Explain Line Power Losses and Voltage Drop
Using numeric examples, explain line power losses, and voltage drop, between half and full load over a single tier 30 km 132kV transmission line (3 cables " 1 per phase) capable of carrying a maximum load of 200 MVA using 200mm^2 aluminum XLPE insulated cables with a resistance of 0.1/km?
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Vi = 132 kV
Resistance of the cable = 0.1 Ohm / km
Resistance total for 30 km = 3 Ohms
Let us say that the load is distributed among 3 phases.
1) Full Load on each line/phase = 200/√3 MVA in each phase (?)
Current on the line (approx) = 200,000 /(√3*132) = 874.77 A each phase
So voltage drop along the line of 30 km = 3 Ohms * 874.77 A
= 2,624.3 kV in each phase
So voltage drop = 2,624.3 *100/132,000 = 2 % along each phase approx.
Power Loss = I^2 R * 3 = 874.77^2 * 3 W *3 phases
= 6.9 MW approx
2) Half load = 100 MVA three phases or 100/√3 each phase
current on the line/phase = 437.38 A
Voltage drop along the line: 3 Ohms * I = 1,282.14 V
Voltage drop % = 1 % approx
Power loss = I^2 R = 1.725 MW
Resistance of the cable = 0.1 Ohm / km
Resistance total for 30 km = 3 Ohms
Let us say that the load is distributed among 3 phases.
1) Full Load on each line/phase = 200/√3 MVA in each phase (?)
Current on the line (approx) = 200,000 /(√3*132) = 874.77 A each phase
So voltage drop along the line of 30 km = 3 Ohms * 874.77 A
= 2,624.3 kV in each phase
So voltage drop = 2,624.3 *100/132,000 = 2 % along each phase approx.
Power Loss = I^2 R * 3 = 874.77^2 * 3 W *3 phases
= 6.9 MW approx
2) Half load = 100 MVA three phases or 100/√3 each phase
current on the line/phase = 437.38 A
Voltage drop along the line: 3 Ohms * I = 1,282.14 V
Voltage drop % = 1 % approx
Power loss = I^2 R = 1.725 MW
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