Economical trapezoid and power saving

Reading: side slope $2:1$ is taken as 2 horizontal to 1 vertical ($z=2$).

Given data

$Q=9$ m³/s, $V=0.75$ m/s, $n=0.025$.

Most economical trapezoid

$$A=\frac QV=\frac{9}{0.75}=12\ \text{m}^2$$ $$A=y^2\left(2\sqrt{1+z^2}-z\right)=y^2(2\sqrt5-2)=2.4721\,y^2\;\Rightarrow\;y=\sqrt{\frac{12}{2.4721}}=2.203\ \text{m}$$ $$b=2y(\sqrt5-2)=2\times2.203\times0.2361=1.040\ \text{m},\qquad R=\frac y2=1.102\ \text{m}$$

Bed slope from Manning's equation:

$$S_t=\left(\frac{nV}{R^{2/3}}\right)^2=\left(\frac{0.025\times0.75}{1.102^{2/3}}\right)^2=3.090\times10^{-4}\quad(\text{1 in }3236)$$

Rectangular channel 3 m wide, 1.2 m deep, same $Q=9$ m³/s

$$A_r=3.6\ \text{m}^2,\quad V_r=\frac{9}{3.6}=2.5\ \text{m/s},\quad P_r=3+2(1.2)=5.4\ \text{m},\quad R_r=\frac{3.6}{5.4}=0.6667\ \text{m}$$ $$S_r=\left(\frac{nV_r}{R_r^{2/3}}\right)^2=\left(\frac{0.025\times2.5}{0.6667^{2/3}}\right)^2=6.707\times10^{-3}\quad(\text{1 in }149)$$

Power

For the same discharge, the power lost per unit length is $\gamma QS_f$ (the head loss per unit length is $S_f$). Over 1 km:

$$P_{trap}=\gamma Q S_t L=9.81\times9\times3.090\times10^{-4}\times1000=27.28\ \text{kW}$$ $$P_{rect}=9.81\times9\times6.707\times10^{-3}\times1000=592.19\ \text{kW}$$ $$\text{Saving}=P_{rect}-P_{trap}=564.9\ \text{kW per km}$$

Equivalently, the rectangular channel needs 6.71 m of fall per km, against 0.31 m for the economical trapezoid.

Answer: economical section $y=2.20$ m, $b=1.04$ m (bed slope 1 in 3236); saving in power $\approx565$ kW per km.