Inverter losses

mountainman

When sizing a battery bank to loads. Does the 85℅ account for inverter idle. My loads are 1100 wh daily. .85 inverter efficiency puts me at 1294 whs .  My 300 pure sine uses 8 watts just being on  x24 hrs.~200 whs. Do I add this also 1294 +200=1494? Or is it already included in the 85℅??

BB.

The 0.85 does include inverter losses... But that is pretty much assuming that the inverter is reasonably loaded when turned on. If, for example, you have a load you run for 5 hours per night, but leave the inverter running 24x7, then you should add in the tare losses too...

For example, say you run a 100 Watt AC load on your 3009 Watt inverter for 5 hours per night and run the inverter 24 hours per day... Very roughly:

• 100 Watts * 0.85 inverter eff * 5 hours per night = 588 Watt*hours at the battery bus
• 8 Watts * 20 hours per night = 160 WH per day
• 588 Watts + 160 Watts = 740 WH per day
• 100 Watts * 5 hours = 500 WH load
• 500 WH load / 740 WH est. battery bus load = 0.68 = ~68% efficiency

You can see that the inverter's efficiency is also a function of its loading (page 2 of this specification--You can see that the inverter is closer to ~92% efficient at 100 Watts):

https://2n1s7w3qw84d2ysnx3ia2bct-wpengine.netdna-ssl.com/wp-content/uploads/2014/02/SureSineENG_R2_1_08.pdf

I use 85% efficiency to allow for less than optimum inverter usage. However, if you operate "outside" an assumed standard loading... Then the 85% is not accurate.

One of the nice things about this MorningStar inverter, is that it has search mode (AC load less than ~8 Watts, the inverter goes into Search mode that consumes ~55 milliamps:

• 0.055 Amps * 12 volts = 0.66 Watts
• 0.66 Watts * 20 Hours per day = 13.2 WH per day (20 hours in standby)
• 500 WH load / (588 WH loaded + 13.2 WH standby) = 0.83 ~ 83% Efficiency

Details do matter--And so does conservation and picking the right loads and equipment.

-Bill

mountainman

This a 24 hr load. unless we are away. 
Then I Turn it off.   A 20 inch box fan a led tv.  and 2 phone chargers. The fan runs 12 hrs a night.  and the tv all day. 
 ~50 watt  constant load. 
I did  a couple weeks test to get an average daily use. 
As an experiment couldn't I hook a dc watt meter from the battery to the inverter.
 And compare dc wh  to ac wh for  more accurate results. and see  the true conversion efficiency for my system?

mountainman

Thanks for the link when my 3 year old....   dreaded word on this forum.  aims  converter that specs 95℅ 1/3 load 90℅ full load .7 amp idle goes out. I'll get me one.

Estragon

If it works for you, that's great!

BB.

Yes, with an AC and DC side Watt*hour meters, you will get more accurate numbers...

However, the power numbers will probably be something like 5% 10% level of accuracy between all of the different measurement devices. And add the issues of waveforms and phase angle measurements---Does not help with accuracy.

Looking at the math and your pattern of energy usage--At least you can figure out what is important to you, and how much you may have to compensate for things like 24x7 AC power (vs size of battery bank, solar array, etc.).

-Bill

Horsefly

Our inverter is the Schneider SW4024. I may have been overly conservative, but in my design I used 85% efficiency for the inverter (which contributed to an overall efficiency of the system of 50%), but I also included 26W continuous tare loss in the energy budget. 26W x 24 hrs/day added 624 watt hours to the overall daily consumption.

mcgivor

Horsefly said:

Our inverter is the Schneider SW4024. I may have been overly conservative, but in my design I used 85% efficiency for the inverter (which contributed to an overall efficiency of the system of 50%), but I also included 26W continuous tare loss in the energy budget. 26W x 24 hrs/day added 624 watt hours to the overall daily consumption.

The SW 2524-230 idle loss is 38W, the SW 4024-230, according to my manual is 40W, not sure if the north American version is any more efficient but that would be 960Wh. Sure adds up but it's the cost of doing business, so to speak.

Horsefly

mcgivor said:

Horsefly said:

Our inverter is the Schneider SW4024. I may have been overly conservative, but in my design I used 85% efficiency for the inverter (which contributed to an overall efficiency of the system of 50%), but I also included 26W continuous tare loss in the energy budget. 26W x 24 hrs/day added 624 watt hours to the overall daily consumption.

The SW 2524-230 idle loss is 38W, the SW 4024-230, according to my manual is 40W, not sure if the north American version is any more efficient but that would be 960Wh. Sure adds up but it's the cost of doing business, so to speak.

That's interesting. I had to go back and look at my manual. For the SW 4024 120/240, the "No-load power draw" is shown as 26W. For the 2524 120/240 it's 21W, and for the 4048 120/240 it's 27W. Not sure why it would be so much more for the 230V version. 

BB.

Things like higher battery bus voltage causes more current draw because of the capacitance of the FETs. Also, possibly because of 50 vs 60 Hz switching (higher frequency (again, more switching across capacitance of FETs, etc.).

-Bill

mountainman

Been looking at the suresine efficiency charts.
Seems at a 50-100 watt load efficiency is slightly higher.  Then at 50℅ it declined. While looking at other  inverters after 50℅ load they also declined 
 So for smallish loads seems it would be more efficient to size your inverter 2-3x  the  size of your max load?

BB.

For larger systems (mutli kWatt off grid systems), the lesser efficiency of smaller loads is a "don't care".

However, this always assumes a "balanced" system design. (more or less, C/8 to C/20 discharge rates for battery bank, C/8 to C/10 solar charging, etc.).

When you have something "different" such high surge loads for short periods of time on a small battery bank (AGM or Li Ion for high surge) and relatively small solar array, then the large AC inverter with (on average) small AC loads, then the Tare losses of the inverter do matter a lot and running the inverter 24x7 becomes a very significant load (sometimes on the order of 1/3 to 1/2 of the entire "solar" load).

For small systems, you have to look closely at all loads (including "vampire loads" that are on all the time, but not in use all the time--Such as wall warts, AC inverters with no loads at times, etc.).

On a 500 WH per day system, a 6 Watt AC inverter (like the MorningStar) on 24x7:

• 6 Watts * 24 hours per day = 144 Watt*Hours per day
• 144 WH / 500 WH = 29% of "system design" is used just to run a 6 Watt Tare load on an inverter 24x7

If you can use the remote on/off or search mode with the MorningStar (8+ Watts of 120 VAC load, inverter "turns on"), 

• 0.025 Amps * 12 volts = 0.3 Watts ("remote off")
• 0.055 Amps * 12 volts = 0.66 Watts ("search mode standby")
• 0.66 Watts * 24 hours per day = 15.84 WH per day (search mode)
• 15.84 WH per day / 500 WH per day system design = 3% of system capacity using "search mode"

Which is why I really like to suggest the MorningStar 300 Watt TSW 12 VDC as there were no smaller inverters with remote/search on-off capability.

https://www.solar-electric.com/morningstar-si-300-115v-ul-inverter.html

Now, some of the Cotek (and similar?) smaller inverters come with search/remote controls (and some even have remote panels):

https://www.solar-electric.com/cotek-se200-112-pure-sine-wave-inverter.html

The right tool for the right job...

-Bill

mountainman

The stand by function interest me.
my radio only pulls 10 watts I wonder how accurate that 8 watt search is. 
Would it sense 10 watts?
 Everything  that runs on ac is 150 watts on the  kil o watt meter. 
 So Im thinking the 300 would be a better match than the 200 cotek?
 With a 5-600 watt p/day design such as mine. Every little bit counts
 With just the inverter adding up to  ~35℅ of my useage.  Seems Im forever turning the inverter off and on trying to conserve.

Would be convenient to  just leave it on.

BB.

The MorningStar does not have any fans... I like that.

Would I bet the farm on 10 Watts vs 8 Watts "searchmode on"? Probably not... However, you could run the remote on off around and place switches near the major loads (radio, etc.) and simply turn on when needed.

You could probably do something "interesting" like a motion detector that "turns on" the inverter when you are around.

Or a simple timer that turns on the inverter when you are around the house (say 5pm to 10pm) and let the search mode manage for other times (larger loads, etc.).

And there is the old twist timer--Twist on for a period of time (say one or two hours) to charge the cell phone/tablet computer/lower power LED light and then shuts off automatically so you don't have to remember (presumably search mode will keep any loads >~8 watts on until they are shut off).

-Bill

Estragon

My MS 300w inverter seems to sense the 8w or whatever pretty well.  Turning on an LED bulb (~10w IIRC) in the bathroom wakes it up.

mountainman

Thanks that's good to know. I see it accepts 2awg .  As far as voltage drop should I shoot for something like  .05 with 6 foot of 2 awg?    Or .12 with 6awg? 

BB.

Generally, a maximum of 0.5 volt drop for 12 volt wiring--I would suggest. Less drop certainly is less losses. Probably a 1% loss (~0.12 volt drop) is the maximum that it is "worth" cabling for:

• 0.5 volts * 50 Amps DC (~600 Watts) = 25 Watt cable loss.

You can use a voltage drop calculator to figure out one way cable run (for this calculator, others use round trip run length) and figure out your cost effective solution:

https://www.calculator.net/voltage-drop-calculator.html

-Bill