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.

Photovoltaic (PV): is the name of a method of converting solar energy into direct current electricity using semiconducting materials that exhibit the photovoltaic effect, a phenomenon commonly studied in physics, photochemistry and electrochemistry. A system employs solar panels composed of a number of solar cells to supply usable solar power. The process is both physical and chemical in nature, as the first step involves the photoelectric effect from which a second electrochemical process takes place involving crystallized atoms being ionized in a series, generating an electric current. Power generation from solar PV has long been seen as a clean sustainable energy technology which draws upon the planet’s most plentiful and 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.

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:

F=P (1+G %) n n=1 per quarter, P = cost (JD), G=governmental taxes.

F=124.8(1+.02)1=127.296 JD

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 **

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: 100w\m^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=2.557 kWp

#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

PBP table

Initial cost =3850

Electricity payment-net cost

Years

-3434.9415.11-2971.2463.72-2503.5467.73-2006.4497.14-1488.5517.95-1064.1424.46

-611.3452.87-22.4588.98

571.61Simple payback period

594.019

1147.6157610

To check if the project is accepted or not: we will use IRR Method.

Y10Y9Y8Y7Y6Y5Y4Y3Y2Y1Now576594.0

1588.9

452.8

424.4

517.9

497.1

467.7

463.7

415.1

-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

−856.288−160.510−4 = −856.288−0

10−irr

IRR%=4.947 IRR>Marr

PROJECT IS ACCEPTED

Names: Lauren Qaisieh Zeina sinnokrot Ruba abu-alrub

Engineering economics

Eng.Nedaa al daher