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Hi , My instructor solved this problem using a circular elemental strip of thickness 'dr'. He told me we get only shear stress in horizontal layers of fluid. He used Newton's law of viscosity to get the value of shear stress at top face of plate. The Newton's law of viscosity is only applied between two plates where there is velocity gradient. But we are also observing linear profile of velocity on the top face of plate. Here my doubt is why can't we apply Newton's law of viscosity for this circular strip? That is, in a circular strip we have a number of layers of fluid with different velocities. So there is velocity gradient between them. But shear stress between is not calculated between layers? Is there no relative motion between those cylinderical fluid layers? What goes wrong in assuming it? Also this problem is solved example in other textbook on fluid mechanics. In that example too they didn't consider shear stress between cylinderical fluid surfaces. I am studying this course for first time. Can anyone please explain elaborately.. Thank you

Edit: I am adding images of how i solved it . So it will avoid any confusion present in my text

enter image description here And the below image is the actual doubt enter image description here

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You are quite correct that there is relative motion along the radial direction as well as normal to the plates. However in viscometers the gap between the plates is orders of magnitude smaller than their radius, so the shear rate in the radial direction is negligibly small compared to the shear rate in the normal direction.

Plate viscometer

The viscosity is defined as:

$$ \eta = \frac{\sigma}{\dot\gamma} $$

where $\sigma$ is the shear stress and $\dot\gamma$ is the shear rate. The shear rate is the change in velocity with distance. Consider the change in velocity with distance as we move in the radial direction. The velocity that the top plate moves at a distance $r$ from the axis is:

$$ v = r\omega $$

and the distance is $r$, so the shear rate in the horizontal is simply:

$$\dot\gamma_h = \frac{r\omega}{r} = \omega $$

Now consider the change in velocity with distance as we move vertically. The bottom plate is fixed, while at a distance $r$ from the axis the top plate is moving with a velocity $v = r\omega$, and the thickness of the liquid layer is $h$ so the shear rate in the vertical direction is:

$$ \dot\gamma_v = \frac{r\omega}{h} $$

So the ratio of the two shear rates is:

$$ \frac{\dot\gamma_v}{\dot\gamma_h} = \frac{r}{h} $$

Viscometers are constructed so that the thickness of the liquid layer $h$ is much less than the radius of the plate, so that means $\dot\gamma_v \gg \dot\gamma_h$ and therefore the force we have to apply is dominated by $\dot\gamma_v$.

Note that commercial viscometers do not use parallel plates because the shear rate in the vertical direction changes with $r$. Instead they use a cone and plate geometry. See for example this answer.

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  • $\begingroup$ Sir thank you for giving your time for answering this question. But Sir can you briefly explain what do you mean by shear rate in radial direction and normal direction. I am not aware of these terminology. I am still learning the subject. Sir we are using the approximation of linear profile in case of small gap in Newton's law of viscosity . Hence, why can't we apply this law for small strip (having small gap ) ? Sir can you show mathematically how the shear rate in radial direction is negligible. I will be happy and satisfied if you mathematically show for radial direction case. Thank you $\endgroup$ Jul 9, 2021 at 7:16
  • $\begingroup$ @HariiPavan I have extended my answer to explain what I mean. $\endgroup$ Jul 9, 2021 at 7:22

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