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How to Convert Capacitor Farads into kVAR & Vice Versa (For Power factor improvement)

How to Convert Capacitor Farads into kVAR & Vice Versa for Power factor improvement?

How to Calculate the Required Capacitor bank value in both kVAR and Farads?
(How to Convert Farads into kVAR and Vice Versa)How to Convert Capacitor Farads into kVAR & Vice Versa For Power factor
How to Convert Capacitor Farads into kVAR & Vice Versa (For Power factor improvement)
In this simple Calculation tutorial, we will find the way “How to Convert Capacitor Farads into kVAR and Vice Versa, usually used in Power Factor improvement Calculation and problems. We will use two simple methods for finding and Conversion for both quantities. For explanation, consider the following simple Example.
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 (i.e, Convert Capacitor Farads into 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
ActualP.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)
= 12kW (1.3333– 0.4843)
= 10.188 kVAR
 
(2) To find the required capacity of Capacitance in Farads (i.e, Convert Capacitor Farads into kVAR) 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 Capacitor kVAR into Farads and Vice Versa
 
Required Capacity of Capacitor in Farads/Microfarads (Convert Capacitor Farads into kVAR)
C = kVAR / (2 π f V2) in microfarad
 
Required Capacity of Capacitor in kVAR (Convert Capacitor kVAR into Farads)
kVAR = C x (2 π f V2)  

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12 comments

  1. thanks buddy ,this information ll use in my project !!!! …. 😀 :)<br /><br />

  2. I have a question on the following lines of your calculations:

    Solution # 2 (Classical Calculation Method)
    Motor Input = P = V x I x Cosθ
    = 400V x 50A x 0.6
    = 12kW
    .
    .
    .
    Required Capacitor kVAR = P (Tan θ1 – Tan θ2)
    = 5kW (1.3333– 0.4843)

    Where did the P=5kW come from? I thought it was calculated to be 12kW. Is this an error?

  3. I got some confusrion for refixing one Philips model HL1642 -600 watts mixie grinder motor wire connection with its Rotary switch connection having overload,INCHERswitch&line indicator .wiring diagrame may be sent to my E mail address for clearing my doubts&refixing the connections correctly…Tnx in anticipation…acn 1-2-15

  4. Hello
    How kvar came to 10.188 kvar.
    If kw =12 and kva = 20, kvar will be 16 for sure.

    Kvar = kva sin theta
    = 400x50x 0.8
    = 16

    hence C will come 318.31 uf

    Please clarify or correct.

    • Dear Jashwant,
      We are not going to calculate the kVAR from kW or kVA here…
      But want to improve the Power factor from 0.6 to 0.9.
      That’s Why we didn’t use that (Which you have mentioned) formula… 🙂

      let me know if you want to know more… Thanks

    • how to calculate the required capacitor to improve power factor by using
      apparent power(kw)=3.66
      voltage(v)=408V
      current(i)=1640A
      VAR=1403
      pf=0.5.
      it is a three phase

  5. i did not understand you have a given exmple 400v and single phase what iis that is it 3phase .still iam new for this subject

  6. This is a right or wrong formula to convert KVAR to microfarad.
    C=KVAR/(2*1.732*F*V^2).
    or any other solution is available?

  7. how calculate
    10.188/2*3.14*50*400*400=202/1000000

  8. Suphot Boonsuksri

    Thank you very much

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