---
title: "Sluice gate profiles and lengths by direct step"
subject: "Hydraulics"
chapter: 8
key: "semester_4/hydraulics:sluice-gate-profiles-length-direct-step"
source: https://www.bctengineer.com/q/semester_4/hydraulics/sluice-gate-profiles-length-direct-step
site: BCT Engineering
---

# Sluice gate profiles and lengths by direct step

Hydraulics · Chapter 8: Non-uniform gradually varied flow (GVF)

- Asked in: 2080 Chaitra, 2+5+5 marks
- Video: [Hydraulic Jumps, WSP, and Direct Step Method- Mirna Ksssem](https://www.youtube.com/watch?v=WRg6o3ezBn0) — Mirna Kassem
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/hydraulics/sluice-gate-profiles-length-direct-step
- Whole chapter: https://www.bctengineer.com/md/semester_4/hydraulics/chapter-8

## Question

A sluice gate is placed in a long rectangular channel with 4 m wide, having bed slope 0.0008 and Manning's n = 0.014. The depth upstream of the gate $y_1$ is 1.85 m and depth downstream of the gate $y_2$ = 0.35 m. Ignoring losses, i) Identify the water surface profiles on both side of the gate. ii) Compute length of the profile for both upstream and downstream case using direct step method. (Take three steps only for each case). [Figure: sluice gate in a channel of slope $S_0$ = 0.0008, upstream depth 1.85 m, downstream depth $y_2$ = 0.35 m]

## Short answer

Upstream: M1 backwater curve, length 1450 m (three direct steps). Downstream: M3 curve from the gate, length 16.1 m to a hydraulic jump (0.411 m to normal depth 1.171 m). Both profiles calculated by direct step method with three steps each.

## Answer

**Given:** $b = 4\ \text{m}$, $S_0 = 0.0008$, $n = 0.014$, $y_1 = 1.85\ \text{m}$ (upstream of gate), $y_2 = 0.35\ \text{m}$ (downstream of gate).

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

$$
y_1 + \frac{V_1^2}{2g} = y_2 + \frac{V_2^2}{2g} \;\Rightarrow\; Q = b\,y_1 y_2\sqrt{\frac{2g}{y_1 + y_2}} = 4(1.85)(0.35)\sqrt{\frac{19.62}{2.2}} = 7.735\ \text{m}^3/\text{s}
$$

**Critical and normal depths** ($q = 1.934\ \text{m}^2/\text{s}$):

$$
y_c = \left(\frac{q^2}{g}\right)^{1/3} = 0.725\ \text{m},\qquad y_n = 1.171\ \text{m}\ (\text{from Manning's equation, } R = A/P)
$$

Since $y_n > y_c$, the slope is **mild**.

#### i) Profiles

- **Upstream of the gate:** $y = 1.85\ \text{m} > y_n$, so an **M1** (backwater) curve, rising towards the gate.
- **Downstream of the gate:** $y = 0.35\ \text{m} < y_c$, so an **M3** curve rising from the vena contracta. The sequent depth of 0.35 m is 1.311 m, which is greater than $y_n = 1.171$ m, so the jump is not at the gate. The M3 curve continues until its conjugate depth equals $y_n$ (at $y_j = 0.411\ \text{m}$), where a **hydraulic jump** to $y_n$ forms.

```text
        M1                     gate
  ...........___----''''''''''|G|
  yn=1.171 ..........         | |  M3  J     yn
                              |_|____/'|----------
```

#### ii) Lengths by direct step

$$
E = y + \frac{V^2}{2g},\quad S_f = \frac{n^2V^2}{R^{4/3}},\quad \Delta x = \frac{E_2 - E_1}{S_0 - \bar S_f}
$$

**Upstream (M1), $y$: 1.85 to 1.25 m in 3 steps of 0.20 m:**

| $y$ (m) | $R$ (m) | $V$ (m/s) | $E$ (m) | $S_f$ | $\bar S_f$ | $\Delta x$ (m) |
|---|---|---|---|---|---|---|
| 1.85 | 0.961 | 1.045 | 1.9057 | 0.00023 | | |
| 1.65 | 0.904 | 1.172 | 1.7200 | 0.00031 | 0.00027 | -348.3 |
| 1.45 | 0.841 | 1.334 | 1.5406 | 0.00044 | 0.00037 | -420.7 |
| 1.25 | 0.769 | 1.547 | 1.3720 | 0.00067 | 0.00055 | -681.3 |

$$
L_{up} = 348.3 + 420.7 + 681.3 \approx 1450\ \text{m (upstream of the gate)}
$$

**Downstream (M3), $y$: 0.35 to 0.411 m in 3 steps:**

| $y$ (m) | $R$ (m) | $V$ (m/s) | $E$ (m) | $S_f$ | $\bar S_f$ | $\Delta x$ (m) |
|---|---|---|---|---|---|---|
| 0.3500 | 0.298 | 5.525 | 1.9057 | 0.03007 | | |
| 0.3704 | 0.313 | 5.220 | 1.7593 | 0.02518 | 0.02763 | 5.46 |
| 0.3908 | 0.327 | 4.947 | 1.6384 | 0.02130 | 0.02324 | 5.39 |
| 0.4113 | 0.341 | 4.702 | 1.5380 | 0.01818 | 0.01974 | 5.30 |

$$
L_{down} = 5.46 + 5.39 + 5.30 \approx 16.1\ \text{m (downstream of the gate, up to the jump)}
$$

**Answer:** M1 upstream (about 1450 m to within 7% of $y_n$) and M3 downstream (about 16 m, then a jump from 0.411 m to 1.171 m).

## Check yourself

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

1. How is discharge found from depths on either side of the gate?
   A. Manning's equation at y1
   B. Critical flow formula
   C. Momentum equation across the gate
   D. Energy equation without loss: Q = b y1 y2 sqrt(2g/(y1+y2))
   Answer: D. Losses are ignored, so specific energy upstream equals that downstream, giving Q = 7.735 m^3/s. (Key point: Discharge from gate)

2. Why is the profile upstream of the gate M1?
   A. Depth 1.85 m is above yn = 1.171 m on a mild slope
   B. Depth is below yc
   C. Depth is between yn and yc
   D. The slope is steep
   Answer: A. On a mild slope with y > yn > yc, the profile is M1 backwater rising toward the gate. (Key point: Upstream profile)

3. Why is the downstream flow an M3 curve ending in a jump?
   A. 0.35 m is above yn so it is M1
   B. 0.35 m is below yc and its sequent depth (1.311 m) exceeds yn, so a jump forms only where the conjugate equals yn
   C. The jump forms right at the gate because 1.311 m is less than 1.171 m
   D. Friction removes supercritical flow
   Answer: B. Sequent depth is more than yn, so tailwater cannot drown the gate; M3 rises until its conjugate equals yn (at 0.411 m). (Key point: Downstream profile)

4. In the sluice gate tables, which friction slope formula is used?
   A. Sf = n^2 V^2 / R^(4/3), with R = A/P
   B. Sf = V^2/gy
   C. Sf = S0
   D. Sf = n q / y
   Answer: A. Manning's formula with hydraulic radius for the 4 m wide rectangular section is used. (Key point: Friction slope used)

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