Question

Determine the minimum surface area in acres of an irrigation reservoir so that the drawdown over...

Determine the minimum surface area in acres of an irrigation reservoir so that
the drawdown over August is no more than 10 ft. the drainage basin feeding into the
reservoir, including the reservoir itself, is 100 mi 2. Assume that 35% of the rain falling on the
basin runs off the reservoir. Assume groundwater flow into and out of the basin is negligible.
Determine the irrigation water needed from the evapotranspiration requirement from beans
growing on 40mi 2 at 35 o North latitude. The reservoir must also supply a continuous 50 cfs of
water for instream flows to protect fish and aquatic life. The evaporation pan coefficient for
this area is 0.800. For the month of August, Temp = 68F; precip = 2.0 inches and Pan
Evaporation = 7.9 inches. Draw a figure and clearly, show the mass balance on the figure.

Hint: use the mass balance equation: inflow-outflow = change in storage
Change in storage = reservoir area * change in height.
Change is height is given so use the mass balance, calculate the change in storages and calculate
reservoir area.

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Answer #1

Precipitation = 2inch=0.16666ft

Let area of reservoir be Ar ft2

Total drainage area(including reservoir area)=100mi2=100*(5280)2 ft2= 2787.84*106 ft2

evaporation from reservoir= 7.9 * 0.8 inch= 6.32inch = 0.52666 ft

total evaporation loss from reservoir in agust= 0.52666Ar ft3

total precipitation inflow in reservoir in agust=0.35*0.166666 * 2787.84*106 =162.6175*106 ft3

total loss in stream from reservoir in agust=50*60*60*24*31=133.92*106 ft3

max permessible drawdown= 10 ft

So,

change in storage from reservoir=total precipitation in reservoir in agust - total loss in agust

drawdown * reservoir area =162.6175*106 - (133.92*106+0.52666Ar )

10* Ar =28.6975 * 106 - 0.52666Ar

  Ar =2.726* 106 ft2 = 62.5842 acre   ANSWER(min area of reservoir needed).

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