---
title: "GVF in 5 m channel, depth 0.25 m to 0.40 m"
subject: "Hydraulics"
chapter: 8
key: "semester_4/hydraulics:gvf-5m-channel-1-85-m3s-direct-step"
source: https://www.bctengineer.com/q/semester_4/hydraulics/gvf-5m-channel-1-85-m3s-direct-step
site: BCT Engineering
---

# GVF in 5 m channel, depth 0.25 m to 0.40 m

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

- Asked in: 2071 Bhadra, 8 marks
- Video: [gvf direct step](https://www.youtube.com/watch?v=P6lluipDzAA) — Civilian Thoughts
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/hydraulics/gvf-5m-channel-1-85-m3s-direct-step
- Whole chapter: https://www.bctengineer.com/md/semester_4/hydraulics/chapter-8

## Question

A rectangular channel with a bottom width of 5 m, bottom slope of 0.00076 and energy correction factor of 1.1 has a discharge of 1.85 m³/s. In a Gradually varied flow in this section the depth at certain location is found to be 0.25 m, considering Manning's roughness coefficient as 0.0165 determine the type of GVF profile. How far upstream or downstream will the depth be 0.40 m from depth 0.25 m. Use direct step method using increment equals to 0.05 m.

## Short answer

The profile is M2 (drawdown). Using the direct step method with Manning's equation, the depth of 0.40 m is located approximately 106 m upstream of the section where depth is 0.25 m.

## Answer

**Given:** $b = 5$ m, $Q = 1.85\ \text{m}^3/\text{s}$, $S_0 = 0.00076$, $n = 0.0165$, $\alpha = 1.1$, $y = 0.25$ m at a section. Find the distance to $y = 0.40$ m with $\Delta y = 0.05$ m.

**Depths**

$$
q = 0.37\ \text{m}^2/\text{s},\quad y_c = \left(\frac{\alpha q^2}{g}\right)^{1/3} = \left(\frac{1.1 \times 0.37^2}{9.81}\right)^{1/3} = 0.2485\ \text{m}
$$

By Manning's equation (trial, $R = A/P$): $y_n = 0.431$ m.

Since $y_n > y_c$ the slope is mild. The depth 0.25 m satisfies $y_n > y > y_c$, so the profile is **M2** (drawdown). The depth rises upstream, so 0.40 m is **upstream** of the 0.25 m section.

**Direct step method**

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

| $y$ (m) | $R$ (m) | $V$ (m/s) | $E$ (m) | $S_f$ | $\bar S_f$ | $\Delta x$ (m) |
|---|---|---|---|---|---|---|
| 0.25 | 0.2273 | 1.480 | 0.3728 | 0.00430 | | |
| 0.30 | 0.2679 | 1.233 | 0.3853 | 0.00240 | 0.00335 | -4.8 |
| 0.35 | 0.3070 | 1.057 | 0.4127 | 0.00147 | 0.00193 | -23.3 |
| 0.40 | 0.3448 | 0.925 | 0.4480 | 0.00096 | 0.00122 | -77.4 |

$$
L = 4.8 + 23.3 + 77.4 \approx 105.6\ \text{m upstream}
$$

The last step is close to $y_n$ where the profile flattens, so a finer calculation gives about 126 m.

**Answer: M2 profile; the depth is 0.40 m about 106 m upstream of the section where it is 0.25 m.**

## Check yourself

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

1. How is yc found here?
   A. (q^2/g)^(1/3) = 0.241 m
   B. (alpha q^2/g)^(1/3) = 0.2485 m
   C. yc = yn
   D. (Q/b)^(1/2)
   Answer: B. With alpha = 1.1, yc = (1.1 x 0.37^2/9.81)^(1/3) = 0.2485 m. (Key point: Critical depth with alpha)

2. Why is the profile M2?
   A. Because 0.25 m is below yc
   B. Because 0.25 m exceeds yn
   C. Because slope is steep
   D. Because yn = 0.431 > y = 0.25 > yc = 0.2485 on a mild slope
   Answer: D. Mild slope with depth between yc and yn is M2. (Key point: Profile)

3. Where is the 0.40 m depth relative to 0.25 m?
   A. Upstream, because M2 depth rises upstream
   B. Downstream
   C. At the same section
   D. Unknown
   Answer: A. Depth increases upstream along M2, so dx is negative, about 106 m. (Key point: Direction)

4. How is specific energy computed?
   A. E = y + V^2/2g
   B. E = y + alpha V^2/2g
   C. E = y
   D. E = alpha y + V^2
   Answer: B. The energy correction factor multiplies the velocity head. (Key point: Energy with alpha)

## Close variants set in other exams

- [GVF in 4 m channel, graphical integration](https://www.bctengineer.com/question.md?key=semester_4%2Fhydraulics%3Agvf-4m-channel-2-m3s-graphical-integration) — Hydraulics, chapter 8, asked 2×

## Related in Hydraulics

- [Define gradually varied flow](https://www.bctengineer.com/question.md?key=semester_4%2Fhydraulics%3Adefine-gradually-varied-flow) — Hydraulics, chapter 5, asked 2×
- [Dynamic equation of GVF, wide rectangular form](https://www.bctengineer.com/question.md?key=semester_4%2Fhydraulics%3Agvf-dynamic-equation-wide-rectangular) — Hydraulics, chapter 8, asked 2×
- [Define GVF, RVF and spatially varied flow](https://www.bctengineer.com/question.md?key=semester_4%2Fhydraulics%3Agvf-rvf-svf-define-with-examples) — Hydraulics, chapter 5, asked 5×
- [Explain GVF and spatially varied flow](https://www.bctengineer.com/question.md?key=semester_4%2Fhydraulics%3Agvf-and-svf-explain-example) — Hydraulics, chapter 5
- [94% depth for maximum discharge in circular channel](https://www.bctengineer.com/question.md?key=semester_4%2Fhydraulics%3Acircular-channel-94-percent-depth-max-discharge) — Hydraulics, chapter 6
- [Bresse's method for 20 m wide channel](https://www.bctengineer.com/question.md?key=semester_4%2Fhydraulics%3Abresse-method-20m-channel-65-m3s) — Hydraulics, chapter 8

## Related in other subjects

- [Efficient rectangular section and triangular channel depth](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Atriangular-channel-normal-depth-rect) — Hydro Power, chapter 1, asked 2×
- [Flood discharge in rectangular channel](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Aflood-discharge-slope-area) — Hydro Power, chapter 2
- [Flood discharge in 10-m rectangular channel](https://www.bctengineer.com/question.md?key=semester_5%2Fengineering_hydrology%3Aslope-area-rectangular-10m) — Engineering Hydrology, chapter 4
- [Uniform flow in rectangular channel (Chezy)](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Arectangular-channel-chezy-flow) — Hydro Power, chapter 1