SOLUTION MANUAL Aircraft Performance, An Engineering Approach 2nd Edition by Sadraey All Chapters 1 to 10 Covered

SOLUTION MANUAL
Aircraft Performance, An Engineering Approach
2nd Edition by Sadraey All Chapters 1 to 10 Covered




SOLUTION MANUAL




1

, Table of Contents
1. Atmosphere.

2. Equations of Motion.

3. Drag Force and Drag Coefficient.

4. Engine Performance.

5. Straight-Level Flight – Jet Aircraft.

6. Straight-Level Flight: Propeller-Driven Aircraft.

7. Climb and Descent.

8. Takeoff and Landing.

9. Turn Performance and Flight Maneuvers.

10. Aircraft Performance Analysis Using Numerical Methods and

MATLAB(R)




2

, Ch. 1

The software package Mathcad is used to solve problems.



1.1. Determine the temperature, pressure and air density at 5,000 m and ISA condition.

There are two methods:
a. Using appendix:
From Appendix A:

- Temperature: 255.69 K
- Pressure: 54,048 Pa
- Air density: 0.7364 kg/m3

b. Calculations:

K J
h = 5000m ISA L1 = 6.5 R1 = 287 Po = 101325Pa
1000m kgK

Sea level: To = (15 + 273)K = 288 K


5000 m: T5 = To − L1h = 255.5 K (Equ 1.6)


5.256
 T5 
P5 = Po  = 54000.3 Pa (Equ 1.16)
 To 

P5 kg
5 = = 0.736 (Equ 1.23)
R1T5 3
m


Same results.




3

, 1.2. Determine the pressure at 5,000 m and ISA-10 condition.


K J
h = 5000m ISA − 10 L1 = 6.5 R1 = 287 Po = 101325Pa
1000m kgK

Sea level: To = (15 + 273 − 10)K = 278 K


5000 m: T5 = To − L1h = 245.5 K (Equ 1.6)


5.256
 T5 
P5 = Po  = 52714.2 Pa (Equ 1.16)
 To 



1.3. Calculate air density at 20,000 ft altitude and ISA+15 condition.



K J
h = 20000ft ISA + 15 L1 = 2 R1 = 287 Po = 101325Pa
1000ft kgK

Sea level: To = [(15 + 273) + 15]K = 303 K To = 545.4R


20000 ft: T20 = To − L1h = 263 K T20 = 473.4R (Equ 1.6)


5.256
 T20  lbf
P20 = Po  = 48143.9 Pa P20 = 1005.5 (Equ 1.16)
 To  ft
2


P20 kg slug
20 = = 0.638 20 = 0.001238 (Equ 1.23)
R1T20 3 3
m ft




1.4. An aircraft is flying at an altitude at which its temperature is -4.5 oC. Calculate:


4