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    • CommentAuthorJT101
    • CommentTimeOct 3rd 2012
     
    Hi all

    I'm a mechanical engineer with some building experience, who's trying to get into eco building, namely passive house.

    I'm trying to do some heat loss calculations (ventilation & fabric) for a large 2 storey Georgian (internal footprint 160m2) triple brick skin house for current (leaky) in Suffolk, and post insulated scenario.

    Wondering if someone can help me out? I have posted some general questions on another post, but thought this one deserved a seperate post.

    How do you do heat loss calculations (fabric) for a solid stone floor

    I came across this formula in on the forums: U = 0.05 + 1.65(P/A) - 0.6(P/A)², which is for a solid floor. It gave me 0.55W/m²K
    It doesn't take into account the floor material or soil temperature which seems odd.
    The floor in this house is solid stone.
    Should I be doing the above calc, or a standard u-value calc involving the conductivity & thickness of the stone as I would do for a wall or window, and just add some factor for direct contact with the soil?

    Many thanks
    • CommentAuthorSprocket
    • CommentTimeOct 3rd 2012 edited
     
    Thermal conductivity of packed dry earth and stone and tiles are all in the same ballpark. So it is reasonable to consider them as one lump.

    The output is a U vale which does incorporate temperature differential. You would still need to scale by your temperature difference and area to get actual loss in watts.

    It's probably good enough under the circumstances. Consider that if you had no stone, just earth, how would you estimate heat loss? What is the insulator and what is the thickness of it? Hard to apply that approach.
    So I think what you have there is a rough approximation (probably of a finite element analysis) for effect of packed dry soil or stone of infinite thickness plus some leakage at the edges.

    On the other hand if you have a floor made up of layers of things with significantly different thermal conductivities then the usual combined U value calcs (plus consideration if the edges are different) is more useful. This is still an approximation of course because heat leaking through the insulation raises the temperature of the soil below so you are back to finite element analysis or a model. However if you have a decent enough insulator in your floor the temperature below will not rise significantly (because it will leak away to surrounding soild fast enough to make that small amount of heat leakage effectively insignificant). In such cases you would typically take ground temeprature below your layers as a fixed 10 degrees C.

    This make sense because in normal consideration of U values we are of course assumming that the amount of heat leaking is not sufficient to raise temperature outside and so change the temperature differential at all. That is not true for a solid earth floor.
    •  
      CommentAuthorfostertom
    • CommentTimeOct 3rd 2012
     
    This touches on a really important but unresolved question - just how does subsoil behave thermally (heat flow straight downward)?

    Complicated by the fact that much (or most) of the loss is in fact by conventional conductive loss in curving path outward under the perimeter and back up to outside air. That being the sum of numerous different-length paths starting from different points on the slab to different points on the ground outside.

    It's the ratio between those two loss types, differing for ea building, that the P/A ratio attempts to address.

    Sprocket, you seem to think that this would be a doddle with finite element analysis? Possibly in a 2D slice thro a typical building cross section (like Therm), but in fact it's a v 3D model, with the plan shape of the slab perimeter making a big difference.

    This wd be mainly dealing with the perimeter conduction loss part of the problem - the straight-down loss still remaining a mystery, for which people (and e.g. Therm) make all sorts of approximations and assumptions. A serious study is badly needed.

    First thing, is that long time-lag is v dominant. We're talking annual cycles of slow heat flow, in fact multi-years till an equilibrium is reached. The rule of thumb is that a 'new' temperature wave propagates at 1m per month. That raises novel, advantageous possibilities - like putting summer heat in over here, and it reaches where it's needed (e.g. an uninsulated floor slab) 6 months later. Pouring heat into uninsulated ground has been proved to work (prototype houses in UK-like weather zones exist) though our boffins immediately prove it will all leak away - but apparently doesn't. The role of ground water flow (if any) thro the ground also needs study - it can whip all the heat away, alternatively (e.g. Foster/Arup's Reichstag scheme) be the heat-transport mechanism. Water flow may be crosswise, or vertical (rise and fall of water table).
    •  
      CommentAuthorfostertom
    • CommentTimeOct 3rd 2012
     
    JT101, a poss solution, that allows your stone floor to remain undisturbed, is to excavate a trench all round your perimeter, down to base of founds (v careful to not undermine, or allow soil beneath to be extruded). Reinstated as a french drain, your rising damp is removed at a stroke. But the trench also contains generous downstanding perimeter insulation, so the block of subsoil beneath your building is largely cut off from conductive loss under its perimeter, to cold outside air. Then the block of subsoil becomes a large thickness of (albeit poor) insulator, as far as straight-down heat flow is concerned. It also becomes an extra bit of temp-stabilising thermal mass available to your interior, the upper part of which will within a year rise to a cosy-floor temp only a little below your maintained internal air temp.

    The insulation in the trench can be say 200 EPS, like underground EWI, and/or quantities of Leca (lighweight expanded clay aggregate) which also makes an ideal french drain fill.
    • CommentAuthordjh
    • CommentTimeOct 3rd 2012
     
    Posted By: JT101I'm a mechanical engineer with some building experience, who's trying to get into eco building, namely passive house

    ....

    How do you do heat loss calculations (fabric) for a solid stone floor

    I'm not an expert but as I understand it there is a standard - BS / I.S. EN ISO 13370: 2007 (Thermal performance of buildings. Heat transfer via the ground. Calculation methods) - for how to do such calculations. And of course if you are interested in Passivhaus, you do it the PHPP way.
  1.  
    We've had this discussion a couple of times before.

    From: http://www.greenbuildingforum.co.uk/newforum/comments.php?DiscussionID=8344

    I wrote:

    "We had a discussion about modeling heat flows into the ground before. Here's a post I made a while ago:

    See page 193 onwards of http://www.nrel.gov/docs/fy08osti/43388.pdf for a description of how BASECAL and BASEIMP (the models used in esp-R and hot2000) work and what the modeling assumptions are in terms of water table, slab thickness etc. These numerical models have been validate against real buildings over multi-year periods.

    Natural Resources Canada has made the BASECALC program available for download on its own: http://canmetenergy-canmetenergie.nrcan-rncan.gc.ca/eng/software_tools/basecalc.html


    See also: http://www.esru.strath.ac.uk/Programs/ESP-r_capabilities/3Dcond.html"

    Paul in Montreal.
  2.  
    This is an interesting question, and would it be similar to a concrete floor?
    • CommentAuthorSprocket
    • CommentTimeOct 3rd 2012
     
    Yes, should be similar for a concrete floor if there is no other insulation.

    All these things are approximations. Even the finite element approach.
    There is so much leeway in the material specs and especially the building specs and air movement etc that it's not worth the effort to model more precisely. If someone has worked out that some rough formula is a close-enough approximation then I am quite happy to use it. By the time you throw it into the mix with losses through walls and roof and windows the small inaccuracies make negligible difference.

    As a friend of mine says, "That'll do, it's close enough for government work!"

    Even if you get the maths precise the conditions are still a huge approximation.
    Some time ago I did a very rough incremental/differential model a swimming pool (ie. software model, not 3D model although it modelled 3 dimensions) just to see what difference various tweaks made. Starting at 30'C inside, 10'C outside it quickly settled into a steady temp gradient. Tried it with no insulation (until recently most pools had no insulation but now it is building regs requirement... long story not necessary here) then added various insulation. Some of the things I tried to add to the model (eg. heat loss at the surface around the oiutside) made very little difference to the overall heat loss. Without any insulation one of the things that made a big difference was heavy rain (another approximation of course - I didn't model the way the water flowed over and through the ground in reality). Anyway, point is that even a small amount of insulation around the pool was enough to let the surrounding soil settle to only just a hairs-breadth over 10'C and that pretty much completely removed any effect of rain.

    Approximations and rules of thumb are plenty good enough for me as long as they are credible.
    • CommentAuthorGaryB
    • CommentTimeOct 3rd 2012
     
    Solid floors in contact with earth are complex as has been previouly stated.

    There is a useful table in the CIBSE Guides for uninsulated floors which you read the length and breadth and it gives you the U-Value. This has been collated from empirical data and is pretty reliable.

    We used to use this before floor insulation became the norm under building regs. We still can use it for large warehouses where the edge effects are negligible over the entire footprint - the average U-value can meet current regs without insulation.

    I'll post a copy or a link when I'm in the office tomorrow.
  3.  
    Posted By: GaryBSolid floors in contact with earth are complex as has been previouly stated.

    There is a useful table in the CIBSE Guides for uninsulated floors which you read the length and breadth and it gives you the U-Value. This has been collated from empirical data and is pretty reliable.

    We used to use this before floor insulation became the norm under building regs. We still can use it for large warehouses where the edge effects are negligible over the entire footprint - the average U-value can meet current regs without insulation.

    I'll post a copy or a link when I'm in the office tomorrow.


    Thanks GaryB, that would be most interesting.

    My house sits on a concrete slab foundation so I'd be interested to know, though I can't see myself installing any insulation there.
    • CommentAuthorJT101
    • CommentTimeOct 3rd 2012
     
    Thanks to all for the very informative comments. This has settled it in my mind. I shall use the formula as I stated , and just add insulation for post work calculations.
    • CommentAuthortony
    • CommentTimeOct 3rd 2012
     
    Heat loss is dependant on the temperature gradient. The temperature of the ground under a house will be close to the average temperature of the house. To get an answer fro heat losses from u values you need to enter temperatures and so under there bout the temperature and the U value are open to debate. I measured mine in my old house a meter down under a solid floor in winter and it was 17C

    Strange that we don't know this stuff in the 21st centuary
    • CommentAuthorpmusgrove
    • CommentTimeOct 4th 2012
     
    I was tempted to insulate the floor in the same manner as the Kiwis which is, I believe, to put vertical insulation in on the circumference of the building down a couple of metres and let the ground under the building reach equilibrium with it. Couldn't convince the BCO though.
    • CommentAuthorCWatters
    • CommentTimeOct 4th 2012
     
    If I've understood this paper correctly vertical insulation does work..

    http://www.ornl.gov/sci/buildings/2012/Session%20PDFs/92_New.pdf

    Not sure I fancy the idea of digging deep trenches close to foundations though.
    •  
      CommentAuthorSteamyTea
    • CommentTimeOct 4th 2012 edited
     
    Paul
    That is probably because they can't work out the heat losses from a slab either.

    The way I see it is as a ratio of absorption area (the floor in your house) and then the emitter area (the other 5 sides, or as many as there are). The ground can be thought of as a semi-infinite heat sink and adjust for rainfall.
    If you work on the worst case, generally February ground temp, as your heating load diminished the ground temp is also increasing, so you would not be far out. Gets around flux variation problems.

    So if we assume a 1 metre cube, then the absorption area is 1, the emitter is 5, so every face in contact the the ground 5m^2, is loosing energy, so 1/5th of the energy in. If we assume 2 of these cubes next to each other then the absorber area is 2, the emitter area is 8, so 1/8th. Once you have a list of different lengths you can plot the data. Then divide the abortion area by the emitter area and you end up with a curve of diminishing exponential growth. This is similar to the Cooling Law. By punching in some real temperatures and times (remember that Watts are units of energy divided by time and Wh are units of energy) you will end up with your losses for any size and shape of slab.

    Edit: May be worth looking at heat sink formula to get a better idea of what is happening. Q(k)=-kA dT/dx

    Chart below shows that as the slab gets longer the losses will get less as any material has the ability to insulate.
      Slab Losses.jpg
    • CommentAuthorTimber
    • CommentTimeOct 4th 2012
     
    I have modelled solid floors in therm before and got some interesting results, although not sure how to turn that slice into a meaningful model/representation of actual heat loss
  4.  
    Thing that bothers me is:

    The 'U' value is as FT says, an average of all the heat flow paths from points on the floor to the outside air. Dominated I s'pose by the flow paths that start from the strip of floor adjacent to the perimiter wall, as these have the shortest path to outside air.

    So a floor with thick perimiter walls standing on it should have longer heat flow paths to outside, and lower floor U, than a house with thin walls, no?

    Our place the walls are about 600-1000mm thick from inside air to outside air, by the time we include the redundant chimney breasts and IWI.

    Not sure how/if the formula accounts for different wall thicknesses.

    Is academic now as we just lifted solid floor and relaid with insulation under.
    •  
      CommentAuthorSteamyTea
    • CommentTimeOct 4th 2012
     
    Posted By: WillInAberdeenThe 'U' value is as FT says, an average of all the heat flow paths from points on the floor to the outside air

    Not quite, it is the sum of all heat losses.
    •  
      CommentAuthorfostertom
    • CommentTimeOct 4th 2012
     
    The wall thickness wd count as slab area, and you take external wall perimeter, because the base of the walls at slab level are pretty much same temp (and same exposure to inside air) as the slab. But it's an inexact estimate so such finesse is inappropriate. Maybe the 'perimeter' line should be half way thro the wall thickness.
    • CommentAuthormike7
    • CommentTimeOct 4th 2012
     
    Posted By: SteamyTea
    Posted By: WillInAberdeenThe 'U' value is as FT says, an average of all the heat flow paths from points on the floor to the outside air

    Not quite, it is the sum of all heat losses.


    The U-value of any element needs to be multiplied by an area and by a temperature difference to get a figure for heat loss (I know you know that). The problem for a solid floor is that the temperature difference for the central area is likely to be between internal and steady deep ground temperature - about 10C in UK , whereas for the perimeter the difference will vary with the time of year. For this reason the convenience of using U values to tot up the heat loss per degree of the whole building falls down a bit if you try to lump in the solid floor with the rest of the calculation.
    • CommentAuthorTimber
    • CommentTimeOct 4th 2012
     
    What about modelling half a slice though the floor (from the centre to the perimeter) and then working out the U-Value of the floor as if it were a circle with the same surface area as the floor? Now I know that this also won't be accurate as the effects at corners and if the floor is a long rectangle, heat loss will vary wildly depending on which path it is taking.

    But given all of that, how far off reality would it really be? In theory, a 3d finite element model should be able to model it realistically (as a computer could), but would this 2d slice method work well enough. Perhaps someone with 3d modelling software and a spare afternoon could do it and let us know? I would happily run the 2d versions and calcs.
    • CommentAuthorSprocket
    • CommentTimeOct 4th 2012
     
    > how far off reality would it really be

    Not enough to worry about IMHO. There are usually bigger (Watts wise) inaccuracies elsewhere.

    > a 3d finite element model should be able to model it realistically

    Perhaps not really. It's still an approximation. All FE models are still approximations.
    But more significantly, none of us build our houses on standard soil in standard climates.

    None of my models (or buildings) have particularly worrying heat loss through the floor. The approximations are close enough and other factors always dominate (usually air changes and then roof).
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