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    • CommentAuthorSeret
    • CommentTimeApr 4th 2013
     
    Indeed, I have confidence that the people who design building heat loss models understand the physics of heat transfer well enough to not miss out any significant sources of losses. I suspect that they understand it a lot better than us here on this forum (myself definitely included!), judging from how we've flailed around the subject in this thread.
    • CommentAuthorEd Davies
    • CommentTimeApr 4th 2013
     
    Posted By: SeretYou're confusing emissivity with transmittance. Low-e coatings are to stop the window itself radiating, not to stop IR radiated from other sources.
    Yes, but…

    Having a low-e value does indeed stop the surface itself radiating. But it also causes the surface to reflect back a larger proportion of the long-wave IR it receives from, e.g., the next sheet of glass closer to the room.
    • CommentAuthorSeret
    • CommentTimeApr 4th 2013
     
    For sure Ed, there's all sorts of things going on in windows. Some energy from either side is reflected, some absorbed, some of that gets re-radiated, some conducted, etc, etc. I don't mind having a headline figure of a u-value that wraps it all up into one usable figure.
    •  
      CommentAuthorfostertom
    • CommentTimeApr 4th 2013
     
    Posted By: Seretit's Thot4-Tcold4 not ΔT4
    Correct, but over limited temp range it's near enough same result.
    •  
      CommentAuthorfostertom
    • CommentTimeApr 4th 2013 edited
     
    Hot emitter body A, cold receiver body C - radiant transfer near enough proportional to temp difference A-C.
    Put a sheet of black paper B between them - B's temp will equilibriate at half way between A and C - A radiates to B and B must re-radiate same amount to C (or burst into flames!). So both legs of the radiation transfer are near enough proportional to half of temp difference A-C.
    So a simple sheet of black paper between will halve the radiant transfer between two bodies, compared to clear line-of-sight.
    Make B a reflective sheet, and it only absorbs one tenth as much from A (reflecting nine tenths), will re-radiate same amount to C, but B's temp will still equilibriate at half way between A and C.

    Are we saying that ordinary float glass really does completely prevent long wave IR radiation through it, as does the sheet of black paper B or the reflective sheet B? And that regardless of reflectivity, coatings etc, a single sheet of intervening glass B will still equilibriate at half way between A and C?
    So the only question is what outward re-radiation (B to C) results from the black paper/reflective sheet/glass sheet's temp difference B-C.
    Will it be half x 0.9 (plain float glass), or half x 0.1 (coated glass), compared to clear line-of-sight radiant transmission?
    • CommentAuthorSeret
    • CommentTimeApr 4th 2013
     
    Posted By: fostertom
    Posted By: Seretit's Thot4-Tcold4 not ΔT4
    Correct, but over limited temp range it's near enough same result.


    Not really, let's look at an example with realistic temperatures. If the cold side is at 5º and the hot side is at 15º the result would be 906,890,480, if you just took ΔT4 you'd get 10,000 which is over 90,000 times smaller. It's absolute temperature in K, not ºC
    • CommentAuthorEd Davies
    • CommentTimeApr 4th 2013
     
    Posted By: fostertomHot emitter body A, cold receiver body C - radiant transfer near enough proportional to temp difference A-C.
    Yes, but not (temp difference A-C)⁴
    •  
      CommentAuthorfostertom
    • CommentTimeApr 4th 2013 edited
     
    Posted By: SeretIf the cold side is at 278K and the hot side is at 288K the result would be 9.0689n8
    And if the cold side is at 278K and the hot side is at 298K the result would be 1.9133n9 - almost exactly twice as much radiant transfer for twice the delta-t.
    I'm talking about the proportionate change, not the absolute value. So
    Posted By: fostertomHot emitter body A, cold receiver body C - radiant transfer near enough proportional to temp difference A-C
    is true - note the word Proportional.
    • CommentAuthorSeret
    • CommentTimeApr 4th 2013
     
    Posted By: fostertom
    I'm talking about the proportionate change, not the absolute value.


    Right I'm with you.

    As for your example given above, assuming only radiative heat transfer the equilibrium temperature of B will be controlled by it's emissivity. You can think of emissivity like insulation. If the heat isn't able to transfer as quickly then the temperature of that medium will rise until the heat transfers quickly enough to match the rate it's being supplied. It won't simply always be half way between A and C, except for one specific value of emissivity.
    •  
      CommentAuthorfostertom
    • CommentTimeApr 4th 2013
     
    Assuming absorbtivity is same as emissivity, which it mostly is, near enough, then radiation absorbed by B from hot source A will only equal radiation emitted (re-radiated) by B to cold target C, when the delta-t's for both transactions are the same. If radiation absorbed by B does not equal radiation simultaneously emitted (re-radiated) by B, then B will steadily either accumulate or disperse heat, hence will either burst into flames or freeze, which it doesn't. Both 'legs' of the temp gradient A to C must be equal, hence B's temp must equilibriate half way between A and C, regardless of its emissivity.
    •  
      CommentAuthordjh
    • CommentTimeApr 4th 2013 edited
     
    Posted By: fostertomAre all our so-precise calcs (esp in e.g. PHPP) in fact way out?

    Well PHPP predictions are frequently checked against the resulting buildings, AIUI. And PHI investigates when there are discrepancies and tweaks the model, I believe.

    Do we have info on the %age of IR that's blocked by different coatings?

    It's not difficult to find diagrams like the one at http://www.commercialwindows.org/lowe.php It seems to show higher transmission than has been mentioned earlier in the thread. But that difference may be accounted for by the difference between radiation that passes straight through glass, in the same way as visible light, and radiation that is absorbed and reemitted at a very similar wavelength.

    Both 'legs' of the temp gradient A to C must be equal, hence B's temp must equilibriate half way between A and C, regardless of its emissivity.

    No, you're ignoring that B is a real physical object with a mass and a specific heat and with conduction and convection to alter its temperature as well. And it can 'see' other objects as well as A and C.
    • CommentAuthorSeret
    • CommentTimeApr 4th 2013
     
    Posted By: fostertomAssuming absorbtivity is same as emissivity, which it mostly is


    Nope, emissivity and surface absorptivity aren't necessarily the same, and there's nothing stopping an object from absorbing and radiating at different rates on different faces. Just polish one and leave one dull or apply a bit of paint, or expose them to different rates of oxidation and you'll have a large effect. Emissivity/absorptivity are a matter of surface finish, they're not a materials property. You can engineer it to be whatever you want.


    then B will steadily either accumulate or disperse heat, hence will either burst into flames or freeze


    It'll only burst into flames if the new equilibrium temperature is too high. As it's temperature rises, it sheds heat faster, so you can reach a new equilibrium. But yes, that might be above melting or combustion point if you're really pumping a lot of heat into something flimsy.
    • CommentAuthorEd Davies
    • CommentTimeApr 4th 2013
     
    Posted By: SeretNope, emissivity and surface absorptivity aren't necessarily the same
    …although they are for any given wavelength.

    Imagine you had two objects next to each other at the same temperature. If one was more absorptive and less emissive that the other then you'd have a net flow of heat between them. One of the laws of thermodynamics prohibits that.
    • CommentAuthorSeret
    • CommentTimeApr 4th 2013 edited
     
    I meant that one object can absorb from one face at a different rate that it emits from another Ed. Such as a pane of glass which had been surface engineered deliberately to do so.
    •  
      CommentAuthorSteamyTea
    • CommentTimeApr 4th 2013
     
    Posted By: Ed DaviesOne of the laws of thermodynamics prohibits that.
    Is that Zeroth Law, it assumes that there is a third body, the atmosphere around the other two and they are all in equilibrium.
    •  
      CommentAuthorfostertom
    • CommentTimeApr 4th 2013 edited
     
    Posted By: djh
    Both 'legs' of the temp gradient A to C must be equal, hence B's temp must equilibriate half way between A and C, regardless of its emissivity.

    No, you're ignoring that B is a real physical object with a mass and a specific heat and with conduction and convection to alter its temperature as well. And it can 'see' other objects as well as A and C.
    Yes I was ignoring all that, to establish basics by simplified thought experiment without complicating features (which could then be added in if you want). No convection/conduction. Allowing enough time for mass and specific heat lags to equilibriate. Assuming only A B and C exist. Surface finish both faces identical. Any objections to that?

    Posted By: SeretAs it's temperature rises, it sheds heat faster, so you can reach a new equilibrium
    That means that if temp difference A-B (driving gain) is higher than B-C (driving loss), heat accumulates in B, whose temp rises still more, so receives less from A and emits more to C, so temp drops back to the 'new equilibrium' which is the temp at which neither heat accumulation nor dispersal is happening, which is temp equidistant between A and C.
    •  
      CommentAuthorSteamyTea
    • CommentTimeApr 4th 2013
     
    As soon as you involve a, temperature difference, a mass and an area of something in thermal calculations you have to stop thinking in a linear fashion. Start thinking in (and calculate in) natural logarithms.
    • CommentAuthorSeret
    • CommentTimeApr 4th 2013 edited
     
    Posted By: fostertomThat means that if temp difference A-B (driving gain) is higher than B-C (driving loss), heat accumulates in B, whose temp rises still more, so receives less from A and emits more to C, so temp drops back to the 'new equilibrium' which is the temp at which neither heat accumulation nor dispersal is happening, which is temp equidistant between A and C.


    No, the temp doesn't drop back, just the rate of increase slows until you reach equilibrium at the new higher temperature.
    •  
      CommentAuthorfostertom
    • CommentTimeApr 4th 2013
     
    That means that B is free to equilibriate at any temp it fancies, while A and C maintain their temps static?

    Remembering that, in equilibrium, radiation flow A to B must equal B to C otherwise B accumulates or sheds nett heat,
    and that (assuming B's absorbtivity equals its emissivity) both radiation flows A to B and B to C are both proportional to their respective delta-t,
    so both delta-t's must be equal,
    then how can B's temp fail to be equidistant between A and C?
    • CommentAuthorSeret
    • CommentTimeApr 4th 2013
     
    Posted By: fostertomThat means that B is free to equilibriate at any temp it fancies, while A and C maintain their temps static?


    Well A and C being static is a condition of the experiment, otherwise everything ends up at the same temperature. The temperature that B equilibrates at is fixed by the emissivity of its faces.


    Remembering that, in equilibrium, radiation flow A to B must equal B to C otherwise B accumulates or sheds nett heat


    Sure, once you reach equilibrium it will be. But if B has differential emissivity then the whole A-B cell will have got hotter than the mid point between A and C. That's exactly why we have low-e coatings on glass: to trap the heat on one side of the system.


    and that (assuming B's absorbtivity equals its emissivity) both radiation flows A to B and B to C are both proportional to their respective delta-t,
    so both delta-t's must be equal,
    then how can B's temp fail to be equidistant between A and C?


    Sure, but if you've chosen B's properties so that it hits that temperature, what does that prove? If you design an experiment so that it achieves certain conditions, then achieving those conditions doesn't prove anything, except that your method and assumptions were correct.

    Or to put it another way: what are you actually trying to say with this thought experiment? I'm suspect I'm rather missing your point...
    •  
      CommentAuthorfostertom
    • CommentTimeApr 4th 2013
     
    Posted By: Seretif B has differential emissivity then the whole A-B cell will have got hotter than the mid point between A and C
    The thought experiment is what happens when emissivity is not differential - if that means that absorbtivity and emissivity are equal. Because AFAIK that is the case with all of: black paper; reflective silver foil; and coatings on glass - in all these cases absorbtivity and emissivity are equal.
    Posted By: SeretThat's exactly why we have low-e coatings on glass: to trap the heat on one side of the system
    To trap heat on one side of the glass does not require absorbtivity to be different from emissivity.
    •  
      CommentAuthordjh
    • CommentTimeApr 5th 2013
     
    • CommentAuthorSeret
    • CommentTimeApr 5th 2013 edited
     
    Posted By: fostertomBecause AFAIK that is the case with all of: black paper; reflective silver foil; and coatings on glass - in all these cases absorbtivity and emissivity are equal.


    I don't know a great deal about glass manufacturing, but I'd be very surprised if the low-e coating was applied to both sides. It would cost twice as much, to no advantage.

    Anyway, it doesn't really matter. The equilibrium temperature of an object is controlled by the emissivity. Consider a zebra crossing in nice bright sun. You can already tell me what will happen, the black stripes will get hotter than the white ones. That's because emissivity can be frequency dependent, and a single body doesn't necessarily absorb and emit at the same rate.

    Bottom line is that your thought experiment would only work the way you wanted it to if it were carefully designed to do so. You couldn't just pick any material for B and get the same results.
    •  
      CommentAuthorSteamyTea
    • CommentTimeApr 5th 2013
     
    Posted By: Seretthe black stripes will get hotter than the black ones.
    You may want to edit that :bigsmile:
    •  
      CommentAuthorfostertom
    • CommentTimeApr 5th 2013
     
    Posted By: SeretI'd be very surprised if the low-e coating was applied to both sides
    O'course it's a single coating but both faces of the coating are operative.
    •  
      CommentAuthorSteamyTea
    • CommentTimeApr 5th 2013
     
    Posted By: fostertomboth faces of the coating are operative.
    I would not be so sure of that.
    • CommentAuthorSeret
    • CommentTimeApr 6th 2013
     
    Posted By: fostertom
    Posted By: SeretI'd be very surprised if the low-e coating was applied to both sides
    O'course it's a single coating but both faces of the coating are operative.


    No, because as we've already established glass is very opaque to long wave IR. It is either reflected or absorbed when it hits the glass. The portion which is absorbed heats the glass, causing it to radiate more from each face. One face will be highly emissive, one won't. The glass will radiate more back towards the source of the radiation than it will to the outside.
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