6.4.1 Visualized Volume
This value is not used for the “Passive House Verification” calculation; it refers to the volume of the visualized 3D model. This will be equal to the gross volume in most cases (when the model is drawn to the exterior dimensions).
6.4.2 Gross Volume
This volume represents the volume of the building measured from the exterior boundary of the thermal envelope. Generally this is the same as the visualized volume. However, if objects are included in the 3D geometry that are not part of the thermal boundary, for shading, etc., the visualized volume may not represent the gross volume. In this case, the gross volume should be calculated.
6.4.3 Net Volume
To determine the net volume, calculate the home’s interior volume (drywall to drywall floor to ceiling, wall to wall) minus volume taken up by interior walls and floor systems. A calculation must be submitted for certification.
6.4.4 Interior Conditioned Floor Area (iCFA)
The reference floor area. The space conditioning criteria (heating and cooling annual demands and peak loads) are per square foot of iCFA. See Section 4.4.1.4 for the definition of iCFA.
6.4.5 Specific Heat Capacity/Thermal Mass
With regard to thermal mass, it’s not just the total mass that matters but the distribution, so that it can interact with the infrared radiation bouncing around the room. The way of figuring this is based on the number of heavy surfaces (0 to six) per room, in an average sense. For example, a two-story building with a first floor concrete slab would have 1/2 of a heavy surface per room on average.
Thermal mass is determined by the equation = [60+n(heavy)*24]*0.176 (BTU/ft2 .F).
Guidelines for Accounting Heavy Surfaces
- Drywall = 0 heavy surfaces - (first ½” included in the baseline)
- 5/8" drywall = 0.34 heavy surfaces
- Double or high-density drywall = 0.5 heavy surfaces
- Two-inch-thick concrete or more, phase-change materials = 1 heavy surface
- Flagstone/tile = 0.5 heavy surfaces
- Example: Concrete floor = 1 heavy surface; Other 5 surfaces = 0.34 each = 1.7 heavy surfaces; Total n (heavy surfaces) = 2.7; Specific capacity = (60+2.7*24) *0.176 = 21.96 BTU/ft2F
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