07 Pressure.pdf

Lesson Objectives

At the end of the lesson you should be able to…

• Define the term pressure in terms of force and area.

Lesson Objectives

At the end of the lesson you should be able to…

• Apply the relationship pressure due to a liquid column = height of column x density of the liquid x gravitational field strength to new situations or to solve related problems.

• Describe and explain the transmission of pressure in hydraulic systems with particular reference to the hydraulic press.

• Explain the effects of atmospheric pressure.

• Describe how the height of a liquid column may be used to measure atmospheric pressure i.e. barometer.

•

Pressure

•

Pressure in Liquids

•

Gas Pressure

What is Pressure??

• It is the

perpendicular force

acting per

unit

area

.

p

= ---

where

p

: pressure

F

: force (in N)

A

: area the force is acting on (in m

)

A

What is Pressure??

• Pressure is a

scalar

quantity

• The S.I. Units for pressure is

Newton per square metre (Nm

)

or

Consider 2 forces…

F

= 10 N

F

= 10 N

Do they exert

the same

pressure?

Consider 2 forces…

F₂ = 10 N F₁= 10 N

A = 1 cm2

A

F

p



01

.

0

01

.

0

10





10

10

10





10

10



Pa

10

1





A

F

p



01

.

0

01

.

0

30

cos

10







10

66

.

8





8

.

66



10

Pa

30°

Factor affecting Pressure: Area

Given a mass of 5 kg, the pressure on a larger area will be lesser than that on a smaller area

5 kg

Factor affecting Pressure: Force

Doubling the force on the same area also doubles the pressure

5 kg

h

5 kg

Applications

• Skiing and surfing

• Studs in soccer boots

Applications of Pressure

• Skis

– A large area of ski reduce the pressure acting on the snow

– Skier will not sink into the snow.

• Studs on soccer boots

– small area of contact with the ground increases the pressure acting on the ground.

Floor vs Bed

Why is sleeping on a bed more comfortable?

Floor Bed

Small total area of contact Large total area of contact Hard, does not conform to body Soft, conforms to body

LOW PRESSURE

Which of these cases is there greater pressure exerted?

(i) an elephant of mass 4000 kg standing on one feet of average size of 0.1m2

(ii) a lady weighing 40kg standing on one stiletto heel of an area of 1cm2

Area of heel of a Stiletto = 1 cm2

= 1 x 10-4 _m2

Pressure = 400 / 1 x 10-4

= 4000 kPa

Killer Heels!

Weight of an Elephant = mg

= 4000 x 10 = 40000 N

Pressure = 40000 / 0.1 = 400 kPa

Example 1:

Weight of woman = mg

A glass block of dimensions 25.0 cm by 10.0

cm by 6.0 cm weighs 100 N. Calculate the

least and greatest pressures it can exert when

resting on a horizontal table.

Dimensions : 25.0 cm x 10.0 cm x 6.0 cm

Weight :100 N

Example 2:

P

= F / A

= 100 / (0.25 x 0.10)

=

4000 Pa

P

= F / A

= 100 / (0.10 x 0.06)

A liquid exerts pressure because of its weight.

The pressure inside a volume of liquid depends on the depth below the surface of the liquid. The

deeper it is, the greater the weight of the overlaying liquid and thus the greater the

pressure.

water spurs out furthest and fastest from the lowest tube

• Pressure at a certain depth from the surface of a liquid is given by

where

p is the pressure in Pa



g is the gravitational field strength in Nkg-1 h is depth in m

Liquid Pressure

How do we obtain

p

=



gh

?

• Consider a column of liquid of density



and base area A as shown.

• Pressure at the base of the liquid column is given by

h

A W

p = F / A

= W / A

• Depends on the

density

and

depth

of the

liquid.

• Does not depend on the

volume

or

cross

sectional area

of the liquid.

same depth pressure at the top of

each tube is equal

pressure at the bottom of each

tube is equal

water spurts out equally fast and reaches the same

distance away from each hole

water

tin

Liquid Pressure

• The pressure at one depth in a liquid

Empty plastic bottle and balloon at the surface

Balloon at 50 feet below the surface

Empty plastic bottle and balloon at the surface

Pressure Changes Underwater

Plastic bottle at 50 feet below the

Example 3:

Given the diagram below and density of

water to be 1000 kgm

, calculate the

density of the methylated spirit.

mercury water Methylated

spirit

12 cm 15 cm

Pressure due to liquid at A and B is the same.

Example 3 (Solution):

mercury water Methylated

spirit

12 cm 15 cm

p

= p



gh

=



gh



(0.15)=(1000)(0.12)



=

800 kgm

side view of a dam

dam water

land

Example 4:

Why is the shape of the

wall

of the dam

slanted towards the bottom?

This thickness of the wall of the dam increases downwards because the deeper it is, the greater

the water pressure. A thicker wall is required to

Find the pressure acting on a shark when it is (i) at the surface

(ii) 15 m in the water.

Density of sea water = 1000 kg m-3 ;

gravitational field strength = 10 Nkg-1 ;

atmospheric pressure, P_o = 1.01  105 _Pa

(i) At the surface, only atmospheric pressure P_o

acts on the shark.

Pressure at the surface = P_o = 1.01 x 105 _Pa.

(ii) At 15 m in the water, ie. h = 15 m, the pressure acting on it is

P = P_o +



= P_o + (1000  10 x 15)

= (1.01  105) + (1.5  105)

= 2.51  105 _Pa

Pressure at the 15 m deep is 2.51  105 Pa.

A vessel with vertical sides and a base

area of 0.050m

contains mercury of

density 13600kgm

and depth 6.0m.

Take atmospheric pressure as 1.01



10

Pa.

Calculate

(a) the weight of mercury in the vessel and

(b) the pressure

due to

the

mercury

on the

base of the tank

(c) the pressure that exerts

on the base

of

the tank.

Did You Get it Right?

(a)

W = mg

= (V



)g

= (Ah)



g

= 0.050 × 6 × 13600 × 10

= 40800 N or 4.08 x 10

N

Example 6 Solution:

Did You Get it Right?

(b) P

= F/A = W/A

= 40800 / 0.05

= 816000 Pa or 8.16 x 10

Pa

or

P

=



gh

= (13600)(10)(6)

= 816000 Pa or 8.16 x 10

Pa

(c)

P

=

P

+

P

=

1.01



10

+ 8.16 x 10

= 917 000 Pa

or = 9.17 x 10

Pa

2.1 Transmission of Pressure In Liquids

Hydraulic systems work by using liquids

under pressure.

Properties used:

•liquids are incompressible.

•if pressure is applied to trapped liquid, the pressure is

P

= P

Since

F

F

A

A

A

F

A

F









Outcome:

A

small effort

applied on a small piston

can lift a

larger load

on a bigger piston.

Applying Principle of Conservation of Energy,

y y

x

x

d

F

d

F







Work Done by Work Done By Smaller Force Larger Force =

Any assumptions?

No friction at the surface in contact with the pistons. i.e. no heat loss as heat or sound.

2.1 Transmission of pressure in liquids

Applying Principle of Conservation of Energy,

y y

x

x

d

F

d

F







The diagram shows a simplified form of a hydraulic press. A force of 15N is exerted on the small piston of area 0.025m2 _{. The large piston has an area of 0.5}

m2. Calculate

(a) the load that can be supported at the large piston. (b) the distance moved by the small piston when the large piston moves 0.15 m.

Pressure at X and Y is the same

P

=

P

A

F

A

F



A

A

F

F



025

.

0

)

5

.

0

)(

15

(



=

300 N

Max load supported at large piston is

300 N

(a)

W

=

W

y y

x

x

d

F

d

F



F

d

F

d



15

)

15

.

0

)(

300

(



=

3 m

Distance moved by small piston is

3 m.

(b)

Applying Princ. Of Conservation of Energy,

Atmospheric Pressure

• The atmosphere is

the layer of air that

surrounds the earth.

• The

weight of

the

layer of

air

exerts a

pressure

on

the

Atmospheric Pressure

• Pressure exerted by

this layer of air at sea

level at 0

C is

1.013



10

Pa

or

Atmospheric Pressure

Atmospheric Pressure

 Because the pressure inside our bodies( 1 atmosphere) is almost the same as the

external pressure and so balances it.

Example 8:

Do mountain climbers experience the same atmospheric pressure as they climb higher?

Air

gets

thinner

at

high altitudes

,

hence the

atmospheric pressure

decreases

.

*Point to Note:

Atmospheric pressure

At high altitudes where

air pressure is lower,

 there is pressure difference

between body and

atmospheric pressure

 breathing is difficult and nose

bleeding may occur

 hence, modern aircraft have

Applications of Atmospheric Pressure

• Drinking with a straw

By sucking, we decrease

the pressure in the straw.

Atmospheric pressure

•

Syringe

To draw liquid into the syringe, its piston is drawn upwards. This decreases the

pressure within the cylinder.

Atmospheric pressure

acting on the surface of the liquid drives the liquid into the cylinder through the nozzle.

Applications of Atmospheric Pressure

•

Suction Caps

To put the suction cap in place, they are

pressed in to force out the air within to create a partial

vacuum.

The greater

Measuring

• Torricelli -- Italian physicist who invented

the

mercury barometer

Torricelli 1608 -1647

Torricelli inverted a glass tube filled with mercury into another container of mercury; the mercury in the tube "weighs" the air in the atmosphere above the container.

A barometer is a device that measures air (barometric) pressure. It measures the weight of the column of air that extends from the instrument to the top of the atmosphere.

vacuum

h _{exerted by} Pressure

Measuring atmospheric pressure

Atmospheric pressure can be measured using a

simple

mercury barometer

.

Trough

Thick Walled Glass Tube

Vacuum

Mercury Barometer

• Pressure P_A = P_{atm =} P_B

• P_B = _Hg  g x h

• Therefore, P_atm = _Hg  g x h • E.g. Atmospheric Pressure is

equal to the liquid pressure exerted by a column of liquid of height, h

Atmospheric Pressure =





g x h

h

vacuum

Pressure exerted by atmosphere

• From experiments, at

atmospheric pressure,

mercury

column

is

found to be

760 mm

Hg

.

Mercury Barometer

h

vacuum

Pressure exerted by atmosphere

• Note:

– if the barometer is titled, the vertical height is taken and not the diagonal

length. h

vacuum

Pressure exerted by atmosphere

Example 9:

Find the pressure (in mmHg) at A, B, C and D as shown in the figure.

*Note:

Example 9 (solution):

P

=

0 mmHg (vacuum)

P

= 960 – 500

=

460 mmHg

P

= 960 – 200

=

760 mmHg (P

)

P

= 960 – 100

• For mercury,

P

= ρ

g

h

1.013 x 10

= (13600) (9.81)

h

h

= (1.013 x 10

) / (13600)(9.81)

= 0.75927…

m

• For water,

P

= ρ

g

h

1.013 x 10

= (1000) (9.81)

h

h

= (1.013 x 10

) / (1000)(9.81)

= 10.32619776…

m

10 m

water air removed

Measuring Atmospheric Pressure

Water Barometer

Water can be used in a barometer instead of

mercury. The glass

Measuring

Gas

• both arms exposed

to the atmosphere

• atmospheric

pressure exerted on

both surfaces is the

same

X Y

mercury

Manometer

The manometer is an instrument used to

pressure at X (gas pressure) = pressure at Z Y X from gas supply Y Y h

Pressure at X can

be

found

by

adding the height

difference

h

to the

atmospheric

pressure at Y

Manometer

The manometer is an instrument used to

Gas pressure at B is given by,

P_B = P_o + Pressure due to liquid column AC P_B = P_o + hρg

where P_o = atmospheric pressure ρ is the density of the liquid used

h is the length of the liquid column AC

The figure shows a

mercury manometer

connected to a large vessel containing some neon gas. Given that

atmospheric pressure is 760 mm Hg. Calculate the pressure of the neon gas in

cm Hg and Pa.

*Given: Density of mercury is 13600 kgm-3

Gravitational field strength is 10 Nkg-1

Example 10:

P_atm

Pneon

P

>

P

P

=

P

+

P

=

760

+

200

=

960 mmHg

P_atm

P_neon

P

= ρ

gh

= (13600)(10)(0.96)

= 130560

=

1.31 x 10

Pa

Example 11:

The figure shows a

mercury manometer

connected to a gas at pressure P_g. Given that

atmospheric pressure, P₀ is 76 cm Hg. Deduce the pressure of the gas in cm Hg and Pa.

*Given: Density of mercury is 13600 kgm-3

Example 11:

P_atm

P_gas

P

<

P

P

=

P

–

P

=

76

-

10

Example 11:

P_atm

P_gas

P

= ρ

gh

= (13600)(10)(0.66)

= 89760

Atmospheric pressure measured by Used in may be Hydraulic

jack Hydraulic brakes Solid pressure

= Weight / area

Hydraulic system Liquid pressure

= hg

Barometer

Gas pressure

Manometer

Pressure = force _area

measured by