# Chapter 9 Fluid Mechanics. Question: The air temperature at an altitude of 10 km is a chilling --- -35 0 C. Cabin temperatures in airplanes flying at.

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Chapter 9 Fluid Mechanics

Question: The air temperature at an altitude of 10 km is a chilling --- -35 0 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.

Fluids – ability to flow – gases and liquids Liquids have definite volume – gases do not. Our atmosphere – “ocean of gas” Density of our atmosphere decreases with altitude Why? Density = m / V The atmosphere exerts pressure – ‘atmospheric pressure’ – caused by the weight of the air. Question: Why are windows on submarines so small?  Pressure varies with depth

Density for solids and liquids are independent of pressure. What about the density of gases? No standard for density of gases…. Why? Compressible Buoyant force – upward force – force opposite of gravity. Objects apparent weight is less in water than in air. Buoyancy is dependent on density. You Tube Demo Buoyancy

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 ) Example: sharp knife vs a dull, flat butter knife

 Barometer Measures atmospheric pressure or “barometric pressure” The SI unit for measuring Pressure = pascal (Pa) = 1 N/m 2 1 m 2 at sea level = 100,000 N --  so 100,000 N/m 2 Average atmospheric pressure at sea level is 101.3kPa Or 1000000 Pa

Conversions: All equivalent units 14.7 pounds per square inch (psi) = 29.92 in Hg = 760 mm Hg = 101.3 kPa = 1.00 atm Example: Convert 232 psi to kPa 232 p.s.i ● 101.3 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

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 doesn’t matter what gas it is, 1 mole = 22.4 L of gas (6.02 x 10 23 particles) at STP (standard temp/press) = 0 0 C and 1 atm of pressure

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.

Charles’s 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

Gay-Lussac’s 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

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

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

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

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 = ? n = 0.25 mol Ideal Gas Law P●V = n●R●T

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