Solution to Water Quality Problems

Solution to Water Quality Problems - 123 Chapter 8...

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Chapter 8 Solutions 8-3 Rate constant Given: BOD 5 = 220.0; L = 320.0 mg/L. Solution: a. Set up Eqn 8-6 BOD t = L o (1 – e -kt ) 220.0 = 320.0 (1 - e -k(5) ) 0.6875 = 1 - e -k(5) -0.3125 = -e -k(5) b. Divide through by -1 and take the natural log of both sides ln (0.3125) = ln (e -k(5) ) -1.163 = -k(5) k = 0.2326 d -1 8-6 Ultimate BOD 5 Given: BOD rate constant = 0.433 d -1 , BOD 5 = 272 mg/L. Solution: a. From Eqn 8-6 BOD t = L o (1 – e -kt ) 272 = L o (1 – e -(0.433)(5) ) 272 = L o (0.885) L o = 307.26 or 307 mg/L 8-9 BOD 5 123
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Given: Oxygen consumption = 2.00 mg/L, 1.00% sample. Solution: a. This an application of Eqn 8-8 with the dilution factor defined by Eqn 8-9 100% BOD 5 = (2.00)(---------) = 200. mg/L 1.00% 8-13 NBOD of glutamic acid Given: Problem 8-1, 150 mg/L of glutamic acid. Solution: a. Calculate the amount of nitrogen that is oxidized. 14 g N (150 mg/L)(-----------------------) = 14.29 mg N/L 147 g C 5 H 9 O 4 N b. Using the relationship from Eqn 8-11: Theo. NBOD = (14.29 mg N/L)(4.57 mg O
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Solution to Water Quality Problems - 123 Chapter 8...

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