ElectricityElectrical Measurements I (year 4) · Electrical Machines I (year 5)

Motor and transformer tests

A machine’s nameplate says what it should do; the test says what it does. With very little energy —without loading it to full power— two or three tests give the equivalent-circuit parameters, the efficiency and the regulation, and allow deciding whether the machine is fit to work.

No-load Short circuit Locked rotor Temperature rise

01Before testing: method, instruments and safety

A test is a planned measurement: the machine is put in a known condition, a few quantities are measured with suitable instruments, and the quantities that cannot be measured directly are calculated from them. The nameplate says what the machine should do; the test says what it does. Operating principles and starting methods are in AC motors and Transformers; this page shows how to check them.

No-load

Rated voltage and nothing connected to the output (the motor, with no load coupled). The machine draws little power, almost all of it to magnetize the core and overcome friction. It gives the iron losses and the excitation current.

Short-circuit or locked rotor

Output shorted (or shaft locked) and reduced voltage, just enough for rated current to flow. It gives the copper losses and the leakage impedance.

Under load

The machine delivers power to a real load or to a brake. It gives the efficiency, the regulation and the temperature rise, but it uses all the energy and needs a suitable load.

Why two «small» tests replace a big one

At full load, a 5 kVA transformer delivers 4 kW at $\cos φ = 0.8$ . In the classic tests the supply provides only 60 W at no load and 110 W on short circuit. Since iron losses depend on voltage and copper losses on current, each test isolates one group of losses with a fraction of the energy. With both, the equivalent circuit is built and everything else is calculated from it: efficiency at any load, regulation, short-circuit current.

Quantity	Instrument	Care when testing
Voltage	True-RMS voltmeter	In parallel, taken at the machine terminals and not at the source: the voltage drop in the leads falsifies the short-circuit test.
Current	Ammeter or clamp meter	In series. Choose the range so that the reading falls in the top third of the scale (see section 06).
Active power	Wattmeter or power analyzer	At no load the power factor is 0.1 to 0.3: use a low-power-factor wattmeter or an analyzer. An ordinary one hardly moves its pointer and reads badly.
Three-phase power	Two wattmeters (Aron connection)	$P = P_{1} + P_{2}$ ; if one reads negative, it is subtracted. See three-phase circuits.
Resistance	Voltmeter-ammeter method, bridge or micro-ohmmeter	With direct current and low current, so as not to heat the winding. Always note the temperature.
Insulation	Megohmmeter	With the machine disconnected and discharged; 500 V DC on low-voltage windings.
Speed	Optical or contact tachometer, stroboscope	Slip comes from it: measure it together with the other quantities, not afterwards.
Temperature	Thermometer, thermocouple or thermal imaging camera	Fix the measuring point; the frame temperature is not the winding temperature.

How each instrument works inside and how it is connected is covered in analog instruments, digital instruments and AC measurements.

Tests are safe only if they are set up properly

No-load test of a transformer: if it is supplied from the low-voltage side, the high-voltage side is left at a raised voltage and unloaded (440 V in the example). Treat it as live.

Short circuit: the voltage is low, but rated current flows. Tight terminals and conductors of adequate cross-section: a loose connection heats up and can catch fire.

Locked rotor: lock the shaft with a firm mechanical device and keep the test to a few seconds; the winding heats quickly.

Always start from zero voltage with a variable autotransformer, bring it back to zero before touching any connection, and verify the absence of voltage with an instrument. The legal framework in Argentina is Law 19,587 and Decree 351/79; the practical details are in occupational health and safety.

02Transformer: no-load and short-circuit tests

Both tests use the same instruments: a variable autotransformer, a voltmeter, an ammeter and a wattmeter. What changes is which winding is supplied and what is done with the other.

Figure 1. Instrument connection, animated. At no load the secondary is left open (only a voltmeter measures it); on short circuit it is closed through an ammeter and the voltage is raised slowly until rated current flows. The wattmeter (W) has its current coil in series and its voltage coil in parallel.

	No-load test	Short-circuit test
Supplied from	The low-voltage side (more convenient).	The high-voltage side (rated current is lower).
The other winding	Open.	Shorted, through a low-resistance ammeter.
Applied voltage	Rated, raised gradually.	Reduced: the voltage that makes rated current flow, a few percent of rated.
Measured	$V_{0}$ , $I_{0}$ , $P_{0}$	$V_{sc}$ , $I_{1 n}$ , $P_{sc}$
Losses that show up	Iron: $P_{Fe} \approx P_{0}$ (copper is negligible because $I_{0}$ is small).	Full-load copper: $P_{Cu} = P_{sc}$ (iron is negligible because the voltage is small).
Parameters obtained	$R_{Fe}$ and $X_{μ}$ : the excitation branch.	$R_{sc}$ and $X_{sc}$ : the leakage impedance.

\cos φ_{0} = \frac{P_{0}}{V_{0} I_{0}} I_{Fe} = I_{0} \cos φ_{0} I_{μ} = I_{0} \sin φ_{0} R_{Fe} = \frac{V_{0}}{I_{Fe}} = \frac{V_{0}^{2}}{P_{0}} X_{μ} = \frac{V_{0}}{I_{μ}}

No-load test. The current splits into a part in phase with the voltage (iron losses,

I_{Fe}

) and another lagging by 90° (magnetizing,

I_{μ}

Z_{sc} = \frac{V_{sc}}{I_{1 n}} R_{sc} = \frac{P_{sc}}{I_{1 n}^{2}} X_{sc} = \sqrt{Z_{sc}^{2} - R_{sc}^{2}}

Short-circuit test. It is measured on the supplied side, so the values are referred to that side. To move them to the other side, multiply or divide by

a^{2}

, where

a

is the turns ratio.

Figure 2. Which part of the equivalent circuit each test reveals, animated, with the example values referred to the high-voltage side. At no load the current hardly goes through the series branch and the voltage falls almost entirely across the excitation branch; on short circuit the opposite happens, because the excitation branch sits in parallel with a minimal impedance.

Example · The parameters of a 5 kVA transformer

Single-phase transformer, 5 kVA, 440/220 V, 50 Hz. No-load test from the low side: 220 V, 1.8 A, 60 W. Short-circuit test from the high side, at rated current: 22 V, 11.36 A, 110 W.

1 · Rated currents. $I_{1 n} = \frac{5000}{440} = 11.36$ A and $I_{2 n} = \frac{5000}{220} = 22.73$ A.

2 · No load. $\cos φ_{0}$ $= \frac{60}{220 \cdot 1.8}$ $= 0.1515$ (φ₀ = 81.3°). $I_{Fe} = 1.8 \cdot 0.1515 = 0.273$ A and $I_{μ} = 1.8 \cdot 0.9885 = 1.779$ A. Then $R_{Fe}$ $= \frac{220^{2}}{60}$ $= 806.7 Ω$ and $X_{μ}$ $= \frac{220}{1.779}$ $= 123.6 Ω$ . Referred to the high side they are multiplied by $a^{2} = 2^{2} = 4$ : 3227 Ω and 495 Ω.

3 · Short circuit. $Z_{sc}$ $= \frac{22}{11.36}$ $= 1.936 Ω$ ; $R_{sc}$ $= \frac{110}{{11.36}^{2}}$ $= 0.852 Ω$ ; $X_{sc}$ $= \sqrt{{1.936}^{2} - {0.852}^{2}}$ $= 1.739 Ω$ . In percent: $ε_{sc}$ $= \frac{22}{440}$ $= 5.0 %$ , $ε_{R} = \frac{110}{5000} = 2.2 %$ and $ε_{X}$ $= \sqrt{{5.0}^{2} - {2.2}^{2}}$ $= 4.49 %$ .

4 · Efficiency at $\cos φ = 0.8$ . At full load: $P_{2} = 5000 \cdot 0.8 = 4000$ W, losses $60 + 110 = 170$ W and $η = \frac{4000}{4170} = 95.9 %$ . At half load the copper losses are divided by four: $60 + 110 \cdot {0.5}^{2} = 87.5$ W, and $η = \frac{2000}{2087.5} = 95.8 %$ . The maximum occurs when copper equals iron: $\sqrt{\frac{60}{110}} = 0.74$ , that is, at 74 % of rated load.

5 · Regulation at full load with $\cos φ = 0.8$ lagging: $ε \approx ε_{R} \cos φ + ε_{X} \sin φ$ $= 2.2 \cdot 0.8 + 4.49 \cdot 0.6$ $= 4.45 %$ . The secondary voltage drops from 220 V to about 210.2 V.

6 · Short circuit at the terminals (with an ideal supply): $\frac{I_{1 n}}{ε_{sc}}$ $= \frac{11.36}{0.05}$ $= 227$ A, twenty times rated. It is the figure used to choose the protection.

Two details that change the result

Resistance depends on temperature. $P_{sc}$ is measured with the machine stabilized and corrected to the reference temperature set by the standard (IEC 60076-1); with a cold winding the copper gives fewer losses than it will in service.

The voltage must be sinusoidal. If the supply is distorted, the iron loses power differently and $P_{0}$ is not the one that applies on the mains. Check the waveform with an oscilloscope before accepting the measurement.

03Induction motor: no-load and locked-rotor tests

The induction motor is a transformer whose secondary rotates. That is why it is tested the same way: at no load you see the excitation branch, and with the rotor locked (slip $s = 1$ , the equivalent of the short circuit) the leakage impedance. The difference is that the motor has mechanical losses that a transformer does not. The rotating field, slip and starting methods are in AC motors.

No-load test

Rated voltage, no load coupled, with the machine running for several minutes until the bearings settle. Voltage, current in the three phases and power are measured. The input power covers three things:

$P_{0}$ $= P_{Fe} + P_{mech} + 3 I_{0}^{2} R_{1}$

iron, friction and windage, and stator copper. To separate iron from mechanical losses, the test is repeated lowering the voltage from $1.1 V_{n}$ to about half, $P_{0} - 3 I_{0}^{2} R_{1}$ is plotted against $V^{2}$ and extrapolated to $V = 0$ : what remains does not depend on voltage and is the mechanical loss.

Locked-rotor test

Shaft locked and reduced voltage, raised from zero until the current is the rated one; quick reading, in a few seconds. With $s = 1$ almost all the current goes through the series branch and the excitation branch can be neglected. $V_{lr}$ , $I_{lr}$ and $P_{lr}$ are measured; the stator resistance, $R_{1}$ , must be measured beforehand with direct current.

Z_{lr} = \frac{V_{lr} / \sqrt{3}}{I_{lr}} R_{lr} = \frac{P_{lr}}{3 I_{lr}^{2}} X_{lr} = \sqrt{Z_{lr}^{2} - R_{lr}^{2}} R_{2}^{'} = R_{lr} - R_{1}

Per-phase values of the equivalent star, with

V_{lr}

the line voltage. Without other data, the leakage reactance is split in halves:

X_{1} \approx X_{2}^{'} \approx X_{lr} / 2

Example · A 5.5 kW three-phase motor

4-pole, 5.5 kW, 380 V, 50 Hz motor, with $I_{n} = 11.5$ A and 1440 rpm on the nameplate. Stator resistance measured with DC: $R_{1} = 1.05 Ω$ per phase (of the equivalent star). No-load test at 380 V: 4.6 A and 340 W. Locked rotor: 65 V, 11.5 A and 650 W.

1 · No load. $S_{0} = \sqrt{3} \cdot 380 \cdot 4.6 = 3028$ VA and $\cos φ_{0}$ $= \frac{340}{3028}$ $= 0.112$ : the 4.6 A current, 40 % of rated, is almost all magnetizing. $Q_{0} = \sqrt{3028^{2} - 340^{2}} = 3008$ var and the no-load reactance is $X_{0} = \frac{380^{2}}{3008} = 48.0 Ω$ ( $X_{1} + X_{m}$ ).

2 · No-load losses. Stator copper: $3 \cdot {4.6}^{2} \cdot 1.05 = 66.7$ W. That leaves $340 - 66.7 = 273.3$ W of iron plus mechanical losses. The extrapolation to $V = 0$ gave 110 W of friction and windage, so the iron loss is $273.3 - 110 = 163.3$ W.

3 · Locked rotor. $Z_{lr}$ $= \frac{65 / \sqrt{3}}{11.5}$ $= 3.263 Ω$ ; $R_{lr}$ $= \frac{650}{3 \cdot {11.5}^{2}}$ $= 1.638 Ω$ ; $X_{lr}$ $= \sqrt{{3.263}^{2} - {1.638}^{2}}$ $= 2.822 Ω$ . Then $R_{2}^{'} = 1.638 - 1.05 = 0.588 Ω$ and $X_{1} \approx X_{2}^{'} \approx 1.41 Ω$ .

4 · Starting at full voltage (extrapolating linearly): $I_{st} \approx 11.5 \cdot \frac{380}{65}$ $= 67.2$ A, that is $5.8 I_{n}$ . The torque comes from the power crossing the air gap, $3 \cdot {11.5}^{2} \cdot 0.588 = 233.4$ W, at synchronous speed $ω_{s}$ $= \frac{2 π \cdot 1500}{60}$ $= 157.1$ rad/s: $T = \frac{233.4}{157.1} = 1.49$ N·m at 65 V, which at 380 V is multiplied by $(380 / 65)^{2} = 34.2$ to give 50.8 N·m. Rated torque is $\frac{5500}{2 π \cdot 1440 / 60} = 36.5$ N·m, so the starting torque is $1.39 T_{n}$ .

What the test estimates and what it does not

The starting current and torque calculated this way are estimates: at full voltage the leakage flux paths saturate and the real current turns out somewhat lower than the extrapolated one. In double-cage rotors, in addition, $R_{2}$ and $X_{2}$ change with the rotor frequency, and with the rotor locked that frequency is the mains frequency. To verify a start, measure it at service voltage with a current recorder.

04Measurements on AC and DC motors

Besides the laboratory tests, the technician measures the motor in service and out of service. This is the list of what is checked on an alternating-current machine and what each value says.

Measurement	How it is done	What it tells
Nameplate and connection	Read voltage, current, speed, $\cos φ$ , insulation class, duty and terminal box.	Whether the motor matches the supply (380 V star or 220 V delta) and the load. See the nameplate.
Line voltage	Between the three phases, at the motor terminals and under load.	Voltage drop in the feeder and voltage unbalance. A small voltage unbalance produces a current unbalance several times larger.
Current in each phase	Clamp meter on each conductor, in normal service.	Overload (against $I_{n}$ ) and unbalance.
Power and $\cos φ$	Two wattmeters or analyzer; $P = \sqrt{3} V I \cos φ$ .	Real motor load and power factor. See AC power.
Winding resistance	Between terminals, with DC, cold and with the temperature noted.	Equal phases (within a few percent), open circuits or shorted turns, and the basis for the temperature-rise test.
Insulation	Megohmmeter between terminals and frame, with the motor disconnected.	Moisture or varnish deterioration. The trend matters, not a single value; as a rough minimum, $(k V + 1)$ MΩ at 40 °C (IEEE 43).
Speed	Tachometer on the shaft.	Slip $s = \frac{n_{s} - n}{n_{s}}$ ; with 1500 rpm synchronous and 1440 rpm, $s = \frac{60}{1500} = 4 %$ .
Temperature	Thermal camera or thermometer on frame and bearings.	Overload, blocked ventilation or worn bearings.

Current unbalance is calculated as the largest deviation from the average, divided by the average. With the readings 7.4 – 7.6 – 9.8 A from the electrical maintenance topic, the average is $\frac{7.4 + 7.6 + 9.8}{3} = 8.27$ A and the maximum deviation $9.8 - 8.27 = 1.53$ A, that is $\frac{1.53}{8.27} = 18.5 %$ : too much for a motor that must run healthy.

DC machines

On a DC machine some things are measured that an AC machine does not have: the armature with its commutator, the field winding and the back electromotive force.

Measurement	How it is done	What it tells
Armature resistance $R_{a}$	Voltmeter-ammeter method with the shaft stationary, at four rotor positions.	If the four readings differ, there are bad brushes, a dirty commutator or damaged bars.
Field resistance	Between the terminals of the excitation winding, with DC.	Open circuits or shorted turns in the field.
Back EMF $E$	Calculated: $E = V - I_{a} R_{a}$ in a motor, $E = V + I_{a} R_{a}$ in a generator.	It is proportional to speed and flux: $E = k Φ n$ .
No-load characteristic (generator)	At constant speed and no load, measure $E$ while varying the field current $I_{f}$ .	The curve rises, bends because of saturation and does not return along the same path because of hysteresis.
Speed regulation (motor)	Measure $n$ at no load and at full load with constant voltage.	How much the speed drops when loaded.

Example · Speed regulation of a shunt DC motor

220 V shunt motor with $R_{a} = 0.8 Ω$ . At full load it draws $I_{a} = 18$ A and runs at 1500 rpm; at no load, $I_{a} = 1.5$ A. With constant field, speed is proportional to the back EMF.

At full load: $E = 220 - 18 \cdot 0.8 = 205.6$ V. At no load: $E_{0} = 220 - 1.5 \cdot 0.8 = 218.8$ V. Then $n_{0}$ $= 1500 \cdot \frac{218.8}{205.6}$ $= 1596$ rpm and the regulation is $\frac{1596 - 1500}{1500} = 6.4 %$ . In practice it is somewhat lower, because armature reaction weakens the flux under load and partly compensates the drop.

A DC motor must never lose its field while the armature is energized

In a shunt or separately excited motor, if the field circuit is cut with voltage on the armature, the flux falls to almost zero and the speed rises without limit until the rotor destroys itself. Before measuring the field resistance, de-energize everything; and never open the excitation circuit while running.

05Load curve and temperature-rise test

The load test is the only one that measures efficiency, speed and power factor directly, and the one that shows the real temperature rise. The motor is loaded with a brake: a Prony brake (band or shoe with a scale), a generator feeding an electrical load, or an electromagnetic brake. Output power comes from torque and speed.

P_{u} = T ω = T \cdot \frac{2 π n}{60} η = \frac{P_{u}}{P_{1}} s = \frac{n_{s} - n}{n_{s}}

T

in N·m,

n

in rpm,

P_{u}

in W. With a 0.25 m arm and a 100 N force, the torque is 25 N·m and, at 1440 rpm,

P_{u} = 25 \cdot 150.8 = 3770

Figure 3. Load curve of the 5.5 kW example motor, animated, calculated from the losses measured in the tests of section 03 (273 W constant and 650 W of copper at full load, proportional to the square of the load). Efficiency peaks near 65 % of load and the power factor grows with load.

Load	Line current	$\cos φ$	Speed	Efficiency	Input power
25 %	5.3 A	0.48	1485 rpm	81.4 %	1.69 kW
50 %	6.9 A	0.70	1470 rpm	86.3 %	3.19 kW
75 %	9.0 A	0.80	1455 rpm	86.6 %	4.76 kW
100 %	11.5 A	0.85	1440 rpm	85.6 %	6.42 kW
125 %	14.2 A	0.87	1425 rpm	84.2 %	8.16 kW

Simplified model built from the tests: constant losses of 273 W (iron plus friction), copper loss of 650 W at rated current proportional to the square of the load, reactive power growing the same way, and useful power equal to the fraction of 5.5 kW. Above 100 % the motor is overloaded; it is shown only to see the trend.

Why a motor should not be oversized

Efficiency is maximum when copper losses equal the constant ones: $x = \sqrt{\frac{273.3}{650}} = 0.65$ of rated load. A motor running at 25 % has $\cos φ = 0.48$ and $η = 81.4 %$ : the installation pays for reactive current and loss energy because a motor was bought «just in case». It is better to choose it to run between 60 % and 100 % of its rating.

Temperature-rise test

The machine is run at its rated load until thermal equilibrium —when the temperature stops rising, with a variation under 1 K per hour on the frame— and the temperature rise of the windings is measured. The temperature of a copper winding is obtained from its resistance, without putting sensors inside: copper changes its resistance with temperature in a known way.

Δ θ = \frac{R_{2} - R_{1}}{R_{1}} (235 + t_{1}) + t_{1} - t_{a}

R_{1}

is the cold resistance at temperature

t_{1}

;

R_{2}

, the hot resistance, measured as soon as the machine stops;

t_{a}

, the ambient temperature at the end of the test. The constant 235 is the one for copper (225 for aluminum). The result is the temperature rise

Δ θ

in kelvin.

Insulation class	Maximum temperature	Permissible rise (by resistance, 40 °C ambient)
A	105 °C	60 K
E	120 °C	75 K
B	130 °C	80 K
F	155 °C	105 K
H	180 °C	125 K

Example · Temperature rise by resistance

A class F motor has, cold and at $t_{1} = 20$ °C, $R_{1} = 2.10 Ω$ between two terminals. After running at full load until equilibrium, $R_{2} = 2.62 Ω$ is measured (extrapolated to the instant of shutdown) at an ambient of $t_{a} = 24$ °C.

$Δ θ$ $= \frac{2.62 - 2.10}{2.10} \cdot (235 + 20) + 20 - 24$ $= 0.2476 \cdot 255 - 4$ $= 59.1$ K. The winding reached $20 + 63.1 = 83.1$ °C; at a 40 °C ambient it would reach $40 + 59.1 = 99.1$ °C. The class F limit is 105 K, so the motor complies, with a margin of almost 46 K.

The winding cools quickly: measure right away

When the machine stops the temperature falls, and so does the resistance. Measure it at 30 s and 1 min intervals for a few minutes and extrapolate the curve back to the instant of shutdown. If it is measured a couple of minutes late without extrapolating, the rise is underestimated. For every 10 K above the class limit, the life of the insulation is roughly halved (rule of thumb).

In a transformer the temperature-rise test is done by the short-circuit method: it is supplied with a power equal to the total losses, $P_{0} + P_{sc}$ , until the oil temperature stabilizes, and then rated current is applied to measure the winding temperature by resistance (IEC 60076-2).

06Visual inspection and selection criteria

Before connecting an instrument, look. Visual inspection takes minutes, costs nothing and finds defects that no electrical test detects, or that make the test dangerous (a broken frame, a low oil level). It is done with the machine disconnected and recorded on a checklist.

Item	What to look for	What it indicates
Nameplate	Legible, complete and matching the installation.	If missing or erased, there is no reference value to test against.
Frame, feet and end shields	Cracks, dents, rust, burnt paint, loose bolts.	Overheating, vibration or poor mounting.
Fan and fins	Obstructions, dirt, damaged blades.	Poor ventilation: the motor will run hotter than in the test.
Shaft and bearings	Turn by hand: play, roughness, noise, grease leaking out.	Worn or poorly lubricated bearing.
Terminal box and connections	Loose or discolored terminals, wrong links; conductors with burnt insulation.	Poor tightening: hot spots. Check the connection against the supply voltage.
Visible winding	Darkened varnish, burnt smell, conductive dust.	Repeated overheating or moisture.
Brushes and commutator (DC)	Wear, sparking, scored or blackened surface, protruding bars.	Poor commutation or brushes of the wrong grade.
Transformer: tank and oil	Leaks, level, oil color, tank deformation.	Overheating, damaged gaskets, moisture ingress.
Transformer: silica gel and bushings	Silica gel color; cracks or dirt on the bushings.	Saturated silica gel no longer protects the oil; a dirty bushing causes discharges.
Grounding	Conductor present, continuous and well connected to the frame.	Without it, an insulation fault leaves the frame live.

How to choose a machine

Criterion	What is defined
Power and duty	The load power with a reasonable margin, and the duty (continuous S1, intermittent, frequent starts) according to IEC 60034-1.
Voltage and frequency	Those of the supply. A 220/380 V motor is connected in delta on a 220 V supply and in star on a 380 V one; a 380/660 V motor, in delta at 380 V and in star at 660 V. Always as the nameplate indicates.
Speed	Poles according to the required speed; with a drive, see variable-speed drives.
Mechanical protection	IP rating (IEC 60529) according to dust, water and contact: a motor outdoors or in a washdown area needs IP55 or higher.
Insulation and environment	Insulation class and ambient temperature; above 40 °C or 1000 m altitude the power must be derated.
Efficiency	IE class (IEC 60034-30-1): a higher-efficiency motor costs more, but the difference is recovered over the running hours.
Mounting	Foot or flange, horizontal or vertical shaft (IM codes of IEC 60034-7).

How to choose an instrument

Criterion	What is defined
Range	The reading should fall in the top third of the scale. At no load (4.6 A) and locked rotor (11.5 A) two ranges are needed, for example 5 A and 15 A.
Accuracy class	A class 0.5 ammeter with a 15 A range has an error of $\pm 0.075$ A: at 4.6 A that is $\frac{0.075}{4.6} = 1.6 %$ . With the 5 A range the error falls to $\frac{0.025}{4.6} = 0.5 %$ .
Waveform	True RMS if there are drives or nonlinear loads. An average-responding instrument reads wrongly.
Wattmeter	Low $\cos φ$ type for no-load tests; otherwise, a power analyzer.
Measurement category	CAT III or CAT IV (IEC 61010-1) for measuring in switchboards and distribution installations.
Calibration	With a valid certificate: without it, the test cannot be documented.

07In the laboratory

Practice 1 · The parameters of a single-phase transformer

Materials: 100 to 500 VA transformer with two windings, variable autotransformer, voltmeter, ammeter, wattmeter (or power analyzer), multimeter for resistance, leads of adequate cross-section.

Steps: 1) read the nameplate and calculate $I_{1 n}$ and $I_{2 n}$ ; 2) measure the resistance of each winding with DC; 3) no-load test from the low side, raising the voltage slowly to rated, and note $V_{0}$ , $I_{0}$ , $P_{0}$ ; 4) short-circuit test from the high side, closing the secondary through the ammeter and raising from zero to $I_{1 n}$ ; note $V_{sc}$ and $P_{sc}$ and bring the voltage back to zero; 5) calculate $\cos φ_{0}$ , $R_{Fe}$ , $X_{μ}$ , $Z_{sc}$ , $R_{sc}$ , $X_{sc}$ , $ε_{sc}$ and the efficiency at 100 % and 50 %.

Accepted if: $R_{sc}$ matches $R_{1} + a^{2} R_{2}$ measured with DC to within 15 % (corrected for temperature) and $P_{0} + P_{sc}$ is less than 10 % of the rated power.

Practice 2 · No-load and locked-rotor tests of a three-phase motor

Materials: 0.37 to 1.5 kW motor, variable three-phase autotransformer, two wattmeters or analyzer, clamp meter, megohmmeter, mechanical device to lock the shaft.

Steps: 1) visual inspection and insulation measurement; 2) stator resistance with DC; 3) no-load at rated voltage, running about 15 minutes; 4) no-load at six decreasing voltages, from $1.1 V_{n}$ to about half; 5) with the shaft locked, raise the voltage from zero up to rated current and read in less than 10 s; 6) calculate parameters, mechanical and iron losses, starting current and torque.

Accepted if: $P_{0} - 3 I_{0}^{2} R_{1}$ against $V^{2}$ gives a straight line, the three no-load currents differ by no more than 10 % from one another and the calculated starting current is 4 to 8 times rated.

Practice 3 · Load curve with a brake

Materials: three-phase motor with Prony brake or brake generator, tachometer, two wattmeters, clamp meter, scale.

Steps: with the motor at rated voltage, load it to 25, 50, 75, 100 and 125 % of rated torque (do not hold 125 % for more than a few minutes). At each point note $V$ , $I$ , $P_{1}$ , the speed and the force on the scale; calculate $P_{u}$ , $η$ , $\cos φ$ and slip; plot against output power.

Accepted if: efficiency has a maximum between 60 % and 90 % of load, $\cos φ$ grows with load and the full-load readings agree with the nameplate within the accuracy of the instruments.

Practice 4 · Temperature rise by resistance

Materials: motor with legible nameplate, micro-ohmmeter or bridge, thermometer, stopwatch.

Steps: note $R_{1}$ and $t_{1}$ cold (the machine has been stopped for several hours); run it at full load until the frame varies by less than 1 K per hour; stop it and measure the resistance every 30 s for 5 minutes; extrapolate the curve back to the instant of shutdown to obtain $R_{2}$ ; note $t_{a}$ ; calculate $Δ θ$ .

Accepted if: $Δ θ$ does not exceed the limit of the insulation class on the nameplate (table in section 05).

Practice 5 · DC motor

Materials: separately excited or shunt DC motor, adjustable DC supply, ammeters, voltmeters, tachometer.

Steps: measure $R_{a}$ with the shaft stationary at four positions; supply rated voltage with the field connected and measure $n$ and $I_{a}$ at no load; load up to rated current and repeat; calculate $E = V - I_{a} R_{a}$ in each case and the speed regulation.

Accepted if: the four $R_{a}$ readings differ by no more than 10 % and the speed calculated from $E$ agrees with the measured one within 5 %.

08Common mistakes

Mistake	Consequence
Short-circuit testing the transformer at rated voltage	A current tens of times rated flows (in the example, 20) and the winding burns in seconds. Always raise the voltage from zero.
Using an ordinary wattmeter in the no-load test	At $\cos φ$ of 0.1 the pointer barely moves and the relative error is huge. Use a low-power-factor one or an analyzer.
Taking the motor's $P_{0}$ as iron loss	It includes friction and windage and the stator copper; they must be subtracted.
Mixing the transformer sides	Values measured on the low and high sides cannot be added: they must be referred to the same side by multiplying by $a^{2}$ .
Keeping the rotor locked too long	Rated current in an unventilated rotor heats the winding very quickly; results change and the motor is damaged.
Measuring winding resistance without noting the temperature	The value is useless for comparison or for calculating the temperature rise.
Measuring hot resistance several minutes after stopping the machine	The winding has already cooled: the rise is underestimated. Extrapolate to the instant of shutdown.
Testing with an unbalanced or distorted supply	Currents and losses are no longer representative; a supply problem is mistaken for a motor defect.
Opening the field of a running DC motor	Runaway: speed rises until the rotor is destroyed.

09Self-assessment

What is obtained from each of the two transformer tests?

The no-load test, at rated voltage, gives the iron losses and the parameters of the excitation branch ( $R_{Fe}$ , $X_{μ}$ ). The short-circuit test, at rated current and reduced voltage, gives the copper losses and the leakage impedance ( $R_{sc}$ , $X_{sc}$ ).

Why does the short-circuit test not destroy the transformer?

Because it is done with reduced voltage, a few percent of rated (5 % in the example), just enough for rated current to flow. At rated voltage the current would be about 20 times larger.

A no-load test reads 220 V, 1.5 A and 45 W. Calculate

\cos φ_{0}

and

R_{Fe}

$\cos φ_{0}$ $= \frac{45}{220 \cdot 1.5}$ $= 0.136$ . $R_{Fe}$ $= \frac{220^{2}}{45}$ $= 1076 Ω$ (referred to the side where it was measured).

A 380 V three-phase motor at no load draws 4.6 A and 340 W. What is its power factor, and why is it so low?

$\cos φ_{0}$ $= \frac{340}{\sqrt{3} \cdot 380 \cdot 4.6}$ $= 0.112$ . Almost all the no-load current is magnetizing (reactive): active power only covers losses, and there is no load to absorb more.

How is the starting current estimated from the locked-rotor test?

The current is measured at reduced voltage and scaled linearly with voltage: $I_{st} \approx I_{lr} \cdot \frac{V_{n}}{V_{lr}}$ $= 11.5 \cdot \frac{380}{65}$ $= 67$ A. It is an estimate somewhat higher than the real one, because of saturation.

A winding's resistance goes from 1.00 to 1.25 Ω. With

t_{1} = t_{a} = 20

°C, what is the temperature rise?

$Δ θ$ $= \frac{1.25 - 1.00}{1.00} \cdot (235 + 20) + 20 - 20$ $= 0.25 \cdot 255$ $= 63.75$ K: it exceeds the limit of class A (60 K) and complies with those of classes E (75 K), B (80 K), F (105 K) and H (125 K).

Why does an oversized motor perform poorly?

Because at low load its power factor and its efficiency fall. The example motor, at 25 % load, has $\cos φ = 0.48$ and $η = 81.4 %$ , against 0.85 and 85.6 % at full load.

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