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  [Àü±â] ¹è¼±³» Line Loss °è»ê¹æ¹ý
  ±Û¾´ÀÌ : ÃÖ°í°ü¸®ÀÚ   °íÀ¯ID : admin     ³¯Â¥ : 07-01-29 23:24     Á¶È¸ : 13728    
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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