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ees3040HW5solns_fa10

# ees3040HW5solns_fa10 - EES 3040 Homework 5 Solutions EES...

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EES 3040 Homework 5 Solutions 1 EES 3040 Homework 5 Solutions Groundwater Well Hydraulics and Water Quality/BOD Reading Assignment: D&M1 Ch 6 (208-271), Ch 8 (pp. 312-316), Ch 13 (564-568). Ch 8 (271-282) D&M2 Ch 7 (264-283), Ch 9 (389-393), Ch 14 (667-671). Ch 9 (332-356) Problems: 1. (4 pts.) D&M1 Ch 8 Problem 8-32 or D&M2 Ch 9 Problem 9-38 Need to find Avg Linear velocity of contaminant and Travel time of contaminant Given: Hydraulic gradient = 0.00086, hydraulic conductivity = 200 m/d, porosity = 0.23, retardation factor = 2.3, travels 100 m. Find the Darcy velocity (v): v = K( h/L) = (200 m/d)(8.6 x 10 -3 m/m) = 0.172 m/d (1 pt) Find the average linear velocity (v’) of the groundwater: v’ = v / η = (0.172 m/d) / (0.23) = 0.748 m/d (1 pt) Noting retardation factor (R): R = v’ / v’ cont so then linear velocity of contaminant = v cont ’ = v’ / R = (0.748 m/d) / (2.3) = 0.325 m/d (1 pt) Travel time of contaminant (t): t = (distance / v cont ’) = (100 m) / (0.325 m/d) = 308 days (1 pt) 2. (6 pts.) GW Well Hydraulics: A well with a 0.30 m diameter has been pumping at a rate of 650 m 3 /day for a long time such that steady-state conditions have been reached. The drawdown at the pumping well is 1.0 m. The aquifer is unconfined and is 13.0 m thick under non-pumping conditions. An observation well 15 m away has a drawdown of 0.40 m. A drawing may be helpful in completing this problem. a.) Determine the hydraulic conductivity of this aquifer. (Answer: K = 57.9 m/d) drawdown = s = H o – h, so then h = H o – s. Since this is an unconfined aquifer, original water table H o = original aquifer thickness = 12.0m So then, h w = 13.0 m – 1.0 m = 12.0 m for r w = (0.30 m)/2 = 0.15 m and h 1 = =13.0 m – 0.40 m = 12.6 m for r 1 = 15 m (0.5 pt) For unconfined aquifer: Q = [

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ees3040HW5solns_fa10 - EES 3040 Homework 5 Solutions EES...

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