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    • CommentAuthoratomicbisf
    • CommentTimeNov 14th 2012
     
    Hi all,

    I've been wondering whether it would be possible/practical/useful to try to determine the heat loss from a house experimentally rather than by u-value calculations etc.

    I was thinking something along the lines of tracking the temperature outside, preferably in cold and not too changeable weather while setting the heating to maintain a constant temperature inside. Then you could measure the fuel use other the time period and possibly adjust for boiler efficiency.

    I have little faith in the EPC we have as it seems just be be a formula and also it would be interesting to compare houses. Especially to compare this house with my parents' one as they are similar in size and less than 1km apart but totally different in construction, as they are often complaining of high energy consumption. This is a 1948/9 steel-framed, system-built semi while theirs is a typical 1901 end-of-terrace with solid walls, the outer layer of Bath stone and the inner of brick.
    • CommentAuthorDarylP
    • CommentTimeNov 14th 2012
     
    Yes, measure the energy used to heat the two dwellings. Compare that with int/ext temp data, but try to do it over a long a time period as possible, to smooth out temperature spikes, and occupancy anomalies.

    I would be interested to see how it compares to your EPC.

    Good luck....:smile:
    • CommentAuthoratomicbisf
    • CommentTimeNov 14th 2012
     
    Posted By: DarylPYes, measure the energy used to heat the two dwellings. Compare that with int/ext temp data, but try to do it over a long a time period as possible, to smooth out temperature spikes, and occupancy anomalies.

    I would be interested to see how it compares to your EPC.

    Good luck....http:///forum114/extensions/Vanillacons/smilies/standard/smile.gif" alt=":smile:" title=":smile:" >


    The trouble with a long period of time that I see is that it may tell you more about the inhabitants than the house itself, for example we're normally out during the day and don't have the heating on then while my mum often works from home and will have the heating on during the day.

    I was thinking of a shorter time period to try and make the comparison better.
    • CommentAuthorDarylP
    • CommentTimeNov 14th 2012
     
    That might be why your parents "are often complaining of high energy consumption"?
    However, if you measure energy in, you then know energy out!

    Cheers..:smile:
    •  
      CommentAuthorSteamyTea
    • CommentTimeNov 14th 2012 edited
     
    You can heat your house up and then see how long it takes to cool, Measuring the temperature difference along the way.
    You need a nice still night, some fan heater/central heating on, someone inside to record the temperature and someone outside to do the same, get a torch (or a couple of iButtons and go have a meal and a night out).
    No opening doors though.

    Suppose you could put a thermometer outside a window and do it that way and you will have at least one really warm evening this year :wink:
    • CommentAuthoratomicbisf
    • CommentTimeNov 15th 2012
     
    Posted By: DarylPThat might be why your parents "are often complaining of high energy consumption"?
    However, if you measure energy in, you then know energy out!

    Cheers..http:///forum114/extensions/Vanillacons/smilies/standard/smile.gif" alt=":smile:" title=":smile:" >


    Indeed, but at the moment we have little idea of what part of the difference is caused by behaviour and what part by thermal efficiency.
    •  
      CommentAuthorSteamyTea
    • CommentTimeNov 15th 2012
     
    Posted By: atomicbisfIndeed, but at the moment we have little idea of what part of the difference is caused by behaviour and what part by thermal efficiency.
    Behavour is somewhere between 25 and 50% in my experience.
    • CommentAuthorEd Davies
    • CommentTimeNov 15th 2012
     
    Steamy, wouldn't your experiment require you to know the thermal mass of the house to be able to calculate anything useful? Even just for comparison the difference between a leaky pile and a light & tight house for rate of temperature drop could be very misleading if you tried to extrapolate to expected energy use long term.

    I think that the hard part of any of these types of experiments is knowing the effect of thermal mass in the layers of insulation. Unless, unrealistically, you can keep both the indoor and outdoor temperatures (and the windspeed, precipitation and radiative temperature of the sky) constant for quite a while the results will be hopelessly fuzzy.

    To find anything useful I think you'd need to monitor indoor and outdoor conditions (more than just temperatures) for quite a while then do quite a sophisticated analysis. Perhaps:

    https://en.wikipedia.org/wiki/Principal_component_analysis
    • CommentAuthoratomicbisf
    • CommentTimeNov 15th 2012
     
    Posted By: Ed DaviesSteamy, wouldn't your experiment require you to know the thermal mass of the house to be able to calculate anything useful? Even just for comparison the difference between a leaky pile and a light & tight house for rate of temperature drop could be very misleading if you tried to extrapolate to expected energy use long term.

    I think that the hard part of any of these types of experiments is knowing the effect of thermal mass in the layers of insulation. Unless, unrealistically, you can keep both the indoor and outdoor temperatures (and the windspeed, precipitation and radiative temperature of the sky) constant for quite a while the results will be hopelessly fuzzy.

    To find anything useful I think you'd need to monitor indoor and outdoor conditions (more than just temperatures) for quite a while then do quite a sophisticated analysis. Perhaps:

    https://en.wikipedia.org/wiki/Principal_component_analysis" rel="nofollow" >https://en.wikipedia.org/wiki/Principal_component_analysis


    I was thinking of thermal mass because the houses will be very different in this respect. Ours will have little apart from the concrete ground floor while theirs will have more in the walls.

    I was hoping it should be possible to get a reasonably useful measure of Watts per Kelvin temperature difference between inside and outside. Obviously not super-accurate, but better than the EPC's "one star out of four" or whatever it is for "system built walls" (coincidentally the same score as masonry non-cavity walls and presumably the same whatever the type of system-building used).
    •  
      CommentAuthordjh
    • CommentTimeNov 15th 2012
     
    Posted By: atomicbisfIndeed, but at the moment we have little idea of what part of the difference is caused by behaviour and what part by thermal efficiency.

    The only simple way to tell, AFAIK, is to control the behaviour. Find out what your mother sets the heating to (i.e. internal temperature that's achieved) and set yours to be the same.

    If you google 'coheating test', you'll find stuff about professional tests using these kinds of experiments.
  1.  
    For what it's worth, I have a very good measure of how much energy I use for heating as I keep the thermostat constantly at the same setting, read my meter weekly and look up the heating (or cooling) degree days for that period and then subtract my "base load" (which I measure when I'm not heating or cooling). Thermal mass and incidental gains (e.g. from the sun) obviously have some effect, but, overall, the measurements are pretty accurate and correlate well with the hot2000 model (on an annual basis) that I have of my house (luckily I have actual measured air leakage figures). It has been interesting to see the consumption per heating degree day decline as different improvements have been made.

    I guess most people in the UK don't keep their houses at a constant temperature, making it much more difficult to estimate accurately. For what it's worth, during last year's heating season I averaged out at 2.75kWh per HDD (basis 18C, internal temperature 21.6C - total of 3601 HDD for the season, heated area 270m2 [90m^2 is the basement])

    Paul in Montreal.
    • CommentAuthoratomicbisf
    • CommentTimeNov 15th 2012
     
    Posted By: Paul in MontrealFor what it's worth, I have a very good measure of how much energy I use for heating as I keep the thermostat constantly at the same setting, read my meter weekly and look up the heating (or cooling) degree days for that period and then subtract my "base load" (which I measure when I'm not heating or cooling). Thermal mass and incidental gains (e.g. from the sun) obviously have some effect, but, overall, the measurements are pretty accurate and correlate well with the hot2000 model (on an annual basis) that I have of my house (luckily I have actual measured air leakage figures). It has been interesting to see the consumption per heating degree day decline as different improvements have been made.

    I guess most people in the UK don't keep their houses at a constant temperature, making it much more difficult to estimate accurately. For what it's worth, during last year's heating season I averaged out at 2.75kWh per HDD (basis 18C, internal temperature 21.6C - total of 3601 HDD for the season, heated area 270m2 [90m^2 is the basement])

    Paul in Montreal.


    Very interesting. We have the heating set to 18C when we are in, but it's switched off when we're out and overnight when asleep. I think most people here just switch it on and turn it up when they feel cold without much rhyme or reason, but I don't like that as I feel it's more comfortable to have a reasonably constant temperature than wild fluctuations.
  2.  
    Why dont you calculate the U values of the fabric of the building (external walls, floor, windows) then use Q=m.Cp. delta T to work out the actual heat loss for each thermal element. Pressure test the building or hazard as guess at the infiltration rate and use Q=1/3.N.V delta T to turn this into Watts.
    This is how we used to do heat loss calcs before EPC, SBEM, etc. This calc method will give you a very good idea of where the heat is going in your house. I set up these simple calcs for a pal who was refurishing a 9,000 sq ft property. By the time he had done the calcs for every room in the house and understood which parts of the house were leaking energy he was fully educated about insulation/draught proofing and it energy benefits. He changed the spec on his original insulation proposals to increase the thermal performance because he could see the reduction in Watts.
    I dont think what you are proposing by doing tests will give you any better idea of your heatloss than doing the above simple calcuations.
    • CommentAuthoratomicbisf
    • CommentTimeNov 15th 2012
     
    Posted By: timevans2000Why dont you calculate the U values of the fabric of the building (external walls, floor, windows) then use Q=m.Cp. delta T to work out the actual heat loss for each thermal element. Pressure test the building or hazard as guess at the infiltration rate and use Q=1/3.N.V delta T to turn this into Watts.
    This is how we used to do heat loss calcs before EPC, SBEM, etc. This calc method will give you a very good idea of where the heat is going in your house. I set up these simple calcs for a pal who was refurishing a 9,000 sq ft property. By the time he had done the calcs for every room in the house and understood which parts of the house were leaking energy he was fully educated about insulation/draught proofing and it energy benefits. He changed the spec on his original insulation proposals to increase the thermal performance because he could see the reduction in Watts.
    I dont think what you are proposing by doing tests will give you any better idea of your heatloss than doing the above simple calcuations.


    I've calculated the U-values for original and improved walls with the Vesma u-value calculator which gives a value of 0.58W/m2K for original downstairs walls (9mm plasterboard, 50mm glassfibre, 70mm cavity, 50mm concrete) and 0.16 for improved (3mm plaster, 13mm plasterboard, 50mm PIR, 70mm PIR, 50mm concrete).

    The trouble is I'm not sure what bearing it has on reality because each element is not composed of simple layers, for example the plasterboard of course is on timber studwork infilled with PIR, and there will be thermal bridging where there floor and ceiling steel beams are bolted to the steel framework in the walls.
    • CommentAuthorEd Davies
    • CommentTimeNov 15th 2012
     
    Posted By: timevans2000Why dont you calculate the U values of the fabric of the building (external walls, floor, windows) then use Q=m.Cp. delta T to work out the actual heat loss for each thermal element.

    Because that doesn't give the actual heat loss, just an estimate of what it might be if the U values are right and if all losses are by conduction (which they aren't).

    I think Paul in Montreal's method is best - looking at heating degree days and energy use over a season. Just that it gets a bit more tricky if you don't keep the internal temperature constant but it should still be possible if you keep detailed records.
    •  
      CommentAuthorSteamyTea
    • CommentTimeNov 15th 2012 edited
     
    If you know the volume of your house, and therefore the mass of air in your house, you can work out the heat differences as it looses temperature.
    This will give you a rapidly decaying line that then flattens out.
    So you should be able to then estimate for any temperature differences what that building will do (within a bit) for those temperature differences.
    I agree that solar, wind and rain will change this, but it is a rough estimate but it takes in the thermal properties of that building, you can't use it as a direct comparison between buildings without a lot more work and monitoring.
    • CommentAuthorEd Davies
    • CommentTimeNov 15th 2012
     
    Posted By: SteamyTeaIf you know the volume of your house, and therefore the mass of air in your house, you can work out the heat differences as it looses temperature.

    A smallish house will probably have not much more than 300 kg of air in it. (Say 100 m² x 2.4 m room height x 1.3 kg/m³.) I think there aren't many houses where that's a significant proportion of the mass, even a significant proportion of the mass inboard of the insulation.
    •  
      CommentAuthorSteamyTea
    • CommentTimeNov 15th 2012 edited
     
    But if you are looking at the thermal losses, measuring the internal temperature drop from an elevated temperature with respect to the ambient and then it can be translated into the heat loss.

    So

    If 300 kg of air is raised to 25°C above ambient and it looses:

    Hour 0 25°C
    Hour 1 15°C
    Hour 2 9°C
    Hour 3 4°C
    Hour 4 1°C
    Hour 5 0.0°C

    That would be an energy loss of:

    Hour 1 0.83 kWh
    Hour 2 0.50 kWh
    Hour 3 0.42 kWh
    Hour 4 0.25 kWh
    Hour 5 0.08 kWh
    Hour 6 0 kWh (estimated)

    Plot this and you would get a curve of about kWh=0.46 * Ln(t) where t is time in hours.
    That would give you a thermal profile for that house. So you could say (for identical conditions or with some certainty boundaries in place) that for every degree C above ambient you are loosing 0.083 kWh overall. It is the overall that is important as this is the cooling curve only.
    The kWh/°C heating curve would be steeper (and positive) as has to include the losses as well as the rise in temperature.
    You could test that one in the same way as you heat the building up.
    This method takes into account all losses (or gains) that are happening as you test it.

    This is just the same as calculating using HDDs but on a shorter time scale, call it HDH (or HDS as we should really be using seconds and the Kelvin scale with limits in place)
    • CommentAuthorEd Davies
    • CommentTimeNov 15th 2012
     
    Naively looking at your numbers you might conclude that the heating required to maintain the original 25 °C would be 830 W (well, a bit more because of the curve in the temperature graph even in the first hour). Actually it'll be a lot more as the house is losing more heat in that hour - not just the heat in the air.

    You can't ignore the heat loss from the plasterboard, floorboards, etc, which will follow (maybe not very closely ¹) the air temperature down. Even the solid stuff inside the insulation will have a mass well in excess of that of the air so the heat loss from the house will, to take the example of your first hour, be well over 0.83 kWh.

    ¹ Sometimes ahead, sometimes behind depending on whether conductive or ventilation heat losses dominate.
    •  
      CommentAuthorSteamyTea
    • CommentTimeNov 15th 2012 edited
     
    I know what you mean about the heat stored in everything but the air, but as we want the air to be at a sensible temperature that is the thing to measure surely. Greater masses, even with lower SHCs take a long time to cool or heat up and over the relatively short timescale of a day (or in my hypothetical example of 6 hours) I think they would have little effect as ventilation losses probably dominate is a real house.

    I was only showing cooling curves as this is about heat loss. If you plot a full heating and cooling cycle, you can look at the time difference between heating and cooling for the same temperature differences, it is how thermal efficiency of ST systems is meant to be calculated for a fixed input.

    Both JSH and myself live in low mass houses, and we were both surprised at how thermally stable they are, seeming more so than higher mass houses that we had both lived in in the past (though we were not measuring them like the sados that we have become). We put this down to better insulation and airtighness.

    I would do this test on my place but I still do not have my heating on, actually 22°C inside and 12°C outside. I do have some loggers stratigraphy placed around the place in anticipation of turning the heating on (have had the fan heater on occasionally and the water is taking more juice to heat up now).

    I still think that if you heat something up and time it cooling down, you can work out the overall energy that is used. It is how they work out time of death with a body. Why the police wear blue rubber gloves when they put the thermometer in. Not very dignified is it :wink:
    •  
      CommentAuthordjh
    • CommentTimeNov 16th 2012
     
    Posted By: SteamyTeaBoth JSH and myself live in low mass houses, and we were both surprised at how thermally stable they are, seeming more so than higher mass houses that we had both lived in in the past (though we were not measuring them like the sados that we have become). We put this down to better insulation and airtighness.

    What matters is the time constant. An insulated house is like a capacitor-resistor filter. The thermal mass is the capacity and the insulation is the resistance. And like a CR filter, it has a time constant that summarises its frequency response.

    I still think that if you heat something up and time it cooling down, you can work out the overall energy that is used.

    But only when you have equilibrium conditions at both ends of the measurement. And you don't when measuring houses except in very special circumstances. The middle of the wall typically takes a really long time to cool down to the same temperature as the surface. Far longer than a day. I was just reading something that suggested for straw bales, it may be as long as months. That's an extreme case, because of the large quantity of hygroscopic transport that goes on inside them.
    • CommentAuthorEd Davies
    • CommentTimeNov 16th 2012 edited
     
    Posted By: SteamyTeaI know what you mean about the heat stored in everything but the air, but as we want the air to be at a sensible temperature that is the thing to measure surely.

    As has been pointed out to you repeatedly on multiple threads, air temperature is only part of the story.

    Still, taking your argument to an extreme, I want my pencil to be at a sensible temperature to make writing comfortable but I wouldn't base calculations of the heatflow out of my house on its heat capacity.

    Posted By: SteamyTeaGreater masses, even with lower SHCs take a long time to cool or heat up and over the relatively short timescale of a day (or in my hypothetical example of 6 hours) I think they would have little effect as ventilation losses probably dominate is a real house.

    Ventilation losses are likely significant but it's not obvious that they dominate to the point that you can reasonably ignore other losses.

    Posted By: SteamyTeaI still think that if you heat something up and time it cooling down, you can work out the overall energy that is used.

    The objective here is to work out how much heat the house loses when it's at a particular temperature difference from the outside. As such you need to take into account the total amount of stuff which is being heated and cooled.

    PS, try moving from a thick stone walled croft house to a static caravan to get some appreciation of the effect of thermal mass.
    •  
      CommentAuthorSteamyTea
    • CommentTimeNov 16th 2012 edited
     
    So Newtons Law of cooling is T=A+S*e^-kt
    and an R-C circuit is V=V(0)*e^-(t/RC)
    So both have the time element already in them. They are actually the same thing really.

    Going back to the OPs original question as to how to estimate the heat loss of a house,I think my method is as valid as the HDD method (as it is the same thing) but you have to do a translation from energy loss in the air to how much energy that air would need to heat up with the current heating system.
    It is quick and dirty admittedly, but easier than just guessing from meter readings as you need to know your base load then, and if you have E7 heating that really throws it as it is a fixed heating load with the temperature fluctuation as the units discharge.
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