Induction motor coverts an electrical power supplied it to into mechanical power. The various stages in this conversion is called Power flow in an induction motor.

The three phase supply is given to the stator is the net electrical input to the motor. If the power factor is Cos Φ and V_{1 , }I_{1} are phase values of supply voltage and current.

P_{in} = V_{1 }I_{1 }Cos_{ }Φ (Power input per phase)

By using phasor Diagram

V_{1 }Cos_{ }Φ = [(-E_{1}) Cos Φ_{1} + I_{1} R_{1} Cos 0^{∘} + j I_{1 }X_{1 }Cos90^{∘}]
P_{in } = [(-E_{1}) Cos Φ_{1} + I_{1} R_{1} Cos 0^{∘} + j I_{1 }X_{1 }Cos90^{∘}] I_{1}
P_{in }= -(E_{1}) [ I_{o} CosΦ_{o }+ (-I^{’}_{2 }CosΦ_{2}) ] + I^{2}_{1 }R_{1}
P_{in }= - E_{1}I_{o} CosΦ_{o }+ E_{1 }I^{’}_{2 }CosΦ_{2 }+ I^{2}_{1 }R_{1}
Stator Core losses = E_{1}I_{o} CosΦ_{o} and Stator Cu losses = I^{2}_{1 }R_{1}Air Gap power (P2) = P_{in }- (Stator Core losses + Stator Cu losses )
P2 = E_{1 }I^{’}_{2 }CosΦ_{2}P2 = ( E^{’}_{2} I^{’}_{2 }CosΦ_{2 }) / s

P2 = [ ( I^{’}_{2 })^{2} R^{’}_{2 }] / s
P2 = P_{c }/ s
P_{c } = Rotor Cu lossGross mechanical Power (Pg) = P_{2 } - P_{c}_{P2 : Pc : Pg = 1 : s :1 - s }
Brake Horse Power (Power available at the shaft) = Pg - [ mechanical losses]

Electrical Engineering (EE) Student at Malaviya National Institute of Technology (MNIT), Jaipur
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