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: 2 : Mechanical Engg. _ ESE MAINS


F

sin

h cp

hcp

=60k= kN/m

c.v

01(a).

Sol: F = PC.G.A

= Ahg

=

52

2

14

3

29100081.9

= 130.8 kN

hcp =hA

sinIh

2G

=

43

252

2

15

452

36

1

43

2

23

= 3m

MBC =

sin

h5F cp

=

5/4

358.130

=

4

1558.130

= 163.5 kN-m

01(b).

Sol:

Consider the control volume as shown in the

diagram. The only external force acting on

the control volume in horizontal direction is

the spring force (F)

Applying linear momentum equation for the

control volume we get,

.v.cinout Vmt

VmVmF

------ (1)

FF

( direction'x'veinactsF

)

0Vm out ( No mass is going out of C.V.)

2in aVVaVVm

0Vmt

( ,0V

free surface velocity)

equation (1) becomes

F = 0 –AV2 + 0

F = AV2 = 22 1002.04

1000 = 10N

k

F =

1000

10= 0.01 m = 1cm

Note:

Even though mass is increasing inside the

C.V. Vmt

is zero because velocity of

that mass (which is equal to free surface

velocity) is zero.

: 3 : Test – 1




1

1 2

2

U

Smooth: Laminar Flow

(x)

U

lessdydu

(x)

moredydu

U

Smooth: Laminar Flow

85 separationpoints

Rough: Turbulent Flow

120

U

Rough: Turbulent Flow

01(c).

Sol: Assumptions:

(i) Pressure is uniform at every cross-section.

(ii) Velocity over each cross-section is uniform

and is equal to average velocity.

Let P1 and P2 be the pressures at section (1)

and (2) respectively.

The linear momentum equation for the

control volume is,

Vmt

VmVmF inout

(P1–P2)A2 = (A2V2)(V2)–(A2V2)(V1)

P1 – P2 = V2(V2 – V1) --------(1)

Applying Bernoulli’s equation between (1)and (2) we get

L

2

22

2

11 hg2

V

g

P

g2

V

g

P

g2

VV

g

VVV

g2

VV

g

PPh

2

2

2

1122

2

2

2

121L

i.e.g2

VVVV2V2h

22

2121

22

L

=g2

VVV2V 2221

21

g2

VVh

221

L

Note: The control volume starts just after thejoint/expansion. Hence pressure force atinlet is P1A2.

01(d).

Sol: Assumption:

Flow over rough surface has turbulent

boundary layer and flow over smooth

surface has laminar boundary layer.

For streamlined objects like aerofoils pressure

drag is minimum (due to small wake) and

skin friction drag contributes mainly to total

drag.

The turbulent boundary layer has more

velocity gradient at surface. Hence it has

more wall shear stress which results in higher

drag force.

For aerofoils (FD)smooth < (FD)rough

In case of blunt objects like sphere both

pressure drag and skin friction drag are

present.

: 5 : Test – 1


The turbulent boundary layer is more resistant

to separation due to high momentum flux.

Hence comparatively it has delayed

separation which results in narrower wake.

Even though skin friction drag is more in case

of flow over rough sphere, overall drag is

reduced because of comparatively much less

pressure drag due to narrower wake.

For spheres, (FD)smooth > (FD)rough

01(e).

Sol: Given:

Vr = 2rsincos; V = –2rsin2

2D continuity equation in cylindrical

coordinate system is,

0Vr

1rV

rr

1r

LHS =

cossinr2rrr

1+ 2sinr2

r1

= cossinr4r1

cossinr4r1

= 0

= RHS

as continuity equation is satisfied flow is

possible

r

Z

Vr2

1rV

rr21

=

cossinr2r2

1sinr2r

rr21 2

=

2sinrr2

1sinr2

rr21 22

= 2.2cosrr2

1sinr4

r21 2

= –2sin2–[1–2sin2]

= –1

0

flow is not irrotational.

02(a).

Sol: Let us consider the coordinate system as

shown in the figure.

The body forces acting on the fluid are

gravity (in downward direction) and inertia

force (in –ve x direction)

After resolving ‘g’ in x and y direction weget

gx = –a–gsin

gy = –gcos

from pressure field equations

asinggxP

x

--------- (1)

and

cosggyP

y --------- (2)

Equations (1) and (2) show that pressure

decreases in x as well as y direction.

PB=PA+l(gcos)0

l

A

B

C

D xy

a

ag sin

g cos

g



PB = lgcos ---------- (1)

i.e. PD = l(gsin+a) --------(2)

PC = PD + l(gcos)

= l(gsin+a) + l(gcos)

i.e. PC = l(gsin+gcos+a) ------(3)

As pressure distribution on face AB has

linear variation.

2BAAB 2

PPF

= 2

2cosg0

i.e. FAB = cosg21 3 ----------(4)

similarly

2DCCD 2

PPF

= 2asingacosgsing21

= a2cosgsing22

3

------------ (5)

FCD–FAB = a2sing22

3

= m(gsin+a) [∵ l3 =m]

FCD – FAB – mgsin = ma

PD=PA+l(gsin+a)0

: 7 : Test – 1


F

Patm P1

Piston

02(b).

Sol: Assumptions:

(i) Pressure at the needle opening is

atmospheric pressure.

(ii) Friction between piston and syringe

surface is neglected.

Applying continuity equation between (1) and

(2) we get,

A1V1 = A2V2

12

2

2

11

2

12 V

DD

VAA

V = 32

2

1015.0

10

= 0.4 m/s

Applying Bernoulli’s equation between (1)and (2) we get,

f

222

211 h

g2V

gP

g2V

gP

(or)

P1–P2 = f2

122 ghVV

2

(or)

P1–Patm = f2

122 ghVV

2

------------(1)

Case-I: Inviscid fluid, hf = 0

P1–Patm = 2122 VV

2

= 232 1014.02

1000

i.e. P1–Patm = 80 Pa

From free body diagram of the piston

F = (P1–Patm)A1

= 201.04

80 = 6.2810–3 N

Case-II: Viscous fluid

The pressure drop (P) across needle is given

by

2DVL32

P

= 23

3

105.0

05.04.010132

= 2560 Pa

gP

h f

equation becomes

P1–Patm = gP

gVV2

21

22

= 80+2560

= 2640 Pa

F = (P1–Patm)A1

= 2640 201.04

= 0.207 N

02(c).

Sol: Applying Bernoulli’s equation between (1)and (2)

f

222

211 h

g2V

gP

g2V

gP



f21

21

22 h

gPP

g2VV

(or)

g2a/Qa/Q 2

12

2 = h–hf

(or) f21

22

2

hha1

a1

g2Q

Q = f22

21

21 hhg2aa

aa

--------(1)

Equation (1) gives actual discharge through

venture meter i.e. discharge with loss. The

ideal or theoretical discharge is obtained

when loss is considered zero.

gh2aa

aaQ

22

21

21th

-----------(2)

The coefficient of discharge for Venturi meter

is defined as

hhh

QQ

C f

th

actd

03(a).

Sol:

Applying Bernoulli’s equation between (1)and (2) we get

2

222

1

211 Z

g2V

gP

Zg2

Vg

P

i.e. 0 + 0 + H = 2

222 Zg2

Vg

P

i.e.

2

2223 Zg2

Vg

gZhPH

[ Hydrostatic law is valid between (2) and

(3) as streamlines are parallel]

2

2

22 Zg2

Vg

gZh0H

2

2

22 Z

g2V

ZhH

V2 = hHg2 ----------(1)

Similarly applying Bernoulli’s equationbetween (3) and (4) we get

4

244

3

233 Z

g2V

gP

Zg2

Vg

P

i.e. 0+0+h = 0+ 0g2

V24

V4 = gh2 ---------(2)

As water level in smaller tank remains same

Qin = Qout

42

22 Vd

4VD

4

2

2

4

dD

VV

From (1) and (2)

2

dD

hHg2

gh2

d

(3)

(1)

D

H

h(2)

(4)

Z2

: 9 : Test – 1


4

Dd

hhH

4

Dd

1hH

4

Dd

1

1Hh

03(b).

Sol: Consider the free body diagram of the block

when it is travelling with terminal velocity

along the incline when the terminal velocity

is reached the block travels with constant

velocity ‘V’ and hence acceleration is zero.

Applying Newton’s law along the incline

A = Wsin

sinWAhV

profilelineargminassu,

hV

dydV

A

hsinWV

Asinhmg

V

----------- (1)

2

33

2.01.030sin81.92.0800101

V

V = 7.848 m/s

Consider the free body diagram of the block

when it is travelling along the rough surface.

Again applying Newton’s 2nd law to the

horizontal motion of the block we get.

–kW = ma

–kmg = ma

a = –kg ----------- (2)

Thus block retards on the rough surface by

deceteration of magnitude g k.

Applying third kinematic equation to the

block.

V2 = u2 +2as

02 = (7.848)2 + 2(–0.29.81) S

S = 15.7 m

Xstopping = S = 15.7 m

03(c) (i) .

Sol: Given:

m = 13.600 kg/m3

= 1050 kg/m3

x = 0.06 m

Wsin–A=ma0

A

Wcos

Wsin

W

W

A

W

k W



1xh m =

1

105013600

06.0

= 0.717 m

V = 717.081.92gh2

= 3.75 m/s

= 13.5 km/hr

03(c) (ii).

Sol:

Given:

s = 850 kg/m3

= 1000 kg/m3

D = 1m

H = 0.8m

for vertical equilibrium

FB = W

gHD4

ghD4

2

s

2

8.01000850

Hh s = 0.68 m

GM = BGImin

=

268.08.0

68.014

164

2

4

= 0.0319 m

gGMK

2T2

= gGM

A/I2

=0319.081.9

14

/1642

24

=

0319.081.916/1

2

= 2.81 seconds

04(a).

Sol: Laminar Boundary Layer:

Ux

48.5xU

x48.5

Re

x48.5x

x

(0.25m) = 61025.0U

48.5

i.e. (0.25m) =

U

1074.2 3

---------(1)

we have

2

xy

xy

2U

y,xu

M

h

H

D

G

B

: 11 : Test – 1


U1074.2

105.12

Umm5.1,m25.0u

3

3

–2

3

3

U1074.2

105.1

U3.0U095.1U34.1

---------- (2)

Solving above equation by trial and error we

get

U = 1.5 m/s

Now

Rex = 61025.05.1xU

= 3.75105

xRe

48.5x

51075.3

25.048.525.0

= 2.24 mm

2fxw V

2C

x

= 2

x

URe

73.021

= 2

55.11000

1075.3

73.021

= 1.34 Pa

Turbulent Boundary Layer:

As free stream velocity remains same

U = 1.5 m/s

6x 1025.05.1xU

Re

= 3.75105

2.052.0

x 1075.3

25.038.0

Re

x38.0x

= 7.29 mm

7

1

xy

Uy,xu

u(0.25m, 1.5mm) =7

1

3

3

1029.7105.1

5.1

= 1.2 m/s

2fxw U

2C

x

=

22.0

x

URe

058.021

2

2.05w 5.110001075.3

058.021

25.0

= 5 Pa

Laminar Turbulent

U 1.5 m/s 1.5 m/s

u (0.25m,

1.5 mm)

1.34 m/s 1.2 m/s

w (0.25 m) 1.34 Pa 5 Pa

(0.25 m) 2.24 mm 7.29 mm

04(b).

Sol: Given:

75.3V

Vumax

We know2

w V -----------(1) (Definition of V)

The head loss by Darcy-Weisbach equation

is given by

hf =gD2

fLV2

------------(2) (Definition of f)

Consider the equilibrium of fluid mass

flowing through the pipe



DLD4

PP w

2

21

P1–P2 =D

L4 w

gD

L4gPP

h w21f

---------- (3)

equating equation (2) and (3)

gD2fLV

gDL4 2

w

2w V

8f --------- (4)

Now, from equation (1) and (4) we get

2V8f

V2

V8f

V

75.3

V8

f

Vu

*V

Vu maxmax

8

f75.31

V

umax

f8

175.31

V

umax

f33.11

04(c).

Sol: Given,

u = umax ( c1 + c2 y + c3 y2)

1. at y = 0 , B = 0

0 = umax (c1 + 0 + 0)

c1 = 0

2. at y = B , u = 0

0 = umax (0 + c2 B + c3 B2)

c2 = –c3 B ------- (1)

3. at y =2

B , u = umax

2

32maxmax 2

Bc

2

Bc0uu

4

Bc

2

Bc1

2

32

Or

4

Bc

2

BBc1

2

33

[∵ from (1) c2 = –c3 B]

4

B

2

Bc1

22

3

23 B

4c

P1 P2

w

wL y

B

: 13 : Test – 1


B

4B

B

4Bcc

232

2

2max yB

4y

B

4uu

i.e.,

2

2

max B

y

B

yu4u

now, udAQ

dy1B

y

B

yu4

B

0 2

2

max

B

0

32

max B3y

B2y

u4

3

B

2

Bu4 max

Bu3

2max

Average velocity, B1

Bu3

2

A

QV

max

i.e, maxu3

2V

05(a).

Sol: Given,

P = 250 kPa,

H = 5 m,

f = 0.02,

L = 10 m,

D0 = 10 cm,

Di = 5 cm,

kentry = 0 ,

Hs = 0, hfs = 0

Case – I :

Water is going out of tank.

Assumption: Pressure at pipe exit is

atmospheric.

Applying Bernoulli’s equation between watersurface in tank and pipe exit we get.

g2

VkhZ

g2

V

g

PZ

g2

V

g

P 22

entryf2

222

1

221

h

2

2

2

21

1

Dg2VLf

g2V

Zg

P

------- (1)

Now, io

2i

20

h DD

DD4

4

P

A4D

= Do – Di

= 10 – 5 = 5 cm

Substituting in equation (1) we get

05.0

1002.01

g2

V5

9810

10250 22

3

V2 = 10.94 m/s

Q = AV2

= 94.1005.01.04

22

Q = 0.232 m3/s

Case – II: Water is going into the tank

Assumption:

5m

P=250 kPa(Gauge)

Do = 10 cm,Di = 5 cm

10 cm



1

2

P = 250 kPa

HP

Hs0

Pump

F1

F2

l2

l1

l1 = l2 = 12 + 6/3 = 15 cm

1. The pump takes water from atmospheric

condition.

2. Suction head of the pump is negligible.

Applying Bernoulli’s equation

between 1 & 2 we get

g2

VKhhZ

g2

V

g

PHZ

g2

V

g

P 2

1exitfdfs2

2

22p1

2

11

i.e.g2

VKhZ

gP

H2

1exitfd2

2P …… (1)

[Note: Even though pipe is well rounded at

joint its exit loss coefficient will be ‘1’because all K.E. in pipe is lost due to mixing

inside the tank]

Since the discharge is same,

V1 = 10.94 m/s

81.92

94.101

05.081.92

94.101002.05

9810

10250H

223

P

= 60.98 m

kW28.16385.0

98.60232.09810gQHP

0

p

05(b).

Sol: Assumptions:

(i) Line of action of drag force passes through

the centre of hemisphere.

(ii) Torque generated by the drag forces on the

two cups aligned to the flow is negligible.

A2

UCF

2

1D1

= 2

2

06.0418

592.1

2

42.0

= 4.453 × 10-3 N

Similarly

AU2

CF 22D

2

= 2

2

06.0418

592.1

2

17.1

= 12.41 × 10-3 N

The torque generated on the shaft of the

anemometer is given by

T = 1122 FF

15.010453.415.01041.12 33

= 1.193 × 10-3 N-m.

: 15 : Test – 1




Vp

constV

V

p

m

Vm

05(c).

Sol:

(i) Kinematic similarity:

The model and the prototype are said to

be kinematically similar if the ratio of

magnitudes of velocities at similar points

is constant and direction of velocities at

those points is same.

The kinematic similarity is obtained when

streamline pattern for the model and

prototype is same.

(ii) Distorted Model:

If length scale ratio for model and

prototype is not same in different

direction then the model is called distorted

model

i.e.vp

vm

Hp

Hm

L

L

L

L

In many situations it is impractical to

maintain same scale ratios in different

direction therefore distorted models have

to be created.

For example if same scale ratio is

maintained for river model in horizontal

as well as vertical direction then vertical

dimension of model river will be very

small and hence unnecessary surface

tension effects will become dominant

which are not present in actual river.

(iii) Incomplete similarity:

For complete similarity all independent

‘’ terms are needed to be matched formodel and prototype. However it is

always not possible to match all the

independent ‘’ terms. If one or more independent ‘’ term are

not matched between model and prototype

then such a similarity is called incomplete

similarity.

For example, the total drag of ship

(friction drag + wave drag) depends on

both Reynolds number (Re) and Froude

number (Fr). However it is not easy to

match Re & Fr simultaneously (∵ Lr =

3/2

r condition has to be satisfied).

Therefore, only ‘Fr’ is matched for modeland prototype to predict wave drag. The

frictional drag is calculated analytically

using boundary layer theory.

ace engineering academy€¦ · ace engineering academy hyderabad|delhi|bhopal ... assumptions: (i...

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