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Room Design Method 1

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SpcFoor

Mechanical
Feb 7, 2011
36
I have a project that I am currently working on and have submitted to both the ahj and architect. The ahj has approved my system and we have started to intall the systems. There were some minor details I over looked in the inital design and once they were pointed out I took care of them. On the inital submittal calcs I used the area/density method. However since some other details are now available I would like to change one of the three calcs. to the room design method. This is a school with a two story classroom and single story kitchen and gym. I have calced the kitchen and gym and the second floor. The second floor is the area where I would like to do the room design method. It has an art room and the opposite end of the building away from the riser. This room has nine heads in it and the most any other room has is six. What do you all think?
 
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I have calced all three areas and the hazard is light in the area where I want to do the room design method
 
What is the ceiling height? Assuming light hazard and you are more than 100 sq ft / head, then you have at least 900 sq ft. If the ceiling is 10' or less, you can just use the QR reduction and calc 900 sq ft. Depending on your spacing and such, the 900 sq ft may be less area than your room design method.

Be sure with room design that you have self closing doors, or you have to get up to 2 heads in the adjacent connecting rooms.

Travis Mack
MFP Design, LLC
 
I have a ceiling that is open from the 1st floor to the second floor which puts it at 23-0. So I cant do the 900 sq ft. All rooms in the adjacent rooms have two heads and or are small enough that is doesnt require two heads.
 
But, if you don't have protected openings as defined in NFPA 13, then you have to take heads in those adjacent rooms as well. As long as you meet all of the criteria to use the room design method, there is no reason that you can't.

Travis Mack
MFP Design, LLC
 
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