# 4.4 Buoyant plume from a steady heat source ```1
Notes on
Instructor: C. C. Mei, 2002
4-4 buoyplum.tex
4.4
Buoyant plume from a steady heat source
[Reference]:
Gebhart, et. al. (Jalluria, Maharjan, Saammakia),
Buoyancy-induced Flows and Transport, 1988,
Hemisphere Publishing Corporation
Let T̃ = T T = temperature variation where Ti nf ty is a constant (no ambient
stratiﬁcation). For a strong enough heat source, we expect boundary layer behavior,
(
(p (
&Agrave;
, u &Agrave; v,
=0
(r
(x
(r
The boundary layer equations are
((ru) ((rv)
+
=0
(x
(r
(4.4.1)
&Atilde;
&quot; (
(u
(u
(u
u
r
+v
= g#(T T ) +
(x
(r
r (r
(r
&Atilde;
( T̃
( T̃
k (
( T̃
u
+v
=
r
(x
(r
r (r
(r
!
(4.4.2)
!
(4.4.3)
The centerline r = 0 is an axis of symmetry,
v=
(u
( T̃
=
=0
(r
(r
(4.4.4)
Far outside the plume r u 0 and T T , (T̃ 0)
(4.4.5)
Rewrite (4.4.3) as
((ruT̃ ) ((rv T̃ )
+
T̃
(x
(r
&Atilde;
((ru) ((rv)
+
(x
(r
&Atilde;
((ruT̃ ) ((rv T̃ )
(
( T̃
=
+
=k
r
(x
(r
(r
(r
!
!
(4.4.6)
2
after using continuity. Now integrating the last equation from r = 0 to r =
Z
( Z
((rv T̃ )
dr
2\$ruT̃ dr + 2\$
(x 0
(r
0
&Atilde;
!
Z
(
( T̃
= k2\$
dr
r
(r
(r
0
Therefore
&macr;
&Atilde;
&macr;
( Z
( T̃
&macr;
2\$ruT̃ dr + 2\$rv T̃ &macr;&macr; = 2\$k r
(x 0
(r
0
!r=
(4.4.7)
r=0
Using the boundary conditions, we get or
Z
0
2\$ruT̃ dr = constant
Note that
Z
0
2\$rdr u'C T̃ = rate of buoyancy ﬂux
= rate of heat ﬂux
= Q(given rate of heat release at x = 0)
Therefore,
Z
Q=
2\$rdr'uC T̃
0
(4.4.8)
This is a boundary condition.
Let the stream function be deﬁned by
ru =
(
,
(r
rv = (
(x
(4.4.9)
(4.4.1) is automatically satisﬁed. From the momentum equation:
&Atilde;
1 (
r (r
!
1 (2
1 ( (
r (x(r r (x (r
&Atilde;
1 (
r (r
!
&quot;
&quot; (
(
= g# T̃ +
r
r (r (r
&Atilde;
1 (
r (r
!#
(4.4.10)
From the energy equation
&Atilde;
1 ( ( T̃
1 ( ( T̃
k (
( T̃
=
r
r (r (x
r (x (r
r (r
(r
!
(4.4.11)
and from the buoyancy ﬂux condition
Z
Q = 2\$'C
0
&Atilde;
!
1 (
rdr
T̃
r (r
(4.4.12)
3
Try a similarity solution with the one-parameter transformation
x a x ,
From (4.4.10),
r = b r ,
= c ,
T̃ = d T 2c4ba = 2c4ba = d = c4b
(4.4.13)
c+d2ba = d2b
(4.4.14)
c+d = 1
(4.4.15)
from (4.4.11)
and from (4.4.12)
From these three equations we get
b
1 d
c
= 1,
= ,
= 1.
a
a
2 a
Th similarity variable can be taken to be
=
r
x1/2
(4.4.16)
and the similarity solution to be
= xF ()
(4.4.17)
T̃ = x1 G()
(4.4.18)
and
After a lot of algebra, and noting
(
1
(
1 r
1 r 1
= 1/2 ,
= 3/2 = 1/2 = (r
x
(x
2x
2x x
2x
we get from (4.4.10)
&Atilde;
F0
&quot;F +
000
and from (4.4.11)
!0
(F &quot;) + g#G = 0
k(G0 )0 + (F G)0 = 0
(4.4.19)
(4.4.20)
.
Before integrating, let us normalize :
= &macr;
, F = F̄ , G = Ḡ.
(4.4.21)
It follows from (4.4.19) that
&quot; 000
F̄ + 3
3
&Atilde;
F̄ 0
&macr;
!0
( F̄ &quot;) + g#&macr;
Ḡ = 0
(4.4.22)
4
where prime denotes d/d&macr;
. Setting = &quot; and
&quot;2
= g#
3
which relates and ,
=
we get
&Atilde;
F̄ 0
F̄ +
&macr;
000
&quot;2
g#4
(4.4.23)
!0
(F̄ 1) + &macr;Ḡ = 0
(4.4.24)
Similar normalization of (4.4.20) gives
k
0 0
(F̄ Ḡ)0 = 0
(&macr;
Ḡ
)
+
2
(4.4.25)
(&macr;
Ḡ0 )0 + P r(F̄ Ḡ)0 = 0
(4.4.26)
. which can be simpliﬁed to
. where
&quot;
= Prandtl Number
(4.4.27)
k
For water &quot; = 102 cm2 /s, k = 1.42cm2 /s, hence P r = 7. For air &quot; = 0.145cm2 /s, k =
0.202cm2 /s, hence P r = 0.75.
We now integrate (4.4.26)to give
Pr =
&macr;Ḡ0 + Pr F̄ Ḡ = constant
Since (x, 0) = 0, we must have F̄ (0) = 0 ; the constant above is zero.
&macr;Ḡ0 + Pr F̄ Ḡ = 0
(4.4.28)
Equation (4.4.28) can be written
F̄
F̄
Ḡ0
d ln Ḡ
= Pr , or
= Pr
Ḡ
&macr;
d&macr;
&macr;
Z
ln Ḡ = Pr
0
&macr;
F̄
d&macr;
+ constant
&macr;
&Atilde;
Ḡ(&macr;
) = Ḡ(0) exp Pr
Z
0
&macr;
!
F̄
d&macr;
&macr;
(4.4.29)
Substituting Eqn. (4.4.29) into Eqn. (4.4.24), the resulting equation for F̄ must be integrated
numerically.
Now let us ﬁnd the boundary condtions for F or F̄ .
5
Eqn. (4.4.8 ) becomes
&Atilde;
Z
Q
1 (
=
dr r
2\$'C
r (r
0
!
Z
G() Z
G Z
r
=
dr x1/2 F 0 =
d(F 0 G) = &quot;
d&macr;
(F̄ 0 Ḡ)
x
r
x
0
0
0
(4.4.30)
Therefore,
Z
0
d&macr;
F̄ 0 Ḡ =
Let us choose
Q
2\$'C&quot;
(4.4.31)
Q
=1
2\$'C&quot;
so that
Z
(4.4.32)
d&macr;
F̄ 0 Ḡ = 1
0
(4.4.33)
is the boundary condition for F̄ and Ḡ. Now (4.4.32) deﬁnes , the scale of G. Note that
larger Q implies larger and smaller . Thus a stronger heat source leads to a greater
centerline temperature and a thinner plume. Also,
u 0 as r hence
F0
1
&quot; F̄ 0
= 2
0, as &macr; u = r =
r
&macr;
The radial velocity is, in general
&Atilde;
1
F0
1
F v = x =
r
r
2
!
Since
v0
as 0,
we must have,
F (0) = 0.
Clearly
F̄ 0
0, and F̄ (&macr;
) = 0 as &macr; 0
&macr;
(4.4.34)
The numerical results by Mollendorf &amp; Gelhart, 1974, are shown in Figs. 4.4.1, for
various Prandtl numbers. A schlierian photograph due to Gebhart (copied from Van Dyke
An Album of Fluid Motion) is hown in Figure ﬁg:plumeVD.
Remark:
&micro;
&para;
1 (
1
F0
x
u=
=
= F 0 1/2
r (r
r x
6
&sup3;
0
&acute;
Along the centerline u(x, 0) = F = constant depending on Pr . Why? Buoyancy acceler0
ation is counteracted by entrainment.
Remark: Let the radius of the plume be a which varies as
a x1/2
This is consistent with the behavior that u x0 , and T̃ x1 , since
a2 uT̃ = Q
On the other hand the mass ﬂux rate is
ua2 x
and the momentum ﬂux rate is
u2 a2 x
hence both approach zero at the source. Thus a plume is the result of energy source, not of
mass or momentum.
```