An autotransformer is used to connect a 12.6-kV distribution line to a 13.8-kV d

Question

An autotransformer is used to connect a 12.6-kV distribution line to a 13.8-kV distribution line. It must be capable of handling 2000 kVA. There are three phases, connected Y - Y with their neutrals solidly grounded.

What must the Nc/Nse turns ratio be to accomplish this connection?

How much apparent power must the windings of each autotransformer handle?

What is the power advantage of this autotransformer system?

(d) If one of the autotransformers were reconnected as an ordinary transformer, what would its ratings be?

Explanation / Answer

given that the transformer is Y-Y Connected ,

hence the primary and secondary must be divided by 3.

turns ratio of transformer is given as below

i.e vH/vL=(Nc+Nse)/Nc =(13.8 KV/3)/(12.6KV/3)

(Nc+Nse)/Nc=13.8/12.6

12.6Nc+12.6Nse=13.8Nc

12.6Nc+12.6Nse-13.8Nc

12.6Nse =1.2Nc

Turns Ratio Nc/Nse=10.5

Power advantage of the transformer Sio/Sw= (Nc+Nse)/Nc= (Nc+10.5Nc)/Nc

= 11.5Nc

=11.5

Hence 1/11.5 power in every tansformer flows in the windings,because 1/3 of overall power is related with every phase ,windings in every transformer should manage

Sw=(200KVA)/(3.11.5)

=200/165

=58 KVA.

voltages on every phase of transformer i.e 13.8kv /3

=7969v

and 12.6 /3

=7275v

Voltage on common winding(Nc)=7275v

lly voltage on series winding(Nse) =7967-7275

=692 volts

hence transformer ratings are i.e 7275/692 volts , with power 58 KVA.

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