91751131 Solucionario Capitulo 15 Paul E Tippens

8/18/2019 91751131 Solucionario Capitulo 15 Paul E Tippens

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Chapter 15. Fluids Physics, 6th Edition

Chapter 15. FLUIDS

Density

15.1. What volume does 0.4 k o! alcohol occupy" What is the #eiht o! this volume"

$

m 0.4 k% &'

()0 k*m

m

V ρ

ρ = = % V = 5.06 + 104 m-

W = DV = ρ gV ' ()0 k*m-/). m*s$/5.06 + 104 m-% W ' -.)$

15$. 2n unkno#n su3stance has a volume o! $0 !t- and #eihs --(0 l3. What are the #eiht

density and the mass density"

-

--(0 l3

$0 !t

W D

V = = % D = 16 l3*!t-

-

$

16 l3*!t

). m*s

D

g ρ = = % ρ ' 5.$( slus*!t-

15-. What volume o! #ater has the same mass o! 100 cm- o! lead" What is the #eiht density o!

lead"

First find mass of lead: m = ρ V = /11.- *cm-/100 cm-% m ' 11-0

Now water:-

-

11-0 11-0 cm

1 *cmw

w

mV

ρ = = = ; V w = 11-0 cm- D = ρ g

D ' /11,-00 k*m-/). m*s$ ' 110,(40 *m-% D = 1.11 + 105 *m-

154. 2 $00m !lask /1 ' 1 + 10- m- is !illed #ith an unkno#n liuid. 2n electronic 3alance

indicates that the added liuid has a mass o! 1(6 . What is the speci!ic ravity o! the

liuid" Can you uess the identity o! the liuid"

V ' $00 m ' 0.$00 ' $ + 10 4 m-% m ' 0.1(6 k

4 -

0.1(6 k

$ + 10 m

m

V ρ −= = ; ρ = 0 k*m

-, en7ene

$0-


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Fluid Pressure

155. Find the pressure in kilopascals due to a column o! mercury 60 cm hih. What is this

pressure in l3*in.$ and in atmospheres"

P = ρ gh = /1-,600 k*m-/). m*s$/0.60 m% P = 0 kPa

$0.145 l3*in.0 kPa %

1 kPa P

=

P ' 11.6 l3*in.$

0 kPa

101.- kPa*atm P = % P ' 0.()0 atm

156. 2 pipe contains #ater under a aue pressure o! 400 kPa. 8! you patch a 4mmdiameter

hole in the pipe #ith a piece o! tape, #hat !orce must the tape 3e a3le to #ithstand"

$ $5 $/0.004 m 1.$5( + 10 m

4 4

D A

π π = = = % %

F P

A=

F = PA = /400,000 Pa/1.$5( + 105 m$% P = 5.0-

15(. 2 su3marine dives to a depth o! 1$0 !t and levels o!!. 9he interior o! the su3marine is

maintained at atmospheric pressure. What are the pressure and the total !orce applied to a

hatch $ !t #ide and - !t lon" 9he #eiht density o! sea #ater is around 64 l3*!t.-

P = Dh ' /64 l3*!t-/1$0 !t% P ' (60 l3*!t$% P = 5-.- l3*in.$

F = PA = /(60 l3*!t$/- !t/$ !t% F = 46,100 l3

15. 8! you constructed a 3arometer usin #ater as the liuid instead o! mercury, #hat heiht o!

#ater #ould indicate a pressure o! one atmosphere"

- $

101,-00 Pa%

/1000 k*m /). m*s

P P gh h

g ρ

ρ = = = %

h ' 10.- m or -4 !t :

$04


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15). 2 $0k piston rests on a sample o! as in a cylinder cm in diameter. What is the aue

pressure on the as" What is the a3solute pressure"

$ $- $/0.0 m 5.0$( + 10 m

4 4

D A

π π = = = %

F mg P

A A

= =

$4

- $

/$0 k/). m*s -.)0 + 10 kPa%

5.0$( + 10 m P = = P ' -).0 kPa

P abs = 1 atm + P gauge = 101.- kPa ; -).0 kPa% P abs = 140 kPa


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1 $

$ 1

h

h

ρ

ρ =

151-. 2ssume that the t#o liuids in the =shaped tu3e in Fi. 15$1 are #ater and oil. Compute

the density o! the oil i! the #ater stands 1) cm a3ove the inter!ace and the oil stand $4 cm

a3ove the inter!ace. @e!er to Pro3. 151$.

-

1 $

$ 1

/1) cm/1000 k*m %

$4 cm

w woil

oil

hh

h h

ρ ρ ρ

ρ = = = ; ρ oil ' ()$ k*m-

1514. 2 #aterpressure aue indicates a pressure o! 50 l3*in.$ at the !oot o! a 3uildin. What is

the ma+imum heiht to #hich the #ater #ill rise in the 3uildin"

$ $ $

$

/50 l3*in. /144 in. *!t %

6$.4 l3*!t

P P Dh h

D= = = % h = 115 !t

The Hydraulic Press

1515. 9he areas o! the small and lare pistons in a hydraulic press are 0.5 and $5 in.$,

respectively. What is the ideal mechanical advantae o! the press" What !orce must 3e

e+erted to li!t a ton" 9hrouh #hat distance must the input !orce move, i! the load is li!ted

a distance o! 1 in." A1 ton ' $000 l3 B

$

$

$5 in.

0.5 in.

o $

i

A %

A= = % % $ = 50 $000 l3

50i F = % F i = 40.0 l3

si = % $ so = /50/1 in.% si = 50 in.

1516. 2 !orce o! 400 is applied to the small piston o! a hydraulic press #hose diameter is 4 cm.

What must 3e the diameter o! the lare piston i! it is to li!t a $00k load"

$ $

$

/$00 k/). m*s % /4 cm

400

o o o oo i

i i i i

F A d F d d

F A d F = = = = %

$06


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d o ' .5 cm

151(. 9he inlet pipe that supplies air pressure to operate a hydraulic li!t is $ cm in diameter. 9he

output piston is -$ cm in diameter. What air pressure /aue pressure must 3e used to li!t

an 100k automo3ile.

$

$

/100 k/). m*s %

/0.16 m

oi o i

o

F P P P

A π = = = % P i = $1) kPa

151. 9he area o! a piston in a !orce pump is 10 in.$. What !orce is reuired to raise #ater #ith

the piston to a heiht o! 100 !t" A 10 in.$/1 !t$*144 in.$ ' 0.06)4 !t$ B

F = PA = &Dh'A; F = /6$.4 l3*!t-/100 !t/0.06)4 !t$% F = 4-- l3

Archimedes’ Principle


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4 -

4 -

0.16 k1.5- + 10 m %

1.5- + 10 m

mV

V ρ = = = ; ρ = 5--- k*m-


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Apparent weight = true weight , buoant fore; m A g = mg , m f g

m A = m , m f ; m f = m 1 m A = ).1( (.$6 ' 1.)1


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@ ' /0.0$(- !t-*s/(.4 al*!t-/60 s*min% 5 = 1$.$ al*min.

$0 al

1$.$ al*minTime = % Time = 1.6- min

15$6. Water !lo#s !rom a terminal - cm in diameter and has an averae velocity o! $ m*s. What

is the rate o! !lo# in liters per minute" / 1 ' 0.001 m-. Go# much time is reuired to !ill

a 40 container" A 2 ' π/0.015 m$ ' (.0( + 104 m$ % & ' 40 ' 0.04 m- B

5 = /A = /$ m*s/(.0( + 104 m$ ' 1.41 + 10- m-*s

-

- -

0.04 m

1.41 + 10 m *sTime = ; Time = "6)7 s

15$(. What must 3e the diameter o! a hose i! it is to deliver o! oil in 1 min #ith an e+it

velocity o! - m*s"

-4 - 0.001 m 1 min 1.-- + 10 m *s

1 min 1 60 s 5

= =

% 5 = /A

$ ,4 -$ ,4 $4 4/1.-- + 10 m *s% 5.66 + 10 m

4 /- m*s

D 5 5 A D

/ /

π

π π

= = = = =

4 $5.66 + 10 m D = ; D = (.5$ + 10- m% D ' (.5$ mm

15$. Water !rom a $in. pipe emeres hori7ontally at the rate o! al*min. What is the emerin

velocity" What is the hori7ontal rane o! the #ater i! the pipe is 4 !t !rom the round"

- $$- $1$ al 1 !t 1 min / !t0.01($ !t *s% 0.0$1 !t

1 min (.4 al 60 s 4

5 A π = = = =

-

$

0.01($ !t *s

0.0$1 !t

5/

A= = % / = 0.1( !t*s

To find the range 4- we first find time to stri.e ground from: = 8gt "

$10


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$

$ $/4 !t0.500 s

-$ !t*s

t

g = = = ; Now range is 4 = / 4t

5ange: 4 = /t = /0.1( !t*s/0.5 s% 4 = 0.40) !t or 4 ' 4.)0 in.

15$). Water !lo#in at 6 m*s throuh a 6cm pipe is connected to a -cm pipe. What is the

velocity in the small pipe" 8s the rate o! !lo# reater in the smaller pipe"

A!/! = A"/"

$$

1 1 1 1$ $

$ $

6 cm/6 m*s

- cm

A / d //

A d

= = =

; /" ' $4.0 m*s

The rate of flow is the same in eah pipe)

Applicatins " #ernulli’s $%uatin

15-0. Consider the situation descri3ed 3y Pro3lem 15$). 8! the centers o! each pipe are on the

same hori7ontal line, #hat is the di!!erence in pressure 3et#een the t#o connectin pipes"

For a hori9ontal pipe- h! = h": P ! + ρ gh! +8 ρ /!" = P " + ρ gh" + 8 ρ /"

"

P ! 1 P " = 8 ρ /"" 1 8 ρ /!

" = 8 ρ &/"" , /!

" '; ρ = 1000 k*m-

P ! , P " = 8/1000 k*m-

A/$4 m*s$

/6 m*s$

B ' $.(0 + 105

Pa% ∆ P = $(0 kPa

15-1. What is the emerent velocity o! #ater !rom a crack in its container 6 m 3elo# the sur!ace"

8! the area o! the crack is 1.$ cm$, at #hat rate o! !lo# does #ater leave the container" A

The pressures are the same at top and at ra.: P ! = P " B

P ! + ρ gh! +8 ρ /!" = P " + ρ gh" + 8 ρ /"

"%

ρ gh! +8 ρ /!" = ρ gh" + 8 ρ /""

Notie that ρ anels out and reall that /! an be onsidered as 9ero)

#etting /! = (- we ha/e: ρ gh! = ρ gh" + 8 ρ /"" or /"

" = "g&h! 1 h" '

/"" = $/). m*s$/6 m% /" ' 10. m*s

$11


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A = /1.$ cm$/1 + 104 m$*cm$ ' 1.$0 + 104 m$%

5 = /A 5 = /10.4 m*s/1.$0 + 104 m$%

5 = 1.41 + 10- m-*s or 5 ' 1.41 *s

15-$. 2 $cmdiameter hole is in the side o! a #ater tank, and it is located 5 m 3elo# the #ater

level in the tank. What is the emerent velocity o! the #ater !rom the hole" What volume

o! #ater #ill escape !rom this hole in 1 min" A Appl Torriellis theorem B

$$ $/). m*s /6 m/ gh= = % / ' ).)0 m*s

$ $5 $/0.0$ m -.1 + 10 m %

4 4

D A 5 /A

π π = = = =

5 = /).)0 m*s/-.1 + 105 m$% 5 = -.11 + 104 m-*s

4 -

-

-.11 + 10 m 60 s 1

1 s 1 min 0.001 m 5

=

% 5 ' 1( *min


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$$

1 $ $

l3 144 in.-0.( 44$1 l3*!t

in. 1 !t P

= =

&Absolute pressure pipe !'

D = ρ g;

--

$

6$.4 l3*!t1.)5 slus*!t

-$ !t*s

D

g

ρ = = =

se onsistent units for all terms: !t, slus


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?#

For gasoline: F * = ρ gh = /60 k*m-/). m*s$/1$0 + 106 m- ' 0.00

W = mg = /0.1 k/). m*s$ ' 0.)0 F * @ W N


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$f ieberg floats- then: F * = mi g and F * = ρ ω gV w

mi g = ρ iV i g; therefore- ρ iV i g = ρ ω gV w

>o that?

-

-

)$0 k*m

0.)-10-0 k*m

w i

i w

V

V

ρ

ρ = = = or ).-J

Thus- 6B)7 perent of the ieberg remains underwater- hene

the e4pression: CThats ust the tip of the ieberg)E


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$ - -

40 4.0 k4.0 k%

). m*s 1.0$ + 10 m

W mm

g V ρ −= = = = = ρ = 4000 k*m-


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154). 2 hori7ontal pipe o! diameter 1$0 mm has a constriction o! diameter 40 mm. 9he velocity

o! #ater in the pipe is 60 cm*s and the pressure is 150 kPa /a What is the velocity in the

constriction" /3 What is the pressure in the constriction"

$ $

$ $ 11 1 $ $ $ 1

$

1$0 mm% /60 cm*s

40 mm

d / d / d / /

d

= = =

% /" = 540 cm*s

P ! +8 ρ /!" = P " + 8 ρ /"

" and P " = P ! + 8 ρ /!" 1 8 ρ /"

"

- $ - $

$ 150,000 Pa H/1000 k*m /0.06 m*s H/1000 k*m /5.4 m*s P = + −

P$ ' 150,000 Pa ; 10 Pa 14,50 Pa ' 1-6,000 Pa% P " ' 1-6 kPa

ho# that the depth h reuired to ive the hori7ontal rane + is iven

3y

$ $

$ $

2 2 4h

−= ±

=se this relation to sho# that the holes euidistant a3ove and

3elo# the midpoint #ill have the same hori7ontal rane" A = 2 , h B

The emergent /eloit / 4 is: $ 4/ gh= and 4 = / 4t and = 8gt "

$$%

$$ 4

4 4 4t t

/ gh gh= = = ; & Now substitute t " into = 8gt " '

$ $ $

H % %$ 4 4 4 g4 g4 g 2 h gh h h= = = − =

%ultipl both sides b ChE and rearrange: h" 1 2h + 4" J = (

Now-$ 4

$

b b ah

a

− ± −= where a = !- b = 2- and =

$

4

4

$1

4

h

G


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$ $ $ $4 *4%

$/1 $ $

2 2 4 2 2 4h h

± − −= = ±

The midpoint is &2"'- and the ± term indiates the distane abo/e and below that point)


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-

1 $

1

0.1--( !t *s

/0.$5 !t

5/

A π = = % /! = 0.60) !t*s

$ $

$ $ 11 1 $ $ $ 1

$

6 in.% /0.60) !t*s

1 in.

d / d / d / /

d

= = =

% /" = $4.5 !t*s

1554. What must 3e the aue pressure in a !ire hose i! the no77le is to !orce #ater to a heiht o!

$0 m"

∆ P = ρ gh = /1000 k*m-/). m*s$/-0 m% ∆P ' 1)6 kPa


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Critical Thin(in& )uestins

1556. 2 livin room !loor has !loor dimensions o! 4.50 m and -.$0 m and a heiht o! $.40 m.

9he density o! air is 1.$) k*m-. What does the air in the room #eiht" What !orce does

the atmosphere e+ert on the !loor o! the room" & ' /4.50 m/-.$0 m/$.40 m%

& ' -4.56 m-% 2!loor ' /4.5 m/-.$ m ' 14.4 m$% %

m

V ρ = W = mg

- -/1.$) k*m /-4.56 m ' 44.6 km V ρ = = % W = /44.6 k/). m*s$% W = 4-(

F = PA = /101,-00 Pa/14.4 m$% F ' 1,460,000

155(. 2 tin co!!ee can !loatin in #ater /1.00 *cm- has an internal volume o! 10 cm- and a

mass o! 11$ . Go# many rams o! metal can 3e added to the can #ithout causin it to sink

in the #ater. A The buoant fore balanes total weight) B

F * = ρ gV = m g + mm g ; mm = ρ V 1 m;

The /olume of the an and the /olume of displaed water are the same)

mm = /1 *cm-/10 cm- 11$ mm ' 6.0


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P A = $116 l3*!t$ + 8($ slus*!t-/-.06 !t*s$

+ /$ slus*!t-/-$ !t*s$/6 !t 1 8($ slus*!t-/0.(64 !t*s$

P2 ' $116 l3*!t$ ; ).-6 l3*!t$ ; -4 l3*!t$ 0.54 l3*!t$% P A = $50) l3*!t

$

$

$ $

l3 1 !t$$0)

!t 144 in. A P

=

% P A = 1(.4 l3*in.

$ &absolute'

91751131 Solucionario Capitulo 15 Paul E Tippens

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Transcript of 91751131 Solucionario Capitulo 15 Paul E Tippens