what is the mechanical power developed by the motor equation
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V = Eb + IaRa
This is known as voltage equation of a motor.
Multiplying both sides by Ia, we get
VIa = EbIa + Ia2Ra
Here VIa = electrical input to the armature,
EbIa = electrical equivalent of mechanical power Pm developed in the armature, and
Ia2Ra = copper loss in the armature.
The power available at the pulley for doing useful work is somewhat less than the mechanical power developed by the armature.
This is evident, since there are certain mechanical/asses (such as bearing and windage field friction and iron losses) that must be supplied by the driving power of the motor.
Condition for maximum power. We know that, the mechanical power developed by the motor,
Pm = VIa – Ia2 Ra
Hence, mechanical power developed by a motor is maximum when back e.m.f. is equal to half the applied voltage. In practice, however, this condition is not realized because in that case current would be much beyond the normal current of the motor. Moreover, half the input would be wasted in the form of heat and taking other losses, such as mechanical and magnetic, into consideration, the efficiency of the motor will be below 50 per cent.
This is known as voltage equation of a motor.
Multiplying both sides by Ia, we get
VIa = EbIa + Ia2Ra
Here VIa = electrical input to the armature,
EbIa = electrical equivalent of mechanical power Pm developed in the armature, and
Ia2Ra = copper loss in the armature.
The power available at the pulley for doing useful work is somewhat less than the mechanical power developed by the armature.
This is evident, since there are certain mechanical/asses (such as bearing and windage field friction and iron losses) that must be supplied by the driving power of the motor.
Condition for maximum power. We know that, the mechanical power developed by the motor,
Pm = VIa – Ia2 Ra
Hence, mechanical power developed by a motor is maximum when back e.m.f. is equal to half the applied voltage. In practice, however, this condition is not realized because in that case current would be much beyond the normal current of the motor. Moreover, half the input would be wasted in the form of heat and taking other losses, such as mechanical and magnetic, into consideration, the efficiency of the motor will be below 50 per cent.
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