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Then you think wrong. A space ship is a closed system...but we'll get back to that. if I have a nozzle at one end of a long pipe and the entire pipe is at 14.7 psi.
lets say the pipe is 60 inch diameter and 500 ft long. and its inside your space ship.
and I have a compressor at the other end of the pipe that compresses air to 100 psi.
if I release the compressed air through the nozzle the reaction of the thrust will be felt by the pipe until the pressure in the pipe reaches 100 psi.
and if the compressor can maintain the 14.7 psi then this can be continuous. Absolutely. But you're missing the "closed system" bit. not only that but the friction between the pipe and the compressed air in the pipes that carries the compresed air back to the nozzle will also be a force in the direction of movement. And here is where you are missing the closed system bit. Based on your description you have what could be considered a linear air accelerator  air goes in the front, is compressed, and is then forced out the back. Thus giving you a vacuum at the front, providing a forward "pull", and a rearfacing nozzel, giving a "push". On earth this system would give you a great deal of thrust  in fact, add a bit of fire and you've got a jet engine. Problem is, if you're spraying that air inside of a space ship that thrust disappears  for two reasons. Firstly, nature abhors a vacuum, and your system is generating one at the front of the pump. Hence, there will be an flow of air towards the front of your air pump equal to the air flow entering your pump. Since that air can only come from within your ship, that vacuum will be replaced by shipboard air, and the movement of that air will exert a force equal to the force produced by the vacuum. Newtons laws  opposite and equal reaction. Secondly, the air coming out of your nozzle has a lot of momentum. That momentum is equal to, but in the opposite direction of, the thrust produced by the nozzle end of your engine. Since your ship is a closed system that air cannot leave the ship, and as such has no option but to transfer that momentum back to the ship. This transfer will occur in the form of impacts between the moving column of air and the walls, etc of your ship. This transfer of momentum back to the ship is in a direction opposite to the thrust of your engine. So just as with the "vacuum thrust", the amount of momentum transfered to the ship will be equal to the momentum of the moving column of air, which in turn is equal to the thrust of your engine. Net effect  zero nozzle thrust. So: Vacuum thrust = 0 Nozzle thrust = 0  Total thrust = 0 according to your way of thinking if there was a hose attached to the place where the nozzle is attached instead of a nozzle and the hose was just laying there on the floor and someone turned on the air valve so that air could come out , the hose would just sit there , and it wouldnt move because you were taught that , and you could just walk up to it and pick it up. Not even close to what I said. The hose will quite efficiently transfer the thrust of the air, resulting in movement of the hose. However, my workshop (equivalent in this case to your spaceship) will stay put as the thrust of that hose is countered because the energy of the air coming out of the hose is transfered to the air and walls of my workshop, producing a net thrust (in the context of my workshop) of zero. So the hose will flow around and knock the unaware in the head. But people standing outside of my shop are 100% safe from being run over by a moving shop  momentum, in the context of the shop  is zero. me , I would want the air pressure to be turned off first. so the hose wouldnt slap me in the head as I approached it.
so if the hose will move then if I held the hose , the ship would move. Only if the contents of that hose are free to leave the ship. If they are not, the momentum of the air will be transfered back to the ship, providing a net zero thrust. the problem here is that when you are driving down "think street" and you see a "do not enter" sign you do not enter , instead you turn onto "do not think street" , I dont see any "do not enter" signs.
so I stay on "think street" LOL, big words for a guy obviously lacking an understanding of basic physics. You cannot generate momentum in a closed system that imparts a net momentum on the closed system itself. You ship is such a closed system. Open the system though  eject that air out of the ship itself  and you'll move along quite nicely. Here's the physics you seem to be missing. Read the pages, then come back and we'll talk: http://en.wikipedia.org/wiki/Closed_systemhttp://en.wikipedia.org/wiki/Thermodynamic_systemhttp://en.wikipedia.org/wiki/Newtons_lawshttp://en.wikipedia.org/wiki/MomentumBryan
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wow , I cant believe all you wrote just to show how little you know about physics. maybe I should say your knowlege of applying physics. lets forget the compressor , because according to you and your knowledge about physics the only item in question is the fact that this system is a closed system. and lets put all the compressed air inside a air tank at one end of the ship. and its still inside the 60 inch diameter pipe and the pipe is still 500 ft long. the air pressure inside the pipe is at 14.7 psi. and suppose the air tank holds a mere 1000 cu ft of 100 psi compressed air inside it. the volume of 14.7 psi air inside the 500 ft long pipe is 8817 cu ft. and the pipe volume capacity is 9817 cu ft. so the tank holds 1/9.8 the pipes area. when I release air from the nozzle there is a thrust or a force that pushes the ship in the opposite direction of the thrust. this force is felt by the pipe just as if it were being applied from outside the pipe at the other end of the pipe. the air inside the pipe at 1 atm presses against every sq inch inside the pipe , not in any particular direction, and continues to do so the entire time that air is being released from the nozzle. so all of the pressure inside the pipe itself cancels out. the only force remaining is the force generated by the thrust of the nozzle. so the stuff you argue with = 0 lb force and the stuff I argue with = a 100 lb force to a 0 lb force. if I use a 1 sq inch nozzle. of course the above does not include the decreasing pressure differential between the 14.7 psi and the final equalized pressure or the initial and final fluid velocity. which greatly increases the force. but it alone tells the story. after the air tank has equalized pressure with the pipe the thrust will stop , until then I win the argument / discussion. or until you can provide some type of solid believable calculations that support your claim. if this pipe were in space and you were inside it and the nozzle was pointed directly away from the sun. and of course your space suit could withstand 100 psi of pressure. would you still believe that you are right? I dont believe that we should attempt space travel until those who learn physics also learn to apply physics without killing everybody on a space ship because of what they have not been taught. However, my workshop (equivalent in this case to your spaceship) LOL your workshop equivalent has a mass of 5.98 × 1024 kg its attached to the earth. but put your workshop in space and see what happens.
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wow , I cant believe all you wrote just to show how little you know about physics. This, coming from the guy promoting a perpetual motion machine in another thread...LOL. lets forget the compressor , because according to you and your knowledge about physics the only item in question is the fact that this system is a closed system. Because that is the problem with your own model. Your space ship is a closed system; matter can neither enter nor leave. You can exert forces within such a system, but the sum of those forces will always be zero at the level of the system itself. ...the air inside the pipe at 1 atm presses against every sq inch inside the pipe , not in any particular direction, and continues to do so the entire time that air is being released from the nozzle.
so all of the pressure inside the pipe itself cancels out.
the only force remaining is the force generated by the thrust of the nozzle. However, none of the above has anything to do with the forces in your closed system. The propulsive force generated where the air leaves the pipe is due to the momentum of the moving air; not friction, etc, within the pipe (in fact, friction in the pipe reduces thrust by slowing the air). Its all newtons laws  you use pressure to accelerate a mass of air in one direction, and you get a an equal force on the opposite that'll push on your tank of compressed air. Momentum, BTW, is a specific physical quantity equal to mass * velocity. Momentum is also conserved in a closed system  you can neither add it, nor take it away. after the air tank has equalized pressure with the pipe the thrust will stop , until then
I win the argument / discussion. Sorry, you don't. Missing half the equation does not equal a valid answer. I'll walk through this slowly; hopefully you'll see where your error is. The pressurized air in the tank represents a stored (potential) energy. Upon letting air out of the tank, that potential energy generates a force which accelerates the air down the pipe. Upon leaving the pipe the air will have a momentum, defined by p = mass * velocity. The ship will have an equal momentum in the opposite direction, due to newtons 3rd law (every action has an opposite, but equal reaction). For the sake of ease, lets assume that 1kg of air is moving at a velocity of 100m/s. This will give the air a momentum of: p(air) = mv p(air) = 1kg*100m/s p(air) = 100kg*m/s Newtons 3rd law dictates our tank (and therefore the ship it is attached to) has the same momentum but in the opposite direction, therefore: p(tank) = p(air) p(tank) = 100kg*m/s Lets assume our ship/tank combo weighs 100kg. In this case this release of air would impart a velocity of: p(tank) = mv, therefore v(tank) = p/m v = 100kg*m/s / 100kg v = 1m/s The negative sign in this case simply means the ship is moving in the opposite direction of the air expelled from the tank. In an open system this would lead to the tank going one way and the air going the other  exactly what you propose is happening. In a fully open system (space) that air would continue moving forever  newtons first law (an object in motion stays in motion). But we're not in a open system. In our closed ship that moving air will eventually hit a bulkhead or hull. When this moving column of air meets the bulkhead/hull that impact creates a force which will be proportional to the momentum of the air (in fact, it will be exactly equal to the force which imparted the air its momentum in the first place). As before, newtons 3rd law dictates this will create an equal, but opposite change in momentum on that bulkhead. So in this case we have the air imparting a momentum to the bulkhead of: p(air) = p(bulkhead); since p(air) = 100kg*m/s; p(bulkhead) also = 100kg*m/s keeping in mind p(tank) = p(air) = 100kg*m/s Now, if the bulkhead and tank were separate entities, you'd have the tank going one way, and the air would blow the bulkhead in the other. However, the bulkhead and tank are connected as they are both part of the ship. Therefore, the total momentum imparted on the ship will be equal to the sum of these two momentums imparted onto these different parts of the ship: So we have: p(ship) = p(tank) + p(bulkhead) p(ship) = 100kg*m/s + 100kg*m/s p(ship) = 0kg*m/s Since velocity is a direct product of momentum, (p=mv) v(ship) = p/m v(ship) = 0kg*m/s/100kg v(ship) = 0m/s Bryan
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WOW !!! so in this closed container ( the pipe ). air will only move in one direction? and that direction just happens to be in the direction of the air comming out of the nozzle... how convienient for you. so after all the air has been released out of the tank there will be a 1000 cu ft block of 100 psi air at the end furtherest from the nozzle. I said a believable calculation. in my world , the air at the other end of the pipe will slowly compress along with the rest of the air in the pipe. because pressure is transmitted equally in a closed system. lets take your assumptions even further. lets place air pressure guages on every square inch of surface area outside the pipe. according to your comical assessment the pressure guages will all keep reading 14.7 psi , except the ones located at the end of the pipe. LOL the bulkhead is 500 ft away... what a load Here's the physics you seem to be missing. Read the pages, then come back and we'll talk:
what will we talk about , your fantasies about physics? This, coming from the guy promoting a perpetual motion machine in another thread...LOL.
exactly .. LOL
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WOW !!!
so in this closed container ( the pipe ). air will only move in one direction? Ahh, I see  your pipe is the closed system, and doesn't exhaust into the ship?. If that's the case, it doesn't help you any. Same problem, only confined to a smaller space. so after all the air has been released out of the tank there will be a 1000 cu ft block of 100 psi air at the end furtherest from the nozzle. I'm not 100% sure I've understood your engine. I am assuming: 1) There is a pipe, sealed on one end 2) There is a pressure tank on the other end 3) You release the air in the pressure tank into the pipe Regardless, the math is the same. In this situation you'll release the air, and it will flow into the pipe until you reach a pressure equilibrium  i.e. the pipe and tank are all at the same pressure; no pressure gradient. The initial movement of the air from the tank into the pipe will create a force that will move the ship forward. However, that force will be exactly countered by an equal and opposite force created by the air coming out of the tank interacting with the pipe itself and any air in the pipe. Net result  no thrust. I said a believable calculation. That's the science of it. Whether you "believe" science or not isn't my problem. Momentum, kinetic energy, potential energy and force are all well established physical concepts. If you can find some math that allows for those to move a closed system, let us know  that would be nobelprize kind of work. in my world , the air at the other end of the pipe will slowly compress along with the rest of the air in the pipe. Nope. The air in the pipe will be compressed equally as air moves in from the tank  gasses naturally fill a void in order to generate an even distribution. You may have a slight imbalance if you release the air quickly enough, but this will only exist for a short period of time. lets take your assumptions even further.
lets place air pressure guages on every square inch of surface area outside the pipe.
according to your comical assessment the pressure guages will all keep reading 14.7 psi , except the ones located at the end of the pipe.
Nope, they wouldn't. I think the issue here is we're sealing up different parts of the system. I was describing the case where the pipe is open to the inside of this ship. In this case the pressure of the whole ship would increase. I now think you're describing a case where the pipe is closed, and thus the pressure in the pipe increases, while the remainder of the ship stays the same. In either case, the net thrust is zero  the initial momentum provided by the movement of the gas from the tank into the pipe or ship will be exactly countered by the interactions of that moving air with the stationary pipe (if the pipe is sealed) or ship (if the pipe is open). Net thrust, in either case, is zero. Newtons laws  every action has an opposite and equal reaction. After all, if it worked the way you think it does, NASA would have no trouble sending people to Mars. Just recycle you propellant back and forth. They don't  strange, isn't it. That the rocket scientists at NASA don't know about this imaginary physics you have created... Bryan
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I think you understood the first time. and lets put all the compressed air inside a air tank at one end of the ship.
and its still inside the 60 inch diameter pipe and the pipe is still 500 ft long.
the tank is inside the 500 ft pipe. there is space between the tank and the pipe. so that the air inside the pipe presses against the entire inside area of the pipe. Regardless, the math is the same. what math are you refering to. the since p(air) = 100kg*m/s; p(bulkhead) also = 100kg*m/s
If you begin with 100 psi tank pressure and 14.7 psi external pressure in the pipe. you dont get an immediate equalization between the two. you would end up with apx 10 psi inside the pipe after the pressures have equalized. its a 1000 cu ft tank inside a apx 10,000 cu ft area so the air in the tank expands inside the pipe and the pressure reduces as it expands. the force that you are referring to is fantasy , the same force is applied to both ends and to every square inch of the inside of the pipe. thereby , no net force from the air acting against the pipe. and I might add that the force caused by the (air pressure) is multiplied by the entire inside area of the pipe. not just the end that you choose. ie.. p(bulkhead) also = 100kg*m/s but go ahead and calculate the force that you think is applied to the end of the pipe , it will be the same force that is applied to the opposite end. thus no net effect. I agree the pipe will not move for long , it will only move as long as the air has a differential pressure that will cause a net force in the opposite direction of air flow. exactly like a jet plane would move if it were inside a larger pipe. and if the jet plane was in the larger pipe in space and it pressed against the end of the pipe , it would transfer its momentum to the pipe , and if it kept on thrusting forward the pipe would continue to move forward. but I suppose you have a answer to that also... and it is a nothinker zero net force , right? and in this scenario the air thrust comming out of the tank is exactly the same as thrust comming out of a jet aircraft.
3/4 inch of dust build up on the moon in 4.527 billion years,LOL and QM is fantasy science.




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lets add up all the forces that apply. as the air is escaping from the tank.
the ends of the tank
114.7 psi > 114.7 psi <
the ends of the pipe
14.7 psi > 14.7 psi <
the rear of the nozzle as the air enters the nozzle
114.7 psi > 14.7 psi <
the front of the nozzle as the air escapes the nozzle
114.7 psi > 14.7 psi <
action = reaction
net result of force to pipe 100 psi <
nothing is gained inside the pipe , nothing is lost inside the pipe except the energy that was put into compressing the air , but the pipe moves.
I would wager that the exact amount of energy that is is held in the compressed air ( due to its compression ) inside the tank is the exact amount of energy that is applied for propulsion of the pipe.
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lets add up all the forces that apply. as the air is escaping from the tank.
the ends of the tank
114.7 psi > 114.7 psi < Pressures are potential energy; as in stored energy. You cannot use them directly to determine the force generated upon the release of that pressure. After all, the same 100psi pressure difference with 1000cuft of air has a lot more stored energy than does 1cuft at the same pressure. Pressure, in this case, is transformed into into momentum thourgh two "conversion" processes. The first is through accelerating the air, giving it kinetic energy. The second is through the expansion of the gas as it leaves a tank  a properly designed nozzle can harness that expansion energy and produce thrust above what the kinetic energy of the air alone provides. The exact amount of force produced is determined by: F=[m*Ve]+[(PtPa)*Ae], where m = mass of gas Ve = exhaust velocity (speed material leaves tank) (PtPa) = pressure difference between tank and air Ae = area of nozzle the are is escaping through You may have noticed that in the above formula you're essentially calculating F(total) = F(from kinetic energy of air) + F(from pressure differential) In your example we can fill in a few numbers. A cuft of air weights approx 34g, so 1000cuft weights 34,000g (34kg). Your pressure difference is 100PSI, which is SI units is 6895kPa. Ve is determined by a fairly complex formula: This becomes more complex when you take into account the pressure decreases as the tank empties, but you get the idea. net result of force to pipe 100 psi < Once again, pressure is stored energy and thus cannot be used to determine the net force generated on an object. You need to calculate the force developed by the mass of the gas, combined with the differential pressure between the inside/outside of the tank to determine the force generated. And once again, we come back to the very simple physical principal you insist on ignoring  newtons 3rd law. As the air leaves the tank, both the tank and that air experience an equal, but opposite force. This force imparts both the tank and air with an equal, but opposite, amount of momentum. In an open system the momentum behind that air is dissipated into the surroundings. Where do your propose it goes in your sealed pipe?The answer, of course, is simple  the momentum of that air is transfered to whatever the air encounters. Namely, the momentum is transfered to the pipe as the air encounters the pipe. Since the momentum of the air is equal, but opposite, to the momentum of the tank, the air will impart and equal, but opposite momentum to the pipe. One equal, plus an equal but opposite, equals zero. A + A = 0 nothing is gained inside the pipe , nothing is lost inside the pipe except the energy that was put into compressing the air , but the pipe moves.
I would wager that the exact amount of energy that is is held in the compressed air ( due to its compression ) inside the tank is the exact amount of energy that is applied for propulsion of the pipe. The amount of ***potential energy*** in the compressed air is the maximum amount which can be "extracted" in the form of thrust. But newtons 3rd law dictates that the force generated will act evenly on both the air coming out of the tank as well as on the tank itself. Since that escaping air is contained in the pipe, it's momentum is transfered to the pipe, thus countering the momentum the tank exerts on the pipe. Giving you a net change in the momentum of the pipe of zero. But hey, ff you're so confident why don't you build it, patent it, and make yourself rich. After all, you've solved the single biggest problem in sending spacecraft outside of lowearth orbit  and in doing so refuted newton himself. Or you could accept reality  you are wrong. Bryan
Last edited by ImagingGeek; 06/02/10 03:45 PM.
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Hi Paul,
One more correction is needed... it would take approximately 16 minutes for light to make the round trip between the Earth and Sun.
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you havent show anything except that you can cut and paste equations. and then dump the same ridiculous sob story on me. what have you shown? nothing. you go to a great extent to put up a equation and explain all the factors involved (at least you think thats all) and then you dont even supply any result from the equation.. is that the way it works in college these days , the test ask you which equasion you use to calculate a given set of factors but you are not required to supply a solution. because that is all you do. supply a mass of useless words beneath a equation. lets see if I can solve the problem this way. The foundational axioms of fluid dynamics are the conservation laws, specifically, conservation of mass, conservation of linear momentum (also known as Newton's Second Law of Motion), and conservation of energy (also known as First Law of Thermodynamics). These are based on classical mechanics and are modified in quantum mechanics and general relativity. They are expressed using the Reynolds Transport Theorem. In addition to the above, fluids are assumed to obey the continuum assumption. Fluids are composed of molecules that collide with one another and solid objects. However, the continuum assumption considers fluids to be continuous, rather than discrete. Consequently, properties such as density, pressure, temperature, and velocity are taken to be welldefined at infinitesimally small points, and are assumed to vary continuously from one point to another. The fact that the fluid is made up of discrete molecules is ignored. For fluids which are sufficiently dense to be a continuum, do not contain ionized species, and have velocities small in relation to the speed of light, the momentum equations for Newtonian fluids are the NavierStokes equations, which is a nonlinear set of differential equations that describes the flow of a fluid whose stress depends linearly on velocity gradients and pressure. The unsimplified equations do not have a general closedform solution, so they are primarily of use in Computational Fluid Dynamics. The equations can be simplified in a number of ways, all of which make them easier to solve. Some of them allow appropriate fluid dynamics problems to be solved in closed form. In addition to the mass, momentum, and energy conservation equations, a thermodynamical equation of state giving the pressure as a function of other thermodynamic variables for the fluid is required to completely specify the problem. An example of this would be the perfect gas equation of state: where p is pressure, ρ is density, Ru is the gas constant, M is the molar mass and T is temperature. [edit] Compressible vs incompressible flow All fluids are compressible to some extent, that is changes in pressure or temperature will result in changes in density. However, in many situations the changes in pressure and temperature are sufficiently small that the changes in density are negligible. In this case the flow can be modeled as an incompressible flow. Otherwise the more general compressible flow equations must be used. Mathematically, incompressibility is expressed by saying that the density ρ of a fluid parcel does not change as it moves in the flow field, i.e., where D / Dt is the substantial derivative, which is the sum of local and convective derivatives. This additional constraint simplifies the governing equations, especially in the case when the fluid has a uniform density. For flow of gases, to determine whether to use compressible or incompressible fluid dynamics, the Mach number of the flow is to be evaluated. As a rough guide, compressible effects can be ignored at Mach numbers below approximately 0.3. For liquids, whether the incompressible assumption is valid depends on the fluid properties (specifically the critical pressure and temperature of the fluid) and the flow conditions (how close to the critical pressure the actual flow pressure becomes). Acoustic problems always require allowing compressibility, since sound waves are compression waves involving changes in pressure and density of the medium through which they propagate. [edit] Viscous vs inviscid flow Viscous problems are those in which fluid friction has significant effects on the fluid motion. The Reynolds number, which is a ratio between inertial and viscous forces, can be used to evaluate whether viscous or inviscid equations are appropriate to the problem. Stokes flow is flow at very low Reynolds numbers, Re ≪ 1, such that inertial forces can be neglected compared to viscous forces. On the contrary, high Reynolds numbers indicate that the inertial forces are more significant than the viscous (friction) forces. Therefore, we may assume the flow to be an inviscid flow, an approximation in which we neglect viscosity completely, compared to inertial terms. This idea can work fairly well when the Reynolds number is high. However, certain problems such as those involving solid boundaries, may require that the viscosity be included. Viscosity often cannot be neglected near solid boundaries because the noslip condition can generate a thin region of large strain rate (known as Boundary layer) which enhances the effect of even a small amount of viscosity, and thus generating vorticity. Therefore, to calculate net forces on bodies (such as wings) we should use viscous flow equations. As illustrated by d'Alembert's paradox, a body in an inviscid fluid will experience no drag force. The standard equations of inviscid flow are the Euler equations. Another often used model, especially in computational fluid dynamics, is to use the Euler equations away from the body and the boundary layer equations, which incorporates viscosity, in a region close to the body. The Euler equations can be integrated along a streamline to get Bernoulli's equation. When the flow is everywhere irrotational and inviscid, Bernoulli's equation can be used throughout the flow field. Such flows are called potential flows. [edit] Steady vs unsteady flow Hydrodynamics simulation of the Rayleigh–Taylor instability [2]When all the time derivatives of a flow field vanish, the flow is considered to be a steady flow. Steadystate flow refers to the condition where the fluid properties at a point in the system do not change over time. Otherwise, flow is called unsteady. Whether a particular flow is steady or unsteady, can depend on the chosen frame of reference. For instance, laminar flow over a sphere is steady in the frame of reference that is stationary with respect to the sphere. In a frame of reference that is stationary with respect to a background flow, the flow is unsteady. Turbulent flows are unsteady by definition. A turbulent flow can, however, be statistically stationary. According to Pope:[3] The random field U(x,t) is statistically stationary if all statistics are invariant under a shift in time. This roughly means that all statistical properties are constant in time. Often, the mean field is the object of interest, and this is constant too in a statistically stationary flow. Steady flows are often more tractable than otherwise similar unsteady flows. The governing equations of a steady problem have one dimension less (time) than the governing equations of the same problem without taking advantage of the steadiness of the flow field. [edit] Laminar vs turbulent flow Turbulence is flow characterized by recirculation, eddies, and apparent randomness. Flow in which turbulence is not exhibited is called laminar. It should be noted, however, that the presence of eddies or recirculation alone does not necessarily indicate turbulent flow—these phenomena may be present in laminar flow as well. Mathematically, turbulent flow is often represented via a Reynolds decomposition, in which the flow is broken down into the sum of an average component and a perturbation component. It is believed that turbulent flows can be described well through the use of the Navier–Stokes equations. Direct numerical simulation (DNS), based on the Navier–Stokes equations, makes it possible to simulate turbulent flows at moderate Reynolds numbers. Restrictions depend on the power of the computer used and the efficiency of the solution algorithm. The results of DNS agree with the experimental data. Most flows of interest have Reynolds numbers much too high for DNS to be a viable option[4], given the state of computational power for the next few decades. Any flight vehicle large enough to carry a human (L > 3 m), moving faster than 72 km/h (20 m/s) is well beyond the limit of DNS simulation (Re = 4 million). Transport aircraft wings (such as on an Airbus A300 or Boeing 747) have Reynolds numbers of 40 million (based on the wing chord). In order to solve these reallife flow problems, turbulence models will be a necessity for the foreseeable future. Reynoldsaveraged Navier–Stokes equations (RANS) combined with turbulence modeling provides a model of the effects of the turbulent flow. Such a modeling mainly provides the additional momentum transfer by the Reynolds stresses, although the turbulence also enhances the heat and mass transfer. Another promising methodology is large eddy simulation (LES), especially in the guise of detached eddy simulation (DES)—which is a combination of RANS turbulence modeling and large eddy simulation. [edit] Newtonian vs nonNewtonian fluids Sir Isaac Newton showed how stress and the rate of strain are very close to linearly related for many familiar fluids, such as water and air. These Newtonian fluids are modeled by a coefficient called viscosity, which depends on the specific fluid. However, some of the other materials, such as emulsions and slurries and some viscoelastic materials (e.g. blood, some polymers), have more complicated nonNewtonian stressstrain behaviours. These materials include sticky liquids such as latex, honey, and lubricants which are studied in the subdiscipline of rheology. [edit] Subsonic vs transonic, supersonic and hypersonic flows While many terrestrial flows (e.g. flow of water through a pipe) occur at low mach numbers, many flows of practical interest (e.g. in aerodynamics) occur at high fractions of the Mach Number M=1 or in excess of it (supersonic flows). New phenomena occur at these Mach number regimes (e.g. shock waves for supersonic flow, transonic instability in a regime of flows with M nearly equal to 1, nonequilibrium chemical behavior due to ionization in hypersonic flows) and it is necessary to treat each of these flow regimes separately. [edit] Nonrelativistic vs relativistic flows Classical fluid dynamics is derived based on Newtonian mechanics, which is adequate for most applications. However, at speeds comparable to the speed of light Newtonian mechanics is inaccurate and a relativistic framework has to be used instead. [edit] Magnetohydrodynamics Main article: Magnetohydrodynamics Magnetohydrodynamics is the multidisciplinary study of the flow of electrically conducting fluids in electromagnetic fields. Examples of such fluids include plasmas, liquid metals, and salt water. The fluid flow equations are solved simultaneously with Maxwell's equations of electromagnetism. [edit] Other approximations There are a large number of other possible approximations to fluid dynamic problems. Some of the more commonly used are listed below. The Boussinesq approximation neglects variations in density except to calculate buoyancy forces. It is often used in free convection problems where density changes are small. Lubrication theory and HeleShaw flow exploits the large aspect ratio of the domain to show that certain terms in the equations are small and so can be neglected. Slenderbody theory is a methodology used in Stokes flow problems to estimate the force on, or flow field around, a long slender object in a viscous fluid. The shallowwater equations can be used to describe a layer of relatively inviscid fluid with a free surface, in which surface gradients are small. The Boussinesq equations are applicable to surface waves on thicker layers of fluid and with steeper surface slopes. Darcy's law is used for flow in porous media, and works with variables averaged over several porewidths. In rotating systems, the quasigeostrophic approximation assumes an almost perfect balance between pressure gradients and the Coriolis force. It is useful in the study of atmospheric dynamics. [edit] Terminology in fluid dynamics The concept of pressure is central to the study of both fluid statics and fluid dynamics. A pressure can be identified for every point in a body of fluid, regardless of whether the fluid is in motion or not. Pressure can be measured using an aneroid, Bourdon tube, mercury column, or various other methods. Some of the terminology that is necessary in the study of fluid dynamics is not found in other similar areas of study. In particular, some of the terminology used in fluid dynamics is not used in fluid statics. [edit] Terminology in incompressible fluid dynamics The concepts of total pressure and dynamic pressure arise from Bernoulli's equation and are significant in the study of all fluid flows. (These two pressures are not pressures in the usual sense—they cannot be measured using an aneroid, Bourdon tube or mercury column.) To avoid potential ambiguity when referring to pressure in fluid dynamics, many authors use the term static pressure to distinguish it from total pressure and dynamic pressure. Static pressure is identical to pressure and can be identified for every point in a fluid flow field. In Aerodynamics, L.J. Clancy writes[5]: To distinguish it from the total and dynamic pressures, the actual pressure of the fluid, which is associated not with its motion but with its state, is often referred to as the static pressure, but where the term pressure alone is used it refers to this static pressure. A point in a fluid flow where the flow has come to rest (i.e. speed is equal to zero adjacent to some solid body immersed in the fluid flow) is of special significance. It is of such importance that it is given a special name—a stagnation point. The static pressure at the stagnation point is of special significance and is given its own name—stagnation pressure. In incompressible flows, the stagnation pressure at a stagnation point is equal to the total pressure throughout the flow field. [edit] Terminology in compressible fluid dynamics In a compressible fluid, such as air, the temperature and density are essential when determining the state of the fluid. In addition to the concept of total pressure (also known as stagnation pressure), the concepts of total (or stagnation) temperature and total (or stagnation) density are also essential in any study of compressible fluid flows. To avoid potential ambiguity when referring to temperature and density, many authors use the terms static temperature and static density. Static temperature is identical to temperature; and static density is identical to density; and both can be identified for every point in a fluid flow field. The temperature and density at a stagnation point are called stagnation temperature and stagnation density. A similar approach is also taken with the thermodynamic properties of compressible fluids. Many authors use the terms total (or stagnation) enthalpy and total (or stagnation) entropy. The terms static enthalpy and static entropy appear to be less common, but where they are used they mean nothing more than enthalpy and entropy respectively, and the prefix "static" is being used to avoid ambiguity with their 'total' or 'stagnation' counterparts. therefore the net force on the pipe = 100 psi <
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the 100 psi < that I use overcomes the nothing that you have.
we can calculate the mass of the air that travels from point a to point b and this will give us the mass that is being moved within the pipe. then we can calculate the velocity at which it moves.
then we can calculate the opposition to pipe movement as zero.
simply because there is no opposition to movement.
see if you can understand this one simple thing.
the only place that any force that can be counted as a force for movement in a direction is where the air is escaping the nozzle.
everything else is canceled out.
what if the tank was just floating inside the pipe and it is located at the middle of the pipe.
nothing to stop its movement.
then the valve is opened.
are you really going to say it would just sit there and not move?
and why wouldnt it move all the way to the end and apply its momentum to the pipe itself?
because you dont want it to. because you know physics but you dont know how to apply physics?
because a equation told you it wont move?
knowing physics is like owning a brand new cadilac with a empty tank of gas. knowing how to apply physics is like owning all the gas stations in the world.
your cadilac is worthless without the gas.
I sometimes wonder how we ever put a man on the moon given the quality of our college graduates and ability to apply what they have learned or supposedly learned.
3/4 inch of dust build up on the moon in 4.527 billion years,LOL and QM is fantasy science.




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you havent show anything except that you can cut and paste equations. and then dump the same ridiculous sob story on me. what have you shown? nothing. Actually, I've shown that physics states your space drive is a physical impossibility. I'd also point out that in "copying & pasting" equations I've given a mathematical basis for my arguments. Verses you, whose evidence consists of: a) whining that I provide mathematical evidence of my claims, b) ignoring basic physical laws such as newtons laws of motion, the conservation of energy, etc, and c) sticking your fingers in your ears and going "neenerneenerneener" every time you hear something you don't like. you go to a great extent to put up a equation and explain all the factors involved (at least you think thats all) and then you dont even supply any result from the equation.. Because I couldn't  to do that I'd have to know the area of the nozzle on the end of the tank, its divergence angle, the volume of the pipe the tank is sealed in, and so forth. Since you did not provide that info, it is not possible for me to complete the calculation. But thanx for confirming those mathematics are beyond your ability to comprehend. After all, anyone who bothered to read what the terms meant would realize that several important variables were not accounted for. therefore the net force on the pipe = 100 psi < Congratulations, you can cutandpaste from the wikipedia. I doubt you even understand what you copied  it was mostly irrelevent to the topic at hand, since fluid dynamics will generally describe the flow of the fluid through a pipe, not once it leaves the pipe. And since most of the energy in a system such as yours are derived by the nozzle  i.e. once the air has escaped the pipe  any calculations as to the energy of the moving fluid in the pipe will underestimate the actual amount of thrust provided. Would this be a bad time to point out I spent a lot of my Phd working out the fluid dynamics of blood flow and how that relates to the distribution of cells within the blood vessel and the forces they experience when interacting with the blood vessels? Even built microfluidic devices that do things like sort cells and mimic our circulatory systems... I'd also point out that I never once said the pipe wouldn't experience 100psi of pressure  the end closest to the tank would experience that pressure immediately after you open the valve. However, none of that changes the original issues, namely: 1) Pressure is not a measure of the amount of energy available for work  its merely a measure of force per unit area due to the presence of a fluid. The actual potential energy available is, as I stated earlier, determined by the volume of gas and the pressure it is at. Therefore saying "the pipe experiences 100psi" is meaningless, as that tells you nothing as to the amount of force generated upon release of the air  for that you need to take in account the mass of the air, and the speed it is moving at  using the equations I outlined in my earlier post. 2) You are ignoring what happens once the air leaves the tank. That moving air has kinetic energy. That energy doesn't magically disappear  it has to go somewhere. That somewhere, in a sealed pipe, is the pipe itself. And since the air is moving in opposition to the force on the tank, guess what the effect of that energy transfer is. Like I said before, build it, show it works, patent it, and show Newton, NASA, etc, to have been wrong. Bryan Bryan
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the 100 psi < that I use overcomes the nothing that you have. No, it does not. PSI is a measure of pressure, not the force accelerating the tank. we can calculate the mass of the air that travels from point a to point b and this will give us the mass that is being moved within the pipe. then we can calculate the velocity at which it moves. Which was the point of the equasions I provided  the very ones you were whining about. then we can calculate the opposition to pipe movement as zero. Assuming it is in space, yes. see if you can understand this one simple thing.
[quote=paul]the only place that any force that can be counted as a force for movement in a direction is where the air is escaping the nozzle. Sorry, that is incorrect. The air leaving the nozzle has kinetic energy (and thus, momentum  the two are closely related). That energy/momentum doesn't magically disappear once the air has moved away from the nozzle  the air will retain that energy/momentum for ever  unless the air encounters something. At which point that energy/momentum will be transfered to the object it encounters. what if the tank was just floating inside the pipe and it is located at the middle of the pipe.
nothing to stop its movement.
then the valve is opened.
are you really going to say it would just sit there and not move? The tank will move, the pipe it is contained in will not. For the sake of simplicity, lets assume the tank, when empty, weights 1kg. Lets also assume it had enough air to accelerate it to a speed of 1m/s. This tank would have: Kinetic energy = 1/2*m*v^2 = 1/2* 1kg * (1m/s)^2 = 0.5J Momentum = mv = 1kg * 1m/2^2 = 1kg*m/s Because of newtons 3rd law, the air that came out of the tank would have the same momentum. Opposite and equal  very important here. So your tank flies forward, and impacts the front of the pipe. When it does this it will transfer that kinetic energy/momentum to the pipe, pushing the pipe forward. But at the same time, you have air carrying energy/momentum in the opposite direction. When that air impacts the back of the pipe it'll transfer that energy/momentum to the back of the pipe, pushing the pipe backwards. Keep in mind  both the tank and the air that came out of it have the same momentum  newtons 3rd law dictates that. So you have an equal amount of momentum being applied to the front and back ends of the pipe. What happens? It should be selfevident! and why wouldnt it move all the way to the end and apply its momentum to the pipe itself? I never said it wouldn't. But why would you expect the air expelled from the tank not to do the same thing? The reason is simple  reality clashes with your preconceptions. so you deny reality. I sometimes wonder how we ever put a man on the moon given the quality of our college graduates and ability to apply what they have learned or supposedly learned.
We put men on the moon by understanding physics. Ever wonder why they didn't use your magical drive to get men there, and instead used big rockets that expel their propellant into space? It isn't cause NASA is dumb, its because your drive is a physical impossibility. Bryan
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Because I couldn't  to do that I'd have to know the area of the nozzle on the end of the tank, its divergence angle, the volume of the pipe the tank is sealed in, and so forth. Since you did not provide that info, it is not possible for me to complete the calculation.
I said the nozzle area is 1 square inch. I said the volue of the tank is 1000 sq ft. I even gave you the pipe volume. I said the fluid is air so SP = 1.29 the DP is 114.7  14.7 = 100 you cant do the math , so you claim that I cant. ohhhhhhh !!!! a college gadget for sure. Actually, I've shown that physics states your space drive is a physical impossibility.
Actually, you havent SHOWN anything.
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I said the nozzle area is 1 square inch. I said the volue of the tank is 1000 sq ft. I even gave you the pipe volume. I said the fluid is air so SP = 1.29 the DP is 114.7  14.7 = 100 You do realize the the exact numbers are meaningless  regardless of how fast or slow the air comes out of the tank, or how much or little of it there there, the momentum of the tank will always be equal, but opposite, to that of the air expelled from the tank. Nonetheless, I did miss your specifications (although even here you didn't give them all). 60" diameter pipe (1.152m), 500' (152.5m) long. You haven't given a divergence angle for the nozzle, so I'll assume it is zero (i.e. there is no cone; just a striaght tube). Volume of pipe = pi*r^2*h = 3.14*(1.152/2)^2*152.5m = 158.95m^3 I'm going to assume your volume of 1000 sq ft is actually cuft. 1000cuft = 28m^3, which is compressed to 10132kPa (100ATA). The force generated upon the release of this air is: F=[m*Ve]+[(PtPa)*Ae] Because we don't have a cone, m*Ve is zero. Therefore, the amount of thrust at a full tank is: F = (PtPa)*Ae F = (10132kPa101kPa)*0.00064516m^2 F = 6471N As the tank empties and the pipe fills, that force will drop. Thrust ends when Pt = Pa. ASCII doesn't allow me to show my work, integrations and all that, but excel gives a total thrust of just over 60kN*s (kilonewton seconds). you cant do the math , so you claim that I cant. Just did the math. And I state again that you cannot; nor do I believe you understand it either. But as I stated at the beginning, the specific numbers do not matter  the force applied to the tank is equal, but opposite to the force applied to the air coming out of the tank. No matter how much force is generated the pipe sits still. Unless you make enough force to rupture the pipe... It isn't rocket science  oh wait, in this case it kinda is... And here we are again  exactly where we were at the beginning of our "discussion"  you still do not have the faintest understanding of newtons laws of motion and law of conservation of energy. And once again I've provided yet another layer of mathematics showing the impossibility of your perpetualmotion engine. Bryan
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Volume of pipe = pi*r^2*h = 3.14*(1.152/2)^2*152.5m = 158.95m^3
1 foot = 0.3048 meters 60 inch / 2 = 30 inch 30^2 = 900 900 * 3.14159 = 2827.431 the cross sectional area = 2827.431 sq inch 500 ft * 12 inch = 6000 inch 2827.433 * 6000 = 16964600.329384883487698274269709 there are 1728 cu in in a cu ft 16964600.329384883487698274269709 / 1728 = 9817 cu ft 9817 cu ft = 277 cu meters somehow you have lost 277  158 = 119 cu meters 119 cu meters lost , how did that happen? And I state again that you cannot I see what happened now ... you used 1.152 instead of 1.524 as the diameter wow , that made a big difference didnt it The force generated upon the release of this air is: F=[m*Ve]+[(PtPa)*Ae]
Because we don't have a cone, m*Ve is zero. Therefore, the amount of thrust at a full tank is: F = (PtPa)*Ae F = (10132kPa101kPa)*0.00064516m^2 F = 6471N one load after another... theres something VERY important when calculating force  YOU NEED MASS  F=MA not F=whatever you want it to be. it looks like your using the initial pressure in the tank (pt)then subtracting the initial pressure in the pipe (pa?)then multiplying that number by the nozzle area (Ae?). so where are you going with this ? Because we don't have a cone, m*Ve is zero so since the mass comes out of a tube as you said , the mass has no velocity? it just covertly sneeks through the nozzle or perhaps it is being transported as they do on star trek !! I said the volue of the tank is 1000 sq ft.
yes , after the strain of trying to coax a little common sence out. but as I recall................... I did use 1000 cu ft earlier at the begining when I mentioned the tank in the pipe and suppose the air tank holds a mere 1000 cu ft of 100 psi compressed air inside it.
you know the mass of air. you know the force , its 100 psi. (believe it or not) all you need now is the acceleration. a=f/m acceleration = force / mass SP ) standard Pressure dry air has a density of 1.2754 kg/m3. this will give you the mass and the initial velocity for your formula. believe me air is light so there wont really be a time to wait as the mass accelerates so we will leave out time for now. now you have the rest of your formula.. And once again I've provided yet another layer of mathematics showing the impossibility of your perpetualmotion engine.
No , you havent provided anything. but you now have the full formula that you obviously already knew , you just didnt want to allow others to see the outcome of actual numbers , you only want them to believe your twisted math and the endless jargon that flows from your knowlege of what other people said.
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Wow this is a new low!!
It's called conservation of momentum. What's so unbelievable about that?
 "The momentum of an isolated system cannot change, ever, for any reason, no matter what happens internally" 
If you're sure it's wrong, then, as ImagingGeek suggested, ACTUALLY BUILD IT AND CLAIM YOUR NOBEL PRIZE!!
As with perpetual motion, this will make you richer than God.
But the advantage here is it's much easier and cheaper. Just get an old PVC pipe, put a pressurized coke bottle inside, rig up a trigger to release the bottle's nozzle, cap the ends, pull the trigger, sit back and wait for the endless stream of money to roll in.




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Now how am I going to prove that paul is wrong ,the only thing I have is words. but the words in physics are supposed to be backed up by the math in physics. AND THE MATH BACKS UP THE WORDS !!! WHY CANT I USE MATH ?
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reactionless propulsion is even being experimented with today. heres a good example , the box can be considered a closed system there is no external forces that apply agaist the box , but the box moves. click here to watch video
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