Density and Specific Gravity

(1)

DENSITY

AND

SPECIFIC GRAVITY

(2)

DENSITY



Maybe we have encountered this tricky

question:

question: ““ Which is heavier, 1 kg of cotton

Which is heavier, 1 kg of cotton

or 1 kg of

or 1 kg of nails?”,or

nails?”,or “Which

“Which is lighter, cork or

is lighter, cork or

plywood?”



A co

A cork

rk is lighter

is lighter than wood

than wood if the

if the volumes of

volumes of

both the cork and the wood are the same.



The preceding statement involves two

important physical quantities: volume and

mass.

(3)

DENSITY



Maybe we have encountered this tricky

question:

question: ““ Which is heavier, 1 kg of cotton

Which is heavier, 1 kg of cotton

or 1 kg of

or 1 kg of nails?”,or

nails?”,or “Which

“Which is lighter, cork or

is lighter, cork or

plywood?”



A co

A cork

rk is lighter

is lighter than wood

than wood if the

if the volumes of

volumes of

both the cork and the wood are the same.



The preceding statement involves two

important physical quantities: volume and

mass.

(4)

DENSITY



 DensityDensity is a measure of how matter is squeezed is a measure of how matter is squeezed

together in a given amount of space. It is defined as an together in a given amount of space. It is defined as an objects mass per unit volume. It is one important property objects mass per unit volume. It is one important property of a substance.

of a substance.

The density can be expressed as The density can be expressed as

ρ ρ = m = m / / VV where: where: ρ ρ = density (kg/m = density (kg/m33₎₎ m = mass (kg) m = mass (kg) V = volume (m V = volume (m33₎₎ 

 The higher the density, the tighter the particles areThe higher the density, the tighter the particles are

packed inside the substance. Density is a physical packed inside the substance. Density is a physical property constant at a given temperature and density can property constant at a given temperature and density can help to identify a

(5)



The density of solids is fairly constant, but for

fluids, it is more variable because fluids are

more compressible, especially gases.



Volumes of solids are usually expressed in cubic

meters (m

3

_{) and those of fluids in liters (L).}



The following may be helpful in converting from

liters to cubic meters and vice versa:

1 L = 0.001 m

3

1 m

3

= 1000 L

1 L = 1000 cm

3

1 mL = 1 cm

3

(6)

U

NITS OF DENSITY



The SI units for density are kg/m

3

. The imperial

(U.S.) units are lb/ft

3

(slugs/ft

3

). While people

often use pounds per cubic foot as a measure

of density in the U.S., pounds are really a

measure of force, not mass. Slugs are the

correct measure of mass. You can multiply

slugs by 32.2 for a rough value in pounds.

(7)

D

ENSITIES OF SOME COMMON SUBSTANCES

Material Density (kg/m3₎ Material Density (kg/m3₎ Air(1atm,200_C) _1.20 _Aluminum _{2.7 x 10}3

Ethanol 0.81 x 103 _{Iron, steel} _{7.8 x 10}3

Benzene 0.90 x 103 _Brass _{8.6 x 10}3 Ice 0.92 x 103 _Copper _{8.9 x 10}3 Water 1.0 x 103 _Silver _{10.5 x 10}3 Seawater 1.03x 103 _Lead _{11.3 x 10}3 Blood 1.06 x 103 _Mercury _{13.6 x 10}3 Glycerin 1.26 x 103 _Gold _{19.3 x 10}3 Concrete 2 x 103 _Platinum _{21.4 x 10}3

(8)

E

XAMPLE

1

An irregular-shaped piece of copper has an unknown mass. When it is placed in a cup of water filled to the brim, 1.26 mL of water overflows. What is the mass of the piece of copper? (density of copper = 8.9 x 103₎

Solution:

The volume of water that overflows is equal to the volume of the copper (1.26 mL).

Convert mL to m3_:

1.26 mL (1L/1000 mL)(0.001 m3_{/1L ) = 1.26 x 10}-6_m3

ρ = m/V; m = ρV

m = (8.9 x 103_{)(1.26 x 10}-6_m3₎

(9)

Find the density of a 0.1 m3_{aluminum having a mass of} 270 kg. Given: V = 0.1 m3 m = 270 kg ρ = ? Solution: ρ = m = 270 kg = 2,700 kg/m3 V 0.1 m3

E

XAMPLE

2

(10)

Find the density of a 51 grams gasoline that

occupies 75 cm

3

.

Given:

V = 75 cm

3

m = 51 g

ρ = ?

Solution:

ρ = m = 51 g = 0.68 g/cm

3

V

75 cm

3

E

XAMPLE

3

(11)

Determine the mass of a solid iron wrecking ball with a radius of 20 cm. Given: r = 20 cm ρ_iron = 7.8 g /cm3 m = ? Solution:

Solve first for the volume of the ball (sphere):

V = 4/3 πr3 = 4/3 π (20cm)3 = 3.35 x 104 cm3

Then, solve for the mass:

m = ρV = (7.8 g/cm3_{)(3.35 x 10}4_cm3₎

m = 2.61 x 105_g

(12)

E

XAMPLE

5

What is the density of a box that is 1.2 m x 2.5 m x 3.0 m and weighs 50 N? Solution: ρ = m ; W = mg V ρ = m = 50N/(9.8 m/s 2₎ _{= 5.10 kg = 0.57 kg/m}3 V 1.2m x 2.5m x 3.0m 9 m3

(13)

R

ELATIVE

D

ENSITY



Another variable, called specific gravity, is

used to identify materials.



Relative density, or specific gravity, is

the ratio of the density of a substance to

the density of a given reference material.

Specific gravity usually means relative

density with respect to water. The term

"relative density" is often preferred in

modern scientific usage.

(14)

R

ELATIVE

D

ENSITY



If a substance's relative density is less than one

then it is less dense than the reference; if greater

than 1 then it is denser than the reference. If the

relative density is exactly 1 then the densities are

equal; that is, equal volumes of the two

substances have the same mass.



If the reference material is water then a substance

with a relative density (or specific gravity) less

than 1 will float in water. For example, an ice

cube, with a relative density of about 0.91, will

float. A substance with a relative density greater

than 1 will sink.

(15)



Specific gravity is often used by

geologists to help determine the mineral

content of a rock sample. In industry,

specific gravity is used to determine the

concentrations of substances in aqueous

solutions.

(16)

E

XAMPLE

1 What is the specific gravity of gold?

Given:

ρ

_gold

= 19,320 kg/m

3

Solution:

specific gravity =

ρ

_gold

ρ

_w

= 19,320 kg/m

3

1000 kg/m

3

(17)

The specific gravity of aluminum is 2.7. Find its

density in kg/m

3

_.

Solution:

Sp.gr = ρ

_aluminum

ρ

_water

ρ

_aluminum

= (Sp. gr)(ρ

_water

) = 2.7(1000 kg/m

3

₎

ρ

_aluminum

= 2,700 kg/m

3

E

XAMPLE

2

(18)

What is the mass in kilograms of 1 liter of kerosene which has specific gravity of 0.8?

Given: V = 1 L Sp. gr = 0.8 m = ? Solution: Sp.gr = ρ_kerosene ρ_water ρ_kerosene = (Sp. gr)(ρ_water) = 0.8(1000 kg/m3_{) = 800 kg/m}3 m = ρV = (800 kg/m3_)(0.001m3_{) = 0.8 kg}

E

XAMPLE

3

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(20)

PRESSURE

Which can cause greater damage to the floor: the boots or the stiletto heels?

The woman wearing the stiletto heels may weigh lighter than the man wearing boots, but her stiletto heels are more damaging to the floor.

This is because when the woman is standing, her weight is concentrated on the small area under the forefoot and heels of each shoe. Force applied on a smaller area produces a larger amount of pressure.

(21)

 How can someone lie on a

bed of nails without getting hurt?

 A nail has a pointed tip,

which concentrates a hammer’s blow onto a small area and produces pressure big enough to part wood fibers and embed the nail.

 However, in the case of the

man lying on a bed of nails, because his weight is distributed on so many nails, the pressure created by each nail on his body is not enough to puncture him.

(22)

PRESSURE



Pressure is a scalar quantity. In standard

units, pressure is measured in newtons

per square meter (N/m

2

_{),also known as}

pascal (Pa). One pascal is the pressure

exerted by a force of 1 N on an area of

1m

2

.

(23)

B

LAISE PASCAL



The unit pascal was

named after Blaise

Pascal (1623 – 1662),

a french physicist and

mathematician, for his

experiment works on

the pressure exerted

by liquids and gases.

(24)

PRESSURE



Pressure is defined as the force acting per unit

(25)

PRESSURE



For a given magnitude of force, the pressure this

force exerts can be increased or decreased by

adjusting the area on which it is applied. Thus to

create a smaller amount of pressure, the area

should be made large.



It therefore makes sense that skis and snow shoes

are spread out and frogs have webbed feet. On the

other hand, a small area causes a bigger amount of

pressure. This is the reason why it hurts a lot when

someone thrusts his or her elbow or leans over

your thigh.

(26)

(27)

H

IGH PRESSURE



Each scissors and knife has a very small

surface area on its cutting edge so that a high

pressure can be exerted to cut something.



Each ice skate has a sharp blade that enables

the skaters to glide smoothly across the

surface of ice. The skater’s weight acting on a

small area of contact gives a high pressure on

ice. Ice will melt under high pressure.

Therefore, when skating, you are actually

gliding through a layer of water.

(28)

L

OW PRESSURE



Snowshoes reduce the pressure acting

on snow by increasing the area over

which the weight is spread. Thus, one

can walk on deep snow without sinking

in.



The large tires of tractors increase the

area of contact with ground and this

reduces the pressure exerted onto the

ground. Therefore, the tractors can move

without sinking into the ground.

(29)

E

XAMPLE

1 A 1.0 m x 1.0 m x 2.0 m block weighs 100 N.

Calculate the pressure under the block

when it is lying on its

a. rectangular surface; and on its

b. square surface.

1.0 m 2.0 m 1.0 m 1.0m 2.0 m 1.0 m

(30)

SOLUTION

The pressure depends on the area underneath the

block.

a. The area of the rectangular surface of the block

is

A = 2.0 m x 1.0 m = 2.0 m

2

P = _F_ = 100 N = 50 N/m

2

_{or 50 Pa}

A

2.0 m

2

b. The area of the square surface is

A = 1.0 m x 1.0 m = 1.0 m

2

P = _F_ = 100 N = 100 N/m

2

or 100 Pa

A

1.0 m

2

(31)

E

XAMPLE

2

Consider a car with a mass of 2,000 kg. The surface area of the part of each tire that is in contact with the floor is 15 cm x 20 cm. What is the pressure under each of the car’s tire?(Assuming that the car’s weight is distributed evenly by the four wheels.)

Solution:

P = _F_ = ¼ W = W = mg = (2,000 kg)(9.8 m/s2_{) =}

A A 4A 4A 4(.15 m x .20 m) P = 1.63 x 105_Pa

(32)

EXERCISE

A box has a mass of 18 kg. It has a height of 0.55

m. and bottom dimensions of 0.70 m x 0.60 m.

a. What force does the box exert on the floor?

b. What pressure does the box exert on the

floor?

Solution:

a. F = mg = 18 kg(9.8 m/s

2

) = 176.4 N

b. P = F =

176.4 N

= 420 Pa

(33)

 The a t m o s p h e r e is a deep ocean of gases piled several

kilometers above us. The weight of all those gases causes a downward pressure on us. The pressure of the atmosphere helps you sip your drink using a straw. By emptying your mouth of air, you create a partial vacuum, allowing the atmosphere to push your drink up the straw and making the liquid move against the flow of gravity.

(34)

A

TMOSPHERIC PRESSURE

 Atmospheric pressure is the force per unit area exerted

into a surface by the weight of air above that surface in the atmosphere of Earth (or that of another planet). In most circumstances atmospheric pressure is closely approximated by the hydrostatic pressure caused by the weight of air above the measurement point.

 Low-pressure areas have less atmospheric mass above

their location, whereas high-pressure areas have more atmospheric mass above their location. Likewise, as elevation increases, there is less overlying atmospheric mass, so that p r e s s u r e d e c r e a s e s w i t h i n c r e a s i n g elevation.

(35)



The pressure of the atmosphere at sea level is

about 1.013 x 10

5

_{Pa. This means that every}

square meter on the Earth’s surface is weighed

down by a force of 101.3 kilonewtons (kN).

This is like being squashed under the weight of

two male elephants.

(36)

U

NITS OF PRESSURE



The pressure at any point in the Earth’s

atmosphere may be expressed in a

non-standard unit called

a t m o s p h e r e

(atm). A

pressure of 1.013 x 10

5

_{Pa is equal to 1 atm.}

1.013 x 10

5

_{Pa = 1 atm}



Other non-standard units of pressure are

b a r ;

millimeter of mercury

(m m Hg),

which is also

known as

t o r r ;

and pound per square inch

(lb/in 2 o r p si).

The bar and millibar (mbar) are

(37)

E

VANGELISTA TORRICELLI

 The unit mm Hg is used in

measuring blood pressure and is derived from the use of the mercury barometer, a pressure gauge invented by the Italian inventor, Evangelista (1608-1647).

 Evangelista Torricelli

discovered Torricelli's Law, regarding the speed of a fluid flowing out of an opening, which was later shown to be a particular case of Bernoulli's principle.

(38)



The following equivalences are useful in the

conversion of pressure measurements:

1 bar = 10

55

Pa

1 millibar = 10

22

Pa

1 atm = 1.013 x 10

55

Pa

= 1.013 bar

= 760 mm Hg

= 14.70 psi (lb/in

22

₎₎

1 kPa = 1000 Pa

1 mm Hg = 1 torr

= 133 N/m

22

(39)

EXAMPLE



The pressure of the atmosphere is 1.013 x

10

55

Pa. How much force does the

atmosphere exert on the roof of a house,

assuming that it is a flat roof with

dimensions of 9.0 m x 12.0 m?

Solution:

F = PA

F = (1.013 x 10

55

_N/m

22

_{)(9.0 m x 12.0 m)}

F = (1.013 x 10

55

N/m

22

)(108m

22

))

F = 1.09 x 10

77

_N

(40)

F F F F F F F F A A₁₁ A A₂₂ A A₃₃ A A₄₄ A A₆₆ A A₅₅

Force acts perpendicula

Force acts perpendicularly on rly on each face. Pressure is theeach face. Pressure is the force acting on a unit area.

force acting on a unit area.

Air

Air pressure pressure on on the the top top face face of of the the cube cube is is duedue to the weight of the air above it. Thus, the force to the weight of the air above it. Thus, the force on the top face is equal to the weight W of the on the top face is equal to the weight W of the air column above it. Hence,

air column above it. Hence, F = W F = W = mg = mg Where:

Where: m m = = mass mass of of air air columncolumn and

and g g = = acceleration acceleration due due toto gravity

gravity But:

But: mass mass m m = = volume volume V V x x densitydensity ρρ and

and volume volume V V = = area area A A x x height height hh Therefore: Therefore: m m = = VVρρ = = AhAhρρ.. Since Since F F = = mgmg = Ah = Ahρρg.g. From From P P = = F F = = AhAhρρgg A A AA We now have We now have P = hP = hρρg.g.

(41)

P

RESSURE DEPENDS ON HEIGHT AND DENSITY



From the equation P = ρgh, since g is

constant for a specific location, air pressure

P depends on

h e i g h t h

and

d e n s i t y

ρ

of

air. As height increases, or ad one goes

higher into the atmosphere, atmospheric

pressure decreases. Hence, atmospheric

pressure is higher in lowlands than on

mountains.

(42)



The human body is used to atmospheric

pressure on the earth’s surface. Airplane

passengers

eventually

feel

pressure

changes in their ears when the airplane

ascends or descends.



The ears are built-in body pressure

measurers. They feel air pressure changes

as painful sensations. Air pressure inside

aircrafts should approximate atmospheric

pressure on the earth’s surface to ensure the

passengers’ comfort and well-being.

(43)

L

IQUID PRESSURE



Do you feel the pressure of water acting against

your eardrums when you swim under water?

Observe that the deeper you swim, the greater

the pressure you feel. What causes this

pressure?



It is caused by the weight of the water above

you that pushes against you. If you swim three

times as deep, the weight of the water above

will also be three times. Therefore the water

pressure will be three times as great. Thus,

(44)

P

RESSURE VARIES WITH DEPTH



Consider

a

beaker

which contains water at

a height h. If you double

the height of water (thus

doubling the weight),

the pressure against the

bottom of the beaker

becomes twice as great.

(45)



Furthermore, if the liquid is two or three

times as dense, liquid pressure is

correspondingly two or three times as

great at any depth.



For instance, if you swim in the sea, which

is denser than ordinary water, the pressure

will be greater proportionally. Thus,

l i q u i d

p r e s s u r e v ar i es w i t h d en s i t y.

(46)

P

RESSURE VARIES WITH DENSITY

 These vessels differ in shape

but contain water of the same height.

 The pressure at the bottom

is independent of the shape of the container provided that the heights of the liquids are exactly the same.

 This means that the higher

the column of liquid, the greater is its pressure: and the more dense the liquid is, the greater too is the pressure.

(47)



This is the formula for hydrostatic pressure. The

height h is always the distance from the surface of

the fluid down to the point that is being studied.

Thus, it is better to use the word

d e p t h

to refer to

the variable h in the equation.



Since both

ρ

and

g

are constants, it is clear that

hydrostatic pressure varies only with the depth of

the fluid column.



A scuba diver, for example, experiences greater

pressure as he/she dives deeper. This increase of

pressure with depth puts limits on scuba diving.

(48)



Another implication of

the equation, P = ρgh,

is that for the same

fluid, pressure at the

same level is the same

everywhere.



This is what we mean

when we say that

“water seeks its own

level”, as shown in the

figure.

(49)



The water pressure at the faucets in our home

depends on the height of the water level at the

reservoir. However, pumps are often used to

help increase the water pressure. The water

level at the reservoir during rainy season is

higher than it is during dry season.



In some areas where there is no water system,

people erect their own water tanks which they fill

by pumping water from deep wells. The higher

the tank, the greater the pressure available at

the water taps. It is clear that the water pressure

does not depend on the diameter of the tank.

(50)

 Suppose a rectangular tank is filled with a liquid of

density ρ to a depth h . The liquid pressure is uniform at

the bottom because all points are at the same depth from the surface the liquid. The pressure is not constant at the sides of the tank, however. The pressure varies from zero at the top, where h is zero, to maximum at the

bottom where the pressure is equal to ρgh.

 To get the force acting on the side, the area is multiplied

by the average pressure. For a rectangular side, the average pressure is the average of the pressure at the top and the pressure at the bottom. Or the average pressure is:

(51)

EXAMPLE



A rectangular tank 4 meters long, 2 meters wide,

and 1 meter high is filled with water. Find the:

a. liquid pressure at the bottom;

b. force on the bottom due to the liquid F

₁

;

c. total force on one side of the tank due to

the liquid F

₂

.

F₁ F₂ 1 m 2 m 4 m

(52)

SOLUTION

a. P = ρgh = (1000 kg/m3)(9.8 m/s2)(1m) = 9,800 N/m2 b. F₁ = PA = (9,800 N/m2)(4m x 2m) = 78,400 N

c. The total force on one side of the tank is the average

pressure multiplied by the area of the side. The average pressure is:

P_ave = ½ ρgh = ½ (9,800 N/m2_{) = 4,900 N/m}2

The area of the side is the area of a rectangular of length 4m and of width 1 m. Hence the force is:

(53)

EXAMPLE



Calculate the pressure on a scuba diver at a

depth of 45.0 m.

Given:

ρ = 1030 kg/m

3

h = 45 m

P = ?

Solution:

P = ρgh

P = (1030 kg/m

3

_)(9.8m/s

2

_{)(45 m)}

P = 454,230 Pa

(54)

E

XAMPLE



A man dove 35 m under the sea. If the seawater

has a density of 1.03 g/cm

3

_{, how much pressure}

in N/m

2

acts on the body of a man?

Given:

ρ = 1.03 g/cm

3

h = 35 m

P = ?

Solution:

P = ρgh

P = (1030 kg/m

3

_{)(9.8 m/s}

2

_{)(35 m)}

P = 353,290 N/m

2

(55)

EXAMPLE



Calculate the pressure of water at the bottom of

a swimming pool which is 1.8 m deep.

Given:

ρ = 1000 kg/m

3

h = 1.8 m

P =?

Solution:

P = ρgh

P = (1000 kg/m

3

_{)(9.8 m/s}

2

_{)(1.8 m)}

P = 17,640 N/m

2

_{or 17,640 Pa}

(56)

EXERCISE

 A swimming pool is 50 m long and 25 m wide. The deep

end has a depth of 2.0 m and the shallow end, 1.22 m. What are the pressures at the bottom of

a. the deep end; and b. the shallow end. Solution:

a. P = ρgh = (1000kg/m3_{)(9.8 m/s}2_{)(2 m) = 19,600 Pa}

(57)

EXERCISE



At a depth of 5.0 m, what is the pressure

experienced by a fish? (Assuming that a fish is

swimming at freshwater)

Given:

h = 5.0 m

ρ

_w

= 1000 k/m

3

Solution:

Neglecting the effect of atmosphere,

P = ρ

_w

gh

P = (1000 k/m

3

_{)(9.8 m/s}

2

_{)(5 m)}

(58)

 Eugene is an expert diver and is exploring an underwater

wreck. If he is at the surface, at what depth will he experience double the pressure that is on him now?

Given:

P_i = 1.01 x 105_Pa

Solution:

For Eugene to experience double the pressure: P_f = 2 P_i Therefore, ∆P = P_f – P_i = 1.01 x 105_Pa

And the change in depth should be:

∆h = ∆P = 1.01 x 105 _{Pa ______}

ρ_wg (1000kg.m3_)(9.8m/s2₎

∆h = 10.31 m

(59)

MEASURING FLUID PRESSURE

 The pressure of fluids is usually measured in terms of

gauge pressure and not absolute pressure. The gauge pressure is the pressure determined by a measuring-device and is equal to the di fference between the p r e s s u r e o f t h e f l u i d a n d t h e p r e s s u r e o f t h e atmos phere.

 For example, when you use a tire gauge, the reading on

the gauge tells the difference between the pressure of the air inside the tire (absolute pressure of the air) and the pressure of the atmosphere. This pressure difference is the pressure gauge.

(60)

MANOMETER

 Another instrument which is used to measure gauge

pressure is the manometer . A manometer may be made of a U-shaped tube containing a liquid, either mercury or water, which we will call the gaug e fl uid .

 One end of the tube, the open end, is exposed to the

atmosphere; the other end, the closed end, is connected to a vessel containing the fluid whose pressure is being measured, or the test fl uid .

 Both the atmosphere and the test fluid exert pressure on

the gauge fluid. The difference between the levels of the gauge fluid in the two arms of the U-tube is the pressure of the test fluid.

(61)

MANOMETER

 At the bottom of the closed

arm, the pressure is the sum of the absolute P_a of the test fluid and the pressure due to the column of gauge fluid in the arm. Since the column has a height of y₁, the pressure exerted by this column is ρgy₁. Thus,

(62)

 At the bottom of the open

arm, the pressure is the sum of the pressure of the atmosphere P_atm and the pressure exerted by the column of gauge fluid with height y₂.

(63)

 The pressures at the

bottom of the closed arm and that at the open arm are equal because they are the pressures at the same point on the U-tube. We can thus equate the expressions for these two pressures:

P_{closed arm} = P_{open arm}

P_a + ρgy₁ = P_atm + ρgy₂

 Rearranging the equation

to isolate Pa, we get: P_a = P_atm + ρgy₂ - ρgy₁ P_a = P_atm + ρg(y₂ - y₁)

(64)

 Thus the difference

between the gauge fluid levels in the two arms of the tube, y₂ – y₁, determines the absolute pressure of the test fluid. If we let y₂ – y₁ = h, we get another expression for absolute pressure.

P_a = P_atm + ρgh

 The difference P_a – P_atm is

the gauge pressure of the manometer.

Equation can be rewritten as: P_a – P_atm = ρgh

(65)

M

ERCURY BAROMETER

 Invented by Evangelista

Torricelli in 1643, this instrument is essentially a glass tube of mercury that is inverted into an open dish. The mercury spills into the dish but not completely because the atmosphere pushes on the mercury, so there is always some mercury left inside the tube.

 The height of the remaining

mercury inside the tube depends on the pressure of the atmosphere.

(66)

 The top of the mercury column

contains only mercury vapor, which has negligible(essentially zero) pressure. At the level of the mercury in the dish, the pressure due to the column of mercury is ρgh, where h is the

height of the mercury column above the mercury dish.

 This same pressure is equal to

atmospheric pressure since the dish is open to the atmosphere, so that in a mercury barometer,

P_atm = ρgh

(67)

 At sea level (at normal

level), the height of the column in a mercury barometer is always equal to 760 mm. Thus the value of atmospheric pressure may be given as 760 mmHg. The use of the mercury barometer ha also resulted to the use of millimeter of mercury as a unit of pressure.

(68)

A

NEROID BAROMETER

 An aneroid barometer , invented in 1843 by French scientist

Lucien Vidie uses a small, flexible metal box called an aneroid cell (capsule), which is made from an a l l o y o f b e r y l li u m a n d c o p p e r . The evacuated capsule (or usually

more capsules) is prevented from collapsing by a strong spring. Small changes in external air pressure cause the cell to expand or contract.

 This expansion and contraction drives mechanical levers such

that the tiny movements of the capsule are amplified and displayed on the face of the aneroid barometer. Many models include a manually set needle which is used to mark the current measurement so a change can be seen. In addition, the mechanism is made deliberately "stiff" so that tapping the barometer reveals whether the pressure is rising or falling as the pointer moves.

(69)

A

NEROID BAROMETER

(70)

A

NEROID BAROGRAPH

 A barograph is a recording aneroid barometer. It

produces a paper or foil chart called a barogram that records the barometric pressure over time.

 Barographs use one or more aneroid cells acting through

a gear or lever train to drive a recording arm that has at its extreme end either a scribe or a pen. A scribe records on smoked foil while a pen records on paper using ink, held in a knob. The recording material is mounted on a cylindrical drum which is rotated slowly by clockwork. Commonly, the drum makes one revolution per day, per week, or per month and the rotation rate can often be selected by the user.

(71)

 Since the amount of

movement that can be generated by a single aneroid is minuscule, up to seven aneroids (so called Vidie-cans) are often stacked "in series" to amplify their motion. It was invented in 1843 by the Frenchman Lucien Vidie (1805 –1866).

(72)

B

OURDON GAUGE

 It is a type of aneroid

pressure gauge

consisting of a flattened curved tube attached to a pointer that moves around a dial. As the pressure in the tube increases the tube tends to straighten and the pointer indicates the applied pressure. It is named after Eugène Bourdon (1808-84), French hydraulic engineer, who invented it.