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* (tanQ1-tanQ2)
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(tanQ1-tanQ2)
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=(1-cos 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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