---
title: "Correction factors for given velocity profile"
subject: "Hydraulics"
chapter: 7
key: "semester_4/hydraulics:energy-momentum-correction-factors-linear-profile"
source: https://www.bctengineer.com/q/semester_4/hydraulics/energy-momentum-correction-factors-linear-profile
site: BCT Engineering
---

# Correction factors for given velocity profile

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

- Asked in: 2068 Magh, 4 marks
- Video: [How to Calculate Energy and Momentum Correction Factors Easily | Energy & Momentum Correction Factor](https://www.youtube.com/watch?v=gFPWbzzLZts) — Hydropower
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/hydraulics/energy-momentum-correction-factors-linear-profile
- Whole chapter: https://www.bctengineer.com/md/semester_4/hydraulics/chapter-7

## Question

For the velocity distribution given in figure below, find energy and momentum correction factors. [Figure: depth 2.4 m; velocity 0.4 m/s at the water surface (over the top 0.4 m), varying linearly to 1.4 m/s at the bed]

## Short answer

Mean velocity V = 0.8167 m/s. Energy correction factor α = 1.48, momentum correction factor β = 1.16.

## Answer

**Reading of the figure (assumed):** flow depth $D = 2.4\ \text{m}$. The top 0.4 m moves at a uniform $0.4\ \text{m/s}$. Below it the velocity increases linearly with depth from $0.4\ \text{m/s}$ to $1.4\ \text{m/s}$ at the bed (over the remaining 2.0 m). Take unit width.

Let $s$ be the depth below the surface:

- $0 \le s \le 0.4$: $u = 0.4$
- $0.4 \le s \le 2.4$: $u = 0.4 + 0.5\,(s - 0.4)$, so $ds = du/0.5$

**Mean velocity** (area under the profile divided by the depth)

$$
\begin{aligned} \int u\,ds &= 0.4(0.4) + \frac{0.4 + 1.4}{2}(2.0) = 0.16 + 1.80 = 1.96 \\ V &= \frac{1.96}{2.4} = 0.8167\ \text{m/s} \end{aligned}
$$

**Integrals of $u^2$ and $u^3$**

$$
\int u^2 ds = 0.4^2(0.4) + \frac{1.4^3 - 0.4^3}{3 \times 0.5} = 0.064 + 1.7867 = 1.8507
$$

$$
\int u^3 ds = 0.4^3(0.4) + \frac{1.4^4 - 0.4^4}{4 \times 0.5} = 0.0256 + 1.9080 = 1.9336
$$

**Correction factors**

$$
\begin{aligned} \alpha &= \frac{\int u^3\,ds}{V^3 D} = \frac{1.9336}{0.8167^3 \times 2.4} = \frac{1.9336}{1.3073} = 1.479 \\ \beta &= \frac{\int u^2\,ds}{V^2 D} = \frac{1.8507}{0.8167^2 \times 2.4} = \frac{1.8507}{1.6007} = 1.156 \end{aligned}
$$

**Answer: Energy correction factor $\alpha \approx 1.48$; momentum correction factor $\beta \approx 1.16$.**

## Check yourself

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

1. How is the mean velocity found from the profile?
   A. Average of 0.4 and 1.4 m/s
   B. Area under the u-depth profile divided by the depth: 1.96/2.4 = 0.817 m/s
   C. Velocity at mid-depth
   D. Surface velocity times 2
   Answer: B. V = integral u ds / D = 1.96/2.4 = 0.8167 m/s. (Key point: Mean velocity method)

2. Energy correction factor alpha uses:
   A. integral u^2 ds/(V^2 D)
   B. V^3 D / integral u^3 ds
   C. integral u^3 ds/(V^3 D)
   D. integral u ds/(V D)
   Answer: C. alpha compares the true kinetic energy flux (u^3) with that of the mean velocity. (Key point: Formula for alpha)

3. Momentum correction factor beta uses:
   A. integral u^2 ds/(V^2 D)
   B. integral u^3 ds/(V^3 D)
   C. integral u ds/(V D)
   D. V^2 D / integral u ds
   Answer: A. Momentum flux involves u^2, giving beta = 1.8507/1.6007 = 1.156. (Key point: Formula for beta)

4. The computed values alpha = 1.48 and beta = 1.16 show that:
   A. beta is larger than alpha
   B. Both equal 1
   C. They are less than 1
   D. alpha exceeds beta, and both exceed 1 due to non-uniform velocity
   Answer: D. Non-uniformity makes both factors greater than 1, with alpha always larger than beta. (Key point: Result ordering)

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