HW 5 Solutions

HW 5 Solutions - 53:050 Natural Environmental Systems...

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53:050 Natural Environmental Systems Solutions to Homework No. 5 – Spring 2011 1. If there is first-order decay, it can be described by the following (as shown in class and in the text): C = C 0 e -kt or ln 0 C C = -kt Substitution of the values given in the problem yields: ln 100 14 . 0 = -(0.2 d -1 )t -6.571 = -0.2t t = 32.9 d As described on p. 57 of the text and derived in class, the half-life is defined by: t 1/2 = k 693 . 0 = 1 - d 0.2 0.693 = 3.465 d t 1/2 = 3.5 d 2. This problem asks how long it will take to achieve a dissolved oxygen (DO) concentration of 6.5 mg/L under the conditions given in the problem. Equation 2-64a can be used to solve this problem: 0 s t s C - C C - C ln = -k a t From Table A-2 in the text, we see that C s = 10.15 mg/L at 15 o C, and = 7.92 mg/L at 28 o C. (a) 2.5 - 10.15 6.5 - 10.15 ln = -(0.034)(t) ln (0.47712) = -0.034t -0.73998 = -0.034t t =21.76 d 22 d (b) k a (28 o C) = 0.038(1.024) 28-20 = 0.046 d
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This note was uploaded on 06/03/2011 for the course 53 50 taught by Professor Parkin during the Spring '11 term at University of Iowa.

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HW 5 Solutions - 53:050 Natural Environmental Systems...

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