Given:
[tex]I_{rms} = 50 A[/tex]
voltage, V = 3600V
step-up voltage, V' = 100000 V
Resistance of line, [tex]R = 100\ohm[/tex]
Solution:
To calculate % heat loss in long distance power line:
Power produced by AC generator, P = [tex]50\times 3600[/tex] W
P = 180000 W = 180 kW
At step-up voltage, V = 100000V or 100 kV
current, I = [tex]\frac{P}{V'}[/tex]
I = [tex]\frac{1800000}{100000}[/tex]
I = 1.8 A
Power line voltage drop is given by:
[tex]V_{drop} = I\times R[/tex]
[tex]V_{drop} = 1.8\times 100[/tex]
[tex]V_{drop} = 180 V[/tex]
Power dissipated in long transmission line [tex]P_{dissipated} = V_{drop}\times I[/tex]
Power dissipated in long transmission line [tex]P_{dissipated} = 180\times 1.8[/tex] = 324 W
% Heat loss in power line, [tex]P_{loss} = \frac{P_{dissipated}}{P}\times 100[/tex]
% Heat loss in power line, [tex]P_{loss} = \frac{324}{180000}\times 100[/tex]
[tex]P_{loss} = 0.18%[/tex]
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