Perfiles aerodinámicos Joukovsky notas suplementarias de aerodinámica
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(2) Luis A. Arriola. Introduction. coordinates and whose angle of attack α is different from zero.. One of the major results obtained in Aerodynamics is the calculation of the ideal flow past a cylinder. Because we do not see flying cylinders, you may wonder why this result is so important. Two reasons will be given here. The cylinder has a simple shape and can be easily tested in a wind tunnel to determine the limitations of ideal fluid theory. Moreover, the cylinder can be mapped into arbitrary shapes by using Figure a Cylinder Figure1.1 Schematic Schematic of aofCylinder appropriate mapping functions. This allows us to determine the velocity and the pressure Figure 1 shows such a cylinder. Note that Figure 1 Schematic of a Cylinder 1 shows such a cylinder. that the isx ′ -coordinate is the direction of the distribution on aDISCUSSION body Figure of arbitrary shape. ′ -coordinate the direction of the the xNote Figure 1 shows such a cylinder. Note that the x′ -coordinate is the direction of the freestream velocity and y’ is normal to the Discussion freestream velocity and y’ is normal to the freestream direction. The angle between the DISCUSSION freestream freestream velocity and y’ is normal to the The freestream direction. The anglethe between the direction. angle between These notes will describe the procedure that maps a cylinder into an airfoil. As an ′ α.isWe xattack the angle of attack α.in terms xcan coordinate axes axes of is the angle can write stream function x x′ attack These notes will describe theaxes procedure coordinate the angle α. ofWe write the the stream function in of terms of x x ′ iscoordinate These notes will describe the procedure that maps a cylinder into an airfoil. As an We can write the stream function in terms that maps cylinder into an Before airfoil.we Asdoanthis, illustration, a Joukowski airfoil awill be considered. we need review the coordinate ( x′ , y′to ) using Equation (2). The result can be written as: ′ ) using Equation the coordinate ( x ′ , ybe (2). The result written as: y ′ )beusing equation 2. of the coordinate ( x ′ ,can illustration, a Joukowski airfoil will illustration, Joukowski airfoil will be considered. Before we do this, weThe needresult to review some resultsarelevant to flow past cylinders. can be written as: considered. Before we do this, we need to R Γ r′ Ψ = U 1 − ln (4) r ′ sin θ ′ + ′ π R 2 r review some results relevant to flow past some results relevant to flow past cylinders. 2 R Γ r′ If the origincylinders. of the coordinates coincides with the origin of a cylinder R, r ′ sin θ ′ + Ψ =ofUradius ln (4) ∞ 1 − 2 To determine Ψ=Ψ( the relationship ( x′ , y′ ) R between x, y ),r ′we have to determine2π (4) the origin of If thethe coordinates withΦ,the origin of a by cylinder radius R, origin ofcoincides thefunction, coordinates then theIf stream function, Ψ, or potential ascoincides described Bertinofand Smith and ( x, y ). To accomplish this, let us introduce the coordinate system ( x ′′ , y ′′ ) whose with the origin of a cylinder of radius R, To determine Ψ=Ψ( x, y ), we have to then theforstream function, Ψ, or potential function, Φ, as described bycoincides Bertin and Smith with the center of the cylinder but is parallel to ( x, y ). (1998) a lifting cylinder be obtained by superimposing a origin uniform flow in the xthen the can stream function, Ψ,Toordetermine potential determine relationship between (between x′ , Ψ=Ψ( x, y ), we havethe to determine the relationship ( x′ , y ′ ) function, Φ, as described by Bertin and ′ ) and y ( x, y ). To accomplish this, let us (1998) for a lifting cylinder can be obtained by superimposing a uniform flow in the xdirection, a doublet at the origin whose strength is: Smith (1998) for a lifting be accomplish this, let us introduce the coordinate and ( cylinder x, y ). To can introduce the coordinate systemsystem ( x ′′ , (yx′′′′ , y ′′ ) whose bywhose superimposing direction, a doublet obtained at the origin strength is: a uniform flow ) whose origin coincides with the center of 2origin coincides in the x-direction, a doublet at the origin with the center the cylinder is parallel theofcylinder butbut is parallel to to ( x,( x,y y).). Β = R U∞ (1) 4 whose strength is: 2 2. ∞. Β = R U∞. Â = R 2 UThe (1)written as: ∞ and a vortex at the origin of strength Γ (gamma). result can be. 2. (1). and aofvortex the origin strength and a vortex at the origin strengthatΓ (gamma). Theofresult can beΓwritten as:. and and. (gamma). The result Γ as: R 2 can be written r Ψ = U ∞ 1 − 2 r sin θ + ln 2Γπ Rr Rr 2 Ψ = U ∞ 1 − 2 r sin θ + ln 2π R r (2) and. (2) (2). 4. Figure 2 Relationship between (x',y') and (x,y). Figure 2. Relationship between (x’,y’) and R2 Γθ (x,y) Figure (3) φ = U ∞ 1 + 2 r cos θ − 2 Relationship between (x',y') and (x,y) 2from πθ Figure (3) 2 that Rr 2 It is notedΓ φ = U ∞ 1 + 2 r cos θ − It is noted (3)from Figure 2 that It is noted from Figure 2 that As a first step rin achieving2πour goal, let us determine theusstream function for afunction x ′′ − ac , y = y ′′ + bc (5) As a first step in achieving our goal, let determine the stream forx a= cylinder ′ ′ , x = x − a y = y ′′ + bc c cylinder whose center is not at the origin of . As a first stepisinnot achieving our goal, let us determine the stream cylinder whose center at the origin of coordinates and whose angle function of attack for α isa different. (5). at the center of thecylinder. cylinder. Again, we we see from −of a ccoordinates , bc are the whose center is not at the origin of coordinates anglethe attack α iscoordinates different whereand− whose are at the center of the Again, see Figure from Fig a c , bc where from zero. 46 | Campus | V. XX | No. 20 | julio-diciembre | 2015 | #2 that: from zero.. #2 that:.
(3) transformation transforms Eq. (9) into an airfoil can be written where − a c , bc are the coordinates we see fromthat Figure ′ + bc Again, (5) x = x ′′at− the ac ,center of the yThe = cylinder. y ′Joukowski that: re − a c , bc are the coordinates at the center of the cylinder. Again, we see from Figure Perfiles aerodinámicos Joukovsky notas suplementarias de aerodinámica 2 that: The Joukowski The Joukowski transformation transformation that transforms that transforms Eq. (9) into Eq.an (9)airfoil into an can airfoil be written can beas: written as: here the αcenter x ′ = xat′′ cos + y ′′ of sinthe α cylinder. Again, we see from Figure hat: − a c , bc are the coordinates that:. where are the theαcenter of x ′ =coordinates x ′′ cos α + yat′′ sin the cylinder. Again, we see from Figure 2 ′ cos α + y ′′ sin α that: x ′y=′ =x ′− x ′′ sin α + y ′′ cos α. The Joukowski a 2 transformation that a2 x1 =equation x(1 + 2 9) ,into an airfoil can y1 =bey (1 − 2 ) transforms r r written as: (6). a2 a2 a2 a2 x ′′αsin+αy ′+ y ′′αcos α ′ sin x ′y=′ =x ′′−cos (11) (1 x1 = x(1 +x1 =2 )x,(1 + 2 ) , (6)y1 = y (1 −y1 =2 )y (1(11) − 2) r schematically in Figure r The resulting shaper is shown 4. rThe actual shape is calc r,mbining we can(5) relate by writing the ′ = desired θ ′ythe − x ′′ sinrelationship, y ′′ cos αi.e.of any α +coordinate (6) arbitrary point B on (6) and θ(6)and yields resulting shape is yshown schematically ( xThe 1 = 2a , 1 = 0 follows: in Figure 4. The actual is calculated as Combining (5) and (6)Combining yieldsythe (5) (6) the desired ′ = desired ′ sinrelationship, ′′ cos der as: (6) shape − x ′and +yields yThe αresulting αi.e. The resulting shape is shape shownis schematically shown schematically in Figure in4.Figure The actual 4. The shape actual is shape calculated is calculate as follows: ′ over, we can relaterelationship, by writing the coordinate of any arbitrary point B on θx ′ =and θ a i.e. ( x + ) cos α + ( y − b ) sin α c c mbining (5) and (6) yields the desired relationship, i.e. The point A (see Figure 3), which has the The point A (Figure 3),coordinate which has the x ′ = ( x ′+ a c ) cosfollows: α + ( y −follows: bc ) sin α linder as: (8) sin( ) sin θ α θ R + = r − b c mbining (5) and (6) yields the desired relationship, i.e. coordinate x ′y=′ =( x−+(axc+) cos α + ( y − b ) sin α (7) (7) a ) sin α + ( y − b ) cos α c c c eover, we can relate θ and θ ′ by writing the of3), any arbitrary on coordinate The point The Acoordinate (see pointFigure A (see Figure which 3), haswhich ,hasB the r the =point acoordinate θ =0 (12) ′cos (7) y=′ =(Rx−sin( (axθ+)we a+cα)can sin +−(θbyθ−) sin cos α′ by (8) ) =+α r( ysin band cc )α θ Moreover, relate ′ x + α ow, if weas:substitute for x ′ , y ′ into Eq. (4), we can determine Ψ=Ψ( x, y ). c c cylinder maps into the trailing edge of the airfoil. writing the coordinate of any arbitrary (7) y ′ = − ( x + a c ) sin α + ( y − bc ) cos α point Substituting equation 11 r =the a , trailing r = a ,edgeequation θof= the 0 airfoil. θ12= into 0 Substituting B on the cylinder as: maps into Eq. (12) (13) into (12) Eq. (11)(1 y , = x 2 a y = 0 yields the coordinate of the trailing edge as: 1 1 ′ ′ if we function substitute for , into Eq. (4), we can determine Ψ=Ψ( ). x y x, y s Now, the stream for a flow past a cylinder whose center is not at the origin ′ (7) y = −R( xsin( + aθc ′)+sin y −θb−c )bcos α (8) (8) ) =+r(sin αα c coordinate of the trailing edge as: 5 , = x 2 a y1 = 0 1 (13) e angle of attack is α.Now, if we substitute maps into maps the into trailing the edge trailing of the edge airfoil. of the Substituting airfoil. Substituting Eq. (12) Eq. into (12) Eq. into (11) Eq. yields (11)theyields for x ′ , y ′ into 5 gives the stream function for a flow past a cylinder whose center is not at the origin equation 4, we can determine Ψ=Ψ( x, y ). Now, if we substitute for x ′ , y ′ into Eq. (4), we cantrailing determine Ψ=Ψ( x, y ). coordinate coordinate of the of theedge trailing as: edge as: 5 ki Transformation hose angle of attack is α. gives the stream function for a flow This past a cylinder whose center is not at the 5 gives the stream origin function for a flow past aattack cylinder whose center is not at the origin and whose angle of is α. ow weTransformation are ready to talk about mapping. As can be seen from Figure 3, the owski 7 Figure 4 Schematic of a Joukowski Airfoil whose angle of attack is α.Joukowski Transformation of the circle under consideration can be written as: Now we are readyNow to talk about As can be seen from Figure 3, the of a Joukowski Airfoil Figure 4.0Schematic we are readymapping. to talk about Next mapping. (theta) is stepped from to 2π using, say, 5° intervals. For a given theta, Eq. (9) g kowski Transformation As can be seen from Figure 3, the equation 2 2 Next (theta) on of the circle under consideration written (9) is stepped from 0 to 2π =circle l 2 + r under + (can 2lrconsideration )be cos( θ r+=δ r)as: of R the (θ can ) andbesince using, say, 5° intervals. For a given theta, written as: Figure 4 Schematic Airfoil = ofr (aθJoukowski ) and since 9 gives3, rthe Now we are ready to talk about mapping. As can be seen equation from Figure 2 2 2 (9) R = l + r + (2lr ) cos(θ + δ ) (9) Next (theta) is stepped from 2π θ using, , say, 5° intervals.yFor= ar given x =0rtocos sin θ theta, Eq. (9) gives ation of the circle under consideration can be written as: with r = r (θ ) and since x1 , y1 follow from Equation 11. Repeating 2 2 2 (10) l = bc 2 + a c2 , 2 tan δ = bc / a c (10) the process will (9) allow to generate the us to generate the de R = l + r + (2lr ) cos( δ) from Eq. (11). Repeating theusprocess will allow x1 , yθ1 +follow x = r cosshape. θ, y = r sin θ of a Joukowski Airfoil desired Figure 4 Schematic. shape. (10) of the Circulation tan δ = bc / a c Determination fromisEq. (11). Repeating the2π process willsay, allow to generateFor the a desired x1Next , y1 follow (theta) stepped from 0 to using, 5°usintervals. given theta, Eq. Although we were able to relate the shape. Determination of the Circulation strength of the doublet to the radius of the r = r (θ ) and since 2 2 2 circle and the free (10)stream velocity, nothing l = bc + a c , tan δ = bc / a c Determination of thewas Circulation said, so far, about Γ. Obviously, we need Although we were able to relate the strength of the doublet to the radius of to determine Γ ifθ we hope to calculate , strength x to= relate r cos y = rtothe sin θ Although we velocity were able and of the doublet over the radius of the the the pressure distribution circle and the free stream velocity, nothing was said, so far, about Γ. Obviously, we the velocity, airfoil. nothing The value of so Γ far, is determined circle and the free stream was said, about Γ. Obviously, we need from Kutta´s condition. According this to determine ΓΓ ifif we we hope to(11). calculate the velocity andtothe pressureusdistribution Figure 3. Schematic of a Circle from Eq. Repeating the process will to generateove th to xdetermine hope to calculate the velocity and the pressure distribution the 1 , y1 follow condition, the flow must leave the trailing allowover 2. l 2 = bc + a c. 2. ,. Figure 3 Schematic of a Circle. airfoil. Thevalue value of determined from Kutta´s condition.condition. According toAccording this condition, airfoil. The ofΓΓis is determined from Kutta´s to this condi shape. the flow must leave the trailing edge smoothly as described by Anderson, Jr (1989). This,. | Campus | V. XX | No. 20 the | julio-diciembre | 2015 | flow must leave the trailing edge smoothly as described 47 by Anderson, Jr (1989). T turn, requires the point A on the cylinder, which maps into the trailing edge, to be a The Joukowski transformation that transforms Eq. (9) into an in airfoil can be written as: Determination of the Circulation in turn, requires the point A on the cylinder, which maps into the trailing edge, to.
(4) L ( x′ , y ′ ) 2the stagnation follows from calculating theIfvelocities at A. (Ifx ′we tagnation apoint follows point from calculating the velocities at A. we consider , y ′ consider )Cat=high Usually, the flowtheseparates angles =of2attack. α should rem + β) π sin(αTherefore, l 1 α +2 β ) ≈ (α + β ) sin(. ρU ∞ c coordinate system 3),Athen thecoordinates: point the coordinates: ordinate system (see Figure 3), (see then Figure the point has the Luis A. Arriola angleAifhas the above theory is to remain 2 αvalid. + βof)This ≈ + Therefore, β ) as setting: sin( (αallows Usually, the flow separates at high angles attack. α should remain. and as a result we have is to remain valid. This setting: + β ) if the (14) r ′ = R , θ ′ = −(αangle θ ′ above = −(α theory + β ) (14) α separates +ofβ ≈ allows βas )attack. (αat +high ) angles and asthe a result Usually, thesin( flow edge smoothly as described by Anderson, Jr we Usually, flow separates at high angles Therefore, α should remain stagnation point. This condition prevents the flow from turning around thehave trailing edge of attack. Therefore, α should remain a small (1989). This, in turn, requires the point A and insures the continuity of the pressure at the trailing The implicationsangle )β )as setting: π) (≈αis(allows C = β2This if remain the sin( above theory valid. on the cylinder, which maps intoedge. angle ifthe thetrailing above theoryofisAtobeing valid. αl + α+toβ+remain e velocity The components are: and as a result we have velocity components are: Cl = 2π (α + β ) This allows setting: edge, to from be a stagnation point. Thisatcondition ′ ) asthe a This stagnation point follows calculating the velocities A.around Ifthe wetrailing consider the (around x ′ , yedge stagnation stagnation point. point. This condition This condition prevents prevents flow the from flow from turning turning the trailing the trailing edge edge tion point. stagnation condition point. prevents This condition thethe prevents from turning the flow around from the turning trailing trailing edge stagnation point. This condition prevents theflow flow around prevents the flow from turning around the edge and insures and insures the continuity the continuity of the the at the trailing edge. edge. The implications The implications of A Ahave being of A being sin(α + β ) ≈ (α + β ) andcoordinate insures thesystem continuity ofFigure theofpressure pressure at the the implications of being (see 3),pressure then theattrailing point A has theascoordinates: trailing edge and insures the continuity and a trailing result we For symmetric airfoils, π (being α + β) sures the continuity and insuresofthe thecontinuity pressure atofthe thetrailing pressure edge. at the The implications edge. The of Aimplications being Cl = of2A offollows the pressure at velocities the trailing edge. The aastagnation a stagnation point follows from calculating calculating the the velocities at A. at A. If consider we consider the (airfoils, ( x′ the ) , y′as ) a result we have x′ y′(and For symmetric stagnation pointpoint follows from from calculating the If we consider the ,, y′ )x′ implications of A being2 a stagnation ′ =then ′at=coordinates: , the r3), R3), θthe − (α If +velocities β )point at A.the nation point a coordinate stagnation follows from point calculating follows from the calculating A. ( (14) the ( x ′ , y ′ ) , y′ ) x ′ consider coordinate system system (see Figure (see then point thevelocities point A has coordinates: coordinate system (see Figure 3), from then the point and asthe a result we have If we ∂Figure Ψ Γ we 2consider R Athehas follows calculating the velocities at 2bπ (α0 + β ) C = (22) c = l ′ ′ U (θ ) = − = −[ U ∞ (1 +∂Ψ2 ) sin θ + (15) Γ R] For symmetric ′ )∞ (1 coordinate A. If∂rwe ′U consider ′(θ ) = − ther ′ (=x ′−,[yU ′+ bc =(15) 0 2π+r ′ 2 ) sinairfoils, θ ] nate system coordinate Figure system 3), then (see point3), A3), has the point the coordinates: ′′ = ′ =+ , Figure (14) (14) rr′′== RR r,′,are: = the RFigure − ((α − β(α)) + β A ) has (14) = the −θ α +coordinates: β symmetric airfoils, ∂then πr ′on rθθ′then 2For r ′ has The(see velocity components system (see the point A Pressure Distribution the Airfoil 2 π (α + β ) C = l which gives the coordinates: For symmetric airfoils, bc = 0 The The velocity components components are: 2 which gives Thevelocity velocity components are: are: ∂ Ψ 1 R ′ ′ ′ ′ (14) (14) rU ′=(rR) ,= R −, θ) cos = −θ(α = −(pressure α + β )(16) ′ + β ) 2θ(14) The distribution follows from Bernoulli’s equation being = rU ∞=(1 2 Pressure on the Airfoil β = 0 R Distribution which gives r ′∂θ ′U ′(r ) = 1∂r ′Ψ = For U ∞ 2symmetric (1 − 2 ) cos θ′ bc = 0(16) Γairfoils, R The′ velocity∂Ψcomponents are: which gives ′ ′ ′ β =0 ∂ r θ r Anderson, Jr. (1989) as ′ U (θ ) = − = −[ U ∞ (1 + 2 ) sin θ + ] (15) 22 2r ′ ′ ′ ∂ π r r 2 ∂ Ψ ∂ Ψ Γ Γ R R elocity components The velocity are:components R +) sin θ)′sin pressure follows from Bernoulli’s equation being de ′′((θθU ′ +The ==′(θ −−)∂=Ψ −=are: U −(15) =∞−(1 +become θΓthe ))(14), [[U [ U+ (15)distribution (15) θ′+ UEq. +∞ (1 U (1 ]] ′ ] coefficient (15) r the conditions indicated in (16) for alift symmetric reduces to:airfoil ′′ =∂− ′′22 ) sin ∂∂rrEq. r ′ ∞ and rrEq. πrr′′ 2πlift r(15) 2π r ′ 2 and and the coefficient a symmetric bc =βairfoil 0=for 2 which gives Pressure Distribution on the Airfoil 0 and the liftEq. coefficient for ato: symmetric airfoil reduces 1 to: 2 1 reduces For the conditions indicated in Eq. (14), Eq. (16) become 2 2 (15) and Jr. (1989) as 1∂Ψ RAnderson, ρ ρ (U ∞ ) P + U = P + ( ) a a ∞ 2 U ′(1r∂)Ψ= 1∂Ψ = UR∞22(1 − ) cosθ ′ (16) on the Airfoil 2PressureΓDistribution R 2 2 (16) ∂ Ψ 1 R rθθ)′ cos ′ −)) cos (1 Cβl ==from (23) (16) (16) ∞ (1 =′∂U ) ==′(′r′)==0′ r== Uθ∞∞′=− (1U cos (16) α′′ +and βθThe + pressure 2−−U )gives ]coefficient distribution follows equation being defin 02παBernoulli’s U r′ = 0 , UU′′((rr)UU (17) rr′′∞22 sin( r ′ 2which θrr ′∂∂θθ ′r ′∂θ−′[ the lift for a symmetric airfoil reduces to: C = 2 πα π 2 R Γ1 l 1 the Airfoil For the conditions ′indicated in equation 2 2 0 = −[ −2 U sin( 2 α + β )P Distribution on =Anderson, +aPressure U∂θΨ ] (U ′Ψ U = 0 , (17) ρ ρBernoulli’s + = P + (U ∞ ) 2 ) ∞ r ∂ Γ Γ R R Jr. (1989) as The pressure distribution follows from equation being d a ∞ For the conditions indicated in Eq.Eq. (14), Eq. (15) and Eq. (16) become (14), equation (15) and equation (16) " " π 2 R For the For conditions the conditions indicated indicated in Eq. in (14), Eq. (14), (15) Eq. and (15) Eq. and (16) Eq. become (16) become ′ +(1 + subscript ′ +designates 2 2 θ θ U ′(θ indicated −′([θU − ) = − in Eq.=U(14), )∞=(15) (1−+and Eq.2 =)(16) sin [ U ] ) sin ] For the conditions Eq. become where an airfoil. The pressure coefficient defined b a (15) (15) β = 0 ∞ 2 and the lift coefficient forThe aπsymmetric airfoil reduces to: from become pressure Cdistribution follows ′ ′ ′ ′ ∂ ∂ π r r r r 2 2 r = 2 πα l . (17) yields Anderson, Jr. (1989) as Γ (1998) Γ Γ is Bernoulli’s equation being defined by Γ Smith 1 1 βα)) + − −sin( +∞ sin( +αβ+) +β]]) + ] ] (17) [[−−0=22= U UUr′′ ==U U 00r′r′,,= =0 (17) (17) (17) 2 [2−U2α Usin( ∞ ∞ 0 ,, U β Uθ′θ′ ==U0U 0θ′==θ′=−−=0 U[− ∞ sin(α + r Jr. (1989) as ρ ρ (U ∞to: + U = P + )2 ( ) π 2π R 2π R2πR" " PaAnderson, (17) Eq. (17) yields 2 R a ∞ and the lift coefficient for a2symmetric airfoil reduces where subscript an airfoil. The pressure coefficient defined by B a designates Cl = 2πα2 2 2 (18) RU Γ = 2(2π1 + α β sin( )) ∞ ∂Ψ ∂Ψ 1R R 1 1 (17) ′(r ) =equation=U Uyields = Smith − 2 ) cos U′(∞ryields )(1=− 2 ) cos Uθ∞′ (1 (1998) θ ′ Pa +(16) Eq. Eq.yields (17) ρ + ρ (U ∞U)a2 2 (U a ) 2 P=a −P∞P (16) Eq.(17) (17) yields ∞ is Eq. (17) yields ′ α + β )) r ′ C = 2πα(18)2= 1 − ( )(24) r ′∂θ ′ Γ = 2(2πRU r r′∂′ θsin( 2 Cpl = 1 ∞ " " (18) 2 U∞ where subscript a designates an airfoil. The pressure coefficient defined by Berti r′ = R ,. (18) the (18)lift RUα ΓΓ==22(Γ(2theorem + βα)) + β ))in Bertin and Smith (1998), 2(2∞πsin( w, using the Kutta-Joukowski described ∞ sin( (18) RU 2π=πRU ∞ sin(α + β )). ρU ∞. an The where(18) subscriptP" a−"2designates P∞ Uairfoil. a a 2 Smith (1998) is " " pressure e conditions For indicated the conditions in Eq. (14), indicated Eq. (15) in Eq. and (14), Eq. Eq. (16) (15) become and Eq. (16) become C = = 1 − ( ) coefficient defined by Bertin and n be determined as: where subscript Theliftpressure coefficient defined by B a designates p an(1998), described in Bertin andin Smith (1998), theBertin Now, Now, using using the the theorem theorem described described in Bertin and Smith and Smith (1998), (1998), the lift the Now, using theKutta-Joukowski Kutta-Joukowski theorem described in andliftSmith Now, using the Kutta-Joukowski Kutta-Joukowski theorem described in Bertin Bertin and Smith (1998), the lift 1airfoil. the 2 U ∞ Smith (1998) is ρU ∞ lift can be determined as: Now, using the Kutta-Joukowski theorem described in Thus, Bertin and (1998), the lift the velocity can can determined be determined as: we Smith need to determine on the airfoil surface. This will be canbe be determined as: as: 2 Smith (1998) is Paof− P∞ Ua 2 can be as, determined as: 2this, β is a small angle and the chord typical airfoils are small. Because of a c , bc Γ Γ Las:= ρU ∞ Γ = 4πρ (U ∞ ) Rsin(α + β ) (19) (19) C = = 1 − ( ) can be determined p ′ ′ 2 2 α β α β = = − − = = + − − + + + U 0 [ 2 U U sin( 0 [ ) 2 U sin( ] ) ] U′ = 0, U r′LθL=== ρ0ρLUU,=∞ ΓρΓU==∞44Γπρ (17) (17) 1of ∞θ)(2U 2 =of ((U 4πρ Rsin( ) is Rsin( α+ + βsmall α)) relating + β )∞anglethe U (19) (19) ∞) ∞β ∞ U πρ Rsin( α β by velocity at the airfoil surface to the velocity U c at the c (19) pical airfoils ras, a c , bc are small. Because this, a and the chord ∞ ∞ 2πR 2πR ρU ∞ (25) , b are small. typical airfoil can be approximatedFor as [see Eq. airfoils (13)]: as, aThus, c c we need 2 to determine the2velocity Pa − Pon U a surface. the airfoil This will be acco 2 2 πρ ∞(19) L = U Γ = U + ρ 4 ( ) Rsin( α β ) = ρUβ∞ Γis=a4πρ ) Rsin(and α ∞+ the β) = 1− ( ) 9C p = ∞ (U ∞ angle Because ofLthis, small 9 9 (19) 9 the relationship follows Thus, from we 1need 2 to determine rfoil can be approximated as [see Eq. (13)]: U∞ chord of the airfoil can be approximated as velocity at the airfoil ρ U by relating the surface to the velocity ∞ velocity on the airfoil surface. This will be U c at the cylin 7) yields Eq. (17) yields[see equation 13]: the velocity the airfoil surface. This will be accomp 9(20)2 c ≈ 4 RThus, we need to determine accomplished by on relating 9the velocity at Γ = ∫to U the = ∫ U c dS U c at the the airfoil surface (20) c ≈ 4R a dS avelocity relationship follows from by relating the(20) velocitycylinder. at the airfoil surface to the velocity U c at the cylinder The relationship(18) follows from Γ = 2(2πRU ∞ sin( Γα=+2(β2)) + β ))we need to determine(18) πRU ∞ sin(αThus, the velocity on the airfoil surface. This will be acco C l defined Asl adefined result, the lift coefficient a result, the lift coefficient C by Anderson, Jr. (1989) can be expressed as: follows from (1989)relationship can beor expressed Γ = ∫ U a dS a = ∫ U c dS c esult, the lift coefficientby defined byJr.Anderson, Jr.by(1989) canthe be expressed C l Anderson, relating velocity atas:the airfoil surface to the velocity U c at the cyli as: using the Now, Kutta-Joukowski using the Kutta-Joukowski theorem described theorem in Bertin described and Smith in Bertin (1998), Smith lift (1998), the lift or andthe Γ = 2using α + β )) (2πRU ∞the sin(Kutta-Joukowski Now, theorem. L Γ = ∫ U a dS a = ∫ UdS c = 2π sin( α + β ) follows from Cl = relationship (21)U a = U c ( c dS c or ) determined canas:be determined as: 1 L 2 ρU c= 2π sin(α + β ) dS a Cl = (21) 1 2 2∞ (21) ρU ∞ c or Γ = U dS = ∫ U c dS c 2 L = ρU ∞ Γ = 4πρ L =(Uρ∞U) 2∞Rsin( Γ = 4α πρ+(Uβ ∞) ) 2 Rsin(α + β (19) ) U a =∫U ca( dSac )(19). (26). dS a the airfoil and the cylinder respectiv where αdSshould are arc lengths around a and dS ally, the flow separates48at high angles of attack. Therefore, remain | Campus | cV. XX a| small No. 20dS| cjulio-diciembre | 2015 | ) U = Uc ( or ly, the flow separates at high angles of attack. Therefore, α should remain a asmall 9 dS a 9 le if the above theory is to remain valid. This allows as setting:.
(5) dS a = [(1 − 2 cos 2θ ) + ( 2 sin 2θ ) ]dS c dS c r r (26) ) 2 2 dS a aPerfiles aaerodinámicos Joukovsky notas suplementarias de aerodinámica or. dxdS 1 = c dx (1 +. Ua = Uc (. ). r2. ) − 2x. r4. (3. Ua = Uc (. ( xdx + ydy). (26). dS a a a or dS a = [1 − 2 2 cos 2θ + ( 2 ) 2 ] 2 2 arc lengths 2 dS cand r r wherewhere around thethe airfoil dS a and dS caare or the cylinder respectively. Now dSc are arca lengths around dSa and dx = dx + − x xdx + ydy ( 1 ) 2 ( ) 2 1 2 airfoil and the a 2rcylinder 2 2 respectively. r 4− 2 xy a dyNow dS a a2 a 2 2 12 = [ 1 + ( − )] dx y x re arc lengths around the airfoil2 and the cylinder respectively. Now 4 = [21 − 2 2 2cos 2θ + ( 2 ) ] r 2 r2 Eq. (4) as dS r (27) (31) (dxUca)22follows ( ) + (dyr ) dS = dx dS a = + (dy ) from c ; c. dx1 = dx(1 +a 22 ) −22 x 4 ( xdx + ydy 2 ) a 2 r2 2 a 2 a2 r 2 2 = 1dx=[1dx + (12+=( y 2(dx )−−x 2)]x − 2 xy ( xdx +dyydy ) dS =nce (dx ) + (dy ) ; dx ra r 1 ) +r(4dy1 )r 4 (27). a. 2. 1. 2. 1. (31). (3. 1. Uc follows from Eq. (4) as. (27). Γ. θ′+ Uas(11). ] (32) The relationship between ( x1 , y1 )Uc y follows givenEq. by (4) Eq. Equation (11) gives ( x, y ) is from c =U θ′ = −[ 2 U ∞ sin 2πR (32) ′) given 2 betweenwith 2 ( x, yby Eq. (8), i.e. θ ( x , y y ) is The relationship a2 2 2 2 1 1a 11 =cos 1, + a2 ( y211. − x ay)]= −r sin 2 xy dy 2 dxθ[equation given by x = r 12 ydy (1 + by ) − (11). 2 xequation (2xdxθ11+r 4gives e ween given Eq. (11)a)gives ( x1 , y1 ) y ( xdx , y1) =isdx with θ ′ given by Eq. (8), i.e. 2 4Equation r 2 a Γ = [ 1 + ( − )] − 2 dx y x xy dy 2 2 r r ′ ′ U c = U θ = −[2 U ∞ sinθ + ] (3 a2 a r4 ) 11 θ ′ = sinθ −1[ r sin θ − bc ] − a 2πR dx1 = dx(1 + r 2 ) − 2 x 4 ( xdx + ydy (33) R (33) 2 θ x = r cos θ , a 22 r yr= r asin 2 ce a a dx1==dx dx[[11+− 2 cos dy dy 22θ ] − 2sin 2θ (28) 2 nce Equation 26, 31 and 32 give Ua and Cp rr 2 ( 2y − x )] − 2rxy 4 r a2 follows equation 25. Note the Note that the range of a 2 Equation 2 (26), (31) and (32) givefrom fromthat Eq. (25). Ua and Cp follows 2 2 since = dx ( − )] − 2 y x xy dy x = r cos θ [,1 + y = r sin θ range of θ ′ is from 0 to 2π. a 2 a r4 dx1x == dx dyθ [1 −θ ,2rcos 2θ ] − sin y2θ= r2sin (28) r cos milarly r r 0 to 2π. θ ′ is from Conclusion 2 2 a a ce dx1 = dx[1 −a 22 cos2 2θ ] − sin 2θa 2 2 dy 2 We can (28)determine the velocity on the ilarly a 2θ ] + sin x dy = r =cos y= θ a(28) r dx dyθ[1,− r cos 2θr sin airfoil(29) surface and the pressure distribution. milarly herefore milarly. refore rly. milarly erefore herefore. fore. erefore. dx1 = dx[12− 2 cos 2θ ] − sin 2θ 2 dy r r r2 r Similarly. (28). 1. over the airfoil just by knowing the velocity x = r cos θ , y = r sin θ CONCLUSION at the cylinder. Consequently, the lift can be a22 a 22 determined too dy1 = dy[1 −a cos 2θ ] + sin 2θ a dx (29) and so, the lift coefficient dx1 = dx[1 − r22 cos 2θ ] − sin 2θ r 22 dy(29) C l . However,(28) this can be done if we map We r a2 r can 2 determine the velocity on the airfoil surface and the pressure distribution a cylinder into a Joukowski airfoil correctly. a dx1 = dx[1 a−2 2 cos 2θ ] − sina22θ 2 dy Therefore (28) here, is an alternative The method presented 22 over the airfoil velocity at the cylinder. Consequently, the lift can be dy12 = dy[1 −a 2 2 rcos dx sin θ r 2 2θ ] + 2just by knowing the (29) a 2 of such a procedure that reduces other a 2θ ) 2 + ( sin r2θ ) 2 ]adS 2 − [1r2−cos dSdya = =[(1dy (30) cos 2θ ]2+ sin 2θ 2 c dx (29) calculations such 1 2 complicated mathematical r r r r determined too and so, the lift coefficient C l . However, this can be done if we map a (30) as the usage of complex numbers.. a22 a2 2 2 2 + ( 2 sin dS a = [(1 − a 2 cos 2θ )cylinder ]dS c into airfoil correctly. (30) The method presented here, is an alternative of a 22 θa)Joukowski dy1 = dy[1 − r2 cos 2θ ] + sinr 2θ 2 dx References (29) r a2 r a2 such a2 procedure that reduces other complicated mathematical calculations such as the 2 Anderson, (1989). to 2 Bertin, J.J., &(29) Smith, M.L. (1998). dy2 1 = dy[Jr, 1a−J.D. cos 22θ ] +Introduction asin 2θ 2 dx 2 2 rcos 2 ( = [(1 −a (Third + θ θ dS a Flight dS 2 ) sin 2 ) ] r 2 2 (30) Edition) New York: Aerodynamics for Engineers (Third 1 c dS a 2 a 22 2of a 2 a 2 complex 2 usage numbers. 2 r r = − θ + [ 1 2 cos 2 ( ) ] (31) Jersey: Prentice Hall. = [(1 −Hill. cos 2θ2 ) + ( 2 sin 2θ ) ]dS c Edition). New dS McGraw (30) dS c a r 2 r 2 r r 2. dS a a22 a 2 22 12 = [1 − 2 a cos 2θ + ( )a ] (31) 2 dSdSas2 ac = [(1 − r 22 cos 2θ ) 2 +r 2( 2 sin 2θ ) 2 ]dS c (30) c follows from Eq. (4) r a2 r a2 REFERENCES 2 dSa 2 a = [(1a−2 2 cos 2θa) 22 +2 (12 2 sin 2θ ) 2 ]dS c (30) dS = [1 − 2 2 rcos2 2θ + ( 2 ) ] 2Γr 1 (31) a U sinrθ ′ +a 2] 2 follows from Eq. (4)dS asdS c aU= (32) θ + ( 2 ) Jr,] J.D. (1989). Introduction [1U−rθ′2= −2[2cos Anderson, to(31) Flight (Third Edition). McGraw Hill. c = ∞2 dS c r r2πR 12. Γ for Engineers (Third Edition). New Jersey: c follows from Eq. (4) ] M.L. (1998). Aerodynamics +12 Smith, dS aas U c = aU2θ′ = −[2UBertin, a 2θ ′2J.J., (32) ∞ sin 2θ + ( 2 ) ] 2πR (31) as[1 − 2 r 2 cos c follows from Eq.dS(4) = r 2 2 Hall Prentice c dS a a 2 1. 12(31) = [1 − 2 2 cos 2θ + ( 2 ) Γ] 2 dS c U c = U θ′r= −[2U ∞ sinθr′ + ] (32) 2θπ′R+ Γ ] ′ U U [ 2 U sin = = − (32) c XXθ | No. 20 |∞julio-diciembre | 2015 | lows from Eq. (4) |as Campus | V. 2πR 12 a. follows from Eq. (4) as. 12. 49.
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