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Hi,
I was wondering how I would maintain a specific pressure (say 50Pa) inside the conditioned room. I know it calls for a lot of fresh air, but how much? Is there any formula to calulate this?
The application is offshore (a refuge area in case of a gas leak, in the plant). So, this is not a clean room, but just the pressurization. Once the air vol is know, I hope the fan should be also strong enough to pump in more volume of air! So, how do I calculate the static?
Regards
Nevin.
Can I refer you to another forum where this is discussed with calculations....
http://www.eng-tips.com/viewthread.cfm?qid=43928
When you have done the calcs - I worked on a few chemistry labs with the same criterea. These wer controlled with outside air damper actuator, VSD on fan and a pressure sensor set at I think around 5 pascals.
The formula mentioned just says about the air leakage through cracks... not the the actual calculation!Quote:
Originally Posted by Brian_UK
I don't think you are reading the thread correctly. The Ashrae formula is in there and is discussed in great detail. But without knowing what your leakage rate of the room is, you will find it hard to calculate your volume flow rate/pressure curve for the fan selection.Quote:
Originally Posted by nevinjohn
Okay, its says about the air volume required, but what about the time required to bring the room back to the design pressure?
Lets say the application is air lock door, designed for 50Pa. When the person enters the air lock, it looses pressure. So, how much time do you think it will need to room back at 50Pa?
Think of a paper bag. If you blow into it, you will pressurise it. If you now put a hole in the bag, your pressure will be lost. As soon as you repair the hole, the bag is pressurised again.Quote:
Originally Posted by nevinjohn
Same thing occurs in your airlock room when you open the door. As long as the airflow into the room is constant, the room will become pressurised as soon as the door is closed again.
Yes, but how can the time delay be calculated for the room to get pressurized back to 50Pa?
How long does it take for the door to close.........
OK, try another route.
If you use an airlock then you have two doors, one direct to outside (unpressurised) and one direct to inside (pressurised).
Fit an electrical interlock between the doors so that only one can be open at one time. Connect the interlock to a pressure sensor that monitors the room air pressure and the air lock pressure. When the two pressures are equal then the inner door can be released.
The AHSRAE orifice equation Q = 2610 x A X (dp)^0.5 gives the air flow through the crack area for a design pressure, say 50Pa. I am trying how to attain this design pressure. Lets say there are two rooms. One with 3000cbft anad another with 1000cb.ftQuote:
Originally Posted by frank
Consider both have the same crack area.
As per the calcualtions, the air flow would be same for both the rooms if both the rooms are designed at 50Pa. So, my question is how to design it for 50Pa.?
Will P1V1 = P2V2 will do???
The easiest way to calculate room pressurization volumes and more like a rule-of-thumb method used by the industry is 0.25 cfm per ft2 of floor area for 39° to 50°F and 0.5 cfm per ft2 of floor area for 55°F and higher room temperature areas. Colder rooms require lower air pressurization volume as they more tightly built.
Added to the calculated air pressurization is the exhausted air volume from the room if any [ process system or, etc…]
Example,
1000 ft2 room floor area and 500 cfm exhausted by stacks [0.25 cfm/ft2 design for pressurization]
1000 ft2 x 0.25 cfm/ft2 = 250 cfm (for air pressurization volume )
500 cfm (exhausted by stacks)
Total: 250 cfm + 500 cfm = 750 cfm
Notes; the air pressurization volume must overcome all the opening, [cracks, conveyors, door infiltration], that can be calculated through ASHRAE formula, also must include the people fresh air intake of 25cfm/per person
The fan used for pressurization comes have external static pressure design in-wg [0, 0.5, ¾, 1, 1.25 in-wg etc..] and should overcome calculated friction losses through ducts, louvers etc..
Regards
Camille Zabbal
Thanks Camille.
Rule of thumb is not accpeted when dealing with clients. I am trying to form a back up calcualation as the basis. One part shall include the leakage air flow as per the ASHARE orifice equation, and the second part shall form the actual basis. I am considering
P (total) / P (atm) = V (total) / V (room)
I found this in BS5588 Part 4
The air supply needed to obtain a given pressure differential is determined by the air leakage out of the space. When air flows through a restriction such as the crack around a door or a window as the result of a pressure differential across the restriction, the relationship between the rate of airflow, the restriction and the pressure differential is given by:
Q = 0.827 x A x (P)^1/N
where
Q is the airflow in M3/s
A is the area of the restriction (m2)
P is the pressure differential (pa)
N is an index that can vary between 1 and 2
For wide cracks such as those around doors and large openings, the value of N can be taken as 2 but for the narrow leakage paths formed by the cracks around windows the more appropriate value of N is 1.6
From Table 2.
Where P = 50 (pa), the value of 1/N, where N = 2 is 7.1
Where P = 50 (pa), the value of 1/N, where N = 1.6 is 11.5
Typical leakage areas from Table 3
Door size 2m high x 0.8m wide, single leaf, rebated, opening into the pressurised space = 0.01m2
Door size 2m high x 0.8m wide, single leaf, rebated, opening out of the pressurised space = 0.02m2
Door size 2m high x 1.6m wide, double leaf, rebated, = 0.03m2
Where more than one door or window exists in the same space to be pressurized, then the sum of the areas in calculated in parallel. (A1 +A2 + A3 etc)
Where the leakage path from the space is via an intermiate space before escaping to an unpressurised space, the sum of the leakage areas should be calculated in series (1/A1 + 1/A2 + 1/A3 etc)