Chapter 4. Transmission Line Parameters ELCT 551: Power System Analysis & Design 1
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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