Economical rectangular condition and trapezoidal bed slope

Condition for the most economical rectangular section

For width $b$ and depth $y$, $P = A/y + 2y$. Setting $dP/dy = 0$ at constant $A$ gives $A = 2y^2$, i.e. $b = 2y$ (width twice the depth) and $R = y/2$. This section carries maximum discharge for a given area.

Bed slope of trapezoidal channel

Data: $b = 6$ m, side slope 2H : 1V ($m = 2$), $Q = 15$ m³/s, $y = 1.2$ m, $n = 0.013$.

$$\begin{aligned} A &= (b + my)y = (6 + 2 \times 1.2) \times 1.2 = 10.08\ \text{m}^2 \\ P &= b + 2y\sqrt{1+m^2} = 6 + 2 \times 1.2\sqrt{5} = 11.367\ \text{m} \\ R &= \frac{10.08}{11.367} = 0.8868\ \text{m} \\ V &= \frac{15}{10.08} = 1.488\ \text{m/s} \end{aligned}$$

Manning's equation, solved for $S$:

$$\begin{aligned} Q &= \frac{1}{n}AR^{2/3}S^{1/2} \\ S &= \left(\frac{Qn}{AR^{2/3}}\right)^2 = \left(\frac{15 \times 0.013}{10.08 \times 0.8868^{2/3}}\right)^2 \\ &= \left(\frac{0.195}{9.304}\right)^2 = 4.39 \times 10^{-4} \end{aligned}$$

Answer: Bed slope $S \approx 0.000439$, i.e. about 1 in 2277.