Mazurek Gravity Staff asked 9 months ago
Solar or photovoltaic (PV) cells convert sunlight to electricity and are commonly used to power calculators, satellites, remote communication systems, and even pumps. The conversion of light to electricity is called the photoelectric effect. It was first discovered in 1839 by Frenchman Edmond Becquerel, and the first PV module, which consisted of several cells connected to each other, was built in 1954 by Bell Laboratories. The PV modules today have conversion efficiencies of about 12 to 15 percent. Noting that the solar energy incident on a normal surface on earth at noontime is about 1000 W/m2 during a clear day, PV modules on a 1-m2 surface can provide as much as 150 W of electricity. The annual average daily solar energy incident on a horizontal surface in the United States ranges from about 2 to 6 kWh/m2.
A PV-powered pump is to be used in Arizona to pump water for wildlife from a depth of 180 m at an average rate of 400 L/day. Assuming a reasonable efficiency for the pumping system, which can be defined as the ratio of the increase in the potential energy of the water to the electrical energy consumed by the pump, and taking the conversion efficiency of the PV cells to be 0.13 to be on the conservative side, determine the size of the PV module that needs to be installed, in m2
Mazurek Gravity Staff answered 9 months ago
Given:
Conversion efficiency of PV modules : 0.12 to 0.15 (12/100, 15/100)
So, for 1000 W/m2 of energy is converted to 150 W for every m2 assuming conversion efficiency of 0.15
Incident energy : Annual dialy Average in US = 2000 Wh/m2 , for 24 hours = 2000 x 24 W/m2 = 48000 W/m2 (worst case scenario is this becuase this is the minimum power incident on a flat surface)
With 6 kWh/m2 –> 6000 x 24 W/m2 –> 144000 W/m2
Calculation:
Potential energy = m x g x h (massxgravityxheight) = (volume x density) x 9.81 x 180
=((400 x 10^-6) x 1000) x 9.81 x 180 (W)
=
In terms of power = 706.32 W x 24 hours = 706.32 x 24 = 16951 Wh = 16.951 kWh
——————————-
For 0.13 efficiency of PV = 16.951 / 0.13 = 130.3975 kWh Energy needs to be generated by PV cell.
Assuming worst case scenario : Minimum size of PV cell needed = 130.3975/2 (2 implies 2 kWh/m2)
= 65.1987 m2 (size of PV cell needed)
If we design with 2 kWh/m2 incident power, we would be able to meet the demand.
This is true if we assume 100 % efficiency for pump.
With 80 % pump efficiency:
Energy required by pump = 16.951 kWh /0.80 = 21.18875 kWh
Solar PV generation should be = 21.18875 / 0.13 = 162.99 (say) 163 kWh
Assuming worst case scenario = Minimum solar PV size required = 163 kWh/ (2 kWh/m2)
= 81.5 m2