Sunday, May 12, 2013
LESSON 9 : FLUIDS and PRESSURE
FLUIDS AND PRESSURE
Fluid is any substance that cannot
maintain its own shape or has no rigidity. Fluid is a term applied to liquids
and gases for they readily and they do not resist shearing stress. Fluids can
flow and alter its shape to conform to the outlines of its container. Both
liquids and gases have many characteristics in common. Liquids are nearly
incompressible, while gases are easily compressed. Liquids tend to have greater
density than gases. The gaseous phase of a substance usually exists at higher
temperature, hence gas molecules are able to break free from one another. Gases are able to escape from an open
container, whereas liquids cannot.
PRESSURE
Pressure
is the force applied per unit area, P = F/A where F is the force applied and A
is the area on which it is applied. Pressure is one of the important concepts
in fluids. The definition is valid for all phases of matter: solid, liquid or
gas. Examples of pressure: record player needle exerts pressure on a disc record, water near the bottom of a
pool exerts pressure on a swimmer’s eardrum, and atmospheric pressure changes
weather conditions. Measurements of pressure are common : Tires must be
inflated to correct pressure, blood pressure should stay within normal range,
and too much pressure in the eye (glaucoma ) can cause blindness.
The SI unit for pressure is Pascal ( Pa
); 1 Pa = 1 newton / m2. An atmosphere
( atm ) is also a unit of pressure, where one atm = the average pressure due to
the weight of the atmosphere at sea
level.
Pressure is as important as the force
creating it. If someone pokes you with his finger, you will certainly feel it.
If, however, a nurse pokes you with a hypodermic needle using the same amount
of force, you just don’t feel it – the needle breaks the skin. The same force
applied to a smaller area creates a larger pressure and has a much different
effect.
Fluids as well as solids, can exert
pressures. Consider the water in a square container. If the water has a mass of
10 kg, its weight of 98 N must be supported by the bottom of the container. If
the bottom has an area of 0.1 m2, then the
pressure due to the weight of the water
on the bottom is 980 N / m2. This
computation is valid only for containers with straight sides. Another example
is atmospheric pressure, which is caused by the weight of air. Atmospheric
pressure is 1.013 x 105
Pa
at sea level which means a column of air 1m on a side extending to the top of
the atmosphere weighs 1.013 x 105 N. Stationary fluids always exert forces
perpendicular to surfaces whether that direction is up or down, left or right.
The reason the force is always perpendicular to the surface is that fluids
cannot withstand shearing or sideways forces and therefore cannot exert
sideways forces.
CONVERSION
FACTORS FOR VARIOUS UNITS OF PRESSURE
Conversion to N / m2 or Pa
|
Conversion to atmosphere, atm
|
||||
1.0 atm
|
=
|
1.013
x 105 N / m2
|
1.0 atm
|
=
|
1.013
x 105 N / m2
|
1.0 dyne / cm2
|
=
|
0.1
N
/ m2
|
1.0 atm
|
=
|
1.013
X 106 dyne/
cm2
|
1.0 kg / cm2
|
=
|
9.8
x 104 N / m2
|
1.0 atm
|
=
|
1.03
kg / cm2
|
1.0 lb / in2
|
=
|
6.9 x 103 N / m2
|
1.0 atm
|
=
|
14.7 lb / in2
|
1.0 mm Hg
|
=
|
133 N / m2
|
1.0 atm
|
=
|
760 mm Hg
|
1.0 cm Hg
|
=
|
1.33 x 103 N / m2
|
1.0 atm
|
=
|
76 cm Hg
|
1.0 cm water
|
=
|
98.1 N / m2
|
1.0 atm
|
=
|
1.03 x 103 cm water
|
1.0 bar
|
=
|
1.0 x 105 N / m2
|
1.0 atm
|
=
|
1.013 bar
|
Problems
1. A woman wearing
high-heeled shoes places about 50 % of its
full weight on the single heel when
walking. If the woman has a mass of 60 kg,
determine the pressure on the ground under one heel if the area of contact is 2.25 cm2. How
would this pressure compare with the pressure exerted the elephant’s foot which is circular with
diameter of 30 cm assuming that it is standing on four legs
and with mass of 3 650 kg.
2. Determine the
pressure in N
/ m2 and atm exerted
by a phonograph needle on a record surface if the needle supports 2.25 g on a circular area
0.48 mm in diameter.
3. Calculate the
force exerted on one side of a 3 m by 5 m wall assuming normal atmospheric
conditions.
4. A hammer exerts a
force of 40 N on a circular surface 2 cm in diameter. Determine the pressure.
BOYLE’S
LAW
Robert Boyle ( 1627 –
1691
)
advanced the study of gases using an air pump made by Robert Hooke. Boyle
observed the relationship between the pressure and the volume of an enclosed
gas at constant temperature. Boyle’s law states that at constant temperature the
pressure exerted by a gas is inversely proportional to the volume in which it
is enclosed.
P a 1/ V ==> PV = k
or PV = constant,
where P is the gas
pressure, V is the volume and the value of
the constant depends on the initial conditions. A complete statement of Boyle’s
law includes the condition that both the temperature and the amount of gas must
be held constant. Alternatively, Boyle’s law can be written in the form P1V1 = P2V2 ; where
the subscripts 1 and 2 refer to the different physical states of the same
sample of gas with the temperature held constant.
This simple
dependence of pressure in a gas on the volume it occupies has many
applications. If the volume of the gas is doubled, the pressure is reduced by
half of its original value. An example is the air pressure in a tire, which results
from putting a large volume of air into the smaller volume of the tire, thus
making the pressure greater than the atmospheric pressure. Another example is
that to drink through a straw, you expand the volume of the air in your mouth, dropping
its pressure and allowing the larger atmospheric pressure to force the fluid up
the straw.
1. A cylinder with height of 20 cm and cross
sectional area of 0.40 m2 has a
close-fitting piston that may be moved to change the internal volume of
the cylinder. Air at 1.013 x 105 N/m2 fills the cylinder.
If the piston is pushed until it is within 8
cm from the end of the cylinder, what is the new pressure of
the air? Assume that the temperature of
the gas remains constant and that the volume of gas in the
gauge is small compared with the volume of
the cylinder.
2. A 150 cm3 gas initially at 500
kPa is allowed to expand until is pressure is reduced to 125 kPa. What is the volume of the gas?
PASCAL’S
PRINCIPLE
One pioneer in the Physics of fluids was the
French philosopher and scientist Blaise Pascal (1623 – 1662). He discovered an
important property of stationary fluids: They can be used to transmit pressure
to a place other than where the pressure is created and is not diminished while
in transit.
This is Pascal’s
principle :
Any pressure applied to a confined
fluid will be transmitted undiminished to all parts of the fluid.
This means that the applied pressure will be added
to whatever pressures already exist in the fluid. One manifestation of Pascal’s
principle is that the total pressure at the bottom of he lake is the sum of the
pressure due to the weight of water plus the atmospheric pressure. According to
Pascal’s principle, atmospheric pressure is transmitted undiminished to the
bottom of the lake. Because water is not rigid, it cannot support the weight of
the atmosphere without transmitting it to all parts of the lake. The effect of
atmospheric pressure often cancels or is negligible, it is tiresome to always
add atmospheric pressure to get total pressure. Gauge pressure is the
pressure above or below atmospheric pressure. Total pressure or absolute
pressure
is gauge pressure plus atmospheric pressure.
Pt = Pg + Patm
where Pt = total
pressure, Pg = gauge pressure, Patm = atmospheric pressure
An easy way to remember
this equation is to recall that a tire gauge reads zero when the tire is flat,
even though a flat tire with a large hole obviously contains air at atmospheric
pressure. The flat tire has a gauge pressure of
zero and a total or absolute pressure of 1 atm. We shall use the convention
that all pressures are gauge pressures unless otherwise specified. Most pressure-measuring
devices yield gauge pressures, and in most circumstances gauge pressure is of
greater concern than total pressure. In the application of Pascal’s principle,
it is important to note that applied pressure, not force, is transmitted
undiminished to all parts of the fluids. Consider a hydraulic system consisting
of two cylinders connected to one another by a tube and filled with a fluid as
shown in the figure below.
The fluid used in the
hydraulic system is incompressible. When a force F1 is exerted on the
small piston, fluid flows from the small cylinder to the large cylinder,
causing the large piston to move upward and exert an upward force to lift a
load connected to it. Pascal’s principle states that the pressure created by
exerting force F1 on the
small piston, P1 = F1 / A1, is
transmitted undiminished to all parts of the fluid. Therefore
P1
= P2 ==> F1 / A1 = F2 / A2
Pascal’s principle holds for gases as
well as for liquids, with some minor modifications due to the change in volume
of a gas when the pressure is changed. In most hydraulic systems a small force
is put into the system and causes a large force to emerge from the other end of
the system. The small cylinder is called the master cylinder and the large
cylinder the slave cylinder. For example, the car brake system have a master
cylinder connected to four slave cylinders, one for each wheel. A small amount
of force exerted by the driver is transformed into a larger amount of braking force
exerted by the slave cylinder on the brake drum of each wheel. The frictional
force exerted by the brakes on the brake drum causes the car to slow down speed
or to stop.
1. A hydraulic jack lifts a 1,200 kg car by
applying a force the small piston. If the diameters of the
large and small pistons are in the
ratio 10 : 1, and the diameter of the large piston is 6 cm,
determine the force exerted by the
operator on the small piston.
2. The large piston of a hydraulic press supports a dentist’s chair and the dentist
wants to lift the
patient by stepping on the pedal
directly on top of the small piston. Find the force exerted by the dentist if the patient plus chair
have a mass of 125 kg and the large piston has radius of 6 cm and
the small piston has radius of 1.2
cm.
Hyperbaric medicine treats
many physical problems through the application of high-pressure air or air-oxygen mixtures. Hyperbaric oxygen therapy uses a special chamber, sometimes called a
pressure chamber, to increase the amount of oxygen in the blood. Some, but not
very many, hospitals have a hyperbaric chamber. Smaller units may be available
in outpatient center. The air pressure inside a hyperbaric oxygen chamber is
about two and a half times greater than the normal pressure in the atmosphere. By
Pascal’s principle, the pressure is distributed throughout the hyperbaric
chamber. This helps your blood carry more oxygen to organs
and tissues in your body. The increased oxygen intake is useful in
treating a variety of problems such as wounds, especially infected wounds, heal more quickly. The therapy may
also be used to treat:
1.
Air or gas embolism
2.
Bone infections (osteomyelitis) that have not improved with other
treatments.
3.
Burns
5.
Certain types of brain or sinus infections
6.
Decompression sickness (for example, a diving
injury)
7.
Gas gangrene
9.
Provide enough oxygen to the lung during a
procedure called whole lung lavage, which is used to clean an entire lung in
patients with certain medical conditions.
10. Radiation
injury (for example, damage from radiation therapy for cancer)
11. Skin grafts
12. Wounds that
have not healed with other treatments (for example, it may be used to treat a
foot ulcer in someone with diabetes or very bad circulation)
PRESSURE
DUE TO WEIGHT OF A COLUMN OF FLUID
P = F /
A =
mg / A
but m
= rV
==> P = rVg / A and V = Ah
==> P = r A h g / A
==> P = h r g
The
pressure due to the weight of a column of fluid depends only of the depth in
the fluid and the density of the fluid. Pressure at the surface of a fluid is
zero since h is zero. As it goes deeper the pressure increases in proportion to
the increase in height.
Another manifestation of how pressure
depends only on depth and density, is found in the intravenous ( IV ) administration of fluids. The
pressure due to the IV fluid at
the entrance of the needle is proportional to h, the height of the surface
above the needle. Paths of the fluid has nothing to do with pressure but is
dependent on h, P = h r g. Applied
pressure can be adjusted by raising or lowering the IV bottle relative to the patient.
Consider a situation
where a medicine dropper is filled with fluid. First air is squeezed out of the
bulb with the tip beneath the fluid surface as shown in Figure 1. The bulb is
then released, and being elastic, returns to its original larger size. The
volume of the gas in the bulb is then increased while the pressure decreases to
a value less than the atmospheric pressure. The atmospheric pressure is then
able to push fluid into the dropper at height h. This situation is analogous to
how a hypodermic syringes are filled with fluid and how we drink soft drink
from a straw. The height h to which the fluid rises in a medicine dropper is
related to the pressure in the bulb, Pbulb , the lower the
pressure in the bulb the higher the fluid rises.
Figure 1
Figure 2
Consider the dropper
in figure 2. The total pressure must be the same at both points mark X, for
they are at the same vertical height. The total pressure at the outer X is
atmospheric, Patm. The
total pressure
At the inner X is the
pressure due to the weight of the fluid plus the pressure of the air in the
bulb which is h r g + Pbulb. The two
pressures must be equal, so
Patm = h r g
+ Pbulb solving for
h
h
= ( Patm – Pbulb) / r g
The pressure, Pbulb is an
absolute pressure. All quantities in the expression is constant except the
pressure in the bulb; the smaller this pressure, the higher h is. Since the
pressure in the bulb cannot be less than zero, there is a limit to how high h
can be. For water, the maximum value
of h
is 10.3 m when Pbulb is zero.
The larger the density of a fluid, the more difficult it is to raise h. The
largest value of h obtainable for mercury is 0.76 m only. The height to which fluid rises is directly related
to the pressure in the bulb, hence the height of a column of fluid can be used
to measure pressure. Many pressure-measuring devices are using this phenomena
to measure pressure.
MEASUREMENT OF
PRESSURE BASED ON PASCAL’S PRINCIPLE AND
P = h r g.
Pascal’s principle states that any
pressure applied to a confined fluid is transmitted undiminished to all parts
of the fluid. Thus, fluid can be use to transmit pressure to a gauge at a
convenient location. As an example, blood pressure can be measured without
putting a gauge into the body. Furthermore, pressure is transmitted undiminished,
so the measurement can be very accurate. In a blood pressure measurement, an
inflatable cuff is placed on the upper arm, and inflated until blood flow is
cut off in the brachial artery. Pressure is created by squeezing the bulb and
is transmitted by air in the tubes (a confined fluid) to the cuff and to the
gauge. The wall of the cuff transmits the pressure to the arm and through it to
the artery. When the applied pressure exceeds the heart’s output pressure , the
artery collapses. The person making the measurement slowly releases the air in
the cuff, lowering its pressure, and listens for flow to resume when pressure
in the cuff becomes lower than the maximum heart output.
Systolic
blood pressure, ( when the heart is contracted ) the maximum blood pressure
is recorded together with the lower pressure called the diastolic pressure (when the heart is relaxed between beats).
Diastolic pressure is the minimum pressure the circulatory system experiences
and can be detected as a change in the sound of the blood flow through the
partially restricted artery. The main point here, based on Pascal’s principle,
is that all these pressures are transmitted undiminished and the pressure read
by the gauge is truly representative of the pressure in the heart. It is important for the cuff to be at the same level as the heart in a
blood pressure measurement. Any effect
due to the weight of the air in
the cuff and connecting tubes is negligible because the density of air is very
small. The gauge can be placed at any convenient location, such as on the wall,
a table, the floor, or anywhere in between, but the cuff must be at the same
level as the heart.
Manometers and analogous
devices are in common used such that pressure units related to the use of these
devices have been developed. A manometer is
often a U-shaped tube filled with fluid usually water or mercury. A single-tube manometer filled with mercury is
commonly used for measuring blood pressure. The pressure in the cuff is h
r g, where h
is the height of the mercury and r is the density of mercury. Common blood
pressure measurements produces value of h ranging from 8 – 300 mm, so h is commonly
recorded in “millimeters of mercury” ( mm Hg ). The mm Hg is almost used universally to
measure blood pressure instead of performing the product of h
r g for each
individual measurement. A normal blood pressure of 100 over 80 means that the systolic pressure
is 120 mm Hg and the diastolic pressure
is 80 mm Hg. If the fluid used in the manometer is water,
h would also be measured in centimeters of water ( cm
H2O ).
Mercury
barometers
measure atmospheric pressure in a manner similar to manometers. A tube closed
at one end is filled with mercury and inverted with its open end in a dish of
mercury. Extreme care must be considered in order not to allow air inside the
tube. If the tube is long enough, part of the mercury will flow out of the tube
thereby leaving a vacuum at the top. In a mercury barometer, the Pbulb is zero (
an absolute pressure of zero is a vacuum ). The pressure due to the weight of
mercury must be equal to the atmospheric pressure. Another way of interpreting this is to say
that since above the mercury is vacuum, the atmosphere is able to raise the
mercury to a height h, where the
pressure due to its weight, P = h r g, equals
atmospheric pressure. Atmospheric pressure varies with altitude and weather conditions. The normal
atmospheric pressure has a value of h equal to 760 mm Hg.
Problems :
1.
A
nurse administers medication in a saline solution to a patient by infusion into
a vein in the
patient’s
arm. The density of the solution is 1000 kg/m3, and the gauge
pressure inside the vein is
2.7
x 103 Pa. How high above the insertion point must
the container be hung so that there is
sufficient pressure to force the fluid into
the patient ?
pressure must the scuba device
deliver air to the diver? How many atm would this pressure be ?
Note :The pressure at the diver’s depth is
greater than atmospheric pressure because of the weight of the water above the diver. If
the air breathed in is not at the same pressure as
the external pressure on
the diver’s chest, the excess pressure will collapse the chest.
Thus, the breathing
apparatus must deliver air to the diver at the pressure of the
surrounding water. ( use P = Patm + h r g )
3.
A tank is filled with water to a depth of 175 cm. What is the pressure at the bottom of
the tank
due to the water alone ? What is the
total pressure ?
4. Determine the maximum force in newtons
exerted by the blood on an aneurysm, or ballooning, by
the aorta, given that the maximum
blood pressure is 140 mm Hg and the area of the aneurysm is 25 cm2.
F = PA = ( h r g ) A = (0.140 m )( 13 600 kg/
m3 )
( 9.8 m /s2 )( 25 cm2
x1 m2/ 10 000 cm2) = 46.648 N
Note that the fluid involve is mercury,
hence we use 13
600
kg/
m3
for the density of mercury
and convert 140 mm to meter. The area 25 cm2 is likewise converted to m2 .
****** This is a
large force for the vessel to withstand, and there is a considerable risk that
the aneurysm will burst.
that the static water pressure at
home is 3 x 105 N /m2, how high above the home is the
surface
of the water in the tower ?
BUOYANT
FORCE AND ARCHIMEDES’ PRINCIPLE
A story has been told
that Archimedes ( 287 – 212 B. C. ) conceived of the principle that bears his
name after King Hiero of Syracuse asked him to determine the actual composition
of the King’s crown, which was alleged to be pure gold. Archimedes was ordered
to do so without damaging the crown.
According to legend, the Greek
scientist’s inspiration came to him as he lay partially submerged in his bath.
On getting into the tub, he observed that the more his body sank into the tub,
the more water ran out over the top. He immediately jumped out of the tub and
rushed through the streets naked, shouting excitedly in a loud voice “Eureka” (“I have found it”).
Archimedes’ principle states that :
A body whether completely or partially submerged
in a fluid, is buoyed upward
by a force that is equal to the weight of
the displaced fluid.
Buoyant
Force = weight of displaced fluid ==> FB = wfl
The principle applies both to liquids
and gases and to objects which are completely or partially submerged. How this principle allowed Archimedes to solve
the problem of the king’s crown is shown in the following problem :
The King’s crown is said to be solid gold but
maybe made of lead and covered with gold.
When it is weighed in air, the scale reads 0.475 kg. When it is submerged in
water, the scale reads 0.437 kg.
Is it
solid gold ? If not, what percentage by
mass is gold?
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