Thermal resistance of oak

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Thi might sound a bit dumb but I don't understand what this is about, can someone please enlighten me :oops:

It's the figures thatI don't understand.

Many thanks

Martin
 
Thanks, Chas.

Martin

It's all to do with heat transfer.

Thermal conductivity = heat flow rate × distance / (area × temperature difference) ie how easily does something conduct heat

Thermal resistance is the reciprocal of thermal conductivity. Another name is thermal insulance (ie insulation). Units in the case of Chas's figures are (sq m X temp difference C)/watts.

The higher the better as far as insulation goes.

My interest stems from being able to calculate the thermal bridging in my roof due to the oak rafters.

Roger
 
AS Roger says the R value of each part of a wall is used to calculate the U value.

By taking a cross section of the structure and using the R value and thickness of each component along with surface resistance and any cavities the U value can be calculated. Any wall, roof or floor construction will have to be below a set U value to meet building regs.

It is also possible from the same data and knowing the average internal and external temps and RH to work out the "dew point" this is where any moisture in the internal air will condence as it passed through the structure, it is important to make sure any vapour barrier is placed before the dew point.

Thats basically it, been 20 odd years since I last did the calcs at college :wink:

Jason
 
To flesh it out a little from a practical perspective, the cottage roof rafters are oak and about 80mm square. Now modern rafters are about 7" x 2' (I think) and Building Regs and insulation values are based around this size of rafter. And therein lies the rub as my meagre 80mm doesn't go very far and to achieve the required U value gets tricky.

The normal approach is to stick thick insulation material in between the rafters. In comparison to the insulation, the rafters will contribute a 'negative' effect since they provide a thermal bridge (which will conduct heat more readily to the outside ..not a good idea) but, with 7" to play with for very thick insulation, this bridging is probably deemed acceptable.

I have 80mm.

From this I have to reserve 25mm (or 50mm..I have seen both values quoted in my research) for an air gap between the sarking felt and the insulation. That leaves me about 65mm worth of Kingspan or equivalent and that isn't enough.

Up until recently, the simple albeit expensive answer was to fix a multi-layer foil insulation material (such as Trioso) spanning across the underside of the rafters. However, there seems to be a bit of a debate as to whether these provide sufficient insulation. The powers-that-be say if it doesn't pass the standard hot box test then it's no good to which the multi-foil manufacturers call foul...that's not a fair test as it only measures heat loss due to convection and conduction and not reflection which is where our product scores.

All very interesting but doesn't help me. One solution would be a combination of Trioso plus kingspan in between the rafters but that seems to increase the cost quite a lot and in discussion with my local BCO I'm not sure that that is still acceptable.

So insulation between the rafters plus a plasterboard with added insulation across the rafters seems to be the best way to go. I want to keep this plasterboard insulation as thin as possible to prevent it eating into the available room space and the guiding parameter that determines this is (at last he's got round to explaining why he wants to know the thermal resistance of oak!) the thermal bridging due to the rafters.

So oak is a better insulator than softwood and so I should be able to reason with my BCO and say 'therefore I can get away with (probably not the optimum words to use :wink: ) a thinner insulation layer in my plasterboard.
 
So oak is a better insulator than softwood and so I should be able to reason with my BCO and say 'therefore I can get away with (probably not the optimum words to use ) a thinner insulation layer in my plasterboard.

NO roger, oak has a lower R value (thermal resistance) so will transmit more heat, not helped by the fact yours are 80mm rather than the usuall 47mm.

If you think about it softwood has a lot more voids so the density is less so its the better insulator than hardwood

You could always strip the roof and add insulation to the top of the rafters :wink: but then planning probably won't allow the increase in ridge height :cry:

Jason
 
jasonB":2dvouoxg said:
NO roger, oak has a lower R value (thermal resistance) so will transmit more heat, not helped by the fact yours are 80mm rather than the usuall 47mm.
Jason

Aaaarghh!

To make things worse in my Building Reg tome it quotes timber battens with a TR of 0.7...which would make oak proportionally even less. Perhaps I should rout all my rafters to a point to get the surface area down :wink:
Actually not such a silly thought, adding beading, I mean. Would that work?
 
Roger,
The lower the wood density, the (generally) the better the insulating properties. Balsa wood is used as an insulator in LNG carriers around the tanks in the ships) LNG is transported at -160 degrees centigrade.
 
Roger

Came across this pdf. Thermal resistivity of softwoods and hardwoods are on pp 3-19/3-20.

Brian
 
CHJ":1sil0qww said:
(R-value )

1.4 per inch for most softwoods

0.7 per inch for most hardwoods.

A 3-1/2" insulated stud wall has a value of about 14


EDIT: I knew I had a reference somewhere

Careful chaps - those figures are in American units.

I earn my living from carrying out computer thermal simulations of fenestration products to determine their thermal properties. (I know - it is a REALLY boring way to make a living!!!).:roll: :roll:

The European Standard for thermal conductivities of common building materials is EN 12524, which gives the following thermal conductivities: -

Hardwood (density 700 kg/m3) = 0.18 W/mK
Softwood (density 500 kg/m3) = 0.13 W/mK

As others have pointed out, the resistivity (R value) is the inverse of the conductivity (you all have calculators so I won't bother converting for you), and the units are mK/W.

If you need conductivities of specific species, then Google can often be your friend, but be careful of the units, as most hits will have American units.

If the values returned are not fairly close to those shown above, then you have the wrong units, and you will get some very strange answers if you try using them in any sort of thermal calcs.

Regards

Gary
 
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