Module 8 Three Phase Systems v3

8/18/2019 Module 8 Three Phase Systems v3

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Three-Phase Systems

EE 102 Circuits 2

Module 8

by:Cesar G. Manalo !r.


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Objectives

Defne a three-phase system.

Describe the wye and delta-type connections in three-phase systems.Solve voltages and currents in three-phase system.Describe a single-phase three-wire system.

Solve voltages and current in single-phase three-wiresystem.


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Introduction

The three-phase (3-φ) ! generators (alternators) has

3 sets o" armature windings called phase windings .#ach o" the 3 windings develop e$actly the samesinusoidal voltages (same magnitude and "re%uency)called phase voltage but are &' electrical degrees

apart.

N S

&' edeg

&' edeg

3-phase 2-pole alternator


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Introduction

n most cases* the winding that produces the +u$ is

the one revolving and the armature winding (thatdevelops phase voltages) is stationary.

The 3 sets o" windings can be wired together to "ormeither a delta(,) connection or a wye( ) connection.

a

c

b

a

c b

E aa’

E cc’

E bb’

c

c

bb

a

a

N S


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Introduction

Y-connected alternator

a

c

ba

c b

E aa’

E cc’

E bb’ N S

c

cbb

a

a

a

b

c

a

b

c

3-phase 2-pole Y-connected alternat


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Introduction


a

c

ba

c b

E aa’

E cc’

E bb’

a

b

c


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Introduction


a

c

ba

c b

E aa’

E cc’

E bb’

/

!

a

c

b

E a

E c E b

/

!

n

Y-connected alternator withcommon point called neutral (n)


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Line-to-line olta!e in 3-"hase Y-connected #lternators

n a balanced system*each o" the threeinstantaneous voltagesa *b * 0 c have e%ualamplitudes but areseparated "rom the other

voltages by a phaseangle o" &' o.

$ #N

$ %N

$ &N

'2 o

'2 o

'2 o

$ #N

$ %N$ &N

#

%

&

$ #%

$ %&

$ N

3-phase alternator with the phases replaced3 sin!le-phase alternators* e-connected 3-phase alternator

"hasor dia!ramo+ phase volta!

a

c

b

E a

E c E

b

/

!

1

2

2 2

-

- -


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$ #N

$ %N

$ &N

'2 o

'2 o'2 o

)( BN AN NB AN AB E E E E E −+=+=

-$ %N

a

bc

#

%

&

$ #%

$ %&

$ N

$ #N

$ N%, o

where $ #N is the phase voltage o" alternator a


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$ #N

)( BN AN NB AN AB E E E E E −+=+=

-$ %N # /

3 # 1 2

( - # / 1 )

AN AN AB E E E 73.13 == BN BN BC E E E 73.13 ==CN CN CA E E E 73.13 ==

a

bc

#

%

&

$ #%

$ %&

$ N

$ #N

$ N%

, o3 o

here+ore. the line-to-line volta!es o+ a w e-connected alternator is '/03 times its phase volta!e/


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a

bc

#

%

&

$ #%

$ %&

$ N

$ #N

$ %N

$ &N

'2 o

'2 o

'2 o

$ #%$ '2 o

'2 o '2 o

here+ore. the line-to-line volta!es o+ a w e-connected alternator is '/03 times its phase volta!e.and the are '2 o out o+ phase +rom each other/

$ %&

"hasordia!ram o+phase and linevolta!es o+ a 3-phase Y-connectedalternator


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#&O1 "1

• The line-to-line voltage o" a 3-phase alternator is '3 4 asmeasured by an ! voltmeter. 5hat is the phase voltage i"measured by the same voltmeter i" the alternator isconnected in wye 6 5rite the polar "orm o" all line-to-line andphase voltages.

Illustrative "roblem ''

Solution78et7 # / line-to-line voltage between 0 /

# 1 phase voltage o" phase 1.

V E E

E E

AB AN

AN AB

9.13273.1

23073.1

73.1

===

=

V E

V E

V E

V E V E

referenceV E

CA

BC

AB

CN

BN

AN

°∠=°−∠=

°∠=°∠=

°−∠=

°∠=

150230

90230

30230

1209.1321209.132

)(09.132

$ #N

$ %N

$ &N

'2 o

'2 o'2 o

$ #%

'2 o

'2 o '2 o

$ %&

$


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Line &urrent in 3-"hase Y-connected#lternators

a

bc

#

%

&

#%

%&

N I

a

I b

I c

' 2 ( o

' 2 ( o

' 2 ( o

I#

I%

I&

Ia

IbIc I #

I &

I % "hase and linecurrent phasordia!ram o+ abalanced-loaded 3-phase alternator

• he current on a line is e ual to the currentthrou!h the phase to which that line isconnected/ hat is. I # I a . I % I b . I & I c /

• *hen the e4ternal load on a 3-phasealternator is 5balanced6. the line and phasecurrents are all '2 o out o+ phase +rom each

# 1

# /1

# !1

#lternator


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Line-to-line olta!e in 3-"hase 7-connected #lternators

• n a ,-connected 3-phase alternator* there is no

neutral pt.• The phase windings are connected one a"ter the

other "orming a delta or closed loop.• The line-to-line voltages e%uals the phase voltages

both in magnitude and in direction because thephases are connected directly to the lines.

$ #

$ %

$ &'2 o

'2 o

'2 o$ #

$ %

$ &

#

%

&

$ #%

$ %&

$

7-connected 3-phase alternators "hasor dia!ram

2

2

2

-

-

-

$ #%

$ %&

$


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Line-to-line &urrent in 3-"hase 7-connected #lternators

$ #

$ %

$ &'2 o

'2 o

'2 o$ #

$ %

$ &

#

%

&

$ #%

$ %&

$

7-connected 3-phase alternators "hasor dia!ram

2

2

2

-

-

-

$ #%

$ %&

$

C CA

B BC

A AB

E E E E

E E

==

=

I#%

I%&I

I#%

I%&

I&#

CACA BC C

BC BC AB B

AB ABCA A

I I I I I I I I

I I I I

33

3

=−==−=

=−=

I#

I%

I&

-I #%

I#

I%

I&

• n a ,-connected balanced 3-phase alternator* theline-currents are &.93 times the phase currents.


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%alanced 3-"hase Loads

a

bc

#

%

&

#%

%&

N

I#

I%

I&

Ia

IbIc

• load is said to be balanced* i" all phase impedances

are e%ual. Z

A

Z B

Z C

3-phase Y-connected%alanced Load. 8 #

8 % 8 &

a

bc

#

%

&

#%

%&

N

I#

I%

I&

Ia

IbIc

Z C

Z B

Z A

3-phase 7-connected

balanced load. 8 # 8 % 8 &

3-phaserecti9er

3-phase inductionmotor


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a

bc

#

%

&

23

N

I#

I%

I&

Ia

IbIc

Z

Z

Z

23 23

3-phase#lternator

%alance3-phaseload.8 : ; 0 L

< =

balance 3-phase load is connected to a 3-phase alternator withline-to-line voltage o" '3 volts. :ind7a) 8oad phase currentsb) 8oad line currentsc) ;hasor diagram o" phase and line currents.

Illustrative "roblem '2


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a

bc

#

%

&

23

N

I#

I%

I&

Ia

IbIc

Z

23 23

3-phase#lternator

%alance3-phaseload.8 : ; 0 L

< =

Z

Z

I#%

I%&

I

8et74 / '3


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%alanced 3-"hase Loads°−∠=−∠= 69.355.269.333055.2 AB I

°−∠=−−∠= 69.12355.2)69.3390(55.2 BC I

°∠=−∠= 116.3155.2)69.33150(55.2CA I

33/,? o

33/,? o

3 3

/ , ?

o

I#%

I%&

I

)( CA ABCA AB A I I I I I −+=−=

)( AB BC AB BC B I I I I I −+=−=)( BC CA BC CAC I I I I I −+=−=

I#

I%

I&

33/,? o

@,/3' o

'


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%alanced 3-"hase "ower

a

bc

#

%

&

#%

%&

N

I#

I%

I&

Ia

IbIc

Z C

Z B

Z A

8 # 8 % 8 &

B #%

I#%

I%&

I#%

'2 o

'2 o '2 o

%&

3 o

I#%

I%&

I

B %&

B

CACAC

BC BC B

AB AB A

I V S

I V S

I V S

===

#pparent power perphase

CACACAC

BC BC BC B

AB AB AB A

I V Q

I V Q

I V Q

φ

φ

φ

sin

sin

sin

=

==

;eactive power perphase

CACACAC

BC BC BC B

AB AB AB A

I V P

I V P

I V P

φ

φ

φ

cos

cos

cos

=

==

;eal power perphase


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CACACAC

BC BC BC B

AB AB AB A

I V Q

I V Q I V Q

φ

φ φ

sin

sinsin

===

;eactive power perphase

CACACAC

BC BC BC B

AB AB AB A

I V P

I V P I V P

φ

φ φ

coscoscos

=

=

=

;eal power perphase

Cor balanced delta-connected load:

P CA BC AB

L P CA BC AB

L P CA BC AB

I I I I I

V V V V V

φ φ φ φ ===

====

====

3

CACAC

BC BC B

AB AB A

I V S

I V S I V S

===

#pparent power perphase

P P P P C B A

P P P P C B A

P P P C B A

I V P P P P

I V QQQQ

I V S S S S

φ

φ

cos

sin

========

====;ASvalueonl notthephase

alue onlnot thephase

P L L P L

L P P P T I V I

V I V Q φ φ φ sin3sin3

3sin3 =

==

P L L P L

L P P P T I V I

V I V P φ φ φ cos3cos3

3cos3 =

==

L LT T T I V Q P S 322 =+=

where:

" phase volta!eL line volta!e

I" phase currentIL line currentS " phase apparentpowerD " phase reactive

power" " phase real

Let:D total reactive power taEen b

load" total real power taEen b loadS total apparent power taEen bload


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a

bc

#

%

&

23

N

I#

I%

I&

Ia

IbIc

Z

Z

Z

23 23

3-phase#lternator

%alance3-phase

load.8 : ; 0 L < =

:ind the per phase apparent* reactive* and true power consumedby the load as shown by the fgure below.

Illustrative "roblem '3

A I I I I P CA BC AB 55.2====

°= 69.33 P φ

V AB V BC V CA V P '3 V

VA I V S

W I V P

VAR I V Q

P P P

P P P P

P P P P

586.5)55.2)(230(

0.48869.33cos)55.2)(230(cos

33.32569.33sin)55.2)(230(sin

===

=°==

=°==

φ

φ

KVAVA P QS

W I V P

VAR I V Q

T T T

P P P T

P P P T

76.15.759,1)24.845()49.563(

99.146369.33cos)55.2)(230(3cos3

99.97569.33sin)55.2)(230(3sin3

2222 ==+=+=

=°==

=°==

φ

φ

Solution7


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3-phase 9B.@ C5 delta-connected induction motor is suppliedby a 3-phase star-connected alternator generating & 4between phases. " the "ull load e ciency and the power "actoro" the induction motor are =' E and .F> respectively* calculateGa) !urrent in each motor phaseb) !urrent in each alternator phase

Illustrative "roblem 'F


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3-phase* > AH* B&> 4 star-connected induction motor has anoutput o" > C5 with an e ciency o" = E and a power "actor o".F>. calculate the line current. " the motor windings are now

connected in delta* what would be the correct voltage o" a 3-phase supply suitable "or the motor6

SeatworE


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Illustrative "roblem'FTwo 3-phase balance loads a e connected in pa allel to a 400-! line. The "i stload is delta-connected with a phase i#pedance o" " # $0%&'0 and the second

is a sta -connected p$ el% esisti&e load o" ( # 2) oh#s pe phase. 'inda) The phaso line c$ ents I A , I B , and I C .

b) The appa ent, eacti&e, and eal powe cons$#ed b% the co#bined load.

c) The ' o" the co#bined load.

Z

Z Z R

R

R

A

B

C

I A

I B

I C

N


26/43

"hase Se uence

N S

"erspective view o+ armature

N S

Side view o+ armature


27/43

"hase Se uence

N S

"erspective view o+ armature

N S

Side view o+ armature (turned ? o )


28/43

"hase Se uence

N S

&' e

deg

N S

# % &"erspective view o+ armature

Side view o+ armature


29/43

"hase Se uence

N S

&' e

deg

N S

# % &


30/43

"hase Se uence

N S

&' e

deg

N S

# % &S$ S$D1$N&$ #%& % %

V a

V b

V c

;hasor representation o"

/! se%uence


31/43

"hase Se uence

&' e

deg

N S

# & %S$ S$D1$N&$ #&% &%# %#&

IJ TKI# 5 1D 1LS #88M51D I#D 1T#I!A 1L#D

V a

V c

V b

;hasor representation o"

!/ se%uence


32/43

"hase Se uence

&' e

deg

N S

# & %

phase se%uencedefnes which o" thethree phase voltages orline voltages comes inse%uence.

There are only twopossible phasese%uence in three-phase

system7 phase se%uence/! or phase se%uence!/.

;hase se%uence /! is

called positive se%uence


33/43

"hase Se uence

&' edeg

N S

# % &

3-"G#S$ F-*I;$

a

bc

#

%

&

#%

%&

N

N#N%N&N

/

!1

3-phase F-wire alternatorwith output connected to

lines #. %. &. H N

3-phase F-wire lines

1

/

!


34/43

"hase Se uence

3-"G#S$ F-*I;$

/

!1

;hase se%uence /!7

&' e

deg

N S

# % &

o P AN V V 0∠=

o

P BN V V 120−∠=o

P CN V V 120∠=

where V P is the line-to-neutral voltage

or the phase voltage


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&hecEpoint '

/

!1

;hase se%uence /!7

&' e

deg

N S

# % &

o P AN V V 0∠=

o

P BN V V 120−∠=o

P CN V V 120∠=



Liven the phase se%uence o" phase voltages 4 1 * 4 /1 * and 4 !1 *

below* what is the phase se%uence o" the line voltages* 4 / * 4 /! *4 ! 6


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"hase Se uence

3-"G#S$ F-*I;$

/

!1

;hase se%uence /!7

&' e

deg

N S

# % &

o P AN V V 0∠=

o

P BN V V 120−∠=o

P CN V V 120∠=

o P AB V V 303 ∠=

o

P BC V V 903 −∠=o

P CA V V 1503 ∠=




37/43

"hase Se uence

3-"G#S$ F-*I;$

/

!1

;hase se%uence !/7

o P AN V V 0∠=

o

P BN V V 120∠=o

P CN V V 120−∠=

o P AB V V 303 ∠=

o

P BC V V 1503 ∠=o

P CA V V 903 −∠=



&' e

deg

N S

# & %


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"hase Se uence

3-"G#S$ 3-*I;$

/

!

;hase se%uence /!7

&' e

deg

N S

# % &

o L AB V V 0∠=

o

L BC V V 120−∠=o

LCA V V 120∠=


39/43

"hase Se uence

3-"G#S$ 3-*I;$

/

!

;hase se%uence !/7

o L AB V V 0∠=

o

L BC V V 120∠=o

LCA V V 120−∠=

&' e

deg

N S

# & %

where V L is the line-to-line voltage


40/43

Sin!le-"hase 3-*ire S stem

n

/N

S

N

S

n

/&F e deg

%n

#n

o P An V V 0∠= o P Bn V V 180∠=

)sin(#a* t vv P An ω = )180sin(#a*o

P Bn t vv += ω

%n

Bn An An An Bn AnnB An AB V V V V V V V V V 22 −==+=−=+=

4 /

F-pole alternatorwith 2 coils # H %

&oils # H %. in series

n

4 n

4 /n

#n #%

2

2

-

-


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:ind the currents and total power dissipation asshown in :ig. & and :ig. ' i" 4 n - 4 /n & 4.

Illustrative "roblem

',

Sin!le-"hase 3-*ire S stem

n

/

n

? & 3 2 NB

? ' 3 2 NB

/

n

n

/

n

? & 3 2NB

? ' 3 2 NB

/

n

? 3 3 2NB ? 3 3 2NB

2

2

-

-

2

2

-

-


42/43

:ind the currents and total power dissipation asshown in :ig. 3 4 n - 4 /n &> 4.

Illustrative "roblem

',

SeatworE

n

/

n

/

n

?

? B 2 N>

?

2

2

-

-


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$N OCS$SSION

Documents

Module 8 Three Phase Systems v3