Questioning the Stefan–Boltzmann Law and Questions About Radiative Heat Loss

[Windy_Skies]

The comments below until this one are a part of a pm exchange.

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Note: The radiative heat transfer equation based on Stefan-Boltzmann 4th power law is erroneous and cannot be relied upon. The only way to possibly measure radiative heat transfer is by measuring the intensity of infrared radiation with electrically sensitive instruments.

The Ultraviolet Catastrophe Illustrated.

The Stefan-Boltzmann law states that radiation intensity scales to the 4th power of temperature, drastically overestimating radiation at high temperatures. If we heat a one cubic meter cube to 2000 C°, we radiate 1483 kW/m2, since a one cubic meter cubic has 6 square meters, we would be radiating 8890 kW, or nearly 9 megawatts of power! Clearly, this is impossible because it would mean heating and melting metal would be physically impossible since it would cool through radiation faster than it can be heated! To heat 7,860 kg worth of steel to 2000 C° in one hour, we need to impart 2032 kWh worth of thermal energy, far less than what we would radiate every second. The Stefan-Boltzmann is wrong and must be modified.

It’s very difficult to be an effective inventor who is also a crackpot. I like this video on that: https://www.youtube.com/watch?v=11lPhMSulSU

And how to address it: https://www.youtube.com/watch?v=Cw97Tj5zxvA

You should go to a physics forum and ask where you are wrong in this quote for example.

My attempt:

You chose a special metal, tungsten for example, that is not melted at 2000 degrees Celcius. Tungsten has an emissivity of 0.46 at 2000 degrees Celcius.

To maintain a 1 m^3 cube of tungsten at that temperature in the vacuum of space, assumed to be at absolute zero, you would use this equation:

P = A \cdot M = A \ \varepsilon\ \sigma T^4

Example calculations, with now also the equation with the temperature of the room taken into account: https://testbook.com/physics/stefan-boltzmann-law

P = eσA (T⁴ – T₀⁴)

Let’s say the cube is in an oven that has been turned off and is now at a temperature of 1950 degrees Celsius, now the energy lost through radiation is:

P = (0.46) (5.67 × 10⁻⁸ W/m²K⁴) (6\ m²) × [(2273.15 K)⁴ – (2223.15 K)⁴] = 355674\ watts

This particular oven is likely very poorly insulated if it can maintain such a temperature difference with such a hot object in it.

The oven will likely be insulated. Let’s assume its outside temperature is at 300 degrees Celsius, it has an emissivity of 0.9, and a surface area of 12 m^2, in a room at 50 degrees Celsius.

P = (0.9) (5.67 × 10⁻⁸ W/m²K⁴) (12\ m²) × [(573.15 K)⁴ – (323.15 K)⁴] = 59404\ watts

[ChristophePochari1]

The Stefan-Boltzmann Constant is in Error..

[ChristophePochari1]

You mention tungsten. Tungsten has an emissivity of 0.74 in its molten state. For a cubic meter block, it would radiate 46,000 kW, or 46,000,000 J/s at 3,400 C. The heat capacity of tungsten is 134 J/kg K, since we are radiating 46,000,000 joules per second, in the first 10 seconds the metal would drop in temperature by 240 K, such rapid cooling is impossible.

[Windy_Skies]

Why are you only referencing Gary Novak. Gary Novak is a complete and utter crackpot. Read what he writes. He thinks all of modern physics is wrong: About Gary Novak, Independent Scientist.

I stumbled onto the misdefinition of energy in physics in 1983. It means the equation for representing kinetic energy is wrong. It has velocity squared within it ( KE=½mv² ), while nothing can move at velocity squared. I developed a simple and unquestionable mathematical proof of the error and conducted physics experiments on the subject. About ninety percent of physics is corrupted by the error. Everything else in physics besides Newton’s laws is corrupted by the resulting shift in standards for contriving instead of measuring.

I did the math on one physics question you had. The physics question you have now you could ask on a physics forum. Here are search results: https://www.csc.com.tw/csc/ts/ena/pdf/no35/pages/2-Analysis%20of%20Temperature%20Dropping%20of%20Molten%20Steel%20in%20Ladle%20for%20Steelmaking.pdf

Problem 97 When a molten metal is cast in a... [FREE SOLUTION] | Vaia

Please go study physics instead of trusting crackpots on the internet.

[Windy_Skies]

@ChristophePochari1

The Stefan-Boltzmann Constant is in Error..

Thermal radiation

In a practical, room-temperature setting, humans lose considerable energy due to infrared thermal radiation in addition to that lost by conduction to air (aided by concurrent convection, or other air movement like drafts). The heat energy lost is partially regained by absorbing heat radiation from walls or other surroundings. Human skin has an emissivity of very close to 1.0.[26] A human, having roughly 2 m2 in surface area, and a temperature of about 307 K, continuously radiates approximately 1000 W. If people are indoors, surrounded by surfaces at 296 K, they receive back about 900 W from the wall, ceiling, and other surroundings, resulting in a net loss of 100 W. These estimates are highly dependent on extrinsic variables, such as wearing clothes.

[ChristophePochari1]

The exposed surface of the molten metal on top of the crucible would instantly solidify if the S-B law were the 4th power as claimed. Imagine a 50mm thick layer over a 1 square meter surface of molten metal, radiating 7600 kW/m2 or 7,600,000 J/s, the first 10 millimeters would cool at a rate of 405 K/s. The lid on the crucible might be magnesium oxide as a refractory metal, if radiated with 7.6 megawatts per square meter with an emissivity of 0.5 would instantly vaporize. In 60 seconds, a 50-millimeter thick magnesium oxide lid placed on top of the molten tungsten bath would reach 2000 C if its starting temperature was 20 C. If this refractory lid were already at 1000 C it would reach 4800 C in 120 seconds which is above its boiling point! This can easily be shown with the calculator below. Such heating is evidently not observed in real life suggesting the 4th power relationship between radiation intensity and temperature cannot hold. Science is an evolutionary process in which discoveries are made by observing discrepancies and paradoxes.

Radiative Warming Calculator

[ChristophePochari1]

Using this radiative cooling calculator, the molten tungsten should cool down to 1000 C in only 140 seconds for the first 20 millimeters of thickness. Clearly, this would make melting it impossible unless heated at the rate of cooling which would require nearly 8 megawatts of thermal power for only 1 square meter of molten metal surface.

Radiative Cooling Calculator

[Windy_Skies]

Do you mind me making this a public topic?

[ChristophePochari1]

No not at all, as long as the debate is kept to the facts and does not become personal or antagonistic.

[Windy_Skies]

People are people are people, so I can’t guarantee that. You can expect the usual antagonism. If you think a comment crosses a line, you can flag it and I will remove it.

[ChristophePochari1]

It doesn’t bother me. I think it would be of great public value to discuss, what I believe and others do as well, the probable deficiencies of the S-B law.

[Windy_Skies]

My physics isn’t good enough to be able to model all these different questions. Let’s just stick with tungsten at 2000 degree Celsius. Your source here is another crackpot, though not as atrociously delusional at the other one, so I can’t trust the model. He himself also says it is simplified.

https://www.researchgate.net/publication/267331267_Tungsten

At room temperature, the thermal conductivity of tungsten is about 175\ W m^{-1} K^{-1}. In the range of temperatures 1200–2800 K, the coefficient of thermal conductivity \lambda \ ,W\ m^{-1}K^{-1}, obeys the following equation:

\lambda = 108.34 - 1.052 \times 10^{-2} \cdot T + \dfrac{23419.9}{T}

where T is temperature, K. At the melting point, the value of \lambda drops from 89.5 to 70.5 W m^{-1} K^{-1} [87]. The variation of the thermal conductivity of tungsten with temperature is shown on the basis of several studies [15–19] in Fig. 3.1.

Taking a shortcut, from the graph we see a thermal conductivity of about 100\ W m^{-1} K^{-1} at 2000 degrees Celsius. I didn’t find a source for its thermal conductivity at very low temperatures, so I am not going to try to calculate the heat loss to space, while in space, at 3 degrees Kelvin.

Let’s take a tungsten cube, and expose one side to either the vacuum of space or to some other environment, one side to a tungsten plate at a constant 2000 degree Celsius, and have the other 4 sides be perfectly thermally insulated.

At room temperature, I’ll say at 25 degrees Celsius, it has a thermal conductivity of about 175\ W m^{-1} K^{-1}, at 2000 degrees Celsius about 100\ W m^{-1} K^{-1}. It’s wrong to do so, but I’ll use the average of about 137.5\ W m^{-1} K^{-1}.

Thermal Conductivity, Stefan Boltzmann Law, Heat Transfer, Conduction, Convecton, Radiation, Physics

P = \dfrac{K \cdot A \cdot\triangle T}{l}

\dfrac{137.5 \cdot 1 \cdot 1975}{1}= 271562.5 \ watts

This is very much simplified, so let’s say this is in a vacuum at room temperature, and the room is infinitely large.

271562.5 watts needs to go into the plate to keep it at 2000 degrees Celsius, and the same 271562.5 watts is lost to the room.

I’m not sure I understand the result, or if it is right. Let’s try to see how hot the outside face of the cube needs to be to radiate at 271562.5 watts .

P = eσA (T⁴ – T₀⁴) => \dfrac{P}{eσA}= T⁴ – T₀⁴ => T = \sqrt[4]{T₀⁴ + \dfrac{P}{eσA}}

\sqrt[4]{298⁴ + \dfrac{271562.5}{0.46 \cdot 5.67×10⁻⁸ \cdot 6}}= 1149\ Kelvin \ / \ 876\ Celsius

867 degrees Celsius is not the same as 25 degrees Celsius, so maybe this equation only holds if the face I now exposed to a vacuum, I instead attach to a constant 25 degrees Celsius heat sink?

[Windy_Skies]

Also please show your work if you do calculations, so it can be checked. You’re probably not including the temperature of the environment in your calculations.

[ChristophePochari1]

The radiative cooling and heating calculator provided seems perfectly accurate, it simply integrates the heat loss as a function of time.

P = ε σ A (T⁴ - T_ambient⁴)

dT/dt = - (P / m c) = - (ε σ A (T⁴ - T_ambient⁴)) / (m c)

dT/dt = - (ε σ A (T⁴ - T_ambient⁴)) / (ρ A d c)

t = ∫ _Tinitial^T_final - (ρ A d c) / (ε σ A (T⁴ - T_ambient⁴)) dT

Symbol	Meaning
P	Radiative power (W)
ε	Emissivity of the tungsten slab (dimensionless)
σ	Stefan-Boltzmann constant (5.67 × 10-8 W/m2K4)
A	Surface area of the slab (m2)
T	Temperature of the tungsten slab (K)
Tambient	Ambient temperature (K)
m	Mass of the tungsten slab (kg)
ρ	Density of tungsten (kg/m3)
V	Volume of the tungsten slab (m3)
d	Thickness of the slab (m)
c	Specific heat capacity of tungsten (J/kg·K)
dT/dt	Rate of temperature change with respect to time (K/s)
t	Time (s)

[ChristophePochari1]

A few additional pieces of evidence that suggest the S-B law is greatly overstating the radiative flux produced by objects.
Gary Novak also points out that ice should radiate 300 W/m2, implying that an ice skating rink should heat the air above it or a table full of ice cubes should heat a room.
MIRVs should also be cooling down faster than they heat up since they reach around 2800 C as they enter the atmosphere. The MIRV is covered with an ablative surface made of phenolic resin. If the surface temperature of the MIRV reaches 2800 C, it should radiate around 4,800 kW, causing the material to rapidly cool beyond what the atmosphere can provide in heat.
The 4th power S-B also predicts unrealistically high radiation from extremely hot plasmas such as plasma torches. A plasma torch is thought to reach as hot as 26000 C, which would radiate 2,700 kW from a 1 cm2 surface. A typical plasma arc in a torch might be 1.5mm wide and 7mm long, giving a surface area of 0.8 cm2, which would mean the plasma torch should radiate 2.1 MW of power, which is 70 times greater than the power usage of a heavy-duty cutting torch of around 30 kW.

[Windy_Skies]

@ChristophePochari1

P = ε σ A (T^4 - T_{ambient} ^4)

Where do you see this as being the function of the calculator? I see a description of what it claims to do, not how it does that. I do see this disclaimer in bold:

Cooling of an infinite slab to deep space (0 K) by radiation
Assuming no conduction, no convection, and no radiation from other sources

Which makes it pretty useless unless you want to cool your tungsten cube in outer space.

Let’s calculate the energy difference between a one meter cube of tungsten at 25 degrees Celsius and at 2000 degrees Celsius: Q=mc \triangle T

Heat capacity c is 132 J/(kg K) (value for solid phase), mass m is 19250 kg, so 19250\ \cdot 132\ \cdot \ 1975 = 5018475000\ J , or 5 GJ.

Here is a graph I made:

I evaluate P=eσA(T⁴–T₀⁴) every second and subtract that from the remainder of the initial 5 GJ, and then convert that back into a temperature value to use for the next second. This should show the cooling of the tungsten cube from its starting temperature at 2273.15 Kelvin in a constant temperature vacuum at 298.15 Kelvin. This looks at the first 100000 seconds / 28 hours. I ignore conduction inside and outside the cube, and also convection and insulation outside the cube. After 28 hours the cube is still at 377 Kelvin / 102 degrees Celsius.

The first 20 seconds:

Temperature	Temperature^4	Seconds	Radiative loss in watts	Starting delta Q in Joules
2273.15	2.67001E+13	0	4177561.803	5018475000
2271.505938	2.66229E+13	1	4165487.559	5014297438
2269.866628	2.65461E+13	2	4153474.29	5010131951
2268.232045	2.64698E+13	3	4141521.556	5005978476
2266.602166	2.63938E+13	4	4129628.922	5001836955
2264.976968	2.63181E+13	5	4117795.958	4997707326
2263.356427	2.62429E+13	6	4106022.236	4993589530
2261.740519	2.6168E+13	7	4094307.334	4989483508
2260.129221	2.60935E+13	8	4082650.832	4985389200
2258.522511	2.60194E+13	9	4071052.315	4981306550
2256.920365	2.59457E+13	10	4059511.372	4977235497
2255.322761	2.58723E+13	11	4048027.595	4973175986
2253.729677	2.57993E+13	12	4036600.58	4969127958
2252.141089	2.57266E+13	13	4025229.928	4965091358
2250.556977	2.56543E+13	14	4013915.241	4961066128
2248.977317	2.55823E+13	15	4002656.128	4957052212
2247.402088	2.55107E+13	16	3991452.198	4953049556
2245.831269	2.54395E+13	17	3980303.066	4949058104
2244.264837	2.53686E+13	18	3969208.349	4945077801
2242.702772	2.5298E+13	19	3958167.67	4941108593
2241.145051	2.52278E+13	20	3947180.652	4937150425

https://www.reddit.com/r/theydidthemath/comments/6p6zs7/request_how_long_would_a_molten_ball_of_iron_take/

The conductive volumetric heat flux dq/dt is given by

dq/dt = ρ.c.dT/dt - ∇.(k∇T).

where ρ is the density, c is the heat capacity, T is the temperature and k is the Boltzmann constant.

This is the first law of thermodynamics and is a form of the second order partial differential equation known as, aptly, the heat equation.There are various methods for solving numerically, but obviously has no analytical solution for arbitrary ΔΤ.

However that’s only the conductive cooling of the air. You’d also need to include black body thermal cooling given by

dQ_rad/dt = 4πr2σT4.

where r is the radius, σ is the Stephan-Boltzman constant and T is the temperature. This equation is a bit similar. If our cannonball is in a vacuum, this is the equation we can use.

[dougselsam]

@Windy_Skies

It’s very difficult to be an effective inventor who is also a crackpot.

Considering that the AWE field has now proven to have involved almost 100% crackpots - results, and especially lack thereof, are undeniable at this poiint. Not sure whether discussing the characteristics of tungsten at 2000 degrees C. could save it, but I’m not seeing how this could be relevant to the discussion. Brings us back to “really smart people” and, what was the term from Barnard, “amazing minds” or something to that effect? The problem with wind energy is, air is invisible so people (especially crackpots) can imagine the air doing whatever they want. Therefore, wind energy has always been, and continues to be, a magnet for crackpots. And airborne wind energy is a neodymium supermagnet!

[Windy_Skies]

@ChristophePochari1

MIRVs should also be cooling down faster than they heat up since they reach around 2800 C as they enter the atmosphere. The MIRV is covered with an ablative surface made of phenolic resin. If the surface temperature of the MIRV reaches 2800 C, it should radiate around 4,800 kW, causing the material to rapidly cool beyond what the atmosphere can provide in heat.

You can ask this on a physics forum, I don’t understand this enough to be useful. This one for example: https://www.physicsforums.com/

From what I understand it, the reentry vehicle is kind of like a hammer and the atmosphere like an anvil at reentry speeds. The reentry vehicle is constantly hammering the air and heating up. Its resultant temperature comes from the balance of the things heating it up and cooling it down.

Here are videos:

Apollo - Atmospheric Entry Phase (1968)

Apollo - Atmospheric Entry Phase (1968)

Planetary Entry- Aerodynamic Heating & Deceleration (NASA)

Planetary Entry- Aerodynamic Heating & Deceleration (NASA)

If that doesn’t explain it enough or is too outdated, you could ask on the physics forum for example.

@ChristophePochari1

Gary Novak also points out

Gary Novak is an insane person. Listen to literally anyone else. Gary Novak is an insane person. Yes, ice radiates heat away. But it also receives radiation, and is insulated by the atmosphere.

Midweek Cuckoo:Gary Novak | moonflake

Let’s start with a quote from him, just to warm you up to the way he thinks:

I was thinking about Einstein’s equation which says that the energy in matter is equal to its mass time the velocity of light squared. It occurred to me that he would have had something if he hadn’t squared the velocity of light, because nothing can move at velocity squared.

Yes indeed, Gary appears to think that the “c squared” is meant to imply that something is actually moving at velocity squared.

[…]

Gary also seems to think that eating transgenic crops will cause us to become genetically similar to the genes used in the crops, which makes me wonder about his qualifications as a biologist. The genetic material of what we eat does not survive our stomach acids! Gary also believes that chlorinated water causes heart disease, that fluoride in drinking water is toxic, that all cancer is caused by nanobacteria or viruses, that there is no scientific method by which carbon dioxide in the atmosphere can cause global warming, that oceans can heat up without transferring that heat to the atmosphere, and that ice ages are caused by hot spots rotating in the earth’s core.

It must be very difficult for Gary to have revolutionized just about every subject in academia, and yet still be known only as a mushroom scientists who can’t get his papers published.