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This image is regarding flow of viscous fluid between two parallel plates. I don't understand how they determine direction of shear stress above and below the element. As the flow is towards right shear-stress should be towards left but at portion above the element it's indicated towards right. Also the shear stress on side faces of element seems to be neglected. [In orginal question we have to find velocity distribution, shear stress distribution and drop of pressure for given length for a flow of fluid between two parallel plates, flow is assumed to be laminar.] (I will be grateful if you have any way to solve the orginal question.)

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  • $\begingroup$ Are you familiar with the Cauchy stress relationship? $\endgroup$ Commented Apr 9 at 10:23
  • $\begingroup$ @ChetMiller I think, no. .. $\endgroup$
    – Qwerty
    Commented Apr 9 at 17:20
  • $\begingroup$ @ChetMiller it will great help if you can provide different perspective (ie cauchy stress relationship based), I tried to read about it but didn't get how it can apply to this scenario. .. thanks 👍 $\endgroup$
    – Qwerty
    Commented Apr 9 at 17:30
  • $\begingroup$ It applies perfectly to this scenario. $\endgroup$ Commented Apr 10 at 1:23
  • $\begingroup$ Are you familiar with dyadic tensor representation? $\endgroup$ Commented Apr 10 at 1:29

2 Answers 2

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The viscous stress must come in equal and opposite pairs because of Newton's third law. The forces on an element are exerted by the elements above and below it much like the friction between surfaces. If the force on the top face of a given element is rightwards, then it means that the element above is exerting this force on the given element. This means that the bottom face of the element above experiences the same force leftwards.

So we see there is really only one force variable here, $F(y)$, for every value of $y$. In other words, at each $y$, there is a pair of equal and opposite forces of magnitude $F(y)$ which act respectively on the elements above and below. So for the fluid element from $y$ to $y+\Delta y$, the force on the top face is $F(y+\Delta y)$ to the right and the force on the bottom face is $F(y)$ to the left. In steady flow, the net force on the element is zero, so the external force on the element must equal $-(F(y+\Delta y) - F(y))$, i.e. it is the (negative) derivative of $F$. After finding $F(y)$, the velocity profile can be determined using the definition of $F/A = \eta \partial v/\partial y$.

For example, in the simple case of Couette flow, the external force on all elements is zero. Therefore, the velocity varies linearly with $y$. If the external force is uniform, such as a river flowing downhill, the velocity will be a quadratic function of $y$.

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  • $\begingroup$ Thank you very much @Vincent Thacker for this nice and clear answer 😊. .. $\endgroup$
    – Qwerty
    Commented Apr 9 at 17:27
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This should help.

Let A be the region inside your fluid parcel and B be the region of fluid surrounding your fluid parcel. Let $\mathbf{n}$ be a unit normal vector at the boundary between A and B, oriented in the direction from A to B. Then the traction vector (force per unit area by the fluid on the B side of the boundary on the fluid on the A side of the boundary) is the dot product of the stress tensor with the unit normal. $$\mathbf{t}=(\sigma_{xx}\hat{\mathbf{x}}\hat{\mathbf{x}}+\sigma_{xy}\hat{\mathbf{x}}\hat{\mathbf{y}}+\sigma_{xy}\hat{\mathbf{y}}\hat{\mathbf{x}}+\sigma_{yy}\hat{\mathbf{y}}\hat{\mathbf{y}})\centerdot (n_x\hat{\mathbf{x}}+n_y\hat{\mathbf{y}})=(\sigma_{xx}n_x+\sigma _{xy}n_y)\hat{\mathbf{x}}+(\sigma_{xy}n_x+\sigma _{yy}n_y)\hat{\mathbf{y}}$$ In your problem, the stress tensor is given by $$\boldsymbol{\sigma}=-p\hat{\mathbf{x}}\hat{\mathbf{x}}+\tau\ \hat{\mathbf{x}}\hat{\mathbf{y}}+\tau\ \hat{\mathbf{y}}\hat{\mathbf{x}}-p\ \hat{\mathbf{y}}\hat{\mathbf{y}}$$So at the upper boundary where $\mathbf{n}=+\hat{\mathbf{y}}$, the traction vector is $$\mathbf{t}=-p\ \hat{\mathbf{y}}+\tau\ \hat{\mathbf{x}}$$And, at the lower boundary where $\mathbf{n}=-\hat{\mathbf{y}}$, the traction vector is $$\mathbf{t}=+p\ \hat{\mathbf{y}}-\tau\ \hat{\mathbf{x}}$$Hope this helps. Note that this rationale has nothing to do with the velocity gradient in the fluid.

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