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A triangular direct runoff hydrograph due to a storm has a time base of 90 hours. The peak flow of 60 m3/s occurs at 20 hours from the start of the storm. The area of catchment is 300 km2. The rainfall excess of the storm (in cm), is
  • a)
    5.40
  • b)
    2.00
  • c)
    3.24
  • d)
    6.48
Correct answer is option 'C'. Can you explain this answer?
Verified Answer
A triangular direct runoff hydrograph due to a storm has a time base o...


∴ Rainfall excess = 3.24 cm
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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.
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A triangular direct runoff hydrograph due to a storm has a time base of 90 hours. The peak flow of 60 m3/s occurs at 20 hours from the start of the storm. The area of catchment is 300 km2. The rainfall excess of the storm (in cm), isa)5.40b)2.00c)3.24d)6.48Correct answer is option 'C'. Can you explain this answer?
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