배선내 Line loss

제어 및 operation2016. 10. 24. 13:01

The only improvement will be on the resistive loss (i^2*R) of the line and substation.

Since improving the power factor will reduce the total line current,

these losses will drop with the reduction ratio squared.

Total Load Power[kVA] = √3 * V[kV] * I[A]

Real Power[kW] = Total Load Power[kVA] * P.F[%]

Line Loss = 3 * I[A]^2 * R [ohm] / 1000

But the resistance of the line an substation (including the transformers) is regularly too small,

then the saving of kW is not that spectacular.

Assuming an hypothetic 480 Volts 3 phase line, driving 1000 amperes LOAD at 0.75 PF and with 0.0050 ohms per phase

total resistance of line and transformers.

Total Load Power[kVA] = √3 * 0.480 * 1000 = 831.36

Real Power[kW] = 831.36 * 0.75 = 623.52 kW

Line Loss = 3 * 1000^2 * 0.0050 / 1000 = 15 kW

If the LOAD Power factor is improved to 0.90

Real Power[kW] = 623.52 kW

Total Load Power[kVA] = 623.52 kW / 0.9 = 692.8

Line Current = 692.8 / (√3 * 0.480) = 833.3

Line Loss = 3 * 833.3^2 * 0.0050 / 1000 = 10 kW

Energy Saving = 10 - 5 kW = 5 kW

RTO care