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THE PROCESS OF DESIGNING A PV SYSTEM

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

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THE PROCESS OF DESIGNING A PV SYSTEM

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  1. THE PROCESS OF DESIGNING A PV SYSTEM

  2. 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

  3. Determine loads demand

  4. 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.

  5. 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.

  6. PV system powering loads: used every day, • the size of the array= daily energy requirement the sun-hours per day

  7. 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.

  8. 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 energy needs with the difference being supplied by utility power.

  9. 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

  10. 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.

  11. Examples

  12. 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 witha multiplier of 5 hours: daily watt-hours = (10 *75 W) * 5 h = 3750 Wh

  13. 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.

  14. 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)

  15. 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

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