9/24/2012
Overview
• Three Phase Power
• Three Phase Analysis
• Per Phase Analysis
06-Three Phase Analysis
• Practical Considerations
ECEGR 450
Electromechanical Energy Conversion
Dr. Louie
Questions
2
Conventions & Assumptions
• How does three phase power compare to single
phase power?
• Voltage: line-to-line in rms
• Current: line in rms
• What is the “neutral” connection in a three phase
system?
• Current direction: source to load
• Balanced three phase
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3
Three Phase Power
4
Three Phase Power
Benefits of three phase power:
• 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
 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
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 S3 : total three phase complex power (VA)
 S: single phase complex power (VA)
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Three Phase Power
Three Phase Power
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
What about instantaneous power?
p3 (t)
imax vmax cos( t
0 ) cos( t
imax vmax cos( t
120 ) cos( t
ia
120 )
imax vmax cos( t
120 ) cos( t
ia
120 )
1
p3 (t)
2
vmaximax [cos(
2
ia
)
cos(2 t
ia
)
cos(2 t
i
120 )
ia
)
cos(2 t
i
120 )]
) cos( 120 )
cos(2 t
i
) cos(120 )
vmaximax [3 cos(
ia
)
cos(2 t
cos(2 t
2
1
cos(2 t
2
3
2
vmaximax cos( )
i
)
250
sin(2 t
sin(2 t
)
i
sin(2 t
2
cos x cos y
i
i
sin x sin y
8
Total instantaneous power
is constant
) sin( 120 )
200
) sin(120 )]
)
2
3
i
sin(2 t
i
3
y)
Three Phase Power
)
i
1
p3 (t)
i
cos(2 t
)
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Power (W)
p3 (t)
1
vmaximax [3 cos(
i
cos(
Three Phase Power
2
cos(2 t
7
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1
)
)
cos(
Next use: cos(x
p3 (t)
ia
ia
)
)]
i
150
100
50
What does this say about the real power?
0
0
0.01
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0.02
0.03
time (s)
0.04
0.05
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Three-Phase Analysis
10
Three-Phase Analysis
• Consider the Y-connected source and load
• Determine Vn’n
• Analysis is easier using admittance, Y
Y
1
Z
• Line current is equal to phase current
n
-
Vn’n
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+
Ina
Y(Van
Inb
Y(Vbn
Vnn' )
Vnn' )
Inc
Y(Vcn
Vnn' )
due to wye connection
n’
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Three-Phase Analysis
Three-Phase Analysis
• Summing the line current
Ina
Inb
Inc
Y(Van
Ina
Inb
Inc
0
Van
Vbn
Vbn
• Since Vnn’ = 0, we can make a hypothetical
connection without affecting the circuit
• Called the Neutral Conductor
• For balanced sources and loads, no current flows
on the neutral conductor
Vcn ) 3YVnn'
• Previously we showed that
Vcn
0
• Therefore
Y(0) 3YVnn'
0
Vnn'
0
Van
Vcn
n
Vbn
Z
n’
Z
13
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In=0
Z
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Three-Phase Analysis
Three-Phase Analysis
Let’s now analyze the current through the a-phase
load
Let’s now analyze the current through the b-phase
load
Van
Van'
Vbn
Vbn'
Ina
Van'
Z
Inb
Vbn'
Z
Vcn
+
Van
-
n
+
Van’
-
Vbn
Z
n’
Z
Z
But
Vcn
+
Van
-
n
+
Van’
-
Vbn
Z
n’
Z
Z
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Three-Phase Analysis
Vcn
Vcn'
Vcn'
Z
Van
Vcn
n
Van(1 120 )
Inc
Vcn'
Z
Van(1 120 )
Z
120 )
Van(1
120 )
Z
Ina(1
120 ) (b-phase current in
terms of a-phase current)
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+
Van’
-
Vbn
Z
• Once Ina is solved for, simply shift it by -/+120
degrees to find Inb and Inc
• Phases can be conceptually decoupled
• No need to solve all three phases
Z
n’
Z
• But
Vcn
Van(1
Vbn'
Z
Three-Phase Analysis
• Let’s now analyze the current through the cphase load
Inc
Vbn
Inb
 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
Ina(1 120 )
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(c-phase current in
terms of a-phase current)
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9/24/2012
Three-Phase Analysis
Per-Phase Analysis
• Balanced Three-Phase Theorem
• Assume:
• Consider the following circuit
• Is it balanced?
 balanced three-phase system
 all loads and sources are Y-connected (or convert
them to Y-connected loads/sources)
 no mutual inductances between phases
then
Z1
Van
 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
Vbn
Vcn
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Z2
Z1
n
Z1
Z2
Z2
20
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Per-Phase Analysis
Per-Phase Analysis
• To analyze this circuit, we first need to find a perphase equivalent
• Need to transform the delta load to Y load
• To analyze this circuit, we first need to find a perphase equivalent
• Need to transform Delta load to Y load
Z1
Van
Z1
n
Vbn
Vcn
Z2
Z1
Z1
Van
Z2
Vbn
Vcn
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Z1
Van
Ina
n
Vbn
Vcn
Z2
Z1
Z1
Z2
Z2
Z2
3
given:
balanced 3-phase source
Van 120 0 V
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Z2
3
Numerical Example
• Analyze a single phase (arbitrarily a-phase)
• Analyze this phase to find the current, power, etc
Z1
Z2
3
Z1
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Per-Phase Analysis
Van
Z1
n
Z2
Z2
3
23
Z1
10
Z2
12 45
find:
Ina, Inb, Inc
total powered delivered to the loads
j2
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Example
Example
Z’2 = 2.82 +j2.82
• Convert to Y
• Redraw circuit
Solution outline
• draw per-phase equivalent
• solve circuit for current
• compute single phase power
• translate per-phase values to 3-phase
Z1
n
Vbn
Vcn
25
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Z2’
Z2’
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Example
• Draw per-phase equivalent
• Solve for Ian and S
Z’2 = 2.82 +j2.82
Ina
Van
120 0 V
Ina
40.32
S
3930.2
35 A
j2822.5 VA
Ina
a
Van
Z1
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Example
• Redraw circuit
• Draw per-phase
equivalent
Z 2’
Z1
Van
Z’
Z1
2
Van
Z||
Z1Z'2
Z1 Z'2
2.417
j1.736
n
27
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Example
Practical Considerations
• Shifting and solving
Inb
40.32
35
120
40.32
Inc
40.32
35
120
40.32
In
S3
155 A
Conductors
85 A
0A
3 3930.19
j2822.51
11790.57
B
j8467.53VA
+
VLL
-
Three phase systems and
components often referred to
by |S3 |
e.g. 100 MVA transformer
C
VLL
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Three phase systems and
components often referred to
by RMS line-line voltages
e.g. 500kV, 13.6kV, 480V
A
29
Vphase ( 3 30 )
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Practical Considerations
Practical Considerations
• Neutral usually bonded to ground at service panel
(see NEC for details)
Phases
(red, blue, black)
Ground
(green)
Phases
(brown, orange, yellow)
• Balanced load assumption can be questionable
 Neutral current nonzero
 Neutral not at ground potential
Neutral
(white)
Neutral
(white)
Ground
(green)
Source: www.electrical-design-tutor.com/
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Summary
• Current (voltages) sum to zero and no current flows on
neutral conductor
• Total S, P, Q = three times single phase S, P, Q
• Three phase systems analysis: convert into wye
connections, solve per-phase
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