ECE201 Lect-61 Series and Parallel Resistor Combinations (2.5, 8.5) Dr. Holbert February 6, 2006.
Chap 4 Three Phase Circuits Karady-Holbert FINAL 2 Slides 97
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Transcript of Chap 4 Three Phase Circuits Karady-Holbert FINAL 2 Slides 97
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EEE 360Chapter 4:
Three-Phase Circuits
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Three-Phase Circuits
Practically all electrical energy generation andtransmission systems use a three-phase circuit.
The three-phase energy is transported through
three or four conductors to large customers. Only the small household and light commercial
loads are supplied by a single phase.
The major advantage of the three-phase systemis the efficiency of power transmission.
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Three-Phase Circuits
Another advantage is generation of constanttorque, which reduces vibration, for rotating
machines. This is particularly important for
industries with large motors.
A three-phase transmission line carries three
times the power of a single-phase line while
requiring practically the same right-of-way.
Polyphase systems (six-phase and twelve-phase) have been tried but proved to be un-
economical for power transmission but are used
for large rectifiers.
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Three-Phase Circuits
Three-Phase QuantitiesBasic Definitions
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Basic Definitions
In this chapter only the balanced th ree-phasesystems are presented.
A three-phase system has three sinusoidal
voltages sources.
In a balanced system, each voltage source has
the same magnitude (V M ) and frequency (ω),
and each is 120° out-of-phase with the other two
240240cos)(
120120cos)(
0cos)(
M M cn
M M bn
M M an
V t V t v
V t V t v
V t V t v
cn
bn
an
V
V
V
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Basic Definitions
When Vbn lags Van by 120° and Vcn lags Vbn by120°, then the system is said to have a positive
phase sequence, or also called an abc phase
sequence.
The phase sequence describes the order in
which the phase voltages reach their maximum
(peak) values with respect to time.
If Vcn and Vbn lag Van by 120° and 240°,respectively, the system has a negative (or acb)
phase sequence.
This class presents only the positive sequence
system.
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Basic Definitions
A balanced three-phase circuit is one in whichthe loads are such that the currents produced by
the voltages are also balanced.
The balanced load currents are described by the
equations below
where the voltage of each phase leads its
corresponding current by an angle of θ
240240cos)(120120cos)(
cos)(
M M c
M M b
M M a
I t I t i I t I t i
I t I t i
c
b
a
II
I
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Basic Definitions
The sum of the balanced voltages and the sumof the balanced currents equal zero
The instantaneous power in each phase is the
product of voltage and current. The total
instantaneous power is
Hence, the instantaneous three-phase power is
constant over time.
0)()()(
0)()()(
t i t i t i
t v t v t v
c ba
cnbnan
)cos(3cos2
3)()()()( rmsrmsM M
c baT I V I V t pt pt pt p
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Basic Definitions
The sinusoidal varying voltages and currents canbe represented by complex phasors.
Vcn
Vbn
Van
Ia
Re
Im
120°
120°
θ
Ib
Ic
120°
120°
Vbn
Vcn
Van + Vbn + Vcn = 0
Ic
Ib
Ia + Ib + Ic = 0
Figure 4.1 Phasor diagram for a
balanced three-phase circuit
with a positive phase sequence
Figure 4.1 presents the
phasor diagram of athree-phase balanced
system.
The phase shift is –120°
The sums of the voltage
vectors and current
vectors are zero.
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Basic Definitions
Similarly the total three-phase complex poweris three times the complex power of any of the
phases: ST = S A + SB + SC = 3 S1
Figure 4.2 shows the sinusoidal voltage and
currents as well as the instantaneous power
variation in time.
The frequency of the voltage and current is 60
Hz, but the power of each phase varies at afrequency of 120 Hz.
The sum of the three instantaneous powers is
constant power .
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Basic Definitions
Figure 4.2 Time-varying voltage, current and power
in a balanced three-phase circuit.
0 5 10 15 20 25 30 35 40-500
0
500
Voltage(kV) Three-Phase Balanced System at 60 Hz, and V Leads I by 45°
0 5 10 15 20 25 30 35 40-1
0
1
Current(kA
)
0 5 10 15 20 25 30 35 40-100
0100
200
300
Time (ms)
Power(
MW)
Va
V b
Vc
Ia
I b
Ic
Pa
P b
Pc
Ptotal
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Three-Phase Circuits
Three-Phase QuantitiesDelta –Wye Load Connections
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Delta-Wye Connections
A balanced three-phase load can be connectedin a delta ( Δ) or in a wye (Y).
Figure 4.3 shows wye-connected loads.
Z Y
Z Y
Z Y
a
c
b
n Load
Z Y
Z Y Z Y
a
b
c
Load
n
Figure 4.3 Wye-connected loads.
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Delta-Wye Connections
Figure 4.4 shows delta-connected loads; the seriesand parallel impedance combination techniques
can not be used for a delta connected load.
Z
Z
Z
a
c
b
Load
Z
Z
Z
a
b
c
Load
Figure 4.4 Delta-connected loads.
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Delta-Wye Connections
The analysis and simplification of a delta-connectedsystem requires the transformation of the delta-
connected impedances to wye-connected impedances.
Figure 4.5 shows the concept of delta –wye impedance
transformation.
Zc
Z b Z a
a b
c
Z 2 Z 3
Z1
a b
c
Y-
Figure 4.5 Delta –
wye impedance transformation
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Delta-Wye Connections
The equations for the delta to wye transformationare
321
32
321
31
321
21
ZZZ
ZZZ
ZZZ
ZZZ
ZZZ
ZZZ
c
b
a
These relations can
be read as the
impedance next to aparticular Y node is
the product of the two
delta impedances
connected to that
node divided by the
sum of the three delta
impedances.
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Delta-Wye Connections
The reverse transformation (Y to Δ) can beperformed by the equations below
For balanced load, when the impedances are
equal
a
ac c bba
b
ac c bba
c
ac c bba
Z
ZZZZZZZ
Z
ZZZZZZZ
Z
ZZZZZZZ
3
2
1
YΔ
ZZ 3
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Three-Phase Circuits
Three-Phase QuantitiesWye Connected Generator
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Wye Connected Generator
Figure 4.6 Three-phase wye-connected generator
n
a
Vab
Ia
Van
VbnVcn
bIb
Icc
Vbc
Vca
Figure 4.6 exhibits a three-phase generator connectedin a wye (Y) configuration.
The generator is represented by three voltage sources.
The magnitudes of the ac
source voltages are the
same.
The phase shift between the
voltages is 120º.
This system has three phase
conductors (a, b and c) and
a grounded neutral (n).
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Wye Connected Generator
The voltage of phase a (Van) is selected as thereference with a phase angle of .
The phasor voltage expressions are
where V P is the phase voltage magnitude between
the phase conductors and neutral, and is called thel ine-to-neutral vo ltage .
An important property of this balanced voltage set is
240
120
P
P
P
V
V
V
cn
bn
an
V
V
V
0cnbnan
VVV
V P is the rms valueof the voltage
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Wye Connected Generator
Figure 4.6 Three-phase wye-connected generator
Figure 4.6 shows the three line-to-neutral voltages aswell as the three l ine-to- l ine voltages (Vab, Vbc, Vca).
The line-to-line voltage is calculated using KVL.
Figure 4.6, the KVL relationfor the n→a→b→n loop is
The line-to-line voltage from
this equation is
0bnaban
VVV
bnanab VVV
n
a
Vab
Ia
Van
VbnVcn
bIb
Icc
Vbc
Vca
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Wye Connected Generator
We substitute for Vbn in terms of Van to define thel ine-to -l ine vo ltage
The line voltage (Vab) leads the corresponding
phase voltage (Van) by 30 degrees.
The line-to-line voltage magnitude is times
larger than the line-to-neutral (phase) voltage.
303
1
120120
an
an
anan
bnanab
V
V
VV
VVV
j e
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Wye Connected Generator
The relations for the three-phase line-to-linevoltages and the line-to-neutral voltages are
The abc phase sequence exhibits the ±120°
phase shift between the three line voltages.
2103303
903303
303
ancnca
anbnbc
anab
VVV
VVV
VV
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Wye Connected Generator
Figure 4.7 Comparison of the relative magnitudes of and the
phase shift between the line, vab(t), and phase, van(t), voltages
t 0 0. 1m s 30m s
0 10 20 30200
100
0
100
200
van t( )
vab t( )
t
m s
Figure 4.7 shows the relative magnitudes of theline and phase voltages and the phase shift
between them. Figure shows that:
• The amplitude of the
line-to-line voltageis larger than the
line-to-neutral
voltage.
• The line-to-neutralvoltage is 30°
behind the line-to-
line voltage.
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Wye Connected Generator
Figure 4.8 Phasor diagram of a
wye-connected balanced system.
30°120°
Vca
Vab
Vbc
Vbn Van
Vcn
–Vbn
Re
Im The Figure 4.8 phasor diagram of the voltages
shows that:
The phase shift between
the line-to-line and line-to-neutral voltages is 30°
(e.g., Van lags Vab by 30°)
The difference of the two
line-to-neutral voltages is
the line-to-line voltage
(e.g., Vab = Van – Vbn)
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Three-Phase Circuits
Three-Phase QuantitiesWye Connected Loads
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Wye Connected Loads A wye-connected generator can be loaded by three
impedances connected in a wye configuration, as seen inFigure 4.9.
Most high-voltage transmission lines use a three-wire
grounded system.
When the neutral point at both the supply and the load aregrounded, the earth interconnects the two neutral points.
Van
Vbn Vcn
Za
Z c Z b
Van
Vbn Vcn
Za
Z c Z b
(b) Four-wire system(a) Three-wire system
Figure 4.9 Three and four wire wye-connected systems
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Wye Connected Loads
The earth acts a conductor and transforms the three-wire
system to a practical four-wire system.
In the distribution level, a true four-wire system is
frequently used. The neutral wire is insulated, although it
is grounded at the transformer sites.
At low voltage the larger industrial complexes use a three-
wire 460 V system for larger motor loads.
The smaller loads are supplied by a 208 V four-wire
system, where
The lighting and small loads are connected between
the phases and the neutral wire at 120 V.
The larger loads are connected between the phases
and supplied by 208 V.
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Wye Connected Loads
Figure 4.10 shows a wye-connected generator loaded bythree impedances connected in a wye.
Three phase conductors (a, b, c) connect the load
impedances (Za, Zb, Zc) to the sources, and the neutral
conductor connects the neutral point of the sources to
the neutral point of the load.
Figure 4.10 Three-phase wye-connected generator with an impedance load
Van
Vcn
Vbn
n
a
b
c
Ic
Ia
Ib
Vab
VbcVca
I0
Za
Zb
Zc
The neutral point of the
sources is grounded to assure
that the potential of the phase
conductors to the groundremains constant.
This is an important safety
consideration.
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Three-Phase Circuits
Three-Phase QuantitiesWye Connected Four-wire
System
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Wye Connected Loads
For the four-wire circuit, the source voltages areconnected directly to the load impedances and the
neutral conductor provides a return path for each phase
current.
Each source line-to-neutral voltage drives a current
through the corresponding impedance.
Figure 4.10 Three-phase wye-connected generator with an impedance load
Figure 4.10 reveals that the
current in each phase can be
calculated by dividing theappropriate line-to-neutral
voltage by the load impedance
in that phase.
Van
Vcn
Vbn
n
a
b
c
Ic
Ia
Ib
Vab
Vbc
Vca
I0
Za
Zb
Zc
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Wye Connected Loads
With a balanced load, the phases do not affect oneanother; it is sufficient to calculate the current in phase a
using Ohm’s Law in the circuit of Figure 4.10
Figure 4.10 Three-phase wye-connected generator with an impedance load
Since Za = Zb = Zc = Z Y, the line
currents in phases b and c areidentical in magnitude to Ia, but
lagging by 120° and 240°,
respectively.
Using this fact and applying KCLat the neutral node yields
a
an
a
Z
VI
0cba0
IIII
Van
Vcn
Vbn
n
a
b
c
Ic
Ia
Ib
Vab
VbcVca
I0
Za
Zb
Zc
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Wye Connected Loads
The complex power in the load of phase a is theproduct of the voltage and the conjugate of the
current:
The total three-phase complex power is threetimes the single-phase power
*
aana IVS
aT SS 3
LoadP3ph_load / 3
Van
pf Load
Ia
This implies that athree-phase circuit
can be represented by
an equivalent single-
phase circuit.
Figure 4.11 Single-phase
equivalent of a three-phasesystem with a balanced load
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Three-Phase Circuits
Three-Phase QuantitiesWye Connected Three-wire
System
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Wye Connected 3-wire System
Three wye-connected sources supply three impedancesalso connected in a wye, as shown in Figure 4.12.
The neutral point (g) of the generator is grounded. But the
supply and load neutral points are not connected together.
This eliminates the direct return path.
Za
Zb
Zc Vngg n
a a
b bc c
Va
Vb Vc
Ia
IbIc
Figure 4.12 Three-wire wye-connected system
The unequal line
currents generate a
voltage difference (Vng)
between the neutralpoint of the generator
and the neutral point (n)
of the load impedances.
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Wye Connected 3-wire System
The system contains three lines, which areg→a→n, g→b→n and g→c→n.
The currents are calculated from the loop
voltage equations (KVL).
As an example in loop
g→a→n→g, the
difference of the line-
to-neutral voltage and
the voltage betweenthe neutral points (Vng)
drives the current
though impedance Za.
Za
Zb
Zc Vngg n
a a
b bc c
Va
Vb Vc
Ia
IbIc
Figure 4.12 Three-wire wye-connected system
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Wye Connected 3-wire System
The loop equations produced line currents are
c
ngc
c
b
ngb
b
a
nga
aZ
VVI
Z
VVI
Z
VVI
Za
Zb
Zc
Vng
g n
a a
b bc c
Va
Vb Vc
Ia
IbIc
Figure 4.12 Three-wire wye-connected system
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Wye Connected 3-wire System
The sum of the currents in node point n is zero.
The node point equation (KCL) for node n is
The voltage
difference
between the
neutral points
is calculated
from this
equation.
0
c
ngc
b
ngb
a
nga
Z
VV
Z
VV
Z
VV
Za
Zb
Zc Vngg n
a a
b bc c
Va
Vb Vc
Ia
IbIc
Figure 4.12 Three-wire wye-connected system
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Wye Connected 3-wire System
After algebraic manipulation, the voltagedifference between the neutral points is
The currents are calculated by substituting theobtained voltage difference in the earlier current
equations:
cba
c
c
b
b
a
a
ng
ZZZ
Z
V
Z
V
Z
V
V111
c
ngc
c
b
ngb
b
a
nga
a
Z
VVI
Z
VVI
Z
VVI
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Wye Connected 3-wire System
For balanced load the equations are simplified by
substituting Za = Zb = Zc = Z Y in the voltage
difference equation
But the sum of the three phase voltages is zero,
which implies that in case of a balanced load the
voltage difference between the neutral points is
also zero.
33
1
111
cba
Y
cba
Y
cba
c
c
b
b
a
a
ng
VVV
Z
VVVZ
ZZZ
ZV
ZV
ZV
V
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Three-Phase Circuits
Three-Phase QuantitiesNumerical Example for Wye
Connected Balanced System
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Balanced Wye System Example
Example: Calculate the currents in a balanced three-phase system, when the wye connected generator
supplies the wye-connected load of 3 kW with a
power factor of 0.8 lagging, and the generator line-to-
line voltage is 480 V. The balanced three-phase Y-Y system is represented
by a single-phase equivalent circuit.
The circuit is energized by the line-to-neutral voltage.
LoadP 3ph_load / 3
Van
pf load
Ia This circuit carries
one-third of the totalthree-phase power.
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The system data are
The line voltage is 480 V, so the line-to-neutral is
The single-phase load power is one-third of the
three-phase load power:
. pf P load load ph lagging80W3000 _ 3
W10003
3
1
ph_load
ph
P P
Balanced Wye System Example
. V12773
V480
anV
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The single-phase complex load power is
The current in phase a is
36.87 A4.51
A70626083
V1277V·A7501000
. j.
. j
*
an
1pha
VSI
Balanced Wye System Example
LoadVan
Ia
V·A7501000)8.0(cosexp80W1000
)(cosexp
1
11
j j .
pf j pf
P load
load
ph
1phS Lagging pf
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With a balanced positive sequence system, thecurrents in phases b and c are simply lagging
phase a by 120° and 240°, respectively
13.83 A514
87276 A514)2401)(87.36 A514(
87156 A514)1201)(87.36 A514(
240
120
.
...e
...e
j
j
ac
ab
II
II
Balanced Wye System Example
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Three-Phase Circuits
Three-Phase QuantitiesNumerical Example for Wye
Connected Unbalanced System
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Unbalanced Wye System Example
Calculate the currents in an unbalanced four-wire Y-Y system.
The loads are
13
712
510
c
b
a
Z
Z
Z
j
j
Van
Vcn
Vbn
n
a
b
c
Ic
Ia
Ib
Vab
Vbc
Vca
I0
Za
Zb
Zc
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The line-to-neutral (phase) voltage is V P = 120 V
The line-to-neutral phasor voltages are
024V120
012V120
0V120
cn
bn
an
V
V
V
Unbalanced Wye System Example
Van
Vcn
Vbn
n
a
b
cIc
Ia
Ib
Vab
Vbc
Vca
I0
Za
Zb
Zc
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Using Ohm’s Law, the phase a current is
626 A7310
A8469
)510(
V120
..
. j .
j a
an
aZ
VI
Unbalanced Wye System Example
Van
Vcn
Vbn
n
a
b
c
Ic
Ia
Ib
Vab
Vbc
Vca
I0
Za
Zb
Zc
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Similarly, the phase b current is
The phase c current is
A63880390789 A6388
3.303.891
120V120
712
120V120
. j...
j
b
bnb
Z
VI
A99476154
0.240 A2319
13
240V201
. j.
.
c
cnc
Z
V
I
Unbalanced Wye System Example
Van
Vcn
Vbn
n
a
b
c
Ic
Ia
Ib
Vab
Vbc
Vca
I0
Za
Zb
Zc
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Using KCL, the neutral conductor current is
As expected, the
neutral current isnot zero for the
unbalanced four-
wire system.
347 A4087
A44450245
)99476154()63880390()8469(
..
. j.
. j.. j.. j.
cba0 IIII
Unbalanced Wye System Example
Van
Vcn
Vbn
n
a
b
c
Ic
Ia
Ib
Vab
Vbc
Vca
I0
Za
Zb
Zc
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Three-Phase Circuits
Three-Phase QuantitiesDelta Connected Generators
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Delta Connected Generators
Figure 4.13 depicts a three-phase generatorconnected in a delta ( Δ).
The voltage between terminals ab, bc and ca
are the line-to-line voltages.
240
120
L
L
L
V
V
V
ca
bc
ab
V
V
V
Figure 4.13 Delta-connected generators
The delta source
line voltages areVab
a
bc
Vca
Vbc
Vab
a
b
c Vbc
Vca
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Delta Connected Generators
For a balanced three-phase system, themagnitudes of these line voltages are equal, and
the phase shift between them is 120º.
The voltage between phase ab is typically
selected as the reference with a zero phaseangle (ψ = 0).
The delta-connected generator can be
converted to an equivalent wye-connectedsupply consisting of three voltage sources, as
illustrated in Figure 4.13.
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Delta Connected Generators
The voltages of the equivalent wye-connectedsources are the line-to-neutral voltages, which
are calculated from the line-to-line voltages
Vab
a
bc
Vca
Vbc
Vab
a
b
cVbc
Vca
Vab
a
b c
Vca
Vbn
Van
a
b
n
Vbc
Vcn
c
Figure 4.13 Delta-connected
generators and the equivalent
wye-connected system.
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Delta Connected Generators
The equations for the calculation of the equivalentline-to-neutral voltages from the line-to-line voltages
are
The practical use of this conversion is that we can
assume either a wye or a delta configuration for the
supply depending on the load.
In the case of a delta load, the delta supply
connection is more advantageous.
In the case of a wye load, the assumption of a wye-
connected source simplifies the calculation.
303030
333
j j j eee ca
cn
bc
bn
ab
an
VV
VV
VV
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Three-Phase Circuits
Three-Phase QuantitiesDelta Connected Loads
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Balanced Delta Load
Figure 4.14 diagrams a delta-connected load in whichthe line-to-line (source) voltages are directly applied to
the load if the line impedance is negligible.
Figure 4.14 Delta connected load.
The calculation of only the
load current ab issufficient in the case of a
balanced load.
The currents in phases bc
and ca will have the sameamplitude and relative
phase angle as Iab, but
will be shifted by ±120º.Ic
Ib
Ia a
b c
Vca
b
c
a
Vab
Vbc Ibc
Iab
Ica
Zbc
Z a b
Z c a
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Delta Connected Loads
The current through impedance Zab is computed usingOhm’s Law:
To determine the line currentIa, we employ KCL at node a:
Substituting the Ohm’s Lawexpressions for the phase
currents yields
ab
ab
abZ
VI
caaba III
ca
ca
ab
ab
aZ
V
Z
VI
Ic
Ib
Ia a
b c
Vca
b
c
a
Vab
Vbc Ibc
Iab
Ica
Zbc
Z a b Z c
a
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Delta Connected Loads
For balanced load and voltage, the equation issimplified
The line current is times larger than the delta
load (phase) current.
303
1
24011
240
240
ab
ab
Δ
ab
Δ
ab
Δ
ab
a
I
I
Z
V
Z
V
Z
VI
j e
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Delta Connected Loads
The voltage Vab and current Iab for Zab obey the
passive sign convention, and the complex power
is thus
The total three-phase complex power is three
times the single-phase power
*
ababab IVS
C
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Delta Connected Loads
The phasor diagram shows
the generated voltages and
currents.
The chart visualizes that in
a delta system there are
two currents (phase and
line) and one voltage (the
line-to-line voltage).
The line current is times
the phase current and the
phase shift is –30º.
Ic
Vca
Ica
Ibc
Vbc
Vab Re
Im
Ib
Ia
Iab
–Ica
30°
f
Figure 4.15 Phasor diagram of a
delta-connected balanced system.
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Three-Phase Circuits
Three-Phase QuantitiesExample for Delta Connected
Balanced Load
Delta Connected Balanced
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Delta Connected Balanced
Load Example
Calculate the line and phase currents in abalanced delta connected three-phase system,
when the sources are loaded by a three-phase
load of 3 kW with a power factor of 0.8 lagging
Vab
a
bc
Vca
Vbc
Vab
a
b
cVbc
Vca
a
b c
Ica
Ia
Ic
Ib
Iab
Ibc
Delta Connected Balanced
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Delta Connected Balanced
Load Example
The system data are
Phase ab line-to-line voltage is the reference
voltage with zero phase shift: Vab = 208 V
Phase ab carries one-third of the total three-
phase load and the line-to-line voltage supplies
this load
V208,8.0W,30003 _ Lload phload V pf P
W10003
W3000
3
3 _
1
phload
ph
P P
Delta Connected Balanced
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Delta Connected Balanced
Load Example
The complex power for each phase is
VA7501000
)8.0(cosexp8.0
W1000
)(cosexp
1
11
j
j
pf j pf
P load
load
ph
1phS
Delta Connected Balanced
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Delta Connected Balanced
Load Example
The load current in the delta ab branch is
For balanced conditions, the other delta branchcurrents are
8736 A016 A6384
V208
VA7501000... j.
j*
ab
1ph
abV
SI
1383 A016
240
87156 A016
120
..
..
abca
abbc
II
IIVab
a
b
c
Vbc
Vca
a
bc
Ica
Ia
Ic
Ib
Iab
Ibc
Delta Connected Balanced
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Delta Connected Balanced
Load Example
The line currents are
Was there an easier way to find Ib and Ic?
53.13 A41.10 A327.8245.6
)361.2527.5()967.5719.0(13173 A4110 A245133410
)60638084()36125275(
8766 A4110 A57290894 )96757190(
)60638084(
j
j j ... j.
. j.. j.
... j.. j.
. j.
bccac
abbcb
caaba
III
III
IIIVca
a
bc
Ica
Ia
Ic
Ib
Iab
Ibc
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Three-Phase Circuits
Three-Phase QuantitiesExample for Delta Connected
Unbalanced Load
Delta Connected Unbalanced
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e ta Co ected U ba a ced
Load Example
A delta connected generator, with a 208 V line-to-line voltage supplies unbalanced delta
connected loads.
Calculate the phase and line currents, and thesupply complex and
active powers.
The figure
shows theequivalent
circuit.
Vab
a
bc
Vca
Vbc
Vab
a
b
c
Vbc
Vca
a
bc
Ica
Ia
Ic
IbIab
Ibc
Delta Connected Unbalanced
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Load Example
The load impedances are
Zab = 23 Ω Zbc = 22 + j 15 Ω
Zca = 25 – j 22 Ω
The reference line voltage is Vab = 208 V
Vab
a
bc
Vca
Vbc
Vab
a
b
c
Vbc
Vca
a
bc
Ica
Ia
Ic
Ib
Iab
Ibc
Delta Connected Unbalanced
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Load Example
The three-phase equivalent circuit shows thateach line-to-line voltage is directly applied to the
corresponding impedance
This implies that the current through eachimpedance can be calculated by dividing the
appropriate
line-to-line
voltage by thecorresponding
impedance
(Ohm’s Law).
Vab
a
bc
Vca
Vbc
Vab
a
b
c
Vbc
Vca
a
bc
Ica
Ia
Ic
Ib
Iab
Ibc
Delta Connected Unbalanced
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Load Example
The line-to-line voltages drive a different currentthrough each impedance.
The line-to-line voltages are
V1.180104
240V208
V1.180104
120V208
V208
j
j
ca
bc
ab
V
V
V
Delta Connected Unbalanced
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Load Example
The phase currents through the impedances arecalculated using Ohm’s Law.
161.3 A246.6 A.9981918.5
2225
V1.180104
3.154 A812.7
A389.3038.7
1522
V1.180104
A044.9 23
V208
j
j
j
j
j
j
ca
ca
ca
bc
bc
bc
ab
ab
ab
Z
VI
Z
VI
Z
VI
Vab
a
b
c
Vbc
Vca
a
bc
Ica
Ia
Ic
Ib
Iab
Ibc
Delta Connected Unbalanced
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Load Example
The currents in the conductors connecting thegenerator to the loads are the line currents.
The line currents are calculated with the node
point equations (KCL).
Vca
a
bc
Ica
Ia
Ic
Ib
Iab
Ibc
6.7 A09.15 A.00296.14
).9981918.5(044.9
j
j
caaba III
Delta Connected Unbalanced
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Load Example
The other two line currents are
3.78 A50.5
A39.512.1
)042.9038.7().9981918.5(
168.1 A43.16
A.39308.16
044.9)389.3038.7(
j
j j
j
j
bccac
abbcb
III
III
Vca
a
bc
Ica
Ia
Ic
Ib
Iab
Ibc
Delta Connected Unbalanced
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Load Example
The complex powers of the phases are
VA858975
3.41VA1299)3.161 A246.6)(120V208(
VA9151343
34.3VA1625 )3.154 A812.7)(120V208(
VA1881) A044.9)(V208(
*
*
j
j
*
cacaca
*
bcbcbc
*
ababab
IVS
IVS
IVS
Delta Connected Unbalanced
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Load Example
The three-phase (supply) complex power is thesum of the complex power of each phase
The total apparent power is 4199 VA
The total real power is 4199 W
8.0VA4199
VA754199
)858975()9151343(1881
j
j j
cabcab3ph SSSS
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Three-Phase Circuits
Three-Phase PowerMeasurement
Three-Phase Power
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Three Phase Power
Measurement
Power measurements generally require
wattmeters.
An analog wattmeter has two inputs: current
and voltage. The current input is connected in series with
the load.
The voltage input is in parallel with the supplyor load.
The wattmeter provides a measurement of the
average power (P ).
Three-Phase Power
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Measurement A voltmeter and an ammeter are used in
conjunction with an analog wattmeter todetermine the rms values of the voltage andcurrent (V rms and I rms).
Digital power meters sample the voltage andcurrent waveforms, i.e., v (t ) and i (t ).
A microprocessor calculates the rms values ofthe voltage and current.
The instantaneous voltage and current readingsare multiplied to calculate the instantaneouspower, p(t ).
The average value of p(t ) gives the real power.
Three-Phase Power
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Measurement
The measured data permit the calculation of thepower factor ( pf ) and the apparent power (|S|).
The magnitude of the complex power, that is, the
apparent power, is
The power factor is
rmsrms I V
*IVS
S
P pf
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Three-Phase Circuits
Three-Phase PowerMeasurement
Four and Three Wire Systems
Four Wire System
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Four-Wire System
Figure 4.21 Measurement connections
for a four-wire, three-phase system.
A
A
V
W
W
Phase a
Phase b
Phase c
Van
Ia
IcV
A
V
W
Ib
Vbn
Vcn
Neutral (n)
P a
P b
P c
In a four-wire system, the loads can beconnected between the phase conductors, or
between the phase conductors and the neutral.
For unbalanced
loading, themeasurement
requires three
wattmeters to
measure each
phase separately
Figure 4.21 shows the
connection diagram
Four Wire System
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Four-Wire System
The measurement system provides a current, avoltage, and a power reading for each of the
three phases
The sum of the
measured threepowers gives the
total three-phase
power
c baT P P P P
Figure 4.21 Measurement connections
for a four-wire, three-phase system.
A
A
V
W
W
Phase a
Phase b
Phase c
Van
Ia
IcV
A
V
W
Ib
Vbn
Vcn
Neutral (n)
P a
P b
P c
Three Wire System
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Three-Wire System
Figure 4.22 Measurement
connections for a three-wire,
three-phase system.
In a three-wire system, the loads are connectedbetween the phase conductors.
The sum of the three line currents is zero, which
permits the measurement of the three-phase
power with just two wattmeters.
Figure 4.22 sketches the
connection diagram for a
three-wire system.
A
A
V
V
W
W
Phase a
Phase b
Phase c
Vab = Va – Vb
Vcb = Vc – Vb
Ia
Ic
P ab
P cb
Three Wire System
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Three-Wire System
The current of phase a and the voltage betweenphases a and b feed one of the wattmeters.
The current of phase c and the voltage between
phases c and b supply the other wattmeter.
The wattmeter multiplies
the voltage and current
values, and determines
the average powervalue.
Figure 4.22 Measurement
connections for a three-wire,
three-phase system.
A
A
V
V
W
W
Phase a
Phase b
Phase c
Vab = Va – Vb
Vcb = Vc – Vb
Ia
Ic
P ab
P cb
Three-Wire System
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Three-Wire System
The readings of the wattmeters can beexpressed in complex notation.
The power is the real value of the voltage timesthe current conjugate.
The reading of the wattmeter that is connectedto current Ia and voltage Vab is
The reading of the wattmeter that is connectedto current Ic and voltage Vcb is
*
ab
*
aa
*
aba
*
aab IVIVIVVIV ReReReabP
*
cb
*
cc
*
cbc
*
ccb IVIVIVVIV ReReRecbP
Three-Wire System
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Three-Wire System
We shall find that the sum of the two wattmeterreadings is the total three-phase real power:
The substitution of the two power values results
in
Simplification of the expression yields
cbabT P P P
*
cb
*
cc
*
ab
*
aa IVIVIVIV ReReT P
*
c
*
ab
*
cc
*
aa IIVIVIV Re
T P
Three-Wire System
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Three-Wire System
The sum of the three line currents is zero suchthat
Substituting the current formula into the power
equation gives
The real value of the components can be
separated individually, which results in:
*
c
*
a
*
bcab IIIIII or
*
bb
*
cc
*
aa IVIVIV Re
T P
cbabc ba
T
P P P P P
P
*
cc
*
bb
*
aa IVIVIV ReReRe
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Three-Phase Circuits
Three-Phase PowerExample for Three-Phase
Transmission
Three-Phase Transmission Example
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Three-Phase Transmission Example
Figure 4.23 Generator supplies a load
through a short transmission line.
Example: A three-phase generator supplies a loadthrough a transmission line.
Figure 4.23 shows the equivalent circuit.
Calculate the generator terminal and excitation
voltages together with the voltage regulation.
The load data are
active power: 40 MW
load voltage: 22 kV
power factor: 0.8
(lagging)
Egen P load /3
Transmission LineGenerator Load
X g Zline
VgenIload
Three-Phase Transmission Example
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The transmission line impedance is
The generator rating is
The latter value ( x g ) is the per unit value of the
generator impedance per phase, which must be
converted from percent to ohms via
915.042.0 j line
Z
125%kV22MVA50 g g g x V S
12.1VA1050
V)(220001.25)(
6
22
g
g
g g S
V x X
Three-Phase Transmission Example
Three-Phase Transmission Example
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Calculate the load current using the power perphase and the line-to-neutral voltage since the
given load voltage is the line voltage
36.9kA31.1
A7871050
)8.0(acosexp
)8.0(3
V000,22
3)W1040(
)(acosexp
3
3
6
j
j
pf j
pf V
P load
load load
load loadI
Three-Phase Transmission Example
Eq. (3.38)
Three-Phase Transmission Example
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The generator excitation voltage is the sum ofthe load line-to-neutral voltage and the voltage
drop on the line and generator impedances.
The calculation
results in:
7.29kV9.26V335,13386,23
)1.12915.042.0)( A7871050(3
V000,22
)(
3
j
j j j
j X g lineloadload
gen ZIV
E
Three Phase Transmission Example
Egen P load /3
Transmission LineGenerator Load
X g Zline
Vgen
Iload
Three-Phase Transmission Example
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The generator terminal voltage is the sum of theload line-to-neutral voltage and the voltage drop
on the line and generator impedances.
The calculation
gives:
kV0.243
2.6kV88.13kV63.086.13
)915.042.0)(9.36kA31.1(3kV22
3
gengen_ll
lineloadload
gen
VV
ZIV
V
j
j
Three Phase Transmission Example
Egen P load /3
Transmission LineGenerator Load
X g Zline
Vgen
Iload
Generator line-to-line voltage
Three-Phase Transmission Example
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Compute the voltage regulation. The generatorregulates the voltage such that in no-load the
load voltage will be the rated generator voltage.
The voltage regulation is
9%%)100(kV22
kV22kV24Reg
load
loadgen_ll
V
VV
Three Phase Transmission Example
Egen P load /3
Transmission LineGenerator Load
X g Zline
Vgen
Iload
Why isn’t the
no-load voltage
l t E ?