Chapter 14: Fluids
In this chapter we will explore the behavior of fluids.
In particular we will study the following:
Static fluids:
Pressure exerted by a static fluid
Methods of measuring pressure
Pascal’s principle
Archimedes’ principle, buoyancy
ΔV
Δm
Fluids
As the name implies, a fluid is defined as a substance that can flow.
Fluids conform to the boundaries of any container in which they
are placed. A fluid cannot exert a force tangential to its surface.
It can only exert a force perpendicular to its surface. Liquids and
gases are classified together as fluids to contrast them with solids.
In crystalline solids the constituent atoms are organized in a rigid
three-dimensional regular array known as the "lattice."
Density :
Consider the fluid shown in the figure. It has a mass Δm and
volume ΔV . The density (symbol ρ ) is defined as the ratio
of the mass over the volume: ρ =
Δm
.
ΔV
SI unit: kg/m 3
If the fluid is homogeneous, the above equation has the form
m
ρ= .
V
Density
Note: density solid > liquid > gas
m
density =ρ =
V
air =
wood =
water =
Al =
Cu =
1.21 kg/m3
550 kg/m3
1000 kg/m3
2700 kg/m3 = 2.7 g/cm3
8960 kg/m3 = 8.9 g/cm3
“Vacuum”
What
is
pressure?
Pressure
F
P=
A
“collisional force”
Units: pascal = Pa = N/m2
€
1 atmosphere = 1 atm
= 1.01×105 Pa
= 760 torr
= 14.7 lb/in2
The collisions of gas molecules
on the wall of the tire keep it
inflated
Uniform ⊥ force on
flat area
Hydrostatic Pressure: Water applies
a force perpendicular to all of the
surfaces in the pool, including the
swimmer, the walls of the pool etc.
Fluids
at
Rest
Static Equilibrium
€

F
∑ =0 ⇒

a=0

∑F = 0
Fbottom − Ftop − mg = 0
pbottom A = ptop A + mg
mg
mg h
= ptop +
A
A h
m
= ptop + gh
V
= ptop + ρ gh Pressure depends on depth
pbottom = ptop +
pbottom
pbottom
NOT horizontal dimensions
pat h = patm + ρgh
pat h > patm for h down
pat h < patm for h up
Example
1. What is the pressure head at the bottom of a 98 ft (30 m) water tower?
2. At what depth is the pressure two-times that of
atmosphere?
pat h = patm + ρ gh
2 patm = patm + ρ gh
patm
1.01 × 10 5 ⋅ Pa
h=
=
ρg
1000 ⋅ kg / m 3 9.8 ⋅ m / s 2
(
= 10.3 ⋅ m = 33.8 ⋅ ft
)(
)
3. What is the maximum height you can suck water up a straw?
p = patm + ρgh
0 = patm + ρgh
patm
h=−
= 10.3 ⋅ m = 33.8 ⋅ ft
ρg
€
4. What is the net force on Grand Coulee dam (width 1200 m - height 150 m)?
pat h = patm + ρgh
dF = (ρgy) dA
D
F=
∫ ( ρgy )W dy
0
= 12 ρ gWD 2
= 1.3 × 1011 N
10
F = 2.9 × 10 lb
Blood
Pressure
Blood pressure of 120/80 is considered
normal… what are these units? How much
pressure is this? WHY?
ρ Hg gh = (13, 600 kg m 3 )(9.8 m s2 )(0.12m)
120mmHg = 1.6 × 10 4 Pa
80mmHg = 1.1 × 10 4 Pa
Difference = 0.5 × 10 4 Pa
What is the pressure difference between your heart and
your feet?
(Density of blood is 1060 kg/m3)
P2 − P1 = ρ gh
= (1060 ⋅ kg m 3 )(9.8 ⋅ m s2 )(1.35m)
= 1.4 × 10 4 Pa
Systolic
Dystolic
Pressure
vs
height:
gasses
Remember: if h is down, pressure goes up;
if h is up, pressure goes down Δp = ρgh
What is the air pressure at 18,000 ft (5,500 m)?
(elevation affects
€pressure- how?)
Assume that the density of air is
proportional to the pressure
(compressible fluid)
 ρ0 
ρ h ph
=
⇒ ρ h = ph  
ρ 0 p0
 p0 
, where at 0 º C & sea level
ρ0 = 1.29 kg/m3 & p0 =1 atm = 1.01×105 Pa
Negative because pressure is
decreasing as you go up.
Δp = ρ gh ⇒ dp = − ρ g dy
 ρ0 
dp = −ph  gdy
 p0 
 p H   ρ0 
ln  = − g( H − 0)
 p0   p0 
ρ 
dp
= −  0  gdy
ph
 p0 
€
pH
∫
p0
p H = p0 e
 ρ0  H
dp
= −   g ∫ dy
ph
 p0  0
ρ
− 0
 p0

 gH

p18,000 ft = 12 patm
€
Measuring
Pressure
Torricelli (1608-1647)
Closed-end Manometer (Hg Barometer)
pat h = 0 + ρgh
pat h = ρgh
€
1mm of Mercury = 1 torr
Open-end Manometer
pat h = patm + ρgh
Δp = ρgh
Absolute Pressure
Gauge Pressure = pg = Δp
patm
1.01 × 10 5 ⋅ N / m 2
hHg =
=
ρ g (13, 550 ⋅ kg / m 3 ) ( 9.8 ⋅ m / s 2 )
= 760 ⋅ mm Hg
€
Question
Is
the
gauge
pressure
at
the
bo.om
of
a
1‐m
high
tube
of
water
on
the
earth
the
same
as
is
on
the
moon?
1. 
2. 
3. 
Yes
No
Some=mes
ΔPmoon = ρ gearth h 

ΔPearth = ρ gmoon h 
gmoon < gearth , so that
ΔPmoon < ΔPearth
Pascal’s
Principle
Blaise Pascal (1623-1662)
Pressure applied to a confined fluid increases the pressure throughout by same amount.
p100 m = patm + ρ gh
p100 m = 1⋅ atm + 9.7 ⋅ atm = 10.7 ⋅ atm
What is the pressure 100m below sea
level?
“Transmitted” throughout the whole
100m
Hydraulic Lever
“Mechanical Advantage”
pout = pin
Fout Fin
=
Aout Ain
Fout = Fin
Incompressible Fluid:
V = din Ain = dout Aout
A
dout = din in
Aout
Work:
With a hydraulic lever, a given force applied over a
given distance can be transformed to a greater
force over a smaller distance
€
Aout
Ain
 A 
A 
Wout = Fout d out =  Fin out  d in in 
Ain  Aout 

= Fin d in = Win
Examples
of
Pascal’s
Principle
Piston B
A person m=75kg stands on circular piston A (diameter=0.40m) of
a hydraulic pump. If you want to lift an elephant weighing 1500kg,
what is the minimum diameter of circular piston B?
P=
€
FA
F
= B
AA AB
AB = AA
2
FB
FA
2
d 
d  F
π B =π A B
 2
 2  FA
 d  FB
dB = 2  A 
 2  FA
= 1.8m
Buoyancy
and
Archimedes’
Principle
Archimedes’ (287?-212 BC)
Buoyancy - lift a rock under water – it’s light
- take it above water – it’s heavy
- many objects float - how?
- Force of gravity points downward
- buoyant force is exerted upward by fluid
Consider a cube in a fluid with density ρ and area A.
Fbottom = Pbottom A = ρghbottom A
1) At top, fluid exerts a€force on cube:
(
)
Ftop = Ptop A = Patm + ρgh top A
DOWNWARD
2) At bottom, fluid exerts a force on cube:
€
Fbottom = Pbottom A = (Patm + ρgh bottom ) A UPWARD
3) Net force due to “fluid” is bouyant force:
Fbuoyant = Fbottom − Ftop = ρ g(hbottom − htop )A
€
= ρ ghA = ρ gV
UPWARD
= m fluid g = weight of displaced fluid