---
title: "Jump after control gate: depth, thrust, energy loss"
subject: "Hydraulics"
chapter: 9
key: "semester_4/hydraulics:jump-downstream-of-gate-thrust-energy-loss"
source: https://www.bctengineer.com/q/semester_4/hydraulics/jump-downstream-of-gate-thrust-energy-loss
site: BCT Engineering
---

# Jump after control gate: depth, thrust, energy loss

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

- Asked in: 2075 Bhadra, 8 marks
- Video: [Open Channel Flow 38 - {How to calculate sequent depth and energy loss in hydraulic jump problem]](https://www.youtube.com/watch?v=eAjApbk4NhA) — Abhijit
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/hydraulics/jump-downstream-of-gate-thrust-energy-loss
- Whole chapter: https://www.bctengineer.com/md/semester_4/hydraulics/chapter-9

## Question

A hydraulic jump is formed in a 4 m wide outlet just downstream of the control gate, which is located at the upstream end of the outlet. The flow depth upstream of the gate is 20 m. If the outlet discharge is 100 m³/s, determine: i) Flow depth downstream of the jump ii) Thrust on the gate; and iii) Energy losses in the jump. Assume the losses through the gate is 5% of velocity head of flow through the gate.

## Short answer

With q=25 m2/s and 5% gate head loss, pre-jump depth is 1.336 m and Fr=5.17, giving downstream depth y3=9.12 m. Gate thrust is about 6.07 MN, and energy loss in the jump is 9.68 m.

## Answer

**Given:** $b = 4$ m, $Q = 100\ \text{m}^3/\text{s}$ ($q = 25\ \text{m}^2/\text{s}$), upstream depth $y_1 = 20$ m. Loss through the gate $= 0.05\,V_2^2/2g$. The jump forms immediately downstream of the gate.

**Flow through the gate** ($V_1 = 25/20 = 1.25$ m/s):

$$
y_1 + \frac{V_1^2}{2g} = y_2 + 1.05\frac{V_2^2}{2g},\qquad V_2 = \frac{25}{y_2}
$$

$$
20.080 = y_2 + \frac{1.05 \times 25^2}{2g\,y_2^2} \;\Rightarrow\; y_2 = 1.336\ \text{m}
$$

$$
V_2 = 18.71\ \text{m/s},\qquad Fr_2 = \frac{18.71}{\sqrt{9.81 \times 1.336}} = 5.17
$$

#### i) Depth downstream of the jump

$$
y_3 = \frac{y_2}{2}\left(\sqrt{1 + 8Fr_2^2} - 1\right) = \frac{1.336}{2}\left(\sqrt{1 + 8(26.73)} - 1\right) = 9.12\ \text{m}
$$

$$
V_3 = \frac{25}{9.12} = 2.74\ \text{m/s}
$$

#### ii) Thrust on the gate

Momentum equation between section 1 (upstream) and section 2 (just downstream of the gate). $F$ is the force of the gate on the water, and the thrust of the water on the gate is equal and opposite:

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

| Term | Value (N) |
|---|---|
| $\gamma b y_1^2/2 = 9810 \times 4 \times 400/2$ | 7 848 000 |
| $\gamma b y_2^2/2$ | 35 012 |
| $\rho Q(V_2 - V_1) = 100\,000 \times 17.46$ | 1 746 473 |

$$
F = 7\,848\,000 - 35\,012 - 1\,746\,473 \approx 6.07 \times 10^6\ \text{N}
$$

#### iii) Energy loss in the jump

$$
\Delta E = \frac{(y_3 - y_2)^3}{4y_2y_3} = \frac{(9.12 - 1.336)^3}{4(1.336)(9.12)} = 9.68\ \text{m}
$$

Check: $E_2 = 1.336 + \dfrac{18.71^2}{2g} = 19.19$ m and $E_3 = 9.12 + \dfrac{2.74^2}{2g} = 9.50$ m, so $E_2 - E_3 = 9.68$ m.

**Answer: (i) $y_3 = 9.12$ m; (ii) thrust on the gate $\approx 6.07$ MN; (iii) energy loss in the jump $\approx 9.68$ m.**

## Check yourself

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

1. How is the depth y2 just after the gate found?
   A. Energy equation with the extra loss 0.05 V2^2/2g (factor 1.05) on the velocity head
   B. Belanger equation from y1
   C. Momentum equation with no loss
   D. Critical depth
   Answer: A. Flow through the gate follows energy conservation with the stated loss, giving y2 = 1.336 m. (Key point: Depth at gate)

2. How is the depth y3 after the jump found?
   A. Energy equation from the gate
   B. Belanger equation using y2 and Fr2 = 5.17
   C. y3 = y1
   D. Manning's equation
   Answer: B. The jump follows momentum, giving y3 = 9.12 m. (Key point: Depth after jump)

3. Which equation gives the thrust on the gate?
   A. Energy equation between 1 and 2
   B. Bernoulli with y3
   C. Continuity between 1 and 2
   D. Momentum equation between 1 and 2, with the gate force F on the water
   Answer: D. The hydrostatic force difference minus F equals the momentum change; the thrust is equal and opposite, about 6.07 MN. (Key point: Thrust on gate)

4. What is the energy loss in the jump?
   A. (y3 - y2)^3/(4 y2 y3), about 9.68 m
   B. (y3 - y2)/2
   C. y3 - y2
   D. V2^2/2g only
   Answer: A. This equals E2 - E3 = 19.19 - 9.50 = 9.68 m. (Key point: Energy loss)

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