Question

An S-hydrograph was prepared for a catchment of 240 km$^2$ using 3-hour unit hydrograph (1 cm rainfall excess). The equilibrium discharge for the S-hydrograph would be ________________ (in m$^3$/s, rounded off to two decimal places).

Hint:

The equilibrium discharge of an S-hydrograph is simply the uniform rate at which the rainfall excess contributes to streamflow: $Q = \dfrac{\text{volume of rainfall excess}}{\text{duration}}$.

Answer 1

Correct Answer: 220

Step 1: Recall formula for equilibrium discharge.
\[ Q_e = \frac{A . R}{T} \] where - $A$ = catchment area in m$^2$, - $R$ = rainfall excess in m, - $T$ = duration of excess rainfall in seconds.
Step 2: Convert given values.
\[ A = 240 \, km^2 = 240 \times 10^6 \, m^2 \] \[ R = 1 \, cm = 0.01 \, m \] \[ T = 3 \, hr = 3 \times 3600 = 10800 \, s \]
Step 3: Substitute into formula.
\[ Q_e = \frac{240 \times 10^6 \times 0.01}{10800} \] \[ = \frac{2.4 \times 10^6}{10800} \approx 222.22 \, m^3/s \] Final Answer: \[ \boxed{222.22 \, m^3/s} \]