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according to the induction principles described in 17; however, the two components are equal and opposite, and no net torque results if the rotor is at rest The resulting mmf is pulsating (ie, changing in amplitude), but not rotating in space, as it would be in a polyphase stator If the rotor is made to turn in either direction, however, the two mmf s will not be equal any longer, because the motion of the rotor will induce an additional mmf, which will add to one of the two mmf s and subtract from the other Thus, a net torque will be established, causing the motor to continue its rotation in the same direction in which it was started In particular, if the rotor is started in the forward direction, the forward mmf, F + , will be greater than the backward mmf, and the motor will continue to rotate in the forward direction Figure 1826 depicts an equivalent circuit for the single-phase induction motor with stationary rotor, where the parameters in the circuit are de ned as follows: Vs = supply voltage RS = resistance of stator winding XS = leakage reactance of stator winding Xm = magnetizing reactance of stator winding XR = leakage reactance of rotor referred to stator at standstill RR = leakage resistance of rotor referred to stator at standstill Eb = voltage induced in the stator winding by the (stationary) pulsating ux in air gap Figure 1827 depicts the equivalent circuit for the same motor with the rotor rotating with slip s Note that the circuit is asymmetrical, because of the different air-gap ux forward and backward components, Ef and Eb , respectively The factors of 05 come from the resolution of the pulsating stator mmf into forward and backward components Note further that the re ected rotor impedance is asymmetrical because of the presence of the slip parameter in the expression for the re ected rotor resistance Further, the circuit model also con rms that the forward induced voltage, Ef , must be greater than the backward voltage, Eb , since the slip is always less than 1 It can be shown that the torque components in the forward and backward directions are given by the expressions Tf = and Tb = Pb s (1812) Pf s (1811)
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Figure 1826 Circuit model for single-phase induction motor with rotor at standstill
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05RR s
Vs + Eb
05XR
05Xm
05RR 2 s
Figure 1827 Circuit model for single-phase induction motor with rotor in motion
where s is the synchronous speed and Pf = Is2 Rf Here, Rf is the resistive component of the forward eld impedance; also, Pb = Is2 Rb (1814) (1813)
18
Special-Purpose Electric Machines
where Rb is the resistive component of the backward eld impedance Since the torque produced by the backward eld is in the opposite direction to that produced by the forward eld, the net torque will consist of the difference between the two: T = T f Tb = Is2 (Rf Rb ) s (1815a)
The mechanical power developed by the motor is Pmech = T m = T s (1 s) = (Pf Pb )(1 s) = Is2 (Rf Rb )(1 s) (1815b)
EXAMPLE 189 Slip in a Single-Phase Induction Motor
Problem
Find the slip of the eld in the forward and backward directions for a single-phase induction machine
Solution
Known Quantities: Motor operating characteristics Find: Forward slip, sf ; backward slip, sb Schematics, Diagrams, Circuits, and Given Data: Motor operating characteristics:
115 V; 60 Hz; 4 poles; 1710 rev/min
Analysis: We rst determine the synchronous speed of the motor:
ns =
120 60 120f = = 1,800 rev/min p 4 ns n 1,800 1,710 = 005 = 5% = ns 1,800
The slip in the forward direction (direction of rotation of the motor) can now be computed: sf =
The slip in the backward direction can be computed as follows, with reference to Figure 1826: sb = 2 sf = 2 005 = 195
EXAMPLE 1810 Analysis of Single-Phase Induction Motor
Problem
Find the input current and generated torque for a single-phase induction motor
Solution
Known Quantities: Motor operating characteristics and circuit parameters
Part III
Electromechanics
Find: Motor input (stator) current, IS ; motor torque, T Schematics, Diagrams, Circuits, and Given Data: Motor operating data:
hp; 110-V; ;
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