how an induction motor works by equation s (and physics)...field induces a current. that current in...

How an Induction Motor Works by Equations (and Physics)

Introduction: Induction motors are the commonest type of motor and account for a very large proportion of heavy-duty motors. Sizes vary from fractional horsepower to several thousand horsepower used for such appli-cations as diesel-electric locomotives. Until relatively recently most such motors had to operate at fixed speeds determined by the available line power frequency. Now the availability of power semiconductors and circuits makes it possible to vary their speed and even in some cases hold them stationary.

All motors require two sets of magnetic fields, one of which might be supplied by a permanent magnet, that in-teract to drive the rotor. One field acts as a sort of environmental field and the second, a time varying field, re-acts with the first to drive the rotor. In induction motors the ‘environmental’ field is supplied by one or more coils in the stator that create a rotating B field in the air gap between rotor and stator.

The distinguishing feature of induction motors is that their rotors have no permanent magnets or any need for current to be driven into the rotor windings from any direct connection. (A direct connection would require brushes, a commutator and more power supply connections, raising cost and decreasing reliability. This is why induction motors are more popular.) Instead the rotor is wound with shorted turns in which the environmental field induces a current. That current in turn produces a field that interacts with the environment to drive the rotor.

The three-phase induction motor is the easiest motor of this type to understand so these notes start with that type. The single-phase induction motor is more subtle and less efficient. It is discussed later.

The magnetic field in the air gap from the voltage applied to the stator: The stator has three sets of windings that are aligned at 120 degrees to each other and are driven by balanced currents that are 120 degrees out of phase. Call the coils A, B, C and we will work in cylindrical coordinates. (See figure below.) The coils are wound in such a way as to generate a field that is a rough stepwise approximation to a cos( )θ distribution. The wind-

ings overlap and each winding slot has two windings in it usually from different phases. (See 4 and 6 pole distri-butions on the handout from class.) Here I assume a two pole distribution because it makes the explanation eas-ier. Such a motor runs at the nominal speed range of 3500 – 3580 RPM. (Four and six pole motors run at 1720 to 1790 and 1140 to 1190 RPM respectively.)

[ ]10 0 2cos( ) cos( ) cos( ) cos( )AB B t r B t t rθ ω θ ω θ ω= = + + −

2 2 410 03 3 2 3cos( ) cos( ) [cos( ) cos( )]BB B t r B t t rπ π πθ ω θ ω θ ω= − − = + − + −

4 4 210 03 3 2 3cos( ) cos( ) [cos( ) cos( )]CB B t r B t t rπ π πθ ω θ ω θ ω= − − = + − + −

4 2102 3 3[3cos( ) cos( ) cos( ) cos( )]A B CB B B B B t t t t rπ πθ ω θ ω θ ω θ ω= + + = − + + + + − + + −

302 cos( )B B t rθ ω= −

Induction Motor Equations ENGN1931F – Spring 2019

2

Let and L R Sω ω ω be the angular velocities of the magnetic field (line frequency), rotor, and slip respectively.

For convenience we assume t that at 0, 0 and =0t θ ϕ= = and therefore PEAK L R St t tθ ϕ ω ω ω− = − = .

The flux in the single-turn coil on the rotor surface is

2

2

302 cos( )ROTOR PEAKB rl d

π

π

ϕ

ϕ

θ ϕ ϕ+

−

′ ′Φ = −∫ .

Substitute and PEAKu d duθ ϕ ϕ′ ′= − = − to get

2

2

2

2

3 30 0 02 2cos( ) sin( ) 3 cos( )

PEAKPEAK

PEAKPEAK

ROTOR SB rl u du B rl u B rl tπ

π

π

π

θ ϕθ ϕ

θ ϕθ ϕ

ω− −

− +

− −− +

Φ = − = =∫

The voltage in the loop is 03 sin( )ROTOR S SdV B rl tdt

ω ωΦ= − =

In the usual treatment of a linear circuit, sin( ) Im(exp( ))S St j tω ω= and the current in the loop is:

( )0 0 2 2 2 2

exp sin( ) cos( )3 Im 3( ) ( )

SROTOR R S S R SR S S

LOOP R R R S R R S R

j tV R t L ti B rl B rlZ R jX R L R L

ω ω ω ωω ωω ω

= = = − + + +

For use later in calculating the heat wasted in the rotor, we will need the mean square value of this current or:

( )

2 2 2 22 01

2 22

9 SR

R S R

B r liR L

ωω

=+

The B field at the leading edge of the rotor coil is:

3 30 02 2 2 2( ( )) cos( ) sin( )PEAK L SB B t r B t rπ πθ θ ϕ ω ϕ ω= − + = − − =

Also at the leading edge of the rotor coil, the positive sign on the voltage implies that the current is out of the

page along the k axis and therefore the force on the wire from the Lorenz force law is in the k r θ× = + direc-tion. The rotor is being pushed in the direction of the rotation of the stator magnetic field. The torque from one side of the rotor is the product of the rotor current, the stator B field and the radius of the rotor itself. The total torque from both sides of the loop is then

2

2 2 20 2 2 2 2

sin ( ) cos( )sin( )9( ) ( )

R S S R S SS

R S R R S R

R t L t tT B r l kR L R L

ω ω ω ωωω ω

= − + +


3

The first term in [ ]2 12sin ( ) 1 cos(2 )S St tω ω= − gives a net positive average torque but it has 100 % variation in

magnitude from no torque to twice the average torque at twice the slip frequency. The second term has no net average torque but would also represent vibration at twice the slip frequency. Figure 1 is a graph showing the two terms and their sum. (The slip frequency is very low, typically 20 – 90 RPM or 0.33 to 1.5 Hz, so this would be really bad vibration.)

Suppose, however, we add a second coil at a right angle to the first. The current in that loop will result in a sim-ultaneous torque that has the same functional form but with all angular arguments moved back by 2

π radians.

i

X

•

•

•

• •

•

X X

STATOR

X

X

- rotor angleϕ

X X

ROTOR

•

- magnetic field angleθ

r j

•

•

Single turn rotor coil

X

X

Coil A – stepwise approximation to cos( )θ field distribution


4

Use the trigonometric identities that 2sin( ) cos( )S St tπω ω− = − and 2cos( ) sin( )S St tπω ω− = to get the total

torque of the two turns as:

2 22 2 2 00 2 2 2

9 29( ) / 1 (2 )

SRS

R S R R S R

B r l fRT B r l k kR L R l f

πωω π τ

= = ⋅+ +

where /R R RL Rτ = is the rotor time constant. This is a constant torque as shown on the median line in Fig. 1.

There are two equivalent forms of the torque expression in this equation. The first is the result of simple substi-tution. The second expression has been rewritten to emphasize how the torque and hence power depends on several properties of the system. It is directly proportional to the square of the stator field 0B (more field in-

duces more current and the torque is a product of field and current), to the volume of the rotor (larger motors, more torque), and to the slip frequency. No slip, no torque because there will be no current in the rotor cage.

Figure 1: Relative torque as a function of slip angle ( Stω ) for a single rotor loop and for two loops at right an-

gle to each other. Net torque from current in phase with the induced voltage and torque component with no net torque, just vibration, from current at 90 deg. phase angle.


5

Notice that the torque equation implies a relationship between the size of the rotor and the mechanical power that it can possibly deliver to the load. The mechanical power is

2 20

2

( ) 36/ 1 (2 )

LINE S SMECHANICAL R

R S R

B r l f f fP TR l f

π πωπ τ

−= ⋅ = ⋅

+

For low slip frequencies, the power of the motor is simply proportional to the product of the square of the gap field strength with the volume of the rotor and the no-load speed of rotation. To increase motor power, general-ly you have to increase the rotor volume or the speed of rotation since the B field strength is limited by the properties of magnet iron and the effective resistance of the rotor is limited by the need to cool the rotor.

Figure 2 shows an evaluation of this torque expression for a 4-pole motor with 0.09Rτ = seconds. The curve

has been normalized to unit peak value to show the relation between torque and shaft speed. Mechanical pow-er is the product of torque with angular velocity and a similarly normalized power curve is superposed. Small motors are sometimes built as single phase induction-run motors and these have no starting torque and have rather less torque for a given stator B-field. (You will do one or two experiments on these motors.)

Obviously, one can increase the torque by adding more loops in pairs until reaching a limit set by the ability to get rid of heat. (The rotor cage has non-zero resistance and that means joule heating anytime there is rotor cur-rent, that is, anytime there is non-zero torque. The more rotor turns, the more heat there is.) The torque and output power increase directly proportionally to the number of pairs of loops. Connecting all the ends of the axial rotor wires by a heavy ring of conductor does not influence the resulting operation very much, so motors generally have a cage-like structure for the loops in which the induced current flows. This is why the commonest induction motors are called ‘squirrel-cage’ motors since the structure looks like what might be used to entertain a small animal in captivity. (Very large motors sometimes have a portion of the ends of the rotor windings exter-nally connected through brushes so you can control the effective resistance of the rotor. Increasing RR lowers the rotor time constant moving the peak torque to lower rotation speed and higher slip. This lowers the maxi-mum torque too, but it substantially increases the starting torque. Usually on a wound-rotor induction motor, the external resistance is optimized for initial starting torque and then the resistor is shorted for optimal steady state operation. This procedure helps control startup current, limiting the inrush current without requiring a high leakage inductance in the stator.)

There is an odd subtle feature of the thermal problem because the rotor resistance changes very rapidly with temperature, increasing by 20 % for a 50 deg. C rise which is typical of rated operation. This causes a similar percent change in the rotor time constant and a readily observable variation in the torque-speed curve during warmup. The change is simply due to the temperature coefficients of resistance of aluminum and copper, the materials from which rotor windings are made. Aluminum is used for squirrel-cage motors with the cages made by die-casting while wound-rotor motors use copper bars.

So far we have assumed that the stator field is uninfluenced by the currents in the rotor cage but that clearly cannot be true. The energy for the rotor has to come from the power mains and that can only happen by chang-


6

ing the fields in the stator. The rotor currents create magnetic fields that add to the stator field both in the air gap and throughout the rotor and stator structure.

Returning to our two-loop model, notice that the first term in the current in each of the two loop equations is what does all the work, that is, the term responsible for the net torque, is 1 sin( )R si tω∝ − for the first loop.

Similarly, the current in the second loop is 2 cos( )R si tω∝ . These currents produce magnetic fields in the gap

that are fixed relative to the rotor itself because that is what the coils are attached to. Their fields are radial in the air gap and 90 degrees rotated relative to each other. The sine and cosine dependence on time means that the sum expresses a radial vector field that rotates relative to the rotor at the slip frequency in the same direc-tion as the rotor is moving.

In the stator reference frame, this field is moving at the line frequency rate, the same angular rate as the stator-induced magnetic field. From the point of view of the stator windings, there is a single magnetic field, the sum of sinusoidal and triangular components, rotating in the air gap that generates voltages in the stator windings. That “back-emf” must match the applied line voltages.

Figure 2: Relative torque versus speed of rotation for a 4-pole 3-phase induction motor having a rotor time constant of 0.09 sec. Power is the product of torque and speed of rotation so vanishes at locked rotor. Single


7

phase, induction-run motors make less efficient use of the stator B-field so have less torque for the same exci-tation MMF.

By Lenz’s law, the rotor field is opposite in sign to the stator-induced field so it tends to reduce that field. This reduces the voltage induced in the stator coils. However, those coils are connected to the line voltages and in-stead of decreasing the terminal voltage, the stator current will increase to generate enough total flux for the back-emf to match the applied voltage. The increase in current is at least partially in phase with the line voltage and supplies the energy both for joule heating the rotor and delivering the mechanical energy. The action is ex-actly analogous to the operation of a transformer, suggesting that we might model the motor with a transformer equivalent circuit and use an energy analysis to determine how the motor will behave from the power applied to it.

Figure 3: Circumferential B-field about the rotor from the slip frequency currents induced in a pair of current loops around the rotor. The results do not depend on whether the loops are inset into rotor slots or not. The direction of rotation of the fields is increasing angle about the rotor or CCW.

The shape of the rotor field is not altogether obvious. Consider a snapshot of the field around the rotor from the current in one of our two loops. The field at that instant is uniform, that is, constant with angle, on each side of the loop, whether or not the loop is inset into a slot in the rotor. If the current in the rotor were constant,


8

the voltage it would induce in a stator coil turn would be a square wave. With two rotor loops there are two overlapping rectangular field shapes that are weighted by the time functions sin( )stω− and cos( )stω , Figure 3

below shows a series of snapshots of the field going counterclockwise around the rotor from one side of one coil at several successive times corresponding to changes in the slip angle Stω . The fields are still constant for ap-

preciable segments of the rotor and do not resemble sinusoids. However, the entire pattern changes continu-ously in a way that shows CCW rotation (increasing angle about the rotor) at the slip frequency.

If another pair of coils is added at right angles to each other and at 45 degrees to the first pair, then snapshots of the field pattern look like Figures 4 and 5. Adding the second pair of rotor loops makes a field pattern that is a stepwise approximation to a triangle wave. As more pairs are added, the approximation gets better and better.

Figure 4: Shape of the circumferential B-field around the rotor from a set of 4 rotor coils equally spaced around the rotor. Only the fields for slip angle ( Stω ) of zero and 180 degrees are shown for clarity. This is a

stepwise approximation of a triangle wave and the more loops are added, the better the approximation.

A consequence of the triangular shape of the rotor B-field is that the stator current has harmonic components at odd multiples of the line frequency even though the line voltage is nearly pure sinusoid. Harmonic currents in the power lines do not, in principle, draw net power from the mains but they do generally increase losses in the mains and in the motor. They are undesirable and much design effort goes into optimizing the stator winding design both to do its fundamental task of creating the sinusoidal stator fields and to minimize the response of the stator to the non-sinusoidal induced currents from the rotor under load.


9

Three effects reduce the harmonic content of the line currents. First, the stator windings are wound as a crude approximation to a cosine distribution. The net induced voltage will more nearly resemble a sinusoid because of the convolution of the winding with the rotor field. That is, any field pattern moving through a sinusoidally wound coil will be roughly sinusoidal.

Second, the drawings of the B fields on the surface of the rotor assume that the field changes abruptly in the stator each time a rotor bar crosses a stator slot. This is why the changes around the surface are steps. Howev-er, as the example rotor that was shown in class was built, its squirrel-cage has an axial angular twist that is slightly more than one stator pitch. This means the effect of the bar going by a stator slot is spread nearly linear-ly in time over almost two slot crossing times. This makes stator flux changes much smoother than depicted here. Smoother transitions reduce both torque fluctuation and harmonic generation in the stator currents.

Figure 5: Shape of the circumferential B-field around the rotor from a set of 4 rotor coils equally spaced around the rotor. Shows 45 deg. increments of slip angle ( Stω ).

Finally, a triangle wave has only odd harmonics. In a 3-phase system, the right connection to the power lines can prevent any line currents at harmonics that are a multiple of 3 times the line frequency. In a wye-wound motor, a harmonic field at an integer multiple of three times the fundamental induces exactly the same voltage varia-tion in each branch of the wye. If the center or star point of the winding is unconnected, then the power lines


10

appear to the motor as a delta connection. With that connection, the 3X-harmonic voltages induce no currents in the power lines; they only affect the voltage of the center point. (This possibility is the same reason that gen-erators are always wye-wound but connect to the primary of delta-wound transmission transformers. Wye-wound motors never have their common point connected to the common point of their source supply.)

Figure 6 shows a triangle wave overlaid by its Fourier series representation with components out to 27 times the fundamental frequency. One cannot see the difference. However, if the 3, 6, 9… harmonics are removed, a bent sinusoid with only about 6 % total harmonic distortion is formed. The figure compares the triangle voltage to both the wave shape of its induced currents and a pure sinusoid at the fundamental.

Figure 6: A triangle wave of the same form as B-field on the perimeter of the rotor and its Fourier series repre-sentation. The voltage and current induced in the stator cannot contain harmonics at multiples of 3 times the line frequency. The second curve shows the result of removing those harmonics from the series while the third curve shows the fundamental component for comparison with a pure sinusoidal waveform.

Transformer-based Motor Models: So far, all analysis has been based on placing the rotor coils on the sur-face of the rotor to make it possible to use the Lorenz force law. When the turns are sunk into slots in the rotor, the wires themselves do not have to sustain the full force of the torque and are much easier to secure to the rotor itself. Removing them from the gap allows for a much narrower gap as well. As with the DC motor, we do


11

not expect the general result to change very much, at least qualitatively. To make calculation easier and to re-move the explicit assumption of an exposed winding, we will develop a transformer model of the motor. By de-riving the model parameters experimentally, we can simulate the line current, torque and efficiency as functions of slip frequency and load by an energy argument. Another motivation for doing actual parametric fits to this model is that the model becomes the basis for simulating operation, control design, and overload response.

A formal approach to solving for the steady state currents and power in a motor begins with the observations that:

• At rest a motor with two rotor turns is a transformer with 5 coils, two of them shorted.

• The self-inductances of the windings are independent of the rotor position

• The mutual inductances of stator to rotor windings are proportional to the sine or cosine of the rotor angle relative to the magnetic axes of the stator coils.

• All stator windings have one resistance and all rotor windings a different resistance that is the same for all rotor loops.

• Flux induced in a coil is M PRML iΦ = and therefore the induced voltage is ( )M PRMdv L idt

= where PRMi

is the current in the inducing coil. (Note: the usual formula for the voltages in a transformer or even for a single coil follows from this if the inductance is independent from time, but that is not the case for a motor.)

With these observations, the steady state equations for a motor with two rotor turns can be written in matrix form. In discussing generator design, we found that for sinusoidally wound stator coils, the mutual inductance between those coils was 1

2M PL L= − where the inductance of one stator coil is PL . The inductance, RTL , is the

total inductance of a rotor winding including both what is fully coupled to the stator and what is rotor leakage current. This is greater than the RL inductance in the equations for motor power and torque. The latter is actu-

ally a leakage inductance and will be part of a parametric model. RL appears here through the relationship be-

tween PL , RTL and MRL based on what fraction of the stator flux is captured by the rotor. With these results,

the motor terminal equations become:


12

1 12 2

2 21 12 2 3 3

4 41 12 2 3 3

2 43 31

2

cos( ) sin( )cos( ) sin( )cos( ) sin( )

cos( ) cos( ) cos( ) 000

A P P P MR R MR R

B P P P MR R MR R

P P P MR R MR RC

MR R MR R MR R RTR

R

v L L L L t L tv L L L L t L t

d L L L L t L tvdt

L t L t L t LvLv

π π

π π

π π

ω ωω ωω ω

ω ω ω

− − − − − − − − − −=

− −= =

12 43 3 2

1

2

sin( ) sin( ) sin( ) 0

0 0 0 00 0 0 0

0 0 0 00 0 0 00 0 0 0

A

B

C

R

MR R MR R MR R RT R

AW

BW

CW

R R

R R

iiii

t L t L t L i

iRiRiR

R iR i

π πω ω ω

⋅ − −

+ ⋅

These equations are non-linear (because of the time-dependent terms in MRL ) and difficult to manipulate into

something useful for comparison to measurements. However, relatively minor manipulation by setting the sta-tor currents to be a set of balanced, three phase sinusoids at the line frequency shows two useful attributes of the solutions:

• All stator voltages and currents are at the line frequency regardless of the frequency in the rotor loops.

• When the rotor moves at its no-load speed, synchronously with the line frequency, the wye-equivalent circuit for one phase is a simple inductor with value 3

2PRM PL L= . (The three phase currents generate

the rotating B field that is responsible for torque. In the parametric model of the motor, they play the role of the primary magnetizing inductance in a standard, fixed transformer, hence the name PRML .)

Figure 7 shows the evolution of a parametric model for one phase of an induction motor. The matrix equations do not include core losses in the stator or resistive losses in its winding. The leakage inductance of the stator is again hidden in the mutual inductance of the rotor and stator. The top section of Fig. 7 shows the inductor PRML

predicted by the matrix equations together with resistors for core and winding loss and some inductance for the leakage inductance of the stator. This is the only model needed for no-load operation.

The middle circuit in Figure 7 shows a possible model for the interaction of the stator and a rotor with one pair of rotor loops. It consists of an ideal transformer loaded by a single R-L circuit representing the rotor loop pair.

The parameter ‘S’ is the fractional slip defined as S S

L L

fSf

ωω

≡ = . The scaled rotor inductance, ROTORL , is explicitly

the leakage inductance on the rotor side of the coupling transformer. The scaled effective resistance depends

on the slip and is given by ROTORRS

. The logic behind interpreting the slip to resistance ratio as the resistive part

of the transformer load begins by considering a design with LROTOR equal zero. If the rotor is locked, S = 1 and the


13

current through the secondary is inversely proportional to RR. Similarly, if the motor runs at synchronous speed, the current is zero because there is no slip field. That condition has S = 0 corresponding to infinite effective re-sistance. Presumably, at intermediate slips, the current is proportional to the difference between synchronous and actual speed, hence to S. Adding the rotor inductance should not change the effective resistance of the cir-cuit.

As additional loop pairs are added to a rotor, the power and torque from each pair add directly so the electrical power, both real and reactive, in the rotor must increase linearly with the number of such pairs. With a fixed voltage, the line voltage across PRML , dividing RROTOR and LROTOR by N increases the total power by a factor of N.

Thus, if this is an adequate model of a motor with two rotor loops, then it can be trivially extended to any num-ber of loop pairs by dividing the rotor parameters by the number of loop pairs. It is not necessary to do this divi-sion explicitly because the final model will not include this factor, but the model parameters will be chosen to match the performance of the fully assembled motor.

The bottom circuit in Figure 7 shows removing the ideal transformer by scaling the R-L load appropriately. There is no way to measure the turns ratio externally so there is little benefit in keeping the ideal transformer in the model. For this final version of a full parametric model to be satisfactory, it must be possible to select its pa-rameters to predict the power delivered to the rotor for both heat and mechanical energy at all values of slip.

To fit parameters to this model, it is first necessary to write expressions for heat and mechanical power in the rotor. Both loops of a right angle pair have the same mean square current so the power to heat the rotor

is,( )

2 2 2 22 0

22

92 SHEAT R R R

R S R

B r lP i R RR L

ωω

= =+

. The equation derived earlier for mechanical power can be rewritten

slightly to make it have a similar form:

( )2 2 2 20

2 2

9( )

L SSMECHANICAL R R

R S R S

B r lP T RR L

ω ωωωω ω

−= ⋅ =

+

. The total power de-

livered to the rotor is the sum of these two and takes the form:

2 2 2 2 2 2 2 2 20 0

2 2 2

9 9( ) 1 ( / )

L LRELECTRICAL HEAT MECHANICAL

R S R S R R R

B r l S B r lR SP P PR L S L R R

ω ωω ω

= + = =+ +

.

This is the form for the power from a voltage source with RMS voltage 03ROTRMS LV B rlω= connected to a resis-

tor of value RRS

through a single-pole, low-pass R-L filter.

The electrical model of Fig. 7 has the same functional dependence of power into the scaled rotor resistance on the driving voltage PRMV . It will correctly predict the same heat and mechanical energy as the physics-based ro-

tor model if the scaled values are chosen properly. It is a per-phase model so it must predict power dissipated in

the resistor ROTORRS′

as equal to one-third the total electrical power. Let PRMV be the RMS voltage across the


14

Figure 7: Electrical equivalent circuits for one of the three phase branches of the motor. At top is the stator excitation equivalent to the stator alone when the rotor is turning synchronously with the line frequency. In the middle is the standard transformer model with the rotor inductance and equivalent load resistance. The bottom circuit shows the rotor circuit impedance transformed to the primary side and broken into two load resistances to separate the mechanical energy from the joule heating of the rotor cage. The only mechanical energy in this model is the electrical power disipated in the slip-dependent resistor at the lower right.

cos( )LINE P LV V tω=

RS LS-LEAK

RCORE LPRM


RS LS-LEAK

RCORE

LPRM

ROTORRS

LROTOR


RS LS-LEAK

RCORE LPRM

( )1 ROTORS RS′−

ROTORL′

ROTORR′


15

magnetizing inductor LPRM, then equating the total power to what is dissipated in the resistor of the electrical model leads to:

( ) ( )

2 2

2 22

3 31 /

ROTORPRM PRMELECTRICAL

ROTORS ROTOR ROTORROTORL ROTOR

RV V SPS RL RR L

Sωω

′= =

′′ ′′ + ′+

This electrical model has the correct dependence of both mechanical and heating power on slip frequency when the time constant is chosen properly. Fig. 7 emphasizes the division of power into its thermal and mechanical components by dividing the equivalent rotor resistance into two parts. Similarly comparison of the expression for rotor heating from field analysis and the transformer model suggests that 0PRM LV B rlω∝ . This is an expres-

sion of the form dVdtΦ

= − where the missing proportionality coefficient is from the number of turns and their

distribution in the stator. Finally choose the parameter ROTORR′ to scale the power for the particular motor.

Figure 8: More drastically simplified model based on assumption that the voltage drop in the stator resistance and stator side leakage inductance is sufficiently small that it does not appreciably change the primary mag-netizing current.

Figure 8 shows an even simpler equivalent circuit than in Fig. 7. It is based on the same simplification made for the common transformer that the voltage drop in the components in series with the line meets two criteria. First, the primary magnetizing current itself produces no significant drop in those components. Second, the drop from the load is sufficiently small that the magnetizing current is not affected. It is easier to fit parameters to this simplified model and power calculations are easier from it. One advantage of the simplification is that the stator leakage inductance simply modifies the value of the scaled rotor inductance and their sum can be ob-tained from a locked rotor measurement. Of course, it is an empirical question whether the model is sufficiently accurate. It probably is at normal rated load.

An empirical fit of parameters is straightforward and is similar to the transformer extraction problem in your first lab. The stator resistance can be measured at DC. The core resistance and LPRM are derived from extrapolat-


RS LS-LEAK

RCORE LPRM

( )1 ROTORS RS′−

ROTORL′

ROTORR′


16

ing the line current at low slip to the no-load, synchronous values. (When the need for accuracy justifies the ef-fort, it is possible to drive the motor with another variable speed motor at the synchronous speed.) Over the normal range of motor use, it may be possible to get the rotor power simply by subtracting the zero slip current from the running current.

There are three ways to determine the rotor parameters, one exploiting locked rotor measurements and the other two separating the electrical measurements into rotor and stator parts. The first way is to use a locked rotor impedance measurement but that is confounded by the change in the value of RCORE when the rotor is sta-tionary. (When running with near zero slip, the rotor magnetic field is changing only at the slip frequency but is changing at the line frequency for a locked rotor. This causes higher losses from eddy currents and hysteresis.) The other method of finding the rotor time constant ( /R ROTOR ROTORL Rτ ′ ′= ) and resistance ( ROTORR′ ) is to fit

them to the slope of the measured power-torque curve.

The electrical model of Fig. 8, only includes the rotor power and within the rotor only the power dissipated in the bottom right side resistor represents the mechanical power and torque. To simplify how a parametric fit would be done, assume that the series resistance of the stator is negligible and that the leakage inductance of the stator is incorporated in the total effective scaled rotor inductance. By inspection, the magnitude squared of the voltage across the total resistance of the rotor is given by

( )

22

21 /PRM

S ROTOR ROTOR

VVL Rω

=′ ′+

and therefore the mechanical power is ( )

( )2

2

13

1PRM

MECHROTORS R

S SVPRω τ

−=

′+.

At low slip, 1

23 MECHROTOR PRM

PR VS

−∂ ′ ≈ ∂ and the maximum power point happens when

1R

S

τω

≈ . Finally, the

mechanical torque is ( )

2

231

PRM

L ROTORS R

V STRωω τ

=′+

.

It is straightforward to use steady-state mechanical measurements to get the two missing rotor parameters.

An alternate way to disentangle the rotor parameters from the electrical measurements is to measure the mo-tor current at zero slip or as close to that as practical. That current is from the magnetizing inductance and core losses. (See Fig. 7 top.) It is approximately constant in the normal slip range so can be subtracted from the measured current with slip. Increase the slip until there is a significant change in current, subtract the no-slip current and convert the difference into an impedance that is directly the reactance of the rotor inductance in

series with ROTORRS′

.

The other piece of information for complete modelling is the moment of inertia of the rotor. The moment of inertia determines how the motor responds to the torque. The dimensions and mass of the rotor and shaft de-termine that value.


17

The Single-phase Induction Motor: Small motors from 1/5 to 3 H.P. are often single phase, having the ad-vantage of being usable where three phase power is not available. Three phase motors are from 80 % to 97 % efficient but single phase motors are generally less efficient because they make poorer use of their “environ-mental” B field. The split-phase motor of your first motor lab is an example of this problem. It is so inefficient that the sale of newly manufactured motors of this type is no longer permitted under US Department of Energy regulations.

The split-phase motor has a single stator coil that is wound sinusoidally in the same way as one of the stator coils of the three phase motor. (The stators of these motors tend to be very small, often with 4- poles. The re-sult is very crude approximations to a sinusoidal field.) The stator B field induced by connecting this motor to the power mains is given by:

[ ]10 0 2cos( ) cos( ) cos( ) cos( )B B t r B t t rθ ω θ ω θ ω= = + + −

.

Notice that this is a stationary magnetic field but it can be regarded as two, counter-rotating fields. A rotor at rest in this field has slip equal to 1 in both directions. There is no applied torque because the two rotating fields are equal strength pulling in opposite directions. In this condition, the rotor current is very high and the rotor heats up rapidly.

However, should we find a way to set the rotor into motion even slowly in one direction, say counterclockwise, the counterclockwise torque from the cos( )Rtθ ω− term increases because the slip is decreasing in that direc-

tion. Similarly the clockwise torque decreases as the clockwise slip is increasing. The difference is now sufficient to accelerate the rotor in the CCW direction until the motor reaches peak torque. Further acceleration then de-creases the torque until the motor reaches a balance between torque and load.

In order to be self-starting, single phase motors are equipped with a second set of stator coils. This “starter” winding is offset from the “run” winding in the same way the three-phase windings are from each other but the spacing is 90 electrical degrees from the pole instead of 120 degrees. This starter winding is driven with a current as closely to 90 degrees out of phase to the run winding as feasible. (Since the power is single phase, the quadrature current has to be generated in a passive circuit, which is a hit or miss problem.) If one coil is driven

by a current propotional to ( )cos tω , its contribution to the radial B field will be ( )0 cos( ) cosrB B tθ ω=

where θ is the angle of rotation from the position of the peak magnetization from that coil. Driving the second coil with a current in quadrature would make the sum of the fields be

( ) ( )0 0cos( ) cos sin( )sin cos( )rB B t t B tθ ω θ ω θ ω= + = − . This is the same rotating field as is generated in

a three phase motor and therefore this motor will behavior much the same way as the three phase motor.

In a split phase motor, the phase shift in the start winding relative to the run winding is generated by making the start winding with a higher resistance (use a smaller wire diameter) and its inductance may be lower. Both possibilities lower the L/R time constant producing a phase shift. However, this can only produce a small phase


18

shift. As a result the phase difference is sub optimal and the winding will heat excessively if left connected in normal operation. The start winding, therefore, is disconnected once the motor is in motion by a centrifugal switch mounted on the end of the motor shaft where the line connections come in.

It is easier if more expensive to change the phase in the start winding by putting a capacitor in series with the start winding. Such motors are said to be “capacitor-start” induction motors. It is straightforward to show that the capacitive reactance has to be higher than the series equivalent inductance of the winding and that one may need some additional resistance in series with the capacitor. The use of such a capacitor markedly improves the starting torque.

It is also common to improve the efficiency as well as the starting torque of a single-phase motor by using a second capacitor that is left in series with the start winding during normal operation. This provides a continuing quadrature current in the second winding so the counterrotating stator field is minimized and the desired rotating field is enhanced. The need for two capactors reflects the fact that the impedance of the winding is very different for the stationary and running conditions. Phase shifts produced by passive circuits only set an exact phase difference for one particular load configuration and do not generally also make fields that are amplitude matched. However, this technique does significantly raise the motor efficiency.

Torque, Speed and Power in the Split Phase Motor: Calculation of the torque and power relations for single phase motors is more difficult than for a three phase motor. The application of such motors does not generally require sophisticated modelling. The principal purpose for considering a model here is to make it possible to understand the measurements you make in a lab on the behavior and efficiency of such a motor.

Split-phase motors have two counterrotating stator fields because the stator field itself is not rotating. The transformer model then is two rotor L/R circuits in parallel with one of them having slip S and the other slip

( )2 S− . Figure 9 shows a straightforward extension of the 3-phase model to account for the split-phase fields.

Figure 9: Transformer model of a split-phase version of a single phase motor. This motor has two rotating B fields that push the rotor in opposite directions. The dotted line marks a division between the dominant rotor rotation direction to the left and the counter torque direction on the right.

LINEV

RS LS-LEAK

2 ROTORRS′

2 ROTORL′

CORER 2 PRML

2(2 )

ROTORRS

′−

2 ROTORL′

2 PRML


19

The locked rotor and zero-slip models derived from Fig. 9 are shown in the upper and lower parts of Figure 10 respectively. There is only a small difference between the two, namely the factor of two on the ROTORL′

component. That factor reduces the current in the high-slip side of the rotor/stator interaction by a factor of 4 from the locked rotor system. While the zero-slip current and power are much higher in the single-phase motor, the torque-slip relation is essentially the same as in the three phase motor.

For your split-phase motor lab, you may assume the free running current and power represent the zero-slip condition. Adding another R-L section for the low-slip interaction in parallel with the zero-slip model reduces the problem of modelling to calculations similar to the three phase power and torque extraction from Fig. 8. That discussion offers two ways to extract the rotor time constant ( /R ROTOR ROTORL Rτ ′ ′= ) and resistance ( ROTORR′ )

come from the measured power and torque curves.

Figure 10: Top is locked rotor model and bottom is S = 0 no-load line frequency synchronous operation.

LINEV

RS LS-LEAK

ROTORR′

ROTORL′

CORER PRML

LINEV

RS LS-LEAK

ROTORR′

2 ROTORL′

CORER PRML

how an induction motor works by equation s (and physics)...field induces a current. that current in...

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