EE 483 Power System 1 Lecture 5: Chapter 4 T.L. Resistance and Inductance

Transmission Line Parameters Bulk electric power is generated at power plants Power plants are located far away from load centers Transmission lines are needed to transmit power from generation to load cites Electric power is transmitted at high voltage to: –Reduce the transmission losses –Reduce cost –Reduce voltage drop –Increase transmission distance and line capacity

Transmission Line Parameters Disadvantage of high voltage transmission: –Transmission cost is proportional to the voltage level –Corona discharge (Ionization of air around the conductor. It is characterized by hissing sound and violet light around the conductor –Corona discharge increases the losses

Transmission Line Voltages High voltage T.L. : 115, 138, 230 kV Extra high voltage T.L. : 345, 500, 765 kV Ultra high voltage T.L. : kV Bundled conductors (more than one conductor for the same phase) is used to reduce the corona power losses and the line inductance

Transmission Line Parameters For each T.L. : Resistance Inductance due to variable magnetic field Capacitance due to electric field Conductance due to leakage current in insulators (neglected)

Types of Conductors  AAC — All aluminum conductor, low cost low weight  AAAC — All aluminum alloy conductor, higher tensile strength  ACSR — Aluminum conductor steel reinforced, layers of aluminum stranded conductors around steel stranded conductors. 24/7: 24 aluminum strands and 7 steel strands ACAR — Aluminum conductor alloy reinforced

Types of Conductors

Conductors are stranded to increase flexibility Layer of strands are spiraled in the opposite direction of the adjacent layer to prevent unwinding

Resistance of Conductors The DC resistance of conductor is given by: Where: : Resistivity of the conductor (m) L : Length of the conductor (m) A : Cross-sectional area of the conductor (m 2 ) In US the L is in feet, A in circular mils (cmil) and  in -cmil/feet

Resistance of Conductors 1 inch = 1000 mils 1 cmil = area of circle with diameter of 1 mil Area in circular inch = [d(inch)] 2 = A CI Area in square inch =  /4[d(inch)] 2 = A SI Area in circular mils = [d(mils)] 2 = A CM Area in square mils =  /4[d(mils)] 2 = A SM 1000 cmils is designated MCM, 1 MCM = 1000 cm

Resistance of Conductors Factors Affecting T.L. Resistance: Temperature Frequency (Skin Effect) Spiraling

Effect of Temperature As the temperature increases the conductor resistance increases For a temperature other than 25 o, the resistance is obtained by: T is temperature at which the conductor will exhibit zero resistance, it is obtained by extrapolation T = 228 for hard drawn aluminum T = 241 for hard drawn copper

Skin Effect It is a nonuniform distribution of the current throughout the cross- sectional area of the conductor Due to the existence of some magnetic flux lines inside the conductor, flux linking a filament inside the conductor is high than flux linking a filament near the surface of the conductor Due to alternating flux an emf is induced on the two filaments

Skin Effect The voltage on the internal filament is higher than the voltage on the external filament This causes the electrons to drift towards the conductor surface Current density close to the conductor surface is higher than the center of the conductor As the frequency increases the current non-uniformity increases

Skin Effect The current non-uniformity causes a reduction in the effective area of the conductor, which increases the resistance To account for skin effect: R ac = k R dc, k ≈ 1.02 at 60 Hz

Spiraling Effect Spiraling causes the stranded conductor length to be more than the given conductor length

Example All aluminum Marigold stranded conductor has a dc resistance of  per 1000 ft at 20 o C and an ac resistance of /mi at 50 o C. If the conductor area is 1113,000 cmil, verify the dc resistance and find the ratio of the ac to dc resistance. Note for All aluminum conductor at 20 o C,  = 2.8x10 -8 .m or 17 .cmil/ft, and T = 228

Problem 4.1 A solid cylindrical aluminum conductor 25 km long has an area of 336,400 circular mils. Obtain the conductor resistance at (a) 20 o C and (b) 50 o C. The resistivity of aluminum at 20 o C is 2.8x10 -8 -m.

Problem 4.3 A three-phase transmission line is designed to deliver MVA at 220-kV over a distance of 63 Km. The total transmission line loss is not to exceed 2.5 percent of the rated line MVA. If the resistivity of the conductor material is 2.8x10 -8 -m, determine the required conductor diameter and the conductor size in circular mils

T.L. Inductance Self inductance:

T.L. Inductance When two coils ( N 1 and N 2 ) are liked by a common flux we have mutual inductance between them where: Where  12 is the mutual flux produced by N 1 but linking with N 2  21 is the mutual flux produced by N 2 but linking with N 1

T.L. Inductance Inductance of a magnetic circuit can be determined as follows: 1.Determine the magnetic field intensity H, using Ampere’s circuital law 2.Determine the magnetic flux density B, ( B = H ) 3.Determine the flux linkage 4.Inductance L = / I

Inductance of a Single Conductor Internal Inductance: Consider a conductor carrying a current I that is uniformly distributed across its cross- section

Inductance of a Single Conductor Assume the radius of the conductor is r To determine the magnetic flux inside the conductor, select a cylinder with radius x, thickness of dx and length of 1 m. H x is constant along the surface of the cylinder, from Ampere’s law:

Inductance of a Single Conductor The current is the area of radius x is: The magnetic field intensity for the current in a circle of radius x is:

Inductance of a Single Conductor The flux passing through the element of thickness dx and length of 1 meter is: The flux linkage/meter caused by the flux in the tabular element is determined by the concept of partial flux linkage as:

Inductance of a Single Conductor

Inductance due to External Flux Linkage: At radius x the magnetic field intensity is:

Inductance of a Single Conductor The total flux linkage the conductor surface and radius D is:

Inductance of a Single Conductor Total Inductance:

Inductance of a Single Phase 2-Wire Line Radius r 1 and r 2 are very small compared with D Assume one conductor is carrying the current to the load and the other is the return path from the load to the supply Supper position can be used to find the loop inductance

Inductance of a Single Phase 2-Wire Line

Example A 60 Hz single phase two-wire overhead transmission line has solid cylindrical copper conductors with 1.5 cm diameter. The conductors are arranged in a horizontal configuration with 0.5 m spacing. Calculate in mH/km: a.The inductance of each conductor due to internal flux linkage only b.The inductance of each conductor due to both internal and external flux linkages c.The total inductance of the line d.The total inductance if the diameter of each conductor is increased by 20% e.The total inductance if the diameter of each conductor is decreased by 20%

Flux Linking a Conductor in a Group of Conductors Consider n solid cylindrical conductor Each conductor carries a phasor current I 1, I 2, I 3,…, I n with a distance D 1x, D 2x, D 3x,…, D nx from a point x

Flux Linking a Conductor in a Group of Conductors

Inductance of a Balanced Three Phase TL with Equal Spacing Inductance per phase is the same as the single phase inductance D s is the geometric mean radius ( GMR ) D s = r e -0.25

Inductance of a Balanced Three Phase TL with Unequal Spacing

It is clear that each line will have its self and mutuals unique to itself. This makes calculation on lines very complicated. To remedy this we transpose the lines

Transposition of Three Phase Lines To equalize the self and the mutual inductance, lines must be transposed

Transposition of Three Phase Lines

D m = the geometrical mean distance = GMD

Problem 4.7 A three-phase, 60-Hz transposed transmission line has a flat horizontal configuration as shown in Figure. The line reactance is ­per kilometer. The conductor geometric mean radius is 2.0 cm. Determine the phase spacing D in meter.

Inductance of Composite Conductors Conductor X composed of n identical filaments, each carries current of I / n Conductor Y composed of m identical filaments, each carries current of - I / m

Inductance of Composite Conductors

Problem 4.5 Find the geometric mean radius of a conductor in terms of the radius r of an individual strand for a.Three equal strands as shown in Figure (a) b.Four equal strands as shown in Figure (b)

Example Calculate the inductance per phase for a three phase fully transposed transmission line if the each conductor is: a.A single solid conductor with radius of 1.5 cm as shown in Fig. b.A bundle of three conductors with a radius of 0.5 cm each as shown in Fig

Example A single phase line is composed of three solid 0.25 cm radius wires. The return circuit is composed of two 0.5 cm radius wires. The arrangement of the conductors is shown in Fig. Find the inductance of each side in H/m

Problem 4.6 One circuit of a single-phase transmission line is composed of three solid 0.5 cm radius wires. The return circuit is composed of two solid 2.5 cm radius wires. The arrangement of conductors is as shown in the Figure. Applying the concept of the GMD and GMR, find the inductance of the complete line in mH/km

Problem 4.10 A single-circuit three-phase transmission transposed line is composed of four ACSR 1,272,000 cmil conductor per phase with horizontal configuration as shown in Figure. The bundle spacing is 45 cm. The conductor code name is pheasant. In MATLAB, use command acsr to find the conductor diameter and its GMR. Determine the inductance per phase per kilometer of the line

Inductance of Three Phase Double Circuit Lines Sometimes HV three-ph lines are made dual circuits, i.e. each circuit is duplicated by another conductor placed symmetrically about a vertical line The dual circuit conductors are fully transposed (i.e. phase ‘ a ’ takes position 1 then position 2 then position 3 in a transposition cycle. The other phases do the same

Inductance of Three Phase Double Circuit Lines The inductance for each phase can be calculated as: Where D s is the geometric mean of the three distances D s1, D s2, and D s3 Here we assume that all conductors are the same size and the GMR = r’

Inductance of Three Phase Double Circuit Lines  D s can be calculated as:

Inductance of Three Phase Double Circuit Lines  D m can be calculated as follows:

Problem 4.11 A double circuit three-phase transposed transmission line is composed of two ACSR 2,167,000 cmil, 72/7 Kiwi conductor per phase with vertical configuration as shown in Figure. The conductors have a diameter of cm and a GMR of cm. The bundle spacing is 45 cm. The circuit arrangement is a 1 b 1 c 1, c 2 b 2 a 2. Find the inductance per phase per kilometer of the line. Find these values when the circuit arrangement is a 1 b 1 c 1, a 2 b 2 c 2.

Problem 4.11