## Water Slope

Consider water is fully flowing through a 0.5 ft diameter pipe at a 5% slope. If the pipe’s roughness coefficient is 0.02, what is the volumetric flow in cubic feet per second?

Hint
Manning’s equation:
$$Q=\frac{K}{n}AR_{H}^{2/3}S^{1/2}$$$where $$Q$$ is the volumetric discharge, $$A$$ is the cross sectional flow area, $$K=1.486$$ for USCS units, $$n$$ is the roughness coefficient, $$R_H$$ is the hydraulic radius, and $$S$$ is the slope. Hint 2 To find hydraulic radius: $$R_H=\frac{A}{P}$$$
where $$A$$ is the area, and $$P$$ is the perimeter.
Manning’s equation:
$$Q=\frac{K}{n}AR_{H}^{2/3}S^{1/2}$$$where $$Q$$ is the volumetric discharge, $$A$$ is the cross sectional flow area, $$K=1.486$$ for USCS units, $$n$$ is the roughness coefficient, $$R_H$$ is the hydraulic radius, and $$S$$ is the slope. To find hydraulic radius: $$R_H=\frac{A}{P}$$$
where $$A$$ is the area, and $$P$$ is the perimeter. Thus, the hydraulic radius is:
$$R_H=\frac{(\pi/4)(D^2) }{(\pi D)}=\frac{D}{4}=\frac{0.5ft}{4}=0.125\:ft$$$To find the volumetric discharge: $$Q=\frac{1.486}{0.02}\cdot \frac{\pi}{4}(0.5)^2(0.125)^{2/3}\sqrt{5/100}$$$
$$Q=74.3\cdot \frac{\pi}{4}(0.25)(0.25)\sqrt{0.05}=0.82\:ft^3/s$$$$$0.82\:ft^3/s$$$