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Monroe L. Weber-Shirk School of Civil and
Environmental Engineering
Fluid Mechanics
EIT Review
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Shear Stress
change in velocity with respect to distance
A
F
2m
N
dy
du
Tangential force per unit area
rate of shear
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P1 = 0 h1
?
h2
Manometers for High Pressures
Find the gage pressure in the center
of the sphere. The sphere contains
fluid with g1 and the manometer contains fluid with g2.
What do you know? _____
Use statics to find other pressures.
1
2
3
=P3
g1
g2
For small h1 use fluid with high density. Mercury!
+ h1g2
- h2g1P1
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Differential Manometers
h1
h3
Mercury
Find the drop in pressure
between point 1 and point
2.
p1 p2 Water
h2
orifice
= p2
p1 - p2 = (h3-h1)gw + h2gHg
p1 - p2 = h2(gHg - gw)
p1
+ h1gw
- h2gHg
- h3gw
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Forces on Plane Areas: Inclined
Surfaces
q
A’
B’
O
O
x
y
c y
c x
R x
R y
Ah F c R g ch
Free surface
centroid
center of pressure
The origin of the y
axis is on the free
surface
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Statics
Fundamental Equations
Sum of the forces = 0
Sum of the moments = 0
A p F c pc is the pressure at the __________________ centroid of the area
y A y
I
A y
A y I y x x
p
2
Line of action is below the centroid
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Properties of Areas
yc
b
a I xc
yc
b
a I xc
A ab=2
c
a y =
3
12 xc
ba I =
2
ab A =
3
c
b d x
+=
3
36 xc
ba I =
2 A R=
4
4 xc
R I
p=
R yc
I xc
0 xyc I =
( )2
272
xyc
ba I b d = -
0 xyc I =
3c
a y =
d
c y R=
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Properties of Areas
3
4 xc
ba I
p= A ab=
4
3c
R y =
a
yc
b
I xc
2
2
R A
p=
4
3c
R y =
4
8 xc
R I
p=
ycR
I xc
0 xyc I =
0 xyc I =
4
16 xc
R I
p=
2
4
R A
p=R
yc
c y a=
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Inclined Surface Summary
The horizontal center of pressure and the
horizontal centroid ________ when the surface
has either a horizontal or vertical axis of symmetry
The center of pressure is always _______ the
centroid
The vertical distance between the centroid and
the center of pressure _________ as the surface
is lowered deeper into the liquid
What do you do if there isn’t a free surface?
y A y
I y x
p
A y
I x x
xy
p coincide
below
decreases
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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 theatmosphere on the other side? Neglect the weight of the gate.
hingewater
F
8 m
4 m
Solution Scheme
Magnitude of the force
applied by the water
Example using Moments
Location of the resultant force
Find F using moments about hinge
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Magnitude of the Force
A p F cr
ab A
abh F r g
m2m2.5πm10m
N 9800
3
r F
b = 2 m
a = 2.5 m
pc = ___
F r = ________
h = _____
hingewater
F
8 m
4 m
Fr
hg
10 m Depth to the centroid
1.54 MN
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Location of Resultant Force
4
3ba I x
h y
y A y
I y x
p
ab y
ba y y p
4
3
y
a y y p
4
2
m12.54
m2.52
y y p
ab A
_______ y y p
________ y
hingewater
F
8 m
4 m
Fr
12.5 mSlant distanceto surface
0.125 m __
p x x b = 2 m
a = 2.5 m
cp
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Force Required to Open Gate
How do we find the
required force?
0hinge M
F = ______ b = 2 m
2.5 ml cp=2.625 m
m5
m2.625 N10x1.54 6
F
tot
cpr
l
l F
F
l tot
hingewater
F
8 m
4 m
Fr
Moments about the hinge
=Fl tot - F r l cp
809 kN
cp
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Example: Forces on Curved
Surfaces
Find the resultant force (magnitude and location)
on a 1 m wide section of the circular arc.
FV =
FH = A p
water 2 m
2 m
3 m W1
W2
W1 + W2
= (3 m)(2 m)(1 m)g + p/4(2 m)2(1 m)g
= 58.9 kN + 30.8 kN
= 89.7 kN
= g(4 m)(2 m)(1 m)
= 78.5 kN y
x
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Example: Forces on Curved
Surfaces
The vertical component line of action goes through
the centroid of the volume of water above the surface.
21V W3
)m2(4W)m1(Fx
water 2 m
2 m
3 m
A
W1
W2
kN89.7
kN30.83
)m2(4kN58.9)m1(
x
Take moments about a vertical
axis through A.
= 0.948 m (measured from A) with magnitude of 89.7 kN
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Example: Forces on Curved
Surfaces
water 2 m
2 m
3 m
A
W1
W2
The location of the line of action of the horizontal
component is given by
y A y
I
yx
p
12
3bh I x
b
h
x I
y
m4.083m4
m1m2m4
m0.667 4
p y
y
x
(1 m)(2 m)3/12 = 0.667 m4
4 m
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Example: Forces on Curved
Surfaces
78.5 kN
89.7 kN
4.083 m 0 . 9
4 8 m
119.2 kN
horizontal
vertical
resultant
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C
(78.5kN)(1.083m) - (89.7kN)(0.948m) = ___ 0
0.948 m
1.083 m
89.7kN
78.5kN
Cylindrical Surface Force Check
All pressure forces passthrough point C.
The pressure forceapplies no moment about point C.
The resultant must pass
through point C.
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Curved Surface Trick
Find force F required to open
the gate.
The pressure forces and force F pass through O. Thus the hinge
force must pass through O!
All the horizontal force iscarried by the hinge
Hinge carries only horizontal
forces! (F = ________)
water 2 m
3 m
A
W1
W2 F
O
W1 + W2 11.23
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Dimensionless parameters
Reynolds Number
Froude Number
Weber Number
Mach Number
Pressure Coefficient
(the dependent variable that we measure experimentally)
Vl R
gl
V
F
2
2C
V
p p
l V W
2
cV M
AV d
2
Drag2C
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Model Studies and Similitude:
Scaling Requirements
dynamic similitude
geometric similitude
all linear dimensions must be scaled identically
roughness must scale
kinematic similitude
constant ratio of dynamic pressures at corresponding
points streamlines must be geometrically similar
_______, __________, _________, and _________
numbers must be the same
Mach Reynolds Froude Weber
C f p M,R,F,W,geometry
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Froude similarity
Froude number the same in model and
prototype
________________________ define length ratio (usually larger than 1)
velocity ratio
time ratio
discharge ratio
force ratio
gl
V F
pm FF
p p
2
p
mm
2
m
Lg
V
Lg
V
p
2
p
m
2
m
LV
LV
m
p
r L
LL r r LV
r
r
r r LV
Lt
2/5
r r r LLL r r r r L AV Q
3 3r r r r r r r 2
r
LF M a L L
t
r = = =
difficult to change g
11.33
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Control Volume Equations
Mass
Linear Momentum
Moment of Momentum
Energy
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Conservation of Mass
cvcs
d t
d Av
021
222111 cscs
d d AvAv
0222111 AV AV
m AV AV 222111
1
2
Q AV AV 2211
v1
A1
V = spatial average of v
If mass in cv
is constant
[M/t]
If density is constant [L3/t]
Area vector is normal to surface and pointed out of cv
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Conservation of Momentum
F M M 1 2
( ) ( )1 1 1 1 1 1V A Qr r = - = -M V V
M V V2 2 2 2 2 2 V A Q
F V V Q Q1 2
F V V Q 2 1 ss p p FFFWF 21
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Energy Equation
l t p h H g
V z
p H
g
V z
p
22
2
222
2
2
2
111
1
1
g
g
g
V K hl
2
2
g
V
D
L f h f
2
2
R
64f
laminar turbulent
Moody Diagram
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z
Example HGL and EGL
z = 0
pump
energy grade line
hydraulic grade line
velocity head
pressure head
elevation
datum
2g
V 2
g
p
2 2
2 2in in out out
in in P out out T L p V p V z h z h h g g
a ag g
+ + + = + + + +
h i i h
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Smooth, Transition, Rough
Turbulent Flow
Hydraulically smooth
pipe law (von Karman,
1930) Rough pipe law (von
Karman, 1930)
Transition function for
both smooth and rough
pipe laws (Colebrook)
51.2
Relog2
1 f
f
D
f
7.3log2
1
g
V
D
L f h f
2
2
(used to draw the Moody diagram)
f
D
f Re
51.2
7.3log2
1
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Moody Diagram
0.01
0.10
1E+03 1E+04 1E+05 1E+06 1E+07 1E+08R
f r i c t i o n f a c t o r
laminar
0.05
0.04
0.03
0.020.015
0.010.0080.006
0.004
0.002
0.0010.0008
0.0004
0.0002
0.0001
0.00005
smooth
l
DC pf
D
0.02
0.03
0.040.05
0.06
0.08
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find head loss given (D, type of pipe, Q)
find flow rate given (head, D, L, type of pipe)
find pipe size given (head, type of pipe,L, Q)
Solution Techniques
Q D gh
L D
D
gh
L
f
f
H
GGG
2 2237
1785 2
2 3
. log.
./
/
DLQ
ghQ
L
gh f f
HG KJ
HG KJ
N
MM
Q
PP
0 66 1 252
4 75
9 4
5 2 0 04
. .
.
.
. .
h f g
LQ
D f
82
2
5
f
D
FH
IK
L
NM
0 25
37
5740 9
2
.
log
.
.
Re
.
Re 4Q
D
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Power and Efficiencies
Electrical power
Shaft power
Impeller power
Fluid power
electric P
water P
shaft P
impeller P
IE
Tw
Tw
gQH p
Motor losses
bearing losses
pump losses
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Manning Formula
1/2
o
2/3
h SR 1
nV
The Manning n is a function of the boundary roughness as well
as other geometric parameters in some unknown way...
RA
P h
A bh P b h 2
Rbh
b h
h
2
Hydraulic radius for wide channels
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