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Watts vs Volt-Amps - huh ??

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  • Sunking
    replied
    PF is no ta huge problem for TSW inverters, but can be a major problem for MSW inverters. However even TSW inverters can have a problem for example running a lot on non linear loads like switching power supplies found in Desktop computers and cheap CFL light bulbs where PF can be .6 or even less. Most TSW inverters can only handle .8 PF at full power.

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  • SunEagle
    replied
    Originally posted by russ View Post
    For residential grid connect PF is a problem for the utility - not the homeowner.
    A bad or low Power Factor is no good for either the utility or the homeowner. Utilities charge a PF penalty to customers to help them reduce waste on their distribution system.

    While a residential connection does not usually have a PF penalty added to their bill by the Utility, a low PF means they have losses in their electrical system. These losses will use up some of the total available kva of their transformer capacity in their inverter.

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  • russ
    replied
    Originally posted by Sunny Solar View Post
    Power factor (PF) is one of those measurements that have no effect on 99.9% of people. Most people buy a AC or refrigerator or microwave oven without ever needing to know their PF.
    Grid power can cope with just about every residence having some devices with far from perfect PF.
    Its only a concern to the following electrical engineers designing big industrial factories and similar, power companies charging those factories,and when you are into using off grid solar power for your residence.
    To most of the people that need to get PF as close to 1 as possible are off grid solar power users. The reasons being that "bad" PF wastes your battery/ies power. Less run time from any given battery capacity.
    If the old standard of assuming PF of .8 then you wont go far wrong in estimating your battery power usage.

    PF is really one of those measurements like Lumens ..If your room is not bright enough you dont go checking the Lumens ,you just go upgrade your existing 60w lamp for a 100w one...You get the idea.
    For residential grid connect PF is a problem for the utility - not the homeowner.

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  • Sunny Solar
    replied
    Power factor (PF) is one of those measurements that have no effect on 99.9% of people. Most people buy a AC or refrigerator or microwave oven without ever needing to know their PF.
    Grid power can cope with just about every residence having some devices with far from perfect PF.
    Its only a concern to the following electrical engineers designing big industrial factories and similar, power companies charging those factories,and when you are into using off grid solar power for your residence.
    To most of the people that need to get PF as close to 1 as possible are off grid solar power users. The reasons being that "bad" PF wastes your battery/ies power. Less run time from any given battery capacity.
    If the old standard of assuming PF of .8 then you wont go far wrong in estimating your battery power usage.

    PF is really one of those measurements like Lumens ..If your room is not bright enough you dont go checking the Lumens ,you just go upgrade your existing 60w lamp for a 100w one...You get the idea.

    Leave a comment:


  • billvon
    replied
    Originally posted by Beanyboy57 View Post
    Is knowing PF critical for me at this point or useful for me to understand?
    It's not that critical. Devices that have big motors in them sometimes have a poor power factor and need bigger inverters to operate. Small wall warts (laptops, cellphone chargers etc) can also have poor power factors, but they are generally smaller loads so the poor power factor matters less. Overall if you just allow some margin for power factor you are generally OK.

    The two areas that you might look into if you want to learn more about power factor are peak rectification and complex impedance. Peak rectification refers to devices that only take power during a small part of the AC waverorm (requiring an inverter that provides a higher peak power) and complex impedance describes loads whose current waveform is not completely in sync with the voltage waveform. Both are more difficult for an inverter to drive, but modern inverters are designed with non-ideal loads in mind.

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  • Beanyboy57
    replied
    PF

    Originally posted by Sunking View Post
    I am OK with that, but before you started educating yourself did you have any clue what PF is? Could you define it and how to calculate it.

    OK since you have some knowledge can you define and know how to calculate it now?
    I am not at that stage of learning yet and perhaps never will be. All I know are the basic formulas for power such as P = IE or P = I(squared) R and Ohm's law E = IR. I am adding to this knowledge daily however.
    As I said I am a new to this and perhaps I will take some time to get to an intermediate level of knowledge which will probably be where I peak in this particular field of study. I know my own capabilities and where my strengths lay.
    My main hobbies are gardening and racing Go-karts, perhaps I know enough about them to teach some of you electrical guys a thing or two!!
    Is knowing PF critical for me at this point or useful for me to understand?

    Leave a comment:


  • Sunking
    replied
    Originally posted by Beanyboy57 View Post
    I have to disagree with you SK,
    I am OK with that, but before you started educating yourself did you have any clue what PF is? Could you define it and how to calculate it.

    OK since you have some knowledge can you define and know how to calculate it now?

    Leave a comment:


  • Beanyboy57
    replied
    Cheers!!

    Originally posted by Sunking View Post
    I chuckled when I first read this post. This is pretty much a DIY forum and the posters looking for help have ZERO electrical knowledge and all this is is GREEK to them and will never understand one bit of it.
    I have to disagree with you SK, just because many of the contributors on this site have little electrical knowledge it doesn't mean we have ZERO knowledge of algebra or mathematics in general. I have to say that since joining this very interesting site I have learned a huge amount, I am taking an online course in Basic concepts in Electricity, I have modified and greatly improved my off-grid setup in my shed and I have a new interest in life now. I can safely say now, that I know and understand the meaning of many of the electrical terms that you commonly use, however when reading this page I realize there is a huge amount that I do not know but that only inspires me to study more (I do have plenty of spare time these days). I thank all of you gurus for your undying patience with us DIY'ers!

    Leave a comment:


  • Sunking
    replied
    Originally posted by randl View Post
    My bad, inetdog and Sunking, sorry.
    No need to apologize, no worries. Be happy.

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  • randl
    replied
    My bad, inetdog and Sunking, sorry. I knew it didn't feel right dividing by 0.8. The "standard" correction should be VA * 0.8.

    randl

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  • Sunking
    replied
    Originally posted by inetdog View Post
    I have always thought that for a given load the VA number was greater than to equal to watts, not the other way around?
    That would be correct
    Watts = VA * PF
    VA = Watts / PF.
    PF = Watts / VA

    PF is a number of 1 or less where 1 is a purely resistive load.

    Leave a comment:


  • inetdog
    replied
    Originally posted by randl View Post
    You could measure the actual power use of each load. If you're calculating a hypothetical load, my electrical engineer friend and my quick read on this topic both tell me that accepted practice is to divide VA by 0.8. If VA=1000 then you would correct to 1249 Watts. This estimate is apparently well accepted but, of course, it could be low for a purely reactive load.
    I have always thought that for a given load the VA number was greater than to equal to watts, not the other way around?

    Leave a comment:


  • randl
    replied
    Originally posted by martin1025
    how can we calculate this and any proper to change ............
    You could measure the actual power use of each load. If you're calculating a hypothetical load, my electrical engineer friend and my quick read on this topic both tell me that accepted practice is to divide VA by 0.8. If VA=1000 then you would correct to 1249 Watts. This estimate is apparently well accepted but, of course, it could be low for a purely reactive load.
    Last edited by russ; 10-26-2012, 10:20 AM. Reason: removed link

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  • randl
    replied
    There's a good, concise explanation of power factor in the August/September 2012 edition of Home Power magazine. It discusses resistive vs. reactive loads, shows an example of in phase and out-of-phase reactive load plots, mentions what sort of loads will have high power factors, etc.

    The trig explanation isn't too helpful for those who don't see the plots...

    Leave a comment:


  • Mike90250
    replied
    Originally posted by Sunking View Post
    I chuckled when I first read this post. This is pretty much a DIY forum and the posters looking for help have ZERO electrical knowledge and all this is is GREEK to them and will never understand one bit of it.
    You can lead a horse to water, but you cannot make it drink! But the info is still valid, and if it gets only a couple people
    asking questions about power factor (why does my 1/2 hp motor draw 1,000 watts) and they figure it out, our work here is done.

    Leave a comment:

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