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Clean Room Pressurization

Clean Room Pressurization

Clean Room Pressurization

(OP)
I am designing the HCAC system for a small clean room (~200ft^2 w/ 8ft ceiling, Class 10,000). The client wants 60 air changes per hour with 0.05 in WG positive pressure (relative to building).

How do I calculate the amount of exhaust air to provide positive room pressurization?

RE: Clean Room Pressurization

Check this thread.
Thread403-31557. I still stick to my stand in that post for if you provide enough air to compensate for leakages there is no pressure built up in the room. However, is your system ciculatory type or once through?

Regards,

RE: Clean Room Pressurization

(OP)
The system is 100% outside air.

Supply Air = 1400 cfm (from 60 ach) + 2 ach exfiltration = 1453 cfm. Where, ach  is the air changes per hour.

Exhaust Air = 1400 cfm

This should maintain a slightly positive room pressure because SA > EA.

RE: Clean Room Pressurization

I'll start with an apology; although I work in the UK (where "British" units originated!), my calculations are generally done in metric units, as are the standards I work to (250 Pa is about 1" water gauge, 1 m3/sec is about 2100 cfm). However, here are some typical figures I use when designing for deliberate positive or negative pressures in hospitals, etc:

Leakage through (closed!) doors (m3/second) at stated pressure difference (Pa) across door (m3/second@Pa):

Single Door:
0.03@5, 0.05@10, 0.06@15, 0.06@20, 0.07@25, 0.07@30, 0.08@40
Double Door:
0.04@5, 0.08@10, 0.10@15, 0.11@25, 0.11@20, 0.12@30, 0.13@40

So, for instance, if the room is reasonably well sealed and the bulk of the air leakage is through the standard gaps (4mm along bottom, 3mm along top and sides, 2mm between double leaves) at doors, then:

If it is a single door and the room pressure is 15 Pa above that on the other side of the (only) single door, the leakage out of the room will be 0.06 m3/sec.  If the supply air is in excess of the extract air by that amount, then the room pressure will be about 15 Pa above the pressure outside the door.

Beware, it is often the case that other "unexpected" leakage paths, etc occur during construction; these must be catered for as well as the doors.

Regards,

Brian


RE: Clean Room Pressurization

As far as my calculations are concerned the supply air is coming out to be 200 cfm more than the return air.

Brian!

As per your calculation the 120 cfm extra air is leaking out of the room. So how the pressure is being maintained?

Or, if I pass only 120 cfm to the room and not taking anything through return (that is cfm in return ducting is zero) still can we be able to maintain the pressure inside?

Note: This is in continuation to my discussion in the other thread I already mentioned. Perhaps I may have some misconception.

Regards,

RE: Clean Room Pressurization

Winwind, As I understand it, the Room Pressure builds from the friction of air leaking out through the openings / gaps present in the structre (under cut below doors ?). If air being supplied is equal to air being exhausted, then we reach a state of equilibrium with respect to the outside. Now, in case we supply more air and exhaust less, the adifferential quantity (supply - exhaust) will tend to escape out of the openings / leakages. It is this frction drop of air trying to leak out through the openings that causes the room pressure to be built up. More the differential, the higher the friction drop, the greater the pressure build up.
You may consider using the requation Q = 2610 x A x dp, where Q = Differential air quantity etween supply & exhaust in cfm / A is the net leakage area in sq ft / dp is pressure differential between room and outside in inches of water column. This equation is from the ASHRAE handbook. Only problem is how to estimate or quantify A. I reckon a close examination of the space and an educated guesstimate are the answers, which will yield a ball park figure.
Anyone having a more accurate method, please help.

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