Introductory fluid mechanics JF207 - JNCASR Desktop

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Introductory fluid mechanics JF207
Lecture#3
Returning to our continuumdescription of fluid motion,we will distinguish between two kinds
of forces exerted on an infinitesimal continuum fluid element (recall that such an element has
to be large compared to the relevant microscopic dimensions - λ
MFP
for dilute gases - for a
continuum description to be valid).These are the body forces and the surface forces.The
former refer to long-ranged forces such as gravity or that arising from an electromagnetic
field,and so on.Such forces,as the name implies,vary slowly with the distance between
interacting elements;for instance,the gravitational force per unit mass acting on any object is
very well represented by the constant vector g (|g| = 9.8m/s
2
) for distances from the earth’s
surface that are less than the order of the earth’s radius (∼ 6.4 x10
6
m).A consequence
of this slow variation is that such forces act equally on all the matter within a sufficiently
small (infinitesimal) element of our fluid continuum,and thus,long-ranged forces manifest as
body forces.Assuming the body force per unit mass at a point x in the fluid,and at time t,
to be F(x,t),the total force on an infinitesimal element around this point with volume dV
is given by ρF(x,t)dV,where ρF is now the force per unit volume.For gravity,this force is
simply the weight of the element,ρgdV;for a conducting fluid in an magnetic field(B),for
instance,there would be an additional Lorentz force (F
L
∝ i∧B,i being the current density.
The second group of forces are short-ranged forces that have a direct molecular origin,and
as a result,are characterized by length scales relevant to the microscopic dynamics.This
would mean that such forces decay extremely rapidly with distance on the macroscopic
scale.Again,considering the example of an dilute gas,one expects such short-ranged forces
to decay on distances of the order of a mean free path.Since any infinitesimal element in
a continuum description is,by definition,much larger than all microscopic scales,the effect
of such forces would be negligible unless two interacting elements are directly in contact;
and then,one would expect these forces to act across the contact surface
1
.In other words,
the short-ranged forces manifest as surface forces,in a continuum description,that act to
transport momentum across the boundaries of an infinitesimal element.In dilute gases,
the momentum transport occurs due to molecules randomly crossing the boundary under
consideration and thereby carrying momentum across in the appropriate direction(often
referred to as the kinetic contribution to the stress);in liquids,the tranport of momentum
can occur without physical translation of molecules via short-ranged forces acting between
pairs of molecules on either side of the boundary and separated by a distance comparable
to the range of the inter-molecular potential (referred to as the potential contribution to
the stress).It must now be clear that the total effect of short-ranged forces acting on a
differential element is decided by its surface area rather than the volume.Hence,it is more
appropriate to consider a plane surface element in the fluid and specify the local short-ranged
force as the total force exerted by the fluid on one side of the element on the other side.If the
area of the element is δA,this may be written as Σ(n,x,t)δA,where Σ is the force per unit
1
Strictly speaking,such forces would be contribute to momentum exchange in a layer of O(λ
MFP
) that
encompasses the contact surface,but since λ
MFP
/V
1
3
→ 0,this layer may be regarded as coincident with
the contact surface itself in a continuum description.
1
n
-n

A


(x,n,t)


(x,-n,t)
Surface force acting
on the element

A
Figure 1:A surface force,Σ(x,n,t) acting on a surface element,δA,with unit normal n.
An equal and opposite force,Σ(x,−n,t) = −Σ(x,n,t) acts on the opposite face.
area (or the stress vector) and n is the unit normal to the element.One has a dependence on
n since one expects the surface force to depend on the orientation of the plane element,and
the latter is determined by n.We will adopt a convention that would amount to a tension
being interpreted as a positive force.Thus,in the above expression for the surface force,
Σ(n,x,t) is the force exerted by the fluid,on the side to which n points,on the fluid on the
side which n points away from.
Now,we proceed to show that it is a stress tensor rather than the stress vector Σ that is
a more fundamental quantity,characterizing the fluid response to an imposed deformation.
In order to do this,we need to extract the n-dependence of Σ.Note that while Σ(n,x,t)
is the force exerted by the fluid on the side to which n points,that exerted by the fluid
on the opposite side should be equal and opposite,since there can be no net force acting
on a surface element (which has zero mass).On the other hand,in accordance with our
notation,this same force can also be written in terms of the unit normal pointing in the
opposite direction(−n) as Σ(−n,x,t).Thus,we must have that Σ(−n,x,t) = −Σ(n,x,t),
implying that the stress vector is an odd function of n.In order to deduce the specific
dependence on n,we consider the tetrahedral volume element shown in figure 2.The three
orthogonal faces have areas δA
1
,δA
2
and δA
3
,and unit (outward) normals −a,−b and −c,
respectively;the fourth inclined face has an area δA and unit normal n.In accordance with
our notation above,the total surface force acting on the tetrahedral element is:
Σ(n,x,t)δA+Σ(−a,x,t)δA
1
+Σ(−b,x,t)δA
2
+Σ(−c,x,t)δA
3
,(1)
where we have assumed the volume element to be small enough(specifically,much smaller
than the characteristic length scales on which Σ varies) that the position coordinate corre-
sponding to all of the faces of the tetrahedron may be approximated as x.Since only the
2
unit normal changes from one term to another in (1),we suppress the dependence on the
arguments x and t in what follows.Further,using Σ(n) = −Σ(−n),and the fact that both
the total volume force and the acceleration of the element must be O(δl)
3
,where δl is a
characteristic linear dimension of the element,we must have that the total surface force on
the given element mush vanish as O(δl) for δl →0.Thus,in the limit,one has
Σ(n)δA−Σ(a)δA
1
−Σ(b)δA
2
−Σ(c)δA
3
= 0.(2)
That is to say,the surface forces,acting on a sufficiently small element,must exist in balance.
Note that the differences between the surface forces acting on the various faces,arising from
the O(δl) difference in the position vectors x specifying the different faces of the tetrahedron,
is O(δl)
3
,and may be balanced by a combination of the body forces and the acceleration of
the element.
Now,the volume of a tetrahedron is again proportional to the scalar triple product (discussed
in an earlier lecture on Cartesian tensors),although,unlike a parallelopiped,there is a
prefactor of 1/6;that is,V
tetrahedron
=
1
6
t
1
 t
2
∧t
3
,where t
1
,t
2
and t
3
are vectors emanating
from a common vertex.Notwithstanding the proportionality factor of 1/6,this may be used
to write the volume of the tetrahedron in figure 2 as δAn δl
2
b,choosing B as the common
vertex;here,we have taken δAn to replace t
2
∧ t
3
in the general formula above,and t
1
to
correspond to the edge of length δl
2
along b.Alternately,taking δA
2
b to correspond to t
2
∧t
3
,
we have the volume being equal to δA
2
δl
2
.Clearly,the two expressions must be equal,which
is only possible if δA(n b) = δA
2
.Similarly,we have the other relations δA(n a) = δA
1
and
δA(n c) = δA
3
from alternate expressions for the tetrahedral volume based on A and C as
the common vertices.Using these relations in (2),we have
Σ(n) = [Σ(a)a +Σ(b)b +Σ(c)c]  n,(3)
after cancelling δA on both sides;or,in index notation:
Σ
i
(n) = {Σ
i
(a)a
j

i
(b)b
j

i
(c)c
j
} n
j
,(4)
Since the vectors n and Σ do not in any way depend on the choice of the axes of refer-
ence ([a,b,c]) the expression within brackets in (4) that relates Σ and n must likewise be
independent of the particular choice of axes.In other words,for given i and j,this expres-
sion must correspond to the ij
th
component of a second-order tensor which we denote by σ.
Thus,the stress vector is a linear functional of n,and one may write:
Σ
i
(n) = σ
ij
n
j
,(5)
relating the stress vector and tensor.One may readily verify this assertion by substituting
(5) in (4) with n (successively) replaced by a,b and c,so as to obtain:
σ
ij
n
j
= (a
j
a
k
+b
j
b
k
+c
j
c
k

ik
n
j
,(6)
which reduces to an identity on noting that the quantity within brackets is just a represen-
tation of δ
jk
in the chosen coordinate system.
3
n
b
a
c

A
1

A
3

A
2
-a
-c
-b
O
A
B
C
Figure 2:The balance of surface forces on a tetrahedral surface element.
In a manner similar to that above,one may consider a balance of the moments of the various
forces acting on the fluid within an infinitesimal volume element.The moment of the surface
forces about a point in the interior of the volume is given by
￿
ǫ
ijk
x
j
σ
kl
n
l
dA.Noting that
the moment arm is O(δl),we have that the moment of the surface forces is O(δl)
3
in the
limit δl →0.On the other hand,the moment of the body forces,given by
￿
ǫ
ijk
x
j
(ρF
k
)dV is
O(δl
4
),and the angular momentum of the element,given by
￿
ǫ
ijk
x
j
ρu
k
dV,is again at least
as small as O(δl)
4
;the angular momentum may,in fact,be reduced to a smaller order ((δl)
5
)
by taking moments about the center-of-mass so that
￿
x
i
dV = 0,and the first non-zero
contribution arises due to the variation of the linear momentum about the origin.In any
case,we note that,for δl → 0,the moment of the surfaces forces must be identically zero,
since it cannot be balanced by the other terms (moment of the body forces and angular
acceleration of the element) in the equation for the conservation of angular momentum.
Thus,for δl →0,the conservation of angular momentum implies
￿
ǫ
ijk
x
j
σ
kl
n
l
dA = 0,(7)
and on applying the divergence theorem,one obtains
￿
ǫ
ijk

∂x
l
(x
j
σ
kl
dV =0,(8)

￿
ǫ
ijk
￿
x
l
∂σ
kl
∂x
l

kj
￿
dV =0.(9)
Noting that the first term is again O(δl) smaller,and that the volume dV is arbitrary,we
4
have that
ǫ
ijk
σ
kj
= 0.(10)
The above relation is satisfied provided the stress tensor,σ is symmetric;that is,σ
ij
= σ
ji
.
In other words,there are only six independent elements of the stress tensor (rather than nine
as is the case for the general second-order tensor).
There are two comments in order here.The first is that we have obviously made a choice
of the point about which the moment of the forces above have been evaluated - this point
was taken as an interior point which made the moment arm of O(δl).The conclusions
with regard to the stress tensor above,however,are independent of this choice.An al-
ternate choice of a fixed exterior point,say x
0
,could be used;in this case,the moment
arm may be written as the sum of a fixed vector from the center-of-mass to the given
exterior point,and a vector of O(δl).The conservation of angular momentum would
then reduce to the form O(δl)
3
.
￿
ǫ
ijk
￿
(x
l
−x
(CM)l
)
∂σ
kl
∂x
l

kj
￿
dV + O(δl)
2
(x
0
−x
CM
) ∧
[Total surface force on the element];leading to the same conclusions.
The second comment is with regard to an assumed absence of body couples acting on the
fluid element under consideration.There are substances such as ferrofluids that experience a
body couple under suitable conditions;a ferrofluid is essentially a suspension of tiny magnetic
particles,and since the microscopic constituents (magnetic dipoles) experience an aligning
torque when the ferrofluid is subject to an ambient magnetic field,there will in general be
a second O(δl)
3
term in the angular momentum conservation,that depends on the total
magnetic couple exerted on the particles within the continuum element under consideration,
and that balances the torque due to the surface forces.In these cases,the stress tensor will
not be symmetric.Referring to figure 3,we note that the volume element under consideration
will have an additional term in the conservation equation due to the contribution from the
microscopic couples.If we denote the orientation of the α
th
magnetic dipole as p
α
,then
the total torque acting on a continuum element,containing N such dipoles,is given by
￿
N
α=1
ǫ
ijk
(x
j
+sp
α
j
)F
α
k
+(x
j
−sp
α
j
)(−F
α
k
),where the forces constituting a microscopic couple
are assumed to be separated by a distance 2s.Assuming the dipole density to be n,so that
N = nδV,and further that a subset (ndV )Ω(p)dp have orientations in the range [p,p+dp],
where Ω(p) is a probability density defined on the unit sphere
2
,one may write the body
couple contribution as
￿
dV
￿
dpnΩ(p)ǫ
ijk
p
j
(2sF
k
).Denoting
￿
dpnΩ(p)p
j
(2sF
k
) as L
i
,the
angular momentum balance should now take the form ǫ
ijk
σ
kj
= −L
i
.
Restricting our consideration to simple fluids,where body couples are absent,we conclude
that the surface forces are entirely characterized by a stress tensor that must be symmetric.
The diagonal elements of the stress tensor are known as normal stresses while the non-
diagonal elements constitute the tangential or shearing stresses.Furthermore,for a fluid
at rest,and with an isotropic microstructure,the stress tensor is expected to be isotropic
since there can be no preferential direction in which the randommolecular motion transports
2
Here,the argument p is taken to denote a dependence on the angles θ and φ,and thus Ω(p) ≡ Ω(θ,φ)
is a probability density defined on the unit sphere such that
￿
π
0
￿
2
0
πdθ sinθdφΩ(θ,φ) = 1
5
p


2s
-F
F
B
Figure 3:The torques on the microscopic constituents,in the presence of an applied magnetic
field,leading to a couple per unit volume.
momentum.Thus,
σ
ij
= −pδ
ij
,(11)
where,in accordance with our convention,p denotes the magnitude of the (compressive) pressure
in a static fluid.The compressive interpretation of the stress in a fluid at rest is consistent
with the inability of simple fluids,under normal conditions,to sustain tensile stresses.Thus,
(11) implies that,under conditions of static equilibrium,there can be no tangential stresses.
This also conforms to our original postulate of what a fluid is - a state of matter that cannot
remain static under the action of shearing stresses.
Having characterized the form of the surface forces,and in particular,the simplified form
that they assume in the absence of fluid motion,we now consider the force balance on a
continuum element under the action of both body and surface forces;this will allow us to
obtain the constraints required on the body forces in order for there to be no motion.This
force balance may be written in the form:
￿
ρF
i
(x,t)dV +
￿
σ
ij
n
j
dA = 0,(12)
and,on using (11),we have
￿
ρF
i
dV −
￿
pn
i
dA = 0.(13)
6
Further,applying the divergence theorem and the arbitrariness of the chosen volume,we
have
∂p
∂x
i
= ρF
i
.(14)
which must be satisfied at every point in the fluid volume.Thus,fluid motion is absent only
when the body force per unit volume,ρF is expressible as the gradient of a scalar.This is,
of course,not true in general,since the Helmholtz decomposition tells us that an arbitrary
vector is in general the sum of the gradient of a scalar potential (φ) and the curl of a vector
potential (Ψ);that is,any vector a may be written as a = ∇φ +∇∧Ψ.
In the simplest case of a constant density fluid acted on by gravity,ρF = g,and (14) reduces
to the familiar statement of hydrostatic equilibrium viz.
∂p
∂x
i
= ρg
i
,(15)
where g
i
is the acceleration vector due to gravity.Of course,(14) with F
i
= g
i
does not
require the density to be a constant.The fluid will continue to be at rest provided the
density varies only in the direction of gravity.This may be seen by taking the curl of (14).
The left-hand-side equals zero,since (∇∧ ∇)p = 0,so that
ǫ
kji

∂x
j
(ρg
i
) = 0,⇒ǫ
ijk
∂ρ
∂x
j
g
i
= 0.(16)
Since ∇ρ is the normal to the constant density surfaces,the above condition implies that
the constant density surfaces must be perpendicular to gravity;in other words,the fluid
density can vary along gravity.A simple illustration of how fluid motion would result on the
violation of this condition is when one tilts a container of water (here,the density variation
is restricted to the interfacial region between water and air,and ∇ρ can be regarded as a
vector perpendicular to the instantaneous interface);tilting causes the water interface to
become inclined to gravity.This immediately leads to fluid motion that acts to restore the
original horizontal orientation
3
.
Restricting our consideration to gravity as the body force,but allowing for the variation of
density as a function of pressure,one may evaluate the pressure variation with height in a
static atmosphere.Since air may be approximated as an ideal gas,one has p = n
a
RT with
n
a
being the number of moles of air,or in terms of the air density,p = ρ
a
(R/M
a
)T where
R = 8314J/K.kmol is the molar gas constant with M
a
≈ 30kg/kmol (a value close to N
2
,
the dominant atmospheric component).Using this relation,and assuming g = −1
3
in (15),
we have
dp
a
dx
3
= −ρ
a
g,(17)

dp
a
dx
3
= −
gM
a
RT
p.(18)
3
The tilted interface leads to the non-coincidence of the constant pressure and constant density surfaces;
a term proportional to δp ∧ δρ,then acts as a (baroclinic) source of vorticity,inducing fluid motion
7
Our analysis will be restricted to the troposphere which accounts for about 80%of the entire
atmospheric mass,and wherein the temperature only varies by about 20% of its value at the
earth’s surface.Thus,an isothermal atmosphere is a reasonable leading-order approximation.
Further,the height of the troposphere ranges from about 20km at the equator to less than
10km near the poles.These distances being considerably smaller than the radius of the
earth,g may also be taken as a constant
4
.With these assumptions,(18) may be easily be
solved to obtain:
p
a
= p
0
a
e

z
H
a
,(19)
where p
0
a
is the pressure at the earth’s surface (z = 0) and H
a
= RT/(M
a
g) is referred
to as the pressure-scale height;that is,the height over which the atmospheric pressure
varies by a significant amount (specifically,by a factor 1/e).Using the values of R and M
a
above,with g ≈ 10m/s
2
,and T = 300K,one finds (RT/M
a
g) ≈ 8km.This is consistent
with our initial assumption of the pressure varying with height much more rapidly than the
temperature (which was taken as constant).
4
The troposphere is largely transparent to the incoming solar radiation whose intensity peaks at a fre-
quency in the mid-visible range.It is the earth’s surface that absorbs the solar radiation,and the troposphere
is,in effect,heated from the bottom by the upwelling (infrared) radiation emitted by the surface.This heat-
ing leads to convection(vertical motion) and mixing - the troposphere being identified by the resulting
characteristic decrease in temperature with height.The tropopause separates the troposphere from the
stratosphere;in contrast to the troposphere,the temperature increases with height in the stratosphere due
to direct absorption of solar radiation(primarily by ozone).Since the surface heating is strongest at the
equator,the vertical motion is strongest in the equatorial regions and the tropopause is thus at the greatest
height close to the equator.
8