Chapter14 Fluids
1/8
Fluids
Objectives:
Distinguish fluids from solids. When mass is uniformly distributed, relate density to mass and volume.
Apply the relationship between hydrostatic pressure, force, and the surface area,fluid density, and the height above
or below a reference level. Distinguish between total pressure (absolute pressure) and gauge pressure.
Identify Pascal’s principle. For a hydraulic lift, apply the relationship between the input area and displacement
and the output area and displacement.
Describe Archimedes’ principle. Apply the relationship between the buoyant force on a body and the mass of the
fluid displaced by the body. For a floating body, relate the buoyant force to the gravitational force and relate the
gravitational force to the mass of the fluid displaced by the body. Distinguish between apparent weight and actual
weight. Calculate the apparent weight of a body that is fully or partially submerged.
Describe steady flow, incompressible flow, nonviscous flow, and irrotational flow. Explain the term streamline.
Apply the equation of continuity to relate the cross-sectional area and flow speed at one point in a tube to those
quantities at a different point. Identify and calculate volume flow rate and mass flow rate.
Calculate the kinetic energy density in terms of a fluid’s density and flow speed. Identify the fluid pressure as being
a type of energy density.
Calculate the gravitational potential energy density. Apply Bernoulli’s equation to relate the total energy density at
one point on a streamline to the value at another point.
Identify that Bernoulli's equation is a statement of the conservation of energy.
Density The density r of any material is defined as the material’s mass per unit volume and usually, where a material sample is
much larger than atomic dimensions,

mass
M
M

and for uniform material  
volume V
V
or

dM
dV
Fluid Pressure A fluid is a substance that can flow; it conforms to the boundaries of its container because it cannot withstand
shearing stress. It can, however, exert a force perpendicular to its surface. That force is described in terms of pressure p:
p
force F
F

and for non - uniform surface p 
area
A
A
or

dF
dA
The force resulting from fluid pressure at a particular point in a fluid has the same magnitude in all directions. Gauge pressure
is the difference between the actual pressure (or absolute pressure) at a point and the atmospheric pressure.
Pressure Variation with Height and Depth Pressure in a fluid at rest varies with vertical position y (depth or height). The
pressure in a fluid is the same for all points at the same level. If h is the depth of a fluid sample below some reference level at
which the pressure is p0, then the pressure in the sample for y measured positive upward,
p2  p1   g h
p2  p1
  g
y2  y1


p2  p1   g h 
p2  p1   g y 
from high pressureto low pressure(goingup)
fromlow pressureto high pressure(goingdeep)
Pascal’s Principle A change in the pressure applied to an enclosed fluid is transmitted undiminished to every portion of the
fluid and to the walls of the containing vessel.
Archimedes’ Principle When a body is fully or partially submerged in a fluid, a buoyant force from the surrounding fluid acts
on the body. The force is directed upward and has a magnitude given by
Fbouyant  m fluid g
where m(fluid) is the mass of the fluid that has been displaced by the body (that is, the fluid that has been pushed out of the
way by the body).
When a body floats in a fluid, the magnitude Fb of the (upward) buoyant force on the body is equal to the magnitude Fg of the
(downward) gravitational force on the body. The apparent weight of a body on which a buoyant force acts is related to its
actual weight by
apparent weight  weight  bouyant force  wapparent  w  Fbouyant
Flow of Ideal Fluids An ideal fluid is incompressible and lacks viscosity, and its flow is steady and irrotational. A streamline
is the path followed by an individual fluid particle.A tube of flow is a bundle of streamlines. The flow within any tube of flow
obeys the equation of continuity:
constant  volume flowrate  area  speed 
Rvolume  A1v1  A2 v2
contant  mass flowrate  density  volume flowrate  Rmass  Rvolume  A1v1  A2 v 2
Bernoulli’s Equation Applying the principle of conservation of mechanical energy to the flow of an ideal fluid leads to
Bernoulli’s equation along any tube of flow:
p  12 v 2  gh  constant  p1  12 v12  gh1  p 2  12 v 22  gh2
Chapter14 Fluids
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Example: Ch14-10 he plastic tube in Fig. 14-30 has a cross-sectional area of 5.00 cm2. The tube is filled with water until the
short arm (of length d=.800 m) is full. Then the short arm is sealed and more water is gradually poured into the long arm. If the
seal will pop off when the force on it exceeds 9.80 N, what total height of water in the long arm will put the seal on the verge
of popping?
p
F
A
A  r 2

p   gh
 h
p
g
F

gA

F
 g r 2 
H total  h  d
Example: Ch14-12 The maximum depth dmax that a diver can snorkel is set by the density of the water and the fact that human
lungs can function against a maximum pressure difference (between inside and outside the chest cavity) of 0.050 atm. What is
the difference in dmax for fresh water and the water of the Dead Sea (the saltiest natural water in the world, with a density of
1.5x103 kg/m3)?
p    g d
 d
p
g
p  0.05 atm  5065Pa
 fresh  998kg / m
3
d  d fresh  d salt 
 1atm  101,325Pa
 salt  1500kg / m 3
and
p  1
1

g   fresh  salt

  0.17m


Example: Ch14-17 Crew members attempt to escape from a damaged submarine 100 m below the surface. What force must be
applied to a pop-out hatch, which is 1.2 m by 0.60 m, to push it out at that depth? Assume that the density of the ocean water is
1024 kg/m3 and the internal air pressure is at 1.00 atm.
F  Foutside  Finside
Foutside   p0  gd  A 
Foutside
p 0  1atm
Finside  p o A  p o LW 
F  Foutside  Finside   p 0  gd A   p o A  gdA  gd LW 
Finside
F   g d LW   7.2 x10 N
5
Example: Ch14-21 Two identical cylindrical vessels with their bases at the same level each contain a liquid of density 1.30x103
kg/m3. The area of each base is 4.00 cm2, but in one vessel the liquid height is 0.854 m and in the other it is 1.560 m. Find the
work done by the gravitational force in equalizing the levels when the two vessels are connected.
W  Mgh  h  h f  hB
M  V  Ah A  h f

h A  hB   h  12 hA  hB   hB
 h A  12 h A  hB   12 h A  hB 
 hf 
 hA  h f
1
2

1
2
hA  hB 
1
2
W  Mgh  Ah A  h f g h f  hB   A h A  hB g  h A  hB   gA h A  hB   0.635 J
4
1
2
1
2
hA
hf
Example: Ch14-24 In Fig. 14-35, water stands at depth D=35.0 m behind the vertical upstream face of a dam of width W=314
m. Find (a) the net horizontal force on the dam from the gauge pressure of the water and (b) the net torque due to that force
about a horizontal line through O parallel to the (long) width of the dam. This torque tends to rotate the dam around that line,
which would cause the dam to fail. (c) Find the moment arm of the torque.
p  gy and
dF  p dA  p (W dy)  gy (W dy)
D
D
1
1 
F   dF   gy (W dy)  gW  y dy  gW  y 2   gWD 2  1.88 x10 9 N
2
2


o
0


d  r  dF  D  y dF  D  y gy (W dy )  gW Dy  y 2 dy
D
3
D
3
3
 Dy
D
y 
D 
3 1 
10
   gW 

  gWD    2.20 x10 N  m
2
3
2
3
6

0


  gW  Dy  y 2 dy  gW 
0
2
hB
Chapter14 Fluids
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Example: ch14-28 A piston of cross-sectional area a is used in a hydraulic press to exert a small force of magnitude f on
the enclosed liquid. A connecting pipe leads to a larger piston of cross sectional area A (Fig. 14-36). (a) What force magnitude
F will the larger piston sustain without moving? (b) If the piston diameters are 3.80 cm and 53.0 cm, what force magnitude on
the small piston will balance a 20.0 kN force on the large piston?
p
F
f

A a
 A
 F f  
a
a
f F F
 A
 r 2
 2
 R
2
2

r
 3.80 
  F    20.0 x10 3 N 
  103N
R


 53.0 

Example: ch14-31 A block of wood floats in fresh water with two-thirds of its volume V submerged and in oil with 0.90V
submerged. Find the density of (a) the wood and (b) the oil
 Vsubmerged 
 block Vsubmerged
2
  Vsubmerged  23 V   block   water    6.7 x10 2 kg / m 3

  block   water 
 water
Vtotal
V
3
 total 
 block Vsubmerged

 oil
Vtotal
 V
  oil   block  total
V
 submerged

  Vsubmerged  0.90V


 1 
2
3
  oil   block 
  7.4 x10 kg / m
 0.90 
Example: Ch14-27 What would be the height of the atmosphere if the air density (a) were uniform and (b) decreased linearly
to zero with height? Assume that at sea level the air pressure is 1.0 atm and the air density is 1.3 kg/m3.
p1  p 0  g  y1  y 0 
p 0  1 atm  1.01x10 5 Pa and
p1  0
 air  constant  1.3kg/m 3
p 0  g  y1  y 0   H   y1  y 0  
p0
1.01x10 5 Pa

 7.9 x10 3 m
g 1.3kg / m 3 9.8m / s 2



 air  linearly decreases with height   0 1 


y

H
y

p1  p 0  g  y1  y 0   p   gy  dp   g dy    0 1   g dy
H


H
y

 dp    0 g 0 1  H  dy
H


y2 
H2
H 
p1  p 0    0 g  y 
  0 g H 
   0 g   
2
H
2
H
2

0


p0 
 0 gH
2
p1  0
 p 
 H  2 0   16.0 x10 3 m
 0 g 
 y
Extend :    0 exp    dp   g dy 
 H
H
 y
 dp   0 g 0 exp  H  dy
H

  H
 y 
 0 
p1  p 0   0 g  H exp      0 gH exp    exp       0 gH exp 1  exp0 
H
H

 0

 H 

 
1 
1  e 
 p 0    0 gH   1    0 gH 

e 
 e 
 p   e   p0   2.7183 
3
 
H   0  
  
  12.5 x10 m  around Troposphere
  0 g  1  e    0 g  1  2.7183 
Chapter14 Fluids
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Example: Ch14-58 The intake in Fig. 14-47 has cross-sectional area of 0.74 m2 and water
flow at 0.40 m/s. At the outlet, distance D=180 m below the intake, the cross-sectional area
is smaller than at the intake and the water flows out at 9.5 m/s into equipment. What is the
pressure difference between inlet and outlet?
p1  12 v12  gh1  p 2  12 v22  gh2


p 2  p1  12  v 22  v12  g h2  h1   D  h2  h1 


p 2  p1  p   v  v12  gD  1.7 x10 6 Pa
1
2
2
2
Example: Ch14-61 A water pipe having a 2.5 cm inside diameter carries water into the basement of a house at a speed of 0.90
m/s and a pressure of 170 kPa. If the pipe tapers to 1.2 cm and rises to the second floor 7.6 m above the input point, what are
the (a) speed and (b) water pressure at the second floor?
A
r 2
v1 A1  v 2 A2  v 2  v1 1  v1 12  3.9m / s
A2
r2
p1  12 v12  gh1  p 2  12 v 22  gh2
h  h2  h1   7.6m,

1
2
2
2
2
1





p 2  p1  12  v12  v 22  g h1  h2   p1  12  v 22  v12  g h2  h1 
p1  170 kPa

p 2  p1   v  v  gh  88 kPa
Example: Ch14-64 In Fig. 14-49, water flows through a horizontal pipe and then out into the atmosphere at a speed v1=15m/s.
The diameters of the left and right sections of the pipe are 5.0 cm and 3.0 cm. (a) What volume of water flows into the
atmosphere during a 10 min period? In the left section of the pipe, what are (b) the speed v2 and (c) the gauge pressure?
2
 d2
d 
Vwater  v1t A1  v1t  r12  v1t   1   v1t  1  6.4m 3
v1 A1  v 2 A2
4
2
2
A
 d1 / 4
d2
 v 2  v1 1  v1
 v1 12  5.4m / s
2
A2
 d2 / 4
d2
p1  12 v12  gh1  p 2  12 v 22  gh2
h1  h2



p 2  p1  12  v12  v 22  g h1  h2 
p1  p 0  1atm  1.01x10 5 Pa
and


p 2  p 0  12  v12  v 22  1.99 x10 5 Pa  1.97atm
gauge p  p 2  1atm  0.97atm
Example: Ch14-70 In Fig. 14-53, water flows steadily from the left pipe section (radius r1=2.00R), through the middle section
(radius R), and into the right section (radius r3=3.00R). The speed of the water in the middle section is 0.500 m/s. What is the
net work done on 0.400 m3 of the water as it moves from the left section to the right section?
A
 r2
r2
v
R2
v1 A1  v 2 A2  v3 A3  v1  v 2 2  v 2 22  v 2 22  v 2
 2
2
A1
 r1
r1
2R  4
v3  v 2
A2
r2
v
R2
 v 2 22  v 2
 2
2
A3
r3
3R  9
 v  2  v  2 
W  K  12 Mv32  12 Mv12  12 V v32  v12  12 V  2    2  
 9   4  

1 1
W  12 Vv 22     2.50 J
 81 16 

   0.998kg / m 3
Chapter14 Fluids
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Example: Ch14-62 A pitot tube (Fig. 14-48) is used to determine the airspeed of an airplane. It consists of an outer tube with a
number of small holes B (four are shown) that allow air into the tube; that tube is connected to one arm of a U-tube. The other
arm of the U-tube is connected to hole A at the front end of the device, which points in the direction the plane is headed. At A
the air becomes stagnant so that vA=0. At B, however, the speed of the air presumably equals the airspeed v of the plane. (a)
Use Bernoulli’s equation to show the speed of the plane.
Outside the Tube :
p A  12  air v A2   air gh A  p B  12  air v B2   air ghB
v A  0 and
hB  h A


p   p A  p B   p   air v
1
2
and

p A  p B  12  air v B2  v A2   air g hB  h A 
2
B
Inside the Tube :
p L   Liquid ghL  p R   Liquid ghR
h  hR  hL  and
p L  p R   Liquid g hR  hL 

p   p L  p R   p   Liquid gh
  Liquid 

  air 
(b) Suppose that the tube contains alcohol and the level difference h is 26.0 cm.
1
2
 air v B2   Liquid gh  v B  2 gh 
What is the plane' s speed relative to the air?  air  1.03 kg/m3 and  Liquid of alcohol is 810 kg/m3.
  Liquid 
  63.3m / s  227.88 km / hr
v B  2 gh 
  air 
63. On a high - altitude aircraft measures a differential pressure of 180 Pa. What is the aircraft's
airspeed if the density of the air is 0.031 kg/m3?
p  12  air v B2
 vB  2
p
 air
 110m / s  396km / hr
Example: Ch14-65 A venturi meter is used to measure the flow speed of a fluid in a pipe. The meter is connected between two
sections of the pipe (Fig. 14-50); the cross-sectional area A of the entrance and exit of the meter matches the pipe’s crosssectional area. Between the entrance and exit, the fluid flows from the pipe with speed V and then through a narrow “throat” of
cross sectional area a with speed v. A manometer connects the wider portion of the meter to the narrower portion. The change
in the fluid’s speed is accompanied by a change Δp in the fluid’s pressure, which causes a height difference h of the liquid in
the two arms of the manometer. (Here Δp means pressure in the throat minus pressure in the pipe.) (a) Find the flow speed V
by applying Bernoulli’s equation and the equation of continuity to points 1 and 2 in Fig. 14-50.
A
a
  12 v A2  12 v a2   p a  p A   g ha  h A 
constant mass flow rate R  v A A  v a a  v a  v A
p A  12 v A2  gh A  p a  12 v a2  gha
ha  h A
2
and
p   p a  p A  
1
2
 v A2  v a2   p
2
  A 2 
 a 2  A2 
 A
 v A2  v a2  v A2   v A   v A2 1      v A2 

2

 a
  a  
 a

p
vA  2
p  a 2 
  a 2  A 2 
Inside the Tube :
p L   Liquid ghL  p R   Liquid ghR
h  hR  hL  and
  Liquid
v A  2 gh
 

p L  p R   Liquid g hR  hL 
p   p L  p R   p   Liquid gh
 a 2 
  2
2 
 a  A 
  Liquid
mass flow rate R  v A A  A 2 gh
 
 a 2 
  2
2
 a  A 
Chapter14 Fluids
6/8
Example: Ch14-80 In an experiment, a rectangular block with height h is allowed to float in four separate liquids. In the first
liquid, which is water, it floats fully submerged. In liquids A, B, and C, it floats with heights h/2, 2h/3, and h/4 above the liquid
surface, respectively. What are the relative densities (the densities relative to that of water) of (a) A, (b) B, and (c) C?
 Vsubmerged 
 block Vsubmerged
  Vsubmerged  V   block   water  997 kg / m 3

  block   water 
 water
Vtotal
 Vtotal 
Vtotal  Ah and Vsubmerged  A xh 
 V

 Ah 
1
   water  
  Liquid   block  total    Liquid   water 
V

A
(
xh
)
 x


 submerged 
1  x  1 / 2, 2 / 3, 1 / 4  x  1 / 2, 1 / 3, 3 / 4
 block Vsubmerged

 Liquid
Vtotal
A
1  1 
 
  2 and
 water  x   1 / 2 
B
 3 and
 water
C
4

 water 3
Example: Ch14-82 What is the acceleration of a rising hot-air balloon if the ratio of the air density outside the balloon to that
inside is 1.39? Neglect the mass of the balloon fabric and the basket.
ma   F  FB  mg
m   inV  and
mg   inV g
 inV a   outV g   inV g
FB   outV g
and
  in a   out   in g
    in 


 g   out  1 g 
a   out
  in

  in

2


a  1.39  1 g  3.82m / s
 out
 1.39
 in
Example: Ch14-71 Figure 14-54 shows a stream of water flowing through a hole at depth h=10 cm in a tank holding water to
height H=40 cm. (a) At what distance x does the stream strike the floor? (b) At what depth should a second hole be made to
give the same value of x? (c) At what depth should a hole be made to maximize x?
The stream of water has horizontal speed.
E1  E 2
 mgh  12 mv02x
 v0 x  2 gh
y  y 0  v0 y t  12 a y t 2

x  x0  v0 x t  12 a x t 2
 x  v0 x t  t 2 gh 
x  2 gh
y  y 0  12 gt 2 
2H  h 
g
2H  h 
2 gh
g
2H  h 
 4hH  h   4  10cm40  10cm  35cm
g
x 2  4hH  h   0  h 2  Hh 
h 
H h  t 
x2
4
 h 
 b  b 2  4ac
2a
 H  H 2  4( x 2 / 4)  H  H 2  x 2

2
2
 H  H 2  x2
 H  H 2  x2
and h  
2
2
H  h  h   h   H  h  40cm  10cm  30cm
h  h 
df
4H H
 0  8hmax  4 H  hmax 

 20cm
dh
8
2
H
 H 
 H  H 
 4hmax H  hmax   4  H    4    H 2  xmax  H  40cm
2
 2 
 2  2 
f  x 2  4hH  h   4h 2  4 Hh 
2
x max
Chapter14 Fluids
7/8
Example: Ch14-83 Figure 14-56 shows a siphon, which is a device for removing liquid from a container. Tube ABC must
initially be filled, but once this has been done, liquid will flow through the tube until the liquid surface in the container is level
with the tube opening at A. The liquid has density 1000 kg/m3 and negligible viscosity. The distances shown are h1=25 cm,
d=12 cm, and h2=40 cm. (a) With what speed does the liquid emerge from the tube at C? (b) If the atmospheric pressure is
1.0x10 5 Pa, what is the pressure in the liquid at the topmost point B? (c) Theoretically, what is the greatest possible height h1
that a siphon can lift water?
2
p C  12 vC2  ghC  p S  12 v S2  ghS  vC2   p S  p C   2 g hS  hC   v S2

p S  p C  p air
and
hS  hC   d  h2
and
v S  vC


vC2  2 g d  h2   vC  2 g d  h2   2 9.8m / s 2 0.12  0.40 m  3.2 m / s
p C  12 vC2  ghC  p B  12 v B2  gh B
5
p C  p air  1.0 x10 Pa
and

Av C  Av B

p B  p C  12  vC2  v B2  g hC  hB 

 vC  v B
and
hC  hB   h1  d  h2 
p B  p air  g h1  d  h2 

 


p B  1.0 x10 5 Pa  1.0 x10 3 kg / m 3 9.8m / s 2 0.25  0.12  0.40m  9.2 x10 4 Pa
 p air
Siphon wor ks as long as p B  0 (else we get backflow)
p B  p air  g h1  d  h2   0 
h1,max 
p air  g h1  d  h2  


p air
 d  h2  h1
g


p air
1.0 x10 5 Pa
 d  h2  
 0.12 m  0.40 m
3
3
2 
g
1
.
0
x
10
kg
/
m
9
.
8
m
/
s





 10.3m  h1
Example: Block on fluids.
mg  FB

 blockVtotal g    waterVsubmerged  g

water

V

submerged
 block
water

 b  Vsubmerged  bVtotal
 water
Vtotal
water




mg  FB  FB , water  FB ,oil    waterVsubmerged  g    oilVsubmerged  g
water
oil




Vsubmerged Vsubmerged
Vsubmerged
Vsubmerged
water
oil
oil
water
Vsubmerged  Vsubmerged  Vtotal 

1 
 1
 1  b  Vsubmerged  1  b Vtotal
Vtotal
Vtotal
Vtotal
Vtotal
water
oil
oil
 blockVtotal g  b Vtotal  water g  1  b Vtotal  oil g
 block  b water  1  b  oil  b water   oil  b oil
  block   oil  b water   oil 
 water  1000kg / m 3 and  block  900kg / m 3 and  oil  875kg / m 3

  oil
b  block
 0.2  20% is submerged in water and remaining 80% is in oil above water.
 water   oil
Extend :  air  1.22kg / m 3
Suppose 20% is sticking out of the oil and water, what is the density of the block?
Chapter14 Fluids
8/8
Example: Lift of airplane:
pF/A
2
2
pTop  12 vTop
 ghTop  pBottom  12 vBottom
 ghBottom
Lift  p  A   pBottom  pTop A
wing thickness  hTop  hBottom   7.0cm  0.07m
wing area  10m 2
and   1.20kg / m3
vTop  85m / s and VBottom  75m / s


2
2
p  pBottom  pTop  12  vTop
 vBottom
 g hTop  hBottom 
85m / s   75m / s   1.20kg / m 9.8m / s 0.07m
p  9.6 x10 N / m   0.82 N / m   9.6 x10 N / m
1
2

p  1.20kg / m
2
3
2
2
2
3
2


2
2
2

Lift  p  A  9.6 x102 N / m 2 10m 2  9.6 x103 N
Viscosity:
v
v
F
 p
(shearing strees)
t
A
x
x 1 v
strain 
 strain rate 

y
t y y
shearing stress F / A
  viscosity coefficient 

strain rate
v/ y
F m
d
c
d'
y
a
F
v
F
p
     v 
y
A

 y
b
for a cylinder :
2
2
 p  R  r 
 p 
     pressure gradient
v   
L
4

 
 L 

2R
dV
 vdA   v2r dr 
dt 
2
2
dV
2  p 
  p  R 4
 p  R  r 
2
2
2r dr  
   
   R  r r dr    

dt
4  L 
8 L  
 L  4 
volume flow rate 


Example: Terminal velocity in fluid:
0   Fy  Fbouyancy  Fvis cos ity  mg


Fbouyancy   fluid 43 r 3 g
Fvis cos ity  6rvT
mg  
 Stoke' s Law
135
 r g
4
ball 3
3
 
  r g    


0   Fy   fluid 43 r 3 g  6rvT  ball 43 r 3 g
6rvT
vT 
4
3
3
ball
2r 2 g
ball   fluid 
9
fluid
L
F
c'