0
\$\begingroup\$

I'm doing some experiments with a transformer, measuring some physical quantities.

First I did an experiment with the transformer in open-circuit. I determined the voltages and currents on both primary and secondary winding.

I know the power factor measures the ratio between active power and apparent power (it was in fact this way that I computed it).

However, what does it exactly mean? Specially in open-circuit? Is it the percentage of power that doesn't get "lost"?

I know that if I had a load connected it would be the percentage of power delivered to the load. But this way I'm not sure.

Then I computed the parameters of the transformer G_m and B_m. I converted them to per-unit values. Does this have any physical significance? Because B_m is more close to 1 than G_m. Does just mean that the inductor parameter is more relevant than the resistor one?

Second I did an experiment with the transformer in short-circuit. I determined the voltages and currents on both primary and secondary winding.

I have the same question about the power factor, what does it mean here? Power delivered to a short-circuit?

Then I computed the parameters of the transformer R_t and X_t. Again, converted them to per-unit values. I have the exactly same questions as before.

Third, I experimented with a load, same thing as before.

Now the power factor must mean the percentage of power delivered to the load, right?

Thanks in advance!

EDIT: Transformer model (valid in pu values).

enter image description here

\$\endgroup\$
4
  • \$\begingroup\$ Transformer...... then you talk about transistor. \$\endgroup\$
    – Andy aka
    Mar 21, 2019 at 8:04
  • \$\begingroup\$ My mistake! I was tired lol. Thank you for the correction \$\endgroup\$ Mar 21, 2019 at 10:09
  • \$\begingroup\$ What is "G_m" and "B_m"? What is "R_t" and "X_t". You should show the equivalent circuit of a transformer and refer to that circuit to avoid confusion: stades.co.uk/index_files/transformer%20equivalent%20circuit.PNG \$\endgroup\$
    – Andy aka
    Mar 21, 2019 at 10:51
  • \$\begingroup\$ Added the model! \$\endgroup\$ Mar 21, 2019 at 11:01

1 Answer 1

0
\$\begingroup\$

enter image description here

This shows the Mutual Inductance which typically draws a VAR power 10% of rated VA so the input current is low P.F.

The short circuit current Isc is then the Rated VA/Zpu so a Zpu=10% of Rated V/I max at full load is 10% of Isc or

Isc = ( V * A )/Zpu

\$\endgroup\$

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.