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How to Calculate the Suitable Capacitor Size in Farads & kVAR for Power factor Improvement

 How to Calculate the Suitable Capacitor Size in Farads & kVAR for Power factor Improvement  (Easiest way ever)

Hi there! With a very important tutorial.. I hope you will find it very useful because I have already spent two days to prepare this article. I think all of those who have sent messages and mails about the topic will never ask again if they follow these simple methods to calculate the proper Size of Capacitor bank in kVAR and micro-farads for power factor correction and improvement in both single phase and three phase circuits. I think it’s too much..
Now let’s begin…
 
Consider the following Examples.
 
Example: 1
A 3 Phase, 5 kW Induction Motor has a P.F (Power factor) of 0.75 lagging. What size of Capacitor in kVAR is required to improve the P.F (Power Factor) to 0.90?
 
Solution #1 (By Simple Table Method)
 
Motor Input = 5kW
From Table, Multiplier to improve PF from 0.75 to 0.90 is .398
Required Capacitor kVAR to improve P.F from 0.75 to 0.90
Required Capacitor kVAR = kW x Table 1 Multiplier of 0.75 and 0.90
= 5kW x .398
= 1.99 kVAR
And Rating of Capacitors connected in each Phase
1.99/3 = 0.663 kVAR
 
Solution # 2 (Classical Calculation Method)
 
Motor input = P = 5 kW
Original P.F = Cosθ1 = 0.75
Final P.F = Cosθ2 = 0.90
θ1 = Cos-1 = (0.75) = 41°.41; Tan θ1 = Tan (41°.41) = 0.8819
θ2 = Cos-1 = (0.90) = 25°.84; Tan θ2 = Tan (25°.50) = 0.4843
Required Capacitor kVAR to improve P.F from 0.75 to 0.90
Required Capacitor kVAR = P (Tan θ1 – Tan θ2)
= 5kW (0.8819 – 0.4843)
= 1.99 kVAR
And Rating of Capacitors connected in each Phase
1.99/3 = 0.663 kVAR
 
Tables (Capacitor sizing in kVAr and Farads for PF correction)
The following tables have been prepared to simplify kVAR calculation for power factor improvement. The size of capacitor in kVAR is the kW multiplied by factor in table to improve from existing power factor to proposed power factor. Check the others Examples below.
Table – from 0.01 to 0.09 (Click image to enlarge)How to Calculate the Suitable Capacitor Size in Farads & kVARTable – from 0.10 to 0.30 (Click image to enlarge)How-to-Calculate-the-Suitable-Capacitor-Size-in-Farads-amp-kVAR-for-Power-factor-Improvement-Easiest-way-everTable – from 0.31 to 0.49 (Click image to enlarge)Calculate the Suitable Capacitor Size in Farads & kVAR for Power factor ImprovementTable – from 0.50 to 0.74 (Click image to enlarge) methods to calculate the proper Size of Capacitor bank in kVAR and micro-farads for power factor correction and improvement in both single phase and three phase circuitsTable – from 0.75 to 1.00 (Click image to enlarge)calculate the proper Size of Capacitor bank in kVAR and micro-farads for power factor correction The whole Table – from 0.10 to 1.0 (Click image to enlarge)How to Calculate the Suitable Capacitor Size in Farads & kVAR for Power factor Improvement (Easiest way ever)
Example 2:
 
An Alternator is supplying a load of 650 kW at a P.F (Power factor) of 0.65. What size of Capacitor in kVAR is required to raise the P.F (Power Factor) to unity (1)? And how many more kW can the alternator supply for the same kVA loading when P.F improved.
 
Solution #1 (By Simple Table Method)
 
Supplying kW = 650 kW
From Table 1, Multiplier to improve PF from 0.65 to unity (1) is 1.169
Required Capacitor kVAR to improve P.F from 0.65 to unity (1)
Required Capacitor kVAR = kW x Table 1 Multiplier of 65 and 100
= 650kW x 1.169
= 759.85 kVAR
 
We know that P.F = Cosθ = kW/kVA . . .or
kVA = kW / Cosθ
= 650/0.65 = 1000 kVA
When Power Factor is raised to unity (1)
No of kW = kVA x Cosθ
= 1000 x 1 = 1000kW
Hence increased Power supplied by Alternator
1000kW – 650kW = 350kW
 
Solution # 2 (Classical Calculation Method)
 
Supplying kW = 650 kW
Original P.F = Cosθ1 = 0.65
Final P.F = Cosθ2 = 1
θ1 = Cos-1 = (0.65) = 49°.45; Tan θ1 = Tan (41°.24) = 1.169
θ2 = Cos-1 = (1) = 0°; Tan θ2 = Tan (0°) = 0
Required Capacitor kVAR to improve P.F from 0.75 to 0.90
Required Capacitor kVAR = P (Tan θ1 – Tan θ2)
= 650kW (1.169– 0)
= 759.85 kVAR
 

How to Calculate the Required Capacitor bank value in both kVAR and Farads?

(How to Convert Farads into kVAR and Vice Versa)

 
Example: 3
 
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 Faradsto 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
 
Example 4
What value of Capacitance must be connected in parallel with a load drawing 1kW at 70% lagging power factor from a 208V, 60Hz Source in order to raise the overall power factor to 91%.
 
Solution:
 
You can use either Table method or Simple Calculation method to find the required value of Capacitance in Farads or kVAR to improve Power factor from 0.71 to 0.97. So I used table method in this case.
P = 1000W
Actual Power factor = Cosθ1 = 0.71
Desired Power factor = Cosθ2  = 0.97
From Table, Multiplier to improve PF from 0.71 to 0.97 is 0.783
Required Capacitor kVAR to improve P.F from 0.71 to 0.97
Required Capacitor kVAR = kW x Table Multiplier of 0.71 and 0.97
= 1kW x 0.783
=783 VAR (required Capacitance Value in kVAR)
Current in the Capacitor =
 
IC = QC / V
= 783 / 208
= 3.76A
 
And
XC = V / IC
= 208 / 3.76 = 55.25Ω
C = 1/ (2 π f XC)
C = 1 (2 π x 60 x 55.25)
C = 48 μF (required Capacitance Value in Farads)
 
Good to Know:
Important formulas which is used for Power factor improvement calculation as well as used in the above calculation
 
Power in Watts
kW = kVA x Cosθ
kW = HP x 0.746 or (HP x 0.746) / Efficiency … (HP = Motor Power)
kW = √ ( kVA2– kVAR2)
kW = P = VI Cosθ … (Single Phase)
kW = P =√3x V x I Cosθ … (Three Phase)
 
Apparent Power in VA
kVA= √(kW2+ kVAR2)
kVA = kW/ Cosθ
 
Reactive Power in VA
kVAR= √(kVA2– kW2)
kVAR = C x (2 π f V2)
 
Power factor (from 0.1 to 1)
Power Factor = Cosθ = P / V I … (Single Phase)
Power Factor = Cosθ =  P / (√3x V x I) … (Three Phase)
Power Factor = Cosθ = kW / kVA  … (Both Single Phase & Three Phase)
Power Factor = Cosθ = R/Z … (Resistance / Impedance)
 
XC = 1/ (2 π f C) … (XC = Capacitive reactance)
IC = V/ XC  … (I = V / R)
 
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)
 
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28 comments

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  2. sir u have biven best examples<br /> and i have one dought how we r going to get 1.169 value<br /> From Table 1, Multiplier to improve PF from 0.65 to unity (1) is 1.169

    • Check in the table… the existing PF is 0.65 and required one is unity (i.e, 1)<br />so the connecting point number between these two numbers (i.e, 0.65 and 1) is 1.169. this is multilier…<br />or<br />you can find it easily by scientific calculator as;<br />θ1 = Cos-1 = (0.65) = 49°.45; Tan θ1 = Tan (41°.24) = 1.169<br />θ2 = Cos-1 = (1) = 0°; Tan θ2 = Tan (0°) = 0<br />(Check solution number

    • Required Capacitor kVAR = P (Tan θ1 – Tan θ2)

  3. pls i need a simple circuit to try and see @fb.com/sarkinfawa

  4. Terrific article. I was examining frequently this weblog and I am <br />amazed! Exceptionally useful details specially the last section <br />I treatment for these types of data significantly.<br /><br />I was seeking this specific information for a pretty very <br />long time. Thank you and good luck.<br /><br />Also visit my web-site: AskNow Psychics Review (<a href="http://www.youtube.com/

  5. Nice Article.<br /><br />Can you also teach us on how to size up the capacitors in capacitor start – capacitor run motor?<br /><br />Thanks..

  6. I am Mechanical Engineer But i love to do Electrical Job to, Love this Page

  7. kalpesh gauswami

    sir …could u plzz help me on this topic – transformer protection with wireless alert (gsm based)??

  8. Abdul Afraaz Khan

    sir, i need your help in calculating the kvar, the power factor and also the capacitor size together with how many step power factor board should i make. iam very new to this so i need your help as soon as possible so that i can quote for the job.
    details are as follows:
    1) volts – 3300v
    amps – 107amps x2
    kw – 525kw

    2) volts – 3300
    amps – 101amps x2
    kw – 485kw

    3) volts – 415v
    hp – 20hp x3

    4) volts – 415v
    hp – 20hp x3

    your help will be highly appreciated.

    thanks
    Abdul

  9. Nemrod P. Baterna

    How to compute required capacitor voltage, Farads for an 8 KW alternator

    Please help, I have an 8 kW generator that don’t produce output voltage, when tested, a capacitor I believe acts as a voltage regulator don’t reads any farads value. “INFINITY”. Any specs/ratings that supposed to be the in the capacitor body was erased due to advance rusting.

    Generator components is still good as it was maintain properly, it’s only this capacitor part need to be replaced. Your help is highly appreciated.

    Sincerely Yours,

  10. i thank you for the valuable information about PF calculations

  11. A friend has an AirFoxx 4000a carpet dryer that burnt up it’s Cap and most of wires. I assume it was a start Capacitor as it’s plastic with epoxy. Was hoping you could help me find out what size is needed to replace it.
    This thing cost around $400 bucks but the company that made them is defunct.
    It is a 120volt 60hz three speed motor. 4 pole 1HP amps are 8,9 and 11 amps.
    There is no info left on the Cap as it was fried real good.
    Any Help appreciated.

  12. Bishan Kumar Bhatnagar

    I want to know how many kvar capacitor bank would be required for 200 kw 3 phase 4 wire load to achieve a pf of 0.9 from.08 lagging for my office

  13. Sir
    Please specify a suitable capacitor specification to be installed for a load of 20 hp as the earlier capacitor is malfunctioning with a power factor of 0.8

  14. sir, I want to know how to calculate a suitable size of a cable for a motor. thanks

    • Fabion Chihanga

      want to know step by step of calculating the capacitance required to bring pf to a certain angle.thank you

  15. An audit for one firm showed that the power factor is almost 70% and that the demand is 1000kw.
    What capacitor size is need to correct power factor to 90%?

  16. converte uf microfare para kvar

  17. Hey Tyrone,Great info for home owners and DIY’s, but I’m wdeonring if you could do a demo on the neutral strap (when it’s needed or not) for us apprentices. I would like to benefit from your experience.Thanks again for your efforts.Regards,Sean

  18. Sir
    How much rating kvar is to be adopted for getting 0.9 pf of 33 hp load of 0.8 pf ( industry)

  19. is the formula is same for calculating kvar of a capacitor for two phase and three phase voltage

  20. great great , it was very helpfull and clear, big thanks to you bro.

  21. i made pf corrector but i have some dought becous i made pf corrector for 230 volt
    what voltage rating cap i chose and its electrolatic

  22. its very faterstic,thx.

  23. My post is part time mali bt im working hv.instrumnt like cb.ct relay .all protaction system now my pay.9360/-p.m.this is resuilt for work in haryana viduit parsarn nigm ltd.

  24. Lokesh rajput

    I am Electrical Engineer and I really like this page…

  25. Ebenezer Ansah

    Dear Sir,

    I would like to ask for your help to solve the question below. Thank you very much. I have been following your articles and I must confess that you are genius. Keep it up. I wish you all the best.
    ************************************************************************************

    Question: A 11/0.433kV, 100kVA distribution transformer of 4.25% impedance is supplying a load of 0.6 power factor. It is required that the power factor be improved by a four – step capacitor bank. Reactive Power of each step of the capacitor bank is 50kVars.

    i. Calculate resonant frequency (in harmonic order) of the circuit at every step of the capacitor bank.

    ii. A measured harmonic content of the above circuit is given in Table-1 below. At what step of the capacitor bank is resonance condition possible and why.

    Harmonic order Fundamental 3rd 5th 7th 9th 11th
    Current (A) 35 16 13 9 7 3

    .

    Calculate the current total harmonic distortion of the circuit.

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