Copper and aluminum are being considered for the cables in a high-voltage transmission line where each must carry a current of \(50 \mathrm{A}\). The resistance of each cable is to be \(0.15 \Omega\) per kilometer. (a) If this line carries power from Niagara Falls to New York City (approximately $500 \mathrm{km}),$ how much power is lost along the way in the cable? Compute for each choice of cable material (b) the necessary cable diameter and (c) the mass per meter of the cable. The electrical resistivities for copper and aluminum are given in Table \(18.1 ;\) the mass density of copper is $8920 \mathrm{kg} / \mathrm{m}^{3}\( and that of aluminum is \)2702 \mathrm{kg} / \mathrm{m}^{3}$

Using the power loss formula P=I²R and given I = 50 A and R=0.15 Ω/km: Power loss in Copper cable per km = \((50)^2 \times 0.15 = 375000 \mathrm{W}\) For a distance of 500 km, total power loss in Copper cable = \(375000 \times 500 = 1.875 \times 10^8 \mathrm{W}\) Similarly, the power loss in Aluminum cable per km = \(375000 \mathrm{W}\) For a distance of 500 km, total power loss in Aluminum cable = \(375000 \times 500 = 1.875 \times 10^8 \mathrm{W}\)

Using the formula R = ρL/A, solve for A, the cross-sectional area (A) of the cable. We are given R = 0.15 Ω/km, and the resistivities ρ (Copper) = 1.72e-8 Ωm and ρ (Aluminum) = 2.82e-8 Ωm. For Copper: \(0.15 = (1.72\times 10^{-8})(1000)/A\) Solving for A, we get A = \(1.72\times 10^{-5} \mathrm{m}^2\) Now, we can find the diameter (d) using the formula \(A = \pi(d/2)^2\). Solving for d, we get d (Copper) = 0.00467 m or 4.67 mm. For Aluminum: \(0.15 = (2.82\times 10^{-8})(1000)/A\) Solving for A, we get A = \(2.82\times 10^{-5} \mathrm{m}^2\) Now, we can find the diameter (d) using the formula \(A = \pi(d/2)^2\). Solving for d, we get d (Aluminum) = 0.00599 m or 5.99 mm.

To find the mass per meter, we first find the volume (V) of a 1 m length cable by using the formula V = A*L. For Copper: Volume (V) = \(1.72\times 10^{-5} \mathrm{m}^2 * 1 \mathrm{m} = 1.72\times 10^{-5} \mathrm{m}^3\) Now using mass density, mass per meter = density * V; Mass per meter (Copper) = (8920 kg/m³) * (\(1.72\times 10^{-5} \mathrm{m}^3\)) = 0.1534 kg/m For Aluminum: Volume (V) = \(2.82\times 10^{-5} \mathrm{m}^2 * 1 \mathrm{m} = 2.82\times 10^{-5} \mathrm{m}^3\) Now using mass density, mass per meter = density * V; Mass per meter (Aluminum) = (2702 kg/m³) * (\(2.82\times 10^{-5} \mathrm{m}^3\)) = 0.0762 kg/m Summary: a) Total power loss in both Copper and Aluminum cable: \(1.875 \times 10^8 \mathrm{W}\) b) Necessary cable diameter: Copper = 4.67 mm, Aluminum = 5.99 mm. c) Mass per meter of the cable: Copper = 0.1534 kg/m, Aluminum = 0.0762 kg/m.