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impi · 26-09-2010, 10:03

Can anyone assist my 'failing brain' how the calculation is done for total amps used when an Inverter is between the battery bank and the item used?
Example: I install a 3000W victron invertor to invert my 12V battery bank to 220 Volts output. I then run a hair dryer (1800 Watt) from my 220 Volt socket in the cabin. How many amps have I taken out of my battery bank after 5 mins:
a) at the inverter
b) at the hair dryer
c) total

**Paul Elliott** · 26-09-2010, 10:23

It's pretty easy to get a ballpark number. It all boils down to one of the variations of Ohm's Law: Watts = Amps x Volts, or P = I x E (I = Amps, P = Power in Watts, E = Volts). This can be rearranged as I = P/E

So, your 1800W hair drier draws 8.18A from the 220V outlet (1800 / 220 = 8.1818...)

If your inverter were 100% efficient (it isn't), the 1800W load would draw 150A from the 12V battery bank (1800 / 12 = 150).

Assume that the inverter is 85% efficient. This is a guess, but probably not far off. The current drawn from the battery will be more, about 176A (150A / 0.85 = 176.47).

This is a lot of current, but since a hair drier isn't used for very long it isn't as bad as it might seem. Assume it takes five minutes to dry the hair (as you asked). That's 1/12th of an hour. The total drain on the battery bank will be 176A / 12 = 14.66... Amp-Hours. It all adds up, but you can probably afford it, especially if it makes someone happy.

MarkJ · 26-09-2010, 10:24

By a brief calculation: drying your wifes hair for 5 minutes will suck your battery dead as a dodo.

Mark

anjou · 26-09-2010, 10:30

Quote:

Originally Posted by Paul Elliott

It all adds up, but you can probably afford it, especially if it makes someone happy.

awww, thats a lovely thing to say.
Im half pissed and feeling rather in tune.

impi · 26-09-2010, 10:51

Hi Paul (Paul Elliott)
Am I understanding this correctly then? We do not consider the amps drawn between the Inverter and the battery bank (except for it's inefficiency) - we simply divide the watts of the appliance by the Invertor voltage and apply an 'inverter' efficiency factor!
Also, could you advise - Does this mean that an inverter 'rests' in neutral mode (NO AMPS drawn from battery bank) when no appliance is plugged in, or does the inverter 'steal' energy by being there?

**David M** · 26-09-2010, 11:13

Inverters will draw a small amount of energy when there is no load on them mostly for running the inverters internal circuitry and it's meters. Smart inverters will provide no AC when it is not needed. You can always choose to manually completely shut it down when you will not be needing AC anytime soon.

**Paul Elliott** · 26-09-2010, 11:19

impi (by the way, are you Finnish?),

Yes, you can calculate using just the load (appliance) power and the battery voltage, and factor in the inverter efficiency. It's really that simple, at least for a rough number.

An inverter will draw some idle current when left on without a load. My inverter goes into a standby mode, where it powers-up the AC outlets briefly to see if anything is plugged in and turned on. If there is a load, the inverter remains on, otherwise it shuts down for a second or two, then tries again. The battery drain is pretty low in this mode, but I haven't measured it.

Other inverters will operate continuously, even if nothing is plugged in. There will be a moderately low "standby" drain on the batteries. Also, an inverter usually has the best efficiency at or near it's rated load. For example, running a cellphone charger on a 2000W inverter is *not* efficient. On the other hand, the overall drain may still be acceptable.

impi · 26-09-2010, 11:22

Fantastic !!! Thanks everyone -This post has really helped me understand the process.
On this topic of INVERTERS, can anyone tell me which is the best SMART INVERTER I could go for?
Are some more efficient than others, and does one get a meter that shows 'inverter efficiency' ?

impi · 26-09-2010, 11:26

No Paul - actually South African - but apparantly roots from Sweden

**Paul Elliott** · 26-09-2010, 11:27

Quote:

Originally Posted by David M

Inverters will draw a small amount of energy when there is no load on them mostly for running the inverters internal circuitry and it's meters. Smart inverters will provide no AC when it is not needed. You can always choose to manually completely shut it down when you will not be needing AC anytime soon.

Exactly. On my boat, when on a passage and I'm watching the amps, we have the occasional "charging party", when I turn on the big inverter and the crew all plug in the chargers for their iPod, cellphone, cameras, my battery-powered drill, etc.

I also carry a much smaller inverter for occasional use, when the big one isn't needed. And, I try to use "native" 12V chargers and small gear. I'm a bit obsessive about these things.

But, as anjou noted, I do try to look at the big picture! Keeping my wife (or crew) happy is a good thing, and if I have to engine-charge an extra ten minutes that's a small price to pay.

**Paul Elliott** · 26-09-2010, 11:30

Quote:

Originally Posted by impi

No Paul - actually South African - but apparantly roots from Sweden

Oh, I was wondering because we have many Finnish relatives and our dog is named "Impi". It means "maiden" in Finnish, I am told.

impi · 26-09-2010, 11:36

OOOOOOOHHHHHH BOY !!!!!

In South Africa Impi is the word used by the Zulu Nation - The word means Warrior.

I was surprised when I purchased a Mitsubishi Pajero - the word pajero means 'little wanker' in some language which of course made me the laughing stock of many of my friends

**noelex 77** · 26-09-2010, 11:36

Quote:

Originally Posted by Paul Elliott

It's pretty easy to get a ballpark number. It all boils down to one of the variations of Ohm's Law: Watts = Amps x Volts, or P = I x E (I = Amps, P = Power in Watts, E = Volts). This can be rearranged as I = P/E

So, your 1800W hair drier draws 8.18A from the 220V outlet (1800 / 220 = 8.1818...)

If your inverter were 100% efficient (it isn't), the 1800W load would draw 150A from the 12V battery bank (1800 / 12 = 150).

Assume that the inverter is 85% efficient. This is a guess, but probably not far off. The current drawn from the battery will be more, about 176A (150A / 0.85 = 176.47).

This is a lot of current, but since a hair drier isn't used for very long it isn't as bad as it might seem. Assume it takes five minutes to dry the hair (as you asked). That's 1/12th of an hour. The total drain on the battery bank will be 176A / 12 = 14.66... Amp-Hours. It all adds up, but you can probably afford it, especially if it makes someone happy.

The current figure is accurate and the AHr figure is accurate, but not really.
Peukert's Formula shows when you discharge a lead acid battery at these sort of amps its capacity plummets.
The bottom line if don’t want to bother with the math’s is that a hair dryer will draw from a battery much more the just Ohms law would suggest.
If the battery bank is not in good condition it may not cope at all.

Short sun dried hair is good.