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I've recently done an experiment on a 415/100 V transformer. Performing the S/C test on the HV side and the Open Circuit test on the LV side and finding winding resistance from applied DC voltage and current.

Therefore I now have results from the experiment and have solved Magnetizing reactance, core loss Resistance, Copper/winding resistance and Leakage reactance equivalent values, depending on which side I transfer the equivalent values to according to my turns ratio (a).

The issue I'm having is that I cannot find a way to split the equivalent leakage reactance into its 2 components (LV winding and HV winding).

Equivalent winding resistances are split into their 2 values from the measured applied DC voltage and current Test.

I was beginning to think that if I had measured the voltage across the S/C side in the S/C test (although it would be small, it would still exist) I would have had enough unknowns to solve for Xl LV side; but because I did not measure that voltage assuming 0 volts, I cannot.

Any ideas? I'm stumped.

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You can split the overall leakage inductance into two parts proportional to the turns ratio squared. For instance, you know that if the transformer was 1:1 the leakage inductance would be shared equally between primary and secondary .

You also know that if the turns ratio was (say) 10:1 and you measured a 1 mH total leakage (referred to the primary) you could attribute all of this 1 mH to the secondary by dividing it by \$(10:1)^2\$. In other words 10 uH on the secondary is equivalent to 1 mH on the primary.

So, whatever your combined leakage is referred to the primary (let's assume it to be 1 mH), split it into two halves and keep 0.5 mH at the primary and transfer 5 uH to the secondary.

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  • \$\begingroup\$ Thanks! ive just found this formulae Xl1=X'l2=(X'eq)/2. and obviously X'eq=Xl1+X'l2. I believe this is exactly what you have just explained. thank you. So because when the leakage inductance is all on one side in the form of equivalent reactance, we can split it in to 2 equal parts (Xl1=X'l2=(X'eq)/2 and X'eq=Xl1+Xl2) does this suggest each winding leaks the same amount of flux? even though the secondary winding reactance/impedance with be reduced by 1/a^2 when transferred back to its respective side? \$\endgroup\$ – Alex Chala Sep 7 '18 at 1:08
  • \$\begingroup\$ Any decently constructed transformer will have the same percentage of leakage on each winding. Clearly you could make a mockery of this by assuming that the secondary is in series with an added external reactance and do the short circuit test and get a silly result but we’re practical folk here and no conjuring is involved. \$\endgroup\$ – Andy aka Sep 7 '18 at 6:20

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