Chapter 4. Transmission Line Parameters

ELCT 551: Power System Analysis & Design

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Topics

• General Information: Design consideration; Resistance; Conductance; Inductance.

• Line Inductance:

• Line Capacitance:

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Solid Cylindrical Conductor

Stranded Conductor

Solid Cylindrical Conductor

Stranded Conductor

1. General Information: Design Considerations

• Elements for Electric Power Transmission Line

Conductors

Insulators

Supporting Structures

Shield Wires

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Conductors

Copper (not often used due to expense)

All aluminum conductor (AAC)

Aluminum conductor, steel reinforced (ACSR)

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AAC

• AAC is composed of strands of aluminum for electrical conductivity

• Conventional strands are circular in cross section, but one type has a trapezoidal shape (more compact)

Aluminum strand

Concentric circular lay Unilay

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ACSR

• ACSR has central strands of steel for mechanical strength, with outer strands of aluminum for electrical conductivity

• Conventional strands are circular in cross section, but trapezoidal shape is available

Steel

Aluminum

Conductors

• ACSR (Aluminum Conductors Steel Reinforced)

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Insulators

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Insulators

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Support Structures and Shield Wires

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765 kV Transmission Line 345 kV Transmission Line

Factors for Transmission Line Design

Electrical Factors

Mechanical Factors

Environmental Factors

Economic Factors

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Electrical Factors

• Lighting:

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Electrical Factors

• Line Sag:

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Major Blackouts

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Immediate causes of the 8/14/03 blackout

• 1:30 Loss of East Lake generator (over-excitation)

• 2:02 Loss of Stuart-Atlanta (tree contact)

• 2:02 MISO system model becomes inaccurate

• 2:14-3:08 Loss of software in FE control center

• 3:05 Loss of Harding-Chamberlain (tree contact)

• 3:32 Loss of Hanna-Juniper (tree contact)

• 3:41 Loss of Star-S. Canton (tree contact)

• 4:06 Loss of Sammis-Star (high overload looked like fault to “zone 3” of the protection system)

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Why so much “tree-contact”?

• Trees were overgrown because right-of-ways had not been properly maintained

• Lines expand and sag due to heat; more prone in summer with high temperature & low winds; more prone with high current.

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BEFORE After

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Mechanical Factor

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• Enough strength of the conductors, insulator strings, and support structures

Environmental and Economic Factors

• Biological effects.

• Lowest cost: construction and maintenance

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Line Resistance

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Line Resistance

• Temperature dependence of resistivity r :

r(T2) = r(T1) (T2+T)/(T1+T)

where T = 228.1°C and r(20 °C) = 2.83 10-8 ohm-m for hard-drawn Aluminum.

• Resistivity and hence line resistance increase as conductor temperature increases (changes is about 8% between 25 C and 50 C)

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Line Resistance

• Because ac current tends to flow towards the surface of a conductor, the resistance of a line at 60 Hz is slightly higher than at dc.

• Because ACSR conductors are stranded, actual resistance, inductance and capacitance needs to be determined from tables.

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Example 4.1

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2. Line Inductance

• Review of Magnetic Theory:

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Magnetic Flux

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Flux linkages and Faraday’s law

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Inductance

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Inductance of Solid Cylindrical Conductor

• To development models of transmission lines, we first need to determine the inductance of a solid cylindrical conductor. To do this we need to determine the wire’s total flux linkage, including

– 1. flux linkages within the wire

– 2. flux linkages outside of the wire

• We’ll assume that the current density within the wire is uniform and that the wire has a radius of r.

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Flux linkage inside

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Flux Linkage outside of the wire

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• Flux linkage per unit length outside wire (out to R):

1 = 1 = rR B(x) dx

= o rR i/(2

x) dx 1 = o i/(2

) ln(R/r) = 2 10-7 i ln(R/r) [Wb/m]

Line Total Flux & Inductance

• Total flux linkage per unit length:

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Inductance Simplification

• Inductance expression can be simplified usingtwo exponential identities:

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External flux linkage of a conductor between D1 and D2

• External flux linkage of a conductor between D1 and D2:

12 = 2 10-7 i ln(D2/D1) [Wb/m]

D1

cond D2

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Flux linkage of a conductor due to current in another

• Flux linkage of cond 1 out to Dk: – Flux linkage of conductor 1 due to

current in conductor k:

1k = 2 10-7 ik ln(Dk/D1k) [Wb/m]

D1k

ik Dk

cond 1 D1 R

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Flux linkage of a conductor due to a group of conductors

• Consider n conductors with i1+…+in=0

1 = 2 10-7{i1[1/4 + ln(D1/r)] + i2ln(D2/D12) +…+ in ln(Dn/D1n)} [Wb/m]

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Flux linkage of a conductor due to a group of conductors

1 = 210-7{i1[1/4+ln(1/r)]

+i2 ln(1/D12) +…+inln(1/D1n)

+i1ln(D1)+i2ln(D2)+…+ in ln(Dn)

- [i1ln(D1)+i2ln(D1)+…+in ln(D1)]}

where the last term in [ ] is equal to zero.

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1 = 210-7{i1[1/4+ln(1/r)] + i2ln(1/D12) +…+ inln(1/D1n) + i1ln(1) + i2ln(D2/D1) +…+ in ln(Dn/D1)}

Now let the point at distance D go to infinity

D1 = D2 = … = Dn

and all terms like ln(D2/D1) ln(1) = 0

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1 = 210-7{i1[1/4+ln(1/r)] + i2ln(1/D12) +…+ inln(1/D1n)} [Wb/m]

For example n=3:

1 = 210-7{i1[1/4+ln(1/r)] + i2ln(1/D12) + i3ln(1/D13)} [Wb/m]

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Equilaterally spaced three-phase line

• Conductors have radius r and spacing D

D

D

D

a b

c

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Equilaterally spaced three-phase line

• Conductors have radius r and spacing D:

a = 210-7 [ialn(1/r')+ibln(1/D)+icln(1/D)]

• ia + ib + ic = 0:

a = 210-7 ia[ln(1/r') - ln(1/D)]

a = 210-7 ia ln(D/r') [Wb/m]

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Equilateral 3 phase line per-phase inductance

• This is the inductance per phase due to balanced three-phase currents. La = a/ia = 210-7 ln(D/r') [H/m]

• For single-phase line: La = a/ia = 410-7 ln(D/r') [H/m]

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Bundled conductors

• Use the distance between bundle centers for phase spacing, use the geometric mean radius of bundle instead of r' :

d

d d

d Ds = (r' d)1/2 for 2 conductor bundle Ds = (r' d2)1/3 for 3 conductor bundle

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Bundled conductor equilateral 3 phase line

d D

d

D

d

Bundle GMR: Ds = (r' d2)1/3

La = 210-7 ln(D/Ds) [H/m]

D

each subconductor: radius = r GMR = r'

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Transposed lines

A

A

B

B

C

C 1

2

3

1

2

3

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Transposed lines

• Positions are numbered 1, 2, 3

• Phases are lettered A, B, C and color coded

• Use the geometric mean spacing Deq= (D12D23D31)1/3

• Often Deq is called equivalent spacing

1 2

3

A B C 1 2 3

1 2 3 C A B

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Average per phase inductance of transposed line

• La = 210-7 ln(Deq/Ds) [H/m]

• Use geometric mean spacing Deq

• Use geometric mean bundle radius Ds

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Average per phase inductance of transposed line

• For equilateral spacing Deq = D

• For single conductor per phase Ds = r' = conductor GMR

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Resistance and Inductance

• For practical stranded conductors, look up the resistance and the conductor GMR from tables supplied by the manufacturer

• For ACSR, see table A.4 in the book for GMR (use in place of r') and the AC resistance for several temperatures

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Suggested procedure for inductive reactance

• Calculate equivalent spacing = Deq (in any units)

• Look up conductor GMR from table and convert to same units as Deq

• Calculate bundle GMR = Ds in same units as Deq

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Suggested procedure for inductive reactance (cont’d)

• Calculate inductance per phase L = 210-7 ln(Deq/Ds) [H/m]

• Calculate inductive reactance in ohms per mile from X = 2 f L [ohm/m] [1609m/mile], where f is frequency in Hz

Review of Line Inductance

• General Equation for any over-head line:

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2

1

N M

2’

1’

Conductor X Conductor Y

GMD

GMR

Review of Line Inductance - 2

• Special cases:

– Each conductor has only one sub-conductor)

– Example:

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Review of Line Inductance - 3

• Special cases:

– Bundled conductors with equal phase spacing

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d D

d

D

d

Bundle GMR: Ds = (r' d2)1/3

La = 210-7 ln(D/Ds) [H/m]

D

each subconductor: radius = r GMR = r'

Review of Line Inductance - 4

• Special cases:

– Bundled conductors with transposed spacing

• Use the geometric mean spacing Deq= (D12D23D31)1/3

• La = 210-7 ln(Deq/Ds) [H/m]

• Use geometric mean spacing Deq

• Use geometric mean bundle radius Ds

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1 2

3

A B C 1 2 3

1 2 3 C A B

3. Line Capacitance

• E field due a line charge q, on a wire of radius r, is directed radially outward

E(R) = q/(2eR) if R > r

= 0 if R < r

• Voltage drop from b to a

vba= RbRa E(R) dR = q ln(Ra/Rb) / (2e)

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Line Capacitance

• If all the charges sum to zero, we can follow a procedure similar to that for the inductances:

voltage of conductor k = vk =

[q1ln(1/Dk1)+…+qkln(1/rk)+…

+qnln(1/Dkn)]/(2e)

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Capacitance of three-phase equilateral line

• Three conductors each equally spaced D and each having radius r

• Balanced three-phase charges

• Air dielectric e = eo

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Capacitance of three-phase equilateral line

• Air dielectric e = eo

va = qa ln(D/r) / (2 eo) = voltage drop with respect to a neutral point p equidistant to three conductors.

C = 2 eo / ln(D/r) [F/m] to neutral point

• For Single-Phase two-wire line C = eo / ln(D/r) [F/m]

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Capacitance calculations

• For transposed non-equilateral lines, use the equivalent spacing Deq just as for inductance

• For bundled conductors, use Dsc =

geometric mean of the conductor radius with the subcond. spacings:

Dsc = (r d)1/2 for 2 conductors

or (r d2)1/3 for 3 conductors, etc.

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Tabulated values

• Capacitive reactances are tabulated in units of Megohm-miles. Divide by the number of miles to get the line capacitive reactance

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Tabulated values

• Or

yc = j w C = j bc = j (1/xc) [siemens/mile]

• So calculate bc = 1/xc then multiply bc by line length to get total line susceptance Bc

• Yc = j Bc [siemens] or [mhos]

Example 4.6: capacitance of single-phase line with stranded conductors

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Example 4.7: capacitance of three-phase line with stranded conductors

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