Consider a metal ball being heated in a furnace. Thermal energy is convected to the ball while simultaneously getting conducted within.

The Fourier Number, in unsteady heat conduction analysis like the one here, is given as the ratio of conductive heat transfer rate in the ball to the rate of energy storage in the ball. I am having trouble with understanding how using this definition can we come up with the expression for Fourier Number which is

$$Fo= \frac{\alpha t}{(L_c)^2}$$

An excerpt from the book I'm referring - Fundamentals of Heat and Mass Transfer by Incropera and Dewitt

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I don't understand, while determining the conductive heat transfer rate, what is $\Delta T$. Is it the difference between the temperatures at the surface and center of the ball? If it is, that temperature difference will vary with time. Is it time averaged temperature difference between the center and the surface?

Furthermore, what is the $\Delta T$ while determining the rate of energy storage? and how is it equal to the $\Delta T $ taken while determining the conductive heat transfer rate? How did they cancel out both the $\Delta T$s at the end to give $\frac{\alpha t}{(L_c)^2}$


1 Answer 1


These are scaling arguments, good for back-of-the-envelope calculations and for broadly classifying process regimes. The actual values may be uncertain by an order of magnitude of more. The important aspect is that $\Delta T$ and $3\Delta T$ and $\Delta T/2$ are all considered to be approximately equal and fundamentally different from $(\Delta T)^2$ or $\ln(\Delta T/T_0)$, for example.

Thus, is doesn't matter if the cross-sectional area is that of a rectangle ($L^2$) or a circle ($\pi L^2/4$), for instance; the important part is the $\sim\!\! L^2$ scaling.

Now to your question. $\Delta T$ is the temperature difference across the solid. Yes, this varies over time. Yes, an energy storage rate $\rho L^3c\Delta T/t$ is strictly meaningful only if the entire solid heats up by a temperature difference $\Delta T$, which is not the same as a temperature difference across the solid. In other words, a spatial gradient in temperature is being equated to a transient difference in temperature. For the reasons given above, this is acceptable for this level of analysis and further can provide insight not available otherwise. What we're checking is whether the Fourier number is, say, $10^{-5}$ or $1$ or $10^5$; it doesn't matter if some temperature estimate was inaccurate by a factor of two. Does this make sense?

  • $\begingroup$ I also recommend rereading the answer you accepted for one of your previous questions. $\endgroup$ Jan 23, 2022 at 19:49
  • $\begingroup$ Oh Thanks it skipped my mind that I've asked a related question before. Yes, the answer does make sense to me. I have just one question though, won't $\Delta T$ and $10\Delta T$ will have different order of magnitudes? For instance, if former is 1 then latter will be 10, differing by one order of magnitude, then how can I consider them approximately equal, even with scaling argument? $\endgroup$ Jan 23, 2022 at 20:19
  • $\begingroup$ If that happens, then the scaling argument might run into problems. That's not the case here, though, so I wouldn't worry about it until it arises. $\endgroup$ Jan 23, 2022 at 21:08
  • $\begingroup$ You mean to say that the $\Delta T$ for conduction and $\Delta T$ for energy storage, would have the same order of magnitude always? That $\Delta T$ for former and $10\Delta T$ for latter is not possible here? $\endgroup$ Jan 24, 2022 at 5:44
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    $\begingroup$ @HarshitRajput Your questions often seem to jump to absolutes and extremes. Consider a simple example, and focus on understanding its behavior: a cube with a temperature difference of $\Delta T$ from one side to the opposite side. The energy to bring the entire cube to the maximum temperature is $\rho L^3c\Delta T/2$ (assuming constant material properties) because the average temperature difference is $\Delta T/2$. The factor of two is discarded because of the approximate nature of the scaling analysis. There’s no $10T$ here. $\endgroup$ Jan 24, 2022 at 14:35

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