Ian Stirling wrote:
Steve wrote:
snip

Can I ask you how you work out these values of heat loss? Please don't
think that I am trying to find fault but rather that I would like to do
calculations like yours for our new property.


Unfortunately, the hard way.


OK, I can google but if you can provide a trusted link or reference that
would give me the necessary information to do the calculations myself
that would be even better.

It seems that there is clearly a diminishing law of returns here, as in
most things to do with physics - once we are in the new place it would
be nice to put some numbers into the equation.



Unfortunately, I have not found a 'nice' solution that works for me -
say javascript/java, pick your construction, specify depths, ...

I did it the hard way.
google for
BS EN ISO 6946:1997 examples

http://www.thenbs.com/BuildingRegs/k...DOMEST_6_6_0_1

looked interesting.

and that will turn up examples of how to calculate the U value, to the
1997 BS standard - which is slightly out of date, but nothing much has
changed, and is fine for indicative purposes.

Basically.
Thermal conductances - measured in w/m/k give the number of watts that
a cube one meter in dimension will pass, per kelvin (or centigrade)
difference across the opposite faces.

For example, stone is about 1.2, rockwool and many other insulators
(polystyrene, compressed straw, wool) about 0.04, kingspan 0.02, wood
0.15.

If you have a tenth of a square meter, it passes a tenth as much heat.
If the wall is a tenth of a square meter thick, then it passes 10 times
the heat.

U value - the number of watts per square meter that a structural element
will pass given a 1C difference.

For monolithic structures, this is just the conductance divided by the
thickness (in meters) - for example 50mm of kingspan has a thermal
conductance of .02 w/w/m / 0.05m = 0.4W/m^2/k

(Or 4 watts per meter square, at a temperature difference of 10C)

For stone, with a conductance of 1.2, you need 3 whole meters (1.2 / 3 =
0.4) to get a U value similarly.

Stone sucks (heat).


Now, it gets more complex.

You can't simply add U values of multiple elements in a wall - as that
would mean that adding another identical element onto the side of this
one would double the heat loss, which is obviously wrong.

You've got to convert into R values, which are the inverse of U values
(R value = 1/U value), then add these, then convert back to U values by
the same manner.

This all works find for homogenous walls.

It gets complex when you consider plasterboard + lath, with insulation
between.

A simple way is to consider these as two seperate structures - in the
last post I made, I used the example of 100m^2 of rockwool, and 8m^2 of
wood - as 1/12th of the area of my example ceiling was wood.

So, simply
((U value of rockwool) * rockwool area ) * ((U value of wood) * wood area)

--------------------------------------------------------------------
total area

= composite U value.


R values can be thought of as the area that will pass one watt, with a
difference of 1C across the structure.

Hope this helps.


Thanks very much for the information and the web link Ian.

I have saved this and bookmarked the page for future use.

The explanation you give has helped me to understand the principle as
well as the actual calculations.

I am now looking forward to spending some time (when I have some :-( )
putting some numbers in...

Cheers

Steve