How Does Viscosity Influence Pressure Drag in 2D Airflow?

AI Thread Summary
Viscosity significantly influences pressure drag in 2D airflow, particularly around airfoils. In inviscid flow, as described by D'Alembert's paradox, there is no drag due to symmetric pressure distribution; however, real-world scenarios involve boundary layers that disrupt this symmetry. The presence of viscosity creates a boundary layer that affects the flow separation and wake formation, leading to pressure differences and drag. When analyzing airfoils at higher angles of attack, the Kutta condition is applied to determine the stagnation point, allowing for the calculation of pressure drag while ignoring friction drag. Ultimately, fluid viscosity alters the net pressure difference between the front and back of an object, impacting overall drag.
Robin04
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Hi!

I'm a bit confused about the source of drag in a 2D flow (actually I'm talking about air). I heard that if the viscosity of a fluid is 0 then there is no drag at all (no pressure and no friction drag). I think I understand why viscosity affects friction drag, but why does it have an effect on pressure drag? For example if we look at an airfoil, at the leading edge near the stagnation point there is a little overpressure because the air slows down near that point and this causes the pressure drag. I don't understand why this overpressure depends on the viscosity of the fluid.

Thank you in advance!
 
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What you are describing is called D'Alembert's paradox. Basically, in potential flow (inviscid flow), the around a cylinder (a circle in 2-D) produces no drag. The pressure at the surface is symmetric in the front and back. Any shape that can be conformally mapped to a circle will give the same answer (so just about any shape you dream up). Of course, we know from measurements that the flow around a cylinder that there is drag. The reason is due to the presence of the boundary layer, which separates and leads to a wake, and the pressure is no longer symmetric around the circle, leading to drag. However, if you ignore viscosity, there is no boundary layer, thus no wake, thus no pressure difference.

Now, when it comes to airfoils, the situation is a little more fun. Clearly you can calculate lift and nonzero (though inaccurate) drag from inviscid theory, but D'Alembert's paradox would seem to imply that it should be zero. When an airfoil moves through a viscous fluid, the boundary layer on the top and bottom typically remains attached along the length and only separates when it meets the sharp trailing edge (or a trailing edge with a sufficiently abrupt corner). If you put the same shape at angle of attack in an inviscid flow, the stagnation point at the trailing edge would be somewhere upstream of that corner on the upper surface and you would measure zero drag. This is overcome by mathematically enforcing the stagnation point at the actual trailing edge (as viscosity would dictate), which is known as the Kutta condition. That way, you can run an inviscid simulation but simulate that one aspect of viscosity in order to get a nonzero answer.

That said, calculating drag on most objects is still rather innaccurate.
 
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Thank you very much for your detailed explanation.

So if I understand it right, in the case of analysing an airfoil at a higher angle of attack in an inviscid flow I get only pressure drag by putting the stagnation point to the trailing edge, and no friction drag due to 0 viscosity?
 
Pressure drag is like the difference in net force between the front and the back.

Fluid that flows around quickly and without resistance makes this force smaller.

Fluid that flows around slowly leaves a smaller pressure behind the object creating a net pressure difference between front and back.
 
Robin04 said:
Thank you very much for your detailed explanation.

So if I understand it right, in the case of analysing an airfoil at a higher angle of attack in an inviscid flow I get only pressure drag by putting the stagnation point to the trailing edge, and no friction drag due to 0 viscosity?

Correct.
 
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