Chapter 6DC and AC Machines

Introduction

• An electrical machine is link between an electrical system and a mechanical system.

• Conversion from mechanical to electrical: generator

• Conversion from electrical to mechanical: motor

Introduction

Machines are called

• AC machines (generators or motors) if the electrical system is AC.

• DC machines (generators or motors) if the electrical system is DC.

DC machines can be divide by:

DC Machines Construction

cutaway view of a dc machine

DC Machines Construction

cutaway view of a DC machine

DC Machines Construction

Rotor of a DC machine

DC Machines Construction

Stator of a dc machine

DC Machines Fundamentals• Stator: is the stationary part of the machine. The

stator carries a field winding that is used to produce the required magnetic field by DC excitation.

• Rotor (Armature): is the rotating part of the machine. The rotor carries a distributed winding, and is the winding where the e.m.f. is induced.

• Field winding: Is wound on the stator poles to produce magnetic field (flux) in the air gap.

• Armature winding: Is composed of coils placed in the armature slots.

• Commutator: Is composed of copper bars, insulated from each other. The armature winding is connected to the commutator.

• Brush: Is placed against the commutator surface. Brush is used to connect the armature winding to external circuit through commutator

DC Machines Fundamentals

In DC machines, conversion of energy from electrical to mechanical form or vice versa results from the following two electromagnetic phenomenaGenerator action:An e.m.f. (voltage) is induced in a conductor if it moves through a magnetic field.Motor action: A force is induced in a conductor that has a current going through it and placed in a magnetic field•Any DC machine can act either as a generator or as a motor.

DC Machines Equivalent Circuit

The equivalent/modelling circuit of DC machine has two components:

Armature circuit: • It can be represented by a voltage source and a

resistance connected in series (the armature resistance). The armature winding has a resistance, RA.

The field circuit: • It is represented by a winding that generates the

magnetic field and a resistance connected in series. The field winding has resistance RF.

DC Motor

Basic Operation of DC Motor

Classification of DC Motor1. Separately Excited DC Motor • Field and armature windings are either connected

separate.

2. Shunt DC Motor• Field and armature windings are either connected in

parallel.

3. Series DC Motor• Field and armature windings are connected in series.

4. Compound DC Motor• Has both shunt and series field so it combines features

of series and shunt motors.

Equivalent Circuit of a DC Motor

Armature circuit - voltage source, EA and a resistor, RA.

The field coils, which produce the magnetic flux are represented by inductor, LF and resistor, RF.

The separate resistor, Radj represents an external variable resistor used to control the amount of current in the field circuit. Basically it lumped together with Rf and called Rf

Equivalent Circuit of DC Motor

F

TF R

VI

AAAT RIEV

F

FF R

VI

AAAT RIEV

AL II

1. Separately Excited DC Motor

2. Shunt DC Motor

FAL III

)( SAAAT RRIEV

LSA III

3. Series DC Motor

)( SAAAT RRIEV

F

TF R

VI FLA III

4. Compound DC Motor

Important terms in DC motor equivalent circuit

• VT – supply voltage• EA – internal generated voltage/back e.m.f.• RA – armature resistance• RF – field/shunt resistance• RS – series resistance• IL – load current• IF – field current• IA – armature current• IL – load current•

n – speed

Speed of a DC Motor

• For shunt motor

• For series motor

1

2

1

2

12

2

1

1

2

1

2

,

A

A

A

A

E

E

n

n

thenIf

E

E

n

n

2

1

1

2

1

2

2

1

1

2

1

2

A

A

A

A

A

A

I

I

E

E

n

n

E

E

n

n

If Constant field excitation, means; if1 = if2 or constant flux; 1 = 2

Flux, ϕ produce proportional to the current produce

Example 1

A 250 V, DC shunt motor takes a line current of 20 A. Resistance of shunt field winding is 200 Ω and resistance of the armature is 0.3 Ω. Find the armature current, IA and the back e.m.f., EA.

SolutionGiven parameters:• Terminal voltage, VT = 250 V• Field resistance, RF = 200 Ω• Armature resistance, RA = 0.3 Ω• Line current, IL = 20 A

Figure 1

Solution (cont..)

the field current,

the armature current,

VT = EA + IARA

the back e.m.f.,

EA = VT – IARA = 250 V – (18.75)(0.3) = 244.375 V

A25.1200

V250

F

TF

FAL

R

VI

III

18.75A

A25.1A20

FLA III

Example 2

A 50hp, 250 V, 1200 rpm DC shunt motor with compensating windings has an armature resistance (including the brushes, compensating windings, and interpoles) of 0.06 Ω. Its field circuit has a total resistance Radj + RF of 50 Ω, which produces a no-load speed of 1200 rpm. There are 1200 turns per pole on the shunt field winding.

Example 2 (cont..)

a) Find the speed of this motor when its input current is 100 A.

b) Find the speed of this motor when its input current is 200 A.

c) Find the speed of this motor when its input current is 300 A.

SolutionGiven quantities:• Terminal voltage, VT = 250 V• Field resistance, RF = 50 Ω• Armature resistance, RA = 0.06 Ω• Initial speed, n1 = 1200 r/min

Figure 2

Solution (cont..) (a) When the input current is 100A, the armature

current in the motor is

Therefore, EA at the load will be A95A5A100

50

V250A100

F

TLFLA R

VIIII

V3.244

V7.5V250

)06.0)(A95(V250

AATA RIVE

Solution (cont..)

• The resulting speed of this motor is

min/r1173

min/r1200250

3.244

11

22

1

2

1

2

V

V

nE

En

E

E

n

n

A

A

A

A

Solution (cont..)(b) When the input current is 200A, the armature

current in the motor is

Therefore, EA at the load will be A195A5A200

50

V250A200

F

TLFLA R

VIIII

V3.238

V7.11V250

)06.0)(195(V250

A

RIVE AATA

Solution (cont..)

• The resulting speed of this motor is

min/r1144

min/r1200250

3.238

11

22

1

2

1

2

V

V

nE

En

E

E

n

n

A

A

A

A

Solution (cont..)(c) When the input current is 300A, the armature

current in the motor is

Therefore, EA at the load will be A295A5A300

50

V250A300

F

TLFLA R

VIIII

V3.232

V7.17V250

)06.0)(295(V250

A

RIVE AATA

Solution (cont..)

• The resulting speed of this motor is

min/r1115

min/r1200V250

V3.232

11

22

1

2

1

2

nE

En

E

E

n

n

A

A

A

A

Example 3

The motor in Example 2 is now connected in separately excited circuit as shown in Figure 3. The motor is initially running at speed, n = 1103 r/min with VA = 250 V and IA = 120 A, while supplying a constant-torque load. If VA is reduced to 200 V, determine

i). the internal generated voltage, EA

ii). the final speed of this motor, n2

Example 3 (cont..)

Figure 3

SolutionGiven quantities

• Initial line current, IL = IA = 120 A

• Initial armature voltage, VA = 250 V

• Armature resistance, RA = 0.06 Ω

• Initial speed, n1 = 1103 r/min

Solution (cont..)

i) The internal generated voltage

EA = VT - IARA

= 250 V – (120 A)(0.06 Ω)

= 250 V – 7.2 V

= 242.8 V

Solution (cont..)

ii) Use KVL to find EA2

EA2 = VT - IA2RA

Since the torque and the flux is constant, IA is constant. This yields a voltage of:

EA2 = 200 V – (120 A)(0.06 Ω) = 200 V – 7.2 V = 192.8 V

Solution (cont..)

• The final speed of this motor

min/r876

min/r1103V8.242

V8.192

11

22

1

2

1

2

nE

En

E

E

n

n

A

A

A

A

Example 4A DC series motor is running with a speed of 800 r/min while taking a current of 20 A from the supply. If the load is changed such that the current drawn by the motor is increased to 55 A, calculate the speed of the motor on new load. The armature and series field winding resistances are 0.2 Ω and 0.3 Ω respectively. Assume the flux produced is proportional to the current. Assume supply voltage as 200 V.

SolutionGiven quantities• Supply voltage, VT = 200 V• Armature resistance, RA = 0.2 Ω• Series resistance, RS = 0.3 Ω• Initial speed, n1 = 800 r/min• Initial armature current, Ia1 = IL1 = 20 A

Figure 4

Solution (cont..) When the armature current increased, Ia2 = 55 A, the back emf

EA2 = V – Ia2 (RA + RS) = 200 – 55(0.2 + 0.3) = 225 V

min/r30050

20

240

225800

2

1

1

212

2

1

1

2

1

2

2

1

1

2

1

2

I

I

E

Enn

I

I

E

E

n

n

E

E

n

n

A

A

A

A

A

A

The speed of the motor on new load

Solution (cont..)

For initial load, the armature current, Ia1 = 20 A and the speed n1 = 800 r/min

V = EA1 + Ia1 (RA + RS)

The back e.m.f. at initial speed

EA1 = V - Ia1 (RA + RS) = 200 – 20(0.2 + 0.3) = 190 V

DC Generator

Generating of an AC Voltage

• The voltage generated in any DC generator inherently alternating and only becomes DC after it has been rectified by the commutator

Armature windings

• The armature windings are usually former-wound. This are first wound in the form of flat rectangular coils and are then puller.

• Various conductors of the coils are insulated each other. The conductors are placed in the armature slots which are lined with tough insulating material.

• This slot insulation is folded over above the armature conductors placed in the slot and is secured in place by special hard wooden or fiber wedges.

Generated or back e.m.f. of DC Generator

• General form of generated e.m.f.,

Φ = flux/pole (Weber)Z = total number of armature conductors = number of slots x number of conductor/slotP = number of polesA = number of parallel paths in armature[A = 2 (for wave winding), A = P (for lap winding)]N = armature rotation (rpm)E = e.m.f. induced in any parallel path in armature

A

PZNE

60

Classification of DC Generator1. Separately Excited DC Generator• Field and armature windings are either connected

separate.

2. Shunt DC Generator• Field and armature windings are either connected in

parallel.

3. Series DC Generator• Field and armature windings are connected in series.

4. Compound DC Generator• Has both shunt and series field so it combines features of series and shunt motors.

Equivalent circuit of DC generator

FAL III

AL II

Separately excited DC generator

F

FF R

VI

AAAT RIEV

Shunt DC generator

F

TF R

VI

AAAT RIEV

FAL III

ASL III

Series DC generator

)( SAAAT RRIEV

Shunt DC generator

F

TF R

VI

AAAT RIEV

Example • A DC shunt generator has shunt field winding

resistance of 100Ω. It is supplying a load of 5kW at a voltage of 250V. If its armature resistance is 0.02Ω, calculate the induced e.m.f. of the generator.

Solution

Given quantities• Terminal voltage, VT = 250V

• Field resistance, RF = 100Ω

• Armature resistance, RA = 0.22Ω

• Power at the load, P = 5kW

Solution (cont..)

A5.2100

V250

F

TF

FLA

R

VI

III

The field current,

A20V250

W5000

TL V

PIThe load current,

The armature current, IA = IL + IF = 20A + 2.5A = 22.5A

The induced e.m.f.,

EA = VT + IA RA = 250V + (22.5)(0.22) = 254.95V

Power flow and losses in DC machines

DC generators take in mechanical power and produce electric power while DC motors take in electric power and produce mechanical power

Efficiency

%xP

P

in

out 100

%xP

PP

in

lossout 100

The losses that occur in DC machine can be divided into 5 categories

1.Copper losses (I2R)2.Brush losses3.Core losses4.Mechanical losses5.Stray load losses

aaa RIP 2

fff RIP 2 Ia = armature current

If = field current

Ra = armature resistance

Rf = field resistance

Core losses – Hysteresis losses and Eddy current losses

Mechanical losses – The losses that associated with mechanical effects.

Two basic types of mechanical losses: Friction & Windage.

Friction losses caused by the friction of the bearings in the machine.

Windage are caused by the friction between the moving parts of the machine and the air inside the motor casing’s

Stray losses (Miscellaneous losses) – Cannot placed in one of the previous categories.

Power Losses

The Power Flow Diagram

For generator

Pout = VTIL

The Power Flow Diagram

For motor

appoutP

Example

A short-shunt compound generator delivers 50A at 500V to a resistive load. The armature, series field and shunt field resistance are 0.16, 0.08 and 200, respectively.

Calculate the armature current if the rotational losses are 520W, determine the efficiency of the generator

Solution

W520Pu W25000A50Vx500Pout

A52200

500If . A552A50A52III Lfa ..

Armature Copper Loss

W441160552RIP 2a

2aca ).().()(

Series Field Copper Loss

W5220080552RIP 22f

2a2cf .).().()(

Shunt Field Copper Loss

W125020052RIP 21f

2f1cf )().()(

Friction + Stray + windage + etc:

W520Pu

Total Losses = W5243152012505220441 .).(

Efficiency, η =losses Totall

Pout

Pout

Pin

Pout

%.@..

1391911305243125000

25000

AC Machine Fundamentals & Induction Machines

The induction machine is the most rugged and the most widely used machine in industry.Like dc machine, the induction machine has a stator and a rotor mounted on bearings and separated from the stator by an air gap.However, in the induction machine both stator winding and rotor winding carry alternating currents.The induction machine can operate both as a motor and as generatorAs motors, they have many advantages. They are rugged, relatively inexpensive and require very little maintenance. They range in size from a few watts to about 10,000 hp. The speed of an induction motor is nearly but not quite constant, dropping only a few percent in going from no load to full load.

INDUCTION MACHINE

The main disadvantages of induction motors are

a. The speed is not easily controlled.b. The starting current may be five to eight times

full-load current.c. The power factor is low and lagging when the

machine is lightly loaded

INDUCTION MOTOR CONSTRUCTION

Two different types of induction motor which can be placed in stator

a) squirrel cage rotorb) wound rotor

Squirrel Cage rotor Wound rotor

Squirrel cage rotor – consists of conducting bars embedded in slots in the rotor magnetic core and these bars are short circuited at each end by conducting end rings. The rotor bars and the rings are shaped like squirrel cage.

Wound rotor – carries three windings similar to the stator windings. The terminals of the rotor windings are connected to the insulated slip rings mounted on the rotor shaft. Carbon brushes bearing on these rings make the rotor terminals available to the user of the machine. For steady state operation, these terminals are short circuited.

Types of rotor

Rotor bars (slightly skewed)

End ring

Squirrel Cage Rotor

Wound Rotor• Most motors use the squirrel-cage rotor because of the

robust and maintenance-free construction.• However, large, older motors use a wound rotor with three

phase windings placed in the rotor slots.• The windings are connected in a three-wire wye. • The ends of the windings are connected to three slip rings. • Resistors or power supplies are connected to the slip rings

through brushes for reduction of starting current and speed control

Induction Motor Components

BASIC INDUCTION MOTOR CONCEPT

A single/three phase set of voltages has been applied to the stator, and single/three phase set of stator currents is flowing. These produce a magnetic field Bs, which is rotating in a counterclockwise direction .

The speed of the magnetic field’s rotation is P

fn e

sync

120

THE CONCEPT OF ROTOR SLIP

The voltage induced in a rotor depends on the speed of the rotor relative to the magnetic field.

Slip speed is defined as the difference between synchronous speed and rotor speed

msyncslip n - n n wherenslip = slip speed of the machinensync = speed of the magnetic fieldsnm = mechanical shaft speed of motor

Slip is the relative speed expressed on a per unit or a percentage basis

100% xn

n s

sync

slip100% x

n

n - n s

sync

msync

In term angular velocity (radians per second, rps)

100% x -

ssync

msync

If the rotor turns at synchronous speed, s = 0 while if the rotor is stationary/standstill, s = 1

synxm s)n- (1 n

synxm s)- (1

THE ELECTRICAL FREQUENCY CONCEPT

Like a transformer, the primary (stator) induces a voltage in the secondary (rotor) but unlike a transformer, the secondary frequency is not necessary the same as the primary frequency.

If the rotor of a motor is locked, then the rotor will have same frequency as the stator.

The rotor frequency can be expressed

er sf f )( msyncr n - n120

P f

A 208V, 10hp, 4 pole, 60Hz, Y connected induction motor has full load slip of 5%.

Calculate,

a. synchronous speed, nsync (Ans:1800rpm)

b. rotor speed, nm (Ans: 1710rpm)

c. rotor frequency, fr at the rated load (Ans: 3 Hz)

d. Shaft torque at the rated load (Ans: 41.7Nm)

Example

The derivation of the induction motor induced-torque equation

m

convind

P

sync

AGind

P

The induced torque in induction motor is

s

RIPAG

222

s

RIPAG

2223

Air gap power

Total Air gap power

a) What is the motor’s slip? (Ans:1.67%)

b) What is the induced torque in the motor in Nm under these conditions? (48.6Nm)

c) What will the operating speed of the motor be if its torque is doubled? (2900 rpm)

d) How much power will be supplied by the motor when the torque is doubled? (29.5kW)

A two pole, 50hz induction motor supplies 15kW to a load at speed 2950 rpm.

Assignment 6.5

Speed control of induction motors

i. Induction motor speed control by pole

ii. changing

iii. Speed control by changing the line frequency

iv. Speed control by changing the line voltage

v. Speed control by changing the rotor

vi. resistance