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.

 

[CHD01]

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]

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]

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]

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]

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]

Thanks, CHD01 - I'd be interested. cbean@eheinc.com

 

[quark]

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]

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]

.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]

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 = V2[sup]2[/sup]/2g

1 lb/sq.ft = 0.1922 inches of wc
1g = 115820 ft/min[sup]2[/sup]

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

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]

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]

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]

Can someone please explain how the constant 2610 was derived?

Q = 2610 x A x (dP)[sup]0.5[/sup]

Thanks,
Freddie

 

[ChasBean1]

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]

ChasBean1,

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

Thanks again,
Freddie

 

[ChasBean1]

It's easy when there's over thirty in a thread! Best of luck, -CB

 

[BikeBill]

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]

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]

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]

CHD01!

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

Best Regards,