Patent Document

CROSS-REFERENCE TO RELATED APPLICATION 
     This application is a continuation of U.S. application Ser. No. 12/867,736 filed Feb. 16, 2011, entitled “Acoustic Turbine,” which is a U.S. National Phase of PCT/US2009/034313 filed Feb. 17, 2009, which claims priority to and the benefit of the filing date of U.S. Provisional Patent Application No. 61/028,974, entitled “Acoustic Production of Kinetic and Electrical Energy,” which was filed on Feb. 15, 2008, the entire disclosures of which are hereby incorporated by reference herein. 
    
    
     TECHNICAL FIELD 
     This application is generally related to the conversion of energy into useful work, and more specifically is related to a method and apparatus for converting heat energy into useful kinetic and electrical energy using acoustic energy. 
     BACKGROUND OF THE RELATED ART 
     Sound fields have been used to levitate objects by taking advantage of the Boltzmann-Ehrenfest principle of adiabatic invariance, which relates the acoustic potential acting on an object positioned in a single-mode cavity to a shift in resonant frequency caused by the presence of the object. In Putterman et al., “Acoustic Levitation and the Boltzmann-Ehrenfest Principle,” J. Acoust. Soc. Am. 85(1), (1989), the torque imposed on the object in a single mode chamber by acoustic energy equals the angular derivative of the experimentally determined frequency shift (Δω). However, the levitation technique disclosed in this article is limited to high frequency sound waves used in single-mode cavities. Further, the created acoustic torque will only align with the cylindrical object being levitated in a horizontal plane, or when θ=90°. These constraints, i.e., high frequency acoustic energy, the single mode cavity and the middle horizontal plane, severely limit the applications of the disclosed levitation technique. 
     In an earlier work, Allen et al., “A Powerful High Frequency Siren,” J. Acoust Soc. Am. 19(5), (1947), a high frequency siren is used to generate chamber pressures of about two atmospheres and an output energy of approximately two kilowatts which can be used for heating purposes. However, this high frequency siren finds little use in most environments where such a sound output would be unacceptable or dangerous. 
     To date, acoustic or sound energy has not been used effectively to impart thrust or force on an object for purposes of providing useful work. While limited examples of levitation are known, the requirement that the object being levitated be disposed in an enclosed chamber renders the technique largely inapplicable. Further, the use of powerful sirens for generating heat is also largely industrially inapplicable and potentially harmful to workers. 
     SUMMARY OF THE DISCLOSURE 
     A method and apparatus that generate kinetic and electrical energy using sound waves is believed to be particularly useful in high efficiency motors and electrical generators. Stated simply, the method and apparatus described herein use sound waves as a catalyst to convert ambient heat energy into kinetic and/or electrical energy. 
     In one embodiment, sound waves at particular frequencies are propagated across one side of a plate or other barrier element, causing flow of fluid (e.g. air) across the surface of the plate which, in turn, causes a reduction in the ambient fluid (air) pressure near the surface of the plate. The difference in fluid pressure on opposite sides of the plate results in net positive thrust on the plate, thereby causing movement of the plate. This movement can be harnessed using, for example, a windmill type of rotor and stator arrangement to generate useful kinetic and electrical energy. 
     In another embodiment, a sonic thrust apparatus includes a barrier element including a first outer surface and a second outer surface. The sonic thrust apparatus further includes an acoustic oscillator disposed in close proximity to the first outer surface of the barrier element. The sonic thrust apparatus further includes a drive mechanism driving the acoustic oscillator to produce sound waves flowing along the first outer surface to create a pressure differential between the first outer surface and the second outer surface and to create a thrust on the second outer surface towards the first outer surface. 
     In yet another embodiment an acoustic turbine includes a rotor having an axis of rotation and at least one lever arm coupled to the rotor and extending way from the rotor. The lever arm includes a thrust element disposed on the lever arm. The thrust element may include a barrier element including a first outer surface and a second outer surface, and may further include an acoustic oscillator disposed in close proximity to the first outer surface of the barrier element. The thrust element may also include a drive mechanism for driving the acoustic oscillator to produce sound waves flowing along the first outer surface to create a pressure differential between the first outer surface and the second outer surface and to create a thrust on the second outer surface towards the first outer surface to rotate the rotor. 
     The operating principle of this system is analogous to the forces that result in lift on an airplane wing, in which faster air flow over the top of the wing, as compared to slower air flow over the bottom of the wing, causes a pressure difference on the opposite sides of the wing which, in turn, causes lift. Consistent with Bernoulli&#39;s law (which predicts the lift of an airplane wing), a sonic oscillator, properly configured, produces sound waves that result in a pressure drop on one side of the plate or other barrier element as compared to the opposite side of the plate or barrier element. This pressure difference results in enough thrust to turn a mechanical gear or other rotating mechanism. This system and method can therefore be used as part of a motor to produce rotation, or as part of a generator to generate electrical energy. 
     The kinetic and electrical energy generated by this system and method is ultimately derived from the heat energy in the ambient fluid (air), as predicted by the ideal gas law, and preliminary research predicts that this energy conversion can be performed with surprisingly high efficiency. 
    
    
     
       BRIEF DESCRIPTION OF THE DRAWINGS 
         FIG. 1  is a perspective view of a sonic thrust element which uses sonic energy to produce thrust on a barrier element in the form of a flat plate; 
         FIG. 2A  is a perspective view of a sonic thrust element that includes a motor system for oscillating the acoustic oscillator in  FIG. 1 ; 
         FIG. 2B  is a cross-sectional view of the sonic thrust element in  FIG. 2A ; 
         FIG. 3  is a perspective view of a sonic motor having multiple sonic thrust elements disposed on a rotor to produce useful kinetic energy; and 
         FIG. 4  is a cross-sectional, side view of a second embodiment of a sonic thrust element which uses sonic energy to produce thrust on a barrier element in the form of a flat plate. 
     
    
    
     DETAILED DESCRIPTION 
     Generally speaking, sound can be used as a catalyst for obtaining torque on a barrier element, such as a flat plate, which thereby causes motion of the plate. When connected to proper mechanical structures, such as a lever arm connected to a rotatable axis, the motion of the plate can be converted into useful kinetic energy to drive a rotating wheel, gear, bar or other element. This motion can also be converted into electrical energy using an electrical rotor and stator configuration as is common in well known generator technology. 
     In a simple form, an acoustic oscillator is used to generate and direct sound waves across the surface of a barrier element in the form of a flat plate. The sound waves, when properly generated, form a low pressure region immediately adjacent to the surface of the flat plate at which the acoustic oscillator is located. Ambient fluid pressure on the other side of the plate results in thrust on the plate, which propels the plate toward the low pressure region, and this thrust can be harnessed to produce useful kinetic and electrical energy. 
       FIG. 1  illustrates an example of a sonic thrust element  5  including a flat plate  10  having an upper surface  12  and a lower surface  14 , and an acoustic oscillator  20  disposed in close proximity to the upper surface  12  of the flat plate  10 . In particular, the acoustic oscillator  20  is illustrated as a square oscillator disposed slightly above, and in the middle of the upper surface of the flat plate  10 . In this case, the oscillator  20  is disposed a finite distance ε above the upper surface  12  of the flat plate  10  and is positioned so as to produce sound waves that flow across the upper surface  12  of the flat plate  10  in a direction generally perpendicular to the normal of the flat plate  10 . The arrows  25 A and  25 B indicate the direction of sound propagation in the system of  FIG. 1 . 
     During operation, electrical or other energy (such as mechanical energy) is provided to drive the acoustic oscillator  20  (via electrical wires or other mechanical mechanism not shown in  FIG. 1 ) so that the acoustic oscillator  20  produces sound waves, having a constant or near constant frequency as described in more detail below. These sound waves propagate across the upper surface  12  of the flat plate  10  on either sides of the oscillator  20 , as illustrated by the arrows  25 A and  25 B. If produced in accordance with the techniques and constraints defined herein, the sound waves produced by the oscillator  20  create a low pressure region immediately adjacent to the upper surface  12  of the flat plate  10 . As the sound waves produced by the oscillator  20  do not affect the fluid pressure immediately adjacent the flat plate  10  on the lower surface  14  of the flat plate  10 , a pressure differential is created on the opposite sides  12  and  14  of the flat plate  10 . This pressure differential results in a thrust (denoted by arrows  26 ) on the lower surface  14  of the flat plate  10 , across the entire surface  14  of the flat plate  10 , and generally in the direction of the upper surface  12  of the flat plate  10 . 
       FIGS. 2A-2B  illustrate another example of a sonic thrust element  7  including a flat plate  10  having an upper surface  12  and a lower surface  14 , and an acoustic oscillator  20  disposed in close proximity to the upper surface  12  of the flat plate  10 . In this case, the flat plate  10  includes an aperture  60  (e.g., a hole, or a slit), and the oscillator  20  is coupled to a motor system  50  via the aperture  60 . The motor system  50  linearly oscillates (drives) the oscillator back and forth along the plane of the upper surface  12  so that the acoustic oscillator  20  produces sound waves, having a constant frequency as described in more detail below. As explained in reference to  FIG. 1 , these sound waves propagate across the upper surface  12  of the flat plate  10  and create a pressure differential on the opposite sides  12  and  14  of the flat plate  10  that results in a thrust on the lower surface  14  of the flat plate  10 , and generally in the direction of the upper surface  12  of the flat plate  10  (i.e., in the direction normal to the surface plane of the flat plate  10 ). 
     The motor system  50  may be implemented in a number of ways. For example, a rotational motor may be used (not shown). The rotational motor may be coupled to a linkage, a cam, etc., to transform the rotational motion of the rotational motor into a linear motion along the upper surface  12  of the flat plate  10 . In some embodiments, the rotational motion and/or the resulting linear motion have an associated frequency of 450 Hz. 
     The motor system  50  may be coupled to the acoustic oscillator  20  in any of a variety of manners (e.g., via bolts, nuts, pins, cams, linkage arms, and so on). In some embodiments, the motor system  50  may be easily detachable from the acoustic oscillator  20  so that different motor systems may be used. Moreover, the motor system  50  may take the form of, or may be powered by mechanical motion collected from other sources, such as sources of waste heat or waste energy. 
     In different embodiments, the different building blocks of the sonic thrust element  7  may be of different shapes and sizes and may be made from a variety of materials. For example, the flat plate  10  may be a square plate, 500 mm on each side. The acoustic oscillator  20  may have a width of 500 mm, and a height and length of 38.1 mm. The aperture in the flat plate  10  may be 88 mm thereby allowing the oscillator  20  to oscillate 88 mm along the flat plate  10 . Both the flat plate  10  and the oscillator  20  may be may be made of aluminum. However, the flat plate  10  and/or the oscillator  20  may be made of other suitable materials, such as carbon fiber, fiberglass, etc. Of course, these materials, and system dimensions are associated with one particular embodiment of the invention, and other sets of dimensions and materials will operate according to the principles described herein. Thus, the invention is not limited to the particular dimensions and materials described herein with respect to  FIG. 2 . 
       FIG. 3  illustrates a motor  27  (referred to herein as a sonic motor) which can harness the thrust on the flat plate  10  of  FIG. 1  to produce useful kinetic energy. In particular, the motor  27  of  FIG. 3  includes a rotor  28  having, in this case, four lever arms  30  rigidly connected to and disposed around a center bar  32 . The bar  32  rotates about a longitudinal axis  34  which defines an axis of rotation for the rotor  28 . As illustrated in  FIG. 3 , thrust elements  5  (described with respect to  FIG. 1 ) are disposed on the ends of the lever arms  30  so that the plates  10  of the thrust elements  5  are disposed extending out from the lever arms  30  in an essentially radial plane with respect to the axis of rotation  34  and/or so that a normal to the surface of the plane of each of the flat plates  10  is disposed in an essentially tangential direction with respect to a circle disposed in a plane perpendicular to the axis of rotation  34  and having a center point on the axis of rotation  34 . 
     While the oscillators  20  of the thrust elements  5  are illustrated as being disposed to extend radially out from the axis of rotation  34 , the thrust elements  5  could be rotated so that the acoustic oscillators  20  extend in any other direction with respect to the axis of rotation  34 . Thus, for example, instead of being disposed horizontally across the plates  10  from side to side, as illustrated in  FIG. 3 , the oscillators  20  could be disposed vertically across the plates  10  (from top to bottom), or an any diagonal direction across the plates  10 . Still further, while each of the oscillators  20  is illustrated as being disposed across a center of one of the flat plates  10 , the oscillator  20  of the thrust elements  5  could be disposed, instead, off-center with respect to an associated flat plate  10 , and even at the edge of the flat plate  10 , if so desired, as long as the oscillator  20  directs sound waves across the top surface  12  of the associated flat plate  10 . Moreover, while the flat plates  10  are illustrated as essentially square plates, the plates  10  could be other shapes as well, such as rectangular, circular, oval, etc. Also, while the surface  14  of the flat plates  10  is illustrated as being flat, this condition may not be necessary, and other surface contours could be used as well. Likewise, while four sonic thrust elements  5  are shown in  FIG. 3  as being connected to the center bar  32  via lever arms  30 , any other number of thrust elements  5 , such as one, two, three, five, etc. could be used instead. 
     In the structure of  FIG. 3 , operation (energization) of the oscillators  20  causes thrust on the flat plates  10 , which thrust is illustrated by the arrows  36  in  FIG. 3 . The thrust  36  on the flat plates  10  places a torque on the lever arms  30  proportional to the lengths of the arms  30 , and imparts a rotational thrust  38  onto the bar  32  around the axis of rotation  34 . The bar  32  may be allowed to rotate and, if desired, may be connected to a gearing mechanism  40  which can be used to harness the rotational kinetic energy imparted to the bar  32  by the sonic thrust elements  5 . If desired, this kinetic energy may be converted to electrical energy with the use of a generator (not shown in  FIG. 3 ). In such an embodiment, the rotor  28  may have an electromagnetic field element which creates a rotating electromagnetic field with respect to (e.g., inside of or around) a stator, which may generate electricity in the stator. As generators are well known, they will not be described in more detail herein. 
     Generally speaking, the operation of the sonic thrust elements  5  of  FIGS. 1, 2 and 3  can be more completely understood on the basis of an analogy to the force imparted to an airplane wing moving with subsonic speed through the air (an ambient fluid). In this case, the shape of the wing causes the pressure above the wing to be less than the pressure below the wing as the wing moves through the air, in the manner predicted by Bernoulli&#39;s law, which in turn causes lift on the wing. In the case of the structures of  FIGS. 1, 2 and 3 , the airplane wing is replaced by a plate  10  having an acoustic oscillator disposed on one side of the plate  10 . The sound waves generated by the oscillators  20  mimic air flowing at the speed of sound over the surface of the plate  10 . According to Bernoulli&#39;s law, the pressure on the surface of the plate  10  over which the sound waves are traveling will be less than the pressure on the opposite side of the plate  10 . Based on the time dependent version of Bernoulli&#39;s law, a thrust on the plate  10  results. However, in order for this thrust to occur as a result of the sound waves produced by the oscillator  20 , a number of conditions must be satisfied based on the mathematical derivations of the physical principles occurring in this system. 
     To begin with, a number of basic variables used in the equations below are defined as:
         L is the length of the plate ( 10 );   S is the width of the plate ( 10 );   H is the height of the oscillator ( 20 ), from the bottom of the oscillator near the plate surface ( 12 ) to the top of the oscillator;   ε is the distance from the bottom of the oscillator to the top of the plate;   α is the length of the rotation arm, defined as the distance between the axis of rotation ( 34 ) and the center of mass of (a) the rotating arm ( 30 ) plus (b) the plate plus (c) the oscillator; and   C is the speed of sound in the ambient fluid.
 
First the wattage output of an acoustic oscillator can be expressed as:
       

     
       
         
           
             
               
                 
                   
                     W 
                     ⁡ 
                     
                       ( 
                       watts 
                       ) 
                     
                   
                   = 
                   
                     Ω 
                     = 
                     
                       
                         
                           256 
                           ⁢ 
                           
                               
                           
                           ⁢ 
                           
                             π 
                             3 
                           
                           ⁢ 
                           δ 
                           ⁢ 
                           
                               
                           
                           ⁢ 
                           
                             SHA 
                             3 
                           
                         
                         
                           2 
                           
                             5 
                             2 
                           
                         
                       
                       ⁢ 
                       
                         1 
                         
                           cm 
                           6 
                         
                       
                     
                   
                 
               
               
                 
                   Eq 
                   . 
                   
                       
                   
                   ⁢ 
                   
                     ( 
                     1 
                     ) 
                   
                 
               
             
           
         
       
         
         
           
             wherein: 
             W is the output of the oscillator in watts; 
             δ is the mass density of the ambient fluid; 
             H is the height of the oscillating diaphragm; 
             S is the width of the oscillating diaphragm (also the width of the plate); and 
             A is the amplitude of the oscillation measured in cm/s.
 
It is well known that a watt is:
 
           
         
         Watt=10 7  dyn-cm/s=10 7  g-cm 2 /s 3    
       
    
     Next, the formula for the optimal thrust (T) on a flat plate due to an acoustic source acting as a catalyst, such as that illustrated in  FIG. 1 , can be expressed as: 
     
       
         
           
             
               
                 
                   T 
                   = 
                   
                     
                       
                         π 
                         ⁢ 
                         
                             
                         
                         ⁢ 
                         δ 
                         ⁢ 
                         
                             
                         
                         ⁢ 
                         
                           A 
                           2 
                         
                         ⁢ 
                         
                             
                         
                         ⁢ 
                         
                           cos 
                           2 
                         
                         ⁢ 
                         
                             
                         
                         ⁢ 
                         
                           ( 
                           
                             2 
                             ⁢ 
                             
                                 
                             
                             ⁢ 
                             π 
                             ⁢ 
                             
                                 
                             
                             ⁢ 
                             ft 
                           
                           ) 
                         
                       
                       
                         ɛ 
                         2 
                       
                     
                     + 
                     
                       
                         8 
                         ⁢ 
                         
                             
                         
                         ⁢ 
                         π 
                         ⁢ 
                         
                             
                         
                         ⁢ 
                         δ 
                         ⁢ 
                         
                             
                         
                         ⁢ 
                         S 
                         ⁢ 
                         
                           
                             LS 
                             π 
                           
                         
                         ⁢ 
                         
                           HA 
                           ⁡ 
                           
                             ( 
                             
                               2 
                               ⁢ 
                               
                                   
                               
                               ⁢ 
                               π 
                               ⁢ 
                               
                                   
                               
                               ⁢ 
                               f 
                             
                             ) 
                           
                         
                         ⁢ 
                         
                             
                         
                         ⁢ 
                         sin 
                         ⁢ 
                         
                             
                         
                         ⁢ 
                         
                           ( 
                           
                             2 
                             ⁢ 
                             
                                 
                             
                             ⁢ 
                             π 
                             ⁢ 
                             
                                 
                             
                             ⁢ 
                             ft 
                           
                           ) 
                         
                       
                       
                         cm 
                         2 
                       
                     
                   
                 
               
               
                 
                   Eq 
                   . 
                   
                       
                   
                   ⁢ 
                   
                     ( 
                     2 
                     ) 
                   
                 
               
             
           
         
       
         
         
           
             wherein: 
             f is the frequency of the sound waves created by the oscillator in hertz; 
             t is the time in seconds; 
             ε is the distance from the bottom of the oscillator to the top of the plate; 
             δ is the mass density of the ambient fluid.
 
However, for the system of  FIG. 1  to operate, the thrust on the plate  10  must point in one direction only, that is, from the lower side  14  of the plate  10  to the upper side  12  of the plate  10  as illustrated in  FIG. 1 . To meet this condition, the second part of the expression on the right side of Eq. (2) must be able to be ignored (i.e., must be much less than the first part), which occurs when:
 
           
         
       
    
                       A     32   ⁢     π     ⁢     S     3   2       ⁢     HL     1   2       ⁢     ɛ   2         ⁢     cm   2       &gt;&gt;           ⁢   f           Eq   .           ⁢     (   3   )                 
wherein &gt;&gt; means at least a magnitude of 10 times greater than.
     This formula was obtained using:   

                   &lt;       cos   2     ⁢   2   ⁢   π   ⁢           ⁢   ft     &gt;=     1   2             Eq   .           ⁢     (   4   )                 
wherein &lt;quantity&gt; means the average value of the quantity over one cycle and setting the sine term in the second part of the expression on the right side of Eq. (2) equal to 1.
 
     Now, at optimal thrust conditions, the fluid pressure on the oscillator side of the plate is near a vacuum, and so, at this condition, the thrust T on the plate is:
 
 T=PLS   Eq. (5)
         wherein:   T is the thrust on the plate;   P is the ambient fluid pressure on the non-oscillator side of the plate;   L is the length of the plate; and   S is the width of the plate.       

     Substituting Eq. (4) into Eq. (2) assuming that the constraint of Eq. (3) is satisfied gives: 
                   T   =       π   ⁢           ⁢   δ   ⁢           ⁢     A   2         2   ⁢           ⁢     ɛ   2                 Eq   .           ⁢     (   6   )                 
and equating the different expressions for the thrust T from Eqs. (6) and (5) gives:
 
                   T   =         π   ⁢           ⁢   δ   ⁢           ⁢     A   2         2   ⁢           ⁢     ɛ   2         =   PLS             Eq   .           ⁢     (   7   )                 
Expressing the right most equality of Eq. (7) as a function of ε (the distance from the bottom of the oscillator to the top surface of the plate) gives:
 
                   ɛ   =           π   /   2       ⁢     δ     1   2       ⁢   A         P     1   2       ⁢     L     1   2       ⁢     S     1   2                   Eq   .           ⁢     (   8   )                 
Now, the torque G produced by a single plate in a rotor system such as that of  FIG. 3  at the optimal thrust conditions is expressed as:
 
                   G   =       PLS   ⁢           ⁢   α     =       π   ⁢           ⁢   δ   ⁢           ⁢     A   2     ⁢   α       2   ⁢           ⁢     ɛ   2                   Eq   .           ⁢     (   9   )                 
From the discussion above, Eqs. (3) and (8) are constraints that must be satisfied when operating the system, wherein Eq. (3) guarantees that the thrust on the plate is always directed from the side of the plate opposite the acoustic oscillator towards the side of the plate with the acoustic oscillator and Eq. (8) is a condition for optimal thrust.
 
     Of course, there is a physical limit to how close the oscillator  20  can be placed to the side  12  of the plate  10 , and thus there is a physical limit to how small ε can be made. A reasonable choice for minimal possible value of ε is believed to be about 3×10 −6  cm (i.e., 300 angstroms) as films of plastics such as collodion can be manufactured at this thickness to be used as windows for low pressure gas retention. Substituting this value for ε into Eq. (8) gives: 
     
       
         
           
             
               
                 
                   
                     
                       π 
                       ⁢ 
                       
                           
                       
                       ⁢ 
                       δ 
                       ⁢ 
                       
                           
                       
                       ⁢ 
                       
                         A 
                         2 
                       
                     
                     
                       2 
                       ⁢ 
                       
                           
                       
                       ⁢ 
                       PLS 
                     
                   
                   ≥ 
                   
                     9 
                     × 
                     
                       10 
                       
                         - 
                         12 
                       
                     
                     ⁢ 
                     
                       cm 
                       2 
                     
                   
                 
               
               
                 
                   Eq 
                   . 
                   
                       
                   
                   ⁢ 
                   
                     ( 
                     10 
                     ) 
                   
                 
               
             
           
         
       
     
     Next, in order to justify neglecting the viscosity terms in the Navier-Stokes equations for the ambient fluid, the following constraint is also determined: 
     
       
         
           
             
               
                 
                   
                     
                       3 
                       ⁢ 
                       
                           
                       
                       ⁢ 
                       δ 
                       ⁢ 
                       
                           
                       
                       ⁢ 
                       
                         C 
                         3 
                       
                     
                     
                       8 
                       ⁢ 
                       
                           
                       
                       ⁢ 
                       
                         π 
                         2 
                       
                       ⁢ 
                       μ 
                       ⁢ 
                       
                           
                       
                       ⁢ 
                       H 
                     
                   
                   &gt;&gt; 
                   
                       
                   
                   ⁢ 
                   
                     f 
                     
                       
                           
                       
                       ⁢ 
                       2 
                     
                   
                 
               
               
                 
                   Eq 
                   . 
                   
                       
                   
                   ⁢ 
                   
                     ( 
                     11 
                     ) 
                   
                 
               
             
           
         
       
         
         
           
             wherein: 
             C is the speed of sound in the ambient fluid; and 
             μ is the absolute viscosity of the ambient fluid.
 
Substituting the ideal gas law (P=ρRT emp ) into Eq. (8) gives:
 
           
         
       
    
     
       
         
           
             
               
                 
                   ɛ 
                   = 
                   
                     
                       
                         
                           π 
                           / 
                           2 
                         
                       
                       ⁢ 
                       
                         δ 
                         
                           1 
                           2 
                         
                       
                       ⁢ 
                       A 
                     
                     
                       
                         ρ 
                         
                           1 
                           2 
                         
                       
                       ⁢ 
                       
                         R 
                         
                           1 
                           2 
                         
                       
                       ⁢ 
                       
                         T 
                         emp 
                         
                           1 
                           2 
                         
                       
                       ⁢ 
                       
                         L 
                         
                           1 
                           2 
                         
                       
                       ⁢ 
                       
                         S 
                         
                           1 
                           2 
                         
                       
                     
                   
                 
               
               
                 
                   Eq 
                   . 
                   
                       
                   
                   ⁢ 
                   
                     ( 
                     12 
                     ) 
                   
                 
               
             
           
         
       
         
         
           
             wherein: 
             ρ is the weight density of the gas; 
             R is the gas constant; and 
             T emp  is temperature.
 
Eq. (12) relates the optimal thrust to the temperature of the gas (ambient fluid). As can be seen, as the density of the gas varies, δ/ρ remains constant, meaning that there is a direct dependence of ε on the temperature T emp  and thus that the energy which produces the thrust in this system comes from the heat in the ambient fluid.
 
           
         
       
    
     A final constraint that must be satisfied is: 
                   f   &gt;&gt;     C       π   ⁢           ⁢   LS                 Eq   .           ⁢     (   13   )                 
This constraint assures that the wavelength of the sound waves are small compared to the dimensions of the plate, so that the sound waves behave like particles traveling near the speed of sound (thereby justifying the airplane wing analogy).
 
     Now, it is possible to describe the thrust on the plate in terms of the wattage output of the oscillator as: 
                   T   =         581.7   ⁢           ⁢     Ω     2   3       ⁢     δ     1   3             ɛ   2     ⁢     S     2   3       ⁢     H     2   3           ⁢         g     2   3       ⁢     cm     16   3           s   2                 Eq   .           ⁢     (   14   )                 
Thus, expressed in terms of the wattage output of the oscillator from Eq. (14), the constraints which must be satisfied to obtain optimal thrust on the plate using sound waves are:
 
     
       
         
           
             
               
                 
                   Constraint 
                   ⁢ 
                   
                       
                   
                   ⁢ 
                   1 
                 
               
               
                 
                     
                 
               
             
             
               
                 
                   
                     
                       
                         5.833 
                         × 
                         
                           10 
                           
                             - 
                             4 
                           
                         
                         ⁢ 
                         
                           PL 
                           
                             1 
                             2 
                           
                         
                       
                       
                         
                           δ 
                           
                             2 
                             3 
                           
                         
                         ⁢ 
                         
                           S 
                           
                             1 
                             6 
                           
                         
                         ⁢ 
                         
                           H 
                           
                             2 
                             3 
                           
                         
                         ⁢ 
                         
                           Ω 
                           
                             1 
                             3 
                           
                         
                       
                     
                     ⁢ 
                     
                       s 
                       
                         
                           g 
                           
                             1 
                             3 
                           
                         
                         ⁢ 
                         
                           cm 
                           
                             2 
                             3 
                           
                         
                       
                     
                   
                   &gt;&gt; 
                   f 
                 
               
               
                 
                   ( 
                   
                     C 
                     ⁢ 
                     
                         
                     
                     ⁢ 
                     1 
                   
                   ) 
                 
               
             
           
         
       
     
     This constraint is derived by solving the right most equality of Eq. (1) for A and substituting this expression of A into Eq. (3). 
     
       
         
           
             
               
                 
                   Constraint 
                   ⁢ 
                   
                       
                   
                   ⁢ 
                   2 
                 
               
               
                 
                     
                 
               
             
             
               
                 
                   ɛ 
                   = 
                   
                     
                       
                         24.12 
                         ⁢ 
                         
                             
                         
                         ⁢ 
                         
                           Ω 
                           
                             1 
                             3 
                           
                         
                         ⁢ 
                         
                           δ 
                           
                             1 
                             6 
                           
                         
                       
                       
                         
                           P 
                           
                             1 
                             2 
                           
                         
                         ⁢ 
                         
                           L 
                           
                             1 
                             2 
                           
                         
                         ⁢ 
                         
                           S 
                           
                             5 
                             6 
                           
                         
                         ⁢ 
                         
                           H 
                           
                             1 
                             3 
                           
                         
                       
                     
                     ⁢ 
                     
                       
                         
                           g 
                           
                             1 
                             3 
                           
                         
                         ⁢ 
                         
                           cm 
                           
                             8 
                             3 
                           
                         
                       
                       s 
                     
                   
                 
               
               
                 
                   ( 
                   
                     C 
                     ⁢ 
                     
                         
                     
                     ⁢ 
                     2 
                   
                   ) 
                 
               
             
           
         
       
     
     This constraint is derived by solving the right most equality of Eq. (1) for A and substituting this expression of A into Eq. (8).
 
Constraint 3:
 
ε≧3×10 −6  cm  (C3)
 
     Assumption based on reasonable expectation of physical limits 
     
       
         
           
             
               
                 
                   Constraint 
                   ⁢ 
                   
                       
                   
                   ⁢ 
                   4 
                 
               
               
                 
                     
                 
               
             
             
               
                 
                   
                     
                       
                         5.817 
                         ⁢ 
                         
                             
                         
                         ⁢ 
                         
                           Ω 
                           
                             2 
                             3 
                           
                         
                         ⁢ 
                         
                           δ 
                           
                             1 
                             3 
                           
                         
                       
                       
                         
                           PLS 
                           
                             5 
                             3 
                           
                         
                         ⁢ 
                         
                           H 
                           
                             2 
                             3 
                           
                         
                       
                     
                     ⁢ 
                     
                       
                         
                           g 
                           
                             2 
                             3 
                           
                         
                         ⁢ 
                         
                           cm 
                           
                             16 
                             3 
                           
                         
                       
                       
                         s 
                         2 
                       
                     
                   
                   ≥ 
                   
                     9 
                     × 
                     
                       10 
                       
                         - 
                         14 
                       
                     
                     ⁢ 
                     
                       cm 
                       2 
                     
                   
                 
               
               
                 
                   ( 
                   
                     C 
                     ⁢ 
                     
                         
                     
                     ⁢ 
                     4 
                   
                   ) 
                 
               
             
           
         
       
     
     This constraint is determined by plugging the ε value of C2 into C3. 
     
       
         
           
             
               
                 
                   Constraint 
                   ⁢ 
                   
                       
                   
                   ⁢ 
                   5 
                 
               
               
                 
                     
                 
               
             
             
               
                 
                   
                     
                       3.80 
                       × 
                       
                         10 
                         
                           - 
                           2 
                         
                       
                       ⁢ 
                       δ 
                       ⁢ 
                       
                           
                       
                       ⁢ 
                       
                         C 
                         3 
                       
                     
                     
                       μ 
                       ⁢ 
                       
                           
                       
                       ⁢ 
                       H 
                     
                   
                   &gt;&gt; 
                   
                       
                   
                   ⁢ 
                   
                     f 
                     
                       
                           
                       
                       ⁢ 
                       2 
                     
                   
                 
               
               
                 
                   ( 
                   
                     C 
                     ⁢ 
                     
                         
                     
                     ⁢ 
                     5 
                   
                   ) 
                 
               
             
           
         
       
     
     The constraint of Eq. (11). 
     
       
         
           
             
               
                 
                   Constraint 
                   ⁢ 
                   
                       
                   
                   ⁢ 
                   6 
                 
               
               
                 
                     
                 
               
             
             
               
                 
                   f 
                   &gt;&gt; 
                   
                     
                       0.564 
                       ⁢ 
                       
                           
                       
                       ⁢ 
                       C 
                     
                     
                       
                         L 
                         
                           1 
                           2 
                         
                       
                       ⁢ 
                       
                         S 
                         
                           1 
                           2 
                         
                       
                     
                   
                 
               
               
                 
                   ( 
                   
                     C 
                     ⁢ 
                     
                         
                     
                     ⁢ 
                     6 
                   
                   ) 
                 
               
             
           
         
       
     
     The constraint of Eq. (13). 
     Now, assuming that the physical parameters (L, S, H, ε, f, etc.) can be chosen such that these conditions (C1-C6) can be met, the rotational torque G that will be produced by a single plate element (i.e., a single sonic thrust element  5  of  FIG. 1 ) attached to an arm is: 
     
       
         
           
             
               
                 
                   G 
                   = 
                   
                     
                       
                         
                           
                             581.7 
                             ⁢ 
                             
                                 
                             
                             ⁢ 
                             
                               Ω 
                               
                                 2 
                                 3 
                               
                             
                             ⁢ 
                             
                               δ 
                               
                                 1 
                                 3 
                               
                             
                             ⁢ 
                             α 
                           
                           ⁢ 
                           
                               
                           
                         
                         
                           
                             ɛ 
                             2 
                           
                           ⁢ 
                           
                             S 
                             
                               2 
                               3 
                             
                           
                           ⁢ 
                           
                             H 
                             
                               2 
                               3 
                             
                           
                         
                       
                       ⁢ 
                       
                         
                           
                             g 
                             
                               2 
                               3 
                             
                           
                           ⁢ 
                           
                             cm 
                             
                               16 
                               3 
                             
                           
                         
                         
                           s 
                           2 
                         
                       
                     
                     = 
                     
                       PLS 
                       ⁢ 
                       
                           
                       
                       ⁢ 
                       α 
                     
                   
                 
               
               
                 
                   Eq 
                   . 
                   
                       
                   
                   ⁢ 
                   
                     ( 
                     15 
                     ) 
                   
                 
               
             
           
         
       
     
     This equation is derived by solving the right most equality of Eq. (1) for A and substituting this expression of A into Eq. (9). 
     From Eq. (15), the power output (PO) of the rotating plate system with a single plate or thrust element is: 
     
       
         
           
             
               
                 
                   
                     P 
                     ⁢ 
                     
                         
                     
                     ⁢ 
                     O 
                   
                   = 
                   
                     
                       
                         .1047 
                         ⁢ 
                         
                             
                         
                         ⁢ 
                         
                           ( 
                           
                             # 
                             ⁢ 
                             
                               R 
                               m 
                             
                           
                           ) 
                         
                         ⁢ 
                         α 
                         ⁢ 
                         
                             
                         
                         ⁢ 
                         PLS 
                       
                       s 
                     
                     = 
                     
                       2 
                       ⁢ 
                       
                           
                       
                       ⁢ 
                       π 
                       ⁢ 
                       
                           
                       
                       ⁢ 
                       
                         R 
                         ps 
                       
                       ⁢ 
                       G 
                     
                   
                 
               
               
                 
                   Eq 
                   . 
                   
                       
                   
                   ⁢ 
                   
                     ( 
                     16 
                     ) 
                   
                 
               
             
           
         
       
         
         
           
             wherein: 
             R ps  is the number of revolutions of the arm per second; and 
             #R m  is the number of revolutions of the arm in one minute such that R ps =#R m /60 s. 
           
         
       
    
     COMPUTATIONAL EXAMPLE 
     An explicit example of the parameters of the system described above is provided below illustrating that all of the constraints (C1-C6) can be simultaneously satisfied. In this example, the ambient fluid is taken to be air at 69 deg F. at standard atmospheric pressure. As such, and as is known for this condition:
 
μ=0.00018 g/cm-s (the absolute viscosity of air).
 
 P= 1.013×10 6  g/cm-s 2  (the ambient air pressure).
 
ρ=1.1781 g/cm 2 -s 2  (the weight density of ambient air).
 
δ=1.2013×10 −3  g/cm 3  (the mass density of ambient air).
 
 C= 33162 cm/s (the speed of sound in ambient air).
 
     The dimensions of the plate and the oscillating diaphragm in this example are taken to be:
 
 L= 60 cm
 
 S= 10 cm
 
 H= 5 cm.
 
Moreover,
 
α=60 cm (the length of the rotation arm);
 
Ω=0.25 watts (the output wattage of the acoustic oscillator); and
 
 f= 16000 hertz=16000/s (the frequency of oscillation of the oscillator).
 
     For this example, it will be assumed that the acoustic oscillator is ¼ percent efficient, which is a reasonable assumption, and that the input wattage to the oscillator is 100 watts, which is reasonably attainable with common acoustic speakers. 
     Substituting these values into the constraints (C1-C6) and solving for these equations provides the following expressions.
 
1.625×10 5 /s&gt;&gt;16000/s  C1:
 
ε=5.454×10 −5  cm  C2:
 
5.454×10 −5  cm≧3×10 −6  cm  C3:
 
2.975×10 −11  cm≧9×10 −14  cm  C4:
 
9.25×10 12 /s 2 &gt;&gt;2.56×10 8 /s 2   C5:
 
1.6×10 4 /s&gt;&gt;7.636×10 2 /s  C6:
 
     Thus, all of the constraints defined by C1-C6 are satisfied, and the system manufactured and operated with these conditions will therefore operate as described. Of course, many other choices of the variables L, S, H, α, and f will work as well, and the operation of the system described herein is certainly not limited to the specific computational example described herein. 
     The described system does not violate the principles of the conversation of energy, and the underlying operation of the system instead, is consistent with the principle of conservation of energy as the energy comes from the heat in the ambient fluid. Generally speaking, the sonic thrust element described herein produces a low pressure region on the side of the plate where the oscillator is located while keeping standard atmospheric pressure on the other side of the plate. By restricting the wavelength of the sound waves to be small compared to the dimensions of the plate, the sound waves behave like particles moving at nearly the speed of sound. The analogy with the flow of air across an airplane wing is thus very strong and Bernoulli&#39;s law in fact predicts a thrust that is close to that of standard atmospheric pressure against a vacuum The pressure differences are directly related to temperature differences, and hence to energy differences. Stated another way, Bernoulli&#39;s law, which predicts the pressure difference on the opposite sides of the plate, combined with the ideal gas law, which predicts a temperature change with pressure change, means that the heat energy in the air is converted into the kinetic energy in the moving plate and lever arm device. Thus, the sound waves are not converted into to kinetic energy directly, but serve as a catalyst for converting heat energy in the ambient fluid (e.g., air) into kinetic energy. 
     It is possible to modify the sonic thrust element  5  of  FIG. 1  to reduce or eliminate the need to account for the finite distance ε between the bottom of the oscillator  20  and the top of the plate  10  in the equations above. In particular, the oscillator  20  can be disposed partially down within the plate  10  to assure sound waves emanate from the oscillator  20  at the same level as the top surface  12  of the flat plate  10 , so that there is no distance between the “bottom of the oscillator” and the top of the flat plate  10 . 
       FIG. 4  illustrates one example of such a configuration. Here, the plate  10  is illustrated as including a top portion  10 A and a bottom portion  10 B connected by side portions  10 C forming a hollow section  102  between the top and bottom portions  10 A and  10 B of the plate  10 . The oscillator  20  extends from above the top portion  10 A of the plate  10  down into the hollow section  102  of the plate  10  through a hole or slit  104  such that the bottom of the oscillator  20  is at the same level as or below the top surface  12  of the top portion  10 A of the plate  10 . A control mechanism  106  is illustrated diagrammatically in  FIG. 4  as being attached to the oscillator  20  to operate the oscillator  20  to produce sound waves of, for example, a constant frequency as defined by the constraints described herein. The mechanism  106  may be electrical or mechanical in nature using any standard oscillator technology. Moreover, if desired, the control mechanism  106  may be at least partially disposed within the hollow section  102  of the plate  10 . 
     Using the sonic thrust element illustrated in  FIG. 4  eliminates the need for Eq. (12) and the constraints (C2) (C3) and (C4) given above for the sonic thrust element  5  of  FIG. 1 . This configuration is thus left with a smaller set of constraints given as: 
                         5.833   ×     10     -   4       ⁢     PL     1   2             δ     2   3       ⁢     S     1   6       ⁢     H     2   3       ⁢     Ω     1   3           ⁢     s       g     1   3       ⁢     cm     2   3             &gt;&gt;   f           (     C   ⁢           ⁢     1   ′       )                   3.80   ×     10     -   2       ⁢   δ   ⁢           ⁢     C   3         μ   ⁢           ⁢   H       &gt;&gt;           ⁢     f   2             (     C   ⁢           ⁢     2   ′       )               f   &gt;&gt;       0.564   ⁢           ⁢   C         L     1   2       ⁢     S     1   2                   (     C   ⁢           ⁢     3   ′       )               
As before, for a single arm rotational device using the thrust element of  FIG. 3 :
 
     
       
         
           
             G 
             = 
             
               α 
               ⁢ 
               
                   
               
               ⁢ 
               PLS 
             
           
         
       
       
         
           and 
         
       
       
         
           
             
               P 
               ⁢ 
               
                   
               
               ⁢ 
               O 
             
             = 
             
               
                 
                   .1047 
                   × 
                   
                     ( 
                     
                       # 
                       ⁢ 
                       
                         R 
                         m 
                       
                     
                     ) 
                   
                   ⁢ 
                   α 
                   ⁢ 
                   
                       
                   
                   ⁢ 
                   PLS 
                 
                 s 
               
               . 
             
           
         
       
     
     Again, if the ambient fluid is taken to be air at 69 deg F. at standard atmospheric pressure, then:
 
μ=0.00018 g/cm-s (the absolute viscosity of air).
 
 P= 1.013×10 6  g/cm-s 2  (the ambient air pressure).
 
ρ=1.1781 g/cm 2 -s 2  (the weight density of ambient air).
 
δ=1.2013×10 −3  g/cm 3  (the mass density of ambient air).
 
 C= 33162 cm/s (the speed of sound in ambient air).
 
     Plugging these values into the constraints (C1′), (C2′) and (C3′) gives the specific constraints: 
     
       
         
           
             
               
                 
                   
                     
                       
                         5.229 
                         × 
                         
                           10 
                           4 
                         
                         ⁢ 
                         
                           L 
                           
                             1 
                             2 
                           
                         
                       
                       
                         
                           S 
                           
                             1 
                             6 
                           
                         
                         ⁢ 
                         
                           H 
                           
                             2 
                             3 
                           
                         
                         ⁢ 
                         
                           Ω 
                           
                             1 
                             3 
                           
                         
                       
                     
                     ⁢ 
                     
                       
                         cm 
                         
                           1 
                           3 
                         
                       
                       s 
                     
                   
                   &gt;&gt; 
                   f 
                 
               
               
                 
                   ( 
                   
                     C 
                     ⁢ 
                     
                         
                     
                     ⁢ 
                     
                       1 
                       ′ 
                     
                   
                   ) 
                 
               
             
             
               
                 
                   
                     
                       
                         9.249 
                         × 
                         
                           10 
                           12 
                         
                       
                       H 
                     
                     ⁢ 
                     
                       cm 
                       
                         s 
                         2 
                       
                     
                   
                   &gt;&gt; 
                   
                     f 
                     
                       
                           
                       
                       ⁢ 
                       2 
                     
                   
                 
               
               
                 
                   ( 
                   
                     C 
                     ⁢ 
                     
                         
                     
                     ⁢ 
                     
                       2 
                       ′ 
                     
                   
                   ) 
                 
               
             
             
               
                 
                   
                     f 
                     &gt;&gt; 
                     
                       
                         
                           1.870 
                           × 
                           
                             10 
                             4 
                           
                         
                         
                           
                             L 
                             
                               1 
                               2 
                             
                           
                           ⁢ 
                           
                             S 
                             
                               1 
                               2 
                             
                           
                         
                       
                       ⁢ 
                       
                         cm 
                         s 
                       
                     
                   
                   ⁢ 
                   
                     
 
                   
                   ⁢ 
                   and 
                   ⁢ 
                   
                     
 
                   
                   ⁢ 
                   
                     
                       P 
                       ⁢ 
                       
                           
                       
                       ⁢ 
                       O 
                     
                     = 
                     
                       
                         
                           1.0606 
                           × 
                           
                             10 
                             5 
                           
                           ⁢ 
                           
                               
                           
                           ⁢ 
                           
                             ( 
                             
                               # 
                               ⁢ 
                               
                                 R 
                                 m 
                               
                             
                             ) 
                           
                           ⁢ 
                           α 
                           ⁢ 
                           
                               
                           
                           ⁢ 
                           LSg 
                         
                         
                           cm 
                           ⁢ 
                           
                               
                           
                           ⁢ 
                           
                             s 
                             3 
                           
                         
                       
                       . 
                     
                   
                 
               
               
                 
                   ( 
                   
                     C 
                     ⁢ 
                     
                         
                     
                     ⁢ 
                     
                       3 
                       ′ 
                     
                   
                   ) 
                 
               
             
           
         
       
     
     A number of examples using this configuration will now be provided, it being understood that other example configuration parameters can be used as well or instead. 
     Example 1 
       α=1000 cm
 
 L= 100 cm
 
 S= 100 cm
 
 H= 10 cm
 
# R   m =100
 
Ω=1 watt.
 
In this example, the constraints (C1′)-(C3′) become:
 
                 5.229   ×     10   4       s     &gt;&gt;   f                   9.249   ×     10   11         s   2       &gt;&gt;     f             ⁢   2                   f   &gt;&gt;       1.870   ×     10   2       s           
and these constraints may be consistently satisfied for f=2×10 3  hertz.
 
     Example 2 
       α=100 cm
 
 L= 20 cm
 
 S= 5 cm
 
 H= 1 cm
 
# R   m =100
 
Ω=0.25 watt.
 
In this example, the constraints (C1′)-(C3′) become:
 
                 2.253   ×     10   5       s     &gt;&gt;   f                   9.249   ×     10   12         s   2       &gt;&gt;     f             ⁢   2                   f   &gt;&gt;       1.870   ×     10   3       s           
and these constraints may be consistently satisfied for f=2×10 4  hertz.
 
     Example 3 
       α=5 cm
 
 L= 2.5 cm
 
 S= 2 cm
 
 H= 0.5 cm
 
# R   m =40
 
Ω=0.001 watt.
 
In this example, the constraints (C1′)-(C3′) become:
 
                 1.169   ×     10   6       s     &gt;&gt;   f                   1.850   ×     10   13         s   2       &gt;&gt;     f             ⁢   2                   f   &gt;&gt;       8.364   ×     10   3       s           
and these constraints may be consistently satisfied for f=10 5  hertz.
 
     While the present invention has been described with reference to specific examples, which are intended to be illustrative only and not to be limiting of the invention, it will be apparent to those of ordinary skill in the art that changes, additions or deletions may be made to the disclosed embodiments without departing from the spirit and scope of the invention.

Technology Category: 5