0.2 x 106
0.6 x 106
2.4 x 106
7.2 x 106
Answer is Wrong!
Answer is Right!
The correct answer is $\boxed{\text{C}}$.
The equilibrium discharge of an S-curve is the maximum discharge that can be sustained by the catchment area. It is calculated by summing
the ordinates of the unit hydrograph for each hour, and then dividing by the time interval between the ordinates. In this case, the time interval is 6 hours, so the equilibrium discharge is:$$Q_e = \frac{\sum_{i=1}^6 Q_i}{6} = \frac{0.2 + 0.6 + 1.2 + 1.8 + 2.4 + 3.0}{6} = 2.4 \times 10^6 \text{ m}^3/\text{hour}$$
Option A is incorrect because it is the equilibrium discharge for a catchment area of 20 km2. Option B is incorrect because it is the equilibrium discharge for a catchment area of 40 km2. Option D is incorrect because it is the equilibrium discharge for a catchment area of 80 km2.