No-Load and Blocked-Rotor Testing

The no-load test first. Take the phase voltage and find the no-load power factor from the three-phase power:

A power factor of 0.15 means \(\phi_0 = 81.4^\circ\) — the current is almost entirely magnetising, exactly as it must be when the shaft is doing no work.

Resolve the no-load current into its two components:

Note what \(R_0\) actually represents here: the whole 520 W of no-load input, which is core loss plus friction and windage plus the 41 W of no-load stator copper loss. Problem 5 separates those three; for the exciting branch alone the lumped value is the conventional answer.

Now the blocked-rotor test. With the rotor held, \(s=1\) and the rotor branch is a low impedance, so the exciting branch may be dropped:

The blocked-rotor power factor is \(1140/(\sqrt3\times83\times20) = 0.396\) — poor, because at standstill the leakage reactance dominates.

Separate resistance from reactance by the impedance triangle:

The equal split is the convention for a wound-rotor or Class A machine; Problem 4 examines what happens when it is not appropriate.

The magnetising reactance follows once \(X_1\) is known, because the no-load impedance is very nearly the series sum \(X_1+X_m\):

The shunt-model value \(X_0 = 46.72\ \Omega\) and the series-model value \(|Z_{nl}| = 46.19\ \Omega\) differ by 1%, which is the price of the very low no-load power factor; the series value is the one to pair with \(X_1\).

The direct-on-line starting current. The blocked-rotor impedance is the standstill impedance, so it need only be scaled to full voltage:

Equivalently, scale the test current in the ratio of the voltages: \(20\times400/83 = 96.4\ \text{A}\) — the same number, and the quicker route in an examination.

The completed parameter set:

Why the test is run at reduced frequency at all. At standstill the rotor frequency equals the supply frequency, and in a deep-bar or double-cage rotor the current then crowds into the top of the bar, raising the apparent resistance well above its running value. Testing at a quarter of rated frequency puts the rotor current where it sits in normal service, at a slip frequency of a few hertz.

Reduce the readings to per-phase quantities:

The rotor resistance, which needs no frequency correction because resistance is not a function of frequency:

The reactance at test frequency, from the impedance triangle:

Scale it to rated frequency. Reactance is proportional to frequency, so the correction is a simple ratio:

This is the only quantity in the test that changes with frequency. The measured power, and therefore \(R_{01}\), is used exactly as read.

Direct-on-line starting current, using the corrected impedance at rated voltage:

The starting torque is the air-gap power at \(s=1\) divided by synchronous speed:

What the uncorrected figure would have given. Using \(X = 0.397\ \Omega\) instead of 1.586 Ω:

A predicted starting current more than twice the true value — enough to specify a starter, a cable and a protective device that are all wrong. The frequency correction is not a refinement.

Start with the DC test, because every later step needs \(R_1\). Direct current between two terminals of a star winding passes through two phases in series:

No reactance, no rotor e.m.f. and no core loss can appear in a DC measurement, which is exactly why it is done first.

The no-load test. Average the three line currents and reduce the voltage to a phase value:

Strip the stator copper loss out of the no-load input to leave the rotational and core losses:

This 371 W is core loss plus friction and windage together; Problem 5 shows how a run at several voltages splits it.

The locked-rotor test, taken at 15 Hz so that the rotor resistance is measured at a realistic slip frequency:

Resolve the locked-rotor impedance and correct the reactance to 60 Hz:

The completed per-phase circuit:

Reduce the stator side to a Thevenin equivalent, as Set 28, Problem 6 requires before any pull-out calculation:

From \(V_{Th} = V_\phi X_m/\sqrt{R_1^2+(X_1+X_m)^2} = 120\times14.03/14.70\) and \(R_{Th} \approx R_1\left(X_m/(X_1+X_m)\right)^2 = 0.243\times(14.03/14.70)^2\).

The slip at pull-out torque:

The pull-out torque. A 4-pole 60 Hz machine has \(N_s = 1800\) rpm, so \(\omega_s = 188.5\) rad/s:

Against a rated torque of roughly \(7.5\ \text{hp} = 5.6\ \text{kW}\) at about 1750 rpm, or 30 N·m, this is a pull-out margin of about 2.2 — entirely normal for a general-purpose cage machine.

Understand what the tests can and cannot see. The blocked-rotor test measures the two leakage reactances in series, and no terminal measurement can distinguish them — there is no accessible node between them. What is measured is only

The rotor resistance is unambiguous, though: \(R_2' = R_{BL}-R_1 = 0.50-0.20 = 0.30\ \Omega\).

The NEMA convention assigns the split by design class, reflecting the rotor-slot geometry each class uses:

Design B, the default for a general-purpose 25 hp machine:

Designs A and C by the same rule:

Note that \(X_m\) moves in the opposite direction to \(X_1\), because their sum is pinned by the no-load measurement.

Now the Thevenin quantities for each case, with \(V_\phi = 460/\sqrt3 = 265.6\ \text{V}\) and \(\omega_s = 188.5\ \text{rad/s}\):

The key observation. Add \(X_{Th}\) and \(X_2'\) in each row:

All three are close to \(X_{BL} = 1.90\ \Omega\), and for the reason that \(X_{Th}\approx X_1\) while \(X_1+X_2' = X_{BL}\) is fixed. The quantity that governs both pull-out results is therefore almost independent of how the split is made.

Slip at maximum torque and pull-out torque, with \(R_2' = 0.30\ \Omega\) throughout:

The pull-out slip varies by 2.5% across the whole range of design classes and the pull-out torque by 3.6% — a spread far smaller than the uncertainty in the test readings themselves.

Where the split does matter. It is not the torque but the magnetising branch and the stator voltage drop:

These affect the no-load current, the power factor at light load and the core-loss estimate. For torque and slip predictions the convention can be applied with a clear conscience; for magnetising-current studies it should not be leaned on.

Identify what the no-load input contains. With the shaft uncoupled the mechanical output is zero, so every watt drawn becomes a loss:

The rotor copper loss is genuinely negligible here: the no-load slip is a few tenths of a per cent, so \(P_{cu2} = sP_g\) amounts to a watt or two.

Remove the stator copper loss from each reading, since it is the one term that depends on the current rather than on the voltage:

Use the different voltage dependence of the two remaining losses. Core loss follows the flux, and hence the voltage, roughly as a square; friction and windage depends on speed, which barely changes:

So plotting \(P'\) against \(V_L^2\) should give a straight line whose intercept is the friction and windage loss — the loss that survives when the flux is taken to zero.

Take the two extreme points to fix the line:

Check the fit at the intermediate points — a curved plot would mean the square law has been broken by saturation:

Agreement to better than a watt, so the straight line is genuine and the extrapolation is trustworthy. A least-squares fit through all five points gives 399.5 W, confirming the two-point estimate.

The core loss at rated voltage is what is left after removing the intercept:

Why the separation is worth the extra readings. The two losses behave differently in every subsequent calculation:

Lumping them together, as Problem 1 was content to do, puts 400 W in the wrong place in the ledger. It leaves the total loss right but the air-gap power — and therefore the torque — wrong.

Extract the series-branch parameters from the blocked-rotor test:

Build the exciting branch from the core loss alone, not from the whole no-load input. Problem 5 established the split:

If the whole 1256 W were put into \(R_0\), the friction and windage would be charged to the machine twice — once at the terminals and again when it is deducted from \(P_m\). Here the shunt branch carries the core loss only.

Solve the series branch at 4% slip:

Add the exciting current to get the line current and the power factor:

Air-gap power, mechanical power and shaft output:

Input power, shaft torque and efficiency:

Verify by the loss ledger, which must close exactly if the exciting branch was built correctly:

The ledger closing to the watt is the strongest available check that the parameters, the slip and the loss split are all consistent.

A sanity check on the prediction itself. The predicted full-load current of 37.9 A is close to the 40 A used in the blocked-rotor test, which is how such a test should be set up; the starting current would be \(230.9/1.429 = 162\ \text{A}\), or 4.3 times full load, and the output of 20.8 kW at 400 V is consistent with a 28 hp frame. Nothing in the answer is physically out of range.