# Can the flow be irrotational if the viscous forces act on fluid?

I tried to answer the question only using the definitions and the Navier-Stokes equation:

$$\rho \frac{Dv}{Dt} = -\nabla P +\rho g -\mu[\nabla \times(\nabla \times v)]$$

In my opinion if the vorticity is zero, then the fluid is irrotational, regardless of presence of the viscous forces, thus $$\mu$$ can have a non-zero value which implies the existence of viskeuze forces, while the $$\nabla \times v = 0$$.

• For the future readers, all the three answers are correct here, but I choose the one which makes more sense to me. – Sam B Aug 10 at 19:43

The answer to the title question is yes, it is possible for the flow to be irrotational if there are nonzero viscous forces acting on fluid.

As an example, let us consider a simple yet physically meaningful example of such flow: radial, spherically symmetric flow in a viscous incompressible fluid (without gravity). A physical realization for such flow is a bubble of gas expanding into the space filled with viscous fluid.

Assuming that the fluid velocity is purely radial $$\mathbf{v}= \hat{\mathbf r}\,v_r(r,t)$$, the continuity equation gives us: $$\mathbf{v}= \frac{\hat{\mathbf r} f(t)}{r^2}.$$ This flow is irrotational and Navier–Stokes equation (with viscous term identically zero) could be solved for the pressure.

So, while viscosity does not enter the equations, the viscous stress tensor is nonzero, and there would be energy dissipation within the fluid volume. Consequently, viscosity would enter the solutions either through energy balance equation or through boundary conditions. For example, at the fluid–gas interface of above mentioned spherical bubble we would have (ignoring gas viscosity and surface tension): $$- p_\text{gas} = -p_\text{fluid}+2 \mu \frac{\partial v_r}{\partial r}$$

• I'm not sure I fully buy the assumption. If you assume the existence of viscosity, then the symmetric rotational flow cannot be at steady-state without an input torque. The torque is supposed to inject energy dissipated by shear stresses. – JZYL Aug 9 at 13:30
• @Jimmy: I hadn't said anything about steady state. And my example is definitely nonstationary. For example, if we consider expansion of gas bubble viscous friction would provide dampening effect by dissipating energy, whereas without viscosity there would be either oscillations or unlimited (but ever slowing) expansion. – A.V.S. Aug 9 at 14:59
• In that case, why would there be bulk viscous damping without shear viscous damping? If you assume incompressibility, then bulk viscous damping should vanish as well (divergence is zero). – JZYL Aug 9 at 15:22
• @Jimmy: what do you mean why? Bulk dissipation power is $\epsilon =\frac {\mu}{2} (\partial_i v_j+\partial_j v_i)^2$. Only vorticity is zero here, square of shear rate tensor is obviously nonzero. – A.V.S. Aug 9 at 15:31
• … and incompressibility means that the trace of the tensor is zero, it says nothing about the square. – A.V.S. Aug 9 at 15:35

I'm not sure your momentum equation is correct. For the x-momentum, we should have:

$$\rho\frac{Du}{Dt}=-\frac{\partial \rho}{\partial x}+\rho g_x+\frac{\partial \tau_{xx}}{\partial x}+\frac{\partial \tau_{yx}}{\partial y}+\frac{\partial \tau_{zx}}{\partial z}$$

For Newtonian fluids,

$$\tau_{xx}=-\frac{2}{3}\mu\nabla \cdot \textbf{V}+2\mu\frac{\partial u}{\partial x}$$ $$\tau_{yx}=\mu(\frac{\partial v}{\partial x}+\frac{\partial u}{\partial y})$$ $$\tau_{zx}=\mu(\frac{\partial u}{\partial z}+\frac{\partial w}{\partial x})$$

If you assume irrotational flow and require the curl to vanish, then the only term left that is related to viscosity is proportional to $$\nabla^2u$$. Shear stress is gone and you would be right, mathematically.

But this is a highly a-physical assumption. Viscosity would result in shear-stress; so irrotationality assumption with a viscous flow isn't a good one.

• Thank you for your answer. I wrote the exact same equation which I found in my book, you can find a screenshot here imgur.com/a/wEPhd60 – Sam B Aug 8 at 15:22
• @SamB This is N-S for incompressible flow. Divergence of the velocity field is zero in this case. Then yes, irrotationality and inviscid are equivalent in this case. – JZYL Aug 9 at 15:23

First, unlike the other answers, I believe your equations are correct thanks to the identity $$\nabla \times \nabla \times \mathbf v = \nabla (\nabla \cdot \mathbf v)-\nabla^2 \mathbf v$$ Irrotational flow means $$\nabla \times \mathbf v=0$$ so in fact as you note correctly Navier-Stokes predicts viscosity plays no role in irrotational flow.

The question of whether the flow is irrotational is answered using the vorticity equation, which predicts the evolution of the vorticity $$\mathbf \omega = \nabla \times \mathbf v$$: $$\frac{D\mathbf \omega}{Dt}= (\mathbf \omega \cdot \nabla) \mathbf u+\nu \nabla^2 \mathbf \omega$$ In many situations the flow is irrotational if and only if $$\mathbf \omega =0$$ (but there are exceptions). Therefore we see that if $$\nu=0$$ (no viscosity) and the vorticity is initially zero, then $$\mathbf \omega =0$$ at all times (this is because $$\mathbf \omega =0$$ is a solution and the solution is unique). However note that if $$\nu \neq 0$$ then in general vorticity will develop even if there was none initially. Of course you can still impose the requirement that the flow is irrotational if you so wish.

As a final remark, I'd like to point out that even though in many cases one may assume that the flow is irrotational, then one may run into issues: for instance you need viscosity to impose the no-slip condition on a boundary. In these situations one assumes that viscosity is relevant near the boundary and the flow is irrotational elsewhere.

• Shouldn't there be a bulk viscosity term in there? I was assuming bulk viscosity is -2/3 of dynamic viscosity. – JZYL Aug 8 at 22:02
• I believe the equations I wrote down are correct. Where do you believe there should be such a term? – John Donne Aug 9 at 5:46
• Take a look at this. This is consistent with the answer given by A.V.S., that there is a bulk dissipation term. Although I'm not certain his conclusion is correct (i.e. if you add viscosity, the rotation cannot be at steady-state without torque). – JZYL Aug 9 at 13:28
• But then the flow you're interested in is not described by Navier-Stokes, and the equation you wrote down is incorrect – John Donne Aug 9 at 16:46
• This is copied verbatim from Fundamentals of Aerodynamics by Anderson. – JZYL Aug 9 at 16:52