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dmcarroll (Mechanical) (OP)
10 Jun 08 11:47
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?
Jabba007 (Mechanical)
10 Jun 08 12:28
In my opinion there are too many variables.  Flow in WILL EQUAL flow out.  Somehow, someway it will.

What's the room FOR may I ask?

Jabba
MintJulep (Mechanical)
10 Jun 08 12:38
The summation of all the leakage paths out of the room will have a pressure vs. flow curve.

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.
dmcarroll (Mechanical) (OP)
10 Jun 08 16:36
Thanks for the responses so far.  It is actually for a lab.  Surely you can calculate something before the room is built.  The system is basically a tank with a continuous bleed valve. Continued help would be appreciated as I can't seem to close in on this problem mentally, but I know somebody has done this before.
Helpful Member!  RossABQ (Mechanical)
10 Jun 08 16:46
Rule of thumb (take it for what it's worth):  starting point for "relatively tight construction" might be 1/2 cfm per sq ft of room.  If you have junky double doors, you may need to double that.  Best bet if you NEED positive pressure is to have dynamic control, a VFD that will ramp up the fan, or control dampers that modulate to maintain it.  The ROT is mostly to size the fan for the added capacity.
Wareagl487 (Mechanical)
10 Jun 08 18:22
If is critical you meet 0.05" wg positive in the room, exactly what RossABQ recommended.  I would add one component to that - a properly sized pressure relief source,  Old school would be either a wall louver or roof mounted relief hood with a barometric damper.  Now IMHO the way to go would be with a space pressure sensor controlling a relief fan with a freq. drive.  The architect needs to help you too if the positive pressure is critical, you will need a vestibule separating the space from the room to not shock the room with the opening and closing of doors.

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.

  
HEC (Mechanical)
10 Jun 08 19:06
You are after around 12Pa, this is not unusual in Pharmaceutical manufacturing facilities, they usually aim for around 12-15Pa across each set of doors. i.e. from outside to first airlock +15Pa, from airlock to production corridoors +15Pa (+30Pa to ambient) and production suites are +15Pa above the corridoor (+45Pa to ambient). One method to achieve this is to, first seal the room as best as is possible, Raven have a good range of door seals. Then take a SWAG (Scientific Wild Arse Guess) at the remaining "openings" from the room. Equate this to an orifice with the appropriate pressure drop, in the pharma example, this would be some to the corridoor 15Pa and some to ambient 30Pa. From this a rough approximation of the make up air is obtained. For the Ventilation system consider a closed circuit with a single fan to recirculate air. The circuit pressure drop will equal the fan pressure rise. Now place a fan blowing in to the circuit rated at the leakage rate (plus a margin) with a pressure rise of 30Pa from ambient. This will effectivly now pressurise the room system.

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


 

Jabba007 (Mechanical)
11 Jun 08 7:25
Everyone makes good points.  It's hard to decide how I would approach it without knowing the criticality of it.  But, essentially .05" is a pretty common number for positive pressurization.  As HEC says, it can probably be accomplished thru balancing the supply and return.  You mention that the system is a tank with a continuous bleed valve... Hmmmm, I am curious to learn more.  I have never heard of this before.  

Curious minds want to know.

Jabba
dmcarroll (Mechanical) (OP)
11 Jun 08 9:53
The idea about it being a tank with a continuous bleed valve was a way that I was trying to think about it in a more simplified mannner with a simple in and out (ruling out all other losses besides the major cracks).  The room will have 6 ac's per hour. Of the 2360cfm supplied only 1860cfm will be returned to acheive the 6 changes.  So there is a continuous flow of 500cfm that will not be returned.

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
quark (Mechanical)
13 Jun 08 5:04
C and n are the variables that are to be evaluated based upon door blower test (and this is the better method).

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)  

DRWeig (Electrical)
13 Jun 08 9:41
Gotta jump on the bandwagon with RossABQ.  You'll never maintain a consistent differential with simply supply / return balance.

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

 
dmcarroll (Mechanical) (OP)
13 Jun 08 10:30
Yes, I made a simple math error when I listed the opening area.  Rest assured I didn't make the mistake in my actual calculations.  I would agree in saying that 500cfm is too large a surplus for the given area to relieve the pressure.  The provisional plan is to install a relief damper so that the door can be opened and overall room pressure can be adjusted.  Thanks all for your help and input.

  
mauricestoker (Mechanical)
17 Jun 08 22:21
I'd suggest looking at ANSI Z9.5 for a good reference on the matter.

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).

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