 # The apparent inconsistency between the above two equa

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The apparent inconsistency between the above two equa-tions can be reconciled in the following way :Suppose that the flux of a motor is decreased by decreasing the field current.Then, followingsequence of events take place :1.Back e.m.f.Eb(=NΦ/K) drops instantly (the speed remains constant because of inertia ofthe heavy armature).2.Due to decrease inEb,Iais increased becauseIa= (VEb)/Ra.Moreover, a small reductionin flux produces a proportionately large increase in armature current.3.Hence, the equationTaΦIa, a small decrease inφis more than counterbalanced by alargeincrease inIawith the result that there is a netincreaseinTa.4.This increase inTaproduces an increase in motor speed.It is seen from above that with the applied voltageVheld constant, motor speed varies inverselyas the flux.However, it is possible to increase flux and, at the same time, increase the speed providedIais held constant as is actually done in a d.c. servomotor.Example 29.20.A 4-pole series motor has 944 wave-connected armature conductors.At acertain load, the flux per pole is 34.6 mWb and the total mechanical torque developed is 209 N-m.Calculate the line current taken by the motor and the speed at which it will run with an appliedvoltage of 500 V.Total motor resistance is 3 ohm.(Elect. Engg. AMIETE Sec. A Part II June 1991)Solution.Ta= 0.159φZIa(P/A) N-m209 = 0.159×34.6×103×944×Ia(4/2);Ia= 20.1 AEa=VIaRa= 50020.1×3 = 439.7 V
1010Electrical TechnologyNow, speed may be found either by using the relation forEborTaas given in Art.Eb=ΦZN×(P/A) or 439.7 = 34.6×103×944×N×2N=6.73 r.p.s.or382.2 r.p.m.Example 29.21.A 250-V shunt motor runs at 1000 r.p.m. at no-load and takes 8A.The totalarmature and shunt field resistances are respectively 0.2and 250.Calculate the speed whenloaded and taking 50 A.Assume the flux to be constant.(Elect. Engg. A.M.Ae. S.I. June 1991)NN=EENΦ×Φ= Φ=

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Electric motor,
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