Solution Manual For Aircraft Performance, An Engineering Approach 2nd Edition By Mohammad H. Sadraey | All Chapters (1-10) | Latest Version A+

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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, 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

, a. Altitude in ISA condition

K
L1 := 6.5⋅ To := 15 °C Sea level: To = 288.15 K
1000⋅ m

ISA Talt := ( −4.5 + 273) ⋅ K Talt = 268.5 K TISA := Talt = 268.5 K


(To − TISA) (Equ 1.6)
TISA To − L1⋅ h h1 := = 3023 m
L1


b. Altitude in ISA+10 condition


ISA + 10 ∆T := 10 TISA := ( −4.5 − ∆T + 273) ⋅ K TISA = 258.5 K


(To − TISA) (Equ 1.6)
TISA To − L1⋅ h h2 := = 4562 m
L1


c. Altitude in ISA-10 condition



ISA + 10 ∆T := −10 TISA := ( −4.5 − ∆T + 273) ⋅ K TISA = 278.5 K


(To − TISA) (Equ 1.6)
TISA To − L1⋅ h h3 := = 1485 m
L1




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