Hydraulic exponent for rectangular section 10/3

Hydraulic exponent

In a prismatic channel, a hydraulic exponent is the power of the depth to which a section factor is proportional. The second hydraulic exponent $N$ (for uniform flow) is defined by

$$K^2\propto y^{N},\qquad K=\frac1nAR^{2/3}=\frac{A^{5/3}}{nP^{2/3}}$$

where $K$ is the conveyance. (The first exponent $M$ is defined by $A^3/T\propto y^M$ for critical flow.)

General expression

Take logarithms and differentiate with respect to $y$:

$$\ln K=\frac53\ln A-\frac23\ln P-\ln n$$ $$\frac{y}{K}\frac{dK}{dy}=\frac53\frac yA\frac{dA}{dy}-\frac23\frac yP\frac{dP}{dy}=\frac N2$$

(since $K\propto y^{N/2}$, $\dfrac yK\dfrac{dK}{dy}=\dfrac N2$). So

$$N=\frac{10}{3}\frac{y}{A}\frac{dA}{dy}-\frac43\frac yP\frac{dP}{dy}$$

Rectangular section (wide channel, $b\gg y$)

$A=by$, $\dfrac{dA}{dy}=b$, so $\dfrac yA\dfrac{dA}{dy}=\dfrac{y\,b}{by}=1$.

$P=b+2y\approx b$, $\dfrac{dP}{dy}=2$, so $\dfrac yP\dfrac{dP}{dy}=\dfrac{2y}{b+2y}\to0$.

$$N=\frac{10}{3}(1)-\frac43(0)=\boxed{\frac{10}{3}}$$

Directly: for a wide rectangle $R\approx y$, so $K=\dfrac{b}{n}y\cdot y^{2/3}=\dfrac bn y^{5/3}$, and $K^2\propto y^{10/3}$ ✓.