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Theme 091011

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Theme 091011

1. 1. SECTION 2. AERODYNAMICS OF BODY OF REVOLUTIONS THEME 9. FUSELAGE GEOMETRICAL PARAMETERS 9.1. General Fuselage is the main part of an airplane structure. It serves for joining of all itsparts in a whole, and also for arrangement of crew, passengers, equipment and freights. The exterior shape of a fuselage is determined by the airplane assigning, range ofspeeds of flight, arrangement of engines and other factors. The airplane fuselages (Fig. 9.1) and engine nacelles have the shape of the bodyof revolutions or close to it. For the airplanes having integral configurations the wingpasses smoothly into the fuselage and the shape of the fuselage cross-section canessentially differ from circular. The fuselages of transport airplanes frequently have tailunit deflected upwards. The noses of modern fighters, as a rule, are rejected downwards. Fig. 9.1. The basic geometrical characteristics of a fuselage The basic geometrical parameters of a fuselage are the following ones. Length of a fuselage l f is the greatest size of a fuselage along its centerline. The area of fuselage midsection S m . f . is the greatest area of fuselage cross-section by a plane, perpendicular to its centerline. The shape of a fuselage cross section essentially influences the interferenceaerodynamic characteristics at the installation of a wing and tail unit. While calculatingthe aerodynamic characteristics of an isolated fuselage it is approximately substituted bya body of revolution with the equivalent area of cross sections. 92
2. 2. Maximum equivalent diameter of a fuselage d m . f . is the diameter of a circle,which area is equal to the fuselage midsection area. d f = d m. f . = 4 Sm. f . π . (9.1) The distinction of a fuselage from the body of revolution is taken into account byan addend in the aerodynamic characteristics depending on design features. Thereforewe shall consider the geometric and aerodynamic characteristics of body of revolutions,connecting them with the specific fuselage below. Fuselage aspect ratio λ f is the ratio of the fuselage length to its maximumequivalent diameter, λf = lf d f . (9.2) In some cases, especially when the fuselage is the body of revolution, it ispossible to allocate nose (head), cylindrical (central) and rear parts (fig. 9.1) and tointroduce the appropriate geometric parameters for them. As the total fuselage lengthl f = l nose + lcil + l rear , then its aspect ratio λ f = λ nose + λ cil + λ rear , (9.3)where λ nose , λ cil , λ rear is the aspect ratio of nose, cylindrical and rear parts, λ nose = l nose d f , λ cil = lcil d f , λ rear = l rear d f . Frequently nose parts of fuselages have bluntness. In some cases, nose can have an inside channel - the engine air intake. The rear part of a fuselage can have a blunt base. Then, the following additional parameters for the description of a nose and rearpart are used: Nose tapering η nose = d nose d f is the ratio of a fuselage nose diameter to itsmaximum equivalent diameter. Tapering of a rear part η rear = d base d f is the ratio of diameter of a base of afuselage to its maximum equivalent diameter. The relative area of the blunt base - S base = d base d 2 . 2 f 92
3. 3. The following angles are used at presence of a cam tail part. An angle of meanline deviation β rear , the deviation of the nose part is determined by the angle β nose(angle β nose is considered as positive at nose part deflection downwards). 9.1. Shape of a nose part and its geometrical parameters The conical shape of a fuselage nose: 1 1 λ nose = , tgβ 0 = ; 2 tgβ 0 2λ nose 1 Wnose = l nose S m . f . - volume of the nose part. 3 1 − η nose 1 − η nose λ nose = , tgβ 0 = ; 2 tgβ 0 2λ nose Wnose = 1 3 ( 2 ) 1 + η nose + η nose l nose S m . f . . The parabolic shape of a fuselage nose: x r = 0 .5 d f x( 2 − x ) , x = , 0 ≤ x ≤ 1; l nose 1 η nose = 0 , tgβ0 = . λ nose [ r = 0 .5 d f ηnose + (1 − ηnose ) x( 2 − x ) ; ] Wnose = 1 15 ( 2 ) 8 + 4η nose + 3η nose l nose S m . f . . The ogival shape formed by arcs of a circle is close to the parabolic one, which β0aspect ratio is equal to λ nose = 0 .5 ctg . Volume of the nose part at λ nose ≥ 2 .5 is 2determined by expression Wnose = 2 3 ( 2 ) 1 + 0 .5η nose l nose S m . f . . The elliptical (ellipsoidal) shape of the nose part. 92
4. 4. x r = 0 .5 d f 1 − ( x − 1) = 0 .5 d f x( 2 − x ) , x = 2 ; l nose 2 Wnose = l nose S m . f . 3 The particular case of an elliptical nose part is the hemisphere. The shapes of rear parts are designed the same as nose ones. THEME 10. FEATURES OF FLOW ABOUT BODYS OF REVOLUTION Lets consider a body of revolution, which is streamlined by undisturbed flow atangle of attack α . The flow can be represented as a result of superposition of two flows- longitudinal with speed V∞ cos α and transversal with speed V∞ sinα . At large anglesof attack the flow is determined by transversal flow, and at small angles of attack - bylongitudinal. The transversal flow is always subsonic at small angles of attack. Conditionally we shall point out three flow modes about body of revolution. 1. Attached flow at small angles of attack ( α = 0 ...5 o ). At small angles of attackthe flow about cross-sections differs only by thickness and status of the boundary layer.The laminar boundary layer of small thickness is in the nose. Further thickness of aboundary layer gradually arises along the length of the body of revolution. Its charactervaries, the boundary layer becomes turbulent. 92
5. 5. 2. At moderate angles of attack ( α ≤ 20 o ) the axis-symmetrical character of flow is upset. The flow about fuselage takes place with separation of the boundary layer at lateral areas. The separated boundary layer is turned into two vortex bundles (Fig. 10.1, a). The location of the point of boundary layer separation depends on the shape and aspect ratio of the nose part, Mach numbers and some other factors. 3. At large angles of attack ( α > 30 o ) the disturbance of a symmetry of a vortex system takes place. Vortex bundles separate from the surface of the body of revolution, not having reached its rear part. At Fig. 10.1. The flow scheme disturbance of flow rotational symmetry the whole system of vortexes on the upper surface of a body of revolution (Fig. 10.1, a) is formed. It results into formation of sizable transversal forces and moments of yaw. At supersonic speed M ∞ > 1 and large angles of attack the internal (hanging) shock waves appear because of a large positive Fig. 10.2. Flow near a cone: a 1-cone; gradient of pressure on leeward fuselage side 2-head shock wave; (Fig. 10.2). 3-vortexes; 4-hanging internal shock At that, the flow structure is similar to waves. the structure of the track behind the cylinderwith circular cross section. The hanging shock waves prevent from the loss of symmetryof the fuselage vortical system. Therefore, at large supersonic speeds transversal forces 92
6. 6. and the moments of a yaw conditioned by non-symmetry of the fuselage vortical areabsent. THEME 11. DEFINITION OF THE AERODYNAMIC CHARACTERISTICS OF A BODY OF REVOLUTION USING PRESSURE DISTRIBUTION It is possible to define aerodynamic forces effecting the body of revolutionknowing the law of pressure distribution along its surface. The pressure in the givenpoint of the surface is determined by two factors: parameters of incoming flow andgeometrical features of the streamlined body. The angle of attack α and number M ∞exert the essential influence onto pressure distribution along outline of the body ofrevolution. If we compare pressure distribution along wing airfoil with pressuredistribution along body of revolution, then it is possible to note the following: therarefaction on the body of revolution is much less than rarefaction on the wing. It takesplace due to spatial flow about the body of revolution. The flow spatiality enfeebles theinfluence of flow compressibility onto character of flow about the body of revolution. Insubsonic flow for the body of revolution the factor of pressure is equal C p incomp Cp = , (11.1) 4 2 1 − M∞ C p incompwhile at flow about the wing C p = , i.e. the factor of pressure C p of the body 2 1− M∞of revolution depends on number M ∞ less than C p for the wing. Lets separately consider the pressure on a lateral surface of the body ofrevolution and the pressure on the blunt base. Lets define the aerodynamiccharacteristics of the body of revolution in body axes, and then by the transitionformulae we shall receive the aerodynamic characteristics in wind axes. Lets assume 92
7. 7. that an overpressure p − p∞ (Fig. 11.1) acts on an elementary site of the lateral surfacedS = r dl dϕ . Fig. 11.1. A nose part of a body of revolution. The function for elementary normal force dY from pressure p − p∞ effectingonto an element of the lateral surface dS looks like this: dY = − ( p − p∞ ) ds cos ϕ cos ϑ = − q∞ C p r dl dϕ cos ϕ cos ϑ , (11.1)taking into account, that dl cos ϑ = dx , we shall receive dY = − q∞C p cos ϕ r( x ) dx . (11.2) Analogously, we shall find a function for elementary longitudinal force dX frompressure p − p∞ dX = ( p − p∞ ) ds sinϑ = q∞ C p r dϕ dl sinϑ , (11.3) drtaking into account, that dl sinϑ = dr and dr = dx , we shall receive dx . dX = q∞C p r r dx dϕ . (11.4) Integrating expressions (11.2) and (11.4) by length of the body of revolution from0 up to l f and by an arc of a circle from 0 up to 2π we shall receive the formulae fornormal and longitudinal force without the account of forces of friction lf 2π Y = − q∞ ∫ r( x) dx ∫ C p ( x , ϕ ) cos ϕ dϕ ; (11.5) 0 0 92
8. 8. lf 2π . X = q∞ ∫ r( x) r( x)dx ∫ C p ( x ,ϕ ) dϕ . (11.6) 0 0 Lets write down the expression for the elementary moment dM z from normaland longitudinal force dM z = − dY x + dX r cos ϕ = ( p − p∞ )ds cos ϕ ( x cos ϑ + r sin ϑ ) ; ⎛ .⎞ (11.7) dM z = q ∞ C p cos ϕ ⎜ x + r r⎟ rdxdϕ ⎝ ⎠also we shall define the function of the longitudinal moment from pressure as lf 2π ⎛ . ⎞ M z = q∞ ∫ ⎝ ⎠ ∫ r( x )⎜ x + r( x ) r( x )⎟ dx C p ( x ,ϕ ) cos ϕ dϕ . (11.8) 0 0 Lets pass in the formulae (11.5), (11.6) and (11.8) from forces to their factors. Asthe characteristic area we shall accept the area of midsection of a body of revolutionS m . f . , and as characteristic length - length of a fuselage l f . We can write down lf π 2 C ya ≈ C y = − Sm. f . ∫ r( x) dx ∫ C p cos ϕ dϕ ; (11.9) 0 0 lf π 2 . Cx = Sm. f . ∫ r( x) r( x) dx ∫ C p dϕ , (11.10) 0 0at α = 0 the factor of pressure C p in the last formula does not depend on the angle ϕ ,and depends only on coordinate x and in this case lf 2π . C xa0 = C x0 = Sm. f . ∫ C p r( x ) r( x )dx ; (11.11) 0 lf π 2 ⎛ .⎞ mz = Sm. f . l f ∫ ⎝ ⎠ ∫ r⎜ x + r r⎟ dx C p cos ϕ dϕ , (11.12) 0 0For thin bodies ( r << 1 ) it is possible not to take into account a moment from &longitudinal force: 92
9. 9. lf π 2 mz ≈ Sm. f . l f ∫ r x dx ∫ C p cos ϕ dϕ . (11.13) 0 0 Dimensionless coordinate of center of pressure location can be defined byformula for an aerodynamic moment relatively to the fuselage nose M z = −Y xc . p . : xc . p . mz m x c . p. = =− ≈− z . (11.14) lf Cy C yа Lets define pressure forces which act onto the blunt base. If the blunt base islocated along the normal to an axis of the body of revolution, there is only longitudinalforce, which at small angles of attack practically does not vary on α . This force iscalled the force of base drag. The force of base drag can be determined at α = 0 . Inthis case, value of ( pbase − p∞ ) along the circle with radius r is a constant( pbase − p∞ ) = const (Fig. 11.2). Fig. 11.2. The blunt base of the fuselage. The function for elementary longitudinal force dX from pressure ( pbase − p∞ )on an element of the base surface dS = 2π r dr looks like this: dX = dX base = − ( pbaase − p∞ )2π rdr = − q∞ C p base 2π rdr . (11.15) Lets pass to the factor of force of base drag rbase 2π C x base = − Sm. f . ∫ C p base r dr . (11.16) 0 Practically, pressure on the blunt base C p base ( r ) ≈ C p base = const and in thiscase it is possible to consider, that 92
10. 10. C x base = − C p base S base . (11.17) Thus, the aerodynamic characteristics of the body of revolution can be calculated,if the pressure distribution along its surface and blunt base is known. 92