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Old Nov 29, 2012, 05:53 PM
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Originally Posted by Crossplot View Post
The objective is to be able to show that the sum of the pressure gradient from centripetal acceleration is the dynamic pressure value at a point on the surface and of equal value to that from the Bernoulli equation from velocity relative to the wing.

If v^2/r can be shown to reduce to v^2/2 then calculating pressure by Bernoulli from velocity relative to the wing makes sense. Otherwise, since relative to the wing does not reflect the actual fluid accelerations, it makes no sense al all.
Now I understand what you're getting at. I think the attached .pdf may answer your question. Thanks for sticking with this, I learned a lot by working through this...
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Old Nov 30, 2012, 03:20 PM
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Shoe

Thanks for the pdf. It will take me some time to get a physical feeling for the effect of some of the symboles.

I have been doing some crude "hand cranking" and found that "r/2' works all over the cylinder. I do not know yet if I have any starting point for the wing,

I am still suggesting that using the Bernoulli equation in traditional streamlines is actualy interigating perpendicular acceleration pressure rather than calculating Bernoulli Principle flow. The velocity changes that we nomally use do not exist in the actual inertial flow.
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Old Dec 02, 2012, 01:23 PM
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Originally Posted by Crossplot View Post
I have been doing some crude "hand cranking" and found that "r/2' works all over the cylinder.
Crossplot,

Multiplying the radial pressure gradient by "r/2" amounts to "bringing an algebra weapon to a calculus fight". Your approach works in this specific case, but only because multiplication by "r/2" transforms the expression for the radial pressure gradient into an expression for the dynamic pressure. For steady, irrotational, uniform-density flow, the increase in dynamic pressure from one location to another is equal to the decrease in (static) pressure.

Just because an analysis leads to an equation that accurately predicts the relationship between flow parameters (in this case the Bernoulli equation), that doesn't mean the analysis is correct. The are an unlimited number of ways you can arrive at the Bernoulli equation (monkeys with typewriters comes to mind), however, the reason the Bernoulli equation accurately predicts the relationship between velocity and pressure in a steady, irrotational, uniform-density flow is because it can be derived from conservation of momentum.

I have updated the .pdf in the post above to address the issues associated with analyzing the flow around an airfoil (or cylinder) from a reference frame where the airfoil is moving through otherwise still air.
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Old Dec 02, 2012, 01:44 PM
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This sub-topic has been much more interesting me that most of the other discussions that have taken place on this thread.
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Old Dec 03, 2012, 10:43 PM
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ShoeDLG -

The pdf update is great. I have not seen a like presentation on the subject before.

Eventully I may have a couple serious questions but with my limmited math skills I will probably miss some important subtle parts of the derivations.

(You would be surprized at the amount of simple algebra I used in years of configurating airplanes !)
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Old Dec 05, 2012, 05:35 PM
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Originally Posted by ShoeDLG View Post
The Bernoulli equation lets you relate the net change in pressure between two points on a streamline to the net change in flow velocity between those points. Again, the equation is derived by integrating all of the infinitesimal pressure changes along the path between the two points… there’s no getting around that..
Getting back to algebra and useing the cylinder for well documented simplicity.
Fluid texts all show the source/sink pattern and the simple formula for flow velocity relative to the surface. (ideal etc.) I reconsile this with the cylinder V and find the actual instantainious velocity relative to the remote still air. ( v is virtually never at right angles to v')

At every location between the stagnation areas the velocity of the flow next to the surface is one V in its source/sink direction. In the direction of its path there is no acceleration and no change in speed. The -V we see at the top of the cylinder indicates direction only and not a difference in speed.

All pressure change must then come from normal accelerations. The primary acceleration for the cylinder is normal to the true flowpath. Lift is determined by acceleration normal to the surface which is a component of that primary acceleration.
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Old Dec 06, 2012, 12:02 AM
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At every location between the stagnation areas the velocity of the flow next to the surface is one V in its source/sink direction. In the direction of its path there is no acceleration and no change in speed. The -V we see at the top of the cylinder indicates direction only and not a difference in speed.
I don't think this is the case. As you watch a cylinder pass by from left to right in the reference frame where the undisturbed air is still, the air at both the forward and aft stagnation points is moving to your right at speed V (the speed of the cylinder). The air at the top and bottom of the cylinder is moving to your left at speed V. Air is most certainly accelerating in the direction of the cylinder's travel. There is also air that is accelerating in the direction of its own instantaneous travel. Perhaps I am misunderstanding what you mean by: "In the direction of its path there is no acceleration and no change in speed".
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Old Dec 06, 2012, 01:39 PM
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As you watch a cylinder pass by from left to right in the reference frame where the undisturbed air is still, the air at both the forward and aft stagnation points is moving to your right at speed V (the speed of the cylinder). The air at the top and bottom of the cylinder is moving to your left at speed V. Air is most certainly accelerating in the direction of the cylinder's travel. There is also air that is accelerating in the direction of its own instantaneous travel. Perhaps I am misunderstanding what you mean by: "In the direction of its path there is no acceleration and no change in speed".
After accelerating to 1 V at the fwd stagnation point, the particle travels a 360 path like a cursive small "e" at 1 V to the rear stagnation point. When it reaches the rear stag point it will be a couple of diameters ahead of where it started at the fwd stag point. The ONLY accelerations have been the normal ones associated with turning.
Each "e" path is unique to only one element . The following element has its own "e" path. What appears as separating when relative to the surface is the difference in location on their own "e" with different vectors. That difference is made up by the thinning of the larger element as in the continuity law.
Bernoulli is only interested in the accelerations of individual elements.

Conditions at the stag points is somewhat imaginary as there is convex flow to turn onto an off of the cylinder.

The first part of http://svbutchart.com should help.
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Old Dec 06, 2012, 05:00 PM
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180°
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Old Dec 06, 2012, 06:39 PM
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180°
180 relative to the cylinder. 360 inertial path, relative to the remote still air.
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Old Dec 07, 2012, 03:10 AM
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Crossplot,

Still not sure I completely understand where you are trying to go with this. Some additional thoughts:

1. Even though the speed of a particle remains constant (equal to V) on the surface of the cylinder (viewed from the reference frame where the remote air is still), the pressure on the surface (the pressure experienced by the particle) does vary with location. You just need to apply the proper form of the Bernoulli equation to relate the pressure to the velocity (in this case the unsteady form of the Bernoulli equation). This is shown in the attached .pdf.

2. There’s nothing imaginary about the “convex” flow near the stagnation point. Just as you can show that crossing “concave” streamlines to get to the top of the cylinder means the pressure is lower than ambient there, you can also show that crossing “convex” streamlines means that the pressure at the stagnation point is higher than ambient there.
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Old Dec 07, 2012, 01:33 PM
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Originally Posted by ShoeDLG View Post
Crossplot,

Still not sure I completely understand where you are trying to go with this. Some additional thoughts:

1. Even though the speed of a particle remains constant (equal to V) on the surface of the cylinder (viewed from the reference frame where the remote air is still), the pressure on the surface (the pressure experienced by the particle) does vary with location. You just need to apply the proper form of the Bernoulli equation to relate the pressure to the velocity (in this case the unsteady form of the Bernoulli equation). This is shown in the attached .pdf.

2. There’s nothing imaginary about the “convex” flow near the stagnation point. Just as you can show that crossing “concave” streamlines to get to the top of the cylinder means the pressure is lower than ambient there, you can also show that crossing “convex” streamlines means that the pressure at the stagnation point is higher than ambient there.
My point all along is to show that Bernoulli can not apply to flow where there is no acceleration. (linear) The basic P(s) = P(t) - P(d) does apply.

1. Pressure changes with the changing normal acceleration as the curvature tightens then relaxes in the "e".
I am having trouble making the pdf work. It may just be just me and am needing a diagram!!

2. I was just trying to justify the use of "stagnation point to stagnation point".
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Old Dec 07, 2012, 02:09 PM
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Your train of thought is difficult to follow because you are mixing reference frames. I'm not sure what you mean by "Bernoulli can not apply", but the relationship:

total pressure = static pressure + dynamic pressure

does not apply in an unsteady flow. Are you having trouble opening the .pdf?
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Old Dec 07, 2012, 04:32 PM
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Originally Posted by ShoeDLG View Post
Your train of thought is difficult to follow because you are mixing reference frames. I'm not sure what you mean by "Bernoulli can not apply", but the relationship:

total pressure = static pressure + dynamic pressure

does not apply in an unsteady flow. Are you having trouble opening the .pdf?
I may have been a little sloppy there, but fluid accelerations are only related to still air and the only way we can locate differences is related to a still cylinder.

I plugged 45deg into the first still air equation and came up with zero.
I do not believe that there is any way that calculus can squeeze pressure change out of fluid that is not changing velocity.
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Old Dec 08, 2012, 02:53 AM
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Crossplot,

I was sloppy with the .pdf in post 791. The first "still air equation" is wrong. The "one half" belongs in front of the "v squared" term inside the parentheses, not outside the parentheses. This equation should be the same as equation (57) from the .pdf file from post 781. Sorry about that. I've corrected the attachment.
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