Fluids
Dr. Robert MacKay
Clark College
Mass Density, r
m
r
V
or
m  rV
Weight Density, D
w
D
V
or
w  DV
Volume
Some Densities
Density
Air
water
steel
Benzene
gold
ice
lead
mercury
g/cm3
.0012
1.00
7.8
0.88
19.3
0.92
11.4
13.6
kg/m3
1.2
1000
7800
880
19300
920
11400
13600
lb/ft3
.075
62.4
487
55
1204
58
705
850
Specific Gravity
rx
Dx
SGx 

rH2O DH2O
Pressure
Force
Pressure =
Area
F
P=
A
F = PA
Quick Quiz 14.1
Suppose you are standing directly behind someone who
steps back and accidentally stomps on your foot with the
heel of one shoe. Would you be better off if that person were
(a) a large professional basketball player wearing sneakers
(b) a petite woman wearing spike-heeled shoes?
Quick Quiz 14.1
Answer: (a). Because the basketball player’s weight is
distributed over the larger surface area of the shoe, the
pressure (F / A) that he applies is relatively small. The
woman’s lesser weight is distributed over the very small
cross-sectional area of the spiked heel, so the pressure is
high.
Figure P14.62
In about 1657 Otto von Guericke, inventor of the air pump, evacuated a
sphere made of two brass hemispheres. Two teams of eight horses each
could pull the hemispheres apart only on some trials, and then "with
greatest difficulty," with the resulting sound likened to a cannon firing (Fig.
P14.62). (a) Show that the force F required to pull the evacuated
hemispheres apart is pR2(P0 – P), where R is the radius of the hemispheres
and P is the pressure inside the hemispheres, which is much less than P0.
(b) Determine the force if
P = 0.100P0 and R = 0.300 m.
Pressure with increasing depth
m=rV & w=mg
Barometer
Patms=1.013x105 N/m2
=1.013x105 Pa
=1.013 bar
= 1 Atmosphere
=1013 mb
= 760 mm Hg
= 29.92 in Hg
= 760 Torr
=14.7 lb/in2
manometer
Pgauge = rgh
Pabs=Patm+Pgauge
Quick Quiz 14.2
The pressure at the bottom of a filled glass of water (ρ =
1 000 kg/m3) is P. The water is poured out and the glass is
filled with ethyl alcohol (ρ = 806 kg/m3). The pressure at
the bottom of the glass is
(a) smaller than P
(b) equal to P
(c) larger than P
(d) indeterminate
Quick Quiz 14.2
Answer: (a). Because both fluids have the same depth, the
one with the smaller density (alcohol) will exert the smaller
pressure.
Archimedes’ Principle
Active Figure 1409
Archimedes’ Principle
p  rf g(h 2  h1 )
FB  pA  rf g(h 2  h1 )A  rf Vg
F B  mf g
The Buoyant force acting on an object
submerged in a fluid equals the weight of fluid
displaced by the object.
Buoyancy
Buoyancy
m= 100 g
400 g
V= m/r = 100 cm3
robj= ?
500g
rf= 1.00 g/cm3
?
Buoyancy
m= 100 g
400 g
V= = m/r = 100 cm3
robj= 5 g/cm3
500g
rf= 1.00 g/cm3
?
Buoyancy
m= 100 g
400 g
V= ?
robj=8.0 g /cm3
500g
r f= ?
?
Buoyancy
m= 100 g
V= 500g/8.0 g /cm3
400 g
V=62.5 cm3
robj=8.0 g /cm3
500g
rf= 100g/62.5 cm3
=1.6 g /cm3
?
Buoyancy
A 20,000,000 tonn ship
Floats in salt water.
(1 tonn=1000kg)
How much water does this ship
Displace?
Active Figure 1410
QUICK QUIZ 14.2
(end of section 14.4)
For a physics experiment, you drop three objects of
equal mass into a swimming pool. One object is a
piece of pine, the second object is a hunk of
copper and the third object is a hunk of lead. The
relationship between the magnitudes of the
buoyant forces on these three objects will be
a) Fcopper > Fpine > Flead,
b) Fpine > Fcopper > Flead,
c) Flead > Fcopper > Fpine or
d) Fcopper > Flead > Fpine.
QUICK QUIZ 14.2 ANSWER
(b). From Archimedes’ principle, the magnitude of the buoyant force will be equal to
the weight of the water displaced. From Table 14.1, lead and copper are more dense
than water and will therefore sink, while pine is less dense than water and will
therefore float. The buoyant force for the pine must equal the weight, mg, of the pine
since these two forces balance. For completely submerged objects, the buoyant force
will be equal to the weight of the water displaced, mwg, and will be less for the denser
lead, because of its smaller volume, than for the copper. In addition, the mass of the
water displaced will be less than the mass of the equal volume of metal displacing it,
so that mw < m. Therefore, the buoyant force on each metal is less than the buoyant
force on the pine.
Buoyancy
A cubical
block of wood has
30.0 cm sides. If
Its density is
600 kg/m3 how much
of it is submerged
when floating in water?
Buoyancy
M=?
A cubical
block of wood has
30.0 cm sides. If
Its density is
600 kg/m3 how much
mass must be placed on its
top to just barely
submerge it?
Quick Quiz 14.6
A glass of water contains a single floating ice cube as in the
figure below. When the ice melts, the water level
(a) goes up
(b) goes down
(c) remains the same
Quick Quiz 14.6
Answer: (c). The ice cube displaces a volume of water that
has a weight equal to that of the ice cube. When the ice cube
melts, it becomes a parcel of water with the same weight and
exactly the volume that was displaced by the ice cube
before.
A barge loaded with steel floats in a lock. If the steel
is then thrown overboard. Does the water level in the
lock a) go up b) go down c) stay at the same
level?
Before
H2O displaced =W(boat)+W(steel)
After
H2O displaced =W(boat)+little more
A barge loaded with steel floats in a lock. If the steel
is then thrown overboard. Does the water level in the
lock a) go up b) go down c) stay at the same
level?
Quick Quiz 14.5
An apple is held completely submerged just below the
surface of a container of water. The apple is then moved to a
deeper point in the water. Compared to the force needed to
hold the apple just below the surface, the force needed to
hold it at a deeper point is
(a) larger
(b) the same
(c) smaller
(d) impossible to determine
Quick Quiz 14.5
Answer: (b). For a totally submerged object, the buoyant
force does not depend on the depth in an incompressible
fluid.
Continuity
Equation of
Continuity
A1x1  A2 x 2
x1
x2
A1  A2
t
t
A1v1  A2 v 2
A1x1  A2 x 2
x1
x2
A1  A2
t
t
A1v1  A2 v 2
Volume rate of flow
Volume
x
Q
 A  Av
t
t
p 2
2
A  pr  D
4
Water flows out of a 1.0 inch hose at a speed of 2.0 ft/s.
How long will it take this hose to fill a 50 gallon drum?
Quick Quiz 14.8
You tape two different sodas straws together end-to-end to
make a longer straw with no leaks. The two straws have
radii of 3 mm and 5 mm. You drink a soda through your
combination straw. In which straw is the speed of the liquid
the highest?
(a) whichever one is nearest your mouth
(b) the one of radius 3 mm
(c) the one of radius 5 mm
(d) Neither – the speed is the same in both straws.
Quick Quiz 14.8
Answer: (b). The liquid moves at the highest speed in the
straw with the smaller cross sectional area.
Example

A 1.0 inch diameter hose is connected to a
0.25 in diameter nozzle. If the water
shoots up in the air to a maximum height
of 15.0 m, what is the flow rate
(gallons/min) in the hose? (231 in3=1
gallon)
QUICK QUIZ 14.5
(end of section 14.5)
You would like to change the opening on the nozzle
of a fire hose so that the water exiting the hose can
reach a height that is four times the present
maximum height the water can reach. To do this,
you should decrease the cross sectional area of the
opening by a factor of
a) 16,
b) 8,
c) 4
d) 2.
QUICK QUIZ 14.5 ANSWER
(d). From the continuity equation, the velocity of the water
exiting the hose is inversely proportional to the cross
sectional area or v  1/A. However, the kinetic energy of
the water that exits the hose will be equal to the potential
energy of the water at its maximum height (when you point
the hose straight up), or
1
2
1 2
mv  mgh, so that h  v  ( ) , or A 
A
2
2
1
h
.
So to increase the height by a factor of four, you must
decrease the area by a factor of 2.
Bernoulli’s Principle

Where the fluid speed is high the internal
pressure is low. [based on the conservation
of energy]
Bernoulli’s Principle
Bernoulli’s Principle
http://www.aerospaceweb.org/question/aerodynamics/
Bernoulli’s Principle
Bernoulli’s Principle
Bernoulli’s Principle
Bernoulli’s Principle
∆m1
A2
v1
A1
F1=P1A1
∆x1 =v1∆t
∆m2
v2
∆m1 = ∆m2
F2=P2A2
∆x2 =v2 ∆t
y2
y1
Bernoulli’s Principle
1 0.0
1 2 0.0
2
P1  rv1  rgh1  P2  rv2  rgh 2
2
2
V1 = 0.0
h1 = h
h2 = 0.0 , P2=P1
Bernoulli’s Principle
1 0.0
1 2 0.0
2
P1  rv1  rgh1  P2  rv2  rgh 2
2
2
1
2
V1 = 0.0
rgh  rv 2
2
h1 = h
2
2gh  v
v  2gh
h2 = 0.0 , P2=P1
Bernoulli’s Principle
1 2
rgh  rv 2
2
2
2gh  v
v  2gh
Bernoulli’s Principle
1 2
1 2
P1  rv1  rgh1  P2  rv2  rgh 2
2
2
Bernoulli’s Principle
1 2
1 2
P1  rv1  rgh1  P2  rv2  rgh 2
2
2
(P1-P2)A=Lift
Surface Tension
g L=F
Water , g
T
0.076 N/m 0°C
0.072 N/m 20°C
0.059 N/m 100 °C
Capillary Action
2 p r g = r p r2 h g
r
h
Viscosity
v
F  A
h
Viscosity
Viscosity (Pa s)
0.0018
0.0010
0.0003
0.0040
Fluid
0°C water
20°C water
100 °C wat
37°C blood
Poiseuille’s Law
Q
2
pr p
8L
Stoke’s Law, terminal velocity
of a sphere