Solutions for Aircraft Performance, An Engineering Approach, 2nd Edition Sadraey (All Chapters included)

Aircraft Performance,
An Engineering Approach
2nd Edition
by Mohammad H. Sadraey


Complete Chapter Solutions Manual
are included (Ch 1 to 10)



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** All Chapters included

, Solutions to problems for
Aircraft Performance: An Engineering Approach, Mohammad Sadraey, 2nd ed.

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
ρ5 := = 0.736 (Equ 1.23)
R1⋅ T5 3
m


Same results.




1

, 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
ρ20 := = 0.638 ρ20 = 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:

2