Fourier number: Definition, Formula, Significance, Application, Examples [with Pdf]

What is fourier number?

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)
Lc = Characteristics length
Lc = `\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

Fourier number significance:

The significances of the Fourier number 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, the higher rate of heat transfer through the body.

3] The larger value of the fourier number indicates,the lower rate of heat transfer through the body.

Fourier number application:

The applications of the fourier number are as follows:-

  1. The Fourier number is used in the analysis of transient/ unsteady state conduction.
  2. The fourier number is also used in analysis transient mass transfer system.

Fourier number examples:

1] A hot object at an initial temperature of 700 K is dipped into the fluid which is at 350 K. The 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_{\infty }` = 350 K
`t_{i}` = 700 K

1] Fourier number:-

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] Now temperature of the object after the interval of 60 seconds is given by, 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 602.82 K.

2] The object with characteristics length of (1 x `10^{-2}`) & 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\times 10^{-5}` m²/s, Bi = `12 \times 10^{-3}`]

Solution:-

Given:-
`L_{c}` = 1 x `10^{-2}`
α = 2 x `10^{-5}` m²/s
`B_{i}` = 12 x `10^{-3}`
`t_{\infty }` = 300 K
`t_{i}` = 800 K
T = 500 K

1] Find the fourier number by using the 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] The time interval to reach the temperature of object to 500 K is given by,

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

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

t = 381.75 Seconds

Therefore the time required to attain a temperature of the object equal to 500 K is 381.75 Seconds.

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