Question

Determine the Reynolds number for a flow of 0.2 m^3/s through a 203 mm inner diameter circular pipe of a fluid with rho=680 kg/m^3 ,μ=3.1*10^(-4) ((N-s)/m^2 ),and σ=0.022 N/m.Is the flow laminar or turbulent?

Answer 1

To find a solution to this problem it is necessary to apply the concepts related to the Reynolds number and its definitions on the type of fluid.

A Reynolds number less than 2000 considers the laminar fluid, while a Reynolds number greater than 4000 is considered a turbulent fluid. (The intermediate between the two values would be a transient fluid)

The mathematical equation that defines the Reynolds number is given by

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

Where

$\rho =$ Density

V= Velocity

D= Diameter

$\mu =$ Viscosity

Our values are given as

$Q = 0.2m^3/s$

$D = 203*10^{-3}m$

$\rho = 680kg/m^3$

$\mu = 3.1*10^{-4}Ns/m^2$

$\sigma = 0.022N/m$

The velocity can be find through the Discharge equation,

Q = VA

Where

V = Velocity

A = Area

Replacing,

$0.2 = V* (2\pi*(\frac{203*10^{-3}}{2})^2)$

$V = 3.08m/s$

Replacing at the Reynolds equation,

$Re = \frac{\rho VD}{\mu}$

$Re = \frac{680*3.08*203*10^{-3}}{3.1*10^{-4}}$

$Re = 1.37*10^6$

Since Reynolds' number is greater than 4000, then we consider this a turbulent fluid.