Question 15.15: The slope of the culvert in Example 15.13 is to be adjusted ...
The slope of the culvert in Example 15.13 is to be adjusted so that normal depth will be at least 1.5 m when the culvert is carrying its design flowrate of 4.7 m³/s (Figure 15.40). What slope would you recommend?
We know that the volume flowrate is given by Eq. 15.53b, Q = (C_0/n)AR^{2/3}_H S^{1/2} _B, with C0 = 1, since we are working in SI units. Solving for the bed slope we obtain S_B =n^2Q^2/C^2 _0 A^2R^{4/3} _H. We can again use Figure 15.6 to write A(y) = wy and RH(y) = wy/(w + 2y). Inserting these into the formula yields
S_B =\frac{n^2Q^2}{(wy)^2\left(\frac{wy}{w+2y} \right)^{4/3} } (A)
Inserting the known values as well as a Manning coefficient n = 0.014 s/m1/3 (notice the units here, which are often omitted in calculations) and width w = 2 m, we obtain
S_B =\frac{(0.014\ s/m^{1/3} )^2(4.7\ m ^3/s)^2}{[(2\ m)(1.5\ m )]^2\left[\frac{2\ m(1.5\ m)}{2\ m+2(1.5\ m)} \right]^{\frac{4}{3}}}=9.5\times 10^{-4}
Thus the bed slope needed is about one-fifth of the slope that produced the slightly supercritical flow in Example 15.13.
