An energy plant produces an output potential of 10000 kV and serves a city 162 k

Question

An energy plant produces an output potential of 10000 kV and serves a city 162 km away. A high-voltage transmission line carries 1340 A to the city. The effective resistance of a transmission line [wire(s)] is 3.27 Ohm/km times the distance from the plant to the city. Consider the money lost by the transmission line heating the atmosphere each hour. Assume the energy plant produces the same amount of power: however, the output electric potential of the energy plant is 20% greater. How much money per hour is saved by increasing the electric potential of the power plant? Answer in units of dollars/hr

Explanation / Answer

The effective resistance of the transmission line (including the resistance of outgoing and returning conductors) is: Re = 3.27ohm/km x 162km = 529.74 ohm The voltage drop across the conductors is: Vd = I x R = 1340A x 529.74ohm =709.851kV The potential (the voltage) delivered to the city is: 10000kV - 709.851kV = 9290.15kV The power dissipated by the transmission line is given by: P=I^2 x R = (1340A)^2 x 529.74ohm = 951201144W The existing load on the plant (in kVA) is: kVAex = 10000kV x 1340A =13400000 kVA The potential is raised 20% to: kVnew = 10000kV x 120% = 12000kV It is stated that the total load (i.e., kVA) remains the same, so the current must decrease (by a factor of 1/1.20). The new current will be: Inew = kVAex/kVnew = 13400000kVA / 12000kV = 1116.66A The power dissipated by the transmission line with reduced current is given by: Pnew = Inew^2 x R = (1116.66A)^2 x 529.74ohm = 660556350W The power saved is given by: Psaved = P - Pnew = 951201144W - 660556350W =290644794W The money per hour saved is: Savings = unknown$/kw x 290644794W

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