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I came across a fact on web that when a fluid flows through a cylinder the shape of its flow is parabolic. But according to me if we have a steady state then the velocity each of the concentric fluid layers(element) shown in the figure must be constant and therefore the force on them must me zero, which implies that the force on each element must be same.

Applying the force due to viscosity

$$F= n A dv/dy$$ where $n$-coefficient of viscosity, A-area of cross section, dv/dy-velocity gradient

If the shape is parabolic then v will be proportional to y^2 and also A will be proportional to y, so overall after writing force equation it will come as a function of y which means it is not a constant.

Can someone please tell me how the shape is parabolic.

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1 Answer 1

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The issue is with your starting point, why would every fluid layer have the same velocity in steady flow? Since you have a non slip boundary condition and if your fluid is actually moving, it is impossible for this assumption to be satisfied. This implies that you have different speed, therefore a non zero and more generally a non constant force.

Check out Poiseuille Flow for more information.

Hope this helps.

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  • $\begingroup$ I meant that velocity across the tube will change but for a particular single cylindrical element along the tube it must be constant therefore the net force on it must be zero $\endgroup$ Jul 17, 2022 at 10:50
  • $\begingroup$ Thankyou I read the article and got my mistake! $\endgroup$ Jul 17, 2022 at 10:56
  • $\begingroup$ Oh ok got it. No problem! Just to make sure you got it right in the end, since cylindrical elements have different speeds, there will be friction forces which go against the velocity gradient and have a tendency of homogenizing the velocity field. Since the problem doesn’t allow this, the parabolic profile is in some sense the best compromise $\endgroup$
    – LPZ
    Jul 17, 2022 at 11:37
  • $\begingroup$ I had another doubt, I recall that I solved a problem involving a river flowing with constant speed and there the velocity of horizontal layers varied directly with distance from base which does not define a parabolic flow, is there some other condition when this happen or in cuboidal tube the flow is not parabolic? $\endgroup$ Jul 17, 2022 at 12:19
  • $\begingroup$ Yes it will depend a lot on the geometry of the tube. The general problem is to solve the 2D Poisson equation on the cross section $\Delta v_z = \nabla p_z/\eta$ with boundary condition $v_z=0$ where $z$ is the axis along the tube $\nabla p_z/$ is the constant pressure gradient along the tube and $\eta$ viscosity. It just so happened that for the disk, you get the parabolic solution. $\endgroup$
    – LPZ
    Jul 17, 2022 at 13:25

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