# Symmetrical Components - Three-phase D-Y transformer with unbalanced voltage and loads

I have a Delta-Wye transformer, N1/N2 = 4, and am trying to solve the circuit below, with 120VAC applied to terminals VAB only. On the secondary I have 1 ohms connected between a-n and 5 ohms between b-n. The other parameters are:

rs = resistance of primary winding A, B, and C

rr = resistance of secondary winding a, b, and c

Llas = Primary leakage inductance on winding A

Llbs = Primary leakage inductance on winding B

Llcs = Primary leakage inductance on winding C

Llar = Secondary leakage inductance on winding a reflected to the primary

Llbr = Secondary leakage inductance on winding b reflected to the primary

Llcr = Secondary leakage inductance on winding c reflected to the primary

Lm = Mutual inductance

Does anyone know how to find the primary line current (IA) using the symmetrical components method? All examples that I could find use transformers balanced loaded. I really appreciate any help!

• home.engineering.iastate.edu/~jdm/ee457/… – Ben Mar 31 at 18:41
• Basically, you need to build a 3x3 matrix with Cross-impedances. The positive and negative sequence impedance will be the same. – Ben Mar 31 at 18:50
• Sorry, but I don't understand how can I build the cross-impedance matrix. How can I calculate Zaa, Zab, Zac..could you please give an example? Thank you!! – Charles Wagner Mar 31 at 18:57
• Are you required to use symmetrical components? If you draw out the winding relationships the problem becomes fairly clear. – relayman357 Mar 31 at 21:15
• Yes, I know how to solve it with the equivalent circuits, but I am required to solve it using symmetrical components =/ – Charles Wagner Mar 31 at 21:30

Ok, here is what I am thinking as an approach. Below I show the transformer connections for a Dyn bank that is standard connected per IEEE (low-side lags by 30° and assumes A-B-C rotation & subtractive polarity transformers).

With your low-side c-phase open circuited, no current can flow in the primary C-phase winding (neglecting magnetizing current). That constraint also prevents current flow in the primary A-phase winding as well in this particular case. So, the only current that can flow on the primary side is in the B-phase winding (the one in middle). As such, the only secondary winding that can have current flow is the b-phase winding.

So, I think you can simply reflect that b-phase load ($$\5\Omega\$$) to the primary and then forget about the transformer. Your problem now being reduced to a primary A-B fault with resistance (adding in your winding impedance data as desired).

Below is an example of sequence network connections for a B-C fault with resistance. You would shift angles for the A-B fault.

Note: Both images are from my lecture notes on symmetrical components.

Additional comments: If the secondary current of one of the 3 two-winding transformations is zero, $$\I_S=0\$$, then the primary current for that particular two-winding transformation is zero as well, $$\I_P=0\$$ (neglecting magnetizing current). Another way to look at it, if the secondary is open-circuited then it's load impedance is $$\\infty\$$...which when reflected to the primary winding by turn ratio squared is still $$\\infty\$$, an open circuit.

Also, the example phase-phase fault calculation in symmetrical components is done in per-unit (e.g. $$\V_{BASE}=13.8\text{kV}, S_{BASE}=100\text{MVA}, \text{and} Z_{BASE} = \frac{13.8^2}{100}=1.904\Omega\$$ so the $$\5\Omega\$$ resistance converted to per-unit is $$\2.625\ \text{pu } \Omega\$$. You can work it easier in your case all in actual units (Volts, Amps, Ohms) without the bother of converting to per-unit.

• Thanks for your help! Unfortunately, it doesn't match the official answer: IA = 3.994 - j6.121 A / IA =(7.309, -56.88°)A – Charles Wagner Apr 1 at 0:10
• Hi Charles, that was just a sample calculation shown in my answer. I did not use your actual values. You will need to work that through yourself. – relayman357 Apr 1 at 0:18
• Oh, ok. However, I think there will be current on windings A and C as well. When I apply 120V to winding B, there will be 120V applied to windings A+C (A+C is in parallel with B), and therefore there will be current going through these windings as well, which will induce a small voltage on terminals b-n and c-n too. – Charles Wagner Apr 1 at 0:27
• I think you need to look a little closer and go through the logic I laid out carefully. See the Additional comments I added at bottom of my answer. – relayman357 Apr 1 at 1:46
• I tried using the values of my problem in your resolution but found the following: V base = 13800; Z base = 1.9044; I base = 4183.7; V_pu = 0.0086956; Z_pu = 2.62550+0.39592j; I_pu = 0.003238-0.000488j; VAB = 120.0; IA = 13.548-2.043j – Charles Wagner Apr 1 at 3:22