1
Lecture Notes on Fluid Dynamics
(1.63J/2.21J)
by Chiang C. Mei, 2002
6.5
Geothermal Plume
6-5g-plume-L.tex
R..A.Wooding, (1963), J Fluid Mech. 15, 527-544.
C. S. Yih, (1965), Dynamics of Nonhomogeneous Fluids, Macmillan.
D. A. Nield and A. Bejan, (1992), Convection in Porous Media. Springer-Verlag.
Consider a steady, two dimensional plume due to a source of intense heat in a porous
medium. From Darcy’s law:
µ
∂p
u=−
(6.5.1)
k
∂x
where k denotes the permeability, and
µ
∂p
w = − − ρg
(6.5.2)
k
∂z
These are the momentum equations for slow motion in porous medium. Mass conservation
requires
(6.5.3)
u x + wz = 0
Energy conservation requires
∂T
∂T
u
+w
=α
∂x
∂z
where
α=
∂2T
∂2T
+
∂x2
∂z 2
K
ρ0 C
(6.5.4)
(6.5.5)
denotes the thermal difusivity.
Equation of state:
ρ = ρ0 (1 − β(T − T0 ))
(6.5.6)
Consider th flow induced by a strong heat source. Let
T − T0 = T ,
p = p o + p
where p0 is the hydrostatic pressure satisfying
−
Eqn. (6.5.2) can be written
∂p0
− ρ0 g = 0.
∂z
µ
∂p
w=−
+ gρ0 βT .
k
∂z
(6.5.7)
2
6.5.1
Boundary layer approximation
Eliminating p from Eqns. (6.5.7) and (6.5.1), we get
µ
(wx − uz ) = gρ0 βTx .
k
Let ψ be the stream funciton such that
u = ψz ,
w = −ψx
then
gρ0 βk (6.5.8)
Tx
µ
For an intense heat source, we expect the plume to be narrow and tall. Let us apply the
boundary layer approximation and check its realm of validity later,
ψxx + ψzz = −
u w,
hence
∂
∂
.
∂x
∂z
ρ0 βk T
ψxx ∼
=−
µ x
or
gρ0 βk T,
ψx ∼
=−
µ
(6.5.9)
which is the same as ignoring ∂p /∂z in Eqn. (6.5.7).
This can be confirmed since u w ∂p /∂x ≈ 0, p inside the plume is the same as that
outside the plume. But
∂p
=0
∂z
outside the plume, hence ∂p /∂z ≈ 0 inside as well.
Applying the B.L. approximation to Eqn. (6.5.4)
uTx + wTz = αTxx
(6.5.10)
Using the continuity equation we get
.
(uT )x + (wT )z = αTxx
Integrating across the plume,
∂
∂z
∞
wT dx = 0
since T = 0 outside the plume. It follows that
∞
∞
wT dx = −ρ0 C
ψx T dx = Q = constant.
ρo C
−∞
(6.5.11)
−∞
−∞
(6.5.12)
3
6.5.2
Normalization
Let us take
WB
(6.5.13)
ū, w = W w̄, T → ∆T θ
H
where H, B, ∆T and W are to be determined to get maximun simplicity. We then get from
the momentum equation,
gρ0 β∆T
w̄ = ψ̄x̄ = −
θ,
µW
x = B x̄, z = H z̄, u =
from the energy equation,
ūθx̄ + w̄θz̄ =
αH
θx̄x̄ ,
W B2
and from the total flux condition,
ρ0 CW B∆
∞
w̄θdx̄ = Q
−∞
Let us choose
gρ0 β∆T
=1
µW
(6.5.14)
αH
=1
W B2
(6.5.15)
ρ0 CW B∆T = Q,
(6.5.16)
and
which gives three relations among four scales, B, H, W, ∆T . Then
w̄ = ψ̄x̄ = −θ,
(6.5.17)
ūθx̄ + w̄θz̄ = θx̄x̄ ,
(6.5.18)
from the energy equation,
and from the total flux condition,
∞
w̄θdx̄ = 1
(6.5.19)
−∞
In addition we require that
w(±∞, z) = 0, θ(±∞, z) = 0
∂w(0, z)
= 0, x = 0.
∂x
From here on we omit overhead bars in all dimensionless equations for brevity.
u(0, z) =
(6.5.20)
(6.5.21)
4
6.5.3
Similarity solution
Now let
x = λa x∗
z = λb z ∗
From Eqn. (6.5.17)
λ
c−a
∂ψ ∗
∂x∗
ψ = λc ψ ∗
θ = λd θ ∗ .
= −λd θ∗ .
For invariance we require,
c − a = d.
From (6.5.19)
−
(6.5.22)
∂ψ ∗ ∗ c−a+a+d
dx λ
= 1.
∂x∗
therefore,
a + d = 0.
(6.5.23)
From Eqn. (6.5.18)
λc+d−a−b = λd−2a .
implying,
c + a − b = 0.
(6.5.24)
Finally
a
3
a
c = , d = − , b = a.
2
2
2
In view of these we introduce the following similarity variables,
η=
x
z 2/3
, ψ = z 1/3 f (η), θ = z −1/3 h(η).
(6.5.25)
Note that at the center line η = 0
and
w = −ψx ∝ z 1/3 f (0)(−)z −2/3 ∼ z −1/3 f (0) ∼ z −1/3
(6.5.26)
θ ∝ z −1/3 h(0)
(6.5.27)
b ∝ z 2/3
(6.5.28)
Thus the velocity and temperature along the centerline decay as z −1/3 and the plume width
grows as z 2/3 .
Substituting these into Eqns. (6.5.17) and (6.5.18), we get, after some algebra
−
and
df
=h
dη
d
d2 h
(f h) = 3 2 .
dη
dη
(6.5.29)
(6.5.30)
5
The boundary conditions are,
f
f (0)
f (±∞), f (±∞)
h(±∞)
=
=
=
=
0
0,
0
0.
(ψ = 0)
(w(0, z) = wmax )
Integrating Eqn. (6.5.30), we get
f h = 3h .
Using Eqn. (6.5.29), we get
f f = 3f .
Integrating again, we get
−6f = f02 − f 2
where f0 = fmax . Let f = −f0 F , then
f0 (1 − F 2 ) = 6F , or
f0 dη
dF
=
2
1−F
6
which can be integrated to give
f0 η
1 1+F
= ln
6
2 1−F
Thus
1+F
1−F
or
1/2
1+F
1−F
= ef0 η/6
= ef0 η/3
Solving for F , we get
F =
ef0 /3 − 1
f0 η
= tanh
f
/3
0
e
+1
6
What is f0 ? Let us use Eqn. (6.5.29)
∞
∞
df
−
h dη =
(f )2 dη = 1
dη
−∞
−∞
since
f = −f0 F = −
f0 η
f02
sech2
6
6
and
h = −f .
(6.5.31)
6
Therefore,
f02
6
2 ∞
sech
f0 η
6
4
−∞
Since
∞
f3
dη = 0
6
∞
sech4 ζdζ = 1.
−∞
sech4 zdz = 4/3.
−∞
we get f0 !
The solution is
and
1/3
9
f0 =
2
(6.5.32)
1/3
1/3
η
9
9
tanh
f=
2
2
6
(6.5.33)
2/3
1/3
9
9
η
2
sech
h = −f = −
2
2
6
(6.5.34)
Computed results are given in Figures.
RemarkChecking the boundary layer approximation.
∂2ψ
∼ z −1 ,
∂x2
∂2ψ
∼ z −5/3
∂z 2
∂2T ∂2T −5/3
∼
z
,
∼ z −7/3
∂x2
∂z 2
hence for large z, B. L. approximation is good.
6.5.4
Return to physcial coordinates
Start from
η=
x̄
z̄ 2/3
(6.5.35)
= f (η)
(6.5.36)
z̄ 1/3 θ = h(η)
(6.5.37)
ψ̄
z̄ 1/3
Then
x/B
η=
=
(z/H)2/3
H 2/3
B
x z 2/3
By eliminating H and ∆T from(6.5.35) and (6.5.37), we get
Qgβ
W =
CB
(6.5.38)
7
From (6.5.36), we get
H
1
W
=
=
2
B
α
α
It follows that
H
1
=
3/2
B
α
Now
ψ̄
z̄ 1/3
Qgβ
CB
Qgβ
C
1/3 H
ψ
ψ z −1/3
=
=
WB H
WB
z 1/3
(6.5.39)
(6.5.40)
8
It can be shown that
1
H 1/3
=
WB
Qgβ
H
B 3/2
1/3
1
= 1/3
α
C
Qgβ
1/3
C
which depends on the fluid properties and the given heat source strength.
Also
z̄ 1/3 θ = h(η) = (Hz)1/3 ∆T T ” = (H 1/3 ∆T )z 1/3 T We can show that
1 1
H 1/3 ∆T = √
ν gβC
1/3
1 Qgβ
Q1/6
=
α
C
ν(αgβ)1/3 C 2/3
(6.5.41)
(6.5.42)
which also depends on the fluid properties and the given heat source strength.