How to find out the quantity of stormwater run-off in a catchment area? Let’s take a mathematical example.
Mathematical Problem:
A catchment area of 20 km2 consists of two-thirds rural and one-third urban areas. The rainfall intensity in this area is recorded as 25 mm/hr. Find the quantity of stormwater run-off in the area in litre per second. ‘K’ for rural area is 0.30 and urban 0.50.
Solution:
Quantity of stormwater run-off
[latex]Q = \frac{KiA}{360}[/latex]
Where,
- K = Coefficient of run-off
- i = Rainfall intensity (in mm/hr) = 25 mm/hr
- A = Catchment area in hector
Rural catchment area = [latex]\left ( \frac{2}{3}\times 20 \right )[/latex] = 13.3333 Km2 = 1333.33 hector.
Urban catchment area = [latex]\left ( \frac{1}{3}\times 20 \right )[/latex] = 6.6666 Km2 = 666.66 hector.
Now,
The quantity of storm water run-off from the rural catchment area
[latex]Q = \frac{KiA}{360}[/latex]
Or, [latex]Q = \frac{0.3\times 25\times 1333.33}{360}[/latex] = 27.777 m3/sec = 27777 litres/sec
The quantity of storm water run-off from the urban catchment area
Q = [latex]Q = \frac{0.50\times 25\times 666.66}{360}[/latex] = 23.125 m3/sec = 23125 liters/sec
Read More: