
sir, plz tell me how much kvar capacitor required for 1000
amp or 500 Kw load ? what is the formula ?
Answers were Sorted based on User's Feedback
Dear Shovan
This is Prakash veer
According to madhu's formulas,
KVAR= KW* (tanQ1tanQ2)
Q1 is intail power factor
Q2 is target power factor
Is This Answer Correct ?  19 Yes  4 No 
Answer / santosh maske
Required data for calculation your demand KVAR
1.Existing P.F of Load (tanQ1)(0.92)
2.Required P.F (Improvement in P.F)(tanQ2)(0.99)
3.Supply Voltage  for reference(Voltage variation if any)
4.Load in KW (exam availabl with you i.e 500KW)
KVAR= KW(tanQ1tanQ2)
KVAR= 500(tan0.92 tan0.99)
KVAR = 500(0.2890)
KVAR = 141.75 KVAR (@ 142KVAR)
if you want to calculate by amp need following data
efficiency of motor, required P.F, KW,maganatising current
of motor,formula find in Electrical Machine Engineering
Is This Answer Correct ?  8 Yes  2 No 
Answer / deepak
kVAR=kVA X sinphi
SO, sin phi=(1cos sq phi) underroot.
SO POWER FACTOR IS REQUIRED.
Is This Answer Correct ?  7 Yes  3 No 
Answer / shovan
Madhu pls explain what is tanphi1 and tanphi2????
Is This Answer Correct ?  5 Yes  1 No 
Answer / elangkumanan
25 kvar = 32amps load taken
12.5 kvar =16amps load taken
l6.25 kvar =8amps load taken
3.125kvar =4amps load taken
1.0775kvar =2amps load taken
1000amps capacitor required to 500kvar
Is This Answer Correct ?  5 Yes  1 No 
250 Kvar Capacitor bank is required for 500Kw load
(i.e,625Kva).
This is because the capacitor rating for any load is equal
to 40% of Kva.
i.e, Capacitor required=40%x Kva Load
Is This Answer Correct ?  7 Yes  5 No 
Answer / saleem iqbal
plz identify your systems existing power factor and
targeted power factor which u want to improve, then apply
this formula
KW X (Tan phi 1  tan phi 2)
Is This Answer Correct ?  2 Yes  0 No 
Reactive power (KVAR) is increase due to run inductive load.as a result power factor is lagging. so have to use capacitive power to reduce this inductive or reactive load. generally we have to use 63% capacitive load to remove the reactive load.i.e.
active power=500KW
so 60% of this will be=(60*500/100) = 300 KVAR
so we have to use 300KVAr capacitive capacitor bank to remove this reactive power .
Is This Answer Correct ?  1 Yes  0 No 
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