fluid mechanics wrap up cee 331 june 27, 2015 cee 331 june 27, 2015
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Fluid Mechanics Wrap UpFluid Mechanics Wrap Up
CEE 331
April 18, 2023
CEE 331
April 18, 2023
ReviewReview
Fluid Properties Fluid Statics Control Volume Equations Navier Stokes Dimensional Analysis and Similitude Viscous Flow: Pipes External Flows Open Channel Flow
Fluid Properties Fluid Statics Control Volume Equations Navier Stokes Dimensional Analysis and Similitude Viscous Flow: Pipes External Flows Open Channel Flow
Shear StressShear Stress
change in velocity with respect to distancechange in velocity with respect to distance
AFAF
2mN
2mN
tU t
U t
Ut
U
dydu dydu
tAU
F t
AUF
AUFt
AUFt
2msN
2msN
dimension of
s1
s1
Tangential force per unit area
Rate of angular deformation
rate of shear
Pressure Variation When the Specific Weight is ConstantPressure Variation When the Specific Weight is Constant
dzdp dzdp
constant zp constant zp
constant z p
constant z p
22
11 z
p z
p 2
21
1 zp
zp
Piezometric head
Center of Pressure: ypCenter of Pressure: yp
Ap ypdAFy Ap ypdAFy
Ap ypdA
Fy
1
Ap ypdAF
y1 sinAyF sinAyF sinyp sinyp
Ap dAy
Ayy
sin
sin1 2
Ap dAyAy
y
sinsin1 2
Ap dAy
Ayy 21
Ap dAy
Ayy 21
Ay
Iy x
p Ay
Iy x
p
Ax dAyI 2Ax dAyI 2
AyII xx2 AyII xx2
yAy
I
Ay
AyIy xx
p 2
yAy
I
Ay
AyIy xx
p 2
Sum of the moments
Transfer equation
y = 0 where p = datum pressurey = 0 where p = datum pressure
Inclined Surface FindingsInclined Surface Findings
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?
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?
yAy
Iy x
p yAy
Iy x
p
Ay
Ixx xy
p Ay
Ixx xy
p coincide
below
decreases
Forces on Curved Surfaces: Horizontal Component
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 ________ of the curved surface
The horizontal component of pressure force on a closed body is always _____
The center of pressure is located on the projected area using the moment of inertia
The horizontal component of pressure force on a curved surface is equal to the pressure force exerted on a horizontal ________ of the curved surface
The horizontal component of pressure force on a closed body is always _____
The center of pressure is located on the projected area using the moment of inertia
projectionprojection
zerozero
Forces on Curved Surfaces: Vertical Component
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
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
C
(78.5kN)(1.083m) - (89.7kN)(0.948m) = ___ 00
0.948 m
1.083 m
89.7kN
78.5kN
Cylindrical Surface Force CheckCylindrical Surface Force Check
All pressure forces pass through point C.
The pressure force applies no moment about point C.
The resultant must pass through point C.
All pressure forces pass through point C.
The pressure force applies no moment about point C.
The resultant must pass through point C.
Uniform AccelerationUniform Acceleration
How can we apply our equations to a frame of reference that is accelerating at a constant rate? _______________________________ _______________________
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.
Use total acceleration including acceleration due to gravity.
Free surface is always normal to total acceleration
Conservation of MassConservation of Mass
cscv
ddtdt
dNAv
cscv
ddtdt
dNAv
N = Total amount of ____ in the systemN = Total amount of ____ in the systemhh = ____ per unit mass = __ = ____ per unit mass = __ N = Total amount of ____ in the systemN = Total amount of ____ in the systemhh = ____ per unit mass = __ = ____ per unit mass = __
cscv
ddtdt
dmAv
cscv
ddtdt
dmAv
cvcs
dt
d Av
cvcs
dt
d Av
massmass11massmass
But dm/dt = 0!But dm/dt = 0!
cv equationcv equation
mass leaving - mass entering = - rate of increase of mass in cvmass leaving - mass entering = - rate of increase of mass in cv
EGL (or TEL) and HGLEGL (or TEL) and HGL
The energy grade line may never be horizontal or slope upward (in direction of flow) unless energy is added (______)
The decrease in total energy represents the head loss or energy dissipation per unit weight
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 __________________
The energy grade line may never be horizontal or slope upward (in direction of flow) unless energy is added (______)
The decrease in total energy represents the head loss or energy dissipation per unit weight
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 __________________
pumppump
coincidentcoincident
reference pressurereference pressure
ltp hHg
Vz
pH
gV
zp
22
22
222
22
111
1
1
ltp hH
gV
zp
Hg
Vz
p 22
22
222
22
111
1
1
Losses and EfficienciesLosses and Efficiencies
Electrical power
Shaft power
Impeller power
Fluid power
Electrical power
Shaft power
Impeller power
Fluid power
electricP electricP
waterP waterP
shaftP shaftP
impellerP impellerP
IEIE
TwTw
TwTw
gQHpgQHp
Motor lossesMotor losses
bearing lossesbearing losses
pump lossespump losses
Linear Momentum EquationLinear Momentum Equation
sspp FFFWF 21 sspp FFFWF 21
21 MMF 21 MMF sspp FFFWMM
2121 sspp FFFWMM 2121
The momentum vectors have the same direction as the velocity vectors
The momentum vectors have the same direction as the velocity vectors
Fp1Fp1
Fp2Fp2
WW
M1M1
M2M2
FssyFssy
FssxFssx
Vector AdditionVector Addition
cs1
cs3
q1
q2cs2
x
y
q3
1M1M
2M2M3M3M
ssFssF
M M M F1 2 3 ss
SummarySummary
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
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
SummarySummary
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
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 EquationsNavier-Stokes Equations
Fa Fa
va 2 ph va 2 ph
a a
ph ph
v2 v2
Navier-Stokes Equation
Inertial forces [N/m3]
Pressure gradient (not due to change in elevation)
Shear stress gradient
h is vertical (positive up)
SummarySummary
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 parametersDimensionless parameters
Reynolds Number
Froude Number
Weber Number
Mach Number
Pressure Coefficient
(the dependent variable that we measure experimentally)
Reynolds Number
Froude Number
Weber Number
Mach Number
Pressure Coefficient
(the dependent variable that we measure experimentally)
ReVl
ReVl
glVFgl
VF
2
2C
Vp
p 2
2C
Vp
p
lV
W2
lVW
2
cV
M cV
M
AVd
2
Drag2C
AVd
2
Drag2C
Froude similarityglVFgl
VFpm FF pm FF
pp
2p
mm
2m
Lg
V
LgV
pp
2p
mm
2m
Lg
V
LgV
p
2p
m
2m
L
V
LV
p
2p
m
2m
L
V
LV
m
pr L
LL
m
pr L
LL rr LV rr LV
rr
rr L
VL
t rr
rr L
VL
t
2/5rrr LLL rrrr LAVQ 2/5rrr LLL rrrr LAVQ
3 3rr r r r r r2
r
LF a L L
tm r= = =3 3r
r r r r r r2r
LF a L L
tm r= = =
difficult to change g
Froude number the same in model and prototype
________________________
define length ratio (usually larger than 1)
velocity ratio
time ratio
discharge ratio
force ratio
Laminar Flow through Circular Tubes
Laminar Flow through Circular Tubes
Velocity
Shear
hpdldra
u
4
22
hpdldra
u
4
22
hpdl
dr
dr
du
2
hpdl
dr
dr
du
2
hpdl
dr
dr
du 2
hpdl
dr
dr
du 2
l
hr l
2
l
hr l
2
l
dhl
40
l
dhl
40
True for Laminar or Turbulent flow
Shear at the wallShear at the wall
Laminar flow
Pipe Flow Energy LossesPipe Flow Energy Losses
R,f
Df
L
DC p
R,f
Df
L
DC p
2
2C
V
ghlp
2
2C
V
ghlp
LD
Vghl
2
2f
LD
Vghl
2
2f
gV
DL
hl 2f
2
g
VDL
hl 2f
2
Dimensional Analysis
Darcy-Weisbach equation
Laminar Flow Friction FactorLaminar Flow Friction Factor
L
hDV l
32
2
L
hDV l
32
2
2
32gD
LVhl
2
32gD
LVhl
gV
DL
hl 2f
2
g
VDL
hl 2f
2
gV
DL
gDLV
2f
32 2
2
gV
DL
gDLV
2f
32 2
2
RVD6464
f
RVD6464
f
Slope of ___ on log-log plot
Hagen-Poiseuille
Darcy-Weisbach
-1
Moody DiagramMoody Diagram
0.01
0.10
1E+03 1E+04 1E+05 1E+06 1E+07 1E+08R
fric
tion
fact
or
laminar
0.050.04
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
lD
C pf
lD
C pf
D
D
0.02
0.03
0.04
0.050.06
0.08
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 TechniquesSolution Techniques
Q Dgh
L DD
gh
L
f
f
F
HGGG
I
KJJJ
2 223 7
1785 2
2 3
. log.
./
/
DLQgh
QL
ghf f
FHG
IKJ
FHG
IKJ
LNMM
OQPP0 66 1 25
24 75
9 4
5 2 0 04
. .
.
.
. .
h fg
LQDf
82
2
5f
D
FH IK
LNM
OQP
0 25
3 75 74
0 9
2
.
log.
.Re .
Re 4QD
Minor LossesMinor 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 coefficient, K, times the velocity head.
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 coefficient, K, times the velocity head.
g
VKh
2
2
g
VKh
2
2
geometry,RepC f geometry,RepC f2
2C
Vp
p 2
2C
Vp
p
2
2C
V
ghlp
2
2C
V
ghlp
g
Vh pl
2C
2
g
Vh pl
2C
2
High ReHigh Re
Swamee Jain Iterative Technique for D and Q (given hl)
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 Calculate minor losses Find new major losses by subtracting minor
losses from total head loss
Assume all head loss is major head loss. Calculate D or Q using Swamee-Jain
equations Calculate minor losses Find new major losses by subtracting minor
losses from total head loss
Q Dgh
L DD
gh
L
f
f
F
HGGG
I
KJJJ
2 223 7
1785 2
2 3
. log.
./
/
minorfl hhh minorfl hhh
42
28
Dg
QKhminor
42
28
Dg
QKhminor
Darcy Weisbach/Moody Iterative Technique Q (given hl)
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
Assume a value for the friction factor. Calculate Q using head loss equations Find new friction factor
minorfl hhh minorfl hhh
42
28
Dg
QKhminor
42
28
Dg
QKhminor
5
2
2
8
D
LQ
gfh f
5
2
2
8
D
LQ
gfh f
Geometric parameters ___________________ ___________________ ___________________
Write the functional relationship
Geometric parameters ___________________ ___________________ ___________________
Write the functional relationship
P
ARh
P
ARh Hydraulic radius (Rh)Hydraulic radius (Rh)
Channel length (l)Channel length (l)
Roughness (e)Roughness (e)
Open Conduits:Dimensional Analysis
Open Conduits:Dimensional Analysis
Re, , ,ph h
lC f
R R
F,M,WRe, , ,p
h h
lC f
R R
F,M,W
glVFgl
VF
R Vl
Open Channel Flow FormulasOpen Channel Flow Formulas
VAQ VAQ
2/13/21oh SAR
nQ 2/13/21
oh SARn
Q
1/2o
2/3h SR
1n
V 1/2o
2/3h SR
1n
V
hSRg
V
2 hSRg
V
2
Dimensions of n?
Is n only a function of roughness?
hSRCV hSRCV
Manning formula (MKS units!)
NO!
T /L1/3
Chezy formula
Boundary Layer ThicknessBoundary 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?
Water flows over a flat plate at 1 m/s. Plot the thickness of the boundary layer. How long is the laminar region?
UxRx
UxRx
U
Rx x
U
Rx x
sm
smxx
/1
)000,500(/101 26
sm
smxx
/1
)000,500(/101 26
0
0.01
0.02
0.03
0.04
0.05
0.06
0.07
0.08
0.09
0.1
0 1 2 3 4 5 6
length along plate (m)
boun
dary
laye
r th
ickn
ess
(m)
-
1,000,000
2,000,000
3,000,000
4,000,000
5,000,000
6,000,000
7,000,000
8,000,000
9,000,000
10,000,000
Rey
nold
s N
umbe
r
laminarturbulentReynolds Number
x = 0.5 m
5/1
5/437.0
Ux
5/1
5/437.0
Ux
Flat Plate:StreamlinesFlat Plate:
Streamlines
U
0 1
2
3
4
2
0
2
2
21U
pp
U
vC p
2
0
2
2
21U
pp
U
vC p
Point v Cp p1234
0 1<U >0>U <0
>p0
>p0
<p0
<p0
Points outside boundary layer!
Flat Plate Drag CoefficientsFlat Plate Drag Coefficients
0.001
0.01
1e+04
1e+05
1e+06
1e+07
1e+08
1e+09
1e+10
Rel
Uln
=Rel
Uln
=
lele
DfCDfC
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
( )[ ] 2.51.89 1.62log /DfC le
-= - ( )[ ] 2.5
1.89 1.62log /DfC le-
= -
( )0.5
1.328
ReDf
l
C =( )0.5
1.328
ReDf
l
C =( )[ ]2.58
0.455 1700Relog Re
Dfll
C = -( )[ ]2.58
0.455 1700Relog Re
Dfll
C = -
( )[ ]2.58
0.455
log ReDf
l
C =( )[ ]2.58
0.455
log ReDf
l
C =
0.20.072ReDf lC -= 0.20.072ReDf lC -=
Drag Coefficient on a Sphere Drag Coefficient on a Sphere
0.10.1
11
1010
100100
10001000
0.10.1 11 1010 102102 103103 104104 105105 106106 107107
Reynolds NumberReynolds Number
Dra
g C
oeff
icie
ntD
rag
Coe
ffic
ient Stokes Law
24ReDC =24ReDC = Re=500000
Turbulent Boundary Layer
More Fluids?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 and Transformation (CEE 655)
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 and Transformation (CEE 655)