This document investigates the optimal tilt and azimuth angles for fixed-mount photovoltaic (PV) systems to maximize peak shaving performance in Atlanta, Georgia. It analyzes hourly solar irradiance and azimuth angle data from 2009 to determine the optimal angles. The results found that a PV system with a tilt angle of 32.6° and azimuth angle of 214° provided the highest peak shaving, supplying 1.13% of Atlanta's peak energy demand. A non-reoriented PV system with tilt of 36.6° and azimuth of 180° supplied 1.07% of peak demand. While the non-reoriented system generated more total daily energy, the reoriented system was more effective for reducing costs during

1. Tilt and Azimuth Angle for Optimal Peak Shaving Performance of
Fixed-mount Photovoltaic Systems
Tianhao Li, and Edward (Ningyuan) Zhang
Georgia Institute of Technology Research Symposium, July, 2013
School of Electrical and Computer Engineering
Georgia Institute of Technology
Atlanta, GA 30332-0250
Abstract—Grid-connected, fixed mount photovoltaic (PV)
systems have the beneficial use of lowering the dependence of
American consumers on electric energy supplied by electric
utility companies during hours of high energy demand. Due to
the nature of PV arrays, it is hypothesized that there is an
optimal panel tilt and azimuth angle that will maximize the peak
shaving effect, which this paper will investigate. To find the
optimal panel tilt and azimuth angles, local hourly irradiance
and solar azimuth angle of a hypothetical PV system installation
in Atlanta, Georgia was found using the typical meteorological
year data (TMY3) from 2009. Programs were written using
MATLAB to determine the optimal panel azimuth and tilt angle
based upon a mathematical model of an energy demand load
curve. It was found that the highest peak shaving performance
of a PV system had a panel tilt angle of 32.6° and a panel
azimuth angle of 214°. At the optimal angles, the PV system
succeeded in providing for 1.13% of total electric energy
demand during Atlanta’s peak demand.
I. INTRODUCTION
Currently, photovoltaic (PV) systems provide less than
one percent of the total energy consumed in the United
States. To meet the total energy demand of the United States,
approximately 10,000 square miles of solar panels would
have to be constructed with the existing PV technology [1].
Although the technology to develop efficient PV systems
with enhanced solar capacity is still lacking, the utilization of
PV systems as an auxiliary energy source has beneficial uses
for American consumers.
One such use of PV systems is to reduce the cost of
electrical energy for consumers during hours of high energy
demand, known as peak demand. The demand for electrical
energy varies throughout a day. Typically, the peak demand
hours are between 12 PM and 6 PM [2]. Electric utility
companies charge consumers extra per kilowatt-hour during
the peak demand. This extra charge is referred to as the “peak
use charge.”
Conveniently, irradiance, which is the density of solar
radiation incident on a given surface, is highest during the
peak demand. PV systems can take advantage of the “peak
shaving” phenomenon, which describes how electric energy
produced by a PV system can lower a consumer’s
dependence on the energy supplied by utility companies
during the peak demand.
The amount of energy generated by a PV system directly
correlates to its peak shaving performance. The orientation of
a PV array is known to affect its ability to absorb solar
energy. Thus, the investigation of this paper is motivated by
the hypothesis that there will be an optimal orientation of the
PV array that will yield the best peak shaving effect.
For a fixed-mount PV system, the PV array orientation is
described by two angles:
1. Panel tilt angle: the angle between the array and the
horizontal axis (usually the ground) and
2. Panel azimuth angle (which is not to be confused with
solar azimuth angle): the angle between the projection of the
normal of the panel surface and the northern direction.
The relationship between the energy output of a PV
system and its orientation is given by the following
equations:
𝑐𝑜𝑠𝜃𝑖 = 𝑐𝑜𝑠𝜃𝑧 𝑐𝑜𝑠𝛽 + 𝑠𝑖𝑛𝜃𝑧 𝑠𝑖𝑛𝛽𝑐𝑜 𝑠(𝛾 − 𝛾𝑠) (1)
𝐼 = 𝐼 𝑑𝑛 𝑐𝑜𝑠(𝜃𝑖) (2)
Where:
𝜃𝑖 = the angle of incidence of the sunlight on the panel
𝜃𝑧 = the solar zenith angle
𝛽 = the panel tilt angle
𝛾𝑠 = the solar azimuth angle
𝛾 = the panel azimuth angle
I = the insolation
𝐼 𝑑𝑛 = the direct normal insolation
Figure 1 illustrates the various factors that affect a PV
system’s ability to absorb solar energy and ultimately its
ability to generate electric energy. A PV system where both
its panel tilt and azimuth angle are variables is referred to as a
reoriented PV system. A reoriented PV system is optimized
for producing the best peak shaving performance. A PV
system that only has its panel tilt angle as a variable is
referred to as a non-reoriented PV system.
Fig.1. Illustration depicting the various angles associated with the orientation
of a PV array [3]. Here, 𝛾 𝑃 and 𝛼 𝑃 + 180° are the panel tilt and azimuth
angle respectively.

2. The objective of this paper is to determine if there is
indeed a certain panel tilt and azimuth angle for grid-
connected, fixed-mount PV systems that will optimize their
peak shaving performance.
II. METHODOLOGY
Atlanta, Georgia was chosen as the location where the
hypothetical PV system will be installed. In order to find the
optimal panel azimuth and tilt angle for the PV system, a
source of local hourly irradiance and solar azimuth angles for
Atlanta was required. These data were obtained via the
typical meteorological year data (TMY3) from 2009 [4].
The interval of the high energy demand hours must also
be determined to ensure that the optimal panel azimuth and
tilt angle were tailored to generate the best peak shaving
performance for a realistic peak demand. The interval used in
this investigation is from 12 PM to 6PM [5]. Only residential
energy consumption was considered in this investigation.
To optimize the insolation, the panel azimuth and tilt
angle must be values that bring Equation 1 as close to one as
possible. The panel tilt angle was determined first for a non-
reoriented PV system. The panel azimuth angle 𝛾, now a
constant, is set to 180° due to the fact that Atlanta is in the
northern hemisphere.
For the case of a reoriented PV system, the panel azimuth
angle was determined first by systematic testing of different
angles using the MATLAB program in Appendix A. The
panel azimuth angle was varied in order to maximize the
overall energy generation of the PV system for the time
interval of 12:00 PM to 6:00 PM. The panel tilt angle was
determined afterwards through algebraic manipulation of
Equation 1. Reoriented PV systems maximize the peak
shaving effect at the cost of the total energy generation during
a day as opposed to non-reoriented PV systems that are
tailored to maximize the energy output for 24 hours. The
solar azimuth angle, 𝛾𝑠, and the solar zenith angle, 𝜃𝑧, were
obtained from the TMY3 data.
The MATLAB program in Appendix A takes in three
columns (the solar zenith angle, the solar azimuth angle, and
the direct irradiance) and 8760 rows (the distinct hours in a
year) of the numerical data in the TMY3 excel file with
respect to each corresponding hour of a day in a year. Every
hourly entry for the year is sampled separately. This allows
the computation of the average values for the solar azimuth
angle, irradiance, and zenith angle.
Using the vectors of the three average values that had
been obtained, the program evaluates a range of angles to
determine the panel tilt angle that maximizes the term
𝑠𝑖𝑛𝜃𝑧 𝑐𝑜𝑠(𝛾 − 𝛾𝑠) of Equation 1 in respect to the six hours
during the peak demand. The program then executes a loop
process to find the panel azimuth angle that maximizes the
result of Equation 1, or the 𝑐𝑜𝑠𝜃𝑖 value.
The comparison of the electric energy output between the
reoriented and non-reoriented PV system is made by
calculating the insolation optimized for the peak demand and
for 24 hours separately. The insolation is found by
multiplying the largest 𝑐𝑜𝑠𝜃𝑖 value with the direct irradiance
attained earlier. The current PV systems are around 19
percent efficient [6]. Therefore to calculate the energy output
of the PV system, the insolation is multiplied by 0.19. The
calculations of the insolation of the two PV systems are done
by the MATLAB program in Appendix B.
Evaluating the peak shaving effect of a PV system
requires a mathematical model of the energy demand curve.
The energy demand curve shown in Graph 1 is modeled by a
sinusoidal curve with a 24 hour period and a peak at 3 PM
every day of the year. The ratio between the maximum and
minimum point is two to one. This indicates that the energy
demand at 3 PM is twice as large as the energy demand at 3
AM. Although this is a very simple mathematical model of
the energy demand of a typical day in Atlanta, it is sufficient
in presenting the peak shaving effect of a PV system.
The annual residential electrical energy consumption of
Atlanta in 2005 was estimated to be around 23,377,516 MWh
[7]. The daily energy consumption is calculated to be a little
over 64,938 MWh. Since the energy demand curve is
approximated by a cosine function and the peak demand is
assumed to be from 12PM to 6PM, the percentage of the
energy demand during the peak hours with respect to the total
energy demand of the day is evaluated by taking the integral
of the following function and dividing it by twice of the
function evaluated from 0 to π:
∫ sin(𝑡) + 3
3𝜋/4
𝜋/4
𝑑𝑡 (3)
The energy demand during the peak demand is evaluated to
be around 32.5 percent of the total energy demand per day.
As a result, the energy demand during the hours of 12 PM to
6 PM in Atlanta is approximately 21,106 MWh.
Fig.2. Energy demand curve with peak demand hours from 12 PM to 6 PM.
An assumption that needed to be made in order to assess
the economical peak shaving effect of the PV system is the
size of the PV system to be installed. It’s not economically
feasible to install a PV system large enough to supply all of
Atlanta’s electrical energy demands during the peak demand.
Thus, it is much more realistic to assume that the PV system
will provide for a percentage of the total electrical energy
demand during the peak demand. Based upon the total
contribution of solar energy in relation to the total energy
2
2.2
2.4
2.6
2.8
3
3.2
3.4
3.6
3.8
4
3:00 7:48 12:36 17:24 22:12 3:00
Time (Hours)
Energy Demand Curve

3. generation in the United States, having the PV system
provide for one percent of the total energy demand during the
peak demand would be a good assumption [1]. A typical PV
system generates 60W per square meter [8]. Therefore the
size of our PV system would be estimated to be around
586,300m2
.
TABLE I. RESULT TABLE
PV System Optimal Panel
Azimuth Angle
(Degrees)
Optimal Panel Tilt
Angle (Degrees)
Non-reoriented 36.6 180
Reoriented 32.6 214
a.
Optimal panel azimuth and tilt angles for a non-reoriented and reoriented PV system. Respective
graphs are shown in Appendix C, D, and E.
Fig.3. Graph depicting total electric energy generated by a non-reoriented
and reoriented PV system per day for one square meter.
TABLE II. VALUES PERTAINING TO PEAK SHAVING PERFORMANCE
III. DISCUSSION
Graphs in Appendix C and D show that for a grid-
connected, fix-mount PV system, there exists an optimal
panel tilt angle of 32.6° and an optimal panel azimuth angle
of 214°. It is expected that there would be an optimal angle
for both the panel tilt and azimuth angle due to the nature of a
PV array where different orientations affect the PV array’s
ability to absorb solar energy.
It is also expected that a reoriented PV system optimized
to generate the most electric energy for the peak demand will
unfortunately output a total electric energy per day that is less
than that of a non-reoriented PV system optimized to
generate the most electric energy for 24 hours. As shown in
Graph 2, the non-orientated PV system produces the greater
amount of total electric energy per day.
1kWh of electric energy costs around 20.3 cents during
the peak demand [9]. With this consideration, the reoriented
PV system can save consumers in Atlanta roughly 20.3
million dollars annually whereas the non-reoriented PV
system saves consumers around 17.1 million dollars annually.
Either way, using PV systems as an auxiliary energy source
proves to be an effective way of reducing peak use charges
from electric utility companies.
Although the cost of installing a PV system, roughly 300
dollars per square meter [10], is very high, new technological
innovations may further reduce the cost of PV systems in the
near future. Additionally, with the promise of more efficient
PV systems with larger solar energy absorption capabilities,
PV systems can become an even more cost effective and
environmentally conscious source of alternative energy.
a
Electric energy is calculated with the assumption that the solar panel efficiency is
19%. Additionally, the attained kWh is for one square meter of the PV array.
0
100
200
300
400
500
600
0 4 8 12 16 20 24
ElectricEnergy(Whr)
Time (Hours)
Total Energy Production Per Day
Reoriented PV
System
Non-oriented PV
System
PV
System
Average
Insolation
Per Hour
During
Peak
Demand
(W/m2
)
Electric
Energy
Generated
During
Peak
Demand
(kWh)*
Electric
Energy
Generated
During Peak
Demand for
Installed PV
System
(kWh)
Percent of
Total
Energy
Demand
During
Peak
Demand
Met By
PV
System
Non-
reoriented
338 0.384 225,139 1.07
Reoriented 359 0.408 239,210 1.13

4. REFERENCES
[1] “Solar.” Internet: http://www.instituteforenergyresearch.org/energy-
overview/solar, [7/19/13].
[2] “Peak Demand Reduction.” Internet:
http://netplusconcepts.com/Commercial.php, [7/19/13].
[3] “The Sun as an Energy Resource.” Internet: http://www.volker-
quaschning.de/articles/fundamentals1/index.php, [7/20/13].
[4] “National Solar Radiation Data Base: 1991-2010 Update.” Internet:
http://rredc.nrel.gov/solar/old_data/nsrdb/1991-
2010/hourly/list_by_state.html, [7/19/13].
[5] A. Chen “Berkeley Lab Researchers Announce OpenADR
Specification to Ease Saving Power in Buildings Through Demand
Response.” Internet: http://newscenter.lbl.gov/news-
releases/2009/04/27/openadr-specification, [7/20/13].
[6] V. Yelundur. “Development of High-efficiency Commercial-ready Si
Solar Cells for Cost-effective PV.” Internet:
http://web.ornl.gov/sci/solarsummit/georgiatech.pdf, [7/20/13].
[7] M.A. Brown. (2008, May). “The Residential Energy and Carbon
Footprints of the 100 Largest U.S. Metropolitan Areas.” Residential
Carbon Footprints of Metropolitan America. [Online]. Available:
https://smartech.gatech.edu/bitstream/handle/1853/22228/wp39.pdf;js
essionid=E9658916A2EDC8B0915D83B52BC5D2E0.smart1?sequen
ce=1 [7/20/13].
[8] “Basics of Solar Energy.” Internet:
http://zebu.uoregon.edu/disted/ph162/l4.html, [7/19/13].
[9] “Electric Service Tariff: Time of Use – Residential Energy Only
Schedule: ‘TOU-REO-7’.” Internet:
http://www.georgiapower.com/pricing/files/rates-and-
schedules/2.20_tou-reo-7.pdf, [7/23/13].
[10] “How Much Does Solar Panel Cost?” Internet:
http://www.solarpanelnexus.com/installsolarpanel/cost.php, [7/21/13].

5. Appendix A
Created by Tianhao Li. 07/21/2013. Georgia Institute of Technology. This program serves the purpose of
processing the data from TMY3 for the Atlanta area. It also calculates the solar insolation for
photovoltaic module when reorientation is and considered. The first part of the program only the hours from
12 PM to 6PM, which is the peak demand interval. The second part of the program evaluates the peak shaving
performance.
function [betaBest,val,valB,ind,suminso,totalinsolation] = pvPower(filename)
[num] = xlsread(filename); high level i/o to import the data from the TMY3 file
length = 8760; number of rows in the TMY3 data sheet
sumvalueA=zeros(1,24); initialization of the values
sumvalueB=zeros(1,24);
sumvalueC=zeros(1,24);
For loop attains the hourly solar irradiance. For example, from 12 PM to 1 PM, the average irradiance for
the whole year is the summation of the hourly values for 365 days. The values are summed separately for
every hour in the 24 hours.
for time = 1:24
for i=time:24:length
sumvalueC(time) = sumvalueC(time) + num(i,4); Summation of irradiance in the TMY3 data
end
end
sumvalueC = sumvalueC./365; takes the daily average
irradiance = sumvalueC; solar irradiance
The following is not affected by the azimuth angle.
for zenith = 1:24
for i=zenith:24:length
sumvalueA(zenith) = sumvalueA(zenith) + num(i,2); apply same technique to acquire zenith angles
end
end
sumvalueA = sumvalueA./365;
zenithA = sumvalueA;
zenithPeak = zenithA(12:17);
cosinePeak = cos((zenithPeak./180)*pi);
sinePeak = sin((zenithPeak./180)*pi);
for azimuth = 1:24
for i=azimuth:24:length
sumvalueB(azimuth) = sumvalueB(azimuth) + num(i,3);
end
end
End of excel data handling
sumvalueB = sumvalueB./365; Daily azimuth angles
azimu = sumvalueB;
azimuPeak = azimu(12:17); Only peak demand hours are concerned
i=1; Initialize the index indicator
for azimuthChange = 180:1:250 For loop ranges from 180 to 250 in order to find the best azimuth angle for
the reoriented PV system. The reason for setting 180 to 250 is because 180 is the best for no peak shaving,
around 1 pm, but for shaving, around 3pm, 227 is the best. 250 goes to the last peak hour, so 250 is large
enough to include the optimal choice.

6. azimuthChangeA = azimuthChange.*ones(1,6); change the size of the angle vector to match with the 6
hours of original solar angles
peakValue(i,:) = irradiance(12:17).*cos(((AzimuthChange1-azimupeak)./180)*pi).*sinepeak;
calculates the second coefficient for cos(beta) The above expression is simply cos(r-
rs)*sin(zenith), which is regarded as the coefficient for cos(beta). Beta is the panel tilt angle we
want to find, and r is the azimuth angle of the panel that we want to find.
i= i+1; ensures the final value for index i will be 72 (the size of the peak value is 72 * 6)
end
secondFactor = peakValue ; This part changes because the peak value is related to the azimuth angle.
firstFactor = zeros(71,6); initializes first vector, with size matched up with 71 tests of azimuth angles
and 6 hours of the peak values
for index = 1:71
firstFactor(index,:) = irradiance(12:17).* cosinePeak; include irradiance in the term to get the
highest insolation.
end
varA = atan(secondFactor./firstFactor); using the formula to sum up A*sin(beta)+B*cos(beta), which gives
sqrt(A^2+B^2)*cos(beta-alpha)
alpha = (varA.*180)./pi;
R = sqrt(secondFactor.^2 + firstFactor.^2); find the largest R with shifting azimuth angle
beta = alpha(:,1)-1; The first column has the lowest alpha values.
betaC = beta;
finalValue = zeros(71,45);
for i=1:45
beta = beta +1;
betaE = (beta*pi)/180;
betaB = [betaE betaE betaE betaE betaE betaE];
k = cos(betaB - varA);
valSum = R.*cos(betaB - varA); fetch six hours insolation
kB = valSum(:,1)+valSum(:,2)+valSum(:,3)+valSum(:,4)+valSum(:,5)+valSum(:,6); daily insolation
finalValue(:,i) = kB;
end
[valB,ind] = max(finalValue); 45 maximum values for insolation;
every column indicating one sample of beta angle
45 indices for 45 degrees range of beta angles
[val,indB] = max(valB); the largest insolation for peak-shaving purpose from 12 PM to 6 PM
ind = ind(indB); get the index for the array of azimuth angles
get the best azimuth angle
betaBest = betaC(ind); the best tilt angle for maximizing insolation
valB = finalValue(1,:);
[valB,ind] = max(valB); valB for non-reoriented PV systems

7. Part II
zenithPeakT = zenithA(6:20); take 24 hours into account
azimuthPeakT = azimu(6:20); no insolation from 1 AM to 5 AM, from 21 PM to 12 AM
cosinePeakT = cos((zenithPeakT./180)*pi);
sinePeakT = sin((zenithPeakT./180)*pi);
peakValueT = irradT.*cos(((180-azimuthPeakT)./180)*pi).*sinePeakT;
secondFactorB = peakValueT; peakValue changes with azimuth angle.
firstFactorB = irradT.* cosinePeakT;
varB = atan(secondFactorB./firstFactorB); find the alpha angle in the combined expression R*cos(beta-alpha)
alphaB = (varB.*180)./pi;
R1 = sqrt(secondFactorB.^2 + firstFactorB.^2); the amplitude for the new cos equation
betaT= min(alphaB)-1;
betaF = zeros(1,15);
for i=1:177 range changed to the 24 hours
betaT = betaT +1;
betaTB = (betaT*pi)/180;
betaF(1,:)= betaTB;
k = cos(betaF - varB);
valSumB(i) = sum(R1.*cos(betaF-varB));
end
[totalInsolation Indexb] = max(valSumB); total insolation without azimuth changes
CALCULATE THE TOTAL INSOLATION FOR REORIENTED PV SYSTEM
newBeta = 36.1683; using the fixed optimal tilted angle from part 1
newBeta = (newBeta*pi)/180;
newAzimuth = 214; the optimal azimuth angle for reoriented PV system
peakValueK = irradT.*cos(((newAzimuth-azimuthPeakT)./180)*pi).*sinePeakT;
secondFactorC = peakValueK;
varC = atan(secondFactorC./firstFactorB); the alpha angle in the combined expression R*cos(beta-alpha)
R3 = sqrt(secondFactorC.^2 + firstFactorB.^2); the amplitude for the new cos equation
newBetaB = zeros(1,15);
newBetaB(1,:) = newBeta;
newInsolation = R3.*cos(newBetaB-varC);
sumInso = sum(newInsolation); total insolation per day for reoriented PV system

8. Appendix B

9. Appendix C

10. Appendix D