Page 186 - Materials Chemistry, Second Edition

Vadose Zone Soil Remediation 169

(d) The vapor flow rate at the exhaust of the extraction pump can be
calculated from Equation (5.7):
 P well   0.9 
Q atm =   P atm   Q well =   1   (0.24)

3
3
= 0.216 m/min= 7.6 ft /min
Discussion:
Using consistent units in Equation (5.6) is very important. In the cal-
culations shown here, the pressure is expressed in N/m , the dis-
2
tance in m, the permeability in m , and the viscosity in N/s/m .
2
2
Consequently, the calculated velocity is in m/s.
Example 5.11: Estimate the Extracted Vapor Flow
Rate of a Soil-Venting Well
A soil-venting well (4-in. diameter) was installed at a site. The pressure in the
extraction well is 0.9 atm, and the radius of influence of this soil-venting well
has been determined to be 50 ft. From Example 5.10, the radial Darcy veloc-
ity right outside the well casing was determined as 177 m/day. Calculate the
radial Darcy velocity at 20 feet away from the center of the venting well by
using Equation (5.4).

Solution:
(a) The radial air flow velocity at 20 ft away from the extraction well
can be found by using Equation (5.4):

 (0.9)(1.013 10 )  1 ( ) 2 
5
×
1
 10 − 12   [(20)(0.3048)] ln[(2/12)/50]  − 0.9 
×
u r =  − 5  0.5
 2(1.8 10 )×  1+  1 ( ) 2  ln[20/(2/12)]
−
1
 0.9  ln[(2/12)/50]
×
÷
−
3
8
= (2.78 10 )( 2.64 10 )( 0.2346)(1.197) 0.5
−
×
−
−
×
= 1.574 × 10 m/s=9.44 10 − 4 m/min= 1.36 m/day
5

(b) For comparison, the radial air flow velocity can also be found as
Q = (2πr H)v = (2πr H)v (assuming that the gas is incompress-
1
2
2
1
ible and one-dimensional radial flow):
Thus r v = r v
2 2
1 1
(2/12 ft)(177 m/day) = (20)(v )
2
v = radial Darcy velocity 20 ft away
2
= 1.48 m/day