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Unformatted text preview: Overhead Conductor
Overhead Spacer Cable
Underground Cable
ThreeConductor Cable
Service Cables Power Systems I 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 Overhead Conductors Where conductor close
proximity is required
In s u la t in g ja c k e t
surrounds each conductor
Plastic spacers keep
conductors from coming in
c o n t a c t w i th o n e a n o t h e r Power Systems I 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 Cables 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 skin effect
at 60 Hz: ac resistance dc resistance Line resistance Power Systems I Transmission Line Parameters Power Systems I Integral of the
flux density that
is normal to a
defined area Magnetic Flux A φ = ∫ B ⋅ da B = µH 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 Image the wire to close at +/ infinity, establishing a
kind of “oneturn 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: Conditions: Power Systems I Inductance of a Single Conductor ⋅ dl = I x → Hx = µI
I
x → Bx = 0 2 x
2π r 2
2π r Power Systems I µ0 I r 3
µ0 I
λint = ∫ dλx =
x dx =
4∫
2π r 0
8π
0 r µ0
→ Lint =
= 0.5 ×10 −7
8π x2
µ0 I
µ0 I 3
x dx
dφ x = Bx dx =
xdx → dλx = 2 dφ x =
2
4
2π r
r
2π r I
I
= x2
π r2 π x Case 1: Points inside of the conductor (x < r) 0 ∫H x Ix
⇒ H=
2π x General: 2πx Inductance of a Single Conductor D1 λext = ∫ dλx = D2 µ0 I
2π µI D
1
d x = 0 ln 2
∫x
2π
D1
D1 D2 → Lext = 2 × 10 −7 ln µ0 I
I x = I → Bx = µ 0 H x =
2π x
µI
µI
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 Inductance of a Single Conductor D2
D1 r + = re −1 4 = DS D
r1 r1 Power Systems I −7 D ȹ
D ȹ
D
ȹ − 7 ȹ − 7 ȹ ȹ ln
L = 2 ×10 ȹ ln −1 4 ȹ = 2 × 10 ȹ ln ȹ = 2 × 10 ȹ ȹ re Ⱥ
ȹ r + Ⱥ
ȹ DS −7 ȹ
ȹ 1
ȹ ln −1 4 + ln D ȹ
L1 = 2 × 10 ȹ ȹ
r1e
Ⱥ
ȹ r1 = r2
L1 = L2 = L L1 = L1(int ) + L1( ext ) = 0.5 × 10 −7 + 2 × 10 −7 ln D
r1 ȹ
ȹ
ȹ
Ⱥ conductors of radii r1 and r2, separated by a distance D L1( ext ) = 2 × 10 −7 ln Inductance of a SinglePhase Line D r2 ȹ 1 ȹ
ȹ ln ȹ L22 = 2 ×10 −7
ȹ r + ȹ
ȹ 1 Ⱥ Power Systems I ȹ 1 ȹ
ȹ ln ȹ
ȹ r + ȹ
ȹ 2 Ⱥ
ȹ 1 ȹ
ȹ ln ȹ
ȹ D Ⱥ 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 1 = 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 ȹ 1n
1 ȹ
ȹ I i ln + ∑ I j
ȹ
ȹ ri+ j =1 Dij ȹ
ȹ Ⱥ j =1 λi = Lii I i + ∑ Lij I j n I1 + I 2 + + I i + + I n = 0 General Case: Total Inductance j≠i −7 1
1
1 ȹ
ȹ λa = 2 × 10 ȹ I a ln + I b ln + I c ln ȹ
r + D
D Ⱥ
ȹ 1
1 ȹ
ȹ λa = 2 × 10 −7 ȹ I a ln − I a ln ȹ
r + D Ⱥ
ȹ 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 Ic Inductance of ThreePhase Lines D
D D Ia Ib −7 1
D12
1
ln
r + 1
D32
ln ln 1 Ⱥ
Ⱥ
D13 Ⱥ
1 Ⱥ
ln
D23 Ⱥ
1 Ⱥ
ln Ⱥ
r + Ⱥ
Ⱥ ȹ 1
1
1 ȹ
ȹ I a ln + I b ln
ȹ
+ I c ln
λa = 2 ×10 ȹ r + D12
D13 ȹ
ȹ Ⱥ
ȹ 1
1
1 ȹ
ȹ
+ I c ln
λb = 2 ×10 −7 ȹ I a ln + I b ln
ȹ + r
D21
D23 ȹ
ȹ Ⱥ
ȹ 1
1
1 ȹ
ȹ
+ I c ln
λc = 2 ×10 −7 ȹ I a ln + I b ln
ȹ r + D31
D32 ȹ
ȹ Ⱥ
= LI Asymmetrical spacing Ⱥ 1
Ⱥ ln + Ⱥ r
1
L = 2 × 10 −7 Ⱥln
Ⱥ D21
Ⱥ 1
Ⱥ
Power Systems I
Ⱥ D31
Ⱥ Inductance of ThreePhase Lines Ic D31
D23 D12 Ia Ib horizontal or vertical configurations are most popular
Symmetry is lost  unbalanced conditions 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 b c Average inductance of each phase will be the same restore balanced conditions by the method of
transposition of lines The practice of equilateral arrangement of phases is not
c o n v e n ie n t position
a
1
b
2
c
3 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
v12 = v12(q1) + v21(q 2 ) =
ln +
ln
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 ȹ
ȹ
ȹ
Ⱥ per mile per phase GMDφ = geometeric mean distance
between conductors
rφ=
conductor radius 0.0389
ȹ GMDφ
log10 ȹ ȹ r
ȹ φ b C C= Equilateral spacing Power Systems I ThreePhase Capacitance 12 i n conductor
R: 0.3263 Ω/mile
GMR: 0.0244 ft
Dia.:
0 .7 2 0 i n 44 i n Power Systems I 44 i n ) = 149.9 Ω mi 1
(0.720 in ) ⋅ 12 = 0.03 ft 0.0389
= 0.177
log10 (4.73 0.03) 1
2 X C = 1 (2π 60 ⋅ 0.177 C= rφ = 1 dia =
2 ȹ 4.73 ȹ
ȹ
10 ȹ ȹ 0.0244 Ⱥ (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 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
u sed
GMRφ : Power Systems I Conductor Bundling Power Systems I 10 ft 30 ft 20 in 20 in
conductor
R: 0.1204 Ω/mile
GMR: 0.0403 ft
Dia.:
1 .1 9 6 i n 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 = 1 dia =
2 Ω mi 1
(1.196 in ) ⋅ 12 = 0.0498 ft
rφ+ = 4 (0.0498) (1.67 ) (1.414 ) (1.67 ) = 0.7568 ft ȹ 39.15 ȹ
ȹ
10 ȹ ȹ 0.7178 Ⱥ (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+ = 1 ⋅ 0.1204 = 0.0301 Ω / mi
4 Example Voltages are expressed as phasetoneutral
Currents are expressed for one phase
The three phase system is reduced to an equivalent singlephase Line parameters: R, L, C, & G depend on the length and the voltage level
short, medium, and long length line models Three types of models All lines are made up of distributed series inductance and
resistance, and shunt capacitance and conductance Transmission lines are represented by an equivalent
circuit with parameters on a perphase basis Power Systems I Transmission Line Modeling circuit equations matrix form ȺVS Ⱥ Ⱥ A B Ⱥ ȺVR Ⱥ
Ⱥ I Ⱥ = ȺC D Ⱥ Ⱥ I Ⱥ
Ⱥ Ⱥ R Ⱥ
Ⱥ S Ⱥ Ⱥ 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 twoport network is the most common
representation
The network is described by the four constants: A, B, C, &
D
Network equations: Power Systems I ABCD TwoPort 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 Modeling of the transmission line parameters The short transmission line model may be used when Power Systems I 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 Short Transmission Line Model VR Load IR A=1
B = Z line ABCD values: D = A=1 C =0 ȺVS Ⱥ Ⱥ1 Z line Ⱥ ȺVR Ⱥ
Ⱥ I Ⱥ = Ⱥ0 1 Ⱥ Ⱥ I Ⱥ
Ⱥ Ⱥ R Ⱥ
Ⱥ S Ⱥ Ⱥ Matrix representation: IS = IR VS = VR + Z line I R Circuit Equations: Power Systems I TwoPort 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 ) = 0.15 Ω + j 2 π × 60 ×1.3263 ×10 −3 ⋅ 40 Find V, S, V.R., and η at the sending end of the line for 40 km, 220 kV transmission line has per phase Power Systems I 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
VR% =
× 100% = 13.6% η =
×100% = 94.4%
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: Modeling of the transmission line parameters The medium transmission line model may be used when Power Systems I 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 C
VS = VR + Z line I R + Y2 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 )I Medium Transmission Line Model R VR Load IR Matrix representation: ABCD values: Circuit Equations: Power Systems I C R YC
2 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 Ⱥ ȺVR Ⱥ
Ⱥ
Z line YC Ⱥ Ⱥ
1 + 2 Ⱥ Ⱥ I R Ⱥ S line R B = Z line
D = 1 + Zline YC
2 ) R R YC
2 R ȺVS Ⱥ Ⱥ 1 + Z line YC
2
= Ⱥ
Ⱥ I Ⱥ
YC 1 + Zline YC
Ⱥ S Ⱥ Ⱥ
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 + TwoPort 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 ) = j 2 π × 60 × 0.0112 × 10 −6 ⋅130 = j 0.549 siemens ( = 4.68 + j 39.2 Ω Z = (r + j ω L ) = 0.036 Ω + j 2 π × 60 × 0.8 ×10 −3 ⋅130 ) C = 0.0112 uF/km F i n d V a n d S a t t h e s e n d i n g e n d o f th e l i n e f o r 130 km, 345 kV transmission line has per phase Power Systems I 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 Ⱥ
Ⱥ
ABCD = Ⱥ
− 3.53 ×10 −7 + j 5.46 ×10 − 4 0.989 + j 0.001284Ⱥ
Ⱥ
Ⱥ
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) 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 The long transmission line model are used when Power Systems I Long Transmission Line Model V(x + Δx) I(x + Δx)
y Δx
Δx
l y Δx I(x)
V(x)
x IR
VR Power Systems I V ( x + Δx) = V ( x) + z Δx I ( x)
I ( x + Δx) = I ( x) + y Δx V ( x + Δx)
V ( x + Δx) − V ( x)
I ( x + Δx) − I ( x)
= z I ( x)
= y V ( x + Δx)
Δx
Δx
dV ( x)
dI ( x)
limit as Δx → 0
= z I ( x) limit as Δx → 0
= y V ( x)
dx
dx VS IS z Δx Long Transmission Line Model VS I + ΔI V+ΔV
Δx 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
γ 2 = 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 = y A1 eγ x − A2 e −γ x
z
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)
= γ 2V ( x)
dx 2
V = A1 eγ x + A2 e −γ x Long Transmission Line Model 1e
Zc ( −e
2 + e−x
2 x yz yz ) 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 (
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 Ⱥ cosh (γ ) Z c sinh (γ )Ⱥ
Ⱥ
ABCD = Ⱥ 1
Ⱥ sinh (γ ) cosh (γ ) Ⱥ
Ⱥ Z c
Ⱥ 1
IS =
sinh (γ ) VR + cosh (γ ) I R
Zc let x →
VS = cosh (γ ) VR + Z c sinh (γ ) I R TwoPort Representation Y’/2 VR ) ( ) Y + 1
cosh (γ ) − 1 1
ȹ γ ȹ
tanh ȹ = (cosh (γ ) − 1) =
=
ȹ
Z c sinh (γ ) Z c
2 Z + ȹ 2 Ⱥ ( → ) + 1 + Z + 4Y + VR + 1 + Z + 2Y + I R
IS = Y + + VS = 1 + Z 2Y VR + Z + I R ( → Z + = Z c sinh (γ ) Y’/2 F i n d th e v a l u e s fo r Z ’ a n d Y ’ VS Represent a long transmission line as a pimodel for
circuit analysis
Z’
The circuit: Power Systems I PiModel 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 ) = 10.88 + j 98.36
Y + 1
ȹ γ ȹ
=
tanhȹ ȹ = j 0.001008
2 Zc
ȹ 2 Ⱥ Zc = z
0.045 + j 0.4
=
= 316.7  j17.76
−6
4 × 10
y Find ABCD for a pi model of the long transmission line 250 km, 500 kV transmission line has per phase Power Systems I 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 ...
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This note was uploaded on 02/01/2012 for the course EEL 4213 taught by Professor Thomasbaldwin during the Fall '11 term at FSU.
 Fall '11
 THOMASBALDWIN

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