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What is bulk viscosity and how does it affect the flow?

Explain the idea of introducing such a term in the Navier-Stokes equation.

What are the consequences if not taken into account?

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closed as off-topic by John Rennie, Emilio Pisanty, ja72, Waffle's Crazy Peanut, Chris White Oct 9 '13 at 4:41

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This is an excellent question and requires more discussion. Therefore, my answer will also have questions in it for others to weigh in.

Bird and Stewart explain this very well in their Transport Phenomena book. In its general form, the viscous stresses may be linear combinations of all the velocity gradients in the fluid: $$\tau_{ij} = \sum_k \sum_l \mu_{ijkl} \frac{\partial v_k}{\partial x_l} $$where $i, j, k$, and $l$ may be 1,2,3. If you observe the equation above, there are 81 quantities $\mu_{ijkl}$ which can be referred to as "viscosity coefficients."

Here is where they start their assumptions.

We do not expect any viscous forces to be present, if the fluid is in a state of pure rotation. This requirement leads to the necessity that $\tau_{ij}$ be a symmetric combination of the velocity gradients. By this we mean that if $i$ and $j$ are interchanged, the combination of velocity gradients remains unchanged. It can be shown that the only symmetric linear combinations of velocity gradients are $$(\frac{\partial v_j}{\partial x_i} + \frac{\partial v_i}{\partial x_j})\&(\frac{\partial v_x}{\partial x} + \frac{\partial v_y}{\partial y} + \frac{\partial v_z}{\partial z}) \delta_{ij}$$

Can this be shown? I have read that the lack of microscopic surface moments ensures that the stress tensor is a symmetric one but I don't quite understand this point.

If the fluid is isotropic-that is, it has no preferred direction-then the coefficients in front of the two expressions above must be scalars so that $$ \tau_{ij} = A(\frac{\partial v_j}{\partial x_i} + \frac{\partial v_i}{\partial x_j}) + B(\frac{\partial v_x}{\partial x} + \frac{\partial v_y}{\partial y} + \frac{\partial v_z}{\partial z}) \delta_{ij}$$

So you can see that the number of "viscosity coefficients" from 81 to 2

Finally, by common agreement among most fluid dynamicists the scalar constant $B$ is set equal to $\frac{2}{3} \mu - \kappa$, where $\kappa$ is called the dilatational viscosity and $B$ is the bulk viscosity or the second coefficient of viscosnity. The reason for writing B in this way is that it is known from kinetic theory that K is identically zero for monatomic gases at low density.

For me this is not a sufficient explanation.I have also seen this refereed to as Stokes hypothesis (which is based on the fact that the thermodynamic pressure of a fluid is equal to its mechanical pressure). enter image description here

I think this needs to be further explored. It is also compound by the fact that it is generally not easy to measure this value experimentally.In addition, the equations of continuum mechanics do not require any fixed relationship between the two coefficients of viscosity.

what are the consequences if not taken into account.

The precise value of the second coefficient of viscosity is not needed for inviscid flows (both $\mu$ and $\kappa$ are assumed zero), for incompressible flows, or when the boundary layer approximations are invoked (normal viscous stresses << shear stresses). Bulk viscosity introduces damping associated with volumetric straining. Its purpose is to improve the modeling of high-speed dynamic events.

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