Aircraft Performance: An Engineering Approach, 2nd Edition – Complete Solutions Manual (Chapters 1-10) by Mohammad H. Sadraey

Aircraft Performance, An
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Engineering Approach
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2nd Edition k




by Mohammad H. Sadraey
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Complete Chapter Solutions Manual are
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included (Ch 1 to 10)
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** Immediate Download
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** Swift Response
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** All Chapters included
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, Solutionsktokproblemskfor
AircraftkPerformance:kAnkEngineeringkApproach,kMohammadkSadraey,k2ndked.

Ch. 1k




TheksoftwarekpackagekMathcadkiskusedktoksolvekproblems.



1.1. Determinekthektemperature,kpressurekandkairkdensitykatk5,000kmkandkISAkcondition.

Therekarektwokmethods:
a. Usingkappendix:
FromkAppendixkA:

- Temperature:k255.69kK
- Pressure:k54,048kPa
- Airkdensity:k0.7364kkg/m3

b. Calculations:

K kk k J
hk k 5000m ISA L1kk 6.5k
k
R1kk 287k Pok k 101325Pa
kgK
k
1000
m

Seaklevel: Tokk (15kk 273)Kk k288kKk


k5000km: T5kk Tok kL1hk k255.5kK (Equk1.6)


5.256
kT5k
P5kk Po  k54000.3kPa (Equk1.16)
kT ok

P5k kkg
5k k0.736k (Equk1.23)
R1T5 3
m


Samekresults.




1

,1.2. Determinekthekpressurekatk5,000kmkandkISA-10kcondition.


k k K J
kk k
hk k 5000m ISAk k 10 L1kk 6.5k R1kk 287k Pok k 101325Pa
1000m kgK

Seaklevel: Tokk (15kk 273kk10)Kk k278kKk


k5000km: T5kk Tok kL1hk k245.5kK (Equk1.6)


5.256
kT5k
P5kk Po  k52714.2kPa (Equk1.16)
kT ok



1.3. Calculatekairkdensitykatk20,000kftkaltitudekandkISA+15kcondition.



k k K J
kk k
hk k 20000ft ISAk k 15 L1kk 2k R1kk 287k Pok k 101325Pa
1000ft kgK

Seaklevel: Tok k [(15kk 273)k k 15]Kk k 303kK Tok k 545.4R


20000kft: T20kk Tok k L1hk k 263kK T20k k 473.4R (Equk1.6)


5.256
kT20k lbf
P20kk Po  k48143.9kPa P20k k 1005.5  
(Equk1.16)
kTk ok  2
ftk

P20k kkg kslug
20k k0.638k 20 k 0.001238k (Equk1.23)
R1T20 3 3
m ftk




1.4. Ankaircraftkiskflyingkatkankaltitudekatkwhichkitsktemperaturekisk-4.5koC.kCalculate:


2

, a. AltitudekinkISAkcondition

k k K
L1kk 6.5k Tok k 15k°C Seaklevel: Tok k288.15kK
1000m

ISA Taltk k (4.5kk 273)K Taltk k268.5kK TISAk k Taltk k268.5kK

To  TISA
k
k


TISAk k Tok k L1h h1k  k3023km (Equk1.6)
L1



b. AltitudekinkISA+10kcondition


ISAk k 10 Tk k 10 TISAk k (4.5k k T k 273)K TISAk k258.5kK

To  TISA
k
k


TISAk k Tok k L1h h2k  k4562km (Equk1.6)
L1



c. AltitudekinkISA-10kcondition



ISAk k 10 Tk k 10 TISAk k (4.5k k T k 273)K TISAk k278.5kK

To  TISA
k
k


TISAk k Tok k L1h h3k  k1485km (Equk1.6)
L1




3