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The so-called SIL (surge impedance loading) of a transmission line is given by P = U2 : Z, where P is the power tansmitted, in MW, U is the line voltage in kV, and and Z is the surge impedance of the line, in ohm. For a 132 kV overhead transmission line, it is typically 400 ohm. The SIL is the power trasnmitted that causes the minimum of losses, that is, it is the most economical condition to operate the line. For example. P = 1322 / 400 = 43.56 MW =43,560 kW The current per phase is I = P / 1.732 U = 43,560 / (1.732 x 132) = 190.5 A If the cable is ACSR (aluminum cable, steel reinforced), it can carry about 1 A/mm2 So a conductor of about 200 mm2 cross section would be a good choice. This calculation just gives a rough idea of the size of the conductor, since the best result is given after several other factors are taken into account, e.g., length of the line, cost of energy loss, cost of cable, cost of steel structures, standardized cable sizes available, etc.

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