J.C.MacSwell Posted February 14, 2015 Posted February 14, 2015 With respect I see nothing steady about the flow in the post83 example, which I have offered to discuss. Ok. Obviously that is not steady, and much more complicated than what would start out as as a constant velocity horizontal flow. I think you have to start by making some simplifying assumptions. when you have hard corners, changing cross sectional area, and varying pressure head before you even get to the airfoil.
studiot Posted February 14, 2015 Posted February 14, 2015 J.C.MacSwell, You seem to know a thing or two about fluids so I don't want to try to "teach my grandfather to suck eggs". The analysis of fluid action can be approached from at least two viewpoints. The Eulerian approach of following a parcel of fluid and calculating its path (position), change of shape, orientation and other properties. The control volume approach of fixing a specific volume in space so that it does not vary in shape or position and calculating the flows of energy, momentum and mass across its boundaries. The point I have been trying to make is how to select the most appropriate for the issue at hand and in particular that these two approaches should not be mixed, as Zet is trying to do. The examples in post83 are best handled by the control volume approach. The entire L shaped duct is made into the control volume. Attempting to use this control volume approach to the open air flow will fail because the volume must extend to infinity, meaning that the energy supply is infinite so the prime statement/assumption 'energy is conserved' is violated.
cxxLjevans Posted February 14, 2015 Posted February 14, 2015 J.C.MacSwell, You seem to know a thing or two about fluids so I don't want to try to "teach my grandfather to suck eggs". The analysis of fluid action can be approached from at least two viewpoints. The Eulerian approach of following a parcel of fluid and calculating its path (position), change of shape, orientation and other properties. The control volume approach of fixing a specific volume in space so that it does not vary in shape or position and calculating the flows of energy, momentum and mass across its boundaries. The point I have been trying to make is how to select the most appropriate for the issue at hand and in particular that these two approaches should not be mixed, as Zet is trying to do. The examples in post83 are best handled by the control volume approach. The entire L shaped duct is made into the control volume. Attempting to use this control volume approach to the open air flow will fail because the volume must extend to infinity, meaning that the energy supply is infinite so the prime statement/assumption 'energy is conserved' is violated. 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.
studiot Posted February 14, 2015 Posted February 14, 2015 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. Larry, how would you start the motion?
cxxLjevans Posted February 14, 2015 Posted February 14, 2015 Larry, how would you start the motion? 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.
studiot Posted February 14, 2015 Posted February 14, 2015 This is exactly where using the apparatus as a control volume comes in handy. Zet specified (I asked him) that there was a vacuum above the top surface of the fluid in the column. So whether the fluid would move at all depends upon the height of the column and gravity. Using a control volume would specify a sensible pressure upon the top surface, opposed by a side pressure from atmosphere when the door was opened.
J.C.MacSwell Posted February 14, 2015 Posted February 14, 2015 If you like I will show how Bernoulli and streamlines should be applied to the example in post 83 OK, let's see how you approach this. Let us know any assumptions you make that you feel might not be obvious.
Zet Posted February 14, 2015 Author Posted February 14, 2015 It would be strange. I don't think different airfoils produce the "same" effect (lesser increase in ...the decrease in the swirl), even if no lifting occurs. I think you should be able to see the differences "in the exit fluid" for each different airfoil. I'd expect some combination of extra turbulence and decreased velocity would be found in the cases with poor airfoil design ...which could be converted (theoretically) into lifting force in the case of a well-designed airfoil. Okay, I misunderstood you when I said it would seem strange to me but not to you that there would be the same lesser increases in the swirl in the rising cases where airfoils of equal masses and volumes but of different shapes rise. Cool. But, I think, you are still saying that (it is not strange to you that) there would be the same overall decrease in the kinetic energy (in all off its various forms, including horizontal and swirling motions) of the fluid (when the differently shaped airfoils of equal mass and equal volume all rise the same distance) to offset the same increases in gravitational potential energy (in the form of these differently shaped risen airfoils)? Yes? No? If I now have your position right ... it still seems strange to me. For me, this is counterintuitive. I would think that when differently shaped airfoils of equal mass and volume all rise the same distance that there would be different amounts of swirl plus horizontal motion and that it wouldn’t all add up in the end to be the same decrease in kinetic energy (overall) in every case. But, that’s just my gut. I wouldn't expect that you could get various poorly-designed airfoils to rise, by using the same amount of 'input' energy, so your second point quoted above should make sense after accounting for the change in input energy needed to get different airfoils to rise. Or maybe I still don’t understand your position. The “input” energy in this system is a certain amount of gravitational potential energy in the L shaped container of fluid. As the fluid drains out of the container (and so there is a decrease in this gravitational potential energy) there is an increase in kinetic energy in the form of the moving fluid. When the moving fluid encounters the airfoil in the horizontal part of the L shaped container, there is a greater decrease in the pressure pushing down on the top of the solid and a lesser decrease in the pressure pushing up on the bottom of the solid. And so, the otherwise neutrally buoyant body will rise (if allowed to do so). The “input” energy at this point (the only option for the rise of the airfoil) is the kinetic energy of the moving fluid (the original gravitational potential energy of the fluid is beyond the scope of the analyses at this point as the fluid moves horizontally across the bottom of the L. And we can assume that this fluid ends up in the same bucket at the same height below the same L shaped container in every case in the end and so there is the same decrease in gravitational potential energy in the form of falling fluid in each case. And so this cannot be the offsetting form of energy for the increase in gravitational potential energy in the rise of differently shaped airfoils). And so, if differently shaped airfoils (all of equal mass and volume) are all in frictionless tracks that allow them to rise the same distance (and if the original height of the L shaped column of fluid and so the subsequent flow of fluid is far beyond that needed to get all of the differently shaped airfoils to all rise that same small little height), then there is the same increase in gravitational potential energy in this form, in the end, in each case, and so ... for energy to be conserved ... there must be the same decrease in another (the “input”) form of energy. It is the flow of the fluid over and under the airfoil (and the subsequent different decreases in pressures) that causes the airfoil to rise. But, I would expect, as it appears perhaps you do too, that the “input” energy (the (gravitational potential energy turned into) kinetic energy) would be different (different amounts of kinetic energy (in all its various motions)) in the end. No? So, perhaps I’m still missing your point. (I do that a lot.) The idea that there would be the same decrease in the overall amount of kinetic energy (in all of its various forms) in the end for each rising differently shaped airfoil case seems counterintuitive ... to me. (But, that my intuitions are wrong, is no surprise either.) ? ...that while these statements are valid, they may raise objections since the statements can't be used to define workable parameters for describing lift, which Bernoulli does do. But for a general understanding how the forces (or the energy) will balance, they seem fairly good. Right. There are two different ... related and overlapping ... analyses here. One is the mechanics of a rising airfoil with no angle of attack. The other is the analysis of how energy is conserved. And while they are two different analyses they both must work in tandem. When an airfoil rises the mechanics of this dynamic must work out in such a way so as energy is conserved. If, mechanically, when an airfoil rises as opposed to when it does not then if in the end there is less overall motion of the fluid (less kinetic energy) then this (mechanical reason) resolves the overlapping conservation of energy analysis where the increase in gravitational potential energy of the rising airfoil (plus the increase in vertical kinetic energy) is offset (logically) by an equal decrease in the overall kinetic energy of the moving fluid. (Whether this is happening in real physical life is another matter and only answered by experimentation, but, for the purposes in this thread, the logic works.) And then, in the next step in this line of analysis where I compared differently shaped airfoils of equal mass and volumes all rising the same difference, I used these two overlapping analysis in the reverse (to find the former). If we know energy must be conserved (which is must in this forum) then we know that when the differently shaped airfoils of equal mass and volume all rise the same distance (and so there is the same increase, in the end, in gravitational potential energy, in this form) that (... even though it seems counterintuitive to me ...) there must be mechanically the same decrease in the overall amount of kinetic energy of the moving fluid (in all of it various motions) in each differently shaped rising case in order for energy to be conserved (since there are no other energy change options). And we can know this to a logical certainty. So, you’re right, I did dodge the direct question of the mechanics of differently shaped airfoils rising and whether or not this would lead to the same lesser overall kinetic energy in the rising versus the non-rising cases. But, logically, I got there just the same. For energy to be conserved, there must be the same lesser amount of kinetic energy (in all it various motions) of the fluid for every risen differently shaped airfoil of equal mass and equal volume to offset the equal increases, in the end, in gravitational potential energy (in the form of these risen bodies). No? And if the Mechanics of Physics are within the bounds of logic (which it must be) then we know this is mechanically what happens ... and we know this to a logical certainty. No? ? --- Would the it be simpler to answer Zet's question if the shape of the container were a U instead of an L? I don’t know. If it is, I’m cool with changing it from an L to a U. The logic and the issue remain the same. ... oscillate forever ... Yep. 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. Yep. 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. Yep. For energy to be conserved, then in the non-rising case the fluid must reach the same height (on the right side) as when it first started (on the left side). And, for energy to be conserved, then in the rising case (where there is an increase in gravitational potential energy in the form of the risen body) there must be an equal decrease in another form of energy. And the only energy change option is the kinetic energy of the moving fluid as it passes over and under the airfoil and thus the height is rises to on the right hand side. The logic and the issue remain the same. Cheers!
cxxLjevans Posted February 14, 2015 Posted February 14, 2015 And, for energy to be conserved, then in the rising case (where there is an increase in gravitational potential energy in the form of the risen body) there must be an equal decrease in another form of energy. And the only 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
studiot Posted February 14, 2015 Posted February 14, 2015 For energy to be conserved, then in the non-rising case the fluid must reach the same height (on the right side) as when it first started (on the left side). Well I've never seen a fluid than can support a vertical free surface like that.
Zet Posted February 14, 2015 Author Posted February 14, 2015 (edited) I'm not sure there is any increase in gravitational potential energy due to rise of the airfoil. Right. If the neutrally buoyant solid starts out in the fluid and then rises in the fluid and remains in the fluid, then there is no increase in gravitational potential energy in both the non-rising and rising cases. There is, rather, an increase in the vertical kinetic energy of the rising solid and displaced fluid downwards in the rising case that does not occur in the non-rising case. This is one way to formulate the question. However, if the non-rising solid and the rising solid (neutrally buoyant when in the fluid when static) emerge from the fluid in the end, then, with the one solid is higher than the other, there is a difference in gravitational potential energies. This is another way to formulate essentially the same question. (I find it easier to talk about differences in gravitational potential energies between the two airfoils in the end, rather than differences in vertical kinetic energies. But, either way, the logic and the issue remain the same.) Edited February 15, 2015 by Zet
MigL Posted February 15, 2015 Posted February 15, 2015 Any elementary text on aerodynamics of flight has a an analysis of the forces involved. one of these forces is drag and it has several components. Even in the case where we consider frictionless surfaces, and so eliminate form-drag, if lift is produced there is always drag-due-to-lift produced. A flat horizontal plate that produces no lift will not produce any drag-due-to-lift as it doesn't affect the fluid in that way. This drag-due-to-lift is produced whether the airfoil rises or not. The constrained airfoil produces the exact same lift, and associated drag-due-to-lift as the rising airfoil. It is in effect the energy analysis that you are doing which is incorrect Zet. If we consider the vertical forces only, you have a force of lift and a force of gravity. If the airfoil is producing lift, that is the NET difference between the two forces, one acting upward and one downward. That means, as the airfoil RISES ( kinetic ) its net potential DECREASES. Unless constrained, an object will always trade off potential for kinetic energy and move when subjected to a force. As studiot has repeated many times, you cannot just 'isolate' gravitational potential in your energy analysis and then wonder why it doesn't add up.
Zet Posted February 15, 2015 Author Posted February 15, 2015 (edited) . That means, as the airfoil RISES ( kinetic ) its net potential DECREASES. So, when the fluid moves over the airfoil there is a certain amount of "dynamic" potential energy. If the airfoil is free to rise, it will. When it rises this potential energy becomes kinetic energy. Okay ... cool. What then happens to this "dynamic" potential energy when the airfoil does not rise ... and after the fluid passes and this potential energy is then gone? The Law of Conservation of Energy states that energy can change forms, but the total amount remains. Are you suggesting that there was an amount of energy in the form of "dynamic" potential energy in the non-rising case and then this amount of energy is then gone (without an offsetting change in another form of energy)? If so, you are suggesting a violation of the Law of Conservation of Energy. ? Edited February 15, 2015 by Zet
MigL Posted February 15, 2015 Posted February 15, 2015 The fluid stops providing lift to the airfoil when either the airfoil stops being thrust forward through the fluid or the fluid stops being thrust past the airfoil ( equivalent ). At this point the airfoil still has the gravitational potential that was being overcome by the lift ( dynamic ? ) potential previously, and translates this into kinetic energy downward. Note that now another source of energy has been removed. That is the thrust of the airfoil or fluid. Note also that a 'falling' airfoil glides ( a well designed glider can move 15-20 times farther horizontally than it does vertically ) as it descends
Essay Posted February 17, 2015 Posted February 17, 2015 Okay, I misunderstood you when I said it would seem strange to me but not to you that there would be the same lesser increases in the swirl in the rising cases where airfoils of equal masses and volumes but of different shapes rise. Cool. But, I think, you are still saying that (it is not strange to you that) there would be the same overall decrease in the kinetic energy (in all off its various forms, including horizontal and swirling motions) of the fluid (when the differently shaped airfoils of equal mass and equal volume all rise the same distance) to offset the same increases in gravitational potential energy (in the form of these differently shaped risen airfoils)? Yes? No? .... No, not the same …because you’d need very different (extra) amounts of initial KE, in order to achieve any lift (increased GPE + vertical KE) for those different airfoils. That may be specified in your OP initial assumptions, but after you introduced "different airfoils" I assumed you’d need a much larger ‘head’ of fluid, compared with a well-designed airfoil, to achieve an equal distance of (or any) lift. This assumption (on my part) explains our misunderstanding about these subsequent modifications, but I think the assumption is necessary for these cases with differently-shaped, less-effective, airfoils. This should also answer your following comments and posts. With the extra (differing) initial KE, then you can expect the "different amounts of swirl plus horizontal motion" will balance with the "same 'differing' decrease in kinetic energy (overall) in every case." ...or words to that effect. ~
Zet Posted February 18, 2015 Author Posted February 18, 2015 Yeah. I tried to set up my thought experiment to avoid issues of “thrust” or “different amounts of initial kinetic energy” and just get to where I’m having a problem (different decreases in pressures on the top and bottom adding to the rise of the airfoils). I tried, ... but I don’t think I succeeded. Airfoils are complicated things, and I probably should stay away from them. I do thank you all for adding to my understanding of the issue. Whether you all believe me or not, I have learned from this discussion. Thank you for all of your time, effort, and ... patience! 2
studiot Posted February 18, 2015 Posted February 18, 2015 (edited) i JCMacSwell post#132 OK, let's see how you approach this. Let us know any assumptions you make that you feel might not be obvious. I did say I would post more detail. This will take several posts and I have been trying to get some stuff together for this. This post will be an introductory post to collect together some preliminaries, for ease of referral. First dimensional analysis is important in Physics, but particularly important in fluids. Some useful dimensions are Now it can clearly be seen that neither pressure nor velocity is energy. So in order to write an equation we need to bring all our terms to a common denominator, preferably a simple one. Engineers have traditionally used 'head' as this common denominator, which has the dimensions of length. Energy is just too complicated. Here are some quantities that have the dimensions of length or head. [math]{\rm{Head}} = \frac{{{\rm{Energy}}}}{{{\rm{Force}}}}{\rm{ = }}\frac{{{\rm{Energy}}}}{{{\rm{Weight}}}}{\rm{ = }}\frac{{{\rm{Pressure x Volume}}}}{{{\rm{Force}}}} = \frac{{{{\left( {{\rm{Velocity}}} \right)}^{\rm{2}}}}}{{{\rm{Acceleration}}}}{\rm{ = }}\frac{{{\rm{Pressure x Volume}}}}{{{\rm{Mass x Acceleration}}}}{\rm{ = }}\left\{ {\frac{{{\rm{Pressure}}}}{{{\rm{Acceleration}}}}{\rm{x}}\frac{{\rm{1}}}{{{\rm{Density}}}}} \right\} = {\rm{Length}}[/math] Now we need some equations of motion to employ these quantities. The simplest form of Bernoulli's equation (in terms of head) as applied to a pipe or duct as in post 83 is [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}}[/math] Note that all these terms have the dimensions of length. This equation is applied to a pipe running full, it does not apply to a pipe only partly full. Further the fluid must act in an incompressible way. However it can be used (with modifications) for a pipe with or without friction, The fluid can have viscoscity or be inviscid And we can introduce machines such as Zet's airfoil in the stream as some sections of the pipe. Each of these adds a term to the equation, increasing or reducing the head at any point. So friction and fluid driven machinery (eg airfoils) introduce a negative head, whilst pumps introduce a positive head, section 1 (subscripts) is taken before the extra and section 2 after it. [math]\frac{{V_1^1}}{{2g}} + \frac{{{P_1}}}{{\rho g}} + {z_1} \pm h = \frac{{V_2^1}}{{2g}} + \frac{{{P_2}}}{{\rho g}} + {z_2}[/math] Where +h is the work added by a pump or -h taken out by a turbine. Compressible fluids, principly gases, have several more terms and are best dealt with by what is known as the enrgy equation. [math]\frac{{V_1^1}}{2} + {P_1}{v_1} + g{z_1} + {U_1} + Q - W = \frac{{V_2^1}}{2} + {P_2}{v_2} + g{z_2} + {U_2}[/math] Finally we will need an estimate of the work done by the airfoil machine against gravity. The work equation for any object lefted in a gravitational field is [math]W = \left( {\frac{1}{2}mV_2^2 - \frac{1}{2}mV_1^2} \right) + \left( {mg{z_2} - mg{z_1}} \right)[/math] We can convert this to a head loss by dividing W by the airfoil weight, which will produce some interesting results when applied to the post83 setup. Edited February 18, 2015 by studiot
cxxLjevans Posted February 19, 2015 Posted February 19, 2015 (edited) [snip] The work equation for any object lefted in a gravitational field is [math]W = \left( {\frac{1}{2}mV_2^2 - \frac{1}{2}mV_1^2} \right) + \left( {mg{z_2} - mg{z_1}} \right)[/math] [snip] 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? Edited February 19, 2015 by cxxLjevans
studiot Posted February 19, 2015 Posted February 19, 2015 Larry, I didn't notice the density appearing in my equation, so how could it affect it? Please bear with me I haven't labelled my variables etc, as I hope that all will become clear in the next post(s). But your post has encouraged me that the interest is still there to complete this discussion.
cxxLjevans Posted February 19, 2015 Posted February 19, 2015 (edited) Larry, I didn't notice the density appearing in my equation, so how could it affect it? Please bear with me I haven't labelled my variables etc, as I hope that all will become clear in the next post(s). But your post has encouraged me that the interest is still there to complete this discussion. 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). Edited February 19, 2015 by cxxLjevans 1
studiot Posted February 19, 2015 Posted February 19, 2015 (edited) Thank you, Larry, for picking that up. Unfortunately I can't now correct post142. The energy equation should, of course contain velocity squared terms. [math]\frac{{V_1^2}}{2} + {P_1}{v_1} + g{z_1} + {U_1} + Q - W = \frac{{V_2^2}}{2} + {P_2}{v_2} + g{z_2} + {U_2}[/math]...........3 I started by using all capitals for quantities, then realised I had two Vs, one for velocity and one for volume and things wnet awry when I sorted that. The work equation does not apply to distributed mass such as fluids, it applies to identifable bodies such as airfoils. Edit I see the error carried right through so the proper Bernoulli equations are without losses/inputs [math]\frac{{V_1^2}}{{2g}} + \frac{{{P_1}}}{{\rho g}} + {z_1} = \frac{{V_2^2}}{{2g}} + \frac{{{P_2}}}{{\rho g}} + {z_2}[/math].............1 and with losses/inputs [math]\frac{{V_1^2}}{{2g}} + \frac{{{P_1}}}{{\rho g}} + {z_1} \pm h = \frac{{V_2^2}}{{2g}} + \frac{{{P_2}}}{{\rho g}} + {z_2}[/math]...............2 I have also now numbered the equations for future reference. Edited February 19, 2015 by studiot
Essay Posted February 20, 2015 Posted February 20, 2015 Yeah. I tried to set up my thought experiment to avoid issues of “thrust” or “different amounts of initial kinetic energy” and just get to where I’m having a problem (different decreases in pressures on the top and bottom adding to the rise of the airfoils). I tried, ... but I don’t think I succeeded. Airfoils are complicated things, and I probably should stay away from them. I do thank you all for adding to my understanding of the issue. Whether you all believe me or not, I have learned from this discussion. Thank you for all of your time, effort, and ... patience! You succeeded in furthering my insight into lift, so thanks; and as a thought experiment, you should be able to make initial conditions all the same, for the various cases ...if you use lots of extra thrust (or whatever force imbalance), which you then account for (minus changes) in the final flow. Then you should prove that changes in lift will balance out with changes in turbulence or drag or.... You might enjoy the points made about flight, in this PBS 'Nature' documentary on OWLS, and how they manage "silent" flight. I think some owls may even have "noise-cancelling" abilities, evolved into their feathers and wing control. ...lots of curvature in those oversized wings. Still hard to imagine their 'striking force' is 12x body weight. Now that is silent but deadly.... ~
cxxLjevans Posted February 20, 2015 Posted February 20, 2015 I did say I would post more detail. This will take several posts and I have been trying to get some stuff together for this. This post will be an introductory post to collect together some preliminaries, for ease of referral. First dimensional analysis is important in Physics, but particularly important in fluids. Some useful dimensions are Now it can clearly be seen that neither pressure nor velocity is energy. So in order to write an equation we need to bring all our terms to a common denominator, preferably a simple one. Engineers have traditionally used 'head' as this common denominator, which has the dimensions of length. Energy is just too complicated. Here are some quantities that have the dimensions of length or head. [math]{\rm{Head}} = \frac{{{\rm{Energy}}}}{{{\rm{Force}}}}{\rm{ = }}\frac{{{\rm{Energy}}}}{{{\rm{Weight}}}}{\rm{ = }}\frac{{{\rm{Pressure x Volume}}}}{{{\rm{Force}}}} = \frac{{{{\left( {{\rm{Velocity}}} \right)}^{\rm{2}}}}}{{{\rm{Acceleration}}}}{\rm{ = }}\frac{{{\rm{Pressure x Volume}}}}{{{\rm{Mass x Acceleration}}}}{\rm{ = }}\left\{ {\frac{{{\rm{Pressure}}}}{{{\rm{Acceleration}}}}{\rm{x}}\frac{{\rm{1}}}{{{\rm{Density}}}}} \right\} = {\rm{Length}}[/math] Now we need some equations of motion to employ these quantities. The simplest form of Bernoulli's equation (in terms of head) as applied to a pipe or duct as in post 83 is [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}}[/math] Note that all these terms have the dimensions of length. This equation is applied to a pipe running full, it does not apply to a pipe only partly full. Further the fluid must act in an incompressible way. However it can be used (with modifications) for a pipe with or without friction, The fluid can have viscoscity or be inviscid And we can introduce machines such as Zet's airfoil in the stream as some sections of the pipe. Each of these adds a term to the equation, increasing or reducing the head at any point. So friction and fluid driven machinery (eg airfoils) introduce a negative head, whilst pumps introduce a positive head, section 1 (subscripts) is taken before the extra and section 2 after it. [math]\frac{{V_1^1}}{{2g}} + \frac{{{P_1}}}{{\rho g}} + {z_1} \pm h = \frac{{V_2^1}}{{2g}} + \frac{{{P_2}}}{{\rho g}} + {z_2}[/math] Where +h is the work added by a pump or -h taken out by a turbine. Compressible fluids, principly gases, have several more terms and are best dealt with by what is known as the enrgy equation. [math]\frac{{V_1^1}}{2} + {P_1}{v_1} + g{z_1} + {U_1} + Q - W = \frac{{V_2^1}}{2} + {P_2}{v_2} + g{z_2} + {U_2}[/math] Finally we will need an estimate of the work done by the airfoil machine against gravity. The work equation for any object lefted in a gravitational field is [math]W = \left( {\frac{1}{2}mV_2^2 - \frac{1}{2}mV_1^2} \right) + \left( {mg{z_2} - mg{z_1}} \right)[/math] We can convert this to a head loss by dividing W by the airfoil weight, which will produce some interesting results when applied to the post83 setup. 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?
studiot Posted February 21, 2015 Posted February 21, 2015 (edited) So the purpose of the collection was to look at zet's post83 'experiment'. It is suprising what can be deduced from a simple static situation in the bent tube. Zet is right to try to simplify, but it takes experience to know what can be left out and what must be included and the idea is to examine this as we go along. I have drawn a square tube of side 'a', since flat surfaces are easy to work with and section area calcuations are simple. The initial setup is shown in Fig1 with elevation (z) reckoned upwards from the base of the horizontal part of the tube. The tube is shown as constant cross section,a x a where a is small compared to the height of the fluid level in the vertical column. This allows a simple representation of the fluid pressure in the lower horizontal part of the tube as the pressure at z=0. Talking of pressure it is worth noting what is meant by 'fluid pressure' and I have inset Fig1a to show this. Consider a small cube within the fluid, shown solid. This cube has a face ABCD on which the cube exerts a fluid pressure PR as shown. The fluid in the adjacent cube, shown dashed, must push back with an equal and opposite pressure PL, or the cube would move. This must, of course be true of the other five faces of the cube. If the face of the cube is against a container, not more fluid, then the container, must supply this pressure instead. These can also be cast in terms of Newton's 3rd law, using force instead of pressure. I will return to these comments later. Now that we have our tube set up, but capped off at section 6 so the fluid can't flow out. To start with let us examine the flow without the airfoil and introduce it later at section 5. Our control volume is from section 1 to section 2, as shown in the green box in Fig2. Since the column is vertical these sections are horizontal. Section 1 is the top free surface of the fluid. The pressure here is atmospheric. In the inset in Fig2a I have returned to my earlier point about the pressure on the walls of the container. The forces to the left (FL) must be equal and opposite to the forces to the right (FR) or again the whole container will move. Again since the fluid is pressing on the container, the container pushes back equally. This may be stating the obvious, but it will become important, along with momentum later. This now leads to Fig3 and a Bernoulli analysis at sections 1 and 2. This is where working in terms of head comes into its own. There are three terms in the simple Bernoulli equation I called equation 1 in post146. At section1 the elevation head is simply z1 Since the velocity is zero the velocity head is zero And the pressure is the ambient (atmospheric) pressure at the free surface. At section 2 The velocity head is still zero The elevation head is now zero, since z2=0 and the pressure head has therefore increased. I have shown this in the equation although I have used vertical section 4 to save a drawing (Remember I said keep 'a' small so we can consider the pressure the same over the whole of vertical sections) This enables us to calculate the pressure along the whole of the horizontal section of fluid. If we now suddenly remove the end cap at section 6 so the fluid flows out, as shown in Fig4. Fig 4 is also about the law of conservation of mass and continuity. Since our fluid is incompressible the same quantity of fluid that flows aout through section 6 must also flow through each other section as shown in red. But since all the sections have the same cross section this means that the velocity has constant magnitude throughout the fluid. If we change thearea at section 5 by introducing the airfoil we must change the velocity to compensate. Unless we know the volumetric flow rate we cannot calculate this changed velocity at section 5 to put into Bernoulli. Fig 5 returns to the static situation before endcap removal for a bit more analysis. Note clearly that the velocity here is zero throughout the fluid. Therefore the momentum is zero throughout the fluid. This comment is important to the airfloil lift, which see the same mechanism about to be discussed, in relation to momentum. I showed in inset Fig 2a that the pressures and forces on the walls of the container balance, and that the wall supply a reaction to the fluid. If we suddenly remove part of those walls in the form of the cap at section 6 we must also remove that reaction. We replace that reaction with the ambient (atmospheric) pressure Pa This explains why the fluid starts to move. The fluid is still at pressure head of (z1 + Pa/rhog) whilst this is opposed by Pa/rhog alone. So the driving head is the difference, z1, as shown inf Fig 6, which allows us to calculate the starting exit velocity of the fluid. Before the cap is removed there is zero momentum. After removal the fluid possesses momentum to the right. So there must be some equal and opposite momentum to the left, because momentum is a conserved quantity. The rate of change of momentum constitutes a force and this force acts on vertical section 3 of the container, pushing it back of the left. This is, of course, how a rocket works. The changes that occur as the level falls require calculus to address and I will leave to a possible later post, along with the application of the Work equation, I have called equation 3, in post 146 to estimating the head loss due to the airfoil. Hopefully this is enough to convince you that because of the interaction with the container walls the analysis of contained fluids is different from that of a semi infinite fluid like an atmosphere. However it is always encouraging when Newton's laws, conservation laws and any specialist laws applicable to the subject in hand, all offer the same answer and fit together nicely. Edited February 22, 2015 by studiot
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