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# 3 Phase Induction Motor Assignment

Added on - 27 Mar 2020

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Induction Motor1|P a g eThree phase induction motors
Induction MotorContentsSolution1 (a)....................................................................................................................................3Solution1 (b)....................................................................................................................................3Solution2..........................................................................................................................................3Solution3 (a)....................................................................................................................................4Solution3 (b)....................................................................................................................................4Solution3 (c)....................................................................................................................................4Solution4 (a)....................................................................................................................................5Solution4 (b)....................................................................................................................................6Solution5 (i).....................................................................................................................................7Solution5 (ii)....................................................................................................................................9Solution5 (iii)...................................................................................................................................9Solution5 (iv).................................................................................................................................10Solution6 (i)...................................................................................................................................10Solution6 (ii)..................................................................................................................................12Solution7 (i)...................................................................................................................................12Solution7 (ii)..................................................................................................................................13Solution7 (iii).................................................................................................................................14Bibliography..................................................................................................................................152|P a g e
Induction MotorSolution1 (a)Practically, It is not possible to catch the sped of rotor with their stator field, and if it happenthere is no e.m.f will be generated, so the difference between speed of rot and stator filed isknown as Slip. It is generally denoted as s in terms of %. Mathematically it is written as%slips=NsNNsx100AndNsNis termed as slip speedAfter calculating motor speed is given asN=Ns(1s)Solution1 (b)As given in question,P = 4, F = 60 Hz, s = 4%We know that,Ns=120FP=120604=1800rpmNow motor speed will beN=Ns(1s)N=1800(10.004)N=1800x0.996=1728rpmSolution2To establish the relationship between we have proceed as follows,Gross output of rotor = Pm=Tgω=Tgx2πNRotor Input = P2=Tgωs=Tgx2πNs......(i)Rotor copper loss = P2– Pm=Tgx2π(NsN).......(ii)Dividing equation (i) and (ii) we getRotorCulossRotorInput=NsNNswhich is nothing but= sWe also know that,Rotor copper loss = s x rotor input (P2) = s x across the air-gap = sP2.....(iii)3|P a g e
Induction MotorIn another way,Rotor Input (P2¿=RotorCopperLosss...........(iv)Again, Rotor Gross Output = Pm=Rotor input (P2) - Rotor Copper loss=Rotor input (P2)- Rotor copper loss =input-s xRotor input=(1-s)input (P2)From above equation (i) (iii) and (iv)P2=Pm=I2R1:(1s):s,P2=Pm=Pcr1:(1s):sSolution3 (a)The synchronous speed of the motor is calculated by the formulaNs=120FPWhere F stands for supplied frequency in HzAnd P stands for number of poles.Solution3 (b)As given in question,P = 6,F = 50 HzSynchronous speedNs=Ns=120FP=120x506=1000rpmSolution3 (c)As given in question, N = 1150Slip speed =NsN= 50Then,Ns=50+1150=1200We know thatNs=120FP,F=NsxP1204|P a g e

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