2
\$\begingroup\$

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

\$\endgroup\$
5
  • \$\begingroup\$ home.engineering.iastate.edu/~jdm/ee457/… \$\endgroup\$
    – Ben
    Mar 31, 2021 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, 2021 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\$ Mar 31, 2021 at 18:57
  • \$\begingroup\$ Are you required to use symmetrical components? If you draw out the winding relationships the problem becomes fairly clear. \$\endgroup\$ Mar 31, 2021 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\$ Mar 31, 2021 at 21:30

1 Answer 1

1
\$\begingroup\$

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

\$\endgroup\$
6
  • \$\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\$ Apr 1, 2021 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\$ Apr 1, 2021 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\$ Apr 1, 2021 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\$ Apr 1, 2021 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\$ Apr 1, 2021 at 3:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.