1. Pressure 1 Pa = 1 Nm-2 Pressure is defined as force per unit area.
Pressure = Force ÷ Area ( P = F/A & F = P x A & A = F/P)
To calculate values of pressure the acting force (Newtons) must be known together with the area on which the force acts.
The unit of pressure (SI) is the Newton per square metre Nm-2
As with stress this unit may also be called the Pascal (Pa)
1 Pa = 1 Nm-2

2. Pressure
If a constant force is applied to two different areas then the applied pressure will change in proportion to the area.
eg. A force of 100N applied to an area of 1 m2 provides a pressure
of 100Nm-2 or 100 Pa.
If the same force were applied to a piston of diameter 50 mm then the
pressure created would be:
Pressure = force/area First calculate area (c.s.a.)
Area of piston = πd2/4 when diameter d = 50 mm
Area = π x 502/4 = mm2
converting to m2 = ÷ = m2 or x10-6
Pressure = F/A = 100N/ m2 = N m-2
Better written as 50.9 kPa

3. Pressure
Try
Calculate the pressure created when a force of 1kN is applied to a
piston of diameter 100mm.
solution
Pressure = force/area Area = c.s.a. of piston = πd2/4
Diameter d = 100mm Area = πx 1002/4 = 7854 mm2 = 7854 x10-6m2
Pressure = 1000N/ 7854 x10-6m2 = Nm-2 or kPa
Force = 1kN
Dia = 100mm
Pressure = F/A

4. Pressure
The basic equation can be transposed and used to find suitable piston diameters to provide a force output.
eg. If a pressure of 100 kPa is used on a piston to provide an output force of 3 ½ kN calculate the minimum diameter of piston.
Force = 3.5kN
Dia = ?
Pressure = 100kPa

5. Pressure
eg. If a pressure of 100 kPa is used on a piston to provide an output force of 3 ½ kN calculate the minimum diameter of piston.
solution
First find piston c.s.a. If P = F/A then A = F/P
when force F = 3500N and pressure P = N/m2
Area = 3500/ = m this is πd2/4
Diameter d = = = m or 211mm
Force = 3.5kN
Dia = ?
Pressure = 100kPa

6. Pressure – Pascals Law
Pascals Law:- named after 17th century French scientist Blaise Pascal
The law states that any pressure created within a sealed system is
constant at each and every point within the system, it is not dependant
upon the shape of the system.
This effect can be put to good use in the principle of the Hydraulic lever
Force F1
Force F2
Cylinder 2
Cylinder 1
Area A2
Area A1

7. Pressure – Pascals Law Thus F1/A1 = F2/A2
The Hydraulic lever has many good examples from car braking
systems to lifting jacks.
As pressure = Force/Area the pressure generated in cylinder A = F1/A1
But this pressure will be the same throughout the system (Pascals Law)
Thus F1/A1 = F2/A2
Force F1
Force F2
Cylinder 2
Cylinder 1
Area A2
Area A1

8. Pressure at Depth
If a liquid, or indeed a gas is held in a container then the weight of the
liquid will exert a pressure on the base of the container.
eg. 1 m3 of water has a mass of 1000kg
the weight of 1000kg = mass x g = 1000 x 9.81 = 9810N
If the base of the container has a surface area of 1 m2 then
Pressure = Force/Area = 9810N/1m2 = N/m2
If the pressure is to calculated at depth of 10m beneath the surface of a
lake of irregular shape then the weight of water cannot easily be found
by the previous method.
The pressure equation can be used in conjunction with density to
obtain a value of pressure.

9. Pressure at Depth
To obtain a value for pressure at depth in fluid the following method is
used. Consider a body at depth h metres in a fluid of density ρ
If the basic equation for pressure is force/area. P = F/A
The force exerted by the fluid = weight of fluid = mass x g (mg)
The mass of the fluid will depend upon the density ( ρ ) of the fluid.
As density = mass/volume then mass = density x volume (m = ρ x v)
The volume of fluid at a particular depth ( h metres) = Area x depth
The basic equation for pressure on the body is Force/area
substitution gives: Pressure = (density x g x Area x depth) ÷ Area
P = (ρ x g x A x h) ÷ A
Thus pressure at depth P = ρ x g x h
Pressure = fluid density (kg/m3) x g x fluid depth (m)

10. Pressure at Depth
eg. Calculate the pressure on an object 30m below the surface of a freshwater pool of density 1000 kg m-3
solution
Pressure at depth = ρgh
when: density ρ = 1000 kg m-3
g = 9.81 ms-2
depth h = 30m
Pressure = x x 30 = Nm-2 or kPa

11. Pressure at Depth
eg. Re calculate the pressure on an object 30m below the surface of the sea. Seawater density 1030 kg m-3
solution
Pressure at depth = ρgh
when: density ρ = 1030 kg m-3
g = 9.81 ms-2
depth h = 30m
Pressure = x x 30 = Nm-2 or kPa

12. Pressure at Depth
Use the link below to calculate the pressure on the body of Herbert
Nitsch and/or Tanya Streeter when breaking the world ‘no limits’ free
dive records. Assume average seawater density is 1020 kgm-3
Note: the average human body has a surface area in the region of
2 m2, use this to calculate the actual forces on the divers when
breaking the records

13. Pressure at Depth Herbert Nitsch = 214m Pressure at depth = ρgh
when: density ρ = 1020 kg m-3
g = 9.81 ms-2
depth h = 214m
Pressure = x x = Nm-2 or MPa
Force = Pressure x area = Nm-2 x 2 m2 = N
Force = 4.28MN ( approx kg or 428 ¼ tonne)

14. Pressure at Depth DO NOT TRY THIS AT HOME !!!!!!!!
Tanya Streeter = 160m
Pressure at depth = ρgh
when: density ρ = 1020 kg m-3
g = 9.81 ms-2
depth h = 160m
Pressure = x x = Nm-2 or 1.6 MPa
Force = Pressure x area = Nm-2 x 2 m2 = N
Force = 3.2MN ( approx kg or 320 tonne)
DO NOT TRY THIS AT HOME !!!!!!!!
Not even in the bath !!!!

15. Atmospheric Pressure
Although the SI pressure unit is the Nm-2 or Pascal it is a very small
unit of pressure as previous examples show.
A more convenient unit is the bar.
This unit is derived from the pressure exerted by the weight of Earths
atmosphere.
On average this pressure is found to be around kPa
This value can range between 96 kPa and 104 kPa at UK sea level.
The bar is taken as Nm-2 or 1 x 105 Pa
thus a pressure of 3 bar = Nm-2 or 300 kPa

16. Atmospheric Pressure
The normal pressure exerted by the weight of Earths atmosphere is
called Atmospheric pressure ( Patmos ) and is averaged at kPa
or bar.
This value will change depending on atmospheric conditions and also
height above or below datum, normally mean sea level.
If a gauge is used to measure pressure within a sealed container then it
will display the pressure as gauge pressure Pgauge
The actual or absolute pressure within the container will be:
Pressure reading from gauge + Atmospheric pressure on the container
The Absolute pressure Pabs = Pgauge + Patmos

17. Atmospheric Pressure
eg. Calculate the absolute pressure within a cylinder if the pressure
gauge reading is 750 kPa and Atmospheric pressure is 1013 mbar
solution
The Absolute pressure Pabs = Pgauge + Patmos
when P gauge = 750 kPa = Pa
Patmos = millibar = bar = x 105 Pa
The Absolute pressure Pabs = Pa Pa = Pa
Pa = kPa = bar abs
Note: even vacuum pressures register as positive values when
expressed as absolute.

18. Atmospheric Pressure
eg. Calculate the absolute pressure within a cylinder if the pressure
gauge reading is -75 kPa (vacuum) and Atmospheric pressure is
1013 mbar
solution
The Absolute pressure Pabs = Pgauge + Patmos
when P gauge = -75 kPa = Pa
Patmos = millibar = bar = x 105 Pa
The Absolute pressure Pabs = Pa Pa = Pa
26300 Pa = kPa = bar abs
Vacuum pressures are often described as negative and sometimes
measure in Torr. (See NPL site for more details.)