Maintaining Pressure in a Room
Maintaining Pressure in a Room
(OP)
I am trying to calculate how to maintain a .05" positive pressure in a room. I know the supply volumetric flow but I am trying to size the return based on the goal of .05" positive pressure. There are 2 doors in the room both with a space under the bottom of the door. (each crack 0.5"x3') I know that the volumetric flow in and out will not be balanced (more in than out) to maintain a positive pressure in the room. So I guess my question: Given the nature of this problem, what rule of thumb, model, or equation could I use to figure this out? What is the room pressure with just the cracks? How do I find the room pressure with the cracks and return of size X? I appreciate your responses.
room dimensions 22'x22'x10'
supply air 2360cfm
crack surface area (for both) 3ft^2
return size X?
room dimensions 22'x22'x10'
supply air 2360cfm
crack surface area (for both) 3ft^2
return size X?





RE: Maintaining Pressure in a Room
What's the room FOR may I ask?
Jabba
RE: Maintaining Pressure in a Room
Adjust the return restriction until you are operating on the .05" point on that curve.
How do you know what that curve will look like before the room exists? Basically you don't.
RE: Maintaining Pressure in a Room
RE: Maintaining Pressure in a Room
RE: Maintaining Pressure in a Room
Thing is, you can't predict the leakage of the room just as Jabba007 stated - so why not let it leak all it wants and then add some "controlled leakage" to achieve the desired pressure.
Not critical, again as RossABQ stated with 1/2 AC/Hr. and call it a day.
Andy W.
RE: Maintaining Pressure in a Room
Another method for determining the required air flow into the system would be to look at the fresh air requirements for the space to be conditioned. this will generally be more than the leakage rate for a well sealed room.
Mark Hutton
RE: Maintaining Pressure in a Room
Curious minds want to know.
Jabba
RE: Maintaining Pressure in a Room
I have been looking at some formulas for volumetric flow through a crack: Q = A*C*(delta P)^n Then further digging through this forum led to this: Q = 2610*A*(delta P)^0.5 This supposedly comes from 51.5 of the ASHRAE 1999 applications book. I don't have this book and can't find the formula anywhere else. If someone does have it, could they maybe provide the explanation that goes along with the book.
http://www.eng-tips.com/viewthread.cfm?qid=43928
According to my text:
C = the flow coefficient, depending on crack type and nature of the flow
n = exponent that depends on the nature of the flow 0.4 < n < 1.0
So my question now is where did 2610 come from? I am fine with the 0.5 as the exponent falls within the stated range, but would still like to know how it was acheived as well.
Ultimately, my goal is to be able to make a model on excel. So when I encounter this again I can enter quantities like room size, number of doors, crack size, supply and return air, etc. and it spits out at least a rough estimate of how I need to design.
I appreciate everyone's input so far. If I figure it out (with your help), I'll gladly share the findings.
-Devin
RE: Maintaining Pressure in a Room
I tried to do a small calculation, in the thread you alluded to, for arriving at the constant 2610. You can use this approximate formula with good success as long as you maitain constant supply flowrate. The observed leakage in actual condition, using a hot wire anemometer, is close to 90% with the type of construction we have.
If you have a leak area only across the doors then you did a mathematical error. You should use a leak area of 0.5 sq.ft instead of 3 sq.ft. 500 cfm seems to be a bit high to flow through a 3' door. You may have to make a provision for bleeding the air from the fan discharge side (if you want to bleed it in a classical way from the return then you may have to use one more fan)
RE: Maintaining Pressure in a Room
In every hospital isolation area I've ever worked (and there have been many, including labs for neoplastics), it's been a simple matter of using a differential pressure sensor designed for room pressurization purposes and modulating the return damper or relief / exhaust damper to maintain a setpoint (in your case, +0.05" WC). With careful tuning, you can maintain temperature control with VAV supply and still keep the room pressure where you need it.
If lab hoods are in there, you need to integrate the controls with the HVAC to accomplish this. The third part of the act is keeping flow velocity across the hood sash within bounds.
Good on y'all,
Goober Dave
RE: Maintaining Pressure in a Room
RE: Maintaining Pressure in a Room
No matter how well you can estimate the leakiness factor, getting the differential volume will come down to the TAB and final adjustments.
I've been using 100 CFM on a standard door (3x7) and 70 CFM differential for new construction/low leakiness for the last ten years (general lab, biosafety, hospital) then making adjustment at balance time. Balance comes down to plus/minus 5 or 10%, so no matter what you design for now with leakiness and duct leakage will generally be overcome by the relatively samll differential it takes to get from 0.01 to 0.05 IWG differential pressure. Once the door opens, or if it stays open (make sure the door hardware includes self-closing) the gnat's-ass design is out the window anyway.
If you're not using hoods and going straight constant volume, its a simple matter of adjusting dampers against a magnahelic. If you're running hoods or VAV, I'd recommend a pressure independent valve or box, like Phoenix or TSI. If running hoods and concerned with energy and NFPA 45, then you'll need to think about running VAV. Unless you're talking about dry lab, immunosuppressed, or non-hazardous compounding, the diferential volume will be inward. If that is a critical issue, then you should start with considering an anteroom/vestibule and perhaps 2-speed fan for door openings.
ANSI Z9.5 writes it up better than I can. If you go to the Phoenix valve web site, they give a good listing of criteria and standard solutions (for their product).