mbrooke
Batteries Included
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- United States
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- Technician
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
While the Scott-T transformer setup is pretty straight forward I'm not sure about an autotransformer version.
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.
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
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
The UK had a need for two phase power? :huh:
Shouldn't the equation for the neutral current "In" at the bottom right corner of the drawing be √(Ia2 + Ib2) instead of √(Ia2 - Ib2)? So if Ia and Ib are equal then the neutral current would be √2≈1.414 higher than each of them.