ECE 441 1
Quadrature-Field Theoryand Induction-Motor Action
• Single-phase induction motor cannot develop a rotating magnetic field
• Needs an “auxiliary” method– That method is another (auxiliary) winding
ECE 441 2
Single-Phase Squirrel-Cage Induction Motor
There are two “Main Poles”
Squirrel-Cage Rotor
Single-Phase Mains Supply
ECE 441 3
Excite the Main WindingStator flux is produced across the air gap – as shown, it is increasing in the downward direction.
The squirrel-cage rotor responds with a mmf in the opposite (upward) direction.
Magnetic axis of the rotor is in line with the magnetic axis of the stator – no rotation!
ECE 441 4
“Main” pole flux (Φ) increasing in the downward direction
Rotor mmf develops in the upward direction
Current “into” the page
Current “out of” the page
ECE 441 5
Cause the rotor to turn clockwise
Rotor conductors cut through the main pole flux.
Current is induced in the rotor bars as shown, producing a magnetic flux perpendicular to the main pole flux. This is known as “Quadrature” flux.
ECE 441 6
The quadrature flux is sustained as the rotor conductors shift their positions – other conductors replace them.
ECE 441 7
Phase Relationship Between the Direct and Quadrature Flux
The “speed” voltage is in phase with the flux that created it, and the flux due to current is in phase with the current that caused it. The instantaneous amplitudes of the direct and quadrature flux are shown above.
ECE 441 8
Resultant Flux
• Determine from
ECE 441 9
Resultant Flux Rotates CW
ECE 441 10
Phase-SplittingSplit-Phase Induction Motor
ECE 441 11
Provides “direct” flux
Provides quadrature flux
Ensures phase difference between winding currents
Start winding
ECE 441 12
Equivalent Circuit
ECE 441 13
Purpose of the “Phase-Splitter”
• Make the current in the Auxiliary Winding out of phase with the current in the Main Winding.
• This results in the quadrature field and the main field being out of phase.
• The locked-rotor torque will be given by
sin
mw aw
lr sp mw aw
i i
T k I I
ECE 441 14
Example 6-1
• The main and auxiliary windings of a hypothetical 120 V, 60 Hz, split-phase motor have the following locked-rotor parameters:– Rmw=2.00 Ω Xmw=3.50 Ω
– Raw=9.15 Ω Xaw=8.40 Ω
• The motor is connected to a 120 V system. Determine
ECE 441 15
Example 6-1 continued
• The locked-rotor current in each winding
2.00 3.50 4.0311 60.2511
9.15 8.40 12.4211 42.553mw mw mw
aw aw aw
Z R jX j
Z R jX j
ECE 441 16
120 029.8 60.3
4.0311 60.2511120 0
9.66 42.612.4211 42.5530
T
mw
mw
T
aw
aw
VI A
ZV
I AZ
Example 6-1 continued
ECE 441 17
Example 6-1 continued
• The phase displacement angle between the main and auxiliary currents
60.3 42.6 17.7mw awi i
ECE 441 18
Example 6-1 continued
• The locked-rotor torque in terms of the machine constant
sin
(29.8)(9.66)sin17.7 87.52
lr sp mw aw
lr sp sp
T k I I
T k k
ECE 441 19
Example 6-1 continued
• External resistance required in series with the auxiliary winding in order to obtain a 30 phase displacement between the currents in the two windings.
ECE 441 20
Example 6-1 continued
• Phasor diagram for the new conditions
' 60.3 30 30.3awi
ECE 441 21
Example 6-1 continued
' '
' '
'
'
030.3
'
30.3aw
aw
T T
aw aw
aw aw Z
Z
aw x aw aw
V VI I
Z Z
Z R R jX
ECE 441 22
Example 6-1 continued
'
'
tan
tan
8.409.15 14.38 9.15 5.23
tan30.3
aw
aw
aw
Z
aw x
aw
x aw
Z
XR R
XR R
R
ECE 441 23
Example 6-1 continued
• Locked-rotor torque for the condition in d
'
'
'
sin
120 09.15 5.23 8.40
7.2 30.29
(29.8)(7.2)sin30 107.1
107.1
lr sp mw aw
T
aw aw
aw
aw
lr sp sp
lr sp
T k I I
VI I
Z jI
T k k
T k
ECE 441 24
Example 6-1 continued
• % increase in locked-rotor torque due to the adding of additional resistance
107.1 87.52100% 22.37%
87.52sp sp
sp
k kX
k
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