Chptr5PrblmSolns - m g 150 3 max = = m Calculate the mass...

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5.1 – 5.33 The solutions for these problems are the solutions for problems 5.1 - 5.33 in the 2 nd edition Solutions Manual . 5.34 On a molar basis the C:N:P ratio for algae is 106:16:1 and on a mass basis it is 41:7.2:1 The mass ratio of the lake water is: 1 4 13.3 03 . 0 12 . 0 40 . 0 P N C = = Therefore, carbon is limiting. 5.35 – 5.53 The solutions for these problems are the solutions for problems 5.34 - 5.52 in the 2 nd edition Solutions Manual .
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5.54 a. From Table 5.11, η = 0.34 for a sand aquifer () () ( ) ( ) () d m 61.2 0.0005 d m 09 . 0 0.34 d d d d = = = = L h v' L h v K η b. t contaminan R v v' = so d m 0.015 6 d m 0.09 t contaminan = = = R v' v and the distance traveled is (0.015 m/d) (365 d) = 5.48 m c. Calculate the maximum contaminant mass dissolved based on the aqueous solubility and density of PCE (use Table 5.14) ( ) () () ( ) g 2,230 m 5.48 m 4.0 m 2.0 0.34
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Unformatted text preview: m g 150 3 max = = m Calculate the mass leaked ( ) ( ) g 59,500 d 365 g 1.63 d mL 100 leaked = = mL m Since the mass leaked > soluble mass, the solubility limit is exceeded and the guideline for the aqueous concentration in the presence of NAPLs (see page 258) is used: C PCE = 0.10 Aq Sol PCE = 15 mg/L d. ( ) ( ) ( ) ( ) d m 0.490 d m 0.09 0.34 m 4 2 m 2 2 3 capture = = = Bv w Q e. ( ) ( ) ( ) yr 22.2 d 8,100 mg g 10 m L 10 L mg 15 d m 0.490 g 59,500 3 3 3 3 PCE PCE = = = = QC m t 5.55 The solution for this problem is the solution for problem 5.53 in the 2 nd edition Solutions Manual ....
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This note was uploaded on 04/03/2008 for the course C E 370 taught by Professor Burgos during the Spring '08 term at Pennsylvania State University, University Park.

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Chptr5PrblmSolns - m g 150 3 max = = m Calculate the mass...

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