LECTURE 6
LECTURE 6
KINEMATICS OF FLUIDS  PART 2
KINEMATICS OF FLUIDS  PART 2
RELATIVE MOTION OF FLUID ELEMENTS
RELATIVE MOTION OF FLUID ELEMENTS
Consider two fluid elements located instantaneously at the close points A and B. We ask what
happens to the relative position of these fluid elements after a short time interval D�t.
The location of the first fluid element after the time D�t can
t+D�t
x3
B'
be expressed as follows
t
B
r�'
xAó� =� xA +� v(t,xA)"t +� O("t2)
r�
A
A'
Since xB =� xA +� � then analogously we have
xBó� =� xA +� � +� v(t,xA +� �)"t +� O("t2),
0
r�=xB-xA
where the vector � describes the relative position of the
r�'=xB'-xA'
x2
fluid elements at the time t.
x1
During a short time interval D�t this vector has changed and can be expressed as
�(t +� D�t) =� xBó� -� xAó� =� �(t) +�[v(t,xA +� �) -� v(t,xA)]D�t +� O(D�t2) =�
=� �(t) +� ��Ń�v(t,x)�ł�D�t +� O(D�t2,D�t |� |2)
�� ��
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 +� D�t) -��(t)
=� lim =� �v(t,x)� +� O(|� |2).
D�t��0
dt D�t
ś�
�� ł�
We have introduced the matrix (tensor) called the velocity gradient ���v�� ij =� vi.
ś�x
j
The velocity gradient �v can be written as a sum of two tensors �v =� D +� R , where
ć�
ś�vi +� ś�v j ��
1
1
D =� [�v +� (�v)T] or dij =� - symmetric tensor,
�� ��
2
�� ��
2 ś�x ś�xi ł�
j
Ł�
and
ć�
ś�vi ś�v j ��
1
1
R =� [�v -� (�v)T] or rij =� -� - skew-symmetric tensor
�� ��
2
�� ��
2 ś�x ś�xi ł�
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.
1
To this end, we note that rij =� -� ��ijk w�k , where w�k are the Cartesian components of the
2
vorticity vector
ś�vk
� =� rotv =� ��ijk ei.
ś�x
1� 3�
4�2�4�j
w�k
Indeed, we have
ć� ć�
ś�v
ś�vg� ś�vg�
ś�vi �� ś�vi ś�vj ��
1 1
j
1 1
-� ��ijk��kb�g� =� -� (d�ib�d� -�d�ig�d� ) =� -� -� =� -� =� rij
�� �� �� ��
2 2 jg� jb�
�� �� �� ��
ś�xb� ś�xb� 2Ł� ś�xi ś�xj ł� 2Ł� ś�x ś�xi ł�
j
1 1 1
Thus, we can write R� =� rij r�j ei =� -� ��ijk r�jw�k ei =� -� ���� =� ����.
2 2 2
d d d
Moreover, we get dt |� |2=� (�,�) =� 2(�, �) =�2(�,R�) =� ���(����) =� 0
dt 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
DEFORMATION OF FLUID ELEMENTS
DEFORMATION OF FLUID ELEMENTS
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 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
ś�vk
1 1(���v)I �� (DSPH )ij =� 1
DSPH =� trD��I =� d�ij ,
3{� 3 3ś�xk
trace
of D
1(���v)��trI =� (���v) �� div v.
Note that trDSPH =�
3
The second part DDEV describes shape changes which preserve the volume.
ć�
ś�vi +� ś�v j �� ś�vk d�ij
1div v��I �� (DDEV)ij =� 1 1
We have DDEV =� D -� -�
�� ��
�� ��
3 2 ś�x ś�xi ł� 3ś�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 D�t the above relation yields
�(t +� D�t) =� �(t) +� DSPH�D�t +� DDEV�D�t +� O(D�t2)
1�4�2�4�3� 1�4�2�4�3�
D��1 D��2
Consider the 2D case when only volumetric part of the deformation exists (see picture).
x2
x2
d 0
�� ł�
We have DSPH =� , tr D =� 2d
ę�
W�(t+D�t)
0 dś�
�� ��
B'
C'
dD�t L2
B C
C B
{
The relative position vector at the time
instant t +� D�t is expressed as
O
A O
A
�(t +� D�t) =� (1+� d D� t)�(t)+� O(D�t2). x1 A'
x1
W�(t)
W�(t)
dD�t L1
The shape of the volume is preserved because the above formula describes the isotropic
expansion/contraction. The volume of the region VolW� (t) =� L1L2 has been changed to
2
VolW� (t +� D�t) =� L1L2(1+� dD�t) =� VolW� (t)(1+� 2d D�t) +� O(D�t2),
VolW� (t +� D�t) -� VolW� (t)
1
and lim =� 2d =� tr D =� ���v
D�t��0
VolW� (t) D�t
{
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
��d11 -� d d12 ł� ��1 (d11 -� d22) d12 ł� -�a� g�
�� ł�
2 2
DDEV =� =� =� .
ę�
ś�
1 1
d12 d22 -� dś� ę� d12 (d22 -� d11)ś� ę� g� a�
2 2 �� ��
�� �� �� ��
x2 x2
The fluid deformation during the short time
C'
interval can be now expressed as g�D�t L2
W�(t+D�t)
�(t +� D�t) =� (I +� D�t DDEV)�(t) +� O(D�t2)
B'
a�D�t L2
{
B
C
B
C
or in the explicit form as
A'
g�D�t L1
A
��x1(t +� D�t) =� (1-�a� D�t)x1(t) +�g� D�t x2(t)
��
O O
��
A x1 x1
x2(t +� D�t) =� g� D�t x1(t) +� (1+�a� D�t)x2(t)
�� a�D�t L1
W�(t)
��
W�(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.
{
{
{
Let s compute again the change of the volume of the fluid region during such deformation.
We get
1-�a� D�t g� D�t
2
VolW� (t +� D�t) =� L1L2 =� L1L2(1-�a�2D�t2 -�g� D�t2) =�
g� D�t 1+�a� D�t
,
=� VolW� (t) +� O(D�t2)
so
VolW� (t +� D�t) -� VolW� (t)
1
lim =� 0.
D�t��0
VolW� (t) D�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).