# Pressure in Fluid Systems. Unit 3 Pressure Pages 43-60  Fluid  Hydraulic System  Pneumatic System  Density  Specific gravity  Buoyant force  Hydrometer.

## Presentation on theme: "Pressure in Fluid Systems. Unit 3 Pressure Pages 43-60  Fluid  Hydraulic System  Pneumatic System  Density  Specific gravity  Buoyant force  Hydrometer."— Presentation transcript:

Pressure in Fluid Systems

Unit 3 Pressure Pages 43-60  Fluid  Hydraulic System  Pneumatic System  Density  Specific gravity  Buoyant force  Hydrometer  Pressure  PSI  Atmospheric Pressure  Absolute pressure  Gage pressure  Manometer

Pressure in a Fluid System  Unit 3 Review  Page 53  #1-15

Fluid  Gas or liquid that conforms to the shape of the container  “Anything that flows”

Hydraulic system  Uses liquid as the fluid Pneumatic system  Uses air or gas as the fluid

Why does a hot air balloon float? Why does motor oil rise to the top of water? Density  Amount of matter in a given amount of substance  = Mass/Volume

Density  SI measured in:  Kg/m 3 or gm/cm 3  English measured in:  Lbm/ft 3 or lb/ft 3

Density  What is the density of gold if you have a 1.036cm 3 piece that had a mass of 20grams?  D=m/v  D=20g/1.036cm 3  D=19.3g/cm 3

Density  What is the density of gold if you have a 3.108cm 3 piece that had a mass of 60grams?  D=m/v  D=60g/3.108cm 3  D=19.3g/cm 3

Other Densities  Platinum  Diamond  Chromium  Tin (white)  Tin (gray)  21.45  3.5-3.53  7.15  7.265  5.769

Density  What is the mass in grams of mercury with a volume of 1cm 3 ?  D = m / v  13.6 g/cm 3 = x / 1cm 3  13.6 g = x

Density  What is the mass in kilograms of balsa wood with a volume of 1m 3 ? .3g / cm 3 = x / 1,000,000cm 3  300,000 g = x  300 kg =x  1m 3 = __cm 3  1m 3 = 100cm x 100cm x 100cm  = 1,000,000 cm 3  D = m / v .3g / cm 3 = x / 1m 3

Specific Gravity  Density of a substance divided by the density of water  Because specific gravity is density/density the units cancel out and is written as a whole number

Specific Gravity  Copper has a density of 8.9g/cm 3  What is its specific gravity?  Specific Gravity = density of substance  = density of water  S.G. = (8.9g/cm 3 ) / (1.0g/cm 3 )  S.G. = 8.9

Buoyant Force  The upward force on a substance from a fluid  Will lead sink or float in water?  Will it sink or float in mercury?

Hydrometer  Instrument that measures density or specific gravity of fluids  Can you drown in quick sand?

Pressure  Force per unit area exerted by a fluid

Force on Airplane Windows  An airplane window has a surface area of 136 square inches.  Air pressure inside the cabin is 12.3 lb/in 2  The force pushing on the window

Pressure  What happens to the pressure as we move away from the earth?  http://www.sciencedaily.com/videos/2006/1201- home_runs_amp_holeinone.htm http://www.sciencedaily.com/videos/2006/1201- home_runs_amp_holeinone.htm

Force on Airplane Windows  An airplane window has a surface area of 144 square inches.  Air pressure inside the cabin is 14.7 lb/in 2  Air pressure outside the window is 6.7 lb/in 2  The force pushing on the window  The force pushing out on the window  Net force on window

Inward force on window  F = P x A  F = (14.7lb/in 2 )(144in 2 )  F = 2116.8 lb  F = P x A  F = (6.7 lb/in 2 )(144in 2 )  F= 964.8lb Outward force on window

Net Force on window  The window is being pushed outward with a net force of 1152 lb.  Net Force = Force out – Force in  Net Force = 2116.8 lb – 964.8 lb  Net Force = 1152 lb

Net Force on window  If the plane rises to a higher altitude and the pressure outside the plane changes to 5.4 lb/in 2  How much stronger will the windows need to be in order to hold the pressure

Inward force on window  F = P x A  F = (14.7lb/in 2 )(144in 2 )  F = 2116.8 lb  F = P x A  F = (5.4 lb/in 2 )(144in 2 )  F= 777.6lb Outward force on window

Net Force on window  The window was originally pushing outward with a net force of 1152 lb.  Therefore it needs to hold 187.2 more pounds of pressure (1339.2 – 1152)  Net Force = Force out – Force in  Net Force = 2116.8 lb – 777.6 lb  Net Force = 1339.2 lb

Pressure  Pressure acts equally in all direction at any point in a fluid and therefore it is a scalar

Absolute vs. Gage Pressure  When we fill a tire to 30lb/in 2 is that the absolute or the gage pressure?  Atmospheric pressure = 14.7 lb/in 2

Absolute Pressure  Total pressure compared to a perfect vacuum Gage Pressure  Pressure measured above atmospheric pressure G.P = Total pressure – atmospheric pressure

Total Pressure  Gage pressure is generally measured “with a gage”

Pressure  Tire gage reads 30lb/in 2  What is the atmospheric pressure?  What is the gage pressure?  What is the total pressure?

Pressure  Tire gage reads 30lb/in 2  What is the atmospheric pressure?  What is the gage pressure?  What is the total pressure?

How does pressure change with depth?  Where is the pressure greater the shallow end or the deep end?  Why?

Pressure increases with depth  There is more water sitting on top of the deep end  There is twice as much weight  Twice as much force  Twice as much pressure

Relationship between pressure and depth

Water Pressure Calculation  Given:  The height of the water in a storage tank is 100 ft above the valve. The weight density of water is 62.4 lb/ft 3  Find:  The pressure at the valve in lb/ft 2

Water Pressure Calculation  P = pw x h  P = (62.4 lb/ft3)(100ft)  P = 6240 lb/ft 2  Given: 1 ft 2 = 144 in 2  Now find:  Pressure in PSI

Water Pressure Calculation  P = pw x h  P = (62.4 lb/ft3)(100ft)  P = 6240 lb/ft 2  Given: 1 ft 2 = 144 in 2  p = (6240 lb/ft 2 )(1ft2/144in 2 ) P = 43.3 lb/in 2 (psi)

Hydraulic lift  Liquids are incompressible  Air compressor increases the pressure to the fluid  Large pushing force is exerted on the lifting piston

Pressure acts like forces  Pressure is a prime mover Measuring Pressures  Manometer – instrument used to measure fluid pressure

Balanced pressure across the valve

Unbalanced pressure across the valve

Pressure on bottom does not depend on the size of the tank

Hydraulic jack?  Large cylinder to a small cylinder  Same pressure = more force in the smaller cylinder  Small to large = allowable force but small increments?

Download ppt "Pressure in Fluid Systems. Unit 3 Pressure Pages 43-60  Fluid  Hydraulic System  Pneumatic System  Density  Specific gravity  Buoyant force  Hydrometer."

Similar presentations