# unbalanced 3 phase generator load

#### GoldDigger

##### Moderator
Staff member
Smart \$...

I strongly contend your assertion that all three windings of the delta-connected generator contribute to the load current is invalid. My simple proof... convert the delta-connected generator into a wye-connected one.

Now if you truly understand Electrical Theory, you must be aware of a very basic tenet, i.e., it is possible to substitute a wye-arrangement for a delta-one, and vice-versa!

Thus, if you substitute a wye-generator for the delta-one, and connect the load to say terminals A and B, terminal C has no connection to the load!

Phil
I think you are misusing that basic tenet. It is absolutely correct that you can substitute a delta for a wye and vice versa without having any effect on what is on the other side of the three terminal points. It certainly can make a difference as to what is going on inside the delta or wye device.

Here is one example: If you have a simple series circuit, coming out of a black box with terminals A and B, you can measure anything you want at those terminals and not be able to tell the difference between a current source with a parallel resistor and a voltage source with a series resistor. But from the point of view of th heat generated within the box, you there will certainly be a difference.
The windings in wye are not the same coils of wire that are used to make a delta of the same voltage, and the way that the current/power comes from the three windings will be different. But from the point of view of the load, there will be no difference.

#### Tom Jones

##### Member
I am not smart or learned enough to understand 3/4 of the arguments being made here. But I have a question about the following assertion: If I have a generator connected 120/208Y instead of the delta that has been discussed here already, and the phases are similarly imbalanced, will there be a similar amount of harmonic (if that is the right term) current in the rotor? Generator manufacturers typically rate their sets at 80% power factor on three phase. Do they have a similar "customary" phase imbalance amount allowed? I have seen many (most?) sets running with a pretty good imbalance and no ill effects although they are usually connected wye. Finally, how would this apply to a set that is connected single phase, either double delta or dog leg/zig zag?
Iceworm... Reur post #54: If you have Stevenson's "Elements of Power Systm Analysis", 2nd edition, the negative-sequence current generates flux that sweeps over the entire rotor, even though the rotor is supplied with DC, at twice line-frequency! A simlar explanation is given in Wagner & Evans, "Symmetrical Components" 1st edition, in the section titled "Rotating Machines!" If neither text you possess contain such information let me know! Regards, Phil Corso

#### Phil Corso

##### Senior Member
Goldigger...

Amazing! Without a word about transforming a delta-circuit consisting of 3-terminals into a wye-circuit with three-terminals, you turned this discussion into a card game where the two-terminal circuit trumps the three-terminal circuit!

I eagerlly await your next trick... how a single-terminal circuit takes all!

Phil

#### Smart \$

##### Esteemed Member
Smart \$...

I strongly contend your assertion that all three windings of the delta-connected generator contribute to the load current is invalid. My simple proof... convert the delta-connected generator into a wye-connected one.

Now if you truly understand Electrical Theory, you must be aware of a very basic tenet, i.e., it is possible to substitute a wye-arrangement for a delta-one, and vice-versa!

Thus, if you substitute a wye-generator for the delta-one, and connect the load to say terminals A and B, terminal C has no connection to the load!

Phil
Actually, wye-delta transform supports my assertion rather than contradicts. Appparently you did not verify your supposition.:happyno:

Let's say as an example, in the delta configuration, each resistance is 10 ohms. Transformed to wye, each resistance is 3⅓.

Now in the delta configuration, with no [external] connection to C, A to B has one 10 ohm resistance (A-B direct) in parallel with two series connected 10 ohm resistances (A-C and C-B, i.e A-B indirect).

R = 1/[1/10 + 1/(10+10)] = 20/3 = 6⅔ ohms

In the wye configuration, with no [external] connection to C, A to B has two 3⅓ resistances in series (A-wye and B-wye).

R = 3⅓ + 3⅓ = 6⅔ ohms​

#### Phil Corso

##### Senior Member
Smart \$/2

Once again... pure obfuscation!!! We are not discussing 3- single phase-loads... just one! Connected to only one-phase of a Delta-connected generator! Did you use the Delta to Wye Transform?

#### ATSman

##### ATSman
240/120V Delta with Stinger Leg

240/120V Delta with Stinger Leg

Most 240/120V Delta systems I have seen have a Stinger leg to ground (one phase winding is center tapped to ground.) A cheap and dirty way of getting single phase 120V source without using a transformer. Could this account for the unbalanced generator loads stated by the OP?

#### Smart \$

##### Esteemed Member
Smart \$/2

Once again... pure obfuscation!!! We are not discussing 3- single phase-loads... just one! Connected to only one-phase of a Delta-connected generator! Did you use the Delta to Wye Transform?
I understand completely, and I did use Y-Δ transform to demonstrate the 'source' impedance.

For reference, go here: http://en.wikipedia.org/wiki/Y-Δ_transform#Demonstration

Note the demonstration actually starts by using the impedance between N1 and N2, with N3 disconnected from the circuit (i.e. not connected to external circuitry).

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#### iceworm

##### Curmudgeon still using printed IEEE Color Books
I am not smart or learned enough to understand 3/4 of the arguments being made here. But I have a question about the following assertion: If I have a generator connected 120/208Y instead of the delta that has been discussed here already, and the phases are similarly imbalanced, will there be a similar amount of harmonic (if that is the right term) current in the rotor? Generator manufacturers typically rate their sets at 80% power factor on three phase. Do they have a similar "customary" phase imbalance amount allowed? I have seen many (most?) sets running with a pretty good imbalance and no ill effects although they are usually connected wye. Finally, how would this apply to a set that is connected single phase, either double delta or dog leg/zig zag?
I suspect you are doing fine. You just are not part of the group trying to see who can fill a quart bottle the fastest. Which, ANAICT (as near as I can tell) that's where the rest of them are.

Assuming you are speaking of small generation (say 100KW or less), I have not seen any evidence that the gen mfg limits the phase imbalance.

Consider 208/120 connected for two legs loaded with phase to neutral, 120V loads, current at generator FLA. Two windings are pumping out full power, and one is zero. The most one is going to get is .67 gen KVA (KW if one is driving resistive loads) The driver is fine, but two windings are getting hot.

Most gen mfg limit DD to 67%. A few limit DD to 50%. I don't know why the difference.

ice

#### iceworm

##### Curmudgeon still using printed IEEE Color Books
Iceworm... Reur post #54:

If you have Stevenson's "Elements of Power Systm Analysis", 2nd edition, the negative-sequence current generates flux that sweeps over the entire rotor, even though the rotor is supplied with DC, at twice line-frequency!

A simlar explanation is given in Wagner & Evans, "Symmetrical Components" 1st edition, in the section titled "Rotating Machines!"

If neither text you possess contain such information let me know!

Regards, Phil Corso
I spent a couple of pleasant hours last night with a cigar and a couple of cups of tea going over Grainger and Stephenson Jr, 1994. Interestingly there is no mention of your statement. There is a discussion about stability, oscillating rotor phase angle, and damper windings. However. the periods concerned were in the seconds range. Nothing in the subcycle to single-cycle range. Makes one wonder if Junior left it out on purpose cause it doesn't matter.

Some discussion:
As noted by mivey, in his reference to Emmanual:
Constant shaft torque from driver to gen
The power function following 1/2 + cos(2wt)
The difference in power around one cycle is two pulses up and down

This power pulsations can only be satisified by a change in the rotor angular momemtum. Which has to be accounted for by accelerating and deacelerating the rotor twice in each revolution. Pretty simple excerise.

One could infer that the continuous rotor phase angle change caused by the accel and deaccel is heating up the rotor through the damper windings. Probably so. But then one might also infer the gen mfg accounts for that in generator capability chart on the rotor heating limit side (high inductive var loadings), or that the additional heating is too small to matter.

Now looking at the issue of DD connection:
Each "cup" (so to speak) is a delta with one open line. All three coils are loaded - just not evenly. Simple exercise to show the line current is made up of: ("<" means phase angle)
positive sequence = .58 <-30
negative sequence .58 <30​

Which is going to give some interesting power pulses. But again, these are two per revolution. Again this is small generation, does it matter

I'm out of my element - can't say. Alternator heat rejection, magnetics design, rotor eddy currents are not anything I know about.

It is however interesting that Grainger and Stevenson Jr don't discuss it other than in passing concerning stability.

Note to Tom Jones:
I promise I am not joining the bottle filling crowd. :roll: They are just too silly.​

ice

#### Phil Corso

##### Senior Member
Ice...

The book I cited was authored by william D. Stevenson, Jr and was published by McGraw-Hill in 1962! LCCC # is 61-17676! Try your University's Library.

Phil

#### iceworm

##### Curmudgeon still using printed IEEE Color Books
Ice...

The book I cited was authored by william D. Stevenson, Jr and was published by McGraw-Hill in 1962! LCCC # is 61-17676! Try your University's Library.

Phil
No, I'd say I'm done with this one. But, perhaps you should consider going to your university library and getting the newer reference I cited.

As I said, I have no idea what changed between 1962 and 1994. And I've got paying work to do.

Just curious: Did you read what I wrote? Or just blow it off as unknowledgable? I'm okay either way. Mostly just curious cause your response appears to translate as, "You don't agree. Go get MY reference" Truely, I am okay either way

Enjoy the day.

So Speakth the Worm

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#### mivey

##### Senior Member
Would I be correct in saying, with respect to the voltage imbalance discussed, that the negative sequence component of the voltage would try to rotate the motor in the opposite direction? That gives some physical meaning to the term.
It causes high induced currents (thus heat) as explained in the Basler paper Smart has linked. I read it a while back and it is a good paper.

#### mivey

##### Senior Member
Smart \$ ...

Rather than play your "Who's on First" game, I can recommend an excellent text on 3-phase generators!

Regards Phil Corso
What text are you recommending?

What Smart\$ has shown is that for a 3-phase transformer delta winding, 2/3 of the load is supplied by the single winding paralleling the load and 1/3 of the load is supplied by each of the other two windings. This can be found in any good transformer book.

In fact for a 30 amp 240 volt single-phase load (7.2 kVA) and no three-phase load and negligible voltage drop, my Westinghouse reference gives the transformer winding loads as 2.4, 2.4, and 4.8 kVA

#### mivey

##### Senior Member
Sorry for dragging you guys back. But I'm feeling a bit dumb this morning. Still working on my first cup of coffee

How does a negative sequence alternator stator current put extra heat in to the rotor? The alternator rotor is strictly DC. What circulating currents? This isn't an induction motor rotor cage. Color me confused. :dunce:

ice
Read the Basler paper. We induce a current by cutting a conductor across a magnetic field. That is the basic operation of rotating a wire loop in a magnetic field to generate current. The induced current direction follows the right hand rule. So a magnetic flux from right to left (index finger) cut by a wire moving down (thumb) induces a current into the page (middle finger).

What the Basler paper shows is that the rotor is rotating in one direction but the magnetic flux is rotating in the other. This cutting of the flux induces currents in the rotor that cause it to heat up.

#### mivey

##### Senior Member
A similar explanation is given in Wagner & Evans, "Symmetrical Components" 1st edition, in the section titled "Rotating Machines!"
Page 23.

also see Blackburn's "Symmetrical Components for Power Systems Engineering" page 267 section 10.4

#### Sahib

##### Senior Member
See the attachment page from one generator manufacturer. They have stated for single phase loading of delta connected gen-set, the capacity is 57% of three phase capacity i.e [1.732*Vline*Iphase/1.732*Vline*Iline]*100=[Iphase/Iline]*100=[Iline/1.732*Iline]*100=100/1.732= 57% concurring with my statement in my post #25.

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#### Smart \$

##### Esteemed Member
See the attachment page from one generator manufacturer. They have stated for single phase loading of delta connected gen-set, the capacity is 57% of three phase capacity i.e [1.732*Vline*Iphase/1.732*Vline*Iline]*100=[Iphase/Iline]*100=[Iline/1.732*Iline]*100=100/1.732= 57% concurring with my statement in my post #25.
Your profile says you are an EE. The manufacturer's published value was most likely determined by an EE. It doesn't surprise me in the least that you arrived at the same conclusion.

As I stated earlier, the 1.732 (√3) factor in the 3? calculations is only valid (depending on the degree of accuracy desired) for balanced 3? calculations. You can prove this to yourself easily by calculating line current for three unequal, delta-connected, single phase loads. A single phase load powered by a 3? source is as extreme as you can get toward an unbalanced condition.

Just the notion that capacity for single phase loading is reduced, yet greater than one-third, logically precludes the power being handled only by the direct-connected winding. The current only has two paths: the direct-connected winding and the two series-connected windings. If we approach each path's current from the perspective each winding being an ideal source (classroom basic theory), 50% of the current will travel through each path. This means each winding will reach its capacity when the load current is twice the capacity of a single winding, which equates to two-thirds the 3? capacity.

However, in the real world, these windings are less than ideal. Essentially, the preceding establishes real world capacity is somewhere between one third and two thirds the full 3? rated capacity. In order to be between those values, some real-world parameter(s) of a power supply must be different than the ideal to cause an other than 50-50 result. Since the voltage between load connection points is a constant for the two current paths, that leaves only two basic parameters: current and impedance...

Are you with me so far?

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#### Sahib

##### Senior Member
Smart \$:
The point I am making is too simple to miss.

Here it is again.

The single phase load capacity of a delta connected generator is Vline*Iine

The rated threephase capacity of the same delta connected generator is 1.732*Vline*Iline.

So the single phase capacity is [
Vline*Iine/1.732*Vline*Iline]*100=57% of three phase capacity.

So it may be seen that it does not matter whether phase currents are balanced or not as far as the above derivation of single phase capacity of 57% of three phase capacity of delta connected generator is concerned.

#### Smart \$

##### Esteemed Member
Smart \$:
The point I am making is too simple to miss.

Here it is again.

The single phase load capacity of a delta connected generator is Vline*Iine

The rated threephase capacity of the same delta connected generator is 1.732*Vline*Iline.

So the single phase capacity is [
Vline*Iine/1.732*Vline*Iline]*100=57% of three phase capacity.

So it may be seen that it does not matter whether phase currents are balanced or not as far as the above derivation of single phase capacity of 57% of three phase capacity of delta connected generator is concerned.
Oh, I get your point. My point is?quite simply?it is wrong.

So entertain me. Answer this (and don't jump off topic )...

What is the wattage capacity of the following circuit and why is it not 3W?