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Chapter 9 Fluid Mechanics. 9-2 Fluid Pressure and Temperature Pressure – a measure of how much force is applied over a given area. Formula: P = F/A Pressure.

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Presentation on theme: "Chapter 9 Fluid Mechanics. 9-2 Fluid Pressure and Temperature Pressure – a measure of how much force is applied over a given area. Formula: P = F/A Pressure."— Presentation transcript:

1 Chapter 9 Fluid Mechanics

2 9-2 Fluid Pressure and Temperature Pressure – a measure of how much force is applied over a given area. Formula: P = F/A Pressure = Force (N) / area (m 2 )

3 Fluids – ability to flow – gases and liquids Our atmosphere – ocean of gas Density of our atmosphere decreases with altitude The atmosphere exerts pressure – atmospheric pressure – caused by the weight (in Newtons) of the air.

4 Barometer Measures atmospheric pressure or barometric pressure The SI unit for measuring Pressure = pascal (Pa) = 1 N/m 2 1 square meter at sea level = 100,000 N -- so 100,000 N/m 2 Average atmospheric pressure at sea level is 101.3kPa Or Pa

5 Conversions: All equivalent units 14.7 pounds per square inch (psi) = in Hg = 760 mm Hg = kPa = 1.00 atm Example: Convert 232 psi to kPa 232 p.s.i kPa 14.7 psi = 1599 kPa Example: Convert 3.50 atm to mm Hg 3.50 atm 760 mm Hg 1 atm = 2660 mm Hg

6 9-4 Properties of Gases Ideal gas Law – relates gas volume, pressure and temperature. For a given Volume of Gas at a given Pressure and a given Temperature there should be a consistent # of molecules/atoms, n Formula : P V = n R T n = number of moles R = universal gas constant = 8.31 J / (mol K) K = SI unit of temperature Kelvin ( 0 C + 273) Avogadro: said that it doesnt matter what gas it is, 1 mole = 22.4 L of gas (6.02 x particles) at STP (standard temp/press) = 0 0 C and 1 atm of pressure

7 Example: How many moles of carbon dioxide gas (CO 2 ) are contained in a 6.2 L tank at 101 kPa and 30 0 C? Given: V = 6.2 L P = 101 kPa T = 30 o C (303 K) And, Constant R = 8.31 kPa x L / mol x K) n = ? Ideal Gas Law -- P V = n R T n = 0.25 mol

8 Boyles Law: The pressure and volume of a gas at constant temperature are inversely proportional. Increase one – decrease the other. Formula: P 1 V 1 = P 2 V 2 Increasing pressure on a gas (compressible) - decreases volume.

9 Boyles Law: A graph of an inverse relationship Example: A volume of gas at 1.10 atm was measured at 326 cm 3. What will be the volume if the pressure is adjusted to 1.90 atm? Given: P 1 = 1.10 atm, V 1 = 326 cm 3, V 2 = ?, P 2 = 1.90 atm Answer V 2 = 189 cm 3 P 1 V 1 = P 2 V 2

10 Charless Law: at constant pressure, the volume of a gas is directly proportional to its Kelvin temperature Increase one, increase the other. Formula: V 1 T 2 = V 2 T 1 proportional to its Kelvin temperature or V 1 = T 1 V 2 T 2 Direct relationship

11 Example: The gas in a balloon occupies 2.25 L at 298 K. At what temperature will the balloon expand to 3.50 L? Given: V 1 = 2.25 L T 1 = 298 K T 2 = ? V 2 = 3.50 L T 2 = 464 K

12 Combined Gas Law P 1 V 1 = P 2 V 2 T 1 T 2 Always convert temperature to Kelvin.

13 Gay-Lussacs Law: The pressure of a gas is directly proportional to the Kelvin temperature if the volume is held constant. Direct relationship P 1 / T 1 = P 2 / T 2

14 Question: The air temperature at an altitude of 10 km is a chilling C. Cabin temperatures in airplanes flying at this altitude are comfortable because of air conditioners rather than heaters. Why? Answer: Airliners have pressurized cabins. The process of stopping and compressing outside air to near sea-level pressures would normally heat the air to a roasting 55 0 C (130 0 F). So air conditioners must be used to extract heat from the pressurized air.


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