91751131 solucionario capitulo 15 paul e tippens
TRANSCRIPT
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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
$0-
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Chapter 15. Fluids Physics, 6th Edition
Fluid Pressure
155. Find the pressure in kilopascals due to a column o! mercury 60 cm hih. What is this
pressure in l3*in.$ 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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Chapter 15. Fluids Physics, 6th Edition
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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Chapter 15. Fluids Physics, 6th Edition
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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Chapter 15. Fluids Physics, 6th Edition
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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Chapter 15. Fluids Physics, 6th Edition
4 -
4 -
0.16 k1.5- + 10 m %
1.5- + 10 m
mV
V ρ = = = ; ρ = 5--- k*m-
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Chapter 15. Fluids Physics, 6th Edition
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( (.$6 ' 1.)1
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Chapter 15. Fluids Physics, 6th Edition
@ ' /0.0$(- !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"
- $$- $1$ al 1 !t 1 min / !t0.01($ !t *s% 0.0$1 !t
1 min (.4 al 60 s 4
5 A π = = = =
-
$
0.01($ !t *s
0.0$1 !t
5/
A= = % / = 0.1( !t*s
To find the range 4- we first find time to stri.e ground from: = 8gt "
$10
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Chapter 15. Fluids Physics, 6th Edition
$
$ $/4 !t0.500 s
-$ !t*s
t
g = = = ; Now range is 4 = / 4t
5ange: 4 = /t = /0.1( !t*s/0.5 s% 4 = 0.40) !t or 4 ' 4.)0 in.
15$). Water !lo#in at 6 m*s throuh a 6cm pipe is connected to a -cm pipe. What is the
velocity in the small pipe" 8s the rate o! !lo# reater in the smaller pipe"
A!/! = A"/"
$$
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
$11
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Chapter 15. Fluids Physics, 6th Edition
A = /1.$ 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
$$ $/). m*s /6 m/ gh= = % / ' ).)0 m*s
$ $5 $/0.0$ m -.1 + 10 m %
4 4
D A 5 /A
π π = = = =
5 = /).)0 m*s/-.1 + 105 m$% 5 = -.11 + 104 m-*s
4 -
-
-.11 + 10 m 60 s 1
1 s 1 min 0.001 m 5
=
% 5 ' 1( *min
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Chapter 15. Fluids Physics, 6th Edition
$$
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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Chapter 15. Fluids Physics, 6th Edition
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Chapter 15. Fluids Physics, 6th Edition
?#
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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Chapter 15. Fluids Physics, 6th Edition
$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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Chapter 15. Fluids Physics, 6th Edition
$ - -
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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Chapter 15. Fluids Physics, 6th Edition
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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Chapter 15. Fluids Physics, 6th Edition
$ $ $ $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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Chapter 15. Fluids Physics, 6th Edition
-
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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Chapter 15. Fluids Physics, 6th Edition
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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Chapter 15. Fluids Physics, 6th Edition
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'