A few months ago Professor Robert G. Brown of Duke University wrote at WUWT and elsewhere that members of Principia Scientific International (PSI) don’t know Mathematics (sic). Brown then defied PSI to prove that the inner core of a hollow metal sphere is not heated by “backradiation” ( a junk science concept alien to thermodynamics experts).
Then, as if to “prove” his claim that “backradiation” is “real” Brown wrote that if we wrapped a hot light bulb up with an aluminum foil, then in a very short time the bulb would be overheated. To Brown the ensuing overheating of the light bulb was by “backradiation” generated from the aluminum foil (applied scientists may be heard sniggering at the very idea!).
However, Brown was unable to show any actual relevant calculations (only a long and boring array of meaningless algebraic gymnastics) to support what he said. Nonetheless, I took very seriously his challenge and I tried to calculate, in the most precise way, what really happens whenever you wrap a light bulb up with an aluminum foil.
What I present herein are typical, sound and established equations that any serious applied physicists, engineers, technicians, etc., will often use in the course of their work in what is euphemistically known as the “real world”. This is perhaps why academics like Brown, residing in their (“unreal world”) ivory towers, so often fail? For instance, it is a routine task for applied scientists and engineers to, say, calculate how much an electric wire may heat surrounding plastic insulation cable, or define the temperature reached by the casing of car engines in close contact with pistons, etc.
Clearly, it comes as a surprise for Mr. Brown (but not for PSI!) that in the fields of applied science it is impossible to find ANY manuals, technical textbooks, etc., using climate science's mysterious “backradiation” to carry out such practical (“real world”) calculations.
“Backradiation” is regarded as a sort of “Arabian Phoenix” in the scientific and technical community. Of course, the academics and global warming believers are saying it exists, but nobody has ever actually found it! Maybe Professor Brown believes otherwise, but I have never met an engineer devising an engine, an industrial plant, or an electric device using “backradiation” to calculate how much heat is passing through the materials!
But let us go back to our calculations.
We can take a small (traditional) light bulb (not new halogen type) with incandescent tungsten filament, 2 centimeters (cm) radius, reaching 100° C Temperature (T) on the glass surface, and very thin aluminum foil (0.01 cm thickness) applied all around the glass surface of the bulb.
Problem: what will be the new temperature reached by the glass surface of the bulb beneath the aluminum foil?
Now, to describe and formalize the above problem in the best mathematical way we can write down a formula like this:
Now, to describe and formalize in the best mathematical way the above problem, we can write down a formula like this:
2π ∫dT = - Q ∫dx/x
No need to be afraid!
This equation simply means, in rigorous mathematical symbolism, that the outflowing (-Q) thermal energy from an unit (1 cm.) zone of a spherical surface 2π (implicit: cm²), of a light bulb whose radius is 2.00 cm., and whose surface temperature is 100° C, is passing through an aluminum foil, whose thickness is 0.01 cm. (2.01 – 2.00 = 0.01).
Integral sign ∫ from “T” bottom to 100 up, simply means: “let us calculate how much the surface temperature of bulb, in close contact with aluminum foil, changes (dT) from an unknown value T (to be discovered) of surface (= 2.00 cm. radius), up to 100° C, corresponding to 2.01 cm. radius.
Not so difficult, is it?
Now, aluminum conductivity on a metric scale is 240 W/m.k, so if we reduce it to a cm. scale, we have to divide per 100, and so we have
K = 2.4 W/cm.k
It is very important to see that the equation above is totally in accordance with 2nd Law of Thermodynamics, and we have – Q = k*… (or Q = -k*…) meaning that heat is flowing from hotter to colder bodies, and NOT the contrary!
As you can see, in the equation above there’s no symbol of “backradiation” at all, only heat flowing from inside the light bulb, to outside.
Now, since – Q = k * A (cm²) *dT/dx (= 0.01cm), we have the following Q rate:
-Q = 2.4/0.01 = 240 joule/sec.°C
Thus, we have:
6.28 ∫dT = 240 ∫dx/x
6.28 * (100 - T) = 240 * (ln 2.01 – ln 2.00) =
628 = 6.28T + (240 * ln (2.01/2.00)) =
628 = 6.28T + (240 * 0.005) =
628 = 6.28T + 1.2 =
T = 626.8/6.28 and finally
T = 99.8° C
So we have that the glass surface of the light bulb decreases its surface temperature very quickly, at a rate of 0.2°C/sec. (100° C – 99.8° C) = 12° C/min., while the aluminum foil surface, in close contact with, it increases it at the same rate, of 0.2° C/sec. But, as the hot electric filament and home electric plant is constantly releasing new electric energy to the glass surface of the bulb, of course there will be no decrease of temperature of the glass surface (that will remain constant at 100° C), but just a rapid increase of temperature of the aluminum foil.
This means that we will record an average increase of surface temperature of aluminum foil around 60° C in just 5 minutes!
But, as in high conductive metals (such as iron, copper, tungsten, silver) the increase of temperature is very rapid, especially when the thickness is very small, then it is experimentally proved that the increase of temperature in aluminum foil will follow a sigmoid function, as the error Gaussian function (erf), namely:
2/√π ∫e ^-t² (dt)
Therefore, it will take much less than 5 minutes for thin aluminum foil to increase the temperature for 50°/60° C.
And then, the aluminum will start to heat the surface of the glass, but not from “backradiation”, but simply by conduction, i.e. the hotter material (aluminum foil) will transmit heat to the boundary of the “colder” material (glass), according to 2nd Law of Thermodynamics.
Now, the question could be: “what would change the surface temperature of the glass light bulb, in close contact with an insulator like polyurethane (and not anymore with a highly conductive metal like aluminum)?”
The answer is not so difficult, as we have learnt from formulas above how to handle this kind of problems.
It is sufficient to replace – in the equations 1) and 2) – the k of aluminum (240), with the k of polyurethane (0.032)
6.28 ∫dT = 0.032 ∫dx/x
Thus, we have:
6.28 * (100 - T) = 0.032 * (ln (2.01/ 2.00))
628 = 6.28T +( 0.032 * 0.005)
628 = 6.28T + 0.00016
T = (628 – 0.00016)/6.28
T = 99.999974° C
This means that the surface of the glass light bulb, in close contact with a thin polyurethane foil, shall “yield” to the polyurethane foil just a very modest amount of energy, corresponding to just 0.000026° C/sec. (100° C – 99.999974° C), namely 0.001° C/min.
But this is not – as anyone can clearly see – the outcome of any fanciful “backradiation”.
The increase in temperature from the aluminum foil surrounding the glass of the light bulb is simply a result of the very high thermal conductivity of aluminum (2.4 W/cm.K). It is the same way as the metal of your car (also with high thermal conductivity) – it too will soon become overheated if you leave your car outside in the hot sun of a summer's day.
But energy ( –Q) – as you can see - is NOT “backradiated” at all because thermal energy is only flowing from hotter to colder bodies (from inside the light bulb to outside). Then the aluminum will start to heat the boundary surface of the glass in close contact with it, but by conductivity, NOT by “backradiation” which doesn’t exist at all!
Conductivity, and NOT “backradiation” is the key-concept to understand regarding the heating mechanism of that aluminum foil.
And again PSI (and applied science) comes to prove this with rigorous, sound, and mathematically indisputable equations. Sadly, for the uninformed academic, Robert Brown, it seems the requisite Math took a little vacation from his statements.