Hydrology miscellaneous

Hydrology miscellaneous

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Hydrology

Hydrology miscellaneous

Direction: One hour triangular unit hydrograph of a watershed has the peak discharge of 60 m³/sec.cm at 10 hours and time base of 30 hours. The f index is 0.4 cm per hour and base flow is 15 m³/sec.

  1. If there is rainfall of 5.4 cm in 1 hour, the ordinate of the flood hydrograph at 15th hour is








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    Rainfall excess = Rainfall – φ index
    = 5.4 – 0.4 = 5 cm
    Base flow = 15 cm³/s

    Ordinate at 15th hour = 5 ×
    60
    		 × 15 + 15 = 240 m³/s.
    30 - 10

    Correct Option: B

    Rainfall excess = Rainfall – φ index
    = 5.4 – 0.4 = 5 cm
    Base flow = 15 cm³/s

    Ordinate at 15th hour = 5 ×
    60
    		 × 15 + 15 = 240 m³/s.
    30 - 10

  1. The catchment area of the watershed is








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    Rainfall depth (excess) = 1 cm (Unit hydrograph)
    Total volume = area under hydrograph

    =
    1
    		 × 60 × 30 × 60 × 60
    2

    = 3.24 × 10-6 m³
    A × 106 × 1 × 10-2 = 3.24 × 106
    ∴ A = 324 km².

    Correct Option: C

    Rainfall depth (excess) = 1 cm (Unit hydrograph)
    Total volume = area under hydrograph

    =
    1
    		 × 60 × 30 × 60 × 60
    2

    = 3.24 × 10-6 m³
    A × 106 × 1 × 10-2 = 3.24 × 106
    ∴ A = 324 km².

  1. The 4-hr unit hydrograph for a catchment is given in the table below. What would be the maximum ordinate of the S-curve (in m³/s) derived from this hydrograph?








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    Correct Option: A

Direction: For a catchment, the S-curve (or S-hydrograph) due to a rainfall of intensity 1 cm/hr is given by Q = 1 – (1 + t) exp (–1) (t in hr and Q in m³/s).

  1. What will be the ordinate of a 2-hour unit hydrograph for this catchment at t = 3 hours ?








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    S1 at 1 hour duration
    = 1 – (1 + t)e–t
    = 1 – (1 + 1)e–1 = 0.264 m³/s
    S2 at 3 hour duration
    = 1 – (1 + 3)e–3
    = 0.80 m³/s

    μ³ =
    S2 - S1
    		 =
    0.80 - 0.264
    		 = 0.27 m³/s.
    22

    Correct Option: C

    S1 at 1 hour duration
    = 1 – (1 + t)e–t
    = 1 – (1 + 1)e–1 = 0.264 m³/s
    S2 at 3 hour duration
    = 1 – (1 + 3)e–3
    = 0.80 m³/s

    μ³ =
    S2 - S1
    		 =
    0.80 - 0.264
    		 = 0.27 m³/s.
    22

  1. What is the area of the catchment ?








  1. View Hint View Answer Discuss in Forum

    Q = 1 – (1 + t)e–t
    Saturation discharge for S-curve will be at t → ∞
    Q = limt→∞ (1 – (1 + t) e-t)
    = 1 – (1 + ∞)e∞ = 1 m³/s
    Saturation discharge = Rainfall intensity × Catchment area
    1 × 60 × 60 = 1 × 10–3 × Area
    ∴ Catchment area = 360 × 10³ m² = .36 km².

    Correct Option: B

    Q = 1 – (1 + t)e–t
    Saturation discharge for S-curve will be at t → ∞
    Q = limt→∞ (1 – (1 + t) e-t)
    = 1 – (1 + ∞)e∞ = 1 m³/s
    Saturation discharge = Rainfall intensity × Catchment area
    1 × 60 × 60 = 1 × 10–3 × Area
    ∴ Catchment area = 360 × 10³ m² = .36 km².