ees3040HW4solns

ees3040HW4solns - EES 3040 Homework 4 Solutions EES 3040...

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EES 3040 Homework 4 Solutions 1 EES 3040 Homework 4 Solutions Hydrology and Groundwater Reading Assignment: D&M1 Chapter 6 (all) or D&M2 Chapter 7, pp. 246-273. Class notes Problems: 1. (6 pts.) Problem 6-1 or 7-1. Hint: Think about given info in terms of mass balance. Given: Lake Kickapoo: length = 12 km, width = 2.5 km, inflow = 3.26 m 3 /s, outflow = 2.93 m 3 /s, total precipitation = 15.2 cm, evaporation = 10.2 cm, and seepage = 2.5 cm. Mass balance on lake: Assume lake is the control volume. Surface area of lake (A): A = (length)(width) A = (12 km)(2.5 km)(1000 m/1 km) 2 = 30,000,000 m 2 or 3.0 x 10 7 m 2 (0.5 pt) Convert rates of input and output into m 3 /month for April: Precipitation = P = (15.2 cm/mo)(1 m/100 cm)(3.0 x 10 7 m 2 ) = 4,560,000 m 3 /month (0.5 pt) Evaporation = E = (10.2 cm/mo)(1 m/100 cm)(3.0 x 10 7 m 2 ) = 3,060,000 m 3 /month (0.5 pt) Seepage = G = (2.5 cm/mo)(1 m/100 cm)(3.0 x 10 7 m 2 ) = 750,000 m 3 /month (0.5 pt) (Note that P, E, and G are multiplied by lake surface area to get volume of water) Inflow = Qin = (3.26 m 3 /s)(86400 s/day)(30 d/mon) = 8,449,920 m 3 /month (0.5 pt) Outflow = Qout = (2.93 m 3 /s)(86400 s/day)(30 d/mon) = 7,594,560 m 3 /month (0.5 pt) Assume surface runoff (R) = 0, transpiration (T) = 0 Calculate change in storage (S) for April: S = P – R – G – E – T + Qin – Qout, or from mass balance ( storage = inputs – outputs), S = (inflow + precipitation) – (outflow + evaporation + seepage) (1 pt) S = (8,449,920+ 4,560,000) – (7,594,560 + 3,060,000 + 750,000) m 3 /month S = 1,605,360 m 3 /month = 1.61 x 10
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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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ees3040HW4solns - EES 3040 Homework 4 Solutions EES 3040...

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