Transmission lines and cables
Transmission lines are classified according to their lengths to:
Short: less than 80 km
Medium: from 80 km to 240 km
Long: longer than 240 km
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Short transmission lines
Transmission lines and cables
Is
IR
Vs
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1
Transmission Line Model
= V R + ZI R
I s
=
I R
IR
VR
Z
V s
Is Two-port Vs
network
A = D = 1 B = Z
V s
= AV R + BI R
C = 0
I s
= CV R + DI R
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VR
Example 1, Solution Z = ( r + jω L)l
= 6+
=
S R
=
Ps
=
V R
=
220∠0 3
= 127∠0
∗
3V R
= 1000∠ − 36.87
V S ( L − L) = 3V S P R
A 220 kV, three phase transmission line is 40 km long. The resistance per phase is 0.15 Ω per km and the inductance per phase is 1.3263 mH per km. Use the short line model to find the voltage and power at the sending end, voltage regulation and efficiency when the line is supplying a three phase load of 381 MVA at 0.8 power factor lagging at 220 kV.
j 20 Ω
The receiving voltage per phase is:
I R
Example 1
=
250kV
3 220 × 1000 × cos(36.8) = 304.8 MW
V S
= V R + ZI R = 144.3∠4.93kV
VR =
250 − 220
η =
3 250 × 1000 × cos( 4.93 + 36.8) = 322.8 MW 16:28
220 304.8 322.8
= 13.6%
= 94.4% 16:28
Medium transmission lines
Example 2 A three phase 60 Hz, completely transposed 345kV, 345kV, 200 km line has two 795,000 cmil 26/2 ACSR conductors per bundle and the following positive sequence line constants: z = 0.032 + j0.35 Ω /km, y = j4.2*10-6 S/km. Full load at the receiving end of the line is 700 MW at 0.99 power factor leading and at 95% of rated voltage. Find the following:
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ABCD parameters of the nominal π circuit Sending end voltage Vs, current Is and power Ps. Percent voltage regulation. Thermal limit. Transmission Transmission line efficiency at full load.
V s
= V R + Z I R +
I s
= I R +
I s
V RY
2
= Y 1 +
+
V RY
V sY
2
YZ = 1 + V R + ZI R 2
2
, subsitute the value of Vs
YZ V R 4
+ 1 +
YZ
I R 2
A = D = 1 +
2
B = Z
C = Y 1 +
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YZ
YZ
4
Example 2, Solution From the table in the previous note, the current carrying capacity is:
d)
Example 2, Solution Y = yl = 8.4 ×10
2*0.9 = 1.8 kA which is less than the actual current
PS
e)
=
η =
730.5 MW 700
730.5
V R
=
I R
=
0.95 × 345 3
b)
3 (0.95 × 345)(0.99)
V RNL
= 356.3 A 356.3 − 327.8
327.8
Example 3
AV r + BI r CV r + AI r A C B A
= 820∠ − 88.8
=
200∠78
Then solve for A, B and C and proceed like the previous example.
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C = 8.277 ×10
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= 1.246∠8.11 kA
V S
Example 3, Solution
Z SC =
3
=
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Z OC =
B = Z = 70.29∠84.78 −4
∠90.08
= 189.2∠0
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=
∠90
700∠ cos 0.99
VR =
V S
−4
−1
= 95.8%
c)
I S
A = D = 0.97∠0.159
Z = zl = 70.29∠84.78
a)
= 8.7%
V S
= AV R + BI R = 199.6∠026.14
I S
= CV R + DI R = 1.241∠15.5 kA
Long transmission lines, cont. V ( x + ∆ x) = V ( x) + ( z∆ x) I ( x) V ( x + ∆ x) − V ( x ) ∆ x
I ( x + ∆ x ) = I ( x ) + ( y∆ x )V ( x + ∆x) I ( x + ∆ x) − I ( x )
= zI ( x)
∆ x
Taking the limit as ∆x approaches zero :
Taking the limit
dV ( x)
dI ( x )
= zI ( x)
dx 2
d V ( x) dx
2
Let : γ
2
= z
dx
dx 2
d 2V ( x)
= zy
d 2V ( x)
= zyV ( x )
dx
= yV ( x + ∆ x )
as ∆x approaches zero :
= yV ( x )
dx
dI ( x)
Long transmission lines
− zyV ( x) =
Ω /m
z = R + j L
0
y = G + jω C S/m
2
− γ V ( x) = 0
2
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Long transmission lines, cont.
Long transmission lines, cont.
V ( x) =
V R + Z C I R
2 V R
I ( x ) =
V ( x) = I ( x ) =
e
γ x
+e
2 γ 1 e x
Z C
− γ x
V R + Z C −e
2
x −γ
e
γ x
Z C
x
+
V R − Z C I R
2 V R
+ I R
e
2
−e
eγ
γ x
+
Z C
dV ( x)
2
e−
γ x
dx I ( x ) =
−γ x
2 x −γ e +e 2
x
x
− I R
I R
V ( x) = cosh(γ x)V R + Z c sinh(γ x ) I R
γ x
V R +
V ( x) = A1eγ
e −γ
I ( x ) =
I R
1 Z c
sinh(γ x )V R
+ cosh(γ x) I R
γ z
Z c =
= γ A1e
γ x
+ A2 e − γ A2e
− γ x
( A eγ − A e γ ) = −
x
1
z y
x
2
γ = zy
is called the propagation constant
γ = α + j β = zI ( x)
y z
( A eγ − A e γ ) = A e −
x
1
x
x γ
1
2
− A2 e
Z c
= V(0) =
A1 +A 2 and I R
= I(0) =
A1 - A 2
B = Z c sinh(γ x ) C = 16:28
4
1 Z c
x) sinh(γ
A1 = 16:28
−γ x
is called the characteristic impedance.
Since VR
A = cosh(γ x) = D
−γ x
V R + Z c I R
2
and A2
=
V R − Z c I R
2
Zc
Example 4, Solution Z C =
0.33∠87.14 4.674 ×10
−6
∠90
Example 4
= 266.1∠ − 1.43
γ l = (0.33∠87.14) × (4.67 ×10−4 ∠90) × 300 = 0.373∠88.57 e
γ l
=e
0.0093
×e
j 0.373
= 1.0094∠0.373
e
−γ l
=e
−0.0093
×e
−
j 0.373
z = 0.0165 + j0.3306 Ω /km, y = j4. 674*10-6 S/km. Calculate the exact ABCD parameters. Compare the exact B parameter with that of the nominal π circuit.
= 0.9907∠ − 0.373
sinh(γ l ) = 0.3645∠88.63
cosh(γ l ) = 0.9313∠0.209
A three phase 60 Hz, completely transposed 765kV, 300 km line has the following positive sequence line constants:
Then from this find the A, B, C and D parameters parameters For example B is calculated as follows: B = 266.1∠ − 1.43 × 0.3645∠88.63 = 97.0∠87.2 Busing
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π
model
= Z = 99.3∠87.14
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Lossless lines
Lossless lines
ABCD Parameters A( x ) = cosh(γ x) = D ( x) A( x ) =
e
j β x
+e
− j β x
2
x ) = cos( β
When line losses are neglected, simpler expressions for the line parameters are obtained.
B = Z c sinh( γ x) β
x) = sinh( j β x) = sinh( γ
e j x
−e
2
− j β x
=
j sin( β x)
C =
1 Z c
sinh( γ x) =
j sin( β x) Z c
For lossless line, R=G=0 and hence:
B = jZ c sin( β x )
z = jω L
Ω /m
y = jω C S/m Wavelength A wavelength is the distance required to change the phase of the voltage or current by 2π 2π. 2π 2π 1 = = λ = β ω LC f LC v= 16:28
5
1 LC
Z c
=
γ = zy Velocity of propagation 16:28
z y =
=
jω L jω C
=
The characteristics impedance is called the surge impedance and is pure real
L C
( jω L)( jω C )
=
jω LC = j β
The propagation constant is pure imaginary
Example 5, solution
Example 5
a) For a lossless line:
β = ω LC = 2π × 60 0.97 × 0.0115 ×10−9 Z C =
λ =
v f
L
=
C =
290.43 Ω
1
v=
=
0.001259 rad/km
2.994 × 105 k m/s
=
LC
A three phase 60 Hz, 500kV, 300 km. The line inductance is 0.97 mH/km and its capacitance is 0.0115 µF/km per phase. Assume a lossless line:
a)
Determine the line phase constant β, the surge impedance Zc, velocity of propagation and the line wavelength.
b)
The receiving end rated load is 800 MW, 0.8 power factor lagging at 500 kV, determine the sending end quantities.
4990 k m
b) The receiving end voltage is: The receiving end current is:
I R
V R
=
800 × 106
=
3 × 500 × 103 × 0.8
500∠0
=
288.67∠0 k V
−1
(0.8) = 1154.7∠ − 36.87 A
3 ∠ − cos
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Surge Impedance Loading Surge impedance loading (SIL) is the power delivered by a lossless line to a load resistance equal to the surge impedance Z c.
I R
=
V R
SIL = 3V R I R
Z c
V ( x) = cos( β x)V R
+ jZ c sin( β x ) I R
V ( x) = cos( β x)V R
+ jZ c sin( β x)
*
=
3
V R Z c
2
Example 5, solution The sending end voltage is: V S I S
V R Z c
V ( x ) = (cos( β x) + j sin(β x ))V R V ( x) 16:28
6
=
V R 16:28
=
=
cos( β l )V R
j
1 Z C
+ jZ C sin( β l ) I R = 356.5∠16.1 kV
sin( β l )V R + cos(β l ) I R
= 902.3∠ − 17.9 A
Complex Power Flow Through Transmission Lines = AV R + BI R
V s I R
=
Let
A = A ∠θ A
V S = V S ∠δ
And
B = B ∠θ B
Voltage Profile under different loading conditions
V R
=
V R
-At no-load, IRNL=0 and
∠0
VNL(x)=cos(β (x)=cos(βl) * VRNL
V S ∠δ − A ∠θ A V R ∠0
S R
B ∠θ B
*
= 3V R I
The no-load voltage increases from VS=cos(β =cos(βl) * VRNL at the sending end to VRNL at the receiving end.
R
The real power at the receiving end of the line is:
P R
=
V S ( L − L ) V R ( L − L ) cos(θ B
2
cos(θ B − θ A )
-From previous slide, voltage profile is constant at SIL.
V S ( L − L ) V R ( L − L ) sin(δ )
-For full load, t he receiving voltage will drop depends on the loading conditions.
− δ ) − A V R ( L − L )
B -For short circuit, VR=0
For a lossless line, B=jX, θA=0, θB=90
P R
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=
X
Complex Power Flow Through Transmission Lines The reactive power at the receiving end of the line is:
Q R
=
V S ( L − L ) V R ( L − L ) sin(θ B
− δ ) −
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7
V S ( L − L ) V R ( L − L ) X
cos(δ ) −
So the maximum power that can be delivered will be
2
Pmax
=
V S ( L − L ) V R ( L − L ) X
This value is called the steady-state stability limit of a lossless line. If an attempt was made to exceed this limit, then synchronous machines at the sending end would lose synchronism with those at the receiving end.
For a lossless line, B=jX, θA=0, θB=90
=
Complex Power Flow Through Transmission Lines
A V R ( L − L ) sin(θ B − θ A )
B
Q R
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V R ( L −L ) X
2
cos( β l )
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Example 6
Power Transmission Capabilities
A three phase power of 700 MW is to be transmitted to a substation located 315 km from a source of power. For a preliminary line design assume the following parameters: Vs = 1 per unit, VR = 0.9 per unit, λ=5000 km, Zc=320 Ω and δ=36.87 a) Based on the the practical line loadability equation equation determine a nominal voltage level for the transmission line.
P R
It is required to transmit 9000 MW to a load center 500 km from the plant based on practical line loadability criteria, Determine the number of three phase, 60 Hz lines to transmit this power with 345 kV and 765 kV lines and surge impedance = 297 and 266 ohm respectively. Assume the sending voltage is 1.0 per unit, the receiving voltage = 0.95 per unit and δ=35.
X
For a lossless line:
P R
Example 7
V S ( L − L ) V R ( L − L ) sin(δ )
For planning and other purposes, it is very useful to express the power transfer formula in terms of SIL.
b) For the transmission transmission voltage obtained obtained in (a) calculate calculate the theoretical theoretical maximum power that can be transferred by the transmission line.
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=
=
V rated
V rated
Z C
P R
sin( β l )
V S ( L − L ) V R ( L − L ) sin(δ )
=
Z C sin( β l )
=
V Spu V Rpu SIL
sin( β l )
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Example 6, solution The line phase constant is:
β l =
2π l
λ
rad =
360 5000
(315) = 22.68o
The practical line loadability:
kV L
=
Pmax
8
P R
V S ( L − L ) V R ( L − L ) (V rated )2 sin(δ )
700 =
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X = Z C sin( β l )
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1 0.9 SIL sin( 22.68)
( Z C )( SIL)
=
=
V Spu V Rpu SIL
sin( β l )
P R
=
V Spu V Rpu SIL
sin( β l )
SIL = 499 MW
sin( 36.87)
(320)(499.83)
= 1167
MW
=
sin( δ )
400 kV
sin(δ )
Line Compensation A transmission line loaded to its surge impedance loading has no net reactive flow into or out of the line and will have a flat voltage profile along its length.
Example
On long transmission lines, light loads less than SIL result in a rise of a voltage at the receiving end and heavy load greater than SIL will produce a large dip in voltage. Shunt reactors are widely used to reduce high voltages under light load or open line conditions.
Can five instead of six 765 kV lines transmit the required power if there are two intermediate substations that divide each line into three 167 km line sections, and if one line section is out of serivce.
If the transmission line is heavily loaded, shunt capacitors, static var control and synchronous motors are used to improve voltage, increase power transfer and improve system stability. 16:28
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Example 8
Shunt Reactors
For the transmission line of example 5: a) Calculate the receiving receiving end voltage when when the line is terminated terminated in an open circuit and energized with 500 kV at the sending end. b) Determine the reactance reactance and the Mvar Mvar of a three three phase shunt reactor reactor to be installed at the receiving end to the keep the no-load receiving voltage at the rated value.
Shunt reactors are applied to compensate for the undesirable voltage effects associated with line capacitance.
I R
V R
=
V S = V R (cos β l +
jX Lsh X Lsh
And
=
sin β l V S V R
Also 16:28
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I S
= I R (−
1 Z C
sin β l )
Z C
− cos β l
X Lsh
For VS = VR
Z C X Lsh
=
sin β l 1 − cos β l
sin( β l ) X Lsh + cos β l )
Z C IS = -IR
Shunt Capacitor Compensation Shunt capacitors are used lagging power factor circuits created by heavy loads.
Example 8, solution The line is energized with 500 kV at the sending end, so the phase voltage is: V S =
The objective is to supply the needed reactive power to maintain the receiving end voltage at a satisfactory level.
500∠0
=
3
288.7 kV
From previous examples, ZC = 290.43, βl = 21.64. W hen the line is open IR = 0 and VR will be: V R ( nl )
V S
=
cos β l
= 310.57 kV
For VR = VS, then:
X Lsh
The reactor rating is:
sin β l 1 − cos β l
Q=
Z C =
(kV Lrated )2 X Lsh
sin 21.64 1 − cos 21.64
=
(500)2 1519.5
290.43 = 1519.5 Ω
= 164.5 M var
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Example 9
Series Capacitor Compensation
The transmission line of example 5 supplies a load of 1000 MVA, 0.8 power factor lagging at 500 kV:
Series capacitors are used to reduce the series reactance between the load and the supply.
a) Determine the Mvar Mvar of the shunt shunt capacitors to be installed at the receiving end to keep the receiving end voltage at 500 kV when the line is energized with 500 kV at the sending end.
This results in improved transient and steady state stability, stability, more economical loading and minimum voltage dip on load buses. P3φ =
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10
=
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V S ( L − L ) V R ( L − L ) X ' − X Cser
sin δ
Example 9, solution From previous examples, ZC = 290.43, βl = 21.64 so the equivalent line reactance for a lossless line is given by: X = Z sin β l = 107.1 Ω C
−
S = 1000∠ cos 1 (0.8) = 800 + j 600 M VA
The receiving end power is:
For the above operating condition, the power angle is obtained from:
800 =
500 500 107.1
δ = 20.04o
sin(δ )
So the net reactive power at the receiving end is: Q R
=
V S ( L− L ) V R ( L − L ) X
cos(δ ) −
V R ( L− L ) X
2
cos( β l ) = 23.15 Mvar
So the required Mvar will be: S C = j23.15 – j600 = -j576.85 -j576.85 Mvar 16:28
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