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Inviscid Drag at subsonic, subcritical Mach # 

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November 18, 2001, 12:33 
Inviscid Drag at subsonic, subcritical Mach #

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This post is somewhat of a continuation of the earlier discussion on the Euler equations, in which I raised the question whether a subsonic compressible inviscid (Euler) flow over a smooth (2D) body like a cylinder should experience any drag. Below the subcritical Mach number (M < 0.5), the flow over a cylinder is shock free, whereas above M > 0.5, the flow accelerates to locally supersonic speed followed by a transonic shock wave. The popular opinion seems to be that the inviscid (Euler) flow over a cylinder below the subcritical Mach number (say M = 0.4) does not undergo any apparent increases in entropy (no shock waves, no boundary layer) and thus the computed drag should be zero. Any nonzero positive drag could only be the result of (heavy) numerical viscosity in the code.
As I stated earlier, I have computed 2D Euler flow over cylinders at subcritical Mach numbers using many different (inviscid) flow solvers, yet the drag is always positive in the compressible range, say M = 0.4, and only approaches zero in the incompressible limit, M = 0, independent of flow resolution. I was wondering if other people have had similar experiences, or if there has ever been a case where the drag was not a function of Mach number and was computed to be (nearly) zero in the subcritical range. In my opinion, it would not make any sense to get zero drag at M > 0, because as the Mach number increases, the flow field becomes more and more asymmetric. Surrounding pressure waves are shifted in the downstream direction, and eventually their fronts will coincide in a transonic shock wave. I believe that zero drag can only be obtained in the incompressible limit (M = 0) where the flow field has perfect symmetry. This case is equivalent to a potential flow solution. Let's say that one accepts the numerical results and the fact that the drag is indeed positive. This raises the question about the entropy generating mechanism, which is not apparent in this type of flow. Classical theory states that in the absence of viscous effects (boundary layers and shocks) a flow should be isentropic (constant entropy). Many textbooks on CFD do not predict any inviscid drag, unless numerical viscosity plays a major role. Here is how I would  intuitively  explain the phenomenon: Entropy, and its increase, is directly tied to how information propagates through a flow field, whether viscous or inviscid. Whenever entropy rises in a flow, you have a loss of information, or the flow of information is disrupted, for example by a boundary layer or a shock wave. Shock waves are the ultimate information destroying mechanism; if you are located behind one, you have no way of telling what goes on in the flow field ahead of you. Similarly, it is difficult to derive any information about an upstream flow field, if you are sitting in a viscous wake. If the wake is turbulent, you are literally surrounded by 'noise'. Although the above entropy generating mechanisms are clearly identifiable, I believe that this 'information' concept can be applied to inviscid shock free (Euler) flow as well. In order to obtain a good numerical solution to an inviscid flow, one has to pay close attention to the eigensystem of the Euler equations. The Euler equations are a set of wave equations. Each wave carries a different signal, one of them being entropy. Different signals travel at different speeds, which are determined by the eigenvalues. The eigenvectors essentially point in the direction of the strongest signal and thus direct the flow of information in an inviscid flow field. At very low Mach numbers (M ~ 0) these eigenvectors point equally in all directions, and any disturbance in the flow field is transmitted throughout the flow almost 'instantly', since the speed of sound is very large compared to the flow speed. Virtually no information is 'lost' in the process, there is no bias in transmission either, and thus the resulting flow is isentropic and dragfree  both from a 'classical' and numerical perspective. As the flow Mach number increases, the eigenvectors shift, and the flow of information obtains a downstream bias. Although information is not necessarily 'lost', there are regions within the flow where the flow of information is abundant, e.g. at a forward stagnation point on a cylinder, and other regions where the flow of information is sparse, e.g. at the rearward stagnation point on a cylinder. It is my opinion that this imbalance of information has to result in an increase in entropy, regardless of its mechanism. And such is the case, at least in all the inviscid computations that I have run over the years, using many different solvers. One may call this entropy and drag generating mechanism 'numerical viscosity', but I believe it is not the kind that approaches zero as the mesh is infinitely refined. As I stated before, the inviscid drag that I experienced seems to be resolution independent, and this would make perfect sense according to my 'theory'. If someone has a better theory that would explain inviscid drag at subsonic subcritical Mach numbers, or if someone thinks that I am "way off" in my computations, I would be happy for any feedback or criticism. Now that I have written almost a 'book' on the subject, I may not be able to reply to each and every response due to time constraints. Thank you for your attention! Axel Rohde 

November 19, 2001, 13:19 
Re: Inviscid Drag at subsonic, subcritical Mach #

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Hi Axel,
If I understand your problem correctly, you are talking about the flow 'around' a cylinder. I am not sure what your code takes into account and what it does not, however it seems clear that the 'impact parameter' of the cylinder is nonzero. Which means that there is actually a colision between the cylinder and the flow. Even if there is no viscosity in the problem, there is a net force acting on the cylinder at all times and the same and opposite force is applied to the 'mass of liquid' (flow). You are solving in the frame of reference where your cylinder is at rest, I guess. Patrick 

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