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Lecture Notes on Fluid Dynamics (1.63J/2.21J) by Chiang C. Mei, 2002
6.5 Geothermal Plume
6-5g-plume-L.tex May 11, 2003
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 k
u = −
∂p ∂x
where k denotes the permeability, and
μ k
w = −
∂p ∂z
− ρg (6.5.2)
These are the momentum equations for slow motion in porous medium. Mass conservation requires ux + wz = 0 (6.5.3)
Energy conservation requires
u
∂T
∂x
∂T
∂z
= α
∂^2 T
∂x^2
∂^2 T
∂z^2
where
α =
K
ρ 0 C
denotes the thermal difusivity. Equation of state: ρ = ρ 0 (1 − β(T − T 0 )) (6.5.6) Consider th flow induced by a strong heat source. Let
T − T 0 = T ′, p = po + p′
where p 0 is the hydrostatic pressure satisfying
−
∂p 0 ∂z
− ρ 0 g = 0.
Eqn. (6.5.2) can be written μ k
w = −
∂p′ ∂z
6.5.1 Boundary layer approximation
Eliminating p′^ from Eqns. (6.5.7) and (6.5.1), we get
μ k
(wx − uz) = gρ 0 βT (^) x′.
Let ψ be the stream funciton such that
u = ψz , w = −ψx
then
ψxx + ψzz = −
gρ 0 βk μ
T (^) x′ (6.5.8)
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,
u w,
∂x
∂z
hence
ψxx ∼= −
ρ 0 βk μ
T (^) x′
or
ψx ∼= −
gρ 0 βk μ
T ′, (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′ ∂z
outside the plume, hence ∂p′/∂z ≈ 0 inside as well. Applying the B.L. approximation to Eqn. (6.5.4)
uT (^) x′ + wT (^) z′ = αT (^) xx′ (6.5.10)
Using the continuity equation we get
(uT ′)x + (wT ′)z = αT (^) xx′.
Integrating across the plume, ∂ ∂z
−∞
wT ′^ dx = 0 (6.5.11)
since T ′^ = 0 outside the plume. It follows that
ρoC
−∞
wT ′^ dx = −ρ 0 C
−∞
ψx T ′^ dx = Q = constant. (6.5.12)
6.5.3 Similarity solution
Now let x = λax∗^ z = λbz∗^ ψ = λcψ∗^ θ = λdθ∗.
From Eqn. (6.5.17)
λc−a
∂ψ∗ ∂x∗
= −λdθ∗.
For invariance we require, c − a = d. (6.5.22)
From (6.5.19)
−
∂ψ∗ ∂x∗^
dx∗^ λc−a+a+d^ = 1.
therefore, a + d = 0. (6.5.23)
From Eqn. (6.5.18) λc+d−a−b^ = λd−^2 a.
implying, c + a − b = 0. (6.5.24)
Finally
c =
a 2
, d = −
a 2
, b =
a.
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
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)
and 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
df dη
= h (6.5.29)
and d dη
(f h) = 3
d^2 h dη^2
The boundary conditions are,
f = 0 (ψ = 0) f ′′(0) = 0 , (w(0, z) = wmax) f (±∞), f ′(±∞) = 0 h(±∞) = 0.
Integrating Eqn. (6.5.30), we get
f h = 3h′.
Using Eqn. (6.5.29), we get f f ′^ = 3f ′′.
Integrating again, we get − 6 f ′^ = f 02 − f 2
where f 0 = fmax. Let f = −f 0 F , then
f 0 (1 − F 2 ) = 6F ′, or
dF 1 − F 2
f 0 dη 6
which can be integrated to give f 0 η 6
ln
1 + F
1 − F
Thus (^) ( 1 + F 1 − F
= ef^0 η/^6
or (^) ( 1 + F 1 − F
= ef^0 η/^3
Solving for F , we get
F =
ef^0 /^3 − 1 ef^0 /^3 + 1
= tanh
f 0 η 6
What is f 0? Let us use Eqn. (6.5.29)
−∞
df dη
h dη =
−∞
(f ′)^2 dη = 1
since
f ′^ = −f 0 F ′^ = −
f 02 6
sech^2
f 0 η 6
and h = −f ′.
Figure 6.5.1: Theoretical solution for a geothermal plume due to Yih
From (6.5.36), we get
H B^2
W
α
α
Qgβ CB
It follows that
H B^3 /^2
α
Qgβ C
Now ψ¯ z ¯^1 /^3
ψ W B
( (^) z H
H^1 /^3
W B
ψ z^1 /^3
Figure 6.5.2: Comparison of theory and experiment. From Nield and Bejan
It can be shown that
H^1 /^3 W B
Qgβ C
H
B^3 /^2
α^1 /^3
C
Qgβ
which depends on the fluid properties and the given heat source strength. Also z¯^1 /^3 θ = h(η) = (Hz)^1 /^3 ∆T T ” = (H^1 /^3 ∆T )z^1 /^3 T ′^ (6.5.41)
We can show that
H^1 /^3 ∆T =
ν
gβC
α
Qgβ C
Q^1 /^6
ν(αgβ)^1 /^3 C^2 /^3
