---
title: "Force on sluice gate, 10 m wide channel"
subject: "Hydraulics"
chapter: 9
key: "semester_4/hydraulics:sluice-gate-force-10m-channel-7-m3s"
source: https://www.bctengineer.com/q/semester_4/hydraulics/sluice-gate-force-10m-channel-7-m3s
site: BCT Engineering
---

# Force on sluice gate, 10 m wide channel

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

- Asked in: 2073 Bhadra, 5 marks
- Video: [PART 1 FLOW THROUGH SLUICE GATE](https://www.youtube.com/watch?v=nIoSLUp6bhg) — AdtU Digital Learning Platform
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/hydraulics/sluice-gate-force-10m-channel-7-m3s
- Whole chapter: https://www.bctengineer.com/md/semester_4/hydraulics/chapter-9

## Question

Figure shows flow through the sluice gate provided in a rectangular channel of width 10 m. If the discharge in the channel is 7 m³/s, determine the force exerted by water in the gate. Take momentum correction factor equals to 1.15. [Figure: sluice gate with upstream depth 2.5 m and downstream depth 0.25 m]

## Short answer

Using beta=1.15, V1=0.28 m/s and V2=2.80 m/s. Momentum between upstream and downstream gives the gate force on water as 283 kN, so the water exerts 283 kN on the gate.

## Answer

**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:

$$
\frac{\gamma b y_1^2}{2} - \frac{\gamma b y_2^2}{2} - F = \rho Q\,\beta\,(V_2 - V_1)
$$

$$
F = \frac{\gamma b}{2}(y_1^2 - y_2^2) - \beta\rho Q(V_2 - V_1)
$$

| Term | Value (N) |
|---|---|
| $\dfrac{9810 \times 10}{2}(2.5^2 - 0.25^2)$ | 303 497 |
| $1.15 \times 1000 \times 7 \times (2.80 - 0.28)$ | 20 286 |

$$
F = 303\,497 - 20\,286 = 283\,211\ \text{N}
$$

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

## Check yourself

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

1. Which equation gives the force on the gate?
   A. Bernoulli equation with a loss
   B. Continuity equation
   C. Manning's equation
   D. Momentum equation between upstream and downstream of the gate
   Answer: D. The net force on the control volume equals the change in momentum flux. (Key point: Principle used)

2. How does the momentum correction factor beta = 1.15 enter?
   A. It multiplies the hydrostatic terms
   B. It multiplies the momentum flux rho Q (V2 - V1)
   C. It multiplies the velocity heads
   D. It is added to the depth
   Answer: B. Beta corrects the momentum flux for non-uniform velocity, reducing F slightly from 286 to 283 kN. (Key point: Role of beta)

3. Which expression is correct for the force of the water on the gate?
   A. F = gamma b (y1^2 + y2^2)/2 - beta rho Q (V2 - V1)
   B. F = gamma b (y1^2 - y2^2)/2 + beta rho Q (V2 - V1)
   C. F = gamma b (y1^2 - y2^2)/2 - beta rho Q (V2 - V1)
   D. F = beta rho Q V2
   Answer: C. The pressure force difference is reduced by the momentum gain, giving about 283 kN downstream. (Key point: Force expression)

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