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)

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

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 )

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

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)

Answer / nagesh.patwari

Kvar = KW * [ tan (cos^-1* Current PF) - tan (cos^-1* Required PF)]

Answer / lohith

kvar=kva*sin

Answer / rk

How did you Calculate Apparent Power=20VA to Reactive power 17.321 KVAR simlarly 12.5VA to 7.5KVAR As the formula is KVAR=KVA/KW

Answer / nag

unity powerfactor

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