Download 05-Three Phase Analysis

1/16/2013
Overview
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05-Three Phase Analysis
Text: 2.4 – 2.7
Three-Phase Sources
Delta and Y Connections
Three-Phase Loads
Three-Phase Power
Three-Phase Analysis
Per-Phase Analysis
ECEGR 451
Power Systems
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Three-Phase Voltages
Van Vbn Vcn
120 deg.
Three Phase
vector diagram
Vcn
150
Vcn 120
100
Voltage (V)
50
0
Van
-50
Van 0
• Power systems use 3phase
• We are concerned with
balanced 3-phase
• Balanced circuit
conditions:
 impedances are equal
for each phase
 voltage source phasors
have equal magnitude
and have a 120 deg.
phase shift
 a, b, c phase rotation
-100
-150
0
0.01
0.02
0.03
time (s)
0.04
Vbn
0.05
Vbn
120
Note: instantaneous
values sum to 0
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Vcn 120
3
Van 0
Vbn
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Benefits of Three-Phase Systems
Three Phase
• Efficient use of conductors over three, single
phases
• Rotating field is needed for some loads (eg three
phase motors)
• Effective for power transfer
• Per-phase analysis can be used in many cases
• Constant power delivery to three phase loads
Three phase generators connected as Y or
a
a
Vc
Va
Vc
c
n
Vb
c
b
Y (Wye)
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120
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Va
Vb
b
Delta
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Three Phase Voltage
Three Phase Voltage
• Phase voltage: the voltage across each source
(Va, Vb and Vc)
• Line voltage: the voltage across the lines of the
sources (also called “Line-to-Line Voltage”)
For Y-connected sources:
Phase Voltages
Van
Vbn
Vb
Vcn
Vc
Line Voltages
Van Vbn Van ( 3 30 )
Vab
Va
by KVL
Vbc
Vbn
Vcn
Vbn ( 3 30 )
Vca
Vcn
Van
Vcn( 3 30 )
a
a
Va
Vc
Vc
Vb
n
a
+
Va
Va
c
Vb
Y
Vc
b
n
Van
-
Vb
c
b
7
c
-
Va
Vbc
Vb
Vca
Vc
Line Voltages
Vab
Do not confuse phase
voltage with
“per-phase voltage” as
discussed later
a
++
Va Van
- + b
Delta-connections:
phase voltages =
line voltages
Vc
-
Va
Va
Vbc
Vb
Vca
Vc
Vb
Vc
0
Van
Vbn
Vcn
0
Vab
Vbc
Vca
0
+
Vb
+
+
Vab
-
-
-
Vca
b
+
V
- bc
-
Va
Vc
•
+
Vc
n
Vb
Va
-
+
c
+
Vb
9
+
b
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Three Phase Current
•
adding vectors
results in 0
Vcn
a
Va
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•
Vbn
Van
a
+
c
Vb
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Voltages (line or phase) sum to zero
Phase Voltages
+
-
b
Three Phase Voltage
For Delta-connected sources:
Vc
Vb
n
c
Three Phase Voltage
-
Vab
Vc
Delta
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Vab
+
Va
Three Phase Current
Phase current: current flowing through the voltage sources
in a three-phase circuit
Line current: current flowing from a three-phase source to
a load
Each set of phase currents and sets of line currents are
balanced:
Phase currents in Y-connected sources:
Vector Diagram
Vcn
Ina
Vbn
Van
 Sum to zero
 Equal in magnitude
 Displaced by 120 degrees
Vcn
n
Inc
120o
Ina
Van
Load
Inb
Inc Inb
Vbn
Van is used as reference
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Three Phase Current
Three Phase Current
Phase currents in Y-connected sources:
Phase currents in Delta-connected sources:
line current
Vector Diagram
Vca
a
Van
Ina
Vbn
n
Vcn
Y-connections:
phase current =
line current
Load
Vab
Vca
Iba
Iac
c
Inc Inb
Vbc
Ica
Load
120o
Iba
Vab
b
Icb
Ibc
Vbc
Vab is used as reference
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Three Phase Current
Summary
Line Current = Phase
Current
Line Voltage = Phase
o
Voltage x 3 30
Line currents in Delta-connected sources:
I12
Iba
Iac
Iba( 3
I34
Icb
Iba
Icb ( 3
30 ) KCL at nodes a, b, c
I56
Iac
Icb
Iac ( 3
30 )
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30 )
Line Current = Phase
o
Current x 3 30
Line Voltage = Phase
Voltage
line current
I12
a
Iba
Iac
c
Va
Vab
Vca
Vbc
Ibc
Vc
Load
I34
n
Va
Vc
Vb
Vb
b
Y (Wye)
I56
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Delta
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Y- Source Conversion
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-Y Source Conversion
a
-
a
Va
c
Vb
+
c
+ b
a
-
+
Each configuration has
identical line-line voltages, line
current, and complex power
delivered to the load
-
+
b
Vb
+
30
3
-
+
-
+
Each configuration has
identical line-line voltages, line
current, and complex power
delivered to the load
b
a
Vc
Va 3 30
Vc 3 30
c
Va
-
+
--
Vc
+
Vc
+
c
+
Va
30
3
--
Vb
3
30
“per-phase” voltage of a delta is
the line-line voltage divided by sqrt(3)
+ b
Vb 3 30
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Example
Example
Convert the shown three-phase source to its Y
equivalent given: Va 480 0
Convert the shown three-phase source to its Y
equivalent given: Va 480 0
a
a
-
+
+
Vc
277 90
-
+
c
277
Va
-
Vb
+
b
c
--
+
•
•
Za
n’
Za
Zc
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Three Phase Loads
Three phase sources are connected to three phase loads in
two common configurations
 Y (wye)
 Delta
Y sources can be connected to delta and/or Y loads
Delta sources can be connected to delta and/or Y loads
Zb
Note: usually we redefine the reference
Voltage to be Van so that Van 277 0
150
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Three Phase Loads
•
277
+ b
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• Circuit analysis is easier if loads are connected as Y
• We can transform balanced Delta connected loads
into balanced Y connected loads mathematically by
ZY
Z
3
 ZY: complex impedance of Y-connected load (Ohms)
 Z : complex impedance of a Delta-connected load
(Ohms)
• Results only apply to terminal conditions
Zc
Zb
Delta
Y
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Conventions & Assumptions
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Three Phase Power
• Voltage: line-to-line in rms
Total power delivered is the sum of power delivered
by each phase
S3 = VanI*na + VbnI*nb + VcnI*nc
= 3VanI*na (due to symmetry)
= 3S
where
• Current: line in rms
• Current direction: source to load
• Balanced three phase
 S3 : total three phase complex power (VA)
 S: single phase complex power (VA)
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Three Phase Power
Three-Phase Analysis
Similarly
P3 = Re{VanI*na} + Re{VbnI*nb} + Re{VcnI*nc}
= 3Re{VanI*na}
= 3P
and
Q3 = Im {VanI*na} + Im {VbnI*nb} + Im {VcnI*nc}
= 3Im{VanI*na}
= 3Q
• Recall that we can make a hypothetical
connection between neutrals without affecting the
circuit
 Neutral Conductor
• For balanced sources and loads, no current flows
on the neutral conductor
Van
Vcn
In=0
Vbn
n
Z
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Z
n’
Z
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Three-Phase Analysis
Three-Phase Analysis
• Phases can be conceptually decoupled
• No need to solve all three phases
• Balanced Three-Phase Theorem
• Assume:
 balanced three-phase system
 all loads and sources are Y-connected (or convert them
to Y-equivalent by dividing delta loads by 3 using perphase voltage of delta sources )
 no mutual inductances between phases
 Solve for a-phase (current or voltage)
 Shift +120o for c-phase, and -120o for b-phase
• We can therefore do a per-phase analysis
then
 all neutrals have the same voltage
 the phases are completely decoupled
 all corresponding network variables occur in balanced
sets of the same sequence as the sources
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Example
Example
Consider the following circuit where the line-line
voltage is 12.47kV, and the load impedance is 1200
+ 300j . Determine the total complex power
consumed by the load.
1. Verify that the circuit is balanced
a
Vc
c
a
Va
Vb
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Z
Z
Vc
b
c
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Va
Z
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Vb
Z
Z
Z
b
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Example
Example
2. Convert sources and loads to Y-connections
2. Convert sources and loads to Y-connections
Van
Vbn
Vcn
a
Vc
c
Va
Vb
12.47k
3
12.47k
3
12.47k
3
7200
30 V
7200
150 V
shift by +30 degrees
for convenience
Van
7200 0 V
Vbn
7200
Vcn
7200 120 V
120 V
7200 90 V
Z
Z
Z
Vcn
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Z
Van
b
n
Vbn
31
Z
Z
Load is already
Y-connected
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Example
Example
3. Draw per-phase equivalent circuit
• Virtual connection between n and n’
Van
7200 0 V
Z
1200
j300
n’
n
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n’
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Example
Example
4. Solve resulting circuit for single phase complex
power and total complex power
4. Solve resulting circuit for single phase complex
power and total complex power
I
S
S3
Van
5.82
14
Z
VanI
40,654 j10,163 VA
3S
121,960
j30, 490 VA
alternatively
S
S3
2
Van
V2
j an
40,654 j10,163 VA
1200
300
3S 121,960 j30, 490 VA
Van
7200 0 V
Z
n
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1200
j300
n’
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