{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

HW9_solution

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

This preview shows pages 1–2. Sign up to view the full content.

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 initial state, Pr = 1.177, Z 0 = 0.2, Z 1 = -0.075, Z = 0.2-0.152(0.075) = 0.189 V = ZRT/P = 94.08 cm 3 /mol m(kg) = MW*V /V = 18.7 kg final state, Pr = 0.212, by looking at the Z chart the conditions are very close to the vapor pressure. By checking the P-H chart or the Antoine equation, the vapor pressure is about 1 MPa, therefore propane is vapor at 0.9 MPa, so use the vapor branch on the Z charts.

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

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 4

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

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online