The 40-min unit hydrograph for a (A) 1.46 mi catchment is given below. Compute t
ID: 1732375 • Letter: T
Question
The 40-min unit hydrograph for a (A) 1.46 mi catchment is given below. Compute the following: a. The volume of direct runoff Vdu (in ft') for the unit hydrograph. b. The depth of direct runoff rau (in inches) for the unit hydrograph. Verify that the unit hydrograph is consistent with 1-in of rainfall excess (i.e., a property of a unit hydrograph) The direct runoff hydrograph Q (t) (in cfs) for a 40-min period with rainfall excess of 0.5 in. Estimate the direct runoff hydrograph Q (t) (in cfs) for a 40-min period with rainfall excess c. d. of 0.5 in, followed by another 40-min period with rainfall excess of 1.5 in. e. Estimate the direct runoff hydrograph Q (t) (in cfs) for the rainfall excess hyetograph shown below Rainfall excess hyetograph 40-min unit hydrograph t (minU (cfs in) t (min) Pe (in) 20 40 60 80 100 120 140 160 180 200 300 500 600 560 400 240 140 60 30 0.5 40 80 120 160 1.0 0.5Explanation / Answer
a.) Volume of direct runoff Vd,U (ft3) = (300+500+600+560+400+240+140+60+30)*20*60 = 3396000 ft3
b.) verification of depth of direct runoff = (volume of direct runoff of UH)/Area of catchment
= (3396000*12*12*12)/(1.46*63360*63360) ( 1 feet = 12 inch, 1 mile = 63360 inch)
= 1.0 hence verified.
c) the direct runoff hydrograph Q(t) in cfs for a 40 min period with rainfall excess of 0.5 inch
= (rainfall excess * areas of catchment)/ time
= (((0.5/12)) * 1.46* 52802)/(40*60)
= 700.64 cfs
d.) similarly from c bit
Q in cfs = (((0.5/12)) * 1.46* 52802)/(40*60) + (((1.5/12)) * 1.46* 52802)/(40*60)
= 700.64 + 2120 = 2820.64 cfs
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