Question

A generator produces 2 MW of electric power at 15 kV. The current is sent to a town 30 km away through aluminum transmission wires (resistivity 2.7 x 10-8 LaTeX: \Omega\cdot mΩ ⋅ m) with a diameter of 6.8 mm. What % of the electric power is lost during transmission?

Answer 1

To solve this problem it is necessary to apply the concepts related to Power in function of the current and the resistance.

By definition there are two ways to express power

$P = I^2R$

P =VI

Where,

P = Power

I = Current

R = Resistance

V = Voltage

In our data we have the value for resistivity and not the Resistance, then

$R = \rho \frac{1}{A}$

$R = 2.7*10^{-8}\frac{1}{\pi r^2}$

$R = 2.7*10^{8}\frac{1}{\pi (6.8*10^-3)^2}$

$R = 1.85*10^{-4}\Omega$

The loss of the potential can mainly be given by the resistance of the cables, that is,

$I = \frac{P}{V}$

$I = \frac{2*10^6W}{15*10^3V}$

$I = 133.3A$

Therefore the expression for power loss due to resistance is,

$P = I^2 R$

$P = 133.3^2 * 1.85*10^{-4}$

$P_l = 3.2872W$

The total produced is $2 * 10 ^ 6 MW$, that is to say 100%, therefore 3.2872W is equivalent to,

$x = \frac{3.2872*100}{2*10^6}$

$x = 1.6436*10^{-4}\%$

Therefore the percentage of lost Power is equivalent to $1.6436 * 10 ^ 4 \%$ of the total