Current Flow With L-L Autotransformer Fault to N

[wwhitney]

Say we have a 208Y/120V system and supply a 208V to 240V autotransformer from lines A-B, with B' the derived line at 240V from A. Then I worked out that B' is 149V from N. (The 32V boost B-B' is 30 degrees out of phase from N-B.)

But what are the current flows if B' faults to N? Is treating a fault like a resistive load reasonable? If so, I expect that I should be able to work out the unknown current phasors in lines A and B just by applying KCL repeatedly (I need to try that). Is there an easier way?

Cheers, Wayne

 

[wwhitney]

If so, I expect that I should be able to work out the unknown current phasors in lines A and B just by applying KCL repeatedly (I need to try that).

OK, that didn't work for me, so looking for some guidance here.

Below is a diagram I made for discussion. The red lines are transformer coils, the wye source on the left, and the autotransformer on the right. The coils are drawn (approximately) with length and direction representing the voltage phasors across the coils. The black lines are just conductors, which I'm happy to assume are impedance-less unless that causes a computational problem. Then I picked 1 ohm for ease of computation for a fault from B' to N.

If I call IA, IB, IB', and IN the currents in the accordingly labeled conductors, with positive sense away from the transformer coils, then IN + IB'=0 by KCL at the 1 ohm fault. IA + IB + IN = 0 by KCL at the transformer wye point. And IA + IB - IB' = 0 by KCL at the autotransformer non-end tap.

But these 3 equations are linearly dependent, the last is just what you get by substituting the first in the second. So I can't solve for IA and IB.

Thanks,
Wayne