I'm given that after a short circuit test a one phase transformer rated at \$2500\$ kVA, \$50000/10000\$ at \$50\$ Hz yields the values of voltage, current and power as follows:

$$720V, 225A,37947W$$.
The transformer has a open circuit current of \$3\$ \$\%\$ of the nominal and the efficiency with nominal charge and power factor \$1\$ is \$97.66\$ \$\%\$

It asked for many things, but where I'm stuck is how can calculate the power of open circuit test?

I'm assuming from the data that the voltage of open circuit is the nominal one (\$50kV\$), the current its the \$3\%\$ of \$I_{1N}\$

$$3\%I_{1N}=0.03\frac{2500kVA}{50kV}=1.5A $$

and the \$P_{0}\$ could be calculated from the \$\eta\$ data, since the power factor is 1 and the charge is \$C=1\$


doing the math this would be \$P_{0}=21.954kW\$ but the reference says this value should be \$13.054kW\$. Then how should be calculated?

  • 1
    \$\begingroup\$ With unloaded secondary, the efficiency figure no longer holds true since you’re not passing any current in the secondary. Do you know your iron hysteresis losses and primary resistance? \$\endgroup\$ – winny Apr 19 '18 at 18:03
  • \$\begingroup\$ No, I dont know the resistance and the losses besides the 37kW values \$\endgroup\$ – riccs_0x Apr 19 '18 at 18:13
  • 1
    \$\begingroup\$ Then you have an unknown variable. I would cheat and assume the resistive losses from the total efficiency figure is split 50-50 on primary and secondary and compute from there. \$\endgroup\$ – winny Apr 19 '18 at 18:26

You're on the right track, but we have to do a little more calculation. The short circuit test data allows us to calculate the resistance. In the schematic below, we apply some voltage \$V_1\$ and measure the current \$I_1\$ on the primary side. What we can then calculate is \$Z_1\$, the equivalent series impedance which is \$R_1 + j X_1\$.


simulate this circuit – Schematic created using CircuitLab

In particular, we have $$R_1 = \frac{P_1}{|I_1|^2}$$ In this case, we have \$R_1=37947/225^2 = 0.75\Omega\$ where \$R_1\$ is the equivalent series resistance on the primary side. If we want to be pedantic, although it's not needed for this problem, we could calculate \$X_1\$ because we know that $$|Z_1|^2 = \left(\frac{|V_1|}{|I_1|}\right)^2 = R_1^2 + X_1^2$$ To translate that to a secondary side equivalent series resistance, we multiply by the square of the turns ratio: \$R_2=R_1\left(\frac{50000V}{10000V}\right)^2=18.739\Omega\$.


simulate this circuit

There are a number of equivalent ways to express efficiency, but in this case the most useful form is $$\eta=\frac{P_{out}}{P_{out}+W_C+W_I}$$ where \$W_C\$ is the copper loss and \$W_I\$ are hysteresis and eddy losses, which are usually combined and called "iron losses" or "core losses". The other key thing to remember is that iron losses are not dependent on load current while copper losses are: \$W_C = |I|^2R\$ Substituting that into the equation gives $$\eta=\frac{P_{out}}{P_{out}+|I_2|^2R_2+W_I}$$ A little algebra lets us solve for \$W_I\$. $$W_I=\frac{P_{out}}{\eta}-P_{out}-|I_2|^2R_2$$ Since we're given efficiency at unity power factor at nominal we can simply plug in the numbers and calculate the result: $$W_I = \frac{2500 kW}{0.9766} - 2500kW - (50A)^2(18.739\Omega)$$ Here, the 50A is the secondary current at nominal (2500VA/50kV). We could equivalently used the primary current at nominal (250A = 2500VA/10kV) and \$R_1\$. Mathematically, they're identical. I calculate \$W_I = 13.054kW\$ which is, directly, the open circuit power.

One can also apply this the other way around. That is, if we are given the open circuit power, we have \$W_I\$ and can, given the efficiency \$\eta\$, calculate \$W_C\$. Since we know that this is at the rated current, which is also how the short circuit test is conducted (if you can forgive the pun) we can extract \$R2\$.

  • \$\begingroup\$ Great idea, only one thing, why the 50A? \$\endgroup\$ – riccs_0x Apr 19 '18 at 21:53
  • 1
    \$\begingroup\$ I added a bit of explanation for that. Let me know if you have any other questions. \$\endgroup\$ – Edward Apr 19 '18 at 21:57
  • \$\begingroup\$ OK, that makes all clear, I never think of calculating the resistance and reflect it to the other side, that’s a nice work. Thanks \$\endgroup\$ – riccs_0x Apr 19 '18 at 22:01
  • 1
    \$\begingroup\$ Reflecting the resistance isn't strictly necessary here. Just habit -- I'm used to seeing the copper loss equivalent series impedance on the secondary side and the iron loss equivalent shunt impedance on the primary side. \$\endgroup\$ – Edward Apr 19 '18 at 22:03
  • \$\begingroup\$ I always get confused with the part of reference to one or other side. Oh well....T[ \$\endgroup\$ – riccs_0x Apr 19 '18 at 22:13
How can be calculated the open circuit power on this transformer?

The open circuit output power of a transformer is always 0. This is because the current is zero (by definition of open circuit).

  • \$\begingroup\$ Yup, You are right I have modified the title and details. \$\endgroup\$ – riccs_0x Apr 20 '18 at 18:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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