# Fluid Mechanics Fluid Mechanics by dfhdhdhdhjr

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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
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

2g
p
g

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
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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