A bar of length L of uniform cross section is perfectly insulated over its lateral surface. The bar is initially at a constant temperature of T0 throughout. For t 0, the two ends of the bar are maintained at zero temperature. Obtain the solution to the problem using Eqs. 5.74 and 5.75. Discuss th

Question

A bar of length L of uniform cross section is perfectly insulated over its lateral surface. The bar is initially at a constant temperature of [latex] T_{0}[/latex] throughout. For t > 0, the two ends of the bar are maintained at zero temperature. Obtain the solution to the problem using Eqs. 5.74 and 5.75. Discuss the nature of the solution specifically with respect its dependence on time.

[latex]A_{n} =\frac{2}{L} \int_{0}^{L}{f(x)sin \left(\frac{n\pi x}{L} \right) } dx[/latex] (5.74)

[latex]T(x,t)=\sum\limits_{1}^{\infty }{e^{-\frac{\alpha n^{2}\pi ^{2} t }{L^{2} } } } \left[\frac{2}{L}\int_{0}^{L}{f(x)sin\left(\frac{n\pi x}{L} \right) dx } \right] sin\left(\frac{n\pi x}{L} \right)[/latex] (5.75)

Accepted Answer

Step 1 In this case [latex] f (x) = T_{0}[/latex] in the interval 0 < x < L. The Fourier coefficients are obtained using Eq. 5.74 as

[latex]A_{n} =\frac{2}{L} \int_{0}^{L}{T_{0} sin \left(\frac{n\pi x}{L} ight) dx}= \frac{2T_{0}}{L} \left\{\frac{-\cos \left(\frac{n\pi }{L} ight) }{\frac{n\pi }{L} } ight\} \mid ^{L}_{0} =\frac{2T_{0} }{n\pi } [1-\cos (n\pi )][/latex]

But cos(nπ) = −1 for odd values of n and cos(nπ) = 1 for even values of n. Hence, [latex] A_{n}[/latex] will be zero whenever n is even and will be [latex]\frac{4T_{0} }{n\pi }[/latex] whenever n is odd. The solution to the temperature field in the bar then becomes, using Eq. 5.75

[latex]T(x,t)=\frac{4T_{0} }{\pi }\sum\limits_{n=1,3,5...}^{\infty }{\frac{sin \left(\frac{n\pi x}{L} ight) }{n} } e^{-\frac{\alpha n^{2}\pi ^{2}t }{L^{2} } }[/latex] (5.76)

Step 2 Solution (5.76) may be recast in nondimensional form by introducing the following nondimensional variables:

Non-dimensional temperature: [latex] heta =\frac{T(x,t)}{T_{0}}[/latex]
Non-dimensional x co-ordinate: [latex]ξ = \frac{x}{L}[/latex]
Non-dimensional time or Fourier number: [latex]F o=\frac{\alpha t}{L^{2}}[/latex]

Then Eq. 5.76 takes the form

[latex] heta (\xi ,F o )=\sum\limits_{n=1,3,5...}^{\infty }\frac{sin (n\pi \xi )}{n\pi } e^{-n^{2} \pi ^{2}F o }[/latex] (5.77)

Step 3 In order to discuss the behavior of the solution we consider the mid-plane temperature (i.e., at [latex]x = \frac{L}{2}[/latex] or ξ = 0.5) variation with Fourier number. Since the result (5.77) is in its nondimensional form, it is applicable to any material. We compare a one-term approximation with the exact in Fig. 5.14. The one-term approximation truncates the series with just the first term and hence is given by

[latex] heta (\xi ,F o )\approx \frac{sin (\pi \xi )}{\pi } e^{-\pi ^{2} F o }[/latex]

It is seen from the plot that a one-term approximation is an excellent representation of the solution for Fo > 0.04 or thereabouts. The one-term approximation is also referred to as the fully developed profile since the shape of the profile in the interval 0 < ξ < 1 remains sinusoidal (and hence invariant) with the amplitude alone varying with Fo.

The Fourier number may be interpreted as the ratio of actual time and a reference or a characteristic time, determined by the size and the material property, the thermal diffusivity. Thus [latex] F o=\frac{t}{t_{ch} }[/latex] where [latex] t_{ch}=\frac{L^{2} }{\alpha }[/latex] is a characteristic time determined by the physical size L of the bar and thermal diffusivity α of the material. The state of affairs for Fo < 0.04 may be termed as the short time solution, i.e., solution for time much smaller than the characteristic time. In this region all or many terms in the Fourier summation contribute significantly to the solution, for a given ξ. However, when the time is larger than the characteristic time the first term alone contributes to the solution. This is because of the fact that exponential terms involving [latex] e^{-{ n^{2}\pi ^{2} F o }} \approx e^{-{10 n^{2} F o }}[/latex] fall off rapidly.

Question 5.5: A bar of length L of uniform cross section is perfectly insu...

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