SINGLE PHASE INDUCTION MOTOR |
Imagine the single phase induction motor is made up of one stator winding and two imaginary rotor windings. One rotor is rotating in forward direction i.e. in the direction of rotating magnetic field with slip s while other is rotating in backward direction i.e. in direction of oppositely directed rotating magnetic field with slip 2 - s.
To develop the equivalent circuit, let us assume initially that the core loss is absent.
Let the stator impedance be Z Ω
Z = R1 + j X1
Where R1 = Stator resistance
X1 = Stator reactance
And X2 = rotor reactance referred to stator
R2 = rotor resistance referred to stator
Hence the impedance of each rotor is r2 + j x2 where
x2 = X2/2
The resistance of forward field rotor is r2/s while the resistance of backward field rotor is r2 /(2 - s). The r2 value is half of the actual rotor resistance referred to stator.
As the core loss is neglected, Ro is not existing in the equivalent circuit. The xo is half of the actual magnetising reactance of the motor. So the equivalent circuit referred to stator is shown in the Fig.1.FIGURE.1 |
Now the impedance of the forward field rotor is Zf which is parallel combination of (0 + j xo ) and (r2 /s) + j x2
While the impedance of the backward field rotor is Zb which is parallel combination of (0 + j xo) and (r2 / 2-s) + j x2.
Under standstill condition, s = 1 and 2 - s = 1 hence Zf = Zb and hence Vf = Vb . But in the running condition, Vf becomes almost 90 to 95% of the applied voltage.
... Zeq = Z1 + Zf + Zb = Equivalent impedance
Let I2f = Current through forward rotor referred to stator
and I2b = Current through backward rotor referred to stator
... I2f = /((r2/s) + j x2) where Vf = I1 x Zf
and I2b = /((r2/2-s) + j x2)
Pf = Power input to forward field rotor
= (I2f)2 (r2/s) watts
Pb = Power input to backward field rotor
= (I2b)2 (r2/2-s) watts
Pm = (1 - s){ Net power input}
= (1 - s) (Pf - Pb ) watts
Pout = Pm - mechanical loss - core loss
... Tf = forward torque = Pf /(2πN/60) N-m
and Tb = backward torque = Pb /(2πN/60) N-m
T = net torque = Tf - Tb
while Tsh = shaft torque = Pout /(2πN/60) N-m
%η = (net output / net input) x 100
2. With core loss
If the core loss is to be considered then it is necessary to connect a resistance in parallel with, in an exciting branch of each rotor is half the value of actual core loss resistance. Thus the equivalent circuit with core loss can be shown as in the Fig. 2.
FIGURE.2 |
= ro║(j xo )
and Zob = Equivalent impedance of exciting branch in backward rotor
= ro║(j xo )
... Zf = Zof ║( r2/s + j x2 )
All other expressions remains same as stated earlier in case of equivalent circuit without core loss.
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