Procedure of deriving Clark UH

Clark's method derives a unit hydrograph by passing the excess rainfall through two effects separately: translation (movement of water to the outlet, represented by a time-area curve) and attenuation/storage (represented by a linear reservoir at the outlet).

Data needed

Catchment area $A$, time of concentration $t_c$, storage constant $K$, and the isochrone map (areas between lines of equal travel time to the outlet).

Procedure

1. Delineate isochrones. Divide the basin into zones by lines of equal travel time to the outlet at interval $\Delta t$ (up to $t_c$). Find the area $A_i$ between successive isochrones by planimeter.
2. Time-area curve. Plot the cumulative area against travel time (or the area histogram).
3. Translation hydrograph. For 1 cm of excess rain, the inflow from each strip to the outlet reservoir is

$$I_i = \frac{2.78\,A_i}{\Delta t}\ \text{m}^3/\text{s}\quad (A_i \text{ in km}^2,\ \Delta t \text{ in h})$$

4. Find $K$. Estimate the storage constant $K$ from the recession limb of an observed hydrograph, $K = -Q/(dQ/dt)$ at the inflexion point, or from regional relations.
5. Route through a linear reservoir $S = KO$:

$$O_2 = C_0 (I_1 + I_2) + C_2 O_1,\quad C_0 = \frac{\Delta t}{2K + \Delta t},\quad C_2 = \frac{2K - \Delta t}{2K + \Delta t}$$

The outflow ordinates form the instantaneous unit hydrograph (IUH) of the catchment. 6. Unit hydrograph of duration $\Delta t$. Take the average of two successive IUH ordinates:

$$U_n = \frac{O_n + O_{n-1}}{2}$$

This is the $\Delta t$-hour unit hydrograph, i.e. the response to 1 cm of excess rain in $\Delta t$ hours. 7. Use. Multiply by excess rainfall depths and superpose to get the flood hydrograph for a storm.

Flow chart

 Isochrone map -> area per strip
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        v
 Inflow I = 2.78 A / dt
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        v
 Route through linear reservoir (K)
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        v
 IUH ordinates O
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        v
 dt-hour UH = average of two IUH ordinates