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

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 = 1963.5 mm2 converting to m2 = 1963.5 ÷ 10002 = 0.0019635 m2 or 1963.5 x10-6 Pressure = F/A = 100N/0.0019635 m2 = 50929.6 N m-2 Better written as 50.9 kPa

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 = 127324 Nm-2 or 127.32 kPa Force = 1kN Dia = 100mm Pressure = F/A

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

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 = 100000 N/m2 Area = 3500/100000 = 0.035 m2 this is πd2/4 Diameter d = = = 0.2111m or 211mm Force = 3.5kN Dia = ? Pressure = 100kPa

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

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

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 = 9810 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.

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)

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 = 1000 x 9.81 x 30 = 294300 Nm-2 or 294.3 kPa

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 = 1030 x 9.81 x 30 = 303129 Nm-2 or 303.1 kPa

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 http://www.impulseadventure.com/freedive/world-record.html http://www.deeperblue.net/ http://www.tanyastreeter.com/ 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

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 = 1020 x 9.81 x 214 = 2141326.8 Nm-2 or 2.14 MPa Force = Pressure x area = 2141326.8 Nm-2 x 2 m2 = 4282653.6 N Force = 4.28MN ( approx. 428265 kg or 428 ¼ tonne)

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 = 1020 x 9.81 x 160 = 1600992 Nm-2 or 1.6 MPa Force = Pressure x area = 1600992 Nm-2 x 2 m2 = 3201984 N Force = 3.2MN ( approx. 320198 kg or 320 tonne) DO NOT TRY THIS AT HOME !!!!!!!! Not even in the bath !!!!

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 101.3 kPa This value can range between 96 kPa and 104 kPa at UK sea level. The bar is taken as 100 000 Nm-2 or 1 x 105 Pa thus a pressure of 3 bar = 300 000Nm-2 or 300 kPa

Atmospheric Pressure The normal pressure exerted by the weight of Earths atmosphere is called Atmospheric pressure ( Patmos ) and is averaged at 101.3 kPa or 1.013 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

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 = 750000 Pa Patmos = 1013 millibar = 1.013 bar = 1.013 x 105 Pa The Absolute pressure Pabs = 750000 Pa + 101300 Pa = 851300 Pa 851300 Pa = 851.3 kPa = 8.513 bar abs Note: even vacuum pressures register as positive values when expressed as absolute.

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 = -75000 Pa Patmos = 1013 millibar = 1.013 bar = 1.013 x 105 Pa The Absolute pressure Pabs = -75000 Pa + 101300 Pa = 26300 Pa 26300 Pa = 26.3 kPa = 0.263 bar abs Vacuum pressures are often described as negative and sometimes measure in Torr. (See NPL site for more details.) http://www.npl.co.uk/server.php?show=nav.1038 http://www.youtube.com/watch?v=ESR_RtlWJL8&feature=related