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.