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

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

Step-by-step 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) ((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* t_c`

`T_l = 0.67* 32.34`

`T_l = 21.6678\ min`

The time to peak i.e

`T_p = (D)/(2)+ T_L \\ \\ T_p = (15)/(2)+ 21.6678 \\ \\ T_p = 29.168 \ min`

`T_r = (T_p)/(5.5) \\ \\ T_r = (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 + (D-t_r)/(4) \\ \\ T_p \ ' = 29.168 + (15-5.30)/(4) \\ \\ T_p \ ' = 31.593`

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

`Q_p = (484 * 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 = (484 * 0.7422)/((31.593)/(60))`

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