Underground Facility Stair Pressurization

[gthomer]

I'm designing a stair pressurization system for an existing facility that is largely underground. My design occupant load is not decided on yet, so I am performing calculations in advance of this decision for two situations/assumptions: 1) all stair doors closed, and 2) some stair doors open. I'm using Chapter 52 of ASHRAE's HVAC Applications Handbook for the closed door calculations, and Chapter 10 of ASHRAE's Principles of Smoke Management for the open door calcs.

Using these texts, I get a believable cfm rating for the fan for the closed door case. But for the open door case, the formulas fall apart and give me nonsense. The problem is that my outside wall area is very small compared to my inside wall area, because only the top floor is above ground. The formulas apparently do not fit my situation.

Does anyone know how to address this stairwell pressurization situation? Are there any good reference books that might be applicable?

 

[friartuck]

You say the equation falls apart

Can you elaborate...after all, you might have the right answer, but not one that you are willing to believe. Possibly the value you have is very high...which is what it would normally be anyway.

You might also try using Colt who provide a design facility.




Friar Tuck of Sherwood

 

[gthomer]

Sorry for the ambiguity.

Specifically, the pressure difference between the stair and the outside, at the bottom of the stair, (DPsob), is 8.7 in. of water. I was shooting for something in the neighborhood of 0.05 in. of water. This pressure difference is eventually used to calculate the flow rate of the fan. But that number is so ridiculous completing the calc was pointless.

The formula in question is 10.9 in ASHRAE's Principles of Smoke Management. It will give you nonsence answers anytime you have an outside wall area that is considerably smaller than your inside wall area.

 

[imok2]

Not sure if this will help you, but I found it in ASHRAE
Q= 2610A. sq root delta”p”

The flow area is frequently the same as the cross-sectional area of the flow path. A closed door with a crack area of 0.11 ft2 and a pressure difference of 0.01 in. of water has an air leakage rate of approximately 29 cfm. If the pressure difference across the door is increased to 0.30 in. of water, the flow is 157 cfm.
Frequently, in field tests of smoke control systems, pressure differences across partitions or closed doors have fluctuated by as much as 0.02 in. of water. These fluctuations have generally been attributed to wind, although they could have been due to the HVAC
system or some other source. To control smoke movement, the pressure difference produced by a smoke control system must be sufficiently large to overcome pressure fluctuations, stack effect, smoke buoyancy, and wind pressure. However, the pressure difference should not be so large that the door is difficult to open.

 

[ChasBean1]

Building on Imok's post, assume five stair levels (five doors leaking) have an estimated open area of 0.55 square feet. Although not with me, Applications Ch. 52 also has an estimated leakage rate per 100 ft2 of wall (although the data there may pertain to hoistways). Add these values.

Say we get 2 ft2 as a result and codes require an average 0.25 in. w.c. pressure.

Q = 2610 A dP^.5

Note that Q is flow (cfm)
A is net open area (ft2)
dP is in inches water column

Flow would be (2610)(2)(.5) = 6,525 cfm

Now open one door. Net open area would be about 22 ft2.

Flow would now be (2610)(22)(.5) = 29,000 cfm

Moral of the story? Design for closed doors... obviously you'd need a ridiculous variance to compensate for open doors to maintain the design pressure. -CB