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VEEKRISH (Mechanical) (OP)
30 Jan 03 3:45
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.
Helpful Member!(4)  quark (Mechanical)
30 Jan 03 5:26
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)1/2. 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 (Mechanical)
30 Jan 03 8:27
VEEKRISH,

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

Thread403-31557

(don't hurt me, quark!)

---KenRad
quark (Mechanical)
30 Jan 03 23:11
Huh! Kenrad

What do I do with that sledge hammer then?

Helpful Member!  ChasBean1 (Mechanical)
31 Jan 03 0:08
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
Helpful Member!  VEEKRISH (Mechanical) (OP)
31 Jan 03 0:18
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 (Mechanical)
2 Feb 03 23:29
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 (Mechanical)
3 Feb 03 10:47
There is a good article "Room Pressure for Critical Environments" published in ASHRAE Journal February 2003. You can free download from "www.ashrae.org"; web site.
nans (Mechanical)
3 Feb 03 14:32
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 (Mechanical)
3 Feb 03 17:09
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 (Mechanical)
4 Feb 03 3:38
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 (Mechanical)
4 Feb 03 12:17
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 (Mechanical)
5 Feb 03 10:45
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 (Mechanical)
5 Feb 03 13:03
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 (Mechanical)
5 Feb 03 22:58
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 (Mechanical)
7 Feb 03 18:05
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 (Mechanical)
8 Feb 03 11:11
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 (Mechanical)
11 Feb 03 9:58
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 (Mechanical)
13 Feb 03 17:50
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 (Mechanical)
14 Feb 03 0:45
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.
CHD01 (Mechanical)
14 Feb 03 10:52
Chasbean1:  In your first post, is not 1200 cfm the total supply from your AC Unit, with 500 cfm your fresh air make-up rate to balance the 500 cfm exhausted through cracks plus outlet damper(s) with 700 re-circulated.   RIght?  If not right, I may be confused.  I envision the AC Unit with outdoor air and re-circulation dampers; plus the room with a weighted damper where the AC dampers control fresh air exchange and the room damper(s) controls room air pressure.

Otherwise I concur with what you say; except I will point out that the ideal gas law still finds a lot of applications with continuous compressible flow.

Thanks for response.

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

ChasBean1 (Mechanical)
17 Feb 03 19:10
CHD01, I meant 1,200 cfm total supply to the room, and 500 cfm total leaving the room via mechanical exhaust, with no recirc (meaning the 700 cfm flow difference is what would leave the room through cracks).
CHD01 (Mechanical)
22 Feb 03 15:19
OK - 100%  fresh air just sounds like a high fresh air rate, although I know we sometimes need an exchnage rate that high. I believe I remember that residential rates are normally about 20% with a recirculation rate of 80%  The 20% flow would then have to equal the mechanical venting plus leakage in order for mass flow to balance.

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

ChasBean1 (Mechanical)
25 Feb 03 13:20
Hi CHD01 - the assumption was that this was a space in a building supplied by a central system that also serves other spaces. The central system might be a 20% OA system, but the room itself should be fed 700 cfm more supply air than exhaust or return air. Note that the 700 cfm is also arbitrary (guess) based on the fact that 0.4 in. w.c. is a pretty high pressure to maintain a room (in fact, the door won't shut unless it has a pretty strong closer, or it would take about 40 lbf to open if it opens into the room).

Best regards, -CB
CHD01 (Mechanical)
26 Feb 03 13:15
Hi ChasBean1:  Boy that explains a lot as to some of the confusion we had initially.  I hear you on the force required, I noticed a slight extra pull in force for the door of an extruder extraction room I designed.

Since you are interested in this, if you don't mind sending your e-mail address over the net I'll send you the spreadsheet I use (I'll even give you my e-mail).  You may use it for your personal use if you like, just don't distribute it for use by others since I wrote it.  Its not that complicated a program nor do I think it would draw that much interest to offer for sale unless I increased its flexibility.  Of course, if you have any upgrades/updates to suggest, let me know.

Quark this offer goes for you too if you have any interest.

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

ChasBean1 (Mechanical)
27 Feb 03 10:44
Thanks, CHD01 - I'd be interested. cbean@eheinc.com
quark (Mechanical)
28 Feb 03 2:23
CHD01!

I am much obliged. My ID is builblock1@yahoo.com.

Some thoughts on the original post.

I have been thinking on the pressure issue since my last post and couldnot come to a common stand point. Actually there are mainly two pressure losses. One is due to expansion and the other is friction. Quantitatively, expansion can cause much pressure drop. If you are keeping a door totally open, the pressure drop from a pressurised room is primarily due to the expansion effect and friction loss is comparatively negligible.

Leakage of air is proportional to pressure difference. Untill and unless you have pressure difference no leakage occurs. So my point is first pressure develops and then leakage starts accordingly. If you know how much pressure you have to build up then you can calculate leakage. As from my previous post, the amount of air required to build up pressure is much smaller than the leak rate it may not be showing up as significant.

It sometimes becomes difficult to loose a concept based on which you did trouble shooting successfully. Yet I am not rigid, for new concept may make my trouble shooting much easier.

I never did (neither will) question anybody's academic or technical credentials(because I too have only Bachelors Degree in ME). I always consider you people in high esteem. All my idea is to put some fun in work.

Yet I disapprove two concepts for one occurance(for example pressure with leakage and without leakage).

One more word if I put this becomes a soap. (It annoys me as much)

Note: To calculate extra air needed ASHRAE equation is sufficient and I totally agree with anybody. But I am confused about the method of pressure buildup.  

CHD01 (Mechanical)
1 Mar 03 12:12
QUARK

I think if micro analyze almost any subject you will find what you might believe to be contradictions.  If you have a open house with many room there is no pressure difference in one sense, yet there is infiltration from leaks.  But how does the infiltration occur - from pressure diffences due to miscellaneous wind currents, from thermal movement, etc.  How can one determine when one affect overrides the other - its all relative.  When I get hung up like that I just try to make a situation as simple as possible without being 100% correct.  Does this help or just make it more frustrating?

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

Waunakee (Mechanical)
3 Mar 03 21:40
.4" static appears quite high. To what are you comparing this to?
If we are talking OA pressure vs builidng pressure, then perhaps you may be looking at .04"WC?
If you are considering .4"wc, then you best have reinforced windows and doors.
The internal static presure compared to the outside from the inside of the room is considerable at that rate, you may wish to consult your building codes.
I did not see any one else questioning the .4" or to what comparitive measurement you are considering.
I also would look at how you attempting to maintain this control. Presure transducers and BAS, or?
Kevin  
quark (Mechanical)
4 Mar 03 3:53
Guys!

Now I am reconfirming my idea that flow through cracks is because of static pressrue buildup inside the room with more positive pressure. The formula can be obtained from Bernoulli's principle.

(P1-P2)/Rho = V22/2g

1 lb/sq.ft = 0.1922 inches of wc
1g = 115820 ft/min2

So V2 = [(2 x 115820)*dP/(0.075 x 0.1922)]1/2
i.e V2 = 4008.6 x dP1/2

So Q = A x V2

There is a coefficient of opening involved based on type of the opening. For square edged narrow opening it is 0.6

So Q = 0.6 x A x V2 and this value almost equals our standard equation. (the inaccuracy may be because of the adjustment in units)

I got a reference for "Fan Engineering by Buffalo Forge Company". If it is available with any of you guys please check and let me know. For me it is almost impossible to get that book.

Now my idea is static pressure is being converted into velocity pressure depending upon the downstream pressure.

Any comments?

CHD01 (Mechanical)
4 Mar 03 11:03
Looks good Quark, flow cannot occur without a pressure drop.  Given the pressure difference, the amount of flow is determined by the resistance to flow and the flow area of the crack.  The resistance to flow is equal to: 1) the entrance loss into the crack, 2) the friction loss through the crack due to flow, and 3) the exit loss from the crack.

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

ChasBean1 (Mechanical)
4 Mar 03 19:18
Quark, or the inaccuracy could be with the coefficient(?). You're right - I think that's probably the basis of it. It's basically takes Point 1 with only pressure energy and zero velocity and translates it to a leak at Point 2 with lower pressure and some velocity. Well done!  -CB
merheb (Aerospace)
22 Apr 03 16:55
Can someone please explain how the constant 2610 was derived?

Q = 2610 x A x (dP)0.5

Thanks,
Freddie
ChasBean1 (Mechanical)
24 Apr 03 12:12
See Quark's post three back. Based on a coefficient (estimate) and the steady flow energy balance, Quark gets to about 2,405 for a coeff. I think his methods for figuring this are accurate.
merheb (Aerospace)
24 Apr 03 21:39
ChasBean1,

Thanks for the info...  I don't know how I missed Quark's post!

Thanks again,
Freddie
ChasBean1 (Mechanical)
25 Apr 03 14:43
It's easy when there's over thirty in a thread! Best of luck, -CB
BikeBill (Mechanical)
1 May 03 10:49
The book mentioned by Quark on 3/4/03, "Fann Engineering" by Buffalo Forge is available from Howden Buffalo.  http://www.howdenbuffalo.com/

They have is as a book only and a book CD combination.

BikeBill

CHD01 (Mechanical)
31 Jul 03 16:39
Quark:

Did you get my program revision for design of air locks?  I get a notice that your yahoo account is discontinued when I try to send you the new program.

I made some improvements, plus I had a typo error in cells D106 and E106 of the new version that you need to correct.  

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

lilliput1 (Mechanical)
31 Jul 03 18:15
From experience in designing isolation rooms, a conservative pressurization CFM to hold 0.01" wg pressurization is 200 CFM. Since pressurization CFM is proportional to the square root of the pressure then CFM2 = CFM1 x (P2/P1)^.5

CFM2 = 200 x (.4/.01)^.5 = 1265 say 1300 CFM.

The bottom of the door should have brush type door sweeps. The ceiling should be gypsum board. Light fixtures and receptacles should be gasketed. The intersection of the wall with the floor, behind the baseboard should be caulked.

But why so high a pressure? If too high, it will be difficult to close door!
quark (Mechanical)
1 Aug 03 0:48
CHD01!

I received your mail and the attachment. Thanks. I will be in touch with you incase I find anything interesting.

Best Regards,

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