---
title: "Turbulence intensity and inlet k, epsilon"
subject: "Applied Computational Fluid Dynamics"
chapter: 5
key: "semester_7/applied_computational_fluid_dynamics:turbulence-intensity-inlet-numerical"
source: https://www.bctengineer.com/q/semester_7/applied_computational_fluid_dynamics/turbulence-intensity-inlet-numerical
site: BCT Engineering
---

# Turbulence intensity and inlet k, epsilon

Applied Computational Fluid Dynamics · Chapter 5: Turbulence Modeling

- Asked in: year not given, 5 marks
- Video: [OpenFOAM Tutorial ( Chap 5 - Section 2): Turbulence Field: k, epsilon, nut  Boundary Conditions](https://www.youtube.com/watch?v=gnoJDPDHESA) — Abecator CFD | Engineering Simulation Academy
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_7/applied_computational_fluid_dynamics/turbulence-intensity-inlet-numerical
- Whole chapter: https://www.bctengineer.com/md/semester_7/applied_computational_fluid_dynamics/chapter-5

## Question

Eight instantaneous velocity readings (m/s) taken at a point in a turbulent duct flow are 10.2, 9.8, 10.5, 9.6, 10.1, 10.4, 9.7 and 10.3. (a) Find the mean velocity, the rms fluctuation and the turbulence intensity. (b) For a CFD inlet of a 0.2 m diameter pipe with mean velocity 12 m/s and turbulence intensity 5%, estimate the turbulent kinetic energy $k$, the turbulence length scale (0.07 D), the dissipation rate $\varepsilon$ and the specific dissipation rate $\omega$ (take $C_\mu=0.09$).

## Short answer

Mean velocity = 10.08 m/s, rms fluctuation = 0.315 m/s, turbulence intensity = 3.1%. Inlet k = 1.5(UI)² = 0.54 m²/s², length scale ℓ=0.014 m, ε = C_μ^{3/4} k^{3/2}/ℓ = 4.66 m²/s³, ω = ε/(C_μ k) = 95.8 s⁻¹.

## Answer

#### (a) Statistics
Mean:
$$
\bar u=\frac{10.2+9.8+10.5+9.6+10.1+10.4+9.7+10.3}{8}=\frac{80.6}{8}=10.075\ \text{m/s}
$$
Fluctuations $u'=u-\bar u$: 0.125, -0.275, 0.425, -0.475, 0.025, 0.325, -0.375, 0.225.

Sum of squares $=0.0156+0.0756+0.1806+0.2256+0.0006+0.1056+0.1406+0.0506=0.795$
$$
u'_{rms}=\sqrt{\frac{0.795}{8}}=0.315\ \text{m/s}
$$
$$
I=\frac{u'_{rms}}{\bar u}=\frac{0.315}{10.075}=0.0313=3.1\%
$$
(If the fluctuations were isotropic, $k\approx\tfrac32u'^2_{rms}=0.149$ m²/s².)

#### (b) Inlet turbulence values
$$
k=\tfrac32(UI)^2=1.5\,(12\times0.05)^2=1.5\times0.36=0.54\ \text{m}^2/\text{s}^2
$$
Length scale: $\ell=0.07D=0.07\times0.2=0.014$ m.

Dissipation rate:
$$
\varepsilon=C_\mu^{3/4}\frac{k^{3/2}}{\ell}=\frac{0.09^{0.75}\times0.54^{1.5}}{0.014}=\frac{0.1643\times0.3968}{0.014}=4.66\ \text{m}^2/\text{s}^3
$$
Specific dissipation rate:
$$
\omega=\frac{\varepsilon}{C_\mu k}=\frac{4.66}{0.09\times0.54}=95.8\ \text{s}^{-1}
$$
(Equivalent: $\omega=\sqrt k/(C_\mu^{1/4}\ell)=0.735/(0.5477\times0.014)=95.8$.)

**Answer:** $\bar u=10.08$ m/s, $u'_{rms}=0.315$ m/s, $I=3.1\%$; inlet $k=0.54$ m²/s², $\varepsilon=4.66$ m²/s³, $\omega=95.8$ s$^{-1}$.

## Check yourself

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

1. For the eight readings, what are the mean and rms fluctuation?
   A. 10.075 m/s and 0.315 m/s
   B. 10.075 m/s and 0.795 m/s
   C. 10.0 m/s and 0.315 m/s
   D. 9.8 m/s and 0.1 m/s
   Answer: A. Mean = 80.6/8 = 10.075; rms = sqrt(0.795/8) = 0.315 m/s. (Key point: Mean and rms)

2. What is the measured turbulence intensity?
   A. 31%
   B. 0.315%
   C. 3.1%
   D. 5%
   Answer: C. I = u'_rms/mean = 0.315/10.075 = 0.0313, i.e. 3.1%. (Key point: Turbulence intensity)

3. For U = 12 m/s and I = 5%, which formula and value give k?
   A. k = (UI)^2 = 0.36
   B. k = 1.5 (UI)^2 = 0.54 m^2/s^2
   C. k = 1.5 U I = 0.9
   D. k = 0.5 (UI)^2 = 0.18
   Answer: B. k = 1.5 (12 x 0.05)^2 = 1.5 x 0.36 = 0.54 m^2/s^2. (Key point: Inlet k)

4. How are epsilon and omega estimated from k and the length scale l = 0.014 m?
   A. epsilon = Cmu^(3/4) k^(3/2)/l = 4.66; omega = epsilon/(Cmu k) = 95.8
   B. epsilon = k/l; omega = k/epsilon
   C. epsilon = Cmu k^2/l; omega = 4.66
   D. epsilon = Cmu^(3/4) k^(3/2)/l = 4.66; omega = epsilon/k = 8.6
   Answer: A. epsilon = 0.1643 x 0.3968/0.014 = 4.66, and omega = epsilon/(Cmu k) = 4.66/(0.09 x 0.54) = 95.8 1/s. (Key point: Epsilon and omega)

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