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Question 19.1: Design of a Transmission Line Design an electrical transmiss......

Design of a Transmission Line
Design an electrical transmission line 1500 m long that will carry a current of 50 A with no more than 5 × 10^5 W loss in power. The electrical conductivity of several materials is included in Table 19-1.

 

Table 19-1 Electrical conductivity of selected materials at T = 300 K*
Material Conductivity (ohm^{-1} · cm^{-1}) Material Conductivity (ohm^{-1} · cm^{-1})
Superconductors Semiconductors
Hg, Nb_{3}Sn Group 4B elements
Yba_{2}Cu_{3}O_{7-x} Infinite (under certain conditions Si 4 × 10^{-6}
MgB_{2} such as low temperatures) Ge 0.02
Metals Compound semiconductors
Alkali metals GaAs 2.5 × 10^{-9}
Na 2.13 × 10^5 AlAs 0.1
K 1.64 × 10^5 SiC 10^{-10}
Alkali earth metals Ionic Conductors
Mg 2.25 × 10^5 Indium tin oxide (ITO) ∼10^{4}
Ca 3.16 × 10^5 Yttria-stabilized zirconia (YSZ) <10^{14}
Group 3B metals Insulators, Linear, and Nonlinear Dielectrics
Al 3.77 × 10^5 Polymers
Ga 0.66 × 10^5 Polyethylene 10^{-15}
Transition metals Polytetrafluoroethylene 10^{-18}
Fe 1.00 × 10^5 Polystyrene 10^{-17} to 10^{-19}
Ni 1.46 × 10^5 Epoxy 10^{-12} to 10^{-17}
Group 1B metals Ceramics
Cu 5.98 × 10^5 Alumina (Al_{2}O_{3}) 10^{-14}
Ag 6.80 × 10^5 Silicate glasses 10^{-17}
Au 4.26 × 10^5 Boron nitride (BN) 10^{-13}
Barium titanate (BaTiO_{3}) 10^{-14}
C (diamond) <10^{-18}
Step-by-Step
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Electrical power is given by the product of the voltage and current or
P = VI = I²R = (50)²R = 5 × 10^5W
R = 200 ohms
From Equation 19-2,
R=ρ\frac{l}{A}=\frac{l}{σA}       (19-2)
A = \frac{l}{R  ⋅  σ} = \frac{(1500 \ m)(100 \ cm/m)}{(200 \ ohms)σ} =\frac{750}{σ}
Let’s consider three metals—aluminum, copper, and silver—that have excellent electrical conductivity. The table below includes appropriate data and some characteristics of the transmission line for each metal.

Any of the three metals will work, but cost is a factor as well. Aluminum will likely be the most economical choice (Chapter 14), even though the wire has the largest diameter. Other factors, such as whether the wire can support itself between transmission poles, also contribute to the final choice.

 

 σ (ohm^{-1} · cm^{-1}) A (cm²) Diameter (cm)
Aluminum 3.77 × 10^5 0.00199 0.050
Copper 5.98 × 10^5 0.00125 0.040
Silver 6.80 × 10^5 0.00110 0.037

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