Spring 07 HW 5 Solutions

Spring 07 HW 5 Solutions - Problem 3-2 The general formula...

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Unformatted text preview: Problem 3-2 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 3-4 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 i-butane 0.0115 58 0.667 0.018244 0.0115 n-butane 0.0233 58 1.3514 0.036964 0.0233 i-pentane 0.0093 72 0.6696 0.018315 0.0093 n-pentane 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 3-7 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 ) = 5-1.3+ 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 3-10 Given the below data:...
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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.

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Spring 07 HW 5 Solutions - Problem 3-2 The general formula...

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