AEE
Unit – 4AC motor Q1) The stator of a 3-phase induction motor has 5 slots per pole per phase. If supply frequency is 50Hz. Calculate a) number of stator poles produced and total number of slots in stator. b) Speed of rotating stator flux.? A1)
Q2) A 3-phase induction motor is wound for 5 poles and is supplied from 50 Hz. Calculate a) synchronous speed b) rotor speed when slip is 4% c) rotor frequency when rotor runs at 500 rpm? A2)
Q3) A 3-phase, 50 Hz,3-pole induction motor has a slip of 4%. Calculate, a) speed of rotor. B) frequency of rotor emf. If the rotor as a resistance of 2ohm, and standstill reactance of 4ohm, calculate the power factor c) at standstill and d) at a speed of 1200rpm? A3)
a) P=2n=2*5=10 poles Total number of slots= 5 slots/pole/phase * 10 poles * 3-phase=150 a) Ns=
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a) Ns= b) rotor speed N=Ns(1-s) =1200(1-0.04) =1152 rpm c)when rotor speed is 500 rpm, slip s=(Ns-N)/Ns= (1200-500)/1200=0.58 rotor current frequency f’=sf=0.58x50=29.17 Hz
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Ns=
R2=2ohm, X2=4ohm Z2=R2+jX2=2+j4=4.47
slip at speed 1200rpm s= Rotor impedance at slip s=0.4 is Z2s=R2+jsX2=2+jx0.4x4=2.56
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Q4) A 3-phase induction motor star connected rotor has an induced emf of 70volts between slip rings at standstill on open circuit. The rotor has a resistance and reactance per phase of 1ohm and 5ohm respectively. Calculate current/phase and power factor when slip rings are short circuited? A4)
Q5) A 3-phase induction motor star connected rotor has an induced emf of 70volts between slip rings at standstill on open circuit. The rotor has a resistance and reactance per phase of 1ohm and 5ohm respectively. Calculate current/phase and power factor when slip rings are connected to star connected rheostat of 2ohm? A5)
Q6) A 6-pole, 3-phase induction motor operates from a supply whose frequency is 50Hz. Calculate i) speed at which the magnetic field of stator is rotating. ii)speed of rotor when slip is 4%. iii)frequency of rotor current when slip is 3%? A6) i)Stator revolves at synchronous speed. So, Ns=120f/P=120 x 50/6=1000rpmii)rotor speed N=(1-s) Ns = (1-0.04) x 1000=960rpmiii)frequency of rotor current f’=sf=0.03 x 50=90rpm Q7) Explain the speed torque characteristics of IM?A7) There is no direct relation in the speed and torque. At, full load the motor runs at a speed N. When mechanical load increases motor speed decreases till the motor torque again becomes equal to the load torque. As long as two torques are equal the motor runs at constant speed. When the supply frequency increases the speed increases and reduces the maximum torque of motor. So, i) F increases, N increases, Tmax decreases.ii) V increases, Tmax increases.
Fig 1(a) Speed Torque Characteristics If the above two features are combined, we see that speed increases and torque kept same. If frequency is increased up to certain level and keeping voltage constant the motor speed increases and torque decreases. But when supply frequency is kept same and voltage is increased, then the motor characteristics in this case are stretched and the value of Tmax increases. Q8) Explain the effect of resistance on speed torque characteristics?A8) It is conventional method, performed by adding a variable rheostat in series with the motor winding. It has many drawbacks which are listed in this section. The torque equation for 3 phase induction motor is
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At the normal operating condition, the slip increases with increasing torque so there is curve is linear. For a fixed load curve, the speed is downward from n1 to n4. From this figure we can also obtain the maximum torque at the starting of the resistance r2”. So, this method has an advantage of achieving maximum torque at the starting period.But the maximum torque is independent of the rotor resistance as per the equation,
Q9) Explain the working of shaded pole IM?A9) This motor works I three operating regions. The poles of the motor are shaded by copper ring which are inductive. They are divided in two equal halves. The operation of such motor takes place at different flux changes:When flux change from zero to positive maximum the copper band is short circuited and maximum current flows through it. This current produces its own flux. This flux opposes the main flux. So, in non-shaded part there is non uniform distribution of flux casing the magnetic axis to shift in middle of the non-shaded part. When flux is still at its maximum and very less induced emf in the shaded portion. The main flux is not affected by the induced emf and hence, uniform flux remains. The flux the decreases from maximum positive to zero, the current is high. The copper band is short circuited and maximum current flows through it. This current produces its own flux. This flux opposes the main flux. So, in shaded part there is non-uniform distribution of flux casing the magnetic axis to shift in middle of the shaded part.
Fig 2 Shaded Pole Induction Motor Q10) Draw the phasor and explain the working of split-phase IM?
A10) In this type of motor an auxiliary winding is connected in addition to the main winding. There is one switch called centrifugal switch (S) which disconnects the auxiliary winding from the main winding. If the starting winding has high resistance than only a phase difference can be created between these windings. The main purpose of creating this phase difference is to get the magnetic field rotated because IM cannot self-start.
Fig 3 Split phase IMFor highly resistive winding the current is almost in phase with the voltage and for highly inductive winding the current lag behind the voltage by large angle. The starting winding is highly resistive so, the current flowing in the starting winding lags behind the applied voltage by very small angle and the running winding is highly inductive in nature so, the current flowing in running winding lags behind applied voltage by large angle. The resultant of these two currents is IT. The resultant of these two currents produces rotating magnetic field which rotates in one direction. In split phase induction motor, the starting and main current get splited from each other by some angle so this motor got its name as split phase induction motor.
Standstill emf/rotor phase=70/ Rotor impedance/phase= Rotor current/phase=40.4/5.09=7.92A Power factor cosφ=0.99
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Rotor resistance/phase=2+1=3 ohm Rotor impedance/phase= Rotor current/phase= (70/ Cosφ=3/5.83=0.514
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From above equation slip is directly proportional to rotor resistance. Hence speed decreases. This method is applicable to wound rotor induction motor.
Fig 1 (b): Torque speed characteristics
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At the normal operating condition, the slip increases with increasing torque so there is curve is linear. For a fixed load curve, the speed is downward from n1 to n4. From this figure we can also obtain the maximum torque at the starting of the resistance r2”. So, this method has an advantage of achieving maximum torque at the starting period.But the maximum torque is independent of the rotor resistance as per the equation,
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A10) In this type of motor an auxiliary winding is connected in addition to the main winding. There is one switch called centrifugal switch (S) which disconnects the auxiliary winding from the main winding. If the starting winding has high resistance than only a phase difference can be created between these windings. The main purpose of creating this phase difference is to get the magnetic field rotated because IM cannot self-start.
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