Electrical Drives and control for Automation-Module IV

THREE PHASE The 3-phase induction motor has a and a . The stator carries a 3-phase winding (called stator winding), while the rotor carries a short-circuited winding (called rotor winding). Only the stator winding is fed from 3-phase supply. The rotor winding derives its voltage and power from the externally energized stator winding through electromagnetic induction and hence the name. The Induction machines can be considered as a transformer with a rotating secondary in the sense that the power is transferred from the stator (primary) to the rotor (secondary) winding only by mutual induction and it can, therefore, be described as a “transformer type” a.c machine in which electrical energy is converted into mechanical energy. Advantages  It has simple and rugged construction.  It is relatively cheap.  It requires little maintenance.  It has high efficiency, good speed regulation and reasonably good power factor.  It has self starting torque. Disadvantages  It is essentially a constant speed motor and its speed cannot be changed easily.  Its starting torque is inferior to d.c motors. CONSTRUCTION A three phase induction motor has 4 main parts: 1. Frame 2. Stator 3. Rotor 4. Shaft and Bearings The stator houses a 3-phase winding which is fed with electric power. The rotor is placed inside the stator and both are separated by an air gap. 1. Frame It is the outerKTUNOTES.IN body of the motor. Its function are to support the stator core and winding, to protect the inner parts of the machine and serve as a ventilating housing or means of guiding the coolant into effective channels. 2. Stator It consists of a steel frame which encloses a hollow, cylindrical core made up of thin laminations of silicon steel to reduce hysteresis and eddy current losses. A number of evenly spaced slots are provided on the inner periphery of the laminations. The insulated conductors are placed in the stator slots and are suitably connected to form a balanced 3-phase star or delta connected circuit. The stator winding is usually 3-phase winding which is supplied from a 3- phase supply mains. The 3-phase stator winding is wound for a definite number of poles as per requirement of speed. When 3-phase supply is given to the stator winding, a rotating magnetic field of constant magnitude is produced. This rotating field induces currents in the rotor by electromagnetic induction.

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3. Rotor The rotor consists of a laminated core, with slots, cut on its outer periphery where windings are placed. The windings may be either of squirrel cage type or wound rotor type. The winding placed in these slots (called rotor winding) may be one of the following two types: (i) Squirrel cage type (ii) type or Wound type

(i) Squirrel cage rotor: It consists of a laminated cylindrical core having parallel slots on its outer periphery. One copper or aluminum bar is placed in each slot. All these bars are joined at each end by metal rings called end rings. This forms a permanently short-circuited winding which is indestructible. The entire construction (bars and end rings) resembles a squirrel cage and hence the name. The short circuited rotor conductors on both sides, constitute a closed path for the rotor current to flow. The rotor winding is not electrically connected to the supply, but by electromagnetic induction from the field created by the stator, a voltage is induced in the rotor conductors and a current is circulated through the end rings. Those induction motors which employ squirrel cage rotor are called squirrel cage induction motors. Most of 3-phase induction motors use squirrel cage rotor as it has a remarkably simple and robust construction However, it suffers from the disadvantage of a low starting torque. It is because the rotor bars are permanently short-circuited and it is not possible to add any external resistance to the rotor circuit to KTUNOTES.INhave a large starting torque.

(ii) Slip ring rotor: It consists of a laminated cylindrical core and carries a 3- phase winding. The rotor winding is uniformly distributed in the slots on the outer periphery of a laminated cylindrical steel core and is usually star-connected. The open ends of the rotor winding are brought out and joined to three insulated slip rings mounted on the rotor shaft with one resting on each slip ring. The three brushes are connected to a 3-phase star-connected rheostat as shown in figure. 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.

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4. Shaft and Bearings Shaft couples mechanical energy developed in the rotor to mechanical load. It is made of carbon steel. Rotor is supported over the shaft through bearings to reduce friction. Rotating Magnetic Field Due to 3-Phase Currents When a 3-phase winding is energized from a 3-phase supply, a rotating magnetic field is produced. This field is such that its poles do not remain in a fixed position on the stator but go on shifting their positions around the stator. For this reason, it is called a rotating field. It can be shown that magnitude of this rotating field is constant and is equal to 1.5 m where m is the maximum flux due to any phase. To see how rotating field is produced, consider a 2-pole, 3-phase winding (3 phase windings are 1200 electrical apart). The three phases X, Y and Z are energized from a 3-phase source and currents in these phases are indicated as Ix, Iy and Iz as shown in fig.

The fluxes produced by these currents (the flux produced by a current is in phase with the current that producesKTUNOTES.IN it) are given by:

where m is the maximum flux due to any phase. The phasor of the three fluxes are shown in figure. It can be proved that this 3-phase supply produces a rotating field of constant magnitude equal to 1.5 m i. At instant 1, t = 0°. Therefore, the three fluxes are given by,

The phasor sum of y and z is the resultant flux, r .

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Electrical Drives and control for Automation-Module IV

ii. At instant 2, t = 30°. Therefore, the three fluxes are given by,

The phasor sum of x, y and z is the resultant flux, r. Phasor sum of x and z,

Phasor sum of 'r and y,

Therefore, resultant flux is displaced 30° clockwise from position 1. iii. At instant 3, t = 60°. Therefore, the three fluxes are given by,

KTUNOTES.IN The resultant flux r is the phasor sum of x and y z 0.

Therefore, the resultant flux is displaced 60° clockwise from position 1. iv. At instant 4, t = 90°. Therefore, the three fluxes are given by,

The phasor sum of x, y and z is the resultant flux r Phasor sum of z and y,

Phasor sum of 'r and x,

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Electrical Drives and control for Automation-Module IV

Therefore, the resultant flux is downward i.e., it is displaced 90° clockwise from position 1. It is proved that a 3-phase supply produces a rotating field of constant value (= 1.5 m, where m is the maximum flux due to any phase). PRINCIPLE OF OPERATION

(i) When 3-phase stator winding is energized from a 3-phase ac supply, a rotating magnetic field is set up which rotates round the stator at synchronous speed, Ns (= 120 f/P), where f is the supply frequency and P is the number of poles on the stator. (ii) The rotating field passes through the air gap and cuts the rotor conductors, which are stationary initially at the instant of starting. When the rotating magnetic field sweeps past the stationary rotor conductors, an emf is induced in the rotor conductors just as emf is induced in the secondary winding of a transformer by the flux set up by the primary current. Since the rotor circuit is short-circuited, currents start flowing in the rotor conductors. (iii) Now the situation is exactly like current-carrying rotor conductors are placed in the magnetic field produced by the stator. Consequently, mechanical force acts on the rotor conductors. The sum of the mechanical forces on all the rotor conductors produces a torque which tends to move the rotor in the same direction as the rotating field. (iv) The factKTUNOTES.IN that rotor is urged to follow the stator field (i.e., rotor moves in the direction of stator field) can be explained by Lenz’s law. According to this law, the direction of rotor currents will be such that they tend to oppose the cause producing them. Now, the cause producing the rotor currents (induced current) is the relative speed between the rotating magnetic field and the stationary rotor conductors. Hence to reduce this relative speed, the rotor starts running in the same direction as that of stator field and tries to catch it. In practice, the rotor can never reach the speed of stator flux (Why cannot 3 phase induction motor run at synchronous speed?). If it did, there would be no relative speed between the stator field and rotor conductors, no induced rotor currents and, therefore, no torque to drive the rotor. The friction and windage would immediately cause the rotor to slow down. Hence, the rotor speed (N) is always less than the stator field speed (Ns). This difference in speed depends upon load on the motor. SLIP The rotor rapidly accelerates in the direction of rotating field. The difference between the synchronous speed Ns of the rotating stator field and the actual rotor speed N is called slip. It is usually expressed as a fraction or percentage of synchronous speed i.e, Fractional slip, s = Synchronous Speed – Rotor Speed Synchronous Speed

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Electrical Drives and control for Automation-Module IV

ROTOR CURRENT FREQUENCY The cause producing rotor current is the relative speed between the rotating magnetic field and the rotor conductor. The frequency of a voltage or current induced due to the relative speed between the rotor conductor and a stator magnetic field is given by the general

NrelativeP formula; Frequency = 120 where Nrelative = Relative speed between stator magnetic field and the rotor conductor. P = Number of poles For a rotor speed N, the relative speed between the rotating flux and the rotor is Ns - N. Consequently, the rotor current frequency f ' is given by;

i.e., Rotor current frequency = Fractional slip x Supply frequency. Therefore it is called slip frequency. EQUATION FOR STATOR AND ROTOR EMFs Let the rotor induced emf at standstill =E2 Stator induced emf = E1 EMF induced in the stator, E1 = 4.44 kw1f  mT1 EMF induced in the rotor at standstill, E2 = 4.44 kw2f mT2 KTUNOTES.IN’ ’ EMF induced in the rotor while running at a slip s, E2 = 4.44 kw2f mT2 = 4.44 kw2 sf mT2 = sE2 where m is the max flux, kw1 and kw2 are the winding factor of stator and rotor respectively and T1 and T2 are no of stator turns / phase and rotor turns / phase respectively. ROTOR CURRENT Figure shows the circuit of a 3-phase induction motor at any slip s. The rotor is assumed to be of wound type and star connected. The rotor e.m.f/phase and rotor reactance/phase are sE2 and sX2 respectively. The rotor resistance/phase is R2 and is independent of frequency therefore, does not depend upon slip. Likewise, stator winding values R1 and X1 do not depend upon slip.

Since the motor represents a balanced 3-phase load, we need consider one phase only, the conditions in the other two phases being similar. Fig (ii) shows one phase of the rotor circuit when the motor is running at slip s. 6 Aswathy Mohandas P Asst.Professor ,EEE,SNGCE

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Electrical Drives and control for Automation-Module IV

Induced emf per phase in rotor winding = sE2 Rotor winding resistance per phase = R2 Rotor winding reactance per phase = 2 f’L2 = 2 sf L2 = s(2 f L2) = sX2 ′ 2 2 Rotor winding impedance per phase, 푍2 = √푅2 + (푠푋2)

ROTOR TORQUE The torque T developed by the rotor is directly proportional to: (i) rotor current (ii) stator flux per pole,  (iii) power factor of the rotor circuit

Torque Under Running Conditions Let the rotor at standstill have per phase induced e.m.f. E2, reactance X2 and resistance R2. Then under running conditionsKTUNOTES.IN at slip s,

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Electrical Drives and control for Automation-Module IV

It can be shown that value of K1 = 3/2 π Ns where Ns is in r.p.s.

At starting, s = 1 so that starting torque is

Note that here Ns is in r.p.s. Condition for Maximum Starting Torque The starting torque will be maximum when rotor resistance/phase is equal to standstill rotor reactance/phase.

Differentiating eq. (i) w.r.t. R2 and equating the result to zero, we get, KTUNOTES.IN

Hence starting torque will be maximum when: Rotor resistance/phase = Standstill rotor reactance/phase Maximum Torque under Running Conditions

In order to find the value of rotor resistance that gives maximum torque under running conditions, differentiate exp. (i) w.r.t. s and equate the result to zero i.e.,

Thus for maximum torque (Tm) under running conditions: Rotor resistance/phase = Fractional slip x Standstill rotor reactance/phase

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Electrical Drives and control for Automation-Module IV

For maximum torque, R2 = s X2. Putting R2 = s X2 in the above expression, the maximum torque Tm is given by,

Slip corresponding to maximum torque, s = R2/X2. It can be shown that:

TORQUE-SLIP CHARACTERISTICS The motor torque under running conditions is given by;

If a curve is drawn between the torque and slip for a particular value of rotor resistance R2, the graph thus obtained is called torque-slip characteristic.

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The following points may be noted carefully: (i) At s = 0, T = 0 so that torque-slip curve starts from the origin. (ii) At normal speed, slip is small so that s X2 is negligible as compared to R2.

Hence torque slip curve is a straight line from zero slip to a slip that corresponds to full-load. (iii) As slip increases beyond full-load slip, the torque increases and becomes maximum at s = R2/X2. This maximum torque in an induction motor is called pull-out torque or break- down torque. 2 2 (iv) When slip increases beyond that corresponding to maximum torque, the term s X2 2 2 2 increases very rapidly so that R2 may be neglected as compared to s X2 .

Thus the torque is now inversely proportional to slip. Hence torque-slip curve is a rectangular hyperbola.

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Electrical Drives and control for Automation-Module IV

EFFECT OF ROTOR RESISTANCE UPON TORQUE - SLIP OR TORQUE - SPEED CHARACTERISTIC The expression for the starting torque of the induction machine shows that, it dependent on the rotor resistance. Whereas the max value of torque is independent of the rotor resistance.

The slip at maximum torque occurs dependent on the rotor resistance (Slip corresponding to maximum torque, s = R2/X2). Therefore, the variation of rotor resistance does not change the magnitude of maximum torque, Tm but merely changes the value of slip at which maximum torque occurs. Larger the rotor resistance the greater the slip at which the maximum torque occurs. Since the starting torque depends on the rotor resistance, as the value of rotor resistance increase the starting torque, Tst also increases.

EFFECT OF CHANGE IN SUPPLY VOLTAGE ON TORQUE – SLIP CHARACTERISTIC KTUNOTES.IN

Therefore, the starting torque as well as maximum torque is very sensitive to changes in the value of supply voltage. These curves show that the slip at maximum torque occurs remains same, (s = R2/X2, independent of supply voltage) while the value of maximum torque comes down with decrease in applied voltage. EQUIVALENT CIRCUIT OF 3-PHASE INDUCTION MOTOR AT ANY SLIP

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Stator circuit: The applied voltage per phase to the stator is V1 and R1 and X1 are the stator resistance and leakage reactance per phase respectively. When the motor is at no-load, the stator winding draws a current I0. It has two components viz., (i) which supplies the no-load motor losses and (ii) magnetizing component Im which sets up magnetic flux in the core and the air-gap. The parallel combination of R0 and Xm, therefore, represents the no-load motor losses and the production of magnetic flux respectively. I0 = Iw + Im Rotor circuit: Here R2 and X2 represent the rotor resistance and standstill rotor reactance per phase respectively. At any slip s, the rotor reactance will be s X2. The induced voltage/phase in the rotor is E'2 = s E2 . Equivalent Circuit of the Rotor Fig (i) shows the equivalent circuit per phase of the rotor at slip s. The rotor phase current is given by;

Mathematically, this value is unaltered by writing it as:

This is shown in Fig (ii) KTUNOTES.IN

This is shown in fig(iii) Deduction of Equivalent circuit The mechanical load on the motor has been replaced by an equivalent electrical resistance RL given by

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Electrical Drives and control for Automation-Module IV

When shifting resistance/reactance from secondary to primary, it should be divided by K2 whereas current should be multiplied by K and voltage should be divided by K. The equivalent circuit of an induction motor referred to primary is shown below.

This deduction has been taken on the assumption that voltage drop in R1 and X1 is small.

TEST ON INDUCTION MOTOR No –load test: KTUNOTES.IN

This test is performed to determine the no load Current I0, no load power factor cos Ø0, windage and friction loss Pwf, no load core loss Pi, no load power input P0, no load resistance R0 and reactance X0. At no-load, the power input is equal to the core loss Pi, stator copper loss Pcu and windage and friction loss Pwf. If the windage and friction losses Pwf are known, and stator 2 copper losses (3I0 R1) are subtracted from power input at no-load Po, the core losses Pi can be determined. Knowing the total core losses Pi, no-load current (Im and Iw respectively), no- load resistance R0 and no-load reactance X0 can be determined.

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Keep the machine on no load and apply rated voltage to the stator. Take the ammeter, voltmeter and wattmeter reading. The ammeter reading will give the no load current I0, wattmeter gives the no load input power P0 and voltmeter gives the applied voltage, V0. V0 = No load voltage (line to line) I0 = No load current (line) P0 = Total input power at no load Stator core loss at no load, Pi = P0 – (Pcu + Pwf) 2 Stator copper loss, Pcu = 3I0 R1

Pi = 3 V0(line)I0(line)CosФ0 No-load power factor, cos Ø0 = Pi √3 V0(line) I0(line) Energy component of no-load current Iw = I0 cos Ø0

Magnetizing component of no-load current, Im = I0SinФ0 No-load reactance, X0 = V0(phase) Im(phase)

No-load resistance, R0 = V0(phase) Iw(phase) :

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This test is performed to determine the short-circuit current Isc with normal voltage applied to the stator; power factor on short-circuit, total equivalent resistance and reactance of the motor as referred to stator (R01 and X01). In this test the rotor is held firmly (rotor windings are short-circuited at slip rings in case of wound rotor motor) and stator is connected across supply of variable voltage. This test is just equivalent to short-circuit test on transformer. Starting with zero voltage across the stator, the applied voltage is gradually increased in steps till the full-load current flows in the stator. The readings of voltmeter, ammeter and wattmeter are noted.

If Vs is the applied voltage (line-to-line value) that causes current Is in the stator winding and Ps is the total input power at short-circuit, then short-circuit current Isc with normal voltage V (line-to –line) applied across the motor is given by Isc = Is V Vs 13 Aswathy Mohandas P Asst.Professor ,EEE,SNGCE

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Electrical Drives and control for Automation-Module IV

Ps = 3 Vs(line)Is(line)CosФs

power factor, cos Øs = Ps . √3 Vs Is 2 Ps= 3 I s(phase) R01 Motor equivalent resistance per phase, as referred to stator, R01 = Ps 2 3 Is Motor equivalent impedance per phase, as referred to stator, Z01 = Vs(phase) Is(phase) 2 2 Motor equivalent reactance per phase, as referred to stator, X01 = √Z01 - R01 Usually the stator reactance per phase, X1 is assumed to be equal to rotor reactance per phase 2 as referred to stator X2’ = X2 / K and so , X1 = X2’= X01 2 The rotor resistance per phase, as referred to stator, can be determined by subtracting R1 from R01. R1 = 1.6 x Rdc (Rdc is obtained from ammeter - voltmeter method) ’ ’ 2 Thus R2 = R01 - R1 (where R2 = R2 / K )

STARTING OF THREE PHASE INDUCTION MOTORS Three phase induction motor are self starting due to the rotating magnetic field. But the motors show tendency to draw very high current at the time of starting. Hence there should be some startingKTUNOTES.IN method to reduce the magnitude of starting current. At start, the speed of the motor is zero and slip is one. So magnitude of rotor induced emf is very large at start. As rotor conductors are short circuited, the large induced emf circulates very high current through rotor at start. When the rotor current is high the stator draws a very high current from the supply. This high starting current in the stator does not harm the motor, since it occurs only for a short duration, but causes voltage drop in the supply lines affecting the performance of other machines connected to the line. Hence it is desirable and necessary to reduce the magnitude of stator current at starting. The common methods used to start induction motors are: (i) Direct-on-line starting (ii) Stator resistance starting (iii) Autotransformer starting (iv) Star-delta starting (v) Rotor resistance starting Methods (i) to (iv) are applicable to both squirrel-cage and slip ring motors. However, method(v)is applicable only to slip ring motors. Direct –on-Line Starting (DOL): In case of small capacity motors having rating less than 5HP, the starting current is not very high and such motors can withstand such starting current without any . Thus there is no need to reduce applied voltage, to control the starting current. Such motors use a

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Electrical Drives and control for Automation-Module IV type of starter which is used to connect stator directly to the supply lines without any reduction in voltage. Hence the starter is known as direct online starter. Though this starter does not reduce the applied voltage, it is used because it protects the motor from various severe abnormal conditions like over loading, low voltage, single phasing etc.

The NO contact is normally open and NC is normally closed. At start, NO is pushed for fraction of second due to which the coil gets energized and attracts the contactor. So stator directly gets supply and keeps contactor in ON position. When NC is pressed, the coil circuit gets opened due to which coil gets de-energised and motor gets switched OFF from the supply. Under overload condition, current drawn by the motor increases due to which there is an excessive heat produced, which increases the temperature beyond limit. Thermal relays get opened due to high temperature, protecting the motor from over load condition. KTUNOTES.IN In this method of starting the motor is started by connecting it directly to 3-phase supply.

Relation between starling and F.L. torques: We know that:

If Ist is the starting current, then starting torque (Tst) is

If If is the full-load current and sf is the full-load slip, then,

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Electrical Drives and control for Automation-Module IV

When the motor is started direct-on-line, the starting current is the short-circuit (blocked rotor) current Isc.

Assume Isc = 5 If and full-load slip sf =0.04. Then,

The starting current is as large as five times the full-load current but starting torque is just equal to the full-load torque. Therefore, starting current is very high and the starting torque is comparatively low. If this large starting current flows for a long time, it may overheat the motor and damage the insulation. Auto-transformer starting: This method connects the induction motor to a reduced voltage at starting and then connecting it to the full voltage as the motor picks up sufficient speed. The tapping on the autotransformer is so set that when it is in the circuit, 65% to 80% of line voltage is applied to the motor. Consequently, starting current is limited to safe value. For large machines (over 25 H.P.), this method of starting is often used. This method can be used for both star and delta connected motors. Relation between starting and F.L. torques. When the motor is direct switched starting current is Ist = Isc. In case of auto-transformer, if a tapping of transformation ratio K (a fraction) is used, then Ist = K Isc,

KTUNOTES.IN

2 The current taken from the supply or by autotransformer is I1 = KI2 = K Isc. Advantages: 1. Low power loss 2. Low starting current 3. Less radiated heat.

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Electrical Drives and control for Automation-Module IV

4. Adjustment of starting voltage by selection of proper tap on the autotransformer Disadvantages: 1. Low power factor 2. Higher cost in case of lower output rating motors

COMPARISON OF SQUIRREL CAGE INDUCTION MOTOR AND SLIP RING INDUCTION MOTOR SQUIRREL CAGE INDUCTION SLIP RING INDUCTION MOTOR MOTOR

Low starting torque High starting torque

Copper losses are less and efficiency is large Copper losses are high and efficiency is low

Low cost High cost

Starting current is high Starting current is high, but it can be reduced using external resistance

Requires low maintenance Requires high maintenance

Speed can be varied by changing the poles Speed can be varied slightly by changing rotor extra resistance

APPLICATIONS OF THREE PHASE INDUCTION MOTORS Squirrel cage induction motors: are used in centrifugal pumps, fans, conveyors, compressors, reciprocating pumps, lathe works etc Slip ring induction motorKTUNOTES.IN: are used in lifts, hoists, cranes, conveyors, pumps, flour mills etc.

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