Guest Post by Willis Eschenbach
For those who enjoy mathematical puzzles, I’m putting this one out there for your pleasure.
Suppose we have a 1 metre by 1 metre by 1 metre concrete block floating in outer space. For the purposes of the puzzle, let’s suppose that there is no longwave background radiation at all.
The block is insulated on four sides, as shown in blue below, with the front and back of the block uninsulated. We’ll further suppose that the insulation is made of Unobtanium, which is a perfect insulator, so no heat at all is lost from the four insulated sides.
Next, let’s assume the emissivity “epsilon” of the concrete block is 0.95. [And as a commenter pointed out, let’s assume that the emissivity and absorptivity across the spectrum are both 0.95 everywhere. Yes, I know this isn’t reality, but it’s a thought experiment.] And we’ll say that the thermal conductivity “k” of the concrete is equal to 0.8 watts per metre per kelvin (0.8 W/m K^-1)
Finally, let’s assume that it gets full-time sunshine on the front side at a rate of 1360 watts per square metre (W/m2). Figure 1 shows the experimental setup.
Figure 1. Setup for the thought experiment. The concrete block (gray) is a one-metre cube. The blue insulation prevents any heat from escaping from the four sides. However, the block is free to gain heat by radiation on the front side, and to lose heat by radiation from both the front and the back sides.
Here’s the puzzle. If the concrete block starts at absolute zero, it will slowly warm up until it is at steady-state, neither warming nor cooling.
So the question is: at steady-state, what will be the temperature T_hot of the hot side and the temperature T_cold of the opposite cold side?
w.
REQUESTS: First, let me ask that when you comment, please quote the exact words you’re discussing. It avoids many problems.
Next, as my high school math teacher would say, please show your work.
Finally, please focus on the question and the answers, and leave out all ad hominems, personal comments, and insults, as well as abjuring any discussion of your opponent’s education, age and species of likely progenitors, improbable sexual habits, or overall intelligence.
Psst! Hey there, nerd at the desk to my right- could you move your left arm so I can see what you’re writing? I have no idea how I got into this exam, and I don’t want to look like a total moron.
The Stephan-Boltzmann Constant = 5.670 374 419 x 10^-8 W m^-2 K^-4 Exact.
See Numerical value 5.670 374 419… x 10-8 W m-2 K-4>NIST Fundamental Physical Constants CODATA 2018
https://physics.nist.gov/cgi-bin/cuu/Value?sigma|category=physchem_in
Willis
can you give me the answer on the back page, i got another class i have to get to.
Stefan-Boltzman, Watson, Stefan-Boltzman.
Assuming an emissivity coefficient = 1 (at steady state the block behaves as a perfect black body, according to the statement), then:
T = 393.54 °K
T = 120.4 °C
Impossible to write the equation here.
That’s what I made it on the hot side and -270.4K on the cold side.
-270.4 K would not be radiating anything. So it would be very similar to having the backside completely insulated, and remove this insulation after 10,000 hour of the sunlight shining on front.
And before removing insulation the back side should be about 120 C. After insulation on back side is removed, the surface rapidly cools. And with insulation removed for 10,000 hours, most of block will be hot {120 C} and it seems backside will be warmer than -270 K, seems likely it’s about 100 K.
Oh, -270.4 K is wrong it should -270.4 C to make any sense.
-270.4 K could be the outside temperature of the unobtanium sides. 😉
Since 0 C = 273.15 K, you have one side at 393.54 K, and the other at 120.4 + 273.15 = 393.55 K. That’s the same to almost 5 significant digits. I assume that’s what you intended?
Front and back the same to almost 5 significant digits?
42
Lol
+1
So are you saying 1 or 43?
..and thanks for the fish?
I think Tyler is correct. Since we know that is the answer to the ultimate question. that no one was able to answer. Judging from the myriad responses below it would appear to meet that criteria. So, I say we name him President of the Galaxy as his reward for the right answer.
Correct – but, what I need to know is, What happened to the CO2 released whilst the concrete was setting – was it left on Earth???!!!
T-hot 260F
T-cold -280F
You have the hot side being colder than the cold side?
Is 260 lower than -280?
Hot side 126dC; cold side 107dC.
126dC?
dC = dangCold?
Space – the Hotter Frontier
One of the heated issues underlying greenhouse theory is whether space is hot or cold.
Greenhouse theory says that without an atmosphere the earth would be exposed to a near zero outer space and become a frozen ice ball at -430 F, 17 K.
Geoengineering techniques that increase the albedo, the ISS’s ammonia refrigerant air conditioners, an air conditioner in the manned maneuvering unit, space suits including thermal underwear with chilled water tubing, UCLA Diviner lunar data and Kramm’s models (Univ of AK) all provide substantial evidence that outer space is relatively hot.
But outer space is neither hot nor cold.
By definition and application temperature is a relative measurement of the molecular kinetic energy in a substance, i.e. solid, liquid, gas. No molecules (vacuum), no temperature. No kinetic energy (absolute zero), no temperature. In the void & vacuum of outer space the terms temperature, hot, cold are meaningless, like dividing by zero, undefined. Same reason there is no sound in space – no molecules.
However, any substance capable of molecular kinetic energy (ISS, space walker, satellite, moon, earth) placed in the path of the spherical expanding solar photon gas at the earth’s average orbital distance will be heated per the S-B equation to an equilibrium temperature of: 1,368 W/m^2 = 394 K, 121 C, 250 F.
Like a blanket held up between a camper and campfire the atmosphere reduces the amount of solar energy heating the terrestrial system and cools the earth compared to no atmosphere.
This intuitively obvious as well as calculated and measured scientific reality refutes the greenhouse theory which postulates the exact opposite even incorrectly claiming the naked earth would be a -430 F ice ball.
Zero greenhouse effect, Zero CO2 global warming and Zero man caused climate change.
–Nick Schroeder February 28, 2020 at 6:58 am
Space – the Hotter Frontier
One of the heated issues underlying greenhouse theory is whether space is hot or cold.–
Space is neither hot or cold. The lunar surface, which very close to vacuum of space, is about 120 C in sunlight and after 2 weeks the surface is about 100 K.
But the top surface of Moon surface is about 0 C just before before the sun goes down- or sun at low angle above horizon is not lunar surface which is level {a rock vertical to surface can be he heated by the sun by a lot {120 C}.
–Greenhouse theory says that without an atmosphere the earth would be exposed to a near zero outer space and become a frozen ice ball at -430 F, 17 K.–
No greenhouse theory doesn’t say this. But people might imagine that Earth without an atmosphere could be similar to the Moon. But the Moon is quite different than Earth and the Moon has slow rotation. And if Moon had enough atmosphere to make any difference {a Mars atmosphere would not make much difference} then the daytime lunar surface temperature would be cooler and night time temperature would be much warmer.
But what is as important as Earth’s dense atmosphere is that 70% of Earth is covered by oceans. Or if Moon had thick atmosphere and an ocean covering 70% of the surface, it would quite similar to Earth.
And like Earth is matters where the Land areas are, within the global ocean.
Or we currently in an Ice Age, this is related to where the land is related to the ocean, in non Ice Age conditions on Earth, Earth has much higher average temperature than 15 C.
Yes, without an atmosphere, the Earth would be hot. Similar to but not identical to the Moon on the Day Side. The Moon’s daylight side remains in Solar Radiation for about 356 hours before It’s Tidally Locked orbit begin to remove the effects of Solar Radiation for a period of 356 hours. The Earth will only heat from Solar Radiation for a period of around 12 hours at the equator before cooling for the next 12 hour period.
So the Earth will not get as hot as the Moon during daylight hours or cool as significantly during the nighttime hours
Nick,
One hundred kilometers up, at the edge of space, there are about one million million million molecules per cubic meter.
Further up at the space station, there are a hundred times less…only ten trillion per cubic meter.
Halfway to the Moon, away from the influence of the atmosphere, there are still a lot of molecules.
Some 7,000,000 per cubic meter.
I do not know where this empty place with no temperature is.
And what does it mean to say at a temperature of absolute zero, there is no temperature?
You said what the temperature is in the same sentence you said there was not one.
If we are doing science, let’s do science.
If you want to demonstrate that 7 million molecules per cubic meter is not a lot of molecules…do that.
But if the point is that with no molecules there is no temperature, and yet there are molecules, what exactly are you saying?
The upper atmosphere of the Earth is called the thermosphere cause it is hot, and it gets hotter when solar activity is high.
This is because the average velocity of the molecules is high, even though there are not a lot of them.
BTW…here is enough air there (at the height that low Earth orbit objects are placed) to have caused the crash of the Spacelab satellite when the atmosphere heated up and expanded more than anticipated back in the 1970s, IIRC.
I am wondering if anyone who knew little to nothing about this subject would be able to learn anything by reading these comments.
I feel like I just got stupider.
Space has no temp because it has no molecules. But it does have some.
Space is hot because stuff in the unfiltered sunshine above the Earth gets real hot on one side?
Huh?
That has nothing to do with the temperature of space.
Even here in Earth we do not count the temperature of an object in direct sun to be an actual air temp.
All measurements are specifically understood to be, and measured in, the shade.
Because otherwise you are measuring a radiation effect, not anything to do with the motion of the molecules.
I am not complaining or anything…you guys comment however you want.
I just do not understand what the point is.
Personally, I like to say things that inform people.
There are 3.3455×10^22 molecules of water in a gram of water. 334,550,000,000,000,000,000,000 (334 sextrillion molecules…
A gram of water is approximately equal to +/- 1/4 teaspoon.
7,000,000 molecules of water takes up 1/4.9357143e+15 (1/4,935,714,300,000,000) 1-5 quadrillionth of a teaspoon. 7,000,000 is a very large number unless you are talking about a very small object
7,000,000 water molecules will fit on the
HeadPoint of a NeedleYour examples would be more apt if you used gas molecules in air.
Nick S
You are making a conceptual error of great importance:
“Greenhouse theory says that without an atmosphere the earth would be exposed to a near zero outer space and become a frozen ice ball at -430 F, 17 K.”
Greenhouse gas theory makes no such assertion. The Greenhouse gas theory (as expounded by the IPCC and Gavin and lots of others who are conceptually misdirected) reads more like this:
“Greenhouse theory says that without any GHG’s in the atmosphere the near-earth average air temperature would the same as the average temperature of the surface of the moon, which has no atmosphere.”
“Greenhouse gas theory” has nothing to say about planets without any atmosphere at all. Greenhouse gas theory is limited to the effects of greenhouse gases. Greenhouse gas theory is subject to the laws of physics, not a progenitor of them.
It the Earth had an atmosphere containing no GHG’s at all, Gavin says (contravening physical laws governing convective heat transfer) the air would be as cold as the surface of the moon. That is how screwed up is the conceptual framework within which NASA/GISS operates.
Thank You.
I have had the same thoughts since I first heard of this CC Propaganda.
As a Nuclear Engineer I learned about flow in a pipe, and Nucleate Boiling. Even with flows of hundreds of feet per second a “film” will persist on the surface of the pipes in coolers or heat exchangers. This film is an insulator, that is, it slows down the transfer of heat from the pipe to the liquid or vice versa. This same effect is present in the atmosphere surrounding the earth. The reason for dimples on golf balls is to minimize this persistent film on the golf ball. Relatively, the Earth is a thousand times smother than the dimpled gulf ball thus the atmosphere is relatively stagnate close to the surface creating an insulator several feet thick of poorly conducting AIR. three to six feet of air is a very good insulator. Period. This is clearly shown in the temperature gradient of atmosphere. There are many charts on the internet. This will be present with or without CO2. It has been much to long since I took advanced thermodynamics courses so someone else can calculate the mu or thermal resistance of this effect.
Uzurbrain
I caution against describing any “stationary” layer or film or clinging when it comes to describing a surface with a flowing fluid (air is also a fluid). The reason is that what you describe is a representation of reality, is a way to think about a problem, not that there is an actual stationary layer, even one molecule thick.
I had the company of a newly graduated young French engineer in Dakar during the design of a concrete cooking stove and the modeling of the equilibrium outside surface temperature was that day’s task. The engineer had been taught the most recent approach to making this calculation which was to assume that there was a stationary “layer” that was conductive only, with no convective heat transfer at all, and that the default thickness was 0.1mm. This “layer” does not exist, it is only a mathematical convenience that approximates what really happens and is, as they say, good enough for government work.
Willis’ proposed problem is interesting for a student: how literally should we take the problem description? The material is given as “concrete”. Heat flow through concrete is like what happens with a porous matrix filled with fluid. If we take it that the 1 metre cube was cut from a long 1×1 m beam, the faces would have a number of sliced stone pieces bound in a limestone matrix. The emissivity of the face is given as 0.95 (average) and that is fine, however a good student will ask what the aggregate size is. The radiation from both faces will be affected by a factor being the aggregate size divided by the length (because the heat conductive properties of the aggregate and the matrix are different). The effective length of the block will be less than one metre if the heat conduction is the average of 0.8 W /sq m/K. In reality (not in effect) the aggregate will mine heat from within the body and radiate it to space more effectively that would a homogenous material (which concrete is not).
For an engineer to ignore the difference, they should first calculate the effect of the granularity and then show that the difference is or is not significant. If it was like a hydro dam containing 76mm aggregate, the difference is probably significant given that the input power was provided to 4 significant digits and the constants are presumed to be perfect.
All gaseous molecules absorb radiation under some circumstances. For instance, N2 and O2 are actually statistically significant absorbers of radiation due to collisional perturbation at the same time as they are impinged upon by a photon. The collision perturbs their magnetic dipole, allowing them to absorb radiation, which they ordinarily wouldn’t do.
There’s a recent study somewhere online, I’ll try to find it. I think I saw it on NoTricksZone.
{ Ten Minutes Later… }
Here it is:
Scientists: Oxygen & Nitrogen ‘Radiatively Important’ Greenhouse Gases With IR Absorption Temps Similar To CO2
https://notrickszone.com/2020/02/10/scientists-oxygen-nitrogen-radiatively-important-greenhouse-gases-with-ir-absorption-temps-similar-to-co2/
Of course, any molecule which can absorb radiation must also be able to emit that radiation.. and given the predominant heat transport means of convection, a N2 or O2 molecule would likely absorb a photon, become excited in one of its available quantum states with some of that energy equipartitioned into translational energy, thus be convectively transported higher in the atmosphere, then release that energy in the form of a photon… so gases actively “pump” energy upward via convection.
And given that the mean free path length of radiation increases with altitude, this means there is more upward terrestrial LW flux than downward. There is no “greenhouse effect” in the atmosphere as the leftists claim it to be. Once again, the leftists and climate loons have turned reality on its head and gone off squawking about the end of the world, when in reality it’s exactly opposite to what they claim. I’ve noticed they tend to do that a lot.
If Earth had no “greenhouse gases”, then no gases could radiatively emit energy to space. Radiative emission to space is the *only* means by which the planet can shed energy. So Gavin Schmidt claiming the atmosphere sans “greenhouse gases” would be cold is, yet again, diametrically opposite to reality. In reality, the gases would heat up via conduction upon contacting the surface and convect upward… but they couldn’t emit that energy to space. The whole of the atmosphere would heat up, with a very small lapse rate.
The problem I have is that the AGW Crowd keeps talking about the absorption od energy in CO2 and ignoring all of the rest of the energy hitting the earth. The percentage of energy from sun in the IR spectrum is 1/10^23 of the energy spectrum emitted by the sun. Even one tenth of 10^23 times the IR energy absorbed is definitely significant.
They also ignore all of the energy from cosmic ray interactions, One source said that the typical cosmic ray impact releases more energy into the atmosphere than all of the power generated on Earth! ! The article also claims that there are hundreds of thousands cosmic ray strikes strikes per hour. What is all of that energy doing? The article was describing Cosmic Rays and had no discussion at all on “Climate Change.” Surely the location of the Earth in our Galaxy affects the number of Cosmic rays hitting us. Also the activity on the Sun also affects the magnetic field around the earth and also affects the number of cosmic rays hitting the Earth. I can hear this activity as background noise while using my Amateur Radio. There are times when this noise is more than an S9 signal. Even the noise from Jupiter rises and falls in a predictable manor. And that “Noise” is strong enough at times to operate a one transistor radio. [Radio JOVE – NASA]
Henrik Svensmark a professor at Danish Technical Univeristy has been investigating the cosmic ray formation of cloud nuclei, after he found out, that the weather changes seen in phase with the solar cycle of approx 11 years could not be explained by TSI wich only vary approx 0.1 pct. He found that the solar wind blows away the cosmic rays and in low solar activity periods it do not. So cosmic rays forms cloud nuclei, which form the clouds, which controls the temperature of the earth.
Sorry, Usurbrain, but your numbers for cosmic ray energy are far, far too high. Here’s the reality:
SOURCE
That’s the energy in a baseball dropped from about 100m (330 feet) …
w.
Oh come on, assume a spherical Earth.
Nick Schroeder wrote, “any substance capable of molecular kinetic energy… placed in the path of the spherical expanding solar photon gas at the earth’s average orbital distance will be heated per the S-B equation to an equilibrium temperature of: 1,368 W/m^2 = 394 K, 121 C, 250 F.”
From the fact that there are actually many such objects which have average temperatures much lower than that you should realize that your statement cannot possibly be correct.
For example, the average temperature of the moon (which has no atmosphere) is much lower than the average temperature of the Earth (which does have an atmosphere), even though they receive the same solar irradiance.
In fact, there is not single, predictable temperature for an object one AU from the Sun, because there are many factors which affect such an object’s temperature.
Differences between the radiation emitted by an object orbiting one AU from the Sun, and the radiation absorbed by that object, can cause the object’s temperature to vary greatly.
One obvious difference is “color.” Solar radiation is mostly much shorter wavelengths than radiation emitted by objects orbiting the Sun. An object which is has high emissivity for long wavelength radiation, but high reflectivity to the shorter wavelengths which dominate solar radiation, will end up being cooler than a similar object which has low emissivity for long wavelengths, but low reflectivity to the shorter wavelengths that dominate solar radiation.
Another difference is that radiation is absorbed by our orbiting object only on the side facing the Sun, but it is emitted in every direction. Manmade satellites are often engineered to make use of that fact. An object which is reflective on the side facing toward the Sun, but black on the side facing away from the Sun, will stay much cooler than an object of uniform color.
Another difference is rotation. Consider the Earth and its Moon. Although the Earth and Moon receive the same level of solar insolation, even if the Earth had no atmosphere, and even if its albedo was identical to that of the moon, the Earth’s average temperature would still be higher than the average temperature of the moon.
That’s because the Moon rotates only 1/27-th as fast as the Earth. That means each side of the Moon is heated by the Sun for 27 times as long as each side of the Earth is heated; and then cools for 27 times as long, as well.
That means that, even if the Earth had no atmosphere, the temperature extremes on the Moon would be much greater than on the Earth. The highs would be higher, and the lows lower.
But since radiative emissions are proportional to the 4th power of surface temperature, there would be more rapid energy loss during the Moon’s higher highs, so the difference between high temperatures on the Moon and Earth would be less than the difference between low temperatures on the Moon and Earth. In other words, the Moon’s average temperature would be lower than the Earth’s average temperature.
Nick Schroeder continued, “Like a blanket held up between a camper and campfire the atmosphere reduces the amount of solar energy heating the terrestrial system and cools the earth compared to no atmosphere.”
That’s not working very well for Venus, is it?
Psst! …. Do you do it by S = UT + 1/2 AT squared….?
Nope. Cuz U is a kluged term to combine the heat transfer resistance against radiation, conduction and convection. This problem deals only with radiant heat transfer, so all you need is emissivity.
If T1 = temp (K) at sunny side and T2 is the temp at opposite side then heat will flow by conduction to the cold side until T1=T2.
Then T2 and T2 will be 255.15K
I would say it would end up at half the sum of the temperature of space and the object that is illuminating it
I am not sure that the thermal conductivity is particularly relevant. That will just act to raise the upstream a bit and lower the downstream a bit
What is the albedo of ‘grey’?
Distance to the illuminating object would not matter?
The Sun is thousands of degrees.
Nothing gets 1/2 that hot in Sunshine, 90 something million miles away.
Nicholas
The distance is unimportant because the flux of the source is not given, just what is impinging on the concrete block, which coincidentally is the same as what arrives at the top of Earth’s atmosphere. By deduction, the distance is probably 93 million miles; however, it could be different if the source were different.
Yes, I understand the problem as stated perfectly well.
I was responding to this sentence:
“I would say it would end up at half the sum of the temperature of space and the object that is illuminating it…”
Which seems to be saying that only the temperature of the source of the incoming energy need be considered in order the known the temp.
And thank you for the reply, Clyde.
Leo
You asked, “What is the albedo of ‘grey’? Probably about the same as the moon.
Answer: the block never existed because it started at absolute zero. But let’s say it did exist at 0.01 K, then it never warmed up because it is surrounded by a perfect insulator and receives no insolation.
It is only surrounded by the perfect insulator on 4 sides.
Oh I see.
What Chair, i mean block?
Willis: What’s the absorption coefficient, alpha, of the surface of the concrete,?…
Yeah, was just gonna ask that. You can’t solve this problem without knowing how much solar irradiance is absorbed. We can assume that the absorption is constant across the incident wavelengths, but the problem should at least provide an absorption a.
Otherwise, I assume a = 0 and then it remains at the initial state.
Joe, Kirschoff’s Law says that at any given frequency, absorptivity = emissivity. So the absorption coefficient = 0.95. But you’re right, I should have specified that the absorptivity and emissivity are invariant across the spectrum … I’ll modify the head post to reflect that.
Good question,
w.
Sorry Willis, but if that was the case spacecraft would roast their electronics.
I’d put some links here but am not sure if that is allowed. Nevertheless, just google spacecraft thermal emissions and absorption materials and I am sure that you’d find examples.
I run a lot of research on this and can confirm they are different. We are getting some very nice materials (with lwoer cost ) soon to be qualified that have emissions >0.9x while their absorption is <.1, sorry can't be more specific. This is key to improving thermal radiator designs.
It’s a thought experiment, Frenchie. You are correct that in the real world, spacecraft are generally highly reflective and often insulated as well. Keeping their temperature down is an ongoing problem.
However, this thought experiment is about concrete. Low-albedo concrete has a visual light absorptivity on the order of 0.7, perhaps more depending on the surface treatment.
w.
Willis
You assumed that there was no dispersion of emissivity with wavelength, i.e. the behavior is the same with visible-light as with IR. So, we have a contradiction. If the absorptivity in the visible region is 0.7, then the albedo has to be 0.3; however, if the absorptivity is 0.95. then the albedo has to be 0.05! You can’t have it both ways unless you change the parameters of the problem by allowing the visible-light behavior to be different from the IR behavior.
Joe, Frenchie, and Willis,
If the absorptivity is 95%, then the reflectivity is 100%-95% or 5%. This is then not a grey body, but a black body, literally, with a reflectance on the order of magnitude of coal, or that mythical Arctic “Dark Water.”
Kirchhoff’s Law is a ratio, not an equality.
In Kirchhoff’s original parlance:
E/A = e
Not E=A.
In modern parlance:
Eν/αν= f(T, ν)
In other words, the ratio between emissive power and absorptivity is equal to specific intensity.
In describing his original formula, Kirchhoff wrote “emissivity”, when he meant “emissive power”. They are not the same.
Only at thermodynamic equilibrium is emissive power and absorptivity equal. If they were always equal, an object could never change temperature.
You’ll note the IPCC assumes in its equations that emissive power = 1 and thus absorptivity = 1, which is clearly unphysical. CO2 is not a hypothetical perfect blackbody.
The correct formula: h = (e (A) (σ)(Ts ^4 – Ta ^4)) / (A * ΔT)
The incorrect IPCC formula, applied to gray bodies: h = ((σ)(T^4)) / (A * ΔT)
Note the lack of e in the IPCC formula… they assume emissive power = 1 and thus absorptivity = 1 at thermal equilibrium, a hypothetical perfect blackbody.
I once owned the textbook to be able to solve this. You may have just copied one of the example problems. But I’ll be darned if I know where it is at the moment. And at least 30 minutes of good hard thought just to narrow down where to look for it.
Maybe the book is John Lienhard’s “A Heat Transfer Textbook”
I’d have to find it to tell you the author.
The chart of thermal conductivities has the formulas at the bottom of the page.
https://www.engineeringtoolbox.com/thermal-conductivity-d_429.html
Hint: Make heat budget equations for the front and back sides, each with heat gains equal to heat losses. Two equations in 2 unknowns (the temperatures).
Roy: +100
ROY the problem is the sides .. he has tried to get around them by say they are insulating but unless they don’t conduct as well they will have a gradient and they will re-radiate thru the back face anyhow. You basically end up with a soldering iron tip situation the front face and the sidewalls make up a larger area than the back face so it’s hotter than you would directly calculate.
Thanks, LdB. I specified that they are perfectly insulating. Because of that, they must also be perfectly non-conducting, because if they could conduct they couldn’t insulate perfectly …
Regards,
w.
That doesn’t stop the material re-radiating as a conduction. So let guess you are trying to use this forumla for the conduction
http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/thercond.html#c3
Note the last part of the statement
What is happening is the material re-radiates each point in the material becomes it’s own heat source. By insulating you stop the external bleed but it doesn’t stop the same thing happening internally. So I am sorry there simply is no way to remove the sidewalls from the calculation.
The only true answer to your question is a rather complicated gradient calculation.
LdB, it’s a thought experiment. Clearly your text doesn’t deal with the wonderful properties of Unobtanium … it neither conducts nor radiates energy.
However, I don’t understand your claim that “the material re-radiates each point in the material becomes its own heat source”. Are you saying that there is radiative heat transfer within the body of a solid object?
Thanks,
w.
Due to the delay these are going to end up in wrong order. However I suggest you read why you need a gradient calculation even on the conduction start by READING THE LINK.
It would appear you want the naive calculation which will pop up above this one which is make the dam thing a large flat sheet because you want to use the flat sheet approximation.
This is a bit like D..slayers not getting the whole re-radation thing, if yuou haven’t worked out the concept by now I am not sure I can help.
These walls are the equivalent of saying that the heat flow is in one direction only, which makes the problem much easier. No finite element analysis required. I’ll pull out my heat transfer book tomorrow and give it a try.
Willis if you do want the naive answer then just make it a sheet 20m x 20m sheet and you want the formula for the middle square meter so the side faces are 10m away and we can ignore the gradient. You don’t need unobtainium then either because you entitled to use the flat sheet approximation.
LdB February 28, 2020 at 9:13 am
Sure, we could do it that way, LdB, or make it 1000m x 1000m. Either your way or the Unobtanium way should give us the same answer … what answer do you get?
Thanks for the attention to detail, and best regards,
w.
Willis,
Yeah, sure, I’m misleading people. Nice!
If CSR = CHF, you would think Kirchoff, Boltzmann, and Planck would have noticed it. But no, when the walls of the cavities they inserted carbon or lamp soot stabilized (CHF=0), B & P discovered the laws of thermal emission. CSR != CHF
But no, they are dumb and you are a genius who figured it all out with your armchair physicist assertions.
Can you walk and chew gum at the same time?
Molecules can move and emit radiation at the same time, and this emission doesn’t slow them down.
“Molecules can move and emit radiation at the same time, and this emission doesn’t slow them down.”
There you violating conservation of energy again.
Greg,
“There you violating conservation of energy again.”
EM radiation is a REFLECTION of Kinetic Energy, not its SPENDING.
Molecular motion disturbs EM fields, causes radiation.
If radiation were to eliminate motion, then you couldn’t get motion at all, and thus no radiation.
I can observe you walking without you slowing down, so to speak.
Great idea, Roy. But Willis wants to include conductive heat flux, otherwise, the solution is too easy (394K).
Hot Side Radiation = Conductive Heat Flux = Cold Side Radiation
HSR => [ CHF ] => CSR
Conduction Formula: q = KA(Th-Tc)/L
Radiation Formula q = ɛσT⁴
Simplify:
Set Absorptivity = Emissivity = 1
Set K = L = A = 1
HSR = σ(Th)⁴
CSR = σ(Tc)⁴
CHF = Th-Tc
Assuming HSR = CHF = CSR:
σ(Th)⁴ = Th-Tc = σ(Tc)⁴
There is only one solution!
Th = Tc = 0
HSR = CHF = CSR = 0
Zoe:
You are demonstrating your complete inability to analyze what is really a very simple problem. You obviously have never taken even an introductory thermodynamics course, because you can’t get even the most basic analysis correct. You can’t do even trivial energy balance (1st Law) calculations.
To satisfy the 1st Law for the cube as a whole in steady-state conditions, HSR + CSR = AbsorbedSolar.
To satisfy the 1st Law for the far end of the cube in steady-state conditions, CHF = CHR.
Those are the constraints of the problem. Many of us here did this exact analysis, as we learned in the first couple of weeks of our first thermo courses. It’s not that hard.
Agreed Ed Bo
It is a first term mechanical engineering thermo problem. It introduces the concept of an albedo for the receiving and re-radiating surface, heat conduction at a given rate and an albedo for the cold side radiating into a space with no return.
So why hide the evidence?
Show it in the book.
Zoe
I am not sure what question you are asking. In order to each the elementary principles of heat transfer there are set problems with many simplifications in order to concentrate the mind on one or two factors. Another task is to calculate the equilibrium temperature of a hollow cube which is insulated and has a window of given dimensions, and a heat source inside with power [Watts].
The question posed by Willis is of this instructional type. The answer has been provided by several contributors. There are on-line calculators for such problems that have all the necessary factors included. These problem usually involve air or water flow and are much more complex than radiative problems.
Remember that the great author of heat transfer textbooks, Adrian Bejan, looked at the GHG-based global warming calculation and said that is as so simple it was not even interesting. His book on Convection Heat Transfer has problems that are extraordinarily difficult that explain things like the development of sets of thunderstorms on hot days and how convective cell structure changes with temperature, and why (meaning the physical and mathematical basis for the changes).
I can add that the temperature response of the atmosphere to an increase in CO2 concentration is very modest, and we are far better off spending our time developing “forever” energy production methods than we are worrying about ending the use of a set of declining resources. We have to have this problem solved within 200 years, 5 or 6 generations.
Zoe says: “Great idea, Roy. But Willis wants to include conductive heat flux, otherwise, the solution is too easy (394K).”
Yes, with no heat conduction through the cube, the front side is 394K. And the back side is 0K. Good work!
With NO conduction, the front is 394K and the back side is 0K.
With a TINY conduction, the front will get slightly cooler and the back will get slightly warmer.
With a bit more conduction, we get the answers provided by multiple people.
With PERFECT conduction, then the front and back are the same.
I forgot to knock off the 5% reflection and just took the 1360 W/m^2 as the flux absorbed at the front surface. Your two equations are a bit messy, but can be solved iteratively in about a dozen steps. I came up with 388.5 K at the front and 222.7 K on the cold side. Anyone who can write the correct equations can easily verify the solution without doing much work.
If you take off the 68 W/m^2 reflected from the front, it changes the front and back temperatures to 383.8 K and 221.4 K, respectively.
To Stan Robertson: I agree (383.3 K and 221.4 K).
Stefan-Boltzman, Watson, Stefan-Boltzman.
Assuming an emissivity coefficient = 1 (at steady state the block behaves as a perfect black body, according to the statement), then:
T = 393.54 °K
T = 120.4 °C
Impossible to write the equation here.
The solar absorptance of rough concrete is 0.60. The total hemispherical emittance at 300K is 0.91.Your problem statement requires the use of a gray body model with an emittance of 0.95.Your result will be meaningless.The essential feature of radiation heat transfer problems involving solar radiation is the non gray behavior of most surfaces .Check an engineering heat transfer textbook.
Will,
I reckon that the hot side to cold side temperature difference has to be large enough to transmit the 1360W. As you’re using 1square metre for the area and 1 metre for the length of transmission then the temperature difference is 1360÷0.8. Which is 1700’C
So if the cold side doesn’t heat then its
Hot = 1427’C
Cold _ – 273’C
Next I will have to work out how the block heats up which is more tricky and will take a bit of tbought
@Red94ViperRT10
Maybe John Leinhard’s excellent “A Heat Transfer Textbook” ?
Most have got the front side because it is relatively easy with SB
I suspect whatever answer you derived is actually wrong Willis because this requires a complex Integral.
The issue is the graphic and question text Willis misleads there are 5 cold sides not 1 🙂
The back side if you ignored the actual 4 sides would be straight forward it is just thermal conductivity thru 1 square meter . The sides create the problem they get progressively colder the further they are away from the hot face creating a gradient along them. The front face to back face conduction is what 98 watts/sec so the side face re-radiation are also going to be significant. Each of the hotter gradient along the sides will also radiate out the back face because it is the furthest away and hence coldest.
So my question is do you really want the hard answer or were you trying to do the naive answer ignoring the sides?
I should say adding a perfect insulator around then sides does not simplify the problem they represent because they conduct. As I stated to Roy you have made a soldering iron tip and it’s a little more difficult than the naive answer.
“The front face to back face conduction is what 98 watts/sec ”
I have a general rule that when people get the units wrong , they probably don’t understand the physics.
So do I. So either you don’t get why it has to have a time component or you don’t get the difference between say KW and KWh.
Firstly it’s kWh not KWh. Second you have the equivalent of kW/h not kWh, I rest my case.
ROFL
In case you are wondering why he is ROFLing…”The kilowatt-hour is a composite unit of energy equal to one kilowatt (kW) of power sustained for one hour. ” in other words kW per hour or kW/h. Even I knew that one.
He was trying to play like some do with English and spelling, that they are more superior. We have a non flattering word for him in the industry, its the old case of those who can’t teach.
Units:
“… in other words kW per hour or kW/h”
The Watt has a time component so there are never W/hr (Watts per hour) in any form.
kWH, often written KWH is one thousand Joules per second for an hour. 1 kWH is 3.6m Joules of energy moved or dissipated or released in the course of one hour.
Any number of Joules per second divided by 3.6 million = kWH which converts the energy component (Joules) and time component (seconds) into kJ and Hr.
Always some to WUWT – learn or teach something every day without finger pointing.
” … The sides create the problem they get progressively colder the further they are away from the hot face creating a gradient along them. …”
—
LDB, the entire block is going to have the same gradient, illuminated side to shadowed side. I don’t know why you’re getting snarled-up on a side-issue. 😉
Hot 125C
Cold -145C
One comment, which may or may not be helpful:
The only part of the block that matters is the face.
The other three sides receive nothing and radiate nothing.
So rather than being 1m deep, it could be infinitely thin.
The only difference that would make is the length of time taken to reach steady state.
Radiation in = radiation out.
What temperature does an object with a known emissivity have to be to emit X watts/m^2
A block has 6 sides
Group problem solving in history.
So correlation IS causation, exactly as AGW hypothesises, and King Arthur proves via witch = duck deduction.
Take that, den1ers!
emissivity of .95, with constant influx of heat on one side (1360 should eventually heat the core to critical temp of planet explosion right ? Blowing out the sunny side. leaving a much cooler half meter thick block which will then start reheating until in half the time cause the reduced thickness remaining block to once again blow off the heated side. ad infinitum eventually exfoliating the block to a thin post card of concrete. cause the ocean stole the heat. and stored it at 4 degree above freezing in the deep.
Alternative answer: The stray CO2 molecule that arrived on the heated face super heated the concrete block driving it to critical temp near infinite Kelvin and it totally vaporized.
Wait the heated side caused the block to start rotating eventually reaching overspin condition and threw off a large mass of material from the prior heated side forming a smaller less dense satellite which began orbiting its reduced spinning source and they lived happily ever after, until the smaller satellite got its faced locked toward the block remanent.
Or something like that.
Willis
Sorry spell checker got your name last time
I went off on a tangent mentally there. The 0.95 emissivity gives the answer to the cold side. So the temperature difference cold side to Absolute Zero has to be
1431’C using the same calculation as before
So my bizarre answer is
Hot = 3131’K
Cold=1431’K
But most likely wildly incorrect
I look forward to seeing the solution
Simple. Use Stefan-Boltzmann equation to compute temperature for given energy flux and albedo.
Then apply Fourier’s Law (not to be confused with Fourier Series, but Fourier had to discover the series to solve the equation) to compute heat conducted through the cube with the given dimesions and thermal conductivity.
I’ll leave the details and final result as an exercise for you students. :-]
120C
44C
Typo, that is 4.4 C
Oh, conductivity is w/m.k not w/m/k
firstly watt = W , not w ; kelvin is K not k .
secondly W/(m.K) is identical to W/m/K , whoever gave you w/m.k is wrong on three counts.
The formula for conductivity is QL/A deltaT. This tells you the units. Wm/m^2K.
W/mK.
If you divide m by K you end up flipping K to become WK/m.
“w/m.k…”
W/m•K is correct.
The • character is produced by holding the Alt key down and typing 0149 on the number pad.
Or full stop when written. It replaces the ‘x’ symbol for multiplication of course, to avoid confusion with variables called ‘x’.
I spent many years studying and working in engineering (Mechanical, also Thermodynamics and fluid dynamics). it is watts / (meter x temperature).
And yes, James Watt was the inventor of the steam engine, I am fully aware they symbol for power and temperature are in upper case.
I also speak French, and m well aware of the use of the alt+numeric keypad number to generate characters form the ASCII table, alt+132 for example.
In fact I am am now a software engineer, I am fully aware of the ASCII table in it’s entirety.
In software it would be W/m*K. The asterisk is used for multiplication in software calculations.
‘^’ is used for the power of, so it can also be written: W*m^-1*K^-1 if you understand software notation.
In fact I know rather a lot about all of this sort of stuff.
Using W/m*K for conductivity is ambiguous at best, but actually wrong.
The rules of algebraic precedence have multiplication and division at the same level of precedence, with operations from left to right. So in that format, it is equivalent to (W/m)*K, and the K ends up in the numerator, which is wrong.
W/m/K is correct, as is W/(m*K).
Operator precedence in the C language is hardly relevant.
10/10/10 is not the same as 10/ 10 x 10
I was referring to the standard algebraic rules of precedence, which have been followed for centuries. There is a reason all serious programming languages follow them.
You say: “10/10/10 is not the same as 10/ 10 x 10”
My point exactly. “W/m*K” is not the same as “W/m/K”. The second is correct. The “K” term must “end up” in the denominator.