1. Quadrature-Field Theory and Induction-Motor Action
Single-phase induction motor cannot develop a rotating magnetic field
Needs an “auxiliary” method
That method is another (auxiliary) winding
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2. Single-Phase Squirrel-Cage Induction Motor
There are two “Main Poles”
Squirrel-Cage Rotor
Single-Phase Mains Supply
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3. Excite the Main Winding
Stator 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!
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4. Current “out of” the page
“Main” pole flux (Φ) increasing in the downward direction
Rotor mmf develops in the upward direction
Current “into” the page
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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.
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6. The quadrature flux is sustained as the rotor conductors shift their positions – other conductors replace them.
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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.
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8. Resultant Flux
Determine from
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9. Resultant Flux Rotates CW
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10. Phase-Splitting Split-Phase Induction Motor
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11. Provides “direct” flux
Start winding
Provides quadrature flux
Ensures phase difference between winding currents
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12. Equivalent Circuit
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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
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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
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15. Example 6-1 continued
The locked-rotor current in each winding
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16. Example 6-1 continued
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17. Example 6-1 continued
The phase displacement angle between the main and auxiliary currents
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18. Example 6-1 continued
The locked-rotor torque in terms of the machine constant
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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.
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20. Example 6-1 continued
Phasor diagram for the new conditions
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21. Example 6-1 continued
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22. Example 6-1 continued
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23. Example 6-1 continued Locked-rotor torque for the condition in d
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24. Example 6-1 continued
% increase in locked-rotor torque due to the adding of additional resistance
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