# unbalanced 3 phase generator load

#### Sahib

##### Senior Member
So you're saying a 3? generator cannot power a single phase load...???

More specifically a 120/240V 3? 4-W 130kW-rated generator cannot power a 40kVA (giving a benefit of a doubt) 240V 1? load.
Unless it is designed for that: sometimes a three phase generator is configured to work as a single phase generator by a manufacturer.

#### iwire

##### Moderator
Staff member
Unless it is designed for that: sometimes a three phase generator is configured to work as a single phase generator by a manufacturer.
All three phase generators in the US can supply single phase loads of various percentages of the rating.

#### Phil Corso

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

PLR stated that the generator was 3-ph, delta. You stated in your 29-Apr (12:28) reply, "One thing many forget is that in a delta configuration, 1/3 of a single-phase load's current is handled by the two indirectly-connected windings."

I've been in this business quite a while, and I'm eager to learn something new! So what I would like you to explain is, how? Specifically, if the single-phase load is connected to, say, phase A-B, then how do the indirectly-connected phases, say, B-C and C-A, carry 1/3 each?

Phil

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#### Sahib

##### Senior Member
All three phase generators in the US can supply single phase loads of various percentages of the rating.

Not only in the US, but all three phase generators in other parts of the world can supply single phase loads of various percentages of the rating with negative sequence current limited: 10 % is the maximum negative sequence current. So what is the percentage of negative sequence current in the OP's case?

#### Sahib

##### Senior Member
Phil:
In delta connection line current=1.732 times phase current. Suppose the load is connected between any two terminals in a delta connected generator. Then VA supplied to load is line currentxvoltage=1.732xphase currentxvoltage. So 1.732 times individual phase supply capacity of the delta connected generator. But this comes with a prize.

#### Smart \$

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

PLR stated that the generator was 3-ph, delta. You stated in your 29-Apr (12:28) reply, "One thing many forget is that in a delta configuration, 1/3 of a single-phase load's current is handled by the two indirectly-connected windings."

I've been in this business quite a while, and I'm eager to learn something new! So what I would like you to explain is, how? Specifically, if the single-phase load is connected to, say, phase A-B, then how do the indirectly-connected phases, say, B-C and C-A, carry 1/3 each?

Phil
I drew up the following for a discussion regarding a transformer with 480V delta-configured secondary, but the principle is the same...

The easiest way to understand is to look at the windings as two separate power sources connected in parallel, each having a different source impedance (z for one winding, 2z for two).

Sahib...

Regards, Phil

#### Smart \$

##### Esteemed Member
Phil:
In delta connection line current=1.732 times phase current. Suppose the load is connected between any two terminals in a delta connected generator. Then VA supplied to load is line currentxvoltage=1.732xphase currentxvoltage. So 1.732 times individual phase supply capacity of the delta connected generator. But this comes with a prize.
That's not correct. See my reply to Phil.

Apparently I don't get the prize....

#### Phil Corso

##### Senior Member
Smart \$

You too, are dead wrong. The line-current is the vector-sum of both legs connected to that line. You can't add them arithmetically!

Phil

#### GoldDigger

##### Moderator
Staff member
Smart \$

You too, are dead wrong. The line-current is the vector-sum of both legs connected to that line. You can't add them arithmetically!

Phil
Phil, you are the one who is right in general but wrong in this particular case.
The voltage between the two connection points (corners of the delta) from the two sources are identical vectorially, since they are normally connected to each other. If you open the corners you still have the same voltage difference at both the two-coil and one-coil side.
As a result, any currents flowing into a (for simplicity resistive) load from the two sources will be in phase also.
In the special case that two vector currents are in exactly the same direction, you are completely justified in adding the amplitudes.

#### Smart \$

##### Esteemed Member
Smart \$

You too, are dead wrong. The line-current is the vector-sum of both legs connected to that line. You can't add them arithmetically!

Phil
The winding current has the same angle (phase) on all three windings, only the magnitude differs.

For example, using the previously posted depictions, what is the current's angle on the two-winding power source? The one-winding power source? Are not the vector sum and arithmetic the same?

#### Smart \$

##### Esteemed Member
Phil, you are the one who is right in general but wrong in this particular case.
...
He's not wrong, he just hasn't thought it through completely... yet.

#### Phil Corso

##### Senior Member
Smart \$...
I want be very clear about your aassertion... that is, the winding to which the load is connected, say A-B, carries two-thirds of the load, while the indirectly-connected windings, B-C and C-A (in series) carry a third of the load!

Sahib...
I thought the SQRT(3) factor appled only to balanced loads? Do you mean to say that Sqrt(3) x 20 = 30?

Phil

#### Phil Corso

##### Senior Member
Goldigger...

I pose a 3rd question to you. Are you saying if the delta were opened so that the 30A load remained connected to A-C, the voltage across winding A-C is only 2/3, of what it normally is, so the load-current now drops to20A?

Phil

#### Phil Corso

##### Senior Member
Smart \$, Sahib; Goldigger...

Phil

#### GoldDigger

##### Moderator
Staff member
Goldigger...

I pose a 3rd question to you. Are you saying if the delta were opened so that the 30A load remained connected to A-C, the voltage across winding A-C is only 2/3, of what it normally is, so the load-current now drops to20A?

Phil
The load current does not change at all, nor does the voltage. The source from which that current is supplied does change though.

And the reason that two thirds of the current comes from the direct winding and one third from the series pair is, as was stated, that the source impedance of the double winding is twice that of the single winding. (To a very good first approximation anyway!)

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#### Smart \$

##### Esteemed Member
Smart \$...
I want be very clear about your aassertion... that is, the winding to which the load is connected, say A-B, carries two-thirds of the load, while the indirectly-connected windings, B-C and C-A (in series) carry a third of the load!

...
That is correct....!!!

PS: Since Golddigger is starting to add nuances, I suppose I should state that each single winding is of equal impedance.

To help, what is a purely resistive load current's phase angle when connected across the missing winding of an open delta compared to the phase angle of each individual winding's voltage?

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#### GoldDigger

##### Moderator
Staff member
PS: Since Golddigger is starting to add nuances, I suppose I should state that each single winding is of equal impedance.
I notice that you have also not specified whether it is a gas or diesel generator, nor whether it is rated for use as a prime mover. :angel:

#### Smart \$

##### Esteemed Member
I notice that you have also not specified whether it is a gas or diesel generator, nor whether it is rated for use as a prime mover. :angel:
That is correct...!!!

#### Phil Corso

##### 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