Fluid Mechanics
EIT Review
Monroe L. Weber-Shirk
School of Civil and
Environmental Engineering
Shear Stress
F

A
du
 
dy
Tangential force per unit area
N
 m 2 
change in velocity with respect to distance
rate of shear
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.
P1 = 0
What do you know? _____
3
?
g1
Use statics to find other pressures.
P1 + h1g2 - h2g1 =P3
For small h1 use fluid with high density.
2
Mercury!
1
g2
h1
h2
Differential Manometers
p1
Water
p2
h3
orifice
h1
h2
Mercury
Find the drop in pressure
between point 1 and point
2.
p1 + h1gw - h2gHg- h3gw = p2
p1 - p2 = (h3-h1)gw + h2gHg
p1 - p2 = h2(gHg - gw)
Forces on Plane Areas: Inclined
Surfaces
Free surface
O
q
FR  ghc A
hc
A’
x
xc
xR
centroid
B’
O
center of pressure
yR
y
yc
The origin of the y
axis is on the free
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
yA
yA
Line of action is below the centroid
Properties of Areas
b
Ixc
Ixc
a
yc
ab
A
=
yc
2
a
b
d
Ixc
A = ab
R
yc
a
yc =
2
a
yc =
3
b+d
xc =
3
A = p R 2 yc = R
ba 3
I xc =
12
I xyc = 0
3
ba 2
=
(b - 2d )
72
ba
I xc =
36
I xyc
p R4
I xc =
4
I xyc = 0
Properties of Areas
Ixc
yc
4R
yc =
3p
p R4
I xc =
8
I xyc = 0
A = p ab
yc = a
p ba 3
I xc =
4
I xyc = 0
p R2
A=
4
4R
yc =
3p
p R4
I xc =
16
p R2
A=
2
R
b
a
yc
Ixc
R
yc
Inclined Surface Summary
The horizontal center of pressure and the
I xy
coincide when the surface x p  x 
horizontal centroid ________
yA
has either a horizontal or vertical axis of
symmetry
 The center of pressure is always _______
below the y  I x  y
p
yA
centroid
 The vertical distance between the centroid and
the center of pressure _________
decreases as the surface
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
applied by the water
hinge
8m
 Location of the resultant force
 Find F using moments about hinge
water
F
4m
Magnitude of the Force
hinge
Fr  pc A
8m
A  ab
h = _____
10 m
Depth to the centroid
water
Fr
F
gh
pc = ___
Fr  gh ab

N
10 m π2.5 m 2 m 
Fr   9800
3
m 

a = 2.5 m
Fr= 1.54
________
MN
b=2m
4m
Location of Resultant Force
hinge
Ix
yp 
y
yA
yh
12.5 m
y  ________
ba 3
yp  y 
4 yab
2
a
yp  y 
4y
m
y p  y  0.125
_______
8m
Slant distance
to surface
Ix 
ba 3
4
Fr
F
4m
A  ab
2.5 m 
yp  y 
412.5 m 
2
x
x p  __
water
a = 2.5 m
cp
b=2m
Force Required to Open Gate
How do we find the
required force?
hinge
8m
Fr
F
Moments about the hinge
 M hinge  0 =Fltot - Frlcp
F
water
4m
Fr lcp
ltot
lcp=2.625 m
2.5 m

1.54 x 10 N 2.625 m 
F
cp
6
5 m 
F = ______
809 kN
b=2m
ltot
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
= (3 m)(2 m)(1 m)g + p/4(2 m)2(1 m)g
water
= 58.9 kN + 30.8 kN
= 89.7 kN
2m
W1
2m
W2
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.
Take moments about a vertical
axis through A.
4(2 m)
xFV  (1 m) W1 
W2
3
4(2 m)
(1 m)58.9 kN  
30.8 kN 
3
x
89.7 kN 
A
3m
water
W1
2m
2m
W2
= 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
bh 3
Ix 
12
W1
3m
b
h
I x  (1 m)(2 m)3/12 = 0.667 m4
y 4m
0.667 m 4
yp 
 4 m   4.083 m
4 m 2 m 1 m 
water
2m
W2
2m
x
Ix
yp 
y
yA
A
y
4.083 m
0.948 m
Example: Forces on Curved
Surfaces
78.5 kN horizontal
89.7 kN vertical
119.2 kN resultant
Cylindrical Surface Force Check
0.948 m
C
1.083 m
78.5kN
89.7kN
All pressure forces pass
through point C.
 The pressure force
applies no moment about
point C.
 The resultant must pass
through point C.

(78.5kN)(1.083m) - (89.7kN)(0.948m) = ___
0
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 is
carried by the hinge
 Hinge carries only horizontal
forces! (F = ________)
W1 + W2
A
water
3m
W1
O
F
2m
W2
11.23
Dimensionless parameters
 Reynolds
 Froude
 Weber
Number
Number
Number
R
Vl

V
F
gl
W
V 2 l

V
c
2Drag
 2p
Cd 
C

 Pressure Coefficient
2
p
2

V
A
V
 (the dependent variable that we measure experimentally)
 Mach
Number
M
Model Studies and Similitude:
Scaling Requirements
 dynamic
similitude
 geometric
similitude
 all
linear dimensions must be scaled identically
 roughness must scale
 kinematic
 constant
similitude
ratio of dynamic pressures at corresponding
points
 streamlines must be geometrically similar
 _______,
Mach __________,
Reynolds _________,
Froude and _________
Weber
numbers must be the same
Cp  f M, R, F,W,geometry
a
f
V
F
gl
Froude similarity

Froude number the same in model and
prototype

________________________
difficult to change g

define length ratio (usually larger than 1)

velocity ratio
Vr  L r
Lr
tr 
 Lr
 time ratio
Vr
Qr  Vr Ar  L r L r L r  L5r / 2
 discharge ratio
3 Lr
3
F
=
M
a
=
r
L
=
L
r
r r
r
r 2
r
 force ratio
tr
Fm  Fp
Vp2
Vm2

g mLm g pLp
2
Vm2 Vp

Lm Lp
Lr 
Lp
Lm
11.33
Control Volume Equations
Mass
Linear
Momentum
Moment of Momentum
Energy
Conservation of Mass
2
If mass in cv

 v  dA   t  d is constant 1
cs
cv
 v
1
cs1
1
 dA1 

2
v 2  dA 2  0
A1
v1
cs 2
Area vector is normal to surface and pointed out of cv
 1V1 A1   2V2 A2  0
1V1 A1   2V2 A2  m

V = spatial average of v
[M/t]
V1 A1  V2 A2  Q If density is constant [L3/t]
Conservation of Momentum
F M
1
M 2
M1 = - ( r 1V1 A1 ) V1 = - ( r Q ) V1
a
f af
F aQfV aQfV
F QaV V f
M 2   2V2 A2 V2  Q V2
1
2
2
1
F  W  F
p1
 Fp  Fss
2
Energy Equation
V12
p2
V22
 z1  1
 Hp 
 z2   2
 H t  hl
g1
2g
g2
2g
p1
2
V
hl  K
2g
hf  f
LV2
D 2g
laminar
turbulent
64
f
R
Moody Diagram
Example HGL and EGL
velocity head

V2
2g
pressure head
p
g
energy grade line
hydraulic grade line
z elevation
pump
z=0
2
in
datum
2
pin
V
pout
Vout
+ zin + a in
+ hP =
+ zout + a out
+ hT + hL
g
2g
g
2g
Smooth, Transition, Rough
h  f
Turbulent Flow
LV2
f
Hydraulically smooth
pipe law (von Karman,
1930)
 Rough pipe law (von
Karman, 1930)
 Transition function for
both smooth and rough
pipe laws (Colebrook)
D 2g
 Re f
 2 log

 2.51
f

1




 3.7 D 

 2 log


  
f
1
 D
2.51

 2 log


f
Re f
 3.7
1




(used to draw the Moody diagram)
Moody Diagram
0.10
0.08
 D
f  Cp 
l  0.06

0.05
0.04
0.03
friction factor
0.05
0.02
0.015
0.04
0.01
0.008
0.006
0.004
0.03
laminar
0.002
0.02
0.001
0.0008
0.0004
0.0002
0.0001
0.00005
0.01
1E+03
smooth
1E+04
1E+05
R
1E+06
1E+07
1E+08

D
Solution Techniques
find
head loss given (D, type of pipe, Q)
2
0.25
8
LQ
4Q
f 
2
hf  f 2
Re 
5

5
.
74

g
D
D
log
 0.9
3.7 D Re
find flow rate given (head, D, L, type of pipe)
L
F
M
NH
IO
KP
Q
F
I
gh
G

178
.  J
Q  2.22 D
logG 
gh J
L
3.7 D
G
J
D
H
L K
find pipe size given (head, type of pipe,L, Q)
L
F
F
LQ I
L I O
D  0.66M
 G J  Q G J P
gh K
gh KP
H
H
M
N
Q
f
5/ 2
2/3

2
1.25
5.2 0.04
4 .75
9 .4
f
f
f
Power and Efficiencies
 Electrical
power
Motor losses
Pelectric  IE
 Shaft
power
Pshaft 
 Impeller
Tw
power
Pimpeller  Tw
 Fluid
bearing losses
power
Pwater  gQHp
pump losses
Manning Formula
1 2/3 1/2
R h So
n
The Manning n is a function of the boundary roughness as well
as other geometric parameters in some unknown way...
V
Hydraulic radius for wide channels
A
Rh 
P
A  bh
P  b  2h
bh
Rh 
b  2h
Drag Coefficient on a Sphere
Drag Coefficient
1000
100
Stokes Law
10
1
0.1
0.1
CD =
1
24
Re
10
102
103
104
Reynolds Number
105
106
Fd  Cd A
107
U 2
2