If a light hazard room is 13 feet x 13 feet, and a k5.6 pendent sprinkler is located seven feet from two of the walls, what is the sprinkler area used for hydraulic calculations?
In short, to determine the minimum flow from a sprinkler in multiplying the area by the required density, is the actual area of the room used (13'-0" x 13'-0" = 169 sqft) or is it computed the same as the maximum area of coverage for a sprinkler by doubling the furthest distance from adjacent walls [(7'-0" x 2) x (7'-0" x 2) = 196 sqft)]?
For a room like this with a K5.6 sprinkler k-factor, it would make a difference between delivering 19.6 gpm to a room (0.1 gpm/sqft x 196 sqft) and 16.9 gpm to a room (0.1 gpm/sqft x 169 sqft).
Posted anonymously for discussion. Discuss This | Submit a Question | Subscribe
Subscribe and learn something new each day:
Top April '22 Contributors
Get 100 Days of Free Sample Questions right to you!
PE PREP SERIES