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FluidMechanics SolutionsforVolume–I_ClassroomPracticeQuestions

Chapter- 1 Properties of Fluids

Common data Q. 04 & 05 04.

Ans: (c)

Sol: D1 = 100 mm , 01. Ans: (c) Sol: For Newtonian fluid whose velocity profile

D2 = 106 mm

Radial clearance, h 

is linear, the shear stress is constant. This



behavior is shown in option (c).

D 2  D1 2

106  100  3mm 2

L = 2m  = 0.2 pa.s

02. Ans: 100 Sol:  

V 0 . 2  1. 5  = 100 N/m2 3  h 3  10

N = 240 rpm



03. Ans: 1 Sol:

2  240 2N = 60 60

 = 8 

WSin30

= 83.77N/m2

30o

W

F  A

W sin 30 

 AV h

100 1 0.1 V  2 2  10 3 V = 1m/s

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r 0.2  8  50  10 3  3 h 3  10

05.

Ans: (b)

Sol: Power, P 

2 2Lr 3 h

2  8   0.2  2  0.05 3  10 3 2



3

= 66 Watt

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:2:



06. Ans: (c) Sol:

ME _ ESE -19 _Vol – I _ Solutions

Bingham plastic  Fluid behaves like a solid until a minimum yield stress beyond

 30

which it exhibits a linear relationship between shear stress and the rate of strain.

18 Slope = constant 6

09. Ans: (b) Sol: V = 0.01 m3

1

 = 0.75  10–9 m2/N 0

1

3

5

du/dy

dP = 2107 N/m2

 Newtonian fluid

07. Ans: (a) Sol:

du  dy

1 1 4   109  9 3  0.75  10

K

 dP dV / V

dV 

u = 3 sin(5y) du  3 cos5y   5  = 15cos(5y) dy  y0.05

K

du  dy y 0.05

 2  107  102  3 = –1.510–4 9 4 10

10. Ans: 320 Pa Sol:  P 

8  8  0.04 32  10 2   1 10 3 10 3 D

P = 320 N/m2

= 0.5  15 cos5  0.05 1   = 0.5  15  cos   0.5  15  4 2   = 7.53.140.707  16.6N/m

2

11. Ans: (d) Sol:



As the temperature is increased, the viscosity of a liquid decreases due to the reduction in intermolecular cohesion.

08. Ans: (d)



Sol:



Ideal fluid 



Newtonian fluid  Shear stress varies



In gases, the viscosity increases with the rise in temperature due to increased

Shear stress is zero.

linearly with the rate of strain.

molecular activity causing an increase in the change of momentum of the molecules,

Non-Newtonian fluid  Shear stress does

normal to the direction of motion.

not vary linearly with the rate of strain.



Thus, statement (I) is wrong but statement (II) is correct.

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:3: 12.

Fluid Mechanics

Ans: (c)

Sol: The surface energy is given by

The plate moves with velocity V

E =   area



V

Surface tension,  is the property of fluid.

2

y

Hence, it is independent of the size of the bubble. Thus, statement (II) is wrong.

From Newton’s law of viscosity, 

Conventional Practice Solutions

du dy

Let A be area of plate

 F1 = 1  Area of plate F1   1 

01. Sol:

V A h y

F2   2  (h-y) 1

h

2

y

F

1

(h-y)

As area increases, surface energy will increase. Thus, statement (I) is correct.

V A y

(i) Shear force on two sides of the plate are equal: F1 = F2 1  VA  2 VA  h y y 1 h  y  2 y

Assumptions:



Thin plate has negligible thickness.



Velocity profile is linear because of narrow gap.

h 1 1  y 2



Given fluid is a Newtonian fluid which

h 1   2  y 2

obeys Newton’s law of viscosity. The force required to pull it is proportional to the total shear stress imposed by the two oil layers. F = F1 + F2 , Where F1 = Force on top sides of plate, F2 = Force on bottom side of plate ineerin ACE Engineeri ng Publications

y

(ii)

2h 1  2

The position of plate so that pull required to drag the plate is minimum. F

 1VA  2 VA ,  y h y [V, A, 1 & 2, h are constant]

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:4:

For minimum force,

dF 0 dy

Assumptions:

–1VA(h –y)–2 (–1) – 2VAy–2 = 0  2 VA  VA  1 2 2 y  h  y

h  y2 y2



The gap between two cylinders is narrow and hence velocity profile in the gap is assumed linear.



  1 2

h y  y

ME _ ESE -19 _Vol – I _ Solutions

No change in properties Torque = Tangential force  radius

1 2

Force = shear stressArea

h 1  1 where y is the distance of the y 2



Where h is the clearance (radial)

thin flat plate from the bottom flat surface.

h

h

y 1

  VA h

10  9.75 2

= 0.125cm = 1.2510–3m

1

Area = DL

2

= 0.12.510-2 02. Ans: 8.105 Pa. S Sol: Torque = 1.2N-m

Speed,

= 7.853910–3m2

N = 90 rpm

Diameter, D1 = 10 cm , H = 2.5cm

D2 = 9.75 cm

Fs 

  r  A h



2N 2  90  3 rad / s  60 60

Torque = Fsr

2.5 cm





rA r h



r 2 A h

1.2 

  3  (0 .05)2  7.8539  10 3 1 .25  10 3

 = 8.105 Pa.s 9.75cm 10 cm ineerin ACE Engineeri ng Publications

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:5:

Fluid Mechanics



The manometer shown in Fig. 2 is for measuring pressure in liquids only.

Chapter- 2 Pressure Measurement & Fluid Statics



The manometer shown in Fig. 3 is for measuring pressure in liquids or gases.

 01. Ans: (a)

The manometer shown in Fig. 4 is an open ended manometer for positive pressure

Sol: 1 millibar = 10 10 = 100 N/m -3

5

2

measurement.

One mm of Hg = 13.610 9.81110 3

-3

= 133.416 N/m2 2

6

05. Ans: 2.2

2

Sol: hp in terms of oil

1 N/mm = 110 N/m 2

4

2

1 kgf/cm = 9.8110 N/m

so ho = smhm 0.85h0 = 13.60.1

02. Ans: (b) Sol:

h0 = 1.6m hp = 0.6+1.6 Local atm.pressure

710 mm

(350 mm of vaccum) 360 mm Absolute pressure

 hp = 2.2m of oil (or) Pp – oil  0.6 – Hg  0.1 = Patm Pp  Patm oil

   Hg  0.1  0.6       oil

Sol: Pressure does not depend upon the volume

13.6 0. 1  0.6 = 2.2 m of oil 0.85 Gauge pressure of P in terms of m of oil

of liquid in the tank. Since both tanks have

= 2.2 m of oil



03. Ans: (c)

the same height, the pressure PA and PB are same.

06. Ans: (b) Sol: h M 

04. Ans: (b)

sw hw 2 sw h w1  h N   h0 s0 s0

Sol:



The manometer shown in Fig.1 is an open ended manometer for negative pressure measurement.

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hM  hN 

9 18  3 0.83 0.83

h M  h N  13.843 cm of oil

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:6: 07. Ans: 2.125 Sol:

FV = x 

FV = gV  1000  10 

I hP  h  Ah 2

ME _ ESE -19 _Vol – I _ Solutions

1m

 2 2 4

FV = 10 kN

D 4  4

 x = 10

64  D 2  2   22  4  2.125m  2 64 2

11. Ans: (d) Sol: Fnet = FH1 – FH2

D D2 FH1     D 1  2 2

08. Ans: 10 Sol: F  ghA



  9810  1.625  1.2 2  0.8 2 4 F = 10kN 09. Ans: 1 Sol:



FH 2   

D D D 2  1  4 2 8

2  1 1  3 D = D 2    = 8  2 8

12. Ans: 2 Sol: Let P be the absolute pressure of fluid f3 at mid-height level of the tank. Starting from 2x

the open limb of the manometer (where 2x

x

Fbottom = g  2x  2x  x FV = gx  2x  2x

pressure = Patm) we write : h  Patm +   1.2 – 2   0.2 – 0.5    0.6   = P 2  or P – Patm = Pgauge

FB 1 FV

= (1.2 - 20.2 – 0.5 0.6 – 0.5 

h ) 2

For Pgauge to be zero, we have, (1.2 – 0.4 – 0.3 – 0.25 h) = 0

10. Ans: 10 Sol:

or

h

0.5 2 0.25

2m

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:7:

Fluid Mechanics

FH = w  6.727  Area (projected)

13. Ans: (b) Sol: The depth of centre of pressure from the

= w  6.727 

free liquid surface is given by hcp = h  h cp  h 

Or,

I xx , c Ah

= w  6.727  4.5

--------(1)

= 932.94 kN

I xx , c

hcp = 6.727 

Ah

From the above relationship, as h increases, I xx ,c Ah

  32 2

= 6.727 

decreases. Thus, at great depth, the

0.10976 R4 R 2  6.727 2 0 .10976  32  2   6.727

= 6.727 + 0.0935 = 6.8205 m from free liquid surface

difference (hcp – h ) becomes negligible. Hence, statement (I) is correct. Also, it is clear from equation (1) that hcp is

= (8 – 6.8205) m from base B = 1.1795 m from base B.

independent of the density of the liquid.

Taking moment about B FA  3 = 932.94  1.1795 Conventional Practice Solutions

 FA = 366.8 kN 02. Sol:

01. Sol:

4 R   h  5  3  3   4  3  = 5  3  = 5 + 1.727 = 6.727 m 3   ineerin ACE Engineeri ng Publications

R h  1.5   2   FH = ghA projected

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:8:

ME _ ESE -19 _Vol – I _ Solutions

R =  g1.5   R  3 2 

Chapter- 3

= 1.5  1.5 3  3

Buoyancy and Metacentric Height

= 27  N





1 3  33 h cp = 3  12 3  3 3 

= 3.25 m from free liquid surface = 3.25 – 1.5 = 1.75 m from A

01. Ans: (d) Sol:  2m

d

 R 2    9  3 27 FB =  3 =   = N 4 4  4 

4m

FB = weight of body

FB will act through the centroid of the 4R quadrant which is at a distance from 3

1.25m

bgVb = fgVfd 640421.25 = 1025(41.25d) d = 1.248m

the vertical line AB. Now, taking moment of the forces about the hinge A, we write

Vfd = 1.24841.25

4R Fs  3  FB   FH  1.75  0 3 where Fs is the force in x-direction on the

Vfd = 6.24 m3

stop at B & Vs is in y-direction (does not contribute in the moment). 27  4R 3Fs = 271.75–  =104(271.75 – 93) 4 3 = 104  27  0.75 = 202.5 kN.m  Fs 

202.5 = 67.5 kN 3

02. Ans: (c) Sol: Surface area of cube = 6 a2

Surface area of sphere = 4 r2 4r2 = 6a2 2  a    3  r Fb,s  Vs

4 3 r 4  r3  3 3  a 3  2  3 r   3   

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2

4 r 3 6   3  2 2 3   r   3  3   

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:9: 03. Ans: 4.76 Sol:

Fluid Mechanics 05. Ans: 1.375

FB = FB,Hg + FB,W

Sol: Wwater = 5 N

WB = FB

Woil = 7 N S = 0.85

x

W – Weight in air

water

FB1 = W – 5 FB2 = W – 7

Hg

(10–x)

W – 5 = 1gVfd…..(1) bgVb= HggHg+wgw

W – 7 = 2gVfd…..(2)

bVb = HgHg+ww

Vfd = Vb W  5  1gVb

SVb = SHgHg+Sww

W  7   2 gVb 2  1   2 gVb

7.6103 = 13.6102(10–x)+102x –6000 = –1260x

Vb 

x = 4.76 cm

2 1000  850 9.81

Vb = 1.359110-3 m3 04. Ans: 11 Sol:

W = 5 + (98101.359110-3) 

W = 18.33 N

FB

W = b g Vb

1.6m

18.33 b 9.81  1.3591  10  3

T

b = 1375.05 kg/m3 Sb = 1.375

FB = W + T W = FB – T

06. Ans: (d) Sol: For a floating body to be stable, metacentre

= fgVfd – T



4 = 10 3  9.81  0. 8 3  10  103 3



should be above its center of gravity. Mathematically GM > 0.

= 21 – 10 W = 11 kN

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: 10 :

ME _ ESE -19 _Vol – I _ Solutions

Shear stress on one side of the plate

07. Ans: (b)

W = FB

Sol:



bgVb = fgVfd bVb = fVfd

Fs = total shear force (considering both sides of the plate)

  0. 6  d 2  2 d  1  d 2  x 4 4 x = 1.2d

 2A   

GM = BM – BG



d d  = 0.052d  2 19.2  d  1. 2 d 4 BG = d – 0.6d = 0.4d BM 

4

I  V 64

2AV y

2 1.5 1.5  2.5  0.1 11  10 3

= 102.2727 N Weight of plate, W = 50 N Upward force on submerged plate,

Thus, GM = 0.052d – 0.4d = –0.348 d

Fv = gV = 900  9.81  1.5 1.5  10–3

GM < 0

= 29.7978 N

 Hence, the cylinder is in unstable condition.

Total force required to lift the plate

08. Ans: 122.475 Sol:

dU dy

V=0.1m/s

= F s + W – FV = 102.2727 + 50 – 29.7978

F

= 122.4749 N

Fs

Fs

09. Ans: (d) Sol:



Statement (I) is wrong because the balloon filled with air cannot go up and up, if it is released from the ground.

W



The thickness of the oil layer is same on either side of plate y = thickness of oil layer 

However, with increase in elevation, the atmospheric pressure and temperature both decrease resulting into a decrease in air density. Thus, statement (II) is correct.

23.5  1.5  11mm 2

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: 11 :

Fluid Mechanics

The centre of buoyancy B will be at a distance

Conventional Practice Solutions

h from O as shown in the figure. 2

01. Ans: (i) 0.33, (ii) 0.5 m

Or,

OB =

h = 0.33L 2

Sol: Given data : Inner diameter of hollow cylinder,

and

OG =

L = 0.5L 2

di = 300 mm Outer diameter of hollow cylinder, do = 600 mm

Now, BM = =

S.G. of wooden hollow cylinder = 0.56 S.G. of oil = 0.85

=

I  4  4 d 0  d i4  2 64   d0  di2  h



d

do M G B

h

L







 d 2i 0.62  0.32 = 16 0.66L 16h 0.0426 = L 2 0

 

Thus, GM = BM – (OG – OB) =

0.0426   0.5L  0. 33L L

=

0.0426  0.17 L L

O

di

For stable equilibrium condition, GM  0. Let 'h' be the depth of immersion of the

Putting GM = 0 for the maximum height of

cylinder in oil and L be the height of the cylinder.

the cylinder, we get

Weight of hollow cylinder = Buoyant force acting on the hollow cylinder Or,

 cyl 

Or,

h

 2   d0  d2i   L   oil  d 20  d 2i  h 4 4

 cyl 0.56 L = L = 0.66 L 0.85  oil

Let us then calculate the maximum height of the cylinder, L for the stable equilibrium condition. ineerin ACE Engineeri ng Publications

0.0426  L2 0.17 

L = 0.5 m

Thus,

h = 0.66 0.5 = 0.33 m

02. Ans: Unstable Sol: Given data:

d = 1...