h = 0 km for your table in (a)], and for T0 = 316 K. Then obtain a Homework 1 AERE355 Fall 2017 Due 9/1(F)

Transcription

1 1 Herk 1 AERE355 Fall 17 Due 9/1(F) SOLUION NOE: If yur slutin des nt adhere t the frat described in the syllabus, it ill be grade as zer Prble 1(5pts) In the altitude regin < h < 1k, e have the flling atspheric equatins: = + λh (19 * ) ; P P = g Rλ (155) ; ρ ρ g (1+ ) Rλ = r P ρ = (157) R (a)(1pts) Write a Matlab cde that ill generate a table ith cluns [ P ρ] h fr the array f altitude values h = :1 :1, give the table, and cent n h it cpares t able A1 n p395 in APPENDIX 1 Slutin: [See 1(a)] Cent: his table is alst identical t able A1 (b)(1pts) Run yur cde fr = 8816 K [ie the teperature at h = k fr yur table in (a)], and fr = 316 K hen btain a plt f the percent change in air density vs altitude assciated ith the increased air surface teperature Slutin: [See 1(b)] his is a (nnintentinal) tricky questin able 1(a) Standard Atsphere able Figure 1(b) Percent change in Air density Fr e have the flling: On a ht day hat tends t happen is that the surface, hich is being ared by the sun, heats the lest level f the atsphere, reducing its density (it is at the sae pressure as its surrundings and its rises) his ill eventually drive cnvectin and ix this arer air vertically Given enugh tie, this ill reduce the ass in the clun f air and therefre reduce the pressure at the surface hese are called "heat ls" and yu can see the fring in the desert areas and they play rles in sea breeze fratin and the nsns

2 Hever, this questin des nt address the cntext f the increased teperature We ill here assue that the surface teperature has increased fr a sufficiently shrt perid, s that the grund pressure reains unchanged hen, fr the 3 ideal gas la, fr = 316 K, the resulting surface density is ρ = 1117kg / (c)(5pts) Based n yur plt in (b) and the definitin f the lift cefficient, explain, quantitatively, explain h the increased teperature influence the lift f a plane flying at an altitude f 6k Include yur reference in relatin t the definitin f the lift cefficient Slutin: Fr C /(5 V L = L ρ S) Hence, fr a plane having a given lift cefficient, its lift is directly prprtinal t ρ At h = 6 k the lift ill decrease by ~33%

3 3 PROBLEM (5pts) [Related t bk Prble 1] An airplane s altieter reading is 5, (a)(1pts) If the utside abient teperature is - C and the plane s true airspeed is 3/s, find the plane s indicated airspeed Slutin: Recall that the altieter uses pressure easureents t arrive at the altieter reading Fr able A1 n p395, fr h = 5, e have P = 54,15N / [Obtained fr a linear interplatin f able A1 e btain ( g / Rλ ) = λh = 54e4 Obtained directly fr P P 1 = 54, I ill use the latter] 3 Hence, fr = C = 53 K, the air density is ρ = P / R = 54,15/[87(53)] = 744kg / Hence, σ ρ / ρ = 744 /15 = 67 Fr (179) n p5, e btain V = V σ = 3 67 = 337 s btain V IAS =, I ill freg the equatins in the bk, and siply use the Matlab cand: crrectairspeed(3,35,54e4,'as','cas') [See als: his gives: EAS AS / V IAS = 43 / s (b)(5pts) Suppse that the altieter reading fluctuates beteen rughly 4,99 and 5,1, and that the ean reading is µ h = 5, Assue that the fluctuatin h has a nral distributin ith σ = 3 Use the Matlab cand nrpdf t arrive at a plt f the prbability density functin (pdf) fr the altieter reading, h Slutin: [See (b)] f(h) Altieter Reading pdf Altitude h () Figure (b) Altieter reading pdf g (c)(5pts) On p17 e have equatin (154): 1 + λ( h h1 ) ln( P / P1 ) = ln Beginning ith this equatin, carry ut Rλ 1 g / Rλ + λh detailed steps t sh that P = P Slutin: In ur situatin, 1 = = K, P1 = P = 11,35N / and h1 = g λ Rλ + h Step 1: ln( P / P ) = ln ; Step : P g λ Rλ + h / P = ; Step 3: + λh P = P g / Rλ (d)(5pts) arrive at the pdf f P : (i) use the Matlab cand nrrnd t generate N=1, siulatins f h (ii) Use these in the expressin in (c) t btain siulated easureents f P(iii) Use the cand histgra t arrive at a plt f an estiate f the pdf f P (iv) Use the cands ean and std t estiate the ean and standard deviatin f P NOE: Recall that R = 87 / K s and λ = 65 K / Slutin: [See (d)] Figure (d) Siulatin-based pdf fr P µ P = 54,1 N / and P = 16 N / σ

4 4 Prble 3(5pts) he plt bel shs bth lift and ent cefficients as functins f the angle f attack fr the NACA 31 airfil Als included are straight-line apprxiatins f these relatins Yu are t use these apprxiatins in this prble Fr: X&ei=WK87UIyzB8OZQW4pYCD&ved=CDIQsAQ&bi=18&bih=637 Figure 1 his plt als includes the CL ( α ZLL ) line (in black)

5 5 (a)(6pts) Estiate the cefficients f the lift del: CL ( α ) = CL + CL α α using the RED line hen plt yur del fr < α < 16 t verify its crrectness Slutin: CL & C L α 4 / 36 = 11 Hence, y del is: ( α) = + 11α C L C(a) Plt f CL(a) = CL + CLa*a a Figure 11(a) Plt f lift del (b)(6pts) Estiate the cefficients f the ent del: C ( α ) = C C α using the BLUE line hen plt + α yur del fr < α < t verify its crrectness Slutin: C 1 and C 3/ 4 = 57 Hence, y del is: C ( α) = 1 57α α C(a) Plt f C(a) = C + Ca*a a Figure 11(a) Plt f ent del (c)(3pts) Cpute the nuerical value f αtri fr yur airfil ent del in (b) Anser: C ( α ) = α 1 75 tri tri (d)(6pts) Use yur equatins fr parts (a) & (b) t estiate the paraeters C and h = h h n f the ent del: C = C + C h L Slutin: Fr (a) e have α = [ CL ( α) ]/ 11 Fr (b) e have C ( α) = 1 57α Substituting the expressin fr α int this gives the del: C = 4 5C L (e)(4pts) Fr yur del in (d) deterine hether the ing has negative pitch stiffness) r psitive pitch stiffness) Explanatin: he ing static argin = h = 5 is greater than zer S the ing has psitive pitch stiffness K n Fr re n this prble, see pp f Andersn Fundaentals f Aerdynaics

6 6 PROBLEM 4(5pts) [Related t bk prble 1] A given ing ent equatin is: C = 8 15C cg L (a)(5pts) Recall fr the discussin bel Figure 5 n p43 that the tri cnditin is defined as that cnditin such that = (ie the tri ent cefficient equals zer) Obtain the crrespnding tri lift cefficient C cg Slutin: C = = 8 15C C = 8 /15 = 53 cgtri L tri L tri (b)(5pts) he (scaled) ing center f gravity f the ing is = xcg hcg = 3 Use (6) n p45 t find the ing neutral c pint h N xn = Assue that the neutral pint equals the aerdynaic center pint c Slutin: Using (6), e have: C = C + C ( h h ) = 8 15C cg ac L gc Hence, h h = 15 h = h + 15 = = 45 gc ac ac gc ac L (c)(1pts) In (7) e see that C L = C L + C Lα α Suppse that the ing lift equals zer hen α = i = ( π /18 ) = 349rad, and it is at tri hen α = α = 5 ( π /18 ) = 873rad Find CL as a functin f α ver the range α 1 Slutin: Fr (a): 53 = C L + C (873) Als, = C L + C (349) L α hese t equatins can be ritten in atrix fr: Hence, (using Matlab): C C L Lα 353 = tri L α 873 C 349 C 1 L Lα 53 = C L CL and Plt f C L vs Angle f Attack CL α hen plt Angle f Attack (deg) Figure 1 Plt f CL versus α (d)(5pts) Fr a ing-alne cnfiguratin as shn in Figure 7 (n p45), explain hy it is nt statically stable Explanatin: Because x > x cg ac

7 7 Appendix Matlab Cde %PROGRAM NAME: h1 %PROBLEM 1 %PAR (a): hg = :1:1; hg = hg'; % Altitude array (k) R = 6371e3; %Earth ean radius () h = R*(R+hG)^-1 * hg; nh = length(h); vec1 = nes(nh,1); g = 987; %Gravity (/s) at sea level R = 87; %Gas cnstant ((/K-s) = 8816; %ep (K) at sea level P = 1135e5; %Pressure (N/) at sea level r = 15; %Density (kg/3) at sea level L = -65; %ep gradiant (K/k) c = g/(r*l); = + L*h; P = P*((/)^(-c*vec1)); r = r*(/)^-(1+c*vec1); frat shrtg able = [hg h P r] %PAR (b): =316; %~11F ep (K) at sea level P = 1135e5; %Assue pressure (N/) at sea level reains cnstant r = P/(R*); %Density (kg/3) at sea level L = -65; %ep gradiant (K/k) c = g/(r*l); = + L*h; P = P*((/)^(-c*vec1)); r1 = r*(/)^-(1+c*vec1); dr=1*(r-r1)/r; figure(1) plt(h,dr) grid xlabel('altitude ()') ylabel('%') title('percent Change in Air Density fr =88K t =316K') %============================================================== %PROBLEM %Part(a): crrectairspeed(3,35,54e4,'as','cas') %PAR (b): hvals=499:1:51; % Altitude Range f Values uh = 5; % Altitude Mean stdh = 3; % Altitude Standard Deviatin f = nrpdf(hvals,uh,stdh); figure() plt(hvals,f,'k') xlabel('altitude h ()') ylabel('f(h)') title('altieter Reading pdf') grid % PAR (d): R=87; L=-65; =8815; P=1135; g=987; N = 1^5; h = nrrnd(uh,stdh,n,1); hnes = nes(n,1); c1 = 1+(L/)*h; c = (-g/(r*l))*hnes; P = P*c1^c; % Pressure Siulated Data Array % % Cpute Estiated pdf: figure(1)

8 histgra(p,'nralizatin','pdf'); grid xlabel('pressure p (N/^)') title('histgra-based & Nral pdfs fr P') upest = ean(p); stdpest = std(p); frat shrtg [upest, stdpest] 8