CP1+B9+Lecture+No.+2+ +Power+System+Modeling

287
NATIONAL ELECTRIFICATION ADMINISTRATION U. P. NATIONAL ENGINEERING CENTER Certificate in Power System Modeling and Analysis Competency Training and Certification Program in Electric Power Distribution System Engineering U. P. NATIONAL ENGINEERING CENTER U. P. NATIONAL ENGINEERING CENTER Power System Modeling Training Course in

Transcript of CP1+B9+Lecture+No.+2+ +Power+System+Modeling

Page 1: CP1+B9+Lecture+No.+2+ +Power+System+Modeling

NATIONAL ELECTRIFICATION ADMINISTRATIONU. P. NATIONAL ENGINEERING CENTER

Certificate in

Power System Modeling and Analysis

Competency Training and Certification Program in Electric Power Distribution System Engineering

U. P. NATIONAL ENGINEERING CENTERU. P. NATIONAL ENGINEERING CENTER

Power System Modeling

Training Course in

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Competency Training & Certification Program in Electric Power Distribution System Engineering

U. P. National Engineering CenterNational Electrification Administration

Training Course in Power System Modeling

Course Outline

1. Utility Thevenin Equivalent Circuit

2. Load Models

3. Generator Models

4. Transformer Models

5. Transmission and Distribution Line Models

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Competency Training & Certification Program in Electric Power Distribution System Engineering

U. P. National Engineering CenterNational Electrification Administration

Training Course in Power System Modeling

� Thevenin’s Theorem

� Utility Fault MVA

� Equivalent Circuit of Utility

Utility Thevenin Equivalent Circuit

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Competency Training & Certification Program in Electric Power Distribution System Engineering

U. P. National Engineering CenterNational Electrification Administration

Training Course in Power System Modeling

Thevenin’s Theorem

Any linear active network with output terminals AB can be replaced by a single voltage source Vth in series with a single impedance Zth

Linear Active

Network

A

B

A

B

+

-

Vth

Zth

The Thevenin equivalent voltage Vth is the open circuit voltage measured at the terminals AB. The equivalent impedance Zth is the driving point impedance of the network at the terminals AB when all sources are set equal to zero.

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Competency Training & Certification Program in Electric Power Distribution System Engineering

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Training Course in Power System Modeling

Utility Fault MVA

Electric Utility Grid

Electric Utilities conduct short circuit analysis at the Connection Point of their customers

FaultIF

Customers obtain the Fault Data at the Connection Point to represent the Utility Grid for their power system analysis

Customer Facilities

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Competency Training & Certification Program in Electric Power Distribution System Engineering

U. P. National Engineering CenterNational Electrification Administration

Training Course in Power System Modeling

Utility Fault MVA

Electric Utility provides the Fault MVA and X/R ratio at nominal system Voltage for the following types of fault:

• Three Phase Fault

Fault MVA3φφφφ X/R3φφφφ

• Single Line-to-Ground Fault

Fault MVALG X/RLG

System Nominal Voltage in kV

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Competency Training & Certification Program in Electric Power Distribution System Engineering

U. P. National Engineering CenterNational Electrification Administration

Training Course in Power System Modeling

Equivalent Circuit of Utility

Positive & Negative Sequence Impedance

From Three-Phase Fault Analysis

1

fTPF Z

VI =

[ ]1

2f

TPFfTPF Z

VIVS ==

[ ]2

3

2

1 ZMVAFault

kVZ ==

φ

Where, Z1 and Z2 are the equivalent positive-sequence and negative-sequence impedances of the utility

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Competency Training & Certification Program in Electric Power Distribution System Engineering

U. P. National Engineering CenterNational Electrification Administration

Training Course in Power System Modeling

Equivalent Circuit of Utility

Zero Sequence Impedance

From Single Line-to-Ground Fault Analysis

021

fSLGF ZZZ

V3I

++=

21 ZZ =

[ ]01

2f

SLGFfSLGF ZZ2

V3IVS

+==

[ ]SLGF

2f

01 S

V3ZZ2 =+ Resolve to real and imaginary

components then solve for Zo

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Competency Training & Certification Program in Electric Power Distribution System Engineering

U. P. National Engineering CenterNational Electrification Administration

Training Course in Power System Modeling

Equivalent Circuit of Utility

Example:

Determine the equivalent circuit of the Utility in per unit quantities at a connection point for the following Fault Data:

System Nominal Voltage = 69 kV

Fault MVA3φφφφ = 3500 MVA, X/R3φφφφ = 22

Fault MVALG = 3000 MVA, X/RLG = 20

The Base Power is 100 MVA

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Competency Training & Certification Program in Electric Power Distribution System Engineering

U. P. National Engineering CenterNational Electrification Administration

Training Course in Power System Modeling

Equivalent Circuit of Utility

Base Power: 100 MVABase Voltage: 69 kVBase Impedance: [69]2/100 = 47.61 ohms

[ ] [ ]Ω

φ

1.36033500

69

MVAFault

kVZZ

2

3

2

21 ====

In Per Unit,

p.u.0.028661.47

1.3603

Z

ZZZ

base

actual21 ====

p.u.0.0286MVA3500

100MVAZZ

FAULT

BASE21 ===

or

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Competency Training & Certification Program in Electric Power Distribution System Engineering

U. P. National Engineering CenterNational Electrification Administration

Training Course in Power System Modeling

Equivalent Circuit of Utility

Solving for the Resistance and Reactance,

R

Z Xθθθθ

1

√[(1 + (X/R)2]X/R

θθθθ

[ ]R/Xtan 1−=θ

θθ

sinZX

cosZR

=

=

( )[ ]

( )[ ]2

1-

1

2

-1

1

Xp.u. 028571.0

22tan sin0.0286X

Rp.u. 1300.0

22tancos0.0286R

==

=

==

= +

-

fV

0.0013+j0.028571+

-01∠

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Competency Training & Certification Program in Electric Power Distribution System Engineering

U. P. National Engineering CenterNational Electrification Administration

Training Course in Power System Modeling

Equivalent Circuit of Utility

For the Zero Sequence Impedance,

.u.p 0.1kV69

kV69Voltage

.u.p 30MVA100

MVA3000SLGF

.U.P

BASE

)actual(SLGF

.U.P

==

==

{ } ( )[ ]{ } ( )[ ] p.u. 0.09987520tan0.1sinZZ2agIm

p.u. 0.00499420tan0.1cosZZ2alRe1-

01

-101

==+

==+

[ ] [ ]1.0

30

0.13

S

V3ZZ2

2

SLGF

2f

01 ===+

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Competency Training & Certification Program in Electric Power Distribution System Engineering

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Training Course in Power System Modeling

Equivalent Circuit of Utility

( ) ( )p.u. j0.0427330.003694

028571.0j1300.02099875.0j004994.0Z

099875.0j004994.0ZZ2

0

01

+=

+−+=

+=+

+

-

+

-

fV

0.0013+j0.028571+

-01∠

+

-

j0.0427330.003694 +0.0013+j0.028571

Positive Sequence

Negative Sequence

Zero Sequence

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Competency Training & Certification Program in Electric Power Distribution System Engineering

U. P. National Engineering CenterNational Electrification Administration

Training Course in Power System Modeling

Equivalent Circuit of Utility

Example:

Determine the equivalent circuit of the Utility in per unit quantities at a connection point for the following Fault Data:

Pos. Seq. Impedance = 0.03 p.u., X/R1 = 22

Zero Seq. Impedance = 0.07 p.u., X/R0 = 22

System Nominal Voltage = 69 kV

Base Power = 100 MVA

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Competency Training & Certification Program in Electric Power Distribution System Engineering

U. P. National Engineering CenterNational Electrification Administration

Training Course in Power System Modeling

Utility Thevenin Equivalent Circuits

The equivalent sequence networks of the Electric Utility Grid are:

+

-

gEr

R1 +jX1+

-

+

-Positive

SequenceNegative Sequence

Zero Sequence

+

-

Equivalent Circuit of Utility

R2 +jX2 R0 +jX0

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Competency Training & Certification Program in Electric Power Distribution System Engineering

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Training Course in Power System Modeling

� Types of Load

� Customer Load Curve

� Calculating Hourly Demand

� Developing Load Models

Load Models

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Competency Training & Certification Program in Electric Power Distribution System Engineering

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Training Course in Power System Modeling

Types of LoadAn illustration:

Sending End

Receiving End

VS = ?

Load2 MVA, 3Ph

85%PF13.2 kVLL

VR = 13.2 kVLL

Line1.1034 + j2.0856 ohms/phase

ISR = ?

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Competency Training & Certification Program in Electric Power Distribution System Engineering

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Training Course in Power System Modeling

Types of LoadAn illustration:

Sending End

Receiving End

VS = ?

Load2 MVA, 3Ph 85% pf lag 13.2 kVLL

VR = 13.2 kVLL

Line1.1034 + j2.0856 ohms/phase

ISR = ?

Constant Power (P & Q)2 MVA = 1.7 MW + j1.0536 MVAR

Constant Current (I∠θ)I = 87.4773 ∠∠∠∠ -31.79o A

Constant Impedance (R & X)Z = 87.12 = 74.0520 + j 45.8948 ΩΩΩΩ

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Competency Training & Certification Program in Electric Power Distribution System Engineering

U. P. National Engineering CenterNational Electrification Administration

Training Course in Power System Modeling

Types of Load

Sending End

Receiving End

VS = ?

Load2 MVA, 3Ph 0.85 pf, lag13.2 kVLL

VR = 13.2 kVLL

Line1.1034 + j2.0856 ohms/phase

ISR = ?

LL KV13.510

V 760.0800,7

)0856.21034.1)(79.314773.87(03

200,13

)(

=

∠=

+−∠+∠=

+=

LLS

o

oo

lineSRRS

V

j

ZIVV

r

rrrr

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Competency Training & Certification Program in Electric Power Distribution System Engineering

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Training Course in Power System Modeling

Types of Load

Sending End

Receiving End

VS = 13.51 kVLL

Load2 MVA, 3Ph 0.85 pf, lag13.2 kVLL

VR = 13.2 kVLL

Line1.1034 + j2.0856 ohms/phase

ISR = 87.48∠∠∠∠-31.79o

MVARjMW

IV ooSS

1010.17256.1

)79.314773.87)(76.0800,7(33 *

+=

∠∠=rr

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Competency Training & Certification Program in Electric Power Distribution System Engineering

U. P. National Engineering CenterNational Electrification Administration

Training Course in Power System Modeling

Types of Load

KW

MWPlosses

6.25

7.17256.1

=

−=

%35.2

%1002.13

2.13510.13

=

×−

=VR

Sending End

Receiving End

VS = 13.51 kVLL

Load2 MVA, 3Ph 0.85 pf, lag13.2 kVLL

VR = 13.2 kVLL

Line1.1034 + j2.0856 ohms/phase

ISR = 87.48∠∠∠∠-31.79o

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Competency Training & Certification Program in Electric Power Distribution System Engineering

U. P. National Engineering CenterNational Electrification Administration

Training Course in Power System Modeling

Types of LoadSending

EndReceiving

End

VS = ?

Load

VR = 11.88 kVLL

Line1.1034 + j2.0856 ohms/phase

ISR = ?

What happens if the Voltage at the Receiving End drops to 90% of its nominal value?

VR =11.88 KVLL

We will again analyze the power loss (Ploss) and Voltage Regulation (VR) for different types of loads

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Competency Training & Certification Program in Electric Power Distribution System Engineering

U. P. National Engineering CenterNational Electrification Administration

Training Course in Power System Modeling

Types of LoadCase 1: Constant Power Load

2 MVA = 1.7 MW + j1.0536 MVAR

o

SRKV

MVAjI

79.311979.97

88.113

0536.17.1

−∠=

−=

r

KV12.224

V 94.08.057,7

)0856.21034.1)(78.311979.97(03

88.11

)(

0

0

=

∠=

+−∠+∠=

+=

j

ZIVV lineSRRS

rrrr

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Competency Training & Certification Program in Electric Power Distribution System Engineering

U. P. National Engineering CenterNational Electrification Administration

Training Course in Power System Modeling

Types of LoadCase 1: Constant Power Load

2 MVA = 1.7 MW + j1.0536 MVAR

KW

WPlosses

722.28

)0134.1)(1979.97(3 2

=

=

%9.2

%10088.11

88.11224.12

=

×−

=VR

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Competency Training & Certification Program in Electric Power Distribution System Engineering

U. P. National Engineering CenterNational Electrification Administration

Training Course in Power System Modeling

Types of LoadCase 2: Constant Current Load

I = 87.4773 ∠∠∠∠ -31.79o A

KV12.190

V 84.08.037,7

)0856.21034.1)(79.314773.87(03

88.11

)(

=

∠=

+−∠+∠=

+=

o

oo

lineSRRS

j

ZIVVrrrr

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Competency Training & Certification Program in Electric Power Distribution System Engineering

U. P. National Engineering CenterNational Electrification Administration

Training Course in Power System Modeling

Types of LoadCase 2: Constant Current Load

I = 87.4773 ∠∠∠∠ -31.78o A

KW

WPlosses

33.25

)1034.1)(48.87(3 2

=

=

%6.2

%10088.11

88.1119.12

=

×−

=VR

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Competency Training & Certification Program in Electric Power Distribution System Engineering

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Training Course in Power System Modeling

Types of LoadCase 3: Constant Impedance Load

Z = 87.12 ∠∠∠∠31.79o ΩΩΩΩ = 74.0520 + j 45.8948 ΩΩΩΩ

KV12.159

KV77.00199.7

79.3112.87

0856.21034.1(79.3112.870

3

88.11

=

∠=

⎥⎦

⎤⎢⎣

⎡∠

++∠∠=

⎥⎦

⎤⎢⎣

⎡ +=

⎥⎦

⎤⎢⎣

⎡+

=

LLS

o

o

oo

Load

LineLoadRS

LineLoad

LoadSR

V

j

Z

ZZVV

ZZ

ZVV

r

r

rrrr

rr

rrr

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Competency Training & Certification Program in Electric Power Distribution System Engineering

U. P. National Engineering CenterNational Electrification Administration

Training Course in Power System Modeling

Types of LoadCase 3: Constant Impedance Load

Z = 87.12 ∠∠∠∠31.79o ΩΩΩΩ = 74.0520 + j 45.8948 ΩΩΩΩ

A 730.78

0856.21034.179.3112.87

77.00199.7

=

++∠∠

=

+=

j

ZZ

VI

o

o

LineLoad

SSR rr

rr

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Competency Training & Certification Program in Electric Power Distribution System Engineering

U. P. National Engineering CenterNational Electrification Administration

Training Course in Power System Modeling

Types of LoadCase 3: Constant Impedance Load

Z = 87.12 ∠∠∠∠31.79o ΩΩΩΩ = 74.0520 + j 45.8948 ΩΩΩΩ

KW

WPlosses

84.18

)0134.1)(73.78(3 2

=

=

%34.2

%10088.11

88.11159.12

=

×−

=VR

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Competency Training & Certification Program in Electric Power Distribution System Engineering

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Training Course in Power System Modeling

Types of Load

18.84 kW2.34 %12.15987.12∠-31.78

ConstantImpedance

25.33 kW2.6 %12.19087.48∠-31.78

ConstantCurrent

28.72 kW2.9 %12.224 2 MVA, 0.85 pf lag

ConstantPower

PlossVRVS*Load

* Sending end voltage with a Receiving end voltage equal to 0.9*13.2 KV

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Competency Training & Certification Program in Electric Power Distribution System Engineering

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Training Course in Power System Modeling

Types of LoadDemandReA= (PA+ IReA Va + Z -1ReA Va

2 )

DemandImA=(QA+ IImA Va + Z -1ImA Va2 )

DemandReC= (Pc+ IReC Vc + Z -1ReC Vc2 )

DemandImC= (Qc+ IImC Vc + Z -1ImC Vc2)

DemandReB= (PB+ IReB Vb + Z -1ReB Vb

2 )

DemandImB = (QB+ IImB Vb + Z -1ImB Vb

2 )

Where:

P,Q are the constant Power components of the Demand

IRe,IIm are the constant Current components of the Demand

Z-1Re,Z-1

Im are the constant Impedance components of the Demand

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Training Course in Power System Modeling

Time Demand (A)1:00 17.762:00 16.683:00 17.524:00 17.405:00 21.006:00 29.887:00 29.648:00 32.289:00 25.92

10:00 21.7211:00 25.2012:00 22.08

Time Demand (A)13:00 20.8814:00 19.8015:00 19.0816:00 19.2017:00 23.0418:00 30.7219:00 38.0020:00 35.0021:00 34.0022:00 27.6023:00 24.8424:00 22.32

24-Hour Customer Load Profile

Customer Load Curve

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Time Demand (A) Per Unit1:00 17.76 0.4672:00 16.68 0.4393:00 17.52 0.4614:00 17.40 0.4585:00 21.00 0.5536:00 29.88 0.7867:00 29.64 0.7808:00 32.28 0.8499:00 25.92 0.682

10:00 21.72 0.57211:00 25.20 0.66312:00 22.08 0.581

Time Demand (A) Per Unit13:00 20.88 0.54914:00 19.80 0.52115:00 19.08 0.50216:00 19.20 0.50517:00 23.04 0.60618:00 30.72 0.80819:00 38.00 1.00020:00 35.00 0.92121:00 34.00 0.89522:00 27.60 0.72623:00 24.84 0.65424:00 22.32 0.587

ΣPU = 15.567

Customer Load Curve• Establishing Normalized Hourly Demand

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Competency Training & Certification Program in Electric Power Distribution System Engineering

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Training Course in Power System Modeling

0.0

0.2

0.4

0.6

0.8

1.0

1.2

Time

Dem

and

(Per

Un

it)

0 2 4 6 8 10 12 14 16 18 20 22 24

Customer Load Curve

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Competency Training & Certification Program in Electric Power Distribution System Engineering

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Training Course in Power System Modeling

0

50

100

150

200

250

300

350

Dem

and (W

)

Customer Energy Bill

Customer Energy Bill Converted to Hourly Power

Demand

Area under the curve = Customer Energy

Bill

0

0.2

0.4

0.6

0.8

1

1.2

Time (24 hours)

Norm

alized

Dem

and (per unit)

Normalized Customer Load Curve

Calculating Hourly Demand

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Competency Training & Certification Program in Electric Power Distribution System Engineering

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Training Course in Power System Modeling

Calculating Hourly Demand

Total Monthly Energy

Total Monthly Energy

Daily EnergyDaily Energy

Hourly Demand

Customer Load Curve

Customer Load Curve ⎟

⎟⎟⎟

⎜⎜⎜⎜

=

∑24

1t

tdailyt

p

pEnergyP

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Competency Training & Certification Program in Electric Power Distribution System Engineering

U. P. National Engineering CenterNational Electrification Administration

Training Course in Power System Modeling

Calculating Hourly Demand

� Example:kWHr Reading (Monthly Bill) = 150 kWHr

Billing Days = 30 days

Daily Energy = 150 / 30 = 5 kWh [24 hours]

Hourly Demand1 = Daily Energy x [P.U.1 / ΣΣΣΣP.U]

= 5 kWh x 0.467 / 15.567

= 0.15011 kW

= 150.11 W

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Competency Training & Certification Program in Electric Power Distribution System Engineering

U. P. National Engineering CenterNational Electrification Administration

Training Course in Power System Modeling

0

50

100

150

200

250

300

350

1:00

3:00

5:00

7:00

9:00

11:0

0

13:0

0

15:0

0

17:0

0

19:0

0

21:0

0

23:0

0

Dem

and

(W)

Calculating Hourly Demand

Hourly Real Demand

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Competency Training & Certification Program in Electric Power Distribution System Engineering

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Training Course in Power System Modeling

( )ttt pfPQ 1costan −=Qt = hourly Reactive Demand (VAR)

Pt = hourly Real Demand (W)

Pft = hourly power factor

� Example:Real Demand (W) = 150.11 W, PF = 0.96 lag

Reactive Demand = P tan (cos-1 pf)

= 150.11 tan (cos-1 0.96)

= 43.78 VAR

Calculating Hourly Demand

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Training Course in Power System Modeling

0

50

100

150

200

250

300

350

1:00

3:00

5:00

7:00

9:00

11:0

0

13:0

0

15:0

0

17:0

0

19:0

0

21:0

0

23:0

0

Dem

and

(W a

nd

VA

R)

Calculating Hourly Demand

Hourly Real & Reactive Demand

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Competency Training & Certification Program in Electric Power Distribution System Engineering

U. P. National Engineering CenterNational Electrification Administration

Training Course in Power System Modeling

Developing Load Models

� Load Curves for each Customer Type� Residential load curves� Commercial load curves� Industrial load curves� Public building load curves� Street Lighting load curves� Administrative load curves (metered)� Other Load Curves (i.e., other types of customers)

� Variations in Load Curves� Customer types and sub-types� Weekday-Weekend/Holiday variations� Seasonal variations

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Competency Training & Certification Program in Electric Power Distribution System Engineering

U. P. National Engineering CenterNational Electrification Administration

Training Course in Power System Modeling

Developing Load Models

� Data Requirements� Customer Data;

� Billing Cycle Data;

� Customer Energy Consumption Data; and

� Load Curve Data.

Distribution Utility Data Tables and Instructions

Converting Energy Bill to Power Demand

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� Generalized Machine Model

� Steady-State Equations

� Generator Sequence Impedances

� Generator Sequence Networks

Generator Models

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Stator:

distributed three-phase winding (a, b, c)

Rotor:

DC field winding (F) and short-circuited damper windings (D, Q)

Axis of a

Axis of c

Axis of b

d-axisq-axis

Phase b winding

Damper winding DDamper

winding Q

Phase a winding

Field winding F

Phase c winding

Generalized Machine ModelConstructional Details of Synchronous Machine

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Primitive Coil Representation

q-axis

phase b

QiQ

vQ

-

+F

D

+v F

-iF

+v D -iD

phase a

phase c

ωωωωm+ Va -

ia

a

c

+Vc

- ic

ibb

+Vb

-

d-axis

θθθθe

dt

dRiv

λ+=

Generalized Machine Model

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Voltage Equations for the Primitive CoilsFor the stator windings For the rotor windings

dt

diRv a

aaa

λ+=

dt

diRv b

bbb

λ+=

dt

diRv c

ccc

λ+=

dt

diRv F

FFF

λ+=

dt

diRv D

DDD

λ+=

dt

diRv Q

QQQ

λ+=

Note: The D and Q windings are shorted (i.e. ).0== QD vv

⎥⎦

⎤⎢⎣

⎡+⎥

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡=⎥

⎤⎢⎣

FDQ

abc

FDQ

abc

FDQ

abc

FDQ

abcp

i

i

R

R

v

v

λλ

Li=λ

Generalized Machine Model

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The flux linkage equations are:

⎥⎥⎥⎥⎥⎥⎥⎥

⎢⎢⎢⎢⎢⎢⎢⎢

⎥⎥⎥⎥⎥⎥⎥⎥

⎢⎢⎢⎢⎢⎢⎢⎢

=

⎥⎥⎥⎥⎥⎥⎥⎥

⎢⎢⎢⎢⎢⎢⎢⎢

Q

D

F

c

b

a

QQQDQFQcQbQa

DQDDDFDcDbDa

FQFDFFFcFbFa

cQcDcFcccbca

bQbDbFbcbbba

aQaDaFacabaa

Q

D

F

c

b

a

i

i

i

i

i

i

LLLLLL

LLLLLL

LLLLLL

LLLLLL

LLLLLL

LLLLLL

λ

λλλλ

λ

or

[ ] [ ][ ] [ ] ⎥

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡=⎥

⎤⎢⎣

⎡λ

λ

FDQ

abc

RRRS

SRSS

FDQ

abc

i

i

LL

LL

Generalized Machine Model

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COIL INDUCTANCES

Stator Self Inductances

emsaa 2cosLLL θ+=

)1202cos(LLL emsbbo++= θ

)1202cos(LLL emscco−+= θ

Stator-to-Stator Mutual Inductances

)1202cos(LMLL emsbaabo−+−== θ

emscbbc 2cosLMLL θ+−==

)1202cos(LMLL emsaccao++−== θ

Generalized Machine Model

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Rotor Self Inductances

Rotor-to-Rotor Mutual Inductances

00

======

QDDQ

QFFQ

FDDFFD L

LLLLLL

COIL INDUCTANCES

QQQQ

DDDD

FFFF

LL

LL

LL

=

=

=

Generalized Machine Model

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)120cos(LLL

)120cos(LLL

cosLLL

eaDDccD

eaDDbbD

eaDDaaD

o

o

+==

−==

==

θ

θ

θ

)120sin(LLL

)120sin(LLL

sinLLL

eaQQccQ

eaQQbbQ

eaQQaaQ

o

o

+−==

−−==

−==

θ

θ

θ

Stator-to-Rotor Mutual Inductances

)120cos(LLL

)120cos(LLL

cosLLL

eaFFccF

eaFFbbF

eaFFaaF

o

o

+==

−==

==

θ

θ

θ

COIL INDUCTANCES

Generalized Machine Model

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q-axis

b-axis

Q iQ vQ

-

+

F D

+ vF -iF

+ vD -iD

d-axis

a-axis

c-axis

ωm

+ Va -ia

a

c+ Vc

-

ic

ibb +V

b-

Equivalent Coil Representation

Rotor coils FDQstationary

Stator coils abcrotating

Generalized Machine Model

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Equivalent Generalized Machine

q-axis

Q vQi

Q

+-

q i

q+-vq

+ vd -

di

m

+ vF -

Fi

F+ vD -

Di

D

d-axis

-

-

Replace the abc coils with equivalent commutated d and q coils which are connected to fixed brushes.

QQQqQqQ

DDDFDFdDdD

DFDFFFdFdF

iLiL

iLiLiL

iLiLiL

+=

++=

++=

λ

λ

λ

QqQqqqq

DdDFdFdddd

iLiL

iLiLiL

+=

++=

λ

λ

Generalized Machine Model

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Transformation from abc to Odqq-axis

b-axis

d-axis

a-axis

c-axis

θe

ib

ia

icω

m

Note: The d and q windings are pseudo-stationary. The O axis is perpendicular to the d and q axes.

q-axis

d-axis

iq

id

qd

Generalized Machine Model

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Equivalence:

1. The resultant mmf of coils a, b and c along the d-axis must equal the mmf of coil d for any value of angle θe.

2. The resultant mmf of coils a, b and c along the q-axis must equal the mmf of coil q for any value of angle θe.

We get Ndid = Kd [Naia cos θθθθe + Nbib cos (θθθθe - 120o)

+ Ncic cos (θθθθe + 120o)]

Nqiq = Kq [-Naia sin θθθθe - Nbib sin (θθθθe - 120o)

-Ncic sin (θθθθe + 120o)]where Kd and Kq are constants to be determined.

Generalized Machine Model

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Assume equal number of turns.

Na = Nb = Nc = Nd = Nq

Substitution gives

id = Kd [ia cos θθθθe + ib cos (θθθθe - 120o) + ic cos (θθθθe + 120o)]

iq = Kq [-ia sin θθθθe - ib sin (θθθθe - 120o) -ic sin (θθθθe + 120o)]

The O-coil contributes no flux along the d or q axis. Let its current io be defined as

io = Ko ( ia + ib + ic )

Generalized Machine Model

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Combining, we get

( ) ( )( ) ( ) ⎥

⎥⎥

⎢⎢⎢

⎥⎥⎥

⎢⎢⎢

+−−−−

+−=⎥⎥⎥

⎢⎢⎢

c

b

a

eqeqeq

ededed

ooo

q

d

o

i

i

i

KKK

KKK

KKK

i

i

i

120sin120sinsin

120cos120coscos

θθθ

θθθ

The constants Ko, Kd and Kq are chosen so that the transformation matrix is orthogonal; that is

[ ] [ ]TPP =− 1

Assuming Kd = Kq, one possible solution is

3

1=oK

3

2== qd KK

Generalized Machine Model

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Park’s Transformation Matrix

[ ] ( ) ( )

( ) ( )

[ ] ( ) ( )

( ) ( )⎥⎥⎥⎥⎥⎥

⎢⎢⎢⎢⎢⎢

+−+

−−−

=

⎥⎥⎥⎥⎥⎥

⎢⎢⎢⎢⎢⎢

+−−−−

+−=

120sin120cos2

1

120sin120cos2

1

sincos2

1

3

2

120sin120sinsin

120cos120coscos

2

1

2

1

2

1

3

2

1

ee

ee

e

eee

eee

P

P

θθ

θθ

θθ

θθθ

θθθ

Generalized Machine Model

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Voltage Transformation

The relationship between the currents is

[ ] abcodq iPi =or

[ ] odqabc iPi 1−=

Assume a power-invariant transformation; that is

qqddooccbbaa iviviviviviv ++=++or

odqTodqabc

Tabc iviv =

Generalized Machine Model

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[ ] odqTodqodq

Tabc iviPv =−1

Substitution gives

[ ]TTabc

Todq Pvv =

[ ] abcodq vPv =Transpose both sides to get

[ ] odqabc vPv 1−=

Note: Since voltage is the derivative of flux linkage, then the relationship between the flux linkages must be the same as that of the voltages. That is,

[ ] abcodq P λλ =

Generalized Machine Model

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Generalized Machine ModelIn summary, using Park’s Transformation matrix,

[ ] abcodq iPi = [ ] odqabc iPi 1−=

[ ] abcodq vPv = [ ] odqabc vPv 1−=

[ ] abcodq P λλ = [ ] odq1

abc P λ=λ −

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Recall the flux linkage equation

⎥⎥⎥⎥⎥⎥⎥⎥

⎢⎢⎢⎢⎢⎢⎢⎢

⎥⎥⎥⎥⎥⎥⎥⎥

⎢⎢⎢⎢⎢⎢⎢⎢

=

⎥⎥⎥⎥⎥⎥⎥⎥

⎢⎢⎢⎢⎢⎢⎢⎢

Q

D

F

c

b

a

QQQDQFQcQbQa

DQDDDFDcDbDa

FQFDFFFcFbFa

cQcDcFcccbca

bQbDbFbcbbba

aQaDaFacabaa

Q

D

F

c

b

a

i

i

i

i

i

i

LLLLLL

LLLLLL

LLLLLL

LLLLLL

LLLLLL

LLLLLL

λ

λλλλ

λ

or

[ ] [ ][ ] [ ] ⎥

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡=⎥

⎤⎢⎣

⎡λ

λ

FDQ

abc

RRRS

SRSS

FDQ

abc

i

i

LL

LL

Generalized Machine Model

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[ ]⎥⎥⎥

⎢⎢⎢

⎡=

cccbca

bcbbba

acabaa

SS

LLL

LLL

LLL

L

( )( )o

o

1202cosLLL

1202cosLLL

2cosLLL

emScc

emSbb

emSaa

−θ+=

+θ+=

θ+=

( )

( )o

o

1202cosLMLL

2cosLMLL

1202cosLMLL

emSacca

emScbbc

emSbaab

+θ+−==

θ+−==

−θ+−==

Recall

where

Generalized Machine Model

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[ ]

( ) ( )( ) ( )( ) ( )⎥

⎥⎥

⎢⎢⎢

−+

+−

+−

+

⎥⎥⎥

⎢⎢⎢

−−

−−

−−

=

1202cos2cos1202cos

2cos1202cos1202cos

1202cos1202cos2cos

eee

eee

eee

m

SSS

SSS

SSS

SS

L

LMM

MLM

MML

L

θθθθθθ

θθθ

Substitution gives

Generalized Machine Model

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Similarly,

Apply Park's transformation to Flux Linkage equation

[ ] [ ][ ] [ ][ ] FDQSRabcSSabc iLPiLPP +=λ

or

[ ][ ][ ] [ ][ ] FDQSRodqSSodq iLPiPLP += −1λ

Generalized Machine Model

[ ] ( ) ( ) ( )( ) ( ) ( )⎥

⎥⎥

⎢⎢⎢

+−++

−−−−

=

120sin120cos120cos

120sin120cos120cos

sincoscos

eaQeaDeaF

eaQeaDeaF

eaQeaDeaF

SR

LLL

LLL

LLL

L

θθθ

θθθθθθ

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The term [ ][ ][ ] 1−PLP SScan be shown

Generalized Machine Model

⎥⎥⎥⎥⎥⎥

⎢⎢⎢⎢⎢⎢

++

++

=

mss

mss

ss

L2

3ML

L2

3ML

M2L

[ ][ ][ ]⎥⎥⎥

⎢⎢⎢

=−

qq

dd

oo

ss

L

L

L

PLP 1

mSSqq

mSSdd

SSoo

LMLL

LMLL

MLL

2

32

3

2

−+=

++=

−=Let

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[ ][ ]⎥⎥⎥

⎢⎢⎢

=

⎥⎥⎥⎥⎥⎥⎥

⎢⎢⎢⎢⎢⎢⎢

=

qQ

dDdF

aQ

aDaFSR

L

LL

L

LLLP

2

300

02

3

2

3

000

Similarly, it can be shown that

Generalized Machine Model

aQqQ L2

3L =aFdF L

2

3L = aDdD L

2

3L =

where

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Finally, we get

QqQqqqq

DdDFdFdddd

oooo

iLiL

iLiLiL

iL

+=λ

++=λ

Generalized Machine Model

[ ][ ][ ] [ ][ ] FDQSRodqSSodq iLPiPLP += −1λ

Substituting, [ ][ ][ ] 1−PLP SS and [ ][ ]SRLP

Note: All inductances are constant.

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The Flux Linkage Equations for the FDQ coils in matrix form is

[ ] [ ] FDQRRabcRSFDQ iLiL +=λ

[ ] [ ]TSRRS LL =Since

we get

[ ] [ ] [ ] FDQRRodqT

SRFDQ iLiPL +=λ −1

Generalized Machine Model

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It can be shown that

[ ] [ ]

⎥⎥⎥⎥⎥⎥⎥

⎢⎢⎢⎢⎢⎢⎢

=−

aQ

aD

aF

TSR

L

L

L

PL

2

300

02

30

02

30

1

⎥⎥⎥

⎢⎢⎢

=

Qq

Dd

Fd

L

L

L

Generalized Machine Model

aFFd L2

3L = aDDd L

2

3L = aQQq L

2

3L =

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Recall that the rotor self- and mutual inductances are constant

[ ]⎥⎥⎥

⎢⎢⎢

=

QQ

DDDF

FDFF

RR

L

LL

LL

L

00

0

0

Upon substitution, we get

QQQqQqQ

DDDFDFdDdD

DFDFFFdFdF

iLiL

iLiLiL

iLiLiL

+=λ

++=λ

++=λ

Note: All inductances are also constant.

Generalized Machine Model

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The Flux Linkage Equation

⎥⎥⎥⎥⎥⎥⎥⎥

⎢⎢⎢⎢⎢⎢⎢⎢

⎥⎥⎥⎥⎥⎥⎥⎥

⎢⎢⎢⎢⎢⎢⎢⎢

=

⎥⎥⎥⎥⎥⎥⎥⎥

⎢⎢⎢⎢⎢⎢⎢⎢

λ

λ

λ

λ

λ

λ

Q

D

F

q

d

o

QQQq

DDDFDd

FDFFFd

qQqq

dDdFdd

oo

Q

D

F

q

d

o

i

i

i

i

i

i

LL

LLL

LLL

LL

LLL

Lo d q F D Q

odq

FD

Q

q-axis

Q vQi

Q+

-

q iq +

-vq

+ vd -

d

idωm

+ vF -

F

iF + vD -

D

i

D

d-axis

Generalized Machine Model

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Transformation of Stator Voltages

Assume Ra = Rb = Rc in the stator. Then,

[ ] abcabcaabc dt

diuRv λ+= 3

Recall the transformation equations

[ ][ ][ ] abcodq

abcodq

abcodq

P

vPv

iPi

λλ =

=

=

Generalized Machine Model

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[ ] [ ] [ ][ ] [ ] [ ]{ }odqodqaabc Pdt

dPiPuRPvP λ+= −− 11

3

Simplify to get

[ ] [ ][ ] [ ] [ ] odqodqodqaodq Pdt

dP

dt

dPPiuRv λ

⎭⎬⎫

⎩⎨⎧+λ+= −− 11

3

Apply Park’s transformation

Generalized Machine Model

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It can be shown that

[ ] ( ) ( )( ) ( ) dt

d

cossin

cossin

cossin

Pdt

d e

ee

ee

eeθ

⎥⎥⎥

⎢⎢⎢

+θ−+θ−

−θ−−θ−

θ−θ−

=−

1201200

1201200

0

3

21

where

mee

dt

dω=ω=

θfor a two–pole machine

m

Pω=

2for a P–pole machine

Generalized Machine Model

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It can also be shown that

[ ] [ ]⎥⎥⎥

⎢⎢⎢

ω

ω−=−

00

00

0001

m

mPdt

dP for a two-pole

machine

Finally, we getooao

dt

diRv λ+=

qmddaddt

diRv λωλ −+=

dmqqaqdt

diRv λωλ ++=

Generalized Machine Model

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Voltage Equation for the Rotor

0

0

=λ+=

=λ+=

λ+=

QQQQ

DDDD

FFFF

dt

diRv

dt

diRv

dt

diRv

Note: No transformation is required.

Generalized Machine Model

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Matrix Form of Voltage Equations

⎥⎥⎥⎥⎥⎥⎥⎥

⎢⎢⎢⎢⎢⎢⎢⎢

⎥⎥⎥⎥⎥⎥⎥⎥

⎢⎢⎢⎢⎢⎢⎢⎢

+

⎥⎥⎥⎥⎥⎥⎥⎥

⎢⎢⎢⎢⎢⎢⎢⎢

+

⎥⎥⎥⎥⎥⎥⎥⎥

⎢⎢⎢⎢⎢⎢⎢⎢

⎥⎥⎥⎥⎥⎥⎥⎥

⎢⎢⎢⎢⎢⎢⎢⎢

=

⎥⎥⎥⎥⎥⎥⎥⎥

⎢⎢⎢⎢⎢⎢⎢⎢

Q

D

F

q

d

o

m

Q

D

F

q

d

o

1

1-

λ

λ

λ

λ

λ

λ

ω

λ

λ

λ

λ

λ

λ

dt

d

i

i

i

i

i

i

R

R

R

R

R

R

v

v

v

v

v

v

Q

D

F

q

d

o

Q

D

F

a

a

a

Q

D

F

q

d

o

The equation is now in the form

[ ] [ ][ ] [ ] [ ] [ ][ ]iGipLiRv mω++=

Generalized Machine Model

Resistance Voltage Drop

Transformer Voltage

Speed Voltage

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[ ]

⎥⎥⎥⎥⎥⎥

⎢⎢⎢⎢⎢⎢

=

QQQq

DDDFDd

FDFFFd

qQqq

dDdFdd

LL

LLL

LLL

LL

LLL

L

dq

F

D

Q

D QFqd

[ ]

⎥⎥⎥⎥⎥⎥

⎢⎢⎢⎢⎢⎢

⎡ −−

=dDdFdd

qQqq

LLL

LL

G

dq

F

D

Q

D QFqd

Generalized Machine Model

Note: All entries of [L] and [G] are constant.

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Summary of Equations

Voltage Equations

( )( )( )( )( )( ) 06

05

4

3

2

1

=λ+=

=λ+=

λ+=

λω+λ+=

λω−λ+=

λ+=

QQQQ

DDdD

FFFF

dmqqaq

qmddad

ooao

piRv

piRv

piRv

piRv

piRv

piRv

Generalized Machine Model

Flux Linkages

( )( )( )( )( )( ) QQQqQqQ

DDDFDFdDdD

DFDFFFdFdF

QqQqqqq

DdDFdFdddd

oooo

iLiL

iLiLiL

iLiLiL

iLiL

iLiLiL

iL

+=λ

++=λ

++=λ

+=λ

++=λ

6

5

4

3

2

1

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Electromagnetic Torque Equation

[ ]

⎥⎥⎥⎥⎥⎥⎥⎥

⎢⎢⎢⎢⎢⎢⎢⎢

⎡−

−=

0

0

0

0

d

q

QDFqdo iiiiiiλλ

[ ] [ ][ ]iGiT Te −=

( )( )[ ]dQqQqDdDqFdFqdqqdd

qddqe

iiLiiLiiLii LL

iiT

−++−−=

+−−= λλ

We get

for a 2for a 2--pole machinepole machine

Generalized Machine Model

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Steady–State Equations

At steady–state condition,1. All transformer voltages are zero.2. No voltages are induced in the damper windings.

Thus, iD = iQ = 0

Voltage Equations

( )

FFF

FdFdddmqaq

qqqmdad

oao

iRv

iLiLiRv

iLiRv

iRv

=

+ω+=

ω−=

=

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Cylindrical-Rotor MachineIf the rotor is cylindrical, then the air gap is uniform, and Ldd = Lqq.

Define synchronous inductance Ls

LS = Ldd = Lqq when the rotor is cylindrical

Voltage and Electromagnetic Torque Equations at Steady-state

qFdFe

FdFmdsmqaq

qsmdad

iiLT

iLiLiRv

iLiRv

=

ω+ω−=

ω−=

Steady–State Equations

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For Balanced Three-Phase Operation

( )( )( )o

c

ob

a

tcosIi

tcosIi

tcosIi

1202

1202

2

+α+ω=

−α+ω=

α+ω=

Apply Park’s transformation [ ] abcodq iPi = , We get

α=

α=

=

sini

cosIi

i

q

d

o

I3

3

0Note:

1. ia, ib and ic are balanced three-phase currents.

2. id and iq are DC currents.

Steady–State Equations

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A similar transformation applies to balance three-phase voltages.

( )( )( )o

c

ob

a

tcosVv

tcosVv

tcosVv

1202

1202

2

+δ+ω=

−δ+ω=

δ+ω=

We get

δ=

δ=

=

sinVv

cosVv

v

q

d

o

3

3

0

Given

Steady–State Equations

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Inverse Transformation

[ ] odqabc iPi 1−=

We get

[ ]tsinitcosii qda ω−ω=3

2

( )[ ]o903

2+ω+ω= tcositcosi qd

Given id and iq, and assuming io = 0,

Steady–State Equations

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Recall the phasor transformation

( ) θ∠↔θ+ω ItcosI2Using the transform, we get

[ ]oo 9003

1∠+∠= qda iiI

assuming the d and q axes as reference. Simplify

qda

qda

jIII

ij

iI

+=

+=33

Steady–State Equations

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Similarly, given vd and vq with vo = 0

[ ]tsinvtcosvv qda ω−ω=3

2

In phasor form,

33qd

a

vj

vV +=

qd jVV +=

( )[ ]o90coscos3

2++= tvtv qd ωω

Steady–State Equations

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Steady-State Operation-Cylindrical

Recall at steady-state

Divide by 3

FdFmdsmqaq

qsmdad

iLiLiRv

iLiRv

ωω

ω

++=

−=

FdFmdsmqaq

qsmdad

iLILIRV

ILIRV

ωω

ω

3

1++=

−=

Steady–State Equations

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Define =ω= sms LX synchronous reactance

=ω= FdFmf iLE3

1Excitation voltage

Phasor Voltage aV

qda jVVV +=

( )( ) ( ) fqdsqda

fdsqaqsda

jEjIIjXjIIR

EIXIRjIXIR

++++=

+++−=

(motor equation)masaaa EIjXIRV ++=

Steady–State Equations

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For a generator, current flows out of the machine

( ) ( )aasaag

gasaaa

VIjXIRE

EIjXIRV

++=

+−+−=

AC

sa jXR +

aI+

-

+

-

aVgE

Equivalent Circuit of Cylindrical Rotor Synchronous Generator

Steady–State Equations

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Salient-Pole MachineIf the rotor is not cylindrical, no equivalent circuit can bedrawn. The analysis is based solely on the phasor diagram describing the machine. Recall the steady-state equations

Divide through by 3FdFmdddmqaq

qqqmdad

iLiLiRv

iLiRv

ωω

ω

++=

−=

3FdFm

ddqaq

qqdad

iLIXIRV

IXIRV

ω++=

−=

Steady–State Equations

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where

== ddmd LX ω direct axis synchronous reactance

== qqmq LX ω quadrature axis synchronous reactance

Define:

FdFm

f iL

E3

ω= = excitation voltage

Steady–State Equations

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We get

From , we getqda jVVV +=

or

fddqqaaa jEIjXIXIRV ++−=

fddqaq

qqdad

EIXIRV

IXIRV

++=

−=

( ) fddqqqda jEIjXIXjIIR ++−+=

( )fddqaqqdaa EIXIRjIXIRV +++−=

Steady–State Equations

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Steady-State Electromagnetic Torque

At steady-state

The dominant torque is the cylindrical torque which determines the mode of operation.

For a motor, Te is assumed to be negative. For a generator, Te is assumed to be positive.

saliency torque

cylindrical torque

( )[ ]qFdFqdqqdde iiLii LLT +−−=

Steady–State Equations

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Since the field current iF is always positive,

)(generator 0 i when 0 (motor) 0 i when 0

q

q

<>><− qFdF iiL

Recall that qda jIII +=

Note: The imaginary component of Ia determinesWhether the machine is operating as a motor or aGenerator.

Steady–State Equations

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What about Id?

fddq

dad

EIXVIRV

+==we get

In general, Ra << Xd. We get

qda jVVV +=

( )( )fdd

fddda

EIXjEIXjIR

+≈++=

Assume . From

0=qI

fddqaq

qqdad

EIXIRV

IXIRV

++=

−=

Steady–State Equations

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constant=+= fddq EIXV

Recall that

FdFm

f iL

E3

ω=

Thus, the excitation voltage depends only on the field current since ωωωωm is constant.

For some value of field current iFo, Ef = Va

and Id = 0.

If the magnitude of Va is constant,

Steady–State Equations

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Operating Modesq-axis

d-axis

Over-excited Motor

Under-excited Motor

Over-excited Generator

Under-excited Generator

Id < 0, Iq > 0 Id > 0, Iq > 0

Id < 0, Iq < 0 Id > 0, Iq < 0

Steady–State Equations

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Over-excitation and Under-excitation

1. If the field current is increased above iFo, thenEf > Va

and the machine is over-excited. Under this condition, Id < 0 (demagnetizing).

2. If the field current is decreased below iFo, then Ef < Va

and the machine is under-excited. Under this condition, Id > 0 (magnetizing).

Steady–State Equations

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Drawing Phasor Diagrams

aaddqqfa

fddqqaaa

qda

qda

IRIjXIXjEV

jEIjXIXIRV

jVVV

jIII

++−=

++−=

+=

+=

A phasor diagram showing Va and Ia can be drawn if the currents Id and Iq are known. Recall

Steady–State Equations

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Over-excited Motor

d-axis

q-axis

fjEqq IX−

aa IR

aVddIjX

δaI

φqjI

dI

Id < 0Iq > 0

Leading Power Factor

Steady–State Equations

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Lagging Power Factor

q-axis

d-axis

aVaa IR

dd IjX

qqIX−

qjI

fjE

φ

δaI

dI

Id > 0Iq > 0

Under-excited Motor

Steady–State Equations

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q-axis

d-axis

aV

aaIR

ddIjXqqIX−

qjI

fjE

φ

δ

aI

dICurrentActual

Id < 0Iq < 0

Over-excited Generator

Lagging Power Factor

Steady–State Equations

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d-axis

Id > 0Iq < 0 aV

aa IRddIjX

qqIX−

qjI

fjE

φ δ

aI

dIActualCurrent

Under-excited Generator

Leading Power Factor

Steady–State Equations

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1. The excitation voltage jEf lies along the quadrature axis.

2. leads jEf for a motorlags jEf for a generator

The angle between the terminal voltage and jEf is called the power angle or torque angle δ.

3. The equation

applies specifically for a motor.

Observations

aVaV

aV

fqqddaaa jEIXIjXIRV +−+=

Steady–State Equations

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4. For a generator, the actual current flows out ofthe machine. Thus Id, Iq and are negative.aI

qqddaaaf

fqqddaaa

IXIjXIRVjE

jEIXIjXIRV

−++=

++−−=or

generator afor

motor afor

gf

mf

EjE

EjE

=

=5. Let

Steady–State Equations

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The generator equation becomes

For a motor, the equation is

6. No equivalent circuit can be drawn for a salient-pole motor or generator.

qqddaaag IXIjXIRVE −++=

qqddaama IXIjXIREV −++=

Steady–State Equations

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Example 1: A 25 MVA, 13.8 kV, 3600 RPM, Y-connected cylindrical-rotor synchronous generator has a synchronous reactance of 4.5 ohms per phase. The armature resistance is negligible. Find the excitation voltage Eg when the machine is supplying rated MVA at rated voltage and 0.8 power factor.

Single-phase equivalent circuit

8.13=aV kV line-to-line97.7= kV line-to-neutral

AC

sjX

aIgE

+

-

+

-

aV

Steady–State Equations

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( ) 208.025 ==aP MW, three-phase67.6= MW/phase

15tan == θaa PQ MVar, three-phase5= MVar/phase

Let o097.7 ∠=aV kV, the reference.

Using the complex power formula*aaaa IVjQP =+

o097.7000,5667,6

* ∠−

=−

=j

V

jQPI

a

aaa

Steady–State Equations

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Apply KVL,

aaSg VIjXE +=( )

V

j

j

o

oo

24.19429,11

766,3791,10

0970,787.36046,15.4

∠=

+=

∠+−∠=

Eg = 11,429 volts, line-to-neutral= 19,732 volts, line-to-line

A

Ajo87.36046,1

628837

−∠=

−=IaWe get

Steady–State Equations

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Example 2: A 100 MVA, 20 kV, 3-phase synchronous generator has a synchronous reactance of 2.4 ohms. The armature resistance is negligible. The machine supplies power to a wye-connected resistive load, 4Ω per phase, at a terminal voltage of 20 kV line-to-line.

(a) Find the excitation voltage

AC

Ω= 4.2SX

aIgE+

-

+

-aV Ω= 4R

Steady–State Equations

Va(L-L) = 20,000 voltsVa(L-N) = 11,547 volts

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Let o0547,11 ∠=aV V, the reference

AmpsR

VI

aa

o0887,24547,11

∠===

aaSg VIjXE +=( )

neutraltolineV

j

j

−−∠=

+=

+=

o96.30466,13

928,6547,11

547,11887,24.2

Applying KVL,

( ) VEg 324,23466,133 ==linetolinekV −−= ,32.23

Steady–State Equations

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(b) Assume that the field current is held constant. A second identical resistive load is connected across the machine terminal. Find the terminal voltage, Va.

Since iF is constant, Eg is unchanged. Thus, Eg = 13,466 V,line-to-neutral.

Ω=ΩΩ= 24//4eqR

referencetheVVLet aa ,0o∠=

o02

1∠== a

eq

aa V

R

VI

Steady–State Equations

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aa

aa

VjV

VVj

2.1

2

14.2

+=

+⎟⎠⎞

⎜⎝⎛=

( )222 2.1 aag VVEgetWe +=22 44.2466,13 aV=

neutraltolineVVa −−= ,621,8

linetolineVVa −−= ,932,14

aasg VIjXE +=Apply KVL,

Steady–State Equations

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neutraltolineVV a −−∠= ,0547,11 o

AmpsR

VI

eq

aa

o0774,52

547,11∠===

( ) 547,1157744.2 += jE g

neutraltolineV

j

−−∠=

+=o19.50037,18

856,13547,11

( ) VE g 241,31037,183 ==linetolinekV −−= 24.31

(c) Assume that the field current iF is increased so that the terminal voltage remains at 20 kV line-to-line after the addition of the new resistive load. Find Eg.

Steady–State Equations

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The equivalent Circuit of Generator for Balanced Three-Phase System Analysis

Ec

Eb

Ea

Ic

Ib

Ia

b

a

c

sa jXR +

aI +

-

aV

+

-gE

Za

Zb Zc

Three-Phase Equivalent Single-Phase Equivalent

Generator Sequence Impedances

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Sequence Impedance of Power System Components

Positive Sequence Negative Sequence Zero Sequence

+

-

Z1

Ia1

E+

Va1

+

-

Ia2

Va2Z2

+

-

Ia0

Va0Z0

2a2a2 ZI - V =1a1a1 ZI– E V = oaoao ZI - V =

From Symmetrical Components, the Sequence Networks for Unbalanced Three-Phase Analysis

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Generator Sequence Impedances

Positive-Sequence Impedance:

Xd”=Direct-Axis Subtransient Reactance

for a salient-pole machine

Xd’=Direct-Axis Transient ReactanceXd=Direct-Axis Synchronous Reactance

Negative-Sequence Impedance:

)"X"X(X qd21

2 +=

for a cylindrical-rotor machine"XX d2 =

Zero-Sequence Impedance:

"X6.0X"X15.0 d0d ≤≤

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Positive Sequence ImpedanceThe AC RMS component of the current following a three-phase short circuit at no-load condition with constant exciter voltage and neglecting the armature resistance is given by

⎟⎟⎠

⎞⎜⎜⎝

⎛ −⎟⎟⎠

⎞⎜⎜⎝

⎛−+=

'

texp

X

E

'X

E

X

E)t(I

ddsdds τ

⎟⎟⎠

⎞⎜⎜⎝

⎛ −⎟⎟⎠

⎞⎜⎜⎝

⎛−+

"

texp

'X

E

"X

E

ddd τ

where E = AC RMS voltage before the short circuit.

Generator Sequence Impedances

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The AC RMS component of the short-circuit current is composed of a constant term and two decaying exponential terms where the third term decays very much faster than the second term.

If the first term is subtracted and the remainder is plotted on a semi-logarithmic paper versus time, the curve would appear as a straight line after the rapidly decaying term decreases to zero.

The rapidly decaying portion of the curve is thesubtransient portion, while the straight line is thetransient portion.

Generator Sequence Impedances

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IEEE Std 115-1995: Determination of the Xd’ and Xd” (Method 1)The direct-axis transient reactance is determined from the current waves of a three-phase short circuit suddenly applied to the machine operating open-circuited at rated speed. For each test run, oscillograms should be taken showing the short circuit current in each phase.The direct-axis transient reactance is equal to the ratio of the open-circuit voltage to the value of the armature current obtained by the extrapolation of the envelope of the AC component of the armature current wave, neglecting the rapid variation during the first few cycles.

Generator Sequence Impedances

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The direct-axis subtransient reactance is deter-mined from the same three-phase suddenly applied short circuit. For each phase, the values of the difference between the ordinates of Curve B and the transient component (Line C) are plotted as Curve A to give the subtransient component of the short-circuit current.

The sum of the initial subtransient component, the initial transient component and the sustained component for each phase gives the corresponding value of I”.

Generator Sequence Impedances

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0.4

0.60.81.0

1.52.0

34568

101214

0 10 20 30 40 50 60Time in half-cycles

Curr

ent

in p

has

e 1 (

per

unit)

+

+

++++

+

++ +++ ++ +++++ + ++ + + + + + + + +

Curve B

Line C

+

+

+

+

+

+

+

++Curve A

Line A

Generator Sequence Impedances

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Phase 1 Phase 2 Phase 3 Ave

(1) Initial voltage 1.0

(2) Steady-state Current 1.4 1.4 1.4

(3) Initial Transient Current 8.3 9.1 8.6

(4) I’ = (2)+(3) 9.7 10.5 10.0 10.07

(5) Xd’ = (1)÷(4) 0.0993

(6) Init. Subtransient Current 3.8 5.6 4.4

(7) I” = (4)+(6) 13.5 16.1 14.4 14.67

(8) Xd” = (1)÷(7) 0.0682

Example: Calculation of transient and subtransientreactances for a synchronous machine

Generator Sequence Impedances

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Negative Sequence Impedance

IEEE Std 115-1995: Determination of the negative-sequence reactance, X2 (Method 1)

The machine is operated at rated speed with its field winding short-circuited. Symmetrical sinusoidal three-phase currents of negative phase sequence are applied to the stator. Two or more tests should be made with current values above and below rated current, to permit interpolation.

The line-to-line voltages, line currents and electric power input are measured and expressed in per-unit.

Generator Sequence Impedances

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Let E = average of applied line-to-line voltages, p.u.I = average of line currents, p.u.P = three phase electric power input, p.u.

IE

Z2 = =Negative Sequence Impedance, p.u.

22 IP

R = =Negative Sequence Resistance, p.u.

22

222 RZX −=

=Negative Sequence Reactance, p.u.Note: The test produces abnormal heating in the rotor and should be concluded promptly.

Generator Sequence Impedances

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Zero Sequence ImpedanceIEEE Std 115-1995: Determination of the zero-sequence reactance, X0 (Method 1)

E A W

V

The machine is operated at rated speed with its field winding short-circuited. A single-phase voltage is applied between the line terminals and the neutral point.Measure the applied voltage, current and electric power.

Field

Generator Sequence Impedances

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Let E = applied voltage, in p.u. of base line-to-neutral voltage

I = test current, p.u.P = wattmeter reading, in p.u. single-phase

base volt-ampere

IE3

Z0 = =Zero Sequence Impedance, p.u.

2

00 EIP

1ZX ⎟⎠⎞

⎜⎝⎛

−=

=Zero Sequence Reactance, p.u.

Generator Sequence Impedances

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Average Machine Reactances

ReactanceTurbo

GeneratorsWater-Wheel Generators

Synchronous Motors

Xd 1.10 1.15 1.20

Xd‘ 0.23 0.37 0.35

Xd” 0.12 0.24 0.30

Xq” 0.15 0.34 0.40

X2 0.12 0.24 0.35

Xq 1.08 0.75 0.90

Xq‘ 0.23 0.75 0.90

Generator Sequence Impedances

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The sequence networks for the grounded-wyegenerator are shown below.

F1

N1

gEr

jZ1+

-

jZ2

F2

N2

jZ0

F0

N0

Grounded-Wye Generator

Generator Sequence Networks

Positive Sequence

Negative Sequence

Zero Sequence

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If the generator neutral is grounded through an impedance Zg, the zero-sequence impedance is modified as shown below.

F1

N1

gEr

jZ1+

-

jZ2

F2

N2

jZ0

F0

N0

3Zg

Grounded-Wye through an Impedance

Generator Sequence Networks

Positive Sequence

Negative Sequence

Zero Sequence

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Ungrounded-Wye Generator

If the generator is connected ungrounded-wye or delta, no zero-sequence current can flow. The sequence networks for the generator are shown below.

F1

N1

gEr

jZ1+

-

jZ2

F2

N2

jZ0

F0

N0

Positive Sequence

Negative Sequence

Zero Sequence

Generator Sequence Networks

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Transformer Models

� Two Winding Transformer

� Short-Circuit and Open-Circuit Tests

� Three Winding Transformer

� Autotransformer

� Transformer Connection

� Three Phase Transformer

� Three Phase Model

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Two-Winding Transformer

+ +

__HVr

XVr

HIr

XIr

HN XN

The voltage drop from the polarity-marked terminal to the non-polarity-marked terminal of the H winding is in phase with the voltage drop from the polarity-marked terminal to the non-polarity-marked terminal of the X winding.

X

H

X

H

NN

VV

=r

rVoltage Equation:

Ideal Transformer

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XXHH ININrr

=

The current that enters the H winding through the polarity-marked terminal is in phase with the current that leaves the X winding through the polarity-marked terminal.

Note: Balancing ampere-turns must be satisfied at all times.

Current Equation:+ +

__HVr

XVr

HIr

XIr

HN XN

Two-Winding Transformer

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Two-Winding Transformer

X

H

V

Va r

r

=

H

X

I

Ia r

r

=

XH VaVrr

=

X

X2

H

H

I

Va

I

Vr

r

r

r

=

a

II X

H

rr

=

Referred ValuesFrom the Transformation Ratio,

X2

H ZaZ =

Dividing VH by IH,

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Practical Transformer1. The H and X coils have a small resistance.2. There are leakage fluxes in the H and X coils.3. There is resistance loss in the iron core.4. The permeability of the iron is not infinite.

+

eH-

φm

NH

+

-

eXNX

iX

vH

iH

vX

ironcore

Two-Winding Transformer

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RH, XH =resistance and leakage reactance of H coil

Rc, Xm =core resistance and magnetizing reactance

+

-

+

-HEv

XEv

HN XN

HIr

XIr

Ideal

HH jXR + XX jXR +

+

-HVr

+

-XVr

H winding X winding

mjXcR

exIv

RX, XX =resistance and leakage reactance of X coil

Two-Winding TransformerEquivalent Circuit

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Two-Winding Transformer

+

-

+

-HEv

XEv

HN XN

HIr

a

I X

r

HH jXR + X2

X2 XjaRa +

+

-HVr

+

-XVar

mjXcR

exIv

Referring secondary quantities at the primary side,

HIr

a

I X

r

HH jXR + X2

X2 XjaRa +

+

-HVr

+

-XVar

mjXcR

exIv

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Two-Winding TransformerThe transformer equivalent circuit can be approximated by

X2

Heq RaRR +=

X2

Heq XaXX +=

HIr+

-HVr

+

-XVar

mjXcR

exIv

eqeq jXR +

HIr+

-HVr

+

-XVar

mjXcR

exIv

eqeq jXR +

Xa1 Ir

Xa1 Ir

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Two-Winding Transformer

Xa1

H IIrr

=

eqeq jXR +

+

-HVr

+

-XVar

For large power transformers, shunt impedance and resistance can be neglected

Xa1

H IIrr

=

eqjX

+

-HVr

+

-XVar

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Tap-Changing Transformera:1

q r

s p

The π equivalent circuit of transformer with the per unit transformation ratio: pqy

a

a2

1−pqy

a

a 1−

pqya

1

Two-Winding Transformer

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Short-Circuit and Open-Circuit Tests

H1

H2

x1

x2

Short-Circuit TestConducted to determine series impedance

With the secondary (Low-voltage side) short-circuited, apply a primary voltage (usually 2 to 12% of rated value) so that full load current flows.

A

V

W

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Short-Circuit and Open-Circuit Tests

Short-Circuit Test

1I

eqeq jXR +

+

-

SCV mjXcR

eI

1sc

e

II

0I

=

≈SCI

2SC

SCeq

I

PR =

SC

SCeq

I

VZ = 2

eq2eqeq RZX −=

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Short-Circuit and Open-Circuit Tests

x1

x2

H1

H2

Open-Circuit TestConducted to determine shunt impedance

With the secondary (High-voltage side) open-circuited, apply rated voltage to the primary.

A

V

W

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Short-Circuit and Open-Circuit Tests

Open-Circuit Test

eqeq jXR +

+

-

OCV mjXcR

eI

eOC II =OCI

OC

2OC

cP

VR = 2

c

2

OC

OC

m R

1

V

I

X

1−⎥

⎤⎢⎣

⎡=

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Short-Circuit and Open-Circuit Tests

Example:

50 kVA, 2400/240V, single-phase transformer

Short-Circuit Test: HV side energized

Open-Circuit Test: LV side energized

Determine the Series and Shunt Impedance of the transformer. What is %Z and X/R of the transformer?

watts617P amps 8.20I volts48V SCSCSC ===

watts186P amps41.5I volts240V OCOCOC ===

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Short-Circuit and Open-Circuit Tests

Solution:

From the short-circuit test

hmo 31.28.20

48Z H,eq ==

hmo 82.142.131.2X2

2

H,eq =−=

From the open-circuit test

( )hmo 310

186

240R

2

L,cq ==

22

m 310

1

240

41.5

X

1⎥⎦⎤

⎢⎣⎡−⎥⎦

⎤⎢⎣⎡= hmo 45X L,m =

( )hmo 42.1

8.20

617R 2H,eq

==

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Short-Circuit and Open-Circuit Tests

Referred to the HV side

hmo 968,30RaR L,c2

H,c ==

hmo 482,4XaX L,m2

H,m ==

%Z and X/R

[ ]hmo 115.2

1000/50

4.2Z

2

BASE ==

%2100x2.115

31.2Z% =⎟

⎠⎞

⎜⎝⎛= 28.1

42.1

82.1R/X ==

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X/R Ratios of Transformers

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Three-Winding Transformer

+

_HVr

HIr

HN +

_XV

rXIr

XN

+

_YV

rYIr

YN

X

H

X

H

NN

VV

=r

r

Y

H

Y

H

NN

VV

=r

r

Y

X

Y

X

NN

VV

=r

r

YYXXHH INININrrr

+=

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ZHX=impedance measured at the H side when the X winding is short-circuited and the Y winding is open-circuited

ZHY=impedance measured at the H side when the Y winding is short-circuited and the X winding is open-circuited

ZXY=impedance measured at the X side when the Y winding is short-circuited and the H winding is open-circuited

From 3 short-circuit tests with third winding open, get

Note: When expressed in ohms, the impedances must be referred to the same side.

Three-Winding Transformer

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HZ

+

-HVr +

-

XVr

XZ

YZ

+

-YVr

XHHX ZZZ +=

YHHY ZZZ += YXXY ZZZ +=

or )ZZZ(Z XYHYHX21

H −+=

)ZZZ(Z XYHYHX21

X +−=

)ZZZ(Z XYHYHX21

Y ++−=

Three-Winding Transformer

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Example: A three-winding three-phase transformer has the following nameplate rating:

H: 30 MVA 140 kV X: 30 MVA 48 kV Y: 10.5 MVA 4.8 kV

Short circuit tests yield the following impedances:

ZHX = 63.37 Ω @ the H sideZHY = 106.21 Ω @ the H sideZXY = 4.41 Ω @ the X side

Find the equivalent circuit in ohms, referred to the H side.

Ω 52.37)41.4()(Z 248

140XY ==

Three-Winding Transformer

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Ω03.66)52.3721.10637.63(Z 21

H =−+=

With all impedances referred to the H side, we get

Ω .66.2)52.3721.10637.63(Z 21

X −=+−=

Ω18.40)52.3721.10637.63(Z 21

Y =++−=

Ω03.66

+

-HVr +

-

XVr+

-YVr

Ω66.2−

Ω18.40

Three-Winding Transformer

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Transformer Polarity

H1 H2

x1 x2

H1 H2

x2 x1

Subtractive Polarity Additive Polarity

V1

V V

V1

Less than V1

Greater than V1

Transformer Connection

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H1 H2

X1 X2

H1 H2

X1X2

Subtractive Additive

“Single-phase transformers in sizes 200 kVA and below having high-voltage ratings 8660 volts and below (winding voltage) shall have additive polarity. All other single-phase transformers shall have subtractive polarity.” (ANSI/IEEEC57.12.00-1993)

Transformer Connection

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Transformer Connection

H1 H2

x1 x2

H1 H2

x1 x2

LOAD

Parallel Connection

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Parallel Connection

� same turns ratio

� Connected to the same primary phase

� Identical frequency ratings

� Identical voltage ratings

� Identical tap settings

� Per unit impedances within 0.925 to 1.075 of each other

Transformer Connection

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Transformer Connection

Three Phase Transformer Bank

H1 H2

x1 x2

H1 H2

x1 x2

H1 H2

x1 x2

WYE-WYE (Y-Y)

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Transformer Connection

Three Phase Transformer Bank

H1 H2

x1 x2

H1 H2

x1 x2

H1 H2

x1 x2

DELTA-DELTA (ΔΔΔΔ-ΔΔΔΔ)

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Transformer Connection

Three Phase Transformer Bank

H1 H2

x1 x2

H1 H2

x1 x2

H1 H2

x1 x2

WYE-DELTA (Y-ΔΔΔΔ)

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Transformer Connection

Three Phase Transformer Bank

H1 H2

x1 x2

H1 H2

x1 x2

H1 H2

x1 x2

DELTA-WYE (ΔΔΔΔ-Y)

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Transformer Connection

Three Phase Transformer Bank

H1 H2

x1 x2

H1 H2

x1 x2

OPEN DELTA – OPEN DELTA

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Transformer Connection

Three Phase Transformer Bank

H1 H2

x1 x2

H1 H2

x1 x2

OPEN WYE - OPEN DELTA

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Three-Phase Transformer

Windings are connected Wye or Delta internally

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Angular DisplacementANSI/IEEEC57.12.00-1993: The angular displacement of a three-phase transformer is the time angle (expressed in degrees) between the line-to-neutral voltage of the high-voltage terminal marked H1 and the the line-to-neutral voltage of the low-voltage terminal marked X1.The angular displacement for a three-phase trans-former with a ΔΔΔΔ-ΔΔΔΔ or Y-Y connection shall be 0o.

The angular displacement for a three-phase trans-former with a Y-ΔΔΔΔ or ΔΔΔΔ-Y connection shall be 30o, with the low voltage lagging the high voltage.

Three-Phase Transformer

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Vector Diagrams

H1

H2

H3

X1

X2

X3

Δ-Δ Connection

X1

X2

X3H1

H2

H3

Y-Δ Connection

H1

H2

H3

X1

X2

X3

Y-Y Connection

X1

X2

X3H1

H2

H3

Δ-Y Connection

Three-Phase Transformer

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0

6

2

4

10

8

IEC Designation for ΔΔΔΔ-ΔΔΔΔ

Dd0 Dd2 Dd4

Dd6 Dd8 Dd10

IEC Designation for Y-YYy0 Yy6

Note: The first letter defines the connection of the H winding; the second letter defines the connection of the X winding; the number designates the angle.

Three-Phase Transformer

IEC Designation

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Three-Phase Transformer

IEC Designation

IEC Designation for Y-ΔΔΔΔ

Yd1 Yd5 Yd7 Yd11

Note: The first letter defines the connection of the H winding; the second letter defines the connection of the X winding; the number designates the angle.

1

7

3

5

11

9

IEC Designation for ΔΔΔΔ-Y

Dy1 Dy5 Dy7 Dy11

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(A-B-C)

1ANVr

1BNVr

1CNVr

1bcVr

1abVr

1caVr

1anVr

1bnVr

1cnVr

1ANVr

1anVr

lagsby 30o

Positive–Sequence Voltages

H1

H2

H3

X1X2

X3A

B

C

c

b

a

N

Three-Phase Transformer

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C a

Positive–Sequence Currents

H1

H2

H3

X1X2

X3A

B

c

b

1AIr

1BIr

1CIr

1cIr

1bIr

1aIr

1baIr

1cbIr

1acIr

1AIr

1aIr

lagsby 30o

1baIr

1cbIr

1acIr1aI

r

1bIr

1cIr

1AIr

1BIr

1CIr

(A-B-C)

Three-Phase Transformer

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Positive Sequence Impedance

XH IIrr

=

111 jXRZ +=

+

-HVr

+

-XVr

Using per-unit values, the positive-sequence equivalent circuit is

Whether a bank of single-phase units or a three-phase transformer unit (core type or shell type), the equivalent impedance is the same.

Note: The negative-sequence impedance is equal to the positive-sequence impedance.

Three-Phase Transformer

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C

2ANVr

2CNVr

2BNVr

(A-C-B)

H1

H2

H3A

BX1

X2

X3c

b

a

N

2cbVr

2baVr

2acVr

2anVr

2bnVr

2cnVr

2ANVr

2anVr

leadsby 30o

Negative–Sequence Voltages

Three-Phase Transformer

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H1

H2

H3

X1X2

X3A

B

C

c

b

a

2AIr

2BIr

2CIr

2cIr

2bIr

2aIr

2baIr

2cbIr

2acIr

2AIr

2aIr

leadsby 30o

2baIr

2acIr

2cbIr

2aIr

2bIr

2cIr

2AIr

2CIr

2BIr

(A-C-B)

Negative–Sequence Currents

Three-Phase Transformer

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Positive– & Negative Sequence Networks

Three-Phase Transformer

1Ir

1Z

+

-

+

-

Primary Side

Secondary Side

2Ir

2Z

+

-

+

-

Primary Side

Secondary Side

Positive Sequence Network

Negative Sequence Network

21 ZZ =

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Transformer Core

3-Legged Core Type

4-Legged Core Type

Shell Type

Note: Only the Xwindings are shown.

Three-Phase Transformer

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Three-Legged Transformer Core

The 3-legged core type three-phase transformer uses the minimum amount of core material. For balanced three-phase condition, the sum of the fluxes is zero.

cφbφaφ

Note: For positive- or negative-sequence flux,

0cba =φ+φ+φ

Three-Phase Transformer

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Zero Sequence Flux

The 3-legged core type three-phase transformer does not provide a path for zero-sequence flux. On the other hand, a bank of single-phase units, the 4-legged core type and the shell-type three-phase transformer provide a path for zero-sequence flux.

3-Legged Core Type

0φ0φ0φ

03φNote: The zero-sequence flux leaks out of the core and returns through the transformer tank.

Three-Phase Transformer

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10 ZZ =

+

-HV

r +

-XV

r

10 ZZ =

+

-HV

r +

-XV

r

Zero Sequence Impedance*Transformer Connection Zero-Sequence Network

*Excluding 3-phase unit with a 3-legged core.

Three-Phase Transformer

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Transformer Connection Zero-Sequence Network

*Excluding 3-phase unit with a 3-legged core.

10 ZZ =

+

-HV

r +

-XV

r

10 ZZ =

+

-HV

r +

-XV

r

Zero Sequence Impedance*

Three-Phase Transformer

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Transformer Connection Zero-Sequence Network

*Excluding 3-phase unit with a 3-legged core.

10 ZZ =

+

-HV

r +

-XV

r

10 ZZ =

+

-HV

r +

-XV

r

Zero Sequence Impedance*

Three-Phase Transformer

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Example: Consider a two-winding three-phase transformer with the following nameplate rating: 25 MVA 69Δ -13.8YG kV (Dyn1) Z=7%. Draw the positive and zero-sequence equivalent circuits. Use the transformer rating as bases.

Positive/Negative Sequence impedance

+

-HV

r +

-XV

r

Z1=j0.07

Zero Sequence impedance

+

-HV

r +

-XV

r

Z0=j0.07

Three-Phase Transformer

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Example: A three-winding three-phase transformer has the following nameplate rating: 150/150/45 MVA 138zG-69zG-13.8Δ kV (Yy0d1).

Draw the positive and zero-sequence equivalent circuits. Use 100 MVA and the transformer voltage ratings as bases.

H-X @ 150 MVA = 14.8% H-Y @ 45 MVA = 21.0% X-Y @ 45 MVA = 36.9%

p.u.10.0)150/100(148.0Z HX ==At the chosen MVA base,

p.u.47.0)45/100(21.0Z HY ==p.u.82.0)45/100(369.0Z XY ==

Three-Phase Transformer

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p.u.125.0)82.047.010.0(Z 21

H −=−+=We get

p.u.225.0)82.047.010.0(Z 21

X =+−=

p.u.595.0)82.047.010.0(Z 21

Y −=++−=

Positive/Negative Sequence Network

HZ

+

-HV

r +

-

XVr

XZ

YZ

+

-YVr

Zero Sequence Network

HZ

+

-HV

rXVr

XZ

YZ

+

-YVr+

-

Three-Phase Transformer

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ABC

abc

abcTY

Core Loss

Three Phase ModelTHREE-PHASE TRANSFORMER AND

3 SINGLE-PHASE TRANSFORMERS IN BANK

Primary Secondary

Admittance Matrix

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2. EPRI Core Loss Model

Three Phase Model

CORE LOSS MODELS

1. Constant P & Q Model

( )( )2

2

VF2

.u.p

VC2

.u.p

EVDBaseSystem

RatingkVAQ

BVABaseSystem

RatingkVAP

ε

ε

+=

+=

22.7F 0.268x10E 00167.0D -13 ===

5.31C 0.734x10B 00267.0A -9 ===|V| in per unit

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Primitive Coils

Primitive Impedances

V6

z11 z33 z44 z55 z66

I1

z12

I5 I6I4I3I2

z23 z34 z45 z56

z22

V1 V5V4V3V2+ + + + + +

- - - - - -

Three Phase Model

z66z65z64z63z62z61

z56z55z54z53z52z51

z46z45z44z43z42z41

z36z35z34z33z32z31

z26z25z24z23z22z21

z16z15z14z13z12z11

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Transformer ModelI1 I2

I3 I4

I5 I6

Three Identical

Single-phase Transformers

in Bank

z11

z33

z55

z12

z34

z56

z22

z44

z66

z66z65

z56z55

z44z43

z34z33

z22z21

z12z11

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Transformer Model

Node Connection Matrix, C

V6

V5

V4

V3

V2

V1

=

Vc

Vb

Va

VC

VB

VA

[V123456] = [C][VABCabc ]

Matrix C defines the relationship of the Primitive Voltages and Terminal Voltages of the Three-Phase Connected Transformer

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Transformer Model

Node Connection Matrix, C

VaVA

VC

VB

IA

IC

IB

Vb

Vc

1

5

3

Ia

Ib

Ic

2

6

4

[V123456] = [C][VABCabc ]

Wye Grounded-Wye GroundedConnection

1

1

1

1

1

1

V6

V5

V4

V3

V2

V1

Vc

Vb

Va

VC

VB

VA

=

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Transformer ModelVaVA

VC

VB

IA

IB

IC

Ia

Ib

Ic

Vb

Vc

1 2

5

6

34

Wye Grounded-Delta Connection

1-1

1

-11

1

-11

1

[V123456] = [C][VABCabc ]

V6

V5

V4

V3

V2

V1

Vc

Vb

Va

VC

VB

VA

=

Node Connection Matrix, C

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Transformer Model

A a

B b

C cN

M

M

M

L1 L2

L1 L2

L1 L2

R1

R1

R1

R2

R2

R2

3 Identical Single-Phase Transformers

connected Wye-Delta

Let,

563412M

64222

53111

ZZMjZZ

ZZLjRZ

ZZLjRZ

==ω==

==ω+=

==ω+=

1

3

5 6

4

2

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Transformer Model

The Primitive Voltage Equations

Z2ZM

ZMZ1

Z2ZM

ZMZ1

Z2ZM

ZMZ1

V6

V5

V4

V3

V2

V1

I6

I5

I4

I3

I2

I1

=

The Inverse of the Impedance Matrix

Z1 Z2 –ZM2

1The Primitive Admittance Matrix

Z1-ZM

-ZMZ2

Z1-ZM

-ZMZ2

Z1-ZM

-ZMZ2

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Transformer Model

YBUS =

YBUS = [C][ Yprim][CT]

2Z1-Z1-Z1-ZMZM

-ZM2Z1-Z1-ZMZM

-Z1-Z12Z1ZM-ZM

-ZMZMZ2

ZM-ZMZ2

ZM-ZMZ2

Z1 Z2 –ZM2

1

A B C a b c

A

B

C

a

b

c

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Transformer Model

Iinj = [C Yprim CT] Vnode

YBUS = [C][Yprim][CT]

1-1

1

-11

1

-11

1

1-1

1-1

-11

1

1

1

The Bus Admittance Matrix

Z1-ZM

-ZMZ2

Z1-ZM

-ZMZ2

Z1-ZM

-ZMZ2

Z1 Z2 –ZM2

1

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2a2yt-a2yt-a2yt-aytayt

-a2yt2a2yt-a2yt-aytayt

-a2yt-a2yt2a2ytayt-ayt

-aytaytyt

ayt-aytyt

ayt-aytyt

Three Phase Model

Define 2

m21

2t zzz

zy

−=

2

1

n

na =

YBUS =

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Three Phase ModelIf the admittances are already in per unit system, then the effective turns ratio ”a” must be

3

1

n

na

2

1 ==

2

m21

2t zzz

zy

−=

ty

ty

ty

3

1− ty

3

1ty

3

1− ty

3

1ty

3

1− ty

3

1ty

3

1− ty

3

1− ty

3

1− ty

3

1ty

3

1ty

3

1ty

3

1− ty

3

1− ty

3

1− ty

3

1− ty

3

1− ty

3

1− ty

3

2ty

3

2ty

3

2ty

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Three Phase ModelSummary

YccYcbYcaYcCYcBYcA

YbcYbbYbaYbCYbBYbA

YacYabYaaYaCYaBYaA

YccYCbYCaYCCYCBYCA

YBcYBbYBaYBCYBBYBA

YAcYabYAaYACYABYAA

[Ybus] =

YSSYSP

YPSYPP

A B C a b c

A

B

C

a

b

c

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Three Phase Model

YSPYPSYSSYPPSECPRI

-YII-YIIYIIYIIDeltaDelta

YIIIYIIITYIIYIIWyeDelta

YIIIYIIITYIYIIWye-GDelta

YIIITYIIIYIIYIIDeltaWye

-YII-YIIYIIYIIWyeWye

-YII-YIIYIIYIIWye-GWye

YIIITYIIIYIIYIDeltaWye-G

-YII-YIIYIIYIIWyeWye-G

-YI-YIYIYIWye-GWye-G

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Three Phase ModelSummary

yt

yt

yt

YI =2yt-yt-yt

-yt2yt-yt

-yt-yt2yt

-ytyt

yt-yt

yt-yt

YII = 1/3

YIII = 1/√√√√3

-ytyt

-ytyt

yt-yt

YIIIT = 1/√√√√3

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Three Phase ModelExample:

Three single-phase transformers rated 50 kVA, 7.62kV/240V, %Z=2.4, X/R=3 are connected Wye(grounded)-Delta. Determine the Admittance Matrix Model of the Transformer Bank. Assume yt = 1/zt

yp.u. = ____ -j ____Zp.u. = ____ +j ____

[Ybus] =

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3-Phase Transformer Impedance Matrix Model

� Distributing Transformer Impedance Between Windings

� Impedance Matrix in Backward-Forward Sweep Load Flow� Wye-Grounded – Wye-Grounded

� Delta-Delta

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Transformer Equations

Consider the winding-to-winding relationship between primary and secondary:

From transformer equations,

PRI

SEC

Va

V=

1PRI

SEC

I

I a=

2PRI

SEC

Za

Z=

PRI

SEC

Na

N=

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Distributing Transformer Impedance Between Windings

� Transformers are typically modeled with series impedance lumped at either end.

� To properly model transformer behavior, series impedance must be modeled in both windings.

� PROBLEM: divide ZT into ZP and ZS given a

'T P SZ Z Z= +

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Distributing Transformer Impedance Between Windings

� ASSUMPTION: Transformer impedance varies as number of wire turns.

Referring ZS to primary side ,

Substituting,

S PZ aZ=

2 3'S S PZ a Z a Z= =

3

3(1 )T P P

P

Z Z a Z

a Z

= +

= +

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Distributing Transformer Impedance Between Windings

To find ZP and ZS,

3

1

(1 )P TZ Za

=+

3(1 )S T

aZ Z

a=

+

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Impedance Matrix in Backward-Forward Sweep Load Flow

� Transformer model involved in� backward summation of current

� forward computation of voltage

� Wye-Grounded – Wye-Grounded

� Delta-Delta

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Wye Grounded – Wye Grounded

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WyeGnd-WyeGnd Backward Sweep

� Secondary to Secondary Winding

� Secondary Winding to Primary Windingif in PU: If not in PU:

� Primary Winding to Primary

_ _1

_ _ 2

_ _ 3

1 0 0

0 1 0

0 0 1

Sec Winding a

Sec Winding b

Sec Winding c

I I

I I

I I

⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥=⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦⎣ ⎦

Pr _ _1 _ _1

Pr _ _ 2 _ _ 2

Pr _ _ 3 _ _ 3

10 0

10 0

10 0

i Winding Sec Winding

i Winding Sec Winding

i Winding Sec Winding

aI I

I Ia

I I

a

⎡ ⎤⎢ ⎥

⎡ ⎤ ⎡ ⎤⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥=⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎣ ⎦ ⎣ ⎦

⎢ ⎥⎢ ⎥⎣ ⎦

Pr _ _1 _ _1

Pr _ _ 2 _ _ 2

Pr _ _3 _ _ 3

1 0 0

0 1 0

0 0 1

i Winding Sec Winding

i Winding Sec Winding

i Winding Sec Winding

I I

I I

I I

⎡ ⎤ ⎡ ⎤⎡ ⎤⎢ ⎥ ⎢ ⎥⎢ ⎥=⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎣ ⎦⎣ ⎦ ⎣ ⎦

Pr _ _1

Pr _ _ 2

Pr _ _ 3

1 0 0

0 1 0

0 0 1

A i Winding

B i Winding

C i Winding

I I

I I

I I

⎡ ⎤⎡ ⎤ ⎡ ⎤⎢ ⎥⎢ ⎥ ⎢ ⎥= ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦

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WyeGnd-WyeGnd Forward Sweep� Primary to Primary winding

� Primary Winding to Secondary WindingIf in PU: If not in PU:

� Secondary Winding to Secondary

Pr _ _1 Pr _ _1 Pr _ _1

Pr _ _ 2 Pr _ _ 2 Pr _ _ 2

Pr _ _3 Pr _ _3 Pr _ _3

1 0 0 *

0 1 0 *

0 0 1 *

i Winding AN i Winding i Winding

i Winding BN i Winding i Winding

i Winding CN i Winding i Winding

V V I Z

V V I Z

V V I Z

⎡ ⎤ ⎡ ⎤⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥= −⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦⎣ ⎦ ⎣ ⎦

_ _1 Pr _ _1

_ _ 2 Pr _ _ 2

_ _ 3 Pr _ _3

10 0

10 0

10 0

Sec Winding i Winding

Sec Winding i Winding

Sec Winding i Winding

aV V

V Va

V V

a

⎡ ⎤⎢ ⎥

⎡ ⎤ ⎡ ⎤⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥=⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎣ ⎦ ⎣ ⎦

⎢ ⎥⎢ ⎥⎣ ⎦

_ _1 Pr _ _1

_ _ 2 Pr _ _ 2

_ _3 Pr _ _3

1 0 0

0 1 0

0 0 1

Sec Winding i Winding

Sec Winding i Winding

Sec Winding i Winding

V V

V V

V V

⎡ ⎤ ⎡ ⎤⎡ ⎤⎢ ⎥ ⎢ ⎥⎢ ⎥=⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎣ ⎦⎣ ⎦ ⎣ ⎦

_ _1 _ _1 _ _1

_ _ 2 _ _ 2 _ _ 2

_ _ 3 _ _ 3 _ _ 3

1 0 0 *

0 1 0 *

0 0 1 *

an Sec Winding Sec Winding Sec Winding

bn Sec Winding Sec Winding Sec Winding

cn Sec Winding Sec Winding Sec Winding

V V I Z

V V I Z

V V I Z

⎡ ⎤ ⎡ ⎤⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥= −⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦

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Delta-Delta Transformer Connection

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Delta-Delta Backward Sweep� Secondary to Secondary Winding

� Secondary Winding to Primary WindingIf in PU: If not in PU:

� Primary Winding to Primary

_ _1

_ _ 2

_ _ 3

1 1 01

0 1 13

1 0 1

Sec Winding a

Sec Winding b

Sec Winding c

I I

I I

I I

⎡ ⎤ −⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥= −⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦⎣ ⎦

Pr _ _1 _ _1

Pr _ _ 2 _ _ 2

Pr _ _3 _ _ 3

10 0

10 0

10 0

i Winding Sec Winding

i Winding Sec Winding

i Winding Sec Winding

aI I

I Ia

I I

a

⎡ ⎤⎢ ⎥

⎡ ⎤ ⎡ ⎤⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥=⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎣ ⎦ ⎣ ⎦

⎢ ⎥⎢ ⎥⎣ ⎦

Pr _ _1 _ _1

Pr _ _ 2 _ _ 2

Pr _ _3 _ _ 3

1 0 0

0 1 0

0 0 1

i Winding Sec Winding

i Winding Sec Winding

i Winding Sec Winding

I I

I I

I I

⎡ ⎤ ⎡ ⎤⎡ ⎤⎢ ⎥ ⎢ ⎥⎢ ⎥=⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎣ ⎦⎣ ⎦ ⎣ ⎦

Pr _ _1

Pr _ _ 2

Pr _ _ 3

1 0 1

1 1 0

0 1 1

a i Winding

b i Winding

c i Winding

I I

I I

I I

⎡ ⎤−⎡ ⎤ ⎡ ⎤⎢ ⎥⎢ ⎥ ⎢ ⎥= − ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦ ⎣ ⎦

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Delta-Delta Forward Sweep� Primary to Primary Winding

� Primary Winding to Secondary Winding If in PU: If not in PU:

� Secondary Winding to Secondary

Pr _ _1 Pr _ _1 Pr _ _1

Pr _ _ 2 Pr _ _ 2 Pr _ _ 2

Pr _ _ 3 Pr _ _ 3 Pr _ _ 3

1 1 0 *

0 1 1 *

1 0 1 *

i Winding AN i Winding i Winding

i Winding BN i Winding i Winding

i Winding CN i Winding i Winding

V V I Z

V V I Z

V V I Z

⎡ ⎤ ⎡ ⎤−⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥= − −⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦⎣ ⎦ ⎣ ⎦

_ _1 Pr _ _1

_ _ 2 Pr _ _ 2

_ _ 3 Pr _ _ 3

10 0

10 0

10 0

Sec Winding i Winding

Sec Winding i Winding

Sec Winding i Winding

aV V

V Va

V V

a

⎡ ⎤⎢ ⎥

⎡ ⎤ ⎡ ⎤⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥=⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎣ ⎦ ⎣ ⎦

⎢ ⎥⎢ ⎥⎣ ⎦

_ _1 Pr _ _1

_ _ 2 Pr _ _ 2

_ _ 3 Pr _ _ 3

1 0 0

0 1 0

0 0 1

Sec Winding i Winding

Sec Winding i Winding

Sec Winding i Winding

V V

V V

V V

⎡ ⎤ ⎡ ⎤⎡ ⎤⎢ ⎥ ⎢ ⎥⎢ ⎥=⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎣ ⎦⎣ ⎦ ⎣ ⎦

_ _ 1

_ _ 2

_ _ 3

1 3 00 0

31 1 5 0

0 03

1 3 00 1

3

a S ec W in d in g

b S ec W in d in g

c S ec W in d in g

V V

V V

V V

∠ −⎡ ⎤⎢ ⎥⎢ ⎥ ⎡ ⎤⎡ ⎤

∠ −⎢ ⎥ ⎢ ⎥⎢ ⎥ = ⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎣ ⎦ ⎣ ⎦∠ −⎢ ⎥⎢ ⎥⎣ ⎦

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Transmission Line Distribution Line

� Series Impedance of Lines

� Shunt Capacitance of Lines

� Nodal Admittance Matrix Model

� Data Requirements

Transmission and DistributionLine Models

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ZccZcbZca

ZbcZbbZba

ZacZabZaa

Y’ccY’cbY’ca

Y’bcY’bbY’ba

Y’acY’abY’aa

Y”ccY”cbY”ca

Y”bcY”bbY”ba

Y”acY”abY”aa

ABC

abc

Transmission and DistributionLine Models

••

••+

-VR

+

-

Z = R + jXL

CY21

CY21

Balanced Three-Phase System

Unbalanced Three-Phase

System

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Conductor Materials� Aluminum (Al) is preferred over Copper (Cu) as a

material for transmission and distribution lines due to:� lower cost

� lighter weight

� larger diameter for the same resistance**This results in a lower voltage gradient at the conductor

surface (less tendency for corona)

� Copper is preferred over Aluminum as a material for distribution lines due to lower resistance to reduce system losses.

Series Impedance of Lines

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Stranding of Conductors

Steel Aluminum

Alternate layers of wire of a stranded conductor are spiraled in opposite directions to prevent unwinding and make the outer radius of one layer coincide with the inner radius of the next.

The number of strands depends on the number of layers and on whether all the strands are of the same diameter. The total number of strands of uniform diameter in a concentrically stranded cable is 7, 19, 37, 61, 91, etc.

Hard-Drawn Copper(Cu)

Aluminum Conductor Steel Reinforced

(ACSR)

Series Impedance of Lines

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Resistance of Conductors� The Resistance of a Conductor depends on the

material (Cu or Al)

� Resistance is directly proportional to Length but inversely proportional to cross-sectional area

� Resistance increases with Temperature

� Skin-Effect in Conductors

LR

Aρ=

R – Resistanceρρρρ – Resistivity of MaterialL – LengthA – Cross-Sectional Area

Series Impedance of Lines

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Conductor Strands O.D. GMR ResistanceType Value Unit (Inches) (feet) (Ohm/Mile)

1 ACSR 6 AWG 6/1 0.19800 0.00394 3.98000 2 ACSR 5 AWG 6/1 0.22300 0.00416 3.18000 3 ACSR 4 AWG 7/1 0.25700 0.00452 2.55000 4 ACSR 4 AWG 6/1 0.25000 0.00437 2.57000 5 ACSR 3 AWG 6/1 0.28100 0.00430 2.07000 6 ACSR 2 AWG 7/1 0.32500 0.00504 1.65000 7 ACSR 2 AWG 6/1 0.31600 0.00418 1.69000 8 ACSR 1 AWG 6/1 0.35500 0.00418 1.38000 9 ACSR 1/0 AWG 6/1 0.39800 0.00446 1.12000 10 ACSR 2/0 AWG 6/1 0.44700 0.00510 0.89500

INDEXSize

Source: Westinghouse T&D Handbook

Resistance of Conductors

Series Impedance of Lines

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Line Inductance

Series Impedance of Lines

Self Inductance: extint LLL +=

Mutual Inductance (between 2 conductors):

1Ir

11z

22z

12z

1

2Ir

1’

2 2’

122111'11 zIzIV +=−

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Carson’s Line

aVr

aIr+

-

aaz

ddz

adz

a a’

d d’

0Vd =r

Local Earth

Remote Earth

Fictitious Return Conductor

dIr

Carson examined a single overhead conductor whose remote end is connected to earth.

The current returns through a fictitious earth conductor whose GMR is assumed to be 1 foot (or 1 meter) and is located a distance Dad from the overhead conductor.

REF

Series Impedance of Lines

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The line is described by the following equations:

' 'aa a a aa a ad dV V V z I z I= − = +r r r r r

' 'dd d d ad a dd dV V V z I z I= − = +r r r r r

Note: , and .a dI I= −r r

0dV =r

' ' 0a dV V− =r r

Subtracting the two equations, we get

( 2 )a aa dd ad aV z z z I= + −r r

a aa aV z I=r r

or

zaa is the equivalent impedance of the single overhead conductor.

2aa aa dd adz z z z= + −

Series Impedance of Lines

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Primitive Impedances:2

(ln 1)aa a a asa

sz r j L r j k

Dω ω= + = + −

2(ln 1)dd d

sd

sz r j k

Dω= + −

2(ln 1)ad

ad

sz j M j k

Dω ω= = −

ra, rd = resistances of overhead conductor and fictitious ground wire, respectively

Dsa, Dsd = GMRs of overhead conductor and fictitious ground wire, respectively

Series Impedance of Lines

Dad = Distance between the overhead conductor and fictitious ground wire

( )( )ohm/meterxfk 71022 −= πω

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Earth Resistance:

Carson derived an empirical formula for the earth resistance.

-31.588 x 10dr f= Ω/mile

-49.869 x 10 f= Ω/km

Note : At 60 Hz,

0.09528dr = Ω/mile

where f is the power frequency in Hz

Series Impedance of Lines

059214.0= Ω/km

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Geometric Mean Radius

For a solid conductor with radius r, r78.0rD 4

1

s ==−

ε

Bundle of Two Bundle of Four

d

dDD scs =

d

d

4 3scs dD09.1D =

Note: Dsc=GMR of a single conductor

Series Impedance of Lines

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Equivalent Impedance:Substitute the primitive impedances into

2aa aa dd adz z z z= + −We get 2

adD( ) lnaa a d

sa sd

z r r j kD D

ω= + +

Define2

adDe

sd

DD

=

We geteD

( ) lnaa a dsa

z r r j kD

ω= + + Ω/unit length

Series Impedance of Lines

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The quantity De is a function of frequency and earth resistivity.

2160 /eD fρ= feet

Typical values of De are tabulated below.

Return Earth Condition

Resistivity(Ω-m)

De(ft)

Sea water 0.01-1.0 27.9-279Swampy ground 10-100 882-2790

Average Damp Earth 100 2790Dry earth 1000 8820Sandstone 109 8.82x106

Series Impedance of Lines

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Three-Phase Line Impedances

aVr

cIr+

-ccz

ddz

a a’

d d’

0Vd =r

dIr

bbz

aaz

b

c

bIraIr

abz

bczcaz b’

c’

bVr

cVr

adzbdz

cdz

REF

+

-+

-

All wires grounded here

Series Impedance of Lines

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The voltage equation describing the line is

aa ab ac ad

ba bb bc bd

ca cb cc cd

da db dc dd

z z z z

z z z z

z z z z

z z z z

a

b

c

d

I

I

I

I

r

r

r

r

'

'

'

'

aa

bb

cc

dd

V

V

V

V

r

r

r

r

'

'

'

'

a a

b b

c c

d d

V V

V V

V V

V V

r r

r r

r r

r r

= =

Since all conductors are grounded at the remote end, we get from KCL

0a b c dI I I I+ + + =r r r r

or

( )d a b cI I I I= − + +r r r r

Series Impedance of Lines

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We can subtract the voltage equation of the ground conductor from the equations of phases a, b and c. The resulting matrix equation is

aa ab ac

ab bb bc

ac bc cc

z z z

z z z

z z z

a

b

c

V

V

V

r

r

r=

a

b

c

I

I

I

r

r

rV/unit length

Self Impedances:

2aa aa ad ddz z z z= − + Ω/unit length

2bb bb bd ddz z z z= − + Ω/unit length2cc cc cd ddz z z z= − + Ω/unit length

Series Impedance of Lines

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Mutual Impedances:

ddbdadabab zzzzz +−−=

ddcdadacac zzzzz +−−=

Ω/unit length

Ω/unit lengthΩ/unit length

ddcdbdbcbc zzzzz +−−=

Primitive Impedances:2

(ln 1)xx xsx

sz r j k

Dω= + − Ω/unit length

2(ln 1)xy

xy

sz j k

Dω= − Ω/unit length

x=a,b,c,d

xy=ab,bc,ca,ad,bd,cd

Series Impedance of Lines

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1. Identical phase conductorsAssumptions:

s sa sb scD D D D= = =2. Distances of the overhead conductors to the

fictitious ground conductor are the same

e ad bd cdD D D D= = =We get

eD( ) lnaa bb cc a d

s

z z z r r j kD

ω= = = + +

Ω/unit lengtheD lnxy d

xy

z r j kD

ω= +xy=ab,bc,ca

Series Impedance of Lines

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Example: Find the equivalent impedance of the 69-kV line shown. The phase conductors are 4/0 hard-drawn copper, 19 strands which operate at 25oC. The line is 40 miles long. Assume an earth resistivity of 100 Ω-meter.

ra=0.278 Ω/mile @ 25oC

Dsc=0.01668 ft @ 60 Hz

s

edaccbbaa D

Dln kj)rr(zzz ω++===

0.016682790ln 121.0j)095.0278.0( ++=

459.1j373.0 += Ω/mileΩ38.58j93.14Z aa +=

10’

a b c

10’

Series Impedance of Lines

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102790

bcab ln 121.0j095.0zz +==683.0j095.0 +=

Ω/mile

Ω33.27j81.3Z ab +=

202790

ac ln 121.0j095.0z +=

Ω97.23j81.3Z ac +=

We get

⎥⎥⎥

⎢⎢⎢

+++

+++

+++

38.58j93.1433.27j81.397.23j81.3

33.27j81.338.58j93.1433.27j81.3

97.23j81.333.27j81.338.58j93.14

Zabc= Ω

Series Impedance of Lines

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aVr cI

r+

-

ccz

ddz

aa’

dd’

0Vd =r

dIr

bbz

aaz

b

c

bIraIr

abz

bcz

cazb’

c’

bVr

cVr adz

bdzcdz

REF

+

-+

-All wires grounded here

Lines with Overhead Ground Wire

wwzwIr

w’

wVr+

-

w

Series Impedance of Lines

wdz

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The primitive voltage equation is

aa ab ac aw ad

ba bb bc bw bd

ca cb cc cw cd

wa wb wc ww wd

da db dc dw dd

z z z z z

z z z z z

z z z z z

z z z z z

z z z z z

'

'

'

'

'

0

0

a a

b b

c c

w

d

V V

V V

V V

V

V

r r

r r

r r

r

r

a

b

c

w

d

I

I

I

I

I

r

r

r

r

r

=

From KCL, we get0a b c w dI I I I I+ + + + =

r r r r r

or( )d a b c wI I I I I= − + + +

r r r r r

V/unit length

Series Impedance of Lines

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It can be shown that

aa ab ac aw

ba bb bc bw

ca cb cc cw

wa wb wc ww

z z z z

z z z z

z z z z

z z z z

a

b

c

w

I

I

I

I

r

r

r

r

a

b

c

w

V

V

V

V

r

r

r

r

=

exx x d

sx

Dz ( r r ) j k ln

Dω= + +

exy d

xy

Dz r j k ln

Dω= + xy=ab,ac,aw,bc,bw,cw

xx=aa,bb,cc,ww

where wV 0=r

Series Impedance of Lines

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Using Kron Reduction technique,

0V1

2

1

II

43

21

ZZZZ

=

where Z1, Z2, Z3 and Z4 are also matrices.

Series Impedance of Lines

131

4211 I)ZZZZ(V −−=

I2 is eliminated and the matrix is reduced to the size of Z1

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Eliminating the ground wire current Iw

⎥⎥⎥

⎢⎢⎢

⎡=

cccbca

bcbbba

acabaa

1

zzz

zzz

zzz

Z

ww4 zZ =

[ ]cwbwaw zzz=2Z =3Z

⎥⎥⎥⎥⎥⎥⎥

⎢⎢⎢⎢⎢⎢⎢

−−−

−−−

−−−

=

ww

wccwcc

ww

wbcwcb

ww

wacwca

ww

wcbwbc

ww

wbbwbb

ww

wabwba

ww

wcawac

ww

wbawab

ww

waawaa

abc

z

zzz

z

zzz

z

zzz

z

zzz

z

zzz

z

zzz

z

zzz

z

zzz

z

zzz

z

We get

Series Impedance of Lines

⎥⎥⎥

⎢⎢⎢

cw

bw

aw

z

z

z

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Example: Find the equivalent impedance of the 69-kV line shown. The phase conductors are the same as in the previous examples. The overhead ground wires have the following characteristics:

rw=4.0 Ω/mile @ 25oC

Dsw=0.001 ft @ 60 Hz

10’a b c

10’

w

15’For the ground wire, we get

sw

edwww D

Dln kj)rr(z ω++=

0.0012790ln 121.0j)095.00.4( ++=

8.1j095.4 +=Ω72j8.163Z ww +=

Ω/mile

Series Impedance of Lines

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aw

edcwaw D

Dln kjrzz ω+==

18.032790ln 121.0j095.0 +=

Ω47.24j81.3ZZ cwaw +==

152790

bw ln 121.0j095.0Z +=

Ω36.25j81.3Zbw +=From a previous example, we got

⎥⎥⎥

⎢⎢⎢

+++

+++

+++

38.58j93.1433.27j81.397.23j81.3

33.27j81.338.58j93.1433.27j81.3

97.23j81.333.27j81.338.58j93.14

Z1= Ω

Ω/mile

Ω/mile

Series Impedance of Lines

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Using the ground wire impedances, we also get

Ω72j8.163Z4 +=⎥⎥⎥

⎢⎢⎢

+

+

+

47.24j81.3

36.25j81.3

47.24j81.3

=2Z = T3Z

Performing Kron reduction, we get

⎥⎥⎥

⎢⎢⎢

+++

+++

+++

11.56j5.170.25j48.67.21j38.6

0.25j48.697.55j71.170.25j48.6

7.21j38.60.25j48.611.56j5.17

Zabc = Ω

Note: The self impedances are no longer equal.

Series Impedance of Lines

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Line TranspositionLine transposition is used to make the mutual impedances identical.

Pos.1

Pos.2

Pos.3

Phase c

Phase a

Phase b

aIr

bIr

cIr

s1 s3s2

Section 2Section 1 Section 3

Note: Each phase conductor is made to occupy all possible positions.

Series Impedance of Lines

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For Section 1⎥⎥⎥

⎢⎢⎢

⎥⎥⎥

⎢⎢⎢

⎡=

⎥⎥⎥

⎢⎢⎢

−−−

−−−

−−−

c

b

a

133132131

123122121

113112111

c

b

a

I

I

I

ZZZ

ZZZ

ZZZ

V

V

V

r

r

r

r

r

r

volts

⎥⎥⎥

⎢⎢⎢

⎥⎥⎥

⎢⎢⎢

⎡=

⎥⎥⎥

⎢⎢⎢

−−−

−−−

−−−

b

a

c

233232231

223222221

213212211

b

a

c

I

I

I

ZZZ

ZZZ

ZZZ

V

V

V

r

r

r

r

r

r

volts

⎥⎥⎥

⎢⎢⎢

⎥⎥⎥

⎢⎢⎢

⎡=

⎥⎥⎥

⎢⎢⎢

−−−

−−−

−−−

a

c

b

333332331

323322321

313312311

a

c

b

I

I

I

ZZZ

ZZZ

ZZZ

V

V

V

r

r

r

r

r

r

volts

Voltage Equations for Each Section

For Section 2

For Section 3

Series Impedance of Lines

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a333222111a I)ZZZ(Vrr

−−− ++=Σb331223112 I)ZZZ(r

−−− +++

c332221113 I)ZZZ(r

−−− +++

a313232121b I)ZZZ(Vrr

−−− ++=Σ

b311233122 I)ZZZ(r

−−− +++

c312231123 I)ZZZ(r

−−− +++

a323212131c I)ZZZ(Vrr

−−− ++=Σb321213132 I)ZZZ(r

−−− +++

c322211133 I)ZZZ(r

−−− +++

The total Voltage Drop at phases a, b, and c are:

Series Impedance of Lines

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Define f1, f2 and f3 as as the ratios of s1, s2 and s3 to the total length s, respectively. We get

a333222111a I)ZfZfZf(Vrr

++=Σ

b313232121 I)ZfZfZf(r

+++

c323212131 I)ZfZfZf(r

+++

a133322211b I)ZfZfZf(Vrr

++=Σ

b113332221 I)ZfZfZf(r

+++

c123312231 I)ZfZfZf(r

+++

a233122311c I)ZfZfZf(Vrr

++=Σb213132321 I)ZfZfZf(r

+++c223112331 I)ZfZfZf(r

+++

Series Impedance of Lines

s

sf 1

1 =

s

sf 2

2 =

s

sf 3

3 =

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Define: 1332321211k ZfZfZfZ ++=

2331221312k ZfZfZfZ ++=

1231322313k ZfZfZfZ ++=

332211s ZZZZ ===

Substitution gives

⎥⎥⎥

⎢⎢⎢

⎥⎥⎥

⎢⎢⎢

⎡=

⎥⎥⎥

⎢⎢⎢

c

b

a

s3k2k

3ks1k

2k1ks

c

b

a

I

I

I

ZZZ

ZZZ

ZZZ

V

V

V

r

r

r

r

r

r

Σ

Σ

Σ

Volts

Series Impedance of Lines

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It can be shown that

s

edas D

Dln ksjs)rr(Z ω++=

⎟⎟⎠

⎞⎜⎜⎝

⎛+++=

31

e3

23

e2

12

e1d1k D

Dlnf

D

Dlnf

D

DlnfksjsrZ ω

⎟⎟⎠

⎞⎜⎜⎝

⎛+++=

23

e3

12

e2

31

e1d2k D

Dlnf

D

Dlnf

D

DlnfksjsrZ ω

⎟⎟⎠

⎞⎜⎜⎝

⎛+++=

12

e3

31

e2

23

e1d3k D

Dlnf

D

Dlnf

D

DlnfksjsrZ ω

Series Impedance of Lines

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Example: Find the equivalent impedance of the 69-kV line shown. The phase conductors are 4/0 hard-drawn copper, 19 strands which operate at 25oC. The line is 40 miles long. Assume s1=8 miles, s2=12 miles and s3=20 miles.

ra=0.278 Ω/mile @ 25oC

Dsc=0.01668 ft @ 60 Hz

⎥⎥⎥

⎢⎢⎢

+++

+++

+++

=

38.58j93.1433.27j81.397.23j81.3

33.27j81.338.58j93.1433.27j81.3

97.23j81.333.27j81.338.58j93.14

Zabc Ω

Without the transposition,

10’

a b c

10’

Section 1

Series Impedance of Lines

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Solving for the mutual impedances, we get

1332321211k ZfZfZfZ ++=

)33.27j81.3(3.0)33.27j81.3(2.0 +++=)97.23j81.3(5.0 ++

Ω65.25j81.3 +=Similarly, we get

2331221312k ZfZfZfZ ++= Ω66.26j81.3 +=

1231322313k ZfZfZfZ ++= Ω32.26j81.3 +=

Series Impedance of Lines

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⎥⎥⎥

⎢⎢⎢

+++

+++

+++

38.58j93.1433.27j81.397.23j81.3

33.27j81.338.58j93.1433.27j81.3

97.23j81.333.27j81.338.58j93.14

Zabc= Ω

⎥⎥⎥

⎢⎢⎢

+++

+++

+++

38.58j93.1432.26j81.366.26j81.3

32.26j81.338.58j93.1465.25j81.3

66.26j81.365.25j81.338.58j93.14

Zabc= Ω

The impedance matrix of the transposed line is

For comparison, the impedance matrix of the untransposed line is

Series Impedance of Lines

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Completely Transposed LineIf s1=s2=s3, the line is completely transposed. We get

⎥⎥⎥

⎢⎢⎢

⎥⎥⎥

⎢⎢⎢

⎡=

⎥⎥⎥

⎢⎢⎢

c

b

a

smm

msm

mms

c

b

a

I

I

I

ZZZ

ZZZ

ZZZ

V

V

V

r

r

r

r

r

r

ΣΣΣ

Volts

s

edas D

Dln ksjs)rr(Z ω++=

)ZZZ(Z 13231231

m ++=m

ed D

Dln ksjsr ω+=

where

Ds, Dm = GMR and GMD, respectively

Ω

Ω

Series Impedance of Lines

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Geometric Mean Distance (GMD)Typical three-phase line configurations

3312312m DDDD =

D12 D23

D31

D12

D23D 31

D12 D23

D31

D12

D23

D31

Series Impedance of Lines

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Example: For the same line assume a complete transposition cycle.

We get the average of the mutual impedances.

⎥⎥⎥

⎢⎢⎢

+++

+++

+++

38.58j93.1421.26j81.321.26j81.3

21.26j81.338.58j93.1421.26j81.3

21.26j81.321.26j81.338.58j93.14

Zabc= Ω

The GMD is10’

a b c

10’

feet 6.12)20)(10(10D 3m ==

Ω21.26j81.3Z m +=The impedance of the transposed line is

Series Impedance of Lines

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Phase to Sequence ImpedancesConsider a transmission line that is described by the following voltage equation:

⎥⎥⎥

⎢⎢⎢

ccbcac

bcbbab

acabaa

ZZZ

ZZZ

ZZZ

⎥⎥⎥

⎢⎢⎢

c

b

a

V

V

V

r

r

r

=⎥⎥⎥

⎢⎢⎢

c

b

a

I

I

I

r

r

r

volts

or

abcabcabc IZVrr

=

From symmetrical components, we have

012abc VAVrr

= and 012abc IAIrr

=

Series Impedance of Lines

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Substitution gives

012abc012 IAZVArr

=or012abc

1012 IAZAV

rr−=

which implies that

AZAZ abc1

012−=

Performing the multiplication, we get

⎥⎥⎥

⎢⎢⎢

−+−

+−−

−−+

=

⎥⎥⎥

⎢⎢⎢

0m0s1m1s2m2s

2m2s0m0s1m1s

1m1s2m2s0m0s

2

1

0

ZZZ2ZZZ

Z2ZZZZZ

ZZZZZ2Z

Z

Z

Z

Note: Z012 is not symmetric.

Series Impedance of Lines

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It can be shown that

)ZZZ(Z ccbbaa31

0s ++=

)ZaaZZ(Z cc2

bbaa31

1s ++=

)aZZaZ(Z ccbb2

aa31

2s ++=

)ZZZ(Z cabcab31

0m ++=

)aZZZa(Z cabcab2

31

1m ++=

)ZaZaZ(Z ca2

bcab31

2m ++=

Series Impedance of Lines

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If the line is completely transposed,

s0s ZZ = m0m ZZ =

0ZZ 2s1s == 0ZZ 2m1m ==

The sequence impedance matrix reduces to

⎥⎥⎥

⎢⎢⎢

+

=

⎥⎥⎥

⎢⎢⎢

ms

ms

ms

2

1

0

ZZ00

0ZZ0

00Z2Z

Z

Z

Z

Note: The sequence impedances are completely decoupled.

Series Impedance of Lines

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For a completely transposed line, the equation in the sequence domain is

⎥⎥⎥

⎢⎢⎢

⎥⎥⎥

⎢⎢⎢

⎡=

2a

1a

0a

2

1

0

2a

1a

0a

I

I

I

Z00

0Z0

00Z

V

V

V

r

r

r

r

r

r

where

s

ma21 D

Dln ksjsrZZ ω+==

2ms

3e

da0DD

Dln ksjsr3srZ ω++=

Ω

Ω

Series Impedance of Lines

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Example: For the same line and assuming a complete transposition cycle, find the sequence impedances of the line. 10’

a b c

10’

Ω21.26j81.3Z m +=

Ω38.58j93.14Z s +=

In the previous example, we got

The sequence impedances are

Ω80.110j55.22Z2ZZ ms0 +=+=

Ω17.32j12.11ZZZZ ms21 +=−==

Series Impedance of Lines

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• Self-capacitance

• Mutual-capacitance

a

b

c

w

CagCbg

Ccg

Cwg

Cab

Cac

Caw

Cbc

Cbw

Ccw

Capacitance of Three Phase Lines

Shunt Capacitance of Lines

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Voltage Due to Charged Conductor

Consider two points P1 and P2which are located at distances D1 and D2 from the center of the conductor.

P1

P2

D1

D2

The voltage drop from P1 to P2 is

1

212 D

Dln

2

qv

πε= Volts

Shunt Capacitance of Lines

+q

xra

rax2

qDE

πεε==

rr

Electric Field of a Long ConductorPermitivity of medium

Electric charge

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Capacitance of a Two-Wire LineThe capacitance between two conductors is defined as the charge on the conductors per unit of potential difference between them.Consider the two cylindrical conductors shown.

qa qb

D

a

aab r

Dln

2

qv

πε=

Due to charge qa, we get the voltage drop vab.

Shunt Capacitance of Lines

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D

rln

2

q

r

Dln

2

qv bb

b

bab πεπε

=−=

Due to charge qb, we also get the voltage drop vba.

orb

bba r

Dln

2

qv

πε=

Applying superposition, we get the total voltage drop from charge qa to charge qb.

D

rln

2

q

r

Dln

2

qv bb

a

aab πεπε

+=

Since qa+qb=0, we get

ba

2a

ab rr

Dln

2

qv

πε= Volts

Shunt Capacitance of Lines

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In general, ra=rb. We getr

Dln

qv a

ab πε= Volts

The capacitance between conductors is

r

DlnV

qC

ab

aab

πε== Farad/meter

The capacitance to neutral is

r

Dln

2C2CC abbnan

πε=== Farad/meter

Shunt Capacitance of LinesSelf-Capacitance

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Mutual Capacitance

In capacitance calculations, the earth is assumed as a perfectly conducting plane. The electric field that results is the same if an image conductor is used for every conductor above ground.

Mirror Conductors below ground

-qb

+qa

+qb

+qc

+qw

-qa-qc

-qw

Dab

Dac

Daw

Haa Hab Hac Haw

Shunt Capacitance of Lines

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The voltage drop from conductor a to ground is

'aa21

a vv =

an

ann

ab

abb

a

aaa D

Hln q...

D

Hln q

r

Hln q(

4

1+++=

πε

)H

Dln q...

H

Dln q

H

rln q

an

ann

ab

abb

aa

aa −−−−

Combining common terms, we get

)D

Hln q...

D

Hln q

r

Hln q(

2

1v

an

ann

ab

abb

a

aaaa +++=

πε

Shunt Capacitance of Lines

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In general, for the kth overhead conductor

k

kkk

bk

bkb

ak

akak r

Hln q...

D

Hln q

D

Hln q(

2

1v +++=

πε

)D

Hln q...

nk

nkn++

Using matrix notation, we get

⎥⎥⎥⎥

⎢⎢⎢⎢

⎥⎥⎥⎥

⎢⎢⎢⎢

=

⎥⎥⎥⎥

⎢⎢⎢⎢

n

b

a

nnncnbna

bnbcbbba

anacabaa

n

b

a

q

q

q

P...PPP

P...PPP

P...PPP

v

v

v

MMMMMMM

Shunt Capacitance of Lines

k

kkkk r

Hln

2

1P

πε=

kj

kjkj D

Hln

2

1P

πε=

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Inversion of matrix P gives

⎥⎥⎥⎥

⎢⎢⎢⎢

+−−−

−−+−

−−−+

=

nnncnbna

bnbcbbba

anacabaa

C...CCC

C...CCC

C...CCC

CMMMMM

Shunt Capacitance of Lines

[ ] [ ][ ]qPv =

Since, q = Cv, ,then [ ] [ ] 1PC −=

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The Shunt Admittance is

⎥⎥⎥⎥

⎢⎢⎢⎢

+−−−

−−+−

−−−+

=

nnncnbna

bnbcbbba

anacabaa

bus

Cj...CjCjCj

Cj...CjCjCj

Cj...CjCjCj

Y

ωωωω

ωωωω

ωωωω

Shunt Capacitance of Lines

The difference between the magnitude of a diagonal element and its associated off-diagonal elements is the capacitance to ground. For example, the capacitance from a to ground is

anacabaaag C...CCCC −−−−=

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Capacitance of a Transposed Line

Pos.1

Pos.2

Pos.3

Phase c

Phase a

Phase b

aq

bq

cq

Section 2Section 1 Section 3

s31 s3

1 s31

Shunt Capacitance of Lines

r

Dln

2

v

qCCC

man

acnbnan

πε====

The capacitance of phase a to neutral is

Farad/meter, to neutral

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fC2

1xc π

=

r

Dln 10 x

f

862.2x m9

c =

r

Dln 10 x

f

779.1x m6

c =

Ω-meter, to neutral

Ω-mile, to neutral

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

Capacitive Reactance

Shunt Capacitance of Lines

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Using matrix notation, we have

Sequence Capacitance

abcabcabc VYIrr

= abcabcabc VCjIrr

ω=

From and , we get012abc VAVrr

= abcabcabc VYIrr

=

012abc012 VACjIArr

ω=or

012abc1

012 VACAjIrr

−= ω

Thus, we have

ACAC abc1

012−=

Shunt Capacitance of Lines

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For a completely transposed line,

ccbbaa0s CCCC ===

acbcab0m CCCC ===Substitution gives

⎥⎥⎥

⎢⎢⎢

+

+

)CC(00

0)CC(0

00)C2C(

0m0s

0m0s

0m0s

=C012

or

0m0s0 C2CC −= 0m0s21 CCCC +==

Shunt Capacitance of Lines

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Example: Determine the phase and sequence capacitances of the transmission line shown. The phase conductors are 477 MCM ACSR 26/7 whose radius is 0.0357 ft. The line is 50 miles long and is completely transposed.Calculate distances

Haa=Hbb=Hcc=80 ft

Hab=Hbc=81.2 ft

Hac=84.8 ftFind the P matrix

a

aa

0ccbbaa r

Hln

2

1PPP

πε===

14’

a b c

14’

40’

Shunt Capacitance of Lines

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For air, 10 x36

1 9-0 π

ε = Farad/meter

Substitution gives

0357.0

80ln 10 x 18P 9

aa =

910 x 86.138= Meter/Farad610 x 29.86= Mile/Farad

ab

ab

0bcab D

Hln

2

1PP

πε==

Similarly, we get

610 x 66.19= Mile/Farad

Shunt Capacitance of Lines

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The P matrix can be shown to be

⎥⎥⎥

⎢⎢⎢

29.8666.1939.12

66.1929.8666.19

39.1266.1929.86

=P

Using matrix inversion, we get the C matrix.

⎥⎥⎥

⎢⎢⎢

−−

−−

−−

34.1254.219.1

54.275.1254.2

19.154.234.12

=C

x 106 mi/F

x 10-9 F/mi

Shunt Capacitance of Lines

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For 50 miles, we get

⎥⎥⎥

⎢⎢⎢

−−

−−

−−

17.627.160.0

27.138.627.1

60.027.117.6

=C x 10-7 F

The capacitances to ground areF43.0CCCC acabaaag μ=−−=

F38.0CCCC bcabbbbg μ=−−=

F43.0CCCC acbccccg μ=−−=Since the line is transposed,

F41.0)CCC(C cgbgag31

0g μ=++=

Shunt Capacitance of Lines

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The self- and mutual capacitances are

F62.0)CCC(C ccbbaa31

0s μ=++=

F105.0)CCC(C cabcab31

0m μ=++=

The sequence capacitances are

0m0s0 C2CC −=

0m0s21 CCCC +==

a

b

cCg0

Cg0Cg0

Cm0

Cm0

Cm0

F41.0 μ=

F725.0 μ=

Shunt Capacitance of Lines

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ZccZcbZca

ZbcZbbZba

ZacZabZaa

YccYcbYca

YbcYbbYba

YacYabYaa

YccYcbYca

YbcYbbYba

YacYabYaa

[Z]

[Y]/2 [Y]/2

[Ikabc]

[IiABC]

[Z]-1+[Y]/2-[Z]-1

-[Z]-1[Z]-1+[Y]/2=

6x1 6x16x6

[Vkabc]

[ViABC]

IiABC Ik

abc

ViABC Vk

abc

Nodal Admittance Matrix Model

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3’ 3’1’

4’

24’

A B C

NPhase Conductor336,400 26/7 ACSR

Neutral Conductor4/0 6/1 ACSRLength: 300 ft.

Example

Nodal Admittance Matrix Model

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Data Requirements

� Phasing

� Configuration

� System Grounding Type

� Length

� Phase Conductor Type, Size & Strands

� Ground/Neutral Wire Type, Size & Strands

� Conductor Spacing

� Conductor Height

� Earth Resistivity

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Ha

VerticalOne Ground Wire

(b)

Dbc

DabDca

Hc HbHg

a

c

b

Hg

TriangularOne Ground Wire

(c)

DabDbc

Dca

Ha Hc Hb

ac

bDca

Horizontal One Ground Wire

(a)

Ha

Dab Dbc

Hb HgHc

a cb

Configuration, Spacing, and Height(Subtransmission Lines)

Distribution Line Models

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Triangular Two Ground Wires

(e)

Dgg

ParallelTwo Ground Wires

(f)

D12

Circuit No. 1

Circuit No. 2

Dgg

Horizontal Two Ground Wires

(d)

Dgg

Line Spacing (Ground Wires)

Distribution Line Models

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A

N

1-Phase (A)

A B

N

V-Phase (AB)

A

N

B

B A

N

V-Phase (BA)

B

N

A

Note:N – Consider

the grounded neutral as Ground Conductor for Hg

Hg

A B C

N

3-Phase (ABC)

A

N

B

C

C A B

N

3-Phase (CAB)

C

N

A

B

B C A

N

3-Phase (BCA)

B

N

C

A

Configuration, Spacing, and Height (Distribution Lines)

Distribution Line Models

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