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Work calculation for polytropic processes

  1. May 24, 2013 #1
    Hi, something is a bit confusing in regards to polytropic processes.Can anyone explain why boundary work in closed system for compressor is calculated as W = ∫pdV and polytropic process after integration is generally W = -(p2V2 - p1V1)/(1-n)...(numbers behind p and V are subscripts)...
    When we talk about control volume, internally reversible process (isn't above one also reversible???) then work is defined as W = -∫Vdp and for polytropic process for compressor work is defined as W = -n*(p2V2-p1V1)/(n-1).
    When I am supposed to use one way of calculation and when other? The results are not same. Aren't they same type of group as control volume compressor should behave during the process of compression as closed system (there is no constant add or removal of mass during compression).
    Thanks in advance!
     
  2. jcsd
  3. May 24, 2013 #2

    jack action

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    Boundary work (W = ∫pdV) is the work done by the expansion of gas, i.e. a variation of a volume (a moving piston inside a cylinder for example). It is referred as a closed system.

    Shaft work (W = -∫Vdp) is any other work that is not boundary work, i.e. where there is no variation of a volume (a turbine spinning inside a fixed housing for example). It is referred as an open system.
     
  4. May 24, 2013 #3
    First of all thanks for your reply.
    Part which I do not understand (below is the task where they are using control volume type of calculation for W = -∫vdp), they are talking about compressor and compressor will change piston location (therefore change specific volume) which holds true in their solution as well. Please see my notes below given task;
    Beginning of task-----------------------------------------------
    An air compressor operates at steady state with air entering at p1=  1 bar, T1=  20 C, and exiting at p2=  5 bar. Determine the work and heat transfer per unit of mass passing through the device, in kJ/kg, if the air undergoes a polytropic process with n=  1.3. Neglect changes in kinetic and potential energy between the inlet and the exit. Use the ideal gas model for air.
    Solution
    Known: Air is compressed in a polytropic process from a specified inlet state to a specified exit pressure.
    Find: Determine the work and heat transfer per unit of mass passing through the device.
    Schematic and Given Data:
    Assumptions:
    1. A control volume enclosing the compressor is at steady state.
    2. The air undergoes a polytropic process with n=  1.3.
    3. The air behaves as an ideal gas.
    4. Changes in kinetic and potential energy from inlet to exit can be
    neglected.

    --------------------------------------------------End of task
    Accompanying solution and its p–v diagram shows polytropic curve from point 1 to point 2 and there is difference in pressure (p is "y-axis") and difference in volume (v is "x-axis"). Now the work
    is defined as an area from curve to pressure axis. Why? As there is differential between point v1 and v2 and curve is known as well as n (polytropic value) why I would not be able to use calculation of work as ∫pdv (by the way I did calculate work for both ways and they are not same and interestingly they are not same for multiplication of value n - work that I got by ∫pdv if I multiply with n I get value of work calculated as -∫vdp.
    What is physical explanation for it? Is it because compressor is not insulated and there is some heat involved, so pressure change is not strictly related to volume only but to heat transfer as well?
    Thanks!
     
  5. May 25, 2013 #4

    jack action

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    Your first clues in the problem statements are the following:

    Matter is entering and exiting your control volume:

    http://upload.wikimedia.org/wikipedia/commons/8/86/First_law_open_system.svg

    This mean you have an open system which, according to Wikipedia, is defined as:

    Why does it differs from a closed system? Again, Wikipedia tells us:

    Because the matter is going in or out with a certain velocity and pressure, it contains an internal energy that can contribute to the work output. This is why we use the difference of enthalpy (which represents the total energy contained in the matter, in any form) between the matter coming in and out to determine how much work was performed by the system. Again, from Wikipedia:

    With closed systems, we only need to consider PV work (which is ∫pdv) as the internal energy does not change since nothing is coming in or out.
     
  6. May 25, 2013 #5
    OK...if you can just confirm did I understood it well as I am having problem physically seeing it.
    "An air compressor operates at steady state with air entering at p1=  1 bar, T1=  20 C, and exiting at p2=  5 bar."
    Are we talking here about boundary system and compressor itself which does not need necessary to include only compressor but area around it as well? That would definitely confuse me even more as work itself is done only inside the compressor.
    My problem as I said it how it is done physically. The compressor (piston based) is getting gas in thru inlet valve (and yes, that is open "control volume" system). But Inlet valve then closes; outlet valve is closed as well. Isn't that now closed system? Let's go further...due to movement of piston, caused by outside work on piston, pressure increases while volume is getting smaller. Let's say compressor is not well insulated, so there is some heat added from outside as well which causes temperature to rise which push pressure even higher (so not only the piston is pushing gas by work, but by heat as well). Internal energy is rising as well and can be now explained thru ΔU = Q -W. Then pressure at some point is high enough so outlet valve is opened and gas is escaping out (at this point we do go from "closed" to "open" system again).
    So, what is not clear to me that physical part of compression how it can be seen as "control volume" (open system)?
    If they did specify this air compressor as rotary based then I would see that as different way of calculation as "piston" in this case is thread-like (constantly rotating and due to shape at same time pushing gas and receiving new one in). Is that maybe case here only they missed to be specific? Sorry if I am becoming pest, but I really struggle to get this one...
    Best regards!
     
  7. May 25, 2013 #6

    jack action

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    Yes. But nowhere I have seen in the statements that the compressor is piston based.

    Although you could probably do the calculations for an open system with a piston based compressor while including the valves, it wouldn't be a steady state case, meaning with a steady, continuous operation. In this case the -∫vdp cannot be simply applied. More complex calculations will be implicated and the answer will obviously be the same as if you would have considered the closed system case of a cylinder with the valves closed because all the work is done at that instant anyway.

    In such problems, when they specified "air entering at" and "air exiting at" or "air passing through the device", you can safely assume it is a rotary based machinery.
     
  8. May 25, 2013 #7
    In all examples before they did have piston based compressor even when they did not specify it in task itself, but in solution they had shown cylinder with piston in it. Based on that I could almost safely assume that compressor is piston based. But again all those task were solved using integral of pdV. This one was first one which used -vdp, and I was not quite sure what was going on and why. The task as it is given (especially knowing it is first where they use different approach) was not quiet clarified. Specifying that compressor is rotary based would be the trigger of that different approach for mass flow.
    Anyway now everything is clear and I would like to thank you so much for help. I was really messed and unsure what was going on.
     
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