Andy Hall wrote:
OK, so you can pretty much use the U value of the insulating material
on the floor for that surface. Heat loss is not going to worse than
represented by that. Strictly, there is a formula taking into
account the perimeter and area of the floor, but in this case it is
not going to make a lot of difference as the floor loss will be small
anyway.
OK, I'll have to find that out, as I didn't buy or install that, but
it shouldn't be too hard to get the U value.
I meant if the building is divided equally in two - one half workshop
other for office etc. and with different temperatures. Obviously if
it is divided differently or temperatures are different you can do the
sums accordingly.
Ah! I see what you mean. Seeing as the two areas will have different
use, it would be beneficial to calculate the insulation requirements
of each? By the way, is the purpose of carrying out these calculations
to establish the minimum thickness of Celotex required, in order that
a standard level of insulation is achieved, or is it to obtain 'values'
which would indicate which heating solution should be implemented?
Sorry if I sound dumb, but this is all new to me and while I certainly
appreciate that a bit of time spent now, calculating temperature
parameters, could save me a lot of time and money, it's very involved
compared to the opinions I have so far been offered by various friends
and associates in the building industry! In summary, is the purpose of
all this so that I can make an informed decision on purchasing the
minimum of spec and amount of Celotex, while not setting myself up for
huge bills resulting from having to overcompensate on a heating solution?
OK, so you have to look at the heat loss through each element of each
surface.
In other words for each side wall, take the total area and subtract
the window areas, for the end wall the doors and so on as you have
done.
Strictly speaking you should take into account all the components of
the wall, e.g. the outer wood, the Celotex and the inner cladding, but
you will find that the Celotex is the dominant one by far (as you
would hope) so for this purpose you can just use that.
45mm Celotex has a U value of 0.5 W/m^2.K
for 55mm it's 0.42
For other thicknesses you can use the table in
http://www.celotex.co.uk/appl/PDF/SOL_PRS.pdf
Just take the reciprocal of the R value.
You can also combine if you want to use multiple thinner sheets. Add
the R values and take the reciprocal of that.
Heat loss for a surface is then given by
Area in sq.m x U value x temperature difference.
Usually you work with -3 degrees outside.
So for an area of 5 sqm, 45mm Celotex and 21 degrees inside it becomes
5 x 0.5 x 18 = 45W.
So for the Celotex covered areas just calculate each piece.
Standard double glazing in wooden frames has a U value of 2.8 W/m^2.K,
or 2.1 if it has low-emissivity glass.
If the doors are substantially glass, then it;'s reasonable for this
purpose to treat them as windows.
So again take the areas and do the sums.
OK, time for the maths bit again...
to convert R to U... U=1/R
So, for 55mm Celotex, R=2.35; U=0.42
Wall A (minus window) = 31.4196m/sq x 0.42 x 18 = 237.5W
Wall B (minus windows) = 28.7016m/sq x 0.42 x 18 = 216.9W
Wall C (minus door) = 8.1016m/sq x 0.42 x 18 = 61.2W
Wall D (minus 2 doors) = 6.4496m/sq x 0.42 x 18 = 48.7W
Gable end E and F = 6.2464m/sq x 0.42 x 18 = 47.2W
Roof elevations G & H = 71.0482m/sq x 0.42 x 18 = 537.1W
For windows and mainly glass doors, U=2.8
3 casement windows = 8.154m/sq x 2.80 x 18 = 410.9W
3 single doors = 4.956m/sq x 2.80 x 18 = 249.7W
Total = 1809.2W
Does this mean I need a heating solution that can produce
1.8kW of heat? Per hour? per day? etc etc!
If you were looking at what is necessary to do Building Regulations
compliance, you ned to be more particular about the other components
because they help a little towards insulation. Here it it doesn't
matter because you are trying to get to a reasonable worst case for
heating requirement. Obviously adjust the outside temperature used
if you are in a colder or exposed position and want to oaccount for
it.
Understood and agreed.
He's probably right. Obviously it would depend on your pattern of
use, how long you want to take into account for the costs (i.e. how
long will you live there) and to some extent taking a punt on energy
costs. You have to run electricity anyway.....
Makes sense, and as I'm hoping to run a profitable business from the
building (usage constraints aside), I would envision that this would
become an overhead of the business and as such would be infinitesimal
compared to accounting for rent (of which I'll have none) when
calculating operating costs and charging such onto clients!
OK. I was trying to understand whether you wanted to leave the rafters
exposed.
Sorry, should have made that clearer.
The laths were in behind purely to make sure that the Celotex was
spaced away from the felt. If you cut the Celotex to be an
interference fit between the rafters you don't really need them.
Interference fit? Held there by friction? Accurate cutting?
Another technique is to put some between the rafters and some over the
top. This reduces cold bridging - i.e. heat conducted through the
rafters, but I am not sure that I would bother for this application.
That would be guilding the lilly
Oh, I see. I thought that it was the "uniform". :-)
It is I suppose... in an environment with lots of 'marketing
types' who are constantly concerned about being 'en vogue', it's
best not to waste valuable creative energy in the mornings by
dressing to show individuality and personal taste! They would
see this as a statement and would spend all day discussing your
choice of attire for the day rather than focussing on making
the 'big decision' about the concept(s) you've spent the last
3 months trying to get them to approve! It's complete irony that
in order to achieve 'closure' (as the American's put it), creatives
have to work with non-creatives!
Worth a try even with these, akthough you may still not reach the
price point of Christian's Seconds outfit.
And would you consider this route?
deano.