ChemicalFluid FloweasyFE ExamCoursework

Reynolds Number for a Viscous Liquid

An oil with density $\rho = 900\ \text{kg/m}^3$ and dynamic viscosity $\mu = 2.0\times10^{-3}\ \text{Pa·s}$ flows at $V = 1.5\ \text{m/s}$ through a pipe of internal diameter $D = 50\ \text{mm}$ .

Find the Reynolds number and state the flow regime.

Given

ρ	900 kg/m^3	Density
V	1.5 m/s	Mean velocity
D	50 mm	Internal diameter
μ	0.002 Pa*s	Dynamic viscosity

Hint 1

Reynolds number is the ratio of inertial to viscous forces.

Hint 2

Re = ρVD/μ, with all quantities in base SI.

Hint 3

Convert 50 mm to 0.05 m before substituting.

Worked solution — try the problem first

$Re = \frac{\rho V D}{\mu}$

Convert the diameter to metres — the only conversion needed:

$D = 50\ \text{mm} = 0.050\ \text{m}$

$Re = \frac{900 \times 1.5 \times 0.050}{2.0\times10^{-3}} = \frac{67.5}{2.0\times10^{-3}} = 33{,}750$

$\boxed{Re = 33{,}750 \ \text{(turbulent)}}$

Regime. For pipe flow, $Re < 2300$ is laminar, $2300 < Re < 4000$ transitional, and $Re > 4000$ turbulent. At 33,750 this is firmly turbulent.

Why it matters. The regime selects the correlation you may use — friction factor, heat transfer coefficient, mixing behaviour all change. Applying a laminar correlation such as $f = 64/Re$ here would be badly wrong.

Reynolds number is dimensionless. If your working leaves units behind, an algebra error has crept in.

Concepts:Reynolds numberFlow regimesDimensionless groups

Sign in to submit

Grading needs an account so your progress, attempts, and daily quota can be tracked. The problem statement and worked solution stay open to everyone.

Create an account