Unit 317 Session 5a principles of heat loss and heat gain

​The principles of heat loss and heat gain in dwellings part 2

​We have looked at calculating heat loss through wall structures however there are some further elements we also need to consider when calculating the heat loss for a whole room or building.

Open Ended

What other factors do you think affect heat loss in a building?

​Glass and glazing

Glass has a fairly high thermal conductivity of around 1.05 W/mK.

The reason why so much heat is lost through the glass is that it is the thinnest part of a building.

The heat loss can only be reduced by increasing the thickness of the glass or using it to form cavities like double and triple glazing.

​Internal design temperatures
The design temperatures within dwellings are based upon the type and usage of the room.
Internal design temperatures should be chosen to ensure satisfactory comfort conditions.

​External design temperatures

The successful design of any central heating system is based on the fact that the dwelling is maintained at a certain specified temperature based upon the prevailing (at the time) external temperature.

It follows that calculations must be based on a realistic external temperature that can be expected for the region during the coldest months.

​Air change rates
'Air change rate: a measure of how many times the air within a defined space (normally a room or a house) is replaced per hour, usually through natural ventilation.'
It is necessary to prevent the air in the room from becoming stale and to prevent the onset of moisture problems and mould growth.

As the air change occurs, the heat in the room is lost by warm air leaving the room and cold air entering.

​Air change U-value = 0.33 W/m3K.

Air change rates per room

​Calculating the heat loss from a dwelling – the tabulation method
The easiest and quickest way to conduct heat loss calculations is by using a table to record all of the figures.
Each room would require its own table.

​The simple room opposite has four outside walls and heat loss through both the floor and roof.

​The U-values required are as follows:
External walls: 0.35
Floor: 0.25
Roof: 0.25 (flat roof)
Windows: 2.9
External door: 2.9 (the same as the windows)
Air change factor: 0.33
Temperature difference:
21°C (internal temperature) -3 (external temperature) = 24 °C

​The calculations revolve around the following elements:

 1 Calculate the air change heat loss. (The room is a kitchen)

 2 Calculate the external wall heat loss.

 3 Calculate the glazing heat losses.

 4 Calculate the floor and roof heat loss.

 

​First, let’s concentrate on Point 1.
Calculate the air change heat loss.

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Point 1 – air change heat loss

The room is 5 m × 3 m × 2.5 m.

This gives a volume of 37.5 m3.

When this figure is multiplied by the temperature difference (24°C), the number of air changes (2) and the air change factor (0.33), the total becomes 594 watts.

This means that because there are two air changes every hour, the room will lose 594 watts of heat during these changes:

37.5 x 24 x 2 x 0.33 = 594 watts

​Point 2 – external wall heat loss

​We now need to calculate the area of the walls of the room.

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​Now look at Point 2.

Point 2 – external wall heat loss

This can be a little involved.

If we laid all the external walls flat, we would end up with the following: 5+ 5 + 3 +3 = 16 m

This figure must now be multiplied by the height of the wall: 16x 2.5
= 40 m2

But this figure is not much use to us as it is.

There are windows and doors that have a greater heat loss than the wall so these MUST be deducted BEFORE we calculate the external wall heat loss.

​So first calculate the area of the windows and doors:

Window: 2 m × 1.5 m = 3

Door: 2 m × 1 m = 2

= 5 m2

This figure can now be deducted from the external wall total:

40 – 5 = 35 m2

So, the heat loss from the external walls is: 35 x 24 x 0.35 = 294 watts

​Point 3 – glazing heat loss

This part of the calculation is made easy by the fact that we have already calculated the areas of the glazing, so it’s a straightforward calculation.

Also, the same U-value is used for both window and door because they are made from the same material.

This isn’t always the case and you must check before doing the calculation:

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​Point 4 – heat loss from the floor and roof/ceiling

​We now need to add the floor and ceiling

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​Point 4 – heat loss from the floor and roof/ceiling

Because these use the same area, the calculation, again, becomes very straightforward.

The area of the room is: 5 m x 3 m = 15 m2

So, now multiply all of the figures together for both floor and ceiling:

Floor: 15 × 24 × 0.25 = 90 watts

Ceiling/roof: 15 × 24 × 0.25 = 90 watts

So, the total heat loss for the room is the sum of all of the calculated elements:

Air loss = 594
+ Walls = 294
+ Door = 208.8
+ Window = 139.2
+ Floor = 90
+ Ceiling = 90

Total = 1416 watts

​Heat-loss for this room is 1416 watts

​Now its your turn, your tutor will give you session 5b for you to complete.