Fourier number: Definition, Formula, Significance (heat transfer)

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.

In this article, we’re going to discuss:

• Fourier number equation:
• Significances of Fourier number in heat transfer:
• Applications:
• Examples with solutions:

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:-

1. It is used in the analysis of transient/ unsteady state conduction.
2. 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.

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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.