Examples – Airfoils & Wings

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

Worked Example #1

Assuming a lift curve slope for the linear range is 0.11 per degree angle of attack, i.e., C_{l_{\alpha}} = 0.11 per deg., then calculate the lift coefficients at 2, 6 and 10 deg. angle of attack for: (a) A symmetric airfoil; (b) A positively cambered airfoil with a zero lift angle of attack of -1.2 deg; (c) A reflexed airfoil with a zero lift angle of attack of 1.7 deg.

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

    \[ C_l = C_{l_{\alpha}} \left( \alpha - \alpha_0 \right) \]

In this case, C_l = C_{l_{\alpha}} \left( \alpha - \alpha_0 \right) = 0.11\left( \alpha - 0 \right) = 0.11 \alpha.

For 2^{\circ}, then C_l = 0.22
For 6^{\circ}, then C_l =  0.66
For 10^{\circ}, then C_l = 1.1


(b) C_l = C_{l_{\alpha}} \left( \alpha - \alpha_0 \right) = 0.11\left( \alpha - (-1.2)\right) = 0.11\left( \alpha + 1.2 \right)

For 2^{\circ}, then C_l = 0.352
For 6^{\circ}, then C_l = 0.792
For 10^{\circ}, then C_l =  1.232


(c) C_l = C_{l_{\alpha}} \left( \alpha - \alpha_0 \right) = 0.11\left( \alpha - 1.7 \right)

For 2^{\circ}, then C_l = 0.033
For 6^{\circ}, then C_l = 0.473
For 10^{\circ}, then C_l = 0.913

Worked Example #2

Given that the pressure coefficient C_p distribution over a 2-dimensional body of chord c is described by C_p = -4(1 - x/c) on the upper surface of the body and C_p = (1-x/c) on the lower surface of the body, then calculate (by integration): (a) The lift coefficient acting on the body; (b) The pitching moment coefficient on the body about its leading edge.

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


(a) The lift coefficient C_l is given by

    \[ C_l = \int_0^1 \Delta C_p d(x/c) =  \int_0^1 (C_{p_{l}} - C_{p_{u}}) d \overline{x} \]

In this case we can see that C_{p_{l}} - C_{p_{u}} = 5 (1 - \overline{x}) so the integral for C_l becomes

    \[ C_l = \int_0^1 5 (1 - \overline{x}) d\overline{x} = 5 \left[ \overline{x} - \frac{\overline{x}^2}{2} \right]_0^1 = 2.5 \]


(b) The pitching moment about the leading edge is

    \[ C_{m_{le}} = -\int_0^1 \Delta C_p \, (x/c) \, d(x/c) =  -\int_0^1 \Delta C_p \, \overline{x} \, d\overline{x} \]

Substituting \Delta C_p= 5 (1 - \overline{x}) then

    \[ C_{m_{le}} =  -\int_0^1  5 (1 - \overline{x}) \overline{x} \, d\overline{x} = 5 \left[ \frac{\overline{x}^2}{2} - \frac{\overline{x}^3}{3} \right]_0^1 = -\frac{5}{6} = -0.833 \]


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

clc
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

The pressure coefficient C_p distribution over a 2-dimensional body of chord c operating at low angle of attack is described by

    \[ C_{p_{u}} = 1 - 300 \left( \frac{x}{c} \right)^2 \quad \mbox{for $0 \le x/c \le 0.1$} \]

    \[ C_{p_{u}} = -2.2277 + 2.2777 \left( \frac{x}{c} \right) \quad \mbox{for $0.1 < x/c \le 1.0$} \]

    \[ C_{p_{l}} = 1 - 0.95 \left( \frac{x}{c} \right) \mbox{for $0.0 \le x/c \le 1.0$} \]

(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):

(b) The lift coefficient C_l is given by

    \[ C_l = \int_0^1 \Delta C_p \, d(x/c) =  \int_0^1 (C_{p_{l}} - C_{p_{u}}) \, d \overline{x} \]

which for this problem we can split into two integrals namely

    \[ C_l =  \int_0^{0.1} (C_{p_{l}} - C_{p_{u}}) d \overline{x} +  \int_{0.1}^1 (C_{p_{l}} - C_{p_{u}}) d \overline{x} = C_{l_{1}} + C_{l_{2}} \]

Taking the first integral gives

    \[ C_{l_{1}} = \int_0^{0.1} (C_{p_{l}} - C_{p_{u}}) d \overline{x} = \int_0^{0.1} \big( \left( 1 - 0.95 \overline{x} \right) - \left( 1 - 300 \overline{x} ^2  \right) \big) d \overline{x} \]

and so

    \[ C_{l_{1}} =  \int_0^{0.1} \left( 300 \overline{x} ^2  - 0.95 \overline{x}  \right) d \overline{x}  = \Bigg[ 100 \overline{x}^3 - 0.475 \overline{x}^2 \Bigg]_{0}^{0.1} = 0.09525 \]

Taking the second integral gives

    \[ C_{l_{2}} = \int_{0.1}^{1} (C_{p_{l}} - C_{p_{u}}) d \overline{x} = \int_{0.1}^{1} \big( \left( 1 - 0.95 \overline{x} \right) - \left(  -2.2277 + 2.2777 \overline{x} \right) \big) d \overline{x} \]

and so

    \[ C_{l_{2}} =  \int_{0.1}^{1} 3.2277 \left(  1-  \overline{x} \right) d \overline{x}  = \Bigg[  3.2277 \left( \overline{x} - \frac{\overline{x}^2}{2} \right) \Bigg]_{0.1}^{1} = 1.307 \]

Therefore,

    \[ C_{l} = C_{l_{1}} + C_{l_{2}} = 0.09525 + 1.307 = 1.4025 \]


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

clc
figure
axis([0.0 1.0 1.0 5.0])
x1 = linspace(0,0.1,500);
x2 = linspace(0.1,1,500);
cpu_1 = 1.-300*x1.^2;
cpu_2 = -2.2277 + 2.2777.*x2;
x = [x1 x2];
cpu = [cpu_1 cpu_2];
cpl = (1.0-0.95*x);
dcp = cpl-cpu;
trapz(x,dcp) % section cl 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 #4

The lift, drag and pitching moment coefficient measurements for a NACA 2412 airfoil as shown in the table below are to be used to calculate certain derived aerodynamic quantities. 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).

\alpha (deg.) C_l C_d C_{m_{1/4}}
-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 1.34 0.015 -0.030

 

(a) Lift curve slope = \displaystyle{\frac{d C_l }{d \alpha}} = C_{l_{\alpha}} = 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 = \alpha_0 = -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 = (C_l/C_d)_{\rm max}, 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

Examine the attached graph, which shows the aerodynamic coefficients for a NACA 66_1-212 airfoil section.
For  the flap up case then estimate the following values for a Reynolds number of 6 \times 10^6 (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^{\circ}.


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


(c) The stall angle of attack. Answer: 15^{\circ}


(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

Modify the MatLab NACA 230-series camberline airfoil generator code given in the e-book plot some of the NACA 231 series reflexed airfoils. Why is reflex camber used on an airfoil?  You should plot the NACA 23112 and NACA 23118 shapes, but you may plot additional shapes in the series if you wish.

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

    \[ \frac{y_{c}}{c} = \frac{k_1}{6} \left[ \left( \frac{x}{c} - m \right)^3 - \frac{k_2}{k_1} \left( 1 - m\right)^3 \, \frac{x}{c} - m^3 \, \frac{x}{c} + m^3 \right] ~\mbox{from $x/c \le m$} \]

and

    \[ \frac{y_{c}}{c} = \frac{k_1}{6} \left[ \frac{k_2}{k_1} \left( \frac{x}{c} - m \right)^3 - \frac{k_2}{k_1} \left( 1 - m \right)^3 \, \frac{x}{c} - m^3 \, \frac{x}{c} + m^3 \right]~\mbox{from $m< x/c \le 1.0$} \]

where m = 0.217, p = 0.15, k_1 =15.793, and k_2/k_1 = 0.00677.

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.

t = 0.12;
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

Lift, drag and pitching moment measurements for a NACA 0012 airfoil are given in the table below. They are to be used to calculate certain derived aerodynamic quantities for further analysis. As part of the analysis you are asked to determine the following: (a) Lift curve slope (in the attached flow regime). (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). (f) Aerodynamic center location.
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 dC_l/d\alpha or C_{l_{\alpha}} = 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 \alpha_0 = -0.02 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 c_l versus the drag coefficient C_d. 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 C_l/C_d. 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 \alpha or C_l. 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

    \[ \frac{x_{cp}}{c} = \frac{1}{4} - \frac{C_{m_{1/4}}}{C_{l}} \]

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 C_{l}. 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

    \[ \frac{x_{ac}}{c} = \frac{1}{4} - \frac{d C_{m_{1/4}}}{dC_{l}} \]

We can obtain the value of dC_{m_{1/4}}/d C_l using

    \[ \frac{d C_{m_{1/4}} }{ d C_l} = \left( \frac{ d C_{m_{1/4}}}{d \alpha}\right) \left( \frac{d \alpha}{d C_{l}}\right) \]

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

    \[ \frac{x_{ac}}{c} = \frac{1}{4} - \frac{0.000687}{0.107} = 0.2436 \]

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

An airplane wing has a semi-span of 9.5m. The root chord of the wing at the centerline of the aircraft is 1.5 m. This constant chord extends to 3 m from the root at the centerline, followed by a linearly tapered part from that point to a tip chord of 1.2 m.  For this wing then calculate: (a) The wing area; (b) The aspect ratio; (c) The mean aerodynamic chord (MAC).

(a) We can split the area calculation of each wing panel into two parts, i.e., S = S_1 + S_2, with the first part ending at a distance y = y_b = 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

    \[ S_1 = 2 (c_r y_b) = 2 \times (1.5 \times 3) = 9.0~\mbox{m$^2$} \]

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

    \[ S_2 = 2 \left( \frac{c_r + c_t}{2} \right) \left( s - y_b \right) = 2 \left( \frac{1.5 + 1.2}{2} \right) \left( 9.5 - 3.0 \right) = 17.55~\mbox{m$^2$} \]

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

    \[ S = S_1 + S_2 = 9.0 + 17.55 = 26.55~\mbox{m$^2$} \]


(b) The aspect ratio A\!R of this wing is

    \[ A\!R = \frac{b^2}{S} =  \frac{4s^2}{S} = \frac{4 \times 9.5^2}{26.55} = 13.6 \]


(c) The MAC is given by

    \[ {\rm MAC} \equiv \overline{\overline{c}} = \frac{2 \displaystyle{\int_{0}^{s} c^2 dy}}{S} \]

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

    \[ \int_{0}^{s} c^2 dy = \int_{0}^{y_b} c^2 dy + \int_{y_b}^{s} c^2 dy \]

The first term is easy because c = c_r so

    \[ \int_{0}^{y_b} c^2 dy = \int_{0}^{y_b} c_r^2 dy = c_r^2 y_b = (1.5)^2 \, 3.0 = 6.75~\mbox{m$^3$} \]

The second term involves the tapered part of the wing, so for y \ge y_b the chord variation is

    \[ c(y) = c_r + \left( \frac{c_t - c_r}{s-y_b} \right) ( y - y_b ) \]

If we let

    \[ A = \left( \frac{c_t - c_r}{s-y_b} \right) \]

then

    \[ c^2  = \left( c_r + A ( y - y_b ) \right)^2 =\left( (c_r - A y_b) + Ay \right)^2 = (c_r - A y_b)^2 + 2 A (c_r - Ay_b) y + A^2 y^2 \]

and if we let

    \[ B = c_r - A y_b \]

then

    \[ c^2  = B^2 + 2 AB y + A^2 y^2 \]

Therefore we have that

    \[ \int_{y_b}^{s} c^2 dy = \int_{y_b}^{s} \left( B^2 + 2 AB y + A^2 y^2 \right) dy = \left[ B^2 y + ABy^2 + \frac{A^2}{3} y^3 \right]_{y_b}^{s} \]

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

    \[ \int_{y_b}^{s} c^2 dy = 11.9~\mbox{m$^3$} \]

and so the MAC is

    \[ \overline{\overline{c}}= \frac{2 \displaystyle{\int_{0}^{s} c^2 dy}}{S} = \frac{ \displaystyle{ 2 \int_{0}^{y_b} c^2 dy + 2 \int_{y_b}^{s} c^2 dy}}{S} = \frac{2 (6.75 + 11.9)}{26.55} = 1.405~\mbox{m} \]

Worked Example #9

A wing has an aspect ratio of 12. The chord of the wing varies smoothly and continuously outward from the centerline of the aircraft according to

    \[ c(y) = 1 - 0.32 \left( \frac{y}{b} \right)~~~~\mbox{in units of meters} \]

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

The area of the wing S is

    \[ S = 2  \int_{0}^{s} c  \, dy \]

The chord for this wing is

    \[ c(y) = 1 - 0.32 \left( \frac{y}{b} \right) = 1 - 0.32 \left( \frac{y}{2s} \right) = 1 - 0.16 \left( \frac{y}{s} \right) \]

and substituting for the chord distribution c(y) gives

    \[ S = 2 \int_{0}^{s} c \, dy = 2 \int_{0}^{s} \left( 1.0 - 0.16 \left( \frac{y}{s} \right) \right) dy = 2 \left[ 1.0 y - \frac{0.08}{s} y^2 \right]_0^s = 2(s - 0.08s) = 1.84s \]

noting that b = 2 s.


The aspect ratio A\!R of this wing is

    \[ A\!R = \frac{b^2}{S} = \frac{4s^2}{S} = \frac{4s^2}{1.82s} = 2.174 s \]

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

    \[ b = 2s = 2 \frac{A\!R}{2.174}  = \frac{2 \times 12}{2.174} = 11.04~\mbox{m} \]

and so the area of this wing is

    \[ S = 1.84 s = 1.84/b/2= 1.84 \times 5.52= 10.157~\mbox{m$^2$} \]

Worked Example #10

A sailplane wing has a span of 15 m. The root chord at the centerline of the aircraft is 1 m and the wing tapers linearly from the root to the tip with a tip chord of 0.5 m. For this wing then calculate: (a) The wing area; (b) The aspect ratio; (c) The mean aerodynamic chord (MAC).

We are given the span of the wing (distance from the left wing tip to the right wing tip) b, the root chord c_r and the tip chord c_t. 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 s. Therefore, with the two wing panels then the area S is

    \[ S = 2 \left( \frac{c_t + c_r}{2} \right) s = 2 \left( \frac{1.0 + 0.5}{2} \right) \frac{15}{2} = 11.25~\mbox{m$^2$} \]

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

    \[ c(y) = c_r + A y \]

where A is a measure of the taper. The boundary conditions are that c(0) = c_r and c(s) = c_t. Therefore,

    \[ c_t = c_r + A s \]

so solving for A gives

    \[ A = \frac{c_t - c_r}{s} \]

and so

    \[ c(y) = c_r + \left( \frac{c_t - c_r}{s} \right) y \]

We can check that this expression is indeed correct by substituting y = 0 and y = s. The area of the wing S will then be

    \[ S = 2  \int_{0}^{s} c dy \]

and substituting for the chord distribution c(y) gives

    \[ S = 2 \int_{0}^{s} c dy = 2 \int_{0}^{s} \left( c_r + A y\right) dy \]

Therefore,

    \[ S = 2 \left[ c_r y + \frac{A y^2}{2} \right]_0^s = c_r s + \frac{As^2}{2} = c_r s + \left(\frac{c_t - c_r}{s} \right) \frac{s^2}{2} = \left( \frac{c_r + c_t}{2} \right) s \]

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


The aspect ratio A\!R of this wing is

    \[ A\!R = \frac{b^2}{S} = \frac{15^2}{11.25} = 20 \]


The MAC is given by

    \[ {\rm MAC} =  \overline{\overline{c}} = \frac{2 \displaystyle{\int_{0}^{s} c^2 dy}}{S} \]

In this case then

    \[ c(y) = c_r + \left( \frac{c_t - c_r}{s} \right) y \]

so

    \[ \int_{0}^{s} c^2 dy = \int_{0}^{s} \left( c_r + \left( \frac{c_t - c_r}{s} \right) y \right)^2 dy = \int_{0}^{s} \left( c_r^2 + 2\left( \frac{c_t - c_r}{s} \right) c_r y + \left( \frac{c_t - c_r}{s} \right)^2 y^2 \right) dy \]

Integrating gives

    \[ \int_{0}^{s} c^2 dy  = \left[ c_r^2 y +  \left( \frac{c_t - c_r}{s} \right) c_r y^2 + \frac{1}{3} \left( \frac{c_t - c_r}{s} \right)^2 y^3  \right]_0^s \]

so

    \[ \overline{\overline{c}} = c_r^2 s +  \left(c_t - c_r \right) c_r s + \frac{1}{3} \left( c_t - c_r \right)^2 s = c_t c_r s  + \frac{1}{3} \left( c_t - c_r \right)^2 s \]

Therefore, the MAC for this wing is

    \[ \overline{\overline{c}} = \frac{2 \left( (0.5 \times 1.0\times 7.5)  + \displaystyle{\frac{(-0.5)^2}{3}} 7.5) \right) }{11.25} = 0.778~\mbox{m} \]

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

    \[ c(y) = c_0\sqrt{ 1 - \left( \frac{y}{s} \right)^2 } \]

where c_0 is the root chord. The area of the wing S is

    \[ S = 2  \int_{0}^{s} c  \, dy \]

and so substituting for the chord in this case gives

    \[ S = 2  \int_{0}^{s} c_0\sqrt{ 1 - \left( \frac{y}{s} \right)^2 }  \, dy  = 2  c_0 \int_{0}^{s} \sqrt{ 1 - \left( \frac{y}{s} \right)^2 }  \, dy \]

This is a standard integral so

    \[ S = 2 c_0 \left[ \frac{1}{2} \left( y \sqrt{1 - \frac{y^2}{s^2} } + s \sin^{-1} \left( \frac{y}{s} \right) \right) \right]_0^s = c_0  \left[ y \sqrt{1 - \frac{y^2}{s^2} } + s \sin^{-1} \left( \frac{y}{s} \right) \right]_0^s = \frac{\pi c_0 s}{2} \]

Therefore, the area of the elliptical wing is

    \[ S = \frac{\pi c_0 s}{2} = \frac{ \pi \times 6.0 \times 15.0}{2} = 141.372~\mbox{ft$^2$} \]


(b) The aspect ratio A\!R for this wing is

    \[ A\!R = \frac{b^2}{S} = \frac{4s^2}{S} = \frac{4 \times 15.0^2}{141.372} = 6.366 \]


(c) The MAC is given by

    \[ {\rm MAC} =  \overline{\overline{c}} = \frac{2 \displaystyle{\int_{0}^{s} c^2 dy}}{S} \]

In this case then

    \[ c(y) = c_0\sqrt{ 1 - \left( \frac{y}{s} \right)^2 } \]

so

    \[ c(y)^2 = c_0^2 \left( 1 - \left( \frac{y}{s} \right)^2 \right) \]

This means then that for this elliptical wing then

    \[ \int_{0}^{s} c^2 dy = \int_{0}^{s} c_0^2 \left( 1 - \left( \frac{y}{s} \right)^2 \right) dy = c_0^2 \int_{0}^{s}  \left( 1 - \left( \frac{y}{s} \right)^2 \right) dy = c_0^2 \left[ y - \frac{y^3 }{3s^2}\right]_0^s = \frac{2}{3} c_0^2 s \]

    \[ \overline{\overline{c}} = \frac{2 \displaystyle{\int_{0}^{s} c^2 dy}}{S} = \frac{ \left( \displaystyle{\frac{4}{3}} \right)c_0^2 s}{S} =  \frac{ \left( \displaystyle{\frac{4}{3}} \right)6.0^2 \, 15.0}{141.372} = 5.093~\mbox{m} \]

Worked Example #12

The drag coefficient C_D of a particular airplane is described by the equation

    \[ C_D = 0.035 + \left( \frac{1.35}{\pi AR} \right) {C_L}^2 \]

where AR 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:

clc
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 AR 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

    \[ C_D = 0.035 + \left( \frac{1.35}{\pi AR} \right) {C_L}^2 = A + B {C_L}^2 \]

This means that

    \[ \frac{C_D}{C_L} = \frac{A}{C_L} + B C_L \]

Differentiating gives

    \[ \frac{d (C_D/C_L)}{d (C_L)} = -\frac{A}{{C_L}^2} + B = 0 \mbox{~~for a maximum or minimum} \]

Therefore, solving for C_L at this condition gives

    \[ C_L = \sqrt{ \frac{A}{B} } = \sqrt{ \frac{0.035 \pi AR}{1.35} } \]

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


For AR = 5 then the C_L for best C_L/C_D is

    \[ C_L = \sqrt{ \frac{0.035 \pi AR}{1.35} } = \sqrt{ \frac{0.035 \times \pi \times 5 }{1.35} } = 0.638 \]

The corresponding C_L/C_D ratio is

    \[ \frac{C_L}{C_D} = \frac{C_L}{C_D} = \frac{C_L}{A + B {C_L}^2} = \frac{C_L}{0.035 + (1.35/\pi/AR){C_L}^2} \]

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

    \[ \frac{C_L}{C_D} = \frac{C_L}{0.035 + (1.35/\pi/AR){C_L}^2} = \frac{0.638} {0.035 + (1.35/\pi/5)0.638^2} = 9.12 \]


For AR = 10 then the C_L for best C_L/C_D is

    \[ C_L = \sqrt{ \frac{0.035 \pi AR}{1.35} } = \sqrt{ \frac{0.035 \times \pi \times  10 }{1.35} } = 0.903 \]

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

    \[ \frac{C_L}{C_D} = \frac{C_L}{0.035 + (1.35/\pi/AR){C_L}^2} = \frac{0.903}{0.035 + (1.35/\pi/10)0.903^2} = 12.89 \]


For AR = 15 then the C_L for best C_L/C_D is

    \[ C_L = \sqrt{ \frac{0.035 \pi AR}{1.35} } = \sqrt{ \frac{0.035 \times \pi \times  15 }{1.35} } = 1.105 \]

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

    \[ \frac{C_L}{C_D} = \frac{C_L}{0.035 + (1.35/\pi/AR){C_L}^2} = \frac{1.105}{0.035 + (1.35/\pi/15)1.105^2} = 15.79 \]


For AR = 20 then the C_L for best C_L/C_D is

    \[ C_L = \sqrt{ \frac{0.035 \pi AR}{1.35} } = \sqrt{ \frac{0.035 \times \pi \times 20 }{1.35} } = 1.276 \]

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

    \[ \frac{C_L}{C_D} = \frac{C_L}{0.035 + (1.35/\pi/AR){C_L}^2} = \frac{1.276}{0.035+ (1.35/\pi/20)1.276^2} = 18.23 \]

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

Consider a flying wing with a wing area of 210 m^2, an aspect ratio of 10, and an Oswald’s efficiency factor of 0.90. The airfoil section on the wing has a profile drag coefficient of 0.015. The mass of the airplane is 50,000 kg. If the aircraft is flying at a density altitude of 3 km and the true airspeed is 230m/s, then calculate the total drag on the aircraft.

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

    \[ L = M g = 50,000 \times 9.81 = 490,500 ~{\mbox N} \]

The lift is given by

    \[ L = \frac{1}{2} \rho V^2 S C_L \]

so the operating lift coefficient of the wing is

    \[ C_L = \frac{L}{ \frac{1}{2} \rho V^2 S} =  \frac{Mg }{ \frac{1}{2} \rho V^2 S} = \frac{490,500 }{0.5 \times 0.90925 \times 230.0^2 \times 210.0} = 0.097 \]

The drag coefficient is given

    \[ C_D = C_{D_{0}} + \frac{{C_L}^2}{\pi \, A\!R e} = 0.015 + \frac{0.097^2}{\pi \times 10.0 \times 0.90} = 0.0153 \]

where the non-lifting part C_{D_{0}} = 0.015, i.e., C_{D_{0}} = C_{d_{0}}.

Therefore, the drag force is

    \[ D = \frac{1}{2} \rho V^2 S C_D = 0.5 \times 0.90925 \times 230.0^2 \times 210.0 \times 0.0153 = 77,271.6~\mbox{N} \]

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 S = 60 m^2, an aspect ratio A\!R = 12, and an Oswald’s efficiency factor e = 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 S. 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.,

    \[ L = W = M \, g = 16,000 \times 9.81 = 156,960~\mbox{N} \]

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

    \[ C_L = \frac{L}{\frac{1}{2} \rho_{\infty} V_{\infty}^2 S} \]

To find C_L, 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

    \[ \rho_{\infty} = 0.90464~\mbox{kg/m$^3$} \]

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

    \[ V_{\infty} = 130.154~\mbox{m/s} \]

Inserting the numbers gives

    \[ C_L = \frac{L}{\frac{1}{2} \rho_{\infty} V_{\infty}^2 S} = \frac{156,960}{\frac{1}{2} \times 0.90464 \times 130.154^2 \times 60.0} = 0.341 \]

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

    \[ D = \frac{1}{2} \rho_{\infty} V_{\infty}^2 S \, C_D \]

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

    \[ C_{D_{\rm wing}} = 0.01 + \frac{{C_L}^2}{\pi \, A\!R \, e} \]

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

    \[ C_{D_{\rm wing}} = 0.01+ \frac{0.341^2}{\pi \times 12 \times 0.9} = 0.01 + 0.00343 = 0.01343 \]

Therefore, the drag force on the wing is

    \[ D_{\rm wing} = \frac{1}{2} \rho_{\infty} V_{\infty}^2 S \, C_D = 0.5 \times 0.9064 \times 130.154^2 \times 60.0 \times 0.01343 = 6,186.32~\mbox{N} \]

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

    \[ \frac{L}{D_{\rm wing}} = \frac{156,960}{6,186.32} = 25.37 \]

(e) The total drag force on the aircraft is

    \[ D = \frac{1}{2} \rho_{\infty} V_{\infty}^2 S \, C_D \]

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

    \[ C_D = 0.03 + C_{D_{\rm wing}} \]

where we are given that for the remainder of the aircraft then C_{D_{0}} = 0.03 (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. Therefore, for the entire aircraft then

    \[ C_D = 0.03 + 0.01343 = 0.04343 \]

Therefore, the drag force on the aircraft is

    \[ D = \frac{1}{2} \rho_{\infty} V_{\infty}^2 S \, C_D = 0.5 \times 0.9064 \times 130.154^2 \times 60.0 \times 0.04343 = 20,005.3~\mbox{N} \]

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

    \[ \frac{L}{D} = \frac{156,960}{20,005.4} = 7.85 \]

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 0.80 at an altitude of 30,000 ft. The aircraft’s drag polar is given by C_D = 0.025 + 0.040 C_L^2. 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: p_{\infty} = 629.6 lb ft^{-2}, T_{\infty} = -47.83^{\circ}F, and \rho_{\infty} = 0.0008903 slug ft^{-3}.

  1. Explain the balance of forces acting on the aircraft.
  2. Calculate the planform area S and aspect ratio AR of the wing.
  3. Calculate the airspeed of the aircraft in knots and ft s^{-1}.
  4. Determine the operating lift coefficient C_L of the wing.
  5. Determine the aircraft’s total drag coefficient C_D at its current flight condition.
  6. What is the zero-lift drag coefficient C_{D_{0}} at these flight conditions?
  7. Determine the induced drag coefficient C_{D_{i}}  at these flight conditions.
  8. Calculate the Oswald’s efficiency factor e for this wing.
  9. Calculate the total drag force acting on the aircraft D and its lift-to-drag ratio L/D.
  10. If the TSFC of each of the four jet engines is 0.5~lb lb^{-1} hr^{-1}, approximately how much fuel will be burned by the aircraft in 30 minutes of flying time?
1. For steady level flight, i.e. cruise, the aircraft is in equilibrium such that the sum of forces in the vertical direction are equal to zero and the sum of forces in the longitudinal direction are equal to zero. This means that for vertical force equilibrium then

    \[ L - W = 0 \]

or

    \[ L = W \]

and for horizontal force equilibrium

    \[ T - D = 0 \]

or

    \[ T = D \]

2. The wing is has trapezoidal panels, so the total wing area is given by

    \[ S = b \left( \frac{ c_{r} + c_{t}}{2} \right)  = 235.0 \left( \frac{35+ 12.25}{2} \right) = 5,552.0~\mbox{ft$^2$} \]

The corresponding aspect ratio of the wing is is

    \[ AR = \frac{b^2}{S} = \frac{235.0^2}{5,552} = 9.95 \]

3. The airspeed is

    \[ V_{\infty} = M \, a = M \, \sqrt{\gamma R \, T} \]

and putting in the numbers gives

    \[ V_{\infty} = 0.80 \sqrt{1.4 \times 1716.59 \times (459.67 - 47.83) } = 795.89~\mbox{ft/s} \]

Because 1~knot = 1.688 ft/s then

    \[ V_{\infty} = 471.5~\mbox{kts} \]

4. The aircraft is cruising in steady, straight and level flight, so L=W and the operating lift coefficient is

    \[ C_{L} = \frac{2W}{\rho V_{\infty}^{2} S} = \frac{2 \times 684,000}{0.0008903 \times 795.89^2 \times 5,552} = 0.437 \]

5. The drag polar is given in the question as

    \[ C_D = 0.025 + 0.040 \, C_{L}^{2} = C_{D_{0}} + K  \, C_{L}^{2} = C_{D_{0}} + C_{D_{i}} \]

so C_{D_{0}} = 0.025. Substituting C_{L} = 0.437 gives

    \[ C_D = 0.025 + 0.040 \times 0.437^{2} = 0.0326 \]

7. The total drag coefficient C_{D} can be decomposed as

    \[ C_{D} = C_{D_{0}} + C_{D_{i}} \]

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

    \[ C_{D_{0}} = 0.025 \]

and the induced (lifting) drag coefficient is

    \[ C_{D_{i}} = K \, C_{L}^{2}= 0.040 \times 0.437^2 = 0.007628 \]

8. Remember that the induced drag coefficient is also given by

    \[ C_{D_{i}} = K \, C_{L}^{2} =  \left( \frac{1}{\pi \, AR \, e} \right) C_{L}^{2} \]

so solving for the Oswald’s efficiency factor gives

    \[ e = \frac{1}{\pi \, K \, AR} \]

and using the known values gives

    \[ e = \frac{1}{\pi \times 0.040 \times 9.95} = 0.80 \]

9. The total drag is then

    \[ D = \tfrac{1}{2}\rho \, V_{\infty}^{2} \, S \, C_{D} \]

and putting in the numbers gives

    \[ D = 0.5 \times 0.0008903 \times 795.89^{2}\times 5, 552 \times 0.0326 = 51,108.22~\mbox{lb} \]

So, the lift-to-drag ratio is

    \[ \frac{L}{D} = \frac{684,000}{51,108.22} = 13.38 \]

10. The fuel flow rate is

    \[ \dot{W}_{f} = ({\rm TSFC}) \, T \]

where T = D. Inserting the numbers gives

    \[ \dot{W}_{f} = 0.50 \times 51,108.22 = 25,554.11~\mbox{lb hr$^{-1}$} \]

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

    \[ W_{f} =  \dot{W}_{f} \, t = 25,554.11 \times 0.5 = 12,777~\mbox{lb} \]

 

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