Question

An aluminum transmission line of 7 km length is designed to carry 100 A current with no more than 2 MW power loss. The required minimum diameter (in mm) of the transmission line is (rounded to the two decimal places) ........... 

Hint:

When calculating the diameter for a power transmission line, ensure to use the correct values for conductivity and resistivity. The formula involves Ohm’s law and the power loss equation.

Answer 1

The power loss in the transmission line due to resistance is given by the formula:
\[ P = I^2 R \] Where:
- \( P = 2 \times 10^6 \, {W} \) (2 MW),
- \( I = 100 \, {A} \),
- \( R \) is the resistance of the transmission line.
The resistance \( R \) of the transmission line is given by:
\[ R = \frac{\rho L}{A} \] Where:
- \( \rho = \frac{1}{\sigma} = \frac{1}{3.77 \times 10^5} \, \Omega \, {cm} \),
- \( L = 7000 \, {m} \),
- \( A = \pi \left(\frac{d}{2}\right)^2 \) is the cross-sectional area of the wire.
Now, substitute the area into the power loss equation:
\[ P = I^2 \times \frac{\rho L}{\pi \left(\frac{d}{2}\right)^2} \] Solving for the diameter \( d \), we get:
\[ d = \sqrt{\frac{4 I^2 \rho L}{\pi P}} \] Substituting the values:
\[ d = \sqrt{\frac{4 \times (100)^2 \times \frac{1}{3.77 \times 10^5} \times 7 \times 10^3}{\pi \times 2 \times 10^6}} \] \[ d \approx 1.12 \, {m} \] Converting to millimeters:
\[ d \approx 1120 \, {mm} \] Thus, the required diameter lies between 1.00 mm and 1.20 mm.