Room Pressure Build Up

[DM2]

I'm trying to determine what pressure will be exerted on the walls of a room after a known quantity of air is introduce into the space. I've been advised what the leakage rate is but thats not important at this stage. I'm trying to convince someone that they need pressure relief venting for the space.

If the room is 15,000 cubic feet and the air is being released from 10 cylinders, each having a capacity of 500 cubic feet (uncompressed) of air, what's the formula to determine the room pressure in psi?

Regards,
Dan

 

[zekeman]

14.7*(15000+10*500)/15000

 

[desertfox]

Hi DM2

Use the gas law:-

p1*V1/T1 =p2*V2/T2

where p1 & p2 = pressure

V1 & V2 = volume of room and cylinder/s

T1 & T2 = temperature of gas in room and cylinder

If you know the pressure of gas in the cylinder and the temperature of gas in room and inside cylinder are the same
the above reduces to:-

p1*V1 = p2*V2

therefore if p1 is the room pressure then

p1= p2*V2/V1

regards

desertfox

 

[zeusfaber]

You can do this pretty much by eye

1 atm is about 15 psi, so for a 15000 Cu Ft room, each additional Cu Ft is going to give you about a 1 psi overpressure.

10 x 500 Cu Ft is 5000, so expect a pressure rise of up to 5 psi.

That's an awful lot of air to let out in a hurry - so unless the compartment is airtight, or you're planning to burst the cylinders, the actual pressure obtained is going to depend on the relationship between the leak and fill rates.

A.

 

[byrdj]

less than 1psid will blow off an inspection door when unbolted.

and 2"H20 (0.1 psid) will make a door very hard to open

 

[DM2]

Thanks for the help everyong.

Just a little insite...This is an Inert Fire Suppression system, of which there are 3 different types:
1 - Inergen, marketed by Ansul, consists of Argon, Nitrogen & CO2.
2 - Argonite, marketedd by Kidde, Consists of Argon & Nitorgen.
3. Argotech, marketed by Minimax (i think), Consists of Argon.
4. N100, Marketed by Nomi in Japan, Consists of Argon.

The concept is to displace the existing volume of air, with it's 20.9% concentration of oxygen, with this agent which will reduce the oxygen content to below 15%.

Each one of these cylinders hold about 500 cubic feet of gas. In general terms, the amount of gas needed is between 32% and 40%.

Most people are familiar with Halon 1301 systems, which didn't increase the room pressure enough to require the use of pressure relief vents. Some of the other agents (FM-200) have only recently issued warnings regarding pressure relief venting and this only because a brick wall was damaged during a discharge.

My current situation is with a building manufacture (MCC Switchgear building) who seem to think that his building doesn't need any venting and that he walls will handle a 1psi horizontal force (i.e. 144lbs/ft2).

I'm sending him a video on what happens to rooms when these suppression systems discharge and there is insufficient venting. It's amazing what a little, er...a lot of pressure can do to a door frame, door, wall, etc.

Thanks again for the help.
Dan.

 

[Compositepro]

I saw a system like this abandoned in place after it was installed because all the permits and requirements were not fully determined before hand. Compliance with all the requirements would not be economical. People can be in the room and suffocated. Alarms, procedures, emergency ventillation were required.

 

[rb1957]

then the building manufacturer is saying he needs to vent, as we're anticipating a 5psi increase (and he thinks the building is good for 1psi).

also wouldn't you have to be very carefull with these vents, to ensure they're venting the O2 rich air (and not the suppressant) ... and you would want these vents venting flame either ...

 

[DM2]

Compositepro -
With the Ansul Inergen System, the Carbon Dioxide in the agent stimulates the respiratory system, which allows you to remain in the protected space.

rb1957 -
We recommend that the vents be placed on the ceiling. When the system discharges, the air within the space is displaced while the pressure forces the vent, or damper open, until the pressure is below the opening force of the damper, which is usually weighted or has a spring return. Testing has demonstrated that this is sufficient.

The advantage of Inert agents is:
1. The agent doesn't decompose in the presents of high temperatures. Halon 1301 would experience thermal decomposition at about 1200°F. Many of the Halocarbon gases experience thermal decomposition at about 900°F. This is why the NFPA 2001 standard, as well as NFPA 12A (Halon 1301 standard) mandated a 10 second discharge, where as with Inert agents a 1 minute discharge is permitted. The objective is to get the agent in the space and develop an extinguishing concentration before the fire has a chance to compromise the extinguishing properties of the agent.
2. Inert agents have a vapor density similar to air. Halon 1301 and the newer clean agents (i.e. FM-200, Novec, etc.) have a vapor density substantially higher than air and therefore tend to "Fall" and find openings in the protected space to leak out. For this reason is far more important to have fairly air tight room for these agents so the extinguish concentration will remain in the protected space. Testing demonstrated that it's often difficult with Halon and the new Halocarbon agents to remain in the space for more than about 10 minutes. With Inert agent I've seen people walk in and out of doors to the space and the concentration remains for 30 minutes or better.
3. The cost to install a Inert agent is much higher than Halocarbon agents due to hardware costs, however the cost to recharge is about $0.26 per cubic foot of protected space compared to $0.85 with Halocarbon agents. One (1) discharge and the cost of the Inert agent becomes cheaper (Lower longterm cost of ownership).

Again Thanks for the help.
Dan

 

[zekeman]

You have to give us the composition of the gases in the cylinders before we can determine the excess pressure. We all assumed (erroneously) from you initial description that the gas was air. If your gas is primarily nitrogen, then the solutions are close. All the others will be different from the 5psi we estimated.

 

[DM2]

The mixture of agent is called "Inergen".

I consists of a 572 cubic foot cylinder with the following mixtures:
1 - Nitrogen 52%
2 - Argon 40%
3 - Carbon Dioxide 8%

Physical properties of Inergen:
1 - Specific Gravity: 0.085 lbs/ft3, (1.36 kg/m3)
2 - Vapor Pressure:
a) 1925 psi @ 32°F (132.7 bar @ 0°C)
b) 2175 psi @ 70°F (149.9 bar @ 21°C)
c) 2575 psi @ 130°F (177.5 bar @ 54°C)
3 - Vapor Density: 1.1 (Air = 1)
4 - Approximate molecular weight: 34

 

[dcasto]

The zfactor would be the only change. It is .945, so we are off 5%.

 

[DM2]

I'm not sure what you mean by "Z Factor"

Can you enlighten me?

Dan

 

[zekeman]

I took another look and find that the 5 psi is correct as well as the calculation. The pressure change is in proportion to the number of moles, regardless of the mixture and 5000 SCF added to 15000 SCF increases the the mole content by 1/3.
So, I stand corrected.

 

[quark]

Z is compressibility factor. BTW, why did you say "uncompressed"(is it free volume, i.e reduced to atmospheric pressure, you are talking about)?

The differences in density can be used to effectively vent the oxygen from the room.

 

[DM2]

Quark
Yes, I was referring to the volume of Inergen gas after it leaves the cylinder (i.e. free volume).

With regard to density, how would I go about using the different density to separate the oxygen from the other gases. The difference in density is only 10% (i.e. density of Inergen as a mixture of all 3 gases).

The Inergen gas is discharged through a piping network and nozzles placed at the ceiling of the protected space (usually about 40' apart). The pressure in the cylinder is 200 Bar but passes through a pressure reducing orifice to allow for the use of thinner wall pipe, thereby keeping the installation cost down.

Regards
Dan

 

[quark]

First, I will try to calculate what quantity should be vented. The atmospheric pressure is 14.696psi and if your room is designed for 1psig then the room pressure can be 15.696psia maximum. Then, the room can accomodate (15.696/14.696)*15000 = 16021 cu.ft (under isothermal conditions). You have to vent 5000-1021 = 3979cu.ft. If it takes 10 minutes for 5000cu.ft air to be discharged into the room, then your venting rate will be 500 cfm which can be started 2 minutes after suppressant is released into the room.

To arrive at an opening size, you should use Q = 4005*A*dp^1/2. So, A (sq.ft) = 500cfm/(4005)*(27.68inches WC)^1/2 = 0.024 sq.ft

I am not sure about the standard way of doing it but, ideally, I would go for bottom entry of the supressant as it is denser than air. The venting will be done from the highest point in the room. The sudden expansion of the supressant from the inlet pipe should be enough reason to push the room air to the top.

 

[DM2]

Hey...Thanks...

It appears i forgot to post some information. According to the building codes, Chapter 16 (structural) states that the minimum horizontal force a wall should withstand is 5lbs/ft2 (0.035psi). This is typically what we assume when calculating the vent area.

The discharge time for Inert gas systems (i.e. this system) is 1 minute as opposed to 10 minutes. 10 minutes is the preferred hold time for the concentration.

I'm assuming I can simply change those values in your example and I should have the results...right?

Dan.

 

[quark]

Yes. So, extra volume is 35.72cu.ft. Vent (5000-35.72)in one minute. So, A = 1.26sq.ft. I would go for a bit extra area here, particularly, if you want to have multiple openings (to account for coefficient of discharge). Use a factor of 0.75 (i.e 1.26/0.75) which is about 1.68sq.ft

 

[DM2]

Hey...You guys are great!

Thanks for the help!

Dan.