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MECH 221 FLUID MECHANICS (Fall 06/07) Tutorial 1

MECH 221 FLUID MECHANICS (Fall 06/07) Tutorial 1. Second TA: CHAU Man Hei (mehei@ust.hk). Leading TA: SIN Ka Fai, Kelvin (meskf@ust.hk). Tel: 2358-8808 ; Office: Rm 1213; Office hour: Wed; 15:00 – 16:30. MECH 221 Fluid Mechanics ( Fall Semester 2006/2007).

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MECH 221 FLUID MECHANICS (Fall 06/07) Tutorial 1

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  1. MECH 221 FLUID MECHANICS(Fall 06/07)Tutorial 1

  2. Second TA: CHAU Man Hei (mehei@ust.hk) Leading TA: SIN Ka Fai, Kelvin (meskf@ust.hk) Tel: 2358-8808 ; Office: Rm 1213; Office hour: Wed; 15:00 – 16:30 MECH 221 Fluid Mechanics(Fall Semester 2006/2007) Instructor: Prof. C. T. Hsu, Mechanical Engineering Email:mecthsu@ust.hk Tel: 2358-7188 ; Office: Rm 2561 Office hour – to be arranged Prerequisites: MATH 100/101 & MATH 150/151 Lecture Time: Tue & Thu; 12:00 – 13:20 Classroom: Rm 1403 Tutorial Time: Wed; 09:00 – 09:50 Classroom: Rm 2503

  3. Assessments #Homework = 10 points Mid term Exam = 30 points Final Exam = 45 points *Others = 15 points # Home works distributed through website every week on Wednesday. Due one week after distribution (collected after tutorial) https://teaching.ust.hk/~mech221/ * Including short quiz, attendance, classroom behavior, etc

  4. Course Focus • Fundamental Concepts • Fluid Statics • Fluid Kinematics, Integral and Differential Equations of Fluid Flows • Conservation of Mass, Momentum and Energy • Dimensional Analysis • Inviscid Flows, Boundary Layer Flows, Pipe Flows, Open Channel Flows

  5. Course Notes • Text Book: Fundamentals of Fluid Mechanics, • 5th or 4th edition B.R. Munson, D.F. • Young and T.H. Okiishi, Wiley and • Sons, 2005 or 2002 • Most materials are available from course web • https://teaching.ust.hk/~mech221/ • Reading the handout may not be sufficient. It is useful to take notes as the instructor explains concepts and elaborates on the handout

  6. Syllabus • 1. Introduction (Chapter 1) Week 1 • 2. Fluid Statics (Chapter 2) Weeks 2-3 • 3. Fluids in Motions (Chapter 3) Weeks 3 -4 • 4. Kinematics of Fluid Motion (Chapter 4) Weeks 4-5 • Integral and Differential Forms of • Equations of Motion (Chapters 5 & 6) Weeks 6-8 • Mid-term Week 8 • 6. Dimensional Analysis (Chapter 7) Week 9 • 7. Inviscid Flows (Chapter 6) Week 10 • 8. Boundary Layer Flows (Chapter 9) Weeks 11-12 • 9. Flows in Pipes (Chapter 8) Weeks 12-13 • 10. Open Channel Flows (Chapter 10) Weeks 13-14 • Summary Review Week 14 • Final According to the University Schedule

  7. Historic Background Prandtl (1875-1953) • Fluid Mechanics is the modern science developed mainly by Prandtl and von Karman to study fluid motion by matching experimental data with theoretical models. Thus, combining Aero/Hydrodynamics with Hydraulics. • Indeed, modern research facilities employ mathematicians, physicists, engineers and technicians, who working in teams to bring together both view points: experiment and theory. Von Karman (1881-1963)

  8. Do you know….? • Tsien Hsue-shen (錢學森) • Father of Chinese Rocketry • Student of von Karman in 1936 From left to right: Ludwig Prandtl, H.S. Tsien, Theodore von Kármán

  9. Fluid Mechanics • Definition • “Fluid”: a substance that deforms continuously when acted on by a shearing stress of any magnitude. • “Mechanics”: the branch of applied mathematics that deals with the motion and equilibrium of bodies and the action of forces, and includes kinematics, dynamics, and statics. • “Fluid mechanics”: a branch of science that studies the mechanics of those free moving particles.

  10. Liquid Gas Mechanics of Particle

  11. Fluid Modeling • Microscopic: • Study the behavior of molecules • VERY complicated!!! • Mesoscopic: • Statistical physics • Macroscopic: • Continuum assumption • Navier-Stokes Equation

  12. Continuum Assumption • …… • A fluid particle is a volume large enough to contain a sufficient number of molecules of the fluid to give an average value for any property that is continuous in space, independent of the number of molecules. • What does “large enough” mean?? • How can we determine??

  13. Continuum Assumption • Knudsen number: Kn = / L  - mean free path L - characteristic length

  14. Continuum Assumption • For continuum assumption: Kn << 1 • Kn < 0.001 -Non-slip fluid flow - B.C.s: no velocity slip - No temp. jump - Classical fluid mechanics • 0.001< Kn < 0.1 -Slip fluid flow - Continuum with slip B.C.s • 0.1< Kn< 10 -Transition flow - No continuum, kinetic gas • 10<Kn-Free molecular flow - Molecular dynamics

  15. D Example 1 • For air duct: • Characteristic scales for standard air: • -> mean free path,  (sea level) ~ 10-7 m • Characteristic length (L): • -> Diameter of the duct (D) = 1 inch (25.4mm) Kn = 10-7/(0.0254) = 3.937x10-6 < 0.001 (Continuum and non-slip fluid flow) Air flow

  16. L h Example 2 • For airplane: • Characteristic scales for standard air: • -> mean free path,  (h=sea level) ~ 10-7 m • Characteristic length (L): • -> Length of the airplane = 10m Kn = 10-7/(10) = 10-8 < 0.001 (Continuum and non-slip fluid flow)

  17. L Example 3 • For micro-channel: • Characteristic scales for standard air: • -> mean free path,  (sea level) ~ 10-7 m • Characteristic length (L): • -> Width of the micro-channel = 1μm = 10-6m Kn = 10-7/(10-6) = 0.1 (Slip fluid flow? Transaction flow? Or others?)

  18. Properties • Thermodynamical • Mean free time - n • Convection time scale - s • Mach number – M • Physical • REV – Representative Elementary Volume • Density -  • Viscosity - µ

  19. Power law: = k (u/ y)m Newtonian fluid: k = µ, m=1 Non-Newtonian fluid: m1 Bingham plastic fluid: = 0 +µu/y : Shearing stress [N/m2] µ: dynamic viscosity [kg/(m.s)] : kinematic viscosity:  = µ/ [m2/s] Viscosity

  20. Dimensional Analysis (MLT) • Primary quantities: • Mass: M • Length: L • Time: T • Example: • Velocity: Length/Time = LT-1 • Momentum: Mass x Velocity = MLT-1 • Density: Mass/Volume = ML-3

  21. Unit conversion • Length • 1 inch = 25.4mm • Volume • 1 L = (10cm)3=0.001m3 • Energy • 1 Btu = 1055.056J • 1 kcal = 4186.8J • 1 kWh = 3,600,000J • Power • 1 hp(UK) = 745.7W • Mass • 1 lbm = 0.4536kg • Force • 1 lbf = 4.448N • Pressure • 1 bar = 100,000Pa • 1 psi = 6894.757Pa • 1 mmHg = 133Pa(γHg = 133kN/m3)

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