Transcript Static Fluids

Static Fluids
• Fluids are substances, such as liquids
and gases, that have no rigidity. A
fluid lacks a fixed shape and assumes
the shape of its container.
• In the liquid state, molecules can flow;
they freely move from position to
position by sliding over one another.
Pressure
• The pressure P acting on a fluid is the force
exerted perpendicularly per unit of the
fluid’s surface area
F
P
A
• Unit of pressure is the N/m2 or Pascal;
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1 N/m2 = 1 Pa (Pascal).
Atmospheric pressure at sea level is
1 atmosphere (atm) = 1.013 x 105 Pa.
1 atm = 14.7 lb/in2.
Pressure in a Liquid.
• A liquid in a container exerts forces
against the walls and bottom of the
container.
• For a liquid in a container, the
pressure the liquid exerts against the
bottom of the container is the weight of
the liquid divided by the area of the
container bottom.
Measuring Pressure
•
A manometer is a U-shaped
tube that is partially filled with
liquid.
•
Both ends of the tube are open
to the atmosphere.
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A container of gas is connected
to one end of the U-tube.
•
If there is a pressure difference
between the gas and the
atmosphere, a force will be
exerted on the fluid in the Utube. This changes the
equilibrium position of the fluid
in the tube.
From the figure:
At point C
Also
Pc  Patm
PB  PB'
The pressure at point B is the pressure
of the gas.
let d  h
PB  PB '  PC  h  D  g
PB  PC  PB  Patm  h  D  g
Pgauge  h  D  g
Pgauge easily remembered as “hot dog”
A Barometer
The atmosphere pushes on the
container of mercury which forces
mercury up the closed, inverted
tube. The distance d is called the
barometric pressure.
From the figure:
and
PA  PB  Patm
let d  h
PA  h  D  g
Atmospheric pressure is equivalent to a column of mercury
76.0 cm tall.
Density
• Density D of a substance is its mass per
unit volume;
M
D
V
• Objects composed of the same substance,
•
whatever the size or mass, have the same
density under the same conditions of
temperature and pressure.
Temperature and pressure affect the
density of substances, appreciably for
gases, but only slightly for liquids and
solids.
Density
• How much a liquid weighs and how
much pressure it exerts depends on its
density.
– For the same depth, a denser liquid exerts a
greater pressure than a less dense liquid.
– For liquids of the same density, the
pressure will be greater at the bottom of the
deeper liquid.
• To convert a density in g/cm3 to kg/m3,
multiply by 1000.
Columnar Fluid Pressure
(sometimes called gauge pressure)
• pressure due to a
column of fluid of height
h and mass density D;
P  hDg
• The pressure of a liquid at
•
rest depends on the
density and depth of the
liquid.
Liquids are practically
incompressible, so except
for changes in the
temperature, the density
of a liquid is normally the
same at all depths.
Columnar Fluid Pressure
• At a given depth, a given liquid
exerts the same pressure against
any surface - the bottom or sides of
its container, or even the surface
of an object submerged in the
liquid to that depth.
• Pressure a liquid exerts depends
only on its density and depth.
• Total pressure (or absolute
pressure) Pabsolute on a submerged
surface equals the pressure the
liquid exerts plus the atmospheric
pressure Po (1 atm = 1.013 x 105
Pa) .
Pabsolute  Po  h  D  g
( peanut  hot dog )
Fluid Pressure
• Pressure of a liquid does not
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depend on the amount of liquid.
Neither the volume or total
weight of the liquid matters.
If you sampled water pressure
at 1 m beneath a large lake
surface and 1 m beneath a small
pool surface, the pressure would
be the same.
The fact that water pressure
depends on depth and not on
volume is illustrated by Pascal
vases.
Water surface in each of the
connected vases is at the same
level.
Occurs because the pressures at
equal depths beneath the
surface are the same.
Fluid
Pressure
• At any point within
a liquid, the forces
that produce
pressure are exerted
equally in all
directions.
• Pressure increases
vertically downward.
• Pressure constant
horizontally.
Forces Exerted By a Fluid
• When the liquid is pressing
against a surface there is a
net force directed
perpendicular to the surface.
• If there is a hole in the
surface, the liquid initially
moves perpendicular to the
surface.
• At greater depths, the net
force is greater and the
horizontal velocity of the
escaping liquid is greater.
Transmission of Pressure:
Pascal’s Principle.
• Pascal’s Principle: A CHANGE
IN PRESSURE IN A CONFINED
FLUID IS TRANSMITTED
WITHOUT CHANGE TO ALL
POINTS IN THE FLUID.
• Ex. Hydraulic lift.
• Hydraulic piston apparatus
uses an incompressible fluid to
transmit pressure from a small
cylinder to a large cylinder.
• According to Pascal’s Principle,
the pressure in the small
cylinder resulting from the
application of F1 to a
frictionless piston is
transmitted undiminished to
the larger piston.
Transmission of pressure:
Pascal’s Principle.
P1 = P2
F2
F1

A1 A 2
• A2 is larger than A1, so the
force exerted by the large
piston is greater than the
force exerted on the small
piston.
• AMA (actual mechanical
advantage) for hydraulic lift:
AMA 
F2
F1
The applied force is
transmitted to the
piston of crosssectional area A2 here.
Apply a force F1
here to a piston
of cross-sectional
area A1.
A2
F2  F1
A1
Transmission of pressure:
Pascal’s Principle.
• The figure shows a hydraulic
system used with brakes. The
force F is applied perpendicularly
to the brake pedal. The brake
pedal rotates about the axis shown
in the drawing and causes a force
to be applied perpendicularly to
the input piston in the master
cylinder. The resulting pressure is
transmitted by the brake fluid to
the output plungers which are
covered with the brake linings.
The linings are pressed against
both sides of a disc attached to the
rotating wheel.
Buoyancy
• Buoyancy: the apparent loss of
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weight of objects when
submerged in a liquid.
Easier to lift objects under
water surface than to lift it
above the water surface.
When submerged, water exerts
an upward force that is
opposite in direction to gravity.
Upward force called the
buoyant force.
Buoyancy
• Forces exerted by liquid
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produce pressure against the
submerged object.
Forces are greater at greater
depths; forces are equal at
the same depth on opposite
sides of the object.
Forces acting upward on the
bottom of the object greater
than those acting downward
on top of the object, simply
because the bottom of the
object is deeper.
Difference in upward and
downward forces is the
buoyant force, B.
Fs refers to the force the
scale exerts on the mass m.
You may also refer to Fs as
the tension in a supporting
string.
Buoyancy
• If the weight of the object
•
•
is greater than the
buoyant force, the object
will sink (as in figure a).
If the weight of the object
is equal to the buoyant
force, the net force on the
object is zero and the
submerged object will
remain at any level (as in
figure b).
If the weight of the object
is less than the buoyant
force, the object will rise
to the surface and float.
Buoyancy
• When an object is submerged in a liquid, the
liquid level will rise.
• Liquid is displaced or moved elsewhere.
• The volume of the liquid displaced is equal to
the volume of the submerged object.
• A completely submerged object always
displaces a volume of liquid equal to its own
volume.
Archimede’s Principle
• When an object is immersed in a fluid, it
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appears to weigh less.
Archimede’s Principle: THE BUOYANT
FORCE EXERTED ON A BODY WHOLLY
OR PARTLY IMMERSED IN A FLUID IS
EQUAL TO THE WEIGHT OF THE FLUID
DISPLACED BY THE BODY.
Archimede’s Principle applies to both
liquids and gases, which are fluids.
Immersed refers to either completely or
partially submerged.
Archimede’s Principle
• Buoyant force: upward force on the object when it is
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immersed in water.
The buoyant force (BF) is the weight of the displaced
fluid - not the weight of the submerged object.
BF = (mass in air - mass in fluid)·gravity
BF = weight in air - weight in fluid
BF = weight of displaced fluid (Dfluid·Vobject submerged·g)
Apparent weight of a submerged object is its weight in
air minus the buoyant force.
Apparent weight = m·g – BF = m·g - Dfluid·Vobject submerged·g
• For an object that is floating or is submerged but
not sinking: mobject·g = Dfluid·Vobject submerged·g
Flotation
• Principle of Flotation: A FLOATING
OBJECT DISPLACES A WEIGHT OF
FLUID EQUAL TO ITS OWN WEIGHT.
• A simple relationship between the
weight of a submerged object and the
buoyant force can be found by
considering their ratio:
Fw object D object

BF
D fluid
Flotation
• Shipbuilding and the principle of
flotation:
– A solid 1-ton block of iron is nearly 8 times
as dense as water, so when it is submerged,
it will displace 1/8 ton of water (not an
amount equal to 8 tons of water).
– Reshape the iron block into a bowl and
submerge, displaces a greater volume of
water. The deeper the bowl is immersed at
the surface, the more water is displaced and
the greater is the buoyant force exerted on
the bowl.
Flotation
• When the weight of the displaced water
equals the weight of the bowl - flotation.
Flotation occurs when the weight of the
bowl equals the buoyant force.
• Every ship must be designed to displace a
weight of water equal to its own weight.
• Submarines:
– Displace a weight of water equal to its
own weight, it remains at a constant
depth.
– Displaces a weight of water greater
than its own weight, rises.
– Displaces a weight of water less than
its own weight, sinks.
Buoyancy in Two Liquids of
Differing Density
• If you have an object submerged in two liquids of
different density, such that the upper portion of
the object is located in the upper liquid and the
lower portion of the object is located in the lower
liquid, the total buoyant force on the object is
equal to the weight of the object
BF = Dfluid·Vobject submerged·g
• Ex. A piece of wood floating partially in water and
partially in oil. The density of the oil is less than
the density of the water.
BFoil  BFwater  Dobject  Vobject  g
(D  V  g)oil  (D  V  g)water  Dobject  Vobject  g
• Gravity cancels out.
(D  V)oil  (D  V)water  Dobject  Vobject
V  A h
(D  V )oil  (D  V )water  Dobject  Vobject
• If d is the height of the object (the 4 cm in
the figure), let y = the portion of the object
in the more dense liquid and d-y = the
portion of the object in the less dense
liquid.
(D  A  d  y )oil  (D  A  y )water  Dobject  Vobject
• This equation can then be solved for the
unknown variable.
Pressure Example
• Water is to be pumped to the top of the Empire State
Building, which is 366 m high. What gauge pressure is
needed to raise the water to a height of 366 m? The
density of water is 1000 kg/m3.
P  hDg
P  366 m  1000
P  3586800
N
m
2
kg
m
3
 9.8
m
s
2
 3586800 Pa
Buoyant Force
• An object weighing 300 N
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in air is immersed in
water after being tied to
a string connected to a
balance. The scale now
reads 265 N. Immersed
in oil, the object appears
to weigh 275 N.
A. Find the density of the
object.
BF  Fw air  Fw H2O
BF  300 N  265 N  35 N
BF  D  V  g
BF
35 N
V

D  g 1000 kg  9.8 m
m3
s2
V  3.5714 x 10 3 m 3
m
D
Fw air  m  g
V
Fw air
300 N
m

 30.61 kg
m
g
9.8 2
s
30.61 kg
kg
D

8571.5
3.5714 x 10 3 m 3
m3
Buoyant Force
• B. Determine the density of the oil.
BF  Fw air  Fw oil
BF  300 N  275 N  25 N
BF  D  V  g
volume of object  volume of fluid displaced
V  3.5714 x 10 3 m3
BF
25 N
kg
D

 714 .9 3
V  g 3.5714 x 10 3 m3  9.8 m
m
s2