AC load and Inverter power calculations

WYsolar

List,

My small solar test project is almost complete. I want it to power various small things, one of them being the occasional use of a shop vac. It draws 7.3A and 120v AC...or 876W

I have a 104Ah of batteries I am feeding with my panels. I want to know how long I can run the vac and not take my batts below 50% discharge. I performed some figuring with what I have learned and came up with the following. I am not sure I did this right so I welcome any corrections:

1) Inverter is a 1000w GoPower Pure sine. It uses 1.4A no-load. I assumed this is DC Amps so the wattage consumed is: 12v x 1.4A = 17W

2) For the vac I used the formula DC Amp = Acwatts /12 *1.1 In this case I get 876W/12 *1.1 = 80A dc used

3) I added the inverter Watts to the vac and get = 876 + 17 = 893 total watts used. 893/12 * 1.1 = 82A dc used total. This load would be 82Ah dc if the vac ran for an hour, right?

4) Fully charged I have 104Ah in my batteries. I don't want to go below 50% so that means I really have 52Ah to use.

5) This is where things get a bit fuzzy. My vac/inverter will consume 82Ah if I run it for an hour. I know I can't do that as I only have 52Ah to use. So...how long can I run it?

6) I took 82 and divided by 60 (minutes in an hour) and I get = 1.4Ah/minute. Now with 52Ah in the batteries I get 52/1.4Ah = 37 minutes of run time on the vac before I reach 50% discharge.

Do these calculations sound right? I would really like to know how to calculate AC load runtimes considering the amount of battery storage I have.

Any advice or corrections are most welcome. Thanks!

Cariboocoot

Re: AC load and Inverter power calculations

Advice: forget it.

In just "simple math" that 7.3 Amps @ 120 VAC is somewhat more than 73 Amps @ 12 VDC which will "flatten" that 104 Amp hour battery instantly. It is unlikely the DC Voltage will remain high enough to even keep the inverter on, much less run the vacuum.

WYsolar

Re: AC load and Inverter power calculations

ok. Sounds like my calculations are way off.

How do you calculate how long an AC load will run based upon the capacity of your DC battery bank?

Cariboocoot

Re: AC load and Inverter power calculations

Convert the AC Watt (hours) to DC Watt (hours) (factor by inverter efficiency). Add in inverter consumption to get total DC Watt (hours). Divide by nominal system Voltage to get approximate Amp (hours) from the battery.

So 876 Watts becomes (876 / 0.85) 1030 Watts DC, plus 17 Watts for the inverter: 1047 Watts. Divide by 12 and you get 87.25 Amps. That's >80% of the battery's Amp hour capacity @ the 20 hour rate. The faster you draw power from a battery the lower its actual capacity. So at this point you're looking at a battery dropping dead very quickly.

In this case the problem is that the current draw is so high that the battery simply can not supply it with a severe Voltage drop. The battery capacity rating is based on about 5 Amps draw steady over 20 hours, not 85 Amps draw.

zoneblue

Re: AC load and Inverter power calculations

And a 1kW inverter wont start a 900W motor. No way, needs maybe 5 times that to get the thing turning.

WYsolar

Re: AC load and Inverter power calculations

Wow.

This is an eye-opener.

To conclude this exercise I now ask:

How would one size a battery bank to run just this one appliance? Suppose it takes 1300W AC just to start the vac, then takes 893W to keep it (and the inverter) running for 0.5 hr everyday? Would a 2x 6v 220Ah "golf cart" battery bank run the vac?

Thanks again for the responses. I sure do take the commercial power grid for granted!

Cariboocoot

Re: AC load and Inverter power calculations

zoneblue wrote: »

And a 1kW inverter wont start a 900W motor. No way, needs maybe 5 times that to get the thing turning.

In this case it probably would as a vacuum cleaner has no load against it upon start. I have measured several motors and found that the main determining factor for start-up surge is how much torque it needs to get moving. Things like compressors and pumps are terrible: factor of 5 is not at all unusual. Saw motors and others that start "no load" can actually start with less than their rated running Watts.

Cariboocoot

Re: AC load and Inverter power calculations

WYsolar wrote: »

Wow.

This is an eye-opener.

To conclude this exercise I now ask:

How would one size a battery bank to run just this one appliance? Suppose it takes 1300W AC just to start the vac, then takes 893W to keep it (and the inverter) running for 0.5 hr everyday? Would a 2x 6v 220Ah "golf cart" battery bank run the vac?

Thanks again for the responses. I sure do take the commercial power grid for granted!

The first thing I'd do is measure it with a Kill-A-Watt and see what it really takes.
If it is in the 800+ Watt running range I would prefer a 24 Volt system to run it from because the current demand will be half what it is on 12 Volts. So you are then looking at drawing 40-ish Amps instead of 80-ish Amps. Right away you have an improvement.

Then you try to keep the current draw in the same neighbourhood as the charge rate: 10%, maybe 20% maximum. For an occasional use item you can use 20%, for something that is run a lot use 10%.

So now 40 Amps becomes 20% of the Amp hour capacity of a 24 Volt battery bank. 40A * 100(%) / 20(%) = 200 Amp hours @ 24 Volts. Round battery size up to nearest available. At that rate the 40 Amps would run for about two hours straight before the batteries were seriously depleted (with no charge input).

How I know this works: I have 232 Amp hour 24 Volt battery bank that has no trouble running the 800+ Watt water pump, 1000+ Watt microwave, or 1200+ Watt septic pump. Even so I run them midday when the panel output is best to avoid depleting stored power.

BB.

Re: AC load and Inverter power calculations

To give you an idea--73-100 amps is probably the average starting current for an automobile. And they have 80 to 100 AH (roughly) batteries... So, even if everything starts and runs--How long would you want to "crank" your engine for your car... I had a friend with a VW Bug and the engine died--He put it in gear and cranked the engine to get to the side of the road and to a safe place to stop--Maybe a 1/2 mile or less?

Anyway, some rules of thumbs. The maximum surge current from a typical lead acid battery bank is about C/0.4... So 1,300 watt surge:

• 1,300 watts * 1/0.85 inverter efficiency * 1/0.4 maximum surge * 1/10.5 volts minimum inverter cutoff = 364 AH @ 12 volt battery bank (lead acid flooded cell)

The above is the design rule of thumb I like to use for a "reliable" installation over typical operating conditions. Obviously, this is a lot more battery bank that you would see on the typical car/pickup truck.

Next is operating power (hours of use)... The maximum draw would probably C/5, the "nominal" maximum draw C/8, and the typical draw for a solar cabin/home would be C/20 (capacity of battery bank divided by hours to take it "dead").

So, if this was a "short term" load (a few hours a day), then I would use C/8 as the average draw rating:

• 876 Watts load * 8 hour discharge * 1/0.85 inverter eff * 1/10.5 inverter cutoff = 785 AH @ 12 volt battery bank

And, that C is defined at 20 hour discharge rate... A battery discharge at 8 hour rate will have less apparent capacity... For this example, lets say the battery bank has a C₈ capacity of 0.8 derating and a 50% maximum battery discharge.

• 785 AH * 12 volts * 1/876 watts * 0.85 inverter eff * 0.80 battery derate * 0.50 maximum discharge = 3.7 hours of "run time" to 50% discharge

To recharge the battery bank, two major checks... One is the minimum charging current and the second is enough hours of sun per day to recharge load.

First charging current of 5% to 13% rate of charge:

• 785 AH * 14.5 volts charging * 1/0.77 panel+controller losses * 0.05 rate of charge = 739 Watt Array minimum
• 785 AH * 14.5 volts charging * 1/0.77 panel+controller losses * 0.10 rate of charge = 1,478 Watt Array nominal
• 785 AH * 14.5 volts charging * 1/0.77 panel+controller losses * 0.13rate of charge = 1,922 Watt array "cost effective maximum"

And lets assume a reasonably sunny climate and a minimum of 4 hours of sun per day (for 9 months a year):
876 Watt * 3.7 hours operation * 1/0.52 system efficiency * 1/4 hours of sun = 1,558 Watt array minimum

So, for a "off grid" home, I would suggest 1,558 to 1,922 Watt solar array for use with that battery bank/load.

So, that is the back of the envelope calculation for your load (as defined) and our basic assumptions. Nothing magic about the above numbers, just a suggested starting point/understanding of the sizing process.

-Bill