GATE-2017 AEROSPACE ENGINEERINGgatepathshala.com/qpapers/GATE2017_AE_COMPLETE_Solutions( CLASSES &...

BEST CLASSES & STUDY MATERIAL FOR AEROSPACE ENGINEERING - GATE GATE PATHSHALA

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GATE-2017 AEROSPACE ENGINEERING – SOLUTIONS

An Effort by GATE PATHSHALA FACULTY

1. Sol. By www.gatepathshala.com (D)

Two vectors determine a plane, and the cross product points in a direction different from both. There are two perpendicular directions. By convention, we assume a right –handed system.

2. Sol. By www.gatepathshala.com (12)

B C D A

2 2 2 2

A B C D

1 2 3 4

22 2 2

1

0 0 0

(( ) ) (( ) ) (( ) ) (( ) )

for path 1 , i.e., from A(0,0) to B(2,0) 0 0

( ) 22

for path 2 , i.e., from

I x y dx x dy x y dx x dy x y dx x dy x y dx x dy

I I I I

y dy

xI x y dx xdx

2

2 22 2

1 000

B(2,0) to C(2,2) 2 0

4 8

x dx

I x dy x y y

http://www.gatepathshala.com/





3. Sol. By www.gatepathshala.com

(0)

4. Sol. By www.gatepathshala.com (C)

5. Sol. By www.gatepathshala.com (B)

6. Sol. By www.gatepathshala.com (B) The pitching moment about the aerodynamic centre of an aerofoil is independent of angle of attack for a given velocity.

7. Sol. By www.gatepathshala.com (C)

8. Sol. By www.gatepathshala.com (0.5448)

o

3.5 3.52 2o

P =1atm (guage)=2atm (absolute)

P 2= 1+0.2M = 1+0.2 1.5 =3.671 p=0.5448atm

p p

9. Sol. By www.gatepathshala.com (B)

00 0 2

1

2 2 2

00

2 2

1 22

1 2 3 4

for path 3,i.e., from C(2,2) to D(0,2) 2 0

( ) ( 2) 2 0 (2 4) 22

for path 4,i.e., from D(0,2) to A(0,0) 0 0

0

Hence, 2 8 2

y dy

xI x y dx x dx x

x dx

I x dy x y

I I I I I

0 12












(B)


(C)

1. The transverse force applied at shear center does not lead to the torsion of

thin-walled beam.

2. The shear center is a center of rotation for a section of thin-walled beam

subjected to pure torsion.


(B)


(C)








Figure shows the variations of (δr)reqd and (δr)free as functions of β. When β is greater than about 150, (δr)free increases rapidly. It((δr)free) equals (δr)reqd at β = βrl and then exceeds (δr)reqd . In this situation, the pedal force would be reverse in direction. This phenomenon is called rudder lock


(686.943 )

2 3

3 32 3 2 2

2 3

In planetary motion : T

T 2.2783 = T = T 365.256

T 1.4953

T 686.9438 Earth days

m m mm e

e e e

m

a

a a

a a


(D)


(D)

Scramjet has a higher specific impulse (change in momentum per unit of propellant) than a rocket engine; it could provide between







1000 and 4000 seconds, while a rocket typically provides around 450 sec or less.


(C)


(D)


(B)

w

In the vaneless region of a centrifugal compressor the angular momentum

is conserved. rC constant


(B)


(B)


(1.360 to 1.370)


(C)












(0.9797)

1

2

5 Hz

2 10 rad/sec

=

0.2 1 9.797989

coulomb damping 0

0

d

c

f

f

k

m

c

c

mx kx N

kx x g

m


(C)


(B)


(D)









(B)

2

2

2

2

2

2

4 6 ( )

Applying Laplace Transform on both sides

'' 4 ' 6 { ( )}

(0) '(0) 4 (0) 6 { ( )}

( ) 2 1 4 ( ) 2 6 ( ) ( )

( ) 4 6 2 9 ( )

( ) 4 6 ( )

d y dyy f t

dt dt

L y L y L y L f t

s L y sy y sL y y L y L f t

s Y s s Y s Y s F s

Y s s s s F s

Y s s s F s

2

2 9

( ) 2 9( )

4 6

s

F s sY s

s s


(0.01)

2

The general solution of the PDE of the form

0 is given by d'Alembert's formula

( ) ( ) 1( , ) ( )

2 2

where,

( ,0) ( )

( ,0) ( )

Given equation is

25 0 ; (

tt xx

x ct

x ct

t

tt xx

u c u

x ct x ctu x t s ds

c

u x x

u x x

u u u

,0) 0.01sin(10 ) , ( ,0) ( )

5

0.01sin(10 ( 5 )) 0.01sin(10 ( 5 ))( , ) 0

2

0.01 sin(10 5.25) sin(10 4.75)(0.25,1)

2

0.01 sin(10 5.25) sin(10 4.75)(0.25,1)

2

(0.25,1) 0.01

tx x u x x

c

x t x tu x t

u

u

u







(C)

22

2

1

22

2

The given differential equation is the most common cauchy differential equation

0

We assume a trial solution given by

Differentiating, we have:

and

( 1)

Suubsti

m

m

m

d y dyx ax by

dx dx

y x

dymx

dx

d ym m x

dx

2 2 1

2

2

2

tuting into the original equation, we have:

( ( 1) ) ( ) ( ) 0

( 1) 0

Given 5 and 4

Substituting the value of and on above equation we get;

(5 1) 4 0

4 4 0

2

For

m m mx m m x ax mx b x

m a m b

a b

a b

m m

m m

m

1 2

2 2

1 2

the one real repeated root,the solution is given by

ln( )

ln( )

m my c x x c x

y c x x c x


(0.543)

0

0

0

0

0

0

0 0

0

AR 10; 2 ; 0.95; 6

For Symmetrical Airfoil, 0

25.19 per radian 0.0905 per degree

211

10 0.95AR

( ) 0.09(6 0 ) 0.543L

a e

aa

a

e

C a







(0.166)

0

0

2

2

0

2 3

2

0

3

2

for ( )

1 for

Momentum thickness, 1

1

1 =

1 =

2 3

1 0.

6 6

yyu y

Uy

u udy

U U

y udy

y y dy

y y

1666


(20)

2

2 2

ˆˆ ˆGiven, 4 2 4

( 1,1,0) and 1 sec

ˆ ˆˆ ˆ ˆ ˆ (4 )(4 4 ) ( 2 )( 2 ) (4 )(4 ) 4 4

ˆ ˆ ˆ ˆ (4 1 1)(4 ) 4 4 4

V txi t yj xzk

P t

dV V V V Va u v w

dt x y z t

a tx ti zk t y t j xz xk xi tyj

a i j i j

ˆ ˆ ˆ 16 4 20

20

a i i i

a


(48.5840)

1 1

2 2

The mach angles ahead and behind the fan

with and corresponding mach numbers are:

180 143 37 1.6616

20 2.9238

o

o

M

M







1

2

2 1

1 2

Corresponding prandtl meyer function values

16.68

48.2654

Flow turning angle, = 48.2654 16.68 31.5854

Fan angle, 37 +31.5854 20 48.584

o

o

o o o

o o o o


(1341.6)

21 1

1

29 , 1.4 , 280 , 0P

T K VP

2

2 1

1

2 2

11 2 2 2

1

2

2

2 1

2

1 1

2 2 1

1 1 2

2

1

1

2 ( 1)29

1

2 ( 1) 2 0.4 55

2 ( 1) 2 1.4 5 0.4

0.4152

( 1)5

2 ( 1)

129 5.8

5

280 5.8 1624 K

5 1.4 287 280 1677ss

P M

P

MM M

M

M

M

M

T P

T P

T

CM C

RT

M

21 2

1

Velocity behind the shock, 1677 0.4152 1.4 287 1624 1341.6 /sC VV m s

RT


(90.478 )

2 1 2 1 2 1 1 1

2

2 21 1 2 2 1 2

1

Reaction force on any duct is given by change of impulse function

( ) ( )

4 .

4

R f f P A mV P A mV

D VAV A V V

D






2

1 2

2 2

1 1 1 2 2 2

2 2

1 2 2 2 1

3 2 2

1

4 8 2 /

8

1 1

2 2

1 ( )

2

1 101 10 1000(8 2 )

2

131

V m s

P V P V

P P V V

P Kpa

2 2

1 1 1

2 1 1 1 2 1

2 2 3

1000 (8 10 ) 24

10.053 /

( ) ( )

= 101 0.08 131 0.08 10 10.053(8 2)4 4

90.478

m AV

m kg s

R P A P A m V V

R N


(4)

m

p

m p

p pm m

p m m pm p

m

p

L=4,

L

For dynamic similarity between model and prototype their

Reynolds no. and force coefficients must be equal

Re = Re

L ρV μρVL ρVL= =

μ μ V L ρ μ

V 1 =4

V

-3

-5

2

2 22

2m m m m

p p p p

m

p

.22 10=0.27415

1000 1.78×10

F α ρV S

F ρ V L 1000 1= = 0.27415

F ρ V L 1.22 4

F=3.8503 4(Nearest Integer)

F


(-0.057)






2 1

2 1

0 0

Sol. 0.057

0.03 0.020.001 per degree

10 0

0.057 per rad

e

e

m m

m

e e

m

C CC

C


(B)

When the landing gear is lowered extra drag is produced which

acts from a point below the CG line and inturn develops a nose down

pitching moment.

L

In order to balance this moment elevator must be deflected up.

And to overcome extra drag and maintain same C throttle

must be increased.


(10.185)

At minimum thrust required condition


(157.691)

2 2

2

1

10.02 0.8

1.0

10.185

Do L LC KC CeAR

AR

AR







2

yield

2 2 2

2

2

1 avg

2

1

2

2

2 2 2

1 2 1 2 yield

330 /

140 / 70 / /

2

35 11025

35 11025

Von Mises Criteria

xx yy xy

xx yy

MN m

MN m MN m x MN m

x

x a b

x a b

2 2 2 2 2 2 2

yield

2 2 2

yield

2 2 2

2 a 2 a

3

35 3(11025 ) 330

a b ab b ab b

a b

x

157.691

x


(D)

1,2

2

3,4

3,4

2

Given roots are

0.02 0.3

2.00 2.5 1

Roots, corresponds to short period oscillation

Damped Frequency, 2.5 1 .........(1)

2

n n

d n

n

i

i i

2

2

2.............................(2)

from (1) and (2)

12.5

2

2.5625 1

Damping Ratio, 0.6246

fr

n

om (2) undamped natural frequency

2 23.201 rad/sec

0.6246

n


(0.272)






0Given, =10 , 10

sin

sin

L

D

C

C

T D W

T D

W W

0

0

cos

cos

cos sin

cos10sin10 0.272

10

L W

LW

T D

W L

T

W


( 0.00)

The pitching moment coefficient of the aircrft at zero AOA is

The pitching moment coefficient of the wing-body at zero AOA is


(A)

Stoichiometric combustion

0.066

0.066 0.066 kg/secf a

f

A

m m


(1.27)

0.02 0.2(0.4 0.3) 0.00moC







0 0

2 1

01

1 2

02 01

02

'

02 02

02 01

'

02 02

02 0

11 45 200 /

165 /

0.8

298 K

1005 J/kgK 1.4

(tan tan )

26585.4428 ( )

324.453 K

0.8

=1.071015

a

c

p

a

p

c

u m s

C m s

T

C

W uC

W C T T

T

T T

T T

T P

T P

1

1

02

01

1.2714

P

P


(215.863)

2

1

05 05

Given, 1

270 / ; 0.9 bar; 290 K

1.6 bar ; 774 K

1147 / ; 1.33

( 1)284.6 /

a a a

p

p

A m

V m s P T

P T

C J kgK

R C J kgK

105

*

*

*

1Ideal expansion 1.850

2

0.864

Nozzle is not choked.a

P

P

P

P P

1205

1

05 05

11.777 1 0.964

2

671.1

e e

a

e

e a

PM M

P

T PT

T P

485.863 / sec

215.863 /

e e e

s e a

V M RT m

F V V Ns Kg






(1138.7)

03

03

04

6

04

04

450 K 44 /

50 / 1005 / K

1 / 1047 / K

( )( )

1 44 10 51 1147 50 1005 450

1138.7336

a

g

g a

v

a p

f p

f v a f p a p

T C MJ kg

m kg s C J kg

m kg s C J kg

m C m m C T m C T

T

T

K


(9174.48)

2

0

0

3

0

10,000 ; 9.8 /

450sec; 11 /

ln

100011 10 0 450 9.8ln

Mass at burnout; 825.51

10,

e

sp e b

sp e

B

e b

B

B

p B

M kg g m s

I V km s V

MV I g

M

V VM

M kg

M M M

000 825.51

Mass at propellant; 9174.48

p

p

M

M kg


(0.446)

2 2

Power, 1800 kW

10 /

1900 / sec 1

a

t

P

m kg s

rad c u

2

3 2

2 2

2

22 2 2

2 1

?

1800 10 10

424.26

0.2232

2 0.44659

a t

d

P m c u u

u

uu r r

d r m









(0.916)

0

4

01 0

3

1

Upward deflection due to UDL

108

10

8

q

q lq l W

EI

Wl

EI

3

2

Downward deflection due to tip mass

3

W

Wl

EI

1 2

3

3

Net deflection upward

10 1

8 3

0.91667

Hence, 0.916

Wl

EI

Wl

EI

k

67


(105)

2

2

2

7 /

?

21 /

56 /

z

x

yy

xy

MN m

MN m

MN m

2 2 2

Equation of circle

h R

Points lying on the circle

,0 ( 7,0)...............(1)

, (21, 56).......(2)

z

yy xy

2 2

2 2

2 2 2

2 2 2

(1) (-7-h) 0 R

(7+h) R R=7+h

(2) (21-h) 56 R

(21-h) 56 (7 h)

h = 63

R = 70





2 2 2

2 2 2

2

h R

63 R 1764

63 42

105; 21(This is already )

so, 105 /

xx

xx

xx yy

xx MN m


(7.25)

0

0

96

sin 45 67.8825

cos 45 67.8825

Let 67.8825

o

x o

x o

M Nm

M M Nm

M M Nm

M

3 3

4

6 4

1 thickness

120 120 118 118

12 12

= 1123518.667

= 1.1235186 10

0

xx yy

xy

mm

I I

mm

m

I

max

6 2

2

( ) 60419556.87( )

Max will occur at -60 ; 60

7250346.825

= 7.250 10 /

= 7.250 /

yx

xx yy

MMy x

I I

My x y x

I

x y

N m

MN m


(0.642) 3 2

1

3 2

2

1

1 1 2 2

300 10

250 10

5 /

2 2

A mm

A mm

q N mm

T A q A q






3 3 3 3

2

2

2 1

10 10 10 2 300 10 5 2 250 10

14 /

14 /

9 /

0.64285

q

q N mm

y N mm

x q q N mm

x

y


(1)

02 2

0

Hence, 1

kx kxmx

kmx kx

m

a


(C)


(A)


(B)


(A)









2 2When she starts from Q and goes towards P, she covers 3 4 5 km to reach P.

Since she travels 15 km towards that direction, the remaining distance she goes is 15 5 10 km.

From figure,

Distance between

2 2Fatema and point P = 10 6 8 km

Hence, she needs to go 8 km towards east to reach point P.


(D)

C = 329

P = 186

M = 295

C P = 83

C M = 217

P M = 63

C P M = 500

C P M = ?

We have,

C P M C+P+M- C P- C M - P M+C P M

500 329 186 295 83 217 63 C P M

C P M 500 447 53


(A)








(D)


(C)

1200 men + 500 women can build 1 bridge in 2 weeks

11200 men + 500 women can build bridge in 1 week..........(1)

2

900 men + 250 women can build 1 bridge in 3 weeks

1900 men + 250 women can build br

3

idge in 1 week.............(2)

Similarly,multilpying (2) by 2, we get

22(900 men + 250 women) can build bridge in 1 week

3

21800 men + 500 women can build bridge in 1 week.............(3)

3

subtracting

(1) from (3)

2 1(1800 men + 500 women 1200 men 500 women) can build bridge in 1 week.

3 2

1600 men can build bridge in 1 week

6

Hence, 3600 men can build 1 bridge in 1 week


(C)


(B)




GATE-2017 AEROSPACE ENGINEERINGgatepathshala.com/qpapers/GATE2017_AE_COMPLETE_Solutions( CLASSES &...

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