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Categories >> Engineering >> Electrical Engineering
 


 

 
Question
sir, plz tell me how much kvar capacitor required for 1000 
amp or 500 Kw load ? what is the formula ?
 Question Submitted By :: Electrical Engineering
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Answer Posted By  
 Answers were Sorted based on User's Feedback
Answer
# 1
Formula : KVAR = KW(tanĂ˜1-tanĂ˜2) 

Is This Answer Correct ?    31 Yes 8 No
madhu253
 
Answer
# 2
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 ?    16 Yes 1 No
prakash veer singh
 
 
Answer
# 3
we need power factor value 

Is This Answer Correct ?    12 Yes 4 No
arun
 
Answer
# 4
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 ?    6 Yes 2 No
santosh maske
 
Answer
# 5
kVAR=kVA X sinphi
SO, sin phi=(1-cos sq phi) underroot.
SO POWER FACTOR IS REQUIRED. 

Is This Answer Correct ?    6 Yes 3 No
deepak
 
Answer
# 6
Madhu pls explain what is tanphi1 and tanphi2???? 

Is This Answer Correct ?    4 Yes 1 No
shovan
 
Answer
# 7
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 ?    3 Yes 1 No
elangkumanan
 
Answer
# 8
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
saleem iqbal
 
Answer
# 9
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
bijoy roy
 
Answer
# 10
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 ?    4 Yes 4 No
malkit singh
 

 
 
 
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