---
title: "Contraction with lowered bed, 10 m³/s"
subject: "Hydraulics"
chapter: 7
key: "semester_4/hydraulics:contraction-lowered-bed-3-5m-wide-10-m3s"
source: https://www.bctengineer.com/q/semester_4/hydraulics/contraction-lowered-bed-3-5m-wide-10-m3s
site: BCT Engineering
---

# Contraction with lowered bed, 10 m³/s

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

- Asked in: 2074 Bhadra, 8 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/hydraulics/contraction-lowered-bed-3-5m-wide-10-m3s
- Whole chapter: https://www.bctengineer.com/md/semester_4/hydraulics/chapter-7

## Question

A 3.5 m wide rectangular channel carries a discharge of 10 m³/s at a depth of 1.75 m. If the width of the channel is reduced to 2.25 m and bed level is lowered by 0.97 m, determine the difference in water level elevation between upstream and contracted section. Assume no energy loss.

## Short answer

Upstream specific energy is 1.886 m; with 0.97 m bed lowering, E2 = 2.856 m. Solving y2 + 1.0067/y2² = 2.856 gives subcritical y2 = 2.720 m. The water surface elevation difference is negligible (≈0.0002 m) because the increased velocity head balances the bed drop.

## Answer

#### Given data
$b_1=3.5$ m, $Q=10$ m³/s, $y_1=1.75$ m; contracted $b_2=2.25$ m, bed lowered by $0.97$ m; no energy loss.
$$
q_1=\frac{10}{3.5}=2.8571\ \text{m}^2/\text{s},\quad V_1=\frac{2.8571}{1.75}=1.633\ \text{m/s},\quad Fr_1=\frac{1.633}{\sqrt{9.81\times1.75}}=0.394
$$
$$
E_1=1.75+\frac{1.633^2}{19.62}=1.8859\ \text{m}
$$
The upstream flow is subcritical.

#### Contracted section
$$
q_2=\frac{10}{2.25}=4.4444\ \text{m}^2/\text{s},\quad y_{c2}=\sqrt[3]{\frac{q_2^2}{g}}=1.2628\ \text{m}
$$
The bed is lowered, so energy above the new bed increases:
$$
E_2=E_1+0.97=2.8559\ \text{m}
$$
Subcritical root of $y_2+\dfrac{q_2^2}{2gy_2^2}=y_2+\dfrac{1.0067}{y_2^2}=2.8559$:
$$
y_2=2.7198\ \text{m}\quad(>y_{c2},\ \text{subcritical})
$$

#### Water level elevations (datum: upstream bed)
- Upstream: $1.7500$ m
- Contracted section: $-0.97+2.7198=1.7498$ m

Difference in water level $=1.7500-1.7498=0.0002$ m, which is practically zero.

**Answer: depth in the contracted section $=2.720$ m; the water surface elevation at the contracted section is the same as upstream (difference ≈ 0.000 m, i.e. less than 1 mm).** The increase in velocity head is exactly balanced by the lowering of the bed.

## Check yourself

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

1. When the bed is lowered by 0.97 m, the specific energy at the contraction is
   A. E1 - 0.97
   B. E1 + 0.97 = 2.856 m
   C. E1
   D. Emin
   Answer: B. Energy above the new, lower bed increases by the drop height. (Key point: Effect of lowering the bed)

2. Why is the subcritical root y2 = 2.72 m chosen?
   A. Upstream flow is subcritical (Fr = 0.39) and E2 is well above Emin
   B. The supercritical root is smaller than zero
   C. Narrowing always gives subcritical flow
   D. The flow is critical
   Answer: A. Approach flow is subcritical and no choking occurs, so the flow remains on the subcritical limb. (Key point: Choice of root)

3. Water surface elevations (datum upstream bed) are 1.750 m upstream and
   A. 2.720 m in the contraction
   B. 1.750 m (-0.97 + 2.7198) in the contraction
   C. 0.97 m in the contraction
   D. 1.263 m in the contraction
   Answer: B. Surface elevation = -0.97 + 2.7198 = 1.7498 m, so the difference is practically zero. (Key point: Water level comparison)

4. Why is the surface level practically unchanged?
   A. Energy is lost in the contraction
   B. The velocity head increase from narrowing balances the bed drop
   C. The flow is critical
   D. The discharge changes
   Answer: B. Narrowing raises velocity and velocity head, which matches the energy gained from lowering the bed. (Key point: Interpretation)

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