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ECE 43200/53201 Power Systems Fall 2020

1

Power-Flow Problem Design Project 1

A final (Word document) written report should be submitted by Design Project due date: TBA

For the nine-bus power system shown in Fig.1, comprising three-generators, three-transformers, six-

transmission lines, and four-loads, consider the following design requirements.

A. Bus Input Data

Consider Sbase= 100 MVA, Vbase = 16.5 kV at bus 1, Vbase = 18 kV at bus 8, Vbase = 13.8 kV at bus 9.

The line-to-line voltages are 230 kV base, 60 Hz at busses 2 through 7.

Bus 1 to which a generator is connected is the slack bus. Buses 8 and 9, to which generators (and

load) are connected, are voltage-controlled (or PV buses). Buses 2 through 7are load (or PQ) buses.

The Table 1 summarizes the power-flow input data and unknowns. You are asked to complete the

last column โUnknownsโ.

Table 1 Bus input data and unknowns

Bus Type V

pu

๐น [ยฐ]

๐ท๐

pu

๐ธ๐

pu

๐ธ๐๐๐๐

pu

๐ธ๐๐๐๐

pu

๐ท๐ pu

๐ธ๐ pu

Unknowns

1 Slack 1.0 0.0 – – – – 0 0

2 Load – – 0 0 – – 0 0

3 Load – – 0 0 – – 1.25 0.5

4 Load – – 0 0 – – 0.9 0.3

5 Load – – 0 0 – – 0 0

6 Load – – 0 0 – – 1.0 0.35

7 Load – – 0 0 – – 0 0

8 Constant voltage 1.025 – 1.63 – 99.0 -99.0 0 0

9 Constant voltage 1.025 – 0.85 – 99.0 -99.0 0.8 0.25

B. Transformer Input Data

The transformer data summarized in Table 2 include per-unit winding impedances and exciting branch

admittances, and the maximum MVA ratings.

Table 2 Transformer input data

Bus-to-Bus R

pu

๐ฟ pu

๐ฎ๐ pu

๐ฉ๐ pu

Maximum MVA pu

1-2 0.0 0.05760 0.0 0.0

5-8 0.0 0.06250 0.0 0.0

7-9 0.0 0.05860 0.0 0.0

ECE 43200/53201 Power Systems Fall 2020

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Fig.1 Single-line diagram of the power system

C. 230 kV Transmission Lines Input Data

โช Tower data: height of all conductors, spacing between phases, and the lines of the transmission system are transposed.

โช Conductor Data: conductor name, radius and DC resistance, sag of all conductors, two bundle conductors.

โช Ground Wires Data: ground wires name, radius and DC resistance, sag of all ground wires, spacing between ground wires, height of ground wires above conductors.

The line data include a maximum 12.0 per-unit MVA rating, for all six lines.

ECE 43200/53201 Power Systems Fall 2020

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Do the following for the power system design project:

I. Determine the actual values of the resistance, inductance, and capacitance of each transmission line using PSCAD/EMTDC. Then, calculate the per-unit (pu) line parameters (R, X, G, and B),

present the results in a table format, and use them to determine the series impedance and shunt

admittance of the linesโ ฮ  (pi) models.

II. Determine the bus admittance matrix YBUS, calculate the power flow on the buses and transmission lines, and solve the network for the unknown voltage magnitudes and phase angles

using PowerWorld.

III. Insert a second line between bus 3 and bus 4, with your own new design line parameters. Determine the new line’s effect on V4, the line loadings, and on the total real power losses.

IV. Determine the lowest per-unit voltage and the maximum line/transformer loading both for the initial case and for the case with the line from bus 7 to bus 4 out of service.

V. Determine the acceptable generation range at bus 8, keeping each line and transformer loaded at or below 100% of its MVA limit, and write your comments in the report.

VI. Determine the effect of adding a 200 MVAR shunt capacitor bank at bus 6 on the power system in Fig.1 and write your comments in the report. Find the MVAR rating of the shunt capacitor

bank that establishes V6 at 1.0 pu.

VII. Open one of the transformers and open one of the transmission lines. This results in overloading the other transformers. Re-dispatch the generators in order to remove the overload.

VIII. The transformer between busses 1 and 2 is now a tap-changing transformer with a tap range between 0.9 and 1.1, a step size of 0.00625, and the tap is on the high voltage side of the

in the reactive power output of generator 1, voltages V2 and V4, and the total real power losses.

Extracredit (10%): Use MATLAB to confirm YBUS, the Jacobian matrix for the power system, and apply

Newton-Raphson method to find the power flow solution.

ECE 43200/53201 Power Systems Fall 2020

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Design Project1: Power Flow Problem

You are asked to contact your instructor to determine all the transmission line data needed

for the project completion.

Transmission Lines System Data

Consider the length of transmission lines as follows:

L23=_______km long, L24= L23 km long, L35=_______km long, L47 = L35 km long, L56 =_______km long,

and L67= L56 km long.

Fig.1 shows a sample of a transmission line, having flat towers and conductors of ACSR type in each

phase, an two ground conductors.

Fig. 1 Sample of a transmission system tower and conductorโs configuration

Tower data:

Tower type: __________________

Height of all conductors: _______[m]

Horizontal spacing between phases: _______[m]

Shunt conductance: 0 [S/m]

The lines of the transmission system are transposed.

Conductor Data: Ground Wires Data: Conductor name: ______________; Ground wires name: 7/16″Utility Grade Steel; Ground wire radius: 0.01106 [m];

Conductor radius: _______[m]; Ground wire DC resistance: 2.865 [ฮฉ /km]; Assume sag of all ground wires: 4.5 [m];

Cond. DC resistance: ________[ฮฉ/km]; Spacing between ground wires: 10.12 [m]; Height of ground wires above conductors: 9.63 [m].

Assume sag of all conductors: 10 [m];

Bundle: ______conductor(s)/phase.

ECE 43200/53201 Power Systems Fall 2020

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AC Transmission Lines Components and Electrical Parameters

Transmission lines are used to transfer large amounts of electrical power over long distances. Most

transmission systems use extra-high voltage ac overhead lines for long-distance energy transport, [1],

[2].

There are also few high-voltage dc (HVDC) transmission systems for point-to-point power transport, long

distance bulk energy transfer, which are highly effective (there is no inductive voltage drop). Typical

examples are the HVDC lines which interconnects Los Angeles with Oregon, and the interconnection

between France and England using underwater cables, [1]. High-voltage ac lines are often used to

connect power plants to a city. A typical example is the high-voltage transmission system between

Hoover Dam and Las Vegas, Nevada. Within a city, sub-transmission lines connect the distribution

substations together, and distribution lines, originating from the substations, supply the energy to the

residential and commercial customers. In congested cities, some transmission lines are replaced by

underground cables.

In this chapter we describe the construction and components of the transmission lines. We address the

most important environmental effects of the transmission lines, that is, the electric and magnetic fields.

The chapter derives equations for calculation of transmission line resistance, inductance, and

capacitance; introduces equivalent circuit for the line and present analysis methods to evaluate line

performance.

The transmission lines commonly consist of three-phases and they can be categorized according to their

voltages as follows, [1]:

Extra-high-voltage lines

โช Voltage: 345 kV, 500 kV, 765 kV โช Interconnection between systems

High-voltage lines

โช Voltage: 115 kV, 230 kV โช Interconnection between substations, power plants

Sub-transmission lines

โช Voltage: 46 kV, 69 kV โช Interconnection between substations and large industrial customers

Distribution lines

โช Voltage: 2.4 kV to 46 kV, with 15 kV being the most commonly used โช Supplies residential and commercial customers

High-voltage DC lines

โช Voltage: ยฑ120 kV to ยฑ600 kV โช Interconnection between regions (e.g., Oregon-California)

ECE 43200/53201 Power Systems Fall 2020

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Fig.1 shows typical extra-high-voltage (EHV) transmission systems, with two different high-voltage

transmission lines [1].

(a) (b)

Fig.1 Transmission line systems with two high-voltage lines using

(a) a more aesthetic steel tube tower; (b) a steel lattice tower.

Fig. 1(b) shows the transmission line with two-conductor bundles per phase, to reduce the corona effect

and generation of (radio and television) noise. The shield conductors at the top of the towers are

grounded to protect each line against lightning.

An EHV transmission line requires towers more than 100 ft. tall and a 200-300-ft-wide corridor. These

lines (see Fig.1) are not an aesthetically pleasing sight, and the general public wants the lines be located

far from their backyard. An additional problem is the people’s aversion to electric and magnetic fields

generated by the lines, although any adverse health effects from these fields has been proven.

These considerations have led to the development of transmission line corridors with several lines remote

from populated areas as much as possible.

Sub-transmission lines are used to interconnect substations and to locally distribute energy within a

city or rural area. These lines are supported with a steel tube (on a concrete foundation) or wood tower

(frequently placed in the ground without a foundation). A single conductor is usually used in each phase

because the lower sub-transmission voltage reduces the corona generation, although multiple conductors

per phase may be used to increase the current-carrying capability of the line. The conductors are

supported by post insulators without a cross-arm, or may be supported by suspension insulators attached

to a cross-arm. Lightning protection might be achieved using one grounded shield conductor placed on top

of the tower. The shield conductor is grounded at each tower. A plate or vertical tube electrode (ground

rod) is used for grounding.

ECE 43200/53201 Power Systems Fall 2020

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Fig.2 shows a typical double-circuit 69 kV sub-transmission line with a wooden pole. This line

carries two three-phase circuits and is located along major city streets and roadways. Note the 12.47

kV distribution line under the 69 kV line and the communication cables under the 12.47 kV circuit.

Fig.2 A typical double circuit sub-transmission line, [1]

Distribution lines supply electricity to the residential and commercial buildings.

(A) In rural areas, there are overhead distribution lines that mostly employ a wood tower with cross-

arm(s). Some utilities use concrete towers. The wood is treated with creosote to protect against rotting.

A simple concrete block foundation or no foundation is used. Small porcelain or plastic post insulators

hold the conductors. The insulator shaft is grounded, to eliminate leakage current from causing the

wood tower to ignite and burn. A simple steel rod is used for grounding. A shield conductor is

seldom used.

Fig.3 shows a distribution line with a 240 V voltage cable connection and pole-mounted transformers

that supplies houses in a neighborhood. The distribution line has four conductors; one of them

(second from the right) is the grounded neutral conductor. A 240/120 V service cable is attached

under the distribution line. The line supplies each transformer through a fuse and a disconnect

switch. The single-phase pole-mounted transformers are installed under the cable connection. The

low voltage 240/120 V line supplies nearby homes. For better utilization of the tower, telephone

and/or cable television lines are often attached to the pole under the transformer.

ECE 43200/53201 Power Systems Fall 2020

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Fig.3 A distribution line with pole-mounted transformers

and service cable connection, [1]

Figures 1 – 3 illustrate that the conductors are suspended on the poles, and the distance between the

poles is called the span, and it varies between 100 to 1500 ft., depending on the voltage. A

conductor, which is hung between two points, sags (from the horizontal position). The sag varies with

the conductor temperature, wind, and icing conditions. In summer, the sag is significantly larger than in

the winter. The sag determines the distance between the conductors and the ground. The National

Electrical Safety Code (ANSI C2-1997, p.232B1) specifies the minimum distance, depending on the line

voltage and the use of the land below the line. As an example, below a 22 kV line the minimum

distance must be more than 14.5 ft. if the land is used for pedestrian traffic.

High wind and icing can damage the conductors, and the lines may clap in stormy weather. In order to

limit line damage, tension towers are used to carry tangential, transverse and vertical loads. The

supporting tower (carrying the conductors) carries only transverse and vertical loads. The advantage of

this construction is that any storm damage may be limited to one section that is terminated with

tension towers at each end.

The tension in the conductor depends on the conductor weight, the ambient temperature, wind, ice,

and the sag at the time of installation. During operation, the maximum tension occurs at wintertime.

The sag calculation must consider the ambient temperature, conductor tensile strength, wind, and ice

(B) In urban areas most new construction uses underground cables placed in concrete ducts, which protect

the cables, and simplify maintenance and replacement of faulty cable sections.

Fig.4 shows a typical section of an urban distribution cable system installation. The figure shows the

concrete duct banks with low-voltage cables. Manholes, placed in the street or easement (e.g., front or

back yard), divide the cable system into sections. In the case of a cable fault, the faulty cable is cut,

pulled out, and replaced. The figure shows the transformer and related switchgear placed in a sidewalk

vault.

ECE 43200/53201 Power Systems Fall 2020

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Fig.4 Underground grid network – cutaway view, [1]

High-voltage 230 kV-69 kV cables are used in large municipalities, such as Chicago and New York,

to supply urban substations located in the middle of the city. These cables carry several hundred

megawatts of power to supply skyscrapers and other urban loads.

1. Components of the AC Transmission Lines

The major components of the transmission line are: the three-phase conductors, the insulators, which

support and electrically isolate the conductors, the tower, which holds the insulators and conductors,

the foundation and grounding, and optional shield conductors, which protect against lightning.

Towers and Foundations

The most frequently used tower types are, [1]:

โช Lattice tower, used for 220 kV and above โช Guyed lattice tower, 345 kV and above โช Tapered steel tube with cross-arm, 230 kV and below โช Concrete tower, for distribution and sub-transmission, and โช Wood tower, for distribution.

Most of the metal towers are stabilized by a concrete foundation and grounded. Buried copper rods or

plates are used for grounding. The grounding resistance determines the lightning strike caused

overvoltage. Overvoltage occurs when the voltage is more than 10%-20% greater than the rated voltage.

Good grounding will limit overvoltage.

Aluminum has replaced copper as the most common conductor metal for overhead transmission.

Although a larger aluminum cross-sectional area is required to obtain the same loss as in a copper

conductor, aluminum has a lower cost and lighter weight. Also, the supply of aluminum is abundant,

whereas that of copper is limited, [3].

Typical phase conductors are, [1]:

ECE 43200/53201 Power Systems Fall 2020

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โช Aluminum Conductor Steel Reinforced (ACSR) โช All Aluminum Conductor (AAC), and โช All Aluminum Alloy Conductor (AAAC) โช Aluminum Conductor Alloy-Reinforced (ACAR)

Typical shield conductors are:

โช Aluminum-Clad Steel (Alumoweld), and โช Extra-High-Strength Steel.

Fig.5 Typical ACSR conductor (54/7 Cardinal), [3]

One of the most common conductor types is aluminum conductor steel-reinforced (ACSR), [2], [3],

which consists of layers of aluminum strands surrounding a central core of steel strands, as shown in

Fig.5. Stranded conductors are easier to manufacture, since larger conductor sizes can be obtained by

simply adding successive layers of strands. Stranded conductors are also easier to handle and more

flexible than solid conductors, especially in larger sizes. The use of steel strands gives ACSR conductors

a high strength-to-weight ratio. For purposes of heat dissipation, overhead transmission-line conductors

are bare (no insulating cover).Most frequently used is ACSR, which has a stranded steel center core and

one to four outside layers of aluminum strands.

(a) (b)

Fig.6 (a) Typical bundled conductorsโ arrangements [1];

(b) three-conductor bundle, 500 KV line [2]

Bundled conductors are used for lines above 220 kV to minimize the electric field strength at the

conductor surface to reduce the corona effect and to increase current-carrying capacity. The bundled

conductor contains 2, 3, or 4 conductors in each phase (Fig.6(a)). The distance (d) between the

conductors is 12-18 inches, which is maintained by aluminum bars (spacers) placed at 20-50 ft.

distances along a span.

Rough estimates of the cost of such lines are available for different transmission voltages.

For example, a 345 kV line costs range from 500,000 dollars per mile (1.6km) in rural areas to over

2 million dollars per mile in urban areas of the state of Minnesota (in 2006, [2]).

ECE 43200/53201 Power Systems Fall 2020

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Underground Cables

In many cities and residential areas, transmission lines are opposed because of environmental and

aesthetical concerns in addition to lack of space for large towers. In these areas, the electrical energy

is transported and distributed by underground cables. The underground cables are insulated

conductors that are buried in the ground or placed in underground concrete cable ducts. Most cables

are placed in concrete ducts, although older installations use directly buried cables. This has been

found to be undesirable because cable life is significantly reduced.

Fig.7 Three-phase distribution cable with solid dielectric

Fig. 7 shows a typical, three-phase distribution cable, with a solid dielectric, that is used in

residential areas. The cable uses three stranded aluminum conductors, which are surrounded by a

semiconducting shield layer that produces a smooth surface and reduces the stranding-caused

electrical field concentration. Solid PEX dielectric provides insulation for each conductor. A

grounded semiconducting layer covers the conductor insulation to assure a cylindrical field

distribution. Fillers are placed between the phases to assure a cylindrical cross-section. A grounded

copper wire surrounding the insulated phase conductors forms the outside shield. Finally, the

weather and water-resistant PVC sheet protects the cable from the environment.

Insulators [1]

Ball-and-socket type insulators composed of porcelain or toughened glass are used for most high-

voltage lines. These are also referred to as “cap-and-pin” insulators. The cross-section of a ball-and-

socket type insulator is shown in Fig.8.

Fig.8 Porcelain insulator (cap-and -pin)

ECE 43200/53201 Power Systems Fall 2020

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The porcelain skirt provides insulation between the iron cap and steel pin. The upper part of the

porcelain is smooth, to promote rain washing and cleaning of the surface. The bottom part is corrugated,

which prevents wetting and provides a longer protected leakage path. Portland cement is used to attach

the cap and pin.

The ball-and-socket insulators are attached to each other by inserting the ball in the socket and securing

the connection by a locking key. Several insulators are connected together to form an insulator string.

Strings of insulators are frequently used for transmission lines. Insulators strings are arranged vertically

on support towers (see Fig.1), and near horizontally on dead-end (tension) towers.

Porcelain post insulators are often used to support sub-transmission lines. These insulators replace the

cross-arm of the towers, as shown in Fig.2. The post insulator consists of a porcelain column, with

weather skirts or corrugation on the outside surface to increase leakage distance. For indoor use, the

outer surface is corrugated. For outdoor use, a deeper weather shed is used. The end fitting seals the

inner part of the tube, to prevent water penetration.

Growing industrial pollution increased the number of flashovers on porcelain insulator strings.

Flashover occurs when an arc bridges an insulator, thereby providing a conduction path between the line

and ground. This motivated the development of pollution-resistant non-ceramic (composite) insulators.

Composite insulators are built with mechanical load-bearing fiberglass rods, which are covered by

rubber weather sheds to assure high electrical strength. End fittings connect the insulator to a tower or

conductor.

All high-voltage composite insulators use rubber weather sheds installed on fiberglass rods. The interface

between the weather-shed, fiberglass rod, and the end fittings is carefully sealed to prevent water

penetration. The most serious insulator failure is caused by water penetration to the interface. The most

frequently used materials are silicon rubber, EPDM (ethylene propylene diene monomer) rubber, and the

alloy of the two. The direct molding of the sheds to the fiberglass rod produces high-quality insulators

and is the best approach, although other methods are used successfully. The rubber contains fillers and

additive agents to prevent discharge-caused tracking and erosion.

Post composite insulators are frequently used on medium- and low-voltage lines. This type of

insulator has a fiberglass core covered by a silicon or EPDM rubber weather-shed.

2. Transmission Line Electrical Parameters

The equivalent transmission line circuits for a balanced system are presented at the end of this section.

These circuits are usually referred to as the positive sequence transmission line equivalent circuits.

The transmission line parameters are the resistance R and the inductance L of the conductors, and the

capacitance C to neutral. These parameters are calculated per unit line length and per phase.

The line carries balanced current and is transposed (see Fig.15). This permits the representation of the

line by a single-phase circuit, which corresponds to phase A. The currents and voltages in phases B

and C are shifted by 120ยฐ and 240ยฐ, respectively. The single-phase equivalent circuit carries one-third

of the line power and is supplied by the line-to-neutral voltage.

The conductors of transmission lines have temperature and frequency dependent electric resistance (R). In

addition, when the transmission lines are energized and carry current, the current produces a magnetic

ECE 43200/53201 Power Systems Fall 2020

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field, and the electric charge creates an electric field around the line. The results of the generated fields

are the line inductance (L), and line capacitance (C), which are distributed parameters all through the

length of the line.

Next, the conductance, G, parameter is discussed briefly which is usually neglected in the analysis.

2.1 Conductance G

Conductance accounts for real power loss between conductors or between conductors and ground, [3].

For overhead lines, this power loss is due to leakage currents at insulators and to corona effect where

the surrounding air is ionized and a hissing sound can be heard in misty, foggy weather [2] .

Insulator leakage current depends on the amount of dirt, salt, and other contaminants that have

accumulated on insulators, as well as on meteorological factors, particularly the presence of moisture.

Corona occurs when a high value of electric field strength at a conductor surface causes the air to

become electrically ionized and to conduct. The real power loss due to corona, called corona loss,

depends on meteorological conditions, particularly rain, and on conductor surface irregularities. Losses

due to insulator leakage and corona are usually small compared to conductor I2R loss. Conductance is

usually neglected in power system studies because it is a very small component of the shunt

admittance, and thus in our future analysis the presence of conductance G will be neglected also.

Fig.9 Transmission line distributed parameter model per-phase [2]

Assuming the three phases of the transmission line to be balanced, the line parameters on a per-phase basis

can be easily be calculated, as shown in Fig.9, where the bottom conductor is neutral (in general,

hypothetical) that carries no current in an ideally balanced three-phase arrangement and under a

For unbalanced arrangements including the effect of the ground plane and the shield wires, computer

programs such as PSCAD/EMTDC and EMTP are available that can also include frequency dependence of

the transmission line parameters, [2].

2.2 Line Resistance, R

The resistance of a transmission line in per unit length of the line (in units ฮฉ/m) is designed to be small to

minimize I2R power losses. These losses go down as the conductor size is increased but the costs of

conductors and towers go up. In the U.S., the Department of Energy estimates that approximately 9% of

the generated electricity is lost in transmission and distribution. Therefore, it is important to keep the

transmission line resistance small.

In a bundled arrangement, it is the parallel resistance of the bundled conductors that is of consideration. The

calculation of the stranded conductor ac resistance is complicated and the results are inaccurate. The

practical method is the use of conductor tables that give the resistance at dc and 60 Hz frequency at various

temperatures.

ECE 43200/53201 Power Systems Fall 2020

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The Table 1 below, [5], shows the technical data of ACSR (aluminum conductor steel reinforced)

conductors.

Table1 Technical Data for ACSR (aluminum conductor steel reinforced) Conductors(*)

Cross-Section

Diameter

Layers

Weight

(Ibs per

1000 ft)

Resistance (ohms/mile)

Code Al (kcmil)

Al (mm2)

Total

(mm2)

Stranding Cond. (in.)

Core (in.)

Strength

(kips)

dc 25ยฐC

ac at 60 Hz 60 Hz

GMR

(ft)

Aluminum Steel 25ยฐC 50ยฐ C 75ยฐC 100ยฐC _ 2776. 1407. 1521. 84 x .1818 19 x .1091 2.000 .546 4 3219 81.6 .0338 .0395 .0421 .0452 .0482 .0667

Joree 2515. 1274. 1344. 76 x .1819 19 x .0849 1.880 .425 4 2749 61.7 .0365 .0418 .0450 .0482 .0516 .0621

Thrasher 2312. 1171. 1235. 76 x .1744 19 x .0814 1.802 .407 4 2526 57.3 .0397 .0446 .0482 .0518 .0554 .0595

Kiwi 2167. 1098. 1146. 72 x .1735 7 x .1157 1.735 .347 4 2303 49.8 .0424 .0473 .0511 .0550 .0589 .0570

Bluebird 2156. 1092. 1181. 84 x .1602 19 x .0961 1.762 .480 4 2511 60.3 .0426 .0466 .0505 .0544 .0584 .0588

Chukar 1781. 902. 976. 84 x .1456 19 x .0874 1.602 .437 4 2074 51.0 .0516 .0549 .0598 .0646 .0695 .0534

Falcon 1590. 806. 908. 54 x .1716 19 x .1030 1.545 .515 3 2044 54.5 .0578 .0602 .0657 .0712 .0767 .0521

Lapwing 1590. 806. 862. 45 x .1880 7 x .1253 1.504 .376 3 1792 42.2 .0590 .0622 .0678 .0734 .0790 .0497

Parrot 1510. 765. 862. 54 x .1672 19 x .1003 1.505 .502 3 1942 51.7 .0608 .0631 .0689 .0748 .0806 .0508 Nuthatch 1510. 765. 818. 45 x .1832 7 x .1221 1.465 .366 3 1702 40.1 .0622 .0652 .0711 .0770 .0830 .0485

Plover 1431. 725. 817. 54 x .1628 19 x .0977 1.465 .489 3 1840 49.1 .0642 .0663 .0725 .0787 .0849 .0494

Bobolink 1431. 725. 775. 45 x .1783 7 x .1189 1.427 .357 3 1613 38.3 .0656 .0685 .0747 .0810 .0873 .0472

Martin 1351. 685. 772. 54 x .1582 19 x .0949 1.424 .475 3 1737 46.3 .0680 .0700 .0765 .0831 .0897 .0480

Dipper 1351. 685. 732. 45 x .1733 7 x .1155 1.386 .347 3 1522 36.2 .0695 .0722 .0788 .0855 .0922 .0459

Pheasant 1272. 645. 726. 54 x .1535 19 x .0921 1.382 .461 3 1635 43.6 .0722 .0741 .0811 .0881 .0951 .0466 Bittern 1272. 644. 689. 45 x .1681 7 x .1121 1.345 .336 3 1434 34.1 .0738 .0764 .0835 .0906 .0977 .0445

Crackle 1192. 604. 681. 54 x .1486 19 x .0892 1.338 .446 3 1533 41.9 .0770 .0788 .0863 .0938 .1013 .0451

Bunting 1193. 604. 646. 45 x .1628 7 x .1085 1.302 .326 3 1344 32.0 .0787 .0811 .0887 .0963 .1039 .0431

Finch 1114. 564. 636. 54 x .1436 19 x .0862 1.293 .431 3 1431 39.1 .0825 .0842 .0922 .1002 .1082 .0436

Bluejay 1113. 564. 603. 45 x .1573 7 x .1049 1.258 .315 3 1255 29.8 .0843 .0866 .0947 .1029 .1111 .0416

Curfew 1033. 523. 591. 54 x .1383 7 x .1383 1.245 .415 3 1331 36.6 .0909 .0924 .1013 .1101 .1190 .0420 Ortolan 1033. 523. 560. 45 x .1515 7 x .1010 1.212 .303 3 1165 27.7 .0909 .0930 .1018 .1106 .1195 .0401

Merganser 954. 483. 596. 30 x .1785 7 x. 1783 1.248 .535 2 1493 46.0 .0987 .0995 .1092 .1189 .1286 .0430

Cardinal 954. 483. 546. 54 x .1329 7 x .1329 1.196 .399 3 1229 33.8 .0984 .0998 .1094 .1191 .1287 .0404 Rail 954. 483. 517. 45 x .1456 7 x .0971 1.165 .291 3 1075 25.9 .0984 .1004 .1099 .1195 .1291 .0385

Baldpate 900. 456. 562. 30 x .1732 7 x .1732 1.212 .520 2 1410 43.3 .1046 .1054 .1156 .1259 .1362 .0417

(*) Electric Power Research Institute (EPRI). Transmission Line Reference Book 345 kV and Above, 2nd Ed., J.J. LaForest, Ed., Palo Alto, CA, 1982, p. 110.

As seen in the first column of the table, the conductors are named after birds. The table gives the geometry

of each conductor, including the cross-sections, number of layers, size of the wires, and so forth. These data

are followed by the weight and mechanical strength of the conductors. The dc resistance is given at 25ยฐC

and the ac resistance is given at several different temperatures. These data are used to determine the

transmission line resistance. From the last column, the geometric mean radius (GMR) is used for the line

inductance calculation (see next section).

For the calculation of the line electrical parameters the conductor diameter, the GMR and the conductor ac

resistance at the appropriate ambient temperature are needed. The temperature dependence of the resistance

has a minor effect on the voltage drop calculation, but significantly affects the line loss. The “Cardinal”

conductors are frequently used on high-voltage lines (see Fig.5). As an example of using the table above, the

line parameters for the Cardinal conductor are extracted. The diameter of this conductor is d = 1.196 in., the

GMR is 0.0404 ft., and the conductor 60 Hz ac resistance at 25ยฐC is 0.0998 ohm/mile and at 75ยฐC is 0.1191

ECE 43200/53201 Power Systems Fall 2020

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ohm/mile. Considering the desert climate in Arizona, the resistance at 25ยฐC corresponds to winter

conditions, if the heating effect of the current is included. The 75ยฐC value represents summer conditions,

including the current and direct sunshine-caused radiation heating.

ACSR conductors such as that shown in Fig.5 are considered hollow, as shown in Fig. 10, because of the

following reasons: the electrical resistance of the steel core is much higher than the outer aluminum

strands, and the skin-effect phenomenon is present at 60-Hz (or 50-Hz) frequency (see details in Appendix

A).

Fig. 10 Cross-section of ACSR conductors, (b) skin-effect in a solid conductor

The line resistance R depends on the length of the conductor l (not accurate for stranded conductors),

the resistivity of the material ฯ (that increases with temperature), and inversely on the effective cross-

sectional area A of the conductor through which the current flows:

๐ = ๐๐

๐ด (1)

The effective area A in Eq. (1) depends on the frequency due to the skin effect, where the current at

60-Hz frequency is not uniformly distributed throughout the cross-section; rather it crowds towards

the periphery of the conductor with a higher current-density J as shown in Fig. 10b for a solid good

conductor (it will be more uniform in a hollow conductor), and decreases exponentially such that at

the skin-depth, ฮด, the current density is decreasing to e-1=0.368 from the conductor’s surface.

The skin depth of a material at a frequency f is [4]:

๐ฟ = 1

โ๐๐๐๐ (2)

where, ๐ = ๐๐๐0 = ๐๐(4๐ ร 10 โ7) H/m, and the conductivity, ๐ =

1

๐ , is the reciprocal of the

conductor’s resistivity.

The average power loss in a conductor with skin effect present may be calculated by assuming that the

total current is distributed uniformly in one skin depth. We may apply this to a conductor of circular

cross section with little error, provided that the conductorโs radius a is much greater than the skin

depth.

โ The resistance at a high frequency where there is a well-developed skin effect is therefore found by considering a slab of width equal to the circumference 2๐a and thickness ๐ฟ. Hence

ECE 43200/53201 Power Systems Fall 2020

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๐ = ๐๐

๐ด =

๐

๐๐ด =

๐

2ฯaฯฮด [ฮฉ] (3)

Example 2-1:

Skin-depth and Conductor’s Resistance Calculation

In Fig.10a, for the aluminum conductor , calculate the skin-depth, ๐ฟ, at 60-Hz frequency, and the resistance for a large conductor diameter, D=30 mm, and a line length l=100 km, at 60Hz.

Solution: The permeability of aluminum can be consider to be that of free space, that is, ๐ = ๐0 = 4๐ ร 10โ7๐ป/๐, and the resistivity is specified as ฯ = 2.65ร 10โ8๐บ๐. Substituting these values in Eq. (2), for a frequency f = 60 Hz, the skin depth is calculated as

๐ฟ = 1

โ๐๐๐๐ =

1

โ๐60(4๐10โ7)( 108

2.65 )

= 0.018747๐ โ 18.75๐๐.

The conductor resistance at 60 Hz, substituting the conductor radius a = D/2 = 15mm in Eq.(3)

๐60๐ป๐ง = ๐๐

๐ด =

๐๐

2ฯaฮด =

(2.65ร10โ8)ร105

2ฯ(15ร10โ3)ร(18.75ร10โ3) = 1.5ฮฉ.

Due to the skin effect, in ACSR conductors, the aluminum thickness T, as shown in Fig.10a, is kept at the order of

the skin depth and any further thickness of aluminum will essentially be a waste that will not result in decreasing the

overall resistance to ac currents, [2]. In case of a type of the ACSR conductor called the Bunting conductor,

the ratio T/D = 0.37, where D = 1.302 inches as listed in the previous Table 1, [5]. Therefore, the skin effect results in a resistance only slightly higher at 60 Hz compared to at dc in such a “hollow” conductor, where the resistance at dc

is listed as 0.0787 ohms/mile versus 0.0811 ohms/mile at 60 Hz, both at the temperature of 25ยฐC.

โข The tables with parameters given in English units (different from SI units) are using conductor cross-sectional area expressed in circular mil (cmil). One inch = 1000 mils, and 1cmil = ฯ/4 sq

mil, [3]

โข For stranded conductors, alternate layers of strands are spiraled in opposite directions to hold the strands together. Spiraling makes the strands up to 1% or 2% longer than the actual

conductor (line) length.

โข Resistivity of metal conductors varies over normal operating temperatures according to ๐๐2 =

๐๐1 ( T2+T

T1+T ), where ๐๐2 and ๐๐1 are resistivities at temperatures T2 and T1 in Celsius degrees,

respectively. T is a temperature constant that depends on the conductor material, and is listed in

tables such as the Table 2 given below [3]

Table 2 Resistivity and temperature constant of conductor metals

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2.3 Line Series Inductance, L

Magnetic Field Generated by Transmission Lines [1]

The current in a conductor produces a magnetic field inside and around the conductor as shown in Fig.11.

The magnetic flux lines produced by current flowing in an isolated, long straight, round conductor are

concentric circles, and the right-hand rule gives the field direction.

Fig.11 magnetic field produced by the current in an conductor

Both ac and dc lines generate magnetic fields. The intensity of the field is proportional with the current.

The dc current generated magnetic field is constant. The dc line generated magnetic field has not been seen

as a health concern because it is constant and is much less than the Earth’s magnetic field, which is about

half of a gauss. However, ac generated magnetic fields, which vary in time according to the system

frequency, have received concern, although their magnitude is also much less than the Earth’s magnetic

field.

The calculation of the line inductance requires the calculation of the magnetic flux inside and outside the

conductor, [1].

External Magnetic Flux

The relation between the magnetic field and the current that generates the field is described by Ampere’s

Law :

โฎ๏ฟฝโโ๏ฟฝ ๐๐โโ โ = โ ๐ผ๐ (4)

where, ๏ฟฝโโ๏ฟฝ is the magnetic field intensity vector measured in amp-turn/m; ๐ผ๐ are the currents within the closed path that generates the field, in amp.

The magnetic field is constant along a flux line, which simplifies the integration. This case is shown in

Fig.12.

Fig.12 Ampere’s law for a magnetic field generated by a

conductor’s current (flowing out of page), [1]

ECE 43200/53201 Power Systems Fall 2020

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In this case, the magnetic field intensity H, produced by the current I flowing through conductor, can be

calculated as ๐ป = ๐ผ

2๐๐ , and the magnetic flux density is ๐ต = ๐0๐ป = ๐0

๐ผ

2๐๐ .

The magnetic flux is the integral of the flux density on a surface S.

ฮฆ = โซ ๏ฟฝโ๏ฟฝ ร ๐๐โโโโ ๐

(5)

As an example, the flux between the conductor surface at rc and point R in Fig.12 is

ฮฆ = โซ ๏ฟฝโ๏ฟฝ ร ๐๐โโโโ ๐

= ๐0 โซ ๐ผ

2๐๐ ๐dr

R

rc = ๐0

๐ผ๐

2๐ ln (

R

rc ) (6)

where, l is the length of the conductor.

Fig.13 External magnetic flux calculation (current into page), [1]

Consider the conductor shown in Fig.13 with a radius rc and carrying current Ic. To calculate the

external flux we select a magnetic flux line at a radius of x. The flux passing through the plane

placed between P1 and P2 is calculated by integrating the flux density between D1 and D2. The

magnetic flux is calculated in a similar manner as shown in Eqs.(5) and (6) and is finally given

by

๐ท12 = ๐0 ๐ผ๐๐

2๐ ln (

D2

D1 ) (7)

This equation will be used to calculate the phase-conductor generated flux and inductance.

The calculation of the magnetic field inside the conductor is based on the magnetic energy stored

in the conductor (beyond the goal of this course) and is finally calculated as

๐ท๐๐๐ก = ๐0 ๐ผ๐๐

2๐

1

4 .

The interesting conclusion is that the internal flux in a conductor carrying a current is independent of

Total Magnetic Flux of the Conductor

The total conductor generated magnetic flux is the sum of the internal and external fluxes. Equation

(7) gives the conductor generated flux passing through the plane placed between two points P1 and P2 as

shown in Fig.14, [1].

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Fig.14 Magnetic flux calculation between the

conductor and an external point

In this case, D1 is equal to the conductor radius, rc. The substitution of D1= rc in Eq.(7) and replacement

of D2 with D gives a general equation for the external magnetic flux:

๐ฑ๐๐๐ = ๐0 ๐ฐ๐๐

2๐ ln (

D

r๐ ) (8)

where, D is the distance from the center of the conductor; rc is the conductor radius; Ic is the

conductor current; and l is the conductor length.

The total conductor generated flux, which is the sum of the internal and external fluxes, is [1]:

๐ฑ๐ = ๐0 ๐ฐ๐๐

2๐ ln (

D

r๐ ) + ๐0

๐ฐ๐๐

2๐

1

4 (9)

Although this formula is suitable for calculations, the equation is modified by substituting 1

4 = ๐๐ (

1

๐ โ

1 4

)

into the above expression. The two natural logarithms are combined together, which results in a

simplified formula. A further simplification is the assumption of an l unit length which gives the flux

in weber per unit length. In the United States Wb/mile is the preferred unit, whereas in Europe Wb/km

is used.

๐ฑ๐

๐ = ๐ฐ๐

๐0

2๐ [๐๐ (

D

r๐ ) +

1

4 ] = ๐ฐ๐

๐0

2๐ ๐๐ (

D

eโ0.25r๐ ) = ๐ฐ๐

๐0

2๐ ๐๐ (

D

GMR ) (10)

where GMR is called geometric mean radius.

The GMR for a solid conductor is GMR = rce -0.25, but for a stranded conductor the GMR values are

given in conductor tables. As an example, the GMR for the Cardinal conductor is 0.0404 ft. from

Table 1, [5].

Equation (10) is used to calculate the flux generated by a conductor. That equation can be interpreted

that the actual conductor is replaced by an ideal conductor (tube) with a radius of GMR. The magnetic

field inside this ideal conductor tube is zero.

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Inductance of a Three Phase Line

In the case of a three-phase line, the current in each phase conductor produces a magnetic field.

The inductance of a three-phase line for a balanced system is usually referred to as positive sequence

inductance. It is assumed that the line is transposed and the line currents are balanced.

Fig.15 shows the concept of line transposition, where each phase occupies each position for one-third of

the line length.

Fig.15 Transposition of a three-phase line

The transposition equalizes the line and permits the calculation of the flux using an average spacing, called

the geometric mean distance (GMD). The GMD is calculated by

๐บ๐๐ท = โ๐ท๐ด๐ต๐ท๐ด๐ถ๐ท๐ต๐ถ 3

(11)

If line transposition is employed, then this implies that DAB, DBC and DAC may be replaced by the GMD.

The approach to compute the line inductance is derived using as example a three-phase line, built

conductors arranged in horizontal position (see Fig. 17). The field lines are concentric circles around

the conductors. Conductor A is exposed to three fluxes: ฮฆAA, the flux generated by the phase A current, ฮฆAB, the flux generated by the phase B current, and ฮฆAC, the flux generated by the phase C current.

(a) (b) (c)

Fig.16 (a) Fluxes produced by currents in phase A (a), phase B (b) and phase C (c)

The current in conductor A produces a flux of ฮฆAA, as shown in Fig.16a between A and F, that can be calculated with Equation (10):

ฮฆAA = ๐ผ๐ด ๐0

2๐ ๐๐ (

๐ท๐ด๐น

๐บ๐๐๐ ) ๐ (12)

where ๐บ๐๐๐ is the conductor GMR obtained from the conductor tables.

The current in conductor B produces a flux of ฮฆAB, as shown in Fig.16b between A and F, that can be calculated with Equation (7):

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ฮฆAB = ๐ผ๐ต ๐0

2๐ ๐๐ (

๐ท๐ต๐น

๐ท๐ด๐ต ) ๐ (13)

Similarly, the current in conductor C produces a flux of ฮฆAC, as shown in Fig.16c between A and F, that can be calculated as

ฮฆAC = ๐ผ๐ถ ๐0

2๐ ๐๐ (

๐ท๐ถ๐น

๐ท๐ด๐ถ ) ๐ (14)

The total flux linkage with conductor A is the sum of the fluxes generated by the phase currents,

that is, the addition of Equations (12), (13), and (14):

๐ฝ๐ = ๐ฐ๐จ ๐0

2๐ ๐๐ (

๐ท๐ด๐น

๐บ๐๐๐ ) ๐+๐ฐ๐ฉ

๐0

2๐ ๐๐ (

๐ท๐ต๐น

๐ท๐ด๐ต ) ๐ + ๐ฐ๐ช

๐0

2๐ ๐๐ (

๐ท๐ถ๐น

๐ท๐ด๐ถ ) ๐ (15)

Similar equations describe the flux linkage with conductors B and C.

For balanced conditions, the sum of the three currents is zero, which permits the elimination of Ic

from Equation (15). The substitution of Ic = -IA – IB results in:

๐ฝ๐ = ๐ฐ๐จ ๐0

2๐ ๐๐ (

๐ท๐ด๐นDAC

๐ท๐ถ๐น๐บ๐๐๐ ) ๐ + ๐ฐ๐ฉ

๐0

2๐ ๐๐ (

๐ท๐ต๐นDAC

๐ท๐ถ๐นDAB ) ๐ (16)

This equation gives the flux linking phase A by calculation of the flux paths through the plane

between conductor A and an arbitrarily selected distant point F. Moving point F to infinity results

in DAF = DBF = DCF. Substitution of these values yields:

๐ฝ๐ = ๐ฐ๐จ ๐0

2๐ ๐๐ (

DAC

๐บ๐๐๐ ) ๐ + ๐ฐ๐ฉ

๐0

2๐ ๐๐ (

DAC

DAB ) ๐ (17)

Similar equations can be derived for phases B and C. Equation (17) shows that if the line is built

with unequal conductor spacing, the voltage drop along the line will be unbalanced even if the

currents are balanced. In other words, if the line is supplied by balanced three-phase voltage,

the phase voltages at the opposite end of the line will be different. That is why utilities commonly

transpose the conductors to avoid these unbalanced voltages. Fig. 15 shows this concept.

If line transposition is employed, then this implies that DAB, DBC and DAC may be replaced by the GMD

given in Equation (11). The substitution of Equation (11) into (17) eliminates the second term on

the right-hand side of Equation (17).

Thus, the simplified flux equation is: ฮฆA = ๐ผ๐ด ๐0

2๐ ๐๐ (

GMD

๐บ๐๐๐ ) ๐ (18)

โ Finally, the line series inductance L of a phase, per unit length, calculated from (18), is

L = ฮฆA

๐ผ๐ด๐ =

๐0

2๐ ๐๐ (

GMD

๐บ๐๐๐ ), [H/m] (19)

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Many high-voltage transmission lines are built with bundled conductors. The application of bundled

conductors reduces the line inductance. The bundled conductors are replaced by an equivalent conductor.

The GMR of this conductor depends on the number of conductors in the bundle and the distance between

the conductors within the bundle.

The equivalent GMRs of the bundles are:

Two-conductor bundle: ๐บ๐๐ = โ๐ ๐บ๐๐๐ (20)

Three-conductor bundle: ๐บ๐๐ = โ๐2๐บ๐๐๐ 3

(21)

Four-conductor bundle: ๐บ๐๐ = โ๐3๐บ๐๐๐ 4

(22)

where d is the distance between the conductors in the bundle, and GMRc is the conductor GMR

obtained from the conductor tables.

Example 2-2

Inductance Calculation of a Three-phase Line:

A three phase line, built with Cardinal conductors arranged as shown in Fig.17, has the following geometrical

data, [1]:

Fig.17Conductor arrangement of the transmission line in Example 2

Phase spacing: DAB = 35 ft; DBC = DAB; DAC = 2DAB.

Conductor height and GMR: H line = 70 ft; GMRC = 0.0404 ft.

The phasor line currents are: ๐ผ๐ด = 1500๐ ๐0ยฐ; ๐ผ๐ต = ๐ผ๐ด๐

โ๐120ยฐ; ๐ผ๐ถ = ๐ผ๐ด๐ +๐120ยฐ.

The coordinates of a distant point F (see Fig.16) are:

DAF = 1000 ft; DBF = DAF; DCF = DAF.

T he length of the line is: l = 100 mi

The flux linking conductor A is determined by calculation of the flux paths through the plane between conductor A and

an arbitrarily selected distant point F. The current in conductor A produces a flux of ๐ฝ๐๐ between A and F, as shown in Fig.16a, that can be calculated with Equation (12) and its absolute value is:

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ฮฆAA = ๐ผ๐ด ๐0

2๐ ๐๐ (

๐ท๐ด๐น

๐บ๐๐๐ ) ๐ = 488.44 ๐๐

The current in conductor B creates a flux of ๐ฝ๐๐ between A and F, as shown in Fig.16b, that can be calculated with Equation (13) and its absolute value is:

ฮฆAB = ๐ผ๐ต ๐0

2๐ ๐๐ (

๐ท๐ต๐น

๐ท๐ด๐ต ) ๐ = 161.85 ๐๐

Similarly, the current in conductor C produces a flux of ๐ฝ๐๐ between A and F, as shown in Fig.16c, that can be calculated with Equation (14) and its absolute value is:

ฮฆAC = ๐ผ๐ถ ๐0

2๐ ๐๐ (

๐ท๐ถ๐น

๐ท๐ด๐ถ ) ๐ = 123.89 ๐๐

The total flux linkage with conductor A is the sum of the fluxes generated by the phase currents, that can be calculated

with the Equation (17):

๐ฝ๐ = ๐ฐ๐จ ๐0

2๐ ๐๐ (

DAC

๐บ๐๐๐ ) ๐ + ๐ฐ๐ฉ

๐0

2๐ ๐๐ (

DAC

DAB ) ๐,

and its absolute value is: | ๐ฝ๐| = 344.53 Wb. Similar equations can be derived for phases B and C. The Equation (17) shows that if the line is built with unequal

conductor spacing, the voltage drop along the line will be unbalanced even if the currents are balanced. In other

words, if the line is supplied by balanced three-phase voltage, the phase voltages at the opposite end of the line will

be different. Utilities commonly transpose the conductors to avoid these unbalanced voltages. If line transposition is

employed, then this implies that DAB, DBC and DAC may be replaced by the GMD. In this case the simplified flux

calculation is given by Equation (18):

ฮฆA = ๐ผ๐ด ๐0

2๐ ๐๐ (

GMD

๐บ๐๐๐ ) ๐ = 337.74 ๐๐.

The inductance of phase A is calculated using the flux-inductance relation

๐ฟ๐ด = ฮฆA

๐ผ๐ด = 0.225 H.

The unit length inductance is given by Equation (19):

L = ฮฆA

๐ผ๐ด๐ =

๐0

2๐ ๐๐ (

GMD

๐บ๐๐๐ ) = 2.25

๐๐ป

๐๐

If the per phase reactance of the line per unit length is needed, this is calculated for a frequency of 60 Hz as

๐๐ฟ = ๐L = (2๐๐) ๐0

2๐ ๐๐ (

GMD

๐บ๐๐๐ ) = 0.85

ฮฉ

mi

ECE 43200/53201 Power Systems Fall 2020

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References

1. George G. Karady and Keith E. Holbert, “Electrical Energy Conversion and Transportation – An Interactive Computer-Based Approach”, IEEE Press, Wiley-Interscience, 2005

2. N. Mohan, “First Course on Power Systems”, MNPERE, Minneapolis, 2006, ISBN 0-9715292-7-2.

3. Duncan J. Glover, Mulukutla S. Sarma, and Thomas J. Overbye, “Power Systems – Analysis and Design”, 4th Ed., Thomson, 2008

4. William H. Hayt, Jr., “Engineering Electromagnetics”, McGrawHill.

5. Electric Power Research Institute (EPRI), “Transmission Line Reference Book: 345kV and above”, 2nd edition.

6. Electric Power Research Institute (EPRI),www.epri.com

PNW Fall 2020 ECE 43200-001 Elmnt Power Syst Engr XLST – Lecture 4 – Transmission Lines

1. Transmission Lines_Notes Part I

2. Project1_Description

3. Project1_Transmission Lines Data