GVF in 4 m channel, graphical integration

Given: $b = 4$ m, $Q = 2.0\ \text{m}^3/\text{s}$, $S_0 = 0.00075$, $n = 0.016$, $\alpha = 1.1$, $y = 0.2$ m at a section. Find the distance to $y = 0.4$ m with $\Delta y = 0.1$ m.

Depths

$$q = 0.5\ \text{m}^2/\text{s},\quad y_c = \left(\frac{1.1 \times 0.5^2}{9.81}\right)^{1/3} = 0.304\ \text{m}$$

By Manning's equation (trial, $R = A/P$): $y_n = 0.525$ m.

The slope is mild ($y_n > y_c$) and $y = 0.2$ m is below $y_c$, so the profile is M3 (supercritical, depth rising downstream).

Graphical integration. From the dynamic equation,

$$\frac{dx}{dy} = \frac{1 - \alpha Q^2T/(gA^3)}{S_0 - S_f},\qquad x = \int_{y_1}^{y_2}\frac{dx}{dy}\,dy = \text{area under the } \frac{dx}{dy}\text{ vs } y \text{ curve}$$

Plot $dx/dy$ against $y$ and find the area under the curve.

• From 0.2 to 0.3 m, the trapezoidal area is $\tfrac{169.4 + 10.9}{2}(0.1) = 9.0$ m; a finer reading of the curve (extra point at 0.25 m, where $dx/dy = 115.8$) gives 10.7 m.
• At $y_c = 0.304$ m, $dx/dy$ changes sign (the denominator of the dynamic equation, $1 - Fr^2$, goes through zero). The M3 curve cannot cross the critical depth gradually; it ends in a hydraulic jump at or before $y_c$. Hence the depth 0.40 m cannot be reached by a continuous profile from 0.20 m, and the 0.3 to 0.4 m "area" has no physical meaning.

Answer: M3 profile. The depth rises from 0.20 m to about 0.30 m (critical) over about 10 m downstream; the curve then ends in a hydraulic jump, so 0.40 m is not reached by gradually varied flow.