Question 11.5: A concrete storm water pipe is laid on a slope of 1 : 100. T......
A concrete storm water pipe is laid on a slope of 1 : 100. The diameter of the pipe is 0.60 m and while conveying the maximum discharge, the depth of flow is not allowed to exceed 0.50 m. Find the capacity of the storm water drain if the Manning n = 0.012.
Given: S_0 = 0.01, Area of cross section of the pipe ={\frac{\pi}{4}}D^{2}={\frac{\pi\times0.60^{2}}{4}}=0.2827\,{\mathrm{m}}^{2} .
Wetted perimeter P = πD, hydraulic radius R = A/P = D/4 = 0.15 m. Assuming the pipe to flow full (but not under pressure), the discharge through the storm sewer is
Q=A\times v =0.2827\times\left\lgroup{\frac{1}{0.012}}\times0.15^{2/3}\times0.01^{1/2}\right\rgroup=0.666\mathrm{~m}^{3}/sThe permissible depth y = 0.50 m. Therefore, y/D = 0.50/0.60 = 0.833. From the graph of the hydraulic properties of circular pipes, Fig. 11.15, corresponding to the above value of y/d, the proportion of flow is 1.02. Therefore, the required flow Q = 1.02 × 0.666 = 0.68 m³/s. This problem can also be solved by working out the geometric properties, A, P and R of the channel for y = 0.5 m and then applying Manning equation directly.
