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VOLTAGE DROP AT XFMR
4

VOLTAGE DROP AT XFMR

VOLTAGE DROP AT XFMR

(OP)
How to calculate voltage drop at transformer secondary if we only know primary voltage and transformer size?

Ex: I have 112.5kVA xfmr, 3-phase (480/208), actual primary voltage is 479.47V.

Thanks!

RE: VOLTAGE DROP AT XFMR

2
You're going to need to know the loading (either kVA or amps), the power factor and the transformer impedance.

Volt Drop (in volts) is approximately equal to I (R pf + X qf)

I is current in amps
R & X are transformer impedance components in ohms
pf = power factor
qf = sqrt(1-pf*pf)

Since you want secondary voltage, I, R & X are secondary quantities.


 

RE: VOLTAGE DROP AT XFMR

For homework, use 479.47 V x (208 V/ 480 V)
For an exact solution, what is the transformer regulation and X/R ratio. What is the impedance and X/R ratio of the feeders, and what is the magnitude and power factor of the load.
In the real world, use the transformer regulation, feeder impedance and loading. Close enough for most work. Be aware that when the feeders are appreciably inductive and/or the load has a poor power factor, the errors will increase.
 

Bill
--------------------
"Why not the best?"
Jimmy Carter

RE: VOLTAGE DROP AT XFMR

I second what magoo2 says but would add that the secondary voltage drop would be from the primary voltage referred to the secondary.  479.47·(208/480) = 207.8

Although your statement of actual primary volts to two decimals is very suspect.  Your basic meter will have a ±1% accuracy (4.8 volts).  Also, every time I've tried to measure a voltage, it varies second by second by at least a volt.  I'd just use 479 volts.
 

RE: VOLTAGE DROP AT XFMR

magoo2 and jgrist; The impedance of a transformer determines the current under short circuit conditions.
Under loading, the regulation of the transformer, (determined primarily but not completely by the resistance of the transformer), is the predominant factor.
Calculating the voltage drop based on the impedance of the transformer will give show greater voltage drop than calculations based on the transformer regulation.

Bill
--------------------
"Why not the best?"
Jimmy Carter

RE: VOLTAGE DROP AT XFMR

waross is right. Regulation is what defines the % change in secondary voltage from no-load to full load and it will vary with the load pf.

Percentage impedance applies to short circuit state.

The regulation detail is not included in the trafo name-plate unlike the percentage impedance.

RE: VOLTAGE DROP AT XFMR

Sorry, guys.  Although the % impedance is obtained by a short circuit test, it is this leakage impedance what determines the secondary voltage under load conditions.  It is what determines the regulation.

With 480 V on the primary, you'd get 208 V on the secondary for an unloaded transformer.  It's a simple ratio as waross points out, so if it's slightly less than 480 V on the primary you get slightly less than 208 V on the secondary.  With load on the secondary, the voltage drop will vary in accordance with the approximate but generally used equation that I gave.   

RE: VOLTAGE DROP AT XFMR

Once quickly.
The current is determined by the impedance of the circuit.
Under short circuit conditions, the impedance of the circuit is the impedance of the transformer and determines the current.
Under normal conditions, the impedance of the circuit is the vector sum of the reactances of both the load and the transformer and the resistances of both the load and the transformer. At normal transformer loading resistance generally predominates and the PU regulation is lower than the PU impedance.
I learned this on Eng-Tips. Thank you to the friends who taught me.
Short circuit, transformer impedance.
Normal load, transformer impedance and load impedance. Resistance often predominates and PU regulation does NOT equal PU impedance.

Bill
--------------------
"Why not the best?"
Jimmy Carter

RE: VOLTAGE DROP AT XFMR

waross,

Go back to the equation

I (R pf + X qf)

Now replace R & X by %R and %X.  Then let load equal 1.0 or 1 per unit.

Under load conditions, power factor dominates, so %R dominates.  Since %R is usually less than %Z, the drop or regulation will be less than %Z.

Does this help?

RE: VOLTAGE DROP AT XFMR

Back to the OP, you also need to know the transformer %Z and the X/R ratio to calculate the calculate what you want.
If you have the transformer equivalent resistance and the inductance (R and X) that would make things easy as magoo2 replied earlier.
Back in school (that was a long time ago), we test transformers to get these parameters: Open circuit and short circuit (or a back-to-back test if you have another similar transformer- Sumpner test). We plot the results in a circle diagram to predict transformer performance at any load condition.
Also, see thread238-147658: Transformer %Z and X/R.
 

RE: VOLTAGE DROP AT XFMR

I think that we are saying the same thing in different words. I think that we both agree that at full load the % regulation will be less than % impedance. I misread or misunderstood one of your posts. Sorry.  

Bill
--------------------
"Why not the best?"
Jimmy Carter

RE: VOLTAGE DROP AT XFMR

Vdrop = IZ
Where all are vector quantities

Z = Transformer impedance, commonly stated as Z as a scalar quantity, commonly in %. The mfg. can probably tell you X/R. From those you can calculate X and R for the transformer.

I = load current
 

RE: VOLTAGE DROP AT XFMR

Thanks for the vector sketches, jghrist. That is what I was trying to express. I guess I wasn't very good at it. Sory for any confusion I may have caused.
The point is that the OP does not have enough information to determine the voltage drop, and calculating the voltage drop of a transformer under load is more complex than many folk realize.

Bill
--------------------
"Why not the best?"
Jimmy Carter

RE: VOLTAGE DROP AT XFMR

(OP)
THANK YOU ALL, I LOVE THIS FORUM!!!

RE: VOLTAGE DROP AT XFMR

jghrist, in your Transformer regulation file where did you get your figures for Z and R?

Or were these pulled from an example that you'd done previously?

Could I also ask that you post the original mathcad file (mcd), please.

RE: VOLTAGE DROP AT XFMR

One assumption with what has been presented so far is that the power factor is lagging.

If the power factor is leading, then the equation becomes

Volt Drop = I(R * pf - X * qf)

The minus sign is the only difference.

RE: VOLTAGE DROP AT XFMR


Vdrop = IZ,
where all are vector quantities

|Vdrop| = |IZ|

...reduces the amount of thinking and number of formulas to remember.

RE: VOLTAGE DROP AT XFMR

alehman,
Unfortunately that's wrong.

The vector IZ is used to get the receiving end voltage and it is the magnitude difference between the sending end and receiving end voltage that we're trying to compute.

jghrist tried to point this out earlier.

RE: VOLTAGE DROP AT XFMR

Quote:

jghrist, in your Transformer regulation file where did you get your figures for Z and R?

Or were these pulled from an example that you'd done previously?

Could I also ask that you post the original mathcad file (mcd), please.
Here's the Mathcad file.  The Z and R are from a power transformer test report.
 

RE: VOLTAGE DROP AT XFMR

jghrist, thanks for the file. Much appreciated.

RE: VOLTAGE DROP AT XFMR

Confused, again. Sending end? Receiving end? Regulation? alehman wrong? Why not primary side, secondary side, resistance, reactance, and all the usual stuff?

Basically, the equation of alehman is correct. We all have learned at school that Vdrop = IZ, using complex variables. But is it (really) so that the voltage drop of a transformer is defined as the magnitude of primary voltage minus the magnitude of secondary voltage? (in suitable units, or assuming 1:1 ratio)

If that is the case, then the equation of alehman only needs a small modification: Vdrop = |V1| - |V1-IZ|, using complex variables (vector quantities). Or is there something that I have missed?

RE: VOLTAGE DROP AT XFMR

Quote:

Confused, again. Sending end? Receiving end? Regulation? alehman wrong? Why not primary side, secondary side, resistance, reactance, and all the usual stuff?

...the equation of alehman only needs a small modification: Vdrop = |V1| - |V1-IZ|, using complex variables (vector quantities). Or is there something that I have missed?
No, you haven't missed anything.  Sending end and receiving end are just generalized circuit analysis terms for V1 (sending end) and V1-IZ (receiving end).  Regulation is just a specific formulation of voltage drop for transformers.  Magoo2's equation for voltage drop is an approximation that is close most of the time and assumes that the voltage angle doesn't change much and that the magnitude of VS is equal to real part of VS (cosine of the voltage angle difference approximately equals 1 if the angle difference is small).  The ABB equation for regulation in my second attachment is a refinement, but still an approximation.

The problem with using Vdrop = |V1| - |V1-IZ| is that the angle of the current depends on the power factor of the load which is in relation to the receiving end voltage, not the sending end voltage.  You don't know the vector value of I until you know the angle of V1-IZ.
 

RE: VOLTAGE DROP AT XFMR

As a further explanation of jghrist's links;
Under steady state short circuit conditions the current is determined by the impedance of the transformer.
Voltage regulation is affected by the resistance of both the load and the transformer, and by the reactance of both the load and the transformer.
For loads with a good power factor the transformer resistance is the predominant factor in transformer regulation. The transformer voltage drop (not regulation) can be expected to increase as the load becomes more inductive and the power factor drops.
The transformer voltage drop will equal current times impedance when the load X:R ratio equals the transformer X:R ratio.

Bill
--------------------
"Why not the best?"
Jimmy Carter

RE: VOLTAGE DROP AT XFMR

To illustrate the difficulty of calculating I·Z, suppose the following Source-Impedance(could be anything, a transformer or a length of cable)-Load:

1. Voltage at the source:  VS = 100 volts at 0°
2. Impedance:  Z = 2 + j2 ohm
3. Load:  100 watts at 80% power factor

What is the vector I?
What is the voltage drop?

RE: VOLTAGE DROP AT XFMR

Does this make sense?

When measured terminal to terminal,
Vdrop = |VS - VR| = |IZ|

When measured terminal to ground at each end,
Vdrop = |VS| - |VR|,
where VR = VS - IZ

In one case it's the magnitude of the difference, in the 2nd case it's the difference of the magnitudes?

The calculations assume constant current load. If you have constant impedance load, the method would be different.

RE: VOLTAGE DROP AT XFMR

Alehman,

Yes, now we're on the same page!

I assume you'll take the current at nominal voltage for your constant current example.

Did you intend to attach the calculation based on jghrist's example?  What you've attached is the equations for |VS| - |VR|.
 

RE: VOLTAGE DROP AT XFMR

Quote:

When measured terminal to terminal,
Vdrop = |VS - VR| = |IZ|
I don't know why the definition would change when dealing with terminal to terminal (Ø-Ø?) voltages.

I think the last equation of alehman's attachment should be:
sqrt [ (VRreal + IR cos Φ + IX sin Φ)² + (VRimg - IR sin Φ + IX cos Φ)² ] - |VR|

RE: VOLTAGE DROP AT XFMR

I disagree with both Alehman's vector againt equation and jghrist correction.

VR doesn't have an imaginary part according to the vector relations.

The negative in the last equation is the result of squaring the j.

But then again I am no expert  and can't compare to the knowledge of either of you two :)

 

RE: VOLTAGE DROP AT XFMR

jghrist,

I apologize. You are correct! Once again I have proved to myself never to judge your comments :)

RE: VOLTAGE DROP AT XFMR

By the way here's the basic equation for regulation I have used in the past, which I would like to throw out there for your comments or corrections

% regulation =  [ ( |V (no load)| – |V(full load)|) / |V(full load)|] x 100

 

RE: VOLTAGE DROP AT XFMR

Thanks Jim. You are correct.

By terminal to terminal, I mean primary terminal to secondary terminal. Thinking about a single-phase xfmr with one leg of each winding grounded, the first equation would be what you see if you place a voltmeter between the ungrounded primary terminal and ungrounded secondary terminal - the magnitude of the vector voltage drop |IZ|.

The second equation would represent measuring the primary to ground voltage magnitude and secondary to ground voltage magnitude separately, and subtracting the quantities. Unarguably the more traditional meaning of voltage drop.

Sorry if I'm making this all too complicated, but it's a  confusing topic (to me anyway).

And to further confuse things, the IEEE Red Book has the following equation:
Vdrop = |VS| + IRcos(phi) + IXsin(phi) - sqrt[|VS|^2 - (IXcos(phi) - IRsin(phi))^2 ]
 

RE: VOLTAGE DROP AT XFMR

zasmat,

Voltage drop when expressed as a % is always taken over the no-load voltage and not over the full load voltage.Full load volatge will vary with the pf of the load.

A lucid explanation and derivation of formulae  for voltage drop is given in IEC standard 60076-8 Application guide on Transformers,clause 7.0.

For most of practical applications voltage drop is IR cos phi + IX sin phi is adequate,where cos phi is the power factor of the load.The effect of second term is so small that in most occasions it can be neglected.

RE: VOLTAGE DROP AT XFMR

prc

How much does the transformer % impedance affect the % regulation ? For example, the large trafos have about 10% impedance. How about their regulation ? (let us assume load pf is 0.8)

All these equation are giving me a headache.

RE: VOLTAGE DROP AT XFMR

Quote:

And to further confuse things, the IEEE Red Book has the following equation:
Vdrop = |VS| + IRcos(phi) + IXsin(phi) - sqrt[|VS|^2 - (IXcos(phi) - IRsin(phi))^2 ]
Note that if VS and I are both 1 pu, then this equation is the same as the exact equation for regulation derived in my second July 11 post attachement.

Regulation is the same as voltage drop when you consider a 1:1 transformer (or take the voltage ratio into account) and realize that V(no load) = VS.
 

RE: VOLTAGE DROP AT XFMR

Edison ,

Let us take a 10% impedance transformer and find out voltage drop for 0.8 pf load. %IR for 100 KVA  to 500 MVA transformer will be 1.2 % to 0.12 %.%IX is square root of impedance square - IR square, all in Pu.

Assuming a IR of 0.005,IX will be 0.0999, so voltage drop at 0.8 pf will be, 0.005X 0.8 + 0.0999X0.6 =0.0603 ie at full load voltage drop at secondary terminals will be 6 % from no-load condition.But at unity pf it will be only 0.5 % as second term (due to sign phi) becomes zero.

RE: VOLTAGE DROP AT XFMR

Thanks prc. So to conclude, the % impedance of the trafo has no direct influence on regulation, am I right ?

RE: VOLTAGE DROP AT XFMR

There is another way, a little more complicate, to calculate the transformer secondary terminal voltage.
This way gives the possibility to follow the primary voltage and load changes to determine this secondary voltage on line.
If we take the simple diagram:
[See the attachment pls.]

Where VS is primary voltage; VL secondary voltage; fi is the angle of load pf =cos (fi) then
VS=SQRT((VL*cos(fi)+sqrt(3)*R*I)^2+(VL*sin(fi)+sqrt(3)*X*I) [for L-L voltage]
But what we need is VL so from this formula using usual algebraic operations we get:
VL=-SQRT(3)*(cos(fi)*R*I+X*I*sin(fi))+SQRT(3*(cos(fi)*R*I+X*I*sin(fi))^2-3*(X*I)^2-3*(R*I)^2+VS^2)
If we put ufi1= SQRT(3)*(cos(fi)*R*I+X*I*sin(fi))/VS and
                Ufi2= SQRT(3)*(sin(fi)*R*I-X*I*cos(fi))/VS then
VL=VS*(-UFI1+SQRT(1-UFI2^2))
VS-VL=VS*(1+UFI1-SQRT(1-UFI2^2)=VS*UFT
If a=Sload/Srated , sqrt(3)*R*Irated/Vsrated=ukr , sqrt(3)*X*Irated/Vsrated=ukx we get:
Ufi1= (ukr*cos(fi)+ukx*sin(fi))
Ufi2= (ukr*sin(fi)-ukx*cos(fi))
Uft=1+ufi1-sqrt(1-ufi^2)
VL=Vsrated*a*(1-uft)
The O.P. examlpe:
Srated=112.5 KVA
Urated=480 V[L-L]
Irated=Srated/sqrt(3)/Urated=112.5*1000/sqrt(3)/480= 135.316A
If ukt%=4% [short-circuit voltage]
Pk=1.4 kw [copper losses]
Ukr%=Pk/Srated*100
Ukr%=1.4/112.5*100=1.244%
Ukx%=sqrt(ukt%^2-ukr%^2)=3.33%
Let say the Uload=0.95*Urated=480*.95=456 V
Let say Iload=2*Irated=2*135.316=270.63 A
Sload=sqrt(3)*456*270.63/1000=213.75 KVA
a= Sload/Srated=213.75/112.5=1.9
And let say [as this sudden increasing load is from a motor starting] pf=0.7[total]
Angle fi=acos(0.7)=0.795 radians=45.57 degrees.
Now:
Ufi1= (ukr*cos(fi)+ukx*sin(fi))= 1.244*cos(0.795)+3.33*sin(0.795))= 3.24968%
Ufi2= (ukr*sin(fi)-ukx*cos(fi))= 1.244*sin(0.795)-3.33*cos(0.795))=- 1.443%
uft=1+a*ufi1/100-SQRT(1-(a*ufi2/100)^2)=1+1.9*3.25/100-sqrt(1-(1.9*1.443/100)^2)= 0.062125
VL=VSׂ*(1-UFT)=480*(1-0.062125)=450.2 V=93.79%
If the load decrease to the rated and voltage is also rated then pf=0.85 and a=1
Angle fi=acos(0.85) =0.55481 radians
Ufi1=1.244*cos(0.55481)+3.33*sin(0.55481) =2.8116
Ufi2=1.244*sin(0.55481)-3.33*cos(0.55481)=-2.1752
Uft=1+2.8116/100-sqrt(1-(2.1752/100)^2)= 0.02835
VL=480*(1-0.02835)=466.4=97.165%
Regards

RE: VOLTAGE DROP AT XFMR

Edison,
We cannot make such a general statement.Percentage impedance is vector sum of IR and IX.Depending on the size of trf IR varies from 1.2 %-0.12 % and IX from 4%-20 %.Voltage drop depends on IR ,IX and pf.At unity pf only IR is involved and zero pf only IX is involved.

RE: VOLTAGE DROP AT XFMR

Quote:

Thanks prc. So to conclude, the % impedance of the trafo has no direct influence on regulation, am I right ?
How on earth did you conclude that?  Wasn't the % impedance a key part of prc's calculation?

RE: VOLTAGE DROP AT XFMR

Sorry, it was an error in my above formulae.
My good old professor told us: "don't take anything for granted, you have to demonstrate it before"
So, I took the load constant a=Sload/Srated from a known hand book. But actually a=Iload/Irated only.
Then I have to introduce KU=VS/VSrated and finally the voltage regulation formula is:
uft=1+a*Ufi1%/100-SQRT(1-(a*Ufi2%/100)^2)
and VL=VSrated*KU*(1-uft)
Now in order to fix the O.P. example:
a=2
uft=1+a*Ufi1%/100-SQRT(1-(a*Ufi2%/100)^2)=1+2*3.25/100-sqrt(1-(2*1.443/100)^2)= 0.0654
VL=VS*KUׂ*(1-UFT) =480*0.95*(1-0.0654) = 426.18 = 88.8%
For the rated VS and rated I, KU=1 and a=1 so the last result is still ok.
Mea culpa!
 

RE: VOLTAGE DROP AT XFMR

In the simplest terms, a transformer and a load consist of the transformer resistance, and the transformer inductive reactance and the load and the load reactance.
The voltage drop depends on all of these factors.
Additional comments;
Load power factor, depends on the load resistance and load reactance.
Transformer impedance, depends on transformer resistance and transformer reactance.
You have a series circuit with transformer resistance and reactance and load resistance and possibly load reactance.
Assuming an infinite source, these values along with the source voltage determmine the other values.

Bill
--------------------
"Why not the best?"
Jimmy Carter

RE: VOLTAGE DROP AT XFMR

In principle you are right, waross. But as these parameters are not constant –as for instance –the induction motor start case , it is very difficult to appreciate the total resistance and reactance in order to calculate the voltage drop up to motor terminals. It is more convenient to state the load current and the resultant pf and from the catalog, data of the transformer [as uk% and pk [kw] –and to calculate, enough accurately, the voltage drop starting down-stream from the transformer. It is from my experience.
Best Regards
 

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