# Rayleigh number: Definition, Significance, Formula, Example [with Pdf]

In this article, we’re going to discuss:

• What is Rayleigh number?
• Rayleigh number significance:
• Numericals on Rayleigh number:

## What is Rayleigh number?

Rayleigh number is defined as the product of grashof number and Prandtl number. The rayleigh number is denoted by the symbol Ra.

As per the definition of rayleigh number,

Where equation for the grashof number is,

G_{r}=\frac{\rho ^{2}gL^{3}\beta \Delta T}{\mu ^{2}}

And the equation for the Prandtl number is,

P_{r}=\frac{\mu C_{p}}{K}

Therefore by putting the equations of Gr and Pr in the above equation of Ra.

Pr = \frac{\rho ^{2}gL^{3}\beta \Delta T}{\mu ^{2}} × \frac{\mu C_{p}}{K}

Pr = \frac{\rho ^{2}gL^{3}\beta \Delta TC_{p}}{\mu K}

Pr = \frac{gL^{3}\beta \Delta T}{(\frac{\mu}{\rho})(\frac{K}{\rho C_{P}})}

Put \frac{\mu}{\rho}=\nu and \frac{K}{\rho C_{P}}=\alpha  in the above equation

Pr = \frac{gL^{3}\beta \Delta T}{\nu \alpha}

This is the required equation for the rayleigh number.

Where,
L = Length
β = Coefficient of volumetric expansion
∆T = Temperature difference
v = Kinematic viscosity
α = Thermal diffusivity

## Rayleigh number significance:

The significances of Rayleigh number are listed below,

1) The rayleigh number states the relation between grashof number and the Prandtl number.
2) Rayleigh number also helps to find the type of fluid flow,
If Ra < 10^9, then the flow is laminar and If Ra > 10^9 then the flow is turbulent.

## Numericals on Rayleigh number:

1) Check the flow for laminar or turbulent,
The properties of fluids are,
Length, = 5 m
Surface temperature, t_{s} = 600 K
Fluid temperature, t_{f} = 400 K
Kinematic viscosity (ν)= 0.2 × 10^(-3) m²/s
Thermal diffusivity (α) = 0.14 × 10^(-5) m²/s

Solution:

Value for coefficient of expansion β is given by,

\beta =\frac{2}{T_{s}+T_{f}}=\frac{2}{600+400}

\beta = 2\times 10^{-3}

Now the rayleigh number is given by,

R_{a}=\frac{gL^{3}\beta (t_{s}-t_{f})}{\nu \alpha }

R_{a}=\frac{9.81\times 3^{3}(2\times 10^{-3}) (600-400)}{(2\times 10^{-3})(0.14\times 10^{-5})}

R_{a}=1.75\times 10^{12}

Here Ra > 10^{9}, therefore the flow is turbulent.

2) In the case of free convection, if the grashoff number is 3×10^{11} and rayleigh number is, 1.8 × 10^{12}, then find the valve for thermal diffusivity if the fluid has kinematic viscosity, v = 0.25 × 10^{-3} m²/s.

Solution:

Given:
Gr = 3× 10^{11}
Ra = 1.8 × 10^{12}
Kinematic viscosity, v = 0.25 × 10^{-3} m²/s

The equation of grashoff number is,

G_{r}=\frac{gL^{3}\beta \Delta T}{\nu ^{2}} — Eq.1

And rayleigh number is given by,

R_{a}=\frac{gL^{3}\beta \Delta T}{\nu \alpha } — Eq.2

Dividing equation 1 by the equation 2, we get,

\frac{G_{r}}{R_{a}}=\frac{\frac{gL^{3}\beta \Delta T}{\nu ^{2}}}{\frac{gL^{3}\beta \Delta T}{\nu \alpha }}

\frac{G_{r}}{R_{a}}=\frac{\alpha }{\nu }

\frac{3\times 10^{11}}{1.8\times 10^{12}}=\frac{\alpha }{0.25\times 10^{-3} }

Thermal diffusivity = α = 4.16 × 10^{-5} m²/s