Fluid Mechanics Paper 1 Solution

FLUID MECHANICS
(FM - I : SOLUTIONS)

1) (c)

2) (a)

    du 
 dy 
 y0

At seperation, du   0
dy 
y0

 0

3) (b)

h  (V1  V2 )2
2g

From continuity,

A1V1 = A2V2
1002 x 5 = 2002 x V2

5  V2
4

h = 0.717 m

4) 60 to 80

5) (b)

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 .a.(v  u)2

= 1000 x 0.015 x (10)2

= 1500 N

6) (b)

7) Sp. Speed for 4 Nozzle

= 20

8) (a)

9) Ans. (c)
According to Newton’s law of viscosity, a Newtonian fluid is one type of fluid, in
which shear stress is proportional to rate of shear strain (i.e. velocity gradient).

 dv 
 dy 


10) Q1  1  cos  3
Q2 1  cos 1

11) Ans. (b)

K  (P2  P1)  (P2  P1)  P2  P1
v1  v2
 dv    v2  v1 
   v1  v1
 v1   

K   80  40  2000MPa
2 1.96 
 2 

12) Ans. (c)

In Laminar flow through a long pipe, the pressure drop (hf )  32.V .L
 g.D2

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Pressure drop per unit length  32.V
 g.D2

hf  1
L D2

Pressure drop per unit length is inversely proportional to the cross – sectional area.

13) Ans. (b)
For a stable floating body the Metacentric height is positive, hence Metacentre should
be above centre of gravity of the floating body.

14) Ans : 0.25 (range 0.24 to 0.26)

Ns  N P

5

H4

Given, Ns1 = Ns2 , H1 = H2,

N1  2, P1 ?
N2 P2

N1 P1  N2 P2

 P1   N1 2   1 2  0.25
P2  N2   2 
 

15) d  20cm, Vmax = 1 m/sec, Vr = 2


V  1 r2 
Vmax  R2 


 1  25 
100 

Vr = 0.75 m/sec

16)

  2xy

d v d u
dx dy

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2y = v , u = -2x

V  U2 V2
 4x2  4y2
 4 16
 20
2 5

17)

QM 2
QN 

hf  FLV 2 = fLQ2
2qd 12.1gd 5

Q   hf x 12.1 x g x d5 1/2
 
 f .L 

QM   fN 1/ 2
QN  fM 
 

 1 1/2  1  0.5
 4  2

18) wz = 0,

wz  1  dv  dv 
2  dx dy 

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wz  1 2 y  2x
2

19) p1  V12  z1  p2  V22  z2
g 2g g 2g

h  v22
2g

h  y  Sm 1
 sp

 0.20  9810  1
 10.8

h = 181.46 m

V2 = 59.66 m/sec

Q A2V2   x (0.15)2 x 59.66
4

= 1.05m3/sec

1 Liter = 10-3 m3

Q = 1054 LPs

20) Ans. 15.7

By Newton’s law of viscosity

  . V
h

 F  .Vh   F
A 

F   ( DL) V
h

F  (0.1)x( x0.1x0.1) (10)
(0.002)

=15.7 N

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21)

hf  (V1  V2 )2  4  2 .m
2g 2g g

V1  0.3  3 m / s
0.1

V2  0.3 1 m / s
0.3

Q  A1V1  A2V2

22) Ans. : 26.4 (26.3 to 26.5)

F   V  0.44x3  26.4N / m2
A y 50x10-3

23) du  dv  0
dx dy

(1)  2x cos y + (-2x) . cos y = 0
(2)  1 (1)  0

24) Q   Q 
D3N m D3 N P

40 x 103 0.8
103 x 1000  403 x N

N  0.8 x 103 x 1000
40 x 10-3 x 403

N = 312.5 rpm

25) Ans. (b)

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m  Q

 dA.U

 . B . dy .U y  ydy

.B
 
0

 .B.U  y2   .B U . 2
  2   2
 0

 1 ..B.U
2

 1 x1x1x10x(10x10-3 )
2

= 0.05 kg/sec

26) Ans. (c)

L

Integrated drag fore (FID ) =  .Bdx
0

Where    
u2 6

(for linear velocity distribution)

Proof :   momentum thickness

   u 1Uu  dy
0 U 


   y 1 y  dy

0  

   y2  y3 
 
 2 3 2 0

  
6

  .u2.

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 1 x (10)2 x  10x10-3 
 
 6 

= 0.167 N say 0.17 N

27 )

P  .L.V
t

Q=AxV

40 x 10-3 =  x (0.2)2 x V
4

V  1.273 m
sec

P  1000 x 1000 x 1.273
11.25

= 113.155 KPa

28)

n  Pd 113.155 x 0.2 x 103
2t  2 x 10 x 10-3

= 1.13 MPa

29 )

d = 300 mm, t = 6 mm

V = 3m/sec , E = 2 x 105 MPa

K = 2000 MPa

P  1000 kg
m3

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P  V K.P  3 x 2000 x 1000

 4242 KN
m2

30) P V x P

1  1
k Et

3x 1000
1 0.3
2000 x 106  2 x 105 x 106 x 0.006

= 3464 . KN/m2

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