# 6 phase forked star transformer

I have a problem i'm trying to get to the bottom of. I am basically trying to find the secondary line voltages of a three phase delta primary, with a six phase forked star connected secondary transformer. The link shows the configuration of the forked star connection, and the phasor diagram to aid it http://www.vias.org/matsch_capmag/img/matsch_caps_magnetics-936.png

My issue is that using the normal equation of root 3 x Vphase/n gives doesnt give me the line voltages for all 6 phases, and looking for line to line voltages I really dont know where to start and there isnt much documentation available on these systems. Any help would be appreciated for me to understand this.

If i had say 11kV primary line voltage (delta) with a turns ratio of 100:1, how would i work out the secondary line voltages of a 6 phase forked star configuration as shown in the link?

Thanks for any help in advance. • Graphically add the vectors, that diagram has already done it for you! – Neil_UK Mar 15 '17 at 7:11
• you'll have to excuse my stupidity, but I don't understand what you mean? – Samuel Winchester-Leigh Mar 15 '17 at 7:51
• It's a case of 'you can't believe it's that simple'. Look up vectors (=== phasors), how they add, Argand diagram, cartesian vectors, wikipedia, no time unfortunately at the moment to pick links, but there's plenty there. – Neil_UK Mar 15 '17 at 8:14
• sorry, euclidean vector rather than cartesian vector. See addition under Argand diagram. – Neil_UK Mar 15 '17 at 9:08
• it is literally the way you sketch it is the way you add it. – Neil_UK Mar 15 '17 at 10:06

I understand your problem and the confusion caused by the reference diagram provided. I always try to align my diagrams with the phasors when I draw them at $\pm 120^\circ$, otherwise I draw them all horizontally to show the ampere-turn balance.

Let us assume the top-left delta is correct. I will then draw the phasors $$\DeclareMathOperator{\kV}{~kV}\DeclareMathOperator{\V}{~V}U_{CB} = 11\kV\angle 0^\circ,$$

$$U_{BA} = 11\kV\angle 120^\circ$$ and $$U_{AC} =11\kV\angle -120^\circ =11\kV\angle 240^\circ.$$

This implies looking at the top-right diagram showing the interconnections and implied core locations by the angle drawn. Therefore from primary to secondary: $${1\over100} U_{CB} = U_{nh} = U_{eg} = U_{di} = 110\V\angle 0^\circ,$$ $${1\over100} U_{BA} = U_{ng} = U_{ci} = U_{bh} = 110\V\angle 120^\circ$$ and $$U_{ni} = U_{fg}= U_{ah} = 110\V\angle 240^\circ$$

The secondary line voltage is then for example $$|U_{bh}-U_{ah}| = |110\V\angle 120^\circ - 110\V\angle 240^\circ| = {\sqrt3\cdot110\V}.$$

(Prepare to get totally lost if you try to use the phasor diagrams provided. :-)

• Please rell me if you would rather use U or E as symbol? – skvery Mar 15 '17 at 13:45
• Hi sorry for the late reply. i would always tend to use U :) – Samuel Winchester-Leigh Mar 16 '17 at 7:04
• As for your reply, thats really appreciated to see it laid out like that, it makes life easier to see it away from the phasor diagrams! – Samuel Winchester-Leigh Mar 16 '17 at 7:05
• although that makes more sense, its only gotten me as far as i got myself :/ I found that each branch of the secondary (nh, ni, ng) = 110V, and that the resultant from that would give me root 3 x 110 giving me 190.52V. I put that to my lecturer and he informed me that looking at the bottom right diagram that would give me the phase voltage of Van in this instance (This is where i am getting confused!) He is basically looking for 6 transmission line voltages, and i have 190.52V! This is why im getting so frustrated lol – Samuel Winchester-Leigh Mar 16 '17 at 7:15
• Quite correct $|U_{an}|= U_{l-l}= \sqrt 3\cdot 110\V$. You can see the symmetry from the LV diagram. – skvery Mar 16 '17 at 10:09