An Approach to Alternative Energy Solutions for the Home

Transcription

1 An Approach to Alternative Energy Solutions for the Home By Robert Rasmussen An Engineering Project Submitted to the Graduate Faculty of Rensselaer Polytechnic Institute in Partial Fulfillment of the Requirements for the degree of MASTER OF MECHANICAL ENGINEERING Approved: Dr. Gutierrez-Miravete, Project Adviser Rensselaer Polytechnic Institute Hartford, Connecticut August, 2011 (For Graduation May 2012)

2 Copyright 2011 By Robert Rasmussen All Rights Reserved ii

3 CO TE TS LIST OF TABLES..iv LIST OF FIGURES. v LIST OF SYMBOLS.. vi ACKNOWLEDGEMENT.vii ABSTRACT. viii 1. INTRODUCTION/BACKGROUND THEORY AND METHODOLOGY PHOTOVOLTAIC ANALYSIS METHODOLOGY ON OBTAINING DATA FOR A GIVEN HOME DATA AND ASSUMPTION FOR THE HOME PHOTOVOLTAIC ANALYSIS RESULTS SOLAR PANEL ARRANGEMENT ON ROOFS ENERGY AND USAGE RESULTS COST RESULTS OVERALL RESULTS CONCLUSION 20 REFERENCES..21 iii

4 LIST OF TABLES Table 1 Chart of Roof s Area and Angles Table 2 - Chart of Solar Insolation Levels for Hartford, CT Area Table 3 - Table of Inefficiencies due to Solar Cell Angle Table 4 Chart of Actual Data for 18 months of Electric Bills Table 5 - Chart of Needed Solar Panel Sizing/Area Table 6 Chart of Cost Results Table 7 Chart of Breakeven Costs iv

5 LIST OF FIGURES Figure 1: An example of a simple semiconductor [1] Figure 2: Construction of a Photovoltaic cell [4] Figure 3: Aerial view of home in Quaker Hill, CT [9] Figure 4: Map of Photovoltaic Solar Resource in US [7] Figure 5: 18 month Electric Charge v. Time Data Figure 6: 18 month Usage v. Time Data Figure 7: 18 month Temperature v. Time Data Figure 8: Break Even Chart for Different Offset Percentages v

6 LIST OF SYMBOLS β = Average daily energy usage (kwh/day) θ = Percentage electric bill will be offset by (dimensionless) µ = Peak Solar Radiation (1 kw/m 2 ) = Solar Insolation level (kwh/m 2 ) η = Efficiency of the system (dimensionless) vi

7 ACK OWLEDGME T The only one deserving of this acknowledgement is my wife Cassie-Ryan who has put up with me and my many hours of schooling to complete this degree. vii

8 ABSTRACT In this day and age, the prices of oil as well as electricity are continuously on the rise with what seems like no chance of stability. This has started to change the thoughts and ideas of many in terms of energy usage and the use of alternative ways to power their homes. In this project a standard home in the State of Connecticut will be analyzed on its annual electricity cost in terms of energy usage. Assumptions will be made based on actual data. With this data, analyses will be done to compare the use of a solar panel system to provide electricity as an alternative form of energy on this home. Solar power will be compared to the standard household that receives electrical power from the grid. This project will focus initially on the home s uses, setup, and location. This project will not only look at providing 100% of the energy needed, but also look into subsidizing the cost of the electrical bill. Initial setup costs, breakeven points, and longevity on solar panel technology will be discussed. Calculations including solar panel output and electrical efficiencies will all be performed. With this data, conclusive analyses will be drawn to show whether or not an alternative approach to obtaining power through a solar panel system is cost effective. All conclusions will be based on the same home in the same area using the same set of initial conditions. viii

9 1. INTRODUCTION AND BACKGROUND For the past ten years, the price of oil has been on a significant rise. The current price of a barrel of oil in 2011 is $ In 2006, the price of a barrel of oil was $ This equates to $65.03 due to inflation. Ten years prior in 1996, oil was at $ This equates to $29.32 taking inflation into account. [13] Oil and energy are such large topics in our country and rightly so. These topics are some of the most important ones even between politicians. It seems as though, no matter what, the United States can no longer rely simply on conventional power plants or any type of commercial plant for that matter. Society needs to learn about the issues of rising energy costs and figure out a way to take these issues into their own hands. Something needs to be done. United States citizens can start on a small scale level by applying the principles of using alternative ways to power their homes and not rely so heavily on the national or even global impact of fluctuating prices. This paper will examine a home in the State of Connecticut and figure out its electrical energy usage and come to a conclusion on how much energy is being consumed yearly along with the monetary impact. Given these sets of parameters, the use of photovoltaics (PV) which is more commonly known as solar panels will be analyzed. Using a cost-effective matrix, this project will figure out what is the cheapest and most efficient way to power a home; by supplementing electricity from the grid, or completely relying on alternative energy and divorcing from power lines altogether, or some combination in between. This project will set out to prove that one home can in fact reduce the consumption of energies currently being used in commercial power generation such as oil, coal, and nuclear. One can see from this paper that this can be proven to be true, along with saving money over time. 1

10 2. THEORY AND METHODOLOGY The method and approach that will be taken in this project is both that from an engineering analysis followed up with a cost-effective approach analysis. The major technology that needs to be discussed before going forward into any type of comparison and analysis is that of electrical solar power technology in the form of solar panels, also known as photovoltaics (PV). This type of technology will be used in analyzing the type of power one can draw from the sun in a given area. Once the basics of photovoltaics are discussed, an analysis of the data for the home under consideration will be discussed. It is at this point that power ratings and assumptions will be made into what the capacity of the PV system needs to be, along with the size required. Once this information is presented, choosing a solar panel system will be performed. This will be done with the use of sizing and power calculations. When the known size of the solar panel that can adequately support this system is chosen, calculations will show which choice is the best choice for the given situation. 2.1 PHOTOVOLTAICS (SOLAR PANELS) Photovoltaic is a term used for the technology of solar cells. It has been derived from the Greek photos meaning light and volt, a unit of measure for electric potential. PVs were first discovered in the late 1830s by Edmond Becquerel who discovered what we know as the photovoltaic effect by figuring out that the silver plates of batteries, when exposed to light, caused the battery voltage to increase. It wasn t until the 1950s that semiconductors, nonmetallic materials with properties that lie between that of a conductive and an insulative material, were developed by engineers and scientists at Bell Laboratories in New Jersey. Stepping forward a few decades, PV cells are now widespread and are used in many applications such as telecommunications, providing power in remote locations, spacecraft, road signs, etc. [3] Many of today s cells use semiconductor technology and rely on what is known as polycrystalline silicon. Polycrystalline silicon consists of small grains of silicon that form solar cell wafers. These wafers are then shaped into PV cells. Other technologies for PVs are monocrystalline silicon, gallium arsenide, thin film PV such as amorphous silicon and cadmium telluride. [3] 2

11 Next will be the more intricate aspect of how semiconductors work. Semiconductors are mostly made of silicon in industry today. There are two types of silicon semiconductors that need to be talked about; n-type and p-type semiconductors. N-type semiconductors are made of silicon that is doped with an impurity to allow free electrons in the material. This doping agent is usually phosphorous. P-type semiconductors are made of silicon that is doped with an impurity to enable the silicon to have a lack of free electrons. Boron is typically the doping agent for a p- type semiconductor. By sandwiching these two types together a p-n junction is created. This p-n junction creates electricity from the sun s rays from the energy that photons impart on the p-n junction. When the sun s rays, or photons, stream onto the p-n junction, this creates more excited electrons. Digging deeper into this technology, we can see that n-type silicon is doped with a material that can allow the crystal structure to have an excess of free electrons. Typically phosphorus is used. For the p-type junction, boron is used due to it only having 3 valence electrons and needing to share with the silicon. This creates electron holes in the crystal structure. In doing this it also leaves more holes, or electron voids. When light in the form of a photon hits the cell, it dislodges an electron and leaves a hole. The electron tends to migrate toward the n-type junction and the hole to the p-type junction. This buildup of electrons and holes create and electric field in the cell. By connecting a wire across the junctions, the electrons will flow from the n-type silicon to the holes in the p-type silicon. This current flow is how electrical energy is produced. This explanation can be seen in Figure 1 below. This is the basic process of how a solar panel or PV works. [3] Figure 1: An example of a simple semiconductor [1] 3

12 The construction of a PV is shown in Figure 2. The outer coating is a protective layer of glass. You then have an anti-reflective coating so as to maximize the amount of incident rays absorbed on the PV. A conductive mesh is put in place to allow for photons to travel in. The rest is the same as Figure 1. This is a simple yet typical example of what today s PV cells look like. Figure 2: Construction of a Photovoltaic cell [4] 2.2 METHODOLOGY ON OBTAINING DATA FOR A GIVEN HOME The home that will be used in this study is located in Southeast Connecticut. Solar tables will be used for this region of the country to obtain solar power from incident rays based on location. For actual calculations, actual solar insolation data will be obtained from tables. The home has Southwest facing roofs. 18 months of electrical bills will be used and analyzed giving important data such as total electrical usage in kilowatt hours (kwh), usage per month, cost per month based on current rates, and a temperature profile of usage versus outside temperature. The time period for this information is between December 2009 and May

13 2.2.1 DATA AND ASSUMPTIONS FOR THE HOME The following is data that will be used throughout this paper. First this paper will take a look at the home under consideration. The home is a 2 story colonial with a detached garage as seen below in Figure 3. Figure 3: Aerial view of home in Quaker Hill, CT [9] As can be seen in the figure, there are three potential roofs that can be used for a solar panel system. Below is a table of the areas of each roof along with the slope and the angle from due south: Roof umber Area (m 2 ) Slope ( ) Angle from Due South ( ) Roof (Southwest) Roof (Southwest) Roof (Southwest) Table 1 The total amount of area that is able to be used is the addition of all 3 roofs, or m 2. The slopes of the roofs vary from around 26 to 35. The position of these roofs are in the southwest direction approximately 75 from due south. They have little to no obstructions such as other 5

14 buildings, trees, utility poles, etc. The angle of the roofs from due south as well as the pitch will come into play later on in the calculations PHOTOVOLTAIC ANALYSIS When analyzing a solar cell for a home, there are many considerations that need to be understood and many assumptions that need to be made. The first thing that needs to be looked at is the sun s power coming in and radiating into the solar cell. This is known as solar radiation or solar insolation. This is measured in kwh/m 2 per day. Values can be found using a solar radiation map as shown in Figure 4 below. Figure 4: Map of Photovoltaic Solar Resource in US [7] As you can see from this map, the values in the United States for estimated solar radiation ranges from kwh/m 2 per day. If you were to look at where Connecticut is, this has a value roughly around 3.5 kwh/m 2 per day. This value is an average and varies throughout the year. In order to arrive at a more precise calculation, more accurate solar radiation data is required. This can be found Table 2 below. Month January February March April May June Solar Insolation (kwh/m 2 /Day) Month July August September October November December Solar Insolation (kwh/m 2 /Day) Table 2 6

15 Looking at Table 2, you can see the large difference in solar radiation values for Hartford, CT throughout the year. As you can see, levels are as low as 1.7 kwh/m 2 per day in January, and as high as 5.58 kwh/m 2 per day in June. Solar insolation, or solar radiation as talked about previously, is simply a measure of the amount of the sun s energy that is radiated on an area of the Earth in a given day. Because there is no sun during the night and minimal sun for most of the day, solar insolation is sometimes referred to as peak sun hours. This value is the amount of solar insolation a certain place or location would receive if you had maximum sunlight for a certain period of time. The maximum, or peak amount of solar radiation that the sun can impart is 1kW/m 2, therefore you can see that the number of peak hours of sunlight is the same numerical value as the solar insolation. You can see this by dividing solar insolation (kwh/m 2 ) by peak solar radiation (1kW/m 2 ) and arriving at hours of peak sunlight. Looking into how to actually calculate power output for a PV cell, the data in Table 4 of the Results section which shows 18 months of electric bills along with each month s corresponding usage data will be used. For the purpose of the data in Table 4, the column labeled Power (kw) shows average power (in kilowatts (kw)) for each month. This number simply reflects an average, and does not show any peak power for the month. When the column is added up and averaged out over the 18 month time period, a value of 0.97 kw is found as the average. This correlates to an average kwh rating per day of kwh. When determining the size of a solar panel, there are many factors that need to be taken into consideration. Some of the questions that need to be answered when choosing solar panels for a home are: how much area is needed for the panels? Or what percentage of the electric bill needs to be subsidized? Or how much up-front cost can be paid? These are all questions that this project will answer either due to the results or based on assumptions. To start out, calculations will be done for various subsidized percentages of the electric bill. This project will later determine cost analysis to see what the best case scenario is. In sizing a solar panel system for the home, Equation 1 is used to size the power output for the system. [14] 2 ( β )( θ )( µ ) ( kwh)( kw / m ) Size power = = = 2 ( Σ)( η) kwh / m kw [1] 7

16 The units for power are kilowatts (kw). In the equation shown, the units for β or average daily usage are kwh. For θ or % offset, this number is the percentage subsidized from the electric bill. µ is the peak solar radiation and is defined as 1 kw/m 2. Ɖ is solar insolation and this is measured in kwh/m 2. η is the efficiency of the system. This number comes from the following factors: Inverter for the system, weather conditions, soiling/dirtying of the panels as well as module inefficiencies. For the solar panel in Southeast Connecticut, the following are values for efficiencies:.95=inverter efficiency,.89=weather conditions,.95=soiling/module inefficiencies. Multiplying all these values, you arrive at a total efficiency of (.95x.89x.95) =0.80 or 80%. % offset is the percentage you want the solar panels to compensate for out of your total power. [5] One last number that goes into your efficiency is that of the angle of the panels. Shown in Table 3 below is efficiency based on both angle from due south and pitch angle. The top row of Table 3 is tilt from the horizon, or pitch angle, and the side column is angle from due south % 96% 100% 100% 93% 82% 66% 30 86% 95% 98% 97% 90% 79% 65% 60 86% 91% 91% 89% 80% 71% 59% 90 86% 85% 81% 76% 67% 58% 48% Table 3 [10] For the purposes of pitch angle for the home, 41 will be used. This is because the angle on a roof can be adjusted to receive maximum sun. For all three roofs as mentioned earlier, they are at an angle of 75 from due south, looking at this chart and interpolating between 60 and 90 for a pitch of 41, you come up with an efficiency of 82.5%. Multiplying this percentage to the 80% previously mentioned for the panel efficiencies, you arrive at an efficiency of 66%. This is quite a drop in efficiency just due to the orientation of the panels. This efficiency value is the final variable that is put into the equation to determine the needed power requirement for a solar panel system. 8

17 Now to give an example, using data from Table 4 from the Results section. Following along on this table, assume 60% compensation in the month of March as shown in the table. This will have a power need of 5.87kW as shown using Equation 1: (22.47)(1)(.6) Size power = = 5. 87kW [1] (3.48)(.66) Continuing on with solar system sizing, most solar panels on the market today are rated at approximately kw/m 2. Using this number we can assume that a roof area of m 2 is needed to support this power. As a note, Table 4 does not show this number; rather it shows an area based on an average over the 18 months. [5] This next section will discuss the cost aspect of the project. Based on research, a solar panel installation which includes, panels, all electrical components, and labor can run anywhere from $7-$9 dollars per Watt. This data was obtained in [5] For this calculation, a value of $10 per Watt will be used. This number will take into account any inflation along with any drop in price. This number is comparable to other prices when researched. Federal and state rebates are also available to help offset the cost of the system. [6] Federal rebates are approximately 30% of the startup cost. This is a number that has been researched and has been found to be accurate around different sites. For the State of Connecticut, they provide a rebate of $1.75/Watt for the first 5kW and $1.25/Watt for the next 5kW. This number is not to exceed $15,000. [12] 9

18 3. RESULTS When deciding what the actual results are to this paper, one must consider what it is that they are trying to come to the conclusion of. The results of this paper give actual data on a real life solar panel system installation on a real life home in southeastern Connecticut. The results show based on assumptions made earlier, the size the system needs to be, how much area on the roof is needed to support the panels, what percentage of the monthly electric bill can be subsidized, cost, and finally break-even points. For the purpose of this project, the main factor is roof area and all the losses associated with the alignment of the solar panels. Cost-effectiveness is also an important factor in the decision making aspect of whether or not alternative power using solar cells is the best for this home situation. 3.1 SOLAR PANEL ARRANGEMENT ON ROOFS As stated earlier, the total area for all 3 roofs is m 2. This area is the more limiting factor when reviewing results that cannot be exceeded. One other roof on the house may be used; however, to utilize this roof as a viable solar panel destination, a lot of money will need to go into tree trimming. This paper ignores this roof in its assumptions and only goes with the three under discussion. 10

19 3.2 ENERGY AND USAGE RESULTS The energy and usage results in this paper are shown in Table 4 below: Date Charge Usage (kwh) Usage (kwh/day) Power (kw) Number of Days Average Temp (F) December-09 $ January-10 $ February-10 $ March-10 $ April-10 $ May-10 $ June-10 $ July-10 $ August-10 $ September-10 $ October-10 $ November-10 $ December-10 $ January-11 $ February-11 $ March-11 $ April-11 $ May-11 $ Average $ Table 4 When looking over the 18 months under observation, Table 4 gives a comprehensive list of averages over this period. Average kwh usage came out to be kwh. This works out to be kwh/day. This table also shows that the home averaged just less than 1 kw over the month. One might see this and wonder why the system isn t sized for 1 kw. Remember that this is just an average and there are peak power times not only throughout the day but also throughout the year. Another piece of data that has been obtained is the average temperature. Figures 5, 6, 7 below show graphical data of cost, usage, and temperature per unit time. 11

20 Figure 5: 18 month Electric Charge v. Time Data 12

21 Figure 6: 18 month Usage v. Time Data 13

22 Figure 7: 18 month Temperature v. Time Data 14

23 These figures and graphs were made to show varying trends in a more graphical approach. This is not overly important to the analysis of this project, but it does provide some information regarding the why? aspect of power usage. In Table 5 shown below, a comprehensive results table clearly shows the size of the system needed for varying power percentage offsets as well as the area of roof required to supply the rated power for the system. Usage (kwh/day) Efficiency Solar Insolation (kwh/m 2 ) 0% 20% 40% 60% 80% 100% Size Needed (kw) Size Needed (kw) Size Needed (kw) Size Needed (kw) Size Needed (kw) Size Needed (kw) Date December January February March April May June July August September October November December January February March April May AVERAGE AVERAGE SIZE (kw) AVERAGE SIZE (m 2 ) Table 5 15

24 In Table 5, you can see again the 18 months under observation. The usage per day is taken from Table 4 shown in the Results section. The efficiency is 66% and was proven to be so earlier in this report. Solar insolation levels for the Hartford, CT area were also an input. The remaining columns in Table 5 are offset power requirements needed. In other words, based on that month s energy requirement, there was a needed power for that month. This was calculated for 0%, 20%, 40%, 60%, 80%, and 100% offset from the power for the month. The average size which is what the solar panel system should be sized for is shown at the bottom of the chart. As shown, for a 20% offset in power, a PV system of 2.55 kw is needed; where for 100% offset kw is needed. This information was calculated using Equation 1. Going back in the report and knowing that for your standard PV panel, you can get approximately kw/m 2, average areas are shown for each percentage offset in Table 5. Going back to the 20%, you can see that m 2 is needed to support this amount of power. For 100% offset, you will need m COST RESULTS Table 6 shown below breaks down the cost analysis for each offset scenario. Chart of Cost Results 0% 20% 40% 60% 80% 100% Cost $ , , , , , Cost with 30% Rebate (Fed) $ , , , , , Cost with CT Rebate $ , , , , , Table 6 Again, using the assumption as stated earlier that 1 Watt was $10.00 for installation cost as well as purchase cost, these are the results that were obtained. What is also shown is both the federal rebate of 30% along with the State of Connecticut s own rebate plan as well. As shown in Table 4, for 100% offset to the home s power, the startup cost is as much as $70, This seems 16

25 like a lot, however as shown later, the payoff will come in time. In the Overall Results section is Table 5. This table is purely a data table for breakeven costs and to support Figure 8. An assumed electric bill cost of $ is used as the monthly savings and helps develop Figure 5 as shown in Appendix C. 3.4 OVERALL RESULTS After gathering all the data, producing all the graphs and charts, final results have been made. Table 5 below is the data obtained to see at what year one would break even based on the data. Year 20% 40% 60% 80% 100% 0 -$13,450 -$26,851 -$41,531 -$56,207 -$70, $11,873 -$25,273 -$39,953 -$54,629 -$69, $10,295 -$23,695 -$38,375 -$53,051 -$67, $8,717 -$22,117 -$36,798 -$51,473 -$66, $7,139 -$20,539 -$35,220 -$49,895 -$64, $5,561 -$18,961 -$33,642 -$48,317 -$62, $3,983 -$17,384 -$32,064 -$46,739 -$61, $2,405 -$15,806 -$30,486 -$45,161 -$59, $827 -$14,228 -$28,908 -$43,584 -$58,259 9 $750 -$12,650 -$27,330 -$42,006 -$56, $2,328 -$11,072 -$25,752 -$40,428 -$55, $3,906 -$9,494 -$24,175 -$38,850 -$53, $5,484 -$7,916 -$22,597 -$37,272 -$51, $7,062 -$6,338 -$21,019 -$35,694 -$50, $8,640 -$4,761 -$19,441 -$34,116 -$48, $10,218 -$3,183 -$17,863 -$32,538 -$47, $11,796 -$1,605 -$16,285 -$30,961 -$45, $13,373 -$27 -$14,707 -$29,383 -$44, $14,951 $1,551 -$13,129 -$27,805 -$42, $16,529 $3,129 -$11,552 -$26,227 -$40, $18,107 $4,707 -$9,974 -$24,649 -$39, $19,685 $6,285 -$8,396 -$23,071 -$37, $21,263 $7,862 -$6,818 -$21,493 -$36, $22,841 $9,440 -$5,240 -$19,915 -$34, $24,419 $11,018 -$3,662 -$18,338 -$33, $25,996 $12,596 -$2,084 -$16,760 -$31, $27,574 $14,174 -$506 -$15,182 -$29, $29,152 $15,752 $1,071 -$13,604 -$28, $30,730 $17,330 $2,649 -$12,026 -$26, $32,308 $18,908 $4,227 -$10,448 -$25, $33,886 $20,485 $5,805 -$8,870 -$23,546 17

26 31 $35,464 $22,063 $7,383 -$7,293 -$21, $37,041 $23,641 $8,961 -$5,715 -$20, $38,619 $25,219 $10,539 -$4,137 -$18, $40,197 $26,797 $12,116 -$2,559 -$17, $41,775 $28,375 $13,694 -$981 -$15, $43,353 $29,953 $15,272 $597 -$14, $44,931 $31,530 $16,850 $2,175 -$12, $46,509 $33,108 $18,428 $3,753 -$10, $48,087 $34,686 $20,006 $5,330 -$9, $49,664 $36,264 $21,584 $6,908 -$7, $51,242 $37,842 $23,162 $8,486 -$6, $52,820 $39,420 $24,739 $10,064 -$4, $54,398 $40,998 $26,317 $11,642 -$3, $55,976 $42,576 $27,895 $13,220 -$1, $57,554 $44,153 $29,473 $14,798 $ $59,132 $45,731 $31,051 $16,376 $1, $60,710 $47,309 $32,629 $17,953 $3, $62,287 $48,887 $34,207 $19,531 $4, $63,865 $50,465 $35,785 $21,109 $6, $65,443 $52,043 $37,362 $22,687 $8, $67,021 $53,621 $38,940 $24,265 $9, $68,599 $55,199 $40,518 $25,843 $11, $70,177 $56,776 $42,096 $27,421 $12, $71,755 $58,354 $43,674 $28,999 $14, $73,333 $59,932 $45,252 $30,576 $15, $74,910 $61,510 $46,830 $32,154 $17, $76,488 $63,088 $48,408 $33,732 $19, $78,066 $64,666 $49,985 $35,310 $20, $79,644 $66,244 $51,563 $36,888 $22, $81,222 $67,822 $53,141 $38,466 $23,790 Table 5 As you can see from the data, the years are to the left and the monetary amount for each offset percentage is in the remaining columns. Where the value is negative, that is an amount that is still owed. Where it becomes positive is the year that you would break even at. Continuing on with the analysis of results, area of the roof is one of the main limiting factors. With an area of m 2, a 60% offset is the maximum percentage that the roof can support. This yields roughly a 7.5 kw system. Now that an offset has been determined, an analysis of the results of cost needs to be addressed. For this size system, with all the government rebates, a startup cost of approximately $41, will be needed. When looking at the 60% offset line (shown below in yellow) Figure 8 below, (derived from Table 5), you can see 18

27 that after approximately 27 years you will break even on the cost of the solar panel system. After that, all your savings is profit. Figure 8: Break Even Chart for Different Offset Percentages This seems like a long time, and actual for the life of a solar panel system, it is. Commercially, systems generally last approximately 25 years. [15] When you roll this factor into the mix, the logical choice would be to choose the 40% offset where you would break even after about 17 years and only have a start-up cost of $26, This choice is the one that makes the most sense for this home. 19

28 4. CONCLUSION When thinking about what conclusions to draw from this analysis, the first thing that comes to mind is is alternative energy really worth it? This question really has no right or wrong answer. For everyone and every situation, there is a different scenario. When looking back at this analysis of the home in southeastern Connecticut, if it was a home that was to be lived in for a long time, that is 20+ years, then yes, choosing a PV system that produces a 40% offset is worth it. However, some of the other options may not be. The variables can be time, money, labor, etc. In the media we seem to always hear about how to become a greener climate, and what you can do to make a change. Well, when it comes down to it, one can make a change. You can in fact lessen your carbon footprint. However there is one stipulation; this change comes at the expense of you! So one must ask themselves, do I value making the Earth a greener place or do I value my money? This is one thing that people must think about when choosing to go green. The other is change. One must make a lot of changes in their lives. If you are one to constantly move about, then maybe this type of system is not for you. Or, the funds just don t support all the up-front costs, and then maybe this type of system is not for you. The final conclusion that has been drawn on is this; there are many, many other sources of power out there that have a lot of power to give. It is time that society does start to utilize ALL its resources, at an affordable rate. What does this mean exactly? Well, it means different things for different people. Everyone has the capability to support themselves and not rely on others for power. Everyone also has the capability to make more of an impact on cleaner energies such as solar power. The final conclusion that can be drawn is this; the price of clean, self-sufficient energy is different for different people. 20

29 REFERENCES 1. CellsWork.aspx July 1, Duffie, John A. Solar engineering of Thermal Processes. Wiley. New York Boyle, Godfrey. Renewable Energy. Power for a Sustainable Future. Oxford University Press, United Kingdom, July 1, July 29, July 19, July 3, July 28, July 30, July 30, July 31, l&currentpageid=1&ee=1&re=1 July 31, August 11, August 17, August 17,