Power System Modelling

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Power System Modeling and Analysis Training Course in

Power System Modeling

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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.

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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.

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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.

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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

a ia

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.

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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.

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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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−1

⎡ 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.

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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.

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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

Note: No transformation is required.

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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

]

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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

2. id and iq are DC currents.

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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

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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 -

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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.

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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.

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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.

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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).

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100

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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101

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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103

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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105

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.

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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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107

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.

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108

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

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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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114

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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116

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.

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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.

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121

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.

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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;?.

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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.

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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.

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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.

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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

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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

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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

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+

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.

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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 +

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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

-

-

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+

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?

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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.

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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;)

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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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164

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

Three Phase Transformer Bank U. P. National Engineering Center National Electrification Administration

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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.

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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.

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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.

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171

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

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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.

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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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175

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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176

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

Note: Only the X windings are shown.

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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

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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.

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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.

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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.

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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.

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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.

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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

U. P. National Engineering Center National Electrification Administration

⎤ 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.

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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.

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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

zaa is the equivalent impedance of the single overhead conductor.

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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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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

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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.

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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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251

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.

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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 )

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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.

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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

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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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Farad/meter

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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

Note: To get the total capacitive reactance, divide xc by the total length of the line.

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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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6x1

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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.

U. P. National Engineering Center National Electrification Administration

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)

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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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