Answer to Question #149122 in Math for Usman

Question

A fluid flow is given by : V = x^2 yi - 2yz^2 j - ( zy^2 - 2z^3 /3 ) k. Prove that it is a case of possible steady incompressible fluid flow.

Answer 1

$$u=x^2 y, v=-2yz^2, w=-zy^2+\dfrac{2}{3}z^3$$

Then

$$\dfrac{\partial u}{\partial x}=2xy, \dfrac{\partial v}{\partial y}=-2z^2 ,\dfrac{\partial w}{\partial z}=-y^2+2z^2$$

For a three-dimensional steady incompressible flow the continuity equation can be written in differential form as

$$\dfrac{\partial u}{\partial x}+\dfrac{\partial v}{\partial y}+\dfrac{\partial w}{\partial z}=0$$

Substitute

$$\dfrac{\partial u}{\partial x}+\dfrac{\partial v}{\partial y}+\dfrac{\partial w}{\partial z}=2xy-2z^2-y^2+2z^2$$

$$=2xy-y^2\not=0$$

Hence the continuity equation for an incompressible flow is not satisfied. Therefore, it is not a possible incompressible flow. 

If a fluid flow is given by : $V=xy^2i-2y^2zj-(zy^2-\dfrac{2}{3}z^2)k,$ then

$$u=x y^2, v=-2yz^2, w=-zy^2+\dfrac{2}{3}z^3$$

$$\dfrac{\partial u}{\partial x}=y^2, \dfrac{\partial v}{\partial y}=-2z^2 ,\dfrac{\partial w}{\partial z}=-y^2+2z^2$$

For a three-dimensional steady incompressible flow the continuity equation can be written in differential form as

$$\dfrac{\partial u}{\partial x}+\dfrac{\partial v}{\partial y}+\dfrac{\partial w}{\partial z}=0$$

Substitute

$$y^2-2z^2-y^2+2z^2=0$$

Hence the continuity equation for an incompressible flow is satisfied. Therefore, it is a possible incompressible flow.