How to Convert Capacitor μ-Farads to kVAR and Vice Versa? – For P.F Correction

How to Convert Capacitor kVAR to μ-Farads & Vice Versa for Power Factor Improvement?

The following simple calculation tutorial shows how to calculate and convert the required capacitor bank value in microfarads and then convert to kVAR and vice versa. We will be using three simple methods to convert the capacitor kVAR in μ-Farads and conversion of microfarad in kVAR.

Let’s see the following examples which show how to find and convert the Required Capacitor bank value in both kVAR and Micro-Farads which is applicable in Power Factor improvement Calculation and capacitor bank sizing.

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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:

Calculate and Convert kVAR in Microfarads

(1) To find the required capacity of Capacitance in kVAR and convert it to micro-farad to improve the P.F from 0.6 to 0.9 (Three Methods)

Solution #1 (Simple Method using the Table)

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 (Classic 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 in kW (Tan θ1 – Tan θ2)

= 12kW (1.3333– 0.4843)

= 10.188 kVAR

Solution # 3 (Using μFarad to kVAR Calculator)

You may directly use the Farad and microfarad to kVAR conversion calculator.

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(2) To find the required capacity of Capacitance in Micro-Farads and convert the Capacitor μ-Farads to kVAR to improve the P.F from 0.6 to 0.9 (Three Methods)

Solution #1 (Simple Method using the Table)

We have already calculated the required Capacity of Capacitor in kVAR, so we can easily convert it into Farads by using this simple formula

C = kVAR / (2 π f V2) in microfarad

Putting the Values in the above formula

= (10.188kVAR) / (2π x 50Hz x 4002V)

= 2.0268 x 10-4

= 202.7 x 10-6

= 202.7μF

Solution # 2 (Classic Calculation Method)

kVAR = 10.188 … (i)

We know that;

IC = V / XC

Whereas XC = 1 / 2π x f x C

IC = V / (1 / 2π x f x C)

IC = V x 2π x f x C

= (400V) x 2π x (50Hz) 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

Solution # 3 (Using kVAR to μFarad Calculator)

The following formulas are used to calculate and convert capacitor kVAR into Farads and Vice Versa

• C = kVAR x 103 / 2π x f x V2                         …     in Farad
• C = 159.155 x Q in kVAR / f x V2                 …     in Farad
• C = kVAR x 109 / (2π f x V2)                       …     in Microfarad
• C = 159.155 x 106 x Q in kVAR / f x V2        …     in Microfarad

Required Capacity of Capacitor in kVAR

• VAR = C x 2π x f x V2x 10-6                               …     VAR
• VAR = C in μF x f  x V2 / (159.155 x 103)          …     in VAR
• kVAR = C x 2π x f x V2  x 10-9                           …     in kVAR
• kVAR = C in μF x f x V2 ÷ (159.155 x 106)        …     in kVAR
• MVAR = C x 2π x f x V2  x 10-12                         …     in MVAR
• MVAR = C in μF x f x V2 ÷ (159.155 x 109)        …     in MVAR

Where:

• C = Capacitance in Microfarad
• Q = Reactive Power in Volt-Amp-Reactive
• f = Frequency in Hertz

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Electrical Technology

1. Unknown says:

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

2. john says:

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?

1. Electrical Technology says:

Thanks for correction :)

2. khan says:

yes bro it is given 12.. and you can see this problem in Vk Mehta book on Power distribution and utilization as well

3. chandran says:

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. Jashwant says:

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

1. Electrical Technology says:

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

2. rathinam says:

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. md rasheed says:

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. rathinam says:

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?

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

8. Suphot Boonsuksri says:

Thank you very much

9. JOHN MULINDI says:

Good reminder of what I learned in my first year of engineering.

10. sang says:

the formula is not right, it should be:

KVAR = Q = Vrated^2/Zc

11. Shaeed says:

I need a power capacitor of UN 380V QN 150 micro F 50 Hz
In 17.9 /10.3 A but I cannot find similar rating in Local market.

Which KVAR and In rating may suitable
Let me know pls

12. Bob Harvey says:

Why does “C = kVAR / (2 π f V2)” in microfarad come out in uF? where does the 1,000,000 come from?

Surely it should be mF? the k of kVAr is only 1000

1. Christian Herzig says:

yes, it is confusing. It should be C = VAR / (2 π f V2)
and the result is in Farad and we have to multiply by 1,000,000 to get uF

13. david rooke says:

Hi hope someone can help me have a 3 phase air compressor need to run it on single phase. Its a 240/415 volts hp 5.5/4 hz 50 Im looking to fit a capacitor what i need to know is what size uf capacitor i need.Thanks for help.

14. Paresh valia says:

If I have AC 5 kva induction motor of 1440 RPM.so i put Capacitor of 0.66 kvar.if my motor RPM change like 5 kva motor of 2800 RPM then any change in capacitore kvar ? I want to know any role of motor RPM to select capacitor size ?

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