Say the drop is 10 PSI: 10/.433=23.1 ft.(h)
v=(2gh)^.5=38.6 ft/sec Area of 2" pipe=((2/24)^2)*PI=.0218 ft^2 38.6 ft/sec * .0218 ft^2 = .84 ft^3/sec, * 7.48 * 60 = 378 GPM. This would be the maximum. It would less proportional to the friction head loss. If it is a short drop to a 90 and right out the wall maybe not too much Pf. Regardless of any 'visual feel' one might get, if System Demand is 400 GPM, one would have to go to plan B in this case. thanks, Brad Casterline, NICET IV Fire Protection Division FSC, Inc. P: 913-722-3473 [email protected] www.fsc-inc.com Engineering Solutions for the Built Environment _______________________________________________ Sprinklerforum mailing list [email protected] http://lists.firesprinkler.org/listinfo.cgi/sprinklerforum-firesprinkler.org
