Did we learn anything?

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rattus

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Did anyone learn anything from the thread on transformer loading? Some of us did:

Apparent power computed at the transformer is less than or equal to that computed at the loads, although with a reasonable load balance the difference is acceptable.

Power, real or reactive, is not a vector quantity.

There really is a phase shift between voltage and current in a wye transformer with a single line to line load of any type.

There really is a basis for the infamous Oregon fudge factor. I do not claim though that it is good practice to use it.

How about the rest of you?
 
Re: Did we learn anything?

There is no basis for the Oregon fudge factor.

An open-delta (primary or secondary)transformer bank can only be loaded to 57.7% of it's 3-phase rating. It cannot be properly balance to any greater loading because of the voltage-current phase shift.

Some people can be confused with cause and effect arguments. The VA rating of two single phase transformers may be added when they are connected in series. Their total VA maybe then be divided by the sum of the winding voltages to determine the current flowing through each winding. But this total VA has nothing to do with the actual loading of a three phase transformer bank.

In the situation rattus is asking about.
I said, the total transformer VA was 2080 and the line voltage was 208, so the winding current was 10A. Working backwards 2080VA/57.7% = 3600VA for the proper 3-phase transformer bank.

rattus said the total VA is 2400VA and the line to line voltage was 208V but the winding current was not simply VA/E but instead was the much more complicated 2*(VA/(V/1.73))or 10A per winding. Working backward (2400VA/2)*3 = 3600VA for the proper 3-phase transformer bank.

I believe all of this came about because some person found a neat way to play with numbers for a unique application rather than using the always applicable 57.7% loading factor.
 
Re: Did we learn anything?

Rattus,

I went back to the original thread about the "Oregon factor" and I think it gives a lower amperage "amps = kVA/(208 * 1.154)" or in essence "amps=kVA/240" not more amperage.

If I understand your argument correctly, I think you found that the 208 volt single phase load will cause an INCREASED amperage on the TRANSFORMER secondary which would be the same as on the feeders to the load). This is the opposite effect of the calculation as related by people who have experienced this test set by the licensing board in Oregon

Jim
 
Re: Did we learn anything?

Originally posted by ccjersey:
Rattus,

I went back to the original thread about the "Oregon factor" and I think it gives a lower amperage "amps = kVA/(208 * 1.154)" or in essence "amps=kVA/240" not more amperage.

Reply: The question is asking what is the ALLOWABLE current based on the xfrmr VA rating. They imply correctly that xfrmr loading should be based on Iphase, not Iload, especially in the case of a severe imbalance.

If I understand your argument correctly, I think you found that the 208 volt single phase load will cause an INCREASED amperage on the TRANSFORMER secondary which would be the same as on the feeders to the load). This is the opposite effect of the calculation as related by people who have experienced this test set by the licensing board in Oregon.

Reply: Again, we are computing the allowable current. I am arguing that to be exact in cases of severe imbalance, one should compute the xfrmr load as Vphase x Iphase, not Vload x I load.

Click to expand...
 
Re: Did we learn anything?

But Jim, this is not an open delta with a balanced load. It is an unbalanced wye. Even so, the 57.7% factor is based on Vphase x Iphase where Iphase is the vector sum of the load currents. As you know, 1/2cos(30) = 0.577. It simply does not apply to this problem.

There is nothing more basic than computing transformer loading as Vphase x Iphase. One cannot argue that this is not the most accurate way to make this computation. We are concerned with the heat in the tranformer, and this is the most direct way to compute it. This is in effect, what Oregon has done. Their fudge factor turns out to be 2/sqrt(3) or 2 x 0.577! It is just as valid as any other fudge factor. It may not be code, it may not be in common use, and it may never be in common use, but it is technically sound.

A couple of electricians have told me privately that they understand and agree with me on this matter. Now why is it so difficult for a Professional Engineer to see this? I think you are stonewalling me.

Just explain why the exact same currents produce more or less heat in the windings as a result of the load configuration? It is that simple, and no amount of formulating will change that.

Rattus
 
Re: Did we learn anything?

rattus, it is the same principal, unless the primary of the transformer bank happens to be connected in a wye configuration (even then I believe it still holds). You acknowledged as much when you analyzed the delta-wye sketch in a different post.

The heat from the windings in the transformer is accurate using my formula and is the same as yours: Wloss = Rwinding * Iwinding^2.

All I have complained about is the reference to 2400 as the loading of this transformer bank. I believe this is gobbly-gook the same as saying the "line voltages in a single phase transformer are 180? apart from neutral so it should be called two-phase". The Oregon factor is a trick with numbers, not good engineering practice.


As I have stated the proper factor of .577 works for all loads. As you have stated your formula is only good for this unique load.
 
Re: Did we learn anything?

Originally posted by jim dungar:
rattus, it is the same principle, unless the primary of the transformer bank happens to be connected in a wye configuration (even then I believe it still holds). You acknowledged as much when you analyzed the delta-wye sketch in a different post.

Reply: I cannot see the connection between an open delta and the two legged wye. One produces three phases, the other does not. One carries a balanced load, the other does not. One can fiddle with cos(30) and come up with all sorts of fudge factors, even 1.154! Furthermore, your formula is Machiavellian. We already know Iphase, why not just multiply it by Vphase?

All I have complained about is the reference to 2400 as the loading of this transformer bank. I believe this is gobbly-gook the same as saying the "line voltages in a single phase transformer are 180? apart from neutral so it should be called two-phase". The Oregon factor is a trick with numbers, not good engineering practice.

Reply: Not true Jim, you claim there is no basis for the OFF, and you must agree by now that apparent power is not a vector, therefore it cannot be added vectorially. We are not adding voltages, and this has nothing to do with 120/240 single phase. If you say the loading is only 2080VA, you are in error by 15%! How can you ignore this error?

Now, the sum of the apparent powers computed at the xfmr may not equal the sum computed at the load, but we all know that, and it is not gobbledy-gook. It is a more accurate indication of the heat generated in the transformers. It is I^2 x R!

As I have stated the proper factor of .577 works for all loads. As you have stated your formula is only good for this unique load.

Reply: Jim, it is not my factor, but you have just admitted that it works for this case, and that means there is a basis for it. I never claimed that it was good practice.
Click to expand...
 
Re: Did we learn anything?

rattus

The formula is a trick of numbers so of course it is possible to design a situation where it looks like it is important. If we were to continue the numbers playing game, using your fomula of 2*Vp*Ip=2*120*10=2400W we should be able to say that based on a 120V line to neutral voltage the transformer can carry 20A.

The loading a transformer sees is the load connected to it. My transformer is not over loaded. Each winding is designed to carry 10A each winding carries 10A.

As an aside, why would you call a proven formula, one that no one disputes, Machiavellian?
 
Re: Did we learn anything?

Originally posted by jim dungar:
rattus

The formula is a trick of numbers so of course it is possible to design a situation where it looks like it is important. If we were to continue the numbers playing game, using your fomula of 2*Vp*Ip=2*120*10=2400W we should be able to say that based on a 120V line to neutral voltage the transformer can carry 20A.

Reply: Jim, of course if one does not understand the meaning of the formula, but it would be rather stupid to draw that conclusion.

The loading a transformer sees is the load connected to it. My transformer is not over loaded. Each winding is designed to carry 10A each winding carries 10A.

As an aside, why would you call a proven formula, one that no one disputes, Machiavellian?

Reply: Jim, I call it Machiavellian because it is far more complicated than taking the product, VpIp. And, in my ignorance, I do not know that this formula applies to this case. I do know that it applies to an open delta with a balanced load. Can you show mathematically that it applies here?
Click to expand...
 
Re: Did we learn anything?

Rattus,

The "Oregon factor", as it has become, is in total error. It allows the calculated line currents of a load attached to a transformer to be too small. It does nothing about the calculated apparent va in the transformer.

The calculation presented by the gentleman in question is :

2000va resistive
208 connected voltage
the factor 1.154 is in the denominator thus

2000va/(208*1.154)= line amps of 8.333 A

But we all know the proper way to calculate line amps for a single phase load is just:

connected load in watts (or va)
divided by the connected voltage. 208V in this case.
2000va/208v=9.615 amps.

On a larger scale, one could undersize the conductors to the connected load, undersize the overcurrent device, and could severely undersize the transformer needed. If the factor was in the numerator, and resulted in a slightly HIGHER current calculated, then I could understand the need for it to fully calculate the transformer heat loading, ie:
(2000va * 1.154)/208v = 11.096 amps.
11.096A * 208V = 2308va xformer loading.
This agrees with the winding loading of
9.615A * 120 V * 2 =2308va.

I do fully agree that the heating in the transformer is of concern. I think that you AND Jim are getting to the same point, just with different applications of the math involved. I also agree that the situation of severely unbalanced Y transformers is FAR more common than we in the field would like to realize.

With all this in mind, I will definitely look harder at transformer loading vs just loads.

Thank You, for the insights.
 
Re: Did we learn anything?

et tu Garrett?

We never heard the exact wording of the Oregon question, but I interpret it to mean
"What is the ALLOWABLE current in the wye transformer with a single line to line load?"

The formula presented is in my opinion,

Imax = VArating/(208V*1.154) = VArating/240V

We are not computing actual currents, we are computing maximum ALLOWABLE currents. With this interpretation of the question, the OFF has a valid basis. Go ask your engineer buddies.

Now, I think it is silly to apply a fudge factor such as this when all you have to do is divide the VA rating by Vphase to obtain this information.

Still, I would like to hear the exact wording of the question.
 
Re: Did we learn anything?

Rattus,

The BIG PROBLEM here is, no one knows just what the exact question is? All involved have been sworn to secrecy to protect "the integrity of the testing procedure".

Sounds like just secrecy to me.

The multiple threads have been very illuminating from the math point of view. I haven't had to think so hard since my college days. (Way to long ago) As we get this straightened out here in Oregon I will try to post the results here.

Thank You,
Garrett

[ March 23, 2005, 01:42 AM: Message edited by: grlsound ]
 
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