Question

The ordinates of a one-hour unit hydrograph (1-hr UH) for a catchment are:
\includegraphics[width=1.0\linewidth]{image55.png} Using superposition, a $D$-hour unit hydrograph is derived. Its ordinates are found to be $3\ \text{m}^3\!/\text{s}$ at $t=1$ hour and $10\ \text{m}^3\!/\text{s}$ at $t=2$ hour. Find the value of $D$ (integer).

Hint:

A $D$-hour unit hydrograph can be obtained from a 1-hr UH by averaging $D$ consecutive ordinates (or equivalently, by the S-curve method and taking a $D$-hour difference, then dividing by $D$).

Answer 1

Step 1: Relation between $D$-hr UH and 1-hr UH (superposition).
The $D$-hr UH ordinate at time $t$ equals the average of $D$ successive ordinates of the 1-hr UH: \[ U_D(t)=\frac{1}{D}\sum_{i=0}^{D-1}U_1(t-i), U_1(\tau)=0\ \text{for }\tau<0. \]

Step 2: Use the value at $t=1$ hour.
\[ U_D(1)=\frac{1}{D}\big(U_1(1)+U_1(0)+\cdots\big) =\frac{1}{D}(9+0+\cdots)=\frac{9}{D}. \] Given $U_D(1)=3\ \Rightarrow\ \dfrac{9}{D}=3 \Rightarrow D=3.$

Step 3: Check with the value at $t=2$ hour.
For $D=3$: \[ U_D(2)=\frac{1}{3}\big(U_1(2)+U_1(1)+U_1(0)\big) =\frac{1}{3}(21+9+0)=\frac{30}{3}=10, \] which matches the given ordinate $\Rightarrow$ value confirmed.
\[ \boxed{D=3} \]