Page 188 - Materials Chemistry, Second Edition

Vadose Zone Soil Remediation 171

(b) The radius of influence can be found by using Equation (5.6):

0.004
Q w = = 0.001
H 4

 π k   P w    P atm  2 
=   µ     ln( R / )   1−   P w     
R I   
w
 π(10 − 12 )   (0.85)(1.013 10 )   1  2 
5
×
=   1.8 10   ln(0.05/)  1−   0.85    
5  
−
×

R I
R I = 16.04 m
Discussion:
Using consistent units is critical for successful calculations in this
example. Specifically, the flow rate is given in m /min, but it needs
3
to be converted to m /s to match the viscosity units in Equation (5.6).
3

Example 5.13: Estimate the Time to Flush One Pore Volume
A soil-venting well was installed in the middle of a plume, and the vapor
extraction rate was found to be 20 ft /m. Assume that the well created a per-
3
fect radial flow with a radius of influence of 50 ft and a thickness of 20 ft. Find
the time needed to flush one pore volume of the capture zone. The porosity
of the subsurface is 0.4, and the volumetric water content is 0.15.

Solution:
(a) The volume of the zone captured by this extraction well = π(R ) H
2
I
= (π) × (50) × (20) = 157,100 ft 3
2
(b) The volume of the air void = (volume of the soil)(ϕ ) = (volume of
a
the soil)(ϕ − ϕ )
w
= (157,100)(0.4 – 0.15) = 39,270 ft 3
(c) Time required to flush one pore volume = (void volume) ÷ (air
flow rate)
= (39,270) ÷ 20 = 1,960 min

Discussion:
1. The flow rate of 20 ft /m is the measured flow rate at the surface.
3
The actual flow rate in the subsurface should be slightly higher
because it is under vacuum.