21/02/2017 LECTURE 18

Material Derivative (Contd…..) Recall, in the last class, we started discussing about material derivative.

For any scalar, vectorial or tensorial quantity Pij ….., the material derivative was given by:

DPij....(,)(,)(,) x t P ij .... x t P ij .... x t vk Dt t xk

The first term indicates local rate change & second term indicates convective rate change.

 Also recall, the Lagrangian strain tensor & Eulerian Strain tensor.  In a similar way, if the spatial gradient of velocity v(,) x t is taken, it will provide you 휕푉푖 velocity gradient tensor 휕푥푗

 Using tensor property :

v11  vvvjj    v  i i     i   xj22  x j  x i    x j  x i  1st part is symmetric tensor & 2nd part is anti-symmetric or skew symmetric tensor.

 The symmetric tensor is the rate of deformation tensor or it is the strain rate tensor.  The skew symmetric tensor is called Vorticity tensor or spin tensor.  So, the velocity gradient term consist of strain rate & rotational rate.  Later on you will see that, if the vorticity tensor is zero, the corresponding velocity flow field is irrotational.

Material Derivative of Volume: Consider a fluid continuum to be in motion. In that fluid continuum, consider an elementary volume having initial configuration as such:

x3, X3 At time

t=t0

(3) dxi

dX3

(1) dxi

dX1 (2) dxi dX2

x2, X2 x1, X1

The elementary configuration’s sides are dX1ê1 , dXê2 , dX3ê3. The box product is given the volume: dU0 dX 1 eˆ 1 dX 2 e ˆ 2. dX 3 e ˆ 3

Volume is parallelepiped is dU0 dX 1 dX 2 dX 3

 As the fluid is moving, this parallelepiped is displaced & deformed. Let the deformed volume at time t = t be as given in the figure (for our convenience, we are right now assuming it to be parallelepiped).  Unlike in the initial configuration, in the later configuration, the sides will be vectors having components in all three directions. The sides are: dx(1),, dx (2) dx (3) xi dxij dX X j (1) (1) xi dxij dX X j (2) (2) xi dxij dX X j (3) (3) xi dxij dX X j

 The volume of the deformed parallelepiped will be dU dx(1) dx (2). dx (3) (1) (2) (3) [Box product]  ijkdx j dx k dx i

As the original initially configured volume had edges dX1, dX2, dX3 & there are particles associated with these edges, we assume in the distorted configuration the same particles occupy (1) (2) (3) the edges dx , dx , dxi respectively. If so, then,

(1) xi dxi  dX1 X1

(2) xi dxi  dX 2 X 2

(3) xi dxi  dX 3 X 3

Therefore the distorted configured volume

(1) (2) (3) dU ijk dx i dx j dx k xxx dU  ikj dX dX dX ijk XXX   1 2 3 1 2 3 xix j  x k  x i ijk J The Jacobian XXXX1  2  3  j dU JdU0 Hence, the material derivative of the volume

D() dUD() JdU DJ 0 dU Dt Dt Dt 0 D() dU (You can easily guess that 0  0 ) Dt

From further operations, it is proved that

퐷퐽 = divergence of velocity vector × Jacobian, 퐷푡 DJ v i.e.  p Dt xp

D() dU v  Jp dU Dt x 0 Therefore, p D() dU v  p dU Dt xp

Material Derivative of property in a Volume: You know that mass is a property associated with volume. If the density of the fluid is ρ then mass is M   dU U i.e. You need to volumetrically integrate density to get mass.

Similarly, there may be several properties in a closed continuum (fluid), where these properties are obtained by volumetric integration.

Let us say that any property (scalar or Vector or Tensor) Bij … is obtained by volumetric integration. i.e, B()(,) t  x t dU ij.... ij .... U

(Note: here we are not saying function of x & t. We say only function of t. this is because volumetric integration done. The material derivative of the property Bij….. will be

DB() t D ij....  [(,)] x t dU  ij.... Dt Dt U D  ((,)) x t dU  ij.... U Dt DD [ (x , t ) dU ( x , t ) ( dU )]  ij.... ij .... U Dt Dt

 v [ (,)vx t   (,)] x t dU   (,) x tp dU ij.... p ij ....  ij .... UUt  xpp  x   v [ (xt , ) v (,)(,)]x t x tp dU  ij.... p ij.... ij .... U t xxpp  [ (x , t ) ( v ( x , t ))] dU  ij.... p ij .... U txp

Applying Gauss Divergence theorem on the second portion of the volumetric integral, you get,

DB() t  ij.... [ (x , t )]dU v n ( x , t ) dA ij....  p p ij .... DtUA t

This equation is very similar to the Reynold’s Transport theorem expression you had studied.

Acknowledgement: The concept of deriving material derivative of volume is referred from the following text: George E. Mase (1970). “Continuum Mechanics”. Schaum’s Outline Series.