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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
1/27/2011 2727
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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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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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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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(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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???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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(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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(c) Static pressure at theexit
kPa00.191p
T
T
p
p
2
1
02
2
02
2
=
=
46 - d
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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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???
RTp
1
1
11
=
=
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222
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75.0C
w2
1ppC
s/m143w
Cww
p
2
11
12p
1
2
1a
2
1t
2
1
=
=
=
+=
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