mec 214 fluid mechanics practicalx

8/13/2019 Mec 214 Fluid Mechanics Practicalx

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&

2

1 2008

214


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TABLE OF CONTENT

WEEK 1

1.0 To find the relation between the pressure and volume of a gas at

constant temperature.

WEEK 2

2.0 To find the relationship between the volume and the absolute

temperature of gas at constant pressure.

WEEK 3

3.0 U-tube manometer for measuring gas pressure

WEEK 4

4.0 Measurement of gas pressure using a pressure gauge

WEEK 5

5.0 To measure the pressure due to the atmosphere

WEEK 6

6.0 Discharge through a small orifice

WEEK 7

7.0 DETERMINATION OF FLUID PROPERTIES

WEEK 8

8.0 VISCOSITY OF WATER

WEEK 9


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9.0 VERIFICATION OF SURFACE TENSION

WEEK 10

10.0 SURFACE TENSION

WEEK 11

11.0 DETERMINATION OF CENTRE OF PRESSURE OF IMMERSED BODY

WEEK 12

12.0 STABILITY AND BUOYANCY OF FLOATINGBODIES

WEEK 13

13.0 FLOW THROUGH A TAPERED PASSAGE

WEEK 14

14.0 DETERMINATION OF VISCOSITY OF FLUID

WEEK 15

15.0 TO MEASURE THE PRESSURE HEAD OF WATER IN A PIPE LINE BY

MEANS OF A U-TUBE


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EEK 1

E

.

B

,

, , .

: L B C .

C B, = .

M

1. (C) (

) .

2. N () ( ).

3. M

4. C () (. )

.


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5. K

. .

6.

7. v

1

A =

( )mml

( )mm

( )c.cv

1. ?

2. H ?

3. B L


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WEEK 2

Experiment

CHARLESS LAW

Object: To find the relationship between the volume and the absolute

temperature of gas at constant pressure.

Apparatus: Flask with neck in the form of a long tube with two right-angle bends,

glass tube, flexible tubing, mercury, copper heater, stirrer,

thermometer.

Theory

The tube AB is graduated to show the volume in the flask and tube.


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WEEK 3

Object:- U-tube manometer for measuring gas pressure.

Apparatus:- U-tubes manometer, gas in a container, mercury, metre rule.

Method:

Fill the u-tube with mercury. Let one end of the u-tube is open to the

atmosphere and hence the surface pressure acting upon the liquid column in

this side of the tube is the atmospheric pressure. Connect the other side to the

container.

Observation:


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In (a) the levels of the liquid in the manometer are the same. Thus the pressure

of the gas P in the container exerted on the left-hand column must exactly

balance the atmospheric pressure \PA exerted on the right-hand column of

liquid, thus

In (b) the pressure of the gas Pb must exceed atmospheric pressure, thus

forcing the liquid in the U-tube round until the excess pressure caused by the

height h of liquid caused the pressure on the two sides to equate. Hence the

gas pressure in the container is:

That is, atmospheric pressure plus the pressure due to the column of liquid Pgh,

where h = column height above xx, P = liquid density, g = 9.81 m/s2

(acceleration due to gravity).

In (c) the gas pressure Pcin the container is below atmospheric pressure. The

greater \


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4

4

M

: .

A: A

M: C .

. ,

, ..


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5

5

: .

A: A , , .

M:

. C . 5.

. , ,

.

= =

A A

A A

M

M

C


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6

6

D

F. 6:

L ,

, . 6.

. ,

I A ,


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C .

I ,

..

C 0.6 0.7.

.


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7

/:

T D K

(1) SAE 40 L

(2) C

U

:

(1) A /

(2) T (

1(3) A

(4) L (SAE 40)

(5) C

(6) S

(7) A M R

F1 F. 2

V :

)

.


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F. 1

9


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I ,

F =

W

=

= A

=

.. =

=

W

().

S

T: F =

G (A)

() :

= F/A

T

.

C

F. 2

(F) :

___1

= M

= G

I

:

____

___3

, (/)

,

_4

,

5

N .

2

.


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= / = ./ 6

C ;

T

E (6), G ()

T:

W

= =

T =


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=()

N/2

T


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F (0) ; :

/P _________(7)

W

= D

P0 = D

= ()

F, E F. (1)

30. N ' . T

. N,

F (2).

W , . A , () . T

.

. A

()

(

2)

I

(S)

()

(V)D L.

1 50 202 30

2 \\ \\ 15 1 2/

3 \\ \\ 10 \\ 1.5/

4 \\ \\ 20 \\ 3.5/

5 \\ \ 25 \\ 3.0/

6 \\ \\ 5 \\ .O/

C . U

(1) R

(2) C

(3) F


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8

T

() T

() T

G

,

() T

P'

1.

2.

3.

()

:

I

1A 1 ,

, , , ,

, , .

__,.

,

.

/


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T :

/ B V V/

3 V = V 2

T / R

/VF S

P V/

/ .

1. A

.

2. I , P'

( )

. F, ,

P' . F

, , '

.

3. I ( )

.

5

F (1) % = 4 % +% + %

.

F % E 1. T %

. T %

''

. E %

.


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9

(1)

T

S 1/2 , ;

, ,

, , .

L

() 3.

T = 9.81"2

C

Y

= /4 ()

() .

C '

, ,

. T ( )

.

F ,

, . (T

.) S . H

. B

. T

. M

, : , , . F

2

. T

A

. I . T

. C


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, . R

: . R .

T :

A ,

, ,

D / S H /(2) M(1)

1

2

3

M........

1.


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2. I ,

. T

.

3. I

.

4. T ''

, , , .

F 0

Y=

4

A () .

1. T

. E %

.

F (): % = % .

2. T

.


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10

1. T

.

2.

S ,

() U E

3.

L =

,

I = AC,

=

,

,

T

.. 2 =

.. = /2

F ,

= ,

, , , A

, , , ,

. N. 1

. F

,

= 2.

_

'

.

_1


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27/51

,

.

. M ,

.

D

.

M .


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()

1.

2.

L =

I = AC,

=

,

,

T

.. 2 =

.. = /2/

F ,

= ,

= ,

=

= (21 + 2) H

2) =

. . 2

. F ,

,

= 2.

_

(2/ +

1


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'Y (21 + 2) _______2

3.

A DABC . A .

C , ,

. T

. P

.

L

.

D .

M /

.

1. R .

2. O . U

. S .

3. I AC

.

4. W .

5. A

( S') ,

..


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31/51

11

1.

T

T

A F. 5

F =

W

A =

.

I.

631

1.1 . T )

A

=

F


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H '

C.G C C.P P WI

C

.


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T CP (

) 0, 0

. T D :

T

= + IG/A

W

= 0 CG

=

W IG= 2 CG.

E D :

F , P

F. = M. W

=

M =

= .

T .T /

, ,

( ) '

'

(


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/

/ = 100 =

75 = 100

= 250

T 63.1

,() ,() () I X M .

1

2

3

4

5

6


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T . T

. W , .. 6 = 90,

,

, . H,

=

. T CP . I

( 7)

148.2. T + (1/2) = 150.0.


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13

1. T

() A

() B

() L

() K .

3.

M MG

W =

2.

M

W

.

.

P

F. 64.1 T

W 0

=

A

J

S

MG =


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=

NOTE

I 0

57.3

M MG

M

W MB =

4. E

T

31:

G

]

,

,

.

.

= MB + BZ GZ

. T I,

. T

. T

F

G, GZ . T

. T

. T

. T


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. A 0

.

R 1 2 3 4 5

G 91 82 75 66 58 0 0 0 0 0 05 0.9 0.8 0.5 0.5 OJ

20 3.6 2.9 2.1 1.8 1.4

30 5.4 4.1 3.2 2.7 2J

4O . 6.9 5.6 4.3 3.5 3J'

50 8.6 6.9 5.4 4.4 3J

55 6.0 4.9 M^

70 7.5 6.2 3

T 300, 250, 200

150 .

M = 2.525

M = 0.2 E

:


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/ = 358

= 203

= 77 D

= 42


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. 9

64.4

C MG : W W =

= 2.5259.81 =

24.77N

J = 0.29.81

1.962N

W

= 4.538

= 4.538 /0 176

_

MG

X 57.3MG


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150

117.5

P G

.

(M)

50


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0 20 40 60 80

M ()

F. 64.6

T MG

GZ (F. 2.16)

T

.F M G, .. MG + . T

' . H,

G , .. GZ . T MG GZ

. F GZ 117.5 MG

. T G GZ

.


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14

T

() A

.

() A

() A

W

F ( )

P/ + Y2/2 + =

P/ =T

+ V2/2

.

T A

). F

.

.

2S^ ^ ?

F. 66.1

B' , :

I

.

=

P

^

/

.

. (A

.

.


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JT

.1

183


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1 2 3 4 5 8 7S

F. 66.2 184


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N ,

H .

H,

.

T

. T

.

T

.


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15

Brief theory. Piezo

employed when pressur

to be measured, since

long tubes, which

conveniently. Furthermo

measured by the piezo

no free atmospheric surfa

overcome by the use of

A U-tube consists

shape, one end of whic

which pressure is to be

remains open to the at

Fig. 2. It contains a li

heavier than the liquid

to be measured.

The pressure head

found from the relation:

*

.

eter tubes cannot be

s in the lighter liquids are

this would require very

cannot be handled

re gas pressures cannot be

eters because a gas forms

e. These limitations can be

-tube manometer.

f a glass tube bent in U-

is connected to a point at

measured and other end

mosphere as shown in

uid (generally mercury)

of which the pressure is

of liquid (h) in a pipe is


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h = h"(S2-\)-

h' Procedure:

1. Connect the U-tube to the pipe carrying liquid (whose pressure is to be measured).

2. Note down the readings of h' and h ".

3. Take number of readings by varying the discharge (say four) and tabulate as

shown in the Table 2.

Table 2. U-tubeObservations

S. No. h' h" Pressure head Intensity of pressure p = wh Remarks

h = h"(S2-S,) (w = sp. wt. of liquid)-h 'S ,

1.

2.

3.

4. '

Specimen calculations : (/)

Conclusions :

Precautions:

1. U-tube should enter the pipe at right angles to the direction in which the fluid flows.

2. The end of the U-tube which is to be connected with the pipe should flush with

its (pipe) inner surface and should not be rough.

3. If large pressures are to be measured, then in U-tube heavier liquids, generally mercury, should

Mean ressure, = .


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mec 214 fluid mechanics practicalx

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