---
title: "Contraction to 1.25 m: is flow possible?"
subject: "Hydraulics"
chapter: 7
key: "semester_4/hydraulics:contraction-possible-flow-specific-energy-4m-channel"
source: https://www.bctengineer.com/q/semester_4/hydraulics/contraction-possible-flow-specific-energy-4m-channel
site: BCT Engineering
---

# Contraction to 1.25 m: is flow possible?

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

- Asked in: 2073 Bhadra, 2+2+3 marks
- Video: [Supercritical Flow Through Contraction](https://www.youtube.com/watch?v=YhRJiAwZAbg) — Bassma Taher
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/hydraulics/contraction-possible-flow-specific-energy-4m-channel
- Whole chapter: https://www.bctengineer.com/md/semester_4/hydraulics/chapter-7

## Question

Water flows in a 4 m wide rectangular channel at a depth of 1.8 m and velocity 1.4 m/s. The channel is contracted to a width of 1.25 m in particular reach. Is the flow possible in given specific energy? If not, what should be the discharge in channel so that flow is possible in the given specific energy? Also determine the depth of flow at contracted section and upstream of contracted section.

## Short answer

Flow is not possible for 10.08 m³/s because E1 = 1.90 m < Emin2 = 2.82 m. The maximum discharge is 5.58 m³/s; at that flow, the contracted section reaches critical depth 1.27 m and the upstream depth is 1.87 m.

## Answer

#### Given data
$b_1=4$ m, $y_1=1.8$ m, $V_1=1.4$ m/s, contracted width $b_2=1.25$ m (no hump).
$$
Q=b_1y_1V_1=4\times1.8\times1.4=10.08\ \text{m}^3/\text{s},\qquad q_1=\frac{10.08}{4}=2.52\ \text{m}^2/\text{s}
$$
$$
E_1=1.8+\frac{1.4^2}{19.62}=1.8999\ \text{m}
$$

#### (a) Is the flow possible?
At the contraction, $q_2=\dfrac{10.08}{1.25}=8.064$ m²/s.
$$
y_{c2}=\sqrt[3]{\frac{q_2^2}{g}}=1.878\ \text{m},\qquad E_{min2}=1.5y_{c2}=2.818\ \text{m}
$$
Since the available energy $E_1=1.900$ m $<E_{min2}=2.818$ m, **the flow is not possible** with this specific energy (the flow would be choked and the upstream level would rise).

#### (b) Discharge possible with the given specific energy
The maximum flow through the contraction occurs with critical flow at $E_{min2}=E_1$:
$$
y_c=\frac23E_1=1.2666\ \text{m},\qquad q_{max}=\sqrt{gy_c^3}=4.465\ \text{m}^2/\text{s}
$$
$$
Q_{max}=q_{max}b_2=4.465\times1.25=5.581\ \text{m}^3/\text{s}
$$
So the discharge should not exceed about 5.58 m³/s.

#### (c) Depths
- **At the contracted section:** critical, $y=y_c=1.267$ m.
- **Upstream:** with $Q=5.581$ m³/s, $q_1=1.3952$ m²/s and $E_1=1.8999$ m (subcritical root):
$$
y_1+\frac{1.3952^2}{19.62y_1^2}=1.8999\;\Rightarrow\;y_1=1.872\ \text{m}
$$
**Answer: not possible for 10.08 m³/s; the discharge must be reduced to $Q\approx5.58$ m³/s; depth at the contraction $=1.27$ m (critical), depth upstream $\approx1.87$ m.**

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