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19 diapositives · 329 Views · Aug 14, 2014

This overview covers essential topics in fluid dynamics, focusing on the relationship between pressure and height. The assignment includes deriving the pressure-height equation and understanding its implications for both ideal gases and fluids. Students should prepare for upcoming quizzes and lab reports, with assignments due soon. Important lessons encompass the behavior of fluids under varying conditions and the role of molecular motion in pressure calculations. Master these concepts to excel in competitive physics coursework.

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  1. Paperwork • Mastering Physics • Course # DRKIDD880131 • Assignment Due tonight • New Assignment up soon… • Lab Reports due next Tuesday 5pm

  2. Schedule Short Term • Today – derive pressure/height, calculate • Friday – Begin Chapter 19 • Monday – Chapter 19 / Solving • Tuesday – Lab #2 • Quiz#2 [Chapter 18, Labs] • Thursday HMWK due 11pm

  3. Pressure vs. Height • Example 18.4 Force = pA + (dp)A dy Thin object, mass m Force = pA For an object in a fluid Pressure on sides of object is the same, so cancels (Book on desk is stationary) Assume pressure felt by top is slightly different than bottom (p+dp)

  4. Pressure vs. Height • Example 18.4 Force = pA + (dp)A dy Thin object, mass m Force = pA For an object in a fluid Pressure on sides of object is the same, so cancels (Book on desk is stationary) Assume pressure felt by top is slightly different than bottom (p+dp) dp can be +, - or even zero. Just much smaller than p for thin object Let’s say this object is stationary – floating in the fluid. What is sum of all forces on object? What are all forces on object? What if “Object” was just a portion of the fluid itself?

  5. Pressure vs. Height • Example 18.4 Force = pA + (dp)A dy mass = rV = rA(dy) Force = pA SF = 0 = pA - [pA + (dp)A] – mg 0 = pA – pA – (dp)A – rVg (dp)A = -rVg (dp)A = -r(Ady)g (dp/dy) = - rg Implications?

  6. Pressure vs. Height • Example 18.4 Force = pA + (dp)A dy mass = rV = rA(dy) Force = pA SF = 0 = pA - [pA + (dp)A] – mg 0 = pA – pA – (dp)A – rVg (dp)A = -rVg (dp)A = -r(Ady)g (dp/dy) = - rg For Ideal Gas r = m/V = pM/(RT) (dp/dy) = - rg

  7. Pressure vs. Height • Example 18.4 Force = pA + (dp)A dy mass = rV = rA(dy) Force = pA Pressure vs. Height Any Fluid (dp/dy) = - rg For Fluid that is an Ideal Gas r = m/V = pM/(RT) (dp/dy) = - pgM/(RT)

  8. Pressure vs. Height • (dp/dy) = - pgM/(RT) • Now need to set up equation to solve • (dp/p) = -(gM/RT)(dy) • Assume a constant temperature (?)

  9. Pressure vs. Height • (dp/dy) = - pgM/(RT) • Now need to set up equation to solve • (dp/p) = -(gM/RT)(dy) • Assume a constant temperature (?)

  10. Pressure vs. Height

  11. Pressure vs. Height Let’s say integration was from sea level (p0=p0, y0 = 0) To a point pF = p, yF = y Need to have known endpoints Then can derive equation for air pressure as a function of height above sea level Happy Equation: Should Check Accuracy Implications? Check at sea level.

  12. Molecular Motion • From Derivation in Section 18.3 • KEAVG=(3/2)NkBT (N is # molecules) • pV = (2/3) KEAVG • pV = NkBT • Where does 3 come from? • Other implications? • Why do people care about root mean square? [vrms] • Hint: zero?

  13. Mean Free PathLifetime (Mean Free Time) • Very Important Concept • Vacuum conditions • Behavior of electrons in solids • Any interacting (or not) particles • Mean distance traveled before collision • lMFP = v tMFP • Lifetime (mean free time) is time between collisions • Why important for electrical conduction?

  14. Lots of Averages… • Gas at temperature 500K • Average KE of One Mole • KEAVG = (3/2)nRT = (3/2)RT • 8.31 J / (mole K) • KEAVG = 6232 J • Say gas is N2 (MM = 0.028kg / mole) • KEAVG = (1/2)m(v2)AVG What is max speed?

  15. Phase Diagram Describes P&T conditions for materials Interesting Points, What is this at STP?

  16. Phase Diagram Supercritical Fluid? Neat – can dissolve things like a liquid & Diffuse through solids like a gas…

  17. Discussion Q18.10Start Gas # molecules = n0 Temperature = T0 pressure = p0 Volume = V0

  18. Discussion Q18.10“Sudden” Hole in wall Gas Initial State # molecules = n0 Temperature = T0 pressure = p0 Volume = V0 Gas Final State # molecules = ? Temperature = ? pressure = ? Volume = ? What Happens here?

  19. Schedule Short Term Today – derive pressure/height, calculate Friday – Begin Chapter 19 Monday – Chapter 19 / Solving Tuesday – Lab #2 Quiz#2 [Chapter 18, Labs] Thursday HMWK due 11pm