An electric power station that operates at 15 kV and uses a 15:1 step-up ideal transformer is producing 320 MW (Mega-Watt) of power that is

Question

An electric power station that operates at 15 kV and uses a 15:1 step-up ideal transformer is producing 320 MW (Mega-Watt) of power that is to be sent to a big city which is located 270 km away with only 2.0% loss. Each of the two wires are made of copper (resistivity = 1.68 × 10-8 Ω.m). What is the diameter of the wires?

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Linh Đan 4 years 2021-08-05T02:55:37+00:00 1 Answers 11 views 0

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  1. Xavia
    0
    2021-08-05T02:57:00+00:00
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    Answer:

    The answer is “0.053 m“.

    Explanation:

    Calculating the voltage sents:

    V=15 \times 15 \ kv=225 \times 10^{5} \ v= 2.25 \times 10^5 \ v\\\\

    Calculating the Power:

    Power = V \times I\\\\320 \ MW= 2.25 \times 10^5 \times I\\\\

    I=\frac{320 \times 10^6}{2.25\times 10^5}=1422.22\ A\\\\

    Given:

     2\%\ \ loss

    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}\\\\
    =\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

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