---
title: "Energy and momentum coefficients, power-law profile"
subject: "Hydraulics"
chapter: 7
key: "semester_4/hydraulics:velocity-distribution-power-law-alpha-beta"
source: https://www.bctengineer.com/q/semester_4/hydraulics/velocity-distribution-power-law-alpha-beta
site: BCT Engineering
---

# Energy and momentum coefficients, power-law profile

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

- Asked in: 2072 Magh, 5 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/hydraulics/velocity-distribution-power-law-alpha-beta
- Whole chapter: https://www.bctengineer.com/md/semester_4/hydraulics/chapter-7

## Question

The velocity distribution in a channel section may be approximated by the equation $u = u_0 (d/d_0)^n$ in which u is the flow velocity at depth d; $u_0$ is the flow velocity at depth $d_0$ and n = a constant. Derive expression for the energy and momentum coefficient.

## Short answer

For u = u₀(d/d₀)ⁿ, the energy correction factor α = (n+1)³/(3n+1) and momentum coefficient β = (n+1)²/(2n+1).

## Answer

Let $d$ be the height above the bed, $d_0$ the full flow depth and $u_0$ the velocity at $d = d_0$ (the surface). Take a unit width, so $dA = dd$ and $A = d_0$.

**Mean velocity**

$$
V = \frac{1}{d_0}\int_0^{d_0} u_0\left(\frac{d}{d_0}\right)^n dd = \frac{u_0}{d_0}\cdot\frac{d_0}{n+1} = \frac{u_0}{n+1}
$$

**Energy coefficient** (kinetic energy correction factor)

$$
\alpha = \frac{\int u^3\,dA}{V^3 A}
$$

$$
\int_0^{d_0} u_0^3\left(\frac{d}{d_0}\right)^{3n} dd = \frac{u_0^3\,d_0}{3n+1}
$$

$$
\alpha = \frac{u_0^3\,d_0/(3n+1)}{\left[u_0/(n+1)\right]^3 d_0} = \frac{(n+1)^3}{3n+1}
$$

**Momentum coefficient** (Boussinesq coefficient)

$$
\beta = \frac{\int u^2\,dA}{V^2 A}
$$

$$
\int_0^{d_0} u_0^2\left(\frac{d}{d_0}\right)^{2n} dd = \frac{u_0^2\,d_0}{2n+1}
$$

$$
\beta = \frac{u_0^2\,d_0/(2n+1)}{\left[u_0/(n+1)\right]^2 d_0} = \frac{(n+1)^2}{2n+1}
$$

**Answer:**

$$
\boxed{\alpha = \frac{(n+1)^3}{3n+1},\qquad \beta = \frac{(n+1)^2}{2n+1}}
$$

Check: for uniform velocity ($n = 0$), $\alpha = \beta = 1$. For $n = 1/7$, $\alpha = \dfrac{(8/7)^3}{10/7} = 1.045$ and $\beta = \dfrac{(8/7)^2}{9/7} = 1.016$, typical of real channels.

## Check yourself

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

1. For u = u0 (d/d0)^n, the mean velocity V is:
   A. u0/(n+1)
   B. u0 (n+1)
   C. u0/n
   D. u0/(2n+1)
   Answer: A. V = (1/d0) integral of u0 (d/d0)^n dd = u0/(n+1). (Key point: Mean velocity)

2. The energy coefficient alpha is defined as:
   A. integral u^2 dA / (V^2 A)
   B. integral u dA / (V A)
   C. integral u^3 dA / (V^3 A)
   D. V^3 A / integral u^3 dA
   Answer: C. Kinetic energy flux uses u^3, so alpha = integral u^3 dA / (V^3 A). (Key point: Energy coefficient definition)

3. The result for alpha is:
   A. (n+1)^2/(2n+1)
   B. (3n+1)/(n+1)^3
   C. (n+1)/(3n+1)
   D. (n+1)^3/(3n+1)
   Answer: D. Integral of u^3 gives u0^3 d0/(3n+1); dividing by V^3 d0 gives (n+1)^3/(3n+1). (Key point: Result for alpha)

4. beta = (n+1)^2/(2n+1). What check confirms the formulae?
   A. For n = 0 (uniform velocity), alpha = beta = 1
   B. For n = 0, beta = 0
   C. For n = 1, alpha = 1
   D. beta is always greater than alpha
   Answer: A. Uniform velocity gives no correction, so both coefficients are 1 at n = 0. (Key point: Result for beta and check)

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