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컨텐츠 | contents |
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스터디 | study |
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| [전기] 배선내 Line Loss 계산방법 |
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| 글쓴이 : 최고관리자 고유ID : admin
날짜 : 07-01-29 23:24
조회 : 14510
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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
[이 게시물은 운영자님에 의해 2013-12-22 22:14:25 참고자료에서 이동 됨]
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