Transformer Stability tests

I assume you've applied a balanced 3-phase 480V source to the HV winding and shorted the LV winding since you don't say otherwise. If you are using a single phase supply or do not have all the LV terminals shorted together then things get somewhat more complex to analyse.

A 6.75% impedance transformer with a primary voltage of 34.5kV will require 2329V on the HV winding to circulate rated current in the secondary. How did you perform the miracle of circulating "a current equivalent to rated current" in the primary using only 480V? And what do you mean by "equivalent to": it is either rated current, or it is not.

Since you applied only 480V instead of 2329V to the HV winding, you would expect current in the ratio of (480/2329)x rated current in the secondary winding, i.e. about 20% of rated current.

To answer your last questions:

It is possible but highly improbable that the nameplate is incorrect.

Impedance is constant - it is independent of applied voltage, although the units it is quoted in are based upon the rated voltage and current of the transformer. It could equally be quoted in ohms.


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One day my ship will come in.
But with my luck, I'll be at the airport!

Dear Sir,
Thank you for your reply.
I am sorry about the few details I input in the thread.

Test condition:
3 phase LV Winding is shorted.
3 phase 480V is applied to the HV Winding

Below is calculation from one of the Nine total installed transformers-please review and advise me:

Rating: 34.5kV / 0.480kV, 2 MVA, %Z=6.53

Base MVA= 2
Base Voltage=0.48kV(480Volts)
Base Current=Base MVA/Base voltage(=Transformer rated current at Secondary)
=2000000 / (1.732*480)
=2405.7A

Base impedance=Zbase=Base Voltage(per phase)/Base current
(480/1.732) / 2405.7 =0.115Ohms

Alternatively,Zbase=Base MVA/[(Base kV)*(BasekV)]
=2/[(0.48)*(0.48)]
=0.115Ohms

Zpu(on 2MVA base)=%Z/100
=6.53/100 =0.0653

Zsc(short circuit impedance in Ohms)=Zpu*Zbase
=0.0653*0.115
=0.0075 Ohms

When we applied 480Volts to the HV Winding(34.5kV side),per phase voltage is measured on the LV Side(480V side)
=(480/1.732) *(0.48/34.5)=3.855V
Therefore current flow on the Short Circuit side=V/Zsc
=3.855/0.0075
=514Amps

The problem lies here
As per calculation we should have 514Amps
But actual current measured at Site is 380Amps

So, our query is-
Is the %Z specified on the name plate based on the rated primary voltage.
Or is there some other reason for getting a lower value of measured Secondary current.

(At the Project, we have two school of thoughts
The first school of thought says-%Z is for the rated primary Voltages and will vary based on the applied Voltage-although I am yet to see a calculation to justify this.
The second school of thought says-%Z is independent of the applied Voltage-I subscribe to the second school of thought-but the actual Site condition puts me in a difficult condition to justify)
Please advise.
Thank you

Scotty is correct - impedance (and the ratio) do not
change, regardless of voltage applied.
Speculations here (not having any direct experience
in the size of transformer in question) are that
1) the construction dictates a certain power level
on input to fully enclose all the windings in the
magnetic field(s) and/or generate the proper
circulating currents, or
2) the supply impedance is entering the equation
and is not reflected in your calculations ( 36 kV
windings will have a considerably higher impedance
than any 480v feed; a considerable mis-match), or
3) something else

I, too, am curious as to how you "maintained the
proper current" at the considerably reduced voltage.
<als>

I agree with your calculated current being about 514A. I got 512A, but via a very different calculation (quicker too):

Rated current at 480V / 2MVA = 2405A

V_sct = voltage applied to HV winding to circulate rated current in shorted LV winding

V_inj = injection test voltage

I_inj = injection test current


V_sct = 34.5x10³ x 0.0653

V_sct = 2253V


V_inj = 480V

I_inj = 2405 x (480/2253)

I_inj = 512A

The most likely reason you're not getting the expected current is the impedance of the interconnects. The LV side shorting link is dominant because of the high current circulating. Unless you have a solid busbar of very substantial cross-section as the shorting link, and near-perfect joints to the LV palms, you will never achieve the theoretical value. [&mu;][&Omega;]'s matter, never mind m[&Omega;]'s! Are you measuring the injection voltage of 480V at the transformer terminals or quoting a nominal value? A slight volt drop under load will also cause the reduction in current you are reporting.



----------------------------------

One day my ship will come in.
But with my luck, I'll be at the airport!

I'm confused by this statement:

When we applied 480 Volts to the HV Winding(34.5kV side),per phase voltage is measured on the LV Side(480V side)=(480/1.732) *(0.48/34.5)=3.855V

Click to expand...

If you shorted the secondary, the per phase voltage measured on the secondary would be zero.

If you short circuit the secondary and apply 480 volts to the primary, you should get

(Rated current)·(Applied voltage)/(Rated voltage)/Zpu

On secondary:

2405.6·480/34500/.0675 = 496A

What actual voltage do you measure on the primary when you get 380A on the secondary?

Dear All,

Scotty-Thanx for the new method of calculation-i learned a few more things.
I believe the impedances of the interconnects are causing the drop.
Also my shorting link is 240mm2 copper cable.The connections of this cable at the busbar may also result in higher impedances.

Secondly
We measure only the input voltage at the source i.e:not from Transformer HV terminal . We have an Auto Trafo connected to the HV terminals of the Xfr and the O/P of the Auto Trafo is measured as the source voltage input.

I am not sure how much of a Voltage drop would occur in the Primary-but we maintain the fixed 480V at the source(Auto Trafo O/P)

Thanx