Transformer short circuit impedance & full impedance when to use which? 6

[TS Liew]

Dear Buddies, The transformer impedance voltage is given in percentage of the total impedance. This is obtained by short circuiting the secondary and inject at the primary to get a full load current on the secondary. This is also often considered the short circuit impedance.
Q1. Which impedance shall I use for calculation during power system analysis in a stable load condition. Some people use the short circuit impedance but I am not sure if this is right?

 

[davidbeach]

Any given transformer has just one impedance; it's a physical property of the transformer. That's the impedance to use.

 

[argotier]

Impedance voltage is NOT a percentage of a total impedance!

It's just the transformer impedance expresed as a relative value (relative to a given base, like: Z% = Z/Zbase). As a relative value (in % or per unit), short-circuit impedance and short-circuit voltage (obtained as you described) are equal (long story short: because the current at wich you tested the transformer is the rated one, so its 100% or 1 pu in relative value).

Now, you want to know when to use the %impedance vs its absolut value in ohms?:

This might help you.

The trick is to know wich base to use, because you can get, for one physical property, a lot of different values.

 

[waross]

Percent impedance is the percent of rated voltage to drive rated current through a short circuited secondary.
To that end it is sometimes referred to as "Percent Impedance Voltage".
For mechanical forces during a short circuit, use the asynchronous current which may be over twice the current calculated with the % imp.
The asynchronous current will approach 2 x √2 or 2.8 times the synchronous short circuit current.
This is the ratio of the peak to peak voltage versus the RMS voltage of a sine wave.
How close it will approach depends on the X/R ratio of the transformer and the point on wave of the switching.
For normal loads, use the transformer regulation percent voltage drop.
For a rigourous solution, consider the Electro Motive Force (EMF or voltage) driving a current through a series circuit comprised of:
The transformer resistance
The transformer inductive reactance
The load resistance
The load reactance, whether inductive or capacitive.
Be aware that the rated regulation is at a specified X/R ratio of the load, and the actual regulation will change depending on the X/R ratio of the load.

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

 

[TS Liew]

David/Waross/Argotier,
Thank you guys very much for your contribution/answer to my question. Your answer has given me greater understanding of the concept involved. I have another thing to add i.e : If I have a string of transformer with different voltages and power along a feeder, then the %Z of the transformers will change base on which base power I select as the base power of the network I want to analyse right?

Say for e.g the circuit above, where both transformer are rated 50MVA given %Z= 10% & 12%. If the 1st the transformer (z=10%) is rated at 100MVA(instead of 50MVA) and I chose to use 100MVA as the base, then the 2nd transformer's impedance would become 6% or 0.06 p.u right?

 

[magoo2]

That is correct. 6% on 100 MVA.

 

[prc]

There are two other impedances, considered some times, during issues connected with transformers.
1)No load impedance- Resistive element based on iron losses and reactive element based on excitation kVA required at rated voltage. This impedance is quite high as excitation KVA is usually less than 0.5 % of rated KVA.
2) Air core reactance of winding- Tis is required for calculation of inrush current and certain overvoltage issues. In this case, core is considered as fully saturated and hence the term air core reactance.

 

[Kiribanda]

TS Liew,
There is something wrong here.
If the second transformer is rated @ 50MVA/12% and therefore @ 100MVA base the impedance is 24% and not 6%.
Pl. refer to basic p.u. equations given in any power systems text book.

 

[TS Liew]

Thank you prc for your input especially on the inrush current aspect.

Thank you Kiribanda for your input, I calculated the 6% or 0.06p.u as below:-

Could you kindly point out what I did wrong? & How to get 24% instead?


TS Liew (Electrical testing, Commissioning, Power system studies)

 

[mparenteau]

This should help you:

https://www.electricalpereview.com/per-unit-example-tips-tricks-watch-electrical-pe-exam/

Mike

 

[prc]

Z,short circuit impedance in ohms, will not change with base MVA.

Zpu= IZ/Rated voltage, where I =rated current corresponding to 50 MVA

When base is 100 MVA, I will be doubled.

z' pu on 100 MVA base = 2IZ/Rated voltage= 2xZpu.

 

[Kiribanda]

TS Liew,
You have already got Z(Base new)=2xZ(Base old).
Below that equation I donot understand what you have derived!!!
Sorry.
My advise is to read a power systems text book to clarify your doubt(s).
From your Thread # 238-468803 too I can see that you have a similar basic question
on basic power system analysis.

 

[xnuke]

TS Liew, the basic problem in your above calculation is that you are using Z[sub]base[/sub] as though it is directly proportional to the MVA rating S[sub]base[/sub], but Z[sub]base[/sub] is inversely proportional to the MVA rating S[sub]base[/sub]:

Z[sub]base[/sub] = V[sup]2[/sup][sub]base[/sub]/S[sub]base[/sub].

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.

 

[waross]

Percent Impedance Voltage;
The voltage required to drive rated current through a short circuited secondary.
If the KVA base is doubled the rated current will be doubled.
Thus the voltage to drive the double rated current through a short circuited secondary will be doubled.
Twice the voltage divided by the rated voltage will be twice as much, hence 24%


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

 

[TS Liew]

Dear buddies, I apologize for the wrong calculation, the above calculation should have been:-

Hence the new Zpu =0.24
Thank you all really much for your help and enlightenment.

TS Liew (Electrical testing, Commissioning, Power system studies)