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# Notes 10 Conductor sizing & an example

By Beatrice Jordan,2014-04-08 20:58
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Notes 10 Conductor sizing & an example

Use of Tables

1.0 Introduction

We have now the ability to compute inductive reactance and capacitive susceptance of a transmission line given its

geometry. To remind you, the most general form of the relevant expressions are:

Dm0llnaInductance (h/m): 2Rb

? D is the GMD between phase positions: m

1/3(1)(2)(3);；D(ddd mababab

? R is the GMR of the bundle b

1/2?Rrd,for 2 conductor bundle;；12b

1/3? ;；rdd,for 3 conductor bundle1213 1/3?;； rddd,for 4 conductor bundle121314

2；，

cacCapacitance (f/m): ln(D/R)mb

? D is the same as above. m

c

R? is Capacitive GMR for the bundle: b

1/2cRrd,for 2 conductor bundle;；12b

1/3;； rdd,for 3 conductor bundle1213

1/3;； rddd,for 4 conductor bundle121314

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We will review conductor tables because many people still use them and because it provides a good way to review some

important and interesting physical properties of conductors that we have not covered and that are not well covered in your book.

2.0 Description of Table A8.1

Table A8.1 on pp 605-607 has 8 columns, with one column, resistance, divided into 4 sub-columns. We describe each one in what follows:

1. Code-word: This is a way of referring to standard conductor sizes that was developed by the Aluminum Association many years ago. They are mostly bird names.

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2. Size: The size is designated in kcmil.

What is a cmil (circular-mil)? A cmil is a unit of measure for area and corresponds to

the area of a circle having diameter of 1 mil,

-3

where 1 mil=10 inches, or 1000 mils=1 in.

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The area of such a circle is πr= π(d/2), or

-3 2-72

π(10in/2)=7.854x10 in.

The area in cmils is defined as the square of

the diameter in mils. Therefore:

2

? 1cmil=(1mil) corresponds to a conductor

-3

having diameter of 1 mil=10 in.

2

? 1000kcmil=1,000,000cmil=(1000mils)

corresponds to a conductor having

diameter of 1000 mils=1 in.

To determine diameter of conductor in inches, take square root of cmils (to get the

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diameter in mils) and then divide by 10:

cmils

3Diameter in inches=. 10

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Example: What is the area in square inches of Conductor Waxwing?

Reference to Table A8.1 indicates it has a size of 266.8 kcmil=266,800 cmils. The

266,800516.53diameter in mils is mils.

Divide by 1000 to get the diameter in inches to be 0.51653 in. The radius is therefore 0.51653/2=.2583in. The area in square

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inches is then πr= π(0.2583)=0.2095in.

Noting the ratio 0.2095/266800=0.7854E-6 gives a constant that, when multiplied by the cmils, gives area in square inches [1, p. 64]:

A=0.7854E-6*cmil in2

3. Stranding Al./St.: The most common

kind of conductor is Aluminum Conductor

Steel Reinforced (ACSR) because it reaps

the benefits of aluminum which are ? High conductivity σ (=1/ρ): 61% of σ copper

as compared to σ which is 25%, σ brasssteel

which is 10%, σ which is 108%) silver

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Abundant supply and low cost ?

? Light

And it reaps the benefit of a few strands of steel, which is increased strength.

A few other kinds of conductors in use are: ? All-Aluminum conductor (AAC)

? All-Aluminum-Alloy Conductor (AAAC) ? Aluminum Conductor Alloy-Reinforced

(AAAC)

(Alumoweld)

See below figure [2] for example of ACSR conductor.

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Examples:

? Waxwing has Al./St. of 18/1 indicating it

has 18 strands of aluminum around 1

strand of steel.

? Partridge has Al./St. of 26/7 indicating it

has 26 strands of aluminum around 7

strands of steel.

4. Number of aluminum layers: The

strands are configured in a number of concentric layers equal to this number.

Example: The conductor “Flicker” has

Al./St. of 24/7. Its core will have 7 strands of steel (6 configured around 1) and 24 strands of aluminum (9 configured around the steel core, 15 configured around the 9).

A general formula for the total number of strands in concentrically stranded cables is

2

# of strands=3n-3n+1

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where n is the number of layers, including the single center strand [1, p. 64]. The above assumes equal diameters for all strands.

For example, for Waxwing, with n=3 (2 aluminum layers), we have 27-9+1=19, which agrees with Table A8.1 indication that the stranding is 18/1.

5. Resistance: The resistance is given

? for DC at 20?C, and

? for AC (60Hz) at 25, 50, and 75?C. The DC resistance is computed according to

l~

RDC (1) A

where ρ is the resistivity (ohm-meters), l is

the conductor length (m), and A is conductor

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cross-sectional area (m).

There are three effects which cause the value used in power system studies to differ from R: DC

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a. Temperature: The resistivity is affected by the temperature and this in turn affects the DC resistance. The influence of

temperature on resistivity is captured by the following equation:

;；，(~(T)~(T20?)1T20? (2)

where all values of temperature T are given in degrees centigrade. Parameter α

ranges from 0.0034-0.004 for copper and from 0.0032-0.0056 for aluminum and about 0.0029 for steel (see note #2 pg 607)

b.Stranding: The effective length is affected by stranding because of spiraling, which tends to increase the DC resistance by about 1 or 2%. So why are conductors stranded instead of being solid [2]?

? They are easier to manufacture since

larger conductors can be obtained by

adding successive layers of strands. ? They are more flexible than solid

conductors and thus easier to handle,

especially in large sizes.

? It reduces skin effect.

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c. Skin effect: This is from the alternating

magnetic field set up by the conductor’s

alternating current. The field interacts with

the current flow to cause more flow at the

periphery of the conductor. This results in

nonuniform current density and therefore

more losses and higher effective resistance.

The increase in resistance from this

influence depends only on frequency and

conductor material.

6. GMR (ft): With a large number of strands, the calculation of the GMR for a stranded

conductor is tedious because we must obtain distances between every combination of strands to compute it, according to:

2nR(dd...d)(dd...d)...(dd...d) s11121n21222nn1n2nn

(see problem 3.7 for an example of this). The availability of this entry in Table A8.1 makes this tedious calculation unnecessary. Note: if you do not have tables, use of r’ instead of R is a good approximation (see s

problem 3.8 for evidence of this).

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Consideration of the expression for R will S

convince you that R will be larger than s

r’(=d), which will cause R to be larger, kkb

and since R goes in the denominator of the b

logarithm, it will cause the inductance to be smaller. So stranding, like bundling, tends to decrease inductance.

7. Inductive ohms/mile X, 1 foot spacing: a

This is the per-phase inductive reactance of a three-phase transmission line when the spacing between phase positions is 1 foot. To see why, note that the per-phase reactance in Ω/m of a non-bundled

transmission line is 2πfl, where a

Dm0llnaΩ/m. Therefore, we can express 2Rs

the reactance in Ω/mile as

)?D1609 meters0m?Xflf22lnLa?R21 miles

D3mf2.022?10ln /mile

Rs

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