Calculate heat loss from actual data?
Discussion
Presumably it's possible to calculate the heat loss in watts from a room, given the room temperature at the start and end of a period of time, and the outside temperature?
I'd like to compare real-life data to theoretical heat loss calcs, to get an idea about radiator sizing etc. The rooms in question is in a flat roof dormer, so no chance of checking insulation levels etc.
The room was heated until 15:30, when the temperature was 20.6 degrees. Door was closed at this time, and room left unoccupied. Smart TRV closed off the radiator, heating stayed off.
By 23:30, the room had dropped to 17.4 degrees. At this point, the heating came back on to make use of cheap rate electricity.
The outside temperature was dropping at a steady rate during this period, from 5.9 degrees to 4.6 degrees.
I'd like to compare real-life data to theoretical heat loss calcs, to get an idea about radiator sizing etc. The rooms in question is in a flat roof dormer, so no chance of checking insulation levels etc.
The room was heated until 15:30, when the temperature was 20.6 degrees. Door was closed at this time, and room left unoccupied. Smart TRV closed off the radiator, heating stayed off.
By 23:30, the room had dropped to 17.4 degrees. At this point, the heating came back on to make use of cheap rate electricity.
The outside temperature was dropping at a steady rate during this period, from 5.9 degrees to 4.6 degrees.
If calculating only for a single room rather than the whole house you’d also need to consider the heat transfer through the walls and floors (if first floor) to the rest of the house as well and the temperature difference over time to that space.
As you’ve then got heat transfer occurring to three different spaces (outside, adjacent rooms and rooms below) you then also need to consider the areas of these elements and their U-values to get meaningfully accurate results from any calculations.
As you’ve then got heat transfer occurring to three different spaces (outside, adjacent rooms and rooms below) you then also need to consider the areas of these elements and their U-values to get meaningfully accurate results from any calculations.
smokey mow said:
If calculating only for a single room rather than the whole house you’d also need to consider the heat transfer through the walls and floors (if first floor) to the rest of the house as well and the temperature difference over time to that space.
As you’ve then got heat transfer occurring to three different spaces (outside, adjacent rooms and rooms below) you then also need to consider the areas of these elements and their U-values to get meaningfully accurate results from any calculations.
I appreciate that heat transfer from adjacent rooms will skew the numbers, but it would likely give me a much better figure than that done during the ASHP pre-installation survey.As you’ve then got heat transfer occurring to three different spaces (outside, adjacent rooms and rooms below) you then also need to consider the areas of these elements and their U-values to get meaningfully accurate results from any calculations.
Because the dormer roof and wall insulation is unknown, the chap doing the calcs said it would be a "best guess". He assumed that the dormer was original to the house (1966), when I know it was done at least 30 years later, and insulation levels were probably higher.
Without knowing the exact build date for the dormer conversion, he said he had no option but to go for the original property build date.
The heat lost, energy in Joules, was stored in the fabric and contents of the room.
Air has a 'specific heat capacity' in joules per kg per K.
Likewise all the other materials in the room, wood plaster and all that.
So you could add all that up and get a figure in Joules per K for the whole room, and that would tell you how many kWh left the room as dropped however many degC.
That's O Level physics, in reality, you don't know how much heat the walls have lost, because they won't all be an even temperature all the way through.
But it can be a useful thought process if you're thinking about rooms with lots of 'mass', there's significant heat stored in the inner leaf of a brick cavity wall for instance.
If you're trying to get at the 'watts per K' of a room, it may be informative to run say a 500W heater in there and see where the temperature goes, when the rest of the house is warm and outdoors is fairly cool but fairly steady.
I used a fan heater for a reality check on a bigger room, observed the temperature remotely. That told me the room was losing way more heat than it should have been calculated from the U values of the walls/windows/ceiling.
Air has a 'specific heat capacity' in joules per kg per K.
Likewise all the other materials in the room, wood plaster and all that.
So you could add all that up and get a figure in Joules per K for the whole room, and that would tell you how many kWh left the room as dropped however many degC.
That's O Level physics, in reality, you don't know how much heat the walls have lost, because they won't all be an even temperature all the way through.
But it can be a useful thought process if you're thinking about rooms with lots of 'mass', there's significant heat stored in the inner leaf of a brick cavity wall for instance.
If you're trying to get at the 'watts per K' of a room, it may be informative to run say a 500W heater in there and see where the temperature goes, when the rest of the house is warm and outdoors is fairly cool but fairly steady.
I used a fan heater for a reality check on a bigger room, observed the temperature remotely. That told me the room was losing way more heat than it should have been calculated from the U values of the walls/windows/ceiling.
I thought it might be fairly simple to get a ballpark number, knowing that the room dropped x number of degrees, with an average temperature difference of y degrees, over z number of hours.
Obviously it's not that simple, as the size of the room and the construction need to be taken into account.
Looking at the other way - how much heat actually needs to be put back in - makes more sense.
At a steady 4 degrees outside temp, the current radiator has brought the room back up to 19.5 degrees with 9 hours of heating (00:3.0 when the hot water cycle finished, to 05:30 when the heating went off for 90 minutes, then on again until now).
That suggests that the radiator would be big enough to maintain temperature if it was zero outside - as long as the flow temperature was high enough.
Currently the rad temp is 44 degrees, so delta T around 25 degrees. The TRV is "cycling" though (room temp now close to target temp), rad temp dropping as low as 37 degrees.
Of course, I don't know how the radiator was running "flat out" for.
At T25, the radiator should be outputting around 430 watts.
Since the room temperature is rising, but the outside temp is 5 degrees above the design temp and the room 1.5 degrees above design temp, it looks like the heat loss calc of 394 watts is in the ball park?
If we get a really cold night, I'll try and keep a 500w oil-filled radiator in the room, with the radiator shut off. I can monitor the actual energy use with a Tapo smart plug.
Obviously it's not that simple, as the size of the room and the construction need to be taken into account.
Looking at the other way - how much heat actually needs to be put back in - makes more sense.
At a steady 4 degrees outside temp, the current radiator has brought the room back up to 19.5 degrees with 9 hours of heating (00:3.0 when the hot water cycle finished, to 05:30 when the heating went off for 90 minutes, then on again until now).
That suggests that the radiator would be big enough to maintain temperature if it was zero outside - as long as the flow temperature was high enough.
Currently the rad temp is 44 degrees, so delta T around 25 degrees. The TRV is "cycling" though (room temp now close to target temp), rad temp dropping as low as 37 degrees.
Of course, I don't know how the radiator was running "flat out" for.
At T25, the radiator should be outputting around 430 watts.
Since the room temperature is rising, but the outside temp is 5 degrees above the design temp and the room 1.5 degrees above design temp, it looks like the heat loss calc of 394 watts is in the ball park?
If we get a really cold night, I'll try and keep a 500w oil-filled radiator in the room, with the radiator shut off. I can monitor the actual energy use with a Tapo smart plug.
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