# Storm Drainage Design Project Hydrographs

Rainfall is the input used to a watershed and the stream flow is also considered as the output of the watershed. A hydrograph also represents how a watershed reacts to rainfall. King1998, in his report said that the response of a watershed to rainfall depends on several factors which affects the shape of a hydrograph. This represents the effect of rainfall on a particular basin. This is a hypothetical response of a watershed to the input of rainfall. This will allow calculations of the response to any input(rainfall). Hydrographs are often affected by factors of soil saturation and the surroundings. The vegetation type and the steepness of the surrounding lands, with the drainage density(Stott,2005). Very dry weather, nornmally creates a crust on river beds and wet winters would increase the discharge. Sand and clay produces a flashy hydrographs but there could be a difference between the two.
be used as a water source of an open channel that would supply water to nearby reservoir. The hydrograph data was used to analyze the water activity in the river after a rainstorm. The study was done for a period of 96 hours. Measurements on the height of water was done on an hourly basis. The results of which are as follows. The water level was almost constant for the first 43 hours. At the start of the 44th hour, the water level started to rise. This stage of the rise of water is called the rising limb.It reached the peak, or the highest water level on the 58th hour. As it started to fall down, it undergoes falling limb. This is when the height of water level starts to recede. The falling limb would come after the rising limb . The peak discharge is when water reaches its highest point and there is the greatest amount of water in the river. The lag time is theperiod of time that is recorded between the peak rainfall and peak discharge.
Design computation of the Channel
Data provided Note: Apply the the Manning’s Formula to get value for Q = 0.8 m3/s breadth b, of a channel with a given value for d.
n = 0.012
S = 1/2000 = 0.0005
d = 0.3
Substitute the data in the equations below:
V = where: v = velocity
Q = Av R = Hydraulic Radius
Q = A S = slope
A = bd n = Manning’s coefficient
R = Q = discharge

Solutions:
A = db
= 0.3(b)
Q = A
R =
0.8 = 0.3b
0.8(0.012) = 0.3b
0.0285 = 0.6b
= 0.3
=
1.447 =
(1.447)3 = b3
3.029 =
3.029 =
3.029 (0.36 + 1.2b + b2) = 0.09b5
1.0904 + 3.6348b + 3.029b2 = 0.09b5
1.0904 + 3.6348b + 3.029b2 – 0.09b5 = 0
b = 3.5799 m.
In order to get the value of the discharge in the river, the first computations will be to solve the value of depth d:
From the given data
v = 4.0m3/s
s = 0.0005
n = 0.012,
Q = Av
v =
v =
4.0 =
4.0(0.012) =
0.048 =

2.17 =
=