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Mechanical room ventilation

  • Monday, April 03, 2017 5:45 PM
    Message # 4710319

    A new replacement hospital looking to put CO2, N2, and N2O manifolds in a small room.  


    I was looking at requirements for mechanical ventilation, and want to use this installation to see if I am fully understanding this requirement, as I haven't had cause to use it recently.  


    Room to contain connected cylinders only:

    24 N2 @ 255 cu' = 6120 cu'

    24 N2O @ 567 cu' = 13608 cu'

    12 CO2 @ 568 cu' = 6816 cu'


    Total 26,544 cu' of gas in use.


    9.3.6.5.3.2 Mechanical exhaust shall be at a rate of 1 CFM of airflow for each 5 cubic feet of liquid designed to be stored in the space and not less than 50 CFM nor more than 500 CFM.


    26544/5 = 5308.  


    My understanding is that the 500 CFM max rate would apply even though gas connected is over 10X noted.


    Is it correct to state that anything over 2500 cubic feet can be ventilated by a 500 CFM system?


    Thanks in advance...

    Bob W.


  • Tuesday, April 04, 2017 6:07 AM
    Reply # 4711100 on 4710319

    Bob,

    Your conclusion turns out to be right, even if your method is not quite.

     

    Remember 9.3.6.5.1 - you are only required to use the largest collection of interconnected vessels for the calculation.  The total quantity of gas stored in the room is not relevant any longer.  

     

    In your example, I assume you have three manifolds: a 12x12 N2, a 12x12 N2O and a 6x6 CO2.

     

    The largest collection of interconnected vessels here is one side of one of these (the manifold controls can be taken to "interrupt" the interconnection for the purpose of this calculation).  Clearly, that is the Nitrous, at 12 x 567 = 6804.  6804/5=1360.  So I need a "motivator" (fan, blower or suchlike) with a capacity of 1360 cuft/min, but since the rule in 9.3.6.5.3.2 is that I never need more than 500 cuft/min, that is where I end up.

     

    Yes, 2500 cuft does equal 500 cuft/min, but be careful how you get there!

  • Tuesday, April 04, 2017 6:00 PM
    Reply # 4712335 on 4710319

    Thanks Mark for the response.  I am glad I asked the question.


    Bob W.

  • Wednesday, April 05, 2017 6:59 AM
    Reply # 4714936 on 4710319
    Steve Bradshaw (Administrator)

    If the limiting factor is the maximum flow of 500 CFM though an exhaust duct then the intent could be to install multiple exhausts which add up to the required exhaust flow.  I hope Keith Ferrari comments - I believe he spoke with someone who was knowledgeable regarding the intent of this ventilation code.

  • Wednesday, April 05, 2017 9:11 AM
    Reply # 4717079 on 4710319
    Steve Bradshaw (Administrator)

    Based on the words "common manifold" in this code, I think many AHJs would base the the exhaust flow on the whole manifold.


    Reference:  NFPA 9.3.7.5.1* For the purposes of this section, the volume of fluid (gas and liquid) to be used in determining the ventilation requirements shall be the volume of the stored fluid when expanded to standard temperature and pressure (STP) of either the largest single vessel in the enclosed space or of the entire volume of the connected vessels that are on a common manifold in the enclosed space, whichever is larger.

  • Wednesday, April 05, 2017 4:45 PM
    Reply # 4718033 on 4710319
    Al Moon (Administrator)

    well omg / I fully understood the requirement for determining the volume of gas within the room ( based on the section # 9.3.7.5.1 that steve listed ):

    But when it can to exhaust rate / volume, I was lost. So the answer can always be 500. I just need to pay attention to the make up air. Is that right?

     

    Last modified: Wednesday, April 05, 2017 4:48 PM | Al Moon (Administrator)
  • Monday, April 10, 2017 8:20 AM
    Reply # 4739384 on 4710319
    Corky Bishop (Administrator)

    10 connected H cylinders of Oxygen or 5 connected G cylinders of Nitrous Oxide would require a 500 cfm fan.

     

    I was very disappointed when this was changed from a dedicated exhaust to being allowed to connect to other area exhausts in the building in 9.3.6.5.3.5.  I believe the wind could overcome an exhaust fan and dispel a broken cylinder lead on a full bottle of Nitrogen or Nitrous Oxide into the nearest restroom.

     

    The remainder of the paragraph says provided that the system does not connect to spaces that contain combustible or flammable materials. Are designers supposed to take methane gas and toilet paper into account for this exclusion?  Doubtful anyone would take it to that extreme.

  • Tuesday, April 11, 2017 6:02 AM
    Reply # 4742866 on 4717079
    Steve Bradshaw wrote:

    Based on the words "common manifold" in this code, I think many AHJs would base the the exhaust flow on the whole manifold.

     

    Reference:  NFPA 9.3.7.5.1* For the purposes of this section, the volume of fluid (gas and liquid) to be used in determining the ventilation requirements shall be the volume of the stored fluid when expanded to standard temperature and pressure (STP) of either the largest single vessel in the enclosed space or of the entire volume of the connected vessels that are on a common manifold in the enclosed space, whichever is larger.


    Steve,

    I think you are correct, AHJ's are rarely trained enough in medical gas to see the fine distinction between largest volume of connected vessels and "whole manifold", but that would be a misinterpretation.  Since the concern is how much gas could reasonably be expected to discharge at one time, the worst case as a practical matter would be something like a pigtail breaking and a single bank emptying.  Failure inside the common controls would of course be worse, as it could discharge both banks, but that is why we have all the relief valves piped to outside.

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