Jump relation in horizontal triangular channel

Geometry (triangular, side slope $z$): $A = zy^2$, $T = 2zy$, centroid depth $\bar y = y/3$.

Specific force (momentum function):

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

Momentum equation for a horizontal channel, $F_1 = F_2$:

$$\frac{Q^2}{gz}\left(\frac{1}{y_1^2} - \frac{1}{y_2^2}\right) = \frac{z}{3}\left(y_2^3 - y_1^3\right)$$ $$\frac{Q^2}{gz}\cdot\frac{y_2^2 - y_1^2}{y_1^2y_2^2} = \frac{z}{3}\left(y_2^3 - y_1^3\right)$$

Put $y_2 = ry_1$:

$$\frac{Q^2}{gz}\cdot\frac{y_1^2(r^2 - 1)}{y_1^4r^2} = \frac{z\,y_1^3}{3}(r^3 - 1)$$ $$\frac{Q^2}{gz^2y_1^5} = \frac{1}{3}\cdot\frac{r^2(r^3 - 1)}{r^2 - 1}$$

Froude number at section 1. For a triangular channel, $V_1 = Q/(zy_1^2)$ and the hydraulic depth is $D_1 = A/T = y_1/2$. The problem defines $Fr_1^2 = V_1^2/(gy_1)$, so

$$Fr_1^2 = \frac{V_1^2}{gy_1} = \frac{Q^2}{gz^2y_1^5}$$

which is exactly the left side of the equation above. Hence

$$\boxed{3Fr_1^2 = \frac{r^2(r^3 - 1)}{r^2 - 1}}$$

Check: for $y_1 = 0.5$, $y_2 = 1.0$ ($r = 2$), the right side is $\dfrac{4 \times 7}{3} = 9.33$, so $Fr_1^2 = 3.11$ and $Fr_1 = 1.76$ (with $Fr_1 = V_1/\sqrt{gy_1}$, which is $\sqrt2$ times smaller than the usual $V_1/\sqrt{gD_1} = 2.49$).