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Average power delivered

  1. Jan 16, 2015 #1
    Hello,

    I have a system of mass m that has an accelerometer strapped to it. The accelerometer is measuring a sinusoidal acceleration with peak amplitude of amax:

    https://www.dropbox.com/s/y49pcrhqbrzk27t/Accel_System.png


    How can I calculate the average power (in Watts) that is being delivered to that system to generate such acceleration?

    Thanks!
     
  2. jcsd
  3. Jan 16, 2015 #2

    A.T.

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    If it's an oscillation, isn't the average delivered power zero?
     
  4. Jan 16, 2015 #3
    I don't think so. No power, no acceleration.

    I think it is something like this, but I am not 100% sure:

    P(t) = F(t) * v(t),

    where F(t) is the force and v(t) is the velocity

    F(t) = m*amax*sin(ω0*t)
    v(t) = ∫ amax*sin(ω0*t) = - amax0*cos(ω0*t)

    P(t) = -m*amax20*sin(ω0*t)*cos(ω0*t)
    P(t) = -m*amax2/(2*ω0)*sin(2*ω0*t)

    I think the average power can be calculated taking the RMS of P(t) Pavg = √(1/T * ∫ P(t)2 * dt).

    What I don't know is what should be the integration period? should it be T = 2*π/ω0 or only half of that because is a square function?

    Thanks!
     
    Last edited: Jan 16, 2015
  5. Jan 16, 2015 #4
    If I am correct, then I am getting:

    Pavg = amax2*m/(2*√2*ω0)

    It would be great if someone could let me know if this is correct.

    Thanks again!
     
  6. Jan 16, 2015 #5

    A.T.

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    The average power can be zero, if positive and negative power cancel.
     
  7. Jan 16, 2015 #6
    To have negative power wouldn't the mass have to be sourcing energy back into the power source?

    What is wrong with my calculation?
     
  8. Jan 16, 2015 #7

    A.T.

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    No, the energy can be dissipated as heat. If the average mechanical power delivered to the oscillating mass wasn't zero, it would accumulate more and more kinetic energy.


    P = F dot v which is negative when F and v oppose each other.
     
  9. Jan 16, 2015 #8
    So you're basically saying the average delivered power by the source is zero; correct? It seems counter-intuitive to me

    There is no loss in this system, so how can the energy be dissipated?

    There is only a mass; there is no friction, no fluid damping, no thermo-elasticity, etc.

    There is not even resistance to the motion (no spring constants), so there cannot even be a transfer of energy between two energy-storing components (typically spring and mass exchanging potential and kinetic energy).

    All there is in this system is mass and a source of energy, so I think the average power cannot be zero. For example, if you want the acceleration to increase in one direction (during the first half of period the sine function), you need to put energy into the system. Similarly, if you want to change its direction (during the second half of the sine function), you need to keep putting energy to decelerate the mass (which has already has energy in the form of 1/2*m*v^2).

    Is this reasoning incorrect? Sorry, I haven't done this in a while.

    Thanks again.
     
    Last edited: Jan 16, 2015
  10. Jan 16, 2015 #9

    A.T.

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    If there is no loss, why is a continuous power supply needed? A loss-less oscillator can oscillate forever, without power supply.

    A decelerating mass doesn't consume energy, it releases energy.
     
  11. Jan 16, 2015 #10
    It is not that the continuos power supply is needed; it is just there, delivering power continuously.

    Anyway, it seems that my understanding of the problem I am dealing with is all wrong.

    Thanks though.
     
  12. Jan 16, 2015 #11

    russ_watters

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    Yes, so whether that is the case or not depends on the details of the system. A spring-mass in gravity would have positive and negative power, for an average of zero and no loss. A wave power station, on the other hand, extracts power with both the up and down stroke. As does alternating current electricity.
    Your order of operations looks wrong (in p(t) is cos on top or bottom?....should be on top), but assuming you just wrote it in calculator order, it may be ok. But for the last bit, you don't need to integrate to find rms manually: it's just 1/sqrt(2).
     
  13. Jan 16, 2015 #12

    russ_watters

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    I wouldn't say your understanding is wong, I'd just say you haven't told us the purpose, so we don't know if it is or not. Calculating average power in this way may be valid -- heck, it is in electricity!
     
  14. Jan 16, 2015 #13
    Thanks Russ. This can be viewed as a wave power station. As a matter of fact, in the problem I am dealing with, the mass is suspended in a fluid, and there is this wave generator sucking and pushing the mass back and forth in an oscillatory manner (viscosity is very small, so losses can be ignored).

    cos is on the top. Notice I do not have parenthesis after the division, so the only term in the denominator is ω0

    Thanks again!
     
    Last edited: Jan 16, 2015
  15. Jan 16, 2015 #14
    Sorry, I didn't want to be too specific because I thought I simple diagram would be enough, but I guess the diagram can be very ambiguous.

    I'm dealing with a component (of mass m) submerged in a fluid, which is used for something known as ultrasonic cleaning. The component has an accelerometer inside that can tell me the acceleration amplitude that it is object is undergoing. So far, I have the peak value of the measured acceleration, and I know the ultrasonic source has a frequency of 40 kHz, but I need to calculate the power at which its emitting to cause such an acceleration.

    A fair assumption I can make is that most of the power is being delivered to the component, and very little is lost due to the fluid's viscosity. That's why I assumed that the best way to calculate the average power of the source was by calculating the instantaneous power of the object and then taking the RMS.

    Yes, I have had a lot of exposure to electrical oscillators, so that's why I assumed it could be calculated this way. First time dealing with a mechanical component though.

    Thanks!
     
  16. Jan 17, 2015 #15

    A.T.

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    If power goes in continuously, and there is no loss, where does all that energy go to? Unless it accumulates in the oscillator, making it oscillate more and more, this would violate energy conservation.
     
  17. Jan 19, 2015 #16
    I agree. I think the way I am postulating the problem is incorrect. The energy in the system will just fluctuate back-and-forth between the mass and the source, so no power would be delivered.

    After extensive searching, I realized that the average power cannot be calculated by using the RMS of the instantaneous power.

    Thanks again.
     
  18. Jan 19, 2015 #17

    A.T.

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    That is one possibility, depending on how the motor applies the force. If the mass is submerged in a fluid it will of course also dissipate energy to that fluid.
     
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