 Solution

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.