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I have one block of material 1, and one block of material 2. They are in contact with each other, sharing an interface of area A.

Material 1 behaves as a lumped capacitance, with temperature T1(t). Material 2 has a temperature distribution of T2(x,t). At time t = 0, T1 > T2, with T2 initially being constant for all x.

How do I determine Q'(t), the rate of heat transfer from material 1 to material 2 due to conduction, in terms of T1(t) and/or T2(x,t)?

Assume that k2, the thermal conductivity of material 2, is known. (Do I need to know k1 or any other constants to solve the problem as well? If so, please let me know and include them in your answer!)

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I know that the usual equation for conductive heat transfer through a single body is

$Q'(x,t) = k*A*\frac{dT(x,t)}{dx}$

So I initially assumed that the rate of heat transfer from material 1 to material 2 was simply

$Q'(t) = k2*A*\frac{dT2(x = 0,t)}{dx}$

However, this doesn't seem quite right. At t = 0, $\frac{dT2}{dx} = 0$ (since T2 is constant for all x at the beginning), so by the formula I suggested, Q'(0) = 0. And yet T1 > T2 at t = 0, so by the laws of thermodynamics, we must have Q'(0) > 0. There's a contradiction here.

What am I not taking into account? I'm assuming there's some factor I need to add to my formula for Q'(t), but I don't know quite enough about conductive heat transfer to know what I'm missing.

Thank you!

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  • $\begingroup$ I discussed solving the heat equation with differing diffusivity in my answer to this post; does that help? $\endgroup$ – Kyle Kanos Dec 29 '19 at 3:04
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If you first consider the case where $T_1$ is constant with time and medium 2 is semi-infinite in thickness (or of finite thickness, but for relatively short times), the solution for the time-dependent temperature profile in medium 2 (Transport Phenomena, Bird, Stewart, and Lightfoot) is given by: $$\frac{T_2-T_{2,\infty}}{T_1-T_{2,\infty}}=\operatorname{erfc}\left({\frac{x}{\sqrt{4\alpha t}}}\right)$$ where erfc is the complimentary error function, $T_{2,\infty}$ is the temperature of medium 2 at time zero (and also at distances far from the boundary) and $\alpha$ is the thermal diffusivity. From this equation, it follows that the heat flux at the boundary between the two media is given by $$q\equiv-k\frac{\partial T_2}{\partial x}=\frac{k}{\sqrt{\pi}\alpha t}(T_1-T_{2,\infty})$$

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This problem is quite practical: to estimate the heat flux when making two objects contact to each other.

The following equation is valid only in block 2.

$Q'(t) = k_2A\frac{dT_2}{dx}|_{(x = 0)}$

From block 1 to block2, the temperature is not continue. So, there is no derivative.

In general, for contact heat transfer, there needs a thermal contact conductance coefficient $h_c$. And the heat flow rate is,

$Q'(t) = h_{c}A(T_1 - T_2)$

In your problem, you didn't provide thermal contact conductance, so the problem cannot be solved.

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