Help me with this simple homework

Discussion in 'Homework Help' started by magodiafano, Feb 28, 2011.

  1. magodiafano

    Thread Starter New Member

    Feb 12, 2011
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    A single phase induction motor is operating at a very light load during a large part of every day an draws 10 A from the supply. A device is propose to increase the efficiency of the motor. During the demostration, the device is placed in parallel with the unloaded motor and the current drawn from the supply drops to 8A. When two devices are placed in parallel, the current drops to 6A. What simple device will cause this drop in current? Is the efficiency of the motor increased by the device?
     
  2. Kermit2

    AAC Fanatic!

    Feb 5, 2010
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    Google up "Power factor correction methods"
     
  3. magodiafano

    Thread Starter New Member

    Feb 12, 2011
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    i tried but it is not very clear.
    I think that the device is the capacitor while the efficiency is increased because adding the capacitor in parallel the power factor will be increased.
    Is it right?
     
  4. Kermit2

    AAC Fanatic!

    Feb 5, 2010
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    That's the basic idea - yes Keep reading because it isn't easy to understand at first.
     
  5. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    It's an arguable point concerning the supposed improved efficiency. You need to perhaps firstly explain in more detail how an increased power factor (i.e. closer to unity) leads to improved efficiency.

    A further comment ...

    From the perspective of the overall configuration - supply + PFC device(s) + motor - one might agree that matters are improved. However, with respect to the motor itself, nothing has changed. The reactive power drawn by the motor (for a given load condition) would be unchanged but is now being partly 'supplied' by the added PFC device(s) rather than solely by the main power supply.
     
    Last edited: Feb 28, 2011
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