[Jonny6001]

I'm working through derivation of the fundamental equations and need to clarify one point regarding elements of the strain rate tensor.

Why do we use half the combination of (dU/dy+dV/dx) etc for the off diagonal terms, rather than d/dy(mu.dU/dy)? I'm finding it difficult to justify this since it's essentially saying that a change in the 'V' velocity is imparting a force on the fluid inside a control volume in the 'X' direction.

I've looked through many text books and websites but have not come across any resource that goes through the derivation entirely. I see how the maths works but not the overall concept of why we need to combine and then split the terms in half.

I look forward to any replies.

[FMDenaro]

Originally Posted by Jonny6001

I'm working through derivation of the fundamental equations and need to clarify one point regarding elements of the strain rate tensor.

Why do we use half the combination of (dU/dy+dV/dx) etc for the off diagonal terms, rather than d/dy(mu.dU/dy)? I'm finding it difficult to justify this since it's essentially saying that a change in the 'V' velocity is imparting a force on the fluid inside a control volume in the 'X' direction.

I've looked through many text books and websites but have not come across any resource that goes through the derivation entirely. I see how the maths works but not the overall concept of why we need to combine and then split the terms in half.

I look forward to any replies.

I am not sure about your question, are you asking why the Newtonian model defines the stress tensor in terms of the symmetric gradient velocity?

[Jonny6001]

Hi FMDenaro,

If we look at the 'X' momentum equation, and the shear stress part:

d(tau_xx)/dx + d(tau_yx)/dy + d(tau_zx)/dz

where d(tau_yx)/dy represent shear stress acting on face normal to 'Y' axis in 'X' direction.

Using the Newton's constitutive law: tau_yx = mu.du/dy

We have d(tau_yx)/dy = d/dy(mu.du/dy) = mu d2y/dy2 (assuming constant viscosity)

This is the formulation that's obtained for incompressible flow. But for compressible flow we end up with:

(tau_yx) = mu.(du/dy + dv/dx)

And (tau_xy) = mu.(du/dy + dv/dx)

Why have we included the dv/dx in the to tau_yx term?

I appreciate that a fluid element deforms in both direction (in 2D) and that the angle that the parallelogram forms with the original vertical side is given by:

(dU/dy).dy.dt

But in my mind, only the vertical sides of the fluid element will be rotated when a shear stress is applied on the top face. I get that the horizontal sides are also rotated, further reducing the angle of the lower left corner of the fluid element. But this requires a shear stress in the 'Y' direction, why does it appear in the 'X' momentum equation?

Apologies if my question is still not clear.

[FMDenaro]

Because in the incompressible flow the divergence-free constrain applies on the velocity field and some terms elide. Therefore you get (mu constant)

Div (2mu S) = mu Lap v

[Jonny6001]

Yes, I appreciate all that. The issue is not with the incompressible version.

Perhaps I'm not explaining my train of though very well. In summary, why does dv/dx appear in the 'X' momentum equation?

I'm looking for a more detailed answer than "because the shear stress is given by (tau_yx) = mu.(du/dy + dv/dx)"
Which is what all the textbooks that I've picked up say.

My previous reply has more detail about the problem that I have with the derivation.

I think the answer is along the lines that the angular deformation of the element is averaged over the X and Y directions (in 2D) to maintain zero net moment on the fluid element. But that's not exactly correct.

Thank you!

[FMDenaro]

The presence of the two derivatives is in the nature of the symmetry of S so that the key of the question is why the Newton law link the stress tensor to the simmetric part of the gradient. To answer, one needs to recall the conservation of the angular momentum for a fluids that leads to the symmetry condition of the stress tensor. That should answer why we have the mutual action of the variation of u and v, along y and x, respectively.