Heat loss
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This article provided a basic description of how to calculate the heat loss for a part of a building structure (say a wall or roof), or for complete buildings. Doing this calculation can greatly help when choosing a boiler, or working out where to best spend money improving insulation.
Causes of heat loss
Mother nature dictates that heat will flow from hotter objects to cooler ones. Generally there is not much we can do about this, but we can change how quickly it happens. Most heat lost from a house is initially via conduction - through walls, floors, and ceilings. These losses will tend to heat the outer surfaces of the building where the heat is then convected away to the atmosphere. The second major loss of heat is through air changes; every time a draft allows cold air in from outside (and conversely warm air out), that cold air will need to be re-heated.
Calculations
It is relatively easy (although a bit tedious!) to do a full set of heat loss calculations in a spread sheet. One works through the property room by room computing a less (or gain) for each room.
Heat loss via conduction
To calculate the rate of loss of heat through a surface we need some basic information about it. The rate of heat loss through a surface will depend on:
- The total area of the surface (A) - the larger it is, the faster will flow through it.
- The temperature difference (ΔT) - the bigger the difference in temperature from one side to the other, the faster the heat flow
- The thermal resistivity of the surface (U) - the more insulating the material, the slower the rate of heat flow.
The rate of heat flow is simply:
F = A x ΔT x U
To get sensible answers, we need to use consistent units for all measurements. So sticking to SI units, the rate of heat loss will be given in Watts. Areas need to be measured in square metres, and the thermal resistance in W/m²K - where K is the temperature difference in Kelvins or degrees Celsius
The area is generally easy to compute, although you may need to break a surface down into a number of sub surfaces if the construction is not consistent over the whole area. So for example an outside wall of 10 m², may include 2 m² of window. These areas will need to have their heat losses computed separately to allow for the different rates of loss through the wall and the window.
The temperature difference is simply the difference between the rooms normal temperature and whatever is the other side of the surface. Now with an outside wall, the chances are the outside temperature will be significantly lower that inside. When computing worst case losses for the depths of winter one would typically use and assumed outside temperature of -3°C. Note that with inside walls, there may actually be a heat gain from the adjoining room rather than a loss - it depends on which room is hotter.
The U value will vary with the building material and the type of construction. Usually you can look these up in a table to find a suitable figure.
Heat loss due to air changes
The simplest way to deal with air changes is to make an assessment of the number of times the complete volume of air in the room will be changed. There are standardised tables for these values which vary for the type of room. However in the absence of a suitable value you can assume 3 changes an hour is a typical worst case for a room with some draft proofing.
If one knows the volume of the room, the number of cubic metres changer per hour is easy to work out. Once you have this you multiply by a standardised constant figure of 0.36 W/m³h
The air change heat loss constant is derived from multiplying the number of cubic meters of air by the mass of 1 cubic meter to convert m³ to kg Then multiplying the mass in kg by the specific heat capacity to get a total in Joules Finally dividing by 3600 to convert figure in J/h to one in J/Sec (Watts)
Worked Example
Lets take a very simple "house" with two rooms:
We will assume it has:
- 9" solid brick walls
- decent argon fill double glazed windows and door
- The partition wall is a plasterboard
- There is a pitched roof over with tiles and felt, and 100mm of loft insulation
- The floor is a concrete slab
- Outside temperature is -3°C
- The left had room is warmest at 21°C, and the smaller room is 18°C
- Air changes in the left room are 1 per hour, and 2 per hour in the right hand one
- and finally the rooms are all 2.2m tall
If we slap all the above figures into a speread sheet, and take the u values from the table at the end, we get:
Room | Surface | Width | Height | Area | Tdelta | u-value | Loss | Air Cng | Vol | Total |
Left Room | Front Wall | 3800 | 2200 | 2.37 | 24 | 2.2 | 125 | 1.00 | 25 | |
Window | 2722 | 2200 | 5.99 | 24 | 1.7 | 244 | ||||
Left wall | 3000 | 2200 | 6.60 | 24 | 2.2 | 348 | ||||
Partition wall | 3000 | 2200 | 6.60 | 4 | 1.8 | 48 | ||||
Rear Wall | 3800 | 2200 | 8.36 | 24 | 1.2 | 241 | ||||
Floor | 3800 | 3000 | 11.40 | 24 | 0.8 | 219 | ||||
Ceiling | 3800 | 3000 | 11.40 | 24 | 0.3 | 82 | ||||
Total | 1307 | 217 | 1524 | |||||||
Right Room | Front wall | 2806 | 2200 | 3.01 | 21 | 2.2 | 139 | 2.00 | 19 | |
Window | 1318 | 1318 | 1.74 | 21 | 1.7 | 62 | ||||
Door | 762 | 1865 | 1.42 | 21 | 1.7 | 51 | ||||
Partion walll | 3000 | 2200 | 6.60 | -4 | 1.8 | -48 | ||||
Back wall | 2806 | 2200 | 6.17 | 21 | 2.2 | 285 | ||||
Right wall | 3000 | 2200 | 6.60 | 21 | 2.2 | 305 | ||||
Floor | 2806 | 3000 | 8.42 | 21 | 0.8 | 141 | ||||
Ceiling | 2806 | 3000 | 8.42 | 21 | 0.3 | 53 | ||||
Total | 989 | 280 | 1269 | |||||||
Total loss | 2793 |
(A Google Docs version of this spreadsheet can be seen here)
Notes:
- Remember to subtract window / door sizes from the front wall area
- We are assuming that the partition wall door is the same u value as the stud wall
- The right hand room gets a nett gain of heat from the left
Conclusions Total heat loss is just under 2.8kW on the coldest of days. If you redo the sums for an more typical average outside temperature of 10°C, then you get a total heat loss of only 1.2kW
Tables of figures
Air changes and typical room temperatures
Room type | Room temp | Air Changes |
---|---|---|
Lounge | 21 | 1 |
Dining Room | 21 | 2 |
Bedroom | 18 | 0.5 |
Hall and Landing | 16 | 1.5 |
Bathroom | 22 | 2 |
Kitchen | 18 | 2 |
u Values
Often you find u values quoted for different building materials. Sometime you will however find a k value given. This is simply the thermal conductivity of the material. This can sometimes be quite handy, since it does not include any implicit concept of the thickness of the material. You can use a k value to calculate a u value simply by multiplying it by the thickness of the material you have and then taking the reciprocal of it.
Common wall and ceilings
Materials | u-Value | Comments |
---|---|---|
Wall - outer 9" solid brick | 2.2 | |
Wall - internal plaster over 4" block | 1.2 | |
Wall - internal PB over stud | 1.8 | No insulation in void |
Floor (ground) - solid concrete | 0.8 | |
Floor - PB + joist + FB flow up | 1.9 | Flow from downstairs to upstairs is faster than the other way |
Floor - PB + joist + FB flow down | 1.5 | |
Roof pitched with felt + 100 insulation | 0.3 | |
Window - wood DG | 2.9 | |
Window - wood - low E | 1.7 | |
Door single glazed | 3 | |
Wall Insulated | 0.6 |
DG Units
Glass specification | Cavity | ||
---|---|---|---|
12 mm | 16 mm | 20 mm | |
float/air/float | 2.9 | 2.7 | 2.8 |
float/argon/float | 2.7 | 2.6 | 2.6 |
float/air/Pilkington K Glass | 1.9 | 1.7 | 1.8 |
float/argon/Pilkington K Glass | 1.6 | 1.5 | 1.5 |
Overall thickness of unit(mm) | 20 | 24 | 28 |
Computing u values for more complex walls
When dealing with complicated wall constructions it can be handy to use the k values of the individual layers of the wall to compute a u value. This also lets you compute an overall u values for complex surfaces made up from several types of materials. You can take the k values of each material and multiply by the thickness of it, and take the reciprocal of the sum of the results.
See an example for a rubble filled wall