Transformer Loading:

[jim dungar]

Re: Transformer Loading:

VA = SQRT(P^2 + Q^2), as shown in my equations.

Where is the Q in your equations.

 

[rattus]

Re: Transformer Loading:

Originally posted by jim dungar:
VA = SQRT(P^2 + Q^2), as shown in my equations.

Where is the Q in your equations.

Jim, You didn't tell us how the neutral current makes a difference. You are sidestepping the question.

Second, we don't need to break the apparent power into real and reactive components. It is simply Vphase x Iphase. You know that!

Third, computing the apparent power your way yields, voila, 2400VA not 2080VA.

Fourth, you have not explained away the fact that a transformer could be overloaded by 15% by computing apparent power at the load instead of at the transformer.

Explain these four points, and I may come over to your side of this argument.

 

[jim dungar]

Re: Transformer Loading:

I have stated my case of identical winding loading and therefore identical transformer loading for the two situations of a single load and two series loads (with no neutral current). Rattus has stated his case for different transformer loadings.

Does anyone else have any comments?

 

[rattus]

Re: Transformer Loading:

Jim,

1) You did not respond to my four points. You are avoiding the real issues. This lessens your credibility.

2) I went to great pains to show that two loads across the phase voltages would create the exact current as a line to line load. So how is the loading any different? The copper and iron losses are the same in both cases. So how come a watt of copper loss is a watt in one case, but only 0.87 watts in the other? That is what you are saying in effect!

I also presented this argument as a load in a black box. That is, we do not know the load configuration. How would you compute the loading in that case? How would you know that it was a line to line load or a line to neutral load. Do you think the transformer is smart enough to know? I am unaware of any instrumentation that would tell us! The transformer gets just as hot either way.

Come on Jim, if you cannot prove your argument with any engineering rule or principle, don't you suspect that you might be wrong?

 

[jim dungar]

Re: Transformer Loading:

I am not avoiding your issues. You are the one saying the transformer loading in the two cases is different, I have always maintained they are identical.

All of my discussion has been on the power loading of the transformer not on the power consumed by the transformer. In both of your examples you have maintained that the loading of each transformer winding is 120V * 10A. The conductor loss of each secondary winding would then be identical (total transformer secondary copper loss = 2*((10A^2)*Rwinding)). The core loss of the transformer is supplied by the primary side but again if the secondaries are identical then the primaries losses are also identical (especially if these are individual single phase transformer connected into a three phase bank).

And as far back as my Mar 09, 19:43 posting I discussed the problem of overloading an open-wye transformer. At that time I stated this single-phase load of 2080VA was the maximum loading possible given each transformer is rated 1200VA.

My only disagreement with you has been your assertion that the transformer sees 2400VA of load (?the transformer thinks it is delivering 2x120xI VA? in first post).

 

[engy]

Re: Transformer Loading:

??discussed the problem of overloading an open-wye transformer. At that time I stated this single-phase load of 2080VA was the maximum loading possible given each transformer is rated 1200VA?

2080W above would be more accurate (yes it?s 2080W & VA for the load, but not the xfmr)
That?s why the 2400VA of transformers can only support 2080W of load.
W in the transformers must equal W of the load. ( Not so for VA)

Screw up the phase relationships in the transformers, then VA?s don?t add up.
V&A is in-phase in the load, not in-phase in the transformers. Something?s gotta give.

Summary

(1) 2080W on 208V single phase. (two phases of three phase system)

Load Watts = 2080
Load VA = 2080

Transformer Amps = 10 (each phase)
Transformer VA = 2400
Transformer W = 2080

(2) 2400W (1200W L-N on each of two phases)

Load Watts = 1200+1200 = 2400
Load VA = 1200+1200 = 2400

Transformer Amps = 10 (each phase)
Transformer VA = 2400
Transformer W = 2400

In both cases the transformers are fully loaded. One setup cannot deliver as much real power as the other. The same basic thing happens for an open delta set of transformers where transformer derating is required. We should probably agree on this before discussing open delta though

 

[rbalex]

Re: Transformer Loading:

To permit the debate to continue, I'd like to provide a framework for the discussion:

From IEEE C37.12.80-2000
"Standard Terminology for Power and Distribution Transformers":

3.4.5 Load (Output).: The apparent power in megavolt-amperes, kilovolt-amperes, or volt-amperes that may be transferred by the transformer.

From IEEE C37.12.00-2000
Standard General Requirements for Liquid-Immersed Distribution, Power, and Regulating Transformers

4.1.6.1 Capability
Transformers shall be capable of
a) Operating continuously above rated voltage or below rated frequency, at maximum rated kVA for any tap, without exceeding the limits of observable temperature rise in accordance with 5.1 1.1 when all of the following conditions prevail:
</font>

• <font size="2" face="Verdana, Helvetica, sans-serif">1) Secondary voltage and volts per hertz do not exceed 105% of rated values.</font>
• <font size="2" face="Verdana, Helvetica, sans-serif">2) Load power factor is 80% or higher.</font>
• <font size="2" face="Verdana, Helvetica, sans-serif">3) Frequency is at least 95% of rated value.</font>

<font size="2" face="Verdana, Helvetica, sans-serif">b) Operating continuously above rated voltage or below rated frequency, on any tap at no load, without exceeding limits of observable temperature rise in accordance with 5.1 1.1, when neither the voltage nor volts per hertz exceed 110% of rated values.

 

[engy]

Re: Transformer Loading:

Anything in that standard that assumes balanced loading, or references unbalanced loading?

 

[jim dungar]

Re: Transformer Loading:

Bob,
Do you agree with my interpretation of those citations as: power transformer load rating is based on VA (and it's multiples)not on watts.

 

[rbalex]

Re: Transformer Loading:

Originally posted by engy:
Anything in that standard that assumes balanced loading, or references unbalanced loading?

There is a reference in C57.12.00, Section 4.2 to IEEE C57.91-1995 that "...provides guidance for loading at other than rated conditions, including
...
c) Loading that results in reduced life expectancy.

Further, in Section 4.3.3 one of the "Unusual Service Conditions is: (j)Specified loading conditions (kVA outputs, winding load power factors, and winding voltages) associated with multiwinding transformers or autotransformers.

Another Guide is IEEE C57.105-1978 Guide for Application of Transformer Connections in Three-Phase Distribution Systems It says in Section 6.2 "...When single-phase and 3-phase loads are combined in Y-connected secondaries, phase difference between these load currents exists even with similar power factors, and simple addition of the two load components, though conservative, is not correct."

Unfortunately, I don't have access to C57.91. Both it and C57.105 are Guides rather than Standards and are not mandatory.

The bottom line is that "standard" three-phase calculations techniques only apply to balanced systems, but most transformers are capable of handling small imbalances. What the tolerance is would require consulting with the manufacturer.

Significantly imbalanced systems require advanced techniques such as symmetrical components to analyze properly.

[ March 14, 2005, 04:15 PM: Message edited by: rbalex ]

 

[rbalex]

Re: Transformer Loading:

Originally posted by jim dungar:
Bob,
Do you agree with my interpretation of those citations as: power transformer load rating is based on VA (and it's multiples)not on watts.

Well, since you asked a direct question, I think the load, i.e., "...[t]he apparent power in ... volt-amperes that [is] transferred by the transformer" in question is 2400VA, 1200VA from each of two windings.

I believe we all already agree that the load and load rating aren?t necessarily the same, especially for significantly imbalanced loads. From what I have seen, we would all still select a "3600VA" transformer if we were forced to use a three-winding transformer in this application.

 

[jim dungar]

Re: Transformer Loading:

The physics of what the "transformer" sees is not dependent on whether it is a single core transformer or a bank of single phase units. However, the total losses of a transformer are dependent on it's construction and/or connection.

The heat generated by the copper/winding watts loss in a transformer is I^2R for each winding and is not related to voltage and therefore VA. In both of these topic situations the current through each winding is identical at 10A, therefore the watts loss of each winding is identical (Ltotal=2*100*Rwinding).

 

[rattus]

Re: Transformer Loading:

Originally posted by jim dungar:
The physics of what the "transformer" sees is not dependent on whether it is a single core transformer or a bank of single phase units. However, the total losses of a transformer are dependent on it's construction and/or connection.

The heat generated by the copper/winding watts loss in a transformer is I^2R for each winding and is not related to voltage and therefore VA. In both of these topic situations the current through each winding is identical at 10A, therefore the watts loss of each winding is identical (Ltotal=2*100*Rwinding).

Jim, I would add that this current, multiplied by the voltage, is the apparent power delivered by the transformer. Throw in PF, and you have the real power delivered.

 

[jim dungar]

Re: Transformer Loading:

Rattus, in your first post you said

It turns out that connecting unbalanced line to line loads on a 3-phase wye yields different loadings depending on the way you compute them ?.The load thinks it is getting 208xI VA, but the transformer thinks it is delivering 2x120xI VA. How would you size the transformer

.

But on March 12 you stated:

It illogical to say it is different for different load configurations. It should be obvious to anyone who knows anything that we are in effect talking of a current rating and that does not change because the line to line voltage is 208V instead of 240V.

My point all along has been it is incorrect to simply add the VA of any
poly-phase transformer bank.

For a three phase transformer bank the total VA is 3 X VA(of the smallest transformer) not the sum of the VA of each unit. For an open-delta (either primary side or secondary side) the resultant total VA is the 3 phase rating X .577 (because of the 120? relationship between the phase voltages) not the sum of the VA of each connected unit. These formulas are based on not exceeding the ampere rating of any one winding.

In your example assuming (2) 1200VA single phase units, the unbalanced transformer rating is VAt=3*1200VA*.577 = 3600*.577 = 2080VA (rounded). Using this formula allows you to easily determine the amount of current in each winding based on the proven equation I=VA/E.

 

[rattus]

Re: Transformer Loading:

Originally posted by jim dungar:
Rattus, in your first post you said

It turns out that connecting unbalanced line to line loads on a 3-phase wye yields different loadings depending on the way you compute them ?.The load thinks it is getting 208xI VA, but the transformer thinks it is delivering 2x120xI VA. How would you size the transformer

.

But on March 12 you stated:

It illogical to say it is different for different load configurations. It should be obvious to anyone who knows anything that we are in effect talking of a current rating and that does not change because the line to line voltage is 208V instead of 240V.

Reply: I see no conflict with these two statements.

My point all along has been it is incorrect to simply add the VA of any
poly-phase transformer bank.

For a three phase transformer bank the total VA is 3 X VA(of the smallest transformer) not the sum of the VA of each unit. For an open-delta (either primary side or secondary side) the resultant total VA is the 3 phase rating X .577 (because of the 120? relationship between the phase voltages) not the sum of the VA of each connected unit. These formulas are based on not exceeding the ampere rating of any one winding.

Reply: Yes Jim, but it still boils down to the the?VA rating of each unit, that is a current limit. It is granted that adding the transformer ratings only tells us the maximum possible load which might be applied. If for example you load the open delta evenly only on the two remaining xfmrs, we can safely deliver 2/3 the rating of the full bank. In the case of the Oregon test, we see that computing the load at the load can lead to significant error.

In your example assuming (2) 1200VA single phase units, the unbalanced transformer rating is VAt=3*1200VA*.577 = 3600*.577 = 2080VA (rounded). Using this formula allows you to easily determine the amount of current in each winding based on the proven equation I=VA/E.

Reply: Jim, this seems to be a lot of trouble when we already know the current and voltage as in this case. It is simply Vphase x Iphase for each xfrmr.

All I am saying is that an exact method of computing xfrmr loading is to take the product Vphase x Iphase for each transformer. This will apply to any loading situation. I do not claim that this method is practical. I know it will not be adopted. It is merely a technical point.

[ March 16, 2005, 10:57 AM: Message edited by: rattus ]