ees3040HW6solns_fa10

ees3040HW6solns_fa10 - 1 EES 3040 Homework 6 Solutions...

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Unformatted text preview: 1 EES 3040 Homework 6 Solutions Reading Assignment : D&M1 Chapter 4, pp. 156-162; Chapter 8, pp. 282-307 D&M2 Chapter 5, pp. 205-215; Chapter 9, pp. 357-384 Problem 1. (6 pts) D&M1 Chapter 8, Problem 8-15 or D&M2 Chapter 9, Problem 9-19 Tannery initial ultimate BOD after mixing Given: Tannery Q w = 0.011 m 3 /s, BOD 5 = 590 mg/L, Creek Q r = 1.7 m 3 /s, BOD 5 upstream of tannery = 0.6 mg/L, k tannery = 0.115 d-1 , k creek = 3.7 d-1 Solution: Calculate the ultimate BOD (L o ) of tannery wastewater using Eqn 8-6: BOD 5 = L o (1- e-kt ) 590 mg/L 590 L o = -------------------- = --------------- = 1,349.2 mg/L 1 - e (-0.115)(5) 1 - 0.56 Thus, in Eqn 8-18, L w = 1,349.2 mg/L (2 pts.) Calculate the ultimate BOD (L o ) of Djanggawul Creek in the same way: 0.6 mg/L 0.6 L o = ----------------- = -------------------- = 0.6 mg/L 1 - e (-3.7)(5) 1 - 9.24 x 10-9 Thus, in Eqn 8-18, L r = 0.6 mg/L (2 pts.) Calculate the initial ultimate BOD using Eqn 8-18: Q w L w + Q r L r L oa = ------------------- Q w + Q r (0.011 m 3 /s)(1,349.2 mg/L) + (1.7 m 3 /s)(0.6 mg/L) 14.84 + 1.02 L oa = ----------------------------------------------------------------- = ------------------ 0.011 m 3 /s + 1.7 m 3 /s 1.711 L oa = 9.269 or 9.3 mg/L (2 pts.) Problem 2. (6 pts) D&M1 Chapter 8, Problem 8-19 or D&M2 Chapter 9, Problem 9-25 Critical point and critical DO Given: L oa = 12 mg/L, DO = saturation = DO s , river temp = 10 C, k d = 0.30 d-1 , k r = 0.40 d-1 Solution: Since the DO in the river is...
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This note was uploaded on 10/29/2011 for the course EES 3040 taught by Professor Chan-hilton during the Fall '10 term at FSU.

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ees3040HW6solns_fa10 - 1 EES 3040 Homework 6 Solutions...

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