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A 220V d.c. shunt motor takes a total current of 80A and runs at 800 r.p.m. The resistance of shunt field is 50Ω and that of armature 0.1Ω The iron and friction losses amount to 1600W. What is the driving power of the motor?

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A 220V d.c. shunt motor takes a total current of 80A and runs at 800 r.p.m. The resistance of shunt field is 50Ω and that of armature 0.1Ω The iron and friction losses amount to 1600W. What is the driving power of the motor?
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1 Answer

  1. This answer was edited.

    Since Ish = 220/ 50 = 4.4 A
    Ia = 80-4.4 = 75.6 Amp
    Eb = V – IaRa
    = 220- (75.6 x 0.1) = 212.44
    Driving Power Eb.Ia = 212.44×75.6
    = 16060 N-m
    Output Power = Driving Power – Iron and Friction Losses
    = 16060-1600
    = 14460 W
    Armature Torque
    Ta = EbIa/2πN if N in r.p.s
    Ta = (60EbIa)/(2πN) if N in r.p.m
    so; Ta = (9.55EbIa)/N if N in r.p.s
    = (9.55 x 212.44 x 75.6)/800
    = 191.7 N-m

    Since the whole of armature torque is not available for doing useful work, a certain part is required to overcome iron & friction losses
    Hence Lost torque = 9.55x (1600/800)
    = 19.1 N-m

    Hence Shaft torque = Ta – Tlost
    = 191.7 – 19.1
    = 172.6 N-m

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