4186720.ppt

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
DC Motor DC Generator

DC Machines Construction
cutaway view of a dc machine

cutaway view of a DC machine

Rotor of a DC machine

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 phenomena
Generator 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.

Classification of DC Motor
1. 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
T
F
R
V
I 
A
A
A
T R
I
E
V 

F
F
F
R
V
I 
A
A
A
T R
I
E
V 

A
L I
I 
1. Separately Excited DC Motor
2. Shunt DC Motor
F
A
L I
I
I 


)
( S
A
A
A
T R
R
I
E
V 


L
S
A I
I
I 

3. Series DC Motor
)
( S
A
A
A
T R
R
I
E
V 


F
T
F
R
V
I  F
L
A I
I
I 

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
1
2
2
1
1
2
1
2
,
A
A
A
A
E
E
n
n
then
If
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.

Solution
Given 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
A
25
.
1
200
V
250






F
T
F
F
A
L
R
V
I
I
I
I
18.75A
A
25
.
1
A
20




 F
L
A I
I
I

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
c) Find the speed of this motor when its

Solution
Given quantities:
• Terminal voltage, VT = 250 V
• Field resistance, RF = 50 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
A
95
A
5
A
100
50
V
250
A
100










F
T
L
F
L
A
R
V
I
I
I
I
V
3
.
244
V
7
.
5
V
250
)
06
.
0
)(
A
95
(
V
250







 A
A
T
A R
I
V
E

Solution (cont..)
• The resulting speed of this motor is
min
/
r
1173
min
/
r
1200
250
3
.
244
1
1
2
2
1
2
1
2






V
V
n
E
E
n
E
E
n
n
A
A
A
A

Solution (cont..)
(b) When the input current is 200A, the armature
A
195
A
5
A
200
50
V
250
A
200










F
T
L
F
L
A
R
V
I
I
I
I
V
3
.
238
V
7
.
11
V
250
)
06
.
0
)(
195
(
V
250








A
R
I
V
E A
A
T
A

Solution (cont..)
min
/
r
1144
min
/
r
1200
250
3
.
238
1
1
2
2
1
2
1
2






V
V
n
E
E
n
E
E
n
n
A
A
A
A

Solution (cont..)
(c) When the input current is 300A, the armature
A
295
A
5
A
300
50
V
250
A
300










F
T
L
F
L
A
R
V
I
I
I
I
V
3
.
232
V
7
.
17
V
250
)
06
.
0
)(
295
(
V
250








A
R
I
V
E A
A
T
A

Solution (cont..)
min
/
r
1115
min
/
r
1200
V
250
V
3
.
232
1
1
2
2
1
2
1
2






n
E
E
n
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

Solution
Given quantities
• Initial line current, IL = IA = 120 A
• Initial armature voltage, VA = 250 V

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
/
r
876
min
/
r
1103
V
8
.
242
V
8
.
192
1
1
2
2
1
2
1
2






n
E
E
n
E
E
n
n
A
A
A
A

Example 4
A 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.

Solution
Given quantities
• Supply voltage, VT = 200 V
• Series resistance, RS = 0.3 Ω
• 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
/
r
300
50
20
240
225
800
2
1
1
2
1
2
2
1
1
2
1
2
2
1
1
2
1
2











I
I
E
E
n
n
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

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/slot
P = number of poles
A = 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
P
ZN
E 

60


Classification of DC Generator
1. 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
F
A
L I
I
I 

A
L I
I 
Separately excited DC generator
F
F
F
R
V
I 
A
A
A
T R
I
E
V 

Shunt DC generator
F
T
F
R
V
I 
A
A
A
T R
I
E
V 


F
A
L I
I
I 

A
S
L I
I
I 

Series DC generator
)
( S
A
A
A
T R
R
I
E
V 


Shunt DC generator
F
T
F
R
V
I 
A
A
A
T R
I
E
V 


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..)
A
5
.
2
100
V
250






F
T
F
F
L
A
R
V
I
I
I
I
The field current,
A
20
V
250
W
5000



T
L
V
P
I
The 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
%
x
P
P
in
out
100


%
x
P
P
P
in
loss
out
100




The losses that occur in DC machine can be
divided into 5 categories
1.Copper losses (I2R)
2.Brush losses
3.Core losses
4.Mechanical losses
5.Stray load losses
a
a
a R
I
P 2

f
f
f R
I
P 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


 app
out
P

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
W
520
Pu  W
25000
A
50
Vx
500
Pout 

A
5
2
200
500
If .

 A
5
52
A
50
A
5
2
I
I
I L
f
a .
. 




Armature Copper
Loss
W
441
16
0
5
52
R
I
P 2
a
2
a
ca 

 )
.
(
)
.
(
)
(
Series Field
Copper Loss
W
5
220
08
0
5
52
R
I
P 2
2
f
2
a
2
cf .
)
.
(
)
.
(
)
( 


Shunt Field
Copper Loss
W
1250
200
5
2
R
I
P 2
1
f
2
f
1
cf 

 )
(
)
.
(
)
(
Friction + Stray + windage
+ etc:
W
520
Pu 
Total Losses = W
5
2431
520
1250
5
220
441 .
)
.
( 




Efficiency, η =
losses
Totall


Pout
Pout
Pin
Pout
%
.
@
.
.
13
91
9113
0
5
2431
25000
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
generator
As 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 rotor
b) 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

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
f
n 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
m
sync
slip n
-
n
n 
where
nslip = slip speed of the machine
nsync = speed of the magnetic fields
nm = mechanical shaft speed of motor
Slip is the relative speed expressed on a per unit or
a percentage basis
100%
x
n
n
s
sync
slip

100%
x
n
n
-
n
s
sync
m
sync


In term angular velocity (radians per second, rps)
100%
x
-
s
sync
m
sync




If the rotor turns at synchronous speed, s = 0
while if the rotor is stationary/standstill, s = 1
synx
m s)n
-
(1
n 
synx
m 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
e
r sf
f  )
( m
sync
r n
-
n
120
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
conv
ind
P

 
sync
AG
ind
P

 
The induced torque in induction motor is
s
R
I
PAG
2
2
2

s
R
I
PAG
2
2
2
3

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

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