UK standards U value 50% higher than that from Therm,

[R K]

Hi

I have modelled a flat roof structure in Therm and got U=0.23 (pfa)

Kingspan modelled the same structure using UK standards and got U=0.32 (pfa),

Kingspan told me: "All the U Values we conduct are  in accordance with BS EN ISO 6946:2017 / I.S. EN ISO 6946:2017 (Building components and building elements. Thermal resistance and thermal transmittance. Calculation methods)."

I attach my model and that of Kingspan.  They have used a very high combined K pf 0.441 W/m2K for the thermal bridge of voids and joist.  I modelled this layer in Therm as two fame cavities and timber separately.  Apparently this is mandated by the UK standards.

Why would Therm give a much lower U value?

Thanks for your thoughts on this very large difference in U values for the same structure.

[archi...@gmail.com]

Hi

From what I can see the different results are due to:

- Kingspan calculation rightly takes into account thermal bridging due for infill panels with different declared thermal conductivity (Flex T 0.027 W/mK Optim-R 0.007 W/mK) on the entire roof surface. This can't be modelled in Therm.

- in your Therm model air cavities between rafters are considered as frame cavities with thermal conductivity of approximately 0.54 W/mK. But in my opinion this is not the case. In fact, according to EN ISO 6946 an unventilated air layer thicker than 15 mm and with upward thermal flux has a thermal resistance of 0.16 m2K/W (table 8 on page 13). In your case layer is 120 mm thick so its equivalent thermal conductivity is L = s/R = 0.12 m / 0.16 m2K/W = 0.75 W/mK.

Ciao!

Fabrizio

[R K]

Thanks Fabrizio for solving this.  I changed L from 0.538 to0.75W/mK but this had almost no impact on the U value.  However, changing the OptimR layer from 0.007 to 0,013 changes the U value to 0.33W/m2K which is inline with the Kingspan result.  So Therm can handle it I think -just take the weighted average conductivity right?  Anyway Chatgpt thinks so:

As to the infill, Kingspan state that they assume 30% infill. If the Optim-R and the infill are effectively in parallel (side-by-side under the same temperature difference) and have the same thickness, then it is valid to combine them into a single “equivalent” layer using an area-weighted conductivity, i.e.

• Fraction of infill: 30%

• Fraction of Optim-R: 70%

With declared conductivities:

• λ_infill = 0.027 W/mK

• λ_Optim-R = 0.007 W/mK

The equivalent conductivity of the mixed layer is:

λ_eq = 0.30 × 0.027 + 0.70 × 0.007
λ_eq ≈ 0.013 W/mK

Because the products are side-by-side and (in this case) of the same thickness, this area-weighted λ is mathematically equivalent to doing an area-weighted average of the U-values for each panel zone. Using this λ_eq as the material conductivity for that layer in THERM is therefore consistent with the underlying physics and should give a valid composite U-value for the roof, subject to the usual approximation that we’re smoothing out the detailed panel pattern.