power_factor_correction example

power_factor_correction example - Q=71.7308 kVA x sin 38.74...

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75 HP Motor PF=0.75 208 Volts 60 Hz POWER FACTOR CORRECTION P=75 HP x 746 W = 55.950 kW S=55.950 kW/0.75 = 74.600 kVA Q=74.600 kVA x sin 41.41 deg. = 49.3426 kVAR's Existing installation After PF correction 75 HP Motor PF=0.75 208 Volts 60 Hz C=? What value of capacitor is needed to increase the power factor to 0.78? P=75 HP x 746 W = 55.950 kW S=55.950 kW/0.78 = 71.7308 kVA
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Unformatted text preview: Q=71.7308 kVA x sin 38.74 deg. = 44.8877 kVAR's Therefore, Qcap. = 49.3426 kVAR's - 44.8877 kVAR's = + 3.4549 kVAR's This is the LEADING reactive power required to reduce the LAGGING motor reactive power to the desired value. Vc Xc Qc 2 Xc=12.5225 Ohms C=211.825 uF (in practice the nearest standard size is selected)...
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This note was uploaded on 03/21/2012 for the course EE 369 taught by Professor Khan during the Fall '10 term at Mohawk College.

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