Wow, lots of posts since I was out doing mundane mechanical stuff like cutting the grass, fixing the mower, finishing cutting the grass, eating dinner... Back to the subject at hand...
Edit:snarkiness removed
![ROFL :ROFLMAO: :ROFLMAO:](data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7)
No worries (whatever you said), I'm enjoying the exchange, and I don't have thin skin when it comes to this stuff
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. Carry on, but don't get in trouble with the mods.
I had a chance to google some things, and now I think I see where theory is meeting the real world.
Like a few others, it's been about 35 years since I was nodding off in front of a blackboard filled with Power Factor correction formulas. Never used it even once in my whole career, so applying this to a real world application is kinda interesting in that geeky kinda way.
So of course,
eridanus is correct when he says the real power is the Watts, not the VA. And as others have found, there is some conflicting info out there on the application of this subject - so here is where the theory and practice meet (from what I've re-learned so far):
As we've discussed, a device with a PF of .5 (CFL) draws 2X the current of an equal wattage device with a PF of 1 (ILB). And 2X the current will have 4X the losses in the distribution system.
What I thought was, that the generator at the power station had to produce Volt-Amps. Well, it does, but as far as I can tell, the Horsepower required to turn it is based on the real power (Watts), not the Volt-Amps. So in theory, (except for compensating for the increased distribution losses), a generator shouldn't burn any more coal producing power for a device with PF1 or PF 0.5 of equal wattage (except for the fact that a generator that can handle higher current will be bigger and heavier and have more friction). However....
Remember those distribution loses? 8% on average for the grid, so the PF 0.5 takes those loses to 4x8=32%, so 24% additional loss. And generators are not 100% efficient either, maybe 90% (no time to google it now). I'd assume a big part of that inefficiency is loss due to current. So the increased current from a PF 0.5 will also increase the losses in the generator. If those losses are 5%, they would also increase to 4X at PF 0.5, to 20%. Add that to the 24% and we are in the ballpark of 1.5X energy consumption for a CFL than the watts rating would indicate.
If someone can put a closer estimate to those numbers, please do. But I fall back to my real world gut check - in the examples I saw, the Utilities bill on a near linear scale, charging 20% more for a .75 PF (.25 away from the ideal of 1.0), and there must be a reason for that.
-ERD50