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Determining change in internal energy and enthalpy

  1. Feb 14, 2016 #1
    1. The problem statement, all variables and given/known data
    A 1 m3 rigid tank has propane at 100 kPa and 300 K. The tank is connected to another 0.5 m3rigid tank which has propane at 250 kPa and 400 K by a ball valve. The valve is opened and both tanks come to a uniform state at 325 K.

    Calculate the change in internal energy and enthalpy of propane in each tank.

    2. Relevant equations
    PV=ZmR(specific)T
    U = u*m [internal energy = specific internal energy * mass]
    H = h*m [same as above, but for enthalpy]
    ΔU = U2 - U1
    ΔH = H2 - H1
    U = Q - W [energy = heat transfer - work]
    W = ∫Pdv


    3. The attempt at a solution

    *note - work is irrelevant here because the tanks are rigid, so no forced volume change

    So this problem had other sub problems where I found the final pressure of the opened system, the initial Z factors, and the initial separated masses of propane.
    I labeled this system as 3 parts - Tank A (the m^3 tank), Tank B (the .5 m^3), and System O (the newly connected, opened valve system).

    Let's just assume for now that my calculations up to this point are correct, I just want to verify if I'm approaching the energy/enthalpy part correctly.

    Po = 139.87 KPa
    Ma = 1.785 Kg
    Mb = 1.69 Kg

    So I went online and found a steam table for superheated propane gas after double checking that the conditions for both tanks fall well within the gas phase on propane's phase diagram data. I found the following data.

    Initial specific internal energy of Tank A = ua1 = 478.2 KJ/Kg
    Initial specific enthalpy of Tank A = ha1 = 534 KJ/Kg

    Initial specific internal energy of Tank B = ub1 = 644.8 KJ/Kg
    Initial specific enthalpy of Tank B = hb1 = 717 KJ/Kg

    *note - the steam tables didn't have the exact conditions I was looking for for a few numbers, but I saw that at constant temperature, even very huge pressure increases didn't have much effect on specific energy/enthalpy properties. So I just used the closest numbers that made sense to use. If this approach is correct then I can just redo it with more exact data via interpolation.

    So my approach was to find the initial internal energy and enthalpy of a tank, and think of these as properties of the molecules in that tank. Then when the tanks are connected, find the final internal energy and enthalpy of those same molecules (so that same mass of propane), at the final pressure and temperature of the newly connected "System O". I would then find the difference, and this would give me the change in energy/enthalpy that the mass from this tank experienced.

    Ua1 = ua1 x Ma = 478.2 KJ/Kg x 1.785 Kg = 853.58 KJ
    Ua2 = 509 KJ/Kg x 1.785 Kg = 908.56 KJ
    Ha1 = ha1 x Ma = 534 KJ/Kg x 1.785 Kg = 953. 19 KJ
    Ha2 = 569.7 KJ/Kg x 1.785 Kg = 1016.91 KJ

    Ub1 = u1b x Mb = 644.8 KJ/Kg x 1.69 Kg = 1089.7 KJ
    Ub2 = 509 KJ/Kg x1.69 Kg = 860.21 KJ
    Hb1= hb1 x Mb = 717 KJ/Kg x 1.69 Kg = 1211.73 KJ
    Hb2 = 569.7 KJ/Kg x 1.69 Kg = 962.79 KJ

    ΔUa = Ua2 - Ua1 = 54.98 KJ
    ΔHa = Ha2 - Ha2 = 63.72 KJ
    (tank A's molecules experienced an increase in temperature and pressure conditions, so this seems reasonable to me)

    ΔUb = - 226.79 KJ
    ΔHb = - 248.9 KJ

    (tank B's molecules experienced a decrease in temperature and pressure conditions, so this seems reasonable to me)

    I'd like to know if there are any immediate, glaring issues with how I thought of this conceptually. Does this approach look sensible / correct? Are there any issues? If you see any issues with my numbers and just point them out that'd be a nice bonus. Any and all help is greatly appreciated, thanks a lot.
     
  2. jcsd
  3. Feb 14, 2016 #2
    It looks like you assumed that the mass of propane in each of the tanks does not change from the initial state to the final state. Is that correct?
     
  4. Feb 14, 2016 #3
    What method did you use to calculate the masses and pressure I came up with slightly different answer when reviewing this? Also to further on Chestermiller I was thinking the final mass in each tank is the total mass multiplied by the ratio of tank volume divided by system volume and compare that to the initially found mass. The total mass of the whole system does not change though
     
  5. Feb 14, 2016 #4
    I don't...think so. The way I imagined this was both tanks initially sealed, each one being a "set" of molecules that share a set of state variables and thermodynamic properties. Then when the valve is opened, the molecules are allowed to migrate around this new, larger volume system. So there are "Tank A molecules" floating around "Tank B" and vice versa, but now both tanks share the same thermodynamic properties and final state. So then I said "well now the Tank A molecules have undergone this process to reach this final state, and the Tank B molecules have undergone their process to reach the same final state, even if they're now both distributed across both tanks." If I imagine it that way, with the total masses within each tank changing, does that blow a hole in my approach here?
     
  6. Feb 14, 2016 #5
    First thing I did was find the Z factors in the initially sealed tanks, using a reduced pressure and temperature Z factor graph. I ended up with a Z of approximately .99 for Tank A and approximately .98 for Tank B.

    Then I found Propane's R constant in my textbook as .1886 KJ/Kg-K

    Then I rearranged the non-ideal gas law (PV=mZR(propane)T) to give me the mass - m = PV/ZRT

    for Tank A, I calculated m = (100 KPa * 1m^3) / (.99 * .1886 KJ/Kg-K * 300 K) = 1.785 kg
    for Tank B, I calculated m = 1.69 kg, the same way as above

    then I imagined the valve opening, and creating one unified system "O", for this new system I had
    To=325 K
    mo = 3.475 kg
    Vo = 1.5 m^3
    Po = ?

    To find Po I just rearranged the non-ideal gas law to give me P=mZRT/V, and used my known "O" properties. (However, since Z wasn't going to change much going off the chart, and have any real effect on calculations, I just averaged the initial z factors for .985)

    Po = [.985 * 3.475 Kg * (.1886 KJ/Kg-K) * 325 K ] / 1.5m^3 = about 139.87 KPa

    Do you see any issues that could be causing the variation?
     
  7. Feb 14, 2016 #6
    If you had thermodynamic tables for propane, then these tables should have had specific volumes tabulated. I don't understand why you used z factors to determine the masses in the tanks. You should have been using the tabulated specific volumes. Once you had the total mass and the total volume, you had the final specific volume after the valve opened. What was that? For that specific volume and temperature 325, you would have the other final properties. Try the problem this way and see what you get. Incidentally, I see no reason why you had to follow molecules from tanks A and B separately, since the molecules from the two tanks are indistinguishable.
     
  8. Feb 14, 2016 #7
    I see what you're saying about the specific volume. I've now gone back and found the mass for tank A that way, and I got 1.776 kg, very close to what I got originally. However, for Tank B I have a problem - for a temperature of 400K, the first time that temperature shows up in the superheated table is by pressure as high as 800 KPa. If I take that specific volume I get a mass of 5.53 kg, very different from my original. Is that not the right way to use the steam table here? I'm using this one by the way.

    https://engineering.purdue.edu/ME200/Thermodynamic_tables_SI_units.pdf [Broken]

    I should also mention that one of the other questions was to find the initial Z factors so I thought I'd use them once I had them. I assumed the problem was trying to lead me in that direction.
     
    Last edited by a moderator: May 7, 2017
  9. Feb 14, 2016 #8
    I see what you mean. I suggest using the tables at 2.5 bars and extrapolating the properties to 127 C. It shouldn't be too bad. Going to 8 bars isn't going to give you right specific volume. Also check the extrapolated specific volume with the volume you calculate with the ideal gas law and also with the z factor. You are going to have to use the extrapolation to get the internal energy and the enthalpy.

    Incidentally, I recognize those tables from Moran et al. It's a great textbook.

    Chet
     
    Last edited by a moderator: May 7, 2017
  10. Feb 14, 2016 #9
    Thank you so much! I'm not very handy at extrapolation on paper yet so I just used google sheets' built in version via dragging data sets. After finding the specific volumes at about 127 C at 2 and 3 bar, I averaged them and got a specific volume of .315 m3/kg, getting a mass of 1.586 kg for tank B. I'm glad, I would have been off by close to 100g.

    So after using the tables to redo my masses, I used the non-ideal gas law to find the final pressure since it was easier than extrapolating properties from the tables. I got 135.3 KPa this time, again pretty close to my original numbers, so that's reassuring.

    After extrapolating the specific internal energies and enthalpies, I found improved energy/enthalpy for Tank B, still somewhat close to my original estimations. I used the table for 100 KPa and 50 C since it's really close to the final conditions and pressure isn't really affecting energy/enthalpy at constant temperature, so now I've calculated the final internal energy and enthalpy for the connected system.

    So when all is said and done the numbers I have to work with are a bit improved from my messier process before, but I'm still confused as to how I should... follow?... the energy change. The problem asks to find the internal energy and enthalpy change of propane in each tank, so am I correct in assuming my original conceptualization was right?

    Basically, how exactly do I find the energy/enthalpy of one of the tanks after it's a connected system if I can't tell how much of the total propane is in it, assuming the propane migrates? I'm guessing I actually need to assume the propane masses stay put in their respective tanks?

    How would you interpret it?
     
  11. Feb 15, 2016 #10
    You find the new amounts of propane in each of the tanks after the equilibration. Then you multiply the new amount in each tank by the new specific enthalpy or new specific internal energy of the combined system. I know it doesn't make much sense to compare in this way, because you are talking about different amounts of mass in each tank, but that's what they ask for. If I were them, I would have just asked for the changes in overall internal energy and enthalpy.
     
  12. Feb 15, 2016 #11
    Unfortunately I think the energy/enthalpy answers I turned in were incorrect (I saw your reply just now, had to turn in this morning in a rush), I think I understand how to have found the new masses now. You'd extrapolate/average specific volume to find it at the final conditions, divide each tank volume by that specific volume to find the mass in each tank. Correct? The masses seem to match up to the total mass pretty well and follow the volume ratio of Tank A to Tank B as I was hoping.

    I'll probably end up finding the correct energy/enthalpy changes at some point when the professor posts the solutions and just do the problem again for practice, but does that seem correct?
     
  13. Feb 16, 2016 #12
    Yes.
    They should match up exactly.
    Yes
     
  14. Feb 16, 2016 #13

    Then again, thank you very much for all your help. I really appreciate it!
     
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