mec 214 fluid mechanics theory x

7/26/2019 Mec 214 Fluid Mechanics Theory x

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EC

EDCA

A

F

C

GEA ECHCA & CAA

EAA ECHAE

EA 2 EEE

HE

1: D 2008

A DA

D ECHAC

E CDE: EC214


2/105

TABLE OF CONTENT

Table of Contents

Week 1

1.0 Fluids

1.1 Definition

1.2 Classification of fluids

1.2.1 Gases

1.2.2 Liquids

1.3 The difference between liquids & gases

1.4 Properties of fluids

Week 2

2.0 Types of fluids

2.1 Ideal and real fluids

2.2 Newtonian fluids

2.3 Non-Newtonian fluids

2.4 Specific gravity of liquids

Week 3


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3.0 Pressure equation

3.1 Pressure of liquids vary with depth

3.2 An orifice tank

4.0 Pressure measuring instrument

4.1 The barometer

4.2 Piezometer

4.3 U-tube manometer

4.4 Inverted U-tube manometer

5.0 Questions and solutions on pressure measurement

6.0 Derivation of equations

6.1 Parallel axes theorem

6.2 Total thrust acting on a vertical plate

WEEK 7 : Understanding the Archimedes principle and its applications

7.1 Buoyancy of floating bodies

7.1 Archimedes principles

7.1.1 Centre of buoyancy

7.1.2 Buoyant force

7..2 Hydrometer

7.3 Floating bodies


4/105

WEEK 8: Analysis and the practical determination of metacentric height

8.1 Determination of metacentric height of a vessel

8.2 Experimental method

WEEK 9: Understanding the principle of conservation of mass

9.1 The principle of conservation of mass

9.2 Conservation of mass

9.2.1 The discharge

9.3 Application of continuity equation

WEEEK 10: Understanding the conservation of energy

10.1 The conservation of energy

10.2 The Bernoullis equation

WEEK 11: To know the momentum equation and its applications

11.1: Introduction and understanding the concept of steady streamline and

stream tube

WEEK 12: To understand the concept of momentum and fluid flow

12.1 Momentum and fluid flow

12.2 Momentum equation for two and three dimensional flow along a

streamline

WEEK 14: FLUID DYNAMICS

14.1 Types of flow

14.2 Determination of laminar and turbulent flow

WEEK 15 : To know the types fluid power machines and their application

15.1 The fluid power machines

15.2 The pumps


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15.2.1 Classification of pump

15.2.2 Centrifugal pump

15.3 Hydraulic turbines

15.3.1 Types of hydraulic turbines

15.3.2 Impulse turbine (pelton wheel)

15.4 Reaction turbine

15.4.1 Types of reaction turbine

15.5 Hydraulic press

15.6 Air Compressors

15.6.1 Types of air compressors

15.6.2 Reciprocating compressor

15.6.3 Rotary compressor


6/105

FD ECHAC

EE 1

1.1.1 D a .

1.2 T .

1.3 Da aa .

1.4 P . (D, , a

a, , , a, , a

b, aa.

FD

L a : a a a a .

1.1 D F a ba a

a , a a b . b

a a a a aab . T a a

b a a.

D aa .

1.22a L: L a a a a

a a ; a a a a.

1.23.1Ga: Ga a a a a

a b . T aa b a a

a a b a a a b .

1.23.2L: .. a, , , , a, ...

1.23.3Ga: O, ab , N ..

1.3

G

1. D , a b a a

b.

1. A a aa a

.

2. A a a

a

2. a. Ca

a a, a a


7/105

a. b a a a

a a.

2b. A a a a

a a

a b a a.

3. A a a

a a a .

3. I a

a a

a a.

4. T a a

a, a

a a . T a

a a a

a ..

4. Ga a a ,

a b a

a a a

.

1.4.1 P .D: T a ba a a a a

ba.

D,

cesubsofvolume

cesubsofmass

tan

tan=

D a ba a b a a;

() Ma

() S

() Ra .

() Ma :

Ma a a ba .

( )3. = mkgv

m

Ta a a


8/105

kTNmxP 15.288,10013.1 25 ==

Wa, 1000 ,3mKg a, 1.23 3mKg

1.42

() S : S , , a .

ggxv

m

v

Mg

volume

weight ====

Ta a: a,331081.9 Nmx ;

A,307.12 Nm

1.43

() Ra S a:

Ra , a a a a ba

aa a .

F a aa a a a( a 4 a a a )

CatH

cesubs

=

402

tan

F a, aa a b a H a a a a

, b .

Ra , b.

Ta a: Wa 1.0; 0.9.

1.44

S V: T a a a a

a ; . a .


9/105

N

mUnit3

=

1.45

P: P a a aa a

a. T b a a a :

. T b a a a a , a a aa, a

a .

. T a b a a a a

aa a (a) a.

AB a aa.

P .

= P

A

1.46 C

V a a a a aa .

I a a a a a a . Ra b

a a aa a a.

X


10/105

1.47 :

A a aa a b b a. Eaa

a a a a

ab a. T a a a

a. T b a

a a aa a baa .

b a a b . A a,

a baa, aa a a a a

a aa .

I ABCD (FIG 1) a a a aa,

P a a aa A a BC . T aa /a a

a a, a b a ( a a), b a

a .

I a , b a a a a , a a a

a. I a , a a a a

. I a a a , a a a ( a a

) a a .

(1) C

(2) A

(3) V1.48 A: A a aa a aa aa

a a. ( Pa )

T a a a a ab a a

. T a a a a a a

a a a. Ta a

a.

C: A a aa a a

( aa a).

A , a b aa b a a a

a .

Sa , a a a b .


11/105

1.49 Sa T: T a a a a ba a

a a a ; a a a a b aa

ba b a a.

1.410 Caa: T a a a b aa. T baa a aa a baa

a aa a ba b a b.

I a ba a aa

aa b a ab , b. I

a aa aa b a ,

a.


12/105

EE 2

2.1T .2.2 Na .

2.3 NNa .

2.4 Sa N a .

2.5 Ea .

2.1 E F FD

T a a, a a a a . T a

b a a a a a a . T

a a . S a a a

, a a aa a. O a, a

a a a a a a aa

a. I a, a , b a a a

b aa b a a a a a a .

F. 2.1:

2.2 Na : T a b N a a N a a

a a a a Na . T a

a a a a a a a

N ba a a.

X


13/105

dy

dvN=

M a a, a .. Ga a

F 1: Vaa a a.

Sa

S

Ra Sa,dy

du

Pa Ba Pa

PPa

Na

Daa

Ia T = 0


14/105

2.3 :

F b N a a a Na a a

.

(1) Pa, a a a a a b

. Ta, a a a a a a.

n

dy

duBA

+=

W A, B a a a.

I =1, aa a a Ba a (.. a )

(2) Pa, a a a a a a

(.. a , a, , b, )

(3) Daa ba a a a a a

a(.. Qa)

(4) T: ba a a

a a a (.. a, .. )

(5) R aa a a

a a a.

(6) Va aa, ba a a a Na

aa b a a , ba a a.

Ea NNa a a ab.


15/105

2.4

1. A, . 0.79

2. B 0.88

3. Cab a 1.59

4. K 0.81

5. M 13.6

6. C 0.85 0.93

7. Lba 0.85 0.88

8. Wa 1.00

F

T a , a, a a a a . T a

b a a a a a a . T

a a .

S a a a , a a aa a. O a, a a a a a a

a aa a.

I a, a , b a a a b aa b

a a a a a a .


16/105

WEEK 3

2.1Ea a a .2.2D a a a a .

2.3Ea a a .

F 3.1

C a a a a 3.1. A a

1. T aa A.

1

2

3

A

B

C


17/105

A

W

Area

Weight

Area

Forceessure ===Pr

B ( )

( )VVolume

MMass=

Ma (M) = Ma(M) aa a ()

M = V

B W(W) = Ma(M) aa a ()

MgW =

( )A

vg

A

Mgessure

==Pr

B (V) = Aa (A) ()

MgW =

( )Avg

AMgPr ==Pessure

B ( ) ( )hheightxAArea)( =VVolume

AhV =

ghA

Ahgessure

==Pr


18/105

3.3

F. 2.3: A O Ta

1

2

3

A

B

C


19/105

WEEK 4

Db a .

Ea

12 T ba.

13 P.

14 U b a.

15 T U b.

3 Db a .

4 Ea

4.1

(1) T Ba:

A aa a a a b a

ba ( 4.1)

F 2

A

a


20/105

T a A a a a b a ba

a a a a b. T a A a a

a. H, a a a b . T

a A a a .

T a ba

b baa a.

L b ab ba; a

ba. L W b a a

a a aa b;

T = a

W = W a

P b a ba

tubeofarea

columnofweight=

a

haxws=

hws=

T b a a ba. T

Ba a a.

T ,m

Kg10x6.13 3

3

9.81x10x13.63

=sw

H, , a = hws

T, 1 , = 1 a 133.3 KN/.

F a a a.

1 a (a) = 101.3N/.

= 1.013ba

= 760

= 10.4 a


21/105

2.41 () P:

T a b a a a

a a. T b P

. 4.2

F. 3: P Tb P

I a a b a a a

b a . I

a, a a .

P A = P

1ghA =

Sa, a 2ghB b ==

1

2

A

B


22/105

2.42() Ub a:

F 4 Ub a

T Ub a, . 4.3, a b a a. T

b Ub a a Q a man a

b , a , b a B

a a b a C a a b a C a a

a b,

P B a B = P C a C

F a b

P Pb = P Pa a A + P 1 P

=1ghA +

F a b.

+= Datessure DC Pr 2h Q

=D A = a

a 20 ghmanC +=

S CB =

21 ghgh manA =+

B C

D

L Q1

2

1

A

F P,


23/105

12 ghghmanA =

() T U b:

F 4.4 Ub a.

T Ub 4.4 a . T

Ub a , a, a a

. S a , a b a

b b.

F a b

ghga manAxx =

F a b

X X

F,

a

a

b

B

A

L,


24/105

( )hbgBxx +=

T ( ) ( )manAB ghabg +=

A a B a a a ,

( )ghmanAB =

I b a a b a a

ghAB = .

I b a a b a a

gh = AB . O

a, b a

a a

a , a a , b a

a a a a a a .


25/105

EE 5

C B .

A aa , , a a b, a V a a

b V a a a. I

aa a a P+, a b a a a

b aa.

strain

pressureinchangemodulusvolumetric

Bulk =

V a a b a . T,

a =V

dv

B =dv

Vd

dv

dv-

=K (T a a a a

a.

T,ulusBulkOriginal

Change

mod

pressureinChange

volume

in volume=

K

=

V

v

I , a => 0

dv

dv-

=K .1.1

a a ba,


26/105

1=V . 1.2

Da,

0=+ dvVd

ddv

=

v-

Sb V a 1.2

ddv

=2

1- 1.3

P a a ba a (1.2) a (1.3) a (1.1)

dK

d= ..1.4

T a K b a (1.4) b a b

a a, a a b a, a a . I a a, a

a b a, a a , a ,

a aaba. T a aaba b a b

a , a a a a a a a

a .

F , aa a b ;

a, baa ( a =1.4)

P a: A a a a ba a a a a.

PV = RT. (1) a a a a a, ; R a

a; a T ab ; R a a; a T ab

a. T a b a .

A a a b. T a a a a

a , a b a a () ab.

I a b a = PRT


27/105

W = /

RT=PV

Ra RT1

x-;1

by =

V

PRT=

C

T a b a

vl, a

R b.

vlRe =

W = .

V = a .

= aa

= a

E a a b a a

a ab a R b aa b a a a

a a , , b a a a

2000=

vd

F a b aa ().

T a R b a . R b a a

a , a b a a a b a

,v

vl V a V

v

1i.e =

=

v


28/105

Ea: F a b

X

X

2

Q

a

A

B

3

5

4

G

Ta a = 3


29/105

K.E =

M = 2+1+3+4 = 10

R = 3+1+2+2 = 8

I =

= M11 + M2R2 + M3R3 +M4R4

= 2 3 + 1 1 + 3 2 + 4 4

= 2 9 + 1 + 3 4 + 4 4

= 18 + 1 + 12 + 16

= 47

M = M1+ M2+ M3+ M4

M =

M = I

I = M

K = (I/M)

M a (a 710)

K.E =

V =

. = ()


30/105

=

K.E (1) = (1)(1)

K.E (2) = (2)(2)

K.E (3) = (3)(3)

K.E (4) = (4)(4)

K.E = K.E1 + K.E2 + K.E3 + K.E4

K.E =

Ea : T a (1) a a a () a ab

a a a .

L P = a

Ma PQ =

S a PQ ab = a (a)

= =

a

S

1

P Q

Z

S


31/105

Ta a a b I =

A 0 =

I = M

M = a

I = = a/3

= = a

= a = a/3

H = a/3

K = a/3

Ea 2:

F I a aa a ab a a .. aa a

2

L = a aa a

Ma PQ = b .. a (a)

Ta a

I = b

I = b

12

I = M = b b M = b

12

K =

12

I = M

12


32/105

Ea 3: F I a aa a 20 10 a 2 ab a a 5 a 20

a .

Ma = b

= 0.01/

Aa = 20

Ma = 20

2 a ab

205

I = 20

Ta a a a

C a a a Z b a .

P a-

zw=P ( = )

T 9.52-wzPQ

T a a

=

=

2

c

dz

dz

2zdaw

I ,02 a = 2

1

2

d

d

awzd

=2

12

2 d

d

zaw

=

22

2

1

2

2 ddaw

= [ ]21222

ddaw


33/105

= ( )( )12122

ddddaw

+

= ( )

+

2

1212

ddddaw

+

2

12 dd aa a .. a a

a b

z

Ta = ( )zddaw

12

= ( ) zwdda

12

A, ( )12 dda a aa a

Ta = aa a a 3 a

D C

T aa a

a a a.


34/105

F. 6

T y

. T

ab a a 00 K a az

D

a yxp

a pab 00.

F stripofareaxpressureDE=

wxpcebdxwx == sin,.

M ab 00 xdp .

= xwxbdx .

= bdxwx2

G

C

O

y

P

C

x

b

bdxxpdp=


35/105

Ta ab 00 = bdxwx2

a a yxP

a

Pab 00

H

= ,xwy.2

-

bdxP a a.

B bdxx2

a aa IAB aa A ab 00

T

ABIxwyP =

.

N

= xwAP ,

x a G b 00

T

=== xA

I

wA

wI

P

wIy AB

x

ABAB

=0-0aboutareaofmoment

0-0aboutareaofmoment

First

Second

L =GI S aa A ab a G aa a a 00

= AK2

W K a a. T aa 2 aa

a:

2-

G xAI +=ABI

-

2-

G

x

xAI

A

y +

=

+=

+= x

x

K

xA

2-2

2xAAK

B

+== xx

x

KxyGC

2


36/105

=

x

KGC

2

WEEK 6

6.1 Sa aa a .

6.2 D a a a a a a b a .6.3 I a a.( ).

Ea 1 T a(I) a a a () a

ab a a a .

L a .

Ma PQ

S a PQ ab XX a (a)2

..2 2 2

Ta a a b I x 2

a X O,

= 3

32 xdxx

a

o

3

x

3

a

I M K2

3

a 3 b M a

F a b

P = a a a

F a b

P = Pb (b+)


37/105

T Pb Pa = (ba) + ( a)

O A a B a a a , Pb Pa = (P a)

I b a a b a P a Pb Pa = .

O a, a Pa a a P,

a a , b a a a a

a a a .

Cb a b .

A aa , , a a b, a V a a

b V a a a. I

aa a a P+, a b a a a

b aa.

strainVolumetricmod

pressureinchangeulusBulk =

V a a b a . T,

a =

B

dv

Vdp=

dv

Vdp-=K (T a a a a

a.

T, a = a

Oa b

K

SP

v

v=

I , a => 0


38/105

dv

VdpK= .1.1

a a ba,

Pv

1= . 1.2

Da,

V + PV = 0

D = ()

P

Sb V a 1.2

= (1) 1.3

P a a ba a (1.2) a (1.3) a (1.1)

dp

dpPK= ..1.4

T a K b a (1.4) b a b a a, a a b a, a

a . I a a, a a b a,

a a , a , a aaba. T a

aaba b a b a , a

a a a a a a a .

F , aa a b ; a,

baa ( a =1.4)

P a: A a a a ba a a a a.

PV = RT. (1) a a a a a, ; R a

a; a T ab ; R a a; a T ab a.

T a b a .


39/105

A a a b. T a a a a

a , a b a a () ab.

I a b a = PRT

W = /

PV =RT

Ra V b 1/; P 1/= RT

P = RT

A

B

3

5

4

G

Ta a = 3

X

P

2


40/105

a : F a b

b

2

d

22

d

Q

G

X


41/105

K.E =

M = 2+1+3+4 = 10

R = 3+1+2+2 = 8

I =

= M11 + M2R2 + M3R3 +M4R4

= 2 3 + 1 1 + 3 2 + 4 4

= 2 9 + 1 + 3 4 + 4 4

= 18 + 1 + 12 + 16

= 47

Ra a

M = M1 +M2 +M3 +M4

M =

M = I

I = M

K = (I/M)

M a (a 710)

K.E =

V =

. = ()

=

K.E (1) = (1)(1)

K.E (2) = (2)(2)


42/105

K.E (3) = (3)(3)

K.E (4) = (4)(4)

K.E = K.E1 + K.E2 + K.E3 + K.E4

K.E =

Ea 3: F I a aa a 20 10 a 2 ab a a 5 a 20

a .

Ma = b

= 0.01/

Aa = 20

Ma = 20

2 a ab

205

I = 20

= 20


43/105

E A A

1. A aa a a a a 33a a 40. Caa

) S

) S

) Ra

2) A a . Q2 b a . Wa b

b a (Pa), (PA) a b a 200 KN/2

a 600 K/3, a a 1000 K/3. Ta a

b 13.6.

Air

Oil

Water

Mercury

3m

2m

500mm

A


44/105

3) Wa b KN/ a a a a 600

a) M a 13.6

b) A 520 K/3

4) Wa b

a) a

b) ab a a a 12 b a.

A a a b 1000 K/3a a 101

KN/2.

5)

A

101 KN/m

P1= 0.87 x 10 Kg/m

Pm= 13.6 x 10 Kg/m3m

4m


45/105

F. Q5 a a a a b . Caa a A

(a b a).

6) T a a a a a

a) 100 a 10 . F a a 20KN. T

a a .

b) T a a a a a a 1:5. T a

a a a 200N a a . D

a a b a b a .

7) Wa b KN/2, a a a a 4000

a) M a 13.6

b) Wa 103K/

3

) O 7.9 KN/3.

8) T a a a 8 a. Wa ,

a a a 1.17?

B

10m

P

A


46/105

F. Q9 a Ja a a a. I a B

150 Ka, aa a P a b a a A.

10) A Ub a aa a . Q10 a

b A a B a a .m-Kg10 -33=

T a Q 13.6 103

K. 3

a B 0.3 a A. Caa = 0.5.

Fig. Q9


47/105

F P

A

a

C

D

Ma Q

a

b

B

++++

++++

++

++++

++

++

++

++

++

++++

++ ++++

++++

++++

++++


48/105

F. 10

A B

1) I a a a 125 a 0.234 /, a b

R b.

Caa a a a a . Ta a

a as

m210x46.1 -5 a a a as

m210x10.1 -6

2) O a 0.9 a 0.17 NS/2 a 75 a

a a 2.7 /.

S a a .

3) Wa a 25 a a a 6 /. D

b aa b a a a a

mskg10x30.1

5-a 3

5-

mkg10 . I a 0.9 a a

ms

kg

10x.69

2- a , a .?


49/105

WEEK 7

UNDERSTANDING THE ARCHIMEDES PRINCIPLE AND ITS

APPLICATION

7.0 BUOYANCY OF FLOATING BODIES

When ever a body is immersed wholly or partially in a fluid it is

subjected to an upward force which tends to lift (buoy) it up. This

tendency for an immersed body to be lifted up in the fluid, due to

an upward force opposite to action of gravity is known as

buoyancy. The force tending to lift up the body under such

condition is known as buoyant force or force of buoyancy or up

thrust. The magnitude of buoyant forces can be determined by

Archimedes principle.

7.1 Archimedes principles:

This states that when a body is wholly or partially immersed in

fluid, it experiences an up thrust which is equal to the weight of

fluid displaced. This is mainly about ships and boats: how they

managed to float upright, and why they sometimes do not.

Consider the equilibrium of a stationary floating object such as a

ship. Two forces act upon it its own weight W which is acting,


50/105

vertically upwards excerted by the surrounding fluid. The upthrust

is governed by the archemedes principle.

7.2 (I) CENTRE OF BUOYANCY

The up thrust on a body always acts through the centre of gravity

of the fluid displaced by the body. This point is called the centre

of buoyancy.

Consider a cylinder in which the upthrust act along the vertical

centre line, because of the symmetry of the cylinder about the

centre line, it would be unreasonable to expect it to act anywhere

else.


51/105

Consider another example say ship. It has a

A much complex shape. The side view has no symmetry, so we

cannot easily tell where the line of action of the upthrust will be. If

we take moment about any point such as G, the centre of gravity

of the ship, we can see there is anticlock wise moment acting. The

ship can be in equilibrium only if the weight and the upthrust are

equal and opposite and also acts along the same restical line.


52/105

7.2 (ii) BOUYANT FORCE

The vertical upward force which the fluid or water exerted on the

body that were immersed in the fluid is known as the buoyant

force.

7.3 HYDROMETER

A given floating body will float at different height in liquids of

different densities for example, because the temperature and

salinity of the sea varies from place to place, its density also

values, and so a ship will float at different depths in different part

of the world. The depth at which a given body floats can be used

to indicate the density of the liquid. An instrument does it is called

a hydrometer.

The hydrometer is usually shaped as shown in the diagrams

below, weighted at the bottom so that it floats upright. The stem

is graduated.

(i) To indicate the density of the liquid


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In the left hand diagram the hydrometer is shown floating in a

liquid of density 0, a volume V0of the hydrometer is submerged

so that V0 of the liquid is displaced. When the hydrometer is

floated in a liquid whose density is grater the diameter floats a

distance H higher, so that submerged volume is now (V0 Ah),

where A is the area of X-section of the stem.

The upthrust on the hydrometer must be equal and opposite to its

weight W.

Therefore, W = e0gV0= e.g. (V0-Ah)

0/= 1- Ah

V0

h = V0 ( 0)

A


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By making the area A rather small, that is, by making the steam

narrow, we obtain a sensitive hydrometer, one which gives a large

change in height in for only a small difference (-0) in densities.

7.4 FLOATING BODIES

Consider a wooden board that float in the position shown below.

It is noted that the centre of buoyancy is below the centre of

gravity. To observe how the body can float stable even with B

below G, consider the situation where the plank has been turned

through an angle.


55/105

From fig (i) the weight W=mg acts through the center of gravity G

and the upthrust R acts through the center of buoyancy B of the

displaced fluid in the same straight line as W. for body tilted at

angle the volume of liquid remains unchanged but the shape of

volume and its center of gravity changes while the center of

buoyancy move from B to B, and a turning moment, WX is

produced.

The metacentric m is the point at which the line of action of the

upthrust R for the as placed position cuts the original vertical

through the center of gravity of the body G.


56/105

The metecentric height is the distance GM for angle of tilt .

Righting moment =WX

Where x= GM.

, provide that the angles of tilt is small so that sin =in rad.

Righting moment =w. GM. Sin.

=w. GM.

Note that comparing fig (ii) and (iv), it can be seen that,

(I) if m lies above G, a righting moment W.GM is produced,

equilibrium is stable and GM is regarded as -ve.

(II) If M lies above G, an overturning moment WGMis produced

equilibrium is unstable and GM is regarded as ve.

(III) If M coincides with G, the body is in neutral equilibrium.

Therefore the stability of a floating body depend on the

metacentric height i.e. the distance G M.


57/105

WEEK 8

DETERMINATION OF METACENTRIC HEIGHT OF A VISSEL.

The metacentric height about the longitudinal axes can be

determined if a load p is moved transversely a distance x across its

deck causing the vessel to move through a small angle. The

overturning movement due to movement of load p=px

the Righting movement =W.Gm sin.

Where Gm= metacentric height and


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W=p+w, (w, weight of vessel and p=load).

Note that:The effect of moving the load p and a distance x is

produce an overturning movement on the vessel and a righting

movement is produced for equilibrium of the vessel in tilled

position.

For small angle sin =

:. Righting movement=W.GM

Where W=total weight of vessel (w+p)

For equilibrium in tilled position, Righting moment = overturning

moment.

WGM=PX.

GM= p x

W sin

Where G.M= metacentric height

=radians.

EXAMPLE 1


59/105

A cylindrical drum with flat ends, whose diameter is 600mm, floats

in fresh water with its vertical, if the total mass m of the drum

(including content) is 225kg, to what depth h will it be

submerged?

Since the cylinder is floating in equilibrium, its weight must be

exactly balanced by the upthrust, and this is equal to the weight of

the water displaced.

Weight of cylinder=weight of water displaced

Mg=water x (volume displaced) g

225xg=1000 x (

x 0.3

2

xh) g.

h=0.796mm or 796mm

EXAMPLE 2


60/105

In an experiment to determine the metacentric weight for a

passenger lines of displacement 28.000 tone (i.e. its mass is

28,000 tones). An object of mass & tone is moved a distance 10m

across the width of the ship. The ship is observed to turn through

an angle of o.380. Find the metacentric height.

For the conversion of angle,

= 0.380

= 0.38 x /180 rod.

The metacentric height, GM= PX.

W

= 8 x 10

28.000x66x10-3

=0.43m

EXAMPLE 3

The crew of a small motor cruiser all moves to one side of the

vessel to wave to a passing boat. The displacement of the cruiser

is 40 tomes. The combined mass of the crew is 400kg and thing


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are initiated distributed evenly over the width of the boat which is

4m. If the metacentric height of

The boat is 0.25m, through what angle does it turn when the crew

all move to one side

The angle, = PX

WGm

= 400X2

40,400X0.25


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= 0.08 rad

:. Q=460.


63/105

WEEK 9:

UNDERSTANDING THE PRINCIPLE OF CONSERVATION OF

MASS

9.1 PRINCIPLE OF CONSERVATION OF MASS

The theory of fluid flow tests on two principles namely the

principle of conservation of mass and principle of conservation of

energy. The principle of conservation of mass stated that matter

can neither be created nor destroyed but can be transformed from

one form to another

This principle can be applied to the flow of fluid. Using the

principle of conservation of mass, it can be said that in a given

period of time, the same mass of fluid that enter a system must be

equal to the amount of fluid flowing out.

9.2 CONSERVATION OF MASS

Consider a flow through a pipe of cross-area A1, density 1, and

valuation V1 at section (1) i.e. point of entry and 2, A2, V2 at

section (2) i.e. at outlet mass flow rate at section (1) = 1A1V1

Mass flow rate at section (2) = 2A2V2


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By conservation of mass, mathematically,

1A1V1= 2A2V2= m- - - (1)

Where m = mass flow rate

For incompressible flow (in general gasses and vapour are

compressible, liquids are incompressible) the density is practical

constant. The equation (1) becomes.

A1V1=A2V2= Q - - - (2)

Where Q = discharge or volume flow rate m3/s.

The equation (2) is called the equation of continuity of flow.

9.3 DISCHARGE

The volume of liquid passing through a pipeline at any given cross

section in unit time is referred to as dischargeor volumetric flow

rate, Q. Assume that the cross section area of the pipe is A and

the fluid flow at uniform velocity V, then the discharge or

volumetric flow rate Q is given by.

Q = AV (m2x m2/s)

The unit is cubic meters per second m3/s


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Give the density of the fluid passing through the pipe of cross

section in unit time, the mass flow rate can found,

Mass flow rate, m = AV = Q (kg/m3

/s)

The unit is kilogram per second (kg/s).

Air is discharged from a compressor outlet at a velocity of 3mk

with a mass flow rate of 7.68 x 10-3

kg/s. if air has density of

1.293kg/m3, determine the volumetric flow rate and the diameter

of the outlet.

SOLUTION: -

Velocity of the air = 3m/s mass flow rate m = 7.68 x 10-3kg/s the

density of air = 1.293kg/m3 Q =?

From the date.

Q = AV.

But m = AV and A = m = 7.88 x 10

-3

V 1.293 x 3

Q = 3


66/105

9.4 APPLICATION OF CONTINIUTY EQUATION.

The continuity equation can be applied to determine the relation

between the flows into and out of a junction consider the figure

below: -

For steady condition

Total inflow to junction = Total out let from junction

1Q1= 2Q2+ 3Q3

For an incompressible fluid, 1= 2= 3 so that Q1= Q2+ Q3or

A1V1= A2V2+ A3V3.

I general if we considered flow toward he junction as positive and

flow from the junction as negative them for steady flow at any

junction the algebraic sum of all mass all mass flows must be zero.


67/105


68/105

WEEK 10

UNDERSTANDING THE CONSERVATION OF ENERGY

10.1 CONSERVATION OF ENERGY

Consider an element of fluid shown below will posses potential

energy due to its height Z above datum and Kinetic energy due to

its velocity V and the same way as any other object.

For an element of weight, mg.

Potential energy = mg Z

Potential energy per unit weight = mgz = Z - - - - - - - - - - (i)

mg

Kinetic energy = mg V2

= mV2

g

Kinetic energy per unit weight = mV2 - - - - - - - - - - --(ii)

Force exerted on AB = PA

After a weight of fluid has flowed along the stream tube, section


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AB will have moved to AB

Volume passing AB = mg /g = m/

Distance AA1= m/A

Work done per unit weight = P/g - - - - - - - - - (c)

The term p/g is called the flow work or pressure energy.

By equation a, b, and c, it can be seen that the 3 terms of

Bernoulli equation are the pressure energy per unit weight, Kinetic

energy per unit weight and the Potential energy per unit weight.

Thus, Bernoullis equation state that for steady flow of a

frictionless fluid along a straight a streamline, the total energy per


70/105

unit weight remain constant from point to point although its

division between the three forms of energy may vary:

Pressure energy K.E per P.E per Total energy

Per unit weight unit weight unit weight per unit weight = constant

P1 + V2+ Z1= H - - - - - - - - (IV)

g 2g

Each of these has the dimension of length or head and they are

often referred to as his pressure head p/g, the velocity head

V2/2g the potential head Z and total head it. Applying this equation

between two points 1 and 2 on stream line, equation d becomes.

P1 + V12 + Z1= P2+ V2

2+ Z2- - - - - - (v)

g 2 g 2

Or total energy per unit weight at 1 = total energy per unit

Weight at 2.


71/105

QUIZ 1.

A jet of 50mm diameter impinges on a curved vane and is deflected

through an angle of 175. The vane moves in the same direction as

that of jet with a velocity of 35 m/s. If the rate of flow is 170 litres

per second, determine the component of force on the vane in the

direction of motion. How much would be the power developed by

the vane and what would be the water efficiency. Neglect friction.

QUIZ 2.

A jet of water , 50 mm in diameter , issues a velocity of 10 m/s

and impinges on a stationary flat plate which destroys its forward

motion. Find the force exerted by the jet on the plate and the work

done.

QUIZ 3.

A jet of water of diameter 75 mm moving with a velocity of 20 m/s

strikes a fixed plate in such a way that the angle between the jet andthe plate is 60. Find the force exerted by the jet on the plate.

i. In the direction normal to the plate , and

ii in the direction of the jet


72/105

EE 11: HE E EA AD ACCA

ACA

11.1 DC AD DEADG F CCE F EAD

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() ; () ; ()

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73/105

a. I b a b a a

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74/105

I a , a a a

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aa b a a; b a a

a a , a

a b a . W a

a


75/105

EE 12: DEADG HE CCE F E AD FD

F

12.1 E AD FD F

I a, a a b a

a M a V:

M = MV.

T a a a , a,

a a a , b a

a a. I aa

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b a a a a . T

a b b a b a a ba (..

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a a a. B N a, a

a a ba b

a . S a a a , a

a a aa a

a ; a .

T a a a a a

ABCD (a b). A a b


76/105

a a a a a

a b a.

mVAVA == 111222 ll

F 12.1 ABCD

. a a a .

T a a a ba

CD a b a

222 2VVAl

Sa a a a AB

a b a

1111 VVAl

T a a a a b

b


77/105

11112222 VVAVVA ll =

O, a a,

N a a

, a a N a b a b a F,

a

T a a ABCD

.

B N a, a a a a

.

12.2 E EA F AD HEE DEA

FAG A EAE

T a ( )12 VVmF = a a

a a , a a a V 1 a 2V

a . T b a a

b 1 a a a a, 2V a a


78/105

a . S b a a

a, a b x a y

a a ( )12 VVmF = a. T, XF a YF a

a ABCD,

F 12.2 E ABCD

directionxinfluidofmomentumofchangeofRateFX =

= directionxinfluidofchangexunitperMass

= ( ) ( )1212 coscos XY VVmVVm =

Sa,


79/105

+=

22

YxFFF

[ ] ( )1212 sinsin YYY VVmVVmF ==

T a b b a ,

Aa, b b a a

. F a , a a b , b

a a V1 a V2 a

a a a

1Zzz VVmF =

QUIZ

D a a a a

b a a .


80/105

EE 13: HE ACA F E EA

13.0 ACA F E EA

13.1 AC F FEE E

DC

A a a a a a

a . W a a a a, a

[ a ]. a b aa b

. T a a b

.

13.1.1 FCE EEED B HE E HE AA AE

1. W a a a ;

2. W a a ;

3. W a .

13.1.2 FCE EEED B HE E HE G AE.

1. W a a ;

2. W a ;

3. W a .

13.1.3 FCE EEED B HE E A AA AE

F a aa a a Na

T F. b a a aa a a

a . L a a b a aa a


81/105

. T a a [a],

a 90 ; b, a a

, b a a a b

a . I , a b a .

T b a [a ]

. [X ],

F =Ra a [ ]

= [Ia a ]

= I .

= [a/] [ b a

a

a].

= aV [V0]

O F= aV2

W = a ; a = aa a = a .

I a b a a a a Xa, a

X .


82/105

13.1.4 FCE EEED A AA FA AE HED CED

HE E

T . b aa a a a 0

a . I a a a a aa a

, a a

= aV

A a (a ), a a

a a ().

L a a a

a. F a , F= aV (V 0)

= aV2

T a a b ; Faa

a F, a

F= F = aV

2

= aV

2

2

F= F = aV2

13.1.5 FCE EEED AA CED AE

. :

C a a aa a () a

b. T a a a ,

aa a.


83/105

T a a a b

:

. C = V (

a a a a a

)

. C a =

A a, a:

F b ( ).

F= aV (V1 V2)

W V1= a = V

V2= a = V

F= aV (V (V ) = aV ( V + V )

O F= aV2(1 + )

Sa, F= aV(V1 V2)

W V1= a = 0

V2= a = V

F= aV (0 V ) = aV2

E

F = F2+ F

2


84/105

E 1:A a, 75 a, a 30/

a a aa a a a . F

b a.

. Da , = 75 = 0.075

V , V = 30/

T b a aa a a b

F = aV2

W, = Ma a = 1000 /3

a = Aa = /4 2= /4 0.075

2= 0.004418 2

F = 1000 0.004418 302

= 3976.2 (A)

E 2:A a a 35/ a a a a

30

0

a . I a aa 25

2

,:

. T b a.

. T a .

. V , V = 35/

Ia a a, = 300

Aa , a = 252= 25 1042

. T b F:


85/105

F = aV2

= 1000 (25 104) 352 300

= 1531.25 (A.)

. T , F:

F= F = 1531.25 300

= 765.625 (A.)

F= F = 1531.25 300

= 1326.1 (A.)

QUIZ 1.

A jet of 50mm diameter impinges on a curved vane and is deflected

through an angle of 175. The vane moves in the same direction as

that of jet with a velocity of 35 m/s. If the rate of flow is 170 litres

per second, determine the component of force on the vane in the

direction of motion. How much would be the power developed by

the vane and what would be the water efficiency. Neglect friction.

QUIZ 2.


86/105

A jet of water , 50 mm in diameter , issues a velocity of 10 m/s

and impinges on a stationary flat plate which destroys its forward

motion. Find the force exerted by the jet on the plate and the work

done.

QUIZ 3.

A jet of water of diameter 75 mm moving with a velocity of 20 m/s

strikes a fixed plate in such a way that the angle between the jet and

the plate is 60. Find the force exerted by the jet on the plate.

i. In the direction normal to the plate , and

ii in the direction of the jet


87/105

Week 14: FLUID DYNAMICS

Fluid dynamic is concerned with the study of fluids in motion.

14.1 TYPES OF FLOW

1. Laminar flow: in this case, fluid particles move along in layers or

laminar with one layer sliding over and adjacent layer. The flow is

governed by Newtons law of viscosity (for one-dimensional flow) it

is also called streamline viscous flow.

2. Turbulent flow: - This occurs when the fluid particle move in very

irregular paths causing an exchange of momentum from one

portion of a fluid to another. This is also called a non viscous flow.

Laminar flow tends occur when the fluid velocity s small or the

fluid velocity is large or both the turbulent flow sets up greater

shear stresses and causes more mechanical energy to be

converted to thermal energy.

3. Steady Flow: This is characterized by a steady mass flow rate and

by that across any section at right angles to the flow all properties

are constant with respect to time. True steady flow is found only in

laminar flow. Steady, turbulent flow is said to exist when the mean

velocity of flow at a section remains constant with time.


88/105

4. Unsteady Flow: This occurs when conditions at any point change

with time. An example is the flow of a liquid being pumped

through a fixed system at an increasing rate.

5. Uniform Flow: This occurs when, at every point the velocity vector

is identical in magnitude and direction at any given instant. That is

du/ds = o in which time is held constant and S is the displacement

in any direction. This equation states that there is no change in

the velocity vector in any direction through out the fluid at any

instant.

6. Non-Uniform Flow: This occurs when the velocity vector vari8es

from place to place at any instant i.e. du/ds o AN example is the

flow of a liquid through a tapered or curved pipe.

To determine whether a particular flow is laminar or turbulent

depends on three particular i.e. velocity, viscosity and density of

the and the cross-sectional area through which the fluid flows.


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14.2 Determination of laminar and turbulent flow

Fig 14.1 long tube carrying water

The sketch shows a long tube along which water flows from left to

right. A thin stream of coloured dye is introduced into the water on

the centre-line of the tube. Would you expect the stream of dye to

be straight, as shown below, or confused, as shown below?

Either is possible

When the water is flowing slowly, the motion is orderly, and all the

particles move in straight lines parallel to the axis of the tube. The


90/105

stream of dye will then be straight. This kind of flow is called

laminar flow.

By contrast, in a fast-moving flow of water the particles

move in a random way, in addition to the general left to right flow.

The dye rapidly gets thoroughly mixed, and after only a short

distance the water becomes a uniform pale colour. This is called

turbulent flow. But what we do mean here by fast and slow? In

particular, at what does the flow change from orderly to random

from laminar to turbulent?

This question was investigated by the British engineer

Osborne Reynolds at Manchester during the early 1880s in one of

the classic experiments of fluid mechanics. He used a thin

filament of coloured dye in water flowing along a glass tube, so

that the pattern traced by the dye could be observed. The water

was stored in a tank in which was allowed to settle for sometime

before each experiment, so that there will be no eddies remaining

in the tank that might affect the behaviour of the water in the

pipe.


91/105

Fig 14.2 Reynolds apparatus

Reynolds observed what we have already discoursed: that there

are two kinds of flow. Furthermore, he showed, when the value of

the dimensionless ratio

ud

Is less than about 2000, the flow is always laminar. (Here u

represent the u represent the mean velocity of the fluid, and

are the density and viscosity, and d is the inside diameter of the

tube.) When the value of this ratio is greater than 2000, the flow

is usually turbulent, unless special precautions are taken.

We have already met the ratio ud


92/105

It is now universally known as Reynolds number. It is denoted by

Re.

Example 1:

Water at 500C flows at mean speed of 0.5m/s along a tube of

5mm diameter. The density of the water is 988 kg/m3 and the

viscosity is 0.548 x 10-3N s/m2. Determine whether the flow is

turbulent or laminar?

Solution

To determine the character of the flow, need to calculate the value

of Reynolds number:

ud

= 988 x 0.5 x (5 x 10-3)

0.548 x 10-3

= 3830

and this is greater then 2000, so the flow is likely to be turbulent.

Example 2:


93/105

A glycerine at temperature 250C is flowing down a pipe of inner

diameter 20mm at a mean speed of 0.5m/s. the density of the

glycerine is 1250kg/m3, and the viscosity is 0.942N s/m2.

Determine whether the flow is laminar or turbulent.

Solution:

We need to calculate the Reynolds number, and this time we find

ud

= 13.3

and this is less than 2000, so the flow is laminar.

Quiz 1:

A pipeline 700mm in diameter carries a flow of methane gas at

150C and gauge pressure 3 bars. The density of the gas is

2.70kg/m3 and its viscosity is 1.15 x 10-5 N s/m2. The mean

velocity of flow of the gas is 0.95 m/s.

Quiz 2

Find the friction factor of the conditions given in questions 1, and

estimate the pressure drop in a 100m length of the pipeline. The


94/105

roughness value of the inner surface of the pipeline is k =

0.28mm.

The characteristic no relating this parameters is refers to as the

Reynolds number and is given by,

Reynolds no. = Vd

Where = density of fluid

V = velocity of fluid

d = cross section diameter of pipe

= dynamic viscosity of fluid

The following condition holds for the Reynolds no for turbulent and

laminar flow.

Reynolds no 2300 means a turbulent flow


95/105

Re = 2100 to 2300 means transition from lamina to turbulent


96/105

EE 15 FD E ACHE

F a b b a a a

a. T b a a a a. I a, a a

a a

a a .

15.1

A a a a a . I

a a b

b aa .

15.1.1 CAFCA F

O ba a aa a b

a a :

1. R a

. Raa

. Aa

. M a

. Ca

2. P a

. Ra

. P

. P

. Daa


97/105

15.1.2 CEFGA

A a a a

, a, a . T a

( ) a baa a ( ba). I a a a a . T

a a a ab . I a

a a aa, ba a

aa ab. I a a

b a a aa a aa. T

/

b a . I

a a a a .

T a a a a a

a b a a , aa

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. A a a a aa

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b a

a .


98/105

FG 15.1 CEFGA

15.2 HDAC BE

Ha b a (a a a

a ba a aa ) .

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a. T a .

P a a b a aa

b. I a , a


99/105

a a a b a b. Wa

a a a

a a a a . A

a a a a a b

. A aaab a

b b .

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a a a a a a a.

15.3 EAC BE

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a a a. Wa a ba

a a a a b a ab . T aa

a a a .

I b a a a a . T

b a a a a . T a

b a , b a

a Ra b.


100/105

FIG 15.2 REACTION TURBINE

15.3.1 E F EAC BE

. F . b a

ba aa a a

a aa .. aa a a. T b a aa

a a a [30 200 ] b a

a a.

T a , a

ba a a . T b a

a, a a a

. T b

a .


101/105

FIG 15.3 P b

. . H a a a aa a b

a b a a a a a a a b a

b . T ba a a a

aaa b a a . W a

a ba a a a , a aa ba a a a a

. T b a a a a

a a a a

ba .


102/105

F 15.4

15.4 H

T a a a a b

aa a . I ba aa a,

a a a a a

a a a .

: T a a a b

a . b.C a a a ,

a a, a

a b, a ab,

.

F 15.5 Ha

L W = W b


103/105

F = F a ,

A = Aa a, a

a = Aa

P b

Aa

=

a

A Paa a, ab b a a a

.

T a =

15.5 A CEA [ a] a a a aa ..

a, a a . I a a

a a , aba , a a

a a, a .

. C a a b a a aa a, ba aaa.

. C a a a a a.

. U a a.

. T a a a a aa a aa.

. Pa a a a a.


104/105

15.5.1 E F A CE

A b a

a. P a a a

a a , a a a a. W a a

a a aa .

15.5.2 ECCAG CE

A a a a a a

b a a a .

Ea a a b a ,

, , b a, a . W a b , a aa

a a b, , , , a

. T a a b a a

b . I a a

a a a. I a a a ,

a I.C. , a ba a I.C. ,

a a aa a b

.P a a a

a . E a

aaab a aa. W a a

ab a aa

aaab.

15.5.3 A A CE

Ra a a a a aa a . A a a a a

a a 1:15. A a a

a a a a 1:15.

Ra a aaba . T

a a a , a a a ,


105/105

a a b b, .

Ra a a a a

ba.

mec 214 fluid mechanics theory x

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