Introductory Chemical Engineering Thermodynamics

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Unformatted text preview: Solutions to Chemical and Engineering Thermodynamics, 3e 5.13 (Also available as a Mathcad worksheet. The Mathcad solution includes graphs.) (a) Use the Clausius-Clapeyron equation H vap ln P2 P1 = R 1 T1 - 1 T2 1 and graphically taking slope, I find H vap ~ 42700 J mol . T (b) The vapor pressure is low enough that the ideal gas approximation should be valid--thus Plotting ln P vs. d ln Pvap ln P vap H vap = = dT T RT 2 either graphically or analytically, we find a f H vap ~ 313,600 J mol 5.14 (a) Start with Eqn. (5.4-6) f = P exp R 1 FV - RT I dPU ln f = F PV - 1I dP S RT z H P K V P z H RT K T W P P 0 0 but 1 d ( PV ) dV dZ dV dP = - = - so P PV V Z V f dZ PV dV Z P 1 = ( Z - 1) - -1 = ( Z - 1) - ln - - dV P Z =1 Z V = RT V 1 V = RT V ln z Z V z F H I K V z FGH IJ K or f 1 = ( Z - 1) - ln Z + P RT V ln V = z FGH RT - P dV V IJ K (Eqn. 5.4-8) (b) Z = 1 + B(T ) RT B and P = 1+ V V V FG H IJ K V V = ln f B B 1 = - ln 1 + + P V V RT V FG H IJ K z LMNFGH RT RT RT B - - V V V V IJ OP dV KQ = B 1 ZB - ln Z + B dV = - ln Z 2 V V V = V z Solutions to Chemical and Engineering Thermodynamics, 3e (c) vdW e.o.s. P= V RT a PV V a - ; Z vdW = = - V -b V2 RT V - b V 2 RT V V = z FGH RT RT RT a V a - P dV = - + 2 dV = RT ln - V V V -b V V - b V V = V = = RT ln V a - V -b V IJ K z FGH IJ K V so ln f vdW P V a - + ( Z - 1) - ln Z V - b RTV Pb a RT P = ln Z - ln Z - - + ( Z - 1) - ln Z RT RT PV RT = ln F H I K = ( Z - 1) - aP A - ln( Z - B) Z Pb . RT ( RT ) (d) Peng-Robinson equation of state. Start with where A = 2 and B = V V = z a FG RT - PIJ dV = z LM RT - RT + OPdV HV K N V V - b V (V + b) + b(V - b)Q L V + d1 + 2 ib OP V a = RT ln - lnM V - b 2 2 b N V + d1 - 2 ib Q M P LV + d1 + 2 ib OP Z a = RT ln - lnM Z - B 2 2 b NV + d1 - 2 ib Q M P V V = [See solution to Problem 4.2 for integral]. Therefore f PR ln P = (Z - 1) - ln Z + ln V + 1+ 2 b Z a - ln Z - B 2 2 bRT V + 1- 2 b a 2 2bRT ln = (Z - 1) - ln( Z - B ) - d i V + d1 - 2 ib LM d MN d i OP i PQ f P will be gotten from V + 1+ 2 b 5.15 (a) liq vap fH 2 S = f H 2 S ; f vap = P f P , where the fugacity coefficient, a f corresponding states. 20 = 0.2237 89.42 255 + 273.15 . TC , H 2 S = 373.2 K Tr = = 0.8002 373.2 ZC , H 2 S = 0.284, which is reasonably close to 0.27 P , H 2 S = 89.42 bar Pr = C Solutions to Chemical and Engineering Thermodynamics, 3e f = 0.765 , fH 2 S = 20 0765 = 153 bar . . . P (b) For a liquid, from Eqn. (5.4-18) Fro m Fig. 5.4-1, f = P vap FfI H PK exp sat LM NM z P P vap V dP RT OP QP a f Pf sat Since Pvap = 6.455 103 Pa at the temperature of interest, we will assume that Also, we will consider the liquid to be incompressible. Thus ~ 1. P vap z P V V dP = RT RT P vap z P dP = V P - P vap RT c h and fH 2 S = P vap exp so that Pressure, Pa fH 2 S, Pa 6,921 9,146 12,960 Reported 6,925 9,175 12,967 LMV c P - P MN RT 7 vap h OP = 6455 expL 0.018( P - 6455) O Pa MN 8.314 10 310.6PQ PQ 3 P = 10 10 . 50 107 . 10 108 . 5.16 (also available as a Mathcad worksheet) (a) There are (at least) two ways to solve this problem. One way is to start from f = P exp or RT ln f = P R 1 FV - RT I dP U S RT z H P K V T W P 0 z FH P 0 V- RT dP P I K RT 8.314 10-6 MPa m3 mol K ( 27315 + 400 )K 0.310748 . = = m3 kg P P(MPa ) 18.01 g mol 10-3 kg g P Solutions to Chemical and Engineering Thermodynamics, 3e From Steam Tables T = 400 C P MPa 0.01 0.05 0.10 0.2 0.3 0.4 0.5 0.6 0.8 1.0 1.2 1.4 1.6 1.8 2.0 $ V m3 kg 31.063 6.029 3.103 1.5493 1.0315 0.7726 0.6173 0.5137 0.3843 0.3066 0.2548 0.2178 0.19005 0.16847 0.15120 $ V - RT P 0.0118 0.00596 0.00448 0.00444 0.00433 0.00427 0.00420 0.00421 0.00413 0.00415 0.00416 0.00416 0.00417 0.00417 0.00417 By numerical integration of this data we find that f ~ -0.0084 MPa m3 kg P f -0.0084 MPa m3 kg ln = = -0.027032 P 0.310748 MPa m3 kg RT ln so f P = 0.97333 and f = 1947 MPa . A second way to use the steam tables is to assume that . steam at 400C and 0.01 MPa is an ideal gas. From the steam tables, at these conditions, we have $ $ H = 3279.6 kJ kg ; S = 9.6077 kJ kg K $ $ $ G = H - TS = 3279.6 - 673.15 9.6077 = -3187.8 kJ kg = -57412.7 kJ kmol = G ( 400 C, 0.01 MPa) = G IG( 400 C, 0.01 MPa) Also G IG( T = 400 C, 2 MPa) - G IG( T = 400 C, 0.01 MPa ) = G IG( T = 400 C, 2 MPa) = -574127 kJ kmol + . = -57412.7 + 8.314 ln 200 = -277603 kJ kmol . Also, from steam tables G (T = 400 C, 2 MPa) = (3247.6 - 673.15 7 .1271) 18.01 = - 2791563 kJ kmol . f G - G IG -2791563 + 277603 . . = exp = exp P RT 8.314 67315 . = 0.9726 f = 0.9726 2 MPa = 1.945 MPa 2 MPa 2 MPa 0.01 z P = 0.01 MPa z V IGdP RT dP P FG H IJ K F H I K Solutions to Chemical and Engineering Thermodynamics, 3e (b) Corresponding states TC = 647.3 K, PC = 22.048 MPa, w = 0.344 Tr = 400 + 273.15 2 = 1.04 ; P = = 0.0907 r 647 .3 22.048 a f From corresponding states chart (actually from Table in Hougen, Watson and Rogatz, Vol. II, p. 601) we have f = 0.983 f = 1.966 MPa P (c) Using the program PR1 we find f = 19.40 bar = 1940 MPa . Comment: The steam table results are probably the most accurate, and the corresponding states results the least accurate. Note that with the availability of the computer program PR1, the P-R e.o.s. is the easiest to use. The results would be even more accurate if the PRSV equation was used. ...
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This homework help was uploaded on 02/06/2008 for the course CBE 103 taught by Professor Rethwisch during the Spring '07 term at University of Iowa.

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