The image above represents reynold’s number.

To compute for reynold’s number, four essential parameters are needed and these parameters are **Fluidization at Minimum Velocity (u _{mf}), Diameter of Bed (d), Density (ρ)** and

**Viscosity (μ).**

The formula for calculating reynold’s number:

Re’_{mf} = ^{umfdρ}/_{μ}

Where:

Re’_{mf} = Reynold’s Number at Minimum Fluidization

u_{mf} = Fluidization at Minimum Velocity

d = Diameter of Bed

ρ = Density

μ = Viscosity

Let’s solve an example;

Find the reynold’s number at minimum fluidization when the fluidization at minimum is 14, the diameter of bed is 8, the density is 12 and the viscosity is 2.

This implies that;

u_{mf} = Fluidization at Minimum Velocity = 14

d = Diameter of Bed = 8

ρ = Density = 12

μ = Viscosity = 2

Re’_{mf} = ^{umfdρ}/_{μ}

Re’_{mf} = ^{(14)(8)(12)}/_{2}

Re’_{mf} = ^{(1344)}/_{2}

Re’_{mf} = 672

Therefore, the **reynold’s number at minimum fluidization **is **672.**

**Calculating the Fluidization at Minimum Velocity when the Reynold’s Number at Minimum Fluidization, the Diameter of Bed, the Density and the Viscosity is Given.**

u_{mf} = ^{Re’mf x μ} / _{dρ}

Where;

u_{mf} = Fluidization at Minimum Velocity

Re’_{mf} = Reynold’s Number at Minimum Fluidization

d = Diameter of Bed

ρ = Density

μ = Viscosity

Let’s solve an example;

Find the fluidization at minimum velocity when the reynold’s number at minimum fluidization is 12, the diameter of bed is 4, the density is 8 and the viscosity is 3.

This implies that;

Re’_{mf} = Reynold’s Number at Minimum Fluidization = 12

d = Diameter of Bed = 4

ρ = Density = 8

μ = Viscosity = 3

u_{mf} = ^{Re’mf x μ} / _{dρ}

u_{mf} = ^{12 x 3} / _{(4)(8)}

u_{mf} = ^{36} / _{32}

u_{mf} = 1.125

Therefore, the **fluidization at minimum velocity **is **1.125.**