Page 238 - Materials Chemistry, Second Edition

Groundwater Remediation 221

• Hydraulic conductivity of the aquifer = 400 gpd/ft 2
• Steady-state drawdown
= 2.0 ft in a monitoring well that is 5 ft from the pumping well
= 1.2 ft in a monitoring well that is 20 ft from the pumping well

Solution:
Inserting the data into Equation (6.1), we obtain

Kb ( 2 h − h ) (400)(30)(2.0 − 1.2)
Q = 1 = = 30.2 gpm
528 log( /)rr 1 528log(20/5)
2
Discussion:
The h –h term can be replaced by s –s , where s and s are the draw-
1
2
1
2
1
2
down values at r and r , respectively.
2
1
6.2.1.2 Steady-State Flow in an Unconfined Aquifer
The equation describing the steady-state flow from a fully penetrating well
in an unconfined aquifer (water-table aquifer) can be expressed as:
Kh 2 − h 2 )
( 2
Q = 1 (AmericanPractical Units)
2 rr
1,055 log( /)
1
1.366 Kh 2 − h 2 )
= ( 2 1 (SIUnits) (6.2)
2 rr
log( /)
1
All the terms are as defined for Equation (6.1).
Example 6.3: Steady-State Drawdown from Pumping
an Unconfined Aquifer
A water-table aquifer is 40-ft (12.2-m) thick. Groundwater is being extracted
from a 4-in. (0.1-m) diameter fully penetrating well.
The extraction rate is 40 gpm (0.15 m /min). The aquifer is relatively sandy
3
and has a hydraulic conductivity of 200 gpd/ft . Steady-state drawdown of 5 ft
2
(1.5 m) is observed in a monitoring well at 10 feet (3.0 m) from the pumping well.
Determine
(a) The drawdown in the pumping well
(b) The radius of influence of the pumping well

Solution:
(a) First let us determine h (at r = 10 ft):
1
1
h = 40 − 5 = 35 ft (or = 12.2 − 1.5 = 10.7 m)
1