CHAPTER 8 – DC MACHINERY FUNDAMENTALS
Summary:
1. A Simple Rotating Loop between Curved Pole Faces
•The Voltage Induced in a Rotating Loop
•Getting
Getting DC voltage out of the Rotating Loop
•The Induced Torque in the Rotating Loop
2. Problems with Commutation in Real Machine
•Armature Reaction
•L di/dt Voltages
•Solutions to the Problems with Commutation
3. The Internal Generated Voltage and Induced Torque
Equations of Real DC Machine
4. The Construction of DC Machine
5 Power Flow and Losses in DC Machines
5.
Dc motor Animation
Magnetic Induction and the DC Generator
Magnetic Induction and the DC Generator
• Faraday’s Law e = N dΦ / dt
– e = the induced voltage in volts (V)
– N = the number of series‐connected turns of wire in turns (t)
– dΦ/dt = rate of change in flux in Webers/second (Wb/s)
• e = B L v
– B = the flux density in teslas (T)
– L = the length of the conductor that is in the magnetic field in meters (m)
in meters (m)
– v = the relative velocity between the wire and the flux, in meters/second (m/s)
2
Magnetic induction in a wire moving in a field.
3
Right‐hand
Right
hand rule for magnetic induction.
rule for magnetic induction.
4
Wire loop rotating in a magnetic field.
Wire loop rotating in a magnetic field.
5
AC generator with slip rings and brushes.
AC generator with slip rings and brushes.
xxx
x
yyy
y
6
DC generator with commutator and brushes.
bc
ad
cb
da
DC generator output waveform.
DC generator output waveform.
9
DC generator with field control.
10
DC generator four‐pole
DC generator four
pole field.
field.
11
DC generator rotor with two coils.
DC generator rotor with two coils.
12
Coil and output waveforms for a two‐winding rotor.
13
15‐4
15
4 Motor Action and the DC Motor
Motor Action and the DC Motor
• F = B L I
– F = the resulting mechanical force in newtons (N)
– B = the flux density in teslas (T)
– L = the effective length of the wire (meters) in the field multiplied by the number of turns
the number of turns
– I = the current in the conductor in amperes (A)
• Ia(start) = (Vt – Vb) / Ra
–
–
–
–
Ia(start) = the armature starting current in amperes (A)
= the armature starting current in amperes (A)
Vt = the applied voltage in volts (V)
Vb = the brush drop in volts (V)
Ra = the armature resistance in ohms (Ω)
the armature resistance in ohms (Ω)
• Ia = (Vt – Vb – Vcemf) / Ra
– Vcemf = the induced counter emf in the armature windings in volts (V).
14
Force on a current‐carrying wire in a magnetic field.
i fi ld
15
Flux compression and resulting force.
16
Simple dc motor.
Simple dc motor.
17
DC motor with electromagnetic field.
18
Segments
a
b
Mica
Bruches
DC Machines ‐ Construction
20
Rotor with several rotor coils and commutator segments.
21
DC Machines ‐ Construction
DC Machines Badariah Bais
KKKF163 Introduction to EE Sem II 2006/07
22
c
b
a
d
DC generator with commutator and brushes.
b
a
c
d
Q v = rω
The Area Under pole faces
AP = π rl
∴ rl =
AP
π
In general
general, the voltage in any real machine will
depend on the same 3 factors:
the flux in the machine
The speed of rotation
A constant representing the
construction of the machine.
The Induced Torque in the Rotating Loop
The resulting total induced torque
in the loop is:
τind = 2 rilB
ilB
The Area Under pole faces
The Area Under pole faces
AP = π rl
∴ rl =
AP
π
the torque expression can be reduced to:
2
τ ind = φ i
π
2
τ ind = φ i
π
In general, the torque in any real machine will depend on
the same 3 factors:
1.The flux in the machine
2.The current in the machine
3.A constant representing the construction of the
machine.
Elements of an armature windings
Elements of an armature windings
The angle between centers of adjacent poles is 180o (electrical) If coil sides are placed 180o electrical apart, the coil is said to be full‐pitch
N
a
a
180oelec
S
b
N
S
b
Elements of an armature windings
g
The most common ways of connecting coils for armature windings:
The most common ways of connecting coils for armature windings:
Lap winding
Wave winding
E d f th
Ends of the coils are connected to the commutator
il
t d t th
t t bars
b
In DC machines most of the coils are full‐pitch.
Elements of an armature windings
Lap winding
Lap winding
Commutator bar
• One coil between adjacent commutator bars
• 1/p of total coils are connected in series
• No. of poles ≡
No of poles ≡ no.
no of brushes ≡
of brushes ≡ no. of parallel paths
no of parallel paths
Elements of an armature windings
g
Elements of an armature windings
Wave winding
•
•
•
•
p/2 coils in series between adjacent commutator
/2 il i
i b
dj
b
bars
½ of all coils between brushes
Regardless of no. of poles, there are always 2 parallel path
The distance between end coils (commutator
(
pitch)) is 2(C±1)/p p
(
)/p
where C is the no. of commutator bars
Elements of an armature windings
g
• The ripple can be further reduced by the use of a e pp e ca be u t e educed by t e use o a
cylindrical iron core and by shaping the pole pieces
– this produces an
approximately
uniform field in the
uniform field in the
narrow air gap
– the arrangement
of coils and core
f il
d
is known as the
armature
• A four
A four‐pole
pole DC generator
DC generator
Rotor with several rotor coils and commutator segments.
44
The Internal Generated Voltage and Induced Torque Equations of Real DC Machine
The voltage in any single conductor under the pole faces was previously shown to be
shown to be
eind per conductor = e = vBl
The voltage out of the armature of a real machine is thus:
The
voltage out of the armature of a real machine is thus E A =
ZvBl
a
Where Z is the total number of conductors and a is the number of current paths.
Q v = rω
A 2πrl
AP = =
P
P
B (2πrl ) 2πrlB ⇒ rlB = Pφ
φ = BAP =
=
2π
P
P
Pφ Zω ZP
EA =
=
φω = Kφω
2π a
2πa
⇒
ZP
K=
2πa
ZrωBl
EA =
a
N
Ia
Armature Reaction
φC
x
If
φr
φC
φm
x
φm
x
x
If
Ia
S
Armature Reaction
φC
x
If
x
φr
φC
φm
x
φm
x
x
x
x
If
N
Ia
Armature Reaction
φC
x
If
φr
φC
φm
x
φm
x
x
If
Ia
S
The Internal Generated Torque Equations of Real DC Machine
The force in any single conductor under the pole faces:
y g
p
Find per conductor = f = iBl
The torque in any single conductor under the pole faces
Tind per conductor = fr = iBlr
The torque out of the armature of a real motor is thus
The torque out of the armature of a real motor is thus:
Tind =
ZI a Blr
a
Where Z is the total number of conductors and a is the number of current paths.
A 2πrl
AP = =
P
P
B (2πrl ) 2πrlB ⇒ rlB = Pφ
φ = BAP =
=
2π
P
P
Tind
ZP
=
φI a = KφI a
2πa
⇒
ZP
K=
2πa
Tind
Pφ ZI a
=
2π a
PZ
=
φI a
2πa
The Power-Flow Diagram:
Classification Of DC Machines
DC Generator
Self Excited
(b)
((c))
Long shunt
Short shunt
Load
L
(a)
Compound Field
Load
L
Shunt Field
Loaad
Series Field
Load
d
Load
Separetly Excited
(d)
DC Mortor
Separetly Excited
Self Excited
Series Field
Shunt Field
Compound Field
Long shunt
Short shunt
(a)
(b)
( )
(c)
(d)
. The Equivalent Circuit of a DC Motor
The internal generated voltage is given by:
Pφ Zω ZP
EA =
=
φω = Kφω
2π a
2πa
the torque induced is
the torque induced is Tind
ZP
=
φI a = KφI a
2πa
The Magnetization Curve of a DC Machine
Separately Excited and Shunt DC Motors
VT = EA + IARA
The Terminal Characteristics of a Shunt DC Motor
E = K eφωm
V = E + I a Ra
V = K eφωm + I a Ra
V
Ra
ωm =
−
Ia
K eφ K eφ
T = K eφ I a
T
Ia =
K eφ
Ra
V
ωm =
−
Tind
2
K eφ (K eφ )
Nonlinear Analysis of a Shunt DC Motor
The total mmf in a shunt dc motor is the field circuit mmf less the mmf due to
armature
t
reaction
ti (AR):
(AR)
Fnet = NFIF - FAR
Th equivalent
The
i l t fi
field
ld current:
t
I F*
FAR
= IF −
NF
EA
n
=
E A0 n0
Example 9.2
A 50HP, 250V, 1200r/min DC shunt motor without compensating windings
has an armature resistance (including the brushes and interpoles) of 0.06
0 06
Ω. Its field circuit has a total resistance R-adj -+ R-F- of 50 Ω, which produces
a no-load speed of 1200r/min. There are 1200 turns per pole on the shunt field
winding, and the armature reaction produces a demagnetising magnetomotive
force of 840 A turns at a load current of 200A. The magnetization curve of this
machine is shown below:
(a) Find the speed of this motor •
when its input current is 200A.
(b) This motor is essentially
identical to the one in Example
9.1 except for the absence of
compensating windings. How
does its speed compare to that
of the previous motor at a load
current of 200A?
VT
Ia = I L − I F = I L −
RF
250
I a = 200 −
= 195A
50
Speed Control of Shunt DC Motors
•Adjusting the field resistance RF (and thus the field flux)
•Adjusting the terminal voltage applied to the armature.
L
Less
common method:
th d
•Inserting a resistor in series with the armature circuit.
Changing the Field Resistance
V −E ↓
IF↓ = VT/RF↑ ⇒
EA↓ (=Kφ↓w), ⇒ I A ↑= T R A
A
Figure above shows a shunt dc motor with an internal resistance of 0.25Ω. It is currently operating with a terminal voltage of 250V and an internal generated voltage of 245V Therefore the armature current
internal generated voltage of 245V. Therefore, the armature current flow is IA = (250V‐245V)/0.25Ω= 20A.
What happens in this motor if there is a 1% decrease in flux? If the flux decrease by 1%, then EA must decrease by 1% too, because EA = Kfw. Therefore, EA will drop to:
EA2 = 0.99 EA1 EA2 0.99 EA1 = 0.99 (245) 0.99 (245) = 242.55V
242.55V
The armature current must then rise to
IA = (250-242.55)/0.25 = 29.8 A
Thus a 1% decrease in flux produced a 49% increase in armature current
Thus,
current.
So, to get back to the original discussion, the increase in current predominates over
the decrease in flux. so, τind>τload , the motor speeds up.
However, as the motor speeds up, EA rises, causing IA to fall. Thus, induced
torque τind drops too, and finally τind equals τload at a higher steady-sate speed than
originally.
Increasing Tind makes Tind > TLoad , and the speed increases Increases speed to increases EA ↑ (=Kφw
↑), The effect of field resistance RF speed
p
control on a shunt motor’s torqueq
speed characteristics.
•over the motor’s normal operating range
•over the entire range from no load to stall conditions
Th Effect
The
Eff t off an Open
O
Field
Fi ld Circuit
Ci
it
Changing the Armature Voltage
This method involves changing the voltage applied to the armature of the motor
without
ith t changing
h
i th
the voltage
lt
applied
li d to
t the
th fi
field.
ld
If the voltage VA is increased, then the IA must rise [ IA = (VA ↑ -EA)/RA]. As IA
increases the induced torque τind =KφΙΑ↑ increases,
increases,
increases making τind > τload , and the
speed of the motor increases.
But, as the speed increases,
But
increases the EA ((=Kφω↑)
Kφω↑) increases
increases, causing the armature
current to decrease. This decrease in IA decreases the induced torque, causing τind =
τload at a higher rotational speed.
Inserting a Resistor in Series with the Armature Circuit
V
RA
ω= T −
τ ind
2
Kφ ( Kφ )
The insertion of a resistor is a very wasteful method of speed control, since the losses in the inserted i
th l
i th i
t d
resistor are very large. For this reason, it is rarely used.
Safe Ranges of Operation for the 2 common methods
Field Resistance Control
Armature Voltage Control
Example 9.3
Figure above shows a 100hp, 250 V, 1200 r/min shunt dc motor with an armature
resistance of 0.03 ohms and a field resistance of 41.67 ohms. The motor has
compensating
ti windings,
i di
so armature
t
reaction
ti can b
be iignored.
d M
Mechanical
h i l and
d core
losses may be assumed to be negligible for the purposes of this problem. The motor is
assumed to be driving a load with a line current of 126A and an initial speed of 1103
r/min. To simplify the problem, assume that the amount of armature current drawn by
the motor remains constant.
(a) If the machine’s magnetization curve is as in Example 9.2, what is the motor’s speed
if the field resistance is raised to 50 ohms?
Ea1 = VT − I a1Ra = 250 − 120 * 0.03 = 246 .4V
Example 9.4
The motor in Example 9.3 is now connected separately excited as shown below.
The motor is initially running with V-A- = 250V, I-A- = 120A, and n= 1103 r/min,
while supplying a constant-torque load. What will the speed of this motor be if
V-A- is reduced to 200V?
The Series DC Motor
A series DC motor is a dc motor whose field windings consist of
relatively
l ti l few
f
turns
t
connected
t d in
i series
i with
ith th
the armature
t
circuit.
i it
Th KVL ffor thi
The
this motor
t is
i VT
= EA + IA (RA + RS)
The induced torque is τind =KφΙΑ
τind = KφΙΑ = ΚcIA2
φ = cIA
Series DC Motor
E = K e K f I aωm
V = E + I a Ra
φ = K f Ia
T = K eφ I a
T = KeK
2
f Ia
V = K e K f I aωm + I a Ra
T
V
Ra
I
=
a
ωm =
−
K
K
e
f
Ke K f I a Ke K f
V
1
Ra
−
ωm =
Ke K f T Ke K f
Speed Control of Series DC Motors.
Unlike with the shunt dc motor, there is only one efficient way to
change the speed of a series dc motor.
motor That method is to
change the terminal voltage of the motor. If terminal voltage is
increased, the speed will increase for any given torque.
. The Compounded DC Motor
A compounded dc motor is a motor with both a shunt and a series
fi ld Thi
field.
This is
i shown
h
b
below:
l
VT = EA + IA (RA + RS)
IA = IIL ‐
IF = VT/RF
IFF = F ± F ‐ F
net
F
SE
AR
IF* = IF ± (NSE/NF) IA – FAR/NF
+ve sign associated with a cumulatively compounded motor
-ve sign
g associated with a differentially
y compounded
p
motor
The Torque-Speed Characteristic of a Cumulatively Compounded DC Motor (CC)