VOLTAGE DROP AT XFMR 4

[anbm]

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!

 

[jghrist]

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.

 

[sibeen]

jghrist, thanks for the file. Much appreciated.

 

[ijl]

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?

 

[jghrist]

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.

 

[waross]

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

 

[jghrist]

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?

 

[alehman]

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.

 

[magoo2]

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|.

 

[jghrist]

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 ? + IX sin ?)² + (VRimg - IR sin ? + IX cos ?)² ] - |VR|

 

[NAZ55]

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

 

[NAZ55]

jghrist,

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

 

[NAZ55]

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

 

[alehman]

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 ]

 

[prc]

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.

 

[edison123]

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.

 

[jghrist]

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.

 

[prc]

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.

 

[edison123]

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

 

[7anoter4]

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=VS?*(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

 

[prc]

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.