
#91
Jan2112, 12:58 AM

P: 1,438

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
[tex]\phi(r,\theta) = U_{\infty}\left(r+\frac{R^2}{r}\right)\cos\theta[/tex] Differentiating to get velocities [tex]v_{r} = U_{\infty}\left(1  \frac{R^2}{r^2}\right)\cos\theta[/tex] [tex]v_{\theta} = U_{\infty}\left(1 + \frac{R^2}{r^2}\right)\sin\theta[/tex] At the surface where [itex]r = R[/itex], [itex]v_r = 0[/itex] [tex]v_{\theta} = 2U_{\infty}\sin\theta[/tex] Clearly, that is not uniform for all [itex]\theta[/itex]. The velocity is symmetric about the [itex]x[/itex]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 [itex]2U_{\infty}\hat{\imath}[/itex] 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. 



#92
Jan2112, 01:17 PM

P: 40

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



#93
Jan2112, 03:13 PM

P: 1,438

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




#94
Jan2112, 04:08 PM

P: 40

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 



#95
Jan2112, 06:28 PM

P: 362

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. 



#96
Jan2112, 07:54 PM

P: 1,438

Keep in mind I said speed. If you look at it, at the stagnation points, [itex]V = U_{\infty}\hat{\imath}[/itex], but at the top and bottom points, [itex]V = U_{\infty}\hat{\imath}[/itex]. 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 [itex]V = 0[/itex], points where [itex]V = \frac{1}{2}U_{\infty}\hat{\imath}[/itex] and any other concoction [itex]V = nU_{\infty}\hat{\imath}[/itex] where [itex]1\le n\le1[/itex] Velocity is absolutely not constant along the circumference in any frame or reference. 



#97
Jan2112, 09:51 PM

P: 40

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. 



#99
Jan2112, 10:32 PM

P: 40

[QUOTE=boneh3ad;3721414]at the top and bottom points, [itex]V = U_{\infty}\hat{\imath}[/itex]. 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!!! 



#100
Jan2112, 11:25 PM

P: 1,438

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. 



#101
Jan2212, 12:20 AM

P: 40

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. 



#102
Jan2212, 10:21 AM

P: 1,438

Earlier, I talked about potentials. The equation Stan keeps quoting is the same equation as mine for [itex]v_{\theta}[/itex], only you have the angle defined differently. My angle was defined in a typical coordinate system with 0 being on the right at the xaxis and increasing counterclockwise. 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 [itex]v_{\theta}[/itex] and [itex]v_{r}[/itex] and plug them into [tex]u = v_r\cos\theta  v_{\theta}\sin\theta[/tex] [tex]u = v_r\sin\theta + v_{\theta}\cos\theta[/tex] Since we already determined that on the surface, [itex]v_{r}=0[/itex], this simplifies to [tex]u = 2U_{\infty}\sin^2\theta[/tex] [tex]v = 2U_{\infty}\sin\theta\cos\theta[/tex] 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: [tex](U_{\infty},0)[/tex] Leaving us with [tex]u = 2U_{\infty}\sin^2\thetaU_{\infty}[/tex] [tex]v = 2U_{\infty}\sin\theta\cos\theta[/tex] 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 [tex]v_r = u\cos\theta + v\sin\theta[/tex] [tex]v_{\theta} = u\sin\theta + v\cos\theta[/tex] This leaves us with (and you can do the algebra if you want) [tex]v_r = U_{\infty}\cos\theta[/tex] [tex]v_{\theta} = U_{\infty}\sin\theta[/tex] Taking the magnitude of this, it is easy to then show that [tex]\bar{V}=U_{\infty}[/tex] So yes, indeed, the fluid parcel is moving with a constant speed of [itex]U_{\infty}[/itex] 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. 



#103
Jan2212, 12:02 PM

P: 40





#104
Jan2212, 06:49 PM

P: 1,438





#105
Jan2312, 05:16 PM

P: 40

The largest percentage of the hundreds of articles and sites firmly define the concept that pressure changes are created by linear variations in the flow tangent and relative to the surface. (in the sense of the venturi) There is additional group that explains that the velocity changes are in response to the pressure variations. The primary flow is from displacement. The source/sink pattern describes the motion effects throughout the entire flowfield. These instantaneous velocity vectors are relative to the remote (still) fluid. When we add the fwd motion of the cylinder we can find the tangent velocity relative to the surface. This tangent velocity, in combination with the surface radius produces the normal acceleration, v^2/R, creating a pressure gradient that must be integrated across the entire flowfield. It is almost luck that all of this can be reduced again to the Bernoulli Equation. We find that we can still apply the Bernoulli Equation to the tangent velocity, relative to the surface. But is this the Bernoulli Principle?? 



#106
Feb1212, 10:16 PM

P: 40

[QUOTE=boneh3ad;3722110] Additionally, your site is quote difficult to follow. [QUOTE]
boneh3ad It appeares that I have a real problem here. If you (or anyone else) have any suggestions my "e" address is on the site. http://svbutchart.com 


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