Tuesday, November 9, 2010

Gord Replies to Comments

UPDATE: Commenter 'Gord,' source of the analysis in the post originally here entitled Why 'Science of Doom' doesn't understand the 1st Law of Thermodynamics, was away from email for a few days and unable to respond to claims in the discussion. I unfortunately misinterpreted that he had dropped out of the discussion and therefore took it down with a "mea culpa." Gord is now back and thus I have put the discussions back up to allow further debate.

Although this debate is irrelevant to AGW for the reasons below, there is merit to continuing the discussions for those interested.

The Science of Doom 'thought experiment' does not prove anything relevant to the Kiehl/Trenberth (K/T) Earth energy budget or the earth/atmosphere system as claimed since

1. The atmosphere cannot behave as both an ideal gas and as a solid PVC blackbody (Science of Doom said elsewhere on this site, "...I realize you are right. As my example is made out of PVC and the atmosphere is made out of gas the example shows nothing useful.")
2. Limitation of conduction and convection by PVC has no relevance to the radiative transfer budget of K/T, which Science of Doom claims is supported by his thought experiment
3. The energy source  ["sun"] should be located outside the sphere
4. The energy source should be intermittent ["day & night"], not continuous. Better yet, the source could be continuous outside the sphere as long as the sphere continously rotates.

The K/T Earth energy budget shows

1. The incoming solar radiation (the only energy source) is 342 Wm-2
2. The earth surface radiation is 390 Wm-2, i.e. 48 Wm-2 more than the only energy source.
3. Another 102 Wm-2 is being lost from the earth surface via evapotranspiration and convection that is not limited by PVC or anything else, and conduction is not a part of the budget. Thus, we now have an energy source of 342Wm-2 creating 492 Wm-2 leaving the earth surface even without limiting convection and conduction as in the SoD example. An additional 30 Wm-2 is reflected from the Earth surface for a new total of 522 Wm-2 leaving the Earth surface.
4. Even if greenhouse gases could behave like a mirror, they cannot add energy to the system, and explain how

342 Wm-2 in becomes 522 Wm-2 out

Why conventional Greenhouse Theory Violates the 1st Law of Thermodynamics

U Mass thinks greenhouse gases can add energy to the system

And this explains why Earth energy budgets that don't incorporate greenhouse gases at all do balance whereas the K/T budget is described by it's creator Kevin Trenberth as "a travesty" that doesn't balance.

201 comments:

  1. "Doomed science" also thinks a cold body can make a hot body hotter

    lol

    ReplyDelete
  2. Thanks for letting me know about this article, it is a beautiful thing.

    (Just to clarify that the original article said the emissivity of the PVC was 0.8. So that is inside surface and outside surface).

    If you refer to the article you cite (which seems fine) you will see the important point:

    "Most solid objects exhibit very low transmission of infrared energy - the majority of incident energy is either absorbed or reflected. By setting transmissivity equal to zero, equation 6 can be restated as follows:

    100% = Emissivity + Reflectivity
    "

    Why not redo your analysis when no energy is transmitted via radiation through PVC.

    80% of energy is absorbed (within a few um of the surface)
    20% is reflected.

    I look forward to the updated analysis!

    ReplyDelete
  3. This example would be relevant to explaining the contribution of geothermal and human produced heat ( which is significant around cities ) , or the temperature of Venus's surface . But what does it have to do with balls radiantly heated from the outside like our earth ?

    ReplyDelete
  4. The thought experiment is absurd for many reasons, including

    1. The atmosphere cannot behave as both a solid PVC blackbody and as an ideal gas
    2. The heat source [sun] should be placed outside the PVC sphere, as 45% of incoming solar radiation is LWIR.
    3. While claiming the post supports the almost entirely radiation-based Trenberth energy budget, SoD wants us to only consider "no energy transmitted via radiation through PVC" [i.e. "the atmosphere"]
    4. SoD continues to ignore his miscalculation of the inner sphere temperature

    ReplyDelete
  5. The analysis is of a heat source inside a PVC sphere.

    I think you have become confused with what it might represent.

    If 30,000W of heat is produced at the center of this hollow sphere your challenge is to work out T1 (inner temperature) and T2 (outer temperature).

    The analysis in your article is clearly flawed because PVC cannot transmit radiation.

    So how does 30,000W leave the outer surface of the PVC sphere?

    (The only mechanism is conduction).

    I look forward to seeing your updated analysis.

    ReplyDelete
  6. While you are reviewing your update, you will probably realize that the outer surface must now be radiating 30,000W.

    If the outer surface is not radiating 30,000W then heat is being accumulated somewhere - which increases the internal temperature - until finally the outer surface temperature rises to the point where radiation = 30,000W.

    So - as a champion of the first of thermodynamics - you will conclude that the equilibrium emission of radiation at the outer surface requires a temperature of 133K.

    q = εσT^4.4πr^2
    if T = 133K, r=13m and ε=0.8

    then q = 30,000 W

    I can only imagine your dilemma from here.

    I think your best course of action is to now play on the "PVC and the atmosphere are not equivalent" idea to muddy the waters and hope that no one else asks you how 30,000W can be conducted through the PVC sphere.

    However, I will commend your bravery if you try to update your analysis without reinventing basic physics.

    ReplyDelete
  7. SoD,
    Yes I know you would like to continue to focus on conduction through an enclosed solid sphere as if that had any relevance whatsoever to the purpose of your post of validating the almost entirely radiation-based Kiehl/Trenberth earth energy budget, but the sad fact is it proves nothing useful or relevant to K/T or the earth/atmosphere system. You don't like the answers when shown the much more analogous radiative transfer calculations. That's fine, continue to live in your fantasy land of the atmosphere is a 3m thick PVC ideal blackbody and we are doomed to global warming due to limited conduction (and convection) thereof.

    ReplyDelete
  8. Well done. Best course of action.

    So it is safe to conclude that you now believe:

    a) "The article by Science of Doom was technically correct but not relevant"?

    OR

    b) Radiation can be transmitted through 3m of PVC?

    ReplyDelete
  9. So, it is safe to summarize & conclude that

    1. conduction has no relevance to K/T or the earth/atmosphere system

    2. limitation of conduction and/or convection has no relevance whatsoever to K/T or the earth/atmosphere system

    3. the only appropriate analogy to the K/T radiative budget is the radiative transfer calculations presented above (sans the extremely important caveat that the energy source should be located outside the sphere)

    4. your post nonetheless claims that K/T understand the 1st law because their radiative transfer budget is somehow analogous to limitation of unrelated conduction and convection through a solid PVC sphere.

    ReplyDelete
  10. I Would say that radiation cannot be transmitted through a 3m PVC sphere therefore you must restate:

    "Incident Energy = Absorbed Energy + Transmitted Energy + Reflected Energy"

    in "Incident Energy = Absorbed Energy + Reflected Energy" since "Transmitted Energy"=0

    This implies that all the power must be radiated by the PCV sphere, which dictates its temperature

    Now, with the conductivity of PVC, you can calculate the total emitted power un the interior surface, which is way above the 30kW. Funny enough, this does not violate the conservation of energy. Would you care to carry out the correct calculation and comment the results for us?

    Best Regards

    ReplyDelete
  11. Whatever happened to the notion of doing simple physical experiments to resolve this basic physics ? 50 years ago I don't think a dispute like this could have gone on so long without somebody doing a relevant experiment or citing one from the previous 150 years .

    I was surprised recently to learn that Ritchie's elegant experiment demonstrating Kirchhoff's ( and Stewart's ) crucial insight that absorption and emission are two sides of the same process 3 decades BEFORE Kirchhoff's work .

    ReplyDelete
  12. When you work out how to do the calculation of heat transfer in the PVC sphere and get to the right answer, I will explain the relevance to the Kiehl and Trenberth summary of the climate system.

    Sharp-eyed observers will probably work it out for themselves.

    ReplyDelete
  13. SoD- did you forget your own comment on this site about this thought experiment?:

    ScienceofDoom said...

    "I did think that your core argument was that a system receiving X W/m^2 couldn't have an internal surface radiating 3X W/m^2.

    I thought that an example showing that this can occur - in an easier to understand environment - would illustrate that your core claim was wrong.

    But I realize you are right. As my example is made out of PVC and the atmosphere is made out of gas the example shows nothing useful.
    ...
    August 3, 2010 7:26 PM"

    and did you forget the comments of physicist Harry Dale Huffman on another thread on this site:

    "I will take any of you seriously when I see you identify obvious flaws once and for all, and have done with them. Trenberth's "Global Energy Budget" has the Earth's surface--the surface, not 4k or 5k or any fraction of a k up in the atmosphere--radiating 390 w/m^2 upwards, greater than the putative only power input, 324 w/m^2 from the Sun. The only reason for this is that 390 w/m^2 is what a blackbody would radiate at the same temperature (288 K) as the "average" temperature of the Earth's surface. But you all should know this is wrong, so why haven't all those who go by that model, including the IPCC (and ScienceofDoom), been laughed off the stage? SOD's "simple example" in "Do Trenberth and Kiehl Understand the First Law of Thermodynamics?" results, in his detailed "scientific analysis", in 1.8 million plus joules per second of energy being created, with the only power source for that being only 30,000 joules per second. That is a physical impossibility and he is an educated idiot, pure and simple. How incompetent is anyone who doesn't confront him with that and refuses to argue with him thereafter? You people keep arguing details that nobody knows the answer to, while letting patent nonsense like Trenberth's model and ScienceofDooms's "simple example" continue to exist, and be promulgated to the public as the scientific consensus and highest standard. You are not helping the world find the truth, you're just vainly arguing difficult and academic questions. This tells any competent physical scientist that "climate science" is a farce. I am not a "climate scientist", I am a physicist, who doesn't have or claim to have all the answers, and who has a few times made minor mistakes on secondary points when trying to do science "on the fly" in forums like this, but who knows when he sees a ridiculous result. I don't make errors about central points. The "consensus" of climate scientists do, in allowing the Trenberth model to represent their "science." It's garbage, and you demean science by not throwing it out, once and for all. "
    October 4, 2010 8:20 AM

    ReplyDelete
  14. The Science of Doom article was "riddled with errors" but you can't identify them?

    You're not going to fix the errors in your own article are you?

    Less charitable readers might think that the reason is it would be even more embarrassing (you would need to create different errors or reach the same conclusion as the Science of Doom article).

    I realize it's because you don't have time for these details.

    I do love your work and will be happy to promote it further from my blog.

    But just let me know if you are going to fix your mistakes and I'll hold off until that time.

    ReplyDelete
  15. SOD will never quit pushing his pseudophysics, I guess.

    Forget, for a minute, about K&T and all the simplistic, silly, radiation cartoons and focus on empirical evidence for the "greenhouse effect," of which there is absolutely NONE! Why is it so much cooler in Atlanta on a summer day, with all the GHGs there (water vapor) than it is in Phoenix? Same latitude and elevation. Where is the GHE?

    Put another way: The direct solar radiation at noon in Atlanta in July is about 900 wm-2. Add the putative "backradiation" (nonsense) and you would have AT LEAST 324 (K&T average). That gives 1224 wm-2, which means that my black shirt should be a scorching 110 C (231 F). Doesn't happen. And the putative "backradiation" should be MUCH higher than this K&T average.

    It is clearly all nonsense.

    ReplyDelete
  16. SoD,

    Thanks very much for the complement and your own comment that your post proves nothing, and for the promos. While you're at it, promo this post on why greenhouse gases can't heat the oceans as opposed to your falacious claims:

    http://hockeyschtick.blogspot.com/2010/08/why-greenhouse-gases-wont-heat-oceans.html

    Back on topic, touche for ignoring your own comment, Harry Dale Huffman's comments, and any relevance to K/T as claimed by your post. Once again, K/T is a radiative transfer budget and thus the only appropriate analogy is the radiative transfer calculations above. It is absurd and silly to focus on conduction through PVC which has absolutely no relevance to the earth/atmosphere system, as stated by yourself. Thanks for admitting that the only-appropriate-analogy radiative transfer calculations above are correct.

    ReplyDelete
  17. I followed your dialog with Science of Doom, however you must be disappointed by SoDs inability to comprehend your points.

    It seems that anything that contradicts his AGW theory is simply ignored.

    His favourite method is to think up some far fetched thought experiment where the readers are led along his path of "faultless logic" to conclude that heat moves from a cold to a hot object.

    Real experiments however scare him off.
    I have asked him to comment on the famous experiment by Wood in SoDs own site.

    He refuses to engage.

    Further I have asked him to solve a simple equation on heat transfer.
    About the level of Physics 101– he refuses.

    He is preoccupied by Gerlich & Tscheuschner.

    This is shown by the number of oblique attacks he makes on them.

    I think he fears that they may be correct.

    He must think that unless he can reinterpret the second law then the whole IPCC case falls.

    This is not the position of more perceptive supporters of the IPCC position and several of them have suggested to him this his loose use of the word HEAT should be avoided

    I describe SoD as “he” because “he” or “she” or “them” refuse to give the slightest details of who they are.
    Most sites(like you own) give some details of the proprietors.
    Its not very brave to mount attacks on named individuals like Gerlich & Tscheuschner from behind a cloak of anonymity.

    ReplyDelete
  18. I only got as far as calculating T1.

    You take the 30kW incident on the inner surface and then calculate T1 so that the surface is emitting 30kW, so the net transfer of energy through the surface is zero.

    What?

    Of course, that means there is now 60kW bouncing around and you will have to do your calculation again!

    ReplyDelete
  19. My response to SOD:

    First let's start with what SOD used as his 30,000 Watt energy source, a light bulb.

    "In the center, we have our super-light-bulb, which radiates 30,000W. It is mysteriously powered, perhaps it is a nuclear device, or just electric with such a thin power cord we can’t detect it – we don’t really care."
    http://scienceofdoom.com/2010/07/26/do-trenberth-and-kiehl-understand-the-first-law-of-thermodynamics/

    The 30,000 Watts will come from a filament in the "super-light bulb"

    In my previous calculation I got (using an emissivity of 1 for the inner sphere) a radiation of 23.87 w/m^2

    SOD wants me to use 0.8 emissivity so the Inner Sphere would only absorb and radiate 0.8 X 23.87 = 19.096 w/m^2

    Using the Stephan Boltzmann Law the temperature T1 = 135.47 K ...NOT the 423K that SOD calculated!

    The inner surface of the sphere has an emissivity of 0.8 so 0.8 X 30,000 = 24000 watts is absorbed by the inner Sphere.

    SOD wants the 0.2 of the orginal 23.87 w/m^2 reflected back towards the filament so 4.77 w/m^2 is reflected back.

    The reflected energy is 0.2 X 30,0000 = 6000 Watts.
    -----------------------------------------
    -----------------------------------------
    I will spend some time explaining what happens with the reflected 4.77 w/m^2 or 6000 Watts.

    Electromagnetic Fields are propagating fields that carry energy from one place to another and are Vector quantities.

    Heat flux
    "Heat flux or thermal flux, sometimes also referred to as heat flux density or heat flow rate intensity is a flow of energy per unit of area per unit of time. In SI units, it is measured in [W·m-2]. It has both a direction and a magnitude so it is a vectorial quantity."
    http://en.wikipedia.org/wiki/Heat_flux

    Scalar "accounting math" does NOT APPLY. One needs to use VECTOR MATHEMATICS when summing VECTOR FIELDS.

    Vector addition of fields
    http://hyperphysics.phy-astr.gsu.edu/hbase/electric/mulpoi.html#c3

    The link above shows how resultant field vectors are calculated.
    Using superposition, many, many sources can analysed at any point in space to produce a SINGLE RESULTANT VECTOR.

    Radiative Heat Transfer equations also show this:

    Heat Radiation
    "Radiation is heat transfer by the emission of electromagnetic waves which CARRY energy away from the emitting object. For ordinary temperatures (less than red hot"), the radiation is in the infrared region of the electromagnetic spectrum. The relationship governing radiation from hot objects is called the Stefan-Boltzmann law:

    Heat Transfer by Radiation using the Stefan-Boltzmann Law

    P = e*BC*A(T^4 - Tc^4)

    Where P = net radiated power (Watts), e = emissivity, BC = Stefan's constant, A = area, T = temperature of radiator and Tc =
    temperature of the surroundings or another body.

    ..when rearranged gives

    P/A = e*BC*T^4 - e*BC*Tc^4 (Watts/m^2)
    http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/stefan.html#c2

    Gord [end of part 1]

    ReplyDelete
  20. [continued]

    This is an obvious subtraction of two Electromagnetic Fields

    It also complies with the Vector subtraction of Electromagnetic Fields which are Vectors.

    The resultant Electromagnetic Field will have a magnitude of P/A and have a direction of propagation in the direction of the larger field produced by the hotter body.
    ------------
    The "super-bulb" filament radiates an unknown X amount w/m^2 but it will be larger than what the inner sphere reflects (4.17 w/m^2) back.

    The Resultant EM Field flux between the filament and the inner sphere surface is a Vector quantity with a Magnitude of X - 4.17 w/m^2 and propagates toward the inner sphere surface.

    There is ZERO w/m^2 propagating to the much hotter filament.

    X - 4.17 w/m^2 is the TOTAL amount of EM Field flux from the filament that can reach the inner sphere surface.

    This is exactly the same as two opposing forces being applied to a to a block of wood, the same Vector addition applies and the resultant force will move the block of wood in the direction of the larger force.

    EM fields work exactly the same way.
    In fact, The Electromagnetic Field is a force, the Electromagnetic Force is one of the four fundamental forces and moves zero mass Photon energy from one place to another.
    ------------
    NOTE:

    Just like the block of wood moves with a resultant vector force and that is the only energy that can be released when the wood block strikes another object, the zero mass Photons move with the resultant EM vector field and X-4.17 watts/m^2 corresponding to 24,000 Watts is all the energy that can be released when it reaches the inner sphere surface.

    ---------
    If both EM field vectors were equal and opposite there would be zero radiation of Photons like this link demonstates.

    Light Interference (including ALL EM fields) is a very common phenomena that demonstrates Vector addition of EM fields.

    Here is an example of complete light cancellation in a reflective soap bubble.

    Cancellation of Light
    http://www.exploratorium.edu/ronh/bubbles/bubble_colors.html
    There is also a good explaination of Constructive and Destructive Interference.

    It should be noted that this is obvously not destruction of energy, it is merely the loss of the ability to move the Photon energy past the bubble surface.

    This is also evident in this equation based on the Law of Conservation of Energy:

    Incident Energy = Absorbed Energy + Transmitted Energy + Reflected Energy
    http://www.optotherm.com/emiss-physics.htm

    This is Energy In = Energy Out

    If absorbed energy and transmitted energy are zero then:
    Incident Energy = Reflected Energy

    Incident Energy - Reflected Energy = 0 (for 100% reflection inside the bubble)
    ---------------------------
    ---------------------------
    So getting back to the SOD example, it should be now be apparent that all that is required to comply with The Law of Conservation of Energy is to use the above equation stated as:

    Incident Energy - Reflected Energy = Absorbed Energy + Transmitted Energy

    Transmitted Energy = 0 and we are left with:

    Incident Energy - Reflected Energy = Absorbed Energy

    Putting the values we already have in we get:

    30,000 watts Incident - 6,000 watts Reflected = 24,000 watts = Energy absorbed by the inner sperical surface that was previously calculated.

    ReplyDelete
  21. [continued]

    Now I will calculate the heat transfer through the PVC Sphere inner to outer surface, only I will do it correctly.

    The outer surface sphere also has an emissivity of 0.8 so 0.8 X 24,000 = 19,200 watts which has to be radiated away by surface re-radiation Plus an additional 4800 watts that is the sum of heat transfer by conduction and energy stored in the PVC that is also radiated.

    The outer sphere area A2 = 2123.72 m^2 and will radiate 19200/2123.72 = 9.04 w/m^2 and the temperature T2 = 112.37 K
    ----
    This is the heat transfer by conduction used by SOD:

    qr = 4pk . (T1 – T2) / (1/r1 – 1/r2) except that this equation is wrong or maybe just typed wrong ?

    The actual equation can be derived easily:

    Ar = 4*pi*r^2

    Q = -k*Ar dT/dr = -k*4*pi*r^2 dT/dr

    Q (1/r^2)dr = -k*4*pi dT

    Integrating both sides gives:

    Q -(1/r) evaluated for r2 and r1 = -k*4*pi (T2-T1)

    Gives Q = k*4*pi (T2-T1)/ (1/r2 - 1/r1) = k*4*pi (T2-T1)(r1*r2)/(r2-r1)

    So for the values of k = 0.19, r1 = 10, r2 = 13, T1 = 135.47 K and T2 = 112.37 K

    Q = 0.19*4*pi (112.37 - 135.47)(10*13)/(13-10)

    Q = 2390 Watts is transfered by conduction.
    ---
    The stored energy in the PVC can now be calculated.

    Stored Energy = 4800 - 2390 = 2410 Watts.
    -----
    Summary:

    Incident Energy = Absorbed Energy + Transmitted Energy + Reflected Energy

    30,000 Watt source = 6000 watts reflected by inner sphere + 0 transmitted energy + 24,000 watts absorbed by inner sphere that has to reach free space.

    24,000 Watts = 19,200 watts absorbed by outer sphere and re-radiated + 2390 watts tranferred by conduction to the outer sphere surface and radiated + 2410 watts stored in the PVC that is also radiated.

    T1 = 135.47 K and T2 = 112.37 K

    And, there is complete compliance with The Law of Conservation of Energy
    ------------------------
    NOTE:

    SOD's use of the conduction equation was done wrong.
    He used 30,000 watts for the value qr when qr is actually the watts tranfered by conduction (2390 watts).
    He then came up with a temperature difference of 290K when it is actually 23.1 K and back calculated the watts at the source to be 1.85 MEGA-WATTS!

    What a HOOT!

    Gord

    ReplyDelete
  22. Thanks Gord!

    SoD, I refer you to your comment above, except that it has now come back to bite you and I ask of you

    ScienceofDoom said...
    The Science of Doom article was "riddled with errors" but you can't identify them?

    You're not going to fix the errors in your own article are you?

    Less charitable readers might think that the reason is it would be even more embarrassing (you would need to create different errors or reach the same conclusion as the [Hockey Schtick] article).

    I realize it's because you don't have time for these details."

    Do let us know once you have fixed the errors in your own article SoD

    ReplyDelete
  23. So, according to Gord, the cavity is accumulating energy at the rate of 6kW or 6kJ/s. This is cool, since given enough pvc and enough time, we can store an arbitrary amount of energy inside a cavity. Amazing.

    Best Regards

    ReplyDelete
  24. Well, this is interesting. I have moderately good reason to believe that SoD is NOT "a climate scientist named Dr. Phillips". In fact, a few months back SoD was being hailed by "skeptics" as somebody to pay attention to because he was really looking into the details of the science behind global warming with a very skeptical eye. Turns out the science was real after all - but SoD's initial stance was very critical, and still is it seems to me.

    But what SoD's site has really brought out is the interesting fact that there are some people out here on the internet who are extremely sure of themselves regarding certain imaginary versions of the laws of physics. MS exhibits an imaginary version of the first law here. Now that SoD has pointed his minions here, I expect a perfect storm, a black hole of imaginary physical laws to form and enter an alternate universe with quite different laws.

    Meanwhile, the rest of us would like to do something about some of the problems in this real world. Ah well.

    ReplyDelete
  25. Arthur,
    Typical alarmist ad hom with absolutely no substance to back it up, and apparently no desire to read or ability to find any fault in the comments immediately above - which provide the details why SoD violates the 1st Law. Ah well.

    ReplyDelete
  26. Um - I counted 11 errors in your first paragraph, and specified several of them (SoD is NOT, as far as I can tell, Dr. Phillips, and certainly not an "alarmist" by any stretch). But most fundamentally your claim:

    "calculates that the 30,000 Watts miraculously turns into 1,824,900 Watts emitted from the inner surface of the PVC sphere. To anyone with a rudimentary understanding of physics, this is an obvious violation of the conservation of energy demanded by the 1st Law of Thermodynamics"

    refers not to the *real* 1st law of thermodynamics, which concerns energy (i.e. Jouels) but to the *imaginary* first law, which somehow constrains radiative and other forms of energy *flow* (watts). Think about it for a bit, and you'll see there might be a teeny tiny problem with your logic here...

    ReplyDelete
  27. Very amusing, for anyone who's had a little training. SOD's example is precisely the sort of simplified 'what if' scenario used to set problems for physics classes-- taking the right approach (i.e. at equilibrium, heat flow through the sphere and radiating from its surface must balance the energy being supplied at the centre) is the key to getting the right analysis and answer-- and that answer (as the elegant original presentation makes very clear) is SOD's. Artificial examples like this are used to illustrate and clarify the use of principles-- which then guide further attempts at more serious applications. A second year course in thermodynamics is a good place to start.

    ReplyDelete
  28. As to Gord's long explanation in the comments here, he is absolutely right that heat flux is a VECTOR quantity. That's why the relevant question at any point on the inner surface of SoD's thought experiment shell is the VECTOR sum of heat flux. When the system is in balance (not heating up or cooling down), that VECTOR sum has to be a steady 23.87 W/m^2 directed OUTWARDS, to match the 30 kW being emitted by the central source.

    That is, in steady state a net 23.87 W/m^2 arrives at the inner surface from the source, and 23.87 W/m^2 leaves (by conduction, since transmission = 0) from that inner surface directed outwards. But what does that tell us about radiation from the inner surface directed inwards to the cavity? Only that radiation directed inwards must balance radiation received (except for the 23.87 W/m^2) - the VECTOR sum of (i.e. difference between) radiative heat absorbed and radiative heat emitted must be 23.87 W/m^2. But the TOTAL value of that radiative heat received could be anything, as long as it balances (minus the 23.87) what it emits.

    That is, the 1st law places *NO CONSTRAINT* on the total radiation emitted by the inner surface.

    ReplyDelete
  29. I was also banned.

    The Science of Doom is really only ...

    The Science of Global Warming

    - you can stay and discuss global warming science and even question it occassionally, but if you point out real errors in the science, you are banned so that others don't hear the evidence.

    ReplyDelete
  30. I see a tiny problem with Gord's analysis as well. Giovi Pele has already pointed it out.

    The heat source inside the sphere is 30,000W.
    The heat leaving the outer surface of the sphere is 24,000W (according to Gord).


    That can happen at the start, but the 6,000J per second of accumulated energy will increase the temperature internally until eventually steady state is reached and 30,000W does leave the sphere.

    The sphere can only radiate into outer space so there is no alternative to the fact that T2 (outer surface temperature) = 133K
    The simple maths for this is already shown in an earlier comment, but repeated here:

    q = εσT^4.4πr^2 (Stefan Boltzmann law x area of the outer surface)

    if T = 133K, r=13m and ε=0.8

    then q = 30,000 W

    Seeing as we are discussing not creating or destroying energy surely even the busiest of people can see that this is correct?

    I look forward to Gord and MS confirming this - or entertaining us with explanations of why the 6,000J per second isn't increasing the temperature of the inner surface of the sphere.

    After that we can progress to the next step.

    ReplyDelete
  31. Perhaps Science of Doom could explain how solar energy coming in at 960 J per second (noon) only increases the Earth surface energy level by 0.0017 J per second or 0.0003C per second.

    Where does the rest of the energy go? Obviously, it flows up and out. How can this outflow be such a high number - twice the rate of the LW back-radiation - LW out-radiation would then have to be 3 times higher than the LW back-in-radiation. Obviously again, there is a mathematical problem here and with the atmospheric radiation theory or at least how it is being "explained" to the general public because the real surface energy balance is quite different than the theory.

    Sorry for the real-Earth example versus a PVC sphere.

    ReplyDelete
  32. None of this will be settled, until there is some empirical evidence. I think I cited the only evidence that is available, earlier. And that evidence shows that the GHE is nonsense.

    ReplyDelete
  33. In fact, the GHE vanishes for exactly the same reason that the REAL greenhouse effect vanishes: convection.

    ReplyDelete
  34. SOD,

    I have already posted the Physics links that shows exactly what happens with the 6000 watts that are reflected including some examples.

    I posted this example of Light Interference inside a soap bubble that is very much like your own example:
    http://www.exploratorium.edu/ronh/bubbles/bubble_colors.html

    They explain, very clearly, what happens within a spherical bubble surface as it becomes more and more reflective.

    "When red is cancelled, it leaves a blue-green reflection. As the bubble thins, yellow is cancelled out, leaving blue; then green is cancelled, leaving magenta; and finally blue is cancelled, leaving yellow. Eventually the bubble becomes so thin that cancellation occurs for all wavelengths and the bubble appears black against a black background."

    Antenna arrays that have nulls in their Radiation Patterns due to cancellation of opposing EM fields is another good example.

    Are you not familiar with the concept of Electromagnetic Field interference like the above or are you arguing that it does not exist?

    In fact, all forces whether they are Mechanical, Gravity or Electromagnetic all display exactly the same Physics of opposing force cancellation.

    Here are Sound Wave Radiation Patterns produced by dipole speakers:
    http://www.kettering.edu/~drussell/Demos/rad2/mdq.html

    The same patern types are produced for Antennas using the Poynting Vectors of Electromagnetic Fields.
    Notice the nulls and peaks calculated by vector mathematics.

    Are you familiar with Vectors and how they are used describe opposing forces?
    --------------
    Do any of the above wave cancellation examples have any energy accumulating?

    This equation based on The Law of Conservation of Energy obviously has no such criteria:

    Incident Energy = Absorbed Energy + Transmitted Energy + Reflected Energy
    http://www.optotherm.com/emiss-physics.htm

    In fact, it clearly shows that the Incident Energy CANNOT increase due to Reflected Energy as You Claim.
    -------------
    You said...

    "That can happen at the start, but the 6,000J per second of accumulated energy will increase the temperature internally until eventually steady state is reached and 30,000W does leave the sphere."

    "I look forward to Gord and MS confirming this - or entertaining us with explanations of why the 6,000J per second isn't increasing the temperature of the inner surface of the sphere."

    I just posted the Physics and actual physical examples that absolutely show why energy does not accumulate.

    Why don't you post the Physics that shows that energy will accumulate and some actual physical examples?

    While you are at it, please post the Physics and any physical example where an Energy Source of 30,000 watts can produce 1.85 MEGA-WATTS!

    Good Luck!

    After that we can progess to the next step.

    Trenberth's Energy Budget Diagram and The fantasy "Greenhouse Effect".

    Including the established Physics Physics they violate and actual measurements that prove they are fairy tale "science".

    ReplyDelete
  35. So Gord doesn't believe in the first law of thermodynamics.

    6,000J per second disappears inside this sphere.

    MS are you with Gord on this? Is this the horse you are backing?

    ReplyDelete
  36. SoD,
    Gord has already explained this twice above and asked you

    "Why don't you post the Physics that shows that energy will accumulate and some actual physical examples?

    While you are at it, please post the Physics and any physical example where an Energy Source of 30,000 watts can produce 1.85 MEGA-WATTS!"

    SoD continues to ignore questions he doesn't like.

    ReplyDelete
  37. Dear MS and Gord

    I can show you how an Energy Source of 30kW can "produce" 1.86MW. Are you willing to listen with an open mind?

    As an aside:

    "When red is cancelled, it leaves a blue-green reflection. As the bubble thins, yellow is cancelled out, leaving blue; then green is cancelled, leaving magenta; and finally blue is cancelled, leaving yellow. Eventually the bubble becomes so thin that cancellation occurs for all wavelengths and the bubble appears black against a black background."

    Antenna arrays that have nulls in their Radiation Patterns due to cancellation of opposing EM fields is another good example"

    Well that's embarassing...What happens in bubbles and in antenna arrays is interference. In interference phenomena, energy does not vanish, but it is redistributed with a different spatial pattern. In this case bubbles behave as something which goes under the name of interferential filter. If we have destructive interference in one direction, that does not mean that light has vanished, but simply that light has formed constructive interference in a different direction. Think of yourself looking at the bubble. If you move your head, you see the displacement of the interference fringes. This means that the light has been redirected. The energy is all there, as it should be. The same goes for the antenna arrays. If you have cancellation along one direction, you have enhanced emission along another one. The result is that the energy is conserved.

    Best Regards

    ReplyDelete
  38. Wow, embarrassing is the only word. Gordon confuses force with energy, thinks that somehow interference destroys energy and that including the ever mysterious Poynting vector will make the bafflegab seem credible.

    MS then climbs on board.

    By the way, it's not at all difficult to determine who writes SoD.

    ReplyDelete
  39. Giovi Pelle writes:

    "I can show you how an Energy Source of 30kW can "produce" 1.86MW. Are you willing to listen with an open mind?"

    If that was possible.Why are we still using Coal,Natural Gas,Geothermal,Nuclear and Hydro power plants.When all you need is a small solar or wind tower to produce the base supply of 30KW and then apply your magic computer formulas.

    This should be a great way to drive all those CO2 "polluters" out of business.

    ReplyDelete
  40. "While you are at it, please post the Physics and any physical example where an Energy Source of 30,000 watts can produce 1.85 MEGA-WATTS!"

    Why are we still using Coal,Natural Gas,Geothermal,Nuclear and Hydro power plants.When all you need is a small solar or wind tower to produce the base supply of 30KW and then apply your magic computer formulas.

    There is indeed a very simple explanation as to how an energy source of 30 kW/s can produce 1.86 MW. The magic has nothing to do with computer formulas. It is a simple, mystical ingredient known as time.

    Referring back to the original Science of Doom article, we see:


    We start with consideration of the complete system and after equilibrium is reached the energy gained will be equal to the energy lost.
    Energy gained, q = 30,000W = Energy lost
    (Note that we are considering energy per second).


    So, if you want to know the answer to the question of how it is possible for a power source generating 30 kW/s to generate 1.86 MW, the answer is wait 62 seconds (30,000 * 62 = 1,860,000).

    The other important concept introduced in the quote above is equilibrium. Obviously, it takes time (there it is again) for this environment to reach a steady state. After 1 second, the source has only generated 30,000 W total. At that point, you are certainly correct in stating that the interior surface cannot be radiating 1.86 MW as we haven't generated that much energy total into the system. However, after some period of time, when we get to the point that the outer sphere is radiating the same amount of energy we're producing, we do reach that point. So, there is no magical energy multiplication. We produced all of that energy (and more) from the source. It just took time.

    So, yes, it is indeed true that "1.8 Megawatts cannot be created from 30 Kilowatts." However, it is easy to see how 1.8 Megawatts can be generated by a source that generates 30 Kilowatts per second. It just takes time.

    ReplyDelete
  41. Arthur,
    1. I was told by a 2nd hand source re "Dr. Phillips," but since I have no direct knowledge whether not "Science of Doom" is "Dr. Phillips," I did a strike through on those references with an update above. Funny that SoD made no mention that he is not "Dr Phillips" in the comments above.
    2. Apparently you don't consider a site called "Science of DOOM" & which repeats the alarmist memes constantly to be alarmist - everyone is entitled to their opinion.
    3. At no point did I say (or Gord) that the 1st law places "A CONSTRAINT" on the total radiation emitted by the inner surface, merely that energy must be conserved. Gord's calculations immediately above clearly show SoD has created massive amounts of energy.
    4. Since you have said nothing in Gord's analysis is flawed and supported it, you have indirectly admitted SoD's analysis is pathetically flawed.
    5. Now, what were the other 8 errors you counted in the first paragraph?

    ReplyDelete
  42. Replies from Gord:

    Re: King of the Road's comment below:

    - The Electromagnetic Field is a force field

    Electromagnetic force
    "The electromagnetic force is one of the four fundamental forces. The other fundamental forces are: the strong nuclear force (which holds quarks together, along with its residual strong force effect that holds atomic nuclei together to form the nucleus), the weak nuclear force (which causes certain forms of radioactive decay), and the gravitational force. All other forces are ultimately derived from these fundamental forces."

    "In physics, the electromagnetic force is the force that the electromagnetic field exerts on electrically charged particles. It is the electromagnetic force that holds electrons and protons together in atoms, and which hold atoms together to make molecules. The electromagnetic force operates via the exchange of messenger particles called photons and virtual photons."

    "The electromagnetic force is the one responsible for practically all the phenomena one encounters in daily life, with the exception of gravity."

    http://en.wikipedia.org/wiki/Electromagnetic_force

    - Interference is NOT destruction of Energy as I posted before:


    Incident Energy = Absorbed Energy + Transmitted Energy + Reflected Energy
    http://www.optotherm.com/emiss-physics.htm

    This is Energy In = Energy Out

    If absorbed energy and transmitted energy are zero then:
    Incident Energy = Reflected Energy

    Incident Energy - Reflected Energy = 0 (for 100% reflection inside the bubble)"

    Note that the above equation IS Based on The Law of Conservation of Energy and Reflection is completely covered.

    --------------
    Re: the Poynting Vector

    I posted this before:

    Electromagnetic Fields are propagating fields that carry energy from one place to another and are Vector quantities.

    Heat flux
    "Heat flux or thermal flux, sometimes also referred to as heat flux density or heat flow rate intensity is a flow of energy per unit of area per unit of time. In SI units, it is measured in [W·m-2]. It has both a direction and a magnitude so it is a vectorial quantity."
    http://en.wikipedia.org/wiki/Heat_flux

    The heat flux IS the Poynting Vector.
    ----
    Power Density and Radiated Power

    The Poynting Vector P is defined as:

    P(vector) = (1/2) E(vector) X H(vector)

    which is a power density with units of W/m2.
    Figure 4.1.3 The w/m2 Varies with Position on the Surface of a Sphere

    "n = unit normal directed outward from the surface."

    "We now continue to calculate the total radiated power from an antenna. It is the number of watts
    per square meter that happens to be at a given point and the direction of the vector is the
    direction of the power flow."

    http://www.engr.psu.edu/cde/courses/ee497c/M4L1.pdf

    ReplyDelete
  43. from Gord, continued:

    Once again, I have NOT said The Law of Conservation is not true, here is what I said...
    "It should be noted that this is obvously not destruction of energy, it is merely the loss of the ability to move the Photon energy past the bubble surface.

    This is also evident in this equation based on the Law of Conservation of Energy:

    Incident Energy = Absorbed Energy + Transmitted Energy + Reflected Energy
    http://www.optotherm.com/emiss-physics.htm

    This is Energy In = Energy Out

    If absorbed energy and transmitted energy are zero then:
    Incident Energy = Reflected Energy

    Incident Energy - Reflected Energy = 0 (for 100% reflection inside the bubble)"

    Reflection is part of the Law of Conservation of Energy that the equation clearly shows.

    SOD is the one that violates the Law of Conservation when the Incident Energy of 30,000 watts magically becomes 1.85 MEGA-WATTS!

    ReplyDelete
  44. from Gord, continued:

    I see that SOD completely ignored the fact that when there is 100% reflection in the soap bubble there is cancellation for all wavelengths and no light leaves the bubble.

    The same occurs for nulls in the Radiation Pattern of Antennas.....complete cancellation due to destructive intereference.

    Other points in the Antenna Radiation pattern have "constructive interference" producing peaks in the Radiation Pattern.

    All these Radiation Patterns are developed with Vector Analysis of the EM fields and verified with actual measurements.
    --------
    I also posted this Heat Transfer link that shows how two opposing EM fields produce ZERO transfer of heat energy confirming the above:

    Radiative Heat Transfer equations also show this:

    Heat Radiation
    "Radiation is heat transfer by the emission of electromagnetic waves which CARRY energy away from the emitting object. For ordinary temperatures (less than red hot"), the radiation is in the infrared region of the electromagnetic spectrum. The relationship governing radiation from hot objects is called the Stefan-Boltzmann law:

    Heat Transfer by Radiation using the Stefan-Boltzmann Law

    P = e*BC*A(T^4 - Tc^4)

    Where P = net radiated power (Watts), e = emissivity, BC = Stefan's constant, A = area, T = temperature of radiator and Tc =
    temperature of the surroundings or another body.

    ..when rearranged gives

    P/A = e*BC*T^4 - e*BC*Tc^4 (Watts/m^2)
    http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/stefan.html#c2

    This is an obvious subtraction of two Electromagnetic Fields

    It also complies with the Vector subtraction of Electromagnetic Fields which are Vectors.

    The resultant Electromagnetic Field will have a magnitude of P/A and have a direction of propagation in the direction of the larger field produced by the hotter body.

    NOTE that when the two fields are equal and opposite the Watts/m^2 goes to ZERO just like what happens in the Soap bubble and Antenna Radiation pattern nulls.

    ReplyDelete
  45. Re: Joe Yangtree's Post where he claims that 30,000 Watts can generate 1.86 Mega-Watts in 62 seconds because it accumulates at 30,000 Watts per second.

    This is absolutely hillarious.

    Why not wait 124 seconds and double the watts to 3.72 Mega-Watts?

    If you wait an entire day (86400 seconds), Joe and SOD would get an astounding 86,000 sec X 30,000 Watts = 2,580,000,000 Watts!

    One of these perpetual motion machine "generators" would easily power everything on the Planet..."It just takes time", as Joe claims.

    ----------------
    Hey Joe and SOD, here is an experiment you can do easily do to verify your claims.

    Take a flashlight and cover the face with a mirror or even some tin foil.

    With the reflective backing in the flashlight and tin foil covering the 0.25 watt bulb you will have 100% reflective cavity.

    Now with 0.25 watts accumulating every second inside cavity you should have 86400 sec X 0.25 Watts = 21,600 watts inside the cavity at the end of the day!

    Neat!....You guys are going to be Rich and Famous!

    ReplyDelete
  46. Hey SOD,

    Any luck coming up with the Physics that shows that energy will accumulate and some actual physical examples?

    How about the Physics and any physical example where an Energy Source of 30,000 watts can produce 1.85 MEGA-WATTS?

    No?

    Gee, what a surprise.

    ReplyDelete
  47. Great is the confusion under the sky. With a curiously high peak at Gord's coordinates.

    "I see that SOD completely ignored the fact that when there is 100% reflection in the soap bubble there is cancellation for all wavelengths and no light leaves the bubble."

    1) At the link you address

    http://www.exploratorium.edu/ronh/bubbles/bubble_colors.html

    The experiment concerns soap films, and not soap bubbles

    2) We are looking at reflected light, therefore if the soap film becomes black, it means we have no reflection and guess what? Energy has not vanished, but instead we have 100% transmission...

    Concernig antennas. Of course antennas have nulls in the pattern. But the intensity integrated over all the solid angle conserves energy.

    Regarding conservation of energy in the PVC experiment, try to follow me closely now:

    1) The light is off, everything is at 0K (to simplify)
    2) At t=0 we turn on the light and the light begins to travel towards the inner sphere wall
    3) At t=r/c, where r is the radius of the cavity, and c the speed of light, the first rays of light reach the cavity wall
    4) 24kW are absorbed, the wall heats up and its thermal emission begins. in the meantime che heat conductivity sink begins to work. Finally 6kW of light are reflected
    5) After t=2r/c the reflected light has crossed the sphere. So the incident intensity over the inner wall is already 30kW+6kW=36kW. We must sum to that all the thermal emission.
    6) The reflected and thermal emission radiation bounce back and fort in the cavity, new energy is always injected by the light bulb, and while the heat sink is getting more and more efficient, energy goes on accumulating in the cavity.
    7) Finally the eat sink is efficient enough and draws the 30kW input. We are at equilibrium.
    8)(1.85MW - 30kW) is the energy ACCUMULATED in the transient process because the heat sink wasn't efficient enough. Now that the heat sink is ok, the excess energy just bounces between the walls.

    Best Regards

    ReplyDelete
  48. To calculate the temperature of the inner surface, T1, we first need to know the energy per second leaving the outer surface. This will allow us to calculate the temperature T2.

    Regardless of Gord's claims to the contrary, the first law of thermodynamics tells us that if the energy source inside the hollow sphere is 30,000W then in equilibrium the energy leaving the outer surface is 30,000W.

    Otherwise energy is being accumulated inside the sphere. If energy is being accumulated it keeps increasing the temperature - as you can see explained in Do Trenberth and Kiehl understand the First Law of Thermodynamics? Part Three – The Creation of Energy?. The article also explains how you can't draw more than 30,000W out of the system, regardless of the claims to the contrary.

    Back to the important point, the first step in the calculation:

    30,000 Joules per second in = 30,000 Joules per second out.

    It doesn't get much clearer than this.

    If you want to claim it doesn't because you don't like the results, go ahead and entertain everyone, but it's pointless me trying to explain the next step if the first law of thermodynamics has already been discarded.

    ReplyDelete
  49. Gord,

    You seem to have misconstrued my position. I didn't say that equilibrium was reached in 62 seconds or that 1.86 MW was "accumulated" anywhere after 62 seconds, just that it is a simple fact that an energy source that produces 30,000 W/s is going to have produced 1.86 MW after 62 seconds. Is this really controversial? That seems to be simple multiplication, and is hardly perpetual motion. If you are introducing X energy per second into a system, then after Y seconds, you will have introduced X*Y energy into the system. Much like SOD, I don’t think that it will be possible to have much of a discussion if this elementary point is not understood.

    After 62 seconds, this 1.86 MW must be somewhere. In the SOD example, some would have already been radiated to "space" and some would have raised the temperature of the sphere. As his calculations show, once equilibrium is reached, 30,000 W/s is being radiated to "space" and 30,000 W/s is still being produced by the power source. At that point, the temperature of the sphere is static and will not change further.

    On to your suggested experiment. As anyone that deals with astronomy knows, there is no such thing as a perfectly reflective surface. Even if both surfaces are 99.999% reflective, light is sufficiently fast to produce billions of "hits" per second for any given photon, with each "hit" giving a 1/100,000 chance of the photon being absorbed. So the odds of a given photon existing for longer than one second in that environment is quite small. Given that the photons will be absorbed by the surrounding material, then we’re back to the original problem that SOD proposed with significantly different parameters.

    You also asked for an example of energy accumulation. Of course, SOD, has already given one example, but another, real world example would be heating water on a stove. Are you claiming that the rise in temperature of the water is not due to the water accumulating energy? Once you turn off the heat, the water will cool as it radiates and conducts energy away.

    Let’s say that we have a stove top that can produce 1600 W/s and we have 1 liter of water we want to boil that is currently at 4 degrees C. To boil it, we have to get it to 100 degrees C, so we need to raise the water by 96 degrees C. That’s 1000g of water (1mL of water = 1g) that we need to raise 96 degrees. By definition, that would take 96,000 calories or ~403,000 J. In your world, how does this work? Can the 1600 W/s stove top ever boil the water? Does the water accumulate energy as its temperature rises? If we’ve transferred 403,000 J to the water from a 1600 W/s stove top, have we created a perpetual motion machine?

    ReplyDelete
  50. Giovi Pelle,

    -The Soap bubble wall is the soap film, all the way around the bubble...it IS THE BUBBLE.

    What did you think the bubble was made of?...Wood?
    -------
    HAHAHA....Where did you dream this statement up from?

    You said...
    "2) We are looking at reflected light, therefore if the soap film becomes black, it means we have no reflection and guess what? Energy has not vanished, but instead we have 100% Transmission..."

    Here is are some quotes from the article:

    "All waves, including light, have a curious property: if two waves combine, the waves can meet each other crest-to-crest, adding up and reinforcing the effect of each other, or they
    can meet crest-to-trough, cancelling each other out so that they have no effect. When they meet crest-to-trough, for every "up" vibration in one wave, there is a corresponding "down" vibration in the other wave. This combination of equal ups and downs causes complete cancellation or interference."

    "When two light waves cancel each other, the result is darkness and this is called "destructive interference."

    "If we were to look at a highly magnified portion of a soap bubble membrane, we would notice that light reflects off both the front (outside) and rear (inside) surfaces of the
    bubble, but the ray of light that reflects off the inside surface travels a longer distance than the ray which reflects from the outside surface. When the rays recombine they can
    get "out of step" with each other and interfere. Given a certain thickness of the bubble wall, a certain wavelength will be cancelled and its complementary color will be seen. Long
    wavelengths (red) need a thicker bubble wall to get out of step than short wavelengths (violet). When red is cancelled, it leaves a blue-green reflection. As the bubble thins, yellow is cancelled out, leaving blue; then green is cancelled, leaving magenta; and finally blue is cancelled, leaving yellow.

    Eventually the bubble becomes so thin that cancellation occurs for all wavelengths and the bubble appears black against a black background."

    http://www.exploratorium.edu/ronh/bubbles/bubble_colors.html

    Do see the word "reflects" used many times?

    Where did you see 100% transmission causes the bubble to go dark when it clearly says "When two light waves cancel each other, the result is darkness and this is called "destructive interference."?

    I assume you have some light bulbs in your house.

    Ever notice that light bulbs have a glass that allows transmission of light out of the bulb.

    When you turn the light on does the interior of the bulb go dark?
    ------------------
    Here is the equation based on The Law of Conservation of Energy.

    Incident Energy = Absorbed Energy + Transmitted Energy + Reflected Energy
    http://www.optotherm.com/emiss-physics.htm

    Where:

    Incident Energy = 30,000 Watts (That's all the Energy from the source and all the rest cannot exceed that amount or Energy is Created)
    Absorbed Energy = 24,000 Watts

    Reflected Energy = 6,000 Watts
    Transmitted Energy = 0 Watts.

    In your step (5) You already have 36,000 watts, a CREATION of 6,000 Watts and a VIOLATION of The Law of Conservation of Energy.

    By step (8) You have CREATED 1.85 MEGA-WATTS - 30,000 Watts = 1.82 MEGA-WATTS from an Incident Energy of 30,000 Watts...A PERPETUAL MOTION MACHINE.

    Why not just take 30,000 Watts from the 1.82 MEGA-WATTS and feed the 30,000 Watt light bulb and you will have a truly PORTABLE PERPETUAL MOTION MACHINE that requires no external
    energy source and will deliver 1.82 MEGA-WATTS (or more) FOREVER.

    Now that's really NEAT....you just solved the Earth's Energy Crisis.

    You and SOD should Patent it really quickly and you will be Rich and Famous!
    -----------------------------------
    By the way, why don't you to post some Physics and any Physical example where an Energy Source of 30,000 watts can produce 1.85 MEGA-WATTS.

    I look forward to your next post.

    ReplyDelete
  51. SOD,

    Here is the equation based on The Law of Conservation of Energy.

    (Do you remember this equation?
    It's from the same article you praised in your Post on Nov 10, 2010 at 11:08PM, where you described the article as...
    "Thanks for letting me know about this article, it is a beautiful thing.")

    Incident Energy = Absorbed Energy + Transmitted Energy + Reflected Energy
    http://www.optotherm.com/emiss-physics.htm

    Where:

    Incident Energy = 30,000 Watts (That's all the Energy from the source and all the rest cannot exceed that amount or Energy is Created)
    Absorbed Energy = 24,000 Watts

    Reflected Energy = 6,000 Watts
    Transmitted Energy = 0 Watts.

    Ever wonder why they have Reflected Energy in the equation?

    Hint: The Reflected energy comes from the Incident Energy source and can only subtract from the Incident Energy source leaving only 24,000 watts available for Absorbtion.

    That's why a Soap bubble with zero absorbtion has to comply with the same equation and The Law of Conservation of Energy:

    Incident Energy = Absorbed Energy + Transmitted Energy + Reflected Energy

    Giving, Incident Energy = Transmitted Energy + Reflected Energy

    And, as the Reflected Energy is increased and Transmitted Energy decreases....

    "When red is cancelled, it leaves a blue-green reflection. As the bubble thins, yellow is cancelled out, leaving blue; then green is cancelled, leaving magenta; and finally blue is
    cancelled, leaving yellow."

    And, finally when there is 100% Reflection, Transmitted Energy out of the bubble goes to ZERO and we have...

    "Eventually the bubble becomes so thin that cancellation occurs for all wavelengths and the bubble appears black against a black background."

    "When two light waves cancel each other, the result is darkness and this is called "destructive interference."

    http://www.exploratorium.edu/ronh/bubbles/bubble_colors.html

    The Physics and Actual Physical examples are very clear.

    But you seem to AVOID talking about the equation except for ("...it is a beautiful thing") and now and you are certainy AVOIDING talking about Soap Bubble example.

    In fact, you have even AVOIDED what I asked you to do in my post to you on Nov 12, 2010 at 8:17 PM:

    "I just posted the Physics and actual physical examples that absolutely show why energy does not accumulate.
    Why don't you post the Physics that shows that energy will accumulate and some actual physical examples?
    While you are at it, please post the Physics and any physical example where an Energy Source of 30,000 watts can produce 1.85 MEGA-WATTS!"

    What a HOOT!

    ReplyDelete
  52. Joe Yangtree,

    SOD gave his calculated example which is not an Actual Physical Example.

    I asked for a ACTUAL PHYSICAL EXAMPLE, like the the link to reflection and destructive interference inside a Soap Bubble that has been actually demonstrated many times.

    I also asked SOD to post the Physics, just like I did when I posted this equation that is based on The Law of Conservation of Energy:


    Incident Energy = Absorbed Energy + Transmitted Energy + Reflected Energy
    http://www.optotherm.com/emiss-physics.htm

    Plus, I posted several other Physics links including The Heat Transfer Equation, the Electomagnetic Force, Heat flux is a Vector, Vector Mathematics for Field calculations...etc.

    So, do you agree that 1.86 MEGA-WATTS can't accumulate inside the inner Sphere?
    ----
    Re: My experiment (that has nothing to do with Astronomy by the way):

    "Hey Joe and SOD, here is an experiment you can do easily do to verify your claims.

    Take a flashlight and cover the face with a mirror or even some tin foil.
    With the reflective backing in the flashlight and tin foil covering the 0.25 watt bulb you will have 100% reflective cavity.
    Now with 0.25 watts accumulating every second inside cavity you should have 86400 sec X 0.25 Watts = 21,600 watts inside the cavity at the end of the day!
    Neat!....You guys are going to be Rich and Famous!"

    Are you saying there will be 21,600 watts inside the cavity at the end of the day or not?

    -------
    Your example deals with stored energy in water where all the energy stored came from an energy source.

    I assume that the 1600 watts/sec energy source in your example (which is 1600 Joules/Sec^2) was actually supposed to be 1600 Joules/sec?

    Mass of the water = 1000 grams
    Delta T = 96 deg C
    Specific Heat of water = 4.18 J/gm X deg C

    Q = mass X delta T X Specific Heat

    Q = 1000 X 96 X 4.18 = 401,280 Joules

    So the water will heat up in 401,280 Joules/ 1600 Joules/sec = 250.8 seconds.

    The water stores 401,280 Joules over 250.8 seconds or 1600 Joules/sec and is exactly the same as the the source provides (1600 Joules/sec)

    This complies with The Law of Conservation of Energy and is not a perpetual motion machine.
    ----------
    Watts are Joules/sec and in SOD's example the 30,000 watt source produces 30,000 Joules/sec.

    The energy produced by the 30,000 watt source is 1,860,000 watts or 1,860,000 Joules/sec.

    The Joules/sec produced is much greater than the source supplies, does NOT comply with The Law of Conservation of Energy and IS A PERPETUAL MOTION MACHINE.

    ReplyDelete
  53. I don't expect to convince Gord of anything. I can't even understand the mishmash of his "explanations".

    What does seem clear is Gord doesn't understand "equilibrium" as a concept.

    Gord writes:

    "Incident Energy = 30,000 Watts (That's all the Energy from the source and all the rest cannot exceed that amount or Energy is Created)
    Absorbed Energy = 24,000 Watts

    Reflected Energy = 6,000 Watts
    Transmitted Energy = 0 Watts.

    In your step (5) You already have 36,000 watts, a CREATION of 6,000 Watts and a VIOLATION of The Law of Conservation of Energy.
    "

    If we want to ask the question what happens in the first second, or the 61st second, or any other point in time, we need to solve the dynamic equations. This is a more involved process and requires that we know the specific heat capacity of the PVC.

    If we want to find out the steady state, or equilibrium conditions the solution is simpler.

    But let me explain the startup from very cold conditions for those who are interested.

    In the first second each square meter of inner surface receives incident radiation of 23.9 J (30,000/area = 30,000/[4.pi.10^2]).

    19.1 J are absorbed and 4.8 J are reflected.

    What happens to this absorbed energy? It goes into heating up the PVC very close to the inner surface.

    Almost nothing is radiated back out because emission of thermal radiation is proportional to T^4. Almost nothing is conducted through to the outer surface - because conduction is proportion to temperature gradient across a material.

    What happens to the reflected radiation? It is incident on another part of the inner surface and some proportion of that is absorbed and some is radiated.

    Much time goes by and the inner surface gets heated up while the heat progressively penetrates deeper into the PVC.

    You can see the dynamic results in a graph presented in Do Trenberth and Kiehl understand the First Law of Thermodynamics? – Part Two. It should give some idea of how the heating takes place through the sphere and how long it takes.

    But the question we actually want to know the answer to is "what is the equilibrium condition?"

    Equilibrium means that the temperature is not changing. And for temperature not to change, energy in = energy out.

    In a separate comment I will explain why the surface radiation is so high and why it can't be used to power anything more than 30,000W, but because so many people have a conceptual problem with this subject I think it is important to focus on first solving the very simple heat transfer equations.

    MS wrote an article that was very flawed.
    Gord has written an analysis that is very flawed.

    They were different from each other. United in rejecting a result by whatever means necessary.

    If you want to understand the solution to this simple problem then you need to understand first of all that the First Law of Thermodynamics says that energy cannot be created or destroyed.

    And yet Gord's groundbreaking analysis of everything under the sun seems unable to grasp this. He has 24,000 Joules per second leaving the system and 6,000 Joules per second added to the system - yet not increasing the temperature (or vanishing).

    This can't happen unless the foundations of 150 years of thermodynamics are wrong.

    If 30,000 Joules per second are input to the system, then in steady state 30,000 Joules per second will leave the system - if less than 30,000 Joules per second are leaving the system then the system will be heating up.

    If anyone apart from Gord wants to ask a question about why this obvious fact is true, please go ahead and ask it. I think it is probably clear to almost everyone.

    ReplyDelete
  54. Gord

    You are indeed a funny guy. Getting down to business:

    1) Soap Bubbles and Soap Films
    a) I said soap film, and not soap bubble because the following picture:

    http://www.exploratorium.edu/ronh/bubbles/bubble_FILM.gif

    clearly shows a planar soap FILM, not a spherical bubble, that's all.
    b) Concerning interference and reflections, I quote you quoting the website http://www.exploratorium.edu/ronh/bubbles/bubble_colors.html:

    "Here is are some quotes from the article:

    "All waves, including light, have a curious property: if two waves combine, the waves can meet each other crest-to-crest, adding up and reinforcing the effect of each other, or they
    can meet crest-to-trough, cancelling each other out so that they have no effect. When they meet crest-to-trough, for every "up" vibration in one wave, there is a corresponding "down" vibration in the other wave. This combination of equal ups and downs causes complete cancellation or interference."

    "When two light waves cancel each other, the result is darkness and this is called "destructive interference."

    "If we were to look at a highly magnified portion of a soap bubble membrane, we would notice that light REFLECTS off both the front (outside) and rear (inside) surfaces of the
    bubble, but the ray of light that REFLECTS off the inside surface travels a longer distance than the ray which REFLECTS from the outside surface. When the rays recombine they can
    get "out of step" with each other and interfere. Given a certain thickness of the bubble wall, a certain wavelength will be cancelled and its complementary color will be seen. Long
    wavelengths (red) need a thicker bubble wall to get out of step than short wavelengths (violet). When red is cancelled, it leaves a blue-green reflection. As the bubble thins, yellow is cancelled out, leaving blue; then green is cancelled, leaving magenta; and finally blue is cancelled, leaving yellow.

    Eventually the bubble becomes so thin that cancellation occurs for all wavelengths and the bubble appears black against a black background."

    http://www.exploratorium.edu/ronh/bubbles/bubble_colors.html

    Do see the word "REFLECTS" used many times?"

    So I actually count REFLECTS 3 times. Furthermore, if we include the part of the post you carefully left out we have:

    "White light is separated into colors as it REFLECTS from the two surfaces of a thin film. Where the two REFLECTIONS interfere constructively, they produce a band of color. Where they cancel each other, that color is subtracted from the spectrum." as comment of Figure http://www.exploratorium.edu/ronh/bubbles/REFLECTION.gif (funny, the figure is named REFLECTION.gif)

    "The alternating bands of light and dark on this soap film are actually bands of color, produced by the REFLECTION and interference of light waves." as comment of Figure http://www.exploratorium.edu/ronh/bubbles/bubble_film.gif which actually shows light REFLECTED by a FILM...

    5 more means a total of 8

    If the film becomes black, it means we have no reflection because all the reflected light undergoes destructive interferences so, where does the light go?
    i) The light magically vanishes. So the energy and we are violating the conservation of energy.
    ii) 0% reflections means 100% transmission.
    Are you willing to affirm that interference destroys energy? If yes, there is not much more to discuss.
    ...

    ReplyDelete
  55. 2) Point (5) of my example
    You are missing something here:

    a) The light emitted by the light bulb at t=0 is partially reflected at t=r/c (6kW) and then at t=r/c+2r/c=3r/c strikes again the inner wall.
    b) The light emitted by the light bulb at t=2r/c strikes the inner wall at t=3r/c
    c) so at time t=3r/c we have two contributions, deriving from the light emitted by the light bulb at different times which sum up. I am not violating conservation of energy, i am just summing up photons emitted by the light bulb at different times. So at t=3r/c we have 30kW (directly emitted by the light bulb at t=2r/c) and 6kW (30kW emitted by the light bulb at t=0, 6kW reflected at t=r/c, traveling across the sphere and getting to the inner wall at 3r/c): total 36kW, without counting thermal emission. As you see the energy builds up because photons emitted AT DIFFERENT TIMES remain trapped until the heat sink is efficient enough.

    Best Regards

    PS "I just posted the Physics and actual physical examples that absolutely show why energy does not accumulate." In your examples energy does not accumulate because in your universe interference destroys energy violating the first principle of thermodynamics...enough said

    ReplyDelete
  56. Many people think that there is something wrong with the high temperature of the inner surface because the value for radiation emitted is so high.

    It seems to many people that energy has been created.

    Over in Do Trenberth and Kiehl understand the First Law of Thermodynamics? Part Three – The Creation of Energy? I try to explain the answer to this question.

    In the comments I produced 2 graphs:
    Energy radiated from the outer surface - against time
    Total energy accumulated in the sphere - against time

    These show that a lot of energy gets stored in the PVC sphere as it is heating up. This is because until the steady state temperature of 423K is reached at the inner surface, less than 30,000W is being radiated from the outer surface. Reaching this equilibrium takes a long time (strictly speaking this equilibrium is never reached, the temperature just gets closer and closer - take a look at the graphs).

    So the first point to note is that a lot of energy has been stored in the PVC sphere during its long heating phase.

    The second point to understand is that each second each square meter of the surface:

    -receives 23.9 J/m² from the internal heat source (30,000J / 1,257 m²)
    -conducts 23.9 J/m² through the wall
    -absorbs 1,452 J/m² radiated from the rest of the inner wall.
    -re-radiates 1,452 J/m².

    The high value of radiation is due to energy just "bouncing around" - not really bouncing of course - absorbed, re-radiated, absorbed, re-radiated.

    So energy is not being created. The high value of radiation is only at all possible because the sphere has had a lot of energy added to it during its heating phase.

    And as soon as you try to power something else using this high temperature, it will simply pull the temperature right back down.

    In steady state the most you can pull out is 30,000W. Of course, a lot of energy has been stored in the PVC sphere so with the appropriately low temperature sink you can pull out all of the stored energy as well.

    ReplyDelete
  57. Just to clarify (from November 16, 2010 11:11 AM) that the numbers that follow this statement:

    "..The second point to understand is that each second each square meter of the surface:.."

    are the values when equilibrium is reached.

    ReplyDelete
  58. SOD,

    You said...
    "Almost nothing is conducted through to the outer surface - because conduction is proportion to temperature gradient across a material."

    HAHAHA...I see SOD has changed his story about how much is conducted through to the outer surface.

    Now he states it is "almost nothing"!

    SOD wrongly used a whopping 30,000 watts for the conduction transfer of energy when it should have been 2390 Watts and back calculated the temperature gradient to be a whopping 290K
    when it should have been 23.1K!

    SOD has so mangled the Conduction fiasco that he can't even remember what he calculated!

    HAHAHA...What a HOOT!
    ----------

    You said...
    "What happens to the reflected radiation? It is incident on another part of the inner surface and some proportion of that is absorbed and some is radiated."

    HAHAHA....another contradiction.

    You also said..
    "In the first second each square meter of inner surface receives incident radiation of 23.9 J (30,000/area = 30,000/[4.pi.10^2]).
    19.1 J are absorbed and 4.8 J are reflected."

    The 19.1 J are absorbed by the inner surface because the emissivity is 0.8 (23.9 X 0.8 = 19.1)!

    Now SOD is saying that the inner surface can absorb MORE than 19.1 J!

    HAHAHA...What a HOOT!
    --------------------
    You said...
    "Equilibrium means that the temperature is not changing. And for temperature not to change, energy in = energy out."
    "What happens to the reflected radiation? It is incident on another part of the inner surface and some proportion of that is absorbed and some is radiated."


    HAHAHA....that's not what the Heat Transfer Physics shows:

    Heat Transfer by Radiation using the Stefan-Boltzmann Law

    P = e*BC*A(T^4 - Tc^4)

    Where P = net radiated power (Watts), e = emissivity, BC = Stefan's constant, A = area, T = temperature of radiator and Tc =
    temperature of the surroundings or another body.

    ..when rearranged gives

    P/A = e*BC*T^4 - e*BC*Tc^4 (Watts/m^2)
    http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/stefan.html#c2

    This is an obvious subtraction of two Electromagnetic Fields

    It also complies with the Vector subtraction of Electromagnetic Fields which are Vectors.

    When T = Tc the Electromagnetic Field (P/A) is ZERO!

    There is ZERO energy being transfered between the parts of the inner surface because the temperature is the same for all the inner surface!

    The reflected energy CANNOT be absorbed by the inner surface and that is also shown by this equation that is based on The Law of Conservation of Energy:

    Incident Energy = Absorbed Energy + Transmitted Energy + Reflected Energy
    http://www.optotherm.com/emiss-physics.htm

    Where:

    Incident Energy = 30,000 Watts
    Absorbed Energy = 24,000 Watts
    (also calculated by the the emissivity of the inner sphere being 0.8...(0.8 X 30,000 = 24,000)
    Reflected Energy = 6,000 Watts
    Transmitted Energy = 0 Watts.

    The Soap Bubble ACTUAL PHYSICAL EXAMPLE also PROVES this because the REFLECTED LIGHT INSIDE THE SOAP BUBBLE IS NOT ABSORBED BY THE BUBBLE SURFACE.
    ----------------------
    Once again, SOD is shown to be a fool trying desperately to make his rants sound "scientific" when they actually violate established Physics and Actual Physical Examples that have been taught and demonstrated countless times.

    SOD has not been able to post ANY Physics or ANY Actual Physical Examples to back up his fantasy and hillarious claims of "energy accumulation" and how 30,000 watts can produce 1.85 MEGA-WATTS.

    He can't post the Physics or Actual Physical Examples because they simply DO NOT EXIST.

    SOD, ever think of about pursuing a career as a Comedian?

    ReplyDelete
  59. Giovi Pelle,

    These two quotes from the Soap Bubble article summarizes the effect of 100% reflection:

    "When two light waves cancel each other, the result is darkness and this is called "destructive interference."
    Eventually the bubble becomes so thin that cancellation occurs for all wavelengths and the bubble appears black against a black background."

    http://www.exploratorium.edu/ronh/bubbles/bubble_colors.html

    However, what YOU SAID WAS...
    "2) We are looking at reflected light, therefore if the soap film becomes black, it means we have no reflection and guess what? Energy has not vanished, but instead we have 100% Transmission..."

    Which is clearly wrong and you also IGNORED my question to YOU:

    "Ever notice that light bulbs have a glass that allows transmission of light out of the bulb.
    When you turn the light on does the interior of the bulb go dark?"

    What a surprise!
    -------------
    Yet you continue to babble:

    "If the film becomes black, it means we have no reflection because all the reflected light undergoes destructive interferences so, where does the light go?
    i) The light magically vanishes. So the energy and we are violating the conservation of energy.
    ii) 0% reflections means 100% transmission.
    Are you willing to affirm that interference destroys energy? If yes, there is not much more to discuss."

    What a HOOT!

    I have already posted a summary for the reflections causing cancellation of light in the Soap Bubble above.

    Now regarding the rest of your rant:

    - Did you not read my previous posts where I have REPEATELY said:

    "It should be noted that this is obvously not destruction of energy, it is merely the loss of the ability to move the Photon energy past the bubble surface.

    This is also evident in this equation based on the Law of Conservation of Energy:

    Incident Energy = Absorbed Energy + Transmitted Energy + Reflected Energy
    http://www.optotherm.com/emiss-physics.htm

    This is Energy In = Energy Out

    If absorbed energy and transmitted energy are zero then:
    Incident Energy = Reflected Energy

    Incident Energy - Reflected Energy = 0 (for 100% reflection inside the bubble)"
    ------------
    I have also posted other Physics links including The Heat Transfer Equation, the Electomagnetic Force, Heat flux is a Vector, Vector Mathematics for Field calculations...etc that all show the same thing.

    If you want to talk about Physics then discuss ANY of the the established Physics links that I posted or Post a Physics link to back-up for your "opinions".

    Now answer my QUESTION THAT YOU IGNORED:

    "Ever notice that light bulbs have a glass that allows transmission of light out of the bulb.
    When you turn the light on does the interior of the bulb go dark?"

    I look forward to your answer.
    -----------------------------------
    PS:

    Re: Your other post where you claim to have posted "Physics" and a "actual physical example".

    Funny, I don't see ANY Physics or ANY "actual physical example".

    All I see is your "ranting opinion" again.

    What a HOOT!

    ReplyDelete
  60. When Gord writes comments like:

    "[citing me]"Almost nothing is conducted through to the outer surface - because conduction is proportion to temperature gradient across a material."

    HAHAHA...I see SOD has changed his story about how much is conducted through to the outer surface.

    Now he states it is "almost nothing"!
    "

    It demonstrates an inability of Gord to understand anything at all.

    As I explained quite carefully, if we are calculating the dynamic conditions at the very start then the heat transfer by conduction will be almost zero.

    This is because T2 (outer surface) is zero and T1 (inner surface) is almost zero. Conducted heat is proportional to temperature difference.

    This is quite different from the equilibrium condition when T1 has heated up to reach 423K and now conducts 30,000 Joules per second.

    Gord is incapable of understanding this, or what "equilibrium" condition means, or what the first law of thermodynamics means.

    As I said before, if anyone apart from Gord wants to discuss it, please ask away.

    ReplyDelete
  61. SOD,

    HAHAHA....HAHAHA....When SOD writes rants like this, he just confirms that he is a fool that has "no clue" about established Physics and can only post his hillarious and totally unsupported "opinions".

    You said...
    "As I explained quite carefully, if we are calculating the dynamic conditions at the very start then the heat transfer by conduction will be almost zero."
    "This is quite different from the equilibrium condition when T1 has heated up to reach 423K and now conducts 30,000 Joules per second."

    I have already posted EXACTLY why your use of Conduction Equation was WRONG and why your calculation of the fantasy 423K inner sphere temperature was WRONG.

    From my Post to YOU and I quote:

    "NOTE:
    SOD's use of the conduction equation was done wrong.
    He used 30,000 watts for the value qr when qr is actually the watts tranfered by conduction (2390 watts).
    He then came up with a temperature difference of 290K when it is actually 23.1 K and back calculated the watts at the source to be 1.85 MEGA-WATTS!"
    --------------
    What "DYNAMIC CONDITIONS" affect the Conduction Equation that you falsely used used to get 30,000 Watts producing 1.85 MEGA-WATTS?

    Where is the fantasy Physics that allows The Law of Conservation of Energy to be VIOLATED and produce your imaginary Perpetual Motion Machine?

    Where is the Actual Physical Example for your delusionary claims?

    How many times have I asked for these?

    All you ever do is "Run For The Hills" every time.

    You can't post them because THEY DON'T EXIST.

    Are you capable of understanding that?

    Why don't you stop your constant babbling and just admit that they DON'T EXIST?

    ReplyDelete
  62. Gord, ok so we have perpetual motion do we? So we have 1850KW, well assume our turbine is running at 90%, we draw 1000KW a second(30kw also enters, and leaves via outter surface), so we have generated 900KW o power... the next second we can only draw 850KW, (because that is all that is left in our reservoir) and generate 765KW o power, in the third second, we are down to only the incoming power, 30KW, and can only produce 27KW... Now if you were putting this energy back into the reservoir, its just going to slow down the rate it loses energy til its at 30KW, this is not perpetual motion, you need to be generating more energy than what you are putting in. As you understand it, boiling water IS perpetual motion... it is not.

    Now if you assume 100% conversion o energy to work, just compressing air, using the heat to expand water is perpetual motion, the universe does not work like that, chaos has a habit o spreading, even mechanically, we have vibrational/noise and heat loses.

    Real life example, try looking at a boiler/steam turbine... you run them hot(they dont heat instantly), but you only use what energy you are putting in, so as that you are not decreasing the pressure, and loosing efficiency. The second you start to draw more than the input, your boiler will cool.

    ReplyDelete
  63. Gord,

    For heaven's sake, just do the calculation.

    ReplyDelete
  64. Anonymous,

    You asked...
    "Gord, ok so we have perpetual motion do we?"

    Answer: Yes you do.

    Funny how you started your rant already assuming that there was 1850 KW generated from a 30 KW source.

    The Law of Conservation of Energy states that Energy cannot be created or destroyed.

    If we only have 30 KW available as a source of energy, how can that produce 1850 KW without Creating Energy?

    Perpetual motion
    "The term perpetual motion, taken literally, refers to movement that goes on forever. However, the term more generally refers to any closed system that produces more energy than it consumes. Such a device or system would be in violation of the law of conservation of energy, which states that energy can never be created or destroyed."

    "Perpetual motion violates either the first law of thermodynamics, the second law of thermodynamics, or both"

    "A perpetual motion machine of the first kind produces energy from nothing, giving the user unlimited 'free' energy. It thus violates the law of conservation of energy."

    http://en.wikipedia.org/wiki/Perpetual_motion

    Producing 1850 KW from a 30 KW source violates The Law of Conservation of Energy and is a perfect example of a Perpetual Motion Machine.
    ------
    You also said...
    "Gord,
    For heaven's sake, just do the calculation."

    Here is the calculation:

    30 KW < 1850 KW....confirming that The Law of Conservation of Energy has been violated because Energy was Created and that a Perpetual Motion Machine has been created.

    Now why don't YOU post the Physics that allows for 1850 KW to be produced from 30KW and post your calculation of how 30 KW = 1850 KW?

    I look forward to your next post.
    ------------
    You said...
    "As you understand it, boiling water IS perpetual motion... it is not."

    Where did you dream that up?

    I think I made myself perfectly clear in my post to Joe Yangtree on Nov 15, 2010 11:13 PM where I said:

    "Mass of the water = 1000 grams
    Delta T = 96 deg C
    Specific Heat of water = 4.18 J/gm X deg C

    Q = mass X delta T X Specific Heat

    Q = 1000 X 96 X 4.18 = 401,280 Joules

    So the water will heat up in 401,280 Joules/ 1600 Joules/sec = 250.8 seconds.
    The water stores 401,280 Joules over 250.8 seconds or 1600 Joules/sec and is exactly the same as the the source provides (1600 Joules/sec)
    This complies with The Law of Conservation of Energy and is not a perpetual motion machine."

    Once again I ask you....Where did you dream that up?

    I look forward to your answer in your next post.
    -------------
    Re: Your examples of Compressing Air and Steam Turbines.

    They both are perfect examples where Energy Out does not exceed Energy In and therefore confirm the validity of The Law of Conservation of Energy.

    However, you assume that 1850 KW can be produced from a 30 KW source without violating The Law of Conservation of Energy.

    So, please post an Actual Physical Example to justify your assumption.

    I look forward to your next post.

    ReplyDelete
  65. Im Mike

    I was the first anonymous, not the second. Just haven't got a profile set up here.

    Ok so my home fire puts out 30KW, i set up a boiler/pressure chamber around my fire(for simplicities sake im going to assume its perfectly insulated) It contains 100litres at 300K, i have the release valve shut, i start up my fire. It delivers 30KW of power a second, so after 62 seconds it has delivered 1850KW of energy into my boiler (1850,000joules, at 4.18J/gm degree K= 418,000J/100lt degree K, it means my water is now 304.4K... need to wait a bit longer) So an hour later my fire has pumped in 108,000,000J! thats a further 258.4K! My steam is now super heated at 562.8K, and i have enough of a differential in pressure that i can run a turbine relatively efficiently(hotter would still be better, its better to expand a little a lot, than a lot a little)

    Thats a total increase in energy contained in the water of 109,850KW of potential energy held under pressure in my boiler. I could open the valve full blast, and run all that energy in one second through my turbine, but i could never get more energy out of it than what ive put in. If i wanted it to run for any length o time, the most a could draw was what i was putting in, or it will cool. And i would be required to use some of that energy to replace the water im taking out of my pressure cylinder as steam.

    I would not be able to heat my water greater than the temperature of the fire, but i can certainly accumulate and store energy.

    You could do it mechanically, with say water also, just pump water up into a reserviour, you could accumulate as much as you want, but you could never get more energy out of it than what has been used lifting the water, you could drop it all back down a much larger pipe, and harness it much more quickly than you stored it though.

    ReplyDelete
  66. Mike,

    You summarized my point about The Law of Conservation of Energy (Energy cannot be created or destroyed) when you said...

    "I could open the valve full blast, and run all that energy in one second through my turbine, but i could never get more energy out of it than what ive put in."

    And also when you said...
    "You could do it mechanically, with say water also, just pump water up into a reserviour, you could accumulate as much as you want, but you could never get more energy out of it than what has been used lifting the water, you could drop it all back down a much larger pipe, and harness it much more quickly than you stored it though."

    "I would not be able to heat my water greater than the temperature of the fire, but i can certainly accumulate and store energy."
    ------------------
    So why did you assume that you can get 1850 KW generated from a 30 KW source?

    Remember, Watts = Joules/sec so 30,000 Joules/sec cannot produce 1,850,000 Joules/sec without violating The Law of Conservation of Energy.

    Ex. 30,000 Joules/sec X 10 seconds = 300,000 Joules and 1,850,000 Joules/sec X 10 seconds = 18,500,000 Joules

    You can't "accumulate and store" more energy than the 30,000 Joules/sec source provides no matter how many seconds of time passes.

    300,000 Joules will ALWAYS be less than 18,500,000 Joules.

    Like I said....

    "Producing 1850 KW from a 30 KW source violates The Law of Conservation of Energy and is a perfect example of a Perpetual Motion Machine."

    (which is exactly what SOD did in his example)

    and...

    "Re: Your examples of Compressing Air and Steam Turbines.
    They both are perfect examples where Energy Out does not exceed Energy In and therefore confirm the validity of The Law of Conservation of Energy.
    However, you assume that 1850 KW can be produced from a 30 KW source without violating The Law of Conservation of Energy.
    So, please post an Actual Physical Example to justify your assumption."

    Which you obviously have not done.
    --------------
    You also forgot to answer this question:

    You said...
    "As you understand it, boiling water IS perpetual motion... it is not."

    Where did you dream that up?

    ReplyDelete
  67. Gord,

    Could you please fix your original calculation. As has been pointed out before there is no transmited readiation in this thought experiment(the PVC shell is opaque).

    It is clear that you are using an incident flux of 24kW on the exterior shell:

    November 12, 2010 5:41 AM
    "Now I will calculate the heat transfer through the PVC Sphere inner to outer surface, only I will do it correctly.

    The outer surface sphere also has an emissivity of 0.8 so 0.8 X 24,000 = 19,200 watts which has to be radiated away by surface re-radiation Plus an additional 4800 watts that is the sum of heat transfer by conduction and energy stored in the PVC that is also radiated.

    The outer sphere area A2 = 2123.72 m^2 and will radiate 19200/2123.72 = 9.04 w/m^2 and the temperature T2 = 112.37 K"


    The outer shell does not recieve 24kW of energy through radiation, the only availiable means of energy input here is conduction.

    Without correcting this error, which was pointed out to you almost immediately, there is really no point to continue as your analysis has no validity in explaining the equilibrium of SoD's thought experiment.

    Neutrino

    ReplyDelete
  68. Neutrino,

    HAHAHAHA...what a HOOT!

    SOD used exactly the same method to calculate the outer surface temperature !!

    I quote from SOD's website:

    "Maths

    The System

    In equilibrium, the outer surface of the sphere has to radiate away all of the heat generated internally. This is the first law of thermodynamics.

    The internal energy source, q = energy radiated from the outer surface

    q = εσT24.4πr22 – the Stefan-Boltzmann equation for emitted energy per m2 x the surface area
    where r2 = radius of the outer surface = 10 + 3 = 13

    If q = 30,000W, r2 = 13m and ε = 0.8

    T2 = q / (εσ.4πr22)1/4 and so T2 = 133K"
    http://scienceofdoom.com/2010/07/26/do-trenberth-and-kiehl-understand-the-first-law-of-thermodynamics/
    -----
    By the way, it should also be noted that there are two mistakes in the above calculation by SOD:

    1) Since emissivity e is 0.8 and q = 30,000 W then the absorbed value is 0.8 X 30,000 = 24,000 Watts

    SOD DIVIDED 30,000 by 0.8 to get 37,500 Watts!

    2) The above equation should have been:

    T2 = (e*q /4*pi*r^2*BC)^0.25 where e = emissivity and BC = Boltzmann Constant = 5.67 X 10^-8

    Plugging in SOD's numbers gives

    T2 = (0.8*30,000/4*pi*13^2 *5.67X10^-8)^0.25 = 118.82 K
    --------------------
    Maybe you should have read SOD's calculation before babbling.

    ReplyDelete
  69. Gord your algebra is a bit off.

    If you accept that q = εσT^4*4πr^2 is accurate then your sovling for T is incorect and SoD's is correct.
    (Although there is a missing set of ()'s but the value he calculated is with the correct ()'s in the equation)

    Moving the emisivity from the demoninator to the numerator changes the equation. This is a basic algebraic manipulation of an equation and should not be a point of contention.

    If you dont see this error then no meaningfull conversation can be had.

    Additionally there is no absorbed value at the outer surface.

    Again you state:
    November 18, 2010 9:19 PM: 1) Since emissivity e is 0.8 and q = 30,000 W then the absorbed value is 0.8 X 30,000 = 24,000 Watts

    There is no absorbed value at the outside surface. The only way that energy can be arriving here is through conduction as the PVC sphere is opaque.

    The outer surface must radiate 30kW or else the system is not be in equilibrium. For a surface of the given area and emisivity the temperature is 133K.

    Neutrino

    ReplyDelete
  70. Neutrino,

    You must have a problem with your reading comprehension.

    The equation used by SOD and myself was the exactly the same.

    I even produced the "quotes" from SOD's website to PROVE IT.

    It's called the Stefan-Boltzmann Equation for heat RADIATION and has NOTHING TO DO WITH CONDUCTION.

    You apparently do not understand what emissivity is or how it is used in the Stefan-Boltzmann Equation.

    Emissivity
    "It is a measure of a material's ability to radiate absorbed energy."
    http://en.wikipedia.org/wiki/Emissivity

    Emissivity = Absorbed Energy / Incident Energy

    When a body has an emissivity (e) of 0.8 it means it can only absorb and radiate 0.8 of the Incident Energy.

    30,000 Incident Watts X 0.8 emissivity = 24,000 Watts absorbed and radiated.
    ---------
    The Stefan-Boltzmann Equation describes what temperature a body will be when it absorbes 100% of the Incident Energy:

    Q1 = σT^4*4πr^2 where Q1 is the Incident Energy (30,000 watts)

    When the Stefan-Boltzmann Equation is written with an emissivity (e) included, it is EXACTLY the same...100% of the Incident Energy is absorbed.

    Here is the Stefan-Boltzmann Equation that SOD used:

    Q2 = e*σT^4*4πr^2

    Q2/e = σT^4*4πr^2 where Q2/e = Q1 = 30,000 watts Incident Energy

    Q2 = ABSORBED ENERGY/ Emissivity = 24,000 Watts / 0.8 = 30,000 Watts Incident Energy!
    ---------

    Instead of using Absorbed Energy (24,000 watts) in the Stefan-Boltzmann Equation, SOD used Incident Energy (30,000 Watts)

    Q2/e = 30,000 Watts/ 0.8 = 37,500 Watts!

    ----------
    You said....
    "The outer surface must radiate 30kW or else the system is not be in equilibrium. For a surface of the given area and emisivity the temperature is 133K."

    37,000 Watts is the value used by SOD for the Outer Surface Radiation to compute the temperature of 133 K !!!

    Proof:

    Q1 = σT^4*4πr^2 where Q1 is the Incident Energy (37,000 Watts), σ = 5.67 X 10^-8, r = 13

    T = (Q1/σ*4πr^2)^0.25

    T = (37,000/ 5.67X10^-8 X 4 X π X 13^2)^0.25

    T = 133K !!!!

    -------

    To use the the Stefan-Boltzmann Equation when 100% of the Incident Energy is not absorbed then you have to compute the Energy that IS ABSORBED.

    Absorbed Energy = Emissivity X Incident Energy = 0.8 X 30,000 Watts = 24,000 Watts.

    Absorbed Energy = σT^4*4πr^2 and the Temperature (T) can then be calculated.

    T = (24,000/ 5.67X10^-8 X 4 X π X 13^2)^0.25

    T = 118.82 K

    Which is what I posted for you before and is the temperature that SOD should have obtained if he understood the Stefan-Boltzmann Equation.

    That is only ONE of the many errors SOD made in his hillarious calculation that resulted in 1.85 Mega-Watts being produced by 30,000 Watts.

    ReplyDelete
  71. Gord,

    There seems to be a conceptual problem here. Not only about emisivity and the Stefan-Boltzmann equation but also how much energy is leaving the outer surface.

    Not much arguement about the SB equation, it is q = εσT^4*4πr^2 and we both agree on that.

    SB equation relates temperature of a surface(T) with how much energy it radiates(q). It is not, as you say, the "Stefan-Boltzmann Equation describes what temperature a body will be when it absorbes 100% of the Incident Energy". Absorbtion is independant of the temperature of the surface, and depends only on its emisivity(emisivity=absorptivity). Emission is dependant on both temperature and emisivity.

    You and SoD did not use the same equation(SB equation rearanged to solve for T):

    Equation that SoD used T = (q/εσ*4πr^2)^0.25 (which is correct)
    Equation that you want to use T = (qε/σ*4πr^2)^0.25 (From your recent post:" 2) The above equation should have been:
    T2 = (e*q /4*pi*r^2*BC)^0.25
    ")

    Flipping the emisivity from denominator to numerator changes the equation, what you are proposing is simply wrong.

    As for all the talk of absorbtion. All I have tried to point out is that there is no absorbtion at the outer surface, so any talk about the absorbed or incident flux is misleading.
    Comments like "Q2/e = Q1 = 30,000 watts Incident Energy" wich implies there is a radiative transfer between inner and outer surfaces are not only confusing but are wrong. Absorbtion is a red herring here because the only input of energy to the outer surface (which is the only one I have been talking about) is through conduction.(and yes SB has nothing to do with conduction, never said it did).


    So, to step back a bit and take another look at the outside surface temperature:

    At equilibrium, which may take a very long time to reach, we know two things about the PVC sphere
    1. 30kW enters the system.
    2. 30kW leaves the system.
    This is by definition, if input does not equal output we do not have equilibrium.

    So the question then becomes, At what temperature will a sphere of 2123m^2 with an emisivity of 0.8 radiate 30kW?

    Plug the numbers into the correct equation above and you will get 133K,

    Now, im not sure if you are objecting to the sphere radiating 30kW, or the standard use of the SB equation to calculate its temperature.

    By definition the sphere is radiating 30kW, its in equilibrium, I cant see how you could argue this point.
    Algebra solves the SB equation for T so this is straight forward as well. We can quickly confirm which solution for T is correct by simply plugging the calculated T's back into the original SB calculation and see how much energy is emited, if its not 30kW you did something wrong.

    Our sphere with surface area 2132m^2 and emisivity 0.8:
    at your T=118.2K it would only be emitting 19.2kW
    at your earlier T=112.37K(from your original analysis) it only emits 15.4kW

    at SoD's T=133K it emits the needed 30kW to be in equilibrium.

    So to try and narrow down where we have a disagreement:
    Equilibrium can be defined as Energy In = Energy Out.
    Knowing the physical characteristics of the outer surface, the input energy and the SB equation we can calculate the temperature.
    Would you agree to that?


    Neutrino

    ReplyDelete
  72. Gord,

    One quick note,

    You seem to be dropping the emisivity and calculating temperatures of what would be a Black Body, not the actual body that we have here in this thought experiment.

    As example you stated that SoD used a flux of 37kW to come up with T=133K. SoD did no such thing, he used a flux of 30kW and an emisivity of 0.8. PVC is not a Black Body and ignoring that, among other things, causes you to come up with incorrect numbers.

    "37,000 Watts is the value used by SOD for the Outer Surface Radiation to compute the temperature of 133 K !!!

    Proof:

    Q1 = σT^4*4πr^2 where Q1 is the Incident Energy (37,000 Watts), σ = 5.67 X 10^-8, r = 13

    T = (Q1/σ*4πr^2)^0.25

    T = (37,000/ 5.67X10^-8 X 4 X π X 13^2)^0.25

    T = 133K !!!!
    "

    133K is the temperature of a Black Body emiting 37kW as the above calculation shows.
    But, it is also happens to be the temperature of a grey body with emisivity of 0.8 emiting 30kW which is what we are talking about here.

    Neutrino

    ReplyDelete
  73. Neutrino,

    MY GOD...you have absolutely no reading comprehensiom and it's clear that you have absolutely no "clue".

    Physics of Emissivity

    "Emitted Energy = Absorbed Energy

    Consider equation 1 for an object in a vacuum at a constant temperature. Because it is in a vacuum, there are no other sources of energy input to the object or output from the
    object. The absorbed energy by the object increases its thermal energy - the transmitted and reflected energy does not. In order for the temperature of the object to remain constant, the object must radiate the same amount of energy as it absorbs.

    Emitted Energy = Absorbed Energy"

    http://www.optotherm.com/emiss-physics.htm

    You CANNOT EMIT MORE ENERGY THAN IS ABSORBED!!!!!

    Did you get that Neutrino?

    --------
    Stefan–Boltzmann law

    "The Stefan–Boltzmann law, also known as Stefan's law, states that the total energy radiated per unit surface area of a black body per unit time (known variously as the black-body
    irradiance, energy flux density, radiant flux, or the emissive power), j*, is directly proportional to the fourth power of the black body's thermodynamic temperature T (also called
    absolute temperature):

    j* = σT^4

    A more general case is of a grey body, the one that doesn't absorb or emit the full amount of radiative flux. Instead, it radiates a portion of it, characterized by its emissivity,
    e.

    j* = eσT^4"
    http://en.wikipedia.org/wiki/Stefan%E2%80%93Boltzmann_law

    Did you UNDERSTAND the part "a grey body, the one that DOESN'T ABSORB OR EMIT THE FULL AMOUNT OF RADIATIVE FLUX. Instead, it RADIATES A PORTION OF IT, characterized by its emissivity, e.????

    In case you don't understand what RADIATIVE FLUX is, it is Watts/Area and in SOD's example Watts/4*pi*r^2.
    ---------------
    How much of the 30,000 Watts can be absorbed and emitted by a body with an emissivity of 0.8?

    Exactly 30,000 X 0.8 = 24,000 Watts!!!

    SOD used 30,000 / 0.8 = 37,000 Watts ABSORBED which IS MORE THAN the 30,000 WATTS of Incident Energy Available (which requires CREATION of Energy)!!!

    Did you get that Neutrino?

    I already PROVED that SOD's calculation used 37,000 Watts ABSORBED to get a temperature of T = 133K !!!

    If the correct amount 24,000 Watts is ABSORBED used you will get a temperature of 118.82 K.

    Did you get that Neutrino or do I have to repeat it again?

    -------------------
    You said...
    "Now, im not sure if you are objecting to the sphere radiating 30kW, or the standard use of the SB equation to calculate its temperature.
    By definition the sphere is radiating 30kW, its in equilibrium, I cant see how you could argue this point."

    SOD had both WRONG!

    As I have repeated about a dozen times now.....

    Here is the Physics link and the equations based on The Law of Conservation of Energy:

    Incident Energy = Emitted Energy + Transmitted Energy + Reflected Energy

    Emitted Energy = Absorbed Energy
    http://www.optotherm.com/emiss-physics.htm

    We have 30,000 Watts as the Incident Energy Source within the INNER SPHERE.

    SOD said the emissivity of the INNER SPHERE was 0.8 so the INNER SPHERE CAN ONLY ABSORB AND EMIT 0.8 X 30,000 = 24,000 WATTS!

    How can the OUTER SPHERE EMIT 30,000 WATTS?...or 37,000 Watts as SOD CLAIMS???

    Come on, ANSWER THE QUESTION and don't forget to include the Physics links to back-up your answer.

    I have already posted the 24,000 Watt calculation before.

    Can't you read or will I have to constantly repeat that again too?

    ReplyDelete
  74. Gord,

    We are not getting anywhere, Ill try and restart from something very basic:

    1) If a system is in equilibrium is it emiting as much energy as it is absorbing? Yes.
    2) Is the system absorbing 30kW? Yes.
    3) Is the system we are talking about in equilibrium? Yes.
    4) Is the system emiting 30kW? Yes.

    Do you disagree with the above questions and the answers given?

    If the system is only emiting 24kW(or any other value less than 30kW, you previously stated 19.2kW and 15.4kW) where is the surplus energy going?
    Is the energy continuing to be stored internally?
    If Yes, then it is not in equilibrium.
    If No, then it is violating the Conservation of Energy.

    Energy In MUST equal Energy Out.
    Energy In is 30kW, how much is the Energy Out?

    So, in an effort to not talk past each other:
    Could you answer the first four questions? Simply Yes/No response will do.

    Neutrino

    ReplyDelete
  75. Neutrino,

    You said...

    "You seem to be dropping the emisivity and calculating temperatures of what would be a Black Body, not the actual body that we have here in this thought experiment.
    "As example you stated that SoD used a flux of 37kW to come up with T=133K. SoD did no such thing, he used a flux of 30kW and an emisivity of 0.8. PVC is not a Black Body and ignoring that, among other things, causes you to come up with incorrect numbers."

    Emissivity = Absorbed Energy/ Incident Energy = 0.8 in SOD's example

    If the Incident Energy is 30 KW how much energy can be absorbed by the body with 0.8 Emissivity?

    Answer: 30,000 Watts X 0.8 = 24,000 Watts.

    SOD DIVIDED the 30,000 Watts (that you say is all the energy available) by the emissivity 0.8 to get 37,000 Watts absorbed by the body!

    Proof:

    Q1 = σT^4*4πr^2 where Q1 is the Incident Energy (37,000 Watts), s = 5.67 X 10^-8, r = 13

    T = (Q1/s*4pr^2)^0.25

    T = (37,000/ 5.67X10^-8 X 4 X p X 13^2)^0.25

    T = 133K !!!!
    -------------------
    I'm sure that you realize that a Mirror has a very low emissivity and is highly reflective.

    The low emissivity means that it absorbs very little energy.

    Let's consider 100 watts of Incident energy being radiated to a mirror with a surface area of 1 sq m.

    If the emissivity of the mirror is 0.01 it can only absorb 100 Watt X 0.01 = 1 Watt

    The temperature of the mirror due to absorbtion of radiated Energy is:

    T = (1 watt / 5.67 X 10^-8)^0.25 = 64.8K

    If we now do the same calculation where we DIVIDE the 100 Watts Incident Energy by the emissivity of 0.01 we get 10,000 Watts being absorbed by the Mirror!

    T = (10,000 Watts/ 5.67 X 10^-8)^0.25 = 648.04 K !!!!
    -------
    If you do SOD's calculation with a lower and lower emissivity, the temperature of the body will increase and increase!

    Why don't you do this and post the results?

    In fact, if the emissivity is dropped to ZERO the energy absorbed by the body will INFINITE and the temperature will be INFINITE regardless of the value of the incident energy as long as the incident energy is not ZERO!!!


    This should make it crystal clear that SOD and you have no idea of how to use the Stefan-Boltzmann Law correctly.

    ReplyDelete
  76. Gord,

    The Stefan Boltzmann Law is:

    1. q = εσT^4*A

    If you think this is wrong just say so and we can end the conversation there.

    If you accept the above as the SB equation then moving forward we can calculate some values.

    You say we have a 1m^2 mirror with emisivity of 0.01 and 100W incident

    2. q = Q*ε
    3. q = 100W*0.01
    4. q = 1W

    So far we agree, now for emission from the mirror:

    Rearanging SB from above to solve for T we get

    5. T = (q/(εσA))^.25
    6. T = (1W/(0.01*σ*1m^2))^.25
    7. T = 205K

    What you calculated(64.8K) is the temperature of a black body 1m^2 radiating 1W. Not your mirror radiating 1W with emisivity of 0.01.
    If Emisivity is not in the equation(or is set to 1 you are dealing with a Black Body)

    As for increasing temperature by decreasing emisivity.
    No the temperature will not go up. If you redo this calculation with a lower emisivity you will get the same T, Why? As ε goes down so does q, so the ratio q/ε stays constant.

    2a. q = Q*ε
    3a. q = 100W*0.00001
    4a. q = 1mW

    5a. T = (q/(εσA))^.25
    6a. T = (1mW/(0.00001*σ*1m^2))^.25
    7a. T = 205K


    Where exactly in the above calculations do you think I have made a mistake?


    Neurtino

    ReplyDelete
  77. Neutrino,

    You said....
    "2. q = Q*ε
    3. q = 100W*0.01
    4. q = 1W
    So far we agree, now for emission from the mirror:"

    WOW, finally YOU are starting to "get it"!

    That is exactly what I have been saying when I said 30,000 Watts X 0.8 emissivity = 24,000 Watts that you have been denying in every post!

    The temperature depends upon the amount of energy ABSORBED!

    That is EXACTLY what SOD HAS NOT DONE!
    --------
    So, cut the CRAP!

    You know perfectly well that I have already taken into account the 100 watts Incident Energy X 0.1 emissivity = 1 watt ABSORBED by the Mirror.

    If you DON'T DO THAT, like SOD didn't do in his example, using SOD's equation where HE ABSOLUTELY DID NOT DO THAT gives:

    T = (q/(εσA))^.25
    T = (100W/(0.01*σ*1m^2))^.25
    T = (10,000 W/ σ*1m^2))^.25

    T = (10,000 / 5.67X10^-8)^0.25 = 648.04 K!!... NOT 205K!!!
    -------------------
    I will do SOD's calculation using SOD's equation AGAIN !

    Incident Energy = 30,000 Watts and emissivity = 0.8.
    Area = 4*pi*r^2 = 4*pi*13^2 = 2123.72 m^2

    HE DID NOT MULTIPLY 30,000 Watts by 0.8 = 24,000 Watts, instead he used 30,000 Watts and DIVIDED it by 0.8 to get 37,000 watts!!!!

    T = (q/(εσA))^.25
    T = (30,000W/(0.8*σ*2123.72m^2))^.25
    T = (37,000 W/ σ*2123.72m^2))^.25
    T = (37,000 / 5.67X10^-8 * 2123.72 m^2)^0.25 = 133 K

    If the emissivity dropped to 0.01 then:

    T = (q/(εσA))^.25
    T = (30,000W/(0.01*σ*2123.72m^2))^.25
    T = (3,000,000 W/ σ*2123.72m^2))^.25
    T = (3,000,000 / 5.67X10^-8 * 2123.72 m^2)^0.25 = 397.29 K

    And, if the emissivity dropped to ZERO (100% reflection) then:

    T = (q/(εσA))^.25
    T = INFINITY!!!!

    Get it now?

    ReplyDelete
  78. Neutrino,

    You said...
    "We are not getting anywhere, Ill try and restart from something very basic:
    1) If a system is in equilibrium is it emiting as much energy as it is absorbing? Yes."

    Correct.

    The inner sphere has an emissivity of 0.8 so it can only absorb 30,000 Watts X 0.8 emissivity = 24,000 Watts.

    Remember your statement:
    You said....
    "2. q = Q*ε
    3. q = 100W*0.01
    4. q = 1W
    So far we agree, now for emission from the mirror:"

    The inner sphere only absorbs 24,000 Watts so it is IMPOSSIBLE for the outer sphere to absorb more than 24,000 Watts and emit 30,000 Watts.
    ------------------
    You said...
    "2) Is the system absorbing 30kW? Yes."

    No, it is aborbing 24,000 watts as per your calculation example above.
    ----------------------
    You said...
    "3) Is the system we are talking about in equilibrium? Yes."

    Yes, I agree.
    ---------------------
    You said...
    "4) Is the system emiting 30kW? Yes."

    No, as you said before "it emiting as much energy as it is absorbing? Yes.", but it only absorbs 24,000 watts so it can only emit 24,000 watts....not 30,000 Watts.

    If it emitted 30,000 Watts it would violate The Law of Conservation of Energy and CREATE 6,000 Watts of Energy.
    ------------
    You asked...
    "If the system is only emiting 24kW(or any other value less than 30kW, you previously stated 19.2kW and 15.4kW) where is the surplus energy going?
    Is the energy continuing to be stored internally?
    If Yes, then it is not in equilibrium.
    If No, then it is violating the Conservation of Energy."

    First, here is what I ACTUALLY said:

    "24,000 Watts = 19,200 watts absorbed by outer sphere and re-radiated + 2390 watts tranferred by conduction to the outer sphere surface and radiated + 2410 watts stored in the PVC that is also radiated."

    And, where did I say anything about 15.4kW?
    ---
    I have posted these equations that are based on The Law of Conservation of Energy MANY, MANY TIMES (why do I have to continually repeat it?):

    Incident Energy = Emitted Energy + Transmitted Energy + Reflected Energy
    Emitted Energy = Absorbed Energy
    http://www.optotherm.com/emiss-physics.htm

    The inner sphere only absorbs 0.8 X 30,000 Watts = 24,000 Watts and that is ALL the energy that can be emitted.

    The inner sphere reflects 0.2 X 30,000 Watts = 6,000 watts which reduces the 30,000 Incident Energy by "destructive interference" (which is NOT destruction of Energy) to 24,000 Watts.

    Exactly what The Law of Conservation of Energy based equation shows and is demonstrated by the Soap Bubble ACTUAL PHYSICAL EXAMPLE that has been demonstrated thousands of times.

    How many times do I have to repeat this ?????

    Now, if you disagree with The Law of Conservation of Energy, the equation that it is based upon it and the the actual physical example (Soap Bubble) that has been demonstrated thousands of times, then POST THE PHYSICS and an ACTUAL PHYSICAL EXAMPLE to support your "opinion".

    Come on, stop your constant babbling and DO IT!
    ------------------
    PS: All the above PHYSICS and the ACTUAL PHYSICAL EXAMPLE was posted by me NUMEROUS TIMES so please stop asking the same questions OVER AND OVER AGAIN.

    The PHYSICS and the ACTUAL PHYSICAL EXAMPLE HAS NOT CHANGED!...DUH!

    ReplyDelete
  79. Gord,

    Lets stick to the simpler mirror until we can agree on that, then move to the sphere.

    Yes, we have always agreed that a surface abosrbs energy at IncidentFlux*Emisivity. No big revalation there.

    So in your example we have 1J accumulated by the mirror every second. And again we are looking for the equilibrium temperature.

    So we go and get the trusty SB equation.

    T is what we want to calculate.
    q is the flux that must be emitted for the mirror to be in balance, As we agreed its 1W.
    ε is the efficiency with which the mirror radiates that 1W.
    A is 1m^2.

    so that gets us back exactly to eq. 5, 6, and 7 from my above post.

    Using a q of anything but 1W is simply wrong. Why? Because that is all the energy in the mirror.(Calculated and both agreed that it is correct)
    Using a ε of anything but 0.01 is simply wrong. Why? Because that is the emisivity of the mirror.(as you defined it in the description)

    So that leaves us with:
    T= 64.8K from q= 1W ε= 1. Incorrect.
    T= 205K from q= 1W ε= 0.01. Correct
    T= 648K from q= 100W ε= 0.01. Incorrect

    Do you agree with this? If not, where is the error.


    And Yes, if we had a surface with emisivity 0 then it will cause a problem, and not just for SoD. Its not Infinity but rather Undefined. Anyways lets deal with the mirror example first then move on to the more complex sphere when we are in agreement.


    Neutrino

    ReplyDelete
  80. Gord,

    Sorry conversation got a bit out of sequence here, the last post of mine from November 22, 2010 6:15 AM
    was about your mirrior, the one you just responded to was from November 20, 2010 2:28 PM
    but some how it wasnt actually added to the page till just this morning.

    Like i said though lets just stick to your mirror till we can agree on it because its a simpler system

    Just to answer one of your questions though because i think it may help with the mirror example as well.

    I made a reference to you stating that 15.4kW was emitted from the outer sphere.
    Yes, you are correct you never actually wrote that.
    But what you did write:
    "The outer sphere area A2 = 2123.72 m^2 and will radiate 19200/2123.72 = 9.04 w/m^2 and the temperature T2 = 112.37 K"
    That temperature is equivilant to saying the sphere(given its area and emisiity) is radiating 15.4kW.
    Dropping the emisivity from the equation when calculating temperature will always get you the wrong answer, unless of course we are talking about a Black Body. This is something you did for the sphere and the mirror.

    By pointing out the above im not argueing what the actual emitted flux is (whether 30kW, 24kW, 19.2kW or even 15.4kW). What I am doing is pointing out that your temperature is not in accordance with what you say the sphere is radiating.
    Just to rephrase it one more way:
    A sphere of 2132m^2 with emisivity 0.8 and emitted flux of 19.2kW does not have a temperature of 112.37K, it does have a temperature of 118.8K.

    So regardless if you are correct about how much energy is arriving at the surface(the mirror or the sphere) you are calculating the wrong temperature based on the number you are using.

    So lets leave the sphere alone for a minute and use the simpler mirror to get the right calculation.
    The one question that I will put to you from this post is:

    Why do you not use the emisivity of an object when you are calculating its temperature?


    Neutrino

    ReplyDelete
  81. Neutrino,

    As I have already pointed out, you are treating the emissivity differently than before.

    You have now correctly used Absorbed Energy = Incident Energy X Emissivity

    Therefore for the Mirror with emissivity of 0.01 the Absorbed Energy = 100 Watts X 0.01 = 1 Watt.

    The mirror can only absorb 1 Watt and only Emit 1 Watt.

    The general Stefan-Boltzmann Equation can now be used to determine the Mirror temperature due to Radiation because we have already determined the effect of the emissivity.

    1)Temperature of Mirror due to Radiation = (Absorbed Energy/ 5.67 X 10^-8)^0.25 = (1 Watt / 5.67 X 10^-8)^0.25 = 64.8 K

    Emitted Energy = Absorbed Energy
    http://www.optotherm.com/emiss-physics.htm

    The above equation based on The Law of Conservation of Energy confirms this.

    If the emissivity of the mirror dropped to 0.000001 the mirror would be more reflective and would absorb LESS of the 100 Watt Incident Energy.

    Absorbed Energy = Incident Energy X Emissivity = 100 watts X 0.000001 = 0.0001 Watts.

    2)The Temperature of the Mirror due to Radiation = (0.0001 Watts / 5.67 X 10^-8)^0.25 = 6.48K
    --------------
    Your method of computing the temperature of the Mirror uses emissivity twice and produces an incorrect Mirror temperature due to Radiation.

    For a mirror with emissivity 0.01 using your incorrect method gives:
    1) Temperature of Mirror due to Radiation = (Absorbed Energy/ (emissivity *5.67 X 10^-8))^0.25 = (1 Watt / (0.01 *5.67 X 10^-8)^0.25 = (100 Watts/5.67 X 10^-8)^0.25 = 204.9K

    The absorbed energy used by you was 100 Watts, where Actual absorbed energy is only 1 Watt!

    For a mirror with emissivity .000001 using your incorrect method gives:
    1) Temperature of Mirror due to Radiation = (Absorbed Energy/ (emissivity *5.67 X 10^-8))^0.25 = (1 Watt / (0.000001 *5.67 X 10^-8)^0.25 = (1,000,000 Watts/5.67 X 10^-8)^0.25 =

    2049.29 K !!

    The absorbed energy used by you was 1,000,000 Watts, where Actual absorbed energy is only 1 Watt!

    Further, the 1,000,000 Watts used to obtain the fantasy 2049.29 K temperature is 10,000 times the value of the 100 Watts Incident Energy!

    The mirror temperature of 2049.29 K is 1776.29 DEG C!!!
    -----------------

    It is very apparent that you don't have a clue about The Law of Conservation of Energy or how to use the Stefan-Boltzmann Law and emissivity correctly as the above calculations PROVE.

    -----------------
    By the way, now that you finally accept that Absorbed Energy = Incident Energy X Emissivity then you obviously agree that the inner Sphere that has an emissivity = 0.8 can only
    absorb 30,000 Watts Incident Energy X 0.8 = 24,000 Watts....RIGHT?

    The 24,000 watts has to reach the Outer Sphere surface which is exactly what I calculated.

    "24,000 Watts = 19,200 watts absorbed by outer sphere and re-radiated + 2390 watts tranferred by conduction to the outer sphere surface and radiated + 2410 watts stored in the PVC that is also radiated."

    I posted the Physics and actual physical examples for "destructive interference" that reduces the Incident Energy (30,000 Watts)- Reflected Energy (6,000 Watts) = Absorbed Energy
    (24,000 Watts) that is the amount absorbed by the inner Sphere that totally complies with The Law of Conservation of Energy.
    -----------
    PS:
    You continually said that the 30,000 Watts Incident Energy has to reach the the outer sphere and be Radiated away.

    In SOD's example he calculated that the 30,000 Watts Incident Energy CREATED 1.85 KW inside the inner sphere.

    According to your statements the created 1.85 KW incident Energy should also reach the outer sphere and be radiated away....RIGHT?

    I look forward to your answers.

    ReplyDelete
  82. Neutrino,

    First an explaination about previous post where the 0.000001 emissivity mirror produced a 1776.29 deg C temperature.

    This was meant to be an example of your previous calculations where the absorbed energy did not change with emissivity and I forgot to specify that.

    Using your latest (current) acceptance of Absorbed Energy = Emissivity X Incident Energy, the absobed energy would be 0.0001 Watts and using the your same method of calculating the mirror temperature would produce the same temperature as the mirror absorbing 1 watt.

    T = 204.9K remained the same despite a reduction of Absorbed Energy from 1 Watt to 0.0001 Watts, which is clearly impossible.
    -----------

    You said...
    "Like i said though lets just stick to your mirror till we can agree on it because its a simpler system".
    "So regardless if you are correct about how much energy is arriving at the surface(the mirror or the sphere) you are calculating the wrong temperature based on the number you are using.
    So lets leave the sphere alone for a minute and use the simpler mirror to get the right calculation."

    I have already posted about the mirror calculations.
    ----------
    You said...

    "The outer sphere area A2 = 2123.72 m^2 and will radiate 19200/2123.72 = 9.04 w/m^2 and the temperature T2 = 112.37 K"
    That temperature is equivilant to saying the sphere(given its area and emisiity) is radiating 15.4kW."

    NOTE: 24,000 watts Incident Energy X 0.8 emissivity = 19,200 Watts ABSORBED.

    T2 = 112.37 K comes from the ABSORBED energy of 19,200 Watts/2123.72 = 9.04 w/m^2.

    T2 = (9.04 w/m^2 / 5.67 X 10^-8)^0.25 = 112.37 K

    How do you get 15,400 watts producing a T2 = 112.37 K???
    -------
    You asked...
    "Why do you not use the emisivity of an object when you are calculating its temperature?"

    I do use it.

    Emissivity = Absorbed Energy watts/ Incident Energy watts

    The value for Absorbed Energy is then used in the Stefan-Boltzmann equation to determine the Body's temperature.

    T = (Absorbed Energy / (Area X 5.67X10^-8))^0.25

    Here is a another great Example you should be familiar with:

    The Earth has a temperature of +15 deg C (288 K).
    The Earth absorbs and radiates 390 w/m^2 (See Trenberth's Earth Energy Budget Diagram)

    T = (390 w/m^2 / 5.67 X 10^-8)^0.25 = 288 K or +15 deg C.

    The Earth also has an Albedo (reflectivity) of A and the Emissivity of the earth = (1 - A) where A = about 0.3 so the Emissivity of the Earth = 0.7

    The Earth's temperature ONLY depends upon how much Energy is ABSORBED by the Earth.

    If we used your hillarious calculation with the improper use of the 0.7 emissivity, the Earth's temperature would be:

    T = (390 w/m^2 / 0.7 X 5.67 X 10^-8)^0.25 = (557.14 w/m^2 / 5.67 X 10^-8)^0.25 = 314.84 K or 41.84 deg C!

    Maybe you should contact Trenberth and explain how he has used Emissivity and the Stefan-Boltzmann Law improperly because your method of calculating the Earth's temperature produces 41.84 deg C???

    ReplyDelete
  83. Gord,

    No, I am not treating Emisivity any different than I have anywhere in these posts.(If I have point it out and I will correct myself)

    Emisivity scales both the absorbtion and emission of surfaces. Using the emisivity to only calculate the absorbtion is just plain wrong. It treats the system as a Grey Body(GB) for absorbtion and as a Black Body(BB) for emission.


    Lets reverse the problem to see if coming at it from a different angle can help.

    Instead of asking what temperature is a mirror, ask how much is emited from a mirror of a given temp, say 205K.
    By the way there is no absorbtion now, this is a new mirror with a 1W internal energy source, radioactive decay! So the effect of ε isnt already factored in as you say.

    Would you use A) q = εσT^4*A or B) q = σT^4*A to calculate the emitted flux given a temperature of 205K?

    If you use B) in this situation you are basically saying that all bodies are BB's and emisivity is meaningless for emission.
    If you use A) here and since you used B) for the original mirror then you are saying that the 1W from the internal source is somehow different from the 1W from the radiant source. In both cases q = 1W.

    The equation for a GB is q = εσT^4*A the mirror(either one) is a GB. Energy is Energy the system doesnt care where it came from.
    Now if you insist that the two systems here are somehow different in a meaningful way please let me know.

    One last note,
    You keep saying things like:(my bold)

    "For a mirror with emissivity .000001 using your incorrect method gives:
    1) Temperature of Mirror due to Radiation = (Absorbed Energy/ (emissivity *5.67 X 10^-8))^0.25 = (1 Watt / (0.000001 *5.67 X 10^-8)^0.25 = (1,000,000 Watts/5.67 X 10^-8)^0.25 =

    2049.29 K !!

    The absorbed energy used by you was 1,000,000 Watts, where Actual absorbed energy is only 1 Watt!
    "

    This makes no logical sense.

    If the system has ε = 0.000001 then q = 0.1mW not 1W. If you change ε it will change q.

    The equation used is T = (q/(εσ*A))^.25. q all by it self is the energy that is emitted which to be in equilibrium is equal to what was absorbed. q/ε is the energy emitted divided by the emisivity, it is not the energy absorbed.

    Not really sure how else to explain this to you.

    Leaving all discussion of the sphere asside until this misunderstanding can be resolved.

    Neutrino

    ReplyDelete
  84. Neutrino,

    You said...
    "No, I am not treating Emisivity any different than I have anywhere in these posts.(If I have point it out and I will correct myself)"

    -------
    Here are examples of where you DID NOT USE Emissivity = (Absorbed Energy/Incident Energy) for the outer shell with 0.8 Emissivity = 24,000 Absorbed Energy / 30,000 Incident Energy!

    From Your Post on Nov 18, 2010 at 9:04 AM:
    "The outer shell does not recieve 24kW of energy through radiation, the only availiable means of energy input here is conduction."

    From Your Post on Nov 19, 2010 at 1:44 AM:
    "Additionally there is no absorbed value at the outer surface.
    Again you state:
    November 18, 2010 9:19 PM: 1) Since emissivity e is 0.8 and q = 30,000 W then the absorbed value is 0.8 X 30,000 = 24,000 Watts
    There is no absorbed value at the outside surface. The only way that energy can be arriving here is through conduction as the PVC sphere is opaque.
    The outer surface must radiate 30kW or else the system is not be in equilibrium. For a surface of the given area and emisivity the temperature is 133K."

    From Your Post on Nov 19, 2010 at 11:10 PM:
    "By definition the sphere is radiating 30kW, its in equilibrium, I cant see how you could argue this point."
    "at SoD's T=133K it emits the needed 30kW to be in equilibrium."

    From Your Post on Nov 20, 2010 at 1:19 AM:
    "SoD did no such thing, he used a flux of 30kW and an emisivity of 0.8. PVC is not a Black Body and ignoring that, among other things, causes you to come up with incorrect numbers."


    From Your Post on Nov 20, 2010 at 2:28 PM:
    "2) Is the system absorbing 30kW? Yes"
    "4) Is the system emiting 30kW? Yes"

    These are ALL examples of where YOU DID NOT accept that the Absorbed Energy = 0.8 Emissivity X Incident Energy = 24,000 Watts!!!
    ----
    You only accepted the fact that Absorbed Energy = Emissivity X Incident Energy when I started to talk about a Mirror with Incident Energy = 100 Watts and an Emissivity = 0.01 producing ABSORBED Energy = 100 W X 0.1 Emissivity = 1 watt

    Here is your Post on Nov 21, 2010 at 3:30 PM where you said...
    "You say we have a 1m^2 mirror with emisivity of 0.01 and 100W incident
    2. q = Q*ε
    3. q = 100W*0.01
    4. q = 1W

    So far we agree, now for emission from the mirror:"

    Then you said in Your Post on Nov 22, 2010 at 6:15 AM:
    "Yes, we have always agreed that a surface abosrbs energy at IncidentFlux*Emisivity. No big revalation there."

    HAHAHAHA...what A HOOT!
    ----------------------------------------
    Continued...

    ReplyDelete
  85. Continuation...
    ----------------------------------------
    I have already explained my mirror example in my previous Post where I said..

    "First an explaination about previous post where the 0.000001 emissivity mirror produced a 1776.29 deg C temperature.
    This was meant to be an example of your previous calculations where the absorbed energy did not change with emissivity and I forgot to specify that."

    ---------------------------
    ---------------------------
    A GREY BODY is one that does not ABSORB the full amount of Incident Energy.

    Instead it will Absorb Energy = Incident Energy X Emissivity

    It is the ABSORBED Energy that produces the Body's temperature.

    The GREY BODY's temperature IS EXACTLY the same as BLACK BODY's temperature that has the Energy = Incident Energy X Emissivity.
    ---
    Emitted Energy = Absorbed Energy or The Law of Conservation of Energy will be violated.

    Which YOUR examples ALL DO because you have used Emissivity TWICE.

    You have calulated Absorbed Energy = Incident Energy X Emissivity and then DIVIDED the Absorbed Energy by Emissivity:

    Absorbed Energy = Incident Energy X Emissivity / Emissivity = Incident Energy !!!
    ---
    That's why your Mirror Calculations ALWAYS produce the SAME HIGH TEMPERATURE no matter how much the ABSORBED ENERGY DROPS!

    You are ALWAYS calculating the TEMPERATURE of the Mirror using only the INCIDENT ENERGY!!!
    ----
    When you were ignoring the Absorbed Energy as in the outer sphere 0.8 X 30,000 = 24,000, you were only using the Incident Energy of 30,000 Watts and then you DIVIDED the Incident Energy by the Emissivity to get 30,000 Incident Energy / 0.8 = 37,000 WATTS!

    No wonder the Temperature was WAY HIGHER than REALITY!
    ---
    I just showed you how the Earth's temperature of +15 deg C (288K) is calculated from ABSORBED Energy of 390 w/m^2 as is evident in Trenberth's Energy Budget Diagram.

    And, when you include DIVISION of the 390 w/m^2 by the Earth's emissivity of 0.7, you get 557.14 w/m^2 which would make the Earth's temperature 41.84 deg C!

    That is absolute PROOF that your calculation method is WRONG.

    ReplyDelete
  86. Gord,

    The mirror and the sphere have radically different geometries, becuase of that they behave differently when thier variables are changed.
    Changing ε in either system will effect the calculations but the effect of those changes is not necesarily the same. I am not saying that ε behaves differently in the two scenarios but that the geometry is different.

    As for the mirror's temperature not changing. If ε goes down both the ability of the mirror to absord and emit are effected. The mirror becomes less efficient at absorbing so q is reduced. Since q is now lower the mirror's need to have a high T(because its now a less efficient emitter) is offset by the fact it has to emit less energy.

    How did I get 15.4kW?
    Easy, I used the SB equation with the flux you said the body was emitting.
    1. q = εσT^4*A
    Solved for T.
    2. T = (q/(εσ*A))^.25
    Its a GB, so not including ε doesnt make sense. And No you didnt already take it into acoount through the absorbtion calculation, SB doesnt care where the energy comes from, all it does is telll you the temerature of a body that is emitting a given flux.
    And again, there is no absorbtion at the outer surface. All the energy, whether you claim its 24kW or I claim its 30kW, arrives at the outer surface via conduction. The solid material between surfaces(PVC) is opaque.

    I think you need to look at the KT diagram again.
    href="http://www.cgd.ucar.edu/cas/Trenberth/trenberth.papers/TFK_bams09.pdf (page 4 is the diagram)
    Total energy absorbed by the earth surface is 333+161 to get 494W/m^2 with the earth surface emitting 396W/m^2. So the temperature of the earth surface(treating it as a BB) comes out to T=289.1K.

    The temperature of the earth surface is only dependant on the emission not the absorbtion. SB doesnt care how much was absorbed since it is an equation about emission. Generally absorbed equals emitted but thats not always the case, the earth surface emits less than it absorbs because it also sheds energy through convection and evaporation.

    Yes the Earth has an albedo of 0.3 (102/341), but thats not the Earth surface, that is for earth surface + earth atmosphere. The albedo for the Earth surface is closer to 0.1 (23/184). So if we wanted to use albedo to get emisivity for the surface we would have to use the 0.1 not the 0.3.

    But... Emisivity is dependant on wavelength, so in the regime of solar radiation(mostly visible light) the surface of the earth has an emisivity somewhere around 0.9. In the regime of surface radiation(mostly infrared) the earth surface has an emisivity of nearly 1.
    If you have some other emisivity for the earth surface in the infrared spectrum feel free to us that, but using the emisivity derived from the atmosphere and the earth surface in a spectrum that the earth is not radiating in is just plain wrong.

    One final attempt to get through to you that emisivity must be in any calculation of a GB. The SB is about emission not absorbtion. SB does not tell us anything about how much energy will be absorbed by an object.
    Calculation of absorbed energy is independant of the SB equation.

    Professor Barton comes up to you and says here is the equation for the energy absorbed by a body:
    3. q =Qε
    Then he goes on to say, here is the equation for emission from a body:
    4. q = εσT^4*A
    He then asks you to combine those two equations and you come up with:
    5. Qε = εσT^4*A
    And solve for T:
    6. T = (Q/(σ*A))^.25 (what i have been using)
    or
    7. T = (Qε/(σ*A))^.25 (what you have been using)

    Problem is 7 doesnt follow algebraicly form 5.
    Only way you can say 7 is correct is if 4 is incorrect, but its not. It is the SB equation. There are no caveats about SB that say when you use ε to scale absorbtion dont use this, it doesnt care where the energy its emitting came from. If you cant accept this I think im at a lose to help you understand your errors.

    Neutrino

    ReplyDelete
  87. (Note I am not the previous anonymous)

    Could you (both) stop messing about please.

    Flux = emissivity * Stefans Constant * T^4

    with emissivity =0.8 , T=133K, that gives

    flux = 0.8 * 5.6704E-8 * 133^4
    = 14.19 Watts/sq m

    Flux is power per area so multiply by the surface area of a 13 m diameter sphere and,
    power = 30 KWatts.
    Conclusion a 13m sphere with e=0.8 and T=133 emits 30KWatts. End of story

    Note that to use this equation properly to find the temperature from the flux you DIVIDE the flux by the emissivity.
    Please also note that power is not the same as energy. I may not have a 30 KWatt source handy, but I have in front of me on my desk, a gadget which produces approximately 10 KWatt from a 1.5 volt battery, and I expect most people do....

    ReplyDelete
  88. Neutrino,

    The reason the mirror temperature does not change is because you are only calculating the temperature based on INCIDENT ENERGY.

    You have calulated Absorbed Energy = Incident Energy X Emissivity and then DIVIDED the Absorbed Energy by Emissivity:
    Absorbed Energy = Incident Energy X Emissivity / Emissivity = Incident Energy !!!

    Example:

    BC = 5.67 X 10^-8
    Area (A)= 1 m^2
    Incident Energy = 100 Watts
    Emissivity (e) = 0.1
    Absorbed Energy (q) = 100 Incident Energy X 0.1 Emissivity = 10 Watts
    q = 10 Watts

    Using your calculation method gives:
    q = e X BC X T^4*A
    T = (q/ e*BC)^0.25
    T = ((q/e) X (1/BC))^0.25
    T = ((q/e) X 1.76 X 10^7)^0.25

    q/e = Incident Energy X Emissivity / Emissivity = 100 X 0.1/0.1 = 100 Watts THE INCIDENT ENERGY

    T = (100 X 1.76 X 10^7)^0.25 = 204.9 K

    If the emissivity is changed to 0.01
    Absorbed Energy (q) = 100 Incident Energy X 0.01 Emissivity = 1 Watt
    q = 1 watt.

    q/e = Incident Energy X Emissivity / Emissivity = 100 X 0.01/0.01 = 100 Watts THE INCIDENT ENERGY

    T = (100 X 1.76 X 10^7)^0.25 = 204.9 K

    No matter what value you use for emissivity the result will ALWAYS BE THE SAME BECAUSE

    Absorbed Energy = Incident Energy X Emissivity / Emissivity = Incident Energy !!!

    Only a BLACK BODY absorbs all the INCIDENT ENERGY and that is what YOU ARE CALCULATING.
    ------
    Like I explained at the very beginning of our discussion:

    The general form for the SB equation is:

    Q1 = BC X T^4*A where Q1 is the INCIDENT ENERGY and all this energy is absorbed by the BLACK BODY(1)

    when emissivity is included the equation is:

    Q2 = e X BC X T^4 where Q2 is the ABSORBED ENERGY

    Q2/e = BC X T^4 (2) where Q2/e is the INCIDENT ENERGY and all this energy is absorbed by the BLACK BODY and is EXACTLY THE SAME AS EQUATION (1)!!

    PROOF:

    Equating (1) and (2) gives

    Q1 = Q2/e

    e = Q2/Q1

    Emissivity e = Absorbed Energy/ Incident Energy

    Absorbed Energy/ Incident Energy = Q2/Q1

    Q2 = ABSORBED ENERGY

    Q2/e = ABSORBED ENERGY X INCIDENT ENERGY/ ABSORBED ENERGY = INCIDENT ENERGY

    Q1 = INCIDENT ENERGY

    Equations (1) and (2) ARE SAME EQUATION and will produce a BLACK BODY TEMPERATURE!!!

    Get it now?
    ---------------------------
    Your Professor Barton example is EXACTLY the same:

    "Professor Barton comes up to you and says here is the equation for the energy absorbed by a body:"
    3. q = Q*e and is the ABSORBED energy

    where Q = Incident Energy and e = Emissivity
    Absorbed Energy = Incident Energy X Emissivity
    ---
    Then he goes on to say, here is the equation for emission from a body:
    4. q = e*BC*T^4*A

    q = Q*e as stated in (3) so Q*e = e*BC*T^4*A gives Q = BC*T^4 where Q = Incident Energy.
    ---
    "He then asks you to combine those two equations and you come up with:
    5. Q*e = e*BC*T^4*A"

    Q = BC*T^4 where Q = Incident Energy.
    ----
    "And solve for T:
    6. T = (Q/(BC*A))^.25 (what i have been using)"...where Q = Incident Energy

    Correct! You have ALWAYS been using Incident Energy that ALWAYS produces a BLACK BODY temperature!
    -------
    "or
    7. T = (Q*e/(BC*A))^.25 (what you have been using)"

    Exactly Correct!

    T is the temperature of a GREY BODY that only Absorbs 0.8 X Incident Energy Q

    Absorbed Energy = Incident Energy X Emissivity = Q*e
    ----------
    Do finally Get It?

    ReplyDelete
  89. Neutrino,

    You said...
    "And again, there is no absorbtion at the outer surface. All the energy, whether you claim its 24kW or I claim its 30kW, arrives at the outer surface via conduction. The solid material between surfaces(PVC) is opaque."

    First, CONDUCTION is method of HEAT TRANSFER from the inner sphere to the outer sphere.

    The PVC also STORES ENERGY otherwise it would not have a TEMPERATURE and that TEMPERATURE produces HEAT RADIATION that has to reach cold space.
    (Stored Energy was ignored by SOD, another HUGE ERROR)

    The Heat Radiation and temperature at the outer sphere is calculated by the SB equation that has an EMISSIVITY of 0.8.

    ****Both SOD and I used the SB equation and EMISSIVITY to do just that, except SOD wrongly divided 30,000 by 0.8 emissivity to get 37,000 watts radiated!****

    The inner sphere has an emissivity of 0.8 and Incident Energy of 30,000 Watts.
    The inner sphere can only absorb 30,000 Watts X 0.8 = 24,000 Watts and THAT AMOUNT HAS TO REACH COLD SPACE.

    The Outer Sphere has an emissivity of 0.8 so it can only Radiate 0.8 X 24,000 = 19,200 Watts.

    The 24,000 - 19,200 = 4800 Watts that has to come from Conduction Heat Transfer AND the STORED ENERGY.

    Using the Conduction equation the watts transferred by conduction was calculated to be 2390 Watts with the remaining 4800 - 2390 = 2410 Watts being STORED ENERGY.

    "24,000 Watts = 19,200 watts absorbed by outer sphere and re-radiated + 2390 watts tranferred by conduction to the outer sphere surface and radiated + 2410 watts stored in the PVC that is also radiated."

    I posted the Physics and actual physical examples for "destructive interference" that reduces the Incident Energy (30,000 Watts)- Reflected Energy (6,000 Watts) = Absorbed Energy
    (24,000 Watts) that is the amount absorbed by the inner Sphere that totally complies with The Law of Conservation of Energy.
    ------
    PS:
    SOD calculated that the 30,000 Watts Incident Energy in the inner sphere magically increased to 1.85 Mega-Watts violating the Law of Conservation of Energy.

    How does that Radiate to cold space?

    ReplyDelete
  90. Anonymous (the previous one),

    See my post to Neutrino for how the SB equation is developed with mathematical proof.
    ----
    Gee, I don't have a 1.5 volt battery that can provide 10 Kilowatts.

    Watts = Volts X Amps

    Amps produced by your battery = 10,000 Watts/1.5 Volts = 6,667 AMPS!

    Buy a hundred of those "batteries" ($100 ?) and you will have a MEGA-WATT power station.

    You will be filthy RICH.

    ReplyDelete
  91. Gord.

    My device is the flash in my camera - 10 joules released in 1 millisecond - think about it - perhaps you will realize the difference between energy and power.

    And I know how the SB equation works...

    Do you disagree at a 13m sphere with emissivity of 0.8 and temperature 133K radiates 30 KWatts?

    ReplyDelete
  92. Gord.

    I finnaly get it,

    I am going to take the advice of the above anon poster and end this pointless conversation now.

    You have absolutely no understanding of what the SB equation represents, its terms or how to use it. As well you are confused about radiation, conduction and the Conservation of Energy. You dont even understand basic Algebra.

    Ive failed to help you see any of your many errors. I have no idea how to get through to you and it seems you are not open to it anyways.

    So unless you want to come to the table in an open exchange, goodbye.

    And thanks to the above anon poster for pointing out my inconsistent use of Power/Flux/Energy, I should have been more carefull and will be in the future.
    A quick heads up, yes Gord does disagree with you he thinks 133K radiates 37kW.
    Thank you again and good luck to you.
    ps. lol camera flash seems about right I couldnt come up with what device you were thingking of.


    Neutrino

    ReplyDelete
  93. Neutrino,

    You said...
    "You have absolutely no understanding of what the SB equation represents, its terms or how to use it. As well you are confused about radiation, conduction and the Conservation of Energy. You dont even understand basic Algebra."

    Really?

    How am I "confused about radiation, conduction and the Conservation of Energy"?

    Gee, I think I posted Physics links and actual physical examples to back up my posts, unlike You, who only posted your unsupported "opinions".

    And, why don't you post where my "basic Algebra" is WRONG???

    1) I showed that your method of using the SB equation ALWAYS produced a temperature based on INCIDENT ENERGY being absorbed by a BLACK BODY.

    2) I showed the mathematical PROOF how your hillarious use of the SB equation was ALWAYS based on INCIDENT ENERGY being absorbed by a BLACK BODY.

    3) I even used your own "Professor Barton" example to PROVE the same thing using "basic Algebra"!

    The same "Professor Barton" example YOU described as "Problem is 7 doesnt follow algebraicly form 5...."!

    Like I said..."Then why don't you post where my "basic Algebra" is WRONG???"

    Kinda hard to dis-prove my "basic Algebra", so you decided to RUN FOR THE HILLS when faced with the PROOF.

    What a HOOT!
    ------
    You said...
    "Quick heads up, yes Gord does disagree with you he thinks 133K radiates 37kW."

    My GOD, where did you dream that little gem of "Wisdumb" up?...or do you suffer from Alzheimer's Disease?

    This is what I ACTUALLY POSTED.

    "SOD used 30,000 / 0.8 = 37,000 Watts ABSORBED which IS MORE THAN the 30,000 WATTS of Incident Energy Available (which requires CREATION of Energy)!!!
    Did you get that Neutrino?
    I already PROVED that SOD's calculation used 37,000 Watts ABSORBED to get a temperature of T = 133K !!!
    If the correct amount 24,000 Watts is ABSORBED used you will get a temperature of 118.82 K.
    Did you get that Neutrino or do I have to repeat it again?"


    What a HOOT!
    -------------
    Adios Amigo, come back when you have something to post besides your unsupported "opinions".

    Thank you and good luck to you.

    ReplyDelete
  94. Anonymous,

    In case you still don't realize it, we are not dealing with stored energy all released in "millisecond" of time with SOD's fantasy production of 1.85 MEGA-WATTS from a 30,000 Watt source in a PVC sphere.

    Conduction of energy, absobtion of energy and storage of energy in the PVC takes TIME and ALL of SOD's example deals with STEADY STATE EQUILIBRIUM PHYSICS like the Stefan-Boltzmann Law is used for.

    It is equivalent to heating a pot of water from a heating element on a stove.

    It is IMPOSSIBLE for the water to release 1.85 Mega-Joules of energy if the source only provides 30,000 Joules of energy and the same goes for SOD's PVC Sphere.

    Get it now?
    -----------
    You said...
    "And I know how the SB equation works..."

    Well that's great.

    If you disagree with the mathematical proof of how the SB equation works that posted for Neutrino, then POST WHERE I HAVE MADE ANY MISTAKES and YOUR CALCULATIONS.

    I look forward to your next post.
    -----------
    You asked...
    "Do you disagree at a 13m sphere with emissivity of 0.8 and temperature 133K radiates 30 KWatts?"

    Ummm, unless you are mentally challenged or can't read my numerous posts you should know what my answer is....YES, I SAY IT IS BS!

    If there is 30,000 watts Incident Energy (SOD's claim, not mine) and the emissivity is 0.8 then the sphere can only absorb and radiate 0.8 X 30,000 = 24,000 Watts.

    Absorbed Energy = Incident Energy X Emissivity

    Do you disagree with that?

    Do you agree that Emitted Energy = Absorbed Energy as this Physics link (that I have posted about a dozen times) says?

    Emitted Energy = Absorbed Energy
    http://www.optotherm.com/emiss-physics.htm

    Did SOD use 24,000 Watts or 30,000 Watts (as you say) for his temperature calculation?

    NO, HE DID NOT

    37,000 Watts is the value used by SOD for the Outer Surface Radiation to compute the temperature of 133 K

    Proof:

    Q1 = BC*T^4*4pr^2 where Q1 is the Incident Energy (37,000 Watts), BC = 5.67 X 10^-8, r = 13

    T = (Q1/BC*4pr^2)^0.25

    T = (37,000/ 5.67X10^-8 X 4 X p X 13^2)^0.25

    T = 133K
    ---------
    Did you not see this analysis before?

    I posted it many times.

    Why don't you print it out and save it, so I don't have to keep repeating it.

    ReplyDelete
  95. I felt bad leaveing one thing on the table. I asked Gord a direct question, where had I treated Emissivity differently, and he actually responded to it(posted:November 23, 2010 12:38 AM ).

    So to clarify:
    Emissivity, as I understand it, is only used in two places.

    The first, when radiation is incident on a surface. Here it is used to calculate how much of that incident is absorbed. q =Qε.
    The second, when radiation is emitted from a surface. This is where StefanBoltzmann is used.q=εσT^4*A
    Both uses are the same really, just scaling the, absorbed or emitted, power.

    As to Gord's accusation I have used Emissivity differently.
    All of his points are:
    1)places I am attempting to show him there is no radiative absorbtion.
    (transfer mechanism from inner surface to outter is conduction)
    2)places I invoked equilibrium to come up with total absorbed and total emitted.

    Neither of these two concepts use Emissivity so im not sure where he thinks I have used it differently for them.

    3)places where I calculate surface temperature from emission or emission from surface temperature.

    Not using Emissivity while calculating emission or temperature seems very odd to me as that is the point of it(its kinda implied in the name!). I have been consistant in using it every time Ive calculated something. Gord just doesnt aggree with the formula or the input numbers used but that has nothing to do with my consistant(or lack thereof) us of the property.

    Anyways since I asked a direct question and Gord actually gave me a direct answer I felt I should respond.


    Neutrino


    ps: Gord if you do want to have a conversation pick one point and stick to it untill the disagreement/misunderstanding is resolved, I suggest algebra since without that everything else you do is wrong from the get go.

    ReplyDelete
  96. Neutrino,

    I will first post this as a reference for my next post to you.
    ------
    Summary of Equations and Results re: The Stefan-Boltzmann Law:

    Emissivity e = Absorbed Energy / Incident Energy

    q = Q*e where q = Absorbed Energy, Q = Incident Energy and e = Emissivity

    q/e = Q (1)

    q = e*BC*T^4*A (2) where q = Absorbed Energy and e = Emissivity and

    q/e = BC*T^4*A where q/e = Absorbed Energy/ Emissivity = Q the Incident Energy.

    Q = BC*T^4*A (3)

    Equations (2) and (3) are the same and give the temperature of a Black Body that absorbs and radiates all the Incident Energy Q = q/e

    To find the temperature of a GREY BODY that only absorbs a fraction of the Incident Energy (Q*e) this is the equation that must be used:

    Q*e = BC*T^4*A (4)

    --------------------------------------
    SOD's method of Calculation:

    SOD calculated the temperature of the outer sphere using the 30,000 Watts from the Energy Source inside the Inner Sphere and used 0.8 as the emissivity of the outer sphere plugged
    into equation (2) q = e*BC*T^4*A which is the same as q/e = BC*T^4*A.

    The value SOD used for q was 30,000 Watts and which was wrong because q is the Absorbed Energy and the Absorbed energy = Incident Energy (30,000) X Emissivity (0.8) = 24,000 Watts.

    Then SOD divided the 30,000 Watts by the Emissivity 0.8 to get 37,000 Watts and then calculated the temperature of the outer sphere using q/e = BC*T^4*A.

    The 37,000 Watts radiated by the outer Sphere exceeded the Energy Source inside the inner Sphere which was only 30,000 Watts, which violates The Law of Conservation of Energy and is impossible.

    The impossible use of 37,000 Watts produced an impossible temperature for the outer Sphere of 133 K.

    SOD then used the Conduction Equation, using the impossible outer sphere temperature 133K and 30,000 Watts of conducted energy to calculate an impossible inner Sphere surface temperature of 423 K.

    SOD then used q = e*BC*T^4*A for the inner sphere where e = 0.8 and the impossible T = 423 K to calculate a hillarious value of q = 1,824,900 Watts when the actual Energy Source is
    30,000 Watts!
    -------------------------------------------------
    Continued...

    ReplyDelete
  97. Continuation...

    My method of Calculation:

    The Energy Source inside the inner Sphere is 30,000 Watts = Q.

    The Inner Sphere Emissivity is 0.8 so it can only absorb 30,000 X 0.8 = 24,000 Watts.

    This complies with the equation based on The Law of Conservation of Energy:

    Incident Energy = Absorbed Energy + Transmitted Energy + Reflected Energy
    http://www.optotherm.com/emiss-physics.htm

    30,000 Watts Incident Energy = 24,000 Watts Absorbed Energy + 0 Watts Transmitted Energy + 6,000 Watts Reflected Energy

    I used the equation for a Grey Body Q*e = BC*T^4*A where Q*e = 24,000 Watts and calculated the inner Sphere temperature to be 135.47 K (SOD calculated 423 K)

    The 24,000 watts absorbed by the inner sphere has to eventually be radiated to cold space as required by The Law of Conservation of Energy.

    Then I used the 24,000 Watts absorbed by the inner Sphere and the emissivity e = 0.8 of the outer sphere to determine the outer Sphere surface temperature using the Grey Body
    equation Q*e = BC*T^4*A where Q*e = 24,000 X 0.8 = 19,200 watts and calculated the outer Sphere temperature to be 112.37 K

    (In SOD's very first calculation, he wrongly used the q/e = BC*T^4*A equation with q/e = 37,000 Watts to get a temperature for the outer Sphere of 133 K)

    The outer Sphere is radiating 19,200 Watts to cold space leaving 24,000 - 19,200 = 4800 Watts to be conducted through the PVC and radiated + Stored Energy in the PVC that is also radiated.

    (SOD totally ignored the Stored energy in his calculations which is another huge mistake)

    Using the Conduction Equation and the temperatures of the inner and outer Spheres, I calculated the Watts transfered by conduction to be 2390 Watts.

    Finally, I calculated the Stored Energy to be 4800 - 2390 = 2410 Watts.
    ---
    Summary of my calculations:

    Incident Energy = Absorbed Energy + Transmitted Energy + Reflected Energy

    30,000 Watt source = 6000 watts reflected by inner sphere + 0 transmitted energy + 24,000 watts absorbed by inner sphere that has to reach free space.

    24,000 Watts = 19,200 watts absorbed by outer sphere and re-radiated + 2390 watts tranferred by conduction to the outer sphere surface and radiated + 2410 watts stored in the PVC that is also radiated.

    T1 = 135.47 K and T2 = 112.37 K

    And, there is complete compliance with The Law of Conservation of Energy

    ReplyDelete
  98. Neutrino,

    - Your use of the SB equation is wrong as is SOD's use of it.
    - Read my mathematical proof again and post where you disagree and your calculations.

    I have already asked you to do this and I'm still waiting for your response.
    ----
    You said...

    "As to Gord's accusation I have used Emissivity differently.
    All of his points are:
    1)places I am attempting to show him there is no radiative absorbtion.
    (transfer mechanism from inner surface to outter is conduction)
    2)places I invoked equilibrium to come up with total absorbed and total emitted.
    Neither of these two concepts use Emissivity so im not sure where he thinks I have used it differently for them."

    What a HOOT!

    - The SB equation deals with Radiative Heat transfer and has nothing to do with Conduction.

    - SOD's first step in his calculation was to use the SB equation with 37,000 watts absorbed and emissivity = 0.8 to determine the outer sphere surface temperature.

    - SOD followed that with a calculation of the watts due to conduction, based on the temperature calculated for the outer Sphere.

    - SOD then used the SB equation to determine the inner Sphere surface temperature.

    - My calculation used the SB equation to calculate the inner and outer Sphere temperatures also based on watts absorbed and emissivity as well.

    - I then calculated the watts due to conduction based on the inner and outer sphere temperatures and the stored energy of the PVC.

    Both myself and SOD have obviously used the SB equation that has nothing to do with conduction and every thing to do with absorbed energy, emissivity and Radiative Heat transfer.

    Also, you only started to use q = Q*e ....Absorbed Energy = Incident Energy * Emissivity when you started talking about mirrors.

    Before that you totally ignored the use of ABSORBED ENERGY.

    Example:
    Your rants about SOD's outer sphere temperature calculation ignored q = Q*e

    There was 30,000 Watts Incident Energy and the outer sphere emissivity is 0.8 so q = 30,000 * 0.8 = 24,000 Watts.

    SOD divided 30,000 Watts by the 0.8 Emissivity = 37,000 Watts which YOU SAID WAS CORRECT....REMEMBER NOW???
    --------------
    PS:

    I have picked the point of SOD's hillarious PVC sphere calculation that:

    - Violates the Law of Conservation of Energy
    - Is a Perpetual Motion Machine that creates energy.
    - Does not use the SB equation properly

    That is why I am posting on this site and why I have posted the Physics Links, Actual Physical Examples and Mathematical Proof as back-up for my posts.

    Why don't you do the same instead of babbling your fantasy un-supported "opinions"?

    ReplyDelete
  99. Gord,

    Only addressing the point that you say SoD(and many other poeple including myself) use the SB equation improperly.

    The SB equation is:
    1. q = εσT^4*A

    Do you dispute that?

    q is the total emitted Power
    ε is the Emisivity
    σ is the StefanBoltzmann constant
    T is the temperature of the surface
    A is the surface area

    Do you dispute any of those?

    Note that q is the emitted power not the absobed power, the SB equation deals solely with the emission of radiation and has nothing to do with absorption(find a reference where it says explicitly that q as written in the above formula is the power absorbed). SB tells us either how much power is radiated given a surface temperature or what temperature is required to emit a certain power, it says absolutely nothing at all about absoption. To find out how much power must be emitted to stay in equilibrium another equation must be used, SB cant do that for you. Once you know what that value is SB can then tell you the temperature.

    Ill head off your apeal to Absorbed = Emitted, thats total absorbed, by any means not just radiative, and total emitted, not just by radiative means. So yes the system must equilibriate, but there is not always a incident radiat power and even it there is its not always equal to the emitted radiant power other excahnges can happen on both sides, convective and conductive.

    So to see if we can both use SB correctly ill repeat the above anon posters question with a twist:
    What is the radiated power(q) of a 2123.7m^2(A) sphere that has 0.8 emissivity(ε) and a temperature of 100.9K(T)?
    Actually do the calculation and get back to me, ill help ya out and set it up even:

    1. q = εσT^4*A
    2. q = 0.8*σ*101K^4*2123.7m^2
    3. q = ???

    Since we agree that 1. is true(do we?) then its just up to simple arithmetic not even any algebra here.

    A few links displaying the form of the SB and stating that it explicitly deals with emission
    en.wikipedia.org/wiki/Stefan%E2%80%93Boltzmann_law
    hyperphysics.phy-astr.gsu.edu/hbase/thermo/stefan.html
    csep10.phys.utk.edu/astr162/lect/light/radframe/sb_tl.html
    planetphysics.org/encyclopedia/StefanBoltzamannLaw.html
    frank.mtsu.edu/~phys2020/Lectures/Part_2__L6-L11/Lecture_6/Stefan-Boltzmann/stefan-boltzmann.html

    even found a calulator, did they program it right?
    www.endmemo.com/physics/radenergy.php


    Neutrino

    ReplyDelete
  100. Gord,

    Tell you what , just use this:

    http://www.endmemo.com/physics/radenergy.php

    type the numbers in and see what you get

    ReplyDelete
  101. Neutrino,

    You said....
    "The SB equation is:
    1. q = e*BC*T^4*A
    Do you dispute that?
    q is the total emitted Power
    e is the Emisivity
    BC is the StefanBoltzmann constant
    T is the temperature of the surface
    A is the surface area
    Do you dispute any of those?"

    Yes, I dispute that q is the total emitted Power.

    Proof: q/e = would BE GREATER THAN "total emitted Power" and Violate The Law of Conservation of Energy.

    q is the ABSORBED power as I have already MATHEMATICALLY PROVEN.

    q/e = the TOTAL EMITTED POWER.

    The equation q = e*BC*T^4*A is q/e = BC*T^4*A (1)

    Your example q = 0.8*BC*101K^4*2123.7m^2 gives a value of q = 10,024 Watts absorbed.

    q/e = 10,024/0.8 = 12,530 Watts of Incident Power.

    The general SB equation is q = BC*T^4*A (2) where all the incident power is absorbed by a Black Body.

    q = 5.67X10^-8 * (101)^4 *2123.7 = 12,530 Watts Incident Power.

    The equation you use q = e*BC*T^4*A IS THE SAME EQUATION as q = BC*T^4*A and will only produce a BLACK BODY TEMPERATURE!!

    I have repeated this about a dozen times now.

    I have showed how your Mirror calculations always produce the same temperature when emissivity is changed because you are ALWAYS using the INCIDENT ENERGY and always getting the SAME TEMPERATURE.

    I showed the MATHEMATICAL PROOF and posted it several times, which you have ignored as well.

    Really, what is your problem?

    ------------
    You said...
    "Note that q is the emitted power not the absobed power, the SB equation deals solely with the emission of radiation and has nothing to do with absorption(find a reference where it says explicitly that q as written in the above formula is the power absorbed)."

    What a HOOT!

    First here is the the Physics link that says EMITTED ENERGY = ABSORBED ENERGY, that I have posted at least a DOZEN TIMES!

    Emissivity in the Infrared
    Physics of Emissivity
    "Emitted Energy = Absorbed Energy"
    http://www.optotherm.com/emiss-physics.htm

    EMITTED ENERGY = ABSORBED ENERGY is based on THE LAW OF CONSERVATION OF ENERGY which you obviously do not understand or refuse to accept.

    What is your problem?

    I am getting really tired of you posting delusional babbling that has been shown to be wrong over and over again!

    So either state WHY you don't accept EMITTED ENERGY = ABSORBED ENERGY, post the Physics to back up your delusional claim "emission of radiation and has nothing to do with absorption" or admit that EMITTED ENERGY = ABSORBED ENERGY !!
    ---------
    continued...

    ReplyDelete
  102. continuation....

    Regarding your links that you posted:

    This one TOTALLY confirms what I have been saying:

    Stefan-Boltzmann law
    P(T) = e*BC*T^4

    "The law is valid only for ideal black objects, the perfect radiators, called black bodies."
    http://planetphysics.org/encyclopedia/StefanBoltzamannLaw.html

    Like I said above and have repeatedly said...
    "The equation you use q = e*BC*T^4*A IS THE SAME EQUATION as q = BC*T^4*A and will only produce a BLACK BODY TEMPERATURE!!"

    Did you get that, Neutrino?
    ---------
    This one also confirms what I have been saying about Grey Bodies.

    Stefan–Boltzmann law

    "The Stefan–Boltzmann law, also known as Stefan's law, states that the total energy radiated per unit surface area of a black body per unit time (known variously as the black-body irradiance, energy flux density, radiant flux, or the emissive power), j*, is directly proportional to the fourth power of the black body's thermodynamic temperature T (also called absolute temperature):

    j* = BC*T^4

    A more general case is of a grey body, the one that doesn't absorb or emit the full amount of radiative flux. Instead, it radiates a portion of it, characterized by its emissivity,
    e.

    j* = e*BC*T^4"
    http://en.wikipedia.org/wiki/Stefan%E2%80%93Boltzmann_law

    Did you UNDERSTAND the part "a grey body, the one that DOESN'T ABSORB OR EMIT THE FULL AMOUNT OF RADIATIVE FLUX. Instead, it RADIATES A PORTION OF IT, characterized by its emissivity, e.????

    Did you get that, Neutrino?

    I also posted the above before!
    ----------
    This one explains absorbed and emitted energy:

    The Stefan–Boltzmann Law

    "Since real objects emit radiation less efficiently than a blackbody at the same temperature (since, by definition, a blackbody is an ideal emitter and absorber of radiation!), we introduce a (unitless) term called the emissivity, cleverly denoted e. The value of the emissivity e ranges from near 0 for an object which is a very poor emitter of radiation (and thus a very poor absorber of radiation, such as a mirror), to a value near 1 for a very good emitter of radiation (very close to a perfect blackbody, for which of course e = 1)."
    http://frank.mtsu.edu/~phys2020/Lectures/Part_2__L6-L11/Lecture_6/Stefan-Boltzmann/stefan-boltzmann.html

    Did you get that, Neutrino?

    ReplyDelete
  103. Neutrino,

    Are you familiar with the AGW claims that the Earth's Black Body temperature would be 5.3 deg C but the Earth's reflection of 30% drops the Earth's temperature to -18 deg C?

    Greenhouse effect
    "If an ideal thermally conductive blackbody was the same distance from the Sun as the Earth, it would have an expected blackbody temperature of 5.3 °C. However, since the Earth reflects about 30%[4] (or 28%[5]) of the incoming sunlight, the planet's actual blackbody temperature is about -18 or -19 °C,[6][7] about 33°C below the actual surface temperature of about 14 °C or 15 °C."
    http://en.wikipedia.org/wiki/Greenhouse_effect

    (By the way that 15 deg C Earth surface temperature is calculated by using 390 w/m^2 in the SB equation T = ((P/A)/BC)^0.25 where P/A = 390 w/m^2)
    ---
    342 w/m^2 of Solar Energy (see Trenberth's Diagram) is used to calculate the 5.3 deg C (278.3 K) Earth Black Body temperature.

    Here is how the calculation was done:

    T = ((P/A)/BC)^0.25 where P/A = 342 w/m^2

    T = (342/ 5.67X10^-8)^0.25 = 278.68 K or 5.68 deg C for the Black Body Earth temperature.

    With a reflection of 0.3 (emissivity = 0.7) the calculation for the Solar energy absorbed by the Earth is 342 X 0.7 = 239.4 w/m^2

    T = (5.67X10-8 X 239.4)^0.25 = 254.91 K or -18.09 deg C for the Earth temperature.

    The Earth can only produce a temperature based on the Energy it absorbs.
    -----------------------
    Notice that they did not use Your equation P/A = e*BC*T^4 which is T = ((P/A)/e*BC )^0.25 and would give:

    T = (342/(0.7 x 5.67X10^-8))^0.25 = 304.67 K or 31.67 deg C instead of -18.09 deg C!

    I wonder why they didn't use your equation?

    Could it be that your equation produced a fantastically high temperature because it increased the Solar Energy absorbed to 342/0.7 = 488.57 w/m^2?

    Unlike you, they know that the Earth can only produce a temperature based on the Energy it absorbs.

    PS: This EXACTLY how SOD managed to get a fantastic 37,000 Watts from 30,000 Watts with a 0.8 emissivity and produce a high temperature of 133K for the outer sphere.

    What a HOOT!

    ReplyDelete
  104. If SOD was correct.Why have those magical heaters not been getting patented?

    It is simply amazing that thousands of highly trained scientists never desire to make these special heaters and become millionaires.

    LOL

    ReplyDelete
  105. OK, here is final definitive proof

    "Yes, I dispute that q is the total emitted Power."

    Gord does NOT know what the SB equation is or what the terms mean. He even declines to use a nice interactive web page which will do the calculation for him.

    No point continuing this further.

    ReplyDelete
  106. Gord says,
    "PS: This EXACTLY how SOD managed to get a fantastic 37,000 Watts from 30,000 Watts with a 0.8 emissivity and produce a high temperature of 133K for the outer sphere."

    That means that everyone, including the poster who thinks that SoD has invented some kind of magical heater, can go to the web page,
    http://www.endmemo.com/physics/radenergy.php
    type in the values 0.8, 133K and the surface area of a 13m radius sphere, and see what the answer really is.

    (Hint: it is not Gord's)

    ReplyDelete
  107. Gord,

    "Yes, I dispute that q is the total emitted Power."
    Then that leaves you at odds with pretty much everyone else. Find a reference that explicitly states that q as I have written it in equation 1. means anything other than the total power emitted.

    Yes the value of q/ε is greater than the value of q. Your 'proof' doesnt prove anything other than that, the value q/ε doesnt mean anything, it isnt the incident, emited, or absorbed anything. What it is is the emitted power divided by ε thats it.
    The SB is about emitted power/flux. Find a reference that explicitly states anything different.

    "The equation you use q = e*BC*T^4*A IS THE SAME EQUATION as q = BC*T^4*A and will only produce a BLACK BODY TEMPERATURE!!"
    No they are not the same equation, if you cant see that see then Im at a lose to help you. One calculates the total emission of a Grey Body the other calculate the total emission of a Black Body. Just plug in the same numbers to both equations and you get a different answer!

    Leaving everything else you posted aside, since the above points form such a major gulf between us anything else said is meaningless.

    BTW, what was your answer for the question posed last time? You never came right out and said it but my guess is that you think its 12.5kW, is that correct? All your talk of incident and absorbed seems a bit confused since in the example there is no input Power(think of it as a sphere that was previously heated and is now slowly cooling as it radiates, no longer in equilibrium).

    A follow up question, what is the total emitted power of the sphere if it is a Black Body?
    (Hint: it has to be different than what you calculate for the original Grey Body)

    So can you please post:
    Emitted power of Grey Body(ε=0.8) is...............
    Emitted power of Black Body is.........................
    Some references showing that q in the equation as I wrote it is not total emitted power as well as references that show the SB is not about emitted power.


    Neutrino

    ReplyDelete
  108. Gord,
    Your comments on the links I posted:

    To say :
    "This one TOTALLY confirms what I have been saying:"
    is a more than a little strange since it starts off with:

    "The Stefan-Boltzmann law, also known as Stefan's law, states that the total energy radiated per unit surface area of a black body in unit time, P is directly proportional to the fourth power of the black body's Thermodynamic temperature T (also called absolute temperature):
    P(T)=εσT^4
    The irradiance P has dimensions of power density (energy per time per square distance), and the SI units of measure are joules per second per square meter, or equivalently, watts per square meter. The SI unit for absolute temperature T is the kelvin. e is the emissivity of the blackbody; if it is a perfect blackbody e = 1. "
    planetphysics.org/encyclopedia/TrivialGroupoid.html

    Its clear that this is saying that P is the total flux which is not what you have been saying.

    To then pick out this comment:
    "The law is valid only for ideal black objects, the perfect radiators, called black bodies."

    To somehow justify your equating εσT^4 with σT^4. Your stating that these are equivalant equations doesnt follow from the quote. The quote is stating that the equation as derived by Boltzmann, σT^4, is for an ideal object known as a BB, that equation has been extended to GB's by adding the ε term, εσT^4. It is not saying that σT^4 = εσT^4.

    Next you want to use this to somehow confirm what you have been saying about GB's:

    "The Stefan–Boltzmann law, also known as Stefan's law, states that the total energy radiated per unit surface area of a black body per unit time (known variously as the black-body irradiance, energy flux density, radiant flux, or the emissive power), j*, is directly proportional to the fourth power of the black body's thermodynamic temperature T (also called absolute temperature):
    j*=σT^4
    A more general case is of a grey body, the one that doesn't absorb or emit the full amount of radiative flux. Instead, it radiates a portion of it, characterized by its emissivity, ε:
    j*=εσT^4"

    en.wikipedia.org/wiki/Stefan%E2%80%93Boltzmann_law

    This one agrees with the first, it establishs that the total emited flux is εσT^4. As to the definition of a GB, its clear and absolutely consistant that a GB only emits a percentage of what a BB emits. A BB emits σT^4 and a GB emits a fraction(ε) of that, εσT^4.

    Finnaly you want this third quote to lock up your view on absorption and emission:

    "Since real objects emit radiation less efficiently than a blackbody at the same temperature (since, by definition, a blackbody is an ideal emitter and absorber of radiation!), we introduce a (unitless) term called the emissivity, cleverly denoted e. The value of the emissivity e ranges from near 0 for an object which is a very poor emitter of radiation (and thus a very poor absorber of radiation, such as a mirror), to a value near 1 for a very good emitter of radiation (very close to a perfect blackbody, for which of course e = 1)."
    frank.mtsu.edu/~phys2020/Lectures/Part_2__L6-L11/Lecture_6/Stefan-Boltzmann/stefan-boltzmann.html

    Not sure what you want this to establish. Yes, emissivity scales both emission and absorption.

    To bad you didnt continue with the quote and get to:

    "We thus write the Stefan-Boltzmann Law in the following form for a real object:
    Real Object Emission: I=E/t*A=εσT^4
    In this equation we have explicitly shown that the total intensity is equal to the total energy radiated at all wavelengths, E , divided by the exposed surface area of the blackbody, A, and the total time over which the emitted radiation is measured, t.
    "

    Once again this confirms that total emission is εσT^4, which is what you dont believe.


    Neutrino

    ReplyDelete
  109. Neutrino,

    HAHAHHA....looks like you can't read and you can't remember YOUR OWN POSTS!

    I'm getting really tired of re-posting everything over and over again.

    You said...
    "Yes, I dispute that q is the total emitted Power."
    Then that leaves you at odds with pretty much everyone else. Find a reference that explicitly states that q as I have written it in equation 1. means anything other than the total power emitted.
    Yes the value of q/ε is greater than the value of q. Your 'proof' doesnt prove anything other than that, the value q/ε doesnt mean anything, it isnt the incident, emited, or absorbed anything. What it is is the emitted power divided by ε thats it."

    Don't you remember your own Post on Nov 23, 2010 at 8:05 AM?

    You said...
    "Professor Barton comes up to you and says here is the equation for the energy absorbed by a body:"
    3. q = Q*e"

    That means q IS THE ABSORBED ENERGY just like you posted! Remember NOW???

    q = Q*e where q = Absorbed Energy, Q = Incident Energy and e = Emissivity
    ----------
    Then you said...

    "Then he goes on to say, here is the equation for emission from a body:
    4. q = e*BC*T^4*A"

    And, q = Q*e as YOU STATED ABOVE and is the ABSORBED ENERGY!

    That's EXACTLY WHAT I SAID in my post to you on Nov 25, 2010 at 11:35 PM
    "q is the ABSORBED power as I have already MATHEMATICALLY PROVEN."

    REMEMBER NOW?
    ---
    So how much is (q/e)?

    q = Q*e so q/e = Q the INCIDENT ENERGY

    q = e*BC*T^4*A" divided by e gives

    q/e = BC*T^4*A"

    q/e = Q

    Q = BC*T^4*A which is the general form of the SB equation that has ALL THE INCIDENT ENERGY being absorbed and emitted by a BLACK BODY and will produce a BLACK BODY TEMPERATURE.

    Like I said in my post to you on Nov 25, 2010 at 11:35 PM:
    "q/e = the TOTAL EMITTED POWER"

    Q = BC*T^4*A IS THE SAME EQUATION as q = e*BC*T^4*A where q = Absorbed Energy

    They BOTH PRODUCE A BLACK BODY TEMPERATURE where ALL THE INCIDENT ENERGY is being absorbed and emitted.
    ----------------
    Your own link that YOU POSTED says the SAME!

    I will REPEAT IT AGAIN!

    Stefan-Boltzmann law
    P(T) = e*BC*T^4

    "The law is valid only for ideal black objects, the perfect radiators, called black bodies."
    http://planetphysics.org/encyclopedia/StefanBoltzamannLaw.html
    --------------
    My God Neutrino, how many times do I have to re-post what YOU have said and re-post your own links???

    You are constantly contradicting yourself to the point of being absolutely delusional.

    Do you suffer from Alzheimers Disease?

    ReplyDelete
  110. Anonymous,

    RE: Your posts Nov 26 at 10:44 AM and 11:57 AM

    I already posted what the calculation on the "nice interactive web page" was:

    q = 0.8*BC*101K^4*2123.7m^2 gives a value of q = 10,024 Watts absorbed.
    ----
    q/e = 10,024/0.8 = 12,530 Watts of Incident Power.

    The general SB equation is q = BC*T^4*A (2) where all the incident power is absorbed by a Black Body.

    q = 5.67X10^-8 * (101)^4 *2123.7 = 12,530 Watts Incident Power.

    The equation you use q = e*BC*T^4*A IS THE SAME EQUATION as q = BC*T^4*A and will only produce a BLACK BODY TEMPERATURE!!

    I have repeated this about a dozen times now.
    ------------
    I posted the Mathematical Proof for this several times.

    If you disagree with the Mathematical Proof, then post where I have made any mistakes and show your calculations.

    We all know that you can babble now let's see if you are just "all foam and no beer".

    I look forward to your next post.

    ReplyDelete
  111. Anonymous - your comment at 8:23 PST has been snipped since it is an ad hom devoid of any scientific content whatsoever. See the comment policy upper right column.

    ReplyDelete
  112. Gord,

    Note:I do need to make one correction to what I have been writting, I should not have used q for both absobed and emitted in my earlier examples, doing that has caused some confusion so from now on I will use q for absorbed and p for emitted.

    I agree this is tedious, continually restating the same thing over and over.
    The reason I am though is because your answers, when you do give them, do not make any logical or mathmatical sense.

    You assert that the SB deals with absorbed rather than emitted. Where is even a single reference that supports this view?
    Then you insist that emitted power is p/ε. Again I ask for a reference that actually states this.

    As for actual sample calculations, you still have not answered them directly. The question was how much power is emitted.

    So,
    Reference that the SB is an equation about absorbed rather than emitted.
    Reference for total emitted power from a GB is anything other than εσT^4*A.
    Answer to how much the sphere radiates, both as a BB and a GB.
    Can you provide me with those, (two links and two numbers thats all)?

    The reason I ask, your answers may help in determining where the misconception is(whether its yours or even maybe mine), but if you do not directly answer them we will just continue on this merry-go-round.


    Neutrino

    ReplyDelete
  113. Ok, Gord is using the SB equation incorrectly.

    The equation gives the TOTAL power emitted by an object. The emissivity is already taken into account. You DO NOT have to multiply or divide or do anything else to the answer afterwards.

    That is all I am going to say, except that a potentially interesting discussion related to this is just starting up at WUWT.

    ReplyDelete
  114. Neutrino,

    You said...
    "I agree this is tedious, continually restating the same thing over and over.
    The reason I am though is because your answers, when you do give them, do not make any logical or mathmatical sense."

    I have already posted the Mathematical Proof several times.

    If you say it does not make any "logical or mathmatical sense" THEN POST WHERE I HAVE MADE ANY MISTAKES AND POST YOUR CALCULATIONS!!!!

    How many times do I have to REPEAT THAT?

    Come on, STOP BABBLING AND DO IT!
    ---------
    You said...
    'Note:I do need to make one correction to what I have been writting, I should not have used q for both absobed and emitted in my earlier examples, doing that has caused some confusion so from now on I will use q for absorbed and p for emitted."
    "You assert that the SB deals with absorbed rather than emitted. Where is even a single reference that supports this view?
    Then you insist that emitted power is p/ε. Again I ask for a reference that actually states this."

    THE LAW OF CONSERVATION OF ENERGY REQUIRES THAT EMITTED ENERGY = ABSORBED ENERGY!!!!

    Once again I find that I have to REPEAT what I have already posted over a dozen times.

    Here is what I posted for you on Nov 25, 2010 at 11:35 PM:

    "First here is the the Physics link that says EMITTED ENERGY = ABSORBED ENERGY, that I have posted at least a DOZEN TIMES!

    Emissivity in the Infrared
    Physics of Emissivity
    "Emitted Energy = Absorbed Energy"
    http://www.optotherm.com/emiss-physics.htm

    EMITTED ENERGY = ABSORBED ENERGY is based on THE LAW OF CONSERVATION OF ENERGY which you obviously do not understand or refuse to accept.

    What is your problem?

    I am getting really tired of you posting delusional babbling that has been shown to be wrong over and over again!

    So either state WHY you don't accept EMITTED ENERGY = ABSORBED ENERGY, post the Physics to back up your delusional claim "emission of radiation and has nothing to do with absorption" or admit that EMITTED ENERGY = ABSORBED ENERGY !!"

    REMEMBER NOW???
    -------------------
    Your own links that YOU HAVE POSTED says the SAME THING!

    I posted this for you on Nov 25, 2010 at 11:37 PM:

    "This one explains absorbed and emitted energy:

    The Stefan–Boltzmann Law

    "Since real objects emit radiation less efficiently than a blackbody at the same temperature (since, by definition, a blackbody is an ideal emitter and absorber of radiation!), we introduce a (unitless) term called the emissivity, cleverly denoted e. The value of the emissivity e ranges from near 0 for an object which is a very poor emitter of radiation (and thus a very poor absorber of radiation, such as a mirror), to a value near 1 for a very good emitter of radiation (very close to a perfect blackbody, for which of course e = 1)."
    http://frank.mtsu.edu/~phys2020/Lectures/Part_2__L6-L11/Lecture_6/Stefan-Boltzmann/stefan-boltzmann.html

    Did you get that, Neutrino?"
    --------------------
    Here, I will repeat it for you a few more times.

    1)EMITTED ENERGY = ABSORBED ENERGY
    2)EMITTED ENERGY = ABSORBED ENERGY
    3)EMITTED ENERGY = ABSORBED ENERGY
    4)EMITTED ENERGY = ABSORBED ENERGY
    5)EMITTED ENERGY = ABSORBED ENERGY

    Start at (1) and read through to (5)
    Repeat as neccessary.

    Do you suffer from Alzheimer's Disease?

    ReplyDelete
  115. Neutrino,

    Read this link at least a dozen times so I don't have to repeat it:
    ----------------------
    Solar Radiation at Earth

    "We need to balance: Total power absorbed = Total power emitted

    Total power absorbed

    Power per unit area x Area of planet facing Sun x Fraction of sunlight absorbed by planet's surface

    1,368 W/m2 x pi x (REarth)2 x (1 - albedo)

    [Note: we use the "Earth as a disk" assumption mentioned earlier to calculate the "area of planet facing Sun".]

    Total power emitted

    Power emitted per unit area x Total surface area of planet

    BC x T^4 x 4pi x (REarth)2

    Where BC is the Stefan-Boltzmann constant, with a value of 5.7 x 10-8 watt / (m2 x K4)

    When we set the power absorbed and the power emitted equations equal to each other, pi and the Earth's radius term cancel out, so we are left with:

    1,368 W/m2 x (1 - albedo) = 4 x BC x T^4

    Or, using Earth's average albedo of 0.31 and solving for T:

    T^4 = [1,368 W/m2 x (0.69)] / 4 X BC

    Which yields T = 254 K ( = -19° Celsius = -3° Fahrenheit)"

    http://www.windows2universe.org/earth/climate/sun_radiation_at_earth.html
    ------------------
    Notice that the equation above:

    1,368 W/m2 x (1 - albedo) = 4 x BC x T^4

    Or, using Earth's average albedo of 0.31 and solving for T:

    T^4 = [1,368 W/m2 x (0.69)] / 4 X BC

    IS EXACTLY THE SAME AS I USED FOR THE SAME CALCULATION

    T^4 = [342 w/m^2 X 0.69]/ 5.67X10^-8

    342 X 0.69 = 235.98 w/m^2 = Absorbed energy

    T = (235.98/5.67X10^-8)^0.25 = 254 K or -19 deg C

    Also, Notice that they said "Total power absorbed = Total power emitted"
    --------
    Here was what said in my post to you on Nov 26, 2010 at 4:47 AM

    "With a reflection of 0.3 (emissivity = 0.7) the calculation for the Solar energy absorbed by the Earth is 342 X 0.7 = 239.4 w/m^2

    T = (5.67X10-8 X 239.4)^0.25 = 254.91 K or -18.09 deg C for the Earth temperature.

    The Earth can only produce a temperature based on the Energy it absorbs."

    The ONLY DIFFERENCE was that I used an emissivity of 0.7 instead of 0.69.
    ------------
    Notice that they did not use Your equation P/A = e*BC*T^4 which is T = ((P/A)/e*BC )^0.25 and would give:

    T = (342/(0.69 x 5.67X10^-8))^0.25 = 305.77 K or 32.77 deg C instead of -19 deg C!
    ------------
    Summary:

    1) The link clearly states "Total power absorbed = Total power emitted"

    2) The link uses exactly the same equation that I used.

    3) Your equation will give THE WRONG TEMPERATURE of +32.77 deg C instead of -19 deg C.

    Start at (1) and read through to (3)

    Read it at least 12 more times so I don't have repeat this post again.

    ReplyDelete
  116. @Gord

    What is really amazing is that you fail to understand how is your approach that violates conservation of energy. You have a 30kW energy source, but the whole sphere emits 24kW. You invoke interference without any plausible reasons, since we are not speaking about point sources or coherent radiation. Let alone the case of an internal rough surface. So in example you accumulate 6kJ/s in the cavity, but you fail to acknowledge that by invoking Gord Interference Theory.

    ReplyDelete
  117. Gord,

    I see we have taken yet another turn on the merry-go-round. You have not responded to anything in the last post.

    As to your 'Proof':
    Two items come to mind that you repeatedly say you have proved.
    A. p/ε is total emitted power
    B. p = εσT^4*A is the same as p = σT^4*A

    Taking only P = σT^4*A as a starting point neither of the above two points hold up.
    1. P = σT^4*A power emitted by a BB
    2. p = εP power emitted by a GB
    3. p = εσT^4*A substitute in for P
    So p is the total power emitted from a GB, your p/ε is the power that the same body would emit if it was a BB.

    Then for your equating tha GB and BB equations.
    1. p = εσT^4*A
    2. P = σT^4*A
    3. p = εP
    4. p ≠ P (Unless ε = 1)
    If p is not equal to P then they are not the same equation.

    These are pretty sloppy proofs but I think get the point accross. What you have done:
    "Proof: q/e = would BE GREATER THAN "total emitted Power" and Violate The Law of Conservation of Energy."
    and:
    "The equation you use q = e*BC*T^4*A IS THE SAME EQUATION as q = BC*T^4*A and will only produce a BLACK BODY TEMPERATURE!!"
    Are not a proofs in any meaningfull sense, they are just statements of your belief,

    I have no problems witht he Law of Conservation of Energy.
    It states Energy can not be created or destroyed. Nowhere have i done either of those two things.

    What you continue to fail to understand is that
    Absorbed = Emitted
    only applies to an equilibrium situation where radiative transfers are the only game in town. If either of those two conditions are not met you can not use that equality.
    There are other methods of input and output, with those the equation becomes:
    Conduction(in) + Convection(in) + Absorbed = Emitted + Convection(out) + Conduction(out)
    So,
    Total In = Total Out
    is what must be conserved while in equilibrium, not just radiation.


    I have tried to answer your questions, could you make an attempt to answer mine?
    Supply some references that explicitly support your claims.
    What is the emission from the sphere?
    (both as it is as a GB and what it would be if it was a BB)
    In case you have forgotten the sphere has the following properties:
    A =2123.7m^2
    T = 100.9K
    ε = 0.8
    there is no incident radiation on it and it is cooling down, slowly radiating away its stored thermal mass. Total In ≠ Total Out, its not in equilibrium.


    I have tried to show you where I believe your conceptual ideas are wrong, do me the favour in return of actual answering my questions.


    Neutrino

    ReplyDelete
  118. Gord,

    quick response to your earth example again.

    Same comments I made last time.

    The planet is in equilibrium and the only exchanges energy through radiation so:
    Absorbed = Emitted
    Is a perfectly acceptable equality.

    The earths emissivity in the visible spectrum is about 0.7, which is what you are using to calculate how much Flux is absorbed, I have no problems with that calculation.

    But..... and huge point here, the earths emissivity in the infrared spectrum is near unity(1)

    Where you go off te tracks is when you use an emissivity of 0.7 to calculate the surface temperature. The emissivity for the earth in the infrared spectrum, which is where it radiates its energy, is 1 not 0.7. The earth is nearly a BB at these wavelengths.

    So doing the calculation with the correct emissivity(1) gets you the standard number. No suprise there.


    I did notice a few things about that link(bold mine):

    "It turns out that there is a mathematical relationship between the temperature of an object and the amount of energy it radiates (the Stefan-Boltzmann law). "
    (note: they are saying the SB is about emission and that its the emission that sets temperature)

    "We need to balance: Total power absorbed = Total power emitted"
    (note: since the only excanges in thier model are radiative this is acceptable)

    "Total power emitted
    Power emitted per unit area * Total surface area of planet
    σ x T^4 * 4π x (Rearth)^2"

    (note: they are treating the earth as a BB for emission which is a good aproximation also saying the total emitted power is given by the SB)

    Nothing on that site contradicts a single thing I have tried to explain to you.

    Still waiting on your answers and a reference supporting your claims..........


    Neutrino

    ReplyDelete
  119. Neutrino,

    I am not going to respond to your posts where I have already posted the same information OVER and OVER again.

    This post will completely prove that your use of the Stefan-Boltzmann Equation is wrong and you are calculating a Black Body temperature.

    RE: Your post on Nov 28, 2010 at 10:30 AM

    You said..
    "The planet is in equilibrium and the only exchanges energy through radiation so:
    Absorbed = Emitted
    Is a perfectly acceptable equality."

    You also said...
    "The earths emissivity in the visible spectrum is about 0.7, which is what you are using to calculate how much Flux is absorbed, I have no problems with that calculation."

    And you said...
    "But..... and huge point here, the earths emissivity in the infrared spectrum is near unity(1)"
    "So doing the calculation with the correct emissivity(1) gets you the standard number. No suprise there."

    So you have agreed that:

    1) -19 deg C is the correct number
    2) Absorbed = Emitted
    3) Use of 0.7 emissivity is OK in the visible spectrum.

    Your only complaint was that the Earth's emissivity = 1 in the Infrared spectrum.
    -----------------
    NO PROBLEM, we don't need to use the Earth's emissivity = 1 in the Infrared spectrum!!!

    We can calculate the total power absorbed by a disk Earth due to Solar radiation using the same parameters as the link I posted with only the 0.69 VISIBLE SPECTRUM Emissivity.

    ALL of which you have already AGREED IS CORRECT!

    1368 w/m^2 = Solar Constant

    Taking the Earth as a disk 1368 w/m^2 /4 = 342 w/m^2 would be absorbed by a Disk Earth....this is the INCIDENT ENERGY.

    Albedo = 0.31 so Emissivity = 0.69

    The total absorbed energy = 342 w/m^ X e = 342 X 0.69 = 235.98 w/m^2....this is amount a GREY BODY ABSORBS.

    Now we can calculate the Earth surface temperature do to Solar Radiation:

    I calculate T = (total absorbed energy / BC )^0.25 = ( 235.98 / 5.67X10^-8 )^0.25 = 253.99 K or -19.01 deg C... THE CORRECT TEMPERATURE!
    --------
    Now calculate the Temperature using YOUR equation:

    T = (total absorbed energy / e*BC )^0.25 = (235.98 / 0.69 * 5.67X10^-8)^0.25 = 278.68 K or 5.68 deg C

    5.68 deg C is the BLACK BODY TEMPERATURE of the Earth with Emissivity = 1 and uses the INCIDENT ENERGY of 342 w/m^2

    PROOF:

    T = (Incident Power / BC )^0.25 = (342 / 5.67X10^-8)^0.25 = 278.68 K or 5.69 deg C !!!!!!!
    ---------------
    There is the absolute PROOF that you have not used the SB equation correctly and will always calculate a BLACK BODY TEMPERATURE.

    ReplyDelete
  120. Giovi Pelle,

    I see that you also cannot read and I have to keep repeating the same Physics Links

    This equation is based on THE LAW OF CONSERVATION OF ENERGY:

    Incident Energy = Absorbed Energy + Transmitted Energy + Reflected Energy
    http://www.optotherm.com/emiss-physics.htm

    Transmitted Energy = 0 Watts

    Incident Energy = Absorbed Energy + Reflected Energy

    Incident Energy = 30,000 Watts THAT'S ALL THE ENERGY AVAILABLE.
    Absorbed Energy = Incident Energy X Emissivity = 30,000 X 0.8 = 24,000 Watts
    Reflected Energy = Incident Energy X Reflectivity = 30,000 Watts X 0.2 = 6,000 Watts.

    REFLECTED ENERGY IS NOT ABSORBED and can only REDUCE THE INCIDENT ENERGY TO COMPLY WITH THE LAW OF CONSERVATION OF ENERGY

    Incident Energy - Reflected Energy = Absorbed Energy.
    30,000 Watts - 6,000 Watts = 24,000 Watts of ABSORBED ENERGY.

    If you want to use Joules the above is:

    Incident Energy - Reflected Energy = Absorbed Energy.
    30,000 Joules/sec - 6,000 Joules/sec = 24,000 Joules/sec of ABSORBED ENERGY.

    Please show me how the 6,000 Joules/sec can increase to a value greater than 6,000 Joules/sec and comply with THE LAW OF CONSERVATION OF ENERGY!
    ---------
    We can use the same equation:
    Incident Energy = Absorbed Energy + Reflected Energy

    To determine the Heat Flux (w/m^2) for all these quantities at the surface of the inner sphere:

    The inner Sphere has a surface area of A = 1256.64 m^2 so the ABSORBED Heat Flux = 24,000 Watts/1256.64 m^2 = 19.10 w/m^2

    The REFLECTED Heat flux at the surface of the inner Sphere has the same area A = 1256.64 m^2 and will have a Heat Flux = 6,000 Watts/1256.64 m^2 = 4.77 w/m^2

    The INCIDENT Heat Flux = 19.1 + 4.77 = 23.87 w/m^2 at the inner sphere Surface.

    All Heat Flux Fields (w/m^2) are VECTOR QUANTITIES.

    Heat flux
    "Heat flux or thermal flux, sometimes also referred to as heat flux density or heat flow rate intensity is a flow of energy per unit of area per unit of time. In SI units, it is measured in [W·m-2]. It has both a direction and a magnitude so it is a vectorial quantity."
    http://en.wikipedia.org/wiki/Heat_flux

    Ihe Incident Heat flux has a Magnitude of 23.87 w/m^2 and a direction of propagation TOWARD the inner sphere surface.

    The Reflected Heat flux has a Magnitude of 4.77 w/m^2 and a direction of propagation OPPOSITE to the Incident Heat Flux.

    The Resultant Heat Flux Vector has a Magnitude of 23.87 - 4.77 = 19.1 w/m^2 and a direction of propagation TOWARD the inner sphere surface.
    There is ZERO w/m^2 propagating in the opposite direction and ZERO energy can be accumulated inside the inner Sphere!

    The Absorbed Heat Flux at the inner Sphere surface is 19.1 w/m^2 and is EXACTLY EQUAL to The Resultant Heat Flux Vector Magnitude of 19.1 w/m^2!

    This totally complies with THE LAW OF CONSERVATION OF ENERGY.

    Please show me how energy can be accumulated inside the inner Sphere without violating THE LAW OF CONSERVATION OF ENERGY and Vector Mathematics!
    ----------
    PS:

    I will not continually repeat my previous posts and Physics links.

    Read before you BABBLE.

    If you just BABBLE without having posting any Physics links to support your "Opinions", I will simply tell you post the supporting Physics before I respond.

    Do you understand that?

    ReplyDelete
  121. Gord,

    And again we go around the merry-go-round.

    Answer the questions put to you.

    A. What is the emitted power form a sphere that has an area of 2123.7m^2, an emissivity of 0.8(our GB) and a temperature of 100.9K?
    B. What is the emitted power from a sphere that has an area of 2123.7m^2, an emissivity of 1(a BB) and a temperature of 100.9K?

    The reason I ask the question is that I think that your answers will be identical. I am not realy sure though, thats why I am asking you directly.
    If your answers are identical you have a serious problem with you conceptual idea of what we are talking about. By definition a GB emits only a portion of the power of a BB, that portion is the term ε. So if your GB doesnt emit less than your BB by exactly the factor ε then you are wrong.
    But since you wont answer the questions I can not be sure if this is were you have the conceptual problem.

    Answer the question.


    Neutrino

    ps: Can we stop with the much more complex Earth model until we agree on the much simpler sphere(or even simpler mirror). And stop misrepresenting what I have said to generate a spurios calculation. The key here is that the earth absorbs in the visible and emits in the infrared(its not hot enough to emit in the visible)
    Using the emisivity from the visible spectrum to calculate the emitted in the infrared spectrum is an obvious error.(SB is about Emission, still waiting on that link that says its not btw.......)

    Known variables(that we both agree on I think):
    1. Q = 342W/m^2
    2. ε(visible) = 0.69
    3 ε(infrared) = 1
    4. q = p (known from equilibrium condition)

    Absorption(in visible):
    4. q = Qε(visible)
    5. q = 342W/m^2*0.69
    6. q = 236W/m^2

    Emission(in infrared):
    7. p = ε(infrared)σT^4
    8. T = (p/(ε(infrared)σ))^.25
    9 . T = (q/(ε(infrared)σ))^.25
    10. T = (236W/m^2/(1*σ))^.25
    11. T = 254K = -19C

    This is the third time I have explained this to you, quite setting up a straw man and answer the questions put to you.

    ReplyDelete
  122. @Gord

    Let's try to look at it from a more fundamental point of view.

    1) Our 30kW source is emitting a photon flux right? Each photon carries a given energy. If we calculate the number of collision per unit time, each collision multiplied for the energy carried by the photon, we get our 30kW
    2) According to your calculations, a given number of photons per unit time (corresponding to 6kW) is reflected.

    Now my question. Will any of the reflected photons be able to strike the inner sphere surface for a second time? I really need you to answer this.

    ReplyDelete
  123. Neutrino,

    I will answer your questions and you can answer mine.

    Emissivity in the Infrared
    Physics of Emissivity
    "Emitted Energy = Absorbed Energy"
    "Incident Energy = Absorbed Energy + Transmitted Energy + Reflected Energy"
    http://www.optotherm.com/emiss-physics.htm

    The above equations are based on The Law of Conservation of Energy.
    ---------
    Your example was stated as:

    "A. What is the emitted power form a sphere that has an area of 2123.7m^2, an emissivity of 0.8(our GB) and a temperature of 100.9K?
    B. What is the emitted power from a sphere that has an area of 2123.7m^2, an emissivity of 1(a BB) and a temperature of 100.9K?"

    My Answer:

    If the Sphere has a temperature of 100.9 K it has absorbed and emitt:

    w/m^2 = BC * T^4

    W/m^2 = (5.67X10^-8 X 100.9^4) = 5.88 w/m^2

    If it's Area = 2123.7m^2 then it has absorbed and will emitt 5.88 w/m^2 X 2123.7 m^2 = 12,487 Watts

    If it's emissivity = 1 then it is a BLACK BODY and it absorbed and will emitt 100% of the 12,487 Watts

    Incident Energy = Absorbed Energy + Transmitted Energy + Reflected Energy

    12,487 Watts Incident Energy = 12,487 Watts Absorbed Energy and (Transmitted Energy + Reflected Energy = 0)

    --------

    If it's emissivity = 0.8 it is a GREY BODY and it absorbed and will emitt 80% of the Incident Energy

    Incident Energy = 12,487/0.8 = 15,609 Watts.

    Incident Energy = Absorbed Energy + Transmitted Energy + Reflected Energy

    15,609 Watts Incident Energy = 12,487 Watts Absorbed Energy + 3,122 Reflected Energy if Transmitted Energy = 0
    -----------------------------------
    Now let's see you do the same calculation for your example.

    Use your SB equation and show all calculations as I have for the Black Body e = 1 and for the Grey Body e = 0.8
    and fill in the amounts for Incident Energy = Absorbed Energy + Transmitted Energy + Reflected Energy for both the Black Body and Grey Body.

    You can assume the Transmitted Energy = 0 as I have done.

    Come on, answer the question.
    -----------------------------
    One more question:

    A Sphere has an emissivity = 0.69
    The Incident Flux = 342 w/m^2

    What will the absorbed Flux be and what will it's temperature be?

    Answer the questions put to you.

    ReplyDelete
  124. Giovi Pelle,

    I have fully replied to your post Nov 28, 2010 at 6:49 AM and included the Physics Links to support my calculations that fully complied with The Law of Conservation of Energy

    I also asked you these simple questions:

    1)Please show me how the 6,000 Joules/sec can increase to a value greater than 6,000 Joules/sec and comply with THE LAW OF CONSERVATION OF ENERGY!

    2)Please show me how energy can be accumulated inside the inner Sphere without violating THE LAW OF CONSERVATION OF ENERGY and Vector Mathematics!

    Answer the questions and we can continue with your next question.
    ---------
    Keep in mind what I told you before as it applies to your next question.

    "I will not continually repeat my previous posts and Physics links.

    Read before you BABBLE.

    If you just BABBLE without having posting any Physics links to support your "Opinions", I will simply tell you post the supporting Physics before I respond.

    Do you understand that?"

    ------------
    I have already posted about Photons, EM fields and the 6KW reflected beginning on Nov 12, 2010 at 5:38 AM and 5:40 AM (READ BEFORE YOU BABBLE).

    And, POST the supporting Physics for your "Opinions".

    ReplyDelete
  125. 1) "Please show me how the 6,000 Joules/sec can increase to a value greater than 6,000 Joules/sec and comply with THE LAW OF CONSERVATION OF ENERGY!"

    6kJ/s stay 6kJ/s. Please notice that's a POWER, not ENERGY . So after one second the accumulated ENERGY 6kJ/s*1/s=6kJ in the cavity. After 10 seconds the accumulated ENERGY is 6kJ/s*10s=60kJ. After 1000s the accumulated energy is 6kJ/s*1000s=6MJ. So after all energy is accumulated in the cavity, since reflected energy is going to stay there forever following your approach.

    2)"Please show me how energy can be accumulated inside the inner Sphere without violating THE LAW OF CONSERVATION OF ENERGY and Vector Mathematics!"

    It can't. Problem is that, as I have shown in point 1, your approach dictates that after time t I have accumulated energy E=6kJ/s*t inside the cavity. So, it is your approach that violates conservation of energy

    3) Concerning the references: "http://www.amazon.com/Classical-Electrodiynamics-Third-David-Jackson/dp/047130932X" I suggest you to begin from page 1.

    4) Now my question. Will any of the reflected photons be able to strike the inner sphere surface for a second time? [yes/no]

    ReplyDelete
  126. Gord,

    Thanks for trying to answer the question, but you seem to have not understood the question.

    There is no incident power anywhere in this example.

    It is SoD's sphere with the power input turned off. I simply chose a temperature that it must pass through while cooling back down. Since 100.9K is less than both of yours and SoD's calculated steady state surface temperature it will at one point be at this temperature regardless of who is correct.

    There is no absorbed only emitted. So stating any value whatsoever for absorbed is simply wrong. The sphere is not in equilibrium so you can not equate absoption and emission. Your appeal to Conservation of Energy is missguided in this scenerio. Yes the sphere can not create energy but it did store a large amount of it in the time it took to equilibriate while the power source was on. It is that stored energy that is slowly being radiated away based solely on its current surface temperature.


    My answers since you asked:

    Emission from the BB
    1. p = εσT^4*A
    2. p = 1*σ*100.9^4*2123.7m^2
    3. p = 12.5kW

    Emission from the GB
    4. p = εσT^4*A
    5. p = 0.8*σ*100.9^4*2123.7m^2
    6. p = 10.0kW

    Incident = Absorbed + Transmitted + Reflected
    7. 0 = 0 + 0 + 0

    There is no incident power. The sphere is losing energy.
    Redo you calculations and actually tell me how much a sphere of the given temperature and emissivities radiates.


    So I ask again, what do the two spheres radiate?

    As you still havent answered my question I will hold off on yours till you do.


    Neutrino

    ReplyDelete
  127. Neutrino,

    Ok, I will answer your questions and then you can do the examples I posted for you.

    The stored heat energy that was absorbed from the the original energy source that caused Each body have a temperature of 100.9 K is Potential Energy.

    Energy Into a Body = Energy Out of a Body.

    Where the Potential Energy is the Energy into the Body (stored Energy) = Kinetic Energy, the Emitted Energy out of the body.

    The absorbed stored energy in Each body is the same because their Temperatures are the same and will be equal to the Emitted Energy.

    Q = BC*T^4*A

    Q = (5.67 X 10^-8 * 100.9^4 * 2123.7 ) = 12,487 Watts Emitted for Each Sphere

    Emitted Energy will decline as the absorbed stored energy and Temperature declines, and can be calculated using Q = BC*T^4*A.

    And this fully complies with "Emitted Energy = Absorbed Energy" that is based on The Law of Conservation of Energy:

    Emissivity in the Infrared
    Physics of Emissivity
    "Emitted Energy = Absorbed Energy"
    http://www.optotherm.com/emiss-physics.htm
    ------------------------
    Now do the examples I asked you to do in my post Nov 29, 2010 at 1:49 PM.

    ReplyDelete
  128. Giovi Pelle,

    Did you read my Posts?

    I posted three Physics Links:

    This equation on the Law of Conservation of Energy:
    Incident Energy = Absorbed Energy + Transmitted Energy + Reflected Energy"
    http://www.optotherm.com/emiss-physics.htm

    Heat flux
    "Heat flux or thermal flux, sometimes also referred to as heat flux density or heat flow rate intensity is a flow of energy per unit of area per unit of time. In SI units, it is measured in [W·m-2]. It has both a direction and a magnitude so it is a vectorial quantity."
    http://en.wikipedia.org/wiki/Heat_flux

    Heat Transfer by Radiation
    http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/stefan.html#c2

    I fully explained why the energy does not accumulate inside the Sphere.

    I also provided this actual Physical Example that has been demonstated countless times as support.

    Cancellation of Light
    http://www.exploratorium.edu/ronh/bubbles/bubble_colors.html

    --------------------
    I'm not interested in your "opinion" so instead of BABBLING "So, it is your approach that violates conservation of energy", either show where my calculations are wrong using the Physics and Physical Example I have posted or post the Physics and links you are using to support your "opinion".

    An actual Physical Example would be nice too.

    Do that or don't waste my time.

    ReplyDelete
  129. Gord,

    Thank you for answering my question, it is very helpfull as it clearly identifies where our conceptual differences lie.

    By definition, a GB is something that only emits a portion of what a BB radiates at a given temperature. That portion is the emissivity.
    That you dont agree with this is clear from your answers and your previous posts, but that is simply just your opinion.

    Below are three quotes from universites describing the SB and how it applies to GB's.
    Each of these quotes clearly state that the emission from a GB is scaled by the emissivity.

    "The Stefan-Boltzmann Law
    The second important law associated with the blackbody emission spectrum and which can be derived from Planck’s equation (6.1) using calculus is the Stefan-Boltzmann Law, which tells us the total intensity of radiation emitted by the surface of the blackbody at all wavelengths, Itotal:

    Ideal Blackbody Emission: I = E/tA = σT^4 (6.6)

    ... Since real objects emit radiation less efficiently than a blackbody at the same temperature (since, by definition, a blackbody is an ideal emitter and absorber of radiation!), we introduce a (unitless) term called the emissivity, cleverly denoted ε. ...
    We thus write the Stefan-Boltzmann Law in the following form for a real object:

    Real Object Emission: I = E/tA = εσT^4 (6.7)
    "
    frank.mtsu.edu/~phys2020/Lectures/Part_2__L6-L11/Lecture_6/Stefan-Boltzmann/stefan-boltzmann.html


    "Stefan-Boltzmann Law
    The energy radiated by a blackbody radiator per second per unit area is proportional to the fourth power of the absolute temperature and is given by

    P/A = σT^4 j/m^2s Stefan-Boltzmann Law
    σ = 5.6703 x 10^-8watt/m^2K^4
    For hot objects other than ideal radiators, the law is expressed in the form:

    P/A = εσT^4
    "
    hyperphysics.phy-astr.gsu.edu/hbase/thermo/stefan.html


    "II. The Stefan-Boltzmann Law
    The total radiated power per unit area of a cavity aperture, summed over all wavelengths, is called its radiant intensity, I(T). It depends on the temperature T, has units of Watt/m2, and is given by a simple formula, given below, .... We note, however, that ordinary hot objects always radiate less efficiently than do cavity radiators; therefore, we associate with them an additional, positive quantity, ε, referred to as the emissivity, which is always less than 1, and depends on the material, the surface, and (usually) the temperature as well. ...

    for a blackbody, or cavity aperture: I(T) = σT^4, σ = 5.670*10^-8W/m^2K^4,(2.1)

    for a real material: I(T) = εσT^4 (2.2)
    "
    panda.unm.edu/Courses/Finley/P262/ThermalRad/ThermalRad.html

    Two of the above links I have already provided for you, did you not read them when I first posted them? One you have even used yourself, did you not read it before you used it?

    There really is no way to read the above and not come to the conclusion that emissivity scales emission.

    So, the ball is in your court. Provide some reason to reject the standard use of the SB as expressed in the above quotes, refrences refuting the above would be a start, or change your conceptual idea of a GB.


    Neutrino

    ReplyDelete
  130. Oh, as for your question.

    Simple answer. 5.5C.
    Given that emissivity scales both absorption and emission there is no suprise at this value really.

    Besides what temperature would you think an uniform object would equilibriate to when placed in a flux that is equivlent to 5.5C?


    Neutrino

    ReplyDelete
  131. "I fully explained why the energy does not accumulate inside the Sphere."

    Sorry, but I didn't get it. And i am not alone here. Would you care to explain one more time?

    ReplyDelete
  132. Neutrino,

    You forgot to do these calculations that I requested in my Nov 29,2010 at 1:49 PM, where the Incident power is not zero.

    "Now let's see you do the same calculation for your example.

    Use your SB equation and show all calculations as I have for the Black Body e = 1 and for the Grey Body e = 0.8
    and fill in the amounts for Incident Energy = Absorbed Energy + Transmitted Energy + Reflected Energy for both the Black Body and Grey Body.

    You can assume the Transmitted Energy = 0 as I have done."
    ----------
    Then we can discuss the SB equation some more.

    ReplyDelete
  133. Furthermore, I want to satisfy your thirst of examples, concerning your fantastic cancellation of light:

    As you will remember, I stated that if the bubble gets black where is very thin, it means that we have destructive interference in reflection, and near total transmission of light on the other side. You can do the calculation for yourself here:

    http://luxpop.com/

    The steps are:

    1) Go at "Light at interface" and read the instructions
    2) Setup a very simple thin film stack
    i)incident angle theta=70 degrees (intereference effect are better seen)
    ii) centre_lamba=500nm
    iii) sweep +\-300nm (visible range)
    iv) 100 sweep points
    v) index of incident material=(1,0) air
    vi)Film stack
    Thick Bubble (400nm,n=(1.33,0))=400.0,1.33,0
    vii)substrate materia n=(1,0) again air
    viii) perform the calculation

    Recover the data, and plot the 3rd(reflectance) and 9th(transmittance) columns againts the wavelenght. You can clearly see the interference effect, for instance constructive interference for yellow-red reflected light (550nm) and destructive interference for blue reflected light. Of course, if you look at the transmitted intensity, you see that when you have destructive interference in reflection, you have constructive interference in trasmission. Now try the calculation with a thin bubble (10nm). You will se that you have essentially no reflection (black) and near 100% transimission. Exactly as i said. Do the calculations for yourself.

    ReplyDelete
  134. Gord,

    I didnt not realize that was a question directed to me,

    To answer the question we would need to know two more things. One, that the sphere is in equilibrium and two, that the incident is from the outside of the sphere(just for simplicity, could do it as internal but then we would just have the original problem again)

    So assuming those two points here is my answers

    A. BB
    1. q = Qε
    2. p = εσT^4*A
    3. p = q
    4. ε = 1

    5. Qε = εσT^4*A
    6. Q = σT^4*A
    7. Q = σ*100.9^4*2123.7m^2
    8. Q = 12.5kW
    9. q = 12.5kW*1 = 12.5kW

    So for BB,
    Incident = 12.5kW
    Absorbed = 12.5kW
    Transmitted = 0W (PVC is opaque)
    Reflected = 0W

    B. GB
    10. q = Qε
    11. p = εσT^4*A
    12. p = q
    13. ε = 0.8

    14. Qε = εσT^4*A
    15. Q = σT^4*A
    16. Q = σ*100.9^4*2123.7m^2
    17. Q = 12.5kW
    18. q = 12.5kW*0.8 = 10.0kW

    So for GB,
    Incident = 12.5kW
    Absorbed = 10.0W
    Transmitted = 0W (PVC is opaque)
    Reflected = 2.5kW

    No suprise here, the GB only absorbs and emits a portion of what the BB does, just as what you would expect from the definition of a GB
    "A more general case is of a grey body, the one that doesn't absorb or emit the full amount of radiative flux. Instead, it radiates a portion of it, characterized by its emissivity, ε:"
    en.wikipedia.org/wiki/Stefan%E2%80%93Boltzmann_law

    I know you do not think that p = εσT^4*A is the SB equation, but as I have been trying to show you it actually is.
    Produce a reference that says otherwise, produce a proof yourself, show where the links I have posted are wrong or accept you were wrong, correct your mistake and move on.


    Neutrino

    ReplyDelete
  135. Giovi Pelle,

    It's called Vector addition of Flux

    Heat flux
    "Heat flux or thermal flux, sometimes also referred to as heat flux density or heat flow rate intensity is a flow of energy per unit of area per unit of time. In SI units, it is measured in [W·m-2]. It has both a direction and a magnitude so it is a vectorial quantity."
    http://en.wikipedia.org/wiki/Heat_flux

    100% Incident Flux --->>>> <<<<----30% Reflected Flux "Equals" 70% Resultant Flux------>>>> Absorbed by the Inner Sphere Surface

    The resultant Vector has a Magnitude of 70% Flux and has a Direction of Propagation TOWARDS the the Inner Sphere Surface

    Incident Energy = Absorbed Energy + Transmitted Energy + Reflected Energy"
    http://www.optotherm.com/emiss-physics.htm

    100% Incident Energy - 30% Reflected Energy = 70% Absorbed Energy by the Inner Sphere (when Transmitted Energy is zero)

    30,000 Watts Incident - 6,000 Watts Reflected = 24,000 Watts Absorbed

    24,000 Watts Inside the Sphere = 24,000 Watts Absorbed by the Inner Sphere Surface.

    There is no build up of Energy inside the Sphere.

    ReplyDelete
  136. Giovi Pelle,

    I remember you babbling:

    "If the film becomes black, it means we have no reflection because all the reflected light undergoes destructive interferences so, where does the light go?
    i) The light magically vanishes. So the energy and we are violating the conservation of energy.
    ii) 0% reflections means 100% transmission.
    Are you willing to affirm that interference destroys energy? If yes, there is not much more to discuss."

    What a HOOT!

    The Physics has not changed.

    Incident Energy = Absorbed Energy + Transmitted Energy + Reflected Energy"
    http://www.optotherm.com/emiss-physics.htm

    100% Reflection still produces:

    Incident Energy = Reflected Energy

    That means Absorbed Energy = 0 and Transmitted Energy = 0

    Cancellation of Light

    "When two light waves cancel each other, the result is darkness and this is called "destructive interference."
    http://www.exploratorium.edu/ronh/bubbles/bubble_colors.html

    ReplyDelete
  137. "It's called Vector addition of Flux

    Heat flux
    "Heat flux or thermal flux, sometimes also referred to as heat flux density or heat flow rate intensity is a flow of energy per unit of area per unit of time. In SI units, it is measured in [W·m-2]. It has both a direction and a magnitude so it is a vectorial quantity."
    http://en.wikipedia.org/wiki/Heat_flux

    100% Incident Flux --->>>> <<<<----30% Reflected Flux "Equals" 70% Resultant Flux------>>>> Absorbed by the Inner Sphere Surface

    The resultant Vector has a Magnitude of 70% Flux and has a Direction of Propagation TOWARDS the the Inner Sphere Surface

    Incident Energy = Absorbed Energy + Transmitted Energy + Reflected Energy"
    http://www.optotherm.com/emiss-physics.htm

    100% Incident Energy - 30% Reflected Energy = 70% Absorbed Energy by the Inner Sphere (when Transmitted Energy is zero)

    30,000 Watts Incident - 6,000 Watts Reflected = 24,000 Watts Absorbed

    24,000 Watts Inside the Sphere = 24,000 Watts Absorbed by the Inner Sphere Surface.
    "


    Impressive...just my last curiosity, since we are speaking about photons: does a reflected photon gets a chance to strike the inner surface more than once?[yes]/[no]

    ReplyDelete
  138. Neutrino,

    HAHAHA....THANK YOU NEUTRINO!

    You still don't realize that what you just calculated PROVES that I am Right.
    ------------
    Are the Temperatures the same for the Black Body and Grey Body .....YES THEY ARE (100.9 K) !

    What value for q = e*BC*T^4 at step (18) did you get? 10 KW !!

    And 10 KW is the ABSORBED Energy !

    (q = ABSORBED ENERGY just like said!)

    What value did you get for q/e ? 12 KW!

    And 12 KW is the INCIDENT ENERGY !

    (q/e = INCIDENT ENERGY just like I said!)

    "So for GB,
    Incident = 12.5kW
    Absorbed = 10.0W
    Transmitted = 0W (PVC is opaque)
    Reflected = 2.5kW "

    EXACTLY THE SAME AS THE BLACK BODY INCIDENT ENERGY!

    "So for BB,
    Incident = 12.5kW
    Absorbed = 12.5kW
    Transmitted = 0W (PVC is opaque)
    Reflected = 0W"

    THATS WHY BOTH EQUATIONS PRODUCED THE SAME BLACK BODY TEMPERATURE !
    -----------------------
    -----------------------
    This should make it CRYSTAL CLEAR.

    Remember when I asked you to calculate this:

    "One more question:

    A Sphere has an emissivity = 0.69
    The Incident Flux = 342 w/m^2

    What will the absorbed Flux be and what will it's temperature be?"
    ---------
    Your reply was:

    Oh, as for your question.

    Simple answer. 5.5C.
    Given that emissivity scales both absorption and emission there is no suprise at this value really.

    Besides what temperature would you think an uniform object would equilibriate to when placed in a flux that is equivlent to 5.5C?"
    ---
    You didn't show your calculation but here is what you did:

    q = e*BC*T^4 so T = (q / e*BC)^0.25

    First you calculated the Absorbed Energy = Incident Energy X Emissivity = Q * e = 342 X 0.8 = 273.6 w/m^2 = q in your equation.

    T = (273.6 / 0.8 x 5.67X10^-8)^0.25 = 278.68 K or 5.53 deg C
    ---
    Now what is the BLACK BODY TEMPERATURE when e = 1 ??

    Q = BC*T^4 where Q = Incident Energy

    T = ( Incident Energy / BC)^0.25 = (342 X 5.67X10^-8)^ = 278.68 K or 5.53 deg C!!!
    -----------------------------------------------
    -----------------------------------------------
    Here is the simple math you have been missing

    Black Body Equation is:

    Q = BC*T^4 where Q = Incident Energy

    Your equation is:

    q = e*BC*T^4 where q = Absorbed Energy

    Q*e = Absorbed Energy

    Q*e = e*BC*T^4

    Q = BC*T^4 the Black Body Equation!!

    YOUR EQUATION AND THE BLACK BODY EQUATION ARE THE SAME AND WILL ALWAYS PRODUCE A BLACK BODY TEMPERATURE!

    ReplyDelete
  139. Gord,

    There is a huge conceptual difference between how you and I apply the SB. The point of asking you the questions was to reveal that difference, So the fact we arrive at a different answer is no suprise.
    Only one of us can be correct though.

    I believe that the SB is about emission, that it has a form of p = εσT^4*A where p is the emitted power. Ive posted links that explicity confirm this.
    You believe that the SB is about absorption and that it has a form q/ε = σT^4*A where q/ε is the absorbed power.(or something like that im not quite sure) You have posted nothing other than you own belief to confirm your opinion.

    Can you respond to why you think the references Ive provided are wrong, or that I have interpreted them incorrectly?
    Can you show me one explicit reference that the SB is not calculating the emitted power from the temperature?


    Neutrino

    ReplyDelete
  140. Neutrino,

    You said...
    "There is a huge conceptual difference between how you and I apply the SB. The point of asking you the questions was to reveal that difference, So the fact we arrive at a different answer is no suprise.
    Only one of us can be correct though."

    Here is the simple math you have been missing

    Black Body Equation is:

    Q = BC*T^4 where Q = Incident Energy

    Your equation is:

    q = e*BC*T^4 where q = Absorbed Energy

    Q*e = Absorbed Energy

    Q*e = e*BC*T^4

    Q = BC*T^4 the Black Body Equation!!

    YOUR EQUATION AND THE BLACK BODY EQUATION ARE THE SAME AND WILL ALWAYS PRODUCE A BLACK BODY TEMPERATURE!

    As long as you keep using q = e*BC*T^4 where q = Absorbed Energy and keep on using Q*e = Absorbed Energy you will always be calculating Q = BC*T^4 the Black Body Equation!!

    And, the temperature T will ALWAYS be the BLACK BODY temperature NOT the GREY BODY temperature.

    That's why EVERY calculation you made including the Mirror temperatures that had different emissivities all produced the same temperature, the Black Body temperature.

    That's why your use of the SB is obviously wrong as I have proven so many times.
    --------------------
    Your rants about " q/e is the absorbed power.(or something like that im not quite sure) You have posted nothing other than you own belief to confirm your opinion." is just BS and you KNOW IT.

    I have ALWAYS said that q = e*BC*T^4 where q = Absorbed Energy and that q/e = Q the INCIDENT ENERGY!

    And, It is PROVEN in the Math above:

    Q*e = Absorbed Energy and q = Absorbed Energy so q/e = Q the INCIDENT ENERGY.

    Your own calculations confirmed that when you calculated the GB having a Incident Power = 12.5 KW, Absorbed Power = 10 KW when the Emissivity e = 0.8.

    q = Absorbed Power = 10 KW divided by the Emissivity e = 0.8 = 12.5 KW the INCIDENT POWER

    So cut the BS.

    Further, The Law of Conservation of Energy requires that Absorbed Energy = Emitted Energy so your babbling about SB being only about emitted power is also just BS and you KNOW IT.
    ----------
    You asked...
    "Can you respond to why you think the references Ive provided are wrong, or that I have interpreted them incorrectly?
    Can you show me one explicit reference that the SB is not calculating the emitted power from the temperature?"

    All you have been doing all along is to calculate the Absorbed Energy = Q*e = q, putting it into the equation q = e*BC*T^4 where q = Absorbed Energy and get the standard Black Body result of Q = BC*T^4.

    All the references posted use q = e*BC*T^4 and that is fine as long as one understands that q = Absorbed Energy and the temperature calculated is ALWAYS the Black Body temperature.

    q = e*BC*T^4 cannot be used to calculate the temperature of a Grey Body.

    The calculation of the Earth that has an e = 0.69 and 342 w/m^2 of Incident Solar Energy is a prime example.

    Using q = e*BC*T^4 where q = Aborbed Energy = 342 X 0.69 = 235.98 w/m^2 will always produce a Black Body temperature for the Earth of 278.68 K or 5.53 deg C.

    The correct method is to use the Absorbed Energy = 235.98 w/m^2 and use that in the standard SB equation Q = BC*T^4 where Q = 235.98 w/m^2 to get the correct Earth temperature of 253.99 K or -19.16 deg C.

    If you want to use the Stefan-Boltzmann equation improperly and deny that The Law of Conservation of Energy requires Absorbed Energy = Emitted Energy, then go ahead.

    Just don't expect me to accept your "delusions" that I have proven to be wrong every single time.

    ReplyDelete
  141. Gord,

    You said:
    "q = e*BC*T^4 cannot be used to calculate the temperature of a Grey Body."
    (note: that q should be a p, SB is about emission)

    That IS the SB equation!!! This is your conceptual problem. Where on earth did you get the idea the SB cant calculate temperature form emission(or emission from temperature)? Provide a reference for your assertion.
    I assume you didnt derive the SB from first principles so I have to also assume your knowledge and use of it are from literiture. What reference led you to your conclusions?


    Neutrino

    ReplyDelete
  142. Gord,

    Another piece of evidence you dont know what you are talking about.
    I'm sure you recognize this link:
    www.optotherm.com/emiss-what.htm

    Its the from the site you keep quoting for Conservation of Energy, I'm wondering if you actually have read the links you are using.

    "What is Emissivity
    All objects and materials do not radiate infrared (thermal) energy equally. Emissivity is a term describing the efficiency with which a material radiates infrared energy. A blackbody has an emissivity of 1.00 and no other material can radiate more thermal energy at a given temperature. An object with an emissivity of 0 emits no infrared energy. Real-world objects have emissivity values between 0 and 1.00. The lower emissivity of most real-world materials reduces the intensity of radiation from the theoretical predictions of Planck’s Law.
    The temperature of an object and its emissivity define how much infrared energy an object will emit. The figure below shows that quartz emits less energy than a blackbody at the same temperature and therefore has an emissivity below 1.00.
    "

    It even has a nice graph that clearly shows the radiation curve of a BB and of a real world substance, with the substance(Quartz) radiating less power than the theoretical BB. Clearly Emissivity scales the power radiated. Even your links support this, why dont you believe it?


    Anyways, I am still waiting on your references that actually support your beliefs....


    Neutrino

    ReplyDelete
  143. The reason I posted the above was to show that there is good evidence to believe emissivity scales emission from a GB.

    My and Gord's analysis of two objects, same temperature and size, only differing in emissivity:

    From my post:November 29, 2010 3:26 PM (my bold)
    "Emission from the BB
    1. p = εσT^4*A
    2. p = 1*σ*100.9^4*2123.7m^2
    3. p = 12.5kW

    Emission from the GB
    4. p = εσT^4*A
    5. p = 0.8*σ*100.9^4*2123.7m^2
    6. p = 10.0kW
    "


    From Gord's:November 29, 2010 10:01 PM (my bold)
    "The absorbed stored energy in Each body is the same because their Temperatures are the same and will be equal to the Emitted Energy.
    Q = BC*T^4*A
    Q = (5.67 X 10^-8 * 100.9^4 * 2123.7 ) = 12,487 Watts Emitted for Each Sphere
    "

    Remember, the quote is from Gord's source not mine, so I would wager a guess that he thinks the site is factually correct or he wouldnt be using it as a reference..

    So we have two options,
    1. The quote is correct and my analysis is in correct.(GB emits less than BB)
    Or
    2. The quote is wrong and Gord's analysis is correct.(GB emits the same as BB)

    Come on Gord, which is it? Your analysis or the quote?


    Neutrino

    ReplyDelete
  144. Neutrino,
    ------------
    Physics of Emissivity

    Incident Energy = Absorbed Energy + Transmitted Energy + Reflected Energy (1)

    Emitted Energy = Absorbed Energy
    Consider equation 1 for an object in a vacuum at a constant temperature. Because it is in a vacuum, there are no other sources of energy input to the object or output from the object. The absorbed energy by the object increases its thermal energy - the transmitted and reflected energy does not. In order for the temperature of the object to remain constant, the object must radiate the same amount of energy as it absorbs.

    Emitted Energy = Absorbed Energy (2)

    "Therefore, objects that are good absorbers are good emitters and objects that are poor absorbers are poor emitters. Applying equation 2, Equation 1 can be restated as follows:

    Incident Energy = Emitted Energy + Transmitted Energy + Reflected Energy (3)

    Setting the incident energy equal to 100%, the equation 3 becomes:

    100% = %Emitted Energy + %Transmitted Energy + %Reflected Energy [4]

    Because emissivity equals the efficiency with which a material radiates energy, equation 4 can be restated as follows:

    100% = Emissivity + %Transmitted Energy + %Reflected Energy (6)

    http://www.optotherm.com/emiss-physics.htm

    Incident Energy = Absorbed Energy + Transmitted Energy + Reflected Energy (1)

    100% = %Absorbed Energy + %Transmitted Energy + %Reflected Energy

    Because Absorptivity equals the efficiency with which a material absorbs energy,

    100% = Absorptivity + %Transmitted Energy + %Reflected Energy

    Equating (6) and the above gives:

    Emissivity = Absorptivity

    Complying with Emitted Energy = Absorbed Energy (2)

    PS:

    You have used Q*e = Absorbed Energy many, many times so cut the BS.

    ReplyDelete
  145. Neutrino,

    You said...
    "That IS the SB equation!!! This is your conceptual problem. Where on earth did you get the idea the SB cant calculate temperature form emission(or emission from temperature)? Provide a reference for your assertion.
    I assume you didnt derive the SB from first principles so I have to also assume your knowledge and use of it are from literiture. What reference led you to your conclusions?"

    My God Neutrino, as I have REPEATELY stated and YOU HAVE VERIFIED in your own calculations:

    q = e*BC*T^4 where q = ABSORBED ENERGY and T is the BLACK BODY TEMPERATURE!

    I have already provided an example where YOU AGREED that the correct Earth temperature is -19 deg C but q = e*BC*T^4 ALWAYS produces the BLACK BODY Earth temperature of +5.53 deg C.

    What is wrong with your Brain and why do you continue to post BS that completely contradicts what YOU HAVE ALREADY CALCULATED?
    ------------
    Absorbed Energy = Emitted Energy and is a requirement of The Law of Conservation of Energy.

    Emissivity is the same as Absorptivity.

    That's why you calculated Q*e = Absorbed Energy many many, times, so stop contradicting yourself.

    This reference completely verifies the FACTS that I have posted and YOUR OWN CALCULATIONS verify.

    Stefan-Boltzmann law
    P(T) = e*BC*T^4

    "The law is valid only for ideal black objects, the perfect radiators, called black bodies."
    http://planetphysics.org/encyclopedia/StefanBoltzamannLaw.html

    So stop the BS.

    ReplyDelete
  146. Neutrino,

    You said...
    "My and Gord's analysis of two objects, same temperature and size, only differing in emissivity:"

    BS!

    Here is what I ACTUALLY CALCULATED:

    "If the Sphere has a temperature of 100.9 K it has absorbed and emitt:

    w/m^2 = BC * T^4

    W/m^2 = (5.67X10^-8 X 100.9^4) = 5.88 w/m^2

    If it's Area = 2123.7m^2 then it has absorbed and will emitt 5.88 w/m^2 X 2123.7 m^2 = 12,487 Watts

    If it's emissivity = 1 then it is a BLACK BODY and it absorbed and will emitt 100% of the 12,487 Watts

    Incident Energy = Absorbed Energy + Transmitted Energy + Reflected Energy

    12,487 Watts Incident Energy = 12,487 Watts Absorbed Energy and (Transmitted Energy + Reflected Energy = 0)

    ---

    If it's emissivity = 0.8 it is a GREY BODY and it absorbed and will emitt 80% of the Incident Energy

    Incident Energy = 12,487/0.8 = 15,609 Watts.

    Incident Energy = Absorbed Energy + Transmitted Energy + Reflected Energy

    15,609 Watts Incident Energy = 12,487 Watts Absorbed Energy + 3,122 Reflected Energy if Transmitted Energy = 0"

    I CALCULATED A HIGHER INCIDENT ENERGY OF 15,609 WATTS FOR THE GREY BODY ABSORBING AND EMITTING 12,487 WATTS RESULTING IN THE SAME 100.9 K TEMPERATURE.
    --------------
    You said...

    "From Gord's:November 29, 2010 10:01 PM (my bold)
    "The absorbed stored energy in Each body is the same because their Temperatures are the same and will be equal to the Emitted Energy.
    Q = BC*T^4*A
    Q = (5.67 X 10^-8 * 100.9^4 * 2123.7 ) = 12,487 Watts Emitted for Each Sphere"

    Remember, the quote is from Gord's source not mine, so I would wager a guess that he thinks the site is factually correct or he wouldnt be using it as a reference.."

    I CALCULATED A HIGHER INCIDENT ENERGY OF 15,609 WATTS FOR THE GREY BODY ABSORBING AND EMITTING 12,487 WATTS RESULTING IN THE SAME 100.9 K TEMPERATURE.

    SO CUT THE BS!
    -------------------
    Either you are suffering from Alzheimers Disease and can't remember what I ACTUALLY POSTED or you are intentionally posting this BS which is LYING.

    So we have two options,

    Which is it?

    ReplyDelete
  147. Gord,

    Im truely confused by your posts,

    From:December 1, 2010 9:17 PM
    "PS:
    You have used Q*e = Absorbed Energy many, many times so cut the BS. "


    Ya I have, and nothing in my last series of posts dealt with that, why did you bring it up?

    From:December 1, 2010 9:22 PM
    "I have already provided an example where YOU AGREED that the correct Earth temperature is -19 deg C but q = e*BC*T^4 ALWAYS produces the BLACK BODY Earth temperature of +5.53 deg C."

    Fourth time Ive had to state this, the earth has an Emissivity of 0.69 in the visible and 1.0 in the infrared. When I agreed that -19C was correct it was for an Earth type sphere, when I stated a temperature of 5.5C we were talking about a sphere of uniform emissivity at 0.69. Not the Earth!

    As for the quote, "The law is valid only for ideal black objects, the perfect radiators, called black bodies."
    Yes the SB as derived (P = σT^4*A) is only for a BB, it is then extended for all GB by adding the term ε(p = εσT^4*A). Where is the BS in that?

    From:December 1, 2010 9:57 PM

    What you actually calculated was both.
    And besides that doesnt even matter, because in both scenarios you came to the conclusion a 100.9K object of 2123.7m^2 is radiating 12.5kW, exactly the same as you calculated for a BB for both scenarios.
    That is the point I was making. The fact you calculated a higher Incident is irrelevant since I was pointing out the emitted.

    The reference I provided dealt with emitted, not absorbed or incident, so I compared it to your calculated emitted. You calculated that a 100.9K object radiated 12.5kW, in both scenarios, regardless of its emissivity.
    Thats contrary to the quote i provided, from your reference no less.

    So, are you wrong or is your reference?


    So to not have any confusion about what we did or didnt say:
    Do you think a object with emissivity of 0.8 radiates less than an object of emissivity 1.0 when they both have the same temperature and dimensions?


    Neutrino

    ReplyDelete
  148. I once asked a question very early in this run around thread.But no answer.

    Could it be because they know that such a "Back-radiation heater" does not exist and can not exist because back-radiation emissions does not create new energy?

    How about this one:

    Why not fill the walls of buildings with pure CO2 to greatly retard heat loss to a minimum?

    Since most modern building already use a wrap to retard heat loss by reflecting it back towards the rooms.It seems like a good idea.So is not being done?

    How come the awesome "heat trapping" greenhouse gas are NOT being used in homes and other obvious applications to keep it warm?

    ReplyDelete
  149. Gee, anonymous, I wonder if it could be ... ummm... because the way a house loses heat is not primarily by radiation but by convection and conduction?

    Of course back radiation creates no energy. Nothing creates energy. On the other hand, no one ever made a claim that implied that DLWR (downward long wave radiation, the correct terminology for "back radiation") does create energy.

    ReplyDelete
  150. You may be interested that over at WUWT a thread has been going for several days discussing a model which is like that used here, but inside out.
    That is it has the heat source (a.k.a. sun) outside the sphere, which is surrounded by a shell which has the property that incoming radiation passes freely, but outgoing radiation is hindered.
    The thread is here: http://wattsupwiththat.com/2010/11/27/people-living-in-glass-planets/

    Most people understood that the model shows the inner surface warming, though arguing that it is much too simple to described a real greenhouse effect. (which is of course the case) A few commentators however were of the same opinion as Gord and felt that energy was being created. There were of course wrong but it might be interesting to the current discussion to see why.

    ReplyDelete
  151. Neutrino,

    You said...
    "Fourth time Ive had to state this, the earth has an Emissivity of 0.69 in the visible and 1.0 in the infrared. When I agreed that -19C was correct it was for an Earth type sphere,
    when I stated a temperature of 5.5C we were talking about a sphere of uniform emissivity at 0.69. Not the Earth!'

    Oh, what a bunch of BS!

    The Earth is a Sphere with Mountains, Valleys, Plains etc. but is modelled as a "Flat Disk".

    There are TWO VERY COLD Poles, a HOT Equator and other regions that receive Solar Energy in unequal amounts that are Averaged Uniformly Over the "Flat Disk".

    The Solar Insolation due to Orbital changes with Seasons is Averaged when the Solar Constant is used.

    The Earth's changing Axis is totally ignored.

    And, THE EARTH'S EMISSIVITY IS ALSO AVERAGED and applied to the "Flat Disk" with an Average Uniform Emissivity = 0.67!
    ---
    For the Earth with Q = Incident Energy = 342 w/m^2 and Emissivity e = 0.69 :

    The equation q = e*BC*T^4 or T = (q/e*BC)^0.25

    where q = ABSORBED ENERGY = Q*e = 342 * 0.69 = 235.98 w/m^2

    Will ALWAYS produce T = (235.98/(0.69 X 5.67X10^-8))^0.25 = 278.68 K or 5.53 deg C the BLACK BODY TEMPERATURE.

    Your equation CANNOT BE USED TO CALCULATE A GREY BODY TEMPERATURE!
    --
    The correct method of calculating the GREY BODY temperature is to use the q = ABSORBED ENERGY = Q*e = 342 * 0.69 = 235.98 w/m^2 in the equation:

    T = (q/BC)^0.25 = (235.98 / 5.67X10^-8)^0.25 = 253.99 K or -19.16 deg C THE CORRECT ANSWER for the GREY BODY whether the the GREY BODY is the Earth or not!

    That is how the Earth's AVERAGE temperature IS ACTUALLY CALCULATED, uses a AVERAGE EMISSIVITY FOR THE EARTH of 0.69 and produces a -19.16 deg C Temperature no matter how much you BABBLE your hillarious delusionary "opinions" to try and AVOID THE FACTS.

    So cut the BS.
    ------------
    You said...
    "As for the quote, "The law is valid only for ideal black objects, the perfect radiators, called black bodies."
    Yes the SB as derived (P = σT^4*A) is only for a BB, it is then extended for all GB by adding the term ε(p = εσT^4*A). Where is the BS in that?"

    The BS is that the quote refers to the equation
    P(T) = e*BC*T^4 !!!

    Stefan-Boltzmann law
    P(T) = e*BC*T^4

    "The law is valid only for ideal black objects, the perfect radiators, called black bodies."
    http://planetphysics.org/encyclopedia/StefanBoltzamannLaw.html

    You can use ANY value of Emissivity e, q = Q*e substituted into T = (q/e*BC)^0.25 gives T = (Q*e/e*BC)^0.25 = (Q/BC)^0.25 and will ALWAYS GIVE THE BLACK BODY TEMPERATURE, NOT THE
    GREY BODY TEMPERATURE!

    Like the quote says "The law is valid only for ideal black objects, the perfect radiators, called black bodies.", IT IS NOT VALID FOR GREY BODIES!

    So stop the BS.
    ------------------------
    continued...

    ReplyDelete
  152. continuation..

    You said...

    "So to not have any confusion about what we did or didnt say:
    Do you think a object with emissivity of 0.8 radiates less than an object of emissivity 1.0 when they both have the same temperature and dimensions?"

    An object with a temperature will only radiate the amount Q = BC*T^4 where Q = ABSORBED ENERGY

    The Law of Conservation of Energy requires that Emitted Energy = Absorbed Energy

    For the Earth with Q = Incident Energy = 342 w/m^2 and Emissivity e = 0.69

    We already know that q = Absorbed Energy = = Q*e = 342 * 0.69 = 235.98 w/m^2

    When used in T = (q/BC)^0.25 = (235.98 / 5.67X10^-8)^0.25 = 253.99 K or -19.16 deg C THE CORRECT ANSWER for the GREY BODY.

    The emissivity was used to calculate the Absorbed Energy that was used to calculate the Temperature T and emissivity is no longer relevant.

    The Body with Temperature -19.16 deg C can only Emitt 235.98 w/m^2 which is the same as it Absorbed.

    The Absorbed Energy = Emitted Energy = 235.98 w/m^2 as required by The Law of Conservation of Energy
    ----
    Now let's calculate the Emitted Energy that corresponds to the to the correct -19.16 deg C or 253.99 K temperature using your equation q = e*BC*T^4:

    q = e*BC*T^4 = 0.67 X 5.67X10^-8 X (253.99 K)^4 = 158.10 w/m^2

    158.10 w/m^2 Emitted Energy DOES NOT EQUAL 235.98 w/m^2 Absorbed Energy and therefore VIOLATES The Law of Conservation of Energy.
    ------------------------------
    You said...
    "The reference I provided dealt with emitted, not absorbed or incident, so I compared it to your calculated emitted. You calculated that a 100.9K object radiated 12.5kW, in both scenarios, regardless of its emissivity.
    Thats contrary to the quote i provided, from your reference no less.
    So, are you wrong or is your reference?"

    The stored heat energy that was absorbed from the the original energy source that caused Each body have a temperature of 100.9 K is Potential Energy.

    (The Absorbed Energy from the original source would have used the Emissivity X Incident Energy to produce the body temperature of 100.9 K so Emissivity is no longer relevant)

    Energy Into a Body = Energy Out of a Body.

    Where the Potential Energy is the Energy into the Body (stored Energy) = Kinetic Energy, the Emitted Energy out of the body.

    The absorbed stored energy in Each body is the same because their Temperatures are the same and will be equal to the Emitted Energy.

    Q = BC*T^4*A

    Q = (5.67 X 10^-8 * 100.9^4 * 2123.7 ) = 12,487 Watts Emitted for Each Sphere

    Emitted Energy will decline as the absorbed stored energy and Temperature declines, and can be calculated using Q = BC*T^4*A.

    And this fully complies with "Emitted Energy = Absorbed Energy" that is based on The Law of Conservation of Energy:

    Emissivity in the Infrared
    Physics of Emissivity
    "Emitted Energy = Absorbed Energy"
    http://www.optotherm.com/emiss-physics.htm

    The Earth's Average Temperature calculation of -19.16 deg C (253.99 K) produced by 235.98 w/m^2 is PROOF for the above.

    ReplyDelete
  153. @gord

    sorry to reiterate my question: does a reflected photon gets a chance to strike the inner surface more than once?[yes]/[no]

    ReplyDelete
  154. Once again Gord you have completely missed the point.

    One comment and one question, thats all I have left for you.

    You continue to appeal to the Law of Conservation of Energy at every turm. I agree(along with everyone else) that you can not violate that Law. But what you continuously fail to understand is that I am not violating it under my paradigm.(and neither are you under yours btw)
    This is why I say that it is a red herring, not because I disagree with it but because given our two sets of analysis we are both consistant with it.

    To deminstrate this lets look at the simplest example we have used, the mirror.
    Q = 100W, A = 1m^2, ε = 0.01:

    You believe that its temperature is 64.8K and:
    1. q = Qε
    2. q = σT^4*A
    So,
    3. q = Qε
    4. q = 100W*0.01
    5. q = 1W (absorbed)
    6. q = σT^4*A
    7. q = σ*64.8K^4*1m^2
    8. q = 1W (emitted)
    (if I have misrepresented you analysis please corect me)

    So emitted equals absorbed, you didnt violate the Law of Conservation of Energy, does that mean you are correct?

    I believe that the temperature is 204.9K and:
    10. q = Qε
    11. p = εσT^4*A
    So,
    12. q = Qε
    13. q = 100W*0.01
    14. q = 1W (absorbed)
    15. p = εσT^4*A
    16. p = 0.01*σ*204.9K^4*1m^2
    17. p = 1W (emitted)

    Again emitted equals absorbed so I didnt violate the Law of Conservation of Energy either. Does that mean I am correct?

    No, being in acordance with the Law of Conservation of Energy does not mean, either of us, that we are correct, it only means that we could possibly be correct.

    So how do we determine who is correct then? Well I have been trying to get you to look at the definition of the SB and that of a GB. But you dont seem to want to do that.

    I have repeatedly posted references(sometimes even your own) that disagree with how you are using the SB and that contravene your idea that a GB emits the same as a BB at equal temperatures.

    So, until you can come up with a reference that supports your view that
    "An object with a temperature will only radiate the amount Q = BC*T^4 where Q = ABSORBED ENERGY"
    we really have nothing left to discuss.
    (note: we are talking about GB's here not ideal BB's)


    Neutrino

    ReplyDelete
  155. Gord,

    You keep misrepresnting or not taking the full context of quotes.


    Here is the qoute in full:(my bold)

    "Stefan-Boltzmann law
    The Stefan-Boltzmann law, also known as Stefan's law, states that the total energy radiated per unit surface area of a black body in unit time, P is directly proportional to the fourth power of the black body's Thermodynamic temperature T (also called absolute temperature):
    P(T) =εσT^4
    The irradiance P has dimensions of power density (energy per time per square distance), and the SI units of measure are joules per second per square meter, or equivalently, watts per square meter. The SI unit for absolute temperature T is the kelvin. e is the emissivity of the blackbody; if it is a perfect blackbody e = 1.
    The constant of proportionality , called the Stefan-Boltzmann constant or Stefan's constant, is non-fundamental in the sense that it derives from other known constants of nature. The value of the constant is
    σ = (2π^5k^4)/(15c^2h^2) = 5.67-4--*10^-8[Ws^-1m^-2K^-4].
    Thus at 100 K the energy flux density is 5.67 W/m2, at 1000 K 56.7 kW/m2, etc.
    The law was discovered experimentally by Jožef Stefan (1835-1893) in 1879 and derived theoretically, using thermodynamics, by Ludwig Boltzmann (1844-1906) in 1884. Boltzmann treated a certain ideal heat engine with the light as a working matter instead of the gas. This law is the only physical law of nature named after a Slovene physicist. The law is valid only for ideal black objects, the perfect radiators, called black bodies. Stefan published this law on March 20 in the article Über die Beziehung zwischen der Wärmestrahlung und der Temperatur (On the relationship between thermal radiation and temperature) in the Bulletins from the sessions of the Vienna Academy of Sciences.
    Derivation
    The Stefan-Boltzmann law can be derived by integrating over all wavelengths the spectral intensity of a black body as given by Planck's radiation law. "


    Pretty selective reading to come away from this and say the SB only applies to a ideal BB. As I said before, its derived from an ideal BB and extended to all GB's by adding ε. That one line you quoted may be slightly ambiguous but read in the overall context its clear the authour believes the SB applies to all GB's(a BB is a GB with emissivity at unity(1) btw)

    Part of your confusion may stem from the fact that both
    P = σT^4 (BB's only)
    P = εσT^4 (all GB's including BB's: also the equation the reference uses)
    are generally referred to as the SB. Is that a perfectly concise scenario? No, but it is the general convention.

    Question:
    If the SB is only valid on a BB then why is ε in the equation that this reference cites as the SB?
    (note: this reference states the SB is about the emitted not absorbed just like all the others I have posted)


    Neutrino

    ReplyDelete
  156. Neutrino,

    I wonder how many times I have to post the very, very simple mathematical proof for your equation:

    The BLACK BODY equation is:

    Q = BC*T^4*A where Q = Incident Power and that is the Absorbed = Emitted Power for the BLACK BODY.

    Your equation is:

    q = e*BC*T^4*A where q = Q*e where Q*e is the Absorbed Power

    Q*e = e*BC*T^4*A

    Q = BC*T^4*A the BLACK BODY equation where Q = Incident Power and that is the Absorbed = Emitted Power for the BLACK BODY.
    -------------
    The Black Body equation is Q = BC*T^4*A the BLACK BODY equation where Q = Incident Power and that is the Absorbed = Emitted Power for the BLACK BODY. (1)

    Q = 100 Watts, A = 1 m^2, e = 0.01, T = 204.9 K

    Here is what you said..

    I believe that the temperature is 204.9K and:
    10. q = Q*e
    11. p = e*BC*T^4*A
    So,
    12. q = Q*e
    13. q = 100W*0.01
    14. q = 1W (absorbed)
    15. p = e*BC*T^4*A
    16. p = 0.01*σ*204.9K^4*1m^2
    17. p = 1W (emitted)


    from (10) q = Q*e therefore e = q/Q

    Subsitute e = q/Q into p = e*BC*T^4*A gives:

    p = (q/Q) * BC*T^4*A

    P*Q/q = BC*T^4*A (2)

    p = 1 watt (what you called emitted power) and q = 1 watt (the absorbed power) so substituting gives

    Q = BC*T^4*A the BLACK BODY equation where Q = Incident Power and that is the Absorbed = Emitted Power for the BLACK BODY.

    Q = 5.67X10^-8 *(204.9K)^0.25 * 1m^2 = 100 Watts.

    THIS IS WHAT YOU ARE ALWAYS CALCULATING, THE BLACK BODY TEMPERATURE EQUATION WHERE ALL THE INCIDENT POWER IS ABSORBED AND EMITTED!!!

    What you are ALWAYS getting wrong is that p what you call emitted is NOT the EMITTED POWER it is ALWAYS equal to q!!!

    The Emitted Power you are actually calculating is 100 Watts!
    ----
    The BLACK BODY EQUATION is Q = BC*T^4*A (1)

    Your equation is P*Q/q = BC*T^4*A (2)

    Gives Q = p*Q/q

    Gives p = q ..... ALWAYS!!!
    -----------
    You said...
    "Again emitted equals absorbed so I didnt violate the Law of Conservation of Energy either. Does that mean I am correct?"

    You obviously ARE violating The Law of Conservation of Energy because you FALSELY stated that p = emitted power = 1 Watt when the ACTUAL emitted power is 100 Watts!
    ------------
    You said...
    "So, until you can come up with a reference that supports your view that
    "An object with a temperature will only radiate the amount Q = BC*T^4 where Q = ABSORBED ENERGY"
    we really have nothing left to discuss.
    (note: we are talking about GB's here not ideal BB's)"

    HAHAHA....HAHAHA....HAHAHA...Oh, that's so Funny!

    The equation Q = BC*T^4 IS THE BLACK BODY EQUATION and so is p = e*BC*T^4*A as I just PROVED for about the 30th TIME!

    They are BOTH BLACK BODY EQUATIONS and ARE NOT GREY BODY EQUATIONS as I have repeately said!

    The Law of Conservation of Energy says Absorbed Energy = Emitted Energy

    I will remind you what you said..
    "You continue to appeal to the Law of Conservation of Energy at every turm. I agree(along with everyone else) that you can not violate that Law."

    What an IDIOTIC and CONTRADICTORY statement!

    Just like your CONSTANT violation of The Law of Conservation of Energy!

    You are hopelessly delusional and are just a waste of my time.

    ReplyDelete
  157. Giovi Pelle,

    I am still waiting for you answer this question:

    1)Please show me how the 6,000 Joules/sec can increase to a value greater than 6,000 Joules/sec and comply with THE LAW OF CONSERVATION OF ENERGY!

    And, POST the supporting Physics for your "Opinions".

    All you said was...
    "6kJ/s stay 6kJ/s. Please notice that's a POWER, not ENERGY . So after one second the accumulated ENERGY 6kJ/s*1/s=6kJ in the cavity. After 10 seconds the accumulated ENERGY is kJ/s*10s=60kJ. After 1000s the accumulated energy is 6kJ/s*1000s=6MJ. So after all energy is accumulated in the cavity, since reflected energy is going to stay there forever following your approach."

    So, either POST the supporting Physics for your "Opinions" or admit that you are wrong.

    I'm still waiting for your response.
    ---------------------
    Did you not understand what I posted for you before?

    "If you just BABBLE without having posting any Physics links to support your "Opinions", I will simply tell you post the supporting Physics before I respond.

    Do you understand that?"

    ReplyDelete
  158. Back on the merry-go-round it seems.

    You do a little bit of algebra, solve for Q get a number of 100W and seem to think this is a suprise.
    If you didnt get 100W then that would be a suprise since Q is the incident and is set to 100W.
    Then you do a bit more algebra and come up with p = q, again thats what it has to be in equilibrium. The suprise would be if p ≠ q.

    Since p is emitted, and q is absorbed and they are equal this seems to confirm the point that my calculations are internally consistant.


    So, still waiting on those references that support you opinion. Having trouble finding them?


    Neutrino

    ReplyDelete
  159. Neutrino,

    They merry-go round is entirely yours.

    The Incident Power of 100 Watts is what you used to calculate the BLACK BODY temperature for your example.

    Q = 100 Watts Incident Power, A = 1 m^2, e = 0.01, T = 204.9 K

    Q = BC*T^4*A is the Black Body equation:

    T = (Q/(BC *A))^0.25 = (100/(5.67 X 10^-8 * 1m^2))^0.25 = 204.9 K
    -----
    Another prime example is your calculation for the Earth temperature:

    Q = 342 w/m^2, e = 0.69

    The BLACK BODY equation is:

    Q = BC*T^4 where Q = Incident Energy 342 w/m^2

    T = (Q/BC)^0.25 = (342/5.67 X 10^-8 )^0.25 = 278.68 K or +5.53 deg C the BLACK BODY EARTH temperature.

    The correct answer for the GREY BODY Earth temperature is:

    q = Absorbed Energy = Incident Energy X Emissivity = 342 w/m^2 X 0.69 = 235.98 w/m^2

    T = (235.98 /5.67 X 10^-8)^0.25 = 253.99 K or -19.16 deg C.
    -----
    This comes from this simple Mathematical Proof that I have posted over a dozen times:

    The BLACK BODY equation is:

    Q = BC*T^4*A where Q = Incident Power and that is the Absorbed = Emitted Power for the BLACK BODY.

    Your equation is:

    q = e*BC*T^4*A where q = Q*e where Q*e is the Absorbed Power

    Q*e = e*BC*T^4*A

    Q = BC*T^4*A the BLACK BODY equation where Q = Incident Power and that is the Absorbed = Emitted Power for the BLACK BODY.

    Emissivity e always cancells in your equation no matter what value is used and ALWAYS produces the BLACK BODY equation Results.

    Every calculation you have done, no matter what value of emissivity you used, produced the BLACK BODY equation Results.

    That is a FACT and if you deny it you are a LIAR.

    ReplyDelete
  160. Gord,


    All you have to do to stop this is provide a reference that says the emission from a GB is not p = εσT^4*A.

    Your 'proofs' do not eshtablish what you think they do but trying to show you that has got us nowhere.

    If I am the Liar with Alzheimers just Babbling my Delusions that you think I am then it should be easy for you to find a reference that actually proves that I am wrong.

    So, where is that reference?


    Neutrino

    ReplyDelete
  161. Neutrino,

    Your Math Skills are absolutely dismal and less than a Grade 5 Student.

    I asked my neighbours Kid who is in Grade 5 to do this Arithmetic problem.

    Q = 100 and e = 0.01 and I asked him to multiply 100 X 0.01 and he came up with the correct answer of 1.

    Then I asked him to divide 1 by 0.01 and he came up with the correct answer of 100.

    I asked him why that was and he correctly said Q X e / e = Q
    ----
    You have already confimed that p = q.

    Your equation is:

    q = e*BC*T^4*A therefore q/e = BC*T^4*A where q = Q*e and Q*e is the Absorbed Power

    Q X e / e = BC*T^4*A

    Q = BC*T^4*A which is what you have been using all along.

    The math does not get any simpler than that.

    Even a Grade 5 Kid understands it.

    Q = BC*T^4*A is the BLACK BODY equation that you have been using.

    You have and will always get the same result every time.

    I have already wasted far too much time on your astounding lack of Math Education.

    If you don't have the Math Skills of a Grade 5 Kid, there is nothing more to discuss.

    End of story.

    ReplyDelete
  162. Gord,

    It is amazing that you can not find a single reference to support your opinion.

    It is also amazing that you do not understand that you are continually proving that the Incident is 100W(the Q in all the equations) not the Emitted.

    If p(emitted) is equal to εσT^4*A then it is trivial that the emitted is 1W. Even your grade 5 Kid can get that right.

    Our disagreement is about the form of the SB, you think its q(absorbed) = σT^4*A and I think its p(emitted)) = εσT^4*A.
    So I ask again, show me the reference that confirms your belief?


    Neutrino

    ps: If the BB equation is p(emitted) = σT^4*A, and if Q = p then your equation would be correct. Do you have any evidence that estabilshes Q = p?

    ReplyDelete
  163. Neutrino,

    You said...
    "Our disagreement is about the form of the SB, you think its q(absorbed) = σT^4*A and I think its p(emitted)) = εσT^4*A.
    So I ask again, show me the reference that confirms your belief?"

    BS! and a LIE!

    I have ALWAYS said q = e*BC*T^4*A where q = Q*e where Q*e is the Absorbed Power !!

    And I have proved that p = q mathematically which also is what The Law of Conservation of Energy says "Absorbed Power = Emitted Power".

    Here is what you posted before:
    "You continue to appeal to the Law of Conservation of Energy at every turm. I agree(along with everyone else) that you can not violate that Law."

    So are you LYING about "I agree(along with everyone else) that you can not violate that Law"???
    ------------
    You said...
    "If p(emitted) is equal to εσT^4*A then it is trivial that the emitted is 1W. Even your grade 5 Kid can get that right."

    The Grade 5 Kid got it right, you got it wrong.

    I asked my neighbours Kid who is in Grade 5 to do this Arithmetic problem.

    Q = 100 and e = 0.01 and I asked him to multiply 100 X 0.01 and he came up with the correct answer of 1.

    Then I asked him to divide 1 by 0.01 and he came up with the correct answer of 100.

    I asked him why that was and he correctly said Q X e / e = Q

    Since p = q as I have proven mathematically and is what The Law of Conservation of Energy says "Absorbed Power = Emitted Power"

    q = e*BC*T^4*A therefore q/e = BC*T^4*A where q = Q*e and Q*e is the Absorbed Power

    Q X e / e = BC*T^4*A

    Q = BC*T^4*A which is what you have been using all along.

    That's why using p = εσT^4*A ALWAYS produces a BLACK BODY temperature as all your calculations show.

    Are you going to LIE about that too?
    ------
    You said...
    "ps: If the BB equation is p(emitted) = σT^4*A, and if Q = p then your equation would be correct. Do you have any evidence that estabilshes Q = p?"

    Q = BC*T^4*A is the Black Body Equation where Q = Incident Energy which is Absorbed and Emitted by the Black Body.

    If p(emitted) = BC*T^4*A and Q = BC*T^4*A then Q = p.

    DUH!
    -------------
    I have already PROVEN that your equation gives the WRONG answer for the Earth temperature when Q = Incident Energy = 342 w/m^2 and Emissivity = 0.69.

    Your equation ALWAYS gives the BLACK BODY temperature of 278.68 K or +5.53 deg C when the ACTUAL GREY BODY EARTH temperature is 253.99 K or -19.16 deg C.

    Are you going to LIE about that too?
    ------------
    Summary:

    1) FACT - your equation ALWAYS produces a Black Body temperature as is proved by simple math that a Grade 5 Kid easily understands yet is way beyond your Math skills.

    2) FACT - that every calculation that YOU HAVE DONE confirms that you ALWAYS have calculated the Black Body temperature.

    3) FACT - You are constantly contradicting yourself when you say that you accept The Law of Conservation of Energy but do not accept Energy Emitted = Energy Absorbed.

    I am tired of wasting my time constantly repeating the above and will no longer respond to your delusionary posts.

    ReplyDelete
  164. @Gord

    Energy Emitted = Energy Absorbed

    Everybody agrees with this. Well, let's see. Let's take a huge battery, which provides the energy to our 30kW source. Therefore we have: Power_in=30kW. Now let's look at the power emitted by our PVC shell: 24kW, therefore:
    Power_out=24kW. So we have: Power_in /= Power_out...that's odd...

    Let's be more specific. Let's take a 30MJ battery attached to our 30kW source. The battery energy is drained in (30MJ)/(30KJ/s)=1000s: the battery is drained in a thousands seconds. Now let's see how much energy is emitted by our PVC in 1000s, and we get 24MJ... we are missing 6MJ, where are they? Would you care to show me, may be with a careful transient analysis?

    Well, sorry to prevent you, but you can call this RANTING, BABBLING, or even worse, UNSUPPORTED BABBLING, it's fine by me, I can sleep with that. Point is that you have 30kW in and 24kW out...and I call it undeniable logic.

    Still waiting for your answer about the reflected photons: do they reach the inner surface once more after the first reflection?([yes][no]). Don't' worry, I don't expect an answer, I don't think you have any idea about what is going on inside the cavity. But please, in case you have a brilliant idea, please enlighten me, or better, me and my babbling.

    ReplyDelete
  165. Gord,

    You just stated: "I have ALWAYS said q = e*BC*T^4*A where q = Q*e where Q*e is the Absorbed Power !!"
    Problem is, no you have not.

    But lets take you on your word that the above quote represents your current belief.
    1. q = Absorbed
    2. Absorbed = Emitted
    3. Q = Incident
    4. T = Temperature

    5. q = Qε
    6. q = εσT^4*A

    I havent changed anything from your above statement, just rewrote everything out to be clear.
    (I did add the Absorbed = Emitted, it is not in the above quote but I do not think you would object to that)

    Now given that 1. through 6. are all true(thats what you just stated, along with Conservation of Energy) Then:
    7. Qε = εσT^4*A
    8. Q = σT^4*A

    This is a trivial result. The only way to reject it is to reject one or more of 1. thorugh 6..

    What this result means is that the Temperature of a GB is only dependant on its Incident, nothing else.
    The above does not mean that the GB is Emitting Q(It does however mean a BB would be Emitting Q at the same temperature).
    What the GB is Emitting, from your own statements, is q(not uppercse Q) where q is equal to Qε or εσT^4*A.

    (note: the only changes I would make to the above, and this does not change the result at all, are:
    1. q = Absorbed
    1a. p = Emitted
    2. p = q
    6. p = εσT^4*A )

    I really am confused as to your position, If the above can be taken seriously then you agree with me but the rest of your post clearly contradicts this.
    So, if the above is your current belief then we are in agreement on the results. If it is not then some references would be helpfull as to why you think that it is wrong.


    Neutrino

    ReplyDelete
  166. Just to elaborate on the last post,

    The example is our mirror, this is what we know about it:
    1. Q = Incident = 100W
    2. ε = Emissivity = 0.01
    3 A = Area = 1m^2
    4. q = Absorbed
    5. p = Emitted
    6. T = Temperature

    This is how the different variables relate to each other:
    7. p = q
    8. q = Qε

    and
    9a. p = εσT^4*A
    10a. q = εσT^4*A
    11a. Qε = εσT^4*A
    12a. Q = σT^4*A
    13a. T = (Q/σA)^.25
    or
    9b. p = σT^4*A
    10b. q = σT^4*A
    11b. Qε = σT^4*A
    12b. T = (Qε/σA)^.25

    If 9a. is true then so is 13a.(and 12a. btw) therefore our mirror has a T of 205K.
    If 9b. is true then so is 12b. therefore our mirror has a T of 65K.

    (note: neither 9a. nor 9b. violate the Conservation of Energy. From 8. both are absorbing 1W.
    Calculate p with T = 205K from 9a. and we get p = 1W, so its not emitting more than its absorbing.
    Calculate p with T = 65K from 9b. and we get p = 1W, so again its not emitting more than its absorbing.)

    Only one of 9a. or 9b. can be correct though, which one do you think it is?
    As always, references to support your answer are required.


    Neutrino

    ReplyDelete
  167. Giovi Pelle,

    I have already posted the Physics links and analysis for 24 KW absorbed several times.
    ----
    I also posted this for you:

    "I'm not interested in your "opinion" so instead of BABBLING "So, it is your approach that violates conservation of energy", either show where my calculations are wrong using the Physics and Physical Example I have posted or post the Physics and links you are using to support your "opinion".

    An actual Physical Example would be nice too.

    Do that or don't waste my time.
    ----------------
    And I posted this for you:

    "I am still waiting for you answer this question:

    1)Please show me how the 6,000 Joules/sec can increase to a value greater than 6,000 Joules/sec and comply with THE LAW OF CONSERVATION OF ENERGY!

    And, POST the supporting Physics for your "Opinions".

    All you said was...
    "6kJ/s stay 6kJ/s. Please notice that's a POWER, not ENERGY . So after one second the accumulated ENERGY 6kJ/s*1/s=6kJ in the cavity. After 10 seconds the accumulated ENERGY is 6kJ/s*10s=60kJ. After 1000s the accumulated energy is 6kJ/s*1000s=6MJ. So after all energy is accumulated in the cavity, since reflected energy is going to stay there forever following your approach."

    So, either POST the supporting Physics for your "Opinions" or admit that you are wrong.

    I'm still waiting for your response.
    ---
    Did you not understand what I posted for you before?

    "If you just BABBLE without having posting any Physics links to support your "Opinions", I will simply tell you post the supporting Physics before I respond.

    Do you understand that?"
    ------------------------------------------------
    I am tired of repeating the above over and over again.

    Do you have some sort of reading disability?

    ReplyDelete
  168. Why is this ridiculous thread still running?

    Gord has incessantly ignored basic science. Although it has been pointed out to him in plain language that
    a) there is NO incident or absorbed radiation on the outer surface of the sphere, he persists in a fantasy version of the SB equations.
    b) the SB equations are totally standard that gives the power RADIATED by an object. Here is a reference
    http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/stefan.html
    there are dozens of identical statements all over the web. Even wikipedia gets it right
    http://en.wikipedia.org/wiki/Stefan–Boltzmann_law
    All you have to do is use the equation exactly as it is written
    http://www.endmemo.com/physics/radenergy.php
    You do not have to modify the answer by dividing by the emissivity or anything else.
    Just use the equation EXACTLY AS IT IS WRITTEN.

    Note to forum mods - why is Gord allowed to insult all other posters on this thread, implying that they suffer from various handicaps?


    PS I am not any previous anons.

    ReplyDelete
  169. Anonymous,

    You said...
    "Why is this ridiculous thread still running?
    "Gord has incessantly ignored basic science."

    Maybe because "rediculous" people make completely FALSE CLAIMS without a shread of Physics back up, like YOU just made.

    Let's see who has incessantly ignored basic science.
    ---
    Regarding your points:

    "a) there is NO incident or absorbed radiation on the outer surface of the sphere, he persists in a fantasy version of the SB equations."

    FACTS:

    1) All objects that have a temperature above absolute zero means that they have ABSORBED ENERGY and will EMITT ENERGY

    2) The Law of Conservation of Energy says Energy Emitted = Energy Absorbed.

    3) SOD's "VERY FIRST CALCULATION" was to use the SB equation to calculate the temperature of the Outer Sphere surface.

    4) SOD used 30,000 Watts as the ABSORBED POWER, A = 2123.73 m^2 and an emissivity of 0.8 to calculate the Outer Sphere surface temperature.
    ---------
    Here is how SOD did the temperature calculation using q = e*BC*T^4*A which can also be written as q/e = BC*T^4*A:

    SOD used q = 30,000 Watts and e = 0.8 so q/e = 37,000 Watts Emitted (which EXCEEDS THE ABSORBED POWER!).

    37,000 = BC*T^4*A (and you should recognize that BC*T^4*A is THE BLACK BODY EQUATION)

    T = (37,000/(5.67X10^-8 X 2123.73))^0.25 = 133K an impossibly high temperature.

    How much would a pure Black Body (emissivity = 1) outer sphere temperature be for 30,000 Watts ABSORBED POWER?

    30,000 = BC*T^4*A (and you should recognize that BC*T^4*A is THE BLACK BODY EQUATION)

    T = (30,000 /(5.67X10^-8 X 2123.73))^0.25 = 125.63K !!!
    ----------
    5) SOD's calculation using what you think is a GREY BODY equation q = e*BC*T^4*A produced a HIGHER TEMPERATURE (133K) than what a BLACK BODY would produce (125.63K)!!

    6) SOD's calculation using what you think is a GREY BODY equation q = e*BC*T^4*A used a HIGHER EMITTED POWER (37,000 Watts) than what a BLACK BODY would EMITT (30,000 Watts)!!

    7) SOD's calculation VIOLATES The Law of Conservation of Energy because EMITTED ENERGY > ABSORBED ENERGY.

    Anonymous, start with Point (1) and tell me if you agree or if you disagree explain why.

    Continue with each of the other points and the calculations shown and tell me if you agree or if you disagree explain why.

    Now SHOW ME where I have "incessantly ignored basic science" or retract your statement.
    ---------------
    continued...

    ReplyDelete
  170. continuation...

    Your next point:

    "b) the SB equations are totally standard that gives the power RADIATED by an object."

    FACTS:

    8) The Law of Conservation of Energy says Energy Emitted = Energy Absorbed therefore Power Radiated = Power Absorbed.

    9) The SB equation q = e*BC*T^4*A where q = Power Radiated and according to The Law of Conservation of Energy q = Power Absorbed.

    10) The SB equation for Black Bodies is Q = BC*T^4*A where Q = Incident Power Emitted and according to The Law of Conservation of Energy Q = Incident Power Absorbed.

    11) Substituting Q = BC*T^4*A into q = e*BC*T^4*A gives q = e * Q which is the Absorbed Power and according to The Law of Conservation of Energy q = Emitted Power

    12) q = e * Q subsituting into q = e*BC*T^4*A gives Q = BC*T^4*A the BLACK BODY EQUATION.

    Anonymous, start with Point (8) and tell me if you agree or if you disagree explain why.

    Continue with each of the other points and the calculations shown and tell me if you agree or if you disagree explain why.

    Now SHOW ME where I have "incessantly ignored basic science" or retract your statement.
    ----------------
    continued...

    ReplyDelete
  171. continuation...

    You said..
    "You do not have to modify the answer by dividing by the emissivity or anything else.
    Just use the equation EXACTLY AS IT IS WRITTEN."

    13) The Law of Conservation of Energy requires that Incident Energy = Absorbed Energy + Transmitted Energy + Reflected Energy

    14) The Law of Conservation of Energy requires that Absorbed Energy = Emitted Energy

    15) The very well known Earth Temperature Calculation for 342 w/m^2 Incident Solar Energy with an Earth Emissivity = 0.69 is done as follows:

    Absorbed Energy of the Earth:
    q = e*Q = 0.69 X 342 = 235.98 w/m^2

    Reflected Energy of the Earth = (1- 0.69)X 342 = 106.02 w/m^2

    Transmitted Energy of the Earth = 0

    Incident Energy 342 w/m^2 = Absorbed Energy 235.98 w/m^2 + Reflected Energy 106.02 w/m^2 + Transmitted Energy 0 w/m^2

    The Reflected Energy cannot heat up the Earth, only the Absorbed Energy can heat up the Earth.

    100% of the Absorbed Energy heats the Earth.

    Temperature of the Earth:

    Absorbed Energy of the Earth = BC*T^4

    Temperature of the Earth = (Absorbed Energy of the Earth / BC)^0.25 = (235.98 / 5.67X10^-8)^0.25 = 253.99K or -19.16 deg C

    This GREY BODY Temperature of the Earth of -19 deg C is widely known and accepted as correct in all the AGW literature and websites.
    It also completely complies with The Law of Conservation of Energy.

    16) Now let's see what Temperature of the Earth is produced by using q = e*BC*T^4 "EXACTLY AS IT IS WRITTEN".

    q = Emitted Energy and according to The Law of Conservation of Energy q = Absorbed Energy.

    Absorbed Energy of the Earth is the same as before:
    q = e*Q = 0.69 X 342 = 235.98 w/m^2

    Temperature of the Earth = (q /(e*BC))^0.25 = (235.98 /(0.69 * 5.67X10^-8)^0.25 = 278.68K or +5.53 deg C!!!

    So using q = e*BC*T^4 "EXACTLY AS IT IS WRITTEN" DID NOT PRODUCE THE CORRECT "GREY BODY" TEMPERATURE OF THE EARTH, WHICH IS -19 DEG C!!

    17) Instead the equation q = e*BC*T^4 PRODUCED THE "BLACK BODY" TEMPERATURE OF THE EARTH WHICH IS +5.53 DEG C!!

    PROOF: A Black Body emitts and absorbs 100% of the Incident Energy = 342 w/m^2

    T = (342 / 5.67X10^-8)^0.25 = 278.68K or +5.53 deg C!

    Anonymous, start with Point (13) and tell me if you agree or if you disagree explain why.

    Continue with each of the other points and the calculations shown and tell me if you agree or if you disagree explain why.

    Now SHOW ME where I have "incessantly ignored basic science" or retract your statement.
    ----------------------------------
    continued...

    ReplyDelete
  172. continuation...

    18) The equation q = e*BC*T^4 cannot be used to calculate a "Grey Body" temperature because it is a "Black Body" equation where T = the temperature of a BLACK BODY and q is the
    Absorbed Energy = Energy Emitted of a "Grey Body".

    T = BLACK BODY TEMPERATURE, NOT A GREY BODY TEMPERATURE.
    q = Emitted Energy = Absorbed Energy = e * Q where Q is the Incident Energy

    Like I proved in (12) q = e * Q subsituting into q = e*BC*T^4*A gives Q = BC*T^4*A the BLACK BODY EQUATION.

    And like I proved in (16) q = e*Q = 0.69 X 342 = 235.98 w/m^2, Temperature of the Earth = (q /(e*BC))^0.25 = (235.98 /(0.69 * 5.67X10^-8)^0.25 = 278.68K or +5.53 deg C!!!...THE BLACK BODY TEMPERATURE.
    -----------------------------------
    19) I will explain what The equation q = e*BC*T^4 is ACTUALLY saying:

    If a Black Body of Temperature T had an emissivity of e what would it radiate?

    Answer: The Black Body Radiation is the Incident Energy X Emissivity = 342 w/m^2 X 0.69 = 235.98 w/m^2 which is correct.

    But a Black Body does not have an emissivity other than e = 1 and a Grey Body does not have a Black Body temperature!

    A Grey Body will always have a temperature less than a Black Body just like the Earth calculation shows.

    Another good example is a white sheet of paper placed next to a black sheet of paper in Sunlight.

    The black sheet of paper will always get hotter than the white sheet of paper.

    The equation q = e*BC*T^4 assumes the white sheet of paper is the same temperature as the black paper (black body temperature T) and then uses the emissivity of the white paper to
    calculate the emitted energy of the white paper.

    If you start with the emitted energy of the white paper and use q = e*BC*T^4 to calculate the temperature T, you will ALWAYS get the temperature of the "Black Body" Black paper.

    That's why you can't use q = e*BC*T^4 "EXACTLY AS IT IS WRITTEN" to calculate the temperature of the white paper.

    To calculate the temperature of the white paper all you have to do is calculate the energy the white paper will absorb q = Q*e = Incident Sunlight Energy X emissivity of the white
    paper and then use q = BC*T^4 to calculate the white paper ACTUAL temperature....just like the ACTUAL Earth temperature is calculated!

    Anonymous, start with Point (18) and tell me if you agree or if you disagree explain why.

    Continue with (19), the calculations shown and explaination and tell me if you agree or if you disagree explain why.

    Now SHOW ME where I have "incessantly ignored basic science" or retract your statement.

    ReplyDelete
  173. @Gord

    1)Please show me how the 6,000 Joules/sec can increase to a value greater than 6,000 Joules/sec and comply with THE LAW OF CONSERVATION OF ENERGY!

    I would be happy to answer, but I really don't get the meaning of this question, if only you could enlighten me...

    2) "Do you have some sort of reading disability?" I can answer to this question: No, but thanks for worrying.

    Now I have one question. In your calculation you have 2390kW transported by conduction to the external surface, and conduction is the only mechanism that can transfer energy there...Nevertheless in your calculation your external surface is happily emitting at 19200W...Where are you getting the missing energy?

    ReplyDelete
  174. Given that Gord just stated:(my bold)
    "9) The SB equation q = e*BC*T^4*A where q = Power Radiated..."
    You would think that there is no disagreement on the form and use of the SB.

    But alas that is not how things play out.
    Same series of posts just a little earlier:(my bold)
    "Here is how SOD did the temperature calculation using q = e*BC*T^4*A which can also be written as q/e = BC*T^4*A:
    SOD used q = 30,000 Watts and e = 0.8 so q/e = 37,000 Watts Emitted...
    "

    Sure the SB, q = εσT^4*A, can be rewritten, q/ε = σT^4*A, but that does not change what q stands for! If q is the Emitted in one then it is still the Emitted in the next.
    That Gord does not get this is a stunning admission on his part.

    There is nothing wrong with making a mistake. But this and other mistakes have been pointed out to Gord by many posters yet he still maintains his flawed and unsupportable positons. Doing that is wrong.

    Gord, if you want to actually have a conversation you need to be able to admit you have made a mistake. If you can not see from the above that you have made a fundemantal basic error then there is not much hope for you. q ≠ q/ε.


    Neutrino

    ReplyDelete
  175. Giovi Pelle,

    The question was first asked by me in my post to you on November 29, 2010 at 4:28 AM.

    Ya might want to read that original post again because you never answered the question.

    I also said in the same post:
    "I will not continually repeat my previous posts and Physics links.
    Read before you BABBLE.
    If you just BABBLE without having posting any Physics links to support your "Opinions", I will simply tell you post the supporting Physics before I respond.
    Do you understand that?"

    I asked you the question Again on Nov 29, 2010 at 2:44 PM
    You never answered the question.

    I asked you the question Again on Dec 4, 2010 at 2:46 AM
    You never answered the question.

    I asked you the question Again on Dec 7, 2010 at 11:15 PM
    You never answered the question.

    Now you say:
    "I would be happy to answer, but I really don't get the meaning of this question, if only you could enlighten me..."

    Well why don't you go back to my original post on November 29, 2010 at 4:28 AM and "READ THE QUESTION" as I originally posted it AND ANSWER THE QUESTION??

    Do you think that might "enlighten" you?
    ----
    You asked:
    "Now I have one question. In your calculation you have 2390kW transported by conduction to the external surface, and conduction is the only mechanism that can transfer energy there...Nevertheless in your calculation your external surface is happily emitting at 19200W...Where are you getting the missing energy?"

    I posted the answer on Nov 12, 2010 at 5:41 AM
    I posted it again on Nov 22, 2010 at 12:00 AM
    I posted it again on Nov 23, 2010 at 6:26 PM
    I posted it again on Nov 25, 2010 at 2:35 AM

    What part of "I will not continually repeat my previous posts and Physics links. Read before you BABBLE." don't you understand?
    -------
    You said...
    "2) "Do you have some sort of reading disability?" I can answer to this question: No, but thanks for worrying."

    HAHAHA...Really?

    ReplyDelete
  176. @Gord

    Wow, You never cease to amaze me. You posted the answer four times, and you got it wrong four times, since, among the many things, you fail to understand that energy reaches the outer surface only by heat conduction. You fail to understand even the tiniest little bit of physics with admirable consistency. I must confess that I begin to suspect that you are acting. Nobody can be that ignorant and arrogant at the same time. I am surprised that MS is not entering the discussion, this shameless display of mistakes is really detrimental for this blog. Nevertheless I think that it's important to go on in the discussion, not for Gord of course, he is a hopeless case, but for someone who might drop in and try to understand something.

    ReplyDelete
  177. I suspect this is pointless, but I will give it another go.

    Gord,
    You invite me to start at your point (1) above. Actually, no, we need a point zero. Remember the advice you were given at school when you had to do a test? READ THE QUESTION!
    So point zero,
    (0) What does the question say. It describes a case with a heat source of 30KW inside a sphere. The RADIATION from the source crosses to the inner surface of the sphere, and is CONDUCTED to the outer surface, The outer surface then RADIATES to space. You are asked to work out the equilibrium conditions i.e the temperature when 30KWatt is RADIATING away from the sphere and 30KWatts is being CONDUCTED between the inner at outer surfaces of the sphere. Got that?
    What is outside the sphere - absolutely nothing - vacuum at 0K - no external heat source - NO INCIDENT RADIATION on the outer surface. The outer surface is only RADIATING - the energy is arriving from inside by CONDUCTION which means that the energy arriving must balance the energy leaving but their behaviors are described by DIFFERENT equations. So what is the first step - you have to use the ABSOLUTELY STANDARD Stefan-Boltzmann equation to calculate what the surface temperature must be if the sphere is RADIATING 30KWatts.
    OK?
    Until you get that there is no point attempting to address your other mistakes. I do not know about the other posters (who have been quite remarkably patient) but I will not comment until you have got point zero.

    You have probably lost sight of the purpose of the question by now - it is simply to show that you can have the inside of an object hotter than the outside while it still remains in balance. The example was a precursor to a more interesting example discussed at SoD (and also at WUWT) which consisted of a heat source OUTSIDE a sphere which is surrounded by a shell which allows free energy transmission in the inward direction but restricts the outward flow - guess what that shows?

    ReplyDelete
  178. Anonymous,

    First you NEVER ASKED A QUESTION!

    You made these STATEMENTS!

    - "Gord has incessantly ignored basic science"
    - "a) there is NO incident or absorbed radiation on the outer surface of the sphere, he persists in a fantasy version of the SB equations."
    - "b) the SB equations are totally standard that gives the power RADIATED by an object"
    - "You do not have to modify the answer by dividing by the emissivity or anything else."
    - "Just use the equation EXACTLY AS IT IS WRITTEN."

    I fully addressed all these in my post to you and asked you to respond to all the points and tell me if you agree and if you disagree to explain why.

    I also asked you to SHOW ME where I have "incessantly ignored basic science" or retract your statement.

    I am STILL WAITING FOR YOU TO DO THAT!
    -------
    Now regarding your current post which I have ALREADY POSTED ANSWERS FOR IN MY LAST POST TO YOU:

    1) All objects that have a temperature above absolute zero means that they have ABSORBED ENERGY and will EMITT ENERGY...GOT THAT?

    - The PVC Sphere has ABSORBED energy from the 30,000 Energy Source.

    - 24,000 Watts was ABSOBED by the Inner Sphere Surface producing a TEMPERATURE of 135.47K.....GOT THAT?

    - There is no Transmission of the EM field but there is TRANSMISSION OF HEAT BY CONDUCTION TO ALL OF THE PVC MATERIAL THAT CAUSES IT'S TEMPERATURE TO RISE...GOT THAT?

    - The TEMPERATURE of the Outer Sphere is less than the TEMPERATURE of the Inner Sphere because ENERGY WAS STORED IN THE PVC MATERIAL and THERE WAS HEAT TRANSFER BY CONDUCTION FROM THE INNER SURFACE TO THE OUTER SURFACE....GOT THAT?

    - READ MY CALCULATIONS BEFORE BABBLING....GOT THAT?

    2) The Law of Conservation of Energy says Energy Emitted = Energy Absorbed...GOT THAT?

    3) SOD's "VERY FIRST CALCULATION" was to use the SB equation to calculate the temperature of the Outer Sphere surface...GOT THAT?

    I also used the SB equation to calculate the temperature of the Outer Sphere surface....GOT THAT?

    continued...

    ReplyDelete
  179. continuation..

    4) SOD used 30,000 Watts as the ABSORBED POWER, A = 2123.73 m^2 and an emissivity of 0.8 to calculate the Outer Sphere surface temperature.

    - The ABSORBED POWER that SOD used for the Outer Sphere surface was the 30,000 Watts of the Source Energy that he said has to reach the Outer Sphere surface as a consequence of The Law of Conservation of Energy and was transferred by CONDUCTION ignoring the FACT that the INNER SPHERE surface could ONLY ABSORB 24,000 Watts so he already VIOLATED The Law of Conservation of Energy....GOT THAT?

    - SOD then WRONGLY used the emissivity of the Outer Sphere surface when he used the 30,000 Watts and DIVIDED it by the Emissivity to get 30,000 / 0.8 = 37,000 Watts which produced an impossibily high Outer Sphere Surface temperature of 133K.....GOT THAT?

    - The 37,000 Watts used by SOD Violated The Law of Conservation of Energy AGAIN (that makes TWO VIOLATIONS)....GOT THAT?

    - SOD also did NOT ACCOUNT for the STORED ENERGY that the PVC material has....GOT THAT?

    - I used the CORRECT 24,000 Watts that the Inner Sphere surface ABSORBED THAT HAS TO REACH the Outer Sphere surface as a consequence of The Law of Conservation of Energy.

    - I correctly used the Emissivity of the Outer Sphere surface and MULTIPLIED the 24,000 Watts (that HAS TO REACH THE OUTER SPHERE SURFACE) by the Emissivity of 0.8 to get the CORRECT 24,000 Watts X 0.8 = 19,200 Watts Radiated by the Outer Sphere surface which produced an Outer Sphere temperature of 112.37K....GOT THAT?

    - Using the correct temperature of the Inner Sphere surface (135.47 K) and the Outer Sphere surface temperature (112.37 K), I calculated the heat transfer by CONDUCTION to be 2390 Watts from the Inner to Outer surfaces....GOT THAT?

    - Finally, the Watts stored in the PVC was calulated to be 24,000 - 19,200 - 2390 = 2410 to comply with The Law of Conservation of Energy....GOT THAT?

    - SOD after VIOLATING The Law of Conservation of Energy Twice and ignoring Stored Energy, proceeded to back calculate the Inner Sphere Surface temperature (using the impossible 133 K temperature of the Outer Sphere) using the 30,000 Watts and came up with a FANTASTIC 423K temperature!....GOT THAT?

    - Next, SOD used the SB equation to calculate a HILLARIOUS 1,824,900 Watts being inside the Inner Sphere where the Source is only 30,000 Watts!....GOT THAT?
    ------------------------------
    Now QUIT AVOIDING MY LAST POST TO YOU.

    Respond to all the points and tell me if you agree and if you disagree to explain why.

    SHOW ME where I have "incessantly ignored basic science" or RETRACT YOUR STATEMENT!

    ReplyDelete
  180. Giovi Pelle,

    You said...
    "Wow, You never cease to amaze me. You posted the answer four times, and you got it wrong four times, since, among the many things, you fail to understand that energy reaches the outer surface only by heat conduction. You fail to understand even the tiniest little bit of physics with admirable consistency. I must confess that I begin to suspect that you are acting. Nobody can be that ignorant and arrogant at the same time. I am surprised that MS is not entering the discussion, this shameless display of mistakes is really detrimental for this blog. Nevertheless I think that it's important to go on in the discussion, not for Gord of course, he is a hopeless case, but for someone who might drop in and try to understand something."

    What a HOOT!...as "ignorant" and "arrogant" as usual and, of course, totally devoid of any facts or any basis in reality as usual.

    First the PVC happens to have a temperature and any body with a temperature will radiate energy.

    That is fundamental Physics and about as simple as it gets but for some reason you can't seem to grasp because of your astonshing ignorance.

    Don't you rememember the Inner and Outer Sphere temperature calculations by SOD and myself?

    SOD's first calculation was to calculate the Outer Sphere temperature using the Stefan-Boltzmann Law and I used the Stefan-Boltzmann Law for that calculation as well.

    The PVC inner surface ABORBS HEAT FROM THE SOURCE and CONDUCTION of HEAT THROUGHOUT the PVC is the reason the PVC HAS A TEMPERATURE AT THE OUTER SURFACE....DUH!

    All my calculations are in my Post on Nov 12, 2010 at 5:38 AM

    A summary of the methods of how I did my calculations and how SOD did his was posted by me on Nov 25, 2010 at 2:34 AM.

    You obviously have not read them or are not mentally capable of understanding them.

    You certainly do you have the ability to READ or understand this:

    "I will not continually repeat my previous posts and Physics links.
    Read before you BABBLE.
    If you just BABBLE without having posting any Physics links to support your "Opinions", I will simply tell you post the supporting Physics before I respond.
    Do you understand that?"
    --------------
    I see that once again, you refuse to answer my question:
    "Please show me how the 6,000 Joules/sec can increase to a value greater than 6,000 Joules/sec and comply with THE LAW OF CONSERVATION OF ENERGY!"

    So, I will no longer waste my time on responding to your "ignorant" and "arrogant" posts utill you provide an answer along with the supporting Physics.

    Since your hillarious CREATION OF ENERGY is IMPOSSIBLE, I expect that this will be the last time I will have to respond to your delusional posts.

    Merry Christmas.

    ReplyDelete
  181. @Gord

    "Finally, the Watts stored in the PVC was calculated to be 24,000 - 19,200 - 2390 = 2410 to comply with The Law of Conservation of Energy....GOT THAT?"

    How on earth can you store a constant power? You can store energy, but you can't store power. 2410W of stored power means that you have a net input of 2410J/s in the PVC. So... since Energy=Power*Time After a time t0 you have Energy=2410J*t0 in the PVC. So, given enough time, you can accumulate an arbitrary energy inside the PVC. Oh... sorry, may be you need a reference for the equation "Energy=Power*Time"...Please Gord go on...you are becoming a living legend at the Physics Department of my university...

    ReplyDelete
  182. Gord

    I know exactly how SoD did his calculation - it is easy to understand - it is basic undergraduate level (or lower) physics - and it is completely and utterly correct. There is no polite way to state this, but I have to say that it is your grasp of physics that is faulty. Where did you 'learn' your physics?

    Why don't you go and talk to someone who understands physics. There is no point in trying to answer your 'questions' because they are (mostly) based on misconceptions. The correct interpretations have been given to you several times.

    For example, this "the INNER SPHERE surface could ONLY ABSORB 24,000 Watts" is incorrect, and you just keep getting more wrong from there.(you keep failing to give a correct answer to the question of what happens to the reflected energy)

    ReplyDelete
  183. Gord,

    The problem is you simply just dont get it.

    I have been trying to point out your improper use of the SB, Giovi Pelle has been trying to show you where you are violating Conservation of Energy, and the above anon poster was showing you your basic lack of understanding about the original problem.

    On top of all that your analysis are not internally consistant. An example of this is your last series of posts.

    You admit that the only form of Energy transfer between the inner and outer surfaces is via conduction:
    "- There is no Transmission of the EM field but there is TRANSMISSION OF HEAT BY CONDUCTION TO ALL OF THE PVC MATERIAL THAT CAUSES IT'S TEMPERATURE TO RISE...GOT THAT?"

    Then go on to state how much is conducted:(my bold)
    "- Using the correct temperature of the Inner Sphere surface (135.47 K) and the Outer Sphere surface temperature (112.37 K), I calculated the heat transfer by CONDUCTION to be 2390 Watts from the Inner to Outer surfaces....GOT THAT?"

    Yet you say the outer sphere is radiating:(my bold)
    "- I correctly used the Emissivity of the Outer Sphere surface and MULTIPLIED the 24,000 Watts (that HAS TO REACH THE OUTER SPHERE SURFACE) by the Emissivity of 0.8 to get the CORRECT 24,000 Watts X 0.8 = 19,200 Watts Radiated by the Outer Sphere surface which produced an Outer Sphere temperature of 112.37K....GOT THAT?"

    So, here is what you are saying:
    1. 24kW absorbed at inner surface
    2. 2390W conducted through to outer surface
    3. 19.2kW emitted from outer surface.

    Besides the fact your numbers are absurd they do not even add up!!!

    You said:
    "- I used the CORRECT 24,000 Watts that the Inner Sphere surface ABSORBED THAT HAS TO REACH the Outer Sphere surface as a consequence of The Law of Conservation of Energy."

    Assuming, for the sake of the arguement, that the 24kW is correct then:
    Why isnt the conduction from inner to outer surface 24kW?
    Why inst the outer surface radiating 24kW?

    Can you not see the inconsistencies in your posrition???


    Neutrino

    ReplyDelete
  184. Anonymous,

    You said..
    "There is no polite way to state this, but I have to say that it is your grasp of physics that is faulty. Where did you 'learn' your physics?"

    I am a Professional Electrical Engineer and have a Consulting Engineering Company in operation for over 22 years.

    I design Microwave, Satellite, Radio, TV, Coaxial, Fiber Optic and Optical Communication Systems.

    What qualifications and education do you have if any ?
    ---
    You said...
    "..the INNER SPHERE surface could ONLY ABSORB 24,000 Watts" is incorrect, and you just keep getting more wrong from there.(you keep failing to give a correct answer to the question of what happens to the reflected energy)"

    I see that you can BABBLE a lot without posting ANY PHYSICS to back-up your delusional "opinions".

    What a HOOT!

    Why don't POST some Physics links for back-up?

    Maybe, because THEY DON'T EXIST???
    --------
    I have already posted why it is only possible for the INNER SPHERE that has an EMISSIVITY of 0.8 to absorb 24,000 Watts several times and I do it ONE MORE TIME FOR YOU.
    ----------
    Physics of Emissivity

    Incident Energy = Absorbed Energy + Transmitted Energy + Reflected Energy (1)

    Emitted Energy = Absorbed Energy
    Consider equation 1 for an object in a vacuum at a constant temperature. Because it is in a vacuum, there are no other sources of energy input to the object or output from the object. The absorbed energy by the object increases its thermal energy - the transmitted and reflected energy does not. In order for the temperature of the object to remain constant, the object must radiate the same amount of energy as it absorbs.

    Emitted Energy = Absorbed Energy (2)

    "Therefore, objects that are good absorbers are good emitters and objects that are poor absorbers are poor emitters. Applying equation 2, Equation 1 can be restated as follows:

    Incident Energy = Emitted Energy + Transmitted Energy + Reflected Energy (3)

    Setting the incident energy equal to 100%, the equation 3 becomes:

    100% = %Emitted Energy + %Transmitted Energy + %Reflected Energy [4]

    Because emissivity equals the efficiency with which a material radiates energy, equation 4 can be restated as follows:

    100% = Emissivity + %Transmitted Energy + %Reflected Energy (6)

    http://www.optotherm.com/emiss-physics.htm
    ----
    Continued..

    ReplyDelete
  185. Continuation...

    Incident Energy = Absorbed Energy + Transmitted Energy + Reflected Energy (1)

    100% = Absorbed Energy/ Incident Energy + Transmitted Energy/Incident Energy + Reflected Energy/Incident Energy

    Equating the above to (6) gives:

    Emissivity = Absorbed Energy/ Incident Energy and since Emitted Energy = Absorbed Energy.....Emissivity = Emitted Energy/ Incident Energy
    =================================
    Absorbed Energy = Emissivity X Incident Energy

    Emissivity = 0.8 for the INNER SPHERE and the Incident Energy = 30,000 Watts

    Absorbed Energy by the INNER SPHERE = 0.8 X 30,000 Watts = 24,000 Watts.

    The PVC has ZERO Transmitted Energy so:

    Incident Energy (30,000 Watts) = Absorbed Energy (24,000 Watts) + Reflected Energy (6,000 Watts) + Transmitted Energy (0 Watts)!!!!!

    Repeating the quote above:
    "The absorbed energy by the object increases its thermal energy - the transmitted and reflected energy does not."

    The Inner Sphere CAN ONLY ABSORB 24,000 Watts!...GOT IT?

    THIS IS THE RESULT BASED ON THE LAW OF CONSERVATION OF ENERGY....NOT BABBLING LIKE YOU CAN ONLY DO.
    =================================
    The Reflected 6,000 Watts reduces the Incident Energy flow by "destructive interference" to 30,000 - 6,000 = 24,000 Watts.

    Incident Energy (30,000 Watts) = Absorbed Energy (24,000 Watts) + Reflected Energy (6,000 Watts) + Transmitted Energy (0 Watts)

    Incident Energy (30,000 Watts) - Reflected Energy (6,000 Watts) = Absorbed Energy (24,000 Watts) + Transmitted Energy (0 Watts)

    I have posted about "destructive interference", EM field being Vector fields, Vector Mathematics, provided Physics links and actual Physical Examples to support this several times and will not repeat them.

    You can READ my previous posts for yourself.
    ---------
    Now QUIT BABBLING and QUIT AVOIDING MY DEC 9, 2010 2:55 AM POST T