Clark's method flood hydrograph, 123 km2

Data and method

Area of basin $A = 123\ \text{km}^2$ (sum of the inter-isochrone areas), $K = 10$ h, $\Delta t = 2$ h. The table has 9 strips of 2 h each (total 18 h), and all strips are used. Although $t_c$ is given as 14 h, the table extends to 18 h, so every strip of the table is used.

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

$$O_2 = C_0 (I_1 + I_2) + C_2 O_1,\qquad C_0 = \frac{\Delta t}{2K+\Delta t},\quad C_2 = \frac{2K-\Delta t}{2K+\Delta t}$$ $$\begin{aligned} C_0 &= \frac{2}{2(10)+2} = 0.0909\\ C_2 &= \frac{2(10)-2}{2(10)+2} = 0.8182\end{aligned}$$

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 $2$-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 $C_2 = 0.8182$ times the previous one (the inflow is zero after $t = 18$ h).

Flood hydrograph

The tabulated outflow is the flood hydrograph produced by 1 cm of excess rain: the IUH (column $O$) and the 2-h unit hydrograph (last column). The IUH peaks at 17.20 m³/s per cm at t = 14 h, and the 2-h unit hydrograph peaks at 16.96 m³/s at t = 16 h. For any storm, multiply the ordinates by the excess rainfall depth in cm (and superpose for successive blocks).