There are a
few complicated experimental formulae for calculating Run-Offs of catchments
which are not strictly accurate and give unreliable results. The rational
method has been evolved in the form of the following general equation for
calculating the flood discharge of Run-Off an area.
Q=
(R.A.P)/36
Where Q is total Run-Off in Cu.Metrs/Second
R is Intensity of Maximum rainfall in Centimeter /Hour.
A
is drainage
area in hectares contributing to Run-Off
P is Factor of imperviousness
Example
Suppose
length of catchment is 1900 ft having width of 40 ft for expected Run-Off in
drainage area ,considering rainfall depth as given by weather department is 5cm
in peak hour of rain intensity.
Length of drainage Area= 1900
Width of
strip causing Run-Off = 40 Ft.
Area of
Run-Off = 1900 x 40 = 76000 Sft
1 Hectares =
107639 Sft
A=
Converting Sft Area into Hectares = 0.70 Hectares
R= Max.
Intensity of Rain fall depth = 05 Cm/Hour
P= This
value varies according to nature of Ground surface.
We have
following factors and we must choose one as per condition of our area, Factor
no. 4 matches with our case.
1- Steep
bare rock = 0.90
2- Rock
steep but wooded = 0.80
3- Plateaus
lightly covered, ordinary ground bare = 0.70
4-Densely built-up
areas of cities with metallic Roads
and paths =
0.70 to 0.90
5-
Residential areas not densely built up with metallic roads =0.50 to 0.70
6-
Residential areas not densely built up with Un-metallic roads =0.20 to 0.50
4-Densely built-up
areas of cities with metallic Roads
and paths =
0.70 to 0.90
P= Average
value (0.80)
Q= (R.A.P)/36
Putting
values in above formula
Q= (5 Cm/Hour
x 0.70 Hectares x 0.70)/36 = 0.068Cubic. Meter
So, we have Discharge of 0.10 Cum = 2.4
Cubic Feet /Sec which is also equal to (0.068x1000) = 68 Liters/Second
Note: Rainfall data will be taken from local departments as per acctual locations in the world.
