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Fluid Mechanics
CHAPTER OUTLINE 14.1 Pressure 14.2 Variation of Pressure with Depth 14.3 Pressure Measurements 14.4 Buoyant Forces and Archimedes’s Principle 14.5 Fluid Dynamics 14.6 Bernoulli’s Equation 14.7 Other Applications of Fluid Dynamics
ANSWERS TO QUESTIONS
*Q14.1 Answer (c). Both must be built the same. The force on the back of each dam is the average pressure of the water times the area of the dam. If both reservoirs are equally deep, the force is the same.
Q14.2 The weight depends upon the total volume of water in the glass. The pressure at the bottom depends only on the depth. With a cylindrical glass, the water pushes only horizontally on the side walls and does not contribute to an extra downward force above that felt by the base. On the other hand, if the glass is wide at the top with a conical shape, the water pushes outward and downward on each bit of side wall. The downward components add up to an extra downward force, more than that exerted on the small base area.
Q14.3 The air in your lungs, the blood in your arteries and veins, and the protoplasm in each cell exert nearly the same pressure, so that the wall of your chest can be in equilibrium.
Q14.4 Yes. The propulsive force of the f ish on the water causes the scale reading to f luctuate. Its aver- age value will still be equal to the total weight of bucket, water, and f ish.
Q14.5 Clap your shoe or wallet over the hole, or a seat cushion, or your hand. Anything that can sustain a force on the order of 100 N is strong enough to cover the hole and greatly slow down the escape of the cabin air. You need not worry about the air rushing out instantly, or about your body being “sucked” through the hole, or about your blood boiling or your body exploding. If the cabin pres- sure drops a lot, your ears will pop and the saliva in your mouth may boil—at body temperature— but you will still have a couple of minutes to plug the hole and put on your emergency oxygen mask. Passengers who have been drinking carbonated beverages may f ind that the carbon dioxide suddenly comes out of solution in their stomachs, distending their vests, making them belch, and all but frothing from their ears; so you might warn them of this effect.
Q14.6 The boat f loats higher in the ocean than in the inland lake. According to Archimedes’s principle, the magnitude of buoyant force on the ship is equal to the weight of the water displaced by the ship. Because the density of salty ocean water is greater than fresh lake water, less ocean water needs to be displaced to enable the ship to f loat.
365
FIG. Q14.
366 Chapter 14
*Q14.7 Answer (b). The apple does not change volume appreciably in a dunking bucket, and the water also keeps constant density. Then the buoyant force is constant at all depths.
Q14.8 The horizontal force exerted by the outside f luid, on an area element of the object’s side wall, has equal magnitude and opposite direction to the horizontal force the f luid exerts on another element diametrically opposite the f irst.
Q14.9 No. The somewhat lighter barge will f loat higher in the water.
Q14.10 The metal is more dense than water. If the metal is sufficiently thin, it can float like a ship, with the lip of the dish above the water line. Most of the volume below the water line is f illed with air. The mass of the dish divided by the volume of the part below the water line is just equal to the density of water. Placing a bar of soap into this space to replace the air raises the average density of the compound object and the density can become greater than that of water. The dish sinks with its cargo.
*Q14.11 Answer (c). The water keeps nearly constant density as it increases in pressure with depth. The beach ball is compressed to smaller volume as you take it deeper, so the buoyant force decreases.
Q14.12 Like the ball, the balloon will remain in front of you. It will not bob up to the ceiling. Air pressure will be no higher at the f loor of the sealed car than at the ceiling. The balloon will experience no buoyant force. You might equally well switch off gravity.
Q14.13 (i) b (ii) c. In both orientations the compound f loating object displaces its own weight of water, so it displaces equal volumes of water. The water level in the tub will be unchanged when the object is turned over. Now the steel is underwater and the water exerts on the steel a buoyant force that was not present when the steel was on top surrounded by air. Thus, slightly less wood will be below the water line on the wooden block. It will appear to f loat higher.
*Q14.14 Use a balance to determine its mass. Then partially f ill a graduated cylinder with water. Immerse the rock in the water and determine the volume of water displaced. Divide the mass by the volume and you have the density. It may be more precise to hang the rock from a string, measure the force required to support it under water, and subtract to f ind the buoyant force. The buoyant force can be thought of as the weight of so many grams of water, which is that number of cubic centimeters of water, which is the volume of the submerged rock. This volume with the actual rock mass tells you its density.
*Q14.15 Objects a and c f loat, and e barely f loats. On them the buoyant forces are equal to the gravita- tional forces exerted on them, so the ranking is e greater than a by perhaps 1.5 times and e greater than c by perhaps 500 times. Objects b and d sink, and have volumes equal to e, so they feel equal-size buoyant forces: e = b = d. Now f has smaller volume than e and g still smaller volume, so they feel smaller buoyant forces: e is greater than f by 2.7 times and e is greater than g by 7.9 times. We have altogether e = b = d > a > f > g > c.
*Q14.16 Answer (b). The level of the pond falls. This is because the anchor displaces more water while in the boat. A f loating object displaces a volume of water whose weight is equal to the weight of the object. A submerged object displaces a volume of water equal to the volume of the object. Because the density of the anchor is greater than that of water, a volume of water that weighs the same as the anchor will be greater than the volume of the anchor.
366 Chapter
368 Chapter 14
Q14.26 (a) Since the velocity of the air in the right-hand section of the pipe is lower than that in the middle, the pressure is higher.
(b) The equation that predicts the same pressure in the far right and left-hand sections of the tube assumes laminar f low without viscosity. Internal friction will cause some loss of mechanical energy and turbulence will also progressively reduce the pressure. If the pres- sure at the left were not higher than at the right, the f low would stop.
*Q14.27 (i) Answer (c). The water level stays the same. The solid ice displaced its own mass of liquid water. The meltwater does the same. You can accurately measure the quantity of H 2 O going into a recipe, even if some of it is frozen, either by using a kitchen scale or by letting the ice f loat in liquid water in a measuring cup and looking at the liquid water level.
(ii) Answer (b). Ice on the continent of Antarctica is above sea level.
SOLUTIONS TO PROBLEMS
Section 14.1 Pressure
P14.1 M = V = ( ) ⎡ ( ) ⎣⎢^
ρiron 7 860 kg m 3 π m
M = 0 111. kg
P14.2 The density of the nucleus is of the same order of magnitude as that of one proton, according to the assumption of close packing:
ρ π
×
( )
− −
m V
27 43 15 3
kg 18 m
kg m 3
With vastly smaller average density, a macroscopic chunk of matter or an atom must be mostly empty space.
P14.3 P
F
A
( × )
− =^ ×
2 2 6 24^106
π
N m 2
P14.4 The Earth’s surface area is 4 π R^2. The force pushing inward over this area amounts to
F = P A 0 = P 0 (^) ( 4 π R^2 )
This force is the weight of the air: Fg = mg = P 0 (^) ( 4 π R^2 ) so the mass of the air is
m
P R
g
( )
0 2 (^ ×^5 )^ ⎡ (^ ×^6 )
2 4 π 1 013.^10 N m^4 π 6 37.^10 m 2 ⎣⎣
⎦ = ×
m s 2 kg
13794_14_ch14_p365-394.indd13794_14_ch14_p365-394.indd 368368 12/28/0612/28/06 3:47:333:47:33 PMPM
Fluid Mechanics 369
Section 14.2 Variation of Pressure with Depth
P14.5 Fel = F fluid or kx =ρ ghA
and h kx gA
ρ
h = ( ) ( × ) ( )
3 3
N m m kg m m s
2 3 2
((. ) ⎡⎣ ( × )⎤⎦
π 1 00 10
m
m
P14.6 (a) P = P 0 + ρ gh = 1 013. × 105 Pa + ( 1 024 kg m (^3) )( 9 80. m s^2 )( 1 1 000 m )
P = 1 01. × 10 7 Pa
(b) The gauge pressure is the difference in pressure between the water outside and the air inside the submarine, which we suppose is at 1.00 atmosphere.
P gauge (^) = P − P 0 = ρ gh = 1 00. × 107 Pa
The resultant inward force on the porthole is then
F = P gauge A = 1 00. × 10 7 Pa ⎡⎣ π ( 0 150. m)^2 ⎤⎦ = 7 09. × 100 5 N
P14.7 Fg = 80 0. kg (^) ( 9 80. m s (^2) ) = 784 N
When the cup barely supports the student, the normal force of the ceiling is zero and the cup is in equilibrium.
F F PA A
A
F
P
g g
= = = (^) ( × )
= = ×
5
5
Pa
..74 × 10 −^3 m 2
P14.8 Since the pressure is the same on both sides, F A
F
A
1 1
2 2
In this case, 15 000 200 3 00
= F^2
or F 2 = 225 N
P14.9 The excess water pressure (over air pressure) halfway down is
P gauge = ρ gh = ( 1 000 kg m (^3) )( 9 80. m s^2 )( 1 20. m) = 1 .1 18 × 10 4 Pa
The force on the wall due to the water is
F = P gauge (^) A = (^) ( 1 18. × 10 4 Pa (^) )( 2 40. m )( 9 60. m) = 2 7. 1 1 × 10 5 N
horizontally toward the back of the hole. Russell Shadle suggested the idea for this problem.
Vacuum
FIG. P14.
FIG. P14.
Fluid Mechanics 371
P14.13 The bell is uniformly compressed, so we can model it with any shape. We choose a sphere of diameter 3.00 m. The pressure on the ball is given by: P = P atm + ρ w gh so the change in pressure on the ball from when it is on the surface of the ocean to when it is at the bottom of the ocean is ∆ P =ρ w gh. In addition:
∆ V V^ ∆ P B
ghV B
ghr B
= −^ = − ρ w^^ = −^4 πρ w B 3
3 , where is thee Bulk Modulus.
kg m 3 m s^2 ∆ V = −
4 π (^) ( 1 030 (^) )( 9 80. (^) ) 1 10 000 1 50 3 14 0 10
3 10
m m a
( )( ) ( ) (^) ( × )
P
2 2 m 3
Therefore, the volume of the ball at the bottom of the ocean is
V − ∆ V = 4 ( ) − = −
π 1 50. m 3 0 010 2. m 3 14 137. m^3 0 010. 22 m 3 = 14 127. m^3
This gives a radius of 1.499 64 m and a new diameter of 2.999 3 m. Therefore the diameter decreases by 0 722. mm.
Section 14.3 Pressure Measurements
P14.14 (a) We imagine the superhero to produce a perfect vacuum in the straw. Take point 1 at the water surface in the basin and point 2 at the water surface in the straw:
P 1 (^) + ρ gy 1 (^) = P 2 (^) +ρ gy 2
1 013. × 10 5 N m 2 + 0 = 0 + (1 000 kg m (^3) )(9 80. m s^2 ) y 2 y 2 = 10 3. m
(b) No atmosphere can lift the water in the straw through zero height difference.
P14.15 P 0 (^) =ρ gh
h P g
= = ×
( × )
0
ρ 9 80
Pa 0.984 10 3 kg m (^3) (( m s^2 )
= 10 5. m
No. The “Torricellian vacuum” is not so good. Some alcohol and water will evaporate.
The equilibrium vapor pressures of alcohol and water are higher than the vapor pressure of mercury.
FIG. P14.
372 Chapter 14
P14.16 (a) Using the def inition of density, we have
h m w (^) A
( )
water water 2 3
g 2 5.00 cm^ g cm
ρ 1 00. 2 20 0. cm
(b) Sketch (b) at the right represents the situation after the water is added. A volume ( A h 2 2 ) of mercury has been displaced by water in the right tube. The additional volume of mercury now in the left tube is A h 1. Since the total volume of mercury has not changed,
A h 2 (^) 2 = A h 1 or h
A
A
2 1 h 2
At the level of the mercury–water interface in the right tube, we may write the absolute pressure as: P = P 0 + ρwater ghw The pressure at this same level in the left tube is given by
P = P 0 (^) + ρHg g h ( + h (^) 2 ) = P 0 +ρwater ghw
which, using equation (1) above, reduces to
ρHg h A ρwater A
1 1 hw 2
or
h h A A = w ( + )
ρ ρ
water Hg 1 1 / 2 Thus, the level of mercury has risen a distance of
h = ( )(^ ) ( ) +
g cm cm g cm
3 (^3) ( 00 ) =^ 0 490.^ cm^ above the original level.
P14.17 ∆ P 0 = ρ g ∆ h = − 2 66. × 103 Pa: P = P 0 (^) + ∆ P 0 = (^) ( 1 013. −0 026 6. (^) ) × 105 Pa = 0 986. × 105 Pa
*P14.18 (a) We can directly write the bottom pressure as P = P 0 + ρ gh , or we can say that the bottom of the tank must support the weight of the water:
PA − P 0 A = m water g = ρ Vg = ρ Ahg which gives again
P = P 0 + ρ gh = 101.3 kPa + (1000 kgm 3 )(9.8 ms^2 ) h = 101.3 kPa + (9.8 kPa m) h
(b) Now the bottom of the tank must support the weight of the whole contents:
P b A − P 0 A = m water g + Mg = ρ Vg + Mg = ρ Ahg + Mg so
P b = P 0 + ρ hg + Mg A Then ∆ P = P b − P = Mg A
(c) Before the people enter, P = 101.3 kPa + (9.8 kPam)(1.5 m) = 116 kPa
afterwards, ∆ P = Mg A = (150 kg)(9.8 ms^2 ) π(3 m) 2 = 52.0 Pa
FIG. P14.
374 Chapter 14
P14.21 At equilibrium (^) ∑ F = 0 or (^) Fapp + mg = B
where B is the buoyant force.
The applied force, Fapp = B − mg
where B = Vol (^) ( ρwater) g
and m = ( Vol ) ρball
So, (^) Fapp = ( Vol ) g ( ρwater −ρball ) = 4 π r g ρ −ρbal 3
3 ( (^) water ll)
Fapp = (^) ( × − ) ( ) −
(^3 ) π. m. m s (^2) ( kg m^3 .. 0 kg m (^3) ) = 0 258. N down
*P14.22 For the submerged object Σ Fy = 0 + B − F (^) g + T = 0 + B = F (^) g − T = 5 N − 3.5 N = 1.5 N
This is the weight of the water displaced. Its volume is the same as the volume V of the object:
B = m water g = ρw V object g = 1.5 N: V object = 1.5 N ρw g
Now the density of the object is
ρobject = m object V object =
m (^) object ρ w (^) g Fg ρ w 1 5 1 5
N N
= = N (1000 kg/m^3 ..
N
= × kg/m 3
P14.23 (a) P = P 0 + ρ gh
Taking P 0 = 1 013. × 105 N m 2 and h = 5 00. cm
we f ind P top= 1 017 9. × 10 5 N m^2
For h = 17 0. cm, we get P bot= 1 029 7. × 10 5 N m^2
Since the areas of the top and bottom are A = ( 0 100. m) 2 = 10 −^2 m 2
we f ind F top (^) = P top (^) A = 1 017 9. × 10 3 N
and F bot = 1 029 7. × 10 3 N
(b) T + B − Mg = 0
where B = ρ w Vg = ( 10 3 kg m (^3) ) (1 20. × 10 − 3 m (^3) )( 9 80. m s^2 ) = 11 1 8. N
and Mg = 10 0 9 80. (. ) =98 0. N
Therefore, T = Mg − B = 98 0. − 11 8. = 86 2. N
(c) F bot (^) − F top = ( 1 029 7. −1 017 9. ) × 10 3 N = 11 8. N
which is equal to B found in part (b).
FIG. P14.
FIG. P14.
Fluid Mechanics 375
P14.24 (a) (b) (^) ∑ Fy = 0 : − 15 N − 10 N + B = 0
B = 25 0. N
(c) The oil pushes horizontally inward on each side of the block.
(d) String tension increases. The oil causes the water below to be under greater pressure, and the water pushes up more strongly on the bottom of the block.
(e) Consider the equilibrium just before the string breaks: − − + = =
N N N
N
oil oil
+ B
B
For the buoyant force of the water we have B Vg V V
= ρ 25 N = ( 1 000 kg m (^3) )( 0 25 (^) block)9 8m s^2 bl
oock = 1 02. × 10 −^2 m^3
For the buoyant force of the oil 50 800 1 02 10 9 8 0
N = ( kg m (^3) ) ( ×^2 m (^3) ) m s^2
f − f
e e
(f) − 15 N + ( 800 kg m (^3) ) f (^) f ( 1 02. × 10 −^2 m (^3) ) 9 8. m s^2 = 0
f (^) f = 0 187. = 18 7. %
*P14.25 (a) Let P represent the pressure at the center of one face, of edge . P = P 0 + ρ gh
The force on the face is F = PA = P 0 A + ρ g ^2 h
It increases in time at the rate dF dt = 0 + ρ g ^2 dh dt = (1030 kgm 3 )(9.8 ms 2 )(0.25 m) 2 (1.9 ms) = 1.20 × 10 3 N/s
(b) B = ρ Vg is constant as both the force on the top and the bottom of the block increase together. The rate of change is zero.
Fg
T B
FIG. P14.24(a)
60 N B
25 N
15 N
oil
FIG. P14.24(e)
B
15 N
oil FIG. P14.24(f)
Fluid Mechanics 377
P14.29 Let A represent the horizontal cross-sectional area of the rod, which we presume to be constant. The rod is in equilibrium:
∑^ Fy =^0 :^ −^ mg^ +^ B^ =^0 = −^ ρ 0 V whole rod^ g^ +ρfluid V immersed^ g
ρ 0 ALg = ρ A L ( − h g )
The density of the liquid is ρ ρ = −
0 L
L h
P14.30 We use the result of Problem 14.29. For the rod f loating in a liquid of density 0 98. g cm 3 ,
ρ ρ ρ
0
0 98 0 0 2 0 98 0
L
L h L L L
g cm cm g cm
3
(^3) .. 98 0 2. g cm cm 0 ( 3 ) =ρ^ L For f loating in the dense liquid, 1 14 1 8 1 14 1 14
g cm cm g cm g cm
3
3 3
ρ L L )) 1 8.^ cm=^ ρ 0 L (a) By substitution, 1 14 1 14 1 8 0 98 0 2 0 98 0 16 1 8
L L
L
cm 556 11 6
cm L =. cm (b) Substituting back, 0 98 11 6 0 2 11 6 0 963
0 0
g cm 3 ( cm − cm ) = cm
ρ ρ g cm 3
(c) The marks are not equally spaced. Because ρ = ρ −
0 L
L h
is not of the form ρ = a + bh , equal-size steps of ρ do not correspond to equal-size steps of h. The number 1.06 is halfway between 0.98 and 1.14 but the mark for that density is 0.0604 cm below the geometric halfway point between the ends of the scale. The marks get closer together as you go down.
P14.31 The balloon stops rising when ( ρ (^) air −ρHe) gV = Mg and ( ρ (^) air −ρHe) V = M
Therefore,
V
M
e
ρ (^) air ρHe − −
V = 1 430 m 3
378 Chapter 14
P14.32 Constant velocity implies zero acceleration, which means that the submersible is in equilibrium under the gravitational force, the upward buoyant force, and the upward resistance force: ∑^ Fy^ =^ may =^0 −^ (^ 1 20.^ ×^10 4 kg^ + m g )^^ +^ ρ wgV +^1100 N=^0 where m is the mass of the added water and V is the sphere’s volume.
1 20 10 1 03 10 4 3
. × 4 + =. × 3 ⎡ (1 50. )^31100
⎣⎢^
kg m π + N 9.8 m s 2 so m = 2 67. × 10 3 kg
P14.33 B = Fg
g
V
ρ ρ gV
ρ ρ
H O sphere
sphere H O
2
2 kg m
glycerin sphere
glycerin
ρ ρ
ρ
g^4 V gV 10
⎠ −^ =
(^500 kg m^3 ) = 1 250 kg m^3
P14.34 By Archimedes’s principle, the weight of the f ifty planes is equal to the weight of a horizontal slice of water 11.0 cm thick and circumscribed by the water line: ∆ B g ∆ V g g
( × ) = ( )
ρwater
50 2 90. 10 4 kg 1 030kg m^3 ( 0 0 110. m) A
giving A^ =^ 1 28.^ ×^10 4 m^2. The acceleration of gravity does not affect the answer.
Section 14.5 F luid Dynamics
Section 14.6 Bernoulli’s Equation
P14.35 Assuming the top is open to the atmosphere, then
P 1 (^) = P 0 Note P 2 (^) = P 0. The water pushes on the air just as hard as the air pushes on the water. F low rate = 2 50. × 10 −^3 m 3 min = 4 17. × 10 −^5 m 3 s.
(a) A 1 (^) >> A 2 so v (^) 1 << v 2 Assuming v 1 = 0, P gy P gy
gy
1 1 2 1 2 2 2 2
2 1
1 2
= ( ) =
ρ v (^) ρ ρ v ρ
v (^) [ (. 0 0 )(16 0. )]1 2 = 17 7. m s
(b) F low rate = =
⎝⎜^
A ( ) = × −
d 2 2
(^25) 4 v 17 7 4 17 10 π
.. m 3 s
d = 1 73. × 10 −^3 m =1 73. mm
FIG. P14.
380 Chapter 14
*P14.38 (a) The mass f low rate and the volume f low rate are constant:
ρ A 1 v 1 = ρ A 2 v 2 π r 12 v 1 = π r 22 v 2 (4 cm) 2 v 1 = (2 cm) 2 v 2 v 2 = 4 v 1
For ideal f low
P 1 (^) gy (^) 1 12 P 2 (^) gy 2 22
4
× + +
ρ ρ v ρ ρ v
. Pa 22
1
2
4
kg m
Pa
( 3 )( )
= × +
v
. ( )(. )( .. ) 5 1 ( ) 2
1 2
1
Pa kg m
Pa 7500 k
v
v gg m 3 =^ 0 825. m s
(b) v 2 = 4 v 1 = 3.30 m/s
(c) π r 12 v 1 = π(0.04 m) 2 (0.825 ms) = 4.14 × 10 −^3 m /s^3
P14.39 The volume f low rate is
125 16 3
1
2 1
cm s
cm 2
3 .
= Av = π⎛⎝ ⎞⎠ v
The speed at the top of the falling column is
v 1
cm s cm cm s
3 2 Take point 2 at 13 cm below:
P gy P gy
P
1 1 1 2 2 2 2 2
0
ρ ρ v ρ ρ v
kg m 3 9 9 8 0 13
(.^ m s^2 ).^ m^ +^ (1 000 kg m^3 )(0 106^.^ m s)^2
= P 00 0 22
2 =^ 2 9 8(^ m s^ )^ 0 13^ m+^0 v. 2. ( .1 106 m s (^) )^2 = 1 60. m s
The volume f low rate is constant:
7 67 2
2 .
.
cm s cm s
cm
π d
d
P14.40 (a) P = = =
⎝⎜^
⎠⎟^
E
t
mgh t
m t gh Rgh
(b) PEL = 0 85 8 5. (^) (. × (^10 5) )( 9 8.)( 87 ) = 616 MW
Fluid Mechanics 381
P14.41 (a) Between sea surface and clogged hole: P 1 (^) 12 gy (^) 1 P 2 (^) 22 gy 2
1 atm + 0 + (1 030 kg m (^3) )( 9 8. m s (^2) )( 2 m) = P 2 + 0 + 0 P 2 = 1 atm +20 2. kPa
The air on the back of his hand pushes opposite the water, so the net force on his hand is
F = PA = (^) ( 20 2 × (^10) )⎛⎝ ⎞⎠ ( × − ) 4
2
. N m 2. m π F = 2 28. N toward Holland
(b) Now, Bernoulli’s theorem is
1 0 20 2 1
atm + +. kPa = atm + (^) (1 030 kg m (^3) ) v 22 + 0 v 2 = 6 26. m s
The volume rate of f low is A 2 2 2 (^2 ) 4 v = (^) (1 2 × 10 −^ ) ( 6 26 ) = 7 08 × 10 − π
. m. m s. m 3 s
One acre–foot is 4 047 m 2 × 0 304 8. m =1 234m^3
Requiring
4 1 74^10 6 20 2
m m s s days
3
. × −^3 =^.^ ×^ =.
*P14.42 (a) The volume f low rate is the same at the two points: A 1 v 1 = A 2 v 2
π (1 cm)^2 v 1 = π (0.5 cm) 2 v 2 v 2 = 4 v 1
We assume the tubes are at the same elevation:
where the pressure is in Pascals
P gy P gy
P P P
1 1
2 1 2 2
2 2
1 2
ρ ρ ρ ρ
ρ
v v
∆ ( vv v
v
v
1
2 1
2
1
2
1
ρ
∆ P kg/m 3
( 0 0 0125.^ m/s) ∆^ P
The volume f low rate is π(0.01 m)^2 ( 0 0125. m/s) ∆ P
= ( 3 93.^ ×^10 −^6 m /s^3 ) ∆^ P^ where^ ∆ P is in pascals
(b) (3 93.^ ×^10 −^6 m /s^3 )^6000 = 0.305 L /s
(c) With pressure difference 2 times larger, the f low rate is larger by the square root of 2 times: (2)^1 ^2 (0.305 Ls) = 0.431 L /s
(d) The f low rate is proportional to the square root of the pressure difference.
Fluid Mechanics 383
Section 14.7 Other Applications of F luid Dynamics
P14.46 Mg = ( P 1 (^) − P 2 (^) ) A for a balanced condition
= × −
A
P
where A = 80 0. m 2 ∴ P 2 = 7 0. × 104 − 0 196. × 10 4 = 6 80. × 104 Pa
P14.47 (a) P 0 (^) gh 0 P 0 0 32
If h = 1 00. m v 3 = 4 43. m s
(b) P + ρ gy + ρ = P + + ρ
2 0 3 v v^2
Since v (^) 2 = v 3 P = P 0 − ρ gy
Since P ≥ 0, the greatest possible siphon height is given by
y P g
≤ = ×
( )( )
ρ
Pa. 10 3 kg m 3 m s^2
m
P14.48 The assumption of incompressibility is surely unrealistic, but allows an estimate of the speed:
P 1 (^) gy (^) 1 1 12 P 2 (^) gy 2 22 2
ρ ρ v ρ ρ v
. atm ... ..
2 2
2
atm + + (^) ( kg m (^3) )
v
v
)) (^) ( × )
.^5
N m kg m
m s
2 3
P14.49 In the reservoir, the gauge pressure is ∆ P = ×
2 00. N− 5 = 8 00. × 104
2.50 10 m 2 Pa From the equation of continuity: A 1 (^) v 1 (^) = A 2 v 2
(^ 2 50.^ ×^10 −^5 m^2 )^ v 1^ =^ (1 00.^ ×^10 −^8 m^2 ) v 2^ v^ 1 =^ (^ 4 00^.^ ×^10 −^4 ) v 2
Thus, v 12 is negligible in comparison to v 22.
Then, from Bernoulli’s equation: P 1 (^) P 2 (^) 1 12 gy 1 (^) 22 gy 2 2
( − ) +^ ρ^ v^ +^ ρ^ =^ ρ^ v +ρ
4 2 2
2
× + + = + (^) ( )
×
Pa kg m 3 v
v
(^4) Pa kg m 3 m s
( ) =.
FIG. P14.
384 Chapter 14
P14.50 Take points 1 and 2 in the air just inside and outside the window pane.
P 1 (^) 1 12 gy (^) 1 P 2 (^) 22 gy 2 2
P 0 0 P 2 2
- = + (1 30. kg m 3 )(11 2. m s) P 2 (^) = P 0 − 81 5. Pa
(a) The total force exerted by the air is outward,
P A 1 − P A 2 = P A 0 − P A 0 + (81 5. N m (^2) )( 4 m )(1 5. m) = 489 NN outward
(b) P A 1 P A 2 (^) 22 A^2
− = ρ v = (^) (1 30. kg m (^3) )( 22 4. m s ) ( 4 m))( 1 5. m ) = 1 96. kN outward
Additional Problems
P14.51 When the balloon comes into equilibrium, we must have
∑^ Fy^ =^ B^ −^ Fg^ , balloon −^ Fg^ , He −^ Fg ,string =^0 Fg , string is the weight of the string above the ground, and B is the buoyant force. Now
F m g F Vg B Vg
g g
, ,
balloon balloon He He air
ρ ρ
and
F m h L g , string = string g
Therefore, we have
ρair Vg m balloon (^) g ρHe Vg m string h L
− − − g = 0
or
h V m m
= L
( ρ (^) air −ρHe ) − balloon string
giving
h =
( 1 29. −0 179. )( kg m 3 ) ( 4 π ( 0 400. m)^3 / 3 ) − 0 250. kkg
0.050 0 kg
( 2 00. m ) = 1 91. m
He
h
FIG. P14.
13794_14_ch14_p365-394.indd13794_14_ch14_p365-394.indd 384384 12/28/0612/28/06 3:48:243:48:24 PMPM
