analysis the effect of unbalance rotor winding torque and speed induction motor
TRANSCRIPT
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
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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
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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)
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=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¿)
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= 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.
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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
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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.
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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
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= 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
Experimental Procedure
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.
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Experiment Results Data
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
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= 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
Experimental Procedure
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.
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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
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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
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Experimental Procedure
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
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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.
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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.
Experiment Results Data
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)
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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
R (%) Nr Slip Pin (Kw)Torsi
(Nm)Ia (A) Ib (A) Ic (A)
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
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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
R (%) Nr Slip Pin (Kw)Torsi
(Nm)Ia (A) Ib (A) Ic (A)
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)
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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
R (%) Nr Slip Pin (Kw)Torsi
(Nm)Ia (A) Ib (A) Ic (A)
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
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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
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= 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
( I ar + a I br + a2 I cr )
= 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
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= 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
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load = 40 %
Slip = 0,41
I ar 1=13
( I ar + a I br + a2 I cr )
= 13
[ 1,57 + (-0,5 + j 0,866) 1,83 + (-0,5 – j 0,866)1,83 ]
= - 0,0866 A
I ar 2=13
( I ar + a2 I br + aI cr )
= 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
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= − 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
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 (0,0783 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
( I ar + a I br + a2 I cr )
= 13
[ 1,8 + (-0,5 + j 0,866) 2,1 + (-0,5 – j 0,866)2,1 ]
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= - 0,1 A
I ar 2=13
( I ar + a2 I br + aI cr )
= 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
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= 0,0915∠12,25
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 (2,855 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
( I ar + a I br + a2 I cr )
= 13
[ 2,1 + (-0,5 + j 0,866) 2,45 + (-0,5 – j 0,866)2,45 ]
= - 0,1166 A
I ar 2=13
( I ar + a2 I br + aI cr )
= 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
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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
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 (2,864x 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
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= 0,356 Nm
Te1 = 0,362 Nm
Te2 = - 0,00606 Nm
load = 100 %
Slip = 0,53
I ar 1=13
( I ar + a I br + a2 I cr )
= 13
[2,3 + (-0,5 + j 0,866) 2,7 + (-0,5 – j 0,866)2,7 ]
= - 0,133 A
I ar 2=13
( I ar + a2 I br + aI cr )
= 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
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= 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
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 (2,857 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
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- For Ra = 4,4 Ω, Rb = 2,4 Ω, Rc = 2,4 Ω
Load = 20 %
Slip = 0,46
I ar 1=13
( I ar + a I br + a2 I cr )
= 13
[ 3 + (-0,5 + j 0,866) 3,3 + (-0,5 – j 0,866)3,3 ]
= - 0,1 A
I ar 2=13
( I ar + a2 I br + aI cr )
= 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
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= − 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
Te = 3 P1 Lm [ Image ( I 11 I 21¿ )+ Image ( I 12 I 22
¿ )]= 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
( I ar + a I br + a2 I cr )
= 13
[ 3,2 + (-0,5 + j 0,866) 3,8+ (-0,5 – j 0,866)3,8 ]
= - 0,2 A
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I ar 2=13
( I ar + a2 I br + aI cr )
= 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)
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= 0,618 Nm
Te1 = 0,618 Nm
Te2 = 0
Load = 60 %
Slip = 0,52
I ar 1=13
( I ar + a I br + a2 I cr )
= 13
[ 3,3 + (-0,5 + j 0,866) 4 + (-0,5 – j 0,866)4 ]
= - 0,233 A
I ar 2=13
( I ar + a2 I br + aI cr )
= 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
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= − 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
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 (2,862 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
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Te1 = 0,72 Nm
Te2 = 0,0267 Nm
Load = 80 %
Slip = 0,53
I ar 1=13
( I ar + a I br + a2 I cr )
= 13
[ 3,55 + (-0,5 + j 0,866) 4,4 + (-0,5 – j 0,866)4,4 ]
= - 0,283 A
I ar 2=13
( I ar + a2 I br + aI cr )
= 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
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= 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
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 (2,866 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
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I ar 1=13
( I ar + a I br + a2 I cr )
= 13
[ 3,8 + (-0,5 + j 0,866) 4,8 + (-0,5 – j 0,866)4,8 ]
= - 0,333 A
I ar 2=13
( I ar + a2 I br + aI cr )
= 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
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= − 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
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 (2,87x 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
( I ar + a I br + a2 I cr )
= 13
[ 2,4 + (-0,5 + j 0,866) 3,5 + (-0,5 – j 0,866)3,5 ]
M. Azhary Siregar. St ([email protected] for FB and Email)
= - 0,366 A
I ar 2=13
( I ar + a2 I br + aI cr )
= 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
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= 0,33289∠345,8
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 (2,874x 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
( I ar + a I br + a2 I cr )
= 13
[ 2,53 + (-0,5 + j 0,866) 3,7 + (-0,5 – j 0,866)3,7 ]
= - 0,39 A
I ar 2=13
( I ar + a2 I br + aI cr )
= 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
M. Azhary Siregar. St ([email protected] for FB and Email)
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
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 (2,877x 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
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= 1,2178 Nm
Te1 = 1,2 Nm
Te2 = 0,0178 Nm
Load = 60 %
Slip = 0,66
I ar 1=13
( I ar + a I br + a2 I cr )
= 13
[ 2,71 + (-0,5 + j 0,866) 4 + (-0,5 – j 0,866)4 ]
= - 0,43 A
I ar 2=13
( I ar + a2 I br + aI cr )
= 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
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= 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
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 (2,882x 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
M. Azhary Siregar. St ([email protected] for FB and Email)
Load = 80 %
Slip = 0,71
I ar 1=13
( I ar + a I br + a2 I cr )
= 13
[ 2,97 + (-0,5 + j 0,866) 4,45 + (-0,5 – j 0,866)4,45 ]
= - 0,493 A
I ar 2=13
( I ar + a2 I br + aI cr )
= 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
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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
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 (2,89x 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
( I ar + a I br + a2 I cr )
= 13
[ 3,3 + (-0,5 + j 0,866) 4,91 + (-0,5 – j 0,866)4,91 ]
M. Azhary Siregar. St ([email protected] for FB and Email)
= - 0,536 A
I ar 2=13
( I ar + a2 I br + aI cr )
= 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
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= 0,501∠356,66
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 (2,898x 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
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- 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
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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
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- For Ra = 3,4Ω Rb = Rc = 2,4Ω
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- For Ra = 4,4Ω Rb = Rc = 2,4Ω
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- For Ra = 5,4Ω Rb = Rc = 2,4Ω
M. Azhary Siregar. St ([email protected] for FB and Email)
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:
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- B. From the results of Experiment
- For Ra = Rb = Rc = 2,4Ω
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M. Azhary Siregar. St ([email protected] for FB and Email)
- For Ra = 3,4Ω Rb = Rc = 2,4Ω
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- For Ra = 4,4Ω Rb = Rc = 2,4Ω
M. Azhary Siregar. St ([email protected] for FB and Email)
- For Ra = 5,4Ω Rb = Rc = 2,4Ω
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M. Azhary Siregar. St ([email protected] for FB and Email)
The results of the calculation load induction motor rotor resistance rotor windings with
balanced and unbalanced, then the curve can be drawn as below:
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M. Azhary Siregar. St ([email protected] for FB and Email)
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.
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