AOE 5104 Final Exam. Thirteen answers, eight points apiece perfect score: 104

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Victoria Patterson
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1 AOE 5104 Final Exam Name: Solution is due on Thusday, Decembe 10 between 9:15 and 9:45 in Room 349 Whittemoe Hall. You may use you notes and the ppt and pdf slides on my website only. Thiteen answes, eight points apiece pefect scoe: 104 1) Explain how we came to the conclusion that the pefect function descibes a flowfield in which thee is a point whee the voticity is infinite and an incompessible and iotational flow (often efeed to as a distubance ) associated with it eveywhee else. Fom hw4, the pefect function is a votex : y x vx = C and v whee is a constant and y = C C = x + y It was shown that this flow is incompessible and iotational eveywhee, except possibly at the point whee = 0, and that the ciculation aound a cicle with its cente at the point whee = 0 is independent of the adius: Γ= πc 0=a constant. Thus, fom Stokes theoem, Γ= Ωin ds, thee must be voticity Γ somewhee, and the aveage voticity inside a cicle of adius is Ω avg =. It follows that Ω avg as π 0 and we conclude that voticity is infinity thee. What to look fo: knowledge of 1) what the pefect function is, including the esults of hw4 incompessibility and iotationality, ) what Stokes theoem is, and 3) the limit as 0 3 pts fo any one, 6 pts fo any two, 8 pts fo all thee ) Why is the pefect function called the pefect function (emembe: thee can be die consequences if you miss this)? The pefect function imitates the esults of bounday-laye theoy pefectly as Re. 8 pts o nothing. 3) Why is thee voticity at the tailing edge in a two-dimensional unsteady flow and not in a steady flow? We took the Kutta condition to be the equiement that the pessues in the meging steams coming off the uppe and lowe sufaces must be equal whethe the flow is steady o unsteady. Fom Benoulli s equations we obtained: closed loop ( φ φ ) ( u u )( u u ) ( u u ) φl φu ul u l u u l + u l u dγ l + u pu pl = + = + = + Δ u = 0 t t t dt d When the flow is steady, Γ dγ = 0 Δ u = 0 dγ When the flow is unsteady, 0 Δ u = dt dt u + u dt What to look fo: 4 pts if they state the Kutta condition coectly, 3 pts if they give Benoulli s equation fo the pessue diffeence coectly, and 1 pt if they finish the explanation. Explanation is in Special Topics: tansient.ppt l u

2 4) In modeling unsteady flow, we moved the voticity being shed at the tailing edge with the paticle velocity. Explain how we came to the conclusion that the voticity moves with the paticle in a flow whee the effects of viscosity ae ignoable and the pessue is continuous. d Fom Special topics: Bounday-Laye Theoy slide 17 Γ = a i d = pid = 0 fo the conditions dt stated. Hee Γ is the ciculation aound a closed loop of paticles, which convects downsteam in the flow. If the loop encicles a egion whee thee is voticity, such as at the tailing edge, then because the ciculation doesn t change as the paticles move downsteam in the flow, it follows fom Stokes theoem that the voticity must be moving with the paticles. 6 pts if they mention the tempoal consevation of ciculation, moe pts if they use Stokes theoem to elate this to the motion of voticity. 5) In thee-dimensional poblems, one often uses the cambe suface to model the flow aound a wing ignoing the thickness. The coodinates of the NACA441 aifoil (the NACA441 aifoil has 4% cambe at 40% of the chod fom the leading edge and 1% thickness) given in the Theoy of Wing Sections by Abbot and von Doenhoff poduce a cambe line that is not vey smooth because the given coodinates have been ounded off. Howeve, a least-squaes fit of the given coodinates of the cambe line smooths the line. The fouth-ode fit is given by: y = x x x x 3 4 Modify you discete-votex method, which was used to model the steady flow ove a flat plate, to model the steady flow ove the cambe line of the NACA441 aifoil. Use cosine spacing fo the x- coodinates and use 15 panels. The following gaph depicts the locations of the nodes, votices, and contol points when only five panels ae used. five-element cambe line with cosine spacing fo a NACA441 aifoil. nodes, o votices, + contol points α flow Othe esults ae shown below, but simila gaphs ae not equied as pat of you solution.

stagnation point on the lowe suface stagnation point on the lowe suface The esults manifest the singulaity at the leading edge, which we accept as pat of this appoximation and which was discussed

3 stagnation point on the lowe suface stagnation point on the lowe suface The esults manifest the singulaity at the leading edge, which we accept as pat of this appoximation and which was discussed when the flat plate was used to appoximate the NACA001 and the symmetic Zhukovskii aifoils. The following is also not equied, but only included to illustate the flowfield. You do not need to take time to make figues like these. Flow field aound the cambe line fo a NACA441 aifoil at 4 o angle of attack. 100 panels and cosine spacing wee used to make the pictue.

4 The following table on the left has the coodinates fom the fit ; the last slightly misses (1, 0). Modify the fit in you solution by eplacing the last y-coodinate with 0. as illustated in the table at the ight. x y x y The following table has the lift (L) and moment about the one-quate chod (M) fo five elements (panels) panels α = 0 o α = 4 o α = L = L = L = 1.4 M = M = M = L = M = L = M = L = 1.31 M = a) Make a table giving the coodinates of the cambe line that you used fo the 15-panel solution. b) Fill in the missing values fo L and M fo the 15- panel solution. Note: to find the fluid velocity on the uppe and lowe sufaces, fist find the velocity at the contol point (whee the no-penetation condition is satisfied), and then add 0.5Δu to find the velocity on the uppe suface and subtact as much to find the velocity on the lowe suface (hee the panels contibute tangential velocity to one anothe because of the cuvatue, unlike the flat plate). c) Plot you values on the gaph below to compae you esults with the expeimental data. Make a few checks with the pogam at the end of this solution, give at least 6 pts if thei pogam agees with finding the discete votices. Thee s an old adage: When some one makes a calculation, no one believes the esults except the one who made the calculation. When some one conducts an expeiment, evey one believes the esults except the one who conducted the expeiment.

5 Re vaies: 3x10 6, 6x10 6, 9x10 6 Δ aifoil with oughened suface aifoil with flap deployed data fo compaison moment about the quate chod aifoil with flap deployed

6 6) Conside the flow aound a sphee of adius a that moves along the z-axis with time-vaying speed, V = V t k, though an incompessible, inviscid fluid that would othewise be at est. We want to A A () solve fo the flowfield in a efeence fame that moves with the sphee, a non-inetial efeence fame, using spheical coodinates: x= sinθ cos φ, y = sinθsin φ, z = cosθ z e e φ VA ( t ) θ e θ A x φ y cosθ As a guess fo the velocity potential Φ, let s assume Φ= C whee C is an abitay function of time. a) Why is Φ assumed to be independent of φ? The flow is symmetic aound the z-axis. 8 pts o nothing b) Show that Φ satisfies the equation govening the flowfield. (Suggestion: fist find the components of the velocity, they ae needed anyway, then substitute the components into the govening equation.) Φ 1 Φ 1 Φ cosθ sinθ Fom hw: v = GadΦ = e + e + e = C e 0 3 C e + e 3 θ sinθ φ cosθ sinθ v = C v C v 0 3 θ = 3 φ = θ φ θ φ

7 ( ) v v 1 v v 1 v θ φ vθ cosθ cosθ cosθ cosθ Div v = cotθ = 6C C 4C + 0 C = θ sinθ φ Continuity Eqn: Φ Div GadΦ = Div = 0 4 pts if they state that the continuity equation is the field equation, but use the wong equation (some students may use cylindical, instead of spheical, coodinates). The latest vesion of hwsoln has the needed fomulas. 8 pts fo the coect answe. c) Satisfy the bounday conditions if that s possible. BC 1) Distubance caused by the motion of the sphee dies out fa fom the body: cosθ sinθ v = C 0 & v C 0 as 3 θ = 3 BC ) No-penetation condition: ( ) 0 ( ) 0 ( ) v v body i n = on = a v VA t k i e = v i e = v = VA t k i e on = a Fom hwsoln: e = sinθ cosφi+ sinθsinφj+ cosθk ki e = cosθ C ava a VAcosθ a VA ava cosθ = V 3 A () t cos θ C = Φ =, v cos & sin = θ v 3 θ = θ 3 a 8 pts fo each BC, 16 pts total d) Is ciculation elevant? Succinctly explain you answe. The voticity is zeo eveywhee in the flow; thus, fom Stokes theoem Γ 0 fo all loops (unlike the two-dimensional case in which a loop may encicle the body and it is not possible to daw the hat used in the deivation of Stokes theoem). The ciculation is not elevant. Be geneous: anything close to this gets 8 pts. e) Find the expession fo the pessue on the suface of the sphee ( = a ). Benoulli s equation fo a solution in a moving efeence fame: Φ vv i p p vi ( VA + ω + vel ) + + = on = a t ρ ρ vv i v v θ = + vi( VAk+ ω + vel ) = + ( veik+ vθeθi k) V + A p p Φ Φ ρ t t 1 eθ ik ik = ( cosθ cosφi+ cosθsinφj sinθk) ik = sin θ, e i k = cosθ θ

8 p p Φ 1 VA 1 = + VA cos θ VA sin θ cos θ + sin θ ρ t 4 a dva a dva 1 5 = cosθ + VA cos θ sin θ cos θ sin θ = cosθ VA cos θ sin θ dt 8 dt 8 a dva 1 5 a dva VA 9 p p ρ cos VA cos sin cos 1 sin dt θ θ 8 θ ρ dt θ 4 θ = = 8 pts fo any of the foms above fom down. f) What is the added mass of the sphee? To find the added mass, we must fist find the dag: suface = a π π p p ds p p a sin θ= 0φ= 0 d d ( ) ( ) F= n = e θ φ θ π π a dva VA 9 = ρa cosθ 1 sin θ ( sinθcosφ sinθsinφ cosθ ) sinθdφdθ dt 4 i+ j+ k θ= 0φ= 0 dv 1 = π ρ k = = π ρ = 3 dt 3 3 A 3 a dag added mass a mass of the fluid displ Pogam to calculate loads 8 pts -- 4 pts fo the elationship between F and p (second line) 3 pts fo ecognizing dag is needed 1 pt fo the coect answe aced by the sphee % NACA 441 cambeline model with a cosine distibution of discete % voticies % cambeline appoximated with a 4th-ode polynomial obtained in % CamFit441.m clc clea alfa = 8 % input the angle of attack alpha = alfa*pi/180; ca = cos(alpha); sa = sin(alpha); n = 15; npts = n + 1 % the numbe elements % the numbe of points % coefficients impoted fom CambeFit441.m fo 4th-ode fit

9 c =[ ]; % cosine distibution of the x-coodinates theta = linspace( 0, pi, npts ); x = *cos( pi - theta ); x = x.*x; x3 = x.*x; x4 = x3.*x; X = [ x', x', x3', x4' ]; y = X*c'; % tweak the tailing edge y(npts) = 0; Could be used to tweak the gaphs % xv(n+1) = 1; (not used) % yv(n+1) = 0; % locations of the discete votices xv = 0.5*x(:npts) *x(1:n); yv = 0.5*y(:npts) *y(1:n); % locations of the contol points xcp = 0.75*x(:npts) + 0.5*x(1:n); ycp = 0.75*y(:npts) + 0.5*y(1:n); % lengths of the panels fo i = 1:n dl(i) = sqt( ( x(i+1) - x(i) )^ + ( y(i+1) - y(i) )^ ); end % vectos nomal to the panels, but not unit vectos % used in lines 55 and 56 nvx = -( y(:n+1) - y(1:n) ); nvy = ( x(:n+1) - x(1:n) ); % unit vectos nomal to the platelets (panels) nx = nvx./dl'; ny = nvy'./dl'; % unit vectos tangent to the platelets ty = -nx; tx = ny; % XXXXXXXXXXXXXXXX calculate the solution XXXXXXXXXXXXXXXXXXXXX % the influence matix fo i = 1:n;

10 R(i) = -ca*nx(i)-sa*ny(i); fo j = 1:n; sq = (xcp(i)-xv(j))^ + (ycp(i)-yv(j))^; % note: each votex has x- and y-components of velocity % associated with it; fo the no-penetation condition, we need % vn = v dot n = 0 at the CPs of the panels: vn = vx*nx + vy*ny % (1/*pi is put in late) A(i,j) = ( -(ycp(i)-yv(j))*nx(i) +(xcp(i)-xv(j))*ny(i) )/sq; end end % the ciculations aound the discete votices in the platelets g = inv(a)*r'; 5 pts fo this % the ciculation aound the aifoil (entie cambeline) cic = *pi*sum(g) % Kutta-Zhukovskii theoem, Dag = 0 lift = *cic % calculate loads fom the pessue distibution % the nomal and tangential components of the velocity at the contol % points (chosen because vn = 0 at the CP on all the panels; so thee % is only tangential velocity thee) % jump in the velocity acoss the cambeline du = g./dl; % ecall the solution fo flow ove a flat plate % the velocity at the CPs: the sum of the feesteam + distubance % associated with the all the votices in the panels fo i = 1:n vx = ca; vy = sa; xp = xcp(i); yp = ycp(i); two points off if this is missed fo j = 1:n sq = ( xp - xv(j) )^ + ( yp - yv(j) )^; vx = vx - g(j)*( yp - yv(j) )/sq; vy = vy + g(j)*( xp - xv(j) )/sq; end % tangential component: ut = v dot t (needed fo the calculation % of the pessue)

11 end ut(i) = vx*tx(i) + vy*ty(i); fo the flat plate: u = cosα du, u = cosα 0.5du uppe lowe cosα = tangential component of the feesteam vel. the tangential component of the vel. associated with votices in the panels is zeo, which is not the case case fo the "cambeed plate" -- we need to do the same steps hee % velocities on the uppe and lowe sufaces uu = ut + 0.5*du(1:n); ul = ut - 0.5*du(1:n); % pessue distibutions on the uppe and lowe sufaces Cpu = 1 - uu.*uu; Cpl = 1 - ul.*ul; % check lift: compae integal of pessue ove the length to *Gamma % components of the foce on each panel Fx = ( Cpl - Cpu ).*nvx'; Fy = ( Cpl - Cpu ).*nvy; Lift = sum( Fy*ca - Fx*sa ) Dag = sum( Fy*sa + Fx*ca ) % moment: the line of action of the foce on the panel goes though % the position of the votex on the panel xvm = xv(1:n); yvm = yv(1:n); % moment about the leading edge, le Mxy = -Fx.*yvm'; Myx = Fy.*xvm; Mle = sum( Mxy + Myx ) % x-coodinate of the cente of pessue xba = Mle/Lift % moment about the quate chod, which is the expeimental quantity % plotted M14 = Mle - 0.5*Lift

Applied Aerodynamics

Applied Aerodynamics Applied Aeodynamics Def: Mach Numbe (M), M a atio of flow velocity to the speed of sound Compessibility Effects Def: eynolds Numbe (e), e ρ c µ atio of inetial foces to viscous foces iscous Effects If