I am a novice. I want to correctly size a solar panel and battery for the following application. I will be powering 2, 6 watt cfl lights and a 40 watt incandescent light. I am planning on using 400 watt inverter from Northern Tool. I need the inverter because of the need for the 40 watt incandescent bulb which will be used as a heat source. The 40 watt bulb will be on 24 hours a day when the temperature is below 32 F. The two 6 watt cfl lights will run a maximum of 5 hours a day year round. I have done research on the internet, but so far some of the information I have obtained is contradictory. I live in southwestern Pennsylvania. What I need to know is the size of the solar panel in watts and the size of the battery Ah.
Using a 120 VAC inverter and a light bulb to keep your pipes from freezing is just a dumb idea.
If you need 40W worth of heat in an enclosed space, just select a bunch of power resistors that add up to that amount, screw them to a heatsink and run them directly from your 12 V battery. For example, a 15-ohm resistor will dissipate about 10W @ 12V, use four of them.
You can also get CFL lamps that run directly from 12V.
Let's say you need a run time of 48 hours, and you have a minimum of 6 hours to recharge.
Your load is drawing a daily average of 3.33 (heater) + 5/24 × 1 (lights) = about 3.5 A. Multiply by 48 hours to get the battery capacity: 170 Ah
Batteries aren't perfectly efficient, so it'll take about 170 / 0.8 = 212.5 Ah to recharge it.
If you want to drive the load and recharge the battery in 6 hours, you need a panel capacity of about 40 A @ 12 V, or 480 W.
This is not an answer to your question, but got too big to be a comment.
About using solar-power to charge a battery, and then convert the battery energy to generate heat aftwerwards... in general, solar panel doesn't have a high convertion ratio, and the heat that you get from an incandescent light isn't good either (since you're generating light, not heat).
Wouldn't it be better to use the solar light directly to heat it to the desired temperature, and use solar energy to heat some water and store it, and use the water as teh heating element during the night?
Solar cell: efficiency of around 20% (http://en.wikipedia.org/wiki/Solar_panel#Efficiencies)
Solar water heating: efficiency around 68% (http://en.wikipedia.org/wiki/Solar_water_heating#Economics.2C_energy.2C_environment.2C_and_system_costs)
Lots of variables/unknowns here. There is no "correct" calculation, but here is an example.
Load : 12W cfl = 12W, 5/24 duty => 12*5/24 * 24 h = 60 Wh
Load : 40W bulb= 40W, 24/24 duty => 40* 24 h = 960 Wh
Total load per 24 hours = 1020 Wh.
Inverter efficiency: 50% (check the data sheet). Inverter loss = 1020 Wh * 50% = 510 Wh, so total load is about 1500 Wh per day.
Hours of autonomy: 48 (No sun for 48 hours). Your system must provide about 1500 Wh for 2 days, so a total load of 3000 Wh.
Load = 1500Wh/24h = 62.5W
Battery: Current I = Load/Voltage = 62.5W/12V = 5.2A DC
Battery Capacity = 5.2A * 48h = 250 Ah
Solar panel must provide enough current to charge a discharged battery plus provide for the load. Assume a maximum of 6 hours useful sun per day. You need 250 Ah to charge the battery in 6 hours = 250 Ah/6h = 41A + 5.2A for the load. Quite a current, and this does not take into account the battery efficiency, maybe 80%.
But this is a back-of-the-envelope calculation, your sun exposure may be better than 5 hours, and even on cloudy days the solar panel will charge some. My suggestion: If you have mains power not too far away, use it.