# 52 Worked Examples: Airfoils & Wings

These worked examples have been fielded as homework problems or exam questions.

Worked Example #1

(a) The equation for the lift coefficient in the linear range is given by

In this case, .

For 2, then = 0.22

For 6, then = 0.66

For 10, then = 1.1

(b)

For 2, then = 0.352

For 6, then = 0.792

For 10, then = 1.232

(c)

For 2, then = 0.033

For 6, then = 0.473

For 10, then = 0.913

Worked Example #2

The pressure distribution in this question is given in analytic form to solve the problem without any numerical integration. But to check, a MATLAB code is given at the end of this answer.

(a) The lift coefficient is given by

In this case we can see that so the integral for becomes

(b) The pitching moment about the leading edge is

Substituting then

Here is some MATLAB code to cross-check the results and plot the pressure distribution.

figure

axis([0.0 1.0 1.0 5.0])

x = linspace(0.0,1.0,100); %set the range of x/c values

cpu = -4.*(1-x);

cpl = 1.*(1-x);

dcp = cpl-cpu;

trapz(x,dcp) % check to find the section cl using the trapezoidal rule

dcpx = -dcp.*x;

trapz(x,dcpx) % check to find the section cl_le using the trapezoidal rule

plot(x,-cpu);hold on

plot(x,-cpl)

xlabel(‘x/c’)

ylabel(‘-C_P’)

legend(‘Upper surface’, ‘Lower surface’)

Worked Example #3

(a) Use MATLAB to plot the pressure coefficient distribution. (b) Calculate the lift coefficient by analytic integration. (c) Calculate the lift coefficient in MATLAB by numerical integration.

(a) The pressure distribution looks like this (see the last part of the answer for the code):

which for this problem, we can split into two integrals, namely

Taking the first integral gives

and so

Taking the second integral gives

and so

Therefore,

(c) The MATLAB code, which uses numerical integration, gives = 1.4025, so the results of the two methods agree.

Worked Example #4

As shown in the table below, the lift, drag, and pitching moment coefficient measurements for a NACA 2412 airfoil are to be used to calculate specific derived aerodynamic quantities. First, determine the values of the following parameters: (a) Lift curve slope; (b) Zero-lift angle of attack; (c) Drag polar (as a plot); (d) The best lift-to-drag ratio; (e) Center of pressure location (as a plot).

(deg.) | |||

-2 | 0.05 | 0.006 | -0.042 |

0 | 0.25 | 0.006 | -0.040 |

2 | 0.44 | 0.006 | -0.038 |

4 | 0.64 | 0.007 | -0.036 |

6 | 0.85 | 0.0075 | -0.036 |

8 | 1.08 | 0.0092 | -0.036 |

10 | 1.26 | 0.0115 | -0.034 |

12 | 1.43 | 0.015 | -0.030 |

14 | 1.56 | 0.0186 | -0.025 |

(a) Lift curve slope = = = 0.0949 per degree, as shown on plot below. The slope is obtained using a least-squares linear fit.

(b) Zero-lift angle of attack = = -2.75 degrees, as also shown on the same plot.

(c) Drag polar (as a plot), as per the plot shown below

(d) The best lift-to-drag ratio = , which is about 115 in this case and not untypical for a two-dimensional airfoil.

(e) Center of pressure location, as shown on the plot below.

Worked Example #5

For the flap up case then estimate the following values for a Reynolds number of (you may also want to annotate the graph):

(a) The zero-lift angle of attack.

(b) The maximum lift coefficient.

(c) The stall angle of attack.

(d) The minimum drag coefficient. Also, why is there a “bucket” in the drag curve?

(e) The lift-to-drag ratio at an angle of attack of 8 degrees.

(f) The minimum drag coefficient with roughness. Explain why the drag of the airfoil increases with the application of roughness.

(a) The zero-lift angle of attack. Answer: -1.5.

(b) The maximum lift coefficient. Answer: 1.45.

(c) The stall angle of attack. Answer: 15

(d) The minimum drag coefficient. Answer: 0.003. There is a “bucket” in the drag curve because this airfoil experiences extended regions of laminar boundary layer flow between certain (low) angles of attack. Such characteristics are not uncommon for certain classes of airfoil sections, especially those used on sailplanes and the like.

(e) The lift-to-drag ratio at an angle of attack of 8 degrees. Answer: 92

(f) The minimum drag coefficient with roughness. Answer: About 0.009. Surface roughness eliminates entirely the run of the laminar boundary layer over the front part of the airfoil and making it turbulent, so increasing skin friction drag. You can think of roughness as equivalent to using some medium grade sandpaper on the surface. Airfoils and wings for airplanes are often tested in the wind tunnel with smooth and rough surfaces, the idea being to simulate the effects of wear and tear on the wing after the airplane has been in operational service. Some airfoils, like the one given in this example, are particularly sensitive to the effects of surface roughness.

Worked Example #6

The camberline for the NACA 3-digit 231-series reflex airfoils is given by

and

where , , , and .

Listed below is a MATLAB code that will draw a NACA 231-series reflexed airfoil. Remember that the purpose of reflex camber is to reduce the pitching moment on the airfoil.

m = 0.217; %location of maximum camber

k1 = 15.793; %constant

k2k1 = 0.006770;

r = 1.1019.*(t^2); %radius of leading edge circle

x1 = linspace(r/3,m,round(m.*500)); %x coordinates nose cicle to m

x2 = linspace(m,1,round((1-m).*500)); %x coordinates m to 1

y_cam_1 = (k1./6).*(((x1-m).^3)-k2k1*((1-m).^3)*x1-(m.^3)*x1+m.^3); %camber line y coord 0 to m

y_cam_2 = (k1./6).*((k2k1*(x2-m).^3)-((k2k1*(1-m).^3)*x2)-((m.^3)*x2)+(m.^3)); %camber line y coord m to 1

x = [x1 x2]; %merged x coordinates

y_cam = [y_cam_1 y_cam_2]; %merged y camber coordinates

dy_cam_1 = (k1./6).*((3.*(x1-m).^2)-k2k1*(1-m).^3-(m.^3)); %derivative of camber line 0 to m

dy_cam_2 = (k1./6).*((3.*k2k1*(x2-m).^2)-k2k1.*(1-m).^3-m.^3).*ones(1,length(x2)); %derivative of camber line m to 1

dy_cam = [dy_cam_1 dy_cam_2]; %merged derivative of camber line

theta = atan(dy_cam); %slope of camber line

y_t = 5.*t.*((0.29690.*sqrt(x))-(0.12600.*x)-(0.35160.*(x.^2)) +(0.28430.*(x.^3))-(0.10150.*(x.^4))); %thickness equation

x_upper = x-(y_t.*sin(theta)); %x coordinates of upper surface

x_lower = x+(y_t.*sin(theta)); %x coordinates of lower surface

y_upper = y_cam+(y_t.*cos(theta)); %y coordinates of upper surface

y_lower = y_cam-(y_t.*cos(theta)); %y coordinates of lower surface

%end points to close off trailing edge

x_end_up = x_upper(end);

x_end_low = x_lower(end);

y_end_up = y_upper(end);

y_end_low = y_lower(end);

dy_cam_005 = (1./6).*k1.*(3.*(0.005-m).^2)-k2k1*((1-m).^3-(m.^3)); %derivative of camber line at x = 0.005

theta_005 = atan(dy_cam_005); %slope of camber line at x = 0.005The shapes of the NACA 23112 and NACA 23118 are shown below.

Worked Example #7

Angle of attack | C_l | C_d | C_m |
---|---|---|---|

0 | 0 | 0.00662 | 0 |

1 | 0.1096 | 0.0067 | 0.0006 |

2 | 0.2182 | 0.00693 | 0.0013 |

3 | 0.3254 | 0.00736 | 0.0024 |

4 | 0.4309 | 0.008 | 0.0038 |

5 | 0.5365 | 0.00881 | 0.0054 |

6 | 0.6509 | 0.00976 | 0.0057 |

7 | 0.7743 | 0.01085 | 0.0057 |

8 | 0.9006 | 0.01203 | 0.0041 |

9 | 0.9957 | 0.01328 | 0.0046 |

10 | 1.0836 | 0.01466 | 0.0046 |

11 | 1.1729 | 0.01627 | 0.0079 |

12 | 1.2585 | 0.01817 | 0.0113 |

13 | 1.3343 | 0.02057 | 0.0157 |

14 | 1.3928 | 0.02328 | 0.0221 |

15 | 1.4322 | 0.02739 | 0.0284 |

16 | 1.4511 | 0.0345 | 0.0315 |

17 | 1.4508 | 0.04615 | 0.0287 |

18 | 1.4004 | 0.06732 | 0.0186 |

19 | 1.2739 | 0.10324 | 0.0001 |

(a) Lift curve slope (in the attached flow regime). The lift-curve-slope is obtained by fitting a straight line (least-squares fit) through the measurements at low angle of attack. In this case or = 0.107 per degree angle of attack.

(b) Zero-lift angle of attack. This is a symmetric airfoil, so the zero lift angle of attack based on the linear fit and so effectively zero degrees as expected.

(c) Drag polar (as a plot). The drag polar is a plot of the lift coefficient versus the drag coefficient . The advantage of this presentation is that a straight line running from the origin of the graph at (0,0) to any point on the polar is the lift-to-drag ratio . The best lift-to-drag ratio is when this line is just tangent to the polar curve.

(d) The best lift-to-drag ratio. Another way finding the lift-to-drag ratio is to plot this ratio as a function of angle of attack or . In this case, the best lift-to-drag ratio is about 75, which is typical for a good airfoil section.

(e) Center of pressure location (as a plot). The center of pressure can be calculated using

The moments are given in the data file about the 1/4-chord. The center of pressure is a function of lift coefficient (and hence also the angle of attack), so it is not a fixed point and it is not a convenient concept to use in aerodynamics. Hence, the use of the center of pressure to resolve the forces and moments is not used much in practice even though the pitching moment is zero about this point.

(f) Aerodynamic center location. By definition, the aerodynamic center is that point about which the moment is independent of . The moments in the data file are given about the 1/4-chord (generally this is the default), so the aerodynamic center is calculated using

We can obtain the value of using

so we just need the lift-curve-slope and the slope of the moment curve (again in the low angle of attack regime). In this case we have

which, unlike the center of pressure, is a fixed point. For most airfoils at low Mach numbers the aerodynamic center is close to 1/4-chord.

Worked Example #8

(a) We can split the area calculation of each wing panel into two parts, i.e., , with the first part ending at a distance = 3 m from the wing root and the second part running to the wing tip. For the first (rectangular) part of the wing panel then

remembering to multiply by 2 because there are two wing panels. For the second (tapered) part then

again with the factor of 2 denoting two wing panels. Therefore, the total wing area is

(b) The aspect ratio of this wing is

(c) The MAC is given by

In this case, we have two parts to the wing panel, first part with constant chord ending at a distance = 3 m from the wing root. We need to evaluate the term

The first term is easy because so

The second term involves the tapered part of the wing, so for the chord variation is

If we let

then

and if we let

then

Therefore we have that

Introducing the limits of integration and evaluating all of the terms gives

and so the MAC is

Worked Example #9

For this wing then calculate the wing span and the wing area.

The area of the wing is

The chord for this wing is

and substituting for the chord distribution gives

noting that .

The aspect ratio of this wing is

Therefore, solving for the span for an aspect ratio of 12 gives

and so the area of this wing is

Worked Example #10

We are given the span of the wing (distance from the left wing tip to the right wing tip) , the root chord and the tip chord . The wing panel is tapered linearly, i.e., each of the left and right wing panel of a trapezoidal shape so the area of one wing panel will be half of the average chord times the semi-span . Therefore, with the two wing panels then the area is

Notice that we can also work with general expression for the chord as a function of distance from the wing root to the right wing tip (a distance equal to the semi-span ) will be

where is a measure of the taper. The boundary conditions are that and . Therefore,

so solving for gives

and so

We can check that this expression is indeed correct by substituting and . The area of the wing will then be

and substituting for the chord distribution gives

Therefore,

i.e., just standard formula for the area of a trapezoid that we have used previously.

The aspect ratio of this wing is

The MAC is given by

In this case then

so

Integrating gives

so

Therefore, the MAC for this wing is

Worked Example #11

A wing of span 30 ft has an elliptical wing planform with a root chord of 6 ft. Calculate for this wing: (a) The wing (planform) area; (b) The aspect ratio of this wing; (c) The mean aerodynamic chord (MAC).

(a) The chord in this case for an elliptical planform shape is

where is the root chord. The area of the wing is

and so substituting for the chord in this case gives

This is a standard integral so

Therefore, the area of the elliptical wing is

(b) The aspect ratio for this wing is

(c) The MAC is given by

In this case then

so

This means then that for this elliptical wing then

Worked Example #12

where is the aspect ratio of the wing. Plot the drag polars for this airplane design for wing aspect ratios of 5, 10, 15 and 20. In each case, determine the best lift-to-drag ratio and the conditions under which this occurs. Assume that the maximum attainable lift coefficient of the wing is 1.7. Comment on your results.

The MATLAB code is listed below:

figure

axis([0.0 0.3 0.0 2.0])

CL = linspace(0.0,1.7,100); %set the range of CL values for AR = 5:5:20 % set the range of AR values

CD = 0.035+1.35*CL.^2/pi/AR;

plot(CD,CL);hold on

end

legend(‘AR=5′,’AR=10′,’AR=15′,’AR=20’);

xlabel(‘C_D’)

ylabel(‘C_L’)

The plot below shows the polars produced by the MATLAB code. Notice the significant effects of aspect ratio on the curve. The lower aspect ratio wing always gives higher induced drag (i.e., “drag due to lift”), so the drag increases more quickly with increasing lift coefficient.

We are asked to find the best lift-to-drag ratio for each wing and also the lift coefficient at which this occurs. It is possible to do that from the graph or numerically within MATLAB, but it can also be done analytically. The drag coefficient is given by

This means that

Differentiating gives

Therefore, solving for at this condition gives

and this will be the lift coefficient to obtain the maximum lift-to-drag ratio.

For then the for best is

The corresponding ratio is

and substituting values gives and a of 0.638 gives the best lift-to-drag ratio for this wing as

For then the for best is

and substituting values with = 0.903 gives the best lift-to-drag ratio for this wing as

For then the for best is

and substituting values with = 1.105 gives the best lift-to-drag ratio for this wing as

For then the for best is

and substituting values with = 1.276 gives the best lift-to-drag ratio for this wing as

The foregoing results clearly show that the best lift-to-drag ratio increases with increasing aspect ratio. Note also that the lift coefficient at which the best lift-to-drag ratio is obtained also increases with increasing aspect ratio.

Worked Example #13

The density of air at a density altitude of 3 km is 0.90925 kg m when using the ISA model. For vertical force equilibrium, the lift on the wing is equal to the weight of the aircraft, , i.e.,

The lift is given by

so the operating lift coefficient of the wing is

The drag coefficient is given

where the non-lifting part , i.e., .

Therefore, the drag force is

giving a lift-to-drag ratio of about 6, which seems reasonable.

Worked Example #14

Consider an aircraft that has a wing with a lifting planform area 60 m, an aspect ratio 12, and an Oswald’s efficiency factor 0.90. The wing has a non-lifting profile drag coefficient of 0.01. The remainder of the aircraft has a non-lifting drag coefficient of 0.03. All force coefficients are based on wing area . The mass of the airplane is 16,000 kg. If the aircraft is flying at a density altitude of 10,000 ft and its true airspeed is 253 kts, then calculate:

(a) The lift force produced by the wing.

(b) The lift coefficient of the wing.

(c) The drag force on the wing.

(d) The lift-to-drag ratio of the wing.

(e) The total drag force on the aircraft.

(f) The lift-to-drag ratio of the aircraft.

(a) With the assumption that the aircraft is flying along in steady unaccelerated flight, the lift force produced by the wing will equal the weight of the aircraft, i.e.,

(b) The lift coefficient of the wing is given by

To find , therefore, we need the density of the air (in this case at 10,000 ft) and the true airspeed in units of m/s. The density can be found from the ISA (assume standard temperature) so

and converting from nautical miles per hour (kts) to m/s (the conversion factor is on the formula sheet) gives

Inserting the numbers gives

(c) The drag force on the wing will be given by

where is the drag coefficient of the wing, which will be given by

where the second part is the induced drag (i.e., drag due to lift). Inserting the numbers gives

Therefore, the drag force on the wing is

(d) Now the lift and drag on the wing are known then the lift-to-drag ratio of the wing is

(e) The total drag force on the aircraft is

where is the net drag coefficient of the aircraft, which will be given by

where we are given that for the remainder of the aircraft then (the non-lifting part) and the second part will be from the wing (which has already been calculated). Notice that all drag coefficients are defined using wing area as a reference so the drag coefficients can be added. For the entire aircraft then

Therefore, the drag force on the aircraft is

(f) Now the lift and drag are known then the lift-to-drag ratio of the entire aircraft is

Worked Example #15

A giant flying-wing drone aircraft, as shown in the figure below, is cruising at in-flight weight of 684,000 lb at a Mach number at an altitude of 30,000 ft. The aircraft’s drag polar is given by . The wing has a span of 235.0 ft, with a root chord of 35.0 ft and a tip chord of 12.25 ft. Assume the following ambient atmospheric conditions: 629.6 lb ft, F, and 0.0008903 slug ft.

- Explain the balance of forces acting on the aircraft.
- Calculate the planform area and aspect ratio of the wing.
- Calculate the airspeed of the aircraft in knots and ft s.
- Determine the operating lift coefficient of the wing.
- Determine the aircraft’s total drag coefficient at its current flight condition.
- What is the zero-lift drag coefficient at these flight conditions?
- Determine the induced drag coefficient at these flight conditions.
- Calculate the Oswald’s efficiency factor for this wing.
- Calculate the total drag force acting on the aircraft and its lift-to-drag ratio .
- If the TSFC of each of the four jet engines is 0.5~lb lb hr, approximately how much fuel will be burned by the aircraft in 30 minutes of flying time?

and for horizontal force equilibrium

The corresponding aspect ratio of the wing is is

and putting in the numbers gives

Because 1 kts (nautical miles per hour) = 1.688 ft/s then

so = 0.025. Substituting gives for the total drag coefficient

Therefore, the minimum (non-lifting) drag coefficient is

7. The induced (lifting) drag coefficient is

8. The induced drag coefficient is also given by

so solving for the Oswald’s efficiency factor gives

and using the give numerical values gives

9. The total drag is then

and putting in the numbers gives

So, the lift-to-drag ratio is

10. The fuel flow rate is

where . Inserting the numbers gives

so during 30 minutes of flight then the approximate weight of fuel burned is

Worked Example #16

The stall airspeed can be estimated using

(1)

We are told that = 3.2 and that = 76.84 lb ft. At 5,000 ft on a standard day we can use the ISA model to determine that = 0.002048 slugs ft. Substituting gives

(2)

which is relatively low and shows the importance of getting a high value of on a wing.