What is the value of using a pressure pan to test outlets?

I had this question recently and my response was:

If the blower door takes the home up to 50 Pa, and the pressure pan reading across the outlet is 49 Pa for example, that says that total of the rest of the leakage path to outdoors is about 7 times larger than the outlet which means the outlet is acting like the primary air containment barrier. Similarly, if the pressure is only 1 Pa, it says that the air barrier actually lies somewhere else and the remaining path to outdoors is 7 times smaller than the tiny hole in the outlet.


It is possible to make some conclusions about what this means but they are obscured by the fact that the major leaks in the house that affect energy are probably 100 to 1000 times larger and therefore 100 to 1000 times more significant than the outlet. It is important to keep hole sizes in perspective to avoid getting distracted by holes that are easy to locate. A large pressure across the outlet says the outlet is not a problem and in all likelihood, the wall is connected to the attic by 20+ feet of cracks between the sheet rock and the top plates. Sealing these top plates will prevent cold air from falling into the walls, cooling them down and affecting comfort even though the reduction in air leakage will be unmeasurable because it exists as a few small holes in the living space such as outlets that are not in themselves the problem nor will fixing them cure much if the wall is connected to the attic as it usually is.


Colin Genge



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David, you could use the Exhaust Fan Flow Meter to measure air flows above 10 cfm through the recessed light, but you would need to pressurize the house to 50 Pa.  For flows below 10 cfm, you would need to make the hole in the Exhaust Fan Flow Meter smaller, say 3 square inches.  The formula to calculate flow in CFM is 1.07 x the area of the hole in square inches x the square root of the pressure in Pa.  The accuracy would be less than the stated 10% accuracy of the Flow Meter, but it would give you a general idea.

If you don't have an Exhaust Fan Flow Meter, you could simply use a shoe box, cut a hole in it and use the formula that Paul mentioned. You can also use the "Hole Flow" feature on the DM-2 gauge to do the math for you and to calculate flow for any size of hole.

Hi Colin, I hope this is not too far off topic.....

I found one of your Linked-In quotes about blower doors to be very fascinating.....

Colin Genge quote"Atmospheric pressure adjusts inside and out almost instantaneously even for a tight house. It would be in the order of a few seconds at most. The enclosure has no ability to contain pressure which is why the test fan flow leaks out as fast as it comes in."

I visualize that many of the building cavities of a house are like a 3-D network of inter-connected containers and conduits.

My thinking is that the building cavities(stud cavities & floor cavities, etc) that are connected to the interior will behave in the same way that you described "The enclosure" above.

Colin, do you think that Once a blower door is cranked up ... that the atmospheric pressure inside any "connected Building Cavities" will adjust almost instantaneously?


I think I can answer this one as well.  The pressure will equalize pretty much instantly in all of the building cavities.  The only exception would be for cavities that are extremely tight -- as in much less than even 1 CFM50 of leakage. 

The reason why pressure equalize quickly is that it takes very little flow to change the pressure in a building cavity by a lot if there is no incoming leakage to create a balance.  For example, an airtight cavity 1 foot wide and 1 foot high and 15 feet long will drop in pressure by 50pa if you can remove just 1 cubic inch of air. So even a cavity that has just 1 CFM50 of leakage path to the inside and 0.1 CFM50 to the outside will take less much less than a second to reach its equilibrium pressure. 


That was so counter-intuitive that I had to create a spreadsheet to verify. Because standard atmospheric pressure is 101,325 Pascals (14.7 psi), a change of 50 Pa requires only a proportional change in volume of about 0.05%.

That would mean a hermetically-sealed 2,500 SF, 20,000 CF house would require evacuation of less than 10 CF of air to produce a pressure differential of 50 Pa.

If the cavities are that tight that their pressures don't move in step with the house pressure ( or close to it) their contribution to the result or ability to affect the result in any way are microscopically small to non-existent so you don't have to worry about them either way.



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