Transformer Reactive power requirement 6

[prashious12]

Hello Everyone,

Is there any standard mentioning the reactive Power Requirement for different % loading for power Transformer??
Precisely, If power factor of load is unity and load is 100% of Transformer rating, How much (approximately or maximum) will be the reactive power consumption of transformer?
Is there any Table mentioning loading and reactive power requirement for transformer?
Or even easy method to calculate the reactive power requirement of transformer provided given load details (Load KW and KVAR), Transformer No-load and Full-load Losses, % impedance etc.

Thanks
Prashious

 

[electricpete]

dpc - I assumed that was a neglgible effect.

Let’s check.
The core loss is 384kw, which is around 0.05% pu. (700mva base).

Working in PU (and again moving core branch in front of primary leakage reactance):
Iresistive = S / V = 0.05% / 1 = 0.05%.
Ireactive = sqrt(Itotal^2 – Iresistive^2) = sqrt(0.6%^2 – 0.05%^2) = 0.6%.
That seems negligible, unless I have made a math error.


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[dpc]

I guess we all have different thresholds of neglect.











David Castor

http://www.cvoes.com

 

[electricpete]

I'm not sure I get your meaning. I believe the 4.1 MVAR error from neglecting magnetizing branch among a total 56+4=60MVAR was significant in the wake of a statement that total absorbed MVAR is calculated to within 1% while neglecting magnetizing branch. I don't see the context where 0.03 MVAR error from neglecting core loss is relevant to anything said before. If you have some disagreement, I'd be interested to know.

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[electricpete]

Sorry I probably should have focused more on the smiley face.
At this moment I can't get the site's smiley faces due to some kind of javascript problems (not kidding). Guess I'll have to settle for this one.
Cheers! ;-)

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[7anoter4]

I beg pardon from our Electric Power Engineers if I am intruding. But , in my opinion only prc is right.
The reactive power does not exist actually-as apparent power [may be also deformant power as I saw somewhere].
The actual unique power is the active power. All other are symbols. One could use these for calculation keeping in mind always that is only a calculation artifice[could be it a French idiom?].
So, it is absurd to speak about "reactive power loss" since reactive power does not exist.

 

[prc]

Sorry, I was sleeping.Let me catch up.I believe the voltage drop in transformer multiplied by the load current will be the MVA lost in transformer.In the words of the legendary GE transformer engineer L F Blume that is the price that we have to give for transforming current from primary to secondary.Similarly,loss of current as excitation current will be the price that we are giving for transferring voltage from one circuit to other.

The exact formula for the voltage drop or regulation in transformer is,

Reguation in pu =kpR+kqX +(kpX-kqR)raised to2/2

where k=actual load as a fraction of the base load on which values of R and X are based ie Normally rated MVA in IEC world.
p=power factor of load (ie cos phi) q=reactance factor of the load (ie sin phi)

When R and X are in %, the division shall be by 200 instead of 2.

So in the original calculation by wilson,MVAR lost will be 0.0025 pu at unity pf and 0.07 pu at full lagging power factor,assuming Z as X.Then the question why MVAR change with power factor?I think it is due to nature of X. It is not a normal reactor,but X contributed by the solenoidal flux between windings.

Can we check the above from actual results from site?The problem with wilson's calculation is MVAR is same at unity pf and full lagging pf. If that is correct, regulation shall be same at all pfs.We know that is not the fact.

 

[jghrist]

prc,
Please read my June 29 post. You are still confusing voltage magnitude drop with loss.

What if the load was capacitive and there was a voltage rise through the transformer? Would the transformer MVA loss be negative?

What if the load were pure reactive? Voltage drop would be IX with no IR. Does that mean there is no copper loss in the transformer? You can push as much current as you want through the windings as long as you are serving only an inductive load, and you won't heat up the copper. Good trick.

 

[electricpete]

Good discussion. I agree with jghrist

the reactive power does not exist actually-as apparent power [may be also deformant power as I saw somewhere].
The actual unique power is the active power. All other are symbols. One could use these for calculation keeping in mind always that is only a calculation artifice

When I said "MVAR losses" I put it in quotes because it is of course not a watts loss, but a term which needs to be considered to balance the reactive power in an identical fashion to the way that we consider real losses when balancing real power.. (rcwilson also used quoates when he first said mvars "lost"). If we subtract reactive power flowing out of one winding of a 2-winding trnasformer to reactive power flowing into the other winding, the difference is found by the sum of all the I^2*X within the reactive elemenrts of transformer equivalent circuit. I call it "mvar losses" because the calculation proceeds identically to a real loss calculation using I^2*R.

Consider S = V conj(I) = ZI*con(I) = Z I^2
S = I^2 Z = I^2 * (R * jX)
S = I^2 R + jI^2X
Comparing this to S = P+jQ, we know that P = I^2 R and Q = I^2 X
The mathematical definitions of P and Q follow a similar pattern, and we can analyse Q balance using the same tools we use for analysing P balance (considering the "losses" from current flow through R for real power and X for reactive).

So, it is absurd to speak about "reactive power loss" since reactive power does not exist.

No, not absurd at all if we want to analyse reactive power balances and flows in a system as discussed above. It is certainly not absurd to talk about reactive power. If you object to the term “loss” then feel free to use your own term.

The problem with wilson's calculation is MVAR is same at unity pf and full lagging pf. If that is correct, regulation shall be same at all pfs.We know that is not the fact.

cwilson used a model including series reactance and neglecting mag reactance, which is reasonably accurate and fairly standard. For a given magnitude of load current, you are correct that the magnitude of the vector voltage drop across will not vary with angle of the load current (lagging or resistive). However the relative angle of the transformer voltage drop in relation to the load will change resulting in generally poorer regulation when feeding an inductive load than when feeding resistive load of same current magnitude (this assumes transformer impedance is primarily reactive).

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[prc]

Thank you jghrist and pete for pointing out my mistake.

I understand IR drop and IX drop will be same irrespective of power factor,depending only on load.

But my logic of reasoning is also correct.But where I went wrong is forgetting to consider the omnipotent 'j'for various parts.

Thank you wilson for the great insight-now I understand how generators supplying the MVARS and transformer improving the pf!!

 

[7anoter4]

I apologize again if my "electrical philosophy" provoked such an inconvenient,
never the less it is not clear if it is a "loss" since this reactive power is returned to the source and the source has to overcome this. Active power is always positive. Reactive could be positive or negative. But this is again "philosophy" and I really sorry.
I have an objection in connection with reactive power loss calculation neglecting core magnetic reactance. I think the magnetic reactance could add another 15% to the total reactive power loss and I'll try to explain.
It is true we have not the noload total current- Io- but only the noload active losses. Also we have not the primary Zp separated from Zs [secondary impedance] but only the total Ztot [see the Steinmetz model].
From no-load part of Steinmetz diagram, neglecting Rp and j*Xp we can calculate Rfe [equivalent resistance of core losses]: Rfe=Up^2/noloadKW Rfe=25^2*1000/13.56=46091.45 ohm
Ife=Up*1000/sqrt(3)/Rfe Ife=25*1000/sqrt(3)/46091.45=0.313 A.
Now if we neglect Ife then Io~Im . The rated current -for 12 MVA apparent transformer power=12/SQRT(3)/25=0.277 kA
Let's say Io=1% as per electricpete example. Io=277*1/100=2.77 A
Xm=Up/sqrt(3)/Io Xm=25000/sqrt(3)/2.77=5210.7 ohm.
Recalculating the equivalent impedance[Rfe||j*Xm] we shall get:
Req= Rfe*Xm^2/(Xm^2+Rfe^2) Xeq= Xm*Rfe^2/(Xm^2+Rfe^2) Req=581.6 ohm Xeq= 5145 ohm.
noloadkw=3*Req*Io^2=3*581.6*2.77^2=13.39 KW
noloadkvar=3*Xeq*Io^2=3*5145*2.77^2=118.43 Kvar
One could find approx. using directly Xm noloadkvar=Up^2/Xm
Noloadkvar=25^2*1000/5210.7=120 kvar.
From uk%=7% and X/R=18.5 we get:
Zshort=7/100*25^2/12=3.646 ohm Xshort=Zshort/SQRT(1+1/18.5^2)=3.6405 Rshort=Xshort/18.5=0.196785
This is-in my opinion-cold copper resistance [20Oc]. For 140 degree C we have to multiply by 1.4323.
and the total [Rp +Rs[reported to primary]=0.196785*1.4323=0.28185 ohm. and the losses [active] at load will be:
copperloadlosses=3*0.28185*277^2=64.88 kw. If we add 13.39 noloadkw total load losses will be: 78.5 kw.
qlosses=3*3.6405*277^2=838 kVAR load reactive losses. Adding 118.43 KVAR no-load total 956.4 kVAR.
:

 

[7anoter4]

Sorry,wrong sketch!

 

[electricpete]

7anoter4 – I appreciate all your contributions on the forums. You provide some of the most detailed calculations around, which is very helpful. I suspect as often occurs in long threads, everyone is saying roughly the same thing, just confused by the fact that so many different people are involved in the discussion and using different terminology. I certainly don’t disagree with anything you said. I don’t think anything that you said contradicts what I said.

I beg pardon from our Electric Power Engineers if I am intruding. But , in my opinion only prc is right

On this particular occasion prc has since revised his thinking (something we all do from time to time).

I apologize again if my "electrical philosophy" provoked such an inconvenient,
never the less it is not clear if it is a "loss" since this reactive power is returned to the source and the source has to overcome this. Active power is always positive. Reactive could be positive or negative. But this is again "philosophy" and I really sorry.

If difference in “philosophy” means difference in “terminology”, then I agree. I explained above the context of the word “loss” 1 Jul 11 13:01 and said that if you object to the word “loss” you are free to use your own term. If you’d like to understand my thought process, then I’d like you to think about filling in the blank: 100mvar flows into the primary winding and 90 mvars flows out the secondary, therefore 10 mvars is ____ in the transformer (and don’t say “lost”). Would “consumed” be better? How about “associated with current flowing in reactive components” (more correct, but a little cumbersome, don’t you think?). You can take your pick and let us know. It really doesn’t matter to me what you pick, and I promise I won’t you’re your term “absurd”. As long as you put the word mvar next to your chosen term, you’d think every power engineer should be able to figure out what you’re talking about, and know that it’s not a watts loss. And if you’re still concerned people might not pick up on that clue, you might even try putting quotes around your chosen term to indicate care is required to avoid misinterpretation of that particular term. But don’t get your hopes up, engineers on eng-tips (myself included) can figure out a remarkably unforeseeable variety of ways to parse/interpret/misinterpet any sentence! ;-). A thermodymamics engineer might object to the use of the term loss to describe even real power, since of course the power is not lost, only transferred. (you knew it meant power lost from the electrical system, but you didn’t say that, so what other conclusion can the thermodynamics engineer possibly draw other than that you don’t understand the principle of conservation of energy). Sorry, I’m getting a little off track (waxing philosophical).

As far as the positive/negative nature of reactive power lost/consumed/associated with passive reactive elements contrasting with the positive-only real power lost in a passive resistive element, I know you already know the math, but imo it supports the analogy of reactive “loss” in an inductive element: By convention S = V I* which means the S which is lost/consumed/associated in a passive inductive passive device is positive while the S which is lost/consumed/associated in a passive capacitive element is negative. The naturally leads to common terminology where we think of capacitive devices as sources of vars and inductive devices as consumers of vars.

I have an objection in connection with reactive power loss calculation neglecting core magnetic reactance. I think the magnetic reactance could add another 15% to the total reactive power loss

I agree, that is a point I made 29 Jun 11 14:42. My calculation came up with 11% assuming 1% no-load current and 8% leakage reactance. I haven’t read your calc in detail but it seems very simlar to mine. You mentioned (calculated?) breaking the primary and secondary reactances which is something I mentioned but did not do. We both made the same assumption that the reactive magnetizing current is practically the same as no-load current has negligible effect on calculating reactive current from no-load current.

I think everyone agrees there are in most circumstances a variety of calculations from simplest/least-accurate to complicated / most-accurate. Using leakage reactance and neglecting magnetizing effect is the first/simplest estimate. We can try to get more and more accurate as we sharpen our pencil. How productive that is depends on the situation.

I don’t see much difference between what you said and what I said.

Although after composing this long response, I looked back noticed you said “I think the magnetic reactance could add another 15% to the total reactive power loss”. That just strikes me as funny, given the way I have interpretted your comments. Maybe I misunderstood you and you were not objecting to my use of the term loss in this context?

I hope you don't take anything I said here too seriously. I may have completely misunderstood what you were saying. Again I certainly do respect your knowledge and your contributions, including detailed responses you provided to my question on several occasions.


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[electricpete]

Corrections in bold:

I promise I won't you're your term "absurd".

should have been:
I promise I won't call your term "absurd".


and
We both made the same assumption that the reactive magnetizing current is practically the same as no-load current has negligible effect on calculating reactive current from no-load current

should have been:

We both made the same assumption that the reactive magnetizing current is practically the same as no-load current because resistive core loss component has negligible effect on calculating reactive current from no-load current


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[electricpete]

Last correction:

the S which is lost/consumed/associated in a passive inductive passive device is positive while the S which is lost/consumed/associated in a passive capacitive element is negative.

should've been:

the Q which is lost/consumed/associated in a passive inductive passive device is positive while the Q which is lost/consumed/associated in a passive capacitive element is negative. (where Q=Im{S})

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[7anoter4]

Thank you very much, electricpete, for your kind appreciations.
But the idea to reduce the core reactive losses [and perhaps active too!] with primary voltage drop it is very interesting.
I neglected it not intentionally but I simply forgot it.
First of all, since the Ife and Im depend on E [e.m.f.] -and this is less than supply voltage Up.
Second, due to this phenomenon the core saturation will be less then lesser losses and since Xm increase Qloss decrease.
So, I have to introduce it in my calculation. The problem remains how to split the total Ztot into Zp and Zs.
For the beginning I'll take as 1/2 Ztot and let's saw the influence. Then, I'll take Zp=Ztot to see an extreme possibility.
Thank you very much indeed!

 

[7anoter4]

They are so many complications in order to appreciate the no-load losses with a reduced supply voltage [E<Up].
We don't know what kind of laminations are, what kind of transformer construction it is -symmetric, 3 columns, 5 columns
triangle connection or star connection with neutral or without and if there is a special construction in order to suppress the 3rd harmonics or other.
I choose a solution neglecting harmonics, air gap influence and hysteresis for a 3 columns transformer-asymmetrical for magnetic flow density B=14000 gauss in noload conditions -25 kV 60 c/s-and I shall take all the drop across the transformer to get E. Total voltage drop is only 200 V and using a magnetic curve of an old type of laminate I got a new
Io=0.875% and Qlosses=940 kVAR. For a half of the total voltage drop will be 950 Kvar. That means neglecting the
magnetic core decreasing current is a good idea and one should not exaggerate extinding the calculation more.

 

[electricpete]

Good points. You proved there is not much error introduced by moving the magnetizing branch is in front of the stator leakage reactance (directly in parallel with the primary voltage).

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[prashious12]

Thanks a lot for such detailed discussion on the point i have put in this forum.
I can conclude from this, there is no IEEE or any other standard which states the typical range of KVAR requirement for particular rating of transformer given with typical % impedance.
We can though calculate from transformer test data. Transformer approximation model also will holds the almost same results as the with comprehensive model.

Thanks a lot to every one who participated in this critical discussion.