Transcript Induction Motors

Induction Motors
Equations, Performance, Electrical Equivalent Circuits
Induction Motor by Bullet Points
• Stator generates rotating, sinusoidal B-Field:
• This field induces current in the rotor cage loops at f  f  f
ur ur
i
• The stator B-Field at each rotor wire is such that  B  $
• Torque pushes in direction of field rotation! (That’s it!!)
• Rotor currents generate triangular B-field rotating in the air gap at slip
speed relative to rotor, so at line rate in reference frame!
• Rotor field reduces field in stator and line current increases to
maintain the stator winding voltage and the gap magnetic field
• Increased line current supplies the mechanical energy and the joule
heating of the rotor.
ur
B  32 B0 cos(  t )$r
S
R
L
R
Relative Torque vs Slip Angle
1
Single Rotor Coil - Resistive Current (Net
Power)
Reactive Torque (No Net Power)
Relative Torque
0.8
Total Torque - Single Turn
0.6
0.4
0.2
0
-0.2
0
60
120
180
Slip Angle (deg.)
240
300
360
B-Field of Rotor vs. Rotor Angle as Function of Slip Cycle Angle
Relative B-Field Strength (a.u.) Offset by 3 for Each Slip Angle
24
21
18
15
0 deg.
45 deg.
12
90 deg.
135 deg.
9
180 deg.
225 deg.
6
270 deg.
315 deg.
3
0
-3
0
60
120
180
240
Angle around rotor from one loop (deg.)
300
360
Shape and Rotation of the Rotor Generated B-Field in the Gap
1
0.8
Relative Magnetic Field Strength (a.u.)
0.6
0.4
0.2
0 deg.
180 deg.
0
-0.2
-0.4
-0.6
-0.8
-1
0
60
120
180
240
300
Angle arount the rotor relative to one turn of the two rotor turns (deg.)
360
Shape and Rotation of the Rotor Generated B-Field in the Gap
1
0.8
Relative Magnetic Field Strength (a.u.)
0.6
0.4
0.2
0 deg.
45 deg.
0
90 deg.
180 deg.
-0.2
-0.4
-0.6
-0.8
-1
0
60
120
180
240
Angle arount the rotor relative to one turn of the two rotor turns (deg.)
300
360
Rotor Induced Triangle Wave in Stator Current Showing Waveshape withHarmonic Multiples of
3 Removed and Compared to Sinusoid
1
0.8
0.6
Relative Current (a.u.)
0.4
0.2
Triangle Wave
0
Fourier to 27th
No multiples of 3
-0.2
Single Sine
-0.4
-0.6
-0.8
-1
0
60
120
180
Cycle Angle (deg.)
240
300
360
Formal Transformer Analogy
• Mutual inductance stator to rotor is time dependent
LP
LM
LM
LMR cos(R t )
LMR sin(R t )  iA   RW
vA


v


LM
LP
LM
LMR cos(R t  23 ) LMR sin(R t  23 )  iB   0
 B
 d 

vC
 
LM
LM
LP
LMR cos(R t  43 ) LMR sin(R t  43 )   iC    0

 dt 
   
2
4
LR
0
vR1  0 
 LMR cos(R t ) LMR cos(R t  3 ) LMR cos(R t  3 )
 iR1   0
2

4

v  0 
 LMR sin(R t ) LMR sin(R t  3 ) LMR sin(R t  3 )
 iR 2   0
0
LR
 R2

0
RW
0
0
0
0
0
RW
0
0
0
0
0
RR
0
0  i A 
 
0  iB 
0   iC 
  
0  iR1 
RR  iR 2 
• The A, B, C voltages are the line voltages
• The rotor voltages are v  R i
• Given rotor frequency, calculate currents and power, subtract rotor and
winding heat to get mechanical power.
• Ugh!
R
R R
Simple Per-Phase Transformer Model
• Know that power flow is constant at constant speed (No torque
variation!)
• Per-phase model with constant impedance that is a function of rotor
speed
• Use basic single-phase transformer model with turns ratio and
secondary impedance dependent on rotor speed
• Stator field is zero-slip model
Electrical Equivalent Circuit of Stator Alone
•
•
•
•
Applies when rotor is turning at zero slip
Derive from locked rotor and zero slip conditions
Accounts for wire loss and stator core loss
Leakage inductance usually larger than for a simple transformer because of air gap and slot shape
Electrical Equivalent Circuit with Ideal Transformer
• S is the “slip” or
S
f SLIP
f NoLoad
Electrical Equivalent Circuit Referred to the Stator
• Basis for calculating efficiency, start inrush, etc.
• Mechanical energy is loss in (1  SS )R ; all else is heat
R
Things Left Out!
• Inrush current
• No-load mechanical drag from cooling, bearing friction, etc.
• Design tradeoffs with cost
• Still to go: single phase operation
• Government efficiency regulation
• Recently – March 2015 – applies to motors to ¼ HP