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CP1 B9 L2 Power System Modelling
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NATIONAL ELECTRIFICATION ADMINISTRATIONU. P. NATIONAL ENGINEERING CENTER
Certificate in
Power System Modeling and Analysis
Competency Training and Certification Program in Electric Power Distribution System Engineering
U. P. NATIONAL ENGINEERING CENTERU. P. NATIONAL ENGINEERING CENTER
Power System Modeling
Training Course in
2
Competency Training & Certification Program in Electric Power Distribution System Engineering
U. P. National Engineering CenterNational Electrification Administration
Training Course in Power System Modeling
Course Outline
1. Utility Thevenin Equivalent Circuit
2. Load Models
3. Generator Models
4. Transformer Models
5. Transmission and Distribution Line Models
3
Competency Training & Certification Program in Electric Power Distribution System Engineering
U. P. National Engineering CenterNational Electrification Administration
Training Course in Power System Modeling
� Thevenin’s Theorem
� Utility Fault MVA
� Equivalent Circuit of Utility
Utility Thevenin Equivalent Circuit
4
Competency Training & Certification Program in Electric Power Distribution System Engineering
U. P. National Engineering CenterNational Electrification Administration
Training Course in Power System Modeling
Thevenin’s Theorem
Any linear active network with output terminals AB can be replaced by a single voltage source Vth in series with a single impedance Zth
Linear Active
Network
A
B
A
B
+
-
Vth
Zth
The Thevenin equivalent voltage Vth is the open circuit voltage measured at the terminals AB. The equivalent impedance Zth is the driving point impedance of the network at the terminals AB when all sources are set equal to zero.
5
Competency Training & Certification Program in Electric Power Distribution System Engineering
U. P. National Engineering CenterNational Electrification Administration
Training Course in Power System Modeling
Utility Fault MVA
Electric Utility Grid
Electric Utilities conduct short circuit analysis at the Connection Point of their customers
FaultIF
Customers obtain the Fault Data at the Connection Point to represent the Utility Grid for their power system analysis
Customer Facilities
6
Competency Training & Certification Program in Electric Power Distribution System Engineering
U. P. National Engineering CenterNational Electrification Administration
Training Course in Power System Modeling
Utility Fault MVA
Electric Utility provides the Fault MVA and X/R ratio at nominal system Voltage for the following types of fault:
• Three Phase Fault
Fault MVA3φφφφ X/R3φφφφ
• Single Line-to-Ground Fault
Fault MVALG X/RLG
System Nominal Voltage in kV
7
Competency Training & Certification Program in Electric Power Distribution System Engineering
U. P. National Engineering CenterNational Electrification Administration
Training Course in Power System Modeling
Equivalent Circuit of Utility
Positive & Negative Sequence Impedance
From Three-Phase Fault Analysis
1
fTPF Z
VI =
[ ]1
2f
TPFfTPF Z
VIVS ==
[ ]2
3
2
1 ZMVAFault
kVZ ==
φ
Where, Z1 and Z2 are the equivalent positive-sequence and negative-sequence impedances of the utility
8
Competency Training & Certification Program in Electric Power Distribution System Engineering
U. P. National Engineering CenterNational Electrification Administration
Training Course in Power System Modeling
Equivalent Circuit of Utility
Zero Sequence Impedance
From Single Line-to-Ground Fault Analysis
021
fSLGF ZZZ
V3I
++=
21 ZZ =
[ ]01
2f
SLGFfSLGF ZZ2
V3IVS
+==
[ ]SLGF
2f
01 S
V3ZZ2 =+ Resolve to real and imaginary
components then solve for Zo
9
Competency Training & Certification Program in Electric Power Distribution System Engineering
U. P. National Engineering CenterNational Electrification Administration
Training Course in Power System Modeling
Equivalent Circuit of Utility
Example:
Determine the equivalent circuit of the Utility in per unit quantities at a connection point for the following Fault Data:
System Nominal Voltage = 69 kV
Fault MVA3φφφφ = 3500 MVA, X/R3φφφφ = 22
Fault MVALG = 3000 MVA, X/RLG = 20
The Base Power is 100 MVA
10
Competency Training & Certification Program in Electric Power Distribution System Engineering
U. P. National Engineering CenterNational Electrification Administration
Training Course in Power System Modeling
Equivalent Circuit of Utility
Base Power: 100 MVABase Voltage: 69 kVBase Impedance: [69]2/100 = 47.61 ohms
[ ] [ ]Ω
φ
1.36033500
69
MVAFault
kVZZ
2
3
2
21 ====
In Per Unit,
p.u.0.028661.47
1.3603
Z
ZZZ
base
actual21 ====
p.u.0.0286MVA3500
100MVAZZ
FAULT
BASE21 ===
or
11
Competency Training & Certification Program in Electric Power Distribution System Engineering
U. P. National Engineering CenterNational Electrification Administration
Training Course in Power System Modeling
Equivalent Circuit of Utility
Solving for the Resistance and Reactance,
R
Z Xθθθθ
1
√[(1 + (X/R)2]X/R
θθθθ
[ ]R/Xtan 1−=θ
θθ
sinZX
cosZR
=
=
( )[ ]
( )[ ]2
1-
1
2
-1
1
Xp.u. 028571.0
22tan sin0.0286X
Rp.u. 1300.0
22tancos0.0286R
==
=
==
= +
-
fV
0.0013+j0.028571+
-01∠
12
Competency Training & Certification Program in Electric Power Distribution System Engineering
U. P. National Engineering CenterNational Electrification Administration
Training Course in Power System Modeling
Equivalent Circuit of Utility
For the Zero Sequence Impedance,
.u.p 0.1kV69
kV69Voltage
.u.p 30MVA100
MVA3000SLGF
.U.P
BASE
)actual(SLGF
.U.P
==
==
{ } ( )[ ]{ } ( )[ ] p.u. 0.09987520tan0.1sinZZ2agIm
p.u. 0.00499420tan0.1cosZZ2alRe1-
01
-101
==+
==+
[ ] [ ]1.0
30
0.13
S
V3ZZ2
2
SLGF
2f
01 ===+
13
Competency Training & Certification Program in Electric Power Distribution System Engineering
U. P. National Engineering CenterNational Electrification Administration
Training Course in Power System Modeling
Equivalent Circuit of Utility
( ) ( )p.u. j0.0427330.003694
028571.0j1300.02099875.0j004994.0Z
099875.0j004994.0ZZ2
0
01
+=
+−+=
+=+
+
-
+
-
fV
0.0013+j0.028571+
-01∠
+
-
j0.0427330.003694 +0.0013+j0.028571
Positive Sequence
Negative Sequence
Zero Sequence
14
Competency Training & Certification Program in Electric Power Distribution System Engineering
U. P. National Engineering CenterNational Electrification Administration
Training Course in Power System Modeling
Equivalent Circuit of Utility
Example:
Determine the equivalent circuit of the Utility in per unit quantities at a connection point for the following Fault Data:
Pos. Seq. Impedance = 0.03 p.u., X/R1 = 22
Zero Seq. Impedance = 0.07 p.u., X/R0 = 22
System Nominal Voltage = 69 kV
Base Power = 100 MVA
15
Competency Training & Certification Program in Electric Power Distribution System Engineering
U. P. National Engineering CenterNational Electrification Administration
Training Course in Power System Modeling
Utility Thevenin Equivalent Circuits
The equivalent sequence networks of the Electric Utility Grid are:
+
-
gEr
R1 +jX1+
-
+
-Positive
SequenceNegative Sequence
Zero Sequence
+
-
Equivalent Circuit of Utility
R2 +jX2 R0 +jX0
16
Competency Training & Certification Program in Electric Power Distribution System Engineering
U. P. National Engineering CenterNational Electrification Administration
Training Course in Power System Modeling
� Types of Load
� Customer Load Curve
� Calculating Hourly Demand
� Developing Load Models
Load Models
17
Competency Training & Certification Program in Electric Power Distribution System Engineering
U. P. National Engineering CenterNational Electrification Administration
Training Course in Power System Modeling
Types of LoadAn illustration:
Sending End
Receiving End
VS = ?
Load2 MVA, 3Ph
85%PF13.2 kVLL
VR = 13.2 kVLL
Line1.1034 + j2.0856 ohms/phase
ISR = ?
18
Competency Training & Certification Program in Electric Power Distribution System Engineering
U. P. National Engineering CenterNational Electrification Administration
Training Course in Power System Modeling
Types of LoadAn illustration:
Sending End
Receiving End
VS = ?
Load2 MVA, 3Ph 85% pf lag 13.2 kVLL
VR = 13.2 kVLL
Line1.1034 + j2.0856 ohms/phase
ISR = ?
Constant Power (P & Q)2 MVA = 1.7 MW + j1.0536 MVAR
Constant Current (I∠θ)I = 87.4773 ∠∠∠∠ -31.79o A
Constant Impedance (R & X)Z = 87.12 = 74.0520 + j 45.8948 ΩΩΩΩ
19
Competency Training & Certification Program in Electric Power Distribution System Engineering
U. P. National Engineering CenterNational Electrification Administration
Training Course in Power System Modeling
Types of Load
Sending End
Receiving End
VS = ?
Load2 MVA, 3Ph 0.85 pf, lag13.2 kVLL
VR = 13.2 kVLL
Line1.1034 + j2.0856 ohms/phase
ISR = ?
LL KV13.510
V 760.0800,7
)0856.21034.1)(79.314773.87(03
200,13
)(
=
∠=
+−∠+∠=
+=
LLS
o
oo
lineSRRS
V
j
ZIVV
r
rrrr
20
Competency Training & Certification Program in Electric Power Distribution System Engineering
U. P. National Engineering CenterNational Electrification Administration
Training Course in Power System Modeling
Types of Load
Sending End
Receiving End
VS = 13.51 kVLL
Load2 MVA, 3Ph 0.85 pf, lag13.2 kVLL
VR = 13.2 kVLL
Line1.1034 + j2.0856 ohms/phase
ISR = 87.48∠∠∠∠-31.79o
MVARjMW
IV ooSS
1010.17256.1
)79.314773.87)(76.0800,7(33 *
+=
∠∠=rr
21
Competency Training & Certification Program in Electric Power Distribution System Engineering
U. P. National Engineering CenterNational Electrification Administration
Training Course in Power System Modeling
Types of Load
KW
MWPlosses
6.25
7.17256.1
=
−=
%35.2
%1002.13
2.13510.13
=
×−
=VR
Sending End
Receiving End
VS = 13.51 kVLL
Load2 MVA, 3Ph 0.85 pf, lag13.2 kVLL
VR = 13.2 kVLL
Line1.1034 + j2.0856 ohms/phase
ISR = 87.48∠∠∠∠-31.79o
22
Competency Training & Certification Program in Electric Power Distribution System Engineering
U. P. National Engineering CenterNational Electrification Administration
Training Course in Power System Modeling
Types of LoadSending
EndReceiving
End
VS = ?
Load
VR = 11.88 kVLL
Line1.1034 + j2.0856 ohms/phase
ISR = ?
What happens if the Voltage at the Receiving End drops to 90% of its nominal value?
VR =11.88 KVLL
We will again analyze the power loss (Ploss) and Voltage Regulation (VR) for different types of loads
23
Competency Training & Certification Program in Electric Power Distribution System Engineering
U. P. National Engineering CenterNational Electrification Administration
Training Course in Power System Modeling
Types of LoadCase 1: Constant Power Load
2 MVA = 1.7 MW + j1.0536 MVAR
o
SRKV
MVAjI
79.311979.97
88.113
0536.17.1
−∠=
−=
r
KV12.224
V 94.08.057,7
)0856.21034.1)(78.311979.97(03
88.11
)(
0
0
=
∠=
+−∠+∠=
+=
j
ZIVV lineSRRS
rrrr
24
Competency Training & Certification Program in Electric Power Distribution System Engineering
U. P. National Engineering CenterNational Electrification Administration
Training Course in Power System Modeling
Types of LoadCase 1: Constant Power Load
2 MVA = 1.7 MW + j1.0536 MVAR
KW
WPlosses
722.28
)0134.1)(1979.97(3 2
=
=
%9.2
%10088.11
88.11224.12
=
×−
=VR
25
Competency Training & Certification Program in Electric Power Distribution System Engineering
U. P. National Engineering CenterNational Electrification Administration
Training Course in Power System Modeling
Types of LoadCase 2: Constant Current Load
I = 87.4773 ∠∠∠∠ -31.79o A
KV12.190
V 84.08.037,7
)0856.21034.1)(79.314773.87(03
88.11
)(
=
∠=
+−∠+∠=
+=
o
oo
lineSRRS
j
ZIVVrrrr
26
Competency Training & Certification Program in Electric Power Distribution System Engineering
U. P. National Engineering CenterNational Electrification Administration
Training Course in Power System Modeling
Types of LoadCase 2: Constant Current Load
I = 87.4773 ∠∠∠∠ -31.78o A
KW
WPlosses
33.25
)1034.1)(48.87(3 2
=
=
%6.2
%10088.11
88.1119.12
=
×−
=VR
27
Competency Training & Certification Program in Electric Power Distribution System Engineering
U. P. National Engineering CenterNational Electrification Administration
Training Course in Power System Modeling
Types of LoadCase 3: Constant Impedance Load
Z = 87.12 ∠∠∠∠31.79o ΩΩΩΩ = 74.0520 + j 45.8948 ΩΩΩΩ
KV12.159
KV77.00199.7
79.3112.87
0856.21034.1(79.3112.870
3
88.11
=
∠=
⎥⎦
⎤⎢⎣
⎡∠
++∠∠=
⎥⎦
⎤⎢⎣
⎡ +=
⎥⎦
⎤⎢⎣
⎡+
=
LLS
o
o
oo
Load
LineLoadRS
LineLoad
LoadSR
V
j
Z
ZZVV
ZZ
ZVV
r
r
rrrr
rr
rrr
28
Competency Training & Certification Program in Electric Power Distribution System Engineering
U. P. National Engineering CenterNational Electrification Administration
Training Course in Power System Modeling
Types of LoadCase 3: Constant Impedance Load
Z = 87.12 ∠∠∠∠31.79o ΩΩΩΩ = 74.0520 + j 45.8948 ΩΩΩΩ
A 730.78
0856.21034.179.3112.87
77.00199.7
=
++∠∠
=
+=
j
ZZ
VI
o
o
LineLoad
SSR rr
rr
29
Competency Training & Certification Program in Electric Power Distribution System Engineering
U. P. National Engineering CenterNational Electrification Administration
Training Course in Power System Modeling
Types of LoadCase 3: Constant Impedance Load
Z = 87.12 ∠∠∠∠31.79o ΩΩΩΩ = 74.0520 + j 45.8948 ΩΩΩΩ
KW
WPlosses
84.18
)0134.1)(73.78(3 2
=
=
%34.2
%10088.11
88.11159.12
=
×−
=VR
30
Competency Training & Certification Program in Electric Power Distribution System Engineering
U. P. National Engineering CenterNational Electrification Administration
Training Course in Power System Modeling
Types of Load
18.84 kW2.34 %12.15987.12∠-31.78
ConstantImpedance
25.33 kW2.6 %12.19087.48∠-31.78
ConstantCurrent
28.72 kW2.9 %12.224 2 MVA, 0.85 pf lag
ConstantPower
PlossVRVS*Load
* Sending end voltage with a Receiving end voltage equal to 0.9*13.2 KV
31
Competency Training & Certification Program in Electric Power Distribution System Engineering
U. P. National Engineering CenterNational Electrification Administration
Training Course in Power System Modeling
Types of LoadDemandReA= (PA+ IReA Va + Z -1ReA Va
2 )
DemandImA=(QA+ IImA Va + Z -1ImA Va2 )
DemandReC= (Pc+ IReC Vc + Z -1ReC Vc2 )
DemandImC= (Qc+ IImC Vc + Z -1ImC Vc2)
DemandReB= (PB+ IReB Vb + Z -1ReB Vb
2 )
DemandImB = (QB+ IImB Vb + Z -1ImB Vb
2 )
Where:
P,Q are the constant Power components of the Demand
IRe,IIm are the constant Current components of the Demand
Z-1Re,Z-1
Im are the constant Impedance components of the Demand
32
Competency Training & Certification Program in Electric Power Distribution System Engineering
U. P. National Engineering CenterNational Electrification Administration
Training Course in Power System Modeling
Time Demand (A)1:00 17.762:00 16.683:00 17.524:00 17.405:00 21.006:00 29.887:00 29.648:00 32.289:00 25.92
10:00 21.7211:00 25.2012:00 22.08
Time Demand (A)13:00 20.8814:00 19.8015:00 19.0816:00 19.2017:00 23.0418:00 30.7219:00 38.0020:00 35.0021:00 34.0022:00 27.6023:00 24.8424:00 22.32
24-Hour Customer Load Profile
Customer Load Curve
33
Competency Training & Certification Program in Electric Power Distribution System Engineering
U. P. National Engineering CenterNational Electrification Administration
Training Course in Power System Modeling
Time Demand (A) Per Unit1:00 17.76 0.4672:00 16.68 0.4393:00 17.52 0.4614:00 17.40 0.4585:00 21.00 0.5536:00 29.88 0.7867:00 29.64 0.7808:00 32.28 0.8499:00 25.92 0.682
10:00 21.72 0.57211:00 25.20 0.66312:00 22.08 0.581
Time Demand (A) Per Unit13:00 20.88 0.54914:00 19.80 0.52115:00 19.08 0.50216:00 19.20 0.50517:00 23.04 0.60618:00 30.72 0.80819:00 38.00 1.00020:00 35.00 0.92121:00 34.00 0.89522:00 27.60 0.72623:00 24.84 0.65424:00 22.32 0.587
ΣPU = 15.567
Customer Load Curve• Establishing Normalized Hourly Demand
34
Competency Training & Certification Program in Electric Power Distribution System Engineering
U. P. National Engineering CenterNational Electrification Administration
Training Course in Power System Modeling
0.0
0.2
0.4
0.6
0.8
1.0
1.2
Time
Dem
and
(Per
Un
it)
0 2 4 6 8 10 12 14 16 18 20 22 24
Customer Load Curve
35
Competency Training & Certification Program in Electric Power Distribution System Engineering
U. P. National Engineering CenterNational Electrification Administration
Training Course in Power System Modeling
0
50
100
150
200
250
300
350
Dem
and (W
)
Customer Energy Bill
Customer Energy Bill Converted to Hourly Power
Demand
Area under the curve = Customer Energy
Bill
0
0.2
0.4
0.6
0.8
1
1.2
Time (24 hours)
Norm
alized
Dem
and (per unit)
Normalized Customer Load Curve
Calculating Hourly Demand
36
Competency Training & Certification Program in Electric Power Distribution System Engineering
U. P. National Engineering CenterNational Electrification Administration
Training Course in Power System Modeling
Calculating Hourly Demand
Total Monthly Energy
Total Monthly Energy
Daily EnergyDaily Energy
Hourly Demand
Customer Load Curve
Customer Load Curve ⎟
⎟⎟⎟
⎠
⎞
⎜⎜⎜⎜
⎝
⎛
=
∑24
1t
tdailyt
p
pEnergyP
37
Competency Training & Certification Program in Electric Power Distribution System Engineering
U. P. National Engineering CenterNational Electrification Administration
Training Course in Power System Modeling
Calculating Hourly Demand
� Example:kWHr Reading (Monthly Bill) = 150 kWHr
Billing Days = 30 days
Daily Energy = 150 / 30 = 5 kWh [24 hours]
Hourly Demand1 = Daily Energy x [P.U.1 / ΣΣΣΣP.U]
= 5 kWh x 0.467 / 15.567
= 0.15011 kW
= 150.11 W
38
Competency Training & Certification Program in Electric Power Distribution System Engineering
U. P. National Engineering CenterNational Electrification Administration
Training Course in Power System Modeling
0
50
100
150
200
250
300
350
1:00
3:00
5:00
7:00
9:00
11:0
0
13:0
0
15:0
0
17:0
0
19:0
0
21:0
0
23:0
0
Dem
and
(W)
Calculating Hourly Demand
Hourly Real Demand
39
Competency Training & Certification Program in Electric Power Distribution System Engineering
U. P. National Engineering CenterNational Electrification Administration
Training Course in Power System Modeling
( )ttt pfPQ 1costan −=Qt = hourly Reactive Demand (VAR)
Pt = hourly Real Demand (W)
Pft = hourly power factor
� Example:Real Demand (W) = 150.11 W, PF = 0.96 lag
Reactive Demand = P tan (cos-1 pf)
= 150.11 tan (cos-1 0.96)
= 43.78 VAR
Calculating Hourly Demand
40
Competency Training & Certification Program in Electric Power Distribution System Engineering
U. P. National Engineering CenterNational Electrification Administration
Training Course in Power System Modeling
0
50
100
150
200
250
300
350
1:00
3:00
5:00
7:00
9:00
11:0
0
13:0
0
15:0
0
17:0
0
19:0
0
21:0
0
23:0
0
Dem
and
(W a
nd
VA
R)
Calculating Hourly Demand
Hourly Real & Reactive Demand
41
Competency Training & Certification Program in Electric Power Distribution System Engineering
U. P. National Engineering CenterNational Electrification Administration
Training Course in Power System Modeling
Developing Load Models
� Load Curves for each Customer Type� Residential load curves� Commercial load curves� Industrial load curves� Public building load curves� Street Lighting load curves� Administrative load curves (metered)� Other Load Curves (i.e., other types of customers)
� Variations in Load Curves� Customer types and sub-types� Weekday-Weekend/Holiday variations� Seasonal variations
42
Competency Training & Certification Program in Electric Power Distribution System Engineering
U. P. National Engineering CenterNational Electrification Administration
Training Course in Power System Modeling
Developing Load Models
� Data Requirements� Customer Data;
� Billing Cycle Data;
� Customer Energy Consumption Data; and
� Load Curve Data.
Distribution Utility Data Tables and Instructions
Converting Energy Bill to Power Demand
43
Competency Training & Certification Program in Electric Power Distribution System Engineering
U. P. National Engineering CenterNational Electrification Administration
Training Course in Power System Modeling
� Generalized Machine Model
� Steady-State Equations
� Generator Sequence Impedances
� Generator Sequence Networks
Generator Models
44
Competency Training & Certification Program in Electric Power Distribution System Engineering
U. P. National Engineering CenterNational Electrification Administration
Training Course in Power System Modeling
Stator:
distributed three-phase winding (a, b, c)
Rotor:
DC field winding (F) and short-circuited damper windings (D, Q)
Axis of a
Axis of c
Axis of b
d-axisq-axis
Phase b winding
Damper winding DDamper
winding Q
Phase a winding
Field winding F
Phase c winding
Generalized Machine ModelConstructional Details of Synchronous Machine
45
Competency Training & Certification Program in Electric Power Distribution System Engineering
U. P. National Engineering CenterNational Electrification Administration
Training Course in Power System Modeling
Primitive Coil Representation
q-axis
phase b
QiQ
vQ
-
+F
D
+v F
-iF
+v D -iD
phase a
phase c
ωωωωm+ Va -
ia
a
c
+Vc
- ic
ibb
+Vb
-
d-axis
θθθθe
dt
dRiv
λ+=
Generalized Machine Model
46
Competency Training & Certification Program in Electric Power Distribution System Engineering
U. P. National Engineering CenterNational Electrification Administration
Training Course in Power System Modeling
Voltage Equations for the Primitive CoilsFor the stator windings For the rotor windings
dt
diRv a
aaa
λ+=
dt
diRv b
bbb
λ+=
dt
diRv c
ccc
λ+=
dt
diRv F
FFF
λ+=
dt
diRv D
DDD
λ+=
dt
diRv Q
QQQ
λ+=
Note: The D and Q windings are shorted (i.e. ).0== QD vv
⎥⎦
⎤⎢⎣
⎡+⎥
⎦
⎤⎢⎣
⎡⎥⎦
⎤⎢⎣
⎡=⎥
⎦
⎤⎢⎣
⎡
FDQ
abc
FDQ
abc
FDQ
abc
FDQ
abcp
i
i
R
R
v
v
λλ
Li=λ
Generalized Machine Model
47
Competency Training & Certification Program in Electric Power Distribution System Engineering
U. P. National Engineering CenterNational Electrification Administration
Training Course in Power System Modeling
The flux linkage equations are:
⎥⎥⎥⎥⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢⎢⎢⎢⎢
⎣
⎡
⎥⎥⎥⎥⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢⎢⎢⎢⎢
⎣
⎡
=
⎥⎥⎥⎥⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢⎢⎢⎢⎢
⎣
⎡
Q
D
F
c
b
a
QQQDQFQcQbQa
DQDDDFDcDbDa
FQFDFFFcFbFa
cQcDcFcccbca
bQbDbFbcbbba
aQaDaFacabaa
Q
D
F
c
b
a
i
i
i
i
i
i
LLLLLL
LLLLLL
LLLLLL
LLLLLL
LLLLLL
LLLLLL
λ
λλλλ
λ
or
[ ] [ ][ ] [ ] ⎥
⎦
⎤⎢⎣
⎡⎥⎦
⎤⎢⎣
⎡=⎥
⎦
⎤⎢⎣
⎡λ
λ
FDQ
abc
RRRS
SRSS
FDQ
abc
i
i
LL
LL
Generalized Machine Model
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COIL INDUCTANCES
Stator Self Inductances
emsaa 2cosLLL θ+=
)1202cos(LLL emsbbo++= θ
)1202cos(LLL emscco−+= θ
Stator-to-Stator Mutual Inductances
)1202cos(LMLL emsbaabo−+−== θ
emscbbc 2cosLMLL θ+−==
)1202cos(LMLL emsaccao++−== θ
Generalized Machine Model
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Rotor Self Inductances
Rotor-to-Rotor Mutual Inductances
00
======
QDDQ
QFFQ
FDDFFD L
LLLLLL
COIL INDUCTANCES
QQQQ
DDDD
FFFF
LL
LL
LL
=
=
=
Generalized Machine Model
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)120cos(LLL
)120cos(LLL
cosLLL
eaDDccD
eaDDbbD
eaDDaaD
o
o
+==
−==
==
θ
θ
θ
)120sin(LLL
)120sin(LLL
sinLLL
eaQQccQ
eaQQbbQ
eaQQaaQ
o
o
+−==
−−==
−==
θ
θ
θ
Stator-to-Rotor Mutual Inductances
)120cos(LLL
)120cos(LLL
cosLLL
eaFFccF
eaFFbbF
eaFFaaF
o
o
+==
−==
==
θ
θ
θ
COIL INDUCTANCES
Generalized Machine Model
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q-axis
b-axis
Q iQ vQ
-
+
F D
+ vF -iF
+ vD -iD
d-axis
a-axis
c-axis
ωm
+ Va -ia
a
c+ Vc
-
ic
ibb +V
b-
Equivalent Coil Representation
Rotor coils FDQstationary
Stator coils abcrotating
Generalized Machine Model
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Equivalent Generalized Machine
q-axis
Q vQi
Q
+-
q i
q+-vq
+ vd -
di
dω
m
+ vF -
Fi
F+ vD -
Di
D
d-axis
-
-
Replace the abc coils with equivalent commutated d and q coils which are connected to fixed brushes.
QQQqQqQ
DDDFDFdDdD
DFDFFFdFdF
iLiL
iLiLiL
iLiLiL
+=
++=
++=
λ
λ
λ
QqQqqqq
DdDFdFdddd
iLiL
iLiLiL
+=
++=
λ
λ
Generalized Machine Model
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Transformation from abc to Odqq-axis
b-axis
d-axis
a-axis
c-axis
θe
ib
ia
icω
m
Note: The d and q windings are pseudo-stationary. The O axis is perpendicular to the d and q axes.
q-axis
d-axis
iq
id
qd
Generalized Machine Model
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Equivalence:
1. The resultant mmf of coils a, b and c along the d-axis must equal the mmf of coil d for any value of angle θe.
2. The resultant mmf of coils a, b and c along the q-axis must equal the mmf of coil q for any value of angle θe.
We get Ndid = Kd [Naia cos θθθθe + Nbib cos (θθθθe - 120o)
+ Ncic cos (θθθθe + 120o)]
Nqiq = Kq [-Naia sin θθθθe - Nbib sin (θθθθe - 120o)
-Ncic sin (θθθθe + 120o)]where Kd and Kq are constants to be determined.
Generalized Machine Model
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Assume equal number of turns.
Na = Nb = Nc = Nd = Nq
Substitution gives
id = Kd [ia cos θθθθe + ib cos (θθθθe - 120o) + ic cos (θθθθe + 120o)]
iq = Kq [-ia sin θθθθe - ib sin (θθθθe - 120o) -ic sin (θθθθe + 120o)]
The O-coil contributes no flux along the d or q axis. Let its current io be defined as
io = Ko ( ia + ib + ic )
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Combining, we get
( ) ( )( ) ( ) ⎥
⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
+−−−−
+−=⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
c
b
a
eqeqeq
ededed
ooo
q
d
o
i
i
i
KKK
KKK
KKK
i
i
i
120sin120sinsin
120cos120coscos
θθθ
θθθ
The constants Ko, Kd and Kq are chosen so that the transformation matrix is orthogonal; that is
[ ] [ ]TPP =− 1
Assuming Kd = Kq, one possible solution is
3
1=oK
3
2== qd KK
Generalized Machine Model
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Park’s Transformation Matrix
[ ] ( ) ( )
( ) ( )
[ ] ( ) ( )
( ) ( )⎥⎥⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢⎢⎢
⎣
⎡
+−+
−−−
−
=
⎥⎥⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢⎢⎢
⎣
⎡
+−−−−
+−=
−
120sin120cos2
1
120sin120cos2
1
sincos2
1
3
2
120sin120sinsin
120cos120coscos
2
1
2
1
2
1
3
2
1
ee
ee
e
eee
eee
P
P
θθ
θθ
θθ
θθθ
θθθ
Generalized Machine Model
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Voltage Transformation
The relationship between the currents is
[ ] abcodq iPi =or
[ ] odqabc iPi 1−=
Assume a power-invariant transformation; that is
qqddooccbbaa iviviviviviv ++=++or
odqTodqabc
Tabc iviv =
Generalized Machine Model
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[ ] odqTodqodq
Tabc iviPv =−1
Substitution gives
[ ]TTabc
Todq Pvv =
[ ] abcodq vPv =Transpose both sides to get
[ ] odqabc vPv 1−=
Note: Since voltage is the derivative of flux linkage, then the relationship between the flux linkages must be the same as that of the voltages. That is,
[ ] abcodq P λλ =
Generalized Machine Model
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Generalized Machine ModelIn summary, using Park’s Transformation matrix,
[ ] abcodq iPi = [ ] odqabc iPi 1−=
[ ] abcodq vPv = [ ] odqabc vPv 1−=
[ ] abcodq P λλ = [ ] odq1
abc P λ=λ −
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Recall the flux linkage equation
⎥⎥⎥⎥⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢⎢⎢⎢⎢
⎣
⎡
⎥⎥⎥⎥⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢⎢⎢⎢⎢
⎣
⎡
=
⎥⎥⎥⎥⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢⎢⎢⎢⎢
⎣
⎡
Q
D
F
c
b
a
QQQDQFQcQbQa
DQDDDFDcDbDa
FQFDFFFcFbFa
cQcDcFcccbca
bQbDbFbcbbba
aQaDaFacabaa
Q
D
F
c
b
a
i
i
i
i
i
i
LLLLLL
LLLLLL
LLLLLL
LLLLLL
LLLLLL
LLLLLL
λ
λλλλ
λ
or
[ ] [ ][ ] [ ] ⎥
⎦
⎤⎢⎣
⎡⎥⎦
⎤⎢⎣
⎡=⎥
⎦
⎤⎢⎣
⎡λ
λ
FDQ
abc
RRRS
SRSS
FDQ
abc
i
i
LL
LL
Generalized Machine Model
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[ ]⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡=
cccbca
bcbbba
acabaa
SS
LLL
LLL
LLL
L
( )( )o
o
1202cosLLL
1202cosLLL
2cosLLL
emScc
emSbb
emSaa
−θ+=
+θ+=
θ+=
( )
( )o
o
1202cosLMLL
2cosLMLL
1202cosLMLL
emSacca
emScbbc
emSbaab
+θ+−==
θ+−==
−θ+−==
Recall
where
Generalized Machine Model
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[ ]
( ) ( )( ) ( )( ) ( )⎥
⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
−+
+−
+−
+
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
−−
−−
−−
=
1202cos2cos1202cos
2cos1202cos1202cos
1202cos1202cos2cos
eee
eee
eee
m
SSS
SSS
SSS
SS
L
LMM
MLM
MML
L
θθθθθθ
θθθ
Substitution gives
Generalized Machine Model
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Similarly,
Apply Park's transformation to Flux Linkage equation
[ ] [ ][ ] [ ][ ] FDQSRabcSSabc iLPiLPP +=λ
or
[ ][ ][ ] [ ][ ] FDQSRodqSSodq iLPiPLP += −1λ
Generalized Machine Model
[ ] ( ) ( ) ( )( ) ( ) ( )⎥
⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
+−++
−−−−
−
=
120sin120cos120cos
120sin120cos120cos
sincoscos
eaQeaDeaF
eaQeaDeaF
eaQeaDeaF
SR
LLL
LLL
LLL
L
θθθ
θθθθθθ
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The term [ ][ ][ ] 1−PLP SScan be shown
Generalized Machine Model
⎥⎥⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢⎢⎢
⎣
⎡
++
++
−
=
mss
mss
ss
L2
3ML
L2
3ML
M2L
[ ][ ][ ]⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
=−
dd
oo
ss
L
L
L
PLP 1
mSSqq
mSSdd
SSoo
LMLL
LMLL
MLL
2
32
3
2
−+=
++=
−=Let
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[ ][ ]⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
=
⎥⎥⎥⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢⎢⎢⎢
⎣
⎡
=
dDdF
aQ
aDaFSR
L
LL
L
LLLP
2
300
02
3
2
3
000
Similarly, it can be shown that
Generalized Machine Model
aQqQ L2
3L =aFdF L
2
3L = aDdD L
2
3L =
where
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Finally, we get
QqQqqqq
DdDFdFdddd
oooo
iLiL
iLiLiL
iL
+=λ
++=λ
=λ
Generalized Machine Model
[ ][ ][ ] [ ][ ] FDQSRodqSSodq iLPiPLP += −1λ
Substituting, [ ][ ][ ] 1−PLP SS and [ ][ ]SRLP
Note: All inductances are constant.
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The Flux Linkage Equations for the FDQ coils in matrix form is
[ ] [ ] FDQRRabcRSFDQ iLiL +=λ
[ ] [ ]TSRRS LL =Since
we get
[ ] [ ] [ ] FDQRRodqT
SRFDQ iLiPL +=λ −1
Generalized Machine Model
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It can be shown that
[ ] [ ]
⎥⎥⎥⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢⎢⎢⎢
⎣
⎡
=−
aQ
aD
aF
TSR
L
L
L
PL
2
300
02
30
02
30
1
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
=
Dd
Fd
L
L
L
Generalized Machine Model
aFFd L2
3L = aDDd L
2
3L = aQQq L
2
3L =
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Recall that the rotor self- and mutual inductances are constant
[ ]⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
=
DDDF
FDFF
RR
L
LL
LL
L
00
0
0
Upon substitution, we get
QQQqQqQ
DDDFDFdDdD
DFDFFFdFdF
iLiL
iLiLiL
iLiLiL
+=λ
++=λ
++=λ
Note: All inductances are also constant.
Generalized Machine Model
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The Flux Linkage Equation
⎥⎥⎥⎥⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢⎢⎢⎢⎢
⎣
⎡
⎥⎥⎥⎥⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢⎢⎢⎢⎢
⎣
⎡
=
⎥⎥⎥⎥⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢⎢⎢⎢⎢
⎣
⎡
λ
λ
λ
λ
λ
λ
Q
D
F
q
d
o
QQQq
DDDFDd
FDFFFd
qQqq
dDdFdd
oo
Q
D
F
q
d
o
i
i
i
i
i
i
LL
LLL
LLL
LL
LLL
Lo d q F D Q
odq
FD
Q
q-axis
Q vQi
Q+
-
q iq +
-vq
+ vd -
d
idωm
+ vF -
F
iF + vD -
D
i
D
d-axis
Generalized Machine Model
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Transformation of Stator Voltages
Assume Ra = Rb = Rc in the stator. Then,
[ ] abcabcaabc dt
diuRv λ+= 3
Recall the transformation equations
[ ][ ][ ] abcodq
abcodq
abcodq
P
vPv
iPi
λλ =
=
=
Generalized Machine Model
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[ ] [ ] [ ][ ] [ ] [ ]{ }odqodqaabc Pdt
dPiPuRPvP λ+= −− 11
3
Simplify to get
[ ] [ ][ ] [ ] [ ] odqodqodqaodq Pdt
dP
dt
dPPiuRv λ
⎭⎬⎫
⎩⎨⎧+λ+= −− 11
3
Apply Park’s transformation
Generalized Machine Model
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It can be shown that
[ ] ( ) ( )( ) ( ) dt
d
cossin
cossin
cossin
Pdt
d e
ee
ee
eeθ
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
+θ−+θ−
−θ−−θ−
θ−θ−
=−
1201200
1201200
0
3
21
where
mee
dt
dω=ω=
θfor a two–pole machine
m
Pω=
2for a P–pole machine
Generalized Machine Model
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It can also be shown that
[ ] [ ]⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
ω
ω−=−
00
00
0001
m
mPdt
dP for a two-pole
machine
Finally, we getooao
dt
diRv λ+=
qmddaddt
diRv λωλ −+=
dmqqaqdt
diRv λωλ ++=
Generalized Machine Model
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Voltage Equation for the Rotor
0
0
=λ+=
=λ+=
λ+=
QQQQ
DDDD
FFFF
dt
diRv
dt
diRv
dt
diRv
Note: No transformation is required.
Generalized Machine Model
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Matrix Form of Voltage Equations
⎥⎥⎥⎥⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢⎢⎢⎢⎢
⎣
⎡
⎥⎥⎥⎥⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢⎢⎢⎢⎢
⎣
⎡
+
⎥⎥⎥⎥⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢⎢⎢⎢⎢
⎣
⎡
+
⎥⎥⎥⎥⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢⎢⎢⎢⎢
⎣
⎡
⎥⎥⎥⎥⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢⎢⎢⎢⎢
⎣
⎡
=
⎥⎥⎥⎥⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢⎢⎢⎢⎢
⎣
⎡
Q
D
F
q
d
o
m
Q
D
F
q
d
o
1
1-
λ
λ
λ
λ
λ
λ
ω
λ
λ
λ
λ
λ
λ
dt
d
i
i
i
i
i
i
R
R
R
R
R
R
v
v
v
v
v
v
Q
D
F
q
d
o
Q
D
F
a
a
a
Q
D
F
q
d
o
The equation is now in the form
[ ] [ ][ ] [ ] [ ] [ ][ ]iGipLiRv mω++=
Generalized Machine Model
Resistance Voltage Drop
Transformer Voltage
Speed Voltage
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[ ]
⎥⎥⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢⎢⎢
⎣
⎡
=
QQQq
DDDFDd
FDFFFd
qQqq
dDdFdd
LL
LLL
LLL
LL
LLL
L
dq
F
D
Q
D QFqd
[ ]
⎥⎥⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢⎢⎢
⎣
⎡ −−
=dDdFdd
qQqq
LLL
LL
G
dq
F
D
Q
D QFqd
Generalized Machine Model
Note: All entries of [L] and [G] are constant.
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Summary of Equations
Voltage Equations
( )( )( )( )( )( ) 06
05
4
3
2
1
=λ+=
=λ+=
λ+=
λω+λ+=
λω−λ+=
λ+=
QQQQ
DDdD
FFFF
dmqqaq
qmddad
ooao
piRv
piRv
piRv
piRv
piRv
piRv
Generalized Machine Model
Flux Linkages
( )( )( )( )( )( ) QQQqQqQ
DDDFDFdDdD
DFDFFFdFdF
QqQqqqq
DdDFdFdddd
oooo
iLiL
iLiLiL
iLiLiL
iLiL
iLiLiL
iL
+=λ
++=λ
++=λ
+=λ
++=λ
=λ
6
5
4
3
2
1
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Electromagnetic Torque Equation
[ ]
⎥⎥⎥⎥⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢⎢⎢⎢⎢
⎣
⎡−
−=
0
0
0
0
d
q
QDFqdo iiiiiiλλ
[ ] [ ][ ]iGiT Te −=
( )( )[ ]dQqQqDdDqFdFqdqqdd
qddqe
iiLiiLiiLii LL
iiT
−++−−=
+−−= λλ
We get
for a 2for a 2--pole machinepole machine
Generalized Machine Model
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Steady–State Equations
At steady–state condition,1. All transformer voltages are zero.2. No voltages are induced in the damper windings.
Thus, iD = iQ = 0
Voltage Equations
( )
FFF
FdFdddmqaq
qqqmdad
oao
iRv
iLiLiRv
iLiRv
iRv
=
+ω+=
ω−=
=
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Cylindrical-Rotor MachineIf the rotor is cylindrical, then the air gap is uniform, and Ldd = Lqq.
Define synchronous inductance Ls
LS = Ldd = Lqq when the rotor is cylindrical
Voltage and Electromagnetic Torque Equations at Steady-state
qFdFe
FdFmdsmqaq
qsmdad
iiLT
iLiLiRv
iLiRv
=
ω+ω−=
ω−=
Steady–State Equations
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For Balanced Three-Phase Operation
( )( )( )o
c
ob
a
tcosIi
tcosIi
tcosIi
1202
1202
2
+α+ω=
−α+ω=
α+ω=
Apply Park’s transformation [ ] abcodq iPi = , We get
α=
α=
=
sini
cosIi
i
q
d
o
I3
3
0Note:
1. ia, ib and ic are balanced three-phase currents.
2. id and iq are DC currents.
Steady–State Equations
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A similar transformation applies to balance three-phase voltages.
( )( )( )o
c
ob
a
tcosVv
tcosVv
tcosVv
1202
1202
2
+δ+ω=
−δ+ω=
δ+ω=
We get
δ=
δ=
=
sinVv
cosVv
v
q
d
o
3
3
0
Given
Steady–State Equations
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Inverse Transformation
[ ] odqabc iPi 1−=
We get
[ ]tsinitcosii qda ω−ω=3
2
( )[ ]o903
2+ω+ω= tcositcosi qd
Given id and iq, and assuming io = 0,
Steady–State Equations
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Recall the phasor transformation
( ) θ∠↔θ+ω ItcosI2Using the transform, we get
[ ]oo 9003
1∠+∠= qda iiI
assuming the d and q axes as reference. Simplify
qda
qda
jIII
ij
iI
+=
+=33
Steady–State Equations
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Similarly, given vd and vq with vo = 0
[ ]tsinvtcosvv qda ω−ω=3
2
In phasor form,
33qd
a
vj
vV +=
qd jVV +=
( )[ ]o90coscos3
2++= tvtv qd ωω
Steady–State Equations
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Steady-State Operation-Cylindrical
Recall at steady-state
Divide by 3
FdFmdsmqaq
qsmdad
iLiLiRv
iLiRv
ωω
ω
++=
−=
FdFmdsmqaq
qsmdad
iLILIRV
ILIRV
ωω
ω
3
1++=
−=
Steady–State Equations
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Define =ω= sms LX synchronous reactance
=ω= FdFmf iLE3
1Excitation voltage
Phasor Voltage aV
qda jVVV +=
( )( ) ( ) fqdsqda
fdsqaqsda
jEjIIjXjIIR
EIXIRjIXIR
++++=
+++−=
(motor equation)masaaa EIjXIRV ++=
Steady–State Equations
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For a generator, current flows out of the machine
( ) ( )aasaag
gasaaa
VIjXIRE
EIjXIRV
++=
+−+−=
AC
sa jXR +
aI+
-
+
-
aVgE
Equivalent Circuit of Cylindrical Rotor Synchronous Generator
Steady–State Equations
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Salient-Pole MachineIf the rotor is not cylindrical, no equivalent circuit can bedrawn. The analysis is based solely on the phasor diagram describing the machine. Recall the steady-state equations
Divide through by 3FdFmdddmqaq
qqqmdad
iLiLiRv
iLiRv
ωω
ω
++=
−=
3FdFm
ddqaq
qqdad
iLIXIRV
IXIRV
ω++=
−=
Steady–State Equations
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where
== ddmd LX ω direct axis synchronous reactance
== qqmq LX ω quadrature axis synchronous reactance
Define:
FdFm
f iL
E3
ω= = excitation voltage
Steady–State Equations
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We get
From , we getqda jVVV +=
or
fddqqaaa jEIjXIXIRV ++−=
fddqaq
qqdad
EIXIRV
IXIRV
++=
−=
( ) fddqqqda jEIjXIXjIIR ++−+=
( )fddqaqqdaa EIXIRjIXIRV +++−=
Steady–State Equations
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Steady-State Electromagnetic Torque
At steady-state
The dominant torque is the cylindrical torque which determines the mode of operation.
For a motor, Te is assumed to be negative. For a generator, Te is assumed to be positive.
saliency torque
cylindrical torque
( )[ ]qFdFqdqqdde iiLii LLT +−−=
Steady–State Equations
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Since the field current iF is always positive,
)(generator 0 i when 0 (motor) 0 i when 0
q
q
<>><− qFdF iiL
Recall that qda jIII +=
Note: The imaginary component of Ia determinesWhether the machine is operating as a motor or aGenerator.
Steady–State Equations
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What about Id?
fddq
dad
EIXVIRV
+==we get
In general, Ra << Xd. We get
qda jVVV +=
( )( )fdd
fddda
EIXjEIXjIR
+≈++=
Assume . From
0=qI
fddqaq
qqdad
EIXIRV
IXIRV
++=
−=
Steady–State Equations
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constant=+= fddq EIXV
Recall that
FdFm
f iL
E3
ω=
Thus, the excitation voltage depends only on the field current since ωωωωm is constant.
For some value of field current iFo, Ef = Va
and Id = 0.
If the magnitude of Va is constant,
Steady–State Equations
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Operating Modesq-axis
d-axis
Over-excited Motor
Under-excited Motor
Over-excited Generator
Under-excited Generator
Id < 0, Iq > 0 Id > 0, Iq > 0
Id < 0, Iq < 0 Id > 0, Iq < 0
Steady–State Equations
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Over-excitation and Under-excitation
1. If the field current is increased above iFo, thenEf > Va
and the machine is over-excited. Under this condition, Id < 0 (demagnetizing).
2. If the field current is decreased below iFo, then Ef < Va
and the machine is under-excited. Under this condition, Id > 0 (magnetizing).
Steady–State Equations
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Drawing Phasor Diagrams
aaddqqfa
fddqqaaa
qda
qda
IRIjXIXjEV
jEIjXIXIRV
jVVV
jIII
++−=
++−=
+=
+=
A phasor diagram showing Va and Ia can be drawn if the currents Id and Iq are known. Recall
Steady–State Equations
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Over-excited Motor
d-axis
q-axis
fjEqq IX−
aa IR
aVddIjX
δaI
φqjI
dI
Id < 0Iq > 0
Leading Power Factor
Steady–State Equations
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Lagging Power Factor
q-axis
d-axis
aVaa IR
dd IjX
qqIX−
qjI
fjE
φ
δaI
dI
Id > 0Iq > 0
Under-excited Motor
Steady–State Equations
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q-axis
d-axis
aV
aaIR
ddIjXqqIX−
qjI
fjE
φ
δ
aI
dICurrentActual
Id < 0Iq < 0
Over-excited Generator
Lagging Power Factor
Steady–State Equations
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d-axis
Id > 0Iq < 0 aV
aa IRddIjX
qqIX−
qjI
fjE
φ δ
aI
dIActualCurrent
Under-excited Generator
Leading Power Factor
Steady–State Equations
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1. The excitation voltage jEf lies along the quadrature axis.
2. leads jEf for a motorlags jEf for a generator
The angle between the terminal voltage and jEf is called the power angle or torque angle δ.
3. The equation
applies specifically for a motor.
Observations
aVaV
aV
fqqddaaa jEIXIjXIRV +−+=
Steady–State Equations
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4. For a generator, the actual current flows out ofthe machine. Thus Id, Iq and are negative.aI
qqddaaaf
fqqddaaa
IXIjXIRVjE
jEIXIjXIRV
−++=
++−−=or
generator afor
motor afor
gf
mf
EjE
EjE
=
=5. Let
Steady–State Equations
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The generator equation becomes
For a motor, the equation is
6. No equivalent circuit can be drawn for a salient-pole motor or generator.
qqddaaag IXIjXIRVE −++=
qqddaama IXIjXIREV −++=
Steady–State Equations
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Example 1: A 25 MVA, 13.8 kV, 3600 RPM, Y-connected cylindrical-rotor synchronous generator has a synchronous reactance of 4.5 ohms per phase. The armature resistance is negligible. Find the excitation voltage Eg when the machine is supplying rated MVA at rated voltage and 0.8 power factor.
Single-phase equivalent circuit
8.13=aV kV line-to-line97.7= kV line-to-neutral
AC
sjX
aIgE
+
-
+
-
aV
Steady–State Equations
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( ) 208.025 ==aP MW, three-phase67.6= MW/phase
15tan == θaa PQ MVar, three-phase5= MVar/phase
Let o097.7 ∠=aV kV, the reference.
Using the complex power formula*aaaa IVjQP =+
o097.7000,5667,6
* ∠−
=−
=j
V
jQPI
a
aaa
Steady–State Equations
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Apply KVL,
aaSg VIjXE +=( )
V
j
j
o
oo
24.19429,11
766,3791,10
0970,787.36046,15.4
∠=
+=
∠+−∠=
Eg = 11,429 volts, line-to-neutral= 19,732 volts, line-to-line
A
Ajo87.36046,1
628837
−∠=
−=IaWe get
Steady–State Equations
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Example 2: A 100 MVA, 20 kV, 3-phase synchronous generator has a synchronous reactance of 2.4 ohms. The armature resistance is negligible. The machine supplies power to a wye-connected resistive load, 4Ω per phase, at a terminal voltage of 20 kV line-to-line.
(a) Find the excitation voltage
AC
Ω= 4.2SX
aIgE+
-
+
-aV Ω= 4R
Steady–State Equations
Va(L-L) = 20,000 voltsVa(L-N) = 11,547 volts
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Let o0547,11 ∠=aV V, the reference
AmpsR
VI
aa
o0887,24547,11
∠===
aaSg VIjXE +=( )
neutraltolineV
j
j
−−∠=
+=
+=
o96.30466,13
928,6547,11
547,11887,24.2
Applying KVL,
( ) VEg 324,23466,133 ==linetolinekV −−= ,32.23
Steady–State Equations
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(b) Assume that the field current is held constant. A second identical resistive load is connected across the machine terminal. Find the terminal voltage, Va.
Since iF is constant, Eg is unchanged. Thus, Eg = 13,466 V,line-to-neutral.
Ω=ΩΩ= 24//4eqR
referencetheVVLet aa ,0o∠=
o02
1∠== a
eq
aa V
R
VI
Steady–State Equations
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aa
aa
VjV
VVj
2.1
2
14.2
+=
+⎟⎠⎞
⎜⎝⎛=
( )222 2.1 aag VVEgetWe +=22 44.2466,13 aV=
neutraltolineVVa −−= ,621,8
linetolineVVa −−= ,932,14
aasg VIjXE +=Apply KVL,
Steady–State Equations
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neutraltolineVV a −−∠= ,0547,11 o
AmpsR
VI
eq
aa
o0774,52
547,11∠===
( ) 547,1157744.2 += jE g
neutraltolineV
j
−−∠=
+=o19.50037,18
856,13547,11
( ) VE g 241,31037,183 ==linetolinekV −−= 24.31
(c) Assume that the field current iF is increased so that the terminal voltage remains at 20 kV line-to-line after the addition of the new resistive load. Find Eg.
Steady–State Equations
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The equivalent Circuit of Generator for Balanced Three-Phase System Analysis
Ec
Eb
Ea
Ic
Ib
Ia
b
a
c
sa jXR +
aI +
-
aV
+
-gE
Za
Zb Zc
Three-Phase Equivalent Single-Phase Equivalent
Generator Sequence Impedances
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Sequence Impedance of Power System Components
Positive Sequence Negative Sequence Zero Sequence
+
-
Z1
Ia1
E+
Va1
+
-
Ia2
Va2Z2
+
-
Ia0
Va0Z0
2a2a2 ZI - V =1a1a1 ZI– E V = oaoao ZI - V =
From Symmetrical Components, the Sequence Networks for Unbalanced Three-Phase Analysis
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Generator Sequence Impedances
Positive-Sequence Impedance:
Xd”=Direct-Axis Subtransient Reactance
for a salient-pole machine
Xd’=Direct-Axis Transient ReactanceXd=Direct-Axis Synchronous Reactance
Negative-Sequence Impedance:
)"X"X(X qd21
2 +=
for a cylindrical-rotor machine"XX d2 =
Zero-Sequence Impedance:
"X6.0X"X15.0 d0d ≤≤
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Positive Sequence ImpedanceThe AC RMS component of the current following a three-phase short circuit at no-load condition with constant exciter voltage and neglecting the armature resistance is given by
⎟⎟⎠
⎞⎜⎜⎝
⎛ −⎟⎟⎠
⎞⎜⎜⎝
⎛−+=
'
texp
X
E
'X
E
X
E)t(I
ddsdds τ
⎟⎟⎠
⎞⎜⎜⎝
⎛ −⎟⎟⎠
⎞⎜⎜⎝
⎛−+
"
texp
'X
E
"X
E
ddd τ
where E = AC RMS voltage before the short circuit.
Generator Sequence Impedances
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The AC RMS component of the short-circuit current is composed of a constant term and two decaying exponential terms where the third term decays very much faster than the second term.
If the first term is subtracted and the remainder is plotted on a semi-logarithmic paper versus time, the curve would appear as a straight line after the rapidly decaying term decreases to zero.
The rapidly decaying portion of the curve is thesubtransient portion, while the straight line is thetransient portion.
Generator Sequence Impedances
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IEEE Std 115-1995: Determination of the Xd’ and Xd” (Method 1)The direct-axis transient reactance is determined from the current waves of a three-phase short circuit suddenly applied to the machine operating open-circuited at rated speed. For each test run, oscillograms should be taken showing the short circuit current in each phase.The direct-axis transient reactance is equal to the ratio of the open-circuit voltage to the value of the armature current obtained by the extrapolation of the envelope of the AC component of the armature current wave, neglecting the rapid variation during the first few cycles.
Generator Sequence Impedances
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The direct-axis subtransient reactance is deter-mined from the same three-phase suddenly applied short circuit. For each phase, the values of the difference between the ordinates of Curve B and the transient component (Line C) are plotted as Curve A to give the subtransient component of the short-circuit current.
The sum of the initial subtransient component, the initial transient component and the sustained component for each phase gives the corresponding value of I”.
Generator Sequence Impedances
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0.4
0.60.81.0
1.52.0
34568
101214
0 10 20 30 40 50 60Time in half-cycles
Curr
ent
in p
has
e 1 (
per
unit)
+
+
++++
+
++ +++ ++ +++++ + ++ + + + + + + + +
Curve B
Line C
+
+
+
+
+
+
+
++Curve A
Line A
Generator Sequence Impedances
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Phase 1 Phase 2 Phase 3 Ave
(1) Initial voltage 1.0
(2) Steady-state Current 1.4 1.4 1.4
(3) Initial Transient Current 8.3 9.1 8.6
(4) I’ = (2)+(3) 9.7 10.5 10.0 10.07
(5) Xd’ = (1)÷(4) 0.0993
(6) Init. Subtransient Current 3.8 5.6 4.4
(7) I” = (4)+(6) 13.5 16.1 14.4 14.67
(8) Xd” = (1)÷(7) 0.0682
Example: Calculation of transient and subtransientreactances for a synchronous machine
Generator Sequence Impedances
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Negative Sequence Impedance
IEEE Std 115-1995: Determination of the negative-sequence reactance, X2 (Method 1)
The machine is operated at rated speed with its field winding short-circuited. Symmetrical sinusoidal three-phase currents of negative phase sequence are applied to the stator. Two or more tests should be made with current values above and below rated current, to permit interpolation.
The line-to-line voltages, line currents and electric power input are measured and expressed in per-unit.
Generator Sequence Impedances
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Let E = average of applied line-to-line voltages, p.u.I = average of line currents, p.u.P = three phase electric power input, p.u.
IE
Z2 = =Negative Sequence Impedance, p.u.
22 IP
R = =Negative Sequence Resistance, p.u.
22
222 RZX −=
=Negative Sequence Reactance, p.u.Note: The test produces abnormal heating in the rotor and should be concluded promptly.
Generator Sequence Impedances
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Zero Sequence ImpedanceIEEE Std 115-1995: Determination of the zero-sequence reactance, X0 (Method 1)
E A W
V
The machine is operated at rated speed with its field winding short-circuited. A single-phase voltage is applied between the line terminals and the neutral point.Measure the applied voltage, current and electric power.
Field
Generator Sequence Impedances
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Let E = applied voltage, in p.u. of base line-to-neutral voltage
I = test current, p.u.P = wattmeter reading, in p.u. single-phase
base volt-ampere
IE3
Z0 = =Zero Sequence Impedance, p.u.
2
00 EIP
1ZX ⎟⎠⎞
⎜⎝⎛
−=
=Zero Sequence Reactance, p.u.
Generator Sequence Impedances
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Average Machine Reactances
ReactanceTurbo
GeneratorsWater-Wheel Generators
Synchronous Motors
Xd 1.10 1.15 1.20
Xd‘ 0.23 0.37 0.35
Xd” 0.12 0.24 0.30
Xq” 0.15 0.34 0.40
X2 0.12 0.24 0.35
Xq 1.08 0.75 0.90
Xq‘ 0.23 0.75 0.90
Generator Sequence Impedances
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The sequence networks for the grounded-wyegenerator are shown below.
F1
N1
gEr
jZ1+
-
jZ2
F2
N2
jZ0
F0
N0
Grounded-Wye Generator
Generator Sequence Networks
Positive Sequence
Negative Sequence
Zero Sequence
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If the generator neutral is grounded through an impedance Zg, the zero-sequence impedance is modified as shown below.
F1
N1
gEr
jZ1+
-
jZ2
F2
N2
jZ0
F0
N0
3Zg
Grounded-Wye through an Impedance
Generator Sequence Networks
Positive Sequence
Negative Sequence
Zero Sequence
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Ungrounded-Wye Generator
If the generator is connected ungrounded-wye or delta, no zero-sequence current can flow. The sequence networks for the generator are shown below.
F1
N1
gEr
jZ1+
-
jZ2
F2
N2
jZ0
F0
N0
Positive Sequence
Negative Sequence
Zero Sequence
Generator Sequence Networks
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Transformer Models
� Two Winding Transformer
� Short-Circuit and Open-Circuit Tests
� Three Winding Transformer
� Autotransformer
� Transformer Connection
� Three Phase Transformer
� Three Phase Model
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Two-Winding Transformer
+ +
__HVr
XVr
HIr
XIr
HN XN
The voltage drop from the polarity-marked terminal to the non-polarity-marked terminal of the H winding is in phase with the voltage drop from the polarity-marked terminal to the non-polarity-marked terminal of the X winding.
X
H
X
H
NN
VV
=r
rVoltage Equation:
Ideal Transformer
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XXHH ININrr
=
The current that enters the H winding through the polarity-marked terminal is in phase with the current that leaves the X winding through the polarity-marked terminal.
Note: Balancing ampere-turns must be satisfied at all times.
Current Equation:+ +
__HVr
XVr
HIr
XIr
HN XN
Two-Winding Transformer
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Two-Winding Transformer
X
H
V
Va r
r
=
H
X
I
Ia r
r
=
XH VaVrr
=
X
X2
H
H
I
Va
I
Vr
r
r
r
=
a
II X
H
rr
=
Referred ValuesFrom the Transformation Ratio,
X2
H ZaZ =
Dividing VH by IH,
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Practical Transformer1. The H and X coils have a small resistance.2. There are leakage fluxes in the H and X coils.3. There is resistance loss in the iron core.4. The permeability of the iron is not infinite.
+
eH-
φm
NH
+
-
eXNX
iX
vH
iH
vX
ironcore
Two-Winding Transformer
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RH, XH =resistance and leakage reactance of H coil
Rc, Xm =core resistance and magnetizing reactance
+
-
+
-HEv
XEv
HN XN
HIr
XIr
Ideal
HH jXR + XX jXR +
+
-HVr
+
-XVr
H winding X winding
mjXcR
exIv
RX, XX =resistance and leakage reactance of X coil
Two-Winding TransformerEquivalent Circuit
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Two-Winding Transformer
+
-
+
-HEv
XEv
HN XN
HIr
a
I X
r
HH jXR + X2
X2 XjaRa +
+
-HVr
+
-XVar
mjXcR
exIv
Referring secondary quantities at the primary side,
HIr
a
I X
r
HH jXR + X2
X2 XjaRa +
+
-HVr
+
-XVar
mjXcR
exIv
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Two-Winding TransformerThe transformer equivalent circuit can be approximated by
X2
Heq RaRR +=
X2
Heq XaXX +=
HIr+
-HVr
+
-XVar
mjXcR
exIv
eqeq jXR +
HIr+
-HVr
+
-XVar
mjXcR
exIv
eqeq jXR +
Xa1 Ir
Xa1 Ir
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Two-Winding Transformer
Xa1
H IIrr
=
eqeq jXR +
+
-HVr
+
-XVar
For large power transformers, shunt impedance and resistance can be neglected
Xa1
H IIrr
=
eqjX
+
-HVr
+
-XVar
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Tap-Changing Transformera:1
q r
s p
The π equivalent circuit of transformer with the per unit transformation ratio: pqy
a
a2
1−pqy
a
a 1−
pqya
1
Two-Winding Transformer
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Short-Circuit and Open-Circuit Tests
H1
H2
x1
x2
Short-Circuit TestConducted to determine series impedance
With the secondary (Low-voltage side) short-circuited, apply a primary voltage (usually 2 to 12% of rated value) so that full load current flows.
A
V
W
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Short-Circuit and Open-Circuit Tests
Short-Circuit Test
1I
eqeq jXR +
+
-
SCV mjXcR
eI
1sc
e
II
0I
=
≈SCI
2SC
SCeq
I
PR =
SC
SCeq
I
VZ = 2
eq2eqeq RZX −=
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Short-Circuit and Open-Circuit Tests
x1
x2
H1
H2
Open-Circuit TestConducted to determine shunt impedance
With the secondary (High-voltage side) open-circuited, apply rated voltage to the primary.
A
V
W
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Short-Circuit and Open-Circuit Tests
Open-Circuit Test
eqeq jXR +
+
-
OCV mjXcR
eI
eOC II =OCI
OC
2OC
cP
VR = 2
c
2
OC
OC
m R
1
V
I
X
1−⎥
⎦
⎤⎢⎣
⎡=
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Short-Circuit and Open-Circuit Tests
Example:
50 kVA, 2400/240V, single-phase transformer
Short-Circuit Test: HV side energized
Open-Circuit Test: LV side energized
Determine the Series and Shunt Impedance of the transformer. What is %Z and X/R of the transformer?
watts617P amps 8.20I volts48V SCSCSC ===
watts186P amps41.5I volts240V OCOCOC ===
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Short-Circuit and Open-Circuit Tests
Solution:
From the short-circuit test
hmo 31.28.20
48Z H,eq ==
hmo 82.142.131.2X2
2
H,eq =−=
From the open-circuit test
( )hmo 310
186
240R
2
L,cq ==
22
m 310
1
240
41.5
X
1⎥⎦⎤
⎢⎣⎡−⎥⎦
⎤⎢⎣⎡= hmo 45X L,m =
( )hmo 42.1
8.20
617R 2H,eq
==
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Short-Circuit and Open-Circuit Tests
Referred to the HV side
hmo 968,30RaR L,c2
H,c ==
hmo 482,4XaX L,m2
H,m ==
%Z and X/R
[ ]hmo 115.2
1000/50
4.2Z
2
BASE ==
%2100x2.115
31.2Z% =⎟
⎠⎞
⎜⎝⎛= 28.1
42.1
82.1R/X ==
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X/R Ratios of Transformers
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Three-Winding Transformer
+
_HVr
HIr
HN +
_XV
rXIr
XN
+
_YV
rYIr
YN
X
H
X
H
NN
VV
=r
r
Y
H
Y
H
NN
VV
=r
r
Y
X
Y
X
NN
VV
=r
r
YYXXHH INININrrr
+=
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ZHX=impedance measured at the H side when the X winding is short-circuited and the Y winding is open-circuited
ZHY=impedance measured at the H side when the Y winding is short-circuited and the X winding is open-circuited
ZXY=impedance measured at the X side when the Y winding is short-circuited and the H winding is open-circuited
From 3 short-circuit tests with third winding open, get
Note: When expressed in ohms, the impedances must be referred to the same side.
Three-Winding Transformer
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HZ
+
-HVr +
-
XVr
XZ
YZ
+
-YVr
XHHX ZZZ +=
YHHY ZZZ += YXXY ZZZ +=
or )ZZZ(Z XYHYHX21
H −+=
)ZZZ(Z XYHYHX21
X +−=
)ZZZ(Z XYHYHX21
Y ++−=
Three-Winding Transformer
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Example: A three-winding three-phase transformer has the following nameplate rating:
H: 30 MVA 140 kV X: 30 MVA 48 kV Y: 10.5 MVA 4.8 kV
Short circuit tests yield the following impedances:
ZHX = 63.37 Ω @ the H sideZHY = 106.21 Ω @ the H sideZXY = 4.41 Ω @ the X side
Find the equivalent circuit in ohms, referred to the H side.
Ω 52.37)41.4()(Z 248
140XY ==
Three-Winding Transformer
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Ω03.66)52.3721.10637.63(Z 21
H =−+=
With all impedances referred to the H side, we get
Ω .66.2)52.3721.10637.63(Z 21
X −=+−=
Ω18.40)52.3721.10637.63(Z 21
Y =++−=
Ω03.66
+
-HVr +
-
XVr+
-YVr
Ω66.2−
Ω18.40
Three-Winding Transformer
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Transformer Polarity
H1 H2
x1 x2
H1 H2
x2 x1
Subtractive Polarity Additive Polarity
V1
V V
V1
Less than V1
Greater than V1
Transformer Connection
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H1 H2
X1 X2
H1 H2
X1X2
Subtractive Additive
“Single-phase transformers in sizes 200 kVA and below having high-voltage ratings 8660 volts and below (winding voltage) shall have additive polarity. All other single-phase transformers shall have subtractive polarity.” (ANSI/IEEEC57.12.00-1993)
Transformer Connection
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Transformer Connection
H1 H2
x1 x2
H1 H2
x1 x2
LOAD
Parallel Connection
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Parallel Connection
� same turns ratio
� Connected to the same primary phase
� Identical frequency ratings
� Identical voltage ratings
� Identical tap settings
� Per unit impedances within 0.925 to 1.075 of each other
Transformer Connection
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Transformer Connection
Three Phase Transformer Bank
H1 H2
x1 x2
H1 H2
x1 x2
H1 H2
x1 x2
WYE-WYE (Y-Y)
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Transformer Connection
Three Phase Transformer Bank
H1 H2
x1 x2
H1 H2
x1 x2
H1 H2
x1 x2
DELTA-DELTA (ΔΔΔΔ-ΔΔΔΔ)
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Transformer Connection
Three Phase Transformer Bank
H1 H2
x1 x2
H1 H2
x1 x2
H1 H2
x1 x2
WYE-DELTA (Y-ΔΔΔΔ)
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Transformer Connection
Three Phase Transformer Bank
H1 H2
x1 x2
H1 H2
x1 x2
H1 H2
x1 x2
DELTA-WYE (ΔΔΔΔ-Y)
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Transformer Connection
Three Phase Transformer Bank
H1 H2
x1 x2
H1 H2
x1 x2
OPEN DELTA – OPEN DELTA
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Transformer Connection
Three Phase Transformer Bank
H1 H2
x1 x2
H1 H2
x1 x2
OPEN WYE - OPEN DELTA
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Three-Phase Transformer
Windings are connected Wye or Delta internally
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Angular DisplacementANSI/IEEEC57.12.00-1993: The angular displacement of a three-phase transformer is the time angle (expressed in degrees) between the line-to-neutral voltage of the high-voltage terminal marked H1 and the the line-to-neutral voltage of the low-voltage terminal marked X1.The angular displacement for a three-phase trans-former with a ΔΔΔΔ-ΔΔΔΔ or Y-Y connection shall be 0o.
The angular displacement for a three-phase trans-former with a Y-ΔΔΔΔ or ΔΔΔΔ-Y connection shall be 30o, with the low voltage lagging the high voltage.
Three-Phase Transformer
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Vector Diagrams
H1
H2
H3
X1
X2
X3
Δ-Δ Connection
X1
X2
X3H1
H2
H3
Y-Δ Connection
H1
H2
H3
X1
X2
X3
Y-Y Connection
X1
X2
X3H1
H2
H3
Δ-Y Connection
Three-Phase Transformer
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0
6
2
4
10
8
IEC Designation for ΔΔΔΔ-ΔΔΔΔ
Dd0 Dd2 Dd4
Dd6 Dd8 Dd10
IEC Designation for Y-YYy0 Yy6
Note: The first letter defines the connection of the H winding; the second letter defines the connection of the X winding; the number designates the angle.
Three-Phase Transformer
IEC Designation
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Three-Phase Transformer
IEC Designation
IEC Designation for Y-ΔΔΔΔ
Yd1 Yd5 Yd7 Yd11
Note: The first letter defines the connection of the H winding; the second letter defines the connection of the X winding; the number designates the angle.
1
7
3
5
11
9
IEC Designation for ΔΔΔΔ-Y
Dy1 Dy5 Dy7 Dy11
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(A-B-C)
1ANVr
1BNVr
1CNVr
1bcVr
1abVr
1caVr
1anVr
1bnVr
1cnVr
1ANVr
1anVr
lagsby 30o
Positive–Sequence Voltages
H1
H2
H3
X1X2
X3A
B
C
c
b
a
N
Three-Phase Transformer
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C a
Positive–Sequence Currents
H1
H2
H3
X1X2
X3A
B
c
b
1AIr
1BIr
1CIr
1cIr
1bIr
1aIr
1baIr
1cbIr
1acIr
1AIr
1aIr
lagsby 30o
1baIr
1cbIr
1acIr1aI
r
1bIr
1cIr
1AIr
1BIr
1CIr
(A-B-C)
Three-Phase Transformer
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Positive Sequence Impedance
XH IIrr
=
111 jXRZ +=
+
-HVr
+
-XVr
Using per-unit values, the positive-sequence equivalent circuit is
Whether a bank of single-phase units or a three-phase transformer unit (core type or shell type), the equivalent impedance is the same.
Note: The negative-sequence impedance is equal to the positive-sequence impedance.
Three-Phase Transformer
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C
2ANVr
2CNVr
2BNVr
(A-C-B)
H1
H2
H3A
BX1
X2
X3c
b
a
N
2cbVr
2baVr
2acVr
2anVr
2bnVr
2cnVr
2ANVr
2anVr
leadsby 30o
Negative–Sequence Voltages
Three-Phase Transformer
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H1
H2
H3
X1X2
X3A
B
C
c
b
a
2AIr
2BIr
2CIr
2cIr
2bIr
2aIr
2baIr
2cbIr
2acIr
2AIr
2aIr
leadsby 30o
2baIr
2acIr
2cbIr
2aIr
2bIr
2cIr
2AIr
2CIr
2BIr
(A-C-B)
Negative–Sequence Currents
Three-Phase Transformer
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Positive– & Negative Sequence Networks
Three-Phase Transformer
1Ir
1Z
+
-
+
-
Primary Side
Secondary Side
2Ir
2Z
+
-
+
-
Primary Side
Secondary Side
Positive Sequence Network
Negative Sequence Network
21 ZZ =
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Transformer Core
3-Legged Core Type
4-Legged Core Type
Shell Type
Note: Only the Xwindings are shown.
Three-Phase Transformer
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Three-Legged Transformer Core
The 3-legged core type three-phase transformer uses the minimum amount of core material. For balanced three-phase condition, the sum of the fluxes is zero.
cφbφaφ
Note: For positive- or negative-sequence flux,
0cba =φ+φ+φ
Three-Phase Transformer
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Zero Sequence Flux
The 3-legged core type three-phase transformer does not provide a path for zero-sequence flux. On the other hand, a bank of single-phase units, the 4-legged core type and the shell-type three-phase transformer provide a path for zero-sequence flux.
3-Legged Core Type
0φ0φ0φ
03φNote: The zero-sequence flux leaks out of the core and returns through the transformer tank.
Three-Phase Transformer
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10 ZZ =
+
-HV
r +
-XV
r
10 ZZ =
+
-HV
r +
-XV
r
Zero Sequence Impedance*Transformer Connection Zero-Sequence Network
*Excluding 3-phase unit with a 3-legged core.
Three-Phase Transformer
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Transformer Connection Zero-Sequence Network
*Excluding 3-phase unit with a 3-legged core.
10 ZZ =
+
-HV
r +
-XV
r
10 ZZ =
+
-HV
r +
-XV
r
Zero Sequence Impedance*
Three-Phase Transformer
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Transformer Connection Zero-Sequence Network
*Excluding 3-phase unit with a 3-legged core.
10 ZZ =
+
-HV
r +
-XV
r
10 ZZ =
+
-HV
r +
-XV
r
Zero Sequence Impedance*
Three-Phase Transformer
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Example: Consider a two-winding three-phase transformer with the following nameplate rating: 25 MVA 69Δ -13.8YG kV (Dyn1) Z=7%. Draw the positive and zero-sequence equivalent circuits. Use the transformer rating as bases.
Positive/Negative Sequence impedance
+
-HV
r +
-XV
r
Z1=j0.07
Zero Sequence impedance
+
-HV
r +
-XV
r
Z0=j0.07
Three-Phase Transformer
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Example: A three-winding three-phase transformer has the following nameplate rating: 150/150/45 MVA 138zG-69zG-13.8Δ kV (Yy0d1).
Draw the positive and zero-sequence equivalent circuits. Use 100 MVA and the transformer voltage ratings as bases.
H-X @ 150 MVA = 14.8% H-Y @ 45 MVA = 21.0% X-Y @ 45 MVA = 36.9%
p.u.10.0)150/100(148.0Z HX ==At the chosen MVA base,
p.u.47.0)45/100(21.0Z HY ==p.u.82.0)45/100(369.0Z XY ==
Three-Phase Transformer
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p.u.125.0)82.047.010.0(Z 21
H −=−+=We get
p.u.225.0)82.047.010.0(Z 21
X =+−=
p.u.595.0)82.047.010.0(Z 21
Y −=++−=
Positive/Negative Sequence Network
HZ
+
-HV
r +
-
XVr
XZ
YZ
+
-YVr
Zero Sequence Network
HZ
+
-HV
rXVr
XZ
YZ
+
-YVr+
-
Three-Phase Transformer
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ABC
abc
abcTY
Core Loss
Three Phase ModelTHREE-PHASE TRANSFORMER AND
3 SINGLE-PHASE TRANSFORMERS IN BANK
Primary Secondary
Admittance Matrix
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2. EPRI Core Loss Model
Three Phase Model
CORE LOSS MODELS
1. Constant P & Q Model
( )( )2
2
VF2
.u.p
VC2
.u.p
EVDBaseSystem
RatingkVAQ
BVABaseSystem
RatingkVAP
ε
ε
+=
+=
22.7F 0.268x10E 00167.0D -13 ===
5.31C 0.734x10B 00267.0A -9 ===|V| in per unit
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Primitive Coils
Primitive Impedances
V6
z11 z33 z44 z55 z66
I1
z12
I5 I6I4I3I2
z23 z34 z45 z56
z22
V1 V5V4V3V2+ + + + + +
- - - - - -
Three Phase Model
z66z65z64z63z62z61
z56z55z54z53z52z51
z46z45z44z43z42z41
z36z35z34z33z32z31
z26z25z24z23z22z21
z16z15z14z13z12z11
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Transformer ModelI1 I2
I3 I4
I5 I6
Three Identical
Single-phase Transformers
in Bank
z11
z33
z55
z12
z34
z56
z22
z44
z66
z66z65
z56z55
z44z43
z34z33
z22z21
z12z11
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Transformer Model
Node Connection Matrix, C
V6
V5
V4
V3
V2
V1
=
Vc
Vb
Va
VC
VB
VA
[V123456] = [C][VABCabc ]
Matrix C defines the relationship of the Primitive Voltages and Terminal Voltages of the Three-Phase Connected Transformer
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Transformer Model
Node Connection Matrix, C
VaVA
VC
VB
IA
IC
IB
Vb
Vc
1
5
3
Ia
Ib
Ic
2
6
4
[V123456] = [C][VABCabc ]
Wye Grounded-Wye GroundedConnection
1
1
1
1
1
1
V6
V5
V4
V3
V2
V1
Vc
Vb
Va
VC
VB
VA
=
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Transformer ModelVaVA
VC
VB
IA
IB
IC
Ia
Ib
Ic
Vb
Vc
1 2
5
6
34
Wye Grounded-Delta Connection
1-1
1
-11
1
-11
1
[V123456] = [C][VABCabc ]
V6
V5
V4
V3
V2
V1
Vc
Vb
Va
VC
VB
VA
=
Node Connection Matrix, C
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Transformer Model
A a
B b
C cN
M
M
M
L1 L2
L1 L2
L1 L2
R1
R1
R1
R2
R2
R2
3 Identical Single-Phase Transformers
connected Wye-Delta
Let,
563412M
64222
53111
ZZMjZZ
ZZLjRZ
ZZLjRZ
==ω==
==ω+=
==ω+=
1
3
5 6
4
2
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Transformer Model
The Primitive Voltage Equations
Z2ZM
ZMZ1
Z2ZM
ZMZ1
Z2ZM
ZMZ1
V6
V5
V4
V3
V2
V1
I6
I5
I4
I3
I2
I1
=
The Inverse of the Impedance Matrix
Z1 Z2 –ZM2
1The Primitive Admittance Matrix
Z1-ZM
-ZMZ2
Z1-ZM
-ZMZ2
Z1-ZM
-ZMZ2
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Transformer Model
YBUS =
YBUS = [C][ Yprim][CT]
2Z1-Z1-Z1-ZMZM
-ZM2Z1-Z1-ZMZM
-Z1-Z12Z1ZM-ZM
-ZMZMZ2
ZM-ZMZ2
ZM-ZMZ2
Z1 Z2 –ZM2
1
A B C a b c
A
B
C
a
b
c
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Transformer Model
Iinj = [C Yprim CT] Vnode
YBUS = [C][Yprim][CT]
1-1
1
-11
1
-11
1
1-1
1-1
-11
1
1
1
The Bus Admittance Matrix
Z1-ZM
-ZMZ2
Z1-ZM
-ZMZ2
Z1-ZM
-ZMZ2
Z1 Z2 –ZM2
1
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2a2yt-a2yt-a2yt-aytayt
-a2yt2a2yt-a2yt-aytayt
-a2yt-a2yt2a2ytayt-ayt
-aytaytyt
ayt-aytyt
ayt-aytyt
Three Phase Model
Define 2
m21
2t zzz
zy
−=
2
1
n
na =
YBUS =
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Three Phase ModelIf the admittances are already in per unit system, then the effective turns ratio ”a” must be
3
1
n
na
2
1 ==
2
m21
2t zzz
zy
−=
ty
ty
ty
3
1− ty
3
1ty
3
1− ty
3
1ty
3
1− ty
3
1ty
3
1− ty
3
1− ty
3
1− ty
3
1ty
3
1ty
3
1ty
3
1− ty
3
1− ty
3
1− ty
3
1− ty
3
1− ty
3
1− ty
3
2ty
3
2ty
3
2ty
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Three Phase ModelSummary
YccYcbYcaYcCYcBYcA
YbcYbbYbaYbCYbBYbA
YacYabYaaYaCYaBYaA
YccYCbYCaYCCYCBYCA
YBcYBbYBaYBCYBBYBA
YAcYabYAaYACYABYAA
[Ybus] =
YSSYSP
YPSYPP
A B C a b c
A
B
C
a
b
c
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Three Phase Model
YSPYPSYSSYPPSECPRI
-YII-YIIYIIYIIDeltaDelta
YIIIYIIITYIIYIIWyeDelta
YIIIYIIITYIYIIWye-GDelta
YIIITYIIIYIIYIIDeltaWye
-YII-YIIYIIYIIWyeWye
-YII-YIIYIIYIIWye-GWye
YIIITYIIIYIIYIDeltaWye-G
-YII-YIIYIIYIIWyeWye-G
-YI-YIYIYIWye-GWye-G
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Three Phase ModelSummary
yt
yt
yt
YI =2yt-yt-yt
-yt2yt-yt
-yt-yt2yt
-ytyt
yt-yt
yt-yt
YII = 1/3
YIII = 1/√√√√3
-ytyt
-ytyt
yt-yt
YIIIT = 1/√√√√3
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Three Phase ModelExample:
Three single-phase transformers rated 50 kVA, 7.62kV/240V, %Z=2.4, X/R=3 are connected Wye(grounded)-Delta. Determine the Admittance Matrix Model of the Transformer Bank. Assume yt = 1/zt
yp.u. = ____ -j ____Zp.u. = ____ +j ____
[Ybus] =
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3-Phase Transformer Impedance Matrix Model
� Distributing Transformer Impedance Between Windings
� Impedance Matrix in Backward-Forward Sweep Load Flow� Wye-Grounded – Wye-Grounded
� Delta-Delta
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Transformer Equations
Consider the winding-to-winding relationship between primary and secondary:
From transformer equations,
PRI
SEC
Va
V=
1PRI
SEC
I
I a=
2PRI
SEC
Za
Z=
PRI
SEC
Na
N=
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Distributing Transformer Impedance Between Windings
� Transformers are typically modeled with series impedance lumped at either end.
� To properly model transformer behavior, series impedance must be modeled in both windings.
� PROBLEM: divide ZT into ZP and ZS given a
'T P SZ Z Z= +
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Distributing Transformer Impedance Between Windings
� ASSUMPTION: Transformer impedance varies as number of wire turns.
Referring ZS to primary side ,
Substituting,
S PZ aZ=
2 3'S S PZ a Z a Z= =
3
3(1 )T P P
P
Z Z a Z
a Z
= +
= +
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Distributing Transformer Impedance Between Windings
To find ZP and ZS,
3
1
(1 )P TZ Za
=+
3(1 )S T
aZ Z
a=
+
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Impedance Matrix in Backward-Forward Sweep Load Flow
� Transformer model involved in� backward summation of current
� forward computation of voltage
� Wye-Grounded – Wye-Grounded
� Delta-Delta
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Wye Grounded – Wye Grounded
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WyeGnd-WyeGnd Backward Sweep
� Secondary to Secondary Winding
� Secondary Winding to Primary Windingif in PU: If not in PU:
� Primary Winding to Primary
_ _1
_ _ 2
_ _ 3
1 0 0
0 1 0
0 0 1
Sec Winding a
Sec Winding b
Sec Winding c
I I
I I
I I
⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥=⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦⎣ ⎦
Pr _ _1 _ _1
Pr _ _ 2 _ _ 2
Pr _ _ 3 _ _ 3
10 0
10 0
10 0
i Winding Sec Winding
i Winding Sec Winding
i Winding Sec Winding
aI I
I Ia
I I
a
⎡ ⎤⎢ ⎥
⎡ ⎤ ⎡ ⎤⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥=⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎣ ⎦ ⎣ ⎦
⎢ ⎥⎢ ⎥⎣ ⎦
Pr _ _1 _ _1
Pr _ _ 2 _ _ 2
Pr _ _3 _ _ 3
1 0 0
0 1 0
0 0 1
i Winding Sec Winding
i Winding Sec Winding
i Winding Sec Winding
I I
I I
I I
⎡ ⎤ ⎡ ⎤⎡ ⎤⎢ ⎥ ⎢ ⎥⎢ ⎥=⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎣ ⎦⎣ ⎦ ⎣ ⎦
Pr _ _1
Pr _ _ 2
Pr _ _ 3
1 0 0
0 1 0
0 0 1
A i Winding
B i Winding
C i Winding
I I
I I
I I
⎡ ⎤⎡ ⎤ ⎡ ⎤⎢ ⎥⎢ ⎥ ⎢ ⎥= ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦
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WyeGnd-WyeGnd Forward Sweep� Primary to Primary winding
� Primary Winding to Secondary WindingIf in PU: If not in PU:
� Secondary Winding to Secondary
Pr _ _1 Pr _ _1 Pr _ _1
Pr _ _ 2 Pr _ _ 2 Pr _ _ 2
Pr _ _3 Pr _ _3 Pr _ _3
1 0 0 *
0 1 0 *
0 0 1 *
i Winding AN i Winding i Winding
i Winding BN i Winding i Winding
i Winding CN i Winding i Winding
V V I Z
V V I Z
V V I Z
⎡ ⎤ ⎡ ⎤⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥= −⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦⎣ ⎦ ⎣ ⎦
_ _1 Pr _ _1
_ _ 2 Pr _ _ 2
_ _ 3 Pr _ _3
10 0
10 0
10 0
Sec Winding i Winding
Sec Winding i Winding
Sec Winding i Winding
aV V
V Va
V V
a
⎡ ⎤⎢ ⎥
⎡ ⎤ ⎡ ⎤⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥=⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎣ ⎦ ⎣ ⎦
⎢ ⎥⎢ ⎥⎣ ⎦
_ _1 Pr _ _1
_ _ 2 Pr _ _ 2
_ _3 Pr _ _3
1 0 0
0 1 0
0 0 1
Sec Winding i Winding
Sec Winding i Winding
Sec Winding i Winding
V V
V V
V V
⎡ ⎤ ⎡ ⎤⎡ ⎤⎢ ⎥ ⎢ ⎥⎢ ⎥=⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎣ ⎦⎣ ⎦ ⎣ ⎦
_ _1 _ _1 _ _1
_ _ 2 _ _ 2 _ _ 2
_ _ 3 _ _ 3 _ _ 3
1 0 0 *
0 1 0 *
0 0 1 *
an Sec Winding Sec Winding Sec Winding
bn Sec Winding Sec Winding Sec Winding
cn Sec Winding Sec Winding Sec Winding
V V I Z
V V I Z
V V I Z
⎡ ⎤ ⎡ ⎤⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥= −⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦
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Delta-Delta Transformer Connection
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Delta-Delta Backward Sweep� Secondary to Secondary Winding
� Secondary Winding to Primary WindingIf in PU: If not in PU:
� Primary Winding to Primary
_ _1
_ _ 2
_ _ 3
1 1 01
0 1 13
1 0 1
Sec Winding a
Sec Winding b
Sec Winding c
I I
I I
I I
⎡ ⎤ −⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥= −⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦⎣ ⎦
Pr _ _1 _ _1
Pr _ _ 2 _ _ 2
Pr _ _3 _ _ 3
10 0
10 0
10 0
i Winding Sec Winding
i Winding Sec Winding
i Winding Sec Winding
aI I
I Ia
I I
a
⎡ ⎤⎢ ⎥
⎡ ⎤ ⎡ ⎤⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥=⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎣ ⎦ ⎣ ⎦
⎢ ⎥⎢ ⎥⎣ ⎦
Pr _ _1 _ _1
Pr _ _ 2 _ _ 2
Pr _ _3 _ _ 3
1 0 0
0 1 0
0 0 1
i Winding Sec Winding
i Winding Sec Winding
i Winding Sec Winding
I I
I I
I I
⎡ ⎤ ⎡ ⎤⎡ ⎤⎢ ⎥ ⎢ ⎥⎢ ⎥=⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎣ ⎦⎣ ⎦ ⎣ ⎦
Pr _ _1
Pr _ _ 2
Pr _ _ 3
1 0 1
1 1 0
0 1 1
a i Winding
b i Winding
c i Winding
I I
I I
I I
⎡ ⎤−⎡ ⎤ ⎡ ⎤⎢ ⎥⎢ ⎥ ⎢ ⎥= − ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦ ⎣ ⎦
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Delta-Delta Forward Sweep� Primary to Primary Winding
� Primary Winding to Secondary Winding If in PU: If not in PU:
� Secondary Winding to Secondary
Pr _ _1 Pr _ _1 Pr _ _1
Pr _ _ 2 Pr _ _ 2 Pr _ _ 2
Pr _ _ 3 Pr _ _ 3 Pr _ _ 3
1 1 0 *
0 1 1 *
1 0 1 *
i Winding AN i Winding i Winding
i Winding BN i Winding i Winding
i Winding CN i Winding i Winding
V V I Z
V V I Z
V V I Z
⎡ ⎤ ⎡ ⎤−⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥= − −⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦⎣ ⎦ ⎣ ⎦
_ _1 Pr _ _1
_ _ 2 Pr _ _ 2
_ _ 3 Pr _ _ 3
10 0
10 0
10 0
Sec Winding i Winding
Sec Winding i Winding
Sec Winding i Winding
aV V
V Va
V V
a
⎡ ⎤⎢ ⎥
⎡ ⎤ ⎡ ⎤⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥=⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎣ ⎦ ⎣ ⎦
⎢ ⎥⎢ ⎥⎣ ⎦
_ _1 Pr _ _1
_ _ 2 Pr _ _ 2
_ _ 3 Pr _ _ 3
1 0 0
0 1 0
0 0 1
Sec Winding i Winding
Sec Winding i Winding
Sec Winding i Winding
V V
V V
V V
⎡ ⎤ ⎡ ⎤⎡ ⎤⎢ ⎥ ⎢ ⎥⎢ ⎥=⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎣ ⎦⎣ ⎦ ⎣ ⎦
_ _ 1
_ _ 2
_ _ 3
1 3 00 0
31 1 5 0
0 03
1 3 00 1
3
a S ec W in d in g
b S ec W in d in g
c S ec W in d in g
V V
V V
V V
∠ −⎡ ⎤⎢ ⎥⎢ ⎥ ⎡ ⎤⎡ ⎤
∠ −⎢ ⎥ ⎢ ⎥⎢ ⎥ = ⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎣ ⎦ ⎣ ⎦∠ −⎢ ⎥⎢ ⎥⎣ ⎦
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Transmission Line Distribution Line
� Series Impedance of Lines
� Shunt Capacitance of Lines
� Nodal Admittance Matrix Model
� Data Requirements
Transmission and DistributionLine Models
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ZccZcbZca
ZbcZbbZba
ZacZabZaa
Y’ccY’cbY’ca
Y’bcY’bbY’ba
Y’acY’abY’aa
Y”ccY”cbY”ca
Y”bcY”bbY”ba
Y”acY”abY”aa
ABC
abc
Transmission and DistributionLine Models
••
••+
-VR
+
-
Z = R + jXL
CY21
CY21
Balanced Three-Phase System
Unbalanced Three-Phase
System
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Conductor Materials� Aluminum (Al) is preferred over Copper (Cu) as a
material for transmission and distribution lines due to:� lower cost
� lighter weight
� larger diameter for the same resistance**This results in a lower voltage gradient at the conductor
surface (less tendency for corona)
� Copper is preferred over Aluminum as a material for distribution lines due to lower resistance to reduce system losses.
Series Impedance of Lines
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Stranding of Conductors
Steel Aluminum
Alternate layers of wire of a stranded conductor are spiraled in opposite directions to prevent unwinding and make the outer radius of one layer coincide with the inner radius of the next.
The number of strands depends on the number of layers and on whether all the strands are of the same diameter. The total number of strands of uniform diameter in a concentrically stranded cable is 7, 19, 37, 61, 91, etc.
Hard-Drawn Copper(Cu)
Aluminum Conductor Steel Reinforced
(ACSR)
Series Impedance of Lines
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Resistance of Conductors� The Resistance of a Conductor depends on the
material (Cu or Al)
� Resistance is directly proportional to Length but inversely proportional to cross-sectional area
� Resistance increases with Temperature
� Skin-Effect in Conductors
LR
Aρ=
R – Resistanceρρρρ – Resistivity of MaterialL – LengthA – Cross-Sectional Area
Series Impedance of Lines
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Conductor Strands O.D. GMR ResistanceType Value Unit (Inches) (feet) (Ohm/Mile)
1 ACSR 6 AWG 6/1 0.19800 0.00394 3.98000 2 ACSR 5 AWG 6/1 0.22300 0.00416 3.18000 3 ACSR 4 AWG 7/1 0.25700 0.00452 2.55000 4 ACSR 4 AWG 6/1 0.25000 0.00437 2.57000 5 ACSR 3 AWG 6/1 0.28100 0.00430 2.07000 6 ACSR 2 AWG 7/1 0.32500 0.00504 1.65000 7 ACSR 2 AWG 6/1 0.31600 0.00418 1.69000 8 ACSR 1 AWG 6/1 0.35500 0.00418 1.38000 9 ACSR 1/0 AWG 6/1 0.39800 0.00446 1.12000 10 ACSR 2/0 AWG 6/1 0.44700 0.00510 0.89500
INDEXSize
Source: Westinghouse T&D Handbook
Resistance of Conductors
Series Impedance of Lines
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Line Inductance
Series Impedance of Lines
Self Inductance: extint LLL +=
Mutual Inductance (between 2 conductors):
1Ir
11z
22z
12z
1
2Ir
1’
2 2’
122111'11 zIzIV +=−
222
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Carson’s Line
aVr
aIr+
-
aaz
ddz
adz
a a’
d d’
0Vd =r
Local Earth
Remote Earth
Fictitious Return Conductor
dIr
Carson examined a single overhead conductor whose remote end is connected to earth.
The current returns through a fictitious earth conductor whose GMR is assumed to be 1 foot (or 1 meter) and is located a distance Dad from the overhead conductor.
REF
Series Impedance of Lines
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The line is described by the following equations:
' 'aa a a aa a ad dV V V z I z I= − = +r r r r r
' 'dd d d ad a dd dV V V z I z I= − = +r r r r r
Note: , and .a dI I= −r r
0dV =r
' ' 0a dV V− =r r
Subtracting the two equations, we get
( 2 )a aa dd ad aV z z z I= + −r r
a aa aV z I=r r
or
zaa is the equivalent impedance of the single overhead conductor.
2aa aa dd adz z z z= + −
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Primitive Impedances:2
(ln 1)aa a a asa
sz r j L r j k
Dω ω= + = + −
2(ln 1)dd d
sd
sz r j k
Dω= + −
2(ln 1)ad
ad
sz j M j k
Dω ω= = −
ra, rd = resistances of overhead conductor and fictitious ground wire, respectively
Dsa, Dsd = GMRs of overhead conductor and fictitious ground wire, respectively
Series Impedance of Lines
Dad = Distance between the overhead conductor and fictitious ground wire
( )( )ohm/meterxfk 71022 −= πω
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Earth Resistance:
Carson derived an empirical formula for the earth resistance.
-31.588 x 10dr f= Ω/mile
-49.869 x 10 f= Ω/km
Note : At 60 Hz,
0.09528dr = Ω/mile
where f is the power frequency in Hz
Series Impedance of Lines
059214.0= Ω/km
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Geometric Mean Radius
For a solid conductor with radius r, r78.0rD 4
1
s ==−
ε
Bundle of Two Bundle of Four
d
dDD scs =
d
d
4 3scs dD09.1D =
Note: Dsc=GMR of a single conductor
Series Impedance of Lines
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Equivalent Impedance:Substitute the primitive impedances into
2aa aa dd adz z z z= + −We get 2
adD( ) lnaa a d
sa sd
z r r j kD D
ω= + +
Define2
adDe
sd
DD
=
We geteD
( ) lnaa a dsa
z r r j kD
ω= + + Ω/unit length
Series Impedance of Lines
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The quantity De is a function of frequency and earth resistivity.
2160 /eD fρ= feet
Typical values of De are tabulated below.
Return Earth Condition
Resistivity(Ω-m)
De(ft)
Sea water 0.01-1.0 27.9-279Swampy ground 10-100 882-2790
Average Damp Earth 100 2790Dry earth 1000 8820Sandstone 109 8.82x106
Series Impedance of Lines
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Three-Phase Line Impedances
aVr
cIr+
-ccz
ddz
a a’
d d’
0Vd =r
dIr
bbz
aaz
b
c
bIraIr
abz
bczcaz b’
c’
bVr
cVr
adzbdz
cdz
REF
+
-+
-
All wires grounded here
Series Impedance of Lines
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The voltage equation describing the line is
aa ab ac ad
ba bb bc bd
ca cb cc cd
da db dc dd
z z z z
z z z z
z z z z
z z z z
a
b
c
d
I
I
I
I
r
r
r
r
'
'
'
'
aa
bb
cc
dd
V
V
V
V
r
r
r
r
'
'
'
'
a a
b b
c c
d d
V V
V V
V V
V V
−
−
−
−
r r
r r
r r
r r
= =
Since all conductors are grounded at the remote end, we get from KCL
0a b c dI I I I+ + + =r r r r
or
( )d a b cI I I I= − + +r r r r
Series Impedance of Lines
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We can subtract the voltage equation of the ground conductor from the equations of phases a, b and c. The resulting matrix equation is
aa ab ac
ab bb bc
ac bc cc
z z z
z z z
z z z
a
b
c
V
V
V
r
r
r=
a
b
c
I
I
I
r
r
rV/unit length
Self Impedances:
2aa aa ad ddz z z z= − + Ω/unit length
2bb bb bd ddz z z z= − + Ω/unit length2cc cc cd ddz z z z= − + Ω/unit length
Series Impedance of Lines
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Mutual Impedances:
ddbdadabab zzzzz +−−=
ddcdadacac zzzzz +−−=
Ω/unit length
Ω/unit lengthΩ/unit length
ddcdbdbcbc zzzzz +−−=
Primitive Impedances:2
(ln 1)xx xsx
sz r j k
Dω= + − Ω/unit length
2(ln 1)xy
xy
sz j k
Dω= − Ω/unit length
x=a,b,c,d
xy=ab,bc,ca,ad,bd,cd
Series Impedance of Lines
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1. Identical phase conductorsAssumptions:
s sa sb scD D D D= = =2. Distances of the overhead conductors to the
fictitious ground conductor are the same
e ad bd cdD D D D= = =We get
eD( ) lnaa bb cc a d
s
z z z r r j kD
ω= = = + +
Ω/unit lengtheD lnxy d
xy
z r j kD
ω= +xy=ab,bc,ca
Series Impedance of Lines
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Example: Find the equivalent impedance of the 69-kV line shown. The phase conductors are 4/0 hard-drawn copper, 19 strands which operate at 25oC. The line is 40 miles long. Assume an earth resistivity of 100 Ω-meter.
ra=0.278 Ω/mile @ 25oC
Dsc=0.01668 ft @ 60 Hz
s
edaccbbaa D
Dln kj)rr(zzz ω++===
0.016682790ln 121.0j)095.0278.0( ++=
459.1j373.0 += Ω/mileΩ38.58j93.14Z aa +=
10’
a b c
10’
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102790
bcab ln 121.0j095.0zz +==683.0j095.0 +=
Ω/mile
Ω33.27j81.3Z ab +=
202790
ac ln 121.0j095.0z +=
Ω97.23j81.3Z ac +=
We get
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
+++
+++
+++
38.58j93.1433.27j81.397.23j81.3
33.27j81.338.58j93.1433.27j81.3
97.23j81.333.27j81.338.58j93.14
Zabc= Ω
Series Impedance of Lines
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aVr cI
r+
-
ccz
ddz
aa’
dd’
0Vd =r
dIr
bbz
aaz
b
c
bIraIr
abz
bcz
cazb’
c’
bVr
cVr adz
bdzcdz
REF
+
-+
-All wires grounded here
Lines with Overhead Ground Wire
wwzwIr
w’
wVr+
-
w
Series Impedance of Lines
wdz
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The primitive voltage equation is
aa ab ac aw ad
ba bb bc bw bd
ca cb cc cw cd
wa wb wc ww wd
da db dc dw dd
z z z z z
z z z z z
z z z z z
z z z z z
z z z z z
'
'
'
'
'
0
0
a a
b b
c c
w
d
V V
V V
V V
V
V
−
−
−
−
−
r r
r r
r r
r
r
a
b
c
w
d
I
I
I
I
I
r
r
r
r
r
=
From KCL, we get0a b c w dI I I I I+ + + + =
r r r r r
or( )d a b c wI I I I I= − + + +
r r r r r
V/unit length
Series Impedance of Lines
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It can be shown that
aa ab ac aw
ba bb bc bw
ca cb cc cw
wa wb wc ww
z z z z
z z z z
z z z z
z z z z
a
b
c
w
I
I
I
I
r
r
r
r
a
b
c
w
V
V
V
V
r
r
r
r
=
exx x d
sx
Dz ( r r ) j k ln
Dω= + +
exy d
xy
Dz r j k ln
Dω= + xy=ab,ac,aw,bc,bw,cw
xx=aa,bb,cc,ww
where wV 0=r
Series Impedance of Lines
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Using Kron Reduction technique,
0V1
2
1
II
43
21
ZZZZ
=
where Z1, Z2, Z3 and Z4 are also matrices.
Series Impedance of Lines
131
4211 I)ZZZZ(V −−=
I2 is eliminated and the matrix is reduced to the size of Z1
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Eliminating the ground wire current Iw
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡=
cccbca
bcbbba
acabaa
1
zzz
zzz
zzz
Z
ww4 zZ =
[ ]cwbwaw zzz=2Z =3Z
⎥⎥⎥⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢⎢⎢⎢
⎣
⎡
−−−
−−−
−−−
=
ww
wccwcc
ww
wbcwcb
ww
wacwca
ww
wcbwbc
ww
wbbwbb
ww
wabwba
ww
wcawac
ww
wbawab
ww
waawaa
abc
z
zzz
z
zzz
z
zzz
z
zzz
z
zzz
z
zzz
z
zzz
z
zzz
z
zzz
z
We get
Series Impedance of Lines
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
cw
bw
aw
z
z
z
241
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Example: Find the equivalent impedance of the 69-kV line shown. The phase conductors are the same as in the previous examples. The overhead ground wires have the following characteristics:
rw=4.0 Ω/mile @ 25oC
Dsw=0.001 ft @ 60 Hz
10’a b c
10’
w
15’For the ground wire, we get
sw
edwww D
Dln kj)rr(z ω++=
0.0012790ln 121.0j)095.00.4( ++=
8.1j095.4 +=Ω72j8.163Z ww +=
Ω/mile
Series Impedance of Lines
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aw
edcwaw D
Dln kjrzz ω+==
18.032790ln 121.0j095.0 +=
Ω47.24j81.3ZZ cwaw +==
152790
bw ln 121.0j095.0Z +=
Ω36.25j81.3Zbw +=From a previous example, we got
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
+++
+++
+++
38.58j93.1433.27j81.397.23j81.3
33.27j81.338.58j93.1433.27j81.3
97.23j81.333.27j81.338.58j93.14
Z1= Ω
Ω/mile
Ω/mile
Series Impedance of Lines
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Using the ground wire impedances, we also get
Ω72j8.163Z4 +=⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
+
+
+
47.24j81.3
36.25j81.3
47.24j81.3
=2Z = T3Z
Performing Kron reduction, we get
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
+++
+++
+++
11.56j5.170.25j48.67.21j38.6
0.25j48.697.55j71.170.25j48.6
7.21j38.60.25j48.611.56j5.17
Zabc = Ω
Note: The self impedances are no longer equal.
Series Impedance of Lines
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Line TranspositionLine transposition is used to make the mutual impedances identical.
Pos.1
Pos.2
Pos.3
Phase c
Phase a
Phase b
aIr
bIr
cIr
s1 s3s2
Section 2Section 1 Section 3
Note: Each phase conductor is made to occupy all possible positions.
Series Impedance of Lines
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For Section 1⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡=
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
−−−
−−−
−−−
c
b
a
133132131
123122121
113112111
c
b
a
I
I
I
ZZZ
ZZZ
ZZZ
V
V
V
r
r
r
r
r
r
volts
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡=
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
−−−
−−−
−−−
b
a
c
233232231
223222221
213212211
b
a
c
I
I
I
ZZZ
ZZZ
ZZZ
V
V
V
r
r
r
r
r
r
volts
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡=
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
−−−
−−−
−−−
a
c
b
333332331
323322321
313312311
a
c
b
I
I
I
ZZZ
ZZZ
ZZZ
V
V
V
r
r
r
r
r
r
volts
Voltage Equations for Each Section
For Section 2
For Section 3
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a333222111a I)ZZZ(Vrr
−−− ++=Σb331223112 I)ZZZ(r
−−− +++
c332221113 I)ZZZ(r
−−− +++
a313232121b I)ZZZ(Vrr
−−− ++=Σ
b311233122 I)ZZZ(r
−−− +++
c312231123 I)ZZZ(r
−−− +++
a323212131c I)ZZZ(Vrr
−−− ++=Σb321213132 I)ZZZ(r
−−− +++
c322211133 I)ZZZ(r
−−− +++
The total Voltage Drop at phases a, b, and c are:
Series Impedance of Lines
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Define f1, f2 and f3 as as the ratios of s1, s2 and s3 to the total length s, respectively. We get
a333222111a I)ZfZfZf(Vrr
++=Σ
b313232121 I)ZfZfZf(r
+++
c323212131 I)ZfZfZf(r
+++
a133322211b I)ZfZfZf(Vrr
++=Σ
b113332221 I)ZfZfZf(r
+++
c123312231 I)ZfZfZf(r
+++
a233122311c I)ZfZfZf(Vrr
++=Σb213132321 I)ZfZfZf(r
+++c223112331 I)ZfZfZf(r
+++
Series Impedance of Lines
s
sf 1
1 =
s
sf 2
2 =
s
sf 3
3 =
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Define: 1332321211k ZfZfZfZ ++=
2331221312k ZfZfZfZ ++=
1231322313k ZfZfZfZ ++=
332211s ZZZZ ===
Substitution gives
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡=
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
c
b
a
s3k2k
3ks1k
2k1ks
c
b
a
I
I
I
ZZZ
ZZZ
ZZZ
V
V
V
r
r
r
r
r
r
Σ
Σ
Σ
Volts
Series Impedance of Lines
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It can be shown that
s
edas D
Dln ksjs)rr(Z ω++=
⎟⎟⎠
⎞⎜⎜⎝
⎛+++=
31
e3
23
e2
12
e1d1k D
Dlnf
D
Dlnf
D
DlnfksjsrZ ω
⎟⎟⎠
⎞⎜⎜⎝
⎛+++=
23
e3
12
e2
31
e1d2k D
Dlnf
D
Dlnf
D
DlnfksjsrZ ω
⎟⎟⎠
⎞⎜⎜⎝
⎛+++=
12
e3
31
e2
23
e1d3k D
Dlnf
D
Dlnf
D
DlnfksjsrZ ω
Series Impedance of Lines
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Example: Find the equivalent impedance of the 69-kV line shown. The phase conductors are 4/0 hard-drawn copper, 19 strands which operate at 25oC. The line is 40 miles long. Assume s1=8 miles, s2=12 miles and s3=20 miles.
ra=0.278 Ω/mile @ 25oC
Dsc=0.01668 ft @ 60 Hz
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
+++
+++
+++
=
38.58j93.1433.27j81.397.23j81.3
33.27j81.338.58j93.1433.27j81.3
97.23j81.333.27j81.338.58j93.14
Zabc Ω
Without the transposition,
10’
a b c
10’
Section 1
Series Impedance of Lines
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Solving for the mutual impedances, we get
1332321211k ZfZfZfZ ++=
)33.27j81.3(3.0)33.27j81.3(2.0 +++=)97.23j81.3(5.0 ++
Ω65.25j81.3 +=Similarly, we get
2331221312k ZfZfZfZ ++= Ω66.26j81.3 +=
1231322313k ZfZfZfZ ++= Ω32.26j81.3 +=
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⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
+++
+++
+++
38.58j93.1433.27j81.397.23j81.3
33.27j81.338.58j93.1433.27j81.3
97.23j81.333.27j81.338.58j93.14
Zabc= Ω
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
+++
+++
+++
38.58j93.1432.26j81.366.26j81.3
32.26j81.338.58j93.1465.25j81.3
66.26j81.365.25j81.338.58j93.14
Zabc= Ω
The impedance matrix of the transposed line is
For comparison, the impedance matrix of the untransposed line is
Series Impedance of Lines
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Completely Transposed LineIf s1=s2=s3, the line is completely transposed. We get
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡=
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
c
b
a
smm
msm
mms
c
b
a
I
I
I
ZZZ
ZZZ
ZZZ
V
V
V
r
r
r
r
r
r
ΣΣΣ
Volts
s
edas D
Dln ksjs)rr(Z ω++=
)ZZZ(Z 13231231
m ++=m
ed D
Dln ksjsr ω+=
where
Ds, Dm = GMR and GMD, respectively
Ω
Ω
Series Impedance of Lines
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Geometric Mean Distance (GMD)Typical three-phase line configurations
3312312m DDDD =
D12 D23
D31
D12
D23D 31
D12 D23
D31
D12
D23
D31
Series Impedance of Lines
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Example: For the same line assume a complete transposition cycle.
We get the average of the mutual impedances.
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
+++
+++
+++
38.58j93.1421.26j81.321.26j81.3
21.26j81.338.58j93.1421.26j81.3
21.26j81.321.26j81.338.58j93.14
Zabc= Ω
The GMD is10’
a b c
10’
feet 6.12)20)(10(10D 3m ==
Ω21.26j81.3Z m +=The impedance of the transposed line is
Series Impedance of Lines
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Phase to Sequence ImpedancesConsider a transmission line that is described by the following voltage equation:
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
ccbcac
bcbbab
acabaa
ZZZ
ZZZ
ZZZ
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
c
b
a
V
V
V
r
r
r
=⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
c
b
a
I
I
I
r
r
r
volts
or
abcabcabc IZVrr
=
From symmetrical components, we have
012abc VAVrr
= and 012abc IAIrr
=
Series Impedance of Lines
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Substitution gives
012abc012 IAZVArr
=or012abc
1012 IAZAV
rr−=
which implies that
AZAZ abc1
012−=
Performing the multiplication, we get
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
−+−
+−−
−−+
=
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
0m0s1m1s2m2s
2m2s0m0s1m1s
1m1s2m2s0m0s
2
1
0
ZZZ2ZZZ
Z2ZZZZZ
ZZZZZ2Z
Z
Z
Z
Note: Z012 is not symmetric.
Series Impedance of Lines
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It can be shown that
)ZZZ(Z ccbbaa31
0s ++=
)ZaaZZ(Z cc2
bbaa31
1s ++=
)aZZaZ(Z ccbb2
aa31
2s ++=
)ZZZ(Z cabcab31
0m ++=
)aZZZa(Z cabcab2
31
1m ++=
)ZaZaZ(Z ca2
bcab31
2m ++=
Series Impedance of Lines
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If the line is completely transposed,
s0s ZZ = m0m ZZ =
0ZZ 2s1s == 0ZZ 2m1m ==
The sequence impedance matrix reduces to
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
−
−
+
=
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
ms
ms
ms
2
1
0
ZZ00
0ZZ0
00Z2Z
Z
Z
Z
Note: The sequence impedances are completely decoupled.
Series Impedance of Lines
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For a completely transposed line, the equation in the sequence domain is
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡=
2a
1a
0a
2
1
0
2a
1a
0a
I
I
I
Z00
0Z0
00Z
V
V
V
r
r
r
r
r
r
where
s
ma21 D
Dln ksjsrZZ ω+==
2ms
3e
da0DD
Dln ksjsr3srZ ω++=
Ω
Ω
Series Impedance of Lines
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Example: For the same line and assuming a complete transposition cycle, find the sequence impedances of the line. 10’
a b c
10’
Ω21.26j81.3Z m +=
Ω38.58j93.14Z s +=
In the previous example, we got
The sequence impedances are
Ω80.110j55.22Z2ZZ ms0 +=+=
Ω17.32j12.11ZZZZ ms21 +=−==
Series Impedance of Lines
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• Self-capacitance
• Mutual-capacitance
a
b
c
w
CagCbg
Ccg
Cwg
Cab
Cac
Caw
Cbc
Cbw
Ccw
Capacitance of Three Phase Lines
Shunt Capacitance of Lines
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Voltage Due to Charged Conductor
Consider two points P1 and P2which are located at distances D1 and D2 from the center of the conductor.
P1
P2
D1
D2
The voltage drop from P1 to P2 is
1
212 D
Dln
2
qv
πε= Volts
Shunt Capacitance of Lines
+q
xra
rax2
qDE
πεε==
rr
Electric Field of a Long ConductorPermitivity of medium
Electric charge
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Capacitance of a Two-Wire LineThe capacitance between two conductors is defined as the charge on the conductors per unit of potential difference between them.Consider the two cylindrical conductors shown.
qa qb
D
a
aab r
Dln
2
qv
πε=
Due to charge qa, we get the voltage drop vab.
Shunt Capacitance of Lines
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D
rln
2
q
r
Dln
2
qv bb
b
bab πεπε
=−=
Due to charge qb, we also get the voltage drop vba.
orb
bba r
Dln
2
qv
πε=
Applying superposition, we get the total voltage drop from charge qa to charge qb.
D
rln
2
q
r
Dln
2
qv bb
a
aab πεπε
+=
Since qa+qb=0, we get
ba
2a
ab rr
Dln
2
qv
πε= Volts
Shunt Capacitance of Lines
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In general, ra=rb. We getr
Dln
qv a
ab πε= Volts
The capacitance between conductors is
r
DlnV
qC
ab
aab
πε== Farad/meter
The capacitance to neutral is
r
Dln
2C2CC abbnan
πε=== Farad/meter
Shunt Capacitance of LinesSelf-Capacitance
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Mutual Capacitance
In capacitance calculations, the earth is assumed as a perfectly conducting plane. The electric field that results is the same if an image conductor is used for every conductor above ground.
Mirror Conductors below ground
-qb
+qa
+qb
+qc
+qw
-qa-qc
-qw
Dab
Dac
Daw
Haa Hab Hac Haw
Shunt Capacitance of Lines
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The voltage drop from conductor a to ground is
'aa21
a vv =
an
ann
ab
abb
a
aaa D
Hln q...
D
Hln q
r
Hln q(
4
1+++=
πε
)H
Dln q...
H
Dln q
H
rln q
an
ann
ab
abb
aa
aa −−−−
Combining common terms, we get
)D
Hln q...
D
Hln q
r
Hln q(
2
1v
an
ann
ab
abb
a
aaaa +++=
πε
Shunt Capacitance of Lines
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In general, for the kth overhead conductor
k
kkk
bk
bkb
ak
akak r
Hln q...
D
Hln q
D
Hln q(
2
1v +++=
πε
)D
Hln q...
nk
nkn++
Using matrix notation, we get
…
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
=
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
n
b
a
nnncnbna
bnbcbbba
anacabaa
n
b
a
q
q
q
P...PPP
P...PPP
P...PPP
v
v
v
MMMMMMM
Shunt Capacitance of Lines
k
kkkk r
Hln
2
1P
πε=
kj
kjkj D
Hln
2
1P
πε=
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Inversion of matrix P gives
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
+−−−
−−+−
−−−+
=
nnncnbna
bnbcbbba
anacabaa
C...CCC
C...CCC
C...CCC
CMMMMM
Shunt Capacitance of Lines
[ ] [ ][ ]qPv =
Since, q = Cv, ,then [ ] [ ] 1PC −=
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The Shunt Admittance is
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
+−−−
−−+−
−−−+
=
nnncnbna
bnbcbbba
anacabaa
bus
Cj...CjCjCj
Cj...CjCjCj
Cj...CjCjCj
Y
ωωωω
ωωωω
ωωωω
Shunt Capacitance of Lines
The difference between the magnitude of a diagonal element and its associated off-diagonal elements is the capacitance to ground. For example, the capacitance from a to ground is
anacabaaag C...CCCC −−−−=
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Capacitance of a Transposed Line
Pos.1
Pos.2
Pos.3
Phase c
Phase a
Phase b
aq
bq
cq
Section 2Section 1 Section 3
s31 s3
1 s31
Shunt Capacitance of Lines
r
Dln
2
v
qCCC
man
acnbnan
πε====
The capacitance of phase a to neutral is
Farad/meter, to neutral
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fC2
1xc π
=
r
Dln 10 x
f
862.2x m9
c =
r
Dln 10 x
f
779.1x m6
c =
Ω-meter, to neutral
Ω-mile, to neutral
Note: To get the total capacitive reactance, divide xc by the total length of the line.
Capacitive Reactance
Shunt Capacitance of Lines
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Using matrix notation, we have
Sequence Capacitance
abcabcabc VYIrr
= abcabcabc VCjIrr
ω=
From and , we get012abc VAVrr
= abcabcabc VYIrr
=
012abc012 VACjIArr
ω=or
012abc1
012 VACAjIrr
−= ω
Thus, we have
ACAC abc1
012−=
Shunt Capacitance of Lines
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For a completely transposed line,
ccbbaa0s CCCC ===
acbcab0m CCCC ===Substitution gives
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
+
+
−
)CC(00
0)CC(0
00)C2C(
0m0s
0m0s
0m0s
=C012
or
0m0s0 C2CC −= 0m0s21 CCCC +==
Shunt Capacitance of Lines
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Example: Determine the phase and sequence capacitances of the transmission line shown. The phase conductors are 477 MCM ACSR 26/7 whose radius is 0.0357 ft. The line is 50 miles long and is completely transposed.Calculate distances
Haa=Hbb=Hcc=80 ft
Hab=Hbc=81.2 ft
Hac=84.8 ftFind the P matrix
a
aa
0ccbbaa r
Hln
2
1PPP
πε===
14’
a b c
14’
40’
Shunt Capacitance of Lines
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For air, 10 x36
1 9-0 π
ε = Farad/meter
Substitution gives
0357.0
80ln 10 x 18P 9
aa =
910 x 86.138= Meter/Farad610 x 29.86= Mile/Farad
ab
ab
0bcab D
Hln
2
1PP
πε==
Similarly, we get
610 x 66.19= Mile/Farad
Shunt Capacitance of Lines
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The P matrix can be shown to be
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
29.8666.1939.12
66.1929.8666.19
39.1266.1929.86
=P
Using matrix inversion, we get the C matrix.
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
−−
−−
−−
34.1254.219.1
54.275.1254.2
19.154.234.12
=C
x 106 mi/F
x 10-9 F/mi
Shunt Capacitance of Lines
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For 50 miles, we get
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
−−
−−
−−
17.627.160.0
27.138.627.1
60.027.117.6
=C x 10-7 F
The capacitances to ground areF43.0CCCC acabaaag μ=−−=
F38.0CCCC bcabbbbg μ=−−=
F43.0CCCC acbccccg μ=−−=Since the line is transposed,
F41.0)CCC(C cgbgag31
0g μ=++=
Shunt Capacitance of Lines
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The self- and mutual capacitances are
F62.0)CCC(C ccbbaa31
0s μ=++=
F105.0)CCC(C cabcab31
0m μ=++=
The sequence capacitances are
0m0s0 C2CC −=
0m0s21 CCCC +==
a
b
cCg0
Cg0Cg0
Cm0
Cm0
Cm0
F41.0 μ=
F725.0 μ=
Shunt Capacitance of Lines
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ZccZcbZca
ZbcZbbZba
ZacZabZaa
YccYcbYca
YbcYbbYba
YacYabYaa
YccYcbYca
YbcYbbYba
YacYabYaa
[Z]
[Y]/2 [Y]/2
[Ikabc]
[IiABC]
[Z]-1+[Y]/2-[Z]-1
-[Z]-1[Z]-1+[Y]/2=
6x1 6x16x6
[Vkabc]
[ViABC]
IiABC Ik
abc
ViABC Vk
abc
Nodal Admittance Matrix Model
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3’ 3’1’
4’
24’
A B C
NPhase Conductor336,400 26/7 ACSR
Neutral Conductor4/0 6/1 ACSRLength: 300 ft.
Example
Nodal Admittance Matrix Model
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Data Requirements
� Phasing
� Configuration
� System Grounding Type
� Length
� Phase Conductor Type, Size & Strands
� Ground/Neutral Wire Type, Size & Strands
� Conductor Spacing
� Conductor Height
� Earth Resistivity
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Ha
VerticalOne Ground Wire
(b)
Dbc
DabDca
Hc HbHg
a
c
b
Hg
TriangularOne Ground Wire
(c)
DabDbc
Dca
Ha Hc Hb
ac
bDca
Horizontal One Ground Wire
(a)
Ha
Dab Dbc
Hb HgHc
a cb
Configuration, Spacing, and Height(Subtransmission Lines)
Distribution Line Models
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Triangular Two Ground Wires
(e)
Dgg
ParallelTwo Ground Wires
(f)
D12
Circuit No. 1
Circuit No. 2
Dgg
Horizontal Two Ground Wires
(d)
Dgg
Line Spacing (Ground Wires)
Distribution Line Models
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A
N
1-Phase (A)
A B
N
V-Phase (AB)
A
N
B
B A
N
V-Phase (BA)
B
N
A
Note:N – Consider
the grounded neutral as Ground Conductor for Hg
Hg
A B C
N
3-Phase (ABC)
A
N
B
C
C A B
N
3-Phase (CAB)
C
N
A
B
B C A
N
3-Phase (BCA)
B
N
C
A
Configuration, Spacing, and Height (Distribution Lines)
Distribution Line Models
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