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## Q. 15.15

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?

## Verified Solution

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.