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CHAPTER 43-PHASE 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 cant 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 3.1 and 3.2 show the inside of induction machine.

Figure 3.1

Figure 3.2Stator ConstructionThe 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 coreStator WindingsStator 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 3.7Wound 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 rotor3.3 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.

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 rpm

n = rotor speed in rpm Slip can also represented in percent. The frequency of the rotor, fr 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 2A 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

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

=

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 4A 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 MOTORThe 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 3.10The 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 = 3V1I1cos

Stator copper loss, Pscl = 3I12R1Core Loss,

Pcl = 3V12/Rm

(always neglected because too small)

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

= Pin - Pscl - PclRotor copper loss,Prcl = 3I22R2Mechanical power/gross output power/converted power,

P mech = Pag Prcl = 3I22R2 /s - 3I22R2

= 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

Figure 3.11Example 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) =

(ii) Y connection, I = IL, V=

P in =3VIcos=

IL=

I = IL=154.59A

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

=

P net output = 29kW

(ii) =

(iii) 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 = 0.4

X1= Stator leakage inductance = 1.4

R2= Rotor resistance = 0.6

X2= Rotor leakage inductance = 2

Rm= no-load loses resistance = 150

Xm= magnetizing reactance = 20

Calculate:

(i) Input power

(ii) Speed of the rotor

(iii) Mechanical power

(iv) Developed torque

(v) Efficiency

Solution

(i) P in =3VIcos

V = 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 I22R2)

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 =

(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 = 0.4

X1= Stator leakage inductance = 1

R2= Rotor resistance = 0.8

X2= Rotor leakage inductance = 3.5

Rm= no-load loses resistance = 150

Xm= magnetizing reactance = 10

Calculate:

(i) Number of poles

(ii) Input power

(iii) Mechanical power

(iv) Developed torque

(v) Efficiency

Solution

(i) n = 855rpm

s = 0.05

ns = 120f/p

, sns = ns n , n = ns sns

= ns(1-s)

ns =

=

ns = 120f/p

p =

(ii)Pin =

Z total =

Pin =

EMBED Equation.3 (iii)Pm = 3(I22R2/s I22R2)

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

Figure 3.12There 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