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3 Phase Induction Motor Assignment

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Added on  2020-03-28

3 Phase Induction Motor Assignment

   Added on 2020-03-28

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Induction Motor1 | P a g eThree phase induction motors
3 Phase Induction Motor Assignment_1
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
3 Phase Induction Motor Assignment_2
Induction MotorSolution1 (a)Practically, It is not possible to catch the sped of rotor with their stator field, and if it happen there is no e.m.f will be generated, so the difference between speed of rot and stator filed is known as Slip. It is generally denoted as s in terms of %. Mathematically it is written as%slips=NsNNsx100And NsN is termed as slip speedAfter calculating motor speed is given as N=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 be N=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 get RotorCulossRotorInput=NsNNs which 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
3 Phase Induction Motor Assignment_3
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 x Rotor 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 formula Ns=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 that Ns=120FP,F=NsxP1204 | P a g e
3 Phase Induction Motor Assignment_4

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