Chapter 15: Fluid Motion
Fluids
 Characteristics
of fluids
• Microscopically molecules of a fluid do not have long-range
order. But liquids do have short-range order unlike gases
• Fluids can flow and conform to the boundaries of a container
• Fluids cannot sustain a shearing stress
Fluids (cont’d)
 Some
useful quantities
 Density for mass m ,volume V
m m


V V
for uniform solid or liquid
unit kg/m3 (at STP for a gas)
in SI unit
 water  1.0 103 kg/m 3 ;  air  1.2 kg/m 3
styrofoam  0.1103 kg/m 3 ; lead  11.35 103 kg/m 3
 Pressure for area
F
dF
p


A A0 dA
A
and normal force F
unit pascal
Pa  1 N / 1 m 2
in SI unit
other useful units:
1 atm  1.01105 Pa  760 torr  14.7 lb / in 2
height of a column of Hg corresponding to this pressure (mm)
Pressure
 Pressure
of a liquid at rest (uniform density)
F1  p1 A
F2  p2 A
fluid level
y1
y2
h
F1  F2   ( y2  y1 ) gA  0
A
imaginary
box
F1
F2
mg A
weight of the imaginary box
p2  p1   ( y2  y1 ) g
 hg
p2  p1  hg
mg
Pressure (cont’d)
 Pressure
p0
of a fluid at rest : gauge pressure
atmospheric
pressure
gauge pressure:
fluid level
y1
y2
h
p2  p1  hg
p1  p0  p1
A
imaginary
box
F1
F2
mg A
gauge
p2  p0  p2
gauge
atmospheric pressure
A simple model for atmospheric pressure
 Pressure
When
of a gas at rest (
  kp )
p2  p1, y2  y1  0,
dp  p2  p1  g ( y1  y2 )  gdy
dp   gdy  kpgdy
dp / dy  kpg
p2  p1e
 kg ( y2  y1 )
Barometer
 Pressure
measurement using a liquid:
(measures absolute pressure)
p0
From
as
h
atmospheric
p0 pressure
p2  p1  g ( y1  y2 )
p2  0, y2  h, p1  p0
p0  gh
If the liquid is mercury, for 1 atm :
p0  1 atm  101.3 103 Pa ,
 Hg  13.6 103 kg/m 3 , g  9.83 m/s
h  p0 / g  0.758 m  760 mm
Manometer
 Pressure
measurement using a liquid :
(measures gauge pressure)
p2  p1  g ( y1  y2 )
p0
p2  p, y2  h,
level 1
pg
pg  p  p0  gh
h
level 2
liquid
tank
manometer
p1  p0 , y1  0
gauge pressure
Pascal’s law
incompressible
 Pressure
applied to an enclosed fluid is transmitted
undiminished to every portion of the fluid and walls
of the containing vessel
p  pext  p0  gh
weight
piston
(area A)
pressure at P
pext
w
liquid
h
P
p
pressure due atmospheric
to weight w: pressure
w/A
p  pext
pressure due to
liquid above P
Pascal’s law (cont’d)
 Hydraulic
lever
p  F1 / A1  F2 / A2
F2  ( A2 / A1 ) F1
x1
F1
A2 / A1  1  F2  F1
x2
x1 A1  x2 A2  x2  ( A1 / A2 )x1
A2
A1
W2  F2 x2  ( A2 / A1 ) F1 ( A1 / A2 )x1
F2
 F1x1  W1
Buoyancy
 Origin
of buoyancy
• Consider a submerged massless object filled with the same fluid
as the fluid that surrounds the object
The object is at rest
Fnet  ( F2  F1 )  m f g  0
buoyancy
P1
F1
mfg
P2
F2
mass of fluid in the object
opposite dir.
buoyant force


Fb  m f g
• Now fill the object with another material
Fnet  ( F2  F1 )  mobj g  0
Fb  m f g
Buoyancy
 Origin
of buoyancy (cont’d)
• Consider a portion of fluid at rest in a container surrounded by an
imaginary boundary represented in dashed line.
• Since the portion of the fluid defined by the surface in dashed line
is at rest, the net force on this portion due to pressure must be
equal to that of the weight of the fluid inside the surface, and
opposite in direction.
dF
dF
dF


 

Fnet   dF  mfluid g  Fb  fluidVg  0
dF
dF
Fnet
X
mg
dF
dF
dF
Fb  mfluid g  fluidVg
buoyant force
• The same argument can be applied when the
imaginary portion of the fluid is replaced by an
object that occupies the same space.
Buoyancy
 Archimedes’s
principle
• The buoyant force on a partly or completely submerged object
is equal to the weight of the displaced fluid:
Fnet  Fb  mobj g  m f g  mobj g
 (  f  obj )Vg
• If
 f  obj , Fnet  0
apparent weight
• If
 f  obj , Fnet  0
Object sinks
W  (  f   obj )Vg  mobj g
Object floats
The object will rise until a part of it comes out above the fluid
Surface when the average density increases to  f
Example
What fraction of an iceberg is submerged in the sea water?
Let’s assume that the total volume of the iceberg is Vi.
Then the weight of the iceberg Fgi is
Fgi   iVi g
Let’s then assume that the volume of the iceberg
submerged in the sea water is Vw. The buoyant force B
caused by the displaced water becomes
B   wVw g
Since the whole system is at its
static equilibrium, we obtain
Therefore the fraction of the
volume of the iceberg
submerged under the surface of
the sea water is
iVi g   wVw g
Vw
i
917kg / m3


 0.890
3
Vi
 w 1030kg / m
About 90% of the entire iceberg is submerged in the water!!!
Example
• A fake or pure gold crown?
Is the crown made of pure
gold?
Tair =7.84 N
T  mg  0
air
Twater =6.86 N
Twater  mg  B  0
Twater  Tair  B  0
B  Tair  Twater   water gVwater
 0.980 N
4
Vwater  Vcrown  1.00 10 m
m  Tair / g  0.800 kg
3
gold=19.3x103 kg/m3
 crown  m / Vcrown  8.00 103 kg/m 3
Example
• Floating down the river
What depth h is the bottom of
the raft submerged?
wood=6.00x102 kg/m3
B  mraft g  0  B  mraft g
mraft g  (  raftVraft ) g
B  mwater g  (  waterVwater ) g  (  water Ah) g
(  water Ah) g  (  raftVraft ) g
 raftVraft
h
 0.0632 m
 water A
A=5.70 m2
Density and Pressure
 Example
• Oil and water
P1  P0  gh1  1.01105 Pa
 (7.00 10 2 kg/m 3 )(9.80 m/s 2 )
(8.00 m)
 1.56 10 Pa
5
Pbot  P1  gh2
 2.06 105 Pa
=0.700 g/cm3
h1=8.00 m
=1025 kg/m3
h2=5.00 m
Ideal fluid flow
 Ideal
fluids in motion
• Incompressible ( density is constant at any position)
• No internal friction (no viscosity)
• Steady (non-turbulent) flow- the velocity at a point is
constant in time.
flow tube
A
flow line
: The path of an individual particle in a
moving fluid
steady flow: A flow whose pattern does not change
with time. Every element passing
through a given point follows the same
flow line
streamline : A curve whose tangent at any point is
in the direction of the fluid velocity at
that point
flow tube : The flow lines passing through the edge
of an imaginary area such as A
flow lines
Continuity equation
 Continuity
equation I (incompressible fluid)
v2
The mass of a moving fluid does not change as it flows.
The volume of the fluid that passes through
area A during a small time interval dt :
dV  Avdt
A2
v2dt
v1
dV / dt  Av
A1
v1dt
dV1  A1v1dt ; dV2  A2v2dt
•In an ideal fluid the density is constant.
•In a time interval dt the mass that flows
into Area 1 is the same as the mass that
flows out of Area 2.
A1v1dt  A2v2dt  A1v1  A2v2
Continuity equation
v2
A2
 Volume
flow rate
dV  Avdt
v2dt
dV / dt  Av
volume flow rate
 Continuity
equation II
(compressible fluid)
v1
1 A1v1dt  2 A2v2dt  1 A1v1  2 A2v2
A1
v1dt
Bernoulli’s equation
 Work
v2
dW  p1 A1ds1  p2 A2ds2
F2  p2 A2
 ( p1  p2 )dV
dV
A2
v1
A1
 Change
ds2  v2dt
dV  A1ds1  A2ds2
ds1  v1dt
y1
in kinetic energy
dK  (1 / 2)  dV (v22  v12 )
 Change
in potential energy
dU  dVg( y2  y1 )
dV
F1  p1 A1
done by pressure
y2
Bernoulli’s equation
 Energy conservation
dW  dK  dU
dW  ( p1  p2 )dV
dK  (1 / 2) dV (v22  v12 )
dU  dVg( y2  y1 )
( p1  p2 )dV  (1 / 2)dV (v22  v12 )  dVg( y2  y1 )
p1  p2  (1 / 2)  (v22  v12 )  g ( y2  y1 )
p1  gy1  (1 / 2) v12  p2  gy2  (1 / 2) v22  const.
Example
 Venturi meter
1 2
1 2
p1  v1  p2  v2
2
2
A1
v2 
v1
A2
1 2 A12
p1  p2  v1 ( 2  1)
2
A2
p1  p2  gh
v1 
2 gh
( A1 / A2 )2  1
h
 Torricelli’s theorem
Example
The velocity of the fluid coming out of a hole in a tank as
shown in the figure can be calculated using Bernoulli’s
equation.
At the top surface the velocity
of the fluid is zero. The pressures
at the top surface and at the hole
are the same, namely, the
atmospheric pressure.
2
2
ptop  gytop  (1 / 2)vtop
 phole  gyhole  (1 / 2)vhole
2
g ( ytop  yhole )  (1 / 2)vhole
 vhole  2 gh
 Siphon
Example
Suppose a U-shaped piece of pipe is completely submerged in
water, filled with water, and then turned upside down under water.
As you slowly pull the top of the U-shaped piece of pipe out of
water, the water does not run out of the pipe. WHY?
 Siphon
Example
Suppose a U-shaped piece of pipe is completely submerged in
water, filled with water, and then turned upside down under water.
As you slowly pull the top of the U-shaped piece of pipe out of
water, the water does not run out of the pipe. WHY?
Air cannot enter the pipe. As the water starts running out of the pipe,
a near vacuum is created in the topmost region of the inverted U. The
pressure here drops to near zero. The atmospheric pressure on the
surface of the water in the bucket pushes the water into the U-shaped
pipe.
 Siphon
Example
If a U-shaped hose or pipe connects a liquid-filled container at a
higher altitude to a container at a lower altitude over a barrier, the
liquid can be siphoned into the container at the lower altitude.
Atmospheric pressure helps to push the liquid over the barrier.
 Siphon
Example
When P1>P2, the fluid can be siphoned from the left to the
right bucket.
Example
• Water garden
A water hose 2.50 cm in diameter is used by a gardener to fill a
30.0-liter bucket. The gardener notices that it takes 1.00 min to fill
the bucket. A nozzle with an opening of cross-sectional area 0.500
cm2 is then attached to the hose. The nozzle is held so that water
is projected horizontally from a point 1.00 m above the ground. Over
what horizontal distance can the water be projected?
30.0 L
volume flow rate 
1.00 min
 1.00 103 cm3  1.00 m 



1
.
00
L
100.0
cm



3
 1.00 min 
3
3

  5.00 10 m /s
 60.0 s 
A1v1  A2v2  A2v0 x (v0 x : the x - component of the initial velocity)
A1v1 5.00 104 m3 / s
v0 x 

 10.0 m/s
4
2
A2
0.500 10 m
y  v0 y t 
1 2
 2t
 2(1.00 m)
gt  t 

 0.452 s; x  v0 x t  4.52 m
2
2
g
9.80 m/s
• Example : A water tank
Consider a water tank with a hole.
(a) Find the speed of the water
leaving through the hole.
P0 
h =0.500 m
y1 =3.00 m
1 2
v1  gy1  P0  gy2
2
v1  2 g ( y2  y1 )  2 gh
 3.13 m/s
(b) Find where the stream hits the ground.
1 2
y  0  y1   gt  v0 y t  t  0.782 s
2
x  v0 xt  v1t  2.45 m
y
x
• Example : Fluid flow in a pipe
Find the speed at Point 1.
A1v1  A2v2
v2 
A2=1.00 m2
A1=0.500 m2
h =5.00 m
A1
v1
A2
1 2
1 2
P0  v1  gy1  P0  v2  gy2
2
2
2
P0 
1 2
1 A 
v1  gy1  P0    1 v1   gy2
2
2  A2 
  A 2 
2 gh
v12 1   1    2 g ( y2  y1 )  2 gh  v1 
 11.4 m/s
  A2  
1  ( A1 / A2 ) 2
Viscosity and turbulence
 Viscosity
Viscosity is internal friction in a fluid, and viscous forces oppose
the motion of one portion of a fluid relative to another.
Viscosity and turbulence
 Drag
If a fluid in laminar flow flows around an obstacle, it exerts a viscous
drag on obstacle. Frictional forces accelerate the fluid backward
against the direction of flow and the obstacle forward in the direction
of flow.
adjacent layers of fluid slide
smoothly past each other and
flow is steady laminar flow
Viscosity and turbulence
 Turbulence
When the speed of a flowing
fluid exceeds a certain critical
value the flow is no longer laminar.
The flow patter becomes extremely
irregular and complex, and it changes
continuously in time. There is no
steady flow pattern. This chaotic flow
Is called turbulence.
Problems
Problem 1
The upper edge of a gate in a dam runs
the water surface. The gate is 2.00 m high
and 4.00 m wide and is hinged along the
horizontal line through its center. Calculate
the torque about the hinge arising from
the force due to the water.
2.00 m
Solution
Denote the width and depth at the bottom of the gate by w and H.
The force on a strip of vertical thickness dh at a depth h is: dF  gh(wdh)
and the torque about the hinge is d  gwh( h  H / 2)dh.
After integrating from h=0 to h=H, you get the torque:
  gwH 3 / 12  2.61  104 N  m.
Problem 2
An object with height h, mass M, and a uniform cross-sectional area A
floats upright in a liquid with density .
(a) Calculate the vertical distance from the surface of the liquid to the
bottom of the floating object in equilibrium.
(b) A downward force with magnitude F is applied to the top of the object.
At the new equilibrium position, how much farther below the surface of
the liquid is the bottom of the object than it was in part (a)?
(c) Calculate the period of the oscillation when the force F is suddenly
removed.
Solution
(a) From Archimedes’s principle gLA  Mg , so L  M /( A).
(b) The buoyant force is: gA( L  x)  Mg  F . With the result of part (a) solving
for x gives: x   F /( gA).
(c) The force is always in the direction toward the equilibrium, namely, a
restoring force Fres  Fbuoy  Mg   gAx. Therefore the “spring constant” is
k  gA and the period of the oscillation is T  2 M / k  2 M /( gA).
Problem 3
You cast some metal of density m in a mold, but you are worried that there
might be cavities within the casting. You measure the weight of the casting
to be w, and the buoyant force when it is completely surrounded by water
V0  B /(  water g )  w /( m g )
to be B. (a) Show that
is the total volume
of any enclosed cavities. (b) If your metal is copper, the casting’s weight
is 156 N, and the buoyant force is 20 N, what is the total volume of any
enclosed cavities in your casting? What fraction is this of the total volume
of the casting?
Solution
(a) Denote the total volume V. If the density of air is neglected, the buoyant
force in terms of the weight is:
B   water gV   water g[( w / g ) / m  V0 ]
Therefore V0  B /(  water g )  w /( m g ).
(b) B /(  water g )  w /( Cu g )  2.52  104 m3. The total volume of the casting is
B /(  water g ), The cavities are 12.4% of the total volume.
Problem 4
A
A U-shaped tube with a horizontal portion of
length contains a liquid. What is the difference
in height between the liquid columns in the
vertical arms (a) if the tube has an acceleration
a toward the right? (b) if the tube is mounted on
a horizontal turntable rotating with an angular
speed  with one of the vertical arms on the
axis of rotation?
A

Solutions

(a) Consider the fluid in the horizontal part of the tube. This fluid with mass
A , is subject to a net force due to the pressure difference between
the ends of the tube, which is the difference between the gauge pressures
at the bottoms of the ends of the tubes. Now this difference is g ( yL  yR ),
and the net force on the horizontal part of the fluid is
g ( yL  yR ) A  Aa,
or ( yL  yR )  (a / g ).
(b) Similarly to (a) consider the fluid in the horizontal part of the tube. As in (a)
the fluid is accelerating. The center of mass has a radial acceleration of
magnitude arad   2 / 2, so the difference in heights between the columns
is ( 2 / 2)(  / g )   22 /( 2 g ).