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Chapter 14
Fluid Mechanics
Fluids
• Fluids (Ch. 6) – substances that can flow (gases,
liquids)
• Fluids conform with the boundaries of any container
in which they are placed
• Fluids lack orderly long-range arrangement of
atoms and molecules they consist of
• Fluids can be compressible and incompressible
Density and pressure
• Density
m
  lim
V  0  V
• SI unit of density: kg/m3
• Pressure (cf. Ch. 12)
P  lim
A0
F
A
• SI unit of pressure: N/m2 = Pa (pascal)
Blaise Pascal
(1623 - 1662)
• Pressure is a scalar – at a given point in a fluid the
measured force is the same in all directions
• For a uniform force on a flat area
F
P
A
Atmospheric pressure
• Atmospheric pressure:
• P0 = 1.00 atm = 1.013 x 105 Pa
Fluids at rest
• For a fluid at rest (static equilibrium) the pressure is
called hydrostatic
• For a horizontal-base cylindrical water sample in a
container
F2  F1  mg
P2 A  P1 A  A( y1  y2 ) g
P2  P1   ( y1  y2 ) g
P  P0  hg
Fluids at rest
• The hydrostatic pressure at a point in a fluid
depends on the depth of that point but not on any
horizontal dimension of the fluid or its container
• Difference between an absolute pressure and an
atmospheric pressure is called the gauge pressure
Pg  P  P0  hg
P  P0  hg
Chapter 14
Problem 12
The tank is filled with water 2.00 m deep. At the bottom of one sidewall is a
rectangular hatch 1.00 m high and 2.00 m wide that is hinged at the top of the
hatch. (a) Determine the force the water causes on the hatch. (b) Find the
torque caused by the water about the hinges.
Measuring pressure
• Mercury barometer
P2  P1   ( y1  y2 ) g
y1  0; P1  P0
y2  h; P2  0
P0  hg
• Open-tube manometer
P2  P1   ( y1  y2 ) g
y1  0; P1  P0
y2  h; P2  P
Pg  P  P0  hg
Pascal’s principle
• Pascal’s principle: A change in the pressure applied
to an enclosed incompressible fluid is transmitted
undiminished to every portion of the fluid and to the
walls of its container
• Hydraulic lever
F1 F2
P 

V  A1x1  A2 x2
A1 A2
A2
A1
x1  x2
F1  F2
A1
W  F1x1  F2 x2
A2
• With a hydraulic lever, a given force applied over a
given distance can be transformed to a greater force
applied over a smaller distance
Archimedes’ principle
• Buoyant force:
For imaginary void in a fluid
p at the bottom > p at the top
B  mf g
Archimedes
of Syracuse
(287-212 BCE)
• Archimedes’ principle: when a body is submerged in
a fluid, a buoyant force from the surrounding fluid
acts on the body. The force is directed upward and
has a magnitude equal to the weight of the fluid that
has been displaced by the body
Archimedes’ principle
• Sinking:
mg  B
• Floating:
mg  B
• Apparent weight:
weight apparent  mg  B
• If the object is floating at the surface of a fluid, the
magnitude of the buoyant force (equal to the weight
of the fluid displaced by the body) is equal to the
magnitude of the gravitational force on the body
Chapter 14
Problem 28
A spherical aluminum ball of mass 1.26 kg contains an empty spherical cavity
that is concentric with the ball. The ball barely floats in water. Calculate (a) the
outer radius of the ball and (b) the radius of the cavity.
Motion of ideal fluids
Flow of an ideal fluid:
• Steady (laminar) – the velocity of the moving fluid at
any fixed point does not change with time (either in
magnitude or direction)
• Incompressible – density is constant and uniform
• Nonviscous – the fluid experiences no drag force
• Irrotational – in this flow the test body will not rotate
about its center of mass
Equation of continuity
• For a steady flow of an ideal fluid through a tube
with varying cross-section
V  Ax  Avt  A1v1t  A2v2 t
A1v1  A2v2
Av  const
Equation of continuity
Bernoulli’s equation
• For a steady flow of an ideal fluid:
Etot  K  U g  Eint
• Kinetic energy
mv 2 Vv 2
K

2
2
• Gravitational potential energy
U g  mgy  Vgy
• Internal (“pressure”) energy
Eint  VP
Daniel Bernoulli
(1700 - 1782)
Bernoulli’s equation
• Total energy
Etot  K  U g  Eint
Vv

 Vgy  VP
2
2
Etot v

 gy  P  const
V
2
2
v1
2
2
 gy1  P1 
v2
2
2
 gy2  P2
Chapter 14
Problem 49
A hypodermic syringe contains a medicine having the density of water. The
barrel of the syringe has a cross-sectional area A = 2.50 × 10-5 m2, and the
needle has a cross-sectional area a = 1.00 × 10-8 m2. In the absence of a force
on the plunger, the pressure everywhere is 1 atm. A force F of magnitude 2.00 N
acts on the plunger, making medicine squirt horizontally from the needle.
Determine the speed of the medicine as it leaves the needle’s tip.
Questions?
Answers to the even-numbered problems
Chapter 14
Problem 6
(a) 1.01 × 107 Pa
(b) 7.09 × 105 N outward
Answers to the even-numbered problems
Chapter 14
Problem 8
225 N
Answers to the even-numbered problems
Chapter 14
Problem 22
3.33 × 103 kg/m3
Answers to the even-numbered problems
Chapter 14
Problem 38
(a) 0.825 m/s
(b) 3.30 m/s
(c) 4.15 L/s