How To Test DC to AC Power Inverter Efficiency - Basic Voltage Drop Results
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- Опубліковано 6 жов 2024
- This is my electrical testing videos below .
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This is my off grid solar projects and power inverters.
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You did the demonstration just like an electrical professor would have..... I could not have done any better myself.....you got all the loses to take into the calculation.....developing heat is your biggest enemy to contend with and you know this......well done Steve.....CHEERS
Thanks Gator. There's so much to electrical I find interesting. Cheers
Great info. I have a small one years ago for camping, for blowing up mattress etc power laptop. 10 percent drop , i wasnt expecting that. Proper cables etc thats a great help bud.
Thanks bud much appreciated 👍
I’ll take your word for it Steve cause you are the smartest friend I have 😂. Thanks Buddy!
HAHAHA I'm think your right Doug 😅😅 Cheers 👍
How's she goin'? Good demonstration of voltage and current losses. To me the key to this comparison is the current draws. Like you mention, if the load stays the same, as the voltage goes down the current goes up. This is a very important point. You need to know what this current will go to so that you can size your wiring correctly. And the longer the distance the more line loss you will encounter in these wires and that means a bigger size cable will be needed. Very interesting these experiments and practical set ups you do Steve. Thanks and take care!!
Thanks Mike , some electrical basics go a long way for those trying to run a 13 amp load on a 16 gauge 150 foot extension cord 😅⭐
Great info there Steve, you take all the guesswork out of it for folks. Cheers
Lloyd I sure like testing stuff . Thanks bud 👍
Good info here Steve. Nice job breaking down the numbers! 10% is significant!
Thanks bud , numbers count big time 👍
I have one of those expensive DC fridges. For me it was worth the extra money because I don't need as much solar power to run it. Not only do you eliminate loss at the inverter, it is also a much more efficient compressor (at least the one I purchased does). It is tough to plunk down that kind of cash, but after five years I don't even notice it, lol.
Yes over time the extra cost doesn't feel so bad LOL
Wow Steve, this was one POWERful video and quite enLIGHTening!! Be safe my YT friend. Cheers.
Thanks buddy , shining some light on subject is a good thing :)
That was a pretty hefty loss when you loaded it up! 😮 Glad I didn’t have to do the math 😁👍
HAHAHA the math was the fun part 🤣
PAPA doesn't Test. Just bought a new light, no idea what it does! All these numbers make me want to drink! PAPA doesn't do math!
LOL no math in Texas eh .
You are on the right track bud, but one big mistake. You cant apply AC watts in your DC math. Also, the kill-a-watt device is redundant here and provides a slightly less accurate reading than doing the math with the voltage and current readings. DC side 12.1 x 8.4 = 101.64. AC side 115.45 x .82 = 94.669. Therefore 94.669/101.64 = 93% efficient. And that's at less than 1/3 of rated capacity!
I work for a utility and see solar applications all the time. The losses get bigger the larger you go. A common solar array I see is 7.2 KW DC/6 KW AC. That equals out to 83% efficient. They'll show the customer the math based on the DC specs instead of AC to pad their numbers.
If you want to go crazy with your understanding of electrical theory, start looking into capacitive and inductive loads and how they affect your watt readings. Your light bulb here is a purely resistive load.
100% correct watts ac to watts dc are not exact as amps AC to amps DC are , the only difference is in how you measure. To present an exact corolation in measurements between ac and dc watts is impossible without more data . I'm presenting a rough expectation at best for an audience that's interested in solar. Cheers 👍
Good video Steve
Thanks bud👍
According to AIMS Corporation, their inverters peak efficiency is at about 40% of their rated maximum load. We know what they do when you get close to 100% of maximum rated load: they get hot and their efficiency drops.
Great info as always. 👍
Thanks Les , I try 🤷♂️
I dunno, I kinda feel like I was tricked into doing math. Very informative video though. I'm seriously thinking of buying a bit of land and building a small off grid place. Use it as a vacation home. If I ever end up doing it, I'll have a bazzillion questions for you! lol
Buy away Bruce , that's a great idea.
nice job man! good explanation! very good this experiment!
Thank you much appreciated 👍
I had a AC DC fridg with frezzer in our boat, it was a full size apartment size It wasan AC with a built in invert. When on shore power it ran on AC, when under way it was on DC, unless I had the on board 10.000 watt generater running.
Sounds like a great setup 👌
Superb entry waooo amazing
Thanks for watching
Awesome!
Did you watch the video Rex?
@@PAPATexas Yes I did!
Thanks!
I've only got one volt meter here. I guess anyone can do electric testing with all those specific tools.
BullyBilly, Colorado.
One is all one needs . You can't do any electrical testing unless you know how to operate the meter.
@11:54 it really is impossible to install or employ too thick of conductors to supply power to your inverter, or to connect to the output side of your inverter. At some point, the cost to benefit ratio drops off of course. As a minimum, I'd use double the thickness of the minimum size for a calculated distance. All this being said: do not use double the minimum rated circuit breakers and fuses of course; as circuit breakers protect equipment, you & your home or vehicle from fires associated with short circuits.
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question, so on the dc side wouldnt you take the volts of the battery x the amps drawn =101.64 watts then minus thst from the watts at the bulb 94 watts= a difference of 7.6 watts. With that considered would the efficeny still be at 90% since you ssid there was a 10% differential?
I'm not sure , never tried that formula . Not sure at all 🤔
You are on the right track. What you are looking for is calculating the gain or loss. This is derived from the ratio of the output/ input. There is no unit for it because both measurements have the same unit. In this case, it makes sense to take the ratio of the powers: 94/101,64 = 0.88. In percentage, it is 88%, resulting in a 12% loss. I think his calculation is wrong. Data collected is good.
@@zegeyealemu1101 yes that is what i was trying to get at lol. from what i understand to get efficiency the way you stated is how i understand it also.
@@zegeyealemu1101 I'm not Einsten or anything but I thought 94/101.64=0.9248 which is approximately 92% efficiency.I stand to be corrected .
@chris Ooops yes, you are correct. The quotient of the duvidion is wrong. The loss is 8%, not 12%, due to division error. Thanks for pointing that out.