part i.a -fundamentals of fluid dynamics
TRANSCRIPT
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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
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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
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#,$ 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
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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& $quation was originall& formulated b&
'aniel ernoulli in *+: #he ernoulli-s rinciple state that as the
speed of a mo/ing fluid (liquid or gas) increases, the pressure withinthat fluid decreases. asicall&, the concept states that the total
energ& in a steadil& flowing fluid s&stem is a constant along the flow
path. 0n increase in the fluid-s speed must therefore be matched b&
a decrease in its pressure.
1n mathematical equation
constant=+ .... E K E P
1
2
2
1k mV mgh =+ V
k
V
mv
V
mgh 1
2
2
1
=+
'i/iding b& constant olume,
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V
k
V
mv
V
mgh 1
2
2
1
=+
k v gh =+ 22
1 ρ ρ
k vh =+ 22
1 ρ γ
k v p =+ 22
1 ρ
k v pv pv p =+=+=+ 2
22
2
11
2
2
1
2
1
2
1 ρ ρ ρ
We can write,
Where: p 3 pressure
/ 3 /elocit&
4 3 densit&
ote: 1n the ernoulli-s concept, it is clear that if
/elocit& increases at one end, it must be
matched b& a decrease in pressure.
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#c$ 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
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( e ) 16789$6"6$#76 $Q70#1" : #he impulse of a force is
equal to the change in the momentum of the mass.
( ) ( )V M V V M Ft o f ∆=−=
Where: Ft – impulse
6f – final momentum
6o – initial momentum
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2. 1mportant 'efinitions in Fluid Flow
(a) ath line is a line made b& a single particle as it mo/es
during a period of time
(a) 9treamline: an imaginar& line within the flow for which the
tangent at an& point gi/es the direction of flow at that point.
(b) 9tream tube: an element of fluid bounded b& streamlines
which enclose or confine the flow.
(c) 9tead& Flow: when the /elocit&, pressure, discharge or
flow rate at a gi/en point in flowing stream of the fluid
remains constant with time or dQ!dt 3 ;.
•P
1
V1 •
P2
V2
•P3V
3
•P
4
V4
•P
5
V5 •P6
V6
Path Line and Streamline
http://../Streamlines%20around%20bodies.ppthttp://../Streamlines%20around%20bodies.ppt
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(d) 7niform Flow: if, at a gi/en instant, the /elocit& remains constant
with respect to a stretch or distance in the flowing stream of
fluid or d!ds 3 ;.
(e) 'ischarge or Flow ead 8oss: the energ& per unit weight lost due to friction (ma?or
loss) or total disturbances (minor loss).
(g) ower is obtained b& multipl&ing the total energ& head (@.$. 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
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$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)
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( 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)
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$5ample 2. 0 liquid ( s32.;;) is flowing in a E; mm diameter pipe. #he total
energ& at a gi/en point is *.C* m!. #he ele/ation of the pipe
abo/e the datum is +.; m and the pressure in the pipe is DE.E a.
%ompute the /elocit& of flow and the >.. of the stream at thatpoint.
9olution:
E; mm
'atum
•
+.; m
p 3 DE.E a
(a) Z p
g
V E ++=
γ 2
2
( ) mm N
m N x
sm
V m 0.3
)/9810(2
/105.65
/81.9247.7
3
23
2
2
++=
smV /712.4=
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(b) AV Q =
( ) ( ) smmQ /712.4050.042π
=
sm xQ /10252.9 33−=
( c) HP watts
E Q P H
/746.
⋅⋅= γ
( )( )( ) HP watts
mm N x sm x
P H /746
47.7/98102/10252.9.
333−
=
horspowr P H 82.1. =
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E@a")l( ;. At a )oint A -'(*( t'( suction )i)( l(ain0 to a )u") is1.7 " ,(lo- t'( )u") an o)(n "ano"(t(* inicat(s a +acuu" of1 "" of "(*cu*!. T'( )i)( is
1 "" in ia"(t(*3 an t'( isc'a*0( is .; ";8s of oil #s 6
.$. 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
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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
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#,$ T'( C'an0( in s)(ci
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( ) 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 =
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$5ample E. 0 +;; mm pipe is connected b& a reducer to a ;; mm pipe.
oints and 2 are at the same ele/ation. #he pressure at is 2;*
a. #he flow is ;.;2 m+!s and the energ& lost between and 2 is
equi/alent to 2;.D a. %ompute the pressure at 2 if the liquid isoil (s 3 ;.;).
9olution:+;; mm dia.
;; mm dia.
• •2 2
'atum
(a)1
1 A
QV =
( ) 2
3
30.04
/028.0
m
sm
π = sm /396.0=
2
2 A
QV =
( ) 2
3
10.04
/028.0
m
sm
π = sm /565.3=
(b)γ
p H L =−21,
3
23
21, /981080.0/1068.20m N xm N x H L =−
m635.2=
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21,22
2
21
1
2
1
22 −+++=++ L H Z p g V Z p g V γ γ
( c) #he energ& equation ( – 2)
635.2
98108.081.92
565.3
98108.0
10207
81.92
396.02
2
2
1
32
+++=++ Z
x
p
x
Z
x
x
x
22 56.299,181
m
N p = or orkPa
m
kN p
22 299.181=
E l C t t' l it ' f t' H t ' if D1
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E@a")l( . Co")ut( t'( +(locit! '(a of t'( H(t3 as s'o-n3 if D16 ""3
D7 6 7 ""3 t'( )*(ssu*( '(a at 1 is ; " of t'( liqui/o-in03 an t'( '(a lost ,(t-((n 1 an 7 is of t'(
+(locit! '(a at 7.
• •
1 2
m p
301 =γ ?et
'atum
*E mm dia. 2E mm dia.
(a) 7sing the continuit& equation:
2211 V AV A =
( ) ( ) 2212 025.04075.04 V V π π =
2
2
1075.0
025.0V V
=
219
1V V =
Solution
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#,$ 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 =−
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$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=
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#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
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$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=
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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
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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
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$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
$le/. ; ('atum)
D; mm dia.
BW.9
D; mm dia.
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$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=
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#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=
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$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
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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 =
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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="
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$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
;
;
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• •• 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=
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• •• 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
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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.
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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
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. 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