MAHALAKSHMImahalakshmiengineeringcollege.com/pdf/eee/Vsem/EE2302/UNIT 2.pdf · gmd e X A N transp =...

R.Thiyagarajan AP/EEE Page 1

MAHALAKSHMI

ENGINEERING COLLEGE

TIRUCHIRAPALLI- 621213.

QUESTION BANK DEPARTMENT: EEE SEMESTER – V SUBJECT CODE: EE2303 SUBJECT NAME: Transmission & Distribution

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 .


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.


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 qA+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

2 [qAln[DA B rA] + [qbln[rA DA B] + [qcln[DB C DA C]

VA C = 1

2 [qAln[DA C rA] + [qbln[DB 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

Dr

+ qBlnrD

+ qClnDD

Va b = 1

2 qAln

Dr

+ qBlnrD

Va c = 1

2 qAln

Dr

+ qBlnDD

+ qClnrD

Va b = 1

2 qAln

Dr

+ qClnrD


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

62qAln

Dr

+ (qb + qc)lnrD

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

2 qAln

Dr

(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 = qA

VA N = 2 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.


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

2[[qA ln [DB C r] + qB ln [r DB C] + qC ln [DA C DA B] ]

VA B first = 1

2[[qA ln [DA B r] + qB ln [r DA B] + qC ln [DA B DA C] ]

VA B ird = 1

2[[qA ln [DA C r] + qB ln [r 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

2[[qA ln [DA C r] + qB ln [DB C DA B] + qC ln [r DA C] ]

VA C se°cnd = 1

2[[qA ln [DA B r] + qB ln [DA C DB C] + qC ln [r DA B] ]

VA C 3rd = 1

2[[qA ln [DB C r] + qB ln [DA B DA C] + qC ln [r 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

6qA ln[(DA B. DA C. DB C ) r

3] + qB ln[ r

3 (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

6qA ln[(DA B. DA C. DB C ) r

3] + qB ln DA B. DA C. DB C

D A B. DA C. DB C

+ qc lnr3

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

182qAln

DA B. DA C. DB C

r3

+ (qB + qC)ln r3

DA B. DA C. DB C

= 1

6qAln

DA B. DA C. DB C

r3


= 1

2qA

GMDr

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

2 qA ln[(GMD) 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 = qA

VA N transp = (2 ) ln gmd

re

XA N transp = 1

4 fln 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.

)m/H(10052.1

]002.07788.0

3.0log102

]002.0

3.0log

4

1[

2

104

)r

Dlog

28(L

6

7

7

ii00c

)(33.0

50010052.15022

l)L(2X

6

cL

3.Derive the expression for inductance unsymmetrical spacing.(AUC NOV 10)

Inductance of Three-Phase Lines with Asymmetrical Spacing


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


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 sub-

segment 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 sub-

segments 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


The above equation can be simplified as

qA

= 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.

)m(00623.0008.07788.0'r),m(008.00002.0

rrS),m(8D 2'eq For round

conductor wire, 'rD'

s


)m/H(10432.100623.0

8log102

'D

Dlog

2L 67

s

eg01

(a). The same current density means the same cross-sectional area. For 2 sub-conductors with spacing of

0.15m,

)m(00439.000564.07788.0'r),m(00564.02

0002.0rrS),m(8D aa

2a

'eq

)m(0257.015.000439.0d'rD a'

sa

)m/H(10148.10257.0

8log102

'D

Dlog

2L 67

sa

eg0a1

Percentage reduction %8.19%100432.1

432.1148.1%100

L

LL%

1

1a1

(b). The same current density means the same cross-sectional area. For 4 sub-conductors (quad) arranged as

a square of side 0.15m,

)m(00311.000399.07788.0'r),m(00399.04

0002.0rrS),m(8D bb

2b

'eq

)m(062.0215.000311.02ddd'rD 34b

'sb

)/(1072.9062.0

8log102

'log

2

7701 mH

D

DL

sb

eg

b

Percentage reduction %1.32%100432.1

432.1972.0%100

L

LL%

1

1b1

5.Derive the expression for loop inductance single phase line.(AUC May 07)

Inductance of a Two-Wire Single-Phase Line


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

GMRA wb

m

ABP =D

DB P

(BB P) = – 0

2 I ln DB P

D wb

m

BAP =D

DA P

(BA P) = – 0

2 I ln DB P

D wb

m

BBP = 0

2 I ln DB P

GMRB wb

m

The total flux linkage of the system at point P is the algebraic summation of λAP and λBP

P = A P + B P = ( A A P + A B p) + ( B A p + B B P)

p = 0

2 Iln DA p

GMRA

DD

A PDB p

GMRB

DD

A P


p = 0

2 Iln D

2

(GMRA)GMRB

wbm

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 DGMR

wbm

Comparing above , it can be seen that the inductance of the single-phase system is

twice the inductance of a single conductor.

GMDA – str ed = n2

Di j

GMDB – str ed = n2

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

GMDA – str ed = n2

Di j

GMDB – str ed = n2

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


GMR n bundle c°nduct = n dn – 1

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

GMR n bundle c°nduct = 1.094 d3 GMR str ered

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