Unit-4
Electrical Machines
The fundamental principle of induction machine is the creation of rotating and sinusoidally distributed magnetic field in the air gap. Three phase balanced power supply is fed to the three-phase stator winding creates a synchronously rotating magnetic field. Due to relative speed between rotating flux and stationary conductors, an emf is induced. The frequency to induced emf is same as supply frequency.
As shown in below fig.a , the stator field is assumed to clockwise. The relative motion of rotor w.r.t stator is anticlockwise. From Right-hand rule, the direction of induced emf in rotor is outwards. By the effect of combined field as shown in fig b the rotor experiences a force tending to rotate it in clockwise direction. Hence rotor rotates in the same direction as the stator field.
The speed of this rotating field is called synchronous speed. If the rotor is initially stationary, its conductors will be subjected to a changing magnetic field, inducing current in the short-circuited rotor at the same frequency. The interaction of air gap flux and rotor mmf produces torque. At synchronous speed rotor cannot have any torque.
There are two main parts of induction motor:
a) Stator
b) Rotor
a) Stator:
It is stationary part of induction motor. It has number of stampings. It is wound with three phase winding which is fed from 3-phase supply. The number of poles here are defined, they are selected according to the speed required. If required speed is greater than we need less number of poles. The stator produces an alternating flux when fed with 3-phase supply which revolves with synchronous speed(Ns = 120f / P). The synchronous speed is inversely proportional to number of poles.
Fig: Stator
b) Rotor: It is rotating part of induction motor. There are two types of rotor
i)Squirrel cage rotor:
The rotor consists of a cylindrical core with parallel slots for carrying the rotor conductor. Each slot has one copper or aluminium bar. Each end of all bars is joined with metal ring. The entire construction resembles a squirrel cage. The rotor is not connected electrically to the supply it has induced current from stator. Almost 90% of induction motors are squirrel cage type. But is has a disadvantage of low starting torque, because the rotor bars are permanently short-circuited.
Fig Squirrel cage rotor
Ii) Phase-wound rotor: It has laminated cylindrical core and the windings are uniformly distributed in the slots which are usually star connected. The other three winding terminals are brought out and connected to three insulated slip rings mounted on shaft with brushes resting on them. The three brushes are connected to a 3-phase star-connected rheostat as shown in Figure below.
Fig: 3-phase star connected
At starting, the external resistances are included in the rotor circuit to give a large starting torque. These resistances are gradually reduced to zero as the motor runs up to speed. The external resistances are used during starting period only. When the motor attains normal speed, the three brushes are short-circuited so that the wound rotor runs like a squirrel cage rotor.
Slip:
The rotor can never have same speed as that of stator. If it occurs then there will not be relative speed between the two, hence no rotor emf, no rotor current and so no torque to maintain rotation. Due to this reason speed of rotor is always less than the stator field. The difference between the synchronous speed Ns and the actual speed N of the rotor is called as slip.
% slip s=
Therefore, rotor speed N=Ns(1-s).
Frequency of rotor current:
When rotor is stationary its frequency is same as the supply frequency. But when rotor revolves its frequency depends upon the slip-speed. Let frequency of rotor current be f’
Ns-N=
But Ns=
Rotor current have same frequency
=
MMF produced by rotor current:
a) For standstill condition:
Let E2=emf induced per phase of rotor at standstill
R2=resistance per phase of rotor.
X2=reactance per phase of rotor at standstill=2
f1L2
Z2=rotor impedance per phase
I2=rotor current per phase
Z2=R2+jX2
=
Power Factor cosφ2=
=
b) Rotor current at slip s:
Induced emf per phase in rotor winding at slip s is E2s=sE2
Rotor winding reactance per phase at slip s will be X2s=2
f2L=sX2
Rotor winding impedance per phase at slip s is Z2s=R2+jX2s=R2+jsX2
Rotor current at slip s, I2s=
Power factor at slip s is cosφ2s=
Que) 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.?
Sol: a) P=2n=2*5=10 poles
Total number of slots= 5 slots/pole/phase * 10 poles * 3 phase=150
c) Ns=
=120x50/10=600rpm
Que) 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?
Sol: a) Ns=
=120x50/5=1200 rpm
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
Que) 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?
Sol: Ns=
=120x50/3=2000 rpm
a) Speed of rotor N=Ns(1-s)=2000(1-0.04)=1920 rpm
b) Frequency or rotor = f’=sf=0.04x50=2Hz
R2=2ohm, X2=4ohm
Z2=R2+jX2=2+j4=4.47
63.43o ohm
c)Power factor cosφ2=0.44(lag)
Slip at speed 1200rpm s=
2000-1200/2000=0.4
Rotor impedance at slip s=0.4 is
Z2s=R2+jsX2=2+jx0.4x4=2.56
oohm
Power factor at 1200rpm is cosφ2s=cos38.66o=0.78(lag)
Q) 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?
Sol: Standstill emf/rotor phase=70/
Rotor impedance/phase=
ohm
Rotor current/phase=40.4/5.09=7.92A
Power factor cosφ=0.99
Que)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?
Sol: Rotor resistance/phase=2+1=3 ohm
Rotor impedance/phase=
=5.83 ohm
Rotor current/phase=(70/
)/5.83=6.93A
Cosφ=3/5.83=0.514
Que) 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%?
Sol: i)Stator revolves at synchronous speed. So, Ns=120f/P=120 x 50/6=1000rpm
Ii)rotor speed N=(1-s)Ns=(1-0.04) x 1000=960rpm
Iii)frequency of rotor current f’=sf=0.03 x 50=90rpm
The below curve can be divided into three regions.
Fig: Torque-Slip Characteristics
i)Motoring Region(0≤s≤1): The induction motor rotates in the same direction as that of the field. Here the speed decreases and torque increases till breakdown torque is reached. In this region air-gap flux is nearly constant. After breakdown torque Tmax , torque decreases and slip increases.
Ii)Generating Region(s<0): In this machine operates as a generator. The rotor rotates at a speed greater than synchronous speed in same direction as that of rotating magnetic field. Due to super synchronous speed the slip becomes negative and creates negative regenerative torque.
Iii)Plugging Region(1≤s≤2): Here slip becomes greater than unity. So, motor rotates in the opposite direction of the rotating magnetic field. Here the stator field is reversed by changing the phase sequence of input supply, due to this the direction of rotating magnetic field also changes. Here torque is positive but speed is negative. So, plugging torque appears as breaking torque. This method of breaking is called plugging.
In general, we can categorise losses as:
i)Constant losses: They remain constant over the normal working range of induction motor. By no-load test we can find these losses. The losses falling in this category are Iron or core loss, Mechanical Loss and Brush Friction loss.
a)Iron or core loss: The core losses depends on the frequency of supply voltage. As the frequency of rotor is always less than that of stator. Hence, core loss in rotor is less compared to that of stator. The core loss can be further classified as hysteresis and eddy current losses.
b)Mechanical and Brush Friction Loss: These are low at starting and increases with increase in speed. In induction motor speed is usually constant hence these losses also remain constant. Mechanical losses occur in bearings and brush friction losses occur in wound rotor induction motor.
Ii)Variable losses: These losses occur due to current flowing in stator and rotor windings. As the load changes, the current flowing in rotor and stator winding also changes and hence these losses also change. Therefore, these losses are called variable losses. The copper losses are obtained by performing blocked rotor test on three phase induction motor.
Efficiency
Rotor Efficiency=
=
3-phase induction moor efficiency=
=
Que) The power input to a rotor of 440V, 50Hz, 6-pole, 3-phase, induction motor is 60kW. The rotor force is observed to make 80 alterations per minute. Calculate i) slip ii) rotor speed iii) rotor cu loss per phase?
Sol: 80 alterations per minute=80/60 cycles per second=1.333Hz=sf
i)slip=1.333/50=0.027 P.u.=2.7%
Ii)Rotor speed N=(1-s) Ns=(1-0.027) x 1000=973.33rpm
Ns=120f/P=120 x 50/6=1000rpm
Iii)rotor cu loss/phase=
=
=0.54kW
Que) The power input to a 3-phase induction motor is 80kW. The stator loss is total 1kW. Find the mechanical power developed and the rotor cu loss per phase if the slip is 3%?
Sol: P2= stator input – stator losses=80-1=79kW
Pm=(1-s)P2==(1-0.03) x 79=76.63kW
Total rotor cu loss=sP2=0.03 x 79=2.37kW
Rotor cu loss per phase=2.37x1000/3=790W
Que) The power input to a rotor of a 400V, 50hz,6-pole, 3-phaseinduction motor is 15kW. The slip is 4%. Calculate i) frequency of rotor current. i)rotor speed. Iii)rotor cu loss iv) rotor resistance/phase if rotor current is 50A?
Sol: i) frequency of rotor current=sf=0.04 x 50=2Hz
Ii) Rotor speed N=(1-s) Ns=(1-0.04) x 1000=960rpm
Ns=120f/P=120 x 50/6=1000rpm
Iii) rotor cu loss=s x rotor input=0.04 x 15=0.6kW
Iv)rotor cu loss/phase=0.6 x 1000/3=200W
602R2=200
R2=0.055Ω
Starting:
Direct on-line starter (Full voltage starting method)
In this induction motor is directly connected to the 3-phase supply. The DOL starter applies full line voltage to the motor terminals.
Fig: D.O.L Starter
If the high input current does not cause any excess voltage drop in circuit than only this starter can be used. The value of armature current in motor is given as
Ia=
The expression for starting torque is given as
Ts starting torque
Tf Full load torque
If per phase rotor current at full load
Is per phase rotor current at starting
sf full load slip
ss starting slip
R2 rotor resistance
Ws synchronous speed of motor
In induction motor torque is given as
T=
The ratio of starting torque to full load torque is given as
x sf
The wiring diagram for a DOL starter is shown above. The working principle of a DOL starter begins with the connection to the 3-phase main with the motor. The control circuit is connected to any two phases and energized from them only. When we press the start button, the current flows through magnetizing coil and control circuit also.
The current energises thecontactor coil and leads to close the contacts, and hence 3-phase supply becomes available to the motor. When we stop it, the current discontinues.ss As the supply to motor breaks the machine stops. . The contactor coil (Magnetizing Coil) gets supply even though we release start button because when we release start button, it will get supply from the primary contacts as illustrated in the diagram of the Direct Online Starter.
The advantages of a DOL starter include:
- Simple and most economical starter.
- More comfortable to design, operate and control.
- Provides nearly full starting torque at starting.
- Easy to understand and troubleshoot.
- DOL starter connects the supply to the delta winding of the motor.
The reduced voltage starting methods are:
Star delta starter: The set up for star delta starter is shown below
Fig: Star delta starter
The starter phases are connected to star by TPDT (triple pole double throw switch). Initially the TPDT switch is at position 1 and when motor attains steady state switch is thrown to position 2. In position 1 terminals are short circuited and at position 2 a, b and c are connected to B, C and A. The expression for starting torque is given as
Ts starting torque
Tf Full load torque
If per phase rotor current at full load
Is per phase rotor current at starting
sf full load slip
ss starting slip
R2 rotor resistance
Ws synchronous speed of motor
In induction motor torque is given as
T=
The ratio of starting torque to full load torque is given as
x sf
Let V1 line voltage
Iss per phase starting current
Iss=
Starting current in stator when connected in delta position
Isd=
Isd=
Iss
Then the torque ratio equation will be
x sf
This shows that the reduced voltage method has an advantage of reducing the starting current but the disadvantage is that all these methods of reduced voltage causes the objectionable reduction in the starting torque.
Speed Control:The speed can be controlled by the following methods.
i) Changing stator Pole: This can be done by Multiple stator windings. In stator two windings are provided which are isolated and have different number of poles. By switching we can control the supply to one winding and hence control the speed.
Ii) Supply voltage control:The torque produced is proportional to 
. The value of 
is very small and R2 is also constant so T α
. But E2α V therefore, T α sV2.This equation shows that if supply voltage is decreased the torque also decreases.
Iii) Adding rheostat in the stator: As we already know Tα
. The value of 
is very small and E2 is also constant soTα
. Now if we increase rotor resistance, R2 torque decreases but to supply the same load torque must remain constant. So, we increase slip, which will further result in the decrease in rotor speed. Thus, by adding additional resistance in the rotor circuit, we can decrease the speed of the three-phase induction motor.
Iv) v/f control: 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.
Construction:
There are two main parts of induction motor:
c) Stator
d) Rotor
d) Stator:
It is stationary part of induction motor. It has number of stampings. It is wound with three phase winding which is fed from 3-phase supply. The number of poles here are defined, they are selected according to the speed required. If required speed is greater than we need less number of poles. The stator produces an alternating flux when fed withsupply which revolves with synchronous speed(Ns = 120f / P). The synchronous speed is inversely proportional to number of poles.
Fig: Stator
e) Rotor: It is rotating part of induction motor.
There are two types of rotor
i)Squirrel cage rotor:
The rotor consists of a cylindrical core with parallel slots for carrying the rotor conductor. Each slot has one copper or aluminium bar. Each end of all bars is joined with metal ring. The entire construction resembles a squirrel cage. The rotor is not connected electrically to the supply it has induced current from stator. Almost 90% of induction motors are squirrel cage type. But is has a disadvantage of low starting torque, because the rotor bars are permanently short-circuited.
Fig Squirrel cage rotor
Ii) Phase-wound rotor: It has laminated cylindrical core and the windings are uniformly distributed in the slots which are usually star connected. The other three winding terminals are brought out and connected to three insulated slip rings mounted on shaft with brushes resting on them. The three brushes are connected to a 3-phase star-connected rheostat as shown in Figure below.
Fig: 3-phase star connected
At starting, the external resistances are included in the rotor circuit to give a large starting torque. These resistances are gradually reduced to zero as the motor runs up to speed. The external resistances are used during starting period only. When the motor attains normal speed, the three brushes are short-circuited so that the wound rotor runs like a squirrel cage rotor.
Principle of working
The fundamental principle of induction machine is the creation of rotating and sinusoidally distributed magnetic field in the air gap. Three phase balanced power supply is fed to the three-phase stator winding creates a synchronously rotating magnetic field. Due to relative speed between rotating flux and stationary conductors, an emf is induced. The frequency to induced emf is same as supply frequency.
As shown in below fig.a , the stator field is assumed to clockwise. The relative motion of rotor w.r.t stator is anticlockwise. From Right-hand rule, the direction of induced emf in rotor is outwards. By the effect of combined field as shown in fig b the rotor experiences a force tending to rotate it in clockwise direction. Hence rotor rotates in the same direction as the stator field.
The speed of this rotating field is called synchronous speed. If the rotor is initially stationary, its conductors will be subjected to a changing magnetic field, inducing current in the short-circuited rotor at the same frequency. The interaction of air gap flux and rotor mmf produces torque. At synchronous speed rotor cannot have any torque.
Torque-Speed Characteristics:
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,
Iii) F increases, N increases, Tmax decreases.
Iv) V increases, Tmax increases.
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.
In Separately excited DC motor field windings are used to excite the field flux. Armature current is supplied to the rotor through brush and commutator. When the motor is excited by a field current if and armature current ia flows, the motor develops a back emf and a torque to balance the load torque at a particular speed. The current if is independent of ia. The armature current has no effect on te field current. The current if is less than ia.
The speed can be controlled by varying voltage in the constant torque region. In the constant power region, field flux should be reduced to achieve speed above the rated speed.
Fig: Armature voltage and field current control
The synchronous generator consists of rotor and stator. The rotor has field poles and stator has armature conductors. When the field poles rotate an induced alternating voltage is produced. The synchronous speed is given by
Ns=
P: Number of Poles
f: frequency
Basic mechanism in operation of these generators is EMI. An emf is induced due to the movement of conductors and flux. When the same poles of rotor and stator are near each other they repel and tend to rotate, and the rotor rotates (in anticlockwise direction). But half a period later stator poles having rotated around interchange their positions. Under this condition the poles attract each other and rotor rotates (in clockwise direction). Hence, due to continuous and rapid rotation of stator poles, the rotor is subjected to a torque which is rapidly reversing. When stator and rotor poles are attracting each other and rotor is not stationary but in CW direction with such a speed that it turns through one pole pitch by time the stator poles are interchanged. It means that if rotor poles change their position along with the stator poles, they also continuously experience a unidirectional torque.
