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I want to convert the pressured air exiting a valve to heat. I know that a vortex tube can be used to do this, but I am wondering if I need to make modifications to the vortex tube to make it more efficient or if there is just another, more efficient, way to generate heat.

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    $\begingroup$ Wait a minute. If compressed air is exiting it's container through a valve to outside atmospheric pressure, it is expanding, so it would be cooling. $\endgroup$ – 0tyranny 0poverty Jan 25 '18 at 15:14
  • $\begingroup$ To my knowledge a vortex tube is already optimized for this. An upstream valve will cool the gas and drop the pressure so eliminate that valve if you can. Or open the valve all the way to there is no obstruction. $\endgroup$ – paparazzo Jan 25 '18 at 15:25
  • $\begingroup$ Sorry I fixed it. $\endgroup$ – layssi Jan 25 '18 at 18:19
  • $\begingroup$ You have not fully fixed it. You are not generating heat. This is an adiabatic device. $\endgroup$ – paparazzo Jan 25 '18 at 18:39
  • $\begingroup$ a vortex tube does indeed generate heat. en.wikipedia.org/wiki/Vortex_tube How efficient this is I have no idea. $\endgroup$ – agentp Jan 25 '18 at 18:41
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One possibility would be to use the air to drive a turbine (probably a radial turbine as they are used in turbochargers). The turbine could be connected to a generator and the generator to a resistor. You will most likely need to cool the bearing of the turbine and the generator as well, so you will generate a lot of heat.

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Assuming that your pressurised air (@4500PSI=320bar) is at room-temperature, 20°C the extraction of the potential pressure energy will result in pretty low temperature at the exhaust of your process.

I used ethermo.us to calculate the thermodynamic states. For a ideal (efficiency=1.0) process the expansion from 320 bar and 20°C to 1 bar (ambient pressure) would result in the condensation of air, which makes everything a lot more complicated, so I expanded only to 12 bar. But even at 12 bar the exhaust air will have -160°C.

I would appreciate if someone could cross-check those numbers and calculations!

Step 1: Mass of compressed air in the container
Volume: $V = 1l$,
density: $\rho_\mathrm{@310bar, 20°C} = 330,33 \frac{kg}{m^3}$ Mass of air in the container, $m_\mathrm{air}=V \cdot \rho = 0,33kg$

Step 2: Specific enthalpy-difference for isentropic expansion. Both states have the same specific entropy ($5\frac{kJ}{kgK}$)
Initial state: $h_\mathrm{@310bar, 20°C} = 248,39 \frac{kJ}{kg}$
Final state: $h_\mathrm{@12bar, -160°C} = 91,86 \frac{kJ}{kg}$

Step 3: Available energy
$\Delta H = \left(h_\mathrm{@310bar, 20°C} - h_\mathrm{@12bar, -160°C} \right) \cdot m_\mathrm{air} = 51,71 kJ $

It is unlikely that a process will extract $52 kJ$ so it is likely to be less. I am afraid (given no errors) that this amount of energy will not justify the technical challenges.


Remark:
The working mechanism of the vortex-tube is analytically described by Polihronov and Straatman1. The author states the following with respect to heating2:

In conclusion: if one needs to convert the energy of pressurized gas into heat by using the vortex tube effect, the best approach is: remove the hot exit of the tube.(...)and drive a heat pump with it to produce heat.

[1]: Polihronov, Jeliazko G.; Straatman, Anthony G. (2012). "Thermodynamics of angular propulsion in fluids". Physical Review Letters.
[2]: Polihronov, Heating using a vortex tube

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  • $\begingroup$ search "vortex tube" on the physics stack exchange for a correct explanation of how it works. $\endgroup$ – niels nielsen Jan 26 '18 at 16:32
  • $\begingroup$ I reformulated the section and added references, thank for nudge! $\endgroup$ – rul30 Jan 26 '18 at 17:24
  • $\begingroup$ @rul30. For scale, my pressured air tank is small, the size of a paintball compressed air tank (maybe around 1 Liter). I want to heat aluminum or copper the size of a post it note. Maybe something like 75mmx75mmx0.1mm Aluminum foil. I want to get it to around 400 degree Celsius for as long as possible. Like a portable heater without the flame. $\endgroup$ – layssi Jan 29 '18 at 14:04
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    $\begingroup$ Thanks @layssi for the info. Sadly it seems that independently of the process you would use to convert the pressure energy, it might only be able to heat on cup of water. $\endgroup$ – rul30 Feb 1 '18 at 13:59
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    $\begingroup$ @layssi. 0,15×4,179×80 = 50,148 kJ (mass of water x heat capacity x temperature difference)...so as long as you find a process with an efficiency of 97% (50,14/51,71) it might work. Good luck, and keep us posted! $\endgroup$ – rul30 Feb 4 '18 at 10:07
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Any engine that can run on compressed air will do this. There are quite a few, but as an example imagine a cylinder that has a bit of compressed air in it. The pressure will push on the piston doing work until it has equalized with the atmospheric pressure. Then a valve opens and perhaps a flywheel or another piston on a crankshaft pushes the piston back and vents the air out to the atmosphere, a new bit of compressed air is admitted and the cycle repeats.

In the absence of a load, the engine will convert all of the pressure energy into heat from friction in the cylinder or bearings as well as air resistance.

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Since the expanding air will try to cool it's immediate surroundings, you need isolate the expanded air supply from the area to be heated. As long as you have a heat sink, you can use an air motor to run a heat pump.

Of course, you can capture heat during the filling of the cylinder too.

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