09 May 2011 ~ 10 Comments

Capacitor sizing for power factor correction

How to sizing capacitor for power factor correction?

Commonly method for power factor correction using the capacitor to generate reactive power ( kVAR ) to reduce apparent power ( kVA ) form inductive load.

This time i want share a simple method to sizing our capacitor to improve power factor for induction motor and utility electricity.This method also can use for capacitor bank sizing.

To properly sizing the amount of capacitor (kVAR) required to correct the lagging power factor,we  must have three (3)  important of information below:

  • kW (kilowatts)
  • Existing Power Factor ( % )
  • Desired Power Factor ( % )

From this information,now we can calculate the capacitor size for power factor correction.The formula to calculate the required kVAR is:

Factor from Table 1 below x kW = kVAR of capacitors required.

Calculation Example :

1 unit air-compressor ( 3 phase 415 VAC ) used an average of 90 kW with an existing power factor of 80%.The desired power factor is 95%. The factor value for this case is 0.421 to raise the power factor from 80% to 95% using table 1.

0.421 x 90 kW = 38 kVAR

So we need capacitor size 38 kVAR to get power factor 95% for 1 unit air-compressor 90 kW.

If working power (kW) or Present Power Factor are not known you can calculate from the following formulas to get the three basic pieces of information required to calculate kVAR.The best range for  desired power factor is around 95% until 100%.

 

 

 

 

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10 Responses to “Capacitor sizing for power factor correction”

  1. haitham hashim 10 May 2011 at 2:14 AM Permalink

    If I had known this site earlier when she entered the Faculty of Electrical Engineering, read the threads started by Mr. Admin and I become a better engineer, thank you, Mr. Adam

  2. Sinnadurai Sripadmanaban 10 May 2011 at 9:45 AM Permalink

    Are pf and efficiency of motors given nameplate of all motors?.
    If no-load current is known(given in nameplate) can we estimate the size of capacitor required a) to start the motor (by reducing starting current)
    b) to correct pf?.

    • lemau 11 May 2011 at 8:28 AM Permalink

      We can estimate the full load ampere using the formula : – kW = ampere x Volt x 1.74

  3. Pieface 13 May 2011 at 8:33 PM Permalink

    Hi
    Nice explanation
    There is something similar explained for capacitor sizing

  4. stefanb2hv 16 May 2011 at 2:03 AM Permalink

    hello, nice article!!! I am really impressed. Actually this is my first time reading your post. But I am very fascinated by the way you explained this simple method in sizing up a to improve power factor for induction motor and utility electricity. Thanks for sharing this information. Really good. I will be subscribing to your feed. Rest assured that I will read your future posts..

    • lemau 16 May 2011 at 11:11 AM Permalink

      HI stefanb2hv..thanks for ur compliment…are u supplier from germany?i like germany stuff.it have a high quality :D

  5. BHARGAV 20 May 2011 at 12:27 PM Permalink

    dear sir,
    thanks for reply
    if i have a only motors HP ratting, then how i can know require KVA of capacitor??
    thanking yoy

    • lemau 20 May 2011 at 8:09 PM Permalink

      no problem bro,
      do you mean for power factor correction for induction motor?
      follow this step :-
      1) Convert hp to kW;formula : 1 hp = 0.746 kW
      2) from kW,you convert to Ampere;formula : I = kW / volt x 1.73
      3) from ampere you can find the kVAR for power factor capacitor..for more detail please read my post about capacitor sizing for power factor.


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