shaiful nizam mohyar universiti malaysia perlis school of microelectronic 2007/2008

SHAIFUL NIZAM MOHYARUNIVERSITI MALAYSIA PERLIS

SCHOOL OF MICROELECTRONIC

2007/2008

Electromechanical Energy Conversion

ElectricalEnergy

MechanicalEnergy

Motor

Generator

2.0 INTRODUCTION

Pelec = vi = Teωm = Pmech

The machine’s magnetic field is the medium of (energy) conversion.

GLOSSARY Torque, T – A force that produces rotation on a axis and

also defined as a linear force multiplied by a radius. In an electric motor, this is the force from the interaction of the magnetic fields produced by the flow of current through the armature and field windings/coils.

Flux, Φ - The magnetic field which is established around an energized conductor or permanent magnet. The field is represented by flux lines creating a flux pattern between opposite poles. The density of the flux lines is a measure of the strength of the magnetic field.

Magnetic field, B - A region of space that surrounds a moving electrical charge or a magnetic pole, in which the electrical charge or magnetic pole experiences a force that is above the electrostatic ones associated with particles at rest.

The voltage in any real machine will depend on three factors: 1. The flux, Φ in the machine. 2. The speed of rotation, ω. 3. A constant representing the construction of the machine.

2.0 INTRODUCTION…contd

Flemming’s Left Hand Rule (Motor Rule)

Use: To determine the direction of a force on a current carrying conductor in a magnetic field.

2.0 INTRODUCTION…contd

The carbon rod is NOT magnetic. When no current flows, the rod is stationary When we turn on the current, the rod

experiences a force that makes it move. The direction of the force is determined by

Fleming' Left Hand Rule

2.1 CONSTRUCTION

Cutaway view of a dc motor

Stator with visible poles

2.1 CONSTRUCTION ….contnd

Rotor of a dc motor.

2.1 CONSTRUCTION ….contnd

General arrangement of a dc machine

2.2 CONSTRUCTION….contnd

The stator of the dc motor has poles, which are excited by dc current to produce magnetic fields.

In the neutral zone, in the middle between the poles, commutating poles are placed to reduce sparking of the commutator. The commutating poles are supplied by dc current.

Compensating windings are mounted on the main poles. These short-circuited windings damp rotor oscillations. .


The poles are mounted on an iron core that provides a closed magnetic circuit.

The motor housing supports the iron core, the brushes and the bearings.

The rotor has a ring-shaped laminated iron core with slots.

Coils with several turns are placed in the slots. The distance between the two legs of the coil is about 180 electric degrees.


The coils are connected in series through the commutator segments.

The ends of each coil are connected to a commutator segment.

The commutator consists of insulated copper segments mounted on an insulated tube.

Two brushes are pressed to the commutator to permit current flow.

The brushes are placed in the neutral zone, where the magnetic field is close to zero, to reduce arcing.


Commutator of a dc motor


The rotor has a ring-shaped laminated iron core with slots.

The commutator consists of insulated copper segments mounted on an insulated tube. Two brushes are pressed to the commutator to permit current flow.

The brushes are placed in the neutral zone, where the magnetic field is close to zero, to reduce arcing.


The commutator switches the current from one rotor coil to the adjacent coil.

The switching requires the interruption of the coil current.

The sudden interruption of an inductive current generates high voltages .

The high voltage produces flashover and arcing between the commutator segment and the brush.

2.1 CONSTRUCTION

I

I

shaftw

h

B

Next slide looks down the shaft

2.2 DC MACHINE OPERATION BASIC

THEORY

Current coming

toward you

Current leaving away from you

B

Shaft


THEORY….contnd

Bdirection by left-hand rule

F

F T = 2hFcos = 2hIwBNcos

Thumb = current Forefinger = BRest = Force


THEORY….contnd

B


THEORY….contnd

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.

2.3 DC MACHINE OPERATION

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.

(a) Rotor current flow from segment 1 to 2 (slot a to b)

Vdc30

NS

Bv

v

a

b

1

2

Ir_dc

(b) Rotor current flow from segment 2 to 1 (slot b to a)

30NS Vdc

a

b

B

v v

Ir_dc


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 counterclockwise rotation.


Vdc30

NS

Bv

v

a

b

Ir_dc


30NS Vdc

a

b

B

v v

Ir_dc


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.


Vdc30

NS

Bv

v

a

b

Ir_dc


30NS Vdc

a

b

B

v v

Ir_dc


2.4 DC Generator Operation 2.4 DC Generator Operation

Fleming’s Right hand rule (Generator Rule) Use: To determine the direction of the induced

emf/current of a conductor moving in a magnetic field.

2.4 DC Generator Operation 2.4 DC Generator Operation

The N-S poles produce a dc magnetic field and the rotor coil turns in this field.

A turbine or other machine drives the rotor.

The conductors in the slots cut the magnetic flux lines, which induce voltage in the rotor coils.

The coil has two sides: one is placed in slot a, the other in slot b.

30NS Vdc

Bv

v

a

b

Ir_dc


30NS Vdc

a

b

1

2vv

B

Ir_dc


2.4 DC Generator Operation2.4 DC Generator Operation....contndcontnd..

In Figure (a), the conductors in slot a are cutting the field lines entering into the rotor from the north pole,

The conductors in slot b are cutting the field lines exiting from the rotor to the south pole.

The cutting of the field lines generates voltage in the conductors.

The voltages generated in the two sides of the coil are added.

30NS Vdc

Bv

v

a

b

Ir_dc


30NS Vdc

a

b

1

2vv

B

Ir_dc



The induced voltage is connected to the generator terminals through the commutator and brushes.

In Figure (a), the induced voltage in b is positive, and in a is negative.

The positive terminal is connected to commutator segment 2 and to the conductors in slot b.

The negative terminal is connected to segment 1 and to the conductors in slot a.

30NS Vdc

Bv

v

a

b

Ir_dc


30NS Vdc

a

b

vv

B

Ir_dc



When the coil passes the neutral zone: Conductors in slot a are

then moving toward the south pole and cut flux lines exiting from the rotor

Conductors in slot b cut the flux lines entering the in slot b.

This changes the polarity of the induced voltage in the coil.

The voltage induced in a is now positive, and in b is negative.

30NS Vdc

Bv

v

a

b

Ir_dc


30NS Vdc

a

b

vv

B

Ir_dc



The simultaneously the commutator reverses its terminals, which assures that the output voltage (Vdc) polarity is unchanged.

In Figure (b)

the positive terminal is connected to commutator segment 1 and to the conductors in slot a.

The negative terminal is connected to segment 2 and to the conductors in slot b.

30NS Vdc

Bv

v

a

b

Ir_dc


30NS Vdc

a

b

1

2vv

B

Ir_dc



2.5 DC Machine Type

There are generally five major types ofDC motors: The separately excited dc motor The shunt dc motor The permanent magnet dc motor The series dc motor The compounded dc motor

The magnetic field produced by the stator poles induces a voltage in the rotor (or armature) coils when the generator is rotated.

This induced voltage is represented by a voltage source.

The stator coil has resistance, which is connected in series.

The pole flux is produced by the DC excitation/field current, which is magnetically coupled to the rotor

The field circuit has resistance and a sourceThe voltage drop on the brushes represented

by a battery

2.6 DC Machine Equivalent Circuit2.6 DC Machine Equivalent Circuit

1. Permanent magnet• The poles are made of permanent

magnets. • No field winding required.• Small size.• Disadvantage is low flux density,

so low torque.

2.6 DC Machine 2.6 DC Machine EquEqu. Circuit. Circuit....contndcontnd..

2. Separately excited The field flux is derived from a separate power source

independent of the generator itself.

B

Field winding

Armature winding


3. Self-excited – shunt machine

• The field flux is derives by connecting the field directly across the terminals of the generator.

B


3. Self-excited - series machine

• field are connected in series with armature

B


3. Self-excited – Compounded dc motor

- both a shunt and a series field

are present


3. Self-excited • Cumulatively compounded

• Differentially compounded

BB

BB


The armature is represented by an ideal voltage source EA and a resistor RA.

The brush voltage drop is represented by a small battery Vbrush opposing the direction of the current flow in the machine.

The field coils, which produce the magnetic flux, are represented by inductor LF and RF.

The separate resistor Radj represents an external variable resistor used to control the amount of current in the field circuit.

Equivalent Circuit of a DC Motor.


The brush drop voltage is often only a very tiny fraction of the generated voltage in the motor.

Therefore, in cases where it is not critical, the brush drop voltage may be left out or approximately included in the value of RA.

Also, the internal resistance of the filed coils is sometimes lumped together with the variable resistor, and the total is called RF , Figure below.

A Simplified Equivalent Circuit eliminating the Brush Voltage Drop and Combining Radj with the Field Resistance .

F

FF R

VI AAAT RIEV

AL II

The Equivalent Circuit of Separately Excited dc Motor.From the above figure,


F

FF R

VI AAAT RIEV

FAL III

Shunt DC motors

From the above figure, The Equivalent Circuit of a Shunt dc Motor

Torque Equation

T = torque of armature (N-m)kA = geometry constant

= flux/pole (Wb) IA = armature current (A)

AA IkT

Geometry Constant

p = number of field polesN = number of active conductors on armature

M = number of parallel paths in armature winding (=p for lap winding, =2 for wave winding)

)(60

),/(2

' rpmM

pNksrad

M

pNk AA

Power Equation

P=power (W) – not counting lossesE = EMF induced in armature (back EMF)

IA = armature current (A)T = torque of armature (N-m) = speed of rotation (rad/s)

Note that Pin = VLIL which will be higher than P because of loss in the field and armature

windings as well as rotational (friction) losses.

TEIP A

EMF Equation

E = EMF induced in armature (V)kA = geometry constant

= flux/pole (Wb) = speed of rotation (rad/s) n = speed of rotation of

armature (rpm)

nkkE AA '

2

60n

Terminal Voltage Equation

VT = voltage at motor terminals

E = EMF induced in armature (V)IA = armature current (A)

RA = armature resistance

AAT RIEV E VT

+

-

+

-

RA

Speed Equation

(applies to shunt connected motor only)

Note that can also be written as kfIf where kf is

/If (normally a constant ratio)

'A

AAT

k

RIVn

Ratio Equation

1

2

1

2

E

E

n

n

Speed-Torque

Torque

Speed

Series

Shunt

Cumulative Compound

Differential Compound

Example 1A 6 pole, 3.0 hp 120V DC lap-wound shunt motor has 960 conductorsin 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 and torque.

AAAIII FLA 8.232.125 VAVRIVE AAT 10275.08.23120

153

62

9606

2

MpN

K A

sradx

V

K

E

KE

A

A

/9.661010153

1023

rpmn 6382

60

kWhp

WhpP 24.2

7463

mNsrad

kWPT 5.33

/9.66

24.2

Example 2A 10hp, 115V Dc series motor takes 40A at its full load speed of 1800rpm. What is the torque at 30A?

srad

n/188

60

18002

60

2

kWhp

WhpP 46.7

74610

mNsrad

kWPT

TP

6.39/188

46.7

AFFAAA IIKKIKT

AF II 2AFA IKKT

025.0

40

6.3922

A

mN

I

TKK

A

FA

mNAIKKTnewAFAnew 2.2230025.0 22

Example 3 (a)A 220V DC shunt motor draws 10A at 1800rpm. The armatureresistance is 0.2 and field winding resistance is 440. (a) What is

thetorque?

AV

R

VI

F

TF 5.0

440

220

AAAIII FLA 5.95.010

VAVRIVE AAT 2182.05.9220

kWAVEIP A 07.25.9218

srad

n/188

60

18002

60

2

mNsrad

kWPT 0.11

/188

07.2

Example 3 (b)A 220V DC shunt motor draws 10A at 1800rpm. The armatureresistance is 0.2 and field winding resistance is 440. (b) What willbe the speed and line current at a torque of 20 N-m (if field current isconstant)?

16.1/188

218

srad

VEK

KE

A

A

AmN

K

TI

IKT

AA

AA

3.1716.1

20

AAAIII FAL 8.175.03.17

VVRIVE AAT 2172.03.17220

rpmxn

sradV

K

E

A

31079.12

60

/18716.1

217

(shunt is constant speed)

shaiful nizam mohyar universiti malaysia perlis school of microelectronic 2007/2008

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