Question

Ordinates of a 2-hr UHG is given below. A storm of 7 cm occurs uniformly over the catchment area in 3 hrs. A base flow of $8m^3/s$ is present in the area . $\varphi$-index is found to be 2.5 mm/hr. What will be the peak flow due to this storm ?
$\begin{array}{cc}\text{Time}& \text{ordinate}\left(m^3/s\right)\\ 0& 0\\ 1& 3\\ 2& 8\\ 3& 6\\ 4& 3\\ 5& 2\\ 6& 0\end{array}$

• A. 14.0
• B. 36.5
• C. 45.5
• D. 66.0

Answer 1

The correct option is C 45.5

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

$\text{Peak of 3- hr UHG}=\frac{9}{\left(\frac{3}{2}\right)}=6m^3/sec$

$\text{Peak flow of storm will be}=(6\times \text{Run off})+\text{Base flow}$

$\varphi -index=\frac{(P-R)}{t}\Rightarrow 0.25=\frac{7-R}{3}$

$R=7-0.75=6.25cm$

$\text{Now, peak flow of storm}=(6\times 6.25)+8$

$=45.5m^3/sec$