## Fluid mechanics, why does the air flow faster over the wing?

 Quote by boneh3ad Either way, you said that the particle slows down in the case of an airfoil. That isn't true. What you just said in this post is exactly what my previous post said.
Yah, I realised tha halfway through my post but it was too late to stop!

But tell me, it is moving at 1V at the stag point and .2V at the top ( all relative to the remote air.) How is that not slowing down?

 Well, -0.2V. It is slowing down in that frame and the speeding back up until it reaches that -0.2V, but in your earlier post you said it sped up over te cylinder, which does not happen in that frame.

 Quote by boneh3ad in your earlier post you said it sped up over te cylinder, which does not happen in that frame.
I certainly hope that it did not come across that way. If it did I apologise.

The cylinder is constant speed, stag point to stag point.

 No it isn't. It is qualitatively the same as the airfoil. Both stagnation points are obviously moving at the same speed as the body and at the top/bottom the flow is moving at twice that speed. It is not constant.

 Quote by boneh3ad No it isn't. It is qualitatively the same as the airfoil. Both stagnation points are obviously moving at the same speed as the body and at the top/bottom the flow is moving at twice that speed. It is not constant.
Oohh yes it is! This is all in friendship but communication is difficult.

I am looking at the inertial path relative to the remote still air only. The flowpath is as a lower case script "e".
Look up the source/sink pattern for the cylinder. While it is an instantaneous picture, all of the surface intercepts are at 1V. The "e" is an integration of those intercepts.

 Listen, the potential flow around a cylinder shows that the the velocity can be described by the potential function of a doublet and a free stream, which give you $$\phi(r,\theta) = U_{\infty}\left(r+\frac{R^2}{r}\right)\cos\theta$$ Differentiating to get velocities $$v_{r} = U_{\infty}\left(1 - \frac{R^2}{r^2}\right)\cos\theta$$ $$v_{\theta} = -U_{\infty}\left(1 + \frac{R^2}{r^2}\right)\sin\theta$$ At the surface where $r = R$, $v_r = 0$ $$v_{\theta} = -2U_{\infty}\sin\theta$$ Clearly, that is not uniform for all $\theta$. The velocity is symmetric about the $x$-axis, but it is not uniform along the surface. This is in the frame of reference of the cylinder, by the way. If you look at it, that means at the stagnation points, the velocity in this frame is zero and at the top and bottom, it is $2U_{\infty}\hat{\imath}$ in Cartesian coordinates. Between these points along the surface the velocity varies smoothly. It is not a constant on the surface. You can find that canonical example in any aerodynamics book or other fluids book that includes potential flow.

 Quote by boneh3ad Clearly, that is not uniform for all $\theta$. The velocity is symmetric about the $x$-axis, but it is not uniform along the surface. This is in the frame of reference of the cylinder, by the way. If you look at it, that means at the stagnation points, the velocity in this frame is zero and at the top and bottom, it is $2U_{\infty}\hat{\imath}$ in Cartesian coordinates. Between these points along the surface the velocity varies smoothly. It is not a constant on the surface.
Absolutly right! However, I have been trying to get across the velocity relative to the remote field. The reason was to show that the Bernoulli Principle did not apply tangent to the surface. Converting the flow relative to the surface to relative to the field is a pretty simple process.

(As a uneducated tech designer, I configurated missles and tactical aircraft using the most basic equations. I never had time for real "rocket science".)

I have absolutely no idea what you are trying to get at then. Based on this quote

 Quote by Stan Butchart While it is an instantaneous picture, all of the surface intercepts are at 1V.
It leads me to believe you are saying the velocity is 1V everywhere on the surface. It isn't.

 Quote by boneh3ad It leads me to believe you are saying the velocity is 1V everywhere on the surface. It isn't.
Right, relative to the surface it is not. However, while it moves along the surface it is moving thru space at 1V because the cylinder is moving.

My reason to continue to pursue this is to try to understand better how to convey concepts. (Which I am failing!)

You might give http://svbutchart.com a shot

 Quote by Stan Butchart The reason was to show that the Bernoulli Principle did not apply tangent to the surface.
I am having trouble seeing what the point you are trying to make with all of this. What do you mean Bernoulli does not apply tangent to the surface?

In a real fluid the particles directly in contact with the surface moves with the surface because of viscosity. Is this what you are talking about? But the velocity of the fluid above the surface is certainly not moving at a constant velocity.

 Quote by Stan Butchart Right, relative to the surface it is not. However, while it moves along the surface it is moving thru space at 1V because the cylinder is moving. My reason to continue to pursue this is to try to understand better how to convey concepts. (Which I am failing!) You might give http://svbutchart.com a shot
But this just isn't true. In the frame where the object is moving, the only places where the flow is moving with speed $U_{\infty}$ are the two stagnation points and the points on top and bottom where $\theta = \pm 90^{\circ}$.

Keep in mind I said speed. If you look at it, at the stagnation points, $V = -U_{\infty}\hat{\imath}$, but at the top and bottom points, $V = U_{\infty}\hat{\imath}$. These are not the same quantities. Also note that in between these points along the circumference, the velocity varies between these two, meaning there are points where $V = 0$, points where $V = \frac{1}{2}U_{\infty}\hat{\imath}$ and any other concoction $V = nU_{\infty}\hat{\imath}$ where $-1\le n\le1$

Velocity is absolutely not constant along the circumference in any frame or reference.

 Quote by RandomGuy88 I am having trouble seeing what the point you are trying to make with all of this. What do you mean Bernoulli does not apply tangent to the surface? In a real fluid the particles directly in contact with the surface moves with the surface because of viscosity. Is this what you are talking about? But the velocity of the fluid above the surface is certainly not moving at a constant velocity.
Expanding on this, Stan originally specified an ideal fluid, meaning zero viscosity.

 Quote by RandomGuy88 I am having trouble seeing what the point you are trying to make with all of this. What do you mean Bernoulli does not apply tangent to the surface? In a real fluid the particles directly in contact with the surface moves with the surface because of viscosity. Is this what you are talking about? But the velocity of the fluid above the surface is certainly not moving at a constant velocity.
The point is to present an understanding to the hundreds that search these pages.

Bernoulli is based upon differences in the speed at different points in the inertial flow. Speed calculated related to different points on the surface does not reflect the actuual inertial flow. Bernoulli Princple does not fit.
However, the Bernoulli Equation with velocity relative to the curved surface does fit when paired with the normal acceleration of turning flow.

If we take the accepted tangential velocity and reconcile it with the cylinder velocity, the inertial velocity will be 1V at all points.

 Quote by Stan Butchart If we take the accepted tangential velocity and reconcile it with the cylinder velocity, the inertial velocity will be 1V at all points.
No it won't. This is an objective fact.

 [QUOTE=boneh3ad;3721414]at the top and bottom points, $V = U_{\infty}\hat{\imath}$. These are not the same quantities. QUOTE] Could you expand on that statement? Another way using the accepted equation for velocity relative to the surface. v=2sinA*V i run a radius line from the center out past the surface at angle A. At the intercept I place a 1V vector running fwd. I reflect the 1V vector accross the radius line. It defines 2 A from horizontal. The base of the 2A triangle is the velocity, tangent and relative to the cylinder. The reflected vector is the inertial vector of the flow relative to far field. It is always 1V. I am not sure why I am finding this concept so difficult!!!
 Because your concept is wrong. There is no frame of reference where the speed is 1V along the entire surface. And to expand on what you referenced, the speed at those points I mentioned is the same but the velocity is not. I was noting the difference.

 Quote by boneh3ad There is no frame of reference where the speed is 1V along the entire surface.
One last shot.
we have a fluid that is moving relative to the remote air. We have a body that is moving relative to the remote air. The body and moving air are moving relative to each other.

I mark the lip of a glass and place it upside down on the table and rotate it. (I should accelerate and decclerate it but that is a detaile.) Now as i rotate it as I move it accross the table. What has the mark traced on the table?

I have resolved it from the source/sink pattern. I have resolved it from the tangental velocity relative tothe surface and i have resolved it from the cylinder surface velocity equation.
I have graphed its path as described in http://svbutchart.com
I have never ever suggested that the constant 1V was relative to the cylinder.

 Quote by Stan Butchart One last shot. we have a fluid that is moving relative to the remote air. We have a body that is moving relative to the remote air. The body and moving air are moving relative to each other. I mark the lip of a glass and place it upside down on the table and rotate it. (I should accelerate and decclerate it but that is a detaile.) Now as i rotate it as I move it accross the table. What has the mark traced on the table? I have resolved it from the source/sink pattern. I have resolved it from the tangental velocity relative tothe surface and i have resolved it from the cylinder surface velocity equation. I have graphed its path as described in http://svbutchart.com I have never ever suggested that the constant 1V was relative to the cylinder.
You have to be careful with your choice of words. Velocity and speed are not the same thing. Going through the math, in the frame you are describing, the speed of a parcel of fluid is, in fact, going to be $U_{\infty}$ along the circumference. The velocity isn't, and is a vector quantity. Additionally, your site is quote difficult to follow. For anyone else on here that has been reading along and thought Stan's explanations were somewhat opaque, I go through the math showing this below.

Earlier, I talked about potentials. The equation Stan keeps quoting is the same equation as mine for $v_{\theta}$, only you have the angle defined differently. My angle was defined in a typical coordinate system with 0 being on the right at the x-axis and increasing counter-clockwise. Otherwise, the two are identical. I am starting from my equation since it will make things easier to change to a Cartesian coordinate system and change frames.

Now, if we want to switch into a Cartesian system, we take my results for $v_{\theta}$ and $v_{r}$ and plug them into
$$u = v_r\cos\theta - v_{\theta}\sin\theta$$
$$u = v_r\sin\theta + v_{\theta}\cos\theta$$

Since we already determined that on the surface, $v_{r}=0$, this simplifies to
$$u = 2U_{\infty}\sin^2\theta$$
$$v = -2U_{\infty}\sin\theta\cos\theta$$

This defines the vector field on the surface of the cylinder in Cartesian coordinates in the frame where we are riding along with the moving cylinder (so it has no velocity and the free stream is moving). Now to change reference frames such that we are viewing the cylinder moving through the stagnant fluid, we just have to add the following:
$$(-U_{\infty},0)$$

Leaving us with
$$u = 2U_{\infty}\sin^2\theta-U_{\infty}$$
$$v = -2U_{\infty}\sin\theta\cos\theta$$

This is the velocity vector along the surface in a Cartesian coordinate system in the inertial frame. The cylinder is moving through stagnant air. We can transform this back into cylindrical coordinates for simplicity using
$$v_r = u\cos\theta + v\sin\theta$$
$$v_{\theta} = -u\sin\theta + v\cos\theta$$

This leaves us with (and you can do the algebra if you want)
$$v_r = -U_{\infty}\cos\theta$$
$$v_{\theta} = -U_{\infty}\sin\theta$$

Taking the magnitude of this, it is easy to then show that
$$|\bar{V}|=U_{\infty}$$

So yes, indeed, the fluid parcel is moving with a constant speed of $U_{\infty}$ in an inertial frame. The problem is, this doesn't tell us a lot about the forces on the cylinder. The important quantity is the speed with respect to the cylinder.