Single phase loading of a three transformer 1

[JBD]

There is a discussion, going on in a different forum (so far it has gone on for a total of 15 pages) concerning the loading of a wye connected transformer.

The only important fact is the transformer in question would have a 208Y/120 connected secondary. With a single 208V 10A resistive load connected to terminals X1 and X2 of the transformer (X3 is not connected).

What the arguement seems to be about is that "the apparent power in the load" is different than "the apparent power delivered by the transformers".

Quote (Rattus)...
In a 120/208V wye, with a single 208V, 10A load, what is the apparent power delivered by the two transformers supplying that load?

Most would say 2.08KVA.

However, if one considers the transformers individually, one sees that each transformer delivers 120Vx10A = 1.2KVA for a total loading of 2.4KVA.

...quote

Is this a trick with numbers or a valid method for analysing transformers?

 

[waross]

This is fun with words.
The real power delivered by the transformers is the same as the real power consumed by the load.
The cause of the different KVAs is the different phase angles of the voltages across the individual transformers compared to the phase angle of the voltage across the load.
One common method of calculating panel and transformer loading is to simply add the load currents of each breaker on each individual phase. We don't consider the voltage of the load. (120 vs. 208)
This gives the total load in amps on each phase and the current in each of the wye connected secondaries. The true KVA of each transformer is then calculated by multiplying the current by the transformer voltage. (120V)
The issue of load KVA vs. transformer KVA is solved by ignoring the utilization voltages of the loads.
Actually we often use the nameplate current of a device, or convert from KVA to current and then use the current in subsequent loading calculations.
Although this effect is not power factor, there is an analogy here to power factor.
By way of example:
You cannot use simple addition to sum the KVAs of 120 volt loads and 208 volt loads because of phase angle differences.
You cannot use simple addition to sum the KVAs of loads of differing power factors because of phase angle differences.
Back to the original post:
What is the KVA of a 10 amp load at 208 volts? 2.08 KVA.
What is the KVA of each of two 120 volt transformers supplying a 208 volt load at 208 volts? 1.2 KVA. (Total 2.4 KVA.)
KVA is used most often to determine heating effects in transformers and conductors. It is also used to help size fuses switches, transformers and breakers.
Use load KVA to determine the proper size of conductors and protection for a load.
Use transformer KVA to determine transformer size, conductor size and protection sizing.
Don't try to add load KVA to determine transformer KVA when you have a phase to phase loads on a wye connection. It doesn't work and you may end up undersizing your transformer. (And feeders)
respectfully
ps
(Let's acknowledge that the current and KVA change with varying voltage, and confine this discussion to transformer vs. load at a fixed voltage.)
war.

 

[jghrist]

You have to look at the angle between the voltage and the current when considering the loading on each phase.

Assume the voltage angle from X0 to X1 is 0° and the voltage from X0 to X2 is -120°. The voltage angle from X2 to X1 is 30°. Current in a resistive load connected between X2 and X1 will be in phase with the voltage, 30°.

Current coming out of one phase will lead the phase voltage by 30°. The real power will be 10A·120V·cos(30°)=1040W. There will be capacitive vars (I leads V) of 10A·120V·sin(30°)=600var. Apparent power is sqrt(1040²+600²)=1200VA.

Current going into the other phase will lag the voltage by 30°. The real power will be the same, 1040W. There will be reactive vars (I lags V) of 600vars. Apparent power is 1200VA.

Total real power for two phases is 2080W. Reactive vars will cancel capacitive vars so total is zero. Total apparent power is sqrt(2080²+0²)=2080VA.

 

[rbulsara]

The kVA capacity of the tranformers utilized or needed will be 1.2 kvA each to deliver 2.08kva of load as the two units are fed by voltages 120 deg out of phase to deliver a single load whose voltage is 30 deg out of phase with any one of the two transformer voltages.

This is akin to open-delta trasfomer ratings,when a single phase load is connected across the two open terminals.

Or put it other way, maximum power deliverd by only two transformers conencted "this way" will only be 0.866 times the total of the ratings. 2.4*0.866=2.08.

Simpler way to look at this that the two 120V are not adding up to 240, but only to 208 so they can't possibly deliver full rated power to a common load.

 

[JBD]

As I said the discussion in the other forum has gone on for many pages. I came here looking comments from other engineers.

Basically one poster says that anyone who does not understand that the apparent load power (2080VA)is different than the apparent transformer power (2400VA)just doesn't correctly understand phasors.

So from the responses is the common opinion that the power drawn by the load and the power supplied by the transformer are the same for this single imbalance?

 

[waross]

Hello again JBD;
I agree with both rbulsara and jghrist.
We have tried to explain the issue in different words.
The KVA of the load is 2.08
The KVA of each transformer is 1.2 KVA.
Because of phase angle differences the KVAs of the two transformers must be added vectorilly.
A vector addition of the two transformer values of 1.2 KVA will result in a total of 2.08 KVA, not 2.4KVA.
rbulsara and jghrist, are we all in agreement here? I think so.
Respectfully

 

[jghrist]

waross,
I agree. It's difficult to explain, however. At first blush, it appears that there is a violation of conservation of energy. There isn't really because the real components of individual phase kVA add up to the total.

 

[JBD]

Thank you for confirming my thoughts.

The leading poster on the other forum said "Apparent power from the transformers may not equal apparent power in the load. Everyone knows that, or should! In this case it is 2400VA vs 2080VA"

 

[rbulsara]

waross:

Yes.

 

[stevenal]

Restated from above: Real power is conserved. Reactive power is conserved. Apparent power in complex quantities (four quadrants)is conserved. The magnitude of the apparent power is not conserved. And yet the sum of the magnitudes of the apparent power can still be useful. Consider sizing a bank of single phase transformers to serve this load. Each has a magnitude kVA rating that must be considered. The real losses and related heating of each transformer depends only on the magnitude of the kVA it is supplying, not the angle. If supplied by a three phase transformer, the sum of the magnitudes of individual winding quantities should be considered.

Another example would be two transformers of equivalent kVA and impedance, but differing ratios in parallel supplying a load within the sum of their ratings. Circulating reactive current, leading on one unit and lagging on the other can cause both units to overload.

 

[mpparent]

Maybe I'm missing something, but the original poster said that the load was connected between x1 and x2. This means the load is across 208V. So how are we arriving at 120V? There's no neutral (no xo mentioned)...so the calculated power is 208*10=2080W. And as others have already said, since this is a resistor, W=kVA.

Mike

 

[stevenal]

120V is across each transformer winding, each of which carries 10A is this example.

 

[rbulsara]

mpparent:

The original post indicates 208 volts is part of a 120/208V Wye which means 3 phase system (seondary of the transfromer).

So the two transformers (or the phase windings of a 3 phase transformers) are seeing 120V each, between the neutral and the line.

 

[mpparent]

Sorry..didn't see the "...two xfmrs supplying the load".

Ha..


Mike

 

[JBD]

Okay, I thought I had it. But, now in a similar discussion at the other site, the problem has been slightly modified. And they are talking about the primary PF being different than the secondary PF depending on how it is wired.
Case #1 (1) single phase 10A resistive load at 208V
Case #2 (2) single phase 10A resistive loads at 120V


quote (winnie)
There is a trick in the question: the power factor of 1. The _load_ has a power factor of 1, meaning that the current through the load is in phase with the voltage difference between terminal A and terminal B of the transformer. But the current through the coils of the transformer is _not_ in phase with the voltage developed across the coils of the transformer.

We have three voltages to consider Vab, Van and Vbn. These are 208V, 120V, 120V respectively. These are each at their own phase angle. I_load is in phase with Vab, and thus not in phase with Van nor with Vbn.

In both the 10A 208V example, and the dual 10A 120V example, you have 10A going through the two secondary coils, and thus 2.5A though each primary coils. This is true in both cases. However the phase angle of the current flow, and thus the power factor, is different between the two cases, and in neither case is the power factor actually unity on all of the supply conductors.
unquote

quote (rattus)
With a resistive line to line load, we see 30 deg phase shifts between the secondary phase voltages and currents. This shift does indeed modify the power factor! That is the fallacy in your argument. Yes, the PF of the load itself is unity, but the PF of the xfrmr and load is 87%.
unquote

 

[jghrist]

As noted in earlier posts, the real power in each winding of Case #1 is 1040W and the apparent power is 1200W. Power factor is 100·1040/1200=87%.

In Case #1, the current is in phase with the voltage and the power factor is unity in the two loaded phases. If the transformer is wye-wye, the power factor will be unity in the primary as well.

 

[stevenal]

Disagree. All the vars produced in one winding are consumed by the other. As I said before, reactive power is a conserved quantity. The total reactive power produced by the transformer is zero. Transformer pf, defined as transformer w/ transformer va is unity.

 

[jghrist]

OK, power factor of the whole transformer is unity, but the pf of one winding is 87% leading and the other winding is 87% lagging.

 

[stevenal]

Sorry jghrist, I didn't intend to disagree with you. When I pushed the submit button, the last thing on my screen was a quote from somebody called rattus, which I completely disagree with. You are right as usual.

 

[jghrist]

As an interesting exercise, I calculated the per line input powers on the primary of a delta-wye transformer with a 10A resistive load between secondary phases (with 120V to neutral secondary):

Phase A 1385.6 W pf=100%
Phase B 346.4 W, +600 var pf=50% lagging
Phase C 346.4 W, -600 var pf=50% leading
Total 2078.3 W pf=100%