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Key

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AE AEROSPACE

ENGINEERINGGATE 2011

MODEL EXAM(KEY)

1/27/2011 22

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(1) (b)

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(2) (b)

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3(c)

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(4) (c)

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(5) (d)

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(6) (b)(B is any Point)

Click to edit Master text stylesSecond level Third level Fourth level Fifth level

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(6) (b)

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(7) (b)

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(8) (c)

Find the Wrong staement on LoadFactor

a)

It is traditionally referred to as g,because of the relation betweenload factor and apparentacceleration of gravity felt on board

the aircraft.

b) Although, it is traditionally referredto as g, it does not take the unit of

acceleration due to gravity (m/s2) ,1/27/2011 1111

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(9) (c)

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(10) -b

Find the wrong staatement on tail wing

a) The conventional configuration with

a low horizontal tail is a naturalchoice since roots of both horizontaland vertical surfaces areconveniently attached directly to the

fuselage.

b) In conventional configuration, theeffectiveness of the vertical tail is

large because interference with the1/27/2011 1313

http://adg.stanford.edu/aa241/stability/taildesign.html

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11 (d)

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(13)=C

k=1.4

T1 = 323 deg K,

p1= 206785 N/m2

C1= 150 m/sM2=1

T2=?

C2=?

Is=?

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a1 = 360.25 m/s

M1 = C1/a1 = 0.416

To1/T1 = 1.034

T01 = 334 K

T02 = T01 ( no heat is added)= 334 K

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To2/T2 = 1.2

T2 = 278 K

a2 = 334.33 K

a2=C2 = 334.33 K

Is = C2/g = 34 s

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(14) (b)

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(15) (d)

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CD

A B0.6 cm

0.

3cm

3600 cm/s

3800 cm/s

2000

cm/s

2100

cm/s

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Solution

B C D A

ABCDA B C D

B C D A

ABCDA B C D

Vcos ds Vcos ds Vcos ds Vcos ds Vcos ds

Vcos ds (3600)(1) ds (2100)(1) ds (3800)( 1) ds (2000)( 1) ds

K (3600)(1)(0.6) (2100)(1)(0.3) (3800)( 1)(0.6) (2000)( 1)(0.3)

K

= + + +

= + + + = = + + +

=

290cm / s (Taking CW as positive)=

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16 - (d)

Kelvins theorem in fluid dynamicsstates that the circulation (defined asthe line integral of the component of

velocity tangential to the closedcontour) in an inviscid andincompressible fluid subject to only

conservative forces is constant.

http://en.wikipedia.org/wiki/Kelvin's_circulation_theorem

http://www.eng.fsu.edu/~dommelen/courses/flm/flm00/topics/vort/node2.htmlhtt ://enc clo edia2.thefreedictionar .com/Kelvin's+circulation+theorem

http://en.wikipedia.org/wiki/Kelvin's_circulation_theoremhttp://www.eng.fsu.edu/~dommelen/courses/flm/flm00/topics/vort/node2.htmlhttp://encyclopedia2.thefreedictionary.com/Kelvin's+circulation+theoremhttp://encyclopedia2.thefreedictionary.com/Kelvin's+circulation+theoremhttp://www.eng.fsu.edu/~dommelen/courses/flm/flm00/topics/vort/node2.htmlhttp://en.wikipedia.org/wiki/Kelvin's_circulation_theorem

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(16)

In fluid mechanics, Helmholtz'stheorems, named afterHermann von Helmholtz, describe the

three-dimensional motion of fluid inthe vicinity ofvortex filaments. Thesetheorems apply to inviscidflows and

flows where the influence ofviscous forces is small and can beignored.

Helmholtzs three theorems are ashttp://en.wikipedia.org/wiki/Helmholtz's_theorems

http://en.wikipedia.org/wiki/Fluid_mechanicshttp://en.wikipedia.org/wiki/Hermann_von_Helmholtzhttp://en.wikipedia.org/wiki/Vortexhttp://en.wikipedia.org/wiki/Inviscid_flowhttp://en.wikipedia.org/wiki/Inviscid_flowhttp://en.wikipedia.org/wiki/Viscosityhttp://en.wikipedia.org/wiki/Viscosityhttp://en.wikipedia.org/wiki/Inviscid_flowhttp://en.wikipedia.org/wiki/Inviscid_flowhttp://en.wikipedia.org/wiki/Vortexhttp://en.wikipedia.org/wiki/Hermann_von_Helmholtzhttp://en.wikipedia.org/wiki/Fluid_mechanics

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(17) (b)

a small aerofoil shaped deviceattached just in front of the wingleading edge to properly direct the

airflow at the front of the wing tomake it to flow more smoothly overthe upper surface while at a high

angle of attack.

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(18) (b)

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(19) - C

In a supersonic flow, a sphereencounters a

Bow shock wave

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2828

20 (d)Limitations of Airys Stress functions

The Airys Stress function is applicable only to plane strain or plane stress problem [3].

The Airys Stress function can only be used if the body force has a special form [3].

Specifically, the requirement is

where is a scalar function of position, F1 & F2 are body forces.

The Airys Stress function approach works best for problems where a solid is subjected to prescribedtractions on its boundary, rather than prescribed displacements [3].

1

1x

F

=2

2x

F

=

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21 (d)

%195

densityinchange%

95.2

1

12

1

2

=

=

=

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22 - C

M 1 = 2.0

T2 = 450K

T1 = ?

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K67.266T

687.1T

T

1

1

2

=

=

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23 - b

http://en.wikipedia.org/wiki/Wave_drag

http://www.answers.com/topic/wave-

drag

http://en.wikipedia.org/wiki/Wave_draghttp://en.wikipedia.org/wiki/Wave_drag

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24) (a) and 25 (d)

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(

26

)

b,2 7

(c)

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28 (a)

( )

( )

( )

( ) ( )

mmdd

d

ddd

FS

FS

pt

pt

yt

yt

n

4.12

10015365

10063652

12730

2

127301

1006365

2

12730

2

12730

22

22

22

2

2

2

2

2

2

2

22

2

2

11

2

2

2121

2

2

2121

1

=

=

=

+

+

=

+

+

=+

+

=+

+

+

=+

+

+

=

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29 (C)

mm42.13d

2

100

d

9000

2

)(

2

2

)FS/(

2

FS

)2/(

2

FS

)2/(

2

pt2

2

21

yt22

21

yt2

2

21

yt

max

=

=

=+

=+

=+

=

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30 - b

222n

2

2

21212n

221n

2

2

21211n

mm

N

d

263522

mmN

d15635

22

=

+

+=

=

+

+

+=

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mm7.12d

100d

16156

-

FS

-

FSm

-

2

pt2n1n

yt

2n1n

yt2n

1n

=

=

=

=

=

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(31) b, 32 - d

T/Theta = kt=GJ/L

k/m kt/I

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Given data :

p01 = 100 kPa

T01 = 288K

Dh = 0.13mDt= 0.3 m

m = 8 kg/s

N = 16200rpm

beta1 = ??? at tip and rootM1t =???

(33) - c

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s

mNDu

smNDu

tt

hh

47.254

60

27.11060

1

1

==

==

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s

mNDu

smNDu

tt

hh

47.254

60

27.11060

1

1

==

==

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( )

11

1

2

22

3

01

0101

7.116

8

0574.0

4

2.1

Cs

m

C

s

kgCAm

mA

DDA

m

kg

RT

p

f

ff

f

htf

==

==

=

=

===

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KT

c

C

TT p

2.281

2

1

2

1

101

=

+=

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kPa92p

T

T

p

p

1

1

1o

1

1o

1

=

=

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34 -d

22 m

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8.25

tan

15.48

tan

7.282

29.165

1

1

1

1

11

2211

22

11

=

=

==

=+=

=+=

t

t

f

t

h

h

f

h

ftt

fhh

uC

u

C

smCuw

s

mCuw

3

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smCuw ftt 63.28221211 =+=

35

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842.0

87.335

1

11

11

==

==

a

wM

smRTa

36

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36 -

36 C

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36 - C

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3

a

a

a

0.7476kg/ m

p 55kPa T 255K

=

==

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KT

T

T

M

T

T

MMa

CMM

smC

smRTa

a

a

a

i

a

a

ai

a

iai

i

aa

07.279

094.1

2

11

687.0

/220

/09.320

0

0

20

=

=

+=

==

==

===

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KTT

kPappIsentropica

kPap

T

T

p

p

a

a

a

a

a

a

a

07.279

41.75Process:1

41.75

001

001

0

100

====

=

=

C i

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Compression

( )

kPap

p

p

KT

T

T

p

p

T

T

05.377

5

99.441

58.1

5

02

01

02

02

01

02

1

1

01

02

01

02

=

=

=

=

=

=

H t Additi

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Heat Addition

KTT

kPapp

1200

05.377

max03

0203

==

==

T bi

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Turbine

kPap

TT

pp

KT

TTcmTTcm

PP

papa

ct

25.226

08.1037

)()(

04

1

03

04

03

04

04

01020403

=

=

=

==

37 b

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37 - b

smC

TTcCC

KT

p

p

T

T

p

p

T

T

j

pj

a

/4.832

)(2

33.692

5045

5

1

0404

5

1

04

5

04

5

=

==

=

=

=

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smp

pRTC

c

ee /21321

12

1

0 =

=

39 C

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39 - C

smRTCa

KT

T

T

c

/6.919

43.2432

1

2

***

*

*

===

=

+=

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smRTC

RTC

/65.919**

1

211

2* 0

==

+

=

40 b

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40 - b

224.3

/75.6614.1259

1

==

===

=

e

ee

ee

e

c

e

c

e

a

CM

smRTaKT

p

p

T

T

579.8

1

1

1

21

*

2

*

)1(2

1

=

+

+

+

=

+

A

A

M

MA

A

e

e

41 a

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41 -a

A single stage turbine has been designed forfollowing parameters:

Inlet temperature : 1000K

Axial velocity : 260 m/s (Ca)

Mean Blade speed : 360 m/s (u)

Nozzle efflux angle : 65 ( 2)

Stage swirl angle : 10 ( 3)

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( )

( )

44.57

tantan1

2.37

tantan1

7222.0

u

C

3

33

2

22

a

=

=

=

=

=

=

42 b

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42 -b

( )

( )

67.1

tantan

291.0R

tantan2

R

23

23

=

+==

=

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3547.32u

w22/u

w

677.1uw

22

,

2

====

==

43 c

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43 c

( )

kgJ217256w

tantanuCw

7222.0

u

C

32a

a

=

+==

=

44 b

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44 - b

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1/30/11 7474

U=150 m/s

w2

w1

C1

C2

85

35

10

75

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1/30/11 7575

w2

w1

Ca2

Ca1

C1

C2

1

1

2

2

U=

15 0

m/s

ct

2 ct1

wt2

wt1

Ca2=75

Ca1 =85

35

10

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1/30/11 7676

T01 = 340 K

Po1 = 185 kPa

(a) Specific work

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(a) Specific work

wt1 = U Ct1 = 150 - 35 =115 m/s

Ct2 = U wt2 = 150 10 = 140 m/s

work, w = U (Ct2 - Ct1)

= 150 (140-35)

= ???

(a) Specific work

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1/30/11 7878

(a) Specific work

wt1 = U Ct1 = 150 - 35 =115 m/s

Ct2 = U wt2 = 150 10 = 140 m/s

work, w = U (Ct2 - Ct1)

= 150 (140-35)

= 15,750 J/kg

(b) Static temperature at

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1/30/11 7979

(b) Static temperature atexit

( )

???

100%Assume,

1

1

1

01

01

0201

0102

==

=

=

=

o

op

p

p

R

RTcw

T

TTcw

TTcw

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1/30/11 8080

1708.1R

100%Assume,

1RTcw

o

1

o01p

==

=

1

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1/30/11 8181

???p

1708.1p

p

1708.1ppR

100%Assume,

1RTcw

02

01

02

01

02o

1

o01p

=

=

==

=

=

45 - d

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45 - d

1/30/11 8282

kPa6.216p

1708.1p

p

1708.1p

pR

100%Assume,

1RTcw

02

01

02

01

02o

1

o01p

=

=

==

=

=

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1/30/11 8585

???Tc2

CTT

???C

CCC

2

p

2

2202

2

2

2a

2

2t

2

2

=

+=

=

+=

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1/30/11 8686

K13.343T

c2

CTT

s/m82.158C

CCC

2

p

2

2202

2

2

2a

2

2t

2

2

=

+=

=+=

(c) Static pressure at the

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1/30/11 8787

(c) Static pressure at theexit

???p

T

T

p

p

2

1

02

2

02

2

=

=

(c) Static pressure at the

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1/30/11 8888

(c) Static pressure at theexit

kPa00.191p

T

T

p

p

2

1

02

2

02

2

=

=

46 - d

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1/30/11 8989

46 d

???Tc2

CTT

???C

CCC

1

p

2

1101

1

2

1a

2

1t

2

1

=

+=

=+=

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1/30/11 9090

79.335T

c2

CTT

s/m92.91C

CCC

1

p

21

101

1

2

1a

2

1t

2

1

=

+=

=

+=

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1/30/11 9393

???

RTp

1

1

11

=

=

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222

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1/30/11 9696

75.0C

w2

1ppC

s/m143w

Cww

p

2

11

12p

1

2

1a

2

1t

2

1

=

=

=

+=

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