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**Pressure in Fluid Systems**

3/25/2017 UNIT 3 Pressure in Fluid Systems

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**Unit 3 Pressure Pages 43-60 Fluid Hydraulic System Pneumatic System**

3/25/2017 Fluid Hydraulic System Pneumatic System Density Specific gravity Buoyant force Hydrometer Pressure PSI Atmospheric Pressure Absolute pressure Gage pressure Manometer

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**Pressure in a Fluid System**

3/25/2017 Pressure in a Fluid System Unit 3 Review Page 53 #1-15

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**Fluid Gas or liquid that conforms to the shape of the container**

3/25/2017 Fluid Gas or liquid that conforms to the shape of the container “Anything that flows”

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**Hydraulic system Pneumatic system**

3/25/2017 Hydraulic system Fluid system that uses liquid as the fluid Pneumatic system Fluid system that uses air or gas as the fluid

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3/25/2017 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

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**Density SI measured in: English measured in: Kg/m3 or gm/cm3**

3/25/2017 Density SI measured in: Kg/m3 or gm/cm3 English measured in: Lbm/ft3 or lb/ft3

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3/25/2017 Density What is the density of gold if you have a 1.036cm3 piece that had a mass of 20grams? D=m/v D=20g/1.036cm3 D=19.3g/cm3

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3/25/2017 Density What is the density of gold if you have a 3.108cm3 piece that had a mass of 60grams? D=m/v D=60g/3.108cm3 D=19.3g/cm3

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**Other Densities Platinum Diamond Chromium Tin (white) Tin (gray) 21.45**

3/25/2017 Platinum Diamond Chromium Tin (white) Tin (gray) 21.45 7.15 7.265 5.769

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**What is the mass in grams of mercury with a volume of 1cm3?**

3/25/2017 Density What is the mass in grams of mercury with a volume of 1cm3? D = m / v 13.6 g/cm3 = x / 1cm3 13.6 g = x

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**What is the mass in kilograms of balsa wood with a volume of 1m3?**

3/25/2017 Density What is the mass in kilograms of balsa wood with a volume of 1m3? 1m3 = __cm3 1m3 = 100cm x 100cm x 100cm = 1,000,000 cm3 D = m / v .3g / cm3 = x / 1m3 .3g / cm3 = x / 1,000,000cm3 300,000 g = x 300 kg =x

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**Density of a substance divided by the density of water**

3/25/2017 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

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**Specific Gravity Copper has a density of 8.9g/cm3**

3/25/2017 Specific Gravity Copper has a density of 8.9g/cm3 What is its specific gravity? Specific Gravity = density of substance = density of water S.G. = (8.9g/cm3) / (1.0g/cm3) S.G. = 8.9

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**The upward force on a substance from a fluid**

3/25/2017 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?

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3/25/2017 Hydrometer Instrument that measures density or specific gravity of fluids Can you drown in quick sand?

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**Force per unit area exerted by a fluid**

3/25/2017 Pressure Force per unit area exerted by a fluid

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**Force on Airplane Windows**

3/25/2017 Force on Airplane Windows An airplane window has a surface area of 136 square inches. Air pressure inside the cabin is 12.3 lb/in2 Find: The force pushing on the window

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**What happens to the pressure as we move away from the earth?**

3/25/2017 Pressure What happens to the pressure as we move away from the earth?

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**Force on Airplane Windows**

3/25/2017 Force on Airplane Windows An airplane window has a surface area of 144 square inches. Air pressure inside the cabin is 14.7 lb/in2 Air pressure outside the window is 6.7 lb/in2 Find: The force pushing in the window The force pushing out the window Net force on window

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**Outward force on window**

3/25/2017 Inward force on window F = P x A F = (6.7 lb/in2)(144in2) F= 964.8lb Outward force on window F = P x A F = (14.7lb/in2)(144in2) F = lb

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**Net Force on window Net Force = Force out – Force in**

3/25/2017 Net Force on window Net Force = Force out – Force in Net Force = lb – lb Net Force = 1152 lb The window is being pushed outward with a net force of 1152 lb.

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3/25/2017 Net Force on window If the plane rises to a higher altitude and the pressure outside the plane changes to 5.4 lb/in2 How much stronger will the windows need to be in order to hold the pressure

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**Outward force on window**

3/25/2017 F = P x A F = (14.7lb/in2)(144in2) F = lb Inward force on window F = P x A F = (5.4 lb/in2)(144in2) F= 777.6lb

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**Net Force on window Net Force = Force out – Force in**

3/25/2017 Net Force on window Net Force = Force out – Force in Net Force = lb – lb Net Force = lb The window was originally pushing outward with a net force of 1152 lb. Therefore it needs to hold more pounds of pressure ( – 1152)

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3/25/2017 Pressure Pressure acts equally in all direction at any point in a fluid and therefore it is a scalar

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**Absolute vs. Gage Pressure**

3/25/2017 Absolute vs. Gage Pressure When we fill a tire to 30lb/in2 is that the absolute or the gage pressure? Atmospheric pressure = 14.7 lb/in2

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**Absolute Pressure Gage Pressure**

3/25/2017 Absolute Pressure Total pressure compared to a perfect vacuum Gage Pressure Pressure measured above atmospheric pressure G.P = Total pressure – atmospheric pressure

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**Gage pressure is generally measured “with a gage”**

3/25/2017 Gage pressure is generally measured “with a gage” Total Pressure

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**Pressure Tire gage reads 38lb/in2 What is the atmospheric pressure?**

3/25/2017 Pressure Tire gage reads 38lb/in2 What is the atmospheric pressure? What is the gage pressure? What is the total pressure?

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**Pressure What is the atmospheric pressure? What is the gage pressure?**

3/25/2017 Pressure Tire gage reads 38lb/in2 What is the atmospheric pressure? What is the gage pressure? What is the total pressure? 14.7 lb/in2 38 lb/in2 38 lb/in lb/in2 = 52.7 lb/in2

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**How does pressure change with depth?**

3/25/2017 How does pressure change with depth? Where is the pressure greater the shallow end or the deep end? Why?

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**Pressure increases with depth**

3/25/2017 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

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**Relationship between pressure and depth**

3/25/2017 Relationship between pressure and depth

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**Water Pressure Calculation**

3/25/2017 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/ft3 Find: The pressure at the valve in lb/ft2

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**Water Pressure Calculation**

3/25/2017 Water Pressure Calculation P = pw x h P = (62.4 lb/ft3)(100ft) P = 6240 lb/ft2 Given: 1 ft2 = 144 in2 Now find: Pressure in PSI

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**Water Pressure Calculation**

3/25/2017 Water Pressure Calculation P = pw x h P = (62.4 lb/ft3)(100ft) P = 6240 lb/ft2 Given: 1 ft2 = 144 in2 p = (6240 lb/ft2)(1ft2/144in2) P = 43.3 lb/in2 (psi)

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**Balanced pressure across the valve**

3/25/2017 Balanced pressure across the valve

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**Unbalanced pressure across the valve**

3/25/2017 Unbalanced pressure across the valve

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

3/25/2017 Pressure on bottom does not depend on the size of the tank

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**Pressure acts like forces**

3/25/2017 Pressure acts like forces Pressure is a prime mover Measuring Pressures Manometer – instrument used to measure fluid pressure

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**Hydraulic lift Liquids are incompressible**

3/25/2017 Hydraulic lift Liquids are incompressible Air compressor increases the pressure to the fluid Large pushing force is exerted on the lifting piston

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**Hydraulic jack? Large cylinder to a small cylinder**

3/25/2017 Hydraulic jack? Large cylinder to a small cylinder Same pressure = more force in the smaller cylinder Small to large = allowable force but small increments?

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3/25/2017 An enclosed fluid under pressure exerts that pressure throughout its volume and against any surface containing it. That's called 'Pascal's Principle', and allows a hydraulic lift to generate large amounts of FORCE from the application of a small FORCE. Assume a small piston (one square inch area) applies a weight of 1 lbs. to a confined hydraulic fluid. That provides a pressure of 1 lbs. per square inch throughout the fluid. If another larger piston with an area of 10 square inches is in contact with the fluid, that piston will feel a force of 1 lbs/square inch x 10 square inches = 10 lbs.

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3/25/2017 So we can apply 1 lbs. to the small piston and get 10 lbs. of force to lift a heavy object with the large piston. Is this 'getting something for nothing'? Unfortunately, no. Just as a lever provides more force near the fulcrum in exchange for more distance further away, the hydraulic lift merely converts work (force x distance) at the smaller piston for the SAME work at the larger one. In the example, when the smaller piston moves a distance of 10 inches it displaces 10 cubic inch of fluid. That 10 cubic inch displaced at the 10 square inch piston moves it only 1 inch, so a small force and larger distance has been exchanged for a large force through a smaller distance.

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