Solution Manual for Aircraft Performance, 1st Edition by Mohammad H. Sadraey 3

AircraftrPropulsionr2E SolutionrManual


Chapter 2 Solutions
r r




Problemr2.1

Givenrthat:



1 2


at=400rm/sr
a1=300rm/s
=1.4



N.S.


Solution:

a) UsingrEquationr2.102-a,

ar2rrrrr ur2 =r rrr tarrr2=rconst.
+r r



r −1 2 r −1

Rearrangingrandrsolvingrforru1,

ru1r =r591.6 mr/rs

InrorderrtorgetrM1:

u
Mr1=r 1rrMra1 1 =r1.972

b) First,rwerneedrtorfindrM2rofrtherflow

ReferrtorAppendixrC,rNormalrShockrTabler(=1.4),

Mr2r r0.58r forr M1=1.972

Then,rwithrtherhelprofrEquationr2.104,




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,AircraftrPropulsionr2E SolutionrManual


*2 (r +1)Mr2 2
M *r2 = r0.615
2rM 2
r+rr(

Mr2−
1)

Problemr2.2r

Solution

a) Fromrtherfigurershown,ranrobliquershockrformsrbetweenrregionr1randr2:rFo

rrM1r=r3.0,r2r=r5o

AccordingrtorFigurer2.23(a),r2r=r23o

M1nr =r M1r sinrr r M1nr r1.172

Usingrthernormalrshockrtable,r

ForrM1nr=r1.172,

p2r r
Mr2n r0.855 and r1.4578
p1

b) Fromrtherfigurershown,randrobliquershockrformsrbetweenrregionr1randr3:rFo

rrM1r=r3.0,r3r=r10o

AccordingrtorFigurer2.23(a),r3r=r27.5o

M1nr =r M1r sinr3r r M1nr r1.385

Usingrthernormalrshockrtable,r

ForrM1nr=r1.385,

p3r
Mr3n r0.7483r and r2.055
p1

D'rq
c) Cd r
1c


Letrusrsubtractrp1rfromrtherthreersurfacesrtorcancelroutrtherbasercontribution,rnamelyrwercanr
write:

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,AircraftrPropulsionr2E SolutionrManual



Dr'r=r(rp2r −r p1 )rrcrtan  2r )+
r r(rp3r −r p1 )rrcrtan  3r ) 


1r 1r  p  p1r 2rr 
Note:rq r rVr =r rr
2 1r
Vr2rr =r V r =r
pr Mr2
2 1r rr p 1 1
1
2 1rr 1 2ar21 1 2 1 1

Therefore,
2(rp2r /r p1r−1)rtan2r +r(rp3r /r p1r−1)rtan3r
Cdr = r0.0359
rM1r2



Problemr2.3




p1=100rkPar 10o
M1=2 2
1
3

Bowrshock
O.S.




Solution:

Anrobliquershockrisrformedronrtherlowerrpanel.rWeruse:rM1r=r

2
r=r10o

torgetrtherwaverangle,rfromrFigurer2.23(a),rr≈r39o

M1nr =r M1r sinrr r M1nr r1.26

Usingrthernormalrshockrtable,



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, AircraftrPropulsionr2E SolutionrManual

p2r r
ForrM1nr=r1.26, Mr2n r0.8071r and r1.6855
p1

Werthenrsolverforrp2,

p2r =r168.55 kPa

Mr2n
Also, Mr2r = rMr2 r1.665
sin(r −rr)

Fromrtherisentropicrtable,rusingrM2r=r1.665:

ptr,2
=r4.6479
p2

TherPitotrtubermeasuresrtherstagnationrpressurerdownstreamrofrarNS,rasrshown,rnamely:
ptr3rrr
Mrrr=r1.665r⎯N⎯ ⎯⎯
S −Table
→r 4.05rr p
r r683.1r kPa
2 tr3
p2

Problemr2.4

Solution:

Anrobliquershockrisrreflectedrfromrarflatrsurfacerwith:

1r=r30o
2r=r9o

FromrFigurer2.23(a),rM1r=r2.6

M1nr =r M1r sinrr r M1nr r1.3

Usingrthernormalrshockrtable,

ForrM1nr=r1.3, Mr2nr r 0.7860

Mr2n
Mr2r = rMr2 r2.193
sin(r −rr)

TherreflectedrobliquershockrhasrM2rasritsrupstreamrMachrnumberrandrtherflowrinrregionr3rha
srtorberparallelrtortherwall,rtherefore,rwerknowrthat:


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