And Now MVaR in words of one sylable or less 6

[JJayG]

Here's one for all of you clever electrical chappies out there.
Imagine trying to explain MVaRs to a new starter, ie. in the simplest way possible.
All analogies welcome (except maybe the horse and barge one).
Points may be awarded for creativity but deducted for over technicality.
This has always been one of those questions that,(along with what is entropy?)has been a source of consternation in power stations since Michael Faraday was a lad.
Go On, you know you want to............

JJ

 

[jghrist]

Electricpete,

As you correctly note, voltage THD is typically low so the distortion power is small. It brings into question how to account for the reactive component of each of the harmonics. For each harmonic frequency, you could say:

S[sub]h[/sub] = V[sub]h[/sub]·I[sub]h[/sub]·[cos(a[sub]h[/sub]) - j·sin(a[sub]h[/sub])]

where a[sub]h[/sub] is the angle between V[sub]h[/sub] and I[sub]h[/sub].

Because the system impedance is usually highly reactive (even more so at higher frequencies), a is usually close to 90° and the real component of the distortion power is low.

 

[electricpete]

Good point. I was a little off-track. For m=n, the term needs to be split into in-phase/quadrature components and put into P and Q buckets. D contains onl m<>n terms.

To summarize the partioning of Im_rms * Vn_rms into buckets:
For m<>n, the term goes into D
For m=n terms, we need to split the current into in-phase and quadrature components which results in two terms: Vn_rms *In_rms_p (goes into the P bucket) and Vn_rms *In_rms_q (goes into the Q bucket)

An interesting question, why bother splitting Q from D…. after all neither one contributes average real power and both add in the same way (SRSS) toward Strue.

The answer is apparently that D is not “convserved” like P and Q are. This is based on the excerpt below from “Electrical Power Systems Quality” by McGraw Hill (which also gives the definition of D as product of ImVn where m <> n).

The reactive power when distortion is present has another interesting peculiarity. In fact, it may not be appropriate to call it reactive power. The concept of var flow in the power system is deeply ingrained in the minds of most power engineers. What many do not realize is that this concept is valid only in the sinusoidal steady state. When dis tortion is present, the component of S that remains after P is taken out is not conserved—that is, it does not sum to zero at a node. Power quantities are presumed to flow around the system in a conservative manner.

This does not imply that P is not conserved or that current is not conserved because the conservation of energy and Kirchoff’s current laws are still applicable for any waveform. The reactive components actually sum in quadrature (square root of the sum of the squares).
This has prompted some analysts to propose that Q be used to denote the reactive components that are conserved and introduce a new quantity for the components that are not. Many call this quantity D, for distortion power or, simply, distortion voltamperes. It has units of voltamperes, but it may not be strictly appropriate to refer to this quantity as power, because it does not flow through the system as power is assumed to do. In this concept, Q consists of the sum of the
traditional reactive power values at each frequency. D represents all cross products of voltage and current at different frequencies, which yield no average power.

P, Q, D, and S are related as follows, using the definitions for S and P previously given in Eqs. (5.1 for S) and (5.5 for P) as a starting point:

S = SRSS(P,Q,D)
Q = Sum Vk Ik sin(thetak)
Therefore D can be determined after P and Q by
D = sqrt(S^2 – P^2 –Q^2

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[PHovnanian]

I don't like McGraw Hill's "conserved' vs "not conserved" terminology as it applies to harmonic currents and voltages.

If one analyzes current harmonics in much the same way one separates fundamental components into positive, negative and zero sequence, you will find that these currents do result in real power being dissipated somewhere in the system. Typically in I^2R conductor losses. And this real power at higher frequencies does come from somewhere. Usually, the non-linear loads that produce the current harmonics also produce the power at those frequencies and feed it back into the system. But there are cases in which generators contribute to harmonic power flows. And this will result in torque harmonics, or vibration.

So just hand waving and saying that these components are not conserved could lead some to ignore their effects on other parts of the power system.

 

[electricpete]

The distinction between Q and D is related to tracing flow of “power”, not power dissipation. Q and D are very similar with respect to power dissipation, i.e. neither Q nor D directly represents power dissipation and both Q and D contribute to S and current losses in other parts of the system.

Now let’s look at the “power flow” aspect.

Let’s say I have two frequencies: f1 and f50 = 50*F1 (never mind that even harmonics are uncommon… it’s just an example)

There are 2 P terms: I1p * V1 and I50p*V50
There are 2 Q terms: I1q * V1 and I50q*V50
There are 2 D terms: I1*V50 and I50 * V1

Consider the above to be at the input to a shunt impedance Zs = R + j*w*L.

Now I want to calculate the output quantities…. So I need to subtract the P, Q, and (?) D consumed in Zs.

I can calculate the real power losses in Zs and the reactive (Q) power consumed in Zs. The output S will be reduced by the amount of real and reactive power consumed.

But how am I going to calculate the D "consumed" in Zs so I can determine the output? For example the term I1*V50 … which frequency to use in computing D consumed in this shunt element?

The answer is it is nonsensical to talk about D consumed. D is an artificial quantity whose primary purpose is to enable us to predict D in presence of harmonic content.

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[electricpete]

Correction:
D is an artificial quantity whose primary purpose is to enable us to predict S in presence of harmonic content.

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[jghrist]

There are 2 P terms: I1p * V1 and I50p*V50
There are 2 Q terms: I1q * V1 and I50q*V50
There are 2 D terms: I1*V50 and I50 * V1

I don't understand the purpose of multiplying the current of one frequency times the voltage of another frequency. The reason for breaking a distorted waveform down into separate frequency components is to be able to analyze each harmonic separately.

 

[electricpete]

The only reason to do it is to compute S.

S = Irms * Vrms
where
Irms = SRSS(I1p, I1q, I50p, I50q)
where p and q are components in-phase and in quadrature with respective voltage harmonic
Vrms = SRSS(V1, V50, V50)

Now plug the expression for Irms and Vrms into the expression for S:
S = sqrt<(V1*I1p)^2+(V1*I1q)^2+(V50*I50p)^2+(V1*I50q)^2 + (V1*I50p)^2+(V1*I50q)^2+(V50*I1p)^2+(V1*I50q)^2)

Now separate those terms into buckets:
S = SRSS(P, Q, D)
where
P = SRSS(V1*I1p, V50*I50p)
Q = SRSS(V1*I1q, V50*I50q)
D = SRSS(V1*I50p,V1*I50q,V50*I1,V1*I50q)
recombine the p and q components within D
D = SRSS(V1*I50,V1*I50)

So, the only purpose of D is to faciliate computing S. I agree there is no other useful purpose of combining voltage and current of different frequencies. That was the point of McGraw Hill's quote above - D is not real in any sense other than the leftover terms in S which are not captured in P and Q.

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[electricpete]

Correction

D = SRSS(V1*I50p,V1*I50q,V50*I1,V1*I50q)
recombine the p and q components within D
D = SRSS(V1*I50,V1*I50)

should have been

D = SRSS(V1*I50p,V1*I50q,V50*I1,V50*I1q)
recombine the p and q components within D
D = SRSS(V1*I50,V50*I1)

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[JJayG]

Many thanks gentlemen, you can now stop arguing amongst yourselves (yeah, like THAT is going to happen). I think we can safely say beyond reasonable doubt that the question will remain contentious for some time. Lets face it if you lot can't explain it, what chance do us mere mortals stand?
Thanks especially to Waross (Bill) for his patient explanation as well as his Fortran program ( who'd have thought a 50 year old Formula Translator language would still be up and running, yeah OK limping a bit, today?
Thanks also to Burnt2x who has definitely won the prize for the answer to give on an interview.............

 

[electricpete]

We can explain it, but you don’t need to go to an EE for what you’re asking for. I had suggested the ME forum.

What was wrong with the analogy I posted 10 Oct 09 17:19? In fact the analogy is mathematically exact as I have shown. But to explain it to your non-technical buddy, you don’t need the math, just the mechanical analogy which is something within the experience of most people.

First you just need an example of a viscous damper. That could be moving a potato masher up and down in a bowl of molasses. Or perhaps the hydraulic piston on a screen door (particularly in the closing direction, may have a check valved bypasss to allow rapid opening, but ignore that). Or could just be moving

Now in scenario 1m, you push the piston using a rigid extension bar. The damper is simply a mechanical heater – converting mechanical energy into heat. The rigid extension bar has no effect on your efforts – your push/pull force/distance is the same as if the bar were absent. (the extension bar is the lossless, inductanceless line).

Finally in scenario 2m, you add a spring in series. Immediately you should realize you are going to have to push a little farther to accomplish the same heating since you have to compress and decompress the spring. The product of force and velocity (apparent power) is higher in scenario 2m than scenario 1m to achieve the same heating. As a result the pusher will probably need to drink more beer.


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