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TL2101 Mekanika Fluida I

TL2101 Mekanika Fluida I. Benno Rahardyan. Pertemuan. Pipes are Everywhere!. Owner: City of Hammond, IN Project: Water Main Relocation Pipe Size: 54". Pipes are Everywhere! Drainage Pipes. Pipes. Pipes are Everywhere! Water Mains. Types of Engineering Problems.

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TL2101 Mekanika Fluida I

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  1. TL2101 Mekanika Fluida I Benno Rahardyan Pertemuan

  2. Pipes are Everywhere! Owner: City of Hammond, INProject: Water Main RelocationPipe Size: 54"

  3. Pipes are Everywhere!Drainage Pipes

  4. Pipes

  5. Pipes are Everywhere!Water Mains

  6. Types of Engineering Problems • How big does the pipe have to be to carry a flow of x m3/s? • What will the pressure in the water distribution system be when a fire hydrant is open?

  7. FLUID DYNAMICSTHE BERNOULLI EQUATION The laws of Statics that we have learned cannot solve Dynamic Problems. There is no way to solve for the flow rate, or Q. Therefore, we need a new dynamic approach to Fluid Mechanics.

  8. The Bernoulli Equation By assuming that fluid motion is governed only by pressure and gravity forces, applying Newton’s second law, F = ma, leads us to the Bernoulli Equation. P/g + V2/2g + z = constant along a streamline (P=pressure g=specific weight V=velocity g=gravity z=elevation) A streamline is the path of one particle of water. Therefore, at any two points along a streamline, the Bernoulli equation can be applied and, using a set of engineering assumptions, unknown flows and pressures can easily be solved for.

  9. Free Jets The velocity of a jet of water is clearly related to the depth of water above the hole. The greater the depth, the higher the velocity. Similar behavior can be seen as water flows at a very high velocity from the reservoir behind the Glen Canyon Dam in Colorado

  10. Closed Conduit Flow • Energy equation • EGL and HGL • Head loss • major losses • minor losses • Non circular conduits

  11. The Energy Line and the Hydraulic Grade Line Looking at the Bernoulli equation again: P/γ + V2/2g + z = constant on a streamline This constant is called the total head (energy), H Because energy is assumed to be conserved, at any point along the streamline, the total head is always constant Each term in the Bernoulli equation is a type of head. P/γ = Pressure Head V2/2g = Velocity Head Z = elevation head These three heads, summed together, will always equal H Next we will look at this graphically…

  12. V is average velocity, kinetic energy Conservation of Energy • Kinetic, potential, and thermal energy head supplied by a pump hp = head given to a turbine ht= mechanical energy converted to thermal hL= downstream Cross section 2 is ____________ from cross section 1! irreversible Point to point or control volume? Why a? _____________________________________

  13. Energy Equation Assumptions • Pressure is _________ in both cross sections • pressure changes are due to elevation only • section is drawn perpendicular to the streamlines (otherwise the _______ energy term is incorrect) • Constant ________at the cross section • _______ flow hydrostatic kinetic density Steady

  14. elevation head (w.r.t. datum) EGL (or TEL) and HGL • The energy grade line must always slope ___________ (in direction of flow) unless energy is added (pump) • The decrease in total energy represents the head loss or energy dissipation per unit weight • EGL and HGL are coincident and lie at the free surface for water at rest (reservoir) • If the HGL falls below the point in the system for which it is plotted, the local pressures are _____ ____ __________ ______ pressure head (w.r.t. reference pressure) velocity head downward lower than reference pressure

  15. Energy equation Energy Grade Line velocity head Hydraulic G L static head pressure head Why is static head important? elevation z pump z = 0 datum

  16. The Energy Line and the Hydraulic Grade Line Lets first understand this drawing: Measures the Total Head 1: Static Pressure Tap Measures the sum of the elevation head and the pressure Head. 2: Pilot Tube Measures the Total Head EL : Energy Line Total Head along a system HGL : Hydraulic Grade line Sum of the elevation and the pressure heads along a system Measures the Static Pressure 1 2 1 2 EL V2/2g HGL Q P/γ Z

  17. The Energy Line and the Hydraulic Grade Line Understanding the graphical approach of Energy Line and the Hydraulic Grade line is key to understanding what forces are supplying the energy that water holds. Point 1: Majority of energy stored in the water is in the Pressure Head Point 2: Majority of energy stored in the water is in the elevation head If the tube was symmetrical, then the velocity would be constant, and the HGL would be level EL V2/2g V2/2g HGL P/γ 2 Q P/γ Z 1 Z

  18. Bernoulli Equation Assumption • _________ (viscosity can’t be a significant parameter!) • Along a __________ • ______ flow • Constant ________ • No pumps, turbines, or head loss Frictionless streamline Steady density Why no a? ____________ point velocity Does direction matter? ____ no Useful when head loss is small

  19. Pipe Flow: Review • We have the control volume energy equation for pipe flow. • We need to be able to predict the relationship between head loss and flow. • How do we get this relationship? __________ _______. dimensional analysis

  20. Example Pipe Flow Problem cs1 D=20 cm L=500 m Find the discharge, Q. 100 m valve cs2 Describe the process in terms of energy!

  21. Flow Profile for Delaware Aqueduct Rondout Reservoir (EL. 256 m) 70.5 km West Branch Reservoir (EL. 153.4 m) Sea Level (Designed for 39 m3/s) Need a relationship between flow rate and head loss

  22. Ratio of Forces • Create ratios of the various forces • The magnitude of the ratio will tell us which forces are most important and which forces could be ignored • Which force shall we use to create the ratios?

  23. Inertia as our Reference Force • F=ma • Fluids problems (except for statics) include a velocity (V), a dimension of flow (l), and a density (r) • Substitute V, l, r for the dimensions MLT • Substitute for the dimensions of specific force

  24. Dimensionless Parameters • Reynolds Number • Froude Number • Weber Number • Mach Number • Pressure/Drag Coefficients • (dependent parameters that we measure experimentally)

  25. Problem solving approach • Identify relevant forces and any other relevant parameters • If inertia is a relevant force, than the non dimensional Re, Fr, W, M, Cp numbers can be used • If inertia isn’t relevant than create new non dimensional force numbers using the relevant forces • Create additional non dimensional terms based on geometry, velocity, or density if there are repeating parameters • If the problem uses different repeating variables then substitute (for example wd instead of V) • Write the functional relationship

  26. Friction Factor : Major losses • Laminar flow • Hagen-Poiseuille • Turbulent (Smooth, Transition, Rough) • Colebrook Formula • Moody diagram • Swamee-Jain

  27. Laminar Flow Friction Factor Hagen-Poiseuille Darcy-Weisbach

  28. Pipe Flow: Dimensional Analysis • What are the important forces?______, ______,________. Therefore ________number and _______________ . • What are the important geometric parameters? _________________________ • Create dimensionless geometric groups______, ______ • Write the functional relationship viscous pressure Inertial Pressure coefficient Reynolds diameter, length, roughness height l/D e/D Other repeating parameters?

  29. Dimensional Analysis • How will the results of dimensional analysis guide our experiments to determine the relationships that govern pipe flow? • If we hold the other two dimensionless parameters constant and increase the length to diameter ratio, how will Cp change? Cp proportional to l f is friction factor

  30. Laminar Flow Friction Factor Hagen-Poiseuille Darcy-Weisbach -1 Slope of ___ on log-log plot

  31. Viscous Flow in Pipes

  32. Viscous Flow: Dimensional Analysis • Two important parameters! R - Laminar or Turbulent e/D - Rough or Smooth Where and

  33. Laminar and Turbulent Flows • Reynolds apparatus inertia damping Transition at R of 2000

  34. v v v Boundary layer growth: Transition length What does the water near the pipeline wall experience? _________________________ Why does the water in the center of the pipeline speed up? _________________________ Drag or shear Conservation of mass Pipe Entrance Non-Uniform Flow Need equation for entrance length here

  35. Images - Laminar/Turbulent Flows Laser - induced florescence image of an incompressible turbulent boundary layer Laminar flow (Blood Flow) Simulation of turbulent flow coming out of a tailpipe Turbulent flow Laminar flow http://www.engineering.uiowa.edu/~cfd/gallery/lim-turb.html

  36. Laminar, Incompressible, Steady, Uniform Flow • Between Parallel Plates • Through circular tubes • Hagen-Poiseuille Equation • Approach • Because it is laminar flow the shear forces can be quantified • Velocity profiles can be determined from a force balance

  37. Laminar Flow through Circular Tubes • Different geometry, same equation development (see Streeter, et al. p 268) • Apply equation of motion to cylindrical sleeve (use cylindrical coordinates)

  38. Laminar Flow through Circular Tubes: Equations a is radius of the tube Max velocity when r = 0 Velocity distribution is paraboloid of revolution therefore _____________ _____________ average velocity (V) is 1/2 umax Vpa2 Q = VA =

  39. Velocity Shear Laminar Flow through Circular Tubes: Diagram Laminar flow Shear at the wall True for Laminar or Turbulent flow

  40. Laminar flowContinue Momentum is Mass*velocity (m*v) Momentum per unit volume is *vz Rate of flow of momentum is *vz*dQ dQ=vz2πrdr but vz = constant at a fixed value of r Laminar flow

  41. Laminar flowContinue Hagen-Poiseuille

  42. The Hagen-Poiseuille Equation cv pipe flow Constant cross section h or z Laminar pipe flow equations

  43. Prof. Dr. Ir. Bambang Triatmodjo, CES-UGM : Hidraulika I, Beta Ofset Yogyakarta, 1993 Hidraulika II, Beta Ofset Yogyakarta, 1993 Soal-Penyelesaian Hidraulika I, 1994 Soal-Penyelesaian Hidraulika II, 1995

  44. Air mengalir melalui pipa berdiameter 150 mm dan kecepatan 5,5 m/det.Kekentalan kinematik air adalah 1,3 x 10-4 m2/det. Selidiki tipe aliran

  45. Minyak di pompa melalui pipa sepanjang 4000 m dan diameter 30 cm dari titik A ke titik B. Titik B terbuka ke udara luar. Elevasi titik B adalah 50 di atas titik A. Debit 40 l/det. Debit aliran 40 l/det. Rapat relatif S=0,9 dan kekentalan kinematik 2,1 x 10-4 m2/det. Hitung tekanan di titik A.

  46. Minyak dipompa melalui pipa berdiameter 25 cm dan panjang 10 km dengan debit aliran 0,02 m3/dtk. Pipa terletak miring dengan kemiringan 1:200. Rapat minyak S=0,9 dan keketnalan kinematik v=2,1x 10-4 m2/det. Apabila tekanan pada ujung atas adalah p=10 kPA ditanyakan tekanan di ujung bawah.

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