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Flame in a centrifuge will go oblique?

  1. Dec 26, 2011 #1
    Flame in a centrifuge will go oblique (nearly parallel to the ground if the device is spinning extremely fast).

    Flame in a centrifuge will not point straight up but go more and more oblique as the rotational speed increases (see the picture)

    Experiment:
    1) Put a candle in a preferably sealed container, or even one that has a few small holes on its top and bottom to avoid starving the flame of oxygen too fast.
    2) Lit the candle, close the jar and rotate it.

    The flame should go more and more oblique as the speed increases (see the picture).

    NOTE: The position of the flame has nothing to do with blowing air on it from one side because the container is sealed.


    attachment.php?attachmentid=42221&stc=1&d=1324933737.jpg

    Source: << Link deleted by berkeman >>

    QUESTION: Will the flame really go oblique?
     

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    Last edited by a moderator: Dec 26, 2011
  2. jcsd
  3. Dec 26, 2011 #2

    berkeman

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    If you are trying to ask a question about how acceleration affects the flame, you need to think about how the flame is formed -- why it normally points up. Do you know?

    And a better-framed question would eliminate the rotational aspect of the question, since that introduces extra complications (like air movement inside the centrifuge). Instead, just ask if the flame is affected in an elevator as it rises up sharply...
     
  4. Dec 26, 2011 #3
    The flame is in a closed container that spins at constant speed about a pivot.
    Air movement inside the spinning container should be the same as in a static container excepting the fact that the static pressure will act oblique not straight down.

    Is this demonstration correct:

    "For high speeds like 100 rot/s the flame will point less than one degree away from the direction of the ray that rotate about the pivot.

    Demonstration:
    g0 = 9.81 m/s2 (perpendicular to the ground)
    ar = 2*pi*f*R (centrifugal acceleration parallel to the ground, to the plane of rotation)
    R = the length of the ray
    f = frequency (rot/s)

    artificial g will be g0 + ar as vectors

    The angle of the artificial g relative to the plane of rotation will be: (360 / (2*pi)) * arctan(9.81/(2*pi*f*R))

    for R = 1 m, if:
    f = 1 rot/s the angle is ~ 57 degree
    f = 10 rot/s the angle is ~ 9 degree
    f = 100 rot/s the angle is ~ 1 degree which means that the flame points nearly parallel to the ground."
     
    Last edited: Dec 26, 2011
  5. Dec 26, 2011 #4

    D H

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    Which way is "up"? It is not up and outward. It's up and inward, toward the rotation axis.

    The flame will get skinnier and taller as the rotation of the gas increases. This is similar to how tornados form.
     
  6. Dec 26, 2011 #5
    The gas does not rotate about the symmetry axis of the container.
    The container, with the flame inside, is at one end of a horizontal bar that has its other end welded to a pivot.

    The flame is inside a machine similar to those used for training jet fighter pilots, to test their resistance when acceleration goes to a few g.

    There is no tornado in the container.
     
    Last edited: Dec 26, 2011
  7. Dec 26, 2011 #6

    AlephZero

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    It's hard to follow your demonstration but the conclusion is correct. This situation sometimes occurs in jet engines if there is a fire in the rotating compressor or turbine. The fire damage is always concentrated along the smallest (inside) radius of the space, not the outside.
     
  8. Dec 26, 2011 #7

    D H

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    Ah, I see. I assumed the axis of rotation was the center of the cylinder.

    Your solution is incorrect. Look at the units. 2*pi*f*R has units of velocity. You should be using ω2r = (2*pi*f)2r.

    Note that I am using r, not R. R is the distance between the rotation axis and the base of the flame while r is the distance between the rotation axis and some point in the flame. This latter quantity decreases as the flame moves toward the axis of rotation. This means the angle with respect to vertical will be greatest at the base of the flame and will decrease as the flame progresses from the base. The flame will start out at some angle with respect to vertical and curve slightly upward along the length of the flame.
     
  9. Dec 26, 2011 #8
    1) Yes ar = (2*pi*f)^2*R and not 2*pi*f*R.
    I will redo the calculations.

    2) If three candles are distributed along the ray R at intervals d then the flame in the middle of the container, considered to be a small flame, will stay at an angle relative to the horizontal plane given by:
    AngleFlame(R) = arctan [ g / ( (2*pi*f)^2*R ) ]
    the next one will stay at:
    AngleFlame(R-d) = arctan [ g / ( (2*pi*f)^2*(R-d) ) ]
    and the third at:
    AngleFlame(R-2d) = arctan [ g / ( (2*pi*f)^2*(R-2*d) ) ]
    (the three flames can be at different heights from the ground level)

    However, I am not sure if we can suppose a long flame can be decomposed in small (infinitesimal) flames to draw the conclusion that a single long flame "will start out at some angle with respect to vertical and curve slightly upward along the length of the flame."
     
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