I am assuming you are talking a light hazard small room that meets the requirements of a small room. In small rooms, you take the area of the room divided by number of heads. That is how I was taught. You would require 4 heads in this room as a minimum. It is 742.5 sq ft. 742.5 / 3 = 247.5. So, you are going to need to space so that you are 9' from the wall in the 16'-6" dimension, but you will need to be 11.25' on center in the 45' dimension if equally spaced.
8.6.2.1.2 The requirements of 8.6.2.1.1 shall not apply in a small room as defined in 3.3.15; the protection area of coverage for each sprinkler in the small room shall be the area of the room divided by the number of sprinklers in the room. > Hello and Merry Christmas, > > Maybe I am tired or something, but I am more confused now than before > for rooms with multiple heads in them. > > If you have a rectangular room 16-6 wide and 45 feet long... You are > spaced 7-6 to head and 9-0 to opposite wall in the narrow dimension; > what is the maximum spacing between the heads going the length of the > room. > And due to lights the spacing will be greater than 45' divided by 4 > heads. > > Thanks, > > Greg McGahan > > > _______________________________________________ > Sprinklerforum mailing list > [email protected] > http://fireball.firesprinkler.org/mailman/listinfo/sprinklerforum > > For Technical Assistance, send an email to: [email protected] > > To Unsubscribe, send an email to:[email protected] > (Put the word unsubscribe in the subject field) > -- _______________________________________________ Sprinklerforum mailing list [email protected] http://fireball.firesprinkler.org/mailman/listinfo/sprinklerforum For Technical Assistance, send an email to: [email protected] To Unsubscribe, send an email to:[email protected] (Put the word unsubscribe in the subject field)
