Thermal Expansion of a trapped fluid with a piston and air chamber

[JonnyRotten]

hi Folks,

I was hoping for some help in a calculation. I have a cylinder with closed ends and a floating piston in the middle. On one side of the piston I have a fluid and on the other side I have air. I will be heating up the cylinder and I would like to calculate how far the piston would travel and also what pressure would be generated. I can calculate how much air pressure will be generated by the piston moving. And I can calculate how far the piston would move due to thermal expansion. At what point would it reach an equilibrium? When the air pressure builds up enough to resist the thermal expansion of the fluid?

Can i use this equation from this post

http://www.eng-tips.com/faqs.cfm?fid=1339

"By equating the change in volume of the fluid to the change in volume of the vessel one gets:αfΔT-βΔP=αvΔT+ΔPD/tE" but also add Volume of piston travel to the right side of the equation?

Any help would be appreciated.

 

[hydtools]

Consider also how air pressure will increase with temperature.
Consider how the cylinder volume will increase with temperature. This reduces effect of fluid expansion.

Ted

 

[BigInch]

Do you really need such accuracy?
1.)If you mean liquid (a fluid can be liquid or gas) and you don't boil the liquid and, if it is almost incompressible and it expands relatively little with temperature, just calculate the pressure using the gas equation.
2.) if the liquid expansion with temperature is significant, just calculate that additional liquid volume then use the gas equation to get pressure.

"People will work for you with blood and sweat and tears if they work for what they believe in......" - Simon Sinek

 

[JonnyRotten]

Thanks for your replies. The liquid expansion is significant. If it was fully trapped, it would reach 10,000psi. I want to add enough of an air chamber so that it only reaches 5,000psi at 121 degreesC. Would P1V1/T1=P2V2/T2 be accurate enough? I was using this for the air chamber.

 

[BigInch]

No. At those pressures you'll have to consider the compressibility factor of the gas, and probably the compression factor (via bulk modulus) of the liquid and the expansion of the container too.

"People will work for you with blood and sweat and tears if they work for what they believe in......" - Simon Sinek

 

[JonnyRotten]

Ok thanks, Could anyone point me in the direction of a set of equations that would allow me to calculate what piston stroke and initial air chamber volume required to ensure that the pressure builds up to 5,000psi? This is obviously a function of initial oil volume.

 

[BigInch]

p1v1 = p2v2 is the initial approximation.
PV = nZRT where Z is evaluated at T the final temperature, which will be really, really hot.

"People will work for you with blood and sweat and tears if they work for what they believe in......" - Simon Sinek

 

[JonnyRotten]

My final temperature is 121 degC. I want to ensure that the pressure in my Oil chamber builds to 5,000psi. Obviously the air chamber pressure will be the same as I have a floating piston separating the two. So using the equations for gases that you have given me, I still need to equate these to the expansion of my oil at 121 degC.

I think what I want to calculate is what delta V is required so that my oil only builds to 5,000psi, instead of the 10,000psi it would reach at a delta V of zero (trapped volume). I can work backwards from there to work out my air chamber volume. I'm not sure how to work out the delta V.

 

[hydtools]

If you compress air to 5000psi, it will be really, really hot. Residual lubricant may ignite. What is the initial pressure and temperature?

Ted

 

[JonnyRotten]

Ah I see. The cylinder will have a constant outside temper of 121 degC in a constantly moving fluid so should dissipate some of the heat. Initial pressure will be atmospheric and temp will be ambient.

 

[JonnyRotten]

At the moment, I am calculating the volume expansion of the fluid due to temperature and then putting that in as a volume decrease in the air chamber and working out the P rise caused by that, using P1V1/T1=P2V2/T2. Then I am just assuming that the oil chamber will be at this pressure.

But I feel that is not correct. It doesn't take into account the pressure of the air chamber acting on the oil chamber. For example, in my calculation spreadsheet, if the oil expansion volume is greater than the total volume of the air chamber (piston stroke volume), then I get a negative answer for pressure. When in reality, the air pressure would ramp up to a point that it would resist the thermal expansion of the oil. This is the part that I am struggling with. I need to equate thermal expansion with compressibilty somehow.

 

[hydtools]

You have to set a limit. The oil cannot expand to more than the total volume of the cylinder less the piston volume. The air volume cannot be less than zero.

Ted

 

[JonnyRotten]

I was hoping to set my limit to approximately 5,000psi (which should be a very small volume in the air chamber). The air compression should be relatively slow - as fast as the oil can expand due to temperature - so the heat should dissipate before it reaches the crazy temperatures that the P1V1/T1=P2V2/T2 calculation gives. Is that fair to say?

 

[BigInch]

The heat won't dissipate unless you take a relatively long time to do the compression. What you may be making instead is some kind of a diesel engine.

"People will work for you with blood and sweat and tears if they work for what they believe in......" - Simon Sinek

 

[prex]

As the pressure builds up as a consequence of heating the liquid, I think you can assume that the gas pressurization is isothermal, including only the effect of gas temperature from initial room temperature to the final 121 °C.
A subset of this problem is treated in faq378-1339: what you have to add is the contribution of gas compression. This one may be evaluated as follows:
P[sub]1[/sub]V[sub]1[/sub]/T[sub]1[/sub]=P[sub]2[/sub]V[sub]2[/sub]/T[sub]2[/sub]
where T[sub]1[/sub]=293 °K and T[sub]2[/sub]=414 °K
V[sub]2[/sub]-V[sub]1[/sub]=P[sub]1[/sub]V[sub]1[/sub]T[sub]2[/sub]/P[sub]2[/sub]T[sub]1[/sub]-V[sub]1[/sub]
[Δ]V=(P[sub]1[/sub]T[sub]2[/sub]/P[sub]2[/sub]T[sub]1[/sub]-1)V[sub]1[/sub]

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[prex]

Sorry, my finger was too fast.
(continued)
[Δ]V/V[sub]1[/sub]=P[sub]1[/sub]T[sub]2[/sub]/T[sub]1[/sub](P[sub]1[/sub]+[Δ]P)-1
and also, if V[sub]i[/sub] is the total (initial) volume (liquid plus gas),
[Δ]V/V[sub]i[/sub]=(P[sub]1[/sub]T[sub]2[/sub]/T[sub]1[/sub](P[sub]1[/sub]+[Δ]P)-1)V[sub]1[/sub]/V[sub]i[/sub]
Now this quantity may be plugged into the equation in the FAQ above equating the various contributions to the relative change in volume.
However also the contributions [α][sub]f[/sub][Δ]T and [β][Δ]P should be factored with V[sub]f[/sub]/V[sub]i[/sub], as the liquid also doesn't occupy the full volume (V[sub]f[/sub] being the liquid initial volume).
Didn't try it, but I guess this should give a reasonable approximation. Of course the solving equation for [Δ]P (or in your case for [Δ]V, as you know the final pressure) is now more complex, due to the strongly non linear nature of the gas law. A numerical approach via Excel or another software should give the result.

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