HW 14 Solutions

HW 14 Solutions - Q = 8.64 x 10-5 day m (1)(100...

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53:050 Natural Environmental Systems Solutions to Homework No. 14 – Spring 2011 1. Problem 13-3 in Davis and Masten. Available space in the landfill = 0.37(11,240,000 m 3 ) = 4,258,800 m 3 Rate of MSW generation = 562,400 people(1.89 kg/cap-d) = 1,062,936 kg/d Volume of daily MSW = 3 kg/m 490 kg/d 1,062,936 = 2,169.3 m 3 /d Remaining Life = /d) m 3 1.9(2,169. m 4,158,800 3 3 (1 yr/365 d) = 2.76 yr 2 . Problem 13-9 in Davis and Masten Leachate flow rate = Q = 1,700 m 3 /yr(1 yr/365 d)(1 d/86,400 s) = 5.39 x 10 -5 m 3 /s Q = K( l h )A h = AK l) Q( h = m/s) 10 x /ha)(3.9 m ha(10,000 12 ) 9 . 0 ( d / m 10 x 5.39 10 - 2 3 -5 = 1.04 m 3. (a) Problem 14-32 in Davis and Masten. Landfill area is 100 ha. The following is taken directly from the authors’ solution manual: Q = K l h A K = 10 -7 sec cm (10 -2 cm m )(86,400 day sec ) = 8.64 x 10 -5 day m l h = 1 10 1 10 + + = 1 when the liner is saturated!
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Unformatted text preview: Q = 8.64 x 10-5 day m (1)(100 hectares)(10 4 hectare m 2 ) = 86.4 m 3 /day 1 What do you think of this? One might argue with the authors definition of l h ! I think, based on other solutions given by the authors, that the following is more correct: Q = 8.64 x 10-5 ( 10 1 10 + )(100)(10 4 ) = 95.0 m 3 /d (b) Travel time. Use the derivation below: Darcy velocity (cm/s) = v = K l h = K( t t h + ) h = head of leachate, t = soil thickness Seepage velocity (cm/s) = v = porosity v Travel time = v t If we substitute the above equations here, we get: Travel time = t) K(h (porosity) t 2 + = s/d) m)(86,400 10 cm/s(1 10 x 1 ) cm/m)(0.45 (100 m) (10 7-2 + (1 yr/365 d) = 128 yr 2...
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HW 14 Solutions - Q = 8.64 x 10-5 day m (1)(100...

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