LESSON 1.8 - PRESSURE
Objectives
On completion of this lesson, you should be able to:
Define: Pressure,
Atmospheric Pressure, The Pascal, Compressibility.
Explain: The effect of depth and density
on pressure exerted by liquids and gases under the
influence of gravity. Explain how the compressibility of
the atmosphere affects the pressure it exerts at various
altitudes.
Calculate: Pressure exerted by a liquid
at a particular depth. The force exerted on a particular
area by a liquid or gas under pressure.
MINI LAB
CHOICE OF
ACTIVITIES
- Can with holes: Drill
three holes at various distances from the bottom
on the side of a large tin can. Fill the can with
water and observe the effect of depth on the
speed of the water leaving through the holes.
- Hydraulic jack: Use a
hydraulic jack to show how a small force on a
small area can be converted to a larger force on
a larger area.
Pressure: The force
acting on a certain area.
Fluid: A material
that can flow: usually a liquid or a gas. Some materials
flow extremely slowly.
Density: The mass
per unit volume of a substance.
g: The
gravitational field vector. The gravitational force on an
object in a vertical direction towards the earth. At the
earth’s surface this is equivalent to 9.81 N/kg.
Pascal: The SI unit
of pressure. 1Pa. is equivalent to 1 N acting on 1 m2.
Some authors prefer to use N/m2 for reporting
pressures in the SI system of units.
Compressibility: The
ability of a material to reduce its volume when exposed
to an increase in pressure. Gases are compressible.
Liquids are generally regarded as incompressible.
Pressure
Pressure is the force that is exerted on a particular
area. A moderate force exerted on a large area creates a
small pressure. A moderate force on a small area creates
a large pressure. The unit of pressure is the Pascal.
This is equivalent to a force of one Newton acting on 1
square meter.
It’s more convenient to report pressures in terms of
kilo-Pascals or even mega-Pascals.
Pressure in fluids
Because liquids and gases can easily change shape,
pressure at any point in a liquid or gas exerts a force
in all directions.
Pressure in fluids
depends on gravity, depth and density
Liquid pressures are often due to gravity. The deeper the
liquid, the greater the pressure.
Scuba divers are well aware of this because they use more
air at higher pressures and have limitations on how long
they can stay down at higher pressures.
For example: At a depth of 10 meters in water, the
pressure is roughly twice that of the atmosphere
The pressure at any point in a liquid depends on the
force of gravity, the density of the liquid and the
height of liquid above the point.
Atmospheric pressure: Normal
atmospheric pressure is roughly 101300 Pa. This is due to
the weight of air above the earth’s surface. Nearly
the earth’s entire atmosphere is located within 30
kilometers of the earth’s surface.
The mass of air directly above a 1 meter-square on the
ground is 10 326 kg. The weight of this air creates a
pressure at ground level of 101300 Pa.
Because air is compressible, there is more air in a cubic
meter close to the earth’s surface than there is
further up. This is why aircrafts that fly at high
altitudes need to be pressurized and oxygen masks are
needed if the pressure is lost. Because air is
compressible, its density changes as its pressure changes
and the formula for calculating air pressure is not as
simple as that for liquids.
Hydraulic equipment
Using pistons with different sizes can cause hydraulic
equipment to create large forces. A hydraulic jack uses a
small piston to create a high pressure in the liquid
connected to a larger piston. This pressure on the larger
piston creates a much larger force that enables the jack
to lift a heavy object. Hydraulic brakes in motor
vehicles and hydraulic pistons in heavy equipment work on
the same principle.
Pressure = Force / Area
In fluids: Pressure = r gh
Where: r = density (kg/m2)
g = gravitational field vector (9.81
N/kg)
h = height or depth of fluid
Example 1.8.1 Pressure
on floor
A man with a mass of 80 kg rests his weight on one of the
heels of his shoes. The area of the heel in contact with
the floor is 16 cm2 or 0.0016 m2. Calculate the pressure exerted by
his heel on the floor.
Solution
The downward force is 80 x 9.81 which is 784.8 Newtons.
The heel exerts a pressure of 784.8 / 0.0016 Newtons per
square meter on the floor.
Pressure = 490,500
Pascals or 490.5 kilopascals.
Some stiletto heels have
an area of about 1 square centimeter.
If a person weighing 50 kg rests on this, the pressure is
roughly 4900 kPa or 4.9 mega-Pascals
Example 1.8.2 Pressure
in liquids
If a tank that is open to the atmosphere contains water
with a density of 1000 kg per cubic meter, the pressure
above atmospheric pressure at a depth of 10 meters is:
1000 x 9.81 x 10 = 981,000
Pascals
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Review
Questions
- What is the pressure
exerted on the floor by a person who weighs 50 kg
and rests all of her weight on a stiletto heel
with an area of 1 square centimeter?
- Why is it a good idea
to let the air out of the tires if a car gets
stuck in soft sand?
- If a probe is used to
measure the pressure at a certain depth in a
water tank, would a more accurate reading be
obtained if the probe was pointing upwards,
sideways or downwards (at the same depth)?
- Why is it easier to
compress gases than liquids?
- What is the pressure
(above atmospheric pressure) at a depth of 30
meters in salt water with a density of 1100
kilograms per cubic meter?
- What is the pressure
above atmospheric pressure at the bottom of a
beaker containing a liquid with a density of 950
kg/m3 if the depth of the liquid is 8
cm?
- If nearly the earth’s
entire atmosphere is located in a region that is
roughly 30 kilometers deep, why is the pressure
at 15 km above the earth’s surface not
roughly half of that at the earth’s surface?
- A hydraulic jack has
a small piston with a diameter of 1cm and a
larger piston with a diameter of 3cm. How much
pressure is exerted on the circular surface of
the larger piston if the smaller piston exerts a
pressure of 300 kPa on the oil connecting the
circular surfaces of the two pistons? (Assume
that the surfaces of the pistons are at the same
depth in the oil)
- A hydraulic jack has
a small piston with a diameter of 1cm and a
larger piston with a diameter of 4cm. How much force
is exerted on the circular surface of the larger
piston if the smaller piston exerts a force of
300 N on the oil connecting the circular surfaces
of the two pistons? (Assume that the surfaces of
the pistons are at the same depth in the oil)
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CHOICE
OF ACTIVITIES
- Can with
holes: Drill three holes at various
distances from the bottom on the side of
a large tin can. Fill the can with water
and observe the effect of depth on the
speed of the water leaving through the
holes.
- Hydraulic
jack: Use a hydraulic jack to show how a
small force on a small area can be
converted to a larger force on a larger
area.
Activity#1.9.1
The Effect of Depth on Pressure
Show how the pressure in the liquid inside a can
varies with the depth of the liquid.
Materials:
1 large can
Procedure
Draw a vertical line on the outside of the can.
Drill or punch 3 holes roughly 50cm (2inches)
apart along this line. (The first hole near the
bottom, the second hole roughly 50cm above this
and the third roughly 50 cm above this.)
Quickly fill the can with water and observe how
the water leaving each of the two lower holes
travels further - at a higher velocity - from the
can than water from the hole above it.
The velocity of the water leaving through a hole
is related to the pressure directly behind the
hole. The variation in velocity illustrates the
variation in pressure with depth.
Activity#1.9.2
Illustrate the Operation of an Hydraulic Jack
Show how a relatively small force acting on a
small area can create a large pressure in a
contained fluid. This pressure can then be used
to create a much larger force by allowing it to
act on a larger area.
Materials:
Hydraulic jack.
Procedure
Show how a hydraulic jack can be used to lift the
side of a motor vehicle.
Point out the sizes of the two pistons in an
hydraulic jack.
Show how the force created by the larger piston
results from the smaller force applied to the
smaller piston.
HANDS-ON
HOMEWORK
Select one or more of the recommended activities for
Lesson 1.9,
collect the items needed and test the procedure
before demonstrating the activity during the next
theory lesson.
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Lesson
1.8 Pressure
- 50 kg has a weight of
50 x 9.81 N = 490.5 N
1 square cm has an area of 1/(100 x 100) = 0.0001
m2.
The pressure is thus 490.5 / 0.0001 = 4,905,000
Pa.
- The weight of the car
can be spread over a larger area and the pressure
on the sand is reduced considerably.
- The pressure is the
same in all directions. There would be no
difference between the probe readings.
- I suppose that the
easiest answer is that gases are compressible and
liquids are usually regarded as incompressible.
The reason for this is that the gas molecules are
not held together with the same forces that hold
liquid particles together and they can move about
more freely and thus move closer to each other or
further apart.
- 1100 x 9.81 x 30 =
323,730 Pa.
- 950kg/m3 x
9.81N/kg x 0.08m = 745.6 Pa.
- The density of the
air drops with pressure. It varies as the air
pressure varies.
- The pressure is the
same: 300 kPa. (300,000 Pa.)
- Area of small piston
in contact with the oil = p x
(1/2)2 = 0.7854 cm2
Area of larger
piston in contact with the oil = p x (4/2)2 = 12.57 cm2
The force on the larger surface will be
multiplied by the ratio of the areas:
Force = (12.57/0.7854) x 300N = 4800 N
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