Economical trapezoid for given velocity

Given data

$V=0.6$ m/s, $z=1.5$, $Q=3$ m³/s, $n=0.003$ (as given in the question).

Area

$$A=\frac QV=\frac{3}{0.6}=5.0\ \text{m}^2$$

Economical section

$$A=y^2\left(2\sqrt{1+z^2}-z\right)=y^2\left(2\sqrt{3.25}-1.5\right)=2.1056\,y^2$$ $$y=\sqrt{\frac{5}{2.1056}}=1.541\ \text{m}$$ $$b=2y\left(\sqrt{1+z^2}-z\right)=2\times1.541\times(1.8028-1.5)=0.933\ \text{m}$$ $$P=b+2y\sqrt{1+z^2}=6.489\ \text{m},\qquad R=\frac AP=\frac y2=0.770\ \text{m}$$

Check: $(b+zy)y=(0.933+1.5\times1.541)\times1.541=5.00$ m².

Bed slope from Manning's equation

$$V=\frac1nR^{2/3}S_0^{1/2}\;\Rightarrow\;S_0=\left(\frac{nV}{R^{2/3}}\right)^2=\left(\frac{0.003\times0.6}{0.770^{2/3}}\right)^2=4.587\times10^{-6}$$

i.e. a slope of about 1 in 218 000.

Answer: depth $y=1.54$ m, bed width $b=0.93$ m, side slope 1.5H:1V, bed slope $S_0=4.59\times10^{-6}$.

Note: $n=0.003$ is unusually small. If the intended value were $n=0.03$ (earth channel), the slope would be $4.59\times10^{-4}$ (1 in 2180).