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

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ircraft Performance,
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2nd Edition by Mohammad H. Sadraey.pdfY!G*EIH(*#Y!(*#Y#*!#!&**#*GESMSS




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


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



** Immediate Download
** Swift Response
** All Chapters included




lutions
ircraft Performance,
forSolutions
Aircraft Performance,
for
AnAircraft
Engineering
Performance,
An Engineering
Approach,
An 2nd
Engineering
Approach,
Edition 2nd
by
Approach,
Mohammad
Edition 2nd
by Mohammad
H.
Edition
Sadraey.pdfY!G*EIH(*#Y!(*#Y#*!#!&**#*GESMSS*(@U(*UE@&*EH*HJE@&EH*@(o8
by Mohammad
H. Sadraey.pdfY!G*EIH(*#Y!(*#Y#*!#!&**#*GESMSS*(@U(*UE@&*EH*HJE@&EH*@(
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,lutions
ircraft Performance,
for10/1/2026,
Aircraft Performance,
6:00:43
An Engineering
AMSolutions
An Engineering
Approach,
for Aircraft
2nd
Approach,
Edition
Performance,
2nd
by Mohammad
Edition
An Engineering
by Mohammad
H. Sadraey.pdfY!G*EIH(*#Y!(*#Y#*!#!&**#*GESMSS*(@U(*UE@&*EH*HJE@&EH*@(o8
Approach,
H. Sadraey.pdfY!G*EIH(*#Y!(*#Y#*!#!&**#*GESMSS*(@U(*UE@&*EH*HJE@&EH*@(
2nd Edition by Mohammad H. Sadraey.pdfY!G*EIH(*#Y!(*#Y#*!#!&**#*GESMSS




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



lutions
ircraft Performance,
forSolutions
Aircraft Performance,
for
AnAircraft
Engineering
Performance,
An Engineering
Approach,
An 2nd
Engineering
Approach,
Edition 2nd
by
Approach,
Mohammad
Edition 2nd
by Mohammad
H.
Edition
Sadraey.pdfY!G*EIH(*#Y!(*#Y#*!#!&**#*GESMSS*(@U(*UE@&*EH*HJE@&EH*@(o8
by Mohammad
H. Sadraey.pdfY!G*EIH(*#Y!(*#Y#*!#!&**#*GESMSS*(@U(*UE@&*EH*HJE@&EH*@(
H. Sadraey.pdfY!G*EIH(*#Y!(*#Y#*!#!&**#*GESMSS*(@U(*UE@&*EH*HJ

,lutions
ircraft Performance,
for10/1/2026,
Aircraft Performance,
6:00:43
An Engineering
AMSolutions
An Engineering
Approach,
for Aircraft
2nd
Approach,
Edition
Performance,
2nd
by Mohammad
Edition
An Engineering
by Mohammad
H. Sadraey.pdfY!G*EIH(*#Y!(*#Y#*!#!&**#*GESMSS*(@U(*UE@&*EH*HJE@&EH*@(o8
Approach,
H. Sadraey.pdfY!G*EIH(*#Y!(*#Y#*!#!&**#*GESMSS*(@U(*UE@&*EH*HJE@&EH*@(
2nd Edition by Mohammad H. Sadraey.pdfY!G*EIH(*#Y!(*#Y#*!#!&**#*GESMSS




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



lutions
ircraft Performance,
forSolutions
Aircraft Performance,
for
AnAircraft
Engineering
Performance,
An Engineering
Approach,
An 2nd
Engineering
Approach,
Edition 2nd
by
Approach,
Mohammad
Edition 2nd
by Mohammad
H.
Edition
Sadraey.pdfY!G*EIH(*#Y!(*#Y#*!#!&**#*GESMSS*(@U(*UE@&*EH*HJE@&EH*@(o8
by Mohammad
H. Sadraey.pdfY!G*EIH(*#Y!(*#Y#*!#!&**#*GESMSS*(@U(*UE@&*EH*HJE@&EH*@(
H. Sadraey.pdfY!G*EIH(*#Y!(*#Y#*!#!&**#*GESMSS*(@U(*UE@&*EH*HJ

,lutions
ircraft Performance,
for10/1/2026,
Aircraft Performance,
6:00:43
An Engineering
AMSolutions
An Engineering
Approach,
for Aircraft
2nd
Approach,
Edition
Performance,
2nd
by Mohammad
Edition
An Engineering
by Mohammad
H. Sadraey.pdfY!G*EIH(*#Y!(*#Y#*!#!&**#*GESMSS*(@U(*UE@&*EH*HJE@&EH*@(o8
Approach,
H. Sadraey.pdfY!G*EIH(*#Y!(*#Y#*!#!&**#*GESMSS*(@U(*UE@&*EH*HJE@&EH*@(
2nd Edition by Mohammad H. Sadraey.pdfY!G*EIH(*#Y!(*#Y#*!#!&**#*GESMSS




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


b. Altitude in ISA+10 condition


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


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


c. Altitude in ISA-10 condition



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


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




3



lutions
ircraft Performance,
forSolutions
Aircraft Performance,
for
AnAircraft
Engineering
Performance,
An Engineering
Approach,
An 2nd
Engineering
Approach,
Edition 2nd
by
Approach,
Mohammad
Edition 2nd
by Mohammad
H.
Edition
Sadraey.pdfY!G*EIH(*#Y!(*#Y#*!#!&**#*GESMSS*(@U(*UE@&*EH*HJE@&EH*@(o8
by Mohammad
H. Sadraey.pdfY!G*EIH(*#Y!(*#Y#*!#!&**#*GESMSS*(@U(*UE@&*EH*HJE@&EH*@(
H. Sadraey.pdfY!G*EIH(*#Y!(*#Y#*!#!&**#*GESMSS*(@U(*UE@&*EH*HJ