# Thermal Expansion of Methanol

Does anyone have coefficients of thermal expansion in F^-1 instead of in C^-1. I have worked out an equation from thermodynamic first principals that derives the throttling equation I then did the first order non-seperable differential equation:

dT/dP = C1*T - C2

to get the following equation:

P2 = {[ln(T1/(T2-(C2/C1)))]/-C1} + P1

Where C1 = V*B/Cp*m and C2 = V/Cp*m

but this equation only gives appropriate solutions for P2 if B is ~0.02 F^-1 or in this range, the tabulated valves for methanol are 0.00135 F^-1 which gives me imaginary numbers when I take the natural log using this B value.

This has been extremely frustrating because I feel like I am very close to solving this blocked in thermal expansion problem to get a pressure but the tabulated values I find are not working with my equation, I am wondering if its because the few websites that state this value are in C^-1 and I am just multiplying by 9/5 to get B = .00135 F^-1.

## Answers and Replies

Mapes
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You should be multiplying by 5/9 to convert from inverse degrees Celsius to inverse degrees Fahrenheit. According to Yaws et al.'s Chemical Properties Handbook, the thermal expansion of methanol at 25°C is

$$0.00116(^\circ\mathrm C)^{-1}=0.000642(^\circ\mathrm F)^{-1}$$

If your equation gives an imaginary number for a physical quantity, it simply isn't applicable for that set of input variables.

You should be multiplying by 5/9 to convert from inverse degrees Celsius to inverse degrees Fahrenheit. According to Yaws et al.'s Chemical Properties Handbook, the thermal expansion of methanol at 25°C is

$$0.00116(^\circ\mathrm C)^{-1}=0.000642(^\circ\mathrm F)^{-1}$$

If your equation gives an imaginary number for a physical quantity, it simply isn't applicable for that set of input variables.

Thank you for catching that, I think I found the error, I took the absolute value of the denominator in the natural log, now I just need to know what a reasonable over pressure value is for a rigid pipe blocked in that is heated from 70 F to 150 F or in that range, I am getting values that are very very high and for some reason as I increase the temperature spread P2 goes down which does not make sense, I feel I made a sign error somewhere or my entire approach is flawed. Thank you for that constant its good to have all your material constants nailed down so you can focus on the flaws in the equations themselves.

I think I have it worked out by taking the absolute value inside the natural log. Would the coefficient of thermal expansion go up or down with increasing temperature (ie when you diverge from 20 C). I ran my numbers again with the absolute value and if I increase the temp without modifying the thermal expansion coefficient the pressure drops as I increase temp but if I slightly lower the expansion coefficient as I step up the temp I see the pressure rise just as I would expect. Going from 70F to ~150 F I am getting pressure spikes of around 500-600 psig that seems reasonable to me now I just need the temperature to follow the pressure correctly.

Mapes
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It might be helpful if you described the problem and showed the steps you used to derive your first equation above.

Attached full calcs

I have attached the full calc files for review. Thank you for the help.

The real problem starts on page 3, the first 2 pages were deriving the joule-thompson/thermal expansion equation for fun. If I remove the mass term and assume m=1 I will get reasonable solutions of P2 = ~600 psig but then the units do not work out because the mass does not = 1 for the given volume and I cant imagine thermal expansion will cause a 800,000 psig rise.

#### Attachments

• Thermal expansion of methanol.pdf
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I have attached the full calc files for review. Thank you for the help.

The real problem starts on page 3, the first 2 pages were deriving the joule-thompson/thermal expansion equation for fun. If I remove the mass term and assume m=1 I will get reasonable solutions of P2 = ~600 psig but then the units do not work out because the mass does not = 1 for the given volume and I cant imagine thermal expansion will cause a 800,000 psig rise.

If I put the temperature units in rankine I get 177000 which is better but still way to high. Maybe I need to step this out every 10 degrees and tweek the thermal expansion coefficient?

Q_Goest
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Hi Ron,
I took a quick look at your attachment but didn't go through in detail. I understand you want to figure out what pressure methanol would go to given it's trapped with no other vapor present. Initial conditions of 70 F and 14.7 psia with final state at 200 F.

Normally, I wouldn't use equations like you are to do this. This is very easy if you have a computerized database for fluid properties. I have a proprietary one that has a few different databases. If you have such a database, all you need to do is input the final temperature, and make the assumption that density remains constant. Database 1 gives me 14,466 psia. Database 2 gives me 22,651 psia. I'd trust database 2.

Of course, at this kind of pressure the container it's in (pipe?) is going to expand considerably, if not rupture, so the final pressure is going to be much lower than calculated.

Hi Ron,
I took a quick look at your attachment but didn't go through in detail. I understand you want to figure out what pressure methanol would go to given it's trapped with no other vapor present. Initial conditions of 70 F and 14.7 psia with final state at 200 F.

Normally, I wouldn't use equations like you are to do this. This is very easy if you have a computerized database for fluid properties. I have a proprietary one that has a few different databases. If you have such a database, all you need to do is input the final temperature, and make the assumption that density remains constant. Database 1 gives me 14,466 psia. Database 2 gives me 22,651 psia. I'd trust database 2.

Of course, at this kind of pressure the container it's in (pipe?) is going to expand considerably, if not rupture, so the final pressure is going to be much lower than calculated. Actually I think I am all wet on my calcs because I tried setting the initital temp to the final temp and P2 should have equaled P1 and it dident so something is wrong. I do not have access to such a data base that is why I am trying to do this thermodynamicly.

So I have 177000 psig, I imagine I would need to iterate the thermal expansion coefficient as the temperature increases (because the beta term is only good for a certian temp). Do you know if the thermal expansion coefficient goes up or down with increase in temperature? Also do you know what equations to use to account for the elasticity of the steel?

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Q_Goest
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Do you know if the thermal expansion coefficient goes up or down with increase in temperature?
You can't use a single value for thermal expansion coefficient. The saturation pressure for methanol is only about 41.2 psia at 200 F, and critical pressure is 1153 psia. The process to increase temperature is taking it well above the point where a single value for thermal expansion is reasonable. What's needed are the fluid's physical properties, and that can only be accessed with a good database.

or

If you have a good equation of state (ex: van der Waal's) then you can determine pressure from the density and temperature. I doubt van der Waals is going to be that accurate though. At any rate, trying to determine pressure from the thermal expansion coefficient just doesn't seem to me to be the right way to go about this problem.

Regarding the vessel or container, if you can describe what you're using, then yes, I can try and explain how you might go about determining the increase in volume given the pressure rise. For example, if you have a pipe, then what is the outside diameter, length, wall thickness and type of material? The amount a pipe expands can be calculated from those variables.

Q_Goest
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Here's the equation of state for density (lbm/ft3) and pressure (psia) given a constant temperature of 200 F.

Density = 9.238E-14*P^3 - 7.922E-09*P^2 + 4.263E-04*P + 4.489E+01
(Note: this is just a curve fit to some data I have.)

Now all you do is finde the intercept where the density is equal to the density at 14.7 psia and 70 F (~49.663 lbm/ft3).

You can then use this equation along with equations for pipe/tank stretch to determine final pressure if you want to take that into account.

You can't use a single value for thermal expansion coefficient. The saturation pressure for methanol is only about 41.2 psia at 200 F, and critical pressure is 1153 psia. The process to increase temperature is taking it well above the point where a single value for thermal expansion is reasonable. What's needed are the fluid's physical properties, and that can only be accessed with a good database.

or

If you have a good equation of state (ex: van der Waal's) then you can determine pressure from the density and temperature. I doubt van der Waals is going to be that accurate though. At any rate, trying to determine pressure from the thermal expansion coefficient just doesn't seem to me to be the right way to go about this problem.

Regarding the vessel or container, if you can describe what you're using, then yes, I can try and explain how you might go about determining the increase in volume given the pressure rise. For example, if you have a pipe, then what is the outside diameter, length, wall thickness and type of material? The amount a pipe expands can be calculated from those variables.

An equation of state is probably better, to evaluate dV/dT instead of using beta, just like you evaluate the joule thompson equation. I fixed my constant of integration and came up with 43,000 psig using a constant beta so im getting closer. I must have dropped a sign somewhere because I had to multiply the ln by -1. I did get 26,000 psig at 150 F. I believe peng robinson would be an adaquate estimate for liquids thats what we typically us in hysys for liquids. Im going to keep whittling away at this. I just got done doing a dynamic pipeline heat up in the middle of winter on mathcad (of course I had to get viscosity vs temp data from the lab and I estimated the rest of the properties) but it worked out very well. After all this is done I will have to take into account the modulus of elasticity to go from 26,000 psig to a real final pressure, do you know how to do that? I also realized im going to have to iterate the heat capacity as the temp goes up because that drastically effects the final pressure.

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Q_Goest
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After all this is done I will have to take into account the modulus of elasticity to go from 26,000 psig to a real final pressure, do you know how to do that? I also realized im going to have to iterate the heat capacity as the temp goes up because that drastically effects the final pressure.
The vessel is going to expand depending on stress and modulus of elasticity. Stress = strain * modulus, so determining stress and given modulus, strain can be determined. With strain, you can find the overall change in dimension.

Regarding heat capacity, again - this is where having a good fluids database comes in. dU = Q, so you don't need Cv as a function of temp and pressure. All you need is the data for dU which is easy.

Engineers shouldn't be spending lots of time calculating fluid properties. There's just no value in doing that. Calculating fluid properties is fine as an excersize to learn something, but once you understand it, it's much more efficient to just use a good database. No engineers in industry should be spending time looking for fluid properties, it's just not worth the money.

The vessel is going to expand depending on stress and modulus of elasticity. Stress = strain * modulus, so determining stress and given modulus, strain can be determined. With strain, you can find the overall change in dimension.

Regarding heat capacity, again - this is where having a good fluids database comes in. dU = Q, so you don't need Cv as a function of temp and pressure. All you need is the data for dU which is easy.

Engineers shouldn't be spending lots of time calculating fluid properties. There's just no value in doing that. Calculating fluid properties is fine as an excersize to learn something, but once you understand it, it's much more efficient to just use a good database. No engineers in industry should be spending time looking for fluid properties, it's just not worth the money.

Unfortunatly we dont have such a data base, I should probably put in a request for one but usually requests for expensive software dont happen. I guess you could consider this a re-learning experence for my PE exam. As far as using dU, I am using thermodynamic equations for dH = TdS + VdP as opposed to dU = TdS + PdV. I can then use an equation of state to solve for dV/dT

The vessel is going to expand depending on stress and modulus of elasticity. Stress = strain * modulus, so determining stress and given modulus, strain can be determined. With strain, you can find the overall change in dimension.

Regarding heat capacity, again - this is where having a good fluids database comes in. dU = Q, so you don't need Cv as a function of temp and pressure. All you need is the data for dU which is easy.

Engineers shouldn't be spending lots of time calculating fluid properties. There's just no value in doing that. Calculating fluid properties is fine as an excersize to learn something, but once you understand it, it's much more efficient to just use a good database. No engineers in industry should be spending time looking for fluid properties, it's just not worth the money.

How did you get a Q only given a deta T with P2 unknown?

Q_Goest
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Hi Ron,
Sorry if I sound like I’m singin’ from a soap box. I’ve heard the complaint about a company not wanting to pay for such software before though and it’s killin’ me. Think about it this way, what’s your charge-out rate? I suspect it’s on the order of $100 per hour unless you work for a small company with a much smaller overhead. Even then, I don’t suppose it’s too much less. So if you can save 2 hours worth of time, you can pay for the database. Look into NIST REFPROP. I’ve used it before and it’s not bad. I use a proprietary one that’s quite a bit more powerful, but the one from NIST has everything a typical user needs. You can purchase it for$200, or $300 with the upgrade. From the looks of all the work you’re putting into this problem, the database would have paid for itself a few times over by now. (note that methanol is included in the database) http://www.nist.gov/srd/nist23.htm [Broken] This database can be linked to EXCEL, so you can write spread sheet programs to do what you need. Not sure what other programs it can be linked to – I see you mentioned mathcad. Here’s how it works. You have the initial state with P and T. You also have a final temperature, so you need a final pressure. What you know about the system is that the two densities are the same (first order estimate). Since you have density and temperature of the final state, you can find any other property of the fluid. That means, you can find not just the pressure, but the internal energy, enthalpy, entropy, thermal conductivity, viscosity, etc… everything about the fluid is known with just the 2 variables, temperature and density. If you want to go a step farther and consider the increase in vessel volume, you can guess at final pressure. This gives you some increase in vessel volume which changes your density. If the density is exactly equal to the mass of the fluid divided by the new, expanded vessel volume, you’re done. If not, do a goal seek to equalize them. That’s the short of it. Calculating the vessel volume is not that difficult. Hoop stress is P*r/t so divide by E and you have strain, then multiply by diameter to get the new diameter. For length, axial stress is ½ the hoop stress, so strain is ½ the strain of the diameter. Multiply by length to get your new length. This is all valid as long as stress doesn't exceed the yield strength of the material. Now if you don’t know the final temperature, you obviously need to know something about the final state. Let’s say you know the total heat input, Q. Since you know the initial state, you can find the initial internal energy, U. Then, given dU = Q for a closed vessel (from first law of thermo), the value for the final internal energy is Ui-Q. Now you go back to your REFPROP database, put in the final internal energy U and density and again, you have the final state, including pressure! It’s that easy! Remember that all you need are 2 known’s such as pressure and temperature or temperature and density or internal energy and density, and you have the state which can be looked up in the REFPROP database. Hope that helps. Last edited by a moderator: Hi Ron, Sorry if I sound like I’m singin’ from a soap box. I’ve heard the complaint about a company not wanting to pay for such software before though and it’s killin’ me. Think about it this way, what’s your charge-out rate? I suspect it’s on the order of$100 per hour unless you work for a small company with a much smaller overhead. Even then, I don’t suppose it’s too much less. So if you can save 2 hours worth of time, you can pay for the database. Look into NIST REFPROP. I’ve used it before and it’s not bad. I use a proprietary one that’s quite a bit more powerful, but the one from NIST has everything a typical user needs. You can purchase it for $200, or$300 with the upgrade. From the looks of all the work you’re putting into this problem, the database would have paid for itself a few times over by now. (note that methanol is included in the database)

This database can be linked to EXCEL, so you can write spread sheet programs to do what you need. Not sure what other programs it can be linked to – I see you mentioned mathcad.

Here’s how it works. You have the initial state with P and T. You also have a final temperature, so you need a final pressure. What you know about the system is that the two densities are the same (first order estimate). Since you have density and temperature of the final state, you can find any other property of the fluid. That means, you can find not just the pressure, but the internal energy, enthalpy, entropy, thermal conductivity, viscosity, etc… everything about the fluid is known with just the 2 variables, temperature and density.

If you want to go a step farther and consider the increase in vessel volume, you can guess at final pressure. This gives you some increase in vessel volume which changes your density. If the density is exactly equal to the mass of the fluid divided by the new, expanded vessel volume, you’re done. If not, do a goal seek to equalize them. That’s the short of it. Calculating the vessel volume is not that difficult. Hoop stress is P*r/t so divide by E and you have strain, then multiply by diameter to get the new diameter. For length, axial stress is ½ the hoop stress, so strain is ½ the strain of the diameter. Multiply by length to get your new length. This is all valid as long as stress doesn't exceed the yield strength of the material.

Now if you don’t know the final temperature, you obviously need to know something about the final state. Let’s say you know the total heat input, Q. Since you know the initial state, you can find the initial internal energy, U. Then, given dU = Q for a closed vessel (from first law of thermo), the value for the final internal energy is Ui-Q. Now you go back to your REFPROP database, put in the final internal energy U and density and again, you have the final state, including pressure! It’s that easy! Remember that all you need are 2 known’s such as pressure and temperature or temperature and density or internal energy and density, and you have the state which can be looked up in the REFPROP database.

Hope that helps.

I will check out that data base, I definatly need a data base of material properties. My bill out rate is 125$/hr and I make about 40$/hr, we used to be a small company (120 people) and I felt good about having hysys because the last company I was at was like 20 people and they had a single user version of hysys that you had to have a hardware key for and I had to check it out of the store room. Anyways since I have spent so much time trying to solve this problem could you help me with deriving this thing from first principles, I have sort of become obsesed with solving this, in fact in trying to solve this problem I have solved the joule thompson equation for T2. Also this will help with my thermodynamics study for the PE.

I started with dH = m*CpdT + V(1-T*beta)dP, at first I assumed dH = 0 which is true for a throttling a gas where there is no energy input which I was able to solve the first order differential equation. But for heat trace failure you have a Q = dH so the equation becomes Q = m*CpdT + V(1-T*beta)dP which I do not know how to solve differentially so I solved it directly using properties of methanol and came up with P = 29,000 psig, however this equation is very weakly dependant upon Q or V so when you account for the volumetric expansion of the steel and the pressure changes little I dont feel I worked the problem correctly when I account for the steel expansion and slight increase in volume the pressure should go down to an order of magnitude of 1000 to 2000 psig. I know that for joule thompson the T*(dV/dT)p term is evaluated using a gas EOS then the rest of the equation can be differentiated but in this case dV/dT is just the volumetric expansion of the fluid (and steel) which makes little difference in the volume.

Please help, I am loosing sleep over this.

Also as a side bar when you do a joule thompson you are typically loosing mass so when you have an equation such as:

T2= (C2/C1) + const/exp(-C1*P) the C1 and C2 terms have mass in them which causes the pressure drop, even if the pressure drop is given you dont know how much mass is lost to give you the new temperature unless you assumed it to be isothermal and used the ideal gas law just to get a rough dm to use in your equation?

Thank you for your help.

Q_Goest
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Hi Ron,
As far as figuring out fluid properties from any kind of basic principals, not sure how you can do that. The van der Waals equation of state for example, is an emperical one such that the constants are derived from measurement.

Looking at your equation:
dH = m*CpdT + V(1-T*beta)dP

I don't understand why you want to use enthalpy (H). Also, your reference to Joule Thompson, which refers to an isenthalpic process, seems a bit odd. As I'd mentioned before, the change in state of the methanol (or any fluid in a constant volume process) is determined from the change in internal energy, not enthalpy. So I guess you could write the equation
dU = m*Cv*dT
In which case you need an equation for Cv in terms of P and T over the range in question. Perry's might have that, but I don't. I've seen such equations for fluids before but never use them.

Hi Ron,
As far as figuring out fluid properties from any kind of basic principals, not sure how you can do that. The van der Waals equation of state for example, is an emperical one such that the constants are derived from measurement.

Looking at your equation:
dH = m*CpdT + V(1-T*beta)dP

I don't understand why you want to use enthalpy (H). Also, your reference to Joule Thompson, which refers to an isenthalpic process, seems a bit odd. As I'd mentioned before, the change in state of the methanol (or any fluid in a constant volume process) is determined from the change in internal energy, not enthalpy. So I guess you could write the equation
dU = m*Cv*dT
In which case you need an equation for Cv in terms of P and T over the range in question. Perry's might have that, but I don't. I've seen such equations for fluids before but never use them.

Also would you not want to include the VdP term, dU = mCvdT + VdP ?

Q_Goest
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Hi Ron,
Sorry if I've mislead you... I think I understand what you're trying to do now. I shouldn't have started talking about expanding pressure vessels, that's just going to complicate things. Pressure vessels and pipes are generally considered fixed volume. For example, CGA and API codes consider cases of pressure vessels and pipes with blocked in fluids where heat is added such that pressure relief valves (PRV's) can be properly sized. Let's neglect any volume increase and just consider how this particular fluid behaves and how the equations should be arranged, given a fixed volume and some heat input (ie: a "blocked in case" as you've called it). For that, the first law reduces to dU = Qin = m Cv dT where Cv is a function of pressure and temperature.

I believe what you're thinking is the PdV term accounts for an expanding pressure vessel. This would be true since the fluid is doing work (W = PdV) by expanding the vessel, but that complicates things unnecessarily. The amount of work done by the fluid in expanding a properly designed vessel or blocked in pipe should be considered negligable. Is that why you were putting the PdV term in - to account for an expanding pressure vessel? If so, my appologies, I'd suggest ignoring this factor and assume the volume change is zero. Generally, these kinds of calculations are done to determine the size of a PRV (sometimes also called a thermal relief valve) needed on the system - is that what you're doing?

Hi Ron,
Sorry if I've mislead you... I think I understand what you're trying to do now. I shouldn't have started talking about expanding pressure vessels, that's just going to complicate things. Pressure vessels and pipes are generally considered fixed volume. For example, CGA and API codes consider cases of pressure vessels and pipes with blocked in fluids where heat is added such that pressure relief valves (PRV's) can be properly sized. Let's neglect any volume increase and just consider how this particular fluid behaves and how the equations should be arranged, given a fixed volume and some heat input (ie: a "blocked in case" as you've called it). For that, the first law reduces to dU = Qin = m Cv dT where Cv is a function of pressure and temperature.

I believe what you're thinking is the PdV term accounts for an expanding pressure vessel. This would be true since the fluid is doing work (W = PdV) by expanding the vessel, but that complicates things unnecessarily. The amount of work done by the fluid in expanding a properly designed vessel or blocked in pipe should be considered negligable. Is that why you were putting the PdV term in - to account for an expanding pressure vessel? If so, my appologies, I'd suggest ignoring this factor and assume the volume change is zero. Generally, these kinds of calculations are done to determine the size of a PRV (sometimes also called a thermal relief valve) needed on the system - is that what you're doing?

Hello,

Yes that is what I was doing (we ended up just putting in a 3/4" x 1" thermal relief valve per API but I wanted to actually calculate the over pressure that could be seen), I was using a thermal expansivity term to calculate the volumetric expansivity. I will still need the volume of the system in order to get the mass but it is a good idea to not consider the PdV term. That being said I loose my P term and I would need an equation for Cv in terms of P in order to solve for P2. I got on the NIST web site and found the dU for my temperature range and density range which really helped (32676 lbf-ft/lbm from 70 to 150 F). I then plugged this into my mathcad sheet and got a pressure of ~1700 psig but this was assuming 1lb for any given volume which is incorrect because your mass changes depending on your volume. I almost feel like im trying to come up with my own equation of state (not really but I feel like it). I ultimatly would like to be able to correctly account for the expansion of the pipe because the pressure should go from like 22000 psia to like 1700 psig from the expansion of the pipe which is a really big deal, of course I was using the thermal expansion coefficient for methanol and I should be using the modulus of elasticity for steel. Also I was using VdP from the dH equation once I changed it to PdV I was able to get a P that was 7900 psig using the Cv and dU from the NIST web site this of course was using a thermal expansion coeficinet of methanol and I need to go back and see what equation you had to relate pressure and modulus of elasticity to get a new volume (expansion) and a new pressure due to that expansion.

Sorry for the delay I have been very busy and had to back burner this but this is definatly great thermo review for the PE and just keeping up with the basics, I was working things out as I wrote this email so I hope it makes sense.

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Hi Ron,
Sorry if I sound like I’m singin’ from a soap box. I’ve heard the complaint about a company not wanting to pay for such software before though and it’s killin’ me. Think about it this way, what’s your charge-out rate? I suspect it’s on the order of $100 per hour unless you work for a small company with a much smaller overhead. Even then, I don’t suppose it’s too much less. So if you can save 2 hours worth of time, you can pay for the database. Look into NIST REFPROP. I’ve used it before and it’s not bad. I use a proprietary one that’s quite a bit more powerful, but the one from NIST has everything a typical user needs. You can purchase it for$200, or \$300 with the upgrade. From the looks of all the work you’re putting into this problem, the database would have paid for itself a few times over by now. (note that methanol is included in the database)
http://www.nist.gov/srd/nist23.htm [Broken]

This database can be linked to EXCEL, so you can write spread sheet programs to do what you need. Not sure what other programs it can be linked to – I see you mentioned mathcad.

Here’s how it works. You have the initial state with P and T. You also have a final temperature, so you need a final pressure. What you know about the system is that the two densities are the same (first order estimate). Since you have density and temperature of the final state, you can find any other property of the fluid. That means, you can find not just the pressure, but the internal energy, enthalpy, entropy, thermal conductivity, viscosity, etc… everything about the fluid is known with just the 2 variables, temperature and density.

If you want to go a step farther and consider the increase in vessel volume, you can guess at final pressure. This gives you some increase in vessel volume which changes your density. If the density is exactly equal to the mass of the fluid divided by the new, expanded vessel volume, you’re done. If not, do a goal seek to equalize them. That’s the short of it. Calculating the vessel volume is not that difficult. Hoop stress is P*r/t so divide by E and you have strain, then multiply by diameter to get the new diameter. For length, axial stress is ½ the hoop stress, so strain is ½ the strain of the diameter. Multiply by length to get your new length. This is all valid as long as stress doesn't exceed the yield strength of the material.

Now if you don’t know the final temperature, you obviously need to know something about the final state. Let’s say you know the total heat input, Q. Since you know the initial state, you can find the initial internal energy, U. Then, given dU = Q for a closed vessel (from first law of thermo), the value for the final internal energy is Ui-Q. Now you go back to your REFPROP database, put in the final internal energy U and density and again, you have the final state, including pressure! It’s that easy! Remember that all you need are 2 known’s such as pressure and temperature or temperature and density or internal energy and density, and you have the state which can be looked up in the REFPROP database.

Hope that helps.

NIST must be getting the constant volume pressure using an equation of state because you cant get pressure using dU = mCvdT, I am guessing once you have the constant volume pressure you can solve for the reduced pressure using dU = mCvdT - PdV using the modulus of elasticity to get dV of course this dV will be very small and you will be dividing by dV which will sky rocket your P term. Man I am so lost and I should not be this lost.

Last edited by a moderator:
Q_Goest
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NIST must be getting the constant volume pressure using an equation of state because you cant get pressure using dU = mCvdT, I am guessing once you have the constant volume pressure you can solve for the reduced pressure using dU = mCvdT - PdV using the modulus of elasticity to get dV of course this dV will be very small and you will be dividing by dV which will sky rocket your P term. Man I am so lost and I should not be this lost.

Let’s just clear one thing up that I think will help. When you have P and T for a fluid, and assuming it is in a single phase (ie: either gas, liquid or supercritical, but not 2-phase) then if you have these two values, you can determine a unique U, H, S and rho for that fluid. We call this the state of the fluid since the fluid can be fully defined in every way from these two values, P and T.

Similarly, any 2 values will suffice. If, instead of P and T, I have P and U, the state of the fluid is defined. If all I have is P and U, I can look up T, H, S and rho.

Another example is if I only have U and rho. Again, having just these two values, I can determine the state of the fluid, including P, T, H and S.

Hopefully that’s clear. What it says is, you don’t need to think about such things as how Cv or Cp vary with temperature. You don’t need to consider how much the liquid ‘expands’ for a given pressure or temperature change. None of that is necessary.

For example, let's say you have the initial state prior to heating. That initial state is given by the pressure and temperature, but it doesn’t have to be. It could be given by rho and T, or P and rho, or rho and U. It only takes 2 values to determine any given state.

Now lets say you have this initial state and want to find the final state. In the case of the blocked in methanol, you already have the final density, because the final density is the same as the initial density, assuming there is no change in volume. With rho known, all you need is one more value and you’re finished. See what value you can find… If you know the final temperature, then you know rho and T which gives you the final state. All you do is take rho and T, plug it into the properties database, and it spits out everything else about the state, including pressure. Alternatively, if you know how much heat went into the fluid, then you can determine the final internal energy from dU = Qin. That is, if you have the initial state and final rho, and you know how much heat (Q) went in, you can find the final internal energy, so you now have rho and U for the final state. Take rho and U, plug them into your database, and you can find all the other parameters, P, T, H, S, etc…

By the way, the same basic process can also be used for a 2 phase fluid. It just may require different values such as vapor fraction instead of one of the other parameters.

Hope that helps…

Let’s just clear one thing up that I think will help. When you have P and T for a fluid, and assuming it is in a single phase (ie: either gas, liquid or supercritical, but not 2-phase) then if you have these two values, you can determine a unique U, H, S and rho for that fluid. We call this the state of the fluid since the fluid can be fully defined in every way from these two values, P and T.

Similarly, any 2 values will suffice. If, instead of P and T, I have P and U, the state of the fluid is defined. If all I have is P and U, I can look up T, H, S and rho.

Another example is if I only have U and rho. Again, having just these two values, I can determine the state of the fluid, including P, T, H and S.

Hopefully that’s clear. What it says is, you don’t need to think about such things as how Cv or Cp vary with temperature. You don’t need to consider how much the liquid ‘expands’ for a given pressure or temperature change. None of that is necessary.

For example, let's say you have the initial state prior to heating. That initial state is given by the pressure and temperature, but it doesn’t have to be. It could be given by rho and T, or P and rho, or rho and U. It only takes 2 values to determine any given state.

Now lets say you have this initial state and want to find the final state. In the case of the blocked in methanol, you already have the final density, because the final density is the same as the initial density, assuming there is no change in volume. With rho known, all you need is one more value and you’re finished. See what value you can find… If you know the final temperature, then you know rho and T which gives you the final state. All you do is take rho and T, plug it into the properties database, and it spits out everything else about the state, including pressure. Alternatively, if you know how much heat went into the fluid, then you can determine the final internal energy from dU = Qin. That is, if you have the initial state and final rho, and you know how much heat (Q) went in, you can find the final internal energy, so you now have rho and U for the final state. Take rho and U, plug them into your database, and you can find all the other parameters, P, T, H, S, etc…

By the way, the same basic process can also be used for a 2 phase fluid. It just may require different values such as vapor fraction instead of one of the other parameters.

Hope that helps…

I have been on the data base and know how it works, what I want to understand is how the data base is comming up with there numbers so I can modify the equations to account for the pipe expansion because ultimatly 22000 psig is going to be reduced to 1000 to 2000 psig when the pipe expands (ie not constant volume) and this is the number I am looking for but I first have to understand how the data base is comming up with the constant volume pressure before I can make that modification.

This is so I can close out a PHA action item, the PSV is already added and the flowrate is very easy to calculate so it wont be the end of the world if I cant get this but it would be nice to understand the thermo of non constant volume thermal expansion.

Q_Goest
Science Advisor
Homework Helper
Gold Member
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.