Fluid Mechanics Wrap Up CEE 331 May 25, 2016
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Transcript Fluid Mechanics Wrap Up CEE 331 May 25, 2016
Fluid Mechanics Wrap Up
CEE 331
May 25, 2016
Review
Fluid Properties
Fluid Statics
Control Volume Equations
Navier Stokes
Dimensional Analysis and Similitude
Viscous Flow: Pipes
External Flows
Open Channel Flow
Shear Stress
AU
F
t
F
A
U
t
du
dy
Ft
AU
dimension of
Tangential force per unit area
U
t
N s
m 2
N
m 2
1
Rate of angular deformation
s
change in velocity with respect to distance
rate of shear
Pressure Variation When the
Specific Weight is Constant
dp
dz
p z constant
p
z constant
Piezometric head
p1
z1
p2
z2
Center of Pressure: yp
y p F ypdA
A
Sum of the moments
1
F yA sin
y p ypdA
p y sin
A
F
1
2
y = 0 where p = datum pressure
yp
y
sin
dA
yA sin A
1
2
yp
y
dA
yA A
Ix
yp
yA
I x y 2 dA
A
Ix Ix y2 A
Ix y2 A Ix
yp
y
yA
yA
Transfer equation
Inclined Surface Findings
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?
Forces on Curved Surfaces:
Horizontal Component
The
horizontal component of pressure force
on a curved surface is equal to the pressure
force exerted on a horizontal ________
projectionof
the curved surface
The horizontal component of pressure force
zero
on a closed body is always _____
The center of pressure is located on the
projected area using the moment of inertia
Forces on Curved Surfaces:
Vertical Component
The vertical component of pressure force on a
curved surface is equal to the weight of liquid
vertically above the curved surface and
extending up to the (virtual or real) free
surface
Streeter, et. al
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
Uniform Acceleration
How
can we apply our equations to a frame
of reference that is accelerating at a constant
rate? _______________________________
Use total acceleration including
_______________________
acceleration due to gravity.
Free surface is always normal to total acceleration
Conservation of Mass
N = Total amount of ____
mass in the system
h = ____
1
mass per unit mass = __
dN
d v dA
dt
t cv
cs
dm
d v dA
dt
t cv
cs
cv equation
But dm/dt = 0!
cs v dA t cv d
mass leaving - mass entering = - rate of increase of mass in cv
EGL (or TEL) and HGL
The energy grade line may never be horizontal or slope
upward (in direction of flow) unless energy is added
(______)
pump
The decrease in total energy represents the head loss or
energy dissipation per unit weight
coincident
EGL and HGL are ____________and
lie at the free surface
for water at rest (reservoir)
Whenever the HGL falls below the point in the system for
which it is plotted, the local pressures are lower than the
__________________
reference pressure
V12
p2
V22
z1 1
Hp
z2 2
H t hl
1
2g
2
2g
p1
Losses and Efficiencies
Electrical
power
Motor losses
Pelectric IE
Shaft
power
Pshaft
Impeller
T
w
power
Pimpeller T
Fluid
bearing losses
w
power
Pwater gQHp
pump losses
Linear Momentum Equation
F W F F
F M M
p1
1
p2
Fss
Fp2
M2
2
M1 M 2 W Fp1 Fp2 Fss
Fssx
The momentum vectors
have the same direction
as the velocity vectors
M1
Fp1
W
Fssy
Vector Addition
M1 M 2 M 3 Fss
q2
cs2
Fss
M3
y
M2
x
M1
cs1
q1
cs3
q3
Summary
Control volumes should be drawn so that the
surfaces are either tangent (no flow) or normal
(flow) to streamlines.
In order to solve a problem the flow surfaces need
to be at locations where all but 1 or 2 of the energy
terms are known
The control volume can not change shape over
time
When possible choose a frame of reference so the
flows are steady
Summary
Control
volume equation: Required to make
the switch from a closed to an open system
Any conservative property can be evaluated
using the control volume equation
mass,
energy, momentum, concentrations of
species
Many
problems require the use of several
conservation laws to obtain a solution
Navier-Stokes Equations
a F
a h p 2 v
Navier-Stokes Equation
h is vertical (positive up)
a Inertial forces [N/m3]
h p Pressure gradient (not due to change in elevation)
2 v Shear stress gradient
Summary
Navier-Stokes
Equations and the Continuity
Equation describe complex flow including
turbulence, but are difficult to solve
The Navier-Stokes Equations can be solved
analytically for several simple flows
Dimensionless parameters
Reynolds
Froude
Weber
Number
Number
Number
Vl
Re
V
F
gl
W
V 2 l
V
c
2Drag
2p
Cd
C
Pressure Coefficient
2
p
2
V
A
V
(the dependent variable that we measure experimentally)
Mach
Number
M
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
time ratio
discharge ratio
force ratio
Vr L r
Lr
tr
Lr
Vr
Qr Vr Ar L r L r L r L5r / 2
3 Lr
Fr = mr a r = r r L r 2 = L3r
tr
Fm Fp
Vp2
Vm2
g mLm g pLp
2
Vm2 Vp
Lm Lp
Lr
Lp
Lm
Laminar Flow through Circular
Tubes
a2 r 2 d
u
p h
4 dl
du
dr
r d
2 dl
du
dr
p h
r d
Velocity
Shear
p h
2 dl
hl True for Laminar or
r
2l Turbulent flow
Laminar flow
Shear at the wall
0
hl d
4l
Pipe Flow Energy Losses
D
f C p
L
Cp
f , R
D
Dimensional Analysis
2 ghl
V2
2 ghl D
f 2
V L
LV2
hl f
D 2g
Darcy-Weisbach equation
Laminar Flow Friction Factor
D 2 hl
V
32 L
hl
Hagen-Poiseuille
32LV
gD 2
LV2
hl f
D 2g
32LV
LV2
f
2
D 2g
gD
64 64
f
VD R
Darcy-Weisbach
Slope of ___
-1 on log-log plot
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
Minor Losses
We
previously obtained losses through an
expansion using conservation of energy,
momentum, and mass
Most minor losses can not be obtained
analytically, so they must be measured
Minor losses are often expressed as a loss
V2
coefficient, K, times the velocity head. h K
High Re
C p f geometry,Re
2p
Cp
V 2
Cp
2 ghl
V2
hl C p
V
2
2g
2g
Swamee Jain Iterative Technique
for D and Q (given hl)
Assume
all head loss is major head loss.
Calculate D or Q using Swamee-Jain
equations
8Q 2
hminor K
Calculate minor losses
g 2 D 4
Find new major losses by subtracting minor
losses from total head loss hl h f hminor
Q 2.22 D
5/ 2
gh f
L
F
I
G
178
. J
logG
gh J
3.7 D
G
J
D
H
L K
2/3
f
Darcy Weisbach/Moody Iterative
Technique Q (given hl)
Assume
a value for the friction factor.
Calculate Q using head loss equations
Find new friction factor
hl h f hminor
hminor K
8Q 2
g 2 D 4
hf f
8
LQ 2
g 2 D 5
Open Conduits:
Dimensional Analysis
Geometric
parameters
Rh
Hydraulic radius (Rh)
___________________
A
P
Channel length (l)
___________________
Roughness (e)
___________________
Write
the functional relationship
l
C p f Re, F,M, W, ,
Rh Rh
Vl
R
F
V
gl
Open Channel Flow Formulas
V
2g
SRh
1 2/3 1/2
V R h So
n
Q VA
Q
1
ARh2 / 3 S o1 / 2
n
V C SRh
Chezy formula
Manning formula (MKS units!)
1/3
Dimensions of n? T /L
Is n only a function of roughness?
NO!
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
9,000,000
Rx
turbulent
Reynolds Number
0.08
1/ 5
0.37 x 4 / 5
U
0.07
0.06
7,000,000
6,000,000
0.05
5,000,000
0.04
4,000,000
0.03
3,000,000
0.02
2,000,000
0.01
1,000,000
0
0
1
2
8,000,000
3
length along plate (m)
4
5
6
Reynolds Number
boundary layer thickness (m)
x
Rx
U
x
1x10 6 m 2 / s(500,000)
1m / s
x = 0.5 m
Flat Plate:
Streamlines
3
U
0
p p0
Cp 1
2
2
2
U
U
v
Cp
p
>p0
0
1
<U >0
>p0
<p0
>U <0
<p0
v2
2
1
4
Point
1
2
3
4
Points outside boundary layer!
Flat Plate Drag Coefficients
0.01
CDf = [1.89 - 1.62log (e / l )]
1 x 10-3
5 x 10-4
2 x 10-4
1 x 10-4
5 x 10-5
2 x 10-5
1 x 10-5
5 x 10-6
2 x 10-6
1 x 10-6
- 2.5
CDf
CDf =
1.328
(Rel )0.5
CDf =
0.455
[log (Re )]
2.58
l
-
e
l
CDf = 0.072Rel- 0.2
1700
Rel
CDf =
0.001
0.455
[log (Re )]
Rel =
Ul
n
+1
0
1e
+0
9
1e
+0
8
1e
+0
7
1e
+0
6
1e
+0
5
1e
1e
+0
4
l
2.58
Drag Coefficient on a Sphere
Drag Coefficient
1000
100
Stokes Law
10
1
0.1
0.1
1
24
CD =
Re
10
102
103
104
Reynolds Number
105
106
107
Re=500000
Turbulent Boundary Layer
More Fluids?
Hydraulic
Engineering (CEE 332 in 2003)
Hydrology
Measurement
Techniques
Model Pipe Networks (computer software)
Open Channel Flow (computer software)
Pumps and Turbines
Design Project
Pollutant Transport
(CEE 655)
and Transformation