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2.1 Remember the construction and

principle operation of a DC motor.
2.2 Apply the principle operation of
DC motor.

LEARNING OUTCOME
2.2 Apply the principle operation of DC motor.
1.2.1 Identify the formula self-excitation DC motor for:
a. back e.m.f. d. losses
b. armature and shaft torque e. efficiency
c. output power
2.2.2 Solve the problems using formula in 2.2.1.
2.2.3 Apply briefly the relationship of motor speed when change the value of:
a. flux (field resistance) b. armature resistance
2.2.4 Describe the characteristic curves base on torque via armature current (Ta/Ia) and speed
via armature current (Na/Ia) for:
a. series motor c. compound motor
b. shunt motor
2.2.5 Draw the various power stages of DC motor.
2.2.6 Show the applications for shunt, series motors and compound motor.

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Significance of Back E.M.F

The presence of back e.m.f. makes the d.c. motor a self-regulating machine i.e., it
makes the motor to draw as much armature current as is just sufficient to develop the
torque required by the load.

Back e.m.f. in a dc motor regulates the flow of armature current i.e., it automatically
changes the armature current to meet the load requirement.

During rotation of the armature, its conductors cut the magnetic flux and emf is
induced in the conductors according to the laws of electromagnetic induction.

The direction of this induced emf can be found by Fleming’s left hand rule and have
direction opposite to the applied voltage.

Induced Counter-voltage (Back EMF)

When the armature of a d.c. motor rotates under the influence of the driving
torque, the armature conductors move through the magnetic field and hence an
e.m.f. is induced in them.

The induced e.m.f. acts in opposite direction to the applied voltage V(Lenz’s law)
and is known as back or counter e.m.f. = Eb .
N = speed (rpm)
𝑁∅𝑍 𝑃
𝐸𝑏 = Volts ∅ = flux per pole (wb)
60 𝐴 Z = no of conductors
P = no of pole pairs
A = Number of parallel
paths
Eb = back emf

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DC
MACHINES

GENERATOR MOTOR

Seperately Self Seperately Self

Excited Excited Excited Excited

Shunt Series Compound Shunt Series Compound

Cumulative Differential Cumulative Differential

Types of Separately-Excited DC Motors

There is no direct connection between the armature and field winding resistance
DC field current is supplied by an independent source
a field circuit is supplied from a separate constant voltage power source.
(such as battery or another generator or prime mover called an exciter)

VF  I F RF
VT  Eb  I A R A  Vbrh

Eb IL  IA

NZØ p
Eb  x (volt )
60 a

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Types of Self-Excited DC Motors

DC SERIES
MOTOR

SELF-
EXCITED DC SHUNT
DC MOTOR MOTOR

DC
COMPOUND
MOTOR

Types of Self-Excited DC Motors

DC Series Motors
The field winding is connected in series with the armature

The resistance of the series field winding Rs is

much smaller than the armature resistance Ra

The resistance of the series field winding Rs is

much smaller than the armature resistance Ra

The flux produced is proportional to the field current but in

this If = ia thus Ø ∝ Ia

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EQUATION
Series DC Motors

Series motors connect

Eb the field windings in
series with the
armature.

VT  Eb  I A R A  RS   Vbrh

I A  IS  IL

Types of Self-Excited DC Motors

DC Shunt Motors
The parallel combination of two windings is connected across a common dc power
supply

The resistance of shunt field winding (Rsh) is always higher than that is armature
winding.
This is because the number of turns for the field winding
is more than that of armature winding

The cross-sectional area of the wire used for field

winding is smaller than that of the wire used for
armature winding

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EQUATION
Shunt DC Motors
IA
ia Ra
RA IL
iL
a field circuit gets its power
from the armature terminals IF if

of the motor.
M Eb RF
Rf VT (dc
supply)

VF  VT  I F RF

VT  Eb  I A R A  Vbrh

IL  I A  IF

EQUATION-summry
SHUNT MOTOR ;
I = Ia + Ish
Vt = Eb + IaRa + Vbrh
Eb = Vt- IaRa -Vbrh
SERIES MOTOR;
I = Ia = Ise
Vt = Eb + Ia{Ra + Rse} + Vbrh
Eb = Vt – Ia(Ra+Rse) -Vbrh

-The back emf is always less than supply voltage (Eb < Vt).
-Ra is very small
-The difference between back emf and supply voltage is very small

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Example 1
A series motor runs at 600rpm when taking 100A from 240 V supply. The resistance
of the armature circuit is 0.12 and that of the series winding is 0.03. Calculate the
back EMF. [225V]

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Example 2
A DC motor takes an armature current of 110A at 480V. The counter voltage of the DC motor is
458V. Calculate the resistance of the armature motor.
[225V]

Types of Self-Excited DC Motors

DC Compound Motors
Compound motor has two field windings; one connected in parallel with the
armature and the other in series with it.
There are two types of compound motor connections
LONG SHUNT CONNECTION
In this the series winding is connected in series with the
armature winding and the shunt winding is connected in Eb

parallel with the armature connection.

SHORT SHUNT CONNECTION
In short shunt compound motor the series winding is connected in
series to the parallel combination of armature and the shunt
winding Eb
This is done to get good starting torque and constant speed
characteristics.

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EQUATION

Equivalent circuit –
Long Shunt Compound DC Motors

R I VF  VT  I F RF
IAi RRA
a a
R
S
f2
IS iL L

VT  Eb  I A ( RA  RS )  Vb
IFi f

M Eb RR F
f1 VT (dc IL  I A  IF
supply)

I A  IS

EQUATION

Equivalent circuit –
Short Shunt Compound DC Motors
IL  IS
IAi RRA R S IS I
a a
Rf2 iL L IL  I A  IF
IF
VT  I L RS
Eb VT (dc
IF 
M
RF supply) RF

VT  Eb  I A R A  I L RS  Vbrh

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TORQUE, OUTPUT POWER AND ROTATION SPEED

Power armature = Armature torque x ω
EbIa = 𝜏a x 2πN rad /sec
60
But Eb in a motor is given by, Eb = ФNZP V
60a

ФNZP x Ia = 𝜏a x 2 π N Where:
60a 60 Po : Mechanical/output Power
N : Speed
ФPZIa 𝜏 : Torque
𝜏a = 2πa N-m

𝜏a = Po x 60
OR
2πN

Speed Control of DC Motor

Many applications require the speed of a motor to be varied over a wide range. One
of the most attractive features of DC motors in comparison with AC motors is the
ease with which their speed can be varied
back emf for a separately excited DC motor:

Eb  Km  VT  I A RA
So
VT  I A R A 2πNm
Speed :  m  
K 60

From the above equation, it is evident that the speed can be varied by using any of the
following methods:
1. Field Control (By Varying Ø )
2. Armature resistance control (By varying Ra)

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DC Motors Characteristic

(i) Torque and armature current characteristics (Ta /Ia ):

It is the curve between armature torque and armature current of a dc motor

(ii) Speed and armature current characteristics (N/Ia ):

It is the curve between speed and armature current.

(iii)Speed and torque characteristics (N/Ta ):

It is the curve between speed and armature torque.

DC Motors Characteristic

DC Series Motor

Torque developed in any dc motor is

T ∝ ØIa

Up to magnetic saturation,
Ø ∝ Ia

So that,

T ∝ Ia2

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DC Motors Characteristic
DC Series Motor

𝐸𝑏 In a DC motor, Speed (N) is directly proportional to back

𝑁 ∝
ϕ emf (Eb) and inversely proportional to Flux per pole (ϕ)

If dc motor has initial values N1, ϕ1 and Eb1 respectively and final values N2, ϕ2and Eb2

𝐸𝑏1 𝑁2 𝐸𝑏2 ϕ1
𝐸𝑏2
= X
𝑁1 ∝
ϕ1
𝑁2 ∝
ϕ2 𝑁1 𝐸𝑏1 ϕ2

DC Motors Characteristic
DC Shunt Motor

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DC Motors Characteristic
DC Compound Motor

DC Motors Characteristic
Comparison of three types of Motor

1) The speed regulation in shunt motor is better than series motor.

2) The starting torque of a series motor is more than other motor.

3) Both shunt and compound motors have definite no load speed,

however series motors have dangerously high speed at no load.

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Example 3
A 4 pole dc motor takes a 50A armature current . The armature has lap
connected 480 conductors. The flux per pole is 20mWb. Calculate the
gross torque developed by the armature of the motor

Given P = 4, a=mp=4, Z=480 Ia=50A

Ф = 20 mWb = 20 x 10-3Wb

Ta = ФPZIa Nm
2πa

Efficiency and Losses

Efficiency is important for several reasons:
1. A less efficient machine will cost more to operate.
2. The losses in the machine are converted to heat, which raises the
operating temperature of the machine.
Unfortunately, not all electrical power is converted to mechanical power
by a motor and not all mechanical power is converted to electrical power
by a generator…
Pout
 x100%
The efficiency of a DC machine is: Pin

Defining the losses in the machine as Ploss:

Pout Pin  Ploss Ploss
 x100%  x100%   (1  ) x100%
Pout  Ploss Pin Pin

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Efficiency and Losses

The Power-flow diagram

On of the most convenient technique to account for power losses in a
machine is the power-flow diagram.

Electrical Mechanical
Input output Power
Power

Pout  Pin  PTL

Electrical power is input to the machine, and the electrical and brush losses
must be subtracted. The remaining power is ideally converted from
electrical to mechanical form at the point labeled as Pconv or Pdev

Efficiency and Losses

The Power-flow diagram

The electrical power that is converted is

Pdev  Eb I A
And the resulting mechanical power is

Pdev   ind m
After the power is converted to mechanical form, the stray losses,
mechanical losses, and core losses are subtracted, and the remaining
mechanical power is output to the load.

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Losses in DC Motor

There are five categories of losses occurring in DC machines.

1. Electrical or copper losses – the resistive losses in the armature and

field windings of the machine.

Armature loss: PCU ( A)  I A2 RA

Field loss: PCU ( F )  I F2 RF

Where IA and IF are armature and field currents and RA and RF are
armature and field (winding) resistances usually measured at normal
operating temperature.

Efficiency and Losses

2. Brush (drop) losses – the power lost across the contact potential
at the brushes of the machine.
PBD  VBD I A

Where IA is the armature current and VBD is the brush voltage drop.
The voltage drop across the set of brushes is approximately constant
over a large range of armature currents and it is usually assumed to be
about 2 V.

Other losses are exactly the same as in AC

machines…

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Efficiency and Losses

3. Core losses – hysteresis losses and eddy current losses. They vary
as B2 (square of flux density) and as n1.5 (speed of rotation of the
magnetic field).

4. Mechanical losses – losses associated with mechanical effects:

friction (friction of the bearings) and windage (friction between the
moving parts of the machine and the air inside the casing). These losses
vary as the cube of rotation speed n3.

5. Stray (Miscellaneous) losses – losses that cannot be classified in

any of the previous categories. They are usually due to inaccuracies in
modeling. For many machines, stray losses are assumed as 1% of full
load.

Efficiency and Losses

The electrical power that is converted is

Pconv  Eb I A
And the resulting mechanical power is

Pconv   indm
After the power is converted to mechanical form, the stray losses,
mechanical losses, and core losses are subtracted, and the remaining
mechanical power is output to the load.

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APPLICATION OF DC MOTORS
MOTORS APPLICATIONS
1. Blowers and fans
2. Centrifugal and reciprocating pumps
3. Lathe machines
DC SHUNT MOTOR 4. Machine tools
5. Miling machines
6. Drilling machines
1. Cranes
2. Hoists
3. Elevators
DC SERIES MOTOR 4. Trolleys
5. Conveyors
6. Electric locomotives
1. Rolling mills
DC COMPOUND 2. Punches
3. Shears
MOTOR 4. Heavy planers
5. Elevators

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Example 4
Calculate the torque in Nm, develop by a DC motor having an armature resistance of 0.25
and running at 750rpm when taking an armature current of 60A from a 480V supply.
[355Nm]

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Example 5
A DC series motor connected across a 460V supply runs at 500rpm when the current is 40A.
The total resistance of the armature and field circuits is 0.6. Calculate the torque of the
armature in Nm. [333Nm]

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Example 6
A series dc machine is consumed a 6.5kW when the 12.5A of armature current is passing
thru the armature and field resistance of 3.3 and 2.0 respectively. Assume stray losses of
1.2kW. Calculate
[520V, 453.75V, 12N-m, 68.8%]
a) terminal voltage, VT
b) back emf, Ea
c) net torque if the speed is at 3560rpm
d) efficiency of the machine

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Example 7
A 600V 150-hp series dc machine operates at its full rated load at 600rpm. The armature and
field resistance are 0.12 and 0.04 respectively. The machine draws 200A at full load.
Assume stray losses 1700W. Determine
a) the armature back emf at full load, Ea
b) developed/mechanical power and developed/mechanical torque [568V, 113.6kW, 1808Nm]

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Example 8
A voltage of 230V is applied to armature of a machines results in a full load armature currents
of 205A. Assume that armature resistance is 0.2. Find the back emf, net power and torque by
assuming the rotational losses are 1445W at full load speed of 1750rpm. [189V, 37.3kW, 203.5Nm]

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Example 9
A 230 V, 10 hp dc shunt motor delivers power to a load at 1200 r/min. The armature current
drawn by the motor is 200 A. The armature circuit resistance of the motor is 0.2  and the
field resistance is 115 . If the rotational losses are 500W, what is the value of the load torque.
[298.4Nm]

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Example 10
A series-connected DC motor has an armature resistance of 0.5  and field winding
resistance of 1.5 . In driving a certain load at 1200 rpm, the current drawn by the motor is
20A from a voltage source of VT = 220V. The rotational loss is 150W. Find the output power
and efficiency. [3450W, 78.41%]

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Example 11
A 6 pole, 3.0 hp 120V DC lap-wound shunt motor has 960 conductors in the armature. It takes
25.0 A from the supply at full load. Armature resistance is 0.75, flux/pole=10.0 mWb, field
winding current is 1.20A. Find the speed. [638rpm]

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Example 12
A 220V DC shunt motor draws 10A at 1800rpm. The armature resistance is 0.2 and field
winding resistance is 440. What is the torque? [11Nm]

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