Chapter 9 Fluid Mechanics
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.
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 Pa
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
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 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