Determine the minimum surface area in acres of an irrigation

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.

Solution

Precipitation = 2inch=0.16666ft

Let area of reservoir be A_r 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= 0.52666A_r ft^₃

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+0.52666A_r )

10* A_r =28.6975 * 10⁶ - 0.52666A_r

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