Control Room Pressure

[esanders]

I'm working on evaluating existing pressurization systems for several control rooms and petrochemical labs. Due to the classification the inside pressure must be maintained at 0.1 in wc, also with all doors open the face velocity across the openings must be at 60 fpm. During meeting the 60 fpm requirement the pressure inside the space is allowed to fall below the 0.1 in w.c.
I calculated the cfm needed to meet the 60 fpm across all door openings but I'm not sure how to go about calculating the cfm required to maintain 0.1 in w.c.
Is meeting the 60fpm airflow adequate?
How do I calculate the amount of air needed to maintain 0.1 in w.c. while the doors are closed?

 

[RossABQ]

There are sections in ASHRAE Fundamentals describing crack losses. There are various rules of thumb including .5 cfm/ft^2 for .05" w.c. difference. If you do some research, you will find that buildings/rtooms aren't as tight as you might think. Is dynamic control going to be used to maintain the pressure? If so, aim high with your capacity for the VAV/pressure control box.

I did this same calc recently for a lab (opposite pressurization) and the doorway velocity wasn't the governing case for volume required. I would have guessed it would be (it had double doors)

 

[quark]

Before saying anything I should, first, wish you good luck. What you are going to acheive is quite a hard task.

Suppose, if you have a door size of 1.2m x 2.1m then flowrate across the door opening at 60fpm would be 1.2*2.1*10.76*60 = 1627 cfm. This should be extra air flow required when the doors are opened. You can divert the room air flow by closing the return damper totally but the speed at which you have to control the things is not possible, in real sense. My practice (in pharma) is to offset the door opening from controls.

During meeting the 60 fpm requirement the pressure inside the space is allowed to fall below the 0.1 in w.c.

Actually, you can get 60fpm in door open condition with a pressure difference of just 0.0002" wc [(60/4005)^1/2], but the moment resistance to flow decreases, fan pumps more volume and then your pressure controller should fight harder.

How do I calculate the amount of air needed to maintain 0.1 in w.c. while the doors are closed?

My practice is to consider 5 mm gap in the bottom, 1 mm for sides and top for a single shutter door. The middle gap, for a double shutter door, is 2 mm. So, the total leak area, for a single door of 1.2mx2.1m, becomes 1.2x0.006+2.1x0.002 = 0.0114 sq.mtr or 0.123 sq.ft.

You can use Q = 2910xAx(dP)^1/2 for calculating leak rate, so Q = 2910x0.123x0.1^1/2 = 114 cfm.

Do let us know when you finish your task.

 

[lilliput1]

Check with Bebco. they make pressurization systems for classified rooms.