If the load is 10Kw and power factor has to improve from 0.5
to 0.8,how can we select the capacitor rating
accordingly.Illustrate by showing formula?
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Answer / guest
at 0.5pf Apparent power is 10/0.5=20VA
reactive power = 17.321kVAR.
At 0.8pf
APPARENT POWER =10/0.8=12.5VA
Reactive power=7.5kVAR
So we need to provide (17.321-7.5)=9.821kVAR reactive power
through capacitor to bring of up 0.8 from 0.5.
So we need a capacitor rated for approximately 10kVAR to
achieve the target.
ANS= CAPACITOR RATING 9.821kVAR (approximately 10 kVAR)
| Is This Answer Correct ? | 10 Yes | 2 No |
Answer / prasad damodar bhoir
kvar= kw (tan phi 1- tan phi 2)
cos phi 1= 0.5, phi 1= cos inverse 0.5= 60
cos phi 2= 0.8, phi 2= cos inverse 0.8= 36.86
kvar= 10 (tan60 - tan 36.86)= 10 (1.73-0.74)
kvar= 9.9
| Is This Answer Correct ? | 4 Yes | 0 No |
Answer / h salim
Load = 10 kW
cos phi 1 = 0.5
cos phi 2 = 0.8
Rating of capacitor Qc = kW(tan phi 1- tan phi 2 )
| Is This Answer Correct ? | 3 Yes | 0 No |
Answer / prakash g.r.
Calculation of Required Kvar
Initial or Present Power Factor =Qi,
Required Power Factor = Qd
KVAr= KW(TANQi-TANQd)
Qi=COS-1PFi = Initial Power Factor Angle
Qd=COS-1PFd = Final Power Factor Angle
| Is This Answer Correct ? | 3 Yes | 0 No |
Answer / ahatesham
at 0.5pf Apparent power is 10/0.5=20VA
reactive power = 17.321kVAR.
At 0.8pf
APPARENT POWER =10/0.8=12.5VA
Reactive power=7.5kVAR
So we need to provide (17.321-7.5)=9.821kVAR reactive power
through capacitor to bring of up 0.8 from 0.5.
So we need a capacitor rated for approximately 10kVAR to
achieve the target.
ANS= CAPACITOR RATING 9.821kVAR (approximately 10 kVAR)
| Is This Answer Correct ? | 2 Yes | 0 No |
Kvar = KW * [ tan (cos^-1* Current PF) - tan (cos^-1* Required PF)]
| Is This Answer Correct ? | 0 Yes | 0 No |
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