I

I

N

N

T

T

R

R

O

O

D

D

U

U

C

C

T

T

I

I

O

O

N

N

T

T

O

O

F

F

L

L

U

U

I

I

D

D

M

M

E

E

C

C

H

H

A

A

N

N

I

I

C

C

S

S

LECTURE NOTES FOR THE COURSE “FLUID MECHANICS I”

(in progress)

written by JACEK SZUMBARSKI

1

M

M

A

A

T

T

H

H

E

E

M

M

A

A

T

T

I

I

C

C

A

A

L

L

P

P

R

R

E

E

L

L

I

I

M

M

I

I

N

N

A

A

R

R

I

I

E

E

S

S

2

VECTORS AND TENSORS IN 3D EUCLIDEAN SPACE E

3

Orthogonal basic unary vectors (versors) : e

1

, e

2

, e

3

i

j

ij

1 if i

j

( , )

0 if i

j

δ







=

= ≡

≠

e e

Any vector in E

3

is expressed as unique linear combinations of the basic versors

1 1

2 2

3 3

i i

a

a

a

a

=

+

+

≡

a

e

e

e

e

- summation (Einstein) convention

1

2

3

[a ,a ,a ]

=

a

- canonical equivalence of E

3

and R

3

I

NNER

(

SCALAR

)

PRODUCT

Let

i i

a

=

a

e

and

j j

b

=

b

e

. We define the inner product of

a

and

b

:

( )

i

j

i

j

i

j ij

i i

,

a b ( , )

a b

a b

δ

⋅ ≡

=

=

=

a b

a b

e e

Note that

i

i

( , )

a

=

a e

hence we can write

i

i

( , )

=

a

a e e

3

V

ECTOR

(

CROSS

)

PRODUCT

We define the operation

×

on the basic vectors:

1

2

3

2

3

1

3

1

2

,

,

× =

× =

× =

e

e

e

e

e

e

e

e

e

,

i

i

no summation!

×

=

e

e

0



,

i

j

j

i

× = − ×

e

e

e

e

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

3

2

3

3 2

1

3 1

1 3

2

1 2

2 1

3

i i

j j

i

j

i

j

a

b

a b

(a b

a b )

(a b

a b )

(a b

a b )

× =

×

=

× =

−

+

−

+

−

a b

e

e

e

e

e

e

e

Practical way of computing the vector product

1

2

3

2

3

1

3

1

2

1

2

3

1

2

3

2

3

1

3

1

2

1

2

3

a

a

a

a

a

a

a

a

a

b

b

b

b

b

b

b

b

b

× =

=

−

+

e

e

e

a b

e

e

e

4

A

LTERNATING SYMBOL

ijk

0 if i

j or i

k or j

k

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

1

∈ = −

,

311

0

∈ =

,

231

1

∈ =

.

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

ijk

j k i

a b

× =∈

a b

e

Another useful operation is the mixed product of three vectors

1

2

3

1

2

3

ijk

i

j k

1

2

3

a

a

a

(

)

b

b

b

a b c

c

c

c

⋅ × =

=∈

a b c

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

ijk

1,i 2, j 3,k

det

a a a

=∈

A

5

2

ND

-

RANK TENSORS IN

E

3

Tensors as bilinear transformations (functionals)

3

3

×

→

E

E

R

Bi-linearity means that

1 1

2

2

1

1

2

2

1 1

2

2

1

1

2

2

T(

, )

T( , )

T(

, ) ,

T( ,

)

T( ,

)

T( ,

) .

α

α

α

α

α

α

α

+

=

+

+

=

+

x

x y

x y

x y

x

y

α y

x y

x y

For two arbitrary vectors

x

and

y

we can write

i i

j j

i

j

i

j

ij i

j

T( , )

T(x

, y

)

x y T( , )

t x y

=

=

=

x y

e

e

e e

The matrix

T

such that

ij

ij

t

 

 

=

T

represents the tensor T in the assumed reference

frame (or with respect to assumed basic versors)

Some operations on tensors:

Addition:

1

2

1

2

1

2

ij

ij

ij

T

T

T

t

t

t

= +

⇒

= +

⇒

= +

T

T

T

Multiplication by a scalar

1

1

1

ij

ij

T

T

t

t

β

β

β

=

⇒

=

⇒

=

T

T

Multiplication of two tensors

1

2

1 2

1 2

ij

ik kj

T

T T

t

t t

=

⇒

=

⇒

=

T

T T

Scalar (Frobenius) product of two tensors

1 2

1

2

ij ij

s

T : T : t t

(double summation !)

=

=

6

2

ND

-

RANK TENSORS IN

E

3

-

CONTINUED

Basic linear functionals

3

→

E

R

:

i

j

ij

( ) :

δ

=

e

e

Tensor product of the basic functionals:

m m

i

j

i

j

i

k k

j

m

m

m

k

i

k

j

k

ik

jm

i

j

(

)( , ) :

( ) ( )

(x

) (y

)

x y

(

) (

)

x y

x y

δ δ

⊗

=

=

=

=

=

=

x y

x

y

e

e

e

e

e

e

e

e

e

e

e

e

Thus we can write

ij i

j

ij

i

j

T( , )

t x y

t (

)( , )

=

=

⊗

x y

x y

e

e

or

ij i

j

T

t

=

⊗

e

e

.

The linear space of the 2

nd

-rank tensors is 9-dimensional.

7

O

RTHOGONAL TRANSFORMATIONS OF COORDINATE SYSTEMS

Assume that different basic vectors are introduced

1

2

3

,

,

′ ′ ′

e e e

(see figure). These vectors can be

expressed by means of the “old” basic vectors.

Consider

i

ik k

z

=

′

e

e

,

m

j

jm

z

=

′

e

e

.

The orthogonality condition for the new base yields

m

ij

ij

i

j

ik

jm

k

ik

jm km

T

T

ik

jk

ij

ij

( )

( , )

z z (

,

)

z z

z z

(

)

(

)

δ

δ

= =

=

=

=

′ ′

=

=

=

I

e e

e e

ZZ

Z Z

.

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

Z

satisfies the relation

1

T

−

=

Z

Z

, i.e.,

it is the orthogonal matrix.

e

1

e

3

e

2

0

8

O

RTHOGONAL TRANSFORMATIONS OF COORDINATE SYSTEMS

-

CONTINUED

Each vector

x

from E

3

can be expressed with respect to both basis, namely

i i

i i

x

x

=

= ′ ′

x

e

e

.

Thus

i i

i

ij

j

j

ji

i

x

x z

x z

=

=

=

′

′

x

e

e

e

,

meaning that

T

1

i

ji

j

ij

j

ij

j

x

z x

(

) x

(

) x

−

=

=

=

′

′

′

Z

Z

and

i

ij

j

x

( ) x

=

′

Z

.

These are the transformation rules for the vectors!

9

O

RTHOGONAL TRANSFORMATIONS OF COORDINATE SYSTEMS

-

CONTINUED

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

ij i

j

ij i

j

T( , )

t x y

t x y

=

= ′ ′ ′

x y

.

We can write



kj

T

km

km

m

m

ij i

j

ij ki

k

mj

k

ki ij

mj

(

)

T

T

m

m

m

k

kj

jm

k

km

k km

t

(

)

T( , )

t x y

t z x z y

x z t z y

x (

) (

)

y

x (

)

y

x t

y

′

=

=

=

=

′

′

′

′

=

=

=

′

′

′

′

′ ′ ′

ZT

ZTZ

x y

ZT

Z

ZTZ





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

T

1

−

=

=

′

T

Z T Z

Z T Z

Thus, we have obtained the transformation rule for the 2

nd

– rank tensors!

10

D

IFFERENT VIEW

: 2

ND

–

RANK TENSORS AS LINEAR MAPPINGS

3

3

→

E

E

Consider the 2

nd

-rank tensor

T

and two vectors

x

and

y

.

We have





i

inner

product

i

ij

j

i

i

w

( , )

T( , )

x t y

x w

=

=

=

x w

x y

.

The vector

w

can be defined as

=

w

y

T

.

The linear transformation

3

3

:

→

E

E

T

is defined by its action on the basic versors as

j

ij i

t

=

e

e

T

Indeed, for any vector

w

we get

j j

j

j

ij

j i

i i

(y

)

y

t y

w

=

=

=

=

=

w

y

e

e

e

e

T

T

T

.

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

T: T( , ) : ( ,

)

→

=

x y

x

y

T

T

,

i

i

T

:

: T( , )

→

=

y

e y e

T

T

.

11

E

IGENVECTORS

,

EIGENVALUES AND TENSOR INVARIANTS

The eigenvalue problem:

1

st

formulation: find

λ

∈

C

and nonzero

w

such that

λ

=

w

w

T

, or

2

nd

formulation: find

λ

∈

C

and nonzero

w

such that

T( , )

( , )

λ

=

x v

x v

for each vector

x

from the space

E

3

.

Equivalently, we have

T

ij

j

i

i

(t v

v )

p ( )

det (

)

0

λ

λ

λ

−

=

⇒

=

−

=

e

0

T

I

.

Thus eigenvalues are the roots of the characteristic polynomial

p

T

(λ).

Tensor

T

is symmetric when

T( , )

T( , )

=

x y

y x

, i.e. when

ij

ji

t

t

=

(check!) or

T

=

T

T

.

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).

12

E

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

1

1

T

1

1

p ( )

det (

)

det (

)

det [ (T

)

]

det

det (

) det

det

det (

) (det )

det (

)

λ

λ

λ

λ

λ

λ

λ

−

−

−

−

=

−

=

−

=

−

=

′

=

⋅

−

⋅

=

⋅

−

⋅

=

−

T

I

ZTZ

I

Z

I Z

Z

T

I

Z

Z

T

I

Z

T

I

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

3

2

T

1

2

3

p ( )

J

J

J

λ

λ

λ

λ

= − +

−

+

where

1

11

22

33

ii

J

tr T : t

t

t

t

=

= ≡

+

+

(“tr” means trace),

2

2

1

2

2

J

[(tr T)

tr T ]

=

−

(calculate for 2D case!),

3

J

det T

=

.

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

rd

-order polynomial)

1

1

2

3

J

λ λ λ

= + +

,

2

1 2

1 3

2 3

J

λ λ λ λ λ λ

=

+

+

,

3

1 2 3

J

λ λ λ

=

.

13

C

AYLEY

-H

AMILTON

T

HEOREM

Any square matrix

A

satisfies its own characteristic polynomial

A

p ( )

det(

)

λ

λ

=

−

A

I

, i.e.

we have

A

p ( )

=

A

0

Proof

For

invertible

square matrix

M

we have

(

)

1

1

T

det

(cof

)

−

−

=

M

M

M

.

Thus

T

(cof

)

det

=

⋅

M

M

M I

.

Let

λ

= −

M

A

I

. Then

T

( ) : [cof (

)]

λ

λ

=

−

B

A

I

is the matrix polynomial of the order not

larger than n -1 (n – dimension of

A

)

n 1

n 1

n 1

n 1

1

0

( )

...

λ λ

λ

λ

−

−

−

−

=

+

+ +

+

B

B

B

B

B

and we have

(

)

n 1

n 1

n 1

n 1

1

0

n

n 1

n 1

1

0

(

)

...

det(

)

(

c

... c

c )

λ λ

λ

λ

λ

λ

λ

λ

−

−

−

−

−

−

−

+

+ +

+

=

=

−

⋅ =

+

+ +

+

A

I

B

B

B

B

A

I I

I

14

C

AYLEY

-H

AMILTON

T

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

k

0

0

c

, k

n 1, n 2, ..,1

c

−

−

−

=

−

+

=

= −

−

=

B

I

B

AB

I

AB

I

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

n

A

, the second one by

n 1

−

A

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

n 1

n 1

1

0

A

0

c

... c

c

p ( )

−

−

=

+

+ +

+

≡

A

A

A

I

A

as stated.

For the matrices with the dimension equal 3 we have

3

2

1

2

3

J

J

J

− +

−

+

=

T

T

T

I

0

.

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

15

P

RODUCT OF ALTERNATING SYMBOLS

Important identity

ijk

k

i

j

i

j

βγ

β

γ

γ

β

δ δ

δ δ

∈ ∈ =

−

Proof

Consider

11

12

13

21

22

23

31

32

33

1 0 0

0 1 0

1

0 0 1

δ

δ

δ

δ

δ

δ

δ

δ

δ

=

=

.

After row’s permutation one gets

i1

i2

i3

j1

j2

j3

ijk

k1

k2

k3

δ

δ

δ

δ

δ

δ

δ

δ

δ

=∈

.

Then, after column’s permutation we obtain

i

i

i

j

j

j

ijk

k

k

k

α

β

γ

α

β

γ

αβγ

α

β

γ

δ

δ

δ

δ

δ

δ

δ

δ

δ

=∈ ∈

.

16

P

RODUCT OF ALTERNATING SYMBOLS

-

CONTINUED

Now, put k = α and apply summation.

The result is as follows

ik

i

i

jk

j

j

ijk

k

kk

k

k

β

γ

β

γ

βγ

β

γ

δ

δ

δ

δ

δ

δ

δ

δ

δ

=∈ ∈

, or





ijk

k

ik

j

k

k

j

i

jk

k

kk

j

i

jk

k

kk

j

3

3

j

i

i

j

i

j

i

j

j

i

j

i

i

j

j

i

(

)

(

)

(

)

3

3

βγ

β

γ

β

γ

β

γ

γ

γ

β

β

β γ

β

γ

β

γ

β

γ

β γ

β γ

β

γ

β γ

δ δ δ

δ δ

δ δ δ

δ δ

δ δ δ

δ δ

δ δ

δ δ

δ δ

δ δ

δ δ

δ δ

δ δ

δ δ

∈ ∈ =

−

−

−

+

−

=

=

−

−

+

+

−

=

−

Exercise: Using index formalism derive the following vector identity

(

)

( , )

( , )

× × =

−

a

b c

a c b

a b c

17

BASIC DIFFERENTIAL OPERATORS (IN CARTESIAN CO.S.)

Gradient of a scalar field

f

f (t, )

=

r

i

1

2

3

i

f

f

f

f

f

,

,

x

x

x

x













∂

∂

∂

∂

∇ =

=

∂

∂

∂

∂

e

(vector)

∇

- nabla operator

Divergence of the vector field

i

i

w (t, )

=

w

r e



formal inner

product

j

3

1

2

1

2

3

j

w

w

w

w

div

x

x

x

x

∂

∂

∂

∂

≡ ∇⋅

=

+

+

=

∂

∂

∂

∂

w

w

(scalar)

Rotation (curl) of the vector field

i

i

w (t, )

=

w

r e

formal vector

prod

3

3

2

1

2

1

1

2

3

2

3

3

1

1

2

j

k

i

uc

jk

i

t

ijk

k

j

i

w

w

w

w

w

w

rot

x

x

x

x

x

x

w

w

x

x





































∂

∂

∂

∂

∂

∂

≡ ∇×

=

−

+

−

+

−

=

∂

∂

∂

∂

∂

∂

∂

∂

=∈

=∈

∂

∂

w

w

e

e

e

e

e



(vector)

18

Gradient of the vector field

i

i

w (t, )

=

w

r e



formal dyadic

produc

i

i

j

j

t

w

Grad

x

∂

≡ ∇

=

⊗

∂

w

w

e

e

(2

nd

–rank tensor)

Divergence of the tensor field

ij

i

j

T

t (t, )

=

⊗

r e

e



formal matrix vector

produc

ij

i

j

t

t

Div T

T

x

−

∂

≡

∇⋅

=

∂

e

(vector)

19

SOME INTEGRAL THEOREMS (INFORMAL REMINDER)

G

REEN

-G

AUSS

-O

STROGRADSKY

(GGO) T

HEOREM

Consider the vector field

( )

=

w

w x

defined in a 3D volume Ω bounded by sufficiently

regular surface ∂ Ω. Then



n

divergence

of

component of

normal to

w

S

( , ) dS

d

Ω

Ω

∂

⋅ =

=

∇⋅

∫

∫

w

w

w n

w n

w

x



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

rotation

of

dS

d

Ω

Ω

∂

×

= ∇×

∫

∫

w

n w

w x



S

TOKES

T

HEOREM

Consider the vector field

( )

=

w

w x

, the closed line (loop)

γ

and sufficiently regular (yet

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

component of

component of

rot

normal to S

tan gent to

S

w

( , ) d

(

, ) dS

τ

γ

γ

⋅ ≡

=

∇×

∫

∫

w

w

w τ

w τ

w n







l

20

K

K

I

I

N

N

E

E

M

M

A

A

T

T

I

I

C

C

S

S

A

A

N

N

D

D

D

D

Y

Y

N

N

A

A

M

M

I

I

C

C

S

S

O

O

F

F

F

F

L

L

U

U

I

I

D

D

M

M

O

O

T

T

I

I

O

O

N

N

21

L

AGRANGIAN AND

E

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, )

=

x

x

ξ

and in

particular

(0, )

=

ξ

x

ξ

.

If we fix

∗

=

ξ

ξ

then the function

(t,

)

∗

=

x

x

ξ

describes some line in E

3

called the

trajectory of the fluid element.

x

1

,

ξ

1

x

3

,

ξ

3

x

2

,

ξ

2

0

ξ

x(t,

ξ)

t=0

t>0

trajectory of a fluid element

Ω

0

Ω

(t)

22

For the fixed time

t

0

≥

the function

(t, )

=

x

x

ξ

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

t = 0

– let’s denote it

Ω

0

- to the region

0

(t)

(t,

)

Ω

Ω

=

x

containing

the same fluid at some later time

t

∗

. Thus, the

region

Ω(t)

is the image of

Ω

0

with respect to

the transformation

(t, )

=

x

x

ξ

; 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

0

.

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

1

(t, )

=

x

x

ξ

and

2

(t

, )

τ

=

+

x

x

ξ

then the following group property holds

2

1

t

[

(t,

]

(

)

τ

τ

τ

=

+

=

=

x

x(

,ξ)

x

, x

ξ)

x

, x

.

x

1

,

ξ

1

x

3

,

ξ

3

x

2

,

ξ

2

0

ξ

x(t,

ξ)

t=0

t>0

trajectory of a fluid element

Ω

0

Ω

(t)

23

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

ξ

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

and the Lagrangian coordinates

ξ

1

,

ξ

2

and

ξ

3

.

The velocity of the fluid element is defined as

t

0

(t

t, )

(t, )

(t, ) : lim

(t, )

t

t

∆

∆

∆

→

+

−

∂

=

≡

∂

x

ξ

x

ξ

x

V

ξ

ξ

(

ξ

– fixed)

Fluid acceleration is defined as

2

2

t

0

(t

t, )

(t, )

(t, ) : lim

(t, )

(t, )

t

t

t

∆

∆

∆

→

+

−

∂

∂

=

≡

=

∂

∂

V

ξ

V

ξ

V

r

ξ

ξ

ξ

a

24

Eulerian viewpoint: the velocity, acceleration and other kinematical or dynamical
quantities are described as functions of time

t

and the position of the fluid element at this

time instant (not at the initial time!), i.e. by the coordinates

x

1

,

x

2

and

x

3

of the vector

x

.

The velocity field is the function of time and space coordinates

(t, )

=

v

v

x

.

The relations between two different viewpoints can be written as

Eulerian to Lagrangian:

(t, )

[t, (t, )]

=

V

ξ

v

x

ξ

Lagrangian to Eulerian:

inverse

transform

(t, )

[t, (t, )]

=

v

x

V

ξ

x



The Eulerian form of the fluid acceleration will be considered later.

25

T

RAJECTORIES OF FLUID ELEMENTS

Lagrange:

t

(t, )

(t, )

∂

∂

=

x

ξ

V

ξ

(

ξ

– fixed parameter). Thus

t

0

(t, )

( , )d

τ

τ

= +

∫

x

ξ

ξ

V

ξ

.

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

d

d

1

2

3

dt

j

j

dt

j

j

x

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

[t,x(t)]

x (0)

(0)

ξ



















=

=

=

⇒

=

=

x

v

x

ξ

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

26

S

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

3

1

2

1

1

2

3

2

1

2

3

3

1

2

3

dx

dx

dx

v (t,x ,x ,x )

v (t,x ,x ,x )

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

1

1

2

3

1

2

2

1

2

3

1

2

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

0

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

0

Ξ

Ξ







=

=

In the above, time

t

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

.

27

S

TREAMLINES OF THE VELOCITY FIELD

-

CONTINUED

The practical method of computing the streamlines is to “freeze” time and “inject” the
marker particles into the “frozen” velocity field. The movement of the marker particle
injected in the point

x

0

is described by the following initial value problem

d

d

0

( )

[t, ( )]

(

0)

τ

τ

τ

τ







=

= =

x

v

x

x

x

where physical time

t

is fixed (“frozen”) and the variable

τ

is the “pseudo-time”. In other

words, the streamlines are the trajectories of the marker particles moving in the frozen
velocity field.

We conclude immediately that fluid element trajectories and the streamlines are
identical if the velocity field does not depend explicitly on time, i.e. if the flow is
stationary.

28

E

XAMPLES

(1) Stationary two-dimensional flow

1

2

2 1

1 2

2

1

(x ,x )

x

x

[ x ,x ]

= −

+

≡ −

v

e

e

Streamlines:

2

2

2

1

2

1

1

2

2

1

2

2

1

dx

dx

x dx

x dx

0

x

x

R , R

0.

x

x

=

⇒

+

=

⇒

+

=

≥

−

We have obtained the family of the concentric circles.

Trajectories:

d

d

1

2

2

1

dt

dt

1

2

x

x ,

x

x

x (0)

R , x (0)

0







= −

=

=

=

The solution is

1

2

x (t)

R cos(t) , x (t)

R sin(t)

=

=

which is the parametric form of the circle

2

2

2

1

2

x

x

R , R

0

+

=

≥

.

29

(2) Nonstationary (or unsteady) flow

1

2

2

1

1 2

2

1

(t,x ,x )

( x

t)

x

[ x

t,x ]

= − −

+

≡ − −

v

e

e

Streamlines:

2

2

2

2

2

1

2

1

1

2

2

1

2

2

1

dx

dx

x dx

(x

t)dx

0

x

(x

t)

C t

R , C

t .

x

t

x

=

⇒

+

+

=

⇒

+

+

= + ≡

≥ −

− −

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

2

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

Trajectories:

d

d

1

2

2

1

dt

dt

1

10

2

20

x

x

t ,

x

x

x (0)

x

, x (0)

x







= − −

=

=

=

The solution

1

10

20

2

10

20

x (t)

(x

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

x (t)

(x

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







=

+

−

−

=

+

+

−

Consider

x

10

= -1

and

x

20

= 0

. Then

x

1

(t) = -1

and

x

2

(t) = -t

so the fluid element moves down the straight

vertical line

x

1

= -1

with the steady velocity equal to

–1

.

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

x

1

x

2

t=0

t>0

tr

a

je

c

to

ry

instantaneous
streamlines

30

S

UBSTANTIAL

(

MATERIAL

, L

AGRANGE

, L

IE

)

DERIVATIVE

Consider a sufficiently regular scalar field

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

1

,x

2

,x

3

)

.

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, )]

=

x

ξ

The rate of change in time of the field

f

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

1

2

1

2

3

1

2

3

3

1

2

3

dx

dx

Df

dF

f

f

f

[t, (t, )]:

(t)

[t, (t, )]

[t, (t, )]

(t)

[t, (t, )]

(t)

Dt

dt

t

x

dt

x

dt

dx

f

f

f

f

f

[t, (t, )]

(t)

v

v

v [t, (t, )]

x

dt

t

x

x

x













∂

∂

∂

=

=

+

+

+

∂

∂

∂

∂

∂

∂

∂

∂

+

=

+

+

+

∂

∂

∂

∂

∂

x

ξ

x

ξ

x

ξ

x

ξ

x

ξ

x

ξ

where we have used the relation

j

j

dx

(t)

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

dt

=

=

x

ξ

31

S

UBSTANTIAL

(

MATERIAL

, L

AGRANGE

, L

IE

)

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

f

f

t

Df

Dt

∂

⋅∇

∂

=

+

v

,

or in the index notation (summation convention is assumed)

j

j

Df

f

f

v

Dt

t

x

∂

∂

=

+

∂

∂

.

▶ 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

f

due to its explicit time dependence at a

fixed space location. It

1

2

3

f

f ( )

f (x , x , x )

=

=

x

we say that

f

is stationary (or steady)

and the local derivative

f / t

∂ ∂

vanishes identically.

▶ The second term is called the

convective derivative

of the field

f

. 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.

32

A

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

j

j

i

i

i

i

i

j

j

Dv

v

v

D

(t, )

v

v

Dt

Dt

t

x

t

x

















∂

∂

∂

∂

=

=

=

+

=

+

∂

∂

∂

∂

v

v

v

a

x

e

e

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

D

(t, )

(

)

Dt

t

∂

=

=

+ ⋅∇

∂

v

v

a

x

v

v

.

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

2

1

2

(t, )

( v )

t

∂

=

+ ∇

+ ×

∂

v

a

x

ω v

where

v | |

=

v

is the velocity magnitude and

= ∇×

ω

v

is the rotation of the velocity field

called

vorticity

.

33

A

CCELERATION

-

CONTUNUED

Proof of the Lamb-Gromeko form

We have (see Appendix)

j

x

ijk

k

i

v

∂

∂

=∈

ω

e

and

2

i

i

v

v v

=

.

Then

j

j

k

k

k

ijk

i

j

i

j

i

i

j

j

i

j

i

j

i

i

2

1

1

2

2

v

v

v

v

(

)

v

x

x

v

v

v

v

v

v

v

v

x

x

x

x

v

v

( v v )

(

)

( v )

x

x

γ

γ

γ

βγ

β

βγ

β

β γ

γ

β

β

γ

β

γ

γ

β γ

β

γ

β

β

β

β

γ

β

γ

γ

β

β β

γ

β

ω

δ δ

δ δ

δ δ

δ δ

















































∂

∂

× =∈

=∈ ∈

=

−

=

∂

∂

∂

∂

∂

∂

=

−

=

−

=

∂

∂

∂

∂

∂

∂

=

−

= ⋅∇ −∇

∂

∂

ω v

e

e

e

e

e

e

v

v

Thus

2

1

2

(

)

( v )

⋅∇ = ∇

+ ×

v

v

ω v

and the Lamb-Gromeko formula follows immediately.

♣

34

R

EYNOLDS

T

RANSPORT

T

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

f (t, )

=

x

. Consider the integral of f calculated over

an arbitrary material volume

Ω(t)

.

(t)

C(t)

f (t, )d

Ω

=

∫

x

x

.

We need to compute the time derivative

(t)

d

C (t)

f (t, )d

dt

Ω

=

′

∫

x

x

.

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

x

1

,

ξ

1

x

3

,

ξ

3

x

2

,

ξ

2

0

ξ

x(t,

ξ)

t=0

t>0

Ω

0

Ω

(t)

35

R

EYNOLDS

T

RANSPORT

T

HEOREM

-

CONTINUED

To calculate the derivative, we will switch from Eulerian variables

1

2

3

[x , x , x ]

=

x

to

Lagrangian variables

1

2

3

ξ = [ξ , ξ , ξ ]

. The integral

C(t)

can be written as

0

0

0

(t)

C(t)

f (t, )d

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

f (t, ) J(t, )d

=

=

=

∫

∫

∫

x

x

x

ξ

ξ

ξ

ξ

ξ

ξ

Ω

Ω

Ω

.

In the above formula we have used the composite function

f

0

0

0

f

f (t, )

f [t, (t, )]

=

=

ξ

x

ξ

,

and also the Jacobi determinant (Jacobian) defined as

1

1

1

1

2

3

2

2

2

1

2

3

3

3

3

1

2

3

x

x

x

x

x

x

x

x

x

J(t, )

det

(t, )

ξ

ξ

ξ

ξ

ξ

ξ

ξ

ξ

ξ

∂

∂

∂

∂

∂

∂

∂

∂

∂

∂

∂

∂

∂

∂

∂

∂

∂

∂





























=

ξ

ξ

.

36

R

EYNOLDS

T

RANSPORT

T

HEOREM

–

CONTINUED

Since the domain

Ω

0

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

0

0

0

0

0

0

f

d

J

C (t)

f (t, )J(t, )d

(t, )J(t, )d

f (t, )

(t, )d

dt

t

t

Ω

Ω

Ω

∂

∂

=

=

+

′

∂

∂

∫

∫

∫

ξ

ξ

ξ

ξ

ξ

ξ

ξ

ξ

ξ

Note that time differentiation of the composite function

f

0

yields

(

)

i

i

i

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

d

0

t

dt

t

x

t

i

t

f (t, )

f [t, (t, )]

f [t, (t, )]

f [t, (t, )]

x (t, )

f

f [t, (t, )]

∂

∂

∂

∂

∂

∂

∂

∂

∂

∂

=

=

=

=

+

⋅

=

=

+ ⋅∇

ξ

x

ξ

ξ

x

ξ

x

ξ

x

ξ

ξ

v

x

ξ





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.

37

Method A

We write the Jacobian using the alternating symbol

3

1

2

i

j

k

x

x

x

ijk

J(t, )

ξ ξ ξ

∂

∂ ∂

∂ ∂ ∂

=∈

ξ

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

1

1

1

i

i

i

x

x

V

t

t

ξ

ξ

ξ

∂

∂

∂

∂

∂

∂ ∂

∂ ∂

∂

=

=

,

2

2

2

j

j

j

x

x

V

t

t

ξ

ξ

ξ

∂

∂

∂

∂

∂

∂ ∂

∂ ∂

∂

=

=

,

3

3

3

k

k

k

x

x

V

t

t

ξ

ξ

ξ

∂

∂

∂

∂

∂

∂ ∂

∂

∂

∂

=

=

The time derivative is

3

3

3

1

2

1

2

1

2

i

j

k

i

j

k

i

j

k

1

1

1

1

1

1

1

1

1

1

2

3

1

2

3

1

2

2

2

2

2

2

2

1

2

3

1

2

3

3

3

3

3

3

3

1

2

3

1

2

3

x

x

V

V

x

x

V

x

x

t

ijk

ijk

ijk

V

V

V

x

x

x

x

x

x

x

x

x

V

V

V

x

x

x

x

x

x

J

ξ ξ ξ

ξ ξ ξ

ξ ξ ξ

ξ

ξ

ξ

ξ

ξ

ξ

ξ

ξ

ξ

ξ

ξ

ξ

ξ

ξ

ξ

ξ

ξ

ξ

ξ

ξ

∂

∂

∂

∂ ∂

∂ ∂

∂ ∂

∂

∂

∂ ∂ ∂

∂ ∂ ∂

∂ ∂ ∂

∂

∂

∂

∂

∂

∂

∂

∂

∂

∂

∂

∂

∂

∂

∂

∂

∂

∂

∂

∂

∂

∂

∂

∂

∂

∂

∂

∂

∂

∂

∂

∂

∂

∂

∂

∂

∂

∂

∂

∂

∂

=∈

+∈

+∈

=

=

+

+



3

2

2

2

1

2

3

3

3

3

1

2

3

j

ij

x

x

x

V

V

V

3

3

i

ij

i 1 j 1

cofactor (i, j) of

(

)

V

[cof ]

ξ

ξ

ξ

ξ

ξ

ξ

ξ

ξ

ξ

∂

∂

∂

∂

∂

∂

∂

∂

∂

∂

∂

∂

∂

∂

∂

= =

∇

=

=

∑∑

J

V

J





38

Consider two square matrices

A

and

B

, and also the product

C = AB

T

. It means that

ik

ij kj

ij kj

k

c

a b

a b

=

≡

∑

,

so we conclude that

ii

ij ij

tr

c

a b

≡ =

C

(trace of the matrix

C

)

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

1

T

T

1

1

1

(cof )

(cof )

det

J

det

−

−

−

=

⇒

=

=

J

J

J

J J

J

J

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

T

1

t

J(t, )

tr

(cof )

(t, )

J(t, ) tr

(t, )

ξ

ξ

−

∂

∂

















=

∇ ⋅

=

∇ ⋅

ξ

V

J

ξ

ξ

V J

ξ

39

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, )]

=

V

ξ

v

x

ξ







and calculate the gradient operator with respect to the Lagrangian variables

k

j

k

j

3

x

x

i

i

ij

k 1

(t, )

V (t, )

v [t, (t, )]

(t, )

ξ

ξ

ξ

∂

∂

∂

∂

∂

∂

=









∇

=

=

∑

V

ξ

ξ

x

ξ

ξ

.

The above formula can be written shortly as

(t, )

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

ξ

∇

= ∇

⋅

V

ξ

x

ξ

J

ξ

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

(

)

t

J(t, )

J(t, ) tr

[t, (t, )]

∂

∂

=

∇

ξ

ξ

v

x

ξ

.

Taking into account that

i

x

i

tr

v

div

∂

∂

∇ =

=

≡ ∇⋅

v

v

v

we finally get the formula

t

J(t, )

J(t, )

[t, (t, )]

∂

∂

=

∇⋅

ξ

ξ

v

x

ξ

40

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,

]

+

=

x(

, ξ)

x , x

ξ)

or

1

2

3

1

1

2

3

2

1

2

3

3

1

2

3

i

i

x t s

,

,

x [ t x (s, ,

,

,x (s, ,

,

,x (s, ,

,

]

ξ ξ ξ

ξ ξ ξ

ξ ξ ξ

ξ ξ ξ

+

=

(

,

)

,

)

)

)

, i = 1,2,3.

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

ξ

j

:

i

i

k

j

k

j

x

x

x

(t s, )

[t, (s, )]

(s, )

ξ

ξ

ξ

∂

∂

∂

+

=

∂

∂

∂

ξ

x

ξ

ξ

,

which can also be written as

ij

ik

kj

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

+

=

J

ξ

J

x

ξ

J

ξ

,

which is equivalent to

(t s, )

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

+

=

J

ξ

J

x

ξ

J

ξ

.

41

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

J(t s, )

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

+

=

ξ

x

ξ

ξ

.

We need to calculate the derivative

t

0

t

0

t

0

J(t

t, ) J(t, )

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

J(t, ) : lim

lim

t

t

t

J[ t, (t, )] 1

J(t, ) lim

t

∆

∆

∆

∆

∆

∆

∆

∆

∆

→

→

→

∂

+

−

−

=

=

=

∂

−

=

ξ

ξ

ξ

x

ξ

ξ

ξ

x

ξ

ξ

Note that

J[ t, (t, )]

∆

x

ξ

is the Jacobian of the “nearly identical” transformation

(t, )

(t

t, )

∆

+

x

ξ

x

ξ

֏

, which can be written shortly as

t

( )

∆

Ψ

x

x

֏

.

The explicit form of this transformation is

2

t

1

2

3

i

i

i

[

( )]

x

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

= +

+

x

∆

Ψ

∆

∆

, i = 1, 2, 3,

42

This, the Jacobi matrix can be calculated as follows

2

i

t

ij

i

ij

j

j

v

[ ] ( t, )

[

( )]

(t, ) t O( t )

x

x

∂

∂

=

= +

+

∂

∂

J

x

x

x

∆

∆

Ψ

δ

∆

∆

or simply

2

( t, )

(t, ) t O( t )

= + ∇

+

J

x

I

v

x

∆

∆

∆

.

Now, it is easy to show (

do it!

) that

di

2

2

3

1

2

1

2

3

v

v

v

v

J( t, ) 1

(t, ) t O( t ) 1

(t, ) t O( t )

x

x

x













∂

∂

∂

= +

+

+

+

= + ∇⋅

+

∂

∂

∂

v

x

x

v

x



∆

∆

∆

∆

∆

Thus, we get

t

0

J( t, ) 1

lim

(t, )

t

→

− = ∇⋅

x

v

x

∆

∆

∆

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, )]

t

∂

=

∇⋅

∂

ξ

ξ

v

x

ξ

.

43

R

EYNOLDS

T

RANSPORT

T

HEOREM

–

CONTINUED

The time derivative

C (t)

′

can be now evaluated as follows

(

)

(

)



0

t

t

t

(t)

(t)

n

t

t

(t)

(t)

(

GGO

normal

Theorem

vel

t)

ocity

(t)

C (t)

f

f

f

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

f

f

f

(t, )d

f

(f ) (t, )d

f d

(f ) d

f d

f v

ds

Ω

Ω

Ω

Ω

Ω

Ω

Ω

∂

∂

∂

∂

∂

∂

∂

∂

∂

∂

⇑

∂

⋅

=









=

+ ⋅∇ + ∇⋅

′

=

+ ⋅∇ + ∇⋅

=

+∇⋅

=

=

+

∇⋅

=

+

∫

∫

∫

∫

∫

∫

∫

v n

v

v

x

ξ

ξ

ξ

v

v

x

x

v

x

x

x

v

x

x

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

f

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.

t

f

0

∂

∂

≡

).

44

P

P

R

R

I

I

N

N

C

C

I

I

P

P

L

L

E

E

O

O

F

F

M

M

A

A

S

S

S

S

C

C

O

O

N

N

S

S

E

E

R

R

V

V

A

A

T

T

I

I

O

O

N

N

D

ENSITY OF FLUID

We define (rather informally) the fluid density as

vol(

)

0

m(

)

lim

vol(

)

Ω

Ω

ρ

Ω

→

=

Thus, the mass of the fluid in the material volume

Ω(t)

can be expressed as

(t)

(t)

m

:

(t, )d

=

∫

x

x

Ω

Ω

ρ

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).

45

P

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

d

dt

(t)

m

0

Ω

=

From Reynolds’ Transport Theorem we get

Re ynolds's

Trans

d

dt

t

(t)

(t)

port Th.

0

d

(

) d

Ω

Ω

ρ

ρ

ρ

∂

∂

⇑









=

=

+ ∇⋅

∫

∫

x

v

x

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

t

(

) 0

ρ

ρ

∂

∂

+∇⋅

=

v

46

P

RINCIPLE OF MASS CONSERVATION

–

CONT

.

Equivalent forms are

t

D

Dt

0

ρ

ρ

ρ ρ

∂

∂

+ ⋅∇ + ∇⋅ =

v

v







→



full

derivative

D

Dt

0

ρ

ρ

+ ∇⋅ =

v

For an incompressible fluid

const

ρ

=

and the above equation reduces to the continuity

equation

div

0

∇⋅ ≡

=

v

v

,

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

t

J(t, )

0

∂

∂

=

ξ

and since

J(0, ) 1

=

ξ

we have

J(t, ) 1

=

ξ

for all times

t

0

≥

.

47

T

IME RATE OF CHANGE OF AN EXTENSIVE QUANTITY

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

H

H(t, )

=

x

. The amount of this quantity contained in the material volume

Ω(t)

is

expressed by the following volume integral

(t)

h(t)

H d

Ω

ρ

=

∫

x

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.

d

d

dt

dt

t

(t)

(t)

D

Dt

t

t

(t)

(t)

( t)

DH

0!

Dt

h(t)

H d

[ ( H)

( H )]d

H

(

) dx

H

H d

H d

Ω

Ω

Ω

Ω

Ω

ρ

ρ

ρ

ρ

ρ

ρ

ρ

∂

∂

⇑

∂

∂

∂

=

∂

=









=

=

+ ∇⋅

=

=

+ ∇⋅

+

+ ⋅∇

=

∫

∫

∫

∫

∫

x

v

x

v

v

x

x





48

DEFORMATION AND STRESS

D

EFORMATION

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

t

.

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

t

∆

.

The location of the first fluid element after the time

t

∆

can be expressed as follows

2

A

A

A

(t,

)∆t O(∆t )

′

=

+

+

x

x

v

x

Since

B

A

=

+

x

x

ρ

then analogously we have

2

B

A

A

(t,

)∆t O(∆t )

′

=

+ +

+

+

x

x

ρ v

x

ρ

,

where the vector

ρ

describes the relative position of

the fluid elements at the time

t

.

t

t+

∆

t

x

1

x

3

x

2

0

A

A'

B

B'

ρ

=x

B

-x

A

ρ

'=x

B'

-x

A'

ρ

ρ

'

49

D

EFORMATION

-

CONTINUED

During a short time interval

t

∆

this vector has changed and can be expressed as

2

2

B

A

A

A

2

(t

t)

(t) [ (t,

)

(t,

)] t O( t )

(t)

(t, )

t O( t , t | | )

′

′









+

=

−

=

+

+ −

+

=

=

+ ∇

+

ρ

x

x

ρ

v

x

ρ

v

x

ρ

v

x ρ

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

2

t

0

d

(t

t)

(t)

lim

(t, )

O(| | )

dt

t

→

+

−

=

= ∇

+

ρ

ρ

ρ

v

x ρ

ρ

∆

∆

∆

.

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

j

x

i

ij

v

∂

∂









∇

=

v

50

D

EFORMATION

–

CONTINUED

The velocity gradient

∇

v

can be written as a sum of two tensors

∇ = +

v

D R

where

T

1

2

[

(

) ]

= ∇ + ∇

D

v

v

or

j

i

ij

j

i

v

v

1

d

2

x

x

















∂

∂

=

+

∂

∂

- symmetric tensor,

and

T

1

2

[

(

) ]

= ∇ − ∇

R

v

v

or

j

i

ij

j

i

v

v

1

r

2

x

x

















∂

∂

=

−

∂

∂

- 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

R

corresponds to the local “rigid” rotation of the

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

D

corresponds to the “real”

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

51

D

EFORMATION

–

CONTINUED

To this end, we note that

1

2

ij

ijk

k

r

ω

= − ∈

, where

k

ω

are the Cartesian components of the

vorticity vector

k

k

ijk

i

j

v

rot

x

ω

∂

=

= ∈

∂

ω

v

e





.

Indeed, we have

j

j

i

i

1

1

2

2

ijk

k

i

j

i

j

ij

i

j

j

i

v

v

v

v

v

v

1

1

(

)

r

x

x

2

x

x

2

x

x

γ

γ

βγ

β γ

γ

β

β

β

δ δ

δ δ

































∂

∂

∂

∂

∂

∂

− ∈ ∈

= −

−

= −

−

=

−

=

∂

∂

∂

∂

∂

∂

Thus, we can write

1

1

1

2

2

2

ij

j

i

ijk

j

k

i

r

ρ

ρ ω

=

= − ∈

= − × =

×

R ρ

e

e

ρ ω

ω ρ

.

Moreover, we get

2

d

d

d

dt

dt

dt

| |

( , )

2( ,

) 2( ,

)

(

)

=

=

=

= ⋅ × =

ρ

ρ ρ

ρ

ρ

ρ Rρ

ρ ω ρ

0

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

1

2

ω

.

52

D

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

d

1

2

dt

=

+

=

+

×

ρ

Dρ

R ρ

Dρ

ω ρ

.

The first terms consists the symmetric tensor

D

, called the deformation rate tensor.

The tensor

D

can be expressed as the sum of the spherical part

D

SPH

and the deviatoric

part

D

DEV

DEV

SPH

=

+

D

D

D

The spherical part

D

SPH

describes pure volumetric deformation (uniform expansion or

contraction without any shape changes) and it defined as



trace
of

k

SPH

SPH ij

ij

k

tr

v

1

1

1

(

)

(

)

3

3

3 x

δ

∂

=

⋅ = ∇⋅

⇒

=

∂

D

D

D

I

v I

D

,

Note that

SPH

1

tr

(

) tr

(

)

div

3

= ∇⋅ ⋅

= ∇⋅ ≡

D

v

I

v

v

.

53

D

EFORMATION

–

CONTINUED

The second part

D

DEV

describes shape changes which preserve the volume.

We have

j

i

k

DEV

DEV ij

ij

j

i

k

v

v

v

1

1

1

div

(

)

3

2

x

x

3 x

















∂

∂

∂

= −

⋅

⇒

=

+

−

∂

∂

∂

D

D

v I

D

δ

and

DEV

SPH

tr

tr

tr

0

=

−

=

D

D

D

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

d

DEV

SPH

dt

=

=

+

ρ

Dρ

D

ρ D

ρ

.

For a short time interval

t

∆

the above relation yields

2

1

2

DEV

SPH

(t

t)

(t)

t

t

O( t )

+

=

+

+

+

ρ

ρ

ρ

ρ

D

ρ

D

ρ





∆

∆

∆

∆

∆

∆

54

D

EFORMATION

–

CONTINUED

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

We have

SPH

d 0

0 d













=

D

,

tr

2d

=

D

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

t

t

∆

+

is

expressed as

2

(t

t)

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

∆

∆

∆

+

= +

+

ρ

ρ

.

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

1

2

Vol (t)

L L

Ω

=

has been changed to

2

2

1

2

Vol (t

t)

L L (1 d t)

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

Ω

Ω

∆

∆

∆

∆

+

=

+

=

+

+

,

and

t

0

Vol (t

t) Vol (t)

1

lim

2d

tr

Vol (t)

t

Ω

Ω

∆

Ω

∆

∆

→

+

−

=

=

= ∇⋅

D

v

O

x

1

x

2

A

B

C

Ω

(t)

B'

C'

A'

{

{

d

∆

t L

2

d

∆

t L

1

Ω

(t+

∆

t)

O

x

1

x

2

A

B

C

Ω

(t)

55

D

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

1

1

2

2

11

12

11

22

12

DEV

1

1

2

2

12

22

12

22

11

d

d

d

(d

d )

d

d

d

d

d

(d

d )

α γ

γ α





































−

−

−

=

=

=

−

−

D

.

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

2

DEV

O( t )

(t

t)

(

t

) (t)

∆

∆

∆

+

= +

+

ρ

I

D

ρ

or in the explicit form as

1

1

2

2

1

2

x (t

t)

(1

t) x (t)

t x (t)

x (t

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.

O

x

1

x

2

A

B

C

Ω

(t)

B'

C'

A'

{

{

α∆

t L

2

α∆

t L

1

Ω

(t+

∆

t)

{

{

γ∆

t L

2

γ∆

t L

1

O

x

1

x

2

A

B

C

Ω

(t)

56

D

EFORMATION

–

CONTINUED

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

We get

2

2

2

2

1

2

1

2

2

1

t

t

Vol (t

t)

L L

L L (1

t

t )

t

1

t

Vol (t) O( t )

Ω

Ω

α ∆

γ ∆

∆

α ∆

γ ∆

γ ∆

α ∆

∆

−

+

=

=

−

−

=

+

=

+

,

so

t

0

Vol (t

t) Vol (t)

1

lim

0

Vol (t)

t

Ω

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)

57

S

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

dA

of the interface

1

2

Ω

Ω

∂ ∩∂

, we have (see figure)

d

dA

=

F

σ

and

2

1

1

2

dA

Ω

Ω

Ω

Ω

→

∂ ∩∂

=

∫

F

σ

The stress vector depends on the point

x

and space

orientation of the surface element

dA

or – equivalently –

of the unit vector

n

perpendicular (normal) to

dA

at the

point

x

.

From the 3

rd

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

( , )

( ,

)

= −

−

σ x n

σ x n

x

1

x

3

x

2

0

Ω

1

Ω

2

dA

n

dF =

dA

58

S

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

j

j

( , )

n x

h

≡

=

n x

,

h

– small number.

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

1

, S

2

and S

3

for

ABC

∆

,

OBC

∆

,

AOC

∆

and

ABO

∆

, respectively. Obviously,

2

S

O(h )

∼

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

j

j

j

j

S

Scos[ ( , )] S ( , )

Sn

=

= ⋅

=

n e

n e

∢

The volume of the tetrahedron is

3

V

O(h )

Ω

∼

.

x

1

x

3

x

2

0

α

n=[n

1

,n

2

,n

3

]

-e

1

-e

2

-e

3

A

B

C

D

59

S

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

fo

vol

surf

rce

force

d

d

dt

Ω

ρ

=

+

∫

v x

F

F



We need to calculate the total surface force

surf

F

.

We have:

on

ABC

∆

:

( , )

( , ) O(h)

=

+

σ x n

σ 0 n

ABC

3

surf

S ( , ) O(h )

∆

=

+

F

σ 0 n

on

OBC

∆

:

1

1

1

( ,

)

( , )

( , ) O(h)

−

= −

= −

+

σ x

e

σ x e

σ 0 e

OBC

3

3

1

1

1

1

surf

S

( , ) O(h )

Sn

( , ) O(h )

∆

= −

+

= −

+

F

σ 0 e

σ 0 e

x

1

x

3

x

2

0

α

n=[n

1

,n

2

,n

3

]

-e

1

-e

2

-e

3

A

B

C

D

60

S

TRESS TENSOR

–

CONTINUED

on

AOC

∆

:

2

2

2

( ,

)

( ,

)

( ,

) O(h)

−

= −

= −

+

σ x

e

σ x e

σ 0 e

AOC

3

3

2

2

2

2

surf

S

( ,

) O(h )

Sn

( ,

) O(h )

∆

= −

+

= −

+

F

σ 0 e

σ 0 e

on

AOB

∆

:

3

3

3

( ,

)

( ,

)

( ,

) O(h)

−

= −

= −

+

σ x

e

σ x e

σ 0 e

AOB

3

3

3

3

3

3

surf

S

( ,

) O(h )

Sn

( ,

) O(h )

∆

= −

+

= −

+

F

σ 0 e

σ 0 e

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



3

2

3

O(h )

O(h )

O(h )

3

vol

j

j

d

d

S[ ( , ) n

( , ) ] O(h )

dt

Ω

ρ

=

+

−

+

∫

v x

F

σ 0 n

σ 0 e







When

h

0

→

the above equation reduces to

j

j

( , ) n

( , ) 0

−

=

σ 0 n

σ 0 e

In general, we can write

j

j

(t, , )

n

(t, , )

=

σ

x n

σ

x e

(summation !)

x

1

x

3

x

2

0

α

n=[n

1

,n

2

,n

3

]

-e

1

-e

2

-e

3

A

B

C

D

61

S

TRESS TENSOR

–

CONTINUED

In the planes oriented perpendicularly to the vectors

e

1

,

e

2

or

e

3

,

the stress vector can be

written as

j

ij

i

(t, , )

(t, )

=

σ

x e

x e

σ

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

j

j

ij

j i

(t, , )

n

(t, , )

(t, )n

=

=

σ

x n

σ

x e

x

e

σ

We define the stress tensor

S

represented by the square matrix

Σ

such that

[ ]

ij

ij

σ

=

Σ

.

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

(t, )

=

x

S

S

.

The stress tensor

S

can be viewed as the linear mapping (parameterized by

t

and

x

)

defined as

3

3

j j

ij

j i

: E

w

w

E

σ

∋ =

∈

w

e

e

֏

S

62

S

TRESS TENSOR

–

CONTINUED

In particular

ij

j i

( )

n

=

=

n

e

σ

σ

S

i.e., the action of

S

on the normal versor

n

at some point of the fluid surface yields

the stress vector

σ

at this point.

Using canonical identification of

E

3

and

R

3

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

n

ar)

product

(

)

( ,

)

= ⋅

≡

n

σ

n Σn n

n Σn





Tangent component can be expressed as

( )

i

n

m

m

ij

j i

km k

i i

ij

j

km k

i

i

n

(

n n )n

[

n

(

n n ) n ]

= −

=

−

=

−

σ

σ

σ

n

e

e

e



τ

τ

σ

σ

σ

σ

σ

or, equivalently (verify!) as

(

)

= × ×

σ

n

σ n

τ

63

P

P

R

R

I

I

N

N

C

C

I

I

P

P

L

L

E

E

O

O

F

F

C

C

O

O

N

N

S

S

E

E

R

R

V

V

A

A

T

T

I

I

O

O

N

N

O

O

F

F

L

L

I

I

N

N

E

E

A

A

R

R

M

M

O

O

M

M

E

E

N

N

T

T

U

U

M

M

2

ND

N

EWTON

’

S

L

AW FOR

C

ONTINUUM

Consider the material volume

(t)

Ω

. The Principle of Momentum Conservation takes the

form

linear momentum

volume force

surface

(t)

(t)

orc

)

e

(t

f

d

dt

d

d

dS

∂

=

+

∫

∫

∫

v x

f x

Σn











Ω

Ω

Ω

ρ

ρ

In the index notation:

d

dt

i

i

ij

j

(t)

(t)

(t)

v d

f d

n dS

∂

=

+

∫

∫

∫

x

x

Ω

Ω

Ω

ρ

ρ

σ

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

d

dt

(t)

(t)

D

d

d

Dt

=

∫

∫

v

v x

x

Ω

Ω

ρ

ρ

64

2

ND

N

EWTON

’

S

L

AW FOR

C

ONTINUUM

–

CONT

.

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

GGO

ij

ij

j

j

(t)

(t)

n dS

d

x

⇑

∂

∂

=

∂

∫

∫

x

Ω

Ω

σ

σ

or in the matrix-vector notation

GGO

(t)

(t)

dS

Div d

⇑

∂

=

∫

∫

Σn

Σ x

Ω

Ω

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

(t)

D

Div

d

0

Dt













−

−

=

∫

v

Σ

f

x

Ω

ρ

ρ

65

2

ND

N

EWTON

’

S

L

AW FOR

C

ONTINUUM

–

CONT

.

Since the choice of the volume

(t)

Ω

is arbitrary, then for the smooth motion the

differential equation follows

D

Div

Dt

=

+

v

Σ

f

ρ

ρ

or in the index notation

ij

i

i

j

Dv

f

Dt

x

∂

=

+

∂

σ

ρ

ρ

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

( )

Div

t













∂ + ⋅∇ =

+

∂

v

v

v

Σ

f

ρ

ρ

or

ij

i

i

j

i

j

j

v

v

v

f

t

x

x

















∂

∂

∂

+

=

+

∂

∂

∂

σ

ρ

ρ

66

S

S

Y

Y

M

M

M

M

E

E

T

T

R

R

Y

Y

O

O

F

F

T

T

H

H

E

E

S

S

T

T

R

R

E

E

S

S

S

S

T

T

E

E

N

N

S

S

O

O

R

R

Σ

Σ

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)

d

=

×

∫

k

r

v x

Ω

ρ

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

k

:

(t)

(t)

( )

d

dt

t

d

D

d

D

D

(

)d

d

d

dt

Dt

dt

Dt

Dt

Ω

Ω

Ω

ρ

ρ

ρ

⇑

=













=

×

=

× + ×

=

×

∫

∫

∫

r

v

k

v

v

r v

x

r v r

x

r

x

From general principles of mechanics we know that

d

dt

=

k

M

The symbol

M

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

material volume

(t)

Ω

67

The total moment

M

is defined as follows

(t)

(t)

d

dS

Ω

Ω

ρ

∂

=

×

+

×

∫

∫

M

r

f x

r Σn

Then, the equation for the angular momentum can be written

(t)

(t)

(

)d

dS

Ω

Ω

ρ

∂

×

−

=

×

∫

∫

r

a f

x

r Σn

,

where

D

Dt

=

v

a

denotes fluid acceleration.

In the index notation, we have

m

ijk

j

k

k

ijk

j

km

ijk

j

km

m

(t)

(t)

(

G

t)

G O

x (a

f )d

x

n dS

(

x

)d

x

Ω

Ω

Ω

ρ

σ

σ

⇑

∂

∂

∈

−

=

∈

=

∈

∂

∫

∫

∫

x

x

The above equation can be transformed as follows

68

ijk

j

k

k

ijk

j

km

m

(t)

km

ijk

j

k

k

j

jm

km

m

(t)

km

ijk

j

k

k

ijk

kj

0 Eq. of

ijk

kj

m

(t)

(t)

(

motion !!!

t)

0

x (a

f )

(x

) d

x

x (a

f ) x

d

x

x ( a

f

) d

d

d

x

Ω

Ω

Ω

Ω

Ω

ρ

σ

σ

ρ

δ σ

σ

ρ

σ

σ

≡

























∂

=

∈

−

−∈

=

∂

∂

=

∈

−

−

−

=

∂

∂

=

∈

− −

−

∈

= −

∈

∂

∫

∫

∫

∫

∫

x

x

x

x

x





It follows that

0

kj

ijk

=

σ

∈

. Using the properties of the alternating symbol (see

Mathematical Preliminaries), we conclude that

32

23

32

23

1jk

kj

13

31

13

31

2 jk

kj

12

21

12

21

3 jk

kj

0

0

0

0

0

0

σ

σ

σ

σ

σ

σ

σ

σ

σ

σ

σ

σ

σ

σ

σ

∈

=

⇒

−

=

⇒

=

∈

=

⇒

−

=

⇒

=

∈

=

⇒

−

=

⇒

=

Thus

ij

ji

σ σ

=

or, equivalently,

T

=

Σ

Σ

, i.e., the

stress tensor is a symmetric tensor

.

69

C

ONCLUSION

The principles of conservation of linear and angular momentum imply the symmetry of the

stress tensor.

Thus, the following statement follows: if the conservation of linear momentum

and symmetry of the stress tensor are postulated, then the conservation of

angular momentum follows automatically.

Yet another way: once the symmetry of the stress tensor is ensured, the fluid

dynamics follows from mass conservation and linear momentum principles.

70

C

C

O

O

N

N

S

S

T

T

I

I

T

T

U

U

T

T

I

I

V

V

E

E

R

R

E

E

L

L

A

A

T

T

I

I

O

O

N

N

S

S

G

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

D

called

the deformation rate tensor and the skew-symmetric part

R

called the (rigid) rotation

tensor.

∇ = +

v

D R

•

Tensor

D

can be expressed as the sum of the spherical part

D

SPH

and the deviatoric part

D

DEV

DEV

SPH

=

+

D

D

D

where

SPH

tr

1

1

(

)

3

3

=

⋅ = ∇⋅

D

D

I

v I

and

j

i

k

DEV

DEV ij

ij

j

i

k

v

v

v

1

1

1

div

(

)

3

2

x

x

3 x

δ

















∂

∂

∂

= −

⋅

⇒

=

+

−

∂

∂

∂

D

D

v I

D

71

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

2

3

0

0

1

2

3

( )

c

c

c

c

...

=

=

+

+

+

+

+

Σ

D

Σ

I

D

D

D

P

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

D

, i.e.

1

2

3

k

k

c

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

=

D

D

D

.

Consider the characteristic polynomial of the tensor

D

3

2

1

2

3

p ( )

det[

]

I

I

I

λ

λ

λ

λ

λ

=

−

= − +

−

+

D

D

I

.

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

3

2

1

2

3

p ( )

I

I

I

= − +

−

+ =

D

D

D

D

D

0

Thus, the 3

rd

power of

D

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

linear combinations of

I

,

D

and

D

2

.

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

nd

order formula

2

0

0

1

2

( )

c

c

c

=

=

+

+

+

Σ

D

Σ

I

D

D

P

72

N

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:

▶

0

c

is a linear function of the invariant

I

1

,

▶

1

c

is a constant,

▶

2

c

0

=

.

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

0

Σ

should correspond to a spherical tensor and

0

0

p

p

= −

= −

⇒

n

I

Σ

Σ

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







1

1

0

0

0

I ( )

2

DE

c

V

3

c

p

(

)

2

p

(

)(

)

2

ζ

µ

ζ

µ

µ

∇⋅

= − ⋅ +

⋅ +

= − +

−

∇⋅

+

D

Σ

Σ

v

Σ

I

I

D

I

v I

D









where

µ

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

ζ

- bulk viscosity (the same unit as

µ

) ; usually

ζ << µ

and

can be assumed zero

.

73

N

EWTONIAN FLUIDS

-

CONTINUED

The constitutive relation written can be written in the index notations

j

k

i

2

3

ij

ij

k

j

i

v

v

v

p (

)

x

x

x

σ

ζ

µ

δ µ

































∂

∂

∂

= − + −

+

+

∂

∂

∂

For an incompressible fluid we have

j

j

v

div

0

x

∂

∇⋅ ≡

≡

=

∂

v

v

and the constitutive relation reduces to the simpler form

j

i

ij

ij

j

i

v

v

p

x

x

σ

δ µ

















∂

∂

= −

+

+

∂

∂

.

74

N

N

A

A

V

V

I

I

E

E

R

R

-

-

S

S

T

T

O

O

K

K

E

E

S

S

E

E

Q

Q

U

U

A

A

T

T

I

I

O

O

N

N

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

nd

Principle of Newton’s dynamics

ij

i

i

j

Dv

f

Dt

x

σ

ρ

ρ

∂

=

+

∂

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

ij

j

k

i

2

3

ij

ij

j

j

j

k

j

j

i

2

j

k

i

2

3

i

i

k

j

j

i

j

2

k

i

1

3

i

i

k

j

j

v

v

v

p

(

)

x

x

x

x

x

x

x

v

v

v

p

(

)

x

x

x

x x

x

x

v

v

p

(

)

x

x

x

x x

σ

δ

ζ

µ

δ µ

ζ

µ

µ

µ

ζ

µ

µ





































































∂

∂

∂

∂

∂

∂

∂

= −

+

−

+

+

=

∂

∂

∂

∂

∂

∂

∂

∂

∂

∂

∂

∂

∂

= −

+

−

+

+

=

∂

∂

∂

∂ ∂

∂

∂

∂

∂

∂

∂

= −

+

+

+

∂

∂

∂

∂ ∂

75

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

grad(div

1

3

)

D

p

(

)

(

)

Dt

ρ

µ∆

ζ

µ

ρ

= −∇ +

+ +

∇ ∇⋅ +

v

v

v

v

f





,

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

N

N

a

a

v

v

i

i

e

e

r

r

-

-

S

S

t

t

o

o

k

k

e

e

s

s

e

e

q

q

u

u

a

a

t

t

i

i

o

o

n

n

1

3

(

)

p

(

) (

)

t

ρ

µ∆

ζ

µ

ρ













∂ + ⋅∇ = −∇ +

+ +

∇ ∇⋅ +

∂

v

v

v

v

v

f

.

For an incompressible fluid the N-S equations simplifies to

(

)

p

t

ρ

µ∆

ρ













∂ + ⋅∇ = −∇ +

+

∂

v

v

v

v

f

,

which is often written in the form

1

(

)

p

t

ν ∆

ρ

∂ + ⋅∇ = − ∇ +

+

∂

v

v

v

v f

where

/

ν µ ρ

=

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

2

/s).

76

The index form of the “incompressible” Navier-Stokes equation is

2

i

i

i

j

i

j

i

j

j

v

v

v

p

1

v

f

t

x

x

x x

ν

ρ

∂

∂

∂

∂

+

= −

+

+

∂

∂

∂

∂ ∂

.

The Navier-Stokes equation is the vector equation (or three scalar equations) with four
unknown fields:
▶ three Cartesian components of the velocity field and
▶ the pressure field.

For an incompressible fluid it is sufficient to add the continuity equation

0

∇⋅ =

v

and

appropriate initial and boundary conditions to obtain a solvable mathematical problem.
However, “solvable” does not mean “easy to solve”.

We need more when the fluid is compressible, as we have one more unknown – the
density

ρ

.

Additional complication comes from the fact that viscosity is temperature dependent!

77

The boundary condition at solid and impermeable surfaces (of the immersed bodies) is
formulated as

=

v

u

at

Ω

∂

,

u

- velocity of the boundary points

The physical meaning of the this conditions is that viscous fluid adheres to a solid
surface, i.e. the velocities of the fluid and of the surface are equal (the no-slip condition)

For the future study …

Another type of the boundary conditions that may be posed for incompressible flow are

inlet/outlet conditions (for internal flows like flows in ducts or pipe systems) and far-field
conditions (for flow in “open” space like a flow around an airplane). Detailed discussion of the
possible forms of such conditions is well beyond the scope of this introductory course. Let us only
mention that the standard far–field condition is typically some approximation of the asymptotic
condition

lim

∞

→∞

=

r

v

v

, where

∞

v

denotes the velocity in the free stream far away from the immersed

body. The situation is more complicated in the case of compressible flows. We have already
mentioned that the momentum and mass conservation equations are not sufficient to describe uniquely
the flow of such fluid. Indeed, we have an additional unknown (density or temperature). We need also
the energy equation which will be derived later on. Also the boundary treatment is more complicated
and usually quite different than for the incompressible flow. Again, more detailed discussion of this
problems will be presented in more advanced courses of Fluid Mechanics and Computational Fluid
Dynamics (CFD). See “Fluid Mechanics” by Kundu & Cohen (Elsevier Academic Press)

78

E

E

U

U

L

L

E

E

R

R

E

E

Q

Q

U

U

A

A

T

T

I

I

O

O

N

N

A

A

N

N

D

D

I

I

T

T

S

S

F

F

I

I

R

R

S

S

T

T

I

I

N

N

T

T

E

E

G

G

R

R

A

A

L

L

S

S

T

HE

E

ULER

E

QUATION

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

(

)

p

t

ρ

ρ













∂ + ⋅∇ = −∇ +

∂

v

v

v

f

Using the Lamb-Gromeko form of the convective acceleration

( )

2

1

2

(

)

v

⋅∇ = ∇

+ ×

v

v

ω v

we can write the Euler equation as follows

( )

2

1

2

1

v

p

t

ρ

∂ +∇

+ × = − ∇ +

∂

v

ω v

f

79

Let us now make the following assumptions:

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

Φ

f

that

f

Φ

= ∇

f

,

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

p

p( )

ρ

=

.

Define the following pressure function

P

:

1

P( ) :

p ( )d

ρ

ρ ρ

ρ

=

′

∫

.

Then

i

i

i

1

1

P[ ( )]

p [ ( )]

p[ ( )]

x

( )

x

( ) x

ρ

ρ

ρ

ρ

ρ

ρ

∂

∂

∂

=

=

′

∂

∂

∂

x

x

x

x

x

⇒

1

P

p

ρ

∇ = ∇

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

P

p /

ρ

=

.

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

p

κ

ρ

≈

, where

P

V

c / c

κ

=

. Then (

check yourself!

)

P

p

P

c T

i

(

1)

κ

κ

ρ

=

=

=

−

(specific enthalpy)

80

B

ERNOULLI

E

QUATION

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

H

0

t

∂ ≡

∂

.

The Euler equation can be written as

(

)

2

1

2

f

v

P

Φ

∇

+ −

= ×

v ω

Choose a streamline and define the tangent unary vector

/ v

=

τ

v

. Next, multiply the above

equation (in the sense of the inner product) by

τ

. The result is

(

) (

)

2

2

1

1

2

2

f

f

d

1

v

P

:

v

P

(

)

0

d

v

Φ

Φ

+ −

= ∇

+ −

⋅ =

⋅ ×

=

τ

v v ω

τ

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

2

1

B

2

f

v

P

C

Φ

+ −

=

The Bernoulli constant

C

B

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

0

× =

v ω

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

0

≡

ω

).

81

In practice, we use the Bernoulli equation in the form of

(

) (

)

2

2

1

1

2

2

f

f

A

B

v

P

v

P

Φ

Φ

+ −

=

+ −

where A and B denote two points on the same streamline.

82

C

AUCHY

-L

AGRANGE

E

QUATION

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

v

Φ

= ∇

v

for some scalar field (the velocity potential)

Φ

v

and

0

= ∇× =

ω

v

,

v

t

t

Φ













∂

∂

= ∇

∂

∂

v

,

2

2

v

v

|

|

Φ

= ∇

.

Then the Euler equation takes the form of

(

)

2

1

v

v

2

t

f

|

|

P

Φ

Φ

Φ

∂

∂

∇

+ ∇

+ −

=

0

meaning that

2

v

1

v

2

f

P

C(t)

t

Φ

Φ

Φ

∂

+ ∇

+ −

=

∂

for some function

C

, which depends only on time.

83

AERO/HYDRODYNAMIC FORCES

S

TRESS AND REACTION FORCE EXTERTED AT AN IMMERSED SURFACE

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

dS

dS

∂

∂

=

=

∫

∫

F

σ

Σn

Ω

Ω

.

For an incompressible fluid we have the constitutive relation





rotation

gradient of

tensor

velocity

p

2

p

2

2

= − +

= − +

∇

−

Σ

I

D

I

v

R

µ

µ

µ

.

Since

1

1

2

2

rot

= − ×

= − ×

R n

n

v

n ω

we can write

p

2

=

= − +

∇ ⋅ +

×

σ

Σn

n

v n

n ω

µ

µ

.

84

The following theorem holds:

If

div

0

=

v

(incompressible flow) and

Ω

∂

=

v

0

then

∇ ⋅ + × =

v n n ω

0

.

Proof

Since

Ω

∂

=

v

0

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

j

j

j

k

k

v

v

n , k 1,2,3

x

Ω

λ

∂

∂

∇

× =

⇒

=

=

∂

n

0

for some real numbers

λ

j

(j = 1, 2, 3).

Next, in the index notation we have

ijk

j

k

i

n

ω

× =∈

n ω

e

,

j

i

j

i

x

v

n

∂

∂

∇ ⋅ =

v n

e

85

After insertion we get



j

j

j

j

i

j

i

i

j

x

x

x

i

ijk

k

j

i

i

ijk

k

j

i

x

x

x

x

x

i

i

j

i

j

j

i

i

j

i

j

i

x

x

x

j

j

i

j

j

j

i

i

div

j

i

j

0

j

j

i

i

(

v

) n

(

v

v ) n

v

(

)

v

n

(

v

v

v ) n

v n

(

v n

v n )

(

n n

n n )

∂

∂

∂

∂

∂

∂

∂

∂

∂

∂

∂

∂

∂

∂

∂

∂

∂

∂

∂

∂

∂

=

∂

=













∇ ⋅ + × =

+∈

=

+∈ ∈

=

=

+

−

=

+

−

=

=

=

−

=

−

=

v

v n n ω

e

e

e

e

e

e

e

0

α

α

αβ

β

α β

β α

β

ω

δ δ

δ δ

λ

λ

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

p

µ

= − −

×

σ

n

n ω

.

Note that

p

0

× =

n n

and

(

)

0

× ⋅ =

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

(p

)dS

Ω

µ

∂

= −

+

×

∫

F

n

n ω

86

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

dS

p dS 2

dS

dS

Ω

Ω

Ω

Ω

µ

µ

∂

∂

∂

∂

=

= −

+

∇ ⋅

+

×

∫

∫

∫

∫

F

σ

n

v n

n ω

.

But







Laplacian of

tensor version

the velocity

of GGO

GGO for

the cross

product

0

dS

Div (

)d

d

(

)d

(

)d

d

dS

Ω

Ω

Ω

Ω

Ω

Ω

Ω

∆

⇑

∂

⇑

=

=

∂

∇ ⋅

=

∇

=

=

= ∇ ∇⋅

− ∇× ∇×

= − ∇×

= −

×

∫

∫

∫

∫

∫

∫

∫

ω

v n

v

x

v

x

v

x

v

x

ω x

n ω

Thus

p dS 2

dS

dS

(p

)dS

Ω

Ω

Ω

Ω

µ

µ

µ

∂

∂

∂

∂

= −

−

×

+

×

= −

+

×

∫

∫

∫

∫

F

n

n ω

n ω

n

n ω

.

87

T

HE INTEGRAL FORM OF THE MOMENTUM EQUATION

(

STEADY MOTION

)

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

ij

i

j

i

j

j

v

v

f

x

x

σ

ρ

ρ

∂

∂ =

+

∂

∂

or, using the (stationary) mass conservation equation

ij

i

j

i

j

i

j

j

j

0

v

v

v

( v )

f

x

x

x

σ

ρ

ρ

ρ

⇓

∂

∂

∂

+

=

+

∂

∂

∂



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

fluid

body

Γ

∪

Γ

=

Ω

∂

ij

i

j

i

j

j

( v v )

f

x

x

σ

ρ

ρ

∂

∂

=

+

∂

∂

n

n

Γ

fluid

Γ

body

Γ

body

Γ

fluid

n

n

88

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

i

j

ij

j

( v v

)

0

x

ρ

σ

∂

−

=

∂

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

j

x

i

j

ij

( v v

)d

0

Ω

ρ

σ

∂

∂

−

=

∫

x

→

i

j

j

ij

j

( v v n

n )dS

0

Ω

ρ

σ

∂

−

=

∫

Equivalently, we can write the vector relation as follows

 

n

j

j

i

i

ij

j

v

i

( v n v

n

)dS

0

Ω

ρ

σ

= ⋅

∂

−

=

∫

v

v n

σ

e

e





,

or simply

momentum flux

through t

n

he boundary

dS

(

) v dS

Ω

Ω

ρ

∂

∂

=

∫

∫

σ

v



89

The obtained relation is nothing else like the integral form of the momentum principle
written for a stationary flow. Note that it contains exclusively the integrals over the
boundary of the control volume (no volume integral are present).

The boundary of the control volume consists of two parts: surface of the body and the fluid
boundary. Thus, we can write

body

fluid

n

dS

(

) v dS

dS

Γ

Ω

Γ

ρ

−

∂

=

−

∫

∫

∫

F

σ

v

σ





,

where the vector

F

is the reaction on the immersed body from the fluid contained in the

control volume.

If we assume that the body is impermeable then

body

n

v

0

Γ

=

and

fluid

fluid

n

(

) v dS

dS

Γ

Γ

ρ

− =

−

∫

∫

F

v

σ

90

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

p

2

µ

= −

+

σ

n

Dn

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

fluid

fluid

fluid

n

(

) v dS

p dS 2

dS

Γ

Γ

Γ

ρ

µ

= −

−

+

∫

∫

∫

F

v

n

D n

Quite often, we can choose

fluid

Γ

in such way that the viscous term is relatively small

and can be neglected. Then

fluid

fluid

n

(

) v dS

p dS

Γ

Γ

ρ

≅ −

−

∫

∫

F

v

n

.

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

p

a

.

Note that for the closed surface

fluid

Γ

we have

fluid

fluid

a

a

p

dS

p

dS

Γ

Γ

=

=

∫

∫

n

n

0

(

why?

)

91

The formula for the full (or net) force can be then written as follows

fluid

fluid

n

a

net

(

) v dS

(p p ) dS

Γ

Γ

ρ

≅ −

−

−

∫

∫

F

v

n

During the tutorial part we will see that the formula in the above form is particularly useful
to calculate the reaction force exerted by a free stream colliding with the solid body.

92

C

C

I

I

R

R

C

C

U

U

L

L

A

A

T

T

I

I

O

O

N

N

,

,

V

V

O

O

R

R

T

T

I

I

C

C

I

I

T

T

Y

Y

A

A

N

N

D

D

S

S

T

T

R

R

E

E

A

A

M

M

F

F

U

U

N

N

C

C

T

T

I

I

O

O

N

N

S

S

C

IRCULATION

Definition: Circulation of the vector field

w

along the (closed) contour

L

is defined as

d

Γ

=

⋅

∫

w



l

L

Kelvin’s Theorem:

Assume that:

•

the volume force field

f

potential,

•

the fluid is inviscid and barotropic

•

the flow is stationary.

Then

: the circulation of the velocity field

v

along any closed material line

L(t)

is constant

in time, i.e.

d

d

dt

dt

(t)

(t)

(t, ) d

0

Γ

≡

⋅ =

∫

v

x



l

L

93

Proof of the Kelvin Theorem:

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

1

P

p

ρ

∇ = ∇

,

Φ

= ∇

f

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

(

)

(P

)

Φ

≡ ⋅∇ = −∇ −

a

v

v

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

0

0

(t)

(t)

(t)

(t, ) d

(t, )

(t, )d

Γ

=

⋅ =

⋅

∫

∫

v

x

V

ξ J

ξ





l

l

L

L

where

(t, )

∂

=

∂

x

J

ξ

ξ

denotes the Jacobi matrix of the transformation between Eulerian and

Lagrangian coordinates.

94

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

0

0

0

0

0

int. of the grad. a

0

0

(t)

(t)

(t)

1

2

0

0

(t)

(

long the closed

t)

(t)

(t

loop

)

d

d

(t, ) d

(t, )

(t, )d

(t, )

(t, )d

dt

dt

(t, )

(t, )d

(t, ) d

(

)(t, ) d

(P

) d

ξ

ξ

Φ

=

⋅ =

⋅

=

⋅

+

+

⋅∇

=

⋅ +

∇

⋅

⋅

=

= − ∇ +

⋅

∫

∫

∫

∫

∫

∫

v

x

V

ξ J

ξ

a

ξ J

ξ

V

ξ

V

ξ

a

x

V V

ξ

















l

l

l

l

l

l

l

L

L

L

L

L

L

L

int. of the grad. along

the closed loo

0

p

0

=

=

∫





where the relation

t

(t, )

(t, )

ξ

∂

= ∇

J

ξ

V

ξ

has been used.

95

V

ORTICITY

As we already know, the vorticity is defined as the rotation of the velocity:

rot

=

≡ ∇×

ω

v

v

Definitions:

▶ A vortex line is the line of the vorticity vector field. At each point of such line, the

vorticity vector is tangent to this line.

▶ The vortex tube is the subset of the flow domain bounded by the surface made of the

vortex lines passing through all point of a given closed contour (the contour

L

on the

picture below)

vortex line

L

vorticity field

96

S

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:

S

d

d

σ

Γ

⋅

=

⋅ =

∫

∫

ω n

v

x



l

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

S

1

and

S

2

.

From the GGO theorem we have

1

2

side

S

S

S

0

0

d

ds

ds

ds

0

Ω

=

= ∇⋅

=

⋅

+

⋅

+

⋅

=

∫

∫

∫

∫

ω x

ω n

ω n

ω n



S

2

S

1

n

2

n

1

S

side

97

Note that the last integral vanishes because the surface

S

side

is made of the vortex lines and

thus at each point of

S

side

the normal versor

n

is perpendicular to the vorticity vector.

Note also that the orientations of the normal versors at

S

1

and

S

2

are opposite (in order to

apply the GGO Theorem, the normal versor must point outwards at all components of
the boundary

Ω

∂

).

Reversing the orientation of

n

at

S

2

, we conclude that

1

2

S

S

ds

ds

⋅

=

⋅

∫

∫

ω n

ω n

98

H

ELMHOLTZ

(3

RD

) 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)

=

a

a

0

l

.

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

t

:

a

∋

∈

x

֏

R

R

R

R

R

R

R

R

F

, i.e.,

t

:

( )

[ (s)]

=

=

x

x s

a

l

F

.

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

Jacobi ma

d

d

d

t

0

ds

trix

ds

ds

(s)

(s)

( (s))

[

/

]( (s))

(s) [

/

]( (s))

(s)

=

=

= ∂ ∂

= ∂ ∂

τ

x

a

x

ξ a

a

x

ξ a

τ



F

99

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

2

D

1

t

Dt

2

( v )

=

= ∂ + ∇

+ ×

a

v

v

ω v

The rotation of

a

can be expressed as

D

t

Dt

(

)

(

)

(

)

(

)

∇× = ∂ ∇× + ∇× × =

− ⋅∇ + ∇⋅

a

v

ω v

ω

ω

v

v ω

In the above, the following vector identity, written for

=

p

ω

and

=

q

v

, is used

(

)

(

)

(

)

(

)

(

)

∇× × = ⋅∇ − ⋅∇ + ∇⋅

− ∇⋅

p q

q

p

p

q

q p

p q

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

/

ρ

ω

as follows

( )



2

D

1

1 D

1

D

1

Dt

Dt

Dt

1

1

1

(

)

(

)

(

)

ρ

ρ

ρ

ρ

ρ

ρ

ρ

ρ

ρ

=− ∇⋅









=

−

= ∇× + ⋅∇ − ∇⋅

+

+

∇⋅ = ∇× +

⋅∇

v

ω

ω

ω

a

ω

v

v ω

ω

v

a

ω

v

100

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

∇× =

a

0

Then, the equation for the vector field

/

ρ

ω

reduces to

( )

D

1

1

Dt

(

)

ρ

ρ

=

⋅∇

ω

ω

v

Define the vector field

c

such that

i

i

j

j

x

c

ω ρ ξ

∂

=

∂

, or equivalently,

Jacobi

matrix

(

/

)

ρ

=

∂ ∂

ω

x

ξ c





.

In the above, the symbol

ξ

denotes the Lagrangian variables.

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

( )

d

d

D

1

Dt

dt

dt

d

dt

L

(

/

)

(

/

) c (

/

)

(

/

) c (

/

)(

/

)

ρ









=

=

∂ ∂

= ∂ ∂

+ ∂ ∂

=

= ∂ ∂

+ ∂ ∂ ∂ ∂

ω

x

ξ c

x

ξ

v

ξ c

x

ξ

v

x

x

ξ c

The right-hand side can be written as

1

R

(

)

(

/

)

(

/

)(

/

)

ρ









=

⋅∇ = ∂ ∂

⋅∇

= ∂ ∂ ∂ ∂ ⋅

ω

v

x

ξ c

v

v

x

x

ξ c

101

Since

L = R

, we obtain

d

dt

(

/

)

(

/

)(

/

)

(

/

)(

/

)

∂ ∂

+ ∂ ∂ ∂ ∂

= ∂ ∂ ∂ ∂

x

ξ

c

v

x

x

ξ c

v

x

x

ξ c

which implies that

d

dt

0

=

c

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

Using the Lagrangian description, we can write

0

(t, )

(0, )

=

=

c

ξ

c

ξ

c

Note that for the initial time

t = 0

the transformation between Lagrangian and Eulerian

descriptions reduces to identity.

t 0

(

/

)

=

∂ ∂

=

x

ξ

Ι

Therefore

0

1

0

0

ρ

=

c

ω

and since

0

(t)

≡

c

c

we get

0

1

1

0

(

/

)

ρ

ρ

= ∂ ∂

ω

x

ξ

ω

.

102

The last equality has the form of the transformation rule for the vectors tangent to

material lines. Since the vector

0

0

/

ρ

ω

is tangent to the vortex line passing through the

point ξ at

t = 0

, it follows that the vector

/

ρ

ω

is tangent to image of this line at some later

time t. But

/

ρ

ω

is also tangent to the vortex line passing through the point

x

, which means

that the vortex lines must be material.

Since the vortex lines are material, so are the vortex tubes. If we define a closed, material

contour lying on the vortex tube’s surface (and wrapped around it), then such a contour

remains on this surface for any time. It follows from the Kelvin Theorem that the circulation

along such contour remains constant. Consequently, the strength of any vortex tube also

remains constant in time. It is important conclusion showing that the vortex motion of the

inviscid, barotropic fluid exposed to a potential force field cannot be created or destroyed.

103

E

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

2

1

1

t

2

( v )

p

ρ

∂ + ∇

+ × = − ∇ +

v

ω v

f

Thus, the application of the rotation operator yields

1

t

(

)

(

p)

ρ

∂ + ∇× × = −∇× ∇ + ∇×

ω

ω v

f

The pressure term can be transformed as follows

2

1

1

1

1

0

(

p)

( )

p

p

p

ρ

ρ

ρ

ρ

ρ

↑

∇× ∇ = ∇

×∇ + ∇×∇ = − ∇ ×∇





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

2

1

t

(

)

(

)

p

ρ

ρ

∂ + ⋅∇ − ⋅∇ = − ∇ ×∇ + ∇×

ω

v

ω

ω

v

f

or, using the full derivative



2

nonpotential

vortex stretching

volume force

baroclinic

term

te

D

1

D

m

t m

t

r

er

(

)

p

ρ

ρ

=

⋅∇

− ∇ ×∇ + ∇×

ω

ω

v

f











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

D

Dt

(

)

=

⋅∇

ω

ω

v

In the 2D case it reduces further to

D

Dt

0

=

ω

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

2

1

1

t

2

( v )

p

ρ

ν ∆

∂ + ∇

+ × = − ∇ +

+

v

ω v

v f

If the rotation operator is applied, we get the equation

t

(

)

(

)

ν ∆

∂ + ⋅∇ − ⋅∇ =

+ ∇×

ω

v

ω

ω

v

ω

f

which reduces to

t

(

)

(

)

ν ∆

∂ + ⋅∇ − ⋅∇ =

ω

v

ω

ω

v

ω

when the field of the volume forces

f

is potential.

106

In the above, the following operator identity has been used

rot

rot (grad div

rot rot )

rot rot

grad div

rot rot

∆

∆

=

−

= −

=

−

=

v

v

v

ω

ω

ω

ω

showing that the vector Laplace and rotation operators commute.

The vorticity equation can be also written equivalently as

D

Dt

(

)

ν ∆

=

⋅∇ +

ω

ω

v

ω

The viscous term describes the diffusion of vorticity due to fluid viscosity. This effect
smears the vorticity over the whole flow domain. Thus, in the viscous case the vortex lines
are not material lines anymore.

107

T

WO

-

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

ψ

such that

y

u

ψ

= ∂

,

x

v

ψ

= −∂

Note that the continuity equation

x

y

u

v

0

∂ + ∂ =

is satisfied automatically. Indeed, we have

x

y

x

u

v

ψ

ψ

0

∂ + ∂ = ∂

−∂

=

xy

y

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:

x

y

u

v

uv

vu

0

ψ

ψ

ψ

∇ ⋅ = ∂ + ∂

= − +

=

v

108

S

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

B

B

B

x

y

AB

A

A

A

B

B

y

x

x

x

y

y

A

A

B

B

A

A

Q

ds

ds

(u n

v n )ds

(u

v

)ds

(

)ds

d

τ

τ

τ ψ τ ψ

ψ

ψ ψ

=

⋅

=

⋅

=

+

=

=

−

=

∂ + ∂

=

= ∇ ⋅ =

−

∫

∫

∫

∫

∫

∫

v n

v n

s

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

n

τ

v

A

B

ψ=ψ

Α

ψ=ψ

B

=ψ

A

+

Q

AB

streamlines

109

S

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

x

y

z

z

(

v

u)

ω

≡ ∇× = ∂ −∂

≡

ω

v

e

e

Then, the streamfunction satisfies the Poisson equation

xx

yy

x

y

(

v

u)

∆ψ

ψ

ψ

ω

≡ ∂

+ ∂

= − ∂ − ∂

= −

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

0

=

v

is automatically satisfied. The complete

description consists of the following equations:

•

Equation of the vorticity transport (2D)

t

x

y

u

v

ω

ω

ω ν ∆ω

∂ + ∂ + ∂ =

•

Equation for the streamfunction

∆ψ

ω

= −

•

Relation between the streamfunction and velocity

y

u

ψ

= ∂

,

x

v

ψ

= −∂

•

Definition of vorticity (2D)

x

y

v

u

ω

= ∂ − ∂

accompanied by appropriately formulated boundary and initial conditions.

110

E

E

N

N

E

E

R

R

G

G

Y

Y

O

O

F

F

A

A

F

F

L

L

U

U

I

I

D

D

F

F

L

L

O

O

W

W

E

E

N

N

E

E

R

R

G

G

Y

Y

E

E

Q

Q

U

U

A

A

T

T

I

I

O

O

N

N

Total energy of the fluid inside the region

Ω

is the sum of the kinetic energy

E

K

and the internal energy

U

.

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

d

K

dt

(U E ) P

P

Q

Q

•

•

Ω

∂Ω

Ω

∂Ω

+

=

+

+

+

111

The right-hand-side terms are:

P

d

Ω

Ω

= ρ ⋅

∫

f v x

- power of the volume forces,

P

dS

dS

∂Ω

∂Ω

∂Ω

=

⋅

=

⋅

∫

∫

σ v

v Σn

- power of the surface (or interface) forces,

Q

qd

•

Ω

Ω

= ρ

∫

x

- power of the internal heat sources (symbol q

denotes their volumetric density) ,

GGO

d T

Q

dS

T

dS

dn

⇑

•

∂Ω

∂Ω

∂Ω

=

= λ

λ∇ ⋅

∫

∫

n

- the heat flux through the boundary

∂Ω

(the

symbol

T

denotes temperature and

λ

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

2

d

1

2

ne

dary

r

t

gy

d

(u

v )d

d

dS

q d

T

dS

Ω

Ω

∂

Ω

Ω

Ω

∂Ω

∂Ω

ρ +

= ρ ⋅

+

⋅

+ ρ

+ λ∇ ⋅

∫

∫

∫

∫

∫

x

f v x

v Σn

x

n















,

where

u

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

2

2

d

1

D

1

Dt

2

2

dt

(u

v )d

(u

v )d

Ω

Ω

ρ +

= ρ

+

∫

∫

x

x

.

113

N

N

e

e

x

x

t

t

,

,

w

w

e

e

t

t

r

r

a

a

n

n

s

s

f

f

o

o

r

r

m

m

t

t

h

h

e

e

s

s

u

u

r

r

f

f

a

a

c

c

e

e

i

i

n

n

t

t

e

e

g

g

r

r

a

a

l

l

s

s

:

:

j

j

j

j

j

i

j

i

j

j

i

j

i

x

x

x

ij i

j

ij i

ij

i

ij

i

1

1

2

2

x

x

x

x

x

ij

i

ij

i

j

ij

i

j

1

1

1

2

2

2

x

x

x

x

ij

i

ij

i

j

ij

i

ji

GGO

dS

v n dS

(

v )d

v d

v d

v d

(

v

v )d

(

v

v )d

v d

(

v

v )d

v d

∂

∂

∂

∂

∂

∂

Ω

Ω

Ω

∂Ω

∂Ω

∂

∂

∂

∂

∂

∂

∂

∂

∂

∂

Ω

Ω

Ω

∂

∂

∂

∂

∂

∂

∂

∂

Ω

Ω

⇑

=

Ω

σ

⋅

= σ

=

σ

=

σ

+ σ

=

=

σ

+ σ

+

+ σ

−

=

=

σ

+ σ

+

+ σ

−

σ

∫

∫

∫

∫

∫

∫

∫

∫

∫

∫

∫

v Σn

x

x

x

x

x

x

x

x

x



j

j

x

i

v d

Div

d

d

∂

∂

Ω

Ω

Ω

=

=

⋅

+

∫

∫

∫

x

Σ v x

Σ : D x

and also

GGO

T

dS

(

T)d

⇑

∂Ω

Ω

λ∇ ⋅

= ∇⋅ λ∇

∫

∫

n

x

.

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

2

D

1

Dt

2

(u

v )

Div

q

(

T)

ρ

+

=ρ ⋅ +

⋅ +

+ρ +∇⋅ λ∇

f v

Σ v Σ: D

114

I

I

N

N

T

T

E

E

R

R

N

N

A

A

L

L

E

E

N

N

E

E

R

R

G

G

Y

Y

A

A

N

N

D

D

V

V

I

I

S

S

C

C

O

O

U

U

S

S

D

D

I

I

S

S

S

S

I

I

P

P

A

A

T

T

I

I

O

O

N

N

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

D

Dt

Div

ρ

=ρ +

v

f

Σ

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

D

Dt

Div

ρ

⋅ =ρ ⋅ +

⋅

v v

f v

Σ v

or, equivalently



2

D 1

Dt 2

v

(

)

Div

ρ

⋅ =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

k

2

d

1

2

dt

E

v d

d

Div

d

Ω

Ω

Ω

ρ

= ρ ⋅

+

⋅

∫

∫

∫

x

f v x

Σ v x



115

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

j

j

j

j

x

x

x

ij

i

ij

i

ij

i

x

i

ij

j

ij

i

Div

d

v d

(

v )d

v d

v

n dS

v d

dS

d

dS

d

∂

∂

∂

∂

∂

∂

Ω

Ω

Ω

Ω

∂

∂

Ω

Ω

∂Ω

∂Ω

Ω

∂Ω

⋅

=

σ

=

σ

− σ

=

=

σ

− σ

=

⋅

−

∇

=

=

⋅

−

∫

∫

∫

∫

∫

∫

∫

∫

∫

∫

Σ v x

x

x

x

x

v Σn

Σ: v x

v Σn

Σ: D x

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

2

d

1

2

dt

v d

d

dS

d

Ω

Ω

Ω

∂Ω

ρ

= ρ ⋅

+

⋅

−

∫

∫

∫

∫

x

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

d

dt

U

ud

qd

T

dS

d

Ω

Ω

Ω

∂Ω

ρ

= ρ

+ λ∇ ⋅

+

∫

∫

∫

∫

x

x

n

Σ : D x





.

116

Observe that both formulae written above contain the term

:

d

Ω

=

∫

Σ : D x

T

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

k

2

3

x

ij

k

ij

ij

[ p (

)

v ]

2 d

∂

∂

σ = − + ζ− µ

δ + µ

.

Hence, we have

k

2

k

0 if

0 (compression)

2

3

x

ij

ij

ij

ij

k

ij

ij

ij

ij

2

3

x

ii

k

ii

ij

ij

0 if

0 (expansion)

positi

2

2

3

ve definite par

r

t

t

d

p

d

(

)

v

d

2 d d

pd

(

)

v d

2 d d

p

(

)(

)

2

>

∇

∂

∂

∂

∂

=

⋅ <

<

∇⋅ >

−

= σ

= − δ

+ ζ − µ

δ

+ µ

=

= −

+ ζ − µ

+ µ

=

− ∇⋅

+

+ ζ − µ ∇⋅

+ µ

v
v

D

Σ : D

v

v

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

0)

dissipation of the machanical energy due to i

energy transfer

2

2

3

d

p

d

(

) (

) d

2

d

− −

− −

<

>

Ω

Ω

Ω

Ω

= − ∇⋅

+ ζ − µ ∇⋅

+ µ

∫

∫

∫

∫

Σ : D x

v x

v

x

D : D x







nt ernal "friction"





.

For incompressible flows we have

0

∇⋅ =

v

, thus

j

j

i

i

j

i

j

i

j

j

j

j

i

i

i

i

j

j

i

i

j

i

j

i

v

v

v

v

1

2

x

x

x

x

v

v

v

v

v

v

v

v

1

2

x

x

x

x

x

x

x

x

d

2

d

d

2

d

d

∂

∂

∂

∂

∂

∂

∂

∂

Ω

Ω

Ω

∂ ∂

∂

∂

∂ ∂

∂

∂

∂

∂ ∂

∂ ∂

∂ ∂

∂

∂

∂

Ω

Ω



























≡

















= µ

= µ

+

+

=

= µ

+

+

= µ

+

∫

∫

∫

∫

∫

i

j

v

x

Σ : D x

D : D x

x

x

x

R

The quantity

R

is called the dissipation rate.

118

F

F

I

I

R

R

S

S

T

T

I

I

N

N

T

T

E

E

G

G

R

R

A

A

L

L

O

O

F

F

T

T

H

H

E

E

E

E

N

N

E

E

R

R

G

G

Y

Y

E

E

Q

Q

U

U

A

A

T

T

I

I

O

O

N

N

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

f

.

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

2

D

1

Dt

2

(u

v )

(p )

ρ

+

= −∇⋅

+ρ ⋅

v

f v

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

2

D

1

Dt

2

(u

v )

p

p

ρ

+

= − ∇⋅ − ⋅∇ +ρ ⋅

v v

f v

119

Since the volume force is potential, the corresponding term in the right-hand side can be
transformed as follows



D

t

0

t

D

(

)

=

ρ ⋅ = ρ ⋅∇Φ = ρ ∂ Φ + ⋅∇Φ = ρ Φ

f v

v

v

Moreover, due to flow steadiness we have



D

t

D

0

t

p

p

p

p

=

⋅∇ = ∂ + ⋅∇ =

v

v

Next, from the mass conservation equation

D

Dt

0

ρ+ρ∇⋅ =

v

we get the following expression for divergence of the velocity field

1 D

Dt

ρ

∇⋅ = −

ρ

v

120

The energy equation can be now written in the following form

D

2

Dt

p

2

D

1

D

1 D

D

Dt

Dt

Dt

(p/

Dt

2

)

(u

v )

p

=−

ρ

ρ

ρ

+

=

ρ−

+ Φ



or

i

2

D

1

Dt

2

( u p

v

) 0

=

+ ρ +

−Φ =





where

i

u p

= + ρ

denotes the mass-specific enthalpy of the fluid.

Thus, the energy equation can be written as

2

D

1

Dt

2

(i

v

) 0

+

−Φ =

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

2

1

2

(i

v

) 0

⋅∇ +

−Φ =

v

121

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

2

1

e

2

i

v

C

const

+

−Φ =

=

In general the energy constant

C

e

can be different for each streamline. If

C

e

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

2

1

B

2

P

v

C

const

+

−Φ =

=

Thus, when the flow is barotropic then

e

B

i P

C

C

const

− =

−

=

i.e., the enthalpy

i

and the pressure potential

P

differ only by an additive constant.

If additionally the fluid is incompressible then its internal energy

u

is fixed and the specific

enthalpy can be defined as

p /

ρ

. Thus, in the incompressible case, the energy and

Bernoulli equations are formally identical.

122

E

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 !)

D

Dt

u

p

ρ

= − ∇⋅

v

We have already used the relation

1 D

Dt

ρ

∇⋅ = −

ρ

v

. Thus, the equation for the internal

energy e can be written as follows

2

p

D

D

D

D

Dt

Dt

Dt

Dt

u

p

(1/ )

p

ρ

= −

ρ = −

ρ = −

υ

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

s

, internal

energy

u

and specific volume

1/

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

D

D

D

D

D

Dt

Dt

Dt

Dt

Dt

T s

u p

p

p

0

=

+

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

2

1

e

2

i

v

C

+

−Φ =

, or equivalently

2

1

2

(i

v

) 0

∇ +

−Φ =

.

Similarly, we call the flow homoentropic if

s

0

∇ ≡

. Thus, when the flow is homoentropic

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

st

Principle of

Thermodynamics can be written in the following form

T ds di (1/ )dp

= −

ρ

then for any stationary flow one has

T s

i (1/ ) p

∇ = ∇ −

ρ ∇

.

124

In the case of a homoentropic flow we get

i

(1/ ) p

P

∇ =

ρ ∇ = ∇

.

Thus, if the flow is homoenergetic and homoentropic, it is automatically barotropic and
the Bernoulli constant C

B

is global. Note that in the case of 2D flows, it implies that the

velocity field is potential (explain why!).

Yet another interesting result can be derived from the Euler equation written in the Lamb-
Gromeko form for the stationary flow

2

1

1

2

( v )

p

ρ

∇

+ × = − ∇ + ∇Φ

v ω

Using the entropy/enthalpy form of the 1

st

thermodynamic principle, we can re-write the

above equation in the following form

2

1

2

T s

( v

i

)

∇ = ∇

+ −Φ + ×

v

ω

We have received the Crocco equation. According to this equation, any nonuniformity in
the spatial distribution of entropy in the homoenergetic flow immediately leads to
vorticity generation.

125

T

T

H

H

E

E

R

R

M

M

O

O

D

D

Y

Y

N

N

A

A

M

M

I

I

C

C

I

I

N

N

E

E

Q

Q

U

U

A

A

L

L

I

I

T

T

Y

Y

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

change of

entropy due

entropy

external heat

exchange

d

d

d

e

dt

dt

dt

i

S

S

S

=

+

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

A

Q

t

dA

t

div d

t div ( )

∆

∆Ω

∆ = −∆

⋅

= −∆

Ω ≈ −∆

∆Ω

∫

∫

q n

q

q x

126

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

e

dS

1 dQ

div ( )

d

dt

T dt

T

=

= −

Ω

q x

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

s

. Then,

according to well-known formula, we have

dS

d

ds

sd

d

dt

dt

dt

Ω

Ω

=

ρ Ω = ρ

Ω

∫

∫

The 2

nd

Principle of Thermodynamics tells us that irreversible processes always lead to

some increase of entropy. Thus

ds

div

d

d

d

0

dt

T

Ω

Ω

Ω

ρζ Ω = ρ

Ω+

Ω >

∫

∫

∫

q

where

i

ds

dt

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

2

div

T

d

div

d

d

T

T

T

Ω

Ω

Ω

 

 

 

⋅∇

Ω =

Ω +

Ω

∫

∫

∫

q

q

q

In the above, the following identity

2

1

1

1

1

div

div

div

T

T

T

T

T

T

 

 

 

 

 

 

=

+ ⋅∇

=

−

⋅∇

q

q q

q

q

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

div

d

dA

T

T

Ω

∂Ω













⋅

Ω =

∫

∫

q

q n

and thus

2

ds

T

d

d

dA

d

0

dt

T

T

Ω

Ω

∂Ω

Ω

⋅

⋅∇

ρ ζ Ω = ρ

Ω+

+

Ω >

∫

∫

∫

∫

q n

q

128

Finally, for an isotropic heat transfer we use the Fourier Law

q

T

= −λ∇

where

0

λ >

is the coefficient of heat transfer.

The inequality can be finally written in the following form (Gibbs-Duhem inequality)

2

dS

T

T

dA

d

dt

T

T

Ω

∂Ω

∇ ⋅

∇

> λ

+ λ

Ω

∫

∫

n

The interpretation can be formulated as follows: if the heat transfer is present in the
flow then there exists a minimal admissible rate of entropy production.

129

S

S

I

I

M

M

I

I

L

L

I

I

T

T

U

U

D

D

E

E

O

O

F

F

I

I

N

N

C

C

O

O

M

M

P

P

R

R

E

E

S

S

S

S

I

I

B

B

L

L

E

E

F

F

L

L

O

O

W

W

S

S

D

D

I

I

M

M

E

E

N

N

S

S

I

I

O

O

N

N

L

L

E

E

S

S

S

S

F

F

O

O

R

R

M

M

O

O

F

F

T

T

H

H

E

E

N

N

A

A

V

V

I

I

E

E

R

R

-

-

S

S

T

T

O

O

K

K

E

E

S

S

E

E

Q

Q

U

U

A

A

T

T

I

I

O

O

N

N

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

1

(

)

p

t

∂ + ⋅∇ = − ∇ +ν∆ +

∂

ρ

v

v

v

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 t

=

ɶ

,

linear dimensions

j

j

x

L x

=

ɶ

,

velocity

V

=

v

vɶ

,

pressure

p

P p

=

ɶ

,

volume force

F

=

f

fɶ

.

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

130

Consequently, we have also dimensionless differential operators

dt

1

t

dt t

T t

∂

∂

∂

=

=

∂

∂

∂

ɶ

ɶ

ɶ

,

j

j

j

j

j

dx

1

x

dx

x

L x

∂

∂

∂

=

=

∂

∂

∂

ɶ

ɶ

ɶ

.

The Navier-Stokes equations can be now written as

2

2

V

V

P

V

(

)

p

F

T

t

L

L

L

∂

ν

+

⋅∇ = −

∇ +

∆ +

∂

ρ

v

v

v

v

f

ɶ

ɶ

ɶ

ɶ

ɶ

ɶ

ɶ

ɶ

ɶ

ɶ

,

or, after multiplication by L/V

2

2

2

L

P

F L

(

)

p

VT t

V L

V

V

∂

ν

+ ⋅∇ = −

∇ +

∆ +

∂

ρ

v

v

v

v

f

ɶ

ɶ

ɶ

ɶ

ɶ

ɶ

ɶ

ɶ

ɶ

ɶ

.

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

St

L

=

,

Euler number

2

V

Eu

P

ρ

=

,

Reynolds number

V L

Re

=

ν

,

Froude number

2

V

Fr

F L

=

.

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

1

1

1

1

(

)

p

St

Eu

Re

Fr

t

∂ + ⋅∇ = − ∇ + ∆ +

∂

v

v

v

v

f

ɶ

ɶ

ɶ

ɶ

ɶ

ɶ

ɶ

ɶ

ɶ

ɶ

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.

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More physically: the similitude numbers tell us how important (or large) are effects
related to flow unsteadiness, pressure forces, viscous forces and gravity forces in
comparison to the inertial forces (related to the convective part of the fluid
acceleration).

For instance, the

Reynolds number

tells us how important are viscous effects – apparently

they become negligible if the Reynolds number is very large. Similarly, the effect of the
volume forces becomes not much important when the Froude number is large, and so on.

However, this general interpretation is correct providing the time and space scaling
quantities are relevant for the physical effect of interest!

Let’s consider a typical example:

In a viscous flow past a wing of an airplane the boundary layer exists in the close vicinity
of the wing’s surface. Typically, the length scale in define as the wing’s chord and the
reference velocity is the velocity of the free stream far from the airplane. Since the
kinematic viscosity of air is of the order 10

-5

(m

2

/s) the Reynolds number is typically very

large, say, of the order of millions. It might suggest that the viscous term can be neglected
and such conclusion is essentially correct for the flow outside the boundary layer. For that
reason, the large scale flow around the airplane can be adequately model by the Euler
equations. However, inside the boundary layer viscous effects are never negligible!.

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The misinterpretations of the large Reynolds number in such case is the result of choice of
an irrelevant length scale: what really matters for the boundary later flow is not a wing’s
chord but rather the boundary layer’s thickness which is smaller by several orders of
magnitude than the wing’s chord!.

Conditions for dynamic similitude of flows:

We say that

two flows are dynamically similar

if:

▶ they are geometrically similar, e.g. the shapes (but usually not dimensions)

of the flow domains are the same,

▶ all similitude numbers computed on the basis of the corresponding scaling

quantities are the same for both flows. It means that the dimensionless
governing equations in both cases are identical,

▶

the dimensionless form of the initial and boundary conditions are identical

.

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The above conditions are very important in

Experimental Fluid Mechanics

. where

investigations are usually carried out with the use of re-scaled models of a real technical
object (a model of an aircraft, a model of a car and so on). On the other hand, these
conditions are very rigorous and it is usually very difficult (or impossible) to meet all
of them simultaneously.

Consider the investigation of the sailing boat model made in the scale 1:10, which is carried
out in the towing basin. The aim of the investigation is to assess a hydrodynamic drag of the
boat.

Clearly, the geometry of the model and the real flows are not strictly similar – the real
object will not be intended to sail inside a 10-times magnified copy of the towing basin! In
fact, the experimentalists assume (reasonably) that the measurements they conduct will give
relevant results because the boat’s model of the boat is relatively small (or, equivalently, the
basin is sufficiently large) and all kinds of “side walls” and “bottom” effects are negligible.

Secondly, it is nearly impossible to keep both Reynolds and Froude numbers the same as in
a real flow – it is actually a kind of self-contradictory demand. Indeed, to keep the same
value of the Froude number, the model should move 10 times slower than the real object,
while keeping the same Reynolds number would require to move the model 10 times faster!
The latter statement assumes that the fluid inside the experimental basin is the same water as

135

in real conditions. It should be noted that - at least in principle – we could play with fluid
viscosity (but water is already the rather low-viscosity fluid!) or with the gravity (towing
basin inside the dropping elevator ?) What is actually done is the splitting of the experiment
into parts devoted to a different regime of the yacht’s motion: for small velocity the frictions
drag dominates and the similitude with respect to viscous effect is crucial, while for the fast
motion the drag is mostly due to gravitational effects (surface waves generated by a moving
boat) and then the Froude number should be kept the same (or close).