Page 101 - Materials Chemistry, Second Edition

84 Practical Design Calculations for Groundwater and Soil Remediation

The specific capacity, defined by Equation (3.12), can also be used to esti-
mate the hydraulic conductivity of an unconfined aquifer.

Example 3.10: Steady-State Drawdown from Pumping
an Unconfined Aquifer

A water-table aquifer is 80 ft (24.4 m) thick. Groundwater is being extracted
from a 4-in. (0.1 m)-diameter fully penetrating well.
The pumping rate is 40 gpm (0.15 m /min). The aquifer is relatively sandy,
3
with a hydraulic conductivity of 200 gpd/ft . Steady-state drawdown of 5 ft
2
(1.5 m) is observed in a monitoring well 10 ft (3.0 m) from the pumping well.
Estimate:
• The drawdown 30 ft (9.1 m) away from the well
• The drawdown in the pumping well

Solution:
(a) First let us determine h (at r = 10 ft):
1
1
h = 80 − 5 = 75 ft (or = 24.4 − 1.5 = 22.9 m)
1
Use Equation (3.14):

2
2
(200)( h 2 − 75 )
40 = ⇒ h 2 = 75.7 ft
1,055 log(30/10)
or
2
2
1.366 [(200)(0.0410)]( h 2 − 22.9 )
(0.15)(1440) = ⇒ h 2 = 23.1 m
log(9.1/3.0)
So drawdown at 30 ft (9.1 m) away = 80 − 75.7 = 4.3 ft (or = 24.4 − 23.1
= 1.3 m)
(b) To determine the drawdown at the pumping well, set r at the
well = well radius = (2/12) ft

2
2
(200)( h 2 − 75 )
40 = ⇒ h 2 = 72.5 ft
1,055 log[(2/12)/10]
So, drawdown in the extraction well = 80 − 72.5 = 7.5 ft

Discussion:
1. In the equation for confined aquifers, the (h − h ) term can be
2
1
replaced by (s − s ), where s and s are the drawdown at r and
2
1
1
2
1
r , respectively. However, no analogy can be made here, that is,
2
(h − h ) cannot be replaced by (s − s ).
2
2
2
2
1
2
2
1