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					                 Fluid Mechanics

                          EIT Review




                             School of Civil and
Monroe L. Weber-Shirk   Environmental Engineering
          Shear Stress

   F                                             N
         Tangential force per unit area         m2 
   A                                              

     du
       change in velocity with respect to distance
     dy
          rate of shear
   Manometers for High Pressures

Find the gage pressure in the center
of the sphere. The sphere contains                     1
fluid with g1 and the manometer
contains fluid with g2.                                g2
What do you know? _____
                      P1 = 0                                     h1
                                          3
                                     g1 ?
Use statics to find other pressures.
                                                            h2
                                                 2
 P1 + h1g2 - h2g1 =P3

For small h1 use fluid with high density.   Mercury!
           Differential Manometers

                      p1      Water                 p2
                                                                   h3
                            orifice
          h1
                                      h2



                                                         Mercury
Find the drop in pressure              p1 + h1gw - h2gHg- h3gw = p2
between point 1 and point              p1 - p2 = (h3-h1)gw + h2gHg
2.
                                       p1 - p2 = h2(gHg - gw)
    Forces on Plane Areas: Inclined
               Surfaces
       Free surface                      O
                                     q       x
         FR  ghc A   hc    A’
                                                 xc
                                                   xR
      centroid
           B’                                           O



                                         The origin of the y
                                yc       axis is on the free
center of pressure         yR
                      y                  surface
                        Statics

 Fundamental     Equations
    Sum of the forces = 0
    Sum of the moments = 0



    F  pc A                              centroid of the area
                pc is the pressure at the __________________


     Ix  y2 A Ix
yp               y     Line of action is below the centroid
        yA      yA
                    Properties of Areas

           b                       a            ba 3
       a             A = ab   yc =       I xc =        I xyc = 0
 Ixc           yc                  2            12

                                   a
                              yc =
       a              ab           3            ba 3
                                                                 ba 2
Ixc                A=                    I xc =        I xyc   =      (b - 2d )
                yc     2           b+d          36               72
                              xc =
           b                         3
           d
                R                               p R4   I xyc = 0
Ixc                  A = p R 2 yc = R    I xc =
               yc                                 4
                      Properties of Areas

      Ixc              yc  p R2         4R          p R4     I xyc = 0
            R           A=         yc =      I xc =
                             2          3p            8

            b
                 a                 yc = a           p ba 3   I xyc = 0
Ixc                     A = p ab             I xc =
                 yc                                   4



            yc             p R2         4R          p R4
R                       A=         yc =      I xc =
                             4          3p           16
         Inclined Surface Summary

 The horizontal center of pressure and the               I xy
  horizontal centroid ________ when the surface x p  x  yA
                       coincide
  has either a horizontal or vertical axis of
  symmetry
 The center of pressure is always _______ the
                                      below            Ix
                                                  yp     y
  centroid                                             yA
 The vertical distance between the centroid and
  the center of pressure _________ as the surface
                          decreases
  is lowered deeper into the liquid
 What do you do if there isn’t a free surface?
           Example using Moments
 An elliptical gate covers the end of a pipe 4 m in diameter. If the
 gate is hinged at the top, what normal force F applied at the
 bottom of the gate is required to open the gate when water is 8 m
 deep above the top of the pipe and the pipe is open to the
 atmosphere on the other side? Neglect the weight of the gate.
Solution Scheme
 Magnitude of the force                                         hinge
  applied by the water                 8m             water

 Location of the resultant force                 F              4m
 Find F using moments about hinge
            Magnitude of the Force

                                                           hinge
Fr  pc A                               8m         water
A  ab                                       Fr
                                               F           4m
h = _____
     10 m      Depth to the centroid
     gh
pc = ___
Fr  gh ab                                  a = 2.5 m
           N
Fr   9800
             10 m π2.5 m 2 m 
             3
           m 
    ________
Fr= 1.54 MN
                                               b=2m
       Location of Resultant Force

     Ix                                                     hinge
yp     y                           8m             water
     yA
                                               Fr
yh
                      Slant distance            F           4m
y  ________
     12.5 m           to surface
          ba 3             ba 3
yp  y              Ix            A  ab
         4 yab               4
                                              a = 2.5 m
         a2
                    yp  y 
                             2.5 m 2
yp  y                                                     cp
         4y                  412.5 m 

y p  y  0.125 m
          _______        x p  __
                               x
                                                b=2m
       Force Required to Open Gate

                                                          hinge
   How do we find the               8m            water
   required force?                           Fr
Moments about the hinge                       F           4m
 M hinge  0 =Fltot - Frlcp
                Fr lcp
         F                                  2.5 m
                 ltot          lcp=2.625 m
                                                               ltot
F
   1.54 x 10 N 2.625 m
            6                                             cp
            5 m 
   F = ______
       809 kN                                b=2m
           Example: Forces on Curved
                   Surfaces
Find the resultant force (magnitude and location)
on a 1 m wide section of the circular arc.

F V = W1 + W 2                                 3m   W1
    = (3 m)(2 m)(1 m)g + p/4(2 m)2(1 m)g
                                         water      2m
    = 58.9 kN + 30.8 kN
    = 89.7 kN                                       W2
                                               2m

FH = p A                                                 x
   = g(4 m)(2 m)(1 m)
   = 78.5 kN                                        y
         Example: Forces on Curved
                 Surfaces
The vertical component line of action goes through
the centroid of the volume of water above the surface.    A
Take moments about a vertical
axis through A.                                          3m   W1
                 4(2 m)
xFV  (1 m) W1         W2                      water         2m
                   3
                     4(2 m)                                   W2
   (1 m)58.9 kN          30.8 kN                    2m
x                     3
               89.7 kN 
= 0.948 m (measured from A) with magnitude of 89.7 kN
        Example: Forces on Curved
                Surfaces
The location of the line of action of the horizontal
component is given by                                     A
       Ix
  yp     y
       yA                  b                             3m       W1

        bh 3                                     water
   Ix                         h                                  2m
        12
                                                                  W2
 I x  (1 m)(2 m)3/12 = 0.667 m4                         2m

 y 4m                                                                 x
          0.667 m 4
 yp                        4 m   4.083 m
      4 m 2 m 1 m 
                                                              y
Example: Forces on Curved
        Surfaces


          0.948 m
                     78.5 kN horizontal
4.083 m
                     89.7 kN vertical

                    119.2 kN resultant
     Cylindrical Surface Force Check
              0.948 m   89.7kN    All pressure forces pass
          C                        through point C.
                                  The pressure force
1.083 m                            applies no moment about
                                   point C.
                                  The resultant must pass
78.5kN                             through point C.



                                       0
(78.5kN)(1.083m) - (89.7kN)(0.948m) = ___
           Curved Surface Trick

 Find  force F required to open
  the gate.                                  A
 The pressure forces and force F
                                            3m   W1
  pass through O. Thus the hinge
  force must pass through O!        water        2m
                                            O
 All the horizontal force is               F    W2
  carried by the hinge
 Hinge carries only horizontal
  forces! (F = ________)
               W1 + W2                           11.23
      Dimensionless parameters
                                            Vl
                                       R
 Reynolds     Number                        
                                          V
                                       F
 Froude    Number                         gl
                                            V 2 l
 Weber    Number                      W
                                             
 Mach    Number                             V
                                       M
                                             c
                                                        2Drag
                                           2p    Cd 
 Pressure Coefficient               Cp 
                                           V 2         V 2 A
    (the dependent variable that we measure experimentally)
  Model Studies and Similitude:
     Scaling Requirements
 dynamic   similitude
   geometric    similitude
     alllinear dimensions must be scaled identically
     roughness must scale

   kinematic    similitude
     constant   ratio of dynamic pressures at corresponding
      points
     streamlines must be geometrically similar

       Mach Reynolds Froude                    Weber
     _______, __________, _________, and _________
      numbers must be the same
            a
    Cp  f M, R, F,W,geometry      f
    V
 F
     gl        Froude similarity                        Fm  Fp


   Froude number the same in model and                  2
                                                        Vm   Vp2
                                                           
    prototype                                         g mLm g pLp
   ________________________
    difficult to change g                               2
                                                       Vm Vp
                                                            2
                                                          
   define length ratio (usually larger than 1)        Lm Lp
                     Vr  L r                                  Lp
   velocity ratio                                      Lr 
                           Lr                                  Lm
                      tr        Lr
 time ratio               Vr
                  Qr  Vr Ar  L r L r L r  L5 / 2
 discharge ratio                                r

                                      3 Lr
 force ratio
                  Fr = M r a r = r r L r 2 = L3r
                                         tr              11.33
  Control Volume Equations

Mass
LinearMomentum
Moment of Momentum
Energy
                   Conservation of Mass

                       If mass in cv                                       2
              
 v  dA   t  d is constant 1
cs                             cv
                                                                v1
 v
cs1
      1   1    dA1     
                        cs 2
                                    2   v 2  dA 2  0   A1

      Area vector is normal to surface and pointed out of cv
 1V1 A1   2V2 A2  0                        V = spatial average of v

  1V1 A1   2V2 A2  m
                                              [M/t]

  V1 A1  V2 A2  Q If density is constant [L3/t]
           Conservation of Momentum

          F M      1    2
                          M

M1 = - ( r 1V1 A1 ) V1 = - ( r Q ) V1


         af af
  M 2   2V2 A2 V2  Q V2

       a a
  F QfV QfV    1           2




   F Qa  f
          V V        2     1            F  W  F   p1    Fp  Fss
                                                              2
               Energy Equation

p1           V12        p2            V22
    z1  1      Hp      z2   2      H t  hl
g1           2g         g2            2g

                                 LV2
                        hf  f
                2
            V
     hl  K
            2g                   D 2g


      laminar                turbulent

          64
       f                    Moody Diagram
          R
                Example HGL and EGL
velocity head         V2
                  
                      2g
                                                p
                             pressure head
                                                g
                                                          energy grade line
                                                         hydraulic grade line


                                                 z elevation
                                   pump
  z=0                                                 datum
                         2                               2
        pin              V
                        in         pout                Vout
            + zin + a in    + hP =      + zout + a out      + hT + hL
        g                2g        g                   2g
     Smooth, Transition, Rough
                                                     LV2
                           h  f
         Turbulent Flow                    f
                                                  D 2g

 Hydraulically smooth          1         Re f      
                                   2 log
                                         
                                                     
                                                     
  pipe law (von Karman,         f         2.51      
  1930)
 Rough pipe law (von                     3.7 D 
                                1
  Karman, 1930)                    2 log
                                         
                                                 
                                                 
                                f          
 Transition function for
  both smooth and rough                D                 
  pipe laws (Colebrook)
                            1
                               2 log     
                                               2.51        
                                       3.7                
                            f                Re f         
                  (used to draw the Moody diagram)
                                   Moody Diagram
                         0.10
                         0.08
     D
f  Cp 
                                                                                  0.05
                                                                                  0.04
      l  0.06                                                                   0.03
                         0.05                                                     0.02

                                                                                            
       friction factor




                                                                                  0.015
                         0.04                                                     0.01
                                                                                  0.008
                                                                                  0.006
                         0.03                                                     0.004
                                                                                            D
                                   laminar
                                                                                  0.002

                         0.02                                                     0.001
                                                                                  0.0008
                                                                                  0.0004
                                                                                  0.0002
                                                                                  0.0001
                                                                                  0.00005
                         0.01                                                     smooth

                           1E+03      1E+04   1E+05       1E+06   1E+07   1E+08
                                                      R
            Solution Techniques
find head loss given (D, type of pipe, Q)
                              0.25                    8 LQ 2
      4Q         f                            hf  f 2
Re 
     D                   LF
                            
                           MH
                                   5.74
                                 0.9
                                         2
                                                  IO
                                                   P  g D5
                     log
                           N
                          3.7 D Re
find flow rate given (head, D, L, type of pipe)
                                                  KQ
                                 F                 I
                          gh
                             logG
                                 G  178 J .
                                                gh J
         Q  2.22 D5/ 2       f

                           L
                                 G D LJ
                                 H
                                  3.7 D
                                                   K
                                                  2/3        f


   find pipe size given (head, type of pipe,L, Q)
                 L F I     2
                                   4 .75
                                        FL I O          5.2 0.04

        D  0.66M G J  Q G J P
                        LQ
                  
                  1.25                     9 .4

                 MH K
                 N      gh f            H KP
                                          gh
                                                 Qf
          Power and Efficiencies

 Electrical   power   Motor losses
  Pelectric  IE
                           bearing losses
 Shaft   power
  Pshaft     Tw
 Impeller   power
                           pump losses
   impeller 
  P           Tw
 Fluid   power
  Pwater  gQHp
                 Manning Formula

          1 2/3 1/2
     V     R h So
          n
The Manning n is a function of the boundary roughness as well
as other geometric parameters in some unknown way...
                                                      A
                                                 Rh 
                                                      P
Hydraulic radius for wide channels
                                                 A  bh
                                                 P  b  2h
                                                       bh
                                                Rh 
                                                     b  2h
                    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
                                                                                   U 2
                      CD =
                                24            Reynolds Number          Fd  Cd A
                                Re                                                  2

				
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posted:1/14/2012
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