Fluid Dynamics - Math 6750 - Fall 2013
Basic principles - Part II
1
Stress tensor
Definition 1. Consider a closed surface S with outward unit normal vector n̂. The stress
vector or traction, denoted by t, is the force per area exerted by the fluid on the plus side of
the surface (the side towards which n̂ is pointing).
The traction at a point is formally defined as
t = lim
∆S→0
∆F
.
∆S
Remark 1. If the volume is under tension, then t · n̂ > 0 and the stress is positive. If the solid
is under contraction, then t · n̂ < 0 and the stress is negative.
To derive general properties of t, we consider the limit as we decrease the material control
volume around a point x to zero while holding the geometry of Vm constant. Let l be a
characteristic linear dimension, so that Vm = l3 . Going back to the conservation of linear
momentum after applying the Reynolds transport theorem, we have
Z
Z
(∂t (ρu) + ∇ · (ρuu) − ρf ) dV =
tdS.
Vm (t)
Sm (t)
Denoting by h·i the mean value of the integrands over the volume or the surface, we have using
the mean-value theorem
h·il3 = h·il2 .
As l → 0, the volume integral of momentum and body-force terms vanishes more quickly than
the surface integral of the stress vector and we have the following proposition.
Proposition 1 (Principle of Stress equilibrium).
!
Z
lim
l→0
tdS
= 0.
Sm (t)
Proof. Taking l → 0 in the averaged equation gives
*Z
+
lim
t→0
tdS
= 0.
Sm (t)
As we let l → 0, the volume shrinks to the point x and the average of the integrand becomes
the value at the point x and the claims holds point wise.
Remark 2. The stress vector depends on the point x and the unit normal vector n̂.
Next, we construct a pyramid around the point x as illustrated in Fig. 1. The goal is to
find an expression for t(n̂) in terms of the component of t on three perpendicular surfaces.
1
e3
C
n̂
x
O
B
e2
A
e1
Figure 1: Cauchy pyramid around a point x.
Proposition 2 (Cauchy stress tensor).
t(n̂) = n̂ · (ei t(ei )) = n · σ(x).
σ is a rank 2 tensor called the stress tensor at the point x.
Proof. We denote by ti the traction vector acting on the perpendicular planes with unit normal
vectors ei . Using Newton’s third law (action-reaction), we note that
t(−ei ) = −ti .
On the pyramid, there are four tractions, the one we are trying to find t(n̂) and the tractions
on the coordinates plane. Applying the conservation of linear momentum to the pyramid and
using the mean-value theorem like in the derivation of the stress equilibrium to the pyramid,
we have
ht(n̂)i∆Sm − hti i∆Si = h·il3 .
∆Si is the projected area of ∆Sm onto the plane perpendicular to ei , and thus the areas are
related by projections of the normal vector n̂:
∆Si = ∆Sm (n̂ · ei ).
Plugging in and dividing by l (∆Sm = l2 ) yields
h·il = ht(n̂)i − hti i(n̂ · ei ).
Taking the limit l → 0, the average reduces to the value at x and we find
t(n̂) = t(ei )(n̂ · ei ) = n̂ · ei ti = n̂ · σ.
2
Remark 3.


σx τxy τxz
• The Cauchy stress tensor is often written as σ = τxy σy τyz , where σi are the
τxz τyz σz
normal stresses and τij the shear stresses.
0 , σ 0 , σ 0 are the eigenvalues of the Cauchy stress
• The principal stresses denoted by σ11
22 33
(rotation to principal axes).
• p̄ = − 13 tr(σ) is the mean (or averaged) normal stress.
• In the principal frame, the Cauchy stress can be rewritten as σ = σ 0 − pI (in a state of
compression), where σ 0 is the deviatoric stress associated with change in shape and pI is
the hydrostatic stress associated with change in volume. In this principal axes, we have

 0
σ11 − 31 σ
0
0
1
0 − 1σ
.
0
0
σ22
p = − σ σ0 = 
3
3
0 − 1σ
0
0
σ33
3
Since tr(σ 0 ) = 0, at least one of the entry of σ 0 has to be positive and one negative, in
another words one of the principal axes corresponds to elongation and one to contraction.
Example 1 (Fluid at rest). We consider an isothermal, stationary fluid, i.e u = 0. The only
surface force is the thermodynamics pressure p (see Thermodynamics) which acts normal to the
surface. By the principle of stress equilibrium, the magnitude of the pressure is independent
of the normal. By convention p is always positive. Remembering that t(n̂) is positive when it
acts inwards, the traction has the form
t(n̂) = −pn̂.
From the previous equation, it follows that the stress tensor in a fluid at rest is purely diagonal
σ = −pI.
Since the stress tensor is diagonal, a fluid at rest can’t sustain any shear stresses.
For a fluid at rest under gravitiy, the conservation of linear momentum becomes
ρg − ∇p = 0.
The last equation reflects the well-known fact (diving, flying a plane) that the pressure increases
with depth under gravity. Assuming that ρ remains constant, the solution is simply
p(z) = p0 + ρgz.
Consider a vertical cylinder of fluid of radius R between z = z1 and z = z2 . The net pressure
force acting on the ends acts upward and has a magnitude
(p(z2 ) − p(z1 ))πR2 .
3
This is equal to the total downward body force ρg(πR2 )(z2 − z1 ) = ρg(πR2 )L. This force is
called the buoyancy force acting on a body immersed in a fluid and is due solely to the increase
of the hydrostatic pressure with depth. If ρs is the density of the solid, then the force due to
gravity is
ρs g(πR2 )L
and the net force is
F = (ρs − ρ)g(πR2 )L.
This is the well-known Archimedes principle. If ρ = ρs , then the body is called neutrally
buoyant.
Knowing that the traction can be expressed as a second rank tensor gives us the conservation of linear momentum equation, however it introduces six new unknowns, the normal and
shear stresses. These quantities can’t be derived from conservation equations, rather they are
educated guesses about the behavior of σ as a function of u and its derivatives.
2
Rotation and rate of shear
We want to represent the rotation and shearing of a fluid element through tensor quantities.
These tensors are going to be fundamental in describing σ, in other words in writing down
constitutive equations for the six component of stress.
We consider a square fluid element ABCD that is being rotated and sheared after a time
δt to a square A0 B 0 C 0 D0 as described in Fig. 2. Let δα be the angle defining the rotation from
the side CD to the side C 0 D0 and δβ be the angle defining the rotation from the side BC to
B 0 C 0 . We remark that ∆x corresponds to a distance travelled over time, while δx is the length
of the side. We know from Calculus that
Z δt
∆x =
u(x(t), y(t))dt,
0
where u is the x-component of u. Since we are considering only short time, the velocity can
be expanded in a Taylor series about (0, 0) yielding
∆x = u(0, 0)δt + . . .
and similarly for ∆y. The goal is to obtain expressions for δα and δβ in terms of u, v.
Proposition 3. In the limit as δx, δy, δt tend to zero, the rate of change of the angles α, β are
α̇ =
∂v
(0, 0)
∂x
β̇ =
∂u
(0, 0).
∂y
(1)
Proof. In class.
Since α is measured counterclockwise, the rate of clockwise rotation of the fluid becomes
with the proposition
1 ∂u ∂v
1
(β̇ − α̇) =
−
.
2
2 ∂y ∂x
4
A0
B0
D0
δβ
∆y
B
δα
A
C0
δy
C
∆x
δx
D
Figure 2: Infinitesimal fluid element being rotated and sheared.
Similarly, the shearing rate at which the sides B 0 C 0 and D0 C 0 are approaching each other is
1
1 ∂u ∂v
(β̇ + α̇) =
+
.
2
2 ∂y ∂x
∂ui
can be decomposed into
Definition 2. The rate of deformation tensor D given by Dij = ∂x
j
its symmetric and antisymmetric parts as follows
∂uj
∂uj
1 ∂ui
1 ∂ui
Dij =
−
+
+
.
(2)
2 ∂xj
∂xi
2 ∂xj
∂xi
The symmetric part is called the rate of strain tensor E = 12 (∇u + ∇uT ), while the antisymmetric part is the vorticity tensor Ω = 21 (∇u − ∇uT ).
Remark 4. The vorticity has 3 independent components corresponding to the curl of u. The
symmetric part has 6 independent components corresponding to the stress at a point. The
rate of deformation tensor is a rank 2 tensor with 9 independent components representing both
rates of rotation and shear of a fluid element.
Exercise 1. Show that the rate of change of the distance δx connecting two points P and Q
can be expressed as
1D
|δx|2 = δx · E · δx + O(|δx|3 ).
2 Dt
HINT: Use a Taylor series expansion and the fact that Ω is antisymmetric.
5
3
Constitutive equations
3.1
Heat flow
Proposition 4 (Fourier’s law of heat conduction). Let K be a positive rank 2 tensor (material
dependent), q be the heat flux and T the temperature. Then
q = −K · ∇T.
For an isotropic material where the heat flux only depends on the magnitude of ∇T and not
on the material’s orientation, K reduces to a diagonal tensor K = kI and
q = −k∇T.
(3)
Remark 5. Fourier’s law says that the heat flux goes from high to low temperature. We already
used Fourier’s law in the derivation of the energy conservation.
3.2
Fluid
The constitutive equations for a fluid are derived from the following observations
• When the fluid is at the rest, the pressure is hydrostatic and equals the thermodynamic
pressure.
• There are not preferred direction in a fluid (isotropy).
• There is no shearing deformation in a rigid-body rotation.
• The stress tensors depends on the rate-of-strain tensor.
From these conditions, we postulate
σ = −pI + τ (E, . . .).
Definition 3. If τ (E) is linear, then the fluid is called Newtonian.
Remark 6. Since σ is symmetric, τ is also symmetric.
Since E and σ are rank 2 tensor, the most general linear relationship is
σ=A:τ
where A is a rank 4 tensor.
Proposition 5 (Newtonian fluid).
σ = (−p + λtr(E))I + 2µE.
µ is called the dynamic viscosity and λ is the second viscosity coefficient.
Proof. In class.
Definition 4. The quantity ν = µ/ρ is called the dynamic viscosity.
6
(4)
liquid is squeezed between the tongue and the palate
and next sheared during the back and forth or sideways
movements of the tongue. However, a study by de
Bruijne, Hendrickx, Anderliesten, and de Looff (1993)
on the break up process in the oral cavity of small
samples of oils of different viscosity and of air bubbles
showed that the effective flow, causing break up, was
purely elongational. Samples of 0.5 g were placed in the
mouth after which they were masticated during a time
varying between 5 and 60 s and next expectorated in an
yield stress higher than 50 Pa will not be broken up and
dispersed by the saliva flow. A stress of about 50 Pa is in
reasonable agreement with the stress at which Shama
and Sherman (1973a) assumed that there is a transition
in the way viscosity is evaluated in the mouth.
During mastication of foods, with a yield stress above
50 Pa, it is likely that they will first be compressed
between the tongue and the palate. Such a deformation
resembles squeezing flow between parallel plates. If the
plates are lubricated this result in a biaxial extensional
Fig. 1. Bounds of shear stress-shear rate associated with oral evaluation of viscosity and superimposed on these bounds shear stress-shear rate data
for various foods (after Shama & Sherman, 1973a).
Figure 3: Relationship between the shear rate and the shear stress for real fluids (left theory,
right from Shama and Sherman (1973)).
Remark 7. The average pressure or mechanical pressure is
2
1
p̄ = − tr(σ) = p − λ + µ ∇ · u.
3
3
Thus in general, the mechanical pressure, which is a measure of the translation of molecules,
is different from the thermodynamic pressure, which is a measure of the total energy.
Definition 5. The quantity κ = λ + 23 µ is called the bulk viscosity.
Proposition 6 (Stokes’ relation). For a monoatomic gas,
κ=0
2
λ = − µ.
3
or
Remark 8.
• In general, the deviation from κ is small and Stokes’ law is used as a constitutive equation.
• For an incompressible fluid, ∇ · u = 0 and thus κ = 0.
Proposition 7 (Netownian incompressible fluid).
σ = −pI + 2µE.
3.3
Equations of state
The equations of state allow to close the system when the energy conservation equation is used.
The kinetic equation of state F (p, ρ, T ) = 0 relates the pressure, density and temperature.
Example 2. For a perfect gas, p = ρRT .
For isothermal flow, T =cst. For incompressible flow, ρ=cst. For barotropic flow, f (p, ρ) = 0.
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The caloric equation of state e = e(T, ρ) relates the internal energy to the temperature and
density.
Example 3. For a perfect gas, e = e(T ) = CV T .
4
Navier-Stokes equations
We summarize the equations for a Newtonian fluid given the material constants µ, λ, k.
Conservation equations:
1. Continuity (conservation of mass):
1 equation - 4 unknowns u, ρ.
Dρ
Dt
+ ρ∇ · u = 0.
2. Eqs of motion (conservation of linear momentum): ρ Du
Dt = ρf + ∇ · σ.
3 equations - 6 unknowns σij (σ = σ T ).
3. Energy: ρ De
Dt = σ : E − ∇ · q.
1 equation - 4 unknowns e, q.
Constitutive equations:
1. Newtonian fluid: σ = (−p + λtrE)I + 2µE.
6 equations - 1 unknown p.
2. Fourier law: q = −k∇T .
3 equations - 1 unknown T .
3. Kinetic equation of state: F (p, ρ, T ) = 0.
1 equation.
4. Caloric equation of state: e = e(θ, ρ).
1 equation.
Substituting the constitutive equations into the conservation equations yield the Navier Stokes
equations. For convenience, we write the equations component wise.
∂t ρ + ∂k (ρuk ) = 0
(5)
ρ∂t uj + ρuk ∂k uj = −∂j p + ∂j (λ∂k uk ) + ∂i [µ (∂j ui + ∂i uj )] + ρfj
(6)
ρ∂t e + ρuk ∂k e = −p∂k uk + ∂j (k∂j T ) + λ(∂k uk )2 + µ(∂j ui + ∂i uj )∂i uj
(7)
p = ρRT
(8)
e = Cv T
(9)
Remark 9. The pressure can be redefined as the dynamic fluid pressure: ∇P = −ρf + ∇p.
Remark 10.
8
• Most fluids are isothermal, i.e. T = cst. Fourier’s law then implies q = 0 and the energy
equation becomes
De
ρ
= σ : E.
Dt
In other words, e can be determined from σ and E.
• More aerodynamics problems at high speed are compressible.
• Most atmospheric problems are not isothermal.
• Most Newtonian fluids satisfy Fourier’s law.
The isothermal and incompressible Navier Stokes equations are simply
ρ
∇·u=0
∂u
+ u · ∇u = −∇P + µ∇2 u.
∂t
(10)
(11)
Remark 11. The pressure is a Lagrange multipliers enforcing the incompressibility condition.
Remark 12.
1. If the left-hand side of Eq. (11) is zero, Eq. (10)-(11) are known as the Stokes equations.
2. If the viscous effects are negligible, then the fluid is called inviscid and the equations are
known as the Euler equations.
5
Boundary conditions
In the case of an infinite domain, we require u → 0 as x → ∞. Imposing boundary conditions
in the presence of solid boundaries is challenging, since discontinuity across surfaces can arise.
On the molecular level, discontinuity can’t arise. There are five types of boundary conditions.
1. Kinematic boundary condition: No net flux of mass through a surface.
BC: u1 · n̂1 = u2 · n̂2 on S a material surface. If S is a solid wall and the reference frame
is the wall, the BC is u · n̂ = 0 at S.
2. Dynamic boundary condition: Tangential velocities are continuous through the surface.
BC: u1 − (u1 · n̂1 = u2 − (u2 · n̂2 on S a material surface. If S is a solid wall, the BC is
known as the no-slip BC: u = (u · û at S.
3. Stress boundary condition: Depending on the problem, different stress BC can be posed.
For example, if the interface is free to move, then the BC is of the form F (xs , t) = 0 on
S (generalization of 1.). Other boundary conditions are the balance of normal stresses
across the surface or motion driven by curvature (capillary flow).
4. Thermal boundary condition: Thermal BC either requires continuity of T at the surface
or no convective heat flux through the surface.
5. Surfactants: e.g. soap concentration on the surface.
9