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GATE 2007 SOLUTIONS

AEROSPACE ENGINEERING IIT Kanpur

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Sol. B

Turbofan engine used in passenger transport aircraft has less specific fuel consumption because of high by pass ratio.

Sol. A

m = 1 kg, 휉 = 0.2 = = √

=

휔 = 5 rad/s

C = 2휉푚휔 = 2 Ns/m

Sol. B

f(휃) = cos휃 sin휃− sin휃 cos휃

Contact- 9740501604

f(훼).f(훽) = cos훼 sin훼− sin훼 cos훼

cos훽 sin훽− sin훽 cos훽

= cos(훼 + 훽) sin(훼 + 훽)− sin(훼 + 훽) cos(훼 + 훽) = f(훼 + 훽)

Sol. A

Sol. D

From Euler iteration formula,

x = x + ∆t × f(x )

Sol. D

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Sol. A

Equation for glider,

L = Wcos훾

D = Wsin훾

tan 훾 = =

We know, tan 훾 = =

Range = height×

For maximum range, 푅푎푛푔푒 = height×

So lift to drag ratio should be maximum.

Sol. D

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Sol. A

J(x) = 푥 − 7x + 30

J’(x) = 2x - 7 = 0

x = 7/2

J”(x) = 2 → positive

So J(x) is minimum at x = 7/2

Sol. C

Longitudinal static stability will be same because of pitching moment is same.

Sol. B

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If propeller is mounted ahead of C.G., it will destabilize the a/c. And longitudinal static stability will decrease.

Sol. C

For steady and level flight,

L = W = 휌푉 푆퐶

T = D = 휌푉 푆퐶

When speed is increases, lift will also increase. For balancing lift 퐶 should be decreases or in other way angle of attack should be decreases by elevator.

Sol. D

For plain strain condition in x-y plane,

휖 = 휖 = 훾 ≠ 0

휖 = 훾 = 훾 = 0

휎 = 휎 = 휎 = 휏 ≠ 0

휏 = 휏 = 0

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Sol. C

Sol. B

Sol. B

Sol. B

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Sol. D

Sol. B

Sol. B

For thin airfoil, critical Mach number is higher than thick airfoil.

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Sol. D

h > h

For satellite A,

Total energy, 퐸 = T + V = 푚푣 -

푟 = 푅 + ℎ

푣 =

퐸 = 푚

-

= -

For satellite B,

Total energy, 퐸 = T + V = 푚푣 -

푟 = 푅 + ℎ

푣 =

퐸 = 푚

-

= -

So, 퐸 > 퐸

푇 < 푇

푉 > 푉

Contact- 9740501604

Sol. D

Both P and Q are square matrices of same size.

(i) For example,

P = 1 01 0 and Q = 0 0

0 1

PQ = 0 00 0

(ii) PQ = 퐼

P = 푄 (|푄| ≠ 0)

(iii)

(P+ Q) = (P+ Q). (P+ Q) = 푃 + QP + PQ + 푄

(iv)

(P− Q) = (P− Q). (P − Q) = 푃 - QP - PQ + 푄

Contact- 9740501604

Sol. B

mg = k훿

1× 9.81 = k× 10 × 10

k = 613.125 N/m

x = 퐴 cos휔 푡 + 퐵 cos휔 푡

At t = 0,

A = 푥 = 10 mm

푥̇ = 0, B=0

So x = 푥 cos휔 푡

휔 = = . = 24.76 rad/s

x = 10 cos(24.76푡)

Sol. B

Earth radius, R = 6.37× 10 m

g = 9.81 m/푠

Contact- 9740501604

h = 6.30× 10 m

r = R + h = 7× 10 m

푉 = = = 7.54 km/s

g =

푉 = √2푉 = 10.66 km/s

∆푉 = 푉 - 푉 = 3.12× 10 m/s

Sol. A

Contact- 9740501604

Curvature, k = =

M = 퐸퐼 =

휌 =

휌 = 휌 + 휌

휌 = +

휌 =

Sol. B

R =

Mass flow rate, 푚̇ = 휌푉퐴 (A = )

Contact- 9740501604

From question, mass flow rate is same

휌퐴 푉 = 휌퐴 푉 = 휌푉퐴 = k

R = × = =

So R ∝

→ R > R

Sol. C

Minimum pressure coefficient does not depend upon dimensions.

Sol. A

L = 휌푉 푆퐶

퐶 is geometric property, it will remains same for same angle of attack. And lift will be different due to different area.

Contact- 9740501604

Sol. D

F = 퐹 sin휔푡

Amplitude, X = /

At resonance, 휔 = 휔

푋 = / = 1 cm...... (1)

At half of resonance frequency, 휔 = 휔 /2

푋 = /[ / ] [ ]

= // [ ]

= 0.5 cm....... (2)

Equations (1)/(2),

= /

퐹0/푘

9/16+[휉]2

= . = 2

휉 = 0.1936

Sol. C

A = 2 10 3

Contact- 9740501604

휆 ,휆 → Eigen values

휆 + 휆 = 5

휆 × 휆 = 6

휆 = 2

휆 = 3

Sol. D

From previous solution,

휆 = 2

휆 = 3

So the eigen values of the matrix 퐴

휆 = 1/2

휆 = 1/3

Contact- 9740501604

Sol. B

r = R + h = 6.37× 10 + 35.90× 10 = 42.27× 10 m

∆푖 = 10.5°

GM = g푅 = 3.9805× 10 푚 /푠

푉 = = . ×. ×

= 3068.72 m/s

∆푉 = 2푉 sin ∆푖/2 = 2× 3068.72 × sin(10.5/2) = 561.58 m/s

훼 = 90 + ∆푖/2 = 95.25°

Sol. C

η = 0.8

W = 73108 N

푉 = 50 m/s

(푅/퐶) = 15 m/s

푅/퐶 = 푉 sin 훾 = 푉 =

(푅/퐶) = = ( )×

Excess brake power = (푅/퐶) × = ×.

= 1371 kW

Contact- 9740501604

Sol. C

Elevator control power, 퐶 = - 휂푉 퐶 휏

= .( )

= - 0.008 per deg

Sol. A

→Two complex roots correspond to Dutch roll stability.

휆 , = - 0.06 ± j 1.5 ; j = √−1

푆 + 2휉 휔 푆 + 휔 = 0 (Dutch roll stability)

Lets,

A푆 + B푆 + C = 0

− = 휆 + 휆 = - 0.12

= 휆 × 휆 = (−0.06) + (1.5) = 2.253

Now, 푆 + 0.12푆 + 2.253 = 0

휔 = 2.253

휔 = 1.501 rad/s

Contact- 9740501604

2휉 휔 = 0.125

휉 = 0.041

Sol. C

At an altitude of 10 km above the sea level,

T = 223 K, 휌 = 0.413 kg/푚 , V (ASI) = 60 m/s

V = V (ASI) = 103.33 m/s

At sea level,

휌 = 1.225 kg/푚

R = 288 J/(kg/K)

Stagnation pressure, 푃 = P + 휌푉

P = 휌푅푇 = 0.413× 288 × 223 = 26524.512 N/푚

So 푃 = 26524.512 + × 0.413 × (103.33) = 2.87× 10 N/푚

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Sol. A

Sol. A

C ∝ l (Tail moment arm)

Sol. B

For vertical tail up side,

< 0

For vertical tail down side,

> 0 (Not desirable)

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Sol. B

x – y + 2z = 0

2x + 3y – z = 0

2x – 2y + 4z = 0

A = 1 −1 22 3 −12 −2 4

|퐴| = 12 – 8 + 2 – 12 + 8 – 2 = 0

|퐴|is zero means the rank of A matrix is less than 3 and also less than the no. of unknown variables. So it has infinite no. of non – trival solutions.

Sol. A

Contact- 9740501604

Skin friction drag in laminar flow is less than in turbulent flow over an airfoil.

The turbulent boundary layer is more energetic due to higher K.E.

The velocity gradients are higher in turbulent boundary layer.

Sol. B

At 0.2 m, boundary layer thickness, 훿 = 5 mm

훿 = ; 푅 =

훿 = = /

훿 = /

...... (1)

훿 = /

..... (2)

Equations (1)/(2),

= /

= ..

/

훿 = 10 mm

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Sol. A

Longitudinal static stability is directly proportional to horizontal tail area.

Elevator control power is also directly proportional to elevator area.

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Sol. B

= ∅ =

For composite shafts,

T = 푇 + 푇

∅ = ∅ = ∅

= ∅

= ∅

T = ∅ + ∅

∅ = +

Sol. A

T = 300 K

Contact- 9740501604

T = 424 K

훾 = 1.4

휂 = 0.81

휂 =

=

휂 =

=

T02T01

−1

0.81 = .

1.413−1

= 2.74

Sol. A

Contact- 9740501604

Sol. B

훼 = 70°

훽 = 37°

For impulsive turbine degree of reaction, °푅 = 0

°푅 = - ∅ (tan훽 − tan훽 )

Flow coefficient, ∅ =

°푅 = - (tan훽 −tan훽 ) = 0

훽 = 훽 = 37°

Sol. C

Contact- 9740501604

Sol. A

→Isentropic efficiency of a compressordoes not remains constant for whole compressor.

→ Flow separation is less critical in centrifugal compressor because of higher turn of flow in radial direction.

→ The pressure ratio of centrifugal compressor is not zero. It has some amount of ratio.

→ Due to complex design, centrifugal compressor is not designed with multi – stages. Axial compressor is designed with multi – stages.

Sol. C

푉 = 푉

푥 = 0

dV∝dx

dV= kdx

= k

Contact- 9740501604

= k

- k = 0

(퐷 – kD).x = 0

푚 – km = 0

m(m - k) = 0

m = 0, k

x = 퐶 푒 . + 퐶 푒 . → (Exponential with time)

x = 퐶 + 퐶 푒 . ..... (1)

At t = 0 → x = 0

0 = 퐶 + 퐶

퐶 = - 퐶

= 퐶 푘푒 .

At t = 0 → V = 푉

푉 = 퐶 푘

퐶 =

퐶 = -

From equation (1),

x = - + 푒 . = (푒 . − 1)

Contact- 9740501604

Sol. B

M = 2000 kg, 퐼 = 280 s, ∆푉 = 500 m/s

∆푉 = c.ln

퐼 = ̇

= ̇̇

=

∆푉 = 퐼 .g.ln

500 = 280× 9.81 × ln

= 1.199

M = 1668.05 kg

M = M - M = 331.95 kg

Sol. B

Contact- 9740501604

. = 0.0280

. = ×[ × × ]

[ × × × × ] = 0.0664

∅ = .

.

= . .

= 0.421

Sol. A

Von – misesyield stress for 2 D,

휎 = 휎 + 휎 − 휎 휎

So Von – misesyield criterion for 2 D,

휎 + 휎 − 휎 휎 ≤ 휎

Contact- 9740501604

Sol. B

Shear stress, 휏 =푉 . ̅.

From below figure,

A = ( − 푧).b

푧̅ = z + . ( − 푧) = . ( + 푧)

Contact- 9740501604

휏 =푉 . ̅.

= 푉 .( ). × .( )

.= 푉푧 . ℎ2

2

퐼푦× 1 −

ℎ2

2

Sol. B

Contact- 9740501604

Sol. B

푚̇ = 719 kg/s

푉 = 1794 m/s

퐴 = 0.635 푚

F = 푚̇ 푉 + 퐴 (푃 − 푃 )

F = 푚̇ 푉 (For optimum expansion)

= 719×1794 = 1289.88 kN

Contact- 9740501604

Sol. A

From figure,

Lets, H’ is the height of inlet.

→Assuming unit depth,

퐴 = H’× 1 = H’

퐴 = 2h× 1 = 2h

푚̇ = 푈 H’ (density is constant)

푚̇ = ∫푢. 푑푦 = ∫ .푑푦 + ∫ .푑푦 = 푈 ℎ

→In above equation h is constant.

From continuity equation,

푚̇ = 푚̇

푈 H’ = 푈 ℎ

So, H’ = h

퐴 = h

퐴 = 2h

From momentum equation, resultant force is equal to change in moment. (Loss of momentum is called drag.)

So, Drag, D = 푀̇ - 푀̇

Contact- 9740501604

푀̇ = 휌푈 ℎ.푈 = 휌푈 h

푀̇ = ∫휌푢푑푦. 푢 = ∫휌푢 푑푦

= ∫ 휌 푑푦 + ∫ 휌 푑푦 = .휌푈 h

Drag, D = 푀̇ - 푀̇ = . 휌푈 h

Sol. A

The flow is incompressible and frictionless. So there is no pressure drop across combustor.

Sol. B

M = 0.85

M = 0.38

훾 = 1.4

Pressure recovery, 휋 = = 0.92

Contact- 9740501604

휋 = 1+휂푑

훾−12 푀푎

2

1+ 훾−12 푀푎

2

휋훾−1훾 =

So, 휂 = 0.813

Sol. A

Sol. C

Contact- 9740501604

Trapezoidal rule,

∫ f(x)dx = [(푦 + 푦 ) + 2(푦 + 푦 + 푦 + ⋯+ 푦 )]

h = 0.2

0 + n× 0.2 = 1

n = 5

x 0 0.2 0.4 0.6 0.8 1

y 1 0.961 0.862 0.735 0.609 0.500

∫ 퐽(푥)푑푥 = ∫ 푑푥 = [(푦 + 푦 ) + 2(푦 + 푦 + 푦 +푦 )]

= . [(1 + 0.500) + 2(0.961 + 0.862 + 0.735 + 0.609)]

= 0.7834

Sol. D

By fuel injection system mixing can be well with air.

Sol. B

Contact- 9740501604

Sol. A

Pressure coefficient, 퐶 = = 1 - 4푠푖푛 휃 = 0

휃 = 30°, 150°, 210° and 330°

Sol. C

x = 푐 /

푥 = c

푥 - c = f(x) = 0

From Newton – Raphson method,

푥 = 푥 – ( )( )

Contact- 9740501604

f(x) = 푥 – c

f(‘x) = 3푥

푥 = 푥 - =

Sol. B

Torsion constant, J = ∫

= = 2휋푟 푡

Contact- 9740501604

Sol. A

From the given statement 푃 > 푃 hence static pressure has to decrease across the wave. This can happen only if there is an expansion wave.

Case 1: 푃 > 푃 In this case static pressure difference between upstream and downstream is higher than design condition. Hence expansion will be more. When this fan hits the solid boundary at B, like reflection will happens. (Note – in solid boundary always like reflection happens just like mirror. i.e., oblique wave reflects as oblique wave and expansion wave reflect as an expansion wave) and for all values of 푃 > 푃 B always seen as expansion wave.

Case 2: 푃 < 푃 Here again the condition given in this problem is 푃 > 푃 . Hence absolute static pressure difference between upstream and downstream is less than that of design condition. This can merely change the width of the fan but expansion will happens to match the downstream conditions. As mentioned earlier B sees an expansion and as it is solid boundary like reflection will occurs.

Hence only option says about compression wave as reflection wave is wrong.

Contact- 9740501604

Sol. C

At stalling, 퐶 reduces drastically.

Sol. C

Contact- 9740501604

y = lim → →

→Using L – Hospital,

y = lim → = 1

Sol. A

Sol. B

Contact- 9740501604

S.M. = 푋 - 푋 =

= -

푋 = 0.3, = - 0.1

S.M. = 푋 - 푋 = - = 0.1

푋 = 0.1 + 0.3 = 0.40

For most forward location,

푋 - 푋 = 0.15

푋 = 0.40 – 0.15 = 0.25

푋 = 0.25 푐̅

Sol. A

For most aft location,

푋 - 푋 = 0.05

푋 = 0.40 – 0.05 = 0.35

푋 = 0.35 푐̅

Sol. D

Contact- 9740501604

= 휕퐶푚휕퐶퐿 = . . = = 0

푋 = 푋 = 0.40

푋 = 0.40 푐̅

Sol. A

Here stiffness matrix has non-zero non-diagonal terms so it is statically coupled and in mass matrix, non-diagonal terms are zero

Contact- 9740501604

Sol. C

m m ω - [m (k + k )+m (k + k )]ω + [k k + k k + k k ] = 0

Here, k = k = k =k

m = m, m = 3m

3푚 휔 - 8km휔 + 3푘 = 0

휔 + 휔 + = 0

휔 = 43푘푚

±√73푘푚

휔 = ± √

Contact- 9740501604

Sol. C

W = 20,000 N, 푃 = 5× 10 J/s

(R/C) = 푉 sin훾 = 푉 =

(R/C) = ( )

4 =

푃 , = 130,000 W

Contact- 9740501604

푃 ∝ 휌

,

, =

휌 = 1.225 kg/푚

휌 = 0.74 kg/푚

푃 , = 푃 , × = 78,530.61 W

At 5km altitude,

(R/C) = , ( ) = 1.42 m/s

Sol. B

(R/C) = 푉 sin훾 = 1.43 m/s

Pitch angle, 훽 = 훼 + 훾

훼 = 4°, 훽 = 5°

훾 = 1°

푉 = ( / ) = 81.93 m/s

휌푉 = 휌 푉

푉 = 푉 = 63.68 m/s

Contact- 9740501604

Sol. C

푎 = 퐶 = = 0.1 per degree

훼 = 8°, u = 20 m/s

L = 35.3 N, 휌 = 1.225 kg/푚 , S = b× 푐 = 1.2× 0.2 = 0.24 푚

L = 휌푢 푆.퐶

퐶 = 0.6 = 훼 =

a = = . = 0.075 per degree

Sol. B

a = 푎0

AR = = = 6

0.075 = 0.100

0.1×

180π

. ×

e = 0.91

Contact- 9740501604

Sol. D

F(s) = ( )( )( )

( )( )( )

= +

1 = A + B..... (1)

10 = 20 A + 2 B.... (2)

A = 4/9

B = 5/9

F(s) = ( )( )( )

= / + /

Sol. B

퐿 ( )( )( )

= 퐿 + 퐿

= 푒 + 푒

Contact- 9740501604

Sol. A

Sol. A

Contact- 9740501604

Sol. C

Inlet condition of compressor,

V = 127 m/s

휌 = 1.2 kg/푚

훾 = 1.4

푃 = 0.9 MPa

Outlet condition of compressor,

V = 139 m/s

푃 = 3.15 MPa

= ..

= 3.5

Sol. B

휂 = 0.89

= = 3.5 . = 1.495

휂 = =

Contact- 9740501604

= = 3.5 . = 1.43

휂 = . .

= 0.8686

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