Power Inverter Question

mathew

Planing to run a ~400W 220V device from a 12V battery with a 500W inverter.
Will the current draw be 2amps or 30amps from the battery?
Or will it draw the full 500/voltage amps??
The battery I have is 100Ah.. so naturaly the 2amps would be favourable for 8 hour operation.
Any help would be great!

DublinDilbert

If you assume the inverter is 100% efficient then;
Pin = Pout
Vin x Iin = Pout
12 x Iin = 400W
Iin = 400/12 = 33A.

But the converter will only be about 85% efficient at best, meaning for 400W out, you'll need to push in about 470W, ->Iin = 39A @ 12 V


Is it a deep cycle 100Ah battery?

Also some mains equipment take in large start up currents, so be careful not to fry the inverter.

mathew

ah.. That creates some problems. There is no way Ill be able to do that so..
Never though of Pin = Pout.. makes perfect sense!!
Ill have to see what I can sort out. It is a deep cycle battery but at 33amps it will never last the 8 hours. Trying to charge it with a 3amp charger will also not be possible in the 10hours I had planned...
Ill have to see if I can find something else. A 12V version of the device might be available.. Ill have to see.

Thanks a million for your help tho!

Red Alert

You will also separately want to see whether your device needs a "pure sinewave" inverter or can make do with the cheaper "modified sine wave" models.

mathew

its gonna need a pure sine wave I thinks.. its a compressor.. so it needs it for the motor I believe...

Looking at getting a 12V DC one instead. It provides 60psi which I think I can cope with. Draws a rich 7amps tho.. so i might have to keep looking... Its probably going to be difficult to find.
Basically I need a compressor that is going to power a piston on and off for 8 hours at night. Then I will use solar during to day to recharge it...
Ambitious I know... but it should be possible

Infact.. Im assuming the same Pin = Pout for charging the battery actually. If I draw 7 amps for 8 hours from a 12 volt battery will I have to replace 720Watts?? As in can I use a 20V 2amp solar panel for 18 hours to recharge it? (I know there isnt 18 hours of sunlight but its the principle Im getting at here)
My formula from the battery manual I was looking it is (Depth of Discharge)Ah / (charger current)A + 2hours = time
That is assuming a 12V charger I believe tho.
I know I will need a charge regulator and all that.. still in the back of the envelope stage here tho.. trying to see if its feasible.

mawk

mathew wrote: »

its gonna need a pure sine wave I thinks.. its a compressor.. so it needs it for the motor I believe...

Looking at getting a 12V DC one instead. It provides 60psi which I think I can cope with. Draws a rich 7amps tho.. so i might have to keep looking... Its probably going to be difficult to find.
Basically I need a compressor that is going to power a piston on and off for 8 hours at night. Then I will use solar during to day to recharge it...
Ambitious I know... but it should be possible

Infact.. Im assuming the same Pin = Pout for charging the battery actually. If I draw 7 amps for 8 hours from a 12 volt battery will I have to replace 720Watts?? As in can I use a 20V 2amp solar panel for 18 hours to recharge it? (I know there isnt 18 hours of sunlight but its the principle Im getting at here)
My formula from the battery manual I was looking it is (Depth of Discharge)Ah / (charger current)A + 2hours = time
That is assuming a 12V charger I believe tho.
I know I will need a charge regulator and all that.. still in the back of the envelope stage here tho.. trying to see if its feasible.

Ill just wade in here and tell you that you wont get 40w from a 20v 2a pv cell. not even nearly. most certainly not in kill o the grange. and super most certainly not for 18 hours a day.

but you might be lucky in that the comperssor you want to run will not have a 100% duty cycle. itl run till the reciever is full and then kick back on for a while when the cylinder has used enough air. depending on cylinder timing and size the duty cycle might be quite low. maybe within the 3 hour capacity of the battery at 33a
what do the cylinders need to drive?


--
edit

is there a reason to not use a petrol engine as opposed to an electric motor? noise? environmental qualms?

backboiler

mathew wrote: »

If I draw 7 amps for 8 hours from a 12 volt battery will I have to replace 720Watts?? As in can I use a 20V 2amp solar panel for 18 hours to recharge it? (I know there isnt 18 hours of sunlight but its the principle Im getting at here)

7 A at 12 V is 84 W (7 * 12).
84 W for 8 hours is about 2.4 MJ (84 * 8 * 60 * 60)
So you have to get a minimum of 2.4 MJ back into the battery.
I seem to remember that the Sun delivers around 1.2 kW/m^2 to the Eath's surface directly underneath it. At out latitude it'll be half that at best, 600 W/m^2. Take your 18-hour day with no cloud cover from sunrise to sunset, that might average out at 400 W/m^2. If we say that a typical PV cell is 12.5% efficient, then you can expect 50 W/m^2.
So, 2,400,000 / 50 = 48,000 s m^2 (that's seconds square metres), meaning that with a square metre of solar panel, you need over 13 hours of direct sunlight to charge your battery. Double the area, half the time, etc.
Now all that is best case agus gan scamall ar bith sa spéir.