Post on 31-Dec-2015
Chapter 6DC Machines
EET103/4
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:
a) DC motorb) DC Generator
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
1.When a conductor moves in a magnetic field, voltage is induced in the conductor.
2. When a current carrying conductor is placed in magnetic field, the conductor experiences a mechanical forces.
DC Machines Fundamentals
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 circuit of DC machines 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
• In a dc motor, the stator poles are supplied by dc excitation current, which produces a dc magnetic field.
• The rotor is supplied by dc current through the brushes, commutator and coils.
• The interaction of the magnetic field and rotor current generates a force that drives the motor.
Basic Operation of DC Motor
• The magnetic field lines enter into the rotor from the north pole (N) and exit toward the south pole (S)
• The poles generate a magnetic field that is perpendicular to the current carrying conductors
• The interaction between the field and the current produces a Lorentz force
• The force is perpendicular to both the magnetic field and conductor
Basic Operation of DC Motor
Basic Operation of DC Motor
• The generated force turns the rotor until the coil reaches the neutral point between the poles.
• At this point, the magnetic field becomes practically zero together with the force.
• However, inertia drives the motor beyond the neutral zone where the direction of the magnetic field reverses.
• To avoid the reversal of the force direction, the commutator changes the current direction, which maintains the counter clockwise rotation.
• Before reaching the neutral zone, the current enters in segment 1 and exits from segment 2
• Therefore, current enters the coil end at slot ‘a’ and exits from slot ‘b’ during this stage
• After passing the neutral zone, the current enters segment 2 and exits from segment 1,
• This reverses the current direction through the rotor coil, when the coil passes the neutral zone
• The result of this current reversal is the maintenance of the rotation
Basic Operation of 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.
Important terms
• 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
Generated or back e.m.f. of DC Motor
• General form of back 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)
EA = back e.m.f.
A
PZNEA
60
Torque Equation of a DC Motor
• The armature torque of a DC motor is given by
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
IA = armature current
Ta = armature torque
)(2
meterNewtonA
PZIT A
a
Equivalent Circuit of DC Motor
F
TF R
VI
AAAT RIEV
F
FF R
VI
AAAT RIEV
AL II
Separately Excited DC Motor
Shunt DC Motor
FAL III
)( SAAAT RRIEV
LSA III
Series DC Motor
)( SAAAT RRIEV
F
TF R
VI FLA III
Compound DC Motor
Speed of a DC Motor
• For shunt motor
• For series motor
1
2
1
212
2
1
1
2
1
2 ,A
A
A
A
E
E
n
nthenIf
E
E
n
n
2
1
1
2
2
1
1
2
1
2
A
A
A
A
A
A
I
I
E
E
E
E
n
n
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 quantities:• 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 r/min 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 r/min. 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
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 is constant ant he 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 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 50 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 250 V.
SolutionGiven quantities• Supply voltage, VT = 250 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..)
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) = 250 – 20(0.2 + 0.3) = 240 V
Solution (cont..) When the armature current increased, Ia2 = 50 A, the back emf
EA2 = V – Ia2 (RA + RS) = 250 – 50(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
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
Generation of an AC Voltage
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.
Lap and wave Windings
There are two types of windings mostly employed:• Lap winding• Wave winding
The difference between the two is merely due to the different arrangement of the end connection at the front or commutator end of armature.
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
Compound 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