Having clad my garage with celotex and plywood, I'm wondering how much heat I need to keep it above freezing during the winter. I've looked for u-values but I keep finding the value for 9" brick walls. I thought a brick wall was about 4" thick. Are you laying walls with the bricks "the wrong way round"? Or is a 9" brick wall, two courses of brick next to each other, i.e. without a cavity?
Does anyone know the u-value for a single 4" thick brick wall?
I had a look at the calculator on the celotex web site but the example it showed used no u-value for plywood. Is this because you need a much greater thickness of wood before it has an insulating effect, or does
3/4" plywood have a u-value?
For the ceiling, I presume you discount the felt on the same basis that it is too thin to have a significant effect?
That's a half-brick thick wall, "one brick" being the stretcher length, i.e. 225 mm or 9 in. (module), 215 mm or 8-5/8 in. (actual). Brick walls can be any multiple of half a brick thick, with half (4-1/2 in.) one (9 in.) and one-and-a-half (13.5 in.) being most common. Look at garden walls and older buildings, e.g. pre-WW2 houses with 9 in. solid walls to see bonded brickwork with both stretchers and headers on show. See also
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Does anyone know the u-value for a single 4" thick brick wall?
That's a bit of a moveable feast, depending on the outside exposure and on the moisture content and density of the bricks. As a guide you could assume 3.3 W/m^2*K for half-brick and 2.4 W/m^2*K for one-brick walls.
A material doesn't have a U-value in its own right, it has a thermal conductivity, k (or k-value) in W/m*K. A material of a particular thickness has a resistance (R-value) which is equal to the thickness (in metres) divided by the k-value. Cavities and insulation layers also have their own R-values. To obtain the U-value for a particular construction you add up the R-values of each of the layers and then add on the air boundary layer resistances for the two faces. The latter - which you have to look up - depend on whether the faces are internal (sheltered) or external (exposed) and whether the heat flow is horizontal, upward or downward. When you have found the overall R-value divide it in to unity to obtain the overall U-value. Simples :-)
This refers to the wall of a house. The outer skin is brick then an air cavity of 50mm then lightweight blocks then 89 mm battens infilled with rockwool and an inner layer of plasterboard. The 88% refers to the proportion of the layer made of rockwool and timber. Ditto the blockwork where 7% is mortar which loses more heat than block so needs to be allowed for. NB Wall ties transmit a lot of heat (in and out), but don't seem to be in this example.
Adding up the thermal resistances of the layers :-
Section passing through blocks and rockwool (i,e, no cold bridging by mortar or battens)
R-value m2K/W external resistance 0.040 brickwork " 0.132 Air cavity 0.180 inner block resistance 0.909 rockwoll " 2.342 plasterboard " 0.130 Total resistance 3.783 U value =3D ONE divided by Total R value, i.e. 1 /
3.783 =3D 0.26
hence the U value of the most well-insulated section is 0.26 but this only about 88% of the wall. You need to repeat this with the sections where the battens bridge the plasterboard and the inner block and also the 7% where mortar bridges the gap between the cavity and either the battens or the rockwool depending on how the battens are arranged. This reduces the overall U value of this example to about 0.32.
When combining the U values of each layer you need to add up the R- values (thermal resistance) and use this to derive the overall U value - but remember to take account of cold bridging caused by battens, mortar and wall ties. Also the amount air leakage and external wind chill is important.
The higher the R-value the better the insulation, whereas with U-value the lower the value the better the insulation.
If you have battened a single skin garage then you can ignore the effect of the cavity and the inner block but the % of the single skin that is mortar is much more significant. Did you remember to line the wall with a DPC before battening and insulating, else wind driven rain will get into the insulation and render it useless.
Plywood has a thermal resistance (R value) of 0.077 (for 10mm). There is a similar example of a timber framed wall on pages 38 and 39.
Basically, you need at least 50mm of PIR insulation (Celotex) which is equivalent to 100mm of rockwool to reduce heat loss by 80%. 100mm of Celotex is the best, but with a cold outer skin you also have the problem of interstitial condensation so you also need a vapour barrier on the WARM side of the insulation if you have used rockwool. I.e use foil backed plasterboard and mastic the joints between boards and between board and batten. If you used celotex then it can be glued right onto the wall with foaming PU glue and then tape the joints with foil tape. PIR insulation does not absorb water and the two foil faces act as DPC to repel wind driven rain and the inner foil face is a vapour barrier. in exposed areas you would need a cavity between brick and insulation to deal with wind-driven rain.
My workshop (12'x17' approx with pitched slate roof, 50mm PIR foil faced foam, 12mm ply lining), has a 2kW fan heater on a separate room stat. I normally leave the stat set on about 6 degrees. Its fairly rare for the heater to kick in. This time of year the outside temp is dropping to close to freezing at worst. The internal temp is staying above 11 degrees at the mo, with nothing more than occasional use. The lights alone can heat it a fair bit over a few hours. In the depths of winter (I recorded temps below -15 on some nights last year!) you might get the heater come on for a handful of 10 min bursts each day.
You can almost ignore the bricks - they are of negligible effect compared to the insulation!
Again if you have insulated the inside of the roof space, then that will dwarf the insulation value of the roof itself.
I had a look at the wikipedia entry you pointed me to. Thanks. yes, my wall is just one row of stretchers, so although it is one brick thick the terminology is a half brick wall, no wonder I was confused!
The only place there are headers are buttresses. There are three of these along the length of the wall.
Thanks. The celotex calculator had these figures and I wondered what they were all about, now I know.
Perhaps that's the easy way to get a rough figure. Work out how much heat is lost through 50mm celotex and use that for my calculation. I guess I also need to work out heat lost through the concrete floor though?
You could guestimate a single brick wall, since a 9" non cavity wall has a u value of 2.2, using a value of 4.4 is likely to be a reasonable approximation.
Celotex type foam has a k value of 0.023 - so 50mm of that gives a u value of 0.023 / 0.05 = 0.46
So that combined with a 4.4 u value brick wall = 1/(1/0.46 + 1/4.4) = overall u value of 0.41 (i.e. not much better that the insulation on its own!)
If you know what the construction is and how much of the slab perimeter is exposed outside you can do some sums...
hts a harder one to crack, as the actual insulation effect of the soil is considerable. Heat is essentially lost via the ground outside the perimeter. Not directly downwards. The crude calculations required by regulations mean that a factor of perimeter length to area is used.
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