- #26

Chestermiller

Mentor

- 21,493

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Would the temperature at the interference be the average of the two temperatures? Was that guess hazardous enough? :-)

Yes. That is correct. So, for the case of two identical slabs at initially different temperatures, the temperature at the interface will be the average of the initial slab temperatures (U

_{A0}+U

_{B0})/2 for all times. What this means is that we can consider each of the two slabs separately. For example, in the case of slab B, its initial temperature is U

_{B0}, and, at time t = 0, the temperature at it surface at x = 0 is suddenly changed to (U

_{A0}+U

_{B0})/2 (and held at this value for all times). The solution to this problem is also the same as for the case of a slab twice as thick 2L, and where the temperatures at its two surfaces x = 0 and x = 2L are suddenly changed to (U

_{A0}+U

_{B0})/2 at time t = 0. Do you know how to solve this kind of problem using the product of a function of x times a function of t? From the solution to this problem, you can find out how long it takes for the temperature anywhere in the slab to approach as closely as you desire to (U

_{A0}+U

_{B0})/2. This will be the amount of time it takes for the temperatures of the two slabs to equilibrate.

Before moving on to the problem of slabs of different properties and different thicknesses, it is worthwhile and instructive to stay with the problem of two identical slabs a little longer, and consider what happens at short times. By short times, I mean before the effective thermal boundary layer δ(t) has had a chance to grow to such an extent that its thickness becomes comparable to that of the slab L. Thus, if δ(t) << L, then far from the boundary, the temperature will still be effectively at the initial temperature of the slab. Thus, the behavior will be essentially the same as if the slab thickness L were infinite. For slab B, the solution to the differential equation for short times in this case is found to be :

[tex]U=U_{B0}+\frac{(U_{A0}-U_{B0})}{2}erfc\left(\frac{x}{2\sqrt{αt}}\right)[/tex]

I hope you are familiar with the complementary error function erfc. At small values of its argument, its value approaches 1, and at values of its argument greater than about 2, its value approaches zero. So, the thermal boundary layer thickness δ is on the order of [itex]δ(t)≈4\sqrt{αt}[/itex].

I'm going to stop here and let you digest what I have said. But, I'd like to continue a little further after you say are comfortable with it.

Chet