WIND ENERGY SYSTEMS - Cd3wd

Chapter 9—Wind Power Plants 9–40
Area of (b) in circular mils = (d in mils) 2
Area of (b) in square mils = (π/4)(d in mils) 2
Number of square mils = (π/4)× number of circular mils
Example. What is the resistance of a copper wire 500 feet long having a diameter of 0.1 inch, if
the resistivity is 10.37 ohm circular mils per foot?
The diameter of the wire is 100 mils (0.1 inch × 1000 mils/inch) and its area in circular mils is d 2 ,
or 10 4 circular mils. Substitution of these values in Eqn. 1 gives
R = ρ l A =10.37500 10 4 =0.5185 Ω
If we have to find the resistance of a rectangular conductor, we simply find the area in
square mils and multiply by 4/π to get the corresponding number of circular mils.
Example. What is the resistance of 1000 feet of a copper conductor 0.1 × 0.2inchincrosssection,
if ρ = 10.37 ohm circular mils per foot?
The cross section of the conductor is 100 × 200 = 2 × 10 4 square mils. Converting to circular
measure gives (4/π)(2 × 10 4 ) circular mils.
R = ρ l A =10.37 1000
(4/π)(2 × 10 4 ) =0.4075 Ω
Resistivity for some common metals is given in Table C.1.
Table C.1. Resistivity ρ at 20 o C.
Selected from CRC Handbook of Tables
for Applied Engineering Science
Ohm α
Material Ohm-meters circular mils Temp.
per foot coefficient
Aluminum wire 2.82 ×10 −8 17.0 0.0039
Brass 7.0 ×10 −8 42.1 0.002
Constantan 49 ×10 −8 295 0.0001
Copper wire 1.724 ×10 −8 10.37 0.00393
Gold 2.35 ×10 −8 14.1 0.004
Iron (pure) 9.7 ×10 −8 58.4 0.00651
Lead 20.6 ×10 −8 124 0.00336
Manganin 44 ×10 −8 265 0.00001
Mercury 98.4 ×10 −8 592 0.00089
Nichrome 100 ×10 −8 602 0.004
Nickel 6.85 ×10 −8 41.2 0.0069
Platinum (pure) 10.5 ×10 −8 63.2 0.00393
Silver 1.59 ×10 −8 9.57 0.0041
Tungsten 5.65 ×10 −8 34.0 0.0046
Wind Energy Systems by Dr. Gary L. Johnson November 21, 2001