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Thermodynamics spring problem

  1. Jun 6, 2016 #1
    Hi guys,
    i just discovered this forum and this is my first post so apologies if i made any mistake with the how im suppose to structure a thread. but nontheless any help would be greatly appreciated!
    cheers

    1. The problem statement, all variables and given/known data

    4. A 250mm diameter insulated cylinder (see figure below) is fitted with a frictionless, leakproof piston, which is attached to an unstressed spring S (spring constant, k = 175kN/m). The cylinder is connected by a closed valve to the mains pipe, in which air flows at a high pressure and a temperture of 27°C. The spring side of the cylinder is open to the atmosphere (Patm = 101kPa).

    The valve is now opened slowly permitting air to move the piston through a distance of 150mm, thus compressing the spring. At this point the valve is closed, trapping air in the cylinder.

    a) Write down the 1st law of thermodynamics for an open system and simplify it for this charging process, which can be assumed to be adiabatic. b) Calculate:

    i) The final pressure of the air in the cylinder.
    ii) The final temperature of the air in the cylinder.
    iii) The mass of air injected into the cylinder.

    answers:
    [636kPa, 340.8K, 0.0479kg]
    3. The attempt at a solution
    i've managed to get the first answer of P2. but im stuck on the 2nd part. am i suppose to use
    P1/T1=P2/T2
    any help would be appreciated!
     
  2. jcsd
  3. Jun 6, 2016 #2

    Simon Bridge

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    Welcome to PF;
    ... that formula would be appropriate if the volume did not change: does it?
    Why not use your answer to (a) to help you here?
     
  4. Jun 6, 2016 #3
    yes the volume does change. but initial volume is 0 because the piston is flushed against the cylinder so i dun really thing that would work..
    how do i use the part (a) answer to help me?
     
  5. Jun 6, 2016 #4

    Simon Bridge

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    You asked about an equation that requires volume to remain the same, but you notice that it changes. Therefore: do not use that equation. The initial volume being zero makes no difference to these facts.

    Read question (a) again.
    1. what is the name of the process?
    2. what was the simplified 1st Law equation you came up with?
     
  6. Jun 6, 2016 #5
    1) i think its an adiabatic system as mentioned in (a)
    2) im not sure if im right but i got -w=m(h2-h2)
     
  7. Jun 6, 2016 #6
    You are working with the open system version of the first law. This is not a steady flow situation, so the equation you wrote down is not applicable. Does the internal energy of the system (i.e., the contents of the cylinder) change? You have mass entering the system, but no mass leaving. Please write down the form of the first law that applies to this situation.
     
  8. Jun 6, 2016 #7
    i see.. so what you mean is the equation should be -w+mihi=m2u2-m1u1?
    im not sure if there is w though but if the spring compresses then there should be work done right?
     
  9. Jun 6, 2016 #8
    Much better. So, if m is the mass that enters, then, in the final state:$$mu=-W+mh_{in}$$
    You are correct that (shaft) work is done on the surroundings. Algebraically, what do you get for the shaft work done (on the piston)? How does the enthalpy and temperature of the air entering the system hin and Tin compare with the enthalpy and temperature of the air in the mains pipe (assuming ideal gas)?
     
  10. Jun 6, 2016 #9
    the enthalpy and temperature of air entering the system should be the same as that in the pipe isnt it?
     
  11. Jun 6, 2016 #10
    Yes.

    From a force balance on the piston, what is the final pressure P in terms of x, k, and patm?
     
  12. Jun 6, 2016 #11
    the final pressure i calculated was 635.8kPa, there shouldnt be an x right since its air? assuming you meant x as in quality..? k would be 1.4 for air and Patm was given in the question at 101kPa.
     
  13. Jun 6, 2016 #12
    Please do it algebraically. There is a good reason I'm asking for this.
     
  14. Jun 6, 2016 #13
    okay correct me if im wrong but what you're asking for is P=Patm+k*x/Areapiston
     
  15. Jun 6, 2016 #14
    Perfect. Now, the final volume is V = Ax. So, again, in terms of these same parameters, what is the final value of PV?
     
  16. Jun 6, 2016 #15
    that would make it P2*Ax?
     
  17. Jun 6, 2016 #16
    Yes. That would be $$PV=\left(P_a+\frac{kx}{A}\right)(Ax)=P_aAx+kx^2=P_aV+kx^2$$
    So now you have the two equations:$$U=mh_{in}-W$$and$$PV=P_aV+kx^2$$
    What do you get if you add these two equations together? (You still haven't answered my question about the algebraic expression for W in terms of k, x, A, and Pa)

    Chet
     
    Last edited: Jun 6, 2016
  18. Jun 6, 2016 #17
    erm im not sure..
     
  19. Jun 6, 2016 #18
    You're not sure how to algebraically add two equations together by adding their left sides together and their right sides together?
     
  20. Jun 6, 2016 #19
    oh.. PV+U=PaV+kx2+mhin−W is this what you mean?
     
  21. Jun 6, 2016 #20
    Yes. Now, on the left hand side of this equation, you have U + PV. What thermodynamic function is that, and what does it represent physically in this problem?

    I need to go to bed now, so I won't be available for a few hours.

    Chet
     
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