CHAPTER 3INDUCTION MACHINE

3.1 INTRODUCTION

Induction motor is the common type of AC motor.

Induction motor was invented by Nicola Tesla (1856-1943) in 1888.

Also known as asynchronous motor.

It has a stator and a rotor mounted on bearings and separated from the stator by an air gap.

It requires no electrical connection to the rotating member.

Such motor are classified induction machines because the rotor voltage (which produce the rotor current and the rotor magnetic field) is induced in the rotor winding rather than being physically connected by wires.

The transfer of energy from the stationary member to the rotating member is by means of electromagnetic induction.

This motor is widely used by the industries because:- Rugged. - Simple construction. - Robust. - Reliable.- High efficiency. - Good power factor. - Require less maintenance - Easy to start.- Rotates itself without external assistant.- Less expensive than direct current motor of equal power and speed.

The weaknesses of this machine are:- Low starting torque if compared to dc shunt motor.- Speed will be reduced when load increased.- Speed can’t be changed without reducing efficiency.

Small single phase induction motors (in fractional horsepower rating) are used in many household appliances such as:- Blenders - Lawn mowers- Juice mixers- Washing machines- Refrigerators

Two phase induction motors are used primarily as servomotor in control system.

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Large three phase induction motors (in ten or hundreds of horsepower) are used in:- Pumps - Fans - Compressors - Paper mills- Textile mills, and so forth.

3.2 INDUCTION MOTOR CONSTRUCTION

Unlike dc machine, induction machine have a uniform air gap.

Composed by two main parts:- Stator- Rotor

Figure 4.1 and 4.2 show the inside of induction machine.

Figure 3.1

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Figure 3.2

Stator Construction

The stator and the rotor are electrical circuits that perform as electromagnets. The stator is the stationary electrical part of the motor. The stator core of a NEMA motor is made up of several hundred thin laminations.

Figure 3.3:Stator core

Stator Windings

Stator laminations are stacked together forming a hollow cylinder. Coils of insulated wire are inserted into slots of the stator core.

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Figure 3.4:Stator winding

Each grouping of coils, together with the steel core it surrounds, form an electromagnet. Electromagnetism is the principle behind motor operation. The stator windings are connected directly to the power source.

Rotor Construction

The rotor also consists of laminated ferromagnetic material, with slot cuts on the outer surface.

The rotor are of two basic types :- Squirrel cage- Wound rotor

Squirrel cage rotor

It consist of a series of a conducting bars laid into slots carved in the face of the rotor and shorted at either ends by large shorting ring.

This design is referred to as squirrel cage rotor because the conductors would look like one of the exercise wheels that squirrel or hamsters run on.

Small squirrel cage rotors use a slotted core of laminated steel into which molten aluminums cast to form the conductor, end rings and fan blades.

Larger squirrel cage rotors use brass bars and brass end rings that are brazed together to form the squirrel cage.

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Skewing the rotor slots help to:- Avoid crawling (locking in at sub-synchronous speeds)- Reduce vibration

Squirrel cage rotor is better than wound rotor because it is:- Simpler - More rugged- More economical- Require less maintenance

Figure 3.5:Squirrel cage Rotor

Figure 3.6 : Rotor core

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Figure 3.7

Wound rotor

Has a complete set of three phase insulated windings that are mirror images of the winding on stator.

Its three phase winding are usually wye connected and ends of three rotor wires are tied to a slip rings on the rotor shaft.

The rotor winding are shorted through carbon brushes riding on the slip rings.

The existence of rheostat enable user to modify the torque speed characteristic of the motor. It is used to adjust the starting torque and running speed.

The three phase rheostat is composed of three rheostat connected in wye with a common lever.

Lever is used to simultaneously adjust all the three rheostat arms. Eg: Movingrheostat to the zero resistance position shorts the resistor and simulates a squirrel cage motor.

Are rarely used because:- More expensive than squirrel cage induction motor.- Larger than squirrel cage induction motor with similar power.- Require frequent maintenance due to wear associates to brushes and slip ring.

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Figure 3.8 Wound rotor induction motor showing rheostat connections

Figure 3.9:Wound rotor

3.3 ROTATING MAGNETIC FIELD

When a three phase stator winding is connected to a three-phase voltage supply, three-phase currents will flow in the winding which induce three-phase flux in the stator.

These flux will rotate at a speed called as Synchronous Speed, ns.

The flux is called as rotating magnetic field.

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The equation is:- where f = supply frequency , p = no. of poles

Rotating magnetic field will cause the rotor to rotate the same direction as the stator flux.

Torque direction is always the same as the flux rotation.

At the time of starting the motor, rotor speed is 0.

The rotating magnetic field will cause the rotor to rotate from 0 speed to a speed that is lower than the synchronous speed.

If the rotor speed is equal to the synchronous speed, there will be no cutting of flux and rotor current equals zero. Therefore, it is not possible for the rotor to rotate at ns.

3.4 SLIP AND ROTOR SPEED

Slip is defined as :

where ns = synchronous speed in rpmn = rotor speed in rpm

Slip can also represented in percent. The frequency of the rotor, fr is:

where s = slipf = supply frequency

Example 1

Calculate the synchronous speed of a 3-phase induction motor having 20 poles when it is connected to a 50 Hz source.

Solution

Example 2

A 0.5 hp, 6-pole induction motor is excited by a 3 –phase, 60 Hz source. If the full-load speed is 1140 rpm, calculate the slip.

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Solution

Example 3

The 6-pole,wound-rotor induction motor is excited by a 3-phase, 60 Hz source. Calculate the frequency of the rotor current under the following conditions:(i) at standstill(ii) motor turning at 500 rpm in the same direction as the revolving field(iii) motor turning at 500 rpm in the opposite direction to the revolving field(iv) motor turning at 2000 rpm in the same direction as the revolving field

Solution

ns = 120f / p = 120(60/6) = 1200 rpm

(i) n=0

=

fr = sf = 1 x 60 = 60Hz

(ii) n = +500

=

fr = sf = 0.583 x 60 = 35 Hz(iii) n = -500

= (s>1 motor is operating as a brake)

fr = sf = 1.417 x 60 = 85 Hz

(iv)n = +2000

=

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fr = sf = -0.667 x 60 = -40 Hz (-ve means that the phase sequence of the voltages induced in the rotor winding is reversed)

Example 4

A 3-phase, 4 pair of poles, 400kW,400V,60Hz induction motor is 780 rpm full-load speed. Determine the frequency of the rotor current under full load condition.

Solution

f rotor = sf

n s =

=

3.5 PER-PHASE EQUIVALENT CIRCUIT OF THREE-PHASE INDUCTION MOTOR

The per-phase equivalent circuit of a three-phase induction motor is just like a single –phase transformer equivalent circuit.

The difference is only that the secondary winding is short-circuited unlike in the transformer it is open-circuited as a load is to be connected later.

Complete Equivalent Circuit For Induction Machine Referred To The Stator Circuit

1,0k 1,0m 1,0m

1,0

k 1

,0k

1,0

m

R1 X1

Rm Xm

X2

R2

s

V input

I1 I2

Figure 3.10

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The subscript ‘1’ is refering to the stator side while ‘2’ is referring to the rotor side

R1, X1, R2, Rm , Xm are value perphase

Input Power, Pin = 3V1I1cosθ

Stator copper loss, Pscl = 3I12R1

Core Loss, Pcl = 3V12/Rm (always neglected because too small)

Power across the air-gap, Pag = 3I22R2 /s

= Pin - Pscl - Pcl

Rotor copper loss, Prcl = 3I22R2

Mechanical power/gross output power/converted power,

P mech = Pag – Prcl

= 3I22R2 /s - 3I2

2R2

= Pag (1-s)

Net power output, Poutput = P mech – P friction & windage loss

For Torque:

Maximum Slip:

3.6 POWER FLOW OF AN INDUCTION MOTOR

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PAG=(3I22R2)/s

= Pin-PSCL-PCL

P CONV = P MECH

=PAG-PRCL

Figure 3.11

Example 5

A 10 poles, 50 Hz, Y connection 3-phase induction motor having a rating of 60kW and 415V. The slip of the motor is 5% at 0.6 power factor lagging. If the full load efficiency is 90%, calculate:(i) Input power(ii) Line current and phase current(iii) Speed of the rotor (rpm)(iv) Frequency of the rotor(v) Torque developed by the motor (if friction and windage losses is 0)

Solution

(i) η =

60

PSCL=3I12R1 PCL=

3V12/RM

PRCL=3I22R2

= sPag

(ii) Y connection, IΦ = IL, VΦ=

P in =3VΦIΦcos=

IL=

IΦ = IL=154.59 A

(iii) n s =

n = n s (1-s) = 600 (1-0.05) = 570 rpm

(iv) fr = sf = 0.05(50) = 2.5Hz

(v) T =

Or

ws=

wm = ws(1-s) = 62.83(1-0.05) = 59.69 rad/s

Example 6

A 3-phase, delta connection, 4 pole, 440V, 60 Hz induction motor having a rotor speed 1200rpm and 50kW input power at 0.8 power factor lagging. The copper losses and iron losses in the stator amount to 2kW and the windage and friction losses are 3kW. Determine:

(i) Net output power(ii) Efficiency(iii) Input current

Solution

(i) ns = 120f/p = 120(60)/4 = 1800 rpm

=

61

P net output = 29kW

(ii) η =

(iii)Δ connection

Example 7

A 3-phase induction motor, delta connection,5 pair of poles, 60 Hz is connected to a 440V source.The slip is 3% and the windage and friction losses are 3kW. The equivalent circuit perphase referred to the stator circuit is:-

R1 = Stator resistance = 0.4ΩX1 = Stator leakage inductance = 1.4ΩR2’ = Rotor resistance = 0.6ΩX2’ = Rotor leakage inductance = 2ΩRm = no-load loses resistance = 150ΩXm = magnetizing reactance = 20Ω

Calculate:(i) Input power(ii) Speed of the rotor(iii) Mechanical power(iv) Developed torque(v) Efficiency

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P input(to stator)

=50kWP input rotor /Pag

=50kW-2kW =48kW

P stator losses =2kW

P rotor losses

=s(P input rotor)= 0.33(48kW)=16kW

Pmech=Pout rotor

=(48-16)kW =32kW

P wind and fric losses

=3kW

P out net

=(32-3)kW=29kW

I1

j20Ω

I2

V2

Solution

1.0k 1.0m 1.0m 1.0

k 1

.0m

(i) P in =3VIcosθV = 440V

Z total =

Pin = 3(440)(29.64)cos(-50.460) = 24907.5W

(ii)

,

n = ns(1-s) , n = 720(1-0.03) = 698.4 rpm

(iii) Pm = 3(I22R2/s – I2

2R2)V2 = 440 – I1Z1

= 440 – (18.87-22.86j)(0.4+j1.4)= 400.45-17.27j V

I2 = A

Pm = 3 [19.942 - 19.942(0.6)] = 23140.5W

(iv) T dev =

Pag = 3I22R2/s = 3(19.94)2(0.6)/(0.03) = 23856.2W

ws =

T =

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0.4Ω j1.4Ω j2Ω

R2/s=0.6/0.03=20Ω

60Hz, 440v,

10 poles,

Δ

S

j10Ω

(v)

Example 8

A 3-phase induction motor, wye connection, 60 Hz is connected to a 220V source.The slip is 5% and rotor speed is 855 rpm. The equivalent circuit perphase is:-

R1 = Stator resistance = 0.4ΩX1 = Stator leakage inductance = 1ΩR2’ = Rotor resistance = 0.8ΩX2’ = Rotor leakage inductance = 3.5ΩRm = no-load loses resistance = 150ΩXm = magnetizing reactance = 10Ω

Calculate:(i) Number of poles(ii) Input power(iii) Mechanical power(iv) Developed torque(v) Efficiency

Solution

1.0k 1.0m 1.0m 1.0

k 1

.0m

(i) n = 855rpms = 0.05ns = 120f/p

, sns = ns – n , n = ns – sns

= ns(1-s)

ns =

=

64

R2/s=0.8/0.05=16Ω

60Hz, 220V,

Y

0.4Ω j1Ω j3.5Ω

S

ns = 120f/p

p =

(ii) Pin =

Z total = Ω

Pin =

(iii) Pm = 3(I22R2/s – I2

2R2)

V2 = – I1Z1

= – (6.53-j12.73)(0.4+j1)

= 111.68-1.438j V

I2 = A

Pm = 3 [6.82 - 6.82(0.8)] = 2108.54W

(iv) T dev =

Pag = 3I22R2/s

ws =

T = (3I22R2/s) / ws =[3(6.8)2(0.8)/(0.05)] / 94.25 = 23.55Nm

(v)

3.7 TORQUE SPEED CHARACTERISTICS

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Figure 3.12

There are 3 regions involve in a 3-phase induction motor:-

(i) Braking/Plugging Braking process occurs at s>0(positive slip). In this case the motor acts as a brake where it rotates in opposite direction respect to the rotor.(2<slip<1).

(ii) MotoringMotoring is the region where induction motor acts as a motor. Slip is reducing from 1 into 0. Slip equals to 0 at synchronous speed,ns.(1<slip<0).

(iii) GeneratingGenerating region is a region where motor acts as a generator. During this time the slip is negative. At this time, the motor acts as a generator.(slip<0)

3.8 DETERMINATION OF CIRCUIT MODEL PARAMETER

The parameter of the equivalent circuit can be determined from the results of a:- No load test- Blocked rotor test-.DC test

The blocked rotor test:

- To determine X1 and X2

- When combines with DC test, it also determines R2

- Test is performed by blocking the rotor so that it cannot turn and measuring the line voltage, line current and three phase power input to the stator- Connection for the test is shown in Figure 4.13

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Figure 3.13 Basic circuit for blocked rotor test and no load test

The no load test:

- To determine magnetizing reactance, Xm, and the combined core, friction and windage losses (these losses are essentially constant for all load condition)- The connection for the no load test are identical to those shown in Figure 12- However, the rotor is unblocked and allowed to run unloaded at rated voltage and rated frequency

DC test:

- To determine R1

- Accomplished by connecting any two stator leads to a variable voltage DC source as shown in Figure 4.14. - The DC source is adjusted to provide approximately rated stator current, and the resistance between two stator leads is determined from voltmeter and ammeter reading

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Figure 3.14 Basic circuit for DC test

Example 9

The following test data were taken on a 7.5hp, four pole, 208 V, 60 Hz Y connected design A induction motor having a rated current of 28A.

DC test : Vdc = 13.6V Idc = 28 A

No-load test: Vt = 208 V f = 60 HzIa = 8.12 A P in = 420 WIb = 8.2 AIc = 8.18 A

Locked rotor test: Vt = 25 V f = 15 HzIa = 28.1 A P in = 920 WIb = 28 AIc = 27.6 A

Sketch the per-phase equivalent circuit for this motor.

Solution

From the DC test,

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From the no-load test,

Pscl = 3I12R1 = 3(8.17A)2(0.243Ω) = 48.7W

Pag = Pin –Pscl = 420W- 48.7W = 371.3W

From the locked-rotor test,

θ = cos-1

R1 + R2 = 0.517(cos 40.4°) = 0.394Ω

DC test, R1 = 0.243Ω, R2 = 0.394Ω – 0.243Ω = 0.151Ω

At 15 Hz, X = 0.517(sin 40.4°)=0.335Ω

At 60 Hz, X = = X1 + X2

For class A induction motor, this reactance is assumed to be divided equally between the rotor and stator,

X1 = X2 = 0.67 Ω

Xm = 14.7 -0.67 = 14.03Ω

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3.9 STARTING OF INDUCTION MOTOR

(i) Direct On Line Starter(DOL)

- A widely-used starting method of electric motors. - The simplest motor starter.- A DOL starter connects the motor terminals directly to the power supply. - Hence, the motor is subjected to the full voltage of the power supply. - Consequently, high starting current flows through the motor. -This type of starting is suitable for small motors below 5 hp (3.75 kW). - Most motors are reversible or, in other words, they can be run clockwise and anti- clockwise. - A reversing starter is an electrical or electronic circuit that reverses the direction of a motor automatically. - Logically, the circuit is composed of two DOL circuits; one for clockwise operation and the other for anti-clockwise operation.- It takes a starting current 6(six) times the full load current. - For large motor the high starting current causes voltage drop in the power system which may trip other motors in the systems.

(ii)Star-Delta Starter

- For star-delta connection the motor windings are connected in star during starting. - The connection is changed to delta when the motor starts running. - The starting current and starting torque of DOL started and start-delta connected motors are as follows:

Example:DOL -6I and 2T

Star-delta - 2I and 2T/3

- Thus it can be seen that the starting current and starting torque are both reduced. - The motor should be capable to start at such reduced torque with load.- The Star Delta starter can only be used with a motor which is rated for connection in delta operation at the required line voltage

(iii)Autotransformer starter

- An Auto transformer starter uses an auto transformer to reduce the voltage applied to a motor during start. - The auto transformer may have a number of output taps and be set-up to provide a single stage starter, or a multistage starter.

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- Typically, the auto transformer would have taps at 50%, 65% and 80% voltage, enabling the motor to be started at one or more of these settings. - As the motor approaches full speed, the auto transformer is switched out of the circuit

Tutorial 3

1. A 3 phase induction machine 373kW, 6 poles is connected to a 440V, 50 Hz, has a full load speed of 950 rpm. If the machine is comprised of 6 poles, calculate the frequency of the rotor current during full load.

2. Determine the synchronous speed of a six pole 460V 60 Hz induction motor if the frequency is reduced to 85 % of its rated value.

3. A 4 pole induction machine is supplied by 60 Hz source and having 4% of full load slip. Calculate the rotor frequency during:(i) Starting

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(ii) Full load

4. A 3-phase induction motor, delta/star connection, 2 poles, 50 Hz is connected to a 410V

source .The rotor speed is 2880 rpm and the windage and friction losses are 600 W. The equivalent circuit perphase referred to the stator circuit is:-

R1 = 0.4Ω X2 = 2 ΩX1 = 2 Ω Rm = 150ΩR2 = 2 Ω Xm = 20Ω

Calculate:(i) Input power(ii) Air-gap power(iii) Mechanical power(iv) Developed torque/torque induced(v) Efficiency

5. A 440V, 50Hz, 10 pole, delta/Y connected induction motor is rated at 100kW. The equivalent parameter for the motor are:

At full load condition , the friction and windage losses are 400W, the miscellaneous losses are 100W and the core losses are 1000W. The slip of the motor is 0.04.(i) Calculate the input power(ii) Calculate the stator copper loss(iii) Calculate the air gap power(iv)Calculate the converted power(v) Calculate the torque induced by the motor(vi)Calculate the load torque(vii) Calculate the starting torque(viii) Calculate the maximum torque and slip(ix) Calculate the efficiency of the motor

6. Squirrel cage and wound rotors are the two common types of rotor used in induction machines. Give four(4) advantages of squirrel cage rotor.

7. A 4 pole induction machine is supplied by 50 Hz source and having 4% of full load slip. Find the rotor frequency during:(i) Starting(ii) Full load

8. A 3-phase, Y-connected, 50 Hz, 4 pair of poles, induction motor having 720 rpm full load speed. The motor is connected to a 415 V supply. The machine has the following impedances in ohms per phase referred to the stator circuit:

R1 = 0.2 Ω X1 = 2.0 ΩR2 = 0.9 Ω X2 = 4.0 Ω

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Xm = 60 ΩIf the total friction and windage losses are 200 W, (i) Find the slip, s.(ii) Find the input power, Pin.

(iii) Find the air gap power, Pag.

(iv)Find the mechanical power, Pm.

(v) Find the torque induced by the motor, τ ind.

(vi)Find the efficiency of the motor.

9. Induction machine is a common type of AC machine. State three weaknesses of the induction machine.

10. A 3-phase, delta-connected, 50 Hz, 2 pair of poles, induction motor having 1455 rpm full load speed. The motor is connected to a 415 V supply. The machine has the following impedances in ohms per phase referred to the stator circuit:

R1 = 0.2 Ω X1 = 0.6 ΩR2 = 0.9 X2 = 0.4 Ω

Xm = 20 Ω

If the total friction and windage losses are 1000 W, calculate:(i) Slip(ii) Input power, Pin

(iii)Air gap power, Pag

(iv) Mechanical power, Pconv

(v) Torque induced by the motor, τ ind

(vi)Efficiency of the motor

11. A 3-phase, Y-connected, 6 poles, 415 V, 50 Hz induction motor having a rotor speed 950 rpm. The input power is 100 kW at 0.85 power factor lagging. The copper and iron losses in the stator are 4 kW and the windage and friction losses are 4 kW. Determine the output power of the motor.

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