Electronic – Calculation of transmission loss of a power plant

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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?

Best Answer

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.