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What is the transmission loss due to heat if my power station produces electricity at a power of \$5 kW\$ and the electricity consumer is \$1 km\$ away?

It is a copper line with a material constant of \$0.017\Omega \frac{mm^2}{m}\$, a voltage of \$1kV\$ and a cross-section of \$10mm^2\$.

My approach, which I don't know if it's true or not, is as follows: $$I_{cable} = \frac{P_{generated}}{U_{grid}} = \frac{5kW}{1kV} = 5A$$ $$R_{cable} = \rho * \frac{l}{A} = 0.017\Omega \frac{mm^2}{m} * \frac{1000m}{10mm^2} = 1.7\Omega$$ $$P_{loss} = (I_{cable}^2R_{cable})* 2 = 85W$$ So that of the \$5000W\$ produced about \$5000W - 85W = 4915W\$ can be consumed.

Is the calculation so possible? If not, then how?

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  • \$\begingroup\$ for a first contribution, using equation markup like a pro, excellent. \$\endgroup\$
    – Neil_UK
    Commented Sep 28, 2018 at 7:08

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Yes, your engineering is correct, the power loss in the cable is \$I^2R\$.

This calculation is a good demonstration for why nationwide power distribution needs transmission voltages in the 100s of kV, and even a district tends to use 10s of kV, to control the power loss in the distribution cables to be reasonable, or avoid unaffordable conductors with the cross section of tree trunks.

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  • \$\begingroup\$ Thank you very much for the detailed and informative answer. In my calculation I have calculated the loss for the outward and return route of the current (\$(I_{cable}^2R_{cable})* 2\$), so that the loss of the cable is doubled. Does the return path also have to be considered? \$\endgroup\$ Commented Sep 28, 2018 at 7:22
  • \$\begingroup\$ @AnneBierhoff You have allowed for the return path correctly. Personally, I would have calculated the total loop resistance as 3.4ohms, and not doubled the power calculation, but that's just a matter of style, if it is indeed a loop of two similar wires for out and return, which it usually is. If the 'return' is for some reason a different conductor type to the 'out', like around a car for instance, that's when calculating the total resistance is the tidier way to do the sums. \$\endgroup\$
    – Neil_UK
    Commented Sep 28, 2018 at 7:49

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