--- title: "Jump with baffle blocks behind sluice gate" subject: "Hydraulics" chapter: 9 key: "semester_4/hydraulics:jump-baffle-blocks-sluice-gate-force" source: https://www.bctengineer.com/q/semester_4/hydraulics/jump-baffle-blocks-sluice-gate-force site: BCT Engineering --- # Jump with baffle blocks behind sluice gate Hydraulics · Chapter 9: Non-uniform rapidly varied flow (RVF) - Asked in: 2081 Chaitra, 8 marks - Video: [important:- hydraulic jump numerical | Force on concrete block by water due to sluice gate](https://www.youtube.com/watch?v=gjHmFp8Ctq8) — Unique Series - Civil Mantraa - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/hydraulics/jump-baffle-blocks-sluice-gate-force - Whole chapter: https://www.bctengineer.com/md/semester_4/hydraulics/chapter-9 ## Question A rectangular channel 5 m wide with manning's roughness n = 0.011 is laid on a slope of 0.0004. A sluice gate is placed at a section of the channel and this gate produces a depth of 4 m just at its upstream and 0.5 m just downstream at vena-contracta. A row of chute or baffle blocks is placed downstream of the gate in order to assist the formation of hydraulic jump after the gate. If the sequent depth of the jump is equal to the normal depth of the channel, determine (i) Head loss in the jump ($\Delta E$) (ii) The force on the chute or baffle blocks ($F_b$). [Figure: sluice gate with $y_1$ = 4 m upstream, $y_2$ = 0.5 m at section 2 downstream, baffle blocks with force $F_b$, section 3 with $y_3 = y_0$, bed slope $S_0$ = 0.0004] ## Short answer Discharge 20.88 m³/s, normal depth 2.102 m. Head loss in the jump ΔE = 1.75 m. The force on the baffle blocks is 30.7 kN, found from momentum balance between vena contracta and tailwater. ## Answer **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): $$ \frac{\gamma b y_2^2}{2} + \rho Q V_2 - F_b = \frac{\gamma b y_3^2}{2} + \rho Q V_3 $$ | Term | Value (N) | |---|---| | $\gamma b y_2^2/2$ | 6 131 | | $\rho Q V_2$ | 174 400 | | $\gamma b y_3^2/2$ | 108 400 | | $\rho Q V_3$ | 41 480 | $$ F_b = 6131 + 174\,400 - 108\,400 - 41\,480 \approx 30\,700\ \text{N} $$ 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.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is the discharge found from the depths upstream of and at the gate? A. Manning's equation with the gate depths B. Momentum equation between the two depths C. Energy equation with no loss, giving Q = b y1 y2 sqrt(2g/(y1+y2)) D. Critical-flow relation Q = b sqrt(g y^3) Answer: C. The flow through the gate is smooth, so energy is conserved between y1 and y2. (Key point: Discharge through gate) 2. Why are baffle blocks needed here? A. The normal depth (2.102 m) is less than the sequent depth (2.428 m) needed for the jump, so blocks supply the missing force B. The normal depth is greater than the sequent depth, so blocks lower it C. To increase the discharge D. To make the flow critical Answer: A. Without blocks the jump needs 2.428 m but only 2.102 m is available; the blocks supply the force difference. (Key point: Role of blocks) 3. How is the head loss in the jump computed? A. E3 - E2 using specific energies B. The difference of specific energies E2 - E3 = 4.056 - 2.303 = 1.75 m C. The difference of depths y3 - y2 D. The difference of velocity heads only Answer: B. Head loss is the specific energy before minus after the jump. (Key point: Head loss in jump) 4. In the momentum equation between 2 and 3, how does the block force Fb act on the water? A. Downstream, adding to the momentum B. Vertically, supporting the water C. Upstream, opposing the flow D. It is neglected Answer: C. The blocks push on the water upstream; the water pushes on the blocks downstream with an equal force, about 30.7 kN. 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