1. Introduction:
Jordan state is very rich in renewable resources especially in solar energy, as
it’s characterized by high solar radiation among regions in the world because it is
located in the earth-sun belt area that has high potential solar energy, which
indicates the potential for installing grid connected Photovoltaic system.
solaris the name of a method of converting(PV):Photovoltaic
that exhibitsemiconducting materialsusingelectricitydirect currentintoenergy
, a phenomenon commonly studied inphotovoltaic effectthe
solarsystem employs. Aelectrochemistryandphotochemistryphysics,
. Thesolar powerto supply usableolar cellsscomposed of a number ofpanels
process is both physical and chemical in nature, as the first step involves
process takes placeelectrochemicalfrom which a secondphotoelectric effectthe
in a series, generating an electricionizedinvolving crystallized atoms being
Power generation from solar PV has long been seen as a clean.current
most plentiful andplanet’senergy technology which draws upon theainablesust
widely distributed (renewable energy) source the sun.
Due to high and reliable solar irradiance in Jordan (5.5 kWh/m2 ·d), a
domestic usage for solar energy in Jordan has high potential for about 330
sunny days per year using solar collectors. Solar irradiance varies with season
and time of the day due to the various sun positions under the unpredictable
weather conditions. One of the scopes of this research is to prove that the
collected solar energy compensates the electrical energy consumption in
residential buildings.
2. In Our project:
A man who pays more than 100 JD per months as a electric bills for his
house decided to install a PV cells but before that he has to look up if this
project is accepted or not .
To check if this project is going to be accepted or not we have to compare
between the photovoltaic and electrical system by selecting a specific
time period to make our study on it, by finding the future worth for
each project then make a comparison between them we will select the
highest one since the project is an investment project.
Procedure:
First of all, we determined our study period which is 10 years (n= per
quarter) and collected the electricity bills. As an example: from
January 2007 to March 2007 the average of electricity demand was
1040 kWh, we calculated the price of the kWh by the following
equation: Kwh*0.12, where 0.12 is the kWh price. So for this period
the cost (JD) =124.8, by using the single cash flow method, then
adding the governmental taxes (2 %) which varies with time
depending on the cost of the bill ,we found the paid value of the bill
(F=future value ) by the following equation :
G=governmental taxes.,n=1 per quarter, P = cost (JD)nF=P (1+G %)
=127.296 JD1F=124.8(1+.02)
And all the rest periods follow the same calculations for the electricity bill.
- then calculate the PV energy production after that we subtract the electricity
production from PV production to calculate the paid cost , If the result of
subtraction was positive means we will pay for the electricity company but if it
was negative it means that we will sell to the electricity company (Profit )
- Cost paid = kWh*.21
- Profit Cost =kWh*.12
Net cost = total cost for the paid and profit
** We can find the payback period as shown in PBP table **
3. PV system:
A photovoltaic system is an integrated assembly of one or more pv modules or
cells and other item
- Modules: the smallest complete environmentally protected assembly or
interconnected PV cells.
Before we install this system we have to make some calculations to know the
capacity of PV and its initial cost.
We will use mono c-Si:
Efficiency = 14-20 %
Typical c-Si module characteristics performance at standard test conditions:
100wm^2, 25C, AM1.5G
- Nominal power: 235 w
-Calculation:
1560 Production [kWh per kWp.year]; given
Capacity (KWp) =demand [kWh per year]/production [kwh per kWp.year]
#of module = capacity (kWp)/ Nominal power
Initial cost (JD) =350(JD per module)* #of module (Module)
PV losses = (14-15) %
Feed-in tariff kWh = .12 “Governmental tariff”
- PV efficiency will be reduced by .5% yearly -
Capacity:
3990/1560=2.557kWp
#of module = 2557/235 = 10.8 - we need to install 11 modules
Initial cost = 350*11=3850 JD
We Complete the calculation using PV Calculator :
http://pvshop.eu/calculator?lang=en
5. To check if the project is accepted or not: we will use IRR Method.
Y10Y9Y8Y7Y6Y5Y4Y3Y2Y1Now
576594.01588.9452.8424.4517.9497.1467.7463.7415.1-3850
MARR=Inflation = 4%
Pw (Marr %) =-cost (now) +
𝐏
( 𝟏+𝐌𝐀𝐑𝐑%)^𝐧
Pw (4%) =-3850+399.135+428.717+415.754
+424.94+425.676+335.409+344.091+430.303+417.34+389.125
PW (4%) =160.5
PW at any value I >Marr
Pw (10%) =-3850+377.364+383.223+351.389+339.526
+321.575+239.567+232.358+274.726+251.918+222.07
PW (10%) =-856.288
−856.288−160.5
10−4
=
−856.288−0
10−irr
IRR%=4.947
IRR>Marr
PROJECT IS ACCEPTED
