DC Machines and Batteries

EEE 3 Lecture 05
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Objectives
• Include DC generators and DC motors in circuit analysis
• Understand the ratings associated with DC Machines
• Find the input, output, and lost power in DC Machines
• Find and relate the speed of DC Machines with electrical
quantities in DC Machines
• Use the model of a battery in circuit analysis
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Outline
• Part 1:
• Overview of Rotating Machines
• Applications of DC Machines
• Review the concepts behind the operation of a basic dc motor
• Review the concepts behind the operation of a basic dc generator
• Peak at the inside of a practical DC Machine
• Part 2:
• Analyzing the operation of Machines
• The model for DC Machines
• Equivalent networks of different types of operations
• The electromechanical relationships
• The ratings of a DC Machine
• Part 3:
• Circuit analysis with DC Machines
• Part 4:
• Batteries and Cells
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PART 1
Overview of DC Machines
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Rotating Machines overview

Electrical Side Mechanical Side

Speed of
Voltage & Rotating Rotation &
Current
Machine Torque

Motor

Generator

5
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Rotating Electrical Machines

• Generator
• Mechanical energy to electrical energy.
• Driven by a “prime mover” – steam turbine, gas engine,
falling/running water

• Motor
• Electrical energy to mechanical energy.
• Supplied with electric power to develop torque.
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Applications of DC Machines
• Mechanical systems requiring the torque-speed characteristic
of DC Machines (may be supplied from the grid)
• Robotic arms
• Disk drives

• Portable Systems & Off-grid System

• E-vehicles
• Mobile devices

* Grid – a.k.a. the power system

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Basic DC Machine Construction

Parts of a Machine
1. Stator
2. Rotor

Machine Windings
1. Armature – where
voltage is induced
2. Field – provides the
magnetic field in the
machine
3. Interpole
4. Compensating
*DC Machine functions as a MOTOR in this video
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DC Machine
q-axis

A 2-pole DC armature
Machine
a

C m d-axis
SF

compensating I shunt field

series field

interpole
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PART 2
Analysis of DC Machines
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Equivalent Steady-State Models of Machine

Parts I a
Ra (motor)

A1 F1

Ia IF A1
+ A2
ωm Te (generator)

(Motor) Eg RF
F1 F2
-
DC Machine
ωm A2 F1
Visual
(generator)
Tm Armature Field
Winding Winding

Note that:

The armature consumes electric power in motors.

The positive direction of the armature current is the one going into the positive terminal of the internal generated voltage.

The armature supplies electric power in generators.

The positive direction of the armature current is the one going into the positive terminal of the internal generated voltage.
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Types of DC Machines
By connecting the machine in different ways, we can arrive at the
different types, such as:
• Separately-excited Shunt
• Armature and field windings have different sources of excitation.
• Self-excited Shunt
• The field winding is connected in PARALLEL with the armature.
• Series field
• The field winding is connected in SERIES with the armature.
• Compound
• Long-shunt
• Short-shunt

NOTE: We will deal only with the shunt DC machines.

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Separately-excited generator
Ra A1 F1

+
Ia IL IF
+ (generator) (generator)
Te RF +
Eg VT Rload VF
-
-
A2 −
F2
ωm Armature Field Winding
Tm (generator) Winding

Eg  mL aFIF Rload

Eg = RaIa + Vt
A1 A2
VF  R F IF
Te  L aFIFIa F1 F2
Tm  Te  Dm DC Machine Visual
www.aliexpress.com
VF
 15

Separately-excited motor
Ra A1 F1

Ia + IL
IF
+ (motor) (motor)
Te + RF +
Eg VT VSource VF
− -
-
A2 −
F2
ωm Armature Field Winding
Tm (generator) Winding

Eg  mL aFIF
VSource
Vt = RaIa +Eg
VF  R F IF A1 A2

Te  L aFIFIa
F1 F2 VF
Tm  Te  Dm DC Machine Visual
www.aliexpress.com
 16

Self-excited generator
Ra A1, F1
+
Ia IF IL
+ (generator)
Te (generator)
VT RLoad
Eg RF
-
A2, F2 −
ωm Armature Field
Tm (generator) Winding Winding

Eg  mL aFIF
Ia  IF  IL RLoad
Eg = RaIa + Vt
Vt  R FIF A1 A2

Te  L aFIFIa
DC Machine Visual
Tm  Te  Dm F1 F2
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Self-excited motor
Ra A1, F1

Ia + IL
IF
+ (motor) (motor)
ωm Te +
(Motor) Eg RF VT - Vsource

-
A2, F2 −
Armature Field
Tm Winding Winding

Eg  mL aFIF IL  IF  Ia Vsource

VT = RaIa +Eg
www.aliexpress.com

VT = RF I F A1 A2

Te  L aFIFIa DC Machine Visual

Tm  Te  Dm F1 F2
 18

Power flows - Motor

Converted Power = EgIa

Output
Input Power
Power

Stray
Mechanical Core Losses
Electrical Losses Losses
Losses

Rotational
Losses
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Power flows - Generator

Converted Power = EgIa

Output
Input Power
Power

Electrical
Mechanical Core Losses
Stray Losses Losses
Losses
Rotational
Losses
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Electromechanical Relationships
1. If saturation is 2. If core saturation is considered, use
neglected, use the magnetization curve.
following expressions:

• Internal generated
voltage, angular speed,
and field current
• Induced torque, field
current, and armature
current
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Terminologies
Symbol Quantity
Ra Armature Resistance
Eg Internal Generated Voltage
VT Terminal Voltage
IF Field Current
Ia Armature Current
IL Line Current
RF Field Resistance
LAF Mutual inductance between armature and field windings
Pin Input power
Pmech,in Input mechanical power
Pconv, Pdev Converted power, Developed power in the armature
Pout Output power
Te Electrical torque
Tm Mechanical torque
D Damping coefficient
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Motor Ratings
Motor should be
operated as a
separately excited
motor

Rated output power Rated speed

Rated terminal voltage

Rated field excitation

what-when-how.com
Full-load Condition:
✓ Rated Output Power
✓ Rated Terminal Voltage
✓ Rated RPM
✓ Rated Armature Current
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Generator Ratings
Rated output power,
Rated speed

Rated terminal (output) voltage,

Rated armature current

Note: For DC Machines, the rotor is

the armature

Rated field excitation, field current

Full-load Condition:
✓ Rated Output Power
✓ Rated Terminal Voltage
✓ Rated RPM
✓ Rated Armature Current
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Generator External Characteristic Curves

(VT vs IL curves)
What happens to the terminal voltage of a DC generator as the load is
increased?
The plot of the terminal voltage versus the load current is called the
external characteristic curve.
Vt
Define: Voltage Regulation separately-excited
VNL  VFL
VR   100
VFL
(in percent) self-excited

IFL IL
Note: The separately-excited generator has a
smaller voltage regulation.
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PART 3
Circuit Analysis with DC machines
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Example 1
A self-excited DC motor is consuming 10kW when its
terminal voltage is 250V. The motor has an armature
resistance of 0.10 Ω, and a field resistance of 50 Ω. Find
the motor’s
a. line current
b. field current
c. armature current
d. internal generated voltage
e. electrical losses

ANS: 40 A; 5 A; 35 A; 246.5 V; 1372.5 W

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Example 1 - Solution
A1, F1

Ia + IL
IF
+ (motor) (motor)
ωm Te +
(Motor) Eg RF VT - Vsource

-
A2, F2 -
Armature Field
Winding Winding

a) Line Current: c) Armature Current:

IA = IL – IF = 40A – 5A = 35A

d) Internal Generated Voltage:

b) Field Current: EG = VT – RAIA = 246.5V
e) Electrical Losses:
PElec, losses = IA2RA + IF2RF = 1372.5 W
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Example 2
A separately-excited DC generator has a no-load terminal
voltage of 125 V when driven at 1800 RPM with the field
current set to 10 Amps. Neglecting saturation, find

a) Eg at 1600 RPM with IF constant at 10 A.

b) the speed in RPM so that Eg = 100 V, with IF constant at 10 A.

c) Eg if the speed and IF are increased to 1850 RPM and

12 A, respectively.

ANS: 111V; 1440 rpm;154 V

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Example 3
A separately-excited DC generator is rated 5 KW and 125 V.
The equivalent armature resistance is 0.2 . Neglecting
saturation of the field winding and assuming that motor
speed and field current are held constant at their values at
full-load conditions…
a) find the terminal voltage Vt at no load conditions.
b) find Vt when the load power is 2.5 KW.

ANS: 133V; 129 V

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Starting a DC motor
For a motor, VT = EG + IARA  IA = (VT – EG)/RA
At starting RA
condition, m=0
which makes EG=0. + IL IA +
RF IF
The starting Vt EG
armature current is
- Te
-
IA,S = VT/RA
ωm

Since RA is very small, IA can be quite large, which will

damage the armature winding.
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Starting a dc motor
At starting condition, armature current is:
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In order to limit the starting current, resistors are connected in

series with the armature at starting condition.
S1 S2 S3 S4
Sa
RA

SF R1 R2 R3 R4
RF IF Ia
Vt Ec
Rh

Te m

Note: The field rheostat Rh is set to zero during the starting period
to maximize the field current (and the electromagnetic torque).
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Starting sequence
1. Switch SF is closed with Rh set to zero.
2. Switch Sa is closed. The electromagnetic torque
and the starting armature current are
Vt
Te = LaFIFIas = kFIas Ias =
Re +R1 +R2 +R3 +R4

3. Te causes the armature to rotate and Ec begins

to build up.

4. Switches S1 to S4 are closed in sequence.

5. The rheostat Rh is increased until the motor

rotates at the desired speed.
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Starting current and speed

Ia
Note: Switching S1 to S4
causes the
discontinuities.
t
m

t
 35

Example 4
Consider a 5 HP, 125 V, 1200 RPM self-excited DC motor. The
efficiency is 85% at full load. The armature and field
resistances are 0.2 Ω and 62.5 Ω respectively.
Determine the external resistance to be connected in series
with the armature of the motor that will limit the armature
current at start-up to twice its full load value.

ANS: 1.69 ohms

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Efficiency, η
Ratio of output power and input power.

η = POUT/PIN
= POUT/(POUT + PLOSSES)
= (PIN – PLOSSES)/PIN
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Losses
• Electrical Losses
• Copper Losses (due to the resistance of the copper wires
in the windings)
• Commutator-Brush Losses
• Core Losses
• Hysteresis Losses
• Eddy Current Losses
• Mechanical Losses
• Bearing Friction and Windage Losses
• Commutator-Brush Friction Losses
• Stray Losses
• Leakage and Armature Reaction Fluxes (assumed as 1%
of output for machines 200 HP or above, otherwise
neglected)
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Power flows - Motor

Developed or Converted Power = PDEV = EgIa

Output
Input Power
Power (Mechanical)
(Electrical)

Stray
Mechanical Core Losses
Electrical Losses Losses (for big
Losses machines)
(I2R)
Rotational
Losses
 39

Power flows - Generator

Developed or Converted Power = PDEV = EGIA

Output
Input Power
(Electrical)
Power
(Mechanical)

Electrical
Core Losses
Stray Mechanical Losses
Losses (I2R)
Losses
(for big
machines)
Rotational
Losses
 40

Example 5
A 10-HP, 230-V shunt DC motor takes a full-load line
current of 40A at rated input voltage. The armature and
field resistances are 0.25 Ω and 230 Ω, respectively. The
brush-contact drop is 2V and the core and frictional losses
are 380W. Calculate the efficiency of the motor. Assume the
stray-load loss is 1% of the rated output.

Ans: 87.6%
 Example 5
Input (40)(230) 9200 W

Field-resistance loss 2302 / 230 230 W

Armature-resistance loss (0.25)(40 – 1)2 380 W

Core loss and friction loss GIVEN 380 W

Brush-contact loss (2)(39) 78 W

Stray-load loss (0.01)(10)(746) 74.6 W

Total losses 1143 W

Power output 9200 – 1143 8057 W

Hence, the efficiency at is given by,

PO 8057
  100%   100%  87.6%W
PI 9200
22 February 2018
 42

Example 6
Consider a 5 HP, 125 V, 1200 RPM self-excited DC motor. The
efficiency is 85% at full load. The armature and field
resistances are 0.2 Ω and 62.5 Ω respectively. Find the
following quantities at full load conditions.
a. The input power
b. The line current
c. The speed of rotation in rad/sec
d. The rotational losses
e. The output torque

ANS: 4388 W; 35.1 A; 125.7 rad/s;

189 W; 29. 7 N.m
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PART 4
Batteries and Cells
 44

Batteries vs. Cells

CELL – electrochemical device which

directly converts electrical energy to
chemical energy or vice versa

BATTERY – an assembly of two or more

cells electrically connected to for a single
enclosed unit.
 45

Charged Battery

Battery discharging

✓ Reactants are available

 The battery has accumulated/stored

net electrical charge

Modern Batteries 2nd Edition

By C. Vincent, Bruno Scrosati
 46

Batteries – Equivalent Circuit

Rint
x1

+
+
+ Ia
Eint Ecell
- (motor)

-
-

x2

Ecell = terminal voltage of the battery

Eint = emf developed because of the redox reaction

inside an electric cell

The equivalent circuit shown is an approximate model.

Modern Batteries 2nd Edition

By C. Vincent, Bruno Scrosati
 47

Batteries

Modern Batteries 2nd Edition

By C. Vincent, Bruno Scrosati
 48

Batteries
• Terminal voltage is dependent on the
• Ohmic resistance of the material
• Electrode losses
• Activation overvoltage – due to the charge transfer
• Concentration overvoltage – due to the accumulation of materials near the
electrode surface
• Area of the electrode/solution interface
• Half-reaction potentials
• Concentration of reactants (number of moles)
• Pressure and Temperature

Modern Batteries 2nd Edition

By C. Vincent, Bruno Scrosati