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Room Pressurization

Room Pressurization

(OP)
Hi,

I am doing a design for a processing room with area 28m x 20 x 6m high. I am asked to keep the pressure of the room to be more than 5 pa. Please I want to know what is the equation to keep the room say at 7 Pa.

They didn't specify how much air changes needed but I am intending to do 12 air change.

Thanks

RE: Room Pressurization

Howdy,
12ac/m is likely OK, but it'll depend on how tight the room is. What is the nature of the process? Are there flammable or combustibles in the process area? What are the requirements for MU air?
7kPa = 28in of water; why such a high pressure?

Sorry, my bad. You clearly started 7pA. I mistook that as 7kPa, which of course is only 0.028 inches of water.

RE: Room Pressurization


Pressure difference can be seen as resistance to flow.
Your air flow will need to ''lose'' 5Pa of head to exit room.

Not very much pressure needed at 5Pa.
Have fresh air ducted to the room, put exhaust behind wall (Corridor, other room, wherever you want)
Put adjustable vent traps in upper part of wall between intake and exhaust. (Sizing needs be done)
Balance system to achieve desired head loss.

Make sure the air speeds remain low, as people don't want it to be ''windy''

A Proud Miner

Ingenieur Minier. QuTbec, Canada.

RE: Room Pressurization

From experience with hospital isolation rooms we found that minimum pressurization CFM for 0.001 in wg range from 200 to 400 cfm per 3' door depending on how good the room is sealed air tight and bottom of doors provided with brush type door seals. Pressure will vary with the square of the CFM so you can prorate the CFM to the pressure you need. Room should be provided with differential pressure alarm that would have time delay to alarm to allow for normal door opening for entry/exit access. Doors should have automatic door closer.

RE: Room Pressurization

(OP)
Thanks all,

The room can be considered as tight. it is processing for milk products.

I was just wondering if there is a law to be followed.

Regards

RE: Room Pressurization

There is no simple equation to solve this problem.

There are many variables with many equations.

RE: Room Pressurization

(OP)
Thanks Willard3.

Ok here is an example if you can please help me to solve it.

Given:

- room area 28m x 20 x 6m high
- 12 air change
- Tight room
- Positive Pressure: 7 Pa

Required:
- Capacity of Supply Fan.
- Capacity of Exhaust Fan

Thanks

RE: Room Pressurization

How many doors? What is the size of each door? The pressure drop is equal to an entrance coefficient x the velocity pressure of air leaking through the opening. So do you have an estimate of what the total size of the openings? What is the flow coefficient through a sharp edge orifice?

RE: Room Pressurization

(OP)
Thanks Lilliput1,

The room has one door 0.8m width x 2m height. Please consider no air leaking.

Thanks.

RE: Room Pressurization

From experience 200 CFM to 400 CFM per door is required to pressurize space to 0.001 in wg
7 Pa = 0.0281305 in wg

Using the 400 CFM as basis:

0.0281305 = .001 x (CFM2/400)^2

CFM2 = 2122 CFM

However the 400 CFM is base on a typical hospital isolation room say max 800 SF vs 6028 SF for you case.
Assuming additional leakage of 200 CFM per additional 800 SF

Add 200 x (6028-800)/800 = 1307

So approximate CFM required is 2122 + 1307 = 3429 say 3500.

You can not say there is no other leakage. I recommend you hire a balancing contractor to measure supply air, return air, exhaust air and space pressurization in a similar room and prorate their findings to get 7 Pa similar to my calculation above. Make sure you coordinate with the architect to specify that all opening be sealed - gaskets at light fixtures, switches, receptacles. Seal opening around piping, walls, conduit, ductwork, etc. Put door closer and brush type door sweep at door bottom. Provide differential pressure alarm with time delay.

RE: Room Pressurization

Saying you have a tight room is not a quantity.

RE: Room Pressurization

Gee,
Here I thought you were pressurizing the room because of flammable gases or other hazards, but it's only a dairy processing room. During my university days I used to work in a dairy, and I can't think of a single reason why pressurization would be essential. Air changes: Yes -- pressurization: No.
Did anyone give you a reason why pressurization is required? (ie to keep bugs, dust, etc out)

If pressurization is required, I would think that an air handling unit (for supply of MU air, heating and/or AC), together with a relief damper set at (say) 7Pa would be more than adequate. The trick here is to ensure that the room is tight enough for the MU fan to maintain 7Pa.

How much MU air is required for this facility? I suspect it will be several thousand cfm. If this is the case I am sure that the MU air fan should be able to maintain 7Pa even if the room is not that tight.
Regards,
GG

ps Only one man-door in a room of 560sq-m? I find that difficult to believe.

RE: Room Pressurization

Lilliput1, I didn't understand the experience rules exactly, do you mean that 0.001 in wg will cause a 400 cfm to leak through the door, I mean if we have 1000 cfm supply then we need to return 600 cfm to the AHU and 400 through the door?
and what does cfm2= 2122 mean in your example calculation, does it mean supply cfm or return cfm?

RE: Room Pressurization

Pi/P2 = (CFM1/CFM2)^2

Solve the above equation to get CFM2 = 2122

So the room will not be a fire trap you should have at least 2 doors.

Also come to think of it I am light in adding only 200 CFM per 800 SF additional floor area because this is based on 0.001 1n wg pressurization, not .0281305 in wg

Again using the above equation to correct the 1307 cfm based on .001 in wg to .0281305 get CFM additional = ((1307)^2 x .0281305)/.001)^0.5 = 6932

Total cfm correcting for (2) doors and the additional floor area beyond 800 SF = 2 x 2122 + 6932 = 11,176 CFM

See also last page of attached article showing authors real world experience.

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