Hydrology miscellaneous
- During a 6-hour storm, the rainfall intensity was 0.8 cm/hour on a catchment of area 8.6 km². The measured runoff volume during this period was 2,56,000 m³. The total rainfall was lost due to infiltration, evaporation, and transpiration (in cm/hour) is
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Rainfall, P = i × t = 0.8 × 6 = 4.8 cm
Run off depth,R = Volume of run off Catchment area = 25600 8.6 × 106
= 0.02976 m
= 2.97 mTotal loss = P - R = 4.8 - 2.97 = 0.34 cm/h r 6 Correct Option: B
Rainfall, P = i × t = 0.8 × 6 = 4.8 cm
Run off depth,R = Volume of run off Catchment area = 25600 8.6 × 106
= 0.02976 m
= 2.97 mTotal loss = P - R = 4.8 - 2.97 = 0.34 cm/h r 6
- The vertical hydraulic conductivity of the top soil at certain is 0.2 cm/h. A storm of intensity 0.5 cm/ hr occurs over the soil for an indefinite period. Assuming the surface drainage to be adequate, the infiltration rate after the storm has lasted for a very long time, shall be
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Hydraulic conductivity, ƒc = 0.2 cm/h
Storm intensity, i = 0.5 cm/h
Infiltration rate = ƒa
If i > ƒc ⇒ ƒa = ƒc
i > ƒc ⇒ ƒa = ƒi
Here i > ƒc
∴ ƒa = ƒc = 0.2 cm/hCorrect Option: B
Hydraulic conductivity, ƒc = 0.2 cm/h
Storm intensity, i = 0.5 cm/h
Infiltration rate = ƒa
If i > ƒc ⇒ ƒa = ƒc
i > ƒc ⇒ ƒa = ƒi
Here i > ƒc
∴ ƒa = ƒc = 0.2 cm/h
- The plan area of a reservoir is 1 km². The water level in the reservoir is observed to decline by 20 cm in a certain period. During this perod the reservoir receives a surface inflow of 10 hectare-meters, and 20 hectare-meters are abstracted from the reservoir for irrigation and power. The pan evaporation and rainfall recorded during the same period at a near by meteorological station are 12 cm and 3 cm respectively. The calibrated pan factor is 0.7. The seepage has from the reservoir during this period in hectare-meters is
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Area = 1 km² = 100 ha
Change in storage = inflow – outflow
⇒ ds = I – Q
I = Surface inflow + rainfall
= 100 + .03 × 100
= 13 ha.m
Q = Irrigation and power + evaporation + seepage loss. = 20 + C p. Ep. A + Seepage loss
= 20 + 0.7 × .12 × 100 + Seepage loss = 28.4 + Seepage loss
ds = – 0.20 × 100 = – 20 (decline, ∴ –ve sign)
∴ – 20 = 13 – (28.4 + Seepage loss)
⇒ Seepage loss = 4.6 ha–mCorrect Option: D
Area = 1 km² = 100 ha
Change in storage = inflow – outflow
⇒ ds = I – Q
I = Surface inflow + rainfall
= 100 + .03 × 100
= 13 ha.m
Q = Irrigation and power + evaporation + seepage loss. = 20 + C p. Ep. A + Seepage loss
= 20 + 0.7 × .12 × 100 + Seepage loss = 28.4 + Seepage loss
ds = – 0.20 × 100 = – 20 (decline, ∴ –ve sign)
∴ – 20 = 13 – (28.4 + Seepage loss)
⇒ Seepage loss = 4.6 ha–m
- The rainfall during three successive 2 hour periods are 0.5, 2.8 and 1.6 cm. The surface runoff resulting from this storm in 3.2 cm. The f index value of the storm is
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R = P – I
I = R if R < Iφ = P - R = (0.5 + 2.8 + 16) - 3.2 t 2 + 2 + 2
= 0.2833 cm/h = 0.566 cm/2h > 0.5 cm
∴ There is no run off for first 2 hours.Infiltration = 0.5 = 0.25 cm/h 2 φindex = P - R = (2.8 + 1.6) - 3.2 = 0.3 cm/h t 4 Correct Option: C
R = P – I
I = R if R < Iφ = P - R = (0.5 + 2.8 + 16) - 3.2 t 2 + 2 + 2
= 0.2833 cm/h = 0.566 cm/2h > 0.5 cm
∴ There is no run off for first 2 hours.Infiltration = 0.5 = 0.25 cm/h 2 φindex = P - R = (2.8 + 1.6) - 3.2 = 0.3 cm/h t 4
- The average rainfall for a 3 hour duration storm is 2.7 cm and the loss rate is 0.3 cm/hr. The flood hydrograph has a base flow of 20 m³/s and produces a peak flow of 210 m³/s. The peak of a 3-h unit hydrograph is
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Run off = Rainfall – losses = 2.7 – (0.3 × 3) = 1.8 cm
Peak of DRH = Peak of flood hydrograph – base flow = 210 – 20 = 190 m³/sPeak of 3h UH = Peak of DRH R = 190 = 105.55 m³/s 1.8 Correct Option: B
Run off = Rainfall – losses = 2.7 – (0.3 × 3) = 1.8 cm
Peak of DRH = Peak of flood hydrograph – base flow = 210 – 20 = 190 m³/sPeak of 3h UH = Peak of DRH R = 190 = 105.55 m³/s 1.8
