I am not an energy Rater .. I am an architect
I am curious if any "Home Energy Pros" make an effort to locate the Neutral Pressure Plane (NPP) of a house that is being blower door tested?
Robert, LOL, simplicity is a must. One thing I have learned in energy auditing is KISS.
Paul Morin gave me this version and John Brooks dug up John Straube's version at http://www.buildingscience.com/documents/digests/bsd-014-air-flow-c...
The "H" is whatever height you select and since we don't know from the start where the NPP is, just use the floor to ceiling. What we want is the pressure per foot so we could actually use 1. I just like to walk readers through just how simple stack effect really is and the many ways we can benefit from understanding it.
I will give your version a try as well.
And for another version.....
Here are a couple of Illustrations (in-progress) that I am adapting from John Klote's paper.
One example is for a single opening
the other example is for two openings
For estimating the height of the NPP ....
Temperature and Season don't make a big difference.
It is the Size and vertical distance between Significant openings that matter most.
The delta-T has a significant impact on the actual pressure difference inside to out (or from floor to ceiling), but not on the location of the NPP.
Nice illustrations. But why would you open the doors and windows on a 35° day? ;-)
Yes the rectangles just under the eaves are the Airlet 100 passive air inlets (there are seven all together, plus a 6" round inlet in the air-sealed laundry room for dryer makeup air and a 4" inlet in the first floor for direct-coupled woodstove combustion air).
There are also two bath exhaust fans (one on each floor) with the ducts dropping 3' before exiting the sidewalls to prevent cold air thermo-siphoning, and a kitchen range hood with a similar down-and-out anti-siphon duct. Both the dryer make up inlet and the woodstove combustion air inlet have similar cold air traps.
There are no upstairs ceiling penetrations other than masonry chimney and plumbing vent - both well air-sealed. The attic hatch is above the thermal/air envelope in the gable wall.
The house tested to 2.13 ACH50 (460 CFM50) with the air inlets taped. That shows an effective leakage area (@ 4 Pa) of 11 square inches, which increases to only 13 square inches when tested with the inlets open - so I don't think they contribute to much passive leakage when the exhaust fans are not running.
I had set the bath fans to run approximately one hour out of three to provide a little more than the ASHRAE 62.2 minimum, and flow tested them for actual CFM (94 CFM for the 110 fan and 74 CFM for the 80).
Since the downstairs has more mechanical openings (including the chimney cleanout door), I would guess that the NPP is somewhat below the first floor ceiling (particularly given 8' ceilings on the 1st floor and 7' ceilings on the 2nd) when no exhaust fans are running, and at the upstairs ceiling when one or more fans are running (baseline ΔP was 2.2 Pa, 4 Pa with one fan running and 8.6 Pa with both bath fans running).
With the static NPP likely below the downstairs air inlets, there's probably little or no flow between lower and upper passive inlets without either fan pressure or some wind pressure. I also installed a sweep on the door at the top of the stairs to pneumatically isolate the two floors.
Robert, thanks for all the detail
How did you calculate "effective leakage area" ?
(I found a ballpark formula of CFM-50 divided by18)
Why do you think "ELA" only increased 2 sq in (13-11)....
after you opened up 7 openings of approx 5 sq in (net free area) each?
CFM50/18 is a ballpark approximation of natural leakage (or more accurately normalized leakage), but the conversion factor (typically 20) depends on the climate zone, height of building, degree of shielding, and "leakiness factor" - as determined by LBNL studies http://www.homeenergy.org/show/article/id/1015.
The formula I use to determine ELA (in SF) is
ELA50 = CFM50/(2400 x √(50/248.88))
ELA4 = (CFM50/NL conversion factor)/(2400 x √(4/248.88))
[ELA @ 4 Pa is the standard]
ELA(sf) = CFM/(2400√(Pa/248.88))
Pa/248.88 = inches water column
The CFM50 went up from 460 to 570 with the Airlets open (24% increase), but when that is converted to ELA at a normalized 4 Pa, it amounts to an 18% increase.
I suspect that's because the combination of the foam dust filter and the mylar reed valve in the Airlet is sufficiently restrictive that it presents less net free area than a simple measurement would show.
With both bath fans running (168 CFM measured flow), I was getting between 6 and 8 CFM per Airlet for a total of about 50 CFM (which was very close to the 62.2 target of 47), which amounted to 30% of the exhaust flow. Including the laundry room inlet and chimney leakage, I measured about 80 CFM, or 47% of exhaust flow - with the remainder from envelope leakage.
Robert, you mentioned "cold air thermo-siphoning"
can you explain how this "works" ?
what is cold air thermo-siphoning?
and how does dropping the duct 3 feet prevent it?
why 3 feet ?
Air increases in density about 2% for every 10°F drop in temperature, so any duct with an outside termination at the same height or higher than the inside termination will allow the denser (winter) cold air to displace the warmer air and thermosiphon into the conditioned space (slowed, but not likely stopped by mechanical dampers, which sometimes get stuck open).
This is a potential problem when exhaust fans terminate through the roof, but even when they terminate at a wall on the same elevation as the fan. It's also a potential problem when inlet ducts come either straight in or in and down.
Indoor air at 70°/40%RH with outside air at 0°F/50%RH will produce a buoyancy (or downward cold air displacement pressure) of 0.0117 pcf. At the same conditions, a 3' chimney would create an upward stack effect pressure of 0.034 psf.
If you calculate actual duct air volume buoyancy vs duct stack effect pressure, they balance out at 3' of duct or "chimney" height.
Whoops! I've got to come up with another rationale for the 3' drop.
Air buoyancy and stack effect are the same thing, so they'll be the same regardless of "chimney" height. I was using two different formulas, with buoyancy based on wet air density and stack effect on dry air density, but the difference is not significant (moist air is actually less dense than dry air).
Any 3' "chimney at a 70° delta-T will create a 1.63 Pa stack pressure, which is the equivalent of the stagnation pressure of a 3.65 mph wind. But I don't know that tells us anything useful.
It's just an old rule of thumb that I learned many years ago and it makes sense to me, so I've used it for 20 years.
So, About a dozen "intentional openings" not including doors and windows
Let's say it is pretty darn cold outside (0 F)
What happens when one bathroom exhaust fan is activated?
and 94 CFM is pushed out of the house
doesn't 94 CFM also enter the house?
Is One exhaust fan powerful enough to raise the NPP all the way to the 2nd floor ceiling?
Can one bathroom exhaust fan make the entire house "negative" (WRTO)?