Replies

1. 
   DanH | Mar 04, 2012 06:35pm | #1

   Well, the simple answer is to divide the Amp-Hour rating by the total current draw.  If the total draw is 15 A DC then the system will run for 100 hours on a 1500 AH battery.  But that's (slightly) complicated by several things.

   First off, you're running an inverter off the battery and running most (all?) of the loads off of 120VAC from this inverter.  The simple jump from (I presume) 12V DC to 120V AC confuses things, since, ignoring conversion losses, the voltage change causes a 10:1 change in current.  That is, a 100 watt light bulb draws about 8.3 amps at 12V but only 0.83 amps at 120V.  But in addition you have conversion losses in the inverter of 10-20%, so that 100 watt 120V AC lamp is really costing you maybe 10 amps at the battery.

   Second, there is a somewhat intangible issue of how long a 1500 AH battery really lasts.  The 1500 AH is an ideal number, measured under certain laboratory circumstances, and the circumstances where the inverter will operate will be different.  In particular, as the battery discharges its voltage drops, and eventually the voltage will drop below a level that the inverter can tolerate.  This probably will occur before the battery is "100% discharged".  At what % of discharge the inverter will cease to function is hard to guess (though the makers of the batteries and inverters probably have charts that produce a good guess).

   Finally, keep in mind that the AH rating of a battery is at a certain discharge rate, and discharging faster or slower than that rate will produce a different result.

   As a SWAG, I'd guess that the battery is good down to about 30% charged/70% discharged -- call it 1000 AH available.  And each 100 watts of 120VAC load, after conversion losses, is about 10 amps.  So that's 100 x 1000 / 10 watt hours, or about 10 KWH.  Add up your total average load and divide into that number.  (For average load on a fridge, eg, figure if it draws say 200 watts 10% of the time then that's 20 watts average.)

   If the battery is not 12V but, say, 48V then scale the above numbers by 4.  That is, a 1500 AH 48V battery would produce about 40 KWH of usable power in the same circumstances.

   And if you're running some lights, etc, directly off of (12V?) battery power, then those don't suffer from inverter losses, but the current draw will be 10 x larger than the draw of a 120V load of the same wattage.  (But a simple way to figure these loads is to take their wattage minus 10% and add that to the 120V load wattage, so you deal with a single number.)

   The inverter manufacturer could probably provide you with better numbers.

2. 
   calvin | Mar 04, 2012 08:35pm | #2

   Man

   if you were confused b/4........................

   No kidding, call the manufacturer-they should have a fair idea according to your use.

   and buy a generator.

3. 
   junkhound | Mar 04, 2012 10:55pm | #3

   Assuning 12 V lead acid battery?

   If ya discharge about any chemical battery to 70%, your 'cycle life' will only be a few dozen cycles. NiMh you can likely get away with over 50%, but definetly NOT with lead acid, which I'm assuming is the 1500 Ahr batt pack.

   Best to discharge to only 30% or so if the owner expect to get any life out of the (expensive?)  battery pack.

   500-72 = 428

   80% efficient inverter, 428*12*.8 = about 4.3 kW-hrs to the 60 hz loads, I'd assume 2kW of other load, so batt will run system for about 2 hrs.  4 or even 6 hours if they have big $$ to replace the battery pack every few power outages.

4. 
   atomsite | Mar 13, 2012 12:41am | #4

   Brated = Bout / DOD *

   Brated = Bout / (DOD * Ct)

   Brated = required battery bank capacity (Ahr)

   Bout = required battery bank output (Ahr)

   DOD = depth of discharge

   Ct = temperature & discharge rate derating factor

   Brated = 1500 Ahr

   Bout = unknown

   DOD = 75%

   Ct = 0.9 (assuming lowest operating temp is +10 deg C & avg discharge rate = C/50)

   Bout = Brated * DOD * Ct = 1500 * 0.75 * 0.9 = 1,012.5 Ahr

   The available battery bank output is 1,013 Ahr (assuming a depth of discharge not exceeding 75%).

   The total battery bank capacity is 1,500 Ahr (which was specified).

5. 
   atomsite | Mar 13, 2012 01:23am | #5

   Bout = (E * ta) / Vsdc

   Bout = battery bank output (Ahr)

   E = daily energy consumption (Whr/day)

   ta = autonomy (in days)

   Vsdc = nominal DC system voltage (VDC)

   Bout = 1,013 Ahr

   E = need to figure this out

   ta = unknown

   Vsdc = 24 VDC (assume 24 volts since this was not specified)

   ta = (Bout * Vsdc) / E

   ta = (1,013 * 24) / E

   To finish the equation we need to know the daily energy consumption (lights, outlets, refrigerator, etc.). Then we have to derate for inverter and wiring losses, etc.

   The load consumption is not accurately specified. How long are the lights on, what is the power consumption of the refrigerator, the refrigerator duty cycle, what loads are plugged into the outlets? Can you give us an estimate of the daily energy consumption is in Wh/day?

   Assumptions:

   Refrigerator = 616 kWhr / yr = 616 / 365 = 1,688 Whr / day

   Outlets = 0 Whr/day (assuming you do not plug anything into them)

   LED lights power consumption = 24 * 3 * 12 = 864 W

   LED lights daily energy consumption (assume lights are on for 10 hrs night only) = 864 * 10 = 8,640 Whr / day

   Total daily energy consumption = 1,688 + 8,640 Whr / day

   Inverter efficiency = 0.9 (typical)

   Wiring losses = .97 (3% typical)

   E = 8,640 / (0.9 * 0.97) = 9,897 Whr / day

   Therefore:

   ta = (1,013 * 24) / 9,897 = 2.5 days

   Days of automony is 2.5 days assuming battery bank system DC voltage is 24 VDC, LED lights are only on for 10 hrs / day, and nothing is plugged into the outlets. Also it's a typical refrigerator.

   This calculation does not account for the PV array load fraction. Which will supply power during the day. Assuming there is no snow on the panels.

   If you could specify the load more accurately then you would get a better estimate. Chances are the days of autonomy are much much less.

   Uh, BTW. Who designed your bimodal system? These are basic calculations for any PV system. These numbers should already be known.