An asphalt pavement drains into a rectangular concrete channel. The catchment surface has an average slope of 1.0%, and the distance from the catchment boundary to the drain is 30 m. The drainage channel is 60 m long, 20 cm wide, and 25 cm deep, and has a slope of 0.6%. For an effective rainfall rate of 50 mm/h, the flowrate in the channel is estimated to be 0.02 m3 /s. The Manning n for an asphalt surface is 0.011 and for a concrete surface is 0.013. Estimate the time of concentration of overland flow for the catchment. (Hint: Use the kinematic-wave equation for overland flow.)

Where [tex]t_{1}[/tex] is time of concentration in minutes, n is Manning’s roughness coefficient, L is travel distance between catchment boundary and drain in metres, [tex]i_{e}[/tex] is effective rainfall intensity in mm/h, [tex]S_{o}[/tex] is the slope of the ground

Substituting L with 30 m as given, [tex]i_{e}[/tex] with 50 mm/h as provided, [tex]S_{o}[/tex] with 0.01 as given, Manning’s coefficient with 0.011 as given then

[tex]t_{1}=\frac {6.99}{(50 mm/h)^{\frac{2}{5}}}*(\frac {0.011*30}{\sqrt {0.01}})^{\frac {3}{5}}[/tex]

[tex]t_{1}=\frac {6.99}{4.78}(\frac {0.33}{0.1})^{\frac {3}{5}}[/tex]

[tex]t_{1}=3 mins[/tex]

The area, A of rectangular channel in terms of channel depth

A=wd where d is depth and w is channel width

Substituting w with 20cm which is 0.2m

A=0.2d

Wetted perimeter of rectangular channel

P=2d+w=20d+0.2

From Manning’s equation of flow

[tex]Q=\frac {1}{n}AR^{\frac {2}{3}}S_{0}^{0.5}[/tex] where q is flow rate, A is cross-sectionall area, R is hydraulic radius, [tex]S_{o}[/tex] is the slope and n is Manning’s coefficient

The flow rate, q is given in the question as 0.02 m3/s and Manning’s coefficient for overland flow which corresponds to concrete is given as 0.013

[tex]0.02\frac {m^{3}}{sec}=\frac {1}{0.013}* \frac {(0.2d)^{\frac {5}{3}* (0.006)^{0.5}}}{(2d+0.2)^{\frac {2}{3}}}[/tex]

[tex]1=\frac {1}{0.013*0.02\frac {m^{3}}{sec}}* \frac {(0.2d)^{\frac {5}{3}* (0.006)^{0.5}}}{(2d+0.2)^{\frac {2}{3}}}[/tex]

[tex]1=\frac {(0.006)^{0.5}}{0.013*0.02}* \frac {(0.2d)^{\frac {5}{3}}}{(2d+0.2)^{\frac {2}{3}}}[/tex]

[tex]1=297.92*\frac {(0.2d)^{\frac {5}{3}}}{(2d+0.2)^{\frac {2}{3}}}[/tex]

d=0.12m

We can now get the flow area which we already stated is 0.2d and now we have d

[tex]A=0.2*0.12=0.024 m^{2}[/tex]

To find the velocity of flow in the drainage channel, we divide discharge by velocity hence

[tex]V_{channel}=\frac {Q_{channel}}{A}[/tex]

[tex]V_{channel}=\frac {0.02}{0.024}=0.83 m/s[/tex]

The time of concentration in the catchment area, [tex]t_{2}[/tex]

[tex]t_{2}=\frac {60}{0.83}=72.3 sec=72.3/60=1.2 mins[/tex]

Total time of concentration, [tex]t_{c} =t_{1}+t_{2}=3+1.2=4.2 mins[/tex]

Therefore, [tex]t_{c}=4.2 mins[/tex]