In The Name of God The Most Compassionate, The Most...

Electric Machines I

In The Name of God The Most Compassionate, The Most Merciful

2017 Shiraz University of Technology Dr. A. Rahideh

2

Table of Contents1. Introduction to Electric Machines

2. Electromagnetic Circuits

3. Principle of Electromechanical Energy Conversion

4. Principle of Direct Current (DC) Machines

5. DC Generators

6. DC Motors


3

Chapter 6Direct Current (DC) Motors

6.1. Different Types of DC Motors

6.2. Basic Relations

6.3. Operating Characteristics of DC Motors


4

Different Types of DC MotorsBase on the magnetic field production, DC Motors are classified as1. Separately excited DC motors: a separate voltage source (from

the armature voltage source) is required for field production.2. Shunt DC motors: field winding is connected in parallel to the

armature winding.3. Series DC motors: field winding is connected in series with the

armature winding.4. Cumulative compound DC motors: both series and parallel

field windings are used and their magnetic fields are added together. It is further divided as long shunt and short shunt.

5. Differential compound DC motors: both series and parallel field windings are used and their magnetic fields are subtracted from each other. It is divided as long shunt and short shunt.

6. Permanent magnet DC motors: field is produced by PM.2017 Shiraz University of Technology Dr. A. Rahideh

5

Specifications of DC MotorsDC motors have the following specifications:

1. Nominal rotational speed

2. Nominal output power (nominal torque can be obtained from the first two items)

3. Efficiency

4. Nominal terminal voltage

5. Speed regulation (SR) 100

fl

flnlSR

No‐load speed Full‐load speed


6

Examples of Mechanical Load

Speed

Torque

Fans, blowers and centrifugal pumps

Speed

Torque

Traction excluding gravity

• In any loads involving the turbulent flow of fluid, the load torque varies as the square of speed.

• It is the windage torque.• The windage is a dominant component

at high speeds for trains and car.

• It is applicable to electric trains and road vehicles.

• It is comprised of the windage, viscous friction, coulomb friction and stiction.

G. K. Dubey, Power Semiconductor Controlled Drives, 1989, Prentice‐Hall, Inc.


7


Speed

Torque

Coiler drives

• It is for applications where the motor is required to operate at constant power.

• Coiler drives are used in steel strip, paper and plastic mills

Speed

Torque

Diesel-electric locomotives

• A diesel‐electric locomotive employs a dc motor fed by a dc generator driven by a diesel engine.

• Instead of d generator, ac generator followed by a rectifier may be used.


8


• The purpose of excavators is to dig earth.• While digging, it may come across a rock,

in this situation the motor should simply stop to prevent damage to the excavator.

• At low speeds, the torque is mainly due to gravity (constant and independent of speed).

• At high speeds, the viscous and windagewill participate in the load torque.

Speed

Torque

Excavators

Speed

Torque

Cranes and Hoists

Low speed High speed


9

Torque DevelopmentThe induced voltage in the armature winding having Z conductors and a parallel paths is obtained as:

is the rotational velocity in rad/s is the magnetic flux of each polep is the number of polesZ is the total number of conductorsa is the number of parallel paths

The electromagnetic power which causes the rotation of rotor isIa is the armature currentTe is the electromagnetic torque

apZEa 2

ee TP aae IEP

ae Ia

pZT 2

kEa apZk2

ae IkT


10

Separately Excited DC MotorsThe schematic diagram of the separately excited dc motor:

The simplified schematic diagram:

IL

N S +

Ia

Field winding Armature winding

_

+

_

If

fV

aI

+

_

tV

fI

+ _

Field winding

Armature circuit


11

Separately Excited DC MotorsThe equivalent circuit of the separately excited dc motor:

ra is the armature winding resistancerf is the field winding resistanceIa is the armature currentIf is the field currentIL is the line currentVt is the motor terminal voltage (armature voltage)Vf is the field winding voltageEa is the induced voltage in the armature winding

La II

aaat rIEV kEa

fff rIV

fI

fV

aI

+

_

tV fI

+ _

LI

aE

ar fr

ae IkT


12

Shunt DC MotorsThe schematic diagram of the shunt dc motor:


IL

N S

Ia


+

_

If

aI

+

_

tV

fI

LI


13

Shunt DC MotorsThe equivalent circuit of the shunt dc motor:

faL III

aaat rIEV kEa

f

tf r

VI fI

aI

+

_

tV

fI

LI aI

+

_

tV

fI LI

aE

ar

fr

ae IkT


14

Effects of Field Winding Open-circuit• Assume during the operation of a shunt or separately excited dc

motor, the field circuit becomes open, then

• Therefore back‐emf decreases • Since the armature voltage source is constant, the armature

current increases significantly• Despite the reduction of flux, the current increase is dominant

and torque increases• Therefore the rotor accelerates• In such a case, the motor should be

disconnected from the voltage source.

0fI resa kE

aI

+

_

tV

fI

LI

res

aaat rIEV

arese IkT

dtdJTT Le


15

Series DC MotorsThe schematic diagram of the series dc motor:


If

IL

N S

Ia


+

_

aI

+

_

tV

fI

LI


16

Series DC MotorsThe equivalent circuit of the series dc motor:

Lfa III

saaat rrIEV

kEa

af II

aI

+

_

tV

fI

LI aI

+

_

tV

fI LI

aE

arsr

ae IkT


17

Compound DC MotorsShort-shunt

The simplified schematic diagram of the short‐shunt compound dc motor:

aI

+

_

tV fI

LI

Series field Shunt field

Cumulative Differential


18

Compound DC MotorsShort-shunt

The equivalent circuit of the short‐shunt compound dc motor:

faL III

Lsaaat IrIrEV

kEa

alDifferenti

CumulativeLsfsh IkIk

f

aaaf r

IrEI

aI

+

_

tV

fI

LI

aE

ar

fr

sr


19

Compound DC MotorsLong-shunt

The simplified schematic diagram of the long‐shunt compound dc motor:

aI

+

_

tV fI

LI

Series field Shunt field

Cumulative Differential


20

Compound DC MotorsLong-shunt

The equivalent circuit of the long‐shunt compound dc motor:

faL III

saaat rrIEV

kEa

alDifferenti

Cumulativeasfsh IkIk

f

tf r

VI

aI

+

_

tV

fI

LI

aE

ar

fr

sr


21

Cumulative vs. Differential Compound DC Motors

• If the direction of the fields produced by the series and shunt field windings are the same, the motor is cumulative;

• Otherwise it is differential.

• A cumulative compound dc motor is a differential compound dc generatorand vice versa.


22

Permanent Magnet DC MotorsThe schematic diagram of the PM dc motor:

The simplified schematic diagram: aI

+

_

tV N

Armature circuit

S

IL

N S

Ia

Permanent Magnet

Armature winding +

_

Permanent Magnet


23

Permanent Magnet DC MotorsThe equivalent circuit of the PM dc motor:

ra is the armature winding resistanceIa is the armature currentIL is the line currentVt is the motor terminal voltage (armature voltage)Ea is the induced voltage in the armature winding

La II aaat rIEV kEa ae IkT

aI

+

_

tV

LI

aE

ar

N S


Neodymium-iron magnet

Alnico magnet

Ferrite magnet

Rare-earth cobalt magnet

0.90 0.75 0.60 0.45 0.30 0.15 0

-H (MA/m)

B (T

)

0.2

0.4

0.6

0.8

1.0

1.2

1.4

24

Common types of Permanent Magnets

rB

cH

Demagnetization curve of four types of permanent magnets


25

Power Flow in DC Motors• In dc motors the input power is electrical (Voltage multiplied by

current)

• In dc motors the output power is mechanical (load torque multiplied by rotational velocity)

aaIEPPPPP lossescorelossesbrushlossesfieldlossesarmatureelec

)( ffLtelecin IVIVPP

LLmechout TPPP

Input electrical power

Ohmic losses due to armature winding

mechlossesmech PPIE aa

Mechanical losses

Brushes lossesOhmic losses due to field windings

Core losses

Output mechanical power


26

Power Flow in DC MotorsImportant note

• At no‐load condition, there is no load torque

• Therefore output power is zero

lossesmech PIE loadnoaa

0LT

mechlossesmech PPIE aa

0mech LTP


27

Operating Characteristics of DC Motors

1. Speed vs. armature current

2. Torque vs. armature current

3. Speed vs. Torque:

aI

ae IT

eT


28

Operating Characteristics of Separately Excited and Shunt DC Motors

1. Speed vs. armature current aI

aaat rIEV

kEa krIV aat

krIV aat

aI

Without ARWith AR a

pZk2


29


2. Torque vs. armature current aIT

apZk2

ae IkT

aI

eT

Without AR

With AR


30


3. Speed vs. torque eT

krIV aat

apZk2

ae IkT

eat T

kr

kV

2)(

kTI e

a

eT

Without ARWith AR


31

Operating Characteristics of Series DC Motors

1. Speed vs. armature current aI

saaat rrIEV

kEa krrIV saat

k

rrIV saat

acI

kc

rrkcIV sa

a

t

aI

Without saturation

With saturation


32


2. Torque vs. armature current aIT

ae IkT

acI

2ae IckT

aI

eTWithout saturation

With saturation


33



kcrr

TkcV sa

e

t ck

TI ea

2ae IckT

kc

rrkcIV sa

a

t

eT

Without saturation

With saturation

Series DC motors should not run without load since

it’s speed increases severely.


34

Operating Characteristics of Compound DC Motors


eT

Differential compound

Shunt & separately excited

Cumulative compound

Series

Differential compound DC motors are not used due to their instability problem.


35

Changing the Direction of Rotation of DC Motors

In order to change the direction of rotation of DC motors,

1. The direction of the field current should be changed while the direction of the armature current should be kept unchanged.or

2. The direction of the armature current should be changed while the direction of the field current should be kept unchanged.

The second method is preferred since the field circuit have higher inductance.


36

DC MotorsExample 1: A voltage of 230 V is connected to the armature of a separately excited DC motor and under this condition the nominal current of 205 A flows in the armature. If the armature resistance is 0.2 ohms,

a) Find the back‐emf.

b) Calculate the output power and torque if the rotational losses are 1445 W and the rotational velocity is 1750 rpm.

Separately Excited


37

DC MotorsSolution 1: separately excited DC motor

a) Calculate the back‐emf ( Ea ).

V230tV 2.0arA205aI

Separately Excited

fV

aI

+

_

tV

fI

+ _

LI

aE

ar fr

aata IrVE 2052.0230 aE V189aE


38

DC MotorSolution 1: separately excited DC motor

b) Calculate the output power and torque

rpm1750n W1445rotP

aaa IEP

Separately Excited

A205aI V189aE

205189aP W38745aP

rotaout PPP 144538745outP W37400outP

out

outPT

6021750

37400

outT Nm5.203outT


39

DC MotorsExample 2: Consider a series DC motor with the following values for nominal voltage, nominal velocity, nominal terminal current, and series field and armature resistances:

a) At nominal condition, calculate the back‐emf ( Ea ).

b) Calculate the developed torque and developed power at nominal condition.

c) If the load varies and the terminal current reduces to 150 A, calculate the speed and developed torque.

Assume the magnetic characteristics is linear.

rpm600nnV600ntV 04.0sr

Series

A200ntI 12.0ar


40

DC MotorsSolution 2: a series DC motor

a) At nominal condition, calculate the back‐emf ( Ea ).

Series

V600ntV rpm600nn A200ntI 04.0sr 12.0ar

aI

+

_

tV

fILI

aE

arsr

saata rrIVE )04.012.0(200600 aE

V568aE


41


b) Calculate the developed torque and developed power at nominal condition.

Series


aI

+

_

tV

fILI

aE

arsr

aaa IEP 200568aP W113600aP

a

aPT

602600

113600

aT

Nm1808aT


42



Series


aI

+

_

tV

fILI

aE

arsr

saata rrIVE 22 )04.012.0(1506002 aE

V5762 aE

11

22

1

2

kk

EE

a

a 11

22

1

2

ˆˆ

nIknIk

EE

a

a

a

a


43



Series


aI

+

_

tV

fILI

aE

arsr

V5762 aE

11

22

1

2

ˆˆ

nIknIk

EE

a

a

a

a

V5681 aE

A2001 aIA1502 aI

2

1

1

212

a

a

a

a

II

EEnn

150200

5685766002 n rpm8112 n


44



Series


aI

+

_

tV

fILI

aE

arsr

A2001 aIA1502 aI

2

1

212

a

aaa I

ITT

Nm10172 aT

11

22

1

2

a

a

a

a

IkIk

TT

11

22

1

2

ˆˆ

aa

aa

a

a

IIkIIk

TT

2

2 2001501808

aT

Nm18081 aT


45

DC MotorsExample 3: Consider the series DC motor of the previous example

a) If the starting current needs to be 150% of nominal current, calculate the external resistance to be connected between the motor and the voltage source

b) With the external resistance calculate the starting torque.

c) If the external resistance remains in the circuit and the terminal current becomes 200 A, calculate the back‐emf and speed.

Assume the magnetic characteristics is linear.


Series

A200ntI 12.0ar


46

DC MotorsSolution 3: series DC motor

a) If the starting current needs to be 150% of nominal current, calculate the external resistance to be connected between the motor and the voltage source.

At starting time, , therefore


Series

A200ntI 12.0ar

extsaaat RrrIEV

aI

+

_

tV

fILI

aE

arsrextR

0aE

extsaat RrrIV

saa

text rr

IVR 04.012.0

2005.1600

extR 84.1extR


47


b) With the external resistance calculate the starting torque.From previous example:


Series

A200ntI 12.0ar

aI

+

_

tV

fILI

aE

ar srextRA200aI

2

a

startastart I

ITT

Nm4068startT

2

2002005.11808

startT

Nm1808aT


48


c) If the external resistance remains in the circuit and the terminal current becomes 200 A, calculate the back‐emf and speed.


Series

A200ntI 12.0ar

aI

+

_

tV

fILI

aE

arsrextR

extsaata RrrIVE 2200600 aE V200aE

2

1

1

212

a

a

a

a

II

EEnn

200200

5682006002 n

rpm2112 n


49

DC MotorsExample 4: Consider a shunt DC motor with the following values for nominal terminal voltage, armature resistance and field resistance:

Assume the no‐load rotational velocity is 1200 rpm, the terminal current at no‐load condition is 3.938 A and the nominal armature current is 40 A.

a) Calculate the armature current at no‐load condition.b) Calculate the developed power at no‐load condition.c) Calculate the efficiency of the motor at nominal condition.d) Calculate the rotational velocity at nominal condition.

V230ntV 3.0ar

Shunt

160fr


50

DC MotorsSolution 4: shunt DC motor

a) Calculate the armature current at no‐load condition.

V230ntV 3.0ar

Shunt

160fr

rpm1200loadnon A938.3)( loadnoLI A40anI

aI

+

_

tV

fI

LI

aE

ar

fr

f

tf r

VI 160230

fI

floadnoLloadnoa III )()(

438.1938.3)( loadnoaI

A438.1fI

A5.2)( loadnoaI


51


b) Calculate the developed power at no‐load condition.

V230ntV 3.0ar

Shunt

160fr


aI

+

_

tV

fI

LI

aE

ar

fr

)()( loadnoaatloadnoa IrVE

V25.229)( loadnoaE

5.23.0230)( loadnoaE

)()()( loadnoaloadnoaloadnoa IEP

5.225.229)( loadnoaP W573)( loadnoaP


52


c) Calculate the efficiency of the motor at nominal condition.

V230ntV 3.0ar

Shunt

160fr


aI

+

_

tV

fI

LI

aE

ar

fr

anatan IrVE V218anE403.0230 anE

ananan IEP 40218anP

W8720anP


53



Note that no‐load developed power is the rotational losses

V230ntV 3.0ar

Shunt

160fr


W573)( loadnoarot PP

W8147outProtanout PPP

W8720anP

5738720outP


54



V230ntV 3.0ar

Shunt

160fr


Ltin IVP

W8147outP

438.140230 inP W6.9530inP

in

out

PP

6.9530

8147 855.0


55


d) Calculate the rotational velocity at nominal condition.

V230ntV 3.0ar

Shunt

160fr


V25.229)( loadnoaE

)( loadnoa

anloadnon E

Enn

rpm1141nn

V218anE

25.2292181200nn


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