10 3 0 5 0.016 10 Vdot 4.43Assume Ideal Gas and P = 1 atm ( 94 459.67) rankine 1.61 10 3 Vdot 31.611 P T ( 68 m 0.7707 s 1atm 459.67) rankine 3 atm ft mol rankine 3 ft 50 sec ndot mol s 3 ndot R T0 P a) T0 Ans. kJ mol K ndot Vdot mol s 3 m 0.943 s Vdot 273.15) K 8.314 10 3 3 ndot R T0 P ICPH T0 T 3.355 0.575 10 R 12 Q ndot R Q 20rankine ndot P Vdot R T0 114 ndot 56.097 mol s Ans. T0 307.594 K T ICPH T0 T 3.355 0.575 10 3 293.15 K 5 0 0.016 10 50.7 K 3 BTU R 7.88 10 Q R ICPH T0 T 3.355 0.575 10 mol K 3 0 5 0.016 10 ndot Q b) T0 R ( 35 273.15)K T 273.15)K mol K 3 m 1.5 sec 3 P Vdot R T0 ndot ICPH T0 T 3.355 0.575 10 3 0 ndot 5 0.016 10 59.325 8.314 10 Q R ICPH T0 T 3.355 0.575 10 35.119 K 3 0 5 0.016 10 Q ndot 17.3216 4.44 First calculate the standard heat of combustion of propane C3H8 + 5O2 = 3CO2(g) + 4H2O (g) H298 Cost 3 393509 J mol 4 241818 6J 2.043 10 2.20 mol s kJ mol K R H298 BTU Ans. sec 3 5 atm m 8.205 10 Vdot ( 25 22.4121 dollars gal mol 80% 115 J mol 104680 J mol kJ Ans. s Estimate the density of propane using the Rackett equation 3 Tc T Vsat 369.8K ( 25 Zc 273.15) K V c Zc Heating_cost 1 Tr 0.276 Tr Vc T Tc Tr cm 200.0 mol 0.806 3 0.2857 Vsat Vsat Cost cm 89.373 mol Heating_cost H298 Heating_cost 0.032 dollars MJ 33.528 Ans. dollars 6 10 BTU 4.45 T0 ( 25 a) Acetylene 273.15) K T ( 500 273.15) K Q R ICPH T0 T 6.132 1.952 10 Q 2.612 10 3 0 5 1.299 10 4J 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 ( 25 Q 273.15)K 30000 a) Acetylene T ( 500 273.15)K J mol Given Q = R ICPH T0 T 6.132 1.952 10 T Find ( T) T 835.369 K T 3 273.15K 0 5 1.299 10 562.2 degC The calculations are repeated and the answers are in the following table: a) Acetylene b) Ammonia c) n-butane d) Carbon dioxide e) Carbon monoxide f) Ethane g) Hydrogen h) Hydrogen chloride i) Methane j) Nitric oxide k) Nitrogen l) Nitrogen dioxide m) Nitrous oxide n) Oxygen o) Propylene 4.47 T0 ( 25 273.15)K T( K) 835.4 964.0 534.4 932.9 1248.0 690.2 1298.4 1277.0 877.3 1230.2 1259.7 959.4 927.2 1209.9 636.3 T T( C) 562.3 690.9 261.3 659.8 974.9 417.1 1025.3 1003.9 604.2 957.1 986.6 686.3 654.1 936.8 363.2 ( 250 a) Guess mole fraction of methane: y 273.15) K Q 11500 0.5 Given y ICPH T0 T 1.702 9.081 10 (1 y 3 2.164 10 y) ICPH T0 T 1.131 19.225 10 Find ( y) y 0.637 Ans. 117 3 6 =Q 0R 5.561 10 6 0R J mol b) T0 ( 100 273.15) K T ( 400 Guess mole fraction of benzene 273.15)K y Q 54000 J mol 0.5 Given y ICPH T0 T ( 1 3 0.206 39.064 10 y)ICPH T0 T y Find y () y c) T0 ( 150 3 3.876 63.249 10 273.15) K 6 13.301 10 =Q 0R 6 20.928 10 0R Ans. 0.245 T ( 250 Guess mole fraction of toluene 273.15)K y Q 17500 J mol 0.5 Given y ICPH T0 T 0.290 47.052 10 ( 1 y 3 15.716 10 y)ICPH T0 T 1.124 55.380 10 Find y () y 0.512 3 6 =Q 0R 18.476 10 6 0R Ans. 4.48 Temperature profiles for the air and water are shown in the figures below. There are two possible situations. In the first case the minimum temperature difference, or "pinch" point occurs at an intermediate location in the exchanger. In the second case, the pinch occurs at one end of the exchanger. There is no way to know a priori which case applies. Pi h a End nc t I e m e a e Pi h nt r di t nc TH1 Se ton I ci I Se ton I ci TH1 Se ton I ci THi Se ton I ci I THi T TH2 TC1 TC1 TCi TCi TC2 118 TH2 T TC2 To solve the problem, apply an energy balance around each section of the exchanger. T H1 Section I balance: mdotC HC1 HCi = ndotH CP dT THi T Hi Section II balance: HC2 = ndotH mdotC HCi CP dT TH2 If the pinch is intermediate, then THi = TCi + T. If the pinch is at the end, then TH2 = TC2 + T. a) TH1 T 10degC TC1 2676.0 TCi For air from Table C.1:A kJ kg 3.355 B 100degC TC2 25degC HCi 100degC HC1 1000degC 419.1 kJ kg HC2 104.8 0.575 10 3 C 0D kJ kg 5 0.016 10 kmol s Assume as a basis ndot = 1 mol/s. ndotH Assume pinch at end: TH2 TC2 THi 110degC Guess: mdotC 1 kg s 1 T Given mdotC HC1 mdotC HCi mdotC THi mdotC ndotH THi TCi HCi = ndotH R ICPH THi TH1 A B C D Energy balances on Section I and HC2 = ndotH R ICPH TH2 THi A B C D II 170.261 degC mdotC Find mdotC THi THi 0.011 kg mol 70.261 degC Ans. TH2 119 TC2 10 degC 11.255 kg s Since the intermediate temperature difference, THi - TCi is greater than the temperature difference at the end point, TH2 - TC2, the assumption of a pinch at the end is correct. b) TH1 TC1 T 10degC 2676.0 TCi kJ kg 100degC TC2 25degC HCi 100degC HC1 500degC 419.1 kJ kg HC2 104.8 1 kmol s THi TCi T TH2 110degC Assume as a basis ndot = 1 mol/s. ndotH Assume pinch is intermediate: Guess: mdotC 1 kg s kJ kg Given mdotC HC1 mdotC HCi mdotC TH2 mdotC ndotH THi TCi HCi = ndotH R ICPH THi TH1 A B C D Energy balances on Section I and HC2 = ndotH R ICPH TH2 THi A B C D II Find mdotC TH2 TH2 5.03 10 3 kg 10 degC mol 48.695 degC mdotC 5.03 kg s Ans. TH2 TC2 23.695 degC Since the intermediate temperature difference, THi - TCi is less than the temperature difference at the end point, TH2 - TC2, the assumption of an intermediate pinch is correct. 4.50a) C6H12O6(s) + 6 O2(g)= 6 CO2(g) + 6 H2O(l) 1 = C6H12O6 , 2 = O2 , 3 = CO2 , 4 = H2O H0f1 1274.4 kJ mol H0f2 120 0 kJ mol M1 180 gm mol H0f3 H0r 393.509 6 H0f3 kJ mol 6 H0f4 b) energy_per_kg mass_glucose H0f4 150 kJ kg H0f1 285.830 mass_person energy_per_kg H0r M3 H0 r 6 H0f2 mass_person kJ mol 44 2801.634 gm mol kJ mol Ans. 57kg mass_glucose M1 0.549 kg Ans. c) 6 moles of CO2 are produced for every mole of glucose consumed. Use molecular mass to get ratio of mass CO2 produced per mass of glucose. 6 275 10 mass_glucose 6 M3 M1 8 Ans. 2.216 10 kg 4.51 Assume as a basis, 1 mole of fuel. 0.85 (CH4(g) + 2 O2(g) = CO2(g) + 2 H2O(g)) 0.10(C2H6 (g) + 3.5 O2(g) = 2 CO2(g) + 3 H2O(g)) -----------------------------------------------------------------0.85 CH4(g) + 0.10 C2H6(g) + 2.05 O2(g) = 1.05 CO2(g) + 2 H2O(g) 1 = CH4, 2 = C2H6, 3 = O2, 4 = CO2, 5 = H2O 6 = N2 kJ mol H0f1 74.520 H0f4 393.509 a) H0c H0c 825.096 kJ mol 2 H0f5 83.820 H0f5 kJ mol 1.05 H0f4 kJ mol H0f2 241.818 0.85 H0f1 H0f3 0 kJ mol kJ mol 0.10 H0f2 1.05 H0f3 Ans. b) For complete combustion of 1 mole of fuel and 50% excess air, the exit gas will contain the following numbers of moles: n3 n3 0.5 2.05mol 121 1.025 mol Excess O2 n4 1.05mol n5 2mol n6 0.05mol 79 1.5 2.05mol 21 n6 11.618 mol Total N2 Air and fuel enter at 25 C and combustion products leave at 600 C. T1 A B C D Q ( 25 T2 273.15) K n3 3.639 n4 6.311 ( 600 n5 3.470 273.15) K n6 3.280 mol n3 0.506 n4 0.805 n5 1.450 n6 0.593 10 3 mol n3 0 n4 0 n5 0 n6 0 10 6 mol n3 ( 0.227) n4 ( 0.906) n5 0.121 5 n6 0.040 10 mol H0c ICPH T1 T2 A B C D R 122 Q 529.889 kJ mol Ans. Chapter 5 - Section A - Mathcad Solutions 5.2 Let the symbols Q and Work represent rates in kJ/s. Then by Eq. (5.8) Work = QH TC TC =1 TH TH 323.15 K Work TC QH 1 250 kJ s kJ s Work or 148.78 Work TH By Eq. (5.1), QH 798.15 K 148.78 kW which is the power. Ans. QC QH QC Work 101.22 kJ s Ans. 5.3 (a) Let symbols Q and Work represent rates in kJ/s TH TC 750 K By Eq. (5.8): But TC 1 Work = QH Work 300 K 0.6 TH Whence 95000 kW QH Work QH QC (b) QH QH 0.35 QC 5.4 (a) TC 303.15 K Carnot 1 TC TH QC Work QH TH Work 6.333 QH Work 5 1.583 10 kW Ans. 5 2.714 10 kW Ans. QC 1.764 4 10 kW Ans. 5 10 kW Ans. 623.15 K 0.55 123 Carnot 0.282 Ans. (b) 0.35 Carnot By Eq. (5.8), Carnot 0.55 TC TH 1 0.636 833.66 K Ans. TH Carnot 5.7 Let the symbols represent rates where appropriate. Calculate mass rate of LNG evaporation: 3 V 9000 molwt m P s 17 gm mLNG mol T 1.0133 bar PV molwt RT 298.15 K mLNG 6254 kg s Maximum power is generated by a Carnot engine, for which Work QH = QC QC TH 512 Work QH QC TH TC QH 1= QC TH TC 1 1 113.7 K QC kJ mLNG kg QC = TC 303.15 K QC 5.8 QC 3.202 Work QH Work 6 10 kW 6 10 kW 8.538 Ans. 6 5.336 Ans. 10 kW Take the heat capacity of water to be constant at the valueCP (a) T1 273.15 K SH2O Sres CP ln Q T2 T2 T2 T1 373.15 K Q CP T2 SH2O 1.305 1.121 kJ kg K Q kJ kg K Sres T1 124 Ans. 4.184 418.4 kJ kg K kJ kg Stotal SH2O Sres Stotal 0.184 kJ Ans. kgK (b) The entropy change of the water is the same as in (a), and the total heat transfer is the same, but divided into two halves. Q Sres 2 Stotal 1 373.15 K 1 323.15 K Sres Sres Stotal SH2O 0.097 1.208 kJ kJ kgK Ans. kgK (c) The reversible heating of the water requires an infinite number of heat reservoirs covering the range of temperatures from 273.15 to 373.15 K, each one exchanging an infinitesimal quantity of heat with the water and raising its temperature by a differential increment. 5.9 P1 T1 P1 V n R T1 (a) Const.-V heating; T2 T1 Q n CV But P2 P1 = 1.443 mol U= Q T2 By Eq. (5.18), 1 T1 Whence 3 0.06m 5 CV 2 R W = Q = n CV T2 Q 15000 J T1 3 10 K S = n CP ln T2 V 500 K n 1 bar S T2 T1 R ln n C V ln T2 T1 P2 P1 S 20.794 (b) The entropy change of the gas is the same as in (a). The entropy change of the surroundings is zero. Whence Stotal = 10.794 J K Ans. The stirring process is irreversible. 125 J K Ans. 5.10 (a) The temperature drop of the second stream (B) in either case is the same as the temperature rise of the first stream CP (A), i.e., 120 degC. The exit temperature of the second stream is therefore 200 degC. In both cases we therefore have: SA SA 8.726 463.15 CP ln SB 343.15 J mol K 473.15 593.15 CP ln SB 7 R 2 6.577 J Ans. mol K (b) For both cases: Stotal SA Stotal SB 2.149 J mol K Ans. (c) In this case the final temperature of steam B is 80 degC, i.e., there is a 10-degC driving force for heat transfer throughout the exchanger. Now SA SA 8.726 463.15 343.15 CP ln Stotal SB J mol K SA SB SB dQ Since dQ/T = dS, 8.512 Stotal dW 5.16 By Eq. (5.8), Ans. J mol K dW = dQ T dW = dQ J mol K 0.214 T =1 353.15 473.15 CP ln Ans. T dQ T T dS Integration gives the required result. T1 Q CP T2 T2 600 K T1 Q 5.82 T 400 K 126 3J 10 mol 300 K S CP ln Work Q Q T2 T1 T Q 5.17 TH1 11.799 Work S Q Work Q Sreservoir S S 2280 3540 Sreservoir T J mol K Sreservoir 0 600 K TC1 J mol K J mol Ans. J mol 11.8 Ans. J Ans. mol K Process is reversible. 300 K TH2 For the Carnot engine, use Eq. (5.8): W QH1 TC2 300 K = TH1 250 K TC1 TH1 The Carnot refrigerator is a reverse Carnot engine. W TH2 TC2 = Combine Eqs. (5.8) & (5.7) to get: TC2 QC2 Equate the two work quantities and solve for the required ratio of the heat quantities: TC2 TH1 TC1 Ans. r r 2.5 TH1 TH2 TC2 5.18 (a) T1 H S (b) C p T2 Cp ln T2 T1 T2 1.2bar T1 R ln P2 mol 4.365 S P1 3J H = 5.82 10 450K H P1 300K 1.582 S = 1.484 127 P2 6bar 3J 10 mol J mol K J mol K Ans. Ans. Cp 7 R 2 3J (c) H = 3.118 10 (d) H = 3.741 10 (e) H = 6.651 10 S = 4.953 mol 3J S = 2.618 mol 3J J mol K J mol K S = 3.607 mol J mol K 5.19This cycle is the same as is shown in Fig. 8.10 on p. 305. The equivalent states are A=3, B=4, C=1, and D=2. The efficiency is given by Eq. (A) on p. 305. Temperature T4 is not given and must be calaculated. The following equations are used to derive and expression for T4. For adiabatic steps 1 to 2 and 3 to 4: T1 V 1 1 = T2 V 2 1 1 T3 V 3 For constant-volume step 4 to 1: P2 T2 = P3 T3 Solving these 4 equations for T4 yields: T4 = T1 7 R 2 T1 ( 200 T4 T1 T3 Eq. (A) p. 306 1 ( 1000 T4 T2 T3 1.4 Cv T2 273.15) K T2 Cp 5 R 2 Cv 1 V1 = V4 For isobaric step 2 to 3: Cp = T4 V 4 T3 ( 1700 873.759 K 1 T4 T1 T3 T2 128 273.15) K 0.591 Ans. 273.15) K 5.21 CV CP P1 R P2 2 bar CP 7 bar T1 298.15 K 1.4 CV With the reversible work given by Eq. (3.34), we get for the actual W: 1 Work 1.35 P2 R T1 1 3J Work 3.6 10 P1 1 But Q = 0, and W = mol Whence S 5.25 P T2 CP ln T1 T 4 R ln T1 Work P2 S 2.914 P1 T2 T1 T2 U = C V T2 471.374 K J mol K CV Ans. 800 Step 1-2: Volume decreases at constant P. Heat flows out of the system. Work is done on the system. W12 = P V2 V1 = R T2 T1 Step 2-3: Isothermal compression. Work is done on the system. Heat flows out of the system. W23 = R T2 ln P3 P2 = R T2 ln P3 P1 Step 3-1: Expansion process that produces work. Heat flows into the system. Since the PT product is constant, P dT T dP = 0 PV= RT P dV = R dT T dP = dT P P dV V dP = R dT (A) V dP = R dT RT dP P 129 In combination with (A) this becomes P dV = R dT R dT = 2 R dT Moreover, P3 = P 1 T1 T3 = P1 T1 T2 V1 P dV = 2 R T 1 W31 = T3 = 2 R T1 T2 V3 Q31 = U31 Q31 = CV = CP W31 = CV T1 2R Wnet W12 = Qin T1 T3 2 R T1 T3 = C P W23 R T1 T3 T2 W31 Q31 7 R 2 W23 W31 Q31 R T2 R T2 ln W12 T2 350 K 1.5 bar P3 P1 T1 W23 T2 3J T1 T2 2.017 10 W31 T2 2.91 W23 P1 R T1 W12 P3 2 R T1 CP 700 K P1 W12 T1 5.82 10 Q31 1.309 10 W31 0.068 Q31 130 10 mol 3J mol 3J mol 4J mol Ans. 5.26 T 403.15 K P1 By Eq. (5.18), S P2 2.5 bar R ln P2 P1 S 7.944 Tres 6.5 bar 298.15 K J mol K Ans. With the reversible work given by Eq. (3.27), we get for the actual W: Work Q 1.3 R T ln Work P2 P1 (Isothermal compresion) Work 4.163 10 3J mol Q here is with respect to the system. So for the heat reservoir, we have Sres Q Sres Tres Stotal S Stotal Sres 13.96 6.02 J Ans. mol K J mol K Ans. 5.27 (a) By Eq. (5.14) with P = const. and Eq. (5.15), we get for the entropy change of 10 moles n 10 mol S n R ICPS 473.15K 1373.15K 5.699 0.640 10 S 536.1 J K 3 0.0 5 1.015 10 Ans. (b) By Eq. (5.14) with P = const. and Eq. (5.15), we get for the entropy change of 12 moles n 12 mol S n R ICPS 523.15K 1473.15K 1.213 28.785 10 S 2018.7 J K Ans. 131 3 8.824 10 6 0.0 5.28 (a) The final temperature for this process was found in Pb. 4.2a to be 1374.5 K. The entropy change for 10 moles is then found as follows n 10 mol S n R ICPS 473.15K 1374.5K 1.424 14.394 10 S 900.86 J K 3 4.392 10 6 0.0 Ans. (b) The final temperature for this process was found in Pb. 4.2b to be 1413.8 K. The entropy change for 15 moles is then found as follows: n 15 mol S n R ICPS 533.15K 1413.8K 1.967 31.630 10 S 2657.5 J 3 9.873 10 6 0.0 Ans. K (c) The final temperature for this process was found in Pb. 4.2c to be 1202.9 K. The entropy change for 18.14 kg moles is then found as follows n 18140 mol S S n R ICPS 533.15K 1202.9K 1.424 14.394 10 6J 1.2436 10 K 3 4.392 10 6 0.0 Ans. 5.29 The relative amounts of the two streams are determined by an energy balance. Since Q = W = 0, the enthalpy changes of the two streams must cancel. Take a basis of 1 mole of air entering, and let x = moles of chilled air. Then 1 - x = the moles of warm air. T0 298.15 K Temperature of entering air T1 248.15 K Temperature of chilled air T2 348.15 K Temperature of warm air x C P T1 x 0.3 T0 ( 1 x)CP T2 T0 = 0 (guess) 132 x Given 1 x = T2 T0 T1 T0 x x Find x () 0.5 Thus x = 0.5, and the process produces equal amounts of chilled and warmed air. The only remaining question is whether the process violates the second law. On the basis of 1 mole of entering air, the total entropy change is as follows. CP 7 2 P0 R Stotal x CP ln Stotal 12.97 P 5 bar T1 ( 1 T0 J mol K 1 bar T2 x)CP ln R ln T0 P P0 Ans. Since this is positive, there is no violation of the second law. 5.30 T1 Tres 303.15 K CV CP R Q= Sres Tres CP ln Stotal S T2 T1 R ln Sres P1 353.15 K Work Q Sres S T2 523.15 K 1800 U J mol Work 5.718 Q J mol K P2 CP Q S P1 Stotal 133 3.42 3 bar 7 2 P2 1 bar R C V T2 T1 Work 3J 1.733 10 2.301 mol J mol K J PROCESS IS POSSIBLE. mol K 5.33 For the process of cooling the brine: kJ T mdot 40 K CP 3.5 T1 ( 273.15 25) K T1 ( 273.15 15) K T2 ( 273.15 30) K T 0.27 258.15 K T kg sec 298.15 K T2 20 303.15 K kg K H S CP ln kJ kg H T2 T1 Eq. (5.26): Wdotideal By Eq. (5.28): 140 S CP T 0.504 mdot H kJ kg K T S Wdotideal Wdotideal Wdot t Wdot 256.938 kW 951.6 kW Ans. t 5.34 E i 110 volt Wdotmech 1.25 hp Wdotelect At steady state: Qdot Wdotelect Qdot T SdotG = Qdot SdotG Wdotelect T 9.7 amp 300 K Wdotelect iE Wdotmech = 1.067 dt U =0 dt dt S =0 dt Qdot Qdot T 134 134.875 W SdotG Wdotmech 0.45 W K Ans. 3 10 W 5.35 i 25 ohm 2 Wdotelect T 10 amp Wdotelect i At steady state: 300 K 3 2.5 10 W dt U =0 dt SdotG 8.333 kmol hr T1 ( 25 Cp R T2 Cp 1 RT dT ln watt K P2 10bar P2 1.2bar 7 5 Cv T2 Qdot T Ans. Cp (a) Assuming an isenthalpic process: S (b) = R Wdotelect SdotG P1 273.15) K Cv 10 Qdot dt S =0 dt 10 watt 7 R 2 Cp SdotG = 3 2.5 5.38 mdot Wdotelect = Qdot T Qdot Qdot T2 T1 298.15 K Ans. Eq. (5.14) P1 T1 T2 7 R ln T1 2 S (c) SdotG (d) T 5.39(a) T1 T H R ln mdot S (0 2 273.15) K P2 P1 SdotG 48.966 Wlost T 500K P1 300K Basis: 1 mol n C p T2 T1 S 6bar T2 n Ws 135 17.628 W K S 371K J mol K Ans. Ans. Wlost P2 5.168 3J 10 1.2bar mol Cp Ans. 7 R 2 1mol H Ws 3753.8 J Ans. S n Cp ln T2 T1 P2 R ln S P1 Eq. (5.27) Wideal H Eq. (5.30) Wlost Wideal Eq. (5.39) SG T S Ws SG T K Ans. 1409.3 J Wlost Wlost J 5163 J Ans. J K Ans. Wideal Wideal Ws 4.698 4.698 Wlost SG (a) 5163J 1409.3J 4.698 J K (b) 2460.9J 2953.9J 493J 1.643 J K (c) 3063.7J 4193.7J 1130J 3.767 J K (d) 3853.5J 4952.4J 1098.8J 3.663 (e) 5.41 3753.8J 3055.4J 4119.2J 1063.8J 3.546 P1 S R ln SdotG QH P2 S P1 mdot S Wdotlost 5.42 P2 2500kPa 1kJ W 0.023 SdotG T SdotG W QH actual K mdot 20 mol sec mol K kJ sec K Ans. 140.344 kW Ans. TH ( 250 TC actual 300K K J kJ 0.468 Wdotlost 0.45kJ T 150kPa J ( 25 0.45 136 273.15)K 273.15)K TH 523.15 K TC 298.15 K TC 1 max actual> max, Since 5.43 QH the process is impossible. 50 kJ Q2 100 kJ T1 350 K T2 250 K QH Q1 Q2 TH T1 T2 SG 0.27 550 K (a) SG (b) Wlost 5.44 0.43 Q1 150 kJ TH max TH Wlost T SG (a) TH 750 MW 1 max Ans. Ans. 81.039 kJ TC ( 20 TC 273.15)K 273.15)K 293.15 K Ans. 0.502 max TH 300 K kJ K 588.15 K TC Wdot QdotH ( 315 TH Wdot T QdotC QdotH Wdot QdotC 745.297 MW max (b) 0.6 QdotC QdotH Wdot QdotH max Wdot QdotC 1.742 (minimum value) QdotH 3 Cp 1 cal gm K T m 165 s QdotC Vdot 137 Cp 1 9 10 W (actual value) 10 MW 3 River temperature rise: Vdot 2.492 gm 3 cm T 2.522 K Ans. 5.46 T1 ( 20 P1 273.15)K T2 P2 5bar ( 27 273.15) K 1atm T3 ( 22 273.15)K First check the First Law using Eqn. (2.33) neglect changes in kinetic and potential energy. 6 H 7 1 7 H 3 R ICPH T1 T2 3.355 0.575 10 ICPH T1 T3 3.355 0.575 10 8.797 10 4 kJ mol 3 5 0 0 0.016 10 0.016 10 5 R H is essentially zero so the first law is satisfied. Calculate the rate of entropy generation using Eqn. (5.23) 6 SG R ICPS T1 T2 3.355 0.575 10 7 1 7 SG P R ICPS T1 T3 3.355 0.575 10 0.013 5.47 a) Vdot 3 kJ mol K Since SG 5 0 3 0.016 10 5 0 0.016 10 R ln P2 P1 0, this process is possible. 3 ft 100000 hr T1 459.67)rankine T2 T 1atm ( 70 ( 70 459.67)rankine ( 20 459.67)rankine Assume air is an Ideal Gas ndot P Vdot R T1 ndot 258.555 lbmol hr Calculate ideal work using Eqn. (5.26) Wideal ndot R ICPH T1 T2 3.355 0.575 10 T Wideal 3 0 R ICPS T1 T2 3.355 0.575 10 1.776 hp 138 5 0.016 10 3 0 5 0.016 10 3 b) Vdot P 3000 m T1 1atm ( 25 273.15) K T hr (5 2 273.15) K T2 (8 273.15) K 0 0.016 10 Assume air is an Ideal Gas P Vdot ndot ndot 34.064 mol s R T1 Calculate ideal work using Eqn. (5.26) Wideal ndot R ICPH T1 T2 3.355 0.575 10 T Wideal 5.48 T1 R ICPS T1 T2 3.355 0.575 10 3 5 5 0 0.016 10 1.952 kW ( 2000 Cp ( ) T T 3 459.67) rankine 3.83 (0 7 0.000306 T2 T rankine ( 300 Hv R 459.67) rankine Tsteam 970 459.67) rankine BTU M lbm ( 212 29 gm mol 459.67) rankine a) First apply an energy balance on the boiler to get the ratio of steam flow rate to gas flow rate.: T2 ndotgas Cp ( ) T Td mdotsteam Hv = 0 T1 T2 Cp ( ) T Td mdotndot T1 mdotndot Hv 15.043 lb lbmol Calculate the rate of entropy generation in the boiler. This is the sum of the entropy generation of the steam and the gas. SdotG = SdotGsteam SdotGgas 139 Calculate entropy generation per lbmol of gas: SdotG ndotgas mdotsteam = Ssteam ndotgas Sgas Hv Ssteam Ssteam Tsteam T2 Sgas C p ( T) dT 1.444 Sgas 9.969 SdotG 11.756 Wlost T 6227 BTU lb rankine 10 3 kg mol lb rankine T1 SdotG mdotndot Ssteam Sgas BTU BTU lbmol rankine Calculate lost work by Eq. (5.34) Wlost SdotG T b) Hsteam Hv Wideal Hsteam Hv Ssteam T Ssteam Tsteam Ssteam BTU Ans. lbmol Wideal 1.444 BTU lb rankine 205.071 BTU lb Calculate lbs of steam generated per lbmol of gas cooled. T2 C p ( T ) dT mn T1 mn Hv 15.043 lb lbmol Use ratio to calculate ideal work of steam per lbmol of gas Wideal mn 3.085 10 3 BTU lbmol Ans. T2 c) Cp ( T) dT Hgas T1 Wideal Hgas T Sgas Wideal 140 9.312 3 BTU 10 lbmol Ans. 5.49 T1 ( 1100 Cp ( ) T T 273.15) K 3.83 (5 2 T2 0.000551 T R K ( 150 Hv 273.15) K 273.15) K 2256.9 Tsteam ( 100 kJ kg M 29 gm mol 273.15) K a) First apply an energy balance on the boiler to get the ratio of steam flow rate to gas flow rate.: T2 Cp ( ) T Td ndotgas mdotsteam Hv = 0 T1 T2 Cp ( ) T Td T1 mdotndot mdotndot Hv 15.135 gm mol Calculate the rate of entropy generation in the boiler. This is the sum of the entropy generation of the steam and the gas. SdotG = SdotGsteam SdotGgas Calculate entropy generation per lbmol of gas: SdotG ndotgas Ssteam = mdotsteam ndotgas Ssteam Sgas Hv Ssteam Tsteam T2 Sgas Cp ( ) T dT T Sgas 6.048 41.835 T1 SdotG mdotndot Ssteam Sgas SdotG 49.708 Wlost 14.8 3 10 J kg K J mol K J mol K Calculate lost work by Eq. (5.34) Wlost SdotG T 141 kJ mol Ans. b) Hsteam Hv Wideal Hsteam Hv Ssteam T Ssteam Tsteam Ssteam 3 6.048 Wideal 10 453.618 J kg K kJ kg Calculate lbs of steam generated per lbmol of gas cooled. T2 Cp ( T) dT T1 mn mn Hv 15.135 gm mol Use ratio to calculate ideal work of steam per lbmol of gas Wideal mn 6.866 kJ mol Ans. T2 c) Cp ( T) dT Hgas T1 Wideal 5.50 T1 a) Hgas ( 830 Sethylene Sethylene Qethylene T Sgas 273.15)K T2 ( 35 21.686 0.09 kJ mol 273.15)K R ICPS T1 T2 1.424 14.394 10 3 Ans. T 4.392 10 ( 25 6 273.15)K 0 kJ mol K R ICPH T1 T2 1.424 14.394 10 3 4.392 10 6 0 kJ mol Qethylene 60.563 Wlost Sethylene T Wideal Qethylene Wlost 33.803 kJ mol Now place a heat engine between the ethylene and the surroundings. This would constitute a reversible process, therefore, the total entropy generated must be zero. calculate the heat released to the surroundings for Stotal = 0. 142 Sethylene QC T =0 Solving for QC gives: QC QC T Sethylene 26.76 kJ mol Now apply an energy balance around the heat engine to find the work produced. Note that the heat gained by the heat engine is the heat lost by the ethylene. QH Qethylene WHE QH QC WHE 33.803 kJ mol The lost work is exactly equal to the work that could be produced by the heat engine 143 Chapter 6 - Section A - Mathcad Solutions 6.7 At constant temperature Eqs. (6.25) and (6.26) can be written: dS = and V dP dH = 1 T V dP For an estimate, assume properties independent of pressure. T V S S 6.8 P1 270 K 1.551 10 3 3m 2.661 P1 P2 1200 kPa 3 K H Ans. Tc 551.7 Zc 408.1 K 1 1 H P1 J kg K Isobutane: 2.095 10 kg V P2 P2 381 kPa T V P2 J kg Ans. 0.282 CP 2.78 Vc cm 262.7 mol 4000 kPa molwt 2000 kPa P1 gm 58.123 mol J gm K 3 Eq. (3.63) for volume of a saturated liquid may be used for the volume of a compressed liquid if the effect of pressure on liquid volume is neglected. 359 T 0.88 360 K Tr 361 T Tc Tr 0.882 0.885 (The elements are denoted by subscripts 1, 2, & 3 2 V V c Zc 1 Tr 131.604 7 V 132.138 3 cm mol 132.683 Assume that changes in T and V are negligible during throtling. Then Eq. (6.8) is integrated to yield: 144 H=T S V 1 P2 T1 S but VP Then at 360 K, H=0 P1 S 0.733 J Ans. mol K We use the additional values of T and V to estimate the volume expansivity: 3 V V3 V V1 1 V V1 T 1.079 cm T mol 4.098835 3 10 T3 T T1 2K 1 K Assuming properties independent of pressure, Eq. (6.29) may be integrated to give S = CP Whence 6.9 T T T S T1 T 6 P P1 V1 P P1 K P2 T molwt CP 298.15 K 250 10 P VP 1 1 bar P2 6 45 10 bar 3 2 10 kPa Ans. 0.768 K 1500 bar 3 1 cm 1003 kg V1 3 By Eq. (3.5), Vave V2 V1 Vave 2 Vave 1 H 134.6 S S T kJ kg P2 970.287 cm P1 V2 U kJ kg K Ans. Q Q H U P1 P1 By Eqs. (6.28) & (6.29), kg Ans. Vave P2 0.03636 P2 3 V2 H V1 exp cm 937.574 kg 5.93 TS 10.84 145 P2 V 2 kJ kg P1 V 1 Ans. Work kJ Ans. Work kg 4.91 U kJ kg Q Ans. 6.10 For a constant-volume change, by Eq. (3.5), T2 T1 36.2 10 T2 P2 6.14 --- 6.16 P2 5 K T1 P1 = 0 T1 1 298.15 K 4.42 10 P1 P2 5 205.75 bar Vectors containing T, P, Tc, Pc, and T2 bar P1 1 323.15 K 1 bar Ans. for Parts (a) through (n): 300 40 308.3 61.39 .187 175 75 150.9 48.98 .000 575 30 562.2 48.98 .210 500 50 425.1 37.96 .200 325 60 304.2 73.83 .224 175 60 132.9 34.99 .048 575 T 35 K P 556.4 bar Tc 45.60 K Pc .193 bar 650 50 553.6 40.73 .210 300 35 282.3 50.40 .087 400 70 373.5 89.63 .094 150 50 126.2 34.00 .038 575 15 568.7 24.90 .400 375 25 369.8 42.48 .152 475 75 365.6 46.65 .140 Tr T Tc Pr P Pc 146 6.14 Redlich/Kwong equation: Pr Tr Eq. (3.53) q z Given z= 1 i q z q Eq. (3.52) zz Find z () Ii HRi R Ti SRi R ln Z i qi Eq. (3.54) 1.5 1 1 14 Z 0.42748 Tr Guess: Z 0.08664 Z ln Z i qi Z i Eq. (6.65b) i qi 1 i qi 1.5 qi Ii Eq. (6.67) The derivative in these i i qi 0.5 qi Ii Eq. (6.68) equations equals -0.5 HRi SRi -2.302·103 J 0.695 -5.461 0.605 -2.068·103 mol -8.767 0.772 -3.319·103 -4.026 0.685 -4.503·103 -6.542 0.729 -2.3·103 -5.024 0.75 -1.362·103 -5.648 0.709 -4.316·103 -5.346 0.706 -5.381·103 -5.978 0.771 -1.764·103 -4.12 0.744 -2.659·103 -4.698 0.663 -1.488·103 -7.257 0.766 -3.39·103 -4.115 0.775 -2.122·103 -3.939 0.75 -3.623·103 -5.523 147 J mol K Ans. 6.15 Soave/Redlich/Kwong equation: 0.08664 1 0.42748 2 0.5 c1 z z= 1 0.480 1.574 0.176 z q Eq. (3.52) Tr Z zz The derivative in the following equations equals: ci q Tri Find z () 0.5 i i 1 14 HRi R Ti Z Ii i qi ln 1 Z i qi Z Eq. (6.65b) i qi 0.5 Tri ci i 1 qi Ii Eq. (6.67) i SRi R ln Z i qi i ci Tri 0.5 qi Ii Eq. (6.68) i Z i qi HRi SRi J mol 0.691 -2.595·103 0.606 -2.099·103 0.774 -3.751·103 0.722 -7.408 0.741 -2.585·103 -5.974 0.768 -1.406·103 -6.02 0.715 -4.816·103 -6.246 0.741 -5.806·103 -6.849 0.774 -1.857·103 -4.451 0.749 -2.807·103 -5.098 0.673 -1.527·103 -7.581 0.769 -4.244·103 -5.618 0.776 -2.323·103 -4.482 0.787 -3.776·103 -6.103 J mol K -4.795 -4.821·103 148 -6.412 -8.947 2 Eq. (3.54) Eq. (3.53) q 1 Given Pr Tr Tr Guess: c Ans. 6.16 Peng/Robinson equation: Guess: Given c 2 0.5 c1 0.37464 Pr Tr Tr z 1 2 0.45724 0.07779 1 1 2 1.54226 0.26992 Eq. (3.53) q Eq. (3.54) Tr 1 z= 1 z q z Eq. (3.52) Z z The derivative in the following equations equals: ci Tri q Find ( z) 0.5 i i 1 14 HRi 1 Ii R Ti Z 2 i qi Z 2 1 i qi i Z ln i qi i ci R ln Z i qi i ci Eq. (6.65b) 0.5 Tri 1 qi Ii i SRi Eq. (6.67) 0.5 Tri Eq. (6.68) qi Ii i Z i qi 2 HRi SRi 0.667 -2.655·103 J -6.41 0.572 -2.146·103 mol -8.846 0.754 -3.861·103 -4.804 0.691 -4.985·103 -7.422 0.716 -2.665·103 -5.993 0.732 -1.468·103 -6.016 0.69 -4.95·103 -6.256 0.71 -6.014·103 -6.872 0.752 -1.917·103 -4.452 0.725 -2.896·103 -5.099 0.64 -1.573·103 -7.539 0.748 -4.357·103 -5.631 0.756 -2.39·103 -4.484 0.753 -3.947·103 -6.126 149 J mol K Ans. Lee/Kesler Correlation --- By linear interpolation in Tables E.1--E.12: 0 h0 equals () HR RTc 1 h1 equals 0 s0 equals () SR R .686 () HR RTc () SR SR R s equals R .950 .471 .705 .024 1.003 1.709 .155 .774 RTc 1 s1 equals .093 .590 HR h equals .591 .675 1.319 .437 .725 .008 .993 .635 .744 Z0 .118 .165 1.265 .184 .705 .699 .770 Z1 .019 .962 h0 .102 1.200 .001 h1 .751 .444 .770 .550 .742 .007 .875 .598 .651 .144 1.466 .405 .767 .034 .723 .631 .776 .032 .701 .604 1.216 .211 .746 Z Z0 .154 Z1 Eq. (3.57) h 150 h0 h1 (6.85) HR ( Tc R) h .711 1.110 .492 .497 .549 .829 .443 .631 .590 .710 s0 .961 .276 .674 .750 .700 s1 .441 .517 .287 0.669 -1.138 SR i HRi si hi Zi Eq. (6.86) .563 .688 ( s R) .589 .491 SR .429 .511 s1 .555 .917 s0 .509 .587 s -0.891 -2.916·103 J mol -7.405 0.59 -1.709 -1.11 -2.144·103 0.769 -0.829 -0.612 -3.875·103 -5.091 0.699 -1.406 -0.918 -4.971·103 -7.629 0.727 -1.135 -0.763 -2.871·103 -6.345 0.752 -1.274 -0.723 -1.407·103 -6.013 0.701 -1.107 -0.809 -5.121·103 -6.727 0.72 -1.293 -0.843 -5.952·103 -7.005 0.77 -0.818 -0.561 -1.92·103 -4.667 0.743 -0.931 -0.639 -2.892·103 -5.314 0.656 -1.481 -0.933 -1.554·103 -7.759 0.753 -0.975 -0.747 -4.612·103 -6.207 0.771 -0.793 -0.577 -2.438·103 -4.794 0.768 -1.246 -0.728 -3.786·103 -6.054 J mol K 151 -9.229 Ans. 6.17 T t 50 273.15 K The pressure is the vapor pressure given by the Antoine equation: T P () t d t 323.15 K 2788.51 t 220.79 exp 13.8858 P ( ) 36.166 50 P P () 1.375 t dPdt 36.166 kPa 1.375 kPa dt K (a) The entropy change of vaporization is equal to the latent heat divided by the temperature. For the Clapeyron equation, Eq. (6.69), we need the volume change of vaporization. For this we estimate the liquid volume by Eq. (3.63) and the vapor volume by the generalized virial correlation. For benzene: Tc T Tc 3 Vc 259 cm mol Pc 562.2 K Tr 0.210 Tr Zc 48.98 bar P Pc Pr 0.575 Pr 0.271 0.007 By Eqs. (3.65), (3.66), (3.61), & (3.63) B0 0.422 0.083 Tr Vvap RT P 1.6 1 By Eq. (3.72), B0 B1 0.941 0.139 0.172 Tr B0 Vliq B1 Pr Vvap Tr V c Zc 1 Tr B1 4.2 1.621 3 4 cm 7.306 10 mol 3 2/7 Vliq 93.151 cm mol Solve Eq. (6.72) for the latent heat and divide by T to get the entropy change of vaporization: S dPdt Vvap S Vliq 100.34 J mol K Ans. 102.14 J mol K Ans. (b) Here for the entropy change of vaporization: S RT dPdt P S 152 6.20 The process may be assumed to occur adiabatically and at constant pressure. It is therefore isenthalpic, and may for calculational purposes be considered to occur in two steps: (1) Heating of the water from -6 degC to the final equilibrium temperature of 0 degC. (2) Freezing of a fraction x of the water at the equilibrium T. Enthalpy changes for these two steps sum to zero: CP t CP x Hfusion = 0 Hfusion 333.4 joule gm J gm K 4.226 t CP t x x Hfusion 6K 0.076 Ans. The entropy change for the two steps is: T2 S T1 273.15 K CP ln ( 273.15 x Hfusion T2 T2 T1 6) K S 1.034709 10 3 J Ans. gm K The freezing process itself is irreversible, because it does not occur at the equilibrium temperature of 0 degC. H1 1156.3 BTU lbm H2 1.7320 BTU S2 lbm rankine 1.9977 BTU 1533.4 S1 6.21 Data, Table F.4: H H2 H1 S S2 H 377.1 BTU lbm S 0.266 lbm BTU lbm rankine S1 BTU Ans. lbm rankine For steam as an ideal gas, apply Eqs. (4.9) and (5.18). [t in degF] T1 ( 227.96 P1 20 psi T1 382.017 K T2 459.67)rankine P2 50 psi T2 810.928 K 153 ( 1000 459.67)rankine molwt 18 lb lbmol R MCPH T1 T2 3.470 1.450 10 molwt H H 372.536 BTU lbm 3 5 0.0 0.121 10 T2 T1 Ans. R MCPS T1 T2 3.470 1.450 10 S 3 5 0.0 0.121 10 ln T2 T1 ln P2 P1 molwt S 0.259 BTU lbm rankine Ans. 6.22 Data, Table F.2 at 8000 kPa: 3 Vliq cm 1.384 gm Hliq 1317.1 J gm 3 Vvap cm 23.525 gm Hvap 2759.9 6 mliq mliq mvap 54.191 kg mliq Hliq Stotal mliq Sliq J gm Svap J gm K 5.7471 J gm K 6 0.15 10 3 cm 2 Vliq Htotal 3.2076 Sliq mvap mvap Hvap 0.15 10 3 cm 2 Vvap 3.188 kg Htotal 154 Ans. Stotal mvap Svap 80173.5 kJ 192.145 kJ K Ans. 6.23 Data, Table F.2 at 1000 kPa: 3 Vliq 1.127 Hliq gm 762.605 J gm Sliq Hvap cm 2776.2 J gm Svap 3 Vvap 194.29 cm gm Let x = fraction of mass that is vapor (quality) x Vvap 70 30 x 2.1382 6.5828 0.5 ( 1 H ( 1 x)Hliq H 789.495 J gm ( 1 2.198 J gm K (Guess) 0.013 S x Hvap gm K Find x () S x)Vliq = x x Given J x)Sliq x Svap J gm K Ans. 6.24 Data, Table F.3 at 350 degF: 3 3 Vliq Hliq 321.76 Vvap mliq 3 lbm 1 Htotal lbm 1192.3 BTU lbm mvap = 3 lbm mvap Vvap = 50 mliq Vliq mliq mliq mvap BTU ft 3.342 lbm Hvap ft 0.01799 lbm mliq 50 Vliq 50 mliq Vliq = 3 lbm Vvap 2.364 lb Vvap 3 lbm mvap mliq mliq Hliq mvap Hvap Htotal 155 0.636 lb 1519.1 BTU Ans. 3 6.25 1 cm 0.025 gm V Data, Table F.1 at 230 degC: 3 Vliq 1.209 cm Hliq gm 990.3 J gm 3 Vvap 71.45 cm Hvap gm J gm 2802.0 x Vvap x H ( 1 x)Hliq x Hvap S x 6.26 x)Vliq 0.552 Vtotal = mtotal Vliq Vtotal 3 0.15 m mtotal ( 1 Vliq x)Sliq S 4.599 x Svap J gm K 0.382 kg mliq mtotal mvap mliq 377.72 gm Ans. 3 Table F.1, 150 degC: Vvap Vtotal Vlv mtotal Vliq Vlv mvap 4.543 3 10 Vtot.liq Vtot.liq 379.23 cm kg mliq Vliq 156 cm 392.4 gm 3 cm 1.004 gm mvap Vvap J gm K mvap Vlv Vliq Vtotal J gm K Vliq Vvap J gm 6.2107 3 Table F.1, 30 degC: mtotal 1991 2.6102 Svap V V=( 1 H Sliq 3 Ans. cm 32930 gm 6.27 Table F.2, 1100 kPa: Hliq 781.124 J gm Hvap Interpolate @101.325 kPa & 105 degC: Const.-H throttling: H2 = Hliq H2 x 6.28 Hvap H2 x Hvap 2779.7 2686.1 J gm J gm Hliq Hliq x Hliq Ans. 0.953 Data, Table F.2 at 2100 kPa and 260 degC, by interpolation: H1 H2 J 2923.5 2923.5 gm J gm S1 6.5640 J gm K molwt 18.015 gm mol Final state is at this enthalpy and a pressure of 125 kPa. By interpolation at these conditions, the final temperature is 224.80 degC and S2 7.8316 J gm K S S2 S S1 1.268 J gm K Ans. For steam as an ideal gas, there would be no temperature change and the entropy change would be given by: P1 2100 kPa P2 125 kPa S P2 R ln P1 molwt S 1.302 J gm K Ans. 6.29 Data, Table F.4 at 300(psia) and 500 degF: H1 1257.7 H2 1257.7 BTU lbm BTU lbm S1 1.5703 BTU lbm rankine Final state is at this enthalpy and a pressure of 20(psia). By interpolation at these conditions, the final temperature is 438.87 degF and S2 1.8606 BTU lbm rankine S 157 S2 S1 S 0.29 BTU lbm rankine For steam as an ideal gas, there would be no temperature change and the entropy change would be given by: P1 P2 300 psi 6.30 lb lbmol P1 S molwt 0.299 BTU lbm rankine Ans. Data, Table F.2 at 500 kPa and 300 degC S1 7.4614 J gm K 1.0912 Hliq 340.564 S2 = S1 = Sliq H2 Hliq The final state is at this entropy and a pressure of 50 kPa. This is a state of wet steam, for which J gm K Sliq 6.31 18 P2 R ln S molwt 20 psi Svap Hvap J gm x Svap x Hvap 7.5947 J gm K 2646.9 Sliq x H2 Hliq J gm S1 Svap Sliq Sliq x 0.98 J gm Ans. xwater 0.031 Ans. xwater 0.122 Ans. 2599.6 Vapor pressures of water from Table F.1: At 25 degC: Psat P xwater 101.33 kPa At 50 degC: Psat xwater 3.166 kPa Psat P 12.34 kPa Psat P 158 6.32 Process occurs at constant total volume: Vtotal 3 ( 0.014 0.021)m Data, Table F.1 at 100 degC: Uliq 419.0 J gm 1.044 cm 1673.0 mvap 0.014 m Vvap Vliq mvap x mass V2 4 4.158 10 3 Vtotal V2 mass mliq gm 3 3 x mass gm cm Vvap gm 0.021 m mliq J 2506.5 3 3 Vliq Uvap cm 1.739 gm mvap (initial quality) This state is first reached as saturated liquid at 349.83 degC For this state, P = 16,500.1 kPa, and U2 Q 1641.7 U2 J gm Uliq Q U1 U1 1221.8 x Uvap J gm Uliq U1 419.868 J gm Ans. 3 6.33 Vtotal 0.25 m Data, Table F.2, sat. vapor at 1500 kPa: 3 V1 cm 131.66 gm U1 2592.4 J gm Of this total mass, 25% condenses making the quality 0.75 Since the total volume and mass don't change, we have for the final state: V2 = V1 = Vliq x= V1 Vvap Vliq Vliq x Vvap (A) Vtotal mass V1 x Whence Vliq Find P for which (A) yields the value x = 0.75 for wet steam 159 0.75 Since the liquid volume is much smaller than the vapor volume, we make a preliminary calculation to estimate: Vvap V1 3 Vvap x 175.547 cm gm This value occurs at a pressure a bit above 1100 kPa. Evaluate x at 1100 and 1150 kPa by (A). Interpolate on x to find P = 1114.5 kPa and Uliq U2 Q 782.41 Uliq J gm x Uvap mass U2 Uvap 2584.9 J gm U2 2134.3 Q Uliq 869.9 kJ U1 J gm Ans. 3 3 Vliq Uvap 6.34 Table F.2,101.325 kPa: Uliq J 418.959 gm Vvap J 2506.5 gm cm 1673.0 gm mliq cm 1.044 gm 0.02 m Vliq 3 3 mvap 1.98 m Vvap mtotal mliq mvap x mvap mtotal 3 V1 Vliq x Vvap Vliq U1 Uliq x Uvap Uliq cm V1 98.326 U1 540.421 x gm 0.058 J gm Since the total volume and the total mass do not change during the process, the initial and final specific volumes are the same. The final state is therefore the state for which the specific volume of saturated vapor is 98.326 cu cm/gm. By interpolation in Table F.1, we find t = 213.0 degC and U2 2598.4 J gm Q mtotal U2 160 U1 Q 41860.5 kJ Ans. 6.35 Data, Table F.2 at 800 kPa and 350 degC: 3 V1 cm 354.34 U1 gm 2878.9 J gm Vtotal 3 0.4 m The final state at 200 degC has the same specific volume as the initial state, and this occurs for superheated steam at a pressure between 575 and 600 kPa. By interpolation, we find P = 596.4 kPa and U2 J gm 2638.7 Q Vtotal V1 U2 Q U1 271.15 kJ Ans. 6.36 Data, Table F.2 at 800 kPa and 200 degC: U1 J gm 2629.9 S1 6.8148 J gm K mass 1 kg (a) Isothermal expansion to 150 kPa and 200 degC U2 J gm 2656.3 Q mass T S2 Also: J gm K S2 Q S1 Work 7.6439 392.29 kJ mass U2 U1 T 473.15 K Ans. Work Q 365.89 kJ (b) Constant-entropy expansion to 150 kPa. The final state is wet steam: J gm K Svap 7.2234 J gm K J gm Uvap 2513.4 J gm Sliq 1.4336 Uliq 444.224 x S1 Svap Sliq x Sliq U2 Uliq x Uvap W mass U2 U2 Uliq W U1 161 0.929 2.367 3J 10 262.527 kJ gm Ans. 6.37 Data, Table F.2 at 2000 kPa: x H1 Hvap 0.94 Hliq x Hvap 2797.2 H1 Hliq J Hliq gm 2.684 10 908.589 3J mass gm J gm 1 kg For superheated vapor at 2000 kPa and 575 degC, by interpolation: H2 3633.4 J gm 6.38 First step: Q mass H2 Q H1 Ans. 949.52 kJ Q12 = 0 W12 = U2 U1 Second step: W23 = 0 Q23 = U3 U2 For process: Q = U3 Table F.2, 2700 kPa: Uliq 977.968 Sliq 2.5924 x1 0.9 U1 S1 Uliq Sliq W = U2 U2 U1 J gm Uvap 2601.8 J gm J gm K Svap 6.2244 J gm K x1 Uvap x1 Svap Uliq U1 Sliq S1 3J 2.439 10 5.861 2 3m 10 2 gm sK Table F.2, 400 kPa: J gm K Svap 6.8943 J gm K J gm Uvap 2552.7 J gm Sliq 1.7764 Uliq 604.237 Vliq cm 1.084 gm 3 3 162 Vvap 462.22 cm gm Since step 1 is isentropic, S2 = S1 = Sliq U2 Uliq x2 Svap x2 Uvap S1 x2 Sliq Svap x2 Sliq 2.159 U2 Uliq Sliq 10 0.798 3J gm 3 V2 Vliq x2 Vvap V2 Vliq 369.135 cm gm V3 = V2 and the final state is sat. vapor with this specific volume. Interpolate to find that this V occurs at T = 509.23 degC and U3 J gm Q 2560.7 401.317 Whence J gm Q Ans. U3 Work U2 Work 280.034 U2 J gm Ans. U1 2605.8 J gm S1 7.0548 Table F.1,sat. vapor, 175 degC U2 2578.8 J gm S2 6.6221 mass T 6.39 Table F.2, 400 kPa & 175 degC: 4 kg Q mass T S2 Q S1 775.66 kJ Ans. ( 175 273.15)K W mass U2 W U1 667.66 kJ 6.40 (a)Table F.2, 3000 kPa and 450 degC: H1 3344.6 J gm S1 7.0854 J gm K Table F.2, interpolate 235 kPa and 140 degC: H2 2744.5 J gm S2 7.2003 163 J gm K Ans. U1 Q J gm K J gm K H H2 S S2 (b) T1 H H1 S1 ( 450 S J gm Ans. J gm K Ans. 600.1 0.115 T2 273.15)K ( 140 273.15)K T1 723.15 K T2 413.15 K P1 3000 kPa P2 235 kPa gm mol 3 5 R ICPH T1 T2 3.470 1.450 10 0.0 0.121 10 Eqs. (6.95) & (6.96) for an ideal gas: Hig molwt molwt 3 R ICPS T1 T2 3.470 1.450 10 Sig Tr1 Tr1 5 ln 0.0 0.121 10 620.6 J gm 647.1 K T1 Sig Pc Pr1 H S S P1 Tc 1.11752 Pr1 Tr2 Pc 0.13602 J gm K Tr2 T2 Tc 0.63846 HRB Tr1 Pr1 R Tc HRB Tr2 Pr2 molwt J Ans. 593.95 gm Hig Sig 0.078 R SRB Tr2 Pr2 J gm K SRB Tr1 Pr1 molwt Ans. 164 Ans. 0.345 220.55 bar P1 0.0605 Pr2 Pr2 The generalized virial-coefficient correlation is suitable here H P2 molwt Hig (c) Tc 18 P2 Pc 0.01066 6.41 Data, Table F.2 superheated steam at 550 kPa and 200 degC: 3 V1 385.19 cm U1 gm 2640.6 J gm S1 J gm K 7.0108 Step 1--2: Const.-V heating to 800 kPa. At the initial specific volume and this P, interpolation gives t = 401.74 degC, and U2 2963.1 J gm S2 7.5782 J gm K Q12 S3 S3 = S 2 Q23 = 0 7.5782 U1 Q12 Step 2--3: Isentropic expansion to initial T. U2 322.5 J gm J gm K Step 3--1: Constant-T compression to initial P. T 473.15 K Q31 T S1 Q31 S3 268.465 J gm For the cycle, the internal energy change = 0. Wcycle = Qcycle = Q12 1 Q31 Q31 = 0.1675 Q12 Wcycle Q12 Ans. 6.42 Table F.4, sat.vapor, 300(psi): T1 T1 459.67) rankine H1 ( 417.35 1202.9 877.02 rankine S1 1.5105 BTU lbm BTU lbm rankine Superheated steam at 300(psi) & 900 degF H2 Q12 1473.6 H2 BTU lbm H1 S2 Q31 1.7591 BTU lbm rankine T 1 S1 165 S3 S3 Q31 S2 218.027 BTU lbm For the cycle, the internal energy change = 0. Wcycle = Qcycle = Q12 1 6.43 Q31 Q31 Q12 Whence Ans. 0.1946 Q12 Wcycle = Data, Table F.2, superheated steam at 4000 kPa and 400 degC: S1 6.7733 J gm K For both parts of the problem: S2 S1 (a)So we are looking for the pressure at which saturated vapor has the given entropy. This occurs at a pressure just below 575 kPa. By interpolation, P2 = 572.83 kPa Ans. (b)For the wet vapor the entropy is given by x S2 = Sliq 0.95 x Svap Sliq So we must find the presure for which this equation is satisfied. This occurs at a pressure just above 250 kPa. At 250 kPa: Sliq 1.6071 S2 Sliq S2 6.7798 J gm K x Svap Svap 7.0520 J gm K Sliq J gm K Slightly > 6.7733 By interpolation P2 = 250.16 kPa Ans. 6.44 (a) Table F.2 at the final conditions of saturated vapor at 50 kPa: S2 7.5947 kJ kg K H2 2646.0 kJ kg S1 S2 Find the temperature of superheated vapor at 2000 kPa with this entropy. It occurs between 550 and 600 degC. By interpolation 166 t1 (degC) 559.16 Superheat: (b) mdot 5 t kg sec H1 ( 559.16 t 212.37)K Wdot kJ kg 3598.0 mdot H2 H1 Ans. 346.79 K 4760 kW Ans. Wdot 6.45 Table F.2 for superheated vapor at the initial conditions, 1350 kPa and 375 degC, and for the final condition of sat. vapor at 10 kPa: H1 kJ kg 3205.4 S1 7.2410 kJ kg K H2 2584.8 kJ kg If the turbine were to operate isentropically, the final entropy would be S2 S1 Table F.2 for sat. liquid and vapor at 10 kPa: kJ kg K Svap 8.1511 kJ kg K kJ kg Hvap 2584.8 kJ kg Sliq 0.6493 Hliq 191.832 x2 S2 Svap Sliq Sliq x2 H' Hliq 0.879 x2 Hvap H' 2.294 H2 H1 3 kJ 10 kg Ans. 0.681 H' H1 Hliq 6.46 Table F.2 for superheated vapor at the initial conditions, 1300 kPa and 400 degC, and for the final condition of 40 kPa and 100 degC: H1 3259.7 kJ kg S1 7.3404 kJ kg K H2 2683.8 kJ kg If the turbine were to operate isentropically, the final entropy would be S2 S1 Table F.2 for sat. liquid and vapor at 40 kPa: 167 Sliq 1.0261 kJ kg K Svap 7.6709 Hliq 317.16 kJ kg Hvap 2636.9 S2 x2 Sliq Svap H2 x2 Sliq kJ kg K kJ kg H' Hliq 0.95 H' 2.522 H1 6.47 Table F.2 at 1600 kPa and 225 degC: H 3 kJ kg P 1600 kPa S 6.5503 J gm K P0 1 kPa 498.15 K VR V 3 cm 132.85 gm 10 Hliq Ans. 0.78 H' H1 V x2 Hvap J gm 2856.3 J gm K Table F.2 (ideal-gas values, 1 kPa and 225 degC) Hig T J gm 2928.7 ( 225 Sig 10.0681 T 273.15)K T R molwt P The enthalpy of an ideal gas is independent of pressure, but the entropy DOES depend on P: HR H Sig Hig R molwt 3 VR 10.96 cm gm HR Reduced conditions: Tr T Tc 72.4 ln P P0 SR J gm S SR Sig 0.11 Sig J Ans. gm K 0.345 Tr Tc 647.1 K Pc 220.55 bar 0.76982 Pr P Pc Pr 0.072546 The generalized virial-coefficient correlation is suitable here B0 0.083 0.422 Tr 1.6 B0 0.558 B1 0.139 0.172 Tr 168 4.2 B1 0.377 By Eqs. (3.61) + (3.62) & (3.63) along with Eq. (6.40) Z 1 B0 B1 R Tc HR Pr Z Tr molwt R SR HRB Tr Pr molwt 3 VR 6.48 9.33 P cm HR gm 53.4 T 1000 kPa (Table F.2) SR 273.15) K 18.015 J gm Hv 2776.2 J gm K Sv 6.5828 762.605 Sl 2.1382 (b) (c) Slv VR T Sl G l 4.445 Vv 453.03 K J gm K T R molwt P Hlv 206.06 gm r 10 Gv 3J Hv Hl Sv Slv gm Hv Vl Slv J gm K Vv Hlv gm 2.014 J Vlv J 3 cm 193.163 gm Hl Ans. gm K gm mol Vv Hl J T cm 194.29 gm Vl (a) Gl 0.077 3 3 cm 1.127 gm Vlv SRB Tr Pr J gm ( 179.88 molwt RT ( Z 1) P molwt VR 0.935 4.445 T Sv G v Hlv r T Sl J gm K 206.01 4.445 J gm K 3 VR cm 14.785 gm Ans. For enthalpy and entropy, assume that steam at 179.88 degC and 1 kPa is an ideal gas. By interpolation in Table F.2 at 1 kPa: 169 J gm Hig 2841.1 J gm Sig 9.8834 J P0 gm K 1 kPa The enthalpy of an ideal gas is independent of pressure; the entropy DOES depend on P: HR Hv SR Sv Sig P R ln P0 molwt Sig Hig HR Sig 64.9 J gm Sig 3.188 Ans. SR 0.1126 J gm K J gm K Ans. (d) Assume ln P vs. 1/T linear and fit three data pts @ 975, 1000, & 1050 kPa. 975 Data: pp 178.79 1000 kPa t 1050 xi 1 yi 273.15 ti P dPdT T Slv 2 ln ppi Slope kPa dPdT Slope K 13 slope ( y) Slope x 22.984 4717 kPa K J gm K Ans. Slv 4.44 0.345 Reduced conditions: T Tc i 182.02 Tc 647.1 K Pc 220.55 bar 0.7001 Pr P Pc Pr 0.0453 Vlv dPdT Tr (degC) 179.88 Tr The generalized virial-coefficient correlation is suitable here B0 0.083 0.422 Tr 1.6 B0 0.664 B1 0.139 0.172 Tr 4.2 B1 0.63 By Eqs. (3.61) + (3.62) & (3.63) along with Eq. (6.40) Z 1 B0 B1 Pr Tr Z 170 0.943 VR RT ( 1) Z P molwt R Tc HR molwt R SRB Tr Pr molwt SR HRB Tr Pr 3 VR cm 11.93 6.49 T HR gm ( 358.43 43.18 J SR gm T 459.67) rankine (Table F.4) Vv BTU lbm Hv 1194.1 BTU lbm rankine Sv 1.5695 330.65 Sl 0.5141 Gl (b) (c) Slv VR Gv Hv Gv BTU lbm rankine T R molwt P r T VR Hl Sv Sl T Sv 89.91 Hlv r Hv BTU lbm 863.45 BTU lbm 1.055 Vv Hlv Vl Slv lbm rankine Vv Hlv BTU T Sl 89.94 150 psi Vlv BTU lbm 3 ft 2.996 lbm Hl P gm mol ft 3.014 lbm Hl (a) Gl 18.015 Ans. gm K 3 3 Vl J 818.1 rankine molwt ft 0.0181 lbm Vlv 0.069 0.235 ft BTU lbm 1.055 BTU lbm rankine 3 Ans. lbm For enthalpy and entropy, assume that steam at 358.43 degF and 1 psi is an ideal gas. By interpolation in Table F.4 at 1 psi: Hig 1222.6 BTU lbm Sig 2.1492 171 BTU lbm rankine P0 1 psi The enthalpy of an ideal gas is independent of pressure; the entropy DOES depend on P: HR Hv P R ln P0 molwt Sig SR HR Hig Sv Sig 28.5 Sig 0.0274 Ans. BTU lbm rankine 0.552 SR Sig BTU lbm BTU lbm rankine Ans. (d) Assume ln P vs. 1/T linear and fit threedata points (@ 145, 150, & 155 psia) 145 Data: pp 355.77 150 psi t 155 xi ti 1 459.67 (degF) 358.43 P T Slv 2 yi ln ppi Slope psi slope ( y) x Slv Vlv dPdT T Tc Tr 0.345 Tc 1.905 1.056 10 psi rankine BTU Ans. lbm rankine Pc 647.1 K Pr 0.7024 3 8.501 dPdT Slope rankine Reduced conditions: Tr 13 361.02 Slope dPdT i P Pc 220.55 bar Pr 0.0469 The generalized virial-coefficient correlation is suitable here B0 0.083 0.422 Tr 1.6 B0 0.66 B1 0.139 0.172 Tr 172 4.2 B1 0.62 By Eqs. (3.61) + (3.62) & (3.63) along with Eq. (6.40) Z 1 HR VR 6.50 B0 R Tc ft Z Tr HR lbm 19.024 For propane: Tc T ( 195 T Tr T Tc 273.15) K Tr BTU lbm SR (Z BTU lbm rankine 1) P0 135 bar P Pc Pr Ans. 0.152 42.48 bar P 468.15 K 1.266 0.0168 Pc 369.8 K P molwt R SRB Tr Pr molwt SR 3 RT VR 0.942 HRB Tr Pr molwt 0.1894 Pr B1 Pr 1 bar 3.178 Use the Lee/Kesler correlation; by interpolation, Z0 V Z Z1 ZRT P 0.1636 V 0.6141 cm 184.2 mol 0.586 R Tc HR1 1.802 SR1 0.717 R J mol K SR1 5.961 HR1 SR HR0 7.674 SR0 1.463 R SR0 12.163 HR H 0.639 Ans. HR1 2.496 R Tc HR0 Z Z1 3 HR0 HR Z0 3J 10 mol 3J 7.948 10 SR0 SR mol R ICPH 308.15K T 1.213 28.785 10 173 13.069 3 10 3J mol J mol K SR1 J mol K 8.824 10 6 0.0 HR S R ICPS 308.15K T 1.213 28.785 10 H 6734.9 J mol Ans. 6.51 For propane: T ( 70 Tr T Tc S Tr 6 8.824 10 15.9 J mol K 0.0 ln P P0 SR Ans. 369.8 K Pc 42.48 bar 0.152 343.15 K P0 101.33 kPa P 1500 kPa Tc T 273.15)K 3 P Pc Pr 0.92793 Pr 0.35311 Assume propane an ideal gas at the initial conditions. Use generalized virial correlation at final conditions. H S H R SRB Tr Pr P P0 ln 6.52 For propane: 1431.3 J mol Ans. S R Tc HRB Tr Pr 25.287 J mol K Ans. 0.152 3 cm Vc 200.0 Zc 0.276 Pc 42.48 bar Tc 369.8 K mol If the final state is a two-phase mixture, it must exist at its saturation temperature at 1 bar. This temperature is found from the vapor pressure equation: P A 1 bar D 1.38551 Given P = Pc exp T () T A () B T Find T) ( B 6.72219 T Tc 1 () T 1.5 C Guess: C 1 T 1.33236 () T 230.703 K 174 () T 3 D T () T 6 2.13868 200 K The latent heat of vaporization at the final conditions will be needed for an energy balance. It is found by the Clapeyron equation. We proceed exactly as in Pb. 6.17. P ( T) A ( T) Pc exp P d P ( T) dT 230.703 K 1 bar Tr Vvap Vvap Hlv 3 ( T) D kPa K Pr 0.815 B1 6 ( T) dPdT 0.024 B1 0.139 Pr Vliq Tr 4.428124 T Tr Tc 0.172 Tr B0 4.2 V c Zc 3 4 cm 10 T Vvap B0 1.6 1 1.847 C 4.428 Pc 0.422 RT P 1.5 ( T) P Pr 0.083 ( T) 1 T B0 B Tr kPa K 0.624 B1 1.109 2 1 Tr 7 3 Vliq mol Vliq dPdT Hlv 75.546 1.879 cm mol 4J 10 mol ENERGY BALANCE: For the throttling process there is no enthalpy change. The calculational path from the initial state to the final is made up of the following steps: (1) Transform the initial gas into an ideal gas at the initial T & P. (2) Carry out the temperature and pressure changes to the final T & P in the ideal-gas state. (3) Transform the ideal gas into a real gas at the final T & P. (4) Partially condense the gas at the final T & P. The sum of the enthalpy changes for these steps is set equal to zero, and the resulting equation is solved for the fraction of the stream that is liquid. For Step (1), use the generalized correlation of Tables E.7 & E.8, and let r0 = H R R Tc 0 and r1 = 175 H R R Tc 1 T1 Tr P1 370 K T1 Tr Tc By Eq. (6.85) r0 H1 P1 Pr 1.001 By interpolation, find: 200 bar r1 3.773 R Tc r0 Pr Pc 3.568 H1 r1 4.708 4J 1.327 10 mol For Step (2) the enthalpy change is given by Eq. (6.95), for which H2 R ICPH T1 T 1.213 28.785 10 3 6 8.824 10 0.0 4J H2 1.048 10 mol For Step (3) the enthalpy change is given by Eq. (6.87), for which Tr H3 230.703 K Tc Tr R Tc HRB Tr Pr H3 232.729 H1 Pr For Step (4), 0.0235 H4 = x Hlv J mol For the process, x 1 bar Pc Pr 0.6239 H2 H1 H3 x H2 0.136 H3 x Hlv = 0 Ans. Hlv 6.53 For 1,3-butadiene: Tc 425.2 K Vc 0.190 cm 220.4 mol Tn 268.7 K 101.33 kPa 3 Pc T Tr 42.77 bar 380 K T Tc Zc 0.267 P 1919.4 kPa T0 273.15 K P0 Tr 0.894 Pr P Pc Pr 176 0.449 Use Lee/Kesler correlation. HOWEVER, the values for a saturated vapor lie on the very edge of the vapor region, and some adjacent numbers are for the liquid phase. These must NOT be used for interpolation. Rather, EXTRAPOLATIONS must be made from the vapor side. There may be some choice in how this is done, but the following values are as good as any: Z0 0.7442 Z1 0.1366 Z Z0 Z1 0.718 3 Vvap ZRT P Vvap HR0 0.689 R Tc HR1 0.892 R Tc HR0 2.436 HR1 3.153 SR0 0.540 R SR1 0.888 R SR0 4.49 SR1 7.383 HR HR Hvap Svap Z 10 3J mol J mol K HR0 3.035 HR1 SR 3J 10 SR0 SR mol 6315.9 J mol Ans. 3 Svap Ans. mol 3J 10 mol J mol K SR1 5.892 R ICPH T0 T 2.734 26.786 10 R ICPS T0 T 2.734 26.786 10 Hvap 1182.2 cm J mol K 3 6 8.882 10 8.882 10 1.624 6 0.0 J mol K 0.0 ln HR P P0 SR Ans. For saturated vapor, by Eqs. (3.63) & (4.12) 2 Vliq V c Zc 1 Tr 3 7 Vliq 177 cm 109.89 mol Ans. 1.092 ln Hn R Tn Pc Hn Tn 0.930 By Eq. (4.13) 1.013 bar Tc H Hn 1 Tr Hvap Sliq Svap 0.38 H Tn 1 Hliq 14003 J mol Tc H Hliq 7687.4 H T Sliq 38.475 6.54 For n-butane: J mol 22449 J Ans. mol J mol K Ans. Tc 425.1 K Vc 0.200 cm 255 mol Tn 272.7 K 101.33 kPa 3 Pc T Tr Zc 37.96 bar 370 K 0.274 P T0 273.15 K P0 Tr T Tc 1435 kPa 0.87 Pr P Pc Pr 0.378 Use Lee/Kesler correlation. HOWEVER, the values for a saturated vapor lie on the very edge of the vapor region, and some adjacent numbers are for the liquid phase. These must NOT be used for interpolation. Rather, EXTRAPOLATIONS must be made from the vapor side. There may be some choice in how this is done, but the following values are as good as any: Z0 V Z 0.1372 ZRT P Z0 V Z1 0.7692 Z Z1 cm 1590.1 mol 3 HR0 2.145 HR1 10 3J mol 178 0.831 R Tc HR1 0.607 R Tc HR0 Ans. 2.937 3J 10 mol 0.742 SR0 0.485 R SR0 4.032 HR HR Hvap Svap SR1 SR1 J mol K HR0 2.733 HR1 mol R ICPH T0 T 1.935 36.915 10 7427.4 J mol SR0 SR 3 R ICPS T0 T 1.935 36.915 10 Hvap 6.942 SR 3J 10 0.835 R Ans. SR1 5.421 3 J mol K 6 11.402 10 11.402 10 Svap J mol K 6 4.197 0.0 0.0 HR P ln SR P0 J mol K Ans. For saturated vapor, by Eqs. (3.72) & (4.12) Vliq V c Zc 1 Tr 3 2/7 Vliq 1.092 ln Hn R Tn Pc By Eq. (4.13) H Hvap Sliq Svap J mol Hn 22514 H Tn 15295.2 Tc 1 Hn 1 Hliq Ans. 1.013 bar 0.930 cm 123.86 mol Tr 0.38 Tn Tc H Hliq 7867.8 J mol Ans. H T Sliq 37.141 J mol K Ans. 179 J mol 6.55 Under the stated conditions the worst possible cycling of demand can be represented as follows: 10, kg/ 000 hr 1/ hr 3 2/ hr 3 Dem and ( hr kg/ ) 1 hr tm e i 6, 000 4, kg/ 000 hr netst age or ofst eam netdepl i eton ofst eam This situation is also represented by the equation: 4000 = 6000 10000 1 = time of storage liquid 2 Solution gives hr 3 The steam stored during this leg is: mprime where 6000 mprime kg hr 4000 kg hr 1333.3 kg We consider this storage leg, and for this process of steam addition to a tank the equation developed in Problem 6-74 is applicable: m1 Hprime H1 Vtank P2 m2 = Hprime Hf2 Vf2 P1 Hfg2 Vfg2 Hfg2 Vfg2 We can replace Vtank by m2V2, and rearrange to get m2 m1 Hprime Hf2 Vf2 Hfg2 Vfg2 V 2 P2 However M1 v1 = m2 V2 = Vtank P1 Hfg2 and therefore 180 = Hprime Vfg2 m2 m1 = V1 V2 H1 Eq. (A) Making this substitution and rearranging we get Hprime Hf2 Vf2 Hfg2 Vfg2 P2 V2 Hfg2 P1 Vfg2 Hprime = H1 V1 In this equation we can determine from the given information everything except Hprime and Vprime. These quantities are expressed by H1 = Hf1 and x1 Hfg1 V1 = Vf1 x1 Vfg1 Therefore our equation becomes (with Hprime = Hg2) Hg2 Hf2 Hfg2 Vf2 Vfg2 P2 V2 P1 Hfg2 Vfg2 = Hg2 Hf1 Vf1 x1 Hfg1 Eq. (B) x1 Vfg1 In this equation only x1 is unknown and we can solve for it as follows. First we need V2: From the given information we can write: 0.95V2 = 1 x2 Vf2 therefore 19 = Then V2 = 1 0.05V2 = x2 Vg2 x2 Vf2 x2 Vg2 Vf2 Vg2 0.05 or 19Vg2 Vf2 Vf2 x2 = = 19Vg2 20 19 Vf2 Vf2 Eq. (C) 1 Vg2 Now we need property values: Initial state in accumulator is wet steam at 700 kPa. P1 700kPa We find from the steam tables Hf1 697.061 kJ Hg1 kg 2762.0 kJ kg 181 Hfg1 Hg1 Hf1 Hfg1 2064.939 kJ kg Vf1 1.108 3 3 3 cm Vg1 gm 272.68 cm Vfg1 gm Vf1 Vfg1 Vg1 Final state in accumulator is wet steam at 1000 kPa. From the steam tables Hf2 762.605 kJ kg Hg2 2776.2 1.127 Hfg2 Hf2 Hfg2 Hg2 cm Vg2 gm 194.29 gm 1000kPa 2013.595 kJ kg 3 3 3 Vf2 kJ kg P2 271.572 cm cm Vfg2 gm Vf2 Vfg2 Vg2 193.163 cm gm Solve Eq. (C) for V2 Vf2 Vg2 V2 19Vg2 0.05 V2 Vf2 Next solve Eq. (B) for x1 1.18595 Guess: x1 10 3 3m kg 0.1 Given Hg2 Hf2 Hfg2 Vf2 Vfg2 P2 V2 x1 x1 Find x1 4.279 Hfg2 P1 10 Vfg2 = Hg2 Hf1 Vf1 x1 Hfg1 x1 Vfg1 4 3 Thus V1 Vf1 V1 x1 Vfg1 V1 m2 = V2 m1 Eq. (A) gives 1.22419 and cm gm mprime = m2 m1 = 2667kg Solve for m1 and m2 using a Mathcad Solve Block: mprime Guess: m1 m2 m1 2 Given m1 m2 m1 3.752 = 4 V1 V2 10 kg m2 m1 = 2667lb m2 3.873 182 4 10 kg m1 m2 Find m1 m2 Finally, find the tank volume Vtank m2 V2 3 Vtank 45.9 m Ans. Note that just to store 1333.3 kg of saturated vapor at 1000 kPa would require a volume of: 3 1333.3kg Vg2 259 m One can work this problem very simply and almost correctly by ignoring the vapor present. By first equation of problem 3-15 m2 m1 Hprime = Hprime Hprime U1 U2 Hg2 = Hprime Uf1 Hprime Uf2 Hprime 2.776 = Hprime Hf1 Hprime Hf2 3 kJ 10 kg Given m2 m1 = Hprime Hprime m1 Hf1 Hf2 m2 3.837 V m2 m1 = 2667lb m2 Find m1 m2 4 10 kg m2 Vf2 0.95 V 6.56 Propylene: 3 45.5 m Ans. 0.140 T Tc 400.15 K P 365.6 K 38 bar Pc 46.65 bar P0 1 bar The throttling process, occurring at constant enthalpy, may be split into two steps: (1) Transform into an ideal gas at the initial conditions, evaluating property changes from a generalized correlation. (2) Change T and P in the ideal-gas state to the final conditions, evaluating property changes by equations for an ideal gas. Property changes for the two steps sum to the property change for the process. For the initial conditions: 183 T Tc Tr Tr P Pc Pr 1.095 Pr 0.815 Step (1): Use the Lee/Kesler correlation, interpolate. H0 H0 2.623 10 S0 4.697 mol J mol K H1 1.623 0.496 R S1 3J 4.124 HR 0.534 R Tc S1 0.565 R S0 H1 0.863 R Tc 3J 10 mol HR SR J mol K H0 H1 3J 2.85 S0 SR 10 mol S1 5.275 J mol K Step (2): For the heat capacity of propylene, A B 1.637 22.706 10 K 3 6.915 10 C K 6 2 Solve energy balance for final T. See Eq. (4.7). (guess) 1 HR = R AT Given 1 B2 T 2 2 Tf 0.908 Find C3 T 3 1 3 Tf T Sig R ICPS T Tf 1.637 22.706 10 Sig 22.774 SR Sig S 184 6.915 10 6 363.27 K Ans. J mol K S 3 1 28.048 J mol K 0.0 ln Ans. P0 P 6.57 Propane: Tc 0.152 Pc 369.8 K 42.48 bar P0 1 bar P 22 bar T 423 K The throttling process, occurring at constant enthalpy, may be split into two steps: (1) Transform into an ideal gas at the initial conditions, evaluating property changes from a generalized correlation. (2) Change T and P in the ideal-gas state to the final conditions, evaluating property changes by equations for an ideal gas. Property changes for the two steps sum to the property change for the process. For the initial conditions: Tr T Tc Tr P Pc Pr 1.144 Pr 0.518 Step (1): Use the generalized virial correlation 3J HR R Tc HRB Tr Pr HR 1.366 10 SR R SRB Tr Pr SR 2.284 J mol K mol Step (2): For the heat capacity of propane, A B 1.213 3 28.785 10 K 8.824 10 C K 6 2 Solve energy balance for final T. See Eq. (4.7). 1 (guess) HR = R AT Given 1 B 2 T 0.967 Find 2 2 C3 T 3 1 Tf Sig R ICPS T Tf 1.213 28.785 10 Sig 22.415 J mol K S SR Sig S 24.699 185 J mol K 1 Tf T 3 3 8.824 10 Ans. 6 Ans. 408.91 K 0.0 ln P0 P 6.58 For propane: T ( 100 T Tc Pc 42.48 bar T 273.15)K Tr 0.152 369.8 K Tc P0 1 bar P 10 bar Pr P Pc Pr 0.235 373.15 K Tr 1.009 Assume ideal gas at initial conditions. Use virial correlation at final conditions. H S R SRB Tr Pr 6.59 H2S: T1 Tr1 Tr1 J mol Ans. J mol K Ans. H ln P P0 801.9 S R Tc HRB Tr Pr 20.639 0.094 P1 400 K T1 Tc 373.5 K Pc 89.63 bar 5 bar T2 600 K P2 25 bar P1 Pr1 Tc Pr1 1.071 T2 Tr2 Pc Tc Tr2 0.056 Pr2 Pr2 1.606 P2 Pc 0.279 Use generalized virial-coefficient correlation for both sets of conditions. Eqs. (6.91) & (6.92) are written 3 5 H R ICPH T1 T2 3.931 1.490 10 0.0 0.232 10 R Tc HRB Tr2 Pr2 HRB Tr1 Pr1 S R ICPS T1 T2 3.931 1.490 10 R SRB Tr2 Pr2 H 7407.3 J mol 3 0.0 5 0.232 10 ln P2 P1 SRB Tr1 Pr1 S 186 1.828 J mol K Ans. 6.60 Carbon dioxide: Tc 0.224 Pc 304.2 K 73.83 bar P0 101.33 kPa P 1600 kPa T 318.15 K Throttling process, constant enthalpy, may be split into two steps: (1) Transform to ideal gas at initial conditions, generalized correlation for property changes. (2) Change T and P of ideal gas to final T & P. Property changes by equations for an ideal gas. Assume ideal gas at final T & P. Sum property changes for the process. For the initial T & P: T Tc Tr Tr P Pr 1.046 Pr Pc 0.217 Step (1): Use the generalized virial correlation R Tc HRB Tr Pr SR HR R SRB Tr Pr 587.999 SR HR 1.313 J mol J mol K Step (2): For the heat capacity of carbon dioxide, A 1.045 10 K B 5.457 3 5 D 1.157 10 K 2 Solve energy balance for final T. See Eq. (4.7). Given 1 (guess) HR = R A T 1 2 1 R ICPS T Tf 5.457 1.045 10 Sig 21.047 SR Sig T Tf T 3 0.0 S 22.36 J mol K 187 302.71 K 5 1.157 10 J mol K S 1 D Tf 0.951 Find Sig B2 T 2 Ans. ln P0 P Ans. 6.61 T0 P0 523.15 K S J mol K 0 A P 3800 kPa 120 kPa For the heat capacity of ethylene: 14.394 10 B 1.424 3 C K 4.392 10 6 2 K (a) For the entropy change of an ideal gas, combine Eqs. (5.14) & (5.15) with D = 0: (guess) 0.4 Given S = R A ln Hig Tf 1.185 Ws 10 3 308.19 K 4.392 10 6 Ans. 0.0 mol Ws (b) Ethylene: 11852 Tc 0.087 Tr0 Tc Tf T0 ln 4J Hig T0 P P0 1 2 R ICPH T0 Tf 1.424 14.394 10 Hig Tr0 C T0 0.589 Find 1 2 B T0 1.85317 Pr0 J mol Ans. 282.3 K P0 Pc Pc 50.40 bar Pr0 0.75397 At final conditions as calculated in (a) Tr T Tc Tr 1.12699 Pr P Pc Use virial-coefficient correlation. The entropy change is now given by Eq. (6.92): 0.5 (guess) Given 188 Pr 0.02381 S = R A ln SRB C T0 Pr Tc Tc 2 T T0 Tr T 1 ln P P0 SRB Tr0 Pr0 T Find Tr T0 1 2 B T0 303.11 K Ans. 1.074 The work is given by Eq. (6.91): Hig R ICPH T0 T 1.424 14.394 10 Hig Ws Hig Ws 6.62 T0 S A 1.208 11567 0.0 HRB Tr0 Pr0 Ans. P0 P 30 bar 2.6 bar For the heat capacity of ethane: B 1.131 6 mol J mol J mol K 4.392 10 4J R Tc HRB Tr Pr 493.15 K 0 10 3 19.225 10 K 3 C 5.561 10 6 2 K (a) For the entropy change of an ideal gas, combine Eqs. (5.14) & (5.15) with D = 0: (guess) Given 0.4 S = R A ln Find Hig B T0 1 2 C T0 2 T 0.745 R ICPH T0 T 1.131 19.225 10 189 T0 3 P P0 1 ln T 367.59 K 5.561 10 6 0.0 Ans. Hig 8.735 Ws 3J 10 mol Ws Hig (b) Ethane: J mol Tr0 Tc Ans. Tc 0.100 T0 Tr0 8735 Pc 48.72 bar Pr0 P0 Pr0 1.6153 Pc 305.3 K 0.61576 At final conditions as calculated in (a) T Tc Tr ( ) T P Pc Pr Tr ( ) 1.20404 T Pr 0.05337 Use virial-coefficient correlation. The entropy change is now given by Eq. (6.83): (guess) 0.5 S = R A ln B T0 SRB Tr Given T0 Tc Pr 2 T T0 Tr Tc ln 1 P P0 SRB Tr0 Pr0 T Find T 1 2 C T0 362.73 K 1.188 The work is given by Eq. (6.91): Hig Hig R ICPH T0 T 1.131 19.225 10 9.034 Ws Hig Ws 8476 10 5.561 10 3J mol R Tc HRB Tr Pr J mol 3 Ans. 190 HRB Tr0 Pr0 6 0.0 Ans. 6.63 n-Butane: T0 S A 425.1 K Pc P0 1 bar 37.96 bar P 7.8 bar For the heat capacity of n-butane: 36.915 10 K B 1.935 T0 Tr0 323.15 K J mol K 0 Tc 0.200 Tr0 Tc 3 C K Pr0 Pc P Pr HRB Tr0 Pr0 2 P0 Pr0 0.76017 6 11.402 10 Pr Pc = 0.05679 0.02634 0.205 HRB0 0.05679 The entropy change is given by Eq. (6.92) combined with Eq. (5.15) with D = 0: (guess) 0.4 S = R A ln Given B T0 SRB T0 Tc Find T Tc Tr Pr Tr 1 2 ln T T0 381.43 K 0.89726 The work is given by Eq. (6.91): Hig R ICPH T0 T 1.935 36.915 10 Hig 6.551 Ws Hig Ws 5680 3 11.402 10 3J 10 mol R Tc HRB Tr Pr J mol P P0 SRB Tr0 Pr0 T 1.18 1 2 C T0 HRB Tr0 Pr0 Ans. 191 6 0.0 Ans. 6.64 The maximum work results when the 1 kg of steam is reduced in a completely reversible process to the conditions of the surroundings, where it is liquid at 300 K (26.85 degC). This is the ideal work. From Table F.2 for the initial state of superheated steam: H1 kJ 3344.6 S1 kg 7.0854 kJ kg K From Table F.1, the state of sat. liquid at 300 K is essentially correct: H2 112.5 kJ kg S2 0.3928 kJ kg K T 300 K By Eq. (5.27), Wideal 6.65 H2 H1 T S2 Wideal S1 1224.3 kJ kg Ans. Sat. liquid at 325 K (51.85 degC), Table F.1: Hliq kJ 217.0 kg Psat 12.87 kPa P1 Sliq 325 K H1 Hliq S1 Sliq cm 1.013 gm Vliq For the compressed liquid at 325 K and 8000 kPa, apply Eqs. (6.28) and (6.29) with 8000 kPa T 3 kJ 0.7274 kg K 460 10 Vliq 1 T Vliq P1 P1 6 K 1 H1 Psat 223.881 S1 Psat 0.724 kJ kg kJ kg K For sat. vapor at 8000 kPa, from Table F.2: H2 2759.9 kJ kg S2 Q Heat added in boiler: 5.7471 H2 kJ kg K H1 T Q 300 K 2536 kJ kg Maximum work from steam, by Eq. (5.27): Wideal H1 H2 T S1 S2 192 Wideal 1029 kJ kg Work as a fraction of heat added: Frac Wideal Frac Q Ans. 0.4058 The heat not converted to work ends up in the surroundings. Q SdotG.surr Wideal T SdotG.system S1 kg 10 SdotG.surr sec S 2 10 kg 50.234 SdotG.system sec kW K 50.234 kW K Obviously the TOTAL rate of entropy generation is zero. This is because the ideal work is for a completely reversible process. 6.66 Treat the furnace as a heat reservoir, for which kg sec kg kW Qdot 50.234 K T Qdot 2536 SdotG kJ T 10 ( 600 SdotG 273.15)K 21.19 kW K T 873.15 K Ans. By Eq. (5.34) T 6.67 Wdotlost 300 K Wdotlost T SdotG For sat. liquid water at 20 degC, Table F.1: H1 kJ kg 83.86 S1 0.2963 kJ kg K For sat. liquid water at 0 degC, Table F.1: H0 0.04 kJ S0 kg 0.0000 kJ kg K For ice at at 0 degC: H2 H0 333.4 kJ kg S2 S0 193 333.4 kJ 273.15 kg K 6356.9 kW Ans. kJ H2 333.44 T S2 293.15 K kg 1.221 mdot 0.5 kJ kg K kg 0.32 t sec By Eqs. (5.26) and (5.28): Wdotideal mdot H2 H1 T S2 Wdotideal Wdot S1 Wdotideal Wdot 42.77 kW 13.686 kW Ans. t 6.68 This is a variation on Example 5.6., pp. 175-177, where all property values are given. We approach it here from the point of view that if the process is completely reversible then the ideal work is zero. We use the notation of Example 5.6: H1 2676.0 S2 0.0 kJ kg kJ kg K S1 Q' 2000 kJ kg K 7.3554 kJ kg H2 T kJ kg 0.0 273.15 K The system consists of two parts: the apparatus and the heat reservoir at elevated temperature, and in the equation for ideal work, terms must be included for each part. Wideal = Happaratus.reservoir Happaratus.reservoir = H2 Sapparatus.reservoir = S2 T' 450 K Given H1 S1 (Guess) kJ 0 = H2 H1 kg T' T Sapparatus.reservoir Q' Wideal = 0.0 Q' T' Q' T S2 S1 T' Find ( ) T' Q' T' 409.79 K (136.64 degC) 194 kJ kg Ans. 6.69 From Table F.4 at 200(psi): H1 BTU 1222.6 S1 lbm Hliq BTU 355.51 lbm Sliq 0.5438 Hliq H2 1.165 10 BTU x Hvap (at 420 degF) lbm rankine Hvap BTU lbm rankine H2 1.5737 1198.3 Svap 1.5454 (Sat. liq. and vapor) BTU lbm BTU x lbm rankine S2 3 BTU lbm Sliq x Svap S2 Hliq 1.505 0.96 Sliq BTU lbm rankine Neglecting kinetic- and potential-energy changes, on the basis of 1 pound mass of steam after mixing, Eq. (2.30) yields for the exit stream: H x S 0.5 H1 H Hvap Sliq H Hliq x Svap Sliq 1.54 BTU lbm 0.994 S Hliq 1193.6 x 0.5 H2 (wet steam) Ans. BTU lbm rankine By Eq. (5.22) on the basis of 1 pound mass of exit steam, SG 6.70 S 0.5 S1 SG 0.5 S2 2.895 10 4 BTU lbm rankine Ans. From Table F.3 at 430 degF (sat. liq. and vapor): 3 3 Vliq ft 0.01909 lbm Vvap ft 1.3496 lbm Uliq 406.70 BTU lbm Uvap 1118.0 VOLliq BTU lbm VOLliq mliq Vliq 195 Vtank mliq 79.796 ft 3 80 ft 3 4180 lbm VOLvap Vtank VOLliq VOLvap VOLvap mvap mvap Vvap mliq Uliq U1 mvap Uvap mliq U1 mvap 0.204 ft 3 0.151 lbm 406.726 BTU lbm By Eq. (2.29) multiplied through by dt, we can write, d mt Ut H dm = 0 (Subscript t denotes the contents of the tank. H and m refer to the exit stream.) m Integration gives: m2 U2 m1 U1 H dm = 0 0 From Table F.3 we see that the enthalpy of saturated vapor changes from 1203.9 to 1203.1(Btu/lb) as the temperature drops from 430 to 420 degF. This change is so small that use of an average value for H of 1203.5(Btu/lb) is fully justified. Then m2 U2 m1 m1 U1 mliq Have m = 0 Have mvap 1203.5 m2 ( mass) m1 Property values below are for sat. liq. and vap. at 420 degF 3 3 Vliq ft 0.01894 lbm Vvap ft 1.4997 lbm Uliq 395.81 BTU lbm Uvap 1117.4 V2 ( mass) U2 ( mass) mass Vtank m2 ( mass) Uliq x mass) ( x mass) Uvap ( 50 lbm (Guess) 196 Uliq BTU lbm V2 ( mass) Vliq Vvap Vliq BTU lbm mass m1 U1 U2 ( mass) Have U2 ( mass) Given mass = mass Find ( mass) mass Ans. 55.36 lbm 6.71 The steam remaining in the tank is assumed to have expanded isentropically. Data from Table F.2 at 4500 kPa and 400 degC: S1 6.7093 3 J gm K S2 = S1 = 6.7093 J gm K V1 64.721 cm Vtank gm 3 50 m By interpolation in Table F.2 at this entropy and 3500 kPa: 3 cm 78.726 gm V2 Vtank m1 6.72 m2 V1 Ans. t2 = 362.46 C Vtank V2 m m1 m2 m 137.43 kg Ans. This problem is similar to Example 6.8, where it is shown that Q= mt Ht H mt Here, the symbols with subscript t refer to the contents of the tank, whereas H refers to the entering stream. We illustrate here development of a simple expression for the first term on the right. The1500 kg of liquid initially in the tank is unchanged during the process. Similarly, the vapor initially in the tank that does NOT condense is unchanged. The only two enthalpy changes within the tank result from: 1. Addition of 1000 kg of liquid water. This contributes an enthalpy change of Hliq mt 2. Condensation of y kg of sat. vapor to sat. liq. This contributes an enthalpy change of y Hliq Thus Hvap = y Hlv mt Ht = Hliq mt y Hlv 197 Similarly, mt Vt = Vliq mt Whence mt Q = Hliq mt H At 250 degC: 209.3 Hliq Vliq mt 6.73 y Vlv mt Hliq H Given: C kJ kg K 3 Vliq cm 1.251 gm 3 kJ 1714.7 kg cm 48.79 gm Vlv 25.641 kg y Hlv Vtank 0.43 kJ kg kJ 1085.8 kg Hlv Q H mt Required data from Table F.1 are: 1000 kg At 50 degC: y y Hlv y Vlv = 0 T1 Q Ans. 832534 kJ 3 0.5 m 295 K kJ kg Hin 120.8 mtank 30 kg Data for saturated nitrogen vapor: 80 0.1640 85 2.287 0.1017 90 T 1.396 3.600 0.06628 95 K P 5.398 bar V 0.04487 100 7.775 0.03126 105 10.83 0.02223 110 14.67 0.01598 198 3 m kg 78.9 At the point when liquid nitrogen starts to accumulate in the tank, it is filled with saturated vapor nitrogen at the final temperature and having properties 82.3 85.0 H kJ kg 86.8 87.7 mvap Tvap Vvap Hvap Uvap 87.4 85.6 By Eq. (2.29) multiplied through by dt, d nt Ut H dm = dQ Subscript t denotes the contents of the tank; H and m refer to the inlet stream. Since the tank is initially evacuated, integration gives mvap Uvap Hin mvap = Q = mtank C Tvap Also, mvap = (A) T1 Vtank (B) Vvap Calculate internal-energy values for saturated vapor nitrogen at the given values of T: 56.006 U ( H 59.041 P V) 61.139 U 62.579 kJ kg 63.395 63.325 62.157 Fit tabulated data with cubic spline: Us lspline ( U) T Uvap () t Tvap Vs interp ( T U t) Us 100 K lspline ( V) T Vvap () t (guess) Combining Eqs. (A) & (B) gives: 199 interp ( T V t) Vs Given Uvap Tvap Tvap mvap 6.74 Hin = mtank C T1 Tvap Vvap Tvap Vtank Find Tvap Tvap 97.924 K mvap 13.821 kg Vtank Vvap Tvap Ans. The result of Part (a) of Pb. 3.15 applies, with m replacing n: m2 U2 H m1 U1 H =Q=0 Whence Also m2 H U2 = m1 H U2 = Uliq.2 x2 Ulv.2 V2 = Vliq.2 x2 Vlv.2 U1 V2 = Vtank m2 Eliminating x2 from these equations gives Vtank m2 H m2 Uliq.2 Vliq.2 Vlv.2 Ulv.2 = m1 H U1 which is later solved for m2 Vtank 3 50 m m1 16000 kg V1 V1 Data from Table F.1 @ 25 degC: 3 Vliq.1 cm 1.003 gm Uliq.1 104.8 kJ kg 3 Vlv.1 cm 43400 gm Ulv.1 2305.1 200 kJ kg Vtank m1 3.125 10 3 3m kg V1 x1 x1 Vliq.1 U1 4.889 10 5 Uliq.1 x1 Ulv.1 U1 Vlv.1 104.913 kJ kg Data from Table F.2 @ 800 kPa: Vliq.2 1.115 720.043 kJ kg Ulv.2 ( 2575.3 Ulv.2 3 cm 1.855 Uliq.2 gm 3 Vlv.2 ( 240.26 Vlv.2 1.115) cm m 0.239 kg gm 720.043) 3 Data from Table F.2 @ 1500 kPa: m1 H U1 Vtank m2 H msteam 6.75 Uliq.2 m2 Vliq.2 m1 H 2789.9 3 kJ 10 kg kJ kg Ulv.2 Vlv.2 m2 Ulv.2 2.086 4 10 kg Vlv.2 msteam 4.855 3 10 kg The result of Part (a) of Pb. 3.15 applies, with Whence kJ kg Ans. n1 = Q = 0 U2 = H From Table F.2 at 400 kPa and 240 degC H = 2943.9 kJ kg Interpolation in Table F.2 will produce values of t and V for a given P where U = 2943.9 kJ/kg. 201 1 384.09 303316 100 384.82 3032.17 1515.61 V2 385.57 t2 200 P2 300 387.08 gm 1010.08 400 i 386.31 3 cm 757.34 15 Vtank 3 1.75 m Vtank massi V2 i 5.77 10 3 T rises very slowly as P increases 0.577 mass 1.155 3 kg 1.733 2 massi 2.311 1 0 0 200 P2 6.76 3 Vtank 400 i Data from Table F.2 @ 3000 kPa: 2m 3 3 Vliq cm 1.216 gm Vvap cm 66.626 gm Hliq 1008.4 kJ kg Hvap 2802.3 x1 0.1 V1 V1 Vliq kJ kg x1 Vvap 7.757 10 3 3m kg Vliq m1 m1 Vtank V1 257.832 kg The process is the same as that of Example 6.8, except that the stream flows out rather than in. The energy balance is the same, except for a sign: Q= mt Ht H mtank 202 where subscript t denotes conditions in the tank, and H is the enthalpy of the stream flowing out of the tank. The only changes affecting the enthalpy of the contents of the tank are: 1. Evaporation of y kg of sat. liq.: y Hvap Hliq 2. Exit of of liquid from the tank: 0.6 m1 kg 0.6 m1 Hliq Thus mt Ht = y Hvap Hliq 0.6 m1 Hliq Similarly, since the volume of the tank is constant, we can write, mt Vt = y Vvap Whence Q= But y= Vliq 0.6 m1 Vliq = 0 0.6 m1 Vliq Vvap Vliq 0.6 m1 Vliq Hvap Vvap Vliq Hliq and H = Hliq 0.6 m1 Hliq 0.6 m1 = H mtank mtank and therefore the last two terms of the energy equation cancel: Q 6.77 0.6 m1 Vliq Hvap Vvap Vliq Hliq Q 5159 kJ Ans. Data from Table F.1 for sat. liq.: H1 100.6 kJ kg (24 degC) H3 355.9 kJ kg (85 degC) Data from Table F.2 for sat. vapor @ 400 kPa: H2 2737.6 kJ kg By Eq. (2.30), neglecting kinetic and potential energies and setting the heat and work terms equal to zero: H3 mdot3 H1 mdot1 H2 mdot2 = 0 203 Also mdot1 = mdot3 Whence mdot3 H1 mdot2 mdot1 mdot3 mdot2 0.484 H1 kg mdot3 mdot2 5 kg sec Ans. sec H3 H2 mdot2 kg sec Ans. mdot1 4.516 6.78 Data from Table F.2 for sat. vapor @ 2900 kPa: H3 2802.2 kJ kg S3 6.1969 kJ kg K mdot3 15 kg sec Table F.2, superheated vap., 3000 kPa, 375 degC: H2 3175.6 kJ kg S2 6.8385 kJ kg K Table F.1, sat. liq. @ 50 degC: 3 Vliq cm 1.012 gm Hliq Psat 12.34 kPa T 209.3 kJ kg Sliq 0.7035 kJ kg K 323.15 K Find changes in H and S caused by pressure increase from 12.34 to 3100 kPa. First estimate the volume expansivity from sat. liq, data at 45 and 55 degC: 3 V V ( 1.015 5 10 1.010) cm T gm 3 3 cm P 10 K 1 Vliq gm V T 3100 kPa 4.941 Apply Eqs. (6.28) & (6.29) at constant T: H1 Hliq S1 Sliq Vliq 1 Vliq P T P kJ kg kJ Psat 204 H1 211.926 S1 Psat 0.702 kg K 10 4 K 1 By Eq. (2.30), neglecting kinetic and potential energies and setting the heat and work terms equal to zero: H3 mdot3 H1 mdot1 H2 mdot2 = 0 Also mdot2 = mdot3 Whence mdot1 mdot2 mdot3 mdot1 mdot3 H3 H2 H1 H2 kg sec mdot1 mdot2 mdot1 1.89 13.11 Ans. kg sec For adiabatic conditions, Eq. (5.22) becomes SdotG S3 mdot3 SdotG 1.973 S1 mdot1 kJ sec K S2 mdot2 Ans. The mixing of two streams at different temperatures is irreversible. 6.79 Table F.2, superheated vap. @ 700 kPa, 200 degC: H3 2844.2 kJ kg S3 6.8859 kJ kg K Table F.2, superheated vap. @ 700 kPa, 280 degC: H1 3017.7 kJ kg S1 7.2250 kJ kg K mdot1 50 kg sec Table F.1, sat. liq. @ 40 degC: Hliq 167.5 kJ kg Sliq 0.5721 kJ kg K By Eq. (2.30), neglecting kinetic and potential energies and setting the heat and work terms equal to zero: H2 Also mdot2 H3 mdot3 Hliq mdot3 = mdot2 mdot1 H1 H3 H1 mdot1 H2 mdot2 = 0 mdot1 H3 mdot2 H2 For adiabatic conditions, Eq. (5.22) becomes 205 3.241 kg sec Ans. S2 mdot3 Sliq SdotG S3 mdot3 SdotG 3.508 mdot2 S1 mdot1 mdot1 S2 mdot2 kJ sec K Ans. The mixing of two streams at different temperatures is irreversible. 6.80 Basis: 1 mol air at 12 bar and 900 K (1) + 2.5 mol air at 2 bar and 400 K (2) = 3.5 mol air at T and P. T1 900 K T2 400 K P1 n1 1 mol n2 2.5 mol CP 600 K 2 bar CP 29.099 (guess) 1st law: T Given n1 CP T T T1 n2 CP T P Given T 5 bar T P R ln T1 P1 T P n2 CP ln R ln T2 P2 P Find ( ) P molwt 28.014 lb lbmol CP Ms 542.857 K 4.319 bar R 7 2 molwt = nitrogen rate in lbm/sec =0 Ans. J K Ans. CP 0.248 BTU lbm rankine = steam rate in lbm/sec Mn J mol K (guess) n1 CP ln P 7 R 2 T2 = 0 J Find ( ) T 2nd law: 6.81 P2 12 bar Mn 206 40 lbm sec (1) = sat. liq. water @ 212 degF entering (2) = exit steam at 1 atm and 300 degF (3) = nitrogen in at 750 degF T3 1209.67 rankine (4) = nitrogen out at 325 degF T4 784.67 rankine H1 180.17 H2 1192.6 BTU lbm S1 lbm BTU lbm rankine (Table F.3) S2 BTU 0.3121 1.8158 BTU lbm rankine (Table F.4) Eq. (2.30) applies with negligible kinetic and potential energies and with the work term equal to zero and with the heat transfer rate given by Ms 3 Given Ms lbm (guess) sec Ms H2 H1 Q = 60 Mn CP T4 Find Ms Ms T 3 = 60 3.933 lbm sec BTU Ms lbm BTU lbm Ms Ans. Eq. (5.22) here becomes SdotG = Ms S2 S4 S3 = CP ln T S1 Mn S4 T4 Q T3 Q T S3 60 BTU Ms lbm 529.67 rankine SdotG Ms S2 SdotG 2.064 S1 Mn CP ln BTU sec rankine T4 T3 Ans. 207 Q T Q 235.967 BTU sec 6.82 molwt 28.014 gm CP mol R 2 molwt 7 CP 1.039 J gm K Ms = steam rate in kg/sec Mn= nitrogen rate in kg/sec kg sec Mn 20 (3) = nitrogen in @ 400 degC T3 673.15 K (4) = nitrogen out at 170 degC T4 443.15 K (1) = sat. liq. water @ 101.33 kPa entering (2) = exit steam at 101.33 kPa and 150 degC H1 419.064 H2 2776.2 kJ kg S1 S2 kJ kg 1.3069 7.6075 kJ kg K (Table F.2) kJ (Table F.2) kg K By Eq. (2.30), neglecting kinetic and potential energies and setting the work term to zero and with the heat transfer rate given by Ms 1 Given Ms kg sec (guess) Ms H2 H1 Ms Find Ms Q = 80 Mn CP T4 1.961 T 3 = 80 kg kJ Ms kg kJ Ms kg Ans. sec Eq. (5.22) here becomes SdotG = Ms S2 S4 S3 = CP ln SdotG Ms S2 SdotG 4.194 S1 Mn S4 T4 T T3 S1 kJ sec K Q T S3 Mn CP ln Ans. 208 T4 T3 298.15 K Q T Q 80 kJ kg Ms 6.86 Methane = 1; propane = 2 T 363.15 K 1 Tc1 0.012 2 190.6 K y2 Tc2 0.5 1 y1 5500 kPa P y1 0.152 Zc1 0.286 Zc2 0.276 369.8 K Pc1 45.99 bar Pc2 42.48 bar The elevated pressure here requires use of either an equation of state or the Lee/Kesler correlation with pseudocritical parameters. We choose the latter. Tpc y1 Tc1 Tpc Ppc Tpr T Tpr Tpc 1.296 44.235 bar Ppr 280.2 K y1 Pc1 Ppc y2 Tc2 y2 Pc2 P Ppc Ppr 1.243 By interpolation in Tables E.3 and E.4: Z0 Z1 0.8010 y1 1 y2 2 0.1100 Z 0.082 Z0 For the molar mass of the mixture, we have: gm molwt molwt y1 16.043 y2 44.097 mol V Vdot D ZRT P molwt V mdot 4A 30.07 3 V Vdot D cm 14.788 gm mdot 3 4 cm 2.07 10 2.964 cm 209 sec Ans. Z Z1 1.4 A Vdot u 0.81 gm mol kg sec u A 30 m sec 2 6.901 cm 6.87 Vectors containing T, P, Tc, and Pc for the calculation of Tr and Pr: 500 425.2 20 42.77 400 304.2 200 73.83 450 552.0 60 79.00 600 617.7 20 21.10 620 T 617.2 Tc 20 P 36.06 Pc 250 190.6 90 45.99 150 154.6 20 469.7 10 33.70 450 430.8 35 78.84 400 374.2 15 Pr P Pc 40.60 1.176 0.468 1.315 2.709 0.815 0.759 0.971 Tr T Tc 50.43 500 Tr 0.948 1.005 Pr 0.555 1.312 1.957 0.97 0.397 1.065 0.297 1.045 0.444 1.069 0.369 Parts (a), (g), (h), (i), and (j) --- By virial equation: 500 425.2 42.77 .190 150 T 20 20 154.6 50.43 .022 500 K P 10 bar Tc 469.7 K Pc 33.70 bar .252 450 35 430.8 78.84 .245 400 15 374.2 40.6 .327 Tr T Tc Pr P Pc 210 1.176 0.97 Tr 0.468 0.397 1.065 Pr 0.297 1.045 0.444 1.069 0.369 B0 0.073 0.422 Eq. (3.65) B1 1.6 0.172 0.139 4.2 Tr DB0 0.675 2.6 Tr Eq. (6.89) DB1 Tr 0.052 0.37 B1 0.321 6.718 4.217 0.306 5.2 Eq. (6.90) 0.443 9.009 10 10 10 3 DB0 3 0.311 0.73 0.056 0.309 0.722 Tr 0.253 B0 Eq. (3.66) 0.845 0.574 DB1 0.522 0.603 0.568 3 0.576 0.51 Combine Eqs. (3.61) + (3.62), (3.63), and (6.40) and the definitions of Tr and Pr to get: Tc B0 Pc B1 VR R HR R Tc Pr B0 Tr DB0 SR R Pr DB0 DB1 (1 B Tr DB1) Eq. (6.88) 211 Eq. (6.87) 3 1.377 10 200.647 94.593 VR 355.907 146.1 1.952 559.501 3 cm 2.469 3 HR 1.226 10 mol 3 J SR mol 2.745 1.746 10 2.256 3 232.454 J mol K 1.74 1.251 10 Parts (b), (c), (d), (e), and (f) --- By Lee/Kesler correlation: By linear interpolation in Tables E.1--E.12: 0 DEFINE: h0 equals 1 () HR () HR h1 equals RTc RTc 0 s0 equals h equals HR RTc s equals SR R 1 () SR R s1 equals () SR R .663 2.008 0.233 .124 Z0 0.208 .050 4.445 5.121 .278 Z1 .088 h0 3.049 h1 2.970 .783 .036 0.671 0.596 .707 0.138 1.486 0.169 1.137 4.381 s0 0.405 5.274 2.675 s1 2.910 0.473 0.557 0.824 0.289 400 304.2 .224 450 T 200 60 552.0 .111 617.7 K .492 600 K P 20 Tc bar 620 20 617.2 .303 250 90 190.6 .012 212 Z s Z1 Eq. (3.57) Z0 s0 HR s1 h0 ( h Tc R) SR Eq. (6.85) ( s R) 3 5.21 10 2.301 0.118 0.235 h1 (6.86) 0.71 Z h 10.207 4 10 4 HR 2.316 10 0.772 4.37 41.291 J mol SR J mol K 5.336 3 10 0.709 34.143 6.88 3 2.358 10 48.289 VR T (Z R P 549.691 1) VR 3 3 1.909 10 cm mol And. 587.396 67.284 The Lee/Kesler tables indicate that the state in Part (c) is liquid. 6.88 Vectors containing T, P, Tc1, Tc2, Pc1, Pc2, 1, and 2 for Parts (a) through (h) 650 562.2 553.6 300 100 304.2 132.9 600 T 60 100 304.2 568.7 350 400 K P 75 150 bar Tc1 305.3 373.5 K Tc2 282.3 190.6 200 75 190.6 126.2 450 80 190.6 469.7 250 100 126.2 154.6 213 K 48.98 .210 .210 73.83 34.99 .224 .048 73.83 Pc1 40.73 24.90 .224 .400 48.72 bar 89.63 50.40 Pc2 45.99 1 bar .100 .087 2 .094 .012 45.99 34.00 .012 .038 45.99 33.70 .012 .252 34.00 50.43 .038 .022 Tpc ( Tc1 .5 Tpr T Tpc .5 Tc2) Ppc ( Pc1 .5 .5 Pc2) .5 P Ppc Ppr 557.9 0.21 218.55 54.41 0.136 436.45 Tpc 44.855 49.365 0.312 293.8 282.05 K Ppc 49.56 67.81 bar 0.094 0.053 158.4 39.995 0.025 330.15 39.845 0.132 140.4 42.215 0.03 1.165 1.373 1.838 1.375 Tpr 1.338 2.026 1.191 1.418 Ppr 1.513 2.212 1.263 1.875 1.363 2.008 1.781 2.369 214 1 .5 2 Lee/Kesler Correlation --- By linear interpolation in Tables E.1--E.12: .6543 1.395 .461 .7706 .1749 1.217 .116 .7527 Z0 .1219 .1929 1.346 .097 .6434 Z1 .7744 .1501 1.510 h0 .1990 1.340 h1 .400 .049 .6631 .1853 1.623 .254 .7436 .1933 1.372 .110 .9168 .1839 0.820 0.172 .890 .466 .658 .235 .729 .242 .944 s0 .430 s1 .704 .224 .965 .348 .750 .250 .361 .095 0 h0 equals s0 equals Z Z0 s s0 ( HR) RTpc ( SR) R 1 h1 equals 0 s1 equals Z1 Eq. (3.57) s1 h Eq. (6.86) 215 ( HR) RTpc ( SR) h equals s equals SR R 1 R h0 HR RTpc h1 Eq. (6.85) HR ( hTpc R) SR ( R) s 0.68 8.213 0.794 2239.984 5.736 0.813 Z 6919.583 4993.974 6.689 0.657 3779.762 HR 0.785 J SR 3148.341 mol 8.183 5.952 mol K Ans. 0.668 2145.752 8.095 0.769 3805.813 6.51 0.922 6.95 Tc J 951.151 3.025 Pc 647.1K At Tr = 0.7: T 220.55bar T 0.7 Tc 452.97 K Find Psat in the Saturated Steam Tables at T = 452.97 K T1 P1 451.15K P2 T2 Psat P1 ( T T1 Psat Pc Psatr T2 957.36kPa Psat T1) P1 Psatr 453.15K 998.619 kPa 1 0.045 P2 Psat 1002.7kPa 9.986 bar 0.344 log Psatr Ans. This is very close to the value reported in Table B.1 ( = 0.345). 6.96 Tc At Tr = 0.7: T Pc 374.2K T T T 471.492 rankine T 0.7 Tc 459.67rankine 40.60bar 11.822 degF Find Psat in Table 9.1 at T = 11.822 F T1 10degF P1 T2 26.617psi 216 15degF P2 29.726psi P2 T2 Psat P1 (T T1 Psat Pc Psatr T1) Psatr P1 Psat Psat 27.75 psi 1 0.047 1.913 bar Ans. 0.327 log Psatr This is exactly the same as the value reported in Table B.1. 6.101 For benzene a) Tc Tn Trn Tc 562.2K Pc Trn 0.210 0.628 Psatrn lnPr0 ( Tr) 5.92714 lnPr1 ( Tr) 15.2518 6.09648 Tr 15.6875 Tr lnPr0 ( Tr) atm Pc Psatrn 353.2K 0.021 6 0.169347 Tr Eqn. (6.79) 13.4721 ln ( Tr) 0.43577 Tr 6 Eqn. (6.80) Eqn. (6.81). lnPr1 Trn lnPsatr ( Tr) 1 Tn 0.271 1.28862 ln ( Tr) lnPr0 Trn ln Psatrn Zc 48.98bar lnPr1 ( Tr) 0.207 Eqn. (6.78) 2 Zsatliq B0 Psatrn Trn 0.083 Zc 0.422 B1 Eqn. (3.73) Eqn. (3.65) 1.6 0.805 0.139 0.172 Trn B1 7 Zsatliq 4.2 Eqn. (3.66) Z0 1 Z0 Trn B0 1 1 T rn Z1 B1 217 Psatrn Eqn. (3.64) 0.974 Z1 1.073 B0 0.00334 Trn Psatrn Trn 0.035 Equation following Eqn. (3.64) Zsatvap Z0 Zlv Zsatvap d Hhatlv Hlv Eqn. (3.57) Z1 dTrn Zsatvap 0.966 Zlv Zsatliq lnPsatr Trn Trn 2 0.963 Hhatlv Zlv Hlv R Tc Hhatlv 6.59 30.802 kJ mol Ans. This compares well with the value in Table B.2 of 30.19 kJ/mol The results for the other species are given in the table below. EstimatedValue (kJ/mol) Table B.2 (kJ/mol) 30.80 30.72 Benz ene 21.39 21.30 isoButane 29.81 29.82 Carbon tetrachlorid e 30.03 29.97 Cy clohex ane 39.97 38.75 nDecane 29.27 28.85 n- ane Hex 34.70 34.41 nOctane 33.72 33.18 Toluene 37.23 36.24 o- lene Xy 6.103 For CO2: Tc a) At Tr = 0.7 Ttr Tt Ttr Tc Pc 73.83bar 216.55K Pt 5.170bar T At the triple point: 304.2K Tt 0.224 0.7Tc T Ptr 0.712 212.94 K Pt Ptr Pc 0.07 lnPr0 ( ) Tr 5.92714 6.09648 Tr 1.28862 ln ( ) 0.169347 Tr Tr Eqn. (6.79) lnPr1 ( ) Tr 15.2518 15.6875 Tr 13.4721 ln ( ) 0.43577 Tr Tr 6 Eqn. (6.80) ln Ptr lnPr0 Ttr lnPr1 Ttr 6 Eqn. (6.81). 218 0.224 Ans. This is exactly the same value as given in Table B.1 b) Psatr 1atm Pc 0.014 ln Psatr = lnPr0 Trn Given Trn Psatr 0.609 Tn Trn Tc Guess: Trn lnPr1 Trn 0.7 Trn Tn Find Trn 185.3 K Ans. This seems reasonable; a Trn of about 0.6 is common for triatomic species. 219 Chapter 7 - Section A - Mathcad Solutions 7.1 u2 325 m R sec 8.314 J mol K molwt 28.9 7 gm CP mol R 2 molwt With the heat, work, and potential-energy terms set equal to zero and with the initial velocity equal to zero, Eq. (2.32a) reduces to H 2 u2 Whence 7.4 But =0 2 T u2 H = CP T 2 T 2 CP 52.45 K Ans. From Table F.2 at 800 kPa and 280 degC: H1 kJ kg 3014.9 S1 7.1595 kJ kg K Interpolation in Table F.2 at P = 525 kPa and S = 7.1595 kJ/(kg*K) yields: H2 kJ 2855.2 kg 3 V2 cm 531.21 gm mdot 0.75 kg sec With the heat, work, and potential-energy terms set equal to zero and with the initial velocity equal to zero, Eq. (2.32a) reduces to: H 2 u2 2 Whence =0 u2 2 H2 H1 m sec u2 By Eq. (2.27), 7.5 A2 mdot V2 u2 565.2 A2 7.05 cm 2 Ans. Ans. The calculations of the preceding problem may be carried out for a series of exit pressures until a minimum cross-sectional area is found. The corresponding pressure is the minimum obtainable in the converging nozzle. Initial property values are as in the preceding problem. 220 H1 3014.9 kJ S1 kg 7.1595 kJ S2 = S 1 kg K Interpolations in Table F.2 at several pressures and at the given entropy yield the following values: 400 425 P 531.21 2855.2 2868.2 450 kPa 507.12 kJ kg 2880.7 H2 485.45 V2 475 465.69 500 mdot 2892.5 2903.9 447.72 0.75 kg sec u2 2 H2 565.2 7.022 m 518.1 sec 494.8 u2 u2 7.05 541.7 mdot V2 A2 H1 A2 2 7.028 cm 7.059 7.127 471.2 Fit the P vs. A2 data with cubic spline and find the minimum P at the point where the first derivative of the spline is zero. i 15 s cspline P A2 pmin pi 400 kPa Pi a2 A( ) P i A2 i interp s p a2 P (guess) 2 Given pmin cm d A pmin = 0 kPa dpmin 431.78 kPa Ans. pmin A pmin 221 Find pmin 2 7.021 cm Ans. 3 cm gm Show spline fit graphically: p 400 kPa 401 kPa 500 kPa 7.13 7.11 A2 7.09 i 2 cm 7.07 A( ) p 2 cm 7.05 7.03 7.01 400 420 440 460 Pi 480 500 p kPa kPa 7.9 From Table F.2 at 1400 kPa and 325 degC: H1 3096.5 kJ kg S1 7.0499 kJ kg K S2 S1 Interpolate in Table F.2 at a series of downstream pressures and at S = 7.0499 kJ/(kg*K) to find the minimum cross-sectional area. 800 294.81 775 P 2956.0 2948.5 302.12 750 kPa H2 2940.8 kJ kg V2 309.82 725 317.97 700 u2 2932.8 2924.9 326.69 2 H2 H1 A2 = 222 V2 u2 mdot 3 cm gm Since mdot is constant, the quotient V2/u2 is a measure of the area. Its minimum value occurs very close to the value at vector index i = 3. 5.561 5.553 V2 5.552 u2 2 cm sec kg 5.557 5.577 At the throat, A2 A2 u2 3 V2 mdot 2 6 cm mdot 1.081 3 kg Ans. sec At the nozzle exit, P = 140 kPa and S = S1, the initial value. From Table F.2 we see that steam at these conditions is wet. By interpolation, Sliq x 7.10 u1 kJ kg K 1.4098 S1 Svap 230 Sliq Svap x Sliq ft sec 7.2479 kJ kg K 0.966 u2 2000 ft sec From Table F.4 at 130(psi) and 420 degF: H1 Btu lbm 1233.6 S1 2 By Eq. (2.32a), H2 H1 1.6310 H H H2 Btu lbm rankine u2 u1 2 H 2 1154.8 78.8 Btu lbm From Table F.4 at 35(psi), we see that the final state is wet steam: Hliq 228.03 Sliq 0.3809 Btu lbm Btu lbm rankine Hvap 1167.1 Svap 1.6872 223 Btu lbm Btu lbm rankine Btu lbm H2 x Hvap S2 u2 x Svap S2 580 x Sliq ( 273.15 By Eq. (2.32a), But T 1.67 BTU lbm rankine 0.039 15)K molwt H= u1 28.9 u2 2 gm CP mol Ans. R 2 molwt 7 2 2 2 Btu lbm rankine = u2 2 Whence H = CP T u2 (quality) SdotG S1 m T2 sec 0.987 S2 Hliq Sliq SdotG 7.11 Hliq 2 T 2 CP Ans. 167.05 K Initial t = 15 + 167.05 = 182.05 degC Ans. 7.12 Values from the steam tables for saturated-liquid water: 3 At 15 degC: V cm 1.001 gm T 288.15 K Enthalpy difference for saturated liquid for a temperature change from 14 to 15 degC: H ( 67.13 1.5 10 K 58.75) J gm t 2K 4 P 4 atm Cp Cp H t 4.19 J gm K Apply Eq. (7.25) to the constant-enthalpy throttling process. Assumes very small temperature change and property values independent of P. 224 T V1 joule 1 P T T 9.86923 cm3 atm Cp 0.093 K The entropy change for this process is given by Eq. (7.26): S T Cp ln T T Apply Eq. (5.36) with Q=0: Wlost T S Wlost 7.13--7.15 P2 350 T1 350 250 T J gm Wlost 80 P1 K 60 60 bar 20 73.83 282.3 126.2 Pc K 369.8 .224 50.40 .087 bar 34.00 .038 42.48 .152 5.457 1.045 1.424 14.394 B 3.280 1.213 10 4.392 0.0 3 K .593 28.785 0.0 C or 1.2bar 304.2 A 1.408 10 1.157 6 10 K 2 3 J gm K 293.15 K 0.413 400 Tc S VP D 8.824 0.0 0.040 0.0 225 5 10 K 2 0.413 kJ kg Ans. As in Example 7.4, Eq. (6.93) is applied to this constant-enthalpy process. If the final state at 1.2 bar is assumed an ideal gas, then Eq. (A) of Example 7.4 (pg. 265) applies. Its use requires expressions for HR and Cp at the initial conditions. 1.151 T1 Tr 1.24 Tr Tc 1.084 P1 Pc Pr 1.981 Pr 1.082 7.13 0.42748 0.08664 Eq. (3.53) Tr 1.765 0.471 Redlich/Kwong equation: Pr 1.19 q Eq. (3.54) 1.5 Tr Guess: z Given z= 1 Z q Find z () i 14 1 q Ii HRi R T1i Z SRi R ln Z i qi i qi z Eq. (3.52) zz ln 1 i Z i qi Z i Eq. (6.65b) i qi 1.5 qi Ii Eq. (6.67) The derivative in these 0.5 qi Ii Eq. (6.68) equations equals -0.5 The simplest procedure here is to iterate by guessing T2, and then calculating it. 280 Guesses T2 302 232 K 385 226 Z i qi 2.681 5.177 0.721 0.773 2.253 kJ 0.521 mol HR 0.956 0.862 Cp HR Cp 2.33 B T1 2 RA C 2 T1 3 1 T1 S Cp ln 302.026 K 232.062 0.480 Guess: P2 P1 SR Ans. 0.42748 1 c1 0.5 2 Tr Eq. (3.54) q Tr 1 Given z= 1 i Ii HRi R ln 0.08664 2 0.176 Eq. (3.53) z 14 T2 T1 2 22.163 1.574 Pr Tr T1 J 31.953 mol K Soave/Redlich/Kwong equation: c D 1 29.947 S Ans. 384.941 7.14 2 31.545 279.971 T2 J mol K 1.59 1.396 T2 T1 T2 4.346 SR R T1i Z ln z q Z i qi Eq. (3.52) Z zz i qi Z i q Eq. (6.65b) i qi 1 ci Tri i 227 0.5 1 qi Ii Eq. (6.67) Find ( z) SRi R ln Z i qi Tri ci i 0.5 Eq. (6.68) qi Ii i The derivative in these equations equals: Tri ci 0.5 i Now iterate for T2: 273 Guesses 300 T2 232 K 384 Z i qi 2.936 0.75 0.79 HR 0.975 0.866 6.126 2.356 kJ 0.526 mol 4.769 J 1.789 mol K SR 1.523 T2 T1 Cp RA 2.679 B T1 2 1 C 2 T1 3 2 1 D T1 2 272.757 T2 HR Cp T1 299.741 T2 231.873 K Ans. 383.554 31.565 S Cp ln T2 T1 R ln P2 P1 SR S 30.028 32.128 mol K 22.18 228 J Ans. 7.15 Peng/Robinson equation: 1 c 2 1 0.37464 1.54226 Pr Tr z q i 14 HRi 2 1 c1 Tr z q Eq. (3.52) z z Find z () 1 Ii R T1i Z 2 ln Z i qi i Z i qi i 2 i qi 1 ci Tri Eq. (6.65b) 0.5 1 qi Ii R ln Z i qi i ci Tri Eq. (6.67) 0.5 Eq. (6.68) qi Ii i The derivative in these equations equals: ci Tri i Now iterate for T2: 270 Guesses T2 297 229 K 383 229 0.5 0.5 Tr Eq. (3.54) q i SRi 0.45724 1 z= 1 Z 0.07779 0.26992 Eq. (3.53) Guess: Given 2 2 Z i qi 3.041 6.152 0.722 0.76 HR 0.95 0.85 2.459 0.6 kJ mol 4.784 J 1.847 mol K SR 1.581 T2 T1 Cp RA 2.689 B T1 2 C 2 T1 3 1 2 D 1 T1 2 269.735 HR Cp T2 T1 297.366 T2 Ans. K 229.32 382.911 31.2 S T2 T1 Cp ln R ln P2 P1 SR 29.694 J 31.865 mol K S Ans. 22.04 7.18 Wdot H1 3500 kW 3462.9 kJ kg Data from Table F.2: 2609.9 H2 kJ kg S1 4.103 kg sec Ans. 7.3439 kJ kg K S2 S1 By Eq. (7.13), Wdot H2 H1 mdot mdot For isentropic expansion, exhaust is wet steam: Sliq x 0.8321 S2 Svap kJ kg K Sliq Sliq Svap x 230 7.9094 0.92 kJ kg K (quality) Hliq H'2 251.453 Hliq x Hvap H2 Hvap Hliq H'2 H1 2609.9 2.421 kJ kg 3 kJ 10 kg H1 H'2 7.19 kJ kg 0.819 Ans. The following vectors contain values for Parts (a) through (g). For intake conditions: 3274.3 kJ kg 6.5597 3509.8 kJ kg 6.8143 kJ 3634.5 kg H1 kJ 3161.2 kg 1389.6 Btu lbm Btu lbm kg K kJ kg K kJ 6.9813 kg K 6.4536 S1 kJ 2801.4 kg 1444.7 kJ kJ kg K kJ 6.4941 kg K 1.6000 1.5677 231 Btu lbm rankine Btu lbm rankine 0.80 0.77 0.82 0.75 0.75 0.80 0.75 For discharge conditions: 0.9441 0.8321 0.6493 Sliq 1.0912 1.5301 kJ 7.7695 kg K kJ kg K 7.9094 kJ 8.1511 kg K kJ kg K kJ kg K 7.1268 Btu 0.1750 1.9200 lbm rankine Btu lbm rankine 0.2200 7.5947 Svap 1.8625 kJ kg K kJ kg K kJ kg K kJ kg K S' 2 = S 1 kJ kg K Btu lbm rankine Btu lbm rankine 289.302 2625.4 kJ kg 80 kg sec 251.453 kJ kg 2609.9 kJ kg 90 kg sec 191.832 Hliq kJ kg kJ kg 2584.8 kJ kg 70 kg sec 340.564 kJ kg 2646.0 kJ kg 65 kg sec 504.701 kJ kg 2706.3 kJ kg 50 kg sec Btu lbm 1116.1 Btu lbm 150 Btu lbm 1127.3 Btu lbm 100 94.03 120.99 Hvap 232 mdot lbm sec lbm sec x'2 S1 Svap H x2 Sliq H'2 H2 Hvap H2 H2 H2 H2 H2 H2 H2 1 H'2 Sliq H1 Hliq Hliq Hliq H2 H1 S2 Sliq Wdot x2 Svap Sliq 3 S2 kJ 2467.8 kg 2471.4 2 7.6873 7.7842 S2 4 2543.4 kJ kg K 7.1022 3 S2 Ans. 6.7127 5 S2 1031.9 Btu 1057.4 lbm 6 S2 7 1.7762 Btu 1.7484 lbm rankine 68030 91230 87653 117544 81672 Wdot H mdot 7.1808 S2 5 7 H 1 2423.9 2535.9 6 Hliq S2 2 4 x'2 Hvap 109523 44836 kW Wdot 60126 12900 17299 65333 87613 35048 46999 233 hp Ans. 7.20 T P0 423.15 K P 8.5 bar 1 bar For isentropic expansion, S J 0 mol K For the heat capacity of nitrogen: A B 3.280 3 0.593 10 K 5 D 0.040 10 K 2 For the entropy change of an ideal gas, combine Eqs. (5.14) & (5.15) with C = 0. Substitute: (guess) 0.5 Given S = R A ln B 1 D T 1 T 2 2 T T0 Find ln P P0 T0 Ans. 762.42 K Thus the initial temperature is 489.27 degC 7.21 T1 CP 32 P1 1223.15 K J mol K P2 10 bar 1.5 bar 0.77 Eqs. (7.18) and (7.19) derived for isentropic compression apply equally well for isentropic expansion. They combine to give: W's C P T1 Ws W's P2 R CP W's 1 P1 H Ws 234 15231 J mol Ws 11728 J mol Ans. Eq. (7.21) also applies to expansion: T2 7.22 H T1 T2 CP Isobutane: Tc 408.1 K Pc T0 P0 5000 kPa P 523.15 K S 0 J mol K A 0.181 36.48 bar 500 kPa For the heat capacity of isobutane: 37.853 10 K B 1.677 T0 Tr0 Ans. 856.64 K Tr0 Tc 3 C K Pr0 1.282 11.945 10 Pr P0 2 Pr0 Pc P Pr Pc 6 1.3706 0.137 The entropy change is given by Eq. (6.92) combined with Eq. (5.15) with D = 0: 0.5 (guess) Given S = R A ln SRB B T0 T0 Tc Find T Tr Pr 1 2 C T0 2 SRB Tr0 Pr0 T0 T Tc T Tr 235 445.71 K 1.092 1 ln P P0 The enthalpy change is given by Eq. (6.91): Hig R ICPH T0 T 1.677 37.853 10 Hig 11.078 H' Hig H' 8331.4 3 6 11.945 10 0.0 kJ mol R Tc HRB Tr Pr HRB Tr0 Pr0 J mol The actual enthalpy change from Eq. (7.16): 0.8 Wdot ndot 700 mol sec ndot H H Wdot H' H 6665.1 J mol Ans. 4665.6 kW The actual final temperature is now found from Eq. (6.91) combined with Eq (4.7), written: 0.7 (guess) Given H = R A T0 1 Tc HRB Find 7.23 B 2 T0 2 T0 Pr Tc 0.875 2 C 1 3 3 3 T0 1 HRB Tr0 Pr0 T T0 T 457.8 K Ans. From Table F.2 @ 1700 kPa & 225 degC: H1 2851.0 At 10 kPa: kJ kg S1 6.5138 x2 0.95 236 kJ kg K Sliq 0.6493 kJ kg K Hliq kJ 191.832 mdot 0.5 Hvap kg kg 2584.8 Wdot sec Svap 8.1511 H H1 2.465 10 (a) Qdot mdot H (b) For isentropic expansion to 10 kPa, producing wet steam: S1 Svap x'2 Hliq H2 Hliq H2 3 kJ H Wdot Sliq H'2 T0 mdot H'2 Hliq H'2 Sliq 2.063 385.848 Qdot kg 0.782 Wdot' 7.24 kJ kg K 180 kW H2 x'2 x2 Hvap kJ kg 12.92 x'2 Hvap P0 673.15 K 8 bar P For isentropic expansion, For the heat capacity of carbon dioxide: A 5.457 B Ans. Hliq kg Ans. 394.2 kW 1 bar S 0 J mol K 3 1.045 10 K kJ sec 3 kJ 10 Wdot' H1 kJ kg 5 D 1.157 10 K 2 For the entropy change of an ideal gas, combine Eqs. (5.14) & (5.15) with C = 0: (guess) 0.5 Given S = R A ln B T0 T0 Find 1 D 2 2 T' 0.693 237 1 T0 T' ln P P0 466.46 K H' H' 9.768 0.0 1.157 10 5 kJ mol 0.75 H 3 R ICPH T0 T' 5.457 1.045 10 Work Work H' H 7.326 Work 7.326 kJ mol Ans. kJ mol For the enthalpy change of an ideal gas, combine Eqs. (4.2) and (4.7) with C = 0: Given H = R A T0 B 2 T0 2 1 Find 2 0.772 1 T D T0 T0 1 T Ans. 519.9 K Thus the final temperature is 246.75 degC 7.25 Vectors containing data for Parts (a) through (e): 500 371 1.2 3.5 450 T1 6 5 376 2.0 4.0 525 P1 10 T2 458 P2 3.0 Cp 5.5 R 475 HS 372 1.5 4.5 550 H 7 4 403 1.2 2.5 [( Cp T2 Cp T1 Ideal gases with constant heat capacities T1) P2 P1 R Cp Eq. (7.22) Applies to expanders as well as to compressors 1 238 0.7 0.803 H 0.649 HS 0.748 0.699 7.26 Cp 7 2 R ndot Guesses: mol 175 T1 sec 0.75 Wdot 550K P1 6bar P2 1.2bar 600kW Given Wdot = 0.065 Wdot .08 ln Wdot kW Find ( Wdot) 0.065 ndot Cp T1 Wdot P2 P1 1 Ans. 594.716 kW Wdot kW 0.08 ln R Cp 0.576 Ans. For an expander operating with an ideal gas with constant Cp, one can show that: T2 P2 P1 T1 1 R Cp 1 T2 433.213 K By Eq. (5.14): S R T2 Cp ln T1 R ln P2 P1 S 6.435 J mol K By Eq. (5.37), for adiabatic operation : SdotG ndot S SdotG 1.126 239 3 10 J K sec Ans. 7.27 Properties of superheated steam at 4500 kPa and 400 C from Table F.2, p. 742. H1 3207.1 S1 6.7093 If the exhaust steam (Point 2, Fig. 7.4) is "dry," i.e., saturated vapor, then isentropicexpansion to the same pressure (Point 2', Fig. 7.4) must produce "wet" steam, withentropy: S2 = S1 = 6.7093 = (x)(Svap) + (1-x)(Sliq) [x is quality] A second relation follows from Eq. (7.16), written: HS) = (0.75)[ (x)(Hvap) + (1-x)(Hliq) - 3207.1] H = Hvap - 3207.1 = ( Each of these equations may be solved for x. Given a final temperature and the corresponding vapor pressure, values for Svap, Sliq, Hvap, and Hliq are found from the table for saturated steam, and substitution into the equations for x produces two values. The required pressure is the one for which the two values of x agree. This is clearly a trial process. For a final trial temperature of 120 degC, the following values of H and S for saturated liquid and saturated vapor are found in the steam table: Hl 503.7 Hv 2706.0 Sl 1.5276 Sv 7.1293 The two equations for x are: xH Hv 801.7 .75 Hl .75 ( Hv Hl) xS The trial values given produce: xH 6.7093 Sl Sv Sl 0.924 xS 0.925 These are sufficiently close, and we conclude that: t=120 degC; P=198.54 kPa If were 0.8, the pressure would be higher, because a smaller pressure drop would be required to produce the same work and H. 240 7.29 P1 5 atm P2 1 atm T1 15 degC 0.55 Data in Table F.1 for saturated liquid water at 15 degC give: 3 V 1001 cm kg Cp 4.190 kJ kg degC Eqs. (7.16) and (7.24) combine to give: Ws H (7.14) Ws 0.223 Eq. (7.25) with =0 is solved for T: H P1) kJ kg T T 7.30 V ( P2 H V ( P2 P1) Cp Ans. 0.044 degC Assume nitrogen an ideal gas. First find the temperature after isentropic expansion from a combination of Eqs. (5.14) & (5.15) with C = 0. Then find the work (enthalpy change) of isentropic expansion by a combination of Eqs. (4.2) and (4.7) with C = 0. The actual work (enthalpy change) is found from Eq. (7.20). From this value, the actual temperature is found by a second application of the preceding equation, this time solving it for the temperature. The following vectors contain values for Parts (a) through (e): 753.15 1 bar 673.15 T0 6 bar 5 bar 1 bar P0 773.15 K 7 bar P 1 bar 723.15 8 bar 2 bar 755.37 95 psi 15 psi 200 150 ndot 0.80 0.75 175 mol sec 0.78 100 0.85 0.5 453.59 0.80 241 S i 0 15 J mol K For the heat capacity of nitrogen: A 3.280 3 0.593 10 B 5 D 0.040 10 K 2 K (guess) 0.5 Given S = R A ln B T0 1 D 22 2 T0 Tau T0 P0 P Find ln 1 P P0 Tau T0 P0 Pi i i i 460.67 431.36 Ti T0 i T i 453.48 K 494.54 455.14 H'i 3 R ICPH T0 Ti 3.280 0.593 10 5 0.0 0.040 10 i 8879.2 7279.8 H' 7103.4 5459.8 9714.4 J mol H H H' 7577.2 6941.7 0.5 5900.5 9112.1 J mol 7289.7 (guess) Given H = R A T0 Tau T0 H 1 B 2 T0 2 2 Find i 242 1 1 D T0 Tau T0 i Hi Ti T0 i i 520.2 492.62 T 1421 819 525.14 K Ans. Wdot ndot H 1326 kW Ans. Wdot 529.34 516.28 7.31 590 1653 Property values and data from Example 7.6: kg H1 3391.6 kJ kg S1 6.6858 kJ kg K mdot H2 2436.0 kJ kg S2 7.6846 kJ kg K Wdot 56400 kW T 300 K Wdotideal 74084 kW Wdotideal t 59.02 sec By Eq. (5.26) mdot H2 H1 Wdot Wdotideal T t S2 S1 Ans. 0.761 The process is adiabatic; Eq. (5.33) becomes: SdotG mdot S2 Wdotlost 7.32 SdotG S1 Wdotlost T SdotG 58.949 kW K 17685 kW Ans. Ans. For sat. vapor steam at 1200 kPa, Table F.2: H2 2782.7 kJ kg S2 6.5194 kJ kg K The saturation temperature is 187.96 degC. The exit temperature of the exhaust gas is therefore 197.96 degC, and the temperature CHANGE of the exhaust gas is -202.04 K. For the water at 20 degC from Table F.1, H1 83.86 kJ kg S1 0.2963 243 kJ kg K The turbine exhaust will be wet vapor steam. For sat. liquid and sat. vapor at the turbine exhaust pressure of 25 kPa, the best property values are found from Table F.1 by interpolation between 64 and 65 degC: kJ kJ Hliq 272.0 Hlv 2346.3 kg kg Sliq 0.8932 kJ kg K Slv 6.9391 kJ kg K 0.72 For isentropic expansion of steam in the turbine: S'3 S2 S'3 6.519 x'3 kJ kg K x' 3 S'3 Sliq H'3 0.811 Hliq H'3 Slv 2.174 H23 H'3 H2 H3 H2 H23 H23 437.996 kJ kg H3 2.345 10 S3 Sliq x3 Slv S3 7.023 kJ kg K For the exhaust gases: ndot 125 mol sec T1 ( 273.15 T2 ( 273.15 T1 673.15 K T2 471.11 K H3 x3 x3 Hliq Hlv 0.883 molwt 18 400)K 3 kJ kg 197.96)K gm mol Hgas R MCPH T1 T2 3.34 1.12 10 Sgas R MCPS T1 T2 3.34 1.12 10 244 3 3 0.0 0.0 T2 0.0 0.0 ln T2 T1 T1 x'3 Hlv 3 kJ 10 kg Hgas 6.687 3 kJ 10 Sgas kmol 11.791 kJ kmol K Energy balance on boiler: mdot ndot Hgas H2 H1 (a) Wdot mdot mdot H3 T 314.302 kW S1 Wdot Wdotideal t sec 293.15 K ndot Hgas mdot H3 H1 T ndot Sgas mdot S3 Wdotideal kg 135.65 kW Ans. Wdot H2 (b) By Eq. (5.25): Wdotideal 0.30971 t Ans. 0.4316 (c) For both the boiler and the turbine, Eq. (5.33) applies with Q = 0. For the boiler: SdotG ndot Sgas Boiler: mdot S2 SdotG S1 0.4534 kW K For the turbine: SdotG mdot S3 Turbine: SdotG 0.156 (d) Wdotlost.boiler 0.4534 Wdotlost.turbine Fractionboiler kW 0.1560 K T kW T K Wdotlost.boiler Wdotideal 245 kW K Ans. S2 Ans. Wdotlost.boiler 132.914 kW Wdotlost.turbine 45.731 kW Fractionboiler 0.4229 Ans. Wdotlost.turbine Fractionturbine Note that: 7.34 t 0.1455 Ans. Fractionturbine Wdotideal Fractionboiler Fractionturbine 1 From Table F.2 for sat. vap. at 125 kPa: H1 kJ kg 2685.2 S1 7.2847 kJ kg K For isentropic expansion, S'2 = S1 = 7.2847 kJ kg K Interpolation in Table F.2 at 700 kPa for the enthalpy of steam with this entropy gives H'2 3051.3 H2 H1 kJ kg H 0.78 H2 H 3154.6 H'2 kJ kg H1 H 469.359 kJ kg Ans. Interpolation in Table F.2 at 700 kPa for the entropy of steam with this enthalpy gives S2 mdot 2.5 kg sec Wdot 7.4586 kJ kg K mdot H 246 Ans. Wdot 1173.4 kW Ans. 7.35 Assume air an ideal gas. First find the temperature after isentropic compression from a combination of Eqs. (5.14) & (5.15) with C = 0. Then find the work (enthalpy change) of isentropic compression by a combination of Eqs. (4.2) and (4.7) with C = 0. The actual work (enthalpy change) is found from Eq. (7.20). From this value, the actual temperature is found by a second application of the preceding equation, this time solving it for the temperature. The following vectors contain values for Parts (a) through (f): 298.15 101.33 kPa 375 kPa 353.15 375 kPa 1000 kPa 303.15 T0 373.15 100 kPa P0 K 500 kPa P 500 kPa 1300 kPa 299.82 14.7 psi 55 psi 338.71 55 psi 135 psi 100 0.75 100 0.70 150 ndot 0.80 mol sec 50 S 0.75 0.5 453.59 16 0.75 0.5 453.59 i J mol K 0 0.70 For the heat capacity of air: A 3.355 0.5 0.575 10 K 3 5 2 (guess) B D 0.016 10 K Given S = R A ln B T0 22 T0 Tau T0 P0 P 1 D Find 1 2 i 247 ln P P0 Tau T0 P0 Pi i i 431.06 464.5 Ti T0 i 476.19 i T K 486.87 434.74 435.71 H'i 3 R ICPH T0 Ti 3.355 0.575 10 5 0.0 0.016 10 i 3925.2 3314.6 5133.2 J 3397.5 mol H' 3986.4 2876.6 5233.6 4735.1 H' H 6416.5 H 4530 J mol 5315.2 4109.4 1.5 (guess) Given H = R A T0 Tau T0 H Wdot 1 B 2 T0 2 ndot H Find i 248 2 1 Tau T0 i D T0 Hi 1 Ti T0 i i 474.68 702 523 511.58 635 474 518.66 T 524.3 1291 Wdot K 304 962 Wdot hp 479.01 1617 1205 476.79 7.36 1250 932 Ammonia: 0 Tc 405.7 K Pc 294.15 K P0 200 kPa J mol K A P 1000 kPa For the heat capacity of ammonia: 3.020 10 K B 3.578 T0 Tr0 0.253 112.8 bar T0 S Ans. kW 227 Tr0 Tc 3 0.186 10 K P0 Pr0 0.725 5 D Pr0 Pc P Pr Pr Pc 2 0.0177 0.089 Use generalized second-virial correlation: The entropy change is given by Eq. (6.92) combined with Eq. (5.15); C = 0: 1.4 (guess) Given S = R A ln B T0 T0 SRB Find T0 Tc 1 D 2 ln 1 2 Pr T P0 SRB Tr0 Pr0 1.437 P Tr 249 T0 T Tc T Tr 422.818 K 1.042 Hig Hig 4.826 H' Hig H' 4652 3 R ICPH T0 T 3.578 3.020 10 0.0 5 0.186 10 kJ mol R Tc HRB Tr Pr HRB Tr0 Pr0 J mol The actual enthalpy change from Eq. (7.17): 0.82 H' H H 5673.2 J mol The actual final temperature is now found from Eq. (6.91) combined with Eq (4.7), written: (guess) 1.4 Given 1 Tc HRB Find B 2 T0 2 T0 Pr Tc 2 1.521 H = R A T0 T R A ln B T0 SRB Tr Pr S 2.347 J mol K T0 T Tc D 2 2 T0 SRB Tr0 Pr0 Ans. 250 1 HRB Tr0 Pr0 Tr S D T0 1 1 T 447.47 K Tr 1.103 1 ln P P0 Ans. 7.37 Propylene: 0 A 365.6 K Pc 303.15 K P0 11.5 bar J 22.706 10 B 1.637 3 K P0 Pr0 0.8292 6.915 10 C K Tr0 Tc P 18 bar For the heat capacity of propylene: mol K T0 Tr0 0.140 46.65 bar T0 S Tc Pr 2 Pr0 Pc P Pr Pc 6 0.2465 0.386 Use generalized second-virial correlation: The entropy change is given by Eq. (6.92) combined with Eq. (5.15) with D = 0: (guess) 1.1 Given S = R A ln B T0 T0 1 2 C T0 2 Tc Find Pr T P P0 SRB Tr0 Pr0 1.069 SRB ln 1 T0 T T Tc Tr 324.128 K Tr 0.887 The enthalpy change for the final T is given by Eq. (6.91), with HRB for this T: Hig R ICPH T0 T 1.637 22.706 10 Hig 1.409 H' Hig 3 6.915 10 3J 10 mol R Tc HRB Tr Pr 251 HRB Tr0 Pr0 6 0.0 H' J mol 964.1 The actual enthalpy change from Eq. (7.17): ndot H' H 0.80 1000 mol Wdot H Wdot ndot H J mol 1205.2 Ans. 1205.2 kW sec The actual final temperature is now found from Eq. (6.91) combined with Eq (4.7), written: (guess) 1.1 Given H = R A T0 B 2 T0 2 T0 Pr Tc 1 Tc HRB 7.38 Methane: C 3 T0 3 1 T T0 0 Tr0 Tc 190.6 K Pc T0 Tc Ans. 308.15 K P0 0.012 45.99 bar 3500 kPa J mol K A 1 327.15 K T0 S 3 HRB Tr0 Pr0 T 1.079 Find 2 P 5500 kPa For the heat capacity of methane: B 1.702 Tr0 9.081 10 K 1.6167 3 C K Pr0 Pr 252 2.164 10 P0 Pc P Pc 6 2 Pr0 Pr 0.761 1.196 Use generalized second-virial correlation: The entropy change is given by Eq. (6.92) combined with Eq. (5.15) with D = 0: (guess) 1.1 Given S = R A ln B T0 T0 1 2 C T0 2 Tc Find Pr T P P0 SRB Tr0 Pr0 1.114 SRB ln 1 T0 T T Tc Tr 343.379 K Tr 1.802 The enthalpy change for the final T is given by Eq. (6.91), with HRB for this T: Hig Hig 1.298 10 H' Hig H' 1158.8 3 R ICPH T0 T 1.702 9.081 10 2.164 10 6 0.0 3J mol R Tc HRB Tr Pr HRB Tr0 Pr0 J mol The actual enthalpy change from Eq. (7.17): 0.78 ndot 1500 H mol sec Wdot H' ndot H H Wdot 1485.6 J mol 2228.4 kW Ans. The actual final temperature is now found from Eq. (6.91) combined with Eq (4.7), written: 1.1 (guess) 253 Given H = R A T0 B 2 T0 2 T0 Pr Tc 1 Tc HRB 7.39 C 3 T0 3 1 3 1 HRB Tr0 Pr0 T 1.14 Find 2 T T0 Ans. 351.18 K From the data and results of Example 7.9, T1 Work T2 293.15 K 5288.3 P1 428.65 K J T mol R ICPH T1 T2 1.702 9.081 10 H 5288.2 S 3 R ICPS T1 T2 1.702 9.081 10 3.201 560 kPa 293.15 K H S P2 140 kPa 6 2.164 10 0.0 J mol 3 6 2.164 10 0.0 ln P2 P1 J mol K Since the process is adiabatic: SG Wideal Wlost H T SG S T S S Wideal Wlost Wideal t Work 254 3.2012 t J mol K 4349.8 938.4 0.823 J mol J mol Ans. Ans. Ans. Ans. 7.42 P1 1atm T1 ( 35 P2 50atm T2 ( 200 T1 473.15 K Cp 273.15) K 308.15 K T2 273.15) K 3.5 R 3 0.65 V Vdot R T1 P1 0.5 ndot m sec Vdot ndot V 19.775 mol sec With compression from the same initial conditions (P1,T1) to the same final conditions (P2,T2) in each stage, the same efficiency in each stage, and the same power delivered to each stage, the applicable equations are: P2 P1 r= 1 N (where r is the pressure ratio in each stage and N is the number of stages.) Eq. (7.23) may be solved for T2prime: T'2 T'2 N R Cp P2 P1 ln ln T1) T1 Eq. (7.18) written for a single stage is: 415.4 K T'2 = T1 ( T2 R1 N Cp Put in logarithmic form and solve for N: P2 P1 N T'2 (a) Although any number of stages greater than this would serve, design for 4 stages. 3.743 T1 (b) Calculate r for 4 stages: N 4 r P2 P1 1 N r 2.659 Power requirement per stage follows from Eq. (7.22). In kW/stage: Wdotr ndot Cp T1 r R Cp 1 Wdotr 255 87.944 kW Ans. (c) Because the gas (ideal) leaving the intercooler and the gas entering the compressor are at the same temperature (308.15 K), there is no enthalpy change for the compressor/interchanger system, and the first law yields: Qdotr Wdotr Qdotr 87.944 kW Ans. Heat duty = 87.94 kW/interchanger (d) Energy balance on each interchanger (subscript w denotes water): With data for saturated liquid water from the steam tables: kJ kJ Hw ( 188.4 104.8) Hw 83.6 kg kg Qdotr mdotw mdotw Hw 1.052 kg sec Ans. (in each interchanger) 7.44 300 290 T1 2.0 1.5 295 K P1 1.2 bar 300 305 1.5 464 6 3.5 547 T2 1.1 5 2.5 455 K P2 6 bar Cp 4.5 R 505 HS 5.5 496 H 8 7 4.0 [( Cp T2 Cp T1 Ideal gases with constant heat capacities T1) P2 P1 R Cp (7.22) 1 256 3.219 0.675 3.729 HS kJ mol 4.745 0.698 HS H 5.959 0.636 4.765 7.47 Ans. 0.793 0.75 The following vectors contain values for Parts (a) through (e). Intake conditions first: 298.15 20 kg 363.15 T1 100 kPa 200 kPa 30 kg P1 333.15 K 20 kPa mdot 15 kg 294.26 1 atm 50 lb 366.48 15 psi 80 lb 2000 kPa 0.75 257.2 5000 kPa 0.70 696.2 5000 kPa 0.75 523.1 20 atm 0.70 0.75 10 K 6 217.3 1500 psi P2 1 sec 714.3 From the steam tables for sat.liq. water at the initial temperature (heat capacity calculated from enthalpy values): 1.003 1.036 V 1.017 4.15 4.20 3 cm gm 4.20 CP 1.002 4.185 1.038 kJ kg K 4.20 By Eq. (7.24) HS V P2 257 P1 H HS 1.906 4.973 HS 2.541 7.104 kJ 5.065 H kg 6.753 1.929 kg 2.756 10.628 kJ 14.17 0.188 By Eq. (7.25) T H V1 T1 P2 0.807 P1 T CP 0.612 K 0.227 1.506 50.82 213.12 Wdot H mdot Wdot 68.15 285.8 101.29 kW Wdot 135.84 hp 62.5 83.81 514.21 689.56 298.338 363.957 T2 T1 T T2 333.762 K 294.487 367.986 t2 t2 T2 K 273.15 t2 t2 t2 T2 K t2 1.8 459.67 t2 1 25.19 2 90.81 degC 60.61 3 4 5 258 70.41 202.7 degF Ans. 7.48 Results from Example 7.10: H kJ 11.57 W kg 11.57 Wideal 300 K H T Wideal T 8.87 kJ kg S kJ kg 0.0090 kJ kg K Wideal S t Ans. t W 0.767 Ans. Since the process is adiabatic. SG S Wlost 7.53 SG T T1 ( 25 T3 S ( 200 Cpv 105 9 3 kJ 10 Wlost 2.7 kJ kg P1 273.15) K J mol K Ans. 1.2bar P3 273.15) K Ans. kg K 5bar Hlv 30.72 P2 5bar kJ mol 0.7 Estimate the specific molar volume of liquid benzene using the Rackett equation (3.72). 3 From Table B.1 for benzene: Tc 562.2K From Table B.2 for benzene: Tn ( 80.0 Zc 0.271 Vc 273.15) K Trn 259 cm mol Tn Tc 2 Assume Vliq = Vsat: V V c Zc 1 T rn 3 7 Eq. (3.72) Calculate pump power Ws V P2 P1 Ws 259 0.053 kJ mol Ans. V cm 96.802 mol Assume that no temperature change occurs during the liquid compression. Therefore: T2 T1 Estimate the saturation temperature at P = 5 bar using the Antoine Equation and values from Table B.2 For benzene from A Table B.2: B Tsat A ln 13.7819 B C degC Tsat P2 kPa Tsat 2726.81 C 217.572 Tsat 415.9 K 142.77 degC Tsat 273.15K Estimate the heat of vaporization at Tsat using Watson's method From Table B.2 Hlv At 80 C: ( 80 Tr1 Hlv2 273.15) K Tc Hlv 1 Tr2 1 Tr1 30.72 kJ mol Tr1 0.628 Tr2 Tsat Tr2 Tc 0.38 Eq. (4.13) Hlv2 26.822 Calculate the heat exchanger heat duty. Q R ICPH T2 Tsat 0.747 67.96 10 Hlv2 Cpv T3 Tsat Q 51.1 kJ mol Ans. 260 3 37.78 10 6 0 0.74 kJ mol 7.54 T1 ( 25 T3 ( 200 Cpv 273.15)K 273.15)K 105 P1 P3 P2 1.2bar 1.2bar 5bar J mol K 0.75 Calculate the compressor inlet temperature. Combining equations (7.17), (7.21) and (7.22) yields: T3 T2 1 P3 1 T2 1 P2 408.06 K T2 R Cpv 273.15K 134.91 degC Calculate the compressor power Ws Cpv T3 T2 Ws 6.834 kJ mol Ans. Calculate the heat exchanger duty. Note that the exchanger outlet temperature, T2, is equal to the compressor inlet temperature. The benzene enters the exchanger as a subcooled liquid. In the exchanger the liquid is first heated to the saturation temperature at P1, vaporized and finally the vapor is superheated to temperature T 2. Estimate the saturation temperature at P = 1.2 bar using the Antoine Equation and values from Table B.2 For benzene from A Table B.2: B Tsat A ln P1 B 13.7819 C degC Tsat kPa Tsat 2726.81 C 217.572 Tsat 358.7 K 85.595 degC Tsat 273.15K Estimate the heat of vaporization at Tsat using Watson's method From Table B.2 At 25 C: Hlv 30.72 kJ mol 261 From Table B.1 for benzene: Tc 562.2K ( 80 Tr1 273.15) K Tr1 Tc Hlv2 Hlv 1 Tr2 1 Tr1 Eq. (4.13) Q 44.393 3 R ICPH T1 Tsat 0.747 67.96 10 Hlv2 Cpv T2 Tsat 7.57 ndot Cp 100 50.6 mol Tr2 Tc 0.638 0.38 Q kJ Tsat Tr2 0.628 Hlv2 37.78 10 30.405 6 kJ mol 0 Ans. kmol hr P1 T1 1.2bar J mol K P2 300K 6bar 0.70 Assume the compressor is adaiabatic. T2 P2 P1 Wdots Wdote T1 (Pg. 77) ndot Cp T2 T2 T1 Wdots 127.641 kW Wdote 3040dollars 380dollars 390.812 K Wdots Wdots C_compressor C_motor R Cp 182.345 kW 0.952 C_compressor kW Wdote 307452 dollars Ans. 0.855 C_motor kW 262 32572 dollars Ans. 7.59 T1 18bar For ethylene: T1 Tr1 Tr1 Tc P2 1.2bar 0.087 P1 375K Tc 282.3K P2 Pr2 A B 1.424 14.394 10 3 Pc C 4.392 10 6 0.357 Pr2 Pc 50.40bar Pr1 P1 Pr1 1.328 Pc 0.024 D 0 a) For throttling process, assume the process is adiabatic. Find T2 such that H = 0. H = Cpmig T2 T1 HR2 Eq. (6-93) HR1 Use the MCPH function to calculate the mean heat capacity and the HRB function for the residual enthalpy. Guess: T2 T1 Given 0 J mol T2 = MCPH T1 T2 A B C D R T2 T1 T2 Pr2 R Tc HRB R Tc HRB Tr1 Pr1 Tc Find T2 T2 365.474 K Ans. Tr2 T2 Tc Tr2 1.295 Calculate change in entropy using Eq. (6-94) along with MCPS function for the mean heat capacity and SRB function for the residual entropy. S R MCPS T1 T2 A B C D ln R SRB Tr2 Pr2 S 22.128 J mol K T2 T1 R SRB Tr1 Pr1 Ans. 263 R ln P2 P1 Eq. (6-94) b) For expansion process. 70% First find T2 for isentropic expansion. Solve Eq. (6-94) with S = 0. Guess: T2 T1 Given 0 T2 P2 J R ln = R MCPS T1 T2 A B C D ln T1 mol K P1 T2 Pr2 SRB R R SRB Tr1 Pr1 Tc T2 Find T2 T2 219.793 K Tr2 T2 Eq. (6-94) Tr2 Tc Now calculate the isentropic enthalpy change, HS. HR2 HS HS HRB Tr2 Pr2 R Tc R MCPH T1 T2 A B C D T2 T1 HRB Tr2 Pr2 R Tc HRB Tr1 Pr1 6.423 R Tc 3J 10 mol Calculate actual enthalpy change using the expander efficiency. H HS H 4.496 3J 10 mol Find T2 such that H matches the value above. Given HS = MCPH T1 T2 A B C D R T2 T1 T2 Pr2 R Tc HRB R Tc HRB Tr1 Pr1 Tc T2 Find T2 T2 268.536 K 264 Ans. 0.779 Now recalculate S at calculated T2 S R MCPS T1 T2 A B C D ln T1 R SRB Tr1 Pr1 R SRB Tr2 Pr2 S 7.77 J mol K T2 R ln P2 Eq. (6-94) P1 Ans. Calculate power produced by expander kJ Ans. P H P 3.147 mol The advantage of the expander is that power can be produced in the expander which can be used in the plant. The disadvantages are the extra capital and operating cost of the expander and the low temperature of the gas leaving the expander compared to the gas leaving the throttle valve. 7.60 Hydrocarbon gas: T1 500degC Light oil: 25degC Cpoil Exit stream: b) T2 T3 Cpgas 150 J mol K J mol K J mol 200degC 200 Hlv 35000 Assume that the oil vaporizes at 25 C. For an adiabatic column, the overall energy balance is as follows. F Cpgas T3 T1 D Hlv Coilp T3 T2 =0 DF 0.643 Solving for D/F gives: DF c) Cpgas T3 Hlv T1 Cpoil T3 T2 Ans. Using liquid oil to quench the gas stream requires a smaller oil flow rate. This is because a significant portion of the energy lost by the gas is used to vaporize the oil. 265 Chapter 8 - Section A - Mathcad Solutions 8.1 With reference to Fig. 8.1, SI units, At point 2: Table F.2, H2 3531.5 At point 4: Table F.1, H4 209.3 At point 1: H1 H4 At point 3: Table F.1, Hliq H4 x3 H3 Hliq 0.96 Sliq For isentropic expansion, x'3 H'3 S'3 Slv x'3 Hlv H3 turbine Ws Ws H3 Hlv x3 Hlv H2 S'3 S2 0.855 H'3 2246 H2 H'3 1.035 3 10 H2 QH 3.322 Ws cycle QH cycle 266 Ans. 0.805 turbine QH H2 H3 Slv 0.7035 x' 3 Sliq Hliq S2 H1 10 0.311 3 Ans. 6.9636 2382.9 2496.9 7.3241 8.2 1.0 (kg/s) mdot The following property values are found by linear interpolation in Table F.1: State 1, Sat. Liquid at TH: H1 860.7 S1 2.3482 P1 3.533 State 2, Sat. Vapor at TH: H2 2792.0 S2 6.4139 P2 3.533 State 3, Wet Vapor at TC: Hliq 112.5 Hvap 2550.6 P3 1616.0 State 4, Wet Vapor at TC: Sliq 0.3929 Svap 8.5200 P4 1616.0 (a) The pressures in kPa appear above. (b) Steps 2--3 and 4--1 (Fig. 8.2) are isentropic, for which S3=S2 and S1=S4. Thus by Eq. 6.82): x3 S2 Sliq Svap Sliq x3 S1 Sliq Svap Sliq x4 0.741 x4 0.241 (c) The rate of heat addition, Step 1--2: Qdot12 mdot ( H2 Qdot12 H1) 1.931 3 10 (kJ/s) (d) The rate of heat rejection, Step 3--4: Hliq x3 ( Hvap Hliq) H3 1.919 H4 10 Qdot34 3 mdot ( H4 (e) Wdot12 699.083 1.22 873.222 mdot ( H3 H2) Wdot23 Wdot41 mdot ( H1 H4) Wdot41 Wdot23 Wdot41 Qdot12 161.617 0.368 Note that the first law is satisfied: Q Qdot12 Q W Qdot34 W Wdot23 0 267 Wdot41 3 10 0 Wdot23 (f) x4 ( Hvap Hliq) Qdot34 H3) Wdot34 0 Hliq H4 H3 (kJ/s) 8.3 The following vectors contain values for Parts (a) through (f). Enthalpies and entropies for superheated vapor, Tables F.2 and F.4 @ P2 and T2 (see Fig. 8.4): 3622.7 6.9013 kJ kg K 3529.6 kJ kg 6.9485 kJ kg K 3635.4 H2 kJ kg kJ kg 6.9875 kJ kg K 6.9145 kJ kg K S2 kJ 3475.6 kg 1507.0 BTU lbm 1.6595 BTU lbm rankine 1558.8 BTU lbm 1.6759 BTU lbm rankine Sat. liq. and sat. vap. values from Tables F.2 and F.4 @ P3 = P4: 191.832 2584.8 kJ kg 251.453 kJ kg 2609.9 kJ kg 191.832 Hliq kJ kg kJ kg 2584.8 kJ kg 2676.0 kJ kg Hvap kJ 419.064 kg 180.17 BTU lbm 1150.5 BTU lbm 69.73 BTU lbm 1105.8 BTU lbm 268 0.6493 kJ kg K 8.1511 kJ kg K 0.8321 kJ kg K 7.9094 kJ kg K kJ 0.6493 kg K Sliq 0.1326 Svap kJ 1.3069 0.3121 8.1511 7.3554 kg K BTU lbm rankine 1.7568 BTU 1.9781 lbm rankine kJ kg K kJ kg K BTU lbm rankine BTU lbm rankine 3 cm 1.010 gm 3 cm 1.017 gm 0.80 1.010 cm gm Vliq 3 cm 1.044 gm 0.0167 ft 0.75 0.75 3 0.75 0.80 turbine 0.78 0.80 pump 0.75 0.78 0.80 3 lbm 3 ft 0.0161 lbm 269 0.75 0.75 80 10000 kPa 10 kPa 100 7000 kPa 20 kPa 70 Wdot 50 3 8500 kPa P1 10 kW 10 kPa P4 6500 kPa 101.33 kPa 50 950 psi 14.7 psi 80 1125 psi 1 psi Vliq P1 Wpump P4 H4 pump S'3 = S2 H3 H2 x'3 S2 Sliq Svap turbine H'3 Sliq H2 Hliq H1 H'3 Hliq H4 Wpump x'3 Hvap Hliq H2 QdotH Wdot Wturbine Wpump H3 H2 H1 mdot QdotC mdot Wturbine QdotH Wdot Answers follow: QdotH mdot1 mdot2 108.64 mdot3 62.13 mdot4 1 70.43 67.29 QdotH kg sec 2 QdotH 3 QdotH 240705 355111 kJ 213277 sec 205061 4 mdot5 mdot6 QdotH 145.733 lbm 153.598 sec 5 QdotH 6 270 192801 BTU 228033 sec 0.332 QdotC 1 160705 0.282 QdotC 255111 kJ Wdot 0.328 QdotC 143277 sec QdotH 0.244 2 3 155061 QdotC 0.246 4 0.333 QdotC 145410 BTU 5 QdotC 152208 6 8.4 sec Subscripts refer to Fig. 8.3. Saturated liquid at 50 kPa (point 4) 3 V4 cm 1.030 gm H4 P4 3300 kPa P1 kJ kg 340.564 50 kPa Saturated liquid and vapor at 50 kPa: Hliq H4 Sliq 1.0912 Hvap kJ kg K 2646.0 Svap 7.5947 By Eq. (7.24), Wpump H1 H1 H4 Wpump kJ kg kJ kg K V 4 P4 343.911 P1 Wpump 3.348 kJ kg kJ kg The following vectors give values for temperatures of 450, 550, and 650 degC: 3340.6 H2 3565.3 7.0373 kJ kg S2 3792.9 7.3282 7.5891 271 kJ kg K S'3 S2 H'3 Hliq QH H2 S'3 x' 3 x'3 Hvap Hliq Sliq Svap Wturbine Sliq H'3 Wturbine H1 H2 Wpump QH 0.914 8.5 0.959 0.314 0.999 x'3 0.297 0.332 Ans. Subscripts refer to Fig. 8.3. Saturated liquid at 30 kPa (point 4) 3 V4 cm 1.022 gm H4 289.302 kJ kg P1 30 kPa Saturated liquid and vapor at 30 kPa: Hliq H4 Hvap 5000 kJ kg 2625.4 P4 7500 10000 Sliq 0.9441 kJ kg K By Eq. (7.24), Svap Wpump 7.7695 V 4 P4 kJ kg K P1 294.381 H1 H4 Wpump H1 296.936 kJ kg 299.491 The following vectors give values for pressures of 5000, 7500, and 10000 kPa at 600 degC 3664.5 H2 3643.7 7.2578 kJ kg S2 3622.7 7.0526 6.9013 272 kJ kg K kPa S'3 S2 H'3 Hliq QH H2 S'3 x'3 x'3 Hvap Hliq Sliq Svap Sliq Wturbine H'3 H2 Wturbine H1 Wpump QH 0.925 8.6 0.895 0.375 0.873 x'3 0.359 0.386 Ans. From Table F.2 at 7000 kPa and 640 degC: H1 kJ 3766.4 S1 kg 7.2200 kJ kg K S'2 S1 For sat. liq. and sat. vap. at 20 kPa: kJ kg Hvap 2609.9 kJ kg kJ kg K Svap 7.9094 kJ kg K Hliq 251.453 Sliq 0.8321 The following enthalpies are interpolated in Table F.2 at four values for intermediate pressure P2: 3023.9 725 P2 750 775 H'2 kPa W12 H'2 579.15 W12 572.442 kJ 565.89 kg 559.572 3040.9 kJ kg 3049.0 800 0.78 3032.5 H1 H2 3187.3 H2 3194 kJ 3200.5 kg 3206.8 273 H1 W12 7.4939 S2 7.4898 7.4851 7.4797 kJ kg K where the entropy values are by interpolation in Table F.2 at P2. x'3 S2 Svap W23 W Sliq H'3 Sliq H'3 W12 Hliq H2 x'3 Hvap Hliq 20.817 W23 W 7.811 kJ kg 5.073 17.723 The work difference is essentially linear in P2, and we interpolate linearly to find the value of P2 for which the work difference is zero: linterp W P2 0.0 kJ kg (P2) 765.16 kPa Also needed are values of H2 and S2 at this pressure. Again we do linear interpolations: linterp P2 H2 765.16 kPa 3197.9 kJ kg H2 3197.9 kJ kg linterp P2 S2 765.16 kPa 7.4869 kJ kg K S2 7.4869 kJ kg K We can now find the temperature at this state by interplation in Table F.2. This gives an intermediate steam temperature t2 of 366.6 degC. The work calculations must be repeated for THIS case: W12 W12 H2 H1 568.5 H'3 Hliq H'3 2.469 x'3 kJ x'3 kg x'3 Hvap 10 Hliq W23 3 kJ W23 kg 274 S2 Svap Sliq Sliq 0.94 H'3 568.46 H2 kJ kg Work W12 Work W23 1137 kJ kg For a single isentropic expansion from the initial pressure to the final pressure, which yields a wet exhaust: x'3 S1 Svap x'3 H'3 H'3 Hliq x'3 Hvap H'3 Sliq 0.903 W' Sliq 2.38 10 W' H1 Hliq 3 kJ 1386.2 kg kJ kg Whence the overall efficiency is: Work overall overall W' 275 0.8202 Ans. 8.7 From Table F.2 for steam at 4500 kPa and 500 degC: H2 3439.3 kJ kg S2 7.0311 kJ kg K S'3 S2 By interpolation at 350 kPa and this entropy, H'3 2770.6 H3 H2 kJ 0.78 kg WI H3 WI 2.918 10 3 kJ kg Isentropic expansion to 20 kPa: S'4 Exhaust is wet: for sat. liq. & vap.: S2 Hliq 251.453 Sliq 0.8321 kJ kg kJ kg K Hvap 2609.9 Svap 7.9094 276 kJ kg kJ kg K H'3 H2 WI 521.586 kJ kg x' 4 S'4 Sliq Svap x'4 H2 H'4 2.317 10 H4 H2 Hliq H'4 0.876 H4 H'4 Sliq x'4 Hvap 2.564 10 Hliq 3 kJ kg 3 kJ kg 3 H5 Hliq V5 V 5 P6 Wpump Wpump 5.841 cm 1.017 P5 20 kPa H6 H5 Wpump H6 257.294 gm P5 kJ kg P6 4500 kPa kJ kg For sat. liq. at 350 kPa (Table F.2): H7 584.270 kJ kg t7 138.87 (degC) We need the enthalpy of compressed liquid at point 1, where the pressure is 4500 kPa and the temperature is: t1 138.87 6 T1 t1 273.15 K t1 132.87 At this temperature, 132.87 degC, interpolation in Table F.1 gives: 3 kJ 558.5 kg Hsat.liq Psat 294.26 kPa Vsat.liq 1.073 Also by approximation, the definition of the volume expansivity yields: 1 1.083 20 Vsat.liq 9.32 10 1.063 3 cm gm K 41 K 277 P1 P6 cm gm By Eq. (7.25), H1 Hsat.liq Vsat.liq 1 T1 P1 H1 Psat 561.305 kJ kg By an energy balance on the feedwater heater: mass H1 H6 H3 H7 mass kg Ans. 0.13028 kg Work in 2nd section of turbine: WII ( kg 1 Wnet mass) H4 WI Wpump 1 kg QH QH 8.8 H2 WII WII 2878 kJ 307.567 kJ Wnet H3 823.3 kJ H 1 1 kg Wnet Ans. 0.2861 QH Refer to figure in preceding problem. Although entropy values are not needed for most points in the process, they are recorded here for future use in Problem 15.8. From Table F.4 for steam at 650(psia) & 900 degF: H2 1461.2 BTU lbm S2 1.6671 BTU lbm rankine S'3 S2 By interpolation at 50(psia) and this entropy, H'3 1180.4 H3 H2 S3 1.7431 BTU lbm WI WI 0.78 H3 1242.2 BTU lbm rankine 278 BTU lbm H'3 WI 219.024 H2 BTU lbm S'4 Isentropic expansion to 1(psia): S2 Exhaust is wet: for sat. liq. & vap.: 69.73 Sliq 0.1326 x'4 S'4 Hvap BTU lbm rankine Sliq 1.9781 BTU lbm rankine H2 H'4 H4 Hvap Hliq 0.944 1047.8 S4 Hliq 931.204 H4 H2 Hliq Sliq S4 H4 H'4 Sliq 0.831 x4 BTU lbm H'4 Svap x' 4 x4 1105.8 Svap BTU lbm Hliq x'4 Hvap 1.8748 Hliq BTU lbm BTU lbm x4 Svap Sliq BTU lbm rankine 3 P5 Wpump P6 H5 1 psi V 5 P6 P5 Wpump H6 650 psi V5 Hliq H5 2.489 ft 0.0161 lbm H6 72.219 BTU lbm Wpump BTU lbm For sat. liq. at 50(psia) (Table F.4): H7 250.21 BTU lbm t7 281.01 S7 0.4112 BTU lbm rankine We need the enthalpy of compressed liquid at point 1, where the pressure is 650(psia) and the temperature is t1 281.01 11 T1 t1 279 459.67 rankine t1 270.01 At this temperature, 270.01 degF, interpolation in Table F.3 gives: Psat 41.87 psi Hsat.liq Vsat.liq 238.96 0.1717 Ssat.liq 0.3960 ft 3 lbm BTU lbm BTU lbm rankine The definition of the volume expansivity yields: 0.01726 1 ft 20 Vsat.liq 4.95 0.01709 10 5 3 P1 P6 H1 257.6 BTU lbm S1 lbm rankine 0.397 BTU lbm rankine 1 rankine By Eq. (7.25) and (7.26), H1 Hsat.liq S1 Vsat.liq 1 Ssat.liq Vsat.liq T1 P1 P1 Psat Psat By an energy balance on the feedwater heater: mass H1 H6 H3 H7 lbm mass 0.18687 lbm Ans. WII 158.051 BTU Work in 2nd section of turbine: WII 1 lbm Wnet WI QH H2 Wnet QH mass H4 H3 Wpump 1 lbm WII Wnet H1 1 lbm QH 0.3112 280 Ans. 374.586 BTU 1.204 3 10 BTU 8.9 Steam at 6500 kPa & 600 degC (point 2) Table F.2: H2 3652.1 kJ kg S2 7.1258 kJ kg K P2 6500 kPa At point 3 the pressure must be such that the steam has a condensation temperature in feedwater heater I of 195 degC, 5 deg higher than the temperature of the feed water to the boiler at point 1. Its saturation pressure, corresponding to 195 degC, from Table F.1, is 1399.0 kPa. The steam at point 3 is superheated vapor at this pressure, and if expansion from P2 to P3 is isentropic, S'3 S2 H'3 3142.6 H3 H2 By double interpolation in Table F.2, kJ kg WI From Table F.1: 0.80 H3 H10 3.244 WI 829.9 281 3 kJ 10 kJ kg kg H'3 WI 407.6 H2 kJ kg Similar calculations are required for feedwater heater II. At the exhaust conditions of 20 kPa, the properties of sat. liq. and sat. vap. are: Hliq kJ 251.453 kg Sliq 0.8321 Hvap Svap kJ kg K 2609.9 7.9094 3 kJ Vliq kg 1.017 cm gm kJ kg K If we find t7, then t8 is the mid-temperature between t7 and t1(190 degC), and that fixes the pressure of stream 4 so that its saturation temperature is 5 degC higher. At point 6, we have saturated liquid at 20 kPa, and its properties from Table F.2 are: tsat 60.09 Tsat H6 Hliq V6 Wpump Wpump V 6 P2 8.238 P6 tsat 273.15 K Vliq P6 20 kPa [Eq. (7.24)] kJ kg H67 Wpump We apply Eq. (7.25) for the calculation of the temperature change from point 6 to point 7. For this we need values of the heat capacity and volume expansivity of water at about 60 degC. They can be estimated from data in Table F.1: 1 Vliq 5.408 1.023 1.012 20 10 3 cm CP gm K 41 CP K 272.0 kJ kg K 230.2 10 4.18 kJ kg K Solving Eq. (7.25) for delta T gives: H67 T67 t7 Vliq 1 Tsat P2 P6 190 t7 T67 CP tsat T67 K t9 2 282 t7 t8 0.678 K t9 5 t7 60.768 t8 130.38 H7 Hliq H67 H7 259.691 kJ kg H8 From Table F.1: t9 T9 125.38 547.9 kJ kg 273.15 t9 K At points 9 and 1, the streams are compressed liquid (P=6500 kPa), and we find the effect of pressure on the liquid by Eq. (7.25). Values by interpolation in Table F.1 at saturation temperatures t9 and t1: Hsat.9 kJ 526.6 kg Hsat.1 kJ 807.5 kg 3 Vsat.9 cm 1.065 gm Vsat.1 cm 1.142 gm Psat.9 234.9 kPa Psat.1 1255.1 kPa 3 3 3 V9 T cm 1.056) gm ( 1.075 V1 1 20 K 9 9 H9 Hsat.9 Vsat.9 1 T1 ( 273.15 Hsat.1 Vsat.1 1 1 T 1 V1 Vsat.1 T 41 9 T9 10 1 T1 P2 1 K P2 Psat.9 Psat.1 1.226 H9 530.9 T1 8.92 190)K H1 ( 1.156 V9 Vsat.9 cm 1.128) gm 10 463.15 K H1 810.089 kJ kg kJ kg Now we can make an energy balance on feedwater heater I to find the mass of steam condensed: mI H1 H9 H3 H10 mI kg 283 0.11563 kg Ans. 31 K The temperature at point 8, t8 = 130.38 (see above) is the saturation temperture in feedwater heater II. The saturation pressure by interpolation in Table F.1 is 273.28 kPa. Isentropic expansion of steam from the initial conditions to this pressure results in a slightly superheated vapor, for which by double interpolation in Table F.2: H'4 2763.2 kJ H4 H2 H4 Then kg H'4 2.941 H2 3 kJ 10 kg We can now make an energy balance on feedwater heater II to find the mass of steam condensed: mII H9 H 7 1 kg H4 mI H10 H8 H8 mII 0.09971 kg Ans. The final stage of expansion in the turbine is to 20 kPa, where the exhaust is wet. For isentropic expansion, x'5 x'5 S2 Svap Sliq H'5 0.889 H5 Then H2 Hliq H'5 Sliq 2.349 H'5 x'5 Hvap Hliq 3 kJ 10 kg H2 H5 2609.4 kJ kg The work of the turbine is: Wturbine Wturbine WI 1 kg 1 kg mI H4 H3 1 kg mI mII H5 H4 936.2 kJ Wturbine QH H2 Wpump 1 kg H 1 1 kg 0.3265 QH 284 QH Ans. 3 2.842 10 kJ 8.10 Tc Isobutane: Pc 408.1 K 0.181 36.48 bar For isentropic expansion in the turbine, let the initial state be represented by symbols with subscript zero and the final state by symbols with no subscript. Then T0 P0 533.15 K S 0 J mol K A 37.853 10 K B Tr0 Tc 450 kPa For the heat capacity of isobutane: 1.677 T0 Tr0 P 4800 kPa 3 11.945 10 C K Pr0 1.3064 Pr 6 2 P0 Pr0 Pc P Pc Pr 1.3158 0.123 Use generalized second-virial correlation: The entropy change is given by Eq. (6.92) combined with Eq. (5.15) with D = 0: 0.8 (guess) Given S = R A ln SRB B T0 T0 Tc Pr 1 2 C T0 2 ln P P0 SRB Tr0 Pr0 T 0.852 Find 1 Tr T T0 T Tc Tr 454.49 K 1.114 The enthalpy change for this final temperature is given by Eq. (6.91), with HRB at the above T: Hig Hig R ICPH T0 T 1.677 37.853 10 1.141 10 4J mol 285 3 11.945 10 6 0.0 Hturbine Hig R Tc HRB Tr Pr Hturbine 8850.6 J mol HRB Tr0 Pr0 Wturbine Hturbine The work of the pump is given by Eq. (7.24), and for this we need an estimate of the molar volume of isobutane as a saturated liquid at 450 kPa. This is given by Eq. (3.72). The saturation temperature at 450 kPa is given by the Antoine equation solved for t degC: VP 450 kPa Avp 14.57100 Bvp Bvp tsat Avp Cvp VP ln kPa 2606.775 tsat Cvp 34 Tsat tsat Tsat 3 Vc 274.068 cm 262.7 mol Zc 0.282 Trsat 273.15 K 307.15 K Tsat Trsat Tc 0.753 2 Vliq Wpump V c Zc 1 T rsat Vliq P0 3 7 Vliq P cm 112.362 mol Wpump 488.8 J mol The flow rate of isobutane can now be found: mdot 1000 kW Wturbine Wpump mdot 119.59 mol sec Ans. The enthalpy change of the isobutane in the cooler/condenser is calculated in two steps: a. Cooling of the vapor from 454.48 to 307.15 K b. Condensation of the vapor at 307.15 K Enthalpy change of cooling: HRB at the initial state has already been calculated. For saturated vapor at 307.15 K: Hig R ICPH T Tsat 1.677 37.853 10 286 3 11.945 10 6 0.0 Hig 4J 1.756 Ha Hig Ha 18082 10 mol R Tc HRB Trsat Pr HRB Tr Pr J mol For the condensation process, we estimate the latent heat by Eqs. (4.12) and (4.13): Tn 261.4 K R Tn 1.092 ln Hn 0.930 Hb Hn 1 Trsat 1 Pc bar mdot Qdotin Wturbine 0.641 1.013 Hn Trn 4J 2.118 10 0.38 Hb Ha 18378 mol J mol Hb Wpump mdot 4360 kW 8.11 Isobutane: Tc Trn Tc Trn Qdotout Qdotout Tn Trn Qdotout 1000 kW Qdotin Qdotin 408.1 K 5360 kW 0.187 Pc 36.48 bar 0.181 Ans. For isentropic expansion in the turbine, let the initial (inlet) state be represented by symbols with subscript zero and the final (exit) state by symbols with no subscript. Then T0 S 413.15 K 0 P0 3400 kPa J mol K 287 P 450 kPa molwt 58.123 gm mol For the heat capacity of isobutane: A 1.677 B T0 Tr0 37.853 10 11.945 10 C K Tr0 Tc 3 K 1.0124 2 P0 Pr0 Pr0 Pc P Pc Pr 6 Pr 0.932 0.123 Use Lee/Kesler correlation for turbine-inlet state, designating values by HRLK and SRLK: HRLK0 1.530 SRLK0 1.160 The entropy change is given by Eq. (6.92) combined with Eq. (5.15) with D = 0: (guess) 0.8 Given S = R A ln B T0 T0 SRB Pr Tc Find 1 2 C T0 1 2 ln P P0 SRLK0 0.809 T T0 T T Tc Tr Tr 334.08 K 0.819 The enthalpy change for this final temperature is given by Eq. (6.91), with HRB at the above T: Hig R ICPH T0 T 1.677 37.853 10 Hig 9.3 3 11.945 10 6 0.0 3J 10 mol Hturbine Hig R Tc HRB Tr Pr Hturbine 4852.6 J mol HRLK0 Wturbine 288 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: 3 Vliq 112.36 cm mol Wpump Vliq P0 P Wpump 331.462 J mol For the cycle the net power OUTPUT is: mdot kg 75 molwt sec Wdot Wdot mdot Wturbine Wpump Ans. 5834 kW The enthalpy change of the isobutane in the cooler/condenser is calculated in two steps: a. Cooling of the vapor from 334.07 to 307.15 K b. Condensation of the vapor at 307.15 K Enthalpy change of cooling: HRB at the initial state has already been calculated. For saturated vapor at 307.15 K it was found in Problem 8.10 as: Tsat Hig 307.15K Tsat Trsat Trsat Tc R ICPH T Tsat 1.677 37.853 10 Hig 2.817 Hig R Tc HRB Trsat Pr Ha 2975 11.945 10 6 0.0 kJ mol Ha 3 0.753 HRB Tr Pr J mol For the condensation process, the enthalpy change was found in Problem 8.10: Hb 18378 J mol Qdotout Qdotout 289 mdot Ha Hb 27553 kW Ans. For the heater/boiler: Qdotin Wdot Qdotout Qdotin Wdot Qdotin 33387 kW Ans. Ans. 0.175 We now recalculate results for a cycle for which the turbine and pump each have an efficiency of 0.8. The work of the turbine is 80% of the value calculated above, i.e., W'turbine 0.8 Wturbine W'turbine 3882 J mol The work of the pump is: W'pump Wdot Wpump W'pump 0.8 mdot W'turbine W'pump Wdot 414.3 J mol 4475 kW Ans. The decrease in the work output of the turbine shows up as an increase in the heat transferred out of the cooler condenser. Thus Qdotout Qdotout Qdotout Wturbine 28805 kW W'turbine mdot Ans. The increase in pump work shows up as a decrease in the heat added in the heater/boiler. Thus Qdotin Qdotin Wdot Qdotin W'pump Wpump mdot 0.134 290 Ans. Qdotin 33280 kW Ans. CP PC 1 bar By Eq. (3.30c): 7 R 2 TC 8.13 Refer to Fig. 8.10. 293.15 K PD PC V C = PD V D 1 VC VD 1 PD = 1.4 5 bar or PD r PC r PC Ans. 3.157 1 Eq. (3.30b): TD QDA = CP TA TC PD QDA PC QDA TA TD CP TD 1500 TA J mol 515.845 K R TC re = VB VA = VC VA = PC R TA PA PD T C PA re T A PC PA re 8.14 2.841 Ans. 3 5 Ratio Ratio = 7 PB PA 1.35 9 Eq. (8.12) now becomes: 0.248 1 1 0.341 1 Ratio 0.396 0.434 291 Ans. 8.16 Figure shows the air-standard turbojet power plant on a PV diagram. 7 TA 303.15 K TC 1373.15 K R CP 2 By Eq. (7.22) WAB = CP TA WCD = CP TC PB R CP 1 = CP TA cr PA PD 2 R CP 1 2 1 = CP TC er PC 7 7 1 where cr is the compression ratio and er is the expansion ratio. Since the two work terms are equal but of opposite signs, cr 6.5 er 0.5 (guess) 2 Given TC er 7 2 1 = TA cr 7 er Find er) ( er 292 1 0.552 By Eq. (7.18): PD TD = TC R CP PC 2 This may be written: TD TC er 7 1 By Eq. (7.11) uE 2 2 2 uD = PD V D 1 PE 1 (A) PD We note the following: er = PD cr = PC PB PA = PC cr er = PE PD PE The following substitutions are made in (A): 1 uD = 0 = 2 R = 7 CP Then molwt uE 2 PE 8.17 TA R 7 TD 1 2 molwt PD 1 bar 305 K PA PE PD V D = R T D 1 cr er 29 2 7 cr er PE PB 1.05bar PD = 1 cr er gm mol uE 843.4 m sec Ans. PD 3.589 bar Ans. 0.8 7.5bar Assume air to be an ideal gas with mean heat capacity (final temperature by iteration): Cpmair MCPH 298.15K 582K 3.355 0.575 10 Cpmair 29.921 J mol K 293 3 0.0 0.016 10 5 R Compressor: TB PB Cpmair TA Wsair R Cpmair PA Wsair TA TB Cpmair Combustion: Wsair 1 8.292 10 3J mol 582.126 K CH4 + 2O2 = CO2 + 2H2O Basis: Complete combustion of 1 mol CH4. Reactants are N mol of air and 1mol CH4. Because the combustion is adiabatic, the basic equation is: HR H298 HP = 0 For H_R, the mean heat capacities for air and methane are required. The value for air is given above. For methane the temperature change is very small; use the value given in Table C.1 for 298 K: 4.217*R. The solution process requires iteration for N. Assume a value for N until the above energy balance is satisfied. (a) TC 1000K HR 57.638 (This is the final value after iteration) N Cpmair N ( 98.15 2 HR 4.896 582.03)K 4.217 R ( 98.15 2 300)K 5J 10 mol The product stream contains: 1 mol CO2, 2mol H2O, 0.79N mol N2, and (0.21N-2) mol O2 1 5.457 2 n A .79 N .21 N i 2 3.470 3.280 1.045 B 3.639 1.450 0.593 0.506 14 294 1.157 10 3 D 0.121 0.040 0.227 5 10 ni 58.638 A ni Ai i i A CpmP HP ni Bi D i 198.517 B CpmP TC 298.15K HP H298 802625 H298 HP 136.223 ni D i i 0.036 D MCPH 298.15K 1000.K 198.517 0.0361 0.0 From Ex. 4.7: HR B 1.292 1.387 10 5 1.3872 10 5 R 6J 10 mol J mol J (This result is sufficiently close to zero.) mol Thus, N = 57.638 moles of air per mole of methane fuel. Ans. Assume expansion of the combustion products in the turbine is to 1(atm), i.e., to 1.0133 bar: PD 1.0133bar PC 7.5bar The pertinent equations are analogous to those for the compressor. The mean heat capacity is that of the combustion gases, and depends on the temperature of the exhaust gases from the turbine, which must therefore be found by iteration. For an initial calculation use the mean heat capacity already determined. This calculation yields an exhaust temperature of about 390 K. Thus iteration starts with this value. Parameters A, B, and D have the final values determined above. Cpm Cpm MCPH 1000K 343.12K 198.517 0.0361 0.0 1.849 10 3 5 1.3872 10 R J For 58.638 moles of combustion product: mol K R Ws TD 58.638 Cpm TC PD Cpm 1 PC TC Ws Cpm TD 343.123 K 295 Ws 6J 1.214 10 mol (Final result of iteration.) Ans. Wsnet Ws Wsair N Wsnet 5J 7.364 10 Ans. mol (J per mole of methane) Parts (b) and (c) are solved in exactly the same way, with the following results: 5 (b) TC N 37.48 Wsnet 7.365 10 (c) 8.18 1200 1500 N 24.07 Wsnet 5.7519 10 me 0.95 line_losses TC 0.35 tm TD TD 5 343.123 598.94 20% Cost_fuel 4.00 dollars GJ Cost_fuel Cost_electricity tm Cost_electricity 0.05 1 me ( cents kW hr line_losses) Ans. This is about 1/2 to 1/3 of the typical cost charged to residential customers. 8.19 TC TH 111.4K 1 Carnot TC Carnot TH Assume as a basis: QH W W QH Hnlv 300K 0.629 QC 2.651 kJ QC W 0.201 mol kJ 0.6 kJ mol Carnot HE 0.377 1kJ HE Hnlv HE 8.206 Ans. 296 QH 1 HE QC 1.651 kJ 8.20 TH ( 27 a) Carnot b) actual TC 273.15)K 1 (6 TC Carnot TH Carnot 0.6 273.15)K 0.07 Ans. actual 0.028 Ans. 2 3 c) The thermal efficiency is low and high fluid rates are required to generate reasonable power. This argues for working fluids that are relatively inexpensive. Candidates that provide reasonable pressures at the required temperature levels include ammonia, n-butane, and propane. 297 Chapter 9 - Section A - Mathcad Solutions 9.2 TH ( 20 TC ( 20 QdotC 0.08 Cost 9.4 TC QdotC Wdot kW hr 253.15 K kJ day TC TH 293.15 K TC 273.15) K 125000 Carnot TH 273.15) K (9.3) 0.6 (9.2) Wdot Wdot Cost 3.797 Carnot 0.381 kW 267.183 dollars yr Ans. Basis: 1 lbm of tetrafluoroethane The following property values are found from Table 9.1: State 1, Sat. Liquid at TH: H1 44.943 S1 0.09142 P1 138.83 State 2, Sat. Vapor at TH: H2 116.166 S2 0.21868 P2 138.83 State 3, Wet Vapor at TC: Hliq 15.187 Hvap 104.471 P3 26.617 State 4, Wet Vapor at TC: Sliq 0.03408 Svap 0.22418 P4 26.617 (a) The pressures in (psia) appear above. (b) Steps 3--2 and 1--4 (Fig. 8.2) are isentropic, for which S3=S2 and S1=S4. Thus by Eq. 6.82): x3 0.971 x4 S1 Sliq Svap Sliq x3 ( Hvap Hliq) H4 Hliq H4 42.118 Q43 59.77 S2 Sliq Svap Sliq x3 x4 0.302 (c) Heat addition, Step 4--3: H3 Hliq H3 101.888 Q43 ( H3 H4) 298 x4 ( Hvap Hliq) (Btu/lbm) (d) Heat rejection, Step 2--1: W21 ( H2 W14 ( H4 0 H3) W32 14.278 H1) W14 2.825 H2) Q43 W14 W32 (f) (Btu/lbm) 71.223 0 W32 (e) ( H1 Q21 W43 Q21 5.219 Note that the first law is satisfied: Q 9.7 Q21 W Q43 W32 Q W14 W TC 298.15 K TH 523.15 K (Engine) T'C 273.15 K T'H 298.15 K 0 (Refrigerator) By Eq. (5.8): Carnot By Eq. (9.3): Carnot By definition: = But TH T'H T'C Wengine 0.43 Carnot 10.926 = QH Given that: 0.6 QH Wrefrig kJ sec 35 QH 7.448 Carnot 0.6 Carnot 20.689 Q'C Q'C Q'C QH Carnot Q'C Carnot T'C Wengine = Wrefrig Whence QH TC 1 kJ Ans. sec 299 Carnot kJ sec Ans. 6.556 9.8 (a) QC kJ sec 4 W QC (c) QC TH TC ( 40 TC 1 5.5 TH TC TH kJ QH W TC = Ans. 2.667 W (b) QH 1.5 kW 227.75 K Ans. sec 273.15)K TH 313.15 K Ans. or -45.4 degC 9.9 The following vectors contain data for parts (a) through (e). Subscripts refer to Fig. 9.1. Values of H2 and S2 for saturated vapor come from Table 9.1. 489.67 600 479.67 0.78 500 469.67 rankine 0.77 459.67 0.76 300 449.67 T2 0.79 0.75 200 107.320 H2 0.22325 104.471 103.015 Btu lbm S2 Btu lbm rankine 0.22647 539.67 rankine S'3 = S2 0.22418 0.22525 101.542 T4 400 0.22244 105.907 QdotC H4 37.978 Btu lbm (isentropic compression) 300 From Table 9.1 for sat. liquid Btu sec The saturation pressure at Point 4 from Table 9.1 is 101.37(psia). For isentropic compression, from Point 2 to Point 3', we must read values for the enthalpy at Point 3' from Fig. G.2 at this pressure and at the entropy values S2. This cannot be done with much accuracy. The most satisfactory procedure is probably to read an enthalpy at S=0.22 (H=114) and at S=0.24 (H=126) and interpolate linearly for intermediate values of H. This leads to the following values (rounded to 1 decimal): 115.5 116.0 Btu lbm H23 H'3 H2 117.2 117.9 H3 H2 H1 116.5 H'3 H23 H4 88.337 kJ kg H23 273.711 30.098 H1 24.084 276.438 36.337 kJ kg H3 279.336 43.414 283.026 50.732 kJ kg 286.918 8.653 mdot 7.361 QdotC H2 mdot H1 6.016 lbm sec Ans. 4.613 3.146 689.6 595.2 QdotH mdot H4 H3 QdotH Btu sec 494 Ans. 386.1 268.6 94.5 100.5 Wdot mdot H23 Wdot 99.2 90.8 72.4 301 kW Ans. 6.697 QdotC 5.25 Wdot 4.256 Ans. 3.485 2.914 TC T2 TH T4 9.793 7.995 TC Carnot TH Carnot TC Ans. 6.71 5.746 4.996 9.10 Subscripts in the following refer to Fig. 9.1. All property values come from Tables F.1 and F.2. T2 ( 4 QdotC H4 273.15)K 1200 142.4 kJ sec kJ kg T4 ( 34 H2 2508.9 S'2 = S2 273.15)K kJ kg 0.76 S2 9.0526 kJ kg K (isentropic compression) The saturation pressure at Point 4 from Table F.1 is 5.318 kPa. We must find in Table F.2 the enthalpy (Point 3') at this pressure and at the entropy S2. This requires double interpolation. The pressure lies between entries for pressures of 1 and 10 kPa, and linear interpolation with P is unsatisfactory. Steam is here very nearly an ideal gas, for which the entropy is linear in the logarithm of P, and interpolation must be in accord with this relation. The enthalpy, on the other hand, changes very little with P and can be interpolated linearly. Linear interpolation with temperture is satisfactory in either case. The result of interpolation is H'3 2814.7 kJ kg H23 H23 H'3 H2 402.368 302 kJ kg H1 H4 H3 H2 H23 H3 QdotC mdot H2 mdot H1 QdotH mdot H4 Wdot mdot H23 QdotC H3 0.507 9.11 T2 kg Ans. sec kJ Ans. sec 204 kW Ans. T2 T4 kg 1404 QdotH Wdot 10 Ans. 5.881 Wdot Carnot 3 kJ 2.911 Carnot 9.238 Ans. Parts (a) & (b): subscripts refer to Fig. 9.1 At the conditions of Point 2 [t = -15 degF and P = 14.667(psia)] for sat. liquid and sat. vapor from Table 9.1: Hliq 7.505 Btu lbm Hvap 303 100.799 Btu lbm H2 Hvap Sliq Btu lbm rankine 0.01733 Svap 0.22714 Btu lbm rankine For sat. liquid at Point 4 (80 degF): H4 37.978 Btu S4 lbm (a) Isenthalpic expansion: QdotC Btu sec 5 H1 0.07892 H2 mdot H1 (b) Isentropic expansion: S1 x1 Sliq Svap H1 Sliq QdotC mdot H2 S1 Hliq x1 Hvap mdot H1 lbm rankine H4 QdotC mdot Btu 0.0759 0.0796 sec Ans. S4 H1 Hliq lbm lbm 34.892 BTU lbm Ans. sec (c) The sat. vapor from the evaporator is superheated in the heat exchanger to 70 degF at a pressure of 14.667(psia). Property values for this state are read (with considerable uncertainty) from Fig. G.2: H2A mdot 117.5 Btu lbm H4 0.262 mdot QdotC H2A Btu S2A 0.0629 lbm rankine lbm sec Ans. (d) For isentropic compression of the sat. vapor at Point 2, S3 Svap and from Fig. G.2 at this entropy and P=101.37(psia) H3 118.3 Btu lbm Eq. (9.4) may now be applied to the two cases: In the first case H1 has the value of H4: H2 a H4 H3 H2 a 304 3.5896 Ans. In the second case H1 has its last calculated value [Part (b)]: H2 H1 H3 b H2 Ans. 3.7659 b In Part (c), compression is at constant entropy of 0.262 to the final pressure. Again from Fig. G.2: H3 138 Btu H3 Wdot lbm Wdot 1.289 QdotC c 9.12 c Wdot H2A mdot (Last calculated value of mdot) BTU sec Ans. 3.8791 Subscripts: see figure of the preceding problem. At the conditions of Point 2 [sat. vapor, t = 20 degF and P = 33.110(psia)] from Table 9.1: H2 105.907 Btu lbm S2 0.22325 Btu lbm rankine At Point 2A we have a superheated vapor at the same pressure and at 70 degF. From Fig. G.2: H2A 116 Btu lbm S2A Btu lbm rankine 0.2435 For sat. liquid at Point 4 (80 degF): H4 37.978 Btu lbm S4 0.07892 Btu lbm R Energy balance, heat exchanger: H1 QdotC H4 H2A 2000 H2 Btu sec H1 mdot 305 27.885 BTU lbm QdotC H2 H1 mdot 25.634 lbm sec For compression at constant entropy of 0.2435 to the final pressure of 101.37(psia), by Fig. G.2: H'3 Wdot mdot 127 Btu lbm 0.75 mdot Hcomp 25.634 lbm sec Hcomp Wdot H'3 Hcomp 396.66 kW H2A 14.667 Btu lbm Ans. If the heat exchanger is omitted, then H1 = H4. Points 2A & 2 coincide, and compression is at a constant entropy of 0.22325 to P = 101.37(psia). mdot Wdot mdot 9.13 QdotC H2 H4 H'3 116 Btu lbm mdot Hcomp 29.443 lbm sec Hcomp Wdot H'3 Hcomp 418.032 kW H2 13.457 Btu lbm Ans. Subscripts refer to Fig. 9.1. At Point 2 [sat. vapor @ 10 degF] from Table 9.1: H2 104.471 Btu lbm S2 0.22418 Btu lbm R S'3 S2 H values for sat. liquid at Point 4 come from Table 9.1 and H values for Point 3` come from Fig. G.2. The vectors following give values for condensation temperatures of 60, 80, & 100 degF at pressures of 72.087, 101.37, & 138.83(psia) respectively. 31.239 H4 37.978 44.943 113.3 Btu lbm H'3 116.5 119.3 306 Btu lbm H1 H4 (a) By Eq. (9.4): 8.294 H2 H'3 (b) H1 H2 H Ans. 5.528 4.014 H'3 H2 0.75 Since H = H3 H2 Eq. (9.4) now becomes H2 6.221 H1 9.14 Ans. 4.146 H 3.011 WINTER TH Wdot 293.15 1.5 QdotH = 0.75 TH TC TH TC Wdot = QdotH TH TC 250 (Guess) Given TH TC Wdot = TH 0.75 TH TC TC Find TC TC 268.94 K Ans. Minimum t = -4.21 degC 307 SUMMER TC 298.15 QdotC 0.75 TH TC TH TC Wdot = TC QdotC TH (Guess) 300 Given Wdot 0.75 TH TC TH 322.57 K TH TC TC Find TH TH = Ans. Maximum t = 49.42 degC 9.15 and 9.16 Data in the following vectors for Pbs. 9.15 and 9.16 come from Perry's Handbook, 7th ed. H4 1033.5 785.3 kJ kg By Eq. (9.8): 9.17 H9 z 284.7 kJ kg H4 H15 H9 1186.7 H15 H15 z 1056.4 0.17 kJ kg Ans. 0.351 Advertized combination unit: TH ( 150 TH 50000 Btu hr WCarnot ( 30 TC 609.67 rankine QC TC 459.67)rankine 489.67 rankine QC 308 TH TC TC 459.67)rankine WCarnot 12253 Btu hr WI WI 1.5 WCarnot 18380 Btu hr This is the TOTAL power requirement for the advertized combination unit. The amount of heat rejected at the higher temperature of 150 degF is QH WI QH QC 68380 Btu hr For the conventional water heater, this amount of energy must be supplied by resistance heating, which requires power in this amount. For the conventional cooling unit, TH ( 120 459.67) rankine WCarnot QC Work TH TC WCarnot TC Work 1.5 WCarnot 9190 Btu hr Btu hr 13785 The total power required is WII 9.18 QH TC 210 WII Work T'H 82165 T'C 260 Btu hr NO CONTEST TH 255 305 By Eq. (9.3): TC TH WCarnot = TC QC I 0.65 TC T'H WI = II TC QC WII = 0.65 9.19 TH T'C QC II I Define r as the ratio of the actual work, WI + WII, to the r Carnot work: T'C 1 1 I r 1.477 Ans. II This problem is just a reworking of Example 9.3 with different values of x. It could be useful as a group project. 309 9.22 TH TC 290K Ws 250K TC Carnot TH 4.063 65% Carnot 6.25 Carnot TC 0.40kW Ans. QC 9.23 QC Ws 3 -3 Q 10 kgm sec H 1.625 2 Ws QC QH 2.025 kW Follow the notation from Fig. 9.1 With air at 20 C and the specification of a minimum approach T = 10 C: T1 ( 10 T4 273.15) K ( 30 T2 273.15) K T1 Calculate the high and low operating pressures using the given vapor pressure equation Guess: PL PH 1bar 2bar PL PL Given ln 4104.67 = 45.327 bar T1 5.146 ln T1 615.0 K T1 K PL PL Find PL bar 2 K 6.196 bar PH PH Given ln 4104.67 = 45.327 bar T4 5.146 ln T4 615.0 K T4 K PH Find PH PH bar 2 K 11.703 bar Calculate the heat load ndottoluene 50 kmol hr T1 ( 100 273.15) K T2 ( 20 273.15) K Using values from Table C.3 QdotC QdotC ndottoluene R ICPH T1 T2 15.133 6.79 10 177.536 kW 310 3 16.35 10 6 0 Since the throttling process is adiabatic: But: Hliq4 = Hliq1 H4 = H1 x1 Hlv1 so: Hliq4 Hliq1 = x1 Hlv T4 and: Hliq4 Hliq1 = Vliq P4 P1 Cpliq ( T) dT T1 Estimate Vliq using the Rackett Eqn. Pc 112.80bar Vc cm 72.5 mol Tn 239.7K 273.15)K Tc Tr 0.723 Tc 0.253 405.7K 3 Zc 0.242 ( 20 Tr Hlvn 23.34 kJ mol 2 Vliq 1 Tr V c Zc 3 7 cm 27.112 mol Vliq Estimate Hlv at 10C using Watson correlation T1 Tn Tr1 Trn 0.591 Trn Tc Tc Hlv Hliq41 Hlvn Vliq PH Hliq41 1.621 1 Tr1 1 0.698 0.38 Trn PL Tr1 Hlv 20.798 R ICPH T1 T4 22.626 kJ mol 100.75 10 Hliq41 x1 kJ mol x1 Hlv 3 192.71 10 0.078 For the evaporator H12 = H2 H1 = H1vap H12 x1 ndot 1 QdotC H12 Hlv H1liq x1 Hlv = 1 H12 ndot 311 19.177 9.258 x1 Hlv kJ mol mol sec Ans. 6 0 Chapter 10 - Section A - Mathcad Solutions 10.1 Benzene: A1 := 13.7819 B1 := 2726.81 C1 := 217.572 Toluene: A2 := 13.9320 B2 := 3056.96 C2 := 217.625 B1 A1 Psat1 ( T) := e T degC (a) Given: x1 := 0.33 Given +C1 A2 kPa Psat2 ( T) := e Guess: T := 100 degC B2 T degC y1 := 0.5 +C2 kPa P := 100 kPa x1 Psat1 ( T) + ( 1 x1) Psat2 ( T) = P x1 Psat1 ( T) = y1 P y1 := Find ( y1 , P) P (b) Given: y1 := 0.33 Given Ans. y1 = 0.545 P = 109.303 kPa Guess: T := 100 degC Ans. x1 := 0.33 P := 100 kPa x1 Psat1 ( T) + ( 1 x1) Psat2 ( T) = P x1 Psat1 ( T) = y1 P x1 := Find ( x1 , P) P (c) Given: x1 := 0.33 Given x1 = 0.169 P := 120 kPa Ans. Guess: P = 92.156 kPa y1 := 0.5 Ans. T := 100 degC x1 Psat1 ( T) + ( 1 x1) Psat2 ( T) = P x1 Psat1 ( T) = y1 P y1 := Find ( y1 , T) T y1 = 0.542 312 Ans. T = 103.307 degC Ans. (d) Given: y1 := 0.33 P := 120 kPa Guess: x1 := 0.33 T := 100 degC x1 Psat1 ( T) + ( 1 x1) Psat2 ( T) = P Given x1 Psat1 ( T) = y1 P x1 := Find ( x1 , T) T (e) Given: Given x1 = 0.173 Ans. T = 109.131 degC Ans. T := 105 degC P := 120 kPa Guess: x1 := 0.33 y1 := 0.5 x1 Psat1 ( T) + ( 1 x1) Psat2 ( T) = P x1 Psat1 ( T) = y1 P x1 := Find ( x1 , y1) y1 x1 = 0.282 z1 := 0.33 x1 = 0.282 y1 = 0.484 Guess: L := 0.5 V := 0.5 Given z1 = L x1 + V y1 (f) Ans. y1 = 0.484 Ans. L+V = 1 L := Find ( L , V) V (g) Vapor Fraction: V = 0.238 Ans. Liquid Fraction: L = 0.762 Ans. Benzene and toluene are both non-polar and similar in shape and size. Therefore one would expect little chemical interaction between the components. The temperature is high enough and pressure low enough to expect ideal behavior. 313 10.2 Pressures in kPa; temperatures in degC (a) Antoine coefficients: Benzene=1; Ethylbenzene=2 A1 := 13.7819 B1 := 2726.81 C1 := 217.572 A2 := 13.9726 B2 := 3259.93 C2 := 212.300 Psat1 ( T) := exp A1 B1 T + C1 B2 Psat2 ( T) := exp A2 T + C2 P-x-y diagram: T := 90 P ( x1) := x1 Psat1 ( T) + ( 1 x1) Psat2 ( T) T-x-y diagram: y1 ( x1) := x1 Psat1 ( T) P ( x1 ) P' := 90 Guess t for root function: t := 90 T ( x1) := root x1 Psat1 ( t) + ( 1 x1) Psat2 ( t) P' , t y'1 ( x1) := x1 Psat1 ( T ( x1) ) x1 Psat1 ( T ( x1) ) + ( 1 x1) Psat2 ( T ( x1) ) x1 := 0 , 0.05 .. 1.0 150 140 130 P ( x1 ) 120 100 T ( x1 ) 110 100 T ( x1 ) 90 P ( x1 ) 50 80 70 0 0 0.5 x1 , y1 ( x1 ) 60 1 314 0 0.5 x1 , y'1 ( x1 ) 1 (b) Antoine coefficients: 1-Chlorobutane=1; Chlorobenzene=2 A1 := 13.7965 B1 := 2723.73 C1 := 218.265 A2 := 13.8635 B2 := 3174.78 C2 := 211.700 Psat1 ( T) := exp A1 P-x-y diagram: B1 T + C1 T := 90 P ( x1) := x1 Psat1 ( T) + ( 1 x1) Psat2 ( T) T-x-y diagram: B2 T + C2 Psat2 ( T) := exp A2 y1 ( x1) := x1 Psat1 ( T) P ( x1 ) P' := 90 Guess t for root function: t := 90 T ( x1) := root x1 Psat1 ( t) + ( 1 x1) Psat2 ( t) P' , t y'1 ( x1) := x1 Psat1 ( T ( x1) ) x1 Psat1 ( T ( x1) ) + ( 1 x1) Psat2 ( T ( x1) ) x1 := 0 , 0.05 .. 1.0 130 160 122.5 115 113.33 107.5 T ( x1 ) 100 T ( x1 ) P ( x1 ) P ( x1 ) 92.5 66.67 85 77.5 20 0 0.5 x1 , y1 ( x1 ) 70 1 315 0 0.5 x1 , y'1 ( x1 ) 1 10.3 Pressures in kPa; temperatures in degC (a) Antoine coefficinets: n-Pentane=1; n-Heptane=2 A1 := 13.7667 B1 := 2451.88 C1 := 232.014 A2 := 13.8622 B2 := 2911.26 C2 := 216.432 B1 T + C1 Psat1 ( T) := exp A1 Psat2 ( T) := exp A2 Psat1 ( T) + Psat2 ( T) 2 P := T := 55 B2 T + C2 P = 104.349 Since for Raoult's law P is linear in x, at the specified P, x1 must be 0.5: x1 := 0.5 y1 := x1 Psat1 ( T) y1 = 0.89 P For a given pressure, z1 ranges from the liquid composition at the bubble point to the vapor composition at the dew point. Material balance: z1 = x1 ( 1 V) + y1 V V ( z1) := z1 := x1 , x1 + 0.01 .. y1 z1 x1 y1 x1 V is obviously linear in z1: 1 x1 y1 V ( z1 ) 0.5 0 0.45 0.5 0.55 0.6 0.65 0.7 z1 316 0.75 0.8 0.85 (b) At fixed T and z1, calculate x1, y1 and P as functions of fraction vapor (V). z1 := 0.5 Guess: x := 0.5 Psat1 ( T) + Psat2 ( T) 2 p := y := 0.5 Three equations relate x1, y1, & P for given V: Given p = x Psat1 ( T) + ( 1 x) Psat2 ( T) y p = x Psat1 ( T) z1 = ( 1 V) x + V y f ( V) := Find ( x , y , p) x1 ( V) := f ( V) 1 y1 ( V) := f ( V) 2 Plot P, x1 and y1 vs. vapor fraction (V) P ( V) := f ( V) 3 V := 0 , 0.1 .. 1.0 150 1 100 x1 ( V) P ( V) 50 0 0.5 y1 ( V) 0 0.5 0 1 0 0.5 1 V V 10.4 Each part of this problem is exactly like Problem 10.3, and is worked in exactly the same way. All that is involved is a change of numbers. In fact, the Mathcad solution for Problem 10.3 can be converted into the solution for any part of this problem simply by changing one number, the temperature. 10.7 Benzene: A1 := 13.7819 B1 := 2726.81 C1 := 217.572 Ethylbenzene A2 := 13.9726 B2 := 3259.93 C2 := 212.300 A1 Psat1 ( T) := e B1 T degC +C1 A2 kPa Psat2 ( T) := e 317 B2 T degC +C2 kPa y1 := 0.70 Guess: T := 116 degC P := 132 kPa (a) Given: x1 := 0.35 Given x1 Psat1 ( T) + ( 1 x1) Psat2 ( T) = P x1 Psat1 ( T) = y1 P T := Find ( T , P) P Ans. T = 134.1 degC P = 207.46 kPa Ans. For parts (b), (c) and (d) use the same structure. Set the defined variables and change the variables in the Find statement at the end of the solve block. (b) T = 111.88 deg_C P = 118.72 kPa (c) T = 91.44 deg_C P = 66.38 kPa (d) T = 72.43 deg_C P = 36.02 kPa To calculate the relative amounts of liquid and vapor phases, one must know the composition of the feed. 10.8 To increase the relative amount of benzene in the vapor phase, the temperature and pressure of the process must be lowered. For parts (c) and (d), the process must be operated under vacuum conditions. The temperatures are well within the bounds of typical steam and cooling water temperatures. 10.9 13.7819 (1) = benzene A := 13.9320 (2) = toluene (3) = ethylbenzene 13.9726 (a) n := rows ( A) Ai Psat ( i , T) := e i := 1 .. n 2726.81 B := 3056.96 3259.93 T := 110 degC 217.572 C := 217.625 212.300 P := 90 kPa zi := 1 n Bi T degC +Ci kPa ki := 318 Psat ( i , T) P Guess: V := 0.5 n zi ki 1 + V ( ki 1) V := Find ( V) V = 0.836 Given i=1 yi := xi := zi ki 1 + V ( ki 1) =1 Ans. Psat ( i , T) T = 110 deg_C Ans. V = 0.575 0.441 y = 0.333 0.226 V = 0.352 0.238 x = 0.345 0.417 0.508 y = 0.312 0.18 V = 0.146 0.293 x = 0.342 0.366 0.572 y = 0.284 0.144 P = 110 kPa (d) T = 110 deg_C Ans. 0.188 x = 0.334 0.478 P = 100 kPa (c) 0.371 y = 0.339 0.29 0.142 x = 0.306 0.552 Eq. (10.16) yi P (b) T = 110 deg_C Eq. (10.17) P = 120 kPa 10.10 As the pressure increases, the fraction of vapor phase formed (V) decreases, the mole fraction of benzene in both phases increases and the the mole fraction of ethylbenzene in both phases decreases. 319 14.3145 2756.22 228.060 10.11 (a) (1) = acetone A := B := C := (2) = acetonitrile 14.8950 3413.10 250.523 n := rows ( A) i := 1 .. n z1 := 0.75 T := ( 340 273.15) degC P := 115 kPa z2 := 1 z1 Bi Ai degC Psat ( i , T) := e Guess: T +Ci kPa ki := Psat ( i , T) P V := 0.5 n zi ki Given i=1 1 + V ( ki 1) V := Find ( V) =1 Ans. V = 0.656 Eq. (10.16) yi := xi := r := Eq. (10.17) zi ki yi P Psat ( i , T) y1 V z1 y1 = 0.805 Ans. x1 = 0.644 Ans. r = 0.705 1 + V ( ki 1) Ans. (b) x1 = 0.285 y1 = 0.678 V = 0.547 r = 0.741 (c) x1 = 0.183 y1 = 0.320 V = 0.487 r = 0.624 (d) x1 = 0.340 y1 = 0.682 V = 0.469 r = 0.639 320 10.13 H1 := 200 bar Psat2 := 0.10 bar P := 1 bar Assume at 1 bar that the vapor is an ideal gas. The vapor-phase fugacities are then equal to the partial presures. Assume the Lewis/Randall rule applies to concentrated species 2 and that Henry's law applies to dilute species 1. Then: y1 P = H1 x1 y2 P = x2 Psat2 P = y1 P + y2 P P = H1 x1 + ( 1 x1) Psat2 x1 + x2 = 1 Solve for x1 and y1: x1 := P Psat2 y1 := H1 Psat2 3 x1 = 4.502 × 10 H1 x1 P Ans. y1 = 0.9 10.16 Pressures in kPa Psat1 := 32.27 Psat2 := 73.14 ( γ 1 ( x1 , x2) := exp A x2 2 ) A := 0.67 z1 := 0.65 ( γ 2 ( x1 , x2) := exp A x1 2 ) P ( x1 , x2) := x1 γ 1 ( x1 , x2) Psat1 + x2 γ 2 ( x1 , x2) Psat2 (a) BUBL P calculation: x1 := z1 x2 := 1 x1 Pbubl := P ( x1 , x2) Pbubl = 56.745 DEW P calculation: y1 := z1 Guess: P' := Given x1 := 0.5 Ans. y2 := 1 y1 Psat1 + Psat2 2 y1 P' = x1 γ 1 ( x1 , 1 x1) Psat1 P' = x1 γ 1 ( x1 , 1 x1) Psat1 ... + ( 1 x1) γ 2 ( x1 , 1 x1) Psat2 x1 := Find ( x1 , P') Pdew Pdew = 43.864 321 Ans. The pressure range for two phases is from the dewpoint to the bubblepoint: From 43.864 to 56.745 kPa (b) BUBL P calculation: y1 ( x1) := x1 := 0.75 x2 := 1 x1 x1 γ 1 ( x1 , 1 x1) Psat1 P ( x1 , 1 x1) The fraction vapor, by material balance is: V := z1 x1 V = 0.379 y1 ( x1) x1 P ( x1 , x2) = 51.892 Ans. (c) See Example 10.3(e). γ 1 ( 0 , 1) Psat1 α 12.0 := α 12.1 := Psat2 α 12.0 = 0.862 Psat1 γ 2 ( 1 , 0) Psat2 α 12.1 = 0.226 Since alpha does not pass through 1.0 for 0<x1<1, there is no azeotrope. 10.17 Psat1 := 79.8 Psat2 := 40.5 ( γ 1 ( x1 , x2) := exp A x2 2 ) A := 0.95 ( γ 2 ( x1 , x2) := exp A x1 P ( x1 , x2) := x1 γ 1 ( x1 , x2) Psat1 + x2 γ 2 ( x1 , x2) Psat2 y1 ( x1) := x1 γ 1 ( x1 , 1 x1) Psat1 P ( x1 , 1 x1) (a) BUBL P calculation: Pbubl := P ( x1 , x2) x1 := 0.05 x2 := 1 x1 Pbubl = 47.971 Ans. y1 ( x1) = 0.196 (b) DEW P calculation: Guess: y1 := 0.05 x1 := 0.1 P' := 322 y2 := 1 y1 Psat1 + Psat2 2 2 ) y1 P' = x1 γ 1 ( x1 , 1 x1) Psat1 Given P' = x1 γ 1 ( x1 , 1 x1) Psat1 ... + ( 1 x1) γ 2 ( x1 , 1 x1) Psat2 x1 := Find ( x1 , P') Pdew Pdew = 42.191 Ans. x1 = 0.0104 (c) Azeotrope Calculation: Guess: x1 := 0.8 y1 := x1 x1 γ 1 ( x1 , 1 x1) Psat1 Given y1 = P P := Psat1 + Psat2 x1 0 2 x1 1 x1 = y1 P = x1 γ 1 ( x1 , 1 x1) Psat1 + ( 1 x1) γ 2 ( x1 , 1 x1) Psat2 xaz1 yaz := Find ( x , y , P) 11 1 Paz 10.18 Psat1 := 75.20 kPa y1 = x1 γ2 γ1 2 = Psat1 and γi = lnγ2 = A x1 Psat1 γ2 ln γ1 A := Psat2 2 A = 2.0998 2 x2 x1 For x1 := 0.6 x2 := 1 x1 323 x2 := 1 x1 ( 2 2 = A x1 x2 ln Whence P Psati x1 := 0.294 Psat2 2 lnγ1 = A x2 Ans. Psat2 := 31.66 kPa At the azeotrope: Therefore xaz1 0.857 yaz = 0.857 1 Paz 81.366 ) ( γ 1 := exp A x2 ) ( 2 x1 γ 1 Psat1 y1 := ) 2 γ 2 := exp A x1 P = 90.104 kPa P 10.19 Pressures in bars: P := x1 γ 1 Psat1 + x2 γ 2 Psat2 Ans. y1 = 0.701 Psat1 := 1.24 x1 := 0.65 A := 1.8 ( γ 1 := exp A x2 Psat2 := 0.89 x2 := 1 x1 ) ( 2 γ 2 := exp A x1 P := x1 γ 1 Psat1 + x2 γ 2 Psat2 y1 = 0.6013 y1 := ) 2 x1 γ 1 Psat1 P Answer to Part (b) P = 1.671 By a material balance, V= z1 x1 For y1 x1 (c) 0V1 Ans. (a) Azeotrope calculation: Guess: x1 := 0.6 y1 := x1 2 γ 1 ( x1) := exp A ( 1 x1) Given y1 = 0.6013 z1 0.65 P := Psat1 + Psat2 2 ( γ 2 ( x1) := exp A x1 ) 2 P = x1 γ 1 ( x1) Psat1 + ( 1 x1) γ 2 ( x1) Psat2 x1 γ 1 ( x1) Psat1 P x1 0 x1 y1 := Find ( x1 , y1 , P) P x1 1 x1 = y1 x1 0.592 y1 = 0.592 P 1.673 324 Ans. 10.20 Antoine coefficients: P in kPa; T in degC Acetone(1): A1 := 14.3145 B1 := 2756.22 C1 := 228.060 Methanol(2): A2 := 16.5785 B2 := 3638.27 C2 := 239.500 P1sat ( T) := exp A1 A := 0.64 x1 := 0.175 ( B1 T + C1 γ 1 ( x1 , x2) := exp A x2 2 z1 := 0.25 ) p := 100 (kPa) ( γ 2 ( x1 , x2) := exp A x1 P ( x1 , T) := x1 γ 1 ( x1 , 1 x1) P1sat ( T) ... + ( 1 x1) γ 2 ( x1 , 1 x1) P2sat ( T) y1 ( x1 , T) := Guesses: x1 γ 1 ( x1 , 1 x1) P1sat ( T) P ( x1 , T ) V := 0.5 L := 0.5 F := 1 T := 100 Given F= L+V z1 F = x1 L + y1 ( x1 , T) V L V := Find ( L , V , T) T T = 59.531 (degC) B2 T + C2 P2sat ( T) := exp A2 p = P ( x1 , T) L 0.431 V = 0.569 T 59.531 y1 ( x1 , T) = 0.307 Ans. 325 2 ) 10.22 x1 := 0.002 y1 := 0.95 B1 := 2572.0 A1 := 10.08 A1 Psat1 ( T) := e x2 := 1 x1 Given P := Guess: A2 := 11.63 B1 Psat2 ( T) := e 0.93 x2 y2 := 1 y1 Psat2 ( T) = x1 γ 1 Psat1 ( T) B2 := 6254.0 A2 T K bar Psat1 ( T) y1 T := 300 K 2 γ 1 := e x2 γ 2 y1 x1 γ 1 y2 T := Find ( T) P = 0.137 bar 326 Ans. B2 T K bar 0.93 x1 2 γ 2 := e T = 376.453 K Ans. Problems 10.25 to 10.34 have been solved using MS-EXCEL 2000 We give the resulting spreadsheets. Problem 10.25 a) BUBL P T=-60 F (-51.11 C) Component methane ethylene ethane xi 0.100 0.500 0.400 b) DEW P T=-60 F (-51.11 C) Component methane ethylene ethane yi 0.500 0.250 0.250 c) BUBL T P=250 psia (17.24 bar) Component methane ethylene ethane T=-50 F T=-60 F T=-57 F (-49.44 C) ANSWER xi Ki yi=Ki*xi Ki yi=Ki*xi Ki yi=Ki*xi 0.120 4.900 0.588 4.600 0.552 4.700 0.564 0.400 0.680 0.272 0.570 0.228 0.615 0.246 0.480 0.450 0.216 0.380 0.182 0.405 0.194 SUM = 1.076 SUM = 0.962 SUM = 1.004 close enough d) DEW T P=250 psia (17.24 bar) Component methane ethylene ethane yi 0.430 0.360 0.210 P=200 psia P=250 psia P=215 psia (14.824 bar) ANSWER Ki yi=Ki*xi Ki yi=Ki*xi Ki yi=Ki*xi 5.600 0.560 4.600 0.460 5.150 0.515 0.700 0.350 0.575 0.288 0.650 0.325 0.445 0.178 0.380 0.152 0.420 0.168 SUM = 1.088 SUM = 0.900 SUM = 1.008 close enough P=190 psia P=200 psia (13.79 bar) ANSWER Ki xi=yi/Ki Ki xi=yi/Ki 5.900 0.085 5.600 0.089 0.730 0.342 0.700 0.357 0.460 0.543 0.445 0.562 SUM = 0.971 SUM = 1.008 close enough T=-40 F T = -50 F T = -45 F (-27.33 C) ANSWER Ki xi=yi/Ki Ki xi=yi/Ki Ki xi=yi/Ki 5.200 0.083 4.900 0.088 5.050 0.085 0.800 0.450 0.680 0.529 0.740 0.486 0.520 0.404 0.450 0.467 0.485 0.433 SUM = 0.937 SUM = 1.084 SUM = 1.005 close enough 327 Problem 10.26 a) BUBL P Component ethane propane isobutane isopentane b) DEW P Component ethane propane isobutane isopentane c) BUBL T Component ethane propane isobutane isopentane d) DEW T Component ethane propane isobutane isopentane T=60 C (140 F) xi 0.10 0.20 0.30 0.40 P=200 psia P=50 psia P=80 psia (5.516 bar) ANSWER Ki yi=Ki*xi Ki yi=Ki*xi Ki yi=Ki*xi 2.015 0.202 6.800 0.680 4.950 0.495 0.620 0.124 2.050 0.410 1.475 0.295 0.255 0.077 0.780 0.234 0.560 0.168 0.071 0.028 0.205 0.082 0.12 0.048 SUM = 0.430 SUM = 1.406 SUM = 1.006 close enough T=60 C (140 F) yi 0.48 0.25 0.15 0.12 P=80 psia P=50 psia P=52 psia (3.585 bar) ANSWER Ki xi=yi/Ki Ki xi=yi/Ki Ki xi=yi/Ki 4.950 0.097 6.800 0.071 6.600 0.073 1.475 0.169 2.050 0.122 2.000 0.125 0.560 0.268 0.780 0.192 0.760 0.197 0.12 1.000 0.205 0.585 0.195 0.615 SUM = 1.534 SUM = 0.970 SUM = 1.010 close enough P=15 bar (217.56 psia) xi 0.14 0.13 0.25 0.48 T=220 F T=150 F T=145 F (62.78 C) ANSWER Ki yi=Ki*xi Ki yi=Ki*xi Ki yi=Ki*xi 5.350 0.749 3.800 0.532 3.700 0.518 2.500 0.325 1.525 0.198 1.475 0.192 1.475 0.369 0.760 0.190 0.720 0.180 0.57 0.274 0.27 0.130 0.25 0.120 SUM = 1.716 SUM = 1.050 SUM = 1.010 close enough P=15 bar (217.56 psia) yi 0.42 0.30 0.15 0.13 T=150 F Ki xi=yi/Ki 3.800 0.111 1.525 0.197 0.760 0.197 0.27 0.481 SUM = 0.986 T=145 F T=148 F (64.44 C) ANSWER Ki xi=yi/Ki Ki xi=yi/Ki 3.700 0.114 3.800 0.111 1.475 0.203 1.500 0.200 0.720 0.208 0.740 0.203 0.25 0.520 0.26 0.500 SUM = 1.045 SUM = 1.013 close enough 328 Problem 10.27 FLASH Component methane ethane propane n-butane T=80 F (14.81 C) zi 0.50 0.10 0.20 0.20 V= 0.855 Ki yi 10.000 0.575 2.075 0.108 0.680 0.187 0.21 0.129 SUM = 1.000 P=250 psia (17.24 bar) Fraction condensed L= 0.145 ANSWER xi=yi/Ki 0.058 0.052 0.275 0.616 SUM = 1.001 Problem 10.28 First calculate equilibrium composition T=95 C (203 F) Component n-butane n-hexane xi 0.25 0.75 P=80 psia P=65 psia P=69 psia (4.83 bar) ANSWER Ki yi=Ki*xi Ki yi=Ki*xi Ki yi=Ki*xi 2.25 0.5625 2.7 0.675 2.6 0.633 0.45 0.3375 0.51 0.3825 0.49 0.3675 SUM = 0.9000 SUM = 1.0575 SUM = 1.0005 Close enough Now calculate liquid fraction from mole balances ANSWER z1= x1= y1= L= 0.5 0.25 0.633 0.347 Problem 10.29 FLASH Component n-pentane n-hexane n-heptane P = 2.00 atm (29.39 psia) T = 200 F (93.3 C) zi 0.25 0.45 0.30 V= 0.266 Ki yi 2.150 0.412 0.960 0.437 0.430 0.152 SUM = 1.000 Fraction condensed L= 0.73 ANSWER xi=yi/Ki 0.191 0.455 0.354 SUM = 1.000 329 Problem 10.30 FLASH Component ethane propane n-butane T=40 C (104 F) V= 0.60 P=110 psia zi Ki yi 0.15 5.400 0.223 0.35 1.900 0.432 0.50 0.610 0.398 SUM = 1.053 Fraction condensed L= 0.40 P=100 psia xi=yi/Ki Ki yi 0.041 4.900 0.220 0.227 1.700 0.419 0.653 0.540 0.373 0.921 SUM = 1.012 ANSWER P=120 psia (8.274 bar) xi=yi/Ki Ki yi xi=yi/Ki 0.045 4.660 0.219 0.047 0.246 1.620 0.413 0.255 0.691 0.525 0.367 0.699 0.982 SUM = 0.999 1.001 Fraction condensed L= 0.80 P=40 psia xi=yi/Ki Ki yi 0.004 9.300 0.035 0.039 3.000 0.107 0.508 1.150 0.558 0.472 0.810 0.370 1.023 SUM = 1.071 ANSWER P=44 psia (3.034 bar) xi=yi/Ki Ki yi xi=yi/Ki 0.004 8.500 0.034 0.004 0.036 2.700 0.101 0.037 0.485 1.060 0.524 0.494 0.457 0.740 0.343 0.464 0.982 SUM = 1.002 1.000 Problem 10.31 FLASH Component ethane propane i-butane n-butane T=70 F (21.11 C) zi 0.01 0.05 0.50 0.44 V= 0.20 P=50 psia Ki yi 7.400 0.032 2.400 0.094 0.925 0.470 0.660 0.312 SUM = 0.907 330 Problem 10.32 FLASH Component methane ethane propane n-butane Component methane ethane propane n-butane Component methane ethane propane n-butane T=-15 C (5 F) zi 0.30 0.10 0.30 0.30 zi 0.30 0.10 0.30 0.30 zi 0.30 0.10 0.30 0.30 Target: y1=0.8 P=300 psia V= 0.1855 Ki yi 5.600 0.906 0.820 0.085 0.200 0.070 0.047 0.017 SUM = 1.079 L= 0.8145 xi=yi/Ki 0.162 0.103 0.352 0.364 SUM = 0.982 P=150 psia V= 0.3150 Ki yi 10.900 0.794 1.420 0.125 0.360 0.135 0.074 0.031 SUM = 1.086 L= 0.6850 xi=yi/Ki 0.073 0.088 0.376 0.424 SUM = 0.960 P=270 psia (18.616 bar) V= 0.2535 L= 0.7465 ANSWER Ki yi xi=yi/Ki 6.200 0.802 0.129 0.900 0.092 0.103 0.230 0.086 0.373 0.0495 0.020 0.395 SUM = 1.000 SUM = 1.000 331 Problem 10.33 First calculate vapor composition and temperature on top tray BUBL T: P=20 psia Component n-butane n-pentane xi 0.50 0.50 T=70 F T=60 F T=69 F (20.56 C) ANSWER Ki yi=Ki*xi Ki yi=Ki*xi Ki yi=Ki*xi 1.575 0.788 1.350 0.675 1.550 0.775 0.450 0.225 0.360 0.180 0.440 0.220 SUM = 1.013 SUM = 0.855 SUM = 0.995 close enough Using calculated vapor composition from top tray, calculate composition out of condenser FLASH P=20 psia (1.379 bar) V= 0.50 Component n-butane n-pentane L= 0.50 T=70 F zi Ki yi 0.78 1.575 0.948 0.22 0.450 0.137 SUM = 1.085 xi=yi/Ki 0.602 0.303 0.905 T=60 F (15.56 C) Ki yi 1.350 0.890 0.360 0.116 SUM = 1.007 ANSWER xi=yi/Ki 0.660 0.324 0.983 Problem 10.34 FLASH T=40 C (104 F) V= 0.60 Component methane n-butane L= 0.40 P=350 psia zi Ki yi 0.50 7.900 0.768 0.50 0.235 0.217 SUM = 0.986 P=250 psia xi=yi/Ki Ki yi 0.097 11.000 0.786 0.924 0.290 0.253 1.021 SUM = 1.038 332 ANSWER P=325 psia (7.929 bar) xi=yi/Ki Ki yi xi=yi/Ki 0.071 8.400 0.772 0.092 0.871 0.245 0.224 0.914 0.943 SUM = 0.996 1.006 close enough Eq. (1) 10.35 a) The equation from NIST is: Mi = ki yi P The equation for Henry's Law is:i Hi = yi P x Solving to eliminate P gives: By definition: Mi Hi ni = ns Ms Eq. (2) Mi Eq. (3) ki xi where M is the molar mass and the subscript s refers to the solvent. = Dividing by the toal number of moles gives: Mi = Combining Eqs. (3) and (4) gives: Hi = xi 1 xs Ms ki If xi is small, then x s is approximately equal to 1 and: Hi = b) For water as solvent: Ms For CO2 in H2O: By Eq. (5): Hi ki Eq. (4) xs Ms gm mol mol 1 Ms ki Eq. (5) 18.015 0.034 kg bar 1 Ms ki Hi 1633 bar Ans. The value is Table 10.1 is 1670 bar. The values agree within about 2%. 10.36 14.3145 Acetone: Psat1 ( ) T Psat2 ( ) T T degC e 14.8950 Acetonitrile 2756.22 T a) Find BUBL P and DEW P values T 50degC x1 0.5 y1 333 kPa 3413.10 degC e 228.060 0.5 250.523 kPa BUBLP x1 Psat1 ( T) 1 x1 Psat2 ( T) BUBLP DEWP 1 DEWP 1 y1 Psat1 ( T) y1 0.573 atm Ans. 0.478 atm Ans. Psat2 ( T) At T = 50 C two phases will form between P = 0.478 atm and 0.573 atm b) Find BUBL T and DEW T values P 0.5atm Given y1 0.5 x1 Psat1 ( T) BUBLT Given x1 1 BUBLT 1 x1 Psat1 ( T) = y1 P DEWT T 50degC x1 Psat2 ( T) = P Find ( T) x1 Guess: 0.5 Find x1 T DEWT Ans. 46.316 degC x1 Psat2 ( T) = 1 51.238 degC y1 P Ans. At P = 0.5 atm, two phases will form between T = 46.3 C and 51.2 C 10.37 Calculate x and y at T = 90 C and P = 75 kPa 13.7819 Benzene: Psat1 ( T) Psat2 ( T) T degC e 13.9320 Toluene: 2726.81 kPa 3056.96 T degC e 217.572 217.625 kPa a) Calculate the equilibrium composition of the liquid and vapor at the flash T and P T 90degC P 75kPa Guess: 334 x1 0.5 y1 0.5 1 x1 Psat1 ( )= y1 P T Given x1 Psat2 ( )= 1 T y1 P x1 x1 Find x1 y1 y1 y1 0.252 0.458 The equilibrium compositions do not agree with the measured values. b) Assume that the measured values are correct. Since air will not dissolve in the liquid to any significant extent, the mole fractions of toluene in the liquid can be calculated. x1 y1 0.1604 x2 0.2919 1 x1 x2 0.8396 Now calculate the composition of the vapor. y3 represents the mole fraction of air in the vapor. Guess: y2 0.5 y3 1 y2 y1 Given 1 y3 y1 y3 P Find y2 y3 y2 0.1 y1 y2 y3 = 1 0.608 y3 y2 x1 Psat2 ( )= 1 T Ans. Conclusion: An air leak is consistent with the measured compositions. 10.38 yO21 ndot yN21 0.0387 10 kmol T1 hr 16.3872 PsatH2O ( ) T e yCO21 0.7288 T2 100degC 3885.70 T degC 230.170 kPa 335 0.0775 yH2O1 25degC P 1atm 0.1550 Calculate the mole fraction of water in the exit gas if the exit gas is saturated with water. PsatH2O T2 yH2O2 yH2O2 P 0.0315 This is less than the mole fraction of water in the feed. Therefore, some of the water will condense. Assume that two streams leave the process: a liquid water stream at rate ndotliq and a vapor stream at rate ndotvap. Apply mole balances around the cooler to calculate the exit composition of the vapor phase. ndotvap Given ndot 2 yO22 Guess: 0.0387 ndot = ndotliq ndot 2 ndotliq yN22 0.7288 yCO22 ndotvap 0.0775 Overall balance ndot yO21 = ndotvap yO22 O2 balance ndot yN21 = ndotvap yN22 N2 balance ndot yCO21 = ndotvap yCO22 CO2 balance yO22 yN22 yCO22 yH2O2 = 1 Summation equation ndotliq ndotvap yO22 Find ndotliq ndotvap yO22 yN22 yCO22 yN22 yCO22 ndotliq yO22 1.276 0.044 kmol hr yN22 ndotvap 0.835 8.724 yCO22 336 kmol hr 0.089 yH2O2 0.031 Apply an energy balance around the cooler to calculate heat transfer rate. HlvH2O Qdot 40.66 kJ T1 mol T1 273.15K T2 ndotvap yO22 R ICPH T1 T2 3.639 0.506 10 3 ndotvap yN22 R ICPH T1 T2 3.280 0.539 10 3 ndotvap yH2O2 R ICPH T1 T2 3.470 1.450 10 HlvH2O ndotliq 0.227 10 5 0 0.040 10 3 3 0 0 0.121 10 10.39 Assume the liquid is stored at the bubble point at T = 40 F Taking values from Fig 10.14 at pressure: xC3 0.05 KC3 3.9 xC4 0.85 KC4 0.925 xC5 0.10 KC5 0.23 The vapor mole fractions must sum to 1. xC3 KC3 xC4 KC4 xC5 KC5 1.004 337 P 18psia 5 1.157 10 Ans. 19.895 kW 273.15K 5 0 ndotvap yCO22 R ICPH T1 T2 5.457 1.045 10 Qdot T2 Ans. 5 10.40 H2S + 3/2 O2 -> H2O + SO2 By a stoichiometric balance, calculate the following total molar flow rates kmol Feed: ndotH2S 10 Products ndotSO2 ndotH2S hr 3 ndotH2S 2 ndotO2 ndotH2O ndotH2S Exit conditions: P T2 1atm 3885.70 16.3872 70degC PsatH2O ( ) T T degC e 230.170 kPa a) Calculate the mole fraction of H2O and SO2 in the exiting vapor stream assuming vapor is saturated with H2O PsatH2O T2 yH2Ovap ySO2 P 1 yH2Ovap yH2Ovap ySO2 Ans. 0.308 Ans. 0.692 b) Calculate the vapor stream molar flow rate using balance on SO 2 ndotvap ndotSO2 ndotvap ySO2 14.461 kmol hr Ans. Calculate the liquid H2O flow rate using balance on H2O ndotH2Ovap ndotH2Oliq ndotH2Ovap ndotvap yH2Ovap ndotH2O ndotH2Ovap 338 ndotH2Oliq 4.461 5.539 kmol hr kmol Ans. hr 10.41 NCL a) 0.01 YH2O yH2O b) P kg Mair MH2O YH2O 1 YH2O PsatH2O ( ) T e 10.42 ndot1 Tdp kmol hr Tdp1 T degC e PsatH2O Tdp1 P 1.606 kPa Ans. ppH2O yH2O P 230.170 Guess: kPa Tdp 32degF 20degC Tdp2 Tdp T 20degC Find ( ) T 57.207 degF Ans. P 10degC y1 MH2O 230.170 18.01 kPa y2 0.023 By a mole balances on the process Guess: ndot2liq mol 1atm 3885.70 16.3872 y1 T 14.004 degC PsatH2O ( ) T gm Ans. yH2O P = PsatH2O ( ) Tdp T 50 29 3885.70 degC Given Tdp 0.0158 ppH2O 1atm Mair 0.0161 yH2O 16.3872 c) 18.01 YH2O kg NCL gm mol MH2O ndot1 ndot2vap ndot1 339 PsatH2O Tdp2 P y2 gm mol 0.012 ndot1 y1 = ndot2vap y2 Given ndot1 = ndot2vap ndot2liq kmol hr ndot2liq ndot2vap 49.441 mdot2liq ndot2liq MH2O 10.43Benzene: mdot2liq Given kmol 10.074 hr kg hr Ans. 13.7819 B1 2726.81 C1 217.572 A2 13.6568 B2 2723.44 C2 220.618 exp A1 B1 T degC Guess: T 0.559 A1 Cyclohexane: Psat2 ( T) Overall balance ndot2liq Find ndot2liq ndot2vap ndot2vap Psat1 ( T) H2O balance ndot2liq exp A2 kPa C1 B2 T degC kPa C2 66degC Psat1 ( T) = Psat2 ( T) T Find ( T) The Bancroft point for this system is: Psat1 ( T) 39.591 kPa Com ponent1 Benzene 2- t Bu anol Acet ti oni le r Com ponent2 Cyclohexane W at er Et hanol T 52.321 degC T (C) 52.3 8 .7 7 65.8 340 P ( Pa) k 39.6 64 .2 60 .6 Ans. Chapter 11 - Section A - Mathcad Solutions 11.1 For an ideal gas mole fraction = volume fraction 3 CO2 (1): x1 0.7 V1 0.7m N2 (2): x2 0.3 V2 0.3m i P 1bar T 12 P 3 ( 25 273.15) K Vi i n n RT S nR 40.342 mol S xi ln xi 204.885 J K Ans. i 11.2 For a closed, adiabatic, fixed-volume system, U =0. Also, for an ideal gas, U = Cv T. First calculate the equilibrium T and P. nN2 4 mol TN2 [75 ( 273.15)K] PN2 30 bar nAr 2.5 mol TAr ( 130 273.15)K PAr 20 bar TN2 348.15 K TAr 403.15 K ntotal nN2 x1 nAr x1 CvAr CpAr CvAr CvN2 3 R 2 R CvN2 nN2 ntotal 0.615 x2 12 nAr ntotal x2 0.385 T T 5 R 2 CpN2 i R Find T after mixing by energy balance: T TN2 Given TAr 2 nN2 CvN2 T (guess) TN2 = nAr CvAr TAr 341 Find T) ( T 273.15 K 90 degC Find P after mixing: PN2 P PAr (guess) 2 Given nN2 nAr R T P = nN2 R TN2 nAr R TAr PN2 PAr P Find ( P) P 24.38 bar Calculate entropy change by two-step path: 1) Bring individual stream to mixture T and P. 2) Then mix streams at mixture T and P. SN2 SAr nAr CpAr ln T TN2 nN2 CpN2 ln Smix ntotal T TAr R R ln SN2 P PAr xi ln xi J K SAr P PN2 11.806 9.547 J K Smix R ln 36.006 J K i S 11.3 SN2 mdotN2 molwtN2 SAr 2 Smix kg sec molarflowtotal 38.27 mdotH2 28.014 molarflowN2 S gm mol mdotN2 molwtN2 molarflowN2 molwtH2 J K 0.5 Ans. kg sec 2.016 molarflowH2 gm mol i mdotH2 molwtH2 molarflowH2 molarflowtotal 342 12 319.409 mol sec y1 molarflowN2 y1 molarflowtotal S R molarflowtotal molarflowH2 y2 0.224 y2 molarflowtotal S yi ln yi J secK 1411 0.776 Ans. i 11.4 T1 T2 448.15 K P1 308.15 K P2 3 bar 1 bar For methane: MCPHm MCPSm 3 MCPH T1 T2 1.702 9.081 10 MCPS T1 T2 1.702 9.081 10 6 2.164 10 3 6 2.164 10 0.0 0.0 For ethane: MCPHe MCPH T1 T2 1.131 19.225 10 MCPSe MCPS T1 T2 1.131 19.225 10 3 6 5.561 10 3 6 5.561 10 0.0 0.0 MCPHmix 0.5 MCPHm 0.5 MCPHe MCPHmix 6.21 MCPSmix 0.5 MCPSm 0.5 MCPSe MCPSmix 6.161 H R MCPHmix T2 S R MCPSmix ln H T1 T2 T1 R ln P2 7228 J mol R 2 0.5 ln ( ) 0.5 P1 The last term is the entropy change of UNmixing J T 300 K S 15.813 mol K Wideal H T Wideal S 2484 J mol Ans. 11.5 Basis: 1 mole entering air. y1 0.21 y2 0.79 t 0.05 T 300 K Assume ideal gases; then H=0 The entropy change of mixing for ideal gases is given by the equation following Eq. (11.26). For UNmixing of a binary mixture it becomes: 343 S R y1 ln y1 S y2 ln y2 By Eq. (5.27): Wideal By Eq. (5.28): Work T 4.273 J mol K Wideal Wideal 25638 t 0 40 mol Ans. 0.942 60 P 0.970 J mol 0.985 20 10 1.000 10 11.16 3J 1.282 Work S 0.913 80 Z bar ln 1 0 end 1 rows ( P) 1 0.885 100 0.869 200 0.765 300 0.762 400 0.824 500 0.910 i 2 end Zi Fi 1 Pi Fi is a well behaved function; use the trapezoidal rule to integrate Eq. (11.35) numerically. Fi Ai i Fi 1 Pi 2 Pi 1 ln i fi exp ln i ln i 1 Ai i Pi Generalized correlation for fugacity coefficient: For CO2: Tc Pc 304.2 K T ( 150 273.15) K Tr T Tc 0.224 73.83 bar Tr P G ( P) exp Pc Tr B0 Tr B1 Tr 344 fG ( P) G ( P) P 1.391 Pi fi bar bar i 10 0.993 9.925 20 0.978 19.555 40 0.949 37.973 60 0.922 55.332 80 0.896 71.676 100 0.872 87.167 200 0.77 153.964 300 0.698 209.299 400 0.656 262.377 500 0.636 317.96 Calculate values: 400 300 fi 0.8 i bar G Pi f G Pi 200 0.6 0.4 bar 0 200 400 100 0 600 0 200 400 Pi Pi bar 600 bar Agreement looks good up to about 200 bar (Pr=2.7 @ Tr=1.39) 11.17 For SO2: Tc 430.8 K Pc T Tr T Tc 600 K P Tr 1.393 Pr 78.84 bar 0.245 300 bar P Pc Pr For the given conditions, we see from Fig. 3.14 that the Lee/Kesler correlation is appropriate. 345 3.805 Data from Tables E.15 & E.16 and by Eq. (11.67): 0 0.672 1.354 1 f GRRT P f 417.9 K a) At 280 degC and 20 bar: Tr ( ) T T Tc T ln GRRT 217.14 bar Tc 11.18 Isobutylene: 0.724 01 Pc ( 280 0.194 40.00 bar P 273.15)K Pr ( ) P Tr ( ) 1.3236 T Ans. 0.323 P Pc 20 bar Pr ( ) 0.5 P At these conditions use the generalized virial-coeffieicnt correlation. f PHIB Tr ( )Pr ( ) T P f P b) At 280 degC and 100 bar: T ( 280 Ans. 18.76 bar P 273.15)K 100 bar Pr ( ) 2.5 P Tr ( ) 1.3236 T At these conditions use the Lee/Kesler correlation, Tables E.15 & E.16 and Eq. (11.67). 1 0.7025 01 f 0.732 0 f 1.2335 P Ans. 73.169 bar 11.19 The following vectors contain data for Parts (a) and (b): (a) = Cyclopentane; (b) = 1-butene Tc Zc T 511.8 420.0 K 0.273 Vc 0.277 383.15 393.15 Pc K P 45.02 40.43 239.3 34 346 0.191 3 258 275 0.196 bar bar cm mol Tn Psat 322.4 266.9 5.267 25.83 K bar T Tr 0.7486 Tr Tc Psat Pc Psatr 0.9361 Psatr 0.117 0.6389 Calculate the fugacity coefficient at the vapor pressure by Eq. (11.68): (a) PHIB Tr Psatr (b) PHIB Tr Psatr 1 2 1 2 0.900 2 1 0.76 Eq. (3.72), the Rackett equation: T Tc Tr 0.749 Tr 0.936 Eq. (11.44): 2 Vsat f f 11.21 V c Zc 1 Tr PHIB Tr Psatr 11.78 20.29 107.546 cm3 133.299 mol 7 Vsat Psat exp Vsat ( Psat) P RT Ans. bar Table F.1, 150 degC: Psat molwt 476.00 kPa 18 3 cm molwt gm Vsat 1.091 Vsat cm 19.638 mol T ( 150 T P 273.15)K 150 bar = 1.084 Ans. 423.15 K 3 Equation Eq. (11.44) with r exp Vsat P RT Psat satPsat = fsat r 1.084 347 r= f fsat gm mol 11.22 The following vectors contain data for Parts (a) and (b): molwt 18 gm mol Table F.2: (a) 9000 kPa & 400 degC; (b) 1000(psia) & 800 degF: T1 ( 400 ( 800 459.67) rankine 3121.2 H1 273.15) K J gm 6.2915 S1 Btu 1389.6 lbm 1.5677 J gm K Btu lbm rankine Table F.2: (a) 300 kPa & 400 degC; (b) 50(psia) & 800 degF: T2 3275.2 H2 J gm 8.0338 S2 Btu 1431.7 lbm 1.9227 J gm K Btu lbm rankine Eq. (A) on page 399 may be recast for this problem as: r (a) exp r= molwt R f2 f1 H2 H1 S2 T1 = 0.0377 (b) r= 0.0377 r S1 f2 0.0542 = 0.0542 f1 Ans. 11.23 The following vectors contain data for Parts (a), (b), and (c): (a) = n-pentane (b) = Isobutylene (c) = 1-Butene: 469.7 Tc 417.9 K 33.70 Pc 40.0 420.0 Zc 313.0 0.194 bar 40.43 0.270 0.252 0.275 0.277 Vc 238.9 239.3 348 0.191 309.2 3 cm mol Tn 266.3 266.9 K T1 200 P 1.01325 300 bar Psat 1.01325 bar 150 Tr 1.01325 0.6583 Tn 0.6372 Tr Tc 0.0301 Psat Pc Pr 0.6355 0.0253 Pr 0.0251 Calculate the fugacity coefficient at the nbp by Eq. (11.68): (a) (b) (c) PHIB Tr Pr 1 1 PHIB Tr Pr 1 0.9572 2 0.9618 PHIB Tr Pr 3 3 3 0.9620 2 Eq. (3.72): Eq. (11.44): 2 Vsat f V c Zc 1 Tr PHIB Tr Pr 0.2857 Psat exp Vsat ( Psat) P R Tn 2.445 f Ans. 3.326 bar 1.801 11.24 (a) Chloroform: Tc Pc 334.3 K 0.882 Trn 0.222 54.72 bar Tn 536.4 K 3 Zc T 0.293 473.15 K Eq. (3.72): Vc cm 239.0 mol Tr T Tc Vsat Tr 2 V c Zc 1 Trn 349 Psat Tn Tc Trn 22.27 bar 0.623 3 7 Vsat cm 94.41 mol Calculate fugacity coefficients by Eqs. (11.68): Pr ( P ) P Pc f ( P) if P Psat ( P) P ( P) if P Psat ( P) P ( P) exp Pr ( P ) B0 Tr Tr ( Psat) Psat exp Psat ( Psat) P B1 Tr Vsat ( P Psat) RT Vsat ( P exp Psat) RT 0 bar 0.5 bar 40 bar 40 f ( P) Psat bar 30 Psat bar 0.8 bar P 20 ( P) 0.6 bar 10 0 0 20 0.4 40 0 20 P P bar bar (b) Isobutane Tc 40 P bar 408.1 K Pc 36.48 bar Tn 261.4 K 0.767 Trn 0.181 3 Zc T 0.282 313.15 K Eq. (3.72): Vc Tr 262.7 T Tc Vsat cm mol Tr 2 V c Zc 1 Trn 350 Psat Tn Tc Trn 5.28 bar 0.641 3 7 Vsat cm 102.107 mol Calculate fugacity coefficients by Eq. (11.68): Pr ( ) P P Pc fP) ( if P Psat ( )P P () P if P Psat ( ) ( sat) P P P () P exp Pr ( ) P P exp B1 Tr Vsat ( Psat) P RT ( sat)Psat exp P Psat B0 Tr Tr Vsat ( P Psat) RT 0 bar 0.5 bar 10 bar 10 Psat bar fP) ( Psat bar 0.8 bar 5 P () P bar 0.6 0 0 5 0.4 10 0 P P bar bar 5 10 P bar 11.25 Ethylene = species 1; Propylene = species 2 282.3 Tc 365.6 Pc K 0.281 Zc Vc 0.289 T 423.15 K P n 2 i 30 bar 1n 50.40 46.65 131.0 188.4 y1 0.140 3 cm mol 0.35 j1n 351 0.087 w bar y2 1 k 1n y1 By Eqs. (11.70) through (11.74) wi wj ij Tc ij 2 1 3 Vc i Vc Vc j ij Tr Tci Tc j Zc ij 1.499 1.317 Tr Tc 157.966 cm3 157.966 188.4 mol 131 0.087 0.114 1.317 1.157 50.345 48.189 Pc 48.189 46.627 282.3 321.261 321.261 Tc 0.114 0.14 2 R Tc ij Vc ij ij Vc Zc j ij Pc T ij ij 13 3 2 Zci Zc 365.6 bar 0.281 0.285 K Zc 0.285 0.289 By Eqs. (3.65) and (3.66): B0i j B0 Bi j B0 Tr B1 i j ij 0.138 0.189 0.189 0.251 B1 R Tc ij B0i j Pc i j B1i j B ij B1 Tr ij 0.108 0.085 0.085 0.046 99.181 cm3 159.43 mol 59.892 99.181 By Eq. (11.64): ij 2 Bi j hatk exp Bi i P RT 20.96 cm3 mol 20.96 0 0 Bj j Bk k 1 2 yi y j 2 i k i fhatk hatk yk P hat 0.957 0.875 352 ij j fhat 10.053 17.059 bar Ans. For an ideal solution , id = pure species Pr P k Pr Pck fhatid idk 0.643 exp 0.95 id idk yk P k Pr 0.595 Tr k B0k k kk 9.978 fhatid 0.873 k k B1k k Ans. bar 17.022 Alternatively, Pr Pr P ij idk Pc exp ij kk Tr B0 k k T 0.873 35 bar 0.012 w 0.43 0.286 0.100 Zc 0.279 0.152 0.276 45.99 0.36 98.6 190.6 305.3 K Pc 48.72 bar 369.8 n P 373.15 K 0.21 Tc 0.95 id kk 11.27 Methane = species 1 Ethane = species 2 Propane = species 3 y k k B1k k Vc 145.5 42.48 i 3 j 1n 200.0 k 1n 1n By Eqs. (11.70) through (11.74) wi ij Vc ij wj Tc ij 2 Vci 1 3 Vc j Tci Tc j Zc Zci ij 13 3 Pc ij 2 Zc j 2 Zc R Tc ij Vc ij ij 353 3 cm mol 1.958 1.547 1.406 Tr T ij 1.547 1.222 1.111 Tr Tc ij 1.406 1.111 1.009 98.6 Vc 120.533 143.378 120.533 145.5 171.308 143.378 171.308 3 cm mol 200 45.964 47.005 43.259 47.005 48.672 45.253 bar 0.056 43.259 45.253 42.428 Pc 0.012 0.056 0.082 0.082 0.126 0.152 190.6 Tc 305.3 265.488 336.006 0.126 0.286 0.282 0.281 241.226 265.488 241.226 0.1 Zc 336.006 K 0.282 0.279 0.278 0.281 0.278 0.276 369.8 By Eqs. (3.65) and (3.66): B0i j Bi j B0 Tr B1 i j B1 Tr ij ij R Tc ij B0i j Pc i j B1i j ij By Eq. (11.64): ij 2 Bi j 0 Bi i Bj j 30.442 107.809 30.442 0 23.482 107.809 23.482 hatk exp P RT Bk k 1 2 yi y j 2 i k i fhatk hatk yk P 0.881 0.775 354 mol 0 ij j 1.019 hat 3 cm 7.491 fhat 13.254 bar 9.764 Ans. For an ideal solution , id = pure species 0.761 P Pck Prk Pr idk 0.718 exp 0.824 Prk Tr idk yk P k id 7.182 fhatid 0.88 GE RT 2.6 x1 = 13.251 bar 9.569 0.759 11.28 Given: k k B1k k kk 0.977 fhatid B0k k 1.8 x2 x1 x2 Substitute x2 = 1 - x1: (a) GE = RT .8 x1 1.8 x1 1 2 x1 = 1.8 x1 x1 3 0.8 x1 Apply Eqs. (11.15) & (11.16) for M = GE/RT: ln 1 = d GE RT dx1 GE RT ln 2 = 1 = 1.8 ln 1 = 1.8 2 x1 GE RT d x1 dx1 2 x1 2 x1 ln 2 = 2.4 x1 1.4 x1 2 2 1.6 x1 3 Ans. 3 1.6 x1 (b) Apply Eq. (11.100): GE = x1 RT 1.8 1 2 x1 x1 x1 1.4 x1 2 2 1.6 x1 1.6 x1 3 This reduces to the initial condition: 355 3 GE RT d x1 GE RT dx1 Ans. (c) Divide Gibbs/Duhem eqn. (11.100) by dx1: x1 d ln 1 dx1 x2 d ln 2 =0 dx1 Differentiate answers to Part (a): d ln 1 =2 dx1 2.8 x1 x1 d ln 1 = 2 x1 dx1 x2 d ln 1 =1 dx1 4.8 x1 4.8 x1 2 3 2 4.8 x1 2.8 x1 x1 d ln 2 = 2 x1 dx1 2 2 x1 4.8 x1 2 These two equations sum to zero in agreement with the Gibbs/Duhem equation. (d) When x1 = 1, we see from the 2nd eq. of Part (c) that d ln 1 When x1 = 0, we see from the 3rd eq. of Part (c) that d ln 2 dx1 dx1 =0 Q.E.D. =0 Q.E.D. (e) DEFINE: g = GE/RT g x1 1.8 x1 ln 1 x1 1.8 ln 2 x1 x1 ln 1 () 0 1.8 2 2 3 x1 0.8 x1 2 x1 1.4 x1 2 1.6 x1 3 1.6 x1 3 ln 2 () 1 356 2.6 x1 0 0.1 1.0 0 g x1 1 ln 1 x1 ln 2 x1 0 ln 1 ( ) 2 ln 2 ( ) 1 3 0 0.2 0.4 0.6 0.8 x1 H H1bar H2bar 0.02715 0.09329 417.4 0.32760 534.5 0.40244 531.7 0.56689 421.1 0.63128 x1 265.6 0.17490 11.32 87.5 347.1 0.66233 0.69984 253 VE 321.7 276.4 0.72792 252.9 0.77514 190.7 0.79243 178.1 0.82954 138.4 0.86835 98.4 0.93287 37.6 0.98233 10.0 357 n x1 rows x1 i1n 0 0.01 1 (a) Guess: a x1 1 2 x1 F x1 3000 x1 1 3000 c a x1 3 x1 1 b a b 3.448 3 10 b 3.202 10 c linfitx1 VE F c x1 250 3 Ans. 244.615 600 400 VEi x1 ( 1 x1) a b x1 c ( x1) 2 200 0 0 0.2 0.4 0.6 0.8 x1 x1 i By definition of the excess properties E V = x1 x2 a b x1 d 3 E V = 4 c x1 dx1 Vbar1 Vbar2 E E = x2 = x1 2 2 c x1 3 (c b) x1 a a 2 2 b x1 b 2 3 c x1 2 (b c) x1 2 (b a) x1 a 2 3 c x1 2 (b) To find the maximum, set dVE/dx1 = 0 and solve for x1. Then use x1 to find VEmax. Guess: x1 0.5 Given 4 c ( x1) x1 3 3 (c Find ( x1) b) ( x1) x1 2 2 (b 0.353 a) x1 Ans. 358 a= 0 VEmax x1 ( 1 x1) a (c) VEbar1 ( ) x1 ( 1 x1 c x1 () a x1 b VEmax 2 b x1 2 x1) a 2 VEbar2 ( ) x1 2 b x1 Ans. 3 c ( 1) x 2( b 536.294 2 c)x1 2 3 c ( 1) x 0 0.01 1 4000 2000 VEbar 1 ( ) x1 x1 VEbar 2 ( ) 0 2000 0 0.2 0.4 0.6 0.8 x1 x1 Discussion: a) Partial property for species i goes to zero WITH ZERO SLOPE as xi -> 1. b) Interior extrema come in pairs: VEbar min for species 1 occurs at the same x1 as VEbar max for species 2, and both occur at an inflection point on the VE vs. x1 plot. c) At the point where the VEbar lines cross, the VE plot shows a maximum. 11.33 Propane = 1; n-Pentane = 2 T B ( 75 273.15)K 276 466 466 809 By Eq. (11.61): P 2 bar y1 n 2 i 0.5 y2 1 1n j 1n y1 3 cm mol B yi y j Bi j i 359 j 3 B 504.25 cm mol Use a spline fit of B as a function of T to find derivatives: 331 b11 980 3 cm 276 b22 mol 809 b12 mol 466 684 235 50 t 558 3 cm 3 cm mol 399 323.15 75 t 273.15 K 348.15 K 100 373.15 3 lspline ( t b11) B11 ( T) vs11 interp ( vs11 t b11 T) B11 ( T) cm 276 mol B22 ( T) cm 809 mol B12 ( T) cm 466 mol 1.92 3.18 3 3 lspline ( t b22) B22 ( T) vs22 interp ( vs22 t b22 T) 3 lspline ( t b12) B12 ( T) vs12 dBdT interp ( vs12 t b12 T) d d B12 ( T) B11 ( T) dT dT d B12 ( T) dT dBdT d B22 ( T) dT cm 3.18 5.92 mol K 3 Differentiate Eq. (11.61): dBdT yi y j dBdTi djBdT i By Eq. (3.38): Z 1 By Eq. (6.55): HRRT By Eq. (6.56): BP RT P R 3 V 13968 cm mol HR 348.037 360 ZR T P V 0.965 dBdT HRRT P dBdT R SRR j Z B T cm 3.55 mol K 0.12 SRR 0.085 J mol SR HR SR 0.71 HRRT R T SRR R J mol K Ans. 11.34 Propane = 1; n-Pentane = 2 T ( 75 273.15)K P 2 bar y1 3 276 ij cm n 466 B 466 2 809 mol 0.5 y2 j1n i1n 2 Bi j Bii B j j By Eqs. (11.63a) and (11.63b): hat1 ( ) y1 exp P B1 1 RT hat2 ( ) y1 exp P B2 2 RT y1 ( 1 2 y1 2 y1) 1 2 12 0 0.1 1.0 1 0.99 0.98 hat1 ( ) y1 0.97 hat2 ( ) y1 0.96 0.95 0.94 0 0.2 0.4 0.6 y1 361 0.8 1 y1 0.0426 0.0817 45.7 0.1177 66.5 0.1510 86.6 0.2107 118.2 0.2624 144.6 0.3472 11.36 23.3 176.6 0.4158 x1 195.7 HE 0.5163 0.6156 85.6 0.9276 43.5 0.9624 22.6 (a) Guess: a x1 1 2 x1 3 x1 0 0.01 1 116.8 0.8650 1n 141.0 0.8181 i 174.1 0.7621 rows x1 x1 191.7 0.6810 F x1 n 204.2 x1 1 x1 1 x1 500 b 100 c 0.01 a b linfit x1 HE F c 539.653 b 1.011 10 c a 913.122 0 100 HEi x1 ( 1 x1) a b x1 c ( x1) 2 200 300 0 0.2 0.4 0.6 x1 x1 i 362 0.8 3 Ans. By definition of the excess properties E H = x1 x2 a d b x1 E dx1 H = 4 c x1 Hbar1 Hbar2 E E 2 = x2 3( c b) x1 a 2 b x1 a 2 = x1 3 2 c x1 b 2 2( b 3 c x1 2( b a)x1 a 2 c)x1 3 c x1 2 (b) To find the minimum, set dHE/dx1 = 0 and solve for x1. Then use x1 to find HEmin. Guess: x1 HE ( ) x1 3 Given x1 0.5 4 c ( 1) x Find ( ) x1 HEmin x1 x1 ( 1 x1) a 3( c x1 ( 1 2 b)( 1) x x1) a 2( b a)x1 x1 2 b x1 c x1 HEmin HEbar1 ( ) x1 HE ( ) ( x1 1 a= 0 HE ( ) x1 x1 204.401 d x1) HE ( ) x1 dx1 d HE ( ) x1 dx1 0 0.01 1 500 HEbar 1 ( ) x1 x1 HEbar 2 ( ) 0 500 1000 0 2 c x1 Ans. 0.512 HEbar2 ( ) x1 (c) b x1 0.2 0.4 0.6 x1 363 0.8 Ans. Discussion: a) Partial property for species i goes to zero WITH ZERO SLOPE as xi -> 1. b) Interior extrema come in pairs: HEbar min for species 1 occurs at the same x1 as HEbar max for species 2, and both occur at an inflection point on the H E vs. x1 plot. c) At the point where the HEbar lines cross, the HE plot shows a minimum. 11.37 (a) y1 (1) = Acetone 0.28 0.307 w n 0.190 2 (2) = 1,3-butadiene y2 T 1 508.2 Tc i y1 j ki j 170 kPa 209 Vc 0.267 1n P 273.15) K 0.233 K Zc 425.2 1n ( 60 220.4 3 cm mol 0 0.307 0.2485 0.082 Eq. (11.70) wi wj 0.2485 2 0.19 0.082 ij 0.126 0.126 0.152 508.2 Tc Tc i Zc Eq. (11.73) Zci j j Zc i 1 ki j Zc 0 K 3 Vc i 2 209 3 j 0.25 0.267 0.276 13 1 Eq. (11.74) Vci j 425.2 0.233 0.25 j 2 Vc Tc 464.851 369.8 Eq. (11.71) Tci j 464.851 Vc 0 214.65 214.65 220.4 200 3 cm mol 0 47.104 45.013 Eq. (11.72) Pci j Zci j R Tci j Vci j Pc 45.013 42.826 bar 42.48 0 Note: the calculated pure species Pc values in the matrix above do not agree exactly with the values in Table B.1 due to round-off error in the calculations. 364 Tri j Tr T Tci j 0.036 0.038 0.656 0.717 Pr 0.717 0.784 Eq. (3.65) P Pci j Pri j 0.038 0.04 0.824 B0i j 0 B0 Tri j 0.74636 B1i j 0.5405 0.27382 0.16178 Eq. (3.66) 0.16178 0.6361 B0 0.6361 0.27382 0.33295 B1 Tri j 0.874 0.558 B1 0.558 0.098 0.34 0.098 Eq. (11.69a) + (11.69b) 0.028 910.278 665.188 n 1 V 1j RT Z P Eq. (6.89) dB0dTri j 665.188 cm3 499.527 mol 3 yi y j Bi j BP RT Z i j B1i j n B i Eq. (3.38) 0.027 R Tci j B0i j Pci j Bi j B Eq. (11.61) 0.028 V Tri j 2.6 365 598.524 mol 1 Z 0.675 B cm 0.963 1.5694 Eq. (6.90) 3 4 cm 10 mol dB1dTri j Ans. 0.722 Tri j 5.2 Differentiating Eq. (11.61) and using Eq. (11.69a) + (11.69b) n n dBdT yi y j i 1j 1 Eq. (6.55) HR Eq. (6.56) SR Eq. (6.54) GR PT B T R dB0dTri j Pci j dBdT i j dB1dTri j 0.727 GR BP 344.051 SR P dBdT HR 101.7 mol K J mol 3 (b) cm V = 15694 mol SR = 1.006 HR = 450.322 J mol K GR = 125.1 3 (c) V = 24255 cm mol GR = 53.3 V = 80972 cm HR = 36.48 mol SR = 0.097 J mol K GR = 8.1 J mol J mol 3 (e) cm V = 56991 mol SR = 0.647 HR = 277.96 J GR = 85.2 mol K 366 J mol J mol 3 (d) J mol J mol HR = 175.666 J SR = 0.41 mol K J mol J J mol J mol Ans. Ans. Ans. Data for Problems 11.38 - 11.40 325 15 308.3 61.39 .187 200 100 150.9 48.98 .000 575 40 562.2 48.98 .210 73.83 .224 50.40 .087 350 T 300 P 35 304.2 Tc 50 Pc 282.3 525 10 507.6 30.25 .301 225 25 190.6 45.99 .012 200 75 126.2 34.00 .038 1.054 1.325 T Tc Tr 2.042 1.023 Tr 0.244 0.817 1.151 Pr 1.063 P Pc Pr 0.474 0.992 1.034 0.331 1.18 0.544 1.585 2.206 11.38 Redlich/Kwong Equation: 0.08664 0.42748 0.02 0.133 Eq. (3.53) 3.234 0.069 Pr Tr 4.559 4.77 0.036 0.081 q 1.5 Tr Eq. (3.54) q 3.998 4.504 0.028 0.04 z 3.847 0.121 Guess: 4.691 2.473 1 367 Given i z= 1 Ii 18 exp Z i fi z q Z ln i qi Eq. (3.52) Z zz i qi Z 1 ln Z i q Find z () Eq. (6.65) i qi i qi i qi Ii Eq. (11.37) i Pi Z i qi fi i 0.925 0.93 13.944 0.722 0.744 74.352 0.668 0.749 29.952 0.887 0.896 31.362 0.639 0.73 36.504 0.891 0.9 8.998 0.881 0.89 22.254 0.859 0.85 63.743 11.39 Soave/Redlich/Kwong Equation c 0.480 1.574 0.176 0.08664 2 1 c1 0.42748 0.5 2 Tr 0.02 0.133 Eq. (3.53) 3.202 0.069 Pr Tr 4.49 4.737 0.036 0.081 q Tr Eq. (3.54) q 3.79 4.468 0.028 0.04 z 3.827 0.121 Guess: 4.62 2.304 1 368 Given i z= 1 Ii 18 exp Z i fi Z z q i qi Eq. (3.52) zz ln Z i qi i i qi q Find z () Eq. (6.65) i qi ln Z 1 Z Z i qi Ii Eq. (11.37) i Pi i qi fi i 0.927 0.931 13.965 0.729 0.748 74.753 0.673 0.751 30.05 0.896 0.903 31.618 0.646 0.733 36.66 0.893 0.902 9.018 0.882 0.891 22.274 0.881 0.869 65.155 11.40 Peng/Robinson Equation 1 c 2 0.37464 1 1.54226 2 0.07779 0.26992 0.45724 2 1 c1 0.5 Tr 0.018 0.12 Eq.(3.53) 3.946 0.062 Pr Tr 5.383 5.658 0.032 0.073 q Tr Eq.(3.54) q 4.598 5.359 0.025 5.527 0.036 4.646 0.108 2.924 369 2 Guess: z 1 Given z = 1 i exp Z fi z i qi i Pi Z 2 1 Eq. (3.52) Z z 1 Ii 18 i z q ln 2 ln Z Z i qi i Z i qi q Find ( z) i i qi i qi i Eq. (6.65) qi Ii Eq. (11.37) fi i 0.918 0.923 13.842 0.69 0.711 71.113 0.647 0.73 29.197 0.882 0.89 31.142 0.617 0.709 35.465 0.881 0.891 8.91 0.865 0.876 21.895 0.845 0.832 62.363 BY GENERALIZED CORRELATIONS Parts (a), (d), (f), and (g) --- Virial equation: 325 T 350 525 308.3 Tc 225 Tr T Tc 304.2 507.6 15 P 190.6 Pr 35 10 25 61.39 73.83 .224 30.25 .301 45.99 Pc .187 .012 P Pc Evaluation of : B0 B0 ( Tr) Eq. (3.65) 370 B1 B1 ( Tr) Eq. (3.66) 0.675 DB0 Eq. (6.89) 2.6 0.722 DB1 5.2 Tr Eq. (6.90) Tr 0.932 Pr B0 exp Tr Eq. (11.60) B1 (a) (d) (f) (g) 0.904 0.903 0.895 Parts (b), (c), (e), and (h) --- Lee/Kesler correlation: Interpolate in Tables E.13 - E.16: .7454 0 1.1842 0.9634 0.9883 1 .7316 .8554 0.210 0.087 1.2071 .7517 0.000 0.038 0.745 01 (b) (c) (e) (h) 0.746 Eq. (11.67): 0.731 0.862 11.43 ndot1 x1 2 kmol hr ndot1 ndot3 ndot2 x1 4 kmol hr 0.333 ndot3 x2 1 x1 ndot1 ndot2 x2 0.667 a) Assume an ideal solution since n-octane and iso-octane are non-polar and very similar in chemical structure. For an ideal solution, there is no heat of mixing therefore the heat transfer rate is zero. b) S t R x1 ln x1 x2 ln x2 ndot3 371 St 8.82 W K Ans. 11.44 For air entering the process: 0.21 xN21 0.79 For the enhanced air leaving the process: xO22 0.5 xN22 0.5 ndot2 50 xO21 mol sec a) Apply mole balances to find rate of air and O2 fed to process Guess: ndotair 40 mol sec ndotO2 10 mol sec Given xO21 ndotair ndotO2 = xO22 ndot2 Mole balance on N2 xN21 ndotair = xN22 ndot2 ndotair ndotO2 ndotair Mole balance on O 2 Find ndotair ndotO2 31.646 mol sec Ans. ndotO2 18.354 mol sec Ans. b) Assume ideal gas behavior. For an ideal gas there is no heat of mixing, therefore, the heat transfer rate is zero. c) To calculate the entropy change, treat the process in two steps: 1. Demix the air to O2 and N2 2. Mix the N2 and combined O2 to produce the enhanced air Entropy change of demixing S12 Entropy change of mixing S23 R xO21 ln xO21 R xO22 ln xO22 Total rate of entropy generation: SdotG SdotG 372 ndotair S12 152.919 W K xN21 ln xN21 xN22 ln xN22 ndot2 S23 Ans. 10 11.50 T 932.1 544.0 30 K GE 273.15K J mol 513.0 50 HE 893.4 J mol 845.9 494.2 Assume Cp is constant. Then HE is of the form: HE = c aT Find a and c using the given HE and T values. a slope ( HE) T a c intercept ( HE) T c 2.155 mol K 3J 1.544 GE is of the form: GE = a T ln J 10 mol T K T bT c Rearrange to find b using estimated a and c values along with GE and T data. GE a T ln B T K T 13.543 c B T 13.559 J mol K 13.545 Use averaged b value 3 Bi b i 1 b 3 13.549 J mol K Now calculate HE, GE and T*SE at 25 C using a, b and c values. HE ( ) T aT HE [25 ( c T K GE ( ) T a T ln TSE ( ) T HE ( ) GE ( ) T T T bT 273.15) ] 901.242 K J Ans. mol ( c GE [25 273.15) ] 522.394 K J Ans. mol TSE [25 ( 373 273.15) ] 378.848 K J Ans. mol Chapter 12 - Section A - Mathcad Solutions 12.1 Methanol(1)/Water(2)-- VLE data: T 333.15 K 39.223 0.5714 42.984 0.2167 0.6268 48.852 0.3039 0.6943 52.784 P 0.1686 0.3681 0.7345 56.652 60.614 0.4461 x1 kPa 0.7742 y1 0.5282 0.8085 63.998 0.6044 0.8383 67.924 0.6804 0.8733 70.229 0.7255 0.8922 72.832 0.7776 0.9141 Number of data points: n Calculate x2 and y2: x2 rows ( ) P 1 n 10 y2 x1 i1n 1 y1 Vapor Pressures from equilibrium data: Psat1 84.562 kPa Psat2 19.953 kPa Calculate EXPERIMENTAL values of activity coefficients and excess Gibbs energy. y1 P 1 x1 Psat1 y2 P 2 GERT x2 Psat2 374 x1 ln 1 x2 ln 2 1 i ln 2 i ln 1 i 2 i i GERTi 1 1.572 1.013 0.452 0.013 0.087 2 1.47 1.026 0.385 0.026 0.104 3 1.32 1.075 0.278 0.073 0.135 4 1.246 1.112 0.22 0.106 0.148 5 1.163 1.157 0.151 0.146 0.148 6 1.097 1.233 0.093 0.209 0.148 7 1.05 1.311 0.049 0.271 0.136 8 1.031 1.35 0.031 0.3 0.117 9 1.021 1.382 0.021 0.324 0.104 10 1.012 1.41 0.012 0.343 0.086 0.2 0.4 0.5 0.4 ln 1 i 0.3 ln 2 i GERT i 0.2 0.1 0 0 0.6 x1 0.8 i (a) Fit GE/RT data to Margules eqn. by linear least squares: VXi Slope Slope x1 i slope ( VX VY) 0.208 VYi GERTi x1 x2 i i Intercept intercept ( VX VY) Intercept 0.683 A12 Intercept A21 Slope A12 0.683 A21 0.475 375 A12 Ans. The following equations give CALCULATED values: 2 1 ( x2) x1 2 ( x2) x1 j exp x2 exp x1 2 X1 j j X1 Y1calc 2 A21 A12 x1 A21 2 A12 A21 x2 X1 1 101 pcalc A12 j .01 j 1 X1 X2 Psat1 j j j j X2 .01 X2 j 2 X1 X2 Psat2 j j j 1 X1 X2 Psat1 j j pcalc j P-x,y Diagram: Margules eqn. fit to GE/RT data. 90 80 Pi kPa 70 Pi 60 kPa pcalc 50 j kPa pcalc kPa 40 j 30 20 10 0 0.2 0.4 0.6 x1 y1 X1 Y1calc i P-x data P-y data P-x calculated P-y calculated 376 i j 0.8 j 1 X1 j Pcalc x1 1 x1 x2 Psat1 i i i i y1calc x2 2 x1 x2 Psat2 i i i x1 1 x1 x2 Psat1 i i i Pcalc i i RMS deviation in P: Pi RMS 2 Pcalc i RMS n 0.399 kPa i (b) Fit GE/RT data to van Laar eqn. by linear least squares: VXi x1 VYi i x1 x2 i i GERTi Slope slope ( VX VY) Intercept intercept ( VX VY) Slope 0.641 Intercept 1.418 a12 a12 1 Intercept 0.705 1 ( x1 x2) 2 ( x1 x2) j a21 1 101 a21 a12 x1 exp a12 1 X2 j j 0.485 Ans. 2 a21 x2 exp a21 1 X1 1 ( Slope Intercept) a21 x2 a12 x1 .01 j 1 2 X1 .00999 j 377 (To avoid singularities) pcalc Pcalc X1 j 1 X1 X2 Psat1 j j j x1 1 x1 x2 Psat1 i i i i X1 Y1calc j j X2 j 2 X1 X2 Psat2 j j x2 2 x1 x2 Psat2 i i i 1 X1 X2 Psat1 j j pcalc y1calc x1 1 x1 x2 Psat1 i i i Pcalc i i j P-x,y Diagram: van Laar eqn. fit to GE/RT data. 90 80 Pi 70 kPa Pi 60 kPa pcalc 50 j kPa pcalc kPa 40 j 30 20 10 0 0.2 0.4 0.6 x1 y1 X1 Y1calc i i j 0.8 j P-x data P-y data P-x calculated P-y calculated RMS deviation in P: Pi RMS Pcalc n 2 i RMS i 378 0.454 kPa 1 (c) Fit GE/RT data to Wilson eqn. by non-linear least squares. Minimize the sum of the squared errors using the Mathcad Minimize function. Guesses: SSE 0.5 12 12 GERTi 21 1.0 21 x1 ln x1 i i 2 x2 i 12 i x2 ln x2 i i x1 Minimize SSE 12 12 x1 exp Pcalc X1 j x1 X1 j X1 Y1calc j x1 21 12 21 x2 x2 12 x1 x1 21 21 .01 j .01 1 X1 X2 Psat1 j j X2 j x1 1 x1 x2 Psat1 i i i i x2 x2 1 101 x2 12 12 x1 2 ( x2) x1 pcalc x2 x1 Ans. 21 12 1 ( x2) x1 1.026 21 21 exp x2 0.476 21 12 j 21 i j X2 j 1 X1 j 2 X1 X2 Psat2 j j x2 2 x1 x2 Psat2 i i i 1 X1 X2 Psat1 j j j pcalc y1calc i x1 1 x1 x2 Psat1 i i i Pcalc i j 379 P-x,y diagram: Wilson eqn. fit to GE/RT data. 90 80 Pi kPa 70 Pi 60 kPa pcalc 50 j kPa pcalc 40 j 30 kPa 20 10 0 0.2 0.4 0.6 x1 y1 X1 Y1calc i i j 0.8 j P-x data P-y data P-x calculated P-y calculated RMS deviation in P: Pi RMS Pcalc n 2 i RMS 0.48 kPa i (d) BARKER'S METHOD by non-linear least squares. Margules equation. Guesses for parameters: answers to Part (a). 2 1 x1 x2 A 12 A 21 exp ( ) A12 x2 2 x1 x2 A 12 A 21 exp ( ) A21 x1 2 380 2 A21 A12 x1 2 A12 A21 x2 1 Minimize the sum of the squared errors using the Mathcad Minimize function. Guesses: A12 SSE A12 A21 0.5 A21 Pi x1 i x2 X1 j X2 X1 Y1calc A21 2 X 1 j X 2 j A 12 A 21 Psat2 j 1 X 1 j X 2 j A 12 A 21 Psat1 pcalc j 1 x1i x2i A 12 A 21 Psat1 i x2 2 x1i x2i A 12 A 21 Psat2 x1 1 x1i x2i A 12 A 21 Psat1 i y1calc 0.435 j j x1 i 0.758 1 X 1 j X 2 j A 12 A 21 Psat1 j i i Pcalc i RMS deviation in P: Pi RMS Pcalc n 2 i 2 2 x1i x2i A 12 A 21 Psat2 i Minimize SSE A12 A21 A21 Pcalc x1 x2 A12 A21 Psat1 i1 i i A12 A12 pcalc 1.0 RMS i 381 0.167 kPa Ans. P-x-y diagram, Margules eqn. by Barker's method 90 80 Pi kPa 70 Pi 60 kPa pcalc 50 j kPa pcalc 40 j 30 kPa 20 10 0 0.2 0.4 0.6 x1 y1 X1 Y1calc i i j 0.8 1 j P-x data P-y data P-x calculated P-y calculated Residuals in P and y1 1 Pi Pcalc 0.5 i kPa y1 y1calc 100 i i 0 0.5 0 0.2 0.4 x1 Pressure residuals y1 residuals 382 i 0.6 0.8 (e) BARKER'S METHOD by non-linear least squares. van Laar equation. Guesses for parameters: answers to Part (b). j X1 1 101 1 x1 x2 a12 a21 .01 j j .00999 exp a12 1 2 x1 x2 a12 a21 exp a21 1 X2 j 1 X1 j 2 a12 x1 a21 x2 2 a21 x2 a12 x1 Minimize the sum of the squared errors using the Mathcad Minimize function. Guesses: a12 SSE a12 a21 0.5 a21 Pi x1 i a12 X1 j x2 i X1 Y1calc j a12 x1 i 2 X 1 j X 2 j a12 a21 Psat2 j 1 X 1 j X 2 j a12 a21 Psat1 j 1 x1i x2i a12 a21 Psat1 i x2 2 x1i x2i a12 a21 Psat2 x1 1 x1i x2i a12 a21 Psat1 i y1calc i i Pcalc i 383 0.83 a21 j pcalc 2 2 x1i x2i a12 a21 Psat2 1 X 1 j X 2 j a12 a21 Psat1 j X2 Pcalc x1 x2 a12 a21 Psat1 i1 i i Minimize SSE a12 a21 a21 pcalc 1.0 0.468 Ans. RMS deviation in P: Pi RMS Pcalc 2 i RMS n 0.286 kPa i P-x,y diagram, van Laar Equation by Barker's Method 90 80 70 Pi kPa 60 Pi kPa pcalc 50 j kPa pcalc j 40 kPa 30 20 10 0 0.2 0.4 0.6 x1 y1 X1 Y1calc i i P-x data P-y data P-x calculated P-y calculated 384 j 0.8 j 1 Residuals in P and y1. 1 Pi Pcalc 0.5 i kPa y1 y1calc 100 i i 0 0.5 0 0.2 0.4 x1 0.6 0.8 i Pressure residuals y1 residuals (f) BARKER'S METHOD by non-linear least squares. Wilson equation. Guesses for parameters: answers to Part (c). j 1 101 1 x1 x2 X1 12 21 .01 j j exp ln x1 x2 2 x1 x2 12 21 exp .01 x2 x1 x2 x1 X1 j 21 12 x2 x1 21 21 21 12 x1 1 j 12 12 ln x2 x1 X2 x2 12 x2 x1 21 Minimize the sum of the squared errors using the Mathcad Minimize function. Guesses: 12 0.5 21 385 1.0 SSE 12 Pi 21 x1 i x1 x2 i1 i i x2 2 x1i x2i i 12 X1 j 1 X1 j X2 j j X2 X1 Y1calc 2 X1 j X2 j 12 21 Psat2 j 1 X1 j X2 j 12 21 Psat1 pcalc x1 1 x1i x2i i j 21 Psat1 12 x2 2 x1i x2i 12 21 Psat2 x1 1 x1i x2i 12 21 Psat1 i y1calc 1.198 21 Psat1 12 j j i 0.348 21 i i Pcalc i RMS deviation in P: Pi RMS Pcalc 2 i RMS n i 386 2 21 Psat2 12 Minimize SSE 21 Pcalc 12 21 12 pcalc 21 Psat1 12 0.305kPa Ans. P-x,y diagram, Wilson Equation by Barker's Method 90 80 Pi kPa 70 Pi 60 kPa pcalc 50 j kPa pcalc 40 j 30 kPa 20 10 0 0.2 0.4 0.6 x1 y1 X1 Y1calc i i j 0.8 1 j P-x data P-y data P-x calculated P-y calculated Residuals in P and y1. 1 Pi Pcalc i 0.5 kPa y1 y1calc 100 i i 0 0.5 0 0.2 0.4 x1 Pressure residuals y1 residuals 387 i 0.6 0.8 12.3 Acetone(1)/Methanol(2)-- VLE data: T 328.15 K 72.278 0.0647 75.279 0.0570 0.1295 77.524 0.0858 0.1848 78.951 0.1046 0.2190 82.528 0.1452 0.2694 86.762 0.2173 0.3633 90.088 0.2787 0.4184 93.206 0.3579 0.4779 95.017 P 0.0287 0.4050 0.5135 96.365 97.646 kPa 0.4480 x1 y1 0.5052 0.5512 0.5844 98.462 0.5432 0.6174 99.811 0.6332 0.6772 99.950 0.6605 0.6926 100.278 0.6945 0.7124 100.467 0.7327 0.7383 100.999 0.7752 0.7729 101.059 0.7922 0.7876 99.877 0.9080 0.8959 99.799 0.9448 0.9336 Number of data points: n Calculate x2 and y2: x2 rowsP) ( 1 x1 n y2 Vapor Pressures from equilibrium data: Psat1 96.885 kPa Psat2 388 68.728 kPa 20 i1n 1 y1 Calculate EXPERIMENTAL values of activity coefficients and excess Gibbs energy. y1 P 1 x1 Psat1 ln 1 i i i GERT x2 Psat2 2 1 i y2 P 2 x1 ln 1 ln 2 i x2 ln 2 GERTi 0.013 0.027 0.568 0.011 0.043 0.544 5.815·10-3 0.052 1.002 0.534 1.975·10-3 0.058 1.58 1.026 0.458 0.026 0.089 6 1.497 1.027 0.404 0.027 0.108 7 1.396 1.057 0.334 0.055 0.133 8 1.285 1.103 0.25 0.098 0.152 9 1.243 1.13 0.218 0.123 0.161 10 1.224 1.14 0.202 0.131 0.163 11 1.166 1.193 0.153 0.177 0.165 12 1.155 1.2 0.144 0.182 0.162 13 1.102 1.278 0.097 0.245 0.151 14 1.082 1.317 0.079 0.275 0.145 15 1.062 1.374 0.06 0.317 0.139 16 1.045 1.431 0.044 0.358 0.128 17 1.039 1.485 0.039 0.395 0.119 18 1.037 1.503 0.036 0.407 0.113 19 1.017 1.644 0.017 0.497 0.061 20 1.018 1.747 0.018 0.558 0.048 1 1.682 1.013 0.52 2 1.765 1.011 3 1.723 1.006 4 1.706 5 389 0.6 0.4 ln 1 i ln 2 i GERTi 0.2 0 0 0.2 0.4 0.6 x1 0.8 i (a) Fit GE/RT data to Margules eqn. by linear least squares: VXi x1 Slope VYi i slope ( VX VY) Slope GERTi x1 x2 i i Intercept Intercept 0.018 intercept ( VX VY) 0.708 A12 Intercept A21 Slope A12 0.708 A21 0.69 A12 Ans. The following equations give CALCULATED values: 2 1 ( x2) x1 2 ( x2) x1 j exp x2 exp x1 X1 j j A12 x1 2 A12 A21 x2 X1 j X1 Y1calc 2 A21 A21 2 1 101 pcalc A12 .01 j j 1 X1 X2 Psat1 j j j 1 X1 X2 j pcalc j .01 X2 j Psat1 j 390 X2 2 X1 X2 Psat2 j j j 1 X1 j P-x,y Diagram: Margules eqn. fit to GE/RT data. 105 100 Pi kPa 95 Pi 90 kPa pcalc 85 j kPa pcalc 80 j 75 kPa 70 65 0 0.2 0.4 0.6 x1 y1 X1 Y1calc i i j 0.8 j P-x data P-y data P-x calculated P-y calculated Pcalc x1 1 x1 x2 Psat1 i y1calc i i i i x2 2 x1 x2 Psat2 i i i x1 1 x1 x2 Psat1 i i i Pcalc i RMS deviation in P: Pi RMS Pcalc n 2 i RMS i 391 0.851 kPa (b) Fit GE/RT data to van Laar eqn. by linear least squares: x1 x2 VXi x1 i VYi i i GERTi Slope slope ( VX VY) Intercept intercept ( VX VY) Slope 0.015 Intercept 1.442 1 a12 Intercept a12 0.693 X1 X2 Pcalc X1 j j 1 X1 X2 j j i X1 j i j i 2 a12 x1 X1 (To avoid singularities) .00999 j Psat1 x1 1 x1 x2 Psat1 i Y1calc 1 j 2 a21 x2 .01 j j Ans. a21 x2 exp a21 1 1 101 Intercept) 0.686 a12 x1 exp a12 1 2 ( x2) x1 pcalc ( Slope a21 1 ( x2) x1 j 1 a21 X2 j 2 X1 X2 j j Psat2 x2 2 x1 x2 Psat2 i i 1 X1 X2 Psat1 j j pcalc i y1calc j 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 95 kPa Pi 90 kPa pcalc 85 j kPa 80 pcalc j kPa 75 70 65 0 0.2 0.4 0.6 0.8 x1 y1 X1 Y1calc i i j 1 j P-x data P-y data P-x calculated P-y calculated RMS deviation in P: Pi RMS Pcalc 2 i n RMS 0.701 kPa i (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 12 0.5 GERTi 21 i 1.0 21 x1 ln x1 i i 393 12 i x2 ln x2 i 2 x2 i x1 i 21 Minimize SSE 12 12 x1 exp pcalc Pcalc X1 j x1 X1 j X1 Y1calc j x1 21 12 21 x2 x2 12 x1 .01 j 1 X1 X2 Psat1 j j X2 x1 21 21 X2 .01 j x1 1 x1 x2 Psat1 i i i i x2 x2 1 101 x2 12 12 x1 2 ( x2) x1 j x2 x1 Ans. 21 12 1 ( x2) x1 0.681 21 21 exp x2 0.71 21 12 j j 1 X1 j 2 X1 X2 Psat2 j j x2 2 x1 x2 Psat2 i i i 1 X1 X2 Psat1 j j j pcalc y1calc i x1 1 x1 x2 Psat1 i i i Pcalc i j 394 P-x,y diagram: Wilson eqn. fit to GE/RT data. 105 100 Pi kPa 95 Pi 90 kPa pcalc 85 j kPa pcalc 80 j 75 kPa 70 65 0 0.2 0.4 0.6 x1 y1 X1 Y1calc i i j 0.8 j P-x data P-y data P-x calculated P-y calculated RMS deviation in P: Pi RMS Pcalc 2 i n RMS 0.361 kPa i (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 exp ( x2) 2 x1 x2 A 12 A 21 exp ( x1) 2 2 A12 2 A21 A12 x1 A21 2 A12 A21 x2 395 1 Minimize the sum of the squared errors using the Mathcad Minimize function. Guesses: A12 SSE A12 A21 0.5 A21 Pi x1 i x2 X1 j X2 X1 Y1calc A21 2 X 1 j X 2 j A 12 A 21 Psat2 j 1 X 1 j X 2 j A 12 A 21 Psat1 pcalc j 1 x1i x2i A 12 A 21 Psat1 i x2 2 x1i x2i A 12 A 21 Psat2 x1 1 x1i x2i A 12 A 21 Psat1 i y1calc 0.672 j j x1 i 0.644 1 X 1 j X 2 j A 12 A 21 Psat1 j i i Pcalc i RMS deviation in P: Pi RMS Pcalc n 2 i 2 2 x1i x2i A 12 A 21 Psat2 i Minimize SSE A12 A21 A21 Pcalc x1 x2 A12 A21 Psat1 i1 i i A12 A12 pcalc 1.0 RMS i 396 0.365 kPa Ans. P-x-y diagram, Margules eqn. by Barker's method 105 100 Pi kPa 95 Pi 90 kPa pcalc 85 j kPa pcalc 80 j 75 kPa 70 65 0 0.2 0.4 0.6 x1 y1 X1 Y1calc i i j 0.8 j P-x data P-y data P-x calculated P-y calculated Residuals in P and y1 2 Pi Pcalc 1 i kPa y1 y1calc 100 i i 0 1 0 0.2 0.4 0.6 x1 Pressure residuals y1 residuals 397 i 0.8 1 (e) BARKER'S METHOD by non-linear least squares. van Laar equation. Guesses for parameters: answers to Part (b). j X1 1 101 1 x1 x2 a12 a21 .01 j j .00999 exp a12 1 2 x1 x2 a12 a21 exp a21 1 X2 j 1 X1 j 2 a12 x1 a21 x2 2 a21 x2 a12 x1 Minimize the sum of the squared errors using the Mathcad Minimize function. Guesses: a12 SSE a12 a21 0.5 a21 Pi x1 i a12 X1 j x2 i X1 Y1calc j a12 x1 i 2 X 1 j X 2 j a12 a21 Psat2 j 1 X 1 j X 2 j a12 a21 Psat1 j 1 x1i x2i a12 a21 Psat1 i x2 2 x1i x2i a12 a21 Psat2 x1 1 x1i x2i a12 a21 Psat1 i y1calc i i Pcalc i 398 0.644 a21 j pcalc 2 2 x1i x2i a12 a21 Psat2 1 X 1 j X 2 j a12 a21 Psat1 j X2 Pcalc x1 x2 a12 a21 Psat1 i1 i i Minimize SSE a12 a21 a21 pcalc 1.0 0.672 Ans. RMS deviation in P: Pi RMS Pcalc 2 i RMS n 0.364 kPa i P-x,y diagram, van Laar Equation by Barker's Method 105 100 95 Pi kPa 90 Pi kPa pcalc 85 j kPa pcalc j 80 kPa 75 70 65 0 0.2 0.4 0.6 x1 y1 X1 Y1calc i i P-x data P-y data P-x calculated P-y calculated 399 j 0.8 j 1 Residuals in P and y1. 1.5 1 Pi Pcalc i 0.5 kPa y1 y1calc 100 i 0 i 0.5 1 0 0.2 0.4 0.6 x1 0.8 i Pressure residuals y1 residuals (f) j BARKER'S METHOD by non-linear least squares. Wilson equation. Guesses for parameters: answers to Part (c). 1 101 1 x1 x2 X1 12 21 .01 j j exp ln x1 x2 2 x1 x2 12 21 .01 x2 x1 x2 x1 X1 j 21 12 x2 x1 21 21 21 12 x1 1 j 12 12 exp ln x2 x1 X2 x2 12 x2 x1 21 Minimize the sum of the squared errors using the Mathcad Minimize function. Guesses: 12 0.5 21 400 1.0 SSE 12 Pi 21 x1 i x1 x2 i1 i i x2 2 x1i x2i i 12 X1 j 1 X1 j X2 j j X2 X1 Y1calc RMS 2 X1 j X2 j 12 j 1 X1 j X2 j 12 21 Psat1 1 x1i x2i i j 21 Psat1 12 x2 2 x1i x2i 12 21 Psat2 x1 1 x1i x2i 12 21 Psat1 i y1calc 0.35 kPa 21 Psat2 pcalc x1 0.663 21 Psat1 12 j j i 0.732 21 i i Pcalc i RMS deviation in P: Pi RMS Pcalc 2 i n i 401 2 21 Psat2 12 Minimize SSE 21 Pcalc 12 21 12 pcalc 21 Psat1 12 Ans. P-x,y diagram, Wilson Equation by Barker's Method 105 100 Pi kPa 95 Pi 90 kPa pcalc 85 j kPa pcalc 80 j 75 kPa 70 65 0 0.2 0.4 0.6 x1 y1 X1 Y1calc i i j 0.8 j P-x data P-y data P-x calculated P-y calculated Residuals in P and y1. 2 Pi Pcalc 1 i kPa y1 y1calc 100 i i 0 1 0 0.2 0.4 0.6 x1 Pressure residuals y1 residuals 402 i 0.8 1 12.6 Methyl t-butyl ether(1)/Dichloromethane--VLE data: T 308.15 K 83.402 0.0141 82.202 0.0579 0.0253 80.481 0.0924 0.0416 76.719 0.1665 0.0804 72.442 0.2482 0.1314 68.005 P 0.0330 0.3322 0.1975 65.096 59.651 0.3880 x1 kPa y1 0.5036 0.2457 0.3686 56.833 0.6736 0.5882 51.620 0.7676 0.7176 50.455 0.8476 0.8238 49.926 0.9093 0.9002 49.720 Psat1 0.4564 53.689 x2 0.5749 0.9529 0.9502 1 y2 x1 49.624 kPa 1 Psat2 y1 85.265 kPa Calculate EXPERIMENTAL values of activity coefficients and excess Gibbs energy. y1 P 1 x1 Psat1 GERTx1x2 y2 P 2 GERT x1 x2 GERT x2 Psat2 n rows ( ) P 403 x1 ln 1 n 14 x2 ln 2 i 1n (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 SSE A12 A21 C A21 C A21 x1 GERTi 0.5 A12 x2 i 0.2 C x1 x2 i i x1 x2 i i i A12 A21 Minimize SSE A12 A21 C 0.336 A21 A12 0.535 0.195 C C Ans. (b) Plot data and fit GeRTx1x2 ( x2) x1 GeRT ( x2) x1 A21 x1 C x1 x2 GeRTx1x2 ( x2)x1 x2 x1 2 ln 1 ( x2) x1 x2 ln 2 ( x2) x1 x1 j A12 x2 2 A12 A12 C x1 3 C x1 A21 2 2 A21 2 A12 A21 C x2 3 C x2 1 101 X1 j .01 j 2 .01 X2 1 j 0 GERTx1x2 i GeRTx1x2 X1 X2 j ln 1 i j 0.1 0.2 ln 1 X1 X2 j j ln 2 i 0.3 0.4 ln 2 X1 X2 j j 0.5 0.6 0 0.2 0.4 0.6 0.8 x1 X1 x1 X1 x1 X1 i 404 j i j i j X1 j i 2 (c) Plot Pxy diagram with fit and data 1 ( x1 x2) exp ln 1 ( x1 x2) 2 ( x1 x2) exp ln 2 ( x1 x2) Pcalc X1 j X1 y1calc 1 X1 X2 Psat1 j j j j j X2 j 2 X1 X2 Psat2 j j 1 X1 X2 Psat1 j j Pcalc j P-x,y Diagram from Margules Equation fit to GE/RT data. 90 Pi 80 kPa Pi 70 kPa Pcalc j 60 kPa Pcalc j kPa 50 40 0 0.2 0.4 0.6 x1 y1 X1 y1calc i i j 0.8 j P-x data P-y data P-x calculated P-y calculated (d) Consistency Test: ln 1 2i ln 1 x1 x2 i i 2 x1 x2 i i GERTi ln 1i 2 i 405 GeRT x1 x2 i i GERTi 0.004 0 GERTi 0 ln 1 2i 0.025 0.004 0 0.5 x1 0.05 1 0 0.5 x1 i 1 i Calculate mean absolute deviation of residuals mean GERT 9.391 4 10 mean ln 1 2 0.021 (e) Barker's Method by non-linear least squares: Margules Equation 2 1 x1 x2 A 12 A 21 C exp ( ) A12 x2 2 A21 A12 C x1 A21 C x2 2 3 C x1 2 2 x1 x2 A 12 A 21 C exp ( ) A21 x1 2 A12 2 3 C x2 Minimize sum of the squared errors using the Mathcad Minimize function. Guesses: A12 0.3 SSE A12 A21 C Pi i A21 C 0.2 x1 x2 A12 A21 C Psat1 i1 i i x1 x2 i A12 A21 0.5 2 x1i x2i A 12 A 21 C Psat2 A12 Minimize SSE A12 A21 C C 0.364 A21 0.521 C 406 2 0.23 Ans. Plot P-x,y diagram for Margules Equation with parameters from Barker's Method. Pcalc X1 j 1 X 1 j X 2 j A 12 A 21 C Psat1 j X2 X1 y1calc j j j 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 Pcalc j 90 Pi 80 kPa Pi 70 kPa Pcalc j 60 kPa Pcalc kPa j 50 40 0 0.2 0.4 0.6 x1 y1 X1 y1calc i i j P-x data P-y data P-x calculated P-y calculated Pcalc x1 i 1 x1i x2i A 12 A 21 C Psat1 i x2 2 x1i x2i A 12 A 21 C Psat2 x1 1 x1i x2i A 12 A 21 C Psat1 i y1calc i i Pcalc i 407 0.8 j Plot of P and y1 residuals. 0.8 0.6 Pi Pcalc i 0.4 kPa y1 y1calc 100 0.2 i i 0 0.2 0 0.5 x1 1 i Pressure residuals y1 residuals RMS deviations in P: Pi RMS Pcalc n 2 i RMS i 408 0.068 kPa 12.8 (a) Data: 0.0523 1.002 0.1299 1.307 1.004 0.2233 1.295 1.006 0.2764 1.228 1.024 0.3482 1.234 1.022 0.4187 x1 1.202 1.180 1.049 0.5001 2 1.129 1 1.092 0.5637 1.102 0.6469 1.076 1.170 0.7832 1.032 1.298 0.8576 1.016 1.393 0.9388 1.001 1.600 0.9813 n 1.120 1.003 1.404 rows x1 GERTi i x1 ln 1 i i 1n n 13 x2 i 1 x1 i x2 ln 2 i i (b) Fit GE/RT data to Margules eqn. by linear least-squares procedure: Xi x1 Yi i GERTi x1 x2 i i Slope slope ( Y) X Intercept intercept ( Y) X Slope 0.247 Intercept 0.286 A12 Intercept A21 Slope A12 0.286 A21 0.534 2 1 ( x2) x1 exp x2 2 ( x2) x1 exp x1 GeRT ( x2) x1 2 A12 A12 2 A21 A12 x1 A21 2 A12 Ans. A21 x2 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 0.4 ln 1 i ln 1 x1 x2 i i ln 2 i 0.3 0.2 ln 2 x1 x2 i i 0.1 0 0 0.2 0.4 0.6 x1 i (c) Calculate and plot residuals for consistency test: GERTi ln 1 2i GeRT x1 x2 i i ln 1 x1 x2 i i 2 x1 x2 i i GERTi ln 1i 2 i 410 0.8 ln 1 2i GERTi 3.314·10-3 -2.264·10-3 -9.153·10-5 -3.14·10-3 0.1 0.098 -0.021 -2.998·10-3 0.026 -2.874·10-3 -0.019 -2.22·10-3 5.934·10-3 -2.174·10-3 0.028 -1.553·10-3 -9.59·10-3 -8.742·10-4 9.139·10-3 2.944·10-4 -5.617·10-4 5.962·10-5 -0.011 9.025·10-5 ln 1 2i 0 0.028 4.236·10-4 0.05 -0.168 0 0.5 x1 1 i 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 31.957 0.1056 33.553 0.0940 0.1818 35.285 0.1829 0.2783 36.457 0.2909 0.3607 36.996 P 0.0455 0.3980 0.4274 37.068 36.978 kPa x1 0.5069 0.5458 y1 0.4885 0.5098 36.778 0.5946 0.5375 35.792 0.7206 0.6157 34.372 0.8145 0.6913 32.331 0.8972 0.7869 30.038 0.9573 0.8916 411 x2 1 Psat1 x1 y2 1 Psat2 27.778 kPa y1 29.819 kPa Calculate EXPERIMENTAL values of activity coefficients and excess Gibbs energy. y1 P 1 y2 P 2 x1 Psat1 GERTx1x2 GERT x2 Psat2 GERT x1 x2 n x1 ln 1 n rows ( ) P x2 ln 2 i 12 1n (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 GERTi A21 x1 i A12 x2 i C x1 x2 i x1 x2 i i i i A12 Minimize SSE A12 A21 C C 1.155 C A21 1.128 A21 A12 0.53 Ans. (b) Plot data and fit GeRTx1x2 ( x2) x1 GeRT ( x2) x1 2 x2 ln 2 ( x2) x1 x1 1 101 A12 x2 C x1 x2 GeRTx1x2 ( x2)x1 x2 x1 ln 1 ( x2) x1 j A21 x1 2 2 A12 2 A21 A12 C x1 3 C x1 A21 2 A12 A21 C x2 3 C x2 X1 j .01 j 412 .01 2 X2 j 1 X1 j 2 1.2 GERTx1x2 i 1 GeRTx1x2 X1 X2 j j 0.8 ln 1 i ln 1 X1 X2 j 0.6 j ln 2 i 0.4 ln 2 X1 X2 j j 0.2 0 0 0.2 0.4 0.6 0.8 x1 X1 x1 X1 x1 X1 i j i j i (c) Plot Pxy diagram with fit and data 1 ( x1 x2) exp ln 1 ( x1 x2) 2 ( x1 x2) exp ln 2 ( x1 x2) Pcalc X1 j X1 y1calc j 1 X1 X2 j j j j Psat1 X2 j 1 X1 X2 Psat1 j j Pcalc j 413 2 X1 X2 j j Psat2 j P-x,y Diagram from Margules Equation fit to GE/RT data. 38 Pi 36 kPa 34 Pi kPa Pcalc 32 j kPa 30 Pcalc j kPa 28 26 0 0.2 0.4 0.6 x1 y1 X1 y1calc i i j 0.8 j P-x data P-y data P-x calculated P-y calculated (d) Consistency Test: ln 1 2i ln 1 x1 x2 i i 2 x1 x2 i i GERTi ln GeRT x1 x2 i i GERTi 1i 2 i 0.004 0 GERTi 0 ln 1 2i 0.025 0.004 0 0.5 x1 0.05 1 0 0.5 x1 i 414 i 1 Calculate mean absolute deviation of residuals mean GERT 6.237 4 10 mean ln 1 2 0.025 (e) Barker's Method by non-linear least squares: Margules Equation 2 1 x1 x2 A 12 A 21 C exp ( ) A12 x2 2 A21 A12 C x1 A21 C x2 2 3 C x1 2 2 x1 x2 A 12 A 21 C exp ( ) A21 x1 2 A12 2 3 C x2 Minimize sum of the squared errors using the Mathcad Minimize function. Guesses: A12 0.3 SSE A12 A21 C A21 0.5 C 0.2 x1 x2 A12 A21 C Psat1 i1 i i Pi x1 i x2 i 2 x1i x2i A 12 A 21 C Psat2 A12 Minimize SSE A12 A21 C 1.114 A21 1.098 C A12 A21 2 0.387 C Ans. Plot P-x,y diagram for Margules Equation with parameters from Barker's Method. Pcalc X1 j 1 X 1 j X 2 j A 12 A 21 C Psat1 j X2 X1 y1calc j j j 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 Pcalc j 415 38 36 Pi kPa 34 Pi kPa Pcalc 32 j kPa Pcalc kPa 30 j 28 26 0 0.2 0.4 0.6 x1 y1 X1 y1calc i i j P-x data P-y data P-x calculated P-y calculated Pcalc x1 i 1 x1i x2i A 12 A 21 C Psat1 i x2 2 x1i x2i A 12 A 21 C Psat2 x1 1 x1i x2i A 12 A 21 C Psat1 i y1calc i i Pcalc i 416 0.8 j Plot of P and y1 residuals. 0.6 0.4 Pi Pcalc i 0.2 kPa y1 y1calc 100 i i 0 0.2 0.4 0 0.5 x1 1 i Pressure residuals y1 residuals RMS deviations in P: Pi RMS Pcalc n 2 i RMS i 417 0.04 kPa 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: A1 16.1154 B1 3483.67 K C1 205.807 K Water: A2 16.3872 B2 3885.70 K C2 230.170 K Psat1 ( T) exp A1 Psat2 ( ) T exp A2 (T B1 273.15 K) C1 (T B2 273.15 K) C2 kPa kPa Parameters for the Wilson equation: 3 V1 a12 775.48 3 cm 75.14 mol 12 ( ) T V2 a21 cal mol cm 18.07 mol 1351.90 a12 V2 exp RT V1 exp x2 1 ( x2 T) x1 2 ( x2 T) x1 21 ( ) T x1 x2 12 ( ) x2 T x1 x2 x1 418 21 ( ) T 21 ( ) T x2 12 ( ) x2 T x2 x1 12 ( ) T 12 ( ) T x1 a21 V1 exp RT V2 21 ( ) T 12 ( ) T x1 exp cal mol 21 ( ) T x1 21 ( ) T P-x,y diagram at T ( 60 273.15) K Guess: P 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) ( 70 kPa x1 1 ( x1 1 Peq x1) ( yeq x) ( x x1 T) Psat1 ( T) Peq x) ( kPa 0 0 20.007 0.05 0.315 28.324 0.1 0.363 30.009 0.15 0.383 30.639 0.2 0.395 30.97 0.25 0.404 31.182 0.3 0.413 31.331 0.35 0.421 31.435 0.4 0.431 31.496 0.45 0.441 31.51 0.5 0.453 31.467 0.55 0.466 31.353 0.6 0.483 31.148 0.65 0.502 30.827 0.7 0.526 30.355 0.75 0.556 29.686 0.8 0.594 28.759 0.85 0.646 27.491 0.9 0.718 25.769 0.95 0.825 23.437 1 1 20.275 419 x 0 0.05 1.0 P,x,y Diagram at T 333.15 K 32 30 28 Peq x) ( kPa Peq x) ( 26 kPa 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 16.1154 B1 3483.67 K C1 205.807 K Water: 16.3872 B2 3885.70 K C2 230.170 K A2 Psat1 ( ) T exp A1 Psat2 ( ) T exp A2 ( T B1 273.15 K) C1 kPa ( T B2 273.15 K) C2 kPa 420 Parameters for the Wilson equation: 3 V1 75.14 a12 775.48 12 ( T) 3 cm cm V2 cal mol 18.07 a21 mol 1351.90 a12 V2 exp RT V1 21 ( T) x1 1 ( x1 x2 T) x1 x2 x2 T-x,y diagram at P x2 x1 21 ( T) 21 ( T) 101.33 kPa Guess: x1 273.15) K Given Teq ( x1) yeq ( x1) x 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 0 0.05 1.0 421 21 ( T) 21 ( T) x2 12 ( T) x2 x1 12 ( T) 12 ( T) x1 a21 V1 exp RT V2 21 ( T) x2 12 ( T) x1 2 ( x1 x2 T) cal mol 12 ( T) exp x2 exp mol Teq x () K yeq x () x 0 0 373.149 0.05 0.304 364.159 0.1 0.358 362.476 0.15 0.381 361.836 0.2 0.395 361.49 0.25 0.407 361.264 0.3 0.418 361.101 0.35 0.429 360.985 0.4 0.44 360.911 0.45 0.453 360.881 0.5 0.468 360.904 0.55 0.484 360.99 0.6 0.504 361.154 0.65 0.527 361.418 0.7 0.555 361.809 0.75 0.589 362.364 0.8 0.631 363.136 0.85 0.686 364.195 0.9 0.759 365.644 0.95 0.858 367.626 1 1 370.349 T,x,y Diagram at P 101.33 kPa 375 Teq( ) 370 x K Teq( ) x K 365 360 0 0.2 0.4 0.6 x yeq x) ( 422 0.8 1 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 16.1154 B1 3483.67 K C1 205.807 K Water: 16.3872 B2 3885.70 K C2 230.170 K A2 Psat1 ( T) exp A1 Psat2 ( T) exp A2 (T B1 273.15 K) C1 (T B2 273.15 K) C2 kPa kPa Parameters for the NRTL equation: b12 500.40 12 ( T) exp b21 1636.57 b12 RT G12 ( T) cal mol 1 ( x1 x2 T) 0.5081 21 ( T ) 2 exp x2 2 exp x1 b21 RT G21 ( T) 12 ( T) exp G21 ( T) x1 x2 G21 ( T) G12 ( T) 12 ( T) 21 ( T ) 2 21 ( T) ( x2 2 ( x1 x2 T) cal mol x1 G12 ( T) ) 2 G12 ( T) 12 ( T) x2 x1 G12 ( T) G21 ( T) 21 ( T) ( x1 x2 G21 ( T) ) 423 2 2 P-x,y diagram at T ( 60 273.15)K Guess: P 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) ( 70 kPa x1 1 ( 1 x1 Peq x1) ( yeq x () x x1 T)Psat1 ( ) T Peq x () kPa 0 20.007 0.05 0.33 28.892 0.1 0.373 30.48 0.15 0.382 30.783 0.2 0.386 30.876 0.25 0.39 30.959 0.3 0.395 31.048 0.35 0.404 31.127 0.4 0.414 31.172 0.45 0.427 31.163 0.5 0.442 31.085 0.55 0.459 30.922 0.6 0.479 30.657 0.65 0.503 30.271 0.7 0.531 29.74 0.75 0.564 29.03 0.8 0.606 28.095 0.85 0.659 26.868 0.9 0.732 25.256 0.95 0.836 23.124 1 1 20.275 0 424 x 0 0.05 1.0 P,x,y Diagram at T 333.15 K 35 Peq ( x) 30 kPa Peq ( x) kPa 25 20 0 0.2 0.4 0.6 0.8 x yeq ( x) 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 B1 3483.67 K C1 205.807 K Water: B2 3885.70 K C2 230.170 K A2 Psat1 ( T) exp A1 Psat2 ( T) exp A2 16.3872 (T B1 273.15 K) C1 (T B2 273.15 K) C2 kPa kPa Parameters for the NRTL equation: b12 500.40 cal mol b21 1636.57 425 cal mol 0.5081 12 ( ) T G12 ( ) T b12 RT 21 ( ) T exp 1 ( x2 T) x1 G21 ( ) T 12 ( ) T 2 exp x2 2 ( x2 T) x1 exp x1 RT exp G21 ( ) T 21 ( ) T x1 x2 G21 ( ) T G12 ( ) 12 ( ) T T 2 2 (2 x 2 b21 x1 G12 ( ) T) G12 ( ) T 12 ( ) T x2 x1 G12 ( ) T G21 ( ) 21 ( ) T T 2 2 (1 x x2 G21 ( ) T) T-x,y diagram at P Guess: T 273.15)K Given P = x1 1 ( 1 x1 T)Psat1 ( ) x1 T ( x1) 2 ( 1 x1 T)Psat2 ( ) 1 x1 T Teq x1) ( yeq x1) ( ( 90 101.33 kPa Find T) ( x1 1 ( 1 x1 x1 Teq x1) Psat1 ( ( ) () Teq x1 ) P 426 21 ( ) T x 0 0.05 1.0 Teq ( x) K yeq ( x) x 0 373.149 0.05 0.32 363.606 0.1 0.377 361.745 0.15 0.394 361.253 0.2 0.402 361.066 0.25 0.408 360.946 0.3 0.415 360.843 0.35 0.424 360.757 0.4 0.434 360.697 0.45 0.447 360.676 0.5 0.462 360.709 0.55 0.48 360.807 0.6 0.5 360.985 0.65 0.524 361.262 0.7 0.552 361.66 0.75 0.586 362.215 0.8 0.629 362.974 0.85 0.682 364.012 0.9 0.754 365.442 0.95 0.853 367.449 1 1 370.349 0 T,x,y Diagram at P 101.33 kPa 375 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: A1 16.1154 B1 3483.67 K C1 205.807 K Water: A2 16.3872 B2 3885.70 K C2 230.170 K Psat1 ( ) T exp A1 Psat2 ( ) T exp A2 ( T B1 273.15 K) C1 kPa ( T B2 273.15 K) C2 kPa Parameters for the Wilson equation: 3 V1 a12 775.48 3 cm 75.14 mol 12 ( ) T V2 a21 cal mol cm 18.07 mol 1351.90 a12 V2 exp RT V1 exp x2 1 ( x2 T) x1 2 ( x2 T) x1 a21 V1 exp RT V2 21 ( ) T 21 ( ) T 12 ( ) T x2 12 ( ) x2 T x1 x1 exp cal mol x1 x2 x1 12 ( ) T 21 ( ) T 12 ( ) T x1 x2 12 ( ) x2 T x2 x1 428 21 ( ) T 21 ( ) T x1 21 ( ) T (a) BUBL P: T ( 60 Guess: P 101.33 kPa Given x1 0.3 x2 1 x1 y1 273.15) K 0.4 y2 1 y1 y1 y2 = 1 y1 P = x1 1 ( x1 x2 T) Psat1 ( T) y2 P = x2 2 ( x1 x2 T) Psat2 ( T) Pbubl Find ( P y1 y2) y1 y2 Pbubl 0.413 y2 0.587 Ans. 273.15) K y1 0.3 y2 1 y1 x1 0.1 x2 1 x1 y1 31.33 kPa (b) DEW P: T ( 60 Guess: P 101.33 kPa y1 P = x1 1 ( x1 x2 T) Psat1 ( T) Given x1 x2 = 1 y2 P = x2 2 ( x1 x2 T) Psat2 ( T) Pdew Find ( P x1 x2) x1 x2 Pdew x1 27.79 kPa x2 0.042 Ans. 0.958 (c) P,T-flash Calculation P Pdew Guess: Pbubl T 2 V Given y1 = y2 = 0.5 x1 y1 ( 60 0.1 0.1 x2 y2 z1 273.15) K x1 1 ( x1 x2 T) Psat1 ( T) P x2 2 ( x1 x2 T) Psat2 ( T) P 429 1 1 x1 x2 = 1 y1 y2 = 1 y1 x1 0.3 x1 ( 1 Eq. (10.15) V) y1 V = z1 x1 x2 y1 Find x1 x2 y1 y2 V) ( y2 V x1 x2 0.08 0.92 y1 y2 0.351 V 0.649 0.813 (d) Azeotrope Calculation Test for azeotrope at: T 1 ( 1 T)Psat1 ( ) 0 T 120 Psat1 ( ) T 2 ( 0 T)Psat2 ( ) 1 T 21.581 121 Psat2 ( ) T 121 273.15)K 2 ( 0 T) 4.683 1 1 ( 1 T) 21.296 0 120 ( 60 0.216 Since one of these values is >1 and the other is <1, an azeotrope exists. See Ex. 10.3(e) Guess: Given P 101.33 kPa x1 y1 x2 y2 0.3 0.3 1 1 y1 x1 y1 P = x1 1 ( x2 T)Psat1 ( ) x1 T y2 P = x2 2 ( x2 T)Psat2 ( ) x1 T x1 y1 x2 = 1 x1 = y1 y2 = 1 x1 x2 y1 Find x1 x2 y1 y2 P) ( y2 Paz Paz 31.511 kPa 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 16.1154 B1 3483.67 K C1 205.807 K Water: 16.3872 B2 3885.70 K C2 230.170 K A2 Psat1 ( T) exp A1 Psat2 ( T) exp A2 (T B1 273.15 K) C1 (T B2 273.15 K) C2 kPa kPa Parameters for the NRTL equation: b12 500.40 12 ( T) G12 ( T) cal mol b21 1636.57 b12 RT exp 1 ( x1 x2 T) 0.5081 b21 21 ( T ) 12 ( T) 2 exp x2 2 exp x1 RT G21 ( T) exp G21 ( T) x1 x2 G21 ( T) G12 ( T) 12 ( T) 21 ( T ) 2 21 ( T) ( x2 2 ( x1 x2 T) cal mol x1 G12 ( T) ) 2 G12 ( T) 12 ( T) x2 x1 G12 ( T) G21 ( T) 21 ( T) ( x1 x2 G21 ( T) ) 431 2 2 (a) BUBL P: T ( 60 Guess: P 101.33 kPa Given y1 P = x1 1 ( x2 T)Psat1 ( ) x1 T x1 0.3 x2 1 x1 y1 273.15)K 0.4 y2 1 y1 y1 y2 P = x2 2 ( x2 T)Psat2 ( ) x1 T y2 = 1 Pbubl Find ( y1 y2) P y1 y2 Pbubl y1 31.05 kPa y2 0.395 (b) DEW P: T ( 60 Guess: P 101.33 kPa Given Ans. 0.605 y1 0.3 y2 1 y1 x1 273.15)K 0.1 x2 1 x1 y1 P = x1 1 ( x2 T)Psat1 ( ) x1 T y2 P = x2 2 ( x2 T)Psat2 ( ) x1 T x1 x2 = 1 Pdew Find ( x1 x2) P x1 x2 Pdew x1 27.81 kPa x2 0.037 0.963 Ans. (c) P,T-flash Calculation P Pdew Pbubl Guess: Given T 2 V y1 = y2 = x1 ( 1 0.5 ( 60 x1 y1 273.15)K 0.1 0.1 x1 1 ( x2 T)Psat1 ( ) x1 T P x2 2 ( x2 T)Psat2 ( ) x1 T P V) y1 V = z1 Eq. (10.15) 432 z1 0.3 x2 y2 1 1 x1 x2 = 1 y1 y2 = 1 y1 x1 x1 x2 y1 Find ( x1 x2 y1 y2 V) y2 V x1 x2 0.06 y1 0.94 y2 0.345 V 0.655 0.843 (d) Azeotrope Calculation Test for azeotrope at: T 1 ( 0 1 T) ( 60 273.15) K 2 ( 1 0 T) 19.863 1 ( 0 1 T) Psat1 ( T) Psat2 ( T) 120 120 2 ( 1 0 T) Psat2 ( T) 20.129 121 Psat1 ( T) 121 4.307 0.235 Since one of these values is >1 and the other is <1, an azeotrope exists. See Ex. 10.3(e). Guess: Given P 101.33 kPa x1 y1 x2 y2 0.3 0.3 1 1 x1 x1 y1 P = x1 1 ( x1 x2 T) Psat1 ( T) y2 P = x2 2 ( x1 x2 T) Psat2 ( T) x1 y1 x2 = 1 x1 = y1 y2 = 1 x1 x2 y1 Find ( x1 x2 y1 y2 P) y2 Paz Paz 31.18 kPa 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: A1 16.1154 B1 3483.67 K C1 205.807 K Water: A2 16.3872 B2 3885.70 K C2 230.170 K Psat1 ( ) T exp A1 Psat2 ( ) T exp A2 ( T B1 273.15 K) C1 kPa ( T B2 273.15 K) C2 kPa Parameters for the Wilson equation: 3 V1 a12 775.48 3 cm 75.14 mol 12 ( ) T V2 a21 cal mol cm 18.07 mol 1351.90 a12 V2 exp RT V1 exp x2 1 ( x2 T) x1 a21 V1 exp RT V2 21 ( ) T 12 ( ) T x2 12 ( ) x2 T x1 x1 exp 2 ( x2 T) x1 21 ( ) T cal mol x1 x2 x1 12 ( ) T 21 ( ) T 12 ( ) T x1 x2 12 ( ) x2 T x2 434 x1 21 ( ) T 21 ( ) T x1 21 ( ) T (a) BUBL T: P x1 Guess: T Given 0.3 x2 1 x1 y1 101.33 kPa 0.3 y2 1 y1 y1 P = x1 1 ( x1 x2 T) Psat1 ( T) ( 60 273.15) K y1 y2 = 1 y2 P = x2 2 ( x1 x2 T) Psat2 ( T) Tbubl y1 Find ( T y1 y2) y2 Tbubl y1 361.1 K y2 0.418 Ans. 0.582 (b) DEW T: P 101.33 kPa y1 0.3 y2 1 x1 Guess: T ( 60 x1 0.1 x2 1 y1 x1 x2 = 1 Given 273.15) K y1 P = x1 1 ( x1 x2 T) Psat1 ( T) y2 P = x2 2 ( x1 x2 T) Psat2 ( T) Tdew x1 Find ( T x1 x2) x2 Tdew x1 364.28 K 0.048 x2 Ans. 0.952 (c) P,T-flash Calculation T Tbubl Tdew Guess: Given P 2 V y1 = y2 = x1 ( 1 x1 y1 0.5 101.33 kPa 0.1 0.1 x1 1 ( x1 x2 T) Psat1 ( T) P x2 2 ( x1 x2 T) Psat2 ( T) P V) y1 V = z1 Eq. (10.15) 435 z1 0.3 x2 y2 1 1 y1 x1 x1 x2 = 1 y1 y2 = 1 x1 x2 y1 Find x1 x2 y1 y2 V) ( y2 V x1 x2 0.09 y1 0.91 y2 0.35 V 0.65 0.807 (d) Azeotrope Calculation Test for azeotrope at: P B1 Tb1 A1 C1 A2 P ln kPa 273.15 K Tb1 370.349 K C2 P ln kPa B2 Tb2 101.33 kPa 273.15 K Tb2 373.149 K 2 ( 0 Tb1) 3.779 1 1 ( 1 Tb2) 16.459 0 1 ( 1 T)Psat1 ( ) 0 Tb2 120 120 P 121 2 ( 0 T)Psat2 ( ) 1 Tb1 19.506 121 P 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 Given 273.15)K x1 0.4 x2 1 y1 y1 y1 P = x1 1 ( x2 T)Psat1 ( ) x1 x1 T y2 = 1 y2 1 x2 = 1 y2 P = x2 2 ( x2 T)Psat2 ( ) y1 x1 T 0.4 436 x1 = y1 x1 x1 x2 y1 Find ( x1 x2 y1 y2 T) y2 Taz Taz 360.881 K x1 0.4546 y1 0.4546 Ans. 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 16.1154 B1 3483.67 K C1 205.807 K Water: 16.3872 B2 3885.70 K C2 230.170 K A2 Psat1 ( T) exp A1 Psat2 ( T) exp A2 (T B1 273.15 K) C1 (T B2 273.15 K) C2 kPa kPa Parameters for the NRTL equation: b12 500.40 12 ( T) G12 ( T) cal mol b21 1636.57 b12 RT exp cal mol 21 ( T) 12 ( T) G21 ( T) 437 0.5081 b21 RT exp 21 ( T) 1 ( x2 T) x1 G21 ( ) T x1 x2 G21 ( ) T G12 ( ) 12 ( ) T T 2 exp x2 2 (2 x 2 ( x2 T) x1 2 21 ( ) T x1 G12 ( ) T) G12 ( ) T x2 x1 G12 ( ) T G21 ( ) 21 ( ) T T 2 exp x1 2 12 ( ) T 2 (1 x x2 G21 ( ) T) (a) BUBL T: P 101.33 kPa x1 0.3 x2 1 x1 Guess: T ( 60 y1 0.3 y2 1 y1 Given 273.15)K y1 P = x1 1 ( x2 T)Psat1 ( ) x1 T y1 y2 P = x2 2 ( x2 T)Psat2 ( ) x1 T y2 = 1 Tbubl y1 Find ( y1 y2) T y2 Tbubl y1 360.84 K 0.415 y2 Ans. 0.585 (b) DEW T: P 101.33 kPa y1 0.3 y2 1 x1 Guess: T ( 90 x1 0.05 x2 1 y1 Given 273.15)K y1 P = x1 1 ( x2 T)Psat1 ( ) x1 T x1 y2 P = x2 2 ( x2 T)Psat2 ( ) x1 T x2 = 1 Tdew x1 Find ( x1 x2) T x2 Tdew 364.27 K x1 0.042 438 x2 0.958 Ans. (c) P,T-flash Calculation Tdew T Tbubl P 2 Guess: Given V x1 y1 0.5 0.1 0.1 x1 1 ( x1 x2 T) Psat1 ( T) y1 = P V) 1 1 y1 x1 x2 = 1 y1 x2 2 ( x1 x2 T) Psat2 ( T) x1 ( 1 0.3 x1 P y2 = z1 x2 y2 101.33 kPa y2 = 1 y1 V = z1 Eq. (10.15) x1 x2 y1 Find ( x1 x2 y1 y2 V) y2 V x1 0.069 x2 y1 0.931 0.352 y2 0.648 V 0.816 (d) Azeotrope Calculation Test for azeotrope at: P B1 Tb1 A1 P ln kPa B2 Tb2 A2 1 ( 0 1 Tb2) P ln kPa 101.33 kPa C1 273.15 K Tb1 370.349 K C2 273.15 K Tb2 373.149 K 2 ( 1 0 Tb1) 14.699 439 4.05 1 ( 1 T)Psat1 ( ) 0 Tb2 120 120 P 121 2 ( 0 T)Psat2 ( ) 1 Tb1 17.578 121 P 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 x1 273.15)K 0.4 x2 1 y1 y1 y1 P = x1 1 ( x2 T)Psat1 ( ) x1 x1 T y2 = 1 1 x1 x2 = 1 y2 P = x2 2 ( x2 T)Psat2 ( ) y1 x1 T Given y2 0.4 x1 = y1 x1 x2 y1 Find x1 x2 y1 y2 T) ( y2 Taz Taz 360.676 K x1 0.4461 y1 Ans. 0.4461 12.20 Molar volumes & Antoine coefficients: 74.05 V 40.73 14.3145 A 18.07 Psat ( T) i 2756.22 16.5785 B 16.3872 exp Ai 3638.27 kPa 273.15 239.500 a 230.170 T ( 65 583.11 161.88 291.27 0 1448.01 469.55 440 273.15) K Ci 0 Wilson parameters: C 3885.70 Bi T K 228.060 107.38 0 cal mol Vj ( i j T) Vi exp ai j RT i j 13 (a) 0.3 p 13 BUBL P calculation: No iteration required. x1 13 x2 ( i x T) exp 1 x3 0.4 ln xj 1 x1 x2 ( i j T) j xp ( p i T) xj p ( p j T) j Pbubl xi ( i x T) Psat ( i T) yi xi ( i x T) Psat ( i T) Pbubl i 0.527 y 0.367 Pbubl Ans. 117.1 kPa 0.106 (b) DEW P calculation: y1 Guess: 0.3 y2 0.4 y3 1 y1 y2 x1 0.05 x2 0.2 x3 1 x1 x2 P Pbubl 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) i x1 x2 x3 xi = 1 Find x1 x2 x3 P Pdew 441 0.035 x 0.19 Pdew 69.14 kPa Ans. 0.775 (c) P,T-flash calculation: z1 0.3 Guess: V z2 Pdew P 0.4 Pbubl T 2 z3 1 z1 338.15 K z2 Use x from DEW P and y from BUBL P as initial guess. 0.5 Given P y1 = x1 ( x T)Psat1 T) 1 ( 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 yi = 1 i i x1 x2 x3 y1 Find x1 x2 x3 y1 y2 y3 V y2 y3 V 0.391 0.109 x 0.345 0.546 y 0.426 V 0.183 442 0.677 Ans. 12.21 Molar volumes & Antoine coefficients: Antoine coefficients: 74.05 V 14.3145 40.73 A 16.5785 18.07 T ( 65 B Psat ( i T) 0 T K i 13 0 b 0.2994 0.5343 0.2994 j 230.170 Bi exp Ai 0.3084 0.5343 0.3084 0.4 bi j RT x3 1 x1 Ci 253.88 1197.41 845.21 13 x2 273.15 0 Gi j l 13 k 13 (a) BUBL P calculation: No iteration required. 0.3 kPa 184.70 631.05 222.64 0 ij x1 239.500 3885.70 NRTL parameters: 0 C 3638.27 16.3872 273.15)K 228.060 2756.22 cal mol 0 exp ij ij x2 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 k Gl j xl j ij G l j xl l l Pbubl xi ( i x T) Psat ( i T) yi i 0.525 y 0.37 Pbubl 115.3 kPa 0.105 443 Ans. xi ( i x T) Psat ( i T) Pbubl (b) DEW P calculation: y1 y2 0.4 y3 1 y1 y2 x1 Guess: 0.3 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 ( xi = 1 P y3 = x3 ( x T)Psat3 T) 3 ( i x1 x2 Find x1 x2 x3 P x3 Pdew 0.038 x 0.192 Pdew 68.9 kPa Ans. 0.77 (c) P,T-flash calculation: z1 z2 0.3 Guess: V 0.5 0.4 Pdew P Pbubl 2 z3 1 z1 T 338.15 K z2 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 yi = 1 xi = 1 i i 444 x1 x2 x3 y1 Find x1 x2 x3 y1 y2 y3 V y2 y3 V 0.391 0.118 x y 0.347 0.426 V Ans. 0.667 0.183 0.534 12.22 Molar volumes & Antoine coefficients: 74.05 V 14.3145 40.73 A 16.5785 18.07 Psat ( i T) 2756.22 B 3638.27 16.3872 T K C 273.15 Wilson parameters: 230.170 P kPa a 161.88 291.27 583.11 0 1448.01 469.55 Vj Vi exp ai j RT (a) 0.3 13 j BUBL T calculation: x1 i x2 0.4 x3 445 101.33kPa Ci 0 ( i j T) 239.500 3885.70 Bi exp Ai 228.060 1 x1 x2 13 107.38 cal mol 0 p 13 ( x T) i exp 1 ln xj ( j T) i j xp ( i T) p xj p ( j T) p j Guess: T 300K y1 0.3 y2 0.3 y3 1 y1 y2 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 ( P= xi ( x T)Psati T) i ( i y1 y2 Find y1 y2 y3 T y3 Tbubl 0.536 y 0.361 Tbubl Ans. 334.08K 0.102 (b) DEW T calculation: y1 Guess: 0.3 y2 0.4 y3 1 y1 y2 x1 0.05 x2 0.2 x3 1 x1 x2 T Tbubl 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 ( xi = 1 i 446 x1 x2 Find x1 x2 x3 T x3 Tdew 0.043 x 0.204 Tdew Ans. 347.4 K 0.753 (c) P,T-flash calculation: z1 0.3 Guess: V z2 0.5 0.2 Tdew T Tbubl 2 z3 1 z1 T 340.75 K z2 Use x from DEW P and y from BUBL P as initial guess. Given P y1 = x1 ( 1 x T) Psat ( 1 T) x1 ( 1 V) y1 V = z1 P y2 = x2 ( 2 x T) Psat ( 2 T) x2 ( 1 V) y2 V = z2 P y3 = x3 ( 3 x T) Psat ( 3 T) x3 ( 1 V) y3 V = z3 xi = 1 i yi = 1 i x1 x2 x3 y1 Find x1 x2 x3 y1 y2 y3 V y2 y3 V 447 0.536 0.125 x y 0.17 0.241 V Ans. 0.426 0.223 0.705 12.23 Molar volumes & Antoine coefficients: Antoine coefficients: 74.05 V 14.3145 40.73 A 16.5785 18.07 P B Psati T) ( 0 T K i 0 13 j b 0.2994 0.5343 0.2994 273.15 0 13 G ( j T) i 13 exp ( j T) i ij cal mol 0 ( j T) i k Ci 253.88 1197.41 845.21 l kPa 184.70 631.05 222.64 0 13 230.170 Bi exp Ai 0.3084 0.5343 0.3084 239.500 3885.70 NRTL parameters: 0 C 3638.27 16.3872 101.33kPa 228.060 2756.22 bi j RT (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 T) i k G ( j T)xl l G ( j T)xl l j l l 448 Guess: T y1 300K y2 0.3 y3 0.3 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= xi ( i x T) Psat ( i T) i y1 y2 Find y1 y2 y3 T y3 Tbubl 0.533 y 0.365 Tbubl Ans. 334.6 K 0.102 (b) DEW T calculation: y1 Guess: 0.3 y2 0.4 y3 1 y1 y2 x1 0.05 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) xi = 1 P y3 = x3 ( 3 x T) Psat ( 3 T) i x1 x2 Find x1 x2 x3 T x3 Tdew 0.046 x 0.205 Tdew 347.5 K 0.749 449 Ans. (c) P,T-flash calculation: z1 0.3 Guess: V z2 Tdew T 0.2 Tbubl 2 z3 1 z1 T 341.011 K z2 Use x from DEW P and y from BUBL P as initial guess. 0.5 Given P y1 = x1 ( x T)Psat1 T) 1 ( 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 yi = 1 i i x1 x2 x3 y1 Find x1 x2 x3 y1 y2 y3 V y2 y3 V 0.537 0.133 x 0.173 0.694 y 0.238 V 0.225 450 0.414 Ans. 3 3 12.26 x1 x2 0.4 1 V1 x1 110 cm V2 mol 90 x1 x2 45 x1 25 x2 By Eq. (12.27): V x1 x2 cm VE x1 x2 mol VE x1 x2 mol 3 3 VE x1 x2 cm x1 V1 7.92 cm mol x2 V2 3 V x1 x2 105.92 cm mol By Eqs. (11.15) & (11.16): Vbar1 V x 1 x2 Vbar2 3 d x2 V x1 x2 dx1 V x1 x2 x1 cm 190.28 mol Vbar1 Ans. 3 d dx1 V x1 x2 Vbar2 49.68 cm mol Check by Eq. (11.11): 3 V x1 Vbar1 x2 Vbar2 V cm 58.63 mol moles V2 3 x1 750 cm V1 moles1 moles2 moles moles2 moles1 x1 moles 1500 cm V2 25.455 mol x2 0.503 1 x1 x2 1.026 0.220 x1 x2 cm mol x1 3 3 VE OK cm 118.46 mol 3 moles1 cm mol 3 3 12.27 V1 105.92 VE cm 0.256 mol 3 By Eq. (12.27), V VE x1 V1 451 x2 V 2 V 88.136 cm mol Vtotal Vtotal V moles 3 2243 cm Ans. For an ideal solution, Eq. (11.81) applies: Vtotal x1 V 1 x2 V2 moles Vtotal 3 2250 cm 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 441579 J (Pg. 457) H3 H H1 H 589 J (Table C.4) 2 ( 285830 J) H2 H3 (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 12.29 53.55 J Ans. 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 H 39.2 kJ H2 Ans. 452 12.30 Calculate moles of LiCl and H2O in original solution: nLiCl nLiCl 0.1 125 kmol 42.39 0.295 kmol 3 20 kmol 42.39 nH2O n'LiCl 0.472 kmol 21.18 nLiCl nH2O Mole ratio, final solution: kmol 6.245 10 mol n'LiCl Mole ratio, original solution: n'LiCl 18.015 nH2O Moles of LiCl added: nLiCl 0.9 125 nH2O nLiCl n'LiCl 8.15 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 H2 Q nLiCl 35 nLiCl H1 kJ mol (Fig. 12.14, n=21.18) n'LiCl H2 32 Q kJ mol (Fig. 12.14, n=8.15) 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 104.8 H1 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 kJ mol 25.5 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 M1 H1 0.2 mol H2 H 0.061 kJ M1 H3 0.2 mol 33.16 kJ mol 10.5 0.2 mol H3 Close enough 0.2 mol M1 1 mol x 0.8 mol x Ans. 0.087 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 104.8 H2 20.756 H3 25.5 H H1 kJ kg kJ mol kJ mol H2 21.01 kJ gm 18.015 kg mol H1 1.509 kJ mol From p. 457 ( H LiCl(s) - H LiCl in 3 mol H2O) From Figure 12.14 H3 0.2 mol 454 H 646.905 J Ans. 12.33 (a) LiCl + 4 H2O -----> LiCl(4H2O) H 0.2 mol H 5.1 kJ 25.5 kJ From Figure 12.14 mol 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 H2 25.5 H kJ mol From p. 457 ( H LiCl(s) - H LiCl in 3 mol H2O) kJ From Figure 12.14 mol 0.2 mol H1 H H2 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 712.58 kJ mol H2 285.83 kJ mol H3 408.61 kJ mol H4 25.5 H (d) 0.2 mol kJ mol H1 From p. 457 for LiCl.H2O From Table C.4 Hf H2O(l) From p. 457 for LiCl From Figure 12.14 H2 H3 H4 H 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 1.472 kJ Ans. H1 25.5 4 H2 H (e) 9 kJ mol () 32.4 0.2 mol From Figure 12.14 kJ From Figure 12.14 mol H1 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 kJ 5 ( 20.756) mol 6 From p. 457 ( H LiCl(s) - H LiCl in 3 mol H2O) H2 kJ 1 () 32.4 mol 6 From Figure 12.14 H3 H (f) kJ mol 25.5 0.2 mol H1 From Figure 12.14 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) H1 H2 kJ 5 ( 712.58) mol 8 5 8 ( 285.83) kJ mol From p. 457 for LiCl.H2O From Table C.4 Hf H2O(l) 456 kJ 3 ( 32.4) mol 8 H3 5 H4 8 H5 H 12.34 ( 408.61) kJ mol From p. 457 for LiCl kJ 25.5 0.2 mol From Figure 12.14 From Figure 12.14 mol H1 H2 H3 H4 H H5 0.403 kJ Ans. BASIS: 1 second, during which the following are mixed: (1) 12 kg hydrated (6 H2O) copper nitrate (2) 15 kg H2O n1 n1 12 kmol 295.61 sec 0.041 15 kmol 18.015 sec n2 kmol sec n2 0.833 6 n1 Mole ratio, final solution: kmol sec n2 26.51 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 H2 H (Table C.4) 6 ( 285.83 kJ) H3 302.9 kJ H1 H2 H3 H4 ( 2110.8 kJ) H H4 457 45.08 kJ 47.84 kJ 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 H3 2 ( 436.805 kJ) H4 ( 439.288 kJ) H H1 (Table C.4) 3 ( 285.83 kJ) Q H2 Q H H3 H H4 (Pg. 457) 19.488 kJ Ans. 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) 2 ( 285.83 kJ) H3 1012.65 kJ Since the process is isothermal, H= H1 H2 Since it is also adiabatic, Therefore, H1 H2 H2 (Table C.4) H3 H=0 H1 H3 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 Ans. 0.1012 458 10 867.85 20 870.06 25 n 862.74 15 12.37 Data: 871.07 Hf 50 872.91 kJ 100 873.82 300 874.79 500 875.13 1000 875.54 Ca + Cl2 + n H2O ---> CaCl2(n H2O) CaCl2(s) ---> Ca + Cl2 HfCaCl2 -------------------------------------------CaCl2(s) + n H2O ---> CaCl2(n H2O) From Table C.4: i Hf HfCaCl2 Htilde 795.8 kJ 1 rows ( n) 65 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) (Table C.4) H1 H2 2 ( 865.295 kJ) H 12.39 795.8 kJ H1 H2 H3 Q H3 871.07 kJ H Q 63.72 kJ Ans. 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 H2 H3 2607.9 kJ 871.8 kJ H298 H1 H298 21.12 kJ 6 ( 285.83 kJ) (Table C.4) (Pb. 12.37) H2 H3 for reaction at 25 degC msoln ( 110.986 msoln 739.908 gm 460 34.911 18.015)gm CP T 12.43 m1 H1 3.28 kJ kg degC 8 25 % m2 m2 m1 47.5 % m2 m3 H3 23 m2 H2 BTU (Fig. 12.17) lbm (Final soln.) H3 m1 H1 (25% soln.) 350 lb H2 msoln CP 16.298 degC Ans. T T m2 BTU lbm H298 T CP T = 0 25 degC 150 lb (H2SO4) m1 Q T 8.702 degC 100 % m1 m3 H298 90 Q BTU lbm (Fig. 12.17) 38150 BTU Ans. 12.44 Enthalpies from Fig. 12.17. x1 H1 20 HE H x2 0.5 BTU 1 H (pure H2SO4) lbm x1 H1 69 H2 x1 108 HE x2 H2 BTU lbm BTU lbm 133 12.45 (a) m1 400 lbm 175 lbm (10% soln. at 200 degF) H1 100 35 % m1 BTU lbm 10 % m2 m1 m3 m2 m1 m2 H2 BTU 152 lbm H3 41 (Fig. 12.19) (Final soln) 27.39 % BTU lbm 461 (pure H2O) BTU Ans. lbm (35% soln. at 130 degF) m2 (50 % soln) (Fig. 12.19) Q m3 H3 m1 H1 Q 43025 BTU Ans. H3 m2 H2 115.826 (b) Adiabatic process, Q = 0. m1 H1 H3 m2 H2 m3 BTU lbm From Fig. 12.19 the final soln. with this enthalpy has a temperature of about 165 degF. 12.46 m1 25 lbm (feed rate) sec 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 x1 m1 m2 Q x2 m2 H2 BTU lbm (Table F.4, 1.5(psia) & 217 degF] lbm m2 7.143 m3 H3 m1 H1 m3 m2 m3 Q sec m1 17.857 20880 BTU sec Ans. 12.47 Mix m1 lbm NaOH with m2 lbm 10% soln. @ 68 degF. BASIS: m2 1 lbm x3 m1 1 lbm (guess) m2 = m3 m1 x2 m2 = x3 m3 m1 0.385 lbm Given m1 0.35 x2 0.1 m3 m1 m2 m1 m3 Find m1 m3 462 m3 1.385 lbm lbm sec From Example 12.8 and Fig. 12.19 H1 BTU lbm 478.7 H3 m1 H1 H2 m3 BTU lbm H3 m2 H2 43 164 BTU lbm 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 8.712 4 10 J Mix 1 mol or 98.08 gm H2SO4(l) with m gm H2O to form a 50% solution. mH2SO4 mH2O 98.08 gm msoln msoln mH2SO4 0.5 mH2SO4 Data from Fig. 12.17: HH2SO4 HH2O Hsoln 0 45 70 BTU lbm [pure acid @ 77 degF (25 degC)] BTU lbm [pure water @ 77 degF (25 degC)] BTU [50% soln. @ 140 degF (40 deg C)] lbm Hmix msoln Hsoln Hmix 18.145 kg Q H298 msoln mH2SO4 HH2SO4 mH2O HH2O BTU lbm Hmix Q 283 463 BTU lbm Ans. 12.49 m1 H1 65 x1 140 lbm H2 BTU lb m2 0.15 0.8 (Fig. 12.17 at 160 degF) BTU lb (Fig. 12.17 at 100 degF) m2 102 x2 230 lbm x3 m3 m1 Q 20000 BTU H3 92.9 BTU lbm H3 m1 x1 m2 x2 x3 m3 Q m1 H1 55.4 % m2 H2 m3 From Fig. 12.17 find temperature about 118 degF 12.50 Initial solution (1) at 60 degF; Fig. 12.17: m1 x1 1500 lbm H1 0.40 98 BTU lbm Saturated steam at 1(atm); Table F.4: m3 m2 x3 m2 m2 m1 x1 m1 m1 125 lbm H2 m2 m2 H3 m2 x3 m2 m1 H1 1150.5 BTU lbm m2 H2 m3 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% This is about as good a result as we can get. 464 H3 m2 5.5 BTU lbm 12.51 Initial solution (1) at 80 degF; Fig. 12.17: m1 x1 1 lbm H1 0.45 95 BTU lbm Saturated steam at 40(psia); Table F.4: m3 m2 m1 H2 m2 x1 m1 x3 m2 m2 m1 m2 0.05 lbm m1 H1 H3 m2 x3 m2 1169.8 BTU lbm m2 H2 m3 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 x1 1 lbm 0.40 H1 77 BTU lbm Saturated steam at 35(psia); Table F.4: BTU lbm H2 1161.1 m3 m1 H3 x3 0.38 m2 m3 1.053 lbm m2 x1 m1 m1 m1 H1 m2 H2 m3 H3 131.2 m2 BTU lbm x3 0.053 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: H1 56 x1 0.35 8 BTU lbm x2 BTU lbm H 1 x1 H2 68 H BTU lbm H H x1 H 1 x2 H2 BTU lbm Ans. 103 12.54 BASIS: 1(lbm) of soln. Read values for H1 & H2 from Fig. 12.17 at 80 degF: BTU lbm H1 4 Q= H2 H=H H x1 H 1 x1 H 1 x2 H2 48 BTU lbm x1 0.4 x2 1 x1 x2 H2 = 0 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: Final solution: x1 x2 2 98.08 2 98.08 15 18.015 3 98.08 3 98.08 14 18.015 Data from Fig. 12.17 at 100 degF: HH2O H1 68 75 BTU lbm BTU lbm HH2SO4 H2 9 101 466 BTU lbm BTU lbm x1 0.421 x2 0.538 Unmix the initial solution: Hunmix x1 HH2SO4 Hunmix 118.185 1 x1 HH2O H1 BTU lbm 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) HfSO3 395720 HfH2SO4 Hrx J mol 285830 J mol J mol 813989 HfH2SO4 HfH2O HfH2O HfSO3 Hrx 1.324 5J 10 mol Finally, mix the constituents to form the final solution: Hmix Q H2 x2 HH2SO4 1 x2 HH2O Hunmix ( 98.08 2 1 lbmol Hrx 14 18.015)lb 137.231 BTU lbm 15 18.015)lb Hmix ( 98.08 3 Hmix Q 76809 BTU Ans. 12.56 Read values for H(x1=0.65), H1, & H2 from Fig. 12.17 at 77 degF: H 125 x1 0.65 BTU H1 lbm x2 0 1 BTU x1 45 H lbm H2 H H 467 BTU lbm x1 H 1 140.8 BTU lbm x2 H2 Ans. From the intercepts of a tangent line drawn to the 77 degF curve of Fig. 12.17 at 65%, find the approximate values: Hbar1 BTU lbm 136 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 m1 42 % 1 lb By a mass balance: x3 = 0.75 m1 m1 12.58 (a) m1 x1 0.75 m1 m2 m2 m1 x3 25 lbm m2 40 lbm 0 x2 m2 1 Ans. 0.786 lbm m3 75 lbm x3 0.25 H3 7 x4 0.42 Enthalpy data from Fig. 12.17 at 120 degF: H1 88 BTU lbm m4 m1 m2 x4 x1 m1 BTU lbm H2 m4 m3 x2 m2 14 140 lbm x3 m3 m4 BTU lbm H4 63 Q m4 H4 BTU lbm (Fig. 12.17) m1 H1 m2 H2 m3 H3 468 Q 11055 BTU Ans. (b) First step: m1 40 lb x1 1 H1 14 m2 75 lb x2 0.25 H2 7 m3 m1 x3 m2 x3 x1 m1 x2 m2 H3 m3 H3 0.511 Q m1 H1 BTU lbm BTU lbm m2 H2 m3 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 H298 1.791 285830 ( 425609 92307)J 5 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 H4 3.88 kJ H H1 H3 44.50 kJ Q H Q 62187 J 469 H2 H3 Ans. H4 68.50 kJ 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: 1 40.00 1 40.00 9 18.015 x1 x1 19.789 % HH2O 45 BTU lbm (Table F.3, sat. liq. at 77 degF) Hsoln 35 BTU lbm (Fig. 12.19 at x1 and 77 degF) HNaOH BTU lbm 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 Cp R molwt HNaOH H H molwt 0.121 HNaOH Hsoln 0.224 16.316 10 K Cp ( 77 x1 HNaOH kJ gm 40.00 3 T Cp 68)rankine 1 H x1 molwt This is for 1 gm of SOLUTION. H 45.259 kJ mol 470 0.245 HNaOH x1 HH2O However, for 1 mol of NaOH, it becomes: H gm mol BTU lbm rankine 480.91 BTU lbm 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 179067 J (Pb. 12.59) (Fig. 12.14 with sign change) H2 74.5 kJ H3 45.259 kJ H4 3.88 kJ H H1 (See above; note sign change) (given) H2 H3 H3 Q H 14049 J Ans. Q 12.61 Note: The derivation of the equations in part a) can be found in Section B of this manual. 0.1 0.2 144.21 0.3 208.64 0.4 262.83 0.5 x1 73.27 302.84 0.6 HE 323.31 0.7 320.98 0.8 279.58 0.85 237.25 0.9 178.87 0.95 100.71 471 kJ kg x2 1 x1 H HE x1 x2 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 3 w w 3 n 3 n a H regress x1 kJ kg b a 735.28 b 3 824.518 c c d d H x1 a b x1 c x1 2 d x1 3 195.199 914.579 kJ kg Using the equations given in the problem statement and taking the derivatives of the polynomial analytically: HEbar1 x1 1 HEbar2 x1 x1 2 x1 H 2 x1 H x1 1 x1 x1 472 b 2 c x1 3 d x1 b 2 c x1 3 d x1 2 2 kJ kg kJ kg 0 500 (kJ/kg) 1000 1500 2000 2500 0 0.2 0.4 0.6 0.8 x1 H/x1x2 HEbar1 HEbar2 12.62 Note: This problem uses data from problem 12.61 0.1 0.2 144.21 0.3 208.64 0.4 262.83 0.5 x1 73.27 302.84 0.6 HE 323.31 0.7 320.98 0.8 279.58 0.85 237.25 0.9 178.87 0.95 100.71 473 kJ kg x2 1 x1 H HE x1 x2 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 3 w w 3 n n 3 a H regress x1 b a 735.28 b 3 kJ kg 824.518 c c d d 195.199 914.579 By the equations given in problem 12.61 x1 a H x1 H H b x1 c x1 x1 x1 1 Hbar1 x1 1 Hbar2 x1 x1 x1 2 H 2 d x1 3 kJ kg x1 2 H x1 x1 1 x1 x1 2 kJ kg 2 kJ b 2 c x1 3 d x1 b 2 c x1 3 d x1 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 Ht = ( 4000 m) H Solving for m: 4000H2 m H3 474 m x1 ( 4000kg)x1 0.9 x1 Eq. (A) . 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( ) 0.9 H3 178.737 H H( ) 0.5 H 303.265 kJ kg kJ kg Define quantities as a function of x 1 Q x1 m x1 4000kg m x1 H x1 m x1 H3 ( 000kg) 1 4 x 0.9 Qtx1 m ( ) 5000 kg 0.5 x1 4000kg m x1 H m x1 H3 Qt0.5) ( 1.836 6 10 kJ Since the heat transfer rate q is constant: q Qtx1 and x1 Q x1 Eq. (C) q 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 x1 0 0.01 0.5 1200 1100 1000 r x1 kg 900 hr 800 700 600 0 1 2 3 4 5 6 x1 hr 12.64 mdot1 20000 lb hr x1 0.8 T1 120degF H1 Enthalpies from Fig. 12.17 x2 0.0 T2 40degF H2 x3 0.5 T3 140degF H3 92 BTU lb 7 70 a) Use mass balances to find feed rate of cold water and product rate. Guess: mdot2 Given mdot1 mdot1 mdot3 mdot2 = mdot3 mdot1 x1 mdot2 x2 = mdot3 x3 476 2mdot1 Total balance H2SO4 balance BTU lb BTU lb 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 mdot3 H3 54.875 BTU lb 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: Substituting -Vdot for dL/dt: L Rearranging this equation gives: L Substituting -dL/dt for Vdot: dx1 L L Eliminating dt and rearranging: x1 dt dx1 x1 ( Vdot)= y1 Vdot dt dx1 dt dx1 dt = x1 y1 Vdot = y1 x1 dx1 y1 477 dL = Vdot y1 dt x1 = dL L dL dt At low concentrations y1 and x1 can be related by: y1 = Psat1 inf1 where: x1 = K1 x1 P dx1 Substituting gives: K1 1 x1 = Integrating this equation yields: Psat1 K1 = inf1 P dL L ln Lf 1 = L0 K1 x1f ln 1 x10 where L0 and x10 are the initial conditions of the system For this problem the following values apply: L0 1mol 600 x10 10 T 130degC Psat1 P exp 16.3872 inf1 Lf L0 exp 6 T degC K1 K1 norg0 L0 1 norg0 1 ln kPa Psat1 230.170 270.071 kPa 15.459 x1f Lf x10 x10 norg0 5.8 inf1 0.999 mole %lossorg 6 10 3885.70 P 1 50 1atm Psat1 K1 x1f 0.842 mole norgf norgf norgf Lf 1 0.842 mole %lossorg norg0 x1f 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 V2 12 V1 exp cal mol a21 a12 RT V1 74.05 V1 0.708 12 3 3 cal 583.11 mol 21 V2 exp cm mol V2 40.73 a21 21 RT cm mol 0.733 ln inf1 ln 12 1 21 ln inf1 0.613 Ans. ln inf2 ln 21 1 12 ln inf2 0.603 Ans. From p. 445 From Fig. 12.9(b) ln inf2 = 0.61 ln inf1 = 0.62 For NRTL equation b12 184.70 cal mol b21 b12 12 G12 cal mol 12 0.3048 b21 12 RT exp 222.64 0.288 21 G12 0.916 G21 21 RT exp 21 0.347 G21 0.9 ln inf1 21 12 exp 12 ln inf1 0.611 ln inf2 From p. 446 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 x1 Psat2 183.4kPa y1 0.253 96.7kPa P 0.456 139.1kPa Check whether or not the system is ideal using Raoult's Law (RL) PRL x1 Psat1 Since PRL<P, 1 1 and PRL x1 Psat2 2 118.635 kPa are not equal to 1. Therefore, we need a model for GE/RT. A two parameter model will work. GE = x1 x2 A21 x1 RT From Margules Equation: ln 1 = x2 ln 2 = x1 Find 1 and 2 Use the values of find A12 and A21. Given A12 1 and A21 2 A12 x1 Eq. (12.10a) A21 2 A12 A21 x2 Eq. (12.10b) A21 ln 1 = 1 x1 2 2 A21 exp 1 2 x1 exp x1 2 x1 2 A21 A12 1 2 x1 Psat2 1.048 A12 0.5 2 A21 2 A12 A12 Find A12 A21 1 x1 y1 P 1 2 at x1=0.253 and Eqs. (12.10a) and (12.10b) to 0.5 ln 2 = x1 A12 2 A21 1.367 1 x1 Psat1 Guess: 2 A12 at x1=0.253 from the given data. y1 P 1 2 A12 x2 1 A21 480 A12 x1 1 x1 A21 0.644 2 A21 2 A12 A21 A12 x1 x1 Eq. (12.10a) Eq. (12.10b) 0.478 a) x1 x1 1 x1 Psat1 P y1 0.5 P x1 1 x1 Psat1 b) 1inf 1 exp A12 12 1.904 3.612 P 0.743 121 Ans. 160.148 kPa exp A21 2inf 120 Psat2 Since 2 x1 Psat2 1inf 1inf Psat1 120 x1 y1 Psat1 2inf Psat2 Ans. 2inf 1.614 121 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 T x1 108.6kPa ( 35 0.389 Psat1 273.15) K Psat2 120.2kPa 73.9kPa Check whether or not the system is ideal using Raoult's Law (RL) PRL x1 Psat1 Since PRL < P, 1 1 and x1 Psat2 2 PRL 91.911 kPa 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 A Find A) ( 1 x1 a) y1 b) P c) 1inf 120 P = x1 exp A 1 A exp A 1 x1 1 x1 exp ( A) 1inf Psat1 2 Psat1 1 x1 exp A x1 2 Psat2 0.677 x1 2 Psat1 2 x1 y1 P x1 1 x1 Psat1 Psat2 x1 1 1inf 120 x1 0.554 2inf 3.201 2 Ans. 2 x1 Psat2 1.968 exp A x1 121 P 110.228 kPa exp ( A) Psat1 2inf Psat2 2inf 121 Ans. 1.968 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= 1 1 1 1= 0 n0 = 1 1= 2 i By Eq. (13.5). yH = yCO = 2 1 2 yH2O = yCO = 2 2 By Eq. (A) and with data from Example 13.13 at 1000 K: T 1000 kelvin 1 G 2 RT 2 Guess: e d Given J mol 1 ( 395790) G de 1 2 ln 2 2 ( 192420 2 2 ln 200240) J mol 2 0.5 e =0 J e mol Find e e 0.3 0.31 0.6 2.082 2.084 G 5 10 2.086 2.088 0.2 0.3 0.4 483 0.5 0.6 0.45308 13.5 (a) H2(g) + CO2(g) = H2O(g) + CO(g) = i= 1 1 1 1= 0 n0 = 1 1= 2 i By Eq. (13.5). 1 yH = yCO = 2 2 yH2O = yCO = 2 2 By Eq. (A) and with data from Example 13.13 at 1100 K: T 1100 kelvin 1 G ( 395960) 2 RT 2 Guess: 1 2 d G de 2 ( 187000 2 2 2 ln 209110) J mol 2 0.5 e Given ln J mol 1 e =0 J mol e Find e e 0.502 Ans. 0.35 0.36 0.65 2.102 2.103 2.104 G 5 10 2.105 2.106 2.107 0.3 0.35 0.4 0.45 484 0.5 0.55 0.6 0.65 (b) H2(g) + CO2(g) = H2O(g) + CO(g) = i= 1 1 1 1= 0 n0 = 1 1= 2 i By Eq. (13.5), 1 yH = yCO = 2 2 yH2O = yCO = 2 2 By Eq. (A) and with data from Example 13.13 at 1200 K: T 1200 kelvin 1 G ( 396020) 2 RT 2 Guess: e d Given G de 1 ln 2 J 2 mol 1 ( 181380 2 2 2 ln 217830) J mol 2 0.1 e =0 J mol e Find e e 0.53988 0.6 0.65 Ans. 0.4 0.41 0.7 2.121 2.122 2.123 G 5 2.124 10 2.125 2.126 2.127 0.35 0.4 0.45 485 0.5 0.55 0.7 (c) H2(g) + CO2(g) = H2O(g) + CO(g) = i= 1 1 1 1= 0 n0 = 1 1= 2 i By Eq, (13.5), yH = yCO = 2 1 2 yH2O = yCO = 2 2 By Eq. (A) and with data from Example 13.13 at 1300 K: T 1300 kelvin 1 G ( 396080) 2 RT 2 Guess: e d Given G de 1 ln 2 J mol 1 2 ( 175720 2 2 2 ln 226530) J mol 2 0.6 e =0 J mol e Find e e 0.57088 Ans. 0.4 0.41 0.7 2.14 2.142 G 5 2.144 10 2.146 2.148 0.35 0.4 0.45 486 0.5 0.55 0.6 0.65 0.7 13.6 H2(g) + CO2(g) = H2O(g) + CO(g) = 1 i= 1 1 1= 0 n0 = 1 1= 2 i By Eq, (13.5), 1 yH = yCO = 2 2 yH2O = yCO = 2 2 With data from Example 13.13, the following vectors represent values for Parts (a) through (d): 1000 3130 1100 T 1200 150 G kelvin 3190 1300 J mol 6170 Combining Eqs. (13.5), (13.11a), and (13.28) gives 2 1 2 2 = 1 2 1 2 = K = exp G RT 2 0.4531 exp 0.5021 G RT Ans. 0.5399 1 0.5709 13.11 4HCl(g) + O2(g) = 2H2O(g) + 2Cl(g) =1 H298 n0 = 6 114408 J mol T 773.15 kelvin G298 487 75948 T0 J mol 298.15 kelvin The following vectors represent the species of the reaction in the order in which they appear: 4 3.156 1 3.639 A 2 A 3 B G 1.267 0.121 D i Bi i Di i 8 5 10 C D 0 4 8.23 10 H298 G B 5 10 0.344 i 0.439 0.227 1 end i A D 0.089 i i Ai 10 1.450 4.442 rowsA) ( 0.151 0.506 B 3.470 2 end 0.623 K T H298 G298 T0 R IDCPH T0 T A B C D R T IDCPS T0 T A B C D 10 4J mol G RT exp K By Eq. (13.5) yO2 = yHCl = 1 yH2O = 6 Apply Eq. (13.28); yHCl 5 5 4 4 0.3508 6 1 yO2 6 yO2 5 4 6 2 0.0397 yCl2 = 6 2 6 (guess) 0.5 4 2 Given yHCl 7.18041 1 yH2O 6 yH2O 488 0.793 Find = 2K 0.3048 2 6 yCl2 yCl2 0.3048 2 6 Ans. 13.12 N2(g) + C2H2(g) = 2HCN(g) n0 = 2 =0 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 A T J mol 42720 G298 B 0.060 T0 C J mol 5 D 0 0.191 10 298.15 kelvin T H298 G298 T0 R IDCPH T0 T A B C D R T IDCPS T0 T A B C D G G 3 0.173 10 923.15 kelvin 39430 H298 3.242 4J 10 K mol exp G RT K 0.01464 By Eq. (13.5), yN2 = 1 yC2H4 = 2 By Eq. (13.28), 2 Given yN2 yN2 0.5 1 1 2 0.4715 1 yHCN = 2 2e 2 = (guess) 2 yC2H4 yC2H4 0.057 Find =K 1 2 0.4715 yHCN 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) n0 = 2.5 =1 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 A T J 68910 mol B 1.424 G298 3 1.601 10 T0 623.15 kelvin 39630 C J mol 0.156 10 6 5 D 0.083 10 298.15 kelvin T H298 G298 T0 R IDCPH T0 T A B C D R T IDCPS T0 T A B C D G H298 G 6.787 10 3J K mol By Eq. (13.5), yCH3CHO = 1 Given yCH3CHO yCH3CHO yH2 = 2.5 By Eq. (13.28), exp 0.5 2.5 1 1.5 = 3K 1 2.5 yH2 0.108 yH2 1.5 2.5 0.4053 G RT K 1.5 3.7064 yC2H5OH = 2.5 2.5 (guess) 0.818 Find yC2H5OH yC2H5OH 2.5 0.4867 Ans. If the pressure is reduced to 1 bar, Given yCH3CHO yCH3CHO 2.5 1 1.5 = 1K 1 2.5 0.1968 yH2 yH2 1.5 2.5 0.4645 490 0.633 Find yC2H5OH yC2H5OH 2.5 0.3387 Ans. 13.14 C6H5CH:CH2(g) + H2(g) = C6H5.C2H5(g) n0 = 2.5 =1 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 J 117440 mol G298 B 923.15 kelvin T 3 4.766 10 T0 A 4.175 J mol 83010 C 1.814 10 6 D 0.083 10 5 298.15 kelvin T H298 G298 T0 R IDCPH T0 T A B C D R T IDCPS T0 T A B C D G H298 G 2.398 10 3J mol By Eq. (13.5), yH2 = K G 1.5 yC6H5C2H5 = 2.5 0.5 2.5 1 yC6H5CHCH2 yC6H5CHCH2 1.5 1 2.5 0.2794 yH2 491 1.36672 1 2.5 2.5 (guess) = 1.0133 K yH2 K RT yC6H5CHCH2 = By Eq. (13.28), Given exp 1.5 2.5 0.5196 0.418 Find yC6H5C2H5 yC6H5C2H5 2.5 0.201 Ans. 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 n0 = 1 = 0.5 By Eq. (13.5), ySO2 = 0.15 1 0.20 yO2 = 0.5 1 0.5 ySO3 = 0.5 1 0.5 From data in Table C.4, H298 J mol 98890 G298 70866 J mol The following vectors represent the species of the reaction in the order in which they appear: 1 5.699 0.5 A 3.639 1 end A i Ai 1.015 0.506 10 3 2.028 D i Bi 2 10 T0 753.15 kelvin i Di i i B 0.227 10 1 end B 0.5415 5 D 1.056 i i T B 8.060 rowsA) ( A 0.801 6 C 0 D 8.995 4 10 298.15 kelvin T H298 G298 T0 R IDCPH T0 T A B C D R T IDCPS T0 T A B C D G H298 G 2.804 10 4J mol K By Eq. (13.28), 0.1 1 Given 0.15 exp 0.5 0.2 G RT K 88.03675 (guess) 0.5 0.5 0.5 492 =K Find 0.1455 By Eq. (13.4), nSO3 = = 0.1455 By Eq. (4.18), H753 H298 J mol R IDCPH T0 T Q Q H753 13.16 C3H8(g) = C2H4(g) + CH4(g) A B 14314 C D J Ans. mol =1 H753 98353 Basis: 1 mole C3H8 feed. By Eq. (13.4) n0 Fractional conversion of C3H8 = By Eq. (13.5), nC3H8 n0 1 yC3H8 = nC3H8 = 1 yC2H4 = 1 = 1 1 1 = yCH4 = 1 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 1.213 1 A 1 end 1.424 i rowsA) ( i Ai 8.824 14.394 10 3 625 kelvin B 4.392 10 6 2.164 1 end C i Bi i Ci i i 1.913 C 9.081 B i (a) T B 1.702 A A 28.785 5.31 10 T0 493 3 C 2.268 298.15 kelvin 10 6 D 0 T H298 G298 T0 R IDCPH T0 T A B C D R T IDCPS T0 T A B C D G H298 G 2187.9 J mol K G RT exp By Eq. (13.28), K 1.52356 (guess) 0.5 2 Given 1 1 Find =K This value of epsilon IS the fractional conversion. Ans. 0.777 2 (b) K 0.85 G R T ln ( K) G 1 K 1 4972.3 J mol 2.604 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. By Eq. (13.5), n0 = 1.5 yC2H6 = 1 yH = 1.5 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 1.131 1 A 19.225 3.249 1 B 0.422 1.424 14.394 5.561 C 0.0 0.0 6 10 5 D 0.083 10 4.392 end i rowsA) ( A i Ai A T 0.0 1 end B C i Bi 3.542 B i Di i i 3 4.409 10 T0 1100 kelvin D i Ci i i 3 10 C 1.169 10 6 D 8.3 3 10 298.15 kelvin T H298 G298 T0 R IDCPH T0 T A B C D R T IDCPS T0 T A B C D G H298 G 5.429 10 3J K mol By Eq. (13.28), 0.5 exp G RT K 1.81048 (guess) 2 Given 1.5 1 By Eq. (13.4), nC2H6 = 1 yC2H6 yC2H6 1 1 0.0899 0.83505 Find =K n= 1 nH2 = nC2H4 = yH2 yC2H4 1 yC2H4 495 0.4551 yH2 1 0.4551 Ans. 13.18 C2H5CH:CH2(g) = CH2:CHCH:CH2(g) + H2(g) (1) (2) =1 (3) Number the species as shown. Basis is 1 mol species 1 + x mol steam. n0 = 1 x By Eq. (13.5), y1 = 1 1 y2 = y3 = x 1 = 0.10 x 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 1.967 1 A 31.630 2.734 1 B 26.786 10 3.249 0.422 0.0 9.873 C 8.882 10 6 D rows ( A) A i i Ai i 4.016 T 950 kelvin G B B i Bi C D i 4.422 10 T0 i Ci 3 C 9.91 298.15 kelvin T H298 G298 T0 R IDCPH T0 T A B C D R T IDCPS T0 T A B C D H298 5 10 1 end i A 0.0 0.083 0.0 end 3 496 i Di i 10 7 D 8.3 3 10 G 3J 4.896 10 K mol G exp ( )( ) 1 0.1 0.1 By Eq. (13.28), K RT x =K 1 Since 0.10 1 x (a) y1 13.19 K = K x 1 0.10 1 y1 (b) x 1 0.843 0.10 6.5894 yH2O 6.5894 7.5894 1 yH2O x 0.0186 ysteam 0.7814 ysteam 0.2 Number the species as shown. Basis is 1 mol species 1 + x mol steam entering. y1 = 1 1 x 235030 =2 n0 = 1 x 1 x 2 = 0.12 y3 = 2 y2 = 0.24 From data in Table C.4, H298 Ans. 0.8682 y2 = 2 y1 Ans. C4H10(g) = CH2:CHCH:CH2(g) + 2H2(g) (1) (2) (3) By Eq. (13.5), 0.53802 J G298 mol 166365 J mol The following vectors represent the species of the reaction in the order in which they appear: 1 1 2 1.935 A 36.915 2.734 B 3.249 26.786 10 0.422 497 3 11.402 C 0.0 8.882 6 10 D 0.0 end i rows ( A) A A B D i Ci i 3 T0 C 2.52 10 6 D 1.66 298.15 kelvin T H298 G298 T0 R IDCPH T0 T A B C D R T IDCPS T0 T A B C D G H298 3J 9.242 10 K mol G RT exp ( 0.12) ( 0.24) By Eq. (13.28), 2 K 1 x 2 Because 0.12 1 x 2 K = K 1 0.12 1 (a) y1 y1 x 2 1 x 2 0.023 (b) ysteam 4.3151 5.3151 0.30066 =K 1 x i Di i 9.285 10 925 kelvin G C i 7.297 T 1 end i Bi i 10 0.083 B i Ai 5 0.0 0.839 4.3151 yH2O 1 yH2O 0.617 ysteam 498 ( 0.24) 0.36 0.812 y1 Ans. Ans. 2 4 10 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 J mol 46110 A B 2.9355 (a) T G298 2.0905 10 T0 300 kelvin 16450 3 J mol C 0 D 5 0.3305 10 298.15 kelvin T H298 G298 T0 R IDCPH T0 T A B C D R T IDCPS T0 T A B C D G H298 G 1.627 P 10 4J P0 1 K mol exp G RT K 679.57 1 From Pb. 13.9 for ideal gases: 1 1 yNH3 P P0 yNH3 = 0.5 2 3 0.5 0.9664 yNH3 2 (b) For 0.9349 Ans. by the preceding equation Solving the next-to-last equation for K with P = P0 gives: 1 K 1.299 K 2 1 1.299 1 K 6.1586 499 Find by trial the value of T for which this is correct. It turns out to be Ans. T = 399.5 kelvin (c) For P = 100, the preceding equation becomes 1 2 1 1 K K 129.9 0.06159 T = 577.6 kelvin Ans. Another solution by trial for T yields (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 For NH3(1): Tc1 Pc1 405.7kelvin T Tc1 Tr1 For N2(2): 100bar Tr1 Tc2 Tr2 583 126.2 1.437 Pc2 126.2kelvin Tr2 4.62 1 P Pc1 Pr1 34.0bar Pr2 100 34.0 0.253 Pr1 0.887 2 112.8bar 0.038 Pr2 2.941 For H2(3), estimate critical constants using Eqns. (3.58) and (3.59) Tc3 1 43.6 21.8 2.016 Pc3 1 3 kelvin T kelvin 20.5 44.2 Tc3 Tr3 42.806 K T Tc3 Pc3 T 2.016 kelvin 0 500 P Pc3 13.62 Pr3 5.061 19.757 bar Pr3 bar Tr3 Therefore, i 13 PHIB Tr1 Pr1 1 0.924 PHIB Tr2 Pr2 2 1.034 PHIB Tr3 Pr3 3 1.029 1 i i 0.5 1.184 i 1.5 The expression used for K in Part (c) now becomes: 1 2 1 K 1 K 0.07292 129.9 1.184 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) =2 Basis: 1 mol CO, 2 mol H2 feed n0 = 3 From the data of Table C.4, 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 300 kelvin 10.815 10 T0 3 C 298.15 kelvin 501 3.45 10 6 D 5 0.135 10 T H298 G298 T0 R IDCPH T0 T A B C D R T IDCPS T0 T A B C D G H298 G 2.439 P 10 P0 1 4J K mol exp G RT K 1.762 4 10 1 By Eq. (13.5), with the species numbered in the order in which they appear in the reaction, y1 = 1 3 y2 = 2 2 2 3 2 By Eq. (13.28), 2 41 y3 (b) 3 y3 2 3 = y3 2 2 P P0 2 K Find 0.9752 0.9291 Ans. By the preceding equation 0.5 3 y3 2 y3 3 (guess) 0.8 3 Given y3 = 0.75 1 Solution of the equilibrium equation for K gives K 3 41 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 3 K 2 2 2 100 K 2.7 10 3 3 41 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 P 528kelvin For CO(1): Tc1 T Tc1 For CH3OH(3): T Tc3 Tr Tc3 Tr1 1.03 P Pc1 Pr1 2.858 3 0.564 Pr1 Pc3 P Pc3 Pr 0.048 34.99bar 3.973 512.6kelvin 1 80.97bar Pc1 132.9kelvin Tr1 Tr 100bar 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 13 PHIB Tr1 Pr1 1.032 1 1.0 1 0.612 0.612 1 i 2 1 i i 503 0.5933 The expression used for K in Part (c) now becomes: 3 K 2 2 100 2 0.593 K 1.6011 10 3 3 41 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 J 178321 G298 mol 130401 J mol The following vectors represent the species of the reaction in the order in which they appear: 1 12.572 1 A 1 i 6.104 B 0.443 10 5.457 13 A T 1.011 i Ai 1151.83 kelvin G B 3 5 D 1.047 10 1.157 1.045 B i Bi i A 3.120 2.637 D i 1.149 10 T0 3 i C 298.15 kelvin T H298 G298 T0 R IDCPH T0 T A B C D R T IDCPS T0 T A B C D H298 504 i Di 0 D 9.16 4 10 G J mol 126.324 Thus K exp G K RT 1.0133 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 J mol 91121 The following vectors represent the species of the reaction in the order in which they appear: 1 1 5.939 A 3.578 1 i B 3.156 13 A 3.020 T 0.795 i Ai 623.97 kelvin G B 3 10 5 D 0.186 10 0.151 0.623 B i A 0.0 16.105 i Bi D i 0.012462 T0 C 298.15 kelvin T H298 G298 T0 R IDCPH T0 T A B C D R T IDCPS T0 T A B C D H298 505 i Di i 0 D 3.5 3 10 G 3J 2.986 Thus 10 K mol exp G K RT 0.5624 Ans. T = 623.97 K 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) yNO2 yNO yO2 0.5 yNO2 = 0.5 = 0.5 =K T From the data of Table C.4, K Given yNO yNO2 G298 G298 exp yNO 298.15 kelvin yNO ( ) 0.21 K RT 10 12 yNO2 0.5 yNO2 = ( ) 0.21 10 6 1.493 7 10 6 6 10 yNO2 yNO = 5 10 yNO K yNO ppm 7.307 J mol 12 = 0.5 See Example 13.9, Pg. 508-510 From Table C.4, 105140 10 6 (a negligible concentration) Ans. 13.26 C2H4(g) + (1/2)O2(g) = <(CH2)2>O(g) H298 J mol (guesses) Find yNO yNO2 This is about 35240 G298 81470 J mol Basis: 1 mol C2H4 entering reactor. Moles O2 entering: nO2 1.25 0.5 Moles N2 entering: nO2 nN2 506 79 21 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: 0.8 1 nO2 n 1.424 0.5 3.639 A 23.463 3.280 4.392 0.0 10 9.296 6 1 kelvin 0.227 D 2 0.0 5 0.5 2 10 kelvin 1 0 0.040 14 A n i Ai B n i n i Ci D i n n0 0.5 i Bi i C y 3 10 kelvin 0.593 0.0 0.0 i 0.506 B 0.385 nN2 C 14.394 n i Di i K y i i K 15.947 i 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 3.629 B 5 0.114 10 kelvin T0 8.816 idcph A T0 C B 1 2 3 3 T0 3 C kelvin 4.904 A ln T0 2 2 1 1 D T0 1 1 2 B T0 C T0 D 130.182 kelvin 1 2 2 idcps 0.417 Given H298 = R A T0 C 3 3 K = exp Find 1 T0 H298 R T0 3 G298 B 2 T0 2 2 1 1 D 1 T0 H298 R T0 0.88244 3.18374 508 idcps 10 6 2 kelvin 298.15 kelvin T0 idcph 3 3 Guess: idcps 2 10 1 T0 idcph 0.0333 0.052 y 0.88244) ( Ans. 0.2496 0.6651 T 13.27 T0 949.23 kelvin Ans. T CH4(g) = C(s) + 2H2(g) (gases only) =1 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 G298 mol 50460 J mol The following vectors represent the species of the reaction in the order in which they appear: 1 9.081 1.702 1 A 2 B 1.771 0.422 3.249 0.0 2.164 i 13 C 0.0 10 6 5 D 0.867 10 0.083 0.0 A i Ai i A 6.567 T B i Bi C i B 923.15 kelvin 7.466 10 T0 3 0.771 10 i Ci D i 3 C 2.164 298.15 kelvin 509 i Di i 10 6 D 4 7.01 10 T H298 G298 T0 R IDCPH T0 T A B C D R T IDCPS T0 T A B C D G H298 G 1.109 10 4J mol By Eq. (13.5), (a) K exp G RT yCH4 = n0 = 1 K 1 2 1 K 4 yCH4 1 2 yH2 1 = 1 4 2 yCH4 1 By Eq. (13.28), 2 0.7893 yCH4 yH2 1 2 0.5659 =K 0.1646 Ans. 0.8354 (b) For a feed of 1 mol CH4 and 1 mol N2, Given 1 2 1 yH2 2 2 (fraction decomposed) 0.7173 K yH2 = 1 2 By Eq. (13.28), 4.2392 n0 = 2 .8 (guess) 2 Find =K 1 (fraction decomposed) yH2 yCH4 2 yN2 2 0.0756 510 1 yCH4 yN2 0.3585 yH2 Ans. 13.28 1/2N2(g) + 1/2O2(g) = NO(g) (1) =0 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 A T 90250 J mol G298 B 0.0725 0.0795 10 T0 2000 kelvin 86550 3 J mol C 0 D 5 0.1075 10 298.15 kelvin T H298 G298 T0 R IDCPH T0 T A B C D R T IDCPS T0 T A B C D G H298 G 6.501 10 4J K1 mol G RT exp 1/2N2(g) + O2(g) = NO2(g) K1 0.02004 = 0.5 (2) 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 3.280 1 A 3.639 1 i 13 0.593 B 0.506 10 4.982 A T 0.297 2000 kelvin 3 5 D 0.227 10 1.195 B i Ai 0.792 D i Bi B 3.925 T0 10 4 298.15 kelvin 511 i Di i i i A 0.040 C 0 D 5.85 4 10 T H298 G298 T0 R IDCPH T0 T A B C D R T IDCPS T0 T A B C D G H298 G 5J 1.592 10 K2 mol exp G K2 RT 6.9373 10 5 With the assumption of ideal gases, we apply Eq. (13.28): yNO (1) yN2 0.5 yNO (2) K1 ( ) 0.7 P0 1 yNO2 yN2 0.5 yNO2 13.29 0.5 0.5 0.5 0.5 = K1 () 0.05 yNO 3.74962 3 10 Ans. 200 yNO2 = 0.5 () 0.7 ( .05) 0 P 0.5 () 0.7 yO2 P P0 yO2 yNO = 0.5 P P0 = () 0.05 0.5 K2 ( .7) 0 ( .05) 0 0.5 K2 yNO2 4.104 10 5 Ans. 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 512 89830 J mol The following vectors represent the species of the reaction in the order in which they appear: 2 3.931 1 5.699 A 3 A B i Ai T 1.015 D 0.783 D i Bi 6.065 10 T0 723.15 kelvin i Di i 3 C D 0 6.28 298.15 kelvin T H298 G298 T0 R IDCPH T0 T A B C D R T IDCPS T0 T A B C D G H298 G 1.538 10 4J K mol By Eq. (13.5), gases only: yH2S = 2 2 n0 = 3 ySO2 = 3 By Eq. (13.28), Given 2 2 1 3 2 G RT K 3 1 Percent conversion of reactants = PC PC = ni0 ni0 ni 100 = i ni0 [By Eq. (13.4)] 100 513 2 3 (guess) Find = 8K 12.9169 (basis) yH2O = 0.5 2 2 exp 5 10 0.121 i B 5.721 3 1.450 i A 10 1.728 3.470 14 0.232 0.801 B 4.114 2 i 1.490 0.767 4 10 Since the reactants are present in the stoichiometric proportions, for each reactant, ni0 = Whence i PC PC 100 13.30 N2O4(g) = 2NO2(g) (a) (b) 76.667 Ans. =1 Data from Tables C.4 and C.1 provide the following values: H298 T0 57200 J mol G298 T 298.15 kelvin A G 3.968 J mol 350 kelvin 0.133 10 3 0 D G RT K C 5 1.203 10 H298 G B 1.696 5080 T H298 G298 T0 R IDCPH T0 T A B C D R T IDCPS T0 T A B C D 10 3J K mol exp 3.911 Basis: 1 mol species (a) initially. Then ya = (a) 1 1 P ya (b) yb = P 5 1 1 P0 ya 1 P0 = 1 K 4P K 1 1 1 2 2 2 0.4241 514 1 K 0.4044 Ans. K 4P K 1 P P0 0.7031 By Eq. (4.18), at 350 K: H H298 R IDCPH T0 T A B C H D 56984 J mol This is Q per mol of reaction, which is 0.7031 0.299 0.4044 Whence Q Q H 13.31 By Eq. (13.32), K= K= K= 1 2 exp 0.1 xB xA xA G exp 0.1 2 xA 1000 xA xA A xA B xA A Whence J mol 1 = xA xA 1 T 2 exp 0.1 xA K = exp 2 xB G RT 298.15 kelvin (guess) .5 Given xA exp 0.1 xA xA = Ans. mol ln b = 0.1 xA 2 xA 1 xB B J 2 2 ln a = 0.1 xB 1 17021 1 xA xA exp 0.1 2 xA 1 = exp G RT xA Find xA Ans. 0.3955 For an ideal solution, the exponential term is unity: Given 1 xA xA = exp G RT xA This result is high by 0.0050. Ans. 515 Find xA xA 0.4005 13.32 H2O(g) + CO(g) = H2(g) + CO2(g) =0 From the the data of Table C.4, H298 T0 A J 41166 G298 mol T 298.15 kelvin B 1.860 28618 J mol 800 kelvin 3 0.540 10 C 5 D 0 1.164 10 T H298 G298 T0 R IDCPH T0 T A B C D R T IDCPS T0 T A B C D G H298 G 9.668 10 3J mol (a) No. Since =0 K exp G RT K 4.27837 , 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), 1 1 1 nCO = 1 = 1 2 NCO2 = nH2 = 1 0.96 1.02 = 0.02 0.941 Let z = w/2 = moles H2O/mole "Water gas". By Eq. (13.5), yH2O = 2z w = 2 2z 2w yCO = 516 1 2 2z yH2 = 1 2 2z yCO2 = 2 By Eq. (13.28) 2z 1 Given 2z (d) 1 z z =K Find z () (guess) 2 z Ans. 4.1 = 1 (gases) 2CO(g) = CO2(g) + C(s) Data from Tables C.4 and C.1: H298 A 172459 J mol B 0.476 G298 0.702 10 120021 3 C J mol 5 D 0 1.962 10 T H298 G298 T0 R IDCPH T0 T A B C D R T IDCPS T0 T A B C D G H298 G 3.074 10 4J K mol G RT exp 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 yCO2 RATIO 2 2z yCO 0.092 yCO2 yCO 1 2 2z yCO 2 5.767 RATIO 10 2.775 3 3 10 No carbon can deposit from the equilibrium mixture. 517 13.33 CO(g) + 2H2(g) = CH3OH(g) (1) =2 This is the reaction of Pb. 13.21, where the following parameter values are given: H298 T 90135 J mol G298 T0 550 kelvin A G B 7.663 H298 10.815 10 3 J mol 24791 298.15 kelvin C 3.45 10 6 5 D 0.135 10 T H298 G298 T0 R IDCPH T0 T A B C D R T IDCPS T0 T A B C D G 3.339 4J 10 K1 mol exp G RT K1 H2(g) + CO2(g) = CO(g) + H2O(g) 6.749 10 4 (2) =0 From the the data of Table C.4, H298 T 41166 J mol G298 T0 550 kelvin J mol 28618 298.15 kelvin The following vectors represent the species of the reaction in the order in which they appear: 1 1 1 3.249 5.457 A 14 B 3.376 1 i 0.422 3.470 A 1.86 B i Ai B 0.557 3 10 1.157 D 0.031 1.450 5.4 10 4 D i Bi 518 C 5 10 0.121 i Di i i i A 1.045 0.083 0 D 1.164 5 10 T H298 G298 T0 R IDCPH T0 T A B C D R T IDCPS T0 T A B C D G H298 G 1.856 10 4J K2 mol G RT exp 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= H2 CO i.j CO2 CH3OH H2O _______________________________________________ j 1 2 -2 -1 -1 1 0 -1 1 0 0 1 By Eq. (13.7) 0.75 yH2 = 21 1 1 yCO = 21 0.05 yCO2 = 0.15 2 2 1 yCH3OH = 21 100 P0 1 2 21 1 1 yH2O = 21 0.1 2 1 21 1 By Eq. (13.40), 1 P 2 0.1 (guesses) Given 11 0.75 0.75 2 21 0.15 21 21 2 2 = 0.15 2 1 2 1 2 0.05 2 = K2 2 P P0 2 K1 1 2 519 Find 1 2 1 0.1186 0.75 yH2 21 1 1 1 yH2 0.6606 yCH3OH yCO yCO2 21 1 2 yH2O 21 0.0528 yH2O 2 1 yCH3OH yCO 0.1555 1 1 yCH3OH 21 yH2 13.34 yCO 2 yN2 3 10 0.15 2 21 0.05 yCO2 8.8812 2 1 21 yH2O yCO2 0.0539 Ans. yN2 0.0116 CH4(g) + H2O(g) = CO(g) + 3H2(g) 0.0655 (1) =2 From the the data of Table C.4, H298 J mol 205813 G298 141863 J mol The following vectors represent the species of the reaction in the order in which they appear: 1 1.702 1 A 1 3 3.470 3.249 0.0 0.0 10 6 D A 7.951 B 0.422 0.121 0.031 5 i 10 C i Bi 14 8.708 10 i Ci D 3 C 520 2.164 i Di i i i i 0.557 3 10 0.083 B i Ai 1.450 0.0 0.0 A B 3.376 2.164 C 9.081 10 6 D 9.7 3 10 T 1300 kelvin 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 G H298 G 1.031 10 5J K1 mol exp G K1 RT 13845 H2O(g) + CO(g) = H2(g) + CO2(g) = 0 (2) This is the reaction of Pb. 13.32, where parameter values are given: H298 A G G 41166 1.860 J G298 mol B 0.540 10 3 28618 C 0.0 J mol D 5 1.164 10 T H298 G298 T0 R IDCPH T0 T A B C D R T IDCPS T0 T A B C D H298 5.892 3J 10 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 yCO2 = 5 By Eq. (13.28), 2 1 Ratio = 5 yH2 yCO 0.1375 Find = K2 Ratio 5 (guess) 0.5 3 Given 3 yH2 = 3 Ratio 1 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 For T = 1300 K, A G 0.476 G298 T B 0.702 10 3 120021 1300 kelvin C 0.0 T H298 G298 T0 R IDCPH T0 T A B C D R T IDCPS T0 T A B C D H298 522 J mol T0 298.15 kelvin D 5 1.962 10 G 4J 5.673 10 K mol exp G RT K 5.255685 10 3 As explained in Problem 13.32(d), the question of carbon deposition depends on: yCO2 RATIO = 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: 5 Ratio Ratio 2 1 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 (1) CO + H2O = CO2 + H2 (2) There are alternative equivalent pairs, but for these: Stoichiometric numbers, i= CH4 H2O i.j CO CO2 H2 j _________________________________________________ j 1 2 -1 0 -1 -1 1 -1 0 1 523 3 1 2 0 For initial amounts: 2 mol CH4 and 3 mol H2O, n0 = 5, and by Eq. (13.7): 2 yCH4 = yCO2 = 3 1 5 yH2O = 21 2 5 yH2 = 21 1 5 1 yCO = 21 31 5 2 5 2 21 2 21 By Eq. (13.40), with P = P0 = 1 bar 3 yCO yH2 = k1 yCH4 yH2O yCO2 yH2 yCO yH2O = k2 From the data given in Example 13.14, G1 J mol 27540 G2 G1 K1 exp K1 K2 RT 1.5 2 2 3 1 1 5 2 3 1 Find 1 2 G2 RT 2 2 2 1000 kelvin 3 31 231 1 T (guesses) 2 1 J mol 1.457 1 1 Given exp K2 27.453 1 3130 21 2 = K1 = K2 2 1 1.8304 2 0.3211 2 2 yCH4 5 1 21 3 yH2O 1 5 524 21 2 yCO 1 5 2 21 2 yCO2 31 yH2 521 yCH4 0.0196 0.0371 yH2 521 yH2O yCO2 2 yCO 0.098 0.1743 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 Water(2): x2 Psat2 = y2 P x1 = 101.33 kPa Psat2 3.166 kPa x2 = y1 P 415kPa y2 P Psat2 (steam tables) Ethylene glycol(3): Therefore, y2 = 1 Psat3 = 0.0 y1 and y3 = 0.0 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, 3 x3 k= y1 P P0 = 2 x2 x3 T y1 x2 Data from Table C.4: k exp 298.15kelvin G298 G298 k RT 525 6.018 72941 12 10 J mol Ans. 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 1 1 1 0 vj ij v i Initial numbers of n0 moles 10 a) Stoichiometric coefficients: Number of components: i 50 50 kmol hr 0 1 0 Number of reactions: 14 1 n0 1 Given values: yA 0.05 yB yC 0.4 yD n0 n0i 12 kmol hr 100 i 0.10 Guess: j 0.4 1 1 kmol hr 2 1 kmol hr Given yA = yC = n01 1 2 n0 1 2 n03 1 2 n0 1 2 n02 yB = 1 n0 yD = 1 n04 n0 2 Eqn. (13.7) 2 1 2 yC yD Find yC yD 1 2 1 1 2 526 44.737 kmol hr 2 2.632 kmol hr (i) nA n01 1 nB n02 1 nC n03 1 nD n04 2 n nA yC (ii) nB nB 42.105 nD nC 5.263 nC 2 2.632 2.632 n nD 52.632 yD 0.8 1 ij v i hr kmol Ans. hr Ans. 40 Initial numbers of n0 moles 0 0 i hr kmol 0.05 2 1 Number of components: hr kmol 1 1 b) Stoichiometric coefficients: vj kmol hr kmol nA 2 40 kmol hr 0 1 0 Number of reactions: 14 1 Given values: yC 0.52 yD yA 0.4 yB 0.4 n0 80 kmol hr 2 1 1 1 Given yA = yC = n01 n0 1 1 n03 n0 2 22 1 1 22 yB = yD = n02 1 22 n0 1 22 n04 n0 527 2 1 kmol hr i 0.04 Guess: 12 n0i n0 2 j 22 Eqn. (13.7) kmol hr yA yB Find yA yB 1 2 1 26 1 kmol hr 2 2 kmol hr 2 yA nA n01 1 2 nA 12 nB n02 1 22 nB 10 nC n03 1 nC 26 nD n04 2 nD 2 hr kmol hr 1 1 vj ij 100 Initial numbers of n0 moles 1 0 0 i Ans. hr kmol 1 Number of components: 0 Number of reactions: 14 n0 Given values: yC 0.3 yD Guess: yA 0.4 yB n0i j 12 100 kmol hr n0 i 0.1 0.4 1 1 kmol hr 2 Given yA = yC = n01 1 2 n0 1 n02 1 2 n0 1 2 2 n03 n0 1 1 0 yB = yD = 2 n04 n0 528 kmol hr 0 1 i 0 1 1 v 0.2 kmol hr kmol 1 c) Stoichiometric coefficients: yB 0.24 2 1 2 Eqn. (13.7) 1 kmol hr yA yB Find yA yB 1 2 kmol hr 37.5 2 0.4 12.5 yB 1 yA 1 kmol 0.2 hr 2 kmol hr kmol nA n01 1 2 nA 50 nB n02 1 2 nB 25 nC n03 1 nC 37.5 nD n04 2 nD 12.5 hr kmol hr kmol hr 1 1 1 d) Stoichiometric coefficients: 1 ij Guess: yA 1 0.2 kmol hr 0 0 0 n0 n0i j n0 12 100 kmol hr i yD yB 60 Initial numbers of n0 moles Number of reactions: 15 0 0.25 40 1 1 v i Given values: yC i 0 0 vj 1 0 Number of components: Ans. 0.20 0.4 529 yE 0.1 1 1 kmol hr 2 1 kmol hr Given n01 yA = 1 n0 n02 yB = 1 n0 1 n0 1 n03 yC = 2 2 n0 1 Eqn. (13.7) 1 n04 yD = 2 n05 2 n0 1 1 yE = yA yB yE Find yA yB yE 1 1 2 20 kmol hr 16 2 kmol hr 1 2 (i) kmol hr kmol 24 hr kmol 20 hr nA n01 1 2 nA nB n02 1 2 nB nC n03 1 nC nD n04 2 nD 16 nE 13.45 (ii) n05 2 nE 16 yA hr yC 0.25 0.2 yE kmol hr kmol 0.3 yD Ans. 0.05 yB 4 0.2 C2H4(g) + H2O(g) -> C2H5OH(g) T0 298.15kelvin P0 T 1bar J mol 1 = C2H4(g) H0f1 2 = H2O(g) H0f2 241818 3 = C2H5OH(g) H0f3 235100 52500 400kelvin G0f1 J mol J mol 530 P 2bar 68460 J mol G0f2 228572 G0f3 168490 J mol J mol H0 H0f1 H0f2 H0f3 H0 45.782 G0 G0f1 G0f2 G0f3 G0 8.378 A A B [( 14.394) ( .450) ( 0.001) 1 210 C [ ( 4.392) () ( 6.002) 010 D [ () ( .121) () 0 0 0 ]10 exp G0 R T0 b) K1 exp H0 1 R T0 K2 mol ( 1.424) ( .470) ( .518) 3 3 a) K 0 K0 K1 K2 1.376 B 6 4.157 3 10 6 C 1.61 10 D 5 1.21 10 4 K298 K0 Eqn. (13.22) K2 Eqn. (13.20) Ans. 29.366 K1 1 IDCPH T0 T A B C D T IDCPS T0 T A B C D exp K400 3 Eqn. (13.21) K298 T0 T kJ mol kJ 9.07 3 Eqn. (13.23) 0.989 K400 10 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 Assuming ideal gas behavior e y3 = e y3 y1 y2 =K e 2 e P P0 e 2 Substituting results in the following expression: e = K400 1 e 2 531 e1 e2 e P P0 e using Solve for a Mathcad solve block. Guess: e 0.5 e 2 Given e = K400 1 e1 e 2 e2 y1 Find e e P0 0.191 e e 1 y1 P 1 e 2 y2 2 e 0.447 y2 e e y3 2 e 0.447 y3 e Ans. 0.105 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 130.680 S0H2O2 S0fH2O2 G0f 136.1064 kJ mol T J mol kelvin 232.95 S0O2 205.152 1bar J mol kelvin J mol kelvin S0H2 H0fH2O2 298.15kelvin P S0O2 S0H2O2 T S0fH2O2 S0fH2O2 G0 f 532 102.882 105.432 kJ mol J mol kelvin Ans. 13.48 C3H8(g) -> C3H6(g) + H2(g) (I) C3H8(g) -> C2H4(g) + CH4(g) (II) T0 P0 298.15kelvin T 1bar 1 = C3H8 (g) H0f1 2 = C3H6 (g) H0f2 19710 3 = H2 (g) H0f3 0 4 = C2H4 (g) H0f4 52510 5 = CH4 (g) H0f5 104680 P 750kelvin J G0f1 mol J mol 1.2bar G0f2 J mol 74520 J mol 0 G0f4 mol 62205 G0f3 J J 24290 68460 G0f5 mol J mol J mol J mol J mol 50460 Calculate equilibrium constant for reaction I: H0I H0f1 H0f2 H0f3 H0I 124.39 G0I G0f1 G0f2 G0f3 G0I 86.495 AI BI [( 28.785) ( 2.706) ( .422) 2 010 CI [ ( 8.824) ( 6.915) () 0 ]10 DI 3 [ () () ( .083) 0 0 010 6 KI1 exp KI2 exp 3.673 BI 5.657 CI 1.909 DI 5 exp mol AI ( 1.213) ( .637) ( .249) 1 3 KI 0 kJ mol kJ 8.3 3 10 10 6 3 10 G0I R T0 H0I 1 R T0 Eqn. (13.21) T0 T Eqn. (13.22) 1 IDCPH T0 T AI BI CI DI T IDCPS T0 T AI BI CI DI 533 KI0 0 KI1 1.348 KI2 1.714 13 10 Eqn. (13.23) KI Eqn. (13.20) KI0 KI1 KI2 KI 0.016 Calculate equilibrium constant for reaction II: H0II H0f1 H0f4 H0f5 H0II 82.67 G0II G0f1 G0f4 G0f5 G0II 42.29 AII ( 1.213) ( 1.424) ( 1.702) BII [ ( 28.785) ( 14.394) ( 9.081) ] 10 CII [ ( 8.824) ( 4.392) ( 2.164) ] 10 DII [ ( 0) kJ ( 0) ] 10 ( 0) KII0 exp G0II R T0 KII1 exp H0II 1 R T0 KII2 exp KII mol kJ mol AII BII 5 Eqn. (13.21) T0 T 10 2.268 10 DII 6 5.31 CII 3 1.913 0 3.897 10 KII1 5.322 10 1 IDCPH T0 T AII BII CII DII T IDCPS T0 T AII BII CII DII KII0 KII1 KII2 Eqn. (13.20) KII2 Eqn. (13.23) 21.328 Assume an ideal gas and 1 mol of C3H8 initially. 1 y1 = y4 = 1 I II I II II 1 I y2 = I 1 y5 = II y3 = I II II 1 I I 1 I y4 y5 y1 = KII 534 II Eqn. (13.7) II The equilibrium relationships are: y2 y3 P0 = KI P y1 P0 P 8 1.028 KII Eqn. (13.28) 6 8 KII0 Eqn. (13.22) 3 Substitution yields the following equations: I 1 I I 1 II I 1 I I II II 1 P0 P = KI II 1 II II I 1 II I 1 I I P0 = KII P II 1 II II 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 1 II 1 I I 1 II 1 I 1 I II P 1 I II P0 P 1 I II 1 I II II II II P0 = KI = KII I II I Find I 0.026 I II 0.948 II II 1 y1 y4 y1 1 I II I II II 1 y2 y5 I 1 I II I 1 I II II II 1 0.01298 y2 0.0132 y3 I y3 I II 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 7K 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 P0 298.15kelvin T 1bar 1 = n-C4H10(g) H0f1 125790 2 = iso-C4H10(g) H0f2 134180 J mol J H0f1 H0f2 H0 8.39 G0 G0f1 G0f2 G0 4.19 C [11.402) ( 11.945) (10 D 3 [( 36.915) ( 7.853) 310 [ () () 0 0 ]10 6 5 exp G0 R T0 b) K1 exp H0 1 R T0 K2 exp 20760 T0 T Eqn. (13.22) 1 IDCPH T0 T A B C D T IDCPS T0 T A B C D 536 mol mol kJ mol 0.258 B 9.38 10 C 5.43 10 D Eqn. (13.21) J mol J kJ A B a) K0 16570 G0f2 mol ( 1.935) ( .677) 1 15bar G0f1 H0 A P 425kelvin 4 7 0 K0 5.421 K1 Ans. 0.364 K2 1 Eqn. (13.23) Ke Eqn. (13.20) K0 K1 K2 Ke Ans. 1.974 Assume as a basis there is initially 1 mol of n-C4H10(g) y1 = 1 y2 = e e y2 a) Assuming ideal gas behavior = Ke y1 e Substitution results in the following expression: 1 = Ke e Solving for Ke yields the following analytical expression for 1 e 1 1 y1 e e y1 0.664 0.336 y2 Ke e e y2 0.336 Ans. b) Assume the gas is an ideal solution. In this case Eqn. (13.27) applies. yi i = P P0 Eqn. (13.27) K i 1 Substituting for yi yields: e 2 =K e1 This can be solved analytically for e to get: 2 e= 2 Ke 1 Calculate i for each pure component using the PHIB function. For n-C4H10: Tr1 1 0.200 Tc1 Tr1 1 Pr1 PHIB Tr1 Pr1 1 1 T Tc1 For iso-C4H10: Tr2 T Tc2 1 2 0.181 Tc2 Tr2 1.041 Pr2 537 425.1kelvin P Pc1 Pc1 37.96bar Pr1 0.395 Pc2 36.48bar Pr2 0.411 0.872 408.1kelvin P Pc2 2 PHIB Tr2 Pr2 Solving for e 2 2 2 yields: e 2 y1 1 e 0.884 y1 0.661 y2 e Ke 1 e y2 0.339 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 T := ( 55 + 273.15) K Margules equations: γ 1 ( x1) := exp ( 1 x1) A12 + 2 ( A21 A12) x1 2 γ 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 γ 1 ( x1) Psat1 + ( 1 x1) γ 2 ( x1) Psat2 y1 ( x1) := x1 γ 1 ( x1) Psat1 Pbubl ( x1) x1 := 0.25 y1 ( x1) = 0.562 x1 := 0.50 Pbubl ( x1) = 80.357 kPa y1 ( x1) = 0.731 x1 := 0.75 (b) Pbubl ( x1) = 64.533 kPa Pbubl ( x1) = 85.701 kPa y1 ( x1) = 0.808 BUBL P calculations with virial coefficients: 3 B11 := 963 cm mol 3 B22 := 1523 cm mol δ 12 := 2 B12 B11 B22 B11 ( P Psat1) + P y22 δ 12 Φ 1 ( P , T , y1 , y2) := exp R T B22 ( P Psat2) + P y12 δ 12 Φ 2 ( P , T , y1 , y2) := exp R T 539 3 B12 := 52 cm mol Guess: x1 := 0.25 P := Psat1 + Psat2 y1 := 0.5 2 y2 := 1 y1 Given y1 Φ 1 ( P , T , y1 , y2) P = x1 γ 1 ( x1) Psat1 y2 Φ 2 ( P , T , y1 , y2) P = ( 1 x1) γ 2 ( x1) Psat2 y2 = 1 y1 y1 0.558 y2 = 0.442 P 63.757 kPa y1 y2 := Find ( y1 , y2 , P) P x1 := 0.50 Given y1 Φ 1 ( P , T , y1 , y2) P = x1 γ 1 ( x1) Psat1 y2 Φ 2 ( P , T , y1 , y2) P = ( 1 x1) γ 2 ( x1) Psat2 y2 = 1 y1 y1 0.733 y2 = 0.267 P 79.621 kPa y1 y2 := Find ( y1 , y2 , P) P x1 := 0.75 Given y1 Φ 1 ( P , T , y1 , y2) P = x1 γ 1 ( x1) Psat1 y2 Φ 2 ( P , T , y1 , y2) P = ( 1 x1) γ 2 ( x1) Psat2 y2 = 1 y1 y1 0.812 y2 = 0.188 P 85.14 kPa y1 y2 := Find ( y1 , y2 , P) P 540 T := 200 K P := 30 bar H1 := 200 bar 14.3 B := 105 y1 := 0.95 3 cm 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: l v fhat1 = H1 x1 fhat1 = y1 φ 1 P By Eq. (11.36): φ 1 := exp B P R T φ 1 = 0.827 Equate the liquid- and vapor-phase fugacities and solve for x1: x1 := 14.4 y1 φ 1 P x1 = 0.118 H1 Ans. Pressures in kPa Data: 0.000 0.0895 0.1981 0.3193 x1 := 0.4232 0.5119 0.6096 0.7135 i := 2 .. rows ( P) (a) 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 It follows immediately from Eq. (12.10a) that: ln γ 1 = A12 Combining this with Eq. (12.10a) yields the required expression 541 (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. γ 1 ( x1 , x2 , A12 , A21) := exp ( x2) A12 + 2 ( A21 A12) x1 2 γ 2 ( x1 , x2 , A12 , A21) := exp ( x1) A21 + 2 ( A12 A21) x2 2 H1 := 50 Guesses: A21 := 0.2 A12 := 0.4 Mininize the sums of the squared errors by setting sums of derivatives equal to zero. Given 0= i 0= i 0= i 2 d H1 Pi x1 γ 1 x1 , x2 , A12 , A21 ... i i i exp ( A12) dA12 + x2i γ 2 x1i , x2i , A12 , A21 Psat2 ( ) ( ) 2 d H1 Pi x1 γ 1 x1 , x2 , A12 , A21 ... i i i exp ( A12) dA21 + x2i γ 2 x1i , x2i , A12 , A21 Psat2 ( ) ( ) 2 d H1 Pi x1 γ 1 x1 , x2 , A12 , A21 ... i i i exp ( A12) dH1 + x2i γ 2 x1i , x2i , A12 , A21 Psat2 ( ( A12 A21 := Find ( A12 , A21 , H1) H 1 ) ) A12 0.348 A21 = 0.178 H 1 51.337 542 Ans. (d) γ1 ( x1 , x2) := exp x22 A12 + 2 ( A21 A12) x1 γ2 ( x1 , x2) := exp x1 A21 + 2 ( A12 A21) x2 2 ( ) exp( A12) + x2i γ2 ( x1i , x2i) Psat2 H1 Pcalc := x1 γ1 x1 , x2 i i y1calc := i ( i ) H1 x1 γ1 x1 , x2 i i i exp ( A 12) i Pcalc i 0.2 0 PiPcalc i (y1iy1calci) 100 0.2 0.4 0.6 0 0.2 0.4 x1 0.6 0.8 i Pressure residuals y1 residuals Fit GE/RT data to Margules eqn. by least squares: i := 2 .. rows ( P) Given 0= i 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 ... A21 x1 ... x1 x2 i i i + A12 x2i 2 0= i 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 A12 A21 := Find ( A12 , A21 , H1) H 1 ... A21 x1 ... x1 x2 i i i + A12 x2i ... A21 x1 ... x1 x2 i i i + A12 x2i A12 0.375 A21 = 0.148 H 1 53.078 γ1 ( x1 , x2) := exp x2 A12 + 2 ( A21 A12) x1 2 γ2 ( x1 , x2) := exp x1 A21 + 2 ( A12 A21) x2 H1 Pcalc := x1 γ1 x1 , x2 + x2 γ2 x1 , x2 Psat2 i i i i exp ( A 12) i i i 2 ( y1calc := i ) ( ( ) H1 x1 γ1 x1 , x2 i i i exp ( A 12) Pcalc i 544 ) Ans. 2 2 0 0.2 PiPcalc i (y1iy1calci) 100 0.4 0.6 0.8 0 0.2 0.4 x1 0.6 0.8 i Pressure residuals y1 residuals 14.5 Pressures in kPa Data: i := 1 .. 7 0.3193 0.4232 0.5119 0.6096 x1 := 0.7135 0.7934 0.9102 1.000 21.63 24.01 25.92 27.96 P := 30.12 31.75 34.15 36.09 x2 := 1 x1 0.5934 0.6815 0.7440 0.8050 y1 := 0.8639 0.9048 0.9590 1.000 Psat1 := P8 (a) It follows immediately from Eq. (12.10a) that: ln γ 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. γ 1 ( x1 , x2 , A12 , A21) := exp ( x2) A12 + 2 ( A21 A12) x1 2 γ 2 ( x1 , x2 , A12 , A21) := exp ( x1) A21 + 2 ( A12 A21) x2 2 H2 := 14 Guesses: A21 := 0.148 A12 := 0.375 Mininize the sums of the squared errors by setting sums of derivatives equal to zero. Given 0 = i 0= ( ) ( ) d dA21 Pi x1 γ 1 ( x1 , x2 , A12 , A21) Psat1 ... i 0= d 2 dA Pi x1i γ 1 x1i , x2i , A12 , A21 Psat1 ... 12 H2 + x2i γ 2 x1i , x2i , A12 , A21 exp ( A21) i i i i H2 + x2 γ 2 x1 , x2 , A12 , A21 i i i exp ( A21) ( ) 2 d 2 dH Pi x1i γ 1 x1i , x2i , A12 , A21 Psat1 ... 2 H2 + x2 γ 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 (d) γ1 ( x1 , x2) := exp x22 A12 + 2 ( A21 A12) x1 γ2 ( x1 , x2) := exp x1 A21 + 2 ( A12 A21) x2 2 546 Ans. ( ) ( ) H2 Pcalc := x1 γ1 x1 , x2 Psat1 + x2 γ2 x1 , x2 i i i i i i i exp ( A 21) ( ) x1 γ1 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 (y1iy1calci) 100 2 3 4 0.2 0.4 0.6 x1 0.8 i Pressure residuals y1 residuals Fit GE/RT data to Margules eqn. by least squares: i := 1 .. 7 y2 := 1 y1 Given 0= i y1 Pi d x1 ln i ... dA12 i x1i Psat1 y2 Pi i + x ln 2i H2 x2 i exp ( A ) 21 547 A21 x1 ... x1 x2 i i i + A12 x2i 2 0= i 0= i y1 Pi d x1 ln i ... dA21 i x1i Psat1 y2 Pi i + x ln 2i H2 x2i exp ( A ) 21 y1 Pi d x1 ln i ... dH2 i x1i Psat1 y2 Pi i + x ln 2i H2 x2 i exp ( A ) 21 A12 A21 := Find ( A12 , A21 , H2) H 2 A21 x1 ... x1 x2 i i i + A12 x2i A21 x1 ... x1 x2 i i i + A12 x2i A12 0.37 A21 = 0.204 H 2 15.065 γ1 ( x1 , x2) := exp x2 A12 + 2 ( A21 A12) x1 2 γ2 ( x1 , x2) := exp x1 A21 + 2 ( A12 A21) x2 2 ( ) ( ) H2 Pcalc := x1 γ1 x1 , x2 Psat1 + x2 γ2 x1 , x2 i i i i i i i exp ( A 21) y1calc := i ( ) x1 γ1 x1 , x2 Psat1 i i Pcalc i i 548 Ans. 2 2 0 PiPcalc 0.2 i (y1iy1calci) 100 0.4 0.6 0.3 0.4 0.5 0.6 0.7 x1 0.8 0.9 1 i 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 i := 2 .. rows ( P) (a) 0.0 0.0932 0.1248 0.1757 0.2000 x1 := 0.2626 0.3615 0.4750 0.5555 0.6718 x2 := 1 x1 0.0 0.1794 0.2383 0.3302 0.3691 y1 := 0.4628 0.6184 0.7552 0.8378 0.9137 Psat2 := P1 It follows immediately from Eq. (12.10a) that: ln γ 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. γ 1 ( x1 , x2 , A12 , A21) := exp ( x2) A12 + 2 ( A21 A12) x1 2 γ 2 ( x1 , x2 , A12 , A21) := exp ( x1) A21 + 2 ( A12 A21) x2 2 H1 := 35 Guesses: A21 := 1.27 A12 := 0.70 Mininize the sums of the squared errors by setting sums of derivatives equal to zero. Given 0= i 0= i 0= i 2 d H1 Pi x1 γ 1 x1 , x2 , A12 , A21 ... i i i exp ( A12) dA12 + x2i γ 2 x1i , x2i , A12 , A21 Psat2 ( ) ( ) 2 d H1 Pi x1 γ 1 x1 , x2 , A12 , A21 ... i i exp ( A12) dA21 i + x2i γ 2 x1i , x2i , A12 , A21 Psat2 ( ) ( ) 2 d H1 Pi x1 γ 1 x1 , x2 , A12 , A21 ... i i i exp ( A12) dH1 + x2i γ 2 x1i , x2i , A12 , A21 Psat2 ( A12 A21 := Find ( A12 , A21 , H1) H 1 ( ) ) A12 0.731 A21 = 1.187 H 1 32.065 550 Ans. (d) γ1 ( x1 , x2) := exp x2 A12 + 2 ( A21 A12) x1 2 γ2 ( x1 , x2) := exp x1 A21 + 2 ( A12 A21) x2 2 ( ) exp( A12) + x2i γ2 ( x1i , x2i) Psat2 H1 Pcalc := x1 γ1 x1 , x2 i i y1calc := i i ( ) H1 x1 γ1 x1 , x2 i i i exp ( A 12) i Pcalc i 0.5 0 PiPcalc 0.5 i (y1iy1calci) 100 1 1.5 2 0 0.1 0.2 0.3 0.4 x1 0.5 0.6 0.7 i Pressure residuals y1 residuals Fit GE/RT data to Margules eqn. by least squares: i := 2 .. rows ( P) Given 0= i y2 := 1 y1 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 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 A12 A21 := Find ( A12 , A21 , H1) H 1 ... A21 x1 ... x1 x2 i i i + A12 x2i ... A21 x1 ... x1 x2 i i i + A12 x2i A12 0.707 A21 = 1.192 H 1 33.356 γ1 ( x1 , x2) := exp x2 A12 + 2 ( A21 A12) x1 2 γ2 ( x1 , x2) := exp x1 A21 + 2 ( A12 A21) x2 2 ( ) ( ) H1 Pcalc := x1 γ1 x1 , x2 + x2 γ2 x1 , x2 Psat2 i i i i exp ( A 12) i i i ( ) exp( A12) x1 γ1 x1 , x2 y1calc := i i i i Pcalc H1 i 552 Ans. 2 2 0 0.5 PiPcalc 1 i (y1iy1calci) 100 1.5 2 2.5 0 0.1 0.2 0.3 0.4 x1 0.5 0.6 0.7 i 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 i := 1 .. 9 (a) 19.30 19.89 21.37 24.95 29.82 P := 34.80 42.10 60.38 65.39 69.36 x2 := 1 x1 0.3302 0.3691 0.4628 0.6184 0.7552 y1 := 0.8378 0.9137 0.9860 0.9945 1.0000 Psat1 := P10 It follows immediately from Eq. (12.10a) that: ln γ 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. γ 1 ( x1 , x2 , A12 , A21) := exp ( x2) A12 + 2 ( A21 A12) x1 2 γ 2 ( x1 , x2 , A12 , A21) := exp ( x1) A21 + 2 ( A12 A21) x2 2 H2 := 4 Guesses: A21 := 1.37 A12 := 0.68 Mininize the sums of the squared errors by setting sums of derivatives equal to zero. Given 0= d dA12 Pi x1 γ 1 ( x1 , x2 , A12 , A21) Psat1 ... i 0= i 0= i i i H2 + x2 γ 2 x1 , x2 , A12 , A21 i i i exp ( A21) ( ) 2 d 2 dA Pi x1i γ 1 x1i , x2i , A12 , A21 Psat1 ... 21 H2 + x2i γ 2 x1i , x2i , A12 , A21 exp ( A21) ( ) ( ) d dH2 Pi x1 γ 1 ( x1 , x2 , A12 , A21) Psat1 ... i i i i H2 + x2 γ 2 x1 , x2 , A12 , A21 i i i exp ( A21) ( A12 A21 := Find ( A12 , A21 , H2) H 2 ) A12 0.679 A21 = 1.367 H 2 3.969 554 2 Ans. (d) γ1 ( x1 , x2) := exp x22 A12 + 2 ( A21 A12) x1 γ2 ( x1 , x2) := exp x1 A21 + 2 ( A12 A21) x2 2 ( ) ( ) H2 Pcalc := x1 γ1 x1 , x2 Psat1 + x2 γ2 x1 , x2 i i i i i i i exp ( A 21) y1calc := ( ) x1 γ1 x1 , x2 Psat1 i i i Pcalc i i 1 PiPcalc 0 i (y1iy1calci) 100 1 2 0 0.2 0.4 0.6 x1 0.8 i 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 ) 21 555 A21 x1 ... x1 x2 i i i + A12 x2i 2 0= i 0= y1 Pi d x1 ln i ... dA21 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 y1 Pi d x1 ln i ... dH2 i x1i Psat1 y2 Pi i + x ln 2i H2 x2i exp ( A ) 21 2 i A12 A21 := Find ( A12 , A21 , H2) H 2 A21 x1 ... x1 x2 i i i + A12 x2i A12 0.845 A21 = 1.229 H 2 4.703 γ1 ( x1 , x2) := exp x2 A12 + 2 ( A21 A12) x1 2 γ2 ( x1 , x2) := exp x1 A21 + 2 ( A12 A21) x2 2 ( ) ( ) H2 Pcalc := x1 γ1 x1 , x2 Psat1 + x2 γ2 x1 , x2 i i i i i i i exp ( A 21) y1calc := i ( ) x1 γ1 x1 , x2 Psat1 i i Pcalc i i 556 Ans. 1 PiPcalc 0 i (y1iy1calci) 100 1 2 0 0.2 0.4 0.6 x1 0.8 i 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 γ 1 := 0.8050 0.8639 0.9048 0.9590 i := 1 .. n Psat2 := 12.30kPa 1.304 1.188 1.114 1.071 1.044 γ 2 := 1.023 1.010 1.003 0.997 x2 := 1 x1 i i γ 1 := i y1 Pi i x1 Psat1 i γ 2 := i y2 Pi i x2 Psat2 i 557 y2 := 1 y1 T := ( 50 + 273.15)K Data reduction with the Margules equation and Eq. (10.5): 1.009 1.026 1.050 1.078 1.105 1.135 1.163 1.189 1.268 i i () () i := 1 .. n GERTi := x1 ln γ 1 + x2 ln γ 2 i i i i Guess: A12 := 0.1 f ( A12 , A21) := A21 := 0.3 n i=1 ( ) 2 GERTi A21 x1i + A12 x2i x1i x2i A12 := Minimize ( f , A12 , A21) A21 A12 = 0.374 A21 = 0.197 ( ) n RMS Error: RMS := i=1 3 RMS = 1.033 × 10 Ans. 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 0.6 0.8 x1 , x1 i Data reduction with the Margules equation and Eq. (14.1): 3 cm B11 := 1840 mol 3 cm B22 := 1800 mol 3 cm B12 := 1150 mol δ 12 := 2 B12 B11 B22 () B11 ( Pi Psat1) + Pi y2 2 δ 12 i Φ 1 := exp i R T 558 γ 1 := i y1 Φ 1 Pi i i x1 Psat1 i () B22 ( Pi Psat2) + Pi y1 2 δ 12 i Φ 2 := exp i R T i := 1 .. n () i i i x2 Psat2 i () GERTi := x1 ln γ 1 + x2 ln γ 2 i i i i Guess: f ( A12 , A21) := A12 := 0.1 A21 := 0.3 n i=1 ( ) 2 GERTi A21 x1i + A12 x2i x1i x2i A12 := Minimize ( f , A12 , A21) A21 RMS := A12 = 0.379 ( 4 Ans. ) n i=1 RMS = 9.187 × 10 A21 = 0.216 2 GERTi A21 x1i + A12 x2i x1i x2i n RMS Error: γ 2 := y2 Φ 2 Pi x1 := 0 , 0.01 .. 1 0.1 GERT i A21 x1+A12 ( 1x1) x1 ( 1x1) 0.05 0 0 0.5 1 x1 , x1 i 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: Tc := 308.3K T := Tn Tr := T Tc Pc := 61.39bar Tn := 189.4K Tr = 0.614 For Redlich/Kwong EOS: σ := 1 ε := 0 := 0.08664 Ψ := 0.42748 1 α ( Tr) := Tr q ( Tr) := 2 Ψ α ( Tr ) Tr α ( Tr ) R Tc β ( Tr , Pr) := Eq. (3.54) Define Z for the vapor (Zv) Given a ( Tr) := Ψ Table 3.1 Table 3.1 Guess: 2 2 Eq. (3.45) Pc Pr Eq. (3.53) Tr zv := 0.9 Eq. (3.52) zv = 1 + β ( Tr , Pr) q ( Tr) β ( Tr , Pr) zv β ( Tr , Pr) ( zv + ε β ( Tr , Pr) ) ( zv + σ β ( Tr , Pr) ) Zv ( Tr , Pr) := Find ( zv) Define Z for the liquid (Zl) Guess: Guess: zl := 0.01 Given Eq. (3.56) ( )( ) 1 + β ( Tr , Pr) zl zl zl = β ( Tr , Pr) + zl + ε β ( Tr , Pr) zl + σ β ( Tr , Pr) q ( T r ) β ( 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) := Iv ( Tr , Pr) := 1 σε 1 σε Zl ( Tr , Pr) + σ β ( Tr , Pr) ln Zl ( Tr , Pr) + ε β ( Tr , Pr) Zv ( Tr , Pr) + σ β ( Tr , Pr) ln Zv ( Tr , Pr) + ε β ( Tr , Pr) 560 Eq. (6.65b) ( ) lnφl ( Tr , Pr) := Zl ( Tr , Pr) 1 ln Zl ( Tr , Pr) β ( Tr , Pr) q ( Tr) Il ( Tr , Pr) Eq. (11.37) ( ) lnφv ( Tr , Pr) := Zv ( Tr , Pr) 1 ln Zv ( Tr , Pr) β ( Tr , Pr) q ( Tr) Iv ( Tr , Pr) Guess Psat: Psatr := Given 1bar Pc lnφl ( Tr , Psatr) = lnφv ( Tr , Psatr) Psatr = 0.026 Zl ( Tr , Psatr) = 4.742 × 10 Psat := Psatr Pc Psat = 1.6 bar 3 Psatr := Find ( Psatr) Zv ( Tr , Psatr) = 0.965 Ans. 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: ω := 0.187 T := Tn Tc := 308.3K Pc := 61.39bar Note: For solution at 0.85T c, set T := 0.85Tc. ε := 0 Tr := T Tc Tr = 0.614 For SRK EOS: σ := 1 Tn := 189.4K := 0.08664 561 Ψ := 0.42748 Table 3.1 1 2 2 α ( Tr , ω ) := 1 + ( 0.480 + 1.574ω 0.176ω ) 1 Tr ( ) ( ) 2 Ψ α Tr , ω Define Z for the vapor (Zv) Given Eq. (3.45) Eq. (3.54) Tr Table 3.1 2 α Tr , ω R Tc a ( Tr) := Ψ Pc q ( Tr) := 2 β ( Tr , Pr) := Guess: Pr Eq. (3.53) Tr zv := 0.9 Eq. (3.52) zv = 1 + β ( Tr , Pr) q ( Tr) β ( Tr , Pr) zv β ( Tr , Pr) ( zv + ε β ( Tr , Pr) ) ( zv + σ β ( Tr , Pr) ) Zv ( Tr , Pr) := Find ( zv) Define Z for the liquid (Zl) Given Guess: zl := 0.01 Eq. (3.56) ( )( ) 1 + β ( Tr , Pr) zl zl = β ( Tr , Pr) + zl + ε β ( Tr , Pr) zl + σ β ( Tr , Pr) q ( T r ) β ( 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) := Iv ( Tr , Pr) := 1 σε 1 σε Zl ( Tr , Pr) + σ β ( Tr , Pr) ln Zl ( Tr , Pr) + ε β ( Tr , Pr) Zv ( Tr , Pr) + σ β ( Tr , Pr) ln Zv ( Tr , Pr) + ε β ( Tr , Pr) 562 Eq. (6.65b) ( ) lnφl ( Tr , Pr) := Zl ( Tr , Pr) 1 ln Zl ( Tr , Pr) β ( Tr , Pr) q ( Tr) Il ( Tr , Pr) Eq. (11.37) ( ) lnφv ( Tr , Pr) := Zv ( Tr , Pr) 1 ln Zv ( Tr , Pr) β ( Tr , Pr) q ( Tr) Iv ( Tr , Pr) Guess Psat: Psatr := Given 2bar Pc lnφl ( Tr , Psatr) = lnφv ( Tr , Psatr) Psatr = 0.017 Psat := Psatr Pc Zl ( Tr , Psatr) = 3.108 × 10 3 Psatr := Find ( Psatr) Zv ( Tr , Psatr) = 0.975 Psat = 1.073 bar Ans. 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%. Acetylene Argon Benzene n-Butane Carbon Monoxide n-Decane Ethylene n-Heptane Methane Nitrogen 14.10 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% (b) Acetylene: ω := 0.187 T := Tn Tc := 308.3K Pc := 61.39bar Note: For solution at 0.85T c, set T := 0.85Tc. For PR EOS: σ := 1 + 2 ε := 1 Tn := 189.4K Tr := T Tc Tr = 0.614 2 := 0.07779 563 Ψ := 0.45724 Table 3.1 2 1 2 2 Table 3.1 α ( Tr , ω ) := 1 + ( 0.37464 + 1.54226ω 0.26992ω ) 1 Tr a ( Tr) := Ψ q ( Tr) := ( ) 2 Eq. (3.45) Pc ( Ψ α Tr , ω ) Tr Define Z for the vapor (Zv) Given 2 α Tr , ω R Tc Eq. (3.54) β ( Tr , Pr) := Guess: Pr Eq. (3.53) Tr zv := 0.9 Eq. (3.52) zv = 1 + β ( Tr , Pr) q ( Tr) β ( Tr , Pr) zv β ( Tr , Pr) ( zv + ε β ( Tr , Pr) ) ( zv + σ β ( Tr , Pr) ) Zv ( Tr , Pr) := Find ( zv) Define Z for the liquid (Zl) Guess: zl := 0.01 Given Eq. (3.56) ( )( ) 1 + β ( Tr , Pr) zl zl = β ( Tr , Pr) + zl + ε β ( Tr , Pr) zl + σ β ( Tr , Pr) q ( T r ) β ( 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) := Iv ( Tr , Pr) := 1 σε 1 σε Zl ( Tr , Pr) + σ β ( Tr , Pr) ln Zl ( Tr , Pr) + ε β ( Tr , Pr) Zv ( Tr , Pr) + σ β ( Tr , Pr) ln Zv ( Tr , Pr) + ε β ( Tr , Pr) 564 Eq. (6.65b) ( ) lnφl ( Tr , Pr) := Zl ( Tr , Pr) 1 ln Zl ( Tr , Pr) β ( Tr , Pr) q ( Tr) Il ( Tr , Pr) Eq. (11.37) ( ) lnφv ( Tr , Pr) := Zv ( Tr , Pr) 1 ln Zv ( Tr , Pr) β ( Tr , Pr) q ( Tr) Iv ( Tr , Pr) 2bar Pc Guess Psat: Psatr := Given lnφl ( Tr , Psatr) = lnφv ( Tr , Psatr) Psatr = 0.018 Zl ( Tr , Psatr) = 2.795 × 10 Psat := Psatr Pc Psat = 1.09 bar Psatr := Find ( Psatr) Zv ( Tr , Psatr) = 0.974 3 Ans. 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%. Acetylene Argon Benzene n-Butane Carbon Monoxide n-Decane Ethylene n-Heptane Methane Nitrogen 14.12 (a) van der Waals Eqn. σ := 0 q ( Tr) := Given 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% ε := 0 Ψ α ( Tr) Tr := Tr := 0.7 1 8 Ψ := β ( Tr , Pr) := Pr Tr zv = 1 + β ( Tr , Pr) q ( Tr) β ( Tr , Pr) α ( Tr) := 1 27 64 zv := 0.9 (guess) zv β ( Tr , Pr) ( zv) 565 2 Eq. (3.52) Zv ( Tr , Pr) := Find ( zv) zl := .01 (guess) 2 1 + β ( Tr , Pr) zl zl = β ( Tr , Pr) + ( zl) Given q ( Tr) β ( Tr , Pr) Eq. (3.56) zl < 0.2 Zl ( Tr , Pr) := Find ( zl) Iv ( Tr , Pr) := β ( Tr , Pr) Zv ( Tr , Pr) Il ( Tr , Pr) := β ( Tr , Pr) Zl ( Tr , Pr) Case II, pg. 218. By Eq. (11.39): lnφv ( Tr , Pr) := Zv ( Tr , Pr) 1 ln ( Zv ( Tr , Pr) β ( Tr , Pr) ) q ( Tr) Iv ( Tr , Pr) lnφl ( Tr , Pr) := Zl ( Tr , Pr) 1 ln ( Zl ( Tr , Pr) β ( Tr , Pr) ) q ( Tr) Il ( Tr , Pr) Psatr := .1 Given lnφl ( Tr , Psatr) lnφv ( Tr , Psatr) = 0 Psatr := Find ( Psatr) Zv ( Tr , Psatr) = 0.839 Zl ( Tr , Psatr) = 0.05 lnφl ( Tr , Psatr) = 0.148 lnφv ( Tr , Psatr) = 0.148 β ( Tr , Psatr) = 0.036 ω := 1 log ( Psatr) ω = 0.302 (b) Psatr = 0.2 Ans. Redlich/Kwong Eqn.Tr := 0.7 σ := 1 ε := 0 := 0.08664 Ψ := 0.42748 .5 α ( Tr) := Tr q ( Tr) := Ψ α ( Tr) β ( Tr , Pr) := Tr Pr Tr Given zv = 1 + β ( Tr , Pr) q ( Tr) β ( Tr , Pr) Zv ( Tr , Pr) := Find ( zv) Guess: zl := .01 566 Guess: zv := 0.9 zv β ( Tr , Pr) zv ( zv + β ( Tr , Pr) ) Eq. (3.52) Given zl = β ( Tr , Pr) + zl ( zl + β ( Tr , Pr) ) zl < 0.2 1 + β ( Tr , Pr) zl q ( Tr) β ( Tr , Pr) Eq. (3.55) Zl ( Tr , Pr) := Find ( zl) Zv ( Tr , Pr) + β ( Tr , Pr) Il ( Tr , Pr) := ln Zl ( Tr , Pr) + β ( Tr , Pr) Zv ( Tr , Pr) Zl ( Tr , Pr) Iv ( Tr , Pr) := ln By Eq. (11.39): lnφv ( Tr , Pr) := Zv ( Tr , Pr) 1 ln ( Zv ( Tr , Pr) β ( Tr , Pr) ) q ( Tr) Iv ( Tr , Pr) lnφl ( Tr , Pr) := Zl ( Tr , Pr) 1 ln ( Zl ( Tr , Pr) β ( Tr , Pr) ) q ( Tr) Il ( Tr , Pr) Psatr := .1 Given lnφl ( Tr , Psatr) = lnφv ( Tr , Psatr) Zv ( Tr , Psatr) = 0.913 Psatr := Find ( Psatr) Zl ( Tr , Psatr) = 0.015 lnφv ( Tr , Psatr) = 0.083 lnφl ( Tr , Psatr) = 0.083 ω := 1 log ( Psatr) 14.15 (a) x1α := 0.1 Guess: ω = 0.058 x2α := 1 x1α A12 := 2 Psatr = 0.087 β ( Tr , Psatr) = 0.011 Ans. x1β := 0.9 x2β := 1 x1β A21 := 2 γ1α ( A21 , A12) := exp x2α A12 + 2 ( A21 A12) x1α 2 γ1β( A21 , A12) := exp x2β A12 + 2 ( A21 A12) x1β 2 γ2α ( A21 , A12) := exp x1α A21 + 2 ( A12 A21) x2α 2 γ2β( A21 , A12) := exp x1β A21 + 2 ( A12 A21) x2β 2 567 Given x1α γ1α ( A21 , A12) = x1β γ1β( A21 , A12) x2α γ2α ( A21 , A12) = x2β γ2β( A21 , A12) A12 := Find ( A12 , A21) A21 (b) x1α := 0.2 Guess: A21 = 2.747 A12 = 2.747 x2α := 1 x1α x1β := 0.9 A12 := 2 Ans. x2β := 1 x1β A21 := 2 γ1α ( A21 , A12) := exp x2α A12 + 2 ( A21 A12) x1α 2 γ1β( A21 , A12) := exp x2β A12 + 2 ( A21 A12) x1β 2 γ2α ( A21 , A12) := exp x1α A21 + 2 ( A12 A21) x2α 2 γ2β( A21 , A12) := exp x1β A21 + 2 ( A12 A21) x2β 2 Given x1α γ1α ( A21 , A12) = x1β γ1β( A21 , A12) x2α γ2α ( A21 , A12) = x2β γ2β( A21 , A12) A12 := Find ( A12 , A21) A21 (c) x1α := 0.1 Guess: A12 = 2.148 A21 = 2.781 x2α := 1 x1α x1β := 0.8 A12 := 2 A21 := 2 x2β := 1 x1β γ1α ( A21 , A12) := exp x2α A12 + 2 ( A21 A12) x1α 2 γ1β( A21 , A12) := exp x2β A12 + 2 ( A21 A12) x1β 2 γ2α ( A21 , A12) := exp x1α A21 + 2 ( A12 A21) x2α 2 γ2β( A21 , A12) := exp x1β A21 + 2 ( A12 A21) x2β 2 568 Ans. Given x1α γ1α ( A21 , A12) = x1β γ1β( A21 , A12) x2α γ2α ( A21 , A12) = x2β γ2β( A21 , A12) A12 := Find ( A12 , A21) A21 14.16 (a) x1α := 0.1 Guess: Given A12 = 2.781 A21 = 2.148 x2α := 1 x1α x1β := 0.9 a12 := 2 Ans. x2β := 1 x1β a21 := 2 2 2 a12 x1α a12 x1β exp a12 1 + x1α = exp a12 1 + x1β a21 x2α a21 x2β 2 2 a21 x2α a21 x2β exp a21 1 + x2α = exp a21 1 + x2β a12 x1α a12 x1β a12 := Find ( a12 , a21) a21 (b) x1α := 0.2 Guess: Given a12 = 2.747 x2α := 1 x1α a12 := 2 a21 = 2.747 x1β := 0.9 Ans. x2β := 1 x1β a21 := 2 2 2 a12 x1α a12 x1β exp a12 1 + x1α = exp a12 1 + x1β a21 x2α a21 x2β 2 2 a21 x2α a21 x2β exp a21 1 + x2α = exp a21 1 + x2β a12 x1α a12 x1β a12 := Find ( a12 , a21) a21 a12 = 2.199 569 a21 = 2.81 Ans. x1α := 0.1 x2α := 1 x1α x1β := 0.8 Guess: a12 := 2 a21 := 2 (c) Given x2β := 1 x1β 2 2 a12 x1α a12 x1β exp a12 1 + x1α = exp a12 1 + x1β a21 x2α a21 x2β 2 2 a21 x2α a21 x2β exp a21 1 + x2α = exp a21 1 + x2β a12 x1α a12 x1β a12 := Find ( a12 , a21) a21 14.18 (a) a := 975 a12 = 2.81 b := 18.4 T := 250 .. 450 A ( T) := a21 = 2.199 Ans. c := 3 a + b c ln ( T) T 2.1 A ( T) 2 1.9 250 300 350 400 450 T 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 1 x x A ( T) ( 1 2 x) = ln x 0.5 x1 ( T) := Find ( x) 570 Eq. (E), Ex. 14.5 x2 ( T) := 1 x1 ( T) UCST := 300 (guess) Given A ( UCST) = 2 UCST := Find ( UCST) UCST = 272.93 LCST := Find ( LCST) LCST = 391.21 LCST := 400 (guess) Given A ( LCST) = 2 Plot phase diagram as a function of T T1 := 225 , 225.1 .. UCST T2 := LCST .. 450 500 T1 400 T1 T2 T2 300 200 0.2 0.3 0.4 0.5 0.6 x1 ( T1 ) , x2 ( T1 ) , x1 ( T2 ) , x2 ( T2 ) (b) a := 540 b := 17.1 T := 250 .. 450 A ( T) := c := 3 a + b c ln ( T) T 2.5 A ( T) 2 1.5 250 300 350 400 450 T Parameter A = 2 at a single temperature. It is a LCST, because A decreases to 2 as T decreases. 571 0.7 0.8 Guess: x := 0.25 Given A ( T) ( 1 2 x) = ln 1 x x x0 x 0.5 Eq. (E), Ex. 14.5 x1 ( T) := Find ( x) LCST := 350 (guess) LCST := Find ( LCST) A ( LCST) = 2 Given Plot phase diagram as a function of T LCST = 346 T := LCST .. 450 450 400 T T 350 300 0.1 0.2 0.3 0.4 0.5 0.6 x1 ( T ) , 1x1 ( T) a := 1500 b := 19.9 T := 250 .. 450 (c) A ( T) := c := 3 a + b c ln ( T) T 3 2.5 A ( T) 2 1.5 250 300 350 400 450 T Parameter A = 2 at a single temperature. It is an UCST, because A decreases to 2 as T increases. 572 0.7 0.8 Guess: x := 0.25 Given A ( T) ( 1 2 x) = ln 1 x x x0 x 0.5 Eq. (E), Ex. 14.5 x1 ( T) := Find ( x) UCST := 350 (guess) A ( UCST) = 2 Given UCST := Find ( UCST) Plot phase diagram as a function of T UCST = 339.66 T := UCST .. 250 350 T 300 T 250 0 0.2 0.4 0.6 0.8 x1 ( T ) , 1x1 ( T) x1α := 0.5 14.20 Guess: x1β := 0.5 Write Eq. (14.74) for species 1: Given 2 2 x1α exp 0.4 ( 1 x1α) = x1β exp 0.8 ( 1 x1β) x1α 1 x1α + x1β 1 x1β =1 x1α := Find ( x1α , x1β) x1β (Material balance) x1α = 0.371 573 x1β = 0.291 Ans. 14.22 Temperatures in kelvins; pressures in kPa. P1sat ( T) := exp 19.1478 P2sat ( T) := exp 14.6511 5363.7 T water 2048.97 T SF6 P := 1600 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 0 100 200 300 400 6 500 6 y1II ( TII) 10 , y1II ( TII) 10 574 600 700 288 286 TI Tstar 284 282 280 650 700 750 800 850 900 6 950 1000 1050 6 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 3885.70 P2sat ( T) := exp 16.3872 T + 230.170 Toluene P := 101.33 Water Find the three-phase equilibrium T and y: T := 25 Guess: P = P1sat ( T) + P2sat ( T) Given y1star := P1sat ( Tstar) P Tstar := Find ( T) y1star = 0.444 For z1 < y1*, first liquid is pure species 2. y1 := 0.2 Given Guess: y1 = 1 Tdew := Tstar P2sat ( Tdew) P Tdew := Find ( Tdew) Tdew = 93.855 For z1 > y1*, first liquid is pure species 1. 575 Ans. Tstar = 84.3 y1 := 0.7 Guess: y1 = Given Tdew := Tstar P1sat ( Tdew) P Tdew := Find ( Tdew) Tdew = 98.494 Ans. 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 P2sat ( T) := exp 16.3872 2910.26 T + 216.432 n-heptane 3885.70 T + 230.170 water P := 101.33 Find the three-phase equilibrium T and y: Guess: T := 50 Given P = P1sat ( T) + P2sat ( T) y1star := P1sat ( Tstar) P Tstar := Find ( T) Tstar = 79.15 y1star = 0.548 Since 0.35<y1*, first liquid is pure species 2. y1 ( T) := 1 P2sat ( T) P Find temperature of initial condensation at y1=0.35: y10 := 0.35 Given Guess: y1 ( Tdew) = y10 Tdew := Tstar Tdew := Find ( Tdew) Tdew = 88.34 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 Tdew 90 T 85 Tstar 80 75 0.3 0.35 0.4 0.45 0.5 0.55 y1path ( T ) P1sat := 75 14.26 Pressures in kPa. P2sat := 110 A := 2.25 ( 2 γ1 ( x1) := exp A ( 1 x1) γ2 ( x1) := exp A x1 ) 2 Find the solubility limits: Guess: x1α := 0.1 Given A ( 1 2 x1α) = ln 1 x1α x1α x1α = 0.224 x1β := 1 x1α x1α := Find ( x1α) x1β = 0.776 Find the conditions for VLLE: Guess: Given Pstar := P1sat y1star := 0.5 Pstar = x1β γ1 ( x1β) P1sat + ( 1 x1α) γ2 ( x1α) P2sat y1star Pstar = x1α γ1 ( x1α) P1sat Pstar := Find ( Pstar , y1star) y1star Pstar = 160.699 Calculate VLE in two-phase region. Modified Raoult's law; vapor an ideal gas. Guess: x1 := 0.1 P := 50 577 y1star = 0.405 P = x1 γ1 ( x1) P1sat + ( 1 x1) γ2 ( x1) P2sat Given P ( x1) := Find ( P) y1 ( x1) := x1 γ1 ( x1) P1sat P ( x1) 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 Pstar PL ( x1) 150 PV ( x1) 125 Pliq Pliq 100 75 50 0 0.2 0.4 0.6 x1 , y1 ( x1) , x1α , x1β x1 := 0 , 0.05 .. 0.2 x1 = PL ( x1) = y1 ( x1) = 0 110 0 0.05 133.66 0.214 0.1 147.658 0.314 0.15 155.523 0.368 0.2 159.598 0.397 578 0.8 1 x1 := 1 , 0.95 .. 0.8 x1 = PL ( x1) = y1 ( x1) = 1 75 1 0.95 113.556 0.631 0.9 137.096 0.504 0.85 150.907 0.444 0.8 158.506 0.414 x1α = 0.224 x1β = 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 n-Pentane: P2sat ( T) := exp 13.7667 2451.88 T + 232.014 n-Heptane: P3sat ( T) := exp 13.8622 2910.26 T + 216.432 P := 101.33 z1 := 0.45 (a) z2 := 0.30 z3 := 1 z1 z2 Calculate dew point T and liquid composition assuming the hydrocarbon layer forms first: Guess: Tdew1 := 100 x2α := z2 x3α := 1 x2α Given P = x2α P2sat ( Tdew1) + x3α P3sat ( Tdew1) z3 P = x3α P3sat ( Tdew1) x2α + x3α = 1 x2α x3α := Find ( x2α , x3α , Tdew1) Tdew1 Tdew1 = 66.602 x3α = 0.706 579 x2α = 0.294 Calculate dew point temperature assuming the water layer forms first: x1β := 1 Tdew2 := 100 Guess: x1β P1sat ( Tdew2) = z1 P Given Tdew2 := Find ( Tdew2) Tdew2 = 79.021 Since Tdew2 > Tdew1, the water layer forms first (b) Calculate the temperature at which the second layer forms: Tdew3 := 100 Given x2α := z2 x3α := 1 x2α y1 := z1 Guess: y2 := z2 y3 := z3 P = P1sat ( Tdew3) + x2α P2sat ( Tdew3) + x3α P3sat ( Tdew3) y1 P = P1sat ( Tdew3) y2 z2 = y3 z3 y1 + y2 + y3 = 1 y2 P = x2α P2sat ( Tdew3) x2α + x3α = 1 y1 y2 y3 := Find ( y1 , y2 , y3 , Tdew3 , x2α , x3α) Tdew3 x2α x3α y1 = 0.288 y2 = 0.388 y3 = 0.324 Tdew3 = 68.437 x2α = 0.1446 x3α = 0.8554 (c) Calculate the bubble point given the total molar composition of the two phases Tbubble := Tdew3 x2α := z2 z2 + z3 x2α = 0.545 580 x3α := z3 z2 + z3 x3α = 0.455 Given P = P1sat ( Tbubble) + x2α P2sat ( Tbubble) + x3α P3sat ( Tbubble) Tbubble := Find ( Tbubble) Tbubble = 48.113 P1sat ( Tbubble) P x2α P2sat ( Tbubble) y2 := P x3α P3sat ( Tbubble) y3 := P y1 = 0.111 y1 := 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 n-Pentane: P2sat ( T) := exp 13.7667 2451.88 T + 232.014 n-Heptane: P3sat ( T) := exp 13.8622 2910.26 T + 216.432 P := 101.33 (a) Guess: Given z1 := 0.32 z2 := 0.45 z3 := 1 z1 z2 Calculate dew point T and liquid composition assuming the hydrocarbon layer forms first: Tdew1 := 70 x2α := z2 x3α := 1 x2α P = x2α P2sat ( Tdew1) + x3α P3sat ( Tdew1) z3 P = x3α P3sat ( Tdew1) x2α + x3α = 1 x2α x3α := Find ( x2α , x3α , Tdew1) Tdew1 Tdew1 = 65.122 x3α = 0.686 581 x2α = 0.314 Calculate dew point temperature assuming the water layer forms first: x1β := 1 Tdew2 := 70 Guess: x1β P1sat ( Tdew2) = z1 P Given Tdew2 := Find ( Tdew2) Tdew2 = 70.854 Since Tdew1>Tdew2, a hydrocarbon layer forms first (b) Calculate the temperature at which the second layer forms: Tdew3 := 100 Given x2α := z2 x3α := 1 x2α y1 := z1 Guess: y2 := z2 y3 := z3 P = P1sat ( Tdew3) + x2α P2sat ( Tdew3) + x3α P3sat ( Tdew3) y1 P = P1sat ( Tdew3) y2 z2 = y3 z3 y2 P = x2α P2sat ( Tdew3) y1 + y2 + y3 = 1 x2α + x3α = 1 y1 y2 y3 := Find ( y1 , y2 , y3 , Tdew3 , x2α , x3α) Tdew3 x2α x3α y1 = 0.24 y2 = 0.503 y3 = 0.257 Tdew3 = 64.298 x2α = 0.2099 x3α = 0.7901 (c) Calculate the bubble point given the total molar composition of the two phases Tbubble := Tdew3 x2α := z2 z2 + z3 x2α = 0.662 582 x3α := z3 z2 + z3 x3α = 0.338 Given P = P1sat ( Tbubble) + x2α P2sat ( Tbubble) + x3α P3sat ( Tbubble) Tbubble := Find ( Tbubble) Tbubble = 43.939 P1sat ( Tbubble) P x2α P2sat ( Tbubble) y2 := P x3α P3sat ( Tbubble) y3 := P y1 = 0.09 y1 := 0.302 0.224 y2 = 0.861 y3 = 0.049 748.4 K 304.2 14.32 ω := 40.51 bar 73.83 Tc := Pc := P := 10bar , 20bar .. 300bar T Tr := Tc T := 353.15K Use SRK EOS From Table 3.1, p. 98 of text: σ := 1 ε := 0 := 0.08664 Ψ := 0.42748 ( )( α := 1 + 0.480 + 1.574 ω 0.176 ω 1 Tr 2 0.5 ) 2 2 2 Ψ α R Tc Eq. (14.31) a := Pc R Tc b := Pc 6.842 kg m5 a= 0.325 s2 mol2 1.331 × 10 4 m3 b= 2.968 × 10 5 mol β 2 ( P) := z2 := 1 b2 P R T q2 := Eq. (14.33) (guess) 583 a2 b2 R T Eq. (14.32) Eq. (14.34) Given z2 = 1 + β 2 ( P) q2 β 2 ( P) ( z2 β 2 ( P) )( z2 + ε β 2 ( P) z2 + σ β 2 ( P) Eq. (14.36) ) Z2 ( P) := Find ( z2) Z 2 ( P) + β 2 ( P) I2 ( P) := ln Z 2 ( P) Eq. (6.65b) For simplicity, let φ1 represent the infinite-dilution value of the fugacity coefficient of species 1 in solution. l12 := 0.088 Eq. (14.103): b1 ( Z2 ( P) 1) ln ( Z2 ( P) β 2 ( P) ) ... b2 0.5 b1 a1 + q2 2 ( 1 l12) I2 ( P) b2 a2 φ 1 ( P) := exp 3 Psat1 := 0.0102bar V1 := 124.5 cm mol Eqs. (14.98) and (14.99), with φsat1 = 1 and (P - Psat1) = P, combine to give: y1 ( P) := Psat1 P φ 1 ( P) P V 1 R T exp 0.1 0.01 y1 ( P) 1 .10 3 1 .10 4 0 50 100 150 P bar 584 200 250 300 0.302 0.038 748.4 K 126.2 14.33 ω := 40.51 bar 34.00 Tc := Pc := P := 10bar , 20bar .. 300bar T Tr := Tc T := 308.15K (K) Use SRK EOS From Table 3.1, p. 98 of text: σ := 1 ε := 0 := 0.08664 Ψ := 0.42748 ( )( ) 2 0.5 α := 1 + 0.480 + 1.574 ω 0.176 ω 1 Tr a := 2 2 Ψ α R Tc R Tc b := Pc Eq. (14.31) Pc b2 P R T z2 := 1 Eq. (14.32) 1.331 × 10 4 m3 b= 2.674 × 10 5 mol 7.298 kg m5 a= 0.067 s2 mol2 β 2 ( P) := 2 q2 := Eq. (14.33) a2 b2 R T Eq. (14.34) (guess) Given z2 = 1 + β 2 ( P) q2 β 2 ( P) ( z2 β 2 ( P) )( z2 + ε β 2 ( P) z2 + σ β 2 ( P) ) Eq. (14.36) Z2 ( P) := Find ( z2) Z 2 ( P) + β 2 ( P) I2 ( P) := ln Z 2 ( P) Eq. (6.65b) For simplicity, let φ1 represent the infinite-dilution value of the fugacity coefficient of species 1 in solution. 585 l12 := 0.0 Eq. (14.103): b1 ( Z2 ( P) 1) ln ( Z2 ( P) β 2 ( P) ) ... b2 0.5 b1 a1 + q2 2 ( 1 l12) I2 ( P) b2 a2 φ 1 ( P) := exp 3 4 Psat1 := 2.9 10 V1 := 125 bar cm mol Eqs. (14.98) and (14.99), with φsat1 = 1 and (P - Psat1) = P, combine to give: y1 ( P) := Psat1 P φ 1 ( P) P V 1 R T exp 10 5 y1 ( P) 10 1 0 50 100 150 P bar Note: y axis is log scale. 586 200 250 300 14.45 A labeled diagram of the process is given below. The feed stream is taken as the α phase and the solvent stream is taken as the β phase. F nF xF1 = 0.99 xF2 = 0.01 R nR xα 1 xα2 = 0.001 Feed Mixer/ Settler S nS xS3 = 1.0 E nE xβ 2 xβ 3 Solvent 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 xF2 := 0.01 xS3 := 1 xα2 := 0.001 xα1 := 1 xα2 Apply mole balances around the process as well as an equilibrium relationship A12 := 1.5 From p. 585 2 γα 2 ( x2) := exp A12 ( 1 x2) A23 := 0.8 2 γβ 2 ( x2) := exp A23 ( 1 x2) Material Balances nS + nF = nE + nR (Total) nS = xβ3 nE (Species 3) xF1 nF = xα1 nR (Species 1) Substituting the species balances into the total balance yields xF1 1 nS + nF = nS + nF xβ3 xα1 Solving for the ratio of solvent to feed (nS/nF) gives nS nF xα1 xF1 xβ3 1 xβ3 xα1 = 587 We need xβ3. Assume exiting streams are at equilibrium. Here, the only distributing species is 2. Then xα2 γα 2 = xβ2 γβ 2 Substituting for γα2 and γβ2 ( ) ( ) ( ) 2 2 xα2 exp A12 1 xα2 = xβ2 exp A23 1 xβ2 Solve for xβ2 using Mathcad Solve Block xβ2 := 0.5 Guess: Given ( ) 2 2 xα2 exp A12 1 xα2 = xβ2 exp A23 1 xβ2 () xβ2 := Find xβ2 xβ2 = 0.00979 xβ3 := 1 xβ2 xβ3 = 0.9902 From above, the equation for the ratio nS/nF is: xα1 xF1 xβ3 1 xβ3 xα1 nSnF := a) nSnF = 0.9112 Ans. b) xβ2 = 0.00979 Ans. 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 γ1α and γ2β. xα1 := 520 6 xα2 := 1 xα1 xβ2 := 10 2 6 10 588 xβ1 := 1 xβ2 γα 1 := γβ 2 := xβ1 3 Ans. 5 Ans. γα 1 = 1.923 × 10 xα1 1 xα1 γβ 2 = 4.997 × 10 1 xβ1 3 14.50 1 - butanenitrile Psat1 := 0.07287bar V1 := 90 cm Psat2 := 0.29871bar V2 := 92 cm mol 3 2- benzene 3 B1 , 1 := 7993 cm mol T := 318.15K i := 1 .. 2 mol 3 B2 , 2 := 1247 cm mol 3 B1 , 2 := 2089 P := 0.20941bar j := 1 .. 2 cm mol B2 , 1 := B1 , 2 x1 := 0.4819 x2 := 1 x1 k := 1 .. 2 y1 := 0.1813 y2 := 1 y1 Term A is calculated using the given data. term_Ai := yi P xi Psati Term B is calculated using Eqns. (14.4) and (14.5) δ j , i := 2 B j , i B j , j Bi , i φhati := exp P 1 Bi , i + 2 R T Bi , i Psati R T φsati := exp y jyk( 2 δ j , i δ j , k) j k term_Bi := φhati φsati Term C is calculated using Eqn. (11.44) fsati := φsati Psati 1.081 1.108 term_A = Vi ( P Psati) fi := φsati Psati exp R T 0.986 1.006 term_B = 589 term_Ci := 1 1 term_C = fsati fi 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 = 2 A 2 1 and 1 x1 = 1 2 A 2 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) := 1 1 A ( T) 2 + 22 A ( T) xr := 0.7 xl := 0.3 xspl1 ( T) := 1 1 A ( T) 2 22 A ( T) From Eq. (E) in Example 14.5, the solubility curves are solved using a Solve Block: 590 1 xr xr xr > 0.5 xr1 ( T) := Find ( xr) 1 xl xl xl < 0.5 xl1 ( T) := Find ( xl) Given A ( T) ( 1 2xr) = ln Given A ( T) ( 1 2xl) = ln Find the temperature of the upper consolute point. T := 300K A ( T) = 2 Tu := Find ( T) 0.3 Given 0.5 Tu = 345.998 K T := 250K .. 346K 360 340 320 300 280 260 240 0.1 0.2 0.4 0.6 0.7 0.8 xl1 xr1 xspl1 xpr1 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 ω 1 := 0.193 Tc1 := 556.4K Pc1 := 45.60bar A1 := 14.0572 B1 := 2914.23 C1 := 232.148 Psat1 ( T) := exp A1 kPa T 273.15 + C 1 K B1 591 2 - n-heptane ω 2 := 0.350 Tc2 := 540.2K Pc2 := 27.40bar A2 := 13.8622 B2 := 2910.26 C2 := 216.432 Psat2 ( T) := exp A2 kPa T 273.15 + C 2 K B2 T := ( 100 + 273.15)K Tr1 := T Tc1 Tr1 = 0.671 Psat1r := Tr2 := T Tc2 Tr2 = 0.691 Psat2r := Psat1 ( T) Pc1 Psat2 ( T) Pc2 Λ 12 := 1.5410 Using Wilson's equation Psat1r = 0.043 Psat2r = 0.039 Λ 21 := 0.5197 γ 1 ( x1) := exp ln x1 + ( 1 x1) Λ 12 ... Λ 12 Λ 21 + 1 x ( 1) x + 1 x Λ 1 x + x Λ ( 1) 1 21 1) 12 1 ( γ 2 ( x1) := exp ln ( 1 x1) + x1 Λ 21 ... Λ 12 Λ 21 + x ( 1) x + 1 x Λ 1 x + x Λ ( 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 γ 1 ( x1) Psat1 ( T) + ( 1 x1) γ 2 ( x1) Psat2 ( T) yi1 ( x1) := x1 γ 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 φhat and φsat. ( φsat1 := PHIB Tr1 , Psat1r , ω 1 ) φsat1 = 0.946 P φhat1 ( P) := PHIB Tr1 , , ω1 P ( c1 φsat2 := PHIB Tr2 , Psat2r , ω 2 ) φ 1 ( P) := c2 φsat1 φsat2 = 0.95 P φhat2 ( P) := PHIB Tr2 , , ω2 P φhat1 ( P) φ 2 ( P) := φhat2 ( P) φsat2 Solve Eqn. (14.1) for y1 and P given x1. Guess: y1 := 0.5 P := 1bar Given y1 φ 1 ( P) P = x1 γ 1 ( x1) Psat1 ( T) ( 1 y1) φ2 (P) P = ( 1 x1) γ 2 ( x1) Psat2 (T) Eqn. (14.1) 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 ) 1.7 bar Pi ( x1 ) 1.6 bar Pii ( x1 ) bar Pii ( x1 ) bar 1.5 1.4 1.3 1.2 1.1 1 0 0.2 0.4 0.6 x1 , yi1 ( x1 ) , x1 , yii1 ( x1 ) P-x Raoult's P-y Raoult's P-x Gamma/Phi P-y Gamma/Phi 594 0.8 Chapter 15 - Section A - Mathcad Solutions 15.1 Initial state: Liquid water at 70 degF. H1 38.05 BTU lbm S1 0.0745 BTU (Table F.3) lbm rankine Final state: Ice at 32 degF. H2 ( 0.02 T (0 7 143.3) BTU S2 lbm 0.0 143.3 491.67 BTU lbm rankine 459.67)rankine (a) 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 Wideal 12.466 Wdotideal H1 T S2 S1 BTU mdot lbm lbm 1 sec Wdotideal mdot Wideal 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 Work Wdot TH 491.67 rankine TH QC QC T TC H2 QC mdot Work 14.018 Wdot TC 14.79 kW t Wdot 0.889 181.37 BTU lbm BTU lbm Work Wdotideal t H1 Ans. 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: H1 1 H2 mdot Wdot HA lbm sec HA 2.484 Wdot HD mdot HB 16.77 kW Wdotideal t lbm mdot Ans. Ans. 0.784 t Wdot sec 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 Sliq 0.0433 BTU lbm BTU lbm rankine Hvap 106.48 Svap 0.2229 597 BTU lbm BTU lbm rankine HA Hvap SA Svap For sat. liquid at 98 degF, P=134.75(psia): HC BTU lbm 44.24 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 BTU lbm 118 HB 121.84 SB 0.228 SD Sliq HB BTU H'B HA HA The entropy at this H is read from Fig. G.2 at P=134.75(psia) lbm BTU lbm rankine xD Svap HD HC Sliq xD SD HD Hliq Hvap 0.094 Hliq xD 0.284 BTU lbm rankine Refrigerent circulation rate: H2 mdot Wdot H1 1 HA lbm sec HA 2.914 Wdot HD mdot HB 47.22 kW Wdotideal t t Wdot THERMODYNAMIC ANALYSIS Wdotlost.compressor Qdotcondenser mdot T mdot HC Wdotlost.condenser mdot T lbm mdot SB T 0.279 (0 7 sec Ans. 459.67)rankine SA HB SC 598 SB Ans. Qdotcondenser Wdotlost.throttle mdot T Wdotlost.evaporator T SD SC mdot SA SD lbm S 2 S1 1 sec 13.152 kW 27.85% Wdotlost.compressor 8.305 kW 17.59% Wdotlost.condenser 14.178 kW 30.02% Wdotlost.throttle 6.621 kW 14.02% Wdotlost.evaporator 4.968 kW 10.52% Wdotideal 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 298.15 K 257190 J 86.513 J K 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 599 nCO2 1 mol nN2 1.881 mol (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 For unmixing the air, define H298 nN2 y1 nair 0.79 y2 1 y1 By Eq. (12.35) with no minus sign: Sunmixing nair R y1 ln y1 Sunmixing 10.174 y2 ln y2 J K For mixing the products of reaction, define y1 nCO2 nN2 Smixing S T y1 nCO2 nCO2 Sunmixing 300 K 0.347 y2 nN2 R y1 ln y1 S298 Wideal 600 T y1 y2 ln y2 S Smixing S Smixing H 1 81.223 Wideal 15.465 J K J K 259 kJ Ans. (b) Adiabatic combustion: Heat-capacity data for the product gases from Table C.1: A nCO2 5.457 nN2 3.280 A 11.627 B 2.16 mol B nCO2 1.045 nN2 0.593 3 10 10 3 mol D nCO2 1.157 nN2 0.040 5 10 D 1.082 5 10 mol T For the products, CP HP = R dT T0 R 298.15 K T0 The integral is given by Eq. (4.7). Moreover, by an energy balance, H298 Guess HP = 0 2 . A 11.627 B 3 2.160 10 K D 5 1.082 10 K Given H298 = R mol A T0 Find 8.796 1 B 2 T0 2 T 601 T0 2 D T0 1 T 1 2622.603 K 2 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 29.701 H H298 H 2.83 S 5 0.0 1.082 10 R mol ICPS H Wideal.cooling 5 10 J 208904 J 0.8078 t Wideal = 29.701 S Wideal.cooling Wideal.cooling t 3 T 246.934 J K S Ans. Ans. The surroundings increase in entropy in the amount: Q H298 Wideal.cooling S Q T 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 S 246.93 J K For sat. liquid and vapor at 1000 kPa, Table F.2: Hliq 762.6 Hvap kJ kg 2776.2 Sliq kJ kg Svap 2.1382 6.5828 kJ kg K kJ Tsat kg K 453.03K (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 0.1 H1 H3 kg s mdot2 H2 mdot1 S3 2 Sliq mdot3 H3 mdot1 Hliq Tsat Given H3 mdot3 H1 mdot1 S3 mdot3 S1 mdot1 mdot3 = mdot1 S3 = Sliq H3 H2 mdot2 = 0 S2 mdot2 = 0 mdot2 H3 kJ s kJ sK Hliq mdot3 = 300 Hliq Tsat mdot1 mdot2 mdot3 Find mdot1 mdot2 mdot3 H3 S3 H3 S3 603 kJ s mdot2 mdot1 H3 0.086 2.767 kg s 10 mdot2 3 kJ S3 kg 0.064 6.563 kg s kJ kg K mdot3 0.15 kg s Ans. 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 S1 Sliq turb xturb xturb Hliq 0.919 H'turb 2.614 10 0.78 Hturb H1 turb H'turb Hturb x'turb H'turb 2.655 10 Sturb Sliq xturb Svap Sturb 6.316 kJ kg K Svap Sliq Hturb Hliq Hvap Hliq 0.94 x'turb Hvap Hliq 3 kJ kg H1 3 kJ kg 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 Hcomp H2 kJ 2993.5 comp kg H'comp H2 Hcomp comp By interpolation: Scomp 7.1803 604 kJ kg K 0.75 3084.4 kJ kg 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 0.15 kg H3 s 2770. kJ kg Given Hcomp H2 mdot2 = H3 mdot3 H1 mdot1 mdot3 = mdot1 Hturb H1 mdot1 H2 mdot2 = 0 mdot2 kJ s H3 Hliq mdot3 = 300 kJ s mdot1 mdot2 Find mdot1 mdot2 mdot3 H3 mdot3 H3 mdot1 mdot3 0.10608 0.14882 kg s kg s mdot2 H3 0.04274 2.77844 kg s 3 kJ 10 kg Ans. Steam at Point 3 is slightly superheated. By interpolation, S3 THERMODYNAMIC ANALYSIS 6.5876 T kJ kg K 300K By Eq. (5.25), with the enthalpy term equal to zero: Wdotideal T mdot3 S3 mdot1 S1 605 mdot2 S2 (assumed) Wdotideal 6.014 kW Wdotlost.turb T mdot1 Sturb Wdotlost.comp S1 T mdot2 Scomp Wdotlost.mixing Wdotlost.turb T mdot3 S3 S2 mdot1 Sturb 2.9034 kW 48.2815% 2.054 kW mdot2 Scomp 34.1565% Wdotlost.comp Wdotlost.mixing 1.0561 kW 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: BTU lbm Hliq 12.090 Svap 0.22525 Hvap BTU Sliq lbm rankine 103.015 0.02744 BTU lbm BTU lbm rankine For sat. liquid at the condenser outlet temperature of 80 degF: H4 37.978 H2 BTU lbm Hvap x1 x1 H1 Hvap S4 S2 H1 Hliq 0.285 606 Sliq S1 Hliq 0.084 BTU lbm rankine H4 S1 Svap 0.07892 x1 Svap BTU lbm rankine Sliq From Example 9.1(b) for the compression step: H BTU lbm 17.48 H3 H2 H H3 120.5 BTU lbm From Fig. G.2 at H3 and P = 101.37(psia): S3 0.231 Wdot BTU lbm rankine mdot Wdot mdot H 1845.1 3.225 lbm hr 10 4 BTU 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 (0 7 QdotC 459.67)rankine 120000 Wdotideal Qdot H4 TH T TC BTU hr QdotC Wdotlost.comp TH ( 10 TC Wdotideal TC T mdot S3 Wdotlost.throttle T mdot S4 T mdot S1 1.533 4 BTU 10 hr S2 H3 mdot Wdotlost.cond 459.67)rankine Qdot S3 S4 607 Qdot 5 BTU 1.523 10 hr Wdotlost.evap 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). BTU 47.53% Wdotideal 15329.9 Wdotlost.comp 5619.4 BTU hr 17.42% Wdotlost.cond 3625.2 BTU hr 11.24% Wdotlost.throttle 4730.2 BTU hr 14.67% Wdotlost.evap 2947.6 BTU hr 9.14% hr The percent values above express each quantity as a percentage of the actual work, to which they sum: Wdot 15.5 32252.3 BTU hr 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 600 500 20 QdotC 400 10 300 0 200 608 BTU sec TC tC Wdotideal 459.67 rankine QdotC TH TC TC For sat. liquid and vapor at the evaporator temperature, Table 9.1: 21.486 18.318 Hliq 107.320 105.907 BTU 15.187 lbm 12.090 104.471 Hvap 103.015 9.026 Sliq Hvap 0.22244 0.04065 H2 lbm 101.542 0.04715 BTU 0.22325 BTU Svap lbm rankine 0.03408 0.02744 0.22418 0.22525 BTU lbm rankine S2 Svap 0.22647 0.02073 For sat. liquid at the condenser temperature: H4 x1 37.978 H1 Hvap BTU lbm Hliq Hliq S4 0.07892 S1 Sliq BTU lbm rankine x1 Svap H1 H4 Sliq From the results of Pb. 9.9, we find: 117.7 118.9 H3 120.1 121.7 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: 123.4 609 8.653 0.227 7.361 0.229 0.231 S3 0.234 BTU lbm rankine mdot H4 Wdotlost.cond S2 H3 mdot T mdot S4 Wdotlost.throttle Wdotlost.evap sec 3.146 T mdot S3 Qdot lbm 4.613 0.237 Wdotlost.comp 6.016 S3 T mdot S1 S4 T mdot S2 T H1 S1 H2 TC Qdot mdot The final term accounts for the entropy change of the refrigerated space (an internal heat reservoir). Wdot mdot H3 H2 36.024 40.844 Wdotideal 20.9 22.419 BTU 41.695 sec 38.325 Wdotlost.comp 21.732 21.379 17.547 30.457 610 BTU sec 11.149 8.754 10.52 9.444 Wdotlost.cond 10.589 BTU sec 11.744 Wdotlost.throttle 7.292 11.826 5.322 10.322 89.818 12.991 95.641 11.268 9.406 Wdotlost.evap BTU sec BTU Wdot sec BTU 94.024 7.369 5.122 sec 86.194 68.765 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 (0 7 459.67)rankine TH TC ( 30 459.67)rankine QdotC Wdotideal QdotC TH TC T Wdotideal TC 2000 BTU sec 163.375 BTU sec For sat. liquid and vapor at the evaporator temperature, Table 9.1: Hliq Hvap H2 18.318 BTU lbm 105.907 Hvap BTU lbm Sliq Svap S2 611 0.04065 0.22325 Svap BTU lbm rankine BTU lbm rankine For sat. liquid at the condenser temperature: H4 37.978 BTU lbm S4 0.07892 BTU lbm rankine From Problem 9.12, H2A H3 BTU lbm 116. H2A S2A 14.667 BTU lbm H3 0.2435 130.67 BTU lbm rankine 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 H2A H1 Hliq Hvap H2 BTU lbm H1 S1 0.109 Sliq x1 Svap S1 Hliq 27.885 0.061 BTU lbm rankine Sliq 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 From Problem 9.12: mdot 25.634 612 lbm sec Wdotlost.comp Qdot H4 T mdot S3 S2A H3 mdot Wdotlost.cond T mdot S4 Wdotlost.throttle Wdotlost.evap T mdot S1 S3 Qdot S4A 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 T mdot S2A Wdot H2A mdot H3 54.31 BTU sec 14.45% 87.08 BTU sec S4 43.45% 23.16% 163.38 Wdotlost.comp Wdotlost.cond Wdotlost.throttle 9.98 BTU sec Wdotlost.evap 45.07 Wdotlost.exchanger 16.16 375.97 S4A BTU sec Wdotideal Wdot S2 BTU sec 2.65% BTU sec 11.99% BTU 4.30% sec 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 0.75 comp kJ kg H1 419.1 H2 2676.0 S1 kJ kg K (sat. liquid) S2 kJ kg 1.3069 7.3554 kJ kg K (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 comp 2757.3 kJ kg 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 614 500.4 kJ kg This enthalpy is a bit larger than that of sat. liquid at 110 degC; find quality and then the entropy: Hliq 461.3 Slv Sliq T Hlv kJ kg K x4 5.8203 S4 kJ kg kJ kg K 300 K x4 Slv Wdotideal Wdotlost.comp Wdot Wdotideal Wdotlost.evap Wdotlost.comp Wdot H1 mdot T S4 mdot T mdot H3 H4 S3 Hliq Hlv S4 mdot H4 Wdotlost.evap 15.8 kJ kg 2230.0 1.5206 T S4 S3 S2 kJ kg K Sliq x4 mdot 1.4185 0.018 0.5 kg sec S1 S1 S2 H2 8.606 kW 21.16% 24.651 kW 60.62% 7.41 kW 18.22% 40.667 kW The figures on the right are percentages of the actual work, to which the terms sum. 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 257.6 S1 0.3970 H2 1461.2 S2 1.6671 H3 1242.2 S3 1.7431 H4 1047.8 S4 1.8748 H5 69.7 S5 0.1326 H7 250.2 S7 0.4112 QH H2 BTU lbm H1 1 lbm Wideal QH 1 BTU lbm rankine 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 T T S1 1 lbm QH TH (From Pb. 8.8) 0.18688 lbm Wlost.turbine S2 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 Wlost.pump.heater T 1 lbm S5 1 lbm m S4 m S7 S5 m S7 S3 1 lbm S1 Q The absolute value of the actual work comes from Pb. 8.8: Wabs.value = 374.61 BTU Wlost.boiler.reservoir 50.43% 224.66 BTU 30.24% Wlost.turbine 98.81 BTU 13.30% Wlost.cond.valve 36.44 BTU 4.90% 8.36 BTU 1.13% Wlost.pump.heater Wideal 742.82 BTU (absolute value) 15.9 The numbers on the right are percentages of the absolute value of the ideal work, to which they sum. 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 1140.0 kJ kg S4 9.359 kJ kg K H5 1009.7 kJ kg S5 8.894 kJ kg K H7 719.8 kJ kg S7 7.544 H9 285.4 kJ kg S9 4.928 H10 796.9 kJ kg S10 617 9.521 kJ kg K kJ kg K kJ kg K H14 1042.1 kJ kg S14 11.015 kJ kg K H15 1188.9 kJ kg S15 11.589 kJ kg K H11 H5 S11 S5 H13 H10 S13 S10 H6 S6 H5 S5 H10 S12 H12 T S10 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 x 0.113 0.25 Q Wlost = T SG T ______________________________________________________________ Wideal = ( m) H fs Wideal H15 ( 1 Wideal 489.001 Wout H12 T ( m) SG = S fs z) H9 z 0.044 H11 x kJ kg K Wout Wlost.a (b) Heat Exchanger II: SG.b SG.b 0.313 kJ kg K T S15 ( 1 z) S9 z S4 S14 ( 1 z) kJ kg (a) Heat Exchanger I: SG.a SG.a H4 ( m) S fs Wlost.b kJ kg S5 S4 T SG.a S7 S6 ( 1 T SG.b 618 S15 Wlost.a x) Wlost.b 13.021 S14 kJ kg S13 ( 1 92.24 kJ kg z) (c) Expander: SG.c 0.157 kJ kg K (d) Throttle: SG.d 0.964 SG.c S12 Wlost.c SG.d kJ kg K S9 z Wlost.d S11 x Wlost.c T SG.c S10 ( 1 T SG.d z 46.241 x) S7 ( 1 x) Wlost.d 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 65.22% 1.478 100.00% Work analysis, Eq. (15.3): kJ/kg Percent of Wout 53.20 10.88% Wlost.a 13.02 2.66% Wlost.b 92.24 18.86% Wlost.c 46.24 9.46% Wlost.d 284.30 58.14% 489.00 100.00% Note that: = Wideal 619 kJ kg 284.304 kJ kg Chapter 16 - Section A - Mathcad Solutions 16.10(Planck's constant) h P 34 6.626 10 Js T 1bar (Boltzmann's constant) 23 J k 1.381 10 K M 6.023 10 mol 3 RT V P 0.025 R ln MkT 5 2 2 Ve NA 2 h 83.800 b) For Krypton: M Sig R ln MkT 2 131.30 M R ln NIST value: 154.84 J mol K Sig 164.08 J mol K J mol K Ans. Ve NA Ans. gm mol NA 3 Sig J mol K 5 2 2 2 mol NA h c) For Xenon 154.84 Sig gm mol 3 2 m NA NIST value: 164.05 Sig 1 mol 3 2 23 NA gm 39.948 a) For Argon: V 298.15K (Avagodro's number) MkT 2 h 2 5 2 Sig Ve NA 164.08 NIST value: 169.68 620 J mol K J mol K Ans. 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, kg·m2 N·m energy [=] [=] s3 s time (b) Electric current is by denition the time rate of transfer of electrical charge. Thus Power [=] Charge [=] (electric current)(time) [=] A·s (c) Since power is given by the product of current and electric potential, then kg·m2 power [=] current A·s3 (d) Since (by Ohms Law) current is electric potential divided by resistance, Electric potential [=] kg·m2 electric potential [=] 3 current A2 ·s (e) Since electric potential is electric charge divided by electric capacitance, Resistance [=] 4 A2 ·s charge [=] Capacitance [=] electric potential kg·m2 1.3 The following are general: ln x = ln 10 × log10 x P sat /kPa = P sat /torr × ( A) 100 kPa 750.061 torr (B) t / C = T /K 273.15 (C ) 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 = 2.3026 a b T /K 273.15 + c + ln 100 750.061 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 mass·velocity2 . Its fundamental units are therefore: E K [=] kg·m2 ·s2 [=] N·m [=] J Potential energy as given by Eq. (1.7) has units of mass·length·acceleration. Its fundamental units are therefore: E P [=] kg·m·m·s2 [=] N·m [=] 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 Second [=] 0.031797 Sc 1 Minute [=] 60 Second [=] 0.19078 Mn 1 Hr [=] 3144.96 Second 1 Hour [=] 3600 Second [=] 1.14469 Hr 1 Dy [=] 31449.6 Second 1 Day [=] (24)(3600) Second [=] 2.74725 Dy 1 Wk [=] 314496. Second 1 Week [=] (7)(24)(3600) Second [=] 1.92308 Wk 1 Mo [=] 3144960 Second 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 κ + N . According to the laboratory report a pure material ( N = 1) is in 4-phase (κ = 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 κ + 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 κ + 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 t t Tw0 Cw + T0 C t = Tw Cw + T C t or (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 ) dτ (C ) In combination with Eq. ( A) this becomes: Ct or dT =K dτ Dene: Ct dT = K Tw0 t (T T0 ) T Cw dτ T T0 Tw0 T t t Cw C βK 1 1 +t t Cw C = T K 1 1 +t Cw Ct αK +K T0 Tw0 +t t Cw C T0 Tw0 +t Cw Ct where both α and β are constants. The preceding equation may now be written: dT = α βT dτ Rearrangement yields: 1 d (α β T ) dT = dτ = β α βT α βT Integration from T0 to T and from 0 to τ gives: α βT 1 ln α β T0 β 624 =τ α βT = exp(βτ ) α β T0 which may be written: When solved for T and rearranged, this becomes: T= where by the denitions of α and β , α α + T0 β β exp(βτ ) Tw C t + T0 C t α = 0 tw Cw + C t β When τ = 0, the preceding equation reduces to T = T0 , as it should. When τ = , it reduces to T = α/β . Another form of the equation for α/β is found when the numerator on the right is replaced by Eq. ( B ): t Tw Cw + T C t α = t Cw + C t β By inspection, T = α/β 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 22 625 2.21 We reformulate the denition of Reynolds number, with mass owrate m replacing velocity u : ˙ m = u Aρ = u ˙ Solution for u gives: u= Whence, Re π2 Dρ 4 ˙ 4m π D2ρ ˙ 4m ˙ 4 m ρD uρ D = = 2ρ µ π Dµ πD µ (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 d (mU ) ˙ +Hm =0 dt and 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 dm dU = m H U Multiply by dt and rearrange: 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: By Eq. (1.3), P= W = RT V b V2 V1 W = RT ln Whence, 2.35 Recall: Whence, By Eq. (2.4), d(P V ) = P d V + V d P d W = V d P d(P V ) V2 V1 RT d ( V b) V b V1 b V2 b and and dW = P dV W= V dP (PV ) d Q = dU d W By Eq. (2.11), U = H P V With d W = P d V Whence, P dV = Q= and dU = d H P d V V d P the preceding equation becomes d Q = d H V d P H V dP 626 . . 2.38 (a) By Eq. (2.24a), m = u Aρ With m , A, and ρ 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, ρ 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 dn . n = dt . dU dn +n Q = ( H U ) dt dt But a material balance yields Then . dU dn +n Q = PV dt dt or H2 H1 + 1 (u 2 u 2 ) = 0 1 22 . . 4m m = u= π ρ D2 Aρ 2.44 (a) By Eq. (2.32a): By Eq. (2.24a): Then u2 2 u2 1 = 4 π 2 . m2 ρ2 1 1 4 4 D1 D2 1 1 ( P2 P1 ) + 2 ρ . Solve for m : and given 4 π 2 π . m = 2ρ( P1 P2 ) 4 . m2 ρ2 2 H2 H1 = 4 4 D1 D2 44 D1 D2 44 D1 D2 4 4 D1 D2 1 ( P2 P1 ) ρ =0 1 /2 . (b) Proceed as in part (a) with an extra term, Here solution for m yields: . m = 2 ρ ( P1 P2 ) ρ 2 C (T2 T1 ) π 4 2 44 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 : πξ πP π πT T = P πV πP 1 V2 = 1 V2 πV πT T P πV πT πV πP + P T π2V πT π P 1 V =ξ + π2V π PπT = ξ π2V π PπT π2V π PπT 1 V 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: πV πP 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 ] dV = dP V Integration from the initial state ( P1 , V1 ) to an intermediate state ( P , V ) for constant 3.7 (a) For constant T , Eq. (3.4) becomes: ln Whence, gives: V = ( P P1 ) V1 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 A(T ) d V = A(T ) exp( P )d P P d V = A(T ) P2 P1 P exp( P )d P [( P1 + 1) exp( P1 ) ( P2 + 1) exp( P2 )] 628 With V1 = A(T ) exp(κ P1 ) and V2 = A(T ) exp(κ P2 ), this becomes: W= 1 [(κ P1 + 1)V1 (κ P2 + 1)V2 ] κ W = P1 V1 P2 V2 + or V1 V2 κ 3.11 Differentiate Eq. (3.35c) with respect to T : T 1δ δ P [(1δ)/δ]1 dT dP =T + P (1δ)/δ dz dz dT P (1δ)/δ d P =0 + P (1δ)/δ dz dz P 1δ δ Algebraic reduction and substitution for d P /dz by the given equation yields: dT 1δ =0 (Mρ g ) + dz δ T P For an ideal gas T ρ/ P = 1/ R . This substitution reduces the preceding equation to: δ1 δ Mg dT = R dz 3.12 Example 2.13 shows that U2 = H . If the gas is ideal, H = U + P V = U + RT For constant C V , and U2 U = RT U2 U = C V (T2 T ) and C V (T2 T ) = RT C P CV R T2 T = = CV CV T Whence, T2 = γ T When C P / C V is set equal to γ , this reduces to: 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 (δ = 0). Here, Eqs. (3.36) and (3.37) reduce to: W = RT1 (1 1) and Q= γ RT1 (1 1) γ 1 Both are indeterminate. The easiest resolution is to write Eq. (3.36) and (3.37) in the alternative but equivalent forms: W= RT1 δ1 T2 1 T1 and Q= (δ γ ) RT1 (δ 1)(γ 1) T2 1 T1 from which we nd immediately for δ = 0 that: W = R (T2 T1 ) and Q= 629 γR (T2 T1 ) = C P (T2 T1 ) γ 1 Isothermal case (δ = 1). Equations (3.36) and (3.37) are both indeterminate of form 0/0. Application of lHˆ pitals rule yields the appropriate results: o W = RT1 ln Note that if P2 P1 y P2 P1 and (δ 1)/δ Q = RT1 ln (δ 1)/δ P2 P1 1 dy =2 δ dδ then P2 P1 ln P2 P1 Adiabatic case (δ = γ ). In this case simple substitution yields: W= P2 P1 RT1 γ 1 (γ 1)/γ 1 and Q=0 Isochoric case (δ = ). Here, simple substitution yields: W =0 and Q= RT1 P2 1 = γ 1 P1 RT1 γ 1 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 630 P1 = 101.33 kPa P2 = 1500 kPa 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 Q = 172.71 kJ or 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 Note also: (c) If n 1 = 0, (d) If in addition Q = 0, Whence, 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 n2 = P2 Vtank RT2 n1 = P1 Vtank RT1 n 2 (C V T2 C P T ) = Q C V T2 = C P T and T2 = CP CV T T2 = γ T (e) 1. Apply the result of Part (d ), with γ = 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 Vtank = 4 × 106 cm3 P2 = 3 bar Appropriate values for R are therefore: R (denominator) = 83.14 bar cm3 mol1 K1 R (numerator) = 8.314 J mol1 K1 Numerically, 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 d n = 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, dU = C V dT or n dU ( H U )dn = d Q H U = P V = RT d Q = mC dT where m is the mass of the tank, and C is its specic heat. nC V dT RT dn = mC dT Thus, 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 or Integration yields: ln T2 T1 = n 2 C V + mC R ln n 1 C V + mC CV T2 = T1 or In addition, n1 = n 2 C V + mC n 1 C V + mC P1 Vtank RT1 and R/CV P2 Vtank RT2 n2 = 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, U = CV T P V = RT Whence, But U= 1 CV CV = = γ 1 C P CV R CV R The isothermal work is then: Whence, W = W = RT ln 3.29 Solve the given equation of state for V : U= Therefore : dV = V1 P2 P1 V= P d V = RT P2 P1 1 dP P Compared with Eq. (3.27) V= 633 1 (PV ) γ 1 RT + B RT P RT dP P2 V2 T (PV ) 3.28 Since Z = P V / RT the given equation can be written: Differentiate at constant T : (PV ) = R θ RT +b RT P V P Whence, By denition [Eq. (3.3)]: T V P 1 V κ RT P2 = T Substitution for both V and the derivative yields: RT θ RT +b RT P κ= P2 Solve the given equation of state for P : RT P= V b+ P T Differentiate: R = + θ V b+ RT V θ RT dθ θ dT T V b+ θ 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 dθ θ 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 + ( + σ )bV + σ b2 V b (V + b)(V + σ b) Substitute V = 1/ρ : Z= 1 a (T )ρ 1 RT 1 + ( + σ )bρ + σ (bρ)2 1 bρ Expressed in series form, the rst term on the right becomes: 1 = 1 + bρ + (bρ)2 + · · · 1 bρ The nal fraction of the second term becomes: 1 = 1 ( + σ )bρ + [( + σ )2 σ ](bρ)2 + · · · 1 + ( + σ )bρ + σ (bρ)2 Combining the last three equations gives, after reduction: Z =1+ b Equation (3.12) may be written: Comparison shows: B =b ( + σ )a (T )b 2 a (T ) ρ + ··· ρ + b2 + RT RT Z = 1 + Bρ + Cρ 2 + · · · a (T ) RT and 634 C = b2 + ( + σ )ba (T ) RT 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= 0.42748 RTc a (T ) = Tr1.5 Pc RT 0.08664 RTc Pc 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 + 3 D P 2 + · · · T Z P Whence, =B T , P =0 Z = 1 + Bρ + Cρ 2 + Dρ 3 + · · · Equation (3.12) with V = 1/ρ : Z ρ Differentiate: = B + 2Cρ + 3 Dρ 2 + · · · T Z ρ Whence, =B T ,ρ =0 3.56 The compressibility factor is related to the measured quantities by: Z= By Eq. (3.39), (a) By Eq. ( A), Thus M PV t PV t = m RT n RT B = ( Z 1)V = ( A) ( Z 1) M V t m dT dm dV t dP dM dZ + t + = T m V P M Z (B) (C ) Max |% δ Z | |% δ M | + |% δ P | + |% δ V t | + |% δ m | + |% δ 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 ), By Eq. (C ), dm dM dV t Z dZ dB + t+ = m M V Z 1 Z B Z dB = Z 1 B dT dP T P Therefore 635 + 2Z 1 Z 1 dm dM dV t + t m M V Max |% δ B | + Z Z 1 |% δ P | + |% δ T | 2Z 1 Z 1 |% δ V t | + |% δ M | + |% δ 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 P V = T a (V b)2 b RT 3/2 (V + b)2 RT 3/2 (V b)2 (V + b)2 ( A) 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) = T In addition, we have the mathematical relation: ( Z / V )T ( P / V )T (C ) 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) Z P = T Combining these three equations gives Z P = T lim For P 0, V , and Eq. ( D ) becomes: P 0 lim For P , V b, and Eq. ( D ) becomes: P Z P Z P = T b a / RT 3/2 RT = T b RT 3.60 (a) Differentiation of Eq. (3.11) gives: Z P = B + 2C P + 3 D P 2 + · whence T If the limiting value of the derivative is zero, then B = 0, and lim P 0 Z P =B T B = B RT = 0 (b) For simple uids, ω = 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 (γ P /γ T )V = Constant. γP γT (a) By Eq. (3.4) applied to a constant-V process: γP γT (b) For an ideal gas P V = RT , and = V = V β κ R V (c) Because a and b are constants, differentiation of Eq. (3.42) yields: γP γT = V 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 ρ , 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 P s. Inappropriate where changes in V are required. (f ) Rackett equation: Saturated liquids; a corresponding states application. (g) Constant β , κ 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/ρ substituted everywhere for V . Subtract 1 from each side of the equation and divide by ρ . Take the limit as ρ 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 σ ) σ 2 ( + σ) ( σ 2 ( = 3Zc + 1) + + 1) + 2 = 3Zc 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 σ 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 σ = 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 σ = 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): Z = 1 + Bρ + Cρ 2 + . . . Eliminate ρ : Z =1+ = 0.077796 Z =1+ = 0.457236 where ρ = P / Z RT C P2 BP + 2 2 2 + ... ZRT Z RT 2 P2 C P2 B Pc Pr ˆ ˆ Pr + C · Pr + . . . + 2 c2 · 2 r 2 + . . . = 1 + B · · Z 2 Tr2 Z Tr R Tc Z Tr RTc Z Tr ( Z 1) Z Tr ˆ ˆ Pr + . . . = B +C · Z Tr Pr Rearrange: ˆ 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: Set V = 1/ρ : whence P= a RT 2 V b V Z= Z rep = Z= a V V b V RT aρ bρ aρ 1 =1+ RT 1 bρ RT 1 bρ bρ 1 bρ Z attr = 638 aρ RT 3.76 Write each modication in Z -form, (a) Z= a V RT V b lim Z = 1 V The required behavior is: (b) Z= lim Z = 1 V a V 2 RT ( V b) lim Z = V Z= V a 1 V b V RT lim Z = 0 V lim Z = 1 The required behavior is: (d ) Z =1 a RT lim Z = 1 The required behavior is: (c) a RT V aρ a =1 RT V RT Although lim Z = 1 as required, the equation makes Z linear in ρ ; i.e., a 2-term virial EOS in V ρ . Such an equation is quite inappropriate at higher densities. 3.77 Refer to Pb. 2.43, where the general equation was developed; For an ideal gas, n= Also for an ideal gas, PV t RT dn = dt and dU = C V dT whence . PV t Q = RT RT 2 ln Integration yields: 3.78 By Eq. (3.4), Integrate: ln PV t RT 2 D2 Vt V2 = ln 2t = ln 2 = ln 2 V1 V1 D1 Note that P V t / R = const. dT dU = CV dt dt P V t dT dT PV t dT = CP CV + RT dt dt RT dt t2 R T2 = CP PV t T1 dV = β dT κ d P V dT dt . dU dn +n Q = PV dt dt t1 . Q dt where β and κ are average values D1 + δ D D1 2 = ln 1 + δD D1 2 = β(T2 T1 ) κ( 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 (ν + 1) + T12 (ν 2 + ν + 1) 3 2 where ν 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 (ν + 1) T1 ν + T1 T2 + T1 = = 2 2 2 and 2 Tam = T12 2 (ν + 2ν + 1) 4 C B T1 (ν + 1) + T12 (ν 2 + 2ν + 1) 4 2 Dene ε as the difference between the two heat capacities: C Pam = A + Whence, ν 2 + ν + 1 ν 2 + 2ν + 1 4 3 ε C P C Pam = C T12 C T12 (ν 1)2 12 Making the substitution ν = T2 / T1 yields the required answer. This readily reduces to: ε= 4.6 For consistency with the problem statement, we rewrite Eq. (4.8) as CP = A + D B T1 (ν + 1) + 2 ν T12 where ν 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 = Whence, T1 (ν + 1) 2 C Pam = A + and 2 Tam = T12 2 (ν + 2ν + 1) 4 4D B T1 (ν + 1) + 2 2 2 T1 (ν + 2ν + 1) Dene ε as the difference between the two heat capacities: ε C P C Pam = This readily reduces to: ε= D T12 D T12 ν 4 1 2 ν + 2ν + 1 ν ν 1 ν +1 2 Making the substitution ν = 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) H298 = 393,509 2H2 (g) + O2 (g) 2H2 O(g) H298 = (2)(241,818) CH4 (g) C(s) + 2H2 (g) CH4 (g) + 2O2 (g) CO2 (g) + 2H2 O(g) H298 = 74,520 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) H298 = (10)(393,509) 11 H2 (g) + 5 1 O2 (g) 11 H2 O(g) 2 H298 = (11)(241,818) 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 = 249,700 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 11 H2 O(g) 11 H2 O(l) C10 H22 (l) + 15 1 O2 (g) 10 CO2 (g) + 11 H2 O(l) 2 4.49 Saturated because the large H298 = 6,345,388 H298 = (11)(44,012) H298 = 6,829,520 J (HHV) H l v overwhelms the sensible heat associated with superheat. Water because it is cheap, available, non-toxic, and has a large H lv . The lower energy content is a result of the decrease in H l v 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 πU = 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 = πU t + π E K + π 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 π Stotal = π S t + π Ssurr Just as πU t for the egg is zero, so is π S t . Therefore, t π Stotal = π Ssurr = Q Q surr = Tξ Tξ Since Q is negative, π Stotal is positive, and the process is irreversible. T = 1 TC H 5.6 By Eq. (5.8) the thermal efciency of a Carnot engine is: 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 π S = C P ln P1 T1 (a) If P2 = P1 , Whence, π S P = C P ln If V2 = V1 , T2 T1 π SV = C P ln T2 T1 R ln 642 T2 P2 = T1 P1 T2 T1 T2 T1 = C V ln Since C P > C V , this demonstrates that (b) If T2 = T1 , ST = R ln Whence, SP > SV . P2 P1 If P2 P1 SV = C P ln This demonstrates that the signs for R ln ST and V2 = V1 , P2 P1 = C V ln P2 T2 = P1 T1 P2 P1 SV are opposite. 5.12 Start with the equation just preceding Eq. (5.14) on p. 170: ig ig dP C dT C dT dS d ln P = P =P P RT RT R For an ideal gas P V = RT , and ln P + ln V = ln R + ln T . Therefore, dT dV dP = + T V P ig dV dT C dT dS = + =P V T RT R Whence, dV dT dP = V T P or ig dT CP + d ln V 1 T R ig ig Because (C P / R ) 1 = C V / R , this reduces to: ig C dT dS + d ln V =V RT R S = R Integration yields: T T0 ig V C V dT + ln V0 RT ********************** 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 : ig ig ig ig C dV C dP dP C dV C dP dS +P =V +P =P RV RP P RV RP R ig Integration yields: 5.13 ig V C C P S + P ln = V ln V0 R P0 R R 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 Whence, C t dTH dTC = H t C C TH TC or d ln TC = d ln TH where t CH t CC Integration from TH0 and TC0 to TH and TC yields: TH TH0 TC = TC0 TH TH0 TC = TC0 or 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 t t W = C H (TH TH0 ) + CC (TC TC0 ) Integration yields: Eliminate TC by the boxed equation of Part (a) and rearrange slightly: W= TH TH0 TH t 1 + CC TC0 TH0 t C H TH0 1 (c) For innite time, TH = TC T , and the boxed equation of Part (a) becomes: T = TC0 From which: T T = (TC0 )1/( Because /( +1) (TH0 ) TC0 TH0 +1 /( +1) + 1) 1 = 1/( T = TH0 T TH0 = TC0 TH0 T = TC0 (TH0 ) and T = (TC0 )1/( TH0 +1) (TH0 ) /( +1)1 + 1), then: 1/( +1) and T TH0 = TC0 TH0 /( +1) Because TH = T , substitution of these quantities in the boxed equation of Part (b) yields: W= 5.14 t C H TH0 TC0 TH0 1/( +1) 1 + 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 CC TC0 TC0 TH0 /( +1) 1 t (a) With d Q C = CC dTC , Eq. ( B ) becomes: TH d QH = t TC CC dTC t d Q H = CC or 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 TC t + CC (TC TC0 ) TC0 dTC t + CC dTC TC t W = CC TH ln or 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: . |W | . r 1 1 = |W | 1r 1r = k Ar 4 (TH )4 or where A= Differentiate, noting that the quantity in square brackets is constant: . . |W | 1 3 |W | dA = + = k ( T H )4 (1 r )r 4 (1 r )2r 3 k (TH )4 dr Equating this equation to zero, leads immediately to: 4r = 3 . |W | k (TH )4 r TC TH 1 (1 r )r 3 4r 3 (1 r )2r 4 r = 0.75 or 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 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 If m 1 = m 2 , 645 T= T = (T1 + T2 )/2 m 1 T1 + m 2 T2 m1 + m2 (1) 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 S t = mC P ln Because T = (T1 + T2 )/2 > T2 T1 T2 S t = 2mC P ln or (2) T T1 T2 S t is positive. T1 T2 , 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. T T0 C P dT T0 T 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. 5.24 By denition, CP Similarly, CP H = dT dT T0 T CP T T= = ln(T0 / T ) ln(T / T0 ) T T0 S = 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 lHˆ 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 | |Qσ | Heat transfer to the house at temperature T Heat transfer from the furnace at TF Heat transfer from the surroundings at Tσ The rst and second laws provide the two equations: |Q| = |Q F | + |Qσ | and 646 |Qσ | |Q| |Q F | =0 Tσ TF T Combine these equations to eliminate | Q σ |, and solve for | Q F |: |Q F | = |Q| With T = 295 K TF = 810 K T Tσ TF Tσ TF T Tσ = 265 K and | Q | = 1000 kJ | Q F | = 151.14 kJ The result is: 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 | |W |refrig = | Q 2 | | Q σ | Equation (5.7) may be applied to both the engine and the refrigerator: Tσ |Qσ | TF |Q F | = = T | Q2| T | Q1| Combine the two pairs of equations: |W |engine = | Q 1 | TF T TF 1 = | Q1| T T |W |refrig = | Q 2 | 1 Tσ T = | Q2| T Tσ T Since these two quantities are equal, | Q1| T Tσ TF T = | Q2| T T or | Q2| = | Q1| Because the total heat transferred to the house is | Q | = | Q 1 | + | Q 2 |, | Q| = | Q1| + | Q1| But | Q1| = | Q F | T TF TF T TF T = | Q1| 1 + T Tσ T Tσ whence |Q| = |Q F | = | Q1| T TF TF T T Tσ TF Tσ T Tσ TF Tσ T Tσ 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 Heat transfer from the tank at temperature T Heat transfer from the house at T Heat transfer to the surroundings at Tσ |Q| |Q | |Qσ | The rst and second laws provide the two equations: |Q| + |Q | = |Qσ | |Q | |Qσ | |Q| =0 T T Tσ and Combine these equations to eliminate | Q σ |, and solve for | Q |: Tσ T T Tσ |Q| = |Q | With T = 448.15 K T = 297.15 K T T Tσ = 306.15 K and | Q | = 1500 kJ | Q | = 143.38 kJ The result is: 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 σ1 | |W |refrig = | Q σ2 | | Q | Equation (5.7) may be applied to both the engine and the refrigerator: Tσ | Q σ1 | = T |Q| Tσ | Q σ2 | = T |Q | |W |engine = | Q | 1 Tσ T Combine the two pairs of equations: = |Q| |W |refrig = | Q | T Tσ T Tσ T = |Q | |Q| = |Q | Tσ T T Tσ T T Since these two quantities are equal, |Q| Tσ T T Tσ = |Q | T T or 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 + Tσ, j dt j 648 Tσ t T . 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 Tσ, 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 Tσ (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: 1 1 T1 T2 SG = | Q | T1 T2 T1 T2 = |Q| For the completely reversible process the net work produced is Wideal : T1 Tσ T1 | W1 | = | Q | and | W2 | = | Q | Wideal = |W1 | |W2 | = Tσ | Q | T2 Tσ T2 T1 T2 T1 T2 This is the work that is lost, Wlost , in the direct, irreversible transfer of heat | Q |. Therefore, Wlost = Tσ | Q | T1 T2 = Tσ 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 = By difference, with Since Tirrev T0 ig CP Srev = 0: P dT ln P T Sirrev = Srev = Tirrev Trev ig CP Trev T0 dT T Sirrev must be greater than zero, Tirrev must be greater than Trev . 650 ig CP P dT ln P T Chapter 6 - Section B - Non-Numerical Solutions νH νS 6.1 By Eq. (6.8), and isobars have positive slope =T P ν2 H ν S2 Differentiate the preceding equation: ν2 H ν S2 Combine with Eq. (6.17): νT νS = P T CP = P P and isobars have positive curvature. 6.2 (a) Application of Eq. (6.12) to Eq. (6.20) yields: νCP νP νCP νP or ν {V T (ν V /ν T ) P } νT = T νV νT = T νCP νP Whence, νV νT For an ideal gas: = P ν2V νT 2 T P = T T R P P νV νT P ν2V νT 2 P ν2V νT 2 and P =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 νP νT νV νT By Eqs. (3.2) and (6.34): V νP νT C P = CV + T Restrict to constant P : dV + = εV V νV νT dP P P νP νT and P C P CV = ε T V Combine with the boxed equation: νV νT = V ε ρ ε ρ 6.3 By the denition of H , U = H P V . Differentiate: νU νT = P νH νT P P νV νT or P 651 νU νT = CP P P νV νT P Substitute for the nal derivative by Eq. (3.2), the denition of β : U T = CP β PV P Divide Eq. (6.32) by dT and restrict to constant P . The immediate result is: U T P T = CV + T P P V V T P Solve for the two derivatives by Eqs. (6.34) and (3.2); substitution gives: U T 6.4 (a) In general, = CV + P β (β T κ P )V κ P T dU = C V dT + T By the equation of state, P= RT V b P dV (6.32) V P T whence = 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 H T Then, = 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 γ 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 R dT d ln(V b) = CV T But from part (b), R / C V = (C P C V )/ C V = γ 1. Then or d ln T = (γ 1)d ln(V b) From which: d ln T + d ln(V b)γ 1 = 0 or T (V b)γ 1 = const. Substitution for T by the equation of state gives P (V b)(V b)γ 1 = const. R 652 or P (V b)γ = const. 6.5 It follows immediately from Eq. (6.10) that: V= G P and S= T 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 and U = H P V = H RT These become: H = (T ) T d (T ) dT and U = (T ) T d (T ) RT dT By Eqs. (2.16) and (2.20), CP = Because H T and CV = P U T V 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 S= T 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 and U = H PV = H PK These become: H = F (T ) + K P T d F (T ) dT and U = F (T ) T By Eqs. (2.16) and (2.20), CP = H T and P 653 CV = U T V d F (T ) dT 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 σ = δ = 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: Then, Z PV RT V= Z= 1 ρ a q b RT 1 exp(qbρ) 1 bρ With the denition, ξ bρ , this becomes: Z= Because ρ = P / Z RT , 1 exp(q ξ ) 1ξ ξ= bP Z RT Given T and P , these two equations may be solved iteratively for Z and ξ . Because b is a constant, Eqs. (6.58) and (6.59) may be rewritten as: 654 ( A) GR = RT ξ 0 ( Z 1) HR = RT ξ 0 dξ + Z 1 ln Z ξ (B) dξ + Z 1 ξ (C ) Z T ξ In these equations, Z is given by Eq. ( A), from which is also obtained: ln Z = ln(1 ξ ) q ξ Z T and = ξ qξ exp(q ξ ) T (1 ξ ) 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: ξ 0 ( Z 1) dξ = exp(q ){ E [q (1 ξ )] E (q )} E (q ξ ) ln(q ξ ) γ ξ where γ is Eulers constant, equal to 0.57721566. . . . ξ and T 0 Z T ξ sξ = q exp(q ){ E [q (1 ξ )] E (q )} ξ 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. B T ( A) B Tc (B) 6.19 Write Eq. (6.75) in log10 form: log P sat = A Apply at the critical point: log Pc = A 655 By difference, 1 1 T Tc log Pr sat = B =B Tr 1 T (C ) If P sat is in (atm), then application of ( A) at the normal boiling point yields: B Tn log 1 = A or A= =B Tc Tn Tn Tc =B Tn 1θ B Tn log Pc With θ Tn / Tc , Eq. ( B ) can now be written: 1 1 Tc Tn log Pc = B Whence, B= Equation (C ) then becomes: Tr 1 log Pc = T Tn 1θ log Pr sat = log( Pr sat )Tr =0.7 = Apply at Tr = 0.7: 3 7 1θ Tn Tr 1 log Pc Tr θ 1θ θ 1θ log Pc By Eq. (3.48), ω = 1.0 log( Pr sat )Tr =0.7 Whence, ω= 3 7 θ 1θ 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 T S and = 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 With C P = a + bT , P CP S T C2 P CP T = P =b T T 2 2 CP CP and P CP T 1 P T CP T S CP T = P T T 1 2 CP CP =1 P 6.84 Division of Eq. (6.8) by d S and restriction to constant T yields: Therefore, =T +V T P S By Eq. (6.25), T H S =T T 656 1 1 = (β T 1) β β P S P a bT = a + bT a + bT Because this quantity is positive, so then is the curvature of an isobar. H S CP T = T 1 βV 2 H S2 Also, = T 1 β2 β S = T 1 β2 2 H S2 Whence, By Eqs. (3.2) and (3.38): V T Whence, β= = P 1 V β P = T V T P S T 1 = T β P β3V 1 β2 β P T and V= P dB R + dT P and T 1 βV β= 1 V RT +B P dB R + dT P Differentiation of the second preceding equation yields: β P = T V P From the equation of state, β P Whence, = T V P 1 V2 dB R + dT P R V P2 = T = T 1 R (β V ) 2 V V P2 V P T RT P2 R β RT R (β T 1) = + 2 2 V P2 VP VP Clearly, the signs of quantity (β T 1) and the derivative on the left are the same. The sign is determined from the relation of β and V to B and d B /dT : βT 1 = T V 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 β 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 (). GR = BP By Eqs. (6.54) and (6.56), S R = P (d B /dT ) and 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 (). V R = B, By Eqs. (3.38) and (6.40), and V R is (). Combine the equations above for G R , S R , and H R : HR = P B T R Therefore, C P dB dT HR T Whence, is (+). HR T =P P (See Fig. 6.5.) P 657 dB d2 B dB T 2 dT dT dT = PT d2 B dT 2 6.89 By Eq. (3.5) at constant T : (a) Work P = dW = P dV = W= W= = 1 κ V 1 P1 ln κ V1 1 1 V 1 P1 d V = ln V d V P1 + ln V1 d V ln κ κ κ V1 V2 V1 ln V d V P1 + 1 ln V1 (V2 V1 ) κ 1 1 [(V2 ln V2 V2 ) (V1 ln V1 V1 )] P1 (V2 V1 ) (V2 ln V1 V1 ln V1 ) κ κ V2 1 + V1 V2 P1 (V2 V1 ) V2 ln V1 κ By Eq. (3.5), ln (b) Entropy V2 = κ( P2 P1 ) V1 ln V1 ln V P1 κ κ dS = β βV d ln V = d V κ κ and 1 d ln V κ d P = S= and By Eq. (6.28), Substitute for d P : V2 V1 κ d S = β V d P P = (c) Enthalpy W = P1 V1 P2 V2 whence By Eq. (6.29), By Eq. ( A), ( A) β (V2 V1 ) κ d H = (1 β T ) V d P d H = ( 1 β T ) V · H= 1 βT 1 dV d ln V = κ κ 1 βT (V1 V2 ) κ 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 6.90 The given equation will be true if and only if H = 134.55 kJ kg1 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 ig Neither C P 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. H ig P = S H ig P + T 658 H ig T P T P T P ig S = CP V T P S T P T S ig = S S ig P P H ig P = T S ig P =T S S ig P T ig CP T T ig 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 ): S P dS = S V dP + V dV P By the chain rule for partial derivatives, S T dS = T P V P V T V T V S T dP + P dV P With Eqs. (6.30) and (6.17), this becomes: dS = 6.93 By Eq. (6.31), P=T (a) For an ideal gas, Therefore RT RT = V V (b) For a van der Waals gas, Therefore P T RT V P= T P CV T V dP + V U V P T and U V P= CP T T = V U V T a RT 2 V b V U V =0 T P T and = V T U V = T A = a (Tc ) · Tc 6.94 (a) The derivatives of G with respect to T and P follow from Eq, (6.10): G T and V= P Combining the denition of Z with the second of these gives: Z P PV = RT RT 659 G P T = T (3/2) A + b) T 1 /2 V ( V 1 2 S = R V b U V and (c) Similarly, for a Redlich/Kwong uid nd: where R V and RT a RT 2= V b V b V dV G P T a V2 Combining Eqs. (2.11) and (3.63) and solving for U gives U = G + T S P V . G T U =GT Replacing S and V by their derivatives gives: 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 2G T 2 = T P U P P 2 G T P = T T 2G T P P 2G P2 P T From the denition of C V and an equation relating partial derivatives: U T CV U T = V U P + P P T T V Combining the three equations yields: C V = T 2G T 2 P P 2G T P T 2G T P 2G P2 +P T P T V Evaluate ( P / T )V through use of the chain rule: P T = V P V V T 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 PT P T Then and C V = T 2G T 2 P P V P and 2G T P = V 2G P2 = T T ( 2 G / T ) P ( 2 G / P 2 )T 2G T P +T +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 l v . Both decrease monotonically as T increases, becoming zero at the critical point. 660 H l v and H sl S sl d P sat = = T V sl V sl dT sl V is assumed approximately constant, then 6.98 By the Clapeyron equation: S sl to If the ratio P sat = A + BT H sl to If the ratio 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: = dT Because Hts v = t Htsl t sat d Pl v dT t Pt RTt2 Hts v Htl v is positive, then so is the left side of the preceding equation. H lv d P sat = T V lv dT 6.100 By Eq. (6.72): But Htl v Pt Htl v Pt Htl v RTt2 RTt2 Z tl v t sat d Pl v dT d Ps sat v Pt Hts v Pt Hts v RTt2 RTt2 Z ts v = d Ps sat v dT V lv = RT P sat Z lv H lv d ln P sat = RT 2 Z l v dT whence (6.73) lv H 1 H lv Tc H l v d ln Pr sat = ·2 = = 2 Z lv lv 2 Z lv Tr RTc Tr Z RT dTr 6.102 Convert αc to reduced conditions: αc 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 ω Tr =1 Thus, αc (L/K) = 5.82 for ω = 0, and increases with increasing molecular complexity as quantied by ω. 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 V T T CP 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 T U = U U V V T The two partial derivatives on the right are found from Eqs. (2.16) and (6.31); thus, T V = U 1 CV P T PT 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 P S = S V S V P This can also be written: P V = S P T V T S S T V 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 662 = T CP CV 1 κV c2 = Substitute into Eq. ( A): V CP MC V κ V CP MC V κ c= or (a) For an ideal gas, V = RT / P and κ = 1/ P . Therefore, cig = RT CP M CV (b) For an incompressible liquid, V is constant, and κ = 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 2γ P1 V1 1 γ +1 γ 1 = γ P1 V1 2 γ +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 J·kg1 . Moreover, P1 V1 = RT1 / M ; whence Pa · m3 · kg1 = u2 throat = γ RT1 M 2 γ +1 With R in units of J·(kg mol)1 ·K1 , RT1 / M has units of J·kg1 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 Z T Because P P = ρ Z RT , ρ ρ= w + ρ Z T = x w + Z T = Z T Whence, x y = Z ρ P P Z RT Z ρ y w y z T ρ T P T ρ T P ρ T and P R = P 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 = ρ Z ρ T ρ T ρ T and P = P ρ P = 2 T Z RT Combining these two equations yields: Z T T =ρ ρ Z ρ T (a) Equation (3.42) with van der Waals parameters becomes: a RT 2 V b V P= Multiply through by V / RT , substitute Z = P V / RT , V = 1/ρ , and rearrange: Z= aρ 1 RT 1 bρ In accord with Eq. (3.51), dene q a /b RT . In addition, dene ξ bρ . Then, Z= Z T Differentiate: 1 qξ 1ξ = ρ Z T ( A) = ξ ξ dq dT By Eq. (3.54) with α (Tr ) = 1 for the van der Waals equation, q = dq = dT Z T Then, In addition, Z ρ Tr2 Tc = (ξ ) ρ =b T 1 d Tr = dT 1 Tr2 Z ξ 664 = T q T = = / Tr . Whence, q 1 = T T Tr qξ T b qb (1 ξ )2 Substitute for the two partial derivatives in the boxed equation: T bρ qξ qbρ = (1 ξ )2 T or qξ = 1 ξ =1 2q Whence, ξ qξ (1 ξ )2 (B) By Eq. (3.46), Pc = R Tc /b. Moreover, P = Zρ RT . Division of the second equation by the rst gives Pr = Zρ bT / Tc . Whence Pr = Z ξ 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 ξ , 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= qξ 1 1+ξ 1ξ ( A) By Eq. (3.54) with α(Tr ) = Tr0.5 for the Redlich/Kwong equation, q = 1.5 q dq = T dT Z ρ Moreover, Z T and T = ρ = / Tr1.5 . This leads to: 1.5 q ξ T (1 + ξ ) bq b 2 (1 + ξ )2 (1 ξ ) Substitution of the two derivatives into the boxed equation leads to a new Eq. ( B ): q= 1+ξ 1ξ 2 1 2.5 + 1.5 ξ (B) As in Part (a), for a given Tr , calculate q , and solve Eq. ( B ) for ξ , 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. 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 = l 1.8604 1.3027 S1 S2 = 0.0921 = l v 7.3598 1.3027 S2 S2 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 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), n1 = Moreover, (γ 1)/γ ) P2 P1 T2 = P1 Vtank RT1 and n2 = P2 Vtank RT2 With γ = 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 By Eq. (6.8), Assume now that RT1 / P1 . Then H =n ˙ ( H )S η V dP = V ( H )S = P P is small enough that V , an average value, can be approximated by V1 = ( H )S = RT1 P1 P RT1 ˙ Ws = n ˙ η P1 and 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 = η For P2 P1 R/C P 1 P sufciently small, the quantity in square brackets becomes: P2 P1 R/C P 1= 1+ P P1 R/C P 1 The boxed equation is immediately recovered from this result. 666 1+ RP C P P1 1 7.41 The equation immediately preceding Eq. (7.22) page 276 gives T2 = T1 π . With this substitution, Eq. (7.23) becomes: π 1 T1 π T1 = T1 1 + T2 = T1 + η η 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: CP ln ln π = R π 1 CP S ln 1 + = η R R 1+ π 1 η π η+π 1 CP SG ln = ηπ R R Whence, 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 (π 1) where π P2 P1 R/C P This equation is a proper basis, because compressor efciency η and owrate n are xed. With all ˙ other variables constant, differentiation yields: dπ d Ws = T1 (π 1) + C P dC P dC P From the denition of π , ln π = Then, and P2 R ln P1 CP whence P2 R 1 dπ d ln π ln = = 2 P1 π dC P dC P CP P2 πR dπ ln = 2 P1 dC P CP π R P2 d Ws ln = T1 π 1 P1 CP dC P = T1 (π 1 π ln π ) When π = 1, the derivative is zero; for π > 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 V ( P / T )V V = = P ( P / V )T T (e) Rearrange the given equation: P T T = V V T P For the nal equality see footnote on p. 266. This result is the equation of (c). V CP 1 · · M CV κ 7.50 From the result of Pb. 7.3: c = With Then V= RT +B P c = PV V P then = T where RT P2 κ= Also, let γ = B γ RT + RT M T CP CV γ RT M BP γ γ = 1+ = ( RT + B P ) RT M RT M RT c= V P 1 V γ 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: ig HS = HS = V dP = RT dP P (const S ) Z RT dP P (const S ) V ig d P = HS ig HS = Z RT d P (const S ) P Z RT d P (const S ) P By extension, and with equal turbine efciencies, 7.52 By Eq. (7.16), H = η( H ) S . W H = . ig = Z H ig W For C P = constant, T2 T1 = η[(T2 ) S T1 ] 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 : 668 T2 = T1 1 + η P2 P1 R/C P P2 P1 1 R/C P From which η= T2 1 T1 P2 P1 R/C P Note that η < 1 1 Results: For T2 = 318 K, η = 1.123; For T2 = 348 K, η = 1.004; For T2 = 398 K, η = 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 ηDiesel 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 rc 1 = r re or Equation (8.7) can therefore be written: ηDiesel ηDiesel = 1 or 1 (1/ r )γ =1 γ 1/ r (rc / r )γ (1/ r )γ rc / r 1/ r 1 =1 γ 1 r γ 1 γ rc 1 rc 1 γ rc 1 γ (rc 1) (b) We wish to show that: γ rc 1 >1 γ (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, Then, R g [1 + θ (x 1)] · ( x 1)2 2! (0 < θ < 1) x a = 1 + a · (x 1) + 1 a · (a 1) · [1 + θ (x 1)]a 2 · (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) γ rc 1 >1 γ (rc 1) and (c) If γ = 1.4 and r = 8, then by Eq. (8.6): 1 8 0.4 ηOtto = 1 ηDesiel = 1 1 8 0.4 rc = 2 ηDesiel = 1 1 8 0.4 rc = 3 ηOtto = 0.5647 and 21.4 1 1.4(2 1) and ηDiesel = 0.4904 31.4 1 1.4(3 1) and ηDiesel = 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, η where, W AB WC D Q BC 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, η= TB T A TA TB + TC TD =1 TC TD TC TD By Eq. (3.30b), TB = T A PB PA (γ 1)/γ and PB PA TA Then, PD PC TD = TC η =1 TC 1 (γ 1)/γ (γ 1)/γ 1 PA PB (γ 1)/γ Multiplication of numerator and denominator by ( PB / PA )(γ 1)/γ gives: η =1 TA TC 671 PB PA (γ 1)/γ = TC PA PB (γ 1)/γ 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 δ/(δ1) ln P = ln K + δP dP = δ1T dT δ = 0 d P /dT = 0 δ = 1 d P /dT = δ d2 P = 2 δ1 dT P 1 dP 2 T T dT P δ d2 P = (δ 1)2 T 2 dT 2 δ dT dP = δ1 T P Sign of d P /dT is that of δ 1, i.e., + ( A) Special cases By Eq. ( A), δ ln T δ1 = δ1 δ1T Constant P Constant T P δP T δ1T Sign of d 2 P /dT 2 is that of δ , i.e., + (B) 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 δ/(δ1) , this requires δ = . 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), δ = γ . In this case Eqs. ( A) and ( B ) become: P γ d2 P = (γ 1)2 T 2 dT 2 γP dP = γ 1T dT We note here that γ /(γ 1) and γ /(γ 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. δ=γ δ=1 δ= P δ=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 0 0 T T Figure 2: The Otto cycle Figure 1: The Carnot cycle P P 0 0 0 0 T T Figure 4: The Brayton cycle Figure 3: The Diesel 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 ν For a Carnot heat pump, ν= |Q H | |W | TH |Q H | = TH TC | Q H | | QC | 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 T H ln TC TH 1 + TH TC 9.5 Differentiation of Eq. (9.3) yields: ερ ε TC = TH TH TC 1 = + 2 (TH TC )2 (TH TC ) TH TC and ερ ε 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, ρ is given by Eq. (9.3). Write this equation for the two cases: ρ= TC TH TC and ρσ = TσC Tσ H TσC Because the directions of heat transfer require that TH > Tσ H and TC < TσC , a comparison shows that ρ < ρσ and therefore that ρ 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 If Carnot = 0. 6 TC TH TC < 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 (a) By denition, h and m H2 O m air or m air = h= Mair pair V t RT 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: P PHsat 2O h pc = h rel P PHsat (h rel /100) 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 γ1 and ln γ2 , 2 y1 P = x1 exp( Ax2 ) P1sat 2 y2 P = x2 exp( Ax1 ) P2sat and 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 x1 = K 1 x 1 + K 2 (1 x 1 ) = 1 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 For water, Ms = 18.015 g mol1 Thus, or Hi = or for xi 0 Hi = 0.018015 kg mol1 . 1 = 1633 bar (0.018015)(0.034) This is in comparison with the value of 1670 bar in Table 10.1. 678 1 Ms k i Chapter 11 - Section B - Non-Numerical Solutions 11.6 Apply Eq. (11.7): ν (nT ) ν ni ¯ Ti νn ν ni =T P ,T ,n j =T ν (n P ) ν ni ¯ Pi T , P ,n j =P P ,T ,n j νn ν ni νm ν ni 11.7 (a) Let m be the mass of the solution, and dene the partial molar mass by: m i ¯ =P T , P ,n j T , P ,n j Let Mk be the molar mass of species k . Then m= ε nk Mk = ni Mi + ε n j M j k and νm ν ni = T , P ,n j ( j = i) j ν (n i Mi ) ν ni Whence, = Mi m i = Mi ¯ T , P ,n j νM t ν mi ˜ (b) Dene a partial specic property as: Mi If Mi is the molar mass of species i , = T , P ,m j mi Mi ni = νM t ν ni ν ni ν mi and T , P ,m j ν ni ν mi = T , P ,m j 1 Mi Because constant m j implies constant n j , the initial equation may be written: 11.8 By Eqs. (10.15) and (10.16), Because V = ρ 1 then dV ¯ V1 = V + x2 d x1 1 dρ dV =2 ρ d x1 d x1 ¯ Mi ˜ Mi = Mi dV ¯ V2 = V x1 d x1 and whence 1 x2 dρ 1 ¯ = V1 = 2 ρ ρ d x1 ρ 1 x2 dρ ρ d x1 = dρ 1 ρ x2 2 d x1 ρ 1 x1 dρ 1 ¯ = V2 = + 2 ρ ρ d x1 ρ With T , P ,m j 1+ x1 dρ ρ d x1 = dρ 1 ρ + x1 2 d x1 ρ 2 ρ = a0 + a1 x 1 + a2 x 1 and dρ = a1 + 2a2 x1 d x1 1 2 ¯ V1 = 2 [a0 a1 + 2(a1 a2 )x1 + 3a2 x1 ] ρ 679 and these become: 1 2 ¯ V2 = 2 (a0 + 2a1 x1 + 3a2 x1 ) ρ 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 (n M ) n1 ¯ For M1 , T , P ,n 2 ,n 3 n n1 Because n = n 1 + n 2 + n 3 , n n1 2n 1 1 3 2 n n = M1 + n 2 n 3 C T , P ,n 2 ,n 3 =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 For the mixture, P = Z RT / V . These two equations combine to give Z = i yi Z i . 11.11 The general principle is simple enough: ¯ ¯ ¯ 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 ) 3 680 or 3 ¯ H2 = 600 + 40 x1 (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 = x 1 =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 681 and 3 2 ¯ V2 = 70 + x1 + 14 x1 (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 ) 3 2 V = 70 + 58 x1 x1 7 x1 Whence, 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 ¯ d V2 2 = 2 x1 + 42 x1 d x1 and 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 = x 1 =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 682 dH ¯ H2 = H x1 d x1 Given that: H = x1 (a1 + b1 x1 ) + x2 (a2 + b2 x2 ) dH = a1 + 2b1 x1 (a2 + 2b2 x2 ) d x1 Then, after simplication, Combining these equations gives after reduction: ¯ H1 = a1 + b1 x1 + x2 (x1 b1 x2 b2 ) ¯ H2 = a2 + b2 x2 x1 (x1 b1 x2 b2 ) and 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 (n M ) ni = T , P ,n j x y = z (n M ) ni + w x w y w y (n M ) V + T , V ,n j z V ni T ,n T , P ,n j Whence, M ¯ ˜ Mi = Mi + n V T ,n V ni or T , P ,n j M ˜ ¯ Mi = Mi n V T ,n V ni T , P ,n j By denition, (nV ) ¯ Vi ni =n T , P ,n j V ni +V or T , P ,n j M ˜ ¯ ¯ Mi = Mi + (V Vi ) V Therefore, n V ni ¯ = Vi V T , P ,n j T ,x ˆ ˆ 11.20 Equation (11.59) demonstrates that ln φi is a partial property with respect to G R / RT . Thus ln φi = ¯ G i / RT . The partial-property analogs of Eqs. (11.57) and (11.58) are: ˆ ln φi P = T ,x ¯ Vi R RT ˆ ln φi T and = P ,x ¯ HiR RT 2 The summability and Gibbs/Duhem equations take on the following forms: GR = RT ˆ xi ln φi and i i 683 ˆ xi d ln φi = 0 (const T , P ) 11.26 For a pressure low enough that Z and ln φ are given approximately by Eqs. (3.38) and (11.36): Z =1+ BP RT and ln φ = BP RT ln φ Z 1 then: 11.28 (a) Because Eq. (11.96) shows that ln γi 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 γ1 = d (G E/ RT ) GE + x2 d x1 RT ln γ2 = 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 Then, d (G E/ RT ) 2 = 1.8 + 2 x1 + 2.4 x1 d x1 whence 3 2 ln γ1 = 1.8 + 2 x1 + 1.4 x1 1.6 x1 3 2 ln γ2 = x1 1.6 x1 and (b) In accord with Eq. (11.11), GE 2 3 2 3 = x1 ln γ1 + x2 ln γ2 = 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 γi and divide by d x1 : x1 d ln γ2 d ln γ1 =0 + x2 d x1 d x1 Differentiate the the boxed equations of part (a): d ln γ1 2 = 2 + 2.8 x1 4.8 x1 d x1 d ln γ2 2 = 2 x1 4.8 x1 d x1 and 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 γ1 d x1 = x 1 =1 684 d ln γ2 d x1 =0 x1 =0 (e) 11.29 Combine denitions of the activity coefcient and the fugacity coefcients: γi fˆi /xi P fi / P or γi = ˆ φi φi 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 SE = and GE T P ,x T T E = CP Working equations are then: E S1 = E H1E G 1 T1 E H2E = H1E + C P T T and and T E E E S2 = S1 + C P 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 E II. For C P = 0 I. E G1 (a) (b) (c) (d) (e) (f) (g) H1E E S1 E CP E S2 H2E E G2 E S2 H2E E G2 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 γ1 = A12 n 2 1 n1 n n2 1 n1 n n2 + A13 n 3 A23 n2n3 n2 = A12 x2 (1 x1 ) + A13 x3 (1 x1 ) A23 x2 x3 Similarly, ln γ2 = A12 x1 (1 x2 ) A13 x1 x3 + A23 x3 (1 x2 ) ln γ3 = A12 x1 x2 + A13 x1 (1 x3 ) + A23 x2 (1 x3 ) (b) Each ln γi is multiplied by xi , and the terms are summed. Consider the rst terms on the right of each expression for ln γi . 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 . x1 = 0: ln γ1 (x1 = 0) = A12 x2 + A13 x3 A23 x2 x3 For pure species 1, x1 = 1: ln γ1 (x1 = 1) = 0 For innite dilution of species 2, x2 = 0: 2 ln γ1 (x2 = 0) = A13 x3 For innite dilution of species 3, x3 = 0: 2 ln γ1 (x3 = 0) = A12 x2 (c) For innite dilution of species 1, GE = GR 11.35 By Eq. (11.87), written with M G and with x replaced by y : i yi G iR Equations (11.33) and (11.36) together give G iR = Bii P . Then for a binary mixture: G E = B P y1 B11 P y2 B22 P G E = P ( B y1 B11 y2 B22 ) or G E = δ12 P y1 y2 Combine this equation with the last equation on Pg. 402: GE T From the last column of Table 11.1 (page 415): S E = Because δ12 is a function of T only: SE = P ,x d δ12 P y1 y2 dT H E = δ12 T By the denition of G E , H E = G E + T S E ; whence, E Again from the last column of Table 11.1: C P = This equation and the preceding one lead directly to: 686 HE T d δ12 dT P ,x E C P = T d 2 δ12 P y1 y2 dT 2 P y1 y2 (G E / RT ) T 11.41 From Eq. (11.95): = P To an excellent approximation, write: H E RT 2 (G E / T ) T or (G E / T ) T = P H E T2 H E (G E / T ) 2 Tmean T P 0.271 785/323 805/298 (G E / T ) = 0.01084 = = 25 323 298 T From the given data: 1060 H E = 0.01082 = 2 3132 Tmean and The data are evidently thermodynamically consistent. 11.42 By Eq. (11.14), the Gibbs/Duhem equation, Given that ¯ M1 = M1 + Ax2 and x1 ¯ ¯ d M2 d M1 =0 + x2 d x1 d x1 ¯ M2 = M2 + Ax1 then ¯ d M1 = A d x1 and ¯ d M2 =A d x1 ¯ ¯ d M2 d M1 = x 1 A + x 2 A = A( x 2 x 1 ) = 0 + x2 d x1 d x1 The given expressions cannot be correct. Then 11.45 (a) For x1 22 M E = Ax1 x2 nd 2 ¯E M1 = Ax1 x2 (2 3x1 ) and Note that at both x1 = 0 (x2 = 1) and x1 = 1 (x2 = 0), In particular, 2 ¯E M2 = Ax1 x2 (2 3x2 ) ¯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 = 2 Ax1 (1 6x1 x2 ) = 2 Ax2 (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 = 2 A and For x1 = 0 d x1 d x1 For (b) For x1 = 1 ¯E d M1 =0 d x1 and ¯E d M2 = 2 A d x1 M E = A sin(π x1 ) nd: ¯E M1 = A sin(π x1 ) + Aπ x2 cos(π x1 ) and ¯E M2 = A sin(π x1 ) Aπ x1 cos(π x1 ) ¯E ¯E d M2 d M1 = Aπ 2 x1 sin(π x1 ) = Aπ 2 x2 sin(π 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 10 A Pb. 11.45 (a) i A . xi MEi MEbar2i xi 0 .. 100 2. xi 2 1 A . xi . xi . 1 MEbar1i xi . 2 3. 1 .01 . i .00001 A . xi . 1 xi xi 2 1.5 MEi MEbar1 1 i MEbar2i 0.5 0 0.5 0 MEi Pb. 11.45 (b) 0.2 0.4 A . sin p .x 0.6 xi 0.8 1 (pi prints as bf p) i MEbar1i A . sin p .x i A.p . 1 xi . cos MEbar2i A . sin p .x i A . p . xi . cos p .x p .x i i 40 30 MEi MEbar1 20 i MEbar2i 10 0 10 0 0.2 0.4 xi 687A 0.6 0.8 1 2 2 3 . xi (n M ) ¯ Mi = ni 11.46 By Eq. (11.7), At constant T and P , M ni = M +n T , P ,n j M xk dM = k T , P ,n j d xk T , P ,x j Divide by dn i with restriction to constant n j ( j = i ): M ni M ni T , P ,n j k xk ni nk xk = n With = T , P ,n j = 1 n M xk = 1 n k =i M xi T , P ,x j (k = i ) M 1 + (1 x i ) xi n M xk xk k xk T , P ,x j (k = i ) 1 ni n n2 T , P ,x j 1 n nj n k n2 = M xi ¯ Mi = M + T , P ,x j nj M xk xk xk ni k M xk T , P ,x j T , P ,x j T , P ,x j For species 1 of a binary mixture (all derivatives at constant T and P ): ¯ M1 = M + M x1 x2 M x1 x1 x2 x2 M x2 = M + x2 x1 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: M x1 dM = x2 d x1 + M x2 x1 d x2 Division by d x1 yields: dM = d x1 M x1 M x2 + 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 = n1 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 1 1 + n 1 + Bn 2 n 2 + Bn 1 ¯E M1 = An 2 + n1 B 1 (n 1 + Bn 2 )2 (n 2 + Bn 1 )2 x1 B 1 + 2 (x2 + Bx1 )2 (x1 + Bx2 ) Conversion back to mole fractions yields: 1 1 + x2 + Bx1 x1 + Bx2 ¯E M1 = Ax2 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 2 ¯E M2 = Ax1 B 1 + 2 (x2 + Bx1 )2 (x1 + Bx2 ) Similarly, (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 M2 = x1 2 d x1 d x1 ¯E d2 M E d M1 = x2 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. β 1 V V T 1 β id = i xi Vi V id = P Vi T xi i xi Vi i The claim is valid only if all the Vi are equal. 690 1 = P i xi Vi xi Vi βi i Chapter 12 - Section B - Non-Numerical Solutions 12.2 Equation (12.1) may be written: yi P = xi πi Pi sat . Summing for i = 1, 2 gives: P = x1 π1 P1sat + x2 π2 P2sat . d π2 d π1 dP π2 + π1 + P2sat x2 = P1sat x1 d x1 d x1 d x1 Differentiate at constant T : Apply this equation to the limiting conditions: For x1 = 0 : x2 = 1 π1 = π1 π2 = 1 For x1 = 1 : x2 = 0 π1 = 1 π2 = π2 = P1sat π1 P2sat or Then, dP d x1 dP d x1 Since both Pi sat and x1 =0 x1 =1 πi = P1sat P2sat π2 Whence, x1 =0 P2sat and 2 ln π1 = Ax2 12.4 By Eqs. (12.15), By Eq. (12.1), dP d x1 dP d x1 x 1 =0 x 1 =1 + P2sat = P1sat π1 P1sat = P2sat π2 are always positive denite, it follows that: dP d x1 Therefore, or d π2 =0 d x1 d π1 =0 d x1 ln and dP d x1 x1 =1 P1sat 2 ln π2 = Ax1 π1 2 2 = A ( x 2 x 1 ) = A ( x 2 x 1 ) = A (1 2 x 1 ) π2 y1 x2 P2sat π1 = = y2 x1 P1sat π2 y1 /x1 y2 /x2 P2sat P1sat = ξ12 r ln(ξ12 r ) = A(1 2x1 ) If an azeotrope exists, ξ12 = 1 at 0 az x1 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 π1 is accompanied by the opposite extremum in ln π2 . Thus the difference ln π1 ln π2 is also an extremum, and Eq. (12.8) becomes useful: d (G E/ RT π1 = ln π1 ln π2 = ln d x1 π2 Thus, given an expression for G E/ RT = g (x1 ), we locate an extremum through: d ln(π1 /π2 ) 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 dA = A12 A21 d x1 Moreover, d 2A =0 2 d x1 and d (G E/ RT ) = A12 A21 d x1 Then, A A12 x1 + A21 x2 where 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 dA d x1 2 dA dA 2 A12 A21 + x1 x2 A2 ( x 2 x 1 ) A 3 d x1 d x1 A 2 = = 2 A12 A21 A3 = A12 A21 2x1 x2 2( x 2 x 1 ) d A 2 + 2 A3 d x1 A A 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(γ1 /γ2 ). The Margules equation is given by Eq. (12.9b), here written: GE = Ax1 x2 RT where d 2A =0 2 d x1 dA = A21 A12 d x1 A = A21 x1 + A12 x2 dA d (G E/ RT ) = A( x 2 x 1 ) + x 1 x 2 d x1 d x1 Then, d 2A dA dA d 2 (G E/ RT ) + x1 x2 2 + (x2 x1 ) = 2 A + (x2 x1 ) 2 d x1 d x1 d x1 d x1 = 2 A + 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 2 A12 3( A21 A12 ) or 692 x1 = (r 2) 3(r 1) r A21 A12 When r = 2, x1 = 0, and the extrema in ln γ1 and ln γ2 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 γ2 curves at x1 = 1 are larger than the intercepts of the ln γ1 curves at x1 = 0. When r = 1/2, x1 = 1, and the extrema in ln γ1 and ln γ2 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 γ1 curves at x1 = 0 are larger than the intercepts of the ln γ2 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 γ1 = d (G E/ RT ) GE + x2 d x1 RT and ln γ2 = 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 x1 x2 x2 x1 d (G E/ RT ) 2 ( A12 A21 ) = A12 A21 D D d x1 Then, and ln γ1 = 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 ln γ1 = A12 1 + D A21 x2 2 A12 x1 A21 x2 2 = A12 A12 x1 + A21 x2 A21 x2 The equation for ln γ2 is derived in analogous fashion. (b) With the understanding that T and P are constant, ln γ1 = (nG E/ RT ) n1 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 2 Differentiation in accord with the rst equation gives: ln γ1 = A12 A21 n 2 ln γ1 = = n1 1 n D ( n D )2 (n D ) n1 n2 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 But [see page 438, just below Eq. (12.10b)]: A12 = ln γ1 1 GE (equimolar) = (ln γ1 + ln γ2 ) 8 RT 12.24 (a) By Eq. (12.6): 1 GE (equimolar) = ( A12 + A21 ) 8 RT and or A21 = ln γ2 1 GE (equimolar) = ln(γ1 γ2 ) 8 RT GE = x1 ln γ1 + x2 ln γ2 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 γ1 = 0.273 x2 and 2 ln γ2 = 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 γ2 d ln γ1 =0 + x2 d x1 d x1 whence 694 x1 d ln γ1 x1 d ln γ1 d ln γ2 = = x2 d x2 x2 d x1 d x1 Integrate, recalling that ln γ2 = 1 for x1 = 0: ln γ2 = ln(1) + x1 0 x1 d ln γ1 d x1 = x2 d x2 0 x1 d ln γ1 d x1 x2 d x2 d ln γ1 = 2 Ax2 d x2 2 (a) For ln γ1 = Ax2 , Whence x1 x1 ln γ2 = 2 A 0 x1 d x1 2 ln γ2 = Ax1 or GE = Ax1 x2 RT By Eq. (12.6), 2 (b) For ln γ1 = x2 ( A + Bx2 ), d ln γ1 2 2 = 2x2 ( A + Bx2 ) + x2 B = 2 Ax2 + 3 Bx2 = 2 Ax2 + 3 Bx2 (1 x1 ) d x2 Whence ln γ2 = 2 A 2 ln γ2 = Ax1 + 3B 2 3 x Bx1 21 x1 0 x1 d x1 + 3 B x1 0 x1 d x1 3 B 2 ln γ2 = x1 A + or x1 0 2 x1 d x1 B 3B 2 Bx1 = x1 A + (1 + 2x2 ) 2 2 3B GE 2 2 Bx1 ) = x1 x2 ( A + Bx2 ) + x2 x1 ( A + 2 RT Apply Eq. (12.6): Algebraic reduction can lead to various forms of this equation; e.g., B GE = x 1 x 2 A + (1 + x 2 ) 2 RT 2 2 (c) For ln γ1 = x2 ( A + Bx2 + C x2 ), d ln γ1 2 2 2 3 = 2x2 ( A + Bx2 + C x2 ) + x2 ( B + 2C x2 ) = 2 Ax2 + 3 Bx2 + 4C x2 d x2 = 2 Ax2 + 3 Bx2 (1 x1 ) + 4C x2 (1 x1 )2 Whence or ln γ2 = 2 A x1 0 x1 d x1 + 3 B ln γ2 = (2 A + 3 B + 4C ) ln γ2 = x1 0 x1 0 x1 (1 x1 )d x1 + 4C x1 d x1 (3 B + 8C ) 2 A + 3 B + 4C 2 2 ln γ2 = x1 A + 2 x1 x1 0 695 0 x1 (1 x1 )2 d x1 2 x1 d x1 + 4C 3 B + 8C 3 8C 3B + 2C B + 3 2 x1 3 4 x1 + C x1 2 x1 + C x1 x1 0 3 x1 d x1 or 2 ln γ2 = 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 ˜ Eliminating 1 + n gives: ˜ Differentiation yields: H= H= ( A) H 1d H 2 x1 d x1 x1 1 d x1 2 = x1 = (1 + n ) 2 ˜ dn ˜ dH = dn ˜ H x1 dHE dH = H E x1 d x1 d x1 E dH ¯ H2E = H E x1 d x1 Comparison with Eq. (11.16) written with M H E , dH ¯ = H2E dn ˜ shows that (b) By geometry, with reference to the following gure, Combining this with the result of Part (a) gives: From which, Substitute: H (1 + n ) ˜ H x1 H d x1 1d H dH = 2 = ˜ ˜ x1 d n dn ˜ x1 d n where Whence, and I= H= ¯ H2E = HI n ˜ H n H2E ˜¯ HE H = x1 x1 696 HI n ˜ dH = dn ˜ and n= ˜ x2 x1 d x1 dn ˜ Whence, ¯ H E x2 H2E x2 ¯ HE H2E = x1 x1 x1 I= ¯ ¯ However, by the summability equation, H E x2 H2E = x1 H1E ¯ I = H1E Then, 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= ˜ A12 n A21 + ˜ 1+n 1+n ˜ n ˜ 1+n ˜ lim (a) It follows immediately from the preceding equation that: lim (b) Because n /(1 + n ) 1 for n , it follows that: ˜ ˜ ˜ H =0 n 0 ˜ H = A12 n ˜ 2 ¯ H2E = x1 [ A21 + 2( A12 A21 )x2 ] (c) Analogous to Eq. (12.10b), page 438, we write: Eliminate the mole fractions in favor of n : ˜ 1 1+n ˜ ¯ H2E = 2 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 lim n 0 ˜ 12.42 By Eq. (12.29) with M H , H=H H t With H t Therefore, CP, = P ,x i dH = A21 dn ˜ xi Hi . Differentiate: H t H0 d( H ) = t t0 C P dt 697 P ,x i H t this becomes P ,x H xi Hi t P ,x P ,x xi C Pi = = CP H= i H0 + t t0 C P dt CP M E = x1 x2 M ( A) dM dME = M( x 2 x 1 ) + x 1 x 2 d x1 d x1 (B) 12.61 (a) From the denition of M: Differentiate: 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 S t = k ln Combining the last three equations gives: N! N1 ! N2 ! N3 ! · · · N! N1 ! N2 ! N3 ! · · · 1 N! 1 St St S = (ln N ! ln = = = N N1 ! N2 ! N3 ! · · · N kN R( N / N A) R From which: ln N ! N ln N N 1 S ( N ln N N N R = 1 ( N ln N N ln Ni ! Ni ln Ni Ni and Ni ) = Ni ln Ni + i i xi N ln xi i ln Ni !) i 1 ( N ln N N xi N ln N ) = i xi N ln xi N ) i xi ln x1 i 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: E dG ¯E G 1 = G E + x2 d x1 Divide through by RT ; Then Given: dene G ln γ1 = G + x2 and GE ; RT dG d x1 GE = A1/ k x! x2 RT and with 698 E dG ¯E G 2 = G E x1 d x1 note by Eq. (11.91) that ln γ2 = G x1 dG d x1 A x 1 Ak + x 2 Ak 21 12 ¯ G iE = ln γi RT G = x 1 x 2 A 1/ k Whence: d A 1/ k dG + A 1/ k ( x 2 x 1 ) = x1 x2 d x1 d x1 and 1 A 1/ k k dA 1 d A 1/ k ( A21 Ak ) = = A(1/ k )1 12 kA d x1 k d x1 2 ln γ1 = x2 A1/ k Finally, ( Ak Ak ) x 1 21 12 +1 kA 2 ln γ2 = x1 A1/ k 1 Similarly, A 1/ k k dG ( A21 Ak )+ A1/ k (x2 x1 ) = x1 x2 12 kA d x1 and ( Ak Ak ) x 2 21 12 kA (b) Appropriate substitition in the preceding equations of x1 = 1 and x1 = 0 yields: ln γ2 = A1/ k = ( Ak )1/ k = A21 21 ln γ1 = A1/ k = ( Ak )1/ k = A12 12 GE = A1/ k = (x1 Ak + x2 Ak )1/ k 12 21 x1 x2 RT g = x1 A21 + x2 A12 (Margules equation) (c) Let g If k = 1, 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 If k = 1, Apply lHˆ 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 , lim ( A12 / A21 )k 0 k Whence For k +, Whence g = A21 lim ln g = ln A21 k except at x1 = 0 where g = A12 lim ( A12 / A21 )k k g = A12 and and lim ln g = ln A12 k 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 γe Pesat = xe γe Pesat = ye P e EtOH Because P is low, we have assumed ideal gases, and for small xe let γe γe . For volume fraction ξe in the vapor, the ideal-gas assumption provides ξev ye , and for the liquid phase, with xe small xe Vel xe Vel xe Vel Vb xb Vb xe Vel + xb Vb l ξe = Ve P volume % EtOH in blood Vb γe Pesat volume % EtOH in gas Vb l sat ξ γ P ξev P Ve e e e Then 12.70 By Eq. (11.95), κ (G E / RT ) κT HE = T RT b blood P ,x E G = x1 ln(x1 + x2 RT κ (G E / RT ) κT 12 ) x2 ln(x2 + x1 d 21 d 12 x2 x1 dT dT = x 2 + x 1 21 x 1 + x 2 12 x d 12 H dT = x1 x2 T x1 + x2 RT ij = Vj d ij = Vi dT H E = x1 x2 12 d 21 dT + x2 + x1 ai j Vj exp RT Vi exp 12 a12 x1 + x2 + 12 21 (i = j ) ai j = RT 2 ai j RT ij (12.24) ai j RT 2 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 ( x 2 + x 1 21 ) 2 (x1 + x2 12 )2 R Because 12 and 21 (12.18) x1 x2 E 21 ) E must always be positive numbers, C P must always be positive. 700 Chapter 13 - Section B - Non-Numerical Solutions 13.1 (a) 4NH3 (g) + 5O2 (g) 4NO(g) + 6H2 O(g) νi = 4 5 + 4 + 6 = 1 ν= n0 = =2+5=7 i i0 By Eq. (13.5), yNH3 = 2 4ε 7+ε yO2 = 5 5ε 7+ε 4ε 7+ε yNO = yH2 O = 6ε 7+ε 2H2 S(g) + 3O2 (g) 2H2 O(g) + 2SO2 (g) (b) νi = 2 3 + 2 + 2 = 1 ν= =3+5=8 n0 = i0 i By Eq. (13.5), yH2 S = 3 2ε 8ε yO2 = 5 3ε 8ε yH2 O = 2ε 8ε ySO2 = 2ε 8ε 6NO2 (g) + 8NH3 (g) 7N2 (g) + 12H2 O(g) (c) νi = 6 8 + 7 + 12 = 5 ν= n0 = =3+4+1=8 i0 i By Eq. (13.5), yNO2 = 3 6ε 8 + 5ε yNH3 = 4 8ε 8 + 5ε yN2 = 1 + 7ε 8 + 5ε yH2 O = 12ε 8 + 5ε C2 H4 (g) + 1 O2 (g) (CH2 )2 O(g) 2 13.2 (1) C2 H4 (g) + 3O2 (g) 2CO2 (g) + 2H2 O(g) (2) The stoichiometric numbers νi, j are as follows: i= C2 H4 O2 (CH2 )2 O CO2 H2 O j νj 1 1 1 2 1 0 0 1 2 2 1 3 0 2 2 0 n0 = =2+3=5 i0 By Eq. (13.7), yC2 H4 = 2 ε1 ε2 5 1 ε1 2 yCO2 = yO2 = 3 1 ε1 3ε2 2 2ε2 5 1 ε1 2 701 5 1 ε 21 yH2 O = y (CH2 )2 2 ε2 5 1 ε1 2 O = ε1 5 1 ε1 2 13.3 CO2 (g) + 3H2 (g) CH3 OH(g) + H2 O(g) (1) CO2 (g) + H2 (g) CO(g) + H2 O(g) (2) The stoichiometric numbers νi, j are as follows: i= CO2 H2 CH3 OH CO H2 O j νj 1 2 1 1 3 1 1 0 n0 = 0 1 1 1 2 0 =2+5+1=8 i0 By Eq. (13.7), yCO2 = 2 ε1 ε2 8 2ε1 13.7 The equation for 5 3ε1 ε2 8 2ε1 yH2 = yCH3 OH = ε1 8 2ε1 yCO = 1 + ε2 8 2ε1 yH2 O = ε 1 + ε2 8 2ε1 G , appearing just above Eq. (13.18) is: H0 G = T ( H0 T0 G ) + R 0 T CP dT RT R T0 T C P dT RT T0 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: P P Ky = ν K (a) Differentiate this equation with respect to T and combine with Eq. (13.14): Ky T = P P P ν Ky H d ln K Ky d K dK = = Ky = RT 2 dT K dT dT Substitute into the given equation for (εe / T ) P : εe T = P K y d εe RT 2 d K y H (b) The derivative of K y with respect to P is: Ky P = ν T P P ν 1 1 K = ν K P 702 P P ν P P 1 ν K y 1 = P P Substitute into the given equation for (εe / P )T : εe P (c) With K y i ( yi )νi , ln K y = K y d εe (ν) P d Ky = T νi ln yi . Differentiation then yields: i 1 d Ky = K y d εe i νi dyi yi d εe ( A) dn d ni yi d εe d εe Because yi = n i / n , 1 n i dn 1 dn i dyi = 2 = n n d εe n d εe d εe But n i = n i0 + νi εe and n = n 0 + νεe dn i = νi d εe and dn =ν d εe Whence, νi yi ν dyi = n 0 + νεe d εe Therefore, Substitution into Eq. ( A) gives 1 d Ky K y d εe νi yi ν n 0 + νεe νi yi = i 1 n 0 + νεe = m = νi2 νi yi i =1 1 n 0 + νεe νi2 νi ν yi i m νk k =1 In this equation, both K y and n 0 + νεe (= n ) are positive. It remains to show that the summation term is positive. If m = 2, this term becomes 2 ( y2 ν1 y1 ν2 )2 ν2 ν1 ν1 (ν1 + ν2 ) + 2 ν2 (ν1 + ν2 ) = 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 νi2 νi yi i =1 m m νk m i k = k =1 ( yk νi yi νk )2 yi yk (i < k ) All quantities in the sum are of course positive. 1 N (g) 22 13.9 + 3 H2 (g) NH3 (g) 2 For the given reaction, ν = 1, and for the given amounts of reactants, n 0 = 2. By Eq. (13.5), By Eq. (13.28), yN2 = 1 (1 2 εe ) 2 εe yNH3 1 /2 3 /2 yN2 yH2 = yH2 = [ 1 (1 2 3 (1 2 εe ) 2 εe yNH3 = εe 2 εe P εe (2 εe ) =K 3 P εe )]1/2 [ 2 (1 εe )]3/2 703 Whence, 1 /2 1 2 εe (2 εe ) = (1 εe )2 3/2 3 2 K P P = 1.299 K P P r εe 2 2 r εe + (r 1) = 0 This may be written: where, r 1 + 1.299 K The roots of the quadratic are: 1 = 1 ± r 1 /2 r 1 /2 εe = 1 ± Because εe < 1, εe = 1 r 1/2 , P P εe = 1 1 + 1.299 K 1/2 P P 13.10 The reactions are written: Mary: 2NH3 + 3NO 3H2 O + 5 N2 2 ( A) Paul: 4NH3 + 6NO 6H2 O + 5N2 (B) Peter: 3H2 O + 5 N2 2NH3 + 3NO 2 (C ) Each applied Eqs. (13.11b) and (13.25), here written: ln K = G / RT and ( fˆi )νi K = ( P )ν i For reaction ( A), G = 3 G fH A 2O 2 G fNH 3 G fNO 3 For Marys reaction ν = 1 , and: 2 5/2 fˆfN fˆf3H 1 2 2O K A = (P ) and 2 fˆf2NH fˆf3NO G A RT ln K A = 3 For Pauls reaction ν = 1, and fˆf6H 1 KB = (P ) 2O fˆf5N 2 and fˆf4NH fˆf6NO ln K B = 3 2 G A RT For Peters reaction ν = 1 , and: 2 KC = ( P ) 1 2 fˆf2NH fˆf3NO and 3 fˆf3H 2O 5 /2 fˆfN ln K C = 2 In each case the two equations are combined: Mary: 1 2 (P ) fˆf3H 2O 5 /2 fˆfN 2 fˆf2NH fˆf3NO 3 704 = exp G A RT G A RT 1 Paul: (P ) fˆf6H 2O fˆf5N G A = exp RT 2 fˆf4NH fˆf6NO 3 2 Taking the square root yields Marys equation. 1 Peter: ( P ) 2 fˆf2NH fˆf3NO 3 fˆf3H 2O 5/2 fˆfN = exp 2 G A RT 1 Taking the reciprocal yields Marys equation. 13.24 Formation reactions: 1 N 22 + 3 H2 NH3 2 (1) 1 N 22 + 1 O2 NO 2 (2) 1 N 22 + 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 = 2π + N r s = 21+520 13.35 (a) Equation (13.28) here becomes: yB = yA P P F =4 0 K=K yB = K (T ) 1 yB Whence, (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 ˆ the pressure dependence, which in the general case appears through the φi 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 705 1 yA = K (T ) yA 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 (1 xA ) PBsat (T ) = (1 yA ) P and F = 2π + N r = 22+21 = 1 (a) Application of Eq. (13.36) yields: (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 yPX = yOX KI = KI P P 0 yEB = yOX K II = K II P P 0 K III = K III (b) These equation equations lead to the following set: yMX = K I yOX yEB = K III yOX (2) yPX = K II yOX (1) (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 = (c) With the assumption that combine to give: (4) C P = 0 and therefore that K 2 = 1, Eqs. (13.20), (13.21), and (13.22) K = K 0 K 1 = exp Whence, 1 1 + K I + K II + K III K = exp G 298 exp RT0 H298 T0 1 T RT0 298.15 500 (8.314)(298.15) H298 1 G 298 The data provided lead to the following property changes of reaction and equilibrium constants at 500 K: Reaction I II III H298 G 298 1,750 1,040 10,920 3,300 1,000 8,690 706 K 2.8470 1.2637 0.1778 (d) Substitution of numerical values into Eqs. (1), (2), (3), and (4) yields the following values for the mole fractions: yOX = 0.1891 13.40 For the given owrates, yMX = 0.5383 n A0 = 10 yPX = 0.2390 and n B0 = 15, nA nB nC nD n yEB = 0.0336 with n A0 the limiting reactant without (II) = n A0 εI εII = n B0 εI = εI εII = εII = n 0 εI εII Use given values of YC and SC / D to nd εI and εII : YC = εI εII n A0 and εI εII εII SC / D = Solve for εI and εII : εII = = 10 6 2 = 15 6 =62 =2 = 17 nA nB nC nD n 2+1 2 SC / D + 1 n A0 YC = SC / D εI = × 10 × 0.40 = 6 10 × 0.40 n A0 YC =2 = 2 SC / D yA yB yC yD =2 =9 =4 =2 = 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 νi G i G P i and from Eq. (6.10), ( G i / P )T = Vi = For the ideal-gas standard state, G P = T νi i RT P Vi T νi i G i P = T i νi Vi = RT / P . Therefore = ν RT P and G ( P2 ) G ( P1 ) = ν RT ln 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 707 xA = 1 Kl + 1 P2 P1 The preceding equation now becomes, P = 1 P= Kl Kl Kl + 1 For K l = 0 1 +1 PBsat + PBsat + Kl Kl 1 +1 1 +1 PAsat For K l = P = PAsat PAsat ( A) P = PBsat (b) Given Raoults law: 1 = x A + x B = yA P= y A / PAsat yB yA sat + PBsat PA P P =P sat + y B PBsat PA 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= For K v = 0 ( K v + 1) PAsat PBsat K v PAsat + PBsat P = PAsat (B) For K v = P = PBsat (c) Equations ( A) and ( B ) must yield the same P . Therefore Kl Kl + 1 Some algebra reduces this to: PBsat + 1 l +1 K PAsat = ( K v + 1) PAsat PBsat K v PAsat + PBsat 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 π r = 2 + 2 2 1 = 1 708 Thus xing T should sufce. 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: ˆ ln νi = ε ln Z ε(nG R/ RT ) ε(n Z ) +1 +n ε ni ε ni ε ni where the partial derivatives written here and in the following development without subscripts are understood to be at constant T , n /ρ (or ρ/ n ), and n j . Equation (6.61) after multiplication by n can be written: ρ2 3 ρ nG R n ln Z + n 2 (nC ) = 2n (n B ) n 2 n RT Differentiate: 3ρ ρ ε(nG R/ RT ) ¯ (n B + n Bi ) + =2 2n n ε ni or By denition, 2 ¯ (2n 2 C + n 2 Ci ) n ε ln Z ln Z ε ni ε ln Z 3 ε(nG R/ RT ) ¯ ¯ ln Z = 2ρ( B + Bi ) + ρ 2 (2C + Ci ) n ε ni 2 ε ni ε (n B ) ε ni ¯ Bi and T ,n j ¯ Ci ε (nC ) ε ni T ,n j The equation of state, Eq. (3.40), can be written: Z = 1 + Bρ + Cρ 2 Differentiate: or n Z = n + n (n B ) ρ ρ ε(n Z ) ¯ (n B + n Bi ) + =1+ n n ε ni 2 ρ ρ + n 2 (nC ) n n 2 ¯ (2n 2 C + n 2 Ci ) ε(n Z ) ¯ ¯ = 1 + ρ( B + Bi ) + ρ 2 (2C + Ci ) ε ni or When combined with the two underlined equations, the initial equation reduces to: ¯ ¯ ˆ ln νi = 1 + ρ( B + Bi ) + 1 ρ 2 (2C + Ci ) 2 The two mixing rules are: 2 2 B = y1 B11 + 2 y1 y2 B12 + y2 B22 3 2 2 3 C = y1 C111 + 3 y1 y2 C112 + 3 y1 y2 C122 + y2 C222 ¯ ¯ Application of the denitions of Bi and Ci to these mixing rules yields: 2 2 ¯ B1 = y1 (2 y1 ) B11 + 2 y2 B12 y2 B22 2 2 2 3 ¯ C1 = y1 (3 2 y1 )C111 + 6 y1 y2 C112 + 3 y2 (1 2 y1 )C122 2 y2 C222 2 2 ¯ B2 = y1 B11 + 2 y1 B12 + y2 (2 y2 ) B22 3 2 2 2 ¯ C2 = 2 y1 C111 + 3 y1 (1 2 y2 )C112 + 6 y1 y2 C122 + 2 y2 (3 2 y2 )C222 709 In combination with the mixing rules, these give: ¯ B + B1 = 2( y1 B11 + y2 B12 ) 2 2 ¯ 2C + C1 = 3( y1 C111 + 2 y1 y2 C112 + y2 C122 ) ¯ B + B2 = 2( y2 B22 + y1 B12 ) 2 2 ¯ 2C + C2 = 3( y2 C222 + 2 y1 y2 C122 + y1 C112 ) In combination with the boxed equation these expressions along with Eq. (3.40) allow calculation of ˆ ˆ ln φ1 and ln φ2 . 14.11 For the case described, Eqs. (14.1) and (14.2) combine to give: yi P = xi Pi sat φisat ˆ φi ˆ If the vapor phase is assumed an ideal solution, φi = φi , and yi P = xi Pi sat φisat φi When Eq. (3.38) is valid, the fugacity coefcient of pure species i is given by Eq. (11.36): ln φi = Therefore, ln Bii P RT and φisat = Bii Pi sat RT Bii ( Pi sat P ) Bii P Bii Pi sat φisat = = ln φisat ln φi = RT RT RT φi For small values of the nal term, this becomes approximately: Bii ( Pi sat P ) φisat =1+ RT φi yi P = xi Pi sat 1 + Whence, yi P xi Pi sat = or Bii ( Pi sat P ) RT 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 γi , Whence, ln γi = ln fˆi ln xi ln f i 1 1 d fˆi 1 d ln fˆi d ln γi = = ˆi d xi xi xi d xi d xi f 1 d fˆi >0 fˆi d xi Combination of this expression with Eq. (14.71) yields: Because fˆi 0, d fˆi >0 d xi (const T , P ) RT d fˆi d ln fˆi d µi = = RT d xi d xi fˆi d xi By Eq. (11.46), the denition of fˆi , d µi >0 d xi Combination with Eq. (14.72) yields: (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 G = 0, in which i 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 For the special case of a binary solution, N = 2, and 711 i 1 ln N = RT ln N N G E (max) = RT ln 2 G E = δ12 P y1 y2 14.17 According to Pb. 11.35, This equation has the form: or δ12 P GE y1 y2 = RT RT 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: δ12 P / RT > 2. Suppose T = 300 K and P = 5 bar. The preceding condition then requires: δ12 > 9977 cm3 mol1 for vapor-phase immiscibility. Such large positive values for δ12 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.) GE = Ax1 x2 RT 14.19 Consider a quadratic mixture, described by: 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: 11 G E > 2 · · · RT = 0.5 RT 22 This is a model-dependent result. Many liquid mixtures are known which are stable as single phases, even though G E > 0.5 RT 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 n1 n1 ln(γ1 )m = C ln γ1 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 γ1 )m ln x1 = C ln(x1 γ1 ) C ln x1 or Differentiate: ln(x1 γ1 )m = C ln(x1 γ1 ) (C 1) ln x1 d ln(x1 γ1 ) C 1 d ln(x1 γ1 )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 γ1 )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): Hil v 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 Hil v P sat dT ln i = 2 P Tbi RT Combination with Eq. (10.1) yields: yi = xi exp T Tbi Hil v 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: ββ xiα γiα = xi γi (i = 1, 2) β β α If phase β is pure species 1 and phase α is pure species 2, then x1 = γ1 = 1 and x2 = γ2α = 1. Hence, β β α x1 γ1α = x1 γ1 = 1 and β β α x2 γ2α = x2 γ2 = 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 = ψ1 = exp (a) Differentiate: sl H1 RTm 1 T Tm 1 T sl H1 d x1 = ψ1 · 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: H sl (Tm B Tm A ) RTm A Tm B xA = exp xB sl HA = where by assumption, sl HB 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 Notice that Tm > T (see Fig. 14.21b). Then x A /x B > 1 if and only if accord with the rule of thumb. sl HA < sl H B , in 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 = ln F1 P y1 + y1 P dy1 = dP Whence, F1 P y1 y1 + y1 ln F1 P ln F1 y1 y1 ln F1 y1 1 y1 F1 y1 + P P dy1 dP dy1 dP 1 P ( A) P This is a general result. An expression for F1 is given by Eq. (14.99): sat V s ( P P1sat ) φ1 exp 1 ˆ RT φ1 F1 From this, after some reduction: ln F1 P = y1 Whence, by Eq. ( A), ˆ ln φ1 P + y1 dy1 = dP V1s RT ln F1 y1 and ˆ ln φ1 y1 P 1 + y1 714 y1 1 V1s + P RT ˆ ln φ1 y1 P = P ˆ ln φ1 y1 P (B) ˆ This too is a general result. If the two-term virial equation in pressure applies, then ln φ1 is given by Eq. (11.63a), from which: ˆ ln φ1 P = y1 1 2 ( B11 + y2 δ12 ) RT dy1 = dP Whence, by Eq. ( B ), and ˆ ln φ1 y1 = P 2 y2 δ12 P RT 2 1 V1s B11 y2 δ12 P RT 2 y1 y2 δ12 P 1 RT y1 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 lHˆ pitals rule gives: o n= Pes P es + c + Pes c + Pes m s 0 2 lim n = lim s 0 = P P + c+P c+P m 2 lims 0 n = or mP c+P which is the Langmuir isotherm. (b) Henrys constant, by denition: Differentiate Eq. ( A): Whence, k= m 2s k lim P 0 m dn = 2s dP es es c c = m cs dn dP e s es c + Pes c + Pes 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 P 0 m dn sinh s = cs dP 715 m d 2n = 2 sinh 2s P 0 d P 2 cs 2m d 3n = 3 sinh 3s lim 3 P 0 d P cs 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 2 With z = 1 + Bn + Cn + · · ·, this becomes: 3 ln( P / n ) = ln k + 2 Bn + Cn 2 + · · · 2 Thus a plot of ln( P / n ) vs. n produces ln k as the intercept and 2 B as the limiting slope (for n 0). Alternatively, a polynomial curve t of ln( P / n ) in n yields ln k and 2 B as the rst two coefcients. 14.37 For species i in a real-gas mixture, Eqs. (11.46) and (11.52) give: g i (T ) µi = ˆ + RT ln yi φi P g ˆ d µi = RT d ln yi φi P At constant temperature, g With d µi = d µi , Eq. (14.105) then becomes: a ˆ d + d ln P + xi d ln yi φi = 0 RT i (const T ) For pure-gas adsorption, this simplies to: a d = d ln P + d ln φ (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 φ = ( Z 1) n = k P · exp P 0 ( Z 1) dP · exp P This equation is the real-gas analog of Eq. (14.109). 716 n 0 (1 z ) dn +1z n 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 bn 1 bn bn 1 bn For bn sufciently small, Whence, by Eq. ( A), exp n k P (1 2bn ) 1 or n ( A) kP 1 + 2bk P which is the Langmuir isotherm. With z = 1 + β n , the adsorption isotherm is: n = k P exp(2β n ) from which, for β n sufciently small, the Langmuir isotherm is again recovered. dP Ad =n P RT 14.41 By Eq. (14.107) with a = A / n , The denition of ψ and its derivative are: ψ A RT and Whence, dψ = dψ = n Ad RT 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, xi = P d d i i yi + Pi yi dP Pi i (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 ), i yi = Pi xi i P = 1 P With P xed, Eq. (C ) can now be written in the simple but approximate form: xi = d i dP P Equation ( A) then becomes: xi dψ = n d or xi δψ = n δ i i where we have replaced differentials by deviations. The deviation in value must be unity. Therefore, yi 1 δ xi = P Pi i i 717 i xi is known, since the true By Eq. (14.132), 1 n= (xi / n i ) i Combine the three preceding equations: yi 1 Pi P i δψ = (xi / n i ) i When xi = yi P / Pi , the Raoults law analogy, is substituted the required equation is reproduced: P δψ = 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: 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 ln γ1 = A12 n 2 Introduce solute-free mole fractions: x2 x2 = x2 1 x1 x2 + x3 Whence, and x3 = x3 1 x1 ln γ1 = A12 x2 (1 x1 )2 + A13 x3 (1 x1 )2 A23 x2 x3 (1 x1 )2 For x1 0, ln γ1 = 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 γ1 = A12 ,2 Whence, Also ln γ1 = A13 ,3 ln γ1 = x2 ln γ1 + x3 ln γ1 A23 x2 x3 ,2 ,3 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 γ1 Whence, or ln H1,2 = ln f 1 + ln γ1 ,2 ln H1,3 = ln f 1 + ln γ1 ,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 α to β , probably one on either side of the minimum in curve II. V2 = ln(x2 γ2 ) RT ¯ 14.44 By Eq. (14.136) with V2 = V2 : Represent ln γ2 by a Taylor series: ln γ2 = ln γ2 |x1 =0 + d ln γ2 d x1 x 1 =0 x1 + 1 d 2 ln γ2 2 2 d x1 x1 =0 2 x1 + · · · But at x1 = 0 (x2 = 1), both ln γ2 and its rst derivative are zero. Therefore, ln γ2 = Also, Therefore, and 1 2 d 2 ln γ2 2 d x1 x1 =0 ln x2 = ln(1 x1 ) = x1 ln(x2 γ2 ) = + ln x2 + ln γ2 = x1 1 1 V2 1 =1+ 2 2 x1 RT Comparison with the given equation shows that: (G E/ RT ) T 14.47 Equation (11.95) applies: 1 2 2 x4 x3 x1 1 1 ··· 4 3 2 1 1 2 d 2 ln γ2 2 d x1 B= d 2 ln γ2 2 d x1 2 x1 · · · x 1 =0 x1 + · · · x 1 =0 1 1 1 2 2 = P ,x 2 x1 + · · · d 2 ln γ2 2 d x1 x 1 =0 HE RT 2 E 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), α12 (b) yi ˆ φi P = xi γi Pi sat φisat Ki γi Pi sat φisat yi · = ˆ P xi φi ˆ γ1 P1sat φ1sat φ2 K1 · sat · = ˆ γ2 P2sat φ1 φ2 K2 α12 (x1 = 0) = γ P1sat φ1 ( P sat ) γ1 P1sat φ1 ( P1sat ) φ2 ( P2sat ) = 1 sat · 1 sat · sat · sat ˆ ˆ P2 P2sat φ1 ( P2 ) φ1 ( P2 ) φ2 ( P2 ) α12 (x1 = 1) = ˆ ˆ φ ( P sat ) P sat φ1 ( P1sat ) φ2 ( P1sat ) P1sat = 1 sat · 2 1 · · sat sat sat φ2 ( P2sat ) γ2 P2 φ1 ( P1 ) φ2 ( P2 ) γ2 P2 The nal fractions represent corrections to modied Raoults law for vapor nonidealities. 719 ˆ (c) If the vapor phase is an ideal solution of gases, then φi = φi for all compositions. ln γi T 14.49 Equation (11.98) applies: = 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 γi (x , T ) =i R γi (x , Tk ) T Tk ¯ HE dT =i R T2 1 1 Tk T γi (x , T ) = γi (x , Tk ) · exp 14.52 (a) From Table 11.1, p. 415, nd: GE T (G E / RT ) T ¯ HiE RT T 1 Tk T 1 Tk = S E = 0 and G E is independent of T . P ,x Therefore (b) By Eq. (11.95), ¯ HiE RT = = P ,x FR ( x ) GE = RT RT HE =0 RT 2 GE = FA ( x ) RT (c) For solutions exhibiting LLE, G E / RT is generally positive and large. Thus α and β 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: α =2 RTU TU = α 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 = π N A d 3 1 + ( K 3 1) 1 eξ/ kT (l 3 K 3 ) e / kT 1 3 From this expression, 1 dB ( K 3 1)ξ eξ/ kT + (l 3 K 3 ) e = kT 2 dT / kT and also for an intermediate temperature Tm : according to which d B /dT = 0 for T Tm = k ln ξ +ξ 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) 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 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 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 E E , HE , and S E . 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, δi j 2 Bi j 1 2 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) δi j > 0. If unlike interactions are stronger than like interactions, | Bi j | > 1 | Bii + B j j | 2 Hence δi j < 0. For identical interactions of all molecular pairs, Bi j = Bii = B j j , and δi j = 0 The rationalizations of signs for H E of binary liquid mixtures presented in Sec. 16.7 apply approximately to the signs of δ12 for binary gas mixtures. Thus, positive δ12 is the norm for NP/NP, NA/NP, and AS/NP mixtures, whereas δ12 is usually negative for NA/NA mixtures comprising solvating species. One expects δ12 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 γi than would otherwise obtain, the values of Hi are in the observed order. 16.9 By Eq. (6.70), H α β = T S α β . For the same temperaature and pressure, less structure or order means larger S . Consequently, S sl , S l v , and S s v are all positive, and so therefore are H sl , H l v , and H s v . H l v H v H l = ( H v H ig ) ( H l H ig ) = H R ,v H R ,l 16.11 At the normal boiling point: Therefore H R ,l = H R ,v H lv At 1(atm), H R ,v should be negligible relative to H l v . Then H R ,l H l v . Because the normal boiling point is a representative T for typical liquid behavior, and because H R reects intermolecular forces, H l v has the stated feature. H l v (H2 O) is much larger than H l v (CH4 ) because of the strong hydrogen bonding in liquid water. ig R R 16.12 By denition, write C lP = C P +C P ,l , where C P ,l is the residual heat capacity for the liquid phase. ig R Also by denition, C P ,l = ( H R ,l / T ) P . By assumption (modest pressure levels) C P C v . P C lP C v + P Thus, H R ,l T P R ,l For liquids, H is highly negative, becoming less so as T increases, owing to diminution of interR molecular forces (see, e.g., Fig. 6.5 or Tables E.5 and E.6). Thus C P ,l 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: R T 1 /3 V t 1 /3 = NA P N 1 /3 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. 723 ...
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