Transcript Document
Flow to a Well
Importance of Wells
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Extraction
Control of salt-water intrusion
Pump and treat
Injection and pumping
Disposal of hazardous waste
Fig. 7.22, Manning
Drawdown depends on:
• K
• T
• S
Basic Setup
Fetter, Figure 5.2
Assumptions
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Confining layer on bottom
Geologic formations are horizontal and infinite
Potentiometric surface: horizontal, static before test
Changes to potentiometric surface due to pumping
Homogeneous, isotropic
Flow is horizontal
Darcy’s law is OK
Constant density and viscosity
Fully penetrating wells
Well has infinitesimal diameter
Two Ways to Go
• Forward solution
– Use equations to predict drawdown
– K, T, S are known
• Backward solution
– Measure heads
– Calculate K, T, S
Radial Symmetry
• Replace x and y with r and q
• q doesn’t matter: drops out of equation
Steady Radial Flow in a Confined Aquifer
h2
dh
Q 2rbK
dr
dh
2rT
dr
h1
b
r2
r1
Steady Radial Flow in a Confined Aquifer
dh
Q 2rT
dr
Q dr
dh
2T r
h2
h1
Q
dh
2T
r2
r1
dr
r
Q
ln r2 ln r1
h2 h1
2T
h2
h1
Q
r2
h2 h1
ln
2T r1
b
r
2
r1
Q
r2
T
ln
2 h2 h1 r1
Steady Radial Flow in an Unconfined Aquifer
dh
Q 2rhK
dr
h2
h1
r2
r1
Steady Radial Flow in a Confined Aquifer
dh
Q 2rhK
dr
h2
Q dr
hdh
2K r
h1
r2
r1
h2
h1
Q
hdh
2K
r2
r1
dr
r
1 2
Q
r2
2
h2 h1
ln
2
2K r1
Q
r2
K
ln
2
2
h2 h1
r1
Unsteady Flow in a Confined Aquifer
h h S h
2
2
x
y
T t
2
2
In polar coordinates:
2 h 1 h S h
2
r
r r T t
Boundary Conditions:
h(r,0) = h0
dh
At r=0, Q 2rT
dr
Unsteady Flow in a Confined Aquifer
2
r S
u
4Tt
Q
h0 h
4T
Q
h0 h
4T
u
u
e
u
u2
u3
u4
...
0.5772 ln u u
2 2! 3 3! 4 4!
We call
u
u
e
u
the well function.
W (u )
u
u
e
u
The Well Function
Fetter, Figure 5.6
The Well Function
2
Q
h0 h
W (u )
4T
r S
u
4Tt
Q
T
W (u )
4 h0 h
4Ttu
S 2
r
The Well Function
Q
h0 h
W (u )
4T
Q
logh0 h log
logW (u )
4T
r 2S
u
4Tt
r 2S 1
t
4T u
r 2S
1
log
log t log
u
4T
Drawdown vs. Time
Fetter, Figure 5.7
Curve Matching
Fetter, Figure 5.8
Jacob Drawdown Method
Q
h0 h
4T
u
u
u
...
0.5772 ln u u
2 2! 3 3! 4 4!
2
3
4
When u gets small,
Q
0.5772 ln u
h0 h
4T
2
Q
r S
h0 h
ln1.78 ln
4T
4Tt
Q
h0 h
4T
4Tt
Q
ln 1.78r 2 S 4T
2.25Tt
ln r 2 S
Jacob Drawdown Method
Q
h0 h
4T
2.25Tt
ln r 2 S
ln x 2.3 log x
2.3Q 2.25Tt
h0 h
log 2
4T r S
2.3Q 2.25T 2.3Q
h0 h
log 2
logt
4T r S 4T
Jacob Drawdown Method
2.3Q 2.25T 2.3Q
h0 h
log 2
logt
4T r S 4T
0
1
Drawdown
2
D(h0-h)
3
4
5
6
0
1
2
log Time
3
Jacob Drawdown
At 0 drawdown, t = t0
2.3Q 2.25T 2.3Q
0
log 2
logt0
4T r S 4T
2.25T
log r 2 S logt0
2.25T 1
2
r S
t0
2.25Tt0
S
r2
Fetter, Figure 5.9
Distance-Drawdown Method
2.3Q 2.25Tt
h0 h
log 2
4T r S
2.3Q 2.25Tt 2.3Q 1
h0 h
log
log 2
4T S 4T r
2.3Q 2.25Tt 2.3Q
log r
h0 h
log
2
4T S 4T
Fetter, Figure 5.10
Leaking Confining Beds
Fetter, Figure 5.3
I. Nonleaky. Straight line: typical decay.
As pumping affects larger and larger area,
the rate of drawdown goes down.
Fetter, Figure 5.4
II. Leaky, w/o storage from confining layer.
As drawdown increases in confined unit,
head difference increases between
unconfined and confined units. Eventually,
drawdown stops. Aquifer may appear
nonleaky for hours or days.
Fetter, Figure 5.4
IV. Like II, but with additional water from
storage in the confining unit. Drawdown is
less at early times, but storage effect
ceases as system reaches steady state.
Fetter, Figure 5.4
Fetter, Figure 5.3
h 1 h e S h
2
r
r y T T t
h0 h
e K'
b'
2
Fetter, Figure 5.3
Assume:
•Water table elevation initially equal to head in aquifer
•Water table doesn’t drop during test
•No loss in storage from confining unit
Hantush-Jacob Formula
Q
r
h0 h
W u,
4T B
2
r S
u
4Tt
Tb'
B
K'
0.5
Fetter, Figure 5.11
Also see Appendix 3
Hantush-Jacob Formula
W(u, r/B)
Tb'
B
K'
Fetter, Figure 5.11
0.5
Curve Matching to Hantush-Jacob
Fetter, Figure 5.12
Hantush-Jacob Formula
Y-axis displacement gives you W(u, r/B) and (h0-h)
Calculate T
X-axis displacement gives you u and t
Calculate S
r / B r /Tb' / K '
0.5
Tb' r / B
K'
r2
2
Fetter, Figure 5.3
Assume:
•Water table elevation initially equal to head in aquifer
•Water table doesn’t drop during test
•Water initially comes from storage in the confining unit
Hantush Solution
Q
h0 h
H u ,
4T
2
r S
u
4Tt
r S'
4B S
0.5
Tb'
B
K'
Fetter, Figure 5.14
0.5
Hantush Solution
Y-axis displacement gives you W(u, ) and (h0-h)
Calculate T
X-axis displacement gives you u and t
Calculate S
r S'
4B S
0.5
16 2Tb' S
K' S'
r2