Why flow routing, time-area method of routing

Why flow routing is required

Flow routing finds the outflow hydrograph from a reach, reservoir, or catchment for a given inflow, accounting for storage. It is required for:

• designing spillways, reservoir capacity, and flood-control works (the outflow peak controls the design);
• flood forecasting and warning at downstream points (arrival time and peak);
• designing flood protection works such as embankments, bridges, and culverts along a river;
• studying how a flood changes (attenuation and lag) as it moves downstream;
• obtaining a catchment hydrograph from rainfall, as in the time-area method.

Time-area method of routing

The time-area method routes excess rain over the catchment surface to the outlet, considering only translation.

1. Draw isochrones at interval $\Delta t$ and find the strip areas $A_1, A_2, \dots, A_N$ ($N\Delta t = t_c$).

 Outlet
   *---- A1 (0 - dt)
   *---- A2 (dt - 2dt)
   *---- A3 (2dt - 3dt)  ... up to tc

2. Divide the effective rainfall into blocks $r_1, r_2, \dots$ of duration $\Delta t$ (cm).
3. Rain on strip $A_i$ reaches the outlet after time $i\Delta t$ at a rate $2.78\,r\,A_i/\Delta t$ (m³/s). The ordinate at time $n\Delta t$ is the sum of the contributions of all blocks:

$$Q_n = \frac{2.78}{\Delta t}\sum_{i=1}^{n} r_i\,A_{n-i+1}$$

For example, $Q_1 = 2.78\,r_1A_1/\Delta t$, $Q_2 = 2.78\,(r_1A_2 + r_2A_1)/\Delta t$. 4. Plot $Q_n$ against time for the direct runoff hydrograph; add base flow for the total flow.

Since storage is neglected, the hydrograph is too peaked. Clark's method corrects this by routing the result through a linear reservoir.