Fluid Flow Concepts and Basic Control Volume Equations by 4ZtQyQU

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									                  External Flows




                             CEE 331
                        September 11, 2012
                              School of Civil and
Monroe L. Weber-Shirk    Environmental Engineering
                Overview

 Non-Uniform   Flow
 Boundary Layer Concepts
 Viscous Drag
 Pressure Gradients: Separation and Wakes
 Pressure Drag
 Shear and Pressure Forces
 Vortex Shedding
         Non-Uniform Flow

 In pipes and channels the velocity
  distribution was uniform (beyond a few
  pipe diameters or hydraulic radii from the
  entrance or any flow disturbance)
 In external flows the boundary layer is
  always growing and the flow is non-
  uniform
        Boundary Layer Concepts

   Two flow regimes
     Laminar boundary layer
     Turbulent boundary layer
           with laminar sub-layer
   Calculations of
     boundary layer thickness
     Shear (as a function of location on the surface)
     Drag (by integrating the shear over the entire surface)
    Flat Plate: Parallel to Flow
                        U          U          U
          to                                      boundary
                                                  layer
    y                                             thickness
U                                      d
          x
                                                  shear

        Why is shear maximum at the leading edge of
        the plate? du
                            is maximum
                   dy
           Laminar Boundary Layer:
             Shear and Drag Force
               d   5                            Ux             nx
                 =                       Re x =           d =5
               x   Re x                         n              U
                                                      square
       Boundary Layer thickness increases with the _______
        root
       ______ of the distance from the leading edge of the plate
                                  Based on momentum and mass
             U 3
t 0  0.332                       conservation and assumed
               x                  velocity distribution
                            U 3
       l         l
Fd  w t 0 dx  w 0.332           dx      Integrate along length of plate
       0         0            x

Fd  0.664w U 3l            On one side of the plate!
        Laminar Boundary Layer:
          Coefficient of Drag
                                    2FD
Fd  0.664w U l    3
                              Cd =    2
                                        = f (Re)
                                   rU A

       2(0.664) w U 3l          Dimensional analysis
Cd 
            U 2lw


       1.328w U 3l              1.328                 1.328
Cd                        Cd                   Cd =
           U 2lw                   Ul                    Re
                                                         Ul
                                                   Rel =
                                                         n
            Transition to Turbulence

     The  boundary layer becomes turbulent when
      the Reynolds number is approximately
      500,000 (based on length of the plate)
     The length scale that really controls the
      transition to turbulence is the
       boundary layer thickness
      _________________________

       Ux       Ud   Red d  d  5      Red = 5 Re x
Re x =    Red =          = = =
       n        n    Re x x x  Re x
                                      Red = 3500
         Transition to Turbulence
                 U            U                  U
                                  d

    y                                                  turbulent
U
          x
                                                       Viscous
                                                       sublayer
    to               This slope (du/dy) controls t0.


                Transition (analogy to pipe flow)
        Turbulent Boundary Layer:
             (Smooth Plates)
                                                                 1/ 5
   d 0.37                        Ux                      
                                                     4/5 
    = 1/ 5                Re x =           d  0.37 x  
                                                         
   x Re x                        n                      U 
                               Grows ____________ than laminar
                                      more rapidly
                        1/ 5
                                Derived from momentum conservation
 t 0  0.029 U  
                 
                    2

                 Ux              and assumed velocity distribution
            x 5/4
                                         1/ 5
        l
                              
Fd  w t 0 dx  0.036 U wl  
                               2
                              
                                                Integrate shear over plate
      0                       Ul 
                                                         2FD    æ eö
Cd = 0.072Rel- 1/ 5      5 x 105 < Rel < 107       Cd =      = f Re,
                                                           2
                                                        rU A    è    lø
                                             Boundary Layer Thickness
                                Water flows over a flat plate at 1 m/s. Plot the thickness of
                                 the boundary layer. How long is the laminar region?
                                  0.1                                                                        10,000,000
                                                     laminar                                                                                             Ux
                                 0.09
                                                     turbulent                                               9,000,000
                                                                                                                                                  Re x =
                                 0.08                Reynolds Number                                         8,000,000                                   n
boundary layer thickness (m) .




                                 0.07                                                                        7,000,000
                                                                
                                                                     1/ 5
                                                                                                                                                        n Re x




                                                                                                                          Reynolds Number
                                 0.06       d  0.37 x   4/5
                                                                
                                                                
                                                                                                             6,000,000                             x=
                                 0.05
                                                               U 
                                                                                                             5,000,000
                                                                                                                                                          U
                                 0.04                                                                        4,000,000                           1x10 6 m 2 / s (500,000)
                                                                                                                                            x
                                 0.03
                                                                                                    nx       3,000,000                                    1m / s
                                 0.02                                                          d =5          2,000,000
                                                                                                    U                                             x = 0.5 m
                                 0.01                                                                        1,000,000

                                   0                                                                         -
                                        0        1               2              3              4   5     6
                                                                      length along plate (m)                                                        Grand Coulee
                Flat Plate Drag Coefficients
       0.01
                                                                                   - 2.5             1 x 10-3
                      CDf = [1.89 - 1.62log (e / l )]                                                5 x 10-4
                                                                                                     2 x 10-4
                                                                                                     1 x 10-4   e
                                                                                                     5 x 10-5
                                                                                                     2 x 10-5   l
CDf                                                                                                  1 x 10-5
                                                                                                     5 x 10-6
                                                                                                     2 x 10-6
                                                                                                     1 x 10-6
                                                                                                      CDf = 0.072Rel- 0.2
                           1.328                    0.455               1700
                CDf =                  CDf =                 2.58   -
                        (Rel )0.5              [log (Re )]
                                                       l
                                                                         Rel                                     0.455
      0.001
                                                                                                     CDf =                 2.58
                                                                                                             [log (Re )]
                                                                                                                     l
            4



                       5



                                   6



                                                7



                                                            8



                                                                               9



                                                                                                 0
         +0



                    +0



                                +0



                                             +0



                                                         +0



                                                                            +0



                                                                                              +1
      1e



                 1e



                             1e



                                          1e



                                                      1e



                                                                         1e



                                                                                           1e


                                                     Ul
                                         Rel =
                                                     n
               Example: Solar Car

 Solar cars need to be as efficient as
  possible. They also need a large surface
  area for the (smooth) solar array. Estimate
  the power required to counteract the
  viscous drag at 40 mph
 Dimensions: L: 5.9 m W: 2 m H: 1 m
 Max. speed: 40 mph on solar power alone
 Solar Array: 1200 W peak

air = 14.6   x10-6   m2/s   air = 1.22 kg/m3
               Viscous Drag on Ships

    The    viscous drag on ships can be calculated
       by assuming a flat plate with the wetted
       area and length of the ship
       FD  Fviscous  Fwave
                                                       0.01
                                                                                                                              - 2.5           1 x 10-3
                                                                    CDf = [1.89 - 1.62log (e / l )]                                           5 x 10-4
                                                                                                                                              2 x 10-4
                                                                                                                                              1 x 10-4 e



       2Fd                         Cd r U 2 A
                                                                                                                                              5 x 10-5
                                                                                                                                              2 x 10-5
                                                                                                                                                       l



Cd 
                                                CDf                                                                                           1 x 10-5
                                                                                                                                              5 x 10-6
                                                                                                                                              2 x 10-6



                      Fviscous   =
                                                                                                                                              1 x 10-6
                                                                                                                                               C Df = 0.072 Re l- 0.2
                                                                       1.328                   0.455                 1700
                                                              CDf =                CDf =                  2.58   -
                                                                      (Rel )0.5            [ log (Re )]               Rel




       U A
                                                                                                    l                                                      0.455
                                                                                                                                              C Df =


                                       2
                                                                                                                                                                      2.58


          2                                           0.001                                                                                            [ log (Re )]
                                                                                                                                                                l




                                                          4



                                                                   5



                                                                               6



                                                                                          7



                                                                                                           8



                                                                                                                          9



                                                                                                                                          0
                                                       +0



                                                                +0



                                                                            +0



                                                                                       +0



                                                                                                        +0



                                                                                                                       +0



                                                                                                                                       +1
                                                      1e



                                                               1e



                                                                           1e



                                                                                      1e



                                                                                                   1e



                                                                                                                      1e



                                                                                                                                      1e
                                                                                                 Ul
                                                                                     Rel =
                                                                                                 n




                                                                                                                      Ul
                      Lr3
   Fwave scales with ____                                     Rel =
                                                                                                                      n
        Separation and Wakes

 Separation   often occurs at sharp corners
   fluidcan’t accelerate to go around a sharp
    corner
 Velocities  in the Wake are ______ (relative
                               small
  to the free stream velocity)
 Pressure in the Wake is relatively ________
                                      constant
  (determined by the pressure in the adjacent
  flow)
         Flat Plate:
        Streamlines

              3
                                                    
    U                        Cp  1
                                       v2    p  p0 
                                          2
          2                                       2 
                                             U 
                                       2
                  4                  U
0         1        Point   v   Cp      p
                   1       0    1     >p0
                   2       <U >0      >p0
                   3       >U <0      <p0
                   4                  <p0
                      Points outside boundary layer!
     Application of Bernoulli
            Equation
p1       v12 p2       2
                     v2                  In air pressure change due to
   h1       h2 
        2g         2g                  elevation is small
            2      2
       p0 U    p v
                                    U = velocity of body relative to fluid
        2g  2g
       U2        v2       p       p0
                            
       2g        2g              

        v2     p  p0 
     1  2  2
               U  2 
                         Cp
        U             
                      Flat Plate:
                 Pressure Distribution
                                                      U 2C p
                                         p  p0 
                     p  p0    C p  2
             2

                 2
            v
   Cp  1                                                                p  p0
                        2                  2 
                    U                U 
              2
           U                                                 2
           >U 3
                                                          Fd  Fd front  Fd rear

           <U 2                                       Fd   p front  prear A

                                                                                     U 2
            0 1                                       
                                               Fd  C p    front
                                                                    Cp   rear
                                                                                     2
                                                                                            A

                                                                                 U 2
                                                    Fd  0.8  1.2                        A
                                                                                     2
1 0.8             0    Cp           -1 -1.2                 Cd = 2
        Drag of Blunt Bodies and
          Streamlined Bodies
   Drag dominated by viscous
                      streamlined
    drag, the body is __________.
   Drag dominated by pressure
    drag, the body is _______.
                       bluff
   Whether the flow is viscous-
    drag dominated or pressure-
    drag dominated depends
    entirely on the shape of the
    body.



                                    Bicycle page at Princeton
           Velocity and Drag: Spheres

       æe                            ö
Cd = f    , Re, M, shape, orientation General relationship for
       èD                            ø submerged objects
                    Spheres only have one shape and orientation!

                  2FD
          Cd 
                 U 2 A


    2FD                        C d U 2 A Where C is a function of Re
      2
        = Cd = f (Re)     FD                    d
   rU A                             2
   How fast do particles fall in
      dilute suspensions?
 Whatare the important
 parameters?
   Initialconditions Acceleration due to gravity
   After falling for some time... drag

 What principle or law could
 help us? Newton’s Second Law...
                Sedimentation:
       Particle Terminal Fall Velocity
                                                  p  particle volume
   F  ma                                  Fb   Ap  particle cross sectional area
Fd  Fb  W  0                                  ρ p  particle density
                                       Fd
   W  ppg                                     ρw  water density
   Fb   p  w g                                g  acceleration due to gravity
                         2                       C D  drag coefficient
                    Vt
 Fd  CD AP  w                                  Vt  particle terminal velocity
                     2
                                            W              2FD
        4                                          Cd 
   p  r   3
                         Ap  r   2

       3                                                  U 2 A
        Particle Terminal Fall Velocity
                  (continued)
          Fd  W  Fb

            Vt 2
CD AP  w            p (  p  w ) g
            2
          2 p (  p   w ) g
 Vt 2                             General equation for falling objects
               C D AP  w
          p       2
             d             Relationship valid for spheres
          Ap 3
          4 gd  p   w                       4 gd  p   w 
 Vt 
   2                                      Vt 
          3 CD         w                        3 CD     w
                    Drag Coefficient on a Sphere
                   1000
Drag Coefficient




                    100                  Stokes Law


                     10

                      1

                    0.1
                          0.1   1   10   102    103   104   105   106   107
                               24                            Re=500000
                          CD =           Reynolds Number
                               Re
                                                 Turbulent Boundary Layer
      Drag Coefficient for a Sphere
              Equations
                                  24                  d 2 g  p   w 
          Laminar flow R < 1 CD =                Vt 
                                  Re                        18
                            4 C @
                                  24    3
Transitional flow 1 < R < 10 D        + 1/ 2 + 0.34
                                  Re Re
                                                       gd ( r p - r w )
 Fully turbulent flow R > 104 CD @0.4           Vt »
                                                          0.3    rw

                 Vt d r
            Re =
                    m
 Example Calculation of Terminal
           Velocity
Determine the terminal settling velocity of a
cryptosporidium oocyst having a diameter of 4 m
and a density of 1.04 g/cm3 in water at 15°C.
                                                     d 2 g  p   w 
 ρ p  1040 kg/m    3
                                              Vt 
 ρw  999 kg/m 3                                           18

 g  9.81 m/s   2
                          Vt   
                                 4x10   6
                                              m  9.81 m/s 2 1040 kg/m 3  999 kg/m 3 
                                                 2


                                                                    kg 
                                                       1.14x10 3      
 d  4x10 6 m                                       18
                                                                   sm
                                                                        

                    kg
   1.14x10 3                               Vt  3.14 x107 m/s
                    sm
                                                Vt  2.7 cm/day
                                                                                   Reynolds
      Pressure Gradients: Separation
               and Wakes

                Diverging streamlines




Van Dyke, M. 1982. An Album of Fluid Motion. Stanford:
Parabolic Press.
    Adverse Pressure Gradients
           Streamlines diverge behind object
 Increasing  pressure in direction of flow
                                           p     V2
 Fluid is being decelerated                 +z+    =C
                                          g      2g
 Fluid in boundary layer has less ______
                                     inertia
  than the main flow and may be completely
  stopped.
 If boundary layer stops flowing then
  separation occurs
          Point of Separation

 Predicting the point of separation on smooth
  bodies is beyond the scope of this course.
 Expect separation to occur where
  streamlines are diverging (flow is slowing
  down)
 Separation can be expected to occur around
  any sharp corners
  (where streamlines diverge rapidly)
    Drag on Immersed Bodies
         (more shapes)
 Figures9.19-21 bodies with drag
 coefficients on p 392-394 in text.
   hemispherical shell       0.38
                                   Why?
   hemispherical shell       1.42
                                   Velocity at
   cube                      1.1  separation point
   parachute                 1.4  determines pressure
                                        in wake.


  Vs             ?        The same!!!
    Shear and Pressure Forces

 Shear   forces
   viscous drag, frictional drag, or skin friction
   caused by shear between the fluid and the solid
    surface
                                    length
   function of ___________and ______of object
                 surface area
 Pressure    forces
   pressure drag or form drag
   caused by _____________from the body
               flow separation
   function of area normal to the flow
            Example: Matrix Power

Cd = 0.32
Height = 1.539 m
Width = 1.775 m
Length = 4.351 m
Ground clearance = 15 cm
100 kW at 6000 rpm
Max speed is 124 mph
Where does separation occur?
Calculate the power required to overcome drag at 60 mph and 120 mph.
 What is the projected area?         (
                                  A @ H - G )W
 A = (1.539m - 0.15m)1.775m = 2.5m2
                Electric Vehicles

 Electric   vehicles are designed to minimize drag.
 Typical    cars have a coefficient drag of 0.30-0.40.
 The   EV1 has a drag coefficient of 0.19.
                         Smooth connection to windshield
                 Drag on a Golf Ball

DRAG ON A GOLF BALL comes mainly from
pressure drag. The only practical way of reducing
pressure drag is to design the ball so that the point
of separation moves back further on the ball. The
golf ball's dimples increase the turbulence in the
                                inertia
boundary layer, increase the _______ of the
boundary layer, and delay the onset of separation.
The effect is plotted in the chart, which shows that
for Reynolds numbers achievable by hitting the
ball with a club, the coefficient of drag is much
lower for the dimpled ball.
Why not use this for aircraft or cars?
 Effect of Turbulence Levels on
              Drag
 Flow over a sphere: (a) Reynolds number =
 15,000; (b) Reynolds number = 30,000,
 with trip wire. Causes boundary layer to become turbulent




         Point of separation
         Effect of Boundary Layer
                 Transition
  Ideal (non     Real (viscous)     Real (viscous)
viscous) fluid    fluid: laminar    fluid: turbulent
                 boundary layer     boundary layer




                                   Increased inertia in
No shear!
                                   boundary layer
          Spinning Spheres

 What  happens to the separation points if we
 start spinning the sphere?
                            LIFT!
                 Vortex Shedding

 Vortices are shed alternately
  from each side of a cylinder
 The separation point and thus the
  resultant drag force oscillates          nd
                                      S
 Frequency of shedding (n) given          U
  by Strouhal number S
 S is approximately 0.2 over a
  wide range of Reynolds numbers
  (100 - 1,000,000)
        Summary: External Flows

   Spatially varying flows
     boundary layer growth
     Example: Spillways

   Two sources of drag
     shear (surface area of object)
     pressure (projected area of object)

   Separation and Wakes
       Interaction of viscous drag and adverse pressure
        gradient
                      Solution: Solar Car
                  0.455                                                                                                             Ul
      CDf =                  2.58                     U = 17.88 m/s                     Rx 
              [log (Rel )]                       
                  2Fd
                                       0.01



                                                      l = 5.9 m
                                                    CDf = [1.89 - 1.62log (e / l )]
                                                                                                              - 2.5           1 x 10-3
                                                                                                                              5 x 10-4




          Cd 
                                                                                                                              2 x 10-4
                                                                                                                              1 x 10-4 e
                                                                                                                              5 x 10-5
                                                                                                                              2 x 10-5
                                                                                                                                       l
                                CDf                                                                                           1 x 10-5
                                                                                                                              5 x 10-6




                                                      air = 14.6 x 10-6 m2/s
                                                                                                                              2 x 10-6




                U 2 A
                                                                                                                              1 x 10-6
                                                                                                                               C Df = 0.072 Re l- 0.2
                                                       1.328                   0.455                 1700
                                              CDf =                CDf =                  2.58   -
                                                      (Rel )0.5            [ log (Re )]
                                                                                    l
                                                                                                      Rel
                                                                                                                              C Df =
                                                                                                                                           0.455
                                                                                                                                                      2.58
                                      0.001                                                                                            [ log (Re )]
                                                                                                                                                l
                                          4



                                                   5



                                                               6



                                                                          7



                                                                                           8



                                                                                                          9



                                                                                                                          0
                                       +0



                                                +0



                                                            +0



                                                                       +0



                                                                                        +0



                                                                                                       +0



                                                                                                                       +1
              C d U 2 A
                                      1e



                                               1e



                                                           1e



                                                                      1e



                                                                                   1e



                                                                                                      1e



                                                                                                                      1e
                                                                                 Ul
                                                                     Rel =




                                                      air = 1.22 kg/m3
                                                                                 n



         Fd 
                    2
        (3x10 )(1.22kg / m ) (17.88m / s ) (11.88m )
            -3          3                             Rel = 7.2 x 106             2                                                                 2

Fd   =2
                             2
                                                       Cd = 3 x 10-3
         Fd =14 N                                A = 5.9 m x 2 m = 11.8 m2
        P =F*U=250 W
        Reynolds Number Check
      Vd
 R
       

R
   3.14 x10     7
                      m/s 4 x10 6 m 999kg/m 3 
                                3 kg
                      1.14x10
                                   sm
R = 1.1 x 10-6



R<<1 and therefore in Stokes Law range
 Solution: Power a Toyota Matrix
         at 60 or 120 mph
        2FD
Cd              f (R )
       U 2 A
       C d U 2 A
FD 
           2
       C d U 3 A
 P
        2
    (0.32)(1.2kg / m3 )(26.82m / s)3 (2.5m2 )
 P=
                        2
P = 9.3 kW at 60 mph
P = 74 kW at 120 mph
Grand Coulee Dam
Turbulent boundary layer reaches surface!
                 Drexel SunDragon IV
                 http://cbis.ece.drexel.edu/SunDragon/Cars.html
   Vehicle ID: SunDragon IV (# 76)
    Dimensions: L: 19.2 ft. (5.9 m) W: 6.6 ft. (2 m) H: 3.3 ft. (1 m)
    Weight: 550 lbs. (249 kg)
    Solar Array: 1200 W peak; 8 square meters terrestrial grade solar cells;
    manf: ASE Americas
    Batteries: 6.2 kW capacity lead-acid batteries; manf: US Battery
    Motor: 10 hp (7.5 kW) brushless DC; manf: Unique Mobility
    Range: Approximately 200 miles (at 35 mph on batteries alone)
    Max. speed: 40 mph on solar power alone, 80 mph on solar and battery
    power.
    Chassis: Graphite monocoque (Carbon fiber, Kevlar, structural glass,
    Nomex)
    Wheels: Three 26 in (66 cm) mountain bike, custom hubs
    Brakes: Hydraulic disc brakes, regenerative braking (motor)
    Pressure Coefficients on a Wing




                      
        v2      p  p0 
Cp  1     2
                         NACA 63-1-412 Flowfield Cp's
        U2      U 2 
     Shear and Pressure Forces:
 Horizontal and Vertical Components
FD    p sin   t 0 cos dA     drag Parallel to the approach velocity

FL    p cos  t 0 sin  dA       lift Normal to the approach velocity
                                       p < p0                       U 2
                                       negative pressure Fd  Cd A
                                                                     2
                                                     A defined as projected
                                                          normal
                                                     area _______ to force!
                                                                     U 2
  U                                                        FL  CL A
                                  lift                                2
                                                          
                                              drag
                       p > p0 positive pressure

								
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