7. Mechanics of fluids
now apply the general principles described in the preceding chapters to
specific problems. In this chapter, we
give a short introduction to the governing equations for fluids, and give
solutions to some simple boundary value problems.
7.1 Summary of governing equations for
have (by definition) the following properties:
A fluid has no
natural reference configuration, so all governing equations are expressed in spatial form. The lack of reference configuration also
means that the response of the fluid can only depend on kinematic variables
that characterize motion only of the current
configuration i.e. the velocity gradient, and measures
derived from it.
A fluid at rest
can support no shear stress the stress state in a stationary fluid is just
a hydrostatic pressure, which may vary with position.
central problem in a fluid mechanics problem is generally to determine the
velocity distribution ,
Cauchy stress distribution and (sometimes) temperature ,
as functions of position and time. The
fluid is characterized by the following physical quantities:
The mass density
The specific internal energy
The specific entropy
The specific Helmholtz free energy
A stress response function
A heat flux response function
Body forces: The fluid is subjected to an external body force per unit mass.
motion is characterized by the usual deformation measures
Strain rate decomposition
is governed by the standard Conservation
Linear momentum conservation
the constitutive law must satisfy the Entropy
Transformations under observer changes:
7.2 General Form for Constitutive
equations for fluids:
now list the general form of the constitutive equations for a fluid that are
consistent with frame indifference and the entropy inequality. In practice, most problems of interest are
approximated using one of several special cases of the general equations. These will be listed separately.
be consistent with frame indifference and the laws of thermodynamics, the
specific free energy, internal energy, Helmholtz free energy, stress response
function and heat transfer function must have the forms
Specific internal energy
Specific Helmholtz free energy
Stress response function
with . Here, can be interpreted as the pressure that would
be measured in the fluid with density and temperature when at rest.
This decomposition is given a-priori, and is constructed to ensure that
the stress state in a fluid at rest supports zero shear stress.
Heat flux response function
is useful also to introduce the specific
heat at constant volume
the equilibrium pressure ,
free energy, entropy and specific heat capacity are related by
addition, the viscous stress and heat flux functions must satisfy
next summarize the reasoning that leads to these conclusions:
Frame indifference shows that the free energy function and the stress and heat transfer
response functions depend only on the symmetric part of the velocity gradient
(i.e. the stretch rate), and must be isotropic functions. To see this, note that frame indifference
for all proper orthogonal tensors . Recall that . If we choose with an arbitrary skew tensor such that ,
and applying frame indifference at time t=0
Then, selecting shows that . Finally, with arbitrary Q, we see that
This implies that the free energy and stress and heat
transfer response functions are all isotropic functions.
Not everyone agrees with the argument that constitutive
behavior must be independent of spin. This
conclusion relies on the assumption that the response functions are invariant
to changes between reference frames that are rotating relative to one
another. If invariance is required only
to transformations between frames that have a fixed relative velocity and
orientation, spin can enter in the constitutive behavior. Some models of turbulent flow do include
The free energy inequality provides the remaining conclusions. Take the time derivative of the free energy
function and substitute into the free energy inequality to see that
(time derivatives are all with fixed x). Note that
and also recall that from mass conservation . Collecting coefficients of rate quantities we
inequality must hold for all possible ,
which can all be independently prescribed.
It follows immediately that
Next, note that
and replace with with it follows that
can now let and since we see that
identities follow from these results, together with the definition . This
definitions of free energy and heat capacity also show that
7.3 Special cases of constitutive
equations for fluids
following special cases of constitutive equations are frequently used in fluid
An Elastic fluid (also known as a barotropic fluid or Eulerian fluid) has free
energy and stress response independent of temperature
Its heat capacity is zero, and heat flow is not
usually considered in applications that use this model. Either the free energy,
or the pressure, must be fit to experiment.
The two are related through the equations listed in the preceding
An Ideal Gas has free energy and stress
where R is
the gas constant, is the specific heat capacity (a constant), and
is an arbitrary constant. Ideal gases are also characterized by the
specific heat at constant pressure and the ratio . An ideal gas is inviscid. Pressure, temperature and density in an ideal
gas are related by the equation of state
. Note that R
in these equations is the individual
gas constant, which has units of Joules/kg/Kelvin, and varies from one gas to
another. It can be computed from the
universal gas constant using the relation where m
is the molecular weight of the gas (the mass of one mole of the gas).
A Compressible Linear (Newtonian) Viscous
Fluid has free energy and stress response function
where are functions of density and temperature that
must be fit to experiment. They are
known as the bulk and shear viscosity of a fluid, respectively.
A Compressible nonlinear (non-Newtonian)
Viscous Fluid has free energy and stress response function
where are the three invariants of ,
and are three functions that must be fit to
Heat Flux: For all these cases, in problems where heat flow is
considered, it is most common to set
k is the thermal conductivity (often
taken to be constant).
7.4 Special forms of field equations for
flow of a viscous fluid is governed by
The mass balance
momentum balance equation
law for a compressible, or incompressible viscous fluid
In problems with
heat transfer, the energy conservation equation.
with any relevant boundary conditions.
For calculations, it is convenient to combine (ii) and (iii) to obtain a
single equation relating the static pressure and velocity, with the following
The Compressible Navier-Stokes Equation is
a convenient form of the linear momentum balance equations for a compressible, linear viscous fluid.
case of Density independent viscosity
this equation can be simplified to
The Vorticity transport equation governs
evolution of vorticity in a fluid with density independent viscosity at
constant temperature. It has the form
This result is not especially useful
for viscous, compressible flows, but simplifies considerably if the fluid is
incompressible, or if the viscosity can be neglected. The result can be derived as follows:
first that . It follows (by direct substitution, noting
that temperature is constant) that
equation can therefore be rearranged as
Taking the curl
of both sides, recalling curl(grad(f))=0,
and expressing the curl of the acceleration in terms of vorticity we then
The Entropy Equation is a useful result for
analyzing gas flows. It states that the
rate of change of specific entropy at a point in a compressible viscous fluid
can be computed from
To see this, start with the identity . It follows that
Next, recall the mass conservation and energy
constitutive equations also show that
7.5 Special forms of field equations for
fluids, we can show the following results
Euler’s equations of motion are the special cases of the linear momentum balance
equation for elastic fluids
acceleration can be written in many other forms, of course)
The Vorticity transport equation for an
elastic fluid reduces to
important conclusion from this result is that if the body force can be derived
from a potential (so its curl vanishes), and the flow is irrotational at time t=0, it remains irrotational.
Bernoulli’s equation is a special result governing steady
flow of an elastic fluid that is subjected to conservative body forces. Conservative body forces can be described by a
potential energy, so that . Then, the Bernoulli equation states that,
along a streamline
Recall that a streamline is a curve that is everywhere
tangent to the velocity vector (for steady flow, the streamlines are coincident
with the trajectories traced by material particles). For the particular case of irrotational flow, this quantity is
constant and independent of position.
This result can be derived as follows. First, recall that
Noting that steady flow implies that ,
the momentum balance equation therefore reads
For irrotational flow ,
and this equation can then be integrated to obtain the Bernoulli equation. Similarly, taking the dot product of both
sides of this equation with the velocity vector and rearranging shows that
therefore along streamlines.
7.6 Incompressible flow
many cases fluids can be approximated as incompressible, which simplifies the
governing equations further.
The mass balance equation reduces to
The Incompressible Navier-Stokes Equation can
be simplified to
quantity is known as the kinematic viscosity.
The Vorticity transport equation reduces
7.7 Incompressible, inviscid, irrotational flow
a flow field is irrotational, the velocity field can be derived from a scalar
is easy to show that of this form is irrotational. The converse can be shown for fluids
occupying a simply connected region the potential may be multiple valued for
multiply connected regions).
If a fluid is incompressible its velocity
field (assuming irrotational flow) can be computed by solving for . In particular
Mass Conservation yields the governing
equation for :
If the fluid is inviscid, then Bernoulli’s equation (generalized to time dependent flow) yields the governing equation for
where is the potential for body force per unit mass,
and f(t) is an arbitrary function of
time. Since the flow is irrotational,
this expression holds everywhere (not just along streamlines).
with appropriate boundary conditions, these equations determine both the
velocity and pressure associated with the flow.
7.8 Stokes Flow
The general Navier-Stokes
equation is nonlinear, because of the terms involving the square of the
velocity in the acceleration. If these
terms can be neglected (as is the case for flows with low values of Reynolds
where V and L are a representative velocity and length), the Navier-Stokes
equation can be approximated as
Velocity fields that satisfy
this equation are known as Stokes Flows. For steady flows, or for particularly slow
flows, the equation can be simplified further by setting .
conditions for fluid flow problems
There are three common types
of fluid mechanics problem:
1. A rigid or elastic solid which moves through a
stationary fluid or gas;
2. A bubble of gas or fluid of a second phase moves
through a surrounding fluid;
3. A fluid or gas flows through rigid or elastic walls of
The following boundary
conditions are most commonly used in these problems:
- Where a viscous
fluid meets a solid boundary, the fluid is usually assumed to have the
same velocity as the solid surface (both normal and tangential components
of velocity are equal). This is the
no slip boundary condition. Normal and tangential components of
traction must also be continuous across the interface.
- Where an inviscid
fluid meets a solid boundary, the normal component of velocity must be
equal in both solid and fluid where the two meet. The tangential component of velocity
may be discontinuous. The normal
component of traction is continuous across the boundary, and the shear traction
- A bubble inside a fluid is usually assumed to be
inviscid, and tractions are required to be continuous across the fluid/gas
interface. For small bubbles, it
may be necessary to account for the effects of surface tension as
well. In this case, the
solid/fluid interface must be modeled as a separate phase, with its own
constitutive behavior. It is above
our pay-grade to discuss this in detail here, but the basic concepts we
develop here for solid and fluid phases should be sufficient for you to be
able to follow texts and papers that work through the thermodynamics of
- In some high-speed compressible viscous flow
problems; or in microfluidics applications, the no-slip boundary condition
is found to be inaccurate. In this
case the tangential component of velocity may be discontinuous across the
boundary. An additional
constitutive law must specify a relationship between the tractions acting
on the interface and the velocity discontinuity.
7.10 Control surface method for fluid flow problems
control surface method is a helpful approach to computing average quantities
such as resultant forces exerted by a fluid flow on a surface.
The method relies on the
conservation laws for a control volume
Linear Momentum Balance
Angular Momentum Balance
Mechanical Power Balance
First law of thermodynamics
Second law of thermodynamics
In addition, the Bernoulli equation can be applied to inviscid, incompressible flows along a
streamline . In this case it has the
where is the body force per unit mass.
1: A jet of incompressible,
inviscid fluid with mass density ,
cross sectional area ,
and speed is incident on an inclined wall, as shown in the
figure. The surrounding atmostpheric pressure is and gravity may be neglected. Calculate the
force acting on the wall.
- Consider a control volume as shown in the figure, with interior R and boundary B.
- Note that
- Note that the fluid is inviscid, and therefore the shear stress
acting on the wall vanishes. Momentum
balance in the j direction thus
- We recognize the last integral as the force exerted by the wall on
the fluid. The force exerted by the
fluid on the wall must be equal and opposite, and so
2: The figure shows an
idealized centrifugal pump. Fluid enters
the rotating blades with radial and tangential velocity and exits with velocity . The tangential velocity of the fluid is
equal to that of the blades at both entry and exit. The blades rotate with angular speed and the mass flow rate through the pump is . Calculate the torque required to rotate the
- Consider a control volume consisting of the
annular region occupied by the blades, but excluding the blades themselves
(only one piece of the CV is shown in the figure)
- Assume axisymmetric flow so fluid velocity is independent of
- Mass conservation gives ,
where z is the out of plane depth of the blades.
- The angular momentum equation gives
The integral on the left represents the net moment
exerted by the blades on the fluid. The
contributions to the integral on the right from the surfaces adjacent to the
blades cancel, leaving only the contributions from the inner and outer surfaces
of the annulus. Evaluating the
- Note that ,
which gives .
Example 3: For
our third example, we work through Von Karman’s classic estimate of the drag
force on a stationary plate inside a steadily flowing stream. The figure shows the problem to be
solved. Ahead of the plate, the fluid
has a uniform horizontal velocity with direction parallel to the edge of the
plate. A boundary layer in which the
fluid flow is slowed develops adjacent to the plate, and as a result the
horizontal velocity has profile at the trailing edge of the plate. The velocity field is assumed to be uniform
outside the boundary layer. We assume that the width of the boundary layer
at the trailing edge of the plate can be
determined somehow. Our goal is then to
determine the drag force on the plate in terms of ,
L and the mass density of the fluid .
choose a control volume that is bounded by the plate, and at the top follows a
streamline outside the boundary layer which terminates at the trailing edge of
the boundary layer. The pressure
outside this control volume is uniform.
No fluid crosses the upper or lower surfaces of the control volume, and
we assume steady-state conditions.
Mass conservation then implies that
Linear momentum balance shows
the plate exerts a force on the control volume (the pressure is uniform, and
there is no shear stress acting on the top surface of the control volume
because the velocity is uniform), so we recognize the integral on the left as
the resultant force exerted by the plate on the control volume. The drag force exerted by the fluid on the
plate must be equal and opposite.
Substituting for h from the first equation into the
second, and noting that a boundary layer must form on both sides of the plate,
gives the following expression for the total drag force
formula is only useful if the velocity profile at the trailing edge of the
plate is known. As a rough guess, we
could take a parabolic velocity distribution which satisfies at y=0
and at . This then yields
use this expression we still need to find the thickness of the boundary layer,
but that’s beyond our pay-grade. An
asymptotic analysis gives ,
where is the fluid viscosity.
7.11 Exact solutions to simple incompressible,
inviscid, irrotational flow problems (potential flow)
The governing equations for
potential flow can be reduced to:
can be generated from a harmonic potential satisfying .
velocity components follow as
The velocity must satisfy at any point where it meets a surface with
normal moving with velocity
The pressure can be computed from the
is a reference pressure at some point where
the velocity and body force potential are zero.
is relatively straightforward to solve the Laplace equation in 3D, many solutions have been found by
guesswork; techniques such as Fourier transforms can be used to solve more
complex problems. In 2D, complex
variable methods are very effective. It
turns out that both the real and imaginary parts of a differentiable function
of a complex number (eg a polynomial function of a complex number z=x+iy) are solutions to Laplace’s
equation. Techniques such as conformal
mapping and analytic continuation are also useful. We don’t have time in this course to discuss
these procedures in detail instead, we just list a few important
surrounding a moving sphere
flow surrounding a rigid sphere with radius a
that starts at the origin, and then moves without rotation with velocity can be computed from the following potential
7.11.2 Flow surrounding a moving cylinder
flow surrounding a rigid cylinder that moves without rotation with velocity can be computed from the following potential
an exercise, you might like to calculate (i) The components of the velocity
field; (ii) the pressure distribution acting on the sphere and cylinder; (iii)
The drag force acting on the sphere and cylinder (the result is surprising at
first, but note that there is no dissipation so at steady-state there can be no
drag; (iv) the total kinetic energy in the fluid.
7.12 Solutions to simple viscous flow problems (Stokes
equations governing Stokes flow are also fairly easy to solve, because the equations
are linear. This means that Fourier
transform techniques and complex variable methods both work. We won’t try to cover these here, but list a
few examples of solutions to Stokes flow problems
7.12.1 Laminar flow of an incompressible viscous fluid
The problem to be solved is illustrated in the figure.
An incompressible fluid with mass
density and viscosity is confined between parallel plates. External boundary constraints at the ends of
the plate induce a (constant) pressure gradient in a direction parallel to the plates. The top plate moves with speed V, and the bottom is stationary.
The velocity field in the
fluid has the form
These results can be derived as
1. Boundary conditions at the top and bottom surfaces of
the plate suggest that the flow velocity must be parallel to the plates, and
independent of . We can assume that
2. The flow is steady, so the general Navier-Stokes
equation reduces to
3. We can easily integrate this equation to see that
4. Boundary conditions at
the plate surfaces imply that . We can then solve for the unknown constants A and B, giving the answer stated.
7.12.3 Rigid sphere moving at steady speed through a
The velocity field
surrounding a rigid sphere that moves without rotation with instantaneous
velocity through a viscous fluid is
For the velocity field can be approximated as
This solution is called a
‘Stokeslet.’ The Stokeslet solution can
also be expressed in terms of the force exerted by the sphere on the fluid for a sphere at the origin we have
Spherical bubble moving at steady speed through a viscous fluid
The velocity field
surrounding a spherical bubble that moves with instantaneous velocity through a viscous fluid is
7.13 Solutions to simple compressible flow problems
Many compressible flow
problems are concerned with flows in air, which can often be approximated as an
ideal gas. The governing equations for
an ideal gas reduce to
Constitutive law: internal energy, free energy and stress response function
R is the gas constant, is the specific heat capacity (a constant),
and is an arbitrary constant. Ideal gases are also characterized by the
specific heat at constant pressure and the ratio .
7.13.1 Sound waves in an ideal gas
amplitude sound waves can be approximated as a small variation in density and
pressure superimposed on a static state.
The velocity field for a small amplitude acoustic wave in still air is
irrotational, and the velocity field and the change in density in the gas
satisfy the equations
the speed of sound. In actual calculations it is often convenient to introduce
a flow potential that satisfies
The flow potential satisfies
the wave equation
results are derived as follows.
- It is usually assumed that heat flow can be
neglected on the time-scales associated with acoustic vibrations. This means that the entropy of the gas
is constant (see the entropy equation).
- For small perturbations, the change in pressure
of the gas is linearly related to its density. We can write
For an ideal gas, it is possible to find a formula for
the sound speed. Recall that
hence, since s=const , and the pressure is related to density by where k
is a constant. It follows that
amplitude perturbations, we approximate the Navier-Stokes equation as
the time derivative of this expression, and assuming body forces to be
independent of time then gives
linearizing the mass conservation equation shows that
these results gives the wave equations.
If the fluid is
at rest at time t=0 its vorticity
vanishes. The vorticity transport
equation then shows that the vorticity must vanish at all times, i.e. the flow
is irrotational. The velocity field
therefore satisfies and hence we can set .
The wave equations can be
solved easily for an infinite region.
Two solutions are:
Plane waves. A plane wave
can be visualized as a propagating ‘wall’ of sound. Let ,
where is a unit vector, and f is an arbitrary
function. This corresponds to a plane
wave travelling in a direction parallel to the unit vector . It is easy to show that this velocity field
satisfies the wave equation for any f(x). The density change follows as ,
and of course the pressure is proportional to the density change.
Monopole. A monopole
is a point source of sound. It would be
produced, e.g. by a vanishingly small radially vibrating sphere. The velocity field has the form
where f(x) is again an arbitrary function.
Dipole. A dipole is two nearby
monopoles emitting equal and opposite signals it would represent, eg, the sound produced by
a sphere moving backwards and forwards along a line. A loudspeaker that is removed from its box behaves
somewhat like a dipole. The displacement field has the form
where is a unit vector parallel to the direction of
the dipole (eg the direction of the line along which the sphere moves).
behind an accelerating piston
is a classic problem that is often used to illustrate the method of
characteristics. The figure shows a
piston in a tube that is filled with an ideal gas with mass density and wave speed (it is more convenient to specify the wave
speed than the pressure, but of course the two are related). At time t=0
the gas and piston are at rest. The
piston then starts to move to the left with constant acceleration a.
The velocity field in the gas
for is given by
wave speed can be calculated from the condition
results can be derived as follows. For one-dimensional flow in the y direction, the Navier-Stokes and mass
conservation equations reduce to
is the sound speed. Solving this equation is a bit above our
pay-grade in this course, but you are probably covering this material in math,
so we’ll work through it anyway. The first step is to write the equations in
let be a left eigenvalue/eigenvector pair of A.
The eigenvalues/vectors are easily shown to be .
It follows that
If we set we see further that
along characteristic lines
such that . Substituting for ,
we see that
These conditions hold along
We can now apply this result
to the piston problem.
The (t,y) characteristic diagram is shown in
the figure. The red characteristic lines
have slope ;
the green ones have slope (we drop the subscript s for clarity)
Note that the red
lines terminate in a region where the gas is stationary, and the wave speed is . Therefore along these lines. It follows that the wave speed in the moving
region of the gas is related to its velocity by
For the green
lines, we know ,
with B some constant. Since each green lines intersects a red one,
we already know the wave speed in terms of v,
and substituting for c shows that .
It follows that v must be constant on the green lines, and therefore they have
constant slope . We can apply this at the point where the
green line intersects the piston (the gas velocity there is equal to that of
the piston) to conclude that
get where D
is a constant of integration and since we know at time we can solve for D and find that
Finally, we can
use this result to solve for the value of corresponding to a general point y at time t in the tube. The solution
the velocity at time t of the point y in the tube follows as giving the solution stated.
The wave speed
follows from . Note that at time the wave speed drops to zero at the
piston. The wave speed can’t be
negative, so after this time the piston separates from the gas and a vacuum
develops between the piston and the advancing gas. After this time the gas/vacuum interface
moves with speed .
7.13.3 Stationary Normal Shock in an ideal gas
Shock waves are surfaces in a compressible fluid or
gas across which fluid properties and state experience a sharp
discontinuity. They occur between
regions where flow is supersonic and subsonic.
The discontinuities across a shock are not arbitrary they are related by the usual conservation laws
and constitutive equations for a fluid.
a stationary shock that separates regions 1 and 2 of a flow. For simplicity, assume that fluid flows in a
direction perpendicular to the plane of the shock. Let ,
denote the (uniform) velocity, pressure, heat
flux, density , specific internal energy and specific entropy just adjacent to
the two sides of the shock.
thermodynamic laws relating these quantities have the following form:
Linear Momentum Balance
Energy Conservation (1st law) .
Entropy Inequality (2nd law)
For an ideal gas, with
negligible heat flow, these can be combined to yield several additional
relations across a normal shock.
Rankine Hugoniot Relations
Prandtl’s relation implies
that flow must be supersonic on one side of the shock and subsonic on the
other. In addition, to satisfy the entropy inequality, the supersonic flow must
be upstream of the shock.
These results can be derived
as follows: Begin by recalling the
conservation laws for a control volume
Linear Momentum Balance
First law of thermodynamics
Second law of thermodynamics
introduce a control volume consisting of a narrow rectangular region enclosing
the shock. The thickness can be reduced
to zero, so the volume integrals can be made arbitrarily small. In addition, recall that the stress in an
ideal gas is hydrostatic. Finally,
evaluating the separate integrals over the two external surfaces of the control
volume and localizing the integrals leads to the conservation relations.
relations can be derived from the energy conservation equation
Recall that ,
so this equation can be rearranged as
The momentum conservation
equation and mass conservation can also be combined to see that
and hence eliminating the
velocity term from the energy equation
which can be rearranged to
give the Rankine Hugoniot relation. The
second identity follows by substituting for an ideal gas.
To show Prandtl’s relation recall
where . The energy equation can then be re-written
Which then gives the first
relation. To show the second note that
We can then use the first
Prandtl relation to see that
where we have used mass
conservation . Finally, recall that
Substituting this into the
preceding result and simplifying gives the second Prandtl formula.