Hydrology miscellaneous Difficult Questions and Answers | Page - 8

Hydrology miscellaneous

The average surface area of a reservoir in the month of June is 20 km². In the same month, the average rate of inflow is 10 m³/s, outflow rate is 15 m³/s, monthly rainfall is 10 cm, monthly seepage loss is 1.8 cm and the storage change is 16 million m³. The evaporation (in cm) in that month is:

1. 46.8
1. 136.0
1. 13.6
1. 23.4

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Surface area = 20 km²
Inflow = 10 m³/s (Q_i) ↓
Rainfall = 10 cm (Q_R) ↓
Outflow = 15 m³/s (Q_o) ↑
Seepage loss = 1.8 cm. (Q_s) ↑
Evaporation loss = ? (Q_E) ↑
Storage change = inflow – outflow = 16 × 10⁶ m³
⇒ 16 × 10⁶ = Q_i +Q_R – (Q_o + Q_s + Q_E)
= (10 × 86400 × 30) + (0.1 × 20 × 10⁶) – [(15 × 86400 × 30) + (1.8 ×10^-2 × 20 ×10⁶) + Q_E]
⇒ 16 × 10⁶ = (2.592 × 10⁶ + 2 × 10⁶) – (12.96 × 10⁶ + .36× 10⁶ + Q_E)
⇒ Q_E = 27.32 × 10⁶ = 1.366 m
20 × 10⁶

= 136.6 cm
Maximum bending stress occurs when wheel load is equidistant from the centre of the beam. The figure shows the position of wheal loads at maximum bending stress condition.

Correct Option: B

Surface area = 20 km²
Inflow = 10 m³/s (Q_i) ↓
Rainfall = 10 cm (Q_R) ↓
Outflow = 15 m³/s (Q_o) ↑
Seepage loss = 1.8 cm. (Q_s) ↑
Evaporation loss = ? (Q_E) ↑
Storage change = inflow – outflow = 16 × 10⁶ m³
⇒ 16 × 10⁶ = Q_i +Q_R – (Q_o + Q_s + Q_E)
= (10 × 86400 × 30) + (0.1 × 20 × 10⁶) – [(15 × 86400 × 30) + (1.8 ×10^-2 × 20 ×10⁶) + Q_E]
⇒ 16 × 10⁶ = (2.592 × 10⁶ + 2 × 10⁶) – (12.96 × 10⁶ + .36× 10⁶ + Q_E)
⇒ Q_E = 27.32 × 10⁶ = 1.366 m
20 × 10⁶

= 136.6 cm
Maximum bending stress occurs when wheel load is equidistant from the centre of the beam. The figure shows the position of wheal loads at maximum bending stress condition.

The Muskingum model of routing a flood through a stream reach is expressed as O₂ = K₀I₂ + K₁I₁ + K₂O₁, where K₀, K₁ and K₂ are the routing coefficients for the concerned reach, I₁ and I₂ are the inflows to the reach, and O₁ and O₂ are the outflows from the reach corresponding to time steps 1 and 2 respectively. The sum of K₀, K₁ and K₂ of the model is

1. –1
1. –0.5
1. 0.5
1. 1

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NA

Correct Option: D

NA

An isolated 3-h rainfall event on a small catchment produces a hydrograph peak and point of inflection on the falling limb of the hygrograph at 7 hours and 8.5 hours respectively, after the start of the rainfall. Assuming, no losses and no base flow contribution, the time of concentration (in hours) for this catchment is approximately

1. 8.5
1. 7.0
1. 6.5
1. 5.5

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NA

Correct Option: D

NA

A 1-h rainfall of 10 cm magnitude at a station has a return period of 50 years. The probability that a 1-h rainfall of magnitude 10 cm or more will occur in each of two successive years is:

1. 0.04
1. 0.2
1. 0.02
1. 0.0004

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Return period of rainfall
T = 50 years
∴ Probability of occurrence once in 50 years,
p = 1 = 0.02
50

Probability of occurrence in each of 2 successive year
= p² = (0.02)² = 0.004

Correct Option: D

Return period of rainfall
T = 50 years
∴ Probability of occurrence once in 50 years,
p = 1 = 0.02
50

Probability of occurrence in each of 2 successive year
= p² = (0.02)² = 0.004

The ratio of actual evapotranspiration to potential evapotranspiration is in the range

1. 0.0 to 0.4
1. 0.6 to 0.9
1. 0.0 to 1.0
1. 1.0 to 2.0

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For soil moisture at field capacities, AET = PET
For soil moisture at permanent willing point, AET ≈ 0
AET/PET have value between 0 and 1.

Correct Option: C

For soil moisture at field capacities, AET = PET
For soil moisture at permanent willing point, AET ≈ 0
AET/PET have value between 0 and 1.