Chapter 5
FLUIDS AND THERMAL
5.1 Hydrostatics
5.2 Fluid flow
5.3 Pascal Law for pressure
5.4 Archimedean Law
5.5 Continuity equation
5.6 Bernoulli equation.
5.7 Diffusion and endosmosis
5.8 Temperature and
thermal expansion
5.9 Calorimetry and heat transfer
Getting point from the Photo ?
Motion of fluids
Part 1
Hydrostatics
Fluid Statics
• By definition, the fluid is at rest.
• Or, no there is no relative motion between
adjacent particles.
• No shearing forces is placed on the fluid.
• There are only pressure forces and no shear.
• Results in relatively “simple” analysis
• Note for the pressure variation in the fluid
• The force per unit area across any surface is
normal to the surface and is the same for all
orientations of the surface.
Learning Check
You are sitting on the area of 1m2, If you change your
seat to area of 0,7m2, How does the pressure you make
to the seat change ??
Increase 1/0.7=1.42
Part 2
Fluid flow
• What do we mean by “fluids”?
– Fluids are “substances that flow”…. “substances
that take the shape of the container”
– Atoms and molecules are free to move .. No long
range correlation between positions.
• What parameters do we use to describe fluids?
– Density
LIQUID:
m

V
incompressible (density almost
constant)
GAS:
compressible (density depends a lot on
pressure)
Part 3
Pascal Law for pressure
Blaise Pascal (1623-1662)
Review Pressure
Pressure
• Pressure is defined as a normal force
exerted by a fluid per unit area.
• Units of pressure are N/m2, which is
called a pascal (Pa). Pa is too small,
in practice, kilopascal (1 kPa = 103 Pa)
and megapascal (1 MPa = 106 Pa) are
commonly used. Other units include
bar, atm, kgf/cm2, mm Hg.
Units
• Standard atmosphere is defined as the
pressure produced by a column of mercury
760 mm in height at 0°C
(29.92 in Hg or of water about 10.3 m )
* (rHg = 13,595 kg/m3) under standard
gravitational acceleration (g = 9.807 m/s2).
• 1 atm = 760 torr and 1 torr (mmHg) = 133.3
Pa
Pressure Units
• Other units include bar, atm, kgf/cm2, mm
Hg.
• 1 bar = 105 Pa
• 1 atm = 101325 Pa = 1.01325 bars
• 1 kgf/cm2 = 9.807 N/cm2 = 9.807  104 N/m2
= 9.807  104 Pa = 0.9807 bar = 0.9679 atm
• 1 atm = 14.696 psi.
• 1 kgf/cm2 = 14.223 psi.
• Mm Hg = 9,8.13,6 =133 Pa
Pressure at a Point:
Pascal’s Law
Pascal’s Law: the pressure at a point in a fluid
at rest, or in motion, is independent of the
direction as long as there are no shearing
stresses present.
* Pressure is the normal force per unit
area at a given point acting on a given
plane within a fluid mass of interest.
Measurement of Pressure:
Barometers
The first mercury barometer was constructed in 1643-1644 by Torricelli. He
showed that the height of mercury in a column was 1/14 that of a water
barometer, due to the fact that mercury is 14 times more dense that water. He
also noticed that level of mercury varied from day to day due to weather
changes, and that at the top of the column there is a vacuum.
Schematic:
Animation of Experiment:
Note, often
pvapor is very
small,
0.0000231 psia
at 68° F, and
patm is 14.7 psi,
thus:
Evangelista Torricelli
(1608-1647)
Learning Check
Torricelli formula
The Pressure in a homogenous, incompressible fluid at rest depends
on the depth of the fluid relative to some reference and is not
influenced by the shape of the container.
Lines of constant Pressure
p = po
h1
p = p1
p = p2
For p2 = gh + po
For p1 = gh1 + po
Compute: P1 and P2 ??
Give P0=760 mmHg, h1=3cm, h=5cm,  =1500kg/m3
Learning Check
A2
F2 
F1  P  F2 A1  F1A 2
A1
Compute: F2 = ??
Give A1/ A2 =3, F1 =45mmHg
Learning test
• What happens with two different fluids??
Consider a U tube containing liquids of
density 1 and 2 as shown:
– Compare the densities of the liquids:
A) 1 < 2
B) 1 = 2
C) 1 > 2
If we use the same liquids in a U tube of twice
the cross-sectional area as the first, compare
the distances between the levels in the two
cases (depth of liquid 2 same in both cases).
A) dI < dII
B) dI = dII
C) dI > dII
2
dI
1
I
2 dII
1
II
Hint
• At the depth of the interface, the pressures
in each side must be equal.
• Since there’s more liquid above this depth on the
left side, that liquid must be less dense!
d2
2
dI
p
C) 1 > 2
1
d1
I
• The pressure depends ONLY on the depth and the density
of the fluid.
p
p
• e.g. consider case I:
d2 
d1 
g 2
g1
d2


p 1
1
d  d 2  d1 
 

p
g  2 1 
B) dI = dII
2 dII

1
II
d1
Part 4
Archimedean Law
Learning Check
H
Calculate D =?
If L=10m, H=3m, W=500000 N, g=10 m/s2
water= 1000 kg/m3
Learning Check
Example: Submarine Buoyancy and
Ballast
• Submarines use both static and dynamic depth
control. Static control uses ballast tanks between
the pressure hull and the outer hull. Dynamic
control uses the bow and stern planes to generate
trim forces.
Submarine Buoyancy and Ballast
Normal surface trim
SSN 711 nose down after accident
which damaged fore ballast tanks
Submarine Buoyancy and Ballast
Damage to SSN
711 (USS San
Francisco) after
running aground
on 8 January
2005.
Submarine Buoyancy and Ballast
Ballast Control Panel: Important station for controlling depth of submarine
Test
• A lead weight is fastened to a large
styrofoam block and the combination
floats on water with the water level
with the top of the styrofoam block as
shown.
–If you turn the styrofoam+Pb
upside down, what happens?
A) It sinks
B)
styrofoam
Pb
C)
Pb
styrofoam
styrofoam
Pb
Hint
Pb
styrofoam
C)
Pb
styrofoam
• If the object floats right-side up, then it also must
float upside-down.
• However, when it is upside-down, the Pb displaces
some water.
• Therefore the styrofoam must displace less water
than it did when it was right-side up (when the Pb
displaced no water).
Part 5
Continuity equation
Learning Check
Write an equation for steady conditions (see ficture below)
???
Part 6
Bernoulli equation
Static P Dynamic P
Hydrostatic P = const
Static Pressure, Dynamic Pressure and Hydrostatic
Pressure have the same unit
Learning Check
Hint:
Use the dynamic pressure
P=F/A=v2/2
Learning Check
Explain this experiment ??