ROOM STATIC PRESSURE CALCULATION 6

[VEEKRISH]

I have a room whose dimensions are : 15 feet x 20 feet. Height is 10 feet. There is one door, whose dimensions are 7 feet x 3 feet.

I need to maintain this room at 0.4 inch static, higher with respect to the outside.

Could you please tell me if there is a mathematical formula available to actually establish the exact air quantity I will have to pump in and exhaust to maintain this diiferential.

 

[quark]

Initially the air volume in room is 3000cu.ft and pressure is atmospheric. (approx. 10000mm of water column) and you want to rise it to a pressure of 0.4inch (10.16 mm of wc - quite high, what is the application?)

volume of air to be present in the room to maintain the above pressure is 10010.16*3000/10000 = 3003.04cft.
This is in static condition. So to maintain your pressure in a leak tight room you have to pump in 3.04cfm air if you want to rise the pressure in one minute(and this is one time requirement). For this you require a fan which has a static pressure of 10.16 mm. Once the pressure is builtup in the room there is no flow into the room as the fan static equals the room pressure.

Now consider there is leakage from the room. The net flow of air through leakage area is calculated by
Q = 2610 x A x (DP)[sup]1/2[/sup]. where A is in Sq.Ft and DP is inches WC. If we want to maintain same pressure in the room, we should supply the amount of air which is leaking out. So in dynamic condition once the pressure is established, you need to supply only the leakage flow rate through the room.

Initial pressure building up depends upon
1. Leakage rate.
2. Room Volume.
3. Difference in supply and Return Air flow rates.

To explain easily I added leakage after stabilisation of pressure. But generally it is a transient condition and increases as the pressure increases and remains constant once the room pressure is stabilised.

If your leakage rate is x cfm at the desired DP then you should initially supply (x+3.04)more cfm initially and then when the pressure establishes it is 'x' cfm only. (Don't worry, this automatically gets balanced)

Room pressure created because of resistance of air passing through narrow cracks seems totally illogical and absurd to me.

 

[KenRad]

VEEKRISH,

For a detailed discussion of this, look at the following thread:

thread403-31557

(don't hurt me, quark!)

---KenRad

 

[quark]

Huh! Kenrad

What do I do with that sledge hammer then?

 

[ChasBean1]

Don't add any confusion. The bolded equation in Quark's post above is your nearest simulation and is from the ASHRAE Handbook of Fundamentals. To help clarify for the equation:

Q = 2610 x A x (DP)1/2

• Q is the room leakage flow rate in cfm,
• 2610 is a conversion factor,
• A is the net open crack area of the room (generally not measurable), and
• DP is the differential pressure (equaling 0.4 in. w.c. in your case).

You're left with two unknowns: Q, which is your air leakage - the offset flow rates in question, and A, which is your net open crack area.

Your key is to design the room with a certain flow offset. Pick a number. I say 700 cfm. Why? It's a 300 ft2 room so if it was office ASHRAE would say supply 300 cfm (general rule). I would guess it's a specialized application and not office because of the pressure requirement. So (guess, based on air exchange, exhaust need, or heat load) supply 1,200 cfm and exhaust 500.

With design cfm delta based on room heat load, exhaust, and air exchange requirements, the rest is up to sealing, caulking, gaskets, and door sweeps.

Your true variable is the net open crack area. Pick sensible design values knowing net open crack area is not measurable and adjust room pressure by sealing the room. For this application, the room will have to be well sealed and the offset will have to be pretty high.

Best regards, -CB

 

[VEEKRISH]

Quark, KenRad & CB - Thanks guys. I now have something to chew. Quark go easy with the sledge hammer!!! You guys really rock this forum. I like it. Thanks again. Regards. Krishnan

 

[CHD01]

My HVAC Manual is not new, but did have methods to calculate leakage based on type of building construction, type of window design, plus door and window dimensions.

Is this not still available? I ask because you all keep saying that leakage cannot be calculated. Granted it can be quite variable, but studies had been done to try an establish reasonable design rates.

The more you learn, the less you are certain of.

 

[sfxf]

There is a good article "Room Pressure for Critical Environments" published in ASHRAE Journal February 2003. You can free download from "

http://www.ashrae.org"

web site.

 

[nans]

Hi SFXF
Tried to find the Room Pressure for critical environments in the ASHRAE website but not successful. Can you pls send the full link or the file to nanda.suryanarayana@amec.com
thanks
nans

 

[CHD01]

Quark & Chasbean1:

This is interesting, I believe I understand both your points - regarding a balance of flow in and out, etc. I think perhaps a way to better explain these differing views is as follows (see if it makes sense to you both):

The equations you all are using (which I have no problem with by the way) is based on volumetric flow! Volumetric flow in and out does not have to be equal - BUT MASS FLOW RATE DOES!!!!

What is happening is that for the room pressure to increase above atmospheric, the volumetric density (lbs/ft3) in the room must increase. This increase requires the air to be compressed by the fan (so the fan static pressure builds with the room pressure). Thus the MASS FLOW of fresh air into the room does equal the MASS FLOW through all open area to atmosphere. Then you must increase the size of the fan to account for any recirculated flow through the fan of course.

Does this not bring into happy harmony both you QUARK and ChasBean1?? The more you learn, the less you are certain of.

 

[quark]

CB, CHD01 and Kenrad!

Thanks for you guys.Infact I am consolidating my thoughts by your inputs. In the beginning of my career I was to control multiple areas (maximum upto 5 rooms with increasing pressure by 15 pascals) under one AHU. Ultimately I used to spend hours in the morning before start up of production. This really annoyed me very much as the area where I have to maintain 75 pascals pressure was the main production area. Somehow I used to pull the matter and thought it was all magic of return damper.

Now it seems clear to me.

In this regard my above post is rather a monologue.

But onething for sure is incase of Isolators, (containment management) as there should absolutely be no leakage, we have to fiddle with the two dampers. My first equation holds good in that condition.

Regards,

 

[ChasBean1]

CHD01: No. The density is the same at the inlet and outlet. The mass flow and volume flow in is equal to the mass flow and volume flow out. Bernoulli's for this dynamic condition reduces to friction (and only friction) which causes the change in pressure. To me, we're "arguing" opinions over a matter that is fact. That said, I'm only a BSME (not MS or Ph.D.) with a nuclear background and I've definitely been conceptually wrong before. I'm at work now and don't have time, - but I'll try to compile a reasonable explaination when I get a chance. Best regards, -CB

 

[quark]

I agree with CB that volume flow in shoud be equal to volume flow out, but under steady state condition. Once the room pressure builds up then the flow remains constant. (Actually before that room volume plays a significant role. Initially flow into room = excess air in the room + flow past room, thus conservation of flow gets satisfied.)

After stabilization of pressure in a room if we increase door opening (more specifically leakage area) pressure drops down. This is because we are letting extra air to flow out.

Leakage occurs because of pressure differential. There is no pressure differential means no flow.

Still I am waiting for CB's post.

Regards,

 

[Yeldud]

ChasBean1 - Nice to see a voice of reason, in a see of confusion... As a fellow BSME'r, i think your BS makes you a qualified BS'r. No Bull!
Regards.

 

[CHD01]

My message regarding mass flow in equaling mass flow out obviously must refer to steady state conditions where free area exists for leakage out. If there is no leakage, then the pressure of the room increases until the fan can no longer bulid static pressure in the room - then flow in goes to zero cfm in from the fan for no leakage.

While the increase in air density is not very much in the room, the density in the room must increase with the increase in room pressure. Thus density in room is slightly greater than outside the room (which is also density of air entering the fan).

Regarding friction loss through free area, at inlet to free area the pressure and air density is greater than it will be at the exit from the free area (free area from cracks). This must be true or there would not be flow. So while the mass flow thru cracks remain constant at crack inlet and outlet - the volumetric flow increases. This is fundamental - look at CRANE and theory of compressible flow that you have all quoted.

While in this example it is not critical, many times it is better it deal with mass flow rather than volumetric flow. Prime example is sizing a relief valve for an air compressor, you need to size the relief valve based on winter conditions - Not On Summer Conditions! The more you learn, the less you are certain of.

 

[ChasBean1]

CHD01 - as an HVAC engineer, my way of thinking is in standard cubic feet per minute (SCFM). I think this industry tends to be guilty of using volume flow and mass flow interchangeably.

We simplify for example by saying if 10,000 cfm is being discharged from a fan, then the sum of all the air outlets and leakage should be 10,000 cfm, neglecting actual volumetric corrections based on the fact that the fan discharge is pressurized and each outlet is essentially atmospheric. We also maintain the equivalent cfm assumption when individual zones are heating the air.

'The density at the inlet and outlet is the same' is faulty - you're right and sometimes I post before I think it through. Your previous post is accurate regarding mass and volume flow. The (true) volume flow could vary but I don't think it's the principal factor in determinining the pressure build-up in a room.

To throw a wrench in the discussion, what if reheated air at 140°F were entering the room while 70°F air was being squeezed out through cracks? Would these thermal changes in volume flow proportionally (or at least relatedly) affect changes in room pressure?

My way of thinking is very simply that air flows from a high to low pressure area. If you consider the room as a large opening in the duct with the air entry point being point 1 and the exit point being point 2, assuming the exit opening is sized equally with the inlet opening, the main reason for pressure build-up in the room is that the friction loss at the point where air is being pushed into point 2 is causing accumulation of air molecules within the room, therefore higher pressure with respect to areas downstream of point 2.

If the opening at point 2 is reduced in size, the mass (or SCFM) flow through the room would still be equal at points 1 and 2 but pressure would build further due to the added restriction (higher friction loss) at the exit opening.

When we close off point 2 completely, representing a perfectly-sealed room, pressure in the room is not a function of the volume flow, mass flow, or friction loss, but is now a static condition equal to the inlet duct pressure. If the room can take it, we can pressurize it to 5,000 psig by using any volume flow of air greater than zero assuming your fan or compressor is equal to the task.

The concept is simple from a common sense standpoint but is complex mathematically for compressible flow with friction. I've posted the Q = 2610 A dP^.5 from ASHRAE 1999 Applications (Ch. 51.5) because I like how it takes a grueling concept with unknowns, scary integrals, internal energy, relative roughness of open area surfaces, etc. and boils it down (with many assumptions, of course) to something usable for this application.

Quark, room pressures generally come to equilibrium pretty quickly (within a second or two in most applications) so if you want to re-think the steady flow energy equation with compressibility, friction, internal energy, etc. during brief transient periods, be my guest!

Now I've gotten too wordy and the more I write the more that can be used against me at a later date... CHD01 - your end cap "the more you learn, the less you are certain of" is definitely true. Best regards, -CB

 

[mwengineer]

I may be butting in a little late here, but how critical is the O.4 static pressure differential? If it is not ctitical, use the ASHRAE equation to estimate the difference in supply/exhaust CFM's then balance the room to that by measuring the delta P while setting up the system.

If the 0.4 is critical, you had better use one several room pressure control units to continually fine tune the in and out.

 

[CHD01]

Yes in this case the differences in density are not critical and flow and mass can essentially be treated the same; sometimes us engineers get turned on by the strangest things and this is one of them. Thus to continue, regarding the reheated air entering the room and hoping I don't confuse myself.

Fundamentally, what should happen is that the hotter makeup air (140 Deg F) entering the room, should make it easier to build up room pressure with less mass flow. The mass flow out of the room for the cooler air (at 70 Deg F) must be the same as mass flow in but the volumetric rate is less. This sounds like a contradiction, but somehow the entering air had to be cooled so there was an energy and/or heat conversion (heat loss or air-conditioning unit). I think this means the input fan to the room might actually be slightly starved for air relative to what it could flow - if flow in actually did equal flow out. But again, your right - the difference in pressure is so small, flow in is essentially flow out. So all this is very theoretical; but real!

Yes Chasbean1 all of what you say is right on - you are completly correct regarding flow from high to low pressure, etc. I got the impression, Quark just had some problems with some earlier applications he had dealt with; and after our exchanges figured out a way to improve upon the model he uses for analysis.


Chasbean1, I see your an HVAC engineer. I've alway liked heat transfer issues and almost took an HVAC job when I was interviewing in my senior year. I'll never forget that opportunity, I interviewed Abbott Laboratories up in Chicago and they met me at the plane with a limo. I almost fell over in shock (this was a few years ago). However, more money won out at another job in the chemical industry plus I was concerned about being viewed as a utility engineer in a pharmaceutical centered industry - but I still wonder what my career would have been like. The more you learn, the less you are certain of.

 

[CHD01]

Quark and Chasbean1:

I have a question:

If you apply the equation (10010.16*3000/10000 = 3003.04cf) or {time = [(ft3 volume)*(dp in psi/14.7)]/[cfm difference in flow]} to the example given to compute increase in air flow required, you got 3 cfm for a required time period of 1 minute with a room volume of 3000 ft3. I agree with this.

This implies a leakage of 3 cfm, if you then take the formula you gave of Q = 2610 x A x dP^0.5 (I agree with this) and solve for the free area for leakage you get a real small number equivalent I think to about 1/4 inch of free area for the entire 3000 ft3 room.

Both these equations need to balance - right?

This seems like a really SMALL leak rate for the real world! Don't you agree?

Also, in this example we've looked at, no one has mentioned the required fresh air rate for occupancy of the room - this could be MUCH MORE than this low leak rate. Depending on how you calculate it, the recommended fresh air exchange rate is: 10-20 cfm per person, or 1 cfm/ft2 of room area, or about 16 cfm if you calculate leakage based on HVAC standards for crack free area, or about 1260 cfm if you elect to maintain a 60 fpm velocity for all openings including doors (unless you use a double door air-lock).

I would take the minmimum required fresh air flow and then use a weighted louver to control room air pressure (remains closed until room static pressure exceeds set-point).

The time to pressure the room should be minimal.

One last thing, in your formula you use a constant of 2610, 10 years or so ago the constant I used was 2800. Do you know if this changed and when?

Thanks


Commments? The more you learn, the less you are certain of.

 

[ChasBean1]

Hi CHD01:

Your question regarding both the equations needing to balance - I don't think so because it's a closed system analysis versus a steady flow system analysis. Quark's post above is fine for tanks. P1V1 = P2V2 (T's cancel). Yes, that will get your increase in room volume from 3,000 ft3 to 3,003 ft3 and give the change in pressure of 0.4 in. w.c. for a non-flowing environment.

3 cfm is a flow rate. If you trust the ASHRAE equation, you'd better seal the room pretty well at this flow rate (allow only 1/4 square inch of net open area) if you want the pressure to build to 0.4 inches. To me, this sounds realistic. You're right, it is a small leak rate for the real world because this is an unrealistically small flow offset between supply and exhaust.

But Pascal's Law and the ideal gas rules don't apply to a steady flow, compressible system with friction loss. We are talking apples and oranges here. The ASHRAE equation won't apply if you completely eliminate leakage and probably won't apply with too high a pressure source like a compressor versus a fan.

If you look back to my first post, I suggested 700 cfm difference. Per the ASHRAE equation, this would allow just under a half a square foot of net open area for leakage, which is still small for the real world, but achievable. The flow offset is picked on gut feel and how good you estimate your drywallers and caulkers to be.