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    • CommentAuthorlineweight
    • CommentTimeSep 13th 2021
     
    I'm trying to come up with a very rough estimate of the proportions of heat loss from various elements of an existing building. In other words, how much heat is being lost through walls vs roof vs floor.

    To do this I'm using this calculator:

    https://www.changeplan.co.uk/u_value_calculator.php

    Which says it calculates according to BS EN ISO 6946:2017.

    Walls and roofs are easy enough to understand but with the ground floor it's more complicated, because the overall U-value given takes into account the perimeter/area ratio.

    I understand in principle that this means the calculation method assumes a higher U-value at the perimeter than it does at the interior of the floor slab. What I am not sure about is what assumptions it makes about what's happening at the perimeter - does it assume that some degree of wall insulation extends around the perimeter of the floor slab?

    In the instance I am looking at, it's a concrete slab in direct contact with the ground, and I know that there is essentially no insulation extending around its perimeter (the external walls themselves sit on the slab and are barely insulated anyway). In addition, the floor level is a fair bit higher than the external ground level on one side, so one edge of the slab is significantly exposed to the outside air.

    Below are the results I get from the calculator. I am unsure what U-value I should best be using for my heat loss estimations.

    I know it's not the "uncorrected" U-value of 1.626 because that would ignore the reality of what is happening in the inner parts of the floor, in contact with the ground but not really affected by perimeter effects.

    However, I am not sure whether the "corrected" value of 0.356 is realistic, because I'm not sure whether the calculation method assumes a standard modern construction where the edges of the slab are insulated somewhat.

    Any thoughts appreciated.
      Screenshot 2021-09-13 at 12.28.29.jpg
  1.  
    That method is from BRE Information Paper 3/90, which like Nessie, many people talk about, but I have never managed to see myself...!

    The assumptions are equally mysterious.

    Legend has it that BRE did lots of finite element models of buildings with concrete slabs, with lots of different variations like shape, soil, water table, edges, etc. They plotted them all on a graph and drew a best-fit line through them, U=0.05+1.65(p/A)-0.6(P/A)² to give a very wet finger estimate.

    I also assume that BRE would have included the resistance of the concrete, and the surface resistances, in their model, but your pic seems to show them being added on separately (double counting?).

    IIUC their method refers to the P/A of the whole slab, however many people seem to look at individual rooms, giving very high or low P/A (like your example?)
    • CommentAuthorlineweight
    • CommentTimeSep 13th 2021 edited
     
    Posted By: WillInAberdeenThat method is from BRE Information Paper 3/90, which like Nessie, many people talk about, but I have never managed to see myself...!

    The assumptions are equally mysterious.

    Legend has it that BRE did lots of finite element models of buildings with concrete slabs, with lots of different variations like shape, soil, water table, edges, etc. They plotted them all on a graph and drew a best-fit line through them, U=0.05+1.65(p/A)-0.6(P/A)² to give a very wet finger estimate.

    I also assume that BRE would have included the resistance of the concrete, and the surface resistances, in their model, but your pic seems to show them being added on separately (double counting?).

    IIUC their method refers to the P/A of the whole slab, however many people seem to look at individual rooms, giving very high or low P/A (like your example?)



    For my particular example, it's a 'slice' through a building that I'm looking at - this means that it's roughly a 6x7m rectangle where two sides are external walls and the other two sides are internal partitions and contiguous with the slab that extends to other parts of the building. So, hence my 12m "perimeter", or at least that's what seemed to make most sense to me.

    It's frustrating that these calculation methods are often rather opaque.

    This seems publicly available at least -

    https://www.bre.co.uk/filelibrary/rpts/uvalue/BR_443_(2006_Edition).pdf
    • CommentAuthorLF
    • CommentTimeSep 13th 2021
     
    Not read fully as should be working but I think there is a lot of benefit in insulating around the slab/footings. The heat loss through building to 14 C soil in the middle of the slab is low so difference is low and driving force for heat loss. Ignore if not valid input.
    • CommentAuthorlineweight
    • CommentTimeSep 13th 2021 edited
     
    Posted By: LFNot read fully as should be working but I think there is a lot of benefit in insulating around the slab/footings. The heat loss through building to 14 C soil in the middle of the slab is low so difference is low and driving force for heat loss. Ignore if not valid input.


    Thanks, I agree that insulating around footings/slab is likely to be of benefit but at this point I'm just trying to quantify what's being lost where.
  2.  
    Most frustrating!

    It's common to just consider the exposed perimeter/area of a part of the building (eg one mid-terrace unit, or one room), but I don't think it's very accurate as that's not the same geometry as the original BRE FE models, from which the formula is extrapolating. If you slice up the building into pieces, calculate their heat losses, then add them all together, I don't think you get the same answer as if you calculate the whole building in one go. This might not be the worst of the inaccuracies!

    If we imagine a square building 14.7m each side, it would have the same P/A as your slice, but would it have the same U? (Who knows!)

    Our house has very thick stone walls with minimal foundations, I've been wondering how to account for the longer heat loss path in this case.

    BRE443 advises not to include the carpet or the screed in the resistances, but is a bit vague!
    • CommentAuthorlineweight
    • CommentTimeSep 13th 2021 edited
     
    I see there's also an updated version of that BRE doc - "2019 for consultation" vs the 2006 one.

    https://www.bregroup.com/wp-content/uploads/2019/10/BR443-October-2019_consult.pdf

    This has quite a bit added to the ground floors section... which I now need to read and attempt to understand.
  3.  
    If it's not too late to comment, I have a question about (for calculation purposes) the expected temperature of the ground below the floor slab. I'm assuming here there must be an expected temperature so that heat loss can be calculated?

    The ground conditions below the floor slab could be an issue here. For example, with very dry ground, heat leaking through the floor slab could heat up the sub-soil below the slab and therefore mitigate heat loss. However, if the slab is on slightly sloping heavy clay ground where there is no water table then the slab is at the mercy of weather conditions. If there is a wet winter and the clay allows gradual percolation then the ground temperature immediately below the slab is compromised and the slab becomes a significant heat leakage element.

    In effect, the ground conditions are a significant factor in floor slab heat loss calculations - is this factor included in any of the calculations?
    •  
      CommentAuthorfostertom
    • CommentTime4 days ago edited
     
    Spot on - the elephant in the room, when it comes to figuring in (whether protecting against, or exploiting) subsoil temperature.

    If it's well-drained e.g. cleanish broken rock subsoil, along with no high (permanent or seasonal) water table, then you can treat the subsoil as a straightforward conductor and storer of heat. Likewise if it's impermeable clay free of fissures providing horizontal watercourses through it.

    However if there's either vertical water movement in the subsoil, whether vertical due to rise and fall of water table, or horizontal due to watercourses through it, then the subsoil as conductor and storer of heat is defeated, dominated or at least complicated by the cooling effect of the water flow. Both these types of flow may be diurnal (after rain) or seasonal or both.

    If what's happening in the subsoil can be known, then all sorts of beneficial use can be made of exploiting subsoil heat storage, which are hardly considered except by enthusiasts. There was a lot of stuff on GBF about this, some years ago.

    But how to tell what's happening in the subsoil? Farmers probably have a fair idea, around their land. Cos that drill boreholes have much experience too. Dowsers? I'm sure there is a body of knowlege that could be collated, incl just looking at the lie of the land with an experienced eye. Personally, I've given up the effort!
    • CommentAuthortony
    • CommentTime3 days ago
     
    I believe that antipodean research some years ago revealed that both wet and dry subsoils stored heat nicely beneath houses and only in the case of flowing water were there significant losses, slowly rising and falling water tables having little influence

    For me the assumptions about subsoil temperatures cause high heat losses through floors to be shown by the calculations and real world heat losses are lower

    Steady state calculations are somewhat flawed as often the assumed steady state is never present - temperature swings daily and seasonally come into play.

    I once met an expert who told me that the temperature below the middle of the slab will be close to the average temperature of the house.
    •  
      CommentAuthorfostertom
    • CommentTime3 days ago edited
     
    Posted By: tonyboth wet and dry subsoils stored heat nicely beneath houses
    Yes - and wet stores more heat because it adds water's thermal mass to the mineral's - and the storage ability of 1 tonne of water is several times greater than 1 tonne of mineral.
    Posted By: tonyonly in the case of flowing water were there significant losses, slowly rising and falling water tables having little influence
    It's a simple matter of the mass flow rate of water - in the extreme case of water table rising and falling completely every 24hrs, the flow rate (rate of water leaving the scene as water table falls) could be quite high.
    Posted By: tonythe temperature below the middle of the slab will be close to the average temperature of the house
    That's true right across the slab - but then getting cooler at depth. That ground temp at depth under open ground is around 2-4m deep and will be the average of annual external air temp - in England perhaps the average of 0C and 20C so 10C.

    Under a slab the surface temp (internal air temp) is maintained at say 20C all year round instead of swinging between 20C and 0C. So you'd think the ground beneath would eventually reach 20C at depth - but not quite, because still the influence of the perimeter reaches inboard. You can say that the 10C is forced deeper.

    The rate of heat loss downward depends on the downward temp gradient - and around the edge a horizontal, outward temp gradient too.

    The point is, that gradient, and that loss, reduces towards the middle of the slab; in the middle of a very large slab, like a warehouse, it may approach zero.
  4.  
    Well, on the london underground, they've certainly reached a point where heat loss to the ground is effectively zero, and something of a problem!

    https://www.ianvisits.co.uk/blog/2017/06/10/cooling-the-tube-engineering-heat-out-of-the-underground/
    •  
      CommentAuthorfostertom
    • CommentTime3 days ago
     
    Yes, that's because surface temp has no influence; the tunnels are the only source of heat or cooling, apart from groundwater flow, and the steady-state temp of the tunnel's place in the long gradient between molten core and av surface temp. The latter gets hotter as you go down - but not as hot as mile-deep mines.
    •  
      CommentAuthordjh
    • CommentTime2 days ago
     
    Posted By: fostertomYes, that's because surface temp has no influence; the tunnels are the only source of heat or cooling, apart from groundwater flow, and the steady-state temp of the tunnel's place in the long gradient between molten core and av surface temp. The latter gets hotter as you go down - but not as hot as mile-deep mines.

    As the excellent article explains, it's actually to do with all the heat generated by the trains themselves, plus their passengers, rather than anything to do with it's position in the planet. The way that the planet is relevant is that the London clay acts as a big insulating blanket around the tunnels and slows down any heat loss or gain from the tunnel.
    •  
      CommentAuthorfostertom
    • CommentTime2 days ago edited
     
    Posted By: djhit's actually to do with all the heat generated by the trains themselves, plus their passengers
    Yes, I include them in
    Posted By: fostertomthe tunnels are the only source of heat or cooling

    Posted By: djhrather than anything to do with it's position in the planet
    Posted By: fostertomthe steady-state temp of the tunnel's place in the long gradient between molten core and av surface temp
    is definitely a third heat/temp input into the equilibrium temp that the surrounding clay has slowly reached.

    Posted By: djhthe London clay acts as a big insulating blanket around the tunnels and slows down any heat loss or gain from the tunnel
    Given long enough timescale for steady-state equilibrium to be reached, insulation ceases to be an influence as then there's no further heat flow. Insulation does affect the length of time it takes to reach that equilibrium tho.
    •  
      CommentAuthordjh
    • CommentTime2 days ago
     
    Posted By: fostertom
    Posted By: djhrather than anything to do with it's position in the planet
    Posted By: fostertomthe steady-state temp of the tunnel's place in the long gradient between molten core and av surface temp
    is definitely a third heat/temp input into the equilibrium temp that the surrounding clay has slowly reached.

    The article nicely explains what the undisturbed equilibrium temperature was ('cool enough to need a jumper' or somesuch) as well as the current temperature.

    Posted By: djhthe London clay acts as a big insulating blanket around the tunnels and slows down any heat loss or gain from the tunnel
    Given long enough timescale for steady-state equilibrium to be reached, insulation ceases to be an influence as then there's no further heat flow. Insulation does affect the length of time it takes to reach that equilibrium tho.

    It's the same as methane emissions. You have to think in terms of the heat emission rates, not the cumulative total heat emitted. And that's exactly the same for insulation as the trains. The timescales here are decades or centuries.
    • CommentAuthorEd Davies
    • CommentTime2 days ago
     
    Posted By: fostertomGiven long enough timescale for steady-state equilibrium to be reached, insulation ceases to be an influence as then there's no further heat flow.
    There will always [¹] be a heat flow.

    [¹] Until the heat death of the Universe :wink:
  5.  
    >>>"It's the same as methane emissions. You have to think in terms of the heat emission rates, not the cumulative total heat emitted. And that's exactly the same for insulation as the trains. The timescales here are decades or centuries."

    That's not correct, and is self contradictory - if it were only the rate of change that were important, then the time scale would be irrelevant, whether days or centuries.

    The tube is creating a cylinder of warmer clay around itself, which is expanding radially. The extent of the cylinder [¹] depends *only* on the cumulative heat that has been emitted ( non-dimensionalised for the physical properties of the clay). No steady state will ever be reached, unless the whole of SE England becomes warmed up to tube temperature, in billions of years from now [²]. However, the rate of heat loss is slowing down because it depends on the temperature gradient across the expanding diameter of the warmed clay. The rate of growth of the cylinder is further slowing down, because of the radial geometry, each additional meter of diameter involves more tonnes of clay than previously.

    [¹] the "cylinder" is a logarithmic temperature distribution so mathematically infinite, so we measure its diameter out to a point of an arbitrary temperature, (1/e) is a good choice.

    [²] the heat death of the tube company might come first.
    •  
      CommentAuthordjh
    • CommentTime2 days ago
     
    Posted By: WillInAberdeenThat's not correct, and is self contradictory

    Whether it's correct or not isn't really the point. Whether it causes Tom to think in a way that's more aligned with the real situation is the point. But please read the comment on the article by 'jb' on 2018-07-26. Or many others that also disagree.

    Your proposed explanation is equally oversimplified and fallacious. There's an actual model at https://www.sciencedirect.com/science/article/pii/S1876610215020949
  6.  
    I regret that "jb" 's comment wouldn't get him/her through a physics exam.

    Had he/she studied high school heat transfer, they would recognise the data given that 79% of the heat generated by the tube is absorbed into the surrounding clay, means that the clay is not yet at thermal equilibrium with the tube.

    Had he/she progressed further, they would recognise this means that both sides of the Fourier equation are non-zero. Radial solutions of this equation around a line source, are cylindrical. All real non-zero solutions in a semi-infinite medium will take infinite time to progress to steady state. Like wot Ed said.

    Heat transfer models use a discrete version of the Fourier equation, so they can produce different results from the continuous version, only if badly formulated.

    If you are "JB", you may still be able to edit that comment.:bigsmile:
    • CommentAuthorEd Davies
    • CommentTime1 day ago
     
    Posted By: WillInAberdeenAll real non-zero solutions in a semi-infinite medium will take infinite time to progress to steady state. Like wot Ed said.
    Agree that it'll never reach a steady state but in practice it'll get so close as not to matter. However, I wasn't writing about that. My point was simply that even close to the steady state condition there will still be a heat flow, contrary to what FT seemed to be implying.
    •  
      CommentAuthorfostertom
    • CommentTime1 day ago
     
    Yes, past the tunnel, e.g. from core to surface, but not in or out of the tunnel (on a macro scale - of course there will be smaller perturbations). Steady state means that no or negligible temp difference remains between the tunnel and its immediate clay surrounding.
    • CommentAuthorEd Davies
    • CommentTime1 day ago
     
    Posted By: fostertomSteady state means that no or negligible temp difference remains between the tunnel and its immediate clay surrounding.
    No, even close to steady state there's still a temperature gradient away from the tunnel and hence a heat flow.

    The heat in question is mostly generated in the tunnel by the motors of the trains, directly or indirectly. It has to go somewhere. Some is removed by ventilation and ground water movement. Some is removed by conduction into the clay which requires a temperature gradient and so sets the ΔT between the tunnel and the main mass of the clay.

    The heat from the core of the Earth is a bit of a red herring in this context but only serves to raise the temperature of the mass of the clay by a small amount which will have the secondary effect of raising the temperature in the tunnel by a similar small amount.
  7.  
    I thought "steady state" meant that there was a situation that wasn't changing. So doesn't that encompass a situation where heat is flowing from inside the tunnels to the surrounding clay - but the rate of flow remains essentially the same. Today the trains generate X amount of heat and 79% of it transfers to the clay. Next week, the same. A year from now, the same. It doesn't necessarily mean that there's no heat flow does it?
    • CommentAuthorEd Davies
    • CommentTime1 day ago
     
    Yes.
  8.  
    Posted By: djhYour proposed explanation is equally oversimplified and fallacious. There's an actual model at https://www.sciencedirect.com/science/article/pii/S1876610215020949


    Interesting paper, one of them lectured long ago when I was training.

    They used a 1D implementation of the Fourier equation to estimate that, after 114years of use, the cylinder of warmer clay around the tube has grown to ~50m radius around the tunnels so far. They neglect that this would reach the surface.

    Despite this, they report that four-fifths of the tube heat is still being conducted away into the soil, so it is still very far off from reaching any kind of steady state (after which point there would be no more heat transfer like FT said). Bad news for tube travel, it's going to get worse.

    They think that changing the present-day heat emission rate by using regenerative braking, would have a negligible (1⁰C) effect, because of all the cumulative heat that has already been stored into the clay over the years.

    Which agrees with what I said, oversimplified or not.

    They tentatively propose using GSHPs to remove the cylinder of cumulative heat, giving 8⁰C improvement, that might be fallacious.


    Going back to housing, the key difference is that the path length from the edges of a ground floor to the outside air, is shallow enough to reach a dynamic nearly-equilibrium within a sensible time.

    Thereafter, the heat loss through this route is far more significant than the heat storage into an ever-expanding region of subsoil. So the BRE formula should work if the edge regions of your floor are similar to BRE's, if only we knew what they were.
  9.  
    Posted By: lineweightI thought "steady state" meant that there was a situation that wasn't changing. So doesn't that encompass a situation where heat is flowing from inside the tunnels to the surrounding clay - but the rate of flow remains essentially the same. Today the trains generate X amount of heat and 79% of it transfers to the clay. Next week, the same. A year from now, the same. It doesn't necessarily mean that there's no heat flow does it?


    Unfortunately it is changing, which is why people are feeling hotter now than 100 years ago. All the heat has to go somewhere, so it is widening the cylinder of warm clay around the tunnels and reducing the temperature gradient. That makes it harder for heat to escape next week, or next year, so each year the train has to get hotter to drive the same amount of heat out into the soil.
    • CommentAuthorEd Davies
    • CommentTime1 day ago
     
    Posted By: WillInAberdeenUnfortunately it is changing…
    So, you're saying it hasn't come close to steady state yet?
    • CommentAuthorlineweight
    • CommentTime1 day ago edited
     
    I don't understand the relevance of the 79% figure to judging whether it has reached "steady state" yet. This is just saying that 21% is removed by ventilation, etc, isn't it?

    If conduction through the surrounding clay was the only means of the heat being removed, then at whatever point it reached steady state or near-enough-steady state, 100% of the heat would still be moving by that mechanism, just as it would be at any point on it's journey towards steady state.
  10.  
    Those MW of heat all have to go somewhere, and at the moment 79% of it is going into warming up the soil that surrounds the tube. More of the soil is warming to a higher temperature each year - so that's not a steady state situation. This makes it harder every year for heat to escape, so the tunnels are not at a steady state temperature either - it's rising each year, a little.

    LU can't carry on doing that indefinitely, without toasting their customers, hence their concern.

    A nearly-steady state would be reached when enough of the soil under London had warmed up to a hot enough temperature that it could no longer warm up very much more. Then, as it could no longer absorb much more sensible heat, it would no longer be able to heat-sink 79% of the tube's electricity bill, maybe only a few % of it. It's a bit subjective when that point will come, but let's hope they have sorted some more active cooling before then.

    Ed: I maybe did mention that possibility, ask me again at the end of the universe and then we can be sure...
   
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