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This means that the battery should handle the current:
5000 / 12 = 416 A (assuming 100% efficiency)
I hope you get the point.
Practically speaking, in this case, the minimum DC voltage of the batteries could be 4*12V
The capacity of each of the 4 batteries is about 200Ah though it depends on the application.
I think you might be confusing several parameters associated with lead acid batteries. Besides as KerimF imples it is not only the current supplied, but also for how long - to which the parameter is labelled as AH
If you use Google you can find many inverter circuits, but - again - this is NOT a project for the novice.
When you find that super strong inverter also find few tons of batteries to power that on 12V.
I'm pretty sure that you can find to buy original "Dry patch on the bottom of the sea" in China.
Dont beleive in all product marks and product names, some inverter are called 3000W and this is not 3KW of power its just name 3000W. Check specification.
Dont beleive in all product marks and product names, some inverter are called 3000W and this is not 3KW of power its just name 3000W. Check specification.
There is nothing about Ohms Law that prohibits the power as you state.
The specifications are clear enough. Blanket condemnation does not enhance your position. You might actually try investigating instead of just stereotyping or blasting others who disagree.
There is nothing about Ohms Law that prohibits the power as you state.
The specifications are clear enough. Blanket condemnation does not enhance your position. You might actually try investigating instead of just stereotyping or blasting others who disagree.
Simply if we use low voltage and high power, as consequence we will have demand of high current, which require good power source and strong wires/stripes.
At high current power source (batteries) suffer trying to supply high current and as we will have voltage drop, this voltage drop will cause more current demand. Batteries under heavy load have voltage drop, to avoid this and to remain battery voltage at high current, we need strong battery bank with lots of batteries, in this case in parallel because we use 12V system, or we can to use 6V or single cell traction batteries in serial/parallel connections.
Because of that usually inverter systems for higher powers use higher voltage, to decrease currents in wires. Additional high DC current is very dangerous can cause electrical flux or overheating some segment and cause fire.
There is no blasting of others.
Ohm Law among other can give you elementar sense for electrical units, to fill how much is 1A, 100A, 12V, 500V,...
On Alibaba you can find 10KW, 20KW @12V is this reasonable ?
Just see what connectors are used and distance between them on inverters from your links. These inverters looks like for power around 300W and with higher power peaks. On one picture they use car battery!.
Additional to all of this, usually deep cycle lead acid should be used in inverter/solar systems. Deep cycle have good possibility to cycle down to 10,5V which starter type cant (can but will be damaged or completely dead), plus deep cycle dont love high starting current or high current at all, this can cause curving of plates. Because of that you need enough number of batteries in battery bank to support needed current for system usage, also which batteries can handle.
In other hand starter batteries (float) are bad solution for this, they handle high starting current but they are not designed for complete discharging specially below 11,9V and also they will suffer under longer high currents load, this can cause curving of lead plates.
Eh, it could be done if you built a good laminated bus for the low voltage DC which had the first DC-DC converter mounted directly on it. The more tricky issue would be to put a bunch of 12V batteries in parallel without having issues with charge balancing.
@kam1787: it seems to me that YOUR arguments are pointless!!!
Please make some calculation on wire size/resistance needed for a 5kVA/12V load. This excluding the efficiency of your inverter and the losses in the conversion.
Just for some back of envelope calculations, your MINIMUM current requirements are of 450A @12V (not counting inverter power losses).
What will be your required wire section?
What will be your voltage drop for 1m of cable? Please take care that being your batteries very big, they should be mounted at some distance from your converter, so we can assume your cable will be at least 2m long.
If you make some calculations you will discover that the voltage drop at such currents will be so "high" that at the input of your inverter you will not get 12V but much less, so your calculations must be changed for a much higher current!
Just as an example take a look at table V of this document:
**broken link removed**
Let's assume you are using a 300mm2 size cable with a length of 2 meters: your DC voltage drop [Flat or vertical (2 cables, single-phase a.c., or d.c., or 3 or 4 cables three-phase)]will be of 0.23V/m.
So you will get a voltage drop of at least 0.5V at the input of your inverter!
Now we have to talk about the losses of your inverter: it is reasonable that at those low voltages the efficiency of your inverter will be no more than 75/80% (I'm VERY generous!) due to internal losses (mosfet RDSon, losses in the transformer, PWM losses to get "pure" sinus wave, ...), so to get real 5000W you really need to juice your inverter with at least 6500W!!!
Now you'll discover that your cable require to pass at least 560A: 6500W/11.5V (remember of the cable voltage drop!!!)!!! So your cable section must be of at least 380mm2.
This section is almost equivalent of a 22mm cable conductor diameter... Take a look at the input connectors of your "fabulous" inverter: are you sure they are enough?
Next time, instead of blindly believing in what you find written on the web, try to make some small calculations... You'll discover that most of what is written is only marketing hype and not technical sound!
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