Answer:
The answer is "0.053 m".
Explanation:
Calculating the voltage sents:
[tex]V=15 \times 15 \ kv=225 \times 10^{5} \ v= 2.25 \times 10^5 \ v\\\\[/tex]
Calculating the Power:
[tex]Power = V \times I\\\\320 \ MW= 2.25 \times 10^5 \times I\\\\[/tex]
[tex]I=\frac{320 \times 10^6}{2.25\times 10^5}=1422.22\ A\\\\[/tex]
Given:
[tex]2\%\ \ loss[/tex]
[tex]I^2R=\frac{2}{100}\times 320 \ MW= 6.4\times 10^6 \\\\I^2R=6.4\times 10^6 \\\\R=\frac{6.4\times 10^6 }{I^2}\\\\[/tex] [tex]=\frac{6.4\times 10^6 }{1422.22^2}\\\\ =3.16\ \Omega \\\\R=\frac{\rho l}{A}=A=\frac{\pi D^2}{4}=\frac{1.68\times 10^{-8}\times 2\times 270\times 10^3}{3.16}=0.002870\\\\D=0.053 m[/tex]
