# Derivation Of Reynolds Number

Derivation of Reynolds number is based on various definitions which are explained below.

## What is Reynolds Number?

Reynolds number which referred to as Re is a dimensionless quantity which is used in fluid mechanics to predict the flow patterns in different fluid flow conditions. It is defined as the ratio of inertial force to viscous force within the fluid. The mathematical representation is given as:

$$R_{e}=\frac{\rho uL}{\mu }=\frac{uL}{\nu }$$

Where,

• ⍴ is the density of the fluid in kg.m-3
• 𝜇 is the dynamic viscosity of the fluid in N.s.m-2
• 𝜈 is the kinematic viscosity of the fluid in m2.s-1
• u is the velocity of a fluid with respect to the object in m.s-1
• L is the characteristic linear dimension in m

Using Reynolds number, it becomes easy to determine whether the fluid flow is laminar or turbulent. Following are the boundary values of a circular pipe that can be used to determine the flow pattern:

• If Re > 2300, then the flow is said to be laminar.
• If 2300 < Re < 4000, then the flow is said to be transient.
• If Re > 4000, then the flow is said to be turbulent.

### Reynolds Number Derivation

The mathematical form of the Reynolds number can be derived as follows:

Re=inertial force/viscous force

$$\frac{ma}{\tau A}$$ $$\frac{\rho V.\frac{du}{dt}}{\mu \frac{du}{dy}.A}\propto \frac{\rho L^{3}\frac{du}{dt}}{\mu \frac{du}{dy}L^{2}}$$ $$\frac{\rho L.\frac{du}{dt}}{\mu}\propto \frac{\rho u_{0}L}{\mu }$$ $$\frac{u_{0}L}{\nu }$$

Where,

• t is the time
• y is the cross-sectional position
• $$u=\frac{dx}{dt}$$ is the flow speed
• τ is the shear stress in Pa
• A is the cross-sectional area of the flow
• V is the volume of the fluid element
• L is the characteristic linear dimension in m
• uis the maximum speed of the object relative to the fluid in m.s-1
• ⍴ is the density of the fluid in kg.m-3
• 𝜇 is the dynamic viscosity of the fluid in N.s.m-2
• 𝜈 is the kinematic viscosity of the fluid in m2.s-1

Following is the Reynolds number representation for the flow in a pipe:

$$R_{e}=\frac{\rho uD_{H}}{\mu }$$ $$=\frac{uD_{H}}{\nu }$$ $$=\frac{QD_{H}}{\nu A }$$

Where,

• DH is the hydraulic diameter of the pipe in m
• Q is the volumetric flow rate in m3.s-1
• A is the pipe’s cross-sectional area in m2
• u is the mean velocity of the fluid in m.s-1
• ⍴ is the density of the fluid in kg.m-3
• 𝜇 is the dynamic viscosity of the fluid in N.s.m-2
• 𝜈 is the kinematic viscosity of the fluid in m2.s-1

### Critical Reynolds number and hydraulic diameter

Critical Reynold number of any fluid flow is defined when the flow pattern changes from laminar to turbulent. It is referred to as Recr. The critical Reynold number is not fixed as it varies from geometry to geometry and also is dependent on flow conditions.

For a circular pipe, the critical Reynolds number is given as Recr=2300 and for non-circular pipes it depends on the hydraulic diameter Dh which is defined as

$$D_{h}=\frac{4A_{c}}{P}$$

Where,

Ac is the cross-sectional area of the pipe

P is the wetted perimeter of the pipe

Hydraulic radius is defined as the cross-sectional area of the channel divided by the wetted perimeter and referred as RH.

$$R_{H}=\frac{A_{c}}{P}$$

Following is the table of hydraulic diameter of non-circular tubes and channels:

 Geometry Hydraulic diameter Circular tube $$D_{H}=\frac{4.\frac{\pi D^{2}}{4}}{\pi D}=D$$ Annulus $$D_{H}=\frac{4.\frac{\pi (D_{out}^{2}-D_{in}^{2})}{4}}{\pi (D_{out}+D_{in})}=D_{out}-D_{in}$$ Square duct $$D_{H}=\frac{4a^{2}}{4a}=a$$ Rectangular duct $$D_{H}=\frac{4ab}{2(a+b)}=\frac{2ab}{a+b}$$ Partially filled rectangular duct $$D_{H}=\frac{4ab}{2a+b}$$

### Reynolds number examples

Example 1: A Newtonian fluid with a dynamic viscosity of 0.40 n.s.m-2 and a specific gravity of 0.93 flows through a 28 mm diameter pipe with a velocity of 2.9 m.s-1.

Answer: The density can be calculated using the specific gravity like

⍴=0.93(1000kg.m-3)

=930kg.m-3

The Reynolds number can be calculated using the formula as

$$R_{e}=\frac{\rho uL}{\mu }=\frac{uL}{\nu }$$

Re=(930kg.m-3) (2.9m.s-1) (28mm) (10-3m.mm-1)/(0.38Ns.m-2)

= 198 (kg.m.s-2)/N

Since Re=198 the flow is laminar.

Example 2: A person turns on the water tap in his kitchen. Water flows out from a copper pipe with 7.00mm diameter at a velocity of 1.00m.s-1. The density of water in the pipe is 1000kg.m-3 and viscosity 0.00135 Pa.s-1. What is the Reynolds number, and what is the type of flow in the pipe?

Answer: The diameter of the pipe=7.00mm=0.007m

Using Reynolds number formula:

$$R_{e}=\frac{\rho uL}{\mu }=\frac{uL}{\nu }$$

Re=(1000kg.m-3) (1.00m.s-1) (0.007m)/0.00135 Pa.s-1

=(1000kg.m-3) (1.00m.s-1) (0.007m)/0.00135 kg.m-1.s-1

=5185

Since Re=5185 the flow is turbulent.

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