0
$\begingroup$

$k$ - thermal conductivity, $h$ - heat transfer coefficient, $\rho C_p$ - Volumetric heat capacity, $T$ - temperature $T_\infty$ - Temperature of exposed cooling surrounding environment

Consider two spheres (say a and b) having the same geometrical dimensions but made of different materials (such that $k_a > k_b$) and being at the same initial temperature. However, one can assume that they are exposed to the same cooling environment of the same $h$ and $T_\infty$. Also, consider that $Bi_a = Bi_b < 0.1$ (so that lumped body analysis can be assumed).

The equation (of the lumped body analysis) suggests that the temperature profiles of both spheres should decrease exponentially with time and that they must overlap each other (as the equation doesn't care for $k$. It only depends on $h$, area, volume, $\rho C_p$). However, my intuition asks whether the thermal conductivity doesn't play any role in the temperature profile, practically. Don't you believe that the T profile of the one with a higher k should decrease faster?

I am wondering if someone can provide both practical and theoretical insight on the temperature profile.

Thanks

$\endgroup$

1 Answer 1

1
$\begingroup$

The lumped-body assumption (i.e., the negligible-Biot-number assumption) implies that the thermal conductivity is irrelevant–in other words, that the thermal conductivity is sufficiently large that the temperature of each body is essentially uniform. This is quite clearly stated in the Wikipedia articles; Incropera & DeWitt is a good textbook reference.

If you wish to consider the influence of the thermal conductivity, don't make that assumption. You're free to solve the transient heat equation in each sphere.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.