2- point blower door test test @ cfm 25 and cfm 50 typical blower door test assumes.65 coefficient...
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![Page 1: 2- Point Blower Door Test Test @ CFM 25 and CFM 50 Typical blower door test assumes.65 coefficient “N” “N” must be between.5 and.75 for accurate test Formula:](https://reader036.vdocuments.us/reader036/viewer/2022082422/56649e5c5503460f94b5411a/html5/thumbnails/1.jpg)
2- Point Blower Door TestTest @ CFM 25 and CFM 50
Typical blower door test assumes .65 coefficient “N”
“N” must be between .5 and .75 for accurate test
Formula:
“N” = (LN Q50/Q25) / (LN 50 Pa /25Pa)
“Q” is flow “LN” is Log Normal
![Page 2: 2- Point Blower Door Test Test @ CFM 25 and CFM 50 Typical blower door test assumes.65 coefficient “N” “N” must be between.5 and.75 for accurate test Formula:](https://reader036.vdocuments.us/reader036/viewer/2022082422/56649e5c5503460f94b5411a/html5/thumbnails/2.jpg)
2- Point Blower Door TestTest House at both CFM 25 and CFM 50 Formula:
“N” = (LN Q50/Q25) / (LN 50 Pa /25Pa)
“N” must be between .5 and .75 for accurate test
Provides additional valuable information Typical blower door test assumes .65 coefficient “N”
A value below .65 suggests large holes A value above .65 suggests smaller holes
![Page 3: 2- Point Blower Door Test Test @ CFM 25 and CFM 50 Typical blower door test assumes.65 coefficient “N” “N” must be between.5 and.75 for accurate test Formula:](https://reader036.vdocuments.us/reader036/viewer/2022082422/56649e5c5503460f94b5411a/html5/thumbnails/3.jpg)
2- Point Blower Door Test Example “N” = (LN Q50/Q25) / (LN 50 Pa /25Pa)
Example test resultsCFM 50 = 2,139CFM 25 = 1,367
“N” = (LN of CFM 50/CFM 25) divided by (LN of 50 Pa / 25Pa)
2,139 divided by 1,367) = 1.56 50 divided by 25 = 2
LN of 1.56 = .44 LN of 2 = .693
.44 divided by .693 = .64
“N” for this example is .64