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The Chézy law relates mean open-channel speed to hydraulic radius and energy slope through a friction coefficient. Equivalently one can use bed shear ${\tau }_b=\rho fu|u|$; for positive uniform flow, balance $\rho gAI={\tau }_b{P}_b$ gives $u=\sqrt{gI/f}\sqrt{A/P_b}$. The absolute value makes drag oppose motion if reverse flows are admitted.

The hydraulic radius is cross-sectional flow area divided by wetted perimeter. A perimeter-averaged boundary shear stress $\tau$ exerts drag per unit fluid volume $\tau \ell /A=\tau /R$. It differs from both actual depth and the radius of a noncircular channel.

Here $s$ is downstream distance, $A$ the wetted cross-sectional area, $u$ the section-averaged downstream velocity, and $M$ the lateral water input per channel length and time. Thus $A_t$ is local water storage and $(Au{)}_s$ the divergence of discharge. Their balance is mass conservation.

The momentum terms $u_t$ and $u{u}_s$ are local and advective acceleration. The bed slope $S$ gives the downstream gravitational acceleration $gS$. Bed shear stress $\tau$ acting over wetted perimeter $l$ produces resisting force per unit mass $\tau l/(\rho A)$; the hydraulic radius is $R=A/l$. Finally $-g{\overline{h}}_s$ is the hydrostatic free-surface pressure-gradient term in the chosen bed-following description. The Saint-Venant approximation depth-averages the flow, assumes a long shallow geometry and hydrostatic pressure, and closes the boundary shear through a roughness law.

To obtain a power-law river kinematic wave, let $P_b$ be the wetted perimeter and $I$ the small bed slope. A uniform-flow force balance per unit river length is $\rho gAI={\tau }_b{P}_b$. With a Chézy friction law ${\tau }_b=\rho f{u}^2$, it gives $u=(gI/f{)}^{1/2}(A/P_b{)}^{1/2}$. Thus the hydraulic radius determines the power of area.

For a channel family with $P_b=P_0{A}^q$, where $q<1$, we have $u=C{A}^{(1-q)/2}$, with $C=(gI/(f{P}_0){)}^{1/2}$. Set $m=(1-q)/2>0$ and $c=(m+1)C$. ThenFor example a broad fixed-width rectangular channel has approximately constant wetted perimeter, giving $m=1/2$; a geometrically similar channel family has $P_b\propto A^{1/2}$, giving $m=1/4$. The exponent is therefore a geometry-dependent consequence of the friction balance, not a universal arbitrary value fixed by Chezy's law alone.

Assume constant mass density, a prismatic channel, approximately uniform cross-sectional velocity $u$, a hydrostatic pressure distribution, small bed slope $S$, and no lateral inflow. Integrating mass conservation over a slice of length $dx$ gives $\rho A_{t}dx=\rho Au{|}_x-\rho Au{|}_{x+dx}$, hence $A_t+(Au{)}_x=0$.

Let $I(A)={\int }_A(H-z)dA$ be the hydrostatic first moment, where $H$ is the actual free-surface depth. The downstream momentum balance isUsing $A_t+(Au{)}_x=0$ reduces its left side to $A(u_t+u{u}_x)+g{I}_x$. For a fixed prismatic cross-section, $dI/dH=A$, so $I_x/A=H_x$. Therefore the exact depth-averaged result isHere hydraulic radius converts perimeter shear stress into force per unit fluid volume. The printed notation $\bar{h}$ yields the requested equation if it denotes this hydraulic pressure head, whose derivative equals $H_x$.

The geometrical definition printed in the next part needs care. Triangular-channel hydrostatic pressure head gives the factor explicitly. For the stated symmetric triangular notch, the actual depth is $H=(\ell /2)\sin \theta$, whereas its printed $\bar{h}=(\ell /4)\sin \theta$ is $H/2$. Indeed $A=H^2\cot \theta$ and $I=AH/3$, giving $I_x/A=H_x=2{\bar{h}}_x$. Thus the equation with precisely the printed geometrical $\bar{h}$ has pressure term $-2g{\bar{h}}_x$. A first-principles derivation cannot remove that factor. The subsequent intended scaling can be carried out algebraically using the printed convention, as below, or with the pressure scale doubled for the actual triangular channel.

For a fixed symmetric triangular channel with depth $H$ and area $A=k{H}^2$, the hydrostatic first moment is $I={\int }_A(H-z)dA=k{H}^3/3=AH/3$. Since $dI/dH=A$, the depth-averaged pressure gradient is $g{I}_x/A=g{H}_x$. A mean geometric depth $H/2$ cannot replace $H$ in that gradient without an extra factor two. This distinction prevents confusion between pressure head, centroid depth, and hydraulic radius.