Solutions Manual for Aircraft Performance: An Engineering Approach, 2nd Edition by Sadraey (All Chapters 1-10) | Step-by-Step Solved Problems

SOLUTION MANUAL j j




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




1

, Table of Contents
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1. Atmosphere.
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2. Equations of Motion.
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3. Drag Force and Drag Coefficient.
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4. Engine Performance.
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5. Straight-Level Flight – Jet Aircraft.
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6. Straight-Level Flight: Propeller-Driven Aircraft.
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7. Climb and Descent.
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8. Takeoff and Landing.
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9. Turn Performance and Flight Maneuvers.
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10. Aircraft Performance Analysis Using Numerical Methods and MATLAB(R)
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Ch. 1 j

2

, ThejsoftwarejpackagejMathcadjisjusedjtojsolvejproblems.



1.1. Determinejthejtemperature,jpressurejandjairjdensityjatj5,000jmjandjISAjcondition.

Therejarejtwojmethods:
a. Usingjappendix:
FromjAppendixjA:

- Temperature:j255.69jK
- Pressure:j54,048jPa
- Airjdensity:j0.7364jkg/m3

b. Calculations:

K jj jJ
hj j 5000m ISA L1jj 6.5j
j
R1jj 287j Poj j 101325Pa
kgK
j
1000
m

Seajlevel:j Tojj (15jj 273)Kj j288jKjT


5000jm: 5jj Toj jL1hj j255.5jK (Equj1.6)


5.256
jT5j
P5jj Po  j54000.3jPa (Equj1.16)
Tj oj

P5jR jkg
5j j0.736j (Equj1.23)
1T5 3
m


Samejresults.




3

, 1.2. Determinejthejpressurejatj5,000jmjandjISA-10jcondition.


j j K J
jj j
hj j 5000m ISAj j 10 L1jj 6.5j R1jj 287j Poj j 101325Pa
1000m kgK

Seajlevel:j Tojj (15jj 273jj10)Kj j278jKjT5j


5000jm: j Toj jL1hj j245.5jK (Equj1.6)


5.256
jT5j
P5jj Po  j52714.2jPa (Equj1.16)
Tj oj



1.3. Calculatejairjdensityjatj20,000jftjaltitudejandjISA+15jcondition.



j j K J
jj j
hj j 20000ft ISAj j 15 L1jj 2j R1jj 287j Poj j 101325Pa
1000ft kgK

Seajlevel: Toj j [(15jj 273)j j 15]Kj j 303jK Toj j 545.4R


20000jft: T20jj Toj j L1hj j263jK T20j j 473.4R (Equj1.6)


5.256
jT20j lbf
P20jj Po  j48143.9jPa P20j j 1005.5  
(Equj1.16)
Tj j oj  2
ftj

P20jR jkg jslug
20j j0.638j 20 j0.001238j (Equj1.23)
1T20 3 3
m ftj




1.4. Anjaircraftjisjflyingjatjanjaltitudejatjwhichjitsjtemperaturejisj-4.5joC.jCalculate:


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