HobbyKing.com New Products Flash Sale
Reply
Thread Tools
Old Nov 18, 2012, 03:05 PM
Registered User
ShoeDLG's Avatar
Germany, BW, Stuttgart
Joined Mar 2012
787 Posts
Quote:
Originally Posted by Crossplot View Post
Multiplying the surface pressure gradient by the surface radius/2 integrates the pressure change built up across the flow field.
I can see how simple multiplication might work in the case of a circular cylinder. I doubt that multiplying the surface pressure gradient on a wing by the local radius of curvature (over 2) would give you the surface pressure.
ShoeDLG is offline Find More Posts by ShoeDLG
Reply With Quote
Sign up now
to remove ads between posts
Old Nov 18, 2012, 05:25 PM
Registered User
Joined Dec 2011
62 Posts
Quote:
Originally Posted by ShoeDLG View Post
Why is it surprising that (in the "still air" reference frame) air has slowed in moving from the stagnation point to the top of the wing?
For someone like yourself that has studied flow it is basic. But how many people in the big wide web world have not said that the air speeds up over the top of the wing?
Crossplot is offline Find More Posts by Crossplot
Reply With Quote
Old Nov 18, 2012, 05:33 PM
Grad student in aeronautics
United States, GA, Atlanta
Joined Oct 2010
447 Posts
Crossplot, you'll have to forgive me for not following you clearly. Also forgive me if this question sounds silly to you.

Consider the reference frame where the wing is fixed. Do you agree that the flow at the stagnation point is slower than the freestream and that the flow moving over the top in a region of low static pressure is faster than the freestream?
DPATE is offline Find More Posts by DPATE
Reply With Quote
Old Nov 18, 2012, 05:36 PM
Registered User
Joined Dec 2011
62 Posts
Quote:
Originally Posted by ShoeDLG View Post
I can see how simple multiplication might work in the case of a circular cylinder. I doubt that multiplying the surface pressure gradient on a wing by the local radius of curvature (over 2) would give you the surface pressure.
Actually this has been giving me a bit of fits. The velocity profile on each side of the "high" point is going to be as less than 1/r. However, the Bernoulli equation, as reduced from centripetal acceleration, remains in general practice throughout the flowfield.
Crossplot is offline Find More Posts by Crossplot
Reply With Quote
Old Nov 18, 2012, 06:10 PM
Registered User
Joined Dec 2011
62 Posts
Quote:
Originally Posted by DPATE View Post
Consider the reference frame where the wing is fixed. Do you agree that the flow at the stagnation point is slower than the freestream and that the flow moving over the top in a region of low static pressure is faster than the freestream?
This is not silly. It is very sticky walking. It very much looks like Bernoulli and I can not swear that ir is not. Air from the moving stream has been accelerated at the stag point to -one V relative to the flow. At the "high" point it is moving +.2-.3 V
in the direction of flow. The flow path ,relative to the stream is difficult to portray. The flow along the surface has expirienced an inertial velocity loss of .8V
What we are interested in is velocitie and acceleration along and normal to that flowpath.
Crossplot is offline Find More Posts by Crossplot
Reply With Quote
Old Nov 18, 2012, 11:25 PM
Registered User
ShoeDLG's Avatar
Germany, BW, Stuttgart
Joined Mar 2012
787 Posts
Crossplot,

Two thoughts.

1. The most common form of the Bernoulli equation (pressure + (1/2)*density*velocity^2 = constant), does not apply in the reference frame where the remote air is still. In that reference frame, the flow is unsteady and you pick up another term: density*(d/dt)potential_function. With the unsteady form of the Beroulli equation, the pressure predicted at the "high point" is lower than at the stagnation point even though the velocity is lower.

2. I still don't see how you would to calculate the pressure at the wing surface just by finding the pressure gradient there. I think you would need to start far from the wing where the pressure is known and integrate the gradient all the way to the wing.
ShoeDLG is offline Find More Posts by ShoeDLG
Reply With Quote
Old Nov 19, 2012, 12:31 PM
Registered User
Joined Dec 2011
62 Posts
Quote:
Originally Posted by ShoeDLG View Post
2. I still don't see how you would to calculate the pressure at the wing surface just by finding the pressure gradient there. I think you would need to start far from the wing where the pressure is known and integrate the gradient all the way to the wing.
Only in such a case as the charactoristics across the field can be predicted from the charactoristics at the surface.

I t may take me some little time to try to reconstruct my thoughts here. --and hope that I have not shot from the hip one too many times!
Crossplot is offline Find More Posts by Crossplot
Reply With Quote
Old Nov 19, 2012, 02:44 PM
Registered User
ShoeDLG's Avatar
Germany, BW, Stuttgart
Joined Mar 2012
787 Posts
The relationship between the characteristics at the surface of an airfoil (the boundary conditions) and the characteristics across the field are governed by the field equation(s). In the case of irrotational, constant-density flow, the field equations simplify to Laplace's Equation:

Laplacian(potential function) = 0

-or-

Divergence(velocity) = 0

The circular cylinder receives so much attention because the boundary conditions on the entire surface can be met by introducing just one singularity (a dipole) at the center of the cylinder. This makes it deceptively simple to relate the field characteristics to the boundary conditions.

If you have an arbitrary surface, like an airfoil, you generally need an infinite number of singularities to satisfy the boundary conditions (although you can often approximate them well with tens of singularities). The flow conditions at any point in the flow around an airfoil depend, not on the boundary conditions at a single point on the surface, but on the boundary conditions at ALL the points on the surface.

So although the pressure gradient at a point on the surface might be determined from the boundary conditions applied only at that point, I think determining the pressure at a point on the surface requires you to solve the field equation(s), and that means applying the boundary conditions over the entire surface (although I have been wrong many times before)
ShoeDLG is offline Find More Posts by ShoeDLG
Reply With Quote
Old Nov 19, 2012, 03:45 PM
Grad student in aeronautics
United States, GA, Atlanta
Joined Oct 2010
447 Posts
Quote:
Originally Posted by ShoeDLG View Post
So although the pressure gradient at a point on the surface might be determined from the boundary conditions applied only at that point, I think determining the pressure at a point on the surface requires you to solve the field equation(s), and that means applying the boundary conditions over the entire surface (although I have been wrong many times before)
A simple example to illustrate (not prove) this is the fact that deflecting a flap will move the LE stagnation point, thereby changing the pressures and velocities in the vicinity of the LE. More specifically, a symmetric airfoil at zero angle of attack but with a deflected flap will have very different upper surface and lower surface pressure distributions.
DPATE is offline Find More Posts by DPATE
Reply With Quote
Old Nov 19, 2012, 06:27 PM
Registered User
Joined Dec 2011
62 Posts
Quote:
Originally Posted by ShoeDLG View Post
(although I have been wrong many times before)
Don't even try to compete in that catagory!
Crossplot is offline Find More Posts by Crossplot
Reply With Quote
Old Nov 26, 2012, 11:24 PM
Registered User
Joined Dec 2011
62 Posts
[QUOTE=ShoeDLG;23309814]So although the pressure gradient at a point on the surface might be determined from the boundary conditions applied only at that point, I think determining the pressure at a point on the surface requires you to solve the field equation(s), and that means applying the boundary conditions over the entire surface

I can't disagree with that at all. However, we continue to determine approximent surface pressure from the Bernoulli equation and the tangential velocity..

I am faced with an appearent paradox.
--Pressure values match Bernoulli Equation values by test. (Surface relative)
--Pressure values are not created from actual inertial flow velocities.
--We match Bernoulli pressures whrn we multiply the surface dp/dr of centripetal acceleration by an equivalent to r/2, right or wrong.(To sum the pressure change across the field)

I am really not qualified to explore the areas away from the "top" areas.
I was thinking maybe that Shoe would like to take a crack at it.
Crossplot is offline Find More Posts by Crossplot
Reply With Quote
Old Nov 27, 2012, 04:41 AM
Registered User
ShoeDLG's Avatar
Germany, BW, Stuttgart
Joined Mar 2012
787 Posts
I think a simple accounting analogy might shed some additional light on why knowing the pressure gradient at the surface of an airfoil alone can’t tell you the pressure there. Suppose you wanted to know how much money you had in your bank account, but you forgot your password (and asking a real person involved a million dollar fee).

If you knew that you added $1,000 to your account every month (the rate of change of your balance), that wouldn’t help you determine your current balance unless there was a point in time at which you did know your balance. For example, if you knew you had $5,000 in your account five months ago, then you could figure out that should now have $10,000 in your account (assuming no withdrawals and a constant monthly rate of deposit).

Similarly, knowing the rate of change of pressure at the surface of an airfoil won’t help you determine the pressure there unless you can start from a point where the pressure is known (typically far from the airfoil), and sum up all the changes as you move toward the airfoil.

If that’s true, how can the Bernoulli equation give you the pressure at the surface knowing only the velocity at the surface? The “traditional” form of the Bernoulli equation is derived by integrating an expression for momentum conservation along a flow streamline under the assumptions of steady, irrotational, uniform density flow. The dynamic pressure (one-half the density times the flow velocity squared) represents the result (integrand) of summing up all of infinitesimal pressure differences as you move along a streamline. When you calculate the static pressure at the surface of an airfoil using the traditional Bernoulli equation (static pressure = total pressure – dynamic pressure), what you are really doing is starting well upstream of the airfoil (where you know the static and dynamic pressure) and integrating the changes in pressure along the streamline until you reach your point of interest on the airfoil. 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.

Quote:
Originally Posted by Crossplot View Post
We match Bernoulli pressures when we multiply the surface dp/dr of centripetal acceleration by an equivalent to r/2, right or wrong.(To sum the pressure change across the field).
I think this approach (integration by simple multiplication of the pressure gradient by r/2) only applies to the special case of a circular cylinder, but agian I could be wrong.
ShoeDLG is offline Find More Posts by ShoeDLG
Reply With Quote
Old Nov 28, 2012, 12:58 AM
Registered User
Joined Dec 2011
62 Posts
Quote:
Originally Posted by ShoeDLG View Post
I think this approach (integration by simple multiplication of the pressure gradient by r/2) only applies to the special case of a circular cylinder, but agian I could be wrong.
Darned good description of Bernoulli. And you are right, the one point that we have nailed down is the top of the cylinder, we have values for v and r all of the way out to ambient so that it summs in a like manner but except in the normal direction. Ps = Pt - Pd still holds along the surface.

My point is that I need help with the rest of the flows. The fact remains that centripetal aceleration occures wherever flow bends and requires specific normal pressure gradients to match measured values.
Crossplot is offline Find More Posts by Crossplot
Reply With Quote
Old Nov 28, 2012, 07:31 AM
Registered User
ShoeDLG's Avatar
Germany, BW, Stuttgart
Joined Mar 2012
787 Posts
Crossplot,

I'm still not sure I understand exactly what your "big picture" objective is.

- If it is to calculate the forces acting on an airfoil purely from examining the boundary conditions on the surface of the airfoil, then I don't think you can "get there from here".

- If it is some other objective then I'm not understanding what that is.
ShoeDLG is offline Find More Posts by ShoeDLG
Reply With Quote
Old Nov 28, 2012, 10:00 PM
Registered User
Joined Dec 2011
62 Posts
Quote:
Originally Posted by ShoeDLG View Post
I'm still not sure I understand exactly what your "big picture" objective is..
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.
Crossplot is offline Find More Posts by Crossplot
Reply With Quote
Reply


Thread Tools

Similar Threads
Category Thread Thread Starter Forum Replies Last Post
Discussion Dymond D47 servo, Which version do you have on hand, how can you tell??? TDboy Hand Launch 5 Apr 21, 2012 02:26 AM
Discussion Is there anything listed that you can't live without? P-51C Life, The Universe, and Politics 44 Aug 05, 2011 02:55 PM
Discussion Fast food that you just can't do without. Bilbobaker Life, The Universe, and Politics 101 Aug 17, 2010 11:30 AM
Joke What Would You do to get that Aircraft You Can't Have ? Chophop Humor 10 Feb 09, 2010 07:00 AM
Discussion HELPIf you have a problem that can't be solved here ccowboyearl Beginner Training Area (Heli-Electric) 2 May 14, 2009 12:31 PM