El-Hawary M.E. Electrical Energy Systems
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the primary side.
B. Find the current drawn by the transformer referred to the primary
side.
C. If the load power factor is changed to 0.9 lagging with the load’s
active power and voltage magnitude unchanged, find the required
primary voltage.
Problem 4.10
Two 2400/600 V single phase transformers are rated at 300 and 200 KVA
respectively. Find the rating of the transformers’ combination if one uses the
following connections:
A. Series-series
B. Parallel-series
C. Parallel-parallel
Problem 4.11
A 30-kVA, 2.4/0.6-kV transformer is connected as a step-up autotransformer
from a 2.4-kV supply. Calculate the currents in each part of the transformer and
the load rating. Neglect losses.
Problem 4.12
A three-phase bank of three single-phase transformers steps up the three-phase
generator voltage of 13.8 kV (line-to-line) to a transmission voltage of 138 kV
(line-to-line). The generator rating is 83 MVA. Specify the voltage, current and
kVA ratings of each transformer for the following connections:
A. Low-voltage windings
∆
, high-voltage windings Y
B. Low-voltage windings Y, high-voltage windings
∆
C. Low-voltage windings Y, high-voltage windings Y
D. Low-voltage windings
∆
, high-voltage windings
∆
Problem 4.13
The load at the secondary end of a transformer consists of two parallel branches:
Load 1: an impedance Z given by
$
4575.0
∠=
Z
Load 2: inductive load with P = 1.0 p.u., and S = 1.5 p.u.
The load voltage magnitude is an unknown. The transformer is fed by a feeder,
whose sending end voltage is kept at 1 p.u. Assume that the load voltage is the
reference. The combined impedance of transformer and feeder is given by:
p.u.08.002.0 jZ
+=
A. Find the value of the load voltage.
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B. If the load contains induction motors requiring at least 0.85 p.u.
voltage to start, will it be possible to start the motors? If not,
suggest a solution.
Problem 4.14
A three phase transformer delivers a load of 66 MW at 0.8 p.f. lagging and 138
KV (line-to-line). The primary voltage under these conditions is 14.34 KV
(line-to-line), the apparent power is 86 MVA and the power factor is 0.78
lagging.
A. Find the transformer ratio.
B. Find the series impedance representation of the transformer.
C. Find the primary voltage when the load is 75 MVA at 0.7 p.f.
lagging at a voltage of 138 KV.
Problem 4.15
A three phase transformer delivers a load of 83 MVA at 0.8 p.f. lagging and 138
KV (line-to-line). The primary voltage under these conditions is 14.34 KV
(line-to-line), the apparent power is 86 MVA and the power factor is 0.78
lagging.
A. Find the transformer ratio.
B. Find the series impedance representation of the transformer.
C. Find the primary apparent power and power factor as well as the
voltage when the load is 75 MVA at 0.7 p.f. lagging at a voltage of
138 KV.
Problem 4.16
A two winding transformer is rated at 50 kVA. The maximum efficiency of the
transformer occurs when the output of the transformer is 35 kVA. Find the rated
efficiency of the transformer at 0.8 PF lagging given that the no load losses are
200 W.
Problem 4.17
The no-load input to a 5 kVA, 500/100-V, single-phase transformer is 100 W at
0.15 PF at rated voltage. The voltage drops due to resistance and leakage
reactance are 0.01 and 0.02 times the rated voltage when the transformer
operates at rated load. Calculate the input power and power factor when the
load is 3 kW at 0.8 PF lagging at rated voltage.
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Chapter 5
ELECTRIC POWER TRANSMISSION
5.1 INTRODUCTION
The electric energy produced at generating stations is transported over
high-voltage transmission lines to utilization points. The trend toward higher
voltages is motivated by the increased line capacity while reducing line losses
per unit of power transmitted. The reduction in losses is significant and is an
important aspect of energy conservation. Better use of land is a benefit of the
larger capacity.
This chapter develops a fundamental understanding of electric power
transmission systems.
5.2 ELECTRIC TRANSMISSION LINE PARAMETERS
An electric transmission line is modeled using series resistance, series
inductance, shunt capacitance, and shunt conductance. The line resistance and
inductive reactance are important. For some studies it is possible to omit the
shunt capacitance and conductance and thus simplify the equivalent circuit
considerably.
We deal here with aspects of determining these parameters on the basis
of line length, type of conductor used, and the spacing of the conductors as they
are mounted on the supporting structure.
A wire or combination of wires not insulated from one another is called
a conductor. A stranded conductor is composed of a group of wires, usually
twisted or braided together. In a concentrically stranded conductor, each
successive layer contains six more wires than the preceding one. There are two
basic constructions: the one-wire core and the three-wire core.
Types of Conductors and Conductor Materials
Phase conductors in EHV-UHV transmission systems employ
aluminum conductors and aluminum or steel conductors for overhead ground
wires. Many types of cables are available. These include:
A. Aluminum Conductors
There are five designs:
1. Homogeneous designs: These are denoted as All-Aluminum-
Conductors (AAC) or All-Aluminum-Alloy Conductors
(AAAC).
2. Composite designs: These are essentially aluminum-
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conductor-steel-reinforced conductors (ACSR) with steel core
material.
3. Expanded ASCR: These use solid aluminum strands with a
steel core. Expansion is by open helices of aluminum wire,
flexible concentric tubes, or combinations of aluminum wires
and fibrous ropes.
4. Aluminum-clad conductor (Alumoweld).
5. Aluminum-coated conductors.
B. Steel Conductors
Galvanized steel conductors with various thicknesses of zinc
coatings are used.
Line Resistance
The resistance of the conductor is the most important cause of power
loss in a power line. Direct-current resistance is given by the familiar formula:
ohms
dc
A
l
R
ρ
=
where
ρ
= resistivity of conductor
l = length
A = cross-sectional area
Any consistent set of units may be used in the calculation of resistance.
In the SI system of units,
ρ
is expressed in ohm-meters, length in meters, and
area in square meters. A system commonly used by power systems engineers
expresses resistivity in ohms circular mils per foot, length in feet, and area in
circular mils.
There are certain limitations in the use of this equation for calculating
the resistance of transmission line conductors. The following factors need to be
considered:
1. Effect of conductor stranding.
2. When ac flows in a conductor, the current is not distributed
uniformly over the conductor cross-sectional area. This is called
skin effect and is a result of the nonuniform flux distribution in the
conductor. This increases the resistance of the conductor.
3. The resistance of magnetic conductors varies with current
magnitude.
4. In a transmission line there is a nonuniformity of current
distribution caused by a higher current density in the elements of
adjacent conductors nearest each other than in the elements farther
apart. The phenomenon is known as proximity effect. It is present
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for three-phase as well as single-phase circuits. For the usual
spacing of overhead lines at 60 Hz, the proximity effect is
neglected.
5.3 LINE INDUCTANCE
The inductive reactance is by far the most dominating impedance
element.
Inductance of a Single-Phase Two-Wire Line
The inductance of a simple two-wire line consisting of two solid cylindrical
conductors of radii
r
1
and
r
2
shown in Figure 5.1 is considered first.
The total inductance of the circuit due to the current in conductor 1 only is given
by:
′
×=
1
7
1
)102(
r
D
lnL
(5.1)
Similarly, the inductance due to current in conductor 2 is
′
×=
2
7
2
)102(
r
D
lnL
(5.2)
Thus
L
1
and
L
2
are the phase inductances. For the complete circuit we have
21
LLL
t
+=
(5.3)
′′
×=
21
7
)104(
rr
D
lnL
t
(5.4)
where
i
ii
r
err
7788.0
41
=
=
′
−
(5.5)
Figure 5.1
Single-Phase Two-Wire Line Configuration.
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We compensate for the internal flux by using an adjusted value for the radius of
the conductor. The quantity r
′
is commonly referred to as the solid conductor’s
geometric mean radius (GMR).
An inductive voltage drop approach can be used to get the same results
)(
2121111
ILILjV
+=
ω
(5.6)
)(
2221122
ILILjV
+=
ω
(5.7)
where V
1
and V
2
are the voltage drops per unit length along conductors 1 and 2
respectively. The self-inductances L
11
and L
22
correspond to conductor
geometries:
′
×=
−
1
7
11
1
)102(
r
lnL
(5.8)
′
×=
−
2
7
22
1
)102(
r
lnL
(5.9)
The mutual inductance L
12
corresponds to the conductor separation D. Thus
×=
−
D
lnL
1
)102(
7
12
(5.10)
Now we have
12
II
−=
The complete circuit’s voltage drop is
112221121
)2( ILLLjVV
−+=−
ω
(5.11)
In terms of the geometric configuration, we have
1
21
7
1
21
7
21
)104(
1
2
11
)102(
I
rr
D
lnj
I
D
ln
r
ln
r
lnjV
VVV
′′
×=
−
′
+
′
×=∆
−=∆
−
−
ω
ω
Thus
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© 2000 CRC Press LLC
′′
×=
−
21
7
)104(
rr
D
lnL
t
(5.12)
where
122211
2LLLL
t
−+=
We recognize this as the inductance of two series-connected magnetically
coupled coils, each with self-inductance L
11
and L
22
, respectively, and having a
mutual inductance L
12
.
The phase inductance expressions given in Eqs. (5.1) and (5.2) can be
obtained from the voltage drop equations as follows:
+
′
×=
−
D
lnI
r
lnIjV
11
)102(
2
1
1
7
1
ω
However,
12
II
−=
Thus,
′
×=
−
1
1
7
1
)102(
r
D
lnIjV
ω
In terms of phase inductance we have
111
ILjV
ω
=
Thus for phase one,
terhenries/me)102(
1
7
1
′
×=
−
r
D
lnL
(5.13)
Similarly, for phase two,
terhenries/me)102(
2
7
2
′
×=
−
r
D
lnL
(5.14)
Normally, we have identical line conductors.
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© 2000 CRC Press LLC
In North American practice, we deal with the inductive reactance of the
line per phase per mile and use the logarithm to the base 10. Performing this
conversion, we obtain
mileperconductorperohmslog
r
D
kX
′
=
(5.15)
where
Hz60at2794.0
10657.4
3
=
×=
−
fk
(5.16)
assuming identical line conductors.
Expanding the logarithm in the expression of Eq. (5.15), we get
r
kDkX
′
+=
1
loglog
(5.17)
The first term is called X
d
and the second is X
a
. Thus
mileperohmsinfactor
spacingreactanceinductivelogDkX
d
=
(5.18)
mileperohmsinspacing
ft-1atreactanceinductive
1
log
r
kX
a
′
=
(5.19)
Factors X
a
and X
d
may be obtained from tables available in many handbooks.
Example 5.1
Find the inductive reactance per mile per phase for a single-phase line with
phase separation of 25 ft and conductor radius of 0.08 ft.
Solution
We first find r
′
, as follows:
ft0623.0
)7788.0)(08.0(
41
=
=
=
′
−
rer
We therefore calculate
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© 2000 CRC Press LLC
mileperohms7274.0
3906.0
25log2794.0
3368.0
0623.0
1
log2794.0
=+=
=
=
=
=
da
d
a
XXX
X
X
The following MATLAB
script implements Example 5.1 based on Eqs. (5.17)
to (5.19)
The answers obtained from MATLAB
are as follows:
Bundle Conductors
At voltages above 230 kV (extra high voltage) and with circuits with
only one conductor per phase, the corona effect becomes more excessive.
Associated with this phenomenon is a power loss as well as interference with
communication links. Corona is the direct result of high-voltage gradient at the
conductor surface. The gradient can be reduced considerably by using more than
one conductor per phase. The conductors are in close proximity compared with
the spacing between phases. A line such as this is called a bundle-conductor
line. The bundle consists of two or more conductors (subconductors) arranged
on the perimeter of a circle called the bundle circle as shown in Figure 5.2.
Another important advantage of bundling is the attendant reduction in line
reactances, both series and shunt. The analysis of bundle-conductor lines is a
specific case of the general multiconductor configuration problem.
% Example 5-1
r=0.08
D=25
r_prime=0.7788*r
Xa=0.2794*(log10(1/(r_prime)))
Xb=0.2794*(log10 (D))
X=Xa+Xb
EDU»
r = 0.0800
D = 25
r_prime = 0.0623
Xa = 0.3368
Xb = 0.3906
X = 0.7274
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© 2000 CRC Press LLC
Figure 5.2 Bundle Conductor.
Figure 5.3 Single-Phase Symmetrical Bundle-Conductor Circuit.
Inductance of a Single-Phase Symmetrical Bundle-Conductor Line
Consider a symmetrical bundle with N subconductors arranged in a
circle of radius A. The angle between two subconductors is 2
π
/N. The
arrangement is shown in Figure 5.3.
We define the geometric mean distance (GMD) by
[][][]
{}
N
NNN
DDD
1
)2(1)2(1)1(1
GMD
++
= (5.20)
Let us observe that practically the distances D
1(N+1)
, D
1(N+2)
, . . . , are all almost
equal in value to the distance D between the bundle centers. As a result,
D
≅
GMD
(5.21)
Also, define the geometric mean radius as
[]
N
N
ArN
1
1
)(GMR
−
′
=
(5.22)
The inductance is then obtained as