is my connection correct ??

[zolar]

yes brother, I know the formula that's why I gave those values,
but generally as I understand now , you mean ( in solar system ) what ever the load is , so there will be 2 values of amps flow ( 1 before the inverter and 1 after the inverter ) and that value stay there as long as the load working.
for example as in my load 400w ; while the load is working so there is a 16.6amps withdrawn from battery due to 24v and its existing between the battery and inverter even though the current will be less ( 1.8 ) after the inverter since the voltage became higher ( 220v )

thanks a lot brother for explanation , and sorry took some of your valuable time , but really I learned a lot from you and others in this helpful nice site.

[SunEagle]

sorry I think I didn't askclearly or I misunderstand ,
if we forget about inverter size ( I mean it could be 2kw less or more ) and suppose I am sure that daily I will go only for that load ( 400w )
when I said 16.6amps and 1.8amps ( cuz I divide 400w /24v between battery and inverter so it will be 16.6 amps ) and after inverter ( I divide 400w / 220v so it will be 1.8 amps ) that's what I am asking about < where is extra amps went after it comes out from battery since I will use only 1.8 amps to generate the load.

sorry brother for not obvious question.

If your load is 400watts at 220v then it will draw 1.8 amps on the wires between the inverter and load. But that same load of 400watts goes through the inverter at 24V (battery voltage) which will draw 16.6 amps on the wires between the battery and inverter.

Because the voltage increases but the wattage stays the same the amps will drop. The formula is Watts = Amps x Voltage. It is just the physics of electricity.

[zolar]

sorry I think I didn't askclearly or I misunderstand ,
if we forget about inverter size ( I mean it could be 2kw less or more ) and suppose I am sure that daily I will go only for that load ( 400w )
when I said 16.6amps and 1.8amps ( cuz I divide 400w /24v between battery and inverter so it will be 16.6 amps ) and after inverter ( I divide 400w / 220v so it will be 1.8 amps ) that's what I am asking about < where is extra amps went after it comes out from battery since I will use only 1.8 amps to generate the load.

sorry brother for not obvious question.

[SunEagle]

thanks a lot gbynum for the link .

thanks a lot suneagle , really nice to hear that info.

so with the load of 400w after the inverter , a cable of 4mm squar more than enough since only 1.8amps will flow between inverter and load,,,, but
where is the rest of amps gone since it will flow 16.6amps between batteries and inverter ? is it lost ?

No it is not lost.

The additional 16.6amps is the possible extra amount that could go through the wires if the load on the inverter is increased to 2kw.

When it comes to electrical circuit wiring you need to size everything at the highest load which makes it safe if you ever need that many watts. If you only sized the DC wires for a couple of amps and for some reason your load increases then you run the possibility of burning up the wires or blowing a fuse. Neither is desirable.

[zolar]

thanks a lot gbynum for the link .

thanks a lot suneagle , really nice to hear that info.

so with the load of 400w after the inverter , a cable of 4mm squar more than enough since only 1.8amps will flow between inverter and load,,,, but
where is the rest of amps gone since it will flow 16.6amps between batteries and inverter ? is it lost ?

[gbynum]

and btw how you are sure of comparing AWG with mm2 ?

you can find a formula (can load into a spreadsheet or use a scientific calculator) and tables here, also.

https://www.rapidtables.com/calc/wire/awg-to-mm.html

[SunEagle]

suneagle,,,,, 1 more question, please
suppose the load is 2kw so the amps going will be 83.3 right?
so that amps in wich cable ?
the cable between batteries and inverter ? or also in cable between inverter and load ?
cuz after inverter the voltage will be 220v , so 2000 devide voltage will be less current ( 2000 / 220 = 9 ).
I hope you understand what I mean

83.3 amps would be in the cables between the 24VDC battery and inverter. If the load was 2kw then the amps in the cables between the inverter and load would be about 9 amps since the voltage is at 220VAC.

[zolar]

suneagle,,,,, 1 more question, please
suppose the load is 2kw so the amps going will be 83.3 right?
so that amps in wich cable ?
the cable between batteries and inverter ? or also in cable between inverter and load ?
cuz after inverter the voltage will be 220v , so 2000 devide voltage will be less current ( 2000 / 220 = 9 ).
I hope you understand what I mean

[zolar]

thanks mike for the link , and thanks suneagle for the info.
and thastinger , I found the inverter manual and I read it carefully and completely but didn't mention anything about grounding not even in safety points
I wonder if what I did by joining the negative ( - ) terminal of the battery to the main house grounding is a correct thing to prevent back voltage from inverter to battery.

[SunEagle]

oh !!!!
but the load only 400W?

It really doesn't matter what the load is to size the wire. It depends on what the inverter can draw if the load is increased by design or by accident. 2000 watts at 24V calculates to 83.3 amps so the wire size should be the next size up which is 100amps. You can always use a smaller wire but you will need to reduce the fuse size so it protects the wire from too many amps.

I have seen people use a smaller wire but after replacing the fuse a couple of times they went to a large wire and corresponding fuse.

[Mike90250]

here's a table

https://www.engineeringtoolbox.com/awg-wire-gauge-d_731.html

[zolar]

and btw how you are sure of comparing AWG with mm2 ?
cuz I said that just with dealing with some cables from both mesurments,
if there is any schedule I can refer to please.
or if you can compare these AWG ( 2 ,4,6,8,10,12,14,16,18) with mm2 please.
thanks

[zolar]

oh !!!!
but the load only 400W?

[SunEagle]

thanks a lot suneagle ,,
the inverter is a 2kw pure sine wave , and and the I am using only neutral and hot conductors ( no ground ), and yes I would like to know tha size with mm2 cuz that's what we are using here, but I can manage comparing with AWG. ( I think 12AWG = 4mm2 ),
thanks again brother

You are correct. 12awg = 4mm2 wire size and a 14awg = 2.5mm2 wire.

Now with a 2000 watt 24V inverter you will need wire that can handle about 100 amps. That is probably a 2AWG copper wire or 35mm2.

[zolar]

thanks a lot suneagle ,,
the inverter is a 2kw pure sine wave , and and the I am using only neutral and hot conductors ( no ground ), and yes I would like to know tha size with mm2 cuz that's what we are using here, but I can manage comparing with AWG. ( I think 12AWG = 4mm2 ),
thanks again brother