Question #149122
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.
1
Expert's answer
2020-12-16T08:30:00-0500
u=x2y,v=2yz2,w=zy2+23z3u=x^2 y, v=-2yz^2, w=-zy^2+\dfrac{2}{3}z^3

Then


ux=2xy,vy=2z2,wz=y2+2z2\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


ux+vy+wz=0\dfrac{\partial u}{\partial x}+\dfrac{\partial v}{\partial y}+\dfrac{\partial w}{\partial z}=0

Substitute


ux+vy+wz=2xy2z2y2+2z2\dfrac{\partial u}{\partial x}+\dfrac{\partial v}{\partial y}+\dfrac{\partial w}{\partial z}=2xy-2z^2-y^2+2z^2

=2xyy20=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=xy2i2y2zj(zy223z2)k,V=xy^2i-2y^2zj-(zy^2-\dfrac{2}{3}z^2)k, then


u=xy2,v=2yz2,w=zy2+23z3u=x y^2, v=-2yz^2, w=-zy^2+\dfrac{2}{3}z^3

ux=y2,vy=2z2,wz=y2+2z2\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


ux+vy+wz=0\dfrac{\partial u}{\partial x}+\dfrac{\partial v}{\partial y}+\dfrac{\partial w}{\partial z}=0

Substitute


y22z2y2+2z2=0y^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. 



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