## Friday, April 06, 2012

### The Same Temperature Problem

Assume a spherical body of unit emissivity, heated from some external (sun) or internal (gravitational collapse), or an electrical resistor, or whatever, which is at a uniform temperature, let's call it Ta.  Now surround the same body with a shell whose emissivity is also unity, at some distance d from the surface of the spherical body with a vacuum in between.  Don't worry about the outer shell emitting to the outside the Bunny got scrith. Call the temperature of the outer shell Tb.   Can it be at temperature Ta?

Let's think about this a bit.  The spherical body emits an amount of energy equal to the amount of energy it is receiving

Ea = 4π r2 σ Ta 4

The shell, also at temperature Ta emits

Eb = 4π( r+d)2 σ Tb 4 = 4π( r+d)2 σ Ta 4
Slight update here for clarity

The ME bunnies now go through a large calculation to figure out how much strikes the sphere and how much misses.  Eli, OTOH, simply pushes r/d >>1.  The net is that Tb < Ta if you want to conserve energy, Eb=Ea.  It's a feature, not a bug.

OK, let's let the outer shell emit to the outside.  Now  we have to divide Eb by two (up and down), but if the hares have it this way we can make the same argument about Ea.  The net of the thing is that radiative equilibrium requires Tb < Ta.

Nick Stokes said...

I guess your second equation should have Tb, not Ta. But isn't the point that the shell emits twice that expression - half up and half down? So in the limit, Ta^4=2*Tb^4?

Which is a stronger version of your conclusion.

Andrew Judd said...

There is no mysterious problem to solve that I can see

1. With no external emissivity the inner sphere is getting hotter and hotter and is heating the external shell to be hotter and hotter

2. If the outer shell is emitting then the hotter heated inner sphere is *still* heating the colder outer shell

No rocket science required

Thomas said...

Are you trying to reproduce Hans Jelbring's lunacy?
http://tallbloke.wordpress.com/2012/02/25/hans-jelbring-stefan-boltzmann-law-and-the-construction-of-a-perpetuum-mobile/

In an equilibrium situation where you have no energy entering or leaving the system Ta=Tb or you violate the second law!

Andrew Judd said...

Thomas

If the inner sphere is being heated you dont have an equilibrium situation. The heated sphere will always emit at an increasingly higher rate than the cooler sphere can ever emit at.

Thomas said...

Andrew, Eli's description of the problem is contradictory. In the text he initially talks about internal heating, but in his equations he doesn't have any. Since he doesn't have any way for energy to leave the system, he can't even have an equilibrium temperature if there is constant energy input.

Anonymous said...

The problem probably is that not all of the power radiated by the outer shell is absorbed by the inner sphere. Rather, just by thinking about the geometry, you note that part of the power radiated by the outer shell is absorbed by (a different part of) the outer shell itself. Now, someone just check that the equilibrium temperature solution is really the solution here. This situation should change with the radiation to vacuum present when the problem becomes a nonequilibrium one.

Speedy Gonzales

Anonymous said...

I'm not sure what is the point of this exercise, but the wrong-headed responses are quite illuminating.

-MO

Tom Curtis said...

Ta = Tb

Emissions from a point on either body do not proceed perpendicular to a tangent with the surface, but equally in all directions. So for any internal point on the shell B, it receives an equal amount of energy from each outer point on the sphere A which is above its horizon. For each ray from an internal point on B that does not intersect with A, it intersects with another point on B.

Ignore conduction. It is then still the case that the temperature of each internal point on B is the same as of every other internal point on B for they all have the same energy balance. Therefore we can ignore all rays not intersecting with A for the energy balance of any given internal point on B, for there is no net energy exchange on any such ray. It follows that any such internal point will increase in temperature until the energy it emits along each ray intersecting with A equals the energy it receives along that ray, ie, until it is the same temperature as the surface of A.

David B. Benson said...

Andrew Judd --- Consider the energy flows of a ordinary, old fashioned, incadescent libht globe (bulb). The electrical energy entering is converted into photons at the resistive element. Some are optical spectrum and escape. Some are IR and heat the glass globe. The glass globe re-emits both inside and outside; to the outside the air is heated and convected away.

And yet after warming up to equilibrium the light globe grows no warmer and the resistive element does not keep growing warmer.

This is similar to the main topic of this thread.

Andrew Judd said...

David

I am not sure what your point is.

As I already pointed out when the shell is not emitting/conducting externally then the sphere is continually heating the outer shell to a higher and higher temperature.

The light bulb is just like the sphere and shell in the later part where where the shell emits/conducts externally.

What is your point??

What are we to learn from this thought experiment and actual observations?

Thomas said...

I think it is time for Eli to step in and state his purpose with his blog post.

EliRabett said...

Consider a surface with two sides. What is the total emission?

Consider a gas. What is the total emission?

Idle bunnies wish to know.

Tom Curtis said...

I assume the same idle bunnies are curious as to how deep a hole is.

EliRabett said...

Depends on the hole