Liquid 100% R134a trapped between 2 valves at 10°C and heated up to 35°C. Let's say volume can't change, what is the equation to calculate final pressure?
Printable View
Liquid 100% R134a trapped between 2 valves at 10°C and heated up to 35°C. Let's say volume can't change, what is the equation to calculate final pressure?
Often wondered about this. A question for a someone who is knowledgeable about Physics I fear. I think if there is 100% liquid the pressure would be enormous!
I think the answer is at the end of this Youtube vid.
Although it shows Enthalpy difference equation, it can be transposed to find the pressure difference.
https://youtu.be/V77rmiRf2qE
Is it something that can be read off a pressure enthalpy chart if you know the density of the liquid at the 10 DegC?
or maybe,
https://www.theleeco.com/support-res...etric%20strain.
.
.
.
.
I thought I could find the solution somewhere in the logp/h but couldn't bind it. This question was from one of my students and I couldn't give him an answer. I hope I find between the help of my colleagues somewhere the right answer.
I will post another one which is not quite clear for me.
I think this is what I was looking for.. now find he cubical thermal expansion for R134a ...found it for water whic was 4 or 5 timesQuote:
Originally Posted by seanf
Found after your help with the correct English expressions this https://pressbooks.bccampus.ca/colle...s-and-liquids/
it's it's 100% full and 100% liquid... then it'll stay liquid no matter the temperature?
so
pressure1 x volume1 / temperature 1
equals
pressure2 x volume2 / temperature 2
(sorry I'm not very good at typing equations!)
I think PV/T might be for ideal gas only.
As simple as this? Never noticed and also never saw a log p/h where the lines of constant volume went up in the liquid phase
I dont know for sure.Quote:
Originally Posted by Peter_1
Seem to be half the charts that came up on a search do have the lines that far to the left.
I think you're spot on SeanQuote:
Originally Posted by seanf
Not so good with the Charts and Math myself, so help me out here.
As I understood the original question (which drew my interest) it was how much pressure liquid would develop if it was trapped and could not expand when heated. (As if Refrigerant was trapped between two valves)
The amount of increase in Temperature would of course be a factor in the pressure,, but I do not think that can be used to determine the final pressure.
I did find this quote. It refers to freezing water, but I think this might be similar.
Freezing force of water: as the water-ice temperature continues to drop to 0°F the forces involved range between 25,000 psi and can continue to reach 114,000 psi.
I found this at https://inspectapedia.com/plumbing/F..._Water_Ice.php
Looking forward to more discussion!
I stopped being lazy (retirement does that to one!) and found this.
http://www.svasd.com/wp-content/uplo...ppedLiquid.pdf
Thanks NH3LVRQuote:
Originally Posted by NH3LVR
Very interesting lecture which I saved for my students (and myself) but there's still no connection to some formula ...seems not that easy to find some explanation for this phenomenaQuote:
Originally Posted by NH3LVR
This is the gas law and not valid for liquids... I guessQuote:
Originally Posted by knighty
After posting, found this where it seems you're right
https://www.eng-tips.com/viewthread....3D%20T2P1%2FT1.
If the volume of the space is 100% liquid and can not change, pressure is not defined, because pressure has to be measured by measuring force (per area). At least an infinite small change (e.g. dV ) has to be thought of. In this small dV, there will be gas at preassure of 9bar (see above), same pressure everywhere inside the liquid. At 10°C it was 5bar.
Maybe you have difficulties to imagine this, because you think of liquid!=gas, but because liquid and gas have 2nd Order Phase Transition and strictly speaking (at >0K) there is no pure liquid or pure gas anyway, this small dV has also measurable realisation (or just find the gas bubble...).
Change of liquid density tabulars dont help here and are not necessary for the same reason.