# Clarification on the Lid-Driven Cavity Problem in CFD

I need some clarifications on The Lid-Driven Cavity Problem.

What does it actually mean? I know cavities are bubbles created when a fluid moves through liquid in low pressure zones, but what does the lid-driven cavity actually mean in context? Are we saying we take the bubbles to be of a square cavity? And also, that the lid is the upper part of the square, where the other three sides are no-slip conditions, correct?

I also need more clarifications on:

initial conditions

boundary conditions

free slip and no-slip boundary conditions

• You should ask another question about non-dimensionalizing the equations and be specific in it about what you need clarification on. It is not directly related to the LDC case and should be its own question. – tpg2114 Feb 19 '19 at 23:05
• Ok will do , gonna edit it out for now – Matias Salgo Feb 19 '19 at 23:07
• The cavity means an empty cavity in (2D) space, not the phenomenon of cavitation of liquids that occurs in multiphase flows. Think of fluid flowing over a surface with some uniform speed, and it meeting a square-shaped ditch at some point (not a very nice analogy, but it's the best I can think of for now). Your question about the initio-boundary conditions is answered here, but can also be deduced from the no-slip conditions. – GodotMisogi Feb 20 '19 at 2:26

The lid-driven cavity does not need to have cavitation (ie. bubbles forming in liquid). It can be a single fluid, which can be liquid or gas, and you will see the characteristic recirculation regions form in the corners. Depending on the Reynolds number, you may see a separation region on the top of the side wall opposite the direction of wall motion (as in the top of the left wall if the top wall is moving left to right). We often call these recirculation bubbles and separation bubbles, but that does not mean there is a gas-in-a-liquid kind of bubble.

Depending on the Reynolds number, the flow may be steady or unsteady. Reynolds numbers under 1000 will be steady and not show the separation region. At Reynolds numbers of 1000 to 5000, there will be a separation bubble that gets progressively stronger, and the corner recirculation bubbles may start to separate. Beyond Reynolds numbers of 3000 to 5000 or so, the flow will be unsteady and high enough Reynolds numbers will begin to transition to something like turbulence.

I spent quite a bit of time working this problem during my PhD thesis and the setup and results are in Chapter 5, Section 3 (and this was published in a journal as well). Check either source for references to the classic literature and datasets.

The initial conditions for this case do not matter really -- a bad choice will just take longer to reach the correct solution (unless it causes the code to blow up or something). Quiescent is just fine -- although if the wall Mach number is large, the sudden impulse could be numerically destabilizing.

In the classic setup, the boundary conditions are no slip walls everywhere, with the top wall moving. If your code is using the compressible form of the governing equations, then you will need to use isothermal boundary conditions for temperature unless you are running at a very low wall Mach number. This is not discussed in the classic papers because they use incompressible codes. But, when the velocity is coupled to the internal energy, the dissipation of the kinetic energy constantly injected at the top wall will continually increase the temperature in the cavity unless an energy sink is applied by keeping the wall temperatures fixed.

• Thank you for the information. As for the initial conditions, I understand that they do not matter too much, but if choosing them for an initial estimate, the variables in question would be only u,v and p? (assuming 2D) – Matias Salgo Feb 20 '19 at 0:22
• @tpg2114 I solved this problem by the finite element method and using the method of the false transient. I wrote code for Mathematica 11.3. Do you have data for comparison so that I can add to my collection on community.wolfram.com/groups/-/m/t/1433064? – Alex Trounev Feb 20 '19 at 1:01
• @MatiasSalgo It depends -- if you are solving the incompressible equations, then it should just be $u$ and $v$. The pressure can be determined from the velocity. If you are solving the compressible equations, then you need initial conditions for $\rho, u, v$ and $T$ or $p$. – tpg2114 Feb 20 '19 at 1:39
• @AlexTrounev Did you solve the lid driven cavity (moving wall) case? Or just the natural convection (heated wall) case? I've been doing some work on the heated wall case with preconditioning, but I haven't published it yet. I'd have to check if I still have the data files for the moving wall case in my thesis... – tpg2114 Feb 20 '19 at 1:41
• Yes, I wrote code for Mathematica 11.3 to solve the lid driven cavity problem. I can publish the code and some pictures. – Alex Trounev Feb 20 '19 at 4:42

I solved the lid driven cavity problem by the finite element method and using the method of the false transient. I wrote code for Mathematica 11.3. In fig. shows the magnitude of the velocity and the streamline of the incompressible flow at $$Re=1000$$.The initial data is set for a fluid at rest. No-slip conditions $$u=v=0$$ are set on the bottom, left, and right border, a constant speed $$u=1, v=0$$ is set on the upper wall.