Memo Date: To: Inst. From: Group A : Subject: Solar cell Lab

Introduction In this lab, solar cell arrays were tested for efficiency and applicability for use in Suburban Ohio. The goal of this lab is to identify the solar cells incident power rate and whether solar panels produce a viable amount of electricity to be a feasible option on new and/or existing homes in the area. The experiment involved using an artificial light source to study the incident power output of two different solar cells at varying angles and lighting conditions. This memo contains the results that provide justification for the conclusion on those characteristics previously mentioned as well as a qualitative and quantitative discussion of the applicability of solar cells.

Results Data was recorded from the lab and is presented below. Table 1 displays the voltages measured from the digital multimeter, the LED binary output reading, and the corresponding decimal number from Task 2. Using a Trimming Potentiometer on the circuit board to vary voltage output values in increments of 0.5 volts (between 0V and 5V), the 8-bit binary number was recorded from the LED displays using equation 1 and then converted to a decimal number using equation 2. In this experiment the power supply voltage was close to 5.00 V with no significant deviation.

Table 1. Binary Voltmeter Calibration Data

Nominal V Input (volts) Actual Measure Voltage Input (DMM volts)

Binary Number

Decimal Number

0.00 0.00 00000000 0

0.50 0.493 00011001 25

1.00 0.999 00110011 51

1.50 1.506 01001100 77

2.00 2.07 01101001 105

2.50 2.50 01111111 127

3.00 3.00 10011000 152

3.50 3.50 10110001 178

4.00 4.00 11001010 194

4.50 4.50 11100011 227

5.00 5.00 11111100 252

Full Voltage 5.10 11111111 255

The following equations were used to calculate the values in Table 1. The sample calculations can be found in the appendix in equations 2.

Binary Output Ndecimal= d7 d6d5d4d3 d2 d1d0

An 8 bit binary number is a combination of 1s and 0s representing the digital states on and off

(1)

Converting Binary Output to Decimal Equivalent

Ndecimal= d7 27 + d6 26 + d5 25 + d4 24 + d3 23+ d2 22 + d1 21 + d0 20

(2)

According to Figure 1 voltage input is directly proportional to the binary output; as the voltage

increased the decimal value increased.

Figure 1: Plot of the Binary Voltmeter Decimal Value (output) vs. the Voltage Input (volts)

Table 2 below describes the energy intensity of an overhead flood light at varying distances from the Silicon Photodiode on the Light Sensor Board. As light hits the silicon chip it generates a current

which the Op-Amp in turn uses to display a voltage with the binary voltmeter. As the distance away from the light sensor decreases both voltage (displayed as LED binary output) and light intensity (Watts/m2) increase. Therefore voltage (decimal value) and light intensity are directly proportional.

Table 2. Light Intensity versus Distance

Distance (inches)

Binary Number

Decimal Number

Light Intensity (Watts/m2)

23 00010010 18 90 19 00011110 30 150 15 0100000 64 320 11 10100110 166 830 7 11110011 243 1215

The following equations were used to calculate the values in Table 2. The sample calculations can be found in the appendix under equation 3.

Intensity of Light Measured

Isolar = 5.0 Ndecimal

Related to the output of the Binary Voltmeter by the following equation and reported in Watts/m2

(3)

According to Figure 2 distance is indirectly proportional to light intensity. As the distance away from the light sensor decreases both voltage light intensity (Watts/m2) increases.

Figure 2: Plot of Light Intensity of an overhead flood light(Watts/m2) vs. the distance (inches) that

source is away from the Light Sensor Board. The power output of Solar Cell A in relation to resistance is presented in Table 3. A light source was placed 7 inches away from the solar cell and voltage was measured directly with the Digital Multimeter for different resistance settings. The power output across Solar Cell A was then calculated using equation 4. Whereas voltage and resistance were directly proportional to one another, powers relationship to these two variables was non linear. Instead, the trend was slightly parabolic with power steadily increasing with voltage and resistance until finally reaching a peak at setting 7. The maximum power output of solar cell A was found to be 0.301 W which corresponded to a resistance of 10 Ohms.

Table 3. Power Output of Solar Cell A as a function of Resistor Value

Resistance Setting (Ohms) Switch Number Voltage (V) Power = V2/R (W)

1 3 0.210 0.044

3 5 0.621 0.129

5 6 0.912 0.166

10 7 1.736 0.301

15 8 1.840 0.226

20 9 1.880 0.177

30 10 1.920 0.123

50 11 1.950 0.076

Switch Number & Resistance for Maximum Power Output Switch # 7 Resistance 10 ohms

The following equation was used to calculate the values in Tables 3-6. The sample calculations can be found in the appendix under equation 4.

Power output from the solar cell

pout=v2/R Power output from the solar cell = Voltage across the solar cell squared the resistance.

(4)

Figure 3 on the following page shows the solar cell producing maximum power output when there is a resistance load of 10 ohms. This marked decrease following resistance values of 10 ohms should be taken into consideration when installing the panel. Power output may be significantly reduced for resistance values greater than 10 ohms. The design of solar cells should therefore seek to use this maximum output as their benchmark for continued performance.

Figure 3: Plot of power Output in watts of Solar Cell A vs. the resistance load (ohms)

Table 4 displays the power output of Solar Cell A under different simulated light conditions. The distance of the overhead light to the Light Sensor on the Solar Meter breadboard was adjusted until the binary code of 1100xxxx corresponding to a bright sky or rather a light intensity of 1000 W/m2, was displayed with the LEDs. The solar meter was replaced with Solar Cell A and the voltage output was recorded with the digital multimeter. Finally the power output in relation to this value was calculated from equation 4. This same methodology was repeated for a cloudy day or a light

intensity of 250 W/m2 (0011xxxx). The power output associated with a bright day is over six times greater than that of a cloudy day.

Table 4. Power Output of Solar Cell A as a function of light conditions

Incident Sunlight Intensity Solar Meter Binary Reading Voltage (V) Power = V2/R (watts)

Bright Sky (1000 W/m2) 11001000 1.75 0.30625

Bright Sky Distance from Light to Solar Cell (inch) h = 8.5 inches

Cloudy Sky (250 W/m2) 00110101 0.68 0.04624

Cloudy Sky Distance from Light to Solar Cell (inch) h = 18 inches

As figure 4 demonstrates solar intensity is directly related to power. The greater the light intensity the greater the power output. Bright clear skies offer no obstruction to solar intensity and therefore produce the greatest power output (0.30625 Watts). Light can still penetrate through cloudy skies (0.04624Watts) but the reduction in power output is significant with a difference in 0.26001 Watts when compared to that of the bright day. Geographical considerations are therefore very relevant when determining the optimal placement of solar cells. In order to obtain the greatest power output, solar cells should ideally be placed in equatorial regions which receive the greatest solar intensity or a region whose climate is fairly mild in order to prevent the reduction in power output associated with a cloudy day.

Figure 4: Plot of the Intensity in (W/m2) of a Bright sky and Cloudy sky vs. the power calculated

produced in watts.

Table 5 displays the the Power Output of Solar Cell A at various sunlight incident angles at maximum light intensity. Resistance was set to the Maximum Power Output switch setting (7) and the light source was adjusted until the LEDs on the Solar Meter breadboard displayed the maximum light intensity or the maximum binary output value. The breadboard was then replaced with Solar Cell A and the Power Output (watts) was calculated from voltage readings taken for four different sunlight incident angles (degrees). Zero degrees is meant to represent noon when light intensity is

greatest. Power output was greatest when the incident angle of sunlight was zero and decreased slightly as the angle increased. This is reasonable, as increasing angles (up to 90 degrees) represent sunlight incident angles that are approaching either sunrise or sunset. Calculated b-values are slightly useful in predicting power output however there appears to be a degree of variation between the predicted and observed values. Error in observed measurements could be due to the noise in the system. As with intensity geographic location can also influence the sunlight incidence angle. Consider the angle at which the Sun's rays strike the Earth's surface. When the sun is directly overhead or 90 degrees from the horizon the incoming solar radiation is most intense. However if it is angled out to 45 degrees the light is spread out over a larger surface area and the resulting intensity is therefore less direct and reduced significantly. Therefore when choosing a location to place the the solar panel, an equatorial region that receives the most direct sunlight intensity or tilting the panel to reduce the total surface area created by the sunlight angle of incidence would be preferable.

With new technology, automatic controls can also be placed on the panel so that it tilts as the sun moves along its trajectory, reducing the potential solar energy lost as a product of the dispersion due to the sunlight angle of incidence. The power output at maximum intensity is comparable to that of a bright day as observed in table 4. Overall Solar intensity appears to play a greater role than the sunlight incident angle in terms of power output. When the power output for various angles is at maximum intensity compared to that of the direct intensity of a cloudy day, the the power output for the maximum intensity is still significantly greater. In other words, even though the sunlight incident angle disperses the intensity it still not enough to alter the power input significantly from a bright day with direct sunlight as when compared to a cloudy day.

Table 5. Variable Incident Light Angle Conditions (Task 3) Solar Cell A

Sunlight Incident Angle (degrees) Voltage ( V ) Power = V2/R (watts) b-component

0 1.720 0.296 0.296

15 1.712 0.293 0.223

30 1.665 0.277 0.043

45 1.577 0.249 0.131

The following equation was used to calculate the b component in Tables 5 and 6

b-component p=P(0) cos () (5)

Figure 5 on the following page shows that as the sunlight incident angle increases the voltage and power decreased. This indicates the most power is produced when the sunlight incident angle is at zero degrees.

Figure 5: Plot of the Sun Incident Angle in degrees vs. the calculated power produced in watts at the

maximum light intensity.

Table 6 displays the Power Output of Solar Cell B at various surface inclination angles (degrees). Solar Cell B is designed slightly differently from A in that it has a fisheye lens that is less sensitive angular variation. Resistance was set to 20 ohms and the height of the lamp was adjusted to

simulate cloudy conditions. Voltage was measured using the digital multimeter and the corresponding Power output across Solar Cell B was calculated for four different surface inclination angles. Power output was greatest at lower surface inclination levels. The b-component was slightly useful for predicting power values however there is still some variation between observed and calculated power values. Again this may have to do with the noise in the electronic circuit and the precision of the measuring instruments. Aside from the power value at 45 degrees, the outputs do not exhibit as much variation when compared to those in Solar Cell A. However with only four values and one possible outlier it is difficult to conclude whether Solar Cell B is less sensitive than A. Power output under bright conditions is nearly double on a bright day regardless of angle.

Table 6. Variable Incident Light Angle Conditions (Task 4) Solar Cell B

Surface Inclination Angle (degrees) Voltage ( V ) Power = V2/R (watts) b-component

0 1.476 0.109 0.109

15 1.475 0.109 0.083

30 1.435 0.103 0.016

45 1.335 0.089 0.047

Figure 6 demonstrates that Power input decreases as the sunlight incident angle shifts away from the vertical angle of 90 degrees under cloudy conditions in Solar Cell B.

Figure 5: Plot of the Sun Incident Angle in degrees vs. the calculated power produced in watts.

Table 7 displays the energy efficiency of Solar Cell A with a solar intensity of 1000 watts/m2 or bright sky conditions. Solar cell efficiency can be calculated by dividing the cells power output at the maximum power point by the amount of light received across its surface area. Solar Cell A has an energy efficiency of 17.4%. This represents the percent of solar energy that is converted into electrical energy.

Table 7. Efficiency of Solar Cell A (Bright Sky Condition)

Active area of Solar Cell A (m2) 0.00176

Light meter reading (Decimal) 200

Solar Intensity (watts/m2) 1000

Incident Power (watts) 1.76

Solar Cell A Output Power (watts) 0.306

Solar Cell A Efficiency (%) 17.4

The following equations as well as equation 4 were used to calculate the values in Tables 7,8 and 9. The sample calculations can be found in the appendix under equation 6, 7 and 8.

Total Active Area

Total active area = Incident Watts per Square Meter x area of solar cell A

(6)

Total incident Power (watts)

Incident Power = Incident Intensity x Area

Incident Power = Incident watts per square meter x area of solar cell.

(7)

Power efficiency of the solar cell

Power Efficiency = pout / pincident

Power efficiency of the solar cell = Power output from the solar cell Power incident from solar cell.

(8)

Table 8 displays the energy efficiency of Solar Cell A with a solar intensity of 250 watts/m2 or cloudy sky conditions. Using the same methodology power efficiency was calculated as 10.5 %. Solar cell A can convert 10.5% of the solar energy absorbed into usable electric energy on a cloudy day. The lower efficiency observed is due in large part to the lower solar intensity (250 watts) received by the cell.

Table 8. Efficiency of Solar Cell A (Cloudy Sky Condition)

Active area of Solar Cell A (m2) 0.00018

Light meter reading (Decimal) 53

Solar Intensity (watts/m2) 250

Incident Power (watts) 0.044

Solar Cell A Output Power (watts) 0.04624

Solar Cell A Efficiency (%) 10.5

Table 9 displays the energy efficiency of Solar Cell B with a solar intensity of 250 watts/m2 or cloudy sky conditions. Solar cell B appears to be less efficient than Solar Cell A it is has only a 4.3% efficiency compared to that of cell A with 10.5% under the same cloudy sky conditions.

Table 9. Efficiency of Solar Cell B (Cloudy Sky Condition)

Active area of Solar Cell B (m2) 0.00431

Light meter reading (Decimal) 53

Solar Intensity (watts/m2) 250

Incident Power (watts) 1.078

Solar Cell B Output Power (watts) 0.046

Solar Cell B Efficiency (%) 4.3

Potential sources of error could have arisen from the noise in the circuit and the precision of the digital multimeter. Since the decimal value is an approximation of the voltage, precision in calculations may be reduced. In addition to this, there was a significant amount of noise which caused readings of LED displays to flicker between two lights. In response the greater value was selected leaving uncertainty in the true value of the binary number and any calculations derived from this.

As stipulated the average energy used per person in an Ohioan home is 33.15 kilowatt-hours per day. Assuming the solar cell would experience 6 hours of sunlight and has an efficiency of 17.4% the active area of solar panels required to provide enough energy to power an already constructed/pre-existing house in Columbus, Ohio is 31.8 m2 (see sample calculations 9-11) on a bright day. On a cloudy day the active area of solar panels required to provide enough energy to power an already constructed /pre-existing house in Columbus, Ohio is 210.5 m2(se sample calculations 12-14) .

Table 10: Financial Projections for Grid-tie solar installations 4

The feasibility of building new homes with solar cells depends greatly on the cost. Costs associated with labor may be greatly reduced if it can be added as an additional project to a construction site. Table 10 displays financial projections associated with system size. The 7,200 - 12,000 system appears to accommodate the 33.15 kilowatt-hours per day but the cost ranges between $11,250 - $21,500. Although this is feasible buyers may not want to make such a large upfront cost.

Conclusions & Recommendations -

This lab was an introduction into the understanding of solar of power and its uses. In experimenting with a strip of solar panels, we as a group were about to understand how solar power is measured. Some this measurements included values for power output with the use of constant values. These constant values include data from light given by the sun on a bright day, as well as a cloudy day. By relating reading with the constant values of light intensity which proportional to the incident of light, we can determine the amount of energy given out in the electrical system. It also important to note that the typical Ohioan home consumes on average 33 Kwh of power, when taking the effectiveness of solar power. Solar panels would about require an active area of about 31.8m^2 per home-210.5m^2 on cloudy day-to supply enough energy to residents.

The calculated results of solar power on bright day with an intensity of 1000Watts/m^2, yields an average voltage 1.75 volts and a power output of .30625 watts. On the other hand on a cloudy day, the intensity is 250Watts/m^2, and yields a voltage of .68 volts and power of .04624 watts. As seen in the data the amount of power disseminated from the solar panel is also proportional to the angle. Larger angles will receive less voltage and power. In terms of bright versus cloudy skies, their efficiencies are 17.4%, and 10.5% respectively. In other words bright skies are more efficient than cloudy skies.

As a an engineering, I would not recommend solar panels in these homes. Solar Panels would require a large amount of space, and most residential home do not have these spaces. The consistency of sunlight plays also plays an important role. In the West Coast, states typically experience between 200-300 sunny days, while states in the East coast experience 60-100 days of sunshine. The geographical location has an impact on the effectiveness of solar power in states like Ohio as it is located in the east. About 1100kwh monthly would be necessary in running the average Ohioan home which would cost between $12,000- $15,000, a rather expensive venture.

References: http://www.currentresults.com/Weather/California/annual-days-of-sunshine.php Solar Cell Lab data.

Conclusion & Recommendations -

This lab enabled the group to understand how the production of solar power and efficiency varies due to multiple factors. This is a problem for determining whether or not a homeowner should install solar cell. However, to decide whether new and/or already existing homes should install solar panels we engineers analyzed the following information:

How many average kilowatt hours does a home in Columbus Ohio use per day? How large solar panel must be to obtain average kilowatts hours per day? How many sun hours do Columbus Ohio homes get per day on average? How much does it cost to install solar cell? How much does it offset the energy bill? (Is it worth installing?)

http://www.currentresults.com/Weather/California/annual-days-of-sunshine.php

This group of engineers determined it important to note there is a particle resistance load required in order for the solar cell to maximize power output. This could vary depending on the solar cell but in this lab the resistance load of 10 ohms arranged for a maximum power output of .301 watts. Sunshine is the most important factor when determining whether or not to install a solar panel. The solar panel needs an abundant of sunlight in order to receive a significant amount of electricity. On a bright sunny day, the sunlight incident intensity is about 1000 Watts/m2 , which produces a voltage of 1.75 and calculated power of .30625 watts. On a cloudy day, the sunlight incident intensity is about 250 Watts/m2 , which produces a considerably less voltage of .68 and calculated power of .04624 watts. Another key factor is the angle at which the sunlight incident is hitting the solar panel. The more of the solar panel that is directly absorbing sunlight the better because larger the power out the solar panel with create. At zero degree sunlight incident on a bright sunny day, sunlight hitting straight down directly on solar panel, the solar panel produced 1.72 volts and a calculated power of .296 watts. While a sunlight incident angle of 45 degrees produced 1.57 volts and a calculated power of .249 watts. Even on a cloudy day the sunlight incident angle affects the power produced. At zero degree sunlight incident on cloudy day the solar panel produces 1.47 volts and a calculated power of .109 watts. However, when the sunlight incident angle is 45 degree, the solar panel produces 1.33 volts and a calculated power of .089 watts. However, it is important to note that when the power output for various angles is at maximum intensity compared to that of the direct intensity of a cloudy day, the power output for the maximum intensity is still significantly greater. What this means is that sunlight intensity is more valuable to have when trying to produce more power than worrying about the sunlight incident angle. The efficiency of the bright sky is 17.4 percent, while the efficiency of a cloudy day is 10.5 percent. This is important to note because the bright sky will produce more power than a cloudy day and in the long run will save a person more money on their electric bill.

As specified in the lab procedures the average energy used per person in an Ohioan home is 33.15 kilowatt-hours per day and the solar panel experience 6 hours of sunlight with an efficiency of 17%. In conclusion, after analyzing the results the active area of solar panels required to provide enough energy to power an already constructed house in Columbus, Ohio on a bright day is 31.8 m2, and on a cloudy day the active area of solar panels required to provide enough energy to power an already constructed house is 210.5 m2 . The sizes required of the solar panels are too large in order to meet the 33.15 kilowatt-hours per day. Already constructed homes do not have enough roof space to hold the solar panels. As an engineer I would not install the solar panels because the solar panel would require being so large in order to reach 33.15 kilowatt-hours per day. To financially install solar panels of these sizes it will cost thousands of dollars. To satisfy 5.525 watt-hours per day would cost a household between $8,000 and $12,000 per person. A family of four is looking to pay around between $32,000 and $48,000 to install solar panels. Overall, Ohio is not suited for solar panels because of the solar panels size requirements and the costs to install.

Conclusion & Recommendations -

This lab demonstrated the factors and concerns in the applicable use of solar power. Practical implementation of solar energy is very important to lowering pollution and combating the effects of climate change. Solar cells only produce pollution in their manufacturing cost. After installation solar cells produce clean free energy. Solar cells do have serious drawbacks that have prevented

them from more widespread use. They have a very high initial cost. With efficiency ratings at or below 20 percent, most solar cells only produce viable amounts of power in certain areas with a great deal of sunlight. The application of solar cells requires many considerations.

Amount of sunlight available to the solar cell The space available for solar panels per application The cost of each installation The cost-benefit ratio of installing the panels.

Using the considerations that the average energy used per person in an Ohioan home is 33.15 kilowatt-hours per day and the solar panel experience 6 hours of sunlight with an efficiency of 17%, the feasibility of installing solar panels in a mid-Ohio home can be deduced. To produce power for one home on a bright sky day, it would require 31.8 square meters of solar panels. That is roughly the size of the average American homes roof. But, under cloudy sky conditions the solar panels size would need to be 210 square meters. That is roughly the size of the average Americans front yard. Mid-Ohio experiences roughly a fifty-fifty split in cloudy to sunny days. The initial cost to install such solar panels would be extremely expensive. Roughly ten thousand dollars per person in the home.

Solar panels are not currently feasible as a main source of energy in mid-Ohio. The amount of overcast days and the extensive size of panels are the main challenges to implementation. I do not recommend the use of solar cells with current technology in mid-Ohio.

1. Current Results. 2015. Days of Sunshine Per Year in Ohio. http://www.currentresults.com/Weather/Ohio/annual-days-of-sunshine.php 2 Grounds-mag. 2015. Average american yard size. http://www.grounds-mag.com/mag/grounds_maintenance_lawn_size/

Conclusion & Recommendations -

The experiments carried out in this lab provided the team with the basic understanding of the requirements of a solar cell to generate energy for the average household. The use of solar energy is very beneficial as it is safe for the environment and also comes from a free source. However, the group learned from this lab that generating electricity from solar cells is influenced by a lot of factors such as lights intensity, the angle at which sunlight hits the solar panels, the size of the solar panels and the availability of sunlight.

On a sunny day, the solar intensity is 1000 watts/m^2 and it is able to generate about 0.306 watts of power. and the solar intensity on a cloudy day is about 250 watts/m^2 and its able to generate only 0.046 watts of power. In order to meet the energy needs of the average ohion( that is; 33.15Kwh per day),A solar panel of 31.8 m^2 with an efficiency of about 17.4% is needed on a sunny day. And 210m^2 panel is required on a cloudy day.

since the weather conditions in ohio fluctuates between sunny and cloudy conditions, a solar panel of about 210 m^2 or more will be needed to provide power for the average home in ohio. this size

http://www.currentresults.com/Weather/Ohio/annual-days-of-sunshine.phphttp://www.grounds-mag.com/mag/grounds_maintenance_lawn_size/

does not seem feasible in that there are no homes big enough to house these kind of panels. Also the average cost of installing solar panels big enough to generate 20 -24Kwh of power in the United States is between $15,000 - $20,000(source: http://www.solarpanelscostguide.com/) .Hence the Mayor will also have to take into consideration the cost involved in installing these solar panels. Taking into consideration the size requirement and the cost of installation of the required solar panels, it is not feasible to install solar panels on the roofs of already constructed/ pre existing houses. However it might be possible to install solar panels on the roofs of new homes if the size and cost problem can be addressed. I would not recommend neither the installation of solar panels on the roofs of pre existing houses nor new ones because it will be very expensive to set up a new house to meet the size requirements and almost impossible to do on an already existing house.

Conclusion & Recommendations - This lab enabled us to understand concepts related to solar energy including solar intensity, electrical resistance, and incidence of sunlight angle as it relates to power output. The results indicated that there are particular conditions that optimize power output and can be used to determine feasibility of solar panel installations. The following recommendations can be used by Mayor Taylor as she decides whether to support funding installation projects.

In order to meet the energy demands of a typical household in Ohio (33.15 kilowatt-hours per day) a solar panel must be at least 31.8m2 to provide optimal power output on a bright day. On average the number of sunny days (either sunny or partly sunny) in Ohio Ranges from 160 to 1801. In addition to this this, the average household size in the Midwest is reported to be 2,265ft2 according to the last government census2. With that in mind the 210.5 m2 size needed for cloudy days would not be preferable. This in turn speaks to the feasibility of the plan. With so many cloudy days in Ohio and the larger surface area required to satisfy these demands questions arise as to both the logistics and the demand. If enough energy can be generated on bright days to make up for the reduction seen on cloudy days than it may be feasible.

On clear day, the sunlight incident intensity is about 1000 Watts/m2 corresponding to a power output of 0.30625 watts. On a cloudy day, the sun's intensity is reduced to 250 Watts/m2 resulting in a decreased power output of 0.04624 watts. The unit would have to produce 5.525 KW/hr on average to supply 33.15 kilowatt-hours per day. If the power observed for a bright sky is maintained consistently than it may produce enough to account for dark photoperiods. Since the team did not examine power output over time, specific recommendations regarding the ability of these solar cells to maintain consistent power output or store enough energy to account for cloudy days can not be made definitively at this time. There are also several other considerations when determining both the design and optimal placement of the panels. Two clear trends emerged from the data. Power output increases both with solar light intensity and the angle of incidence that is most direct. Therefore it is essential that the placement of solar panels are such that it reduces the solar dispersion caused by the angle of incidence. In Ohio the sun is always in the southern half of the sky. Therefore panels should be placed facing the south to receive direct sunlight and tilted such that it reduces the angle of sunlight incidence. If roofs do not face south than this can be compensated for by using more panels or

http://www.solarpanelscostguide.com/

larger collectors. It may be possible to place these panels on a fixed mount or a southward facing wall. Expenses and efficiency associated with those modifications would have to be assessed before recommending this course to current homeowners. It is advisable that new development focus on southern facing roofs. As evidenced by this data a resistance value of 10 ohms is the most ideal as it maximized power output. Adjusting resistor values to anything greater than this will result in a marked decrease in power output. Therefore when choosing the panel, selecting for the higher efficiency unit may be best especially if cloudy days become an issue. Again this will up the cost but may solve any issues associated with low intensity days. It appears that this is feasible however significantly more relevant information is needed before any positions can be supported. For example, what is the size of most houses in the area? If they can house a set of solar panels, is the roof in the proper condition. If not being housed on the roof do they have the room to install these units elsewhere? Are there members in the community that can afford the upfront investment? Cost is a serious investment with totals ranging between $15,000 and $40,000 when all things are considered3. It is highly recommended that the Mayors team conduct a survey in the local community to gauge interest. Regardless of tax incentives there will still be a rather large upfront cost. If no one is interested in paying or even learning more about different energy options than the campaign platform should be shifted to something the community is interested in. Until more information on this becomes available it is recommended that this is tabled until community interest is established and long term energy outputs can be assessed. References 1. Current Results. 2015. Days of Sunshine Per Year in Ohio. http://www.currentresults.com/Weather/Ohio/annual-days-of-sunshine.php 2. U.S. Census 2010.Median and Average Square Feet of Floor Area in New Single-Family Houses Completed by Location. https://www.census.gov/const/C25Ann/sftotalmedavgsqft.pdf 3. Energy Informative. 2015.How Much Solar Panels Cost. http://energyinformative.org/solar-panels-cost/ 4.SolarPanelCostGuide.com. 2015. Complete Solar Panel Cost Guide. http://www.solarpanelscostguide.com/

Appendix

Sample Calculations

Decimal Equivalent Ndecimal= d7 27 + d6 26 + d5 25 + d4 24 + d3 23+ d2 22 + d1 21 + d0 20

(2)

Example: 10011000

http://www.currentresults.com/Weather/Ohio/annual-days-of-sunshine.phphttps://www.census.gov/const/C25Ann/sftotalmedavgsqft.pdfhttp://energyinformative.org/solar-panels-cost/http://energyinformative.org/solar-panels-cost/http://www.solarpanelscostguide.com/

Ndecimal= 1 128 + 0 64 + 0 32 + 1 16 + 1 8 + 0 4 + 0 2 + 0 1

Ndecimal= 152

Intensity I=5 Ndecimal (3)

Binary Number = 00010010

I=518 = 90 (W/m2)

Power P=v2/R (4)

v = 0.21, R=1

P=(0.21volts)2/1=0.044 watts

b-component p=P(0) cos () (5)

p= 0.103 cos(30)

p= 0.0159

Total Active Area Total active area = Incident Watts per Square Meter x area of solar cell A

(6)

Incident Watts per Square Meter = 1000 (W/m2) area of solar cell A = 0.00176 (m2) Total active area = 1.76

Total incident Power (watts)

Incident Power=Incident Intensity x Area

(7)

Incident Power=1000W/m2 x 0.00176m2

Incident Power= 1.76 Watts

Power efficiency of the solar cell

Power efficiency = pout / pincident (8)

Power efficiency = 0.306 Watts/1.76 Watts

Power efficiency= 0.174 x 100

Power efficiency= 17.4%

Incident Radiation Incident Radiation = Solar Cell Efficiency x Solar Intensity (watts/m2) 0.174 x 1000 W/m2 =174 W/m2 = 0.174 KW/m2

(9)

average output power over a day

average output power over a day = Total Kilowatt-hours needed per day. /Hour 33.15 KWh/6h = 5.525 KW

(10)

Solar Cell Area (Bright day)

Solar Cell Area= Rate of Energy Transfer /Incident Radiation 5.525 KW/ 0.174 KW/m2 = 31.8 m2

(11)

Incident Radiation Incident Radiation = Solar Cell Efficiency x Solar Intensity (watts/m2) = 0.105 x 250 W/m2 =26.25 W/m2 = 0.02625 KW/m2

(12)

Average output power over a day

average output power over a day = Total Kilowatt-hours needed per day. /Hour 33.15 KWh/6h = 5.525 KW

(13)

Solar Cell Area (Cloudy day)

Solar Cell Area (cloudy day) = Rate of Energy Transfer / Incident Radiation 5.525 (KW) /0.02625 (W/m2) = 210.5 m2

(14)