Part I.A -Fundamentals of Fluid Dynamics

8/8/2019 Part I.A -Fundamentals of Fluid Dynamics

1/38

FLUID DYNAMICS

PART 1 FUNDAMENTALS OF FLUID DYNAMICS

1.Basic Equations of Flui D!na"ics#a$ CONTINUITY E%UATION & T'( )*inci)l( of cons(*+ation of "ass

"a! ,( stat( as follo-s& t'( "ass of /ui )assin0 an! s(ction )(* unit ti"( isconstant. T'( (quation -'ic' *(sults f*o" t'is a))lication is 2no-n as t'(equation of continuity or continuity equation.

a.1 Fo* co")*(ssi,l( /ui #0as$3 (quation of continuit! is3

41A151 6 47A757 # 208s$ o* 91A151 6 97A757 #N8s$

-'(*(& 4 : (nsit! of /ui #208";$

9 : s)(ci


2/38

a.7 Fo* Inco")*(ssi,l( /ui #liqui$3 (quation ofcontinuit! is

*(uc( to3

A151 6 A757 #";8s$

sinc( 4 an 9 "a! ,( assu"( constants.

Fo* an! s(ction&

RT

pg =γ

='(*(&

an RT

p= ρ

Not(& Un(* a.1 an a.73 fo* ci*cula* )i)(s3

2

2

21

2

144

V DV D π π =

o* 2

2

1

21

V D

DV

=

t consQV A tan==⋅ Q is discharge or flow rate


3/38

#,$ ENER>Y E%UATION& T'( (n(*0! of t'( /o-in0 st*(a" of

/ui )(* unit

ti"( )assin0 an! u)st*(a" s(ction is t'( sa"( as t'((n(*0! )(* unit ti"(

)assin0 an! o-nst*(a" s(ction )lus t'( loss of (n(*0!,(t-((n t'( t-o

s(ctions.

,.1 Fo* inco")*(ssi,l( /ui #liqui$3 t'( (quation is3

21,22

2

21

1

2

1

22 −+++=++ L H Z

p

g

V Z

p

g

V

γ γ

='(*(& g V 2

2

? 5(locit! '(a # " o* N?"8N$

γ

p? P*(ssu*( '(a #" o* N?"8N$

Z ? El(+ation '(a #" o* N?"8N$

g

V

2

2

? T'( 2in(tic (n(*0! )(* unit -(i0't

γ

p Z an ? Constitut( t'( )ot(ntial (n(*0! )(* unit -(i0't

L H ? T'( '(a loss ,(t-((n s(ctions 1 an 7


4/38

b.2 For compressible fluid (gas) under isothermal conditions, the equation is,

21,22

1

1

2

211

1

1

2

1 ln2

ln2

−+++=++ L H Z p p

g

V Z p

p

g

V

γ γ

Where: p – absolute pressure ( !m2 )

"#$: #he %oncept of $nerg& $quation was originall& formulated b&

'aniel ernoulli in *+: #he ernoulli-s rinciple state that as the

speed of a mo/ing fluid (liquid or gas) increases, the pressure withinthat fluid decreases. asicall&, the concept states that the total

energ& in a steadil& flowing fluid s&stem is a constant along the flow

path. 0n increase in the fluid-s speed must therefore be matched b&

a decrease in its pressure.

1n mathematical equation

constant=+ .... E K E P

1

2

2

1k mV mgh =+ V

k

V

mv

V

mgh 1

2

2

1

=+

'i/iding b& constant olume,


5/38

V

k

V

mv

V

mgh 1

2

2

1

=+

k v gh =+ 22

1 ρ ρ

k vh =+ 22

1 ρ γ

k v p =+ 22

1 ρ

k v pv pv p =+=+=+ 2

22

2

11

2

2

1

2

1

2

1 ρ ρ ρ

We can write,

Where: p 3 pressure

/ 3 /elocit&

4 3 densit&

ote: 1n the ernoulli-s concept, it is clear that if

/elocit& increases at one end, it must be

matched b& a decrease in pressure.


6/38

#c$ If (@t(*nal (n(*0! is a( to t'( st*(a" ,(t-((n )oints 1

an 7 as fo* instanc( ,! a )u")3 t'( co")l(t( (n(*0!(quation ,(co"(s&

21,22

2

211

2

1

22 −+++=+++ L p H Z

p

g

V E Z

p

g

V

γ γ

where: $p 3 e5ternal energ& head added

(d) If (n(*0! is 0i+(n u) ,! t'( s!st(" to a tu*,in( ,(t-((n)oints 1 an 73 t'( *i0't si( of t'( (n(*0! (quation"ust inclu( a t(*" E T to *()*(s(nt

t'( '(a 0i+(n u).

21,22

2

21

1

2

1

22 −++++=++ LT H E Z p

g

V Z

p

g

V

γ γ

where: $# 3 energ& gi/en up


7/38

( e ) 16789$6"6$#76 $Q70#1" : #he impulse of a force is

equal to the change in the momentum of the mass.

( ) ( )V M V V M Ft o f ∆=−=

Where: Ft – impulse

6f – final momentum

6o – initial momentum


8/38

2. 1mportant 'efinitions in Fluid Flow

(a) ath line is a line made b& a single particle as it mo/es

during a period of time

(a) 9treamline: an imaginar& line within the flow for which the

tangent at an& point gi/es the direction of flow at that point.

(b) 9tream tube: an element of fluid bounded b& streamlines

which enclose or confine the flow.

(c) 9tead& Flow: when the /elocit&, pressure, discharge or

flow rate at a gi/en point in flowing stream of the fluid

remains constant with time or dQ!dt 3 ;.

•P

1

V1 •

P2

V2

•P3V

3

•P

4

V4

•P

5

V5 •P6

V6

Path Line and Streamline

http://../Streamlines%20around%20bodies.ppthttp://../Streamlines%20around%20bodies.ppt


9/38

(d) 7niform Flow: if, at a gi/en instant, the /elocit& remains constant

with respect to a stretch or distance in the flowing stream of

fluid or d!ds 3 ;.

(e) 'ischarge or Flow ead 8oss: the energ& per unit weight lost due to friction (ma?or

loss) or total disturbances (minor loss).

(g) ower is obtained b& multipl&ing the total energ& head (@.$. A .$.), which ma& be in m! b& QB to obtain m!s.

'i/iding b& *CD gi/es the horsepower, or

746.

E Q P H

⋅⋅

=

γ


10/38

$5ample . 0 fluid in a pipe 2;; mm in diameter with a mean /elocit& of

+.;E m!s. #he pressure at the center of the pipe is +E a, and the

ele/ation of the pipe abo/e a reference datum is C.D; m. %ompute

the total head in meters if the liquid is (a) water, (b) molasses (s 3.E;),

( c) gas (B3D.EC !m+).

9olution:

3 +.;E m!s2;; mm

'atum

p3 +E a

Z p

g

V E ++=

γ 2

2

(a)

( )

( ) mm N

m N x

sm

sm E 60.4

/9810

/1035

/81.92

/05.33

23

2

2

++=

•

C.D; m

m E 642.8= (water)


11/38

( b) Z p

g

V E ++=

γ 2

2

( )( ) mm N xm N x

sm sm E 60.4

/98105.1/1035

/81.92/05.3

3

23

2

2

++=

m E 453.7=

(c ) ( )( )

mm N

m N x

sm

sm E 60.4

/54.6

/1035

/81.92

/05.33

23

2

2

++=

m E 76.356,5= (gas)

(molasses)


12/38

$5ample 2. 0 liquid ( s32.;;) is flowing in a E; mm diameter pipe. #he total

energ& at a gi/en point is *.C* m!. #he ele/ation of the pipe

abo/e the datum is +.; m and the pressure in the pipe is DE.E a.

%ompute the /elocit& of flow and the >.. of the stream at thatpoint.

9olution:

E; mm

'atum

•

+.; m

p 3 DE.E a

(a) Z p

g

V E ++=

γ 2

2

( ) mm N

m N x

sm

V m 0.3

)/9810(2

/105.65

/81.9247.7

3

23

2

2

++=

smV /712.4=


13/38

(b) AV Q =

( ) ( ) smmQ /712.4050.042π

=

sm xQ /10252.9 33−=

( c) HP watts

E Q P H

/746.

⋅⋅= γ

( )( )( ) HP watts

mm N x sm x

P H /746

47.7/98102/10252.9.

333−

=

horspowr P H 82.1. =


14/38

E@a")l( ;. At a )oint A -'(*( t'( suction )i)( l(ain0 to a )u") is1.7 " ,(lo- t'( )u") an o)(n "ano"(t(* inicat(s a +acuu" of1 "" of "(*cu*!. T'( )i)( is

1 "" in ia"(t(*3 an t'( isc'a*0( is .; ";8s of oil #s 6

.$. Co")ut( t'(total '(a at )oint A -it' *(s)(ct to a atu" at t'( )u").

Solution&

• 0

'atum

.2; m

(a) A

QV =

( ) 2

3

10.04

/030.0

m

sm

π = sm /82.3=

Z p

g

V E ++=

γ 2

2

(b) ( )

( ) m

m N x

m N x

sm x

sm20.1

/981085.0

/10015.24

/81.92

/82.33

23

2

2

−−

+= N m N /33.3 −−=

p 3 Bmhm 3 (+.D 5 G; !m+)(;. m) 3 2C.;E a


15/38

E@a")l( . A +(*tical ci*cula* stac2 ; " 'i0' con+(*0(s unifo*"l!f*o" a ia"(t(*

of " at t'( ,otto" to . " at t'( to). Coal 0as -it' a unit-(i0't of .1

N8"; (nt(*s at t'( ,otto" of t'( stac2 -it' a +(locit! of ;. "8s. T'( unit

-(i0't of t'( 0as inc*(as(s unifo*"l! to .G N8"; at t'( to).Co")ut( t'(

"(an +(locit! (+(*! . " u) t'( stac2.

Solution&

*.E m

*.E m

*.E m

*.E m

D m

E.; m

coal gas

(a) %hanges in the diameter 3

30

56−

2

+

C

E

mm 5.7/25.0=

#herefore for e/er& *.E m height, the change

1n diameter is ;.2E m. 9tarting from the bottom

#he diameters changes as follows:

' 3 D.;; m

'2 3 E.*E m

'+ 3 E.E; m

'C 3 E.2E m'E 3 E.;; m


16/38

#,$ T'( C'an0( in s)(ci


17/38

( ) 52

4

2

3

2

2

22 54

59.625.54

12.650.54

65.575.54

18.505.364

71.4 V xV xV xV x x

=

=

=

=

π π π π π

smV /02.32 =

smV /026.33 =

smV /066.34 =

smV /139.35 =


18/38

$5ample E. 0 +;; mm pipe is connected b& a reducer to a ;; mm pipe.

oints and 2 are at the same ele/ation. #he pressure at is 2;*

a. #he flow is ;.;2 m+!s and the energ& lost between and 2 is

equi/alent to 2;.D a. %ompute the pressure at 2 if the liquid isoil (s 3 ;.;).

9olution:+;; mm dia.

;; mm dia.

• •2 2

'atum

(a)1

1 A

QV =

( ) 2

3

30.04

/028.0

m

sm

π = sm /396.0=

2

2 A

QV =

( ) 2

3

10.04

/028.0

m

sm

π = sm /565.3=

(b)γ

p H L =−21,

3

23

21, /981080.0/1068.20m N xm N x H L =−

m635.2=


19/38

21,22

2

21

1

2

1

22 −+++=++ L H Z p g V Z p g V γ γ

( c) #he energ& equation ( – 2)

635.2

98108.081.92

565.3

98108.0

10207

81.92

396.02

2

2

1

32

+++=++ Z

x

p

x

Z

x

x

x

22 56.299,181

m

N p = or orkPa

m

kN p

22 299.181=

E l C t t' l it ' f t' H t ' if D1


20/38

E@a")l( . Co")ut( t'( +(locit! '(a of t'( H(t3 as s'o-n3 if D16 ""3

D7 6 7 ""3 t'( )*(ssu*( '(a at 1 is ; " of t'( liqui/o-in03 an t'( '(a lost ,(t-((n 1 an 7 is of t'(

+(locit! '(a at 7.

• •

1 2

m p

301 =γ ?et

'atum

*E mm dia. 2E mm dia.

(a) 7sing the continuit& equation:

2211 V AV A =

( ) ( ) 2212 025.04075.04 V V π π =

2

2

1075.0

025.0V V

=

219

1V V =

Solution


21/38

#,$ T'( (n(*0! (quation #1 : 7$

21,22

2

211

2

1

22 −+++=++ L H Z

p

g

V Z

p

g

V

γ γ

; (atm.)

g

V

g

V

g

V

205.0

230

2

9

12

2

2

2

2

2

+=+

302

5.1281

1 2

2

2

2 −=−−⋅ g

V

g

V

302

0376.12

2 −=−− g

V

m g

V 913.28

2

2

2 =−


22/38

$5ample *. 1n the figure, a E; mm pipe line leads downhill from a reser/oir

and discharges into air. 1f the loss of head between 0 and is CC m,

compute the discharge.

$le/. CD mAW.9

B

?et$le/. ; ('atum)

E; mm dia.

9olution:

$nerg& equation (0 – )

! A L ! ! !

A A A H Z

p

g

V Z

p

g

V −+++=++ ,

22

22 γ γ

negl. ; (atm.) ; (atm.)

m g

V m ! 44

246

2

+=

m g

V ! 22

2

=

s

mV ! 264.6=


23/38

#hen, ! !V AQ =

( )

=

smmQ 264.6050.0

42π

s

mQ

3

012.0=

ote:

g

V A

2

2

is negligible since the water surface in the large reser/oir

will drop down /er& slowl&.

$le/. CD mAW.9

$le/. ; ('atum)

E; mm dia.

B

?et


24/38

$5ample . 0 pump draws water from a 2;; mm suction pipe and

discharging through a E; mm pipe in which the /elocit& is

+.DE m!s. #he pressure is +C.C* a at 0. #he E; mm pipe

discharges into the air at %. #o what height h abo/e can the water

be raised, 1f is .; m abo/e 0 and 2; > is deli/ered to thepumpI 0ssume that the pump operates at *;J efficienc& and the

frictional loss between 0 and % is +.; m.

C

B

A

•

•

•2;; mm

E; mm

W.9.'atum

.; m

h

9olution

(a) 7sing %ontinuit& $q.

! ! A A V AV A =

( ) ( ) ( )658.315.0

4

20.0

4

22 π π = AV

( )658.320.0

15.0 2

= AV

s

mV A 058.2=


25/38

A AV AQ = ! !V A=

( )058.220.04

2

= xπ

s

m3065.0=

(b) 7sing the energ& equation (0 – %)

" A L" " "

p A A A H Z

p

g

V E Z

p

g

V −+++=+++ ,

22

22 γ γ

( ) ( )( ) 0.380.1081.92

658.309810

1047.34

81.92

058.2 232

++++=++−+ h x E x

x p

; (atm.)

p E h +−= 78.8

C

B

A

•

•

E; mm

'atum.; m

h

•

2;; mm


26/38

Where:

J eff. (> of pump)746

p E Q ⋅⋅=

γ

( ) ( ) ( )

746

9810065.02070.0

p E =

m E p 38.16= ( $nerg& added b& the pump to the s&stem)

therefore

38.1678.8 +−=h

mh 60.7=

$5ample G #he D; mm pipe conducts water from reser/oir 0 to a pressure


27/38

$5ample G. #he D; mm pipe conducts water from reser/oir 0 to a pressure

turbine, which discharges through another D; mm pipe into tailrace

. #he head losses are:

From 0 – : E2!2g

From 2 – : ;.22!2g

1f the discharge is ;.*; m+!s, what > is being gi/en up b& the

water turbine.

$le/. CD mAW.9

$le/. ; ('atum)

D; mm dia.

BW.9

D; mm dia.


28/38

$le/. D; mAW.9

D; mm dia.

$le/. ; ('atum)BW.9

D; mm dia.

(a) 7sing $nerg& $quation (0 – )

! A L ! ! !

T A A A H Z

p

g

V E Z

p

g

V −+++=−++ ,

22

22 γ γ

negl. ; negl. ;

++=− g

V

g

V E T 22.025060

22

Where:

A

QV =

( ) 2610.0

4

708.0

π =

s

m423.2=

−=

g

V E T

22.560

2

#hen,

−=

81.92

423.22.560

2

x E T m444.58=


29/38

#herefore

746

E Q

HP

⋅⋅

=

γ

( )

hp s

m N

mm

N

s

m

HP −

=

746

444.589810708.03

3

hp HP 131.544=

ote : s

m N − s

# = $atts=


30/38

$5ample ;. 1n a test to determine the discharge coefficient of a E; mm b&

2 mm enturi 6eter the total weight of water passing through the

meter in E minutes was +;; . 0 mercur&water differential gage

connected to inlet and throat showed an a/erage mercur& difference

of +D; mm. 'etermine the meter coefficient.

9olution:

(a) 6eter coefficient,

.

.

th

act

Q

Q" =

(b) 0ctual 'ischarge,

t%m

Vo&'mQact =.

t

$

Qact γ =.

min/60min5

/98103100 3

. s x

m N N

Qact =

s

m xQact

33

. 10053.1 −=

E; mm dia.

2 mm

+D; mm>g

H2O

• •

1 2


31/38

E; mm dia.

2 mm

+D; mm>g

H2O

• •

1 2

(c ) $nerg& equation ( – 2), >8 neglected (theoretical /alues)

21,22

2

21

1

2

1

22 −+++=++ L H Z

p

g

V Z

p

g

V

γ γ

γ γ

21

2

1

2

2

22

p p

g

V

g

V

−=−

;

where :

2211 V AV A =

( ) ( ) 22

1

2012.0

4050.0

4V V

π π =

2

2

1015.0

012.0V V

=

21 0576.0 V V =


32/38

and ( ) ( ) ( ) ( )γ γ

21 6.13360.00.1360.0 p p

=−+

( ) ( ) ( ) ( )γ γ

21 6.13360.00.1360.0 p p

=−+

m p p

536.421 =−γ γ

#herefore,

m g V

g V 536.4

22

2

1

2

2 =−

( )m

g

V

g

V 536.4

2

0576.0

2

2

2

2

2 =−

m g

V 536.4

2997.0

2

2 =

s

mV 45.9

2 =

9o, 22V AQth =

( ) ( )45.9012.04

2π =thQ

s

m xQth

3

310069.1

−=

Finall&,

3

3

10069.1

10053.1−

−

= x

x"

985.0="


33/38

$5ample . 0 pitot tube in a pipe in which air (Ba 3 2 !m+) is flowing is

connected to a manometer containing water as in the figure shown.

1f the difference in water le/els in the manometer is G; mm, what is the

/elocit& of flow in the pipe, assuming a tube coef. %p 3 ;.GGI

9olution:

21,22

2

21

1

2

1

22 −+++=++ L H Z p

g

V Z

p

g

V

γ γ

(a) $nerg& equation ( – 2), >8 3 ; ( for theoretical /alues)

• •• 2 +

G; mm

>2"

air

;

;


34/38

• •• 2 +

G; mm

>2"

air

21,22

2

211

2

122

−+++=++ Laa

H Z p

g

V Z

p

g

V

γ γ

; ;

aa

p p

g

V

γ γ

12

2

1

2−=

%onsidering the manometer reading starting from point 2 proceeding to point +

aa

w

a

p p

γ γ

γ

γ

32 0090.0

0 =+−−a

p

γ

1=


35/38

• •• 2 +

G; mm

>2"

air

aa

w

a

p pγ γ

γ γ

32 0090.00 =+−−

a

w

aa

p p

γ

γ

γ γ

090.012 =−

a

pγ

1=

0ssuming the specific weight of air, Ba 3 2 !m+

( )

12

9810090.012 =−aa

p p

γ γ m575.73=

then,aa

p p g V

γ γ

12

2

1

2 −=

575.732

2

1 = g

V

s

mV 994.371 =

Finall&,

s

m xV act 994.3799.0=

1 xV " V pact =

s

mV act 614.37=

EXERCISE PROBLEMS


36/38

EXERCISE PROBLEMS

.0 Fluid flowing in a pipe +; cm in diameter has a uniform /elocit& of

C m!s. #he pressure at the center of the pipe is C; a, and the ele/ation

of the pipe-s center line abo/e an assumed datum is C.E m. %ompute the total energ& per unit weight of the flowing fluid if it is (a) oil (s 3 ;;)

(b) gas (B 3 .E; !m+).

2.0 liquid of specific gra/it& .*E flows in a *E mm pipe. #he total energ&

at a point in the flowing liquid is ;; K!. #he ele/ation of the pipe abo/e

a fi5ed datum is +.; m and the pressure in the pipe is G; a. 'etermine

the /elocit& of flow and the power a/ailable at that point.

+.oint 0 in the suction pipe is m below the pump. 1t is mounted with an

open manometer which reads a /acuum of 2; cm of mercur&. #he pipe is

; cm in diameter and the flow is +E liters!s of water. %ompute the total energ& at point 0 with respect to a datum through the pump.


37/38

C. 0 cit& requires a flow of .E m+!s for its water suppl&. 'etermine the

diameter of the pipe if the /elocit& of flow is to be .; m!s.

E. 0 pipe line consists of three successi/e lengths of E;; mm, C;; mm,

and +;; mm pipes. With continuous discharge of +;; liters!s of oil

(;.*E) compute the mean /elocit& in each pipe.

D. 0 +;; mm pipe is connected b& a reducer to a ;; mm pipe. oints and

2 are along the same ele/ation. #he pressure at point is 2;; a. #he

flow is +; liters!s and the energ& lost between and 2 is equi/alent to 2; a. %ompute the pressure at point 2 if the liquid flowing is water.

*.%ompute the /elocit& of the ?et if the larger diameter is ;; mm and the

smaller diameter is +; mm. #he pressure head at point is +; m of the

flowing water and the head lost between points and 2 is E J of the /elocit& head of the ?et.

1 mm 3 mm • V2

• V1

1n figure C; liters!s of sea water (s3 ;+) is flowing from


38/38

. 1n figure, C; liters!s of sea water (s3.;+) is flowing from

to 2, and the pressure at is GE a while at 2 the pressure

is – 2; a. oint 2 is D m higher than . %ompute the

energ& lost in a between and 2 if ' 3 +;; mm and

'2 3 ;; mm.

!1

!2

•

V2

•

V1

1

2

D m

Part I.A -Fundamentals of Fluid Dynamics

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Transcript of Part I.A -Fundamentals of Fluid Dynamics