---
title: "GVF in 12 m wide channel, 2 steps"
subject: "Hydraulics"
chapter: 8
key: "semester_4/hydraulics:gvf-30m3s-12m-wide-two-steps"
source: https://www.bctengineer.com/q/semester_4/hydraulics/gvf-30m3s-12m-wide-two-steps
site: BCT Engineering
---

# GVF in 12 m wide channel, 2 steps

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

- Asked in: 2068 Bhadra, 7+1 marks
- Video: [Methods of solving GVF | Direct step Method | Hydraulics | Unit:8 | Prashant YT | BE Civil | PU,PoU|](https://www.youtube.com/watch?v=2MaZez93EzU) — Prashant YT
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/hydraulics/gvf-30m3s-12m-wide-two-steps
- Whole chapter: https://www.bctengineer.com/md/semester_4/hydraulics/chapter-8

## Question

A rectangular channel conveying a discharge of 30 m³/sec is 12 m wide with a bed slope 1 in 6000 and having Manning's n = 0.025. The depth of flow at a section is 1.5 m. Find how far upstream or downstream of this section the depth of flow will be 2 m. Find also the types of profile. Use direct step method for calculation and take only two steps for calculation.

## Short answer

The profile is M2 (drawdown). Using the direct step method with two steps, the depth of 2 m is found approximately 640 m upstream of the section where depth is 1.5 m.

## Answer

**Given:** $Q = 30\ \text{m}^3/\text{s}$, $b = 12$ m, $S_0 = 1/6000$, $n = 0.025$, $y = 1.5$ m at a section. Find the distance to $y = 2$ m with 2 steps.

**Depths**

$$
q = 2.5\ \text{m}^2/\text{s},\quad y_c = \left(\frac{2.5^2}{9.81}\right)^{1/3} = 0.860\ \text{m},\quad y_n = 3.034\ \text{m}\ (\text{Manning, } R = A/P)
$$

**Profile type:** $y_n > y_c$ (mild), and $y_n > y > y_c$ (zone 2), so the profile is **M2** (drawdown). The depth rises upstream, so the 2 m section is **upstream** of the 1.5 m section.

**Direct step method** (2 steps, $\Delta y = 0.25$ m):

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

| $y$ (m) | $R$ (m) | $V$ (m/s) | $E$ (m) | $S_f$ | $\bar S_f$ | $S_0 - \bar S_f$ | $\Delta x$ (m) |
|---|---|---|---|---|---|---|---|
| 1.50 | 1.200 | 1.667 | 1.6416 | 0.001360 | | | |
| 1.75 | 1.355 | 1.429 | 1.8540 | 0.000850 | 0.001105 | -0.000938 | -226.1 |
| 2.00 | 1.500 | 1.250 | 2.0796 | 0.000570 | 0.000710 | -0.000543 | -415.4 |

$$
L = 226.1 + 415.4 \approx 641.6\ \text{m upstream}
$$

(A finer calculation gives about 676 m.)

**Answer: M2 profile; the depth is 2 m about 640 m upstream of the section where it is 1.5 m.**

## Check yourself

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

1. What profile is the 1.5 m to 2 m reach (yc 0.860, yn 3.034)?
   A. M1
   B. M3
   C. M2
   D. S2
   Answer: C. Mild slope, y between yn and yc: M2. (Key point: Profile)

2. Where is the 2 m depth relative to the 1.5 m section?
   A. Downstream
   B. Upstream
   C. At the same place
   D. Cannot say
   Answer: B. M2 depth rises upstream toward yn. (Key point: Direction)

3. How is depth divided for two steps?
   A. 1.5 to 2 m in equal steps of 0.25 m
   B. One step of 0.5 m
   C. Steps of 0.1 m
   D. Steps of 1 m
   Answer: A. Two equal steps: 1.50, 1.75, 2.00 m. (Key point: Number of steps)

4. What is the total length?
   A. About 226 m
   B. About 6.4 km
   C. About 64 m
   D. About 640 m upstream
   Answer: D. 226.1 + 415.4 is about 641.6 m upstream. (Key point: Result)

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