Sunday, August 10, 2014

KAVR into Farad; Capacitor KVAR conversion into Farad

Below we study how to Convert Farads into kVAR and Vice Versa. It usually comes in the mind that capacitor used for motors or higher capacity inductive load is in KVAR but capacitor rating is in farad . Then how we can convert KVAR into Farad.


This can be explained by some very simple examples as below:-

Example 1:
A Single phase 400V, 50Hz, motor takes a supply current of 50A at a P.F (Power factor) of 0.6. The motor power factor has to be improved to 0.9 by connecting a capacitor in parallel with it. Calculate the required capacity of Capacitor in both kVAR and Farads.
Solution.:
(1) To find the required capacity of Capacitance in kVAR to improve P.F from 0.6 to 0.9 (Two Methods)

Solution #1 (By Simple Table Method)
Motor Input = P = V x I x Cosθ
                              = 400V x 50A x 0.6
                              = 12kW
From Table, Multiplier to improve PF from 0.60 to 0.90 is 0.849
Required Capacitor kVAR to improve P.F from 0.60 to 0.90
Required Capacitor kVAR = kW x Table Multiplier of 0.60 and 0.90
= 12kW x 0.849
= 10.188 kVAR

Solution # 2 (Classical Calculation Method)
Motor Input = P = V x I x Cosθ
                              = 400V x 50A x 0.6
                              = 12kW
Actual P.F = Cosθ1 = 0..6
Required P.F = Cosθ2 = 0.90
θ1 = Cos-1 = (0.60) = 53°.13; Tan θ1 = Tan (53°.13) = 1.3333
θ2 = Cos-1 = (0.90) = 25°.84; Tan θ2 = Tan (25°.50) = 0.4843
Required Capacitor kVAR to improve P.F from 0.60 to 0.90
Required Capacitor kVAR = P (Tan θ1 - Tan θ2)
= 5kW (1.3333– 0.4843)
= 10.188 kVAR

(2) To find the required capacity of Capacitance in Farads to improve P.F from 0.6 to 0.9 (Two Methods)
Solution #1 (Using a Simple Formula)
We have already calculated the required Capacity of Capacitor in kVAR, so we can easily convert it into Farads by using this simple formula
Required Capacity of Capacitor in Farads/Microfarads
C = kVAR / (2 π f V2) in microfarad
Putting the Values in the above formula
 = (10.188kVAR) / (2 x π x 50 x 4002)
= 2.0268 x 10-4
= 202.7 x 10-6
= 202.7μF
Solution # 2 (Simple Calculation Method)
kVAR = 10.188 … (i)
We know that;
IC = V/ XC
Whereas XC = 1 / 2 π F C
IC = V / (1 / 2 π F C)
IC = V 2 F C
= (400) x 2π x (50) x C
IC = 125663.7 x C
And,
kVAR = (V x IC) / 1000 … [kVAR =( V x I)/ 1000 ]
= 400 x 125663.7 x C
IC = 50265.48 x C … (ii)
Equating Equation (i) & (ii), we get,
50265.48 x C = 10.188C
C = 10.188 / 50265.48
C = 2.0268 x 10-4
C = 202.7 x 10-6
C = 202.7μF

Good to Know:
These are the main Formulas to Convert Farads into kVAR and Vice Versa
Required Capacity of Capacitor in Farads/Microfarads
C = kVAR / (2 π f V2) in microfarad
Required Capacity of Capacitor in kVAR

kVAR = C x (2 π f V2)