Question

30,000 kW at power factor 0·8 lagging is being transmitted over a 220 kV, three-phase transmission line. The length of the line is 275 km and the efficiency of transmission is 90%. Calculate the weight of copper required. Also calculate the weight of copper had the power been transmitted over a single-phase transmission line for the same line voltage and losses. Assume that the resistance of 1 km long conductor and 1 sq. cm is 0·l73 Ω and specific gravity of copper is 8·9.

Answer 1

Solution:

(i) Total power transmitted = 30000 kW

Power factor = 0.8

We know that for a three phase system total power,

P = 3.ViIcos

therefore line current is,

30000 x 10 P 98.415.A V3.VICOS 3 x 220 x 103 x 0.8

Transmission Efficiency is 90%, therefore total losses W in line is 10% of transmitted power.

$W= 30000\times \frac{10}{100}=3000kW$

We know that total copper losses in the line,

$W= 3.I_{L}^{2}R=3.I_{L}^{2}\left ( \frac{\rho l}{a} \right )$

where l is length and a is cross-sectional area.

therefore,

pl 3000 x 103kW 3.1Z а

$a=\frac{3.I_{L}^{2}.\rho l}{3000\times 10^{3}}$

therefore cross-sectional area,

$a=\frac{3\times 98.415^{2}\times 0.173\times 275}{3000\times 10^{3}}=0.46cm^{2}=0.46\times 10^{-4}m^{2}$

Now volume of copper required,

3al 3 x (0.46 x 104) x (275 x 103) = 38.015m3 Volume

Specific gravity of copper=8.9

density of copper=8.9X1000=8900 kg/cubic meters

Weight of copper required,

${\color{Red} Weight=Volume\times density=38.015\times 8900=338333 Kg}$

(ii) for single phase 2 wire system,

Total power,

$P=V_{L}I_{L}cos\phi$

therefore line current is,

$I_{L} =\frac{P}{V_{L}cos\phi}=\frac{30000\times 10^{3}}{220\times 10^{3}\times 0.8}=170.45A$

Transmission Efficiency is 90%, therefore total losses W in line is 10% of transmitted power.

$W= 30000\times \frac{10}{100}=3000kW$

We know that for two wire,total copper losses in the line,

$W= 2.I_{L}^{2}R=2.I_{L}^{2}\left ( \frac{\rho l}{a} \right )$

where l is length and a is cross-sectional area.

therefore,

$2.I_{L}^{2}\left ( \frac{\rho l}{a} \right )=3000\times 10^{3}kW$

$a=\frac{2.I_{L}^{2}.\rho l}{3000\times 10^{3}}$

therefore cross-sectional area,

$a=\frac{2\times 170.45^{2}\times 0.173\times 275}{3000\times 10^{3}}=0.921cm^{2}=0.921\times 10^{-4}m^{2}$

Now volume of copper required,

$Volume=2al=2\times (0.921\times 10^{-4})\times (275\times10^{3})=50.655 m^{3}$

Specific gravity of copper=8.9

density of copper=8.9X1000=8900 kg/cubic meters

Weight of copper required,

${\color{Red} Weight=Volume\times density=50.655\times 8900=450829 Kg}$