---
title: "GVF in 4 m channel, graphical integration"
subject: "Hydraulics"
chapter: 8
key: "semester_4/hydraulics:gvf-4m-channel-2-m3s-graphical-integration"
source: https://www.bctengineer.com/q/semester_4/hydraulics/gvf-4m-channel-2-m3s-graphical-integration
site: BCT Engineering
---

# GVF in 4 m channel, graphical integration

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

- Asked in: 2070 Bhadra, 8 marks
- Video: [GVF Computation (Advanced Numerical and Graphical Integration Method)](https://www.youtube.com/watch?v=NHpG7JCWzNI) — Effective Learning Center
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/hydraulics/gvf-4m-channel-2-m3s-graphical-integration
- Whole chapter: https://www.bctengineer.com/md/semester_4/hydraulics/chapter-8

## Question

A rectangular channel with a bottom width of 4 m, bottom slope of 0.00075 and energy correction factor of 1.1 has a discharge of 2.0 m³/s. In a Gradually varied flow in this section the depth at certain location is found to be 0.2 m, considering Manning's roughness coefficient as 0.016 determine the type of GVF profile. How far upstream or downstream will the depth be 0.40 m from depth 0.20 m. Use Graphical Integration Method using increment equals to 0.1 m.

## Short answer

The profile is M3 (supercritical). By graphical integration, the depth increases from 0.20 m to near critical depth (0.304 m) over about 10 m downstream, where the curve ends in a hydraulic jump; thus 0.40 m is not reached by gradually varied flow.

## Answer

**Given:** $b = 4$ m, $Q = 2.0\ \text{m}^3/\text{s}$, $S_0 = 0.00075$, $n = 0.016$, $\alpha = 1.1$, $y = 0.2$ m at a section. Find the distance to $y = 0.4$ m with $\Delta y = 0.1$ m.

**Depths**

$$
q = 0.5\ \text{m}^2/\text{s},\quad y_c = \left(\frac{1.1 \times 0.5^2}{9.81}\right)^{1/3} = 0.304\ \text{m}
$$

By Manning's equation (trial, $R = A/P$): $y_n = 0.525$ m.

The slope is mild ($y_n > y_c$) and $y = 0.2$ m is below $y_c$, so the profile is **M3** (supercritical, depth rising downstream).

**Graphical integration.** From the dynamic equation,

$$
\frac{dx}{dy} = \frac{1 - \alpha Q^2T/(gA^3)}{S_0 - S_f},\qquad x = \int_{y_1}^{y_2}\frac{dx}{dy}\,dy = \text{area under the } \frac{dx}{dy}\text{ vs } y \text{ curve}
$$

| $y$ (m) | $V$ (m/s) | $R$ (m) | $S_f$ | $Fr^2 = \alpha V^2/gy$ | $dx/dy$ (m/m) |
|---|---|---|---|---|---|
| 0.20 | 2.500 | 0.1818 | 0.01553 | 3.504 | 169.4 |
| 0.30 | 1.667 | 0.2609 | 0.00427 | 1.038 | 10.9 |
| 0.40 | 1.250 | 0.3333 | 0.00173 | 0.438 | -573.1 |

Plot $dx/dy$ against $y$ and find the area under the curve.

- From 0.2 to 0.3 m, the trapezoidal area is $\tfrac{169.4 + 10.9}{2}(0.1) = 9.0$ m; a finer reading of the curve (extra point at 0.25 m, where $dx/dy = 115.8$) gives **10.7 m**.
- At $y_c = 0.304$ m, $dx/dy$ changes sign (the denominator of the dynamic equation, $1 - Fr^2$, goes through zero). The M3 curve cannot cross the critical depth gradually; it ends in a **hydraulic jump** at or before $y_c$. Hence the depth 0.40 m cannot be reached by a continuous profile from 0.20 m, and the 0.3 to 0.4 m "area" has no physical meaning.

**Answer: M3 profile. The depth rises from 0.20 m to about 0.30 m (critical) over about 10 m downstream; the curve then ends in a hydraulic jump, so 0.40 m is not reached by gradually varied flow.**

## Check yourself

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

1. Why is the profile M3?
   A. y = 0.2 m is above yn
   B. Mild slope (yn 0.525 > yc 0.304) with y = 0.2 m below yc
   C. Slope is steep
   D. y is between yn and yc
   Answer: B. Mild slope with depth below yc is zone 3: M3, supercritical with depth rising downstream. (Key point: Profile type)

2. What does the distance equal in graphical integration?
   A. Value of dx/dy at the end depth
   B. Average Sf times dy
   C. Area under the dx/dy versus y curve between the two depths
   D. Slope of the curve
   Answer: C. x is the integral of dx/dy dy, i.e. the area under the curve. (Key point: Graphical integration)

3. Why can the area from 0.3 to 0.4 m not be used?
   A. The curve is below the axis
   B. The depth step is too large only
   C. Sf is zero
   D. dx/dy changes sign at yc because 1 - Fr^2 passes through zero, so the curve cannot cross yc gradually
   Answer: D. M3 cannot reach critical depth gradually; a hydraulic jump ends it, so 0.40 m is not reached by GVF. (Key point: Singularity at yc)

4. What is the conclusion for the 0.2 to 0.3 m portion?
   A. About 10 m downstream, then a jump
   B. About 500 m upstream
   C. About 100 m downstream
   D. Zero
   Answer: A. Finer integration gives about 10.7 m before the profile ends in a jump. (Key point: Result)

## Close variants set in other exams

- [GVF in 5 m channel, depth 0.25 m to 0.40 m](https://www.bctengineer.com/question.md?key=semester_4%2Fhydraulics%3Agvf-5m-channel-1-85-m3s-direct-step) — Hydraulics, chapter 8, asked 2×

## Related in Hydraulics

- [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 gradually varied flow](https://www.bctengineer.com/question.md?key=semester_4%2Fhydraulics%3Adefine-gradually-varied-flow) — Hydraulics, chapter 5, 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
- [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
- [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

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