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When object is heated and the heat source is removed, how long does it take to cool down and how it does it change the air temperature around it. If We know the temperature of the heated object and the temperature of environment, the property of material how dose the air temperature changes 1 meter from the object during 1 hour.
Thanks!

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The main question asks how a heated object raises the temperature of the air around it, therefore I take a more mechanistic approach here. You can certainly quantify the temperature rise but depending on the physics, that can be fairly easy or exceedingly difficult.

Mechanistically, there are three ways the hot object can heat the surrounding air: conduction, convection, and radiation. If we ignore active convection, i.e. movement of the air around the object due to outside influences, the temperature of the object will largely determine the mode of heat transport.

The Nusselt number, $$\frac{hL}{k},$$ is the ratio of resistance to conductive heat transfer to resistance to convective heat transfer. The larger the Nusselt number, the more important convection will be. If we're specifically thinking about heat transfer to a point one meter away from the surface, then we can just plug in 1 m for the length-scale and think about the ratio of the convective heat transfer coefficient, $h$, and the thermal conductivity, $k$. In general, $h$ must be estimated but has values from 0.5-1000 W/m^2-K depending on the thermal differences involved. However, the thermal conductivity of air is between 0.01 and 0.1 W/m-K, so in almost all situations convection will dominate conduction. With convection, it's standard to simply use Newton's law of cooling to determine the temperature change.

But what about radiation? This situation is a lot more complex, particularly if you must consider both radiation and convection. The overall ratio of radiative heat transport to convective heat transport is given by the Boltzman number, $$\text{Bo} = \frac{1}{\text{Ra}} = \frac{\rho C_P L}{\sigma \epsilon T_0^3},$$ (also known as the inverse of the Radiation number in some places). Note that I added in the emissivity, $\epsilon$, which wasn't included in the link because they were assuming a black body. The smaller the Boltzman number, the more important radiation is going to be. Again, the characteristic length here is probably appropriately considered as one meter.

But this only thinks about the object emitting radiation and not the air absorbing that radiation one meter away. To think about that, you need to calculate the emission spectrum of the object (often assuming it's grey) and then you need to figure out how much of that radiation is going to actually be absorbed by the air.

Then if you want to figure out the final temperature of the air and how long it takes to get to that temperature, you need to calculate the view factor from the surface of your object to the patch of air you wish to consider (this is greatly simplified by thinking about a spherical object or an infinite cylinder or plane). And this is all just to figure out the heat fluxes. Once you get those, you need to plug them into the general form of the general conservation equation for heat and then solve the first-order non-linear differential equation.

This is why people generally use computers to answer these types of questions if radiation is considered.

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  • $\begingroup$ Unfortunately I didn't have enough reputation to link to emissivity :-/ en.wikipedia.org/wiki/Emissivity $\endgroup$ – JeffP Mar 12 '18 at 20:53
  • $\begingroup$ +1 you have covered this question about as well as anyone can expect given what was asked. That said; in general I usually don't give such detailed answers to questions like this one; where the asker provided very little and seems as though they were looking for an exact equation with very few input details... This actually reads a lot like how I answer questions too. I'm honestly skimming this over thinking "did I just pass out for an hour, make a new account with a fake name and answer a question?!" Also added your link for emissivity. $\endgroup$ – JMac Mar 12 '18 at 21:06
  • $\begingroup$ @JMac, thanks for adding the link! Normally I answer questions like this on Quora but since I had a question to ask about thermodynamics today, I figured I'd try my hand at answering here. My philosophy on questions like this is that if this was asked as a homework problem (usually with real numbers), would somebody be able to to turn in my answer for credit. If the answer is no, but I still feel I answered the question as completely as possible, I think I've contributed an appropriate answer :-). $\endgroup$ – JeffP Mar 12 '18 at 22:17

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