2.00AJ / 16.00AJ Exploring Sea, Space, & Earth: Fundamentals of Engineering Design Spring 2009

Transcription

1 MIT OpenCourseWare AJ / 16.00AJ Exploring Sea, Space, & Earth: Fundamentals of Engineering Design Spring 2009 For information about citing these materials or our Terms of Use, visit:

2 FUNdaMENTALs for Design Analysis: Fluid Effects & Forces Prof. A. H. Techet 2.00a/16.00a Lecture 4 Spring 2009

3 Water & Air Courtesy of the U.S. Navy. Courtesy of NASA. Hydrodynamics v. Aerodynamics Air Water is almost 1000 times denser than air!! = 1.2 kg / m Water Density Density 3 3! = 1025 kg / m (seawater) µ = 1.82" 10 N # s / m! 5 2 3! = 1000 kg / m (freshwater) Dynamic Viscosity Dynamic Viscosity Kinematic Viscosity Kinematic Viscosity! = µ " = $ # 5 2 / m / s µ = 1.0" 10 N # s / m! = #! 3 2 " m / s Image by Leonardo da Vinci. Fluid

4 Hydrostatic Pressure

5 Pressure under water Pressure is a Force per Area (P = F/A) Pressure is a Normal Stress???? Pressure is isotropic. How does it act on these 2D shapes?

6 Pressure increases with depth Hydrostatic Pressure Pressure on a vertical wall: dp dz = "! g Absolute Pressure p =! g ( z h) p! a "! +z Gauge Pressure p g = " g( h! z) The NET pressure force acts at the CENTER of PRESSURE FOR MORE DETAILS WITH THE DERIVATION OF THE HYDROSTATIC EQUATION, SEE THE READING ON PRESSURE POSTED ON THE CLASS WEBPAGE

7 Pressure on a sphere at depth? F "" = p! nˆ ds S Pressure acts normal to the surface. By convention pressure is positive in compression. The total force is the integration of the ambient pressure over the surface area of the sphere.

8 Archimedes Principle Weight of the displaced volume of fluid is equal to the hydrostatic pressure acting on the bottom of the vessel integrated over the area. W z M F B L F g D x Pressure, p BUOYANCY FORCE --> F z = p*a = (ρ g D) * (L W) = ρ g* (L D W) = ρ g * Volume

9 Center of Buoyancy Center of buoyancy is the point at which the buoyancy force acts on the body and is equivalently the geometric center of the submerged portion of the hull. To calculate the center of buoyancy, it is first necessary to find the center of area! 1) Calculate the Area of the body: A =! y( x) dx 3) Calculate the coordinates for the 2) Find the 1 st moment of the area: M M xx yy =! y( x) x dx Center of Buoyancy: M =! x( y) y dy yy x = A and y = M xx A

10 Stability? A statically stable vessel with a positive righting arm. self-righting Statically unstable vessel with a negative righting arm.

11 Fluids in Motion

12 Fluids Follow Basic Laws Conservation of Mass Conservation of Momentum Conservation of Energy

13 Flows can be described simply Streamlines are lines everywhere parallel to the velocity (no velocity exists perpendicular to a streamline) Streaklines are instantaneous loci of all fluid particles that pass through point x o Pathlines are lines that one single fluid particle follows in time In Steady flow these are all the same! Steady flow does not change in time.

14 Conservation of Mass What goes in must come out! Control Volume: nozzle Mass: density * volume m =!" U Area A Area a V Volume: area * length! = A" L Length: velocity * time L = U! " t! m = "AU # $t To conserve Mass: m in = m out!au " #t!# " $# m in =!av #t!# " $# m out

15 Conservation of Momentum Newton s second law states that the time rate of change of momentum of a system of particles is equal to the sum of external forces acting on that body.! F i = d dt { m V} P da P + dp Forces: Gravity (hydrostatic) Pressure!!!! x = P Shear (viscous/friction) External Body F da P x "( )! F = " dpda = + dp da = dv m dt dv m dt

16 Pressure Along a Streamline Bernoulli s Equation (neglecting hydrostatics) Atmospheric pressure P1 +! v1 = P2 +! v2 = Const 2 2 General eqn. on streamline: 1! 2 2 P + v = C P atm =P static v 1 2 Stagnation pressure: 1 Po = Patm +! v 2 Stagnation point (high pressure) 2 Streamline (C = constant) NB: The body surface can also be treated as a streamline as there is no flow through the body.

17 Aero/Hydro-dynamic Forces Lift V α Drag L =! V " CL " S D =! V " CD " S

18 Aero/Hydro-dynamic Forces Lift V α Drag L =! V " CL " S D =! V " CD " S Dynamic Pressure (like in Bernoulli s equation!!)

19 Aero/Hydro-dynamic Forces Lift V α Drag L =! V " CL " S D =! V " CD " S Empirical Force Coefficients

20 Aero/Hydro-dynamic Forces Lift V α Drag L =! V " CL " S D =! V " CD " S Wing Planform Area

21 Aero/Hydro-foil Geometry Leading Edge Velocity b, span t, thickness Trailing Edge c, chord S = Planform Area S = b*c for rectangular foil Aspect Ratio: AR = b 2 / S

22 Lift on a hydrofoil Courtesy of Spyros A. Kinnas. Used with permission.

23 LIFT COEFFICIENT v. AOA Coefficient of Lift for a NACA 0012 Airfoil as A function of angle of attack NACA 0012 First two numbers indicate camber Double zeros indicate symmetric foil

24 LIFT COEFFICIENT v. AOA Coefficient of Lift for a NACA 0012 Airfoil as A function of angle of attack NACA 0012 Last two numbers indicate thickness of foil as % of total chord length

25 LIFT COEFFICIENT v. AOA Coefficient of Lift for a NACA 0012 Airfoil as A function of angle of attack NACA 0012 slope = 2π (forαin radians) Stall AOA For a symmetrical foil: C l = 2!" (α in radians!) F LIFT = 1 2!U 2 C l S

26 Maneuvering with a Rudder Images removed due to copyright restrictions. Please see: Fig and 14-2 in Gillmer, Thomas Charles, and Bruce Johnson. Introduction to Naval Architecture.Annapolis, MD: U.S. Naval Institute Press, 1982.

27 Images removed due to copyright restrictions. Please see Fig. 14-5, 14-6, and 14-9 in Gillmer, Thomas Charles, and Bruce Johnson. Introduction to Naval Architecture. Annapolis, MD: U.S. Naval Institute Press, 1982.

28 Turning Moment on a Vehicle Vehicle Forward speed Single Motor Turn Moment Arm, r M = r x T M T, thrust Sum of the Moments (Torques) about the CG equals the time rate of change of angular momentum.. Σ M CG = Ιθ Ι = moment of Inertia of the vehicle about CG* *can calculate in Solidworks! Vehicle Forward speed Turning a Vehicle with a Rudder Moment Arm, r M M = r x L Lift on the Rudder: L = 1 / 2 ρv 2 (2 π α) A A = area of rudder α in radians Coefficient of Lift: C L = 2 π α α LIFT, L

29 Viscous Drag Skin Friction Drag: C f laminar turbulent Form Drag: C D due to pressure (turbulence, separation) Streamlined bodies reduce separation, thus reduce form drag. Bluff bodies have strong separation thus high form drag.

30 Friction Drag The transfer of momentum between the fluid particles slows the flow down causing drag on the plate. This drag is referred to as friction drag. Friction Drag Coefficient: C f = 1 2 F 2!U Aw units " 2 [ MLT ] " 2 [ MLT ] (non-dimensional coefficient) A = Wetted Area w THIS IS A SHEAR FORCE THAT COMES FROM SHEAR STRESS AT THE WALL!

31 Flat Plate Friction Coefficient Turbulent boundary layer F = 1 2 ρu2 C f A w Drag coefficient, C D Laminar boundary layer Transition at Re x = 5 x 10 5 A w = wetted area Turbulent boundary layer Reynolds number, Re L Variation of drag coefficient with Reynolds number for a smooth flat plate parallel to the flow. Figure by MIT OpenCourseWare.

32 Viscous flow around bluff bodies (like cylinders) tends to separate and form drag dominates over friction drag.

33 Drag on Bodies DRAG ACTS INLINE WITH VELOCITY Image removed due to copyright restrictions. Please see

34 Form Drag Drag Force on the body due to viscous effects: F D = Where C D is found empirically through experimentation A is profile (frontal) area 1 2! Image removed due to copyright restrictions. Please see 2 U CD A C D is Reynolds number dependent and is quite different in laminar vs. turbulent flows Form Drag or Separation Drag or Pressure Drag (same thing!)

35 Flow Separating from a Cylinder Y Separation point 'Negative' velocities (separated flow) d Wake High pressure Stagnation pressure Y Low pressure Figure by MIT OpenCourseWare.

36 Classical Vortex Shedding Image removed due to copyright restrictions. Please see Alternately shed opposite signed vortices

37 Vortex shedding results from a wake instability Images removed due to copyright restrictions. Please see Fig in Homann, Fritz. "Einfluß großer Zähigkeit bei Strömung um Zylinder." Forschung auf dem Gebiete des Ingenieurwesens 7 (January/February 1936): 1-10.

38 Reynolds Number? Non-dimensional parameter that gives us a sense of the ratio of inertial forces to viscous forces Re = ( ) (! 2 2 ) inertial forces U L!UL = = (viscous forces µul µ ) ( ) ν water = ρ/µ = 10 6 m 2 /s L=1-10 mm L = 9.5m Courtesy NOAA. Speeds: around of 2*L/second Speeds: in excess of 56 km/hr Image from Wikimedia Commons,

39 Reynolds NumberDependence Regime of Unseparated Flow A Fixed Pair of Foppl Vortices in Wake Two Regimes in which Vortex Street is Laminar Transition Range to Turbulence in Vortex Vortex Street is Fully Turbulent Laminar Boundary Layer has Undergone Turbulent Transition and Wake is Narrower and Disorganized Re-establishment of Turbulent Vortex Street R d < < R d < < R d < < R d < 300 Transition to turbulence 300 < R d < 3*10 5 3*10 5 < R d < 3.5* *10 6 < R d Regimes of fluid flow across smooth circular cylinders (Lienhard, 1966). Figure by MIT OpenCourseWare. From Blevins

40 Drag Coefficient: Cylinder C D VD Re = v Drag coefficient for a smooth circular cylinder as a function of Reynolds number. Figure by MIT OpenCourseWare.

41 Drag Coefficient: Sphere C D Theory due to Stokes Re = VD v Golf ball Drag coefficient of a smooth sphere as a function of Reynolds number. Drag coefficient, C D Spheres with roughness Rough spheres x 10-5 Smooth spheres x x ε = 150 x 10-5 D 0 2 x Reynolds number, Ud/v 4x 10 6 Figure by MIT OpenCourseWare.

42 Trade-off between Friction and Pressure drag c = Body length inline with flow t = Body thickness Images removed due to copyright restrictions. Please see Fig and 7.15 in White, Frank M. Fluid Mechanics. Boston, MA: McGraw-Hill, 2007.

43 2D Drag Coefficients For 2D shapes: use C D to calculate force per unit length. Use a strip theory type approach to determine total drag, assuming that the flow is uniform along the span of the body. Images removed due to copyright restrictions. Please see Table 7.2 in White, Frank M. Fluid Mechanics. Boston, MA: McGraw-Hill, 2007.

44 Trade-off between Friction and Pressure drag c = Body length inline with flow t = Body thickness Images removed due to copyright restrictions. Please see Table 7.2 in White, Frank M. Fluid Mechanics. Boston, MA: McGraw-Hill, 2007.

45 3D Drag Coefficients Images removed due to copyright restrictions. Please see Table 7.3 in White, Frank M. Fluid Mechanics. Boston, MA: McGraw-Hill, 2007.

46 More 3D Shapes Images removed due to copyright restrictions. Please see Table 7.3 in White, Frank M. Fluid Mechanics. Boston, MA: McGraw-Hill, 2007.

47 Calculating Drag on a Simple Structure Drag total = Drag sphere +2*Drag motors Example: Spherical vehicle with motors Vehicle Forward speed Drag Drag sphere = 1 / 2 ρv 2 C D A A = front area = π r 2 C D = drag coefficient (depends on velocity) C D = (laminar) or 0.2 (turbulent) Drag motors ~ 1 / 2 ρv 2 C D A ** A = front area = π r 2 C D = drag coefficient (depends on aspect ratio) Assuming L/D = 2 C D =0.85 **This neglects the propellers which add some drag to the vehicle Use linear superposition to find total drag on a complex shape

48 Fluid Forces Force on a surface ship is a function of X?? 1) Fluid properties: density (ρ) & viscosity (µ) 2) Gravity (g) 3) Fluid (or body) velocity (U) 4) Body Geometry (L) F = f (!, µ,g,u, L) Dimensional Analysis: 1) Output variable (Force) is a function of N-1 input variables (ρ, µ, g, U, L). Here N=6. 2) There are M=3 primary dimensions (units) for the variables listed above [Mass, Length, Time] 3) We can determine P=N-M non-dimensional groups (P=3). 4) How do we find these groups?

49 Fluid Forces Force on a surface ship is a function of X?? 1) Fluid properties: density (ρ) & viscosity (µ) 2) Gravity (g) 3) Fluid (or body) velocity (U) 4) Body Geometry (L) F = f (!, µ,g,u, L) Dimensional Analysis: 1) Output variable (Force) is a function of N-1 input variables (ρ, µ, g, U, L). Here N=6. 2) There are M=3 primary dimensions (units) for the variables listed above [Mass, Length, Time] 3) We can determine P=N-M non-dimensional groups (P=3). 4) How do we find these groups?

50 Dimensional Analysis F = f (!, µ,g,u, L) F ρ U L µ g [kg m/s 2 ] [kg/m 3 ] [m/s] [m] [kg/m/s] [m/s 2 ] Mass [M] Length [L] Time [t] Step 1) Make all variables containing mass non-dimensional in mass by dividing through by density: F/ρ and µ/ρ

51 Dimensional Analysis F! = f (µ, g,u, L)! F/ρ [m 4 /s 2 ] ρ/ρ [--] U [m/s] L [m] µ/ρ [m 2 /s] g [m/s 2 ] Mass [M] Length [L] Time [t] Step 2) Rewrite Matrix deleting density column and mass row

52 Dimensional Analysis F! = f (µ, g,u, L)! F/ρ [m 4 /s 2 ] U [m/s] L [m] µ/ρ [m 2 /s] g [m/s 2 ] Length [L] Time [t] Step 3) Non-dimensionalize all variables containing time, using velocity: F/ρU 2 and µ/ρu and g/u 2

53 Dimensional Analysis F!U = f ( µ 2!U, g U, L) 2 F/ρU 2 [m 2 /s 0 ] U/U [--] L [m] µ/ρu [m 1 /s 0 ] g/u 2 [m -1 /s 0 ] Length [L] Time [t] Step 4) Rewrite Matrix

54 Dimensional Analysis F!U = f ( µ 2!U, g U, L) 2 F/ρU 2 [m 2 /s 0 ] L [m] µ/ρu [m 1 /s 0 ] g/u 2 [m -1 /s 0 ] Length [L] Step 5) Non-dimensionalize all variables containing length: F/ρU 2 L 2 and µ/ρul and gl/u 2

55 Dimensional Analysis F!U 2 L = f ( µ 2!UL, gl U ) 2 F/ρU 2 L 2 L/L µ/ρul gl/u 2 [m 0 ] [--] [m 0 ] [m 0 ] Length [L] Step 5) Your equation is non-dimensional! Yea! But does it make sense??

56 Classical non-dimensional parameters in fluids F!U 2 L = f ( µ 2!UL, gl U ) 2 C F = F 1!U 2 2 L 2 Force Coefficient (can be found through experiments and is considered an empirical coefficient, L 2 is equivalent to Area of object) Fr = U 2 gl Re =!UL µ Froude Number Reynolds Number (important for all forces in air or water) (in fluids typically only important when near surface of ocean/water)