Fluid Mechanics
Z. Suo
BASIC EQUATIONS IN FLUID MECHANICS
Division of labor. Subject to forces or a change in temperature, a fluid
deforms. Understanding this deformation is the object of fluid mechanics.
The deformation of a body of fluid is often inhomogeneous. Fluid
mechanics regards the body as a sum of many small pieces. Each small piece
(called a fluid element, a material particle, or a material point) consists of a large
number of molecules, but is much smaller than the body. Each small piece
evolves in time through a sequence of homogeneous states. The small pieces
communicate with one other, and the body as whole undergoes inhomogeneous
deformation.
Fluid mechanics divides the labor into two distinct tasks. First, analyze
the homogeneous deformation of a small piece by constructing a rheological
model, also known as constitutive law, equations of state, or rheological equation
of state. Second, analyze the inhomogeneous deformation of the body by
describing various means of communication among the small pieces:
compatibility of geometry, balance of forces, conservation of mass, and
conservation of energy.
The following notes begin with the homogeneous deformation of a small
piece, and end with inhomogeneous deformation of a body.
Homogeneous State
Name each thermodynamic state of a substance using values of
two thermodynamic properties. We will consider a pure fluid, a substance
that consists of a large number of molecules of the same species. A fluid can be
either a liquid or a gas. When the temperature is low and the pressure is high,
the substance is a liquid, and molecules touch one another but frequently change
neighbors. When the temperature is high and the pressure is low, the substance
is a gas, and molecules fly in space and collide with one another and with the wall
of the container; between collisions, on average a molecule flies a distance longer
than the size of the molecule.
A pure substance can be in many thermodynamic states, and has many
thermodynamic properties. Examples of thermodynamic properties include
temperature, pressure, volume, energy, entropy, smell, color, electric field, and
polarization. When the substance is in a particular thermodynamic state, the
value of every thermodynamic property is fixed. We name each state of a
substance by the values of its properties.
How many properties do we need to differentiate states of a substance?
The answer is two. How do we know? The answer depends on empirical
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evidence and on the limit of our attention. One property is too few: if we hold
the value of the temperature we can still change the state of the substance by
changing the pressure. Three properties are too many: once the temperature and
pressure are fixed, the thermodynamic state of the substance is fixed if we neglect
minor effects of, say, the electric field and magnetic field.
The choice of two properties, rather than any other number of properties,
is also motivated by a pedagogical consideration. Indeed, fluid mechanics may be
taught by choosing zero thermodynamic property—that is, by restricting to
incompressible fluids and neglecting the effect of temperature. This way, one can
focus on the Navier-Stokes equation and analyze the effects of viscosity and
inertial under various initial and boundary conditions. But this approach breaks
the link with thermodynamics and obscures the structure of the theory.
Thus, we assume that the thermodynamic state of a fluid is capable of two
independent variations. When we specify the values of two properties, the
thermodynamic state of the fluid is fixed, so are all its other thermodynamic
properties. We can use any two properties as coordinates of a plane. A point in
the plane represents a thermodynamic state of the substance. Given the values of
two properties, we look up values of all other properties in thermodynamic tables
and plots, in textbooks or online.
Thermodynamic relations.
Volume, energy and entropy of a
substance are proportional to the amount of the substance. Let v, u and s be the
volume, energy, and entropy per unit mass. (The density relates to the volume
per unit mass as ρ = 1 / v .) We have assumed that the thermodynamic state of
the substance is capable of two independent variations. Once the volume and
energy is specified, the entropy is fixed. This relation is described by a function
of two variables, s u,v . This function has been tabulated for many substances.
( )
In fluid mechanics we are interested in two other thermodynamic
properties: the temperature T and the pressure p. These two properties relate to
the partial derivative of the function s u,v :
( )
1 ∂s ( u,v)
p ∂s ( u,v)
,
.
=
=
T
∂u
T
∂v
The two relations express the temperature and pressure as functions of volume
and energy.
Recall an identity in calculus:
∂s u,v
∂s u,v
ds =
du +
dv .
∂u
∂v
Combining the above expressions, we write
du + pdv
.
ds =
T
( )
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Thus, the variation in entropy relates to four properties: u, v, p and T. Because
these four quantities can be readily measured, this relation is commonly used to
determine the function s u,v in experimental measurements.
( )
Stress. We pull a gum by a force, and define the tensile stress as the
force divided by the cross-sectional area of the gum in the deformed state. We
now generalize this definition of stress to multiaxial loading.
Consider a piece of fluid in the shape of a unit cube, with faces parallel to
the coordinate planes. Now consider one face of the block normal to the axis x j .
Acting on this face is a force. Denote by σ ij the component i of the force acting
on this face. The nine quantities σ ij are components of the stress.
We adopt the following sign convention. When the outward normal vector
of the area points in the positive direction of axis x j , we take σ ij to be positive if
the component i of the force points in the positive direction of axis x i . When the
outward normal vector of the area points in the negative direction of the axis x j ,
we take σ ij to be positive if the component i of the force points in the negative
direction of axis x i .
The nine components of stress form a three-by-three matrix. The matrix
is symmetric, σ ij = σ ji . You can confirm this relation by the balance of moment
acting on the cube.
Stress is a tensor. Consider a plane of unit normal vector n. Define the
traction t as force acting on the plane divided by the area of the plane. Consider a
tetrahedron formed by the plane and the three coordinate planes. The balance of
forces acting on the tetrahedron requires that
ti = σ ij n j .
The stress is a linear map that maps one vector (the unit normal vector) to
another vector (the traction). In linear algebra, such a linear map is called a
tensor.
Rate of extension. Consider two material particles in a fluid moving
apart. We define the rate of extension by the relative velocity between the two
material particles divided by the distance between them. Let D11 , D22 and D33 be
the rates of extension along the three coordinates.
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Rate of dilation.
A piece of fluid consists of a fixed number of
()
molecules. The volume of the piece is a function of time, V t . Define the rate of
dilation by
dV
.
Vdt
We now relate the rate of dilation to the rates of extension. Consider a simple
case that the piece of fluid is of a rectangular shape, sides l1 ,l2 ,l3 in the current
state. The rates of extension of the sides are
D11 =
dl
dl1
dl
, D22 = 2 , D33 = 3 .
l1dt
l2dt
l3dt
The volume of the piece is V = l1l2l3 . The volume changes at the rate
dl
dl
dl
dV d
=
l1l2l3 = l2l3 1 + l3l1 2 + l1l2 3 .
dt dt
dt
dt
dt
(
)
The rate of dilation is
dV
= D11 + D22 + D33 .
Vdt
Even though we have derived this result using a piece of fluid of
rectangular shape, the conclusion is correct for a piece of fluid of any shape, so
long as the deformation is homogeneous. We will show that the three rates of
extension are components of a tensor, called the rate of deformation. The sum
D11 + D22 + D33 is the trace of the tensor, an invariant of the tensor. Thus, at a
given time, the volume of the piece is V, the trace of the rate of deformation is
Dkk , and the piece changes its volume at the rate VDkk .
Rate of shear. Consider three material particles P, A and B in a piece of
fluid undergoing homogeneous deformation. At a give time, line PA is normal to
line PB, but the two lines rotate relative to each other. We define the rate of shear
by the one half of the rate at which the angle between the two lines changes. Let
D23 , D31 and D12 be the rates of shear with respects to the three coordinates.
Rate of deformation is a tensor. In defining the rates of shear, D23 ,
D31 and D12 , we have introduced a factor of 2. We do so to make the three rates
of shear, along with the three rates of extension, D11 , D22 and D33 , constitute the
components of a symmetric, second-rank tensor Dij .
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To see that the six rates of deformation are indeed the components of a
tensor, we abstract the definition as follows. Consider two points A and B, of
coordinates x A and x B , and velocities v A and v B . The piece of liquid is
undergoing a homogeneous deformation. That means that the velocity field is
linear in space:
(
)
vB − v A = L xB − x A .
The linear map L maps one vector x B − x A to another vector v B − v A . This linear
map is called the velocity gradient, and is a tensor. The deformation is
homogeneous when the velocity gradient L is independent of position in space.
Using a picture, we can readily confirm that the rate of deformation
relates to the velocity gradient as
1
L + L ji .
2 ij
Thus, the rate of deformation is a tensor.
Consider a line of material particles in the piece. In the current state, the
line is in the direction of a unit vector n, and is of length l. Write x B − x A = ln .
Dij =
(
)
(
)
The line of material particles elongates at a rate dl / dt = v B − v A ⋅ n .
(
By
)
definition, the rate of extension of the line of material particles is Dn = dl / dt / l .
( )
Thus, Dn = n ⋅ Ln , or
Dn = Dij ni n j .
Consider two lines of material particles in the piece. In the current state,
the two lines are two unit vectors normal to each other, m and n. By definition,
( ( )
(
))
the rate of shear with respect to the two vectors is Dmn = m ⋅ Ln + n ⋅ Lm / 2 .
This expression is the same as
Dmn = Dij mi n j .
Rheological state. The thermodynamic state of a substance is capable
of two independent variations. We name each thermodynamic state of the
substance by the values of two thermodynamic properties, for example, the
temperature and the density. Once the temperature and density are specified, the
values of pressure, energy and entropy are fixed.
To these five thermodynamic properties, we have just added six
components of stress and six components of rate of deformation. We now
assume that two thermodynamic properties, along with six out of twelve
components of stress and rate of deformation, specify a rheological state of the
substance. For example, in addition to the temperature and density, we use the
values of the six components of stress to specify a rheological state.
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Incidentally, the pressure due to the applied forces in general differs from
the pressure determined from the thermodynamic relations.
Rheological equations of state. A thermodynamic equation of state
relates thermodynamic properties. In particular, once we have chosen two
thermodynamic properties as independent variables, any other thermodynamic
property is a function of the two independent properties. We have shown all
these functions are derived from a single function, s u,v . We now need to relate
( )
the six components of stress to the six components of rate of deformation by
constructing six rheological equations of state.
Newton’s law of viscosity. Specifically, we wish to generalize
Newton’s law of viscosity. Let us recall this law. Two rigid plates sandwich a
layer of fluid, thickness h and area A. When we apply a shear force F, one plate
slides relative to the other plate at a velocity V. Newton’s law of viscosity says
that the velocity V is linear in the force F. In modern words, the shear stress is
τ = F / A , the shear rate is γ = v / h , and Newton’s law of viscosity is τ = ηγ ,
where η is the viscosity. The viscosity is material specific, and depends on
pressure and temperature. We write Newton’s law of viscosity as
σ 12 = 2η D12 .
We further assume that the fluid is isotropic, so that the same viscosity applies in
all directions. For example, we write
σ 13 = 2η D13 ,
σ 23 = 2η D23 .
What happens to a fluid in extension or under hydrostatic compression? We next
generalize Newton’s law of viscosity to any linear, isotropic, viscous fluid. We do
so in consistence with thermodynamics.
The energy of an isolated system is conserved. A unit mass of a
fluid is subject to external forces, and is connected to a reservoir of energy. The
fluid, the external forces, and the reservoir together constitute a composite
thermodynamic system. The composite system is an isolated system—it
exchanges neither energy nor matter with the rest of the world.
The external forces give a state of stress σ ij . As the fluid deform at the
rate of deformation Dij , the external forces changes its potential energy at the
rate −vσ ij Dij .
As the reservoir transfers heat to the fluid, the energy of the
reservoir changes at the rate −vQ . The energy of the composite system is the sum
of the parts. The conservation of energy requires that the energy of the isolated
system should be constant:
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du
− vσ ij Dij − vQ = 0 .
dt
The entropy of an isolated system never decreases. The fluid
changes entropy at the rate ds/dt. The reservoir itself is in a state of
thermodynamic equilibrium, held at a constant temperature TR . When the
reservoir changes energy at the rate −vQ , the entropy of the reservoir changes at
the rate −vQ /TR . The mechanism that applies the external forces does not
change entropy. The entropy of the composite system is the sum of the parts.
Thermodynamics requires that the entropy of the isolated system should never
decrease:
ds vQ
−
≥0.
dt TR
The equality holds when the isolated system is in thermodynamic equilibrium,
whereas the inequality holds when the isolated system is not in thermodynamic
equilibrium.
Thermodynamic condition in terms of independent changes. In
the above thermodynamic condition, the change in entropy and the transfer of
energy transfer are not independent changes. We now express this condition in
terms of seven independent changes: Dij and Q. Recall the thermodynamic
(
)
relation ds = du + pdv /T and the geometric relation dv / dt = vDkk . We obtain
that
(σ
ij
"1 1 %
+ pδij Dij + $$ − '' Q ≥ 0 .
# T TR &
)
This thermodynamic condition holds for arbitrary rate of deformation Dij and
rate of transfer Q, a total of seven independent variables.
Thermodynamic equilibrium.
When the equality in the
thermodynamic condition holds for arbitrary rate of deformation Dij and rate of
transfer Q, the composite system is in thermodynamic equilibrium. This
condition of equilibrium leads to
σ ij + pδij = 0 , T = TR .
The first expression corresponds to a total of six independent equilibrium
equations, σ 12 = σ 23 = σ 31 = 0 and σ 11 = σ 22 = σ 33 = − p . For the fluid, the external
forces, and the reservoir to equilibrate, the fluid must be in a hydrostatic state,
the stress due to the applied forces must match the thermodynamic pressure of
the fluid, and the temperature of the fluid must equal the temperature in the
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reservoir. That is, the composite in equilibrium is fully specified by the
temperature T and the pressure p of the fluid.
Generalized Newton’s law of viscosity. When the inequality in the
thermodynamic condition holds for arbitrary rate of deformation Dij , the
composite system is not in thermodynamic equilibrium. We satisfy this inequality
by prescribing a kinetic model that linearly relates two symmetric tensors,
σ ij + pδij and Dij . For an isotropic liquid, the linear relation between two
symmetric tensors takes the general form
!
$
1
σ ij + pδij = 2η # Dij − Dkkδij & + ζ Dkkδij ,
3
"
%
where η and ζ depend on two independent thermodynamic properties, but are
independent of the rate of deformation. The pressure p relates to the
thermodynamic equation of state. This kinetic model provides the six rheological
equations of state that connect the two tensors, the stress and the rate of
deformation.
The kinetic model generalizes Newton’s law of linear, isotropic, viscous
flow. To see this generalization clearly, we motivate the model in a different way.
The trace of the rate of deformation, Dkk , describes the rate at which the piece of
liquid changes its volume. The deviatoric part of the rate of deformation,
eij = Dij − Dkkδij / 3 , describes the rate at which the piece of liquid changes its
shape. The mean stress is σ m = σ kk / 3 , and the deviatoric stress is sij = σ ij − σ mδij .
Note
the
(
identity
σ ij Dij = sij eij + σ m Dkk ,
and
write
the
inequality
as
)
sij eij + σ m + p Dkk ≥ 0 . This inequality suggests a kinetic model:
sij = 2ηeij , σ m + p = ζ Dkk .
The thermodynamic inequality holds when both η and ζ are nonnegative. The
constant η represents the shear viscosity that resists the change in shape, and
the constant ζ represents the dilation viscosity that resists the change in volume.
The dilation of a fluid is viscoelastic. Let us examine the pair of
equations in the kinetic model separately. Whereas sij = 2ηeij generalizes
Newton’s law of viscosity, σ m + p = ζ Dkk is a viscoelastic model of the Kelvin type.
The thermodynamic pressure related to specific volume and temperature through
a thermodynamic equation of state, p = f v,T . This expression describes the
( )
elasticity of the fluid in dilation. Recall the expression for the rate of dilation
Dkk = dv / dt / v , we rewrite σ m + p = ζ Dkk as
(
)
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dv
.
vdt
This expression represents a spring and a dashpot in parallel, subject to an
applied stress σ m .
( )
σ m + f v,T = ζ
Note that in general σ m ≠ − p . When a stress σ m is suddenly applied and
held at a constant, the fluid creeps and the thermodynamic pressure of the fluid
approaches this applied stress. The time needed is estimated by ζ / B , where B is
bulk elastic modulus of the fluid. Take representative values for water at the
room temperature, ζ = 10−3 Pa ⋅ s and B = 109 Pa , the relaxation time ζ / B = 10−12 s .
For many substances, as the temperature drops, viscosity increases steeply, but
the elastic modulus does not. It is conceivable that viscoelastic dilation can be
important.
Inhomogeneous State
A body of fluid is a collection of small pieces of fluid. As the body evolves,
each piece evolves through a sequence of homogeneous states, as described in the
previous section. Different pieces communicate through the compatibility of
geometry, balance of forces, conservation of mass, and conservation of energy.
Time derivative of a function of material particle. At time t, a
material particle X moves to position x (X , t ) . The velocity of the material
particle is
V=
( ).
∂x X,t
∂t
Here, the independent variables are the time and the coordinates of the material
particles when the body is in the reference state.
We can also use x as the independent variable. To do so we change the
variable from X to x by using the function X (x , t ), and then write
v (x, t ) = V (X, t ).
Both v (x , t ) and V (X, t ) represent the same physical object:
the velocity of
material particle X at time t. Because of the change of variables, from X to x, the
two functions v (x , t ) and V (X, t ) are different. We indicate this difference by
using different symbols, v and V.
This practice, however, is not always convenient. Often we simply use the
same symbol for the velocity, and then indicate the independent variables:
v (x , t ) and v (X , t ). They represent the same physical object: the velocity of
material particle X at time t.
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Let G (X , t ) be a function of material particle and time. For example, G
can be the temperature of material particle X at time t. The rate of change in
temperature of the material particle is
∂G (X , t )
.
∂t
This rate is known as the material time derivative.
We can calculate the material time derivative by an alternative approach.
Change the variable from X to x by using the function X (x , t ), and write
g(x, t ) = G(X, t )
Using chain rule, we obtain that
∂G (X , t ) ∂g(x , t ) ∂g(x , t ) ∂x i (X , t )
.
=
+
∂t
∂t
∂x i
∂t
Thus, we can calculate the material time derivative from
∂G (X , t ) ∂g(x , t ) ∂g(x , t )
=
+
vi (x , t ).
∂t
∂t
∂x i
A shorthand notation for the time derivative of a function of material particles is
( )
( )
( )
Dg ∂G X,t ∂g x,t ∂g x,t
=
=
+
vi x,t .
Dt
∂t
∂t
∂xi
( )
In particular, the acceleration of a material particle is
∂v (x, t ) ∂vi (x, t )
ai (x, t ) = i
+
v j (x, t ).
∂t
∂x j
Compatibility of geometry. The rate of deformation relates to the
gradient of velocity as
1 " ∂v ∂v %
Dij = $$ i + j '' .
2 # ∂x j ∂xi &
This relation expresses the compatibility of geometry.
Balance of forces. Consider a small volume in the body. The balance
of forces now also include the inertial force:
∂σ ij
Dvi
.
+ bi = ρ
∂x j
Dt
Conservation of mass. Consider a fixed volume in space. The
conservation of mass requires that the increase in the mass in the volume equals
the mass flowing into the volume:
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( )
∂ρ ∂ ρ vk
+
=0.
∂t
∂x k
Conservation of energy. Let q be the flux of energy due to conduction,
i.e., the energy conducted per unit area per unit time. Consider a fixed volume in
space. The conservation of energy requires that the increase in the energy in the
volume equals work done by the stress, the energy transferred from the reservoir,
and the heat from neighboring parts of the fluid:
∂ ρ u ∂ ρ uvk
∂q
+
= σ ij Dij + Q − i .
∂t
∂x k
∂xi
( ) (
)
Using the conservation of mass, we simplify the conservation of energy as
∂q
Du
ρ
= σ ij Dij + Q − i .
Dt
∂xi
Increase of entropy. Under isothermal conditions, the free energy.
The free energy does not change.
Ds Q
∂ # qk &
% (≥0.
ρ
− +
Dt TR ∂x k %$ T ('
(
)
Note that Ds / Dt = Du / dt + pDv / Dt /T and Dv / Dt = vDkk .
Rewrite the
thermodynamic inequality as
"1 1 %
∂ "1%
+ pδij Dij + $$ − '' Q + qk
$ '≥0 .
∂x k # T &
# T TR &
The inequality holds when the composite system is not in thermodynamic
equilibrium, whereas the equality holds when the composite system is in
thermodynamic equilibrium. The thermodynamic condition holds for three
independent and arbitrary fields: the velocity v, the rate of heat Q, and the heat
flux q. Consequently, each of the three terms must be nonnegative.
(σ
)
ij
Thermal conduction. We satisfy the inequality as follow. The first
term is nonnegative if we adopt the model of viscosity. The second term is
nonnegative if we require that the heat transfer in the direction from high
temperature to low temperature. The last part of the above expression is
nonnegative if we adopt Fourier’s model of thermal conduction:
∂T
.
qi = −κ
∂xi
The thermal conductivity κ depends on the two thermodynamic variables.
Some other thermodynamic relations. Starting from the function
s u,v and the ration ds = du + pdv /T , we can derive many more functions an
( )
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many more relations by mathematical manipulations.
Of course, such
mathematical manipulations will never yield any new information. They do,
however, produce relations that are convenient under various experimental
conditions.
It is common that s increases with u when v is fixed. Thus, we can invert
the function s u,v and get anther function u s,v . Rewrite ds = du + pdv /T
( )
( )
(
)
as
du = Tds − pdv .
Recall an identity in calculus:
du =
( ) ds + ∂u ( s,v) dv .
∂u s,v
∂s
∂v
A comparison of the above two expressions gives that
∂u s,v
∂u s,v
, p=−
.
T=
∂s
∂v
These two relations express temperature and pressure as functions of the entropy
and volume.
Define the enthalpy by
h = u + pv .
Rewrite du = Tds − pdv as
dh = Tds + vdp .
Write
∂h s, p
∂h s, p
, v=
.
T=
∂p
∂s
Define the Helmholtz free energy by
f = u − Ts .
Rewrite du = Tds − pdv as
df = −sdT − pdv .
Write
∂f T ,v
∂f T ,v
,
.
s=−
p=−
∂T
∂v
Define the Gibbs free energy by
g = u − Ts + pv .
Rewrite du = Tds − pdv as
dg = −sdT + vdp .
Write
∂g T , p
∂g T , p
, v=
.
s=−
∂p
∂T
( )
( )
( )
( )
( )
(
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References
G.K. Batchelor, An Introduction to Fluid Mechanics. Cambridge University Press,
1967.
L.D. Landau and E.M. Lifshitz, Fluid Mechanics, 2nd ed. Butterworth Heinemann,
1987.
Z.G. Suo, Notes on thermodynamics, http://imechanica.org/node/288
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