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Transformer Percent Impedance 1

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deltawhy

Electrical
Jun 1, 2011
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Hello,
I have a question regarding transformer %Z. The things that are known are as follows:
1. %Z is acquired through testing the transformer via a short circuit test.
2. %Z is the percentage of rated primary voltage that when applied, produces the full load current through the secondary.
3. %Z is a representation of a transformers leakage reactance (the flux produced by the primary windings that does not link with the secondary).

Now, if those statements above are true, can someone please shed light on the following:
1. Can you consider %Z to be the "impedance" of the transformer during steady state conditions (ie. does %Z only apply during transient events when the transformer is essentially short circuited)?
 
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Your understanding is correct. The impedance that you calculate during short circuit tests can be applied to other conditions. Remember that load conditions are quite different that short circuit conditions. The current magnitude is higher under short circuits and it's power factor is determined by the X/R ratio of the transformer.

Under short circuit conditions, the load (i.e., a bolted fault) is negligible so the only impedance in the circuit is the transformer s-c impedance.

Under load conditions, the load impedance dominates and controls the amount of current and the resulting power factor.

Under load conditions, we're usually concerned with how the impedance affects the load voltage. The most commonly used equation is:

Voltage Drop = I * (R * cos theta + X * sin theta)

This incorporates the transformer R & X as well as the load current.
 
Think of %Z as an inductor. It's inductance is constant whether under steady state or transient conditions. The only time when %Z changes is due to the action of the tapchanger (and this will depend on the trfr design) and change in frequency, for example with harmonics.

Regards.
 
I am afraid ltawhy ,your 3 sentence is not entirely correct.
First of all the "impedance" includes not only the reactance[series reactance that means inductive reactance]
but the resistance of the circuit-depending on temperature and frequency also.
Second this reactance it is of leakage magnetic flux of both-primary and secondary winding. You may understand this
using Steinmetz circuit diagram.
 
 http://files.engineering.com/getfile.aspx?folder=ac49277a-fd18-4c47-af28-823964628ab5&file=Transformer_Steinmetz_Circuit.jpg
Thanks for the responses.
I have gone over the transformer impedance models in my old power text over and over, and have made marginal headway. Actual impedance can be calculated using %Z and the transformer base impedance. This is constant under any condition.
Here is where I reach a contradiction, and if someone could point out the solution I would very much appreciate it:

When I use an old arc welder as a ballast for an induction heater, I can achieve two levels of impedance, one when I short the secondary windings and one where I open the secondary windings. I notice the impedance is much higher when the secondary is open. I can only assume (either correctly or incorrectly) this occurs from the magnetizing reactance.

Please elaborate if possible.

Thanks
 
The open circuit or no-load condition gives you the high impedance case where you see the magnetizing impedance.

The short circuit impedance gives you the nameplate impedance in % or ohms. This is orders of magnitude lower than the magnetizing impedance.
 
You must consider the complete circuit. A transformer fed circuit will consist of the transformer resistance, the transformer reactance, the load resistance and the load reactance. Consider conductors as part of the load.
The impedance is a combination of resistance and reactance.
The transformer impedance is valid only under short circuit conditions, when only the transformer resistance and reactance are in the circuit.


Bill
--------------------
"Why not the best?"
Jimmy Carter
 
Leakage impedance is the series impedance seen by the load current.In a 1:1 voltage transformer,secondary voltage will be reduced by the voltage drop across the series impedance.This voltage drop is the price to be paid for transferring current from one circuit to other.

The shunt magnetising impedance across input voltage, draws a current from primary current to magnetise the core so that secondary will have an induced voltage= primary voltage.Secondary current will be primary current minus the current through shunt magnetisng impedance.This loss in current as magnetising current is the price to be paid for transferring voltage from one circuit to other.
 
Deltawhy, Your last post was correct. You've stumbled onto one of the means to determine magnetizing impedance - the open circuit impedance test, primary side. I would bet that your old welder is an autotransformer type.
 
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