2- point blower door test test @ cfm 25 and cfm 50 typical blower door test assumes.65 coefficient...

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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: “N” = (LN Q50/Q25) / (LN 50 Pa /25Pa) “Q” is flow “LN” is Log Normal

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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:

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:

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:

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