Walker Renewable

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Arapahoe Community College *

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101

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Physics

Date

Dec 6, 2023

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docx

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10

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Purpose: The purpose of this lab is to learn how to build a solar collector and to also measure the results and to determine the efficiency of the thermal collector. Another purpose is to learn about solar collectors and renewable energy sources. Hypothesis: If the solar thermal collector is built correctly and placed in the right elements then the collector will operate with above 50 percent efficiency. Experiment: In this experiment I built a solar collector and placed it in sunlight. Then I funneled a liter of water through the solar collector continually filling up a cup on the output end. When that cup became full I poured the water back into the top collector. Every five minutes I recorded the temperature of the top part and recorded the information. I was able to use all this information to figure out the efficiency of the solar collector. For the second exercise, I used a multimeter and a silicone solar cell to analyze its collection of the Sun’s rays. I angled the solar cell horizontal to the Earth, perpendicular to the Earth and perpendicular to the sun to evaluate how those different angles showed different wattage.
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Exercise 1: Building and Analyzing a Solar Thermal Collector Data Table 1. Measurements for the Solar Thermal Collector. Mass (g) of 50 mL of Density of water (g/mL) Beginning temperature Ending temperature Beginning air temperature Ending air temperature
water of water of water 48.5 .97 21.3 42.5 20.8C 26.4 Data Table 2. Water Temperature. Time (min) Temperature (°C) Weather changes 0 21.3 Slight breeze 5 25.5 No change 10 27.9 No change 15 29.8 No change 20 31.4 No change 25 33.0 Slight breeze 30 34.5 More wind 35 35.9 No change 40 37.4 No Change 45 38.5 No Change 50 39.4 Slight breeze 55 40.8 No Change 60 41.5 No change Questions A. Graph the data from Data Table 2 . Place temperature on the Y-axis and time on the X-axis. 0 10 20 30 40 50 0 5 10 15 20 25 30 35 40 45 Temperature of Water Time(minutes) Temperature(Celcius)
Evidence: This graph shows the change in the temperature of water from the solar collector over an hour, every five minutes the temperature was evaluated and recorded and represents every five minutes the temperature was taken. B. Calculate the amount of energy the water absorbed. To do this, you will use the following equation: Q = ΔTmc p . Q is the amount of energy absorbed by the water, ΔT is the change in water temperature, m is the mass (g) of the water heated, and c p is the specific heat capacity of water, which is 4.184 joules/g/ºC. You will need to use the density calculation to determine the mass of the liter of water by multiplying density by mL. Finally, convert this to watt-hours. One joule is equal to 0.00028 watt-hours. m=.97(1000)= 970g ΔT=42.5-21.3= 21.2 ºC Q=(970)(21.2)(4.184)=86,039.776J 86,039.776(0.00028)=24.091 watt-hours Evidence: I weighed a ml of water on a scale and it weighed .97 grams. There are 1000ml of water in a liter so I multiplied .97 grams per ml by 1000ml to come up with 970 grams in a liter. The change in temperature is going to consist of the final reading minus the initial temperature of the water. 42.5 degrees Celcius was the final reading minus 21.3 degrees celcius(initial reading) meant there was a 21.2 degree celcius change in temperature. To find the amount of energy absorbed by the water (Q), I needed to multiply the grams in a liter of water times the change in temperature times the specific heat capacity of water(provided in the problem). C. Determine the percent efficiency of your solar collector. To do this, use one of the appropriate values for solar insolation (how much solar energy strikes a square meter in an hour): Sunny, summer: 800 watt-hours per m 2 (Wh/m 2 ) Sunny, winter: 400 Wh/m 2 Partly cloudy, summer: 400 Wh/m 2 Partly cloudy,winter: 200 Wh/m 2 Next, multiply the appropriate value above by the area of your solar collector (0.07 m 2 ) to determine the watt-hours received by your collector over the course of one hour. For example, on a sunny summer day, a 0.5 m 2 solar collector would receive 400 watt-hours of solar energy in one hour, 0.5 m 2 x 800 Wh/m 2 . Divide the energy that the water absorbed
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by the watt-hours the collector received in one hour and multiply by 100 to determine percent efficiency. 400(.07)=28 24.091/28=0.8604 .8604(100)=86.04 The solar collector operated with 86.04 percent efficiency. D. What are the sources of error that exist for calculating the percent efficiency of your solar water collector? The sources of error that exist for calculating the percent efficiency of my solar water collector is trying to pick a value to use for a decently warm fall day. I chose to use Sunny winter/ partly cloudy summer for mine. Another error can be making sure no water is lost during the experiment and eliminating wind. E. How would you make the collector more efficient? I could make the collector more efficient by spending more time making sure the solar collector is sealed as much as possible. I would shorten the tubing to allow for better flow of water as well.
Exercise 2 : Analyzing a Silicon Solar Cell Data Table 3. Voltage, Amperage, and Wattage Data from a Silicon Solar Cell. Solar cell parallel to Earth Solar cell perpendicular to Earth Solar cell perpendicular to Sun’s rays Voltage .55 .59 .59 Amperage .0069 .0155 .1537 Wattage .003795 .009145 .090683 Data Table 4. Calculated Values from Using a Silicon Solar Cell. Maximum Wattage/Power (W) Area of PV cell in m 2 Power density (W/m 2 ) 0.090683 0.0004m 2 226.7075 Questions A. What difference, if any, was there among the different positions of the solar cell? What does this tell you about the optimal way to position solar panels for highest electricity production? There was a significant difference between the different positions, parallel to the earth was the least efficient, perpendicular to the Earth is more efficient but the most efficient was perpendicular to the Sun’s rays. This shows that solar panels should be positioned perpendicular to the Sun’s rays for best electricity production. B. How many solar cells of the type you worked with would it take to light a 100 W bulb for one hour? (Assume that the bulb is used during the middle of the day, or that you have a way to store the energy the solar cell produces during the day for use at night or when it is cloudy.) How many square meters would that require? This would require .4411m 2 to power a 100 watt lightbulb for an hour. 100/ (.090683)=1,102.743(0.0004)=.4411 Evidence: It takes 100 watts to power a 100 watt lightbulb for an hour. The maximum wattage of the solar cell I used was .090683 W when perpendicular to the sun’s rays. I took 100 watts divided by maximum wattage and then multiplied the result by the area of our solar cell to discover how many square meters of our solar cell would be needed to power the lightbulb.
C. How much does it cost a homeowner to use electricity for 30 years if the electricity is purchased from the electric power company? Hint: An average house in the United States uses 31.5 kilowatt hour (kWh)/day and electricity costs $0.10/kWh. This does not account for the rises in electricity prices that will occur over the next 30 years. 31.5(365)=11,497.5 11,497.5(.1)= 1,149.75 1,149.75(30)=34,492.5 It would cost a home owner $34,492 to use electricity that is purchased from the electric power company over 30 years. Evidence: To find the cost over 30 years I took the 31.5 kilowatt hours a day and multiplied it by 365 days to represent a year. Then I took that result and multiplied it by .1 to represent 10 cents for every kWh to find the price spent for 1 year. Then, I used that result and multiplied it by 30 years. D. An average house in the United States uses 31.5 kWh/day. How may square meters of solar cells are need to supply all of the electricity? Use the average solar insolation of 4.2 kWh/ m 2 per day and an efficiency of 15% for the solar panels, which means that they are able to convert 15% of the solar radiation that strikes them into electricity. 31.5/(4.2x.15) 31.5/.63=50 m 2 50m 2 of solar cells are needed to supply the electricity for the average house. Evidence: For this problem, I divided 31.5kWh/day by the result of multiplying 4.2kWh/m 2 and .15 representing 15 percent efficiency. E. What are the limitations and advantages of using solar systems to providing electricity? Some of the limitations can be the initial setup cost which some cannot afford to pay. Another limitation of using solar systems is the access to sunlight. It is beneficial to live somewhere with more sunlight year round but in Colorado there is less sunlight during the winter. Advantages are using clean renewable energy provided by the sun. Another advantage is the low cost of electricity due to the solar systems over time. Reflection: This experiment was extremely informative and I discovered that the solar collector was actually 86 percent efficient. I was pleased with the results and I accept my hypothesis.
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Source: Solar . Energy.gov. (n.d.). https://www.energy.gov/solar Solar energy . Education. (n.d.). https://education.nationalgeographic.org/resource/solar-energy/ Encyclopædia Britannica, inc. (2023, October 30). Solar energy . Encyclopædia Britannica. https://www.britannica.com/science/solar-energy