Design Stand-Alone Photovoltaic System Case Study

Miss. L. Raja Rajeswari, Dr. C. V. Krishna Bhanu / International Journal of Engineering
Research and Applications (IJERA) ISSN: 2248-9622 www.ijera.com
Vol. 3, Issue 2, March -April 2013, pp510-515

Design of Stand-Alone Photovoltaic System – A Case Study
Miss. L. Raja Rajeswari *, Dr. C. V. Krishna Bhanu**
* (Department of EEE, Gayatri Vidya Parishad College of Engineering, Visakhapatnam, AP. )
* (Department of EEE, Gayatri Vidya Parishad College of Engineering, Visakhapatnam, AP. )

ABSTRACT Lnm Average night time load
In this paper the design of a stand-alone PV Photo voltaic
photovoltaic system has been explained in two (Rb)mi Tilt factor for direct solar radiation
methods; method-I, considering the solar radiation Rd Tilt factor for daily diffuse solar
data of the location and the load, and method-II, radiation
considering only the load. The various constraints Rr Tilt factor for daily ground reflected
considered in the preliminary phase and the later solar radiation
stages have been thoroughly discussed along with a XC Critical radiation
case study pertaining to a college. The paper β Tilt angle
describes various parameters required for the δm Declination angle
design of a photovoltaic system taking the load  Latitude of the location
conditions and economic constraints in the end.
Cost analysis of conventional and stand alone  Solar radiation utilizability
mi
photovoltaic system for both the methods is carried Ƞb Average energy efficiency of the battery
out. Ƞc Cabling efficiency
Keywords – photovoltaic system, PV system, solar ȠT Maximum power point tracking
energy, inverter efficiency
Ƞ vr Voltage regulator efficiency
I. INTRODUCTION Ƞw PV array wiring efficiency
The ever increasing demand for electrical ω mi Hour angle
energy due to rapid industrialization and increasing ω sm Sunset hour angle
population brings in a need to go for sustainable ω stm Sunrise hour angle
energy sources i.e. renewable energy sources like PV Photovoltaic
solar, wind, ocean energy, etc. Solar energy is
ultimate source of energy which is abundant in nature
and is available irrespective of location. III. METHOD-I
The complete analytical methodology has
This paper attempts to design a standalone photo been presented in Method-I. Solar radiation data, tilt
voltaic system in two methods. The Method-I is an factor and instantaneous solar radiation the location
analytical method which takes solar radiation data have been used to estimate the stand-alone PV
into consideration to design the PV system and the system.
Method-II does not consider the data from solar
radiation but still can give the design of the PV A. ESTIMATION OF ENERGY REQUIREMENT
system. The Method-II can be considered to be more
The load demand must be met by the energy output of
simple and practical. Cost analysis of conventional the PV array installed. So, for a certain load, the
system and stand alone system for both analytical required daily output from the PV array (Epl)m should
method and practical method are carried out and be equal to
economical solution is recommended.
(Epl)m = (Ldm + Lnm / ηb)/(ηw ηT ηvr ηc )
II. LIST OF NOTATIONS USED (1)
Hbm Direct solar radiation
Hdm Diffuse solar radiation where Ldm is the average daytime load, and Lnm is the
Hgm Ground reflected part of global solar average night time load, ηb is the average energy
radiation efficiency of the battery, ηw is the PV array wiring
HTm Monthly average daily solar radiation efficiency, ηT is the maximum power point tracking
(Ib)mi Hourly direct solar radiation efficiency, ηvr is the voltage regulator efficiency,and
(Id)mi Hourly diffuse solar radiation ηc is the cabling efficiency.
(Ig)mi Total solar radiation
(ITb)mi Instantaneous solar radiation Annual daily average energy Epl, required by the load
is given by
Ldm Average day time load

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1 12 ( Ldm  Lnm /  b) like Sine wave inverter, PWM inverter, etc are
Epl  
12 m 1 ( w T vr c)
(2) available in market which can be selected depending
upon requirement.
B. REQUIREMENT OF PV ARRAY AREA
Inverter o/p power rating = 1.5 x Total maximum load
PV array must be capable of covering load and power
(8)
conditioning unit losses on a yearly basis for
calculating minimum PV array area (Amin) which can
F. SOLAR RADIATION UTILIZABILITY
be estimated from the equation given by
The PV battery systems are based on an important
design tool, namely solar radiation utilizability which
 12 ( Ldm  Lnm /  b ) 
  ( w T vr c ) 
is very vital in designing of PV battery systems. This
A min  INT 
1
 A (3) solar radiation utilizability (Φ) is defined as the
 A HTm ( e ) ( d ) 
12
fraction of total radiation incident on the array that
  m
 exceeds a specified intensity called critical level. The
 1 
critical level (Ic)mi is a radiation level at which the rate
where ΔA is the incremental step in array area, which of electrical energy production is just equal to the
depends on the array configuration (e.g., it can be the hourly load (Ld)mi which is given by
area of an additional module or sub-array). The
( Ld )mi
maximum PV array area (Amax) can be estimated as a ( Ic)mi  (9)
multiple (say two times) of Amin. For a given value of A(( e)m( d )( w)(T )( vr )( c))
PV array area A, between Amin and Amax, it is possible
to compute the energy flow in a PV system. The solar radiation utilizability (Φ)mi is defined as

C. PV SIZING ( IT )mi  ( Ic)mi
The number of PV panels required for the system is  mi  (10)
estimated from PV array area „A‟ required and area of ( IT )mi
the module chosen for the design.
A The average daily utilizability is defined as a fraction
Number of panels required = (4) of the monthly total solar radiation that is above a
Area of Module critical radiation level.
D. BATTERY SYSTEM SIZING IV. METHOD-II
The total capacity of battery required is obtained from
supply watt hour and days of autonomy i.e. number of In Method-II, first a load chart is developed based on
days the battery alone should supply power to load. the load and consumption of energy. Different
This is taken based on number of cloudy days. appliances, their wattage, number of appliances and
duration of usage are required to develop the load
The supply watt hour is calculated by: chart. The load chart is prepared by multiplying the
number of appliances with wattage of each appliance
Supply Wh = Load Wh x Battery Autonomy (5) to get maximum watts. This maximum watt is
multiplied with duty factor to get average watts. This
The battery watt hour is calculated from: is multiplied with number of hours of usage to get
watt hours. For different appliances the maximum
Supply Wh watts, average watts and total watt hours are
Battery Wh = (6) aggregated individually for calculation purpose.
(DOD  BEF  ACEF )
A. PV SIZING
And, battery ampere hour is obtained from: PV sizing means determining the number of panels
required to support the load. Initially the total power
BatteryWh that is required to be generated from panels is
Battery Ah = (7)
VAT calculated from the formula

E. INVERTER SIZING Load Wh
Peak PV watts (Total Wp) = (11)
Most of the electrical appliances work on alternative (PSH)  SYSEF
current and the electricity generated from PV system
is direct current in nature. Hence inverters are used in Here load Wh is obtained from the load chart. PSH is
PV system to convert DC electricity to AC and make the average daily peak sun hour in design month for
it suitable for AC applications. Inverters are placed selected tilt and orientation of the PV array. This is
between PV arrays and the load. Various Inverters

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the amount of time where useful solar energy is 23:00 Television 2 x 150 5 1500
present. This value varies from location to location. Total 27720W 138600
23:00 Fans 336 x 60 8 161280
PSH – average daily peak sun hours in design month to Lights 25 x 40 8 8000
for selected tilt and orientation of PV array. 6:00 Total 21160W 169280
Table 2: Daily Load Data of G.V.P.C.O.E. Girls
SYSEF is the overall system efficiency i.e. product of Hostel
AC system efficiency, Battery efficiency and DC
system efficiency. A. ESTIMATION OF ENERGY REQUIREMENT
From Equation 2,
SYSEF = DCEF x BEF x ACEF (12) 307880
(123020  )
The number of modules required is estimated ( Epl ) m  0.80 = 568062Wh
from the formula: (0.97  0.95  0.99  0.98)

Number of PV modules = Total Wp / Module Wp B. REQUIREMENT OF PV ARRAY AREA
(13) From Equation 3,
 568062 
The battery and inverter sizing are similar to Method- A min  INT  1.98 = 618.13m
2

I.  1.98  919 
1 12
V. CASE STUDY AND CALCULATIONS
Where,
12 1
 HTm( e)m( d ) = 919 (Refer Appendix-A)
Girls Hostel Block in Gayatri Vidya Parishad College
of Engineering & Technology in Visakhapatnam, C. PV SIZING
Andhra Pradesh, INDIA has been considered for From Equation 4,
making the model calculations. A panel or 280Wp 618.13
Number of panels required = = 312
manufactured by EMMVEE with area of 1.98m 2 is 1.98
considered for the purpose of calculation.
Therefore, 312 panels of 280Wp are required.
(i) METHOD-I
The collected data and the calculations are as below: D. BATTERY SYSTEM SIZING
The total hostel load is as From Equations 5, 6 and 7,
Appliance Qty Wattage (W)
Tube lights 144 40 Supply Wh = 430890  2 = 861780
Fans 361 60
Geysers 10 2000 861780
Television 2 150 Battery Wh = = 1877516Wh
(0.60  0.85  0.90)
Grinder (large) 2 1000
Table 1: Hostel Load
1877516
Battery Ah = = 19557Ah
The daily load data of G.V.P.C.O.E. Girls Hostel has 96
been tabulated as shown below: E. INVERTER SIZING
Time QtyxWatts Energy From Equation 8,
Appliance Hrs Inverter o/p power rating = 1.5 x 27720 = 41580W
(Hrs) of appliance (Wh)
6:00 Lights 32 x 40 3 3840
to Fans 61 x 60 3 10980 F. SOLAR RADIATION UTILIZABILITY
9:00 Geysers 10 x 2000 2 40000 From Equation 9,
Grinders 1 x 1000 1 1000 123020
( Ic)mi 
Total 25940W 55820 721 0.13  0.90  0.97  0.95  0.99  0.98
9:00 Fans 25 x 60 6 9000 = 1631.16
to Lights 4 x 40 6 960
16:00 Television 2 x 150 6 1800 (IT)mi = (Ib)mi(Rb)mi + (Id)mi Rd + (Ig)mi Rr (14)
Total 1960W 11760
16:00 Fans 361 x 60 2 43320 (Ig)mi = (Ib)mi + (Id)mi (15)
to Lights 144 x 40 2 11520 ( Ib)mi  ( Id )mi  (Cos mi  Cos sm)
18:00 Television 2 x 150 2 600  (a  bCos mi ) 
Hgm 24 (Sin sm   smCos sm)
Total 27720W 55440
18:00 Fans 361 x 60 5 108300 (16)
to Lights 144 x 40 5 28800
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 (Cos mi  Cos sm) has been done. Conventional cost analysis is same in
( Id )mi  Hdm  (17) Method-I and Method-II, whereas the stand-alone PV
24 ( Sin sm   smCos sm)
system cost analysis differs. 60% of the load is

a  0.409  0.5016 Sin( sm - ) (18) considered to be supplied from state electricity board
3 (SEB), and 40% from diesel powered generators due
 to the power outages at the location selected.
b  0.6609  0.4767 Sin( sm - ) (19)
3
A. CONVENTIONAL SYSTEM
here,
Non peak hour unit cost = Rs.5.97/-
 mi = 300
Peak hour unit cost = Rs.6.97/-
 sm = 89.686 Diesel generation cost per unit = Rs.12/-
 a = 0.6574, b = 0.4124 Non peak hour units = 292.3KWh
 (Ig)mi = 92.8547, (Id)mi = 18.7707 & (Ib)mi =
292.3
74.6839 Non peak hour KWAh =  324.77 KWAh
0.9
and hence Non peak hour units from S.E.B. = 194.86KWAh
(IT)mi = 7179 Non peak hour units from S.E.B. cost per day =
Rs.1163/-
From Equation 10, Non peak hour units from diesel generator =
7179  1631 129.91KWAh
 mi  = 0.772 Diesel generation cost per day = Rs.1558/-
7179 Non peak hour cost per year = 2722.2 x 365
= Rs.993615/-
The solar radiation utilizability  =0.772 is 77%, i.e. Peak hour units = 154KWAh
77% of the energy from the PV arrays is available for Peak hour units from S.E.B. = 92.4KWAh
charging the batteries during day time. Peak hour units from diesel generator = 61.6KWAh
S.E.B. units cost per day = Rs.647/-
(i) METHOD-II Diesel generation cost per day = Rs.739.2/-
A. PV SIZING Peak hour cost per year = 1358 x 365
The load chart we considered for the calculations is = Rs.504868/-
shown below in Table 3. KVA cost per year = Rs.75000/-
Tota Total cost of the system = Rs.1573483/-
H
W l Total
Qt o
Load att Max avg. Wh B. STAND-ALONE PV SYSTEM OF METHOD-I
y u
s .Wat watts Total watts of the system = No. of panels x peak
rs
t power of each panel
Tube 14 = 364 x 280
40 5760 5760 7 40320
lights 4 = 101920W
36 2166 1 32490 Cost per one watt = Rs.165/-
Fans 60 21660
1 0 5 0 Total cost of the system = 101920 x 165
20 2000 = Rs.16816800/-
Geyser 10 20000 3 60000
0 0 The payback period can be estimated by dividing the
Televisio 15 30 1 Total cost of the stand alone PV system by Total cost
2 300 4800 per year of conventional system.
n 0 0 6
4772 43002 16816800
Total
0
47720
0   10.68
1573483
Duty factor = 1.0 for all appliances In this case the payback period is 10 years, 8 months.
Table 3: Load Chart Annual savings after payback period is Rs.1573483/-
Load Wh 430020
Total Wp = =
5.6  0.072 5.6  0.72 C. STAND-ALONE PV SYSTEM OF METHOD-II
= 106651W Cost per watt = Rs.165/-
106651 Total maximum watts = No. of panels x peak power
Number of PV panels = = 381 of each panel
280 = 381 x 280
381 panels of 280Wp are required. = 106680W
Total cost of system = 106680 x 165
VI. COST ANALYSIS = Rs.17602200/-
The cost analysis for the conventional system and The payback period can be estimated by dividing the
stand alone photovoltaic system for both the methods Total cost of the stand alone PV system by Total cost

513 | P a g e

per year of conventional system. of the panel to be 20 years as claimed by the industry,
17602200 the remaining 50% of the life, electrical energy
= = 11.18 generated will be at a very small cost because of
1573483
Therefore payback period is 11 years, 3 months. batteries and the maintenance.
Annual savings after payback period is Rs.1573483/- Keeping in view, the present power shortages, and
also frequent interruption of power supply particularly
D. COMPARISON BETWEEN METHOD I & II
for institutions in rural areas (like ours), it is
A tabular column has been constructed using all the economical to utilize solar energy for meeting
above data for comparison between the two methods electrical energy requirement. This will improve
employed so as to get a clear picture of the reliability of power supply at a reasonably low cost
advantageous method based on resources. This tabular
which benefits the organization.
column is as shown in Table 4.
REFERENCES
Sl. Descripti
Method I Method II
No. on
Total Journal Papers:
1 430890Wh 430020Wh
Energy
No. of PV Mohan Kolhe, “Techno-economic optimum sizing
2 panels 364 381 of a stand-alone solar photo voltaic system”, IEEE
required Transactions on energy conservation, Vol.24, No.2,
Total June 2009, Pg. 511-519.
3 Capacity of 19557Ah 19517Ah
Battery (Ah)
Total Conferences:
Inverter
4
capacity
41580W 48000W Proceedings of Work shop on “Off-grid Solar PV
required Components and Systems” by National Centre for
Promotion of Photovoltaic Research and Education
Total Cost
5 Rs.16816800/- Rs.17602200/- (NCPRE) at IIT Bombay from April 11th to 13th, 2012
of System
at New Delhi.
Payback 10 years, 8 11 years, 3
6
Period months months
Books:
Annual
Savings
Solar Engineering of Thermal Processes – 2nd
7 after Rs.1573483/- Rs.1573483/-
edition by John A. Duffie , William A. Beckman,
payback
New York; Wiley Publications; 1991
period
Table 4: Comparison between Method I & Method A. Mani, Handbook of solar radiation data for
II India, 1980 New Delhi, India; Allied Publishers
Private Limited 1981
VII. CONCLUSIONS
The objective of this paper was to design a Planning and Installing Photovoltaic Systems – 2nd
standalone photovoltaic system on radiation data edition by EARTHSCAN Publications
(available for the location) to optimize the space
requirement and cost of installation. A standalone Websites Referred:
photovoltaic system for meeting the electrical energy
and for the energy requirement for Girls Hostel is www.leonics.com
presented as a case study for explaining the www.sunpossible.com
methodology adapted. www.powermin.nic.in
www.wikipedia.com
From the work presented, it can be concluded that the www.emmvee.com
design based on monthly average daily energy
generated (Method I) gives a requirement of 364
panels whereas the calculation based on thumb rule
(Method II) gives 380 panels.

The initial investment is very high; the rate of return
is less, i.e. 10 years, 8 months for Method I and 11
years, 3 months for Method II. Assuming the life span

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APPENDIX-A

Hb Hd Hg Rb Htm= A Htm*A
m m m HbmR *ƞd
(K bm+H
Wh dmRd
/ m2) +
HgmR
r
Jan 7000 1416 8416 1.20 9863.9 0.13 1171.8
6 1 2 3
Feb 7272 1490 8762 1.13 9764.0 0.13 1151.1
7 4 1 8
Mar 6589 1863 8452 1.06 8869.0 0.13 1053.6
4 5 2 4
Apr 5870 2167 8037 0.99 8011.3 0.13 951.75
8 5 2
May 5117 2640 7757 0.95 7484.7 0.13 889.19
2 9 2
Jun 2811 2965 5776 0.93 5541.0 0.13 678.23
2 7 6
Jul 2027 3100 5127 0.94 4955.2 0.13 615.45
8 8
Aug 2487 3046 5533 0.97 5431.4 0.13 669.69
8 1 7
Sep 3559 2552 6111 1.03 6210.4 0.13 760.15
7 6
Oct 5220 1829 7049 1.11 7623.1 0.13 898.77
2 6 1
Nov 6149 1555 7704 1.18 8853.0 0.13 1051.7
7 6 2 4
Dec 6650 1397 8047 1.22 9567.8 0.13 1136.6
8 1 2 6
Sum Sum
92175.33 11028.28
Average Average
7681.28 919.02

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Design Stand-Alone Photovoltaic System Case Study

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