External Flow-compressible flow solver-lift/drag
 

Hello,

I would like to follow up to the question dealing with the external flow computations-lift and drag.

(1) If I use a compressible flow solver (density based) to obtain a steady-state solution, lift, drag, etc., around the wing flying at Mach=0.2 can I 'scale' this solution (velocity, Re number, etc. scaling) to obtain lift and drag for this wing at Mach=0.01? (2) If yes how do I apply the scaling? (3) If yes can I use the same compressible code with preconditioning method to obtain solution at Mach=0.01 and prove that the scaling works?

Thanks, Tom

Re: External Flow-compressible flow solver-lift/dr
 

(1). The lift and drag coefficients in this range (incompressible flow) are not functions of Mach number,so you can just use it through the definition. (2). The Reynolds number will always affect the results, whether it is laminar, transitional or turbulent. You will get different answers as a function of Reynolds number. (3). From Mach number point of view, the flow at Mach 0.2 is the same as the flow at Mach 0.01, both are incompressible flows.(even if the density is not strictly constant, the effect is negligible.) (4). I have no idea about the code you mentioned, so there is no way I will be able to answer the code related questions. Normally, a density-based method will have convergence problem and oscillations in the solution, at very low Mach numbers.

Re: External Flow-compressible flow solver-lift/dr
 

Hi,John,

Excuse me,I donot know what's your mean,you said that the coef. isn't the function of Mach number,but on other hand, you also said the the coef.is the function of Re number for incompressible flow again,we know,the Re number is the linear function of velocity,and the mach number is directly connected with the velocity,why?

Pendy

Re: External Flow-compressible flow solver-lift/dr
 

Since in case of incompressible flow "pressure" is not a thermodynamic variable, "MACH" no. is not a right nondimensional parameter to characterize the flow. Up to Mach no. 0.3 Reynolds no. is the one which characterizes the flow. That is what John mean by saying that "from Mach no. point of view flow is same at Mach no. 0.001 and at 0.2". Hope this will clarify the doubt.

GS

Re: External Flow-compressible flow solver-lift/dr
 

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Re: External Flow-compressible flow solver-lift/dr
 

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Re: External Flow-compressible flow solver-lift/dr
 

I agree with everything said so far. But if I have a CFD solution (compressible flow solver) to a flow around a wing traveling at v=68 m/sec (~Mach=0.2) ~Re=6.8e6, what can I say, if anything, about this wing traveling at v1=10m/sec and Re=50,000? How can I get a new solution (or an approximation) at v1=10m/sec and Re=50,000 without having access to an incompressible flow solver?

Thanks, Tom.

Re: External Flow-compressible flow solver-lift/dr
 

(1). You will have to read the text book and accept it first. This will let you to develop the feeling related to the non-dimensional parameters. (2). First of all, the skin friction coefficient is normally expressed as a function of Reynolds number. You can pick up any fluid mechanics book and search for the chart. You need to accept it as a fact, that is, it is based on the experiment. And these charts are good for low Mach number range, or incompressible flows. Therefore, you don't see the Mach number in it as a separate parameter. (3). But as the Mach number starts to increase, the skin friction coefficient also starts to change. This is especially the case for transonic and supersonic flows. That is also from the testing data. (4). In your case, the Mach number effect can be neglected in low speed flow (low Mach number). Then the skin friction coefficient will be a function of the Reynolds number only. Now, for the pipe flows, or flow over a flat plate, the skin friction coefficient formula (as a function of Reynolds number) is available (based on the previous test data), so, you can just use the chart to find the value. (5). Obviously, in your case, you will have to compute the new case, because you don't have the formula as a function of Reynolds number, unless it is a pipe flow or flat plate flow. (6). You could assume that your wing behaves like a flat plate, and use the flat plate formula to account for the shift. (it is normally called, the Reynolds number correction.) The lift is probably less sensitive to the Reynolds number, so, you can forget about it for now.