Question

The ordinates of a 2-hour unit hydrograph for a catchment are given as
$\begin{array}{cccccc}\text{Time(h)}& 0& 1& 2& 3& 4\\ \text{Ordinate}& 0& 5& 12& 25& 41\\ \left(m^3/s\right)& & & & & \end{array}$
The ordinate (in $m^3/s$) of a 4-hour unit hydrograph for this catchment at the time of 3 h would be

• A. 15

Answer 1

The correct option is A 15
Lagging the ordinate of a 2-hr unit hydrograph by a 2-hr and adding it with 2-hr unit hydrograph.
$\begin{array}{ccccc}\text{Time}& \text{A}& \text{B}& \text{A+B}& \frac{A+B}{2}\\ & \text{Ordinate}& \text{Lagged}& & \\ & & \text{Ordinate}& & \\ 0& 0& & 0& 0\\ 1& 5& & 5& 2.5\\ 2& 12& 0& 12& 6\\ 3& 25& 5& 30& 15\\ 4& 41& 12& 53& 26.5\end{array}$


Hence ordinate of resulting hydrograph at 3 hrs would be,
$=15m^3/s$