Solar Energy

This is one of those topics where we expect the situation in Oregon to be significantly different than in England. This is based on a couple of ideas that we have. One is that England is much cloudier than Oregon, which is true. The other, less widely known, is that England is further north than we are which also gives them less available sunshine.

1) What are the differences between the three ways that David suggests we might gather energy from the sun? Is the amount of sunshine available for each of these processes the same or different?

2) You may have heard about "passive solar" design for houses. Look it up if you need to. Which of the three strategies for gathering solar energy is used in passive solar design?

3) How would you decribe the difference between the type energy gathered by solar hot water panels and PV panels?

4) How much sunshine is available (average over the year in kW/m² or W/m²) for us to gather here in Central Oregon (based on our discussions in class)? Show me that you can figure this out for yourself by calculating what the average solar energy available in Vermont using the same NREL maps we used in class. Look closely at the units and convert what is on the map to the units requrested.

5) Given the efficiencies listed on this wiki table active flat panel systems how many square meters of solar panel would you need to have to provide for your daily hot water needs (12 kWh/day) when you are receiving 100 W/m² of sunshine on a winter day in Central Oregon? How many square meters would you need on a summer day when the sunshine is delivering energy at 300 W/m²?

6) Why is Oregon's net metering law so important to the cost effective development of solar generated electricity? Be sure you think about and discuss what times of year we can best generate solar electricity and when we most need it.

7) This is actually just a giant unit conversion problem....go slow and be sure to track the units from what you know (kg of oil/m²) to where you are going which is kWh/m² annually. So here are some ag figures for biofuels production. In the Columbia basin one can harvest 2500 kg of canola seed per acre from an irrigated field in the spring. An acre is 4050 m². The 2500 kg of canola seed will produce 1000 kg of canola oil which has an energy density of 40 MJ/kg. Remember that 3.6 MJ = 1 kWh. How many kWh/m²are being generated annually? From this now determine the average power capacity assuming that this energy was generated continuously throughout the whole year. This number should be less than 2 W/m² to be consistent with our text.