# 3 phase and 2 phase in the same system

#### mbrooke

##### Senior Member
Would this be a viable system? Is this called a Taylor system? Does the scott-T autotransformer do any phase displacement or shifting? Could a neutral be added for the 3 phase?

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#### winnie

##### Senior Member
I don't know about the practicality, in the sense that I've no experience with how well that system would serve loads that need it. But it looks to me that:

a) X2 of the middle transformer could be grounded/used as a neutral for an ordinary 'high leg' delta system
b) You could add a 'zig-zag' or 'grounding' transformer to derive a neutral for the 3 phase system, and end up with a 240/139V three phase system
c) The 'two phase' system is quite asymmetric; if you were to ground X2 of the middle transformer than the A and B lines (phase 2) would both be 120V to ground, but the C line would be 208V to ground and the 'D' line (the unlabeled other side of phase 1) would be 32V to ground

-Jon

#### mbrooke

##### Senior Member
I don't know about the practicality, in the sense that I've no experience with how well that system would serve loads that need it. But it looks to me that:

a) X2 of the middle transformer could be grounded/used as a neutral for an ordinary 'high leg' delta system
b) You could add a 'zig-zag' or 'grounding' transformer to derive a neutral for the 3 phase system, and end up with a 240/139V three phase system
c) The 'two phase' system is quite asymmetric; if you were to ground X2 of the middle transformer than the A and B lines (phase 2) would both be 120V to ground, but the C line would be 208V to ground and the 'D' line (the unlabeled other side of phase 1) would be 32V to ground

-Jon
Look so- I'll take your word for it.

#### pv_n00b

##### Senior Member
While the Scott-T transformer setup is pretty straight forward I'm not sure about an autotransformer version.

#### winnie

##### Senior Member
While the Scott-T transformer setup is pretty straight forward I'm not sure about an autotransformer version.
The existing 'high leg' delta already gives the 90 degree phase difference (A to B is 120 degrees from B to C which is 120 degrees from C to A, a normal 3 phase delta. But the _midpoint_ to C is 90 degrees out of phase with A to B). Since you already have the phase difference, you only need to adjust voltage.

-Jon

#### gar

##### Senior Member
191212-1522 EST

winnie has provided a good discussion.

Had the original question been posed as a phasor diagram it would be immediately obvious what was going on. Note: only two transformers are required for the source delta (an open delta).

.

#### 310 BLAZE IT

##### Senior Member
I am working on replacing a two phase system in Philadelphia (one of the only places it still exists) and wanted to point out that there in fact were two different types: 4-wire and 5-wire,with the 5 wire using a common neutral.

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#### mbrooke

##### Senior Member
How does a 5 wire system work? Why 5 wires not 4 or 3?

#### Hv&Lv

##### Senior Member
How does a 5 wire system work? Why 5 wires not 4 or 3?
Look at your two phase system on the picture you posted. Then add a neutral.

#### mbrooke

##### Senior Member
Look at your two phase system on the picture you posted. Then add a neutral.
What are the voltages from each phase to neutral?

#### winnie

##### Senior Member
How does a 5 wire system work? Why 5 wires not 4 or 3?
5 wire has 2 phases 90 degrees apart, but with a common center tap. Relative to this center neutral, you have 4 'hot' wires at 0, 90, 180, and 270 degrees. (Lets not revisit the 180 degree phase difference being a different phase or the same phase )

My guess is that 5 wires lets you feed both 240V 2 phase loads such as motors, and also provides 120V L-N for receptacles and lights (much as modern 208/120V wye three phase is used).

3 wire was also used, with your two phases 90 degrees apart and a common wire.

-Jon

#### Hv&Lv

##### Senior Member
Wait...
split phase? Two phase??

:rotflmao::rotflmao:

#### mbrooke

##### Senior Member
5 wire has 2 phases 90 degrees apart, but with a common center tap. Relative to this center neutral, you have 4 'hot' wires at 0, 90, 180, and 270 degrees. (Lets not revisit the 180 degree phase difference being a different phase or the same phase )

My guess is that 5 wires lets you feed both 240V 2 phase loads such as motors, and also provides 120V L-N for receptacles and lights (much as modern 208/120V wye three phase is used).

3 wire was also used, with your two phases 90 degrees apart and a common wire.

-Jon
Are 120 and 240 the only voltages? I'd imagine 90* gives something strange like the square root of two.

#### winnie

##### Senior Member
Are 120 and 240 the only voltages? I'd imagine 90* gives something strange like the square root of two.
Yes, you also have the L-L voltage which would be root 2 * the L-N voltage. I suspect that this voltage was not used, in much the same way that the 208V 'wild leg' L-N voltage in a delta system wouldn't (shouldn't? isn't commonly?) used.

-Jon

1

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#### mbrooke

##### Senior Member
Yes, you also have the L-L voltage which would be root 2 * the L-N voltage. I suspect that this voltage was not used, in much the same way that the 208V 'wild leg' L-N voltage in a delta system wouldn't (shouldn't? isn't commonly?) used.

-Jon
Don't think I've seen anything take 170 volts.

#### mbrooke

##### Senior Member
The UK had a need for two phase power? :huh:

#### Tony S

##### Senior Member
The UK had a need for two phase power? :huh:
Not normally as far as I know but they are used on 25kV railway supplies. (The LH drawing.)

I did the drawing as part of an article on transformer configuration.

#### synchro

##### Senior Member
Shouldn't the equation for the neutral current "In" at the bottom right corner of the drawing be √(Ia[SUP]2[/SUP] + Ib[SUP]2) [/SUP]instead of √(Ia[SUP]2[/SUP] - Ib[SUP]2[/SUP])? So if Ia and Ib are equal then the neutral current would be √2≈1.414 higher than each of them.

#### Tony S

##### Senior Member
Shouldn't the equation for the neutral current "In" at the bottom right corner of the drawing be √(Ia[SUP]2[/SUP] + Ib[SUP]2) [/SUP]instead of √(Ia[SUP]2[/SUP] - Ib[SUP]2[/SUP])? So if Ia and Ib are equal then the neutral current would be √2≈1.414 higher than each of them.
No it’s minus but at times you have to transpose Ia with Ib.