Critical depth in four channel sections

Condition for critical flow:

$$\frac{Q^2}{g} = \frac{A^3}{T}$$

For $Q = 6\ \text{m}^3/\text{s}$: $\dfrac{Q^2}{g} = \dfrac{36}{9.81} = 3.670\ \text{m}^5$.

i) Circular, $D = 1.5\ \text{m}$

With $\theta$ (rad) the angle subtended at the centre by the water surface:

$$A = \frac{D^2}{8}(\theta - \sin\theta),\quad T = D\sin\frac{\theta}{2},\quad y = \frac{D}{2}\left(1 - \cos\frac{\theta}{2}\right)$$

Trial and error (bisection) on $A^3/T = 3.670$ gives $\theta = 4.651\ \text{rad}$:

$$A = 1.589\ \text{m}^2,\quad T = 1.093\ \text{m},\quad \frac{A^3}{T} = \frac{4.011}{1.093} = 3.670 \checkmark$$ $$y_c = 0.75\,(1 - \cos 2.3254) = 1.264\ \text{m}$$

ii) Rectangular, $b = 3\ \text{m}$

$$q = \frac{6}{3} = 2\ \text{m}^2/\text{s},\qquad y_c = \left(\frac{q^2}{g}\right)^{1/3} = \left(\frac{4}{9.81}\right)^{1/3} = 0.742\ \text{m}$$

iii) Trapezoidal, $b = 2.5\ \text{m}$, $z = 2$ (2H : 1V)

$$A = (2.5 + 2y)y,\quad T = 2.5 + 4y$$

Solve $\dfrac{[(2.5 + 2y)y]^3}{2.5 + 4y} = 3.670$. Trial: $y = 0.691$ gives $A = 2.683$, $T = 5.265$, $A^3/T = 19.32/5.265 = 3.670$.

$$y_c = 0.691\ \text{m}$$

iv) Triangular, $z = 1$ (1H : 1V)

$A = y^2$, $T = 2y$, so $\dfrac{y^6}{2y} = \dfrac{Q^2}{g}$:

$$y_c = \left(\frac{2Q^2}{g\,z^2}\right)^{1/5} = \left(\frac{2 \times 36}{9.81}\right)^{1/5} = 1.490\ \text{m}$$