Question

Q2. 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. (TOTAL 25 MARKS)

Answer 1

Given that,

The annual average solar energy incident on a horizontal surface daily ranges from about 2 to 6 kWh/m². Let us take this as 6kWh/m².

Conversion Efficiency of the PV cells is 0.13.

Total Electrical energy we can obtain = 6 kWh/m² * Efficiency * Total Area of the PV module

= 0.78 kWh/m² * Area of the PV module ----------( 1 )

Let us now consider the PV powered pump.

Since most of the calculation involving in finding the power of the pump is represented in terms of horsepower. I will follow the same.

Water horsepower =

TDH = Total Dynamic Head = Vertical distance liquid travels (in feet) + friction loss from pipe (since friction loss is not given we will assume it to be 0) = 180 * 3.28 feet (conversion from meter to feet, 1m = 3.28 feet)

TDH = 590.4 feet

Q = flow rate of liquid in gallons per minute = gallons per minutes ( liter is converted into gallon by the conversion formula 1 liter = 0.264 gallons and day is converted into minutes)

Q = 0.07338112566 gallons per minutes

SG = specific gravity of liquid (this equals 1 if you are pumping water) = 1

Water horse power = hp

(Note : Total power required by PV pump can be calculated by the relation between power and work done which is Power = Work done/time = mass * acceleration due to gravity * height / time. Power calculated in this way is given at the end and it is found to be equal to the Power calculated in the above way)

Water horsepower is the power required to lift a given volume of the liquid to given height ideally. But no pumps are ideal, Since we are told to assume a reasonable efficiency, let us assume the efficiency of the pump to be 0.5 (50%).

Then the actual power required = Water horse power / efficiency of the pump = 0.02188091747 hp (horse-power)

Total power required in watts = Power in hp * 746 (1hp = 746 watt) = 16.32316 watts --------------( 2 )

Energy spent by pump in a day to lift 400 liters of water= power * time (time is equals to 24 hours here) = 391.7559 Watt hour --------------( 3 )

By the law of conservation of energy, equation (1) is equal to equation (3) hence,

0.78 k Watt hour/m² * Area of the PV module = 391.7559 Watt hour

Hence the total Area of the PV module required = .

The area of PV module is low because the water needs to be lifted is low (which is just 400 liters per day).

Had we considered solar energy incident of the surface to be 2 kWh/m² then following the same procedure as above one would get total Area of the PV module as 1.50675 m².

EXTRA :

Power can also be calculated in watt directly rather than calculating in horsepower then converting it to watt.

Power = Work done per unit time = mass * acceleration due to gravity * height / time

Here 400 liters of water is lifted to a height of 180 meter.

For simplicity, let us take mass of 1 liter is approximately equals to 1 kg ( 1 liter is exactly equals to 1 kg at 4 ⁰C).

The acceleration due to gravity = 9.8 m/s².

Time is equals to 1 day = 24 * 60 * 60 second = 86400 second

Power =

Since PV pump is not ideal, let us assume it is 50 % efficient.

Total Power required = which is same as equation (2).