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Dissipating Power

  1. Apr 16, 2015 #1
    1. The problem statement, all variables and given/known data
    I need clarification of how something is worded in my homework description. I have two impedance loads; one of them dissipates 5kW of power, which I understand to be positive power. Then, I have a second impedance load that draws 8kVA of power. Does the term "draw" specify that power is negative or positive? I've never known this term to be used when describing circuit power.

    2. Relevant equations
    Dissipating power is positive and providing power is negative, but I am unfamiliar with "draw"

    3. The attempt at a solution
    Seen above
     
  2. jcsd
  3. Apr 16, 2015 #2

    SteamKing

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    It's not clear what you mean by "positive" or "negative" power.

    In this instance, "draw" means that the impedance load is consuming 8kVA of power from the generator.

    The load which dissipates 5kW or power suggests that this energy is possibly converted into heat, which then radiates away.
     
  4. Apr 16, 2015 #3

    gneill

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    Staff: Mentor

    Following on along the lines of SteamKing's response, something that "draws power" is generally a passive load that does not produce power via voltage or current generators. Now, it may be a reactive load and thus be a complex impedance. Assigning a sign to a complex number is problematical since the imaginary component may be positive or negative while the real component is positive.

    In order to distinguish Real Power from Reactive Power and Apparent Power, different units are used (although they all boil down to power per unit time, or Joules/second).

    If you look up "Power Triangle" you'll see the labels on the different legs of the triangle.

    Real Power is specified in Watts (or kW, MW, GW).

    Reactive Power is specified in VAR, or Volt-Amperes Reactive (or kVAR,...)

    Apparent Power (on the hypotenuse of the Power Triangle) is given in VA (or kVA, MVA, ...).
     
  5. Apr 16, 2015 #4
    Ok, thank you, that helps with understanding that draw implies consumption.

    One other question... So, I know that VA is the unit for apparent power, and VA is also the unit for complex power.
    For 8kVA, there's no explained distinction whether its apparent or complex power. Since there are no imaginary values, and it's not written in a polar form, should I assume that "8kVA" implies apparent power?
     
  6. Apr 16, 2015 #5

    NascentOxygen

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    Staff: Mentor

    Apparent power is the magnitude of the complex power. It's the same measurement.
     
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