“Reactive Power – A Strange Concept”

L V Nagarajan 

It was very interesting to read a paper by R.Fetea and A. Petroianu (of University of Cape Town), on the above subject. (refer: http://www.el.angstrom.uu.se/kurser/water05/Reactive_Power.pdf).

Reactive Power as a concept is really strange, but is very necessary for power system management. So we have to live with it. But we need some kind of reconciliation with many valid points raised by the authors of the above paper. Let me try the same in the following paragraphs. 

Let us start from their first two equations:

v = Vmax Cos(ωt)

 i = Imax Cos(ωt – θ)

Actual Instantaneous power = vi = Vmax Cos(ωt) Imax Cos(ωt – θ)

Let us now take the expression for current:

 I = Imax Cos(ωt – θ)

    = {Imax cos θ cos ωt + Imax sin θ sin ωt}

    = {Imax cos θ cos ωt – Imax sin θ cos (ωt – π/2)}  

We may now recognize the two terms in the above expression as:  in-phase and 90-deg-lagging components of current, I.

We say,

Active component:      Ia = Imax cos θ cos ωt

Reactive component: Ir = Imax sin θ cos (ωt – π/2)

Let,

Root Mean Square value of v, V= (Vmax/√2)

Root Mean Square value of  i,  I = (Imax /√2)

Then,

(instantaneous) Active power, p = v Ia

 i.e.,    p =  Vmax cos(ωt) Imax cos θ cos ωt

                 = (Vmax/√2) (Imax/√2) cos θ (2cos2 ωt)

                 = VI cos θ (2cos2 ωt)

                 = P (1 + cos 2ωt),        where P = VI cos θ

This is a positive sinusoidal function with an average value of P, the active power, transferred through the circuit.

(instantaneous) Reactive Power, q = v Ir

i.e.,  q = Vmax cos(ωt) Imax sin θ sin ωt

              = (Vmax/√2) (Imax/√2) sin θ (2 sin ωt cos ωt)

              = VI sin θ (sin 2ωt)                                        

              = Q (sin 2ωt), where Q = VI sin θ

This is a sinusoidal function with an average value of zero. But still we say a reactive power of value Q is transferred through the circuit!! Why at all?

Assume a power source and a load connected as below

If the load R is purely resistive,

            V —————————- R

there will be only active power flow through the circuit, as voltage and current will be in phase.

Suppose we add an inductive load XL. Now the current will lag the voltage by an angle, and hence there will be some reactive power flow also.  

            V —————————- R+ XL

Though average of this reactive power will be zero, the source will still have to ‘supply’ this reactive power flow also.

Suppose we now add a capacitative load Xc, to exactly compensate this inductance.

            V —————————- R + XL +  Xc

Then the source need not ‘supply’ any reactive power, as the same is ‘compensated’ by the capacitance. (i.e, the lag by inductance is nullified by the lead created by capacitance). However there will be reactive power flow between XL and Xc. Sometimes it is said, Xc ‘produces’ lagging reactive power to ‘supply’ XL

It is exactly in this way, we handle the reactive power Q, even though its average is always zero.

Actual instantaneous Power

                          = Active power p + Reactive power q

                          = P (1 + cos 2ωt) + Q (sin 2ωt)

                          = P + P (cos 2ωt) + Q (sin 2ωt)

The average values of both second and third terms above are zeros over a voltage cycle. Hence P become actual average power transmitted over the circuit.

As phase angle θ varies from 0 to 90, the value of ‘active’ power P reduces from maximum value of VI to zero and ‘reactive power’, Q increases from zero to VI. However all through, the magnitudes of AC voltage and AC current remain the same and hence the value of VI. For this reason VI is known as apparent power, S.

 S2 = VI2(cos2 θ + sin2 θ)

       = (VI cosθ) 2 +  (VI sinθ) 2  = P2 +  Q2  

S = √ (P^2 + Q^2) = VI, is an important parameter known as VA which is widely used for specifying power ratings of electrical devices such as generators, transformers and even major loads. Now, in phase component of S, S cosθ is same as active power P. The cross phase component of S, S sinθ is called reactive power, Q, which does not appear very strange now.

Now we may use complex algebra to revisit the same concepts as above. 

Phasor Representation

By using Euler’s Formula, any sinusoidal function can be expressed as
V = Vmax cos(ωt- θ) = Real Part of [Vmax ej(ωt- θ)]

                                           = Re[Vmax e-jθ  ej(ωt) ]

[Vmax e-jθ] is known as V’, the phasor representation of V(ωt), represented as V∟-θ

Now, V = Re [V’ ej(ωt) ]

The time-dependency has been effectively factored out, in the phasor representation as it deals with only the static quantities of amplitude and phase angle.

By phasor representation as above, we may write,

V’ = V∟0 and I’ = I∟-θ

Apparent Power S’= V.I* = V.{I e-jθ }* = VI∟θ

Alternately, if V’ = V∟-θ1 and I’=∟-θ2

Apparent Power S’= V.I* = V e-jθ1.{I e-jθ2 }*

                                      = VI∟(θ2- θ1) = VI∟θ,

where θ = (θ2- θ1), the actual phase difference.

 (i.e)    S’ = VI e-jθ   = VI (cos θ + j sin θ)

                  = VI cos θ + j VI sin θ

                   = P + j Q

As we see the real part of S’ is ‘Active Power’, VI cos θ; and imaginary part of S’ is Reactive Power, VI sin θ. In this representation the Reactive Power does not seem as strange as earlier. 

Let us again consider only a resistive load, R. As in this case current will be in phase with voltage, θ = 0. Hence,

 S’ = V’. I’* = V∟0 . I∟0 = VI cos 0 + j VI sin 0

(i.e)      S∟0 = P + j0, where P = VI

Impedance Z’ = V∟0 / I∟0  = V/I cos 0 + j V/I sin 0

                             = R + j0

Now consider only an inductive load of XL   In this case the current will lag voltage by an angle of 90 degrees, i.e., π/2. Now,

S’ = V’ I’* = V∟0. I∟π/2  = VI cos π/2 + j VI sin π/2

(i.e)      S∟π/2 = 0 + jQ, where Q = VI

Impedance Z’ = V∟0 /  I∟-π/2  = V/I cos π/2 + j V/I sin π/2

                              = 0 + j XL

Now consider an combined load of R and XL . In this case there will be a phase difference of, say, θ, lagging. Hence,

S’ = V’ I’* = V∟0. {I∟- θ}*  = VI cos θ + j VI sin θ

(i.e)       S∟θ = P + j Q, where P = VI cos θ and Q = VI sin θ

Impedance = Z’ = V∟0 /  I∟- θ  = V/I cos θ + j V/I sin θ

                                 = R + j XL

               We recognize R = V/I cos θ and XL = V/I sin θ    

When a capacitor is added to the load, we have

                Impedance Z’ = R + j XL – j Xc

Here also we observe the mutually nullifying effect of XL and Xc. Hence we are able to represent combined resistive and ‘reactive’ loads conveniently as a phasor or a complex number, known as impedence, Z.

It appears that Reactive Power concept, though somewhat strange, is very useful. By this concept, complicated trigonometric functions have been reduced to simple(!) complex algebra.

Yet another set of strange things about reactive power is its direction of flow and sign. If you still have appetite for further confusion you may refer to my blog:

 https://lvnaga.wordpress.com/2010/02/19/electrical-power-and-power-factor/   and

https://lvnaga.files.wordpress.com/2010/02/direction-of-flow-of-active-and-reactive-power.doc

We may meet again later.

L V Nagarajan

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