HW9_solution - Homework 9 solution 6.1 a) Pr = 0.88, the T...

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Homework 9 solution 6.1 a) Pr = 0.88, the T limit where Z 0 > 0.95 is Tr = 1.75, T > 220K b) Pr = 0.41, the T limit where Z 0 > 0.95 is Tr = 1.4, T > 425K c) Pr = 0.47, the T limit where Z 0 > 0.95 is Tr = 1.5, T > 775K Therefore, it can be seen that there isn’t a single temperature above which the ideal gas law is valid; the temperature depends on the fluid. 6.2 i) a) methane MW = 16, Tr = 1.574, initial state, Pr = 2.172, Z 0 = 0.85, Z 1 = 0.22, Z = 0.85+0.011(0.22) = 0.85 V = ZRT/P = 212 cm 3 /mol m(kg) = MW*V /V = 3.02 kg final state, Pr = 1.086, Z 0 = 0.93, Z 1 = 0.04, Z = 0.93 + 0.011(0.04) = 0.93 V = ZRT/P = 463.9 cm 3 /mol m(kg) = MW*V /V = 1.38 kg 3.02-1.38=1.64 kg lost b) methane MW = 16. Initial state Using PREOS.xls, T (K) 300 Z V fugacity P (MPa) 10 cm 3 /gmol MPa 0.833735 207.9502 8.229178 m(kg) = MW*V /V = 3.078 kg final state T (K) 300 Z V fugacity P (MPa) 5 cm 3 /gmol MPa 0.901785 449.8462 4.506731 m(kg) = MW*V /V = 1.423 kg 3.078-1.423=1.66 kg lost ii) a) propane, MW = 14, Tr = 0.811
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This note was uploaded on 05/10/2008 for the course CHBE 2110 taught by Professor Gallivan during the Spring '08 term at Georgia Institute of Technology.

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HW9_solution - Homework 9 solution 6.1 a) Pr = 0.88, the T...

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