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Overhead Conductor Overhead Spacer Cable Underground Cable Three-Conductor Cable Service Cables

Power Systems I

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Transmission Lines


ACSR Aluminum Conductor with inner Steel Reinforced strands ACAR Aluminum Conductor with inner Al allow Reinforced strands ACSR/AW Aluminum Conductor with inner Alumoweld Steel Reiforced strands Aluminum - current carrying member Steel - structural support

Power Systems I

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Overhead Conductors


Where conductor close proximity is required Insulating jacket surrounds each conductor Plastic spacers keep conductors from coming in contact with one another

Power Systems I

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Overhead Cable


Power Systems I

Cables


Underground transmission and distribution cables Semiconducting material surrounds the conductor to grade the electric field Plastic jacket provides insulation and protection Neutral strands for an outer shell for protection and return currents

Power Systems I

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Cables


u

u

u

ρ = conductor resistivity l = conductor length A = conductor crosssectional area

Rnew = Rold

Rac = 1.02 ⋅ Rdc

ρl Rdc = A

TAl = 228°C

T + told

increased resistance at conductor temperature rises wiring is rated for 65°C, 75°C, or 90°C T + t new ambient temperature is 20°C

Temperature effects

n

skin effect at 60 Hz:

ac resistance

u

n

dc resistance

u

Line resistance

Power Systems I

l

l

Transmission Line Parameters


Power Systems I

φ = ∫ B ⋅ da

Integral of the flux density that is normal to a defined area A

B = µH

Magnetic Flux

Integral of the scalar product of a closed path ie and the magnetic field equals the encircled current

B

A

H

φ

A=

B=

H=

F = ∫ H ⋅ d l = ie

Γ

Γ=

Ampere’s circuital law

Review of Magnetics and Inductance


I

I

I

Power Systems I

φ ∑ ∫ B ⋅ da ∑ ∫ µH ⋅ da ∑ = = =

λ L= I

Inductance

i =1

λ = ∑ φi

N

Flux Linkage

Review of Magnetics and Inductance


infinite straight wire is an approximation of a reasonably long wire

u

u

u

u

u

Image the wire to close at +/- infinity, establishing a kind of “one-turn coil� with the return path at infinity Straight infinitely long wire of radius r Uniform current density in the wire. Total current is Ix Flux lines form concentric circles (i.e. H is tangential) Angular symmetry - it suffices to consider Hx

Assumptions:

u

Conditions:

Power Systems I

l

l

Inductance of a Single Conductor


→ Hx =

µ0 I I → = x x B x 2 2 2π r 2π r

Power Systems I

r

µ0 I r 3 µ0 I x dx = λint = ∫ dλx = 4 ∫ 2π r 0 8π 0

µ0 → Lint = = 0.5 ×10 −7 8π

x2 µ0 I 3 µ0 I x dx xdx → dλx = 2 dφ x = dφ x = Bx dx = 2 4 2π r 2π r r

Ix I = π r 2 π x2

0

Case 1: Points inside of the conductor (x < r)

⋅ dl = I x

l

x

Ix ⇒ H= 2π x

General:

l

∫H

2πx

Inductance of a Single Conductor


D1

λext = ∫ dλx =

D2

µ0 I 2π

µ 0 I D2 1 d x = ln ∫D x D1 2π 1

D2

→ Lext = 2 × 10 −7 ln

µ0 I I x = I → Bx = µ 0 H x = 2π x µ I µ I dφ x = Bx dx = 0 dx → dλx = dφ x = 0 dx 2π x 2π x

Case 2: Points outside of the conductor (x > r)

Power Systems I

l

Inductance of a Single Conductor

D2 D1


r ′ = re −1 4 = DS

D r1

r1

Power Systems I

−7

D  D D  −7  −7   L = 2 ×10  ln −1 4  = 2 × 10  ln  = 2 × 10  ln  re   r′   DS

−7

  1  L1 = 2 × 10  ln −1 4 + ln D    r1e r1 = r2 L1 = L2 = L

L1 = L1(int ) + L1( ext ) = 0.5 × 10 −7 + 2 × 10 −7 ln

D r1

  

conductors of radii r1 and r2, separated by a distance D

L1( ext ) = 2 × 10 −7 ln

l

Inductance of a Single-Phase Line

D

r2


 1  ln  L22 = 2 ×10 −7  r1′ 

Power Systems I

 1  ln   r2′   1  ln   D

λ1 = L11 I1 − L12 I1 I1 = − I 2 → λ2 = − L21 I 2 + L22 I 2

L12 = L21 = −2 ×10 −7 (ln D ) = 2 ×10 −7

L11 = 2 ×10

−7

λ1 = L11 I1 + L12 I 2 λ2 = L21 I1 + L22 I 2

From the 2 conductor case:

Flux Linkage - Self and Mutual Inductances


Power Systems I

λi = 2 ×10 −7

j≠i

n   1 1  I i ln + ∑ I j    ′ r D = 1 j i ij  

j =1

λi = Lii I i + ∑ Lij I j

n

I1 + I 2 + L + I i + L + I n = 0

General Case:

Total Inductance

j≠i


−7

1 1 1  λa = 2 × 10  I a ln + I b ln + I c ln  r′ D D  1 1  λa = 2 × 10 −7  I a ln − I a ln  r′ D  D −7 λa = 2 × 10 I a ln r′ D L = 0.2 ln DS

I a + Ib + Ic = 0

Symmetrical spacing

Power Systems I

l

Ic

Inductance of Three-Phase Lines

D D

D

Ia

Ib


−7

1 D12 1 ln r′ 1 D32 ln

ln

1   D13  1  ln D23  1  ln r ′ 

 1 1 1    + I c ln λa = 2 ×10  I a ln + I b ln r′ D12 D13    1 1 1   + I c ln λb = 2 ×10 −7  I a ln + I b ln ′ r D21 D23    1 1 1   + I c ln λc = 2 ×10 −7  I a ln + I b ln r′ D31 D32   = LI

Asymmetrical spacing

 1  ln ′  r 1 L = 2 × 10 −7 ln  D21  1  Power Systems I  D31

l

Inductance of Three-Phase Lines

Ic

D31 D23

D12

Ia

Ib


horizontal or vertical configurations are most popular Symmetry is lost - unbalanced conditions

u

c a

a b

b c

a

b

a

c

Each phase occupies each position for the same fraction of the total length of the line

Power Systems I

u

b

c

Average inductance of each phase will be the same

restore balanced conditions by the method of transposition of lines

u

u

The practice of equilateral arrangement of phases is not convenient

position a 1 b 2 c 3

l

l

Transposition

c a

b


D2

Power Systems I

q =Cv

Capacitance

D1

v12 = vD1 − vD2 = − ∫ E ⋅ dl

Electric field

D=εE

Electric field

A

q e = ∫ D ⋅ da

Gauss’s law

R

Review of Electric Fields

D Gaussian Surface

h


q1 D ln 2πε 0 r

q D2 q ln dx = 2πε 0 x 2πε 0 D1 R

h

Power Systems I

q D ln v12 = πε 0 r

q1 D q2 r ln + ln v12 = v12(q1) + v21(q 2 ) = 2πε 0 r 2πε 0 D

q2 r v21(q 2 ) = ln 2πε 0 D

v12 (q1) =

D1

v12 = ∫

D2

Infinite Straight Wire

D

q C= v 2π ε C= D ln r

Infinite wire of radius r


c

C

C

a

n

   

)

per mile per phase

GMDφ = geometeric mean distance between conductors conductor radius rφ=

0.0389  GMDφ log10   r  φ

b

C

C=

Equilateral spacing

Power Systems I

l

Three-Phase Capacitance


12 in

conductor R: 0.3263 Ω/mile GMR: 0.0244 ft Dia.: 0.720 in

44 in

Power Systems I

44 in

)

) = 149.9

Ω mi

)PLSKV

(0.720 in ) ⋅ 121 = 0.03 ft 0.0389 = 0.177 log10 (4.73 0.03)

1 2

X C = 1 (2π 60 ⋅ 0.177

C=

rφ = 12 dia =

 4.73   10   0.0244 

(60) log

60 = 0.326 + j 0.639 Ω / mi

Z a = (0.3263) + j 0.2794

= 56.8 in = 4.73 ft

GMDφ = 3 d12 d 23 d13 = 3 (45.6 ) (88) (45.6 )

Calculate the resistance, inductive reactance, and capacitive reactance per phase and rated current carrying capacity for the overhead line shown. Assume the line operates at 60 Hz

Example


D12 3 conductors

D12

2 conductors

D13

D12 4 conductors

D13

D14

Commonly used to reduce the electric field strength at the conductor surface Used on overhead lines above 230 kV Conductors are connected in parallel Typical bundled conductor configurations

Power Systems I

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Conductor Bundling


n

i=2

rφ′ = n rφ ⋅ ∏ D1i

Equivalent radius

i=2

GMRφ′ = n GMRφ ⋅ ∏ D1i

n

The use of bundled conductors effects the impedance of the line, the GMRφ , the GMDφ , and the equivalent radius GMDφ : the distance between the center of each bundle is used GMRφ :

Power Systems I

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Conductor Bundling


Power Systems I

10 ft

30 ft

20 in

20 in conductor R: 0.1204 â&#x201E;Ś/mile GMR: 0.0403 ft Dia.: 1.196 in

Calculate the resistance, inductive reactance, and capacitive reactance of the overhead line shown. Assume the line operates at 60 Hz

Example


1 2

Power Systems I

X C = 1 (2π 60 ⋅ 0.177

)

) = 116.85

0.0389 C= = 0.0227 log10 (39.15 0.7568)

rφ = 12 dia =

Ω mi

)PLSKV

(1.196 in ) ⋅ 121 = 0.0498 ft rφ′ = 4 (0.0498) (1.67 ) (1.414 ) (1.67 ) = 0.7568 ft

 39.15   10   0.7178 

(60) log

60 = 0.0301 + j 0.485 Ω / mi

Z a = (0.0301) + j 0.2794

GMRφ = 4 (0.0403) (1.67 )(1.414 ) (1.67 ) = 0.7178 ft

GMDφ = 3 D12 D23 D13 = 3 (31.6 ) (60 )(31.6 ) = 39.15 ft

R′ = 14 ⋅ 0.1204 = 0.0301 Ω / mi

Example


Voltages are expressed as phase-to-neutral Currents are expressed for one phase The three phase system is reduced to an equivalent single-phase

Line parameters: R, L, C, & G

u

u

depend on the length and the voltage level short, medium, and long length line models

Three types of models

u

All lines are made up of distributed series inductance and resistance, and shunt capacitance and conductance

u

u

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Transmission lines are represented by an equivalent circuit with parameters on a per-phase basis

Power Systems I

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Transmission Line Modeling


circuit equations

matrix form

u

u

VS   A B  VR   I  = C D   I    R  S 

I S = C VR + D I R

VS = A VR + B I R

All transmission line models may be described as a twoport network The ABCD two-port network is the most common representation The network is described by the four constants: A, B, C, & D Network equations:

Power Systems I

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ABCD Two-Port Network


The line length is less than 50 miles (80 km), or The line voltage is not over 69 kV

VS

RL

XL VR

The shunt capacitance and conductance are ignored The line resistance and reactance are treated as lumped parameters

Circuit of the short model

u

u

Modeling of the transmission line parameters

u

u

The short transmission line model may be used when

Power Systems I

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Short Transmission Line Model


Gen.

IS

VS

= VR + I R Z

VS = VR + I R ( R + j ω s L)

IS = IR

Z=R+jωL

Circuit analysis of the short line model

Power Systems I

l

Short Transmission Line Model

VR

Load

IR


Matrix representation:

ABCD values:

u

u

Circuit Equations:

Power Systems I

l

D = A=1

C =0

A=1 B = Z line

VS  1 Z line  VR   I  = 0 1   I    R  S 

IS = IR

VS = VR + Z line I R

Two-Port Representation


R = 0.15 Ω/km L = 1.3263 mH/km

(

381 MVA load at 0.8 lagging pf at 220 kV

)

(

)

Z = 6 + j 20 Ω 220,000∠0° VR = = 127,000∠0° 3 S R ( 3φ ) = 381∠ cos −1 0.8 = 381∠36.9° = 304.8 + j 228.6 MVA

Z = (r + j ω L ) l = 0.15 Ω + j 2 π × 60 ×1.3263 ×10 −3 ⋅ 40

u

Find V, S, V.R., and η at the sending end of the line for

u

40 km, 220 kV transmission line has per phase

Power Systems I

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Short Transmission Line Example


3 VR*

381× 106 ∠ − 36.9° = 1000∠ − 36.9° A = 3 × 127,000∠0°

Power Systems I

250 - 220 304.8 × 100% = 13.6% η = ×100% = 94.4% VR% = 220 322.8

= 322.8 + j 288.6 = 433∠41.8° MVA

S S (3φ ) = 3 ⋅ VS ⋅ I S* = 3 ⋅ (144,330∠4.93°)(1000 ∠ − 36.9°)

VS − LL = 3 ⋅ VS = 250 kV

= 144,330∠4.93°

VS = VR + Z I R = 127,000∠0° + (6 + j 20)(1000∠ − 36.9°)

IR =

S R* ( 3φ )

Short Transmission Line Example


The line length is greater than 50 miles (80 km) The line length is less than 150 miles (250 km)

VS

YC/2

RL

XL

YC/2

VR

Half of the shunt capacitance is considered to be lumped at each end of the line The line resistance and reactance are treated as lumped parameters

Circuit model:

u

u

Modeling of the transmission line parameters

u

u

The medium transmission line model may be used when

Power Systems I

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Medium Transmission Line Model


Gen.

IS

VS

½ YC

IS C

R

R

R

Z line YC 4

R

YC 2

YC 2

Z line YC 2

S

line R

VS = VR + Z line I R + Y2C VR

( ) )V + Z I = (1 + = (I + V ) + V )V + (1 + = Y (1 +

½ YC

Z=R+jωL

Z line YC 2

Circuit analysis of the short line model

Power Systems I

l

)I

Medium Transmission Line Model

R

VR

Load

IR


Matrix representation:

ABCD values:

u

u

Circuit Equations:

Power Systems I

l

C

R

R

R

Z line YC 4

(

A = 1+ C = YC 1 +

Z line YC 2 Z line YC 4

(

)

Z line YC 2

)I

R

Z line  VR   Z line YC   1 + 2  IR 

S

line R

B = Z line D = 1 + Zline2 YC

)

YC 2

YC 2

R

VS   1 + Z line2 YC I  =  Z line YC Y 1 +  S  C 4

IS

Z line YC 2

VS = VR + Z line I R + Y2C VR

( ) )V + Z I = (1 + = (I + V ) + V )V + (1 + = Y (1 +

Two-Port Representation


R = 0.036 Ω/km L = 0.80 mH/km

(

270 MVA load at 0.8 lagging pf at 325 kV

)

(

)

325,000∠0° VR = = 187,600∠0° 3 S R (3φ ) = 270∠ cos −1 0.8 = 270∠36.9° = 216 + j162 MVA

Y = ( j ω C ) l = j 2 π × 60 × 0.0112 × 10 −6 ⋅130 = j 0.549 siemens

(

= 4.68 + j 39.2 Ω

Z = (r + j ω L ) l = 0.036 Ω + j 2 π × 60 × 0.8 ×10 −3 ⋅130

u

)

C = 0.0112 uF/km

Find V and S at the sending end of the line for

u

130 km, 345 kV transmission line has per phase

Power Systems I

l

l

Medium Transmission Line Example


3 VR*

270 ×106 ∠ − 36.9° = = 480∠ − 36.9° A 3 ×187,600∠0°

(

)

Power Systems I

= 421.5∠ − 25.58°

(480∠ − 36.9°)(0.989 + j 0.001284)

I S = C VR + D I R = (187,600∠0°) − 3.53 ×10 −7 + j 5.46 ×10 − 4 +

= 199,160∠4.02°

(480∠ − 36.9°)(4.68 + j39.2)

0.989 + j 0.001284 4.68 + j 39.2   ABCD =   −7 −4 3 . 53 10 5 . 46 10 0 . 989 0 . 001284 − × + × + j j   VS = A VR + B I R = (187,600∠0°)(0.989 + j 0.001284 ) +

IR =

S R* (3φ )

Medium Transmission Line Example


Power Systems I

= 325

345 0.989 + j 0.001284 - 325

×100% = 7.3%

= 218.9 + j124.2 MVA pf = 0.87 VR ( NL ) − VR ( FL ) VS ( FL ) / A − VR ( FL ) VR% = ×100% = × 100% VR ( FL ) VR ( FL )

S S (3φ ) = 3 ⋅ VS ⋅ I S* = 3 ⋅ (199,160∠4.02°)(421 ∠ − 25.58°)

VS − LL = 3 ⋅ VS = 345 kV

Medium Transmission Line Example


The line length is greater than 150 miles (250 km)

u

u

u

Accuracy obtained by using distributed parameters The series impedance per unit length is z The shunt admittance per unit length is y

Modeling of the transmission line parameters

u

The long transmission line model are used when

Power Systems I

l

l

Long Transmission Line Model


V(x + ∆x)

I(x + ∆x) y ∆x ∆x l

y ∆x

I(x) V(x) x

IR VR

Power Systems I

V ( x + ∆x) = V ( x) + z ∆x I ( x) I ( x + ∆x) = I ( x) + y ∆x V ( x + ∆x) V ( x + ∆x) − V ( x) I ( x + ∆x) − I ( x) = z I ( x) = y V ( x + ∆x) ∆x ∆x dV ( x) dI ( x) limit as ∆x → 0 = z I ( x) limit as ∆x → 0 = y V ( x) dx dx

VS

IS

z ∆x

Long Transmission Line Model


VS

I+∆I

V+∆V ∆x

V

I

x

VR

Load

IR

d 2V ( x) dI ( x) d 2 I ( x) dV ( x) =z =y 2 2 dx dx dx dx d 2V ( x) d 2 I ( x) = z ( y V ( x) ) = y ( z I ( x) ) 2 2 dx dx 2 γ = z y propagation constant Power Systems I

Gen.

IS

Long Transmission Line Model


(r + jωL )(g + jωC )

Power Systems I

@x=0⇒

(

VR + I R Z c A1 = 2

(

(

)

VR − I R Z c A2 = 2

)

)

1 dV ( x) γ I ( x) = = A1 eγ x − A2 e −γ x = yz A1 eγ x − A2 e −γ x z dx z 1 Zc = z y I ( x) = A1 eγ x − A2 e −γ x characteristic impedance Zc

γ = α + jβ = z y =

d 2V ( x) 2 V ( x) γ = 2 dx V = A1 eγ x + A2 e −γ x

Long Transmission Line Model


1 e Zc

)

VR +

VR + Z c

−x y z

yz

e

yz

x yz

ex

) y z)I

+e 2

yz

−x y z

− e−x 2

Power Systems I

c

R

R

V ( x) = cosh x y z VR + Z c sinh x y z I R

(

−e 2

+ e−x 2

x yz

yz

( 1 I ( x) = sinh (x y z )V + cosh (x Z

I ( x) =

V ( x) =

ex

VR + Z c I R x y z VR − Z c I R − x y z V ( x) = e e + 2 2 V Z +I V Z −I I ( x) = R c R e x y z − R c R e − x y z 2 2

IR

IR

Long Transmission Line Model

eθ + e −θ cosh θ = 2

eθ − e −θ sinh θ = 2

Hyperbolic Functions


Power Systems I

γ = zy Zc =

z y

 cosh (γ l ) Z c sinh (γ l )  ABCD =  1  sinh (γ l ) cosh (γ l )    Zc

1 IS = sinh (γ l ) VR + cosh (γ l ) I R Zc

let x → l VS = cosh (γ l ) VR + Z c sinh (γ l ) I R

Two-Port Representation


Y’/2

′ ′

)

)

(

)

→ Z ′ = Z c sinh (γ l )

Y’/2

VR

Y′ 1 cosh (γ l ) − 1 1 γ l tanh  = (cosh (γ l ) − 1) = =  Z c sinh (γ l ) Z c 2 Z′  2 

Z′Y′ Z′Y′ ′ I S = Y 1 + 4 VR + 1 + 2 I R

(

VS = 1 + Z 2Y VR + Z ′I R

(

Find the values for Z’ and Y’

VS

Represent a long transmission line as a pi-model for circuit analysis Z’ The circuit:

Power Systems I

l

l

l

Pi-Model of a Long Transmission Line


z = 0.045 + j 0.4 Ω/km Y = j 4. 0 uS/km

γ = zy = (0.045 + j 0.4 )(4 × 10 −6 ) = 7.104 × 10 −5 + j 0.001267 Z ′ = Z c sinh (γ l ) = 10.88 + j 98.36 Y′ 1 γ l tanh =  = j 0.001008 2 Zc  2 

Zc =

0.045 + j 0.4 z = = 316.7 - j17.76 −6 4 × 10 y

Find ABCD for a pi model of the long transmission line

u

250 km, 500 kV transmission line has per phase

Power Systems I

l

l

Long Transmission Line Example


)

′ ′

)

Power Systems I

C = Y ′ 1 + Z 4Y = j 0.00100

(

B = Z ′ = 10.88 + j 98.36

(

Z ′ = 10.88 + j 98.36 Y′ = j 0.001008 2 ′ ′ A = D = 1 + Z 2Y = 0.9504 + j 0.0055

Long Transmission Line Example


Lecture Power system