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Finite element analysis of the induction motor

N. Bianchi L. Alberti

June 1, 2006

1 IntroductionThis example shows the computation of a threephase induction motor bymeans of finite element method. Two analyses are carried out: at firstthe noload test, then the lockedrotor test are simulated. At noload,magnetostatic simulations are carried out varying the magnetizing current.In the lockedrotor test, the frequency is changed so as to compute thedependence of the rotor parameters on the operating frequency.

From these two tests, the parameters of the equivalent circuit of theinduction motor are obtained, from which the motor performance can bepredicted. Among the others, the power and the torque produced, the stator

current and the power factor as a function of the rotor slip.Let us refer to the circuit sketched in Fig. 1, where

Rs () stator inductance,Ls (H) stator leakage inductance,Rr () rotor resistance referred to the stator,Lr (H) rotor leakage inductance,Lm (H) magnetizing inductance.

The parameters Ls and Rs have to be known from measurement or fromanalytical calculation.

R (s)r

L (s)r

L (im)m

L s R s

Eim

is ir

Figure 1: Equivalent circuit of the induction motor

Department of Electrical Engineering, University of Padova, [email protected]

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In this circuit, the magnetizing inductance Lm is considered to be a non

linear function of the magnetizing current Im and the rotor parameters L2and R2 are considered to be a nonlinear function of the rotor slip.

2 Definitions and preprocessing

The structure of the induction motor under analysis is shown in Fig. 2 whichalso highlights the materials of the various blocks. Particular care has to begiven to the materials used in the simulations. In particular

Copper The stator slot is filled by several wires. However, it is consideredthat the slot is completely filled by copper, considering an equiva-

lent conductive bar. In this bar the total current of all the wires isassigned. In the magnetodynamic computation, this yields a nonuniform distribution of the stator current within the slot, accordingto the operating frequency. In order to avoid this mistake, the con-ductivity of the copper can be equated to zero, yielding an infinitepenetration thickness.

Iron The iron nonlinear characteristic (the BH curve) has to be used inthe noload simulation. However the higher values of the curve are notalways available, and they are interpolated during the finite elementanalysis. It is convenient to assign the high values of flux density B and

magnetic field H when they are not given, extending the curve witha slop close to the 0. An example of BH curve is given in Table 1.At last, in the magnetodynamic simulations of the lockedrotor testsa linear characteristic for the iron can be adopted, so as to limit theresearch to the effect of the frequency on the rotor parameters.

Aluminum The rotor bars are completely filled by Aluminum. Its mainparameter is the conductivity, which determines the rotor resistance.It is convenient to set its value considering the operating temperatureof the rotor, thus a reasonable value is Al=15 MS/m (correspondingto 120 K temperature rise).The resistive effect of the two external rings can be taken into accountby means of a further reduction of the conductivity Al as

Al,eq =Al

1 + kring(1)

where the factor kring is defined as

kring =2Qr

(2p)2DringLstk

SbarSring

(2)

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Table 1: Example of BH curveB H B H B H

(T) (A/m) (T) (A/m) (T) (A/m)

0 0 1.58 2000 1.80 100000.3 200 1.64 3000 1.805 110000.6 400 1.68 4000 1.810 12000

0.90 600 1.72 5000 1.815 140001.05 700 1.74 6000 1.82 150001.20 800 1.76 7000 2.00 870001.35 900 1.78 8000 2.25 1480001.50 1000 1.79 9000 2.50 215000

Figure 2: Materials in the simulation

It is useful to group the various objects together, so as to facilitate thepostprocessing of the solution. The convention adopted by the authors isdescribed hereafter:

Rotor All the parts of the rotor are fixed to belong to the group numbered10, as also shown in Fig. 3.

Stator All the parts of the stator except for the slots are fixed to belongto the group 1000, as shown in Fig. 4.

Slots The stator slots are numbered in counterclockwise direction start-ing on the right-hand side. They are labelled Slot1, Slot2, ..., andnumbered 1001, 1002, ..., as shown in Fig. 4. These labels allow torecognize each single slot for both the automatic assignment of the slotcurrent and the computation of the corresponding flux linkage once thesolution has been carried out.

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Figure 3: All otor parts belong to group 10

Figure 4: Stator (group 1000) and numbers assigned to the stator slots

2.1 Slot matrix

An useful tool for processing the stator windings is the slot matrix. Itis formed by m matrix vectors (where m is the number of phases) whoselength is equal to the number Qs of stator slots. The element of thesevectors describes how the different phases fill the stator slots, assuming avalue ranging from +1 to 1.

For instance, referring to the aphase and the jth slot, it is

kaj = +1 the jth slot is completely filled by conductors of the aphase;

kaj = 0 there are no conductors of the aphase within the jth slot;

kaj = 1 the jth slot is completely filled by conductors of the aphase, but withnegative polarity.

An example of slot matrix is reported in Table 2. It refers to the 24slotstator shown in Fig. 4, and a singlelayer fullpitch winding.

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Table 2: Example of slot matrixslot no. ka kb kc slot no. ka kb kc slot no. ka kb kc

1 0 1 0 9 0 0 1 17 1 0 02 0 1 0 10 0 0 1 18 1 0 03 0 1 0 11 0 0 1 19 1 0 04 0 1 0 12 0 0 1 20 1 0 05 1 0 0 13 0 1 0 21 0 0 16 1 0 0 14 0 1 0 22 0 0 17 1 0 0 15 0 1 0 23 0 0 18 1 0 0 16 0 1 0 24 0 0 1

2.2 Automatic setting of slot currentsThe conventions assumed above above allows an easy automatic analysis ofthe machine. The slot currents are set as follows. The variable Isim meansthe rms value of the current fixed for the simulation. The example refers toLUA code for FEMM tool.

-- series conductors per slot

nc = 82

-- slot number (simulated)

Qs = 24

-- slot matrix

ka= { 0, 0, 0, 0, 1, 1, 1, 1, 0, 0, 0, 0,

0, 0, 0, 0, -1, -1, -1, -1, 0, 0, 0, 0}

kb= {-1, -1, -1, -1, 0, 0, 0, 0, 0, 0, 0, 0,

1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0}

kc= { 0, 0, 0, 0, 0, 0, 0, 0, -1, -1, -1, -1,

0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1}

-- phase currents (real and imaginary part)

Ia_re = sqrt(2) * Isim

Ib_re = sqrt(2) * (-Isim/2)

Ic_re = sqrt(2) * (-Isim/2)

Ia_im = 0

Ib_im = sqrt(2) * (-sqrt(3) * Isim/2)

Ic_im = sqrt(2) * (sqrt(3) * Isim/2)

for q = 1, Qs, 1 do

Islot_re = nc * (Ia_re*ka[q] + Ib_re*kb[q] + Ic_re*kc[q])

Islot_im = nc * (Ia_im*ka[q] + Ib_im*kb[q] + Ic_im*kc[q])

modifycircprop("Islot" .. q, 1 , Islot_re)

modifycircprop("Islot" .. q, 2 , Islot_im)

end

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3 Noload test

The simulations are carried out at zero frequency, assuming to work in therotor reference frame. This corresponds to a rotor slip s equal to zero. Thusa magnetostatic field problem is solved. Stator currents are imposed andthe nonlinear behavior of the magnetizing inductance is computed. A seriesof simulations are carried out varying the stator current.

The nonlinear characteristic of the iron has to be used in the simulation.Fig. 5 shows the flux plot during the noload test.

Figure 5: Map of flux density under noload test

0 0.5 1 1.5 2 2.5 30

0.2

0.4

0.6

0.8

1

Noload test

Stator current (A)

Fluxlinkage

(Vs)

Inductance

(H)

Flux linkage (Vs)Magnetizing inductance (H)

Figure 6: Flux linkage versus stator current

Fig. 6 shows the phase flux linkage and the magnetizing inductance as afunction of the stator current. The effect of the saturation is evident.

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0 0.5 1 1.5 20

100

200

300

400

500Noload test

Stator current (A)

Phase

tophase

voltage

(V)

Figure 7: Voltage versus stator current

0 0.5 1 1.5 2 2.5 30

0.5

1

1.5

2

2.5Noload test

Stator current (A)

Magneticenergy(J)

Wm

= int H.dB

0.5 * int A.J

Figure 8: Energy and coenergy versus stator current

0 0.5 1 1.5 2 2.5 30.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1Noload test

Stator current (A)

Magn

etizinginductance(H)

Lmag

from

Lmag

from Wm

Lmag

from A.J

Figure 9: Magnetizing inductance versus stator current

Referring to the aphase, the magnetizing inductance is computed as

Lm =aIa

(3)

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where Ia is the current and a is the flux linkage.

Fig. 7 shows the phasetophase noload voltage that can be computedfrom the noload test. In the same figure the measured voltage is shownusing circles. An appreciable agreement between simulated and measuredvoltage is observed. The higher difference is found in the saturation region.It is attributed to the difference in the magnetic characteristic that is usedin actual motor and in simulation.

Fig. 8 shows the magnetic energy versus the stator current. It is worthnoticing that different computation yield different results in the saturationregion. The computation of the whole magnetic energy

Wm = vol

B

0

H dBdvol (4)

can not be used. Its value becomes lower and lower with the saturation andcan not more used for the computation of the magnetizing inductance. Onthe contrary, the apparent energy

WAJ =

vol

A Jdvol (5)

can be used to estimate the apparent magnetizing inductance of the machine.It can be also observed that

WAJ = Wm + W

m (6)

where Wm is the magnetic coenergy

Wm =

vol

H0

B dHdvol (7)

A numerical comparison is given in Table 3. At low currents it is observedthat Wm = W

m =1

2WAJ, while this is not hold at higher currents.

As confirmation of this remark, Fig. 9 shows the magnetizing inductancecomputed using three different ways. Using the fluxlinkage or the energy(5) yield a correct value of its apparent inductance:

Lm =2

3

WAJ

I20

(8)

On the contrary, using the magnetic energy (4), the inductance obtaineddoes not make sense.

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Table 3: Magnetic energy comparisonIm Wm W

m WAJ(A) (J) (J) (J)

0.20 0.056 0.056 0.1120.38 0.201 0.202 0.4050.56 0.411 0.436 0.8480.74 0.605 0.738 1.3440.92 0.743 1.082 1.8261.1 0.843 1.449 2.293

1.28 0.920 1.830 2.7521.46 0.986 2.222 3.2091.64 1.044 2.621 3.6661.82 1.100 3.026 4.1272.00 1.154 3.438 4.593

4 Lockedrotor tests

These tests are carried out assuming a lockedrotor and imposing a fixedcurrent within the stator slots. A series of simulations is carried out atvarious frequencies so as to weigh up the dependence of the rotor parameterson the operating frequency.

In these simulations the iron is assumed to be linear, so as to allow the

superposition of the effect to be applied. The corresponding magnetizinginductance of the equivalent circuit of Fig. 1 is fixed to its linear value. Thisassumption does not affect the computation since the main purpose of thistest is to compute the rotor parameters.

From each finite element analysis the following quantities are computed:

f (Hz) FrequencyIs (A) Currenta (Vs) aphase flux linkage (real and imaginary part)b (Vs) bphase flux linkage (real and imaginary part)c (Vs) cphase flux linkage (real and imaginary part)

T (Nm) Torque computed from Maxwell stress tensorPjr (W) Rotor Joule lossesWm (J) Magnetic energyWm (J) Magnetic coeneryWAJ (J) Integral ofA J

As an example, Fig. 10 and Fig. 11 show the flux plots during the lockedrotor test at two different frequency: at 10 Hz and at 50 Hz, respectively.It is worth noticing the higher shielding effect of the rotor current at thehigher frequency.

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Figure 10: Map of flux density under locked rotor test (10 Hz)


Fig. 12 shows the magnetic energy computed in different ways. In this

case, the motor works in linear conditions, so that the computation yieldthe same result.Fig. 13 shows the computed torque as a function of the working fre-

quency. It is worth noticing that the torque computed here refers to aconstant current source, fixed in all the simulation, and then it has not tobe confused with the torque obtained at fixed voltage to plot the mechanicalcharacteristic of the machine.

The Maxwell stress tensor is computed along a line within the airgapof the motor. The torque is also proportional to the power transferred fromthe stator to the rotor. It is equal to the ratio between the rotor Joule losses

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0 10 20 30 40 500

0.01

0.02

0.03

0.04

0.05

0.06Locked rotor test

Rotor frequency (Hz)

Energy

(J)and

coenergy

(J)

int H.dBint B.dHint A.J

Figure 12: Energy versus frequency

0 10 20 30 40 500

0.01

0.02

0.03

0.04

0.05



Torque(Nm)

from Maxwell tensorfrom rotor losses

Figure 13: Torque versus frequency

0 0.2 0.4 0.6 0.8 10

20

40

60

80

100

120Rotor parameters

slip

Resist

ance()

Reactance()

Figure 14: Rotor parameters obtained from locked rotor test

Pjr and the synchronous speed, given by

T =Pjr

2f/p(9)

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where f is the frequency adopted in the simulation and p is the pole pair.

Then 2f/p is the synchronous speed of each simulation.Finally, Fig. 14 shows the rotor parameters that are obtained from the

lockedrotor simulations corresponding to each operating frequency f. Theequivalent parameters are computed as

Req =Pjr3I2

Leq =2

3

WmI2

(10)

where I is the rms current used in the simulation.

R (s)2

L (s)2

L mR (s)eq

L (s)eq

Figure 15: Equivalent circuits corresponding to the lockedrotor simulation

Referring to Fig. 15, from the equivalence of the two circuits, it results

Lr = LmLeq(Lm Leq) (Req/)

2

(Lm Leq)2 + (Req/)2

Rr = ReqLm + LrLm Leq

(11)

At last, let us remark that the leakage inductance of the stator andthe rotor can not be separated. The meaning of Lr is the total leakageinductance, of the twodimensional model, referred to the stator.

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5 Induction motor performance

From the noload tests and the lockedrotor tests, the parameters of theequivalent circuit of the induction motor are obtained. In particular themagnetizing inductance (see Fig. 9) and the rotor parameters (see Fig. 14).

The equivalent circuited is completed adding the elements that are notconsidered in the finite element analysis: the stator winding resistance andthe end winding leakage inductance. Then, fixing a value of the phasevoltage, for instance V=230 V, the computation of the motor performanceis carried out.

0 0.2 0.4 0.6 0.8 10

500

1000

1500

2000

2500

3000

3500From equivalent circuit

slip

Mechandelectricpower(W)

Figure 16: From equivalent circuit: input and output power versus slip

0 0.2 0.4 0.6 0.8 10

1

2

3

4

5


slip

Torque(Nm)

Current(A)

Torque (Nm)Current (A)

Figure 17: From equivalent circuit: torque versus slip

Fig. 16 shows the electrical power (input power) and the mechanicalpower (output power) of the motor versus the rotor slip. A good agreementbetween the simulated power (solid line) and the measured power (triangles)is evident, even though the iron losses and the mechanical losses are not

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0 0.2 0.4 0.6 0.8 10.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9


slip

cos

Figure 18: From equivalent circuit: power factor versus slip

0 0.2 0.4 0.6 0.8 10

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8From equivalent circuit

slip

Efficiency

Figure 19: From equivalent circuit: efficiency versus slip

considered in the computation. As a consequence, this justifies the lowervalues of the predicted mechanical power and of the predicted torque.

Fig. 17 shows the torque produced by the motor and the stator currentas a function of the rotor slip. Once again, the agreement between simulatedvalues (solid line) and measurements (triangles and circles) is evident.

Fig. 18 shows the power factor (cos ) versus the rotor slip. The mea-sured values are reported using triangles. Finally, Fig. 19 shows the motorefficiency (without considering iron and mechanical losses) versus the rotorslip.

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6 A second example: a 3.7kW 2pole induction

motor

This section reports the simulation of a 3.7kW 2pole induction motor,adopting the same procedures illustrated above. However, in this case, wetake advantage of the symmetry of the machine and only a half of motoris studied. This is possible by imposing an odd periodicity (or an antiperiodicity) along the two boundaries of the machine as illustrated in Fig. 20.

Figure 20: Antiperiodicity along the two boundaries

6.1 Noload test

The simulations of noload test are carried out at zero frequency. Fig. 21shows the corresponding flux plot.

Figure 21: Map of flux density under noload test

Fig. 22 shows the phase flux linkage and the magnetizing inductanceas a function of the stator current. Fig. 23 shows the phasetophase noload voltage. In the same figure the measured voltage is shown using circles.Fig. 24 shows the magnetic energy versus the stator current. At last, Fig. 25shows the magnetizing inductance computed from fluxlinkage, energy (5)and the magnetic energy (4). As above the last computation does not makesense.

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0 5 10 15 20 250

0.2

0.4

0.6

0.8

1

1.2

1.4Noload test

Stator current (A)

Fluxlinkage

(Vs)

Inductance

(H)

Flux linkage (Vs)Magnetizing inductance (H)

Figure 22: Flux linkage versus stator current

0 5 10 15 200

100

200

300

400

500Noload test

Stator current (A)

Phase

tophasevoltage

(V)

Figure 23: Voltage versus stator current

0 5 10 15 20 250

5

10

15

20Noload test

Stator current (A)

Magneticenergy(J)

Wm

= int H.dB

0.5 * int A.J

Figure 24: Energy and coenergy versus stator current

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0 5 10 15 20 250

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

0.45Noload test

Stator current (A)

Magnetizinginductance(H)

Lmag

from

Lmag

from Wm

Lmag

from A.J

Figure 25: Magnetizing inductance versus stator current

6.2 Lockedrotor tests

As above these tests are carried out assuming a lockedrotor and imposinga fixed current within the stator slots, but varying the operating frequency.The lamination iron is assumed to be linear. Fig. 26 shows the flux plotsduring the lockedrotor test at 50 Hz.


0 10 20 30 40 500

0.1

0.2

0.3

0.4

0.5

0.6



Energy(J)and

coenergy(J)

int H.dBint B.dHint A.J

Figure 27: Energy versus frequency

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0 10 20 30 40 500

0.05

0.1

0.15

0.2

0.25

0.3



Torque(Nm)

from Maxwell tensorfrom rotor losses

Figure 28: Torque versus frequency

0 0.2 0.4 0.6 0.8 10

10

20

30

40

50

60Rotor parameters

slip

Resistance()

Reactance()

Figure 29: Rotor parameters obtained from locked rotor test

Fig. 27 shows the magnetic energy computed in different ways. Fig. 28shows the torque computed by means of the Maxwell stress tensor and therotor Joule losses as a function of the working frequency, using a constantstator current. Finally, Fig. 29 shows the rotor parameters that are obtainedfrom the lockedrotor simulations corresponding to each operating frequencyf.

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6.3 Induction motor performance

From the noload tests and the lockedrotor tests, the parameters of theequivalent circuit of the induction motor are obtained. This equivalent cir-cuited is completed adding the stator winding resistance and the end wind-ing leakage inductance. Then, the computation of the motor performance iscarried out for any given stator voltage, i.e. V=230 V, .

0 0.2 0.4 0.6 0.8 10

2000

4000

6000

8000

10000

12000

14000

16000


slip

Mechandelectricpower(W)

Figure 30: From equivalent circuit: input and output power versus slip

0 0.2 0.4 0.6 0.8 10

10

20

30

40


slip

Torque(Nm)

Current(A)

Torque (Nm)Current (A)

Figure 31: From equivalent circuit: torque versus slip

Fig. 30 shows the electrical power (input power) and the mechanicalpower (output power) of the motor versus the rotor slip. The measuredpower is reported by means of triangles, highlighting a good prediction.

Fig. 31 shows the torque produced by the motor and the stator currentas a function of the rotor slip. The measured currents are reported by meansof circles while the measured torque by means of triangles.

Fig. 32 shows the power factor (cos ) versus the rotor slip. The mea-sured values are reported using triangles. Finally, Fig. 33 shows the motor

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0 0.2 0.4 0.6 0.8 10.3

0.4

0.5

0.6

0.7

0.8

0.9


slip

cos

Figure 32: From equivalent circuit: power factor versus slip

0 0.2 0.4 0.6 0.8 10

0.2

0.4

0.6

0.8


slip

Efficiency

Figure 33: From equivalent circuit: efficiency versus slip

efficiency (without considering iron and mechanical losses) versus the rotorslip.

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