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Resolve the Thermo Paradox

  1. Nov 8, 2005 #1

    Q_Goest

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    Imagine two infinitely large air tanks connected by a long, straight pipe. Tank A is at a higher pressure than tank B, so air is flowing from one to the other. For the sake of argument, we will say there is no heat flux or work into or out of the pipe.

    Now we put a perfectly insulated cylinder inside tank A that is fitted with a piston which is free to move, though there can be no heat transfer through the cylinder walls so the air inside the cylinder undergoes an adiabatic expansion/compression as the piston moves. There is air inside this cylinder also which is at the same temperature as the air in tank A, so the state of the air inside the cylinder and the state of the air inside tank A is the same. Since the piston is free to move, the pressure of the air inside the cylinder and the pressure of the air outside the cylinder is ALWAYS the same. If air pressure dropped outside the cylinder, the piston would move to allow the air inside to be at the same pressure as the air outside.

    The cylinder is now moved from tank A to the inlet of this pipe and allowed to travel down the pipe to tank B. As it does, the pressure in the pipe drops and the piston moves out of the cylinder so that pressure inside and outside the cylinder remains the same. When it gets to tank B, the pressures are the same.

    Question: Will the air inside the cylinder and the air in tank B be at the same state when the cylinder arrives at B? Note that the pressure will be the same, but that does not necessarily mean the temperature will. Why or why not?

    Observations:
    - The air inside the cylinder represents a control mass that undergoes an adiabatic expansion and the air is doing work on the piston but the pressure inside the cylinder and outside the cylinder are the same.
    - A mass of air flowing through the pipe can also have a control surface drawn around it, and that control mass can be made equal to the control mass of the air inside the cylinder.
    - The control mass of air flowing through the pipe also experiences an adiabatic expansion as it goes from tank A to tank B, but it is doing no work.
    - If we write the first law for the air in the cylinder, we obtain dU = W
    - If we write the first law for the air in the pipe, we obtain dH = 0

    Both the air in the pipe and air in the cylinder undergo seemingly equivalent expansions as they progress from tank A to tank B, they are both adiabatic, and both maintain the same pressure as they expand going down the pipe. But from the perspective of the first law, the end states are different. Are they different, the same or is there an error in the observations?
     

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  3. Nov 9, 2005 #2

    Q_Goest

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    Just curious, is it clear why this an apparant paradox?

    If it is clear, is it simply too difficult to reconcile?
     
  4. Nov 9, 2005 #3

    GCT

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    will consider when I have the time, right now I'm way too busy.
     
  5. Nov 10, 2005 #4

    GCT

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    where did you obtain this problem?
     
  6. Nov 10, 2005 #5

    Q_Goest

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    I made it up. Hopefully it provides some insight into the pV and work issue we were discussing in the general physics area.
     
  7. Nov 10, 2005 #6

    Q_Goest

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    If instead of putting a cylinder fitted with a piston into the pipe, we put a frictionless piston into the pipe that could seal on the ID of the pipe, and a second identical piston just a few inches behind it, what process would the gas between the two pistons experience as it flowed from tank A to tank B? Would the internal energy be converted to work, or would it be isenthalpic?
     
  8. Nov 12, 2005 #7

    GCT

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    well from a quick consideration an isenthalpic process indicates a relationship between the work of expansion/compression with that of internal energy (for this adiabatic process). It's the expansion/compression which differentiate the change in internal energy and temperature in this case. I don't think though that the process has to be perfectly isenthalpic for the temperature difference, it is an adiabatic process after all. I think that in order for this situation to be considered isenthalpic you're gonna need more stringent conditions. My point is that work is converted into internal energy, whether isenthalpic or not.

    I've worked out some equations from a similar experiment, which I may want to discuss later.
     
  9. Nov 12, 2005 #8

    Q_Goest

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    Hey there GCT. Thanks for the responce, thought this one was going to drop off the edge of the world and never come back! lol

    Maybe this paradox is too tough to resolve for the Physics Forum folks. <proding other PF'ers>

    Would you agree that adiabatic flow through a pipe is isenthalpic? Why or why not?

    Would you agree that a perfectly insulated cylinder fitted with a piston results only in the conversion of internal energy to work as the piston is moved? Why or why not?

    Not sure I understand what you mean here. (which process?) Please clarify.
     
  10. Nov 13, 2005 #9

    GCT

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    well H=U+PV, there's no heat input, does the work of expansion/compression contributes to the change in internal energy whether isenthalpic or not. I understand what you're trying to get at here, I was thinking the same thing after reading up on the kelvin's version of the joules/thompson experiment, the diagram shows (Atkins) a complete compression by the left handed piston and a complete expansion...so you begin to wonder on if that was not the actual case (partial compression, then a greater expansion) and if that makes any difference. Right now, I'm quite busy, so I hadn't researched it too much.
     
  11. Nov 15, 2005 #10

    Q_Goest

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    Hey GCT, I'd like to give you my perspective on this paradox. I find it pretty interesting, but not beyond explanation, and I think the explanation can help others here in understanding thermo and the first law in particular just a bit better.

    I had a guy show me a form of the first law equation that I've found tremendously useful. I'd already graduated from college with a BS in mechanical engineering, and had worked at a large aerospace company as a thermodynamics analyst for 8 years when I moved to the company I work for now and this guy with a PhD showed me a new way of looking at the first law. That was 9 years ago, and I've found this form of the equation to be tremenously useful, much more so than the more simple U (or E) = Q + W .

    Attached is a diagram showing a generic control volume and the first law written out in a slightly different way than usual. The only real difference is that enthalpy is added to the equation which makes all the difference in the world when trying to understand and apply it. In words, the change in internal energy inside some control volume is equal to the heat, work and enthalpy going into and coming out of the CV. We can of course add PE and KE if we want to be rigerous, but those terms are generally neglected since they are almost always zero or exceedingly small. Note also that heat, work or enthalpy into a CV is generally considered positive, and out of a CV it is generally considered negative.

    Applying a control volume such as this to any portion of the pipe with a gas flowing through it, we find the following:
    dU = 0 because there is no energy stored inside the CV. The CV remains constant.
    dW = 0 because no work is put into or removed from the CV. Generally, work is considered energy removed from the fluid as work, not pV energy.
    dQ = 0 since we defined this process as adiabatic earlier. Certainly a fluid can gain or reject heat as it flows through a pipe, but we will only look at the examples given above.
    H in = mass flow in times specific enthalpy. Note the enthalpy is "… the total energy of a body (fluid) is its internal energy plus the extra energy it is credited with by having a volume V at pressure p." (from MIT Web site)
    H out = similarly mass flow out times specific enthalpy.

    So applying the first law using the CV described on the attached and the first law equation derived from that, we find the flow through a pipe is isenthalpic.

    If we apply the control volume to the cylinder expanding with a piston on it, we find a different outcome though. By putting the CV only around the expanding fluid, we find the change in internal energy is equal to the work that leaves the CV. This is a very different outcome. For example:

    For isenthalpic expansion of air from 500 psig and 70 F to 0 psig, the temperature of the air drops to 54.7 F (from computerized database I use).

    But if the air is expanded and allowed to do work from 500 psig and 70 F to 0 psig, the temperature of the air drops to 50.7 F.

    So there's a significant difference between the two processes. The question is, why should there be a difference, the control volumes seem identical?

    The difference is because the gas flowing through the pipe has molecules/atoms bouncing off the walls of the pipe which resist the flow and create a pressure drop. Note that the molecules bouncing off the walls do so by a completely elastic collision since we previously defined this as an adiabatic process. This process can be viewed as work done by the gas on itself.

    The cylinder fitted with a piston has no such work, it only does work on the piston, so the internal energy leaves the gas and it does this because the wall is receeding away from the gas as the molecules bounce off the moving piston. Momentum and energy is given up by the molecules and put into the motion of the piston.

    In conclusion, one could look at the pV energy term in enthalpy as a work term that is already part of the fluid in a given state, or one can look at the pV energy term as work which is exchanged with the internal energy of the fluid. But if we say it is a work term, we must be very specific to say it is work done by the gas on the gas itself as it expands. I'm not convinced there's any benefit in looking at it either way, so long as one understands the concepts behind it. I rather like the way the MIT web site has it defined though.
     

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  12. Nov 16, 2005 #11

    GCT

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    hmm, very nice, thanks for the info I'll be studying it more throughouly when I have the time.
     
  13. Nov 17, 2005 #12

    Q_Goest

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    Correction. Temp drops to -271 F. My mistake.
     
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