open delta transformer

[7anoter4]

On an open delta three phase transformer in order to obtain a high leg voltage [208 V for instance] we take a full phase-let’s say a_b and a half of the other phase winding [a_n]
In my opinion, a half secondary winding transformer presents only 1/4 impedance [instead of 1/2 as it seems to be].
The short-circuit impedance of a transformer viewed from secondary side it is as follows:
Zab=Z'pab+Zsab
where Z'pab = secondary referred primary impedance Zpab*ksp
ksp=(no.turns.sec[ws]/no.turns.primary[wp])^2
Zsab=Rsab+Xsab.i complex secondary impedance.
Xsab=kxs*ws^2
where kxs it is a constant which depends on frequency and magnetic circuit geometry.
If we split the secondary winding wsnew=ws/2 then Xan=Xsab/4 and kspan=(ws/2/wp)^2=kspab/4
Only Rsan=Rsab/2 but it will be negligible so Zan≈Zab/4=ZT/4.

 

[7anoter4]

On the other hand the secondary half bobbin height it is also a half [Lc/2] so
Xs2=2*π*f*μo*(ws/2)^2*Lmt*(d/3+s)/(Lc/2)
where Lmt is the average turn length,d the bobbin thickness and s the air space width.
Xs2 of secondary half will be only a half of the entire secondary winding.
The new total impedance will be then:
ZT2=Zp*ws^2/wp^2/4+Zs/2
If Zp*ws^2/wp^2≈Zs then
ZT1=2xZs and ZT2=3/4xZs
ZT2/ZT1=0.75/2=0.375 [less then 0.5 but more then 0.25].

 

[waross]

In North America, open delta installations are often to support a small three phase load along with a large single phase load.
You may see a 5 KVA transformer paired with a 25 KVA transformer.
A 5 KVA paired with a 50 KVA is not unusual.
In such a case the fault current of a wild leg ground fault will be predominantly limited by the impedance of the smaller transformer.

Bill
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Jimmy Carter