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A 4 hour unit hydrograph of a basin with area of 1728 km2 can be approximated as a triangle with base period of 48 hours. The peak ordinate of hydrograph is
- a)400m3/s
- b)300m3/s
- c)200m3/s
- d)100m3/s
Correct answer is option 'C'. Can you explain this answer?
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Calculation of Peak Ordinate of Hydrograph
Given:
Area of basin = 1728 km2
Base period of hydrograph = 48 hours
Unit hydrograph duration = 4 hours
Unit hydrograph shape = Triangle
We know that the peak flow of a hydrograph is given by the formula:
Peak flow = (P × A)/T
where,
P = peak ordinate of unit hydrograph
A = catchment area
T = time base of unit hydrograph
Here, A = 1728 km2 and T = 4 hours.
Calculation of P
To calculate P, we need to find the area of the triangle unit hydrograph. Since the base period of the hydrograph is 48 hours and the unit hydrograph duration is 4 hours, the triangle is made up of 12 equal parts.
The area of the triangle can be calculated as:
Area of triangle = (base × height)/2
where,
base = 12 × 4 hours = 48 hours
height = peak ordinate of unit hydrograph
Since the unit hydrograph shape is a triangle, the peak ordinate occurs at the midpoint of the base. Therefore, the height of the triangle is half of the peak ordinate.
Hence, the area of the triangle can be expressed as:
Area of triangle = (48 × P/2)/2
= 12P
Now, we know that the catchment area is 1728 km2 and the time base of the unit hydrograph is 4 hours. Therefore, the peak flow can be calculated as:
Peak flow = (P × A)/T
= (P × 1728)/4
= 432P
Equating this with the area of the triangle, we get:
Area of triangle = Peak flow
12P = 432P
P = 200 m3/s
Therefore, the peak ordinate of the hydrograph is 200 m3/s.
Given:
Area of basin = 1728 km2
Base period of hydrograph = 48 hours
Unit hydrograph duration = 4 hours
Unit hydrograph shape = Triangle
We know that the peak flow of a hydrograph is given by the formula:
Peak flow = (P × A)/T
where,
P = peak ordinate of unit hydrograph
A = catchment area
T = time base of unit hydrograph
Here, A = 1728 km2 and T = 4 hours.
Calculation of P
To calculate P, we need to find the area of the triangle unit hydrograph. Since the base period of the hydrograph is 48 hours and the unit hydrograph duration is 4 hours, the triangle is made up of 12 equal parts.
The area of the triangle can be calculated as:
Area of triangle = (base × height)/2
where,
base = 12 × 4 hours = 48 hours
height = peak ordinate of unit hydrograph
Since the unit hydrograph shape is a triangle, the peak ordinate occurs at the midpoint of the base. Therefore, the height of the triangle is half of the peak ordinate.
Hence, the area of the triangle can be expressed as:
Area of triangle = (48 × P/2)/2
= 12P
Now, we know that the catchment area is 1728 km2 and the time base of the unit hydrograph is 4 hours. Therefore, the peak flow can be calculated as:
Peak flow = (P × A)/T
= (P × 1728)/4
= 432P
Equating this with the area of the triangle, we get:
Area of triangle = Peak flow
12P = 432P
P = 200 m3/s
Therefore, the peak ordinate of the hydrograph is 200 m3/s.
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A 4 hour unit hydrograph of a basin with area of 1728 km2 can be approximated as a triangle with base period of 48 hours. The peak ordinate of hydrograph isa)400m3/sb)300m3/sc)200m3/sd)100m3/sCorrect answer is option 'C'. Can you explain this answer?
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