8.4.14 Solving Problems Involving Photovoltaic cells, Solar Heating Panels and Seasonal and Regional Variation in Solar Radiation.

Objective: Solve problems involving specific applications of photovoltaic cells and solar heating panels.

Table of Contents
1. Problem 1 and its solution: involving the principle behind the photovoltaic cells
2. Problem 2 and its solution: involving a specific application of photovoltaic cells
3. Problem 3 and its solution: involving a specific application of solar heating panels
4. Problem 4 and its solution: involving a specific application of variation in solar radiation.
5. Problem 5 and its solution: involving a specific application of solar heating panels and variation in solar radiation.



solar-cell_p-n.jpg
NP silicon cell.

http://www.pv.unsw.edu.au/images/future-students/solar-cell_p-n.jpg

1. The diagram above shows an actual NP silicon solar cell. Explain the steps by which incident sunlight would provide energy to light the bulb.

Solution:
First, note that silicon atoms form a covalent network structure (or “bunch up”) in the P-layer, represented by the yellow area. As incident light strikes the surface of the N-type semiconductor (the thin layer of silicon, represented by the blue area) the photons from the light pass through the N-layer and transfer energy to electrons orbiting the silicon atoms, which in turn dislodge them into the N-layer. Positively charged silicon ions are left behind in the P-layer, and the attraction between the separated dislodged electrons and positively charged ions create potential difference, acting as a battery. Current flows in the circuit and light the bulb.




solar-powered-car.jpg
Solar powered car using photovoltaic cells

http://www.economicshelp.org/blog/wp-content/uploads/2008/07/solar-powered-car.jpg

2. You are designing a solar car with a total roof area for solar cells of 6.4m2. Calculate the electrical power available, assuming total cell efficiency of 17% and a constant light intensity of 980 Wm^-2.

Solution:
Power input = (light intensity)*(roof area)
= (980 Wm^-2)*(6.4m^2)
= 6.3 x 10^3 W
= 6.3 kW
Since Efficiency = (Power output)/(Power input),
Power output = Efficiency*(Power input)
= 0.17*(6.3 kW)

1.1 kW of electrical power is available for the operation of the solar car.



solar_hotwater_residential.gif
solar heating panels/solar hot water system.

http://www.njsolarsolutions.com/Solar-Hot-Water/pics/solar_hotwater_residential.gif

3. A solar hot water system receives solar energy at the rate of 10.5 MJm^-2 per day. If the collector area is 4.8m^2, collector efficiency is 0.7 and the water volume is 325dm^3,
a) Calculate the total water energy gain per day;
b) Estimate the temperature rise of the water in the tank during the day.

Solution:
Water energy gain per day = (rate at which the system receives solar energy)*(collector area)
= (10.5 MJm^-2)*(4.8m^2)
= 50 MJ
= 50000 kJ
We use th equation Q = mc∆T.

(Q=heat gained by the system; m=mass of water; T=temperature)
Noting that density of water is 1.0 kg dm^-3,
m = (density of water)*(volume)
= (1.0 kg dm^-3)*(325dm^3)
= 325 kg
∆T = Q/(mc) = 50000 kJ/(325kg*4.18 kJ oC^-1 kg^-1)
= 37 oC.
The temperature of the water increases by 37 oC during the day.



asdfads.gif
In the diagram above, solar radiation intensity decreases in the following order: red, light red, yellow, light yellow, green, light green, blue and light blue. The diagram represents the maximum amount of solar radiation (in W m^-2) received on a horizontal surface at sea level.

http://www.ldeo.columbia.edu/~kushnir/MPA-ENVP/Climate/lectures/energy/analema.gif

Questions 4-5 are based on the above diagram.

4. What is the difference between the solar radiation intensity received at latitudes of 15 oN and 30 oS on June? What accounts for this difference?
Solution:
Solar radiation intensity at 15 oN = 450 W m^-2
Solar radiation intensity at 30 oS = 250 W m^-2
Difference = (450 - 250) W m^-2 = 200 W m^-2

The reason for such difference is that on June, the Northern hemisphere tilts towards the sun while the Southern hemisphere tilts away from the sun: although the Northern hemisphere receives the same amount of sunlight as the Southern hemisphere, it receives it in a smaller area, which results in a substantially larger intensity of solar radiation.
fig17_3.jpg
Varying distribution of solar radiation at different tilt angles. (a) the Northern hemisphere tilts towards the sun. (b) No tilt (c) The Southern hemisphere tilts towards the sun.
L60_solar_panels.jpg
Solar heating panels installed at certain tilt angle.

http://www.earth.rochester.edu/fehnlab/ees215/fig17_3.jpg http://mynasadata.larc.nasa.gov/images/L60_solar_panels.jpg

5. Suppose you are living in Vancouver, Canada, situated at about 49 oN Latitude. You plan to install solar heating panels on top of your roof. Considering the varying intensities of solar radiation at different seasons, how would you adjust the angle of tilt of the solar panels to ensure that they receive maximum amount of solar radiation?

Solution:
Note that the solar heating panels would receive maximum amount of solar radiation when they are situated perpendicular to the sunlight rays, for maximum solar radiation falls in the smallest area possible.

From May until August, during which the Northern hemisphere tilts towards the sun as shown in case (a) of the above diagram, solar radiation falls nearly directly towards the Northern hemisphere, which means that the solar radiation will fall directly on top of the roof of your house. Therefore, the heating panels must lie flat on the roof to receive maximum amount of solar radiation.

From September until December, during which the Northern hemisphere tilts away from the sun as shown in case (c) of the above diagram, solar radiation falls obliquely towards the Northern hemisphere, which means that the solar radiation will fall obliquely on top of the roof of your house. Therefore, the heating panels must tilt at angles so that they are perpendicular to the incoming solar radiation.

Sources


"Energy flux from the sun." The Earth's orbit and the seasons. 1 Apr. 2009. <http://www.earth.rochester.edu/fehnlab/ees215/fig17_3.jpg>

Griel, Sharon. "Photovoltaic cells." The Solar Power. 3 Mar. 2009 <http://www.pv.unsw.edu.au/images/future-students/solar-cell_p-n.jpg>

Kushnir, Yochanan. "Solar Radiation and the Earth's Energy Balance." The Earth Radiation Budget. 1 Apr. 2009. <http://www.ldeo.columbia.edu/~kushnir/MPA-ENVP/Climate/lectures/energy/analema.gif>

"Solar Hot Water System." The Alternative Resources. 2 Apr. 2009 <http://www.njsolarsolutions.com/Solar-Hot-Water/pics/solar_hotwater_residential.gif>

Walding, Richard. New Century Senior Physics. Oxford: Oxford UP, 1999.