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Old Dec 08, 2012, 06:01 AM
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Originally Posted by Crossplot View Post
I do not believe that there is any way that calculus can squeeze pressure change out of fluid that is not changing velocity.
If you look at the equation relating pressure to velocity in the reference frame where the remote air is still and the cylinder is moving with steady velocity V, the pressure experienced by a fluid element depends not only on the fluid element's speed, but also its direction (through the term that represents the product of the cylinder's speed times the fluid element's speed in the cylinder's direction of travel: V*v_x). This equation was derived by integrating the expression for momentum conservation under the following conditions:

-Uniform density
-Flow is unsteady in the reference frame where the remote air is still, but is steady in a reference frame that is moving with the cylinder

If the fluid element's speed and direction of travel remain constant, then the pressure it experiences will remain constant. If the fluid element's speed remains constant, but its direction of travel changes, then (in an unsteady flow), then the pressure it experiences will change.
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Old Dec 08, 2012, 06:25 PM
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Originally Posted by ShoeDLG View Post
If the fluid element's speed and direction of travel remain constant, then the pressure it experiences will remain constant. If the fluid element's speed remains constant, but its direction of travel changes, then (in an unsteady flow), then the pressure it experiences will change.
That is the way that I see it. The change in the direction of travel is the normal acceleration that produces pressure change.
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Old Dec 09, 2012, 04:15 AM
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This brings up the question I think you have been trying to get at: is the pressure experienced by a fluid element on the surface of the cylinder determined purely by the normal acceleration it is experiencing?

You can do a thought experiment that will strongly suggest this is not the case. The pressure (in coefficient form) over the surface of a circular cylinder is shown in the first attachment (for the case of inviscid, uniform density flow).

Nothing too surprising here... In the frame of reference of the cylinder, an element of air at the surface accelerates along its path, giving rise to the pressure variation shown (in accordance with the steady Bernoulli equation). In the frame of reference where the remote air is still, an element of air at the surface accelerates normal to its path, giving rise to the pressure variation shown (in accordance with the unsteady Bernoulli equation). Although the analysis is different, the pressure at the surface is the same regardless of the reference frame you choose for your analysis (see the attachment in post 791).

Now suppose you take the circular cylinder and sand a small region at the top and bottom perfectly flat. When an element of air at the surface flows over this region, it will be moving in a straight line (in both reference frames). The normal acceleration of a fluid element moving in a straight line is zero. What is the pressure experienced by the fluid element as it passes over the flat region?

Does the absence of normal acceleration (absence of any flow curvature) mean that the pressure in this region is unchanged compared to ambient pressure? If that's the case then the pressure will be the same as ambient (pressure coefficient of 1.0). This is shown by the black line in the second attachment.

Does the absence of flow curvature instead mean that pressure experienced by a fluid element will remain roughly unchanged as it passes over the flat region? This is shown by the blue line in the second attachment.

Experience (and intuition) certainly suggests that sanding a small flat in the surface of a cylinder or airfoil will not cause a dramatic shift in the surface pressure there.

Is the pressure experienced by a fluid element on the surface of the cylinder determined purely by the normal acceleration it is experiencing? No. While the rate of pressure change may be determined by normal acceleration, the pressure itself is not. The pressure at any point on a cylinder (or airfoil) is determined by the conditions everywhere on the surface. When you deflect a flap, the pressure doesn't just change on the flap, it changes over the entire airfoil.

It is the velocity of a fluid element at the surface that determines the pressure it experiences, not its acceleration. A fluid element's velocity represents the sum total (the integration) of the accelerations it experienced as it traveled from where the pressure was ambient to its current location on the cylinder (or airfoil) surface. If you are very careful, you can connect the sum of the accelerations the fluid element experienced to the sum of the pressure changes it experienced. Just knowing the acceleration or rate of pressure change the fluid element is experiencing right now is of no help in knowing the pressure it is experiencing.

There's another way to see that the pressure experienced by a fluid element is not purely a result of the normal accelerations it is experiencing (or has experienced). Viewed in the reference frame where the remote air is still, the pressure experienced by a fluid element far in front of an approaching cylinder is ambient pressure. As the cylinder approaches, the fluid element accelerates in a straight line (no normal acceleration) in the direction of the cylinder's travel. When the the cylinder and the fluid element finally meet (bringing the fluid element to the forward stagnation point), the fluid element has experienced no normal acceleration, but the pressure it is experiencing has increased by one half the fluid density times the cylinder's speed squared. A fluid element does not have to experience normal acceleration to experience a change in pressure.
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Old Dec 09, 2012, 02:57 PM
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In the frame of reference of the cylinder, an element of air at the surface accelerates along its path, giving rise to the pressure variation shown (in accordance with the steady Bernoulli equation). ).

Now suppose you take the circular cylinder and sand a small region at the top and bottom perfectly flat. Does the absence of flow curvature instead mean that pressure experienced by a fluid element will remain roughly unchanged as it passes over the flat region?

While the rate of pressure change may be determined by normal acceleration, the pressure itself is not. The pressure at any point on a cylinder (or airfoil) is determined by the conditions everywhere on the surface.

It is the velocity of a fluid element at the surface that determines the pressure it experiences, not its acceleration. A fluid element's velocity represents the sum total (the integration) of the accelerations it experienced as it traveled from where the pressure was ambient to its current location on the cylinder (or airfoil) surface. If you are very careful, you can connect the sum of the accelerations the fluid element experienced to the sum of the pressure changes it experienced.

As the cylinder approaches, the fluid element accelerates in a straight line (no normal acceleration) in the direction of the cylinder's travel. A fluid element does not have to experience normal acceleration to experience a change in pressure.
When we calculate pressure with flow relative to a cyl or wing we ascribe velocity and acceleration from velocities that are made up primarily from the motion of the wing. ??

For flat areas, you are right pressures are projected everywhere. There is the same effect on the top back of most wings. As they flatten out the pressure field tapers.

The rate of change from normal acceleration is its own buildup across the field. If it were not providing a smooth transition point to point there would be chaos.

How do we get velocity change without acceleration?
The point of this exercize is that the air, related to the remote air, around a cylinder experiences no linear acceleration between the stag points.
Note that normal accelerations are massive and dominent.

Linear accelerations are basic Newton. Normal accelerations are to all intents linear accelerations.
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Old Dec 09, 2012, 03:23 PM
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The rate of change from normal acceleration is its own buildup across the field.
Can you explain what you mean by this?

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The point of this exercize is that the air, related to the remote air, around a cylinder experiences no linear acceleration between the stag points.
This is an interesting feature of inviscid flow around a cylinder, but I don't see how this sheds any light on the bigger picture of whether momentum exchange is involved when a wing generates lift.
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Old Dec 09, 2012, 11:44 PM
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Can you explain what you mean by this?

This was meant to indicate the changing rate of acceleration, across the field, normal to the surface, The actual rate of change along the surface is the result of changing rates of curvature in the flow relative ro the remote air.

>This is an interesting feature of inviscid flow around a cylinder, but I don't see how this sheds any light on the bigger picture of whether momentum exchange is involved when a wing generates lift.
This is tough. The normal accelerations contain vertical components that sum up to lift.

I have an undefencible feeling that since the flow over the wing removes pressure (downward weight) that the situation is in essence bouyancy as with balloons and boats ! (No repy suggested!)
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Old Dec 10, 2012, 12:51 PM
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The normal accelerations contain vertical components that sum up to lift.
Ah, how simple and tidy life would be if this were the case. However, if true, this statement means that every lifting wing flying anywhere near a horizontal surface (like the ground) is violating Newton's Second Law. The sum of the forces acting on the air would not be equal to the air's rate of momentum change.
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Old Dec 10, 2012, 01:17 PM
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Perhaps Newton got it wrong

As in, you can pour a liquid into another container of liquid without that poured liquid affecting say some sediment at the bottom of the container.

i.e. the falling liquid dissipates it's energy before hitting another solid. It's doesn't always disturb the sediment.

See what comes of thinking and peeing at the same time.
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Old Dec 10, 2012, 04:29 PM
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Ah, how simple and tidy life would be if this were the case. .
Lets start with what we know. The flow must accelerate towards the surface in order to change its velocity vector. v^2/R is classic physics. The pressure change at the surface is equal to the sum of the total mass accelerations, across the flow, normal to a suface element. Lift pressure is the vertical component of that pressure area and is equal to the downward vertical component of the normal mass acceleration.

What happens in the way of momentums from a passing wing is at best a thesis. If one cannot resolve the thesis with what we do know then one must suspect the thesis.
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Old Dec 11, 2012, 03:46 AM
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I agree the pressure distribution on the surface of an airfoil (or cylinder) can be connected to the pressure changes associated with flow curvature (normal acceleration experienced by fluid elements). However, there’s a straightforward example that should convince you there’s more to the story than simple equality between lift and vertical momentum exchange.

Suppose you have two airfoils both moving at the same speed and generating the same lift. One of the airfoils is many chords above the ground, but the other is only a chord or so above the ground. Because the air cannot flow into or out of the ground, the ground’s proximity has the effect of straightening the streamlines (reducing their curvature), thus reducing the magnitude of the accelerations experienced by an element of air as the wing passes by. If an airfoil can generate the same amount of lift with less streamline curvature, then there MUST be more going on than just the momentum exchange associated with air flowing along curved streamlines.
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Old Dec 11, 2012, 04:00 AM
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Low pressure on top, higher pressure underneath. Yet the low pressure air causes downwash into the higher pressure air ?

Still can't get my head around that.

http://www.google.co.uk/search?num=1....1.immBMYVbRzY
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Old Dec 11, 2012, 01:42 PM
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....should convince you there’s more to the story than simple equality between lift and vertical momentum exchange.
I cannot stick my neck out beyond what happens near the wing surface. Everything else falls under conjecture rather than fact. In spite of all of the good reading, ground effect mechanism is not 100% totally obvious, downwash seems totally left over circulation with no obvious connection to reaction from lift and for ground pressure mechanism I don't have a clue!

eflightray - now there is a page!!
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Old Dec 12, 2012, 05:05 PM
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Ground effect defined- Just tip over a piece of plywood - note the slow contact with the floor
Downwash - - just like dishwater - - it is a result of work having been performed
Why try to make the obvious confusing.--
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Old Dec 12, 2012, 05:16 PM
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... Why try to make the obvious confusing.--
Attempting to comprhend the "obvious" is what makes us human.
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Old Dec 12, 2012, 06:27 PM
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