1. Photovoltaics Tyler Gerlach Abel de Vos

2. Question: How many solar panels does it take to heat the water that is cycled through the Linfield Pool every two weeks? (Note: This is not including the number of solar panels that would be needed to keep the pool at 80 degrees as heat is lost to the air over time. We will be learning this later in the semester.)

3. Assumed: - Hottest month of the year - Solar panels move to face the sun throughout the day - 1 foot of pool water is replaced every two weeks by tap water - Tap water is at an average of 55 degrees - The deeper flat section of the pool is 13 ft. deep and the shallow end is 4 ft. deep - The area of the deeper section and the area of the shallow section are roughly the same - The slope from 4-13 ft is constant, and the average height of a slope is always in the middle, 8.5 ft in this case. - Therefore the average depth of the pool is 8.5 ft. - The efficiency of the solar panel we use is 18.3% - The efficiency of the pool heater is 80%

4. We want to see how much energy is needed to heat the pool back up to 80°F after we add 1 foot of 55°F water. The average depth of the pool is about 8.5 ft Before one ft of water is added, the water is 80°F So the amount of water in the pool is 7.5ft at 80°F The water added = 55°F = 1 foot so the average = 1/8.5 * 55 + 7.5/8.5 * 80 = 77.05°F 80°F - 77.05°F = 2.95°F = ΔT Data

5. Amount of water The pool contains 282,000 gallons of water 1/8.5(the drop) = 2/17 (we do this to remove the decimal) 2/17 * 282,000 gallons =33176.46 gallons of water drop every two weeks Average of 77.05°F after adding of this water. 8.337lb = 1 (gallon of water) 8.337lb/gallon * 33176.46 gallon = 276592 lb This is the mass that has to be refilled every two weeks

6. Amount of energy needed Qto raise temp.= mass × c × ΔT c = 1 Btu / lb°F (Given by Prof. Heath) mass = 2351034 lb ΔT = 2.95°F So: 2351034 lb * 1 Btu / lb°F * 2.95°F = 6935550.3 Btu 1 kWh = 3413 Btu So 6935550.3/3413 = 2032.1 kWh Since the heater is 80 % efficient: 2032.1kWh/0.80 = 2540.1 kWh needs to come into the heater to heat the pool.

7. Solar Panels In the hottest month of the year is 8.6 kWh / square meter. The efficiency of our solarpanel is 18.3% 2540.1 kWh = input on the panels * 0.183 So 2540.1 /0.183 = 13880.5 kWh is needed on the panels To get this in one day, 13880.5 kWh / 8.6 kWh/m2 = 1614 m2 of panels.

8. Factors we did not use in our calculations: Heat is lost to air Rest of the year is colder o Efficiency of the solar panels may include this Efficiency of natural gas heater

9. Heat vs. Electricity - Passive solar energy is about capturing the sun's energy as heat. - Photovoltaic cells produce electricity, so using them to produce heat is incredibly inefficient. The best way to heat the pool using solar energy would be to redesign the architecture of the building that houses the pool keeping in mind the requirements of energy flow in a passive solar energy system, however depending on how much energy is required, there may even be supplemental - Passive solar energy is often used for heat - Photovoltaic cells create electricity which makes them better suited for lighting While there are examples of success with photovoltaic pool heating systems, because of their inefficiency they are not very realistic and can be costly when truly done right. The best way to set up an off grid pool heating system would be to redesign the building that houses the pool to be highly energy efficient with passive solar heating

10. Sources: http://www.swimmingpool.info/pool-heater.html http://www.grapesolar.com/index.php?action=products.gss100t s