HW 6 Solutions

HW 6 Solutions - 53:050 Natural Environmental Systems Solutions to Homework No 6 Spring 2011 1 Problem 4-5 from Davis and Masten First draw the

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53:050 Natural Environmental Systems Solutions to Homework No. 6 – Spring 2011 1. Problem 4-5 from Davis and Masten. First, draw the mass-balance diagram: Rappahannock River: Q Rap = 3.00 m 3 /s and C Rap = 0 mg/L Tin Pot Run: Q TP = 0.05 m 3 /s and C TP = ? Detection limit = 1.0 mg/L Q TP = 0.05 m 3 /s, C TP = ? Q Rap = 3 m 3 /s Q mix = 3 + 0.05 = 3.05 m 3 /s C Rap = 0 C mix = 1 mg/L Mass Balance: Accumulation = In – OUT + Reactions Accumulation = 0 (assume steady-state) No Reactions So, 0 = Q Rap C Rap + Q TP C TP – Q mix C mix 0 = 3(0) + 0.05(C TP ) – 3.05(1) Solving for C TP gives C TP = 61 mg/L Mass rate = Q TP C TP = 0.05 m 3 /s(61 mg/L)(10 3 L/m 3 )(10 -6 kg/mg)(86,400 s/d) = 263.5 kg/d 2. Problem 4-16 from Davis and Masten. The general mass-balance equation given in class is: ± = reactions C Q - C Q dt d(CV) out out in in Since this is a batch system with no flow in and now flow out, the above equation reduces to: kCV - dt d(CV) = or kC - dt dC = integrating gives: C = C 0 e -kt The problem states that C 0 = 2 mg/L, k = 1.0 d -1 , and t = 8/24 = 0.333 d C = 2e -1.0(0.333) = 1.43 mg/L To return to the original C = 2.0 mg/L, we need to add 2.0 – 1.43 = 0.57 mg/L. 1
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Mass required = mg/kg 10 ) L/m )(1,000 m mg/L(4,000 57 . 0 6 3 3 = 2.28 kg 3. Problem 4-18 in Davis and Masten. Actually, this problem can be worked easily by recognizing the definition of half life – the amount of time it takes for half the material remaining to be removed. Half Life
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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 6 Solutions - 53:050 Natural Environmental Systems Solutions to Homework No 6 Spring 2011 1 Problem 4-5 from Davis and Masten First draw the

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