---
title: "Contraction and bed rise, 3.5 m channel (1.2 m depth)"
subject: "Hydraulics"
chapter: 7
key: "semester_4/hydraulics:contraction-bed-rise-3-5m-wide-1-2m-depth"
source: https://www.bctengineer.com/q/semester_4/hydraulics/contraction-bed-rise-3-5m-wide-1-2m-depth
site: BCT Engineering
---

# Contraction and bed rise, 3.5 m channel (1.2 m depth)

Hydraulics · Chapter 7: Energy and Momentum Principles in Open channel flow

- Asked in: 2071 Bhadra, 6 marks
- Video: [#8 numerical on channel transition || width contraction || open channel flow ||](https://www.youtube.com/watch?v=32bMKbMF7hE) — Numerical's Civil
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/hydraulics/contraction-bed-rise-3-5m-wide-1-2m-depth
- Whole chapter: https://www.bctengineer.com/md/semester_4/hydraulics/chapter-7

## Question

A 3.5 m rectangular channel carries discharge of 4 m³/s of water at a depth of 1.2 m. If the width is reduced to 2.0 m and bed raised by 0.15 m, determine the depth of flow at reduced section and upstream of the reduced section.

## Short answer

Flow chokes; depth at reduced section becomes critical at 0.74 m. Upstream depth rises to 1.22 m due to backwater.

## Answer

#### Given data
$b_1=3.5$ m, $Q=4$ m³/s, $y_1=1.2$ m; contracted $b_2=2.0$ m, bed raised $\Delta z=0.15$ m.
$$
q_1=\frac{4}{3.5}=1.1429\ \text{m}^2/\text{s},\quad V_1=\frac{1.1429}{1.2}=0.952\ \text{m/s},\quad E_1=1.2+\frac{0.952^2}{19.62}=1.2462\ \text{m}
$$

#### Critical conditions at the contracted section
$$
q_2=\frac{4}{2}=2\ \text{m}^2/\text{s},\quad y_{c2}=\sqrt[3]{\frac{2^2}{9.81}}=0.7415\ \text{m},\quad E_{min2}=1.5y_{c2}=1.1123\ \text{m}
$$
Energy available at the contracted section if the upstream depth is unchanged:
$$
E_2=E_1-\Delta z=1.2462-0.15=1.0962\ \text{m}<E_{min2}=1.1123\ \text{m}
$$
The available energy is less than the minimum needed, so the flow is **choked**: critical flow occurs in the reduced section and the upstream depth must rise.

#### Depth at the reduced section
$$
y_2=y_{c2}=0.742\ \text{m}
$$

#### Upstream depth
Upstream energy must be $E_1'=E_{min2}+\Delta z=1.1123+0.15=1.2623$ m.
$$
y_1'+\frac{q_1^2}{2gy_1'^2}=y_1'+\frac{0.06657}{y_1'^2}=1.2623\;\Rightarrow\;y_1'=1.217\ \text{m}
$$
The upstream depth rises from 1.20 m to 1.217 m.

**Answer: depth at the reduced section $=0.74$ m (critical); depth upstream $=1.22$ m.**

## Check yourself

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

1. Here E2 = 1.0962 m and Emin2 = 1.1123 m. What does this show?
   A. The flow is choked; critical flow occurs in the reduced section
   B. The flow is subcritical and unaffected
   C. The flow is supercritical
   D. The width is too large
   Answer: A. Available energy is less than the minimum required, so the upstream depth must rise. (Key point: Choking test)

2. The depth at the reduced section is
   A. 1.2 m
   B. 1.0962 m
   C. 0.742 m, the critical depth for q2 = 2 m^2/s
   D. 1.217 m
   Answer: C. When choked, the contracted section holds critical depth: yc2 = (4/9.81)^(1/3) = 0.742 m. (Key point: Depth at the reduced section)

3. The new upstream specific energy is
   A. E1 - Δz
   B. Emin2 + Δz = 1.2623 m
   C. Emin2 - Δz
   D. E1
   Answer: B. Upstream energy must supply the minimum energy plus the bed rise: 1.1123 + 0.15. (Key point: Upstream energy)

4. Solving y + 0.06657/y^2 = 1.2623 gives an upstream depth of about
   A. 1.20 m
   B. 1.50 m
   C. 1.217 m
   D. 0.74 m
   Answer: C. The subcritical root is 1.217 m, a small rise from 1.20 m. (Key point: Upstream depth)

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