10" of vacuum is not very much, 1atm is 9.8m of water or 385" of water so a 10" vacuum is (385-10)/385 or 97.4% of normal atmospheric pressure, so 1000 (s)cfm ( or standard cubic feet per minute) is only 1000/.974 = 1026 (a)cfm (or actual cubic feet per minute) at the specified reduced pressure. How much air 10" of vacuum will move is then dependent of the size of the fan and the ducting, as Howard alludes to. For example most gas systems run with a gas velocity or 15-25 m/s, to transport 1026 acfm at a velocity of 20m/s you would require
1026/35.29/60 = 0.48m3/s
0.48 / 20 = 0.24m2 (x-sectional area of ducting)
square root of (0.24 * 4 / 3.142) * 100 = 17.4cm diameter of ducting (c. 6")
using
http://www.engineeringtoolbox.com/press ... d_852.html you can calculate the specific pressure drop that will be experienced. Based on 1000cfm air, 3.28' (1m) length, 6" id duct, and 14.3psi (97.4% of 1atm 14.7psi) you get a pressure drop of .00631psi.
.00631 psi = .00631 / 14.7 * 385 = .17" of water.
So your system could have up to 10/.17 = 58m of equivalent 6" ducting before you would exceed the 10" of available suction and then flow would reduce (note fittings, bends, gates etc all have an equivalent lenght, ie a 90degree bend is equivalent to 5-8 pipe diameters)
For interest the same flow in a 4" pipe has a specific presure drop of .047 psi or 1.2"/meter, so you would see flow reductions after only 8m of ducting. Pressure drop at equivalent volumetric rate is a function of d^5! You see the need for large duct diameter in a HVLP system!
Regards
Fitz (woodworker by night, chemical and process engineer by day)