**Several Passive House projects undergoing certification have requested a way to estimate the additional R-value of an adjoining unconditioned spaces (such as garage).**

A good example is a garage which is part of the building but outside of the Passive House Thermal Envelope. This garage will increase the insulation value of the thermal envelope by buffering it from the exterior conditions. Other example of unconditioned spaces would be store rooms, conservatories, mud room, laundry space, and attic knee wall spaces. Roof spaces and crawl spaces are not appropriately dealt with by the method below.

Typically the first method folks propose is some way to use the temperature zone X feature built into PHPP (areas Tab) to do this. That would only work if you knew the answer. In other words, you could change the temperature zone X factor to get the right answer. But since you actually don’t know the answer before you start this is not immediately useful.

There are three options I’m aware of to estimate the additional R-value of an adjoining unconditioned spaces:

- Estimate the additional R-value is Zero and use that. That’s the most conservative choice. You ignore all the other benefit of that garage space. That’s obviously the least favourable to your model, so people don’t want to do that typically.
- Estimate the additional R-value using ISO 6946:2007 Section 5.4.3 Other Spaces which is intended for an unheated garage or entry space and recommends modifying the thermal resistance of the surface between the conditioned and unconditioned space with an additional homogeneous layer, R
_{u}. The largest assumption is that the unheated space is not insulated. If the insulation is to be taken into account a heat balance per ISO13789 is recommended in the standard as a more accurate method. - Use a heat balance per ISO 13789 or a full transient heat transfer model to estimate the heat transfer across the wall and enter it into PHPP by adjusting the Temperature Zone X. This method is fairly complicate and less conservative and I do not recommend this complexity for projects unless they are attempting research.

Obviously I recommend option 2 above using ISO 6946:2007 section 5.4 to estimate the additional R-value. As you can see the ISO 6946 method is quite simple and given by the equation below:

R_{u} is the additional thermal resistance to be added to the surface as a line item in the U-values sheet of PHPP.

A_{i} is the area of the internal wall of the thermal envelope adjoining the unconditioned space

A_{e} is the area of the a unconditioned space walls, and roof (and suspended floor but neglect slab area) surfaces facing the exterior conditions

U_{e} is the U-value of these surfaces (W/m^{2}K)

V is the volume if the unconditioned space (m3)

n is the air change rate of the unconditioned space in Air Changes per Hour (ACH). It’s the average air changes per hour of that space over the essentially heating period or cooling period, whichever you’re concerned about. If you measured the unconditioned space air leakage rate with a blower door and the blower door said it was 6 ACH at 50 Pa, it’s the air change rate is not 6 ACH. To convert the measured air leakage at 50 Pa it is common to divide by 20 to obtain a reasonable estimate of the air change rate. It depends on wind speed, exposure, and a bunch of other things, but 20 is a reasonable number. BR442, Appendix A Table A5 Typical air change rates for unheated spaces provides some typical value to be used for n in the design phase. ISO 6946 suggests n=3 ACH for unknown and this is definitely conservative unless you have intentional ventilation opening. I’d recommend using n=1 ACH for normal construction including insulated garage doors and n=3 for a standard garage door.

Once the R_{u} value has been calculated a specific U-value in PHPP can be setup and the additional R_{u} entered in the U-value table as another layer using a fictional thickness and the resultant Thermal Conductivity. For Example if R_{u} was 0.4 m^{2}K/W and a fictional thickness of 100mm or 0.1m was used the Thermal conductivity would be 0.1m/0.4m^{2}K/W or 0.25 W/mK.