EKT 103EKT 103EKT 103EKT 103

CHAPTER 4CHAPTER 4

AC Machine

AC MachineC ac e

: Alternating current (ac) is the primary source of electrical energy.

: It is less expensive to produce and transmit alternating current (ac) than direct current (dc)current (ac) than direct current (dc). : For this reason, and because ac voltage is induced into the

armature of all generators, ac machines are generally more practicalpractical.

: May function as a generator (mechanical to electrical) or a motor (electrical to mechanical)

AC GeneratorAC Generator This process can be described in terms of Faraday's law

when you see that the rotation of the coil continually changes the magnetic flux through the coil and therefore generates a voltage

AC MotorAC Motor a current is passed through th il ti tthe coil, generating a torque on the coil. Since the current is alternating, the motor will run smoothly only at the frequency of the sine wave.frequency of the sine wave.

Generator and MotorGenerator and Motor

How Does an Electric Generator Work?

Classification of AC Machines

i) S h M hi:Two major classes of machines;

i) Synchronous Machines: Synchronous Generators: A primary source of

electrical energy.electrical energy. Synchronous Motors: Used as motors as well as

power factor compensators (synchronous condensers).ii) A h (I d ti ) M hiii) Asynchronous (Induction) Machines: Induction Motors: Most widely used electrical motors in

both domestic and industrial applicationsboth domestic and industrial applications. Induction Generators: Due to lack of a separate field

excitation, these machines are rarely used as generators.

Synchronous Machine

Synchronous Machine

: Origin of name: syn = equal, chronos = time: Synchronous machines are called synchronous because: Synchronous machines are called synchronous because

their mechanical shaft speed is directly related to the power systems line frequency.

: the rotating air gap field and the rotor rotate at the same speed, called the synchronous speed.

: Synchronous machines are ac machine that have a field: Synchronous machines are ac machine that have a field circuit supplied by an external dc source. DC field winding on the rotor,g , AC armature winding on the stator

Synchronous Machine

: Synchronous machines are used primarily as generators of electrical power, called synchronous generators or alternators.

: They are usually large machines generating electrical power at hydro nuclear or thermal power stationsat hydro, nuclear, or thermal power stations.

: Synchronous motors are built in large units compare to induction motors (Induction motors are cheaper for smaller ratings) and used for constant speed industrial drives

: Application as a motor: pumps in generating stations, electric clocks timers and so forth where constant speed is desiredclocks, timers, and so forth where constant speed is desired.

Synchronous Machine

The frequency of the induced voltage is related to the rotor speed by:

where P is the number of magnetic poleswhere P is the number of magnetic polesfe is the power line frequency. Typical machines have two-poles, four-poles, and six-poles

Synchronous Machine

Construction

Energy is stored in the inductanceAs the rotor mo es there is a change in As the rotor moves, there is a change in the energy stored

Either energy is extracted from the magnetic field (and becomes mechanical energy motor)

Or energy is stored in the magnetic field gy gand eventually flows into the electrical circuit that powers the stator generator

Synchronous Machine

Construction

DC field windings are mounted on the (rotating) rotor - which is thus a rotating electromagnet

AC windings are mounted on the (stationary) stator resulting in three-phase AC stator voltages and currents

:The main part in the synchronous machines arei) Rotorii) Stator

Synchronous MachineRRotor: There are two types of rotors used in synchronous

machines:machines: i) cylindrical (or round) rotors ii) salient pole rotors) p

: Machines with cylindrical rotors are typically found in higher speed higher power applications such as t b t U i 2 4 l th hiturbogenerators. Using 2 or 4 poles, these machines rotate at 3600 or 1800 rpm (with 60hz systems).

: Salient pole machines are typically found in large (many: Salient pole machines are typically found in large (many MW), low mechanical speed applications, including hydrogenerators, or smaller higher speed machines (up to 1 2 MW)1-2 MW).

: Salient pole rotors are less expensive than round rotors.

Synchronous Machine Cylindrical rotor

D 1 mTurbine

L 10 mSteam

Uniform air gap

Stator winding

N High speed

d-axis

Stator

Uniform air-gap

Rotor

3600 r/min2-pole 1800 r/min4-pole Direct-conductor cooling (using q-axis

Rotorwinding

S

Direct-conductor cooling (using hydrogen or water as coolant)

Rating up to 2000 MVA

Turbogenerator

Synchronous Machine Cylindrical rotor

Stator

Cylindrical rotor

Synchronous Machine Salient Pole1. Most hydraulic turbines have to turn at low speeds

(between 50 and 300 r/min)

2 A l b f l i d th t2. A large number of poles are required on the rotor

Non-uniformN

d-axis

D 10 m

Non uniform air-gap

N

S Sq axis

TurbineH d ( t )

S S

N

q-axis

Hydro (water)

Hydrogenerator

Synchronous Machine Salient Pole

Stator

: Synchronous machine rotors are simply rotating electromagnets built to have as many poles as are produced by the stator windings.

: Dc currents flowing in the field coils surrounding each pole magnetize the rotor poles.

: The magnetic field produced by the rotor poles locks in with a rotating: The magnetic field produced by the rotor poles locks in with a rotating stator field, so that the shaft and the stator field rotate in synchronism.

: Salient poles are too weak mechanically and develop too much wind resistance and noise to be used in large, high-speed generators driven by steam or gas turbines. For these big machines, the rotor must be a solid, cylindrical steel forging to provide the necessary strength.

: Axial slots are cut in the surface of the cylinder to accommodate the field windings.

: Since the rotor poles have constant polarity they must be li d ith di t tsupplied with direct current.

: This current may be provided by an external dc generator or by a rectifierby a rectifier. : In this case the leads from the field winding are connected to insulated

rings mounted concentrically on the shaft. : Stationary contacts called brushes ride on these slip rings to carry

current to the rotating field windings from the dc supply. : The brushes are made of a carbon compound to provide a good: The brushes are made of a carbon compound to provide a good

contact with low mechanical friction. : An external dc generator used to provide current is called a brushless

exciter .

Synchronous Machine

Stator: The stator of a synchronous machine carries the armature or load

winding which is a three-phase winding.: The armature winding is formed by interconnecting various g y g

conductors in slots spread over the periphery of the machines stator. Often, more than one independent three phase winding is on the stator An arrangement of a three-phase stator winding is shown instator. An arrangement of a three-phase stator winding is shown in Figure below. Notice that the windings of the three-phases are displaced from each other in space.

Synchronous Machine

Construction :Stator

Synchronous MachineMagnetomotive Forces (MMFs) and Fluxes Due to Armature and Field Windings

Flux produced by a stator winding

Synchronous Machine

Magnetomotive Forces (MMFs) and Fluxes Due to Armature d Fi ld Wi diand Field Windings

Synchronous Machine

Magnetomotive Forces (MMFs) and Fluxes Due to Armature d Fi ld Wi diand Field Windings

Two Cycles of mmf around the Statory

Synchronous Generator

Equivalent circuit model synchronous generatorq y g

: If the generator operates at a terminal voltage VT while supplying a load corresponding to an armature current Ia then;load corresponding to an armature current Ia, then;

: In an actual synchronous machine, the reactance is much greater than the armature resistance, in which case;

: Among the steady-state characteristics of a synchronous generator, its voltage regulation and power-angle characteristics are the mostits voltage regulation and power angle characteristics are the most important ones. As for transformers, the voltage regulation of a synchronous generator is defined at a given load as;

Synchronous Generator

Phasor diagram of a synchronous generator

The phasor diagram is to shows the relationship among the voltages within a phase (E,V, jXSIA and RAIA) and the current IA in the phase.

Unity P.F (1.0)

Synchronous Generator

Lagging P.F

Leading P.F.

Synchronous Generator

Power and TorqueI t t ll th h i l i i tIn generators, not all the mechanical power going into a synchronous generator becomes electric power out of the machine

The power losses in generator are represented by difference betweenThe power losses in generator are represented by difference between output power and input power shown in power flow diagram below

Synchronous Generator

LossesR tRotor

- resistance; iron parts moving in a magnetic field causingcurrents to be generated in the rotor body

- resistance of connections to the rotor (slip rings)Stator

- resistance; magnetic losses (e g hysteresis)resistance; magnetic losses (e.g., hysteresis)Mechanical

- friction at bearings, friction at slip ringsSt l d lStray load losses

- due to non-uniform current distribution

Synchronous Generator

The input mechanical power is the shaft power in the generator given by equation:

The power converted from mechanical to electrical form internally is given by

The real electric output power of the synchronous generator can be expressed in line and phase quantities as

and reactive output power

Synchronous Generator

In real synchronous machines of any size, the armature resistance RA is more than 10 times smaller than the synchronous reactance XS (Xs >> RA). Therefore, RA can be ignored

Synchronous Motor

Synchronous Motor

Power Flow

Example : Synchronous Generator.

: A three-phase, wye-connected 2500 kVA and 6.6 kV generator operates at full-load. The per-phase armature resistance Ra and the synchronous reactance Xd are (0 07+j10 4)synchronous reactance, Xd, are (0.07 j10.4). Calculate the percent voltage regulation at

(a) 0.8 power-factor lagging, and(b) 0 8 f t l di(b) 0.8 power-factor leading.

Solution.

Induction Machine: The machines are called induction machines because of the

rotor voltage which produces the rotor current and the rotor magnetic field is induced in the rotor windings.

: Induction generator has many disadvantages and low efficiency Therefore induction machines are usuallyefficiency. Therefore induction machines are usually referred to as induction motors.

Induction Machine

Three-phase induction motors are the most common and f tl t d hi i i d tfrequently encountered machines in industry simple design, rugged, low-price, easy maintenance wide range of power ratings: fractional horsepower to 10 MW run essentially as constant speed from no-load to full load Its speed depends on the frequency of the power source

not easy to have variable speed controlnot easy to have variable speed control requires a variable-frequency power-electronic drive for optimal

speed control

ConstructionConstructionSlip rings

Cutaway in a typical woundtypical wound-rotor IM. Notice the brushes and the slip rings

Brushes

ConstructionA i d i h i An induction motor has two main partsi) a stationary stator

consisting of a steel frame that s pports a hollo consisting of a steel frame that supports a hollow, cylindrical core

core, constructed from stacked laminations, , , having a number of evenly spaced slots, providing the space for

the stator winding

Stator of IM

ConstructionConstructionii) a revolving rotor

composed of punched laminations stacked to create a composed of punched laminations, stacked to create a series of rotor slots, providing space for the rotor winding

conventional 3-phase windings made of insulated wire (wound rotor) similar to the winding on the stator(wound-rotor) similar to the winding on the stator

aluminum bus bars shorted together at the ends by two aluminum rings, forming a squirrel-cage shaped circuit (squirrel cage)(squirrel-cage)

ConstructionConstruction Two basic design types depending on the rotor design

i l d ti b l id i t l t d squirrel-cage: conducting bars laid into slots and shorted at both ends by shorting rings.

wound-rotor: complete set of three-phase windings p p gexactly as the stator. Usually Y-connected, the ends of the three rotor wires are connected to 3 slip rings on the rotor shaft In this way the rotor circuit ison the rotor shaft. In this way, the rotor circuit is accessible.

Construction1. Squirrel cage the conductors would look like one of

the exercise wheels that squirrel or hamsters run on.

Construction2. Wound rotor have a brushes and slip ring at the end

of rotor

Notice theNotice the slip rings

Rotating Magnetic Field Balanced three phase windings, i.e.

mechanically displaced 120 degrees form h th f d b b l d th heach other, fed by balanced three phase

source A rotating magnetic field with constant

it d i d d t ti ith dmagnitude is produced, rotating with a speed

)(120 rpmP

fn esync =Where fe is the supply frequency andP is the no. of poles and nsync is called

Py

P is the no. of poles and nsync is calledthe synchronous speed in rpm(revolutions per minute)

Synchronous speedy p

P 50 Hz 60 Hz

2 3000 3600

4 1500 18004 1500 1800

6 1000 1200

8 750 900

10 600 72010 600 720

12 500 600

Principle of operation This rotating magnetic field cuts the rotor windings and produces an

induced voltage in the rotor windingsD t th f t th t th t i di h t i it d f b th Due to the fact that the rotor windings are short circuited, for both squirrel cage and wound-rotor, and induced current flows in the rotor windingsTh t t d th ti fi ld The rotor current produces another magnetic field

A torque is produced as a result of the interaction of those two magnetic fields

kB BWhere ind is the induced torque and BR and BS are the

ind R skB B = ind q R S

magnetic flux densities of the rotor and the stator respectively

Induction motor speed At what speed will the IM run?

Can the IM run at the synchronous speed, why? If rotor runs at the synchronous speed, which is the same speed

of the rotating magnetic field, then the rotor will appear stationary to the rotating magnetic field and the rotating magnetic field will not cut the rotor. So, no induced current will flow in the rotor and no rotor magnetic flux will be produced so no torque is generated and the rotor speed will fall below the synchronous speed

When the speed falls, the rotating magnetic field will cut the rotor windings and a torque is produced

Induction motor speedInduction motor speed

So, the IM will always run at a speed lower thanSo, the IM will always run at a speed lower than the synchronous speed

The difference between the motor speed and the psynchronous speed is called the Slip (s)

nnn =

Where n = slip speed

msyncslip nnn =

Where nslip= slip speednsync= speed of the magnetic fieldnm = mechanical shaft speed of the motorm p

The Slippmsync nns

=Where s is the slip

syncns =p

Notice that : if the rotor runs at synchronous speed

s = 0s = 0

if the rotor is stationary

s = 1

Slip may be expressed as a percentage by multiplying the b b 100 ti th t th li i ti d d tabove eq. by 100, notice that the slip is a ratio and doesnt

have units

Induction Motors and TransformersInduction Motors and Transformers

Both IM and transformer works on the principle of p pinduced voltage Transformer: voltage applied to the primary windings produce an

induced voltage in the secondary windingsinduced voltage in the secondary windings Induction motor: voltage applied to the stator windings produce

an induced voltage in the rotor windingsThe difference is that in the case of the induction motor the The difference is that, in the case of the induction motor, the secondary windings can move

Due to the rotation of the rotor (the secondary winding of the IM), the induced voltage in it does not have the same frequency ofthe induced voltage in it does not have the same frequency of the stator (the primary) voltage

FrequencyFrequency The frequency of the voltage induced in the rotor y g

is given by

120rP nf =

Where fr = the rotor frequency (Hz)

120

P = number of stator polesn = slip speed (rpm)

( )P n n ( )120

s mr

P n nf

P sn

=120

se

P sn sf= =

FrequencyFrequency What would be the frequency of the rotors

i d d lt t d ?induced voltage at any speed nm?

er sff = When the rotor is blocked (s=1) , the frequency

f th i d d lt i l t th l

er ff

of the induced voltage is equal to the supply frequency

On the other hand if the rotor runs at On the other hand, if the rotor runs at synchronous speed (s = 0), the frequency will be zerobe zero

TorqueTorque While the input to the induction motor is electrical power,

its output is mechanical power and for that we shouldits output is mechanical power and for that we should know some terms and quantities related to mechanical powerA h i l l d li d t th t h ft ill Any mechanical load applied to the motor shaft will introduce a Torque on the motor shaft. This torque is related to the motor output power and the rotor speedp p p

and ).( mNPoutload = )/(602 sradnmm =

m 60

Horse powerHorse power Another unit used to measure mechanical power is the p

horse power It is used to refer to the mechanical output power of the

tmotor Since we, as an electrical engineers, deal with watts as a

unit to measure electrical power, there is a relationunit to measure electrical power, there is a relation between horse power and watts

wattshp 7461 =p

ExampleExample

A 208-V, 10hp, four pole, 60 Hz, Y-connectedA 208 V, 10hp, four pole, 60 Hz, Y connected induction motor has a full-load slip of 5 percent1. What is the synchronous speed of this motor?2. What is the rotor speed of this motor at rated load?3. What is the rotor frequency of this motor at rated

l d?load?4. What is the shaft torque of this motor at rated load?

SolutionSolution1. 120 120(60) 1800

4e

syncfn rpm

P= = =

2.

4P

(1 )(1 0 05) 1800 1710

m sn s n=

3.

(1 0.05) 1800 1710 rpm= =

0.05 60 3r ef sf Hz= = =

4. out outload P P n = = 260

10 746 / 41.7 .1710 2 (1/ 60)

mmn

hp watt hp N m

= =

1710 2 (1/ 60)

Equivalent CircuitEquivalent Circuit The induction motor is similar to the transformer with the

exception that its secondary windings are free to rotateexception that its secondary windings are free to rotate

As we noticed in the transformer, it is easier if we can combine these two circuits in one circuit but there are some difficultiestwo circuits in one circuit but there are some difficulties

Equivalent CircuitEquivalent Circuit When the rotor is locked (or blocked), i.e. s =1, the

largest voltage and rotor frequency are induced in thelargest voltage and rotor frequency are induced in the rotor.

On the other side, if the rotor rotates at synchronous speed, i.e. s = 0, the induced voltage and frequency in the rotor will be equal to zero.

Where ER0 is the largest value of the rotors induced voltage

0RR sEE =Where ER0 is the largest value of the rotor s induced voltage

obtained at s = 1(locked rotor)

Equivalent CircuitEquivalent Circuit The same is true for the frequency, i.e.

It is known thater sff =

fLLX 2

So, as the frequency of the induced voltage in

fLLX 2==

the rotor changes, the reactance of the rotor circuit also changes

Where X is the rotor reactance2 rrrrr LfLX == Where Xr0 is the rotor reactance

at the supply frequency (at blocked rotor)

2 re

rrrrr

Lsf

f

= 0rsX=

Equivalent CircuitEquivalent Circuit

Then, we can draw the rotor equivalent circuit asThen, we can draw the rotor equivalent circuit as follows

Where ER is the induced voltage in the rotor and RR isWhere ER is the induced voltage in the rotor and RR is the rotor resistance

Equivalent CircuitEquivalent Circuit Now we can calculate the rotor current as

E

0

( )R

RR R

R

EIR jX

sE

= +

Dividing both the numerator and denominator by s so

0

0( )R

R R

sER jsX

= +g y

nothing changes we get0R

REI R=

Where ER0 is the induced voltage and XR0 is the rotor reactance at

0( )R

RR

R jXs+

Where ER0 is the induced voltage and XR0 is the rotor reactance at blocked rotor condition (s = 1)

Equivalent CircuitEquivalent Circuit Now we can have the rotor equivalent

circuit

Equivalent CircuitEquivalent Circuit Now as we managed to solve the induced voltage and g g

different frequency problems, we can combine the stator and rotor circuits in one equivalent circuitWhereWhere

22 0e ff RX a X=

22

2

e ff R

R

R a RII

a

==

1 0

e ff

e ff R

S

aE a E

Na

==e ff

R

aN

=

Power losses in Induction machines

Copper lossespp Copper loss in the stator (PSCL) = I12R1 Copper loss in the rotor (PRCL) = I22R2

C l (P ) Core loss (Pcore) Mechanical power loss due to friction and windage How this power flow in the motor? How this power flow in the motor?

Power flow in induction motorPower flow in induction motor

Power relationsPower relations

3 cos 3 cosP V I V I = =3 cos 3 cosin L L ph phP V I V I = =2

1 13SCLP I R=( )AG in SCL coreP P P P= +

23P I R22 23RCLP I R=conv AG RCLP P P=

( )out conv f w strayP P P P+= + convindm

P =

Equivalent CircuitEquivalent Circuit

We can rearrange the equivalent circuit asWe can rearrange the equivalent circuit as follows

Actual rotorResistance

i l t tActual rotor resistance

equivalent to mechanical load

Power relationsPower relations

3 cos 3 cosP V I V I = =3 cos 3 cosin L L ph phP V I V I = =2

1 13SCLP I R=( )AG in SCL coreP P P P= +

23P I R

conv RCLP P= + 2 223 RI s=RCLPs

=22 23RCLP I R=

conv AG RCLP P P= 2 22 (1 )3 R sI s= (1 )RCLP s

s=

( )fP P P P= +

s s(1 )conv AGP s P=

convi d

P = (1 ) AGs P=( )out conv f w strayP P P P+ + indm

(1 ) ss

Power relationsPower relations

A GP co n vPA G co n v1 1-s

R C LPR C Ls

: :AG RCL convP P P 1 : : 1-s s

ExampleExampleA 480-V, 60 Hz, 50-hp, three phase induction motor is drawing 60A at 0.85 PF lagging. The stator copper losses are 2 kW, and the rotor copper losses are 700 W. The f i ti d i d l 600 W th lfriction and windage losses are 600 W, the core losses are 1800 W, and the stray losses are negligible. Find the following quantities:following quantities:

1. The air-gap power PAG.2. The power converted Pconv.3. The output power Pout.4. The efficiency of the motor.

SolutionSolution1. 3 cosin L LP V I =

3 480 60 0.85 42.4 kW= =AG in SCL coreP P P P=

2.42.4 2 1.8 38.6 kW= =

conv AG RCLP P P= 70038.6 37.9 kW

1000= =

3. &60037 9 37 3 kW

out conv F WP P P= = =37.9 37.3 kW

1000

SolutionSolution37.3 50 hptP = = 50 hp0.746outP

100%outP = 4.

100%

37.3 100 88%

inP =

= =100 88%42.4

ExampleExampleA 460-V, 25-hp, 60 Hz, four-pole, Y-connected induction motor has the following impedances in ohms per phase referred to the stator circuit:following impedances in ohms per phase referred to the stator circuit:R1= 0.641 R2= 0.332X1= 1.106 X2= 0.464 XM= 26.3 Th t t l t ti l l 1100 W d d t b t tThe total rotational losses are 1100 W and are assumed to be constant. The core loss is lumped in with the rotational losses. For a rotor slip of 2.2 percent at the rated voltage and rated frequency, find the motors

1. Speed2. Stator current3. Power factor

4. Pconv and Pout5. ind and load6 Effi i6. Efficiency

SolutionSolution1. 120 120 60 1800 rpm

4e

syncfn

P= = = p4sync P

(1 ) (1 0.022) 1800 1760 rpmm syncn s n= = =0 332R2. 22 20.332 0.4640.022

15.09 0.464 15.1 1.76

RZ jX js

j

= + = += + = j

2

1 11/ 1/ 0.038 0.0662 1.76f M

ZjX Z j

= =+ + 1 12.94 31.1

0.0773 31.1= =

SolutionSolutiontot stat fZ Z Z= +

0.641 1.106 12.94 31.111.72 7.79 14.07 33.6

jj

= + + = + =

1

460 03 18.88 33.6 A

14 07 33 6V

IZ

= = =

3.4.

14.07 33.6totZ cos33.6 0.833 laggingPF = =3 cos 3 460 18.88 0.833 12530 Win L LP V I = = =in L L

2 21 13 3(18.88) 0.641 685 WSCLP I R= = =

12530 685 11845 WP P P= = =12530 685 11845 WAG in SCLP P P= = =

SolutionSolution(1 ) (1 0.022)(11845) 11585 Wconv AGP s P= = =

& 11585 1100 10485 W10485

out conv F WP P P= = =

5.

10485= 14.1 hp746

=11845 62.8 N.m18002

AGind

P = = =18002 60ind

sync 10485 56.9 N.m1760

outload

P = = =

6.

17602 60load

m 10485100% 100 83 7%outP = = =100% 100 83.7%12530inP

= = =

Induction Motor Power and Torque

The output power can be found asPout = Pconv PF&W Pmisc

Th i d d t d l d tThe induced torque or developed torque:

ExampleExampleA two-pole, 50-Hz induction motor supplies 15kW to a p , ppload at a speed of 2950 rpm.1. What is the motors slip?2. What is the induced torque in the motor in N.m under these

conditions?3. What will be the operating speed of the motor if its torque is

?doubled?4. How much power will be supplied by the motor when the

torque is doubled?

SolutionSolution1. 120 120 50 3000 rpm

2e

syncfn

P= = = p2

3000 2950 0.0167 or 1.67%3000

sync

sync m

Pn n

sn = = =

2.3000syncn

no givenf WP +Q

3

assume and

15 10 48.6 N.m

conv load ind load

convi d

P PP

= =

= = = 48.6 N.m2295060

indm

SolutionSolution3. In the low-slip region, the torque-speed curve is

linear and the induced torque is direct proportionallinear and the induced torque is direct proportional to slip. So, if the torque is doubled the new slip will be 3.33% and the motor speed will be

4

(1 ) (1 0.0333) 3000 2900 rpmm syncn s n= = =

4.2(2 48.6) (2900 ) 29.5 kW60

conv ind mP

== =

60