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So, I was thinking about trying to build some kind of body cooling suit and started looking into the wonderful world of refrigeration systems. I've managed to get myself side-tracked pretty badly because I can't figure out why I'm right/wrong about this situation. It's a pretty simple concept.

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A fan blows air into a funnel, the air speeds up out of the funnel, as the air speeds up it loses pressure, because it loses pressure, it loses temperature, I thought perhaps that flow would be obstructed by air rushing in from the other side but this can be fixed rather easily with a check valve. I didn't think it would be much but when you have a pretty extreme funnel the results get pretty dramatic, presuming my calculations are correct, which is usually a long shot. I just don't understand if or where I'm going wrong, although I'm pretty sure I'm going wrong.

I'm aware in refrigeration systems they use a different type of expansion device this is just for my own understanding.

Calculations are below, I just used an online calculator:

Online Calculator

Gravity = 9.81 m/s

Fluid Density = 1.27 kg/m^3

Position 1

Pressure = 101,325 pascals

Height = 1 m

Speed = 15 m/s

Pipe Diameter = 0.1m

Temperature = 293.15 Kelvin

Position 2

Pressure = 12,171 pascals

Height = 1 m

Speed = 375 m/s

Pipe Diameter = 0.02m

Temperature = 293.15/8.325 = 35.21 Kelvin

Pressure change of 89,154 pascals

The Temperature out of the funnel is -237.94 Celsius! (This seems insane to me)

So to get the temperature at position 2 because temperature was not included on the online calculator, I just figured the pressure had been divided by 8.325 so I did the same to the temperature. Don't know if that's right. If anyone could enlighten me I'd be really appreciative!

-David.

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  • $\begingroup$ How do you get 12 Pa at position 2? If you blow to the atmosphere, you should have atmospheric pressure at the output. Considering this, you would actually need compressed air at roughly 0.9 bar to make it work. And that air having room temperature when compressed could surely cool something just when simply expanded. $\endgroup$
    – Tomáš Létal
    Commented Sep 2, 2022 at 17:50
  • $\begingroup$ The calculator gave me 12,171 pascals, I mean, I don't really understand why it would be at atmospheric pressure, isn't air that's moving faster at a lower pressure? And the air that's coming out is apparently at 375m/s, which again sounds off but I digress. $\endgroup$
    – David Nolan
    Commented Sep 2, 2022 at 18:01
  • $\begingroup$ If you start at atmospheric pressure, the air will not move anywhere. Movement of the air is caused by the pressure difference. In your case, you are somehow maintaining just 12.171 Pa at the outlet, which cannot be the atmosphere, instead, you would need confined space with low vacuum. $\endgroup$
    – Tomáš Létal
    Commented Sep 2, 2022 at 18:29
  • $\begingroup$ Is the movement of the air not caused by the fan? Maybe that's a silly question but the pressure differences the fan creates. I agree the figures are ridiculous I just don't understand why the calculator is so off. See the comment below for the values I was responsible for entering myself, the rest were calculated by the website I linked, Maybe I'm miss-applying the formula? $\endgroup$
    – David Nolan
    Commented Sep 2, 2022 at 18:48

1 Answer 1

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Pressurized air cools off as it expands. We don't need a calculator to tell us that.

To do it correctly, what we need are units that match reality, you can put anything into an online calculator. Yours has you putting air into a vacuum, and no fan will draw a vacuum like that. A regular fan will have pressure differences measured in cm of water, not in significant fractions of atmospheric pressure.

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  • $\begingroup$ Yes, I think you're right, Can you tell me which units I imputed incorrectly, the vacuum unit was one the calculator spit out. I simply entered the diameter of the pipes, density, pressure at position 1, heights, gravity and speed at position 1. and this is what it came out with. Is the calculator wrong perhaps? $\endgroup$
    – David Nolan
    Commented Sep 2, 2022 at 18:44

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