Geoscience Reference
In-Depth Information
to be depth-dependent, which seems reasonable. Substitution into Eq. (7.13) (with
A
=
0) then yields
d
2
V
d
z
2
d
W
d
t
k
ρ
c
P
W
V
=
(7.39)
which, upon rearranging, becomes
1
W
d
W
d
t
k
ρ
c
P
1
V
d
2
V
d
z
2
=
(7.40)
Because the left-hand side of this equation is a function of
z
alone and the
right-hand side is a function of
t
alone, it follows that each must equal a constant,
say,
c
1
.However, substitution of Eq. (7.38) into the boundary conditions (i) and (ii)
yields, respectively,
W
(
t
)
=
e
i
ω
t
(7.41)
and
V
(
z
)
→
0 s
z
→∞
(7.42)
Boundary condition (i) therefore means that the constant
c
1
must be equal to i
ω
(differentiate Eq. (7.41)tocheck this). Substituting Eq. (7.41) into Eq. (7.40)gives
the equation to be solved for
V
(
z
):
d
2
V
d
z
2
i
ωρ
c
P
V
k
=
(7.43)
This has the solution
V
(
z
)
=
c
2
e
−
qz
+
c
3
e
qz
(7.44)
where
q
=
(1
+
i)
ωρ
c
P
/
(2
k
) (remember that
√
i
=
(1
+
i)
/
√
2) and
c
2
and
c
3
are
constants. Equation (7.37), boundary condition (ii), indicates that the positive
exponential solution is not allowed; the constant
c
3
must be zero. Boundary
condition (i) indicates that the constant
c
2
is
T
0
; so, finally,
T
(
z
,
t
)isgivenby
T
(
z
,
t
)
=
T
0
exp(i
ω
t
)exp
−
(1
+
i)
ωρ
z
c
P
2
k
ωρ
c
P
2
k
ωρ
c
P
2
k
=
T
0
exp
z
exp
i
z
−
ω
t
−
(7.45)
For large
z
this periodic variation dies out. Thus, temperatures at great depth are
unaffected by the variations in surface temperatures, as required by boundary
condition (ii).
At a depth of
2
k
ωρ
c
P
L
=
(7.46)
the periodic disturbance has an amplitude 1
/
eofthe amplitude at the surface. This
depth
L
is called the skin depth. Taking
k
=
2.5Wm
−
1
◦
C
−
1
,
c
P
=
10
3
Jkg
−
1
◦
C
−
1
and
ρ
=
2.3
×
10
3
kg m
−
3
,which are reasonable values for a sandstone, then for the
daily variation (
ω
=
7.27
×
10
−
5
s
−
1
),
L
is approximately 17 cm; for the annual
