Chapter6_A

# Chapter6_A - Chapter 6 Section A Mathcad Solutions 6.7 At...

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6.8 Isobutane: T c 408.1 K := Z c 0.282 := C P 2.78 J gm K := P 1 4000 kPa := P 2 2000 kPa := molwt 58.123 gm mol := V c 262.7 cm 3 mol := Eq. (3.63) for volume of a saturated liquid may be used for the volume of a compressed liquid if the effect of pressure on liquid volume is neglected. T 359 360 361 K := T r T T c := T r 0.88 0.882 0.885 = (The elements are denoted by subscripts 1, 2, & 3 VV c Z c 1T r () 2 7 ⎯⎯⎯⎯⎯⎯⎯ := V 131.604 132.138 132.683 cm 3 mol = Assume that changes in T and V are negligible during throtling. Then Eq. (6.8) is integrated to yield: Chapter 6 - Section A - Mathcad Solutions 6.7 At constant temperature Eqs. (6.25) and (6.26) can be written: dS β V dP = and dH 1 β T V dP = For an estimate, assume properties independent of pressure. T 270 K := P 1 381 kPa := P 2 1200 kPa := V 1.551 10 3 m 3 kg := β 2.095 10 3 K 1 := S β V P 2 P 1 := H1 β T V P 2 P 1 := S 2.661 J kg K = Ans. H 551.7 J kg = Ans. 144

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P 2 1500 bar := β 250 10 6 K 1 := κ 45 10 6 bar 1 := V 1 1003 cm 3 kg := By Eq. (3.5), V 2 V 1 exp κ P 2 P 1 () := V 2 937.574 cm 3 kg = V ave V 1 V 2 + 2 := V ave 970.287 cm 3 kg = By Eqs. (6.28) & (6.29), HV ave 1 β T P 2 P 1 := U HP 2 V 2 P 1 V 1 := H 134.6 kJ kg = Ans. U 5.93 kJ kg = Ans. S β V ave P 2 P 1 := QT S := Work UQ := S 0.03636 kJ kg K = Ans. Q 10.84 kJ kg = Ans. Work 4.91 kJ kg = Ans. HT S V P + = but H0 = Then at 360 K, S V 1 P 2 P 1 T 1 := S 0.733 J mol K = Ans. We use the additional values of T and V to estimate the volume expansivity: VV 3 V 1 := V 1.079 cm 3 mol = TT 3 T 1 := T2 K = β 1 V 1 V T := β 4.098835 10 3 × K 1 = Assuming properties independent of pressure, Eq. (6.29) may be integrated to give SC P T T ⋅β V ⋅∆ P = PP 2 P 1 := P2 10 3 × kPa = Whence T T 1 C P S β V 1 P + molwt := T 0.768 K = Ans. 6.9 T 298.15 K := P 1 1 bar := 145
Pr P Pc := Tr T Tc := ω .187 .000 .210 .200 .224 .048 .193 .210 .087 .094 .038 .400 .152 .140 := Pc 61.39 48.98 48.98 37.96 73.83 34.99 45.60 40.73 50.40 89.63 34.00 24.90 42.48 46.65 bar := Tc 308.3 150.9 562.2 425.1 304.2 132.9 556.4 553.6 282.3 373.5 126.2 568.7 369.8 365.6 K := P 40 75 30 50 60 60 35 50 35 70 50 15 25 75 bar := T 300 175 575 500 325 175 575 650 300 400 150 575 375 475

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## This note was uploaded on 04/18/2011 for the course CEM 497 taught by Professor Ku during the Spring '11 term at Punjab Engineering College.

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Chapter6_A - Chapter 6 Section A Mathcad Solutions 6.7 At...

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