Jump after control gate: depth, thrust, energy loss

Given: $b = 4$ m, $Q = 100\ \text{m}^3/\text{s}$ ($q = 25\ \text{m}^2/\text{s}$), upstream depth $y_1 = 20$ m. Loss through the gate $= 0.05\,V_2^2/2g$. The jump forms immediately downstream of the gate.

Flow through the gate ($V_1 = 25/20 = 1.25$ m/s):

$$y_1 + \frac{V_1^2}{2g} = y_2 + 1.05\frac{V_2^2}{2g},\qquad V_2 = \frac{25}{y_2}$$ $$20.080 = y_2 + \frac{1.05 \times 25^2}{2g\,y_2^2} \;\Rightarrow\; y_2 = 1.336\ \text{m}$$ $$V_2 = 18.71\ \text{m/s},\qquad Fr_2 = \frac{18.71}{\sqrt{9.81 \times 1.336}} = 5.17$$

i) Depth downstream of the jump

$$y_3 = \frac{y_2}{2}\left(\sqrt{1 + 8Fr_2^2} - 1\right) = \frac{1.336}{2}\left(\sqrt{1 + 8(26.73)} - 1\right) = 9.12\ \text{m}$$ $$V_3 = \frac{25}{9.12} = 2.74\ \text{m/s}$$

ii) Thrust on the gate

Momentum equation between section 1 (upstream) and section 2 (just downstream of the gate). $F$ is the force of the gate on the water, and the thrust of the water on the gate is equal and opposite:

$$\frac{\gamma b y_1^2}{2} - \frac{\gamma b y_2^2}{2} - F = \rho Q(V_2 - V_1)$$ $$F = 7\,848\,000 - 35\,012 - 1\,746\,473 \approx 6.07 \times 10^6\ \text{N}$$

iii) Energy loss in the jump

$$\Delta E = \frac{(y_3 - y_2)^3}{4y_2y_3} = \frac{(9.12 - 1.336)^3}{4(1.336)(9.12)} = 9.68\ \text{m}$$

Check: $E_2 = 1.336 + \dfrac{18.71^2}{2g} = 19.19$ m and $E_3 = 9.12 + \dfrac{2.74^2}{2g} = 9.50$ m, so $E_2 - E_3 = 9.68$ m.

Answer: (i) $y_3 = 9.12$ m; (ii) thrust on the gate $\approx 6.07$ MN; (iii) energy loss in the jump $\approx 9.68$ m.