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B Method to calculate work while moving two walls

  1. Sep 26, 2016 #1
    Hi guys,

    It is not some homework, so I think it is better to post here. I would like to know if my method is correct for the work of the walls x and y.

    I suppose no mass, no friction and the gas keeps its temperature constant.

    An external device modify the shape of a triangle but it keeps constant the area. The external device turns the walls x clockwise around the dot O and the wall y counterclockwise around the dot O. The wall x rotates at the same angle as the wall y. At the same time, the external device moves the wall z up. Inside the triangle there is a gas and the pressure of the gas is higher than outside. There are gaskets between x/z and y/z: the gas doesn't escape.

    tr1.png

    I noted:

    tr3.png

    At start, a=b=√2/2 m and c=1 m
    Difference of pressure P=1 Pa
    Area of the triangle = 1*1/2 = 0.5 m² = constant
    Calculation for an angle of 1e-1 rad
    Depth of the device = 1 m


    Energy from x and y :

    The area is constant so a*b=0.5, so a=0.5/b
    Pythagoras's theorem: c²=a²+b²
    b=c*cos(α):

    c²=0.25/(c²*cos²(α))+c²*cos²(α)
    <=> (c²-c²*cos²(α))*(c²*cos²(α))=0.25
    <=> c⁴*(1-cos²(α))cos²(α)=0.25
    <=> c=(0.25/((1-cos²(α))*cos²(α)))^0.25

    There are two walls but I need to divide by two (the moment), so the work is:

    ## \int_{\frac{pi}{4}}^{\frac{pi}{4}+0.1} (\frac{0.25}{(cos^2(x)*(1-cos^2(x)))^{0.25}} dx ##

    Is it ok for you ?
     
  2. jcsd
  3. Sep 26, 2016 #2

    CWatters

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    Not sure I follow that but I have some comments...

    So no change in volume or pressure? Work is being done against what exactly?

    I can see that the centre of mass of the gas moving vertically so I suppose it's KE (and possibly PE) is changing which may require work but otherwise I don't see it.
     
  4. Sep 26, 2016 #3
    No, the volume is constant and the pressure too.

    The external device needs to give an energy to move the z wall.

    I supposed there is no mass, even the gas, I would like to simplify the calculations.

    I drawn the triangle at start and at the end (pink):

    tr4.png
     
    Last edited: Sep 26, 2016
  5. Sep 26, 2016 #4

    CWatters

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    Not that I can see. Apart from some effects due to turbulence the z wall is not moving relative to the gas because the volume is constant.

    Work = pressure * change in volume.

    No change in pressure or volume means no work.

    Consider a bicycle pump mounted on a bike. The handle of the pump might well be moving at 30mph due to the bikes motion but so is the rest of the pump. The volume in the pump isn't changing so it takes no work to move the handle of the pump at 30mph like this. Your situation appears identical.
     
  6. Sep 26, 2016 #5
    "No change in pressure or volume means no work."

    I agree, the external device recovers an energy from the walls x and y but it needs to give the same energy to move the wall z. There is a difference of pressure, inside the triangle there is the pressure 2P and outside there is P, so there are forces on the walls x, y and z. I'm interesting of the energy recovered from the walls x and y.

    The length of three sides of the triangle is changing.
     
  7. Sep 26, 2016 #6

    CWatters

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    I don't see how that changes anything.

    The energy flowing through the system from z to x and y is independent of anything going on in the system. The more you push in the more you get out but the amount you can push in is entirely up to the external system. You can't calculate it any way I can think of.
     
  8. Sep 26, 2016 #7

    Nidum

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    I'm not convinced that the constant volume stipulation is even achievable with that geometry .

    Personally I don't think that the problem as given is worth pursuing further .
     
  9. Sep 26, 2016 #8

    CWatters

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    It's not achievable indefinitely but see no reason why it can't be constant over a range.
     
  10. Sep 26, 2016 #9
    " I'm not convinced that the constant volume stipulation is even achievable with that geometry ."

    The geometry can respect the volume constant with that:

    The area is constant so a*b=0.5, so a=0.5/b
    Pythagoras's theorem: c²=a²+b² (a, b, c are not x, y, z)
    b=c*cos(α):

    The length of the z wall is : 2*0.5/(sqrt(2)/2-x)

    "You can't calculate it any way I can think of." Three walls are independent, I can rotate x and y and move z.
     
  11. Sep 26, 2016 #10
    If the area isn't changing (hard to see how that is possible), then there is no work done.
    The differential work is given by
    ##dU = PdV##
    if all other things are constant. if dV is also 0, then no work.
     
  12. Sep 26, 2016 #11
    "If the area isn't changing (hard to see how that is possible), then there is no work done." I agree, but I want the work from x and y not x +y +z.

    Yes, the area doesn't change because I rotate clockwise the wall x, I rotate counterclockwise the wall y and I move up the wall z, the external device moves the wall z to keep constant the area, the device is under control.
     
  13. Sep 26, 2016 #12

    CWatters

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    For small displacements of wall z the length of the wall can be considered constant. The volume swept out by wall z can be calculated by multiplying it's length by displacement. A notional Energy Input can be calculated using pressure * swept volume.

    You can do the same for walls x and y giving a notional energy output.

    Since the swept volumes are the same by design you just end up proving energy in equals energy out for any small displacement.

    The fact that the shape of the swept volumes is different (rectangular for z vs triangular for x an y) makes no diffeence
     
  14. Sep 26, 2016 #13
    x, y, and z are pneumatically connected. The work is not well defined. Basically, if you have two sides pushing on opposite sides of a massless wall, all the work is external to the wall, and depends on the pushers.
     
  15. Sep 26, 2016 #14
    "Basically, if you have two sides pushing on opposite sides of a massless wall, all the work is external to the wall, and depends on the pushers." I don't understand, the walls don't pushing themselves. The difference of pressure (inside/outside) give a force on the walls. I rotate independently the wall x, independently the y, independently the wall z. But sure, the external device controls the angle of rotation and the translation to keep constant the volume.

    "The fact that the shape of the swept volumes is different (rectangular for z vs triangular for x an y) makes no diffeence" I agree.

    I would like to know if my calculations are correct for the walls x and y. I don't know what you called small displacements but I would like to calculate the work for an angle of 0.1 rd so the length d= 6.77 cm
     
  16. Sep 26, 2016 #15
    Consider a pneumatic lift with incompressible fluid. When I push one piston down, the other one goes up. Now, if I put a car on top of one piston, it takes more work for me to push down on the other piston than without a car. The details of the pneumatic lift are irrelevant to the calculation.
     
  17. Sep 26, 2016 #16
    "Consider a pneumatic lift with incompressible fluid. When I push one piston down, the other one goes up. Now, if I put a car on top of one piston, it takes more work for me to push down on the other piston than without a car. The details of the pneumatic lift are irrelevant to the calculation."

    I can't understand why I couldn't calculate the work of a wall in rotation. If you're right, I can't calculate the work from a pneumatic actuator ? With an actuator, I can suppose the pressure constant (it is easier to calculate), the cylinder gives an energy : dP*S*x, with S the surface, dP the difference of pressure, and x the length.
     
    Last edited: Sep 26, 2016
  18. Sep 26, 2016 #17
    Maybe we need to know more about the actual machine so we can help you ask the right question. Right now it sounds a little contrived so we can't come up with reasonable assumptions.
     
  19. Sep 27, 2016 #18

    CWatters

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    The walls x an y must be connected to wall z at the corners. Indeed something must provide a force between z and xy that counters the pressure difference or the seals will open or the system explode. When calculating the force on wall z you have to subtract these forces. You are left with a net zero force on wall z. Walls x an y pull upwards on z with a force exactly equal to the pressure differential. The result is it takes no force to move wall z.

    See diagram.
    Missing Force.jpg
     
    Last edited: Sep 27, 2016
  20. Sep 27, 2016 #19
    "or the system explode." An external device controls the walls. The walls x, y and z are independent. That external device needs to give an energy to move in translation the z wall and recover the same energy from x and y. I would like to know if my calculations are correct for the walls x and y because it is not only a question of physics it is a question about maths, I would like to know if my integrals are correct.
     
  21. Sep 27, 2016 #20

    sophiecentaur

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    The shape of the container is not relevant here - the prism shape just gets in the way of understanding, I think.
    You could get exactly the same situation in principle by having a cylinder withe pistons at each end. As piston A moves left, so does piston B (moved by the same actuator). Clearly, nothing will happen to the energy situation except some increase in total KE. Doing it infinitely slowly will eliminate even that. If you move the cylinder, instead, the internal situation is identical and KE change is zero. A case of 0=0, I think.
     
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