The size of a typical parking space in the USA is between 7.5 to 9 feet wide and

Question

The size of a typical parking space in the USA is between 7.5 to 9 feet wide and 10 to 20 feet long. Assuming PV panels are covering 8 feet by 18 feet of a parking space in a parking lot that can accommodate 300 cars. If a polycrystalline panel with 18% efficiency is used to cover all 300 parking spots in the lot, how much electric power in kWp (kilo watt peak)) can the PV covering the whole parking lot (300 parking spots) generates, assuming solar irradiation on the panel surface is 1,000 W/m^2 [please enter the power in kW units!]? kW

Explanation / Answer

Given area covered by solar panels is A = 8*18 = 144 ft2 = 2.4384*5.4864 = 13.37303776 m2

Given solar irradiation on panels is G = 1000 W/m2

Given conversion efficiency of solar polycrystalline solar panels eff = 0.18;

Electrical Power in kWp = A*G*eff = 13.37303776*1000*0.18 = 2.408 kWp ------------------------------------------- (Answer)

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