Transformer Voltage Drop

[joan271273]

When I energize a 34.5-6.9kV Power transformer unloaded, the voltage in the secondary is not exactly 6.9kV. Is that basically due to voltage drop across the Xtr Impedance?.
Can someone clarify this calculation for me.

 

[DanDel]

With no secondary load, the measured secondary voltage should be in direct ratio to the primary winding and the tap setting. With a secondary load, the voltage will be lower since the current flow with cause a voltage drop across the transformer impedance.

 

[jghrist]

If the transformer is not loaded, the voltage not being 6.9 kV is not due to voltage drop across the impedance. Most likely it is because the voltage on the high side is not exactly 34.5 kV, the transformer is not on the neutral tap, or there are errors in the VTs or meters.

Voltage drop is load current times impedance, but you have to consider the vector relationship. With a resistive load, the IZ drop will be nearly at 90 deg to the voltage because the impedance is nearly all inductive. If it were all inductive, then if the load voltage was 1 pu at an angle of 0 deg and the load was 1 pu, then the IZ drop for Z=6% would be 0.06 pu at an angle of 90 deg. The source voltage would be sqrt(1^2 + 0.06^2) = 1.0018

 

[bigamp]

It is normal to specify a transformer to have slightly higher than nominal voltage ratio (typically 4.5% higher)in order to ensure that the transformer produces your required nominal voltage at full load.

For example, if a 6.6kV transformer is to be powered from a 33kV system, the transformer ratio would be specified to be 33kV/6.9kV at the nominal "zero" tap. At no load the transformer voltage would be 6.9kV with 33kV applied. At fullish load and depending on power factor the voltage would be near enough 6.6kV. Not sure if this helps you.

The formula for determining transformer regulation for any given load is from "The J&P Transformer Book". For LAGGING load is as follows:

Reg = a(Xsin(p)+ Zcos(p) + ((a^2)(Xcos(p)- Rsin(p))^2)/200

Where:

Reg = regulation in %
a = actual load current divided by transformer rated load current
X = transformer reactance in % on transformer base
R = transformer resistance in % on transformer base
p = phi, the load power factor angle (cos(p)= power factor)

Actual voltage = (1-Reg(%)) x no laoad voltage

In your case, if you are looking for 6.9kV as nominal voltage, the actual voltage ratio on your transformer may well be 34.5kV/7.2kV. In any case, the ratio of transformation at no load and at all the various taps should be stated on the transformer rating plate.

Regards

 

[hknair]

Hi,

Your question is itself answering your query.The rating of transformer is mentioned at no load only.rated regulation at ratedload cuurent at rated pf is also mentioned upon the name plates of every transformer.You can also calculate this regulation under site conditions by conducting Sc and OC tests.If your transformer is feeding a radial load then various type of compensators are also available.The principle of this compensator is either to provide a load shedding or to change the tap(automatic voltage regulators) which ever is avilable.Anyhow you cannot purchase a transformer which supplies the noload voltage at fullload.And even if you purchase one mind that under light load conditions the load is subjected to the over voltage subject to the regulation your transformer is having.

 

[wilkin]

BS171 OR ANY POWER TRANSFORMER SPEC GIVES A TOLERANCE BETWEEN THE DECLARED TRANSFORMER RATIO AND THE TESTED TRANSFORMER RATIO AT NO LOAD.THIS VALUE IS SMALL AND CHECKED WITH ACCURATE INSTRUMENTS.(1%)

 

[jbartos]

Suggestion to the original posting joan271273 (Electrical) Jan 15, 2003 marked ///\\When I energize a 34.5-6.9kV Power transformer unloaded, the voltage in the secondary is not exactly 6.9kV.
///Is it higher than 6.9kV or lower than 6.9kV?\\ Is that basically due to voltage drop across the Xtr Impedance?.
///The power transformer is stated "unloaded." Normally, the voltage drop across the Xtr is analyzed for the loaded transformer.\\Can someone clarify this calculation for me.
///A little clarification of your posting is needed.\\\