Analysis the Effect of Unbalance Rotor Winding Torque and Speed Induction Motor

ANALYSIS THE EFFECT OF UNBALANCE ROTOR WINDING TORQUE AND

SPEED INDUCTION MOTOR

Windings on the rotor can be equipped with an external 3 phase resistance may not be

balanced. And also, the broken bars on the shell of the rotor can cause the motor winding

resistance is not balanced. The figure below shows a coil winding rotor which has a

resistance that is not balanced.

Image of induction motors with unbalanced rotor windings

On the subject had previously been on the explain that if there is a series of three

phases that have a resistance that is not balanced then there will be components in the

circuit is symmetrical. Then there will be symmetrical components in the circuit of the rotor

above the amount shown in the following equation:

I ar 0=13

( I ar + I br + I cr )

I ar 1=13

( I ar + a I br + a2 I cr )

I ar 2=13

( I ar + a2 I br + aI cr ) = 0

M. Azhary Siregar. St ([email protected] for FB and Email)

mailto:[email protected]

From the equation above, gained large rotor currents (I2) for the positive sequence (I21) and

negative sequence (I22), and also in getting on the rotor voltage (V2) of:

V ar=¿ -Zar I ar ; V br=¿ -Zbr I br ; V cr=¿ -Zcr I cr

In the first approach, all rotor currents have a frequency f2 = Sf1 at steady state

(fixed). Forward motion of magnetic force, generated by Iar1, Ibr1, Icr1, interacting normally

with the windings.

I 21 R2−V 21=− j ω1 Sψ21;ψ21=L2 I21+Lm I 11

I 11R1−V 11=− j ω1 ψ11 ;ψ11=L1 I 11+Lm I 21

I 11=− j ω1 Lm I 11+V s

R1+ j ω1 L1

Component of magnetic force that retreated from the spinning rotor currents of the

stator located on n1 speed '. The n1’ 'is:

n1 '= n – Sf 1

p =

f 1

p ( 1 - 2S )

Makai this will induce electromotive force at a frequency f1’ = f1 (1-2S). Style

retreat is produced by I22, which will produce the opposite torque.

I 22 R2−V 22=− j ω1 Sψ22;ψ22=L2 I22+Lm I 12

I 12 R1=− j ω1(1−2S )ψ12;ψ12=L1 I 12+Lm I 22

I 12=− jω1(1−2S )Lm I 22

R1+ j ω1(1−2S)Ls



Where:

L1 = Ls + Lm; L2 = Lr + Lm

Where

I21 = forward rotor current (Ampere)

I22 = backward rotor current (Ampere)

I11 = forward stator current (Ampere)

I12 = backward stator current (Ampere)

R2 = rotor resistance (Ω)

R1 = stator resistance (Ω)

V (21) = fordward rotor voltage (Volt)

V (22) = backward rotor voltage (Volt)

Vs = Voltage source (Volt)

L1 = stator mutual inductance (H)

L (2) = rotor mutual inductance (H)

Lr = rotor inductance (H)

Ls = stator inductance (H)

Lm = Magnetic Inductance (H)

ω1 = velocity of flow phase angle frequency (rad / s); ω1 = 2 π F1

ψ_21 = positive sequence rotor flux linkage (Wb)

ψ_22 = negative sequence rotor flux linkage (Wb)

S = Slip

Known torque is:

Te = Pg

ωs ; ωs =

ω1

P 1 (Rad/s)



=3P 1 I2

2 R2

ω1 S

Where P1 = Number of pole pairs

From equeation I 21 R2−V 21=− j ω1 Sψ21;ψ21=L2 I21+Lm I 11 , rotor in short circuit if the

voltage on the rotor (V2) = 0, then:

I 21 R2=− j ω1 Sψ21

By adding his I 21¿ (I21 conjugate) to the equation above then:

I 21¿ I 21 R2=− j ω1 Sψ21 I 21

¿

By combining the real and imaginary numbers above obtained equation:

I 21¿ I 21 R2+ jω1 Sψ21 I 21

¿

So Pg are:

Real = 3 (I 21¿ I 21 R2 )

Imaginer = 3 j ω1 Sψ21 I 21¿

Substituting equation Te = Pg

ωs

to the equation above, we obtain:

Te = 3P 1 Imag(ψ 21 I21¿)

By lowering the equation above, we obtain:

Te = 3P 1 Imag [ ( L2 I 21+Lm I 11) I 21¿ ]

= 3P 1 Imag (Lm I11 I 21¿)



= 3P 1Lm Imag ( I11 I 21¿)

The equation above also applies to the equation I 22 R2−V 22=− j ω1 Sψ22;ψ22=L2 I22+Lm I 12 which

produces torque opponent.

So the torque expression is:

T e= 3 P1 Lm [ Image ( I 11 I 21¿ )+ Image ( I12 I 22

¿ )]= T e1+T e2

Where torque is the sum of forward (Te1) and reverse torque (Te2). To Te1 using

symmetrical component sequence "1" and to Te2 use symmetrical component sequence "2".

Pictures of the slip torque curve

Reverse torque component is positive (motoring) of 1-2s <0 or S> 0.5 and negative

(breaking) at S <0.5. In the beginning, the torque component will move backwards

(motoring). And also for S = 0.5, turning torque is 0, because the induction electromotive

force to this situation occurred at a frequency f1 = f1 (1-2S), if S = 0.5, the F1 '= 0 and does

not occur induction in these circumstances. Behind Torque Torque is also called George or

monoaksial.



Experiments are done at the Laboratory of Electrical Engineering Conversion USU.

To get results from the influence of an unbalanced rotor resistance is carried out experiments

by using a balanced rotor resistance. Results from the experiment can be compared between

the prisoners of a balanced rotor with an unbalanced.

Previous attempts to obtain necessary parameters - the parameters of the induction motor in

order to analyze the effect of rotor imbalance.

Tools used in this experiment is

1. Three-phase induction motor

type: rotor windings

motor specifications: - AEG TYP C AM 112MU 4RI

- Δ / Y 220/380 V 10.7 / 6.2 A

- 2.2 Kw, 0.67 cosφ

- 1410 rpm, 50 Hz

- isolation B

2. Ampere meter

3. Volt Meter

4. Prisoners Slide

5. Watt Meter 3φ

6. AC and DC voltage source



DC Resistance Experiment

A. DC Resistance Experiments On The stator winding

Experiment series

Experimental Procedure

Stator winding connections are made ties Y. to be measured are two of the three stator

windings.

The series of stator windings connected to supply DC voltage

DC supply voltage is increased until at a certain value.

When the voltage showing on the amount of 15.4 volts, the appointment of voltmeters

and ammeters gauge recorded

If you've completed a series of removable.



Experiment Results Data

Rdc =

VI (Ω)

Phasa V (volt) I (Ampere)

U – V 12,89 4,2

Data Analysis

For the above data was obtained:

Rdc =

VI

= 12,89

4,2

= 3,07 Ω

Because the relationship of the stator is Y, then the RDC is

Rdc = 3,07

2

= 1.535 Ω

Rac = 1.2 x 1.535



= 1.842 Ω

So the stator resistance is

Rs = 1842 Ω

B. DC Resistance Experiments on Rotor winding

Experiment series

Image of DC resistance in the rotor experiment


Rotor winding connections made Y relationship, which will be measured are two of

the three rotor windings.

Circuit rotor windings connected to supply DC voltage Raise the DC supply voltage

slowly, until at a certain value.

When the voltage showing on the amount of 3.5 volts, the appointment of voltmeters

and ammeters gauge recorded

if you've finished circuit is removed.




Phasa V (volt) I (Ampere)

K – M 2,38 3,4

Data Analysis

For the above data was obtained:

Rdc =

VI (Ω)

= 2,383,4

= 0.7 Ω

Because the relationship on the rotor is Y, then the RDC is

Rdc = 0.72

= 0.35Ω

Rac = 1.2 x 0.35



= 0,42 Ω

So the rotor resistance is

Rr = 0.42 Ω

Suspended Rotor Experiment (Block Rotor)

Experiment series

From the data obtained in the measurement of rotor motors in a state of suspended or

short circuit then calculated Xs and Xr '. The series of measurements when connected briefly

shown in the figure below


The procedure is performed to obtain data on short circuit is:

Induction motor is coupled with direct current machine

All switches in an open state, the voltage regulator in the state minimum.



Switch S1 is closed, PTAC1 be increased so that the induction motor started to spin

slowly.

Switch S3 then closed, PTDC2 raised until the appointment of ammeters A3 reaches

the amplifier current price nominal direct current machine

Switch S2 is closed and PTDC1 raised so that the engine block direct current

induction motor and the rotation stops. Then the appointment of measuring devices

A1, W and T recorded

The measurement is repeated several times to get the best value.

Suspended Rotor Experiment Results Data

Vbr (Volt)IBR ( Ampere ) PBR ( Watt ) F1 (Hz) Fbr (Hz)

98

6,2 575 50 50

Data Analysis

From the above data is obtained:

Zbr=V Br

√ 3 I br

= 98

√3 x 6 .2

= 9.125 Ω



θBr=cos−1[ PBr

√3 xV Br x IBr]

= cos−1[ 5731052.4 ]

= 57o

X br=F1

FBr

x [ SinθBr x Zbr ]

= 5050

x [sin57o x 9.125 ]

= 7.6528 Ω

X s=0.5 x Xbr

= 3.8264 Ω ; Ls = X1

ω1 = 0,01218 H

X r=0.5 x Xbr

= 3.8264 Ω ; Lr = X2

ω1 = 0,01218 H

Experiment Zero Cost

The series of experiments




The procedure is performed to obtain data required are:

All switches open, the voltage regulator at minimum position.

Switch S1 is closed, PTAC1 increased slowly until the voltage of 350 volts.

When a voltage is 350 volts, the meter reading is recorded large ammeters and

wattmeters each phase.

Once recorded, the series is removed.

Data Analysis

ω1 =2 π 50=314

V 0 ( Volt ) P0 ( watt ) I 0 (Ampere)

350 300 3,33



X m=V 1

I nl √3−X1

(Ω)

= 350

3.33√ 3 −3.8264

= 56.85 Ω

Lm = Xm

ω1

= 0,181 H

Experimental ”ANALYSIS THE EFFECT OF UNBALANCE ROTOR WINDING

TORQUE AND SPEED INDUCTION MOTOR”

Experiment series

Chain series of experiments as Figure 4.5 above.

Create a relationship outside of detainees in relation Y.

Connect the external resistance to the rotor terminals, respectively - each prisoner out

for the price of 2 Ohm.

Close PTAC1 S1 which connects with the stator terminal and then raise PTAC1 until

the specified nominal voltage.



Close switch S and then raise up to A3 show the current PTDC1 amplifier

par.

Prisoner R made the maximum for which data are determined and S2 closed, then

note the appointment of A4, A5, A6, W, and T and n.

Lower the resistance R in stages according to the data that is specified, and

maintained for a constant voltage in V1. Note the appointment of A4, A5, A6, W, and

T and n.

Repeat procedure 4 to 7 with no one detainee zoom out with the values specified,

which is 3 Ω, 4 Ω, and 5 Ω, to obtain the stator resistance is not balanced.

Experiments carried out until the current asynchronous machine A1, A4, A5, A6, does

not exceed the nominal current.

Experiment completed.


Vin = 300 Volt (LL)

Vs = 300√ 3

= 173,2 Volt (LN)

Ra = 2,4 Ω

Rb = 2,4 Ω

Rc = 2,4 Ω

F1 = 50 Hz ; ω1 =2 π 50=314

P1=2

R (%) Nr Slip Pin (Kw)Torsi

(Nm)Ia (A) Ib (A) Ic (A)



20 1150 0,23 0,2 0,625 0,7 0,7 0,7

40 1100 0,26 0,26 0,75 0,8 0,8 0,8

60 1050 0,3 0,32 0,775 1 1 1

80 1000 0,33 0,35 1 1,1 1,1 1,1

100 950 0,36 0,45 1,125 1,3 1,3 2,3

Vin = 300 Volt (LL)

Vs = 300√ 3

= 173,2 Volt (LN)

Ra = 3,4 Ω

Rb = 2,4 Ω

Rc = 2,4 Ω

F1 = 50 Hz ; ω1 =2 π 50=314

P1=2



20 900 0,4 0,45 2,4 1,34 1,6 1,6

40 880 0,41 0,48 2,6 1,57 1,83 1,83

60 850 0,43 0,52 2,7 1,8 2,1 2,1



80 790 0,47 0,61 2,8 2,1 2,45 2,45

100 700 0,53 0,62 2,9 2,3 2,7 2,7

Vin = 300 Volt (LL)

Vs = 300√ 3

= 173,2 Volt (LN)

Ra = 4,4 Ω

Rb = 2,4 Ω

Rc = 2,4 Ω

F1 = 50 Hz ; ω1 =2 π 50=314

P1=2



20 800 0,46 0,6 2,75 3 3,3 3,3

40 750 0,5 0,62 3 3,2 3,8 3,8

60 720 0,52 0,63 3,25 3,3 4 4

80 700 0,53 0,65 3,6 3,55 4,4 4,4

100 500 0,66 0,74 4 3,8 4,8 4,8

Vin = 300 Volt (LL)



Vs = 300√ 3

= 173,2 Volt (LN)

Ra = 5,4 Ω

Rb = 2,4 Ω

Rc = 2,4 Ω

F1 = 50 Hz ; ω1 =2 π 50=314

P1=2



20 660 0,56 0,52 2,8 2,4 3,5 3,5

40 550 0,63 0,55 3,2 2,53 3,7 3,7

60 500 0,66 0,58 3,4 2,71 4 4

80 430 0,71 0,63 4,2 2,97 4,45 4,45

100 350 0,76 0,67 4,8 3,3 4,91 4,91

Data Analysis

- for Ra = 2,4 Ω, Rb = 2,4 Ω, Rc = 2,4 Ω

load = 20 %

Slip = 0,23

Te = 3P 1ω1

I 22 R2

S

= 3x 2314

0,72 2,40,23

= 0,0977 Nm



load = 40 %

Slip = 0,26

Te = 3P 1ω1

I 22 R2

S

= 3x 2314

0,82 2,40,26

= 0,112 Nm

load = 60 %

Slip = 0,3

Te = 3P 1ω1

I 22 R2

S

= 3x 2314

12 2,40,3

= 0,152 Nm

load = 80 %

Slip = 0,33

Te = 3P 1ω1

I 22 R2

S

= 3x 2314

1,12 2,40,33

= 0,168 Nm

load = 100 %

Slip = 0,36

Te = 3P 1ωs

I 22 R2

S



= 3x 2157

1,32 2,40,36

= 0,215 Nm

- for Ra = 3,4 Ω, Rb = 2,4 Ω, Rc = 2,4 Ω

load =20 %

Slip = 0,4

I ar 1=13


= 13

[ 1,34 + (-0,5 + j 0,866) 1,6 + (-0,5 – j 0,866)1,6 ]

= - 0,0866 A

I ar 2=13

( I ar + a2 I br + aI cr )

= 13

[ 1,34 + (-0,5 - j 0,866) 1,6 + (-0,5 + j 0,866)1,6 ]

= - 0,0866 A

So I 21 = - 0,0866 A , I 22 = - 0,0866 A

I 11=− j ω1 Lm I 21+V s

R1+ j ω1 L1

= − j 314 x0,181 x (−0,0866)+173,2

1,842+ j314 x0,19318

= 173,2+ j 3,91

1,842+ j 60,65



= 173,24∠1,2960,68∠ 88,26

= 2,854 ∠-86,97 A

I 12=− jω1(1−2S )lm I 22

R1+ j ω1(1−2S) L1

= − j 314 (1−2 x0,4 ) 0,181x (−0.0866)

1,842+ j 314 (1−2x 0,4 ) 0,19318

= j 0,98

1,842+ j 12,13

= 0,98∠90

12,27∠81,36

= 0,07987∠8,64

Te = 3 P1 Lm [ Image ( I 11 I 21¿ )+ Image ( I 12 I 22

¿ )]= T e1+T e2

= 3 x 2 x 0,181 [ image (2,85 x 0,0866 ∠-86,97 - 180) + image (0,07987 x 0.0866 ∠8,68 -

180)]

= 1,086 [(0,2468 sin -266,97) + (0,007 sin -171,32)]

= 0,266 – 0,00114

= 0,264Nm

Te1 = 0,266 Nm

Te2 = - 0,00114 Nm



load = 40 %

Slip = 0,41

I ar 1=13


= 13

[ 1,57 + (-0,5 + j 0,866) 1,83 + (-0,5 – j 0,866)1,83 ]

= - 0,0866 A

I ar 2=13


= 13

[ 1,57 + (-0,5 - j 0,866) 1,83 + (-0,5 + j 0,866)1,83 ]

= - 0,0866 A

So I 21 = - 0,0866 A , I 22 = - 0,0866 A

I 11=− j ω1 Lm I 21+V s

R1+ j ω1 L1

= − j 314 x0,181 x (−0,0866)+173,2

1,842+ j314 x0,19318

= 173,2+ j 3,91

1,842+ j 60,65

= 173,24∠1,2960,68∠ 88,26

= 2,854 ∠-86,97 A

I 12=− jω1(1−2S )lm I 22

R1+ j ω1(1−2S) L1



= − j 314 (1−2 x0,41 )0,181 x (−0.0866)

1,842+ j 314 (1−2x 0,41 )0,19318

= j 0,886

1,842+ j 10,91

= 0,886∠90

11,06∠80,41

= 0,0783∠9,59


¿ )]= T e1+T e2

= 3 x 2 x 0,181 [ image (2,85 x 0,0866 ∠-86,97 - 180) + image (0,0783 x 0.0866

∠9,59 - 180)]

= 1,086 [(0,2468 sin -266,97) + (0,00678 sin -170,41)]

= 0,266 – 0,00112

= 0,265 Nm

Te1 = 0,266Nm

Te2 = - 0,00112Nm

load = 60 %

Slip = 0,43

I ar 1=13


= 13

[ 1,8 + (-0,5 + j 0,866) 2,1 + (-0,5 – j 0,866)2,1 ]



= - 0,1 A

I ar 2=13


= 13

[ 1,8 + (-0,5 - j 0,866) 2,1 + (-0,5 + j 0,866)2,1 ]

= - 0,1 A

So I 21 = - 0,1 A , I 22 = - 0,1 A

I 11=− j ω1 Lm I 21+V s

R1+ j ω1 L1

= − j 314 x0,181 x (−0,1)+173,2

1,842+ j314 x 0,19318

= 173,2+ j 5,68

1,842+ j 60,65

= 173,29∠1,8760,68∠ 88,26

= 2,855 ∠-86,39 A

I 12=− jω1(1−2S )lm I 22

R1+ j ω1(1−2S) L1

= − j 314 (1−2 x0,43 )0,181 x (−0.1)1,842+ j 314 (1−2 x 0,43 ) 0,19318

= j0,795

1,842+ j 8,49

= 0,795∠ 90

8,689∠77,75



= 0,0915∠12,25


¿ )]= T e1+T e2

= 3 x 2 x 0,181 [ image (2,855 x 0,1 ∠-86,39 - 180) + image (0,0915 x 0.1 ∠12,25 - 180)]

= 1,086 [(0,2855 sin -266,39) + (0,00915 sin -167,75)]

= 0,31 – 0,0022

= 0,3078 Nm

Te1 = 0,31 Nm

Te2 = - 0,0022 Nm

load = 80 %

Slip = 0,47

I ar 1=13


= 13

[ 2,1 + (-0,5 + j 0,866) 2,45 + (-0,5 – j 0,866)2,45 ]

= - 0,1166 A

I ar 2=13


= 13

[ 2,1 + (-0,5 - j 0,866) 2,45 + (-0,5 + j 0,866)2,45 ]

= - 0, 1166 A

So I 21 = - 0,1166 A , I 22 = - 0,1166 A



I 11=− j ω1 Lm I 21+V s

R1+ j ω1 L1

= − j 314 x0,181 x (−0,1166 )+173,2

1,842+ j314 x 0,19318

= 173,2+ j 6,62681,842+ j 60,65

= 173,38∠2,2

60,68∠ 88,26

= 2,864 ∠-86,06 A

I 12=− jω1(1−2S )lm I 22

R1+ j ω1(1−2S) L1

= − j 314 (1−2 x0,47 ) 0,181x (−0.1166 )

1,842+ j 314 (1−2 x0,47 ) 0,19318

= j0,397

1,842+ j 3,64

= 0,397∠90

4,08∠63,41

= 0,0973∠26,59


¿ )]= T e1+T e2

= 3 x 2 x 0,181 [ image (2,864x 0,1166 ∠-86,06 - 180) + image (0,0973 x 0.1166 ∠29,59-

180)]

= 1,086 [(0,334 sin -266,06) + (0,0113 sin -150,41)]

= 0,362 – 0,00606



= 0,356 Nm

Te1 = 0,362 Nm

Te2 = - 0,00606 Nm

load = 100 %

Slip = 0,53

I ar 1=13


= 13

[2,3 + (-0,5 + j 0,866) 2,7 + (-0,5 – j 0,866)2,7 ]

= - 0,133 A

I ar 2=13


= 13

[ 2,3 + (-0,5 - j 0,866) 2,7 + (-0,5 + j 0,866)2,7 ]

= - 0,133 A

So I 21 = - 0,133 A , I 22 = - 0,133 A

I 11=− j ω1 Lm I 21+V s

R1+ j ω1 L1

= − j 314 x0,181 x (−0,133)+173,2

1,842+ j314 x 0,19318

= 173,2+ j 7,5591,842+ j 60,65



= 173,36∠2,5

60,68∠ 88,26

= 2,857 ∠-85,76A

I 12=− jω1(1−2S )lm I 22

R1+ j ω1(1−2S) L1

= − j 314 (1−2 x0,53 )0,181 x (−0.133)

1,842+ j 314 (1−2 x 0,53 ) 0,19318

= − j 0,4535

1,842− j 3,64

= 0,4535∠270

4,08∠−63,15

= 0,111∠333,15


¿ )]= T e1+T e2

= 3 x 2 x 0,181 [ image (2,857 x 0,133 ∠-85,76 - 180) + image (0,111 x 0.133 ∠333,15-

180)]

= 1,086 [(0,38 sin -265,76) + (0,0147 sin 153,15)]

= 0,4115 + 0,00721

= 0,4187 Nm

Te1 = 0,4115 Nm

Te2 = 0,00721 Nm



- For Ra = 4,4 Ω, Rb = 2,4 Ω, Rc = 2,4 Ω

Load = 20 %

Slip = 0,46

I ar 1=13


= 13

[ 3 + (-0,5 + j 0,866) 3,3 + (-0,5 – j 0,866)3,3 ]

= - 0,1 A

I ar 2=13


= 13

[ 3 + (-0,5 - j 0,866) 3,3 + (-0,5 + j 0,866)3,3 ]

= - 0,1 A

So I 21 = - 0,1 A , I 22 = - 0,1 A

I 11=− j ω1 Lm I 21+V s

R1+ j ω1 L1

= − j 314 x0,181 x (−0,1)+173,2

1,842+ j314 x 0,19318

= 173,2+ j 5,68

1,842+ j 60,65

= 173,29∠1,87860,68∠88,26

= 2,8558 ∠-86,382A

I 12=− jω1(1−2S )lm I 22

R1+ j ω1(1−2S) L1



= − j 314 (1−2 x0,46 ) 0,181x (−0.1)1,842+ j 314 (1−2 x 0,46 ) 0,19318

= j0,45

1,842+ j 4,85

= 0,45∠ 90

5,19∠69,2

= 0,0867∠20,8


¿ )]= T e1+T e2

= 3 x 2 x 56,86 [ image (2,8558 x 0,1 ∠-86,382 - 180) + image (0,0578 x 0.1 ∠20,8 -

180)]

= 341,16 [(0.28558 sin -266.382) + (0,00578 sin -159,2)]

= 0,31 – 0,00223

= 0,307 Nm

Te1 = 0,31 Nm

Te2 = - 0,00223 N

Load = 40 %

Slip = 0,5

I ar 1=13


= 13

[ 3,2 + (-0,5 + j 0,866) 3,8+ (-0,5 – j 0,866)3,8 ]

= - 0,2 A



I ar 2=13


= 13

[ 3,2 + (-0,5 - j 0,866) 3,8 + (-0,5 + j 0,866)3,8 ]

= - 0,2 A

So I 21 = - 0,2 A , I 22 = - 0,2 A

I 11=− j ω1 Lm I 21+V s

R1+ j ω1 L1

= − j 314 x0,181 x (−0,2)+173,2

1,842+ j314 x 0,19318

= 173,2+ j 11,361,842+ j 60,65

= 173,57∠3,7560,68∠ 88,26

= 2,86 ∠-84,51A

I 22=− j ω1(1−2 S)Lm I 22

Rs+ j ω1(1−2 S)Ls

= − j 157 (1−2 x0,5 )56,85 x (−0.1666)

1,842+ j 157 (1−2 x0,5 )0,19318

= 0

Te = 3 P1 Lm [ Image ( I 11 I r 1¿ )+ Image ( I 22 I r2

¿ )]= T e1+T e2

= 3 x 2 x 0,181 [ image (2,86 x 0,2 ∠-84,51 - 180) ]

= 1,086(0.572 sin -264,51)



= 0,618 Nm

Te1 = 0,618 Nm

Te2 = 0

Load = 60 %

Slip = 0,52

I ar 1=13


= 13

[ 3,3 + (-0,5 + j 0,866) 4 + (-0,5 – j 0,866)4 ]

= - 0,233 A

I ar 2=13


= 13

[ 3,3 + (-0,5 - j 0,866) 4 + (-0,5 + j 0,866)4 ]

= - 0,233 A

So I 21 = - 0,233 A , Ir2 = - 0,233 A

I 11=− j ω1 Lm I 21+V s

R1+ j ω1 L1



= − j 314 x0,181 x (−0,233)+173,2

1,842+ j314 x 0,19318

= 173,2+ j 13,241,842+ j 60,65

= 173,705∠ 4,4260,68∠ 88,26

= 2,862 ∠-83,84 A

I 12=− jω1(1−2S )lm I 22

R1+ j ω1(1−2S) L1

= − j 314 (1−2 x0,52 )0,181 x (−0.233)

1,842+ j 314 (1−2 x 0,52 )0,19318

= − j0,53

1,842− j 2,42

= 0,53∠270

3,04∠−52,72

= 0,1743∠322,72


¿ )]= T e1+T e2

= 3 x 2 x 0,181 [ image (2,862 x 0,233 ∠-83,84 - 180) + image (0,1743 x 0.233 ∠322,72

- 180)]

= 1,086 [(0.6668 sin -263.84) + (0,0406 sin 142,72)]

= 0,72 + 0,0267

= 0,747 Nm



Te1 = 0,72 Nm

Te2 = 0,0267 Nm

Load = 80 %

Slip = 0,53

I ar 1=13


= 13

[ 3,55 + (-0,5 + j 0,866) 4,4 + (-0,5 – j 0,866)4,4 ]

= - 0,283 A

I ar 2=13


= 13

[ 3,55 + (-0,5 - j 0,866) 4,4 + (-0,5 + j 0,866)4,4 ]

= - 0,283 A

So I 21 = - 0,283 A , I 22 = - 0,283 A

I 11=− j ω1 Lm I 21+V s

R1+ j ω1 L1

= − j 314 x0,181 x (−0,283)+173,2

1,842+ j314 x 0,19318

= 173,2+ j 16,081,842+ j 60,65

= 173,94∠5,3

60,68∠ 88,26



= 2,866 ∠-82,96 A

I 12=− jω1(1−2S )lm I 22

R1+ j ω1(1−2S) L1

= − j 314 (1−2 x0,53 )0,181 x (−0.283)

1,842+ j 314 (1−2x 0,52 )0,19318

= − j 0,965

1,842− j 3,64

= 0,965∠270

4,08∠−63,16

= 0,2365∠333,16


¿ )]= T e1+T e2

= 3 x 2 x 0,181 [ image (2,866 x 0,283 ∠-82,96 - 180) + image (0,2365 x 0.283 ∠333,16

- 180)]

= 1,086 [(0.811 sin -262.96) + (0,067 sin 153,16)]

= 0,874 + 0,03285

= 0,9068 Nm

Te1 = 0,874 Nm

Te2 = 0,03285 Nm

Load = 100 %

Slip = 0,66



I ar 1=13


= 13

[ 3,8 + (-0,5 + j 0,866) 4,8 + (-0,5 – j 0,866)4,8 ]

= - 0,333 A

I ar 2=13


= 13

[ 3,8 + (-0,5 - j 0,866) 4,8 + (-0,5 + j 0,866)4,8 ]

= - 0,333 A

So I 21 = - 0,333 A , I 22 = - 0,333 A

I 11=− j ω1 Lm I 21+V s

R1+ j ω1 L1

= − j 314 x0,181 x (−0,333)+173,2

1,842+ j314 x 0,19318

= 173,2+ j 18,9251,842+ j 60,65

= 174,23∠ 6,23560,68∠ 88,26

= 2,87 ∠-82,025 A

I 12=− jω1(1−2S )lm I 22

R1+ j ω1(1−2S) L1



= − j 314 (1−2 x0,66 ) 0,181x (−0.333)

1,842+ j 314 (1−2 x 0,66 ) 0,19318

= − j6,056

1,842− j 19,41

= 6,056∠270

19,5∠−84,57

= 0,31∠354,57


¿ )]= T e1+T e2

= 3 x 2 x 0,181 [ image (2,87x 0,333 ∠-82,025 - 180) + image (0,31 x 0.333 ∠354,57-

180)]

= 1,086 [(0.9557 sin -262.025) + (0,103 sin 174,57)]

= 1,0278 + 0,01

= 1,0378 Nm

Te1 = 1,0278 Nm

Te2 = 0,01 Nm

- load = 20 %

Slip = 0,56

I ar 1=13


= 13

[ 2,4 + (-0,5 + j 0,866) 3,5 + (-0,5 – j 0,866)3,5 ]



= - 0,366 A

I ar 2=13


= 13

[ 2,4 + (-0,5 - j 0,866) 3,5 + (-0,5 + j 0,866)3,5 ]

= - 0,366 A

So I 21 = - 0,366 A , I 22 = - 0,366 A

I 11=− j ω1 Lm I 21+V s

R1+ j ω1 L1

= − j 314 x0,181 x (−0,366)+173,2

1,842+ j314 x0,19318

= 173,2+ j 20,8

1,842+ j 60,65

= 174,44∠6,84860,68∠88,26

= 2,874∠-81,412 A

I 12=− jω1(1−2S )lm I 22

R1+ j ω1(1−2S) L1

= − j 314 (1−2 x0,56 ) 0,181x (−0.366)

1,842+ j 314 (1−2x 0,56 ) 0,19318

= − j2,5

1,842− j 7,28

= 2,5∠270

7,51∠−75,8



= 0,33289∠345,8


¿ )]= T e1+T e2

= 3 x 2 x 0,181 [ image (2,874x 0,366 ∠-81,412 - 180) + image (0,33289 x 0.366 ∠

345,8- 180)]

= 1,086 [(1,052sin -261.412) + (0,1218 sin 165,8)]

= 1,13 + 0,0324

= 1,1624 Nm

Te1 = 1,13 Nm

Te2 = 0,0324 Nm

Load = 40 %

Slip = 0,63

I ar 1=13


= 13

[ 2,53 + (-0,5 + j 0,866) 3,7 + (-0,5 – j 0,866)3,7 ]

= - 0,39 A

I ar 2=13


= 13

[ 2,53 + (-0,5 - j 0,866) 3,7 + (-0,5 + j 0,866)3,7 ]

= - 0,39 A

So I 21 = - 0,39 A , I 22 = - 0,39 A



I 11=− j ω1 Lm I 21+V s

R1+ j ω1 L1

= − j 314 x0,181 x (−0,39)+173,2

1,842+ j314 x 0,19318

= 173,2+ j 22,161,842+ j 60,65

= 174,61∠7,3

60,68∠ 88,26

= 2,877∠-80,96 A

I 12=− jω1(1−2S )lm I 22

R1+ j ω1(1−2S) L1

= − j 314 (1−2 x0,63 )0,181 x (−0.39)1,842+ j 314 (1−2 x 0,63 ) 0,19318

= − j5,76

1,842− j 15,77

= 5,76∠ 270

15,87∠−83,33

= 0,363∠353,33


¿ )]= T e1+T e2

= 3 x 2 x 0,181 [ image (2,877x 0,39 ∠-80,96 - 180) + image (0,363 x 0.39 ∠353,33-

180)]

= 1,086 [(1,122sin -260.96) + (0,1415 sin 173,33)]

= 1,2 + 0,0178



= 1,2178 Nm

Te1 = 1,2 Nm

Te2 = 0,0178 Nm

Load = 60 %

Slip = 0,66

I ar 1=13


= 13

[ 2,71 + (-0,5 + j 0,866) 4 + (-0,5 – j 0,866)4 ]

= - 0,43 A

I ar 2=13


= 13

[ 2,71 + (-0,5 - j 0,866) 4 + (-0,5 + j 0,866)4 ]

= - 0,43 A

So I 21 = - 0,43 A , I 22 = - 0,43 A

I 11=− j ω1 Lm I 21+V s

R1+ j ω1 L1

= − j 314 x0,181 x (−0,43)+173,2

1,842+ j314 x 0,19318

= 173,2+ j 24,431,842+ j 60,65

= 174,91∠8,0260,68∠ 88,26



= 2,882∠-80,24A

I 12=− jω1(1−2S )lm I 22

R1+ j ω1(1−2S) L1

= − j 314 (1−2 x0,66 ) 0,181 x (−0.43)1,842+ j 314 (1−2 x 0,66 ) 0,19318

= − j 7,82

1,842− j 19,4

= 7,82∠270

19,48∠−84,57

= 0,401∠354,57


¿ )]= T e1+T e2

= 3 x 2 x 0,181 [ image (2,882x 0,43 ∠-80,24 - 180) + image (0,401 x 0.43 ∠354,57-

180)]

= 1,086 [(1,24sin -260.24) + (0,1724 sin 174,57)]

= 1,327 + 0,0177

= 1,3447 Nm

Te1 = 1,327 Nm

Te2 = 0,0177 Nm



Load = 80 %

Slip = 0,71

I ar 1=13


= 13

[ 2,97 + (-0,5 + j 0,866) 4,45 + (-0,5 – j 0,866)4,45 ]

= - 0,493 A

I ar 2=13


= 13

[ 2,97 + (-0,5 - j 0,866) 4,45 + (-0,5 + j 0,866)4,45 ]

= - 0,493 A

So I 21 = - 0,493 A , I 22 = - 0,493 A

I 11=− j ω1 Lm I 21+V s

R1+ j ω1 L1

= − j 314 x0,181 x (−0,493)+173,2

1,842+ j314 x 0,19318

= 173,2+ j 28,021,842+ j 60,65

= 175,45∠9,2

60,68∠ 88,26

= 2,89∠-79,06 A



I 12=− jω1(1−2S )lm I 22

R1+ j ω1(1−2S) L1

= − j 314 (1−2 x0,71 )0,181 x (−0.493)

1,842+ j 314 (1−2 x 0,71 )0,19318

= − j 11,768

1,842− j 25,47

= 11,768∠270

25,53∠−85,86

= 0,46∠355,86


¿ )]= T e1+T e2

= 3 x 2 x 0,181 [ image (2,89x 0,493 ∠-79,06 - 180) + image (0,46 x 0.493 ∠355,86 -

180)]

= 1,086 [(1,424sin -259.06) + (0,226 sin 175,86)]

= 1,518 + 0,0163

= 1,5343 Nm

Te1 = 1,518 Nm

Te2 = 0,0163 Nm

Load = 100 %

Slip = 0,76

I ar 1=13


= 13

[ 3,3 + (-0,5 + j 0,866) 4,91 + (-0,5 – j 0,866)4,91 ]



= - 0,536 A

I ar 2=13


= 13

[ 3,3 + (-0,5 - j 0,866) 4,91 + (-0,5 + j 0,866)4,91 ]

= - 0,536 A

So I 21 = - 0,536 A , I 22 = - 0,536 A

I 11=− j ω1 Lm I 21+V s

R1+ j ω1 L1

= − j 314 x0,181 x (−0,536)+173,2

1,842+ j314 x0,19318

= 173,2+ j 30,461,842+ j 60,65

= 175,85∠ 9,9760,68∠ 88,26

= 2,898∠-78,3 A

I 12=− jω1(1−2S )lm I 22

R1+ j ω1(1−2S) L1

= − j 314 (1−2 x0,76 ) 0,181x (−0.536)

1,842+ j 314 (1−2x 0,76 ) 0,19318

= − j15,84

1,842− j 31,54

= 15,84∠270

31,59∠−86,66



= 0,501∠356,66


¿ )]= T e1+T e2

= 3 x 2 x 0,181 [ image (2,898x 0,536 ∠-78,3 - 180) + image (0,501 x 0.536 ∠356,66-

180)]

= 1,086 [(1,553sin -258,3) + (0,268 sin 176,66)]

= 1,65 + 0,017

= 1,667 Nm

Te1 = 1,65 Nm

Te2 = 0,017 Nm

Experimental Analysis Results Table

- Ra = Rb = Rc = 2,4 Ω

R (%) SlipTorsi

(Nm)

20 0,23 0,097

40 0,26 0,112

60 0,3 0,152

80 0,33 0,168

100 0,36 0,215



- Ra = 3,4 Ω , Rb = Rc = 2,4 Ω

R (%) SlipTe1

(NM)

Te2 (Nm)Te (Nm)

20 0,4 0,266 -0.00114 0,264

40 0,41 0,266 -0,00112 0,265

60 0,43 0,31 -0,0022 0,3078

80 0,47 0,362 -0,00606 0,356

100 0,53 0,4115 -0,00721 0,4187

- Ra = 4,4 Ω , Rb = Rc = 2,4 Ω

R (%) SlipTe1

(NM)

Te2 (Nm)Te (Nm)

20 0,46 0,31 -0,00223 0,307

40 0,5 0,618 - 0,618

60 0,52 0,72 0,0267 0,747



80 0,53 0,874 0,03285 0,9068

100 0,66 1,0278 0,01 1,0378

- Ra = 5,4 Ω , Rb = Rc = 2,4 Ω

R (%) SlipTe1

(NM)

Te2

(Nm)

Te

(Nm)

20 0,56 1,13 0,0324 1,1624

40 0,63 1,2 0,0178 1,2178

60 0,66 1,327 0,0177 1,3447

80 0,71 1,518 0,0163 1,5343

100 0,76 1,65 0,017 1,667

Curve Effect of Rotor Resistance Of un Balanced Torque And speed Induction

Motor

- A. Calculation of Results

for Ra = Rb = Rc = 2,4Ω



- For Ra = 3,4Ω Rb = Rc = 2,4Ω



- For Ra = 4,4Ω Rb = Rc = 2,4Ω



- For Ra = 5,4Ω Rb = Rc = 2,4Ω



The results of the calculation of loading experiments winding rotor induction motor with

rotor resistance is balanced and not balanced, then the curve can be drawn as below:



- B. From the results of Experiment

- For Ra = Rb = Rc = 2,4Ω





- For Ra = 3,4Ω Rb = Rc = 2,4Ω



- For Ra = 4,4Ω Rb = Rc = 2,4Ω



- For Ra = 5,4Ω Rb = Rc = 2,4Ω





The results of the calculation load induction motor rotor resistance rotor windings with

balanced and unbalanced, then the curve can be drawn as below:





Then it can be concluded that:

If the rotor is not balanced then the prisoners with the same load torque is

large and the rotation will decrease.

The more unbalanced rotornya then with the same load, the greater the torque

and rotation decreases.

Motor with unbalanced stator resistance can degrade the performance of the

engine induction.



Documents

Analysis the Effect of Unbalance Rotor Winding Torque and Speed Induction Motor