MAHALAKSHMI
ENGINEERING COLLEGE
TIRUCHIRAPALLI- 621213.
QUESTION BANK
SEMESTER – V
SUBJECT NAME: Transmission & Distribution
DEPARTMENT: EEE
SUBJECT CODE: EE2303
UNIT 2
TRANSMISSION LINE PARAMETERS
PART-A
1. On what a factor does the skin effect depends? (AUC MAY 2009)
The skin effect depends upon the 1.type of the material 2.frequency of the current 3.and diameter of
conductor& shape of conductor. It increases with the increase of cross-section, permeability and supply
frequency.
2. What is the bundling of the conductors.(AUC MAY 2009)
It is a conductor made up of 2 or more sub conductors and is used as one phase conductors.
3. Define skin effect.(AUC NOV 2010)
The steady current when flowing through the conductor, does not distribute uniformly, rather it has the
tendency to concentrate near the surface of the conductor. This phenomenon is called skin effect.
4. What are advantages of using bundled conductors? (AUC NOV 2010)
Reduced reactance.
Reduced voltage gradient.
Reduced corona loss.
Reduced Interference.
5. What is the need of transposition?(AUC NOV 2011)
Transposition means changing the positions of the three phases on the line supports twice over the
total length of the line .the line conductors in practice ,are so transposed that each of the three possible
arrangements of conductors exit for one-third of the total length of the line .
R.Thiyagarajan AP/EEE
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6. Define The Term Disruptive Voltage (AUC NOV 2011)
When the ac transmission voltage goes above the particular value, the corona effect will occur. The
corresponding value of voltage is called as critical disruptive voltage.
7. Briefly explain ACSR.(AUC MAY 2007)
ACSR is one of the most used conductors in transmission lines. It consists of alternate layers of
stranded conductors, spiraled in opposite directions to hold the strands together, surrounding a core of
steel strands. Figure.shows an example of aluminum and steel strands combination.
8. What is meant by proximity effect? (AUC NOV 2007)
The alternating magnetic flux in a conductor caused by the current flowing in a neighbouring conductor
gives rise to a circulating current which cause an apparent increase in the resistance of the conductor
.This phenomenon is called as proximity effect
9 . Why skin effect is absent in dc system?
The steady current when flowing through a conductor distributes itself uniformly over the whole cross
section of the conductor .That is why skin effect is absent in dc system.
9. What is the effect of skin effect on the resistance of the line?
Due to skin effect the effective area of cross section of the conductor through which current flow is
reduced. Hence the resistance of the line is increased when ac current is flowing.
10. On what factors the skin effect depend?
Nature of the material, Diameter of the wire , Frequency and shape of the wire.
11.
Define symmetrical spacing.
In 3 phase system when the line conductors are equidistant from each other then it is called
symmetrical spacing.
12.
What is the necessity for a double circuit line?
To reduce the inductance per phase and to increase the efficiency.
R.Thiyagarajan AP/EEE
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11. Mention the factors governing the inductance of a line.
Radius of the conductor and the spacing between the conductors.
12. Define a neutral plane.
It is a plane where electric field intensity and potential is zero.
13. Define voltage regulation.
Voltage regulation is defined as the change in voltage at the receiving (or load) end when the full-load
is thrown off, the sending-end (or supply) voltage and supply frequency remaining unchanged.. % voltage
regulation= ((Vs-Vr)/Vr)*100 where Vs is the voltage at the sending end Vr is the receiving end voltage.
PART-B
1.Derive the equation for capacitance three phase unsymmetrical spaced over head line.
(AUC MAY 09) (May 07)(Nov 07)
Capacitance of a Three-Phase Line:
Consider a three-phase line with the same voltage magnitude between phases, and assuming a
balanced system with abc (positive) sequence such that q A+qB+qC=0. The conductors have radii rA, rB, and rC,
and the space between conductors are DAB, DBC, and DAC (where DAB, DBC, and DAC > rA, rB, and rC).Also, the
effect of earth and neutral conductors is neglected.
The expression for voltages between two conductors in a single-phase system can be extended to
obtain the voltages between conductors in a three-phase system. The expressions for VAB and VAC are
VA B =
1 [q ln[D
A
AB
2
rA] + [qbln[rA
VA C =
1 [q ln[D
A
AC
2
rA] + [qbln[DB C
DA B] + [qcln[DB C
DA C]
DA B] + [qcln[r C
DA C]
If the three-phase system has triangular arrangement with equidistant conductors such that
DAB=DBC=DAC=D, with the same radii for the conductors such that rA=rB=rC=r (whereD > r), the expressions for
VAB and VAC are
Va b =
1
2
qAln D + qBln r + qCln D
r
D
D
Va b =
1
2
qAln D + qBln r
r
D
Va c =
1
2
qAln D + qBln D + qCln r
r
D
D
Va b =
1
2
qAln D + qCln r
r
D
R.Thiyagarajan AP/EEE
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Balanced line voltage with the sequence abc, expressed in terms of the line to neutral voltages are
VA B = 3 VA N 30° and VA c = – VC A = 3 VA N 30°
Where V AN is the line to neutral voltage . therefore VAN can be expressed in terms of VAB and VAC as
VA N = V A B + VA C
3
And thus substituting VAB and VAC from the equation we have
VA N =
1
6
2qAln D + (qb + qc)ln r
r
D
Under balanced condition qA+ qB+qC=0 or –qA= (q B+ qc) then the final expression for the line to neutral voltage
is
V A N = 1 qAln D
r
2
(V)
The positive sequence capacitance per unit length between phase A and neutral can now be obtained. The
same result is obtained for capacitance between phases B and C to neutral
CA N = qAA N = 2
V
ln D
r
(F/m)
Capacitance of Stranded Bundle Conductors
The calculation of the capacitance in the equation above is based on
1. Solid conductors with zero resistivity (zero internal electric field)
2. Charge uniformly distributed
3. Equilateral spacing of phase conductors
In actual transmission lines, the resistivity of the conductors produces a small internal electric field and
therefore, the electric field at the conductor surface is smaller than the estimated. However, the difference is
negligible for practical purposes.
Because of the presence of other charged conductors, the charge distribution is nonuniform, and
therefore the estimated capacitance is different. However, this effect is negligible for most practical
calculations. In a line with stranded conductors, the capacitance is evaluated assuming a solid conductor with
the same radius as the outside radius of the stranded conductor. This produces a negligible difference.
Most transmission lines do not have equilateral spacing of phase conductors. This causes differences between
the line-to-neutral capacitances of the three phases. However, transposing the phase conductors balances the
system resulting in equal line-to-neutral capacitance for each phase and is developed in the following manner.
Consider a transposed three-phase line with conductors having the same radius r, and with space
between conductors DAB, DBC, and DAC , where DAB, DBC, and DAC > r.
R.Thiyagarajan AP/EEE
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Assuming abc positive sequence, the expressions for VAB on the first, second, and third section of the
transposed line are
VA B Sec°nd =
1 [[q ln [D
A
BC
2
VA B first =
1 [[q ln [D
A
AB
2
r] + qB ln [r
VA B
1 [[q ln [D
A
AC
2
r] + qB ln [r
ird
=
r] + qB ln [r
DB C] + qC ln [DA C
DA B]]
DA B] + qC ln [DA B
DA C]]
DA C] + qC ln [DA B
DB C]]
Similarly the expression for VAC on the first, second, and third section of the transposed line are
VA B first =
1 [[q ln [D
A
AC
2
VA C se°cnd =
VA C 3rd =
r] + qB ln [DB C
1 [[q ln [D
A
AB
2
1 [[q ln [D
A
BC
2
r] + qB ln [DA C
r] + qB ln [DA B
DA B] + qC ln [r
DA C]]
DB C] + qC ln [r
DA C] + qC ln [r
DA B]]
DB C]]
Taking the average value of the three sections, we have the final expressions of VAB and VAC in the transposed
line
V A B = VA B first + VA B Sc°nd + VA b 3rd
3
= 1
6
qA ln[(DA B. DA C. DB C )
3
3
r ] + qB ln[ r
(DA B. DA C. DB C) ] + qc ln DA C. DA C. DB C
DA C. DA C. DB C
V A C = VA C first + VA C Sc°nd + VA C 3rd
3
= 1
6
qA ln[(DA B. DA C. DB C )
3
r ] + qB ln
DA B. DA C. DB C
D A B. DA C. DB C
3
+ qc ln
r
DA C. DA C. DB C
For a balanced system where –qA =(qB+ qc) the phase to neutral voltage VAN is
V A N transp = VA B transp + VA C Transp
3
=
1
18
= 1
6
3
r
2qAln DA B. DA3 C. DB C + (qB + qC)ln
D
.
D
A
B
A C. DB C
r
qAln DA B. DA3 C. DB C
r
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= 1
2
qA GMD
r
Where GMD = (DA B. DB C. DC A = geometrical mean distance for a three-phase line.
For bundle conductors, an equivalent radius re replaces the radius r of a single conductor and is
determined by the number of conductors per bundle and the spacing of conductors. The expression of re is
similar to GMR bundle used in the calculation of the inductance per phase, except that the actual outside
radius of the conductor is used instead of the GMR phase. Therefore, the expression for VAN is
V A N transp = 1 qA ln[(GMD)
2
re]
where
re=(d n_1 r)1/n=equivalent radius for up to three conductors per bundle (m)
re=1.09 (d3r)1\4=equivalent radius for four conductors per bundle (m)
d=distance between bundle conductors (m)
n=number of conductors per bundle
Finally, the capacitance and capacitive reactance, per unit length, from phase to neutral can be evaluated as
CA N = qAA N transp =
V
(2 )
XA N transp = 1 ln GMD
re
4 f
ln gmd
re
m
2.A 50Hz single-phase line consists of two parallel conductors 30cm apart. If each conductor has a
diameter of 4mm, calculate the inductive reactance of a 500m length of the line.
Solution:
Using equation 8 in lecture note, noting that we assume the conductor is a round conductor (given as
“diameter of 4mm” in the question). Equation 8 is about how to calculate the inductance of ONE conductor in a
circuit. The inductance for the line is the sum of the inductance of the “go” and the “return” conductors.
Lc
( 0
8
4
log
D ii
)
r
10 7 1
[
2
4
log
0
2
0.3
]
0.002
0.3
2 10 7 log
]
0.7788 0.002
1.052 10 6
XL
2 ( Lc ) l
2 2 50 1.052 10 6 500
0.33 ( )
( H / m)
3.Derive the expression for inductance unsymmetrical spacing.(AUC NOV 10)
Inductance of Three-Phase Lines with Asymmetrical Spacing
R.Thiyagarajan AP/EEE
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It is rather difficult to maintain symmetrical spacing as shown in Fig. while constructing a transmission line.
With asymmetrical spacing between the phases, the voltage drop due to line inductance will be unbalanced
even when the line currents are balanced. Consider the three-phase asymmetrically spaced line shown in Fig.
in which the radius of each conductor is assumed to be r . The distances between the phases are denoted by
Dab, Dbcand Dca. We then get the following flux linkages for the three phases
Let us define the following operator
Note that for the above operator the following relations hold
Let as assume that the current are balanced. We can then write
Substituting the above two expressions in we get the inductance of the three phases as
R.Thiyagarajan AP/EEE
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It can be seen that the inductances contain imaginary terms. The imaginary terms will vanish only when
Dab= Dbc= Dca. In that case the inductance will be same as given by
Transposed Line
The inductances that are given in are undesirable as they result in an unbalanced circuit configuration. One
way of restoring the balanced nature of the circuit is to exchange the positions of the conductors at regular
intervals. This is called transposition of line and is shown in Fig. In this each segment of the line is divided into
three equal sub-segments. The conductors of each of the phases a, b and c are exchanged after every subsegment such that each of them is placed in each of the three positions once in the entire segment. For
example, the conductor of the phase-a occupies positions in the sequence 1, 2 and 3 in the three subsegments while that of the phase-b occupies 2, 3 and 1. The transmission line consists of several such
segments.
1.
In a transposed line, each phase takes all the three positions. The per phase inductance is the average
value of the three inductances calculated in above equ. We therefore have
This implies
we have a + a2 = - 1. Substituting this in the above equation we get
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The above equation can be simplified as
q
A
=
q
2 x
(c)
Defining the geometric mean distance ( GMD ) as
H/m
Symmetrically spaced conductors. Comparing these two equations we can conclude that GMD can be
construed as the equivalent conductor spacing. The GMD is the cube root of the product of conductor
spacings.
4.A 3-phase fully-transposed overhead line is to be designed with a geometric mean spacing between
phases of 8m. Calculate the inductance per km, if a single round conductor with a cross-sectional area
of 200mm2 is used for each phase. What will be the percentage reduction in inductance if (AUC DEC10)
(a) The phase conductors are split into 2 (twin) sub-conductors, with a spacing of 150mm?
(b) The phase conductors are split into 4 (quad) sub-conductors, arranged at the corner of a square of side
150mm?
Assume a constant design current density.
Solution:
Under the balanced three-phase situation, analysis is carried out for a single-phase
system and positive-sequence inductance L1 is used as the series impedance of the line.
D eq '
8 (m), S
'
conductor wire, D s
r2
r
0.0002
0.008 (m), r ' 0.7788 0.008
0.00623 (m) For
round
r'
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L1
0 log D eg
2
D's
2 10 7 log
8
0.00623
1.432 10 6
( H / m)
(a). The same current density means the same cross-sectional area. For 2 sub-conductors with spacing of
0.15m,
D eq '
Dsa '
L1a
ra 2
8 (m), S
ra ' d
0.0002
2
ra
0.00439 0.15
0 log D eg
2
D sa '
0.00564 (m), ra ' 0.7788 0.00564 0.00439 (m)
0.0257 (m)
2 10 7 log
8
0.0257
1.148 10 6
L1a L1
100%
L1
Percentage reduction %
( H / m)
1.148 1.432
100%
1.432
19.8%
(b). The same current density means the same cross-sectional area. For 4 sub-conductors (quad) arranged as
a square of side 0.15m,
D eq '
8 (m), S
D sb '
4r ' d
b
L1b
0
2
log
rb 2
d d
Deg
Dsb '
0.0002
4
rb
0.00311 0.153
2
2 10
Percentage reduction %
0.00399 (m), rb ' 0.7788 0.00399 0.00311 (m)
7
log
8
0.062
2
9.72 10
L1b L1
100%
L1
7
0.062 (m)
( H / m)
0.972 1.432
100%
1.432
32.1%
5.Derive the expression for loop inductance single phase line.(AUC May 07)
Inductance of a Two-Wire Single-Phase Line
R.Thiyagarajan AP/EEE
Page 10
Now, consider a two-wire single-phase line with solid cylindrical conductors A and B with the same
radius r, same length l, and separated by a distance D, where D > r, and conducting the same current I, as
shown in Fig.. The current flows from the source to the load in conductor A and returns inconductor B (IA=-IB).
The magnetic flux generated by one conductor links the other conductor. The total flux linking
conductor A, for instance, has two components: (a) the flux generated by conductor A and (b) the flux
generated by conductor B which links conductor A.
As shown in Fig. , the total flux linkage from conductors A and B at point P is
λ AP= λAAP+ λABP
λ BP= λBBP+ λBAP
Fig 10: External magnetic flux around conductors in a two-wire single-phase line.
Where
λ AAP=flux linkage from magnetic field of conductor A on conductor A at point P
λ ABP=flux linkage from magnetic field of conductor B on conductor A at point P
λ BBP=flux linkage from magnetic field of conductor B on conductor B at point P
λBAP =flux linkage from magnetic field of conductor A on conductor B at point P
The expressions of the flux linkages above, per unit length, are
AAP
=
0
2
I ln DA P A
GMR
wb
m
DB P
ABP
=
(BB P)
= –
I ln DB P
D
2
wb
m
(BA P)
= –
I ln DB P
D
2
wb
m
D
0
DA P
BAP
=
D
BBP
=
0
2
I ln DB P B
GMR
0
wb
m
The total flux linkage of the system at point P is the algebraic summation of λAP and λBP
P
=
p
=
AP
0
2
+
Iln
BP
=(
DA p
A
GMR
AAP
+
D
AP
D
R.Thiyagarajan AP/EEE
)+(
ABp
DB p
B
GMR
BAp
+
)
BBP
D
AP
D
Page 11
2
p =
Iln
0
2
D
(GMRA)GMRB
wb
m
If the conductors have the same radius, rA=rB=r, and the point P is shifted to infinity, then the total flux linkage
of the system becomes
Fig 11: Flux linkage of (a) conductor A at point P and (b) conductor B on conductor A at point P. Single-phase
system.
=
0
2
Iln
D
(GMRA)
wb
m
and the total inductance per unit length becomes
L1 ase sys m =
I
=
0
Iln
D
GMR
wb
m
Comparing above , it can be seen that the inductance of the single-phase system is
twice the inductance of a single conductor.
2
GMDA – str
ed
=n
GMDB – str
ed
=n
2
Di j
Di j For a line with stranded conductors, the inductance is determined using a new GMR
value named GMR stranded, evaluated according to the number of conductors. If conductors A and B in the
single-phase system, are formed by n and m solid cylindrical identical sub conductors in parallel,
respectively,then
2
GMDA – str
ed
=n
GMDB – str
ed
=n
2
Di j
Di j
Generally, the GMRstranded for a particular cable can be found in conductor tables given by the manufacturer.
If the line conductor is composed of bundle conductors, the inductance is reevaluated taking into account the
number of bundle conductors and the separation among them. The GMRbundle is introduced to determine the
final inductance value. Assuming the same separation among bundle conductors, the equation for GMRbundle,
up to three conductors per bundle, is defined as
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n–1
GMR n bundle c°nduct = n d
GMR str ered
Where
n=number of conductors per bundle
GMR stranded=GMR of the stranded conductor
D=distance between bundle conductors
For four conductors per bundle with the same separation between consecutive conductors, the GMR bundle is
evaluated as
3
GMR n bundle c°nduct = 1.094 d GMR str ered
R.Thiyagarajan AP/EEE
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