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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!

enter image description here

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  • \$\begingroup\$ home.engineering.iastate.edu/~jdm/ee457/… \$\endgroup\$ – Ben Mar 31 at 18:41
  • \$\begingroup\$ Basically, you need to build a 3x3 matrix with Cross-impedances. The positive and negative sequence impedance will be the same. \$\endgroup\$ – Ben Mar 31 at 18:50
  • \$\begingroup\$ 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!! \$\endgroup\$ – Charles Wagner Mar 31 at 18:57
  • \$\begingroup\$ Are you required to use symmetrical components? If you draw out the winding relationships the problem becomes fairly clear. \$\endgroup\$ – relayman357 Mar 31 at 21:15
  • \$\begingroup\$ Yes, I know how to solve it with the equivalent circuits, but I am required to solve it using symmetrical components =/ \$\endgroup\$ – Charles Wagner Mar 31 at 21:30
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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).

enter image description here

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.

enter image description here

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.

enter image description here

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  • \$\begingroup\$ Thanks for your help! Unfortunately, it doesn't match the official answer: IA = 3.994 - j6.121 A / IA =(7.309, -56.88°)A \$\endgroup\$ – Charles Wagner Apr 1 at 0:10
  • \$\begingroup\$ 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. \$\endgroup\$ – relayman357 Apr 1 at 0:18
  • \$\begingroup\$ 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. \$\endgroup\$ – Charles Wagner Apr 1 at 0:27
  • \$\begingroup\$ 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. \$\endgroup\$ – relayman357 Apr 1 at 1:46
  • \$\begingroup\$ 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 \$\endgroup\$ – Charles Wagner Apr 1 at 3:22

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