360hw1_14

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MAE 360. Aerodynamics. Homework Assignment 1. Due September 4. Instructions: Homework solutions are expected to be presented in a professional manner. In addition to the plots, gures, calculations and codes that are requested, you must include an explanation of the work you have done. MATLAB code should be included as an appendix to your solution document. All plots and gures must have appropriate captions, titles, labels, legends, etc. The following data describe the surfaces of two di/erent airfoils. x and z are made dimensionless by dividing by the airfoil chord. NACA 0012 NACA 2412 x z u z 0.0028 0.0064 0:0064 0.0139 0.0188 0:0188 0.0359 0.0301 0:0301 0.0682 0.0399 0:0399 0.1102 0.0480 0:0480 0.1609 0.0541 0:0541 0.2191 0.0580 0:0580 0.2837 0.0596 0:0596 0.3530 0.0590 0:0590 0.4257 0.0566 0:0566 0.5000 0.0525 0:0525 0.5743 0.0471 0:0471 0.6470 0.0410 0:0410 0.7163 0.0343 0:0343 0.7809 0.0275 0:0275 0.8391 0.0209 0:0209 0.8898 0.0147 0:0147 0.9318 0.0093 0:0093 0.9641 0.0050 0:0050 0.9861 0.0020 0:0020 0.9972 0.0004 0:0004 x z u z 0.0022 0.0067 0:0061 0.0121 0.0201 0:0174 0.0332 0.0334 0:0266 0.0649 0.0460 0:0336 0.1067 0.0573 0:0385 0.1577 0.0668 0:0412 0.2165 0.0737 0:0421 0.2819 0.0777 0:0414 0.3523 0.0786 0:0395 0.4258 0.0764 0:0367 0.5006 0.0718 0:0331 0.5752 0.0654 0:0289 0.6481 0.0575 0:0244 0.7175 0.0487 0:0199 0.7820 0.0394 0:0156 0.8401 0.0301 0:0116 0.8906 0.0213 0:0080 0.9323 0.0136 0:0050 0.9644 0.0073 0:0027 0.9862 0.0029 0:0010 0.9972 0.0006 0:0002 1. Using MATLAB, create a plot of each airfoil cross section. Be sure that the x and z axes are scaled with respect to each other. Discuss the similarities and di/erences between the two airfoil shapes. 2. The ideal pressure coe¢ cients are known at each point and are given below. Plot the pressure coe¢ cients vs. wing x-location for both upper and lower surfaces of the airfoils. Be sure to indicate which data are for the upper surface and which are for the lower surface. Recall that the standard for plotting pressure coe¢ cient is negative up. Recall also that point number 1 is the rearmost point on the lower surface and the points are numbered clockwise. Compare and contrast the two pressure distributions. 3. Using MATLAB, estimate the lift and drag coe¢ cients for the two airfoils. The angle of attack, , is 4.5 . Find the lift and drag per unit span if (a) the chord length is 0.5 m, =1:225 kg/m 3 , and V 1 = 100 m/s, and (b) the chord length is 1.0 ft., =0:0015 slugs/ft 3 and V 1 = 300 mi/hr. Do your values seem to make sense? Explain. point no. 1 2 3 4 5 6 7 8 9 10 C p 0012 0:3353 0:2334 0:1668 0:1185 0:0811 0:0508 0:0256 0:0040 0:0147 0:0309 C p 2412 0:3436 0:2551 0:2032 0:1701 0:1476 0:1316 0:1195 0:1097 0:1014 0:0945 point no. 11 12 13 14 15 16 17 18 19 20 C p 0012 0:0437 0:0515 0:0516 0:0399 0:0108 0:0435 0:1348 0:2834 0:5257 0:8836 C p 2412 0:0900 0:0923 0:0978 0:1052 0:1202 0:1509 0:2098 0:3202 0:5290 0:8853

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Page 1: 360hw1_14

MAE 360. Aerodynamics.Homework Assignment 1. Due September 4.

Instructions: Homework solutions are expected to be presented in a professional manner. In addition to the plots,figures, calculations and codes that are requested, you must include an explanation of the work you have done.MATLAB code should be included as an appendix to your solution document. All plots and figures must haveappropriate captions, titles, labels, legends, etc.

The following data describe the surfaces of two different airfoils. x and z are made dimensionless by dividing bythe airfoil chord.

NACA 0012 NACA 2412

x zu z`0.0028 0.0064 −0.00640.0139 0.0188 −0.01880.0359 0.0301 −0.03010.0682 0.0399 −0.03990.1102 0.0480 −0.04800.1609 0.0541 −0.05410.2191 0.0580 −0.05800.2837 0.0596 −0.05960.3530 0.0590 −0.05900.4257 0.0566 −0.05660.5000 0.0525 −0.05250.5743 0.0471 −0.04710.6470 0.0410 −0.04100.7163 0.0343 −0.03430.7809 0.0275 −0.02750.8391 0.0209 −0.02090.8898 0.0147 −0.01470.9318 0.0093 −0.00930.9641 0.0050 −0.00500.9861 0.0020 −0.00200.9972 0.0004 −0.0004

x zu z`0.0022 0.0067 −0.00610.0121 0.0201 −0.01740.0332 0.0334 −0.02660.0649 0.0460 −0.03360.1067 0.0573 −0.03850.1577 0.0668 −0.04120.2165 0.0737 −0.04210.2819 0.0777 −0.04140.3523 0.0786 −0.03950.4258 0.0764 −0.03670.5006 0.0718 −0.03310.5752 0.0654 −0.02890.6481 0.0575 −0.02440.7175 0.0487 −0.01990.7820 0.0394 −0.01560.8401 0.0301 −0.01160.8906 0.0213 −0.00800.9323 0.0136 −0.00500.9644 0.0073 −0.00270.9862 0.0029 −0.00100.9972 0.0006 −0.0002

1. Using MATLAB, create a plot of each airfoil cross section. Be sure that the x and z axes are scaled withrespect to each other. Discuss the similarities and differences between the two airfoil shapes.

2. The “ideal”pressure coeffi cients are known at each point and are given below. Plot the pressure coeffi cientsvs. wing x-location for both upper and lower surfaces of the airfoils. Be sure to indicate which data arefor the upper surface and which are for the lower surface. Recall that the standard for plotting pressurecoeffi cient is “negative up.” Recall also that point number 1 is the rearmost point on the lower surface andthe points are numbered clockwise. Compare and contrast the two pressure distributions.

3. Using MATLAB, estimate the lift and drag coeffi cients for the two airfoils. The angle of attack, α, is 4.5◦.Find the lift and drag per unit span if (a) the chord length is 0.5 m, ρ = 1.225 kg/m3, and V∞ = 100 m/s,and (b) the chord length is 1.0 ft., ρ = 0.0015 slugs/ft3 and V∞ = 300 mi/hr. Do your values seem to makesense? Explain.

point no. 1 2 3 4 5 6 7 8 9 10Cp 0012 0.3353 0.2334 0.1668 0.1185 0.0811 0.0508 0.0256 0.0040 −0.0147 −0.0309Cp 2412 0.3436 0.2551 0.2032 0.1701 0.1476 0.1316 0.1195 0.1097 0.1014 0.0945

point no. 11 12 13 14 15 16 17 18 19 20Cp 0012 −0.0437 −0.0515 −0.0516 −0.0399 −0.0108 0.0435 0.1348 0.2834 0.5257 0.8836Cp 2412 0.0900 0.0923 0.0978 0.1052 0.1202 0.1509 0.2098 0.3202 0.5290 0.8853

Page 2: 360hw1_14

point no. 21 22 23 24 25 26 27 28 29 30Cp 0012 0.7645 −1.1228 −1.7589 −1.5149 −1.2611 −1.0673 −0.9139 −0.7851 −0.6722 −0.5713Cp 2412 0.7276 −0.7932 −1.6456 −1.5388 −1.3699 −1.2333 −1.1185 −1.0120 −0.9060 −0.7937

point no. 31 32 33 34 35 36 37 38 39 40Cp 0012 −0.4803 −0.3982 −0.3239 −0.2560 −0.1929 −0.1326 −0.0728 −0.0112 0.0546 0.1283Cp 2412 −0.6782 −0.5744 −0.4810 −0.3941 −0.3114 −0.2310 −0.1509 −0.0691 0.0163 0.1081

point no. 41 42Cp 0012 0.2168 0.3353Cp 2412 0.2126 0.3436

4. Based on the data and plots above, see if you can explain why the lift coeffi cients are different.