Find the diameter required for reinforced concrete pipe laid at a slope of 0.001 and required to carry a uniform flow of \(19.3 \mathrm{ft}^{3} / \mathrm{sec}\) when the depth is \(75 \%\) of the diameter.

The Manning formula, used to calculate the velocity of the flow in open channels is \(V= \frac{1}{n} * R^{2/3} * S^{1/2}\), where:\nV is the velocity,\nR is the hydraulic radius (the ratio of the cross-sectional area of flow to the wetted perimeter),\nn is the roughness coefficient, with value of 0.014 for reinforced concrete according to the Manning's roughness coefficients,\nS is the slope of the hydraulic grade line (or the energy line).

It is given that the uniform flow is 19.3 \(ft^{3}/sec\) and the slope (S) is 0.001. Given that the flow depth is 75% of the diameter (D), we can calculate hydraulic radius R with the formula \(R= \frac{A}{P}\) where A is the cross-sectional area of the flow and P is the wetted perimeter. For a circle, \(A= \pi * D^{2}/4\) and \(P= \pi * D\). Given that the depth of the flow is 75% of the diameter, only 75% of the circle's area is filled. Thus, the cross-sectional area of flow becomes \(A= 0.75 * \pi * (\frac{3}{4}D)^{2}\); and the wetted perimeter becomes \(P= 0.75 * \pi * D\). Thus, hydraulic radius \(R= \frac{0.75 * \pi * (\frac{3}{4}D)^{2}}{0.75 * \pi * D} = \frac{3}{4}D\). Now, substitute these values into the Manning's equation.

Rearrange the Manning formula to compute for D after substituting R, n and S. Now \(V = \frac{1}{n} * (\frac{3}{4}D)^{2/3} * S^{1/2}\) becomes \(D = (\frac{V * n}{S^{1/2}})^{3/2} * \frac{4}{3}\). Substituting the given data and the known value of n gives the diameter of the circular pipe.