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Rasmussen, Robert
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)
© Copyright 2011
By
Robert Rasmussen
All Rights Reserved
ii
COTETS
LIST OF TABLES………………………………………………………………………………..iv
LIST OF FIGURES………………………………….……………………………………………v
LIST OF SYMBOLS………………………………………………………………………..……vi
ACKNOWLEDGEMENT……………………………………………………………………….vii
ABSTRACT………………………………………………………………….…………………viii
1. INTRODUCTION/BACKGROUND……………………………………………………........1
2. THEORY AND METHODOLOGY……………………………………………………..……2
2.1. PHOTOVOLTAIC ANALYSIS……………………………………………………..…...2
2.2. METHODOLOGY ON OBTAINING DATA FOR A GIVEN HOME………..…….....4
2.2.1. DATA AND ASSUMPTION FOR THE HOME…………………………………5
2.2.2. PHOTOVOLTAIC ANALYSIS………………………………………………......6
3. RESULTS………..………………………………………………………………………......10
3.1. SOLAR PANEL ARRANGEMENT ON ROOFS.……………………………………..10
3.2. ENERGY AND USAGE RESULTS………………………………………………..…..11
3.3. COST RESULTS……………………………………..…………………………………16
4. OVERALL RESULTS………………………………..……………………………………...17
5. CONCLUSION………………………………………………………………………………20
REFERENCES…………………………………………………………………………………..21
iii
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
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
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/m2)
∑ = Solar Insolation level (kWh/m2)
η = Efficiency of the system (dimensionless)
vi
ACKOWLEDGMET
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
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
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 $86.84. In 2006, the price of a barrel of oil was $58.30. This equates to
$65.03 due to inflation. Ten years prior in 1996, oil was at $20.46. 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
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
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 ptype 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
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 2011.
4
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 (m2) Slope (˚) Angle from Due South (˚)
Roof 1
26.01
28.8
75 (Southwest)
Roof 2
17.84
34.2
75 (Southwest)
Roof 3
26.76
34.7
75 (Southwest)
Table 1
The total amount of area that is able to be used is the addition of all 3 roofs, or 70.61 m2. 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
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.
2.2.2 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/m2 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 2.2-6.8 kWh/m2 per day. If you were to look at where Connecticut is, this has a
value roughly around 3.5 kWh/m2 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
Solar Insolation
(kWh/m2/Day)
January
February
March
April
May
June
1.70
2.43
3.48
4.07
5.14
5.58
Month
Solar Insolation
(kWh/m2/Day)
July
August
September
October
November
December
5.38
5.04
4.13
2.91
1.81
1.42
Table 2
6
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/m2 per day in January, and
as high as 5.58 kWh/m2 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/m2, 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/m2) by peak solar radiation
(1kW/m2) 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 23.24 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]
Size power
( β )(θ )( µ ) ( kWh)(kW / m 2 )
=
=
= kW
kWh / m 2
(Σ)(η )
7
[1]
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/m2. Ɖ is solar insolation and this is measured
in kWh/m2. 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.
0
15
30
41
60
75
90
0
86% 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
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:
Size power =
(22.47)(1)(.6)
= 5.87 kW
(3.48)(.66)
[1]
Continuing on with solar system sizing, most solar panels on the market today are rated at
approximately 0.108 kW/m2. Using this number we can assume that a roof area of 54.35 m2 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 2008. [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
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 70.61 m2. 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
3.2
ENERGY AND USAGE RESULTS
The energy and usage results in this paper are shown in Table 4 below:
Date
December-09
January-10
February-10
March-10
April-10
May-10
June-10
July-10
August-10
September-10
October-10
November-10
Charge
$128.66
$141.30
$111.68
$122.24
$117.67
$130.61
$127.60
$133.88
$129.89
$143.81
$111.54
$112.18
Usage (kWh) Usage (kWh/day) Power (kW) Number of Days Average Temp (F)
741
22.45
0.94
33
44.9
797
23.44
0.98
34
26.6
607
21.68
0.90
28
28.5
674
22.47
0.94
30
35.7
645
21.50
0.90
30
49.9
727
23.45
0.98
31
53.7
708
23.60
0.98
30
64
747
24.90
1.04
30
72.3
720
23.23
0.97
31
75.7
808
26.06
1.09
31
70.8
604
21.57
0.90
28
62.2
608
20.97
0.87
29
50
December-10
$127.51
705
21.36
0.89
33
40.4
January-11
$150.19
847
25.67
1.07
33
27.7
February-11
$136.00
808
26.93
1.12
30
22.1
March-11
$116.72
676
24.14
1.01
28
31.6
April-11
$111.97
644
22.21
0.93
29
40.8
May-11
$127.16
746
22.61
0.94
33
53.3
Average
$126.70
711.78
23.24
0.97
30.61
47.23
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 711.78 kWh. This works out to
be 23.24 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
Figure 5: 18 month Electric Charge v. Time Data
12
Figure 6: 18 month Usage v. Time Data
13
Figure 7: 18 month Temperature v. Time Data
14
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.
Date
December09
January-10
February-10
March-10
April-10
May-10
June-10
July-10
August-10
September10
October-10
November10
December10
January-11
February-11
March-11
April-11
May-11
AVERAGE
0%
20%
40%
60%
80%
100%
(kWh/m )
Size
Needed
(kW)
Size
Needed
(kW)
Size
Needed
(kW)
Size
Needed
(kW)
Size
Needed
(kW)
Size
Needed
(kW)
Solar
Insolation
Usage
(kWh/day)
Efficiency
22.45
23.44
21.68
22.47
21.50
23.45
23.60
24.90
23.23
0.66
0.66
0.66
0.66
0.66
0.66
0.66
0.66
0.66
1.42
1.70
2.43
3.48
4.07
5.14
5.58
5.38
5.04
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
4.79
4.18
2.70
1.96
1.60
1.38
1.28
1.40
1.40
9.58
8.36
5.41
3.91
3.20
2.77
2.56
2.81
2.79
14.38
12.54
8.11
5.87
4.80
4.15
3.84
4.21
4.19
19.17
16.71
10.81
7.83
6.40
5.53
5.13
5.61
5.59
23.96
20.89
13.52
9.78
8.00
6.91
6.41
7.01
6.98
26.06
21.57
0.66
0.66
4.13
2.91
0.00
0.00
1.91
2.25
3.82
4.49
5.74
6.74
7.65
8.99
9.56
11.23
20.97
0.66
1.81
0.00
3.51
7.02
10.53
14.04
17.55
21.36
25.67
26.93
24.14
22.21
22.61
0.66
0.66
0.66
0.66
0.66
0.66
1.42
1.70
2.43
3.48
4.07
5.14
0.00
0.00
0.00
0.00
0.00
0.00
4.56
4.58
3.36
2.10
1.65
1.33
9.12
9.15
6.72
4.20
3.31
2.67
13.68
13.73
10.08
6.31
4.96
4.00
18.24
18.30
13.43
8.41
6.61
5.33
22.80
22.88
16.79
10.51
8.27
6.66
23.24
0.66
3.41
AVERAGE
SIZE (kW)
AVERAGE
2
SIZE (m )
0.00
2.55
5.10
7.66
10.21
12.76
0.00
23.71
47.43
71.14
94.85
118.56
2
Table 5
15
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
12.76 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 0.108 kW/m2,
average areas are shown for each percentage offset in Table 5. Going back to the 20%, you can
see that 23.71 m2 is needed to support this amount of power. For 100% offset, you will need
118.56 m2.
3.3
COST RESULTS
Table 6 shown below breaks down the cost analysis for each offset scenario.
Chart of Cost Results
Cost
Cost
with 30%
Rebate
(Fed)
Cost
with CT
Rebate
0%
20%
40%
60%
80%
100%
$0.00
25,524.53
51,049.06
76,573.58
102,098.11
127,622.64
$0.00
17,867.17
35,734.34
53,601.51
71,468.68
89,335.85
$0.00
13,450.38
26,850.75
41,531.13
56,206.51
70,881.88
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 $70,000.00. This seems
16
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 $126.70 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
-$70,882
1
-$11,873
-$25,273
-$39,953
-$54,629
-$69,304
2
-$10,295
-$23,695
-$38,375
-$53,051
-$67,726
3
-$8,717
-$22,117
-$36,798
-$51,473
-$66,148
4
-$7,139
-$20,539
-$35,220
-$49,895
-$64,570
5
-$5,561
-$18,961
-$33,642
-$48,317
-$62,993
6
-$3,983
-$17,384
-$32,064
-$46,739
-$61,415
7
-$2,405
-$15,806
-$30,486
-$45,161
-$59,837
8
-$827
-$14,228
-$28,908
-$43,584
-$58,259
9
$750
-$12,650
-$27,330
-$42,006
-$56,681
10
$2,328
-$11,072
-$25,752
-$40,428
-$55,103
11
$3,906
-$9,494
-$24,175
-$38,850
-$53,525
12
$5,484
-$7,916
-$22,597
-$37,272
-$51,947
13
$7,062
-$6,338
-$21,019
-$35,694
-$50,370
14
$8,640
-$4,761
-$19,441
-$34,116
-$48,792
15
$10,218
-$3,183
-$17,863
-$32,538
-$47,214
16
$11,796
-$1,605
-$16,285
-$30,961
-$45,636
17
$13,373
-$27
-$14,707
-$29,383
-$44,058
18
$14,951
$1,551
-$13,129
-$27,805
-$42,480
19
$16,529
$3,129
-$11,552
-$26,227
-$40,902
20
$18,107
$4,707
-$9,974
-$24,649
-$39,324
21
$19,685
$6,285
-$8,396
-$23,071
-$37,747
22
$21,263
$7,862
-$6,818
-$21,493
-$36,169
23
$22,841
$9,440
-$5,240
-$19,915
-$34,591
24
$24,419
$11,018
-$3,662
-$18,338
-$33,013
25
$25,996
$12,596
-$2,084
-$16,760
-$31,435
26
$27,574
$14,174
-$506
-$15,182
-$29,857
27
$29,152
$15,752
$1,071
-$13,604
-$28,279
28
$30,730
$17,330
$2,649
-$12,026
-$26,701
29
$32,308
$18,908
$4,227
-$10,448
-$25,124
30
$33,886
$20,485
$5,805
-$8,870
-$23,546
17
31
$35,464
$22,063
$7,383
-$7,293
-$21,968
32
$37,041
$23,641
$8,961
-$5,715
-$20,390
33
$38,619
$25,219
$10,539
-$4,137
-$18,812
34
$40,197
$26,797
$12,116
-$2,559
-$17,234
35
$41,775
$28,375
$13,694
-$981
-$15,656
36
$43,353
$29,953
$15,272
$597
-$14,079
37
$44,931
$31,530
$16,850
$2,175
-$12,501
38
$46,509
$33,108
$18,428
$3,753
-$10,923
39
$48,087
$34,686
$20,006
$5,330
-$9,345
40
$49,664
$36,264
$21,584
$6,908
-$7,767
41
$51,242
$37,842
$23,162
$8,486
-$6,189
42
$52,820
$39,420
$24,739
$10,064
-$4,611
43
$54,398
$40,998
$26,317
$11,642
-$3,033
44
$55,976
$42,576
$27,895
$13,220
-$1,456
45
$57,554
$44,153
$29,473
$14,798
$122
46
$59,132
$45,731
$31,051
$16,376
$1,700
47
$60,710
$47,309
$32,629
$17,953
$3,278
48
$62,287
$48,887
$34,207
$19,531
$4,856
49
$63,865
$50,465
$35,785
$21,109
$6,434
50
$65,443
$52,043
$37,362
$22,687
$8,012
51
$67,021
$53,621
$38,940
$24,265
$9,590
52
$68,599
$55,199
$40,518
$25,843
$11,167
53
$70,177
$56,776
$42,096
$27,421
$12,745
54
$71,755
$58,354
$43,674
$28,999
$14,323
55
$73,333
$59,932
$45,252
$30,576
$15,901
56
$74,910
$61,510
$46,830
$32,154
$17,479
57
$76,488
$63,088
$48,408
$33,732
$19,057
58
$78,066
$64,666
$49,985
$35,310
$20,635
59
$79,644
$66,244
$51,563
$36,888
$22,213
60
$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 70.61 m2, 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 $41,500.00 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
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 $26,850.00. This choice is the one that makes the most
sense for this home.
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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.
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REFERENCES
1. http://kids.actewagl.com.au/education/energy/RenewableEnergy/SolarEnergy/HowSolar
CellsWork.aspx July 1, 2011
2. Duffie, John A. Solar engineering of Thermal Processes. Wiley. New York. 1980.
3. Boyle, Godfrey. Renewable Energy. Power for a Sustainable Future. Oxford University
Press, United Kingdom, 2004.
4. www.mrsunsolar.com/eletricinfo.php July 1, 2011
5. http://www.findsolar.com/index.php?page=rightforme July 29, 2011
6. www.solarpowerauthority.com July 19, 2011.
7. http://www.nrel.gov/gis/solar.html July 3, 2011
8. www.solarpanelsplus.com/solar-insolation-levels July 28, 2011
9. www.maps.bing.com July 30, 2011
10. http://www.solar-estimate.org/showfaq.php?id=58 July 30, 2011
11. http://www.solar-estimate.org/ July 31, 2011
12. http://www.dsireusa.org/incentives/incentive.cfm?Incentive_Code=CT10F&State=federa
l&currentpageid=1&ee=1&re=1 July 31, 2011
13. http://www.fintrend.com/inflation/inflation_rate/Historical_Oil_Prices_Table.asp August
11, 2011
14. http://pvcdrom.pveducation.org/SUNLIGHT/AVG.HTM August 17, 2011
15. http://solarpanelsathome.org/saving-money-with-a-residential-solar-power-system
August 17, 2011
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