Thanks for the responses. As David mentioned it was the volume^1.87 that was throwing me for a loop. As Rconner suggested the 7.57 coefficent likely includes the cg term and would allow for my unit balance.
When calculating pressure drop in long pipes I've used the following formula.
dp = 7.57*(q^1.85)*(L)*(10^4)) / ((d^5)*p) (1)
where
dp = pressure drop (kg/cm2)
q = air volume flow at atmospheric conditions (FAD) (m3/min)
L = length of pipe (m)
d = inside diameter of pipe (mm)...
