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 heat lost will need to be replaced.
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 out measurements. Therefore 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 ... TBC
Tables of figures
Air changes
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.
Materials u-Value
Wall - outer 9" solid brick 2.2 Wall - internal plaster over 4" block 1.2 Wall - internal PB over stud 1.8 Floor (ground) - solid concrete 0.8 Floor - PB + joist + FB flow up 1.9 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 glaze 3 Wall Insulated 0.6
DG Units - http://www.glassmasterltd.com/uvalues.html
float/air/float float/argon/float float/air/Pilkington K Glass float/argon/Pilkington K Glass Overall thickness of unit(mm) 12 mm 16 mm 20 mm 2.9 2.7 1.9 1.6 20 2.7 2.6 1.7 1.5 24 2.8 2.6 1.8 1.5 28
"Difficult" Walls
Rubble fill
From one of Hugo's posts:
Outside surface resistance = 0.04m^K/W Limestone: conductivity = 1.13W/mK; divide the thickness by this value to give resistance. Mortar (and presumably loose fill): conductivity = 0.84W/mK [1] Rockwool: conductivity = 0.05m/0.038W/mK = 1.32m^K/W [2] Plasterboard resistance = 0.06m^K/W. Inner surface resistance = 0.13m^K/W
Add up all the resistances, then take the reciprocal to give your U-value.
[1] It depends on the proportions of wall to rubble-fill, but I would have thought about 75% stone to 25% mortar and crap, which gives an average conductivity of about 1.04W/mK.
[2] If you want a precise figure, you should also include for the studs in the dry-lining.
R = 0.04 + 0.56/1.04 + 1.32 + 0.06 + 0.13 = 2.09m^2K/W. Therefore U-value = 1/2.09 = 0.48W/m^2K. To work out heating, probably best to use 0.6-0.8 unless you can put actual values to the thicknesses of the leaves of the wall.