In A Challenge for Bryan I put up a simple heat transfer problem and asked for the equations. Bryan elected not to provide these equations. So I provide the answer, but also attempt some enlightenment for people who don’t think the answer can be correct.
As DeWitt Payne noted, a post with a similar problem posted on Wattsupwiththat managed to gather some (unintentionally) hilarious comments.
Here’s the problem again:
Spherical body, A, of radius ra, with an emissivity, εa =1. The sphere is in the vacuum of space.
It is internally heated by a mystery power source (let’s say nuclear, but it doesn’t matter), with power input = P.
The sphere radiates into deep space, let’s say the temperature of deep space = 0K to make the maths simpler.
1. What is the equation for the equilibrium surface temperature of the sphere, Ta?
The condition of case A, but now body A is surrounded by a slightly larger spherical shell, B, which of course is itself now surrounded by deep space at 0K.
B has a radius rb, with an emissivity, εb =1. This shell is highly conductive and very thin.
2a. What is the equation for the new equilibrium surface temperature, Ta’?
2b. What is the equation for the equilibrium temperature, Tb, of shell B?
The reason for the “slightly larger shell” is to avoid “complex” view factor issues. Of course, I’m happy to relax the requirement for “slightly larger” and let Bryan provide the more general answer.
The reason for the “highly conductive” and “thin” outer shell, B, is to avoid any temperature difference between the inside and the outside surfaces of the shell. That is, we can assume the outside surface is at the same temperature as the inside surface – both at temperature, Tb.
This kind of problem is a staple of introductory heat transfer. This is a “find the equilibrium” problem.
How do we solve these kinds of problems? It’s pretty easy once you understand the tools.
The first tool is the first law of thermodynamics. Steady state means temperatures have stabilized and so energy in = energy out. We draw a “boundary” around each body and apply the “boundary condition” of the first law.
The second tool is the set of equations that govern the movement of energy. These are the equations for conduction, convection and radiation. In this case we just have radiation to consider.
For people who see the solution, shake their heads and say, this can’t be, stay on to the end and I will try and shed some light on possible conceptual problems. Of course, if it’s wrong, you should easily be able to provide the correct equations – or even if you can’t write equations you should be able to explain the flaw in the formulation of the equation.
In the original article I put some numbers down – “For anyone who wants to visualize some numbers: ra=1m, P=1000W, rb=1.01m“. I will use these to calculate an answer from the equations. I realize many readers aren’t comfortable with equations and so the answers will help illuminate the meaning of the equations.
I go through the equations in tedious detail, again for people who would like to follow the maths but don’t find maths easy.
Energy in, Ein = Energy out, Eout : in Watts (Joules per second).
Ein = P
Eout = emission of thermal radiation per unit area x area
The first part is given by the Stefan-Boltzmann equation (σTa4, where σ = 5.67×10-8), and the second part by the equation for the surface area of a sphere (4πra²)
Eout = 4πra² x σTa4 …..[eqn 1]
Therefore, P = 4πra²σTa4 ….[eqn 2]
We have to rearrange the equation to see how Ta changes with the other factors:
Ta = [P / (4πra²σ)]1/4 ….[eqn 3]
If you aren’t comfortable with maths this might seem a little daunting. Let’s put the numbers in:
Ta = 194K (-80ºC)
Now we haven’t said anything about how long it takes to reach this temperature. We don’t have enough information for that. That’s the nice thing about steady state calculations, they are easier than dynamic calculations. We will look at that at the end.
Probably everyone is happy with this equation. Energy is conserved. No surprises and nothing controversial.
Now we will apply the exact same approach to the second case.
First we consider “body A”. Given that it is enclosed by another “body” – the shell B – we have to consider any energy being transferred by radiation from B to A. If it turns out to be zero, of course it won’t affect the temperature of body A.
Ein(a) = P + Eb-a ….[eqn 4], where Eb-a is a value we don’t yet know. It is the radiation from B absorbed by A.
Eout(a) = 4πra² x σTa4 ….[eqn 5]- this is the same as in case 1. Emission of radiation from a body only depends on its temperature (and emissivity and area but these aren’t changing between the two cases)
– we will look at shell B and come back to the last term in eqn 4.
Now the shell outer surface:
Radiates out to space
We set space at absolute zero so no radiation is received by the outer surface
Shell inner surface:
Radiates in to A (in fact almost all of the radiation emitted from the inner surface is absorbed by A and for now we will treat it as all) – this was the term Eb-a
Absorbs all of the radiation emitted by A, this is Eout(a)
And we made the shell thin and highly conductive so there is no temperature difference between the two surfaces. Let’s collect the heat transfer terms for shell B under steady state:
Ein(b) = Eout(a) + 0 …..[eqn 6] – energy in is all from the sphere A, and nothing from outside
= 4πra² x σTa4 ….[eqn 6a] – we just took the value from eqn 5
Eout(b) = 4πrb² x σTb4 + 4πrb² x σTb4 …..[eqn 7] – energy out is the emitted radiation from the inner surface + emitted radiation from the outer surface
= 2 x 4πrb² x σTb4 ….[eqn 7a]
And we know that for shell B, Ein = Eout so we equate 6a and 7a:
4πra² x σTa4 = 2 x 4πrb² x σTb4 ….[eqn 8]
and now we can cancel a lot of the common terms:
ra² x Ta4 = 2 x rb² x Tb4 ….[eqn 8a]
and re-arrange to get Ta in terms of Tb:
Ta4 = 2rb²/ra² x Tb4 ….[eqn 8b]
Ta = [2rb²/ra²]1/4 x Tb ….[eqn 8b]
or we can write it the other way round:
Tb = [ra²/2rb²]1/4 x Ta ….[eqn 8c]
Using the numbers given, Ta = 1.2 Tb. So the sphere is 20% warmer than the shell (actually 2 to the power 1/4).
We need to use Ein=Eout for the sphere A to be able to get the full solution. We wrote down: Ein(a) = P + Eb-a ….[eqn 4]. Now we know “Eb-a” – this is one of the terms in eqn 7.
Ein(a) = P + 4πrb² x σTb4 ….[eqn 9]
and Ein(a) = Eout(a), so:
P + 4πrb² x σTb4 = 4πra² x σTa4 ….[eqn 9]
we can substitute the equation for Tb:
P + 4πra² /2 x σTa4 = 4πra² x σTa4 ….[eqn 9a]
the 2nd term on the left and the right hand side can be combined:
P = 2πra² x σTa4 ….[eqn 9a]
And so, voila:
T’a = [P / (2πra²σ)]1/4 ….[eqn 10] – I added a dash to Ta so we can compare it with the original value before the shell arrived.
T’a = 21/4 Ta ….[eqn 11] – that is, the temperature of the sphere A is about 20% warmer in case 2 compared with case 1.
Using the numbers, T’a = 230 K (-43ºC). And Tb = 193 K (-81ºC)
Explaining the Results
In case 2, the inner sphere, A, has its temperature increase by 36K even though the same energy production takes place inside. Obviously, this can’t be right because we have created energy??.. let’s come back to that shortly.
Notice something very important – Tb in case 2 is almost identical to Ta in case 1. The difference is actually only due to the slight difference in surface area. Why?
The system has an energy production, P, in both cases.
- In case 1, the sphere A is the boundary transferring energy to space and so its equilibrium temperature must be determined by P
- In case 2, the shell B is the boundary transferring energy to space and so its equilibrium temperature must be determined by P
Now let’s confirm the mystery unphysical totally fake invented energy.
Let’s compare the flux emitted from A in case 1 and case 2. I’ll call it R.
- R(case 1) = 80 W/m²
- R(case 2) = 159 W/m²
This is obviously rubbish. The same energy source inside the sphere and we doubled the sphere’s energy production!!! Get this idiot to take down this post, he has no idea what he is writing..
Yet if we check the energy balance we find that 80 W/m² is being “created” by our power source, and the “extra mystery” energy of 79 W/m² is coming from our outer shell. In any given second no energy is created.
The Mystery Invented Energy – Revealed
When we snapped the outer shell over the sphere we made it harder for heat to get out of the system. Energy in = energy out, in steady state. When we are not in steady state: energy in – energy out = energy retained. Energy retained is internal energy which is manifested as temperature.
We made it hard for heat to get out, which accumulated energy, which increased temperature.. until finally the inner sphere A was hot enough for all of the internally generated energy, P, to get out of the system.
Let’s add some information about the system: the heat capacity of the sphere = 1000 J/K; the heat capacity of the shell = 100 J/K. It doesn’t much matter what they are, it’s just to calculate the transients. We snap the shell – originally at 0K – around the sphere at time t=100 seconds and see what happens.
The top graph shows temperature, the bottom graph shows change in energy of the two objects and how much energy is leaving the system:
At 100 seconds we see that instead of our steady state 1000W leaving the system, instead 0W leaves the system. This is the important part of the mystery energy puzzle.
We put a 0K shell around the sphere. This absorbs all the energy from the sphere. At time t=100s the shell is still at 0K so it emits 0W/m². It heats up pretty quickly, but remember that emission of radiation is not linear with temperature so you don’t see a linear relationship between the temperature of shell B and the energy leaving to space. For example at 100K, the outward emission is 6 W/m², at 150K it is 29 W/m² and at its final temperature of 193K, it is 79 W/m² (=1000 W in total).
As the shell heats up it emits more and more radiation inwards, heating up the sphere A.
The mystery energy has been revealed. The addition of a radiation barrier stopped energy leaving, which stored heat. The way equilibrium is finally restored is due to the temperature increase of the sphere.
Of course, for some strange reason an army of people thinks this is totally false. Well, produce your equations.. (this never happens)
All we have done here is used conservation of energy and the Stefan Boltzmann law of emission of thermal radiation.