---
title: "Direct step, 16 m wide channel, 40 m³/s"
subject: "Hydraulics"
chapter: 8
key: "semester_4/hydraulics:gvf-profile-40m3s-16m-wide-direct-step"
source: https://www.bctengineer.com/q/semester_4/hydraulics/gvf-profile-40m3s-16m-wide-direct-step
site: BCT Engineering
---

# Direct step, 16 m wide channel, 40 m³/s

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

- Asked in: 2074 Bhadra, 10 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/hydraulics/gvf-profile-40m3s-16m-wide-direct-step
- Whole chapter: https://www.bctengineer.com/md/semester_4/hydraulics/chapter-8

## Question

A rectangular channel carrying a discharge of 40 m³/sec is 16 m wide having slope 1/5000 and Manning's coefficient n = 0.024. The depth of flow in a particular section is 1.5 m. Find how far upstream or downstream of this section the flow depth is 2.5 m. Determine the type of flow profile and using direct step method calculate the length of profile taking 3 steps for calculation.

## Short answer

The profile is M2 on a mild slope (normal depth 2.671 m, critical depth 0.860 m), with depth increasing upstream. Direct step using three equal steps gives the 2.5 m depth about 4.3 km upstream of the 1.5 m section.

## Answer

**Given:** $Q = 40\ \text{m}^3/\text{s}$, $b = 16\ \text{m}$, $S_0 = 1/5000 = 0.0002$, $n = 0.024$, $y = 1.5$ m at a section. Find the distance to $y = 2.5$ m.

**Depths**

$$
q = 2.5\ \text{m}^2/\text{s},\quad y_c = \left(\frac{2.5^2}{9.81}\right)^{1/3} = 0.860\ \text{m},\quad y_n = 2.671\ \text{m}\ (\text{Manning, } R = A/P)
$$

Since $y_n > y_c$ the slope is mild, and $y_n > y > y_c$ for both 1.5 m and 2.5 m. The profile is **M2** (drawdown). For M2 the depth increases upstream, so the 2.5 m depth is **upstream** of the 1.5 m section.

**Direct step method** (3 equal steps, $\Delta y = 0.333$ 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) | $S_f$ ($\times 10^{-4}$) | $\bar S_f$ ($\times 10^{-4}$) | $S_0 - \bar S_f$ | $\Delta x$ (m) |
|---|---|---|---|---|---|---|---|
| 1.500 | 1.263 | 1.667 | 1.6416 | 11.7 | | | |
| 1.833 | 1.492 | 1.364 | 1.9281 | 6.3 | 9.0 | -0.000700 | -409.2 |
| 2.167 | 1.705 | 1.154 | 2.2345 | 3.8 | 5.0 | -0.000300 | -1012.9 |
| 2.500 | 1.905 | 1.000 | 2.5510 | 2.4 | 3.1 | -0.000110 | -2870.7 |

$$
L = 409.2 + 1012.9 + 2870.7 \approx 4293\ \text{m upstream}
$$

Because the last step approaches $y_n$, where the profile is asymptotic, three steps give a lower value than a finer calculation (about 5.15 km).

**Answer: M2 profile; the depth is 2.5 m about 4.3 km upstream of the section with 1.5 m depth (3-step direct step).**

## Check yourself

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

1. What is the profile (yc = 0.860, yn = 2.671, y = 1.5 to 2.5)?
   A. M1
   B. M3
   C. S2
   D. M2
   Answer: D. Mild slope with yn > y > yc is the M2 drawdown curve. (Key point: Profile type)

2. Where is the 2.5 m depth relative to the 1.5 m section?
   A. Upstream, since M2 depth increases upstream
   B. Downstream
   C. At the same section
   D. Not determinable
   Answer: A. M2 depth rises toward yn upstream, so the deeper section lies upstream. (Key point: Direction)

3. Which relation gives the length of each step?
   A. dx = dy/S0
   B. dx = (E2 - E1)/(S0 - mean Sf), with E = y + V^2/2g
   C. dx = (V2 - V1)/g
   D. dx = mean Sf/S0
   Answer: B. Energy difference over (S0 - mean Sf) gives each step. (Key point: Step formula)

4. Why do 3 steps give 4.3 km when a finer calculation gives about 5.15 km?
   A. Manning's n is wrong
   B. Q is too large
   C. The last step near yn is long and nonlinear, so coarse steps underestimate
   D. Steps are in the wrong direction
   Answer: C. Near yn, S0 - Sf tends to zero and dx varies sharply, so a coarse step underestimates. (Key point: Accuracy)

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