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Product Description:

Grid-tied systems have the luxury of a free storage device, the electric grid. When arrays produce more power than they are consuming, a building’s electric metre spins backwards. In small-scale applications, the PV system simply slows the rotation of the metre, saving the owner money and reducing greenhouse gas emissions. Grid-tied systems also eliminate several components of stand-alone systems that reduce the overall efficiency and increase the cost.

Now that yearly AC power demand (kWh/yr) and the insolation of a location have been determined, the AC power rating (kW) of the grid-tied PV/ inverter system can be calculated. Insolation is given in units of kilowatt-hours per square metre per day (kWh/m2-d) and because the amount of energy that hits the earth’s surface from the sun is roughly 1 kW/m2, units of kWh/m2-d can be equated to hours of sunlight per day.  e.g.:  From the table above, Perth receives a yearly average of 6 kWh/m2-d. This is the same as saying it receives sunlight for 6 hours a day at 1kW/m2.

To find:             PAC (kW)

Equation:          PAC (kW) = Delec / (Iavg x 365 d/yr)

Where:             PAC = AC power of PV/ inverter system (kW)

                        Delec = electric demand of building (kWh/yr)

                        Iavg = average insolation of location (kWh/m2-d)

 

Photovoltaic modules operate at lower efficiencies as their temperatures increase. In sizing an array, we need to consider the effect temperature will have on the performance of the array. A list of average daily high temperatures for locations across Australia is located here.

To find:             Tcell

Equation:          Tcell = Tamb + [(NOCT - 20°) / 0.8] x S

Where:             Tcell = cell temperature (°C)

                         Tamb = ambient temperature (°C)

                         NOCT = Nominal Operating Cell Temperature, 45°C

                         0.8 = solar irradiance, 0.8 kW/m2

                         S = wind speed, 1 m/s

 

As the temperature of the cell increases beyond 25°C, the power produced by the cell drops by approximately 0.4% / °C.  An equation to determine the amount of power that is lost from an array using the location’s average daily high temperature appears thus:

To find:             Ploss

Equation:          Ploss = .004 x (Tcell - 25°)

                        Temperature derate = (1 - Ploss)

Where:             Ploss = power lost due to temperature (kW)

                        Tcell = cell temperature (°C)

                        .004 = temperature derating constant, 0.4% / °C

i.e. for every increase of 1°C above 25°C, there is an efficiency drop of 0.4%

 

Imperfections in the manufacturing process can cause some PV panels to be rated slightly higher than others in an array. When these unequal panels are linked together the power output will be slightly less than the down rated power multiplied by the number of panels. This is referred to as ‘mismatch’ and can account for up to a 3% reduction in the power output. An additional 3% decrease in efficiency can be caused by mismatch due to dust or dirt on the panels. Inverter inefficiencies also contribute to the loss of power when converting from DC to AC power.  85-92% efficient inverters are common in PV system applications. For these calculations, a 90% efficiency is assumed.

To find:             PDC, STC

Equation:          PDC, STC = PAC / (dirt x mismatch x inverter x temp. derate)           

Where:             PDC, STC = PV array’s DC rated power (kW)

                        PAC = PV array’s AC rated power (kW)

Inefficiencies:

• dirt = .97
• mismatch = .97
• inverter = .90
• temp. derate = .88

 

The PDC, STC value that has just been calculated is the amount of power that the PV array needs to produce in order to supply adequate AC power in the building. This value is also the relevant kW rating value used to purchase PV modules.  The equation below can be used to determine the amount of PV panel area that is required to produce this power. The value will vary according to the panel’s efficiency.

To find:            A (m2)

Equation:          A (m2) = PDC, STC/ PV eff.

Where:             A = area of PV panels (m2)

                        PDC, STC= PV array’s DC rated power (kW)

                        PV eff. = PV panel efficiency (decimal)

 

The result of this equation may then be divided by the area of an individual PV module and rounded up to the nearest whole number. This will be the total number of PV modules required in the array. In a grid-connected system, arrays are generally wired in series to increase the voltage through the wires and minimise losses. By operating at a high voltage, current, which is inversely proportional to voltage, is reduced. This allows for the use of smaller diameter wires which can help to save money.

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