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Change in u and h is for a piston cylinder device.

  1. Jun 3, 2013 #1
    1. The problem statement, all variables and given/known data
    0.03m^3 of air is contained in the spring-loaded piston-cylinder device. The spring constant is 875N/m and the piston diameter is 25cm. When no force is exerted by the spring on the piston the state of the air is 2250kPa and 240C. This device is now cooled until the volume is halved. Determine the change in internal energy and specific enthalpy.


    2. Relevant equations
    du=Cv(T2-T1)
    dh=Cp(T2-T1)
    Pv=RT
    P2A = PatmA - F spring
    3. The attempt at a solution
    I'm having trouble getting the P2.
    1. The problem statement, all variables and given/known data



    2. Relevant equations

    http://imgur.com/sRT21x6
    http://minus.com/lE1w4b9sdcKqd
    3. The attempt at a solution
     
    Last edited: Jun 3, 2013
  2. jcsd
  3. Jun 3, 2013 #2

    rude man

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    << When no force is exerted by the spring on the piston the state of the air is 2250kPa and 240C. >>

    What's holding the piston in at p = 2250 kPa if the spring is not exerting any force to it?
     
  4. Jun 3, 2013 #3
    I have no idea. I guessed that it was the weight of the piston maybe but there was no mention of it.
     
  5. Jun 3, 2013 #4

    rude man

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    That would make sense.

    In that case, you have enough info to compute p2 and of course V2 = V1/2 given so you can also compute T2. Assuing ideal gas law of course.

    Then there's the 1st law of thermodynamics and your other stated equations.

    I don't see dh = Ch(T2-T1) though. What's Ch?
     
  6. Jun 3, 2013 #5
    That's the thing though, when I plug in the numbers I get p2 = 95.9Kpa which is not even close to the correct answer of 2244.54 Kpa

    Oh and sorry I meant Cp and not Ch.
     
  7. Jun 3, 2013 #6

    rude man

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    Show us how you plugged in the numbers to get p2.

    OK on the dh.

    BTW I confirmed the answer is correct for p2.
     
    Last edited: Jun 3, 2013
  8. Jun 3, 2013 #7
    Fs = kx where x= V2-V1 / A = .306m

    Fs = 875(.306) =267.75 N

    A = D^2/4

    P2 = 101375 - 267.75/ (.049)

    P2 = 95.9 kPa
     
  9. Jun 3, 2013 #8

    rude man

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    where did this come from? It's not p1.
    P2 = 95.9 kPa[/QUOTE]
     
  10. Jun 4, 2013 #9
    [/QUOTE]

    That is the P atmosphere. Why would it be P1?
     
  11. Jun 4, 2013 #10

    rude man

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    That is the P atmosphere. Why would it be P1?[/QUOTE]

    Because that was the pressure when the spring was relaxed.

    p2 = p1 - pressure of the spring pulling up on the piston. Remember, we decided the piston had the mass needed to keep it at p1 initially?
     
  12. Jun 4, 2013 #11
    I think I get it now but why do we neglect the effect of atmospheric pressure on the piston?
     
  13. Jun 4, 2013 #12

    rude man

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    Because it's included in the force that keeps the piston down in state 1. I assumed it was just the weight of the piston but the atosphere of course adds to it. But no matter - p changed from p1 to p2 and the difference is entirely accounted for by the spring stretching. Atmospheric pressure is included in both p1 and p2 as is the weight of the piston. Obviously, considering how high p1 is, atmospheric pressure is << pressure due to the weight of the piston.
     
  14. Jun 4, 2013 #13
    Thank you. I actually get it now.
     
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