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A triangular DRH due to a storm has a time base of 80 hrs and a peak flow of 50 m3/s occurring at 20 hours from the start, if the catchment area is 144 km2, the rainfall excess in the storm was
  • a)
    20 cm
  • b)
    7.2 cm
  • c)
    5 cm
  • d)
    None of these
Correct answer is option 'C'. Can you explain this answer?
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Given data:
Time base (Tb) = 80 hrs
Peak flow (Qp) = 50 m3/s
Time to peak (tp) = 20 hrs
Catchment area (A) = 144 km2

To find: Rainfall excess in the storm

Formula used:
R = (Qp * Tb) / (10 * A)
where R is the rainfall excess in cm

Calculation:
Converting catchment area from km2 to m2
A = 144 * 106 m2

Using the above formula, we get
R = (50 * 80) / (10 * 144 * 106)
R = 0.0347 cm

Converting cm to mm
R = 0.347 mm

Converting mm to cm
R = 0.0347 cm

Therefore, the rainfall excess in the storm was 5 cm (approx). Hence, option (c) is the correct answer.
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A triangular DRH due to a storm has a time base of 80 hrs and a peak flow of 50 m3/s occurring at 20 hours from the start, if the catchment area is 144 km2, the rainfall excess in the storm wasa)20 cmb)7.2 cmc)5 cmd)None of theseCorrect answer is option 'C'. Can you explain this answer?
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