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Popular Science - Weaponology - zahoor ahmad Pakistan - Member - New Member

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power factor

12/17/2008 2:54 AM

what is power factor give me full detail

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#1

Re: power factor

12/17/2008 3:01 AM
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#2

Re: power factor

12/17/2008 3:44 AM

The power factor (P.F.) is = REAL POWER / TOTAL APPARENT POWER

In an Ideal Sinusoidal Signal Both current and voltage waveforms are assumed to be IDEAL SINUSOIDAL waveforms. If the phase difference between the input voltage and the current waveforms is defined as the phase lag angle or displacement angle,

REAL POWER : P = VRMS IRMS Cosφ

TOTAL APPARENT POWER: S = VRMS IRMS

Then, by definition: P.F. = (P/S)Cos φ.

Check the below article : www.st.com/stonline/products/literature/an/4042.pdf

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#5
In reply to #2

Re: power factor

12/18/2008 1:51 AM

thanks for such good answere.

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#3

Re: power factor

12/17/2008 5:32 AM

ITS A HUGE TOPIC what actually u want to know ?pls clarify...

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#4

Re: power factor

12/17/2008 11:02 PM

it is a factor which gives the exact value of useful power by the load when multiplied with apparent power.

p=vi cos(angle)

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#6

Re: power factor

01/21/2011 4:41 PM

Lets take an example of to understand somewhat the concept of power factor.

We have a 500 VA UPS. It runs 2 computers each having Load of 0.9 A and P.F 0.6 at 220 VAC system. When we add a third computer, UPS fails to provide power to these three computers. UPS Fails Why?

For a Single Computer at P.F = 0.6

Real Power = 220 x 0.9 x 0.6 = 119 W

Apparent Power = 220 x 0.9 = 198 VA

When both computers were on we have

Real Power = 238 W

Apparent Power = 396 VA

When the Third Computers was added then

Real Power = 357 W

Apparent Power = 594 VA

Since Apparent Power of UPS is 500 VA and at this P.F i-e 0.6, 594 VA are required so UPS fails to fulfill the requirement.

Now a Capacitor is added to improve the Power Factor from 0.6 to 0.95 then Current Reduces to 0.5693 A.

Now for a single system

Real Power = 220 x 0.5693 x 0.95 = 119 W

Apparent Power = 220 x 0.5693 = 125.246 VA

Now If three Computers are added to the system then

Real Power = 220 x 0.5693 x 0.95 = 357 W

Apparent Power = 220 x 0.5693 = 375.738 VA

System will run quite smoothly as Requirement reduces from 594 VA to 376 VA.

By improving the P.F

1.We reduce the size of Transformers, cables etc.

2.We are very efficiently utilizing the System Power.

Also see the video

http://www.youtube.com/watch?v=AI_e3dF14eg

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