# Is a radiation heat pump possible

1. Dec 10, 2008

### philrainey

is it possible to get a cold surface radiating more heat to a hot surface than the hot surface to the cold?

1. The problem statement, all variables and given/known data
Find the net radiation between two surfaces per square metre,1mm apart
one surface flat and the other sigsawed with valleys 100 mm across tops and 100mm deep. Both surfaces perfect black bodies with a perfect vaccuum inbetween.The flat surface been at 273 degrees kelvin and the sigsawed one at 283 degrees kelvin. I'm told radiation at different angles from the surface it radiates from is proportional in intensity to the cosine of the angle to the normal. I think some of the radiation emitted by the sigsawed surface will hit the opposite side of the valley and be reasorbted by it. All the radiation emitted by the flat surface will hit the sigsawed surface and be asorbed by it.

2. Relevant equations
total radiation emitted from a square metre of surface area=(56.7*10^-9)*(surface temperture)^4

3. The attempt at a solution
=314 watts
I can't do advanced maths to work out the average intenisty from the angles that would hit the flat surface from the sigsawed one so I divided up each side of the valley into points and found the average of the cosine of the angle to the normals. I get 0.6 of the radiation emitted by a equilvient flat surface of 1 square meter
radiation from sigsawed surface to flat=0.6*(56.7*10^-9)*(283)^4
=218 watts

=96 Watts from the cold surface to the hot

2. Dec 10, 2008

### cesiumfrog

Use wikipedia to learn about the second law of thermodynamics.

3. Dec 11, 2008

### philrainey

Quoting the second law of thermodynamics means nothing more to me than quoting the bible. It is only considered a law till a exception is found. Carnots max effiency is based on a isothermal cycle been the most efficent heat pump. What if it isn't. What if it is not a law but only a belief?

Last edited: Dec 11, 2008
4. Dec 11, 2008

### Crazy Tosser

Disproving the Second Law would not be easy, nor is it necessary. You can just look at it's vague definition to see that the Law is most likely not a cause, but an effect of a more fundamental law yet to be discovered or formulated.

Exceptions to the Second Law happen from time to time on a molecular scale, when a slow molecule gives up energy to a fast molecule in a collision, but on the macroscopic scale it's as likely for a cold object to transfer energy to a hot one as it is for all air molecules in your room to gather in a corner. It's possible, but extremely improbable.

5. Dec 12, 2008

### cesiumfrog

If you'd bothered to learn thermodynamics, you'd realise that if your device worked then it would be trivial to obtain unlimited energy from it. I suggest you stop discussing this perpetual motion possibility here because you don't want anyone else to steal your idea, fame, and fortune.

Alternatively you might take it as evidence that you've made a mistake and do some double checks, for example, whether you accounted for how the increased surface area compensates the decreased radiation intensity at an angle and, you know, actually do the math or the experiment rather than just guessing numbers.

No, it would be necessary (compare the Clausius phrasing with the OP). And indeed, the law has been discovered to simply derive from statistics (and is called the Fluctuation theorem if your interest isn't limited to the macroscopic).

Last edited: Dec 12, 2008
6. Dec 12, 2008

### philrainey

perhaps I can calculate it properly using a speadsheet if someone can tell me what the anti differential of sinx is? I can break the sides of the valley into little evenly spaced points and work out the range of the angles of the radiation from the points thats hit the cold plate and use those angles in the anti differential equation to find the proportion of total radiation that the sigsawed plate radiates that hits the cold plate? Take the average of these points.Would this be correct?

7. Dec 12, 2008

### cesiumfrog

8. Dec 13, 2008

### Staff: Mentor

The difference between the second law of thermo and the Bible is that one has been tested and the other has not. It is absolutely nothing like a belief. Science is not a guessing game.

I don't have time to figure out the proof right now, but I'll give you a hint about where your error might be: if you have to guess about anything in a calculation and you come up with an incorrect answer, it is a good bet that your error lies in that guess.

Also, the term "radiation heat pump" is a misnomer. It implies a similarity to other heat pumps, where there is none. Heat pumps are well-understood thermodynamic devices that operate with an input of energy. That's the "pump" part of "heat pump". What you describe here is an attempt to passively (without an input of energy) move heat from one place to another. In other words: no pump.

Last edited: Dec 13, 2008
9. Dec 13, 2008

### Staff: Mentor

It wouldn't be the average, it would be the integral. This is a compound integral problem. Sometimes the integral is an average (such as with the area under a line), but here it is not.

10. Dec 13, 2008

### philrainey

you are right taking the average of ten points is not the exact way of doing it but my maths is very limited so I was just trying to find a way I can roughy do it without hard calculus that I can't remeber how to do. Someone with reasonable maths could do it exact perhaps easy but I can't which is why I would like someone with math skills to help me. I have done a calculation which must be wrong as it says it works which is unlikely.
I divided one wall of a valley into ten and worked out the angle with the cosine rule that misses the top of the opposite side of the valley for each point( I'll call angle P), so the range of the angle of radiation for the point is 0 degrees to the angle I worked out by the cosine rule. I'm working on the basis the radiation changes in intensity distribution according to the sine of this angle(cosine to the normal but sine to the valley side) Told the anti differential of sinx is -cosx. Enter the angle P into the -cosP equation minus -cos0.divide it by the area under the sine wave of 180 degrees which I figour is 2.Add the ten fractions together and divide by ten (take the average because I can't to the proper maths)and this I say is the fraction of radiation emitted by the sigsawed surface that hits the flat surface.The area of the sigsawed surface is 2.2square metres 31% of the radiation from the sigsawed surface hits the flat.work out the radiation emitted by the sigsawed surface using the Stefan-Boltzmann equation.=(56.7*10^-9)*2.2*(temperture^4)*0.31

11. Dec 13, 2008

### philrainey

=248 watts compared to the 314 watts emitted by the colder flat plate. Please help me to rid this idea out of my head by finding me wrong but please don't quote the "law"at me. I must be wrong made a mistake.Save me from myself. I'll go back to cleaning the workshop.

12. Dec 13, 2008

### philrainey

my calculation is wrong because it only considers two dimensions but the things is three dimensional with radiation also radiating at angles along the valley

13. Dec 13, 2008

### philrainey

one would need a hundred little four sided piamids as the shape for the hot surface and the calulation would still be incorrect because of the radiation going at varies angles not perpeniduar to the sides. perhaps the area would be for each triangle in the piamid dimple 0.1*0.1*0.5 4 sides per dimple so *4 *100=2 square metres the calculation is too hard for me but I think the result not far different. If it was a cone and a circle surface perhaps it would be correct?

14. Dec 13, 2008

### philrainey

I'm happy to right it off now I think it dosn't work .with a cone too much of the area is around the circle base where the most radiation lands on the flat cold surface is at it's most intense.

15. Dec 14, 2008

### philrainey

I have a much better shape that is much easier to calculate.Try calculating this one the net radiation between two spheres one within the other a perfect vaccuum inbetween.outer Big sphere is 0.13m diameter at 1500 kelvin. the little sphere is .01m diameter at 1550 kelvin. what is the net radiation and in which direction.

16. Dec 14, 2008

### Staff: Mentor

The Stefan-Boltzman equation describes exactly that scenario, Philrainey. You can use it directly to find the answer.

17. Dec 15, 2008

### philrainey

so the little sphere area is0.000314metres square at 1550 kelvin=(56.7*10^-9)*(1550^4)*0.000314=102.7 watts.
Total radiation from the big sphere area of0.053066 metres square=(56.7*10^-9)*(1500^4)*.053066=
15232 watts radiating from all the surface of the colder big sphere
but only 7.7% of this radiation hits the little sphere as the rest misses it and hits the big sphere again on opposite side.15232*.077=1172watts so the net radiation is 1172-102.7=1069.3 watts from the colder big sphere to the smaller sphere

18. Dec 15, 2008

### philrainey

is the surface area of a sphere =<r^2>*4*3.14

19. Dec 15, 2008

### Staff: Mentor

You're not using the equation correctly. The only area that is considered is the area of the small sphere. The equation describes the radiation from a surface to it's surroundings. If the larger sphere is .13 m or 13 km in diameter, it doesn't make any difference.

Use the last form of the equation shown here: http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/stefan.html

20. Dec 15, 2008

### philrainey

I accept I'll be wrong but a have trouble accepting the logic that area of the outer sphere has no effect on the net radiation as the small sphere will always emit the same at a given temperture and all of it hits the big sphere but the bigger the big sphere the more area you have emitting radiation at the same amount per square metre of which the highest intensity (if the intensity at a given angle is proportional to the cosine of the angle to the normal) will hit the middle of the big sphere. but then again the angle widgth that hits the small sphere will decrease so maybe it cancels each other out. Then if I manage to calculate the way I have been and get rid of mistakes I should get the same answer as them.