Chapter 6
DC 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
DC Motor
DC Generator
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 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.
DC Motor
Basic Operation of DC Motor
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
1. Separately Excited DC Motor
VF
IF 
RF
IL  IA
VT  E A  I A RA
2. Shunt DC Motor
VT
IF 
RF
IL  I A  IF
VT  E A  I A RA
3. Series DC Motor
VT  E A  I A ( RA  RS )
I A  IS  IL
4. Compound DC Motor
VT
IF 
RF
I A  IL  IF
VT  E A  I A ( RA  RS )
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
n 2 E A 2 1


n1 E A1  2
If  2  1 , then
n2 E A2

n1 E A1
If Constant field excitation,
means; if1 = if2 or constant
flux; 1 = 2
• For series motor
n2 E A 2 1


n1 E A1  2
n2 E A 2 I A1


n1 E A1 I A 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..)
IL  I A  IF
the field current,
V T 250V
IF 

 1.25A
RF 200
the armature current,
I A  IL  IF
 20A  1.25A
 18.75A
VT = EA + IARA
the back e.m.f.,
EA = VT – IARA = 250 V – (18.75)(0.3) = 244.375 V
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.
Solution
Given 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
VT
I A  IL  IF IL 
RF
250V
 100 A 
50
 100 A  5 A  95 A
Therefore, EA at the load will be
E A  VT  I A RA
 250V  (95 A)(0.06 )
 250 V  5.7 V
 244.3V
Solution (cont..)
• The resulting speed of this motor is
n2 E A 2

n1 E A1
E A2
n2 
 n1
E A1
244.3V

1200 r / min
250 V
 1173 r / min
Solution (cont..)
(b) When the input current is 200A, the armature
current in the motor is
VT
I A  IL  IF IL 
RF
250V
 200 A 
50
 200 A  5 A  195 A
Therefore, EA at the load will be
E A  VT  I A RA
 250V  (195 A)(0.06 )
 250 V  11.7 V
 238.3 V
Solution (cont..)
• The resulting speed of this motor is
n2 E A 2

n1 E A1
E A2
n2 
 n1
E A1
238.3V

1200 r / min
250 V
 1144 r / min
Solution (cont..)
(c) When the input current is 300A, the armature
current in the motor is
VT
I A  IL  IF IL 
RF
250V
 300 A 
50
 300 A  5 A  295 A
Therefore, EA at the load will be
E A  VT  I A RA
 250V  (295 A)(0.06 )
 250 V  17.7 V
 232.3 V
Solution (cont..)
• The resulting speed of this motor is
n2 E A2

n1 E A1
E A2
n2 
 n1
E A1
232.3 V

1200 r / min
250 V
 1115 r / min
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
Solution
Given 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
n2 E A2

n1 E A1
E A2
n2 
 n1
E A1
192.8 V

1103 r / min
242.8 V
 876 r / min
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
• 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
n2
E A 2 1


n1
E A1  2
EA2 = V – Ia2 (RA + RS)
n2
E A2 I1
= 200 – 55(0.2 + 0.3)


n1
E A1 I 2
= 225 V
E A2 I1
n 2  n1 

E A1 I 2
225 20
 800 

240 50
The speed of the motor on new load  300 r / min
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 formerwound. 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.,
ZN
P
E

60
A
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
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
Separately excited DC generator
IL  IA
VF
IF 
RF
VT  EA  I A RA
Shunt DC generator
IL  I A  IF
VT
IF 
RF
VT  EA  I A RA
Series DC generator
IL  IS  I A
VT  E A  I A ( RA  RS )
Shunt DC generator
IL  I A  IF
VT
IF 
RF
VT  EA  I A RA
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..)
I A  IL  IF
The field current,
V T 250V
IF 

 2.5A
RF 100
The load current,
P 5000W
IL 

 20A
VT
250V
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
Pout

x100%
Pin
Pout  Ploss

x100%
Pin
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
Pf  I R f
2
f
Ia = armature current
If = field current
Pa  I Ra
2
a
Ra = armature
resistance
Rf = field resistance
Power Losses
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.
The Power Flow Diagram
Pout = VTIL
For generator
The Power Flow Diagram
Pout   app
For motor
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
Pu  520W
Pout  500Vx 50A  25000W
500
If 
 2.5A
200
Ia  If  IL  2.5A  50A  52.5A
Armature Copper
Loss
Series Field
Copper Loss
Shunt Field
Copper Loss
Pca  (Ia )2 Ra  (52.5)2 (0.16)  441W
Pcf 2  (Ia )2 Rf 2  (52.5)2 (0.08)  220.5W
Pcf 1  (If ) Rf 1  (2.5) (200)  1250W
2
Friction + Stray + windage
+ etc:
Total Losses =
2
Pu  520W
( 441  220.5  1250  520)  2431.5W
Efficiency, η =
Pout
Pout

Pin
Pout  Totall losses
25000

 0.9113 @ 91.13%
25000  2431.5
AC Machine Fundamentals &
Induction Machines
INDUCTION MACHINE
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.
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
Types of 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.
Squirrel Cage Rotor
Rotor bars (slightly skewed)
End ring
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
nsync
120 f e

P
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
nslip  nsync - nm
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
n sync - n m
n slip
s
x 100%
s
x 100%
n sync
n sync
In term angular velocity (radians per second, rps)
 sync -  m
s
x 100%
 sync
If the rotor turns at synchronous speed, s = 0
while if the rotor is stationary/standstill, s = 1
nm  (1 - s)n synx
 m  (1 - s) synx
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
f r  sf e
P
fr 
(n sync - nm )
120
Example
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)
The derivation of the induction motor inducedtorque equation
The induced torque in induction motor is
2 R2
Air gap power
PAG  I 2
s
Total Air gap power
PAG
2 R2
 3I 2
s
 ind 
 ind
Pconv
m
PAG

 sync
Assignment 6.5
A two pole, 50hz induction motor supplies 15kW to
a load at speed 2950 rpm.
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)
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