Page 395 - APPLIED PROCESS DESIGN FOR CHEMICAL AND PETROCHEMICAL PLANTS, Volume 1, 3rd Edition

Ejectors and Mechanical Vacuum Systems 363

Mixture avg mo! wt = 26.2/0.603 =43.4 Example 6-7: Total Weight of Mixture
Molecular weight correction (From Figure 6-18) = 1.18
Calculate the total weight of mixture to be handled
when evacuating 25 pounds per hour of air from 28.5
26 2 inches vacuum with a mixture temperature at 80°F.
Air equivalent (at 80°F) = · = 22.2 lbs/hr
1.18 Barometer = 30 inches Hg.
Temperature correction (Figure 6-17, using air curve) = 0.999 Absolute pressure = 30.0 - 28.5 = 1.5 in. Hg abs
70°F air equivalent for mixture = 22.2/0.999 = 22.2 lbs/hr
P,. = 1.034 in. Hg abs (water vapor at 80°F)
Pa = 1.50 - 1.034 = 0.466 in. Hg abs (air)
This is the value to compare with a standard manufac- Wm = 25 + 0.62(25) (1.034)/0.466 = 59.4 lbs mixture/hr
turer's test or performance curve at 70°F.
The air bleed is used to maintain a constant condition.
However, a control valve may be used instead. Control or Total Volume of a Mixture
hand valves in the lower pressure vapor lines to an ejector
are not recommended, as they must be paid for in system The total fixed volume of a mixture of gases and vapors
pressure drop and ejector utility requirements. at a given condition is the same as the volume of any one
component (gas laws), and its pressure is composed of the
sum of the individual partial pressures of each component.
Non-Condensables Plus Water Vapor Mixture
This is also a frequent process situation. To determine Example 6-8: Saturated Water Vapor-Air Mixture
the 70°F air equivalent, the non-condensable are deter-
mined as in Example 6-5 and the water vapor as in Exam- An air-water vapor mixture is saturated with water
ple 6-2. The total for the mixture is the sum of these two vapor at 80°F and 2 in. Hg abs total pressure. The air in
values. the mixture is 60 lbs/hr. Determine the volume at these
conditions.
Total Pressure of System at Suction to Ejector
Vapor pressure water at 80°F = 1.034 in. Hg abs
P = P n + P'vt + P',2 + · · · (6-3) Partial pressure air= 2.0 - 1.034 = 0.966 in. Hg abs
Weight of mixture:
Air-Water Vapor Mixture Percent Curves Wm = 60 + 0.62(60) (1.034)/0.966 = 99.8 lbs/hr
Weight of water vapor = 39.8 lbs/hr (from above)
Volume of air under its condition in the mixture:
For saturated air-water vapor systems, Figures 6-20A, B,
C, and D are useful in solving for the pounds of water
vapor per pound of air (Dalton's Law, Equation 6-2). Vol of air = vol of mixture = V
WRT . H
, use 111. g (6- 5)
70. 73 Pa
Example No. 6-6: Use of Water Vapor-Air Mixture
Curves Gas constant, R = 1544/MW = 1544/29 = 53.3 (6-6)

A system handles 50 lbs/hr of air that is saturated with 5 53 3 460 80
+
· ) (
water vapor at 3 inch Hg abs and 95°F. Find the total V = 0( 70.73 (0.966) ) = 25, 300 cu ft/hr
amount of water vapor.
= 422 cu fl/min at 80°F
For 3 inch Hg abs and 95°F saturation, the fraction of
water vapor from Figure 6-20 is 0. 77 lbs water vapor /lb air. As a check:
From steam tables, specific volume of water vapor at
Total water vapor= (50) (0.77) = 38.5 lbs/hr 80°F and 1.034 in. Hg abs = 633.8 cu ft/lb
Total mixture = 50 + 38.5 = 88.5 lbs/hr
Volume = (39.8) (633.8) = 25,200 cu fl/hr
Weight of Air and Water Vapor Mixture = 421 cu ft/min
As an alternate method, total mols could be calculated
" = \" + 0.62 (W .)(P") and converted to volume at 80°F and 2 mm Hg abs.
''V m 'Va (6- 4)
(text continued on pag, 366)