CE 408 Water Resources and Supply Homework 1 Date of assignm

CE 408 Water Resources and Supply Homework #1 Date of assignment: 1. (5 pts) lIf capacity of a reservoir is 43,240 acre-flyr, what would be the constant annual yi 2. (10 pts) A mean draft of 120 million gallons per January 19, 2018 January 24, 2018 Total 50 pts rate of this reservoir? Report the yield in the unit of gallon per min (Epm). catchment area (watershed). At the stream line, the reservoir is estimated to be 6,500 acres. The day (mgd) is to be developed from a 220 mi free the I rainfall is 31 in, the mean annual runoff is 12 in, and the mean annual evaporation (from face) is 43 in. (1) Find the net gain or loss in storage this represents. (2) Compute e vol 3. (10 pts) An impounding reservoir is expected to provide a constant draft of 550 million gallons per square miles per year. The following record of monthly mean flow values (million gallons per square mile storage required using the tabular method. Inflow 33 10 8 5 25 38 105 85 45 30 4. (15 pts) The table below shows the monthly flow data of the Little Weiser River in Idaho. (1) Determine the required active reservoir capacity to produce a yield of 5,000 acre- /month. (2) Determine amount of water needs to be withdrawn from the reservoir in each low flow period. Report volume in acre-ft. Use the graphic method and show all steps of your work. Use of Excel is recommended. You may do it by hand, but you have to use graphic paper and to scale. Hand drawing of straight lines will not be acceptable. 520 2,100 4,410 2,750 3.370 10 5,990 7.560 5,800 8,590 21,960 30,790 14,320 70 709 12 5,170 2003 15,680 19,630 90 710 650 780 870 10 12 10 870 2002 1,800 8,370 5,200 7,210 6,540 8,300 12 14,270 2004 16,290 9,290 od should be considered to be 90% certain that location of a (l0pts) What designretum peri structure will not be flooded in the next 25 years? 5

Solution

Hi,

Due to constraints, solving first problem in the sequence.

Capacity of reservoir = 43240 acre-ft/year = 43240 * 43560 ( ft3)/year

= 1883534400 ft3/year

Constant Annual Rate of Reservoir in lb/m = [1883534400 ft3 * 62.42796 ( lb/ft3 ) / [ 365 * 1440 min]

= 3728.60 lb/min

Now we know that 1 gallon = 8.34 lb

Therefore converting the answer to gallons per min format , we get

= 3728.60 * ( 1 / 8.34 ) gpm

= 447.07 gpm

Constant Annual Rate of Reservoir in gpm is 447.07 gpm

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