HW3 Solution

Zachary Lovering
Oct. 31, 2008
MAE 2
HW#3
Homework 3 Solution
Problem 5.23
Description:
Consider a finite wing with an area and aspect ratio of 21.5 and 5, respectively (this is comparable to
the wing on a Gates Learjet, a twin-jet executive transport). Assume the wing has a NACA 65-210 airfoil,
a span efficiency factor of 0.9, and a profile drag coefficient of 0.004. If the wing is at a 6° angle of
attack, calculate and .
Given:
Wing area, 21.5
Aspect ratio, 5
Airfoil, NACA 65-210
Span efficiency factor, 0.9
Profile drag coefficient, 0.004
Angle of attack, 6°
Find:
Lift coefficient,
Drag coefficient,
Assumptions:
1. Steady flight
2. Incompressible flow
3. Finite wing
Equations:
. / (5.65)
(Slope of a straight line)
(5.61)
(5.58)
Analysis:
The lift curve slope is given in equation 5.65 as
1 57.3 /
As noted on pg. 333, “… and are theoretically different but are in practice approximately the same
value for a given wing.” So, we set . To find , the slope of the infinite wing lift curve can be
evaluated at any two points on the linear section of the curve. This curve in found in Appendix D for the
NACA 65-210 wing section. Selecting two points on this curve, the slope is given by

Additionally, from this same curve, the angle of attack at zero lift, , can be found. With these values
known, the finite wing lift coefficient, , is given in equation 5.61 as,
1 57.3 /
The finite wing drag coefficient is given in equation 5.58 as,
Since is known and is simply the profile drag, can be found.
Numerical Substitution:
Selecting the following two points on the linear portion of the lift slope curve for the NACA 65-210
airfoil,
Next the slope is calculated to be,
Using the equation for found in the analysis,
Using the equation for found in the analysis,
1.05 at 8°
0 at 1.5°
1.05 0
8° 1.5°
1.05
9.5°
0.11
0.11
1
6° 1.5°
57.3 0.11
1
1
0.95
0.825
1.4458
0.57
0.004 0.57
0.95
0.027

Problem 5.26
Description:
Consider a light, single-engine airplane such as the Piper Super Cub. If the maximum gross weight of the
airplane is 7780 , the wing area is 16.6 , and the maximum lift coefficient is 2.1 with flaps down,
calculate the stalling speed at sea level.
Given:
Gross weight, 7780
Wing area, 16.6
Maximum lift coefficient, , 2.1
Find:
Stalling speed,
Assumptions:
1. Sea level
2. Steady level flight
Equations:
,
(5.71)
Analysis:
The stalling speed may be calculated using equation 5.71.
Numerical Substitution:
Sea level density is found in Appendix A as,
Using the equation from the analysis,
2
, 1.225
27780
1.225
16.6 2.1
15560 ·
42.7035
19.1

Problem 9.4
Description:
Consider a turbojet-powered airplane flying at a standard altitude of 40000 at a velocity of 530 /
. The turbojet engine has an inlet and exit areas of 13 and 10 , respectively. The velocity and
pressure of the exhaust gas at the exit are 1500 / and 450 / , respectively. Calculate the thrust
of the turbojet.
Given:
Standard altitude, 40000
Air speed, 530 /
Inlet Area, 13
Exit Area, 10
Exhaust velocity, 1500 /
Exhaust pressure, 450 /
Find:
Thrust,
Assumptions:
1. Steady flow
2. Conditions at the inlet are the free stream conditions
Equations:
(9.25)
Analysis:
Using the thrust equation (Eq. 9.25),
To solve for the air mass flow rate, , the continuity equation is used.
Substituting the continuity equation into the thrust equation yields,
where and are found from Appendix A at the altitude of 40000 .
Numerical Substitution:
At a standard altitude of 40000 , from Appendix A it is found that
393.12
5.8727 · 10
The velocity must be converted into consistent units,

530
777
Using the equation for thrust found in the analysis,
Note that,
5.8727 · 10
1
5280
1 3600
777
13 1500
450 393.12 10 ·
4289 567 4858
·
777

Problem 9.6
Description:
A small ramjet engine is to be designed for a maximum thrust of 1000 at sea level at a velocity of
950 /. If the exit velocity and pressure are 2000 / and 1.0 , respectively, how large should
the inlet be?
Given:
Maximum thrust, 1000
Air velocity, 950 /
Exit velocity, 2000 /
Exit pressure, 1.0
Find:
Inlet area,
Assumptions:
1. Sea level
2. Steady flow
3. Conditions at the inlet are the free stream conditions
Equations:
(9.25)
Analysis:
Using the thrust equation (Eq. 9.25),
Since the air pressure ahead of the engine, , and exit pressure, , are equal, the thrust equation
reduces to,
Finally, substituting in the continuity equation,
and solving for yields,
Numerical Substitution:
The density at sea level is 0.002377 / . Using the equation found for inlet area from the
analysis,
1000
0.002377
950
2000
950

·
1000
2371
· 0.42

Problem 9.16
Description:
For the turbojet engine operating under the conditions given in Example 9.3, calculate the propulsive
efficiency, as defined in Prob. 9.15.
Given:
Air velocity, 500 /
Exhaust velocity, 1600 /
Find:
The propulsive efficiency,
Assumptions:
1. Steady flow
2. Thrust from pressure difference is negligible
3. Conditions at the inlet are the free stream conditions
Equations:
/
Analysis:
The propulsive efficiency was defined in Problem 9.15 as,
Numerical Substitution:
Converting the air velocity to consistent units,
Using the equation from the analysis,
500
733.33
2
1 /
1
5280
1 3600
2
1600
1
733.33
2
3.1818
0.63