Wednesday, February 29, 2012

Miracles of 2LoT

I've been arguing over at Tallbloke's. It's one of those posts where a sceptic does an elementary analysis and makes elementary errors which contradict "consensus" science. A new scientific discovery is announced. Being Galileos, they don't have to check their work.

The sceptic here is Hans Jelbring. He looks at a simple problem, two concentric spheres without heat sources, and checks their radiation balance to find what the temperature difference should be. Consensus science says, of course, that there should be none, but he found one, and then spent time working out the resulting perpetual motion machine. I'm not sure what was the point of that, but Trenberth was mentioned.

I have sometimes done these analyses myself, being intrigued when what looks like a problem determined by geometry turns out to have a solution constrained by the Second Law of Thermodynamics (2LoT). Given the complexity, that can look like a miracle.


The concentric problem.

When you have concentric convex shapes, the radiation from the inner one ends up on the outer one. A really rookie mistake you can make is to expect the converse. Since the outer surface is larger, the inner body ends up being very hot. Hans managed to make that error here (case II, pipes). But he avoided it in the main post, which concerned spheres. There he noted, correctly, that he needed to calculate how much of the radiation from any one outer point impinged on the central body, and how much missed.

So he drew a plot which you can see there, but which I'll modify to the one at right. There is a small surface element at dA on S2, the outer sphere. Some of its emission, in a cone angle α impinges on S1. The rest misses and ends up back on S2.

I've shown dA at the bottom, but all locations are equivalent, and the total incident on S1 is got by summing the various dA's. I'll assume the spheres are black. Some trig relations:

R1 = R2 sin α, r = R2 cos α

The total emitted by dA is given by the Stefan Boltzmann relation $$F= dA\ \sigma \ T_2^4 $$ To get the fraction within the impinging cone, we want the part of that that would impinge on the surface S (if S1 wasn't there).

Hans made here a common error of assuming that the radiation from dA is as for a sphere, uniform in all directions. Then you can just divide the area of S by the area of the hemisphere of which it is part. But dA is flat, and does not radiate uniformly. In fact, in its own plane it doesn't radiate at all (think of seeing a disc side on).


There is a Law and theory applicable here. It's called Lambert's cosine law, and says that the intensity of radiation is proportional to cos θ, the angle from the normal.

So that tells how the radiation incident on S should be summed. An integral is needed. You can imagine a ring element formed by an increment in θ. Its surface area will be \(dS = 2\pi\ r^2\ sin \theta d\theta\). And if the impinging radiance is \(I(\theta) = I_0\ cos \theta\) W/m2, then the total on S will be $$ I_0\int_\alpha^0 cos\theta dS = 2\pi\ r^2\ I_0 \int_\alpha^0 cos\theta\ sin\theta\ d\theta = \pi\ r^2\ I_0\ (1 - cos^2 \alpha) = \pi\ r^2\ I_0\ sin^2 \alpha$$ We can relate I0 to F by using this formula for the hemisphere, which catches all the radiation. The lower integration limit is then not α, but π/2. So $$ F = \pi r^2 I_0$$ So the power dP transferred from dA to S1 is $$dP = F\ sin^2 \alpha = dA\ \sigma\ T_2^4\ sin^2 \alpha$$ Integration over dA is just summation, so the total power P2 from S2 to S1 is $$ P2 = 4\pi R2^2\ T_2^4 sin^2 \alpha = 4\pi R1^2 T_2^4 $$ which exactly matches the Stefan-Boltzmann emission from S1 if \(T_2=T_1\).

Heat sources

The discussion on the Tallbloke thread was quite interesting, though Hans seems to have dropped out. DocMartyn posed the problem - what if S2 was a conducting shell in space and S1 had a heat source (Pu) generating 300W. He simplified with S2 as 2 sq m and S1 as 1 sq m. He asked what the temperature of S1 would be. It's actually enough to work out the fluxes from each body.

The above reasoning is useful here. We can say that 300W has to be radiated out to space, and the 150 W/m2 sets the temperature of S2. It means also 150W/m2 is radiated inward. An amount P of this is absorbed by S1, which then radiates in total 300+P W.

To get P, imagine that the 300 W was now generated within S2 rather than S1. Surprisingly perhaps, this does not change the temperature of S2. It still radiates 300W outward and inward, of which P arrives at S1.

But now we know that S1, with no source, is at the same temperature as S2. So it radiates 150 W outward (its area is 1 sq m). And P is what comes in, so P=150.

So the answer to the original problem is that the flux from S1 (with source) is 300+150=450 W/m2, and so the temperature is T1=298.48 °K. If T2 is the temperature of the shell, and T0 the temperature corresponding to 300 W emission only (ie S1 without shell), then \(T1^4=T0^4+T2^4\)

An interesting aspect of this reasoning is that nothing was said about spheres. If you allow the bodies to be conductive enough to keep their temperature uniform, they could have been any reasonable shape, though you have to account carefully if S1 isn't convex.

I think DocMartyn chose his numbers with a common shell model of the greenhouse effect in mind.

Update
There's a fairly simple generalization which doesn't require the bodies to be spherical or concentric (though they need to be convex). Each has to be at a uniform temperature, which will generally require perfect conductivity.

Suppose that S1 and S2 have respective areas A1 and A2, and there is a power source P watts. Start by assuming that is on S2. Then it must radiate P outwards, creating an emittance P/A2 W/m2. That forces (S-B) a temperature T0. All this is steady-state, black-body.

Then S1 must also be at T0, and so also emits P/A2 W/m2. For power balance, this is also what it receives.

Now suppose P shifts to body S1. S2 still has to emit P W, so is still at the same temperature. So the environment of S1 hasn't changed, and it still receives P/A2 W/m2, or P*A1/A2 W. But with the extra source, it must now emit P*(1+A1/A2) W, or emittance P*(1/A1+1/A2)W/m2. That determines its temperature.



22 comments:

  1. Oh dear. Hans doesn't do a very good job of understanding you. What a waste of everyone's time.

    Though it *could* be useful. The people who comment there could read your words, realise that you're right, and realise that therefore Hans and Tallbloke are clueless.

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    Replies
    1. If you are going to engage in Lambert's law, you also have to make some assumptions about the surface of the inner sphere and how the absorption and reflection vary with angle.

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    2. Eli,
      yes, the assumption (by HJ and I) was that both spheres are black, ie black bodies.

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  2. If you have a black body in front of another black body I cannot see (pun intended) how, at thermal eqilibrium, you would be able to dectect its presence.

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  3. I haven't read tallblokes guest post but I used to chase these guys away all the time from tAV. They just don't get that you ain't gonna prove AGW wrong with some basic theory about radiation or whatever. I've posted one once just to see how people reacted - it didn't go well for them. SOD left a comment asking why I would put a thing like that up if I didn't endorse it.

    More often than not, they are not technical people in general. I had one guy insist that in converting units, there was a division by 1 and that was where all of radiation theory went wrong. I told him over and over that he was in error.

    At least you set the record straight. Have you dropped a link to this over there?

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  4. Anon,

    Indeed so - if it's isothermal then it's featureless. That's another way of seeing why you don't really have to analyze the geometry to know the answer.

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  5. Jeff,
    Yes, it can be frustrating. There was another post by Stephen Wilde there on no backradiation etc. It got me looking for the earliest measurements I could find, and this 1837 paper by Pouillet turned up.

    I hadn't posted a link as the discussion there seemed to have gone quiet, but your post showed up there.

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  6. Nick,

    Is there an easy way to write equations on blogs? I've worked with LaTeX a little but it is such a pain in the butt that i usually resort to looking around the web for a screen grab.

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  7. Jeff,
    I've installed the MathJax version of Latex. It works well in posts and comments. There's some overhead - it taken a little time to render the Latex, and adds a bit to the size of every page.

    There are some WYSIWYG Latex editors; I haven't tried them.

    But I find that you can go a long way with the symbols in html, with <sub> and sup as well. It's trappy, though, as I found at TB's. Blog hosts often restrict their use in comments.

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  8. If one body is cooler than the other you can ignore its radiation to the warmer one because all that happens is that a standing wave is set up corresponding to the amount of radiation the cooler one could emit. You then deduct this amount which the cooler one could emit from that which the warmer one could emit. The difference represents that portion of the warmer one's radiation which is absorbed and converted to thermal energy in the cooler one.

    That's what actually happens physically. I know you can get the same result by assuming radiation from the cooler one is absorbed in the warmer one, but it isn't and the warmer one's absorptivity for radiation from the cooler one is zero.

    This is why radiation from a cooler atmosphere cannot transfer thermal energy to a warmer surface. It can only slow the rate of radiative cooling by forming standing waves which do not transfer any energy between the atmosphere and the surface. Other cooling processes like evaporation and diffusion followed by convection will increase to compensate. Carbon dioxide molecules can play no greater role than water vapour molecules. Hence their effect in regard to any greenhouse conjecture would be less than 4% of the effect of water vapour. Work out the consequences for yourself.

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  9. Doug,
    You've been saying this everywhere, but it's totally wrong. One thing is that you never in the real world get a standing wave in 3D over millions of wavelengths. Secondly, while an exact standing wave won't transmit power, it takes only a slight realignment of phase to do so.

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    Replies
    1. oh hell, the whole point of thermal radiation is that it is incoherent.

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  10. Very nice. If you add a third sphere?

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  11. Dallas,
    Yes, interesting thought. The "Update" formula extends. The outer shell S3 emits P at emittance P/A3; The second receives irradiance P/A3 from S3 (again it gets the same as it would get at isothermal), total power P+A2*P/A3. And the innermost, S1, receives that emittance P*(1/A2+1/A3) that S2 emits inward (same as outward), which is power P*A1*(1/A2+1/A3). So balancing, S1 emits that plus the source P, with an emittance of P*(1/A1+1/A2+!/A3).

    For N spheres, P*(1/A1+....+1/AN)

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  12. Nick,
    Thanks for the good work. The Jelbring stuff is the new skydragon. Argue!

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  13. Anon writes "If you have a black body in front of another black body I cannot see (pun intended) how, at thermal eqilibrium, you would be able to dectect its presence."

    Admiral Ackbar would know.

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  14. Nick, Once you have n spheres, then you can adjust for imperfection by making the spheres progressively more oblique. One thing I have noticed, is that the non-uniform distribution of water produces forcing response more like oblique spheres than concentric spheres. You math is much less rusty than mine, http://paoc.mit.edu/labweb/notes/chap5.pdf so you might have more fun with the potential temperature profile of moist air :)

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  15. Nick, I still cannot understand how the heat fluxes for your solution can possibly balance.
    Initially the sphere has an output of 300 W and a flux of 300 W/m2.
    When enveloped by the shell, the shell must radiate 150W/m2.
    In your final steady state you state that the inner sphere emits 450 W/m2.
    We know that the sphere cannot radiate to itself, so the total 450 W must be directed to the shell, so the inner surface of the shell gets a minimum of 225 W/m2. The shell will also radiate some fraction to itself.
    So what are the heat fluxs?
    Outer shell efflux is known to be 150W/m2.
    Inner shell efflux is known to be 150W/m2.
    What is the influx from the sphere to the inner shell?
    What is the influx from the inner shell to the inner shell?

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  16. DocM,
    I think it is:
    Inner sphere S1 emits 450 W, which all lands on the shell S2.
    It receives 150 W from the shell, and generates 300.

    S2 emits 300W out, and 300 W in. Of the 300 in, 150 W lands on S1 and the remaining 150W lands on S2.
    So the balance in W for S2 is
    emits 300+300
    receives 450 (from S1) plus 150 (from S2)
    And for S1
    emits 300W+150W
    receives 150W

    I think it is easier to think of balance in W, then divide by respective areas for W/m2.


    Maybe the simplified numbers create confusion, since 150W has 2 different roles. Suppose S2 is bigger - surface 2.5 m2. Then it still has to emit 300 W out and in. But a smaller fraction of the 300 in goes to S1; only 120W. So 180W lands on S2. Then
    S1 receives 120W, emits 300+120W=420W
    S2 emits 600W, receives 420+180.

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  17. Indeed,

    "Inner sphere S1 emits 450 W"

    The sphere, 1 m2, has a steady state temperature of 298.4 K. It radiates 450 W so the inner shell receives 225 W/m2 from the sphere

    "the remaining 150W lands on S2"

    it also gets 75W/m2 from the shell; a total of 300 W/ms, thus the inner shell, at steady state has an influx 0f 300 W/m2 and the inner surface must radiate at 300 w/m2. So the twice what the outgoing flux should be.

    The influx and the efflux of the inner surface and sphere surface MUST balance.

    "S2 emits 300+300'

    The material in the shell does not know it has two surfaces, the inner surface is getting 300W/m2 and must heat up until it radiates at 300 W/m2; thus it should be at 269.5 K.

    However, your solution can work if the rate the shell material conducts heat is far greater that the rate at which it radiate heat. In that case, heat absorbed at the inner surface can either reradiate (slow) or migrate into the body (fast) and will have an identical likelihood of eventually being emitted on either face.

    If the rate of heat transfer in the shell is slow, then there will be a steady state temperature gradient in the shell. The Earth itself is an example of this.
    If you have ever visited a hot steel rolling mill you would see this in action. The white hot steel cools quickly, by radiation and not by heat transfer into the air. you can observe the cooling and note that the inner core is much hotter than the outer surface.
    Here is a video of a hot steel forge. Not how the yellow hot surface of the pigs is transformed into the white hot center on compression.

    http://www.youtube.com/watch?v=tLRkOupbARM


    Using a guillotine forge cutter on ingots, you ofter observe a white/red center in the two halves even though the exterior surface is dull.

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  18. Doc,
    "However, your solution can work if the rate the shell material conducts heat is far greater that the rate at which it radiate heat. "

    Your original spec at Tallblokes said the shell was 1000 atoms thick (could be hard to make). That's about 0.2 micron. I said there that 300W flowing through would produce a 0.00003C temp drop. That was working on 1μ, at 0.2 it's less than .00001C. That compares with the 228K or whatever needed to radiate 300W.

    In this problem, as stated, spherical symmetry means you don't have to worry about lateral conduction. Nor do you have to worry about the heat distribution within the sphere. That depends on the unstated conductivity anyway.

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  19. "There was another post by Stephen Wilde there on no backradiation etc. It got me looking for the earliest measurements I could find, and this 1837 paper by Pouillet turned up."

    Kevin McKinney has written an essay on The History of Climate Science - William Charles Wells over at SkepticalScience concerning Wells' Essay On Dew that dates to 1814 and may be of interest.

    ReplyDelete