7. Mechanics of fluids




We 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 fluids


Fluids have (by definition) the following properties:

(1)   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.

(2)   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.


The 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.


Its motion is characterized by the usual deformation measures

*  Velocity Gradient  

*  Strain rate decomposition  

*  Vorticity  

*  Vorticity-Spin relation  

* Velocity-acceleration relations





and is governed by the standard Conservation laws:


  Mass Conservation  

  Linear momentum conservation  

  Energy conservation  


Finally the constitutive law must satisfy the Entropy Inequality:  


Transformations under observer changes:

*  Velocity Gradient  

*  Stretch rate  

*  Spin  

*  Cauchy Stress  



7.2 General Form for Constitutive equations for fluids: 


We 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.


To 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 entropy  

* 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  

It is useful also to introduce the specific heat at constant volume  


Then, the equilibrium pressure , free energy, entropy and specific heat capacity are related by


In addition, the viscous stress and heat flux functions must satisfy



We 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 requires that


for all proper orthogonal tensors .  Recall that .  If we choose  with  an arbitrary skew tensor such that , then , and applying frame indifference at time t=0 then gives


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 spin.


·         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 see that


This 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


We can now let  and since  we see that



·         The remaining identities follow from these results, together with the definition  .  This implies that


In addition, since




The definitions of free energy and heat capacity also show that






7.3 Special cases of constitutive equations for fluids


The following special cases of constitutive equations are frequently used in fluid mechanics:


* 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 section.


* An Ideal Gas has free energy and stress response function


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 experiment.



Heat Flux: For all these cases, in problems where heat flow is considered, it is most common to set


where k is the thermal conductivity (often taken to be constant).




7.4 Special forms of field equations for viscous fluids


The flow of a viscous fluid is governed by

(i)                 The mass balance equation

(ii)               The linear momentum balance equation

(iii)             The constitutive law for a compressible, or incompressible viscous fluid

(iv)             In problems with heat transfer, the energy conservation equation.

together 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 results


* The Compressible Navier-Stokes Equation is a convenient form of the linear momentum balance equations for a compressible, linear viscous fluid.



* For the 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:  

(i)                 Recall first that .  It follows (by direct substitution, noting that temperature is constant) that



(ii)               The Navier-Stokes equation can therefore be rearranged as


(iii)             Taking the curl of both sides, recalling curl(grad(f))=0, and expressing the curl of the acceleration in terms of vorticity we then obtain




* 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 conservation equations


The constitutive equations also show that






7.5 Special forms of field equations for elastic fluids


For elastic 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



* The Vorticity transport equation for an elastic fluid reduces to 


An 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


and therefore  along streamlines. 





7.6 Incompressible flow


In 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


The quantity  is known as the kinematic viscosity.


* The Vorticity transport equation reduces to 





7.7 Incompressible, inviscid, irrotational flow (potential flow)


If a flow field is irrotational, the velocity field can be derived from a scalar potential


(it 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 the pressure


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).


Together 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 number , 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 .



7.9 Boundary 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 some sort.   


The following boundary conditions are most commonly used in these problems:

  1. 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.


  1. 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 vanishes.



  1. 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 interfaces.


  1. 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


The 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


* Mass Conservation:  

* 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 form


where  is the body force per unit mass.



Example 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.




  1. Consider a control volume as shown in the figure, with interior R and boundary B.
  2. Note that  
  3. Note that the fluid is inviscid, and therefore the shear stress acting on the wall vanishes.  Momentum balance in the j direction thus shows that


  1. 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  



Example 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 pump.


  1. 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)
  2. Assume axisymmetric flow
     so fluid velocity is independent of  
  3. Mass conservation gives
    , where z
    is the out of plane depth of the blades.
  4. 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 integrals gives


  1. 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 .  


We 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 that


Only 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



The 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


To 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:

* The solution can be generated from a harmonic potential satisfying

* The 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 Bernoulli equation


where  is a reference pressure at some point where the velocity and body force potential are zero.


It 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 solutions.







7.11.1 Flow surrounding a moving sphere


The 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


The flow surrounding a rigid cylinder that moves without rotation with velocity  can be computed from the following potential



As 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 flow)


The 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 between plates


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 follows. 

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 viscous fluid


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




7.12.3 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


* Mass conservation


 * Navier Stokes


* Energy conservation


* Entropy equation


* Constitutive law: internal energy, free energy and stress response function


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 .



7.13.1 Sound waves in an ideal gas


Small 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




is 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



These 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


and hence, since s=const ,  and  the pressure is related to density by  where k is a constant.  It follows that


·         For small amplitude perturbations, we approximate the Navier-Stokes equation as


Taking the time derivative of this expression, and assuming body forces to be independent of time then gives


·         Note that


and linearizing the mass conservation equation shows that


Combining 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).


7.13.2 Flow behind an accelerating piston


This 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


The wave speed can be calculated from the condition



These results can be derived as follows. For one-dimensional flow in the y direction, the Navier-Stokes and mass conservation equations reduce to


where  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 matrix form


Now, 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 lines satisfying  


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  

·         Integrating, we 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 is


Finally, 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.  


Consider 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.


Conservation and thermodynamic laws relating these quantities have the following form:


* Mass Balance  

* 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


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

Mass Conservation:  

Linear Momentum Balance  

First law of thermodynamics



Second law of thermodynamics


Then, 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.


The Rankine-Hugoniot 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 that


where .   The energy equation can then be re-written as


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.