This preview shows pages 1–4. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Problem 32 The general formula for this molecule is C n H 2n . Because the pressure is low we assume this is an ideal gas. 3 3 (100 460) 0.103 10.73 42.1 14.7 m m RT RT PV nRT PV RT M M V P P lb psi ft R lb M ft R lb mol psi lb mol = = = = + = =  Knowing the general formula: M(C n H 2n ) = 12 2 n n + = 14n = 42.1 lb/lbmol Thus: n = 3, C 3 H 6 the molecule is propene Problem 34 Assuming the total number of moles will be 1 lbmol and the gas is an ideal gas: m n m n M M = = i i i t t i i i t i t t PV n RT PV n RT V PV n RT n PV n RT V n = = = = componet Mole fraction MW Mass lb wt% vol% Methane 0.6904 16 11.0464 0.302144 0.6904 Ethane 0.0864 30 2.592 0.070897 0.0864 Propane 0.0534 44 2.3496 0.064267 0.0534 ibutane 0.0115 58 0.667 0.018244 0.0115 nbutane 0.0233 58 1.3514 0.036964 0.0233 ipentane 0.0093 72 0.6696 0.018315 0.0093 npentane 0.0085 72 0.612 0.01674 0.0085 Hexane 0.0173 86 1.4878 0.040695 0.0173 C7+ 0.0999 158 15.7842 0.431734 0.0999 total mass 36.56 Problem 37 void Bulk V Porosity V = Assuming the gas obeys ideal gas law (pressure is low) and temperature remains constant. Initial state Final state Initial state: P 1 = 14.7 psia (760mm Hg) V 1 = V vessel  (V bulk  V void ) = 51.3+ V void Final State: P 2 = 6.47 psia (334.7 mm Hg) V 2 = 5 +V 1 1 1 2 2 1 1 2 2 14.7 (3.7 ) 6.47 (8.7 ) 0.235 void void void PV nRT PV nRT PV PV V V V cc = = = + = + = Porosity = 0.235/1.3 = 0.1806 or 18.06 % Problem 310 Given the below data:...
View
Full
Document
This note was uploaded on 09/01/2011 for the course PGE 312 taught by Professor Peters during the Fall '08 term at University of Texas at Austin.
 Fall '08
 Peters

Click to edit the document details