Lecture Notes
EE 330 Power System Analysis I
Lecture Notes
EE 330 Power System Analyses I
Chapter 5: Line Model & Performance
Lecturer: Dr Ibrahim Rida
Electrical Engineering Department
University of Hail
First Semester 101 – October, 2010
1. Introduction
Transmission lines are represented by an equivalent model with circuit parameters on a
“per-phase” basis.
Terminal voltage: line-to-neutral
Current: one phase
Therefore 3-phase system reduces to an equivalent single-phase system. The model is
used to calculate voltage (V), current (I) and power (S) flows. Depending on the length,
lines can be: short, medium, and long.
The electrical power is transmitted over the lines at approximately the speed of light. To
account for wave propagation effects, an equivalent π model is developed for long lines.
To maintain voltage level along transmission lines within operationally acceptable range
during both loaded and unloaded conditions, line compensation is used.
2. Short Line Model
Short line is considered for lines less than 80 km (50 miles) long, and voltage level less
than 69 kV. In this model, the capacitance can be ignored. Only L and R is considered,
so, line impedance is calculated as given in equation (5.1), as:
Z=
(r + j ω L) l
=
R + jX
where l is the line length.
See Figure 5.1 for short line model (per phase), where VS and IS are the voltage and
current on the sending end, while VR and IR are the voltage and current on the receiving
end.
If a three-phase load with apparent power (3-phase SR) is connected at the end of the
transmission line, the receiving end current is obtained by (as in (5.2)):
October, 2010
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Lecture Notes
IR =
EE 330 Power System Analysis I
S R* ( 3 ph )
3VR*
The phase voltage at the receiving end is:
Vs = VR + Z IR
Since the capacitance (C) is neglected, the sending-end and receiving-end currents are
equal, i.e.: IS = IR.
We can view the model as a two port network shown in Figure 5.2 with equations (5.5)
and (5.6), or in matrix form as (5.7), where constant ABCD is given in (5.8).
VS = AVR + BI R
I S = CV R + DI R
Or, in the matrix form:
⎡VS ⎤ ⎡ A B ⎤ ⎡V R ⎤
⎢ I ⎥ = ⎢C D ⎥ ⎢ I ⎥
⎦⎣ R ⎦
⎣ S⎦ ⎣
For short lines:
A = 1, B = Z, C = 0, D = 1
Voltage Regulation (V.R.) is defined as the percentage change in voltage at the
receiving end of the line (expressed as percent of full-load voltage) in going from no-load
to full-load, as given in (5.9):
Percent VR =
VR ( NL ) − V R ( FL )
VR ( FL )
× 100
At no load, IR = 0, so VR(NL) is as given in (5.10) and since A=1, VR(NL) = VS.
V.R. is dependent on the load power factor:
• Poorer at low lag PF loads
• Negative at leading PF loads (capacitive loads)
See Figure 5.3
If VS is known, the sending end power is calculated from (5.11): SS(3ph) = 3 VS IS*
The total loss is given in (5.12): SL(3ph) = SS(3ph) – SR(3ph)
and the transmission line efficiency is given by (5.13):
η=
PR ( 3 ph )
PS (3 ph )
See Example 5.1.
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Lecture Notes
EE 330 Power System Analysis I
3. Medium Line Model
Transmission lines between 80 km (50 miles) to 250 km (150 miles) are referred to
medium length lines. See Figure 5.4. Line charging of these lines is appreciable and must
be considered. Half of the shunt C is lumped at each end of the line; such a model is
known as nominal π model, shown in Figure (5.4).
Z is the total series impedance of the line as calculated from (5.1) and Y is the total shunt
admittance as given in (5.14):
Y = (g + j ω C) l
where g - the shunt conductance/m representing the leakage current of insulators and due
to corona. Under normal conditions, g = 0.
Applying KCL, IL is given as in (5.15):
Y
I L = I R + VR
2
From KVL the VS is given as in (5.16), so in equation (5.17) VS is obtained as:
⎛ ZY ⎞
V S = ⎜1 +
⎟VR + ZI R
2 ⎠
⎝
The sending-end current (IS) is given as in (5.18) or (5.19),
Y
I S = I L + VS
2
The constant ABCD for medium lines are given as in (5.20) and (5.21):
⎛ ZY ⎞
A = ⎜1 +
B=Z
⎟
2 ⎠
⎝
⎛ ZY ⎞
⎛ ZY ⎞
C = Y ⎜1 +
D = ⎜1 +
⎟
⎟
4 ⎠
2 ⎠
⎝
⎝
In general, ABCD constants are complex and since the π model is a symmetrical 2-port
network, A=D. Also, the determinant of the transmission matrix is unity as in (5.22):
AD – BC = 1
Finally, the relation between the sending and receiving ends in the matrix form is given
as in (5.23) for VR and IR:
⎡VR ⎤ ⎡ D
⎢ I ⎥ = ⎢− C
⎣ R⎦ ⎣
October, 2010
− B ⎤ ⎡Vs ⎤
A ⎥⎦ ⎢⎣ I s ⎥⎦
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Lecture Notes
EE 330 Power System Analysis I
Example:
A three-phase completely transposed 345 kV, 200 km line has the following positive
sequence line constants: z = 0.032 + j0.35 Ω/km , y = j 4.2 μs/km. Full load at the
receiving end of the line is 700 MW at 0.99 leading PF and 95% of rated voltage.
Assuming a medium-length line, determine the following:
(a) ABCD parameters of the nominal π circuit.
(b) Sending-end voltage Vs, current Is, and real power Ps
(c) Transmission line efficiency at full load.
Solution:
(a) The total series impedance and the shunt admittance values are:
Z = zl = (0.032 + j 0.35) (200) = 6.4 + j70 = 70.29 ∟84.78˚ Ω
Y = yl = (j4.2 x10-6) (200) = 8.4x10-4 s
A = D =1+
ZY
(70.29∠84.78o )(8.4 × 10−4 ∠90o )
=1+
= 0.9706 + j 0.00269 = 0.9706∠0.159o
2
2
B = Z = 70.29 ∟84.78˚ Ω
C = (8.4 x 10-4 ∟90˚)(1+0.01476 ∟174.78˚) = 8.277 x 10-4 ∟90.08˚ s
(b) The receiving-end voltage and current quantities are:
VR = 0.95 × 345 = 327.8 kVLL
327.8 o
∠0 = 189.2∠0o kVLN
3
700∠ cos −1 0.99
IR =
= 1.246∠8.11o A
3 (0.95 × 345)(0.99)
VS = AVR + BIR
VS = (0.9706∟0.159˚)(189.2∟0˚)+(70.29∟84.78˚)(1.246∟8.11˚)
= 183.6∟0.159˚ + 87.55∟92.89˚ = 179.2 + j87.95
= 199.6 ∟26.14˚ kV(L-N)
VS ( L − L ) = 3 × 199.6 = 345.8 kV
VR =
IS =
IS =
=
=
CVR + DIR
(8.277 x 10-4∟90.08˚)(189.2∟0˚)+(0.9706∟0.159˚)(1.246∟8.11˚)
0.1566 x 10-4∟90.08˚ + 1.209∟8.27˚ = 1.196 + j0.331
1.241∟15.5˚ kA.
The power delivered to the sending end is:
PS = 3 x VS x IS x cos(θv – θi)
PS = 3 x 199.6 x 1.241 x cos(26.14˚ – 15.5˚) = 730.3 MW
October, 2010
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Lecture Notes
EE 330 Power System Analysis I
(c) The no-load receiving-end voltage is:
V
345.8
VR ( NL ) = S =
= 356.3 kVL − L
A 0.9706
356.3 − 327.8
100% = 8.7%
VR =
327.8
(d) The full-load line losses are:
PS – PR = 730.5 – 700 = 30.5 MW
700
P
η = R 100% =
100 = 95.8%
730.5
PS
See also Example 5.2 and Example 5.3.
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Lecture Notes
EE 330 Power System Analysis I
4. Long Line Model
The length of line is more than 250 km (150 miles), and we will use the distributed
parameters as in Figure 5.5.
For a small segment of line Δx at a distance x from the receiving end of the line, the
phasor voltage is as in (5.24) and (5.25). Taking limit as Δx --> 0, we have (5.26). From
Kirchoff’s current law, we have (5.27) and (5.28), and taking Δx --> 0, we have (5.29).
Differentiating (5.26) and together with (5.29), we have (5.30).
d 2V ( x)
Let γ = zy, so we have (5.32):
− γ 2V ( x) = 0
2
dx
2
The solution is given in (5.33): V ( x) = A1eγx + A2e −γx
where γ is the propagation constant as in (5.34). The real part of γ is α (called
attenuation constant) and the imaginary is β (known as phase constant (Rad/m)):
γ = α + jβ = zy = (r + jωL)( g + jωC )
From (5.26) I(x) is derived as in (5.36):
(
)
1
A1eγx − A2e −γx
ZC
where ZC is the characteristic impedance of the line; it is given as in (5.37):
z
ZC =
y
I ( x) =
To find constant A1 and A2, we use the boundary conditions:
• x=0 (to find VR and IR), where V(x) = VR and I(x) = IR and with (5.33) and (5.36),
we find A1 and A2 as in (5.38):
V + ZC I R
A1 = R
2
V − ZC I R
A2 = R
2
• Substituting (5.33) and (5.36), the general expression for V(x) and I(x) is as in
(5.39) and (5.40), or (5.41) and (5.42). The hyperbolic functions (sinh and cosh)
can be recognized in these equations, hence the general expression for V(x) and
I(x) can be written as in (5.43) and (5.44).
• For x = ℓ (to find VS and IS), where V(ℓ) = VS and I(ℓ) = IS , the relation between
the sending end and receiving end of the line is given as in (5.45) and (5.46):
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Lecture Notes
EE 330 Power System Analysis I
VS = cosh γl VR + Z C sinh γl I R
IS =
1
sinh γl VR + cosh γl I R
ZC
Therefore, the equations in the matrix form in terms of the constant ABCD is as given in
(5.47):
⎡VS ⎤ ⎡ A B ⎤ ⎡VR ⎤
⎢V ⎥ = ⎢C D ⎥ ⎢ I ⎥
⎦⎣ R ⎦
⎣ R⎦ ⎣
Where the values of the ABCD constants are given in equations (5.48) and (5.49):
A = cosh γl
C=
1
sinh γl
ZC
B = Z C sinh γl
D = cosh γl
Note that A=D and AD-BC = 1.
So the accurate equivalent π model is as in Figure 5.6. Writing the expression in (5.17)
and (5.19), we obtain equations (5.50) and (5.51):
⎛ Z 'Y ' ⎞
VS = ⎜1 +
⎟VR + Z ' I R
2 ⎠
⎝
⎛ Z 'Y ' ⎞
⎛ Z 'Y ' ⎞
I S = Y ' ⎜1 +
⎟VR + ⎜1 +
⎟I R
4 ⎠
2 ⎠
⎝
⎝
where the new terms Z’ and Y’ are given by equations (5.53) and (5.54):
sinh γ l
Z ' = Z C sinh γ l = Z
γl
γl
γ l Y tanh 2
Y' 1
=
=
tanh
2 ZC
2
2 γl
2
See Example 5.4:
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Lecture Notes
EE 330 Power System Analysis I
5. Voltage and Current Waves
From phasor value of voltage at any point along the line as given by (5.33) and γ = α +
jβ; then the time domain expression for the instantaneous voltage v(t, x) can be written as
in (5.55):
v(t , v) = 2ℜA1eαx e j (ωt + βx ) + 2ℜA1e −αx e j (ωt − βx )
This equation shows that the voltage at any point along the line is composed of two
components, as given in (5.56):
v(t, x) = v1(t, x)+v2(t, x)
Where, as in (5.57) and (5.58):
v1 (t , v) = 2ℜA1eαx e j (ωt + βx ) = 2 A1eαx cos(ωt + βx)
v2 (t , v) = 2ℜA2e
−αx
e
j (ωt − βx )
= 2 A2e
−αx
is the incident wave and
cos(ωt − β x) is the reflected wave.
This is called the traveling wave.
Similarly, the current can be expressed, as the voltage, as the sum of incident wave and
reflected wave.
The velocity of wave propagation, as in (5.60) is:
v=
ω 2π f
=
β
β
The wavelength λ is the distance which results in a phase shift of 2π radian, so:
λ=
2π
β
If the line losses are ignored, i.e. g = 0 and r = 0, the real part of the propagation constant
will be zero, α =0, then from (5.34), we have β reduces, as in (5.62):
β = ω LC
The characteristic impedance, ZC, becomes, as in (5.63):
ZC =
October, 2010
L
,
C
which is called the surge impedance.
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Lecture Notes
EE 330 Power System Analysis I
For the lossless line, the velocity of propagation and wavelength are given in (5.64) and
(5.65):
1
ν=
LC
1
λ=
f LC
And when the internal flux linkage of conductor is neglected, GMRL = GMRC, the values
of v and λ can be approximated, as in (5.66) and (5.67):
v≈
λ≈
1
μ0ε 0
1
f μ0ε 0
Where, μ0 = 2π 10-7 is the permeability and ε0 = 8.85 10-12 is the permittivity.
Substituting these values, we find that the velocity of the wave will be approximately the
velocity of light, that is c = 3 108 m/s.
At 60 Hz, the wavelength λ = 5000 km; whereas, the surge impedance, Zc, equals as in
(5.68):
1 μ0 GMD
GMD
= 60 ln
ZC ≈
ln
2π ε 0 GMRC
GMRC
For typical transmission lines the surge impedance varies from approximately 400Ω (for
69 kV) to 250Ω (double circuit 765 kV).
For lossless line, γ = jβ and the rms voltage and current from (5.43) and (5.44) become as
in (5.71) and (5.72):
VS = cos β l VR + jZ C sin β l I R
IS = j
1
sin β l VR + cos β l I R
ZC
This is easier to be used for hand calculation. For more accurate computer-based
calculations, equations (5.47 – 5.49) can be used.
The boundary or terminal conditions can be obtain from (5.71) and (5.72), for example
for open-circuit IR = 0, the no-load receiving end voltage is given by:
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Lecture Notes
EE 330 Power System Analysis I
VR ( NL ) =
VS
cos β l
At no load, line current is entirely due to the line charging capacitive current and the
receiving end voltage is larger than the sending end voltage. As ℓ is longer, cos βℓ is
smaller, so VR(NL) is larger.
For a solid short circuit at the receiving end, VR = 0, then (5.71) and (5.72) reduce to:
VS = jZ C sin βl I R
I S = cos β l I R
6. Surge Impedance Loading
Surge Impedance Loading (SIL) is a useful measure of transmission line capacity as it
indicates the loading at minimum line’s requirement for reactive power. Example: for
230 kV lines, the SIL = 150 MW, while for 765 kV line, the SIL = 2000 MW.
The SIL is when the line is loaded with impedance equal to its characteristic impedance
ZC. SIL is calculated using (5.77).
3V
SIL = 3V I = R
ZC
2
*
R R
Since VR =
VL (rated )
3
, the SIL (in MW) also can be calculated using (5.78).:
(kV
SIL =
)
2
L ( rated )
ZC
MW
VR
and VR in (5.70), with VR ZC, the voltage and current
ZC
relationships will be as given in (5.79) and (5.80):
Substituting IR in (5.69) with
V ( x) = (cos β x + j sin β x)VR
or V ( x) = VR ∠β x
I ( x) = (cos β x + j sin β x) I R
or
I ( x) = I R ∠β x
These equations indicate that in the voltage and current at any point along a lossless line
under surge impedance loading (SIL) are constant in magnitude and are equal to the
receiving end values.
Since QS = QR = 0 (Zc is purely resistive), the reactive losses in the line inductance are
totally offset by the reactive power supplied by the shunt capacitance or:
October, 2010
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Lecture Notes
EE 330 Power System Analysis I
ωL|IR|2 = ωC|VR|2.
From this we have Z C =
VR
L
=
, as given in equation (5.63).
IR
C
For loads larger than SIL, shunt capacitors are needed to minimize voltage drop along the
line. For loads less than SIL, shunt inductors are needed. Generally the transmission line
full-load is much higher than SIL. See Figure 5.11 for various loading condition. See
Example 5.5.
7. Complex Power Flow through Transmission Lines
From (5.5) and expressing constant ABCD and VS and VR in polar forms
( A = A ∠θ A , B = B ∠θ B , B = B ∠θ B , VS = V ∠δ ) , IR can be written as in (5.81). 3-phase
SR is expressed as in (5.82) and with IR, the expression for receiving end power will be,
using L-N voltages, as follows (5.83):
2
V V
A VR
∠(θ B − θ A )
S R ( 3φ ) = 3 S R ∠(θ B − δ ) − 3
B
B
Using the line-to-line voltages power will be (5.84):
VS ( L − L ) VR ( L − L )
A VR ( L − L )
2
∠(θ B − δ ) −
B
B
The 3-phase PR and QR are written as in (5.85) and (5.86):
S R ( 3φ ) =
PR (3φ ) =
QR ( 3φ ) =
VS ( L − L ) VR ( L − L )
B
VS ( L − L ) VR ( L − L )
B
cos(θ B − δ ) −
sin(θ B − δ ) −
A VR ( L − L )
∠(θ B − θ A )
2
cos(θ B − θ A )
B
A VR ( L − L )
2
B
sin(θ B − θ A )
Since IS can be written as in (5.88), then 3-phase PS and QS are written as in (5.89) and
(5.90).
PS ( 3φ ) =
QS (3φ ) =
A VS ( L − L )
2
cos(θ B − θ A ) −
B
A VS ( L − L )
B
2
sin(θ B − θ A ) −
VS ( L − L ) VR ( L − L )
B
VS ( L − L ) VR ( L − L )
B
cos(θ B + δ )
sin(θ B + δ )
The transmission line losses are as in (5.91) and (5.92):
PL ( 3φ ) = PS (3φ ) − PR ( 3φ )
QL (3φ ) = QS ( 3φ ) − QR ( 3φ )
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Lecture Notes
EE 330 Power System Analysis I
See Example 5.9 (g) in using “pwcirc(ABCD)” to see receiving end power circle
diagram for assessing the performance characteristic of the transmission line (plotting 3phase QR vs PR for fixed receiving end voltage and varying sending end voltage).
For a lossless line: B = jX’, θA=0, θB = 90˚ and A=cos βl, and the real power transferred
over the line is given by (equation (5.93) and (5.94)):
P3φ =
Q3φ =
VS ( L − L ) VR ( L − L )
X'
VS ( L − L ) VR ( L − L )
Where, X’ = ZC sin βl
X'
sin δ
cos δ −
VR ( L − L )
X'
2
cos β l
For a system operating at constant voltage, the power transferred is proportional to the
sine of the power angle δ. The maximum power that can be transmitted under stable
steady-state condition occurs for an angle of 90˚. However, to maintain adequate margin
of stability, the operating angle is usually limited to 35˚ to 45˚.
8. Power Transmission Capability
The power transfer ability of a line is limited by:
1- Thermal loading limit
2- Stability limit.
The thermal limit is specified by the current-carrying capacity of the conductor as
specified in manufacturer’s data.
For planning and other purposes, it is very useful to express power transfer in terms of
SIL, and construct the line loadability curve. For a lossless line X’ = ZC sin β l, and from
equation (5.93) power can be given as (5.96):
⎛ VS ( L− L ) ⎞⎛ VR ( L− L ) ⎞⎛ V 2 ⎞ sin δ
⎟⎜
⎟⎜ rated ⎟
P3φ = ⎜
⎜ Vrated ⎟⎜ Vrated ⎟⎜⎝ Z C ⎟⎠ sin β l
⎝
⎠⎝
⎠
The first two terms in the equation are the PU voltage (VSpu, VRpu); the third term is the
SIL. Then power can be given as (5.97):
P3φ =
=
VSpu VRpu SIL
sin β l
VSpu VRpu SIL
October, 2010
⎛ 2π ⎞
sin ⎜
⎟
⎝ λ ⎠
sin δ
sin δ
Page 12 of 14
Lecture Notes
EE 330 Power System Analysis I
Try function loadabil(L,C,f )
See Example 5.6
9. Line Compensation
Figure 5.11 (page 182) shows the voltage profile of a long line for various loading
conditions:
• Loaded with SIL, no Qout and Qin and has a flat voltage profile along the line.
• Light loads less than SIL, rise of voltage at the receiving end. Shunt reactors is
used to reduce the voltage for this case.
• Heavy loads larger than SIL, produce a large dip in voltage. Shunt capacitor,
static VAR control, synchronies condenser are used to improve voltage, increase
power transfer and improve system stability.
Shunt Reactors
Consider a reactor of reactance XL,sh connected at the receiving end of a transmission line
as shown in Figure 5.7. The receiving end current is as given in equation (5.98);
substituting into (5.71), the sending end voltage can be computed as:
⎞
⎛
Z
VS = VR ⎜⎜ cos βl + C sin βl ⎟⎟
X L ,sh
⎠
⎝
Note that VS and VR are in phase, e.g. no real power is transmitted over the line.
sin β l
X L ,sh =
ZC
Solving for XL,sh (5.99):
VS
− cos β l
VR
sin β l
X L ,sh =
ZC
For VS = VR (5.100):
1 − cos βl
The relation between IS and IR can be given by substituting (5.98) into (5.72):
⎛ 1
⎞
I S = ⎜⎜ −
sin β l X L ,sh + cos β l ⎟⎟ I R
⎝ ZC
⎠
When VS = VR (5.101) substituting for XL,sh from (5.100) it can be found that: IS = -IR
It is worth to not that, when VS = VR, voltage at the mid-span is given by:
VR
Vm =
βl
cos
2
Also, the current at the mid-span is zero
See Example 5.7.
October, 2010
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Lecture Notes
EE 330 Power System Analysis I
Shunt Capacitors
Shunt capacitors are used to maintain the receiving end voltage. Capacitors are connected
either directly to the bus bar or the TRF tertiary windings. Given VS and VR, equation
(5.85) and (5.86) can be used to calculate the required MVar of capacitors needed. See
Example 5.9(f) that uses “shntcomp(ABCD)”.
Series Capacitors
Series capacitors are normally installed in the midpoint of the line to improve transient
and steady state stability. The reactive power production varies with the line loading. The
power transfer as in (5.103), see Figure 5.8. The percentage of compensation is XCser/
X’ is in the range of 25% to 70%. Special protective device is needed to bypass short
circuit not to pass the capacitor. See Example 5.9(f) with “sercomp(ABCD)” and
“srshcomp(ABCD)”. See Example 5.8.
10. Line Performance Program
Run “lineperf” and see Example 5.9.
October, 2010
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