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Solar panels.

  • #1

Homework Statement



First of all I apologise if this is in the wrong place.

Anyway we have been given the task of finding out the efficiency of a solar panel, where energy from a light bulb is used to heat the water in the panel. I was just wondering if someone would just check that my physics is correct.



Homework Equations



1. E=mcΔT
2. Energy= power x time
3. Energy efficiency= Useful energy out/total energy in

The Attempt at a Solution



This is my idea, since we know both the mass and specific heat capacity of the water, by recording the change in temperature of the water over a set period of time the energy required can be worked out using equation 1. If we then work out the power output of the bulb during that time by equation 2. we can work out the efficenicy of the panel using the 3rd equation.
 

Answers and Replies

  • #2
malty
Gold Member
162
0
Hey GL:),

Measuring the energy of the solar panel sounds ok, just remember that the specific heat capacity is also dependent on temperature somewhat.

The light bulb is being used to supply energy to the solar panel, but what are you using the solar panel to power??

In my opinion it would be simpler to just measure the power output of the bulb, and to measure the power output of the device the solar panel is supplying.

The effeciency of the solar panel will just be Power of Device/ Power of Light Bulb, this way is of course neglecting energy losses between light bulb,water and device which should be minimal.

EDIT: I just realised that I should have asked, "How exactly will this solar panel work?"
 
Last edited:
  • #3
Chi Meson
Science Advisor
Homework Helper
1,789
10
Malty, this is the kind of panelthat is just supposed to heat water, I believe. So the change in temperature of the water IS the end result of the exchange of energy between bulb and panel.

Great Lad:
One decision needs to be made clear: the "energy in," is it all the energy released by the bulb? Or is it just the light that is incident on the panel? If so, you have to consider how far away the bulb is, and recognize that the intensity decreases as an inverse square of the distance to the panel. (intensity is power/4πr^2).
 
  • #4
Yeah Chi Meson that is the sort of panel I have to use.

I was referring to all of the energy giving out by the bulb though I know not all of it will be transferred to the water, I was just hoping I could just sort of mention that as a way to improve the experiment and be done with it :smile:

Is there any way I could ensure that more light is absorbed by the panel, apart from placing it as close as possible? The only thing I can thing of is placing a load of mirriors around the equipment, hopefully reflecting as much light as possible to the panel. I don't think I have to explain too much about these things but just show that I have taken them into consideration.

Oh and thank you for posting, I didn't expect such a speedy response.
 
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  • #5
Chi Meson
Science Advisor
Homework Helper
1,789
10
Is there any way I could ensure that more light is absorbed by the panel, apart from placing it as close as possible? The only thing I can thing of is placing a load of mirriors around the equipment, hopefully reflecting as much light as possible to the panel.
Both of those are valid and important considerations. Another is to be aware of the kind of surface that absorbs light best. Hint: not white.
 
  • #6
mgb_phys
Science Advisor
Homework Helper
7,774
12
One other thing to remember is that the solar panel isn't at it's most efficent absorbing the long wavelenghts given off by the lightbulb.
But the water will efficnetly absorb almost all of the bulbs energy.
 
  • #7
Well glad thats done and dusted. Thanks everyone for the help.
 

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