Question 12.9: The channel in Fig. 12–21 consists of an unfinished concrete......
The channel in Fig. 12–21 consists of an unfinished concrete section (n = 0.014) and overflow regions on each side that contain light brush (n = 0.050). If the bottom of the channel has a slope of 0.0015, determine the steady volumetric flow when the depth is 2.5 m, as shown.
Fluid Description. We have steady uniform flow, and we will assume the water is an incompressible fluid.
Analysis. For an approximate solution, the cross section is divided into three composite rectangles, Fig. 12–21. The flow through the entire cross section is thus the sum of the flows through each composite rectangle. For the calculation, note that the wetted perimeter does not include the liquid boundary between the rectangles because they are not part of the channel’s wall or bed surface. Since n = 0.014 s /m^{1/3} for unfinished concrete, and for light brush n = 0.050 s /m^{1/3}, as stated, the Manning equation, written in the form of Eq. 12–20, becomes
= (0.0015)^{1/2} [\frac{[(5 m) (1 m)]^{5/3}}{0.050 (1 m + 5 m)^{2/3}} + \frac{[(2 m) (2.5 m)]^{5/3}}{0.014 (1.5 m + 2 m + 1.5 m)^{2/3}} + \frac{[(5 m) (1 m)]^{5/3}}{0.050 (1 m + 5 m)^{2/3}}]
= 20.7 m^3/s