Jump with baffle blocks behind sluice gate

Given: $b = 5$ m, $n = 0.011$, $S_0 = 0.0004$. At the gate $y_1 = 4$ m upstream and $y_2 = 0.5$ m at the vena contracta. The sequent depth after the jump equals the normal depth $y_3 = y_0$.

Discharge (energy equation across the gate, no loss):

$$Q = b\,y_1y_2\sqrt{\frac{2g}{y_1 + y_2}} = 5(4)(0.5)\sqrt{\frac{19.62}{4.5}} = 20.88\ \text{m}^3/\text{s}$$ $$q = 4.176\ \text{m}^2/\text{s},\quad V_2 = \frac{20.88}{5 \times 0.5} = 8.352\ \text{m/s},\quad Fr_2 = 3.77$$

Normal depth by Manning's equation (trial, $R = A/P$): $y_0 = y_3 = 2.102$ m, $V_3 = 20.88/(5 \times 2.102) = 1.986$ m/s.

(Without blocks, the sequent depth of 0.5 m would be $y_2' = 2.428$ m. The blocks reduce the required depth to 2.102 m.)

i) Head loss in the jump

$$E_2 = 0.5 + \frac{8.352^2}{2g} = 0.5 + 3.556 = 4.056\ \text{m}$$ $$E_3 = 2.102 + \frac{1.986^2}{2g} = 2.102 + 0.201 = 2.303\ \text{m}$$ $$\Delta E = E_2 - E_3 = 4.056 - 2.303 = 1.75\ \text{m}$$

ii) Force on the blocks

Momentum equation between sections 2 and 3 for the control volume (the force of the blocks on the water, $F_b$, acts upstream; bed friction and the gravity component are neglected):

$$\frac{\gamma b y_2^2}{2} + \rho Q V_2 - F_b = \frac{\gamma b y_3^2}{2} + \rho Q V_3$$ $$F_b = 6131 + 174\,400 - 108\,400 - 41\,480 \approx 30\,700\ \text{N}$$

The force on the blocks by the water is equal and acts downstream.

Answer: (i) $\Delta E = 1.75$ m; (ii) $F_b \approx 30.7$ kN.