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So I know the most "prominent" conditions of carburetor icing is anywhere from 20 to 70 °F with visible moisture. There is also a tendency to have icing in the carburetor in the venturi throat due to a drop in temperature due to fuel vaporization.

The problem I'm having is if fuel is vaporized, wouldn't that be increasing the temperature? I don't understand how the temperature would drop.

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Good question! There's two parts to the answer.


The first part is the ideal gas law.

The formula of the ideal gas law is $PV = nRT$, where

  • P = Pressure
  • V = Volume
  • n = amount of substance
  • R = the ideal gas constant, which can be ignored for our purposes
  • T = Temperature

Since we're ignoring $R$, we're left with $PV = nT$.

Now, say that you send some air through the throat of a venturi. Pressure decreases greatly (that's the point), volume stays close enough that we can leave it alone, the amount of molecules remain unchanged, so to keep the equation in balance, the temperature drops.


The second part is the latent heat of vaporization

In order for a liquid to change state and become a gas, a certain amount of heat is required. Consider a firefighter spraying water on a fire. The liquid water is heated by the fire and evaporates into steam. The energy of the fire is decreased by the amount of energy that went into the steam and which is now floating away into the atmosphere.

Now, replace the fire with your carburetor, and replace the water with fuel. The heat energy in your carburetor is transferred to the fuel in order to cause it to vaporize. The fuel gets hotter (but who cares, it's about to get burned in the cylinders anyway) and the carburetor gets cooler.

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  • $\begingroup$ I'm no mathematician, but I took the presumptive step of changing "N" to "n" in your simplified (without "R") equation. If that's incorrect, please roll it back. $\endgroup$ – FreeMan Feb 9 '17 at 21:27
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    $\begingroup$ The fuel does not get hotter when evaporating. This is exactly why cooling by evaporating works: you can remove a lot of energy without increasing the temperature. It's not a magic heat engine that makes hot fumes and cold remaining liquids, which is how your answer describes it now. $\endgroup$ – Sanchises Feb 10 '17 at 8:25
  • $\begingroup$ @Sanchises - I think we may be saying the same thing using different words. I agree that the fuel can absorb heat energy without an increase in temperature, if that energy is used to vaporize the fuel. If you're talking about something else, please feel free to edit my answer or add your own. $\endgroup$ – Steve V. Feb 10 '17 at 16:11
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    $\begingroup$ @SteveV.: No, you're not saying the same thing. The fuel doesn't get hotter, it gets colder. And it's the cold vaporized fuel which then cools the carburetor in turn. The reason is that vaporization is adiabatic; there's no significant heat exchange during vaporization. $\endgroup$ – MSalters Feb 10 '17 at 23:33
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    $\begingroup$ Hi Steve, the second part is by far more important. Maybe you might want to add some expressions from this Wikipedia article and the concept of latent heat to improve the answer. $\endgroup$ – Peter Kämpf Feb 11 '17 at 9:31
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Sorry - disagree with the previous answer.

There's really just one effect. Molecules generally attract each other, until they come very close. If there's (virtually) no heat, then the molecules will settle at this close distance. You now have a solid.

Heat the solid a bit, and the molecules will start moving, and will no longer be at these closest distances, but they'll generally stay fairly close still - but they'll have enough energy to change neighbors.

Heat the liquid even further, and you'll see some of the molecules fly off entirely, and disappear in the gas phase. They still attract each other, so the speed of each individual molecule is constantly varying.

Now these change-over points are interesting. At these points, we see that with heat added, the average speed of molecules doesn't change. The extra energy goes to increasing the distances between molecules.

What happens in a carburetor is fairly simple. The fuel line is under normal pressure, and the fuel molecules are packed in a fluid. As the fuel enters the carburetor, it encounters a lower pressure which allows the molecules to disperse. As this dispersion increases the distance, the attraction between molecules slows down the escaping molecules - this is evaporative cooling.

Exactly how much cooling you have depends on the attraction as function of distance, which depends on the exact molecules. And aviation fuel is a mixture of many different chemicals, so that's rather complex. But on average, it's predictable enough for engineering purposes. But yes, you might need to heat the carburetor, else you'll get droplets of fuel instead of a vapor.

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  • $\begingroup$ Well, it looks like one of us is mistaken, because I agree with everything you've written up to, but not including As this dispersion increases the distance, the attraction between molecules slows down the escaping molecules - this is evaporative cooling. Both of our answers are lacking in references, so I'll try to find some evidence for mine. If you look for some evidence for yours maybe together we can meet at the truth. $\endgroup$ – Steve V. Feb 11 '17 at 2:36
  • $\begingroup$ See physics.stackexchange.com/questions/286013/… - this is fairly basic thermodynamics. $\endgroup$ – MSalters Feb 11 '17 at 12:19
  • $\begingroup$ Physics are based on models. Models are like opinions: everyone has one. $\endgroup$ – user3528438 Feb 8 '18 at 15:27

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