INTRODUCTION TO FLUID MECHANICS
INTRODUCTION TO FLUID MECHANICS
LECTURE NOTES FOR THE COURSE  FLUID MECHANICS I
(in progress)
written by JACEK SZUMBARSKI
MATHEMATICAL PRELIMINARIES
MATHEMATICAL PRELIMINARIES
1
VECTORS AND TENSORS IN 3D EUCLIDEAN SPACE E3
Orthogonal basic unary vectors (versors) : e1, e2, e3
ńł1 if i = j
(ei,ej) = �ij a"
�ł
0 if i `" j
ół
Any vector in E3 is expressed as unique linear combinations of the basic versors
a = a1e1 + a2e2 + a3e3 a" aiei - summation (Einstein) convention
a = [a1,a2,a3] - canonical equivalence of E3 and R3
INNER (SCALAR) PRODUCT
Let a = aiei and b = bjej. We define the inner product of a and b:
a�"b a" a,b = aibj (ei,ej) = aibj�ij = aibi
( )
Note that (a,ei) = ai hence we can write a = (a,ei)ei
2
VECTOR (CROSS) PRODUCT
We define the operation � on the basic vectors:
e1 �e2 = e3 , e2 �e3 = e1 , e3 �e1 = e2,
ei �ei = 0 , ei �ej = -ej �ei
no summation!
Assuming linearity with respect to both arguments, we extend this operation to all vectors in
the space E3
a�b = aiei � bjej = aibj ei �ej = (a2b3 - a3b2)e1 + (a3b1 - a1b3)e2 + (a1b2 - a2b1)e3
Practical way of computing the vector product
e1 e2 e3
a2 a3 a1 a3 a1 a2
a�b = a1 a2 a3 = e1 - e2 + e3
b2 b3 b1 b3 b1 b2
b1 b2 b3
3
ALTERNATING SYMBOL
0 if i = j or i = k or j = k
ńł
�ł
"ijk = 1 if {i, j,k} is an even permutation of {1,2,3}
�ł
�ł
-1 if {i, j,k} is an odd permutation of {1,2,3}
ół
For instance we have "213= -1 , "311= 0 , "231=1.
The vector product of a and b can be nicely written as follows
a�b ="ijk a bkei
j
Another useful operation is the mixed product of three vectors
a1 a2 a3
a�"(b�c) = b1 b2 b3 ="ijk aibjck
c1 c2 c3
Determinant of the matrix A (dim A = 3): det A ="ijk a1,ia2,ja3,k
4
2ND-RANK TENSORS IN E3
Tensors as bilinear transformations (functionals) E3 � E3 R
T(ą1x1 +ą2x2,y) = ą1T(x1,y) +ą2T(x2,y) ,
Bi-linearity means that
T(x,ą1y1 + ą2y2) = ą1T(x,y1) +ą2T(x,y2) .
For two arbitrary vectors x and y we can write
T(x,y) = T(xiei,yjej) = xiyjT(ei,ej) = tijxiyj
The matrix T such that T�ł ij = tij represents the tensor T in the assumed reference
�ł łł
�ł
frame (or with respect to assumed basic versors)
Some operations on tensors:
2
Addition: T = T1 + T2 �! T = T1 + T2 �! tij = t1 + tij
ij
Multiplication by a scalar T = � T1 �! T = � T1 �! tij = � t1
ij
Multiplication of two tensors T = T1T2 �! T = T1T2 �! tij = t1 t2
ik kj
2
Scalar (Frobenius) product of two tensors s = T1 : T2 := t1 tij (double summation !)
ij
5
2ND-RANK TENSORS IN E3 - CONTINUED
Basic linear functionals E3 R: ei(ej) := �ij
Tensor product of the basic functionals:
(ei "ej)(x,y) := ei(x)ej(y) = ei(xkek)ej(ymem) =
= xkymei(ek)ej(em) = xkym�ik� = xiyj
jm
Thus we can write T(x,y) = tijxiyj = tij (ei "ej)(x,y) or T = tij ei "ej.
The linear space of the 2nd-rank tensors is 9-dimensional.
6
ORTHOGONAL TRANSFORMATIONS OF COORDINATE SYSTEMS
Assume that different basic vectors are introduced
e3
e1, e2 , e2 (see figure). These vectors can be
2
2 3
expressed by means of the  old basic vectors.
Consider e2 = zikek , e2 j = zjmem.
i
0
The orthogonality condition for the new base yields
(I)ij = �ij = (e2 ,e2 j) = zikzjm(ek,em) = zikzjm�km =
i
e2
.
e
1
= zikzjk = (ZZT)ij = (ZTZ)ij
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 Z-1 = ZT, i.e.,
it is the orthogonal matrix.
7
ORTHOGONAL TRANSFORMATIONS OF COORDINATE SYSTEMS - CONTINUED
Each vector x from E3 can be expressed with respect to both basis, namely
x = xiei = x2 e2 .
i i
Thus x = xiei = x2 zij ej = x2 j zji ei ,
i
meaning that
xi = zjix2 j = (ZT)ijx2 j = (Z-1)ijx2 j
and
x2 = (Z)ijxj.
i
These are the transformation rules for the vectors!
8
ORTHOGONAL TRANSFORMATIONS OF COORDINATE SYSTEMS - CONTINUED
Consider the tensor T and its representation with respect to both basis (reference frames)
T(x,y) = tijxiyj = t2 x2 y2 j.
ij i
We can write
T(x,y) = tijxiyj = tijzkix2 zmjy2 = x2 zkitijzmj y2 =
m m
k k
(ZT)kj
= x2 (ZT)kj(ZT)jm y2 = x2 (ZTZT)km y2 = x2 t2 y2
m m m
k k k km
tkm
2
(ZTZT )km
The matrix representing the tensor T in the new base is given as
T2 = Z T ZT = Z T Z-1
Thus, we have obtained the transformation rule for the 2nd  rank tensors!
9
DIFFERENT VIEW: 2ND  RANK TENSORS AS LINEAR MAPPINGS E3 E3
Consider the 2nd-rank tensor T and two vectors x and y.
We have T(x,y) = xi tijyj = xiwi = (x,w).
inner
wi
product
The vector w can be defined as w = Ty.
The linear transformation T : E3 E3 is defined by its action on the basic versors as
Tej = tijei
Indeed, for any vector w we get
w = Ty = T(yjej) = yjTej = tijyjei = wiei.
Equivalence between 2-rank tensors and linear mappings can be established as follows
T T: T(x,y):= (x,Ty) , T T: Ty := T(ei,y)ei.
10
EIGENVECTORS, EIGENVALUES AND TENSOR INVARIANTS
The eigenvalue problem:
1st formulation: find  "C and nonzero w such that Tw =  w, or
2nd formulation: find  "C and nonzero w such that T(x,v) =  (x,v) for each vector x
from the space E3.
Equivalently, we have
(tijv - vi)ei = 0 �! pT() = det(T - I) = 0.
j
Thus eigenvalues are the roots of the characteristic polynomial pT().
Tensor T is symmetric when T(x,y) = T(y,x), i.e. when tij = t (check!) or T = TT.
ji
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).
11
EIGENVECTORS, 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
pT() = det(T2 - I) = det(ZTZ-1 - I) = det [Z(T - I)Z-1] =
= det Z�"det (T - I)�"det Z-1 = det Z�"det(T - I)�"(det Z)-1 = det(T - I)
We are mostly interested in 3D case. Then, we can write
pT() = -3 + J12 - J2 + J3
where
J1 = trT := tii a" t11 + t22 + t33 ( tr means trace),
1
J2 = [(trT)2 - trT2] (calculate for 2D case!),
2
J3 = detT.
The following relations hold between invariants and the eigenvalues (Viete s formulas
for 3rd-order polynomial)
J1 = 1 + 2 + 3 , J2 = 12 + 13 + 23 , J3 = 123.
12
CAYLEY-HAMILTON THEOREM
Any square matrix A satisfies its own characteristic polynomial pA() = det(A - I), i.e.
we have pA(A) = 0
Proof
For invertible square matrix M we have M-1 = det M (cof M)T. Thus
( )-1
M(cof M)T = det M�"I.
Let M = A - I. Then B() := [cof (A - I)]T is the matrix polynomial of the order not
larger than n -1 (n  dimension of A)
B() = n-1Bn-1 + n-1Bn-1 +...+ B1 + B0
and we have
(A - I) n-1Bn-1 + n-1Bn-1 +...+ B1 + B0 =
( )
= det(A - I)�"I = (n + cn-1n-1 +...+ c1 + c0)I
13
CAYLEY-HAMILTON THEOREM - CONTINUED
The above equality is satisfied for any number  so the corresponding matrix coefficients



at both sides should be the same. Thus
-Bn-1 = I
-Bk-1 + ABk = ckI , k = n -1, n - 2,..,1
AB0 = c0I
Let s multiply (from the left side) the first equation by An, the second one by An-1 and so
on (then the last equation remains unchanged) and sum up all equations. The left-hand side
of the obtained equation is zero since all terms will cancel out in pairs! Thus we get
0 = An + cn-1An-1 +...+ c1A + c0I a" pA(A)
as stated.
For the matrices with the dimension equal 3 we have -T3 + J1T2 - J2T + J3I = 0.
This relation will be used in the section devoted to the constitutive relations in fluid
mechanics.
14
PRODUCT OF ALTERNATING SYMBOLS
Important identity "ijk"k�ł = �i�� -�ił�
jł j�
Proof
�11 �12 �13 1 0 0
Consider �21 �22 �23 = 0 1 0 =1.
�31 �32 �33 0 0 1
�i1 �i2 �i3
After row s permutation one gets � �j2 �j3 ="ijk.
j1
�k1 �k2 �k3
�ią �i� �ił
Then, after column s permutation we obtain � � � ="ijk"ą�ł .
ją j� jł
�ką �k� �kł
15
PRODUCT OF ALTERNATING SYMBOLS - CONTINUED
Now, put k = ą and apply summation.
�ik �i� �ił
The result is as follows �jk � � ="ijk"k�ł , or
j� jł
�kk �k� �kł
"ijk"k�ł = �ik(� �kł -�k�� ) -�i� (� �kł - �kk � ) +�ił (� �k� - �kk �j� ) =
j� jł jk jł jk
3 3
= � �ił -�i�� -�i�� + 3�i�� +� �ił - 3� �ił = �i��jł -� �ił
j� jł jł jł j� j� j�
Exercise: Using index formalism derive the following vector identity
a�(b�c) = (a,c)b - (a,b)c
16
BASIC DIFFERENTIAL OPERATORS (IN CARTESIAN CO.S.)
Gradient of a scalar field f = f (t,r)
�ł łł
"f "f "f "f
"f = , , = ei (vector)
�ł
"x1 "x2 "x3 śł "xi
�ł �ł
" - nabla operator
Divergence of the vector field w = wi(t,r)ei
"w1 + "w2 + "w3 = "wj (scalar)
div w a" "�"w =
"x1 "x2 "x3 "x
formal inner
j
product
Rotation (curl) of the vector field w = wi(t,r)ei
�ł łł �ł łł �ł łł
"w3 - "w2 e1 + "w1 - "w3 e2 + "w2 - "w1 e3 =
rot w a" "� w =
�ł �ł �ł
"x2 "x3 śł "x3 "x1 śł "x1 "x2 śł
formal vector �ł �ł �ł �ł �ł �ł
product
(vector)
"wj
="ijk "wk ei ="ijk ek
"xj "xi
17
Gradient of the vector field w = wi(t,r)ei
"wi
Grad w a" "w = ei "ej (2nd  rank tensor)
"x
j
formal dyadic
product
Divergence of the tensor field T = tij(t,r)ei " ej
"tij
DivT a" "�"T = ei (vector)
"x
formal matrix-vector
j
product
18
SOME INTEGRAL THEOREMS (INFORMAL REMINDER)
GREEN-GAUSS-OSTROGRADSKY (GGO) THEOREM
Consider the vector field w = w(x) defined in a 3D volume &! bounded by sufficiently
regular surface " &!. Then
(w,n) dS = "�"w dx
+" +"
divergence
"&! &!
w�"n=wn
of w
component of
w normal toS
We have analogous (dual) theorem with vector products, namely n� wdS = "� w dx
+" +"
rotation
"&! &!
of w
STOKES THEOREM
Consider the vector field w = w(x), the closed line (loop) ł and sufficiently regular (yet
arbitrary) surface S spanned (like a soap bubble) by this line. Then
(w, �) dl = ("� w,n) dS
+"
+"
ł S
w�"�a"w�
component of
component of
rot w normal toS
w tangent toł
19
KINEMATICS AND DYNAMICS OF FLUID
KINEMATICS AND DYNAMICS OF FLUID
MOTION
MOTION
20
LAGRANGIAN AND EULERIAN VIEWS ON THE FLUID MOTION
Fluid element is defined as an individual and
x3,�3
infinitely small portion of the fluid. Each fluid
trajectory of a fluid element
element is characterized by its instantaneous
t>0
&!(t)
location (or position) vector x, which is the
function of time t and the initial position � of
x(t,�)
t=0
the element, i.e. its location at the time instant
�
0
t = 0. Thus we have x = x(t,�) and in
&!0
particular � = x(0,�).
x2,�2
x1,�1
If we fix � = �" then the function x = x(t,�") describes some line in E3 called the
trajectory of the fluid element.
21
x3,�3
For the fixed time t e" 0 the function x = x(t,�)
describes the transformation of the region filled
trajectory of a fluid element
with the fluid at the time t = 0  let s denote it
t>0
&!(t)
&!0 - to the region &! (t) = x(t,&!0) containing
the same fluid at some later time t". Thus, the
region &!(t) is the image of &!0 with respect to
x(t,�)
t=0
the transformation x = x(t,�); we call &!(t) the
�
0
material volume because all the time it consists
&!0
of the same fluid elements, i.e. those which are
belong originally to &!0.
x2,�2
x1,�1
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 x1 = x(t,�) and x2 = x(t +� ,�)
then the following group property holds
x2 = x(t +� ,�) = x[� ,x(t,�)] = x(� ,x1).
22
Any fluid motion can be described using either Lagrangian or Eulerian
viewpoint.
Lagrangian viewpoint: each fluid element is identified uniquely by its position at t = 0, i.e.
by the vector �. All kinematical and dynamic quantities are described as functions of time
and the Lagrangian coordinates �1, �2 and �3.
The velocity of the fluid element is defined as
x(t + "t,�) - x(t,�) "x
V(t,�) := lim a" (t,�) (�  fixed)
"t0
"t "t
Fluid acceleration is defined as
V(t + "t,�) - V(t,�) "V "2r
a(t,�) := lim a" (t,�) = (t,�)
"t0
"t "t
"t2
23
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 x1, x2 and x3 of the vector x.
The velocity field is the function of time and space coordinates v = v(t,x).
The relations between two different viewpoints can be written as
Eulerian to Lagrangian: V(t,�) = v[t,x(t,�)]
Lagrangian to Eulerian: v(t,x) = V[t, �(t,x)]
inverse
transform
The Eulerian form of the fluid acceleration will be considered later.
24
TRAJECTORIES OF FLUID ELEMENTS
t
"
Lagrange: "t x(t,�) = V(t,�) (�  fixed parameter). Thus x(t,�) = � +
+"V(� ,�)d� .
0
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
ńłdt j
x = v (t,x1,x2,x3) , j =1,2,3.
ńłdt
x = v[t,x(t)]
j
�ł �ł
�!
�ł �ł
x (0) = �j
x(0) = �
�ł
j
�ł
ół
ół
Typically, the above Initial Value Problem has to be solved numerically (e.g. using the
Runge-Kutta methods)
25
STREAMLINES 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
dx1 dx2 dx3
= =
v1(t,x1,x2,x3) v2(t,x1,x2,x3) v3(t,x1,x2,x3)
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
ńłś1(t,x1,x2,x3,C1,C2) = 0
�ł
ś (t,x1,x2,x3,C1,C2) = 0
2
ół
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.
26
STREAMLINES 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 x0 is described by the following initial value problem
d
ńłd�
x(� ) = v[t,x(� )]
�ł
�ł
x(� = 0) = x0
�ł
ół
where physical time t is fixed ( frozen ) and the variable � 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.
27
EXAMPLES
(1) Stationary two-dimensional flow v(x1,x2) = -x2e1 + x1e2 a" [-x2,x1]
dx1 = dx2 �! x1dx1 + x2 dx2 = 0 �! x1 + x2 = R2 , R e" 0.
2
Streamlines:
2
-x2 x1
We have obtained the family of the concentric circles.
d d
ńł
x1 = -x2 , x2 = x1
dt
Trajectories: �łdt
�ł
x1(0) = R , x2(0) = 0
�ł
ół
The solution is
x1(t) = R cos(t) , x2(t) = Rsin(t)
2
which is the parametric form of the circle x1 + x2 = R2 , R e" 0.
2
28
(2) Nonstationary (or unsteady) flow v(t,x1,x2) = (-x2 - t)e1 + x1e2 a" [-x2 - t,x1]
Streamlines:
dx1 = dx2
2
�! x1dx1 + (x2 + t)dx2 = 0 �! x1 + (x2 + t)2 = C + t2 a" R2 , C e" -t2.
-x2 - t x1
Again: the family of concentric circles but the pattern of the streamlines moves down
along the 0x2 axis with the steady speed equal  1 (see figure).
x2
d d
ńł
x1 = -x2 - t , x2 = x1
dt
Trajectories: �łdt
�ł
t=0
x1(0) = x10 , x2(0) = x20
�ł
ół
ńłx1(t) = (x10 +1)cos(t) - x20 sin(t) -1
The solution �ł
x1
x2(t) = (x10 +1)sin(t) + x20 cos(t) - t
ół
instantaneous
streamlines Consider x10 = -1 and x20 = 0. Then x1(t) = -1 and
x2(t) = -t so the fluid element moves down the straight
t>0
vertical line x1 = -1 with the steady velocity equal to  1.
The trajectories in the unsteady flow can be quite different that the streamlines!
29
trajectory
SUBSTANTIAL (MATERIAL, LAGRANGE, LIE) DERIVATIVE
Consider a sufficiently regular scalar field f = f(t,x) = f(t,x1,x2,x3).
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,x(t,�)]
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
Df dF(t) = "f "f "f
[t,x(t,�)]:= [t,x(t,�)]+ [t,x(t,�)]dx1 (t) + [t,x(t,�)]dx2 (t) +
Dt dt "t "x1 dt "x2 dt
�ł �ł
dx3
"f "f "f "f "f
+ [t,x(t,�)] (t) = + v1 + v2 + v3 [t,x(t,�)]
�ł
"x3 dt "t "x1 "x2 "x3 �ł
�ł łł
dx
j
where we have used the relation (t) = v [t,x(t,�)] , j =1,2,3.
j
dt
30
SUBSTANTIAL (MATERIAL, LAGRANGE, LIE) 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
Df "f
= + v�""f ,
Dt "t
convective
derivative
local
derivative
or in the index notation (summation convention is assumed)
Df "f
= + vj "f .
Dt "t "xj
ś% 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 f = f (x) = f (x1,x2,x3) we say that f is stationary (or steady)
and the local derivative "f / "t vanishes identically.
ś% 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.
31
ACCELERATION
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
�ł
Dvi "vi "vi �ł
Dv "v "v
a(t,x) = = ei = + v ei = + v
�ł �ł
j j
�ł �ł
Dt Dt "t "x "t "x
j j
�ł łł
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
Dv "v
a(t,x) = = + (v�"")v.
Dt "t
An alternative way of expressing the fluid acceleration is the Lamb-Gromeko form
"v
a(t,x) = + "(1 v2) + �� v
2
"t
where v =| v | is the velocity magnitude and � = "� v is the rotation of the velocity field
called vorticity.
32
ACCELERATION - CONTUNUED
Proof of the Lamb-Gromeko form
"
We have (see Appendix) � ="ijk "x vk ei and v2 = vivi.
j
Then
"v "v
j
�� v ="k�ł �kv�eł ="k�ł"ijk j v� eł = (�i��jł -�ił�j� ) v� eł =
"xi "xi
�ł �ł
�ł "vj "v �ł "v�
"vł
j
= �i�� v� -�ił� v� �ł eł = v� - v� �ł eł =
�ł �ł
�ł
jł j�
�ł �ł
�ł
"xi "xi łł "x� "xł
�ł
�ł łł
�ł �ł
"vł
"
= v� - (1 v� v� )�ł eł = (v�"")v - "(1 v2)
�ł �ł
2
�ł
"x� "xł 2
�ł łł
Thus
(v�"")v = "(1 v2) + �� v
2
and the Lamb-Gromeko formula follows immediately. c&
33
REYNOLDS TRANSPORT THEOREM
We will prove the mathematical result known as the
x3,�3
Reynolds Transport Theorem, which plays the
fundamental role in derivation of differential forms
t>0
of the conservation principles in Continuum
&!(t)
Mechanics.
x(t,�)
Consider any sufficiently regular scalar field t=0
f = f (t,x). Consider the integral of f calculated over
�
0
an arbitrary material volume &!(t).
&!0
C(t) = f (t,x)dx.
x2,�2
+"
x1,�1
&! (t)
d
We need to compute the time derivative C2 (t) = f (t,x)dx.
+"
dt
&! (t)
NOTE: This task is nontrivial since the integration domain is itself time dependent!
34
REYNOLDS TRANSPORT THEOREM - CONTINUED
To calculate the derivative, we will switch from Eulerian variables x = [x1,x2,x3] to
Lagrangian variables � = [�1,�2,�3]. The integral C(t) can be written as
C(t) = f (t,x)dx = f[t,x(t,�)]J(t,�)d� = f0(t,�)J(t,�)d�.
+" +" +"
&! (t) &!0 &!0
In the above formula we have used the composite function f0
f0 = f0(t,�) = f[t,x(t,�)],
and also the Jacobi determinant (Jacobian) defined as
"x1 "x1 "x1
�ł łł
"�1 "�2 "�3
�ł śł
�ł"x "x2 "x2 śł
2
J(t,�) = det (t,�).
"�1 "�2 "�3
�ł śł
"x3 "x3 "x3
�ł śł
"�1 "�2 "�3
�ł śł
�ł �ł
35
REYNOLDS TRANSPORT THEOREM  CONTINUED
Since the domain &!0 is time-independent (it is actually the initial form of the material
volume &!(t) at the time t = 0), we can move the differentiation operator under the sign of the
integral and get
"f0 (t,�)J(t,�)d� + f0(t,�) (t,�)d�
d "J
C2 (t) = f0(t,�)J(t,�)d� =
+" +" +"
dt "t "t
&!0 &!0 &!0
Note that time differentiation of the composite function f0 yields
" d " " "
f0(t,�) = f[t,x(t,�)] = f[t,x(t,�)]+ f[t,x(t,�)]�" xi(t,�) =
"t dt "t "xi "t
=Vi(t,�)=vi[t,x(t,�)]
"
= f + v�""f [t,x(t,�)]
( )
"t
This part was easy! We need to calculate the time derivative of the Jacobian which has
appeared in the second integral in the formula for C2 (t). This is much more complicated & .
Basically, we have two methods.
36
Method A
We write the Jacobian using the alternating symbol J(t,�) ="ijk "x1 "x2 "x3
"�i "�j "�k
Note that partial derivatives with respect to time and Lagrangian variables commute, hence
"x1 " "x1 "V1 "x2 " "x2 "V2 "x3 " "x3 "V3
" " "
= = , = = , = =
"t "�i "�i "t "�i "t "�j "�j "t "�j "t "�k "�k "t "�k
The time derivative is
"
J ="ijk "V1 "x2 "x3 +"ijk "x1 "V2 "x3 +"ijk "x1 "x2 "V3 =
"t "�i "�j "�k "�i "�j "�k "�i "�j "�k
"V1 "V1 "V1 "x1 "x1 "x1 "x1 "x1 "x1
"�1 "�2 "�3 "�1 "�2 "�3 "�1 "�2 "�3
"x2 "x2 "x2 "V2 "V2 "V2 "x2 "x2 "x2
= + + =
"�1 "�2 "�3 "�1 "�2 "�3 "�1 "�2 "�3
"x3 "x3 "x3 "x3 "x3 "x3 "V3 "V3 "V3
"�1 "�2 "�3 "�1 "�2 "�3 "�1 "�2 "�3
3 3
"
= Vi [cof J]ij
""
"�j
i=1 j=1
cofactor (i,j) of J
("� V)ij
37
Consider two square matrices A and B, and also the product C = ABT. It means that
cik =
"a bkj a" aijbkj,
ij
k
so we conclude that
trC a" cii = aijbij (trace of the matrix C)
Moreover, from the construction of the inverse Jacobi matrix we have
1
J-1 = (cof J)T �! (cof J)T = detJ J-1 = J J-1
det J
Hence, the formula for the time derivative of the Jacobi determinant can be written as
follows
"
�ł �ł
J(t,�) = tr "� V�"(cof J)T łł(t,�) = J(t,�) tr "� V�"J-1łł(t,�)
"t
�ł �ł �ł �ł
38
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
V(t,�) = v[t,x(t,�)]
Lagrange
Euler
and calculate the gradient operator with respect to the Lagrangian variables
3
" " k
�ł
"� Vłł ij (t,�) = Vi(t,�) = vi[t,x(t,�)]"x (t,�).
"
"�j "xk "�j
�ł �ł
k=1
The above formula can be written shortly as
"� V(t,�) = "v[t,x(t,�)]�"J(t,�)
Thus, the time derivative of the Jacobian can be re-written in the following form
"
J(t,�) = J(t,�) tr"v [t,x(t,�)].
( )
"t
"
Taking into account that tr"v = vi = divv a" "�"v
"xi
"
we finally get the formula J(t,�) = J(t,�)"�"v[t,x(t,�)]
"t
39
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 x(t + s,�) = x[t,x(s,�)] or
xi(t + s,�1,�2,�3) = xi [t,x1(s,�1,�2,�3),x2(s,�1,�2,�3),x3(s,�1,�2,�3)] , i = 1,2,3.
Let s differentiate the above formula with respect to the Lagrangian coordinate �j:
"xi (t + s,�) = "xi [t,x(s,�)] "xk (s,�),
"�j "�k "�j
which can also be written as
[J]ij(t + s,�) = [J]ik[t,x(s,�)] [J]kj(s,�),
which is equivalent to
J(t + s,�) = J[t,x(s,�)] J(s,�).
40
Then, from the fundamental property of matrix determinant, we have
J(t + s,�) = J[t,x(s,�)] J(s,�).
We need to calculate the derivative
" J(t + "t,�) - J(t,�) J(t,�) J["t,x(t,�)]- J(t,�)
J(t,�):= lim = lim =
"t0 "t0
"t "t "t
= J(t,�)"t0 J["t,x(t,�)]-1
lim
"t
Note that J["t,x(t,�)] is the Jacobian of the  nearly identical transformation
x(t,�) Ź x(t + "t,�), which can be written shortly as x Ź� (x).
"t
The explicit form of this transformation is
[� (x)]i = xi + vi(t,x1,x2,x3)"t + O("t2) , i = 1, 2, 3,
"t
41
This, the Jacobi matrix can be calculated as follows
"vi
"
[J]ij("t,x) = [� (x)]i = �ij + (t,x)"t + O("t2)
"xj "t "x
j
or simply J("t,x) = I + "v(t,x)"t + O("t2).
Now, it is easy to show (do it!) that
�ł �ł
"v1 + "v2 + "v3 (t,x)"t + O("t2) =1+ "�"v(t,x)"t + O("t2)
J("t,x) =1+
�ł
"x1 "x2 "x3 �ł
�ł łł
div v
J("t,x) -1
Thus, we get lim = "�"v(t,x)
"t0
"t
and  after returning back to the Lagrangian variables - the formula for the time derivative
of the Jacobian is obtained
"
J(t,�):= J(t,�) "�"v [t,x(t,�)].
( )
"t
42
REYNOLDS TRANSPORT THEOREM  CONTINUED
The time derivative C2 (t) can be now evaluated as follows
"
C2 (t) = f + v�""f + f "�"v [t,x(t,�)]J(t,�)d� =
( )
"t
+"
&!0
" "
�ł"t
= f + v�""f + f "�"v (t,x)dx = f +"�"(f v)łł(t,x)dx =
( )
"t
+" +"
�ł �ł
&! (t) &! (t)
" "
= f dx + "�"(f v) dx = f dx + f vn ds
"t "t
+" +" +" +"
Ń!
&! (t) &! (t) &! (t) "&! (t)
v�"n
GGO
normal
Theorem
velocity
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 a" 0).
43
PRINCIPLE OF MASS CONSERVATION
PRINCIPLE OF MASS CONSERVATION
DENSITY OF FLUID
We define (rather informally) the fluid density as
� = lim0 m(&! )
vol(&! )
vol(&! )
Thus, the mass of the fluid in the material volume &!(t) can be expressed as
m&! (t) := �(t,x)dx
+"
&! (t)
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).
44
PRINCIPLE OF MASS CONSERVATION
The mass of the fluid in an arbitrary material volume &!(t) does not change in time (&!(t)
contains permanently the same fluid elements).
d
Thus, we can write dt m&! (t) = 0
From Reynolds Transport Theorem we get
d "
�ł"t
0 = � dx = � + "�"(� v)łłdx
dt
+" +"
�ł �ł
Ń!
&! (t) &! (t)
Reynolds's
Transport Th.
Since the volume &!(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
"
� + "�"(� v) = 0
"t
45
PRINCIPLE OF MASS CONSERVATION  CONT.
D�
"
Equivalent forms are � + v�""� + � "�"v = 0 + � "�"v = 0
"t
Dt
D�
full
Dt
derivative
For an incompressible fluid � = const and the above equation reduces to the continuity
equation
"�"v a" div v = 0,
which describes the geometric constrain imposed on the velocity field (volume
conservation).
In Lagrangian description, the continuity equations is equivalent to the following condition
(why?) for the Jacobi determinant
"
J(t,�) = 0
"t
and since J(0,�) = 1 we have J(t,�) =1 for all times t e" 0.
46
TIME 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 &!(t) is
expressed by the following volume integral
h(t) = � Hdx
+"
&! (t)
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
d d "
h(t) = � Hdx = ["t (� H) + "�"(� H v)]dx =
dt dt
+" +"
Ń!
&! (t) &! (t)
Reynolds
Trans.Th.
" " D
�ł
= H � + "�"(� v)łłdx + � H + v�""H dx = � Hdx
( )
Dt
"t "t
+" +" +"
�ł �ł
&! (t) &! (t) &! (t)
= 0! DH
=
Dt
47
DEFORMATION AND STRESS
DEFORMATION
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+"t
x3
B'
"t can be expressed as follows
t
B
xA2 = xA + v(t,xA)"t + O("t2)
�'
�
A
A'
Since xB = xA + � then analogously we have
xB2 = xA + � + v(t,xA + �)"t + O("t2),
0
�=xB-xA
where the vector � describes the relative position of
�'=xB'-xA'
the fluid elements at the time t .
x2
x1
48
DEFORMATION - CONTINUED
During a short time interval "t this vector has changed and can be expressed as
�(t + "t) = xB2 - xA2 = �(t) +[v(t,xA + �) - v(t,xA)]"t + O("t2) =
= �(t) + �ł"v(t,x)�łł"t + O("t2,"t |� |2)
�ł �ł
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
d� �(t + "t) -�(t)
= lim = "v(t,x)� + O(|� |2).
"t0
dt "t
We have introduced the matrix (actually it represents the tensor) called the velocity
gradient
"
�ł łł
"v�łij = vi
"x
�ł
j
49
DEFORMATION  CONTINUED
The velocity gradient "v can be written as a sum of two tensors
"v = D + R
�ł
"vi "v j �ł
1
1
where D = ["v + ("v)T] or dij = + - symmetric tensor,
�ł �ł
2
�ł �ł
2 "x "xi łł
j
�ł
�ł
"vi "v j �ł
1
1
and R = ["v - ("v)T] or rij = - - skew-symmetric tensor
�ł �ł
2
�ł �ł
2 "x "xi łł
j
�ł
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.
50
DEFORMATION  CONTINUED
1
To this end, we note that rij = - "ijk �k , where �k are the Cartesian components of the
2
vorticity vector
"vk
� = rotv = "ijk ei.
"x
j
�k
Indeed, we have
�ł �ł
"v
"vł "vł
"vi �ł "vi "vj �ł
1 1
j
1 1
- "ijk"k�ł = - (�i�� -�ił� ) = - �ł �ł �ł �ł
- = - = rij
2 2 jł j�
�ł �ł �ł �ł
"x� "x� 2�ł "xi "xj łł 2�ł "x "xi łł
j
1 1 1
Thus, we can write R� = rij �j ei = - "ijk �j�k ei = - ��� = ���.
2 2 2
d d d
Moreover, we get dt |� |2= (�,�) = 2(�,dt�) =2(�,R�) = ��"(���) = 0
dt
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
1
fluid and the local angular velocity is equal �.
2
51
DEFORMATION  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
d 1
� = D� + R� = D� + ���.
dt 2
deformation
rigid rotation
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 DSPH and the deviatoric
part DDEV
D = DSPH + DDEV
The spherical part DSPH describes pure volumetric deformation (uniform expansion or
contraction without any shape changes) and it defined as
"vk
1 1("�"v)I �! (DSPH )ij = 1
DSPH = trD�"I = �ij ,
3 3 3"xk
trace
of D
1("�"v)�"trI = ("�"v) a" div v.
Note that trDSPH =
3
52
DEFORMATION  CONTINUED
The second part DDEV describes shape changes which preserve the volume.
�ł
"vi "v j �ł "vk
1div v�"I �! (DDEV)ij = 1 1
We have DDEV = D - + �ij
�ł �ł -
�ł �ł
3 2 "x "xi łł 3"xk
j
�ł
and trDDEV = trD - trDSPH = 0
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
� = D� = DSPH� + DDEV� .
dt
For a short time interval "t the above relation yields
�(t + "t) = �(t) + DSPH�"t + DDEV�"t +O("t2)
"�1 "�2
53
DEFORMATION  CONTINUED
Consider the 2D case when only volumetric part of the deformation exists (see picture).
x2
x2
d 0
�ł łł
We have DSPH = , tr D = 2d
�ł
&!(t+"t)
0 dśł
�ł �ł
B'
C'
d"t L2
B C
C B
{
The relative (wrt the origin) position
vector at the time instant t + "t is
expressed as
O
A O
A
x1 A'
x1
&!(t)
&!(t)
�(t + "t) = (1+ d " t)�(t)+ O("t2).
d"t L1
The shape of the volume is preserved because the above formula describes the isotropic
expansion/contraction. The volume of the region Vol&! (t) = L1L2 has been changed to
2
Vol&! (t + "t) = L1L2(1+ d"t) = Vol&! (t)(1+ 2d "t) + O("t2),
Vol&! (t + "t) - Vol&! (t)
1
and lim = 2d = tr D = "�"v
"t0
Vol&! (t) "t
54
{
DEFORMATION  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
�łd11 - d d12 łł �ł1 (d11 - d22) d12 łł
�ł-ą ł
łł
2 2
DDEV = = =
�ł
śł.
1 1
d12 d22 - dśł �ł d12 (d22 - d11)śł �ł ł ą
2 2 �ł �ł
�ł �ł �ł �ł
x2 x2
The fluid deformation during the short time
C'
ł"t L2
interval can be now expressed as
&!(t+"t)
�(t + "t) = (I + "t DDEV)�(t) + O("t2)
B'
ą"t L2
{
B
C
B
C
or in the explicit form as
A'
ł"t L1
A
ńłx1(t + "t) = (1-ą "t)x1(t) +ł "t x2(t)
�ł
O O
�ł
A x1 x1
x2(t + "t) = ł "t x1(t) + (1+ą "t)x2(t)
ą"t L1
�ł &!(t)
ół
&!(t)
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.
55
{
{
{
DEFORMATION  CONTINUED
Let s compute again the change of the volume of the fluid region during such deformation.
We get
1-ą "t ł "t
2
Vol&! (t + "t) = L1L2 = L1L2(1-ą2"t2 -ł "t2) =
ł "t 1+ą "t
,
= Vol&! (t) + O("t2)
so
Vol&! (t + "t) - Vol&! (t)
1
lim = 0.
"t0
Vol&! (t) "t
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)
56
STRESS 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.
x3
Thus, for an infinitesimal piece dA of the interface
dF = dA
"&!1 )""&!2, we have (see figure)
&!2
n
dF = � dA and F&! &!1 = �dA
+"
&!1
2
"&!1)""&!2
0
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
x2
x1
point x.
From the 3rd principle of Newton s dynamics (action-reaction principle) we have
�(x,n) = -�(x,-n)
57
d
A
STRESS TENSOR - CONTINUED
Consider small tetrahedron as depicted in the figure below.
x3
The front face "ABC belongs to the plane
which is describes by the following formula
C
(n,x) a" njxj = h , h  small number.
n=[n1,n2,n3]
The areas of the faces of the tetrahedron are S,
S1, S2 and S3 for "ABC, "OBC, "AOC and
-e2
-e1
"ABO, respectively. Obviously, S <" O(h2)
0
Moreover, the following relations hold for
j = 1,2,3:
ą B
A
Sj = Scos[""(n,ej)] = S�"(n,ej) = Snj
-e3 D
x2
x1
The volume of the tetrahedron is V&! <" O(h3).
58
STRESS TENSOR - CONTINUED
The momentum principle can be written for the fluid contained inside the tetrahedron
volume as follows
d
x3
�vdx = Fvol + Fsurf
+"
dt
&!
total volume total surface
force
force
time derivative
C
of the momentum
n=[n1,n2,n3]
We need to calculate the total surface force Fsurf .
-e2
-e1
We have:
0
on "ABC: �(x,n) = �(0,n) + O(h)
"ABC
Fsurf = S�(0,n) + O(h3) ą B
A
-e3 D
on "OBC:
x2
x1
�(x,-e1) = -�(x,e1) = -�(0,e1) + O(h)
"OBC
Fsurf = -S1 �(0,e1) + O(h3) = -Sn1�(0,e1) + O(h3)
59
STRESS TENSOR  CONTINUED
on "AOC: �(x,-e2) = -�(x,e2) = -�(0,e2) + O(h)
"AOC
Fsurf = -S2 �(0,e2) + O(h3) = -Sn2 �(0,e2) + O(h3)
on "AOB: �(x,-e3) = -�(x,e3) = -�(0,e3) + O(h)
"AOB
Fsurf = -S3 �(0,e3) + O(h3) = -Sn3 �(0,e3) + O(h3)
When the above formulas are inserted to the equation of
x3
motion we get
C
d
�vdx = Fvol + S[�(0,n) - nj �(0,ej)] + O(h3)
+"
n=[n1,n2,n3]
dt
&!
O(h3)
O(h2)
-e2
-e1 O(h3)
0
When h 0 the above equation reduces to
�(0,n) - nj�(0,ej) = 0
ą B
A
In general, we can write
-e3 D
x2
�(t,x,n) = nj�(t,x,ej) (summation !)
x1
60
STRESS TENSOR  CONTINUED
In the planes oriented perpendicularly to the vectors e1, e2 or e3, the stress vector can be
written as
�(t,x,ej) =�ij(t,x)ei
This, the general formula for the stress vector takes the form
�(t,x,n) = nj�(t,x,ej) =�ij(t,x)njei
We define the stress tensor S represented by the square matrix Ł such that Ł = �ij.
[ ]
ij
The stress tensor depends on time and space coordinates, i.e. what we actually have is the
tensor field S =S (t,x).
The stress tensor S can be viewed as the linear mapping (parameterized by t and x)
defined as
S : E3 w = wjej Ź �ijwjei "E3
61
STRESS TENSOR  CONTINUED
In particular S (n) =�ijnjei = �
i.e., the action of S on the normal versor n at some point of the fluid surface yields
the stress vector � at this point.
Using canonical identification of E3 and R3 we can simply write
� = Łn
It is often useful to calculate the normal and tangent stress components at the point of
some surface.
Normal component is equal �n = (n�"Łn)n a" (n,Łn) n
inner (scalar)
product
Tangent component can be expressed as
�� = � -�nn =�ijnjei -(�kmnknm)niei = [�ijnj -(�kmnknm)ni]ei
(� )i
�
or, equivalently (verify!) as �� = n�(��n)
62
PRINCIPLE OF CONSERVATION OF LINEAR MOMENTUM
PRINCIPLE OF CONSERVATION OF LINEAR MOMENTUM
2ND NEWTON S LAW FOR CONTINUUM
Consider the material volume &! (t). The Principle of Momentum Conservation takes the
form
d
�vdx = �fdx + ŁndS
dt +" +" +"
&! (t) &! (t) "&! (t)
linear momentum volume force surface force
d
In the index notation: dt �vidx = �fidx + �ijnjdS
+" +" +"
&! (t) &! (t) "&! (t)
The formula for time derivative of the volume integral of an extensive quantity gives
Dvdx
d
�vdx = �
dt
+" +"
Dt
&! (t) &! (t)
63
2ND NEWTON S LAW FOR CONTINUUM  CONT.
The next step is to transform (using the GGO integral theorem) the surface integral to
the volume integral
"�ij
�ijnjdS = dx
+" +"
"xj
Ń!
"&! (t) &! (t)
GGO
or in the matrix-vector notation
ŁndS = DivŁdx
+" +"
Ń!
"&! (t) &! (t)
GGO
Finally, the integral form of the momentum conservation can be re-written as
ńł �ł
Dv
� - DivŁ - �f dx = 0
�ł żł
+"
Dt
ół �ł
&! (t)
64
2ND NEWTON S LAW FOR CONTINUUM  CONT.
Since the choice of the volume &! (t) is arbitrary, then for the smooth motion the
differential equation follows
Dv
� = DivŁ + � f
Dt
or in the index notation
Dvi "�ij
� = + � fi
Dt "xj
Writing the time derivative in the extended form, the above equation takes the form of
�ł �ł
"v
� + v�"" v�ł = DivŁ+ � f
( )
�ł
"t
�ł łł
or
�ł
"vi "vi �ł "�ij
�ł �ł
� + vj = + � fi
"t "x "xj
�ł �ł
j
�ł łł
65
SYMMETRY OF THE STRESS TENSOR Ł
SYMMETRY OF THE STRESS TENSOR Ł
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)
k = r��vdx
+"
&! (t)
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:
�ł �ł
dk D d Dv Dv
= � (r � v)dx = � r � v + r � dx = � r � dx
�ł �ł
+" +" +"
Ń!
dt Dt dt Dt Dt
�ł łł
&! (t) &! (t)
dr=v &! (t)
dt
dk
From general principles of mechanics we know that = M
dt
The symbol M denotes the total moment of external forces (a torque) acting of the
material volume &! (t)
66
The total moment M is defined as follows
M = r � �f dx + r � ŁndS
+" +"
&! (t) "&! (t)
Then, the equation for the angular momentum can be written
r � �(a - f )dx = r � Łn dS ,
+" +"
&! (t) "&! (t)
Dv
where a = denotes fluid acceleration.
Dt
In the index notation, we have
"
� "ijk x (ak - fk)dx = "ijk x �kmnmdS = ("ijk x �km)dx
j j j
+" +" +"
Ń!
"xm
&! (t) "&! (t) &! (t)
GGO
The above equation can be transformed as follows
67
�ł łł
0 = � "ijk x (ak - fk)-"ijk " (x �km)śłdx =
�ł
j
+"
"xm j
�ł �ł
&! (t)
�ł "�km -� �km dx =
łł
= "ijk �ł�x (ak - fk) - x
śł
j j jm
+"
"xm
�ł �ł
&! (t)
"�km ) dx - "ijk �kj dx = - "ijk �kj dx
= "ijk x (�ak - fk -
j
+" +" +"
"xm
&! (t) &! (t) &! (t)
a" 0 Eq.of motion!!!
It follows that "ijk �kj = 0. Using the properties of the alternating symbol (see
Mathematical Preliminaries), we conclude that
"1jk �kj = 0 �! �32 -�23 = 0 �! �32 = �23
"2 jk �kj = 0 �! �13 -�31 = 0 �! �13 = �31
"3jk �kj = 0 �! �12 -�21 = 0 �! �12 = �21
Thus �ij = � or, equivalently, Ł = ŁT , i.e., the stress tensor is a symmetric tensor.
ji
68
CONCLUSION
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.
69
CONSTITUTIVE RELATIONS
CONSTITUTIVE RELATIONS
GENERAL CONSIDERATIONS
The constitutive relations for the (simple) fluids is the relations between stress tensor Ł
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 DSPH and the deviatoric part
DDEV
D = DSPH + DDEV
1 1("�"
where DSPH = trD�"I = v)I
3 3
�ł
"vi "v j �ł "vk
1div 1 1
and DDEV = D - v�"I �! (DDEV)ij = + �ij
�ł �ł -
�ł �ł
3 2 "x "xi łł 3"xk
j
�ł
70
The general constitutive relation for a (simple) fluid can be written in the form of the
matrix  polynomial
Ł =P (D) = Ł0 + c0I + c1D + c2D2 + c3D3 +...
where the coefficients are the function of 3 invariants of the tensor D, i.e.
ck = ck[I1(D),I2(D),I3(D)].
Consider the characteristic polynomial of the tensor D
pD() = det[D - I] = -3 + I12 - I2 + I3.
The Cayley-Hamilton Theorem states that the matrix (or tensor) satisfies its own
characteristic polynomial meaning that
pD(D) = -D3 + I1D2 - I2D + I3 = 0
Thus, the 3rd power of D (and automatically all higher powers) can be expressed as a
linear combinations of I, D and D2.
Hence, the most general polynomial constitutive relation is given by the 2nd order formula
Ł =P (D) = Ł0 + c0I + c1D + c2D2
71
NEWTONIAN 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:
ś% c0 is a linear function of the invariant I1,
ś% c1 is a constant,
ś% c2 = 0.
If there is no motion we have the Pascal Law: pressure in any direction is the same. It
means that the matrix Ł should correspond to a spherical tensor and
0
Ł0n = -p �! Ł0 = -pI
The constitutive relation for the Newtonian fluids can be written as follows
Ł = -p�"I +ś ("�"v)�"I + 2� DDEV = -pI + (ś - 2 �) ("�"v) I + 2� D
3
c1
I1(D)
Ł0
Ł0 c0
where
� - (shear) viscosity (the physical unit in SI is kg/m�s)
ś - bulk viscosity (the same unit as �) ; usually ś << � and can be assumed zero.
72
NEWTONIAN FLUIDS - CONTINUED
The constitutive relation written can be written in the index notations
�ł
�ł
"vk łł "vi "vj łł
2
�ł śł
�ij = -p + (ś - �) �ij + � +
�ł
3
"xk śł "x "xi �ł
�ł śł
�ł śł
j
�ł �ł
�ł
For an incompressible fluid we have
"v
j
"�"v a" div v a" = 0
"xj
and the constitutive relation reduces to the simpler form
�ł
"vi "v j łł
�ł śł
�ij = -p�ij + � + .
"x "xi �ł
�ł śł
j
�ł
73
NAVIER-STOKES EQUATION
NAVIER-STOKES EQUATION
Let us derive the equation of motion of Newtonian fluids. Earlier, we have derived the
general form from the 2nd Principle of Newton s dynamics
Dvi = "�ij + � fi
�
Dt "x
j
We have to calculate the explicit form of the first term in the right-hand side of the above
equation:
�ł
"�ij
�ł �ł
"vk "vi "v j �ł
"p " "
2
= - �ij + (ś - �) �ij + � + =
�ł �ł
�ł
3
�ł �ł
"x "x "x "xk �ł "x "x "xi łł
j j j j j
�ł łł
�ł
�ł �ł
"v
�ł �ł
"vk "2vi
"p " "
j
2
= - + (ś - �) + � + � =
�ł �ł
3
�ł �ł
"xi "xi �ł "xk �ł "x "x "xi �ł "xj łł
j j
�ł łł
�ł �ł
"vk "2vi
"p "
1
= - + (ś + �) + �
3
"xi "xi �ł "xk �ł "xj"x
j
�ł łł
74
After the obtained formula is inserted into the equation of motion, we get
Dv
1
� = -"p + �"v + (ś + �) " ("�"v) + � f ,
3
Dt
grad(div v)
or, writing the fluid acceleration in the full form, we obtain the Navier-Stokes equation
Navier-Stokes equation
�ł
"v
1
� + (v�"")vłł = -"p + �"v + (ś + �)"("�"v) + � f
.
�ł śł
3
"t
�ł �ł
For an incompressible fluid the N-S equations simplifies to
�ł
"v
� + (v�"")vłł = -"p + �"v + � f ,
�ł śł
"t
�ł �ł
which is often written in the form
"v 1
+ (v�"")v = - "p +� "v +f
"t �
where � = � / � is called the kinematic viscosity of fluid (the SI unit is m2/s).
75
The index form of the  incompressible Navier-Stokes equation is
"vi "vi "2vi
1 "p
+ v = - +� + fi.
j
"t "xj � "xi "xj"xj
The Navier-Stokes equation is the vector equation (or three scalar equations) with four
unknown fields:
ś% three Cartesian components of the velocity field and
ś% the pressure field.
For an incompressible fluid it is sufficient to add the continuity equation "�"v = 0 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 � .
Additional complication comes from the fact that viscosity is temperature dependent!
76
The boundary condition at solid and impermeable surfaces (of the immersed bodies) is
formulated as
v = u at "&! , 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 v = v" , where v" denotes the velocity in the free stream far away from the immersed
r "
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)
77
EULER EQUATION AND ITS FIRST INTEGRALS
EULER EQUATION AND ITS FIRST INTEGRALS
THE EULER EQUATION
The Euler equation are the special case of the Navier-Stokes equation for a hypothetic
inviscid (i.e. possessing zero viscosity) fluid:
�ł łł
"v
� + (v�"")vśł = -"p + � f
�ł
"t
�ł �ł
Using the Lamb-Gromeko form of the convective acceleration
1
(v�"")v = " v2 + �� v
( )
2
we can write the Euler equation as follows
"v 1
1
+ " v2 + �� v = - "p +f
( )
2
"t �
78
Let us now make the following assumptions:
(1) Volume force field is potential, i.e. there exist such scalar field Śf that f = "Śf ,
(2) Fluid is barotropic, i.e. its pressure is the function of density (or vice-versa) p = p(�).
1
Define the following pressure function P: P(�):= p2 (�)d� .
+"
�
" 1 "� 1 " 1
Then P[�(x)] = p2 [�(x)] = p[�(x)] �! "P = "p
"xi �(x) "xi �(x) "xi �
Note that for the incompressible fluid (� constant) we simply have P = p / � .
A less trivial example is an isentropic motion of an ideal gas where p H" �� , where
� = cP / cV. Then (check yourself!)
� p
P = = cPT = i (specific enthalpy)
(� -1)�
79
BERNOULLI EQUATION
"H
Assume that the flow is stationary, i.e. for any physical quantity H we have a" 0.
"t
1
The Euler equation can be written as " v2 + P -Śf = v ��
( )
2
Choose a streamline and define the tangent unary vector � = v / v. Next, multiply the above
equation (in the sense of the inner product) by �. The result is
d 1
1 1
v2 + P -Śf := " v2 + P -Śf �"� = v�"(v ��) = 0
( ) ( )
2 2
d� v
Hence, the function under the gradient is constant along the streamline:
1
v2 + P -Śf = CB
2
The Bernoulli constant CB can be  in general  different for different streamlines, unless
v �� = 0 (for instance, the flow may be irrotational, i.e. such that the vorticity � a" 0).
80
In practice, we use the Bernoulli equation in the form of
1 1
v2 + P -Śf = v2 + P -Śf
( ) ( )
2 2
A B
where A and B denote two points on the same streamline.
81
CAUCHY-LAGRANGE EQUATION
Now, the flow can be unsteady but we assume that the velocity field is potential. Then
v = "Śv
for some scalar field (the velocity potential) Śv and
"v
� = "� v = 0 , = "�ł " Śv �ł , v2 =|"Śv |2.
�ł �ł
"t "t
�ł łł
Then the Euler equation takes the form of
" 1
" Śv + |"Śv |2 +P -Śf = 0
( )
"t 2
2
"Śv + 1 "Śv + P -Śf = C(t)
meaning that
2
"t
for some function C, which depends only on time.
82
AERO/HYDRODYNAMIC FORCES
STRESS AND REACTION FORCE EXTERTED AT AN IMMERSED SURFACE
The flow-induced force is defined as the surface integral of the surface stress vector
F = �dS = ŁndS.
+" +"
"&! "&!
For an incompressible fluid we have the constitutive relation
Ł = -pI + 2� D = -pI + 2� "v - 2� R .
rotation
gradient of
tensor
velocity
1 1
Since Rn = - n�rotv = - n��
2 2
we can write � = Łn = -pn + 2� "v�"n + � n��.
83
The following theorem holds:
If div v = 0 (incompressible flow) and v = 0 then
"&!
"v �"n + n�� = 0.
Proof
Since v = 0 then the boundary "&! is the izosurface for all components of the velocity
"&!
field and the gradients of these components must be perpendicular (normal) to "&! .
Thus, we can write
" vj
"v �n = 0 �! = j nk , k =1,2,3
j
"&!
"xk
for some real numbers j (j = 1, 2, 3).
Next, in the index notation we have
" vi
n�� ="ijk nj�k ei , "v�"n = nj ei
" x
j
84
After insertion we get
"
"v�"n + n�� = ("" vi +"ijk �k)nj ei = ("" vi +"ijk"ką� "xą v� )nj ei =
x x
j j
�ł
" " " "
= vi + (�ią�j� -�i�� ) v� łł nj ei = ("" vi + v - vi)nj ei =
"x ją "xą x "xi j "x
�ł j śł j j
�ł �ł
" "
= v nj ei = ("" v nj - v ni)ei = (j ni nj - j nj ni)ei = 0
"xi j xi j "x j
j
= div v=0
Using the above result in the formula for the stress vector, we finally obtain the formula
� = -pn - � n��.
Note that pn�n = 0 and (n ��)�"n = 0 so the first term corresponds to the normal
stress while the second one  to the tangent stress at the boundary surface "&!.
The total aerodynamic force can be calculated from the integral formula
F = - (pn + � n��)dS
+"
"&!
85
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
F = �dS = - pndS+ 2� "v�"ndS+ � n��dS .
+" +" +" +"
"&! "&! "&! "&!
But
"v�"ndS = Div("v)dx = "v dx =
+" +" +"
Ń!
"&! &! &!
Laplacian of
tensor version
the velocity
of GGO
= "("�"v)dx - "�("� v)dx = - "��dx = - n��dS
+" +" +" +"
Ń!
=�
=0
&! &! &! "&!
GGO for
the cross
product
Thus
F = - pndS- 2� n��dS+ � n��dS= - (pn + �n��)dS.
+" +" +" +"
"&! "&! "&! "&!
86
THE INTEGRAL FORM OF THE MOMENTUM EQUATION (STEADY MOTION)
n
Consider a stationary flow. The momentum equation in the
index form reads
�fluid �body
"vi = "�ij + � fi
� v
j
n
"x "x
j j
or, using the (stationary) mass conservation equation
�fluid
"�ij
"vi + vi
"
� v (�v ) = + � fi
j j
n
"x "x "xj
j j
Ó!
0
�body
n
Two terms in the left-hand side of the above equation can
be collected into one term, namely "&! = �body *" �fluid
"�ij
"
(�viv ) = + � fi
j
"x "x
j j
87
Assume next that the volume forces can be neglected. Then the momentum equation
reduces to
"
(� viv -�ij) = 0
j
"x
j
In the next step, we calculate the volume integral and apply to it the GGO theorem. The
result is
"
(� viv -�ij)dx = 0 (� viv nj -�ijnj)dS = 0
"x j j
+" +"
j
&! "&!
Equivalently, we can write the vector relation as follows
(� v nj vi ei - �ijnjei )dS = 0 ,
j
+"
"&!
v
vn=v�"n
�
or simply �dS = (�v)vn dS
+" +"
"&! "&!
momentum flux
through the boundary
88
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
�dS = (�v)vn dS - �dS,
+" +" +"
� "&! �fluid
body
-F
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 vn � = 0 and
body
-F = (�v)vn dS - �dS
+" +"
�fluid �fluid
89
Consider now an incompressible flow. The surface stress vector is expressed as
� = -pn + 2� Dn
and the formula for the reaction force can be written as follows
F = - (�v)vn dS- pndS+ 2� DndS
+" +" +"
�fluid �fluid �fluid
Quite often, we can choose � in such way that the viscous term is relatively small
fluid
and can be neglected. Then
F E" - (�v)vn dS - pndS.
+" +"
�fluid �fluid
Sometimes the part of the body surface is in the contact with some other motionless fluid
(typically  the surrounding air) having a uniform pressure pa.
Note that for the closed surface �fluid we have pa ndS = pa ndS = 0 (why?)
+" +"
�fluid �fluid
90
The formula for the full (or net) force can be then written as follows
Fnet E" - (�v)vn dS- (p - pa)ndS
+" +"
�fluid �fluid
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.
91
CIRCULATION, VORTICITY AND STREAMFUNCTIONS
CIRCULATION, VORTICITY AND STREAMFUNCTIONS
CIRCULATION
Definition: Circulation of the vector field w along the (closed) contour L is defined as
� = w�"dl
+"
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
� (t) a" v(t,x)�"dl = 0
dt dt
+"
L(t)
92
Proof of the Kelvin Theorem:
Since the flow is barotropic and the volume force field is potential, we can write
1
"P = "p , f = "Ś
�
Thus, the acceleration (which consists of the convective part only) can be expressed as
a a" (v�"")v = -"(P -Ś )
In order to evaluate the time derivative of the circulation along the material line, it is
convenient to use Lagrangian approach. Thus, the circulation can be expresses as
� (t) = v(t,x)�"dl = V(t,�)�"J(t,�)dl0
+" +"
L(t) L0(t)
"x
where J(t,�) = denotes the Jacobi matrix of the transformation between Eulerian and
"�
Lagrangian coordinates.
93
Then, the time derivative of the circulation is evaluated as follows
d d
v(t,x)�"dl = V(t,�)�"J(t,�)dl0 = a(t,�)�"J(t,�)dl0 +
+" +" +"
dt dt
L(t) L0(t) L0(t)
+ V(t,�)�""� V(t,�)dl0 = a(t,x)�"dl + "� (1 V�"V)(t,�)�"dl0 =
2
+" +" +"
L0(t) L(t) L0(t)
=0
int. of the grad.along the closed loop
= - "(P +Ś )�"dl = 0
+"
L(t)
=0
int.of the grad.along
the closed loop
where the relation "tJ(t,�) = "� V(t,�) has been used.
94
VORTICITY
As we already know, the vorticity is defined as the rotation of the velocity:
� = rot v a" "�v
Definitions:
ś% 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.
ś% 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)
vorticity field
vortex line
L
95
STRENGTH OF THE VORTEX TUBE
It is defined as the flux of vorticity through a cross-section of
S2
the tube. Using the Stokes Theorem we can write:
n2
v�"dx = �
+"��"nd� = +"
S l
Sside
We see that the strength of the vortex tube is equal to the
circulation of the velocity along a closed contour wrapped
n1
around the tube.
S1
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 &! located
between two cross-section S1 and S2 .
From the GGO theorem we have
0 = "�"�dx = ��"nds + ��"nds = 0
+" +"��"nds + +" +"
&! S1 S2 Sside
= 0
96
Note that the last integral vanishes because the surface Sside is made of the vortex lines and
thus at each point of Sside the normal versor n is perpendicular to the vorticity vector.
Note also that the orientations of the normal versors at S1 and S2 are opposite (in order to
apply the GGO Theorem, the normal versor must point outwards at all components of
the boundary "&! ).
Reversing the orientation of n at S2, we conclude that
��"nds
+"��"nds = +"
S1 S2
97
HELMHOLTZ (3RD ) THEOREM
Assume that:
ś% the flow is inviscid and barotropic,
ś% 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 l0 :a = a(s).
At some later time instant t > 0, the shape of the material line follows from the flow




mapping Ft :R a Ź x "R , i.e., l : x = x(s) = Ft[a(s)].
R R
R R
R R
The corresponding transformation of the tangent vector can be evaluated as follows
d d d
�(s) = x(s) = Ft(a(s)) = ["x / "�](a(s)) a(s) = ["x / "�](a(s))�0(s)
ds ds ds
Jacobi matrix
98
Let s now write the acceleration in the Lamb-Gromeko form:
D
a = v = "tv + "(1 v2) + �� v
Dt 2
The rotation of a can be expressed as
D
"�a = "t("� v) + "�(�� v) = � - (��"")v +("�"v)�
Dt
In the above, the following vector identity, written for p = � 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 � / � as follows
D 1 1 D 1 D 1
�ł łł
� = � - � � = "�a + (��"")v -("�"v)��ł +
( )
� � �
Dt Dt
�ł
�2 Dt
=-�"�"v
1 1 1
+ �"�"v = "�a + (� ��"")v
� �
99
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 � / � reduces to
D 1 1
� = (� ��"")v
( )
�
Dt
"xi cj , or equivalently, � = � ("x / "�) c.
Define the vector field c such that �i = �
"�j
Jacobi
matrix
In the above, the symbol � denotes the Lagrangian variables.
The left-hand side of the above equation can be transformed as follows
D 1 d d
L = � = �ł("x / "�)cłł = ("x / "�)dt c + ("v / "�)c =
( )
�
Dt
dt
�ł �ł
d
= ("x / "�)dt c + ("v / "x)("x / "�)c
The right-hand side can be written as
1
R = (� ��"")v = �ł("x / "�) c�""łł v = ("v / "x)("x / "�)�"c
�ł �ł
100
Since L = R, we obtain
d
("x / "�) c + ("v / "x)("x / "�)c = ("v / "x)("x / "�)c
dt
d
which implies that dt c = 0
Thus, c is constant along trajectories of the fluid elements.
Using the Lagrangian description, we can write c(t,�) = c(0,�) = c0
Note that for the initial time t = 0 the transformation between Lagrangian and Eulerian
descriptions reduces to identity.
("x / "�) = �
t=0
1
Therefore c0 = �0 and since c(t) a" c0 we get
�0
1 1
� = ("x / "�) �0.
� �0
101
The last equality has the form of the transformation rule for the vectors tangent to
material lines. Since the vector �0 / �0 is tangent to the vortex line passing through the
point � at t = 0, it follows that the vector � / � is tangent to image of this line at some later
time t. But � / � 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.
102
EQUATION 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
1
"tv + "(1 v2) + �� v = - "p + f
�
2
Thus, the application of the rotation operator yields
1
"t� + "�(�� v) = -"�(�"p) + "�f
The pressure term can be transformed as follows
1 1 1 1
"�(� "p) = "(� )�"p + "�"p = - "� �"p
�
�2
ę!
0
Note: the above term vanishes identically when the fluid is barotropic since the gradients
of pressure and density are in such case parallel.
103
The equation of the vorticity transport can be written in the form
1
"t� + (v�"")� - (��"")v = - "� �"p + "�f
�2
D 1
or, using the full derivative � = (��"")v - "� �"p + "�f
Dt
�2
nonpotential
vortex stretching
volume force
baroclinic
term
term
term
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.
104
For the barotropic (in particular  incompressible) motion of inviscid fluid, the vorticity
equation reduces to
D
� = (��"")v
Dt
D
In the 2D case it reduces further to � = 0
Dt
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
1
"tv + "(1 v2) + �� v = - "p +� "v + f
�
2
If the rotation operator is applied, we get the equation
"t� + (v�"")� - (��"")v =� "� + "�f
which reduces to "t� + (v�"")� - (��"")v =� "�
when the field of the volume forces f is potential.
105
In the above, the following operator identity has been used
rot "v = rot (graddivv - rot rotv) = -rot rot� = graddiv� - rot rot� = "�
showing that the vector Laplace and rotation operators commute.
The vorticity equation can be also written equivalently as
D
� = (��"")v +� "�
Dt
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.
106
TWO-DIMENSIONAL INCOMPRESSIBLE FLOW. STREAMFUNCTION.
The streamfunction is a very convenient concept in the theory of 2D incompressible flow.
The idea is to introduce the scalar field � such that
u = "y� , v = -"x�
Note that the continuity equation
"xu + "yv = 0
is satisfied automatically. Indeed, we have
"xu + "yv = "xy� - "yx� = 0
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:
"� �"v = u"x� + v"y� = -uv + vu = 0
107
STREAMFUNCTION 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=�A+QAB
B B B
B
QAB =
+"v�"nds = +"v�"nds = +"(u nx + v ny)ds =
A A A
�
n B B
=
streamlines v y x
+"(u� - v�x )ds = +"(� "x� +�y "y� )ds =
A A
B
A
=
+""� �"ds =�B -�A
�=�ę
A
The volumetric flux through the line segment is equal to the difference of the
streamfunction between the endpoints of this segment.
108
STREAMFUNCTION 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
� a" "� v = ("xv - "yu)ez a" �ez
Then, the streamfunction satisfies the Poisson equation
"� a" "xx� + "yy� = -("xv - "yu) = -�
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 v = 0 is automatically satisfied. The complete
description consists of the following equations:
" Equation of the vorticity transport (2D) "t� + u"x� + v "y� =� "�
" Equation for the streamfunction "� = -�
" Relation between the streamfunction and velocity u = "y� , v = -"x�
" Definition of vorticity (2D) � = "xv - "yu
accompanied by appropriately formulated boundary and initial conditions.
109
ENERGY OF A FLUID FLOW
ENERGY OF A FLUID FLOW
ENERGY EQUATION
ENERGY EQUATION
Total energy of the fluid inside the region &! is the sum of the kinetic energy EK
and the internal energy U.
The energy changes during the fluid motion due to:
ś% external volume and surface forces,
ś% heat conducted through the boundary "&!
ś% heat produced by internal sources.
The time derivative of the total energy of the fluid in &! can be written as
" "
d
(U + EK) = P&! + P"&! +Q&!+Q"&!
dt
110
The right-hand-side terms are:
P&! =
+"�f �"vdx - power of the volume forces,
&!
P"&! = ��"vdS = v�"ŁndS - power of the surface (or interface) forces,
+" +"
"&! "&!
"
Q&! =
+"�qdx - power of the internal heat sources (symbol q
&!
denotes their volumetric density) ,
"
dT
Q"&! =  dS = "T�"ndS - the heat flux through the boundary "&! (the
+" +"
dn Ń!
"&! "&!
GGO
symbol T denotes temperature and  is the
coefficient of heat conduction).
111
Using the quantities defined above we get
d 1
v�"ŁndS + "T�"n dS,
dt 2
+"�(u + v2)dx = +"�f �"vdx + +" +"�qdx + +"
&! &! "&! &! "&!
total energy power of the internal heat
power of the heat flux through
volume forces sources in &!
surface forces the boundary "&!
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
d 1 D 1
dt 2 Dt 2
+"�(u + v2)dx= +"� (u + v2)dx.
&! &!
112
Next, we transform the surface integrals:
Next, we transform the surface integrals:
" " "
v�"ŁndS = �ijvinj dS = (�ijvi)dx = �ij vi dx + vi dx =
"x "x ij "x
+" +" +" +" +"�
j j j
Ń!
"&! "&! &! &! &!
GGO
" 1 " 1 "
= �ij vi dx +
2 2
"x ij x "xi j ij x "xi j
+" +"� ("" vi + v )dx + +"� ("" vi - v )dx =
j j j
&! &! &!
" 1 " 1 " 1 "
= �ij vi dx + vi dx - �ji "x jvi dx =
2
"x ij x "xi j 2 ij "x 2
+" +"� ("" vi + v )dx + +"� +"
j j j
&! &! &! &!
=�ij
= DivŁ�"vdx + Ł: Ddx
+" +"
&! &!
and also "T�"n dS =
+" +""�"("T)dx.
Ń!
"&! &!
GGO
Assume that all physical fields involved are sufficiently regular. Since the volume &! can be
arbitrary the partial differential equations  called the energy equations  follows
D 1
� (u + v2) =�f �"v + Div Ł�"v + Ł:D+�q +"�"("T)
Dt 2
113
INTERNAL ENERGY AND VISCOUS DISSIPATION
INTERNAL ENERGY AND VISCOUS DISSIPATION
It is interesting to consider separately the balance of kinetic and internal energy. To this
end, consider the momentum equation
D
� v =�f + DivŁ
Dt
The momentum equation can be multiplied by the velocity vector v. We get
D
� v�"v = �f �"v + DivŁ�"v
Dt
D
or, equivalently � (1 v�"v ) =�f �"v + DivŁ�"v
Dt 2
v2
In the next step, the above equation is integrated in the volume &! which yields the following
integral expression for the temporal rate of change of the kinetic energy
d 1
�v2 dx = DivŁ�"vdx
dt 2
+" +"�f �"vdx + +"
&! &! &!
Ek
114
Let s transform the second integral in the right-hand side. We can write (GGO again!)
" " "
DivŁ�"vdx = �ij vi dx = (�ij vi)dx - vi dx =
"x "x ij "x
+" +" +" +"�
j j j
&! &! &! &!
"
= vi �ijnj dS- vi dx = v�"ŁndS- Ł:"vdx =
ij "x
+" +"� +" +"
j
"&! &! "&! &!
= v�"ŁndS- Ł:Ddx
+" +"
"&! &!
The rate of change of the kinetic energy can be expressed as follows
d 1
�v2 dx = v�"ŁndS- Ł:Ddx
dt 2
+" +"�f �"vdx + +" +"
&! &! "&! &!
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
"T�"ndS+ Ł:Ddx.
dt
+"�udx = +"�qdx + +" +"
&! &! "&! &!
U
115
Observe that both formulae written above contain the term
T := Ł:Ddx
+"
&!
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
2
�ij =[-p+(ś- �)"" vk ]�ij +2�dij .
3 xk
Hence, we have
2 "
Ł:D = �ij dij = -p�ijdij + (ś - �) vk �ij dij + 2�dij dij =
3 "xk
2 "
= -pdii + (ś - �) vk dii + 2�dij dij = -p"�"v +
3 "xk
> 0if "�"v<0 (compression)
< 0if "�"v>0 (expansion)
=trD2
2
+ (ś - �)("�"v)2 + 2� D: D
3
positive-definite part
116
The  energy transfer term takes the following form
2
2
Ł: Ddx = -
3
+" +"p"�"vdx + (ś - �)+"("�"v) dx + 2�+"D: Ddx .
&! &! &! &!
internal-to -mechanical irreversible mechanical-to-internal energy transfer, i.e.
(if < 0) or reverse (if > 0) dissipation of the machanical energy due to internal "friction"
energy transfer
For incompressible flows we have "�"v = 0, thus
"vi "v j "vi "v j
1
Ł:Ddx = 2�
�ł �ł�ł �ł
2 "x "xi "x "xi
+" +"D:Ddx = �+"�ł + �ł�ł + �łdx =
j j
�ł łł�ł łł
&! &! &!
"vj
"vi "vi "v j "vj "v "vi "vi "vj
�ł �ł
1
= � + + 2"xij "xi �ł dx = � + dx a" R
�ł �ł �ł �ł
2 "x "xj "xi "xi "xj "x "xi
+"�ł +"
j j
�ł łł �ł łł
&! &!
The quantity R is called the dissipation rate.
117
FIRST INTEGRAL OF THE ENERGY EQUATION
FIRST INTEGRAL OF THE ENERGY EQUATION
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 f = "Ś .
We begin with the differential energy equation, which in the case of an ideal fluid reduces to
D 1
� (u + v2) = -"�"(pv)+�f �"v
Dt 2
By expanding the pressure term, this equations can be re-written equivalently as
D 1
� (u + v2) = -p"�"v - v�""p+�f �"v
Dt 2
118
Since the volume force is potential, the corresponding term in the right-hand side can be
transformed as follows
D
�f �"v = �v�""Ś = �("tŚ + v�""Ś) = � Ś
Dt
=0
Moreover, due to flow steadiness we have
D
v�""p = "tp + v�""p = p
Dt
=0
Next, from the mass conservation equation
D
� +�"�"v = 0
Dt
we get the following expression for divergence of the velocity field
1 D
"�"v = -� Dt�
119
The energy equation can be now written in the following form
p
D 1 D 1 D D
(u + v2) = � - p + Ś
�
Dt 2 Dt Dt
�2 Dt
D
=-Dt(p/�)
D 1
or ( u + p � + v2 -Ś) = 0
Dt 2
= i
where i = u + p � denotes the mass-specific enthalpy of the fluid.
Thus, the energy equation can be written as
D 1
(i + v2 -Ś) = 0
Dt 2
Since the flow is stationary, the above equation is equivalent to
1
v�""(i + v2 -Ś) = 0
2
120
Using the same arguments as in the case of the Bernoulli Eq., we conclude that along each
individual streamline
1
i + v2 -Ś = Ce = const
2
In general the energy constant Ce can be different for each streamline. If Ce 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
1
P + v2 - Ś = CB = const
2
Thus, when the flow is barotropic then
i - P = Ce - CB = 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 / �. Thus, in the incompressible case, the energy and
Bernoulli equations are formally identical.
121
ENTROPY 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
� u = -p"�"v
Dt
1 D
We have already used the relation "�"v = - �. Thus, the equation for the internal
�
Dt
energy e can be written as follows
p
D D D D
u = -�2 Dt� = -p (1/�) = -p �
Dt Dt Dt
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/ �. The corresponding form of this principle reads
Tds = du + pd�
122
For the thermodynamic process inside individual fluid element one can write
D D D D D
T s = u + p � = -p �+ p � = 0
Dt Dt Dt Dt Dt
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
1 1
i + v2 -Ś = Ce , or equivalently "(i + v2 -Ś) = 0.
2 2
Similarly, we call the flow homoentropic if "s a" 0. Thus, when the flow is homoentropic
then the entropy is uniformly distributed in the flow domain. Since the 1st Principle of
Thermodynamics can be written in the following form
Tds = di -(1/�)dp
then for any stationary flow one has T"s = "i -(1/�)"p.
123
In the case of a homoentropic flow we get
"i = (1/�)"p = "P.
Thus, if the flow is homoenergetic and homoentropic, it is automatically barotropic and
the Bernoulli constant CB 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
1
"(1 v2) + v�� = -�"p + "Ś
2
Using the entropy/enthalpy form of the 1st thermodynamic principle, we can re-write the
above equation in the following form
T"s = "(1 v2 +i -Ś) + v��
2
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.
124
THERMODYNAMIC INEQUALITY
THERMODYNAMIC INEQUALITY
Consider the material volume &!. 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.
d d d
We have dtS = Si + Se
dt dt
irreversible change of
change of entropy due
entropy
external heat
exchange
Consider a small portion of the volume "&! surrounded by the surface "A.
The amount of heat exchanged by this volume during a small time interval "t can be
expressed as
"Q = -"t q�"ndA = -"t divqd&! H" -"tdivq(x)"&!
+" +"
"A "&!
125
The increments of entropy of the volume "&! which occurs in the time interval "t due to the
heat exchange is
dSe = 1 dQ divq(x)
= - d&!
dt T dt T
It is convenient (and natural) to introduce the mass-specific density of entropy s. Then,
according to well-known formula, we have
dS d dsd&!
=
+"�sd&! = +"� dt
dt dt
&! &!
The 2nd Principle of Thermodynamics tells us that irreversible processes always lead to
some increase of entropy. Thus
dsd&!+ divq
d&! > 0
+"� śd&! = +"� dt +"
T
&! &! &!
dsi
where ś = represents a mass-specific density of entropy sources corresponding to all
dt
irreversible processes.
126
The heat flux integral term can be transformed as follows
divq q q�""T
d&! = d&!
�ł �ł
+" +"div�ł T �łd&! + +"
T
T2
�ł łł
&! &! &!
In the above, the following identity
1 1 1
div�ł q �ł = divq + q�""�ł 1 �ł = divq - q�""T
�ł �ł �ł �ł
T T T T
T2
�ł łł �ł łł
has been used. We also have (the GGO theorem)
q q�"n
dA
�ł �ł
+"div�ł T �łd&! = +"
T
�ł łł
&! "&!
and thus
dsd&!+ q�"n q�""T
dA + d&! > 0
+"� śd&! = +"� dt +" +"
T
T2
&! &! "&! &!
127
Finally, for an isotropic heat transfer we use the Fourier Law
q = -"T
where  > 0 is the coefficient of heat transfer.
The inequality can be finally written in the following form (Gibbs-Duhem inequality)
2
dS "T�"ndA + "T
> 
+" +" T d&!
dt T
"&! &!
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.
128
SIMILITUDE OF INCOMPRESSIBLE FLOWS
SIMILITUDE OF INCOMPRESSIBLE FLOWS
DIMENSIONLESS FORM OF THE NAVIER-STOKES EQUATION
DIMENSIONLESS FORM OF THE NAVIER-STOKES EQUATION
The Navier-Stokes equation for an incompressible flow can be written in the following form
"v 1"p + �"v +f
+ (v�"")v = -
"t �
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:
v,
time t = T t
linear dimensions x = Lx ,
v
j j
velocity v = V v,
v
v
pressure p = Pp,
v
volume force f = Ff .
In the above, all symbols with wave are dimensionless quantities.
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Consequently, we have also dimensionless differential operators
dx
v
" dv " 1 " " " 1 "
t
j
= = , = = .
"t dt "v T "v "x dx "xj L "xj
t t v v
j j
The Navier-Stokes equations can be now written as
V "v V2 (v�"")v = - P �V"v + Ff
v
v v v v,
+ v v "p + v
v
T "v L �L
t
L2
or, after multiplication by L/V2
L "v P � FL
v
v v v v.
+ (v�"")v = - "p + "v + f
v v v v
VT "v V L
t
�V2 V2
In the above equation, four nondimensional combinations of the scaling quantities have
appeared.
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We can define the following similarity numbers
VT
Strouhal number St = ,
L
�V2 ,
Euler number Eu =
P
V L
Reynolds number Re = ,
�
V2 .
Froude number Fr =
FL
Finally, the Navier-Stokes equation can be written in dimensionless form as follows
1 "v 1 1 1
v
v v v v
+ (v�"")v = - "p + "v + f
v v v v
St Eu Re Fr
"v
t
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 (m2/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:
ś% they are geometrically similar, e.g. the shapes (but usually not dimensions)
of the flow domains are the same,
ś% 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,
ś% 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
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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).
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