A single lake 5 square miles in area with an average depth of 50 ft when full. C

Question

A single lake 5 square miles in area with an average depth of 50 ft when full. Currently, it contains 60% of its total water capacity . A single stream flows into the lake at a rate of 200 cubic feet per second, a local water utility withdraws water from the lake for public consumption at an average rate of 129 X 10^5 gallons per day. assume all other potential inputs and outputs are negligible.

A) Estimate the amount of time it will take to fill the lake to capacity (in days)

B) If the rate of withdrawal by the water utility increases by 50% how long will it take for the lake to fill?

C) If the utility seeks to keep the lake at 60% capacity by withdrawing the equivalent stream input, what is the mean residence time of the water? What is the turnover rate?

D) What assumptions did you implicitly or explicitly use to make these estimates?

Explanation / Answer

Volume of the lake is Area * depth = (5 square miles* 5280 ft per mile) *50 ft deep = 1,320,000 cubic feet

1 cubic foot = 7.48052 gallons

129 * 10^5 gallons / 7.48052 = 1,724,479.047 cubic feet per day.

200 cubic feet per second * 3600 seconds per hour * 24 hours per day = 17,280,000 cubic feet per day.

60% * 1,320,000 cubic feet = 792,000 cubic feet starting out.

So, with x = number of days, the relationship to fill the lake is given by 1,320,111=792,000 +17,280,000*x - 1,724,479.047*x

solving for x gives x = 0.0339501 days

2)

the rate of withdrawl is 1,724,479.047 cubic feet per day

increase of 50% is 1,724,479.047 + (.50)*1,724,479.047 =2,586,718 cubic feet per day

New relationship is 1,320,111=792,000 +17,280,000*x - 2,586,718 *x

solving for x = x = 528111/14693282 ~~ 0.035942 days

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