Hydrology miscellaneous
- The direct runoff hydrograph of a storm obtained from a catchment is triangular in shape and has a base period of 80 hours. The peak flow rate is 30 m³/sec and catchment area is 86.4 km². The rainfall excess that has resulted the above hydrograph is
-
View Hint View Answer Discuss in Forum
Volume of run off = Area of hydrograph
= 1 × 30 × (80 × 60 × 60) 2
= 4320000 m³/s
Area of catchment = 86.4 km² = 86.4 × 106 m²
Rainfall excess = run off= Volume of run off Area of catchment = 4320000 = 0.05 m 86.4 × 106 Correct Option: A
Volume of run off = Area of hydrograph
= 1 × 30 × (80 × 60 × 60) 2
= 4320000 m³/s
Area of catchment = 86.4 km² = 86.4 × 106 m²
Rainfall excess = run off= Volume of run off Area of catchment = 4320000 = 0.05 m 86.4 × 106
- 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
-
View Hint View Answer Discuss in Forum
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 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
-
View Hint View Answer Discuss in Forum
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
- 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
-
View Hint View Answer Discuss in Forum
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
- A watershed got transformed from rural to urban over a period of time. The effect of urbanization on storm runoff hydrograph from the watershed is to
-
View Hint View Answer Discuss in Forum
NA
Correct Option: C
NA
