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A triangular DRH due to a 6-h storm in a catchment has a time base of 100 h and a peak flow of 40 m3/s. The catchment area is 180 km2. The 6-h unit hydrograph of this catchment will have a peak flow in m3/s of
  • a)
    10    
  • b)
    20
  • c)
    30    
  • d)
    None of these
Correct answer is option 'A'. Can you explain this answer?
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Given data:
- Triangular DRH due to a 6-h storm in a catchment
- Time base = 100 h
- Peak flow = 40 m3/s
- Catchment area = 180 km2

To find: Peak flow in m3/s of 6-h unit hydrograph of the catchment

Solution:
1. Calculate the total volume of runoff generated by the storm using the DRH:
Total volume of runoff = (Time base / 2) x Peak flow
= (100/2) x 40
= 2000 m3

2. Calculate the average rainfall intensity over the catchment during the storm:
Average rainfall intensity = Total volume of runoff / Catchment area
= 2000 / (180 x 106)
= 0.0111 m/h

3. Use the formula for the 6-h unit hydrograph:
Peak flow of 6-h unit hydrograph = (Average rainfall intensity x Catchment area) / 10
= (0.0111 x 180 x 106) / 10
= 1998 m3/h
= 1998 / 3600 m3/s
= 0.555 m3/s
≈ 10 m3/s

Therefore, the peak flow in m3/s of 6-h unit hydrograph of the catchment is 10 m3/s (option A).
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A triangular DRH due to a 6-h storm in a catchment has a time base of 100 h and a peak flow of 40 m3/s. The catchment area is 180 km2. The 6-h unit hydrograph of this catchment will have a peak flow in m3/s ofa)10 b)20c)30 d)None of theseCorrect answer is option 'A'. Can you explain this answer?
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