Copper and aluminum are being considered for a high-voltage transmission line th

Question

Copper and aluminum are being considered for a high-voltage transmission line that must carry a current of 60 A. The resistance per unit length is to be 0.155 0/km. The densities of rapper and aluminum are 8960 and 2700 kg/m^3, respectively. Compute the magnitude J of the current density for a copper cable. Compute the magnitude J of the current density for an aluminum cable. And compute the mass per unit length lambda for a copper cable. And compute the mass per unit length lambda for an aluminum cable.

Explanation / Answer

Given

copper and aluminum wires of

carrying current i = 60 A, resistance per unit length is = 0.155 ohm/km = 0.155*10^-3 ohm/m

densities of

copper wire d_C= 8960 kg/m3

Aluminum d_A = 2700 kg/m3

the resitivity of copper is 1.68*10^-8 ohm m

       aluminum is 2.12*10^-8 ohm m

we know the relation of rho = R*A/l , given R/l for each

   area of cross section of the copper wire is A = rho*(R/l)

   pir^2 = rho(R/l)

       = 1.68*10^-8 (0.155*10^-3) m2
       = 2.604*10^-12 m2
for aluminum is

   pir^2 = rho(R/l)

       = 2.12*10^-8 (0.155*10^-3) m2

       =3.286*10^-12 m2

now the current density is J = I/A = I/pir^2

   J_C = 60 /(2.604*10^-12) = 2.30415*10^13 A/m2

   J_A = 60 /(3.286*10^-12) = 1.82593*10^13 A/m2

now mass per unit length is density = m/(pi*r^2*l)

       d *pi*r^2 = m/l

for copper m/l = d_c*pi*r^2 = 8960*2.604*10^-12 = 2.333184*10^-8 kg/m

for aluminum is m/l = d_a*pi*r^2 = 2700*3.286*10^-12 = 8.8722*10^-9 kg/m

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