Question

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

Answer 1

Answer:

$\mathbf{Q_p =682 \ \ ft^3/s}$

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 $t_c = 0.0078L^{0.77} S^{-0.385}$

where;

L = 6870

S = 5/264

$t_c = 0.0078(6870)^{0.77} (\dfrac{5}{264})^{-0.385}$

$t_c =32.34$ min

Since 60 min = 1 hour

32.34 min will be (32.34*1)/60

= 0.539 hour

Lag time $T_l = 0.67\times t_c$

$T_l = 0.67\times 32.34$

$T_l = 21.6678\ min$

The time to peak i.e

$T_p = \dfrac{D}{2}+ T_L \\ \\ T_p = \dfrac{15}{2}+ 21.6678 \\ \\ T_p = 29.168 \ min$

$T_r = \dfrac{T_p}{5.5} \\ \\ T_r = \dfrac{29.1678}{5.5} \\ \\ T_r = 5.30 \ min$

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

Then;

$T_p \ ' = T_p + \dfrac{D-t_r}{4} \\ \\ T_p \ ' = 29.168 + \dfrac{15-5.30}{4} \\ \\ T_p \ ' = 31.593$

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

$Q_p = \dfrac{484 \times A}{T_p \ '}$

where

484 = peak factor

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

A = 0.7422 mile²

$T_p \ ' = 31.593/60$

∴

$Q_p = \dfrac{484 \times 0.7422}{\dfrac{31.593}{60}}$

$\mathbf{Q_p =682 \ \ ft^3/s}$