EQUAL FRICTION METHOD EXAMPLEDuct system for general office ;

Total air quantity5400 cfm18 air terminals300 cfm eachOperating pressure for all terminals-0.15 in.wgRadius elbowsR/D = 1.25

Find ;Initial duct velocity, area, size and friction rate in the ductTotal equivalent length of duct run, with highest resistance Total static pressure required at Fan discharge

DUCT LAYOUT

SELECT AN INITIAL VELOCITYRECOMMENDED MAX DUCT VELOCITIES FOR LOW VELOCITY SYSTEMS

Velocity dipilih 1700 fpm

Duct area : 5400 cfm/1700 fpm = 3.18 sq.ft

Duct size (TABLE 6)

Duct size : 22 in x 22 in

Circular equiv. diameter

Initial friction rate

Friction rate ; 0.145 in wg per 100 ft of eauivalent length

Percent section area in branches

Percent of cfm = air quantity in duct section/ total air quantity

It appears that the duct run from the Fan to terminal 18 has the highest resistance

Friction of rectangular Elbow

The total friction loss in the ductwork frfom the Fan to terminal 18 :

Loss= total equiv length x friction rate

= 229 ft x (0.145/100 ft) = 0.332 in. wg

Total pressure at Fan DischargeThe sum of the terminal operating pressure and the loss in the ductwork

0.332 in. wg + 0.15 in.wg = 0.482 in.wg

Friction losses