Force on sluice gate, 10 m wide channel

Given: $b = 10$ m, $Q = 7\ \text{m}^3/\text{s}$, upstream depth $y_1 = 2.5$ m, downstream depth $y_2 = 0.25$ m, $\beta = 1.15$.

Velocities

$$V_1 = \frac{7}{10 \times 2.5} = 0.28\ \text{m/s},\qquad V_2 = \frac{7}{10 \times 0.25} = 2.80\ \text{m/s}$$

Momentum equation between sections 1 and 2 (horizontal bed, friction neglected). $F$ is the force of the gate on the water (acting upstream); the force of the water on the gate is equal in size and acts downstream:

$$\frac{\gamma b y_1^2}{2} - \frac{\gamma b y_2^2}{2} - F = \rho Q\,\beta\,(V_2 - V_1)$$ $$F = \frac{\gamma b}{2}(y_1^2 - y_2^2) - \beta\rho Q(V_2 - V_1)$$ $$F = 303\,497 - 20\,286 = 283\,211\ \text{N}$$

Answer: Force of water on the gate $\approx 283\ \text{kN}$ (about 286 kN if $\beta = 1$).