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Equation of Continuity

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Equation Of Continuity


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Equation Of Continuity

1) For an incompressible fluid flow, two components of velocity (u,v,w) are given by u=x2+2y2+3z2, v=x2yy2z+zx. Determine the third component w so that they satisfy the equation of continuity. Also, find the zcomponent of acceleration.

[2018, 10M]

Using the continuity equation in cartesian coordinates:

ux+vy+wz=0
2x+x22yz+wz=0
wz=2yz2xx2
Integrating wrt z, we get:
w=yz22xzx2z+f(x,y)

The z component of acceleration:
az=(q)w+ωt=uwx+vwy+wωz

Therefore,

az=(x2+2y2+3z2)(fx2z2xz)+(x2yy2z+xz)(fy+z2)+(yz22xzx2z+f(x,y))(2yz2zx2)

2) Show that (x2a2)cos2t+(y2b2)sec2t=1 is a possible form for the boundary surface of a liquid.

[2007, 12M]


3) Show that: u=2xyz(x2+y2)2, v=(x2y2)z(x2+y2)2, w=yx2+y2 are the velocity components of a possible liquid motion. Is this motion irrotational?

[2002, 15M]

Here, we have:

ux=2yz1(x2+y2)2x2(x2+y2)2x(x2+y2)4=2yzx2+y24x2(x2+y2)3=2yzy23x2(x2+y2)3 vy=z2y(x2+y2)22(x2+y2)2y(x2y2)(x2+y2)4=2yzx2+y2+2(x2y2)(x2+y2)3=2yz3x2y2(x2+y2)3

and

w/z=0

Hence, the equation of continuity u/x+v/y+w/z=0 is satisfied and so the liquid motion is possible.

Furthermore, we have

Ωx=wyvz=x2y2(x2+y2)2x2y2(x2+y2)2=0Ωy=uzwx=2xy(x2+y2)2+2xy(x2+y2)2=0Ωz=vxuy=2xz(3y2x2)(x2+y2)32xz(3y2x2)(x2+y2)3=0 w/y=v/z,u/z=w/x,v/x=u/y

and hence the motion is irrotational.


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