Aircraft Performance, A m m
n Engineering Approach
m m
2nd Edition m
by Mohammad H. Sadraey
m m m
Complete Chapter Solutions Manualar
m m m m
e included (Ch 1 to 10)
m m m m m
** Immediate Download
m m
** Swift Response
m m
** All Chapters included
m m m
, Solutionsmto mproblemsmfor
Aircraft mPerformance:mAnmEngineering mApproach,mMohammad mSadraey,m2ndmed.
Ch. 1m
ThemsoftwarempackagemMathcadmismusedmtomsolvemproblems.
1.1 . Determine mthemtemperature,mpressure mand mairmdensity matm5,000 mmmand mISAmcondition.
Theremaremtwo mmethods:
a. Using mappendix:
FrommAppendix mA:
- Temperature:m255.69 mK
- Pressure:m54,048 mPa
- Airmdensity:m0.7364 mkg/m3
b. Calculations:
K mm J
h m=m5000 ISA L1 m=m 6.5
m m
R1 m=m 287 Po m=m101325Pa
m
m m
1000 kgK
m
Seamlevel To m =m (15 m+m 273)Km =m 28
:m5000mm: 8 mKmT5 m=m To m −m L1h m=m 255. (Equm1.6)
5 mK
5.256
mT5 m
P5 m=m Po =m 54000.3 mP (Equm1.16)
m To m a
P5 m mkg
5 m= =m0.736 (Equm1.23)
R1T5 3
m
Samemresults.
1
,1.2 . Determine mthempressure matm5,000 mmmand mISA-10 mcondition.
m m K mm J
h m=m5000 ISAm −m1 L1 m=m 6.5 R1 m=m 287 Po m=m101325Pa
m 0 1000m m
kgK
Seamlevel To m=m(15 m+m273m−m10)Km=m278m
:m5000mm: KmT5 m=m To m −m L1h m=m 245.5 mK (Equm1.6)
5.256
mT5 m
P5 m=m Po =m 52714.2 mP (Equm1.16)
T
m o m a
1.3 . Calculate mairmdensity matm20,000 mftmaltitudemand mISA+15 mcondition.
m m K mm J
h m=m20000f ISAm+m1 L1 m=m2 R1 m=m 287 Po m=m101325Pa
t 5 1000ft m
kgK
Seamlevel To m =m [(15 m+m 273)m +m 15]Km =m30 To m=m545.4R
: 3 mK
20000mf t T20 m=m To m −m L1h m =m 26 T20 m=m473.4 (Equm1.6)
: 3 mK R
5.256
mT2 0 m lbf
P20 m=m Po =m 48143.9 mP P20 m=m1005.5 (Equm1.16)
T
mmm o m a ft m
2
P20 m mkg slu
20m =m0.638 20 =m0.001238m (Equm1.23)
R1T20 3 g
= m
3
ftm
2
, 1.4 . An maircraftmismflyingmatman maltitudematmwhich mitsmtemperature mism-4.5 mo C.mCalculate:
3
n Engineering Approach
m m
2nd Edition m
by Mohammad H. Sadraey
m m m
Complete Chapter Solutions Manualar
m m m m
e included (Ch 1 to 10)
m m m m m
** Immediate Download
m m
** Swift Response
m m
** All Chapters included
m m m
, Solutionsmto mproblemsmfor
Aircraft mPerformance:mAnmEngineering mApproach,mMohammad mSadraey,m2ndmed.
Ch. 1m
ThemsoftwarempackagemMathcadmismusedmtomsolvemproblems.
1.1 . Determine mthemtemperature,mpressure mand mairmdensity matm5,000 mmmand mISAmcondition.
Theremaremtwo mmethods:
a. Using mappendix:
FrommAppendix mA:
- Temperature:m255.69 mK
- Pressure:m54,048 mPa
- Airmdensity:m0.7364 mkg/m3
b. Calculations:
K mm J
h m=m5000 ISA L1 m=m 6.5
m m
R1 m=m 287 Po m=m101325Pa
m
m m
1000 kgK
m
Seamlevel To m =m (15 m+m 273)Km =m 28
:m5000mm: 8 mKmT5 m=m To m −m L1h m=m 255. (Equm1.6)
5 mK
5.256
mT5 m
P5 m=m Po =m 54000.3 mP (Equm1.16)
m To m a
P5 m mkg
5 m= =m0.736 (Equm1.23)
R1T5 3
m
Samemresults.
1
,1.2 . Determine mthempressure matm5,000 mmmand mISA-10 mcondition.
m m K mm J
h m=m5000 ISAm −m1 L1 m=m 6.5 R1 m=m 287 Po m=m101325Pa
m 0 1000m m
kgK
Seamlevel To m=m(15 m+m273m−m10)Km=m278m
:m5000mm: KmT5 m=m To m −m L1h m=m 245.5 mK (Equm1.6)
5.256
mT5 m
P5 m=m Po =m 52714.2 mP (Equm1.16)
T
m o m a
1.3 . Calculate mairmdensity matm20,000 mftmaltitudemand mISA+15 mcondition.
m m K mm J
h m=m20000f ISAm+m1 L1 m=m2 R1 m=m 287 Po m=m101325Pa
t 5 1000ft m
kgK
Seamlevel To m =m [(15 m+m 273)m +m 15]Km =m30 To m=m545.4R
: 3 mK
20000mf t T20 m=m To m −m L1h m =m 26 T20 m=m473.4 (Equm1.6)
: 3 mK R
5.256
mT2 0 m lbf
P20 m=m Po =m 48143.9 mP P20 m=m1005.5 (Equm1.16)
T
mmm o m a ft m
2
P20 m mkg slu
20m =m0.638 20 =m0.001238m (Equm1.23)
R1T20 3 g
= m
3
ftm
2
, 1.4 . An maircraftmismflyingmatman maltitudematmwhich mitsmtemperature mism-4.5 mo C.mCalculate:
3