Why does heating my house require 280 watts per degree Celsius above ambient?

Previously I explained how I learned that for each degree Celsius the outside temperature falls below 20 °C, it takes 280 watts of heating to keep my house at 20 °C.

In order to provide this heating, I burn gas which last winter resulted in the emission of around 17 kg of carbon dioxide per day – around 2.5 tonnes in all.

I would really like to reduce this shameful figure, but I have only finite resources. In order to act I need to know where best to spend my money.

In this article I will explain how I came to understand the relative significance of the windows, roof and walls in this heat loss.

Windows

It is easier to estimate the heat loss from windows than it is from walls.

This is because walls are opaque and (without expert knowledge) it is not obvious what the wall is made of. Moreover, different walls in the house can have different construction and thickness. However, being transparent, one can see directly the type and construction of windows.

The heat flow through a window(or wall) is characterised by a U-value. This states the amount of heat which flows across 1 square metre of the window when there is one degree Celsius of temperature difference across the window.

The units are for U-values are watts per metre squared per degree Celsius (W/m²/°C) or watts per metre squared per kelvin  (W/m²/K). These two units are equal to each other.

Roughly speaking U-values for windows are [Link]:

• Old single-glazed windows: 6 W/m²/°C
• Old double-glazed windows: 4 W/m²/°C
• New double-glazed windows: 1.5 W/m²/°C
• The best triple-glazed windows: 1.0 W/m²/°C

I proceeded as follows:

• I made a list of the 21 windows, skylights and glazed doors in in my house.
• I measured their area – width × height in metres.
• I multiplied their area by their U-value to get the transmission per degree Celsius through that window.
• I then added them all up.

For each window in the house I multiplied the area by the estimated U-value to get the heat transmitted per degree Celsius of temperature difference. I colour-coded the column to highlight which windows were the worst. Adding up all the windows came to 75.7 watts per degree Celsius. If I replaced all the windows with the best available I might be able to reduce this to 24.0 watts per degree Celsius.

The estimated total transmission through all the windows and doors came to about 76 watts per degree Celsius. I concluded that:

• Firstly,  I could see which windows lost the most energy – they are colour-coded red, amber, and green in the figure above. There are no surprises – the largest area windows lose the most energy.
• Secondly, I could see that if I replaced all the old windows with modern ones (U = 1.5 W/m²/°C), I might hope to reduce the window losses by roughly half their current value, to around 36 watts per degree Celsius. If I spent a lot – on triple-glazed windows and used insulating blinds, I might hope to achieve U = 1.0 W/m²/°C and reduce the losses to 24 watts per degree Celsius.
• Thirdly, since the house as a whole is losing 280 watts per degree Celsius, I could see that windows and doors account for about a quarter of the energy lost from the house.
• And finally, logically, the remaining 75% of the losses (280 – 76 = 204) must be going the through the roof, walls, and floors or lost in draughts.

Roof and Walls 

By analysing the thermal transmission of the windows and doors (transmission = 76 watts per degree Celsius), I concluded that roof and walls must be transmitting about 204 watts per degree Celsius.

• Is this estimate reasonable?

To answer this question I embarked on yet another tedious and difficult exercise.

• The tediousness arises because I need to add up all the areas of the roof and walls, subtract the areas of the windows and skylights, and then estimate the U-value,
• The difficulty arises because I don’t know the materials from which the walls of the house are constructed!

Most of the walls date from the 1930’s (I think) and are probably solid brick. A 1970’s extension is probably not much better thermally, but I don’t know. However, the extension we built 10 years ago was built to building regulations at the time and I have a pretty good idea of the appropriate U-value.

So I made measurements of the wall areas. And then I assumed (link) that:

• The old walls had a U-value of 2 W/m²/°C – a value appropriate for a double-skin solid brick wall.
• The new walls had a U-value of 0.3 W/m²/°C – a value specified by current building regulations.

For each wall or roof, I multiplied the area by the estimated U-value to get the heat transmitted per degree Celsius of temperature difference. I colour-coded the column to highlight which were the worst. Adding it up came to about 229 watts per degree Celsius. If I clad all the walls to achieve a U-value of 0.3 watts per metre squared per degree Celsius, I might be able to reduce this to 54 watts per degree Celsius.

With these assumptions I estimated the heat transmission through the roof and walls. As shown in the table above, I arrived at an estimate of 229 watts per degree Celsius. This should be compared with estimate of 204 watts per degree Celsius that I arrived by analysing:

• My gas meter readings
• The average weekly temperature
• The estimated properties of the windows.

Given all the uncertainties, I take this as confirmation that within about 10% uncertainty, I can understand the thermal properties of my house.

Summary

Currently my house loses 280 watts for each degree Celsius the external temperature falls below ambient. Of those 280 watts,

• roughly 76 watts flow through the windows and doors
• the remaining 204 watts flow through the walls, floors and roof.

With modern double-glazing I could reasonably hope to reduce the glazing losses from 76 watts to around 36 watts, or possibly even lower with triple-glazing and thermal blinds.

Cladding the entire house I could hope to reduce the losses from around 204 watts to around 50 watts.

• What should I do?

In the next article I will discuss my strategy.