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Homework+Solutions+Chapter+11+2nd+Assignment

Homework+Solutions+Chapter+11+2nd+Assignment - PETE 2032...

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PETE 2032 Homework Solutions – Chapter 11 Second Assignment 11-9 Use the correlation of Fig 11-6 to calculate the density of a reservoir liquid at its bubble point of 1763 psia at a reservoir temperature of 250 o F. The composition of the well stream is as follows. Compare your answer with laboratory measurement of 42.12 lb/ft 3 Component Composition, Mole fraction H 2 S 0.0879 CO 2 0.0270 N 2 0.0009 C 1 0.2112 C 2 0.0763 C 3 0.0703 i-C 4 0.0147 n-C 4 0.0428 i-C 5 0.0171 n-C 5 0.0237 C 6 0.0248 C 7 + 0.4033 1.0000 Properties of Heptanes plus Specific Gravity 0.8500 MW 215 lb/lb-mole
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PETE 2032 Homework Solutions – Chapter 11 Second Assignment Solution to 11-9 ρ of C 3 + = 101.2278/2.0109 = 50.34 lb/ft 3 W2 = 2.3196/103.5473 = 2.24 wt % W1 = 3.3883/108.1239 = 3.13 wt % W H 2 S = 2.9953/108.1239 = 2.77 wt % From Fig 11-6 Pseudo-Liquid Density = 47.6749 lb/ft 3 Pressure Adjustment + 0.5428 (Fig 11-3) Density at 60 o F & p = 48.2177 lb/ft 3 Temp Adjustment - 5.3799 (Fig 11-4) Density of reservoir liquid 42.8378 lb/ft 3 H 2 S Adjustment - 0.2833 (Fig 11-7) Density of Reservoir liquid 42.5546 lb/ft 3 1763 psia & 250 o F Component Composition, Mole fraction z M z*M ρ Vol Z*M/ ρ H 2 S 0.0879 34.076 2.9953 50.0355 0.0599 CO 2 0.0270 44.010 1.1883 N 2 0.0009 28.013 0.0252 C 1 0.2112 16.043 3.3883 C 2 0.0763 30.070 2.2943
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