Pressure rise due to temperature (Flowsimulation)

Simulation Generality
1

As part of a thesis topic, I want to analyze the pressure rise of a gas in a tank when it is immersed in water at 90°C and atmospheric pressure.

In short:
Pressure variation of a gas contained in a closed volume following the transfer of heat provided by a hot fluid (water at 90° at high flow).

 

At the beginning, gas = 0°C and 3 bar (liquid + gas). At the end of the reaction, gas = 90°C and 6 bar (gas).

 

I manage to carry out the (external) simulation well. But only the temperature rises as predicted. The pressure is only around 4.1 bars.

 

Could anyone take a look at my work? (Appendix)

 


pression-temperature.zip
2

Hello

 

I'm trying to take a look at the day, for now by looking quickly I don't see any big errors in the definition of the calculation.

 

@+

 

3

Hello

 

How do you get the pressure of 6 bars at the end of the reaction?

I don't have SW so I can't look at your model and, anyway, I don't know this tool... On the other hand, the result of Flowsimulation seems pretty good to me.

Sorry but I basically come back to my ideal gas equation P.V=n.R.T. Here, V, n and R are constant so P=a.T. At t0, P0 = a.T0 with P0 = 3 bars and T0 = 273 °K. A tf, Pf = a.Tf with Tf = 273+90=363 °K. As a result, Pf = (P0/T0). Tf = 4 bar approx. (3.99 exactly). So, if you have a result at 4.1 bars, it would be consistent.

 

 

4 Likes
4

Thank you for your help.

 

 

Pounding:

 

By regaling a thermodynamic diagram of R134a (in the appendix):

We can see that starting from t=0 °C and P = 3 bars  to reach T=90°C (all this at constant volume), we reach a pressure of  6 bars.

 

That's why I'm surprised that flowsimulation only finds 4.1. ;-)

 

 

Do you have any idea where this can come from?


r134a-700694.gif
5

Hi @ s_pirlet

as in concidere that the diagram is good ;-)

and that the gas used is R 134 A

 

See temperature/pressure correspondence table for some refrigerants

http://www.energieplus-lesite.be/index.php?id=11279

is at   0° =  1.91bar

is at  40° =  9.11 bar

is at  60° = 15.73 bar

 

so at 90° you should be at 33 bar ?????

 

or I am saying a outright which is possible

your part does not deform?????????????

@+ ;-)

6

Hello

 

OK, I understand your mistake better. In fact, it's your starting point that is wrong because you took it in the "liquid+gas" zone (inside the black curve) and not in the gaseous state only (right part of the graph). So, if you recalibrate your starting point and follow the green line, you fall back on about 4 bars. I've put all this back on the graph.

 

On the other hand, if you have to use this graph for your dissertation, be careful because it also tells you that your initial hypothesis is not entirely accurate. Indeed, if you look closely, you see that you can't be purely gaseous at 0° and 3 bars. Either you keep 3 bars but your starting temperature will be slightly higher or you keep 0° but the starting pressure must be lower.

 


r134a-2.gif
1 Like
7

http://www.techno-froid.com/index.php?option=com_content&task=view&id=28&Itemid=32

see this page Thank you as well as the page you previous link

explain to me HELP I don't understand +

@+

8

http://fr.wikipedia.org/wiki/1,1,1,2-t%C3%A9trafluoro%C3%A9thane

 

 

@+ ; '((((

9

Hi @gt22

 

The data you found are probably more likely to be obtained at constant enthalpy and not at constant volume, which explains the difference in values.

In a refrigerated circuit, the opposite phenomenon is used to the one we are interested in here. In other words, the pressure (compression or expansion) of the gas is used to increase or lower its temperature.

10

Hello everyone and thank you very much for your help.

 

 

I think I expressed myself badly, which seems to have created confusion.

 

I am trying to raise the pressure of a liquid+gas mixture (R134a) at a constant volume.

 

My initial point is around 0°C and 3 bars (liquid+gas) to reach about 90°C and 6bars (gas only). (cf thermo curve).

 

When I carry out my simulation on Flowsimulation, the simulator gives me a value of 90°C and 4.1 bars....

 

Could you help me interpret this result?

 

11

your liquid and your gas are R 134 A: so liquid and gaseous state ?

enclosed in a volume of x cm3 

3 Bar Pressure

to 0° celcius

@+

12

Hello

 

Indeed, I had forgotten this detail. As a result, if there is a phase transition, all this is no longer valid.

On the other hand, given the consistency of the results obtained by taking only a gaseous state, it seems quite clear to me that FlowSimulation made the same error of initial condition.

So, I think you need to look for ways to better define your starting point (liquid/gas proportion, enthalpy, ...) so that the change of state is properly taken into account.

13

See this link

with a thermodynamic property calculator

 

This applet is used to calculate the thermodynamic properties of some ideal gases and vapors.

By using this software package, you agree to the terms of the license (see licenceS4E2.pdf file).

License Software package "Thermoptim-Light Calculator"

http://direns.mines-paristech.fr/Sites/Thopt/fr/co/applet-calculateur.html

 

@+ ;-)