of 18
Chapter 11
THE PLANE PLUME
11.1 Generalities
Plumes are an important subject in civil and environmental engi-
neering, because of the frequent need to dispose of waste heat and/or
combustion products. In a uniform ambient fluid at rest, similarity
solutions exist for round and plane plumes for both laminar and tur-
bulent flow. The new property of such flows is that momentum is
continuously added to the fluid, so that the velocity on the center-
line increases, or at least does not decrease, with increasing distance
from the source. If the initial momentum flux is not negligible, the
flow may behave like a jet for a time, and then like a plume, as the
acquired momentum begins to dominate the motion. The problem of
a plume in a stably stratified ambient fluid in also important. Finally,
the plume in a crossflow is an even more difficult problem than the
jet in a crossflow, a problem already discussed in section x.
11.1.1
Dimensional preamble
Plumes are usually treated within the Boussinesq approximation,
which assumes that variations in density can be ignored everywhere
except in the energy equation and in the driving buoyancy term in
the momentum equation. In the engineering literature, the approx-
559
560
CHAPTER 11.
THE PLANE PLUME
imation is often stated as if it were self-evident. In what follows, a
more rigorous argument will be attempted.
Recall the result of taking the limit
M
→
0 in the equations
of motion for a perfect gas in SECTION 1.2.3. The equations of
continuity, momentum, energy, and state have the form
Dρ
Dt
+
ρ
div
~u
= 0
(11.1)
ρ
D~u
Dt
=
−
grad
p
+
ρ
~
F + div
τ
(11.2)
ρc
p
DT
Dt
=
−
div
~q
+
ρQ
(11.3)
ρ
T = constant
(11.4)
where the last two equations assume a calorically and thermally per-
fect fluid (
h
=
c
p
T
and
p
=
ρRT
, respectively, with
c
p
and
R
con-
stant). To these are added Newton’s hypothesis for viscous stress and
Fourier’s hypothesis for heat flow,
τ
=
μ
(grad
~u
+ grad
~u
∗
) =
μ
def
~u
(11.5)
~q
=
−
k
grad T
(11.6)
where the two constants
μ
and
k
are assumed to depend only on the
state of the fluid.
The steady laminar plane plume is defined in FIGURE 11.1.
The flow is driven by a line heat source at the origin, with
E
the
energy input per unit time per unit length. The fluid is characterized
by the temperature and density in the uniform ambient region, and
by four secondary state variables, the viscosity, the heat conductivity,
the specific heat at constant pressure, and the volume coefficient of
expansion. The coordinates are labelled for consistency with other
flows already considered, with
x
increasing upward. The acceleration
of gravity
g
is directed downward, in the negative
x
-direction.
There are altogether eight parameters, listed below in the form
of dimensional statements. For the present, the argument will proceed
11.1. GENERALITIES
561
Figure 11.1: Steady laminar plane plume. (Caption
provided by B. Coles)
562
CHAPTER 11.
THE PLANE PLUME
without benefit of equations.
(Change sub 0 to sub
α
?)
[
E
]
=
ML
T
3
[
ρ
0
]
=
M
L
3
[
μ
0
]
=
M
LT
[
T
0
]
=
Θ
[
k
0
]
=
ML
T
3
Θ
(11.7)
[
g
]
=
L
T
2
[
c
p
]
=
L
2
T
2
Θ
[
β
]
=
1
Θ
.
Let the first four statements be interpreted as defining equations for
M
,
L
,
T
,
Θ
and solved, to obtain
M
=
(
ρ
5
0
ν
9
0
E
3
)
1
/
2
,
L
=
(
ρ
0
ν
3
0
E
)
1
/
2
,
T
=
ρ
0
ν
2
0
E
,
Θ
=
T
0
.
(11.8)
In passing, define
U
=
L
T
=
(
E
ρ
0
ν
0
)
1
/
2
.
(11.9)
and note that, as usual,
UL
ν
0
= 1
.
(11.10)
11.1. GENERALITIES
563
The relationships (11.8) can now be inserted in the last four of equa-
tions (11.7) to obtain four dimensionless groups,
P
1
=
k
0
T
0
E
, P
2
=
ρ
3
0
ν
5
0
g
2
E
3
, P
3
=
ρ
0
ν
0
c
p
T
0
E
, P
4
=
β T
0
(11.11)
in which the four physical parameters
k
0
,
g
,
c
p
, and
β
are isolated.
Within limits, a different group of four statements might have been
selected initially, with the same result, since the four groups (11.11)
can be multiplied or divided arbitrarily by one another. This opera-
tion is evidently an ad hoc illustration of Buckingham’s Π theorem
(
ref
), which states that
p
dimensional parameters and
q < p
dimen-
sional units imply
p
−
q
dimensionless groups. One of these, obtained
by dividing the third group in (11.11) by the first, is familiar;
Π
3
=
P
3
P
1
=
ρ
0
ν
0
c
p
k
0
=
ν
0
κ
0
=
Pr .
(11.12)
Another, a global Froude number, follows on rearrangement of
P
2
;
Π
2
=
P
−
1
/
2
2
=
1
g
(
E
3
ρ
3
0
ν
5
0
)
1
/
2
=
1
g
(
E
ρ
0
ν
0
)(
E
ρ
0
ν
3
0
)
1
/
2
=
U
2
gL
=
Fr
2
.
(11.13)
A third amounts to a global Rayleigh number,
Π
1
=
P
1
/
2
2
P
3
P
4
P
1
=
g
(
ρ
3
0
ν
5
0
E
3
)
1
/
2
ν
0
κ
0
βT
0
=
g
(
ρ
3
0
ν
9
0
E
3
)
1
/
2
βT
0
κ
0
=
gβT
0
L
3
κ
0
ν
0
=
Ra .
(11.14)
The ratio Π
1
/
Π
3
also has a name (see Eckert for comment),
Ra
Pr
=
gβT
0
L
3
ν
2
0
=
Gr .
(11.15)
Finally, a fourth parameter is
P
4
itself,
Π
4
=
P
4
=
βT
0
.
(11.16)
Note from equation (x) of section x that Π
4
= 1 for a perfect gas.
For water at 20
◦
C
, Π
4
=
xxxx
.
(Do the moving observer argu-
ment.)
564
CHAPTER 11.
THE PLANE PLUME
The similarity argument for the equations of motion is es-
sentially independent of the dimensional argument just completed,
which did not use equations except to establish the relevance of the
eight quantities in equations (11.7). The two most common sources
of buoyancy forces are heating
(why is cooling not allowed?)
as
in the sketch, or programmed density differences, as for flow of fresh
or hot water into salt or cold water. The first case will be considered
here and the second, which is a jet-plume relaxation, in section x.
(
Should also do thermal.
)
The equations of motion for the laminar plane plume, with the
boundary-layer and Boussinesq approximations, are
∂u
∂x
+
∂v
∂y
= 0
,
(11.17)
ρ
0
(
∂uu
∂x
+
∂uv
∂y
)
=
βρ
0
(
T
−
T
0
)
g
+
μ
0
∂
2
u
∂y
2
,
(11.18)
ρ
0
c
p
(
∂u
(
T
−
T
0
)
∂x
+
∂v
(
T
−
T
0
)
∂y
)
=
k
0
∂
2
(
T
−
T
0
)
∂y
2
.
(11.19)
Suitable boundary conditions for
u
(
x, y
) and
T
(
x, y
) are
u
(
x,
±∞
) = 0
(11.20)
T
(
x,
±∞
) = 0
(11.21)
and the symmetry condition
ψ
(
x,
0) =
v
(
x,
0) = 0
(11.22)
with null conditions on higher derivatives as needed.
Two integrals can be found immediately by integrating (11.18)
and (11.19) from
−∞
to
∞
in
y
;
ρ
d
d
x
∞
∫
−∞
uu
d
y
=
β ρ g
∞
∫
−∞
(
T
−
T
0
) d
y
(11.23)
ρ c
p
d
d
x
∞
∫
−∞
u
(
T
−
T
0
) d
y
= 0
(11.24)
11.1. GENERALITIES
565
from which, in the second case,
ρ c
p
∞
∫
−∞
u
(
T
−
T
0
) d
y
=
E
= constant
.
(11.25)
Equation (11.23) describes the budget for the changing momentum,
which increases with increasing
x
if
T > T
0
. The integral in equation
(11.24), when expressed in terms of (
ρ
−
ρ
0
), is often referred to
as “buoyancy flux”, for reasons that are not apparent to me; it is
obviously an energy flux. The constant
E
has the units of energy per
unit time per unit length passing any station
x
= constant and is
here identified with the power per unit length at the heat source at
the origin.
The momentum equation (11.18) and energy equation (11.19),
together with the integral conservation law (11.25),
ρ
0
c
p
∞
∫
−∞
u
(
T
−
T
0
) d
y
=
E ,
(11.26)
are the substance of the similarity formulation. The affine transfor-
mation operates on a large number of quantities; there are eight
parameters (11.7) and five variables
x
,
y
,
u
,
v
,
T
.
(Recall the rule
“transform everything in sight.” Note that
T
transforms like
566
CHAPTER 11.
THE PLANE PLUME
T
0
. Discuss
ψ
, boundary conditions.)
Put
x
=
a
̂
x
y
=
b
̂
y
ψ
=
c
̂
ψ
μ
0
=
d
̂
μ
0
ρ
0
=
e
̂
ρ
0
κ
0
=
f
̂
κ
0
(11.27)
c
p
=
p
̂
c
p
β
=
q
̂
β
T
=
r
̂
T
T
0
=
r
̂
T
0
E
=
s
̂
E
g
=
t
̂
g
Four alphabetic invariants are obtained, constituting a complicated
but faithful image of the equations of the problem
(where is
T
11.1. GENERALITIES
567
replaced by
T
−
T
0
?)
;
bce
ad
= 1
,
(11.28)
b
3
eqrt
cd
= 1
,
(11.29)
bcep
af
= 1
,
(11.30)
cepr
s
= 1
.
(11.31)
If equation (11.30) is divided by equation (11.28), the result is
dp
f
= 1
(11.32)
which requires invariance for one of the dimensionless parameters;
c
p
μ
0
k
0
=
ν
0
κ
0
=
Pr
=
̂
Pr .
(11.33)
The remaining three equations, say (11.28), (11.29), and (11.31), can
be solved for
b
,
c
,
r
to establish dimensionless forms for
y
,
ψ
, (
T
−
T
0
);
b
5
e
2
qst
a
2
d
3
p
= 1
,
(11.34)
c
5
e
3
p
a
3
d
2
qst
= 1
,
(11.35)
r
5
a
3
d
2
e
2
p
4
qt
s
4
= 1
.
(11.36)
Hence choose an ansatz of the form
A
(
c
p
ρ
0
βgEν
2
0
x
3
)
1
/
5
ψ
=
f
[
B
(
βgE
ρ
0
c
p
ν
3
0
x
2
)
1
/
5
y
]
=
f
(
η
)
,
(11.37)
568
CHAPTER 11.
THE PLANE PLUME
D
(
βgc
4
p
ρ
4
0
ν
2
0
x
3
E
4
)
1
/
5
(
T
−
T
0
) =
θ
(
η
)
(11.38)
with boundary conditions
f
(0) =
f
′
(
±∞
) =
f
′′
(
±∞
) =
θ
(
±∞
) =
θ
′
(
±∞
) = 0
.
(11.39)
(Note from the form of
η
that
δ
varies like
ν
3
/
5
and like
x
2
/
5
.
What about
u
c
and
Fr
?)
Substitution in (11.34)-(11.36)
1
yields
5
3
ABf
′′′
+
ff
′′
−
1
3
f
′
f
′
−
5
3
A
2
B
2
D
θ
= 0
,
(11.40)
5
3
AB
Pr
θ
′′
+
fθ
′
+
f
′
θ
= 0
,
(11.41)
∞
∫
−∞
f
′
θ
d
η
=
AD
(11.42)
with one parameter, the Prandtl number.
(Check the other inte-
gral
(11.23).)
Equation (11.41) can be integrated once to obtain
5
3
AB
Pr
θ
′
+
fθ
= 0
.
(11.43)
The velocities
u
and
v
are (note
βgE/c
p
ρ
0
recurs)
u
=
B
A
(
β
2
g
2
E
2
x
c
2
p
ρ
2
0
ν
0
)
1
/
5
f
′
(
η
)
(11.44)
and
v
=
1
A
(
βgν
2
0
E
c
p
ρ
0
x
2
)
1
/
5
(
2
5
η f
′
−
3
5
f
)
.
(11.45)
1
Original ms is unclear about these equation numbers.
11.1. GENERALITIES
569
Provided that
η f
′
→
0 as
η
→ ∞
, the entrainment velocity at the
edge of the plume is
v
(
x,
∞
) =
−
3
5
1
A
(
βgν
2
0
E
c
p
ρ
0
x
2
)
1
/
5
f
(
∞
)
.
(11.46)
The notation will eventually put
f
(
∞
) =
C
. A constant local Froude
number can be formed from the centerline velocity, which varies like
x
1
/
5
, and the plume thickness, which varies like
x
2
/
5
. A maximum-
slope thickness for the plume can be defined in the same way as for
the plane jet in SECTION X, namely
4
= 2
f
(
∞
)
f
′
(0)
.
(11.47)
A second relationship is obtained by putting
η
=
4
/
2 when
y
=
δ/
2,
in the argument of
f
in equation (11.37);
4
=
B
(
βgE
ρ
0
c
p
ν
3
0
x
2
)
1
/
5
δ .
(11.48)
Elimination of
δ
gives
δ
=
2
B
f
(
∞
)
f
′
(0)
(
c
p
ρ
0
ν
3
0
x
2
βgE
)
1
/
5
.
(11.49)
Finally, from equation (11.44) at
y
= 0,
u
c
=
B
A
(
β
2
g
2
E
2
x
c
p
ρ
2
0
ν
0
)
1
/
5
f
′
(0)
.
(11.50)
It follows that
(must be small)
Fr
2
=
u
2
c
gδ
=
B
3
2
A
2
[
f
′
(0)]
3
f
(
∞
)
(
βE
c
p
ρ
0
ν
0
)
(11.51)
which is constant
(of what order?)
and formally independent of
T
0
,
k
0
, and
g
. The dimensionless combination in parentheses can be
recognized as the ratio of the fourth to the third of the dimension-
less groups (11.11). (
How does
T
−
T
0
vary with
x
? Who solved the
problem first? Proceed to entrained flow, with and without wall. If
source is cold and there is a wall, is the outcome a wall plume? Do
thermal. Discuss Grashof number, Richardson number. See Schmidt
and Beckmann.
)
570
CHAPTER 11.
THE PLANE PLUME
11.1.2 The plane turbulent plume
In the turbulent case, the boundary-layer approximation for
the equations of motion for a plane plume, with the usual stream
function (
u
=
∂ψ/∂y, v
=
−
∂ψ/∂x
) is
ρ
0
(
ψ
y
ψ
xy
−
ψ
x
ψ
yy
) =
ρ
0
β g
(
T
−
T
0
) +
∂τ
∂y
,
(11.52)
ρ
0
c
p
[
ψ
y
(
T
−
T
0
)
x
−
ψ
x
(
T
−
T
0
)
y
]
=
∂q
∂y
.
(11.53)
The integral conservation law is again
(mention buoyancy flux)
ρ
0
c
p
∞
∫
−∞
u
(
T
−
T
0
)d
y
=
E .
(11.54)
In these equations,
τ
=
−
ρ
0
u
′
v
′
,
(11.55)
q
=
−
ρ
0
T
′
v
′
.
(11.56)
The second of these is usually defined without the factor
c
p
.
To implement an affine transformation, list the variables and
11.1. GENERALITIES
571
parameters in a table, together with their dimensions;
x
=
a
̂
x
L
y
=
b
̂
y
L
ψ
=
c
̂
ψ
L
2
/
T
ρ
0
=
d
̂
ρ
0
M
/
L
3
c
p
=
p
̂
c
p
L
2
/
T
2
Θ
β
=
q
̂
β
1
/
Θ
(11.57)
T
=
r
̂
T
Θ
T
0
=
r
̂
T
0
Θ
E
=
s
̂
E
ML
/
T
3
g
=
t
̂
g
L
/
T
2
τ
=
m
̂
τ
M
/
LT
2
q
=
n
̂
q
M
/
T
3
.
From the six parameters
ρ
0
, c
p
, β, T
0
, E,
and
g
, characteristic
scales appear as
M
=
(
E
6
g
9
ρ
0
)
1
/
5
,
L
=
(
E
2
g
3
ρ
2
0
)
1
/
5
,
T
=
(
E
g
4
ρ
0
)
1
/
5
,
Θ
=
T
0
(11.58)
together with two dimensionless groups,
β T
0
,
c
5
p
ρ
2
0
T
5
0
g
2
E
2
.
(11.59)
572
CHAPTER 11.
THE PLANE PLUME
The second of these is also a group for the laminar problem
(there-
fore basic?)
.
Affine transformation of equations (11.52) - (11.54) yields the
following alphabetical relations:
c
2
d
abm
= 1
,
(11.60)
bdqrt
m
= 1
,
(11.61)
cdr
an
= 1
,
(11.62)
cdpr
s
= 1
.
(11.63)
Equations (11.62) and (11.63) can first be combined in the form
(why?)
anp
s
= 1
(11.64)
and (11.62) discarded. An equivalent condition to (11.62) is that
q
transforms like
ρψ
x
(
T
−
T
0
). If it is also assumed that
τ
transforms
like
ρψ
2
y
, the relation implied is
mb
2
c
2
d
= 1
.
(11.65)
When this is compared with (11.60), it appears that
a
b
= 1
.
(11.66)
(
Question by Hall: is this above the line or below the line?
Answer: below the line.
) Hence the combination
y/x
is invariant
under the transformation, and is evidently the proper independent
variable. When
b
is replaced by
a
throughout, three relations remain,
in addition to (11.64);
c
2
d
a
2
m
= 1
,
adqrt
m
= 1
,
cdpr
s
= 1
.
(11.67)
11.1. GENERALITIES
573
These can be solved for
c, m
and
r
. The result is
c
3
dp
a
3
qst
= 1
,
r
3
a
3
d
2
p
2
qt
s
2
= 1
,
m
3
p
2
d
2
q
2
s
2
t
2
= 1
.
(11.68)
The proper similarity variables, with constants for future use,
are therefore
A
ψ
x
(
ρ
0
c
p
β g E
)
1
/
3
=
f
(
B
y
x
)
=
f
(
η
)
,
(11.69)
C
(
T
−
T
0
)
x
(
ρ
2
0
β g c
p
2
E
2
)
1
/
3
=
θ
(
η
)
,
(11.70)
τ
(
c
p
2
ρ
0
β
2
g
2
E
2
)
1
/
3
=
g
(
η
)
,
(11.71)
q
c
p
x
E
=
h
(
η
)
.
(11.72)
When these are substituted in equations (11.52) - (11.54), the
result is
f f
′′
+
A
2
B
2
C
θ
+
A
2
B
g
′
= 0
,
(11.73)
f
′
θ
+
fθ
′
+
ACh
′
= 0
,
(11.74)
∞
∫
−∞
f
′
θdη
=
AC .
(11.75)
Note from equation (11.74) that there is a relation between
f, θ
and
h
independent of position in the dimensionless flow;
fθ
+
ACh
= constant = 0
(11.76)
since
f
and
h
are zero on the plane of symmetry. In physical variables,
this is
ψ
(
T
−
T
0
)
−
x
T
′
v
′
= 0
.
(11.77)
The mean-velocity components in the turbulent plane plume
are
u
=
B
A
(
βgE
c
p
ρ
0
)
1
/
3
f
′
,
(11.78)
574
CHAPTER 11.
THE PLANE PLUME
v
=
1
A
(
βgE
c
p
ρ
0
)
1
/
3
(
η f
′
−
f
)
.
(11.79)
Hence the velocity in the plane of symmetry,
u
c
, is constant. The
entrainment velocity is
v
(
x,
∞
) =
−
1
A
(
βgE
c
p
ρ
0
)
1
/
3
f
(
∞
)
(11.80)
and is also constant. In the plane of symmetry the temperature (
T
c
−
T
0
) and the turbulent heat transfer vary like 1
/x
, while the Reynolds
shearing stress is constant. (
Why not
τ
∼
u
2
c
,
T
−
T
0
∼
T
c
−
T
0
?
Should do without boundary-layer approximation.
)
Note from (11.78) that
u
c
is independent of
x
;
u
c
=
B
A
(
βgE
C
p
ρ
o
)
1
/
3
f
′
(0)
(11.81)
and from (11.70) that
T
c
−
T
0
=
1
Cx
(
E
2
ρ
2
0
βgc
2
p
)
1
/
3
θ
(0)
(11.82)
so that the Reynolds shearing stress (11.71) and the Reynolds heat
transfer (11.72) can be written
τ
=
A
2
B
2
[
f
′
(0)]
2
ρ
0
u
2
c
g
(
η
)
(11.83)
and
q
=
AC
Bf
′
(0)
θ
(0)
ρ
0
u
c
(
T
c
−
T
0
)
h
(
η
)
.
(11.84)
(
Why not do this in the beginning?
)
From handout:
Laminar plume: nice work. No evidence of instability. Velocity in
plane of symmetry is increasing with
x
. Thickness goes like
ν
2
/
3
.
Sparrow. If (similarity) theory and experiment do not agree, I choose
to believe the theory and look for sources of experimental error. This
11.1. GENERALITIES
575
policy might be useful for the turbulent problem, where there is no
theory.
Sparrow. Note turbulent case; independent variable is
y/x
, and uses
up
0
.
6
: this flow is not a boundary layer. The entrained flow for the
round turbulent plume uses Legendre polynomials.
Kotsovinos. Good on relaxation from jet to plume, given some initial
momentum. K’s plume not very wide; tank is small. See also picture
of jet.
Anwar. Does not comment on development into pair of counter-
rotating vortices.