# Prandtl number: Definition, Significance, Equation, Examples [with Pdf]

Contents:

## What is the prandtl number?

Prandtl number is defined as the ratio of kinematic viscosity (v) to the thermal diffusivity (α). Prandtl number is denoted by the symbol Pr.

By definition of the Prandtl number,

## Prandtl number formula:

The prandtl number is given by,

Pr=\frac{\upsilon }{\alpha }

Where equations of kinematic viscosity (V) and thermal diffusivity (α) are given by,

\nu =\frac{\mu }{\rho } and \alpha =\frac{K}{\rho \times C_{P}}

Put these values in the equation of Pr,

\therefore Pr=\frac{\frac{\mu }{\rho }}{\frac{K}{\rho \times C_{P}}}

Pr=\frac{\mu C_{P}}{K}

Where,
μ = Dynamic viscosity
Cp = specific heat of the fluid
K = thermal conductivity

In the case of boundary layer flow, the Prandtl number is given by,

Pr=(\frac{\delta }{\delta t})^{3}

Where,
δ = hydrodynamic boundary layer thickness
δt = thermal boundary layer thickness

The value for the Prandtl number depends on the properties of fluids at given working conditions.

## Physical significance of prandtl number in heat transfer:

The significance of Prandtl number in heat transfer are as follows:

1) The number gives the relationship between kinematic viscosity and thermal diffusivity of the fluid
2) In the case of boundary layer flow, the Prandtl number relates the velocity boundary layer thickness and thermal boundary layer thickness.
3) The smaller value of the Prandtl number indicates, the rate of thermal diffusion is higher than momentum diffusion. And in this case, δth > δH
4) The higher value of the Prandtl number indicates the rate of thermal diffusion is lower than the rate of momentum diffusion. In this case, δth < δH

## Prandtl number solved examples:

1) The water is flowing over the heated plate. The water has prandtl number of 6. Find the relation between velocity boundary layer thickness and thermal boundary layer thickness.

Solution:

Given: Pr = 6

The Prandtl number is given by,

Pr=(\frac{\delta }{\delta th})^{3}

6=(\frac{\delta }{\delta th})^{3}

1.817=\frac{\delta }{\delta th}

δ = (1.817) δth

Therefore the velocity boundary layer thickness is 1.817 times of thermal boundary layer thickness.

2) For the fluid passing over the heated plate the hydrodynamic boundary layer thickness at a certain point is 1.5 m. find the thickness of the thermal boundary layer if the fluid has the following properties.
Dynamic viscosity = 0.001 Pa.s
Specific heat, Cp = 1.2 KJ/Kg.K
Thermal conductivity, K = 1.1 W/m.K

Solution:

Given:
μ = 0.001 Pa.s
Cp = 1.2 KJ/Kg.K = 1.2 × 10³ J/Kg.K
K= 1.1 W/m.K
δ = 1.5 mm

The Prandtl number for the given fluid is given by,

Pr=\frac{\mu C_{P}}{K}

Pr = 1.09

Now the thickness of the thermal boundary layer is given by,

Pr=(\frac{\delta }{\delta th})^{3}

1.09=(\frac{1.5}{\delta _{th}})^{3}

δth = 1.457 mm 