According to the popular interpretation of 'Bernoulli's Principle', moving air should always be associated with a lower static gas pressure than resting air, but clearly this can not be correct: If one considers a pipe with air resting in it, then the static pressure on the inside of the wall (due to the random motion of molecules) is given through the ideal gas law as p=n*k*T (where n is the volume number density of air, k the Boltzmann constant and T the temperature). Now consider the air in the pipe not resting but moving uniformly through it, i.e. assume that the amount of air leaving the pipe at one end is exactly replaced by the amount of air entering the pipe at the other end. Because the total amount of air is the same as for the stationary case, the density n is the same as well and hence also to static gas pressure p=n*k*T (assuming that the gas temperature is unchanged). The walls of the pipe experience therefore the same static pressure whether or not the air is moving in it (in contradiction to the popular view based on Bernoulli's Principle).
I don't know where and how you came up with this "popular" obviously incorrect interpretation of Bernoulli's principle... Daniel.
Well, just do a corresponding Google search and you will find a lot of sites claiming this (e.g. http://nasaui.ited.uidaho.edu/curriculum/velocity.htm ).
Look up the definition of total pressure. Static pressure is only one component of total pressure. For anything to flow in an enclosed vessel, i.e. a pipe, there must be a pressure differential.
You are quite right. The popular interpretation is wrong or, at best incomplete, as are most explanations of Bernouilli's principle in explaining airplane wing lift. Bernouilli's principle is simply a statement about conservation of energy in a dynamic fluid. Pressure is a measure of energy/unit volume and represents potential or stored energy. If the pressure changes, then the energy density changes. If the quantify of fluid is constant, the energy has to go somewhere. It does. It results in change in kinetic energy of the fluid. Furthermore, motion is relative. One cannot say that a volume of gas is 'moving' and another 'at rest' without defining a reference frame. So it cannot at all be correct to say that moving air has a lower static gas pressure than resting air, because we can't say, in absolute terms, which one is moving. One can only speak about changes in pressure and velocity. Essentially Bernouilli's law says that for a given quantity of fluid in thermal isolation and not subject to external forces (ie. external pressures), if that fluid experiences an increase in kinetic energy (ie. speed), the pressure of that fluid must decrease so that the increase in kinetic energy of fluid is equal to the drop in pressure x volume. AM
Andrew Mason: Add to that the all-important condition that this relation between pressure&velocity holds between two points on the SAME streamline.
If the flow is uniform as assumed (i.e. it has the same speed at both ends of the pipe) there can not be a pressure differential as the latter would correspond to a force and hence accelerate the flow (assuming the air to be non-viscous). As an example, imagine for instance a pipe (with a sufficiently large diameter to prevent friction) moving longitudinally through the stationary air. Clearly no pressure differential is necessary here for the air to flow through the pipe, and obviously it would be wrong to assume that in this case the static pressure on the walls of the pipe would be reduced (as Bernoulli's Principle would suggest).
Good post AM. Thomas2, I agree that the examples you cited on that web site do over simplify the explanation by saying flow equals lower static pressure than non flow conditions. All of those examples do require viscous flow to work, which is a limitation of Bernoulli. If that is what you are saying, then I misread your original post.
It is true that not all sources explicitly state the connection between pressure and flow speed in this blunt way, but even serious scientific references dealing with aerodynamics imply in fact the same thing through the simplifying assumptions that are being made for the corresponding physical models: all corresponding derivations actually assume not only zero viscosity but also incompressibility. See for instance the NASA webpage www.grc.nasa.gov/WWW/K-12/airplane/bern.html and note the statement (halfway down the page) 'Assuming that the flow is incompressible, the density is a constant' (which obviously must be the case). Now have a look at the page dealing with the concept of stream lines around an airfoil http://www.grc.nasa.gov/WWW/K-12/airplane/stream.html . I quote from this page: 'Since the streamline is traced out by a moving particle, at every point along the path the velocity is tangent to the path. Since there is no normal component of the velocity along the path, mass cannot cross a streamline. The mass contained between any two streamlines remains the same throughout the flowfield. We can use Bernoulli's equation to relate the pressure and velocity along the streamline. Since no mass passes through the surface of the airfoil (or cylinder), the surface of the object is a streamline'. Now taking these statements together, this obviously results in the logical conclusion that the density n along the whole of the airfoil is constant i.e. equal to original density of the undisturbed air. Since one certainly can assume the temperature T to be constant as well, this means the static pressure p=n*k*T is constant along the airfoil. According to these assumptions, aerodynamic lift should therefore be impossible altogether (obviously, the dynamic pressure could not contribute either with this 'model' as the velocity is always tangential to the airfoil).
I think density remains constant along the airfoil. Lifting is not an exclusive propierty of compressible flows. In incompressible flow there is a decoupling of energy transport and momentum transport. In fact one can solve [tex] \nabla \cdot \overline{v}=0[/tex] [tex] \overline{v}\nabla\overline{v}=-\frac{1}{\rho}\nabla P+\nu\nabla^2\overline{v}[/tex] [tex] \overline{v}\cdot\nabla T=\alpha\nabla^2 T[/tex] and obtain a lift component with constant density. The temperature will reshape itself according to the momentum transport. At low Mach numbers making use of the equation of state has little sense.
I can't really see how (in the absence of any heat sources or sinks) the temperature of the air could change along a streamline given the assumption of incompressibility (which is usually made for small velocities). Since therefore neither the density n nor the temperature T changes along a streamline, the pressure p=n*k*T should consequently be constant as well.
I can't really see how the temperature of the air is NOT going to change. As I have said before, the expression P=nkT has NO sense for incompressible aerodynamics. Your substance is incompressible, and this formula is given for compressible substances. In incompressible flow, the pressure reshapes itself into the fluid field in order to yield the continuity of mass, and it is decoupled of temperature variations, which are a CONSEQUENCE of the transport of Internal Energy due to convective and diffusive phenomena (See the last equation I wrote: it is the Navier-Stokes Internal Energy Equation). Various effects enhance a change of temperature (spatial variations): i) Although initial temperatures of surrounding air and airfoil were the same, heat transfer by convective effects will change the temperature of the airfoil. If you're so lucky to have a thermal resistance which makes possible to control the airfoil temperature to be the same of the surrounding air (namely [tex]T_{\infty}[/tex]) then the energy equation would yield a constant temperature along the flow field. Let's see. Try to solve [tex] \overline{v}\cdot\nabla T=\alpha\nabla^2 T[/tex] with [tex] T=T_{\infty} [/tex] at [tex] x\rightarrow\infty[/tex] and [tex] x=x_{airfoil}[/tex] The solution of this problem is [tex] T=T_{\infty} [/tex] everywhere. ii) Of course this last result is far away from reality, because I have neglected viscous dissipation, which surely has a strong effect inside the boundary layer and will change the temperature field. Pay attention: No matter the temperature will be constant in our "imaginary" experiment I have modelled theoretically above, the pressure is NOT the same along the flow field. The proper deflection of the flow over the airfoil is going to reorder pressure gradients to yield the conservation of mass. As a conclusion, the equation of state in an incompressible fluid is NOT P=nkT, but [tex] \rho=constant[/tex], which is very different. The density is NOT a function of the thermodynamic state. Moreover, fluid-mechanical variables (velocity & pressure) are DECOUPLED of fluid-thermal variables (Temperature).
Hey Thomas, maybe this will help a bit. The assumptions for Bernoulli's are: Ref: http://www.princeton.edu/~asmits/Bicycle_web/Bernoulli.html Also: Ref: http://142.26.194.131/aerodynamics1/Basics/Page3.html So the incompressible (constant density) assumption is only valid for velocity below the speed of sound. (Still, that gives us plenty of velocity to play with!) From Cambridge University Press: Ref: http://www.fluidmech.net/tutorials/bernoulli/compressible-bernoulli.htm What all this says is that for the incompressible case, internal energy and density remain constant, so you are safe to say that temperature does not change for the incompressible flow case. You are correct in saying the temperature does not change. This is true for a gas, and it's true for a liquid of course as well. For the compressible flow case, temperature does change. But that isn't your question. The real question is, "Why does the flow across a wing result in a lower pressure?" (hope I got that right! lol) Imagine a given mass of gas moving through a converging/diverging nozzle. Imagine just one gram of air and the volume it would take up, and you are traveling along with this one gram of air. When static, you might imagine this volume as being a cube shape. As this gram of air accelerates past the smallest point in the nozzle, the volume (in order to maintain constant density) stretches in the direction of motion. In other words, the flow accelerates and the cube shape this volume took before must take on a rectangular shape as it accelerates. So: i) the length of the cube in the direction of motion increases, and ii) the length of the cube in the direction perpendicular to the motion decreases. The total pressure of this deformed cube stays constant, as does the density. That's what Bernoulli's tells us. But the total pressure is made up of 3 different forms, the static, dynamic, and head pressure. For a streamline going past a wing, we can ignore changes in head pressure since we will assume horizontal flight. The static pressure then decreases as dynamic pressure increases as the small cube of air accelerates. But this is with respect to a stationary observer. If one were to stand in the center of the converging/diverging nozzle, you might sense the velocity increase because the dynamic pressure on you could be felt just as you feel the push of the wind. Similarly, that stationary observer would sense a lower static pressure (ie: perpendicular to the direction of motion). A physical explanation for the lower static pressure for the stationary observer might best be explained by examining the cause of the pressure. Pressure is caused by the change in momentum of the molecules. As the molecules bounce off the surface of a wing or internal surface of the nozzle, they impart a small force, and it's the sum of these forces which results in pressure. In the case for the moving fluid, the impacts in the direction of motion perpendicular to the direction of flow are spread out. I don't know how to explain that any better, I guess it's a visualization you have to think about for a while. But anyway, the static pressure decreases, and static pressure is the pressure in the direction perpendicular to the direction of motion. ______________________________________________________ "Beware of those who stand on soap boxes. They are not there to help you but to impress themselves." ~ Anonymous.
This is the answer to the OP. Static pressure isn't the only kind of pressure present in the pipe when the air is moving, Thomas2. When the air is stationary, it is. So the total pressure when the air is moving is equal to the static pressure when the air is stationary (which is kinda the whole point of Bernoulli's equation). Thomas2, you are falling into a common trap where you assume that if an equation has all the variables you are looking for (the ideal gas law in this case), you can apply it whereever you want. This is not the case. In this case, the ideal gas law may provide all the variables you want - but it doesn't have all the variables you need. Andrew_Mason, for a question about Bernouli's principle in a pipe, lift on a wing is not relevant (though I do know that that's where Thomas2 is going). Lets try to keep lift out of it for now. edit: Its also interesting that you're making separate, simultaneous claims, Thomas2: that Bernoulli's equation is wrong and also just that its interpreted (or applied) wrong.
To say the truth I am a bit lost in this thread. I haven't got the point about what is being discussed here. So don't take my previous comments seriously. I think I have misunderstood what Thomas meant. Anyway, all what I posted I think is true (although maybe not related with the subject of this thread by the way).
First of all, the 'incompressibility' of a gas is clearly a very relative property. All gases are in principle compressible as there is a lot of empty space between the molecules. It is just that for certain densities and temperatures one needs large forces to compress it, i.e. the compression will in many cases only be small and consequently one may treat it then as 'incompressible'. Obviously, the density can in this case be taken as constant, but the fact that the ideal gas law p=nkT is restricted by the condition n=const. does not mean that it it is not valid anymore (see for instance http://www.grc.nasa.gov/WWW/K-12/airplane/eqstat.html where the ideal gas law is in fact used in connection with aerodynamics). You mentioned that viscous dissipation should lead to a variation of the temperature along the streamline of an airfoil, but a) the air is usually assumed to be non-viscous for these basic considereations of aerodynamic lift and b) dissipation of energy would anyway lead to an increase of the temperature and hence the static gas pressure, whereas what you want to produce lift is a pressure decrease.
I repeat again I think I am out of the aim of this thread. If you are treating with Incompressible substances, the equation of state is not valid and not needed to solve the flow field. If you are treating with Gas dynamic at low Mach Numbers but allowing a short compressibility, the equation of state is another constitutive equation in order to solve the Navier Stokes equations. I don't know how basic is the lifting force, but it is CLOSELY related to the viscous behavior of the fluid.
Clausius2: You might consider viewing the following thread prior to engaging in a discussion with Thomas2: https://www.physicsforums.com/showthread.php?t=64384
In the meanwhile I have looked into some of the issues that have come up in the course of this discussion in some more detail. Despite looking through several aerodynamics theory books , I could actually not find any indication that the pressure profile along an airfoil should be temperature related (as Clausius suggested above). As mentioned by me already, this raises the question how the static pressure can change if the density stays the same (which it should under the assumption of an incompressible flow). I also found some apparent confusion in the literature regarding the definition of 'viscosity' and 'friction' in this context. In aerodynamics theory, an inviscous fluid is claimed to result in a flow without drag on objects (see http://www.centennialofflight.gov/essay/Theories_of_Flight/Ideal_Fluid_Flow/TH7.htm , http://astron.berkeley.edu/~jrg/ay202/node95.html ). This assumption actually equates friction with viscosity, which is clearly incorrect. Viscosity is generally defined as the presence of 'shear stress' i.e. a friction between parallel streamlines. However, even if the viscosity is zero in this sense (i.e. if the molecules of a gas are assumed not to be interacting with each other), there will be a friction and a resulting momentum loss by molecules hitting the object and being elastically reflected (this is easy to show from the energy and momentum conservation laws for an elastic collision of two objects). Even in an ideal non-viscous gas, an object will therefore lose momentum due to friction and hence experience drag (a plane surface whose normal has an angle alpha to the gas flow experiences a normal force proportional to cos(alpha) and thus the drag is proportional to cos^2(alpha) whereas the lift is ~sin(alpha)*cos(alpha) ).