AS 310: Aircraft Performance

Exercise 4: Drag and Applications

The first part of this week’s assignment is to revisit our reciprocating engine powered (i.e. propeller type) aircraft from last week.

1. Selected Aircraft (from last week’s module):

Aero Commander 720 Shrike

Make sure to review your data and results from last week and any feedback that you may have received on your work, in order to prevent continuing with faulty data.

2. Main Wing Airfoil type & on-line database designator (from last week’s module): NACA 23012

3. Aircraft Maximum Gross Weight [lbs] (from last week’s module): 6750 lbs

4. Wing Span [ft] (from last week’s module): 49 ft 1 in = 49 ft

5. Average Chord Length [ft] (from last week’s module): 5.2 ft

6. Wing Area ‘S’ [ft2] (from last week’s module): 255 ft^2

7. Find the Aspect Ratio ‘AR’ for your selected aircraft wing. (Use the wing span and average chord length from last week’s module/from above. See also page 63 in your textbook.):

Span = 49 ft , average chord = 5.2 ft

AR = span/average chord = 49 / 5.2 = 9.42

8. CLmax for your airfoil (from last week’s module): CLmax = 1.5413 AOA = 15.75

9. Standard sea level Stall Speed ‘Vs’ for your aircraft [kts] (from last week’s calculation):

L = W = 6750 lbs Cl Max =1.5413 ,  =1 S =255 ft^2

V^2 = 295 * 6750 / 1.5413/ 255 V = 71.18 KTAS

This document was developed for online learning in ASCI 309. File name: Ex_4_Drag&Applications Updated: 07/11/2015 2

Find the appropriate drag polar curve for your airfoil selection (2. above; from last week’s module). You can utilize any officially published airfoil diagram for your selected airfoil or use again the Airfoil Tool at

http://airfoiltools.com/search .

Concentrate for this exercise on the Cl/Cd (coefficient of lift vs coefficient of drag) plot, i.e. the so called drag polar. Use again only the curve for the highest Reynolds-number (Re) selected (i.e. remove all checkmarks, except the second to last, and press the “Update plots” tab).

How to find the minimum Cd

10. From the polar plot, find the CDmin value for your airfoil, i.e. the lowest value that the coefficient of drag ‘Cd’ (bottom scale in the online tool depiction) reaches. (Tip: for a numerical breakdown of the plotted curve, you can again select the “Details” link and directly read the lowest CD value in the table – third column, labeled “CD”): 0.00587

What we’ve just found (…with some degree of simplification…) is the parasite drag coefficient for our airfoil, i.e. the drag that exists due to skin friction and the shape of our airfoil, even when

This document was developed for online learning in ASCI 309. File name: Ex_4_Drag&Applications Updated: 07/11/2015 3 little or no lift is produced. However, this value will only represent the airfoil, i.e. main wing portion of our aircraft; therefore, let us for the remainder of our calculations assume that our aircraft is a Flying Wing type design and the total CDP for the aircraft is the same as the CDmin that we’ve just found.

Let us also assume that we are at standard sea level atmospheric conditions and that our wing has an efficiency factor of e = 0.82.

A. Prepare and complete the following table for your aircraft (with the data from 1. through 8. above). Start your first row with the Stall Speed ‘Vs’ (from 7. above) and start the second row from the top with the next higher full twenty knots above that stall speed. Then increase speed with every subsequent row by another 20 knots until reaching 300 kts. You are again encouraged to utilize MS® Excel as shown in the tutorial video and can also increase your table detail. However, the below depicted, and above described, interval is the minimum required for this assignment.

Equations for Table:

W 2 q = CL = qS CDi =[1/ (πeAR)] CL 2 CD = CDP + CDi CD = CDP + [1/ ( e AR)] CL Dp = CDp q S

2 Di = CDi q S = [1/ ( e AR)] CL q S Dt = Di + Dp = CD q S

AR = 9.42, L = W = 6750 lbs Cl Max =1.5413 ,  =1 S =255 ft^2 ,Stall speed @ standard sea level = V = 71.18 KTAS, e = 0.82

V q CDP CDI DP DI DT CL CD CL / CD

(KTAS) (psf) CDP CDI (lb) (lb) (lb) 71.18 17.17489 1.541238 0.00587 0.097937 0.103807 14.84721 25.70823 428.9227 454.6309 60 12.20339 2.169118 0.00587 0.193987 0.199857 10.85337 18.26664 603.6601 621.9268 80 21.69492 1.220129 0.00587 0.061379 0.067249 18.14356 32.47403 339.5588 372.0328 100 33.89831 0.780882 0.00587 0.025141 0.031011 25.18109 50.74068 217.3176 268.0583 120 48.81356 0.542279 0.00587 0.012124 0.017994 30.1364 73.06658 150.915 223.9816 140 66.44068 0.398409 0.00587 0.006544 0.012414 32.09272 99.45173 110.8763 210.3281 160 86.77966 0.305032 0.00587 0.003836 0.009706 31.42665 129.8961 84.8897 214.7858 180 109.8305 0.241013 0.00587 0.002395 0.008265 29.16105 164.3998 67.07335 231.4731 200 135.5932 0.195221 0.00587 0.001571 0.007441 26.23477 202.9627 54.32941 257.2921 220 164.0678 0.161339 0.00587 0.001073 0.006943 23.23698 245.5849 44.90034 290.4852 240 195.2542 0.13557 0.00587 0.000758 0.006628 20.45485 292.2663 37.72876 329.9951 260 229.1525 0.115515 0.00587 0.00055 0.00642 17.99258 343.007 32.14758 375.1546 280 265.7627 0.099602 0.00587 0.000409 0.006279 15.86272 397.8069 27.71909 425.526 300 305.0847 0.086765 0.00587 0.00031 0.00618 14.03874 456.6661 24.1464 480.8125

Answer the following questions from your table.

I) Determine the minimum total drag ‘Dmin’ [lbs] (i.e. the minimum value in the total drag ‘DT’ column): 210.3281 lbs

II) Determine the airspeed at which this minimum drag occurs ‘VDmin’ [kts] (i.e. the speed associated with the row in which ‘Dmin’ was found): 140 kts

III) Compare parasitic ‘DP’ and induced ‘DI’ drag at VDmin. What is special about this point in your table? The difference between Dp and Di is at its minimum of about 11 while the differences for others is a lot bigger than this number.

IV) Determine the maximum CL/CD value in your table (i.e. the maximum value in the CL/CD column) and the speed at which it occurs. 32.09272 @ 140 kts

V) Compare your results in IV) with II) and comment on your findings.

The maximum CL/CD occurs at 140 kts, and minimum drag occurs at 140 kts. While the drag is at its minimum, lift-drag ratio will be at its maximum. Thus, the airspeed when drag is at it minimum, the lift drag ratio will be at maximum achieving the most optimum efficiency.

VI) Explain which values in your table will directly allow glide performance prediction and how (Tip: Reference again the textbook discussion pp. 61-63). At 60 ktas, the lift-drag ratio is at its minimum. In other words, the airspeed generates minimum positive lift with very small amount of climb. Thus it is best for glide performance.

B. If the gross weight of your aircraft is decreased by 10% (e.g. due to fuel burn), how would the stall speed change? Support you answer with calculation as well as written assessment. (Remember, stall speed references and discussions can be found pp. 43-45 in your textbook.)

When gross weight is 6750 lbs at standard sea level, L = W = 6750 lbs Cl Max =1.5413 ,  =1 S =255 ft^2

V^2 = 295 * 6750 / 1.5413/ 255 V = 71.18 KTAS

When gross weight decreased by 10% 6750 lbs * (1-10%) = 6075 lbs

V^2 = 295 * 6075 / 1.5413/ 255 V =67.53 ktas. When the gross weight decreased by 10 %, stall speed decreased also. This is due to the fact that it does requires less lift than before.

For the second part of this assignment use the given figure below (Figure 1.13 from Aerodynamics for Naval Aviators [1965]) to answer the following questions. (This assignment is designed to review some of the diagram reading skills required for your midterm exam; therefore, please make sure to fully understand all the diagram information and review book, lecture, and/or tutorials if necessary.):

Figure 1.13 from Aerodynamics for Naval Aviators (1965). C. What is the Angle of Attack at Stall for the aircraft in Figure 1.13? 20 degree

D. What Angle of Attack is associated with Best L/D? 6 degree

E. What would be the best Glide Ratio for this aircraft? 12.4 degree

F. What is the maximum coefficient of lift (CLmax) value? 1.5 at stall angel.

This document was developed for online learning in ASCI 309. File name: Ex_4_Drag&Applications Updated: 07/11/2015