Use of 1.154 in single phase calc on three phase

[jim dungar]

There is no question that there are ways to develop the 1.154 multiplier.

My point is, it serves no useful purpose in three phase systems.

It takes several facts and jumbles them together into a single conclusion. It does not help to size a three phase transformer. It does not help to determine the watts loss of the three phase transformer. It does not help determine the input power to the three phase transformer. It is simply a game with numbers, kind of like saying that breathing air is a leading cause of accidents because almost every one in involved in one was breathing at the time.

 

[rattus]

An easier way:

An easier way:

It is not a fudge factor, it is derived from understanding the formula for Power, and it is correct:

King, your math is correct, however the question on the Oregon exam involves apparent power, not real power, therefore your explanation does not quite fit the situation, and it is overly complicated.

Simply put, apparent power in the load is,

Pal = Vll x Iload

and, apparent power delivered by the two transformers is,

Pax = 2 x Vp x Iload

The ratio of the two powers is,

Pax/Pal = 2 x Vp/Vll = 2xVp/(Vpx1.732) = 2/1.732 = 1.154

Or you could rationalize the expression to

2xsqrt(3)/3

 

[kingpb]

Jim,

Your point is valid. I'm not sure why this factor would ever show up on any test, except maybe a math test. You wouldn't size a transformer this way, because if you had a large single phase load, and a dedicated transformer, then why would you use 3 phase, just use a single phase. If your using three phase it's going to be sized much larger then the load anyway, so who cares.

However the OP asked a question and then people wanted to know where the factor came from, and the answer was provided. Now, whether the factor is useful, well, that is an entirely different question!

 

[jim dungar]

Jim,
However the OP asked a question and then people wanted to know where the factor came from, and the answer was provided. Now, whether the factor is useful, well, that is an entirely different question!

Actually part of the original question was if there was a valid reason to use the factor.

Is there a good reason for using the 1.154 multiplier

My answer is no, because it provides useless information. Every one has spent time deriving the factor but no one has said how or why it is used.

 

[rattus]

Actually part of the original question was if there was a valid reason to use the factor.

My answer is no, because it provides useless information. Every one has spent time deriving the factor but no one has said how or why it is used.

Jim, this factor could be used to provide a more conservative estimate of transformer loading for this unbalanced case. You have made it clear that you won't use it, and I doubt that anyone else will either. I would think that in a normal situation, the imbalance would be so small that it would not matter. One can imagine though, a wye service loaded with a single line to line load. Then it would apply.

 

[steve66]

To use this "fudge" factor, you first have to assume that a transformer can dissipate its full rating while using only two of its coils. So for a 30KVA 3 phase transformer, you would have to assume two windings could each supply 15KW each.

If you are following the NEC, I think that might be a problem with the overcurrent protection, since OCP would be sized assuming each winding can only provide 10KW each.

I prefer to just limit each winding to 1/3 of the total transformer rating. A 30KVA transformer gets no more than 10KVA per winding. If you stick with this rule, you don't need the fudge factor.

BTW: I hope everyone realizes that:

P (1 phase LL) = 2 * V phase * I phase * cos Theta

Is not the same power that is disipated in the load.

 

[jim dungar]

One can imagine though, a wye service loaded with a single line to line load. Then it would apply.

Rattus,
I can easily imagine many examples of this single phase imbalance.
Now, if you are trying to describe how much of the available transformer loading has been used then say that. Do not say that the load is one amount of watts but the transformer thinks it is delivering a different amount of watts.


All the fudge factor does is show that at a given amount of current a 100% loaded 1-phase 240V transformer needs to be 1.154 times larger than a 100% imbalanced 3-phase wye 208V one. Or that through the magic of physics 2 fully loaded single phase transformers connected in a series 1ph 2w connection can supply a larger load then if they are connected in a 3ph open wye.

 

[rattus]

Rattus,
I can easily imagine many examples of this single phase imbalance.
Now, if you are trying to describe how much of the available transformer loading has been used then say that. Do not say that the load is one amount of watts but the transformer thinks it is delivering a different amount of watts.


All the fudge factor does is show that at a given amount of current a 100% loaded 1-phase 240V transformer needs to be 1.154 times larger than a 100% imbalanced 3-phase wye 208V one. Or that through the magic of physics 2 fully loaded single phase transformers connected in a series 1ph 2w connection can supply a larger load than if they are connected in a 3ph open wye.

Jim, VA not watts! 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.

And, real power, in watts, from the transformer equals real power dissipated in the load. In this case, 2080W. We all know that, don't we?

Simply put, the two active transformers should be sized to handle 1200VA each instead of 1040VA. The third transformer can be omitted.

A transformer can safely deliver no more than its rated apparent power regardless of the configuration or load connections. I never said otherwise.

I cannot follow the rest of your argument, but you seem to be making this thing overly complicated, and you seem to be saying things I did not say.

 

[RayS]

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.

thank you rattus! I just read that thread, and now it makes some sense.

 

[e57]

WOW - I AM SORRY I ASKED!!!!!!

1st - Oregon has a problem.... Not sure if they have the exact wording of a question that drags you into the neither-world - but if they do - SOMEone has too much time on thier hands and may be selling this question to his flunkies...... (the answer to trick question #3 is ---- $50)

No-one is going to size a 208 wye transformer that way as a single phase! A feeder from it- maybe? Multiple feeders from it? I would have to think about it.....

BTW a single phase 208 as a diagram (two windings center-taped for grounded) would look remarkably like an Edison... Would it even be 208..... Really? (without that 3rd phase to off-set potential)

 

[rattus]

You are welcome:

You are welcome:

thank you rattus! I just read that thread, and now it makes some sense.

The original question was poorly worded, and I think we never saw the exact wording, so we had to read between the lines. We may still have it wrong.