Froude number ratio for alternate depths

Alternate depths $y_1$ and $y_2$ have the same specific energy and the same discharge per unit width $q$.

Express $E$ in terms of Froude number

For a rectangular channel, $Fr=\dfrac{V}{\sqrt{gy}}=\dfrac{q}{\sqrt g\,y^{3/2}}$, so

$$\frac{V^2}{2g}=\frac{Fr^2\,y}{2}$$ $$E=y+\frac{Fr^2y}{2}=\frac{y\,(2+Fr^2)}{2}$$

Equal specific energy

$$\frac{y_1(2+F_1^2)}{2}=\frac{y_2(2+F_2^2)}{2}\;\Rightarrow\;\frac{y_1}{y_2}=\frac{2+F_2^2}{2+F_1^2}\qquad(1)$$

Ratio of Froude numbers

Since $q$ is the same for both depths,

$$\frac{F_2}{F_1}=\frac{q/(\sqrt g\,y_2^{3/2})}{q/(\sqrt g\,y_1^{3/2})}=\left(\frac{y_1}{y_2}\right)^{3/2}$$ $$\left(\frac{F_2}{F_1}\right)^{2/3}=\frac{y_1}{y_2}\qquad(2)$$

Result

From (1) and (2):

$$\boxed{\left(\frac{F_2}{F_1}\right)^{2/3}=\frac{2+F_2^2}{2+F_1^2}}$$

Hence proved.