MAE SUMMER 2015 HOMEWORK 1 SOLUTION

Transcription

1 MAE 04 - SUMMER 205 HOMEWORK SOLUTION Problem :. Plot the drag coefficient, c D, the aerodynamic efficiency, AE, and the center of pressure, x cp, of the airplane as a function of the angle of attack. By combining together the plots for the polar curve and the c L as a function of α, we obtain the c D data. The result is shown in Figure. Figure : Drag coefficient of the airplane as a function of the angle of attack. The aerodynamic efficiency is defined as AE = c L /c D. By combining the c L and c D plots we get AE as a function of α, represented in Figure 2. Consider the sketch presented in Figure 3 for the geometry of the problem. The origin of the (x, y) axis is located at the front of the airplane. Applying the definition of center of pressure: M cp = N (x cp x c/4 ) + Mc/4 = 0.

2 Figure 2: Aerodynamic efficiency of the airplane as a function of the angle of attack. By using the definition of the aerodynamic coefficients, defining the average chord length as c = S/b, and using the relation c N = c L cos(α) + c D sin(α), we find that the location of the center of pressure is x cp = 0.6 d S b cm c/4 c N, Figure 4 represents the location of the center of pressure as a function of the angle of attack. y N U8 η x M c/4 c.p. ξ α Figure 3: Sketch for the geometry of the problem. 2

3 Figure 4: Center of pressure as a function of the angle of attack. 2. Calculate the stalling velocity of the airplane. What are α, c D, c Mc/4 and AE at this flight condition? 2W c L,max.8 V stall = 74.2 m/s, ρsc L,max α stall 38, c D,stall.2, c Mc/4,stall 0.09, AE stall.5. ** The angle of attack α stall is very large. Since the engines are fixed to the airplane, the thrust at high angles of attack have a non-zero component parallel to the lift. Now that we know that α is large, we need to take into account this component of the thrust in order to calculate the actual V stall. However, this is out of the scope of this problem, and for simplicity, here on I will use the approximation of small α (all the text in blue), and I will consider it as good in my grading. If you are interested in knowing the actual solution, keep reading. Consider the force balance plotted in Figure 5. Equilibrium of forces renders W = L + T sin(α), D= T cos(α), 3

4 U8 α T L D W Figure 5: Force balance for high angles of attack. and thus: 2 ρv 2 S [c L + c D tan(α)] = W, 2W V stall = ρs [c L + c D tan(α)] max The stall doesn t happen now at the maximum c L but at the maximum c L +c D tan(α), which is the coefficient of total force counteracting the airplane s weight. Figure 6 depicts the dependence of c L + c D tan(α) with α. It can be seen that this coefficient of force doesn t reach a maximum for the range of angles of attack provided. This means that the airplane can reach very high angles of attack and thus very low speeds without stalling. Potentially, if the curve doesn t reach a maximum, in a flight with lower weight, the airplane could stay vertically with zero speed and balancing the Figure 6: Coefficient of total force counteracting the airplane s weight. 4

5 Figure 7: Airplane flying at α = π 2 rad. weight with just the thrust provided by the engines. And just to finish, Figure 7 shows a cool picture of a MIG-29 performing the cobra maneuver in which it stays vertical, with zero velocity, and balanced just by the engines thrust 3. Under the conditions described in the previous part, calculate the lift, drag, moment with respect to the front of the airplane M 0 and traction acting on the airplane. L stall = W = 230, 535 N, D stall = 2 ρv stallsc 2 D,stall 5, 700 N, M 0,stall = N 0.6 d + M c/4 = 2 ρv stallsc [ dc ] (c L,max cos(α) + c D,stall sin(α)) + c Mc/4 M 0,stall 2, 788, 496 N m, T stall = D stall 5, 700 N. 4. Calculate the maximum velocity of the airplane. What are α, c L, c Mc/4 and AE at this flight condition? 5

6 The airplane has two engines, and thus the maximum thrust is T max = 2T. 2T max c D,min 0.05 V max = m/s, ρsc D,min α cd,min 5, c L,cD,min 0.4, c Mc/4,c D,min 0.7, AE cd,min Under the conditions described in the previous part, calculate the lift, drag, moment with respect to the front of the airplane and traction acting on the airplane. L cd.min = 2 ρv 2 maxsc L,cD,min = 43, 596 N W, D cd,min = T max = 58, 400 N, M 0,cD,max = N 0.6 d + M c/4 = [ 2 ρv maxsc d ( cl,cd,min c cos(α) + c D,min sin(α) ) ] + c Mc/4 M 0,cD,min 2, 754, 348 N m, T cd,min = T max = 58, 400 N. 6. What is the angle of attack that maximizes AE? Calculate the lift to drag ratio, velocity, lift L, drag D, moment with respect to the front of the airplane M 0 and thrust T for that angle of attack. α AEmax 0, AE max 6.3, 2W V AEmax = 37.4 m/s, ρsc L,AEmax L AEmax = W = 230, 535 N, D AEmax = W AE max 36, 332 N, M 0,AEmax = N 0.6 d + M c/4 = 0.6 d (L AEmax cos(α) + D AEmax sin(α)) + 2 ρv AE 2 max Scc Mc/4 M 0,AEmax 2, 74, 243 N m, T AEmax = D AEmax 36, 332 N. 6

7 7. When the airplane is at a height h = 4, 000m. both engines are turned off (T = 0N.) and the airplane is glided to the ground. Calculate the maximum distance that the airplane is able to glide at maximum lift to drag ratio. If x is the traveled distance: AE = x h = tan(α) x AE max = AE max h 25, 38 m. 8. Specify the units of all the previous results. All the results in the previous questions have the correct units. Bonus: Is there anything wrong with questions 4. and 5.? The velocity is supersonic, and thus the theory used doesn t hold. 7

8 Problem 2:. What is the relation between the lift L and normal force N on the airfoil? Justify why we can integrate along the reference system (x 2, y 2 ) instead of (x, y) to calculate L. The integral of c p on the surface of the airfoil, projected on (x 2, y 2 ) will render the normal and axial forces. However, we are interested in the lift force Given that α is very small: L = N cos(α) A sin(α) L N. 2. Integrate c p along the surface of the airfoil to calculate L. Hint: use the coordinate system (x 2, y 2 ). c c l c n = c c p,u (x 2 )dx 2 + c p,l (x 2 )dx 2 = { } 2x 2 /c 2α + 2x 2 /c + 2x 2/c α2 + 2x 2 /c + 2α 2x 2 /c + 2x 2 /c 2x 2/c α2 dx 2 = + 2x 2 /c c 2x 2 /c 4α + 2x 2 /c dx 2 Making the change of variable x 2 = c 2 ξ: c l 2α ξ= ξ= ξ dξ = 2πα + ξ L = 2 ρu 2 Sc l ρu 2 Sπα 3. Calculate the coefficient of moment at the leading edge of the airfoil, c ml.e.. 8

9 Using the reference system (x 2, y 2 ): c m,l.e. = c 2 c 2 [ c p,u (x 2 ) c p,l (x 2 ) 4α Making the change of variable x 2 = c 2 ξ: c m,l.e. = α ξ= ξ= 2x 2 /c + 2x 2 /c ] ( x 2 + c 2 ( x 2 + c ) dx 2 2 ξ + ξ (ξ + ) dξ = α 2 π ) dx 2 = 4. Calculate the coefficient of moment at the c/4 point, c mc/4. c m,c/4 = c m,l.e. + 4 c l c m,c/4 = 0 9