Page 276 - Materials Chemistry, Second Edition
P. 276
Groundwater Remediation 259
Example 6.18: Determine the Required Injection
Pressure of Air Sparging
Three air-sparging wells were installed into the plume of the aquifer
described in Example 6.17. The injection air flow rate into each well is 5 ft /
3
min. The height of the water column above the air injection point is 10 ft. The
aquifer matrix consists mainly of coarse sand. Determine the minimum air
injection pressure required. Also, for the purpose of comparison, determine
the air injection pressure if the aquifer formation is clayey.
Solution:
(a) Use Equation (6.33) to convert the water-column height to pres-
sure units as:
lb m ft lb f
= 62.4
P hydrostatic =ρg h hydrostatic 32.2 (10ft) 2
ft 3 s 2 32.2 lb m − ft/s
= 624 lb f = 4.33 lb f = 4.33 psi
ft 2 in 2
Note: The density of water at 60°F is 62.4 lb /ft . In other words,
3
m
the specific weight of water is 62.4 lb /ft . The water-column
3
f
height of 33.9 ft at 60°F is equivalent to a pressure of 1 atm or
14.7 psi.
(b) From Table 2.2, pore radius of fine-sand media is 0.05 cm. Use
Equation (2.11) to determine the height of capillary rise:
0.153 0.153
= = = 3.06 cm = 0.1 ft
h c
r 0.05
Use the discussion in part (a) to convert the capillary rise to the
capillary pressure:
0.1 ft
= (14.7 psi) = 0.04 psi
P capillary
33.9 ft
(c) Use Equation (6.31) to determine the minimum air injection
pressure:
+
= 4.33 0.04 = 4.37 psig
P injection = P hydrostatic + P capillary
(d) If the aquifer formation is clayey, then the pore radius is 0.0005
cm (from Table 2.2). Use Equation (2.11) to determine the height
of the capillary rise:
0.153 0.153
= = = 306 cm = 10 ft
h c
r 0.0005