In the Godavari basin, consider a storm occurrence.Duration = 6 hoursEffective rainfall = 15 cmPeak flow = 930 m3/sConsider another storm of the same duration with effective rainfall = 7.5 cm.Find out the corresponding peak flow (in m3/s). (Answer up to the nearest integer)Correct answer is '465'. Can you explain this answer?

Question

Answer

Area is same

Answer

Given Information:
- Duration of the storm = 6 hours
- Effective rainfall for the first storm = 15 cm
- Peak flow for the first storm = 930 m3/s

Calculating Peak Flow for the Second Storm:
To find the corresponding peak flow for the second storm, we can use the Rational Method, which is commonly used in hydrologic analysis.

The Rational Method equation is given by:
Q = CiA
Where:
- Q is the peak flow (in m3/s)
- C is the runoff coefficient
- i is the rainfall intensity (in cm/h)
- A is the catchment area (in hectares)

Step 1: Calculating the Rainfall Intensity for the Second Storm:
The effective rainfall for the second storm is given as 7.5 cm, and the duration is 6 hours. To calculate the rainfall intensity, we divide the effective rainfall by the duration:
Rainfall intensity (i) = Effective rainfall / Duration
= 7.5 cm / 6 hours
= 1.25 cm/hr

Step 2: Calculating the Runoff Coefficient:
The runoff coefficient depends on various factors such as land use, soil type, and antecedent moisture conditions. Since this information is not given, we assume a typical value of 0.5 for the runoff coefficient.

Step 3: Calculating the Peak Flow for the Second Storm:
Using the Rational Method equation, we can now calculate the peak flow for the second storm:
Q = CiA
= (0.5) * (1.25 cm/hr) * A

To convert the units to m3/s, we need to convert the rainfall intensity from cm/hr to m/s and the catchment area from hectares to square meters.

1 cm/hr = 0.0278 m/s
1 hectare = 10,000 m2

Substituting the values:
Q = (0.5) * (1.25 cm/hr) * A
= (0.5) * (1.25 cm/hr) * (A hectares * 10,000 m2/hectare)
= (0.5) * (1.25 cm/hr) * (A * 10,000 m2)
= 0.625 * (A * 10,000) m3/s

Step 4: Finding the Corresponding Peak Flow:
Since the catchment area (A) is not given, we cannot calculate the exact peak flow. However, we are asked to find the answer to the nearest integer.

For the first storm, the peak flow is given as 930 m3/s. Let's assume the corresponding catchment area for the second storm is X hectares.

Using the peak flow equation for the first storm:
930 m3/s = (0.5) * (1.25 cm/hr) * (X * 10,000 m2)
Cancelling out the units and solving for X:
X = 930 / (0.5 * 1.25 * 10,000)
X ≈ 74.4 hectares

So, for a catchment area of approximately 74.4 hectares, the corresponding peak flow for the second storm would be:
Q = 0.625 * (

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