# Free energy in sunlight

1. Apr 9, 2007

### cesiumfrog

Thermodynamics question: How much work can be obtained from sunlight?

It seems like a solar panel within the sun would not continuously generate electricity (else it could purely convert heat to work) but instead would emit as much black-body radiation as it absorbs. Back here the light is obviously more ordered: the photons only vary in direction by about a degree (rather than coming from all directions) and the frequency-to-density-ratio is much higher (far from thermal equilibrium for this region of space) but it's still less ordered than laser light (so seems improbable that even a perfect theoretical device could obtain 100% efficiency from sunlight). How would you go about calculating the maximum efficiency?

2. Apr 9, 2007

### Staff: Mentor

Perpendicular to the sun's rays, the solar flux is 340 w/m^2.

http://en.wikipedia.org/wiki/Earth's_energy_budget

The best solar cells are about 40% efficient. That means that 40% of the energy that hits them is converted to electricity and the remaining 60% is either reflected (not much) or reradiated as heat (most of it).

For almost any energy conversion device, but specifically the two types we are talking about here, whether a solar cell or thermodynamic device, essentially all of the inefficiency comes from waste heat. Perhaps ironically/coincidentally, at 40%, a solar cell comes in in the ballpark of the most efficient single-stage thermodynamic energy conversion systems. The principle benefit of a system such as a solar tower running a steam cycle would be first-cost.

The efficiency calculations, anyway, are straightforward: efficiency is useful work out (from an electric meter) divided by work in (converted from a fuel meter or photosensor in this case).

Last edited: Apr 9, 2007
3. Apr 10, 2007

### cesiumfrog

Sorry, I seem not to have phrased my question clearly.
I'm not talking here about comparisons between two specific technologies.
That's how you measure the real efficiency of a particular implementation, but I was asking how to calculate the maximum theoretical efficiency of an ideal sunlight (to electricity, say) converter. Think "Carnot efficiency for solar cells"; how many of those 340W/sq.m. are provably unusable regardless of how far technology advances?

4. May 10, 2007

### marcus

do you allow concentration of the sunlight by parabolic mirror?

at what temperature do you dump heat?

suppose you dump heat at 400 K
and you concentrate the sunlight to 5400 K
would not the theoretical efficiency be 5000/5400?

I know you are talking about ideal photovoltaic conversion (not a steam engine )
but I tend to think the same theoretical limit applies.
the reason sunlight is such a highgrade form of energy is that it's color temperature is 5000-some kelvin.
(that's the basic reason you can get 40 percent in a practical cell)
a photovoltaic is not all that different from a thermocouple (a solid-state heat engine) except that it is using 5000 kelvin photons.
Let me know if my answer is just plain dumb. I'm saying the carnot limit applies, in a certain sense, to photovoltaics as well as other stuff.

Last edited: May 11, 2007
5. May 11, 2007

### Xezlec

I don't mean to be picky, but do you by chance have a reference for that 40% figure? I always thought 15% or so was the figure (don't remember where I heard that), and I'd be interested to hear more about the subject.

6. May 11, 2007

### cesiumfrog

Yeah, seems the maximum thermodynamic efficiency for converting the sun's heat (5800K) into work here (300K) is 94.8%. But that would require some very impressive engineering, to set up a reversible heat pump 1AU long..

I found a paper ("Thermodynamic limitations to solar energy conversion", Peter Wurfel, Physica E 2002) on the topic. It claims the maximum theoretical efficiency from sunlight is 93%. This is lower on the grounds that a black body radiating into space is an entropy producing process (since the surface of the body is exchanging heat with the EM field without being in proper thermodynamic equilibrium with the field, unlike say in the interior of the body).

The paper then discusses various ways of processing the sunlight, and recognising that these are inevitably all irreversible processes too, claims the maximum efficiency (and this requires maximal sunlight concentration) is 86%.

7. May 11, 2007

### Staff: Mentor

I pulled it out of the air, but.....
http://en.wikipedia.org/wiki/Solar_cell

8. May 11, 2007

### Staff: Mentor

Cesium - sorry, I guess this thread dropped down fast and I missed your latest post there....

But see, that's the thing - it is different for every type of process. If you want to do a thermodynamic process, you can use the temperature of the source and sink to find carnot efficiency, but that doesn't have anything to with the temperature of the sun. If we're talking direct light->heat, the efficiency doesn't depend on delta-T because there is no heat of rejection: radiative heat transfer is 100% efficient.

Solar cells themselves are not strictly limited in conversion by anything other than the wavelength of light they absorb. "40% efficient" means they convert 40% of the total solar energy that hits them to electricity, but what they actually do is convert 100% of a specific wavelength or small range of wavelengths to electricity. So it is entirely an electrochemical limitation, not a thermodynamic limitation.
http://yarchive.net/space/spacecraft/solar_cell.html

9. May 11, 2007

### cesiumfrog

I don't really understand what you're trying to say (obviously the temperature of the sun is relevent, and the efficiency can not be 100%), but I recommend reading the published paper (though I haven't gone in detail through its entirety yet). It almost sounds like you're implying that "non-thermodynamic" processes (meaning, things with fancy technical mechanisms that don't "seem" like simple "heat engines") are not limited to "thermodynamic" efficiency (not limited from reducing entropy of the universe), which is silly.

Last edited: May 11, 2007
10. May 11, 2007

### Ratiocinator

11. May 11, 2007

### Staff: Mentor

No. What is the efficiency of an electric heater? 100% and it doesn't matter what the temperature is.
It doesn't have anything to do with "fancy chemical mechanisms" - a solar cell is electrochemical. It converts radiative energy directly to electricity. Thermodynamic devices - heat engines, as you correctly called them - convert heat energy to mechanical energy. A solar cell (or an electric heater) is not a heat engine.

A solar water heater is another example. It's efficiency is limited only by it's insulation and as such the theoretical maximum efficiency is 100%. All of the captured energy is converted to heat in the water.

12. May 12, 2007

### vanesch

Staff Emeritus
Well, you can use the EM field as your "reversible energy transporter". In fact, you can do much better than this, because what ultimately counts, is not *earth's* temperature, but the temperature of the "black night sky". As such, you can use optical (reversible) means to have a black body "see" only the sun's surface (with parabolic mirrors), until that body reaches the sun's surface temperature and optical equilibrium is set up, and you can have another black body "see" only the night sky, and it will reach a few Kelvin. Now, we have two heat reservoirs at said temperature, and we can hence install a perfect Carnot engine between them.

This is not a very FAST converter, but it is "efficient" in the sense that once equilibrium is reached, it doesn't "extract" any energy from the sun (because it radiates it back).

13. May 12, 2007

### Xezlec

But heat is obviously the special case. Heat is like pure entropy. You're talking about converting sunlight to electricity, which is all nice and ordered. How can the efficiency be 100%? I thought I had understood that solar cells are much less efficient at capturing photons than chlorophyll.

And what little I thought I knew (from my solid state electronic devices class) about solar cells' operation does not allow any way I can see for the efficiency to be 100%. If the photon's energy is even a tiny bit above the bandgap, the extra energy is wasted, right? And the probability of a photon actually hitting the right atom to even transfer its energy is not 100%, right?

I thought my 15% number was specifically for the wavelengths that the cell was sensitive to.

14. May 12, 2007

### cesiumfrog

....

Russ, I said "technical", not "chemical".

Solar cells will never "convert 100% of a specific wavelength [..] to electricity" because they, like all things, are still subject to thermodynamic limitations.

Regards your solar water heater example: If it does a good job of absorbing heat from the warm sunlight (Kirchoff's law basically states that a good black body is a good black body..), it will unfortunately also do a good job of radiating heat away from heated water, which limits the percentage of solar energy transferred to the water, bounding efficiency even with otherwise perfect insulation.

Read the peer-reviewed literature, Russ. The paper I referenced above does also specifically discuss the theoretical upper limits of electrochemical solar cell efficiency.

Last edited: May 12, 2007
15. May 12, 2007

### cesiumfrog

Interesting argument. If the idea actually works, and it was implemented to operate very slowly, that might *extract* energy from the surface of the sun (and leak some of it to empty space) with very high efficiency (~5799/5800). However, it would be extracting energy from the incidence *sunlight* with very low efficiency (~0%). But (keeping in mind the point of the exercise) if you tried to extract a greater rate of useful work (say aiming for a few percent efficiency in obtaining work from incident *sunlight*) then (since the hot reservoir temperature decreases slightly, and is no longer in exact equilibrium with the surface of the sun) additional entropy will be created as energy is radiated from the sun to the reservoir, which suggests you would no longer even be able to claim such high efficiency for extracting work from the surface of the sun (I suppose you'd be looking at something like ~5600/5700).

Last edited: May 12, 2007
16. Jul 27, 2010

### jrf1994

Here is my question: Compared to solar panels how efficient are plants at capturing solar energy?

According to this article solar efficiency in plants vary according to the mineralization of the soil. http://www.aglabs.com/newletters/harvesting_solar_energy.html" [Broken]

Unfortunately the author doesn't give any ranges of efficiency. Any ideas?

Last edited by a moderator: May 4, 2017
17. Jul 28, 2010

### K^2

Under perfect lighting and atmospheric conditions, photosynthesis can be up to around 6% efficient. Sorry, I do not have the source anymore. I found that number a long time ago for some estimates I was making.

At any rate, it's not even as efficient as ordinary solar cells, let alone high efficiency ones.

18. Jul 28, 2010

### jrf1994

19. Jul 28, 2010

### dipstik

i have seen themo efficiency calcs similar to marcus in journals talking about theoretical limits to solar efficiency.

using the ratio of temperatures.

20. Aug 1, 2010

### Antiphon

The terrestrial AM1.5G insolation spectrum has about 1000 Watts per square meter.

The thermodynamic limit for photovoltaic conversion (a Carnot cell in effect) is about 62%.

The limit was given by Quiessar and Shockley in 1961.

21. Aug 1, 2010

### Staff: Mentor

Yes, I'd say that that 340 W/sq m I posted 3 years ago is not the right/relevant number. It's the average over the surface area of the earth, not the peak for perpendicular rays.

22. Aug 2, 2010

### Antiphon

I wish the iPhone interface would show the date of a post. It only shows the time which is pretty useless for avoiding old threads.

23. Aug 2, 2010

### Staff: Mentor

Ditto for Blackberry - annoys me too.

24. Aug 2, 2010

### mikelepore

More research in the manufacturing area is very important. The most important issues are the cost of manufacturing the solar devices and also the external costs of that manufacturing (pollution, workplace hazards.). It may be justified to use even a very low efficiency device if the manufacturing process can be made more simple. For equal area on both cases, a 1 cent device that's 1 percent efficient would be better than a 10 dollar device that's fifty percent efficient.