Open-Circuit and Short-Circuit Tests

The performance of a transformer can be calculated from its equivalent circuit, and these tests are very economical and convenient, because they furnish the required information without actually loading the transformer.

The equivalent circuit of Chapter 44 contains four parameters, here referred to the primary:

• the equivalent resistance \(R_{01}\);

• the equivalent leakage reactance \(X_{01}\);

• the core-loss conductance \(G_0\), or equivalently the resistance \(R_0\);

• the magnetising susceptance \(B_0\), or equivalently the reactance \(X_0\).

• The shunt branch parameters are determined by this test.

• The core loss and the magnetising current depend on the applied voltage only and are practically unaltered by the load current — the point established in Chapter 41 and used ever since.

• Hence rated voltage is supplied, generally to the LV winding, keeping the HV winding open.

• Since \(I_0\) is very small compared with the full-load current, the loss in \(R_1\) is neglected.

• Thus the whole of \(W_0\) drawn from the source is dissipated as heat in the core.

Either test can in principle be done from either side, but there is a strongly preferred choice in each case, and it is a practical matter of instruments and safety.

Whichever side is used, the parameters obtained refer to that side, and must be transferred by \(K^{2}\) before being combined with quantities on the other. Failing to do so is the commonest error in these calculations.

• The input voltage is reduced to a small fraction of rated value and the secondary terminals are short-circuited.

• A current then circulates in the secondary winding.

• Since only a small fraction of rated voltage is applied, the flux in the core — and hence the core loss — is very small.

• Hence the power input on short circuit is dissipated as heat in the windings.

• Since the applied voltage is very small, of the order of 5 to 8 %, the magnetising branch can be eliminated from the equivalent circuit altogether.

The open-circuit wattmeter reads the total iron loss. If the two components are wanted separately, the test is repeated.

The requirement of equal flux density is essential and easily overlooked. If the voltage were held constant while the frequency changed, \(B_m\) would change too, and the two unknowns could not be separated — the measurement would confound a change of frequency with a change of flux. Chapter 42 worked a numerical example of this separation.

The two wattmeter readings are the two losses of Chapter 44, and the impedance from the short-circuit test gives the percentage drops of Chapter 45. Everything follows.

The tests just described explain the convention directly, because each measures a loss governed by a different quantity.

• Copper losses \(\left(I^{2}R\right)\) depend on the current passing through the transformer winding — which the short-circuit test measures at rated current.

• Iron losses — core and insulation losses — depend on the voltage — which the open-circuit test measures at rated voltage.

• Total losses therefore depend on voltage and current, expressed together in volt-amperes, and not on the load power factor.

• Hence the transformer rating is expressed in VA or kVA, not in W or kW.

• Four parameters describe the transformer: \(R_{01}\), \(X_{01}\), and the shunt branch \(R_0\) (or \(G_0\)) and \(X_0\) (or \(B_0\)).

• The open-circuit test applies rated voltage with the far winding open. Iron loss and magnetising current depend on voltage alone, so \(W_0\) is the iron loss and \(R_0 = V_1^{2}/W_0\), \(X_0 = V_1/I_\mu\).

• The short-circuit test applies 5 to 8 % of rated voltage with the far winding shorted. Flux and hence iron loss are negligible, so \(W_{sc}\) is the full-load copper loss and \(Z_{01} = V_{sc}/I_{sc}\), \(R_{01} = W_{sc}/I_{sc}^{2}\), \(X_{01} = \sqrt{Z_{01}^{2} - R_{01}^{2}}\).

• Supply the LV side for the OC test (rated voltage is easier) and the HV side for the SC test (rated current is smaller).

• Use a low-power-factor wattmeter for the OC test, and hold rated voltage and frequency.

• Two OC tests at different frequencies but the same flux density separate hysteresis from eddy loss, since \(W_0/f = A + Bf\).

• Efficiency at any load follows from \(W_0\) and \(W_{sc}\); regulation at any power factor from \(v_r\) and \(v_x\).

• The rating is in kVA because copper loss follows current and iron loss follows voltage, neither depending on power factor — and the manufacturer cannot know the load.

• Forgetting which side the parameters refer to. OC on LV gives LV-side values; transfer by \(K^{2}\) before combining with SC values from the HV side.

• Computing \(R_0\) as \(V_1/I_0\). It is \(V_1/I_w\), equivalently \(V_1^{2}/W_0\).

• Treating \(W_0\) as copper loss or \(W_{sc}\) as iron loss. Each test isolates the other one.

• Running the SC test at rated voltage. That is a short circuit in earnest — 20 times rated current.

• Running the OC test below rated voltage. \(X_0\) depends on the flux density, so the value would be wrong.

• Using an ordinary wattmeter for the OC test. At a power factor near 0.2, a low-power-factor instrument is required.

• Doing the two-frequency test at constant voltage. The volts per hertz must be held constant so that \(B_m\) does not change.

• Forgetting that \(W_{sc}\) is the full-load copper loss. Only if the test ran at rated current; otherwise scale by the square of the current ratio.

• Omitting the cross-check. \(\sqrt{v_r^{2}+v_x^{2}}\) must equal \(V_{sc}/V_1\).

• Quoting a transformer rating in kW. Its limits are current and voltage, neither of which involves power factor.

Always note which winding each test was performed on, and check that the SC current is the rated value for that side.

1. P46.1 An OC test on the 200 V LV side gives 200 V, 0.8 A, 70 W. Find \(\cos\phi_0\), \(I_w\), \(I_\mu\), \(R_0\) and \(X_0\).

   Show answer

   \[\cos\phi_0 = \frac{70}{160} = 0.4375, \qquad \sin\phi_0 = 0.8992\]

   \[I_w = 0.350~\mathrm{A}, \qquad I_\mu = 0.7194~\mathrm{A}\]

   \[R_0 = \frac{(200)^{2}}{70} = 571.4~\Omega, \qquad X_0 = \frac{200}{0.7194} = 278.0~\Omega\]

2. P46.2 An SC test gives 30 V, 15 A, 250 W on a 600 V winding. Find \(Z_{01}\), \(R_{01}\), \(X_{01}\) and the percentage impedance.

   Show answer

   \[Z_{01} = \frac{30}{15} = 2.000~\Omega, \qquad R_{01} = \frac{250}{225} = 1.111~\Omega\]

   \[X_{01} = \sqrt{4.000 - 1.235} = \sqrt{2.765} = 1.663~\Omega\]

   \[\%Z = \frac{30}{600}\times100 = 5.00\,\%\]

3. P46.3 A 10 kVA transformer has \(W_0 = 120\) W and \(W_{sc} = 200\) W. Find the efficiency at full load, 0.9 pf.

   Show answer

   \[\eta = \frac{9000}{9000 + 120 + 200} = \frac{9000}{9320} = 96.57\,\%\]

4. P46.4 For P46.3, find the load for maximum efficiency and that efficiency.

   Show answer

   \[x = \sqrt{\frac{120}{200}} = 0.7746 \quad\Longrightarrow\quad 7.746~\mathrm{kVA}\]

   \[\eta_{\max} = \frac{(7746)(0.9)}{6971 + 240} = \frac{6971}{7211} = 96.67\,\%\]

5. P46.5 For the transformer of P46.2 on a 600 V winding rated 15 A, find \(v_r\), \(v_x\) and the regulation at 0.85 lagging.

   Show answer

   \[v_r = \frac{(15)(1.111)}{600}\times100 = 2.778\,\%, \qquad v_x = \frac{(15)(1.663)}{600}\times100 = 4.158\,\%\]

   \[\sin\phi = 0.5268\]

   \[\text{Regulation} = (2.778)(0.85) + (4.158)(0.5268) = 2.361 + 2.190 = 4.551\,\%\]

   Check: \(\sqrt{v_r^{2}+v_x^{2}} = 5.00\,\%\), equal to \(V_{sc}/V_1\) \(\checkmark\)

6. P46.6 An SC test is run at 12 A instead of the rated 15 A, giving 160 W. Find the full-load copper loss.

   Show answer

   \[W_{sc,FL} = (160)\left(\frac{15}{12}\right)^{2} = (160)(1.5625) = 250~\mathrm{W}\]

   Copper loss varies as the square of the current, so the reading must be scaled up.

7. P46.7 Why is the OC test performed on the LV side and the SC test on the HV side?

   Show answer

   Each test is placed on the side that makes the awkward quantity manageable.

   OC test needs rated voltage, and the LV winding has the lower voltage — far easier and safer to supply 220 V than 2200 V. The current is small either way, so nothing is lost.

   SC test needs rated current, and the HV winding has the lower current — 22.7 A rather than 227 A. Only a few percent of rated voltage is needed, so the supply is easy to obtain, and shorting the LV winding requires only a thick link.

   Since the two windings differ in voltage and current in opposite senses, the two tests naturally fall on opposite sides. Whichever side is used, the parameters obtained refer to that side and must be transferred by \(K^{2}\) before being combined.

8. P46.8 Explain quantitatively why the SC test measures copper loss almost free of iron loss.

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   Because the applied voltage is only about 5 % of rated, and from Chapter 41 \(B_m \propto V\), so the flux density is 5 % of normal.

   Iron loss follows \(P_h \propto B_m^{1.6}\) and \(P_e \propto B_m^{2}\), so it falls to between \((0.05)^{1.6} = 0.0083\) and \((0.05)^{2} = 0.0025\) of its rated value.

   For a machine with 500 W of iron loss that is 1.3 to 4.1 W — against a short-circuit wattmeter reading of some 700 W. The contamination is a fraction of a percent, well below the instrument's own accuracy.

   The converse argument applies to the OC test: at 5 % of rated current the copper loss is \((0.05)^{2} = 0.25\,\%\) of its full-load value.

9. P46.9 Why must the two-frequency loss-separation test hold the flux density constant?

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   Because the relation being fitted, \(W_0 = Af + Bf^{2}\), assumes \(A\) and \(B\) are constants — and they are constants only at a fixed \(B_m\). In full, \(P_h = k_hB_m^{1.6}f\) and \(P_e = k_eB_m^{2}f^{2}\).

   If the voltage were held fixed while the frequency changed, then \(B_m \propto V/f\) would change too, and both coefficients would vary from one reading to the next. The measurement would confound a change of frequency with a change of flux, and the two unknowns could not be separated.

   The remedy is to hold the volts per hertz constant — halve the frequency and halve the voltage — so that \(B_m\) is identical in both experiments and only \(f\) varies.

10. P46.10 Explain fully why a transformer is rated in kVA rather than kW.

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    Because its two limits are thermal, and neither involves power factor.

    The winding may not exceed its rated current, or the \(I^{2}R\) loss overheats it. The core may not exceed its rated voltage, or the flux density saturates it and the iron loss and magnetising current run away. The product of these two limits is a volt-ampere limit.

    The losses correspond exactly: copper loss is fixed by current, iron loss by voltage. Neither depends on the load's power factor, so the machine is equally fully loaded regardless of what it is feeding.

    The manufacturer also cannot know the load. It may be resistive, inductive, capacitive or mixed, giving a different power factor in every installation. A rating in kW would be meaningless without also stating that power factor.

    The same 50 kVA transformer delivers 50 kW at unity power factor, 40 kW at 0.8, and no real power at all into a purely reactive load — while being fully loaded in every case.