Clark's 2-h IUH, 110 km2

Data and method

Area of basin $A = 110\ \text{km}^2$ (sum of the inter-isochrone areas), $K = 12$ h, $\Delta t = 2$ h. The table has 9 strips of 2 h each (total 18 h = $t_c$).

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

$$O_2 = C_0 (I_1 + I_2) + C_2 O_1,\qquad C_0 = \frac{\Delta t}{2K+\Delta t},\quad C_2 = \frac{2K-\Delta t}{2K+\Delta t}$$ $$\begin{aligned} C_0 &= \frac{2}{2(12)+2} = 0.0769\\ C_2 &= \frac{2(12)-2}{2(12)+2} = 0.8462\end{aligned}$$

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 $C_2 = 0.8462$ times the previous one (the inflow is zero after $t = 18$ h).

Clark's 2-h IUH (2-h unit hydrograph)

The 2-h unit hydrograph is the last column of the table, obtained by averaging two successive IUH ordinates. Its peak is 13.85 m³/s at t = 16 h; the IUH peak is 13.96 m³/s per cm at t = 14 h.