Why does viscosity effect the normal component of stress in a Newtonian fluid?

Question

The constitutive law for a Newtonian fluid is

$$ \boldsymbol{\tau} = 2\mu \mathbf{D} + \lambda\left(\nabla\cdot\mathbf{v}\right)\mathbf{I}$$

where $\mu$ is the dynamic viscosity. Assuming we have a flow field that has a form

$$ \mathbf{v} = x\hat{\mathbf{i}} + 0\hat{\mathbf{j}} + 0\hat{\mathbf{k}}$$

then the normal stress in the $x$ direction is found to be

$$ \tau_{11} = 2\mu + \lambda$$

My question is: Why is dynamic viscosity affecting the normal stress at all? I am probably missing something really basic here but viscous forces come into picture only when there is movement between fluid layers and there is sliding. For the given velocity field, I cannot see any sliding of fluid layers that can cause viscosity to affect the stress. There is only stretching.

It may have something to do with the velocity field I chose and that it is not an admissible one. I am not sure if the validity of the constitutive law depends on whether the velocity field is admissible.

Nevertheless, how can viscosity affect normal components of stress?

Answer 1

The usual picture you see in wkipedia and other sources is indeed over-simplified. Viscosity resists more general velocity gradients, not just pure shear flows $\nabla_x v_y\neq 0$. For example, if you have a compressible fluid undergoing non-isotropic scaling expansion $$ \vec{v} = (\alpha x,\beta y,\gamma z) $$ then shear viscosity will try to equalize the expansion rates $\alpha\simeq \beta \simeq \gamma$. Bulk viscosity will resist the overall expansion of the fluid.

You can view you example $$ \vec{v} = (x,0,0) = \frac{1}{3} \left( (2x,-y,-z)+(x,y,z) \right) $$ as a linear combination of anisotropic shear flow, and pure expansion. Shear viscosity counteracts the first, and bulk viscosity the second term.

Answer 2

While @Thomas's answer does tackle the effects of general viscosity terms on normal stress, it doesn't discuss the effect that dynamic viscosity (or "shear" viscosity) has on the normal stress of the flow field. And this comes down to Mohr's law, where normal stresses can be "transformed" into shear stresses on a different plane. This is typically a mechanics of materials topic, but doesn't normally come up in fluids courses and explains dynamic viscosity impact on normal stresses via normal strain.

From Mohr's law, if you have a purely normal stress 2D stress tensor, orienting the axes 45 degrees will give a shear stress.

To give a mental image as to why this is, imagine you have two fluid cubes, one water and the other tar. If you try to deform both fluid cubes using only normal forces (imagine squishing/stretching them in one direction), which one will be easier? Clearly*, the water will be easier since it has significantly less viscosity than tar. If we constrain the "squishing" deformation to be incompressible, then the bulk viscosity will have no effect since:

$$\text{Incompressibility:} \quad \frac{\partial u_j}{\partial x_j} = 0$$

and

$$\text{Bulk Viscosity Stress:} \quad K \frac{\partial u_j}{\partial x_j}$$

Therefore, only the dynamic viscosity $\mu$ will have the effect.

Another "proof" from the previous exercise: If you were to perform the deformation mentioned above, energy would be expended to do the work of deformation. Since fluid continuously deforms under stress and we're assuming incompressible fluid, no mechanical energy will be stored by the fluid (in other words, it's not a spring. Air springs operate on principle of compressibility). If we set adiabatic boundary conditions on the fluid cube, that means that the energy must be going into heat. And what's the primary way fluids turn kinetic energy into heat? Viscous action.

_{*I normally hate statements like these in textbooks, but in this case I do hope that the result is fairly obvious.}