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






RE: VOLTAGE DROP AT XFMR
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 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
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"Why not the best?"
Jimmy Carter
RE: VOLTAGE DROP AT XFMR
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
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
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
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
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
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
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
Bill
--------------------
"Why not the best?"
Jimmy Carter
RE: VOLTAGE DROP AT XFMR
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
RE: VOLTAGE DROP AT XFMR
RE: VOLTAGE DROP AT XFMR
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
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"Why not the best?"
Jimmy Carter
RE: VOLTAGE DROP AT XFMR
RE: VOLTAGE DROP AT XFMR
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
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
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
RE: VOLTAGE DROP AT XFMR
RE: VOLTAGE DROP AT XFMR
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
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
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
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
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
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
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
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
I apologize. You are correct! Once again I have proved to myself never to judge your comments :)
RE: VOLTAGE DROP AT XFMR
% regulation = [ ( |V (no load)| – |V(full load)|) / |V(full load)|] x 100
RE: VOLTAGE DROP AT XFMR
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
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
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
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
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
RE: VOLTAGE DROP AT XFMR
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
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
RE: VOLTAGE DROP AT XFMR
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
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
Best Regards