ELE B7 Power Systems
Engineering
Power System Components’ Modeling
Power System Components
ELE B7


The main components of a power system are
generators, transformers and transmission lines.
In this lecture, we shall discuss the models of these
components that will be used subsequently in power
system studies.
Slide # 1
ELE B7
Section I: Transmission Lines
Slide # 2
Transmission lines- CONSTRUCTION





ELE B7
Three-phase conductors, which
carry the electric current;
Insulators, which support and
electrically isolate the conductors;
Tower, which holds the insulators
and conductors;
Foundation and grounding; and
Optional shield conductors, which
protect against lightning
Slide # 3
Transmission lines- VOLTAGE LEVELS


ELE B7
Overhead Transmission lines (OTL) are operating at different
voltage levels:
Distribution: 6.3, 11, 13.8, 22, 33, 69 kV
Supplies residential and commercial customers

Subtransmission: 69, 110, 132 kV
Interconnection between substations and large
industrial customers

Transmission: 132, 220, 400 kV
Interconnection between substations, power
plants

EHV transmission: 500, 735, 765 kV
Interconnection between systems

UHV (experimental): 1200, 1500 kV
Transco 220 KV
Slide # 4
Transmission Line- TYPES OF CONDUCTORS


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ELE B7
Transmission line conductors can be made of copper or aluminum
However, aluminum conductors have completely replaced copper
for overhead lines because of the much lower cost and lighter
weight of an aluminum conductor compared with a copper
conductor of the same resistance.
Symbols identifying different types of aluminum
conductors are as follows:
AAC
all-aluminum conductors
AAAC all-aluminum-alloy conductors
Aluminum outer strands
2 layers, 30 conductors
ACSR aluminum conductor, steel-reinforced
ACAR aluminum conductor, alloy-reinforced
Steel core strands,
7 conductors
ACSR
Slide # 5
Transmission Line- PARAMETERS

A transmission line has four parameters :

Resistance,
Inductance,
Capacitance, and

Conductance.




ELE B7
The conductance, exists between conductors or between
conductors and the ground, accounts for the leakage current at
the insulators of overhead lines and through the insulation of
cables.
Since leakage at insulators of overhead lines is negligible, the
conductance between conductors of an overhead line is usually
neglected.
Slide # 6
Transmission Line Parameters-RESISTANCE
ELE B7

It is very well known that the dc resistance of a wire is given by:
Ω
where ρ is the resistivity of the wire in Ω - m, l is the length in
meter and A is the cross sectional area in m2

The line resistance increases by:





Stranding
Temperature
Skin effect
AC resistance higher than DC
Accurate value from Tables
Slide # 7
Transmission Line Parameters-RESISTANCE
ELE B7
Example:
A three-phase overhead transmission line is designed to deliver 190.5MVA at 220kV over a
distance of 63km, such that the total transmission line loss is not to exceed 2.5% of the rated
line MVA. Given the resistivity of the conductor material to be 2.84*10-8 ohm-meter.
Determine the required conductor diameter and the conductor size. Neglecting power loss
due to corona and insulator leakage and other reactive loss.
Solution:
The total transmission line loss is:
PLoss  190.5 * 2.5%  4.7625 MW
IL 
190.5( MVA )
 500 A
3 ( 220 kV )
d  1.894 cm  0.7456 in  556 ,000cmil
PLoss  3 * I L 2 * R
R
4.7625 MW
3( 500 )
2
 * l 2.84  10 8 * 63  10 3
A

R
6.35
A  2.81764  10 4 m 2
 6.35 
1 in = 1000 mils
1 cmils = sq mil
Slide # 8
Transmission Line Parameters-INDUCTANCE
ELE B7
• The inductance per phase of 3-phase equilateral spaced solid
conductors is given by:
D
L  2  10 (ln ) H / m
r
7
where,
r
D is the distances between the conductor
r   r e 1 / 4, and r is the radius of the conductor
D
D
D
• For inequilaterally spaced conductors, the inductance per phase is:
7
L  2  10 ln
where,
Deq
r
H /m
Deq  3 D12 D23 D31
Slide # 9
Transmission Line Parameters-INDUCTANCE
ELE B7
• The inductance per phase of 3-phase stranded conductors is:
L  2  10 7 ln
GMD
H /m
GMR
where,
- GMD is the Geometrical Mean Distance
- GMR is the Geometrical Mean Radius
• The inductance per phase of 3-phase
bundled conductors is:
L  2  10
GMR  2 Ds  d
d
7
GMD
ln
H /m
GMR
GMR  3 Ds  d 2
D31
D23
D12
GMD  Deq  3 D12 D23 D31
GMR  1.091 4 Ds  d 3
d
d
d
d
d
d
d
Slide # 10
Transmission Line Parameters-INDUCTANCE
ELE B7
• Transposition:
When the conductors of a three-phase line are not spaced
equilaterally, the flux linkages and the inductance of each phase are
not the same. A different inductance in each phase results in
unbalance circuit.
Balance of the three phases can be restored by exchanging the
positions of the conductors at regular intervals along the line as shown
below
Pos.1
D12
Cond. a
Cond. b
Pos.3
Cond. c
Cond. c
D23
D23
D23
Cond. b
D12
D12
Pos.2
D31
Cond. c
Cond. a
Cond. b
Cond. a
Such an exchange of the conductor positions is called transposition.
Slide # 11
Transmission Line Parameters-INDUCTANCE
ELE B7
Slide # 12
Transmission Line Parameters-CAPACITANCE
ELE B7
• Capacitance of a Three-Phase Transposed Line:
2 o
Cn 
ln( Deq / r )
F /m
Phase a
GMD  Deq  3 D12 D23 D31
Phase
b
Phase
c
• Capacitance of Three-Phase Bundled Line:
2 o
Cn 
ln( Deq / D bS )
F /m
where,
 ε0=8.85×10-12
 r is the equivalent radius of the conductor
d
Dsb  2 r  d
Dsb
 rd
3
d
d
2
d
d
Dsb
 1.091 r  d
4
3
d
d
d
Slide # 13
Transmission Line Parameters-CAPACITANCE
ELE B7
EXAMPLE:
A 400kV, 60Hz, three phase with bundled conductor line, two sub-conductors per-phase
as shown in the Figure. The center to center distance between adjacent phases is 12 m and
distance between sub-conductors is 45 cm. The radius of each sub-conductor is 1.6 cm.
Find the capacitance per phase per km.
D31
D12
a’
a
d
D23
b’
b
d
c’
c
d
SOLUTION:
D bS  r  d  1.6  45  8.485cm  0.08485m
Deq  3 Dab Dac Dbc  3 12 *12 * 24  15.119m
2 o
Cn 
 0.0107 F / km
b
ln( Deq / D S )
NOTE
Use r  for L
Use r for C
Slide # 14
Section II: Transmission Line
Performance
Transmission line representation

Series impedance
z  r  j x  / mi


Z  z  R j X

Shunt admittance
y  g  j b S / mi

ELE B7
Y  y  G  j B S
G is almost always ignored
R is sometimes ignored for analysis purposes (lossless line), but never in the real world!
Slide # 16
Transmission line models
ELE B7
Short line (up to 50 miles long)
+
VS
–
IS
IR
Z
+
VR
–
Ignore shunt
admittance
Medium-length line model (50-150 miles long)
+
VS
–
IR
IS
Z
½Y
½Y
+
VR
Nominal 
circuit
–
Slide # 17
Two-Port Network- ABCD parameters

ABCD parameters
+
VS
–

IR
IS
VS  AVR  BI R
I S  CVR  DI R

ELE B7
Two-port
Network
or
+
VR
–
VS   A B  VR 
 I   C D   I  (1)
 R
 S 
Applies to linear, passive, two-port networks
A, B, C and D depend on transmission line parameters
Slide # 18
ABCD Parameters – Example 1

Find the ABCD parameters
for the nominal  circuit
+
VS
IS
ELE B7
IR
Z
½Y
+
½Y
VR
–
–
KVL equation:
Y VR 

VS  VR  Z  I R 

2 

 YZ 
 VS  1 
 VR  Z I R
2 

( 2)
KCL equation:
Y Vs Y VR
IS 

 IR
2
2

Y 2Z 
 YZ 


VR  1 
 I S  Y 
I R

4 
2 


(3)
By inspection,
A  D  1
YZ
pu,
2
B  Z ,
 YZ 
C  Y 1 
 S
4 

Slide # 19
ABCD Parameters
ELE B7

Series impedance Z:
A  D  1,

Shunt admittance Y:
A  D  1,

Why use ABCD?
–
–
+
VS
–
B  Z ,
B  0,
C 0
C Y S
Easier than circuit analysis for hand calculations
Easier to concatenate elements
IR
IS
 A1
C
 1
B1 
D1 
 A2
C
 2
B2 
D2 
+
VR
–

+
VS
–
IR
IS
 A1
C
 1
B1   A2
D1  C2
B2 
D2 
+
VR
–
Slide # 20
ABCD Parameters – Example 2
ELE B7
A 345 kV transmission line has z = 0.05946 + j0.5766 /mi, y =
7.3874×10-3 S/mi and  = 222 mi. Calculate:
a)
ABCD parameters, assuming nominal  circuit model
Z  z  13.2  j128 ,
A  D  1  YZ

C  Y 1  YZ
b)
2
Y  y  j1.64  103 S
 0.895110.69 pu,
  1.554  10
4
3
B  Z  128.6884.11 
90.33 S
Receiving-end no-load voltage
VS  AVR  BI R
VR 
IR  0
VS
345 3

 222.5  0.69 kV
A 0.895110.69
VR LL  385.4 kV (12% above nominal)
Slide # 21
The long transmission line
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

ELE B7
Impedance and admittance parameters are
distributed rather than lumped
At 60 Hz, effects of distributed parameters are
significant for long lines (>150 miles)
Need a new model to accurately represent “long”
transmission lines
1.
2.
Derive exact model for a generic transmission line
represented as a two-port network (ABCD matrix)
Determine relationship to nominal  model
Slide # 22
The long transmission line

ELE B7
The exact network equations in ABCD format:
cosh( ) Z c sinh( ) 
V
 S 

 I    sinh(  )
cosh( ) 
 S
 Zc

where ,
A
C

V R 
I 
 R
cosh( ) Z c sinh(  ) 
B  

 sinh( )

cosh( ) 
D 
 Zc

  zy m 1 (propagati on constant)
Zc 
z
 (character istic impedance)
y
Slide # 23
The long transmission line

Relationship to nominal  circuit model:
 cosh(   )
V S  
 I    sinh(   )
 S
Zc


V S  
I   
 S  Y

Z c sinh(   ) 
 V R 
cosh(   )   I R 

Y 'Z '
2
Y 'Z

'1 
4

1


Y ' Z '

1
2 
Z'
'


V R 
I 
 R
tanh(   2 )
Y'

2
Zc
Z '  Z c sinh(   ),

ELE B7
Exact network equations
in ABCD format
Equivalent  circuit equations
(Z’ and Y’ are the series
impedance and shunt
admittance of the equivalent
 circuit that models the
terminal behavior exactly)
In terms of the “usual” Z = z and Y = y:
Z ' Z
sinh(   )

and
Y ' Y tanh(   2 )

2
2
 2
Slide # 24
The long transmission line – Example 3

For the line in Example 2, calculate:
a.
Propagation constant and characteristic impedance
 
b.
zy  2 . 069  10  3  87 . 06  m 1 ,
Zc 
ELE B7
z
 280 . 11   2 . 94 
y
Equivalent  circuit parameters
Z' Z
sinh(   )
 1 . 2422  84 . 32 

( 4 % lower )
Y ' Y tanh(   2 )

 0 . 8346  10  3  89 . 9 
 2
2
2
Y '  1 . 6693  10  3  89 . 9 
( 2 % higher )
Slide # 25
Loadability

ELE B7
Power transfer capability of a transmission line may
be limited by any one of the following:
1.
2.
3.
4.
5.
Conductor temperature & sag requirements
Voltage profile
Stability considerations
Real power losses
Reservation requirements
Slide # 26
Temperature / sag limitations

Operate within conductor or insulator temperature
rating
–
–
–

Heat gain (I2R, other heat sources)
Heat dissipation (wind, conduit)
Summer vs. winter ratings; continuous vs. emergency
ratings
Meet minimum sag requirements
–

ELE B7
Heat causes conductors to stretch, which reduces ground
clearance for overhead lines
For most short transmission lines, temperature / sag
limitations dictate transfer capability
Slide # 27
Voltage profile

Voltage regulation (typically, VR  10%)
VR %  

V R NL  V R FL
V R FL
 100 ,
VR NL
VS

A
Operating range
0 .95  V R  1 .05 pu

ELE B7
and
0 .95  V S  1 .05 pu
Voltage profile is a consideration for all lines
Slide # 28
Stability considerations

Steady-state stability limit
PMAX
V RV S
sin  ,
P  PS  PR 
X
PMAX 
V RV S
X
or
  V  V
s
PMAX
3
3
Real Power
For a loss-less 3-phase line (R=G=0),
ignoring distributed effects:

ELE B7
R
V RV S
X
0
30
60
90
120
150
180

–
For  > 90o, synchronism between sending and receiving end cannot
be maintained
–
Lines are operated at  < 35o to prevent transient instability during
system disturbances
Long lines are typically stability-limited
Slide # 29