PV Solar: Homeower Analysis (1)

Part 1 – Sizing Your System
In my last post I questioned the affordability of renewable energy and the homeowners’ place in the energy market. By becoming the producer as well as the consumer, the homeowner has found a place in the market and helps make renewables more affordable. This will begin a series of posts where I compare the cost of fossil fuel to renewables. The objective of this analysis is to determine whether a renewable energy alternative such as solar, can be as affordable as the fossil fuel generated electricity transmitted to my home. The renewable energy alternative for this analysis is a hypothetical photovoltaic (PV) solar system installed in my home.
I’ll begin by trying to determine the size of the solar powered system I want install. Note, the size of the system isn’t determined by the size of the home, but rather by the amount of electricity that is consumed in kilowatt hours and how much of that consumption I would like to eliminate. I’ve decided to size the system to produce approximately the same amount of electricity I consumed last year. I used my monthly energy statements to determine the actual electricity use for the year. The use includes lighting, cooling, dishwasher, oven, refrigerator, freezer, the washer in laundry room, and all plugins (computers, televisions, etc.). I excluded the use of natural gas which includes heating, water heater, clothes dryer and stove top.
Electricity use is usually billed in kilowatt hours (kWh, W h): 1 kW·h = 1000 W·h, which is a unit of energy equal to 3,600,000 joules. My total electricity use for the year was 8455 kWh, this is an average use of 704.59 kWh/month. What I want to determine now is the amount of power the system needs produce per day to eliminate my electricity bill. To do this I’ll divide the average monthly electricity usage by 30 days (704.59 kWh/30 = 23.49kWh), which tells me the system needs to produce at least 23.49 kWh per day to eliminate the bill.

Next I need to determine the average number of hours of full sunlight that is available, in this area, on a daily basis. To do this I’ll use a Solar Insolation Map which shows the average daily solar radiation in kWh/m2/day (Collector Orientation: Flat Plate Tilted South at Latitude). Sun hours are the average number of hours per day of usable solar radiation. In 1 hour under ideal conditions, 1 square meter receives the equivalent of approximately 1 kWh of solar energy. The map shows that my area receives approximately 5-6 hours of optimal sunlight per day. This means, on average, I have 5.5 hours during the day to produce the 23.49 kWh of power, needed to eliminate the electric bill.
To determine the size of the system I need to purchase, the amount of power needed per day is divided by the number of sun hours (23.49kWh/5.5 h = 4.271 kW). The calculation shows I’ll need at least a 4.271 kW system (1 kW = 1000 watt hour = 1000 W) which is the same as a 4271 W system. Due to module and inverter inefficiencies along with other power losses, the calculated system size of 4.271 kW is multiplied by a factor of 1.2 to give a more real-world estimate (4.271 kW x 1.2 = 5.13 kW) of 5.13 kW.
Ok, now that I have determined that I need a system approximately 5.13 kW in size, I will need to price out various options to come up with a cost. In my next post I’ll pick up from here and start the process of pricing the system.
This is a learning experience for me, so if you see any error or omissions, please share and leave a comment.
