Contents:

## What is the Fourier number?

**The Fourier number is the dimensionless quantity used in the calculation of unsteady-state heat transfer. The Fourier number is the ratio of the rate of heat conduction to the rate of heat stored in a body. It is derived from the non-dimensionalization of the heat conduction equation.**

## Fourier number equation:

The Fourier number for heat transfer is given by,

`F_{O}=\frac{\alpha\ \tau }{L_{C}^{2}}`

Where,

α = Thermal diffusivity

𝜏 = Time (second)

`L_{c} = \text{Characteristics length} = \frac{\text{Volume}}{\text{Area}}`

The Fourier number in the mass transfer is given by,

`F_{O}=\frac{D. \tau }{L_{C}^{2}}`

Where D = Mass diffusivity

## Significances of Fourier number in heat transfer:

The significances are as follows:-

1] The Fourier number indicates the relation between the rate of heat conduction through the body and the rate of heat stored in the body.

2] The larger value of the Fourier number indicates, a higher rate of heat transfer through the body.

3] The lower value of the Fourier number indicates the lower rate of heat transfer through the body.

## Applications:

The applications are as follows:-

- It is used in the analysis of transient/ unsteady state conduction.
- It is also used in the analysis of transient mass transfer systems.

## Examples with solutions:

1] A hot object at an initial temperature of 700 K is dipped into the fluid which is at 350 K. The whole system has a biot number of 0.015.

Find the temperature of the object after the interval of 60 seconds. The necessary data for the analysis is given below:-

(Characteristic length: Lc = 8 x 10^{-3}

Thermal diffusivity, α = 2.3 x 10^{-5} m²/s)

**Solution:-**

**Given:-**L

_{c}= 8 x 10

^{-3}

α = 2.3 x 10

^{-5}m²/s

B

_{i}= 0.015

t = 60 seconds

T

_{∞}= 350 K

T

_{i}= 700 K

**1] Fourier number:-**

The Fourier number is given by,

`F_{o}=\frac{\alpha.t}{L_{C}^{2}}`

`F_{o}=\frac{(2.3\times 10^{-5})\times 60}{(8\times 10^{-3})^{2}}`

F_{o} = 21.56

**2] Temperature of the object after the interval of 60 seconds using lumped system analysis**:-

`\frac{T-T_{\infty}}{T_{i}-T_{\infty}}=e^{-Bi.F_{o}}`

`\frac{T-350}{700-350}=e^{-0.015\times 21.56}`

**T = 603.28 K**

∴ The temperature of the object after the time interval of 60 seconds is 603.28 K.

2] The object with characteristics length of (1 x 10^{-2}) & a temperature of 800 K is kept in cold fluid at a temperature of 300 K.

Calculate the time interval to become the temperature of the object equal to 500 K. [Assume α = 2 x 10^{-5} m²/s, Bi = 12x 10^{-3}]

**Solution:-**

**Given:-**L

_{c}= 1 x 10

^{-2}

α = 2 x 10

^{-5}m²/s

B

_{i}= 12 x 10

^{-3}

T

_{∞}= 300 K

T

_{i}= 800 K

T = 500 K

**1] Fourier number using a method of lumped system analysis**:-

`\frac{T-T_{\infty}}{T_{i}-T_{\infty}}=e^{-Bi.F_{o}}`

`\frac{500-300}{800-300}=e^{-0.015\times Fo}`

F_{o} = 76.35

**2] Time interval to reach temperature of the object to 500 K**:-

`F_{o}=\frac{\alpha.t}{L_{C}^{2}}`

`76.35=\frac{(2\times 10^{-5})t}{(1\times 10^{-2})^{2}}`

**t = 381.78 Seconds**

Therefore the time required to attain a temperature of the object equal to 500 K is 381.78 Seconds.