Surface profiles on critical slope

On a critical slope, $S_0 = S_c$ and the normal depth equals the critical depth, $y_n = y_c$. The NDL and CDL coincide, so only two zones exist and zone 2 is absent.

$$\frac{dy}{dx} = S_0\,\frac{1 - (y_c/y)^{10/3}}{1 - (y_c/y)^3}$$

Zone 1: $y > y_n = y_c$ (C1 curve)

Numerator $> 0$ and denominator $> 0$, so $dy/dx > 0$ (depth rises downstream).

• Downstream end: $y \to \infty$, so $dy/dx \to S_0$: the surface becomes horizontal.
• Upstream end: $y \to y_c$. The numerator and denominator both tend to zero, and the limit (with $x = y_c/y \to 1$) is

$$\lim_{x\to 1}\frac{1 - x^{10/3}}{1 - x^3} = \frac{10/3}{3} = \frac{10}{9}$$

so $dy/dx \to 1.11\,S_0$ (with the Chezy-type exponent 3 it would be exactly $S_0$). The profile approaches the critical line at a finite, nearly horizontal angle (not vertically), and is unstable with surface undulations because $Fr \approx 1$.

Zone 3: $y < y_n = y_c$ (C3 curve)

Numerator $< 0$ and denominator $< 0$, so $dy/dx > 0$.

• Upstream end: starts from a control at small depth (e.g. below a gate).
• Downstream end: $y \to y_c$ with the same finite slope $\approx 1.11\,S_0$: it approaches the critical line at a gentle angle, without a jump.

Zone 2 would lie between $y_n$ and $y_c$, which are equal, so no C2 curve exists. Uniform flow at $y_n = y_c$ itself is unstable and is avoided in design.