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ÄÁÅÙÃ÷ | contents |
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½ºÅ͵ð | study |
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[Àü±â] ¹è¼±³» Line Loss °è»ê¹æ¹ý |
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±Û¾´ÀÌ : ÃÖ°í°ü¸®ÀÚ °íÀ¯ID : admin
³¯Â¥ : 07-01-29 23:24
Á¶È¸ : 14475
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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] = ¡î3 * 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] = ¡î3 * 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 / (¡î3 * 0.480) = 833.3
Line Loss = 3 * 833.3^2 * 0.0050 / 1000 = 10 kW
Energy Saving = 10 - 5 kW = 5 kW
[ÀÌ °Ô½Ã¹°Àº ¿î¿µÀÚ´Ô¿¡ ÀÇÇØ 2013-12-22 22:14:25 Âü°íÀÚ·á¿¡¼ À̵¿ µÊ]
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