Transformer Impedance and voltage drop

[Ftx]

I am trying to calculate voltage drop on my transformer...
Primary voltage 16000V ,Secondary voltage 120/240
TX= 100 KVA %z=3.3 1phase

Can anyone help me how to calculate voltage drop on transformer.
The result of my calculation is given me 3.28% voltage drop, However my Supervisor has Voltage drop on Xfmr at 5.26%

 

[xnuke]

At full load secondary current, the voltage drop is equal to the transformer impedance rating. That's how percent systems representation works. Your answer is closest.

xnuke
"Live and act within the limit of your knowledge and keep expanding it to the limit of your life." Ayn Rand, Atlas Shrugged.
Please see FAQ731-376 for tips on how to make the best use of Eng-Tips.

 

[Ftx]

But He has done it with calculator , see picture.
I don't know how he got 5.26%

 

[Skogsgurra]

Why not ask him?

It looks like he is either adding cable impedance - or he is calculating voltage drop for a much higher load. There is something said about 230 kVA just above the 5.26%

Are you interpreting the task correctly? Or is it your supervisor that hasn't?

Gunnar Englund

http://www.gke.org

--------------------------------------
Half full - Half empty? I don't mind. It's what in it that counts.

 

[bacon4life]

The voltage drop depends on the R/X ratio of the transformer and on the power factor of the load. Your answer is closer.

 

[waross]

To expand on bacon's post;
The transformer impedance is comprised of the transformer internal resistance and the transformer inductive reactance.
For voltage drop, use the transformer % regulation at the appropriate power factor. It will be less than the % impedance.
For loads of high power factor the transformer internal resistance predominates.
You must know either the transformer % regulation or the X/R ratio to calculate the voltage drop under normal loads.
The % impedance is only valid for short circuit calculations where the transformer impedance is the only impedance in the circuit.

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

 

[Ftx]

My supervisor is not here at the moment , and he is using special calculator.
So base on this information :
Primary voltage 16000V ,Secondary voltage 120/240
TX= 100 KVA %z=3.3 x/R=3.1
Do you guys get 3.28% voltage drop as well?
This is only voltage drop on Transformer...
Thank you

 

[waross]

That calculator shows the voltage drop for a 100 KVA transformer with a 230.52 KVA load.
That's 230.52% of full load current. Is this for calculating voltage dip during motor starting?
If you convert your supervisors figure to 100% loading it comes out as 2.28% voltage drop under full load.
That is probably correct at some power factor.
The voltage drop calculated by % impedance times rated voltage is a directed value. It is predominantly inductive. It can not be subtracted numerically from the rated voltage. It must be considered as a vector.

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

 

[Mbrooke]

Is the 230kva a value taking advantage of a transformer's overload ability? I somehow have the feeling (though I could be wrong), your supervisor is interested about what the voltage drop will be above the nameplate kva rating.

 

[Ftx]

Yes Waroos, this calculation is for voltage drop on initial stage .
There are loads attached to Transformer , But I don't know Which Kva he used! 100kva or 230KVa.
I am getting only 3.6% on transformer . I believe I am using either wrong voltage or Kva..

 

[Ftx]

I think you are right Mbrooke. He might done it opposite way.
He has from
TX-------------------Point A Cable-----------------------point B cable.
5.26% --------- 1.26%dropA + 5.26------------- 2% drop on cableB + 5.26
Voltage drop--------=6.52% Total drop on point A---------=7.26%