Page 98 - Materials Chemistry, Second Edition

Plume Migration in Aquifer and Soil 81

For example, if a well produces 50 gpm and the drawdown in the well
is 5 ft, the specific capacity of this pumping well is 10 gpm/ft (i.e., it will
produce 10 gpm for each foot of available drawdown). A rough estimate on
transmissivity (in gpd/ft) can be obtained by multiplying the specific yield
(in gpm/ft) by 2,000 for confined aquifers and 1,550 for unconfined aquifers
[2]. The hydraulic conductivity (in gpd/ft ) can then be determined by divid-
2
ing the transmissivity with the aquifer thickness (in ft).

Example 3.7: Steady-State Drawdown from Pumping a Confined Aquifer
A confined aquifer (30 ft or 9.1 m in thickness) has a piezometric surface 80
ft (24.4 m) above the bottom confining layer. 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 (3.10):
(200)(30)( h 2 − 75)
40 = ⇒ h 2 = 76.7 ft
528 log(30/10)
or
2.73 [(200)(0.0410)](9.1)( h 2 − 22.9)
(0.15)(1, 440) = ⇒ h 2 = 23.4 m
log(9.1/3.0)

So drawdown at 30 ft (9.1 m) away = 80 − 76.7 = 3.3 ft (or = 24.4 −
23.4 = 1.0 m)
(b) To determine the drawdown at the pumping well, set r at the
wellbore = well radius = (2/12) ft:
(200)(30)( h 2 − 75)
40 = ⇒ h 2 = 68.7 ft
528 log[(2/12)/10]
So, drawdown in the extraction well = 80 − 68.7 = 11.3 ft