chisam14 Posted November 7, 2007 Report Share Posted November 7, 2007 I have two single phase, 120V motors in parallel. One is a 3/4 hp(10.4A) and the other is a 1/4hp(4A). I need to know what size capacitor to put in parallel with them to help aid power factor. The cap will be controlled by a microprocessor. If anyone knows the size or formulas for calculating the size please let me know. Link to comment Share on other sites More sharing options...
marke Posted November 8, 2007 Report Share Posted November 8, 2007 To determine the amount of correction to apply, you first need to determine the level of the magnetizing current. If you run the motors open shaft, (No shaft load) then the current drawn will essentially be the magnetizing current. If you apply static correction, then the correction current should be 80% of the magnetizing current. If you apply bulk correction, then you can correct higher than that. Best regards, Mark Empson | administratorSkype Contact = markempson | phone +64 274 363 067LMPForum | Power Factor | L M Photonics Ltd | Empson family | Advanced Motor Control Ltd | Pressure Transducers | Smart Relay | GSM Control | Mark Empson Website | AuCom | Soft Starters Link to comment Share on other sites More sharing options...
AB2005 Posted November 8, 2007 Report Share Posted November 8, 2007 The cap will be controlled by a microprocessor. Why you want to use microprocessor? According to your question, you want to connect capacitor at motor terminal (static correction) then there is no need of microprocessor. You calculate the capacitor as Marke replied. "Don't assume any thing, always check/ask and clear yourself". Link to comment Share on other sites More sharing options...
marke Posted November 13, 2007 Report Share Posted November 13, 2007 Once you determine the amount of reactive current that you need, you can calculate the impedance using ohms law. Z = V/I When you have calculated the impedance, you can calculate the capacitance as z = 1/(wc) therefore C = 1/(2 x pi x F x Z) or you can multiply the volts time the amps, divide by 1000 to get the required KVAR. Best regards, Mark Empson | administratorSkype Contact = markempson | phone +64 274 363 067LMPForum | Power Factor | L M Photonics Ltd | Empson family | Advanced Motor Control Ltd | Pressure Transducers | Smart Relay | GSM Control | Mark Empson Website | AuCom | Soft Starters Link to comment Share on other sites More sharing options...
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