Insulation thickness

[david3456]

I want to convert the part of the garage that is integral to the house to a utility room. It is currently a double garage and a quarter of it is under the main house. There is a pillar in the middle supporting the corner of the house so it is a matter of separating it off by filling in two gaps.

The idea is two layers of plasterboard (outer, fire retardant & sealed), then a layer of plywood for mechanical protection on the garage side. I am looking for guidance on the gap for insulation between the two layers of plasterboard.

If I look at celotex xr4000, it quotes a conductivity of 0.022, which for 75mm thickness, converts to 0.29, better than the required 0.35. Is this the correct maths? The boards will obviously add a bit more but studding will take some away.

 

[endecotp]

look at celotex xr4000, it quotes a conductivity of 0.022, which for 75mm thickness, converts to 0.29, better than the required 0.35. Is this the correct maths? The boards will obviously add a bit more but studding will take some away.

Yes, 0.022 / 0.075 = 0.29.

The contribution of the studs is sometimes approximated as 15%, which would increase that to 0.85 x 0.29 = 0.34.

Whether 0.35 is the right target value, and whether a stud wall is suitable to separate a garage from a utility room, are questions that maybe someone else will comment on.

 

[david3456]

Whether 0.35 is the right target value, and whether a stud wall is suitable to separate a garage from a utility room, are questions that maybe someone else will comment on.

Thank you. I know about the need for 100mm of masonary at the base of the wall but I expect there are a few more issues to uncover.

 

[chappers]

I would have thought that a target u-value of 0.28 would be required as this is a new thermal element.

 

[Doggit]

I get a bit confused with thermal conductivity figures, but I think you've got it the wrong way round. You seem to have divided, where you should have multiplied. Having tried to use Celotex for external wall insulation, I would have needed 85mm PIR on a 9" brick wall, to get a u value of 0.3 for current building regs.

Check out http://www.celotex.co.uk/technical-services/online-u-value-calculator. You can create an account, and then use the calculator.

 

[endecotp]

if you're referring to my arithmetic, no, I'm confident that's right.

 

[Doggit]

Your maths is correct, it's just the way that it's been applied that the problem - but correct me if I'm wrong of course. I think you're trying to do 0.022/0.75 which make 0.029 rather than 0.29. I know that thermal conductivity (0.022) doesn't relate to the building regs U value which needs to be 0.3 after the insulation has been added to the other components of a wall. Building control now want 130mm of PIR insulation for a roof in a loft conversion, so to my mind, using 75mm of celotex, isn't going achiever the required figure.

Check out http://www.ccfltd.co.uk/Trade-Support/Calculating_U_Values to see the difference between thermal conductivity, thermal resistance, and thermal transmitance. I know the general principle, but I can never work out how they achieve the figures; which is why I use celotexes calculator

 

[noseall]

A typical spec' would be to fill the 100mm void with Celotex and then add a further 25mm across the studs. I haven't seen a thermally relevant stud wall spec' (whether that be a dormer, garage wall or ceiling, raked ceiling etc) whereby they hadn't shown insulation going across the studs as well as between.

Thicknesses of Celotex specified going across the timbers vary from 25mm up to 50mm.

 

[chappers]

according to the celotex calculator 120mm between only would achieve 0.27 for a garage partition but that would involve a 125mm stud.
75mm between and 25 over would achieve 0.23 and you could go lower to achieve 0.28

 

[endecotp]

Your maths is correct, it's just the way that it's been applied that the problem - but correct me if I'm wrong of course.

You're wrong.

I think you're trying to do 0.022/0.75 which make 0.029 rather than 0.29.

NO.

I know the general principle, but I can never work out how they achieve the figures; which is why I use celotexes calculator

I do know how to get the right figures, which is what I've done above. Here it is with units, in case that helps you understand it:

0.022 W/mK / 75mm = 0.29 W/m^2K

 

[tony1851]

You have to work out the thermal resistance of each layer first. (thermal resistance = thickness [in m.]÷thermal conductivity)

Starting on garage-side and adding up thermal resistances of each element; 12.5 plasterboard = 0.05
12.5 plywood = 0.09
75 PIR = 3.4
12.5 plasterboard = 0.05
______
total resistance = 3.59

The U-value is the reciprocal of this, = 1/3.59 = 0.278.

There are small additions to the total resistance due to the internal- and external surfaces of the wall (around 0.12 internally); there will be a reduction
due to the studs because their conductivity will be far greater than the PIR - separate figures need to be done for the U-value through the stud-part of the wall,
and the figures averaged out.
To prevent patern-staining on the interior, put 25mm PIR across the studs, and then the plasterboard.

 

[Doggit]

Nicely demonstrated Tony, but I've never managed to work out how to convert .022W/mk into thermal resistance in the first place.

 

[endecotp]

That's what Tony shows in the first line of his last post.