Introduction to power system Chapter two

1. ….PER UNIT[P.U]
CHAPTER- TWO
Constants of Transmission line

2. …Con’t
• Transmission line of electric power is done by 3-phase, 3-wire
overhead lines.
• An electric transmission line can be represented by
a) Resistance
b) Inductance
c) Capacitance
• With constantly distributed along its length.
• These are constant or parameters of the line.

3. Inductors and Inductance
• An inductor is a device which stores energy in a magnetic field.
• The inductance L of an inductor is the ratio of its total magnetic flux
linkages to the current (I) through the inductor or
• Mutual inductance between two circuits is defined as the flux
linkages of one circuit due to the current in the second circuit per
ampere of current in the second circuit.
• If the current I2 produces λ12 flux linkages with circuit 1, the mutual
inductance is

4. Magnetic Field Intensity Due to Current Conductors
• Current carrying conductor produces a magnetic field which is in the form of
closed circular loops around the conductor.
• The relation of the magnetic field direction to the current direction can be easily
remembered by means of the right hand rule.
• Let us consider a long current carrying conductor with radius R as shown in
Figure below.
• We will consider here that the current is uniformly distributed across the section of
the conductor. The flux linkages here will be both due to internal flux and
external flux.
• The magnetic field intensity due to the current distribution inside the conductor is
calculated as follows:
1. Consider a cylinder with radius r < R: -The current enclosed by the cylinder
will be:
where I is the current through the conductor.

5. …Can't
• Therefore, the magnetic field intensity at a
distance r due to this current, using
Ampere’s Law is
• which means that the magnetic field
intensity inside the conductor is directly
proportional to the distance from the
center of the conductor.
2. Now consider a cylinder with radius r > R.
Applying Ampere’s Law is
 which means H is inversely
proportional to r outside the
conductor.
 The variation of H as a function of r is
shown in above Figure.

6. Inductance of Two-Wire Transmission Line
• Inductance is the flux linkages per ampere.
• So the objective is to find out the flux linkages to this system of
conductors.
• There are two flux linkages:
a. Due to internal flux, and
b. Due to external flux.

7. Internal flux linkages:
• In order to determine the internal flux linkages, we start with the
magnetic field intensity H at any distance r < R.
This flux density as we see is varying with r.
 We can assume this to be constant over an infinitesimal
distance dr.
 The flux lines are in the form of circles concentric to the
conductor.
 Therefore, the flux lines passing through the concentric
cylindrical shells of radii r and r + dr.
d = B . Area normal to flux density = Bdrl

8. …Con’t
 From this it is clear that the flux linkage due to internal flux is independent of the
size of the conductor.

9. External flux linkages:
• These flux linkages are due to the flux lines outside the conductor.
• There will be no flux line that encloses both the conductors.
• This is because for any distance r > D the total current enclosed is zero.
• The magnetic field intensity H due to one conductor at any distance
R ≤ r < D
• The flux density B can be considered uniform over a distance dr.

10. …Con’t
 The radius R’ is that of a fictitious (its far from the true)conductor assumed to have no internal flux
linkages.
 But with the same inductance as the actual conductor with radius R. The quantity 𝑒 Τ
−1
4= 0.7788.
 The multiplying factor of 0.7788 to adjust the radius in order to account for internal flux
linkages applies only to solid round conductors.

11. …Con’t
READING ASSIGNMENT
Flux Linkages of one conductor in a group of conductors

12. Inductance of three phase[unsymmetrically] transmission line
• Consider a single circuit 3-φ system in Figure having three conductors a, b and c
carrying currents Ia, Ib and Ic respectively.
• The three conductors are unsymmetrically placed i.e., a ≠ b ≠ c and each has a
radius of R meters.
• The flux linkage of conductor a due to Ia, Ib and Ic from equation

13. …Con’t
 It is clear from the expressions for inductances of conductors a, b and c that the three inductances are unequal.
 They contain imaginary term which is due to the mutual inductance.
 In case the transmission line is transposed i.e., each conductor takes all the three positions of the conductors.
 Each position for one third length of the line we will see later.
 The average value of the inductance

14. Transposition of power line and Composite conductor
• When the conductors of a three-phase line are not spaced equilaterally, the problem of finding the
inductance becomes more difficult.
• The flux linkages and inductance of each phase are not the same. A different inductance in each
phase results in an unbalanced circuit.
• Balance of the three phases can be restored by exchanging the positions of the conductors at
regular intervals a long the line so that each conductor occupies the original position of every other
conductor over an equal distance. Such an exchange of conductor positions is called transposition.
• Transposition results in each conductor having the same average inductance over the whole
cycle.

15. …Cont.
• Modern power lines are usually not transposed at regular intervals
although an interchange in the positions of the conductors may be
made at switching stations in order to balance the inductance of the
phases more closely.
• So that each conductor occupies the original position of every other
conductor over an equal distance.

16. …Con’t
• To find the average inductance of one conductor of a transposed line , we first
determine the flux linkages of a conductor for each position it occupies in the
transposition cycle and then determine the average flux linkages.
• To find the phasor expression for the flux linkages of a in position 1 when b is in
position 2 and c is in position 3 , we obtain

17. …Con’t

18. INDUCTANCE
CALCULATIONS FOR
COMPOSITE
&
BUNDLED CONDUCTORS

19. …Cont.
• The composite conductors consist of two groups of conductors each
having m and n number of strands respectively as shown in Fig.
• The current is assumed to be equally divided amongst the strands.
• One group of conductors act as a ‘go’ conductor for the single-phase
line and the other as the ‘return’.
• The current per strand is I/m ampere in one group and – I/n ampere
in the other.

20. …Cont.
• The flux linkage of strand 1 in conductor A is
given by

21. …Cont.

22. …Cont.
• The mnth root of the product of the mn distances between m strands
of conductor A and n strands of conductor B is called geometric mean
distance (GMD) and is denoted as Dm and,
• The m^2th root of m^2 distances i.e., the distances of the various
strands from one of the strands and the radius of the same strand,
the distances of such m groupings constitute m^2 terms in the
denominator, is called the geometric mean radius (GMR) or self GMD
and is denoted as Ds.
• The expression for inductance of conductor A consisting of m strands
from the above equation becomes.

23. Concept of Geometric Mean Distance(GMD)
• Geometric mean distance is a mathematical concept used for the calculation of
inductance.
• By definition the geometric mean distance of a point with respect to a number of
points is the geometric mean of the distances between that point and each of the
other points.
• The geometric mean distance of point P with
respect to five points on the circle is

24. …Cont.
• In case, the number of points on the circle are increased to infinity, it can be seen intuitively that,
the geometric mean distance between the point P and the infinite points on the circle will be the
geometric mean of all the distances and will correspond to the distance between the point P and
center of the circle.
• The GMD between two circular areas will be the distance between the centers of the two areas.
• The GMD method does not apply strictly to non-homogeneous conductors such as ACSR(
Aluminum conductor steel reinforced) or when the current is not uniformly distributed over the
section of the conductor.
• An approximate value of inductance for ACSR conductors can be calculated by assuming
negligible current in the steel strands.

25. Examples
1) Determine the inductance of a 3-phase line operating at 50Hz and conductors
arranged as follows. The conductor diameter is 0.8cm.
2) A conductor consists of seven identical strands each having a radius of r.
Determine the factor by which r should be multiplied to find the self GMD of
the conductor.

27. INDUCTANCE of DOUBLE CIRCUIT 3-PHASE LINE
• The double circuit line consists of three conductors in each circuit (Fig. below).
• The three conductors corresponds to three phases, a, b, c and a′, b′, and c′.
• Conductors a and a′ are electrically parallel and constitute one phase.
• Similarly conductors b, b′ and c, c′ form other phases. This means there are two conductors (strands) per
phase.
• Since the conductors are not symmetrically placed, to calculate the inductance of the line, the conductors
should be transposed.

28. …Cont.
• The GMD of the conductors in phase ‘a’ with the conductors in other
two phases in position 1,

29. …Cont.
 Here the conductors of two phases are placed diagonally opposite rather than in the same horizontal plane, in
all the three positions.
 By doing this the self GMD of the conductors is increased whereas the GMD reduced, thereby the inductance
per phase in lowered.

30. Bundled Conductors
• For voltages in excess of 230 kV, it is in fact not possible to use a
round single conductor.
• Instead of going in for a hollow conductor it is preferable to use more
than one conductor per phase which is known as bundling of
conductors.
• A bundle conductor is a conductor made up of two or more sub-
conductors and is used as one phase conductor. It is found that the
increase in transmission capacity justifies economically the use of two
conductor bundles on 220 kV lines.

31. …Cont.
The following are the advantages in using bundle conductors:
a) Reduced reactance.
b) Reduced radio interference.
c) Reduced surge impedance ( ).
d) Reduced corona loss.
e) Reduced voltage gradient.
 The reactance of the bundle conductors is reduced because the self GMD of the conductors is
increased.
 The basic difference between a composite conductor and bundled conductor is that the sub
conductors of a bundled conductor are separated from each other by a distance of almost 30cms
or more and the wires of a composite conductor touch each other.

32. skin effects and proximity effect
• When direct current flows in the conductor, the current is uniformly distributed across the section of the conductor.
• Whereas, the flow of alternating current is non-uniform, with the outer filaments of the conductor carrying more
current than the filaments closer to the center.
• This results in a higher resistance to alternating current than to direct current and is commonly known as skin
effect.
• This effect is more, the more is the frequency of supply and the size of the conductor.
• A conductor could be considered as composed of very thin filaments.
• The inner filaments carrying currents give rise to flux which links the inner filaments only.
• Where as, the flux due to current carrying outer filaments enclose both the inner as well as the outer filaments.

33. Cause of skin effects….
• The flux linkages per ampere to inner strands is more as compared to
outer strands.
• Hence the inductance/impedance of the inner strands is greater than
those of outer strands which results in more current in the outer
strands as compared to the inner strands.
• This non-uniformity of flux linkage is the main cause of skin effect.

34. …Cont.
• The alternating magnetic flux in a conductor caused by the current
flowing in a neighboring conductor gives rise to circulating currents
which cause an apparent increase in the resistance of a conductor.
• This phenomenon is called proximity effect.

35. Electric field and potential
difference

36. Electric field and potential difference
• The flow of current through a conductor gives rise to a magnetic field and
• The charging of conductor results in an electric field. A charge if brought in the
vicinity of this electric field experiences a force.
• The intensity of this field at any point is defined as the force per unit charge and
is termed as electric field intensity designated as E.
• The units of this field are newton per coulomb or volts per meter.
• The direction of electric field intensity is the same as the direction of the force
experienced by the unit charge.

37. …Cont.
• Consider the field produced by a thin line of charge as shown in Fig. below.
• Let a positive charge ρL coulomb per meter be uniformly distributed along the infinitesimally thin
line of infinite length.
• It is required to find out electric field intensity E at P
due to infinite line charge.
• Take an infinitesimal charge ρL*dy which could be
considered a point charge.
• The electric field intensity dE at P due to this charge is
given by
• Therefore, total intensity at P due to infinite line of charge

38. Potential Difference Between Two Points Due to Line
of Charge
• The potential at any distance r from the charge is the work done in moving a unit positive charge from
infinity to that point.
• The potential difference between two points at distances r1 and r2 is the work done in moving a unit positive
charge from r2 to r1 as shown in Fig. or
• It is the line integral of the electric field intensity between points r2 and r1.
• Here Er is taken as negative because the unit charge is to be moved
against the direction of the electric field intensity Er.
• Now substituting for Er from equation

39. Capacitance of single and three phase Transmission line
• Capacitor: It is an electrical device which consists of two conductors separated by a dielectric
medium and is used for storing electrostatic energy.
• Capacitance: The capacitance of a capacitor is the ratio of the charge on one of its conductors to
the potential difference between the conductors.
• Consider a 1-φ transmission line. Let a fixed potential V be applied between the conductors. so that
the charge per unit length of each conductor is ρL coulomb per meter.
• The length of the line is very large as compared with the distance of separation h of the conductors,
and radius r of each conductor is very small as compared to the distance of separation.

40. …Cont.
• It is to be noted that the charge ρL coulomb/meter is distributed on the
surface of the conductor which is non-uniformly distributed over the
surface.
• such that it has higher density on the adjacent sides of the conductors. This
charge distribution can be considered as a line charge as in the previous
section.
• The surface of the conductor represents an equipotential surface with
circular cross-section and radius equal to r.
• So the objective will be to find out the equivalent line charge distribution
for a system of two conductors with operating voltage V, distance of
separation h and radius of the equipotential surface r.
• This equivalent charge distribution, as can be seen from Fig. below.

41. …Cont.
• Since the charge is assumed to be uniformly distributed
over the surface of the conductor.
• This could be considered as concentrated along the axis
on conductor.
• The electric field intensity at point P due to ρL is
Total electric field intensity at P

42. …Cont.
• The potential difference between the conductors
• And the capacitor will be

43. Comparison…
• Compare this expression with the expression for inductance equation of a single phase transmission
line.
• Equation for inductance contains a constant term corresponding to the internal flux linkages.
• Whereas, since charges reside on the surface of the conductor, similar term is absent in the
capacitance expression.
• As a result of this, the radius in the expression for capacitance is the actual outside radius of the
conductor.
• Whereas, for inductance equation the radius is the self GMD of the conductor.
• The concept of self GMD is applicable for inductance calculation and not for the capacitance.

44. …Cont.
• Sometimes it is required to know the capacitance between one
conductor and a neutral point between them which will be defined as
the charge on one of the conductors per unit of voltage difference
between the neutral and the conductor.
• This means the capacitance of one conductor with respect to the
neutral plane is two times the capacitance of the single-phase line

45. Capacitance of three phase Transmission line
• For an un-transposed line the capacitances between conductor to neutral of the
three conductors are unequal.
• In transposed lines the average capacitance of each conductor to neutral is the
same as the capacitance to neutral of any other phase.
• The three positions of the conductors are shown in Fig.

46. …Cont.
• Let a point P be at a large distance D from the system such that Da,Db
and Dc are approximately same.
• It is required to find out the potential of conductor a due to charges ρa,
ρb and ρc per unit length of the conductors.
• Since it is a 3-phase balanced system, taking ρa as the reference
charge,

47. …Cont.
• The potential of conductor ‘a’ with respect to point P due to the
charge on the conductor itself,
• Similarly, the potential of conductor ‘a’ due to the charges ρb and ρc
respectively are

48. …Cont.

49. …Cont.
• Since the conductors b and c also occupy the same three positions
as occupied by conductor a, the average voltage of the conductors is
same and,
• Therefore, the capacitance is also the same.

50. Capacitance of Double Line
• Normally two configurations of conductors are used:
i. Hexagonal spacing, and
ii. Flat vertical spacing.
• First of all an expression of capacitance for hexagonal
spacing derived.

51. Hexagonal Spacing
• Since the conductors of the same phase are connected in parallel the charge per
unit length is the same Fig below.
• Also, because of the symmetrical
arrangement the phases are balanced and the
conductors of each phase are also balanced if
the effect of ground is neglected.
• Therefore, the transposition of conductors is
not required.

52. …Cont.
• Assume a point P very far from the system of conductors such that the
distances of the conductors from P are almost same.
• It is to be noted here that point P corresponds to almost zero potential.
• The potential of conductor a with respect to point P due to the charge
on the conductor itself and the charges on conductors b, c, a′, b′ and c′
is given by

53. …Cont.

54. …Cont.
• The above equation represents an expression for the capacitance of
conductor a alone,
• Whereas, there are two conductors per phase a and a′.
• Therefore, the capacitance of the system per phase will be twice the
capacitance of one conductor to neutral, i.e.,
 Since the conductors of different phases are symmetrically placed,
the expression for capacitance for other phases will also be the same.

55. Flat Vertical Spacing
• The conductors of different phases are not symmetrically placed;
• Therefore, the derivation of capacitance expression will require the
transposition of conductors as shown in Fig below.

56. …Cont.
• It is required to find out average voltage of conductor a in the three
different positions due to the charge on conductor a and the conductors
b, c, a′, b′ and c′.
• For this we again assume a point very far from the system of
conductors such that Da ~− Db ~− Dc ~− Da′ ~− Db′ ~− Dc′.
• Since point P is at a very large distance from the system of
conductors, the potential of point P is approximately zero.
• The potential of conductor a in position 1.

57. …Cont.

58. …Cont.
• The capacitance C per phase of the system

59. C
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CHAPTER-THREE
Performance of Transmission lines