2008-Solution for hw2

2008-Solution for hw2 - 400/10 60 600 K K K = = 2 0.01(20...

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32 2 at A B C 1. Pressure head (m) 65 15 88 2. Hydraulic head (m) 90 90 88 3,4. Contours and direction are shown in the Figure 5. Perpendicular distance 2 2 (1500) (750) 1299 x ft = - = 6. Hydraulic gradient= 90 88 0.0015 1299 - = 3.7 1. q=0.5in/day=0.042ft/day 2. 1/ 5280 ι = 3. 0.042 5280 221.8 / 1 q K ft day = = = 4. At 4 0 , 5 2 62.43 3.229 10 / pcf lb s ft γ μ - = = - 5. 5 2 9 11 2 (221.8 / )(3.229 10 / ) 1 [ ] 62.43 86400 1 (1.33 10 )[ ] 125 1.062 10 ft day lb s ft day k K pcf s Darcy darcys ft - - - - = = = = 3.9
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1. 2 2 300 600 400 1300 300 /15 600/
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Unformatted text preview: 400 /10 60 600 / K K K + + = = + + + 2. 0.01 (20 1)(2.5/1300) 0.26 / 6.24 / K orK m hr or m day = = 3. 2 2 1300 6.24 60 600 / 4.04 / K K m day = + = 3.10 1. 3 2 0.02 0.48 / ( )( ) ft q or ft day hr ft = 2. Between point a and b 40(90 ) 90 40(90 ) 0.48 90 88.92 b b b h q h or orh ft-=-= = 3. Between point b and c (1)( ) ( 80) 18.58 b h x x q x x ft +-+ = =...
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This note was uploaded on 04/19/2010 for the course CE 421 taught by Professor Shu-guangli during the Spring '05 term at Michigan State University.

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2008-Solution for hw2 - 400/10 60 600 K K K = = 2 0.01(20...

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