M1 profile upstream of a free outfall

M1 profile

On a mild slope ($y_n > y_c$), the M1 profile is the backwater curve in zone 1, where $y > y_n > y_c$. Its depth increases downstream. It is asymptotic to the normal depth line upstream and approaches a horizontal water surface downstream. It occurs upstream of dams, weirs and reservoirs. (The profile in the numerical part, upstream of a free outfall, is the drawdown curve M2, because the depth falls from $y_n$ to $y_c$.)

Numerical part

Given: $b = 10\ \text{m}$, $Q = 50\ \text{m}^3/\text{s}$, $S_0 = 0.0006$, free outfall. Concrete lining: $n = 0.014$ assumed.

Depths ($q = 5\ \text{m}^2/\text{s}$):

$$y_c = \left(\frac{5^2}{9.81}\right)^{1/3} = 1.366\ \text{m},\qquad y_n = 2.169\ \text{m}\ (\text{Manning, } R = A/P)$$

Since $y_n > y_c$ the slope is mild, and the profile is M2. The depth is $y_c$ at $4y_c = 5.46$ m upstream of the fall, which is the control section. The point asked about is 2000 m upstream of the fall, i.e. $2000 - 5.46 = 1994.5\ \text{m}$ upstream of the control.

Bresse's method (three depth steps)

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

$\dfrac{y_n}{S_0} = 3614.5\ \text{m}$ and $1 - (y_c/y_n)^3 = 1 - 0.6298^3 = 0.7502$.

Three equal depth steps from $y_c$ upwards (trial for the end depth):

The last row is the depth at which the distance equals 1994.5 m upstream. A fine numerical integration gives 2.135 m, which confirms it.

Answer: Depth of flow 2 km upstream of the fall $\approx 2.14\ \text{m}$ (very close to $y_n = 2.17$ m).