Lecture Electromechanical energy conversion: Polyphase Induction Machines - Nguyễn Công Phương

Nguyễn Công Phương

ELECTROMECHANICAL ENERGY
CONVERSION
Polyphase Induction Machines


Contents
I. Magnetic Circuits and Magnetic Materials
II. Electromechanical Energy Conversion
Principles
III. Introduction to Rotating Machines
IV. Synchronous Machines
V. Polyphase Induction Machines
VI. DC Machines
VII.Variable – Reluctance Machines and Stepping
Motors
VIII.Single and Two – Phase Motors
IX. Speed and Torque Control
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Polyphase Induction Machines
1. Introduction to Polyphase Induction Machines
2. Currents and Fluxes in Polyphase Induction
Machines
3. Induction – Motor Equivalent Circuit
4. Analysis of the Equivalent Circuit
5. Torque and Power by Use of Thevenin’s

Theorem
6. Parameter Determination from No – Load and
Blocked – Rotor Tests
7. Effects of Rotor Resistance
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Introduction to Polyphase
Induction Machines (1)
• Induction motor: alternating current is supplied to the stator
directly & to the rotor by induction or transformer action
from the stator.
• Two kinds of rotor:
– Wound rotor (relatively uncommon)
– Squirrel-cage rotor (the most commonly used)

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Introduction to Polyphase
Induction Machines (2)
• The rotor speed: n (r/min).
• The synchronous speed of the stator field: ns
(r/min).
• The rotor slip: ns – n (r/min)

ns  n
(%)
• The fractional slip: s 
ns
•  the rotor speed: n = (1 – s)ns
• The mechanical angular velocity: ωm = (1 – s)ωs
• The slip frequency: fr = sfe
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Introduction to Polyphase
Induction Machines (3)
• The rotor terminals of an induction motor are
short circuited.
• The rotating air – gap flux induces slip –
frequency voltages in the rotor windings.
• The operating speed can never equal the
synchronous speed.
• The rotor currents produce a rotating flux wave
which rotate at sns (r/min) with respect to the
rotor.
• The rotor speed: n (r/min)
• The speed of the rotor’s flux wave:
sns + n = sns + ns(1 – s) = ns
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Introduction to Polyphase
Induction Machines (4)
• The speed of the stator field: ns (r/min).
• The speed of the rotor field: ns (r/min).
•  the rotor currents produce a field which
rotates at synchronous speed and hence in
synchronism with that produced by the stator
currents.
•  the stator & rotor fields are stationary with
respect to each other, & produce a steady
torque, called asynchronous torque.
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Introduction to Polyphase
Induction Machines (5)

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Polyphase Induction Machines
1. Introduction to Polyphase Induction Machines
2. Currents and Fluxes in Polyphase Induction
Machines
3. Induction – Motor Equivalent Circuit

4. Analysis of the Equivalent Circuit
5. Torque and Power by Use of Thevenin’s
Theorem
6. Parameter Determination from No – Load and
Blocked – Rotor Tests
7. Effects of Rotor Resistance
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Currents and Fluxes in Polyphase
Induction Machines (1)
Resultant flux – density
wave

Rotor – mmf
wave

a c

b a

  90o

c b

Rotation
Resultant flux – density
wave

Rotor – mmf
wave

a c

a

b a

  90o  

c b

a

Rotation

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Currents and Fluxes in Polyphase
Induction Machines (2)
Flux-density wave

Instantaneous bar – voltage
magnitudes

s

1 2 3 4 5 6 7 8 9

11 12 13 14 15 16 1

(1  s )s

s

Instantaneous bar – current
magnitudes
1 2 3 4 5 6 7 8 9

11 12 13 14 15 16 1


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(1  s )s
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Currents and Fluxes in Polyphase
Induction Machines (3)
Rotor – mmf wave
Flux-density wave
s



1 2 3 4 5 6 7 8
9

11 12 13 14 15 16 1

(1  s )s

Fundamental component of
rotor – mmf wave

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Polyphase Induction Machines
1. Introduction to Polyphase Induction Machines
2. Currents and Fluxes in Polyphase Induction
Machines
3. Induction – Motor Equivalent Circuit
4. Analysis of the Equivalent Circuit
5. Torque and Power by Use of Thevenin’s
Theorem
6. Parameter Determination from No – Load and
Blocked – Rotor Tests
7. Effects of Rotor Resistance
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Induction – Motor Equivalent
Circuit

a

R1


Vˆ1



Iˆ1

X1

Iˆc
Rc

Iˆ

Iˆ2
Iˆm


Eˆ 2

X2
R2

s



Xm
b

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Polyphase Induction Machines
1. Introduction to Polyphase Induction Machines
2. Currents and Fluxes in Polyphase Induction
Machines
3. Induction – Motor Equivalent Circuit
4. Analysis of the Equivalent Circuit
5. Torque and Power by Use of Thevenin’s
Theorem
6. Parameter Determination from No – Load and
Blocked – Rotor Tests
7. Effects of Rotor Resistance
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Analysis
of the Equivalent Circuit (1)

a

Pgap

R
n I 2
s
2
ph 2

R1



Iˆ1

Protor  n I R2
Pmech  Pgap  Protor



Rc

1 s
s

 Pmech  (1  s ) Pgap

 Protor  sPgap

Iˆm



X2

R2
s

Eˆ 2


Xm

a
R1


Vˆ1


Iˆ2

b

2 R2
 n ph I 2
 n ph I 22 R2
s

 n ph I 22 R2

Iˆ

Iˆc

Vˆ1

2
ph 2

X1

Iˆ1

X1

Iˆc
Rc

Iˆ

Iˆ2
Iˆm



R2

Eˆ 2



Xm
b

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X2

1 s
R2
s
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Ex. 1

Analysis
of the Equivalent Circuit (2)

A three – phase, two – pole, 60-Hz induction motor is operating at 3502 r/min with
an input power of 15.7 kW and a terminal current of 22.6 A, the stator – winding
resistance is 0.20 Ω/phase. Calculate the power dissipated in rotor?

Pstator  3I12 R1  3  22.62  0.2  306 W
Pgap  Pinput  Pstator  15.7  0.3  15.4 kW
ns 

120
120

fe 
60  3600 r/min
poles
2

ns  n 3600  3502
s

 0.0272
ns
3600

Protor  sPgap  0.0272  15.4  419 W
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Analysis
of the Equivalent Circuit (3)
Pmech  mTmech
m  (1  s )s
 Pmech  (1  s )sTmech
Pmech  (1  s ) Pgap
 Tmech 
Pgap

Pgap
s

R2
n I
s
2
ph 2

 Tmech 

Pshaft

n ph I 22 ( R2 / s )

s
 Pmech  Prot

Tshaft  Tmech  Trot
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Ex. 2

Analysis
of the Equivalent Circuit (4)

Given a three – phase, six – pole, Y – connected, 220-V (line-to-line) 60-Hz 7.5-kW
induction motor, R1 = 0.249Ω/phase, R2 = 0.144, X1 = 0.503, X2 = 0.209, Xm = 13.25,
Prot = 403W (independent of load), s = 2%. Compute the speed, output torque and
power, stator current, power factor, and efficiency?

a

 R2


jX

2  jX m
s

Z ab  
R2
 jX 2  jX m
s

R1



Iˆ1

Vˆ1

 0.144


j
0.209

 j13.25

0.02


0.144
 j 0.209  j13.25
0.02



X1

Iˆ2
Iˆm



X2

Eˆ 2

R2
s



Xm
b

 5.43  j 3.11

220
Vˆ1
3
ˆI 

 15.93  j10.14  18.88  32.5o A
1
R1  jX 1  Z ab 0.249  j 0.503  5.43  j 3.11
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Ex. 2

Analysis
of the Equivalent Circuit (5)

Given a three – phase, six – pole, Y – connected, 220-V (line-to-line) 60-Hz 7.5-kW
induction motor, R1 = 0.249Ω/phase, R2 = 0.144, X1 = 0.503, X2 = 0.209, Xm = 13.25,
Prot = 403W (independent of load), s = 2%. Compute the speed, output torque and
power, stator current, power factor, and efficiency?

Iˆ1  15.93  j10.14  18.88  32.5o A

pf  cos( 32.5o )  0.844 lagging

ns 

120

120
fe 
60  1200 r/min
poles
6

n  (1  s )ns  (1  0.02)1200  1176 r/min

s 

2
2
2
e 
2 f e  2  60  125.7 rad/sec
poles
poles
6

m  (1  s )s  (1  0.02)125.7  123.2 rad/sec
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Analysis
of the Equivalent Circuit (6)

Ex. 2

Given a three – phase, six – pole, Y – connected, 220-V (line-to-line) 60-Hz 7.5-kW
induction motor, R1 = 0.249Ω/phase, R2 = 0.144, X1 = 0.503, X2 = 0.209, Xm = 13.25,
Prot = 403W (independent of load), s = 2%. Compute the speed, output torque and
power, stator current, power factor, and efficiency?
a
Iˆ  18.88  32.5o A
1

Z ab  5.43  j 3.11 

R1



Pgap  3I12 Rab

Iˆ1

X1

Vˆ1

 3  18.82  5.41  5740W



Pshaft  Pmech  Prot

Iˆ2
Iˆm



X2

Eˆ 2

R2
s



Xm
b

 (1  s ) Pgap  Prot  (1  0.02)5740  403  5220W
Tshaft 

Pshaft
m



5220
 42.4Nm
123.2
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Analysis
of the Equivalent Circuit (7)

Ex. 2

Given a three – phase, six – pole, Y – connected, 220-V (line-to-line) 60-Hz 7.5-kW
induction motor, R1 = 0.249Ω/phase, R2 = 0.144, X1 = 0.503, X2 = 0.209, Xm = 13.25,
Prot = 403W (independent of load), s = 2%. Compute the speed, output torque and
power, stator current, power factor, and efficiency?

Pshaft  5220W

Pin  3Re{Vˆ1 Iˆ1*}

 3Re{127(18.88  32.5o )}
 6060W



Pshaft
Pin



5220
 86.1%
6060
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Polyphase Induction Machines
1. Introduction to Polyphase Induction Machines
2. Currents and Fluxes in Polyphase Induction
Machines
3. Induction – Motor Equivalent Circuit
4. Analysis of the Equivalent Circuit
5. Torque and Power by Use of Thevenin’s
Theorem
6. Parameter Determination from No – Load and
Blocked – Rotor Tests
7. Effects of Rotor Resistance
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Torque and Power by Use of
Thevenin’s Theorem (1)
Vˆ1,eq  Vˆ1
Z1,eq

jX m
R1  j ( X 1  X m )

jX m ( R1  jX 1 )

R1  j ( X 1  X m )

a
R1



Iˆ1

X1

Iˆ2
Iˆm

Vˆ1


Eˆ 2

Iˆ2 
Tmech 

Z1,eq  jX 2  R2 / s
2
ph 2

n I ( R2 / s )
s

R2
s



Xm

 R1,eq  jX 1,eq

Vˆ1,eq



X2

b
a



R1,eq

Vˆ1,eq

X 1,eq

Iˆ2


Eˆ 2

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R2
s





n phV1,2eq ( R2 / s )
1

s ( R1,eq  R2 / s ) 2  ( X 1,eq  X 2 ) 2

X2

b

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Torque and Power by Use of
Thevenin’s Theorem (2)
Tmech

n phV1,2eq ( R2 / s )
1

,
s ( R1,eq  R2 / s ) 2  ( X 1,eq  X 2 ) 2

s

ns  n
ns

Motor

600

400

200

0

Braking region

Motor region

Generator region

Torque

-200

Generator

-400

-600

-800

-1000

-1200

2

1.5

1

0.5

0

-0.5

-1

-1.5

Fractional slip s

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