Hydraulic exponent for triangular section 16/3

Hydraulic exponent

The second hydraulic exponent $N$ (for uniform flow) is defined by $K^2\propto y^N$, where the conveyance is

$$K=\frac1nAR^{2/3}=\frac{A^{5/3}}{nP^{2/3}}$$

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

General expression

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

Triangular section (side slope $z$)

$$A=zy^2,\qquad \frac{dA}{dy}=2zy,\qquad \frac yA\frac{dA}{dy}=\frac{y\cdot2zy}{zy^2}=2$$ $$P=2y\sqrt{1+z^2},\qquad \frac{dP}{dy}=2\sqrt{1+z^2},\qquad \frac yP\frac{dP}{dy}=1$$ $$\frac N2=\frac53(2)-\frac23(1)=\frac{10}{3}-\frac23=\frac83$$ $$\boxed{N=\frac{16}{3}}$$

Directly: $R=\dfrac{zy}{2\sqrt{1+z^2}}\propto y$, so $K\propto y^2\cdot y^{2/3}=y^{8/3}$ and $K^2\propto y^{16/3}$ ✓. This is exact for a triangle for all depths because $A\propto y^2$ and $P\propto y$.