energy for remote farm house

[alternate energy]

I would like to install solar power system for my farm house

following are the appliances using 220VAC, insolation is about 5.9  in my area

4. flourescent lights 40w for 8 hours
   
4. energy saver lights 24w for 8 hours
   
2. fans 100w for 10 hours
   
   

I am new to this technology and confused how many solar panels , batteries of 200ah , charge controller opf how many amperes and inverters of how many watts will be used, with 3 sunless days, although professional are available but their prices are too high, can some guide me for the calculation for above energy requirement

thanks

[wdyasq]

First, you need to get things in its' own units.

40W X 8hr = 320Whr; 24W X 8hr = 192Whr; 100W X 10hr = 1000Whr Total - 1,500Whr/day

For 3 days of running you will need 4.5kWh of power. At 12V and no losses that is ~375AH. I will use a 50% pull-down of the batteries. That will be ~800 AH at 12V. That is 8, 200 AH 6V Batteries in series/parallel wireing.

If using solar, "PV", it will take 300W of panels at 5 hours of sun per day just 'to keep even'. 400W would allow one to recharge the 3 day depleted batteries in 3 days.

All this is with no losses and 100% inverter efficiency. I might suggest higher than 12V for the system batteries if you are planning on expanding.

Good luck,

Ron

[Waterlogged]

I come up with:

4. x40x8=1280
   
4. x24x8=768
   
2. x100x10=2000
   
   

total= 4048 Watt Hours per day.

Figure about 3 times the previous amount.

I got to go.

Rod

[jimovonz]

I get just over 4kWh/day too. I would double this figure to allow for panel/inverter/battery and other inefficiencies (For instance if you assume 80% for each you get 51% system efficiency. This may be a bit harsh but as I said, there are other inefficiencies as well. It is far better if your system over performs compared to it under performing...) If you use 5.9 full sun hours equivalent for your insolation you'd need 8000/80/5.9=16.9 or 17 panels (rounded up to an even 18 for a 24v system). Of course if 5.9 is your average insolation over the year then you can expect it to be quite a bit lower over winter if you live far from the equator. You can get more detailed data for your site if you follow the link/instructions I posted here: http://www.fieldlines.com/story/2004/10/6/164945/703.

With this sized system, using a tracker may be worth your while. Your insolation data probably relates to a fixed horizontal surface (but may also relate to a north facing (southern hemisphere, south facing otherwise), optimally inclined surface) and would be quite a bit better for a surface that tracks the suns movement. With 18 panels nominally rated at 80w you'd have a absolute max charge current of 18*80/24=60A (24V system). You'd only aproach this with a good MPPT controller on an exceptionally clear/cool/sunny day.

If you want three days storage using 6V, 200ah batteries (discharged to 50%) then you need 3*8000/100/6=40 batteries being 4X10 serries/parallel for a 24V system.

[wdyasq]

I guess I would have the same number if I read it the same way....

A lot of power just for lights and fans.

Ron

[Clifford]

You can significantly cut down the size of your system by making sure you turn off your lights when you exit a room.

Consider Compact Florescent Lights...   5 - 10 watts each.  I guess it depends on how bright you need the house to be.

Everything you've listed, lights & fans would be relatively easy to convert to 12/24V, and thus you could eliminate the inverter.

My recomendation is that if you are doing a "do-it yourself" installation, to plan to expand the system over time, and to do some experimentation.