Question

Ordinates of a 2-hr UHG is given below. The peak flow of a 3-hr UHG will be ____ $m^3/s$.
$\begin{array}{cc}\text{Time}& \text{ordinate}\left(m^3/s\right)\\ 0& 0\\ 1& 3\\ 2& 8\\ 3& 6\\ 4& 3\\ 5& 2\\ 6& 0\end{array}$

• A. 24
• B. 12
• C. 9
• D. 6

Answer 1

The correct option is D 6

T = 3 hrs

D = 2 hrs

$\frac{T}{D}=1.5$

Then, method of superposition cannot be used.

Use S-curve method.

$\begin{array}{cccccc}\text{Time}& \text{2 hr}& \text{Addition}& S_A& S_B\text{(delay}& (S_A-S_B)\\ & \text{UHG}& & & \text{T hrs)}& \\ 0& 0& -& 0& -& 0\\ 1& 3& -& 3& -& 3\\ 2& 8& 0& 8& -& 8\\ 3& 6& 3& 9& 0& 9\to \text{Max}\\ 4& 3& 8& 11& 3& 8\\ 5& 2& 9& 11& 8& 3\\ 6& 0& 11& 11& 9& 2\\ 7& 0& 11& 11& 11& 0\\ 8& 0& 11& 11& 11& 0\end{array}$

$\text{Peak ordinate 3 - hr UHG}=\frac{(S_A-S_B)}{T/D}$

$=\frac{9}{3/2}$

$=6m^3/sec$