Bresse's method for 20 m wide channel

Given: $b = 20\ \text{m}$, $Q = 65\ \text{m}^3/\text{s}$ ($q = 3.25\ \text{m}^2/\text{s}$), $S_0 = 0.0001$, $n = 0.025$, $\alpha = 1.1$, $y = 2\ \text{m}$ at a section.

Critical and normal depth

$$y_c = \left(\frac{\alpha q^2}{g}\right)^{1/3} = \left(\frac{1.1 \times 3.25^2}{9.81}\right)^{1/3} = 1.058\ \text{m}$$

Manning's equation with $R = A/P$, solved by trial: $y_n = 4.024\ \text{m}$ ($A = 80.5\ \text{m}^2$, $R = 2.87$ m, $V = 0.81$ m/s).

Type of profile

$y_n > y > y_c$, and the slope is mild ($y_n > y_c$), so the profile is M2 (drawdown). On an M2 curve the depth increases upstream, so the depth of 2.6 m lies upstream of the 2 m section.

Bresse's method

For a wide channel with constant Chezy $C$:

$$x = \frac{y_n}{S_0}\left[u - \left(1 - \left(\frac{y_c}{y_n}\right)^3\right)\phi(u)\right],\qquad u = \frac{y}{y_n}$$ $$\phi(u) = \frac{1}{6}\ln\frac{u^2 + u + 1}{(u-1)^2} + \frac{1}{\sqrt3}\tan^{-1}\frac{2u + 1}{\sqrt3}$$

Here $\dfrac{y_n}{S_0} = 40\,238\ \text{m}$ and $1 - (y_c/y_n)^3 = 1 - (0.2629)^3 = 0.9818$.

Distances relative to the 2 m section:

• 2.0 m to 2.3 m: $-12\,709 - (-12\,228) = -481$ m
• 2.3 m to 2.6 m: $-13\,521 - (-12\,709) = -812$ m

$$x_{2.0 \to 2.6} = -1292\ \text{m}$$

The negative sign means upstream. As a check, the direct step method with the actual rectangular geometry gives about 1220 m with the same two 0.3 m steps and about 1276 m with a fine integration, so the Bresse result is reasonable.

Answer: M2 profile; the depth is 2.6 m about 1.29 km upstream of the section where the depth is 2 m.