Question

The Reynolds number, rho VD/mu, is a very important parameter in fluid mechanics. Verify that the Reynolds number is dimensionless, using both the FLT system and the MLT system for basic dimensions, and determine its value for methane flowing at a velocity of 4 m/s through a 2-in-diameter pipe.

Answer 1

Re = 1 10⁴

Step-by-step explanation:

Reynolds number is

Re = ρ v D /μ

The units of each term are

ρ = [kg / m³]

v = [m / s]

D = [m]

μ = [Pa s]

The pressure

Pa = [N / m²] = [Kg m / s²] 1 / [m²] = [kg / m s²]

μ = [Pa s] = [kg / m s²] [s] = [kg / m s]

We substitute the units in the equation

Re = [kg / m³] [m / s] [m] / [kg / m s]

Re = [kg / m s] / [m s / kg]

RE = [ ]

Reynolds number is a scalar

Let's evaluate for the given point

Where the data for methane are:

viscosity μ = 11.2 10⁻⁶ Pa s

the density ρ = 0.656 kg / m³

D = 2 in (2.54 10⁻² m / 1 in) = 5.08 10⁻² m

Re = 0.656 4 2 5.08 10⁻² /11.2 10⁻⁶

Re = 1.19 10⁴