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x = x ξ ( y , t ' ) , t ,

where the parameter t ' along it lies in the interval 0 t ' t . If we regard the motion as having been proceeding for all time, then the origin of time is arbitrary and t' can take negative values - t ' t .

The flow field illustrated in 4.13 by Aris is assigned as an exercise.

Dilatation

We noticed earlier that if the coordinate system is changed from coordinates ξ to coordinates x , then the element of volume changes by the formula

d V = ( x 1 , x 2 , x 3 ) ( ξ 1 , ξ 2 , ξ 3 ) d ξ 1 d ξ 2 d ξ 3 = J d V 0

If we think of ξ as the material coordinates, they are the Cartesian coordinates at t = 0 , so that d ξ 1 d ξ 2 d ξ 3 is the volume d V o of an elementary rectangular parallelepiped. Consider this elementary parallelepiped about a given point ξ at the initial instant. By the motion this parallelepiped is moved and distorted but because the motion is continuous it cannot break up and so at some time t is some neighborhood of the point x = x ( ξ , t ) . By the above equation, its volume is d V = J d V o and hence

J = d V d V o = ratio of an elementary material volume to its initial volume.

It is called the dilation or expansion . The assumption that x = x ( ξ , t ) can be inverted to give ξ = ξ ( x , t ) , and vice versa, is equivalent to requiring that neither J nor J - 1 vanish. Thus,

0 J

We can now ask how the dilation changes as we follow the motion. To answer this we calculate the material derivative D J / D t . However,

J = ε i j k x i ξ 1 x j ξ 2 x k ξ 3 = x 1 ξ 1 x 1 ξ 2 x 1 ξ 3 x 2 ξ 1 x 2 ξ 2 x 2 ξ 3 x 3 ξ 1 x 3 ξ 2 x 3 ξ 3

Now

D D t x i ξ j = ξ j D x i D t = v i ξ j

for D / D t is differentiation with ξ constant so that the order can be interchanged. Now if we regard v i as a function of x 1 , x 2 , x 3 ,

v i ξ j = v i ξ j x m ξ j

The above relation can now be applied to differentiation of the Jacobian.

D J D t = ε i j k D D t x i ξ 1 x j ξ 2 x k ξ 3 + ε i j k x i ξ 1 D D t x j ξ 2 x k ξ 3 + ε i j k x i ξ 1 x j ξ 2 D D t x k ξ 3 = ε i j k v i x m x m ξ 1 x j ξ 2 x k ξ 3 + ε i j k x i ξ 1 v j x m x m ξ 2 x k ξ 3 + ε i j k x i ξ 1 x j ξ 2 v k x m x m ξ 3 = ε i j k v 1 x 1 x i ξ 1 x j ξ 2 x k ξ 3 + ε i j k x i ξ 1 v i x 2 x j ξ 2 x k ξ 3 + ε i j k x i ξ 1 x j ξ 2 v 3 x 3 = v 1 x 1 J + v 2 x 2 J + v 3 x 3 J = v J

where we made use of the property of the determinant that the determinant of a matrix with repeated rows is zero. Thus,

D ln J D t = v

We thus have an important physical meaning for the divergence of the velocity field. It is the relative rate of dilation following a particle path. It is evident that for an incompressible fluid motion,

v = 0 for incompressible fluid motion.

Use of a stream function to satisfy the mass-conservation equation (batchelor, 1967)

In the cases of flow of an incompressible fluid, and of steady flow of a compressible fluid, the mass-conservation equation reduces to the statement that a vector divergence is zero, the divergences being of u and ρ u respectively. If we impose the further restriction that the flow field either is two-dimensional or has axial symmetry, this vector divergence is the sum of only two derivatives, and the mass-conservation equation can then be regarded as defining a scalar function from which the components of u or ρ u are obtained by differentiation. The procedure will be described here for the case of an incompressible fluid.

Assume first that the motion is two-dimensional, so that u = ( u , v , 0 ) and u and v are independent of z . The mass-conservation equation for an incompressible fluid then has the form

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Source:  OpenStax, Transport phenomena. OpenStax CNX. May 24, 2010 Download for free at http://cnx.org/content/col11205/1.1
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