Real Power Vs Apparent Power

• Sep 10, 2013

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Dear All,
From Calculation point of view ,
Power Factor=Real Power / Apparent Power
Since,Apparent Power = Vrms * Irms ,
Power Factor =cosangle

therefore,Real Power=Vrms * Irms * cosangle

So according to formula Real Power also depends on Phase Angle.Since residential electric bill are based on Real Power,so from above formula they also charge for Phase angle,so by improving power factor the bill will get reduce?

But according to several discussions & comments they are saying bill is independent of power factor?


Thanks & waiting for knowledge enhancement,:smile:

 

• Sep 10, 2013

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For domestic use, they don't bill differentiate for active and reactive power, bill for aparent power only. For industrial is imposed a maxim reactive power so if there are a lot of inductiv or capacitive loads, need to compensate power factor to be minim 0.9 (in most country). So, for home, no advantage to improve power factor to 1 from billing point of view but there are an advantage that consist in reducing cable loses (reducing Irms); these losees are proportional with Irms^2 so, for example 10% reducing of Irms result in 21% loses reducing; these losese is counted by billing device... so may be not neglijable.

 

• Sep 10, 2013

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Dear iop95,
As Irms increases due to power factor, indirectly the Apparent Power increases,thus bill increases?



Thanks & waiting for knowledge enhancement,:smile:

 

• Sep 10, 2013

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For a constant active load if you increase power factor (reduce reactive power) will reduce Irms; keep account that Irms=(Iactive^2 + Ireactive^2)^1/2, so power factor close to 1 mean Ireactive go to zero, so remain only active component, hence Irms = Iactive.

 

• Sep 10, 2013

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Dear iop95,
thanks but,can you explain with some example like,
For 1KW Load, Power Factor = 0.8
Change Power Factor =0.95


Thanks & waiting for knowledge enhancement,:smile:

 

• Sep 10, 2013

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For 1kW load, 230Vrms voltage for example and PF=0.8, aparent power (S) = 1/0.8=1.25kVA, Irms=1250/230 = 5.43A; if PF increase to 0.95, S=1/.95=1.052kVA, Irms=1052/230=4.57A; difference is 18.8% that mean loses reduction of 35.3%.

 

• Sep 10, 2013

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Dear iop95,
Lose reduction?
Irms=1/2(Irmsreal 2 + Irmsreactive 2)?


Thanks & waiting for knowledge enhancement,:smile:

 

• Sep 10, 2013

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Cable lose is Pj=r*Irms^2; for a givem circuit, r (cable resistance) presume to be constant, so if Irms reduce by 18.8% mean that Pj will reduce with 1.88^2, 35.3%.

 

• Sep 11, 2013

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Mr.Cool

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why would your typical residential power factor be initially so low at 0.8?

 

• Sep 11, 2013

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Dear Mr.Cool,
that was just for example point of view for different power factor 0.8 to 0.95,to understand the concept,



Thanks & waiting for knowledge enhancement,:smile:

- - - Updated - - -

 

• Sep 11, 2013

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Mr.Cool

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ok, just wondering. using power factor in your home IS legitimate ** IF **
- your power factor is initially 0.8 to 0.85 ON THE AVERAGE (not just the single day you took a measurement)
- your caps can bring PF to 0.99%
- your utility billing will reward you for improved power factor
- you do the cost benefit analysis to determine pay back period (taking into consideration life expectancy of your caps)

 

• Sep 11, 2013

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Dear iop95,
How come 1.88?
Difference is 0.86 which is 16% of 5.43Irms?
Irms=[(Irmsreal^2 * Irmsreactive^2)^1/2] ?


Thanks & waiting for knowledge enhancement,:smile:

- - - Updated - - -

- - - Updated - - -

Dear Mr.Cool,
But even though we correct our power factor,no doubt losses will get reduce & ultimately durability of electronics load get increased,the bill to be paid remains same?


Thanks & waiting for knowledge enhancement,:smile:

 

• Sep 11, 2013

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No, it's 1.188 times current reduction that mean 1.411 times loses reduction.

 

• Sep 11, 2013

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Means Only capacitor can satisfy the power factor correction,then why people go for ir1150 is it for Dc to Ac application?


Thanks & waiting for knowledge enhancement,:smile:

 

• Sep 11, 2013

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Power factor correction usually is linked with inductive loads, such as electromotors, heat pumps compressors, elevator motor, .... specially in factories and industries. But adequate capacitor is already mounted with electromotor.



See this video:
https://www.youtube.com/watch?v=V_ooTeO_Tq8



Search existing EDABoard threads there is interesting threads about this topic.




Best regards,
Peter

 

• Sep 11, 2013

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ir1150 is a PFC correction circuit that is in the front of DC converter. Any rectifier with filter capacitor looks like an inductive load and such PFC ic try to correct PF by "imposed" current to be in phase with voltage (PF = 1).

 

• Sep 11, 2013

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tpetar

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For home usage there is no benefit from that.

Industry is other story.


In home worst device what someone can use with PF 0,7-0,8 is REL transformer welding device, but this is not 24h per day. :smile:



Best regards,
Peter

 

• Sep 11, 2013

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Any rectifier with filter capacitor looks like an inductive load

Click to expand...

Not particularly, phase angle of it's fundamental current is slightly capacitive (e.g. 10 - 20 degree). In other words, it's displacement power factor is > 0.95.

Power factor correction for rectifier supplies is wanted (and even required in newer power quality standards) to improve the distortion power factor, the non-sinoidal current waveform. For an introduction see https://en.wikipedia.org/wiki/Power_factor

 

• Sep 12, 2013

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Means using only capacitor will not provide us the required inphase voltage & current i.ePower Factor =1?
But using external capacitor even though we are using capacitor at the input of every electronic load so what effect it will create,because external capacitor initially tries to acquuire current & slowly voltage across of it increases(Current leads & voltage lags)but, what about input capacitor which is again parallel to external capacitor, What will be the effect?


Thanks & waiting for knowledge enhancement,:smile:

 

• Sep 12, 2013

• #20

FvM

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Capacitors can compensate reactive apparent power (e.g. from a motor or fluorescent lamp ballast choke), not harmonic currents (distortion power factor) from electronic loads.

As said, you rarely need to care about power factor for home power supply. You have to think about it if you operate a factory and as a manufacturer of electronic devices.