cxxLjevans

In the above W equation, I assume {V,z}_1 is the airfoil {velocity,height} at the start of some time period, and {V,z}_1 is the airfoil {velocity,height} at the end of some time period. Is that right? But I thought for the fluid, V_1 was fluid velocity at entrance to the wind tunnel and V_2 was the fluid velocity at exit from the wind tunnel. So V for the airfoil depends on time, but V for the fluid depends on distance along the wind tunnel. So, how can that be?

Good point. That equation does just describe the work on the airfoil body. On the other hand, the energy of airfoil and fluid would require some density terms. After all: [math]m={\rho Vol_foil} where Vol_foil is the volume of the airfoil. Anyway, the point is as the airfoil rises, it gravity potential increased but the fluid must fall and equal amount, and since the densities are the same, there's no change in the sum of gravitational potential energies of airfoil and fluid. OOPS. I now realize that the two points at which the fluid velocities and pressures height were measured take that into account because at the airfoil is in between these two points and aren't displacing the airfoil volume at either of these points. Now I realize why someone (maybe you) said density doesn't matter). Sorry for noise Also, density: [math]\rho was removed from the previous equation: [math]\frac{{V_1^1}}{{2g}} + \frac{{{P_1}}}{{\rho g}} + {z_1} = \frac{{V_2^1}}{{2g}} + \frac{{{P_2}}}{{\rho g}} + {z_2} = H = {\rm{A constant}} from some reason, and that equation describes energy in the fluid. Also, the above equation should have some velocity squared terms instead of just velocity terms according to: http://en.wikipedia.org/wiki/Hydraulic_head where it's called the velocity head. However, you did have velocity squared in an earlier equation: Hence, it was probably just a typo(using math markup is difficult).

But if the fluid has the same density as the airfoil, then as the airfoil rises the fluid falls the same amount, and wouldn't that just cancel the 2nd term: [math] \left( {mg{z_2} - mg{z_1}} \right)[/math] in the W equation, if it were included in the W equation?

Since the body has the same density as the fluid (one of the assumptions) I'm not sure there is any increase in gravitational potential energy due to rise of the airfoil. Am I missing something? -Larry

Start with the fluid height in the left vertical channel X meter higher than that in the right vertical channel. This height difference could be enforced by a door of some sort. The experiment starts when that door is instantly removed. I'd guess that whatever energy is used to raise the airfoil would be reflected in the difference in height in the opposite channel after the opposite channel reached its maximum height during an oscillation.

Would the it be simpler to answer Zet's question if the shape of the container were a U instead of an L? In this case, the fluid would simply oscillate between the left and right vertical columns of the U until something stopped it. I, and I think Zet, would assume, with frictionless flow, it would oscillate forever if the arifoil were constrained vertically; however, if it were allowed to rise, then, again I'm guessing, the energy used to raise the airfoil would dampen the ozcillation until ist stopped or the arifoil reached its vertical limit.

The Euler equation: http://www.grc.nasa.gov/WWW/k-12/airplane/eulereqs.html assume viscosity can be neglected: The Euler equations neglect the effects of the viscosity of the fluid So are you saying the Euler equatons would not predict lift for an airfoil. But then why does a further quote from same page: For some problems, like the lift of a thin airfoil at low angle of attack, a solution of the Euler equations provides a good model of reality. indicates that it would? Hmm, maybe your saying that with Euler turbulence and vortices would be present and they would produce lift? Sorry, I don't understand. Could you explain more? -regards, Larry Hi Zet, I also, initially, could not get the java applet to work. I had to make some adjustments to my FileFox browser prefserences (or something having to do with enableing java applets, but I forget what exaclty) but eventually I got it to work. I've looked at the code (which you can if you download) briefly, and I found code that mentioned viscosity, but viscosity of different forms. I've yet to figure out how to allow the airfoil to rise; hence, I don't know whether it can answer the question. I tried emailing the author listed on the web page (I think it was Thomas Benson) ;however, my mailer said there was no longer such an email address I've also looked at the SU2 docs, but, agian, it willl talk some time to figure out how to use it and graph the results to make the more understandable. I did ask for help here: http://www.cfd-online.com/Forums/cfd-freelancers/148415-effect-rising-hydrofoil-fluid.html and got a "private" reply where some guy offered help for a "reasonable price". I'm not sure I want to pay, but paying might be less work than going on forever here I probably try my hand using su2 (since I've been a c++ programmer for some time) and if it gets too hard, I might pay the money. -regards, Larry

If the angle of attach is positive, I can see why the velocity below would decrease and the pressure would increase. However, if the angle of attach was 0, and the airfoil was symmetrical, I would think the pressure above would exactly mirror the pressure below. At: http://www.grc.nasa.gov/WWW/k-12/airplane/foil3.html when a symmetric airfoll (an ellipse) is specified with 0 angle of attack, the pressures above an below look the same. The velocities both look same too. Could you explain more the reasoning behind your pressures? Is your airfoil not symmetric or is the angle of attack positive? TIA. -Larry

Would there be drag if there was no viscosity and no turbulence and no vortices created? I think in this thought experiment, those were the assumptions.

Couldn't the missing variables simply be given a name, such as VolFlowRate, and the a solution found using in terms of that name? If not, then pick some value for the missing varable and solve the problem. Maybe you mean the volumetric flow rate at the start of the channel before encountering the airfoil. I guess you could just set that to 1 in some aribitrary units and set the density to 1 (of both airfoil and fluid), and viscosity to 0. IOW, for every term in the Bernoulli equatoin, pick a value and solve the problem. Why wouldn't that work? OOPS. Bernoulli s for steady state. Hmm. So maybe that wouldn't work. Maybe just using some CFD code, such as: http://su2.stanford.edu/ could answer the problem; however I suspect it would require a lot of time to learn how to do that ; however, since this thread seems to be going on for so long, maybe resorting to using a CFD to answer the question would be desirable. -regards, Larry

[snip] > To do this look at my diagram. I have drawn two > streamlines BC and AD and two more (B'C' and A'D') exactly > 1 metre alongside so they form a square box section stream > tube. The streamlines are above and below a 1 metre > section of airfoil. So everything in the third dimension > is measured per metre. [snip] > Both sections have the same area so > > U1x = U2x > > This is the obstruction does not slow the fluid > horizontally, in accordance with D'Alembert. What does > happen is a vertical downward velocity Uy is imparted to > the fluid. How? Maybe this: http://en.wikipedia.org/wiki/Lift_%28force%29#Flow_deflection_and_Newton.27s_laws provides an explanation for the downward velocity; however, I haven't read it closely. I'd guess a downward flow is caused by a positive angle of attack, like that experienced by your flattened hand angled up and held outside of a speeding car window. In addition, this page: http://www.grc.nasa.gov/WWW/K-12/airplane/shed.html shows a "downwash" which looks like it *may* correspond to your Uy at the exit to your stream tube. However, my question (and I suspect Zet's ) is what happens when there's 0 angle of attack and the only lift is provided by the difference in the static presssure between the upper and lower surfaces of the airfoil. Bernoulli said there'd be a change in pressure caused by the fluid moving faster over the top and that produces the lift. And (like Zet), I'm wondering where the energy comes from causing that rise? Does the fluid flowing under the wing slow down or decrease in pressure or both in order to compensate for the energy required to raise the foil? > > The mass flowrate equals the fluid density times the > volumetric flowrate, > > and the momentum flowrate equals the mass flowrate times > the horizontal or vertical velocity. > > At section1 there is zero vertical momentum in the fluid > but at section 2 there is a downwards momentum flow of > > (pUxA2)Uy > Now, if we assume the flow conditions used to derive Bernoulli's equation: * Velocity head * Elevation head * Pressure head remain constant (where these terms are as defined here: http://en.wikipedia.org/wiki/Hydraulic_head ) then, since the Velocity head has increased (since Ux remains constant but Uy increases from 0) the pressure head must decrease to compensate. Hence, the airfoil is sucked backward (I guess you could say that is part of the drag). > But a momentum flow is the rate of change of momentum and > this is the definition of a force. > > The force exerted on the airfoil is the lift force and > this is equal to in magnitude but opposite in direction to > this force on the fluid. > > So the lift force = Fy = -(pUxA2)Uy > What about the difference in pressure between the upper surface of the air foil and the lower surface of the airfoil? I thought this was the reason given by the Bernoulli equation to explain lift. Indeed, the page: http://en.wikipedia.org/wiki/Lift_%28force%29#Increased_flow_speed_and_Bernoulli.27s_principle claims: For any airfoil generating lift, there must be a pressure imbalance, i.e. lower average air pressure on the top than on the bottom. Bernoulli's principle states that this pressure difference must be accompanied by a speed difference. -regards, Larry

I thought friction was caused by viscosity and 0 viscosity (and 0 friction, IIUC) is used to derive Euler's equation: http://en.wikipedia.org/wiki/Inviscid_flow#Reynolds_number Since Euler assumed frictionless flow was OK, I can see why Zet assumed it was.

That's what I would guess; however, if the energy comes from the horizontal thrust, then that begs the question of how a horizontal force causes a change in vertical potential energy. IOW, how can something pushing horizontally cause movement of something vertically? OTOH, if the energy comes from the fluid, then how is that energy substracted from the fluid. Does the fluid drop, decreasing is potential energy? Does its speed decrease? Does its static pressure decrease? If so, then how?

Can't the problem be simplified by: 1) making the fluid incompressible (as in a hydrofoil instead of an airfoil) 2) making the hdrofoil density the same as fluid density 3) making the fluid viscosity = 0 (no energy loss due to friction) IIUC, 1) would avoid any change in fluid energy due to compression (in air, change in internal energy is P dV). Also, 3) would avoid change in fluid energy due to frictional heating. 2) would make the gravitational potential energy constant. How is the loss of the free stream energy, mentioned here: gravitational/potential energy gain in the second case is accounted for as a loss of the free stream energy, accomplished. Does the velocity of the free stream decrease with respect to its value before encountering the foil. If so, then this is, essentially, what Zet said in post#46 which says: when the airfoil rises there must be a decrease in the horizontal velocity of the wind -regards, Larry

Could you please provide a very simple example (hopefully by adding to my sketchy example ) which is complete enough to calculate an answer? Also, maybe you could at least outline the calculation steps needed in this example to get an answer. I'd be happy to do the calculations myself then. TIA. -regards, Larry

Did I misinterpret MigL's statement? If so, could you clarify what MigL meant by: The fluid has a certain amount of energy and it imparts some of it to the airfoils ( heating, lift, etc. ). Also, if the energy does not come from the air, then does it only come from the thrust? To put concrete numbers on this, assume: 1) force of gravity, g, is 1.0 meter/(second^2) 2) thrust, T, is 2.0 (kilogram*meter)/(second^2) 3) wt. of airfoil is 1.0 kilogram 4) the airfoil and rocket engine have no drag 5) the fluid is incompressible 6) density of fluid = 1 (in "appropriate" units [ this needed to calculate the change in p in the bernoulli equation A on: http://en.wikipedia.org/wiki/Bernoulli%27s_principle ] ). 7) speed of airfoil at start time is 1.0 meter/sec 8) the airfoil cross section is an isosceles triangle with width=1 meter and height of .1 meter. 9) the airfoil is within a wind tunnel stetching 1 meter above and below the base of the triangle (this data is needed to calculate the speed up in fluid flow over top of triangle, AFAICT). 10) density of airfoil is 1.1 * density of fluid. ( this is needed in order for energy to be used in lifting the airfoil. If the airfoil density = fluid density then there would be no change in gravity potential no matter what the height of airfoil. ) What is the difference in height of airfoil after 1 sec.? Since there's no drag, and since no energy comes from the fluid, I guess part of the thrust is used to accellerate the airfoil horizontally and part is used to lift the airfoil vertically. But how much is used horizontally and how much is used vertically?

From: The fluid has a certain amount of energy and it imparts some of it to the airfoils ( heating, lift, etc. ). I gather that the energy used to lift the airfoil a certain distance, d, comes from a decrease in the fluids energy (via decrease in pressure or slowing down). OK, so how much comes from slowing down the air speed and how much from decrease in pressure? BTW, I've had the same questions as Zet and I applaud him in being persistent in getting an answer he (and hopefully I) can understand. Thanks Zet.