10.4 amp air compressor

brewer · August 2019

I'm a beginner on learning how much draw certain products use. Just got a 10.4 amp air compressor that I want to use on my simple 12 6v battery, 24 volt system I'll only use it when it's sunny.

Is this a reasonably low draw compressor?

Thanks.

mcgivor · August 2019

What voltage is the compressor, what capacity are the batteries? These details are nessersary to begin a discussion.

brewer · August 2019

120V How do I figure battery capacity? I have 12 6V batteries. 4 rows of 4 in parallel.

Thanks

icarus · August 2019

Remember starting current for motor loads is 10-100 times higher than running current, so the battery and the inverter need to be able to handle that starting current. Compressors are often even harder to start, if they have to start against significant head pressure. Tony

mike95490 · August 2019

brewer said:

120V How do I figure battery capacity? I have 12 6V batteries. 4 rows of 4 in parallel.

Thanks

Ouch. Massive parallel (more than 2 banks) batteries.

for 12 batteries, you have 2 in series, and then 6 parallel sets.

read up about about parallel issues here http://www.smartgauge.co.uk/batt_con.html

to figure capacity, you need the voltage & Ah of the battery, like a golf cart battery is often 6v, 200ah

if you had those as your bank, each 12v string (2 in series) is 12v 200ah

6 of those in parallel gives you 12V , 1200ah. That would properly require about 120A of charge to keep the batteries healthy,

That battery bank also can be measured in Watt Hours, and would be 14,400 watt hours total.

120A @ 14V = 1680 harvested watts of PV power

If you had configured that to be a 24V bank you still have 14,400wh but 24V , 600ah, needing only 60A of charge current (Just a single charge controller) and still needing 1680 harvested watts of PV power, but you can use smaller wire gauge.

Now, getting to starting that compressor @ 120VAC

10A @ 120V (roughly) is 1200 watts. Add .6 power factor, and you are at about 2,000 watts that the inverter HAS to supply out of phase because of the inductive motor. At 24VDC, that's about 85A coming out of the battery, constant (till the compressor shuts off)

Starting that mess, 10x is a good number, so your inverter is going to have to be pretty stout to handle the starting load.

(that's why I have a 6kw inverter to run a 1/2 hp pump on a 48V system, I cannot afford to not pump 4,000 gallons daily for trees and my livestock water)

Your cables have to handle close to 850A dc, while starting, without voltage sag, or your inverter shuts down from undervoltage, not overload. The shorter the inverter cables, the better

Double those amps for trying this at 12V and you see it's going to be nightmarish.

jonr · August 2019

If it doesn't have one, adding an "unloader" will help some. But I don't know - maybe every compressor has one. Another approach is a smaller compressor, perhaps with a very large tank (if you need lots of CFM). I see a Dewalt at 2.6 amps.

mcgivor · August 2019

Working with 12VDC to 120VAC is a 10:1 ratio, in other words 10.4A at 120VAC is 104A at 12V, start or inrush is significantly higher which may well overwhelm the batteries which are of unknown capacity at this stage. To add to the equation induction motors generally have a poor power factor which may not affect the battery itself but will have influence on the generation source, the inverter.

Realistically it would be better to use a generator to power the intermittent load of a compressor, but that's only an option, 12V systems should be limited to ~1000W maximum of non inductive loads.

BB. · August 2019

Compressors are "difficult" loads... They typically have high starting surge requirements, and actually do a lot of work (compressing a gas to much higher pressures and lots of volume of air moved). All these make for a very difficult load for solar/battery power.

The other big issue is that Amps and Amp*Hours are a partial power/energy unit. You need to know the working voltage too. As the guys above talk about above:

Power = Voltage * Current (power is a rate, like miles per hour)
Energy = Voltage * Current * Time (energy is an amount, like total miles driven)

If you have a 10 Amp load:

10 amps * 12 volts = 120 Watts
10 amps * 24 colra = 240 Watts
10 amps * 120 volts = 1,200 Watts
10.4 amps * 120 volts = 1,248 Watts

So, the running load for your compressor is 1,248 Watts... (that is something 1.5 HP--Not a small load). The size inverter needed to run that load would probably be something like:

1,248 Watts * 1/0.50 Power factor starting * 5 starting current = 12,480 VA (similar to, but not Watts) starting load
12,480 VA / 2.0 amount of "surge rating" for good AC inverter = 4,160 VA (Watt) minimum AC inverter Rating

Different inverters respond to heavy starting loads differently... Some just output the surge current and maybe lose a bit of output voltage. Others may simply shut down if too much starting current. So, look at 4.2 kW as the minimum rating for you needs... If you have other loads running at the same time (lights, other tools, refrigerator, etc.), you would need a larger inverter--Like 6 kWatt would not be unreasonable.

Then there is the battery bank... You have a typo in your post... 12x 6 volt batteries or 4x4 batteries in series parallel--Or 16 batteries... Lets say you really have 4s x 3p bank and the batteries are "golf cart" type at ~200 AH each. such a battery bank would be 24 volts @ 600 AH.

Using our rule of thumb for the maximum AC inverter supported by an XX AH by YY Voltage bank... Our typical rule of thumb is a 500 Watt inverter for every 100 AH @ 24 volts:

600 AH (@ 24 volt) bank * 500 Watts per 100 AH = ~3,000 Watt suggested maximum inverter

If you were running your system full time (shop, building a house, etc.), you would need roughly a 2x larger battery bank to support a ~6 kWatt AC inverter (reliably over time)... And if you are full time a 10% minimum solar array (note for this size inverter you would go to 600 AH @ 48 volt battery bank and 48 volt inverter--But the math works out the same at 1,200 AH and 24 volts--Same amount of energy battery bank):

1,200 AH * 24 volts * 1/0.77 panel+controller deratings * 0.10 rate of charge = 3,740 Watt minimum solar array

If this is for your place in Mexico, assume at least 5 hours of sun per day. Your 3,740 Watt solar array would produce (minimum, sunny day) something like:

3,740 Watt array * 0.52 off grid system efficiency * 5.0 hours per day = 9,724 WH per day (this is not a "small system")

lapRunning just your compressor, that would run roughly:

9,724 WH per day * 1/1,248 Watt load = ~7.8 hours per sunny day

And your battery bank would be suggested to supply 2 days of energy, maximum 50% discharge (for longer life), WH per cloudy day:

1,200 AH * 24 volts (or 600 AH @ 48 volts suggested) * 0.85 AC inverter eff * 1/2 days storage * 0.50 max discharge = 6,120 WH per cloudy day suggested

If this is energy to build your home... This system, after you are done, the right size for your home, or 2x (or larger) than you need?

For construction of a home, typically, a genset (and/or gasoline driven portable compressor) is probably more cost effective during construction, and design/build your off grid solar power system sized for you completed home.

If you are living on site, a small off grid power system (maybe a 1,000 WH per day for lights, laptop computer, cell phone, RV water pump, etc.) is a better option (for overnight quiet time) and use solar+genset to charge during the day when work is done/more energy needed.

If you are running a full size refrigerator (vs a portable smaller propane fridge or blocks of ice), then you are probably looking at a minimum of 3.3 kWH (3,300 WH) of off grid power per day.

I will stop here... Questions, answers? The above is a bunch of guesses for a full time off grid system for long term use... May not apply to your needs.

-Bill

brewer · August 2019

Thank you for all the info. I'll now be researching it all. At this time I'll run the genny for the compressor.

10.4 amp air compressor

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