u-values for brick
Fred
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?
TIA
Tabby
Andy Wade
> I thought a brick wall was about 4" thick.
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 http://en.wikipedia.org/wiki/Brickwork.
> 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.
> [...] 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?
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 :-)
--
Andy
ad
> On 14/11/2010 10:47, Fred wrote:
>
> > I thought a brick wall was about 4" thick.
>
> 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.
>
> > 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.
>
> > [...] 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?
>
> 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 :-)
>
> --
> Andy
If you look at pages 34 and 35 of Approved Doc L (available online)
you can see the calculation.
Taken from the document :-
Layer Material Thickness (mm) R-value U-value
1 Brick 102 0.132
0.77
2 Air cavity 50 0.18
3a Light block(93%) 100 0.909 0.11
3b Mortar(7%) 100 0.114 0.88
4a Rockwool(88%) 89 2.342 0.038
4b Timber battens(12%) 89 0.685
0.13
5 Plasterboard 12.5 0.050 0.25
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 :-
1. 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 = ONE divided by Total R value, i.e. 1 /
3.783 = 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.
John Rumm
>
> 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
Not much...
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.
> 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?
You can almost ignore the bricks - they are of negligible effect
compared to the insulation!
> For the ceiling, I presume you discount the felt on the same basis
> that it is too thin to have a significant effect?
Again if you have insulated the inside of the roof space, then that will
dwarf the insulation value of the roof itself.
--
Cheers,
John.
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Fred
<spamb...@maxwell.myzen.co.uk> wrote:
>That's a half-brick thick wall, "one brick" being the stretcher length,
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.
>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.
Thanks. The celotex calculator had these figures and I wondered what
they were all about, now I know.
Fred
<see.my.s...@nowhere.null> wrote:
>You can almost ignore the bricks - they are of negligible effect
>compared to the insulation!
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?
John Rumm
> On Sun, 14 Nov 2010 17:59:01 +0000, John Rumm
> <see.my.s...@nowhere.null> wrote:
>
>> You can almost ignore the bricks - they are of negligible effect
>> compared to the insulation!
>
> 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
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!)
> guess I also need to work out heat lost through the concrete floor
> though?
If you know what the construction is and how much of the slab perimeter
is exposed outside you can do some sums...
The Natural Philosopher
50mm celotex is practically full spec insulation.
Brick k factor is an average of about 1 W meter, so on 4" brick (100mm)
is about a U value of 10 or so,
Celotex is about .020 k so 50mm is a U value of 1 more or less.
Combined its 1/11 or around 0.9 So adding the celotex reduces heat loss
by about 11 times.
As aginst 10 for the celotex alone.
The Natural Philosopher
> I also need to work out heat lost through the concrete floor
> though?
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.