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Unformatted text preview: PGE 323 Homework #1 Key January 30, 2006 1. Darcy’s Law Calculations a) Uniform Permeability i) DARCY UNITS L = 2 ft = 60.96 cm d= 2 in = 5.08 cm θ = 90 o ρ = 0.8 g/cm 3 μ = 0.7 cp k= 100 md = 0.1 D Start with the equation 1.16: ( 29 A B kA q P P gLsin L =  + ρ θ μ For this problem, flow is going up, so the change in the elevation is positive (h out minus h in). Thus, the gravity term gLsin ρ θ is positive. You can also reason this out by realizing that the dip angle is a positive 90 degrees and the sine of 90 degrees is positive. Finally, you can reason it out by thinking that gravity is opposing upward flow since it is acting vertically downward. All of these points of view lead to the same correct answer. 2 3 3 2 6 2 g cm 0.8 *980 *60.96cm (0.1D)(20.3)cm cm cm s q (1.02atm 1.36atm ) 0.0139 dynes (0.7cp)(60.96) cm sec 10 cm =  + = ii) SI UNITS L = 2 ft = 0.6096 m d= 2 in = 0.0508 m θ = 90 o ρ = 800 kg/m 3 μ = 0.0007 Pas k= 100 md = 1 E13 m 2 2 2 3 3 2 1E 13m *0.002m kg m m q (103367Pa 137823Pa 800 *9.8 *0.6096m) 1.37E 8 0.0007Pa s*0.6096m sec m s =  + = b) Harmonic Mean Permeability Start with equation 1.221 for harmonic mean permeability (flow in series): 1 N i i i 1 x 1 k L k = ∆ = & For just two sections, the sum in this equation expands to the following: 1 1 1 2 1 2 x x 1 1 15.24cm 15.24cm k 0.075D L k k 30.48cm 0.05D 0.15D ∆ ∆ = + = + = Then insert this permeability into equation 1.17....
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This note was uploaded on 05/15/2011 for the course PGE 323 L taught by Professor Johns during the Spring '10 term at University of Texas at Austin.
 Spring '10
 Johns

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