HW 10 Solutions

HW 10 Solutions - 1 100 ) = 200 mg/L 3. NBOD of 100 mg/L...

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53:050 Natural Environmental Systems Solutions to Homework No. 10 – Spring 2011 1. (a) Problem 9-7 in Davis and Masten. BOD t = BOD L (1 – e -kt ) equation 9-6 with BOD L substituted for L 0 250 = BOD L (1 – e -0.233(5) ) BOD L = 0.233(5) - e - 1 250 = 0.688 250 = 363 mg/L (b) BOD 5 at 28 o C k T = k 20 (θ) T-20 equation 9-7 k 28 = 0.233(1.056) 28-20 = 0.360 d -1 BOD 5 = 363(1 – e -0.360(5) ) = 303 mg/L 2. Problem 9-11 in Davis and Masten. Here we are just dealing with sample dilution. BOD 5 = (mg/L DO decrease)(dilution factor) BOD 5 = 2(
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Unformatted text preview: 1 100 ) = 200 mg/L 3. NBOD of 100 mg/L alanine. The following is from my solution to No. 9 of HW 9: C 3 H 7 O 2 N + 3O 2 → 3CO 2 + 2H 2 O + NH 3 NH 3 + 2O 2 → NO 3-+ H + + H 2 O NH 3-N produced from 100 mg/L alanine = 100 g/mol) alanine(89 mol 1 g/mol) N(14-NH mol 1 3 = 15.7 mg/L NBOD = 15.7 mg/L g/mol) N(14-NH mol 1 g/mol) (32 O mol 2 3 2 = 71.8 mg/L O 2 1...
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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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