I am converting a garage to a usable room and am researching different options when it comes to insulation. I know more insulation is better, however there is a point where adding more insulation will never reduce the heating bills by enough to pay for itself.

Assuming the external dimensions of the room is 5.5m long, 3.5m wide and 2.5m high, this would give a surface area of:

  • 5.5m * 2.5m * 2 walls = 27.5 m2
  • 3.5m * 2.5m * 2 walls = 17.5 m2
  • 3.5m * 5.5m * 2 (1 roof and 1 floor) = 38.5 m2

The total surface area would be 83.5 m2

If 75 m2 had a 0.33 w/m2k and 8 m2 has a 1.5 w/m2k (windows and doors), would I be right to assume that this would give a 37 w/k for the entire room?

If the difference between the internal and external temperature was 10 degrees C, I assume that I'd need 370 watts to heat the room. If the walls/ceiling/floor had a 0.22 w/m2k, then without listing all the calculations I'm assume this would be 270 watts, therefore a saving of 120 watts or about £0.03 of electric per hour.

Firstly, I was wondering if someone could please confirm if I'm at least on the right track please?

Secondly, any recommendations on how I can work out the division of lost between the walls, floor and ceiling please?

  • You're on the right track. Ecneral's approach is better in that it includes more details and gives visibility into the individual pieces. Even from what you've done, it's easy to see that windows are a killer in so far as heat loss/gain or energy efficiency is concerned.
    – SteveSh
    May 10 at 19:49
  • Windows always are. But a screen inside and a camera outside are not quite the same thing as an actual window... so we tolerate them.
    – Ecnerwal
    May 10 at 19:55
  • Thanks guys, but anyone know how to determine any weighting I should give the ceiling compared to the floor, etc please?
    – Jay
    May 11 at 19:59

I think the simplest way to evaluate this is with a spreadsheet, where you collect the area of each "type" and can plug in numbers for different insulation values for each type, and view the effect on that part and on the total.

i.e if all the windows are the same sort or will be when purchased, you can lump that area & U value.

If there are different types, separate them into different rows in the spreadsheet.

The walls get a row, unless the walls differ, then they get as many rows as there are different areas of wall-types.

The ceiling gets a row.

The floor gets a row, but is often tricky to characterize since insulation value and exterior temperature is harder to know for sure on the floor.

...because that's the way I do it.

Other then "metric R" tacked on the end, this is all in "English" units mostly used only in the USA now, but it's easy to do in the units applicable to your area. A few numbers are input, most are calculated. It's easy to see the difference that changing one part makes to that part and the overall result. For instance, increasing my ceiling insulation can be done, but will have little impact on the overall use as it's already less than half what the walls are, and likewise less than half what the windows are.

enter image description here

When you get heavily insulated, you also need to pay attention to the heat lost (or exchanged) via ventilation, with or without heat recovery, as it becomes a significant part of the overall load.


Maybe this site can help. An easy first step is to open and edit an example. Layers can be switched on/off by a mouse click.


In a few minutes it reveals problems with insulations/moisture or the order of layers. Toggling between R- value/U- value is done by clicking on the tool icon next to the result.

The screen shot is an example from another stack exchange question.enter image description here

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