BCO refused roof - U value too low.

[sm_jamieson]

My roof insulation was based on the knauf insulation spec for between-joist-only insulation, the example of which was 225mm wool between 225mm joists, with 6.2% bridging due to the joists, which gave a sufficient U-value. After refusal, I checked mine and it was 10% bridging due to joists being thicker although further apart. So, I need to fix it. Also, I could replace some of the wool with celotex so that the bridging is OK. I need to keep the roof a certain thickness, thus not wanting to add celotex under or over the joists. How do I calculate U values when there are parallel conductivity paths (joist / wool) and serial (if I add a thin layer of celotex) in the same structure ? I need calcs to be able to prove that my fixed insulation spec is OK. Thanks, Simon.

[Christian McArdle]

Work with conductivities, which is the reciprocal of the uValues. Then you can multiply them by the area, add them up and take the reciprocal again to get the final value.

Finally, are you sure you can't get anything under the joists? Cold bridging has deleterious effects in addition to the overall energy loss, which is all the BCO will be interested in.

In particular, with cold bridging, you will get cold lines on the plasterboard which will discolour and attract mould. You would be far better, if at all possible, to install even a little insulation underneath, even if it is only 10mm or 12mm to eliminate this possibility and provide a uniform warm surface to the plaster.

Replacing rockwool with celotex is unlikely to be a bad idea, even with the underboarding. The building regs minimum is a minimum. It would be to your own benefit to improve on it.

Christian.

[The Natural Philosopher]

You can sum U values over parallel paths easily enough.

U value = sum (U value times area for all the bits)/total area.

Thats is basically saying that for a given section there will be a certain heat loss in watts per degree C..and summing all those losses gives you an overall watts per degree C of the whole roof..dividing that by the total area nets you an 'average' U value..don't forget that if you e.g. put a skin of plasterboard on the bottom, and a skin of e.g. chipoard on the top, you need to do the other calculation for U values in series.. which is that 1/Uvalue final = 1/U value plasterboard +

1/Uvalue chipboard + 1/Uvalue calculated as above for beams and rockwool.

I can't be arsed to do the calculations for you..but they should be available online.

[sm_jamieson]

What about conductivities in series ? I guess these operate in the reciprocal sense to the reciprocal rule, if that makes sense.

I also need calcs for the roof, so I am seeing a structural engineer anyway next week. I will see if he can get the roof thinner and allow for the celotex. I would still need to work out the thickness of celotex to get the U-value acceptable. By the way, the reason for the roof thickness fuss, is that I need the pitch over

12.5 degrees for the tiles, I need headroom at the eves and I have a mono-pitch roof on the party wall to line up with. And ... I'm keeping the roof open nearly to the apex.

Simon.

[Christian McArdle]

Actually, that's what I accidentally described, due to a brain fart. To calculate the overall uValue, just multiply the %age of the material by the uValue of that part and sum them all together. Only take reciprocals for the series calculation.

Christian.

[Tony Bryer]

Not that easy unfortunately: you need to do R/upper and R/lower calculations. Example on our website at

[sm_jamieson]

room ?? But, it leads me to part L appendix B which has a nice example of studs / insulation / sheathing that I can apply. The sums aren't to bad really, so no need to buy superheat ! Thanks, Simon.

[sm_jamieson]

Hah, BCO is wrong. He probably looked at the cold bridging and got scared I calculated for my case and the one from the knauf website. My figure for the case on the website came up with 0.18 which matched their figure, so sums OK. My figure for my case came up with 0.199. Less than 0.2. And that's without including the plasterboard or boundary layer. So, I can prove it if I want. If he disagrees, he will have to show me his calcs !

But, I take on board the cold lines factor. But, 225mm of wood is a relatively good insulator I would have thought, so not sure this would be a problem in practice.

But I need to keep on good terms, I may have to confront him with legalities if he wants me to pay for the electrical inspection. They want a note in the spec that "a test certificate to BSxxxx will be given to building control". I'll have to be careful how I word that bit.

Simon.

[Hugo Nebula]

On 31 Jan 2006 03:38:51 -0800, a particular chimpanzee named sm [e-mail] randomly hit the keyboard and produced:

Calcs! You think the BCO hasn't got more to do than calculate the heat loss through your roof. He'll have just looked up your spec on the Knauf website, noted that their spec is based on 38mm wide trusses at 600mm centres, and that yours are obviously thicker and closer, and asked you to prove your spec.

BTW, have you allowed for ventilation? Either you need to leave a

50mm gap over the insulation, or counterbatten over. Maybe he's assumed that the thickness of insulation will need to be reduced by 50mm.