THE PROCESS OF DESIGNING A PV SYSTEM

Potential of Solar Energy in Palestine

it has about 3000 sunshine hours per year

annual average of solar radiation amounting to 5.4 kWh/m2 – day (19.44MJ) on horizontal surface, which classified as a high

in December, it amounts to 2.63 kWh/ m2 - day.

In June: 8.4 kWh/m2 - day

Determine loads demand

Example: if a refrigerator is rating of 768 kWh /year,

=768 / 365 = 2.104 kWh /day.

2.104 kWh /day * 1000 = 2104 Daily Watt Hours.

Example : A house has the following electrical appliance usage:

One 18 Watt fluorescent lamp with electronic ballast used 4 hours per day.

One 60 Watt fan used for 2 hours per day. One 75 Watt refrigerator that runs 24 hours per day with compressor

run 12 hours and off 12 hours. The system will be powered by 110 Wp PV module.

1. Determine power consumption demands Total appliance use=(18 W x 4 hours) +(60 W x 2 hours)+(75 W

x 24 x 0.5 hours)  = 1,092 Wh/day Total PV panels energy needed = 1,092 x 1.3 (the energy lost in the system)

= 1,419.6 Wh/day.

PV system powering loads: used every day,

the size of the array= daily energy requirement

the sun-hours per day

2. Size the PV panel Total Wp of PV panel capacity needed =

1,419.6 / 3.4   (For Thailand) = 417.5 Wp

  Number of PV panels needed=417.5/110 = 3.8 modules

                                                                         Actual requirement = 4 modules

          So this system should be powered by at least 4 modules of 110 Wp PV module.

For systems designed for non-continuous use (such as weekend cabins),:

PV system powering loads: used

weekly, weekly energy requirement= energy

requirement* 7days

the size of the array= daily energy requirement

the sun-hours per day * 7days

Generally, gridconnected systems are designed to provide from 10 to 60% of the energyneeds with the difference being suppliedby utility power.

Area of the PV system:

Example Suppose we need an output of 1.4kWh per day, and the

insolation at a site is 4.44kWh/m2per day and system efficiency 100% .

It follows that the PV array rating= =1.4kWh ÷ 4.44kWh/m2 = 0.315m2

Area of PV : A rating of r kilowatts peak (r kWp) means r kW per 1.0

kW/m2 of insolation. It has dimensions kW ÷ (kW/m2) = m2.

For example, 2.4kWp means the array gives 2.4kW of power when the solar irradiance falling on it is 1.0kW/m2. In other words, it is equivalent to 2.4m2 of collector area at 100% efficiency.

Another example: 30Wp is equivalent to 0.03m2 of collector area at 100% efficiency.

Examples

In order to account for the average daily solar exposure time. Each area is assigned an “area exposure time factor,” which depending on the location may vary from 2 to 6 hours.

Example :calculate daily watt-hours (Wh) for a solar panel

array consisting of 10 modules with a power rating of 75 Watt in an area located with a multiplier of 5 hours:

daily watt-hours = (10 75 W) 5 h 3750 Wh

Example 2 a homeowner uses 10,000 kWh per year and wants to

offset half of that power consumption, 5,000 kWh per year, with a solar electric system. in unshaded conditions, a 1kwatt solar electric system with the panels facing polar south and mounted at 45 degrees will generate about 1,200 kWh in a typical year .

system power =5,000 kWh per year /1,200 kWh per year= 4.2

= 4.2 x 1,000 = 4,200 watt system.

Therefore, a system big enough to generate 5,000 kw-hours per year would need to be about 4,200 watts.

For Homeowners - if you were to install a 5 kW (5,000 watts) PV System:

* 5,000 W x 5 hrs/day* = 25 kWh (kilowatt hours)/day DC* 25 kWh/day - 5 kWh DC** = 20 kWh useable AC/day* (or 7,300 kWh/year)

For Business Owners - if you were to install a 25 kW (25,000 watts) PV System:

* 25,000 W x 5 hrs/day* = 125 kWh (kilowatt hours)/day DC* 125 kWh/day - 25 kWh DC** = 100 kWh useable AC/day* (or 36,500 kWh/year)

For example,

a system of five 200-watt panels has a rated peak output of 1,000 watts. Using the rated power, Estimate the total electricity production of the system based on the sun and shade conditions for your location about 5 hours

System power = Sun hours /day x Rated power x System efficiency x 365 = kWh per year

= 5 x 1000 x 1.3 x 365= 2372500 watt/ year =

= 2372 kw/year