% Impedance of Power Transformer ?

[squelch]

can someone give an additional info/discussion/inputs regarding this ? 

[GuDo]

ang alam ko depende sa windings mo, gauge ng wire na ginamit pati ang core type..

[???]

Alam ko is fault,protection and load calculation.

[tony]

transformers have turns ratio, and this ratio comes into play in calculating for regulation.....iimpedance can either be primary reffered or secondary reffered:

when reffered to secondary:

Zp = [(Np/Ns)^2]Zs or Zs = [(Np/Ns)^2]Zp as i recall,

here are some google finds: http://www.google.com.ph/search?ix=uca&sourceid=chrome&ie=UTF-8&q=%25+Impedance+of+Power+Transformer

[kinyo]

% Impedance of transformer is essentially what is also called per-unit impedance of the transformer ... where the unit impedance is the impedance of the load at full capacity

%Impedance = (transformer impedance)/(full-load impedance)  *  100%

it also represents the voltage regulation of the transformer ... for example, if %Impedance is rated 10%, it means the voltage at the output will drop 10% from no-load to full-load

 

[???]

Say for example ang %Z nya ay 1.9%

25kva single phase 13.2kv to 220V

What's the meaning?

[robbietan]

a very good explanation is here: http://www.transformerworld.co.uk/impedance.htm

Percentage Impedance (Z%)
 

The impedance of a transformer is marked on most nameplates -  but what is it and what does the Z% figure mean?

Definition
The percentage impedance of a transformer is the volt drop on full load due to the winding resistance and leakage reactance expressed as a percentage of the rated voltage.

It is also the percentage of the normal terminal voltage required to circulate full-load current under short circuit conditions

Measuring Impedance
The impedance is measured by means of a short circuit test.  With one winding shorted, a voltage at the rated frequency is applied to the other winding sufficient to circulate full load current - see below:
 

The percentage impedance can then be calculated as follows:

Z%  =  Impedance Voltage   x  100
Rated Voltage

Changing the Impedance Value
The most economical arrangement of core and windings leads to a 'natural' value of impedance determined by the leakage flux.  The leakage flux is a function of winding ampere turns and the area and length of the leakage flux path.  These can be varied at the design stage by changing the volts per turn and the geometric relationship of the windings.

The Effect of Higher and Lower Impedances
The impedance of a transformer has a major effect on system fault levels.  It determines the maximum value of current that will flow under fault conditions.

It is easy to calculate the maximum current that a transformer can deliver under symmetrical fault conditions.  By way of example, consider a 2 MVA transformer with an impedance of 5%.  The maximum fault level available on the secondary side is:

                    2 MVA x 100/5 = 40 MVA

and from this figure the equivalent primary and secondary fault currents can be calculated.

A transformer with a lower impedance will lead to a higher fault level (and vice versa)

The figure calculated above is a maximum.  In practice, the actual fault level will be reduced by the source impedance, the impedance of cables and overhead lines between the transformer and the fault, and the fault impedance itself.

As well as fault level considerations, the impedance value also:

determines the volt drop that occurs under load - known as 'regulation'
affects load sharing when two or more transformers operate in parallel

[kinyo]

Say for example ang %Z nya ay 1.9%

25kva single phase 13.2kv to 220V

What's the meaning?

on the secondary side, the unit impedance is (220^2)/25000 = 1.936 ohms

if the transformer is given a rating of 1.9% %Z ... it means that, as seen on the secondary, the internal impedance of the transformer is (0.019)*(1.936) = 0.036784 ohms

the transformer can now be represented as an ideal voltage source (220V) in series with 0.036784 ohms for purposes of analysis for secondary currents

so at full-load, the current is 220/(0.036784 + 1.936) = 111.51753 A

the voltage on the 1.936 ohm is 111.51753 * 1.936 = 215.898 V

% voltage regulation = (220 - 215.898)/(215.898)  * 100% = 1.899% (same as %Impedance rating)

hth  

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1.9 % of secondary voltage will drop at full load.

220V x .019 = 4.18 V will drop in secondary voltage at full load

25kVA / 220 V = 113.6 A

Z = 4.18 V / 113.6 A = 36.8 mohm at 25kVA load

25kVA is the transformer rated capacity. we can put full load at 80% with its rated capacity depende na yon sa inyo, but usually 80% load of rated capacity as a full load is a good practice.
 

[???]

Ok nagkasundo.


Ito pa napulot ko rin

I  X  ( 100 / %Z ) = kAIC

ano naman meaning neto?


gwapo points sa inyo.

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KAIC - Kilo Ampere Interrupting Capacity

its a fuse, mcb or hv breaker current interupting capacity during fault event. to quantify kaic depende na sa equipment.