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    • CommentAuthormike7
    • CommentTimeDec 18th 2017 edited
     
    For the sort of heating situation people here are likely to be interested in there's an easy way to estimate radiant heat transfer without getting involved with fearsome-looking maths using Kelvins to the power of four and lots of noughts somewhere or finding the online calculator at eng. toolbox.

    Provided that the source emissivity is 0.9, the answer is 6 Watts per square metre for each degree C difference between the source and ambient - roughly.
    How rough? It depends also on the mean of the source and the ambient temperatures. If that mean is 35C, eg say 50C source and 20C ambient, it gives a very accurate answer of 6 x 30 = 180W/m2

    If the mean is not 35C the result will be lower or higher by 1% for every degree C lower or higher. Thus in WillinAberdeen's warm ceiling idea the source was 30C and ambient 20, so the mean is 25C, ie 10C lower than 35C, and therefore the answer will be 10% lower than the 60W estimate, ie 54W.

    The whole thing assumes that the emissivity is 0.9, which is a fair guesstimate for many likely surfaces but if you insist on chrome plated radiators for example, then there would be a large adjustment to be made pro rata for a very different emissivity.
    • CommentAuthorgyrogear
    • CommentTimeDec 19th 2017
     
    great, many thanks !

    https://www.youtube.com/watch?v=BWpYQjuJ0u0

    gg
    :devil:
    •  
      CommentAuthorfostertom
    • CommentTimeDec 19th 2017 edited
     
    Brilliant mike. I've enjoyed a running battle for years, decade even, here on GBF on this topic, about the difference between

    a) the fourth power of the difference between two absolute temps, and

    b) the difference between the fourth powers of two absolute temps.

    a) produces large and wild numbers for the supposed radiant heat transfer between two bodies;
    b) approximates to nicely linear (heat transfer proportional to delta-t) over small ranges of absolute temps (e.g. 263K to 313K).

    What you've done is reduce b) to a v pragmatically useful rule of thumb. Invaluable. Thanks.
    •  
      CommentAuthordjh
    • CommentTimeDec 19th 2017
     
    Excellent idea. :bigsmile:

    For anybody that does want to find the Engineering Toolbox calculator, it's at https://www.engineeringtoolbox.com/radiation-heat-transfer-d_431.html

    There are also a lot of other things such as radiator sizing etc https://www.engineeringtoolbox.com/heat-emission-radiators-d_272.html
  1.  
    Thanks Mike for suggesting as a rule of thumb: U= 6 W/m2K for radiant heat transfer at moderate temperatures.

    U is the conductivity. (Or should it be 'radiativity' here?)

    Here are some bits I came across during previous thread:

    SAP / BR443 gives the thermal resistance downwards from a sheltered surface as R=0.17, which nicely ties up with
    1/R = U = 6 W/m2K


    Or some "fearsome maths", never be a feartie! Using:  hot temp = T ,  cold temp = t,  temp difference = d = (T -t)

    U = εσ(T^4 – t^4)/d    

    binomial series:
    T^4-t^4
    = T^4-(T-d)^4
    = T^4 - T^4(1 + 4(d/T) - 6(d/T)^2 + ….  )

    Drop the terms beyond (d/T)^2 which are very small so:
    T^4-t^4 = 4 T^3 d
    And so
    U = 4 εσ T^3

    Plug in ε = 0.9; σ=5.67E-8 and T=310K gives:
    U = 6 W/m2K

    Or try different Kelvin temps to give the corresponding U's.

    Previous thread: http://www.greenbuildingforum.co.uk/newforum/comments.php?DiscussionID=15448&page=1#Item_0
  2.  
    Posted By: WillInAberdeenU = εσ(T^4 – t^4)/d

    binomial series:
    T^4-t^4
    = T^4-(T-d)^4
    = T^4 - T^4(1 + 4(d/T) - 6(d/T)^2 + ….  )

    Drop the terms beyond (d/T)^2 which are very small so:
    T^4-t^4 = 4 T^3 d
    And so
    U = 4 εσ T^3

    Plug in ε = 0.9; σ=5.67E-8 and T=310K gives:
    U = 6 W/m2K

    Or try different Kelvin temps to give the corresponding U's.


    OMG I'm going to get that other bottle of wine !!:shocked::confused:
    • CommentAuthorgoodevans
    • CommentTimeDec 20th 2017
     
    I recon that 6 Watt rule of thumb is a little on the low side for wall and floor emitters - if you add the effects of convection I think 11W per degree difference is closer to the mark for ideal UFH conditions. However you have to use the surface temp of the emmitter - for radiators it will be very close to the water temp - for UFH the surface temp is affected by the floor covering.

    emitters on the ceiling may be almost purely radiative (except for heating the room above) so the 6W rule would apply here.
    •  
      CommentAuthorSteamyTea
    • CommentTimeDec 20th 2017
     
    So no one has taken view factor and reciprocity into account then.
    • CommentAuthorgyrogear
    • CommentTimeDec 21st 2017
     
    Posted By: SteamyTeaSo no one has taken view factor and reciprocity into account then


    by all means please feel free to enlighten us
    :shamed:

    gg
    •  
      CommentAuthorfostertom
    • CommentTimeDec 21st 2017
     
    I guess view factor means occlusion and/or angularity (to add more jargon) but what's reciprocity? the fact that both the hotter and the cooler body emit at each other and it's the (nett) difference that is usually of interest?
    • CommentAuthormike7
    • CommentTimeDec 21st 2017 edited
     
    Posted By: WillInAberdeen


    Or some "fearsome maths", never be a feartie! Using:  hot temp = T ,  cold temp = t,  temp difference = d = (T -t)

    U = εσ(T^4 – t^4)/d



    Its not that the maths itself is all that fearsome, possibly it is at least as much the notation used that is off-putting for those whose maths education stopped short of yours - and mine. I read somewhere (R Dawkins?) that sales drop by 10% for every formula printed in a book .... so I purposely expressed the rule of thumb in words, and it helps that the rule is simple enough to do this. Doing it with the correction for other temperatures gets a bit wordy, but still I'd guess it's more widely acceptable than writing the formula.

    Anyway, prolonging the fun a bit more :- as any green builder will know, another way of looking at it is to find the slope of the W=f(T^4) curve using differential calculus, which gets us dW/dT =4f(T^3). Hands up all whose eyes have not glazed over already...

    ...pressing on, the rate of change of the slope would be the second differential d2W/dT2 which is 12f(T^2). The percentage change in W as T varies would be 100 x 12f(T^2)/4f(T^3) which is 300/T i.e. spot on 1% for T=300K and close enough at my chosen 35C (308K). That the figures of 6W and 1% are so round and memorable is a sure sign God had a hand in this, or would be if I wasn't an atheist.

    Note to goodevans - you're quite right that it would be too low, but the rule is intended to deal only with the radiation component of heat transfer.
    •  
      CommentAuthorSteamyTea
    • CommentTimeDec 21st 2017
     
    Posted By: mike7R Dawkins?
    Was Simon Mitton
    • CommentAuthormike7
    • CommentTimeDec 21st 2017
     
    Thanks ST. Re Simon Ditton Wiki says :"Earlier in his career, while employed by the Cambridge University Press, he was the editor in question when "Stephen Hawking famously put the success of his bestseller A Brief History of Time down to advice from his editor that for every equation in the book the readership would be halved. As a result the book included only a single equation, E = mc2."
    I got it from Hawking's book.
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