HW 11 Solutions

HW 11 Solutions - 53:050 Natural Environmental Systems...

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53:050 Natural Environmental Systems Solutions to Homework No. 11 – Spring 2011 1. (a) Problem 9-23 in Davis and Masten. Here we will be using Equations 9-26 and 9-30, respectively, to calculate k r and k d . k r = 5 . 1 5 . 0 h ) u ( 90 . 3 = 5 . 1 5 . 0 0 . 3 ) 5 . 0 ( 90 . 3 = 0.531 d -1 at 20 o C k d = k + h u η = 0.20 + 0 . 3 5 . 0 (0.40) = 0.267 d -1 at 20 o C (b) Problem 9-24 in Davis and Masten. k r = 5 . 1 5 . 0 h ) u ( 90 . 3 = 5 . 1 5 . 0 0 . 4 ) 5 . 2 ( 90 . 3 = 0.771 d -1 at 20 o C k d = k + h u η = 0.20 + 0 . 4 5 . 2 (0.50) = 0.513 d -1 at 20 o C (c) The differences are expected because both k r and k d depend on the ratio, h u , which increased as a result of the flood (velocity, u, increased greater than the height, h). In addition, the bed-activity coefficient increased. Thus, you would expect both k r and k d to increase as a result of the flood. 2. Problem 9-27 in Davis and Masten. What DO will result if L w is reduced by 50% due to wastewater treatment? It is given that L r and D o = 0. We are also told that the critical DO = 4 mg/L, therefore: D c = 10.83 – 4 = 6.83 mg/L Consider the following equations: L o = r w r r w w Q Q L Q L Q + + since L r = 0, a 50% reduction in L w gives a 50% reduction in L o t c = o d d r o d r d r L k ) k - (k D - 1 k k ln k - k 1 since D o = 0, t c does not change. D c = ( 29 c r c r c d t k - o t k - t k - d r o d e D e - e k - k L k + since D o = 0 and t c does not change, and L o is reduced by 50%, then D c is reduced by 50%. Therefore, New minimum DO = 10.83 – 0.5(6.83) = 7.42 mg/L 1
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3. Problem 9-30 in Davis and Masten. Use the data in the “corrected” table. The following solution is taken from the authors’ solutions manual. What is the DO at Edinkira? t =
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HW 11 Solutions - 53:050 Natural Environmental Systems...

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