gul Posted August 4, 2009 Report Share Posted August 4, 2009 sir, dears, i have three phase induction motor and the name plate parameters are as under 1) horse power=10 2) power in va=7.5kva 3) rated current=14a 4) rated voltage =420v now according to three phas eequation the power VA= rated voltage*current*1.732 = 420*14*1.732=10184VA=10.184KVA THIS VALUE IS DIFFERENT FROM THE NAME PLATE VALUE i.e power 7.5kva so explain this difference factor? Link to comment Share on other sites More sharing options...
marke Posted August 4, 2009 Report Share Posted August 4, 2009 Hello gul The rating in 2) should read 7.5KW, not KVA KW = KVA x eff x pf. Best regads, Mark. 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...
JEGAN Posted September 3, 2009 Report Share Posted September 3, 2009 Dear gul, Induction motor should be KW rating. formula is KW = 1.732 * v * A * effi * pf jegan. Link to comment Share on other sites More sharing options...
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