calculating voltage drop in a transformer

Thread Starter

ben22

Joined Jul 10, 2013
25
What they call no load losses, I would call core loss. What they call full load losses, I would call copper loss, or I squared R loss.
But wouldn't no-load losses be the core losses [I-core squared x R-core] plus the "copper losses" of the entire "core current" (I-core + I-magnetizing) flowing through the primary winding [I-o squared x R-primary]?

Also, do you know if, since we are talking about about a 13.2 kV primary, 240 V secondary transformer, I can assume that the R and X, and thus Z, values of the windings are in the same ratio (55 to 1) as the coils?
 
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But wouldn't no-load losses be the core losses [I-core squared x R-core] plus the "copper losses" of the entire "core current" (I-core + I-magnetizing) flowing through the primary winding [I-o squared x R-primary]?
Because the magnetizing current (more properly the exciting current) is so small compared to normal load current, we usually ignore the copper losses caused by the exciting current. This is determined by the "open circuit test".

Likewise, when the "short circuit test" is performed, the core loss is assumed to be negligible compared to the copper loss.

These two tests allow us to determine the core and copper losses independent of one another as a practical matter. It's true that there is some tiny amount of copper loss when performing the open circuit test, and some tiny amount of core loss when performing the short circuit test, but they are ignored.

Also, do you know if, since we are talking about about a 13.2 kV primary, 240 V secondary transformer, I can assume that the R and X, and thus Z, values of the windings are in the same ratio (55 to 1) as the coils?
The ratio for impedances is as the square of the turns ratio, not directly as the turns ratio.

Theoretically, a transformer has minimum loss (and maximum efficiency) when the copper and core losses are equal at a particular operating point.

Distribution transformers (pole pigs) are seldom operated at full load for extended periods of time, but they are energized 24 hours a day. So they are designed with low core loss relative to their copper loss. Other transformers may have different design criteria. Thus, the winding DC resistances may not be apportioned in the same way for different transformers, similarly the reactances; in particular they may not be apportioned as the square of the turns ratio.

That's why the manufacturer gives a value for the transformer impedance. This number takes everything into account.
 

Thread Starter

ben22

Joined Jul 10, 2013
25
Likewise, when the "short circuit test" is performed, the core loss is assumed to be negligible compared to the copper loss.
But in my real example of a 10 kVA transformer, the core loss is more than 1/2 the amount of copper loss. This does not seem to be negligible! I am missing something here?

Even in a 500 kVA transformer they give the core loss as ~17% the size of the copper loss.

On another note, can I calculate the total impedance of the transformer from the % impedance given in the data thus?

full load current in the secondary (41.667 amps) divided by the % impedance = 2.03% gives short circuit current of 2,052.5 amps across the secondary. Secondary voltage of 240 V divided by this gives the total actual impedance of the transformer: 0.11693 ohms.
 
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But in my real example of a 10 kVA transformer, the core loss is more than 1/2 the amount of copper loss. This does not seem to be negligible! I am missing something here?

Even in a 500 kVA transformer they give the core loss as ~17% the size of the copper loss.
Have a look at the "Short circuit test" section of this page:

http://www.electrical4u.com/open-and-short-circuit-test-on-transformer/

The secondary is shorted for this test. Then a variac is used to apply a voltage much less than rated voltage to the primary, slowly raising the applied voltage until the primary draws the rated current. You can imagine what would happen in you shorted the secondary with rated voltage applied to the primary--lots of sparks and molten copper if it weren't for a fuse or circuit breaker! So, with a short on the secondary it takes much less than the rated voltage to get the normal rated current in the transformer. It's because the voltage on the primary is so much less than normal that the volt-seconds seen by the core are also much less than normal. This then means that the maximum magnetic flux in the core is much less than under normal operating conditions, and that in turn means that the core loss under these conditions is much less than normal--so much less that it is negligible and ignored.

This is how the copper loss is measured without core loss being added to the measurement.

In a similar way the open circuit test measures the core loss without the copper loss being also measured. With no load, the exciting current is so low (compared to full load current) that the I squared R losses in the copper are negligible.

On another note, can I calculate the total impedance of the transformer from the % impedance given in the data thus?

full load current in the secondary (41.667 amps) divided by the % impedance = 2.03% gives short circuit current of 2,052.5 amps across the secondary. Secondary voltage of 240 V divided by this gives the total actual impedance of the transformer: 0.11693 ohms.
Don't forget that it's the impedance referred to the secondary; at the primary it would be 353.7 ohms. The transformer specs you gave in post #15 show an impedance of 1.46%--where did 2.03% come from?
 

Thread Starter

ben22

Joined Jul 10, 2013
25
Ok I understand what you mean about how those values are derived from these tests, thanks for explaining.

Don't forget that it's the impedance referred to the secondary; at the primary it would be 353.7 ohms. The transformer specs you gave in post #15 show an impedance of 1.46%--where did 2.03% come from?
In post #15 I copied that value wrong (1.46% is for a 5 kVA), it should have been 2.03%, sorry about that.

Given what you said above, it would seem that the "total impedance" value, derived from the short-circuit test, does not actually take into account the resistance associated with core losses. But when I look at the total (core plus copper) losses according to the manufacturer (its 202 watts) and I plug into the power loss formula P-loss = I squared x R, where I is max secondary current or 41.667 amps, then I get 0.11635 ohms, which is awfully close to the 0.11693 value calculated from the % impedance of 2.03%.

So should we assume then that this % impedance value takes into account both core losses and copper losses?

If I'm trying to figure out the total voltage drop of the transformer, then, can I just refer it all to the secondary? So let's say the secondary current is 10 amps, will total voltage coming out of the secondary be 240 - 1.1693 V, or 238.8307 V? This seems low, especially given the values for % regulation given by the manufacturer...

Is it possible to calculate VD in the primary and in the secondary separately? I think this means I need to calculate it in the core as well...not really sure how to do this, but this is ideally what I want to do - to be able to work backwards from those values given by the manufacturer to get primary voltage, secondary voltage and primary current.
 
Are all the values in post #15 for a 5 KVA transformer?

Normally, I wouldn't expect the core loss to affect the transformer impedance much because it's a shunt loss. Imagine just how much IR drop you would expect in the primary due to exciting current, given that the primary wire is large enough to carry the full load current without very much loss under even that condition. The much smaller exciting current wouldn't cause much loss in the resistance of the primary winding.

Some of the impedance loss is due to leakage inductance, and that is a lossless phenomenon, so calculations involving only real power loss won't give the full impedance value. Leakage inductance may be the factor causing the impedance to be larger than you would calculate using only the copper loss, rather than the core loss accounting for the increase in the impedance. Better just use the manufacturer's number.

I also notice that two values for regulation are given, for different power factors. We don't know exactly how the manufacturer got those values.

If you want to separate the total impedance loss into the primary and secondary parts, you will have to know more detailed transformer parameters, such as the AC resistance of primary and secondary separately. For large transformers, the conductors may be large enough for skin effect to come into play.

If you know the actual transformer you're going to use, consult the manufacturer for detailed information on transformer parameters, and just how the various loss factors are measured.
 

Thread Starter

ben22

Joined Jul 10, 2013
25
Are all the values in post #15 for a 5 KVA transformer?
no load loss should be 47, full load loss should be 155, total power losses 202. The rest are accurate.

So should we assume then that this % impedance value takes into account both core losses and copper losses?
So do you think this is accurate?

If I'm trying to figure out the total voltage drop of the transformer, then, can I just refer it all to the secondary? So let's say the secondary current is 10 amps, will total voltage coming out of the secondary be 240 - 1.1693 V, or 238.8307 V? This seems low, especially given the values for % regulation given by the manufacturer...
Lets forget about the % regulation values given by the manufacturer for a minute, would you say this is an accurate way to calculate the voltage coming out of the secondary under a 10-amp load?
 
Lets forget about the % regulation values given by the manufacturer for a minute, would you say this is an accurate way to calculate the voltage coming out of the secondary under a 10-amp load?
You should be aware that you need to take the word "accurate" with a grain of salt when it comes to grid frequency transformers. Let's say that it is reasonably accurate.
 

Thread Starter

ben22

Joined Jul 10, 2013
25
You should be aware that you need to take the word "accurate" with a grain of salt when it comes to grid frequency transformers. Let's say that it is reasonably accurate.
So finally, if I want to get the best approximation for voltage drop given the values I have from the manufacturer, then instead of using % impedance or the no-load and full-load losses, do you recommend I use the % voltage regulation numbers ? Since 1.0 PF regulation is 1.57% (output voltage is 236.29 V at full load), then I can increase voltage up to 240 at no load in a linear relationship as current decreases?

And I could take into account power factor by again assuming a linear relationship given the 0.8 PF voltage regulation value at full load...
 
The way the impedance measurement is made, core loss is near zero, and doesn't figure into the measurement.

The measurement of regulation is made with full rated voltage applied to the primary, so core loss is near maximum. As load is applied, the core loss decreases slightly, so the drop in output voltage is slightly compensated for by the decrease in core loss.

You have suggested that you might linearly vary certain parameters with applied load. Be aware that core loss does not decrease linearly with applied voltage. If you decrease the applied voltage by one half, core loss will not decrease by half. It will decrease much more than that; the variation with applied voltage is highly non-linear.

Probably the best thing for you is to use the regulation parameter, adjusted for your load PF.
 

Thread Starter

ben22

Joined Jul 10, 2013
25
The way the impedance measurement is made, core loss is near zero, and doesn't figure into the measurement.

The measurement of regulation is made with full rated voltage applied to the primary, so core loss is near maximum. As load is applied, the core loss decreases slightly, so the drop in output voltage is slightly compensated for by the decrease in core loss.

You have suggested that you might linearly vary certain parameters with applied load. Be aware that core loss does not decrease linearly with applied voltage. If you decrease the applied voltage by one half, core loss will not decrease by half. It will decrease much more than that; the variation with applied voltage is highly non-linear.

Probably the best thing for you is to use the regulation parameter, adjusted for your load PF.
This transformer will be used in a power distribution circuit, so applied voltage will be regulated by the generator to remain at or near a setpoint. Current will vary as the load varies. My understanding is the core loss is constant as the load current changes but voltage stays the same, that is why the no-load power loss given as a separate value.

Does this change your recommendation at all?
 
This transformer will be used in a power distribution circuit, so applied voltage will be regulated by the generator to remain at or near a setpoint. Current will vary as the load varies. My understanding is the core loss is constant as the load current changes but voltage stays the same, that is why the no-load power loss given as a separate value.

Does this change your recommendation at all?
Core loss is not perfectly constant with changes in load current. The flux in the core depends of the voltage seen by the core. The resistance of the primary wire causes a slight decrease in the voltage seen by the core as the load current in the primary increases. The core loss decreases more rapidly for small decreases in primary voltage than a linear relation would suggest, because transformers are operated somewhat into saturation, which leads to a non-linear relation between voltage seen by the core, and the flux in the core.

I'd stick with regulation.

You might find this thread informative:

http://forum.allaboutcircuits.com/threads/confused-about-transformer-saturation.37358/
 

Thread Starter

ben22

Joined Jul 10, 2013
25
Core loss is not perfectly constant with changes in load current. The flux in the core depends of the voltage seen by the core. The resistance of the primary wire causes a slight decrease in the voltage seen by the core as the load current in the primary increases. The core loss decreases more rapidly for small decreases in primary voltage than a linear relation would suggest, because transformers are operated somewhat into saturation, which leads to a non-linear relation between voltage seen by the core, and the flux in the core.
Also power loss in the core is equal to V squared / R, so its proportional to V squared rather than V, right?

This is another reason why I would like to calculate the voltage drop separately in the primary and secondary windings, but still not seeing any way to do that well wit current the values given by the manufacturer....oh well.
 

Thread Starter

ben22

Joined Jul 10, 2013
25
So...seems like accurately modeling a power distribution transformer is actually quite hard,damn near impossible when you do not have all the X and R values for every part of the machine.

Thank you for your help and explanations!
 
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