http://wattsupwiththat.com/2009/11/17/the-steel-greenhouse/

http://climateandstuff.blogspot.co.uk/2013/04/the-copper-greenhouse-new-test.html

http://climateandstuff.blogspot.co.uk/2013/03/the-copper-greenhouse.html

http://climateandstuff.blogspot.co.uk/2013/03/a-cool-object-reduces-energy-loss-from.html

http://climateandstuff.blogspot.co.uk/2013/03/does-thermal-radiation-travel-from-cool.html

http://climateandstuff.blogspot.co.uk/2012/05/cool-body-can-transfer-measurable-heat.html

The iron greenhouse energy budget |

The tests trying to produce some replication of Willis's Iron greenhouse have been repeated using a multichannel thermocouple probe recorder measuring to 0.01°C (accuracy 1°C) and somewhat modified test setup.

The sensor has been modified to make it more responsive (less copper in the plate than in the original cone. Fine wire thermocouple used to reduce the heat conduction.

The insulated hot box uses much thicker insulation and the double sided grey sprayed copper plate temperature is monitored using another fine wire thermocouple. The temperature between the 2 IR windows and outside the window is now measured. A small fan is used to provide a continuous stream of ambient air between the hot box and the sensor to prevent conduction and convection effects upsetting the sensor reading. The heating voltage is maintained to 9.254 volts +-3mV ensuring constant power input to the hot body.

The object of the experiment is not to replicate EXACTLY the steel greenhouse thought experiment. For a starter the hot object is only "surrounded" on one side and the major loss of heat from the object is via conduction through the insulation. There is obviously conduction and convection occurring in the experiment which are prevented by using a vacuum in the thought experiment.

Inside of Hotbox |

Table-top setup |

### Grey painted Plate Insertion:

What is expected is a significant increase in the hot body temperature when the grey plate is inserted. The Sensor should report a drop in temperature until the system has reached equilibrium It should then be at the same level as before the grey plate was inserted - I.e. the radiation from the hot box should be constant before and after grey plate insertion.### Reflective plate insertion

The rise in temperature of the hot body should now be significantly hotter than either no plate or grey plate (100% of forward facing IR should now be reflected back onto the hot body causing the temperature to rise until the additional energy balance can be restored.

The sensor should show a drop in heat for the time that the reflective plate is in position.

Method

- During the test the voltage applied to the resistors heating the hot body is monitored and maintained within +-3mV of the nominal (giving a power variability of 0.065%)
- The test setup was run monitoring temperatures for about 10 hours.
- The recorded results were then analysed over a period when the ambient was most stable (after sunset).
- To allow for ambient vatiation the measured forced ventilation temperature in front of the IR window was smoothed and then subtracted from the hot body temperature.

Results

The first temperature rise is with a reflective plate and the second rise is with a grey painted copper plate inserted between hot body and IR window. The low temperatures are when no intermediate plate is inserted.

The air outside the IR window plot (green) is the temperature as measured from the forced airflow. The corrected temperatures refer to measured temperature less the air temperature.

The temperature of the Hot Body shows results as expected - maximum temperature from reflective plate lower temperature from grey plate lowest temperature from no plate.

The measured temperature (= IR output) from the sensor does not agree with expected result..

With the grey plate not equalling the hot body temperature there would be expected lower IR emission so perhaps this could explain the lower temperature compared to no plate.

With the reflective plate the IR output does not go to zero. possibly the plate warms and the insulation is insufficient. The IR sensor needs to be improved - possibly an IR thermometer? But will these then read the temperature of the IR windows?

So the results show

A definite increase in hot body temperature if a reflective plate is used (5.5°C)

A definite increase in hot body temperature if a grey plate is used (3.5°C)

If you believe that backradiation or reflection cannot add energy to the hot body from which the radiation originates then these results alone disprove this belief.

Good start.

ReplyDeleteIt is a bit hard to see your configuration (need more pictures). However, I would not place aluminum on the inside of the enclosures - too much reflection. Black felt seems a more logical choice for this experiment.

You show the copper sheets being inside the same chamber as the heater. That allows heating by conduction. I would use an IR window to separate them so that only radiation would cause changes in temperature. Also arrange a way to change the plate without opening the hot chamber. This should make it easy to show the cooling effect of removing the sheets.

I would also make sure that the copper sheets extend into the insulation on all 4 sides. Also, don't allow your aluminum linings to conduct heat from one side of the sheet to the other. just allow conduction thru the sheet.

Hopefully, these suggestions will provide a better than a 5.5°C increase.

The aluminium was placed on the inside of the chamber after someone suggested that the insulation is "transparent" to IR. I now realise that placimg it behind the hot plate was not wise - it absorbs heat by conduction and cools the hot plate.

DeleteUse of LDPE wrap does not prevent conduction and also impedes IR see:

http://climateandstuff.blogspot.co.uk/2013/03/does-thermal-radiation-travel-from-cool.html

where I show the effect of the material.

The chamber is not at all like the thought experiment so lining with an IR absorber (heats chamber) or an IR reflector (heats absorbers in chamber and the chamber)is perhaps not easy to decide.

The copper sheets do NOT separate the chamber into 2 sections - they freely float in the centre but are the same size as the hot plate. To seal them into the chamber walls would be possible but not easy.

What is required is a vacuum!

A few comments/questions.

ReplyDelete1) 2000 s is about 1/2 hr, so these experiments run about 5 hours, correct?

2) Are the small downward spikes in the sensor temperature @ 15000s & 25000s due to opening the box to change the plates?

3) is seems that the "heat sink" is a bit counter-productive. It will increase surface area (and hence conduction/convection) but the metal may have a poor emissivity (depending on the finish). It seems a thin sheet of metal with a coat of paint attached to the heating elements would increase the emissivity and decrease conduction/convection to the environment.

4) What did you expect for the sensor temperature? With no plate in place, the sensor should "see" the hot heat sink and receive the maximum IR, reaching a high temperature. With the plates in the way, it sees the cooler plates, and should be cooler. I think this is what your data shows.

You say:

"The object of the experiment is not to replicate EXACTLY the steel greenhouse thought experiment."Unfortunately, those who don't understand science and who don't want to think will use this as an excuse to invent all sorts of excuses. I suspect that even if you could exactly build the system exactly as described in the experiment, the uber-skeptiocs would still wriggle around with excuses as to why it doesn't show the effects of back-radiation.

If the "steel greenhouse" is a real effect then adding an extra shell would simply double the radiation again.

Delete235 out at "new" outer shell plus 235 in = 470. Original shell must now radiate 470 out and 470 in and planet needs to radiate 940 !

Add another and the planet now radiates 1880 - and another 3760 and another 7520 etc etc.

At this rate a candle could power a power station.

You're not quite right. With a second shell, the orginal shell would radiate 470 out and 470 in, and the planet would need to radiate 470+235=705. With a third shell, the planet would need to radiate 940.

DeleteI know it seems counterintuitive to many, but consider this: if you look at the system as a whole, there is 235 of generated power, and 235 of power out to the void of the universe. That is all. Whatever is happening inside that system cancels out.

I often explain that first-law-of-thermodynamics energy balancing is like balancing your bank accounts. You could earn $235 a week and spend $235 a week (to the "outside universe") and the level of your total bank accounts would stay constant, no matter how much shuffling you did between your checking and savings accounts. And the radiative exchange between these shells is like shuffling money between your own accounts. It won't make you rich...

In the physical world, if you look at high end "dewar" flasks (glorified thermos jugs) for things like holding liquid nitrogen, they have multiple layers of metal shells in the vacuum chamber (separated by plastic sheets to stop conduction). These are equivalent to the multiple shells in "steel greenhouse" thought experiment. And yes, they really work to reduce energy transfer between inside and outside.

No matter whose maths is right the concept is still a claim that trapping energy by layering shells around other shells is increasing the internal energy beyond the power available and that does NOT happen in reality.

DeleteYou are clearly claiming that the original 235 internal energy source combines with the 235 back radiation to enable the surface to heat to the temperature where it emits 470 ! All from a 235 source in the first place !

But you say Rosco created a strwaman "You've set up a strawman -- nobody does the calculations in the way you say is incorrect."

But that is exactly what YOU say IS correct - you can't have it both ways.

You CLEARLY say it is correct to simply sum radiative fluxes !!

The only way the "planet" can emit 470- or whatever - due to the sum of the 235 internal + 235 backradiation is for the surface to heat to the temperature equivalent to 470 or about 301 K - 28 degrees C.

So YOU claim that the temperature of the surface of the planet - originally at a temperature of about 254 K or minus 18 degrees C (235 Watts per square metre) can be increased to 301 K or plus 28 degrees simply by adding a steel shell to trap the radiation.

And you claim Rosco has created a strwaman !!!

It may sound OK when you think of ludicrous though bubble conjectures where the arithmetic is OK and silly claims about photons not knowing about the state of the emitter or the target - actually that is total BS anyway - a photon is emitted with precisely the energy level caused by its emission.

The whole scenario looks completely ludicrous when you consider the reality that it is the TEMPERATURE of the object that drives the power of the radiative flux.

And unlike thought bubbles we can measure temperatures - thefordprefect tried to achieve this but ran into a problem he hadn't originally considered - how do you actually establish an original boundary position.

Can he prove it is backradiation or is it simply possible that better insulation simply allows the source to heat to higher levels than when energy is "wasted" - the lid on the pot effect that is primarily a conductive/convective effect.

For the steel greenhouse to be real the TEMPERATURE MUST INCREASE - 1.2 times the original K temperature for doubling the power flux.

If you think placing a steel shell around an ice cube will cause it to heat to 30 degrees C then I think you should tell Westinghouse to stop wasting their time building freezers.

If it you say it is correct here in the "steel greenhouse" thought bubble then I think reality (as well as Rosco) has shown reasonable evidence to prove you're wrong.

Just to show how wrong this concept is you only need to look at what Captain Curt says is reality.

DeleteHis claim is that a second shell causes the first to radiate 470 in and out.

The new outer shell radiates 235 in and out.

And the planet now radiates 705.

Lets add 'em up.

The 235 in and out at the second shell cancel.

Likewise the 470 in and out at the original.

This leaves the 705 out from the planet.

Or 705 out from planet plus 470 out from original shell plus 235 from second shell gives a total of 1410 out.

1410 out minus 470 in from original shell minus 235 in from outer shell still leaves 705 out.

Where does the extra 470 come from ????

Amazing !

Arithmetic may work like this but physics doesn't !

Just what physical world are YOU looking at Captain Curt ???

1. the experiments ran for 8 to 10 hours but error occurred during some of the tests - power to hot plate outside the +-3mv range - solar heating of room - incorrectly fitted lid to box.

ReplyDeleteThe plots shown are those where such other influences were smallest.

2. yes the spikes are caused when the plates are inserted.

3. The heatsink is black anodised to increase emissivity. Unfortunately it is mounted with fins in direction that will increase convection. However the inside of the box should reach an equilibrium temperature limiting the effect of convection.

It would have been better in retrospect to have a much lower thermal mass as the hot body but I was worried that the resistors would provide uneven heating. Perhaps I will change this to a grey copper plate for the next tests

4. The sensor should show some sort of correlation with the temperature from emitters inside the hot box. I had hoped that the reflective plate should emit very little IR (perhaps this is shown for the 1st 2000 seconds. But eventually the reflective plate emits the same as the black body plate.

I agree that the slayers will always find something wrong - but if carried out sufficiently carefully then perhaps potential slayers may be convinced.

A simpler experiment would perhaps be to suspend a heated source in a IR transparent chamber. measure the source final temperature. Drop a painted aluminium foil cover (close fitting) over the chamber and record the final temperature of the heated body.

Tested a PET bottle with the thermal camera but surprisingly it is opaque to IR.

A plastic drink beaker (polystyrene) seems to be transparent to IR (virtually invisible in the camera!). So I think this will be my next test as it is a step closer to the steel greenhouse.

I was contemplating some sort of "Saran Wrap pipes" (polyethylene) for the IR transparent chamber. Build the chamber slightly larger than the heater. Then blow air around that chamber. Then have another saran wrap shell (so the air blows through the annulus). Then put the heat shield right outside that wall. Perhaps surround the heat shield by another saran wrap shell. Basically concentric tubes to isolate all the parts.

ReplyDeleteThere is still air involved, but each part is pretty well thermally isolated from the other parts and from the room. Energy will not be conducted from the heater to the shield. It could ONLY get there by IR. And energy could ONLY get back by radiation.

As an extension, ice and/or dry ice could be arranged around the whole thing. The whole thing could also be made with polystyrene. It would be good to keep the parts as close together as possible, of course, to minimize the IR exchange with the room and maximize the exchange with the inner heater.

You have a good experiment. Don't let the slayers influence the design. My recent communications with them indicate that they fully understand, and accept, the physics involved. They appear to just enjoy being contrary and wasting other people's time.

ReplyDeleteI agree that the LDPE wrap won't prevent conduction, but it should reduce it a lot. In your configuration, conduction is like a short circuit in a electrical circuit. Basically, if the heater and the plate are at different temperatures, then convection will try to make them the same temperature and erase the effect you are trying to observe.

I checked your comments on another page explaining why you want the plates and heater to be the same size and in the same chamber. However, if your heat source is a blackbox with an IR window, and the plate is in a convectively isolated space, I think you will get a larger temperature change.

The electric current is only capable of raising the temperature of the body to a certain maximum temperature. (On Venus, the Sun can only raise the surface temperature to about 150K very roughly. On Uranus it cannot raise the temperature above absolute zero with direct radiation into the depths – because no radiation can get there.) Your object will be cooling at the same time that the current is trying to heat it, so it never reaches its maximum temperature. But, as I wrote in my paper over a year ago (March 2012) about “Radiated Energy and the Second Law of Thermodynamics” radiation from a cooler object is only able to slow that portion of the rate of cooling of a warmer object which is itself by radiation. So, when you introduce your grey sheet you do in fact reduce the rate of cooling of the electrical heated object, and the new equilibrium temperature is closer to the maximum temperature to which the current could heat it.

ReplyDeleteHowever, what determines Earth’s mean surface temperature is the temperature plot already established in the troposphere. It’s just a matter of where that intersects the surface. That then becomes the “supporting” temperature, basically a minimum that the surface would cool down to in the early pre-dawn hours on a calm night. Then the Sun can indeed add more energy the next day in some regions, because the supporting temperature may be only 278K to 280K for example. The Sun is not shining with a fixed low flux equivalent to 255K as a mean night and day value. In case you haven’t noticed, it shines more strongly upon you in the middle of the day. So temporary thermal energy explains weather variations, and oxygen and nitrogen molecules provide most of the slowing of surface cooling (by non-radiative processes) so we are not too cold in the evening, whilst radiation plays a small part in this too, mostly water vapour being the key player. But moist regions are not the warmest: they are the coolest because the primary determinant of surface temperatures is the underlying support temperature. That is lower in moist regions because the temperature gradient is less steep, and thus the plot intersects the surface at a lower temperature.

Please explain what you

ReplyDeletethinkyour experiment has to do with planetary atmospheric, surface, crust, mantle and core temperatures. Uranus is thousands of degrees hot down in the depths of its atmosphere. There must be net outward radiation from those depths, but the energy is known not to be coming from any internal energy generation. The energy in fact originally came from the Sun, but no Solar radiation can possibly get down there. Hence it got there by the non-radiative process which is explained inmy paper, and which you will not read about elsewhere yet,, and which you will not understand until you understand the process described in statements of the Second Law of Thermodynamics. In that process, as that Law says, thermodynamic equilibrium evolves spontaneously towards a state of maximum accessible entropy. Are you with me? That state must be isentropic, which means it must display a temperature gradient. Then, when new Solar radiation is absorbed in the uppermost layers of the Uranus atmosphere it disturbs the equilibrium, and the only way that equilibrium can be re-established is by some of that extra thermal energy moving downwards, that is from cooler to warmer regions.DJC

ReplyDeleteThe object was to show the steel greenhouse effect of willis

There is only vague similarities to the earth system.

The earth is not significantly internally heated. But the sun effectively does this.

The amosphere is transparent (almost) to shorter wavelengths where most of the suns power resides. The GHGs do not absorb significantly the solar power (the LWIR is way down in the tail of the Planck curve). The emitted wavelengths from the warmed earth are peaking neart the GHG absoption bands so the earths LWIR will get absorbed and re-emitted/conducted to the non ghg molecules.

The sun is therefore like the power supplied to my hot plate.

The plate being very conductive copper will be transmitting 50% out and 50% towards the hot plate similar to GHGs (but as a BB not discrete bands).

The problem you have is demonstrating that it is any sort of back radiation that is causing any temperature increase or is it merely a case of the better the insulation the more efficient the heating element works - less "wasted" energy - and this is the real cause of increased temperature.

ReplyDeleteReducing "heat" loss by improving insulation - i.e. reducing conductive/convective heat loss is nothing new and nothing to do with whether or not back radiation produces any measurable thermal effect.

Until you can establish the maximum temperature the element can induce in "perfect" conditions you actually have no starting point for any calculations.

My solution to this problem was to see if summing radiative fluxes works as climate science claims it does.

To this end I used spotlights to heat an object - the obvious object to heat was a thermometer bulb.

I placed 2 spotlights so that I was certain they would heat the thermometer to a consistent temperature - one to 30 degrees C and one to 36 degrees C - I was trying for 35 but got fed up with all the tinkering so accepted 36.

Now I have real data that can be reproduced and I can calculate the flux emitted by the thermometer bulb at each temperature by the SB equation. Logically this implies each spotlight is supplying a flux equivalent to this.

Summing the fluxes in the manner used by climate science does NOT give the real temperature observed in the experiment.

The problem for climate science is the theory does not match reality and here is why.

At 30 C - 303 K - the thermometer bulb emitted flux calculates to ~478 W/sq m.

At 36 C - 309 K - the thermometer bulb emitted flux calculates to ~517 W/sq m.

Add these and you get a result of 995 W/sq m with a SB calculated temperature of 363 K or 90 degrees C.

This theoretical result should immediately tell you the climate scientists have got it wrong - 30 C + 36 C = 90 C ?????

I haven't yet supplied all of the puzzle.

Ambient air temperature was a consistent 18 C - 291 K which is equivalent to ~406 W/sq m - I'm rounding here as nothing is really precise but you get the idea.

So the way I see it :-

Spotlight 1 provides an EXTRA 72 W/sq m above ambient conditions of 406 W/sq m.

Spotlight 2 provides an EXTRA 111 W/sq m above ambient conditions of 406 W/sq m.

Adding up fluxes in a correct manner gives 406 + 72 + 111 for a total of 589 W/sq m.

589 W/sq m gives a SB temperature of ~319 Kelvin or ~46 degrees C.

My final experimental result showed 46 degrees C.

This experiment is easily reproducible - the temperatures do NOT matter as long as you record them honestly and ambient conditions do not change very much.

The method I show for summing radiative fluxes works time and again whilst the way climate science does it fails time and again.

I trust a simple reproducible experiment that yields results which combine well with theory - there is no supposition necessary at all - just observable results.

These results also suggest that a thermal response of an increase in temperature requires a higher power of radiative flux so "back" radiation of lower power does not appear capable of inducing any thermal response and results claimed for such may be explainable by other means.

If you can see any flaw in this experiment please tell me - email is roscomac@dodo.com.au.

Rosco:

DeleteYou've set up a strawman -- nobody does the calculations in the way you say is incorrect. (You cannot add temperatures.)

The way you advocate as the correct method is in fact the standard method used in climate science.

There are many possible issues to critique in climate science, but this is not one of them.

I have not set up any straw man!!

DeleteTo deny that they actually do what I say they do is to deny reality !!!

There are endless examples where a sum of two radiative fluxes that I describe as incorrect is shown as a simple explanation of the greenhouse effect.

Example include the University of Washington lecture notes on the greenhouse effect and Trenberth et al Energy Budget.

Both of these show the sum of radiative fluxes producing a certain temperature response.

The Washington University claim 239.7 W/sq m solar + 239.7 W/sq m "back" radiation combine to produce a temperature of 303 K

Here is the URL

http://www.atmos.washington.edu/2002Q4/211/notes_greenhouse.html

Trenberth et al do exactly the same thing - they claim 161 Solar + 333 "back" combine to balance the surface radiating 494 W/sq m at a temperature of ~305 Kelvin.

I cannot see why you claim that this evidence is not real - they make the claim that I say is wrong and they clearly do not sum fluxes in the manner I show.

Are you the guy who heated the light bulb on WUWT - because THAT experiment WAS a joke.

No workable data at all plus absolutely no evidence that all the energy didn't come from the only real energy source - the light bulb filament.

Besides who says I add temperatures ??

DeleteAt 30 degrees C the SB calculated flux is ~478 W/sq m.

I think I added fluxes not temperatures.

Tell you what - prove me wrong ! Show some proof of what you claim is expert knowledge of thermodynamics by a valid reproducible experiment with actual data that doesn't rely on misinterpretations of deluded reality !

Rosco30 June 2013 22:41:00 BST

Delete"Reducing "heat" loss by improving insulation - i.e. reducing conductive/convective heat loss is nothing new and nothing to do with whether or not back radiation produces any measurable thermal effect."

------------

tfp:

the plate is a blackbody the thermal conductivity of copper is excellent. Measurement is significant only when temperatures are stable. I fail to see how a conductive black body in a substantially isothermal sealed enclosure can effect the conduction or convection to ambient. It howevere does sit between the heater and the IR window, and I would therefore suggest that the only difference it can make is to the IR transmission out of the box.

----------

Rosco

Until you can establish the maximum temperature the element can induce in "perfect" conditions you actually have no starting point for any calculations.

----------------

tfp

the heater temperature in this setup has a maximum temperature of about 60°C At this temperature the energy in per second = the energy loss per second.

If I had doubled the insulative properties of the box then the temperature at balance would be higher. If the insulation was 100% effective then the temperature of the heater would be effectively infinite - i.e. no limit

The really amazing thing about people like "Cap'n" Curt is they can't even see they are completely inconsistent !!

ReplyDeleteHe says "You've set up a strawman -- nobody does the calculations in the way you say is incorrect. "

BUT a few comments above he is actually advocating that creating extra energy out of nothing with his interpretation of how adding steel or what ever metal greenhouse shells "works" in reality.

Show me proof that this works - it has never been shown - never.

Wow - people can design methods of reducing energy loss !

Curt - I hate to break it to you but that isn't increasing energy fluxes and has nothing to do with your claim of your "proof" of the steel greenhouse effect.

I mean - come on - this is typical delusional behaviour.

Smear someone with something they actually do not even claim.

Deny reality by claiming that climate scientists do not show methods that I say they do when there is clear evidence they do in their own published documents - how wrong is that ????

I didn't make this stuff up - look at THEIR documents.

Look up multi layer insulation on the web.

DeleteIt works

It is the same principle as steel greenhouses

If you really believe you can generate extra energy by a proposition such as the one proposed in the Steel greenhouse theory then why hasn't this been engineered into a useful application ?

DeleteOf course multi layer insulation can slow down energy loss in either direction and keep things warmer longer or reduce the energy required to keep things cold - that has been known for millenia !

It does NOT demonstrate any "back radiative" effect at all !

Because Inuit people build shelters out of the only building material known to them doesn't mean they think the ice is warming them by radiation - they know it is better to sheltered from a hostile environment than not.

If the proposal works as climate scientists claim then the sum of the radiation from a dozen ice cubes in a sealed insulated container where the initial temperature was say 0 C or 273 K should heat it to - lets do the math they way climate science says -

Ice cube at minus 18 C initially - 255 K - radiating at 239.7 W/sqm. - check out the link to the Washington University lecture - http://www.atmos.washington.edu/2002Q4/211/notes_greenhouse.html

12 x 239.7 = 2876.4 W/sqm equivalent to a Stefan-Boltzmann calculated temperature of 474.58 K or plus 201 degrees C.

If, as University professors claim it is valid for a sum of 239.7 + 239.7 then it must apply universally - maths is like that - if it works fro 2X it works for 200 X etc.

It is all based on the assumption that the principle of superposition apples to wave motion and that 2 waves in equal phase will combine to produce double the amplitude.

This has never, to my knowledge, actually been demonstrated.

Not only that but there is actual experimental data to show it does not.

There is sound experimental data that shows that waves do NOT "superpose" - I am referring to the well proven mechanism shown to combine the sound pressure levels using the dB scale - a logarithic summ NOT a linear sum.

Doubling the sound pressure - that is superimposing 2 equal waves - does NOT increase the "loudness" by a factor of 2 - we know that is fact - it increases by a steady 3 dB for each doubling.

If sound waves do NOT superimpose in the linear manner described by the hypothesis of superposition why should I accept radiation would ??

Especially when there is an abundance of observational evidence that says it doesn't ?

As I said - the Washington University Professor states plainly in written format that radiation levels of 239.7 + 239.7 - ie 2 ice cubes - combine to cause 303 K or plus 30 degrees C - it is what THEY say - you can see it everywhere including the Trenberth et al Energy Budget used extensively by the IPCC.

You cannot deny it is what they claim - the undeniable evidence is right there in their own published works.

When I made my original comment I was simply trying to show why you had an insurmountable problem with your experiment - how to establish a necessary boundary condition - which can only be established in a true vacuum but there is always conductive energy loss while you have an external energy supply.

However I now realise you cannot really be serious if you believe :-

"If I had doubled the insulative properties of the box then the temperature at balance would be higher. If the insulation was 100% effective then the temperature of the heater would be effectively infinite - i.e. no limit"

The first line shows you understand your boundary condition problem !!

How much insulation effect is required to establish the TRUE power of the heater element ?? Because you NEED to know this to PROVE any "bsck radiation" thermal effect !!!!

I cannot believe a rational human being believes 100% insulation leads to infinite power from a finite power input!

I mean, why is science bothering with nuclear fusion at all - 100% insulation is the answer.

Rosco

DeleteI cannot believe a rational human being believes 100% insulation leads to infinite power from a finite power input!

tfp

power input is constant power lost to the heater is constant at zero - there is NO power change

The power input has to go somewhere but as stated it can go nowhere so the source heats up - it will heat up until the power loss is equal to the power in. i.e. it will heat without limits in this hypothetical situation.

I DID NOT SAY INFINITE POWER - JUST INFINITE TEMPERATURE

Rosco:

Deletef you really believe you can generate extra energy by a proposition such as the one proposed in the Steel greenhouse theory then why hasn't this been engineered into a useful application ?

Of course multi layer insulation can slow down energy loss in either direction and keep things warmer longer or reduce the energy required to keep things cold - that has been known for millenia !

It does NOT demonstrate any "back radiative" effect at all !

-----------

tfp

your first 2 paragraphs contradict:

why hasn't this effect been utilised followed by "of course MLI works"!!!

If you do not belieive back radiation is required for MLI to work, then can you provide a suitable explanation on its operation please?

I DID NOT SAY INFINITE POWER - JUST INFINITE TEMPERATURE

DeleteRemember all of the scientists that progressed the work on radiation and quantum theory understood the principle that the radiatiob flux is proportional to the temperature.

No matter how the arithmetic seems to be valid the FINAL ultimate truth is that an object emitting a certain flux SIMPLY HAS TO BR AT THE CORRESPONDING STEFAN_BOLTZMANN TEMPERATURE.

Flux (W/sqm) = sigma X T^4

Look at the flaw in Curt's logic below

"The 235 power output sets the temperature of the shell, given the emissivity (=1.0 for a pure blackbody). This gives a temperature for the shell of ~255K (-18C)."

So the shell is at ~255 - OK he can do the math so far.

"So we have so far established that the shell must be outputting 235+235=470, and this is at equilibrium! So it must be receiving 470 as well to maintain this equilibrium, and the only place it can get it is radiation from the sphere."

BUT - what is the shell at:-

~255 K emitting 235 W/sq m OR

~302 K emitting 470 W/sq m ?

It can't be both.

Infinite temperature requires infinite energy input - look at the equation - doubling the energy only increases the absolute temperature by the fourth root of 2 or ~1.2

rosco:

DeleteInfinite temperature requires infinite energy input - look at the equation - doubling the energy only increases the absolute temperature by the fourth root of 2 or ~1.2

----------

tfp

As stated the energy loss =0 gives infinite temp. Energy loss = 0 implies no conduction no convection and no radiation loss. Where does all the constant power go if there is no heat loss?

------

rosco:

BUT - what is the shell at:-

~255 K emitting 235 W/sq m OR

~302 K emitting 470 W/sq m ?

---------------

tfp

it is emitting 235 OUT as stated

It seems to me you want to believe your "science" instead of what REAL scientists established - the radiation emitted is proportional to the temperature.

DeleteYou mention multi layer insulation but liquid nitrogen does not remain liquid without the work performed by the refrigeration cycle. Turn of the power and ultimately it will end up at ambient temperature.

No one has ever demonstrated your hypothesis of infinite temperature by perfect insulation.

The observations of cavity radiation were based on the hypothesis that any radiation entering would not escape the cavity.

Curt says it is elementary but:

In the 7th paragraph he says the shell is emitting 235 - agreed ? His claim is there in black and white -

"The 235 power output sets the temperature of the shell, given the emissivity (=1.0 for a pure blackbody). This gives a temperature for the shell of ~255K (-18C)."

But in the 9th paragraph he says the shell is emitting 470 - agreed ? His claim is there in black and white -

"So we have so far established that the shell must be outputting 235+235=470, and this is at equilibrium!"

Can't you see this is just nonsensical ?

It can't be both at once - this is fairy tale stuff.

If it is real demonstrate it.

OK, let's break it down into baby steps, because obviously Rosco and Joker have never performed even an elementary energy balance exercise, as you would get in the first week or two of an introductory undergraduate thermodynamics course. And you will see that no energy is "created" by the shells of a "steel greenhouse" - all analysis depends on conservation of energy.

ReplyDeleteOne of the first things you learn in such a thermodynamics course is that you can perform an energy balance on any system you define. The art is to define your systems (often called control masses) for the maximum possible insight. For the single shell steel greenhouse, we will define three systems:

1. The whole system: sphere and shell together

2. Just the shell

3. Just the sphere

Also, you can learn a lot just by examining these systems at equilibrium, typically the simplest case. If your analysis doesn't work at equilibrium, it won't work anywhere else.

For each case, look at all of the energy/power flows in and out of the system. And yes, they do add up - that is the implication of the 1st law. In this case, we will talk in terms of power (really power flux density).

For the sphere and shell together, at equilibrium, you have 235 power input from the radioactive decay inside the sphere, and 235 power output from the external surface of the shell radiating to space. You have energy balance, and therefore equilibrium, so no change in internal energy of the system, and no change in temperature.

The 235 power output sets the temperature of the shell, given the emissivity (=1.0 for a pure blackbody). This gives a temperature for the shell of ~255K (-18C).

Next, we look at the shell alone as our new system to analyze. We already know that it is radiating 235 from its outer surface. Its inner surface must also be radiating 235 inward toward the sphere. Actually it would be slightly more as there would be a thermal gradient between the surfaces. But for a thin shell of highly thermally conductive material, this is negligible, so we will call it 235. (Many people miss that the inner surface must radiate at least as much as the outer surface.)

So we have so far established that the shell must be outputting 235+235=470, and this is at equilibrium! So it must be receiving 470 as well to maintain this equilibrium, and the only place it can get it is radiation from the sphere.

Now we look at the sphere itself. It is receiving 235 from the internal radioactive decay, and 235 from the shell. So to maintain equilibrium, it must be outputting 470. Therefore its surface temperature at equilibrium must be high enough to radiate 470 outward. For a blackbody, this would be a temperature of about 29C or 302K.

For each system, the energy balance works out. This is the type of problem given to starting thermodynamics students, and this is the answer expected.

If you had two shells, the outer shell would radiate 235 outward, 235 inward, and receive 470 from the inner shell. Balance. The inner shell would radiate 470 outward, 470 inward (940 total), receive 235 from the outer shell, and 705 from the sphere (940 total). Balance. The sphere would receive 235 from the radioactive decay and 470 from the inner shell (705 total) and radiate outward 705. Balance. The surface of the sphere has even a higher temperature.

(continued in next post)

Seriously I dont fail to "get" the arithmetic I just do not believe what you claim happens. I wouldn't believe it from anyone !

DeleteIf this really happens then prove it by demonstration - an ice cube at minus 18 radiates ~235 and more as it melts.

Show the doubling effect you claim by reproducible experiment. But you claim not only a mere doubling but a never ending increase equivalent to the original for each extra shell put in place.

I bet you can't.

I still cannot get over that people buy this stuff !

This can continue for a long time -- not forever, as there are some 2nd law limitations (e.g. you cannot concentrate sunlight to achieve a higher temperature than the surface of the sun). But you can get a wide variety of temperatures depending on the radiative barriers between the sphere and cold space.

ReplyDeleteThis properly done energy balance analysis shows that the effect is not creating any additional power as it increases temperatures internal to the system.

This principle has been engineered into useful devices. High end vacuum flasks have multiple layers of foil in the vacuum gap (separated by sheets of non-thermally conductive material) to provide this exact kind of effect. In such a flask, a very small power input to the material inside the flask would produce a very high temperature because all these separate radiative shells provide no good thermal path out. Even low end Thermos flasks provide one or two shells for this purpose.

DeleteDidn't you see the complete nonsense of your analysis??

Here are YOUR OWN statements:-

"The 235 power output sets the temperature of the shell, given the emissivity (=1.0 for a pure blackbody). This gives a temperature for the shell of ~255K (-18C)."

"Next, we look at the shell alone as our new system to analyse. We already know that it is radiating 235 from its outer surface. Its inner surface must also be radiating 235 inward toward the sphere."

"So we have so far established that the shell must be outputting 235+235=470, and this is at equilibrium! So it must be receiving 470 as well to maintain this equilibrium, and the only place it can get it is radiation from the sphere."

"Now we look at the sphere itself. It is receiving 235 from the internal radioactive decay, and 235 from the shell. So to maintain equilibrium, it must be outputting 470. Therefore its surface temperature at equilibrium must be high enough to radiate 470 outward. For a blackbody, this would be a temperature of about 29C or 302K."

WELL GENIUS - WHICH IS IT ??

Is the shell radiating at 235 to “space” at minus 18 C; or,

Is the shell radiating 470 at plus 29 C or 302 K ??

Because you claim it is doing BOTH at the same time in order that it can supply the 235 that heats the sphere beyond what its internal power is capable of.

Remember that it is the temperature of the object that determines its radiation.

You have completely shredded your own argument.

You have completely proven that YOU DO add radiative fluxes as I claimed earlier does not work as well as demonstrating what nonsense the idea is !!

What makes you think the shell has to radiate at 470 for there to be 235 both “up” and “down” ??

A point source radiating at 235 radiates 235 in all directions – you DO NOT multiply 235 x 4 pi steradians

This is nonsense – arrant nonsense !!

If the shell is radiating 235 because its temperature is minus 18 C it is radiating 235 in all directions – only fools try to claim because it is up and down it must be double.

Your logic defies the inverse square law :-

“Any point source which spreads its influence equally in all directions without a limit to its range will obey the inverse square law. This comes from strictly geometrical considerations. The intensity of the influence at any given radius r is the source strength divided by the area of the sphere.”

You say placing a solid shell doubles the flux at whatever radius you nominate – I say Bull****.

Actually you do multiply by 4 pi steradiand to determine the TOTL power output from a point source with a given flux power.

DeleteBut this is TOTAL power not W/sq metre.

This is the fundamental Flaw of climate science and a correct experiment proves it beyond doubt.

Steel greenhouses are fantasy !

"But you can get a wide variety of temperatures depending on the radiative barriers between the sphere and cold space."

ReplyDeleteThis statement demonstrates your complete lack of scientific understanding.

First - it is simply absurd to apply physical properties such as hot or cold to something that by its very definition has no physical properties at all - vacuum space !

But even if it were valid why do YOU ASSUME the space at the location of the Earth's orbit is "COLD" ??

The Earth is moderately close to a moderate star that is continuously emitting powerful radiation !

We have experts tell us that radiation is of the order of 1370 watts per square metre - all the time - the only means to avoid it is in the shadow of a planet !

(they use this and the inverse square law as one method of calculating the sun's temperature )

We have experts tell us that radiation is capable of heating the Moon's unprotected surfaces to in excess of 390 Kelvin !

What bit of that sounds "COLD" to you.

Sure it is believed the trace radiation of deep space corresponds to a temperature of a few Kelvin but the space for hundreds of thousands of kilometres areound Earth cannot be classed as "COLD" by any stretch of the imagination !

To claim things like this shows a lack of credibility or curiousity.

Don't believe everything you read - question everything !

Surely the fact that above Earth's atmosphere there is radiation capable of inducing temperatures higher than 390 Kelvin PROVES the atmosphere helps SHIELD us from this fearsome power ??

ReplyDeleteIt is only when climate science divides it by 4 to account for their "radiative balance" that the mistakes happen.

Consider 235 W/sqm is equivalent to ~243 Kelvin - minus 18 degrees C.

4 x 235 = 940 W/sqm is equivalent to ~359 Kelvin - plus 86 degrees C.

Under the noonday sun on a clear day is the asphalt on the road to Riyadh in Saudi Arabia at minus 18 degrees C or closer to plus 85 degrees C ??

And is it the Sun heating the asphalt to near 86 degrees C or is it the almost non-existant water vapour and 0.04% CO2 ??

I've seen asphalt melt in outback Australia in summer when it can be extremely hot and very dry - dangerous enough that you can be seriously harmed in a few hours without adequate hydration.

Singapore on the other hand has near continuous saturated humidity and the temperatures are almost 10 - 15 cooler during the day.

Averaging radiation fluxes is nonsense as is summing them as demonstrated as A++ Class science as claimed by Curt.

Sorry - typo alert !

DeleteOnviously should be Consider 235 W/sqm is equivalent to ~253 Kelvin - minus 18 degrees C - not 243 !

And yes 253 K is ~minus 19 C - just using Curt's figures.

Delete"This properly done energy balance analysis shows that the effect is not creating any additional power as it increases temperatures internal to the system."

DeleteSo let me get this straight.

For each shell added the sphere with 235 W/sq m internal energy can radiate 235 W/sq m plus "x" by 235 W/sq m where x is the number of shells added ?? And this means that this

"effect is not creating any additional power as it increases temperatures internal to the system."

How precisely do you increase temperature without increasing power ???

Never Mind - let's continue :-

The melting point of Lead is about 327.5 degrees C or about 600.5 Kelvin.

So if I have a sphere of lead and provide it with a continuous 235 W/sq m and place 31 shells around it BY YOUR SCIENCE it will then radiate 235 + 31 x 235 = 7520 W/sqm.

To do this it must be at a temperature of 603.47 Kelvin and will melt.

Wonder why those blacksmiths didn't figure that one out ?

All that work slaving over a furnace at temperatures high enough to peel skin when all they needed was minus 18 degrees C and 31 shells.

You are kidding - right ?

Show me any actual proof that this happens and I'll happily admit I'm wrong !!!!

You'll be inline for a Nobel when you do it !

Rosco, you seem to have discovered an amazing form of steel that only radiates from one side! I want some of that, and I'm willing to pay good money for it! Lots of other people would be too.

ReplyDeleteBut first I have some questions about this steel you say you have.

How do you know which side radiates and which side doesn't? Does it come labeled? Because I want to use it for a heat sink design I'm presently working on. I need to be sure that I orient the radiating side outward. If I can do this, it would be so much more effective at dissipating heat than the radiating-from-both-sides steel I'm using now. But if I orient it in the wrong direction, I'd be screwed!

By the way, which half of the sphere radiates and which doesn't? What is the best way to orient the sphere?

I thought I had broken the problem down into small enough baby steps, but I guess I was mistaken. The radiation power flux density as a function of temperature is given in watts per square meter of surface area. A shell has two surfaces -- the inner surface and the outer surface. So for each square meter of shell cross-sectional area, there are two square meters of surface area. This is literally as simple as 1+1=2, but you cannot understand it.

The sphere, on the other hand, only has an outer surface. You need to contemplate this fundamental geometrical issue.

You need also to understand that in the real world on earth, there are also conductive and convective modes of heat transfer, so you don't have pure radiative transfers. But still, multiple radiative barriers are commonly used to reduce heat transfer.

And you still do not even comprehend the difference between power and temperature. The fundamental equation you would learn in the first week of a beginning heat transfer course is:

dT/dt = Q/Cp

where dT/dt is the rate of change of temperature T with respect to time t, Q is the net (signed) power transfer into the system, and Cp is the thermal capacitance of the system. If you put a steady power into the system and limit the losses to less than the input, the temperature of the system will steadily increase. In a real system, the losses, especially the radiative ones, will increase with temperature until they match the power input. Your whole mode of analysis is simply fundamentally wrong.

Captain Curt says :

Delete"I thought I had broken the problem down into small enough baby steps, but I guess I was mistaken. The radiation power flux density as a function of temperature is given in watts per square meter of surface area. A shell has two surfaces -- the inner surface and the outer surface. So for each square meter of shell cross-sectional area, there are two square meters of surface area. This is literally as simple as 1+1=2, but you cannot understand it."

It is just fantastic when a fool provides you with the ammunition to destroy their own argument. The funny thing is he doesn't realise he has been doing this from the start.

Willis Eschenbach proposed this complete BS as follows:

The sphere has internal radioactive decay causing it to emit 235 W /sq. m - can we up it to 240 as the math is easier ? He also proposed that placing the shell would cause the radiation from the sphere to heat the shell to the point where it emitted 240 (235) out and 240 (235) back. Remember - let's say 240.

Let us consider initially - sphere - no shell.

The total power is 240 W/sq. m x 4 x pi x r^2 sq. m - let's say 240 x SAsp where SAsp is surface area of sphere.

Now we insert the shell and as Cap'n Curt rightly points out the surface area of the shell is double - that is 2 x SAsp.

Now the shell is intercepting the total power which is 240 x SAsp.

But Captain Curt insists it radiates at 240 over 2 x SAsp - 240 x SAsp out and 240 SAsp back.

Now I may be as stupid as his condescending language has been designed to convey but even I can see this is twice the energy available and hence a gross violation of the laws of thermodynamics !!!

So initially the shell must radiate at 120 out and 120 “back” hence Captain Curt is either simply wrong or an alchemist of the first order.

But you guys want to talk "back radiation" don't you ?

OK – let’s do that.

Do we agree that as proposed initially by Willis the shell is receiving all energy from the sphere and the 1st law of thermodynamics is valid and because of the double surface area the shell must radiate 120 initially and not the 240 that the sphere is else we break the first law ?

I mean that is a no brainer - right ??

240 out from sphere plus 120 "back" from shell = 360 - this is why I wanted to up it to 240 - my tiny brain can't handle ever increasing numbers of decimals.

So now we have a total power of 360 X SAsp out from the sphere. This means the shell must now radiate 180 out and 180 "back" to maintain a balance in accordance with the first law - because as the good Captain points out there is that pesky factor of 2.

(A note to the wise – we have already broken most of the laws of physics with this little thought bubble do far so what do a few more matter ?)

Of course this now means we suddenly have not only the 360 out but also the 180 "back" and a simple sum demands it must now be 540 out.

You can't add the 180 "back" to the original 240 out because 240 out only results in 120 "back" NOT 180 !!

So now we have 540 x SAsp out and the shell must radiate 270 out and 270 "back" to maintain a balance in accordance with the first law - because as the good Captain points out there is that pesky factor of 2.

Hmmm. - where do we stop because we seem to have missed the "equilibrium" point the good Captain insists exists ???

At 360 out from the sphere with 180 out to space and 180 "back" there is insufficient energy leaving the “system” to cover the original 240 radioactive decay.

This is a runaway system – perpetual motion – or as I prefer to call it - total BS.

We continue below because of limit to size of post.

OK OK – you want to add the “back radiation” to the original 240 each time – I’m not exactly sure how that works but let’s do it.

DeleteSo 240 out initially gives 120 “back”. We still agree on this, right ?

This now gives 360 out – 180 “back” and this now means 240 + 180 = 420 out ? – But it’s what we agreed.

420 out results in 210 “back” and this now gives 240 + 210 = 450 out.

450 out results in 225 “back” and this now gives 240 + 225 = 465 out.

465 out results in 232.5 “back” and this now gives 240 + 232.5 “back = 472.5 - Oh hell there’s those damn decimals but were getting close to the magic “equilibrium” of 480 out from the sphere = 240 out and 240 “back” at the shell.

Even my tiny brain can see this is tending to a limit of 480 which would be the good Captains “equilibrium”.

See what happens in this fantasy – each iteration gets closer to the magic 240 out and 240 “back” the good Captain insists is sound science “equilibrium”.

But it is nothing like what Captain Curt proposes as sound “science”.

But I find this an extraordinary proposal that is just nonsense – but it gets to the right answer - so it must be right hey ?

If Cap'n Curt promised to double our money for free we'd rightly call him a fraud !

DeleteBut he can double energy at the turn of a sentence !

The one thing I find amazing is the confidence that people have in something that has never been shown to be true - NEVER.

ReplyDeleteIf it had been shown to be anything more than a thought bubble there would be a published document describing the reproducible experiment - not simply opinion.

No matter what - theory is only theory until proven !

There is no such experimental evidence - if there is show me the proof and if I am satisfied I will happily admit I am wrong !!

AS usual when the "evangelical" fail to supply actual evidence they resort to insult.

Your "science" says that YOU can add an extra 235 W/sq metre to the sphere for each extra shell you add.

Do you really believe you can increase the temperature of a sphere from 255 K or minus 18 C to 302 K or plus 29 C by simply fitting a shell in this manner??

If so demonstrate it !

You can claim anything you like but if you can't show it is real it probably isn't.

Lets take a blowtorch to your shell and cut a section out such that we have a flat steel plate.

This "plate" at 255 K is radiating what ?

235 W/sq m ? - or

235 up plus 235 down? or

235 up plus 235 down plus 235 left edge plus 235 right edge plus 235 front edge plus 235 rear edge?

I think it is radiating 235 !

I also think no matter what sarcasm you spit you are not as smart as you think are.

Remember Richard Feynman said -

"It doesn't matter how beautiful your theory is. It doesn't matter how smart you are. If it doesn't agree with experiment it is wrong."

As I have said time and again -

I have shown an experiment that does not support your claims about adding radiative fluxes. My experiment is easily reproducible and produces consistent results.

You have shown nothing but thought bubbles and opinion.

I will happily admit I am wrong the minute you show me my experiment is wrong and you have one showing your claim actually works that I can perform confirming your results.

Until then you have NO REAL evidence - only opinion that has no evidentiary value !

I am simply amazed how you cannot see through the nonsense of your claims.

DeleteI guess you really are that dumb.

The proposal is radiation and the manner in which the Steel Greenhouse can double the flux !

It is absurdly easy to prove your description is wrong !

I have been waiting for your cognitive abilities to grasp the simple obvious truth that falsifies your BS so I'll explain it in terms even you

should be able to grasp.

The other claim that back radiation can cause extra heating is another issue.

You claim your analysis is sound - I say you are obviously wrong !

Initially the total power emitted by the sphere is 235 Watts per square metre x 4 x pi x r^2 in units of Watts or Joules per second spread equally in 3 dimensional space. This is the ONLY energy available.

Let's ignore the geometrical differences implicit in the requirement that the sphere must have a larger area - lets ignore the inverse square law.

We have 4 x pi x r^2 x 235 Watts total being radiated to the shell -

BUT

YOU claim there is double this total at the shell - 235 x 4 x pi x r^2 out to space plus 235 x 4 x pi x r^2 "back radiated".

Get it yet ?

For YOUR initial claim YOU NEED

235 x 4 x pi x r^2 "out to space" PLUS 235 x 4 x pi x r^2 "back radiation".

There is no way around it - YOUR INITIAL CLAIM IS WRONG !

Therefore the shell can only radiate 117.5 each way - not 235 each way as you claim.

I mean even you with your astounding grasp of grade 2 arithmetic must understand that because your "science" results obviously - EVEN TO A GENIUS LIKE YOU - A CLEAR CREATION OF ENERGY OUT OF NOTHING - it is not going to happen !!

Violation of well established physics principles right there !!

This alone absolutely falsifies your analysis - even you cannot possibly argue with this.

You are proven wrong beyond doubt !

What then of YOUR energy balance ?

Like all of the proponents of this nonsense you just make stuff up !

I repeat: You simply have no idea how to perform energy balances. You do not understand the difference between power (which is not conserved) and energy (which is conserved). You don't even really understand the difference between power and temperature. You desperately need to take an introductory thermodynamics course before you are ready to discuss these matters.

ReplyDeleteIt is trivial to show that your analysis fails completely. You say that the outer shell would only radiate 117.5 outward at equilibrium. Let's test that idea. Take the complete sphere+shell system. Its power input is 235. You say that its power output is 117.5. This is not even close to equilibrium -- the internal energy of the system is increasing at a rate of 117.5, which means its temperature would be increasing.

If you had ever taken a thermodynamics course, you would understand that this is the very first stage of the analysis that you should perform. It would show that your initial idea is completely wrong, and you need to modify your analysis. This should have been the very first step in your analysis -- that for a system in equilibrium with a 235 power input, you need a 235 power output. Anything else is wrong from the start; there is no point in continuing.

You seem to think that my analysis showing 235 out and 235 in from the shell is creating energy. It is not, as I carefully performed energy balances on each component of the system to show that energy was conserved. But perhaps you need an analogy to understand it. Performing energy balances is an accounting exercise, just like balancing a bank account, using "conservation of money".

Let's say you had a single bank account, and every week you deposited your $235 paycheck, and every week you withdrew $235 to spend. The balance of your bank account would be steady over time; it would be in equilibrium. This case is analogous to the sphere alone.

Now you open a second bank account. You still deposit your $235 each week in your first account, but you will spend your money from the second account. So each week you transfer $470 from your first account to the second to make sure you can cover large expenses. But you don't, so every week you spend $235 from the second account and transfer $235 back to the first account.

Have you created any money in doing these transfers? No! Are they perfectly valid with money still conserved? Yes! By your logic, these transfers between accounts would be creating money, and you should be transferring money as rapidly as you can between your bank accounts to make yourself rich!

Back to the physical realm: Your claim is that putting a radiative barrier around a radiating body with a separate power source could not make that body increase in temperature. It is trivial to demonstrate that this is wrong, which I did in a simple experiment I described over at WUWT. Your analysis cannot explain those real-world results.

http://books.google.com.au/books?id=dQGC0ifkE34C&pg=PA24&lpg=PA24&dq=concentric+sphere+black+body&source=bl&ots=Zh6N1e35jc&sig=m-7nVWch4_zv-l3ISR5k7bluSUQ&hl=en&sa=X&ei=_ldYUd7EFZLU9ATVzYHoDw&redir_esc=y#v=onepage&q=concentric%20sphere%20black%20body&f=true

DeleteThe above is a link to a text book where the answer to the proposal is that the radiative flux from the shell is half that of the initial radiative flux - as I said above.

You will note no mention is made of increasing the initial flux above its original value in this solution.

As you can clearly see in this example Q is the initial rate of energy loss by the blackbody – ie the sphere.

So Q = 235.

Q’ is the rate of energy loss by the blackbody – i.e. the sphere - after being surrounded by the shell.

And Q’’ is the rate of energy loss by the shell.

As you can see from the solution Q’/Q is equal to the ratio of the radius of the Shell squared over the sum of the square of each radius i.e. - R^2/(R^2 + r^2).

In our little thought experiment it is assumed the radius of each is approximately equal.

This gives a relation where Q’(radiation from the Shell) = Q (initial radiation from Sphere) x 1^2/( 1^2 + 1^2) or Q/2. That is the shell radiates at 117.5 not 235.

You mocked this assessment. I have shown a balid reference where it is stated as a valid solution in a recognized textbook - Problems and Solutions on Thermodynamics and Statistical Mechanics (Major American Universities Ph.D. Qualifying Questions and Solutions).

Please do not mock that Google Books mistakenly refer to Mechanichs - it is a typo by Google.

I have searched high and low for an authoritative reference supporting your analysis.

Can you please provide such a reference in order that I may read it ?

The above is problem number 1026 in the text book.

DeleteI find this result surprising as well - I simply quote it.

If R is almost equal to r then it is implied that Q' = 1/2 Q.

Note you claim the same ratio but you double Q from 235 to 470.

If R = 2r then it is implied that Q' = 4/5 Q. - 235 x 4/5 = 188.

If R = 3r then it is implied that Q' = 9/10 Q. - 235 x 9/10 = 211.5

If R = 4r then it is implied that Q' = 16/17 Q.

There is a problem in the same book - number 1023 - which cites the same answer for a close fitting shell - Q' = 1/2 Q.

This example states this is at equilibrium.

Many people cite this as proof of your assessment. Equally many cite it as evidence of the opposite.

I have seen another reference which I am still trying to relocate.

As I said I would like to read your assessment reference.

Different problem. No constant internal power source. The fact that you cannot tell the difference tells me all that I need to know about your skills.

ReplyDeleteIn this (poorly stated) problem is the constraint that the sphere itself is maintained at a constant temperature T. So of course, its temperature will not be increased by the presence of the shell, and the shell will be at lower temperature.

It gets incredibly tiresome trying to discuss this with people who don't understand such basics as telling the difference between systems with no power source, a constant power source, and a power source that maintains a constant temperature. That includes several posters on this thread.

I repeat for the umpteenth time -- you need to take a real thermodynamics class before you can understand what is going on properly. At least go out and buy an introductory engineering thermodynamics textbook. Virtually any will do. They will explain early on how to define your systems and subsystems (aka "control masses") and perform equilibrium energy balances on them. I double-checked my analysis with my undergraduate text: Reynolds & Perkins, Engineering Thermodynamics.

As I have said before, the key to this problem is the energy balance on the total sphere+shell system. For it to be in equilibrium with a constant 235 power input, it must be radiating a constant 235 power output from the outer shell. Everything else follows from there.

How do YOU propose the temperature is held constant in light of continual radiation loss ? How is this different from the original proposal of the Steel Greenhouse ? Isn't the idea of the internal power supply to simply supply energy that maintains the temperature initially ?

"The 235 power output sets the temperature of the shell, given the emissivity (=1.0 for a pure blackbody). This gives a temperature for the shell of ~255K (-18C)."

DeleteIn your analysis you make these statements:-

This is the power output by the Shell I assume.

Then you say -

"Next, we look at the shell alone as our new system to analyse. We already know that it is radiating 235 from its outer surface. Its inner surface must also be radiating 235 inward toward the sphere."

I have no argument that an object at ~255 K will radiate ~235 W/sq m in all directions - this accords with the SB equation.

"So we have so far established that the shell must be outputting 235+235=470, and this is at equilibrium! So it must be receiving 470 as well to maintain this equilibrium and the only place it can get it is radiation from the sphere."

Now you have implied that the shell is no longer radiating at ~235 which is the SB calculated flux of an object at ~255 K but it is now radiating double that.

Why is - for the shell - 470 W/sq metre = sigma x ~302^4 suddenly wrong in your analysis. The math is right - 470 is the flux emitted by an object at 302 K.

You clearly say "So we have so far established that the shell must be outputting 235+235=470, and this is at equilibrium!"

If a plate of steel is at ~255 K does it emit 235 or 470 ? Or as a steel plate really has six surfaces perhaps it is six times or do we need to consider all the area components because that is no longer a flux.

Because YOU say both - you cannot deny that.

Which is it - does an object at a certain temperature emit a flux as calculated by the SB equation or not ?

Also this explanation of MLI seems to disagree with your analysis.

"The principle behind MLI is radiation balance. To see why it works, start with a concrete example - imagine a square meter of a surface in outer space, at 300 K, with an emissivity of 1, facing away from the sun or other heat sources. From the Stefan-Boltzmann law, this surface will radiate 460 watts. Now imagine we place a thin (but opaque) layer 1 cm away from the plate, thermally insulated from it, and also with an emissivity of 1. This new layer will cool until it is radiating 230 watts from each side, at which point everything is in balance. The new layer receives 460 watts from the original plate. 230 watts is radiated back to the original plate, and 230 watts to space. The original surface still radiates 460 watts, but gets 230 back from the new layers, for a net loss of 230 watts. So overall, the radiation losses have been reduced by half by adding the additional layer."

Oh wait - a different problem again I bet - no internal power supply - yet they say -

"The original surface still radiates 460 watts, but gets 230 back from the new layers, for a net loss of 230 watts. So overall, the radiation losses have been reduced by half by adding the additional layer."

How does it radiate 460 and the system lose 230 without some internal energy supply - it would soon radiate all energy away ?

So why don't they explain this in the explanation.

This is now the third separate example where the claim is made that the radiation output is halved by the shell.

So if the shell is simply emitting 235 in all directions and you do not add the flux up to the flux down - that is the shell is really only emitting 235 as an object at 255 K does it does not require 470 from the sphere to emit 235 to space - it simply requires 235.

DeleteWhy is this wrong ? You yourself say the shell is emitting 235 at ~255 K.

Stefan-Boltzmann say 470 is emitted by objects at ~302 K.

You simply do not know how to do these things at all.

ReplyDeleteI think I prefer the real examples given by real textbooks than your example!

You have a shell at a temperature of ~255 K emitting ~235 W/sqm and then out of nowhere you double the flux of the shell because you can't understand that a temperature of ~255 K emits ~235 W/sqm in 3 dimensions - you do not add "up" to "down" to double the flux because that is impossible because if you double the flux you raise the temperature of the shell.

The concept that the shell ever radiates 235 is your construction – none of the textbook examples I see seems to do this.

I prefer:-

The sphere has a 235 internal power supply.

The sphere accordingly radiates 235 in 3 dimensions.

The shell intercepts this.

The shell radiates at 117.5 in all directions.

To maintain this 117.5 radiation it requires the other half of the 235 it intercepts from the sphere – because it has no energy input other than the 235 from the shell.

It gets 235 from the sphere, the sphere radiates 117.5 to space and is accordingly at a temperature of ~213 Kelvin.

There is balance in this analysis.

Sphere has internal source of 235 and consequently radiates 235 from the sphere to the shell.

The shell absorbs 117.5 and heats to 213 K. This causes the Shell to radiate at 117.5. - 117.5 + 117.5 = 235.

This means 177.5 radiated in all directions – 117.5 to space and 117.5 back to the sphere.

This is a correct balance in accordance with real science not BS made up by – well you know.

There is a real balance here without any need to make up an imaginary temperature increase caused by energy that never existed.

Balance 235 out from sphere – 117.5 back to sphere – 117.5 from shell to space equals net from sphere. The Sphere remains at 255 K and the shell is at 213 K.

This is consistent with the 3 examples quoted by me.

Let’s do this the real manner quoted by every text book I’ve seen.

Let “To” be the external temperature, r = radius of sphere and R = radius of shell.

The rate of energy loss from the sphere before being surrounded by the shell is given by :-

Q = 4 x pi x r^2 x sigma (T^4 – To^4) - this is real science - Look it up !

The rate of energy loss from the sphere after being surrounded by the shell is given by :-

Q’ = 4 x pi x r^2 x sigma (T^4 – Ti^4) where Ti is the temperature of the Shell

The rate of energy loss by the shell is given by :-

Q’’ = 4 x pi x R^2 x sigma (Ti^4 – To^4).

If you say To = 0, and R is approximately equal to r – which is one of the requirements of the Steel Greenhouse it is trivial to show 2 x Ti^4 = T^4.

In this example this result is T = ~255 K and the shell is ~213 K.

213 K x the fourth root of 2 is ~255 K.

Your analysis is not a balance in any shape or form.

Your whole proposal is absurd !!

Except that you forgot the constant 235 W/m2 that powers the whole system. You have just had what my kids would call an EPIC FAIL.

ReplyDeleteYou are solving a different problem, and the worst thing is, you can't even get it after it's already been pointed out to you.

You said –

ReplyDelete“Except that you forgot the constant 235 W/m2 that powers the whole system.”

Are you really that stupid ?

When I derived the result from first principles I used the value for the temperature associated with the 235 W/sqm – 255 K to obtain the 213 K for the shell.

Can't you see that ?

Umm - that's an initial condition, not an ongoing power flow. Every time I think you can't make another basic mistake, you surprise me.

DeleteYou are effectively arguing that, given an initial temperature, it doesn't matter what a house's furnace does after that when analyzing the effects of insulation.

You said above –

ReplyDelete“In this (poorly stated) problem is the constraint that the sphere itself is maintained at a constant temperature T. So of course, its temperature will not be increased by the presence of the shell, and the shell will be at lower temperature.”

This is one of the stupidest statements I have ever read and it shows you know no real science at all!!

How do you make this leap ? By what mechanism do you propose its temperature is held constant ???

Did you look at the book I quoted ? Did you notice the title?

Problems and Solutions on Thermodynamics and Statistical Mechanics (Major American Universities Ph.D. Qualifying Questions and Solutions)

Compared to your shining intellect this is obviously a book written by morons – I’ll give them your name so you can correct their errors !

The author dealt with the Steel Greenhouse proposition from first principles -you simply make a fundamental mistake assumption.

There is a fundamental equation that describes radiant energy transfer – you may have heard of it. It is quoted in EVERY Physics book I have ever read.

“Then the net rate of radiation from a body at temperature T with surroundings at temperature T’ is

H(net) = A e x sigma(T^4 – T’^4)

In this equation a positive value of H means a net heat flow out of the body. Equation (17.26) shows that for radiation, as for conduction and convection, the heat current depends on the temperature difference between two bodies."

If T > T’ the net flow is from the body to the surroundings.

If T = T’ there is no net flow.

If T < T’ the net flow is from the surroundings to the body.

What about this don’t you get ?

The result you dismiss is derived by sound science from first principles.

Your absurd proposal is derived from you’re an assumption that is demonstrably wrong -

AND

YOU claim a result that all of the textbooks I have ever seen including the quote above CLEARLY SAY DOES NOT HAPPEN.

You are arguing against more than a century of accepted science.

You are an idiot – only a moron would argue they are right when all of real science says they are wrong.

How exactly is the solution derived from first principles solving a different problem ???

You clearly do not know what you are talking about!

You have offered no real science only your opinion!

Like so many people who think they know it all you simply make things up.

Only an idiot could think energy from a supply at ~255 K radiating against a shell can induce a temperature of ~303 K when the above mentioned equation clearly says that energy does NOT flow from a lower temperature to a warmer one.

In what way is any of this different from the "Steel Greenhouse" proposal ?

You cannot even see that the result derived from first principles requires the same temperature difference between the sphere and the shell as your FAILED analysis - the ratio of 1:2 for the fourth power of the temperatures.

You have this same ratio - you arrive at 470 W.sqm - ~303 Kelvin versus 235W/sqm - ~255Kelvin at the shell.

You analysis is simply wrong.

You make an assumption that is just wrong - the shell cannot radiate at 235 W/sqm as you claim.

Get some help from a real scientist - you are arguing that well established science is wrong.

No sophistry on your part overcomes your failure to understand you are wrong !

I simply cannot believe you cannot understand this !

As a final post on this issue I’ll explain a simple fact about why your experiment did not establish the result you claim.

ReplyDeleteFirst the equation I quote above says the net flow of heat is from hot to cold.

It “qualifies” this by including area considerations

H(net) = A x e x sigma(T^4 – To^4).

Inside your bulb is a filament that is heated to over 3000 Kelvin – reference example http://hypertextbook.com/facts/1999/AlexanderEng.shtml

At 3000 Kelvin it is emitting in 4,592,700 Watts per square metre at all frequencies of the EM spectrum – with significant amounts of IR – more than visible. And as we all know a Watt is a joule per second.

The area of this filament is tiny so obviously the number of joules per second is small.

Your first problem was measuring the temperature of the glass of the bulb. This is definitely NOT the source of any energy – remember the little equation above – it is heated by the filament.

You then enclose the bulb in all sorts of enclosures and measure the temperature of the glass of the bulb?

Because the temperature rises you rush to the conclusion you have proven back radiation.

Well you may think you have but you have actually proven nothing at all.

Enclosing the bulb as you did simply prevented the filament fulfilling its design purpose – emitting energy as light to triumph over the dark without overheating.

Coincident with that light is significant IR.

The filament is pumping energy out at 4.6 MW/square metre and you are trapping a lot of that.

Of course the temperature will increase – even Neanderthals knew that would work when they moved their fire inside a cave because it worked better in a confined space.

The problem you have is PROVING that the FINAL temperature you recorded is NOT SIMPLY the temperature the filament is capable of inducing on its own in the circumstances.

You have no proof of this at all – remember that pesky equation that says there is NO NET FLOW OF “HEAT” FROM A COOLER SOURCE TO A WARMER ONE?

If there is no net flow then the filament is most likely causing ALL the heating effect.

Unless you can PROVE otherwise your experiment simply confirms something that has been known for thousands of years – putting a lid on a pot simply increases efficiency.

The fact that thousands of people will read the pseudoscience you and Anthony Watts espouse as “proof” of back radiation when it is nothing of the sort and is in direct opposition to the laws of physics is a travesty both should be ashamed of.

This comment has been removed by the author.

DeleteThe speed at which you can generate errors is simply breathtaking!

DeleteThe control case for my test was the glass shell, which is far better at reducing conductive/convective losses than the metal shells, because its thermal resistance is hundreds of times higher than the metal shells. So it is already the "pot lid".

But with either metal shell in its place, the bulb temperature is significantly higher. Conductive/convective losses from the bulb are higher, and the power input from the bulb has not increased (and actually decreased a bit). So what is the difference? It has to be the radiation from the metal shell.

The glass shell is almost totally transparent to the visible and near infrared radiation from the bulb and so lets it pass through to the room. The foil shell reflects almost all of it back, and the black metal shell absorbs and re-radiates it back. That is the difference from the control case.

Your insistence that the bulb surface temperature I measured is not important is completely misguided. First, the bulb in the glass is a physical object subject to the laws of thermodynamics. You say that radiation from a cooler object can NEVER contribute to increasing the temperature of a warmer object. The experiment proves that it is possible, in cases where there is another power source for the warmer object.

Besides, the bulb surface is analogous to the earth's surface. No one is claiming that the earth's surface is the source of its own power -- that comes almost entirely from the sun (and a tiny bit from the core).

Also, I did show that the filament temperature increased. The reduced current through the filament at constant voltage means that the filament resistance increased. For a metal such as tungsten, this means that the temperature increased. If you go through the math, this was a 3 to 4K increase in temperature of the filament. I have since done refinements of the experiment, and gotten an 11K increase in filament temperature.

If your case is that all I did was trap the radiative energy that is ultimately from the filament, then you are making the same case as those saying that greenhouse gases are trapping energy that is ultimately from the sun. If you want to call this a "pot lid", you are using the same metaphor as those saying that radiatively active gases act as a greenhouse roof. You have actually conceded the case!

Let's take these a piece at a time. First, look at the radiative heat transfer equation you show.

ReplyDeleteH(net) = A e x sigma(T^4 – T’^4)

Of course I've seen it and used it, in my undergraduate and graduate thermodynamics and heat transfer courses, in my professional industrial career, and in my teaching at major universities. It is important to realize what it does and does not say. That is what you need to understand.

Note that in my solution, this equation is completely satisfied. Look at the drawing at the top of this post. Since it talks about power flux density, we will rearrange this equation to:

H/A(net) = e.sigma(T^4 - T'4)

With blackbodies, e=1.0 and can be removed. Also, we'll take the temperature terms out of parentheses so the terms in the equation match the drawing components:

H/A(net) = sigma.T^4 - sigma.T'^4

In terms of the sphere at T and the shell at T', we have:

H/A(net) = 470 - 235 = 235

So we have 235 W/m2 net radiative power flux density from the warmer sphere to the cooler shell. Completely consistent with this equation.

In terms of the shell at T' and space at 0K, we have:

H/A(net) = 235 - 0 = 235

So we have 235 W/m2 net radiative power flux density from the shell to the space OUTSIDE the shell. Again, completely consistent with the equation.

Now, your analysis has for the sphere and shell:

H/A(net) = 235 - 117.5 = 117.5 W/m2 net from sphere to shell

And you have from shell to space:

H/A(net) = 117.5 - 0 = 117.5 W/m2 net from shell to space

So the use of this equation by itself does not distinguish between my analysis and yours. And in both cases, the temperatures differ by a factor of the 4th root of two.

To settle the question, we have to go to energy balance analysis. The net radiative transfer from sphere to shell is the only power flow out of the sphere. For the sphere to be in thermal equilibrium, this power out must match the power in. The power flux density input is given as 235 W/m2, so the net output must also be 235 W/m2 for thermal equilibrium. My solution provides this; yours does not.

To get this 235 W/m2 NET radiative output, the sphere's surface temperature must be high enough to radiate 470 W/m2. That is ~303K for a blackbody.

Nothing in this analysis says that power flows from the cooler shell to the warmer sphere. Starting with the sphere alone at ~255K radiating 235 W/m2 to space AND having a constant 235 W/m2 internal power source, we put the shell in place around the sphere. Let's say that the shell is at 0K to start, so initially the net from sphere to shell is 235 W/m2. But as the blackbody shell absorbs this radiation (because that is what a blackbody does), its temperature starts to increase from 0K. As it increases, it starts radiating (because that is what a blackbody does).

And using the equation YOU cite, this means that the NET radiative output from the sphere is reduced. Given that the power input is constant, the sphere now has more power coming in than leaving, so its internal energy, and therefore temperature, increase. This is true even though the NET power exchange between sphere and shell is ALWAYS from the warmer sphere to the cooler shell. My analysis does NOT contradict this equation.

(continued)

ReplyDeleteThe warming of the sphere will continue until its temperature reaches a value where the NET power transfer from sphere to shell matches the constant power input.

Considering the sphere and shell together, the shell will warm from its initial 0K until its NET power transfer to space matches the constant input power to the system. So at equilibrium, the shell will radiate 235 W/m2 from its OUTER surface to space. This requires a temperature of ~255K for the shell.

Your analysis fails this simple energy balance. You have for the sphere: 235 in and 117.5 out. Not close to equilibrium, so it will warm. You have for the sphere plus shell: 235 in and 117.5 out. Not close to equilibrium, will warm.

What you are missing in just looking at the net radiative transfer equation you cite is that this just expresses what happens between these two bodies, and tells you nothing about what happens in the interactions between these bodies and the rest of the universe. Even in simplified problems, there are many possibilities, all of which I have dealt with in problems and labs. Some of these are:

Case 1: These two bodies are completely isolated from the rest of the universe (no other power transfers). In this case, the two bodies will equilibrate at a temperature in between the initial temperatures T and T'.

Case 2: These two bodies also radiate to the cold of space, but that is the only other power transfer. In this case, both bodies will equilibrate at ~0K with the cold of space, but the sphere will cool from its initial temperature T to this temperature more slowly with the shell around it than it would without.

Case 3: These two bodies also radiate to the cold of space, but one of them (i.e. the sphere) has a thermostatically controlled power input to maintain it at its initial temperature T. Obviously, this body will maintain its temperature - by definition of the problem. The other body will reach equilibrium base on these conditions. In the case of the sphere/shell problem we are discussing, this would reach equilibrium with 117.5 power input to the sphere, 117.5 NET radiative power from sphere to shell (235 gross from sphere to shell, 117.5 gross from shell to sphere), and 117.5 gross/net from the shell's outer surface to space.

Case 4: These two bodies also radiate to the cold of space, but one of them (i.e. the sphere) has a constant power input sufficient to maintain its temperature at the initial value in the absence of the other body. This is the case we have been discussing, as shown at the top of the post. In this case, the system would reach equilibrium with 235 power input to the sphere (given), 235 NET radiative power from sphere to shell (470 gross from sphere to shell, 235 gross from shell to sphere), and 235 gross/net from the shell's outer surface to space.

My quibble with the problem you cite is that it does not clearly state whether it is case 2, 3, or 4, all of which lead to different answers. (At every level of my education, including graduate school, I have had to deal with poorly constrained problems like these.) Since it does not state a constant power input, Case 4 can probably be ruled out. Just from being testwise, Case 2 is unlikely, because that would make it a trick question, with both ending up at 0K. Although if I had given the problem worded this way to my students, I would have to give full credit to a student who gave this answer if he could back it up with proper analysis.

That leaves Case 3 as the most likely scenario the question writer meant. From the answer, that's what they seemed to want. It's also possible that a Case "2.5" was intended, with no other power source, but the sphere's thermal capacitance was huge compared to the shell. In this case, the system would reach a "pseudo equilibrium" in the conditions they state quite quickly, but then would slowly decay to ~0K as in Case 2.

(continued)

ReplyDeleteAll of my analysis has been strictly standard. My thermo and heat transfer professors drummed in to my head: energy balance, energy balance, energy balance. It is the first thing you assess in any problem -- you have not been using it at all.

As I told you before, I have already backed up my analysis with experimentation, placing a blackbody shell around a body with an internal power source. Even though the shell always had a lower temperature than the internal body, the temperature of the internal body was higher with the black shell in place than not (with the control being a radiatively transparent shell). With the transparent shell, the body surface temperature was ~105C; with the black shell, it was ~124C.

For those looking in, the results can be found at:

http://wattsupwiththat.com/2013/05/28/slaying-the-slayers-with-watts-part-2/

The experimental results are consistent with my analysis, but not yours.

Here's another real-world example: A few years ago, I designed a power electronics system. One of the big challenges in such a design is making sure that the power transistors do not overheat -- you must design an effective heat removal system. We did a lot of analysis and modeling, and figured out what heat sinking was needed. We found that a thick metal plate whose flat outside surface was directly exposed to the room ambient was sufficient to remove the heat while keepin the temperature of the power transistors within specifications. I also specified that there be no cosmetic metal shell on this side of the product, because I wanted the heat sink to be able to radiate directly to the room.

When the first prototype was being tested, the technicians came to me, alarmed because the power transistors were overheating. I went to the lab to look at the system, and found that my instructions that the metal shell not go over the heat sink metal plate had been ignored. I got them to remove the cover from that side, and the power transistors ran 20C cooler for the same (constant) electrical power input under our test conditions.

So that's another example where the presence of a (highly conductive) metal shell around a body with an internal power source leads to a higher temperature of the inside body. Your analysis would have been useless in diagnosing the problem. And getting this right was absolutely essential to having a properly working system.

By the way, you keep saying that my statement that the shell radiates 235 W/m2 from the outer surface is simply an assumption. It is not! It is calculated using the 1st Law of Thermodynamics (conservation of energy) applied to the sphere-plus-shell system. Using the 1st Law, for a system in thermal equilibrium, power output must match power input. The problem states that there is a constant 235 W/m2 power input to the system. Therefore, at equilibrium, there must be a constant 235 W/m2 power output from the system. Since the only power transfer mechanism to the rest of the universe is radiative, the outer surface must transfer 235 W/m2 to space. No assumptions, just the logical application of fundamental physical laws.

You have said above"

"The 235 power output sets the temperature of the shell, given the emissivity (=1.0 for a pure blackbody). This gives a temperature for the shell of ~255K (-18C)."

ReplyDelete"Next, we look at the shell alone as our new system to analyse. We already know that it is radiating 235 from its outer surface. Its inner surface must also be radiating 235 inward toward the sphere."

"So we have so far established that the shell must be outputting 235+235=470, and this is at equilibrium! So it must be receiving 470 as well to maintain this equilibrium, and the only place it can get it is radiation from the sphere."

I simply cannot believe you cannot see you have arbitrarily doubled the flux at the shell in these statements.

Here is what you say in case you missed it for God knows how many times :-

"So we have so far established that the shell must be outputting 235+235=470, and this is at equilibrium!”

Anyway let’s consider the following.

If the shell were double the radius of the sphere it would radiate at one quarter of the sphere’s original flux.

235 x 4 x pi x1^2 Watts out from the sphere = 235/4 x 4 x pi x 2^2 times area for shell = balance.

Sphere originally at 253.72963858051897709704191572713 Kelvin – 235 W/sq m

Shell at 179.41394802829681914881881858825 Kelvin - 58.75 W/sq m

If the sphere was 1 metre radius the total Joules per second out from the sphere is 235 (W/sq m) x 4 x pi x 1^2 (sq m).

I am right on this am I not ?

This has to be the maximum number of Joules per second the shell can intercept when it is first placed.

This surely has to be 235 (W/sq m) x 4 x pi x 1^2 (sq m) – OUT FROM SPHERE - intercepted over four times the area for a maximum of 58.75 W/sq m and equivalent to 179.41394802829681914881881858825 Kelvin.

We can agree on that can’t we ?

Why is there any need to suddenly increase the output from the sphere to 470 to keep the output from the shell to space to a value equivalent to 235 x 4 x pi x 1^2 as originally output by the sphere ?

It already equals that in total – ¼ of the flux x 4 times the surface area.

235 x A = 58.75 x 2^2A.

Now let’s say the shell is 1.1 times the sphere radius.

Still 235 x 4 x pi x1^2 for original sphere.

235 x 4 x pi x1^2 intercepted over an area of 4 x pi x 1.1^2. Can we agree on that?

(1/1.1)^2 is equal to 0.8264462809917355371900826446281.

This means the shell is emitting 194.2148760330578512396694214876 Watts per square metre at a temperature of 241.92171816932995830304055179298 Kelvin.

235 x A = 194.2148760330578512396694214876 x 1.1^2 x A

Now let’s say the shell is 1.01 times the sphere radius.

(1/1.01)^2 is 0.98029604940692089010881286148417

which means the shell is emitting 230.36957161062640917557102244878 Watts per square metre at a temperature of 252.47042664615565921050042208223 Kelvin.

Do I need to go on ? We are tending to a limit of 1. Why is there suddenly any need to arbitrarily double the flux if for all real scenarios there is no need to maintain balance ?

Can’t you see the fundamental error you simply keep making ?

You arbitrarily double the output flux.

What is the total surface area of a spherical shell of radius R?

ReplyDeleteIf you had to buy enough paint to paint all the surfaces of a spherical shell with a 10-meter radius, and each liter of paint could cover 10 square meters, how much paint would you buy?

Who cares ?

ReplyDeleteYou have defied ALL REAL SCIENCE throughout this whole discussion and you are still trying sophistry to cover your mistake.

Let's consider your paint analogy but the right way around - because you are STILL making the same mistake time and again.

And let's not change the figures - let's say you need 235 litres for each square metre.

YOU HAVE ENOUGH PAINT TO PAINT THE SPHERE INITIALLY IF YOU BUY - 235 litres per square metre x 4 x pi x r^2.

This is the initial premise.

But because the shell has 2 sides you can only do one of the following :-

Paint ONE side at the required rate of 235 litres per square metre; or,

Paint both sides at 117.5 litres per square metre.

But somehow you magically claim that suddenly you really have 235 x 4 x pi x r^2 plus 235 x 4 x pi x r^2 litres of paint - out of nowhere !

This is so stupid it is beyond belief !!!

You have consistently made the mistake of thinking because you have 235 Watts per square metre over a certain area you can arbitraily double the area and still have the same 235 Watts per square metre when all simple logic says that is wrong.

As I have said all along I will happily admit I am wrong if I am and will happily say so in any forum.

But I am not wrong.

I have said you doubled the flux when you should have halved it.

I showed how a textbook on Thermodynamics for students who want to qualify for PhD admission says you are wrong.

I gave the REAL example of a sphere and shell where the shell was double the sphere's radius and how the increased area simply had to mean the flux was 1/4 of the original and how the only balance possible for a consistent 235 out was 1/4 flux over 4 times the area.

I then showed as the ratio of the radius decreased the limit is 1 NOT 2 as you claim.

I still cannot believe that you are still claiming you are right when even your own examples show you have made a fundamental mistake !

Be an adult ! Admit you made an error and make sure you get it right in future.

You cannot even understand the elementary school concept that a shell has an inside and an outside. You would only have bought half as much paint as needed. You don't have the conceptual basis to start examining these types of problems.

ReplyDeleteRadiation as a function of temperature is per unit of SURFACE AREA. A shell has a surface area of 8 * pi * r^2. If you had ever done any real thermal design, you would know that more surface area means more heat transfer.

How would one surface of a shell "know" that there is another surface, so it would have to cut its own radiation in half -- that is what you are implying.

Why does the sphere not also cut its own radiation in half per unit of surface area, because the opposite side must radiate?

Do you really not understand the difference between a problem where you are told that there is a body "at temperature T" which means there is a constant temperature, and a body with a constant power input, which means that the temperature is free to vary in the problem?

Do you really believe that a system with constant 235 power input and constant 117.5 power output is at thermal equilibrium. Where does the other 117.5 go?

You wouldn't make it out of the first week of a thermodynamics course.

You are obviously a complete idiot.

DeleteYou cannot even see that when you say "A shell has a surface area of 8 * pi * r^2" you are admitting that you arbitrarily double the energy.

You are so stupid you deny real science for your delusional fantasy.

You provided the paint anology which didn't work in your favour so your back to claiming "authority".

As I said Real Science says that if you increase the radius of the shell to twice the radius of the sphere you quarter the flux.

235 x 4 x pi x r^2 === 235/4 x 4 x pi x 2r^2.

If you cannot see the equality there I am amazed you have the common sense to get out of bed in the morning.

Let's work through it once more

r Sphere = 1 metre R Shell = 2 metre.

235 x 4 x pi x 1^2 = 2953.1 Joules per second - lets check units - W/sq m x Area = W. W = J/sec.

This number of Joules per second is all you have initially - that is the initial starting proposition.

2953.1/(4 x pi x R^2) = Watts per square metre for the shell.

As R^2 is now 4 this evaluates to 2953.1/(16 x pi). Which is 58.75 Watts per square metre.

58.75 Watts per square metre over 4 times the area is the same energy as 235 Watts per square metre over area of 1.

You are the most unbelievable idiot I have ever had the misfortune to get into a discussion with.

You are wrong - every textbook ever written on Thermodynamics and radiation physics says you are wrong.

Have you never seen this equation

H(net) = A e x sigma(T^4 – T’^4) ???

What does it say if T is greater than T' ?? What does it say if T = T' ??

Can't you see you are preserving the Watts per square metre NOT ENERGY.

You are trashing the First Law by saying you have double the ENERGY !

I simply cannot believe you can't see that !

There is no point going on.

I say you are wrong !

Examples where we double the radius say you're wrong !

I showed how a textbook on Thermodynamics for students who want to qualify for PhD admission says you are wrong. !

You insist you're right.

Well good luck with that - I think anyone who really thinks about this will see you are defying the first law of thermodynamics by creating energy out of nothing to maintain the flux.

For your information 235 out from sphere is exactly balanced by 117.5 out from shell plus 117.5 back to sphere.

H(net) = A e x sigma(T^4 – T’^4) says that because the shell is at a cooler temperature than the sphere the net energy transfer is from the sphere.

If you cannot understand that; and,

the fact that a flat plate of steel at ~255 K radiates 235 in all directions - you do not double it because it has two sides; and,

the fact that the text example found the Shell temperature from first principles using real science while you use pseudo science with the mistaken assumption you make

Then there can be no resolution to this.

But remember YOU claim it is OK to create energy out of nothing - not me I say there is no need because at every second there is a total of 235 W/sq m x 4 x pi x r^2 squarw metres emitted by the sphere.

Then you magically double it.

I'm through arguing - I might as well talk to the wall!

I’ll try this once more.

ReplyDeleteThe following quote is from - University Physics by Young and Freedman, 13 Edition, published in 2012, Page 575.

“Radiation and Absorption

While a body at absolute temperature T is radiating, its surroundings at temperature Ts are also radiating, and the body absorbs some of this radiation. If it is in thermal equilibrium with its surroundings, T = Ts and the rates of radiation and absorption must be equal. For this to be true, the rate of absorption must be given in general by . then the net rate of radiation from a body at temperature T with surroundings at temperature Ts is

Hnet = A ξ σ (T4 - Ts4) (17.26)

In this equation a positive value of H means that there is a net heat flow out of the body. Equation (17.26) shows that for radiation, as for conduction and convection, the heat current depends on the temperature difference between two bodies.”

This is verbatim from the text.

When the shell is first brought into place its temperature will be Ts – it has no energy itself and has not yet absorbed any from the sphere.

So let’s use Ts as the value of the temperature of the shell and it is free to increase and T as the as the value of the temperature of the sphere.

Note there are NO constraints on T except equation (17.26).

As the shell absorbs radiation its temperature will increase BUT during ALL of this time the value of Hnet in equation (17.26) will be positive and this means the energy transfer is from the sphere to the shell.

The minute it stops being positive ALL heating effect on the shell also stops.

When finally the temperature of the shell increases to the value where it equals the temperature of the sphere then equation (17.26) says there is NO net heat flow.

The heating effect simply stops.

Until this point the sphere has NOT changed its temperature - equation (17.26) demands this – because if it did the shell would NEVER equilibrate at any temperature.

But this isn’t sound enough science for you – at equilibrium you claim the process continues until YOUR preferred equilibrium is reached.

Let me say it again in simple English – the only way the shell can heat the sphere is for the shell to reach a higher temperature than the sphere -

This is completely indisputable – for the sphere to heat beyond its initial temperature with the initial 235 radiation the value of the shell’s temperature - Ts - will have to exceed the value of the sphere’s temperature- T.

And that NEVER happens – not even in your analysis!

Simple logic and actual science says so.

You are mistaken – if you can’t see that I am simply amazed.

This is completely pointless. You do not even have the basic conceptual framework to begin to analyze these problems. You pick out one equation but have no idea how it fits in to the larger picture.

ReplyDeleteEven just looking at the interaction between sphere and shell, you claim that the sphere transmits power to the shell without changing temperature. In other words, you are creating energy out of nothing. You say your equation demands this -- it does no such thing. It merely states what the net radiative power transfer is between two bodies of given temperatures at an instant.

To use this information in a system, you must also consider both the effect that this transfer has on the temperatures of the bodies, and also what other power sources and sinks there are. In the case we are discussing, the sphere has a separate power source and the shell radiates to space. You MUST account for both of these.

In an above post, I spent (wasted, it would seem) a lot of time laying out variations on this problem showing how they differ. I also explained at length that the net power flow (using the equation you champion) is always from warmer sphere to cooler shell in any of these scenarios, even as the sphere warms because the constant internal power is greater than the net outward radiative flow (even though that flow is from sphere to shell).

I've mainly been participating in this discussion to prepare for future classes so I can anticipate mistakes beginning students will make. You've provided more examples than I possibly could have dreamed of. Go on believing what you want, but please, please, do not design anything that could possibly cause harm, because you have absolutely no concept what you are doing. I am done here.

You appear to have lost all reason.

Delete“In an above post, I spent (wasted, it would seem) a lot of time laying out variations on this problem showing how they differ. I also explained at length that the net power flow (using the equation you champion) is always from warmer sphere to cooler shell in any of these scenarios, even as the sphere warms because the constant internal power is greater than the net outward radiative flow (even though that flow is from sphere to shell).”

You are now making stuff up – how do you propose your original equilibrium ever occurs if what you say immediately above happens – “because the constant internal power is greater than the net outward radiative flow”

So are you now saying that the heating will never stop?

Anyone that believes that fitting 4 shells around an object initially at the temperature of minus 18 degrees C will cause that object to heat to over 100 degrees C is simply delusional.

Remember 235 W/sq m is equivalent to ~254 K or ~minus 18 C.

Not one of your arguments has any real basis in science.

I am simply amazed you can continue to argue the same thing despite changing your position (and demolishing your own argument) more often than most change their underwear.

Rosco:

ReplyDelete...how do you propose your original equilibrium ever occurs if what you say immediately above happens – “because the constant internal power is greater than the net outward radiative flow”

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tfp:

He is saying that the power in the core is 235.

You said "For your information 235 out from sphere is exactly balanced by 117.5 out from shell plus 117.5 back to sphere."

i.e. the loss to space was 117.5

BUT we said the core generates 235 there is an inbalace and the system is not at equilibrium (235 generated 235 lost to space)

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Rosco:

So are you now saying that the heating will never stop?

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tfp:

No - heating stops when energy lost to space = energy generated at core (i.e. the outside of the shell is emitting 235)

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Rosco:

Anyone that believes that fitting 4 shells around an object initially at the temperature of minus 18 degrees C will cause that object to heat to over 100 degrees C is simply delusional.

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tfp

Agreed. But an ice cube does not have an internal source of constant power.