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Answer:
[tex]\mathbf{Q_p =682 \ \ ft^3/s}[/tex]
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 [tex]t_c = 0.0078L^{0.77} S^{-0.385}[/tex]
where;
L = 6870
S = 5/264
[tex]t_c = 0.0078(6870)^{0.77} (\dfrac{5}{264})^{-0.385}[/tex]
[tex]t_c =32.34[/tex] min
Since 60 min = 1 hour
32.34 min will be (32.34*1)/60
= 0.539 hour
Lag time [tex]T_l = 0.67\times t_c[/tex]
[tex]T_l = 0.67\times 32.34[/tex]
[tex]T_l = 21.6678\ min[/tex]
The time to peak i.e
[tex]T_p = \dfrac{D}{2}+ T_L \\ \\ T_p = \dfrac{15}{2}+ 21.6678 \\ \\ T_p = 29.168 \ min[/tex]
[tex]T_r = \dfrac{T_p}{5.5} \\ \\ T_r = \dfrac{29.1678}{5.5} \\ \\ T_r = 5.30 \ min[/tex]
Since D = 15 min is not equal to [tex]T_r[/tex], then we hydrograph apart from [tex]T_r[/tex] duration lag time.
Then;
[tex]T_p \ ' = T_p + \dfrac{D-t_r}{4} \\ \\ T_p \ ' = 29.168 + \dfrac{15-5.30}{4} \\ \\ T_p \ ' = 31.593[/tex]
Now, we need to determine the peak discharge [tex]Q_p[/tex] by using the formula:
[tex]Q_p = \dfrac{484 \times A}{T_p \ '}[/tex]
where
484 = peak factor
Recall that A = 475 acres, to miles, we have:
A = 0.7422 mile²
[tex]T_p \ ' = 31.593/60[/tex]
∴
[tex]Q_p = \dfrac{484 \times 0.7422}{\dfrac{31.593}{60}}[/tex]
[tex]\mathbf{Q_p =682 \ \ ft^3/s}[/tex]