A very long bar of a material of thermal conductivity k = 2.5 W/m K has a square cross section of side L = l = 0.1 m. Heat is generated at a uniform rate of G = 10^4 W/m^3 inside the bar. All the surfaces of the bar are maintained at a temperature of T = 0. Determine the temperature at x = y = 0; x

Question

A very long bar of a material of thermal conductivity k = 2.5 W/m K has a square cross section of side L = l = 0.1 m. Heat is generated at a uniform rate of [latex] G = 10^{4} W/m^{3}[/latex] inside the bar. All the surfaces of the bar are maintained at a temperature of T = 0. Determine the temperature at x = y = 0; x = 0.05, y = 0; x = 0, y = 0.05; and x = 0.05, y = 0.05 where all lengths are in m.

Accepted Answer

Evaluation of the desired temperatures are based on the solution presented above. With the data specified in the problem, reference temperature due to heat generation is calculated as

[latex]T_{ref} =\frac{GL^{2} }{2k}=\frac{10^{4} imes 0.1^{2} }{2 imes 2.5}=20k[/latex]

The function [latex]\phi (x)[/latex] is then given by Eq. 5.94 as

[latex]\phi (x)=\frac{GL^{2} }{2k}\left[1-\left(\frac{x}{L} ight) ^{2} ight] [/latex] (5.94)

[latex]\phi (x)=20\left[1-\left(\frac{x}{0.1} ight)^{2} ight][/latex]

The temperatures required to be calculated at different locations specified in the problem require the summation of Fourier series to obtain [latex]ψ(x, y)[/latex]. Consider the location x = y = 0 that is at the center of the square cross section. Both the cos and cosh functions are unity at this point. Hence the function [latex]ψ(0, 0) = \sum\limits_{0}^{\infty }{a_{n} }[/latex] where an are given by Eq. 5.98. For convergence to 5 digits after the decimal point 3 terms in the series are adequate. These are shown in the table below

[latex]a_{n}=-\frac{4(-1)^{n} }{(\lambda _{n}L )^{3} } \frac{GL^{2} }{2k} \frac{1}{\cos h(\lambda _{n}l )}[/latex] (5.98)

n	[latex]\lambda _{n}L[/latex]	[latex]a_{n}[/latex]
0	1.570796	-8.226191
1	4.712389	0.013734
2	7.853982	-0.000128
Sum of three terms=−8.21258

Hence the temperature at x = 0, y = 0 is obtained as

[latex]T (0, 0) = ψ(0, 0) + \phi(0) = −8.21258 + 20 = 11.787417 ≈ 11.79 °C[/latex]

Similarly temperatures at other locations specified in the problem are obtained. The cosine and hyperbolic cosine factors are to be used appropriately. As an example, the calculations are shown for x = y = 0.05 in the following tabulation:

n	[latex]\lambda _{n}L[/latex]	[latex]a_{n}[/latex]	Fourier terms
0	1.570796	-8.226191	-7.704980
1	4.712389	0.013734	-0.051691
2	7.853982	-0.000128	0.002301
3	10.995574	0.000002	0.000174
4	14.137167	0.000000	-0.000017
5	17.278760	0.000000	-0.000002
Sum of five terms = −7.75421

Hence the temperature at x = 0.05, y = 0.05 is obtained as
[latex]T (0.05, 0.05) =ψ(0.05, 0.05) + \phi(0.05)[/latex]

[latex]=-7.75421+2\left[1-\left(\frac{0.05}{0.1} ight)^{2} ight] \approx 7.25^{\circ } C[/latex]

It is left as an exercise to the reader to show that the temperatures at the other two points specified in the problem are the same and equal to 9.17°C. We show in Fig. 5.19a, b the temperature variations along two directions. First direction is along x, y = 0 or y, x = 0. The second direction is along direction x = y that passes through origin. Note that the coordinates are non dimensionalized with L as the characteristic length. r in Fig. 5.19b is defined as [latex]r=\sqrt{x^{2} +y^{2} }=x\sqrt{2} =y\sqrt{2}[/latex]

Question 5.7: A very long bar of a material of thermal conductivity k = 2....

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