48vdc 120vac setup

robocop

have 4 pcs 500w panel each / 4kw power inverter and 60amp or less max mppt /16pcs lipo 3.2v 280ah batt.(bank of 48v 280ah)

house 120v  grid  2 refrigerators total about 600w  1 tv 200w  sometimes air fryer 1200 w  and a total of led lights of 65w

 living in the tropics .
Will this set up be enough for off grid.

Appreciate  any help on this

Bob

BB.

Bob,

The math...

Battery bank capacity. Assume 80% to 20% battery cycling (6-% of capacity, allow for aging and less than perfect charging). For 2 days.

• 48 volts * 280 AH * 0.60 capacity usage * 1/2 days of storage (bad weather) * 0.85 inverter eff = 3,427 Watt*Hours per day storage
• 3,427 WH / 24 hours = 142 Watts continuous load (just to give you an idea)

Nominally during good weather you will only be using the battery bank at night, and using solar to run loads plus charging during the day... The 24 loading is support during bad weather and such (worst case). Do you have a genset for backup power? (want to use a genset?).

To run the loads you ask, a 2,000 Watt inverter is probably the minimum. The 4 kWatt inverter is "better", but you need to look at "tare losses" (power used by the inverter just being turned on. Some inverters use 10-20 Watts, others use upwards of 40 Watts--Which is a fair amount of power usage in this size system--I.e., 142 Watts of 24 hour average load support power - 40 Watts from "poor inverter design" = 100 Watts of average battery load support). 

For your refrigerators, need to know their power consumption. Efficient refrigerators tend to use around 1,000 to 1,500 WH per day. Or just about all of your 2 day of 3,427 WH battery storage (weather, backup genset, etc.).

Next, 2,000 Watts of solar for Willemstad, Curaçao. Fixed array facing south, 12 degree tilt:

https://pvwatts.nrel.gov/pvwatts.php

Month	Solar Radiation ( kWh / m2 / day )	AC Energy ( kWh est monthly production with 2kWatt array)
January	5.83	199
February	6.08	187
March	6.19	209
April	5.89	192
May	5.83	195
June	5.84	191
July	5.88	198
August	6.26	211
September	6.33	205
October	5.78	194
November	5.46	179
December	5.48	185
Annual	5.90	2,345

Your "worst month" is December at 5.48 hours of sun. Assuming AC system efficiency of 61% for lithium solar AC power:

• 2,000 Watt array * 0.61 system eff * 5.48 hours of average sun per day (December) = 6,686 Watt*Hours of average solar per day (December)

You are in a very sunny location... The "conservative" design would assume using 50% to 65% of "predicted" power--Allowing for bad weather, and such:

• 6,686 WH per day (Dec Average) * 0.50 "reliable solar fudge factor" = 3,328 WH per day

You have lots of solar--And can easily run your refrigerator and lighting loads. And probably run the air fryer during the day time (and not bad weather).

One thing to look at is your battery bank and BMS (battery management system) specifications. Typically I would guess that your bank would support 100 Amps to 280 Amps max continuous power (need to check specs). For example:

• 100 amps * 48 VDC battery bank * 0.85 inverter eff = 4,080 Watts max supported AC loads (just a conservative guess)

You did not mention "water pumping"... That can be a big issue for smaller systems too.

Anyway, at first look, I would guess that your system would support your daily loads quite nicely... Especially if you can shift loads to middle of a sunny day (washing machine, vacuum, cooking, etc.). And may need to conserve if you have bad weather predicted (when solar harvest is poor to non-exist ant during heavy clouds/rain).

-Bill

robocop

Hi Bill   
There is no waterpump and 1 refrigerator consume 3.0A  the other 2.3A

BB.

3 amps at what voltage? 48 VDC, 120 VAC, 230 VAC or what?

Also, refrigerators typically run for 50% of the time and are off for 50% of the time....

For example:

• 3 amps * 48 VDC * 50% duty cycle * 24 hours per day = 1,728 Watt*Hours per day

If this is 120 or 230 VAC, it gets a bit more complicated... There is the "power factor" of the AC current vs the AC voltage... Just measuring current with an Amp Meter is not really "good enough". You need a power meter that measures both voltage and current at the same time.

Power Factor is a number that ranges from 0.0 to 1.0 and typically is around 0.6 to 0.95 for an electric induction motor. For a PF of 0.8 it would look like:

• 3 amps * 120 VAC * 0.8 PF * 50% duty cycle * 24 hours per day = 3,456 Watt*Hours per day

And during cool ambient temperatures, the duty cycle may be under 0.50 (50%) and during hot weather, the duty cycle may be over 50% (motor runs longer during hot weather).

-Bill

littleharbor2

robocop said:

Hi Bill   
There is no waterpump and 1 refrigerator consume 3.0A  the other 2.3A

Hopefully those numbers are the max draw when in defrost mode. My fridge says it draws 6.5 amps but when running it is only drawing 0.75 amps varified with a Kill-a-watt meter.

robocop

 ok Bill thanks for the information

robocop

Sorry it was littleharbort2  thnks to you