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Why isn't Reactive Power defined as Q = S - P ?

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  1. Jan 20, 2016 #1
    The instantaneous power in an AC circuit is given by: p = i(t)⋅v(t) = VrmsIrmscosφ + VrmsIrmscos(2ωt -φ).

    The average power P = VrmsIrmscosφ is often a useful quantity to know. For example, it can tell me the work being done by a motor.

    The apparent power S = VrmsIrms is also a useful quantity to know. For example, it can tell me the maximum work the motor could potentially do (when the current and voltage are in phase φ=0).

    However... the reactive power Q = VrmsIrmssinφ = sqrt(S2 - P2) doesn't really tell me anything useful. All I know is that it is some unintuitive measure of the difference between the apparent and average power. Why isn't reactive power simply defined as Q = S - P instead?
     
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  3. Jan 20, 2016 #2

    Hesch

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    Because P and Q have perpendicular directions. Thus Pythagoras must be used.
    upload_2016-1-20_17-14-21.png
     
  4. Jan 20, 2016 #3
    The definition you give is fine; the reactive power is the energy stored in the system.
     
  5. Jan 20, 2016 #4
    This is circular reasoning. We defined it to be that way!
     
  6. Jan 20, 2016 #5

    Hesch

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    Q = S - P →
    S = P + Q , which doesn't make sense.

    A phase shifted current, I / φ, can be represented by the complex value , I / φ = I*cos φ + j*I*sin φ.

    Thus S = V * I = V * ( I*cos φ + j*I*sin φ ) = P + jQ ,
    j = 1 / 90°

    So the only "definition" comes from the complex representation of the phase shifted current, I / φ.
     
  7. Jan 20, 2016 #6

    jim hardy

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    When you study generators
    peaks of reactive current and 'real' current occur with rotor directly or perpendicular to stator winding
    so it was kinda natural for the old timers' thought processes to go with that complex representaion
    figuring it out the first time from physical hands-on is different from trying to back into it from algebra.
     
  8. Jan 21, 2016 #7
    S = VrmsIrms
    P = VrmsIrmscosφ
    Q = VrmsIrms - VrmsIrmscosφ

    Then S = P + Q would make sense. This definition for reactive power is more intuitive because you can easily tell how much power in Watts is being stored, and how much power you are using for useful work. For example, 100VAr would mean you are missing out on 100W of extra useful power.

    You're right that it comes out naturally with: VI* = S = P + jQ as long as you define reactive power to be Q = VrmsIrmssinφ. The single complex number contains a lot of information: P = Re{VI*}, Q = Im{VI*}, S = mag{VI*}, φ = arg{VI*} which is probably why Q was defined this way. On the other hand, you're now stuck with an ugly definition for reactive power. For example, 100VAr does not tell you how much power is being stored in Watts.

    Do you have an example (or link to one) of this?
     
  9. Jan 21, 2016 #8

    Hesch

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    How would you calculate:

    100 [W] + 100 [VAr] = 200 [unit ?] ?
     
  10. Jan 21, 2016 #9

    jim hardy

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  11. Jan 21, 2016 #10

    cnh1995

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    Cosider a box floating in space in the absence of gravity.
    images?q=tbn:ANd9GcQ1sW3msIIspdmpLHaidthYPq4WO7htWvFErlwPd7Ii8GadbHVq.jpg
    If you give some velocity 'v' to the block at an angle Φ, it will move horizontally with velocity vcosΦ and vertically with velocity vsinΦ. If you know the vertical velocity vsinΦ and horizontal velocity vcosΦ and you are asked to find the total velocity v, would you say it is vsinΦ+vcosΦ? No, because velocity is a vector, you MUST consider it's direction and therefore, the angle it makes with the reference. Hence, you'll have to use Pythagoras theorem here for vector addition. Similarly, voltage and current are PHASORS. Hence, you must consider phase angles as Hesch has demonstrated in #2 and #5. VrmsIrmscosΦ is the power dissipated in the circuit, hence, VrmsIrmssinΦ turns out to be the reactive power in the circuit. Apparent power is the phasor addition of the powers and not simple arithmatic addition.
    Also, you can verify this by drawing some waveforms. Take a simple RL ac circuit with some convenient component values and plot the waveforms of active power(power dissipated in R) and reactive power(power absorbed by L)separately for a complete cycle and then calculate their corrosponding average values.
     
    Last edited: Jan 21, 2016
  12. Jan 22, 2016 #11
    Once, I supervized installation of meteting equipment on Power Substation 220KV in Bashkortostan (Russia). Our device showed that nearby plant consumes only 0.5MW of active power but generates 10MVAR of reactive power (due to heating inductors that were switched off just before the measurement).
    As for the question then let's look at the speed in horizontal and vertical directions. No one would mechanically sum them. Formula
    V=SQRT(Vh*Vh+Vv*Vv)
    is almost obvious.
    The situation with active and reactive powers is essentially the same. They are (in specific meaning) orthogonal to each other.
    Sorry, haven't read previous answers carefully.
     
  13. Jan 22, 2016 #12
    But is it not true that power is a scalar quantity and does not therefore have a phase angle, as do voltage and current?
     
  14. Jan 22, 2016 #13
    If {Vn} and {In} are series of samples (voltage and current) then
    P - active power is a sclar multiplication P=ΣVn*In
    and thus is a scalar value while voltage and current are vectors. Though RMS values of both voltage and current are scalar values (length of respective vectors).
     
  15. Jan 22, 2016 #14

    sophiecentaur

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    Yes. Power must be a scalar quantity because Energy is scalar and power is energy per unit time. Imo, "reactive Power" is really an oxymoron which has been introduced into the vernacular for convenience. In a purely reactive component, energy is stored and not dissipated so the 'lost power' must be there because of resistive components in the supply impedance. If people stick with volts and current for their calculations then they can't go wrong and will get the right answers (arithmetic permitting).
     
  16. Jan 22, 2016 #15
    Agree, so it seems incorrect to show vector diagrams of P and Q.
     
  17. Jan 22, 2016 #16

    sophiecentaur

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    I'm not a Power Engineer so I would have to be more careful with my 'shorthand', which is what that diagram is. I would do any calculations the 'formal' way and stand a better chance of getting the right answer. A PQ vector diagram has got to be wrong but it 'may' be forgiveable if the people who use it are aware that they are being sloppy.
     
  18. Jan 22, 2016 #17

    jim hardy

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    Phasors aren't vectors i've been told by folks who are better at maths than i..

    I'd have to go back to Steinmetz - i think it was he who came up with idea of complex numbers to represent AC current....
     
  19. Jan 24, 2016 #18
    Assume a voltage Vcos(wt) across an impedance of R + jX = Z∠φ

    The total instantaneous power is:
    p(t) = v(t).i(t) = Vcos(wt).Icos(wt - φ) = VI.cos(wt).[cos(wt)cosφ + sin(wt)sinφ] = (VI/2)cosφ[1 + cos(2wt)] + (VI/2).sinφ.sin(2wt)

    The expression above is the total power consumed, and it does not tell you the power consumed by the resistor, or the power consumed by the reactance. The power consumed by the resistor is actually NOT (VI/2)cosφ[1 + cos(2wt)], nor is the power consumed by the reactance (VI/2).sinφ.sin(2wt).

    To find the power consumed by the individual resistor and reactance, you have to find the voltage across these individually (e.g. by voltage divider).
    vR(t) = (VR/Z).cos(wt - φ) = Vcosφcos(wt - φ)
    vX(t) = (VX/Z).cos(wt + 90 - φ) = Vsinφcos(wt + 90 - φ)

    Hence:
    pR = vR(t).i(t) = Vcosφcos(wt - φ).Icos(wt - φ) = (VI/2).cosφ.[1 + cos(2wt - 2φ)]
    pX = vX(t).i(t) = Vsinφcos(wt + 90 - φ).Icos(wt - φ) = (VI/2).sinφ.sin(2φ - 2wt)

    Note that adding these two terms together gives the total instantaneous power above (after some further trig identities).

    The average power consumed is P = VrmsIrmscosφ.
    The peak power consumed by the reactance is Q = VrmsIrmssinφ.

    Although both of these quantities are scalars and mean different things (P is the average power, and Q is the peak power consumed by the reactance), you can artificially construct a vector of magnitude VrmsIrmsand angle φ. The x-component of this vector will be the average power P, and the y-component will be the reactive power Q.

    I could also define another scalar quantity, Q = S - P, and call it the "reactive power", but it would not be the peak power consumed by the reactance. If you want the "reactive power" to be the peak power consumed by the reactance, then it must be equal to VrmsIrmssinφ.
     
  20. Jan 24, 2016 #19

    sophiecentaur

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    If we are trying to get to the bottom of something as fundamental as this, we could at least get the basics right. How can any power be "consumed" by a reactance? The Current and Volts are in quadrature. The reason that Power Factor is relevant is that the I and V are greater when there are reactive elements in the circuit. The excess current is passing through the Resisitive components of the supply circuit, which 'consume' extra Power. Also, the extra Volts may be relevant.
    There is more power consumed when the PF is not unity but there is no way it can be dissipated in reactances. Reactances can only store energy and return it (twice) per cycle. The 'Peak' power involved, (rate that energy is being stored, instantaneously) can be given a value but it is not very meaningful
     
  21. Jan 24, 2016 #20
    Power is consumed by a reactance half the time, and released half the time. Just because on the average the power consumed is 0 doesn't mean a reactance doesn't draw power for half the cycle. The peak power it draws is equal to Q.

    That value is the "reactive power" Q agree?
     
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