Fun with Neutral Pressure Planes (NPP)

Replies to This Discussion

Permalink Reply by John Brooks on April 30, 2012 at 1:36pm

Bud, I think the key to how a tube or a chimney chase relates to the NPP is to look at the point (altitude) where the "tube" "connects" to the outdoors

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Permalink Reply by John Brooks on April 30, 2012 at 1:54pm

Here is an Illustration of a tube as the only significant opening

I say the NPP will be at elevation 12 feet ...

the point where the tube "connects" to the outdoors

this is assuming that the enclosure is heated, the house has already burped and the tube is full of 70 degree air...

 

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Permalink Reply by Robert Riversong on September 5, 2012 at 2:17pm

If that were true, then a chimney extending a couple feet above the roof peak would move the NPP outside the house, and that isn't the case. It merely moves it upward according to the chimney air flow and how much that reduces the interior pressure.

But why is there a fish in the house?

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Permalink Reply by Bud Poll on April 30, 2012 at 2:23pm

Once you follow a chase or tube from one end or the other, adding or subtracting the stacks of air, you end up with the stack pressure across that path.  I would assume we could also use the equation, as we need to assume the inside (path) and outside (path) temperatures.  Once we have the pressure across the path we can divide it in half and view half as pushing air in and the other half as pushing air out, and a NPP in the middle.  I'm shaky on how the pressures set up across a uniform path so I tend to view it as a large chamber with one opening high and the other low.

Bud

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Permalink Reply by John Brooks on May 1, 2012 at 5:27am

Bud, did you notice the "red herring" in my previous illustration?

I continue to believe that the barometric pressure at the point(s) where our enclosures "CONNECT" to the outdoors is the key.

This is where the pressure is applied or resisted.

I am attaching a John Straube quote and illustration.

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Permalink Reply by Bud Poll on May 1, 2012 at 8:46am

Assuming the outside section of tube is filled with outside temperature air and the inside section of tube is filled with inside air, start at the bottom of the outside tube and subtract the air pressure as you move up to the opening.  Since the air in the tube is the same temperature as outside, when you reach the opening the pressure inside the tube will be the same as outside. 

You now have your reference pressure, also the NPP.  If you calculate up or down from the opening on the inside, inside the tube or not, you increase of decrease a different amount than outside, thus the pressures above and below the NPP.

Bud

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Permalink Reply by Robert Riversong on September 5, 2012 at 2:18pm

Yes, but that's true only of a hermetically-sealed cylinder, not a leaky house.

Oh! Red Herring. I get it (I guess). I couldn't identify the fish.

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Permalink Reply by John Brooks on September 9, 2012 at 4:50am

Robert, Humor me here.....

If the tube were the only "significant opening"....

Imagine that the house is extremely tight (< 1 ACH-50)

and the cross sectional area of "the tube" is say 4 times greater than all the other openings combined.....

would you agree that the NPP is near elevation 12 feet?

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Permalink Reply by Robert Riversong on September 9, 2012 at 9:40am

It's an interesting thought experiment (does it have a real world application?).

I don't think you'd be correct even if the house was hermetically sealed, because this is not a horizontal opening but a chimney.

A 4:1 leak area ratio would still allow some stack effect flow, somewhat slowed by the U-turn in the "chimney" and the restriction of the other openings, but that would effectively reduce house interior pressure such that, assuming the other leaks to be balanced high and low, the NPP would rise.

Assuming 10' effective chimney height, that tube would create a pressure differential of about 2.5 Pa.

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Permalink Reply by John Brooks on September 9, 2012 at 9:49am

the real world application ...would be for locating and sizing the intentional openings (intake ports & exhaust ports) of a house to manipulate the location of the NPP

I am extrapolating my thought experiment from the John Klote paper

http://fire.nist.gov/bfrlpubs/fire91/PDF/f91013.pdf

in particular Example 3

Permalink Reply by Robert Riversong on September 9, 2012 at 10:01am

I suspect that, with such a tight house, a 2.5 Pa draw would move the NPP to the ceiling.

I don't follow how this example applies to locating and sizing intake and exhaust ports, if those are the ports of an HRV/ERV which, if barometrically balanced would not alter the natural NPP. Or are you referring to passive "holes" in the envelope?

In any case, such horizontal "ports" wouldn't be chimneys unless there was a significant offset and channel flow within the envelope.

Permalink Reply by John Brooks on September 9, 2012 at 11:26am

My take-away from Klote-Example 3

(and the entire Klote Paper)

is that the height of the NPP is strongly influenced by...

the elevation of the most significant Opening(s)

In a "tight house" I would think that the Significant openings ARE the Intentional Openings.

The intentional openings might be ....

Passive openings like Aldes Air inlets

or the termination of an exhaust fan (off or on)...

or the intake portal for "supply ventilation"

or an open basement window

or an open clerestory window

or they might be the exterior connection points of an HRV/ERV

I realize that wind and mechanical effects can overshadow stack effects

however, it seems to me that it would be better to work "with" stack (or container) effects rather than "against" them.

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