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College of Engineering and Computer ScienceMechanical Engineering Department

Mechanical Engineering 390Fluid Mechanics

Spring 2008 Number: 11971 Instructor: Larry Caretto

Solutions to Exercise Four Bernoullis Equation Part II

1 Water flows through the pipe contraction shown in thefigure at the right. For the given 0.2m difference inmanometer level! determine the flow rate as a functionof the diameter of the small pipe! D. "#ro$lem andfigure %.%0 from Munson et al.! Fluid Mechanicste&t.'

Apply the ernoulli e!uation "or incompressible# in$isci%"lo&s# sho&n belo bet&een t&o points '1( an% '2( along astreamline in the center o" the pipe)

( )

( )02

2

1

2

212

12 =

+

+ g

VV

g

pp

zz

*he measurement tube at point 1 is "acing the "lo& so it &ill measure the stagnation pressure)*his means that the ele$ation %i""erence sho&n in the %iagram 'h + 0)2 m( times the speci"ic

&eight o" the "lui% is the %i""erence bet&een the stagnation pressure at point '1( 'p1,-12.2( an%

the static pressure# p2at point '2() /e thus ha$e the "ollo&ing interpretation o" the height%i""erence)

ghV

pppV

pghh

+=+==22

2

1

122

2

1

1

/e can substitute this epression "or p2into our ernoulli e!uation along &ith the "act that at thecenter o" the pipe 2+ 1) *his gi$es the ernoulli e!uation as

( ) ( ) ( )

02

202

2

1

2

21

2

1

12

1

2

212

12 =

+

+

+=

+

+

g

VV

g

pghV

p

g

VV

g

ppzz

/e see that the terms in p1an% -12cancel lea$ing the "ollo&ing result)

ghVg

Vh 20

2 2

2

2==+

*he "lo& rate is simply -2A2+ -222.3# &ith 2+ )

( ) 22

2

222

556.181.92.02

4

2

4

2 Ds

m

s

mmghDghAVAQ =====

*he "lo& rate &ill ha$e units o" m4.s &hen is in meters)

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2 ( #itotstatic tu$e is used to measure the velocit) of helium in a pipe. *he temperatureand pressure are +0oF and 2, psia. ( water manometer connected to the #itotstatic tu$eindicates a reading of 2.% in. Determine the helium velocit). -s it reasona$le to considerthe flow as incompressi$le E&plain. "#ro$lem %.%+ from Munson et al.! Fluid Mechanicste&t. Figure $elow from solutions manual.'

*he "lo& can be assume% incompressible i" the ach number# a + -.c# &here c + '=>*(1.2is the

soun% spee%# is less than 0)4) /e can assume that the "lo& is incompressible# then compute the$elocity an% ach number# a) I" a ? 0)4 our ans&er &ill be correct)

A itotstatic tube measures the %i""erence bet&een the static pressure# ps# &here the "lui% has a

$elocity# -# an% the stagnation pressure# p0+ ps, -2.2# &hich occurs &hen the original $elocitt#

-# is re%uce% to ero)) 'See the %iscussion o" the itotstatic tube in the tet an% lecture presentations "or morein"ormation)( /e can obtain the $elocity# -# "rom the%i""erence in static an% stagnation pressures by sol$ing

the e!uation p0+ ps, -2.2 "or -) /hen the pressure

%i""erence is measure% by a manometer# as sho&n at theright 'the top part o" the %iagram# lin=ing each si%e &ith

the "lo&ing "lui% &as truncate% in copying(# the measure% $alue o" p0@ ps+ 'mano@ "lui%(h) In this

case &here the manometer "lui% is &ater an% the "lo&ing "lui% is helium &e can neglect thespeci"ic &eight o" heilum an% &rite p0@ ps+ manoh so the e!uation "or $elocity becomes)

fluid

mano

fluid

s hppV

22

0=

=

/e can compute the %ensity o" the "lui%# eliuim# "rom the i%eal gas la& using the $alue o" > +2)33103"tBlb".slugB> "or helium "oun% in *able 1)7 in the insi%e "ront con$er) /e can use thepressure o" 2 psia %irectly since it is alrea%y an absolute pressure# but &e ha$e to con$ert thetemparature o" 30o< to a >an=ine temperaure by the e!uation > + o< , 39)7# so &e use atemperature o" 399)7 > in both the %ensity calculation an% the ach number calculation belo&)

( )3

4

4

2

2

2

1080.5

67.49910446.2

14425

ftslugsx

RRslug

lbftxft

in

in

lb

RTP

f

f

=

==

/e can chec= the assumption that the speci"ic &eight o" the helium &as neglegible compare% tothe manometer "lui%# &ater# as "ollo&s:

3

2

23

4

3

38.62

1

174.321080.54.62

ft

lb

ftslug

slb

s

ft

ft

slugsx

ft

lbg

fff

HemanoHemano =

==

*his gi$es an error o" only 0)04D) Ether gases# &ith higher molecular &eights# &houl% gi$e alarger# but still neglegible# error)

Subsituting the %ensity "oun% abo$e an% the speci"ic &eight o" &ater# the manometer "lui%# as2)3 lb"."t

4in the e!uation "or $elocity gi$es)

( )

===

3

4

3

1080.5

12

13.2

4.622

2

ft

slugsx

in

ftin

ft

lb

hV

f

fluid

mano

203 ft/s

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/e can no& chec= the compressibliity assumption by computing the ach number) *he $alue o"= "or elium is also "oun% in *able 1)7 on the insi%e "ront co$er)

( ) ( )

063.0

1

67.499

10446.2

66.1

203

2

4

=

===

slb

ftslug

RRslug

lbftx

s

ft

kRT

V

c

VMa

f

f

*his $alue is &ell belo& the upper limt o" a + 0)4 "or &hich &e sai% &e coul% consi%er a gas "lo&

incompressible so &e conclu%e that the helium flow can be considered incompressible.

% Water flows from a large tan/ as shownin the figure at the right. (tmosphericpressure is 1+., psia and the vaporpressure is 1.0 psia. -f viscous effectsare neglected! at what height! h! willcavitation $egin *o avoid cavitation!should the value of D1$e increased or

decreased *o avoid cavitation! shouldthe value of D2$e increased ordecreased. E&plain. "#ro$lem andfigure %.0 from Munson et al.! FluidMechanicste&t.'

Ca$itation is li=ely to occur at the thin %iameter point &here 1+ 1 in) 'oint 2 is open to theatmosphere so ca$itation &ill not occur here) *o %etermine i" ca$itation &ill occur &e ha$e to "in%the pressure at the point &here 1+ 1 in) *o %o this &e ha$e to =no& the o$erall "lo& rate) /ecan "in% this by applying ernoulliFs e!uation bet&een the top o" the large tan= 'point 0( an% theeit 'point 2() *his gi$es

( ) ( )

02

2

2

2

020

20 =

+

+

g

VV

g

ppzz

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/e can use the pre$ious result that -22+ 2gh an% the %e"inition o" +g to re&rite the pre$ious

result as "ollo&s)

( ) 01/22

1/

2

1/

2

/ 2

1

2

2

1

2

1

2

212

2

2

1

2

21

2

2

2

1

2

2

2

21=+=

+=

+=

+ hAA

pgh

g

AApV

g

AAp

g

VAAV

g

p

Sol$ing this e!uation "or the gage pressure# p1# an% noting that the area ratio is the same as the

%iameter ratio s!uare% gi$es)

( )hDDp 41

4

21 /1=

I" &e a%% the atmospheric pressure to both si%es o" the e!uation &e &ill obtain the absolutepressure at point 1# &hich must be geater than the $apor pressure to a$oi% ca$itation))

( ) vatmatmabs pphDDppp >+=+= 44

21,1 /1

Substituting the gi$en %ata that 1+ 1 in# 2+ 2 in# patm+ 13)4 psia# p$+ 1)0 psia# an% using aspeci"ic &eight o" 2)3 lb"."t

4"or &ater gi$es the "ollo&ing numerical result)

psiapsiah

in

in

ft

lb

lb

inpsia

in

ft f

f

60.13.14

1

21

4.621

144

4

3

2

2

2

>+

fthpsiahft

psia96.17.12

5,6

Note than &hen an ine!uality is multiplie% or %i$i%e% by a negati$e number the %irection o" theine!uality is re$erse%)

*o %etermine the e""ects o" 1an% 2&e can sol$e ( ) vatm pphDD >+ 44

2 /1 "or h) Note that

since 2H 1&e &ill be %i$i%ing by a negati$e number so &e ha$e to change the %irection o" theine!uality)

( ) ( )1//1 4

142

41

42

=