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Course Outline 1. Utility Thevenin Equivalent Circuit 2. Load Models 3. Generator Models 4. Transformer Models 5. Transmission and Distribution Line Models
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Utility Thevenin Equivalent Circuit
Thevenin’s Theorem
Utility Fault MVA
Equivalent Circuit of Utility
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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 A Linear Active Network
+
A Zth
Vth B
-
B
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. U. P. National Engineering Center National Electrification Administration
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Utility Fault MVA Electric Utilities conduct short circuit analysis at the Connection Point of their customers
Electric Utility Grid IF Fault
Customer Facilities
Customers obtain the Fault Data at the Connection Point to represent the Utility Grid for their power system analysis
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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 MVA3φ
X/R3φ
• Single Line-to-Ground Fault
Fault MVALG
X/RLG
System Nominal Voltage in kV U. P. National Engineering Center National Electrification Administration
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Equivalent Circuit of Utility Positive & Negative Sequence Impedance From Three-Phase Fault Analysis
I TPF =
Z1 =
V
[V ]
2
f
S TPF
Z1
= V f I TPF =
f
Z1
Where, Z1 and Z2 are the equivalent positive2 sequence and kV = Z 2 negative-sequence Fault MVA 3φ impedances of the utility
[ ]
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Equivalent Circuit of Utility Zero Sequence Impedance From Single Line-to-Ground Fault Analysis
I SLGF =
3V f Z1 + Z2 + Z0
2Z1 + Z0 =
[ ]
3 Vf
S SLGF = V f I SLGF =
[ ]
3Vf
2
2Z 1 + Z 0
Z1 = Z2
2
SSLGF
Resolve to real and imaginary components then solve for Zo
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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 MVA3φ = 3500 MVA,
X/R3φ = 22
Fault MVALG = 3000 MVA,
X/RLG = 20
The Base Power is 100 MVA U. P. National Engineering Center National Electrification Administration
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Equivalent Circuit of Utility Base Power: 100 MVA Base Voltage: 69 kV Base Impedance: [69]2/100 = 47.61 ohms
[kV ]
2
Z1 = Z2 =
[69 ]
2
Fault MVA 3 φ
=
3500
= 1.3603 Ω
In Per Unit,
or
Z actual 1.3603 = = 0.0286 p.u. Z1 = Z2 = Z base 47 . 61 100MVA BASE Z1 = Z2 = = 0.0286 p.u. 3500 MVA FAULT U. P. National Engineering Center National Electrification Administration
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Equivalent Circuit of Utility Solving for the Resistance and Reactance, √[(1 + (X/R)2]
Z
X
θ
θ
R
1
θ = tan −1 [ X / R ] X/R R = Z cos θ X = Z sin θ
R 1 = 0.0286 cos [tan -1 (22 )] = 0 . 00 13 p.u. = R 2
X 1 = 0.0286 sin [tan = 0 . 028571
-1
(22 )]
p.u. = X 2
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+ 0.0013+j0.028571 +
V f 1∠0 -
-
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Equivalent Circuit of Utility For the Zero Sequence Impedance,
SLGF P .U . = Voltage
P .U .
2Z 1 + Z 0 =
3000 MVA SLGF ( actual ) 100 MVA BASE
= 30 p .u .
69 kV = = 1 . 0 p .u . 69 kV
[ ]
3Vf
2
S SLGF
3[1.0 ] = = 0 .1 30 2
[ } = 0.1sin [tan
] (20 )] = 0.099875
Re al {2 Z 1 + Z 0 } = 0.1cos tan -1 (20 ) = 0.004994 Im ag {2 Z 1 + Z 0
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p.u. p.u.
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Equivalent Circuit of Utility 2 Z 1 + Z 0 = 0.004994 + j0.099875
Z 0 = (0.004994 + j0.099875) − 2(0.0013 + j0.028571) = 0.003694 + j0.042733 p.u. +
+
+
0.0013+j0.028571 +
V f 1∠0
0.0013+j0.028571
0.003694 + j0.042733
-
-
Positive Sequence
-
Negative Sequence
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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
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Equivalent Circuit of Utility The equivalent sequence networks of the Electric Utility Grid are: +
+
R2 +jX2
R0 +jX0
+
r + Eg
R1 +jX1
-
-
-
Positive Sequence
Negative Sequence
-
Zero Sequence
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Load Models
Types of Load
Customer Load Curve
Calculating Hourly Demand
Developing Load Models
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Types of Load An illustration: Sending End
VS = ?
Line
1.1034 + j2.0856 ohms/phase ISR = ?
Receiving End
VR = 13.2 kVLL Load 2 MVA, 3Ph 85%PF 13.2 kVLL
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Types of Load An illustration: Sending End
Line
1.1034 + j2.0856 ohms/phase
VS = ?
ISR = ?
Constant Power (P & Q) 2 MVA = 1.7 MW + j1.0536 MVAR
Receiving End
VR = 13.2 kVLL Load 2 MVA, 3Ph 85% pf lag 13.2 kVLL
Constant Current (I∠θ) I = 87.4773 ∠ -31.79o A
Constant Impedance (R & X) Z = 87.12 = 74.0520 + j 45.8948 Ω U. P. National Engineering Center National Electrification Administration
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Types of Load Sending End
Line
1.1034 + j2.0856 ohms/phase
VS = ?
ISR = ?
r r r r VS = VR + I SR ( Z line )
Receiving End
VR = 13.2 kVLL Load 2 MVA, 3Ph 0.85 pf, lag 13.2 kVLL
13,200 o = ∠0 + (87.4773∠ − 31.79o )(1.1034 + j 2.0856) 3 = 7,800∠0.760o V
r VSLL = 13.510 KVLL
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Types of Load Sending End
Line
1.1034 + j2.0856 ohms/phase
VS = 13.51 kVLL
ISR = 87.48∠ ∠-31.79o
Receiving End
VR = 13.2 kVLL Load 2 MVA, 3Ph 0.85 pf, lag 13.2 kVLL
r r* 3VS I S = 3(7,800∠0.76o )(87.4773∠31.79o ) = 1.7256 MW + j1.1010 MVAR
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Types of Load Sending End
Line
1.1034 + j2.0856 ohms/phase
VS = 13.51 kVLL
ISR = 87.48∠ ∠-31.79o
Plosses = 1.7256 − 1.7 MW
Receiving End
VR = 13.2 kVLL Load 2 MVA, 3Ph 0.85 pf, lag 13.2 kVLL
= 25.6 KW 13.510 − 13.2 VR = × 100% 13.2 = 2.35% U. P. National Engineering Center National Electrification Administration
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Types of Load Sending End
VS = ?
Line
1.1034 + j2.0856 ohms/phase ISR = ?
Receiving End
VR = 11.88 kVLL Load
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
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Types of Load Case 1: Constant Power Load 2 MVA = 1.7 MW + j1.0536 MVAR r 1.7 − j1.0536 MVA I SR = 311.88KV = 97.1979∠ − 31.79o
r r r r VS = VR + I SR ( Z line )
11.88 0 = ∠0 + (97.1979∠ − 31.78)(1.1034 + j 2.0856) 3 = 7,057.8∠0.940 V = 12.224 KV U. P. National Engineering Center National Electrification Administration
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Types of Load Case 1: Constant Power Load 2 MVA = 1.7 MW + j1.0536 MVAR
Plosses = 3(97.19792 )(1.0134) W = 28.722 KW 12.224 − 11.88 VR = × 100% 11.88 = 2.9%
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Types of Load Case 2: Constant Current Load I = 87.4773 ∠ -31.79o A
r r r r VS = VR + I SR ( Z line )
11.88 o = ∠0 + (87.4773∠ − 31.79o )(1.1034 + j 2.0856) 3 = 7,037.8∠0.84o V = 12.190 KV
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Types of Load Case 2: Constant Current Load I = 87.4773 ∠ -31.78o A
Plosses = 3(87.482 )(1.1034) W = 25.33 KW 12.19 − 11.88 VR = × 100% 11.88 = 2.6%
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Types of Load Case 3: Constant Impedance Load Z = 87.12 ∠31.79o Ω = 74.0520 + j 45.8948 Ω
r VR r VS
r r ⎡ Z Load ⎤ r ⎥ = VS ⎢ r ⎣ Z Load + Z Line ⎦ r r r ⎡ Z Load + Z Line ⎤ r = VR ⎢ ⎥ Z Load ⎣ ⎦
11.88 o ⎡ 87.12∠31.79o + (1.1034 + j 2.0856 ⎤ = ∠0 ⎢ ⎥ o 87 . 12 ∠ 31 . 79 3 ⎣ ⎦ = 7.0199 ∠0.77o KV
r VSLL = 12.159 KV
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Types of Load Case 3: Constant Impedance Load Z = 87.12 ∠31.79o Ω = 74.0520 + j 45.8948 Ω
r r VS r I SR = r Z Load + Z Line 7.0199 ∠0.77 = 87.12∠31.79o + 1.1034 + j 2.0856 = 78.730 A o
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Types of Load Case 3: Constant Impedance Load Z = 87.12 ∠31.79o Ω = 74.0520 + j 45.8948 Ω
Plosses = 3(78.732 )(1.0134) W = 18.84 KW 12.159 − 11.88 VR = × 100% 11.88 = 2.34%
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Types of Load Constant
Power Constant
Current Constant Impedance
Load
VS*
VR
Ploss
2 MVA, 0.85 pf lag
12.224
2.9 %
28.72 kW
87.48 ∠-31.78
12.190
2.6 %
25.33 kW
87.12 ∠-31.78
12.159
2.34 % 18.84 kW
* Sending end voltage with a Receiving end voltage equal to 0.9*13.2 KV U. P. National Engineering Center National Electrification Administration
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Types of Load ReA
Va2 )
ImA
Va2 )
DemandReA= (PA+ IReA Va + Z
-1
DemandImA=(QA+ IImA Va + Z
-1
DemandReB= (PB+ IReB Vb + Z -1ReB Vb2 ) DemandImB = (QB+ IImB Vb + Z -1ImB Vb2 ) DemandReC= (Pc+ IReC Vc + Z DemandImC= (Qc+ IImC Vc + Z
-1
2) V ReC c
-1
2) V ImC c
Where:
P,Q are the constant Power components of the Demand IRe,IIm are the constant Current components of the Demand Z-1Re,Z-1Im are the constant Impedance components of the Demand U. P. National Engineering Center National Electrification Administration
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Customer Load Curve 24-Hour Customer Load Profile Time Demand (A) 1:00 17.76 2:00 16.68 3:00 17.52 4:00 17.40 5:00 21.00 6:00 29.88 7:00 29.64 8:00 32.28 9:00 25.92 10:00 21.72 11:00 25.20 12:00 22.08 U. P. National Engineering Center National Electrification Administration
Time Demand (A) 13:00 20.88 14:00 19.80 15:00 19.08 16:00 19.20 17:00 23.04 18:00 30.72 19:00 38.00 20:00 35.00 21:00 34.00 22:00 27.60 23:00 24.84 24:00 22.32 Competency Training & Certification Program in Electric Power Distribution System Engineering
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Customer Load Curve • Establishing Normalized Hourly Demand Time Demand (A) Per Unit 1:00 17.76 0.467 2:00 16.68 0.439 3:00 17.52 0.461 4:00 17.40 0.458 5:00 21.00 0.553 6:00 29.88 0.786 7:00 29.64 0.780 8:00 32.28 0.849 9:00 25.92 0.682 10:00 21.72 0.572 11:00 25.20 0.663 12:00 22.08 0.581
Time Demand (A) Per Unit 13:00 20.88 0.549 14:00 19.80 0.521 15:00 19.08 0.502 16:00 19.20 0.505 17:00 23.04 0.606 18:00 30.72 0.808 19:00 1.000 38.00 20:00 35.00 0.921 21:00 34.00 0.895 22:00 27.60 0.726 23:00 24.84 0.654 24:00 22.32 0.587
ÎŁPU = 15.567 U. P. National Engineering Center National Electrification Administration
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Customer Load Curve
Demand (Per Unit)
1.2 1.0 0.8 0.6 0.4 0.2 0.0
0
2
4
6
8
10
12
14
16
18
20
22
24
Time
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Calculating Hourly Demand 350 300
Customer Energy Bill
Demand (W)
250 200 150
1.2
N o rm a lize dD e m a n d(p e ru n it)
100 1
0.8
50
Area under the curve = Customer Energy Bill
0.6
0
0.4
0.2
0 Time (24 hours)
Normalized Customer Load Curve U. P. National Engineering Center National Electrification Administration
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Calculating Hourly Demand Total Total Monthly Energy Energy Monthly Daily Energy Energy Daily
Customer Customer Load Load Curve Curve
Hourly Demand ⎛ ⎜ pt ⎜ Pt = Energy daily 24 ⎜ ⎜ ∑ pt ⎝ 1
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⎞ ⎟ ⎟ ⎟ ⎟ ⎠
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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 / ÎŁP.U] = 5 kWh x 0.467 / 15.567 = 0.15011 kW = 150.11 W U. P. National Engineering Center National Electrification Administration
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Calculating Hourly Demand 350 300
200 150 100 50
23:00
21:00
19:00
17:00
15:00
13:00
11:00
9:00
7:00
5:00
3:00
0 1:00
Demand (W)
250
Hourly Real Demand U. P. National Engineering Center National Electrification Administration
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Calculating Hourly Demand
(
−1
Qt = Pt tan cos pf t
)
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 U. P. National Engineering Center National Electrification Administration
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Calculating Hourly Demand Demand (W and VAR)
350 300 250 200 150 100
23:00
21:00
19:00
17:00
15:00
13:00
11:00
9:00
7:00
5:00
3:00
0
1:00
50
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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 U. P. National Engineering Center National Electrification Administration
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Developing Load Models Converting Energy Bill to Power Demand
Data Requirements Customer Data; Billing Cycle Data; Customer Energy Consumption Data; and Load Curve Data. Distribution Utility Data Tables and Instructions U. P. National Engineering Center National Electrification Administration
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Generator Models
Generalized Machine Model
Steady-State Equations
Generator Sequence Impedances
Generator Sequence Networks
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Generalized Machine Model Constructional Details of Synchronous Machine Axis of b q-axis
d-axis Phase b winding
Phase c winding
Field winding F
distributed threephase winding (a, b, c)
Axis of a
Rotor: Damper winding D
Damper winding Q
Axis of c
Stator:
Phase a winding
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DC field winding (F) and shortcircuited damper windings (D, Q)
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Generalized Machine Model Primitive Coil Representation b +V
-
phase b
iQ + Q
θe
D iD -+
v
F
-
-
+
v
iF
F
b
d-axis
D
ib
vQ
q-axis
ωm +V c -
ic c
phase c U. P. National Engineering Center National Electrification Administration
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phase a
+ Va -
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Generalized Machine Model Voltage Equations for the Primitive Coils For the stator windings
For the rotor windings
dλa v F = R F iF + v a = R a ia + dt dλb v D = R D iD + v b = R b ib + dt dλc v Q = R Q iQ + v c = R c ic + dt Note: The D and Q windings are shorted (i.e. v D
⎡ v abc ⎤ ⎡ Rabc ⎢v ⎥=⎢ ⎣ FDQ ⎦ ⎣
⎤ ⎡ i abc ⎤ + ⎥ ⎢ ⎥ R FDQ ⎦ ⎣i FDQ ⎦
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⎡ λ abc ⎤ p⎢ ⎥ λ ⎣ FDQ ⎦
dλF dt dλD dt dλQ
dt
= v Q = 0 ).
λ = Li
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Generalized Machine Model The flux linkage equations are:
or
⎡λa ⎤ ⎡Laa ⎢λ ⎥ ⎢L ⎢ b ⎥ ⎢ ba ⎢ λc ⎥ ⎢Lca ⎢ ⎥=⎢ ⎢λF ⎥ ⎢LFa ⎢λD ⎥ ⎢LDa ⎢ ⎥ ⎢ ⎢⎣λQ ⎥⎦ ⎢⎣LQa
Lab
Lac
LaF
Lbb Lcb LFb LDb
Lbc Lcc LFc LDc
LbF LcF LFF LDF
LQb LQc LQF
⎡ λ abc ⎤ ⎡ [L SS ] ⎢λ ⎥=⎢ ⎣ FDQ ⎦ ⎣[L RS ] U. P. National Engineering Center National Electrification Administration
LaD LaQ ⎤ ⎡ia ⎤ LbD LbQ ⎥⎥ ⎢⎢ib ⎥⎥ LcD LcQ ⎥ ⎢ic ⎥ ⎥⎢ ⎥ LFD LFQ ⎥ ⎢iF ⎥ LDD LDQ⎥ ⎢iD ⎥ ⎥⎢ ⎥ LQD LQQ ⎥⎦ ⎢⎣iQ ⎥⎦
[LSR ]⎤ ⎡ i abc ⎤ [L RR ]⎥⎦ ⎢⎣i FDQ ⎥⎦ Competency Training & Certification Program in Electric Power Distribution System Engineering
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Generalized Machine Model COIL INDUCTANCES Stator Self Inductances L aa = L s + L m cos 2θ e
L bb = L s + L m cos( 2θ e + 120 o )
Lcc = Ls + Lm cos( 2θ e − 120 o ) Stator-to-Stator Mutual Inductances
Lab = Lba = −M s + Lm cos(2θ e − 120o ) Lbc = Lcb = −M s + Lm cos2θe Lca = Lac = −M s + Lm cos(2θe + 120o ) U. P. National Engineering Center National Electrification Administration
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Generalized Machine Model COIL INDUCTANCES Rotor Self Inductances
LFF = LFF LDD = LDD LQQ = LQQ Rotor-to-Rotor Mutual Inductances
L FD = L DF = LFD L FQ = L QF = 0 L DQ = L QD = 0 U. P. National Engineering Center National Electrification Administration
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Generalized Machine Model COIL INDUCTANCES Stator-to-Rotor Mutual Inductances
LaF = LFa = LaF cosθ e LbF = LFb = LaF cos(θ e − 120o ) LcF = LFc = LaF cos(θ e + 120 ) o
LaQ = LQa = − LaQ sin θ e
LaD = LDa = LaD cosθe LbD = LDb = LaD cos(θe − 120o ) LcD = LDc = LaD cos(θe + 120o )
LbQ = LQb = − LaQ sin( θ e − 120 o ) LcQ = LQc = − LaQ sin( θ e + 120 o ) U. P. National Engineering Center National Electrification Administration
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Generalized Machine Model Equivalent Coil Representation q-axis b-axis
Q iQ vQ + ib
b
Stator coils abc rotating
+V
b-
Rotor coils FDQ stationary
F + V c -
a
ic c
D
iF iD + vD + v F + Va a-axis
d-axis
ia
ωm
c-axis U. P. National Engineering Center National Electrification Administration
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Generalized Machine Model Equivalent Generalized Machine Replace the abc coils with equivalent commutated d and q coils which are connected to fixed brushes. q-axis
Q
λ F = LFd i d + LFF i F + LFD i D λ D = LDd id + LDF i F + LDD i D λQ = LQq iq + LQQ iQ
vQ +
- i Q
q
vq +
- i q
ω m
d
F
D
+ vd -
i + vF F -
i + vD D -
i d
d-axis
λ d = L dd i d + L dF i F + L dD i D λ q = L qq i q + L qQ iQ
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Generalized Machine Model Transformation from abc to Odq q-axis
b-axis
q i q
ib c-axis ω m
q-axis
d
d-axis
ic
id ia
d-axis
θe
a-axis
Note: The d and q windings are pseudo-stationary. The O axis is perpendicular to the d and q axes. U. P. National Engineering Center National Electrification Administration
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Generalized Machine Model 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 θe + Nbib cos (θe - 120o) + Ncic cos (θe + 120o)] Nqiq = Kq [-Naia sin θe - Nbib sin (θe - 120o) -Ncic sin (θe + 120o)] where Kd and Kq are constants to be determined. U. P. National Engineering Center National Electrification Administration
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Generalized Machine Model Assume equal number of turns. Na = Nb = Nc = Nd = Nq Substitution gives id = Kd [ia cos θe + ib cos (θe - 120o) + ic cos (θe + 120o)] iq = Kq [-ia sin θe - ib sin (θe - 120o) -ic sin (θe + 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 ) U. P. National Engineering Center National Electrification Administration
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Generalized Machine Model Combining, we get ⎡io ⎤ ⎡ ⎢ ⎥ ⎢ ⎢id ⎥ = ⎢ ⎢ iq ⎥ ⎢ ⎣ ⎦ ⎣
Ko K d cos θ e
Ko
K d cos (θ e − 120 )
− K q sin (θ e − 120 )
− K q sin θ e
⎤ ⎡ia ⎤ ⎥ K d cos (θ e + 120 ) ⎥ ⎢⎢ib ⎥⎥ − K q sin (θ e + 120 )⎥⎦ ⎢⎣ic ⎥⎦ Ko
The constants Ko, Kd and Kq are chosen so that the transformation matrix is orthogonal; that is
[P ]− 1
=
[P ]T
Assuming Kd = Kq, one possible solution is
K
o
=
1 3
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Generalized Machine Model Park’s Transformation Matrix
[P ] =
[P ]−1
=
⎡ ⎢ ⎢ 2⎢ 3⎢ ⎢ ⎢ ⎣ ⎡ ⎢ ⎢ 2⎢ 3⎢ ⎢ ⎢ ⎣
1
1
2
2
cos θ e
cos (θ e − 120 )
− sin θ e
− sin (θ e − 120 )
1
cos θ
2 1 2 1 2
cos (θ e − 120 ) cos (θ e + 120 )
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⎤ ⎥ 2 ⎥ cos (θ e + 120 ) ⎥ ⎥ ⎥ − sin (θ e + 120 )⎥ ⎦ 1
⎤ ⎥ ⎥ − sin (θ e − 120 )⎥ ⎥ ⎥ − sin (θ e + 120 )⎥ ⎦ − sin θ e
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Generalized Machine Model Voltage Transformation The relationship between the currents is
i odq = [P ]i abc or
i abc = [P ] iodq −1
Assume a power-invariant transformation; that is
vaia + vbib + vcic = voio + vd id + vqiq or
T abc abc
v i
=v
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T odq odq
i
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Generalized Machine Model Substitution gives
v
T abc
[P]
−1
i odq = v T odq
v
T odq odq
=v
i
[P]
T abc
T
Transpose both sides to get
v odq = [P ]v abc
v abc = [P ] v odq −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,
λ
odq
= [P ]λ
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Generalized Machine Model In summary, using Park’s Transformation matrix,
i odq = [P ]i abc
i abc = [P ] iodq −1
v odq = [P ]v abc
v abc = [P ] v odq
λ odq = [P ]λ abc
λ abc = [P ] λ odq
−1
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−1
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Generalized Machine Model Recall the flux linkage equation
or
⎡λa ⎤ ⎡Laa ⎢λ ⎥ ⎢L ⎢ b ⎥ ⎢ ba ⎢ λc ⎥ ⎢Lca ⎢ ⎥=⎢ ⎢λF ⎥ ⎢LFa ⎢λD ⎥ ⎢LDa ⎢ ⎥ ⎢ ⎢⎣λQ ⎥⎦ ⎢⎣LQa
Lab
Lac
Lbb Lcb LFb LDb
Lbc Lcc LFc LDc
LQb LQc
⎡ λ abc ⎤ ⎡ [L SS ] ⎢λ ⎥=⎢ ⎣ FDQ ⎦ ⎣[L RS ] U. P. National Engineering Center National Electrification Administration
LaF LaD LaQ ⎤ ⎡ia ⎤ LbF LbD LbQ ⎥⎥ ⎢⎢ib ⎥⎥ LcF LcD LcQ ⎥ ⎢ic ⎥ ⎥⎢ ⎥ LFF LFD LFQ ⎥ ⎢iF ⎥ LDF LDD LDQ⎥ ⎢iD ⎥ ⎥⎢ ⎥ LQF LQD LQQ ⎥⎦ ⎢⎣iQ ⎥⎦
[LSR ]⎤ ⎡ i abc ⎤ [L RR ]⎥⎦ ⎢⎣i FDQ ⎥⎦ Competency Training & Certification Program in Electric Power Distribution System Engineering
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Generalized Machine Model Recall
where
⎡ Laa [LSS ] = ⎢⎢ Lba ⎢⎣ Lca
Lab Lbb Lcb
Lac ⎤ Lbc ⎥⎥ Lcc ⎥⎦
L aa = L S + L m cos 2 θ e
( (cos 2 θ
Lbb = L S + L m cos 2 θ e + 120 o
) )
o L cc = L S + L m − 120 e L ab = Lba = − M S + L m cos 2 θ e − 120 o
(
Lbc = Lcb = − M S + L m cos 2 θ e
(
Lca = L ac = − M S + L m cos 2 θ e + 120 o U. P. National Engineering Center National Electrification Administration
) )
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Generalized Machine Model Substitution gives ⎡ LS [LSS ] = ⎢⎢− M S ⎢⎣− M S
− MS LS − MS
− MS ⎤ − M S ⎥⎥ LS ⎥⎦
cos(2θ e − 120 ) cos(2θ e + 120 )⎤ ⎡ cos 2θ e ⎥ cos 2θ e + Lm ⎢⎢cos(2θ e − 120 ) cos(2θ e + 120 ) ⎥ ⎢⎣cos(2θ e + 120 ) cos 2θ e cos(2θ e − 120 )⎥⎦
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Generalized Machine Model Similarly, ⎡ ⎤ − L aQ sin θ e L aF cos θ e L aD cos θ e ⎢ ⎥ [L SR ] = ⎢ LaF cos (θ e − 120 ) LaD cos (θ e − 120 ) − LaQ sin (θ e − 120 )⎥ ⎢ LaF cos (θ e + 120 ) L aD cos (θ e + 120 ) − LaQ sin (θ e + 120 )⎥ ⎣ ⎦
Apply Park's transformation to Flux Linkage equation
[P ]λ abc or
= [P ][L SS ]i abc + [P ][L SR ]i FDQ
λ odq = [P ][L SS ][P ] i odq + [P ][L SR ]i FDQ −1
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Generalized Machine Model The term [P ][LSS ][P ]−1 can be shown ⎡ ⎢Ls − 2M ⎢ = ⎢ ⎢ ⎢ ⎢⎣
s
3 Ls + M s + Lm 2
⎤ ⎥ ⎥ ⎥ ⎥ 3 Ls + M s + Lm ⎥ ⎥⎦ 2
Let L oo = L S − 2 M L dd = L S + M
S
L qq = L S + M
S
S
3 Lm 2 3 − Lm 2 +
[P ][Lss ][P ]
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⎡ Loo ⎢ =⎢ ⎢ ⎣
Ldd
⎤ ⎥ ⎥ Lqq ⎥⎦
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Generalized Machine Model Similarly, it can be shown that ⎡ ⎢ 0 ⎢ ⎢ 3 [P][LSR ] = ⎢ LaF 2 ⎢ ⎢ 0 ⎢⎣
⎤ 0 ⎥ ⎥ ⎡ ⎥ ⎢ 0 ⎥ = ⎢ LdF ⎥ ⎢⎣ 3 LaQ ⎥ ⎥⎦ 2
0 3 LaD 2 0
LdD
⎤ ⎥ ⎥ LqQ ⎥⎦
where
LdF =
3 LaF 2
LdD =
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3 LaD 2
L qQ =
3 L aQ 2
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Generalized Machine Model and [P ][LSR ]
Substituting, [P ][LSS ][P ]
−1
λ odq = [P ][L SS ][P ] i odq + [P ][L SR ]i FDQ −1
Finally, we get
λ o = Loo io λ d = Ldd i d + LdF i F + LdD i D λ q = Lqq i q + LqQ iQ Note: All inductances are constant. U. P. National Engineering Center National Electrification Administration
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Generalized Machine Model The Flux Linkage Equations for the FDQ coils in matrix form is
λ FDQ = [L RS ]i abc + [L RR ]i FDQ Since we get
[LRS] =[LSR]
T
λFDQ = [LSR ] [P] i odq + [LRR ]i FDQ T
−1
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Generalized Machine Model It can be shown that
[LSR ]T
⎡ ⎢ ⎢ [P ]−1 = ⎢⎢ ⎢ ⎢ ⎢⎣
L Fd =
3 L aF 2
0 0 0
3 LaF 2 3 LaD 2 0
L Dd =
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⎤ 0 ⎥ ⎥ ⎡ ⎥ ⎢ 0 ⎥ =⎢ ⎥ ⎢ 3 LaQ ⎥ ⎣ ⎥⎦ 2
3 L aD 2
LFd LDd
L Qq =
⎤ ⎥ ⎥ LQq ⎥⎦
3 L aQ 2
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Generalized Machine Model Recall that the rotor self- and mutual inductances are constant
⎡ LFF LFD 0 ⎤ ⎥ ⎢ [LRR] = ⎢LDF LDD 0 ⎥ ⎥ ⎢ 0 0 L QQ ⎦ ⎣
Upon substitution, we get
λ F = LFd id + LFF iF + LFDiD λ D = LDd id + LDF iF + LDDiD λ Q = LQqiq + LQQiQ
Note: All inductances are also constant. U. P. National Engineering Center National Electrification Administration
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Generalized Machine Model The Flux Linkage Equation
o d q F ⎡λo ⎤ o ⎡ Loo ⎢λ ⎥ d ⎢ Ldd LdF ⎢ d⎥ ⎢ ⎢λq ⎥ q ⎢ Lqq ⎢ ⎥= ⎢ LFd LFF ⎢λF ⎥ F ⎢ ⎢λD⎥ D ⎢ LDd LDF ⎢ ⎥ ⎢ LQq ⎢⎣λQ ⎥⎦ Q ⎢⎣
q-axis
Q i Q
vQ +
q iq vq + F
d
ωm
id
+ vd -
iF
D
D LdD LFD LDD
Q ⎤ ⎡io ⎤ ⎥ ⎢i ⎥ ⎥⎢ d ⎥ LqQ⎥ ⎢iq ⎥ ⎥⎢ ⎥ ⎥ ⎢iF ⎥ ⎥ ⎢iD⎥ ⎥⎢ ⎥ LQQ⎥⎦ ⎢⎣iQ⎥⎦
d-axis
i + vF - + vD D
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Generalized Machine Model Transformation of Stator Voltages Assume Ra = Rb = Rc in the stator. Then,
v abc = Ra [u3 ]i abc
d + 位 abc dt
Recall the transformation equations
i odq = [P ] i abc
v odq = [P ] v abc
位
odq
= [P ] 位
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Generalized Machine Model Apply Park’s transformation
[P ]v abc
= [P ]R a [u 3 ][P ] i odq −1
d + [P ] dt
{[P ]
−1
λ odq
}
Simplify to get
v odq = R a [u 3 ]i odq + [P ][P ]
−1
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Generalized Machine Model It can be shown that
⎡ 2⎢ d −1 [P] = ⎢ 3 dt ⎢⎣
0 0 0
− sinθe
− cosθe
⎤ dθe ⎥ − sin(θe −120) − cos(θe −120)⎥ dt − sin(θe +120) − cos(θe +120)⎥⎦
where
dθ e = ωe = ω m dt P = ωm 2 U. P. National Engineering Center National Electrification Administration
for a two–pole machine
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Generalized Machine Model It can also be shown that
⎡ 0 d −1 [P] [P] = ⎢⎢ 0 dt ⎢⎣ 0 Finally, we get
0 0 ωm
0 ⎤ − ωm ⎥⎥ 0 ⎥⎦
for a two-pole machine
d v o = R aio + λo dt d v d = R a i d + λ d − ω mλ q dt d v q = R aiq + λ q + ω mλ d dt
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Generalized Machine Model Voltage Equation for the Rotor
vF vD vQ
d = R F iF + 位F dt d = R D iD + 位D = 0 dt d = R Q iQ + 位Q = 0 dt
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Generalized Machine Model Matrix Form of Voltage Equations ⎡ vo ⎤ ⎡Ra ⎤ ⎡io ⎤ ⎡λo ⎤ ⎡ ⎢v ⎥ ⎢ ⎥ ⎢i ⎥ ⎢λ ⎥ ⎢ R a ⎢ d⎥ ⎢ ⎥⎢ d⎥ ⎢ d⎥ ⎢ ⎢vq ⎥ ⎢ ⎥ ⎢iq ⎥ d ⎢λq ⎥ Ra ⎢ ⎢ ⎥=⎢ ⎥ ⎢ ⎥ + ⎢ ⎥ + ωm ⎢ RF ⎢vF ⎥ ⎢ ⎥ ⎢iF ⎥ dt ⎢λF ⎥ ⎢ ⎢vD ⎥ ⎢ ⎥ ⎢iD ⎥ ⎢λD ⎥ ⎢ RD ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ RQ ⎥⎦ ⎢⎣iQ ⎥⎦ ⎢⎣ ⎢⎣λQ ⎥⎦ ⎢⎣vQ ⎥⎦ ⎢⎣
-1 1
⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦
⎡λo ⎤ ⎢λ ⎥ ⎢ d⎥ ⎢λq ⎥ ⎢ ⎥ ⎢λF ⎥ ⎢λD ⎥ ⎢ ⎥ ⎢⎣λQ ⎥⎦
The equation is now in the form
[v ] = [R ][i ] + [L ] p [i ] + ω m [G ][i ] Resistance Voltage Drop
Transformer Voltage
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Generalized Machine Model d d ⎡ Ldd q ⎢ ⎢ [L] = F ⎢LFd D ⎢LDd ⎢ Q ⎢⎣
q
F LdF
D LdD
LFF
LFD
Lqq LDF LDD LQq
Note: All entries of [L] and [G] are constant.
Q
⎤ LqQ ⎥⎥ ⎥ ⎥ d q F D Q ⎥ LQQ ⎥⎦ d ⎡ − Lqq − LqQ ⎤ ⎥ q ⎢L L L dF dD ⎢ dd ⎥ ⎥ [G] = F ⎢ ⎥ D⎢ ⎢ ⎥ ⎥⎦ Q ⎢⎣
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Generalized Machine Model Summary of Equations Flux Linkages
Voltage Equations
(1) (2) (3) (4) (5) (6)
vo = Ra io + pλ o vd = Ra id + pλ d − ωm λ q vq = Ra iq + pλ q + ωm λ d vF = RF iF + pλ F vD = Rd iD + pλ D = 0 vQ = RQiQ + pλ Q = 0
(1) (2) (3) (4) (5) (6)
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λ o = Looio λ d = Ldd id + LdF iF + LdDiD λ q = Lqqiq + LqQiQ λ F = LFd id + LFF iF + LFDiD λ D = LDd id + LDF iF + LDDiD λ Q = LQqiq + LQQiQ
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Generalized Machine Model Electromagnetic Torque Equation
Te = −[i] [G][i] T
[
= − io id iq iF iD We get
Te = −(− λq id + λd iq )
[
⎡ 0 ⎤ ⎢− λ ⎥ ⎢ q⎥ ⎢ λd ⎥ iQ ⎢ ⎥ ⎢ 0 ⎥ ⎢ 0 ⎥ ⎥ ⎢ ⎢⎣ 0 ⎥⎦
]
= − (Ldd − Lqq ) id iq + LdF iF iq + LdDiDiq − LqQiQid
]
for a 2-pole machine U. P. National Engineering Center National Electrification Administration
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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
vo = Ra io
vd = Ra id − ω m Lqq iq vq = Ra iq + ω m (Ldd id + LdF iF ) v F = R F iF U. P. National Engineering Center National Electrification Administration
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Steady–State Equations Cylindrical-Rotor Machine If 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 v = R i − ω L i d
a d
m
s q
vq = Raiq − ωm Ls id + ωm LdF iF Te = LdF iF iq U. P. National Engineering Center National Electrification Administration
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Steady–State Equations For Balanced Three-Phase Operation
ia = 2 I cos (ωt + α )
( ) i = 2 I cos (ωt + α + 120 ) Apply Park’s transformation i odq = [P ]i abc , We get ib = 2 I cos ωt + α − 120 o o
c
io = 0
id = 3 I cos α
Note: 1. ia, ib and ic are balanced three-phase currents.
iq = 3 I sin α U. P. National Engineering Center National Electrification Administration
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Steady–State Equations A similar transformation applies to balance threephase voltages. Given
va = 2 V cos(ωt + δ)
( ) 2 V cos(ωt + δ + 120 )
vb = 2 V cos ωt + δ − 120 vc = We get
o
o
vo = 0 vd =
3 V cos δ
vq =
3 V sin δ
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Steady–State Equations Inverse Transformation Given id and iq, and assuming io = 0,
i abc = [P ] i odq −1
We get
[
2 ia = id cos ωt − iq sin ωt 3
[
(
]
2 = id cos ωt + iq cos ωt + 90o 3 U. P. National Engineering Center National Electrification Administration
)]
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Steady–State Equations Recall the phasor transformation
2 I cos (ω t + θ ) ↔ I∠ θ Using the transform, we get
[
1 Ia = id ∠0o + iq ∠90o 3
]
assuming the d and q axes as reference. Simplify
iq id Ia = + j 3 3 I a = I d + jI q
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Steady–State Equations Similarly, given vd and vq with vo = 0
[
]
2 va = vd cos ωt − vq sin ωt 3 2 o = v d cos ωt + vq cos (ωt + 90 ) 3
[
In phasor form,
]
vq vd Va = + j 3 3
=Vd + jVq U. P. National Engineering Center National Electrification Administration
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Steady–State Equations Steady-State Operation-Cylindrical Recall at steady-state
vd = Ra id − ω m Ls iq vq = Ra iq + ω m Ls id + ω m LdF iF Divide by 3
Vd = Ra I d − ωm Ls I q 1 Vq = Ra I q + ωm Ls I d + ωm LdF iF 3
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Steady–State Equations Xs = ωmLs = synchronous reactance 1 E f = ωm LdFiF = Excitation voltage 3 Phasor Voltage V a Define
V a = Vd + jVq
= Ra I d − X s I q + j (Ra I q + X s I d + E f ) = Ra (I d + jI q ) + jX s (I d + jI q ) + jE f
V a = Ra I a + jX s I a + E m U. P. National Engineering Center National Electrification Administration
(motor equation)
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Steady–State Equations For a generator, current flows out of the machine
( )
( )
V a = Ra − I a + jXs − I a + E g E g = Ra I a + jXs I a + V a R a + jX
+
+
Eg
AC
s
Ia
-
Va -
Equivalent Circuit of Cylindrical Rotor Synchronous Generator U. P. National Engineering Center National Electrification Administration
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Steady–State Equations Salient-Pole Machine If the rotor is not cylindrical, no equivalent circuit can be drawn. The analysis is based solely on the phasor diagram describing the machine. Recall the steady-state equations
vd = Raid − ω m Lqqiq vq = Raiq + ω m Ldd id + ω m LdF iF Divide through by
3
Vd = Ra I d − X q I q Vq = R a I q + X d I d + U. P. National Engineering Center National Electrification Administration
ω m LdF iF 3
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Steady–State Equations where
Xd =ωmLdd =
Xq =ωmLqq = Define:
Ef =
ωm LdF 3
iF
direct axis synchronous reactance quadrature axis synchronous reactance
= excitation voltage
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Steady–State Equations We get
Vd = Ra I d − X q I q Vq = Ra I q + X d I d + E f
From
Va =Vd + jVq , we get V a = Ra Id − Xq Iq + j(Ra Iq + Xd Id + Ef )
or
= Ra (Id + jIq ) − Xq Iq + jXd Id + jEf
V a = Ra I a − X q Iq + jXd Id + jEf U. P. National Engineering Center National Electrification Administration
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Steady–State Equations Steady-State Electromagnetic Torque At steady-state
[
Te = − (Ldd − Lqq ) id iq + LdF iF iq
]
saliency cylindrical torque torque 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. U. P. National Engineering Center National Electrification Administration
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Steady–State Equations Since the field current iF is always positive,
â&#x2C6;&#x2019; LdFiF iq < 0 > 0 Recall that
when iq > 0 (motor) when iq < 0 (generator)
I a = Id + jIq
Note: The imaginary component of Ia determines Whether the machine is operating as a motor or a Generator. U. P. National Engineering Center National Electrification Administration
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Steady–State Equations What about Id? Assume From we get
Iq = 0
.
Vd = Ra I d − X q I q Vq = Ra I q + X d I d + E f
Vd = Ra I d Vq = X d I d + E f
In general, Ra << Xd. We get
V a = V d + jV q
= R a I d + j (X d I d + E f ≈ j (X d I d + E f ) U. P. National Engineering Center National Electrification Administration
)
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Steady–State Equations If the magnitude of Va is constant,
Vq = X d I d + E f = constant Recall that
Ef =
ω m L dF 3
iF
Thus, the excitation voltage depends only on the field current since ωm is constant. For some value of field current iFo, Ef = Va and Id = 0. U. P. National Engineering Center National Electrification Administration
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Steadyâ&#x20AC;&#x201C;State Equations Operating Modes q-axis Over-excited Motor
Under-excited Motor
Id < 0, Iq > 0
Id > 0, Iq > 0 d-axis
Id < 0, Iq < 0
Id > 0, Iq < 0
Over-excited Generator
Under-excited Generator
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Steadyâ&#x20AC;&#x201C;State Equations Over-excitation and Under-excitation 1. If the field current is increased above iFo, then Ef > 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). U. P. National Engineering Center National Electrification Administration
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Steady–State Equations Drawing Phasor Diagrams A phasor diagram showing Va and Ia can be drawn if the currents Id and Iq are known. Recall
I a = I d + jI q V a = Vd + jVq V a = Ra I a − X q I q + jX d I d + jE f V a = jE f − X q I q + jX d I d + Ra I a U. P. National Engineering Center National Electrification Administration
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Steady–State Equations Over-excited Motor Id < 0 Iq > 0
− X q Iq
Ra I a
q-axis
jEf
Va
jXd Id
δ
Ia
jIq
Id
φ
d-axis
Leading Power Factor U. P. National Engineering Center National Electrification Administration
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Steady–State Equations Under-excited Motor Id > 0 Iq > 0
Ra I a
q-axis
Va
jX d I d
− XqIq jIq
δ
jEf
Ia
φ
Id
d-axis
Lagging Power Factor U. P. National Engineering Center National Electrification Administration
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Steady–State Equations Over-excited Generator q-axis − XqIq
jEf Id < 0 Iq < 0
jXd Id
Ra I a
δ
φ
Va Actual Current
Id
jIq Lagging Power Factor U. P. National Engineering Center National Electrification Administration
d-axis
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Steady–State Equations Under-excited Generator
jXd Id
Id > 0 Iq < 0
Ra I a Va
jEf − Xq Iq
φ δ Actual Current
Leading Power Factor U. P. National Engineering Center National Electrification Administration
Id d-axis
jIq
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Steady–State Equations Observations 1. The excitation voltage jEf lies along the quadrature axis. 2. V a leads jEf for a motor V a lags jEf for a generator The angle between the terminal voltage Va and jEf is called the power angle or torque angle δ. 3. The equation
V a = Ra I a + jXd Id − X q Iq + jEf applies specifically for a motor. U. P. National Engineering Center National Electrification Administration
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Steady–State Equations 4. For a generator, the actual current flows out of the machine. Thus Id, Iq and I a are negative.
V a = −Ra I a − jXd Id + Xq Iq + jEf or
jEf = V a + Ra I a + jXd Id − Xq Iq 5. Let
jE f = E m
for a motor
jE f = E g
for a generator
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Steady–State Equations The generator equation becomes
E g = V a + Ra I a + jXd Id − X q Iq For a motor, the equation is
V a = Em + Ra I a + jXd Id − Xq Iq 6. No equivalent circuit can be drawn for a salient-pole motor or generator. U. P. National Engineering Center National Electrification Administration
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Steadyâ&#x20AC;&#x201C;State Equations 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 jXs power factor. Single-phase + + equivalent circuit Ia
Eg
AC
Va = 13.8 kV = 7.97 kV
line-to-line line-to-neutral
U. P. National Engineering Center National Electrification Administration
Va -
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Steady–State Equations Pa = 25(0.8) = 20 MW, three-phase = 6.67 MW/phase Qa = Pa tan θ = 15 =5 Let
MVar, three-phase MVar/phase
V a = 7.97∠0o kV, the reference.
Using the complex power formula * a a
Pa + jQa = V I Ia =
Pa − jQa V
* a
6,667 − j5,000 = 7.97∠0o
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Steady–State Equations Ia = 837 −
We get
j628 A
= 1,046∠ − 36.87o A Apply KVL,
Eg = jXS I a + V a
(
)
= j 4.5 1,046∠ − 36.87o + 7,970∠0o = 10,791 + j3,766 = 11,429∠19.24o V Eg = 11,429 volts, line-to-neutral = 19,732 volts, line-to-line U. P. National Engineering Center National Electrification Administration
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Steady–State Equations 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
X S = 2.4Ω +
Eg
AC
-
+
Ia
Va -
U. P. National Engineering Center National Electrification Administration
R = 4Ω Va(L-L) = 20,000 volts Va(L-N) = 11,547 volts Competency Training & Certification Program in Electric Power Distribution System Engineering
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Steady–State Equations V a = 11,547∠0o V, the reference V a 11,547 o Ia = = = 2,887∠0 Amps R 4 Applying KVL, E g = jXS I a + V a Let
= j 2.4(2,887) + 11,547 = 11,547 + j 6,928
= 13,466∠30.96o V line − to − neutral
E g = 3 (13,466 ) = 23,324 V = 23 .32 kV, line − to − line U. P. National Engineering Center National Electrification Administration
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Steady–State Equations (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.
Req = 4Ω // 4Ω = 2Ω
Let Va = Va ∠ 0 o , the reference
Va 1 o Ia = = Va∠0 Req 2 U. P. National Engineering Center National Electrification Administration
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Steady–State Equations Apply KVL,
E g = jX s I a + V a ⎛1 ⎞ = j 2.4⎜ Va ⎟ + Va ⎝2 ⎠ = Va + j1.2Va
We get Eg = Va + (1.2Va ) 2
2
13 , 466 = 2 . 44 V a 2
2
2
Va = 8,621 V , line − to − neutral Va = 14,932 V , line − to − line U. P. National Engineering Center National Electrification Administration
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Steady–State Equations (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.
V a = 11 , 547 ∠ 0 o V , line − to − neutral
V a 11,547 Ia = = = 5,774 ∠ 0 o Amps Req 2 E g = j 2.4(5774 ) + 11,547 = 11,547 + j13,856 = 18,037 ∠50 .19 o V line − to − neutral
E g = 3 (18 , 037 ) = 31, 241 V = 31 .24 kV line − to − line U. P. National Engineering Center National Electrification Administration
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Generator Sequence Impedances The equivalent Circuit of Generator for Balanced Three-Phase System Analysis a Ia
Za
R a + jX
s
Ea
+
Eb Zb
Ec
Zc
Ib
Ia
Eg b
Ic
-
+
Va -
c
Three-Phase Equivalent U. P. National Engineering Center National Electrification Administration
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Sequence Impedance of Power System Components From Symmetrical Components, the Sequence Networks for Unbalanced Three-Phase Analysis +
+
+
Ia1 Va1
Ia2
Z1
Va2
Ia0
Z2
Va0
Z0
+
E -
-
V a1 = E â&#x20AC;&#x201C; I a1 Z 1
-
V a2 = - I a2 Z 2
Positive Sequence Negative Sequence U. P. National Engineering Center National Electrification Administration
V ao = - I ao Z o Zero Sequence
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Generator Sequence Impedances Positive-Sequence Impedance: Xd”=Direct-Axis Subtransient Reactance Xd’=Direct-Axis Transient Reactance Xd=Direct-Axis Synchronous Reactance Negative-Sequence Impedance:
X2 = 12 (X d "+ X q " ) for a salient-pole machine for a cylindrical-rotor machine X2 = X d " Zero-Sequence Impedance:
0.15X d " ≤ X 0 ≤ 0.6X d " U. P. National Engineering Center National Electrification Administration
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Generator Sequence Impedances Positive Sequence Impedance The 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
⎛ −t ⎞ E ⎛ E E ⎞ ⎟⎟ ⎟⎟ exp⎜⎜ + ⎜⎜ − I( t ) = X ds ⎝ X d ' X ds ⎠ ⎝τ d' ⎠ ⎛ E ⎛ −t ⎞ E ⎞ ⎟⎟ exp⎜⎜ ⎟⎟ + ⎜⎜ − ⎝ X d" X d' ⎠ ⎝τ d" ⎠ where E = AC RMS voltage before the short circuit. U. P. National Engineering Center National Electrification Administration
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Generator Sequence Impedances 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 the subtransient portion, while the straight line is the transient portion. U. P. National Engineering Center National Electrification Administration
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Generator Sequence Impedances IEEE Std 115-1995: Determination of the Xdâ&#x20AC;&#x2122; and Xdâ&#x20AC;? (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. U. P. National Engineering Center National Electrification Administration
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Generator Sequence Impedances The direct-axis subtransient reactance is determined 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 Iâ&#x20AC;?. U. P. National Engineering Center National Electrification Administration
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Current in phase 1 (per unit)
Generator Sequence Impedances 14 12 + 10 +++ Curve B + ++ 8 ++ ++ ++ 6 ++ Line ++ +++ 5 ++ + 4 3+
2.0 1.5 1.0 0.8 0.6 0.4 0
C ++
++ + ++
+ +
Line A
+ + + + + +
Curve A
10
20
30
40
50
60
Time in half-cycles U. P. National Engineering Center National Electrification Administration
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Generator Sequence Impedances Example: Calculation of transient and subtransient reactances for a synchronous machine 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
(5) Xd’ = (1)÷(4)
0.0993
(6) Init. Subtransient Current 3.8 (7) I” = (4)+(6) (8) Xd” = (1)÷(7) U. P. National Engineering Center National Electrification Administration
10.07
13.5
5.6
4.4
16.1
14.4
14.67 0.0682
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Generator Sequence Impedances Negative Sequence Impedance IEEE Std 115-1995: Determination of the negativesequence 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 perunit. U. P. National Engineering Center National Electrification Administration
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Generator Sequence Impedances 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.
E Z2 = =Negative Sequence Impedance, p.u. I P R 2 = 2 =Negative Sequence Resistance, p.u. I 2
X2 = Z2 â&#x2C6;&#x2019; R 2
2
=Negative Sequence Reactance, p.u. Note: The test produces abnormal heating in the rotor and should be concluded promptly. U. P. National Engineering Center National Electrification Administration
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Generator Sequence Impedances Zero Sequence Impedance IEEE Std 115-1995: Determination of the zero-sequence reactance, X0 (Method 1) 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 V voltage, current and electric power. Field U. P. National Engineering Center National Electrification Administration
E
A
W
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Generator Sequence Impedances Let E = applied voltage, in p.u. of base line-toneutral voltage I = test current, p.u. P = wattmeter reading, in p.u. single-phase base volt-ampere
3E Z0 = =Zero Sequence Impedance, p.u. I X0 = Z0
⎛P⎞ 1−⎜ ⎟ ⎝ EI ⎠
2
=Zero Sequence Reactance, p.u. U. P. National Engineering Center National Electrification Administration
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Generator Sequence Impedances Average Machine Reactances Turbo Water-Wheel Synchronous Reactance Generators Generators Motors Xd 1.10 1.15 1.20 Xq
1.08
0.75
0.90
X d‘
0.23
0.37
0.35
X q‘
0.23
0.75
0.90
X d”
0.12
0.24
0.30
X q”
0.15
0.34
0.40
X2
0.12
0.24
0.35
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Generator Sequence Networks Grounded-Wye Generator The sequence networks for the grounded-wye generator are shown below. F1
r + Eg
F0
F2
jZ1 jZ0
jZ2
-
N2
N1
Positive Sequence
Negative Sequence
U. P. National Engineering Center National Electrification Administration
N0
Zero Sequence
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Generator Sequence Networks Grounded-Wye through an Impedance If the generator neutral is grounded through an impedance Zg, the zero-sequence impedance is modified as shown below. F1
r + Eg
F0
F2
jZ1
jZ0
jZ2
3Zg
-
N2
N1
Positive Sequence
Negative Sequence
U. P. National Engineering Center National Electrification Administration
N0
Zero Sequence
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Generator Sequence Networks 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
r + Eg
F0
F2
jZ1 jZ0
jZ2
-
N2
N1
Positive Sequence
Negative Sequence
U. P. National Engineering Center National Electrification Administration
N0
Zero Sequence
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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 Ideal Transformer The voltage drop from the polaritymarked terminal to the non-polaritymarked 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. N N Voltage Equation:
r VH NH r = NX VX
U. P. National Engineering Center National Electrification Administration
+
r VH _
r IH
H
X
r + IX r VX _
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Two-Winding Transformer +
r VH _
r IH
NH N X
r IX
+
r VX _
Current Equation:
r r NH IH = N X IX
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 satisfied at all times. U. P. National Engineering Center National Electrification Administration
must
be
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Two-Winding Transformer Referred Values From therTransformation Ratio,
VH a= r VX r IX a= r IH
r r V H = aV X r IH
r IX = a
Dividing VH by IH,
r r VH 2 VX r =a r IH IX
ZH = a2 Z X
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Two-Winding Transformer Practical Transformer 1. 2. 3. 4.
The H and X coils have a small resistance. There are leakage fluxes in the H and X coils. There is resistance loss in the iron core. The permeability of the iron is not infinite. Ď&#x2020;m
iH vH
iX
+
+
eH
eX
-
NH
NX
vX
-
iron core U. P. National Engineering Center National Electrification Administration
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Two-Winding Transformer Equivalent Circuit v RH + jX H I ex +
r VH
r IH R c
jX m
H winding
N H N X R X + jX X +
v EH
+
-
-
v EX
Ideal
r IX
+
r VX -
X winding
RH, XH =resistance and leakage reactance of H coil RX, XX =resistance and leakage reactance of X coil Rc, Xm =core resistance and magnetizing reactance U. P. National Engineering Center National Electrification Administration
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Two-Winding Transformer Referring secondary quantities at the primary side, RH + jX H a 2 R X + ja 2 X X NH N X v +
r VH
I ex r IH R jX m c
-
RH + jX H +
r VH
r IH R c
v I ex
r IX a
+
+
+
-
-
-
r v aV X EH
v EX
a 2 R X + ja 2 X X jX m
U. P. National Engineering Center National Electrification Administration
r IX a
+
r aV X -
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Two-Winding Transformer The transformer equivalent circuit can be approximated by
Req + jX eq +
r VH
v Iex
r IH R c
jX m
R eq = R H + a 2 R X
r + 1 a IX r
aV X
-
X eq = X H + a 2 X X
-
r r IH V H Rc +
U. P. National Engineering Center National Electrification Administration
v Iex
Req + jX eq 1 a
jX m
r IX
+
r aV X -
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Two-Winding Transformer For large power transformers, shunt impedance and resistance can be neglected
R eq + jX eq +
r VH -
r r I H = a1 I X
jX eq +
r aV X
r VH
-
-
U. P. National Engineering Center National Electrification Administration
+
r r I H = a1 I X
+
r aV X -
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Two-Winding Transformer Tap-Changing Transformer a:1 q s
r
1 y pq a
p
The π equivalent circuit of transformer with the per 1− a y pq 2 unit transformation ratio: a U. P. National Engineering Center National Electrification Administration
a −1 y pq a
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Short-Circuit and Open-Circuit Tests Short-Circuit Test Conducted to determine series impedance With the secondary (Low-voltage side) shortcircuited, apply a primary voltage (usually 2 to 12% of rated value) so that full load current H1 x1 flows. W
A V
H2 U. P. National Engineering Center National Electrification Administration
x2
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Short-Circuit and Open-Circuit Tests Short-Circuit Test Req + jX eq +
VSC
I SC
Ie
Rc
Ie ≈ 0
I1 jX m
I sc = I 1
-
PSC Req = 2 I SC
Z eq
VSC = I SC
U. P. National Engineering Center National Electrification Administration
X eq = Z − R 2 eq
2 eq
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Short-Circuit and Open-Circuit Tests Open-Circuit Test Conducted to determine shunt impedance With the secondary (High-voltage side) opencircuited, apply rated voltage to the primary.
W
x1
H1
x2
H2
A V
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Short-Circuit and Open-Circuit Tests Open-Circuit Test Req + jX eq +
I OC
VOC
Ie
I OC = I e
jX m
Rc
2 OC
V Rc = POC
2
⎡ I OC ⎤ 1 1 = ⎢ ⎥ − 2 Xm Rc ⎣VOC ⎦
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Short-Circuit and Open-Circuit Tests Example: 50 kVA, 2400/240V, single-phase transformer Short-Circuit Test: HV side energized
VSC = 48 volts
I SC = 20.8 amps
PSC = 617 watts
Open-Circuit Test: LV side energized
VOC = 240 volts
I OC = 5.41 amps
POC = 186 watts
Determine the Series and Shunt Impedance of the transformer. What is %Z and X/R of the transformer? U. P. National Engineering Center National Electrification Administration
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Short-Circuit and Open-Circuit Tests Solution: From the short-circuit test
Z eq ,H
48 = = 2.31 ohm 20.8
R eq ,H =
617 = 1 .42 o hm 2 (20 .8 ) 2
X eq ,H = 2.31 − 1.42 = 1.82 ohm 2
From the open-circuit test
Rcq ,L
2 ( 240 ) =
= 310 ohm
186 2 2 1 ⎡ 5.41 ⎤ ⎡ 1 ⎤ = ⎢ − ⎥ ⎢ 310 ⎥ Xm ⎣ 240 ⎦ ⎣ ⎦ U. P. National Engineering Center National Electrification Administration
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Short-Circuit and Open-Circuit Tests Referred to the HV side
Rc ,H = a 2 Rc ,L = 30 ,968 ohm
X m ,H = a 2 X m ,L = 4 ,482 ohm %Z and X/R
Z BASE =
[2.4 ]2 50 / 1000
= 115.2 ohm
⎛ 2.31 ⎞ %Z = ⎜ ⎟ x100 = 2% ⎝ 115.2 ⎠ U. P. National Engineering Center National Electrification Administration
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X/R Ratios of Transformers
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Three-Winding Transformer +
r VH
r IH
NX
NH
r IX
+
r VX _
_
NY
r VH NH r = NX VX
r IY
r VY _
r VH NH r = NY VY r r r NH IH = N X IX + N Y IY
U. P. National Engineering Center National Electrification Administration
+
r VX NX r = NY VY
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Three-Winding Transformer From 3 short-circuit tests with third winding open, get 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 Note: When expressed in ohms, the impedances must be referred to the same side. U. P. National Engineering Center National Electrification Administration
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Three-Winding Transformer ZH
ZX
+
+
r VH
ZY
VY
-
Z HX = Z H + Z X Z HY = Z H + Z Y or
+ r -
r VX -
Z XY = Z X + Z Y
Z H = 12 ( Z HX + Z HY − Z XY ) Z X = 21 ( Z HX − Z HY + Z XY ) Z Y = 12 ( − Z HX + Z HY + Z XY )
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Three-Winding Transformer 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 立 @ the H side ZHY = 106.21 立 @ the H side ZXY = 4.41 立 @ the X side Find the equivalent circuit in ohms, referred to the H side. 140 2
Z XY = (
48
) ( 4.41 ) = 37 .52 立
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Three-Winding Transformer With all impedances referred to the H side, we get
Z H = 21 ( 63.37 + 106.21 − 37.52 ) = 66.03 Ω Z X = 12 ( 63.37 − 106.21 + 37.52 ) = −2.66. Ω ZY = 21 ( −63.37 + 106.21 + 37.52 ) = 40.18 Ω 66.03 Ω
− 2.66 Ω
+
+
r VH
40.18 Ω
U. P. National Engineering Center National Electrification Administration
+ r
VY -
r VX -
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Transformer Connection Transformer Polarity V1
V1 H1
H1
H2
V
H2
V
Less than V1
Greater than V1 x1
x2
Subtractive Polarity U. P. National Engineering Center National Electrification Administration
x2
x1
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Transformer Connection H1
H2
H1
Subtractive
H2 Additive
X1
X2
X2
X1
â&#x20AC;&#x153;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.â&#x20AC;? (ANSI/IEEEC57.12.001993) U. P. National Engineering Center National Electrification Administration
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Transformer Connection Parallel Connection H1
H2
x1
x2
H1
H2
x1
x2
LOAD U. P. National Engineering Center National Electrification Administration
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Transformer Connection 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
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Transformer Connection H1
H2
x1
x2
H1
H1
H2
x1
H2
x1
x2
x2
WYE-WYE (Y-Y)
Three Phase Transformer Bank U. P. National Engineering Center National Electrification Administration
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Transformer Connection H1
H2
x1
x2
H1
H2
x1
H1
H2
x1
x2
x2
DELTA-DELTA (Î&#x201D;-Î&#x201D;)
Three Phase Transformer Bank U. P. National Engineering Center National Electrification Administration
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Transformer Connection H1
H2
x1
x2
H1
H1
H2
x1
H2
x1
x2
x2
WYE-DELTA (Y-Î&#x201D;)
Three Phase Transformer Bank U. P. National Engineering Center National Electrification Administration
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Transformer Connection H1
H2
x1
x2
H1
H1
H2
x1
H2
x1
x2
x2
DELTA-WYE (Î&#x201D;-Y)
Three Phase Transformer Bank U. P. National Engineering Center National Electrification Administration
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Transformer Connection H1
H2
x1
x2
H1
H2
x1
x2
OPEN DELTA â&#x20AC;&#x201C; OPEN DELTA
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Transformer Connection 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 U. P. National Engineering Center National Electrification Administration
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Three-Phase Transformer Angular Displacement ANSI/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 transformer with a Δ-Δ or Y-Y connection shall be 0o. The angular displacement for a three-phase transformer with a Y-Δ or Δ-Y connection shall be 30o, with the low voltage lagging the high voltage. U. P. National Engineering Center National Electrification Administration
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Three-Phase Transformer Vector Diagrams H2
H3
H1
Δ-Δ Connection H2
X2 X1
X3
X1 H1
H2
X2
H3
Y-Δ Connection H2
X2
X3 X2
X1 X1 H1
X3
H3
Y-Y Connection U. P. National Engineering Center National Electrification Administration
H1
H3
X3
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Three-Phase Transformer IEC Designation 0
IEC Designation for Î&#x201D;-Î&#x201D; Dd0 Dd2 Dd4 Dd6
Dd8
10
2
8
4
Dd10
IEC Designation for Y-Y Yy0 Yy6
6
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. U. P. National Engineering Center National Electrification Administration
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Three-Phase Transformer IEC Designation 11
IEC Designation for Y-Î&#x201D; Yd1 Yd5 Yd7 Yd11
1
9
3
IEC Designation for Î&#x201D;-Y Dy1 Dy5 Dy7 Dy11
7
5
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. U. P. National Engineering Center National Electrification Administration
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Three-Phase Transformer Positiveâ&#x20AC;&#x201C;Sequence Voltages B
H2 N
A C
r VBN1
H1
H3
(A-B-C) r
X3
r Vab1
Van1
r VAN1
X2
X1
r VCN1
U. P. National Engineering Center National Electrification Administration
r Vca1
b
c
r Vbn1
a
r r r AN1 Vbc1 Van1 lags V o r Vcn1
by 30
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Three-Phase Transformer Positiveâ&#x20AC;&#x201C;Sequence rCurrents B r
H2
IB1 r IA1 A r IC1 H1
C
r IA1
r IB1
Iba1
X1
H3
r Iac1
(A-B-C)
r IC1
r Ia1 r Iba1
U. P. National Engineering Center National Electrification Administration
b r X2 r Ib1 Icb1 r Ic1 r c X3 Ia 1
r Icb1
r Ib1
a
r r Ia1 lags IA1 by 30o
r Iac1 r I1 CompetencycTraining & Certification Program in Electric Power Distribution System Engineering
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Three-Phase Transformer Positive Sequence Impedance Whether a bank of single-phase units or a threephase transformer unit (core type or shell type), the equivalent impedance is the same. Using per-unit values, the positive-sequence equivalent circuit is
Z1 = R1 + jX1 +
r VH -
r r IH = IX
+
r VX -
U. P. National Engineering Center National Electrification Administration
Note: The negativesequence impedance is equal to the positivesequence impedance. Competency Training & Certification Program in Electric Power Distribution System Engineering
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Three-Phase Transformer Negativeâ&#x20AC;&#x201C;Sequence Voltages B
A Cr
H2 N H1
H3
VCN2
r Vcn2
(A-C-B)
r Vac2
r VAN2
r VBN2
X2
X1
r Van2
U. P. National Engineering Center National Electrification Administration
X3
b
c a
r Vcb2 r Vbn2 r Vba2
r r Van2 leads VAN2 by 30o
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Three-Phase Transformer Negativeâ&#x20AC;&#x201C;Sequence r Currents B r
IB 2 r IA2 A r H IC2 1
C
Iba2
H2
H3
r IC 2
b r X2 r Ib2 Icb2 r Ic2 r c X3 Ia 2
X1
r Ic 2
r Iac2
r Iac2
(A-C-B)
r IA 2
r IB 2
r Iba2
r Ia2
U. P. National Engineering Center National Electrification Administration
a
r Icb2
r Ib 2
r r Ia2 leads IA2 by 30o
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Three-Phase Transformer Positiveâ&#x20AC;&#x201C; & Negative Sequence Networks Z2
Z1 + Primary Side
-
r I1
+
+
Secondary Side
-
Positive Sequence Network
Primary Side
-
Z1 = Z2
U. P. National Engineering Center National Electrification Administration
r I2
+ Secondary Side
-
Negative Sequence Network
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Three-Phase Transformer Transformer Core
3-Legged Core Type
Shell Type 4-Legged Core Type U. P. National Engineering Center National Electrification Administration
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Three-Phase Transformer 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. Note: For positive- or negative-sequence flux,
φa
φb
φc
U. P. National Engineering Center National Electrification Administration
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Three-Phase Transformer 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.
3φ0
φ0
φ0
φ0
Note: The zerosequence flux leaks out of the core and returns through the transformer tank.
3-Legged Core Type U. P. National Engineering Center National Electrification Administration
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Three-Phase Transformer Zero Sequence Impedance* Transformer Connection
Zero-Sequence Network
Z0 = Z1
+ r
+ r
VH
VX
-
-
Z0 = Z1 + r
+ r
VH
VX
-
-
*Excluding 3-phase unit with a 3-legged core. U. P. National Engineering Center National Electrification Administration
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Three-Phase Transformer Zero Sequence Impedance* Transformer Connection
Zero-Sequence Network
Z0 = Z1
+ r
+ r
VH
VX
-
-
Z0 = Z1 + r
+ r
VH
VX
-
-
*Excluding 3-phase unit with a 3-legged core. U. P. National Engineering Center National Electrification Administration
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Three-Phase Transformer Zero Sequence Impedance* Transformer Connection
Zero-Sequence Network
Z0 = Z1
+ r
+ r
VH
VX
-
-
Z0 = Z1 + r
+ r
VH
VX
-
-
*Excluding 3-phase unit with a 3-legged core. U. P. National Engineering Center National Electrification Administration
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Three-Phase Transformer Example: Consider a two-winding three-phase transformer with the following nameplate rating: 25 MVA 69Î&#x201D; -13.8YG kV (Dyn1) Z=7%. Draw the positive and zero-sequence equivalent circuits. Use the transformer rating as bases. Positive/Negative Sequence impedance
Zero Sequence impedance Z0=j0.07
Z1=j0.07 + r
+ r
+ r
+ r
VH
VX
VH
VX
-
-
-
-
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Three-Phase Transformer Example: A three-winding three-phase transformer has the following nameplate rating: 150/150/45 MVA 138zG-69zG-13.8Î&#x201D; kV (Yy0d1). H-X @ 150 MVA = 14.8% H-Y @ 45 MVA = 21.0% X-Y @ 45 MVA = 36.9% Draw the positive and zero-sequence equivalent circuits. Use 100 MVA and the transformer voltage ratings as bases. At the chosen MVA base,
Z HX = 0.148 ( 100 / 150 ) = 0.10 p.u. Z HY = 0.21( 100 / 45 ) = 0.47 p.u. Z XY = 0.369 ( 100 / 45 ) = 0.82 p.u. U. P. National Engineering Center National Electrification Administration
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Three-Phase Transformer We get
Z H = 21 ( 0.10 + 0.47 − 0.82 ) = −0.125 p.u. Z X = 12 ( 0.10 − 0.47 + 0.82 ) = 0.225 p.u. Z Y = 21 ( −0.10 + 0.47 + 0.82 ) = −0.595 p.u. Zero Sequence Network
Positive/Negative Sequence Network
ZH + r
VH -
ZX ZY
ZH
+
+ r
r VX
-
-
VY
U. P. National Engineering Center National Electrification Administration
+ r
VH -
ZX ZY
+
+ r
r VX
-
-
VY
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Three Phase Model THREE-PHASE TRANSFORMER AND 3 SINGLE-PHASE TRANSFORMERS IN BANK Primary A B C
Secondary a b c
abc T
Y
Admittance Matrix
U. P. National Engineering Center National Electrification Administration
Core Loss
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Three Phase Model CORE LOSS MODELS 1. Constant P & Q Model 2. EPRI Core Loss Model
( (
) )
2 kVA Rating 2 CV A V + Bε Pp .u . = System Base 2 kVA Rating 2 FV Q p .u . = D V + Eε System Base
A = 0.00267 D = 0.00167
B = 0.734x10 -9 E = 0.268x10 -13
U. P. National Engineering Center National Electrification Administration
C = 13.5 F = 22.7
|V| in per unit
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Three Phase Model I1
I2
I3
I4
I5
I6
+ V1 + V2 + V3 + V4 + V5 + V6
Primitive Coils
z12
z23
z11
-
z11
z12
z13
z14
z15
z21
z22
z23
z24
z25 z26
z31
z32
z33
z34
z35
z36
z41
z42
z43
z44
z45
z46
z51
z52
z53
z54
z55
z56
z61
z62
z63
z64
z65
z66
z34
z22
-
z45
z33
-
z56
z44
-
z55
-
z66
-
z16
U. P. National Engineering Center National Electrification Administration
Primitive Impedances
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Transformer Model Three Identical Single-phase Transformers in Bank z11
z12
z21
z22
I1
I2 z11
z12
z22
I3
I4 z33
z33
z34
z43
z44
z34
z44
I5 z55
z56
z65
z66
U. P. National Engineering Center National Electrification Administration
I6 z55
z56
z66
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Transformer Model Node Connection Matrix, C V1
VA
V2
VB
V3
VC
V4
=
Va
V5
Vb
V6
Vc
[V123456] = [C][VABCabc ] Matrix C defines the relationship of the Primitive Voltages and Terminal Voltages of the Three-Phase Connected Transformer U. P. National Engineering Center National Electrification Administration
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Transformer Model Va
VA IA
VC
VB
1
2 3
IC 5
Ia 4 Ib Vb
6 Ic
IB
Vc
Wye Grounded-Wye Grounded Connection
Node Connection Matrix, C V1
1
VA
V2
[V123456]
=
[C][VABCabc ]
V3 V4 V5 V6
U. P. National Engineering Center National Electrification Administration
1
=
VB
1
VC 1
Va
1
Vb 1
Vc
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Transformer Model 6
VA IA
VC VB
1
Ia 2 Ib
3
IB 5
Va
4
IC
Vb Ic Vc
Wye Grounded-Delta Connection
Node Connection Matrix, C V1
VA
1
V2
[V123456] = [C][VABCabc ]
V3 V4 V5 V6
U. P. National Engineering Center National Electrification Administration
1
=
VB
-1
VC
1 1
-1
Va Vb
1 -1
1
Vc
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Transformer Model R1
A
3 Identical Single-Phase B Transformers connected Wye-Delta C Let,
R2
M 1 L1
R1
a L2 2
M 3 L1
R1 N
b L2 4
M
5 L1
R2
L2 6
R2
c
Z 1 = R1 + jωL1 = Z 3 = Z 5 Z 2 = R2 + jωL2 = Z 4 = Z 6 Z M = Z 12 = jωM = Z 34 = Z 56
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Transformer Model V1
Z1 ZM
I1
V2
ZM Z2
I2
V3 V4
=
Z1 ZM
I3
ZM Z2
I4
V5
Z1 ZM
I5
V6
ZM Z2
I6
The Primitive Voltage Equations
The Inverse of the Impedance Matrix
The Primitive Admittance Matrix
Z2
-ZM
-ZM
Z1
1
Z2
-ZM
Z1 Z2 â&#x20AC;&#x201C;ZM2
-ZM
Z1
U. P. National Engineering Center National Electrification Administration
Z2
-ZM
-ZM
Z1
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Transformer Model YBUS = [C][ Yprim][CT] A
B
C
Z2
a
b
-ZM
ZM
Z2
YBUS =
1 Z1 Z2 â&#x20AC;&#x201C;ZM2
-ZM ZM
A ZM
B
-ZM
C
Z2
ZM
ZM
2Z1
-Z1
-Z1
-Z1
2Z1
-ZM
a b
-Z1
-Z1
2Z1
c
-ZM ZM
U. P. National Engineering Center National Electrification Administration
-ZM
c
-ZM
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Transformer Model The Bus Admittance Matrix
Iinj = [C Yprim CT] Vnode YBUS = [C][Yprim][CT] 1 1
Z2
-ZM
-ZM
Z1
1
1 Z1 Z2 â&#x20AC;&#x201C;ZM2
1 -1
-1 1 -1
1
U. P. National Engineering Center National Electrification Administration
1 1
Z2
-ZM
-ZM
Z1
-1
1 1
Z2
-ZM
-ZM
Z1
-1
1 -1
1
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Three Phase Model Define
z2 yt = z1 z 2 â&#x2C6;&#x2019; z m2
n1 a= n2
yt
-ayt yt
YBUS =
-ayt ayt
-ayt
ayt
yt
ayt
ayt
2a2yt
-a2yt
-a2yt
-a2yt
2a2yt
-a2yt
-a2yt
-a2yt
2a2yt
-ayt ayt
ayt
-ayt
U. P. National Engineering Center National Electrification Administration
-ayt
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Three Phase Model If the admittances are already in per unit system, then the effective turns ratio ”a” must be
n1 1 = a= n2 3
−
yt
1 3
−
yt
z2 yt = 2 z1 z 2 − z m
1
yt −
1 3 1 3
1
yt yt
3 1
−
3 1 3
U. P. National Engineering Center National Electrification Administration
1
yt
yt
yt yt
−
1 3
yt
yt
3 2 yt 3 1 − yt 3 1 − yt 3
3 1 3
yt 1
yt −
3 1
yt yt
3 1 1 − yt − yt 3 3 2 yt − 1 yt 3 3 2 1 yt − yt 3 3
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Three Phase Model Summary
[Ybus] = YPP YPS
A
B
C
a
b
c
A
YAA
YAB
YAC
YAa
Yab
YAc
B
YBA
YBB
YBC
YBa YBb YBc
C
YCA
YCB
YCC YCa
YCb Ycc
a
YaA
YaB
YaC
Yaa
Yab
Yac
b
YbA
YbB
YbC
Yba
Ybb
Ybc
c
YcA
YcB
YcC
Yca
Ycb
Ycc
YSP YSS U. P. National Engineering Center National Electrification Administration
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Three Phase Model PRI
SEC
YPP
YSS
YPS
YSP
Wye-G
Wye-G
YI
YI
-YI
-YI
Wye-G
Wye
YII
YII
-YII
-YII
Wye-G
Delta
YI
YII
YIII
YIIIT
Wye
Wye-G
YII
YII
-YII
-YII
Wye
Wye
YII
YII
-YII
-YII
Wye
Delta
YII
YII
YIII
YIIIT
Delta
Wye-G
YII
YI
YIIIT
YIII
Delta
Wye
YII
YII
YIIIT
YIII
Delta
Delta
YII
YII
-YII
-YII
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Three Phase Model Summary
2yt -yt -yt
yt yt
YI =
yt
-yt YIII = 1/â&#x2C6;&#x161;3
-yt
yt -yt
yt
YII = 1/3 -yt 2yt -yt -yt -yt 2yt
yt -yt
U. P. National Engineering Center National Electrification Administration
YIIIT = 1/â&#x2C6;&#x161;3
yt
yt -yt yt
-yt
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Three Phase Model Example: 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
Zp.u. = ____ +j ____
yp.u. = ____ -j ____
[Ybus] =
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3-Phase Transformer Impedance Matrix Model
Distributing Transformer Impedance Between Windings
Impedance Matrix in BackwardForward Sweep Load Flow Wye-Grounded â&#x20AC;&#x201C; Wye-Grounded Delta-Delta
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Transformer Equations Consider the winding-to-winding relationship between primary and secondary: From transformer equations,
VPRI =a VSEC
I PRI 1 = I SEC a U. P. National Engineering Center National Electrification Administration
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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
ZT = Z P + Z S ' U. P. National Engineering Center National Electrification Administration
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Distributing Transformer Impedance Between Windings
ASSUMPTION: Transformer impedance varies as number of wire turns.
Z S = aZ P Referring ZS to primary side ,
ZS ' = a ZS = a ZP 2
3
Substituting,
ZT = Z P + a Z P 3
= (1 + a3 ) Z P U. P. National Engineering Center National Electrification Administration
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Distributing Transformer Impedance Between Windings To find ZP and ZS,
1 ZP = ZT 3 (1 + a ) a ZS = Z T 3 (1 + a ) U. P. National Engineering Center National Electrification Administration
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Impedance Matrix in BackwardForward Sweep Load Flow
Transformer model involved in backward summation of current forward computation of voltage
Wye-Grounded â&#x20AC;&#x201C; Wye-Grounded
Delta-Delta
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Wye Grounded â&#x20AC;&#x201C; Wye Grounded
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WyeGnd-WyeGnd Backward Sweep
Secondary to Secondary Winding ⎡ I Sec _ Winding _1 ⎤ ⎡1 0 0 ⎤ ⎡ I a ⎤ ⎢ ⎥ ⎢ ⎥ ⎢I ⎥ I 0 1 0 = Sec _ Winding _ 2 ⎢ ⎥ ⎢ ⎥⎢ b⎥ ⎢ I Sec _ Winding _ 3 ⎥ ⎢⎣ 0 0 1 ⎥⎦ ⎢⎣ I c ⎥⎦ ⎣ ⎦
Secondary Winding to Primary Winding
if in PU:
⎡ I Pr i _ Winding _1 ⎤ ⎡1 0 0 ⎤ ⎡ I Sec _ Winding _1 ⎤ ⎢ ⎥ ⎢ ⎥ ⎥ ⎢I = I 0 1 0 Pr i _ Winding _ 2 Sec _ Winding _ 2 ⎢ ⎥ ⎢ ⎥ ⎥⎢ ⎢ I Pr i _ Winding _ 3 ⎥ ⎢⎣ 0 0 1 ⎥⎦ ⎢ I Sec _ Winding _ 3 ⎥ ⎣ ⎦ ⎣ ⎦
⎡1 ⎢ ⎡ I Pr i _ Winding _1 ⎤ ⎢ a ⎢ ⎥ ⎢ I Pr i _ Winding _ 2 ⎢ ⎥ = ⎢0 ⎢ I Pr i _ Winding _ 3 ⎥ ⎢ ⎣ ⎦ ⎢0 ⎣⎢
If not in PU:
Primary Winding to Primary
0 1 a 0
⎤ 0⎥ ⎥ ⎡ I Sec _ Winding _1 ⎤ ⎢ ⎥ 0 ⎥ ⎢ I Sec _ Winding _ 2 ⎥ ⎥ ⎥ ⎥ ⎢I 1 ⎥ ⎣ Sec _Winding _ 3 ⎦ a ⎦⎥
⎡ I A ⎤ ⎡1 0 0 ⎤ ⎡ I Pr i _ Winding _1 ⎤ ⎥ ⎢ I ⎥ = ⎢0 1 0 ⎥ ⎢ I ⎢ B⎥ ⎢ ⎥ ⎢ Pr i _ Winding _ 2 ⎥ ⎢⎣ I C ⎥⎦ ⎢⎣0 0 1 ⎥⎦ ⎢⎣ I Pr i _ Winding _ 3 ⎥⎦
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WyeGnd-WyeGnd Forward Sweep
Primary to Primary winding ⎡VPr i _ Winding _1 ⎤ ⎡1 0 0 ⎤ ⎡VAN ⎤ ⎡ I Pr i _ Winding _1 * Z Pr i _ Winding _1 ⎤ ⎢ ⎥ ⎢ ⎥ ⎥ ⎢V ⎥ − ⎢ I = V 0 1 0 * Z ⎢ Pr i _ Winding _ 2 ⎥ ⎢ ⎥ ⎢ BN ⎥ ⎢ Pr i _ Winding _ 2 Pr i _ Winding _ 2 ⎥ ⎢VPr i _ Winding _ 3 ⎥ ⎢⎣0 0 1 ⎥⎦ ⎢⎣VCN ⎥⎦ ⎢ I Pr i _ Winding _ 3 * Z Pr i _ Winding _ 3 ⎥ ⎣ ⎦ ⎣ ⎦
Primary Winding to Secondary Winding
⎡1
If in PU: ⎡VSec _ Winding _1 ⎤
⎡1 0 0 ⎤ ⎡VPr i _ Winding _1 ⎤ If not in PU: ⎢ ⎡VSec _ Winding _1 ⎤ ⎢ a ⎢ ⎥ ⎢ ⎢ ⎥ ⎥ ⎢ ⎥ ⎢VSec _ Winding _ 2 ⎥ = ⎢ 0 1 0 ⎥ ⎢VPr i _ Winding _ 2 ⎥ VSec _ Winding _ 2 ⎥ = ⎢ 0 ⎢ ⎢VSec _ Winding _ 3 ⎥ ⎢⎣ 0 0 1 ⎥⎦ ⎢VPr i _ Winding _ 3 ⎥ ⎢ ⎣ ⎦ ⎣ ⎦ ⎢V ⎥ ⎢ ⎣
Sec _ Winding _ 3
Secondary Winding to Secondary
⎦
⎢0 ⎣⎢
0 1 a 0
⎤ 0⎥ ⎥ ⎡VPr i _ Winding _1 ⎤ ⎢ ⎥ 0 ⎥ ⎢VPr i _ Winding _ 2 ⎥ ⎥ ⎥ ⎥ ⎢V 1 ⎥ ⎣ Pr i _ Winding _ 3 ⎦ a ⎦⎥
⎡Van ⎤ ⎡1 0 0 ⎤ ⎡VSec _ Winding _1 ⎤ ⎡ I Sec _ Winding _1 * Z Sec _ Winding _1 ⎤ ⎥ ⎢ ⎥ ⎢V ⎥ = ⎢0 1 0 ⎥ ⎢V − I * Z ⎢ bn ⎥ ⎢ ⎥ ⎢ Sec _ Winding _ 2 ⎥ ⎢ Sec _ Winding _ 2 Sec _ Winding _ 2 ⎥ ⎢⎣Vcn ⎥⎦ ⎢⎣0 0 1 ⎥⎦ ⎢⎣VSec _ Winding _ 3 ⎥⎦ ⎢⎣ I Sec _ Winding _ 3 * Z Sec _ Winding _ 3 ⎥⎦ U. P. National Engineering Center National Electrification Administration
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Delta-Delta Transformer Connection
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Delta-Delta Backward Sweep
Secondary to Secondary Winding
Secondary Winding to Primary Winding
⎡ I Sec _ Winding _1 ⎤ ⎡ 1 −1 0 ⎤ ⎡ I a ⎤ ⎢ ⎥ 1⎢ ⎥ ⎢I ⎥ 0 1 1 = − I ⎢ Sec _ Winding _ 2 ⎥ 3 ⎢ ⎥⎢ b⎥ ⎢ I Sec _ Winding _ 3 ⎥ ⎢⎣ −1 0 1 ⎥⎦ ⎢⎣ I c ⎥⎦ ⎣ ⎦
If in PU:
⎡ I Pr i _ Winding _1 ⎤ ⎡1 0 0 ⎤ ⎡ I Sec _ Winding _1 ⎤ ⎢ ⎥ ⎢ ⎥ ⎥⎢ ⎢ I Pr i _ Winding _ 2 ⎥ = ⎢0 1 0 ⎥ ⎢ I Sec _ Winding _ 2 ⎥ ⎢ I Pr i _ Winding _ 3 ⎥ ⎢⎣0 0 1 ⎥⎦ ⎢ I Sec _ Winding _ 3 ⎥ ⎣ ⎦ ⎣ ⎦
⎡1 ⎢ ⎡ I Pr i _ Winding _1 ⎤ ⎢ a ⎢ ⎥ ⎢ I Pr i _ Winding _ 2 ⎢ ⎥ = ⎢0 ⎢ I Pr i _ Winding _ 3 ⎥ ⎢ ⎣ ⎦ ⎢0 ⎢⎣
If not in PU:
Primary Winding to Primary
⎡ I a ⎤ ⎡ 1 0 −1⎤ ⎡ I Pr i _ Winding _1 ⎤ ⎥ ⎢ I ⎥ = ⎢ −1 1 0 ⎥ ⎢ I ⎢ b⎥ ⎢ ⎥ ⎢ Pr i _ Winding _ 2 ⎥ ⎢⎣ I c ⎥⎦ ⎢⎣ 0 −1 1 ⎥⎦ ⎢⎣ I Pr i _ Winding _ 3 ⎥⎦ U. P. National Engineering Center National Electrification Administration
0 1 a 0
⎤ 0⎥ ⎥ ⎡ I Sec _ Winding _1 ⎤ ⎢ ⎥ 0 ⎥ ⎢ I Sec _ Winding _ 2 ⎥ ⎥ ⎥ ⎥ ⎢I 1 ⎥ ⎣ Sec _ Winding _ 3 ⎦ a ⎥⎦
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Delta-Delta Forward Sweep
Primary to Primary Winding
Primary Winding to Secondary Winding
⎡VPr i _ Winding _1 ⎤ ⎡ 1 −1 0 ⎤ ⎡VAN ⎤ ⎡ I Pr i _ Winding _1 * Z Pr i _ Winding _1 ⎤ ⎢ ⎥ ⎢ ⎥ ⎥ ⎢V ⎥ − ⎢ I V 0 1 1 * Z = − ⎢ Pr i _ Winding _ 2 ⎥ ⎢ ⎥ ⎢ BN ⎥ ⎢ Pr i _ Winding _ 2 Pr i _ Winding _ 2 ⎥ ⎢VPr i _ Winding _ 3 ⎥ ⎢⎣ −1 0 1 ⎥⎦ ⎢⎣VCN ⎥⎦ ⎢ I Pr i _ Winding _ 3 * Z Pr i _ Winding _ 3 ⎥ ⎣ ⎦ ⎣ ⎦ ⎡1 ⎢a ⎡VSec _ Winding _1 ⎤ ⎡1 0 0 ⎤ ⎡VPr i _ Winding _1 ⎤ ⎡ ⎤ V ⎢ ⎢ ⎥ ⎢ ⎥ Sec _ Winding _1 ⎥⎢ ⎢ ⎥ ⎢ ⎢VSec _ Winding _ 2 ⎥ = ⎢0 1 0 ⎥ ⎢VPr i _ Winding _ 2 ⎥ V ⎢ Sec _ Winding _ 2 ⎥ = ⎢ 0 ⎢VSec _ Winding _ 3 ⎥ ⎢⎣0 0 1 ⎥⎦ ⎢VPr i _ Winding _ 3 ⎥ ⎣ ⎦ ⎣ ⎦ ⎢VSec _ Winding _ 3 ⎥ ⎢ ⎣ ⎦ ⎢0 Secondary Winding to Secondary ⎣⎢
If in PU:
If not in PU:
⎡ 1∠ − 3 0 ⎢ 3 ⎡V a ⎤ ⎢ ⎢ V ⎥ = ⎢ 1∠ − 1 5 0 ⎢ b⎥ ⎢ 3 ⎢⎣ V c ⎥⎦ ⎢ ⎢ 1∠ − 3 0 ⎢ 3 ⎣
0 0 0
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⎤ 0⎥ ⎥ ⎥ 0⎥ ⎥ ⎥ 1⎥ ⎦
0 1 a 0
⎤ 0⎥ ⎥ ⎡VPr i _ Winding _1 ⎤ ⎢ ⎥ 0 ⎥ ⎢VPr i _ Winding _ 2 ⎥ ⎥ ⎥ ⎥ ⎢V 1 ⎥ ⎣ Pr i _ Winding _ 3 ⎦ a ⎦⎥
⎡ V S e c _ W in d in g _ 1 ⎤ ⎢ ⎥ ⎢ V S e c _ W in d in g _ 2 ⎥ ⎢ V S e c _ W in d in g _ 3 ⎥ ⎣ ⎦
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Transmission and Distribution Line Models
Series Impedance of Lines
Shunt Capacitance of Lines
Nodal Admittance Matrix Model
Data Requirements
Transmission Line U. P. National Engineering Center National Electrification Administration
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Transmission and Distribution Line Models +•
-•
Z = R + jXL 1 YC 2
•+
1 YC 2
VR
•-
Balanced Three-Phase System
A B C
Unbalanced Three-Phase System
Zaa
Zab
Zac
Zba
Zbb
Zbc
Zca
Zcb
Zcc
a b c
Y’aa
Y’ab
Y’ac
Y”aa
Y”ab
Y”ac
Y’ba
Y’bb
Y’bc
Y”ba
Y”bb
Y”bc
Y’ca
Y’cb
Y’cc
Y”ca
Y”cb
Y”cc
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Series Impedance of Lines 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. U. P. National Engineering Center National Electrification Administration
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Series Impedance of Lines Stranding of Conductors 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. Steel
Aluminum
Hard-Drawn Copper
Aluminum Conductor Steel Reinforced
(Cu)
(ACSR)
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Series Impedance of Lines 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
L R=ρ A
R – Resistance ρ – Resistivity of Material L – Length A – Cross-Sectional Area
Resistance increases with Temperature
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Series Impedance of Lines Resistance of Conductors Conductor Size Type Value Unit INDEX 1 ACSR 6 AWG 2 ACSR 5 AWG 3 ACSR 4 AWG 4 ACSR 4 AWG 5 ACSR 3 AWG 6 ACSR 2 AWG 7 ACSR 2 AWG 8 ACSR 1 AWG 9 ACSR 1/0 AWG 10 ACSR 2/0 AWG
Strands 6/1 6/1 7/1 6/1 6/1 7/1 6/1 6/1 6/1 6/1
O.D. (Inches) 0.19800 0.22300 0.25700 0.25000 0.28100 0.32500 0.31600 0.35500 0.39800 0.44700
GMR Resistance (feet) (Ohm/Mile) 0.00394 3.98000 0.00416 3.18000 0.00452 2.55000 0.00437 2.57000 0.00430 2.07000 0.00504 1.65000 0.00418 1.69000 0.00418 1.38000 0.00446 1.12000 0.00510 0.89500
Source: Westinghouse T&D Handbook
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Series Impedance of Lines Line Inductance Self Inductance: L = L int + L ext
Mutual Inductance (between 2 conductors): z 11 1 r I1 z 2
r I2
1’ 12
2’
z 22
V 1− 1' = I 1 z 11 + I 2 z 12 U. P. National Engineering Center National Electrification Administration
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Series Impedance of Lines Carson’s Line Carson examined a single overhead conductor whose remote end is connected to earth.
Local Earth REF
+ r
Va -
z aa
a
r Ia
r Id
Remote Earth
z ad
r Vd = 0
d
a’
zdd
d’
Fictitious Return Conductor
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. U. P. National Engineering Center National Electrification Administration
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Series Impedance of Lines The line is described by the following equations:
r r r r r Vaa ' = Va − Va ' = zaa I a + zad I d r r r r r Vdd ' = Vd − Vd ' = zad I a + zdd I d
r r r r r Note: I a = − I d , Vd = 0 and Va ' − Vd ' = 0. Subtracting the two equations, we get or
r r Va = ( zaa + zdd − 2 zad ) I a r r zaa Va = zaa I a
= zaa + zdd − 2 zad
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Series Impedance of Lines Primitive Impedances:
2s − 1) zaa = ra + jω La = ra + jω k (ln Dsa 2s zdd = rd + jω k (ln − 1) Dsd 2s zad = jω M = jω k (ln − 1) Dad ω k = (2π f )(2 x10 −7 ) ohm/meter
ra, rd = resistances of overhead conductor and fictitious ground wire, respectively Dsa, Dsd = GMRs of overhead conductor and fictitious ground wire, respectively Dad = Distance between the overhead conductor and fictitious ground wire U. P. National Engineering Center National Electrification Administration
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Series Impedance of Lines Earth Resistance: Carson derived an empirical formula for the earth resistance. -3 立/mile r = 1.588 x 10 f d
= 9.869 x 10-4 f
立/km
where f is the power frequency in Hz Note : At 60 Hz,
rd = 0.09528
立/mile
= 0.059214 U. P. National Engineering Center National Electrification Administration
立/km
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Series Impedance of Lines Geometric Mean Radius For a solid conductor with radius r, Ds Bundle of Two
= rÎľ
â&#x2C6;&#x2019;
1 4
= 0.78 r
Bundle of Four d
d d
Ds = Dsc d
Ds = 1.09 4 Dsc d 3
Note: Dsc=GMR of a single conductor U. P. National Engineering Center National Electrification Administration
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Series Impedance of Lines Equivalent Impedance: Substitute the primitive impedances into We get
zaa = zaa + zdd − 2 zad
D ad 2 zaa = ( ra + rd ) + jω k ln Dsa Dsd D ad 2 De = Define Dsd We get
De zaa = (ra + rd ) + jωk ln Dsa
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Ω/unit length
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Series Impedance of Lines The quantity De is a function of frequency and earth resistivity.
De = 2160 Ď / f
feet
Typical values of De are tabulated below. Return Earth Condition Sea water Swampy ground Average Damp Earth Dry earth Sandstone
Resistivity (Ί-m)
De (ft)
0.01-1.0 10-100 100 1000 109
27.9-279 882-2790 2790 8820 8.82x106
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Series Impedance of Lines Three-Phase Line Impedances r Ia r Ib r Ic
a +r
b
Va
-
+r
Vb +r Vc -
REF
c
z aa
a’
zbb
zab z ca b’
z cc
zbc z ad
r Vd = 0 d
r Id
zdd
U. P. National Engineering Center National Electrification Administration
zbd
c’
z cd
All wires grounded here
d’
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Series Impedance of Lines The voltage equation describing the line is
r r r Vaa ' Va − Va ' r r r Vbb ' Vb − Vb ' r = r r Vcc ' Vc − Vc ' r r r Vdd ' Vd − Vd '
=
zaa zba
zab zbb
zac zbc
zad zbd
zca zda
zcb zdb
zcc zdc
zcd zdd
r Ia r Ib r Ic r Id
Since all conductors are grounded at the remote end, we get from KCL or
r r r r I a + Ib + Ic + I d = 0 r r r r I d = −( I a + I b + I c )
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Series Impedance of Lines We can subtract the voltage equation of the ground conductor from the equations of phases a, b and c. The resulting matrix equation is
r Va r Vb r Vc
=
zaa zab zac
zab zbb zbc
zac zbc zcc
r Ia r Ib r Ic
V/unit length
Self Impedances:
zaa = zaa − 2 zad + zdd
Ω/unit length
zcc = zcc − 2 zcd + zdd
Ω/unit length Ω/unit length
zbb = zbb − 2zbd + zdd
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Series Impedance of Lines Mutual Impedances:
z ab = z ab − z ad − z bd + z dd z bc = z bc − z bd − z cd + z dd z ac = z ac − z ad − z cd + z dd
Ω/unit length Ω/unit length Ω/unit length
Primitive Impedances:
2s − 1) z xx = rx + jω k (ln Dsx
Ω/unit length
2s z xy = jω k (ln − 1) Dxy
Ω/unit length
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x=a,b,c,d
xy=ab,bc,ca,ad,bd,cd
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Series Impedance of Lines Assumptions: 1. Identical phase conductors
Ds = Dsa = Dsb = Dsc 2. Distances of the overhead conductors to the fictitious ground conductor are the same
De = Dad = Dbd = Dcd We get
De zaa = zbb = zcc = (ra + rd ) + jω k ln Ds De Ω/unit length z xy = rd + jω k ln Dxy xy=ab,bc,ca U. P. National Engineering Center National Electrification Administration
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Series Impedance of Lines 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
10’ a
10’ b
c
De z aa = z bb = z cc = ( ra + rd ) + jωk ln Ds 2790 = ( 0.278 + 0.095 ) + j0.121 ln 0.01668 Ω/mile = 0.373 + j1.459 Z aa = 14.93 + j 58.38 Ω U. P. National Engineering Center National Electrification Administration
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Series Impedance of Lines z ab = z bc = 0.095 + j0.121 ln 2790 10 = 0.095 + j0.683 Z ab = 3.81 + j 27.33 Ω z ac = 0.095 + j0.121 ln 2790 20
Ω/mile
Z ac = 3.81 + j 23.97 Ω We get
⎡14.93+ j58.38 3.81+ j27.33 3.81+ j23.97 ⎤ Zabc= ⎢ 3.81+ j27.33 14.93+ j58.38 3.81+ j27.33 ⎥ ⎥ ⎢ ⎢⎣ 3.81+ j23.97 3.81+ j27.33 14.93+ j58.38⎥⎦ U. P. National Engineering Center National Electrification Administration
Ω
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Series Impedance of Lines Lines with Overhead Ground Wire r a b +r
c
Ia r Ib r Ic r Iw
z aa
a’
zbb
zab z ca
z cc
zbc
Va +r z ww z ad Vb +r w Vc +r zbd Vw z cd z wd r Vd = 0 REF d r Id
b’ c’ w’
All wires grounded here
d’
zdd
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237
Series Impedance of Lines The primitive voltage equation is
r r Va − Va ' r r Vb − Vb ' r r Vc − Vc ' r 0 − Vw ' r 0 − Vd '
=
zaa zba
zab zbb
zac zbc
zaw zbw
zad zbd
zca
zcb
zcc
zcw
zcd
zwa zda
z wb zdb
zwc zdc
zww zdw
zwd zdd
r Ia r Ib r Ic r Iw r Id
V/unit length
From KCL,rwe get r or
r r r I a + Ib + Ic + I w + Id = 0 r r r r r I d = −( I a + Ib + Ic + I w )
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Series Impedance of Lines It can be shown that
r Va r Vb r Vc r
Vw
=
zaa zba zca zwa
zab zbb zcb zwb
zac zbc zcc zwc
zaw zbw zcw zww
De zxx = ( rx + rd ) + jωk ln Dsx De z xy = rd + jω k ln Dxy U. P. National Engineering Center National Electrification Administration
r Ia r Ib r Ic r
Iw
where
r
Vw = 0
xx=aa,bb,cc,ww
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Series Impedance of Lines Using Kron Reduction technique,
I1 I2
V1
Z1 Z2 = Z3 Z4 0
where Z1, Z2, Z3 and Z4 are also matrices. â&#x2C6;&#x2019;1
V1 = (Z1 â&#x2C6;&#x2019; Z2Z4 Z3 )I1 I2 is eliminated and the matrix is reduced to the size of Z1 U. P. National Engineering Center National Electrification Administration
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Series Impedance of Lines Eliminating the ground wire current Iw
⎡ z aa ⎢ Z 1 = ⎢ z ba ⎢⎣ z ca
We get
z abc
z ab z bb z cb
z ac ⎤ ⎡zaw ⎤ ⎢z ⎥ ⎥ z bc ⎥ Z2 = ⎢ bw ⎥ ⎢⎣zcw ⎥⎦ z cc ⎥⎦
⎡ z aw z wa ⎢ z aa − z ww ⎢ z bw z wa ⎢ = z ba − ⎢ z ww ⎢ ⎢ z ca − z cw z wa ⎢⎣ z ww
z aw z wb z ab − z ww z bw z wb z bb − z ww z cw z wb z cb − z ww
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Z 3 = [zaw zbw zcw ]
Z 4 = z ww z aw z wc ⎤ z ac − ⎥ z ww ⎥ z bw z wc ⎥ z bc − z ww ⎥ z cw z wc ⎥⎥ z cc − z ww ⎥⎦
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Series Impedance of Lines 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: w rw=4.0 Ω/mile @ 25oC Dsw=0.001 ft @ 60 Hz
15’
For the ground wire, we get
z ww
Z ww
De = ( rw + rd ) + jωk ln Dsw
10’ a
10’ b
c
2790 = ( 4.0 + 0.095 ) + j0.121 ln 0.001 = 4.095 + j1.8 Ω/mile = 163.8 + j72 Ω
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Series Impedance of Lines z aw = z cw Z aw
De Ω/mile = rd + jωk ln Daw
2790 = 0.095 + j0.121 ln 18.03 = Z cw = 3.81 + j 24.47 Ω
Z bw = 0.095 + j0.121 ln 2790 Ω/mile 15 Z bw = 3.81 + j 25.36 Ω From a previous example, we got
⎡14.93+ j58.38 3.81+ j27.33 3.81+ j23.97 ⎤ ⎢ 3.81+ j27.33 14.93+ j58.38 3.81+ j27.33 ⎥ Z1= ⎢ ⎥ ⎢⎣ 3.81+ j23.97 3.81+ j27.33 14.93+ j58.38⎥⎦ U. P. National Engineering Center National Electrification Administration
Ω
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Series Impedance of Lines Using the ground wire impedances, we also get
⎡ 3 .81 + j 24 .47 ⎤ ⎢ 3 .81 + j 25 .36 ⎥ T Z = Z2 = ⎢ 3 ⎥ ⎢⎣ 3 .81 + j 24 .47 ⎥⎦
Z 4 = 163.8 + j72 Ω
Performing Kron reduction, we get
⎡17.5 + j56.11 Zabc = ⎢ 6.48 + j 25.0 ⎢ ⎢⎣ 6.38 + j 21.7
6.48 + j 25.0 17.71+ j55.97 6.48 + j 25.0
6.38 + j 21.7 ⎤ 6.48 + j 25.0 ⎥ Ω ⎥ 17.5 + j56.11⎥⎦
Note: The self impedances are no longer equal. U. P. National Engineering Center National Electrification Administration
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Series Impedance of Lines Line Transposition Line transposition is used to make the mutual impedances identical. r Ia Phase c r Pos.1 Ib Phase a r Pos.2 Ic Phase b
Pos.3
Note:
s1
s2
s3
Section 1
Section 2
Section 3
Each phase conductor is made to occupy all possible positions.
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Series Impedance of Lines Voltage Equationsr for Each ⎡V a ⎤ ⎡ Z 11 − 1 ⎢r ⎥ ⎢ V = ⎢ Z 21 − 1 For Section 1 ⎢ rb ⎥ ⎢V c ⎥ ⎢⎣ Z 31 − 1 ⎣ ⎦ r ⎡V c ⎤ ⎡ Z 11 − 2 ⎢r ⎥ ⎢ For Section 2 ⎢V a ⎥ = ⎢ Z 21 − 2 r ⎢V b ⎥ ⎢⎣ Z 31 − 2 ⎣ ⎦ r ⎡V b ⎤ ⎡ Z 11 − 3 ⎢r ⎥ ⎢ For Section 3 ⎢Vrc ⎥ = ⎢ Z 21 − 3 ⎢V a ⎥ ⎢⎣ Z 31 − 3 ⎣ ⎦ U. P. National Engineering Center National Electrification Administration
Section r Z 12 − 1 Z 13 − 1 ⎤ ⎡ I a ⎤ ⎢r ⎥ ⎥ Z 22 − 1 Z 23 − 1 ⎥ ⎢ I b ⎥ r Z 32 − 1 Z 33 − 1 ⎥⎦ ⎢⎣ I c ⎥⎦ r Z 12 − 2 Z 13 − 2 ⎤ ⎡ I c ⎤ ⎢r ⎥ ⎥ Z 22 − 2 Z 23 − 2 ⎥ ⎢ I a ⎥ r ⎢ Z 32 − 2 Z 33 − 2 ⎥⎦ ⎣ I b ⎥⎦ r Z 12 − 3 Z 13 − 3 ⎤ ⎡ I b ⎤ ⎢r ⎥ ⎥ Z 22 − 3 Z 23 − 3 ⎥ ⎢ I c ⎥ r ⎢ Z 32 − 3 Z 33 − 3 ⎥⎦ ⎣ I a ⎥⎦
volts
volts
volts
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Series Impedance of Lines The total Voltage Drop at phases a, b, and c are:
r r ΣVa = ( Z 11−1 + Z 22 − 2 + Z 33 −3 )I a r + ( Z 12 −1 + Z 23 − 2 + Z 31−3 )I b r + ( Z 13 −1 + Z 21− 2 + Z 32 − 3 )I c r r ΣVb = ( Z 21−1 + Z 32 − 2 + Z 13 −3 )I a r + ( Z 22 −1 + Z 33− 2 + Z 11− 3 )I b r + ( Z 23 −1 + Z 31− 2 + Z 12 − 3 )I c r r ΣVc = ( Z 31−1 + Z 12 − 2 + Z 23 −3 )I a r + ( Z 32 −1 + Z 13 − 2 + Z 21−3 )I b r + ( Z 33 −1 + Z 11− 2 + Z 22 − 3 )I c
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Series Impedance of Lines Define f1, f2 and f3 as as the ratios of s1, s2 and s3 to the total length r s, respectively. We get r
ΣVa = ( f 1 Z 11 + f 2 Z 22 + f 3 Z 33 )I a r
+ ( f 1 Z 12 + f 2 Z 23 + f 3 Z 31 )I b
r ΣVb r ΣVc
r + ( f 1 Z 13 + f 2 Z 21 + fr3 Z 32 )I c = ( f 1 Z 21 + f 2 Z 32 + f 3 Z 13 )I a r + ( f 1 Z 22 + f 2 Z 33 + f 3 Z 11 )I b r + ( f 1 Z 23 + f 2 Z 31 + f 3 Z 12 )I c r = ( f 1 Z 31 + f 2 Z 12 + f 3 Z 23 )I a r + ( f 1 Z 32 + f 2 Z 13 + f 3 Z 21 )I b r + ( f 1 Z 33 + f 2 Z 11 + f 3 Z 22 )I c
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s1 f1 = s s2 f2 = s s3 f3 = s
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Series Impedance of Lines Define:
Z k 1 = f 1 Z 12 + f 2 Z 23 + f 3 Z 13 Z k 2 = f 1 Z 13 + f 2 Z 12 + f 3 Z 23 Z k 3 = f 1 Z 23 + f 2 Z 13 + f 3 Z 12 Z s = Z 11 = Z 22 = Z 33
Substitution gives
r ⎡ ΣV a ⎤ ⎡ Z s ⎢ r⎥ ⎢ ⎢ΣVrb ⎥ = ⎢ Z k 1 ⎢ ΣVc ⎥ ⎢⎣ Z k 2 ⎣ ⎦
Z k1 Zs Zk3
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r Z k 2 ⎤⎡I a ⎤ ⎢r ⎥ ⎥ Z k 3 ⎥ ⎢ I b ⎥ Volts r Z s ⎥⎦ ⎢⎣ I c ⎥⎦ Competency Training & Certification Program in Electric Power Distribution System Engineering
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Series Impedance of Lines It can be shown that
De Z s = ( ra + rd )s + jωks ln Ds ⎛ De De De ⎞ ⎟⎟ Z k 1 = rd s + jωks ⎜⎜ f 1ln + f 2 ln + f 3 ln D12 D23 D31 ⎠ ⎝ Zk2
⎛ De De De ⎞ ⎟⎟ = rd s + jωks ⎜⎜ f 1ln + f 2 ln + f 3 ln D31 D12 D23 ⎠ ⎝
Z k3
⎛ De De De ⎞ ⎟⎟ = rd s + jωks ⎜⎜ f 1ln + f 2 ln + f 3 ln D23 D31 D12 ⎠ ⎝
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Series Impedance of Lines 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
Without the transposition,
10’ a
10’ b
c
Section 1
⎡14.93 + j58.38 3.81+ j27.33 3.81+ j23.97 ⎤ Zabc = ⎢⎢ 3.81+ j27.33 14.93 + j58.38 3.81+ j27.33 ⎥⎥ Ω ⎢⎣ 3.81+ j23.97 3.81+ j27.33 14.93 + j58.38⎥⎦ U. P. National Engineering Center National Electrification Administration
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Series Impedance of Lines Solving for the mutual impedances, we get
Z k 1 = f 1 Z 12 + f 2 Z 23 + f 3 Z 13 = 0.2( 3.81 + j 27.33 ) + 0.3( 3.81 + j 27.33 ) + 0.5( 3.81 + j 23.97 ) = 3.81 + j 25.65 立 Similarly, we get
Z k 2 = f 1 Z 13 + f 2 Z 12 + f 3 Z 23
= 3.81 + j 26.66 立
Z k 3 = f 1 Z 23 + f 2 Z 13 + f 3 Z 12
= 3.81 + j 26.32 立
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Series Impedance of Lines The impedance matrix of the transposed line is
⎡14.93+ j58.38 3.81+ j25.65 3.81+ j26.66 ⎤ Zabc= ⎢ 3.81+ j25.65 14.93+ j58.38 3.81+ j26.32 ⎥ Ω ⎢ ⎥ ⎢⎣ 3.81+ j26.66 3.81+ j26.32 14.93+ j58.38⎥⎦ For comparison, the impedance matrix of the untransposed line is
⎡14.93+ j58.38 3.81+ j27.33 3.81+ j23.97 ⎤ Zabc= ⎢ 3.81+ j27.33 14.93+ j58.38 3.81+ j27.33 ⎥ Ω ⎢ ⎥ ⎢⎣ 3.81+ j23.97 3.81+ j27.33 14.93+ j58.38⎥⎦ U. P. National Engineering Center National Electrification Administration
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Series Impedance of Lines Completely Transposed Line If s1=s2=s3, the line is completely transposed. We r r get ⎡ ΣV ⎤ ⎡ Z Z Z ⎤ ⎡I ⎤
a s r ⎢ ⎥ ⎢ ⎢ΣVrb ⎥ = ⎢ Z m ⎢ Σ Vc ⎥ ⎢ Z m ⎣ ⎣ ⎦ where
m
Zs Zm
a r ⎢ ⎥ ⎥ Z m ⎥⎢I b ⎥ r Volts Z s ⎥⎦ ⎢⎣ I c ⎥⎦ m
De Z s = ( ra + rd )s + jωks ln Ds Ω Z m = ( Z 12 + Z 23 + Z 13 ) 1 3
De Ω = rd s + jωks ln Dm
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Series Impedance of Lines Geometric Mean Distance (GMD) Typical three-phase line configurations D12
D23 D12
D31 D12
D23
D
D31 D23
D31
D
31
23
D12
Dm = 3 D12D23D31 U. P. National Engineering Center National Electrification Administration
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Series Impedance of Lines Example: For the same line assume a complete transposition cycle. 10’
The GMD is
Dm = 3 10( 10 )( 20 ) = 12.6 feet
a
10’
b
c
We get the average of the mutual impedances.
Z m = 3.81 + j 26.21 Ω The impedance of the transposed line is
⎡14.93+ j58.38 3.81+ j26.21 3.81+ j26.21 ⎤ Zabc= ⎢ 3.81+ j26.21 14.93+ j58.38 3.81+ j26.21 ⎥ Ω ⎢ ⎥ ⎢⎣ 3.81+ j26.21 3.81+ j26.21 14.93+ j58.38⎥⎦ U. P. National Engineering Center National Electrification Administration
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Series Impedance of Lines Phase to Sequence Impedances Consider a transmission line that is described by the following voltage equation:
or
r ⎡Va ⎤ ⎡ Z aa ⎢r ⎥ ⎢ V ⎢ rb ⎥ = ⎢ Z ab ⎢Vc ⎥ ⎢⎣ Z ac ⎣ ⎦
Z ab Z bb Z bc
r r Vabc = Z abc I abc
r Z ac ⎤ ⎡ I a ⎢r ⎥ Z bc ⎥ ⎢ I b r Z cc ⎥⎦ ⎢⎣ I c
⎤ ⎥ ⎥ ⎥ ⎦
volts
From symmetrical components, we have
r r Vabc = AV012
and
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r r I abc = AI 012
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Series Impedance of Lines Substitution gives or
r r AV012 = Z abc AI 012 r r −1 V 012 = A Z abc A I 012
which implies that
Z 012 = A −1 Z abc A Performing the multiplication, we get
⎡ Z 0 ⎤ ⎡ Z s 0 + 2 Z m0 ⎢Z ⎥ = ⎢ Z − Z m1 ⎢ 1 ⎥ ⎢ s1 ⎢⎣ Z 2 ⎥⎦ ⎢⎣ Z s 2 − Z m 2
Z s2 − Z m2 Z s0 − Z m0 Z s 1 + 2 Z m1
Z s 1 − Z m1 ⎤ Z s 2 + 2 Z m 2 ⎥⎥ Z s 0 − Z m0 ⎥⎦
Note: Z012 is not symmetric. U. P. National Engineering Center National Electrification Administration
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Series Impedance of Lines It can be shown that
Z s 0 = 31 ( Z aa + Z bb + Z cc )
Z s 1 = 31 ( Z aa + aZ bb + a 2 Z cc ) Z s 2 = 31 ( Z aa + a 2 Z bb + aZ cc ) Z m 0 = 13 ( Z ab + Z bc + Z ca ) Z m 1 = 13 ( a 2 Z ab + Z bc + aZ ca )
Z m 2 = 31 ( aZ ab + Z bc + a 2 Z ca ) U. P. National Engineering Center National Electrification Administration
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Series Impedance of Lines If the line is completely transposed,
Z s0 = Z s
Z m0 = Z m
Z s1 = Z s 2 = 0
Z m1 = Z m 2 = 0
The sequence impedance matrix reduces to
⎡ Z 0 ⎤ ⎡Z s + 2 Z m ⎢Z ⎥ = ⎢ 0 ⎢ 1⎥ ⎢ ⎢⎣Z 2 ⎥⎦ ⎢⎣ 0
0 Zs − Zm 0
0 ⎤ 0 ⎥⎥ Z s − Z m ⎥⎦
Note: The sequence impedances are completely decoupled. U. P. National Engineering Center National Electrification Administration
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Series Impedance of Lines For a completely transposed line, the equation in the sequence domain is r r
V a0 ⎡Z 0 r V a 1 = ⎢⎢ 0 r ⎢⎣ 0 Va2
where
0 Z1 0
0 ⎤ ⎡ I a0 ⎢r ⎥ 0 ⎥ ⎢ I a1 r Z 2 ⎥⎦ ⎢⎣ I a 2
Dm Z 1 = Z 2 = ra s + jωks ln Ds Z 0 = ra s + 3rd s + jωks ln
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De
⎤ ⎥ ⎥ ⎥ ⎦
Ω 3
Ds Dm
2
Ω
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Series Impedance of Lines Example: For the same line and assuming a complete transposition cycle, find the sequence impedances of the line.
10’
a
In the previous example, we got
10’
b
c
Z s = 14.93 + j 58.38 Ω
Z m = 3.81 + j 26.21 Ω The sequence impedances are
Z 0 = Z s + 2 Z m = 22.55 + j110.80 Ω Z 1 = Z 2 = Z s − Z m = 11.12 + j 32.17 Ω U. P. National Engineering Center National Electrification Administration
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Shunt Capacitance of Lines Caw w
b Cbw
Cab a
Cbc
Ccw
Cac c
Cag
Cbg Ccg
Cwg â&#x20AC;˘ Self-capacitance â&#x20AC;˘ Mutual-capacitance
Capacitance of Three Phase Lines U. P. National Engineering Center National Electrification Administration
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Shunt Capacitance of Lines Voltage Due to Charged Conductor Consider two points P1 and P2 which are located at distances D1 and D2 from the center of the conductor. The voltage drop from P1 to P2 is Electric charge
v 12
D2 = ln 2πε D1 q
Volts
D1
P1 P2
D2
q +
x
ˆ ar
r r D = E=
ε
q 2πε x
aˆ r
Permitivity of medium Electric Field of a Long Conductor
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Shunt Capacitance of Lines Capacitance of a Two-Wire Line The 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 Due to charge qa, we get the voltage drop vab.
v ab
qa D ln = 2Ď&#x20AC;Îľ ra
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Shunt Capacitance of Lines Due to charge qb, we also get the voltage drop vba.
v ba
qb D ln = 2 πε rb
or
v ab
qb qb rb D ln = =− ln 2πε rb 2πε D
Applying superposition, we get the total voltage drop from charge qa to charge qb.
v ab
qa rb qb D ln + = ln 2πε ra 2πε D
Since qa+qb=0, we get
v ab
qa D2 ln = 2πε ra rb
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Shunt Capacitance of Lines Self-Capacitance In general, ra=rb. We get
v ab
qa
D = ln πε r
Volts
The capacitance between conductors is qa πε = C ab = Farad/meter D Vab ln r The capacitance to neutral is
C an = C bn = 2C ab
2πε = D ln r
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Shunt Capacitance of Lines 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.
Dab
+qb
+qa
Daw Dac
Haa Hab
-qa
Hac
-qb
+qw
+qc
Haw -qc -qw
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Shunt Capacitance of Lines The voltage drop from conductor a to ground is
va = 21 vaa' H aa H ab H an 1 = ( q a ln + qb ln + ... + q n ln 4πε ra Dab Dan ra Dab Dan − q a ln − qb ln − ... − q n ln ) H ab H an H aa Combining common terms, we get
H aa H ab H an ( q a ln + qb ln + ... + q n ln ) va = Dab Dan 2πε ra 1
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Shunt Capacitance of Lines In general, for the kth overhead conductor
H bk H kk H ak ( q a ln vk = + qb ln + ... + q k ln Dak Dbk rk 2πε H nk + ... + q n ln ) Dnk 1
Using matrix notation, we get
Pab Pbb M Pnb
…
⎡v a ⎤ ⎡ Paa ⎢v ⎥ ⎢ P ⎢ b ⎥ = ⎢ ba ⎢M⎥ ⎢ M ⎢ ⎥ ⎢ ⎣v n ⎦ ⎣ Pna
Pac Pbc M Pnc
... Pan ⎤ ⎡q a ⎤ H kk 1 ln Pkk = ⎥ ⎢ ⎥ ... Pbn ⎥ ⎢qb ⎥ 2πε rk H kj 1 M M ⎥⎢ M ⎥ ⎥ ⎢ ⎥ Pkj = 2πε ln D kj ... Pnn ⎦ ⎣q n ⎦
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Shunt Capacitance of Lines
[v ] = [P ][q ] Since, q = Cv, ,then
[C ] = [P ]
−1
Inversion of matrix P gives
⎡+ C aa ⎢− C ba ⎢ C= ⎢ M ⎢ ⎣− C na
− C ab + Cbb M − C nb
− C ac − Cbc M − C nc
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... − C an ⎤ ⎥ ... − Cbn ⎥ M M ⎥ ⎥ ... + C nn ⎦
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Shunt Capacitance of Lines The Shunt Admittance is
Ybus
⎡ + jω C aa ⎢ − jωC ba ⎢ = ⎢ ⎢ ⎣ − jω C na
− jω C ab
− jω C ac
+ jωC bb
− jω C bc
− jω C nb
− jω C nc
... − jωC an ⎤ ⎥ ... − jω C bn ⎥ ⎥ ⎥ ... + jω C nn ⎦
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
C ag = C aa − C ab − C ac − ... − C an
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Shunt Capacitance of Lines Capacitance of a Transposed Line Pos.1 Pos.2 Pos.3
qa
Phase c
qb
Phase a
qc
Phase b
1 3
1 3
s
Section 1
1 3
s
Section 2
s
Section 3
The capacitance of phase a to neutral is
C an = C bn = C cn
qa 2Ď&#x20AC;Îľ = = Dm v an ln r
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Farad/meter, to neutral
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Shunt Capacitance of Lines Capacitive Reactance 1 xc = 2πfC
Dm 2.862 9 x 10 ln xc = f r Dm 1.779 6 xc = x 10 ln f r
Ω-meter, to neutral Ω-mile, to neutral
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274
Shunt Capacitance of Lines Sequence Capacitance Using matrix notation, we have
r r r r I abc = jωCabcVabc I abc = YabcVabc r r r r From Vabc = AV012 and I abc = YabcVabc, we get r r A I 012 = jω C abc A V012 r r or −1 I 012 = jωA Cabc AV012 Thus, we have
C012 = A −1C abc A U. P. National Engineering Center National Electrification Administration
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Shunt Capacitance of Lines For a completely transposed line,
Cs0 = Caa = Cbb = Ccc C m 0 = C ab = C bc = C ac Substitution gives
C012 or
0 0 ⎤ ⎡( Cs0 − 2Cm0 ) ⎥ ⎢ = 0 ( C + C ) 0 s0 m0 ⎥ ⎢ ⎢⎣ 0 0 ( Cs0 + Cm0 )⎥⎦ C0 = Cs0 − 2Cm0
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Shunt Capacitance of Lines 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. 14’ 14’ Calculate distances a b c Haa=Hbb=Hcc=80 ft
Hab=Hbc=81.2 ft Hac=84.8 ft Find the P matrix
40’
H aa Paa = Pbb = Pcc = ln 2πε 0 ra 1
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Shunt Capacitance of Lines 1 -9 = x 10 ε For air, 0 36π
Farad/meter
Substitution gives
80 Paa = 18 x 10 ln 0.0357 = 138.86 x 10 9 Meter/Farad = 86.29 x 10 6 Mile/Farad 9
Similarly, we get
H ab Pab = Pbc = ln 2πε 0 Dab = 19.66 x 10 6 Mile/Farad 1
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Shunt Capacitance of Lines The P matrix can be shown to be
⎡86.29 19.66 12.39 ⎤ P = ⎢19.66 86.29 19.66 ⎥ x 106 mi/F ⎢ ⎥ ⎢⎣12.39 19.66 86.29 ⎥⎦ Using matrix inversion, we get the C matrix.
⎡ 12.34 − 2.54 − 1.19 ⎤ ⎢− 2.54 12.75 − 2.54 ⎥ x 10-9 F/mi C= ⎢ ⎥ ⎢⎣ − 1.19 − 2.54 12.34 ⎥⎦ U. P. National Engineering Center National Electrification Administration
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Shunt Capacitance of Lines For 50 miles, we get C=
⎡ 6.17 − 1.27 ⎢− 1.27 6.38 ⎢ ⎢⎣ − 0.60 − 1.27
− 0.60 ⎤ − 1.27 ⎥⎥ 6.17 ⎥⎦
x 10-7 F
The capacitances to ground are
Cag = Caa − Cab − Cac = 0.43 μF Cbg = Cbb − Cab − Cbc = 0.38 μF Ccg = Ccc − Cbc − Cac = 0.43 μF Since the line is transposed,
Cg0 = 13 (Cag + Cbg + Ccg ) = 0.41 μF U. P. National Engineering Center National Electrification Administration
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Shunt Capacitance of Lines The self- and mutual capacitances are
C s0 = 13 ( C aa + Cbb + Ccc ) = 0.62 μF C m0 = 13 ( C ab + Cbc + C ca ) = 0.105 μF The sequence capacitances are Cm0
C0 = C s0 − 2C m0 = 0.41 μF
C1 = C 2 = C s 0 + C m0
b Cm0
a
Cm0 c
Cg0
Cg0
Cg0
= 0.725 μF U. P. National Engineering Center National Electrification Administration
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Nodal Admittance Matrix Model [Z]
IiABC ViABC
Zaa
Zab
Zac
Zba
Zbb
Zbc
Zca
Zcb
Zcc
Yaa
Yab
Yac
Yba
Ybb
Ybc [Y]/2
Yca
Ycb
Ycc
[IiABC] abc]
[Ik
6x1
=
[Y]/2
Ikabc Vkabc Yaa
Yab
Yac
Yba
Ybb
Ybc
Yca
Ycb
Ycc
[Z]-1+[Y]/2
-[Z]-1
[ViABC]
-[Z]-1
[Z]-1+[Y]/2
[Vkabc]
6x6
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Nodal Admittance Matrix Model 1’
3’
3’
Example
A
B
C 4’
Phase Conductor 336,400 26/7 ACSR Neutral Conductor 4/0 6/1 ACSR Length: 300 ft.
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N
24’
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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 U. P. National Engineering Center National Electrification Administration
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Distribution Line Models a
Dca Dab Dbc a
b
Ha
Hb
b
b Dab c
Hc
Horizontal One Ground Wire (a)
Dab
Dca
Hg
Hg
Dbc
a
c Dbc
Dca
Hc Hb Ha
Vertical One Ground Wire (b)
Hg
Ha
c Hc
Hb
Triangular One Ground Wire (c)
Configuration, Spacing, and Height (Subtransmission Lines) U. P. National Engineering Center National Electrification Administration
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Distribution Line Models Dgg
Dgg
Dgg
D12 Circuit No. 1 Horizontal Two Ground Wires (d)
Triangular Two Ground Wires (e)
Circuit No. 2
Parallel Two Ground Wires (f)
Line Spacing (Ground Wires)
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Distribution Line Models A
B
C
N
B
A B C
C
A
N
A
N
B
B
A
A B
C A B
N
3-Phase (CAB) A B A
N N
V-Phase (AB)
B
N
3-Phase (BCA)
N
A
N
3-Phase (ABC) A
C
B C
N N
V-Phase (BA)
Hg
Note: N â&#x20AC;&#x201C; Consider the grounded neutral as Ground Conductor for Hg
1-Phase (A)
Configuration, Spacing, and Height (Distribution Lines) U. P. National Engineering Center National Electrification Administration
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