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# A DC permanent magnet motor draws 5A from a 200V DC source, under no-load conditions, and...

A DC permanent magnet motor draws 5A from a 200V DC source, under no-load conditions, and draws 50A from the same source, and has a speed of 2500rpm at full load. Ra = 1.4Ω. Determine:

a) the rotation loss* (W)

b) the no-load speed (rpm)

c) the output torque (Nm)

d) the efficiency of the motor at full load.

(*Assume the rotational loss is proportional to speed!)

a) Since the motor is operating under no load condition, net mechanical output power is zero. Hence the gross power developed by the armature must supply the core loss and friction losses of the motor.

Therefore, total rotation loss = core loss + friction loss

= Source voltage x armature current + (armature current)2 x armature resistance

= 200 V x 5 A + (5 A)2 x 1.4

= 1035 W

b) Full load speed = 2500 rpm = 2500x2/60 rad/s 261.8 rad/s

Source voltage = armature current x armature resistance + ke x speed v [ ke = back EMF constant of motor ]

In case of full load: 200 = 50 x 1.4 + ke x 261.8 => ke = 0.4966 V/(rad/s)

In case of no load : 200 = 5 x 1.4 + 0.4966 x no-load speed 388.7 rad/s = (388.7x60/2) rpm 3712 rpm

c) Electrical power due to back emf = mechanical power due to rotation

=> back EMF constant x rotational speed x armature current = Torque x rotational speed

=> No load Torque = back EMF constant x armature current = 0.4966 V /(rad/s) x 5 A = 2.483 Nm

Full load torque = 0.4966 V /(rad/s) x 50 A = 24.83 Nm

d) Input power at full load = source voltage x armature current = 200 V x 50 A = 10000 W

Output power at full load = Torque x rotational speed = 24.83 Nm x 261.8 rad/s 6500.5 W

Efficiency = Output power/Input power = 6500.5 W / 10000 W x100% 65 %

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