a rural kansas watershed that is ungauged has an area of 475 acres and a main channel length of 6870 feet with an average slope of 100 feet per mile. what is the peak flow in ft3/s for a 15-minute unit hydrograph using the scs unit hydrograph method

Explanation:

Given that:

Area = 475 acres

The length of the channel (L) = 6870 feet

The average water shield slope (S) = 100 feet/mile

Since; 1 mile = 5280 feet

Burst duration D = 15 min

∴

100 feet/mile = 100/5280

The average water shield slope (S) = 5/264

Using hydrograph method:

The time of concentration

where;

L = 6870

S = 5/264

min

Since 60 min = 1 hour

32.34 min will be (32.34*1)/60

= 0.539 hour

Lag time

The time to peak i.e

Since D = 15 min is not equal to , then we hydrograph apart from duration lag time.

Then;

Now, we need to determine the peak discharge by using the formula:

where

484 = peak factor

Recall that A = 475 acres, to miles, we have:

A = 0.7422 mile²

∴

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