Max Single Phase Load?

[davelectric]

I am installing a 45 kva xfmr 480 to 240 delta with a 120v center tap. I've been told that the max center tap 120v load can not exceed 5 percent of total xfmr capacity (2,250). Could you explain? Wouldn't the max be 15kva on that phase or 7.5 nuetral 120v loads combined C1 plus C2?

 

[charlie b]

Re: Max Single Phase Load?

I think you are right. I believe there is no code requirement that the load ?can not exceed 5 percent. . . .? The three secondary windings can each handle 15 KVA, for the total of 45 KVA. One half of the center-tapped winding should therefore be able to handle one half of the 15 KVA.

 

[jtester]

Re: Max Single Phase Load?

Many transformer manufacturers limit single phase current, read unbalanced current, to 5% or less. It is the nature of a delta-delta bank to circulate unbalanced currents through all of the transformers, and increase loading, without increasing output. Another way that utilities look at it is if all 3 transformers have the same impedance, the lighting transformer, provides 1/3 of the 3 phase power and 2/3 of the single phase power. Each of the power transformers provides 1/3 of the 3 phase and 1/3 of the single phase. These are loads inside the transformer bank, not loads flowing in the line conductors outside the bank.

I know that adds up to 3/3 of the three phase and 4/3 of the single phase, but that is how it is. These aren't Code limitations, but electrical circuits, and manufacturers requirements.

Manufacturers will let you load high leg delta transformers to more than 5% single phase load, but you must give up significant 3 phase capacity.

Now all you who don't believe the 4/3 single phase load can start the discussion.

 

[charlie b]

Re: Max Single Phase Load?

Since when does 2/3 plus 1/3 plus 1/3 add up to 4/3?

Actually, what you need to do is to eliminate the duplication (i.e., double-counting) within your math. Part of the lighting transformer?s ?2/3 of the single phase load? and part of each power transformer?s ?1/3 of the single phase load? are one and the same. You can?t count it twice.

So the correct addition of loads would be: 2/3 plus 1/3 plus 1/3 minus 1/6 minus 1/6, for a total of 3/3.

 

[charlie]

Re: Max Single Phase Load?

Charlie, I don't understand. Where do you get the ability to subtract the 1/6th twice? When we load three transformers in a bank, we load them with the single phase spread in the manner of 1/3, 2/3, and 1/3. Additionally, we will use test data and break it apart in the same manner.

 

[jim dungar]

Re: Max Single Phase Load?

As mentioned above, the 5% limitation is due to circulating currents in a closed-delta (laws of physics). This limitation only exists in closed-delta single-core transformers. Most utilties use single phase transformers to create 240/120V 3P4W systems so they do not have the same problems as contractor purchasing a single dry-type transformer.

 

[davelectric]

Re: Max Single Phase Load?

Jim that is exactly what a the Westinghouse engineer told me. I don't understand why the phase couldn't be loaded to more than 5% as long as don't exceed the 15kva.

 

[charlie b]

Re: Max Single Phase Load?

Originally posted by charlie: Charlie, I don't understand. Where do you get the ability to subtract the 1/6th twice?

You don?t get to count more than 100% of the power generated by the three windings. The final answer cannot be 4/3. There is a double-counting somewhere. I merely took one half of the ?extra 1/3? and assigned it to each of the other two windings. I assigned 1/6 as the double-counted load amount from each power transformer. I then subtracted the two double-counted amounts of 1/6 from the 4/3, to get back to 3/3, or 100%.

It?s just my attempt to rationalize an irrational statement: that the three windings supply more than 100% of the power that they are supplying. I don?t know if I can back this up with math, for I?m a bit rusty on hard-core trigonometry. But here is my reasoning:

When you say that the two power transformers are each supplying 1/3 of the single phase load, you must necessarily be talking about single phase 240 volt circuits. There is no center tap on the power winding to give you 120 volts.

Draw a triangle with Point A at the top, Point B at the bottom right, Point C at the bottom left. Consider a load connected between Points A and B. A single electron leaving Point A towards a single phase, 240V load will return to Point B, and will join with other single and three phase currents. But from there, that single electron might flow through one winding back towards Point A, or it might flow through another winding towards Point C. For that matter, the electron that left Point A in the first place might have originated from Point B or from Point C. It is in this sense that the three windings are ?sharing? the single phase loads.

But no matter how you slice it, ?power? is the one quantity that must always add up. You can?t always add one amp of current to two amps of current, and get three amps as the result. That is because of phase angle differences between the two currents. But one watt plus two watts will always be three watts, regardless of phase angles.

 

[jtester]

Re: Max Single Phase Load?

Charlie

The single phase load is actually as I said. If I could draw a transformer bank I could show you. I'll try to help the reader understand.

1. Visualize or draw a delta secondary. Ground the center of one leg.
2. The phases from the grounded leg are labeled A and C. The high leg is B.
3. All of the single phase current must come from phases A and C.
4. There are 2 parallel paths between phases A and C. They are 1.) the transformer with the grounded secondary, and 2.) the other two transformers in series.
5. If impedances are equal and we have a delta-delta bank, the single phase current in Phase A-C will split 2/3 through the grounded transformer, and 1/3 in the two in series.
6. From that you see that the grounded transformer provides 2/3 of the single phase power, and the power transformers each provide 1/3 of the single phase.

Ah the mysteries of modern science.

 

[charlie]

Re: Max Single Phase Load?

Charlie, this is no different than three phase load on an open delta bank where 57.735% (inverse of the radical 3) is placed on each transformer. It doesn't add up and logic would apply the load to each transformer at 50%.

 

[charlie b]

Re: Max Single Phase Load?

Originally posted by jtester: I'll try to help the reader understand.

I am afraid that your logic is flawed, and that your mathematical result is inaccurate. It was a valiant, though futile attempt. No matter what else happens, you cannot produce more than 100% of what you produce. Here is where your reasoning falls short:

3. All of the single phase current must come from phases A and C.

This is true if, and only if, you are referring to currents going to single phase 120 volt loads.

4. There are 2 parallel paths between phases A and C. They are 1.) the transformer with the grounded secondary, and 2.) the other two transformers in series.

The two ungrounded (power) transformers are NOT in series!

By definition, two items are in series if the current from the first has no other place to go than into the second. That is not the case here. Using your notation (I drew the picture you described), current going from Point C to Point B (i.e., through one transformer winding) might go from Point B to Point A (i.e., through the other transformer winding), or if might go out to a load. As soon as you connect any three-phase load, or connect any single phase load between Point B and one of the other points, you will have more than two items connected at Point B. That is no longer a series connection.

6. From that you see that the grounded transformer provides 2/3 of the single phase power, and the power transformers each provide 1/3 of the single phase.

Let us, just for the sake of discussion, presume that this three-phase transformer will only have single phase, 120 volt loads. Thus, there are no wires connected to Point B, other than the two power transformer windings. That will put the two in series, as you have described earlier. Your description of current splitting 2/3 and 1/3 now becomes accurate. In fact, you are describing a ?current divider? circuit. However, you have one incorrect, key word in your statement 6. You need to replace the word ?each? with ?together.? Your item 6 should read, ?From that you see that the grounded transformer provides 2/3 of the single phase power, and the power transformers TOGETHER provide 1/3 of the single phase.?

Each of the two power transformers will supply 1/6 of the single phase, and together they supply 1/3. So the total is 2/3 plus 1/6 plus 1/6 equals 3/3.

But I reiterate that this logic vanishes completely, the moment you connect a load at Point B.

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• <font size="2" face="Verdana, Helvetica, sans-serif">Aside to Forum Members who are Engineers or Other Lovers of Science and Math: </font>

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• <font size="2" face="Verdana, Helvetica, sans-serif">Do you think the ?Principle of Superposition? applies here? Can we treat the single-phase 120V, the single phase 240 volt, and the three phase 240 volt loads as individual, linear components, and then add up their effects to get the true total? Would that make the rule of 2/3 and 1/6 and 1/6 apply to the single phase 120v loads, even if there are loads connected at Point B?</font>

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Ah the mysteries of modern science.

Mystery Resolved!

 

[charlie b]

Re: Max Single Phase Load?

Originally posted by charlie: Charlie, this is no different than three phase load on an open delta bank where 57.735% (inverse of the radical 3) is placed on each transformer.

Part of the confusion here is the conversation use of the word ?load.? Does that word mean ?current? or ?power??

Each winding might account for 57.735% of the current. But you are adding two currents that are not in phase with each other. That is why 57.735% plus 57.735% adds up to 100%.

But power does not add up in that fashion. You are not adding KVA at one phase angle to KVA at a different phase angle. You add KVA to KVA to get KVA. So long as you have a balanced three-phase load on the open delta bank, the two windings will each supply 50% of the power.

 

[charlie]

Re: Max Single Phase Load?

Each winding might account for 57.735% of the current. But you are adding two currents that are not in phase with each other. That is why 57.735% plus 57.735% adds up to 100%.

That must be the same in the closed delta configuration with single phase load (current?). I know that we determine the single and three phase loads and then split them as I have indicated.

It is times like this that I wish I had the engineering background to make a good argument.

 

[coulter]

Re: Max Single Phase Load?

Guys -

Sheesh - I've got litle vector symbols scribbled all over the place and I'm having a little trouble with a few things. I generally think pretty well in vectors, but this one is tough.

Here are a few things I am seeing that are causing me to stumble:

1. <snip>But no matter how you slice it, ?power? is the one quantity that must always add up. You can?t always add one amp of current to two amps of current, and get three amps as the result. That is because of phase angle differences between the two currents. But one watt plus two watts will always be three watts, regardless of phase angles. <snip>

Well, I wouldn't exactly see it that way. For power out to load, voltage is a vector, current is a vector, both have magnitude and phase angle. So when you multiply the two together, then the result is a vector. Now add two power vectors, and the result is not a simple addition of the magnitudes. Now for heat disapated in the windings, heat is scalar, porportional to the current in the winding and should be alble to add all three

So, help me out here to get a handle on this vector monster. So, the following discussion restricts the loads to single phase, resistive, 240V connected to phases A-C. Also, no three phase loads or single phase loads connected to phase B. And, please deal with the 120V loads later too.

JT -If I am understanding you correctly, the current through the series ungrounded windings is one-half of the current through the grounded winding. Okay, that makes sense, the impedance is double cause there are two windings in series.

So the Iab = Ibc - 1/2(Ica). And Pab = Pbc = 1/2Pca. Now, I've been paying attention to you two, so it looks like the power put out by winding A-B has a phase angle of +60 deg and the power output of B-C has a phase angle of -60 deg relative to the phase angle of C-A. So some of the power pumped out by A-B is getting sucked up by B-C, in fact I would say half. Again from the previous conversations, power delivered to the load is split 2/3 from grounded winding, so Po (output to load) = Pca +1/2Pab +1/2 Pbc Okay that all matches all up

Second place I am seeing a difference of definition is in the terms concerned with power out from each winding, power delivered to load, power disipated as transformer heat. And I got some confusion here.

Translating Charlie's comments, I think he is saying, half of the power put out by winding A-B gets sucked up by the opposite phase angle component of B-C. So for any output Po, the transformer is heating winding C-A by an amount porportional to 2/3Po, heating A-B and B-C proportional to 1/3Po. Since these are just heat and are scalar, we ought to able to just add them for total heat, which is porportional to 2/3Po + 1/3Po + 1/3Po. So for any load sinking Po, the transformer is heated as though it were putting out 4/3Po.

How am I doing? This sucker is brutal.


***********************************
Do you think the ?Principle of Superposition? applies here? Can we treat the single-phase 120V, the single phase 240 volt, and the three phase 240 volt loads as individual, linear components, and then add up their effects to get the true total?
**********************************
yes but i'm not doing it.

carl

 

[jtester]

Re: Max Single Phase Load?

Let me take a shot at Charlie the utility guy's 57.7% and see if Charlie b. is talking the same power that we are.
The power out of a closed delta bank at full load, Pc, is VxIcx1.732 where Ic is the phase wire current in the closed delta example. The current thru each xfmr, It, = Ic/1.732. Remember each xfmr is fully loaded at this point.

The power out of an open delta bank at full load, Po, is VxIox1.732 where Io is the open delta phase wire current. Looking at the legs that are open, It=Io. Remember same size transformer still fully loaded, except the phase current can't split in the open leg.

Compare currents It=Ic/1.732=Io

Substituting Ic/1.732 for Io gives
Po=Vx(Ic/1.732)x1.732 = VxIc
Notice this is the closed delta power without the 1.732.
Po = Pc/1.732 or Po = .577xPc
Hence the statement that the capacity of the open delta system is 57.7% of the closed delta system.

An example, if you have 3 10 kva transformers in a closed delta feeding 30 kva of 3 phase load, you have 30 kva of fully loaded transformation and 30 kva of capacity. If you take one transformer out and create an open delta, you have 20 kva of fully loaded transformation but only 17.32 kva of capacity.

 

[charlie b]

Re: Max Single Phase Load?

Coulter: You posted your comments after I left work, and I keep my related text books at work. I cannot properly address your comments until I get a chance to read my text books. But I can respond to a couple of your statements, and perhaps we can work this out together.

First, you are incorrect in saying that the product of two vectors is a vector. There are two ways to multiply vectors. One is called the ?Cross Product.? It?s result is a vector. The other is called a ?Dot Product.? Its result is a scalar. What I don?t wish to assert without first looking at the book is whether the product of voltage and current is a ?Cross Product? or a ?Dot Product.?

Secondly, you are incorrect in saying that the multiplication of a voltage vector and a current vector results in a vector, if you are working with the load, and a scalar (i.e., number), if you are working with the conductors. They are both vectors, or they are both scalars. I still need to work out which is true. In the conductor, the current sees only pure resistance, so that the voltage dropped along the conductor is in phase with the current through the conductor. You are simply multiplying two vectors of the same phase angle.

Third, you can use trigonometry to do the multiplication. For those who fear that word, I?ll only talk about it, and not show the actual trig. Keep in mind that the vector addition stuff is merely a simplification, to allow us to get the answers without going through the trig. But here is what the trig tells us:

If ?a? represents the phase angle between voltage and current, and ?w? represents the frequency (we base that on 60 Hertz in the USA), and if Vm and Im represent the peak values for voltage and current, then the power of a single phase circuit, as a function of time, is given by:

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• <font size="2" face="Verdana, Helvetica, sans-serif">P(t) = ? (Vm Im) cos (a) [1+ cos (2wt)] + ? (Vm Im) [sin (a) sin (2wt)]</font>

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This bizarre thing is essentially the sum of a constant, a cosine (at twice 60 Hertz) with one magnitude, and a sine (again, at twice 60 Hertz) with different magnitude. That is no longer a ?sine wave.? I am not at all sure it can be represented by a vector in the same graph as the voltage and current vectors.

This does not resolve the issue. But perhaps it might address some of your confusion.

 

[charlie b]

Re: Max Single Phase Load?

Jtester: Thanks for reminding me where the ?huh factor? comes from on an open delta. It is not a matter of, ?Gee, why is it 57% and not 50%.? Rather, it is a matter of, ?Gee, why is it 57.7% and not 66.7%.?

If you start with a delta transformer, then open one leg, intuition might suggest that the remaining 2/3 of the windings should be able to give you 2/3 (i.e., 66.7%) of power. Intuition fails here. You really get only 57.7%, as you have explained very nicely.

That said, I still submit that, given a balanced load, the two remaining windings will share that load equally. That is, each will get 28.85%.

 

[jtester]

Re: Max Single Phase Load?

Can't produce more than 100% of what you produce. Charlie, you are correct, but we are talking load on individual transformer legs adding up to 4/3 of the single phase load served. You have already seen that in the open delta example, the capacity is less than the load seen on the transformers.

The two undgrounded power transformers are not in series. Charlie, again under many circumstances that is true, but to make modeling the system manageable, power engineers use the following assumptions when they tout this idea.
Single phase 240 volt loads can be spread around all legs of the delta. Those that balance each other out will look like equivalent 3 phase loads and can be considered such. When finally circuited, the remaining ones that don't balance, must be hooked across phase A&C. The final assumption related to single phase loading is that all 120 volt loads must be balanced, no neutral current either. Not exactly the real world, but not far off either. We're exchanging basic ideas, not computer programs.

Once you understand the assumptions that make this problem manageable, I think you will agree that the two parallel paths are the lighting transformer, and the two power transformers in series.

I have some work by Westinghouse that shows how the load divides based on power factor of the load, and which phase, A,B, or C, you hook the lighting transformer up to. Those calcs are quadratic equations and are generalizations of work done by J.C. Neupauer back in the 1950's. Clearly my statement is a generalization, but I believe that it will hold water.

Those interested, I will fax a copy of exerpts from Westinghouse and from GE transformer manuals.

 

[steve66]

Re: Max Single Phase Load?

You are correct, Charlie B, that you cannot use basic vector math formulas when one wave is wt and another is 2wt. You would have to use trig identities and write out the formulas longhand keeping the frequencies, and hoping the 2wt terms cancel out.

More on Charlie B's formula:

The last term with the sin(2wt) is the apparent power (its integral over one cycle is zero).

The first term with (1+cos(2wt)) is the real power. It contains two parts. The constant (1/2 Vm Im cos (a) )* 1 is the dc power. The other factor (1/2 Vm Im cos (a) cos(2wt) is a pulsating real power (i.e it is delivered to the load in pulses, but not returned to the source like the apparent power).

It is a basic concept that the power in an AC circuit is at twice the frequency of the applied frequency. But you may be able to use trig identies to get sin^2 wt and cos^2 wt terms instead of the double frequency terms. At the very least, you could break it up into a fundemental wt component and a 2wt harmonic, do the math, and see what happens.

 

[charlie b]

Re: Max Single Phase Load?

Originally posted by jtester: Charlie, you are correct, but we are talking load on individual transformer legs adding up to 4/3 of the single phase load served.

It isn?t 4/3. Let me go back to my earlier comment on your earlier post:

Originally posted by jtester: 6. From that you see that the grounded transformer provides 2/3 of the single phase power, and the power transformers each provide 1/3 of the single phase.

The power transformers do not EACH provide 1/3. Instead, they TOGETHER provide 1/3. Therefore, each provides 1/6.

The sum is 2/3 plus 1/6 plus 1/6. The result is not 4/3, it is 3/3.

You have already seen that in the open delta example, the capacity is less than the load seen on the transformers.

I don?t quite understand what you mean by ?the capacity is less than the load.? It almost sounds like you are saying that the power received by the load is different than the power supplied by the two transformer windings. But I?m sure you did not mean that.

I would put it this way: When you lose one third of the windings (i.e., to create the open delta configuration), the remaining two thirds of the windings cannot supply 2/3 of the power that the original three windings could supply.