Comparing Efficiency of Solar Energy Trapping Methods

[SMUDGY]

I am to design a laboratory experiment to compare the energy conversion rates of two standard methods for solar energy to be 'trapped' for domestic use. See below*. I need to compare the two systems and find out which is more efficient. The energy source should be a standard 150W light bulb connected to the 230V mains.

I know to work out efficiency between the two methods use the equation: input energy/output energy x 100. (Is that correct)

How would i set up the solar cell to measure the power output of the cell and how would i set up the experiment to measure the rate of heating of water by the bulb.

* The two standard methods for solar energy to the 'trapped' are as follows:

1. To use a solar cell (photovoltaic cells), this converts the radiation energy to electrical energy. (First system)

2. To use a solar panel (an array of pipes on the roof), which absorb the solar radiation to heat water.

 

[Curious P]

G'Day
For the efficiency you would use output energy / input energy (not the other way around).

To be honest I don't really get what you mean by "The energy source should be a standard 150W light bulb connected to the 230V mains. "

But I'll help where I can.
For method one, you would use either some kind of maximum power point tracker (MPPT) or multimeter/voltmeter to see what the voltage and current (and therefor power) being supplied by your PV cells.
The efficiency of this would be given by Voltage (preferably at max power point) times Current (also MPP) (side note: This Voltage times Current = Power with units of Watts) divided by the input power (solar radiation), this of course varies depending on location, time of day, clouds, etc. But 1kW/m2 is usually a nice round number for estimating.
Additionally, if your MPPT has some nice software, it will track the efficiency of your panel/system in real time.

For 2., you'd need to use the equation deltaQ=mC(deltaT)
where delta Q is the change in energy, m = mass of water, C = specific heat capacity and delta T is the change in temperature

You'll need to keep an eye on your units, C for water is around 4.18 and the units are Joules per gram.Kelvin (so keep mass in grams), Q is in Joules and Power (what you want for energy out) is Joules/second

If you plan on lighting your light with this , you'll need a way to convert the heated water to electricity (such as using the steam to turn a turbine/generator, where you will get some more efficiency losses. Likewise for the solar cells, they generate DC current and if you want to power the light from a socket you would need to convert the electricity to AC using an inverter (more efficiency losses)

 

[Curious P]

Just to add on for method 2:
For flowing water, you can incorporate the water flow rate (in mL per second) (1g = 1mL roughly), to help convert the energy to power

 

[dec697]

I'm doing a similar experiment. I was just wondering in experiment 2 do you have to time anything, and if so, what?

Also, would you need to account for the surface area and keep both the cells and the panels the same size (makes sense to me) and is it simply input power/output power x 100 in exp. 1 or do we need to do intensity or something??

 

[dec697]

i mean either experiment, and I am not to sure how you would go about setting up exp. 2

 

[dec697]

don't worry about any of the above questions, i have it sorted now,

although i would like to check if you can just put thermometer into one of the pipes on the solar panels??