1 PHASE INDUCTION MOTOR

CHAPTER 3

INDUCTION 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.

·         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

           

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.

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.

·         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

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: Moving

rheostat 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.

            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, n_s.

·         The flux is called as rotating magnetic field.

·         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 n_s.

3.4 SLIP AND ROTOR SPEED

·         Slip is defined as :

                                   

where n_s = synchronous speed in rpm

                                    n = rotor speed in rpm

·         Slip can also represented in percent.

·         The frequency of the rotor, f_r is:

where s = slip

                                    f = 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.

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

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

(i) n=0

=

f_r  = sf = 1 x 60 = 60Hz

(ii) n = +500

=

f_r  = sf = 0.583 x 60 = 35 Hz

(iii) n = -500

= (s>1 motor is operating as a brake)

f_r  = sf = 1.417 x 60 = 85 Hz

(iv)n = +2000

=

f_r  = 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

Figure 3.10

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

R₁, X₁, R₂, R_m , X_m are value perphase

Input Power,               P_in = 3V₁I₁cosθ

Stator copper loss,       P_scl = 3I₁²R₁

Core Loss,                   P_cl = 3V₁²/R_m               (always neglected because too small)

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

        

                                                =   P_in - P_scl  - P_cl

Rotor copper loss,       P_rcl = 3I₂²R₂

Mechanical power/gross output power/converted power,

P _mech = P_ag – P_rcl

 = 3I₂²R₂ /s - 3I₂²R₂

                                                = P_ag (1-s)

Net power output, P_output = P _mech – P _{friction & windage loss}

For Torque:

           

Maximum Slip:

3.6 POWER FLOW OF AN INDUCTION MOTOR

				P CONV = P MECH =PAG-PRCL

 

					PRCL=3I22R2              = sPag
	PSCL=3I12R1		PCL= 3V12/RM

 

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 =

(ii) Y connection, I_Φ = I_L, V_Φ=

P _in =3V_ΦI_Φcos=

I_L=

I_Φ = I_L=154.59A

           

(iii)  n _s =

     

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

(iv)  f_r = sf = 0.05(50) = 2.5Hz

(v) T =

Or

     w_s=

     w_m = w_s(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)         n_s = 120f/p = 120(60)/4 = 1800 rpm

  =

P input(to stator) =50kW

Pmech=Pout rotor =(48-16)kW =32kW

P out net =(32-3)kW =29kW

P input rotor /Pag =50kW-2kW =48kW

 

 

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 = 0.4Ω

X₁        = Stator leakage inductance = 1.4Ω

R₂’       = Rotor resistance  = 0.6Ω

X₂’       = Rotor leakage inductance  = 2Ω

R_m       = no-load loses resistance = 150Ω

X_m       = magnetizing reactance = 20Ω

Calculate:

(i) Input power

(ii)  Speed of the rotor

(iii) Mechanical power

(iv) Developed torque

(v) Efficiency

j2Ω

j1.4Ω

0.4Ω

 Solution

I1

I2

V2

60Hz, 440v, 10 poles, Δ

j20Ω

R2/s=0.6/0.03=20Ω

(i)         P _in =3VIcosθ

V = 440V

Z _total =

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

(ii)                 

            ,

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

(iii)       P_m        =          3(I₂²R₂/s – I₂²R₂)

V₂ =          440 – I₁Z₁

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

      =          400.45-17.27j V

I₂   =          A

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

(iv)       T _dev =

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

w_s  =         

T    =         

(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 = 0.4Ω

X₁        = Stator leakage inductance = 1Ω

R₂’       = Rotor resistance  = 0.8Ω

X₂’       = Rotor leakage inductance  = 3.5Ω

R_m       = no-load loses resistance = 150Ω

X_m       = magnetizing reactance = 10Ω

           

Calculate:

(i) Number of poles

(ii) Input power

(iii) Mechanical power

(iv) Developed torque

(v) Efficiency

Solution

	0.4Ω		j1Ω		j3.5Ω

 

           

60Hz, 220V, Y

j10Ω

R2/s=0.8/0.05=16Ω

(i)         n = 855rpm

s = 0.05

n_s = 120f/p

,      sn_s = n_s – n ,    n = n_s – sn_s

                                                   = n_s(1-s)

                                                n_s =  

                                                   =

n_s = 120f/p

p =

(ii)        P_in =

Z _total =Ω

           

            P_in =

(iii)       P_m        =          3(I₂²R₂/s – I₂²R₂)

             V₂       =           – I₁Z₁

      =          – (6.53-j12.73)(0.4+j1)

      =          111.68-1.438j V

I₂   =          A

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

(iv)           T _dev =         

P_ag =          3I₂²R₂/s

w_s  =         

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

(v)       

3.7 TORQUE SPEED CHARACTERISTICS

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)        Motoring

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

(iii)       Generating

Generating 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

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

The no load test:

- To determine magnetizing reactance, X_m, 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

                        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 Hz

                                                Ia = 8.12 A                  P _in = 420 W

                                                Ib = 8.2 A                  

                                                Ic = 8.18 A

Locked rotor test:                   Vt = 25 V                    f = 15 Hz

                                                Ia = 28.1 A                  P _in = 920 W

                                                Ib = 28 A                   

                                                Ic = 27.6 A

Sketch the per-phase equivalent circuit for this motor.

Solution

From  the DC test,

From the no-load test,

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

P_ag =  P_in –P_scl = 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

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.

- 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

(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 Ω

X₁        = 2 Ω                           R_m       = 150Ω

R₂        = 2 Ω                           X_m       = 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 Ω

R₂        =          0.9 Ω                           X₂        =          4.0 Ω

X_m       =          60 Ω               

If the total friction and windage losses are 200 W,

(i)     Find the slip, s.

(ii)  Find the input power, P_in.

(iii) Find the air gap power, P_ag.

(iv) Find the mechanical power, P_m.

(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 Ω

R₂           =          0.9                               X₂        =          0.4 Ω

X_m       =          20 Ω

           

If the total friction and windage losses are 1000 W, calculate:

(i) Slip

(ii) Input power, P_in

(iii)Air gap power, P_ag

(iv) Mechanical power, P_conv

(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.