3phase circuit
TRANSCRIPT
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Three-phase power circuits
Lecture Notes: EEE
Instructor: Dr. U D Dwivedi
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Introduction
Almost all electric power generation and most of the power
transmission in the world is in the form of three-phase ACcircuits. A three-phase AC system consists of three-phase
generators, transmission lines, and loads.
There are two major advantages of three-phase systems over
a single-phase system:1) More power per kilogram of metal from a three-phase
machine;
2) Power delivered to a three-phase load is constant at all
time, instead of pulsing as it does in a single-phase system.
The first three-phase electrical system was patented in 1882
by John Hopkinson -British physicist, electrical engineer,Fellow of the Royal Society.
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Generation of three-phase voltages
and currents
A three-phase generatorconsists of three single-
phase generators with
voltages of equal
amplitudes and phasedifferences of 1200.
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Generation of three-phase voltages
and currents
Each of three-phasegenerators can be connected
to one of three identical loads.
This way the system would
consist of three single-phasecircuits differing in phase
angle by 1200.
The current flowing to each
load can be found as
VI
Z=
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Generation of three-phase voltages
and currents
Therefore, the currents flowing in each phase are
0
0
0
0
120120
240240
A
B
A
V
I IZ
VI I
Z
VI I
Z
= =
= =
= =
(1)
(2)
(3)
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Generation of three-phase voltages
and currents
We can connect the negative (ground) ends of the threesingle-phase generators and loads together, so they share
the common return line (neutral).
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_
__
++
+
n
a
bc
c
b
a
V 0
V -120V -240
Wye Connected
Source
Wye Connected Source
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a
bc
a
b
c
+
+
+
_
_
_Delta
Source
Delta Source
Vab = | Vab| 0
Vbc
= Vab
-120
Vca= Vab-240
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Zl
Zl
Zl
ZL
ZL ZL
a
n
bc
A
B C
N
Wye Wye System
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Zl
Zl
Zl
a
bc
A
B C
ZL
ZLZ L
+
+
+
_
_
_
Delta Delta System
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Zl
Zl
Zl
a
bc
A
B C
+
+
+
_
_
_
ZL
ZL
ZL
Delta Wye System
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_
__
++
+
n
a
b
c
c
b
a
V 0
V -120V -240
A
B C
ZZ
Z
IaA ICA
IBC
IAB
Wye Delta System
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Generation of three-phase
voltages and currents
As long as the three loads are equal, the return current in
the neutral is zero!
Such three-phase power systems (equal magnitude, phase differences
of 1200, identical loads) are called balanced.
In a balanced system, the neutral is unnecessary!
Phase Sequenceis the order in which the voltages in the individual
phases peak.
abc acb
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Voltages and currents
There are two types of connections in three-phase circuits: Yand .
Each generator and each load can be either Y- or -connected. Anynumber of Y- and -connected elements may be mixed in a power system.
Phase quantities- voltages and currents in a given phase.
Line quantities voltages between the lines and currents in the lines
connected to the generators.
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Voltages and currents
1. Y-connection
Assuming
a resistiveload
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Voltages and currents
1. Y-connection (cont)
0
0
0
0
120
240
an
bn
cn
V V
V V
V V
=
=
=
0
0
0
0120
240
a
b
c
I II I
I I
=
=
=
Since we assume a resistive load:
(6)
(7)
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Voltages and currents
1. Y-connection (cont 2)
The current in any line is the same as the current in the corresponding phase.
L
I I
=
Voltages are:
0
0
0 1 3 3 30 120
2 2 2 23 1
32 2
303
aab bV V V V V V jV V j V
j V
V
V
= = = = +
= + =
(8)
(9)
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Voltages and currents
1. Y-connection (cont 3)
3LL
V V
=
Magnitudes of the line-to-line voltages and the line-to-neutral voltages are related as:
In addition, the line voltages are
shifted by 300with respect to the
phase voltages.
(10)
In a connection with abc
sequence, the voltage of a
line leadsthe phase voltage.
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Van
Vbn
Vcn- Vbn
Vab
Vab = Van - Vbn
Vab = 3 Van 30o
30
o
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( )
=
=
=
=
=
30||3
2
3
2
1||||
)120sin120(cos1||
120||0||
p
pp
p
pp
bnanab
V
jVV
jV
VV
VVV
=
=
210||3
90||3
pca
pbc
VV
VV
VoltageLine== ||3 pL VV
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Voltages and currents
2. -connection
0
0
0
0
120
240
ab
bc
ca
V V
V V
V V
=
=
=
0
0
0
0
120
240
ab
bc
ca
I I
I I
I I
=
=
=
assuming a
resistive load:
(3.14.1) (11)
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Voltages and currents
2. -connection (cont)
LLV V=
The currents are:
0 0
0
1 30 240
2 2
3 3 3 132 2 2 2
3 30
ab caa I I I I I I I
I
j I
I j I I j
= = +
= = =
=
3LI I=
(12)
(13)
(14)The magnitudes:
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Voltages and currents
For the connections with the abcphase sequences, the current of a
line lagsthe corresponding phase current by 300(see Figure below).
For the connections
with the acbphase
sequences, the line
current leadsthe
corresponding phase
current by 300.
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Power relationships
For a balanced Y-connected load with the impedance Z= Z0:
0
0
( ) 2 sin
( ) 2 sin( 120 )
( ) 2 sin( 240 )
an
bn
cn
v t V t
v t V t
v t V t
=
=
=
and voltages:
The currents can be found:
0
0
( ) 2 sin( )
( ) 2 sin( 120 )
( ) 2 sin( 240 )
a
b
c
i t I t
i t I t
i t I t
=
=
=
(3.17.1)
(3.17.2)
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Power relationships
The instantaneous power is:
( ) ( ) ( )p t v t i t=
Therefore, the instantaneous power supplied to each phase is:
0 0
0 0
( ) ( ) ( ) 2 sin( ) sin( )
( ) ( ) ( ) 2 sin( 120 ) sin( 120 )
( ) ( ) ( ) 2 sin( 240 ) sin( 240 )
a an a
b bn b
c cn c
p t v t i t VI t t
p t v t i t VI t t
p t v t i t VI t t
= =
= =
= =
(15)
(16)
Since
[ ]1
sin sin cos( ) cos( )2
= + (17)
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Power relationships
Therefore
[ ]0
0
( ) cos cos(2 )
( ) cos cos(2 240 )
( ) cos cos(2 480 )
a
b
c
p t VI t
p t VI t
p t VI t
=
=
=
The total power on the load
( ) ( )( ) 3 cos( )tot a b cp t p tp t Vp t I + + ==
The pulsing components cancel each other because of
1200phase shifts.
(18)
(19)
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Power relationships
The instantaneous
power in phases.
The total power
supplied to the load isconstant.
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Power relationships
Phase quantities in each phase of a Y- or -connection.
23 cos 3 cosP V I I Z = =
23 sin 3 sinQ V I I Z = =
23 3S V I I Z = =
Real
Reactive
Apparent
Note: these equations are valid for balanced loads only.
(20)
(21)
(22)
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Power relationships
Line quantities: Y-connection.
Power consumed by a load: 3 cosP V I =
Since for this load 3L LLI I and V V = =
Therefore: 3 cos3
LLL
VP I =
Finally: 3 cosLL LP V I =
(23)
(24)
(25)
(26)
Note: these equations are valid for balanced loads only.
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Power relationships
Line quantities: -connection.
Power consumed by a load: 3 cosP V I = (27)
Since for this load 3L LLI I and V V = = (28)
Note: these equations were derived for a balanced load.
Therefore: 3 cos3
LLL
IP V =
Finally: 3 cosLL LP V I =
(29)
(30)
Same as for a Y-connected load!
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Power relationships
Line quantities: Y- and -connection.
3 sinLL LQ V I =
3 LL LS V I=
Reactive power
Apparent power
(31)
(32)
Note: is the angle between the phase voltage and the phase
current the impedance angle.
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Analysis of balanced systems
We can determine voltages, currents, and
powers at various points in a balanced circuit.
Consider a Y-connected generator and load via
three-phase transmission line.
For a balanced Y-connected system,
insertion of a neutral does not change the
system.
All three phases are identical except of 1200
shift. Therefore, we can analyze a single
phase (per-phase circuit).
Limitation: not valid for -connections
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Analysis of balanced systems
A -connected circuit can be analyzed via the transform of impedances by theY-transform. For a balanced load, it states that a -connected load consistingof three equal impedances Zis equivalent to a Y-connected load with the
impedances Z/3. This equivalence implies that the voltages, currents, and
powers supplied to both loads would be the same.
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Analysis of balanced systems: Ex
Example 3-1:
for a 208-V three-
phase ideally balanced
system, find:
a) the magnitude of
the line current IL;b) The magnitude of
the loads line and
phase voltages VLL
and VL;
c) The real, reactive,
and the apparentpowers consumed
by the load;
d) The power factor of
the load.
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Analysis of balanced systems: Ex
Both, the generator and the load are Y-
connected, therefore, its easy to construct
a per-phase equivalent circuit
a) The line current:0 0 0
0
0
120 0 120 0 120 07.94 37.1
(0.06 0.12) (12 9) 12.06 9.12 15.12 37.1L
L load
VI A
Z Z j j j
= = = = =
+ + + + +
b) The phase voltage on the load:
0 0 0 0(7.94 37.1 )(12 9) (7.94 37.1 )(15 36.9 ) 119.1 0.2L L LV I Z j V = = + = =
The magnitude of the line voltage on the load:
3 206.3LL LV V V= =
Btw: is the load inductive or capacitive??
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Analysis of balanced systems: Ex
c) The real power consumed by the load:
03 cos 3 119.1 7.94 cos 36.9 2270load
P V I W
= = =
The reactive power consumed by the load:
03 sin 3 119.1 7.94 sin 36.9 1702 var loadQ V I = = =
The apparent power consumed by the load:
3 3 119.1 7.94 2839loadS V I VA = = =
d) The load power factor:
0cos cos 36.9 0.8loadPF lagging= = =
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One-line diagrams
Since, in a balanced system, three
phases are similar except of the
1200phase shift, power systems
are frequently represented by a
single line showing all threephases of the real system.
This is a one-line diagram.
Such diagrams usually include allthe major components of a power
system: generators, transformers,
transmission lines, loads.
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Using the power triangle
If we can neglect the impedance of the transmission line, an important
simplification in the power calculation is possible
If the generator voltage in the system is known,
then we can find the current and power factor at
any point in the system as follows:
1. The line voltages at the generator and the loads
will be identical since the line is lossless.
2. Real and reactive powers on each load.
3. The total real and reactive powers supplied to
all loads from the point examined.
4. The system power factor at that point using thepower triangle relationship.
5. Line and phase currents at that point.
We can treat the line voltage as constant and use the power triangle method to
quickly calculate the effect of adding a load on the overall system and power factor.