VI. DC Machines
Introduction
DC machines are used in applications requiring a wide range
of speeds by means of various combinations of their field
windings
Types of DC machines:
•
•
•
•
Separately-excited
Shunt
Series
Compound
1
DC
Machines
Motoring
Generating
Mode
Mode
In Motoring Mode: Both armature
and field windings are excited by DC
In Generating Mode: Field winding
is excited by DC and rotor is rotated
externally by a prime mover coupled
to the shaft
1. Construction of DC Machines
Basic parts of a DC machine
2
Construction of DC Machines
Copper commutator segment
and carbon brushes are used
for:
(i) for mechanical rectification
of induced armature emfs
(ii) for taking stationary armature
terminals from a moving member
Elementary DC machine with commutator.
(a) Space distribution of air-gap flux density
in an elementary dc machine;
Average gives us a DC voltage, Ea
Ea = Kg f r
Te = Kg f a
= Kd If a
(b) waveform of voltage between brushes.
3
Electrical Analogy
(a) Space distribution of air-gap flux density
in an elementary dc machine;
(b) waveform of voltage between brushes.
2. Operation of a Two-Pole DC Machine
4
Space distribution of air-gap flux density, Bf
in an elementary dc machine;
One pole spans 180º electrical in space
B f  B peak sin( )
Mean air gap flux per pole: avg / pole  Bavg A per
Aper pole: surface area spanned
by a pole
pole
 0 B peak sin( )dA
 0 B peak sin( )rd
For a two pole DC machine, avg / pole  2 B peak r
Space distribution of air-gap flux density, Bf
in an elementary dc machine;
One pole spans 180º electrical in space
B f  B peak sin( )
Flux linkage a:
with 0= 0
ea 
 a  Navg / pole cos( )
 : phase angle between the magnetic axes of
the rotor and the stator
a  Navg / pole cos( r t )
da
  r Navg / pole sin( r t )
dt
For a two pole DC machine: Ea 
2

 (t )   r t   0
Ea 
1


0 ea ( t )dt
 r Navg / pole
5
In general:
Kg: winding factor
E a  K g f  r
f: mean airgap flux per pole
r: shaft-speed in mechanical rad/sec
 r  2
nr
nr: shaft-speed in revolutions per minute (rpm)
60
DC machines with number of poles > 2
f elec 
P nr Pnr

2 60 120
avg / pole 
Ea 
P: number of poles
2
2 B peak r
P
P 2
N r avg / pole  K g f  r
2
A four-pole DC machine
Schematic representation of a DC machine
6
Typical magnetization curve of a DC machine
Torque expression in terms of mutual inductance
Te 
dM fa
1 2 dL f 1 2 dLa
if
 ia
 i f ia
2
d
2 d
d
Te  i f ia
dM fa
M fa  M̂ cos 
d
ˆi i
Te  M
f a
Alternatively, electromagnetic torque Te can be derived from power conversion equations
Pmech  Pelec
Ea  K g f  m
Te m  Ea I a
Te m  K g f  m I a
Te  K g f I a
7
In a linear magnetic circuit
Te  K g K f I f I a
where
f  K f If
Field-circuit connections of DC machines
(a) separately-excited
(c) shunt
(b) series
(d) compound
8
Separately-excited DC machine circuit in motoring mode
Pmech

Pout

Pf & w

 
 

internal electromechanical power
or gross output power
E a  K g f  m
Kg 
E a  Vt
p Ca
2 a
output power produced
friction and windage
p : number of poles
Ca : total number of conductors in armature winding
a : number of parallel paths through armature winding
Te produces rotation (Te and m are in the same direction)
Pmech  0, Te  0 and ωm  0
Separately-excited DC machine circuit in generating mode
E a  Vt
Te and m are in the opposite direction
Pmech  0, Te  0 and ωm  0
Generating mode
– Field excited by If (dc)
– Rotor is rotated by a mechanical prime-mover at m.
– As a result Ea and Ia are generated
9
3. Analysis of DC Generators
Separately-Excited DC Generator
also
Vt  E a  I a ra
where
E a  K g f  m
Vt  I L R L
where
I L  Ia
Terminal V-I Characteristics
Terminal voltage (Vt) decreases slightly as load current increases
(due to IaRa voltage drop)
10
Terminal voltage characteristics of DC generators
Series generator is not used due to poor voltage regulation
Shunt DC Generator (Self-excited DC Generator)
– Initially the rotor is rotated by a mechanical prime-mover at m
while the switch (S) is open.
– Then the switch (S) is closed.
11
When the switch (S) is closed
Ea= (ra + rf) If
Load line of electrical circuit
Self-excitation uses the residual magnetization & saturation properties
of ferromagnetic materials.
– when S is closed Ea = Er and If = If0
– interdependent build-up of If and Ea continues
– comes to a stop at the intersection of the two curve
as shown in the figure below
Solving for the exciting current, If
E a  K g f  m
where
f  K f I f
Ea  K d If  m
where
Kd  K g K f
Integrating with the electrical circuit equations
K d i f  m   La  Lf 
di f
 ra  rf  i f
dt
Applying Laplace transformation we obtain
K d  m I f ( s )   La  Lf sI f ( s )  ra  rf I f ( s )   La  Lf I f0
So the time domain solution is given by
if ( t )  I f0
 r  r  K dm
  a f
La  Lf
e 

 t

12
if ( t )  I f0
 r  r  K dm
  a f
La  Lf
e 

 t

Let us consider the following 3 situations
(i) (ra + rf) > Kdm
lim if ( t )  0
t 
Two curves do not intersect.
(ii) (ra + rf) = Kdm
if ( t )  I f 0
Self excitation can just start
(iii) (ra + rf) < Kdm
Generator can self-excite
Self-excited DC generator under load
Ia= If + IL
Vt= Ea - ra Ia = rf If
Ea= ra Ia + rf If
Ea= (ra + rf) If + ra IL
Load line of electrical circuit
13
Series DC Generator
also
Vt  E a  I a ( ra  rs )
where
E a  K g f  m
Vt  I L R L
and
I L  Ia  I s
Not used in practical, due to poor voltage regulation
Compound DC Generators
(a) Short-shunt connected compound DC generator
also
Vt  E a  I a ra  I s rs
where
E a  K g f  m
Vt  I L R L
and
IL  Is
and
Ia  If  Is
14
(b) Long-shunt connected compound DC generator
also
Vt  E a  I a ra  rs 
where
E a  K g f  m
Vt  I L R L
and
I L  Ia  If
and
Ia  Is
Types of Compounding
(i) Cumulatively-compounded DC generator (additive compounding)
Fd  Ff  Fs
 

field
mmf
shunt
field
mmf
series
field
mmf
for linear M.C. (or in the linear region of the magnetization curve, i.e. unsaturated magnetic circuits)
 d  f   s
(ii) Differentially-compounded DC generator (subtractive compounding)
Fd  Ff  Fs
for linear M.C. (or in the linear region of the magnetization curve, i.e. unsaturated magnetic circuits)
 d  f   s
Differentially-compounded generator is not used in practical, as it exhibits
poor voltage regulation
15
Terminal V-I Characteristics of Compound Generators
Above curves are for cumulatively-compounded generators
Magnetization curves for a 250-V 1200-r/min dc machine.
Also field-resistance lines for the discussion of self-excitation are shown
16
Examples
1.
A 240kW, 240V, 600 rpm separately excited DC generator has an armature
resistance, ra = 0.01 and a field resistance rf = 30. The field winding is
supplied from a DC source of Vf = 100V. A variable resistance R is connected
in series with the field winding to adjust field current If. The magnetization
curve of the generator at 600 rpm is given below:
If (A)
1
1.5
2
2.5
3
4
5
6
Ea (V)
165
200
230
250
260
285
300
310
If DC generator is delivering rated load and is driven at 600 rpm determine:
a) Induced armature emf, Ea
b) The internal electromagnetic power produced (gross power)
c) The internal electromagnetic torque
d) The applied torque if rotational loss is Prot = 10kW
e) Efficiency of generator
f) Voltage regulation
2.
A shunt DC generator has a magnetization curve at nr = 1500 rpm as shown
below. The armature resistance ra = 0.2 , and field total resistance rf = 100 .
a) Find the terminal voltage Vt and field current If of the generator when it
delivers 50A to a resistive load
b) Find Vt and If when the load is disconnected (i.e. no-load)
17
Solution:
a)
b)
NB. Neglected raIf voltage drops
3.
The magnetization curve of a DC shunt generator at 1500 rpm is given below,
where the armature resistance ra = 0.5 , and field total resistance rf = 100 ,
the total friction & windage loss at 1500 rpm is 400W.
a) Find no-load terminal voltage at 1500 rpm
b) For the self-excitation to take place
(i) Find the highest value of the total shunt field resistance at 1500 rpm
(ii) The minimum speed for rf = 100.
c) Find terminal voltage Vt, efficiency  and mechanical torque applied to
the shaft when Ia = 60A at 1500 rpm.
18
Solution:
a)
b) (i)
b) (ii)
4. Analysis of DC Motors
DC motors are adjustable speed motors. A wide range of torquespeed characteristics (Te-m) is obtainable depending on the motor
types given below:
•
•
•
•
Series DC motor
Separately-excited DC motor
Shunt DC motor
Compound DC motor
19
DC Motors Overview
(a) Series DC Motors
(b) Separately-excited DC Motors
20
(c) Shunt DC Motors
(d) Compound DC Motors
21
DC Motors
(a) Series DC Motors
The back e.m.f:
E a  K g f  m
Electromagnetic torque:
Te  K g f I a
Terminal voltage equation:
Vt  E a  I a ( ra  rs )
Ia  I s
Assuming linear equation:
 f  K f Is
Te  K g f I a
  f  K f Is
Te  K g K f I s I a
 Ia  I s
Te  K d I a2
K g f  K d I a
m 
Ea
V  I a ra  rs 
 t
K g f
K g f
 E a  K g f  m , Vt  E a  I a ( ra  rs )
m 
Vt  I a ra  rs 
K d Ia
 K g f  K d I a
E a  K d I a  m  Vt  I a ra  rs 
Ia 
Te 
 E a  K g f  m
Vt
K d  m  ra  rs 
K d Vt 2
K d  m  ra  rs 
2
 Te  K d I a2
22
Te 
K d Vt 2
K d  m  ra  rs 2
thus
Te 
1
 m2
Note that: A series DC motor should never run no load!
Te  0

m  
overspeeding!
(b) Separately-excited DC Motors
The back e.m.f:
E a  K g f  m
Electromagnetic torque:
Te  K g f I a
Terminal voltage equation:
Vt  E a  I a ra
Assuming linear equation:
f  K f If
23
Vt  E a  I a ra
Vt  K g  f  m 
Te ra
K g f
Vt
Te ra
 m 
K g f
K g f
2
ra
Vt

K g f
K g f
2

m 

 E a  K g f  m ,
Te
Te  K g f I a
 m   0  K l Te
No load (i.e. Te = 0) speed:  0 
Slope:
Kl 
ra
K g f 2
Vt
K g f
very small!
Slightly dropping m with load
(c) Shunt DC Motors
The back e.m.f:
E a  K g f  m
Electromagnetic torque:
Te  K g f I a
Terminal voltage equation:
Vt  E a  I a ra
Assuming linear equation:
f  K f If
24
Vt  E a  I a ra
Vt  K g  f  m 
Te ra
K g f
Vt
Te ra
 m 
K g f
K g f
2
ra
Vt

K g f
K g f
2

m 

 E a  K g f  m ,
Te
Te  K g f I a
 m   0  K l Te
Same as separately excited motor
No load (i.e. Te = 0) speed:  0 
Slope:
Kl 
Vt
K g f
ra
K g f 2
very small!
Slightly dropping m with load
 m   0  K l Te
Note that: In the shunt DC motors, if suddenly the field terminals are
disconnected from the power supply, while the motor was running,
overspeeding problem will occur
E a  K g f  m
so
 f  0  m  
Ea is momentarily constant, but f will decrease rapidly.
overspeeding!
25
Motor Speed Control Methods
(a) Controlling separately-excited DC motors
Shaft speed can be controlled by
i.
Changing the terminal voltage
ii. Changing the field current (magnetic flux)
i.
Changing the terminal voltage
 m   0  K l Te
Te  K g f I a
0 
Vt
K g f
Vt    0  ,
Vt  E a  I a ra
Te 
26
ii.
Changing the field current
 m   0  K l Te
Te  K g f I a
f  K f I f
0 
Vt
K g f
(linear magnetic circuit)
I f   f  ,  0 ,
Te 
(b) Controlling series DC motors
Shaft speed can be controlled by
i.
Adding a series resistance
ii. Adding a parallel field diverter resistance
iii. Using a potential divider at the input (i.e. changes the
effective terminal voltage)
27
i.
Adding a series resistance
For the same Te produced
Ea drops, Ia stays the same
E a  Vt  I a ( ra  rs  rt )
For the same Te,
f is constant
 but m drops since Ea = Kg f m..
New value of the motor speed, m is
given by
m 
Ea
K g f
rt  
ii.
 Te  K g f I a
 E a  K g f  m
Ea , m 
Adding a parallel field diverter resistance
When we add the diverter resistance
Is drops i.e. Is < Ia.
Ea remains constant,
For the same Te produced, Ia increases
Ia 
Vt  E a
ra  ( rs || rd )
 rs || rd  rs
When series field flux drops, the motor
speed m = Ea / Kg f should rise, while
driving the same load.
m 
rd  
Ea
K g f
I s ,
 E a  K g f  m
  f  K f Is
I a ,
f  ,
m 
28
iii.
Using a potential divider
Let us apply Thévenin theorem to the
right of Vt
VTh 
R2
Vt
R1  R2
RTh  R1 || R2
This system like the speed control by
adding series resistance as explained in
section (i) where rt  RTh and Vt  VTh.
For the same Te produced
Ea drops rapidly, Ia stays the same
E a  VTh  I a ( ra  rs  RTh )
For the same Te,
f is constant
 but m drops rapidly since Ea = Kg f m..
New value of the motor speed, m is
given by
Ea
 Te  K g f I a
m 
K g f
 E K  
a
Vt  
g
f
m
E a  ,  m  
If the load increases, Te and Ia increases and Ea decreases, thus motor speed m drops down more.
Ex1:
A separately excited DC motor drives the load at nr = 1150 rpm.
a)
Find the gross output power (electromechanical power output)
produced by the DC motor.
b) If the speed control is to be achieved by armature voltage control and
the new operating condition is given by:
nr = 1000 rpm and Te = 30 Nm
find the new terminal voltage Vt while f is kept constant.
29
Soln:
a)
Gross output power is given by
Pconv  Te m  Ea I a
For Ia = 36 A
Ea  VT  I a ra  125  36  0.37  112 V
Thus
Pconv  Te m  Ea I a  112  36  4.03 kW
Note that induced torque is found to be Te = 33.4 Nm
b)
New gross output power is found to be
Pconv2  Te 2 m 2  Ea 2 I a 2  30  2  1000 / 60  3.14 kW
As Ea = Kg f m and f is constant
Ea 2 
nr 2
1000
E a1 
112  97.1 V
nr 1
1150

Ea1  m1 nr 1


Ea 2  m 2 nr 2
Then, new armature current is found to be
Ia 2 
Pconv 2 3.14k

 32.3 A
Ea 2
97.1
Consequently, the new terminal voltage is given by
Vt  Ea 2  I a 2 ra  97.1  32.3  0.37  109 V
30
Ex2: A 50 hp, 250 V, 1200 rpm DC shunt motor with compensating windings has
an armature resistance (including the brushes, compensating windings, and
interpoles) of 0.06 . Its field circuit has a total resistance Radj + RF of 50 ,
which produces a no-load speed of 1200 rpm. The shunt field winding has
1200 turns per pole.
a) Find the motor speed when its input current is 100 A.
b) Find the motor speed when its input current is 200 A.
c) Find the motor speed when its input current is 300 A.
d) Plot the motor torque-speed characteristic.
Solution:
At no load, the armature current is zero and therefore Ea = VT = 250 V.
a) Since the input current is 100 A, the armature current is
I A  IL  IF  IL 
VT
250
 100 
 95 A
RF
50
Therefore E A  VT  I A RA  250  95  0.06  244.3V
and the resulting motor speed is:
n2 
EA2
244.3
n1 
1200  1173rpm
E A1
250
31
b) Similar computations for the input current of 200 A lead to n2 = 1144 rpm.
c) Similar computations for the input current of 300 A lead to n2 = 1115 rpm.
d) At no load, the torque is zero.
For the input current of 100 A:  ind
For the input current of 200 A:  ind
For the input current of 300 A:  ind
The induced torque is  ind 
2443  95

 190N - m
2 1173 / 60
2383 195

 388N - m
2 1144 / 60
2323  295

 587 N - m
2 1115 / 60
EA I A

The torque-speed
characteristic of the motor is:
Ex3: A 100 hp, 250 V, 1200 rpm DC shunt motor with an armature resistance of
0.03  and a field resistance of 41.67 . The motor has compensating
windings, so armature reactance can be ignored. Mechanical and core losses
may be ignored also. The motor is driving a load with a line current of 126 A
and an initial speed of 1103 rpm. Assuming that the armature current is constant
and the magnetization curve is
a) What is the motor speed if the field
resistance is increased to 50 ?
b) Calculate the motor speed as a function
of the field resistance, assuming a
constant-current load.
c) Assuming that the motor next is
connected as a separately excited and
is initially running with VA = 250 V,
IA = 120 A and at n = 1103 rpm while
supplying a constant-torque load,
estimate the motor speed if VA is
reduced to 200 V.
32
Soln:
shunt
a) For the given initial line current of 126 A, the initial armature current will be
I A1  I L1  I F 1  126 
250
 120 A
41.67
Therefore, the initial generated voltage for the shunt motor will be
E A1  VT  I A1 RA  250  120  0.03  246.4V
After the field resistance is increased to 50 Ω, the new field current will be
IF 2 
250
 5 A
50
The ratio of the two internal generated voltages is
E A 2 K22 2 n2


E A1 K11 1n1
Since the armature current is assumed constant, EA1 = EA2 and, therefore
n2 
1n1
2
The values of EA on the magnetization curve are directly proportional to the flux.
Therefore, the ratio of internal generated voltages equals to the ratio of the fluxes within
the machine. From the magnetization curve, at IF = 5A, EA1 = 250V, and at IF = 6A,
EA1 = 268V. Thus:
n2 
1n1 E A1n1 268
1103  1187rpm


250
2
EA2
b) A speed vs. RF characteristic is shown on the
right
33
separately excited
c) For a separately excited motor, the initial generated voltage is
E A1  VT 1  I A1 RA
Since
E A 2 K22 2 n2


E A1 K11 1n1
and since the flux  is constant
E n
n2  A 2 1
E A1
Since the both the torque and the flux are constants, the armature current IA is
also constant. Then
n2 
VT 2  I A2 RA
200  120  0.03
1103  879rpm
n1 
250  120  0.03
VT 1  I A1 RA
Ex4: A 250-V series dc motor with compensating windings. and a total series
resistance ra + rs of 0.08 . The series field consists of 25 turns per pole.
with the magnetization curve (at 1200 rpm) shown below.
a) Find the speed and induced torque of
this motor for when its armature current
is 50 A.
b) Calculate the efficiency of the motor
34
Soln:
a)
To analyze the behavior of a series motor with saturation. pick points along the
operating curve and find the torque and speed for each point. Notice that the
magnetization curve is given in units of magnetomotive force (ampere-turns) versus
Ea for a speed of 1200 r/min. so calculated Ea values must be compared to the
equivalent values at 1200 r /min to deternine the actual motor speed.
For Ia = 50 A
Ea  VT  I a ra  rs   250  50  0.08  246 V
Since Ia = Is = 50 A. the magnetomotive force is
Fs  NI s  25  50  1250 A - turns
From the magnetization curve at F = 1250 A-turns. Ea0 = 80 V. To get the
correct speed of the motor.
nr 
Ea
246
nr 0 
1200  3690 rpm
Ea 0
80
To find the induced torque supplied by the motor at that speed, recall that
Pconv = Ea Ia = Tem. Therefore,
Te 
b)
Ea I a
m

246  50
 31.8 Nm
2  3690 / 60
Efficiency is given by the ratio of the output and input power values. Thus,
% 
246  50
Pout
T
E I
 100  e m  100  a a  100 
 100  98.4%
250  50
Pin
Vt I t
Vt I t
35