Thermal Expansion of Methanol

  • Thread starter rppearso
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  • #26
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Hi Ron,
Glad to hear the database worked out for you. As for how it works, to the best of my knowledge, there are simply a whole bunch of equations behind it that determine the various properties given the specific state of the fluid as that state is defined by 2 parameters. I guess it’s something like putting equations to all the data in steam tables.

Regarding how to incorporate the case of an expanding vessel (whether it’s just ‘stretching’ or you have a spring loaded piston or other way of removing work) that gets just a bit more tricky. I suppose such a case has application to things such as Sterling engines perhaps. At any rate, there has to be some way of determining the final state, be it final temperature or total heat transferred or some other parameter.

If I had to do this, I’d put a spreadsheet together and have an input so I could ‘guess’ at the final pressure. That guess at final pressure would go into determining total work removed from the system, be it from the vessel stretching or from a piston moving through some distance. Given the guess at pressure, one could then calculate the final volume (which determines final density) and work removed (Wout).

Now let’s say you have as a given, the total heat input. The first law now reduces to:
dU = Qin – Wout

You have Qin (a given) and now with the guess you made, you have Wout. At that point, you can calculate the final internal energy:

U2 – U1 = Qin – Wout
So
U2 = U1 + Qin – Wout

So now you have U2 and rho (remember that the guess at final pressure also gave you final volume so you have mass divided by volume) which gives you the final state.

The final state calculated from this will give you the final pressure which isn’t likely to match the final pressure that you guessed – but that doesn’t matter.

Now you have another block in your spreadsheet that finds the difference between the pressure you guessed and the pressure calculated from the above first law equation. That block should be 0 if you guessed correctly. So all you need to do now is use the solver to set that cell to 0 by changing the pressure you guessed at. When the pressure you guess at and the actual pressure calculated by the first law are the same, you’ve accurately calculated the final state, including the final volume of the vessel and work removed.
So in other words, dU = mCvdT - PdV, I know what the dT term is and if I guessed a PdV I would know what dU was but thats with a guessed PdV, instead of guessing you could solve explicitly for P by rearranging the equation. I guess you could use the tabulated dU from NIST but would that number necessarily stay constant if the volume were changing?
 
  • #27
Q_Goest
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So in other words, dU = mCvdT - PdV
Cv is for a constant volume process in which PdV is zero (ie: no volume change, so dV = 0). If the process were constant pressure, with work being done, would you use Cp instead? If the process were somewhere between constant volume and constant pressure, what would you use?

What I’m getting at, is there’s no easy way to determine heat input with just an initial state and a final temperature without considering what work is being done. The less work, the closer you are to a constant volume process. The more work, the closer to a constant pressure process. So you don’t have enough information yet to determine the final state if all you know is temperature. The system could be a cylinder with a piston on it such that only the weight of the piston acts against pressure. Such a system would be constant pressure. Or the piston could have an infinitely rigid spring attached such that it’s a constant volume process. Or it could be somewhere between the two so you have no special way of determining heat and work.

If the only thing you know about the final state is the temperature, the work done by the fluid during this process needs to be determined. Consider how you might determine work done if you had a spring attached to a piston. The work done would be PdV but we could also write an equation for work knowing how pressure changes depending on the movement of the piston which is a function of your spring. You might want to write out the equation for the work done in this case to help understand how work can be determined. Note that this equation will be the same one for a blocked in section of pipe. As long as the stresses in the pipe are below yield, the material acts as a spring.

Once you have the equation for work as a function of a linear spring and a piston, you can now make a guess at the final pressure and determine work caused by that final pressure. Again, the rest of the process generally follows what I wrote before. With a final pressure you have final volume, final rho, everything needed for a final state (U2). With that, you have U1, U2 and work so you can then calculate the heat (Q) needed to obtain that final state.

I hope that works. I can't know unless I work it out myself, so I may have missed something somewhere. If you work it out, I'd be interested in seeing how you did it.
 
  • #28
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Cv is for a constant volume process in which PdV is zero (ie: no volume change, so dV = 0). If the process were constant pressure, with work being done, would you use Cp instead? If the process were somewhere between constant volume and constant pressure, what would you use?

What I’m getting at, is there’s no easy way to determine heat input with just an initial state and a final temperature without considering what work is being done. The less work, the closer you are to a constant volume process. The more work, the closer to a constant pressure process. So you don’t have enough information yet to determine the final state if all you know is temperature. The system could be a cylinder with a piston on it such that only the weight of the piston acts against pressure. Such a system would be constant pressure. Or the piston could have an infinitely rigid spring attached such that it’s a constant volume process. Or it could be somewhere between the two so you have no special way of determining heat and work.

If the only thing you know about the final state is the temperature, the work done by the fluid during this process needs to be determined. Consider how you might determine work done if you had a spring attached to a piston. The work done would be PdV but we could also write an equation for work knowing how pressure changes depending on the movement of the piston which is a function of your spring. You might want to write out the equation for the work done in this case to help understand how work can be determined. Note that this equation will be the same one for a blocked in section of pipe. As long as the stresses in the pipe are below yield, the material acts as a spring.

Once you have the equation for work as a function of a linear spring and a piston, you can now make a guess at the final pressure and determine work caused by that final pressure. Again, the rest of the process generally follows what I wrote before. With a final pressure you have final volume, final rho, everything needed for a final state (U2). With that, you have U1, U2 and work so you can then calculate the heat (Q) needed to obtain that final state.

I hope that works. I can't know unless I work it out myself, so I may have missed something somewhere. If you work it out, I'd be interested in seeing how you did it.

I think I see what you are saying, basically calculate dV using the modulus of elasticity (which is the springiness of the metal so to speak) and that gives you dV and you can get dU from the "methanol tables" and Q is the one known (mCpdT), then solving for P is cake. You asked a really good question earlier and I think what I am trying to do is equivalent to deriving the "steam tables" except in this case it would be "methanol tables" and that got me thinking how are the steam tables (or any other table for that matter, air tables, etc) derived? Would this be a graduate level thermo question? I defiantly want to work this out but I want to make sure I'm not going down a rabbit hole like I did with the attached calc's. If you think what I wrote is correct I will proceed with working out the problem.
 
  • #29
stewartcs
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You asked a really good question earlier and I think what I am trying to do is equivalent to deriving the "steam tables" except in this case it would be "methanol tables" and that got me thinking how are the steam tables (or any other table for that matter, air tables, etc) derived?
For most substances, the relationships among thermodynamic properties are too complex to be expressed by simple equations. Therefore, properties are frequently presented in the forms of tables. Some thermodynamic properties can be measured easily, but others cannot and are calculated by using the relations between them and measurable properties. The results of these measurement and calculations are presented in property tables - such as steam.

Since programs such as REFPROP from our friends at NIST are now available, tables are somewhat obsolete (other than for academic purposes). I wouldn't recommend making a table, I'd just buy REFPROP for $200 bucks. I'm sure the time you would spend on making a table is more valuable than that!

Hope that helps.

CS

PS
Sorry for butting in Q!
 
  • #30
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For most substances, the relationships among thermodynamic properties are too complex to be expressed by simple equations. Therefore, properties are frequently presented in the forms of tables. Some thermodynamic properties can be measured easily, but others cannot and are calculated by using the relations between them and measurable properties. The results of these measurement and calculations are presented in property tables - such as steam.

Since programs such as REFPROP from our friends at NIST are now available, tables are somewhat obsolete (other than for academic purposes). I wouldn't recommend making a table, I'd just buy REFPROP for $200 bucks. I'm sure the time you would spend on making a table is more valuable than that!

Hope that helps.

CS

PS
Sorry for butting in Q!
I guess I am interested in how they came up with the tables as an academic excersize since derivation of the steam tables was never covered in 300 level thermo (which is why im thinking this is 500 level thermo stuff, which is ok)
 
  • #31
Q_Goest
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Hi CS. Feel free to butt in all you want. :smile: I think Ron already bought REFPROP database though (per post 24).

Ron, I assume you bought REFPROP, is that right?

What I'm saying about using Cp is that you can't use it. You can't use Cp nor Cv because both of these are for specific cases (either constant pressure or constant volume). So I would recommend using the database.

I understand you have the initial state, volume, and mass. The only thing you know about the final state is temperature. You don't have volume either since we are assuming that will change depending on pressure. But pressure depends on heat input.

Here’s my suggestion. Calculate vessel volume from the simplified description of a cylinder with a piston and a spring holding back the piston. You can then equate your actual hardware to this model since even if it is a blocked in pipe, you have a linear spring rate dependant on the pipe's modulus of elasticity. So you should be able to create a graph of volume versus pressure for any given model, and that graph will vary depending on the spring rate you choose.

You should also be able to create a graph of pressure versus volume given your known temperature. These two graphs should cross at a single point which is the final volume and pressure for the system at your given final temperature.

Rather than actually graph it, I’d suggest making a spreadsheet using REFPROP and take a guess at the pressure. This will give you the final fluid density which you can then calculate the volume from. This volume however, has to exactly equal the volume of your simplified model with the piston and spring for the given pressure. When the methanol volume and the cylinder volume are equal, you'll have found the solution for the final state of the fluid, including the final pressure. Put a cell in your spreadsheet that takes the difference between the two volumes and make that equal zero by changing the guess at pressure.

Once you do this, you will have found the final state and you could also determine heat input if that’s important to you.
 
  • #32
204
3
Hi CS. Feel free to butt in all you want. :smile: I think Ron already bought REFPROP database though (per post 24).

Ron, I assume you bought REFPROP, is that right?

What I'm saying about using Cp is that you can't use it. You can't use Cp nor Cv because both of these are for specific cases (either constant pressure or constant volume). So I would recommend using the database.

I understand you have the initial state, volume, and mass. The only thing you know about the final state is temperature. You don't have volume either since we are assuming that will change depending on pressure. But pressure depends on heat input.

Here’s my suggestion. Calculate vessel volume from the simplified description of a cylinder with a piston and a spring holding back the piston. You can then equate your actual hardware to this model since even if it is a blocked in pipe, you have a linear spring rate dependant on the pipe's modulus of elasticity. So you should be able to create a graph of volume versus pressure for any given model, and that graph will vary depending on the spring rate you choose.

You should also be able to create a graph of pressure versus volume given your known temperature. These two graphs should cross at a single point which is the final volume and pressure for the system at your given final temperature.

Rather than actually graph it, I’d suggest making a spreadsheet using REFPROP and take a guess at the pressure. This will give you the final fluid density which you can then calculate the volume from. This volume however, has to exactly equal the volume of your simplified model with the piston and spring for the given pressure. When the methanol volume and the cylinder volume are equal, you'll have found the solution for the final state of the fluid, including the final pressure. Put a cell in your spreadsheet that takes the difference between the two volumes and make that equal zero by changing the guess at pressure.

Once you do this, you will have found the final state and you could also determine heat input if that’s important to you.
I have not bought REFPROP yet I have just been using the NIST data base
 
  • #33
204
3
check it out,

Here is what I did for the thermal expansion, see what you think, I still ended up with 9300 psig even including the modulus of elasticity and I would like a reality check on that number because I was expecting something like 2000-3000 psig but its in the ball park and I have been surprised before.

Thank you for the help this has been great working on this problem along with the Joule thompson problem.

My next endevor is to get into line integration to solve for the surface areas of wierd surfaces for fire relief valves.
 

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  • #34
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Sorry the dT should be 80 not 540. This will give you a pressure of 1500 psig which exceeds the pressure of A spec piping. This is so sweet that I was finally able to get a reasonable answer.
 

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