Fluid Dynamics - Math 6750 - Fall 2013
Basic principles - Part I
1
The continuum approximation
Microscopic
individual molecules
many-body problems
few thousands molecules
short time
laws of dynamics
ε ∼ mean free path
Macroscopic
infinitely divisible without change of character
continuum
limit of the mean of a quantity over the molecular fluctuations
large physical dimension
laws of mechanics and probability
L: physical length of the fluid flow
Any macroscopic quantity can be defined as an appropriate spatial average of molecular quantities over a small enough volume V . For example, the velocity u is
Z
1
u=
wdV.
V V
The above continuum approximation is valid if
ε V 1/3 L.
On the microscopic scale, all transports phenomena stems from statistical mechanics.
On the macroscopic level, transports mechanisms can be split into transport by mean of a
continuum variable (e.g. u) (also called convective processes) and transports resulting from
molecular fluctuations (also called diffusive processes). We will see that molecular fluctuations
lead to surface contributions and stresses.
Example 1 (Billiard-ball gas).
1
2
Tensor Calculus (Cartesian)
The cartesian basis vectors are ei , i = 1, 2, 3, for example e1 = [1 0 0]T . A vector is given by
u = ui ei . Summation over the range of an index is understood when repeated dummy indices
appear in the same multiplicative term, also called Euler summation.
We start with a few basic facts about the Kronecker Delta and the permutation symbol.
Definition 1 (Kronecker Delta).
(
1
δij =
0
if i = j
if i =
6 j
.
Definition 2 (Permutation symbol).


0 when any two indices are equals
εmnr = 1 when m, n, r are 1, 2, 3 or an even permutation of 1, 2, 3


−1 when m, n, r are an odd permutation of 1, 2, 3
Proposition 1.
1. δmm = 3
δmn δmn = 3.
2. εijk εilm = δjl δkm − δjm δkl
εijk εijk = 6.
Exercise 1. Use the definitions to prove the proposition.
Vector operations like dot and cross products are defined on the basis vector.
Definition 3 (Operations). The dot and cross products of basis vectors are
ei · ej = δij
and ei × ej = εijk ek .
Remark 1. From the above definitions, it is easy to see that for two vectors u, v
u · v = ui ei · vj ej = ui vj δij = ui vi
and u × v = εijk ui vj ek .
Exercise 2. Show the identity for the cross product of u and v.
We will use the next fact about δij and derivatives in many derivations.
Proposition 2. Let x be a position vector with coordinates xi . Then
δij =
2
∂xi
.
∂xj
.
2.1
Tensor algebra
Vectors are examples of rank one tensors. Matrices are example of second-rank tensors, generally just called tensors. A rank zero tensor is simply a scalar.
Definition 4 (Rank two tensor). A second-rank tensor T represents a linear vector function
associating with each vector u another vector v, i.e u = T · v. The dyadic form of a tensor T
is T = Tij ei ej = Tij ei ⊗ ej .
There are three basics operations on tensors. The inner and double inner product (·, :) are
contractions and reduce the rank of tensor. The tensor or outer product (⊗) increases the rank
of the tensor. More specifically, let u, v be vectors and T, U be tensors. Then we have
u · v = ui ei · vj ej = δij ui vj = ui vi
rank 0,
u · T = ui ei · Tjk ej ek = ui Tjk δij ek = ui Tik ek
rank 1,
T · U = Tij ei ej · Skl ek el = Tij Skl ei δjk el = Tij Sjl ei el
rank 2,
u ⊗ v = ui vj ei ej
rank 2,
u ⊗ T = ui Tjk ei ej ek
rank 3,
T : U = Tij Uji
rank 0.
Remark 2. eij = ei ⊗ ej form the unit basis for the second rank tensors with respect to the
scalar product :. It is a nine dimensional vector space.
Remark 3. The cross product is not a general tensor operation as it is only defined between
two rank one tensors.
Definition 5. A tensor is symmetric if Tij = Tji . A tensor is antisymmetric (also called skew
symmetric) if Tij = −Tji .
Definition 6. A tensor is isotropic if its components are invariant with respect to all possible
rotations of the coordinate system. That is if the quantity represented by the tensor is a
function of position only.
Remark 4. All tensors of rank 0 are isotropic, while all tensors of rank 1 are not (vectors have
a direction).
Theorem 1 (Isotropic tensors). The only isotropic tensors of order 2, 3, 4 are of the form
rank 2 :
αδij
rank 3 :
αεijk
rank 4 :
αδij δpq + β(δip δjq + δiq δjp ) + γ(δip δjp − δiq δjp ),
where α, β, γ are scalars.
3
2.2
Tensor calculus
All of the previous operations are algebra operations. Since tensors represent quantities that are
changing in space and time, we can define tensor calculus operations like we did in multivariable
Calculus. The main operator is the nabla or del operator, which we define next. For the
moment, we will define everything with respect to the Cartesian coordinates system and leave
the generalization to spherical and cylindrical coordinates for later.
Definition 7 (Del operator).
∇ = ei
∂
= ei ∂i .
∂i
We can now apply tensor operations between ∇ and a general tensor field. The most important
ones are the gradient and the divergence.
Definition 8 (Gradient). The gradient of tensor field A of any rank is
∇A = ∇ ⊗ A.
Example 2. Let φ be a scalar field and u a vector field. Then
∂φ
∂φ
=
ek = ∂k φek ,
∂x
∂xk
∇u = ei ∂i uj ej = ∂i uj ei ej .
∇φ =
Definition 9 (Divergence). The divergence of a tensor field A of any rank is
div = ∇ · A.
Example 3. Let u be a vector field and T be a tensor field. Then
div(u) = ∇ · u = ei ∂i · uj ej = ∂i uj δij = ∂i ui ,
div(T) = ∇ · T = ei ∂i · Tjk ej ek = ∂i Tjk δij ek = ∂i Tik ek .
→
−
←
−
←
−
Remark 5. ∇u is sometimes denoted ∇u to avoid confusion with ∇ defined as: u ∇ =
ui ei ∂j ej = ∂j ui ei ej .
Since ∇ is a rank one tensor, we can take its cross product with a vector field.
Definition 10 (Curl). The curl (or rotation) of a vector field u is
curl(u) = ∇ × u = ei ∂i × uj ej = ∂i uj εijk ek .
Definition 11. A vector field is irrotational if curl(u) = 0.
Taking the inner product of ∇ with itself is called the Laplace operator ∆.
Definition 12 (Laplace operator).
∆ = ∇2 = ∇ · ∇.
4
Example 4. The Laplacian of a scalar field φ(x) is ∇ · (∇φ) = ∆φ.
The next proposition summarizes different relations between algebra and calculus operations.
Proposition 3. Let φ be a scalar field and u be a vector field. Then
1. ∇ · (φu) = ∇φ · u + φ∇ · u.
2. ∇ × (∇φ) = 0.
3. ∇ · (∇ × u) = 0.
4. ∇ × (∇ × u) = ∇(∇ · u) − ∇2 u.
Exercise 3. Using the definitions and index notations prove the proposition.
The divergence theorem from Calculus can be extended to tensors.
Theorem 2 (Divergence thm). Let n̂ be the outward unit normal vector to the surface S
enclosing the volume V . Let T be a second rank tensor. Then
Z
Z
∇ · TdV.
n̂ · TdS =
S
V
Proof. Exercise.
Exercise 4. Proof the general divergence theorem.
5
3
Frames and Kinematics
There are two basic coordinate systems that might be employed: Eulerian and Lagrangian
coordinates. Basic conservation equations are easier to derive in Lagrangian coordinates, while
most problems are solved in Eulerian coordinates.
Eulerian
spatial
fluid dynamics
variables: x position in space, t time
fluid velocity: u(x, t)
follows flow through a control volume fixed in space
Lagrangian
material, referential
elasticity
variables: α initial configuration, t time
fluid velocity: v(α, t)
follows identifiable pieces of matter
Definition 13 (Flow lines).
1. The flow is steady if u is independent of t, i.e u(x, t) = u(x) or ∂t u = 0.
2. A streamline is a line in the fluid whose tangent is everywhere parallel to u.
3. A streakline is a line on which lie all those fluid elements that at some earlier passed
through a certain point in space (e.g. drug release).
4. A flow flield is 2D when u(x, t) is everywhere perpendicular to a certain direction and
independent of displacements parallel to that direction.
5. A flow field is axisymmetric when u expressed in cylindrical coordinates is independent
of the azimuthal angle θ.
Remark 6.
• A streamline x = x(s), y = y(s), z = z(s) satisfies
dx/ds
dy/ds
dz/ds
=
=
u
v
w
for all t.
• The path of a particle coincide with the streamline only if the flow is steady.
Definition 14 (Circulation and Vorticity).
H
1. The circulation Γ is defined as Γ = u · dγ, where the contour integral is taken counterclockwise.
2. The vorticity ω is the curl of u: ω = ∇ × u.
Definition 15. A material volume, Vm (α, t), is a volume of fluid whose boundaries are defined
such that there is zero flux of a specified extensive property (e.g. mass). A control volume,
V (x), is a fixed and constant volume, whose location is fixed.
6
Remark 7. The material volume is a function of time, since the boundaries of the volume will
change with the fluid. It is also a function of initial position α. The control volume is not a
function of time, only of location x.
Consider the transformation x(α, t) of the spatial position x as a function of the Lagrangian
variable α and time. We will assume that, given a location x, we can backtrack and find the
fluid particle’s initial location α. This means that the function x(α, t) is invertible. In other
words, we can find a function α = α(x, t) that is continuous and single-valued. The necessary
and sufficient condition for invertability is for a non-vanishing Jacobian to exist. We recall
that the Jacobian is


∂x1
1
∂x
 ∂α
∂x2
J=
= det  ∂α
1
∂α
∂x3
∂α1
∂x1
α2
∂x2
α2
∂x3
α2
∂x1
∂α3
∂x2 
∂α3  .
∂x3
∂α3
J represents the dilatation of an infinitesimal volume dV as it follows the motion and the
change in dV is given by the usual change of coordinates formula (V0 initial volume).
Proposition 4.
dV = JdV0 .
(1)
The next proposition given without proof is known as the Euler’s expansion formula.
Proposition 5.
DJ
= J∇ · u.
Dt
4
(2)
Material Derivative
We want to understand how a field variable f changes as particles move along the flow. In the
Lagrangian frame, f is observed to change by ∆f when t changes by ∆t. From the Eulerian
point of view, f is considered a function of x, y, z, t. However as the fluid flows, x, y, z change
by ∆x, ∆y, ∆z and thus the change in f is by differential calculus
∂f
∂f
∂f
∂f
∆t +
∆x +
∆y +
∆z.
∂t
∂x
∂y
∂z
Since this change equals the change in the Lagrangian frame, we have after dividing by ∆t
∆f
∂f ∆x ∂f ∆x ∂f
∂f ∆z
=
+
+
∆y +
.
∆t
∂t ∆t
∂x ∆t
∂y
∂z ∆t
(3)
The limit of the left-hand side of Eq. (3) as ∆t → 0 represent the total change in f in the
Lagrangian frame or the change for an observer following the fluid and watching a particular
parcel of fluid. It is denoted by Df
Dt and called the material (or convective) derivative. As ∆t →
∆x ∆y ∆z
0 the fractions ∆t , ∆t , ∆t on the right-hand side of Eq. (3) tend to the velocity components
u, v, w, respectively. In vector form, we have the following definition.
Definition 16. The material derivative of a field variable f is
∂f
Df
=
+ (u · ∇)f.
Dt
∂t
7
(4)
∂f
Remark 8. ∂f
∂t , ∂x , . . . are the usual partial Eulerian derivatives. Equation (4) expresses the
Lagrangian rate of change in terms of the Eulerian derivatives.
Exercise 5. Express the Lagrangian acceleration a =
5
Du
Dt
in terms of its Eulerian derivatives.
Reynolds’ Transport Theorem
Consider f a fluid property and follow a specific parcel of fluid for a short time ∆t. Then in
the Lagrangian variables α, t f = f (t) is a function of t only as the control volume moves with
the fluid. The rate of change of f over the material volume Vm is defined as
"
!#
Z
Z
Z
1
D
f (t)dV = lim
f (t + ∆t)dV −
f (t)dV
.
∆t→0 ∆t
Dt Vm
Vm (t+∆t)
Vm (t)
Theorem 3 (Reynolds).
D
Dt
Z
Z
f dV =
Vm
Vm
∂f
+ ∇ · (f u) dV.
∂t
(5)
Proof. In class.
6
Conservation of Mass
If a given fluid parcel of mass m is followed along the fluid its shape and size can change, but
its mass will not. No mass is created: Any imbalance in the mass flux in and out of a
control volume is compensated by an accumulation of mass inside.
Let Vm be a Rmaterial volume and ρ be the density of the fluid. The total mass of fluid contained in Vm is Vm ρdV . Mathematically, the conservation of mass is (Lagrangian derivative)
D
Dt
Z
ρdV
= 0.
(6)
∂ρ
+ ∇ · (ρu) = 0.
∂t
(7)
Vm
Theorem 4 (Conservation of mass in Eulerian).
Proof. In class.
Theorem 5 (Conservation of mass in Lagrangian).
d
(Jρ) = 0.
dt
(8)
Proof. In class.
Remark 9. The conservation of mass equation is also known as the continuity equation.
Definition 17. A fluid is incompressible if ρ is constant, i.e.
8
Dρ
Dt
= 0.
Proposition 6 (Continuity for an incompressible fluid). For an incompressible fluid, the continuity equation becomes
∇ · u = 0.
(9)
Exercise 6. Use the conservation of mass in Eulerian form to prove the continuity equation for
an incompressible fluid.
7
Conservation of Momentum
The conservation of momentum is an application of Newton’s 2nd law to a fluid element that
we follow along the fluid in a Lagrangian frame.
7.1
Linear momentum
Consider the a material volume Vm . Newton’s 2nd law states that the rate at which the
momentum of the fluid mass is changing is equal to the sum of the external forces acting on
the mass. In other words,
Z
D
ρu = Sum of the forces acting on Vm .
(10)
Dt Vm
There are two types of forces:
1. Body forces such as gravitational or electromagnetic force;
2. Surface forces such as pressure forces or viscous stresses. A surface force acts from the
fluid outside the control volume on the fluid inside and vice versa.
Let f be the resultant of all body forces per unit mass and t beR the resultant of the surface
force per unit area. Then the total acting force on a volume Vm is Vm ρf dV and the net surface
R
force acting on the surface Sm is Sm tdS. Therefore, Eq. 10 becomes
D
Dt
Z
Z
Z
ρudV =
Vm
tdS +
Sm
ρf dV.
(11)
Vm
Remark 10. t is called the traction (or stress) vector.
There are nine components of stress at any given point, one normal component and two
shear components on each coordinate planes, illustrated in Fig. 1 in which the stress components act at a point as the length of the cube goes to zero. We will see later that this means
that the stress at a point may be represented by a tensor of rank 2 in the following form.
Definition 18. σ so that t(n̂) = n̂ · σ = σ T · n̂ for any traction across a surface with outward
unit normal n̂ is called the Cauchy stress tensor.
Remark 11. The conservation of angular momentum (next section) gives that σ is symmetric
and therefore σ has only 6 independent components.
9
e2
σ23
σ21
σ22
σ13
σ11
σ33
σ12
e1
σ31
σ32
e3
Figure 1: Nine components of stress as they act as a point in a fluid.
Theorem 6 (Cauchy’s equation of motion - Conservation of linear momentum).
∂u
ρ
+ u · ∇u = ρf + ∇ · σ
∂t
(12)
or
ρ (∂t ui + uj ∂j ui ) = ρfi + ∂j σji
(13)
Proof. In class.
7.2
Angular momentum
In a material volume Vm the rate of change of the angular momentum equals the sum of the
torques acting on the volume, i.e
Z
D
(x × ρu) dV = Sum of toques acting on Vm .
(14)
Dt Vm
There are four sources of torque:
1. Moment of the surface forces;
2. Moment of the body forces;
3. Body couples per unit mass c. A body couple is not due to a body force and creates
rotation without translation. For most fluids, except ferrofluids, c = 0.
4. Surface couples per unit area r. In practice, r = 0.
10
Again, let f be the resultant of all body forces per unit mass and t be the resultant Rof the surface
force per unit area. Then the total moment of the surface forces acting on Sm is Sm (x × t)dS
R
and the total moment of the body forces acting on Vm is Vm (x × ρf )dV . Therefore with the
definition of the Cauchy stress, Eq. 14 becomes
Z
Z
Z
D
(x × ρf )dV.
(15)
(x × (n̂ · σ)) dS +
(x × ρu)dV =
Dt Vm
Vm
Sm
Exercise 7. Show that
Z
Z
(x × (∇ · σ) − ε : σ) dV.
(x × (n̂ · σ)) dS =
V
S
HINT: Use the permutation symbol definition of the cross product and the general divergence
theorem.
Theorem 7 (Angular momentum conservation).
ε:σ=0
(16)
σ = σT .
(17)
and
Proof. Exercise.
Exercise 8. Prove the angular momentum theorem.
HINT: The second statement follows from writing out the contraction and using the definition
of the permutation symbol.
8
Conservation of Energy
The conservation of energy is the application of the first law of thermodynamics to a fluid
element in a Lagrangian frame. The energy of the fluid consists of two parts: intrinsic and
kinetic energy. Let e be the internal energy per unit mass and let 1/2u · u be the kinetic energy
per unit mass. The first law of thermodynamics applied to an element of fluid as it flows says
that the rate of change of the total energy equals to the sum of the rate at which work is being
done on the fluid by external forces and the rate at which heat is being added by conduction.
In other words,
Z 2
D
ρu
+ ρe dV = Sum of the rate of work done by forces and of heat added in Vm .
Dt Vm
2
(18)
Since work is force dotted with distance, rate of work is force
dotted
with
rate
of
distance.
R
This implies that the rate of work of the external forces is Vm ρf · u and that of the surface
R
forces is Sm t · udS. Using the divergence theorem and the definition of the Cauchy stress, the
surface integral can be rewritten as
Z
Z
Z
t · udS =
n̂ · (σ · u)dS =
∇ · (σ · u)dV.
Sm
Sm
Vm
11
Let q be the conductive heat flux leaving Vm through the surface. Thus the net of amount of
heat leaving the fluid per unit time is
Z
Z
∇ · qdV.
q · n̂dS =
Vm
Sm
Therefore, Eq. (18) becomes
Z Z
Z
Z
D
ρ 2
ρe + u dV =
ρf · udV +
∇ · (σ · u)dV −
∇ · qdV.
Dt Vm
2
Vm
Vm
Vm
(19)
Theorem 8 (Thermal energy balance).
ρ
De
= σ : E − ∇q
Dt
(20)
where E = 12 (∇u + ∇uT ) is the rate-of-strain tensor.
Remark 12. The first term represents the contribution to the internal energy due to the mean
motion.
Proof. We apply the Reynolds transport theorem (5) to the left-hand side
Z Z ρ 2
∂
ρu2
ρu2
D
ρe + u dV =
ρe +
+ ∇ · (ρe +
)u dV
Dt Vm
2
2
2
Vm ∂t
2
2
2
2
Z u
u
u
u
=
ρ∂t
+2 +
+ e ∂t ρ + ρu · ∇
+2 +
+ e ∇ · (ρu) dV
2
2
2
2
Vm
Z
Z 2
D u2
u
ρ
=
+ e dV +
+ 2 (∂t ρ + ∇ · (ρu)) dV.
2
2
Vm Dt
Vm
By the conservation of mass (7), the second term is zero. Since Vm is arbitrary, the integrand
is zero and we have
D u2
+ e = ∇ · (σ · u) + ρf · u − ∇q.
ρ
Dt 2
We use index notation to simplify the first term
∇ · (σ · u) = ei ∂i · (σjk ej ek · ul el ) = ei ∂i · (σjk uk ej ) = δij ∂i (σjk uk ) = ∂i σik uk + σik ∂i uk
1
1
= (∇ · σ) · u + σik ∂i uk + σki ∂i uk = u · (∇ · σ) + σ : E.
2
2
Together we have
ρ Du2
De
+ρ
= u · (∇ · σ) + σ : E + ρf · u − ∇q.
2 Dt
Dt
Taking the dot product of the conservation of mass (7) with u yields
ρ
Du
ρ Du2
·u=
= ρf · u + u · (∇ · σ).
Dt
2 Dt
The last equation is the mechanical energy balance. Combining it with the above energy
equation yields the desired result.
12