Page 185 - Materials Chemistry, Second Edition

168 Practical Design Calculations for Groundwater and Soil Remediation

Calculate the steady-state flow rate entering the well per unit well screen
length, vapor flow rate in the well, and the vapor rate at the extraction pump
discharge by using the following additional information:

• Permeability of the formation = 1 darcy
• Well screen length = 20 ft
• Viscosity of air = 0.018 cP
• Temperature of the formation = 20°C

Strategy:
We need to perform a few unit conversions first:
1 atm = 1.013 × 10 N/m 2
5
1 darcy = 10 cm = 10 m 2
−12
−8
2
1 poise = 100 centipoise = 0.1 N/s/m 2
(So, 0.018 centipoise = 1.8 × 10 poise = 1.8 ×10 N/s/m )
−5
2
−4
Solution:
(a) The velocity at the wellbore, u , can be found by using Equation
w
(5.5):
 k      P RI  2 
u w =   2µ    R ln( P w w R I    1−   P w     
 
R /)
w
 110 − 12   (0.9)(1.013 10 )   1  2 
×
5
×
=  − 5    1−   
×
 2(1.8 10 )   [(2/12)(0.3048)] × ln[(2/12)/50]   0.9  
−
−
8
−
×
= (2.78 10 )( 3.15 10 )( 0.2346)
5
×
×
= 2.05 10 m/s= 0.123 m/min= 177 m/day
3
−

(b) The vapor flow rate entering the well per unit screen interval can
be found by using Equation (5.6):
Q w = 2π
w
H Ru w
3
= 2π[(2/12)(0.3048)m](0.123 m/min) = 0.039 m/min/m

(c) The vapor flow rate in the well = (Q /H) × H = (0.039 m /min/m)
3
w
[(20 ft)(0.3048 m/ft)]
= 0.24 m /min = 8.4 ft /min
3
3