Efficiency versus load for Conext SW 4024

Ram

Greetings - I've benefited hugely by browsing around on this forum, and decided to join today to learn and hopefully contribute as well!
Briefly, I have a residential solar setup here in Bangalore, India, consisting of 6xSolarWorld 270W mono panels, and 4x 200AH flooded batteries. For about a year I have been trying to finalize my inverter purchase but after blowing up a few, pulling out my hair over the poor quality of a few others (including Phocos) and trying to pool some money to help a local manufacturer improve his design, I am back to square one. Considering ending my misery by buying a Conext SW which is sold and supported in India. (Waiting for the 4048 but I will pose the question for the 4024.)
In this post I have a specific question about efficiency when operated in pure off-grid mode. The datasheet of course gives the peak efficiency of 92%, but I am trying to get some feel of the efficiency at a few specific intermediate and [very] low loads. The inverter will run for the most part at about 500-700W, and will run at ~2.7kW only when I run my dishwasher. (I felt it would be simpler to have one inverter sized for the peak load though that peak only occurs for a couple of hours a day.) Therefore:
Q1: As I ramp up the load from zero, efficiency would improve till it hit the sweet spot, after which it would slightly decline. At what load would I expect to see an efficiency of greater than 85%?
Q2: At what load does the Conext SW4024 exhibit its stated peak efficiency?
Anyone who is able to share some measurements from personal experience?
Thanks! - Ram

BB.

According to page 109 of the manual:

http://www.solar-electric.com/lib/wi...120-manual.pdf

• 40 Watts Tare losses for 4024 and 27 Watts for 4048 [No-load power draw (Inverter On)]

If you assume that all losses are Tare losses (not true--But at least it gives you an approximate answer):

• 40 Watts / 0.15 losses = 267 Watts total inverter operational power
• 267 Watts * 0.85 to load eff = 227 Watt load for estimated 85% efficiency (based on Tare losses)

If you want to model the efficiency without more data... Assume that the peak efficiency is around 90% at 4,000 Watts (that is for the 24 volt inverter model), that means the losses are:

• 4,000 Watts * 0.10 losses = 400 Watts
• 400 Watts - 40 Watts Tare = 360 Watts "non Tare loses".

Assume those loses are I²R loses, then the formula for losses may look like:

• Max AC current = 4,000 Watts / 240 VAC = 16.7 Amps
• Losses = 40 Watts Tare + (XYZ Watts Load*1/240 VAC*1/16.8 amps AC max current)² * 360 Watts

Then 267 Watt load losses estimated losses would be:

• 40 Watts Tare + [(227 Watts * 1/240 VAC ) * 1/16.8 Amps]² * 360 Watts = 40 Watts + 1.6 Watts = 41.1 Watts total losses
• 227 Watt load / (227 Watts + 41.6 Watts estimated losses) = 84.7% efficiency (estimated when I²R losses are taken into account)

The above is a bunch of guesswork from a less than ideal set of date from the Mfg.... But it is probably close enough to answer your questions (for example, we do not know the efficiency at 100% load, only at some unknown "optimum" efficiency load).

If I am correct that the losses are mostly I²R for operational losses (vs fixed Tare losses), then you can see that the efficiency will fall again as you get higher current levels (the squared current function).

Also, i would be using VA or Amps for the I²R losses rather than the load's Watt usage (i.e., if you have poor power factor, the losses will be higher because of the higher current for power power factor/inductive loads).

-Bill

PS: Here is a company that graphs their inverter output efficiency:

http://www.morningstarcorp.com/wp-co...NG_R2_1_08.pdf

You can see that for them, at rated output (300 amps continuous), their efficiency is ~85%. At at 2x rated load (surge current) falls to ~70% efficiency.

If you wanted to be "conservative", perhaps use 85% efficiency.

Also, I see from the MorningStar graph that their out efficiency seems to be pretty linear--Perhaps remove the "square" function from my equation--It would match better:

-BB

zoneblue

Bill, your approach there is consistant with the analysis that poster "northguy" here did a couple years back. After in depth monitoring (true rms datalogging) he concluded that the gross conversion efficiency was basically flat, and the fall off in net efficiency at low power levels was primarily a result of the tare. His inverter was an XW, and as i recall the actual peak was 89%.

If you were to graph this youd get, theoretically this. Attachment not found.

Typically you get an additional but smaller fall off at very high loads. Presumably thats a product of heating and or magnetic saturation.

zoneblue

OP is right to be particular about inverter selection. We see far too many people choosing inverters with no thought to their impact on system resources. Heres an example of a guy with only three panels and an SW.

http://www.altestore.com/blog/2014/1...-as-you-think/

His loads were only 200W, and his 4000W invertor is known to have the highest tare in the industry.
And he wondered why it didnt work. The inverter tare was basically eating everything.

Here, since installing the VFX, which is second best in class in terms of tare, it still feels like another mouth to feed compared to the Steca 1100W unit that preceded it.

Ram

BB, zoneblue, thanks for your detailed responses! Another useful article is here:
http://www.homepower.com/article/?file=HP113_pg36_Perez
which is almost flat but for a slight taper which causes the author to postulate that the peak actually occurs at about 20-30% of peak load rating.
This helps me sharpen the question to the following two subquestions:
(i) How significant are "iron" (core/eddy/hysteresis losses)?
(ii) More important - are there any components that may "turn on" when the inverter goes from a no-load state to supplying a load, causing the quantum of fixed overheads to be higher than estimated above? I have seen this happen in an inverter I worked with. Then for example, the analysis would change slightly to:

• 4,000 Watts * 0.10 losses = 400 Watts
• 400 Watts - 40 Watts Tare - X watts other turn-on overleads = 360 - X Watts "load-dependent losses".

The implication would then be that the I2R losses are lower than initially estimated, but at lower power the efficiency drops because of the increased overheads, and the 85% point would be hit at a higher load level.
Would be interesting to measure inverter current draw at any one very low load to see if there are any such overheads that kick in from a no-load to low-load condition!
Best - Ram

Ram

There is one more interesting factor we should look at for the SW, and this could be linked to the point I made about additional overheads kicking in with a load. The Addendum for the SW released Dec 2014 says that whether you set Load Shave to zero or not, whether you engage AC Support or not, upto 2 amps can be drawn from grid to prevent the inverter from injecting current into the grid. This changes what is stated in the user manual where the figures (and text) state zero draw from the grid in these two modes. 2 amps is of course a lot of current and can seriously jeopardize efficiency, if as someone in these forums pointed out (wrt the XW), any of that grid draw just gets dissipated away in monitoring the grid status rather than in supplying the load. I suppose I could just turn off the AC so it operates pure off-grid though. (Someone at Schneider India advises me that the XW can work with zero grid draw but the SW "cannot" and I am researching the truth of that before I buy the SW, but that is probably off-topic for this specific thread.)

BB.

A lot of those questions are "answered" by the design engineers to meet specifications/cost points. Best you can do is look at the various products and see which will meet your needs/cost point.

And, people design for particular environments. Some will derate over ~25C (room temperature)--and some will not derate until the room gets much hotter.

There are even some devices that have parallel switching devices... Turn on one section at low output power and turn on all three at high output power (saves switching losses when powering low power loads).

In other cases, you may want a small/lower power inverters for small loads that run 24x7... And a larger inverter to run the well pump an hour per day.

Unless you are trying to design an inverter--You sort of have to look at the specifications and your (and your neighbor's) experiences--And go from there.

Another engineering saying... Price, Function, Reliability--Pick any two to optimize.

-Bill

LorenAmelang

The Conext SW4024 manual linked above (975-0638-01-01 Rev D 2-2015) is the first I've seen with the "40 W" spec:
DC Input            - SW 2524 120/240 - SW 4024 120/240 - SW 4048 120/240
No-load power draw (Inverter On) - 38 W - 40 W - 27 W

The currently posted manual (975-0638-01-01 Rev E 7-2015) shows:
No-load power draw (Inverter On) - 21 W - 26 W - 27 W

Which is much more in line with the competition. Though still high compared to my old Trace SW4024 at 16 Watts... 

Does anyone know what they changed? Hardware? Firmware? Or just the spec? Has anyone with a pre-July 2015 inverter done a firmware update and noticed the lower idle consumption?

The Conext SW4024 has moved to the top of my list of possible replacements for my suddenly fried Trace. I doubt it will start the 1 Hp Code K table saw the Trace could start, but at least it can do generator support. Hopefully it will start the 3/4 Hp water pump...  But most of the actual power goes into the ~75 Watt ozone water treatment system, with its nasty power factor. The 105 pound trace was good about reflecting the apparent power back. I fear no modern inverter will be as efficient.

Dave Angelini

Starting those loads you mention is more a function of having a large 1000+ AH battery and 4/0 cables in good condition.
The firmware has changed a few times and this inverter is being used in grid self consumption uses quite a bit world wide.