monroe l. weber-shirk s chool of civil and environmental engineering fluid mechanics eit review
TRANSCRIPT
Monroe L. Weber-Shirk
School of Civil and
Environmental Engineering
Fluid MechanicsFluid Mechanics
EIT ReviewEIT Review
Shear StressShear Stress
change in velocity with respect to distancechange in velocity with respect to distance
AFAF
2mN
2mN
dydu dydu
Tangential force per unit area
rate of shear
P1 = 0P1 = 0 hh11
??
hh22
Manometers for High PressuresManometers for High Pressures
Find the gage pressure in the center of the sphere. The sphere contains fluid with 1 and the manometer contains fluid with 2.
What do you know? _____
Use statics to find other pressures.
Find the gage pressure in the center of the sphere. The sphere contains fluid with 1 and the manometer contains fluid with 2.
What do you know? _____
Use statics to find other pressures.
11
22
33
=P3=P3
1
2
For small h1 use fluid with high density. Mercury!Mercury!
+ h12+ h12 - h21- h21P1P1
Differential ManometersDifferential Manometers
h1
h3
Mercury
Find the drop in pressure between point 1 and point 2.
p1p2Water
h2
orificeorifice
= p2= p2
p1 - p2 = (h3-h1)w + h2Hg
p1 - p2 = h2(Hg - w)
p1p1 + h1w+ h1w - h2Hg- h2Hg- h3w- h3w
Forces on Plane Areas: Inclined Surfaces
Forces on Plane Areas: Inclined Surfaces
A’
B’B’
OO
OO
xx
yy
cycy
cxRx
RyRy
AhF cR ch
Free surfaceFree surface
centroid
center of pressurecenter of pressure
The origin of the y axis is on the free surface
StaticsStatics
Fundamental Equations Sum of the forces = 0 Sum of the moments = 0
Fundamental Equations Sum of the forces = 0 Sum of the moments = 0
ApF c ApF c pc is the pressure at the __________________centroid of the area
yAy
IAy
AyIy xx
p 2
yAy
IAy
AyIy xx
p 2
Line of action is below the centroid
Properties of AreasProperties of Areas
yc
baIxc
yc
b
aIxc
A ab=2c
ay =
3
12xc
baI =
2ab
A =
3c
b dx
+=
3
36xc
baI =
2A Rp=4
4xc
RI
p=R
ycIxc
0xycI =
( )2
272xyc
baI b d= -
0xycI =
3c
ay =
d
cy R=
Properties of AreasProperties of Areas
3
4xc
baI
p=A abp=
43c
Ry
p=
ayc
b
Ixc
2
2R
Ap
=43c
Ry
p=
4
8xc
RI
p=
ycR
Ixc
0xycI =
0xycI =
4
16xc
RI
p=
2
4R
Ap
=Ryc
cy a=
Inclined Surface SummaryInclined 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?
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
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.
hingewater
F
8 m
4 m
Solution SchemeMagnitude of the force applied by the waterMagnitude of the force applied by the water
Example using MomentsExample using Moments
Location of the resultant forceLocation of the resultant force
Find F using moments about hingeFind F using moments about hinge
Magnitude of the ForceMagnitude of the Force
ApF cr ApF cr
abA abA
abhFr abhFr
m 2m 2.5πm 10m
N 9800
3
rF m 2m 2.5πm 10
m
N 9800
3
rF
b = 2 m
a = 2.5 m
pc = ___
Fr= ________
h = _____
hingehingewaterwater
FF
8 m
4 m
FrFr
hh
10 m Depth to the centroid
1.54 MN
Location of Resultant ForceLocation of Resultant Force
4
3baI x
4
3baI x
hy hy
yAy
Iy x
p yAy
Iy x
p
abyba
yy p
4
3
aby
bayy p
4
3
ya
yy p 4
2
y
ayy p 4
2
m 12.54
m 2.5 2
yy p
m 12.54
m 2.5 2
yy p
abA abA
_______ yy p _______ yy p
________y ________y
hingehingewaterwater
FF
8 m
4 m
FrFr
12.5 m12.5 mSlant distance to surfaceSlant distance to surface
0.125 m0.125 m __px __px xxb = 2 m
a = 2.5 m
cp
Force Required to Open GateForce Required to Open Gate
How do we find the required force?How do we find the required force?
0hingeM 0hingeM
F = ______ b = 2 m
2.5 mlcp=2.625 m
m 5
m 2.625N 10 x 1.54 6
F
m 5m 2.625N 10 x 1.54 6
F
tot
cpr
l
lFF
tot
cpr
l
lFF
ltot
hingehingewaterwater
FF
8 m
4 m
FrFr
Moments about the hinge=Fltot - Frlcp=Fltot - Frlcp
809 kN809 kN
cpcp
Example: Forces on Curved Surfaces
Example: Forces on Curved Surfaces
Find the resultant force (magnitude and location) on a 1 m wide section of the circular arc.
FV =
FH = ApAp
water 2 m
2 m
3 m W1
W2
W1 + W2W1 + W2
= (3 m)(2 m)(1 m) + p/4(2 m)2(1 m)= (3 m)(2 m)(1 m) + p/4(2 m)2(1 m)= 58.9 kN + 30.8 kN= 58.9 kN + 30.8 kN= 89.7 kN= 89.7 kN
= (4 m)(2 m)(1 m)= (4 m)(2 m)(1 m)= 78.5 kN= 78.5 kN y
x
Example: Forces on Curved Surfaces
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
)m 2(4W)m 1(Fx
21V W
3)m 2(4
W)m 1(Fx
water 2 m
2 m
3 m
A
W1
W2
kN 89.7
kN 30.83
)m 2(4kN 58.9)m 1(
x
kN 89.7
kN 30.83
)m 2(4kN 58.9)m 1(
x
Take moments about a vertical axis through A.Take moments about a vertical axis through A.
= 0.948 m (measured from A) with magnitude of 89.7 kN= 0.948 m (measured from A) with magnitude of 89.7 kN
Example: Forces on Curved Surfaces
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
yAy
Iy x
p yAy
Iy x
p
12
3bhI x 12
3bhI x
b
h
xI xI
yy
m 4.083m 4m 1m 2m 4
m 0.667 4
py m 4.083m 4m 1m 2m 4
m 0.667 4
py
y
x
(1 m)(2 m)3/12 = 0.667 m4(1 m)(2 m)3/12 = 0.667 m4
4 m4 m
Example: Forces on Curved Surfaces
Example: Forces on Curved Surfaces
78.5 kN78.5 kN
89.7 kN89.7 kN
4.083 m
0.94
8 m
119.2 kN119.2 kN
horizontalhorizontal
verticalvertical
resultantresultant
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.
Curved Surface TrickCurved 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 = ________)
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 = ________)
water 2 m
3 m
A
W1
W2FFOO
W1 + W2W1 + W2 11.23
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)
VlRVlR
glVFgl
VF
2
2C
Vp
p 2
2C
Vp
p
lV
W2
lV
W2
cV
M cV
M
AVd
2
Drag2C
AVd
2
Drag2C
Model Studies and Similitude:Scaling Requirements
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
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 sameMach Reynolds Froude Weber
C fp M, R, F,W,geometrya fC fp M, R, F,W,geometrya f
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
glVFgl
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 M a L L
tr= = =3 3r
r r r r r r2r
LF M a L L
tr= = =
difficult to change g
11.33
Control Volume EquationsControl Volume Equations
MassLinear MomentumMoment of MomentumEnergy
MassLinear MomentumMoment of MomentumEnergy
Conservation of MassConservation of Mass
cvcs
dt
d Av
cvcs
dt
d Av
021
222111 cscs
dd AvAv 021
222111 cscs
dd AvAv
0222111 AVAV 0222111 AVAV
mAVAV 222111 mAVAV 222111
1122
QAVAV 2211 QAVAV 2211
vv11AA11
V = spatial average of vV = spatial average of v
If mass in cv is constantIf mass in cv is constant
[M/t][M/t]
If density is constant [L3/t]If density is constant [L3/t]
Area vector is normal to surface and pointed out of cvArea vector is normal to surface and pointed out of cv
Conservation of MomentumConservation 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 Qa f a fF V V Q Qa f a f1 2
F V V Q 2 1a fsspp FFFWF 21 sspp FFFWF 21
Energy EquationEnergy Equation
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
gV
Khl 2
2
g
VKhl 2
2
g
V
D
Lfh f
2
2
g
V
D
Lfh f
2
2
R64
f R64
f
laminar turbulent
Moody Diagram
zz
Example HGL and EGL
z = 0
pump
energy grade line
hydraulic grade line
velocity head
pressure head
elevation
datum
2g
V2
2g
V2
pp
2 2
2 2in in out out
in in P out out T L
p V p Vz h z h h
g ga a
g g+ + + = + + + +
2 2
2 2in in out out
in in P out out T L
p V p Vz h z h h
g ga a
g g+ + + = + + + +
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
51.2
Relog2
1 f
f
D
f
7.3log2
1
D
f
7.3log2
1
g
V
D
Lfh f
2
2
g
V
D
Lfh f
2
2
(used to draw the Moody diagram)
f
D
f Re
51.2
7.3log2
1
f
D
f Re
51.2
7.3log2
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
FHGIKJ
FHGIKJ
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
Power and EfficienciesPower 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
TT
TT
QHpQHp
Motor lossesMotor losses
bearing lossesbearing losses
pump lossespump losses
Manning FormulaManning Formula
1/2o
2/3h SR
1n
V 1/2o
2/3h SR
1n
V
The Manning n is a function of the boundary roughness as well The Manning n is a function of the boundary roughness as well as other geometric parameters in some unknown way...as other geometric parameters in some unknown way...
RAPh
A bh
P b h 2
Rbh
b hh 2
Hydraulic radius for wide channelsHydraulic radius for wide channels
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 =
2
2UACF dd
2
2UACF dd