electrical machines 1...course ilos 1. describe the construction of the transformers. 2. explain the...

ELECTRICAL MACHINES 1 EPMN301

1

Course Grading System

Marks

Final Exam Semester Work- 20 Mid-term- 15 Quizzes

- 15 Assignments- 10 Lab

2

Course Schedule

Lecture: Sunday 11:00 - 1:00 (9301) Tutorial: Sunday 4:00-7:00 pm (9301) Office Hours: Tuesday & Wednesday 11-12 Quizzes: W2-W4-W6-W10-W12 Assignments: W3-W5-W11-W14 Lab: W9-W13 Mid-term: W8 Email: [email protected] Scholar page: https://goo.gl/pkSTxw

3

mailto:[email protected]

https://goo.gl/pkSTxw

Course ILOs

1. Describe the construction of the transformers.

2. Explain the theory of operation of the transformers.

3. Calculate the parameters of the transformers using the different tests.

4. Predict the performance of the transformers.

5. Discriminate between the construction and applications of special transformers.

6. Identify the general constructional features of DC machines.

7. Formulate the EMF and torque equations of DC machine.

8. Analyze the operation of different types of DC motors and generators.

9. Asses the performance of different types of DC motors.

10. Implement different methods of, starting, speed control and braking of DC motors.

11. Work effectively as a team member.

12. Use practical work in laboratory.

4

References

A.E. Fitzgerald & Charles Kingsley, Electric Machinery, 7th Edition.

Stephan J. Chapman, Electric Machinery Fundamentals, 5th Edition.

5

Course Contents

Revision: Basic Principles Transformers:o Construction, theory of operation & Equivalent circuit. o Per-Unit System, Tests, Efficiency & Regulation.o Special Transformers.o Three-phase transformers & Parallel operation.

DC Machines:o DC Machines fundamentalso EMF & Torque Equationso Types of DC Generators & DC Motorso Equivalent circuito External Characteristics of DC Machineso Starting & Speed control of DC Motors

6

Revision: Basic Principles

AC Circuits Three-phase Circuits

7


Phasor Representation

1 sin( )mv V tω=

2 sin( )m ov V tω θ= +

8



1 0rmsV V= ∠

2 rms oV V θ= ∠

2m

rmsVV =

9



1V

2V

Phasor Diagram

1 1 0V V= ∠

2 2 oV V θ= ∠ oθ

10



1V

2V

1 2TV V V= +

oθ

1V

TV

11



oV V θ= ∠

V A jB= +

Polar form

Rectangular form

V

oθx

y

cos oA V θ=

sin oB V θ=

Real

Img

1tanoBA

θ −=

2 2V A B= +ojV V e θ=

12



1 1 1V A jB= + 2 2 2V A jB= +

1 2TV V V= ±

1 2 1 2( ) ( )TV A A j B B= ± + ± T T TV V θ= ∠

2 21 2 1 2( ) ( )TV A A B B= ± + ± 1 1 2

1 2

tanTB BA A

θ − ±=

±

1 1 1V V θ= ∠ 2 2 2V V θ= ∠

13



1 1 1V A jB= + 2 2 2V A jB= +

1 2TV V V= ×

1 1 2 2[ ] [ ]TV V or Vθ θ= ∠ × ÷ ∠

T T TV V θ= ∠

1 1 1V V θ= ∠ 2 2 2V V θ= ∠

1 2 1 2TV V or V θ θ= × ÷ ∠ ±

1 2TV V or V= × ÷

1 2Tθ θ θ= ±

14


Loads: R-L Load

V I(R j L)ω= +

LV I(R j X )= +

V IZ= Impedance

LZ R j X Z Φ= + = ∠

1tan LRω−Φ =2 2

LZ R X= +

V 0 VIZ Z

ΦΦ

∠= = ∠−

∠current lags voltage

VΦ

I

15


Loads: R-L Load

RV IR=

V IZ=

L LV IX=Φ

R

ZLX

Φ

I×

16


Loads: R-C Load

1V I(R j )Cω

= −

CV I(R j X )= −

CZ R j X Z Φ= − = ∠−

1 1/tan CRω−Φ =

2 2CZ R X= +

V 0 VIZ Z

ΦΦ

∠= = ∠

∠−current leads voltage

VΦ

I

17


Loads: R-C Load

RV IR=

V IZ=C CV IX=

Φ

R

Z CXΦ I×

18


Loads: R-L-C Load

L CV I [R j( X X )]= + −

L CZ R j( X X ) Z Φ= + − = ∠

1tan L CX XR

− −Φ =2 2

L CZ R ( X X )= + −L CX X> veΦ = +

L CX X< veΦ = −V 0 VI

Z ZΦ

Φ∠

= = ∠∠±

inductive load

capacitive load

19


Loads

inductive load capacitive load

Pure inductive load Pure capacitive load

L CX X>

L CX X<

inductive load

capacitive load

Resistive load

20


AC Power

p( t ) v( t ) i( t )= ×

( ) 2 sin( )v t V tω=

( ) 2 sin( )i t I tω= ±Φ

( ) [cos( ) cos(2 )]p t VI tω= Φ − ±Φ

cosavgP VI= Φ Active Power (W)

21


Active and Reactive Power

( ) [cos( ) cos(2 )]p t VI tω= Φ − ±Φ

For pure resistive load:

0Φ =

( ) [1 cos(2 )]p t VI tω= −

avgP VI=

22



For pure inductive load:

90oΦ = −

( ) cos(2 90 )op t VI tω= −

( ) sin(2 )p t VI tω=

avgP zero=

( ) [cos( ) cos(2 )]p t VI tω= Φ − ±Φ

23



For pure capacitive load:

90oΦ =

( ) cos(2 90 )op t VI tω= +

( ) sin(2 )p t VI tω= −

avgP zero=

( ) [cos( ) cos(2 )]p t VI tω= Φ − ±Φ

24


AC Power: Inductive Load

2 cosP I R VI= = Φ

2S VI I Z= =2 sinL LQ I X VI= = Φ

Φ

RV IR=

V IZ=L LV IX=

Φ

Apparent Power (VA)

Active Power (W)

Reactive Power (VAR)

I×

25


AC Power: Capacitive Load

RV IR=

V IZ= C CV IX=ΦΦ

2 cosP I R VI= = Φ

2S VI I Z= =

2 sinC CQ I X VI= = Φ

Apparent Power (VA)

Active Power (W)


I×

26


Complex Power & Power Factor

Φ

*S P jQ VI= ± =

Leading Load

* *( )I I I= ∠ Φ = ∠±Φ

Lagging Load

S

P

QΦ

QS

Pcos sinS VI jVIφ φ= ±

cosPpower factorS

= = Φ

27


PPower FactorS

=

28

Revision: Basic Principles29


sin( )mv V tω=

2 Tπ ω=2T πω

= (sec)

3-phase AC Supply

30

The three-phases are called: A-B-C or R-S-T or R-Y-B

Balanced 3-phase supply: the three voltages are equal in magnitude and are 120o out of phase.

sin( )a mv V tω=

sin( 120 )ob mv V tω= −

sin( 240 )oc mv V tω= −

sin( 120 )oc mv V tω= +

0aV V= ∠

120bV V= ∠−

120cV V= ∠



A-B-C A-C-B

Positive sequence Negative sequence

0aV V= ∠

120bV V= ∠−

120cV V= ∠

0aV V= ∠

120bV V= ∠

120cV V= ∠−

32


abV

bcV

caVanV

bnV

cnV

Line-to-neutral

Line-to-line

ab an bnV V V= − bc bn cnV V V= − ca cn anV V V= −

0anV V= ∠

120bnV V= ∠−

120cnV V= ∠

33

3 30ab an bnV V V V= − = ∠

3 90bc bn cnV V V V= − = ∠−

3 150ca cn anV V V V= − = ∠

3 ( 30)LL LnV V θ= ∠ +



abV

bcV

caVLine-to-line

0abV V= ∠ 120bcV V= ∠− 120caV V= ∠

35


Balanced Star Load (ZA = ZB = ZC)

abV

bcV

caV anV

bnV

cnV

aI

bI

cI

ab bc ca LV V V V= = = an bn cnV V V Vφ= = =3LVVφ =

a b c LI I I I= = = an bn cnI I I Iφ= = =LI Iφ =

VI

Zφ

φφ

=

36


Balanced Delta Load (ZA = ZB = ZC)

ab bc ca LV V V V= = = LV Vφ =

a b c LI I I I= = = ab bc caI I I Iφ= = =

abV

bcV

caV caIabI

bcI

aI

bI

cI

37


b bc abI I I= −

c ca bcI I I= −

caIabI

bcI

aI

bI

cI

3a b c LI I I I Iφ= = = =

ab bc caI I I Iφ= = =

3 30a ab caI I I Iφ= − = ∠−

38

YY

YD

DD

YD


Star-Star Balanced System0a b cI I I I zero+ + = =

n NV V zero= =


Star-Star Balanced System

Single phase equivalent circuit


Star-Delta Balanced System

3YZZ ∆=


Star-Delta Balanced System

Single phase equivalent circuit

3YZZ ∆=



Star

Delta3-phase power = 3 x per phase power

3LV Vφ= LI Iφ =

LV Vφ= 3LI Iφ=

3 cos 3 cosL LP V I V Iφ φ= Φ = Φ

3 sin 3 sinL LQ V I V Iφ φ= Φ = Φ

3 3 L LS V I V Iφ φ= =

Active Power (W)


Apparent Power (VA)

PPower FactorS

=

44

electrical machines 1...course ilos 1. describe the construction of the transformers. 2. explain the...

Documents