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:-
Lc = 8 x 10-3
α = 2.3 x 10-5 m²/s
Bi = 0.015
t = 60 seconds
T∞ = 350 K
Ti = 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}}`
Fo = 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:-
Lc = 1 x 10-2
α = 2 x 10-5 m²/s
Bi = 12 x 10-3
T∞ = 300 K
Ti = 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}`
Fo = 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.
