Answer Happy

Note-
thin symmetric airfoil mean angle of attack zero left is zero.
For equation 5-51, coordinate transformation formula and other related formulas from chapter 5 to find An, circculation(y), lift coefficient, induced angle of attack(y), CL, Cd,i.
So, to solve this task, i have a hint for you to do this in easy way step by step in matlab. please follow this instruction by gather some formulas from above.
please help me to do this alteast how it begin with. thank you a lot and i will surly upvote.
Problem 4. Write your own Matlab code to apply the analysis of Chapter 5 to an untwisted rectangular wing of AR = 6 flying at an angle of attack a =10 shown in Fig. 2. Assume that the wing section is the thin symmetric airfoil (such as NACA 0012) for the entire wing, 1=1, and c=1. Repeat the following calculations twice to produce two sets of results for N=19 and N=39 in Eq. (5.51), respectively. In order to solve for the N unknowns in the equation, A. (n=1,2,...,N), chose N different spanwise locations yor shown in Fig. 1 specified by: b yo. =-+nay 2 (n=1,2,1,N) O (4) where N is an odd number and Ay=b/(N+1). Obtain the corresponding e. by the coordinate transformation and then construct a set of linear equations to solve for all A, (n=1,2,...,N). a 1. Print out your numerical results of 4, (n=1,2,.,N) in a table. 2. Derive the formulas and then plot (y), (y) and a (y) vs. y along the span for -b/25 y<b/2. 3. Calculate C, and Co. At the end of your report, attach a printout of your Matlab code to the end of you PDF file as an appendix. In addition, upload a separate Matlab .m file of your source code for inspection AN وام 명이 You your How You y sy Figure 2. The top view of a rectangular wing.
Consider the transformation e y = cos e (5.46) where the coordinate in the spanwise direction is now given by 8, with 0 <ST. In terms of 0, the elliptic lift distribution given by Equation (5.31) is written as (0) = lo sine (5.47) Equation (5.47) hints that a Fourier sine series would be an appropriate expression for the general circulation distribution along an arbitrary finite wing. Hence. assume for the general case that re) = 25V A, sin ne (5.48) dr where as many terms N in the series can be taken as we desire for accuracy. The coefficients A, (where n = 1.....N) in Equation (5.48) are unknowns; however, they must satisfy the fundamental equation of Prandtl's lifting-line theory; that is. the A, 's must satisfy Equation (5.23). Differentiating Equation (5.48), webtain dr de de = 261. nA cos no (5.49) dy de dy dy Substituting Equations (5.48) and (5.49) into (5.23), we obtain 2b LinA,cos ne (0) A, sin nd + a2_.6) + -de (5.50) πε(Θ) Jo cos - cos Blo The integral in Equation (5.50) is the standard form given by Equation (4.26). Hence, Equation (5.50) becomes 1 # no N a(0) 25 sin né. ΣΑ, sin nθ + αι. - (Θ)) + Σ Α, c(@o) sin en (5.51) Examine Equation (5.51) closely. It is evaluated at a given spanwise location: hence, 0is specified. In turn, b, c), and (0) are known quantities from the geometry and airfoil section of the finite wing. The only unknowns in Equa- tion (5.51) are the A's. Hence, written at a given spanwise location (a specified es), Equation (5.51) is one algebraic equation with N unknowns, A1, A2,.... A. However, let us choose N different spanwise stations, and let us evaluate Equa- tion (5.51) at each of these N stations. We then obtain a system of N independent
algebraic equations with N unknowns, namely, A1, Az.....Ay. In this fashion. actual numerical values are obtained for the A, numerical values that ensure that the general circulation distribution given by Equation (5.48) satisfies the fundamental equation of finite-wing theory. Equation (5.23). Now that (@) is known via Equation (5.48), the lift coefficient for the finite wing follows immediately from the substitution of Equation (5.48) into (5.26): 2 262 C = BsLady [(y)dy = sinnᎾ sin Ꮎ dᎾ (5.52) In Equation (5.52), the integral is for n = 1 sin no sine de - { /2 10 forn #1 Hence, Equation (5.52) becomes be Gi=1175 = AAR (5.53) Note that C, depends only on the leading coefficient of the Fourier series expan- sion. (However, although C, depends on A, only, we must solve for all the A's simultaneously in order to obtain A.) The induced drag coefficient is obtained from the substitution of Equa- tion (5.48) into Equation (5.30) as follows: ha CD = (5.54) -1/2 vs L Fa:(9) dy 5 ( 4, sinne ) a:(0) sinó de no 25 The induced angle of attack a (8) in Equation (5.54) is obtained from the substi- tution of Equations (5.46) and (5.49) into (5.18), which yields 1 a) 4V-2 ਨ। L (dr/ddy - Σελ. 1 cos ne -do cose - coses (5.55) The integral in Equation (5.55) is the standard form given by Equation (4.26). Hence, Equation (5.55) becomes sin nd, α, (θ) = ΣΗΑ, (5.56) sin In Equation (5.56), is simply a dummy variable which ranges from 0 tot across the span of the wing: it can therefore be replaced by 8, and Equation (5.56) = 4
can be written as sin no α (0) - ΣΠΑ, sine (5.57) 1 Substituting Equation (557) into (5.54), we have CIA, ) (įna, sinne) de A, sinne пA (5.58) Examine Equation (5.58) closely; it involves the product of two summations. Also note that, from the standard integral, form sin mo sin ko = / (5.59) for m=k Hence, in Equation (5.58), the mixed product terms involving unequal subscripts (such as A, A2, A2A1) are, from Equation (5.59). equal to zero. Hence, Equa- tion (5.58) becomes {1,2 CD = C. (Σ) - ΛΑΣ που 212 S NA? = TARA - =XAR A 1:[+9] (5.60) Substituting Equation (5.53) for C, into Equation (5,60), we obtain Ср. CZ (1 + 8) TAR (5.61) where 8 = n(A/A). Note that 8 > 0; hence, the factor 1 + 8 in Equa- tion (5.61) is either greater than I or at least equal to 1. Let us define a span efficiency factor, e as e = (1+8)- Then Equation (5.61) can be written as Col C CAR (5.62) where e < 1. Comparing Equations (5.61) and (5.62) for the general lift distri- bution with Equation (5.43) for the elliptical lift distribution, note that 8 = 0 and e = 1 for the elliptical lift distribution. Hence, the lift distribution which yields minimum induced drag is the elliptical lift distribution. This is why we have a practical interest in the elliptical lift distribution,
Elliptic wing Rectangular wing T Tapered wing Figure 5.18 Various planfors for straight wings.
Numerical solution of fundamental equation (for general wing shapes) Consider the fundamental equation of Prandtt's lifting-line theory. r(y) a(y)= 1.p (dr/dy) dy IV.cy.) 5+246)+ K y - y 421-12 . Using this equation, we could theoretically solve for the circulation distribution 'yo) and eventually the lift, induced drag, etc. for any given wing shape if we know the chord distribution cy), the angle-of-attack a(y), and angle-of-zero-lift at-obo). ho For a general wing shape, it will not be possible to perform the integration in the fundamental equation analytically. Let us suppose that we could use a Fourier sine series approximation to r(in terms of the coordinate 6) to represent the circulation distribution over a wing of arbitrary shape. The coefficients A, are to be determined later. ()= 25v 4 sin(ne) Numerical solution of fundamental equation (for general wing shapes) As we increase the number of terms in the sum N, we expect the result to become more accurate Plugging in the Fourier series form of r into the fundamental equation (and writing everything terms of 8, instead of yo) we arrive at a different form of the fundamental equation 25 a(0) ΣΑ,sin (πθ.)+α(Θ)-ΣΑ. sin(no) sin() Rearranging the above equation, we get 2b Σ sin(ne) sin (0) 4. =a(0)-2-(0) πείθ.) και . 3.-203 |ac(0) Sin(ne)+n Now let:
. Now let: sin (no, he)= 2b sin (no,)+n ac(0) sin(0) Numerical solution of fundamental equation (for general wing shapes) . Then we get: N 21 (0)4, = a (0.)-a,-(0) n=1
. To solve for the coefficients A, numerically, we evaluate the previous equation at N spanwise locations of (i = 1,2,..,N), thereby obtaining Nequations for N unknowns. . In matrix form, (0)h(0 (C) (2) hy(0) A h (2) 4, a(0.)- a (0) a(z)-a (0) --- LO [(e) h(a) h, (C) LA] a(@x)-2.(0) La 0] OM [H]yv [z]xx1 = : NxN Nxl Nxl Problem configuration AIN arله y 2 . . . We consider: Rectangular wing Aspect ratio AR = 6 Thin, symmetric airfoil along the entire wing Uniform chord with c = 1 Uniform angle-of-attack (no geometric twist) with a = 10° V = 1 . . .
Objective and workflow . Objective: Using the numerical formulation of the fundamental theorem with the Fourier sine series approximation to obtain (y), 46), a(y), CL, and Cpx for the given wing with N = 19 and N = 39 . COS Workflow (for your code): 1. Generate a set of N point along the span, denoted yon, where we apply the fundamental equation. The equation is specified in the prompt. 2. Transform these coordinates from yo to @, by inverting equation (5.46); Yo = -cos en 3. Generate the matrices and vectors shown previously by evaluating the fundamental equation at each point of 4. Solve for the coefficients An 5. Using these results, compute and/or plot the required quantities Hint = The aspect ratio is defined as AR = b2/S. Since the aspect ratio is given, this will allow you to obtain a numerical value for the span. • The wing uses a thin, symmetric airfoil for the entire wing. What does that mean for the angle-of-zero-lift a -o? Use the following lines at the beginning of your code cle; clear: close all; This will essentially reset the workspace every time you run your code, minimizing potential errors. You can solve a matrix equation using either z-H\bor z-inv (H)*b Make sure your arrays are properly sized • Use a double "for" loop to generate the matrix H The problem statement asks you to "derive" certain equations. If the equations are provided to you, either in the textbook or the notes, you can just write the equations as is.