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Yap Wen Jiun March 9, 2000 23:04

Help:lift and drag
For a viscous incompressible flow past an airfoil, using the x-momentum equation and a suitable control volume the drag and lift can be calculated using the the law that the force acted on the airfoil by the control volume equal to the force acted on the control volume by the airfoil. Then taking the conservation of momentum of either the x- and y-direction, the drag which is the force in x-direction (the inlet velocity is in x-direction) can be formulated as D=Sum (pressure force + density * velocity^2) at inlet - sum (pressure force + density * velocity^2) at outlet and lift, L is formulated with the almost same formulation. As this formula is directly derived from the Navier-Stoke equation for viscous, newtonian flow, it should be valid (a little error tolerance) for whatever control volume should be as long as the airfoil is surrounded by the control volume. Am I right ? But I found different answer using this formula for both the drag and lift for different control volume in PHOENICS. Is this formula a very cruel approximation ? What is its limitation ? If so, why is this formula can be found in almost all the books for aerodynamics?

I can give more details about how I use this formula (in PHOENICS) if needed. So please help me.

Andrea Panizza March 10, 2000 05:52

Re: Help:lift and drag
I can see three possible errors in your reasoning: 1) if upper and lower surface of your control volume are not stream-surfaces, then you are forgetting to take into account the momentum flux across them. The y-component of V transports x-momentum across upper and lower surfaces, so if you forget to include this in the balance, then you are not correctly calculating drag. 2)If that is not the case, then note that you are not taking into account the contribution of viscous stresses T to momentum flux. The reason why you do not find them in the formulas on your textbooks is that for Newtonian fluids is easy to show that the surface integral of T tends to zero as the control volume increases indefinitely in size. So, when using small control volumes, taking into account viscous stresses may solve your troubles. 3)Lastly, problems may arise if you are studying flow in ducts (e.g. wing in windtunnel). In this case, obviously you cannot choose a control volume of arbitrary size: also, if you choose a control volume that exactly fits the test chamber, you will need to take into account viscous stresses, because on the walls of the chamber you have boundary layer. Best choice would be to use big control volumes that end a little before the walls , so for high Re you may safely neglect the contribution from T.

I hope any of those helps: I think that if you decide to always neglect T, it is reasonable to expect a certain dependence of D and L from the control volume, but expecially for Re going to infinity, this dependence, assuming control surfaces are all far away from the wing and do not enter boundary layers on the walls, shold become extremely weak. NACA used your formula for a long time and , as far as I know, with very good results.

Yap Wen Jiun March 10, 2000 07:48

Re: Help:lift and drag
The upper and lower surface I selected is not very far from the airfoil, less than that of the airfoil chord length. The chord length is 4.5m and the distance between these 2 surfaces is 4.5m and the airfoil is situated in the middle. So they are probably not stream surfaces. But how can the Y-component of the V transport a "X-momentum" ? This problem should affect Lift but not Drag, right ?

I haven't study the viscous stresses in very detail. But what I wanted to argue is that the viscous stresses is still a kind of momentum loss, so the difference between the upstream and the downstream momentum, which is the momentum loss should already include this effect ?? And if it is very important and necessary to include this effect, can you suggest how can I modify my formula to include them ? Thanks

John C. Chien March 10, 2000 11:24

Re: Help:lift and drag
(1). Remove the airfoil, and compute the flow field, the check the lift and the drag. you should get zero lift and zero drag. (2). Put in a finite flat plate at zero angle of attack, compute the flow field and check the lift and the drag. You should get zero lift and finite drag. Check the drag against text book data. (3). If you can pass these two validation tests, then repeat the airfoil case at zero angle of attack first, and check the lift and the drag. (4). In item-1 and item-2, you should vary the size of control volume to verify the results.

Andrea Panizza March 10, 2000 18:33

Re: Help:lift and drag
I too believe that Su and Sl, upper and lower surfaces of your control volume, are not stream surfaces. Now suppose Su is plane and horizontal: the versor normal to Su is y (I mean the vertical versor: if yours is a 3D problem, then this would be z). So, when you calculate the flow of rho*u (the x-component of momentum rho*V) across Su, you write sum(rho*u*<V.y>)=sum(rho*u*v) ( < . > = dot product). In other words, as V crosses the horizontal surface Su, it transports the x-momentum rho*u . Think of it this way: at outlet you have a smaller momentum flux than at inlet, so in time dt mass exiting from outlet is less than mass entering from inlet. This mass, with associated momentum, must exit from somewhere since you are studying incompressible flow, so it exits across Su and Sl. I would suggest that you include momentum flow Q across Su and Sl in your calculations, before including viscous stresses T: in my experience the error which comes from not including Q is bigger than the error which comes from neglecting T. For example, if Su and Sl are plane and horizontal, the terms you need to add to your formula for drag D are

-sum(rho*u*v) at Su + sum(rho*u*v) at Sl.

Good luck.

adam April 2, 2000 17:13

Re: Help:lift and drag
Going for the world record bicycle length jump once again. Need help choosing best angle for launch ramp, and help figuring out the effects of wind drag on me at different speeds. Would like to have an prediction of distance at different speeds.

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