Hydrology miscellaneous

Hydrology miscellaneous

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Hydrology

Hydrology miscellaneous
  1. During a 3-hour storm event, it was observed that all abstractions other than infiltration are negligible. The rainfall was idealized as 3 one hour storms of intensity 10 mm/hr, 20 mm/hr and 10 mm/hr respectively and the infiltration was idealized as a Horton curve, f = 6.8 + 8.7 exp (–t) (f in mm/hr and t in hr). What is the effective rainfall ?








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    f = 6.8 + 8.7 e–t
    t = 0, f = 6.8 + 8.7 = 15.5 > 10 mm
    t = 0 to 1
    Infiltration = ¹∫0(6.8 + 8.7 e-t)dt

    = 6.8t +
    8.7 e-t
    		¹
    - 1 0

    = 12.3 mm > 10 mm
    ∴ No excess rainfall t = 1 to 2
    Total infiltration = ²∫1(6.8 + 8.7 e-t)dt
    = 6.8t +
    8.7 e-t
    		²
    - 1 1

    = 8.82 < 20 mm
    ∴ Rainfall excess = 20 – 8.82 = 11.17 mm
    t = 2 to 3
    Total infiltration = ³∫2(6.8 + 8.7 e-t)dt
    = 7.54 > 10 mm
    Rainfall excess = 10 – 7.54 = 2.45
    Total rainfall excess = 10 + 11.17 + 2.45 = 13.63 mm.

    Correct Option: D

    f = 6.8 + 8.7 e–t
    t = 0, f = 6.8 + 8.7 = 15.5 > 10 mm
    t = 0 to 1
    Infiltration = ¹∫0(6.8 + 8.7 e-t)dt

    = 6.8t +
    8.7 e-t
    		¹
    - 1 0

    = 12.3 mm > 10 mm
    ∴ No excess rainfall t = 1 to 2
    Total infiltration = ²∫1(6.8 + 8.7 e-t)dt
    = 6.8t +
    8.7 e-t
    		²
    - 1 1

    = 8.82 < 20 mm
    ∴ Rainfall excess = 20 – 8.82 = 11.17 mm
    t = 2 to 3
    Total infiltration = ³∫2(6.8 + 8.7 e-t)dt
    = 7.54 > 10 mm
    Rainfall excess = 10 – 7.54 = 2.45
    Total rainfall excess = 10 + 11.17 + 2.45 = 13.63 mm.

  1. Two observation wells penetrated into a confined acquifer and located 1.5 km apart in the direction of flow, indicate head of 45 m and 20m. If the coefficient of permeability of the acquifer is 30m/day and porosity is 0.25, the time of travel of an inert tracer from one well to another is








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    Discharge velocity,
    V = hydraulic gradient (i) × Coeff. of permeability (k)

    i =
    hƒ
    		 =
    45 - 20
    		 = 0.01667
    L1.5 × 10³

    k = 30 m/d
    ∴ V = 30 × 0.01667 = 0.5 m/d
    Actual velocity, Vs =
    V
    		 =
    0.5
    		 = 2 m/d
    η0.25

    Time of travel =
    distance
    		 =
    1.5 × 10³
    		 = 750
    Vs2

    Correct Option: C

    Discharge velocity,
    V = hydraulic gradient (i) × Coeff. of permeability (k)

    i =
    hƒ
    		 =
    45 - 20
    		 = 0.01667
    L1.5 × 10³

    k = 30 m/d
    ∴ V = 30 × 0.01667 = 0.5 m/d
    Actual velocity, Vs =
    V
    		 =
    0.5
    		 = 2 m/d
    η0.25

    Time of travel =
    distance
    		 =
    1.5 × 10³
    		 = 750
    Vs2