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Thermal Expansion of Methanol

  1. May 3, 2008 #1
    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.
     
  2. jcsd
  3. May 3, 2008 #2

    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

    [tex]0.00116(^\circ\mathrm C)^{-1}=0.000642(^\circ\mathrm F)^{-1}[/tex]

    If your equation gives an imaginary number for a physical quantity, it simply isn't applicable for that set of input variables.
     
  4. May 3, 2008 #3
    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.
     
  5. May 4, 2008 #4
    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.
     
  6. May 4, 2008 #5

    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.
     
  7. May 5, 2008 #6
    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.
     

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    Last edited: May 5, 2008
  8. May 5, 2008 #7
    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?
     
  9. May 5, 2008 #8

    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.
     
  10. May 5, 2008 #9
     
    Last edited: May 5, 2008
  11. May 5, 2008 #10

    Q_Goest

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    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.
     
  12. May 6, 2008 #11

    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.
     
  13. May 6, 2008 #12
    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.
     
    Last edited: May 6, 2008
  14. May 6, 2008 #13

    Q_Goest

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    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.
     
  15. May 6, 2008 #14
    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
     
  16. May 6, 2008 #15
    How did you get a Q only given a deta T with P2 unknown?
     
  17. May 6, 2008 #16

    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

    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.
     
  18. May 7, 2008 #17
    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.
     
  19. May 8, 2008 #18

    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.
     
  20. May 30, 2008 #19
    Also would you not want to include the VdP term, dU = mCvdT + VdP ?
     
  21. May 30, 2008 #20

    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?
     
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