Hydrology miscellaneous
Direction: The drainage area of a watershed is 50 km². The φ index is 0.5 cm/hour and the base flow at the outlet is 10 m³/s. One hour unit hydrograph (unit depth = 1 cm) of the watershed is triangular in shape with a time base of 15 hours. The peak ordinate occurs at 5 hours.
- The peak ordinate (in m³/s/cm) of the unit hydrograph is
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Volume of water = Area under curve
= 1 × 15 × Q0 × 3600 2
Area × Depth = Volume
Depth = 1 cm (UH)(50 × 106) × (1 × 10–2) = 1 × 15 × 3600 × Q0 2
∴ Q0 = 18.52 m³/s.Correct Option: B
Volume of water = Area under curve
= 1 × 15 × Q0 × 3600 2
Area × Depth = Volume
Depth = 1 cm (UH)(50 × 106) × (1 × 10–2) = 1 × 15 × 3600 × Q0 2
∴ Q0 = 18.52 m³/s.
Direction: A four hour unit hydrograph of a catchment is traingular in shape with base of 80 hours. The area of the catchment is 720 km². The base flow and f-index are 30 m³/s and 1 mm/h, respectively. A storm of a 4 cm occurs uniformly in 4 hours over the catchment.
- The peak flood discharge due to the storm is
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Rainfall excess, R = P – φt
= 4 – 0.1 × 4 = 3.6 cm
Peak of DRH = Peak of UH × R
= 50 × 3.6 = 180 m³/s
Peak of flood hydrograph = Peak of DRH + Base flow
= 180 + 30 = 210Correct Option: A
Rainfall excess, R = P – φt
= 4 – 0.1 × 4 = 3.6 cm
Peak of DRH = Peak of UH × R
= 50 × 3.6 = 180 m³/s
Peak of flood hydrograph = Peak of DRH + Base flow
= 180 + 30 = 210
- The peak discharge of four hour unit hydrograph is
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Area of UH = Area of catchment × unit depth
= 1 × (80 × 60 × 60) × QP 2 
Volume of rain fall = A × 1 cm
= 720 × 106 × 0.01 m∴ 1 × 80 × 60 × 60 × QP 2
= 720 × 106 × 0.01
∴ Qp = 50 m³/sCorrect Option: B
Area of UH = Area of catchment × unit depth
= 1 × (80 × 60 × 60) × QP 2
Volume of rain fall = A × 1 cm
= 720 × 106 × 0.01 m∴ 1 × 80 × 60 × 60 × QP 2
= 720 × 106 × 0.01
∴ Qp = 50 m³/s
Direction: The ordinates of a 2-h unit hydrograph at 1 hour intervals starting from time t = 0, are 0, 3, 8, 6, 3, 2 and 0 m³/s. Use trapezoidal rule for numerical integration, if required.
- A storm of 6.6 cm occurs uniformly over the catchment in 3 hours. If f-index is equal to 2 mm/h and base flow is 5 m³/s, what is the peak flow due to the storm?
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Effective rainfall = rainfall – φ index
φ index = 2 mm/hour = (2 × 3) + 10–1 (3 hours)
(10–1 to convert mm to cm)
= 0.6
Effective rainfall = 6.6 – 0.6 = 6 cm
Peak discharge = Effective rainfall × maximum ordinate of UH + Base flow
= 6 × 8 + 5 = 53 m³/s.Correct Option: C
Effective rainfall = rainfall – φ index
φ index = 2 mm/hour = (2 × 3) + 10–1 (3 hours)
(10–1 to convert mm to cm)
= 0.6
Effective rainfall = 6.6 – 0.6 = 6 cm
Peak discharge = Effective rainfall × maximum ordinate of UH + Base flow
= 6 × 8 + 5 = 53 m³/s.
- What is the catchment area represented by the unit hydrograph?
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Volume of water = area under graph= 
1 × 1 × 3 
+ (3 × 1) + 1 × 5 × 1 + 1 × 2 × 1 + (6 × 1) + 1 × 3 × 1 + (3 × 1) + 1 × 2 × 1 + (2 × 1) + 1 × 2 × 1 2 2 2 2 2 2
= 22 m³ × 3600
(3600 is multiplied to convert hour to seconds)
= 79200 m³
Depth = 1 cm (UH)V = depth area Area = 79200 = 7.92 km² 1 Correct Option: C
Volume of water = area under graph= 1 × 1 × 3 + (3 × 1) + 1 × 5 × 1 + 1 × 2 × 1 + (6 × 1) + 1 × 3 × 1 + (3 × 1) + 1 × 2 × 1 + (2 × 1) + 1 × 2 × 1 2 2 2 2 2 2
= 22 m³ × 3600
(3600 is multiplied to convert hour to seconds)
= 79200 m³
Depth = 1 cm (UH)V = depth area Area = 79200 = 7.92 km² 1
