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8/18/2019 Module 8 Three Phase Systems v3
1/43
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
Y-connected alternator
a
c
ba
c b
E aa’
E cc’
E bb’
a
b
c
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Introduction
Y-connected alternator
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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Line-to-line olta!e in 3-"hase Y-connected #lternators
$ #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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Line-to-line olta!e in 3-"hase Y-connected #lternators
$ #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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Line-to-line olta!e in 3-"hase Y-connected #lternators
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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Line-to-line olta!e in 3-"hase Y-connected #lternators
#&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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%alanced 3-"hase Loads
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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%alanced 3-"hase Loads
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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%alanced 3-"hase "ower
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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%alanced 3-"hase "ower
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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%alanced 3-"hase "ower
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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%alanced 3-"hase "ower
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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%alanced 3-"hase "ower
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$#ed b% the co#bined load.
c) The ' o" the co#bined load.
Z
Z Z R
R
R
A
B
C
I A
I B
I C
N
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"hase Se uence
N S
"erspective view o+ armature
N S
Side view o+ armature
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"hase Se uence
N S
"erspective view o+ armature
N S
Side view o+ armature (turned ? o )
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"hase Se uence
N S
&' e
deg
N S
# % &"erspective view o+ armature
Side view o+ armature
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"hase Se uence
N S
&' e
deg
N S
# % &
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"hase Se uence
N S
&' e
deg
N S
# % &S$ S$D1$N&$ #%& % %
V a
V b
V c
;hasor representation o"
/! se%uence
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"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
8/18/2019 Module 8 Three Phase Systems v3
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"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
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"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
/
!
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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∠=
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∠=
where V P is the line-to-neutral voltage
or the phase voltage
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 ∠=
where V P is the line-to-neutral voltage
or the phase voltage
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"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 −∠=
where V P is the line-to-neutral voltage
or the phase voltage
&' 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∠=
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"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
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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
-
-
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: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