# Phase shift in Star-Delta Transformers

Thread Starter

#### Siddhartha Mahavira

Joined Oct 22, 2015
2
Hi Everyone !!

We all know that the per-phase analysis of a star-delta transformer requires converting the delta-side into an equivalent-star connection. Upon doing this, we find that the new equivalent secondary delta-side phase lags by 30 degrees with respect to the primary star-side phase. We usually represent this by multiplying the term e^(j30deg) to the equivalent per-phase transformation-ratio a:1 from primary to secondary; ie, the new transformation ratio is a.e^(j30deg) : 1 .

Now, if we were to transfer an impedance from the secondary-side to the primary-side, applying the transformation rule of Z(referred_to_primary) = (primary-to-secondary transformation-ratio)^2 x Z(secondary), we would get :
Z(referred_to_primary) = (a.e^(j30deg))^2 x Z(secondary)
= a^2 x e^(j60deg) x Z(secondary).

This implies that even a purely resistive balanced-load on the delta side would reflect as an inductive load on the primary side, which is not possible. Calculations based on this would then go on to give erroneous results (as I have myself tried out and confirmed).

So where lies the contradiction ?! Please help !! I can furnish further details as well as a specific problem.

Thanks !!

#### crutschow

Joined Mar 14, 2008
27,210
The phase of the voltage between phases changes in the transformation but not the phase between voltage and current (the power factor remains constant as would be expected).

• Siddhartha Mahavira
Thread Starter

#### Siddhartha Mahavira

Joined Oct 22, 2015
2
Yes, the power factors on the primary- as well as the secondary- sides remain the same actually. Direct calculations as well as complex (active and reactive) volt-ampere balance both show that. But the transformed per-phase circuit involving star:delta doesn't show this. Here's why..
for the equivalent per-phase star:star :
V(primary) = a.e^(j30deg) x V(secondary) = a.V(secondary).e^(j30deg)
where a.e^(j30deg) is the equivalent primary-to-secondary transformation ratio. Then,
I(primary) = (1/(a.e^(j30deg)) x I(secondary) = (1/a).I(secondary).e^(-j30deg)
Thus, Z(primary) = V(primary)/I(primary) = a^2 x V(secondary)/I(secondary) x e^(j60deg)
= a^2 x Z(secondary) x e^(j60deg)
Which shows that the reflected primary impedance always has an additional 60 degree angle with respect to the actual secondary impedance.

#### crutschow

Joined Mar 14, 2008
27,210
Does not compute. Similar threads