HW_5_solution

# HW_5_solution - PGE 312 Physical and Chemical Behavior of...

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PGE 312 Physical and Chemical Behavior of Petroleum Fluids I Homework 5, Fall 2005 Problem 3-2 Pressure (P) = 14.7psia Temperature(T) = 100 o F = 560 o R Density (ρ) = 0.103 lb/ft3 R = 10.73 psia-ft 3 /lbmol- o R The general formula for this compound is C n H 2n Therefore, the molecular weight of the compound is: M = 12n + 2n = 14n (1) Applying the ideal gas law, RT M m nRT PV * Therefore, P RT RT PV m M * Applying the values of ρ, R, T and P, 42.102 7 . 14 560 * 73 . 10 * 103 . 0 M from equation (1) M = 14n = 42.102 Therefore, n = 3.0 From the general formula for this compound ,C n H 2n , The compound is C 3 H 6 Propene.

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Problem 3-7 P 1 = 760mmHg P 2 = 334.7mmHg V 1 = (1.3*Φ)+(5.0 - 1.3) = 1.3Φ+3.7 V 2 = 5 +V 1 = 1.3Φ+3.7+ 5.0 = 1.3Φ+8.7 Appl ying Boyle’s law, 2 2 1 1 V P V P 760*(1.3Φ+3.7) = 334.7*(1.3Φ+8.7) 2.271*(1.3Φ+3.7) = (1.3Φ+8.7) 1.652Φ = 0.2973 Φ = 0.18
Problem 3-11 Tank dimensions: Diameter (D) = 80ft Height (h) = 25ft Oil height (h o ) = 15ft Barometric Pressure (P i ) = 30inHg = 14.73psia Flow rate (q) in the tank = 15000bbl/day = 0.975ft 3 /s Calculating the Initial Volume to air: 3 2 2 50286 ) 15 25 ( * 2 80 * ft h r V i i Calculating the Final Volume to air: ) ( 6 . 5028 * 2 80 * 3 2 2 ft h h h r V f f f f 1oz/in 2 = .0625psia

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HW_5_solution - PGE 312 Physical and Chemical Behavior of...

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