Force on blocks after sluice gate jump

Given: $b = 6$ m, upstream depth $y_1 = 6$ m, depth of the stream just after the gate $y_2 = 1.2$ m, depth after the jump $y_3 = 3.1$ m.

Flow rate

Energy equation across the gate (no loss):

$$y_1 + \frac{V_1^2}{2g} = y_2 + \frac{V_2^2}{2g} \;\Rightarrow\; Q = b\,y_1y_2\sqrt{\frac{2g}{y_1 + y_2}}$$ $$Q = 6(6)(1.2)\sqrt{\frac{19.62}{7.2}} = 43.2 \times 1.651 = 71.3\ \text{m}^3/\text{s}$$

Force on the blocks

$$V_2 = \frac{71.31}{6 \times 1.2} = 9.905\ \text{m/s},\qquad V_3 = \frac{71.31}{6 \times 3.1} = 3.834\ \text{m/s}$$

Without blocks the sequent depth of 1.2 m would be $\dfrac{1.2}{2}\left(\sqrt{1 + 8(2.887)^2} - 1\right) = 4.336$ m. Since the actual depth after the jump (3.1 m) is smaller, the blocks are needed to hold the jump in place.

Momentum equation between sections 2 and 3 (horizontal bed, friction neglected). The blocks push on the water in the upstream direction with force $F_b$:

$$\frac{\gamma b y_2^2}{2} + \rho Q V_2 - F_b = \frac{\gamma b y_3^2}{2} + \rho Q V_3$$ $$F_b = \frac{\gamma b}{2}(y_2^2 - y_3^2) + \rho Q(V_2 - V_3)$$ $$F_b \approx 192\,500\ \text{N} \approx 192\ \text{kN}$$

The water pushes on the blocks with an equal force in the downstream direction.

Answer: $Q = 71.3\ \text{m}^3/\text{s}$; force on the blocks $\approx 192\ \text{kN}$.