Norwegian university of Science and Technology
NTNU
Faculty of Engineering Science technolog
Dept. of Energy and Process Engineering
TEP 4160 AERODYNAMICS
Exercise 11
Use the Vortex Lattice program to solve the following problem:
An airplane that weighs m=1000kg has a strainght unswept wing with an area of A=15m2, an aspect ratio of AR = 7.5 and a taper ratio of 0.8. The wing uses the NACA 0012 airfoil. The built in angle of attack for the root section is a = 0o with respect to the airplane central axis and the built in angle of attack for the tip is a = -3o. If the plane is to fly at a steady velocity of V=170km/h at constant altitude where r=1.17m3, compute the required lift coefficient for the wing, CL.
Choose an angle of attack where you know potential flow should apply and use e.g. 5 panels in the streamwise direction. Play with the number of panels in the spanwise direction. How many spanwise vortices do you need to get a solution that is basically independent of the number of panels?
Next, run VortexLattice for two angles of attack and do a final calculation that gives the required lift coefficient. What is the required angle of attack (a)?
For this angle of attack, find the following parameters:
a) The spanwise load distribution (Cl [y]/ CL). Repeat for at least one other angle of attack and compare. Explain any differences / similarities. (Answers may be found e.g. in Eqs. 7.48 to 7.51 ...... )
b) The vortex induced drag coefficient (Cd). (Assume e=0.95 in Eq. 5.48)
c) The effective angle of attack (aeff) (The effective angle of attack is the angle the plane would need to have to produce the required lift if the wing had the theoretical lift slope dCL/d a = 2p.)
d) Repeat the calculations for twist angles of g = 0, 1 and 2 degrees at the same angle of attack. How do the new load distributions differ from the first.
e) Compare the load distributions to the optimum elliptic distribution, , where Cl [y=0] is the lift coefficient on the centre line and b is the wing span.
Norwegian university of Science and Technology
NTNU
Faculty of Engineering Science technolog
Dept. of Energy and Process Engineering
TEP 4160 AERODYNAMICS
Solution to Exercise 11
First we convert the velocity from km/h to m/s: U = 170km/h = 47.2m/s. Hence the lift coefficient needed for steady flight is .
In order to be able to run VortexLattice we need to determine the input parameters. The sweep and dihedral angles are zero for the straight wing and the built in twist is g = +3o. From the area and the aspect ratio we can determine the wing span and the root chord length.
Since this is a symmetric airfoil we expect it to operate in the linear lift range for e.g. a = 5o. We then generate the "Rot.dat" file with 6 coordinate pairs for the straight root camber line, e.g.
0,0
0.3,0
0.6,0
0.9,0
1.2,0
1.571,0
Running the program with a range of spanwise panels gives e.g.
Ny=5: CL = 0.313
Ny=10: CL = 0.308
Ny=15: CL = 0.307
Ny=19: CL = 0.306
The last value is also obtained for 30 panels (not an option in the student version of the program), and so we conclude that 19 spanwise vortices will be adequate.
We now run the program also for a = 5o which gives CL = 0.713. A linear interpolation between the two gives the required angle of attack to be a = 7.41o. (A check by running the program for this angle of attack confirms that we get CL = 0.502.)
a) Figure 1(a) and (b) show the normalized lift distributions for a = 7.41o and a = 2o.
Figure 1 Normalized lift distributions for two different angles of attack
It is obvious from the two figures that the normalized lift distributions are the same. This is because the influence matrix is independent of the angle of attack and the circulation distribution (and therefore the lift coefficient) is linearly dependent on G and the angle of attack (see text book, Eq. 7.48). Therefore when scaled by any lift coefficient (in this case the value on the centre line), the distributions will be the same.
b) The vortex induced drag coefficient is written . With e=0.95, AR=7.5 and CL = 0.502 we then get .
c) The lift coefficient for the 3D wing may be written . Using the values of CL from the initial calculations we find that when a is measured in degrees. Hence the wing produces zero lift at . The two-dimensional wing will also produce zero lift at the same angle of attack, but the slope will be different, the theoretical slope being 2p(=0.1097 when a is measured in degrees). (See page 231 in the text book.) Therefore the same wing with infinite span is expected to have the lift distribution from which we find that the required lift coefficient of CL = 0.502 is obtained for a = 5.50o.
d) and e) The figure below shows the load distributions for g = 0, 1, 2 and 3 degrees. Also included for reference is the elliptic distribution which is the optimal load distribution. It is apparent that by increasing the twist, the load on the wing tip is reduced and at the same time the load near the centre is increased. We see that we may use the wing twist to optimise the lift distribution. For for g = 3 degrees, the load distribution is very close to the elliptic distribution. Since this is the optimal distribution, which has a wing efficiency of e = 1, we expect e to be close to, but not quite, equal to 1. Hence the suggested value of e = 0.95 in question b).
We may also use the taper ratio to load or unload the tip area of the wing. For a straight, untwisted wing the optimum tip ratio is close to 0.3. This was used for the high altitude reconnaisance airplane U-2 which needed to have a very efficient wing to be able to produce very high lift coefficient when operating at high altitude where the density is low.
Figure 2 Effect of wing twist
Figure 3 Lockheed U-2 reconnaisance aircraft
Side 3