رژیم های مختلف سرعت

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فصل ۱- برخی مطالب پایه 

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رژیم های مختلف سرعت سه شدي 1 مه مسد م] ا ES aS ee FIGURE 131 Block diagram categorizing the types of aerodynamic ows. 

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عنم و فيضم بر 1 (surge) 

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AN{,=—p, ds, cos @~7, ds, sin 6 dA, pu ds, sin +7, ds, cos 0 7 ‎ds, + | (p,cos = 7; sin 8) ds,‏ )8 توي + ق وم يم) ‎Le‏ ‎ ‎TE ‎(p, sin 64+ 7, cos 0) | (prsin 0+ 7; cos 0) ds; ue ‎ 

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Tas 2 1 | ‏ای‎ ‎ena ce a: Ret, Clr 0.20.0.00) gms | 

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۳ ۳ ares 2 9 Lift coefficient: Drag coefficient: Normal force coefficient: Axial force coefficient: Moment coefficient: 

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Ar sttking the NoseCone is smoothly diverted over the top and around the sides of the traler Auch loss turbulence is created, reducing ar drag also Fedues the effect of crosswinds. 2 2 ‘Without trailer mounted NoseCone Ar striking flat front of box is forced up over to and ut toe dee. coating trbuint a wc {sipulled along by the box effectively increasing {hd size of tho unt Being pulled 

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‎Thee, A= ronal ea‏ 4 - هه ‎vm ‎a 0310 ‎m 0207 ‎ ‎ ‎& ot & SB ox ™ ‎ ‎B vse ‎“ont, A = ‏اه‎ ‎ ‎ ‎ ‎ ‏اله - ف ‎‘Stramlinedtody,‏ ‏عدو كك عمد ‎Ferm aveaae) ‏ع اج مده تقد بره نوكيو ‎Senivaie, A Woral area ‎ ‎ ‎ ‎ 

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Effect of Surface Roughness (s ماو هس 

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Northrop T-38 jet trainer. Fraps tll down (48°) Flaps 20° Flaps 0") 1 ‏بيد‎ ‎a a a Flight Mash number, Ma 

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Aerodynamic tools Review of vector relations for convenience in expressing the aerodynamic tools} Basie flow equations Useful concepts for the (containing the fundamental implementation of the physics of flows} basic low equations Commu equation] ‘Substantial derivative Momentum equation ‘Streamline Eneray equation [ Vortwity ما Circulation ۳ Velocity potential ‘The solution of practi aerodynamic ‏ام‎ 

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Gradient of a Scalar Field vergence of a Vector Field % ie Direction ofthe مالا جر ‎Cartesian: Vein yz) lt‏ ويه ‎J pat the point ts.)‏ / سانا الع امل مار سر | مج دور رح دم ور ‎mM‏ 2 ‎ae ae‏ ی تا ‎‘Spherical: V= V(r, 0,0) = Vie,+ Veer + Veo.‏ بر عام ۲ تست ‎1a 13 1‏ ع ۲۱-۱۱ 52-5 ‎vp‏ ‎Spherical Pin,‏ 

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Curl of a Vector Field V=V(x, ¥, 2) = V(r, 6, 2) =V(r, 6, ®) Cartesian: Va Vit Vit Vek Spherical ree (rin Bes, aa a ar 38 2 0 0 

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Surface Integrals 1 p dS =surface integral of a scalar p over the open surface S (the result is a vector) IJ A dS =surface integral of a vector A over the 6 open surface S (the result is a scalar) i Ax dS = surface integral of a vector A over the ‎open surface S (the result is a vector)‏ ع ‎ ‎Line Integrals Consider a vector field ‎(7.4.0) ‎ ‏ماهد (ه ور داه ۸ ‎ ‎Volume Integrals ‎pd¥=volume integral of a scalar p over the volume (the result isa scalar) ‎ ‎ ‎ ‎ff a av = volume integral of a vector A over the volume 1 (the result is a vector) ‎ ‎ ‎ ‎ ‎ ‎ ‎ 

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2.2.11 Relations Between Line, Surface, and Volume Integrals Stokes’ theorem: $s ‏هه‎ as divergence theorem: fpaas=4fh waar gradient theorem: “هم طلا عه م فإ 7 5 

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AN APPLICATION OF THE MOMENTUM EQUATION: DRAG OF A TWO-DIMENSIONAL BODY ffeov-asu- ff ras, —* ', For a constant pressure, J (pd), =0 ركه ‎fhov‏ او ‎jac pau dy‏ - د زول سس ‏عدم | | اوه ام له ام ‎fl‏ 2 ‏]هه ‎ ‎ ‎ ‎ 

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Example 2.1. Consider an incompressible flow, laminar boundary layer growing along the surface of a flat plate, with chord length ¢, as sketched in Fig. 2.19. The definition of a boundary layer was discussed in Sec. 1.10 and illustrated in Fig. 1.28 ‘The significance of a laminar flow is discussed in Chap. 15; it is not relevant for this example. For an incompressible, laminar, flat plate boundary layer, the boun- dary-layer thickness, 8, at the trailing edge of the plate is 8 ce VRe, ‏و‎ ‎Re, === and the skin friction drag coefficient for the plate is Mea DY 18 1 quell) boundary of ۱ ‏سس‎ vole 2 000 profile =a) here.] Let us assume that the velocity profile through the boundary layer is given by a power-law variation ‏سا‎ دا 0 (3) Calculate the value of n, consistent with the information given above. 

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Solution. From Ea, (2.75), و | یه من ول لاما “عمو where the integral is evaluated at the trailing edge of the plate. Hence, ۳ ] Inserting the laminar boundary-layer result for C as well as the assumed variation of velocity, both given above, we can write this integral as PG Gy TO Carrying out the integration, we obtain ۳ هه ‎(Ge) 3‏ م2 ‎mal‏ 1328 1 1 ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎no 02s ‎۶ ‎1 ‎or ‎ ‎Sse seta sna ‏و کر‎ lige ea Pro ora fa ple ‎or 0.2656n?~0.6016n + 0.1328 =0 oa ‎ ‎Using the quadratic formula, we have ‎n=2 of 028 0 

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Streamli St ۳3 Pane line for a 1 تا For steady flow, streamlines and pathlines are the same. dsxV=0 جك زو ده و ازجا ‎bik‏ ‏عه ‎Jac dy‏ وه =i(wedy ede) +f(ude— was) +h(vdx—udy) =0 wdy—vdz=0 udz—wdx=0 vdx—udy=0 (2) Equation of «stent in ‘ho -aieasional caren pace. (6) Sketch of steantabe ie differential equations for the streamline “ thee deal nee 

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Example >? Consider the velocity field given by w=y/(x?-+y?) and v= —x/(x'+y"). Calculate the equation of the streamline passing through the point (0,5), Solution dy/dx=v/u=—x/y, and ydy=-xdx Integrating, we obtain ‏عجر دقر‎ where c is a constant of integration. For the streamline through (0, 5), we have S=0+¢e or ‘Thus, the equation of the streamline is عسل Note that the streamline is a circle with its center at the origin and a radius of 5 units. 

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حرکت عمومی یک المان سیال ۱- حرکت انتقالی ‎Granslational Motion)‏ ۲ تقییر > ‎Enlargement: Dilatation)‏ ۳ چرخش جسم صلب ‎Rigid-Body Rotation)‏ ؟- تغبير ‎Deformation J‏ مواق ‎anna‏ | دم ‎[EEE] fee ke‏ ‎a(R +92)‏ )8+ 5( 4 وتو وب »كز ۷+ . + ‎(R48)‏ + ام ‎ ‎ ‎ ‎ ‎ ‎ ‎ 

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a vorticity exit o,j+ ok we i XS ANGULAR VELOCITY, VORTICITY, AND STRAIN 

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e/a Ln Example 2.1, the low was incompressible. From steady How sven by Ea (245), repeated below, we hae foram isompresibe ow, where» conan, en) Equation (2) wil provide an eaprssion for» a flows: [ “سام mere, rm Example 21, we wate hat «es This caution holes at any x sation lon he pte, ot ist at x= Testor pe 2.3. For the velocity feld given in Example 22, caleulate the vorticity. Seaton ‏ره ره‎ | 2 م ا ها دص ‎lr yo‏ مير رمات یدعب شاد 01+ 9404-0 The flow fed is ierotational at every pot except at he origin, where x°+ y= Example 24, Consider the hounday-ayer veloc profie ved ia Example 2, samt) Va=(y/ 87° lth ow rotational tata! ‎tow he otationaty condition i gven by Ee‏ همه و هم ‎mame,‏ )1( ‎Does his reton bod for the vicous ound ye ow in Example 21 Let came ‏ع‎ From the boundary ye sea pole sv by ‎ae ‎sere” ‎we aban ‎en 

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1 2 Integrating with respect co y; we have Cyne, es und C, can be a funetion of x Evaluating Eg. (E.4) at the 1 we obtain C30. Hence, ره where C) is a const wall, where »=0 and In tum, we obtain by differentiation و م (Note: » is finite inside a boundary layer; the steamlines within a boundary ste deflected upward. However this “displacement” effet is usually small compared to the running eneth in the x direction, and vis of small magnitude in comparison to w Both of these statements will be verified in Chaps. 16 and 17.) Recasting Eq. (1) in the same general form as Eq. (ES), we have a" Hence, y From Eqs. (E.S) and (E.6), we ean write conc ۳۹ هه ‘Therefore, the irotationality condition does no! hold; the Now is rotational 

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Example 25, For the velocity fled given in Example 22, calulate the circulation ‘ound a cll path of agus Sim Assome hat and © given in Example 22 fe in units of meters er second Solution, Since we ate dealing wih a iculae path, itis easier to work this problem im polar coordinates, where x=rcos ®.y =rsin ‏رک و‎ Vim w.cos 8+ Bsn and Vy= usin #4 e088 Therelore, ‎sine‏ فم ان ‎ ‎0 tebe) ‎ ‏مس مه (- )مجه مل اا د ۲ ‎[Note that we never used the 5-m diameter of the circular path; in this case, the value of I'is independent of the diameter of the path ‎ ‎ ‎CIRCULATION ‎From Stokes’ theorem ‎raf vee [five ‎ ‎ ‎d= (0 xV)-dS=—(VxV)-nds ‎(¥xV) a ‎ ‎ 

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STREAM FUNCTION |, consider two-dimensional steady flow Mass flow = Ad = pV An = pu Ay + po(—Ax) Letting cd approach ab, Eq. (2.135) becomes in the limit 

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مه ۳ ‎war‏ ما Point (x + dx, y + dy) Streamline 2 5 Streamline | 

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PROBLEMS 24. Consider a body of arbitrary shape. If the pressure distribution over the surface of the body is constant, prove that the resultant pressure force on the body is 2er0 [Recall that this fact was used in Eq. (2.68),] 22. Consider an airfoil in « wind tunnel (ie. a wing that spans the entire test section) Prove that te lift per unit span can be obtained from the pressure distributions on the top and bottom walls of the wind tunnel (ie, from the pressure distributions fom the walls above and below the airfoil) 23, Consider a velocity field where the x and y components of velocity are given by us ex/(x"+y") and v= oy/(x°+y"), where ¢ is @ constant, Obtain the equations of the streamlines 24, Consider a velocity field where the x and y componente of velocity are given by = oy/(x°+y") and v= ~cx/(x"+)"), where ¢ i & constant. Obtain the equations (of the streamlines 25. Consider a velocity field where the radial and tangential components of velocity are V,=0 and V,=cr, respectively, where ¢ is a constant. Obtain the equations of the streamlines 2.6. Consider a velocity field where the x and y components of velocity are given by ue ex and ‏جه‎ cy, where ¢ is a constant. Obtain the equations of the streamlines, 2.7, The velocity field given in Prob, 2.3 is called source flow. For source flow, calculate: (a) The time rate of change of the volume of a fluid element per unit volume (b) The vorticity Hint: Ibis simpler to convert the velocity components to polar coordinates and deal ‘with a polar coordinate system. 28. The velocity field given in Prob, 2.4 called vortex flow. For vortex low, calculate (a) The time rate of change of the volume of a fluid element per unit volume (b) The vortciy Hint: Agu, for convenience use polar coordinates, 229, te the flow field given in Prob, 25 ireotational? Prove your answer. 2.10, Consider a flow field in polar coordinates, where the stream function is given as = (7,0), Starting with the concept of mass flow between two streamlines, derive Eqs. (2.1394 and b), 2.11, Assuming the velocity fied given in Prob. 2.6 pertains to an incompressible low, calculate the stceam function and velocity potential. Using your results, show that lines of constant @ ate perpendicular to lines of constant ¥. 

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فصل ۲- جربان پتانسیل 

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General noalifting bodies; source ppanel numerical technique ‘Basic aspects of inviscid, incompressible flow Flow over Semintinite Nonliftin and ova flow over" shaped acylinder bodies 1 Lifting flow over aevlinder Katte Jouowske theorem سس Some elementary Flows Uniform flow Sources and sinks Laplace’s equation General philosophy and tse in solving problems لم Synthesis of superposition of elementary ows Ventuti Lowspeed ‘wind tunnel Pitot tube, airspeed Bernoulli's equation f |, Some simple applications: 

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VELOCITY POTENTIAL for an irrotational flow, £=VxV=0 Consider the following vector identity: if @ isa scalar function, then _. ۱ ۲۳۳۸0 cylindrical coordinates, y, ‏گ- ۷ فك‎ ۷-۲ 2 ‏لا‎ and in spherical coordinates, $= (x y, 2) b= 4,42), b= ۵۲ ۵, ). 3 2% کب بر بو ‎ay? az‏ ad ab, ab 

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RELATIONSHIP BETWEEN THE STREAM FUNCTION AND VELOCITY POTENTIAL For a streamline, ¥(x, y)= constant. ay = ax+% ay =o Ox ay ۵ =-vdx+udy=0 (2) at Ax] ycone ۷ Similarly, for an equipotential line, (x, y) = constant. Along this line, 32 dp= 6 ‏ین‎ 9۴ dy = 0 oy dé =udx+vdy=0 (2) ‏كدت‎ ‎x) scone 9 (2) ‏عمجم‎ ‎AX] y—con (dy / dx) 

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CONDITION ON VELOCITY FOR INCOMPRESSIBLE FLOW GOVERNING EQUATION FOR IRROTATIONAL, INCOMPRESSIBLE FLOW: LAPLACE’S EQUATION ۷۰۷ 0 ۷-۲ ۲۰۲۵0 Cartesian coordinates: & = (x, y, 2) Vb =0 Spherical coordinates: @ = 6(r, 0, P) ‎OP) + & (sin 9 2P) 4-2 (1 38) |.‏ 9 مروع) فالب ید ‎sn 958) +5 (sin 38) +5 (55 9) °‏ | ‎ ‎ ‎ ‎ ‎ 

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irrotationality condition 1. Any irrotational, incompressible flow has a velo potential and stréam function (for two-dimensional flow) that both satisfy Laplace's equation. 2. Conversely, any solution of Laplace's equation represents the velocity potential or stream function (two-dimensional) for an irrotational, incompressible flow. Note that Laplace’s equation is a second-order linear partial differential 2quation. رك ۰۰ برقع 

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‎-Boundary Conditior‏ شرایط مرزی ‎the shapeof the body Boundary condiiont at infinity and on a body; inviscid flow Y» = f(x), then ‎Wurtace = By =y, = Const ‏)2( هل ‎dx \u) suetace‏ ‎ ‎30 _ ‎u= ‎ ‎ ‎ ‎ ‎‘Wall Boundary Conditions ۷۰۵۰۱۲۵۱۰۱0 ‎ ‎ ‎ 

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روش عمومی حل مسائل Reflecting on our previous discussions, the general approach to the solution of irrotational, incompressible flows can be summarized as follows: 1. Solve Laplace's equation for $ [Eq. (3.40)] or ¥ [Eq. (3.46)] along with the proper boundary conditions [such as Eqs. (3.47) and (3.48)]. These solutions are usually in the form of a sum of elementary solutions (to be discussed in the following sections). 2. Obtain the flow velocity from V=V¢ or u=ay/ay and 3. Obtain the pressure from Bernoulli’s equation, p+!pV’ p» and Vj. are known freestream conditions. ay/ax. =potipVx, where 

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UNIFORM FLOW: OUR Fist] aw $= Var cos 6 Y= Vor sin 6 = Veds 1 0 Ve ds= = Vil -O(h) + Val +0(K) =0 r=0 Equation is true for any arbitrary closed curve in the uniform flow. ۱ r=-4 ‏لال ی‎ 4 49-۷۰۰0 | c ۰ ELEMENTARY FLOW $= Vax +const 

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SOURCE FLOW: OUR SECO! ELEMENTARY FLOW flow across the surface of the cylinder is sree Aw Sink flow V-V=0 everywhere except at the origin, where it is infini Thus, the origin is a singular point, 7 ‎2anV,‏ نو م ‎€ ‎۶ | ‎ ‎ ‎v,=0 ‎=v inr+s(0) 7 ‎ ‎۸ ‎=——] ‏و9‎ on nr ‎ ‎ ‎ie = V,=0 => =const+ f(r) ‎ 

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> v= Lor aw. V,=0 ™® w=const+f(4) The equation of the streamlines A r=-[[xv-as=0 = >— @=const i, 5 | satisfy Laplace’s simply by substitution into V¢ =0 and Viy =0 

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COMBINATION OF A UNIFORM FLOW WITH A SOURCE AND SINK Uniform stream Source veterind v2 8 lay A ee ee A A r 27۳ ‏كاوه - #2 لقي روي م صو‎ ‏ا‎ ‏او ۷۷ قم‎ 2 sin @ Voc sin 0 = 0 

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EXAMPLE 81. ‘An offshore powerplant cooling-water intake sucks in 1500 fein water 30 ft deep, asin Fig. ۳0۱۰ 1 the tidal velocity approaching the intake is 0.7 fl, (a) how far downstream does the in- take effect extend and (b) how much width ۲ of tidal low is entrained into the intake? Solution Recall from Eq, (4.131) that the sink strength mis related to the volume flow Q and the depth Jy into the paper ‘Therefore from Fig. 8.5 the desired lengths @ and L are 26 ‏»مر‎ Ans. (a) Ans. (b) 

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Superposition of a uniform flow and a source-sink pair; flow over a Rankine oval. A W= Voor sin 6+ 5 (0\~ 02) = const Thus, the stagnation streamline is given by / =0, ic. Vor sin 0+*-(0,~63)=0 27 oA w= Var sin O45 6 27 2) A w= Vor sin ote (8 OA=OB= 

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۱ ‏مه‎ ‎DOUBLET FLOW: OUR THIRD 0 ar r—1 cos 1 ELEMENTARY FLOW Source A 1 G@) Source-sink pair : 40+ 400, | r—lcos@ — ‘we have a singularity of strength (20-90); 

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NONLIFTING FLOW OVER A CIRCULAR CYLINDER Lay t Vt eh vireo Fay Mer 608 8) بنج ٩0 ‏کر‎ vosine 2 gaye sino ne ۱ (:-4) Vic cos 9-0 x sin@ ۷-۷ ‏ص و‎ 11 2 ‏مگ‎ ‏متوعيلا دن‎ of (r, @)=(R,0) and (R, 7). Let R?=k/27V,. ‏بي‎ ‎<1 =0. y= (Vorsin (1 = ۷-0 22۷ ۶ و 

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Rear hal of evince سم ۷۶0 Vy =-2V.. sin 0 5 ‏2)--ه‎ C,=1-4sin’ 6 atl (Cpa ~ Cpu) ax edo pte & + (Cpu Epa) dy 

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Example Consider the nonlifting flow over a circular cylinder. Calculate the locations on the surface of the cylinder where the surface pressure equals the freestream pressure, Solution. When p = po, then C, =0. From Eq, (3.101), 30°, 150°, 210°, 330° These points, as well as the stagnation points and points of minimum pressure, are illustrated in Fig. 3.25. Note that at the stagnation point, where C, = 1, the pressure is Pot qos the pressure decreases to ‏مر‎ in the first 30° of expartsion around the body, and the minimum pressure at the top and bottom of the cylinder, consistent with C,=—3, is po-34e0 Hence, 

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| ‏ی‎ 0 as VORTEX FLOW: OUR FOURTH, ELEMENTARY FLOW | ۱ ‏هرد ] ]که رم | | مد‎ | سیخ | a | Jfirxvias-iexv ds [te tt ‏((م)‎ ۱ in the limit as r->0, we have ۲ ‏ا‎ | 1 ‏را‎ | 2aC =|VxV| ds Ze ۱ 2 ‏بعرم | يده‎ 256 tC | gs j= tS | However,as 0, dS 0. Therefore, in the limit as r>0, pos | ‏موعدم‎ ‎| ] ‏ورگ‎ ۲ V-ds=~V,(2ar) | arr ۱ 

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oO w y= constant و P = Pam 

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Kelvine’s Theorem 

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0 ] of Basle, Plane Potential Flows Description of ‘Velocity Flow Field Yelocity Potential tream Funetion Components! Ueosa Uniform flow at = Ulecosa + ysina) — = Ulycosa ~ xsina) Using dangle a With the x axis: Source or sink tl ne m > 0 source m <0 sink, Free vortex r>0 ‘counterclockwise 3 ۳-9 lockisise motion Doublet ۳7 2۳ "Nelocky components re ‏ده مج له نع وه 1 الم‎ oa a ae 3810 6 

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LIFTING FLOW OVER A CYLINDER Nonlifting flow Vortex of Lifting flow over over a evlinder strength P ‏ی‎ ‎9 r | yay tds ۳ ‏انم‎ ‎Const = ~5— = (Vorsin a(t ‎if r= R, then y=0‏ معي يك )سيدا ‎ ‎ ‎2 ۲. =» o=aresin( ~ Fe) Vacos 0=0 ‎ ‎a) ‎4۹۷۰ ‎R® cr ‏مومس(‎ 

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Go P< etek سل ‎oa as oe‏ الع ۲ ‎(ce) > tna‏ , ۳ ی .7/2— ‎0=a/2 or‏ مس گس 27 ۴ 1 ‎ry.‏ + ‎oe‏ ی اب 4۳۷۰ 47۷ ‎af" : 5‏ 5 و ‎The velocity on the surface of the cylinder i‏ هرب 4 > ات ۵ ۷-2 .۰ ۵ وزو 16 >« r = Vp=—2V-n sin @-—— V=Vy=~-2V..sin 8 97 ‏أ‎ ce ccf 5 ‏و‎ ‎r‏ رده ‎OF RV “(ssa ]‏ مس ‎ ‎ ‎nw) ‏0ب اب‎ Cuz cos 0 ‎ ‎ ‎ 

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FORCES ACTING ON THE CYLINDER 

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2 5 50 [i cvee-tf Gace ‏معدي كاماد ود‎ A 2 jh oes Te ‏مومسم و‎ x=Reos@ dx=—Rsin 0d0 9 (oe Kutta-Joukowski theorem, Ler 5 9 - q=-= I G, sin 6 do German mathematician M. Witheim Kutta (1867-1944) 2 ‏ول‎ Nikole leukowski (1847-1921), a ip ‏ری | كدو‎ 00 | “لاوم ع لآ 

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تام) ال ‎Pretoer's‏ 

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® C, strongly depends on rate of rotation. ® The effect of rate of rotation on C, is small. ® Baseball, golf, soccer, tennis players utilize spin. ® Lift generated by vp ‎rotation is called The‏ و ی ‎Magnus Effect. ‎ ‏و 

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Example Consider the lifting flow over a circular cylinder. The lift coefficient is 5. Calculate the peak (negative) pressure coefficient. Solution. Examining Fig. 3.27, note that the maximum velocity for the nonlifting flow over a cylinder is 2V.. and that it occurs at the top and bottom points on the cylinder. When the vortex in Fig. 3.27 is added to the flow field, the direction of the vortex velocity is in the same direction as the flow on the top of the cylinder, but opposes the flow on the bottom of the cylinder. Hence, the maximum velocity for the lifting case occurs at the top of the cylinder and is equal to the sum of the nonlifting value, -2V,., and the vortex, -I'/27R. (Note: We are still following the usual sien convention: since the velocity on the top of the cylinder is in the opposite irection of increasing or he polar coordinate system, he vel ‘here are negative.) Hence, mes ‏یت‎ en ‘The lf coefficient and Fare related through Eq, (3.138) Hence, 0 لمع و ‎Substituting Eq. (E.2) into (F.1), we have‏ ,2.7960 حلا تت رلا حال ‎Me 5 (63)‏ Substituting Eq. (E.3) into Ea. (3.38), we obtain هس 

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Example For the flow field in Example 3.10, calculate the location of the stagnation points and the points on the cylinder where the pressure equals freestream static pressure. Solution. From Eq. (3.123), the stagnation points are given by From Example 3.10, Thus, 7 كمه 203.4 

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ای عا سر قد 315377 ‏سروح‎ OTE) ۳5 ‎and 29620‏ امد ‎Also,‏ ‎ ‎2588 and 1780" ‎“Thee ate four points onthe cise cylinder where p= po. These are sketched in Fig. 3.31 alongwith he stagnation point locations. Ac shown in Example 3.10, he Imnimum presure occurs the op ofthe elinde and equal 0 pe ~6824u. A Theat minimum presure ‏سید‎ athe bottom ofthe einer where #= 3/2 Tas ‎ ‏اتلد مد ‎ ‏تس ‏2 ‎۳ ‎Hence, a the Batom ofthe ead, p= #504540 ‎ ‎ ‎“To find the foeations whete p= pe, fist construct a formula for the pressure Sofie on the esnde race: ‎Ths, ‎ ‎From Example 510.1) RV ‎ ‏مس ‏۱ص دوز ‏مرس ‎ ‎| check on this equation ca be obtained by calling C; a #~90" and sing iTit ares wih he esl obtained in Example 310 For 8907, we have ‎,0367-3189-4=[=582] ‎“Tiss the same seta in Example 3.10; the eguation chek “Fond the valuts of? where pp. se C0 ‏انهم م تقد دودو ‎ ‎ 

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NONLIFTING FLOWS OVER ARBITRARY BODIES: THE NUMERICAL SOURCE PANEL METHOD Ads db =F nr 27 > ooun=[ Ads a ln ae Ina — Antara low’ Source seat on tat Flow over the bo of body, with Mis) of given shape 9 calculated to make the body surface a streamline 

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| 9 +1 2 -3 

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9 Res = 186.107 سيو / 2 Soper ‎Teo 7‏ ۳ پا ی اس انا ‎Pa‏ ‎So‏ + ما 25 7 ‎Ta‏ 2 ‎ ‏موه ‏دام 

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PROBLEMS Note: All the following problems assume an inviscid, incompressible flow. Also, standard sea level density and pressure are 1.23 kg/m? (0.002377 slug/) and 1.0110" N/m? (2116 1b/f¢), respectively ‎For an irrotational low, show that Bernoulli's equation holds between any points‏ لق ‎in the flow, not just along a streamline.‏ ‎32. Consider @ venturi with ‏له‎ throat-to-inlet area ratio of 0.8, mounted on the side of fan aieplane fuselage. The airplane is in flight at standard sea level. If the static pressure at the throat is 21001b/ft, calculate the velocity of the airplane, ‎44. Consider a venturi with a small hole drilled in the side of the throat. This hole is connected via a tube t0 a closed reservoir. The purpose of the venturi is to create {vacuum in the reservoir when the venturi is placed in an airstream. (The vacuum js defined as the pressure difference below the outside ambient pressure.) The venturi has a throat-to-inlet area ratio of 0.85. Calculate the maximum vacuum obtainable in the reservoir when the venturi is placed in an airstream of 90 m/s at standard sea level conditions. ‎34. Consider a low-speed open-circuit subsonic wind tunnel with an inlet-to-throat area fatio of 12. The tunnel is turned on, and the pressure difference between the inlet (he settling chamber) and the test section is read as a height difference of 10cm ‘on a Ustube mercury manometer. (The density of liquid mercury is 1.36% 10" kg/m.) Calculate the velocity of the air in the test section ‎138. Assume that a Pitot tube is inserted into the tet-section flow of the wind tunnel in Penh ٩4 The tunnel test section is completely sealed from the outside ambient pressure. Calculate the pressure measured by the Pitot tube, assumi Dron te tunel ‏موجه مان‎ ee MOSS HE ‎346. A Phot tube onan airplane ty airplane flying at standard sea level reads 1,07 x 10° N/m’, Wt is the velocity of the airplane? ‏ین‎ Wt 3. Ata given pont onthe ura ft ‏م‎ Surface ofthe wing ofthe airplane in Prob. 3.6, the flow Nelo 6 0m/s Calculate the peste cefient ahs point 34, Consider uniarm ow with velocity VS with ‏.ا ياوا‎ Show tha his foi physically posible incompressible flow and that it is irrotational, ee ws. ‎Show that a source flow is a physieal isa physically posible incompressible Row everywhere ‘xcept atthe ogi. Also show that it inotatonl vetuneres 

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310, Prove thatthe velocity potential and the stream function fora uniform flow, Eqs. (2.53) and (3.55), respectively, satisfy Laplace's equation ‎Prove that the velocity potential and the stream function for # source flow, Eqs‏ لد ‎and (3.72), respectively, satisfy Laplace's equation.‏ )267( ‎342, Consider the Now over a semi-infinite body as discussed in Sec. 3.11 If Vs is the ‘locity ofthe uniform stream, and the stagnation point is t upstream ofthe source (a) Draw the resulting semi-infinite body t0 scale on graph paper. ‎(b) Plot the pressure coefficient distribution over the body, ie, plot Cy versus distance along the centerline of the body ‎33. Derive Eq. (3.81). Hint: Make use of the symmetry of the flow field shown in Fig 3.1; i, start with the knowledge that the stagnation points must lie on the exis ligned with the diection of V.. ‎3.14. Derive the velocity potential fora doublet; ie, derive Eq. (3.88), Hint: The easiest ‘method isto start with Eq. (3.87) forthe stream function and extract the velocity ‎potential ‎Consider the nonlifting fow over a cirular cylinder. Derive an expression for the ‎pressure coeficient at an arbitrary point (r 6) in this flow, and show that it reduces ‎{0 Eq, (3.101) on the surface of the cylinder, ‎346, Consider the nonlifting flow over a circular cylinder of a given radius, where V.=201h/s. If Ve is doubled, ie, Va= 40/5, does the shape ofthe streamlines change? Explain, ‎317, Consider the lifting Now over a circular eylinder of a given radius and with a given sirculation. If Vis doubled, keeping the circulation the same, does the shape of the streamlines change’ Explain ‎‘348. The lit on a spinning circular cylinder in a frestream with a velocity of 30 m/s and at standard sea level conditions is 6 N/m of span. Calculate the circulation around the cylinder. ‎3619. A typical World War I biplane fighter (such as the French SPAD shown in Fig. 3.45) has « number of vertical interwing steuts and diagonal bracing wices. Assume for a given aieplane thatthe (otal length for the vertical struts (summed together) is 25, and that the struts are cylindrical with a diameter of 2in. Assume also that the total Tength of the bracing wites is 804, with a cylindrical diameter of in CCaleulate the drag (in pounds) contributed by these struts and bracing wites when the aieplane is ying at 120 mi/h at standard sea level. Compare this component of rag with the total zero-lift drag forthe airplane, for which the total wing area is 23010 and the zetolift drag coefficient is 0.036 ‎ ‎ ‎ ‎ ‎