FM1_lecture_notes

LECTURE NOTES FOR THE COURSE “FLUID MECHANICS I”

(in progress)

written by JACEK SZUMBARSKI

VECTORS AND TENSORS IN 3D EUCLIDEAN SPACE E

Orthogonal basic unary vectors (versors) : e

, e

1 if i

( , )

0 if i







= ≡

≠

e e

Any vector in E

is expressed as unique linear combinations of the basic versors

1 1

2 2

3 3

i i

≡

- summation (Einstein) convention

[a ,a ,a ]

- canonical equivalence of E

and R

NNER

(

SCALAR

)

PRODUCT

Let

i i

and

j j

. We define the inner product of

and

( )

j ij

i i

a b ( , )

a b

⋅ ≡

a b

e e

Note that

( , )

a e

hence we can write

( , )

a e e

ECTOR

(

CROSS

)

PRODUCT

We define the operation

on the basic vectors:

× =

no summation!

× = − ×

Assuming linearity with respect to both arguments, we extend this operation to all vectors in
the space E

3 2

3 1

1 3

1 2

2 1

i i

j j

a b

(a b

a b )

(a b

a b )

(a b

a b )

× =

−

a b

Practical way of computing the vector product

× =

−

a b

LTERNATING SYMBOL

ijk

0 if i

j or i

k or j

1 if {i, j, k} is an even permutation of {1,2,3}

1 if {i, j, k} is an odd permutation of {1,2,3}











∈ =

−

For instance we have

213

∈ = −

311

∈ =

231

∈ =

The vector product of a and b can be nicely written as follows

ijk

j k i

a b

× =∈

a b

Another useful operation is the mixed product of three vectors

ijk

j k

(

)

a b c

⋅ × =

=∈

a b c

Determinant of the matrix A (dim A = 3):

ijk

1,i 2, j 3,k

det

a a a

=∈

RANK TENSORS IN

Tensors as bilinear transformations (functionals)

→

Bi-linearity means that

1 1

, )

T( , )

, ) ,

T( ,

)

T( ,

)

T( ,

) .

x y

α y

x y

For two arbitrary vectors

and

we can write

i i

j j

ij i

T( , )

T(x

, y

)

x y T( , )

t x y

x y

e e

The matrix

such that

 

 

represents the tensor T in the assumed reference

frame (or with respect to assumed basic versors)

Some operations on tensors:

Addition:

= +

⇒

= +

⇒

= +

Multiplication by a scalar

⇒

Multiplication of two tensors

1 2

ik kj

T T

t t

⇒

T T

Scalar (Frobenius) product of two tensors

1 2

ij ij

T : T : t t

(double summation !)

RANK TENSORS IN

CONTINUED

Basic linear functionals

→

( ) :

Tensor product of the basic functionals:

m m

k k

(

)( , ) :

( ) ( )

) (y

)

x y

(

) (

)

x y

δ δ

⊗

x y

Thus we can write

ij i

T( , )

t x y

t (

)( , )

⊗

x y

ij i

⊗

The linear space of the 2

-rank tensors is 9-dimensional.

RTHOGONAL TRANSFORMATIONS OF COORDINATE SYSTEMS

Assume that different basic vectors are introduced

′ ′ ′

e e e

(see figure). These vectors can be

expressed by means of the “old” basic vectors.

Consider

ik k

′

The orthogonality condition for the new base yields

jm km

( )

( , )

z z (

)

z z

(

)

(

)

= =

′ ′

e e

Z Z

We conclude that the transformation of the basis preserves orthonormality of the basic
vectors if and only if the transformation matrix

satisfies the relation

−

, i.e.,

it is the orthogonal matrix.

RTHOGONAL TRANSFORMATIONS OF COORDINATE SYSTEMS

CONTINUED

Each vector

from E

can be expressed with respect to both basis, namely

i i

= ′ ′

Thus

i i

x z

′

meaning that

z x

(

) x

(

) x

−

′

and

( ) x

′

These are the transformation rules for the vectors!

RTHOGONAL TRANSFORMATIONS OF COORDINATE SYSTEMS

CONTINUED

Consider the tensor T and its representation with respect to both basis (reference frames)

ij i

T( , )

t x y

= ′ ′ ′

x y

We can write

ij i

ij ki

ki ij

(

)

k km

(

)

T( , )

t x y

t z x z y

x z t z y

x (

) (

)

x (

)

x t

′

′ ′ ′

ZTZ

x y

ZTZ

The matrix representing the tensor T in the new base is given as

−

′

Z T Z

Thus, we have obtained the transformation rule for the 2

– rank tensors!

IFFERENT VIEW

: 2

–

RANK TENSORS AS LINEAR MAPPINGS

→

Consider the 2

-rank tensor

and two vectors

and

We have

inner

product

( , )

T( , )

x t y

x w

x w

x y

The vector

can be defined as

The linear transformation

→

is defined by its action on the basic versors as

ij i

Indeed, for any vector

we get

j j

j i

i i

)

t y

Equivalence between 2-rank tensors and linear mappings can be established as follows

T: T( , ) : ( ,

)

→

x y

: T( , )

→

e y e

IGENVECTORS

EIGENVALUES AND TENSOR INVARIANTS

The eigenvalue problem:

formulation: find

∈

and nonzero

such that

, or

formulation: find

∈

and nonzero

such that

T( , )

( , )

x v

for each vector

from the space

Equivalently, we have

(t v

v )

p ( )

det (

)

−

⇒

−

Thus eigenvalues are the roots of the characteristic polynomial

(λ).

Tensor

is symmetric when

T( , )

x y

y x

, i.e. when

(check!) or

If the tensor T is symmetric then its all eigenvalues are real and the eigenvectors
corresponding to different eigenvalues are orthogonal (the proof can be found in
standard algebra textbooks).

IGENVECTORS

EIGENVALUES AND TENSOR INVARIANTS

CONTINUED

The characteristic polynomial is invariant, i.e. it is the same in all orthogonal reference
frames. Indeed, according to the transformation rule we have

p ( )

det (

)

det (

)

det [ (T

)

]

det

det (

) det

det

det (

) (det )

det (

)

−

′

⋅

−

⋅

−

⋅

−

ZTZ

I Z

We are mostly interested in 3D case. Then, we can write

p ( )

= − +

−

where

tr T : t

= ≡

(“tr” means trace),

[(tr T)

tr T ]

−

(calculate for 2D case!),

det T

The following relations hold between invariants and the eigenvalues (Viete’s formulas
for 3

-order polynomial)

λ λ λ

= + +

1 2

1 3

2 3

λ λ λ λ λ λ

1 2 3

λ λ λ

AYLEY

-H

AMILTON

HEOREM

Any square matrix

satisfies its own characteristic polynomial

p ( )

det(

)

−

, i.e.

we have

p ( )

Proof

For

invertible

square matrix

we have

(

)

det

(cof

)

−

Thus

(cof

)

det

⋅

M I

Let

= −

. Then

( ) : [cof (

)]

−

is the matrix polynomial of the order not

larger than n -1 (n – dimension of

)

n 1

( )

...

λ λ

−

+ +

and we have

(

)

n 1

(

)

...

det(

)

(

... c

c )

λ λ

−

+ +

−

⋅ =

+ +

I I

AYLEY

-H

AMILTON

HEOREM

CONTINUED

The above equality is satisfied for any number

λλλλ

so the corresponding matrix coefficients

at both sides should be the same. Thus

n 1

k 1

, k

n 1, n 2, ..,1

−

= −

−

Let’s multiply (from the left side) the first equation by

, the second one by

n 1

−

and so

on (then the last equation remains unchanged) and sum up all equations. The left-hand side
of the obtained equation is zero since all terms will cancel out in pairs! Thus we get

n 1

... c

p ( )

−

+ +

≡

as stated.

For the matrices with the dimension equal 3 we have

− +

−

This relation will be used in the section devoted to the constitutive relations in fluid
mechanics.

RODUCT OF ALTERNATING SYMBOLS

Important identity

ijk

βγ

δ δ

∈ ∈ =

−

Proof

Consider

1 0 0

0 1 0

0 0 1

After row’s permutation one gets

ijk

=∈

Then, after column’s permutation we obtain

ijk

αβγ

=∈ ∈

RODUCT OF ALTERNATING SYMBOLS

CONTINUED

Now, put k = α and apply summation.

The result is as follows

ijk

βγ

=∈ ∈

, or

ijk

(

)

(

)

(

)

βγ

β γ

δ δ δ

δ δ

δ δ δ

δ δ

δ δ δ

δ δ

∈ ∈ =

−

Exercise: Using index formalism derive the following vector identity

(

)

( , )

× × =

−

b c

a c b

a b c

BASIC DIFFERENTIAL OPERATORS (IN CARTESIAN CO.S.)

Gradient of a scalar field

f (t, )













∂

∇ =

∂

(vector)

∇

- nabla operator

Divergence of the vector field

w (t, )

r e

formal inner

product

div

∂

≡ ∇⋅

∂

(scalar)

Rotation (curl) of the vector field

w (t, )

r e

formal vector

prod

ijk

rot





































∂

≡ ∇×

−

∂

=∈

∂

(vector)

Gradient of the vector field

w (t, )

r e

formal dyadic

produc

Grad

∂

≡ ∇

⊗

∂

–rank tensor)

Divergence of the tensor field

t (t, )

⊗

r e

formal matrix vector

produc

Div T

−

∂

≡

∇⋅

∂

(vector)

SOME INTEGRAL THEOREMS (INFORMAL REMINDER)

REEN

-G

AUSS

-O

STROGRADSKY

(GGO) T

HEOREM

Consider the vector field

( )

w x

defined in a 3D volume Ω bounded by sufficiently

regular surface ∂ Ω. Then

divergence

component of

normal to

( , ) dS

Ω

∂

⋅ =

∇⋅

∫

w n

We have analogous (dual) theorem with vector products, namely

rotation

Ω

∂

= ∇×

∫

n w

w x

TOKES

HEOREM

Consider the vector field

( )

w x

, the closed line (loop)

and sufficiently regular (yet

arbitrary) surface S spanned (like a soap bubble) by this line. Then

component of

rot

normal to S

tan gent to

( , ) d

(

, ) dS

⋅ ≡

∇×

∫

w τ

w n

AGRANGIAN AND

ULERIAN VIEWS ON THE FLUID MOTION

Fluid element is defined as an individual and

infinitely small portion of the fluid. Each fluid

element is characterized by its instantaneous

location (or position) vector x, which is the

function of time t and the initial position ξ of

the element, i.e. its location at the time instant

t = 0. Thus we have

(t, )

and in

particular

(0, )

If we fix

∗

then the function

(t,

)

∗

describes some line in E

called the

trajectory of the fluid element.

x(t,

ξ)

t=0

t>0

trajectory of a fluid element

Ω

(t)

For the fixed time

≥

the function

(t, )

describes the transformation of the region filled
with the fluid at the time

t = 0

– let’s denote it

Ω

- to the region

(t)

(t,

)

Ω

containing

the same fluid at some later time

∗

. Thus, the

region

Ω(t)

is the image of

Ω

with respect to

the transformation

(t, )

; we call

Ω(t)

the

material volume because all the time it consists
of the same fluid elements, i.e. those which are
belong originally to

Ω

Note:

two, originally different, fluid elements cannot drop into the same point where the

velocity vector is not zero, i.e., only one trajectory can go through such point. These
condition can be described mathematically as follows. If

(t, )

and

, )

then the following group property holds

[

(t,

]

(

)

,ξ)

, x

ξ)

, x

x(t,

ξ)

t=0

t>0

trajectory of a fluid element

Ω

(t)

Any fluid motion can be described using either Lagrangian or Eulerian
viewpoint.

Lagrangian viewpoint: each fluid element is identified uniquely by its position at

t = 0

, i.e.

by the vector

. All kinematical and dynamic quantities are described as functions of time

and the Lagrangian coordinates

and

The velocity of the fluid element is defined as

t, )

(t, )

(t, ) : lim

(t, )

∆

→

−

∂

≡

∂

(

– fixed)

Fluid acceleration is defined as

t, )

(t, )

(t, ) : lim

(t, )

∆

→

−

∂

≡

∂

RAJECTORIES OF FLUID ELEMENTS

Lagrange:

(t, )

∂

(

– fixed parameter). Thus

(t, )

( , )d

= +

∫

We have obtained direct integral formula which can be calculated numerically (e.g. using
the trapezoidal integration rule)

Euler: we have the following initial value problem

v (t,x ,x ,x ) , j 1,2,3.

[t,x(t)]

x (0)

(0)













⇒

Typically, the above Initial Value Problem has to be solved numerically (e.g. using the
Runge-Kutta methods)

TREAMLINES OF THE VELOCITY FIELD

The streamline: line l such that for every point P on l the velocity vector at the point P is
tangent to l.

The tangency condition can be written as

v (t,x ,x ,x )

The above equalities can be view as the differential equivalent of the “edge” description of
the 2-parameter family of lines in 3-dimensional space, namely

(t,x ,x ,x ,C ,C )







In the above, time

is treated as the fixed parameter. In other words, the pattern of

streamlines is determined for each time instant separately and – in general – the form
of

the streamlines at different time instants is not the same

XAMPLES

(1) Stationary two-dimensional flow

2 1

1 2

(x ,x )

[ x ,x ]

= −

≡ −

Streamlines:

x dx

R , R

⇒

≥

−

We have obtained the family of the concentric circles.

Trajectories:

x ,

x (0)

R , x (0)







= −

The solution is

x (t)

R cos(t) , x (t)

R sin(t)

which is the parametric form of the circle

R , R

≥

(2) Nonstationary (or unsteady) flow

1 2

(t,x ,x )

( x

[ x

t,x ]

= − −

≡ − −

Streamlines:

x dx

t)dx

C t

R , C

t .

⇒

= + ≡

≥ −

− −

Again: the family of concentric circles but the pattern of the streamlines moves down
along the 0x

axis with the steady speed equal –1 (see figure).

Trajectories:

t ,

x (0)

, x (0)







= − −

The solution

x (t)

1)cos(t) x sin(t) 1

x (t)

1)sin(t) x cos(t) t







−

Consider

= -1

and

= 0

. Then

(t) = -1

and

(t) = -t

so the fluid element moves down the straight

vertical line

= -1

with the steady velocity equal to

–1

The trajectories in the unsteady flow can be quite different that the streamlines!

t=0

t>0

instantaneous
streamlines

UBSTANTIAL

(

MATERIAL

, L

AGRANGE

, L

)

DERIVATIVE

Consider a sufficiently regular scalar field

f = f(t,x) = f(t,x

)

For an observer moving with a given fluid element the value of this field is a time dependent
quantity described by the composite function

F(t) : f [t, (t, )]

The rate of change in time of the field

seen by such observer moving with the fluid is

called the substantial (material, Lagrange, Lie or full) derivative of the field f.

Mathematically, we have

[t, (t, )]:

(t)

[t, (t, )]

(t)

[t, (t, )]

(t)

[t, (t, )]

(t)

v [t, (t, )]













∂

where we have used the relation

(t)

v [t, (t, )] , j 1,2,3.

UBSTANTIAL

(

MATERIAL

, L

AGRANGE

, L

)

DERIVATIVE

CONTINUED

Since the arguments at both sides are the same, we have obtained the scalar field which can
be written using “nabla” operator as

convective

derivative

local

derivative

∂

⋅∇

∂

or in the index notation (summation convention is assumed)

∂

▶ The first term in the right-hand side of the above definition is called a

local derivative

It “measures” the rate of change of the field

due to its explicit time dependence at a

fixed space location. It

f ( )

f (x , x , x )

we say that

is stationary (or steady)

and the local derivative

f / t

∂ ∂

vanishes identically.

▶ The second term is called the

convective derivative

of the field

. It is generally

nonzero even if the field f is stationary. It measures the rate of change due to the
movement of the observer. This part of the substantial derivative vanishes identically if
the field f is uniform in space, i.e. its instantaneous value at each point is the same.

CCELERATION

Consider the acceleration of fluid elements in Eulerian description. In order to calculate the
acceleration we need to differentiate the velocity along the trajectories of fluid elements.
We have

(t, )

















∂

We see that the acceleration is the vector field. In popular notation, the convective part of
the acceleration is written using the formal inner product of the velocity field and nabla
operator

(t, )

(

)

∂

+ ⋅∇

∂

An alternative way of expressing the fluid acceleration is the Lamb-Gromeko form

(t, )

( v )

∂

+ ∇

+ ×

∂

ω v

where

v | |

is the velocity magnitude and

= ∇×

is the rotation of the velocity field

called

vorticity

CCELERATION

CONTUNUED

Proof of the Lamb-Gromeko form

We have (see Appendix)

ijk

∂

=∈

and

v v

Then

ijk

(

)

( v v )

(

)

( v )

βγ

β γ

β β

δ δ

















































∂

× =∈

=∈ ∈

−

∂

−

∂

−

= ⋅∇ −∇

∂

ω v

Thus

(

)

( v )

⋅∇ = ∇

+ ×

ω v

and the Lamb-Gromeko formula follows immediately.

♣

EYNOLDS

RANSPORT

HEOREM

We will  prove  the  mathematical  result known  as  the
Reynolds’  Transport  Theorem,  which  plays  the
fundamental  role  in  derivation of differential  forms
of  the  conservation  principles  in

Continuum

Mechanics

Consider any sufficiently regular scalar field

f (t, )

. Consider the integral of f calculated over

an arbitrary material volume

Ω(t)

(t)

C(t)

f (t, )d

Ω

∫

We need to compute the time derivative

(t)

C (t)

f (t, )d

Ω

′

∫

NOTE: This task is nontrivial since the integration domain is itself time dependent!

x(t,

ξ)

t=0

t>0

Ω

(t)

EYNOLDS

RANSPORT

HEOREM

CONTINUED

To calculate the derivative, we will switch from Eulerian variables

[x , x , x ]

Lagrangian variables

ξ = [ξ , ξ , ξ ]

. The integral

C(t)

can be written as

(t)

C(t)

f (t, )d

f [t, (t, )]J(t, )d

f (t, ) J(t, )d

∫

Ω

In the above formula we have used the composite function

f (t, )

f [t, (t, )]

and also the Jacobi determinant (Jacobian) defined as

J(t, )

det

(t, )

∂





























EYNOLDS

RANSPORT

HEOREM

–

CONTINUED

Since the domain

Ω

is time-independent (it is actually the initial form of the material

volume

Ω(t)

at the time

t = 0

), we can move the differentiation operator under the sign of the

integral and get

C (t)

f (t, )J(t, )d

(t, )J(t, )d

f (t, )

(t, )d

Ω

∂

′

∂

∫

Note that time differentiation of the composite function

yields

(

)

V (t, ) v [t, (t, )]

f (t, )

f [t, (t, )]

x (t, )

f [t, (t, )]

∂

⋅

+ ⋅∇

This part was easy! We need to calculate the time derivative of the Jacobian which has
appeared in the second integral in the formula for

C (t)

′

. This is much more complicated … .

Basically, we have two methods.

Method A

We write the Jacobian using the alternating symbol

ijk

J(t, )

ξ ξ ξ

∂

∂ ∂

∂ ∂ ∂

=∈

Note that partial derivatives with respect to time and Lagrangian variables commute, hence

∂

∂ ∂

∂

∂ ∂

∂

∂ ∂

∂

The time derivative is

ijk

ξ ξ ξ

∂

∂ ∂

∂

∂ ∂ ∂

∂

=∈

+∈

i 1 j 1

cofactor (i, j) of

(

)

[cof ]

∂

= =

∇

∑∑

Consider two square matrices

and

, and also the product

C = AB

. It means that

ij kj

a b

≡

∑

so we conclude that

ij ij

a b

≡ =

(trace of the matrix

)

Moreover, from the construction of the inverse Jacobi matrix we have

(cof )

det

−

⇒

J J

Hence, the formula for the time derivative of the Jacobi determinant can be written as
follows

J(t, )

(cof )

(t, )

J(t, ) tr

(t, )

−

∂

















∇ ⋅

V J

Finally, we need to get back to the Eulerian variables. To this end, we use the relation
between Lagrangian and Eulerian definitions of the fluid velocity

Lagrange

Euler

(t, )

[t, (t, )]

and calculate the gradient operator with respect to the Lagrangian variables

k 1

(t, )

V (t, )

v [t, (t, )]

(t, )

∂









∇

∑

The above formula can be written shortly as

(t, )

v[t, (t, )] (t, )

∇

= ∇

⋅

Thus, the time derivative of the Jacobian can be re-written in the following form

(

)

J(t, )

J(t, ) tr

[t, (t, )]

∂

∇

Taking into account that

div

∂

∇ =

≡ ∇⋅

we finally get the formula

J(t, )

[t, (t, )]

∂

∇⋅

Method B

This method is based upon the group property of the transformation of the material
volume at initial time

t = 0

to the volume (consisting of the same fluid particles) at some

later time

t > 0

. We can write

t s

[ t

(s,

]

, ξ)

x , x

ξ)

x t s

x [ t x (s, ,

,x (s, ,

]

ξ ξ ξ

(

)

, i = 1,2,3.

Let’s differentiate the above formula with respect to the Lagrangian coordinate

(t s, )

[t, (s, )]

(s, )

∂

which can also be written as

[ ] (t s, ) [ ] [t, (s, )] [ ] (s, )

which is equivalent to

(t s, )

[t, (s, )] (s, )

Then, from the fundamental property of matrix determinant, we have

J(t s, )

J[t, (s, )] J(s, )

We need to calculate the derivative

J(t

t, ) J(t, )

J(t, ) J[ t, (t, )] J(t, )

J(t, ) : lim

lim

J[ t, (t, )] 1

J(t, ) lim

∆

→

∂

−

∂

−

Note that

J[ t, (t, )]

∆

is the Jacobian of the “nearly identical” transformation

(t, )

t, )

∆

, which can be written shortly as

( )

∆

The explicit form of this transformation is

[

( )]

v (t,x ,x ,x ) t O( t )

= +

∆

, i = 1, 2, 3,

This, the Jacobi matrix can be calculated as follows

[ ] ( t, )

[

( )]

(t, ) t O( t )

∂

= +

∂

∆

or simply

( t, )

(t, ) t O( t )

= + ∇

∆

Now, it is easy to show (

do it!

) that

J( t, ) 1

(t, ) t O( t ) 1

(t, ) t O( t )













∂

= +

= + ∇⋅

∂

∆

Thus, we get

J( t, ) 1

lim

(t, )

→

− = ∇⋅

∆

and – after returning back to the Lagrangian variables - the formula for the time derivative
of the Jacobian is obtained

( )

J(t, ) : J(t, )

[t, (t, )]

∂

∇⋅

∂

EYNOLDS

RANSPORT

HEOREM

–

CONTINUED

The time derivative

C (t)

′

can be now evaluated as follows

(

)

(

)

(t)

(

GGO

normal

Theorem

vel

ocity

(t)

C (t)

[t, (t, )]J(t, )d

(t, )d

(f ) (t, )d

f d

(f ) d

f d

f v

Ω

∂

⇑

∂

⋅









+ ⋅∇ + ∇⋅

′

+ ⋅∇ + ∇⋅

+∇⋅

∇⋅

∫

v n

Note that the last equality has been obtained by the use of the Green-Gauss-Ostrogradsky
(GGO) Theorem. We see that the rate of change of

C(t)

is the sum of two components.

The first component appears due to the local time variation of the integrated function

and it appears even if the fluid is in rest (no motion). In contrast, the second term is
entirely due to the fluid motion and it assumes nonzero value even if the field f is
stationary (i.e.

∂

≡

ENSITY OF FLUID

We define (rather informally) the fluid density as

vol(

)

lim

vol(

)

Ω

→

Thus, the mass of the fluid in the material volume

Ω(t)

can be expressed as

(t)

(t, )d

∫

Ω

NOTE: The concept of a density of a continuous medium can be introduce formally in
terms of the measure theory (the Radon-Nikodym Theorem, see for instance “Probability
and Measure” by Billingsley, also in Polish).

RINCIPLE OF MASS CONSERVATION

The mass of the fluid in an arbitrary material volume

Ω(t)

does not change in time (

Ω(t)

contains permanently the same fluid elements).

Thus, we can write

(t)

Ω

From Reynolds’ Transport Theorem we get

Re ynolds's

Trans

(t)

port Th.

(

) d

Ω

∂

⇑









+ ∇⋅

∫

Since the volume

Ω(t)

is arbitrary and the motion is assumed continuous, the upper

equality can hold if and only of the integrand vanishes identically.

This way we get the differential equation of mass conservation

(

) 0

∂

+∇⋅

RINCIPLE OF MASS CONSERVATION

–

CONT

Equivalent forms are

ρ ρ

∂

+ ⋅∇ + ∇⋅ =

→

full

derivative

+ ∇⋅ =

For an incompressible fluid

const

and the above equation reduces to the continuity

equation

div

∇⋅ ≡

which describes the geometric constrain imposed on the velocity field (volume
conservation).

In Lagrangian description, the continuity equations is equivalent to the following condition
(why?) for the Jacobi determinant

J(t, )

∂

and since

J(0, ) 1

we have

J(t, ) 1

for all times

≥

IME RATE OF CHANGE OF AN EXTENSIVE QUANTITY

Consider an extensive physical quantity, characterized by its mass-specific density

H(t, )

. The amount of this quantity contained in the material volume

Ω(t)

expressed by the following volume integral

(t)

h(t)

H d

Ω

∫

The examples are: the Cartesian components of the linear momentum, kinetic and internal
energy. We need to know how to evaluate the time derivative of such integrals.

Using the Reynolds’ theorem and the differential equation of mass conservation we can
write

(

)

Re ynolds

Trans. Th.

(t)

( t)

h(t)

H d

[ ( H)

( H )]d

(

) dx

H d

Ω

∂

⇑

∂









+ ∇⋅

+ ⋅∇

∫

DEFORMATION AND STRESS

EFORMATION

Consider two fluid elements which are located at close points A and B at the time instant

We ask what happens to the relative position of these two fluid elements during a short time
interval

∆

The location of the first fluid element after the time

∆

can be expressed as follows

(t,

)∆t O(∆t )

′

Since

then analogously we have

(t,

)∆t O(∆t )

′

+ +

ρ v

where the vector

describes the relative position of

the fluid elements at the time

∆

-x

'=x

-x

EFORMATION

CONTINUED

During a short time interval

∆

this vector has changed and can be expressed as

(t) [ (t,

)

(t,

)] t O( t )

(t)

(t, )

t O( t , t | | )

′









−

+ −

+ ∇

x ρ

∆

∆ ∆

In the above, we have dropped the lower index “A” at the location vector corresponding to
the first element.

The rate of change of the vector describing the relative position of two close fluid elements
can be calculated

(t)

lim

(t, )

O(| | )

→

−

= ∇

x ρ

∆

We have introduced the matrix (actually it represents the tensor) called the velocity
gradient

∂









∇

EFORMATION

–

CONTINUED

The velocity gradient

∇

can be written as a sum of two tensors

∇ = +

D R

where

[

(

) ]

= ∇ + ∇

















∂

- symmetric tensor,

and

[

(

) ]

= ∇ − ∇

















∂

−

∂

- skew-symmetric tensor

We will show that the change of the relative position of the fluid elements due to the
action of the antysymmetric tensor

corresponds to the local “rigid” rotation of the

fluid.
Next, we will show that the action of the symmetric part

corresponds to the “real”

deformation, i.e. it is responsible of the change in shape and volume.

EFORMATION

–

CONTINUED

To this end, we note that

ijk

= − ∈

, where

are the Cartesian components of the

vorticity vector

ijk

rot

∂

= ∈

∂

Indeed, we have

ijk

(

)

βγ

β γ

δ δ

































∂

− ∈ ∈

= −

−

= −

−

∂

Thus, we can write

ijk

ρ ω

= − ∈

= − × =

R ρ

ρ ω

ω ρ

Moreover, we get

| |

( , )

2( ,

) 2( ,

)

(

)

= ⋅ × =

ρ ρ

ρ Rρ

ρ ω ρ

i.e., the distance between two (arbitrary) fluid elements is fixed and there is no shape
deformation.

The skew-symmetric part of the velocity gradient describes

pure rigid rotation

of the

fluid and the

local angular velocity is equal

EFORMATION

–

CONTINUED

The rate of change of the relative position vector (or – equivalently – the velocity of the
relative motion of two infinitely close fluid elements) can be now expressed by the formula

deformation

rigid rotation

Dρ

R ρ

Dρ

ω ρ

The first terms consists the symmetric tensor

, called the deformation rate tensor.

The tensor

can be expressed as the sum of the spherical part

SPH

and the deviatoric

part

DEV

DEV

SPH

The spherical part

SPH

describes pure volumetric deformation (uniform expansion or

contraction without any shape changes) and it defined as

trace
of

SPH

SPH ij

(

)

(

)

3 x

∂

⋅ = ∇⋅

⇒

∂

v I

Note that

SPH

(

) tr

(

)

div

= ∇⋅ ⋅

= ∇⋅ ≡

EFORMATION

–

CONTINUED

The second part

DEV

describes shape changes which preserve the volume.

We have

DEV

DEV ij

div

(

)

3 x

















∂

= −

⋅

⇒

−

∂

v I

and

DEV

SPH

−

To  explain  the  geometric  interpretation  of  both  parts  of  the  deformation  rate  tensor,
consider  the  deformation  of  a  small,  initially  rectangular  portion  of  a  fluid  in  two
dimensions.  Assume there is no rotation part and thus we can write

DEV

SPH

Dρ

ρ D

For a short time interval

∆

the above relation yields

DEV

SPH

(t)

O( t )

∆

EFORMATION

–

CONTINUED

Consider the 2D case when only volumetric part of the deformation exists (see picture).

We have

SPH

d 0

0 d













The relative (wrt the origin) position
vector at the time instant

∆

expressed as

(1 d t) (t) O( t )

∆

= +

The shape of the volume is preserved because the above formula describes the isotropic
expansion/contraction. The volume of the region

Vol (t)

L L

Ω

has been changed to

Vol (t

L L (1 d t)

Vol (t)(1 2d t) O( t )

Ω

∆

and

Vol (t

t) Vol (t)

lim

Vol (t)

Ω

∆

Ω

∆

→

−

= ∇⋅

Ω

(t)

{

∆

t L

∆

t L

Ω

(t+

∆

Ω

(t)

EFORMATION

–

CONTINUED

Assume now that the spherical part of the deformation rate tensor is absent. The deviatoric
part of this tensor in a 2D flow can be written as follows

DEV

d )

α γ

γ α





































−

The fluid deformation during the short time
interval can be now expressed as

DEV

O( t )

(

) (t)

∆

= +

or in the explicit form as

x (t

t) x (t)

t x (t)

x (t

t x (t) (1

t) x (t)







= −

+ +

∆

α ∆

γ ∆

∆

γ ∆

α ∆

Note the presence of shear, which manifests in the change of the angles between the
position vectors corresponding to different fluid elements in the deforming region.

Ω

(t)

{

α∆

t L

α∆

t L

Ω

(t+

∆

{

γ∆

t L

γ∆

t L

Ω

(t)

EFORMATION

–

CONTINUED

Let’s compute again the change of the volume of the fluid region during such deformation.

We get

Vol (t

L L

L L (1

t )

Vol (t) O( t )

Ω

α ∆

γ ∆

∆

α ∆

γ ∆

α ∆

∆

−

Vol (t

t) Vol (t)

lim

Vol (t)

Ω

∆

Ω

∆

→

−

We conclude that this time the instantaneous rate of the volume change is zero.

Thus, instantaneously, the deviatoric part of the deformation describes pure shear (no
expansion/contraction)

TRESS TENSOR

According to

Cauchy hypothesis

, the surface (or interface) reaction force acting between

two adjacent portions of a fluid can be characterized by its surface vector density called the
stress.

Thus, for an infinitesimal piece

of the interface

Ω

∂ ∩∂

, we have (see figure)

and

Ω

→

∂ ∩∂

∫

The stress vector depends on the point

and space

orientation of the surface element

or – equivalently –

of the unit vector

perpendicular (normal) to

at the

point

From the 3

principle of Newton’s dynamics (action-reaction principle) we have

( , )

( ,

)

= −

−

σ x n

Ω

dF =

TRESS TENSOR

CONTINUED

Consider small tetrahedron as depicted in the figure below.

The front face

ABC

∆

belongs to the plane

which is describes by the following formula

( , )

n x

≡

n x

– small number.

The areas of the faces of the tetrahedron are S,
S

, S

and S

for

ABC

∆

OBC

∆

AOC

∆

and

ABO

∆

, respectively. Obviously,

O(h )

∼

Moreover, the following relations hold for
j = 1,2,3:

Scos[ ( , )] S ( , )

= ⋅

n e

∢

The volume of the tetrahedron is

O(h )

Ω

∼

n=[n

]

-e

TRESS TENSOR

CONTINUED

The momentum principle can be written for the fluid contained inside the tetrahedron
volume as follows

time derivative

of the momentum

total surface

total volume

vol

surf

rce

force

Ω

∫

v x

We need to calculate the total surface force

surf

We have:

ABC

∆

( , )

( , ) O(h)

σ x n

σ 0 n

ABC

surf

S ( , ) O(h )

∆

σ 0 n

OBC

∆

( ,

)

( , )

( , ) O(h)

−

= −

σ x

σ x e

σ 0 e

OBC

surf

( , ) O(h )

∆

= −

σ 0 e

n=[n

]

-e

TRESS TENSOR

–

CONTINUED

AOC

∆

( ,

)

( ,

)

( ,

) O(h)

−

= −

σ x

σ x e

σ 0 e

AOC

surf

( ,

) O(h )

( ,

) O(h )

∆

= −

σ 0 e

AOB

∆

( ,

)

( ,

)

( ,

) O(h)

−

= −

σ x

σ x e

σ 0 e

AOB

surf

( ,

) O(h )

( ,

) O(h )

∆

= −

σ 0 e

When the above formulas are inserted to the equation of
motion we get

O(h )

vol

S[ ( , ) n

( , ) ] O(h )

Ω

−

∫

v x

σ 0 n

σ 0 e

When

→

the above equation reduces to

( , ) n

( , ) 0

−

σ 0 n

σ 0 e

In general, we can write

(t, , )

x n

x e

(summation !)

n=[n

]

-e

TRESS TENSOR

–

CONTINUED

In the planes oriented perpendicularly to the vectors

the stress vector can be

written as

(t, , )

(t, )

x e

This, the general formula for the stress vector takes the form

j i

(t, , )

(t, )n

x n

x e

We define the stress tensor

represented by the square matrix

such that

[ ]

The stress tensor depends on time and space coordinates, i.e. what we actually have is the
tensor field

(t, )

The stress tensor

can be viewed as the linear mapping (parameterized by

and

)

defined as

j j

j i

: E

∋ =

∈

TRESS TENSOR

–

CONTINUED

In particular

j i

( )

i.e., the action of

on the normal versor

at some point of the fluid surface yields

the stress vector

at this point.

Using canonical identification of

and

we can simply write

σ Σn

It is often useful to calculate the normal and tangent stress components at the point of
some surface.

Normal component is equal

inner (scal

ar)

product

(

)

( ,

)

= ⋅

≡

n Σn n

n Σn

Tangent component can be expressed as

( )

j i

km k

i i

km k

(

n n )n

[

(

n n ) n ]

= −

−

or, equivalently (verify!) as

(

)

= × ×

σ n

EWTON

’

AW FOR

ONTINUUM

Consider the material volume

(t)

Ω

. The Principle of Momentum Conservation takes the

form

linear momentum

volume force

surface

(t)

orc

)

∂

∫

v x

f x

Σn

Ω

In the index notation:

(t)

v d

f d

n dS

∂

∫

Ω

The formula for time derivative of the volume integral of an extensive quantity gives

(t)

∫

v x

Ω

EWTON

’

AW FOR

ONTINUUM

–

CONT

The next step is to transform (using the GGO integral theorem) the surface integral to
the volume integral

GGO

(t)

n dS

⇑

∂

∫

Ω

or in the matrix-vector notation

GGO

(t)

Div d

⇑

∂

∫

Σn

Σ x

Ω

Finally, the integral form of the momentum conservation can be re-written as

(t)

Div













−

∫

Ω

EWTON

’

AW FOR

ONTINUUM

–

CONT

Since the choice of the volume

(t)

Ω

is arbitrary, then for the smooth motion the

differential equation follows

Div

or in the index notation

∂

Writing the time derivative in the extended form, the above equation takes the form of

( )

Div













∂ + ⋅∇ =

∂

















∂

We will accept the following definition of the angular momentum (wrt the origin of the
coordinate system) of the fluid enclosed in the material volume

(t)

Ω

(t)

∫

v x

Ω

Note:

the above formula is not the most general one. There exist fluid models (called micropolar

fluids) where the total angular momentum contains additional contribution due to internal rotational
degrees of freedom of the fluid particles.

Let’s calculate the time derivative of the angular momentum

(t)

( )

(

Ω

⇑













× + ×

∫

r v

r v r

From general principles of mechanics we know that

The symbol

denotes the total moment of external forces (a torque) acting of the

material volume

(t)

Ω

The total moment

is defined as follows

(t)

Ω

∂

∫

f x

r Σn

Then, the equation for the angular momentum can be written

(t)

(

Ω

∂

−

∫

a f

r Σn

where

denotes fluid acceleration.

In the index notation, we have

ijk

(t)

(

G O

x (a

f )d

n dS

(

Ω

⇑

∂

∈

−

∈

∂

∫

The above equation can be transformed as follows

ijk

(t)

ijk

(t)

ijk

0 Eq. of

ijk

(t)

(

motion !!!

x (a

f )

) d

x (a

f ) x

x ( a

) d

Ω

δ σ

≡

























∂

∈

−

−∈

∂

∈

−

∂

∈

− −

−

∈

= −

∈

∂

∫

It follows that

ijk

∈

. Using the properties of the alternating symbol (see

Mathematical Preliminaries), we conclude that

1jk

2 jk

3 jk

∈

⇒

−

⇒

∈

⇒

−

⇒

∈

⇒

−

⇒

Thus

σ σ

or, equivalently,

, i.e., the

stress tensor is a symmetric tensor

ENERAL CONSIDERATIONS

The constitutive relations for the (simple) fluids is the relations between stress tensor Σ
and the deformation rate tensor D. It should be postulated in such form so that the stress
tensor is automatically symmetric. Let’s remind two facts:

•

The velocity gradient can be decomposed into two parts: the symmetric part

called

the deformation rate tensor and the skew-symmetric part

called the (rigid) rotation

tensor.

∇ = +

D R

•

Tensor

can be expressed as the sum of the spherical part

SPH

and the deviatoric part

DEV

SPH

where

SPH

(

)

⋅ = ∇⋅

v I

and

DEV

DEV ij

div

(

)

3 x

















∂

= −

⋅

⇒

−

∂

v I

The general constitutive relation for a (simple) fluid can be written in the form of the
matrix “polynomial”

( )

...

where the coefficients are the function of 3 invariants of the tensor

, i.e.

c [ I ( ),I ( ),I ( )]

Consider the characteristic polynomial of the tensor

p ( )

det[

]

−

= − +

−

The Cayley-Hamilton Theorem states that the matrix (or tensor) satisfies its own
characteristic polynomial meaning that

p ( )

= − +

−

+ =

Thus, the 3

power of

(and automatically all higher powers) can be expressed as a

linear combinations of

and

Hence, the most general polynomial constitutive relation is given by the 2

order formula

( )

EWTONIAN FLUIDS

The behavior of many fluids (water, air, others) can be described quite accurately by
the linear relation. Such fluids are called

Newtonian fluids

For Newtonian fluids we assume that:

▶

is a linear function of the invariant

▶

is a constant,

▶

If there is no motion we have the Pascal Law: pressure in any direction is the same. It
means that the matrix

should correspond to a spherical tensor and

= −

⇒

The constitutive relation for the Newtonian fluids can be written as follows

I ( )

(

)

(

)(

)

∇⋅

= − ⋅ +

⋅ +

= − +

−

∇⋅

v I

where

- (shear) viscosity (the physical unit in SI is kg/m·s)

- bulk viscosity (the same unit as

) ; usually

ζ << µ

and

can be assumed zero

EWTONIAN FLUIDS

CONTINUED

The constitutive relation written can be written in the index notations

p (

)

δ µ

































∂

= − + −

∂

For an incompressible fluid we have

div

∂

∇⋅ ≡

≡

∂

and the constitutive relation reduces to the simpler form

δ µ

















∂

= −

∂

Let us derive the equation of motion of Newtonian fluids. Earlier, we have derived the
general form from the 2

Principle of Newton’s dynamics

∂

We have to calculate the explicit form of the first term in the right-hand side of the above
equation:

(

)

(

)

x x

(

)

x x

δ µ





































































∂

= −

−

∂

= −

−

∂

∂ ∂

∂

= −

∂

∂ ∂

After the obtained formula is inserted into the equation of motion, we get

grad(div

)

(

)

(

)

µ∆

= −∇ +

+ +

∇ ∇⋅ +

or, writing the fluid acceleration in the full form, we obtain the

(

)

(

) (

)

µ∆













∂ + ⋅∇ = −∇ +

+ +

∇ ∇⋅ +

∂

For an incompressible fluid the N-S equations simplifies to

(

)

µ∆













∂ + ⋅∇ = −∇ +

∂

which is often written in the form

(

)

ν ∆

∂ + ⋅∇ = − ∇ +

∂

v f

where

ν µ ρ

is called the kinematic viscosity of fluid (the SI unit is m

/s).

ULER

QUATION

The Euler equation are the special case of the Navier-Stokes equation for a hypothetic
inviscid (i.e. possessing zero viscosity) fluid:

(

)













∂ + ⋅∇ = −∇ +

∂

Using the Lamb-Gromeko form of the convective acceleration

( )

(

)

⋅∇ = ∇

+ ×

ω v

we can write the Euler equation as follows

( )

∂ +∇

+ × = − ∇ +

∂

ω v

Let us now make the following assumptions:

(1) Volume force field is potential, i.e. there exist such scalar field

that

= ∇

(2) Fluid is barotropic, i.e. its pressure is the function of density (or vice-versa)

p( )

Define the following pressure function

P( ) :

p ( )d

ρ ρ

′

∫

Then

P[ ( )]

p [ ( )]

p[ ( )]

( )

( ) x

∂

′

∂

⇒

∇ = ∇

Note that for the incompressible fluid (ρ constant) we simply have

p /

A less trivial example is an isentropic motion of an ideal gas where

≈

, where

c / c

. Then (

check yourself!

)

c T

(

−

(specific enthalpy)

ERNOULLI

QUATION

Assume that the flow is stationary, i.e. for any physical quantity H we have

∂ ≡

∂

The Euler equation can be written as

(

)

∇

+ −

= ×

v ω

Choose a streamline and define the tangent unary vector

/ v

. Next, multiply the above

equation (in the sense of the inner product) by

. The result is

(

) (

)

(

)

+ −

= ∇

+ −

⋅ =

⋅ ×

v v ω

Hence, the function under the gradient is constant along the streamline:

+ −

The Bernoulli constant

can be – in general – different for different streamlines, unless

× =

v ω

(for instance, the flow may be irrotational, i.e. such that the vorticity

≡

AUCHY

-L

AGRANGE

QUATION

Now, the flow can be unsteady but we assume that the velocity field is potential. Then

= ∇

for some scalar field (the velocity potential)

and

= ∇× =













∂

= ∇

∂

= ∇

Then the Euler equation takes the form of

(

)

∂

∇

+ ∇

+ −

meaning that

C(t)

∂

+ ∇

+ −

∂

for some function

, which depends only on time.

AERO/HYDRODYNAMIC FORCES

TRESS AND REACTION FORCE EXTERTED AT AN IMMERSED SURFACE

The flow-induced force is defined as the surface integral of the surface stress vector

∂

∫

Σn

Ω

For an incompressible fluid we have the constitutive relation

rotation

gradient of

tensor

velocity

= − +

∇

−

Since

rot

= − ×

R n

n ω

we can write

= − +

∇ ⋅ +

Σn

v n

n ω

The following theorem holds:

div

(incompressible flow) and

Ω

∂

then

∇ ⋅ + × =

v n n ω

Proof

Since

Ω

∂

then the boundary

Ω

∂

is the izosurface for all components of the velocity

field and the gradients of these components must be perpendicular (normal) to

Ω

∂

Thus, we can write

n , k 1,2,3

Ω

∂

∇

× =

⇒

∂

for some real numbers

(j = 1, 2, 3).

Next, in the index notation we have

ijk

× =∈

n ω

∂

∇ ⋅ =

v n

After insertion we get

ijk

div

(

) n

(

v ) n

(

)

(

v ) n

v n

(

v n

v n )

(

n n

n n )

∂













∇ ⋅ + × =

+∈

+∈ ∈

−

v n n ω

αβ

α β

β α

δ δ

Using the above result in the formula for the stress vector, we finally obtain the formula

= − −

n ω

Note that

× =

n n

and

(

)

× ⋅ =

n ω n

so the first term corresponds to the normal

stress while the second one – to the tangent stress at the boundary surface

Ω

∂

The total aerodynamic force can be calculated from the integral formula

)dS

Ω

∂

= −

∫

n ω

Interestingly enough, the above formula for the force F can be derived

without

the

assumption that the velocity is zero at the boundary (however in such case the formula
for the stress vector is not true!). Indeed, we have

p dS 2

Ω

∂

= −

∇ ⋅

∫

v n

n ω

But

Laplacian of

tensor version

the velocity

of GGO

GGO for

the cross

product

Div (

(

Ω

∆

⇑

∂

⇑

∂

∇ ⋅

∇

= ∇ ∇⋅

− ∇× ∇×

= − ∇×

= −

∫

v n

ω x

n ω

Thus

p dS 2

)dS

Ω

∂

= −

−

= −

∫

n ω

HE INTEGRAL FORM OF THE MOMENTUM EQUATION

(

STEADY MOTION

)

Consider a stationary flow. The momentum equation in the
index form reads

∂

∂ =

∂

or, using the (stationary) mass conservation equation

( v )

⇓

∂

Two terms in the left-hand side of the above equation can
be collected into one term, namely

fluid

body

∪

Ω

∂

( v v )

∂

fluid

body

fluid

Assume next that the volume forces can be neglected. Then the momentum equation
reduces to

( v v

)

∂

−

∂

In the next step, we calculate the volume integral and apply to it the GGO theorem. The
result is

( v v

Ω

∂

−

∫

→

( v v n

n )dS

Ω

∂

−

∫

Equivalently, we can write the vector relation as follows

( v n v

)dS

Ω

= ⋅

∂

−

∫

v n

or simply

momentum flux

through t

he boundary

(

) v dS

Ω

∂

∫

Consider now an incompressible flow. The surface stress vector is expressed as

= −

and the formula for the reaction force can be written as follows

fluid

(

) v dS

p dS 2

= −

−

∫

D n

Quite often, we can choose

fluid

in such way that the viscous term is relatively small

and can be neglected. Then

fluid

(

) v dS

p dS

≅ −

−

∫

Sometimes the part of the body surface is in the contact with some other motionless fluid
(typically – the surrounding air) having a uniform pressure

Note that for the closed surface

fluid

we have

fluid

∫

(

why?

)

IRCULATION

Definition: Circulation of the vector field

along the (closed) contour

is defined as

⋅

∫

Kelvin’s Theorem:

Assume that:

•

the volume force field

potential,

•

the fluid is inviscid and barotropic

•

the flow is stationary.

Then

: the circulation of the velocity field

along any closed material line

L(t)

is constant

in time, i.e.

(t)

(t, ) d

≡

⋅ =

∫

Proof of the Kelvin Theorem:

Since the flow is barotropic and the volume force field is potential, we can write

∇ = ∇

= ∇

Thus, the acceleration (which consists of the convective part only) can be expressed as

(

)

≡ ⋅∇ = −∇ −

In order to evaluate the time derivative of the circulation along the material line, it is
convenient to use Lagrangian approach. Thus, the circulation can be expresses as

(t)

(t, ) d

(t, )

(t, )d

⋅ =

⋅

∫

ξ J

where

(t, )

∂

denotes the Jacobi matrix of the transformation between Eulerian and

Lagrangian coordinates.

Then, the time derivative of the circulation is evaluated as follows

int. of the grad. a

(t)

(

long the closed

(t)

loop

)

(t, ) d

(t, )

(t, )d

(t, )

(t, )d

(t, )

(t, )d

(t, ) d

(

)(t, ) d

) d

⋅ =

⋅

⋅∇

⋅ +

∇

⋅

= − ∇ +

⋅

∫

ξ J

V V

int. of the grad. along

the closed loo

∫

where the relation

(t, )

∂

= ∇

has been used.

TRENGTH OF THE VORTEX TUBE

It is defined as the flux of vorticity through a cross-section of
the tube. Using the Stokes’ Theorem we can write:

⋅

⋅ =

∫

ω n

We  see  that  the  strength  of  the  vortex  tube  is  equal  to  the
circulation  of  the  velocity  along  a  closed  contour  wrapped
around the tube.

The  above  definition  does  not  depend  on  the  choice  of  a

particular contour. Indeed, since the vorticity field is divergence-free, the flux of the
vorticity is fixed along the vortex tube. To see this, consider the tube segment

Ω

located

between two cross-section

and

From the GGO theorem we have

side

Ω

= ∇⋅

⋅

∫

ω x

ω n

side

ELMHOLTZ

) T

HEOREM

Assume that

▶ the flow is inviscid and barotropic,
▶ the volume force field is potential.

Then:

the vortex lines consist of the same fluid elements, i.e. the lines of the vorticity field

are material lines.

Proof:

We need the transformation rule for the vectors tangent to a material line.

Let at initial time t = 0 the material line be described parametrically as

(s)

At some later time instant t > 0, the shape of the material line follows from the flow
mapping

∋

∈

, i.e.,

( )

[ (s)]

x s

The corresponding transformation of the tangent vector can be evaluated as follows

Jacobi ma

trix

(s)

( (s))

[

]( (s))

(s) [

]( (s))

(s)

= ∂ ∂

ξ a

Let’s now write the acceleration in the Lamb-Gromeko form:

( v )

= ∂ + ∇

+ ×

ω v

The rotation of

can be expressed as

(

)

(

)

(

)

(

)

∇× = ∂ ∇× + ∇× × =

− ⋅∇ + ∇⋅

ω v

v ω

In the above, the following vector identity, written for

and

, is used

(

)

(

)

(

)

(

)

(

)

∇× × = ⋅∇ − ⋅∇ + ∇⋅

− ∇⋅

p q

q p

p q

Next, one can calculate the Lagrangian derivative of the vector field

as follows

( )

1 D

(

)

(

)

(

)

=− ∇⋅









−

= ∇× + ⋅∇ − ∇⋅

∇⋅ = ∇× +

⋅∇

v ω

100

From the equation of motion and assumed flow properties that the acceleration field is
potential and thus

∇× =

Then, the equation for the vector field

reduces to

( )

(

)

⋅∇

Define the vector field

such that

ω ρ ξ

∂

, or equivalently,

Jacobi

matrix

(

)

∂ ∂

ξ c

In the above, the symbol

denotes the Lagrangian variables.

The left-hand side of the above equation can be transformed as follows

( )

(

)

(

) c (

)

(

) c (

)(

)









∂ ∂

= ∂ ∂

+ ∂ ∂

= ∂ ∂

+ ∂ ∂ ∂ ∂

ξ c

The right-hand side can be written as

(

)

(

)

(

)(

)









⋅∇ = ∂ ∂

⋅∇

= ∂ ∂ ∂ ∂ ⋅

ξ c

101

Since

L = R

, we obtain

(

)

(

)(

)

(

)(

)

∂ ∂

+ ∂ ∂ ∂ ∂

= ∂ ∂ ∂ ∂

ξ c

which implies that

Thus, c is constant along trajectories of the fluid elements.

Using the Lagrangian description, we can write

(t, )

(0, )

Note that for the initial time

t = 0

the transformation between Lagrangian and Eulerian

descriptions reduces to identity.

t 0

(

)

∂ ∂

Therefore

and since

(t)

≡

we get

(

)

= ∂ ∂

103

QUATION OF THE VORTICITY TRANSPORT

In fluid mechanics the vorticity plays a very important role, in particular in understanding
of the phenomenon of turbulence. In this section we derive the differential equation
governing spatial/temporal evolution of this field.

Recall that the equation of motion of an inviscid fluid can be written in the following form

( v )

∂ + ∇

+ × = − ∇ +

ω v

Thus, the application of the rotation operator yields

(

)

(

∂ + ∇× × = −∇× ∇ + ∇×

ω v

The pressure term can be transformed as follows

(

( )

↑

∇× ∇ = ∇

×∇ + ∇×∇ = − ∇ ×∇

Note: the above term vanishes identically when the fluid is barotropic since the gradients
of pressure and density are in such case parallel.

104

The equation of the vorticity transport can be written in the form

(

)

(

)

∂ + ⋅∇ − ⋅∇ = − ∇ ×∇ + ∇×

or, using the full derivative

nonpotential

vortex stretching

volume force

baroclinic

term

t m

(

)

⋅∇

− ∇ ×∇ + ∇×

The change of the vorticity appears due to the following factors:

•

Local  deformation  of  the  pattern of vortex  lines  (or vortex  tubes)  known  as  the  “vortex
stretching”  effect.  This  mechanism  is  believed  to  be  crucial  for  generating
spatial/temporal  complexity  of  turbulent  flows.  The  vortex  stretching  term  vanishes
identically for 2D flows.

•

Presence of baroclinic effects. If the flow is not barotropic then the gradients of pressure
and density field are nonparallel. It can be shown that in such situation a torque is
developed which perpetuates rotation of fluid elements (generates vorticity).

•

Presence of nonpotential volume forces. This factor is important e.g. for electricity-
conducting fluids.

105

For the barotropic (in particular – incompressible) motion of inviscid fluid, the vorticity
equation reduces to

(

)

⋅∇

In the 2D case it reduces further to

We conclude that in any 2D flow the vorticity is conserved along trajectories of fluid
elements.

If the fluid is viscous, the vorticity equation contains the diffusion term. We will derive this
equation assuming that the fluid is incompressible. Again, we begin with the Navier-Stokes
equation in the Lamb-Gromeko form

( v )

ν ∆

∂ + ∇

+ × = − ∇ +

ω v

v f

If the rotation operator is applied, we get the equation

(

)

(

)

ν ∆

∂ + ⋅∇ − ⋅∇ =

+ ∇×

which reduces to

(

)

(

)

ν ∆

∂ + ⋅∇ − ⋅∇ =

when the field of the volume forces

is potential.

107

DIMENSIONAL INCOMPRESSIBLE FLOW

. S

TREAMFUNCTION

The streamfunction is a very convenient concept in the theory of 2D incompressible flow.
The idea is to introduce the scalar field

such that

= ∂

= −∂

Note that the continuity equation

∂ + ∂ =

is satisfied automatically. Indeed, we have

∂ + ∂ = ∂

−∂

The streamfunction has a remarkable property: it is constant along streamlines.

To see this, it is sufficient to show that the gradient of the streamfunction is always
perpendicular to the velocity vector (why?). It is indeed the case:

∇ ⋅ = ∂ + ∂

= − +

108

TREAMFUNCTION AND THE VOLUMETRIC FLOW RATE

Consider a line joining two points in the (plane) flow domain. We will calculate the
volumetric flow rate (the volume flux) through this line.

We have

(u n

v n )ds

)ds

(

)ds

τ ψ τ ψ

ψ ψ

⋅

−

∂ + ∂

= ∇ ⋅ =

−

∫

v n

The volumetric flux through the line segment is equal to the difference of the
streamfunction between the endpoints of this segment.

ψ=ψ

=ψ

streamlines

109

TREAMFUNCTION AND VORTICITY

There exists a relation between the streamfunction and vorticity. Since the flow is 2D,
the vorticity field is perpendicular to the flow’s plane and can be expressed as

(

≡ ∇× = ∂ −∂

≡

Then, the streamfunction satisfies the Poisson equation

(

∆ψ

≡ ∂

+ ∂

= − ∂ − ∂

= −

Two dimensional motion of an incompressible viscous fluid can be described in terms of the
purely kinematical quantities: velocity, vorticity and streamfunction. The pressure field is
eliminated and the continuity equation

div

is automatically satisfied. The complete

description consists of the following equations:

•

Equation of the vorticity transport (2D)

ω ν ∆ω

∂ + ∂ + ∂ =

•

Equation for the streamfunction

∆ψ

= −

•

Relation between the streamfunction and velocity

= ∂

= −∂

•

Definition of vorticity (2D)

= ∂ − ∂

accompanied by appropriately formulated boundary and initial conditions.

110

Total energy of the fluid inside the region

Ω

is the sum of the kinetic energy

and the internal energy

The energy changes during the fluid motion due to:
▶ external volume and surface forces,
▶ heat conducted through the boundary

∂Ω

▶ heat produced by internal sources.

The time derivative of the total energy of the fluid in

Ω

can be written as

(U E ) P

•

Ω

∂Ω

Ω

∂Ω

111

The right-hand-side terms are:

Ω

= ρ ⋅

∫

f v x

- power of the volume forces,

∂Ω

⋅

∫

σ v

v Σn

- power of the surface (or interface) forces,

•

Ω

= ρ

∫

- power of the internal heat sources (symbol q

denotes their volumetric density) ,

GGO

d T

⇑

•

∂Ω

= λ

λ∇ ⋅

∫

- the heat flux through the boundary

∂Ω

(the

symbol

denotes temperature and

is the

coefficient of heat conduction).

112

Using the quantities defined above we get

power of the

internal heat

power of the

heat flux through

sources in

volume forces

surfa

total e

ce forces

the boun

dary

v )d

q d

Ω

∂

Ω

∂Ω

ρ +

= ρ ⋅

⋅

+ ρ

+ λ∇ ⋅

∫

f v x

v Σn

where

denotes the mass-specific internal energy of the fluid.

In  order  to  derive  the  differential  energy  equation,  we  have  to  transform  all  surface
integrals into the volume integrals.

Consider  first  the  derivative  on  the  left  side.  On  the  basis  of  the  Reynolds  transport
theorem and the mass conservation principle, we also have

v )d

Ω

ρ +

= ρ

∫

113

ij i

GGO

v n dS

(

v )d

v d

(

v )d

(

v )d

v d

(

v )d

v d

∂

Ω

∂Ω

∂

Ω

∂

Ω

⇑

Ω

⋅

= σ

+ σ

−

+ σ

−

∫

v Σn

v d

Div

∂

Ω

⋅

∫

Σ v x

Σ : D x

and also

GGO

(

T)d

⇑

∂Ω

Ω

λ∇ ⋅

= ∇⋅ λ∇

∫

Assume that all physical fields involved are sufficiently regular. Since the volume

Ω

can be

arbitrary the partial differential equations – called the energy equations – follows

v )

Div

(

=ρ ⋅ +

⋅ +

+ρ +∇⋅ λ∇

f v

Σ v Σ: D

114

It is interesting to consider separately the balance of kinetic and internal energy. To this
end, consider the momentum equation

Div

=ρ +

The momentum equation can be multiplied by the velocity vector v. We get

Div

⋅ =ρ ⋅ +

⋅

v v

f v

Σ v

or, equivalently

D 1

Dt 2

(

)

Div

⋅ =ρ ⋅ +

⋅

v v

f v

Σ v

In the next step, the above equation is integrated in the volume

Ω

which yields the following

integral expression for the temporal rate of change of the kinetic energy

v d

Div

Ω

= ρ ⋅

⋅

∫

f v x

Σ v x

115

Let’s transform the second integral in the right-hand side. We can write (GGO again!)

Div

v d

(

v )d

v d

n dS

v d

∂

Ω

∂

Ω

∂Ω

Ω

∂Ω

⋅

− σ

⋅

−

∇

⋅

−

∫

Σ v x

v Σn

Σ: v x

v Σn

Σ: D x

The rate of change of the kinetic energy can be expressed as follows

v d

Ω

∂Ω

= ρ ⋅

⋅

−

∫

f v x

v Σn

Σ : D x

If the above equality is subtracted from the integral form of the balance of the total energy,
we get the integral formula for the temporal rate of change of the internal energy, namely

Ω

∂Ω

= ρ

+ λ∇ ⋅

∫

Σ : D x

116

Observe that both formulae written above contain the term

Ω

∫

Σ : D x

but with opposite signs! This term describes the transfer of mechanical energy into
internal energy (or vice versa).

Let’s look at the structure of this term in details. The stress tensor for the linear fluids is
expressed as

[ p (

)

v ]

2 d

∂

σ = − + ζ− µ

δ + µ

Hence, we have

0 if

0 (compression)

0 if

0 (expansion)

positi

ve definite par

(

)

2 d d

(

)

v d

2 d d

(

)(

)

∇

∂

⋅ <

∇⋅ >

−

= σ

= − δ

+ ζ − µ

+ µ

= −

+ ζ − µ

+ µ

− ∇⋅

+ ζ − µ ∇⋅

+ µ

v
v

Σ : D

D : D

117

The “energy transfer” term takes the following form

int ernal to mechanical

irreversible mechanical to int ernal energy transfer, i.e.

(if

0) or reverse (if

dissipation of the machanical energy due to i

energy transfer

(

) (

) d

− −

Ω

= − ∇⋅

+ ζ − µ ∇⋅

+ µ

∫

Σ : D x

v x

D : D x

nt ernal "friction"

For incompressible flows we have

∇⋅ =

, thus

∂

Ω

∂ ∂

∂

∂ ∂

∂

∂ ∂

∂

Ω



























≡

















= µ

∫

Σ : D x

D : D x

The quantity

is called the dissipation rate.

118

If the fluid is ideal (i.e., inviscid and not heat-conducting) then an algebraic relation between

kinematic and thermodynamic parameter - known as the first integral of the energy equation

- can be derived. The derivation procedure is similar to that for the Bernoulli Equation. The

main difference is that the flow need not to be barotropic – we assume only flow

steadiness and potentiality of the volume force field, i.e. that

= ∇Φ

We begin with the differential energy equation, which in the case of an ideal fluid reduces to

v )

(p )

= −∇⋅

+ρ ⋅

f v

By expanding the pressure term, this equations can be re-written equivalently as

v )

= − ∇⋅ − ⋅∇ +ρ ⋅

v v

f v

120

The energy equation can be now written in the following form

1 D

(p/

)

v )

=−

ρ−

+ Φ

( u p

) 0

+ ρ +

−Φ =

where

u p

= + ρ

denotes the mass-specific enthalpy of the fluid.

Thus, the energy equation can be written as

) 0

−Φ =

Since the flow is stationary, the above equation is equivalent to

) 0

⋅∇ +

−Φ =

121

Using the same arguments as in the case of the Bernoulli Eq., we conclude that along each
individual streamline

const

−Φ =

In general the energy constant

can be different for each streamline. If

is the same for

all streamlines then the flow is called homoenergetic.

Let us recall that if the flow is barotropic then along each streamline we have

const

−Φ =

Thus, when the flow is barotropic then

i P

const

− =

−

i.e., the enthalpy

and the pressure potential

differ only by an additive constant.

If additionally the fluid is incompressible then its internal energy

is fixed and the specific

enthalpy can be defined as

p /

. Thus, in the incompressible case, the energy and

Bernoulli equations are formally identical.

122

NTROPY OF A SMOOTH FLOW OF IDEAL FLUID

We will show that if the flow is smooth (i.e., all kinematic and thermodynamic fields are
sufficiently  regular)  then  the  entropy  of  the  fluid  is  conserved  along  trajectories  of
fluid  elements.  To  this  end,  let’s  consider  the  equation  of  internal  energy  derived  in  the
earlier  section.  For  the  inviscid  and  not  heat-conducting  fluid,  this  equation  reduces  to
(explain !)

= − ∇⋅

We have already used the relation

1 D

∇⋅ = −

. Thus, the equation for the internal

energy e can be written as follows

(1/ )

= −

ρ = −

Let us remind that the first principle of thermodynamics can be expressed in terms of
complete differentials of three parameters of thermodynamic state: entropy

, internal

energy

and specific volume

υ = ρ

. The corresponding form of this principle reads

T ds du pd

+ υ

123

For the thermodynamic process inside individual fluid element one can write

T s

u p

υ = −

υ+

υ =

In the above, the equation for the internal energy has been used. We see that entropy of the
fluid is fixed along trajectories, as stated. We will show in the Fluid Mechanics III course
that this statement is no longer valid if strong discontinuities (called shock waves) appear.

We have already introduced the concept of homoenergetic flows. In such flows we have

global

−Φ =

, or equivalently

) 0

∇ +

−Φ =

Similarly, we call the flow homoentropic if

∇ ≡

. Thus, when the flow is homoentropic

then the entropy is uniformly distributed in the flow domain. Since the 1

Principle of

Thermodynamics can be written in the following form

T ds di (1/ )dp

= −

then for any stationary flow one has

T s

i (1/ ) p

∇ = ∇ −

ρ ∇

125

Consider the material volume

Ω

. The temporal rate of change of total entropy of the

fluid inside this volume can be split in two parts: one part appears due to irreversible

processes while the other is related to the heat transfer.

We have

irreversible

change of

entropy due

entropy

external heat

exchange

Consider a small portion of the volume

∆Ω

surrounded by the surface

∆A

The amount of heat exchanged by this volume during a small time interval

∆t

can be

expressed as

div d

t div ( )

∆

∆Ω

∆ = −∆

⋅

= −∆

Ω ≈ −∆

∆Ω

∫

q n

q x

126

The increments of entropy of the volume ∆Ω which occurs in the time interval ∆t due to the
heat exchange is

1 dQ

div ( )

T dt

= −

Ω

q x

It is convenient (and natural) to introduce the mass-specific density of entropy

. Then,

according to well-known formula, we have

Ω

ρ Ω = ρ

Ω

∫

The 2

Principle of Thermodynamics tells us that irreversible processes always lead to

some increase of entropy. Thus

div

Ω

ρζ Ω = ρ

Ω+

Ω >

∫

where

ζ =

represents a mass-specific density of entropy sources corresponding to all

irreversible processes.

127

The heat flux integral term can be transformed as follows

div

Ω

 

 

 

⋅∇

Ω =

Ω +

Ω

∫

In the above, the following identity

div

 

 

 

+ ⋅∇

−

⋅∇

q q

has been used. We also have (the GGO theorem)

div

Ω

∂Ω













⋅

Ω =

∫

q n

and thus

Ω

∂Ω

Ω

⋅

⋅∇

ρ ζ Ω = ρ

Ω+

Ω >

∫

q n

129

The Navier-Stokes equation for an incompressible flow can be written in the following form

(

)

∂ + ⋅∇ = − ∇ +ν∆ +

∂

v f

In  order  to  make  comparisons  between  various  flow  it  is  necessary  to  introduce  the
dimensionless  form  of  the  Navier-Stokes  Equation.  To  this  end  we  choose  reference  or
scaling quantities for:

time

T t

linear dimensions

L x

velocity

vɶ

pressure

P p

volume force

fɶ

In the above, all symbols with wave are dimensionless quantities.

130

Consequently, we have also dimensionless differential operators

dt t

T t

∂

L x

∂

The Navier-Stokes equations can be now written as

(

)

∂

⋅∇ = −

∇ +

∆ +

∂

or, after multiplication by L/V

F L

(

)

VT t

V L

∂

+ ⋅∇ = −

∇ +

∆ +

∂

In the above equation, four nondimensional combinations of the scaling quantities have
appeared.

131

We can define the following similarity numbers

Strouhal number

V T

Euler number

Reynolds number

V L

Froude number

F L

Finally, the Navier-Stokes equation can be written in dimensionless form as follows

(

)

∂ + ⋅∇ = − ∇ + ∆ +

∂

Note that the coefficient at dimensionless convective acceleration is equal 1. The remaining
terms in this equations are multiplied by reciprocals of the similitude numbers. We can say
that each similitude number is a measure of significance of a corresponding term in
comparison with the convective acceleration term.