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Adiabatic Processes and Ideal Gases

  1. May 18, 2006 #1
    Hey.

    Im doing this question where Air (said to be ideal gas) is compressed from a pressure of 0.1 MPa and Temp 300K to pressure .5 MPa and temperature 480K. We are supposed to determine whether the process is possible to do adiabatically (sorry if that is the wrong spelling).

    I hadnt done a question like this before so i began by assuming it was an adiabatic process. by doing this and using thermodynamic data from Moran and Shapiro (our text book) i got the work done on the system to be approx. 130 kJ/kg

    I then went on to try and use the ideal gas equation to determine the work required for this process (using the definition, work = integral (pressure) dV)

    now, to do this i had to make the assumption that for an ideal gas an Adiabatic process is also polytropic (pV^n = const.)

    My notes say that polytropic processes do not have to be adiabatic, but it does not describe whether or not adiabatic processes have to be polytropic (my assumption).

    My question is,

    If a process is adiabatic, does it mean it is also polytropic?

    PS.

    Using my above method i get a rather nice answer, being that it is not possible, but i do not enjoying making assumptions.

    cheers,

    bart
     
  2. jcsd
  3. May 18, 2006 #2

    siddharth

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    Why do you need to find the work done?

    An adiabatic process is one in which no heat is gained or lost by the system (ie, Q=0). So, just apply the definition here.

    Isn't this question very straight-forward?
     
  4. May 18, 2006 #3
    when you apply the definition and the enrgy balance equation all you can calculate is the work. i understand what an adiabatic process is, but i was unsure how to determine whether it is possible for this process so i began by looking at how the process being adiabatic might affect it.

    i apologise, maybe i was not clear but we are supposed to determine if it is possible to do adiabatically. maybe a better question is why wouldnt this process be possible to do adiabatically.

    we are covering the second law of thermodynamics at the moment, but all my notes on the topic are for cycles, which this is not. so im am not sure how it may be applied here.

    cheers,

    bart
     
  5. May 18, 2006 #4

    Andrew Mason

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    The adiabatic process is also polytropic because the adiabatic condition:
    [tex]PV^\gamma = \text{constant}[/tex]

    applies. In the case of the adiabatic process, [itex]n = \gamma = C_p/C_v[/itex]. So, find out the value of [itex]\gamma[/itex] for air and see if the relation applies.

    AM
     
  6. May 19, 2006 #5

    siddharth

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    You can first calculate the change in internal energy and then apply the fist law to calculate the work. So, it's obvious to me that you can have an adiabatic process between the 2 states, and the corresponding work, energy and change in entropy can be calculated. Note that this adiabatic process may be irreversible also.

    If you are specifically trying to see if the process can be both reversible and adiabatic, remember that the change in entropy in a reversible adiabatic process is 0. So, calculate the change in entropy between the two states. If it's non zero, then the process can't be reversible adaibatic.
     
  7. May 19, 2006 #6

    Andrew Mason

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    [itex]\gamma = 1.4[/itex] for air.

    Since PV=nRT and

    [tex]PV^\gamma = K[/tex]

    [tex]V = (KP)^{-1/\gamma} = nRTP^{-1}[/tex]

    So in terms of temperature and pressure:

    [tex]TP^{1/\gamma - 1} = \text{constant}[/tex]

    Plugging in the values given and :

    [tex]300 * .1^{-.286} = 580 \approx 480 * .5^{-.286} = 585 [/tex]

    So the answer should be yes.

    AM
     
    Last edited: May 19, 2006
  8. May 19, 2006 #7

    siddharth

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    When you use [itex] PV^{\gamma} = K [/itex], you are assuming that the process is reversible (or fully resisted), which need not be true, right?
     
  9. May 19, 2006 #8

    Andrew Mason

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    One can expand a gas adiabatically but non-reversibly (free expansion). This is because there is no work done in a free expansion. But when a gas is compressed the incremental work done to it is Pdv where P is the pressure of the gas itself, not the external pressure.

    So I am having difficulty seeing how you could have an adiabatic compression (ie without exchanging heat with the surroundings) without it being fully resisted. Can you give an example?

    AM
     
  10. May 19, 2006 #9

    siddharth

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    That's what I don't understand actually. The work done is usually [tex] P_{ext} dV [/tex], right? And in a fully resisted process, [tex] P_{gas} = P_{ext} [/tex], at every stage because the process is reversible.


    Ok. Let's say you have a completely insulated tank, filled with air at some low pressure. On the top is a movable and insulated piston open to the atomsphere which is at a higher pressure than the air inside. There is a latch present which prevents the piston from falling initially.

    When the latch is removed, the piston will fall and I think the resulting process will be an irreversible adiabatic process, which won't be fully resisted.
     
    Last edited: May 19, 2006
  11. May 20, 2006 #10

    Andrew Mason

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    In compression, the work done on the gas would be [itex]P_{gas}dV[/itex]. In expansion, the work done by a gas on its surroundings would be [itex]P_{ext}dV[/itex]. If the compressing pressure on the gas is greater than the internal pressure of the gas, the compressing surface has to accelerate. This imparts additional kinetic energy in the gas which is eventually results in additional heat being added and we know this cannot all be converted back to mechanical work. So it is non-reversible. But you could say it is also not adiabatic since additional heat is added (by converting external work into heat).

    This is not a reversible process. You are right. But the falling (frictionless) piston does not result in immediate thermodynamic equilibrium. The piston, having kinetic energy, compresses the gas beyond the pressure sufficient to balance the weight of the piston. This causes the gas to push the piston back up and the piston oscillates up and down.

    The energy of the resulting oscillating compression wave inside the gas eventually dissipates as heat. So, again, there is effectively heat added to the gas. In my view, one could argue that such a process is non-adiabatic.

    The whole problem is one of defining 'adiabatic'. In my view, the term 'adiabatic' should be used only to reversible processes because 'adiabatic non-reversible' processes effectively transfer additional heat to the gas.

    AM
     
  12. May 20, 2006 #11
    what i did in the end for this question was look at the entropy change in going from state 1 to state 2.

    this turned out to be negative.

    i then looked at the formula for a closed system;

    s2-s1 = integral (delta Q x 1/T) + entropy production

    for an adiabatic process, integral (delta Q x 1/T) = 0, this would mean that the entropy production would be less than zero which is not possible.

    from this i concluded the process is not able to be done adiabatically. now, i had to assume it was a closed system which i believe is a fair assumption.

    does that seem ok?
     
  13. May 20, 2006 #12

    Andrew Mason

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    The approach is right to begin with. The change in entropy of a gas undergoing a reversible adiabatic process is necessarily 0. If you determine the entropy change to be non-zero, you will know that it is not a reversible adiabatic process. You don't have to go any further. How did you determine that the change in entropy was negative? Can you show us your work?

    AM
     
  14. May 20, 2006 #13

    Andrew Mason

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    In significant figures, I get the change in entropy to be 0.

    Since ds = dQ/T and dQ = dU + dW where dW is the work done by the gas:

    (1)[tex]\Delta S = \int_{T_{i}}^{T_{f}} dQ/T = \int_{T_{i}}^{T_{f}} dU/T + \int_{T_{i}}^{T_{f}} dW/T[/tex]

    Since d(PV) = PdV + VdP = nRdT and dW = PdV, dW = nRdT - VdP = nRdT - nRTdP/P

    [tex]dS = dQ/T = (dU + dW)/T = nC_vdT/T + nRdT/T - nRdP/P = nC_pdT/T - nRdP/P[/tex]

    [tex]\Delta S = \int_{T_{i}}^{T_{f}} dS = nC_p\int_{T_{i}}^{T_{f}} dT/T - nR\int_{P_{i}}^{P_{f}} dP/P[/tex]

    [tex]\Delta S = nC_p\ln\left(\frac{T_f}{T_i}\right) - nR\ln\left(\frac{P_f}{P_i}\right)[/tex]

    C_p for air is: 3.5R

    In this case:

    [tex]\Delta S = nR(3.5*\ln(480/300) - \ln(.5/.1)) = nR(1.6-1.6) = 0[/tex]

    AM
     
    Last edited: May 20, 2006
  15. May 21, 2006 #14

    siddharth

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    Yeah Andrew, that's perfectly right.

    I understand what you say about it being practically not possible to achieve a adiabatic compression process, which isn't fully resisted.
     
  16. May 21, 2006 #15
    Im doing Thermodynamics as part of an engineering course so its more equatoins and tables of thermodynamical data (this leaves me feeling rather empty, thus i try to understand the origins of the equations as much as i can).

    I just found a mistake in one of my calculations because some of my units were in kJ/kgK and some were in J/kgK

    revised, i use the equation;

    s2-s1 = s0(T2)-s0(T1)-Rln(p2/p1) = 2.17760-1.70203-(8.314/28.97)ln(0.5/0.1) = 0.01368 kJ/kgK

    By the procedure i was using before, this means that entropy production would be greater than 0 and the process possible. Though i wonder how the entropy production occurs.

    I got the s0 values from a table and used that to calculate the entropy change (the s0 values are just difficult to calculate integrals that were determined experimentally). Im not sure you guys would have the same text book we use here, but it was Fundamentals of Engineering Thermodynamics by Moran and Shapiro table A-22

    The reason i look at entropy production is because the question does not say the process has to be reversible. it simply asks if it is possible to be done adiabatically.

    Thats why i thought i should show that for the process to be adiabatic entropy production was greater than or equal to 0. Otherwise i dont think the solution would have been complete.
     
    Last edited: May 21, 2006
  17. May 21, 2006 #16

    Andrew Mason

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    The value for R is 8.31451 m^2·kg/s^2·K·mol Where do you get the 10^3/28.97? Are you using table S values for air or a diatomic gas? What units are they in? If you apply significant figures, your answer is actually 0.

    In any event, since the adiabatic condition, [itex]P_iV_i^\gamma = K = P_fV_f^\gamma[/itex] applies to within significant figures, it can be done adiabatically and reversibly. Entropy change is also 0 to within significant figures which also means it is adiabatic.

    I should also point out that an entropy change of a system can be negative. It is the entropy of the system and its surroundings (universe) that cannot decrease. In this case, you do not have a closed system. The gas can is being compressed due to work supplied from its surroundings. In the example discussed above by siddharth of a piston dropping and compressing a gas, the gas could have a negative entropy change.

    AM
     
    Last edited: May 21, 2006
  18. May 22, 2006 #17
    We are taught what is probably not standard notation. But in the equations i have been given R is not the universal gas constant, rather it is the gas constant for the gas. This is the universal gas constant divided by the molar mass of the gas.

    In this case

    R= (8.314 kJ/kmol K) / (28.97 kg/kmol) - i accidently forgot to remove the x10^3 when iedited it, sorry

    The values in my previous post are all for kJ and kg, and as i said, i still get a positive number, not zero or even one that rounds to zero.

    You didnt say if you found the same textbook i was using, but if you did i am using Equation 6.21a in chapter 6.

    In regards to your last point, i do realise that entropy change of the system does not need be positive in general, BUT, in this case, where we also have the definition for entropy change as;

    s2-s1 = integral (ds/T) + entropy production

    simplifying (with ds=0) to;

    s2-s1 = entropy production

    surely entropy change has to be greater than or equal to 0, otherwise it implies negative entropy production which (according to the countless reminders all through all my texts and notes) is not possible.
     
    Last edited: May 22, 2006
  19. May 22, 2006 #18

    Andrew Mason

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    Your pressures are .5 and .1 mPa respectively. Not .500 and .100. So how can you be more accurate than 2 figures? If you are accurate to two significant figures, the answer is 0. An answer in more than two significant figures is simply wrong.

    If you had pressures of .518 mPa and .100 mPa respectively, the entropy change would be 0.000 J/K-mole.

    AM
     
    Last edited: May 22, 2006
  20. May 22, 2006 #19
    The value i get is 0.01368 kJ/kg K which equals 13.68 J/kgK and in 1 significant figure that is 10 J/kgK - surely this would be considered non zero in one significant figure.

    Im not trying to argue, but what is the more correct answer?
     
  21. May 23, 2006 #20

    Andrew Mason

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    I am not sure where you get such a large value for change in entropy. I get less than 1 J/K-mole. Perhaps you should check to see if your units of KJ/K-mole are correct. I note you are using the gas constant 8.314 which has units of J/K-mole so I am not sure how you end up with KJ.

    I can show you why this is very very close to an adiabatic compression in which the condtion [itex]PV^\gamma = K[/itex] applies.

    Using PV=nRT,

    1. [tex]P_iV_i = nRT_i[/tex]

    2. [tex]P_fV_f = nRT_f[/tex]

    Dividing 1. by 2. we get:

    [tex]P_iV_i/P_fV_f = T_i/T_f[/tex]

    [tex]V_i/V_f = T_iP_f/T_fP_i[/tex]

    Plugging in values:

    [tex]V_i/V_f = 300*.5/480*.1 = 3.125[/tex]

    So, letting the final volume be 1 unit:

    [tex]P_iV_i^\gamma = .1*3.125^{1.4} = .1 *4.929 \approx .5[/tex] to sig. figs.

    [tex]P_fV_f^\gamma = .5*1^{1.4} = .5[/tex]

    AM
     
    Last edited: May 23, 2006
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