Hydraulic jump in 90° triangular channel

For a 90° triangular channel the side slopes are 1H : 1V, so $z = 1$.

Derivation

Geometry at depth $y$: $A = y^2$, $T = 2y$, and the centroid of the area is at $\bar y = y/3$ below the surface.

Specific force:

$$F = \frac{Q^2}{gA} + A\bar y = \frac{Q^2}{gy^2} + \frac{y^3}{3}$$

For a jump in a horizontal channel, $F_1 = F_2$:

$$\frac{Q^2}{g}\left(\frac{1}{y_1^2} - \frac{1}{y_2^2}\right) = \frac{y_2^3 - y_1^3}{3}$$ $$\frac{Q^2}{g}\cdot\frac{y_2^2 - y_1^2}{y_1^2y_2^2} = \frac{y_2^3 - y_1^3}{3}$$ $$\boxed{Q^2 = \frac{g}{3}\cdot\frac{y_1^2y_2^2\,(y_2^3 - y_1^3)}{y_2^2 - y_1^2}}$$

Flow rate

With $y_1 = 0.5$ m and $y_2 = 1.0$ m:

$$Q^2 = \frac{9.81}{3}\cdot\frac{(0.25)(1)(1 - 0.125)}{1 - 0.25} = 3.27 \times 0.29167 = 0.9538$$ $$Q = 0.977\ \text{m}^3/\text{s}$$

Froude numbers

The hydraulic depth is $D = A/T = y/2$.

$$V_1 = \frac{Q}{y_1^2} = \frac{0.9766}{0.25} = 3.906\ \text{m/s},\qquad Fr_1 = \frac{3.906}{\sqrt{9.81 \times 0.25}} = 2.49$$ $$V_2 = \frac{0.9766}{1.0} = 0.977\ \text{m/s},\qquad Fr_2 = \frac{0.977}{\sqrt{9.81 \times 0.5}} = 0.44$$

Answer: $Q \approx 0.98\ \text{m}^3/\text{s}$; $Fr_1 \approx 2.49$ (supercritical) and $Fr_2 \approx 0.44$ (subcritical).