CE 473/573 Groundwater
Fall 2007
Details in the analyses of wells
For steady flow in a leaky confined aquifer with azimuthal symmetry
1 d
r dr
dh
e
r
+ = 0,
dr
T
where e = −K ′ (h − h0 )/b′ is the leakage, K ′ is the hydraulic conductivity of the confining
layer, b′ is the thickness of the confining layer, and h0 is the head outside the confined
aquifer. Inserting this definition of the leakage and expanding the first term give
d2 h 1 dh
K′
+
+
(h0 − h) = 0.
dr 2
r dr
Kbb′
Call λ2 = Kbb′ /K ′ and use the drawdown s = h0 − h:
d2 s 1 ds
s
+
− 2 = 0.
2
dr
r dr λ
Put r ′ = r/λ. The chain rule gives
ds dr ′
1 ds
ds
= ′
=
.
dr
dr dr
λ dr ′
Then
d2 s
1 ds
−s=0
2 + ′
′
r dr ′
dr
or
d2 s
2
′ ds
r
− r ′ s = 0,
2 +r
′
′
dr
dr
which is the modified Bessel’s equation of order zero.
′2
For steady flow in a confined aquifer with azimuthal symmetry and no leakage
1 ∂
S ∂s
=
T ∂t
r ∂r
∂s
r
.
∂r
In class we argued that we should seek a solution of the form
s
=f
Q0 /T
Sr 2
Tt
.
Use the chain rule to compute the derivatives
∂s
=
∂t
∂s
=
∂r
∂s
=
r
∂r
1 ∂
∂s
r
=
r ∂r
∂r
Then
Q0 df ∂u
Q0 Sr 2 df
Q0 df
=−
=
−
u
T du ∂t
T T t2 du
T t du
Q0 df ∂u
2Q0 Sr df
=
T du ∂r
T T t du
2Q0 df
u
T du
2Q0 d
df ∂u
2Q0 S d
df
u
=
u
T du
du ∂r
T T t du du
d
df
df
u
+u
=0
du du
du
Put g = udf /du. Then separating and integrating give
g=u
df
= C1 e−u .
du
Integrate again (using x as a dummy variable)
f = C2 − C1
Z
∞
u
e−x
dx
x
Since f (∞) = 0, C2 = 0. Then since the flow into the well must be Q0 , C1 = −1/4π.
Finally,
Z ∞ −x
e
Q0
Q0
s=
dx =
W (u),
4πT u
x
4πT
where W (u) is the well function.