On summer day, the solar irradiation in Athens, Georgia provides 735 W/m2 at noon time with ambient temperature 26C. Find the temperature derate factor for the Hanwha 300W single crystal silicon solar module at noon. The NOCT temperature is 45C and the temperature coefficient of Pmax is -0.39% per C
Solar irradiance (SI) = power per unit area received from the Sun = 735 W/m2 = 0.735 kW/m2
Ambient temperature Ta= 26oC
Nominal Operating Cell Temperature TNOCT = 45oC
Temperature coefficient of Pmax = -0.39% per oC
Temperature of solar cell can be determined from following expression:
Tcell = Ta + 1.25 *SI* ( TNOCT - 20) = (26 + 1.25*735*(45-20))oC = 48.97oC [SI has to be expressed in kW/m2 to use in this expression]
Assuming the rated power of the module is measured at 25 degrees Celsius (per standard practice),
power derate factor at temperature Tcell is given by
Pmax x Temperature coefficient x (Tcell - 25) = 300 W x (-0.39%) /oC x (48.97 - 25)oC = -28.045 W
i.e. The module will produce 28.045 W less power than the rated power for the given conditions.
On summer day, the solar irradiation in Athens, Georgia provides 735 W/m2 at noon time with...
Near the equator, the Sun is approximately straight overhead at noon, and around 1000 W/m2 solar irradiance reaches Earth. Around this time of day, assuming ambient temperature of 40C, and a 10m x 10m x 2m-tall greenhouse with ¼”-thick glass walls that reflect away 20% of the incoming sunlight, and only thinking of conductive losses, estimate the maximum temperature reached in the interior of the greenhouse. Hint: when are losses and sources balanced? glass conductivity is 1.05 W/(m k)