---
title: "Force on blocks after sluice gate jump"
subject: "Hydraulics"
chapter: 9
key: "semester_4/hydraulics:sluice-gate-jump-blocks-force-6m-channel"
source: https://www.bctengineer.com/q/semester_4/hydraulics/sluice-gate-jump-blocks-force-6m-channel
site: BCT Engineering
---

# Force on blocks after sluice gate jump

Hydraulics · Chapter 9: Non-uniform rapidly varied flow (RVF)

- Asked in: 2076 Baisakh, 8 marks
- Video: [NCH5.7 Energy and momentum equation on sluice gate followed by a concrete block](https://www.youtube.com/watch?v=VGEaJ5MteLQ) — Aaradhya Elite Academy
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/hydraulics/sluice-gate-jump-blocks-force-6m-channel
- Whole chapter: https://www.bctengineer.com/md/semester_4/hydraulics/chapter-9

## Question

A sluice across a rectangular prismatic channel 6 m wide discharges a stream 1.2 m deep. What is the flow rate when the upstream depth is 6 m? The conditions downstream cause a hydraulic jump to occur at a place where concrete blocks have been placed on the bed. What is the force on the blocks if the depth after the jump is 3.1 m?

## Short answer

Energy equation across the sluice gate gives Q = b y1 y2 sqrt(2g/(y1+y2)) = 71.3 m3/s. Momentum between the jet and post-jump sections, with the blocks providing upstream force, gives F_b = 192 kN on the blocks.

## Answer

**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 $F_b$:

$$
\frac{\gamma b y_2^2}{2} + \rho Q V_2 - F_b = \frac{\gamma b y_3^2}{2} + \rho Q V_3
$$

$$
F_b = \frac{\gamma b}{2}(y_2^2 - y_3^2) + \rho Q(V_2 - V_3)
$$

| Term | Value (N) |
|---|---|
| $\dfrac{9810 \times 6}{2}(1.2^2 - 3.1^2)$ | $-240\,443$ |
| $1000 \times 71.31 \times (9.905 - 3.834)$ | $+432\,900$ |

$$
F_b \approx 192\,500\ \text{N} \approx 192\ \text{kN}
$$

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 $\approx 192\ \text{kN}$.**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. How is Q found from y1 = 6 m and y2 = 1.2 m?
   A. Q = b y1 y2 sqrt(2g/(y1+y2)) from the no-loss energy equation
   B. Q from Manning's equation
   C. Q = b yc sqrt(g yc)
   D. From the momentum equation
   Answer: A. Energy is conserved across the gate, giving 71.3 m^3/s. (Key point: Flow rate)

2. Why are blocks needed to hold the jump?
   A. The actual depth after the jump (3.1 m) is less than the sequent depth (4.336 m)
   B. The actual depth is greater than 4.336 m
   C. Fr2 is below 1
   D. The flow is subcritical at the gate
   Answer: A. With too low a tailwater the jump would be swept away, so the blocks supply the extra force. (Key point: Why blocks are needed)

3. How is Fb found?
   A. Energy equation between sections 2 and 3
   B. Continuity
   C. Momentum equation: Fb = gamma b (y2^2 - y3^2)/2 + rho Q (V2 - V3)
   D. Fb = gamma A y3
   Answer: C. The hydrostatic and momentum terms are balanced by Fb, giving about 192 kN. (Key point: Force on blocks)

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