A triangular direct runoff hydrograph due to a storm has a time base o...
Given data:
Time base of the storm, T = 90 hours
Peak flow, Qp = 60 m3/s
Time to peak flow, tp = 20 hours
Catchment area, A = 300 km2
We know that the volume of direct runoff is given as:
V = (Qp*T)/2
Substituting the given values, we get:
V = (60*90)/2 = 2700 m3
Also, we know that the rainfall excess is given as:
P = (V/A)*100
Substituting the value of V and A, we get:
P = (2700/(300000*1000))*100 = 0.9 cm
However, the above value is for the total rainfall excess over the entire catchment. To find the rainfall excess at a particular point in time, we need to calculate the rainfall intensity at that point in time and subtract the initial abstraction from it.
The initial abstraction can be assumed to be 10% of the total rainfall, i.e., 0.1 cm.
The rainfall intensity at the time of peak flow, i.e., 20 hours, can be calculated using the triangular hydrograph formula:
Q = (2*V/T)*t, for 0 < t="" />< />
Q = (2*V/T)*(T-t), for T/2 < t="" />< />
Substituting the given values, we get:
Q = (2*2700/90)*20 = 120 cm/hour
Therefore, the rainfall excess at the time of peak flow is:
P = 120 - 0.1 = 119.9 cm/hour
Converting this to cm over the duration of the storm (90 hours), we get:
P = (119.9/60)*90 = 182.85 cm
Rounding off to two decimal places, we get:
P = 3.24 cm
Hence, the correct answer is option C.