Calculation of the Infiltration Credit
The infiltration credit that can be used to reduce the required installed fan flow requires estimating the infiltration using the blower door test result. This calculation can be reduced to a few inputs using certain assumptions. This section provides this reduced equation for the infiltration credit. For a more detailed stepbystep discussion, see the end of this page.
The infiltration rate at operating conditions, measured in CFM, can be estimated as:
$$Q_{inf} = 0.052\ast wsf\ast S\ast Q_{50}$$
Where:
\(\mathbf{wsf}\) = a weather factor specific to a geographic location
\(\mathbf{S}\) = a factor accounting for the height of the house, determined from Table A1
\(\mathbf{Q_{50}}\) = the blower door test result in CFM50 (cubic feet per minute at 50 Pa)
\(\mathbf{Q_{inf}}\) = infiltration in CFM
Table A1. SFactors for Various House Heights
Stories 
1 
1.5 
2 
2.5 
3 
SFactor 
1.00 
1.18 
1.32 
1.44 
1.55 
In ASHRAE Standard 62.22013, the infiltration rate \(\mathbf{Q_{inf}}\) is the same as the infiltration credit, and can be fully subtracted from the fan flow rate.
Use of ASHRAE Standard 62.22013, Appendix A
ASHRAE Standard 62.22013 includes an appendix that details an alternative compliance method intended for existing homes that did not meet the ASHRAE 62.2 local exhaust requirements when built. The strategy is to evaluate how much local exhaust deficit there is in each room that should have local exhaust, based on intermittent fan requirements, and to increase the continuous primary fan flow rate to account for this deficit. This section provides guidance on how to determine the increase to the primary fan flow rate to comply with ASHRAE 62.22013.
Per ASHRAE 62.22013
 Each bathroom should have a 50 CFM fan, if used on demand.

Note: Half baths do not require ventilation per ASHRAE Standard 62.22013 – only bathrooms with a shower and/or tub require local exhaust.

 Each kitchen should have a 100 CFM fan, if used on demand.
For each of these rooms that does not meet the stated local exhaust requirements
1. Calculate the deficit. If there is a fan that exhausts to the outside but does not have the required flow, the deficit is only the difference between the required flow and the measured flow.
2. Reduce the deficit by 20 CFM for each of these rooms that have an operable window (if allowed by the authroity having jurisdiction). Sum up all of the individual deficits.
3. Divide by 4.
4. Add the result to the required primary continuous fan flow rate.
Example #1
 Kitchen has no exhaust to outside but has an operable window.
 Bathroom #1 has no exhaust but has an operable window.
 Bathroom #2 has a fan that exhausts to outside but moves only 32 CFM.
Deficit for kitchen is 10020 = 80 CFM (20 CFM credit for operable window)
Deficit for bathroom #1 is 5020 = 30 CFM (20 CFM credit for operable window)
Deficit for bathroom #2 is 5032 = 18 CFM
Sum of deficits is 80+30+18 CFM = 128 CFM
Increase required primary fan flow rate by 128/4 = 32 CFM
Example #2:
 Kitchen has a fan to outside that moves only 60 CFM and an operable window.
 Bathroom #1 has a fan that moves only 20 CFM.
 Bathroom #2 has a fan that moves only 32 CFM.
Deficit for kitchen is 1006020 = 20 CFM (20 CFM credit for operable window)
Deficit for bathroom #1 is 5020 = 30 CFM
Deficit for bathroom #2 is 5032 = 18 CFM
Sum of deficits is 20+30+18 CFM = 68 CFM
Increase required primary fan flow rate by 68/4 = 17 CFM
Detailed StepbyStep Process for Determining Infiltration Credit
This process determines the infiltration credit using only a blower door result, three house characteristics (floor area, volume, number of abovegrade stories), and a factor used to account for local weather.
The calculations that are required are for the equivalent leakage area (ELA), normalized leakage (NL), and infiltration (I) at normal operating conditions.
1) Calculation of ELA
$$ELA = \frac{Q_{50}}{50^{n}}\Delta P^{n}\sqrt{\frac{\rho }{2\Delta P}}$$
Where:
\(\mathbf{Q_{50}}\) = blower door leakage at 50 Pa [\(ft^{3}/min\) @ 50 Pa (or CFM50)]
\(\mathbf{n}\) = house leakage curve exponent
\(\mathbf{\Delta P}\) = reference pressure difference between inside and outside (Pa)
\(\mathbf{\rho}\) = density
By assuming that n = 0.65 (experimental average value for residential houses), ΔP = 4 Pa (typical reference value for ELA), and the density is a constant of \(1.2 kg/m^3\), and by converting all metric units to consistent inchpound (IP) units, the ELA can be rewritten as:
$$ELA = 0.000381\ast Q_{50}$$_{ }
(with \(\mathbf{Q_{50}}\) measured as CFM50, ELA has units of \(ft^2\))
2) Calculation of NL
$$NL=\frac{1000\ast ELA}{A_{floor}}\ast \left ( \frac{H}{H_{r}} \right )^{0.4}$$
Where:
\(\mathbf{A_{floor}}\) = floor area of the house (\(ft^2\))
\(\mathbf{H}\) = height of the house above grade (ft)
\(\mathbf{H_{r}}\) = reference height of one story = 8 ft
The normalized leakage was developed assuming that the volume is 2.5 meters (8.2 ft) multiplied by the floor area. Using this assumption, substituting for ELA, and by assuming that the height of one story above grade is 8 ft, the NL can be rewritten as:
$$NL = \frac{3.1242\ast Q_{50}}{V}\ast \left ( stories \right )^{0.4}$$
Where:
\(\mathbf{V}\) = volume of the house (\(ft^3\))
(TECHNICAL NOTE: The height of 2.5 m (8.2 ft) was used for determining the constant in order to be consistent with ASHRAE Standard 62.22013 in sections that the user does not need to input information; however, a height of 8 ft was used for the story factor. An analysis of the impact of the use of 8 ft for the story factor instead of 2.5 m (8.2 ft) shows less than a 1% error, which was considered acceptable in the name of simplicity for the user.)
3) Calculation of infiltration at normal operating conditions
$$I = NL\ast wsf$$
Where:
\(\mathbf{wsf}\) = a weather factor specific to a geographic location
In this equation I is in air changes per hour (ACH). The weather factor can be found in a table in ASHRAE Standard 62.22013.
Once the infiltration I is determined, it can be converted to CFM using the volume of the house.
$$Q_{inf} = \frac{I\ast V}{60}$$
Where:
\(\mathbf{Q_{inf}}\) = infiltration in CFM
\(\mathbf{60}\) = conversion from hours to minutes
The infiltration rate at operating conditions, measured in CFM, can then be estimated as
$$Q_{inf} = 0.052\ast wsf\ast \left ( stories \right )^{0.4}\ast Q_{50}$$
In ASHRAE Standard 62.22013, there is no default infiltration. The measured infiltration rate based on the blower door test can be subtracted from the fan requirement, including any adjustment for deficits, in full.
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