Testing of Single-Phase Induction Motors

Start with the blocked-rotor test, because it alone fixes the rotor parameters. At standstill the forward and backward branch resistances are both \(R_2'/2\), so they add to \(R_2'\) and the magnetising branch, being much the larger impedance, can be dropped:

Split the two blocked-rotor quantities. The resistance splits by subtraction because \(R_{1m}\) was measured with d.c.; the reactance splits by the usual equal-share assumption:

Now the no-load test, which supplies the magnetising reactance. Work through the power factor:

Extract \(X_m\) from \(X_0\). At no load the slip is nearly zero, so \(R_2'/2s\to\infty\) and the forward branch is effectively \(j\tfrac12X_m\); the backward branch, however, is at \(2-s\approx2\) and still contributes \(j\tfrac12X_2'\):

Ignoring the backward branch here would give 83.5 Ω, about 7% high. The backward field is small at no load but it is not zero.

The rotational loss is what remains of the no-load input after the copper loss. The copper term includes the backward branch resistance \(R_2'/2(2-s) \approx R_2'/4\):

This 74.8 W is the sum of core loss and friction and windage. A single no-load reading cannot separate them; Problem 5 shows the test that can.

Collect the equivalent circuit:

Blocked-rotor test on the main winding. This is the test that fixes the rotor, so it comes first:

No-load test on the main winding:

The magnetising reactance. Collecting the no-load reactances in terms of the blocked-rotor value, \(X_1+\tfrac12X_2' = \tfrac12X_{bm}+\tfrac14X_{bm} = 0.75X_{bm}\):

The factor of 2 multiplies the whole bracket, because \(\tfrac12X_m\) is what appears in the circuit. Applying it only to \(X_{nL}\) would give 69.17 Ω.

The rotational loss from the no-load input:

Blocked-rotor test on the auxiliary winding, with the main winding open. Only the auxiliary branch is energised, so the reading gives its own impedance:

The two are equal to within the precision of the readings, so \(X_{ba}\approx0\): the auxiliary winding of a split-phase motor is deliberately wound with fine wire and few turns, making it nearly a pure resistance at standstill. That is the whole basis of the phase split.

Refer the auxiliary winding to the main winding. The measured resistance is the winding's own resistance plus the rotor resistance seen through the turns ratio \(a = N_a/N_m\):

The auxiliary winding is equivalent to half the effective turns of the main winding, which is consistent with its being a high-resistance, low-reactance starting winding rather than a running winding.

Impedance under blocked-rotor conditions:

Resistance under blocked-rotor conditions, taken from the wattmeter and the current, not from the impedance:

The rotor resistance is the difference. At standstill the forward and backward halves of the rotor resistance add up to the whole of \(R_2'\), so only the stator part has to be removed:

The leakage reactances follow from the same test:

The no-load data, so far unused, gives the magnetising reactance and the rotational loss:

(a) The blocked-rotor triangle:

A blocked-rotor power factor of \(7.5/12.5 = 0.6\), which is typical: at standstill the machine is dominated by leakage reactance.

(b) Separate stator from rotor:

(c) The no-load reactance, then the magnetising reactance:

(d) The rotational loss and the no-load power factor:

Note that of the 180 W drawn at no load, more than half is copper loss — 91.9 W — and a good part of that belongs to the backward field, which is still turning at nearly twice synchronous speed relative to the rotor.

Sanity check on the whole set. Feed the parameters back into the equivalent circuit at \(s=1\): the two branch resistances are \(5.1/2 = 2.55\ \Omega\) each, giving \(2.4+2.55+2.55 = 7.5\ \Omega\) — the measured \(R_e\), as it must be. The parameters are self-consistent.

The principle of the test. The no-load input contains three things: copper loss, core loss and friction-and-windage. Copper loss can be calculated at each point; core loss is proportional to the square of the flux and therefore to \(V^2\); friction and windage depends only on speed, which barely changes because the machine stays close to synchronism at every voltage. So

A straight line in \(V^2\) whose intercept is what we want: friction and windage is the loss that survives when the flux is taken to zero.

The effective no-load resistance is the same at every point:

Strip the copper loss out of each reading:

The top row reproduces Problem 4's 88.1 W, as it should — the rated-voltage point is the same test.

Fit the straight line. Taking the first and last rows, which are furthest apart and so give the best-conditioned slope:

Every intermediate row agrees: at 170 V, for instance, \(kV^2 = 26.3\) W and \(66.3-26.3 = 40.0\) W. The points really do lie on a line, which is the evidence that the model is right.

The core loss at rated voltage is then the remainder of the rated-voltage rotational loss:

Why the separation matters. Friction and windage is a constant drag that must be subtracted from gross mechanical power at every load, whereas core loss varies with the applied voltage and is properly charged to the electrical side. Lumping them together, as a single no-load reading forces you to do, makes any efficiency calculation at reduced voltage wrong — and it hides a worn bearing, which shows up as a rising intercept while the slope stays put.