Clark IUH, 157 km2 basin

Data and method

Area of basin $A = 157\ \text{km}^2$ (sum of the inter-isochrone areas), $K = 9.5$ h, $\Delta t = 1$ h. The strips add up to 157 km², equal to the given area.

Step 1: Translation (time-area) inflow. For 1 cm of excess rain over each strip, the inflow to the imaginary linear reservoir is

$$I = \frac{2.78\,A_i}{\Delta t}\ \text{m}^3/\text{s per cm}$$

Step 2: Storage (linear reservoir) routing. With $S = KQ$ and the continuity equation, the outflow is

$$O_2 = C_0 (I_1 + I_2) + C_2 O_1,\qquad C_0 = \frac{\Delta t}{2K+\Delta t},\quad C_2 = \frac{2K-\Delta t}{2K+\Delta t}$$ $$\begin{aligned} C_0 &= \frac{1}{2(9.5)+1} = 0.0500\\ C_2 &= \frac{2(9.5)-1}{2(9.5)+1} = 0.9000\end{aligned}$$

Check: $C_0 + C_0 + C_2 = 1.0000$ (must equal 1).

Computation

Inflow ordinates are taken at the end of each strip (time $= i\Delta t$); inflow at $t=0$ and outflow at $t=0$ are zero. Column $O$ is the IUH ordinate (routed outflow per cm), and the $1$-h unit hydrograph ordinate is $U_n = (O_n + O_{n-1})/2$.

Beyond the last row the hydrograph is a recession, with each ordinate equal to $C_2 = 0.9000$ times the previous one (the inflow is zero after $t = 7$ h).

Result

The IUH of the catchment is given by the column $O$. Peak IUH ordinate = 33.86 m³/s per cm at t = 6 h. The 1-h unit hydrograph (last column) peaks at 33.01 m³/s at t = 7 h.