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.

Power factor correction is the one of famous topic for all electrical person at around the world.Many discussion and debate about how to sizing the capacitor for power factor correction.

**How to sizing capacitor for power factor correction?**

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%.

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

Thanks for ur compliment

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?.

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

Hi

Nice explanation

There is something similar explained for capacitor sizing

hi pieface,thanks for share nice info :D

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..

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

dear sir,

thanks for reply

if i have a only motors HP ratting, then how i can know require KVA of capacitor??

thanking yoy

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.

Above calculation is for global compensation only when a PFI Panel is to be installed at site for whole load(s).

Please BE CAREFUL when installing shunt capacitor with motors (sometimes called individual compensation). You should select the Kvar as per following formula:

Qc (Kvar) = 0.9 x Io x Voltage (U) x 1.732

Io is the no load current of a motor which shall be 20% to 30% of heavy motors.

Thanks,

Muhammad Waqar

Lahore Pakistan

m.wrj@msn.com

I got that I need a Supco Hard Start Kit Capacitor of 24???? I have a small common house freezer that does NOT already have any capacitor. I want to install one to help it get started. All I know is it is rated at 1/6th Horse Power or 124kW, 1.2 amp, 240volt on the sticker on the back of the unit, I do not know the efficiency percent etc. I found these equations difficult can anyone check over my answers for if what I got below is wrong or right. Or if you are interested to do the math and tell what i should get in capacitor size?

I have included all my workings outer the math. New at this so not sure what I’m doing or if its right.

Power Factor = 0.222

TWO WAYS BELOW TO GET kW

number time by 1.73×Rated amps 1.2×motor voltage 240× power Factor 0.222

kW = 1.73×1.2×240×0.222

kW = 0.110

kW = HP .124 x .746 / .84 as %age (a guess) motor efficiency

kW = 0.110

kVa = 1.73×1.2amp×240v/1000

kVa = 0.49824

Power Factor = kW .124(1/6th Horsepower) / kVa 0.49824

Power Factor = 0.222

Answer:

PF 0.222 x kW 0.110 = 24kVAR???

Answer = 24kVAR

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