D C MOTOR LECTURE electrical engineering.pptx

DC Machines
• If the conversion is from mechanical to
electrical energy, the machine is said to act as a
generator.
•If the conversion is from electrical to
mechanical energy, the machine is said to act
as a motor.
•When a conductor moves in a magnetic
field, voltage is induced in the conductor.
(Generator action)
•When a current carrying conductor is placed in
a magnetic field, the conductor experiences a
mechanical force. (Motor action)

RULES
 Fleming’s Left Hand Rule:“Hold out your left hand with
forefinger, second finger and thumb at right angles to
one another. If the forefinger represents the direction
of the field, and the second finger that of the current, then
thumb gives the direction of the motion or force.”
FORE FINGER = MAGNETIC FIELD
900
900
900
MIDDLE FINGER= CURRENT
T
H
U
M
B
=
M
O
T
I
O
N
FORCE = B IAl

Right-Hand Screw Rule:Turn the current
vector i toward the flux vector B. If a
screw is turned in the same way, the
direction in which the screw will move
represents the direction of the force f.

The simplest DC machine
segments
brushes

Parts
(i) Magnetic Frame orYoke
(ii) Pole-cores and Pole-shoes
(iii) Pole Coils or Field Coils
(iv) Armature Core
(v) Armature Windings
(vi) Commutator
(vii) Brushes and Bearings

ArmatureWindings:-
Two types of windings mostly employed for
the armatures of DC machines are known as
Lap Winding and Wave Winding.
 In a lap winding, the number of parallel paths
(a) is always equal to the number of poles (b)
and also to the number of brushes.
In wave windings, the number of parallel
paths (a) is always two and there may be two
or more brush positions.

Armature Winding in a DC Machine

Lap Winding of a DC Machine
• Used in high current
low voltage circuits
•Number of parallel
paths
equals number of
brushes
or poles

Wave Winding of a DC Machine
• Used in high voltage
low current circuits
•Number of parallel
paths
always equals 2

Pole-pitch:-
The periphery of the armature divided by the
number of poles of the generate.
i.e. the distance between two adjacent poles
It is equal to the number of armature
conductors (or armature slots) per pol. If there
are 400 conductors and 4 poles, then pole
pitch is 400/4= 100 conductor.
Coil span or Coil pitch:-
It is the distance, measured in terms of
armature slots (or armature conductors),
between two sides of a coil.
 It is, in fact, the periphery of the armature
spanned by the two sides of the coil

Back Pitch:-
 The distance, measured in terms of the armature
conductors. which a coil advances on the back of the
armature is called back pitch and is denoted byYb.
 It is equal'to the number difference of the conductors
connected to a given segment of the commutator.
Front Pitch:-
 The number of armature conductors or elements
spanned by a coil on the front (or commutator end of
an armature) is called the front pitch and is
designated by Yf
 Alternatively, the front pitch may be defined as the
distance (in terms of armature conductors) between
the second conductor of one coil and the first
conductor of the next coil which are connected
together to commutator end of the armature.

Resultant Pitch:-
 It is the distance between the beginning of one coil
and the beginning of the next coil to which it is
connected as shown in Fig.
 As a matter of precaution, it should be kept in mind
that all these pitches, though normally stated in
terms of armature conductors, are also sometimes
given in terms of armature slots or commutator bars.
Commutator Pitch (Yc):-
 It is the distance (measured in commutator bars or
segments) between the segments to which the
two ends of a coil are connected as shown in Fig.
From these figures it is clear that for lap winding,Yc
is the difference of Yb and Yf whereas for wave
winding it is the sum of Yb and Yf

D C MOTOR LECTURE electrical engineering.pptx

EMF Equation
 =flux/pole in weber.
φ
Z=total number of armature conductors =
No. of slots*No. of conductors/slot.
P=No. of poles.
A= No. of parallel paths in armature.
N=armature rotation in rpm
E=EMF induced in any parallel path in
armature.
Generated EMF=e.m.f. generated in one of
the parallel paths.

 Average EMF generated/conductor= d /dt volt
φ
Now, flux cut/conductor in one revolution,
d = * P web
φ φ
Number of revolution per second =N/60;
Then, time for one revolution, dt=60/N second
 Hence according to Faraday's laws of electromagnetic
induction,
 EMF generated/conductor=d /dt= PN/60 Volt
φ φ
 For a wave-wound generator:-
No. of parallel paths is 2
No. of conductors (in series) in one path=Z/2
EMF generated/conductor=d /dt= PN/60 *Z/2 Volt
φ φ
 For a lap-wound generator :-
No. of parallel paths =P
No. of conductors (in series) in one path=Z/P
EMF generated/conductor=d /dt= ZN/60Volt
φ φ

In general, generated EMF is
EMF generated/conductor = PN/60* Z/A
φ
Volt
Where,
A=2 for wave-winding.
A= P for lap-winding.

Magnetization Curve
m
a
a K
E 


•Flux is a non-linear
function of field current
and
hence Ea is a non-linear
function of field current
•For a given value of flux
Ea
is directly proportional to
m

Different connections of DC
machines:-
(a) Separately excited DC machine.
(b) Series DC machine.
(c) Shunt DC machine.
(d) Compound DC machine.

Separately excited DC machine
The armature and field winding are electrically separate from
each other.
The field winding is excited by a separate DC source.
Total input power = Vf If +VT Ia

Field Coil
Armature
RA
Vf
+
-
Separately Excited DC Machine

Self excited machines
 Shunt machine:-
 The armature and field winding are connected in
parallel.
 The armature voltage and field voltage are the same..
 Total current drawn from the supply, IL = If + Ia
 Total input power = VT * IL

Shunt Field Coil Armature
RA
Shunt Excited DC Machine

 Series DC machine:-
 The field winding and armature winding are connected in series.
 The field winding carries the same current as the armature
winding.
 A series wound motor is also called a universal motor. It is
universal in the sense that it will run equally well using either an
ac or a dc voltage source.
 Reversing the polarity of both the stator and the rotor cancel out.
Thus the motor will always rotate the same direction irregardless
of the voltage polarity.

Series Field Coil
Armature
RA
Series Excited DC Machine

 Compound DC machine:-
 If both series and shunt field windings are used, the
motor is said to be compounded. In a compound
machine,
 The series field winding is connected in series with
the armature, and the shunt field winding is connected
in parallel.Two types of arrangements are possible in
compound motors:-
 Cumulative compounding - If the magnetic fluxes
produced by both series and shunt field windings are
in the same direction (i.e., additive), the machine is
called cumulative compound.
 Differential compounding - If the two fluxes are in
opposition, the machine is differential compound.
 In both these types, the connection can be either short
shunt or long shunt.

Shunt Field Coil Armature
RA
Compound Excited DC Machine
Series Field Coil
he shunt and series field aid each other it is called a cumulatively
ited machine
he shunt and series field oppose each other it is called a differentia
ited machine

Torque equation of dc motor
 E is the supply voltage,
 Eb is the back emf produced and
 Ia, Ra are the armature current and armature
resistance

Then the voltage equation is given by.
E = Eb + Ia.Ra -------------(2)
But keeping in mind that our purpose is to derive the
torque equation of dc motor we multiply both sides of
equation (2) by Ia.
E I
∴ a = Eb Ia + Ia
2
.Ra-------------(3)
Now Ia
2
.Ra is the power loss due to heating of the
armature coil, and the true effective mechanical power
that is required to produce the desired torque of dc
machine is given by
Pm = Eb.Ia -------------(4)
The mechanical power Pm is related to the

Pm = Tg. -------------(5)
ω
Where is speed in rad/sec. Now equating equation (4) & (5)
ω
we get
Eb.Ia = Tg.ω
Now for simplifying the torque equation of dc motor we
substitute
Eb = P ZN / 60*A -------------(6)
φ
Where, P is no of poles
is flux per pole
φ
Z is No. of conductors
A is No. of parallel paths
and N is the speed of the D.C. Motor.
Hence, = 2π N/60 -------------(7)
ω
Substituting equation (6) and (7) in equation (4), we get:
Tg = P.Z. .I
φ a / 2πA

The torque we so obtain, is known as the electromagnetic torque
of dc motor, and subtracting the mechanical and rotational losses
from it we get the mechanical torque.
T
∴ m = Tg - mechanical losses.
This is the torque equation of dc motor. It can be further
simplified as:
Tg = ka I
φ a
Where, ka = P.Z/2πA
Which is constant for a particular machine and therefore the
torque of dc motor varies with only flux and armature current I
φ a
The Torque equation of a dc motor can also be explained
considering the figure below.

Here we can see Area per pole Ar = 2π.r.L/P
B = /A
φ r
B = P. /2πrL
φ
Current / conductor Ic = Ia / A
force per conductor = f
∴ c = B L Ia/A
Now torque Tc = fc.r = BLIa.r/A
∴ Tc = P.I
φ a/2πA
Hence the total torque developed of a dc machine is
Tg = P.Z. .I
φ a /2π.A
This torque equation of dc motor can be further
simplified as:
Tg = ka I
φ a
Where, ka = P.Z/2π.A
Which is constant for a particular machine and therefore the
torque of dc motor varies with only flux and armature
φ
current Ia

Significance of back e.m.f
 The back e.m.f. is always less than supply voltage (Eb < V). But is very
small hence under normal running conditions, the difference between
back e.m.f. and supply voltage is very small.

Working at no-load and on-load
In the case of a dc series motor, at the
time of starting there'll be no back emf.
So the full current will be applied and the
motor will run in dangerous speed.This
will damage the motor. So it is always
started in loaded condition.

Losses
Copper Losses or Electrical Losses
Core Losses or Iron Losses
Brush Losses
Mechanical Losses
Stray-Load Losses

Copper Losses or Electrical Losses
 These losses are the winding losses because these occurs in the winding of the
machine.The copper or electrical are present because of the resistance of the
winding. Currents flowing through these windings produce ohmic losses (i.e. I2
R
losses).The windings that may be present addition to Armature winding are the
field windings, interpole and compensating windings.
Armature current losses = Ia
2
Ra, where Ia is Armature current and Ra is Armature
Resistance.These losses are about 30% of total full-load losses.
 Copper losses in the shunt field of a shunt machine =Ish
2
Rsh where Ish is the current
in the shunt field and Rsh is the resistance of the shunt field winding.The shunt
regulating resistance is included in Rsh.
 Copper loss in the series field of a series machine = Ise
2
Rse where Ise is the current
through the series field winding and Rse is the resistance of the series field
winding.
 In a compound machine, both shunt and series field losses occur.There losses are
about 20% of full load losses.
 Copper loss in interpole windings =Ia
2
Ri where Ri is resistance of interpole
winding.

Core Losses or Iron Losses
These losses also called magnetic losses.
These are of two types viz. Hysteresis and
Eddy-current losses. Since DC machines
are usually operated at constant speed
and constant flux density, these losses are
almost constant.These are about 20% of
full-load losses.

Brush Losses
 There is a power loss at the brush contacts between
the copper commutator and the carbon brushes. In
practice, thin loss depends upon the brush contact
voltage drop and the Armature current Ia. It is given
by
PBD=VBDIa
The voltage drop across a set of brushes is
approximately constant over a large range of
Armature currents. Unless stated otherwise, the
brush voltage drop is usually assumed to be about
2V.The brush droploss is, therefore, taken as 2Ia

Mechanical Losses
The losses associated with mechanical
effects are called mechanical losses.
They consist of bearing friction loss and
windage loss.Windage losses are those
associated with overcoming are friction
between the moving parts of the machine
and the air inside the machine for
cooling purposes.These losses are
usually very small.

Stray-Load Losses
 Stray-load loss consists of all losses not covering above.
These are the miscellaneous losses that result from
such factors as (i) the distortion of flux because of
Armature reaction, (ii) short circuit currents in the coil,
undergoing commutation etc.These losses are very
difficult to determine.The indeterminate nature of the
stray-load loss makes it necessary to assign reasonable
value. For most machines stray losses are taken by
convention to be 1% of full load output power.
So friends this was all about various losses occurs in a
DC Machine and here I think every kind of loss was
covered. If any loss is still left, feel free to add a
comment.

Armature Reaction(AR)
• AR is the magnetic field produced by the
armature current
•AR aids the main flux in one half of the
pole and opposes the main flux in the
other half of the pole
•However due to saturation of the pole
faces the net effect of AR is demagnetizing

Effects of Armature Reaction
• The magnetic axis of the AR
is 900
electrical (cross) out-
of-phase with the main flux.
This causes commutation
problems as zero of the flux
axis is changed from the
interpolar position.

Minimizing Armature Reaction
•Since AR reduces main flux, voltage in
generators and torque in motors
reduces with it. This is particularly
objectionable in steel rolling mills that
require sudden torque increase.
•Compensating windings put on pole
faces can effectively negate the effect
of AR.These windings are connected
in series with armature winding.

D C MOTOR LECTURE electrical engineering.pptx

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