流體力學講義_王曉剛_義守大學
TRANSCRIPT
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0. 4
1. ..7-dimensionsunits.7-viscosity.8-CouettePoiseuille..15
-..18-stress field..19- ..23
2. Fluid Statics.24- ..24-..32- .38-..47- buoyancy..50
3. - Bernoulli equation..51
- streamline.51
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- .59-.67-.71
4. Fluid Kinematics.80-velocity field..80-acceleration field..84- control volume
system..87
-Reynolds transport theorem905.finite control volume.98
-.98- 105-116
6. differential..124- kinematics124-130-..140- inviscid flow.144-viscous flow..154
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0
meteorology
oceanographyhydrology
aerodynamics
ArchimedesHero of
Alexandria
~400
Da Vinci 15
jethydraulic jump
Mariotte 16
Newton 17
viscosity of linear fluid-
newtonian fluidperfect
or frictionless fluid
bernoulliEulerLagrange
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Laplace
Eulerdifferential
integral
hydraulics
PittoWeberHagen
PoiseuilleDarcy
19 experimental
hydraulics theoretical
hydrodynamics
Froude
Rayleighdimensional analysis
Reynolds
Reynolds number
NavierStokes
Navier-Stokes
20
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Prantdlboundary layer
theory-
20 von Karman
Taylor
~1990
computational fluid dynamics, CFD CFD
PHOENIX, FLUENT, CFD2000
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1.1 dimensionsunits
principle dimensions
SI units
Mass {M}/ Force {F}
(F = MLT-2)
kilogram (kg)/(N)
Length {L} meter (m)
Time {T} (s)
Temperature {} (K)
force of 1 newton (N) = 1 (kg.m/s2)
energy of 1 joule (J) = 1 (N.m)
power of 1 watt (W) = 1 (J/s)
pressure of 1 pascal (Pa) = 1 (N/m2)
viscosity = kg/m.s
specific heat = J/kg.K = m2/s2.K
Bernoullis equation
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ghuppo ++=2
2
1
{N/m2
} = {N/m2
} + {kg/m3
. m2
/s2
} + {kg/m3
. m/s2
. m}
(kg/m3. m2/s2= kg/m.s2= kg . m/s2. 1/m2= N/m2)
density (kg/m3) = mass/unit volume
specific volume v(m3/kg) = 1/
specific weight = g =
specific gravity s.g. = /H2O
s.g.Hg= 13.6
Hg= (13.6)(1000 kg/m3) = 13.6 X 103kg/m3
1.2 viscosity
:?
shear stress
(deformation angle force)
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(fluid deforms continuously)
hookes law
deformation rate
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P
U
U 0no-slip condition u = u(y) u(y) = Uy/b
velocity gradientdu/dy = U/b
t AB
b
a =tan
a = Ut
b
tU=
deformation rate
tt
0lim
=&
dy
du
b
U==&
= P/A
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& dydu
dy
du= (1.1)
muviscosity
dynamic viscositykg/m.s
nu= /= kinematic
viscosity(m2/s)
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(1.1)newtonian fluid
non-newtonian fluid
momentum exchange
long change
cohesion force
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1.3 CouettePoiseuille
1.Couette
2.pressure dropPoiseuille
du/dyvelocity profile or velocitydistribution
frictional force
V
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control volumea
= 0 x
p = const. In x-direction (Why?)
1= 2= y
.constdy
du==
byayu +=)(
1. u = 0 at y = 0 (no-slip condition)2. u = V at y = h (no-slip condition)
(,)
yh
Vyu )()( =
.consth
V==
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dp/dx 0+x p1> p2,
dp/dx < 0
lrrpp 2)(2
21 =
dr
durpp =
=
2)( 21
l
)4
(1
)()(2
21 rcpp
yu
=l
u = 0 at y = R,4
2R
c=
))((4
1)(
2221 rRpp
yu
=l
= -dp/dx, (dp/dx = pressure gradient)
@r=0221
max )(41)0( Rppuu
l==
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max22120
2
1)(
8
12)(uR
pp
R
rdrruu
R
=
==
l
HW
1.4
viscometer
torque meter
y
yU /
=
=rU )2(/
rhr
T
A
rT
A
F
===
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hr
Ty32
=
1.5 stress field
forces
1.body forces- gravitational force,dVgr
EM force
2. surface forces- stress forcepressure force
An Ft
Fnt n
shear stressA
FtA
t
0lim
=
normal stressA
FnA
n
0lim
=
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A
FxA
xx
0lim
=
A
FyA
xy
0lim
=
A
FzA
xz
0lim
=
xyx: x
y: y
18
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)( xxxx = )( yyyy = )( zzzz =
)( xyyx = )( yxxy =
xyyx = )()( yxxy =
6
x 0, y 0 z 0
6
zzzyzx
yzyyyx
xzxyxx
6
tensor
non-viscous flow or inviscid
flowfrictionless
0.....==== yzyxxy
pzzyyxx === hydrostatic
pressure
dy
duyx =
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1.6
1.system2.control volume
finiteintegral
infinitesimal
differential
sysEWQ =
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1. lagrangian
2.eulerian
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fluid statics
shear stress
hydrostatic
pressure
2.1 pressure distribution
V
sin0 sxpzxpF syy ==
2cos0
zyxgsxpyxpF szz
==
cossy= sinsz=
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sy pp = zgpp sz 2
1+=
0x 0y 0z
pppp zyx ===
scalarvector- Pascals law
)(3
1zzyyxxp ++=
pressure field
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y
zyxy
pzx
y
y
ppzx
y
y
ppFy
=
+
= )
2()
2(
zyxx
pFx
= zyx
z
pFz
=
zyxkz
pj
y
pi
x
pkFjFiFF
zyxpress
)(rrrrrrr
+
+
=++=
gradient
kz
jy
ix
gradrrr
+
+
==
kz
fscalarj
y
fscalari
x
fscalarfscalerfscalargrad
rrr
+
+
==)()()(
)()(
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:
zyxpFpress =r
directional derivative
f Pbr 1=b
r
f PbrDbf df/ds
sPfQf
dsdffD
sb )()(lim
0==
Qbr C C
bsPkszjsyisxsrrrrrr
+=++= )()()()( 0s
chain rule
dsdz
zf
dsdy
yf
dsdx
xf
dsdffDb
++
==
bkds
sdzj
ds
sdyi
ds
sdx
ds
srd rrrrr
=++=)()()()(
)(fbds
dffDb ==
r
ar
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)(fa
a
ds
dffDa == r
r
222
32),,( zyxzyxf ++= P(2,1,3) kiarvr
2=
kzjyixfrrr
264)( ++= P kjifrrr
668)( ++=
789.15
4)668(
5
)2()( =++
=== kji
kif
a
a
ds
dffDa
rrrrr
r
r
f Par
1. f P f P
cos)(cos)( ffbfDb == r
br(f) Dbf (f)
2. f x,y,z)=C=const. P f P normal vector
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HW: 22 531500),,( yxzyxp =
P(-0.2,0.1,1)
xyz
zyxpzyxkz
pj
y
pi
x
pFpress =
++
= )(
rrrr
pzyx
F
volumed
Fd
f
presspress
press ===
rr
r
)(
driving action p
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-T heat flux
gfgravrr
=
)(2
2
2
2
2
2
z
V
y
V
x
Vfviscous
+
+
=
rrrr
conservation of momentum
...... 2 +++=+++== Vgpffffa viscousgravpressi
i
rr
rrrrr
hydrostatic pressure
0=ar 02 = Vr Why?
gp r+=0
{} +
{} = 0
p surface of constantpressure
0)()( =+++
+
+
kgjgigkz
pj
y
pi
x
pzyx
rrrrrr
0== yx gg ggz =
0=xp 0=y
p gzp =
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x y z
p = p(x,y,z) = p(z)
=2
112 gdzpp
incompressible = const
)( 1212 zzgpp =
compressible
gRT
pg
z
p==
==2
11
22
1ln
Tdz
Rg
pp
pdp
=
RT
zzgpp
)12
12
(exp
HW:
T(K) = (288 0.006507 z(m))
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2.2
absolute pressure
gage pressure
p = 29.92 in Hg = 760 mm Hg = 101.325 kPa
1 Pa = 1 N/m
barometer
atmvapor pghpgh =+
mercury Hg
mmmg
ph atm 76076.08.910006.13
1001.15
==
==
1 bar
/ equal level/equal pressure
principle -
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p1= p2= p3
hydraulic jackhydraulic brake
B,C Cp
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C
B
C
B
C
B
A
A
pA
pA
F
F=
=
AB>> AC
piezometer
A
atmA pghp += 11 absolute
11ghpA = gage
U-U-tube manometer
U/
A
2211 ghghpA =+ gage
1
22 ghpA
U
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A,B
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ghgphgp BA 211 )( ++=++ ll
ghppBA )( 12 =
2 < 1
flow nozzle U
Q
BA ppKQ = K
212 )( ghpp BA =
inclined-tube manometer
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sin22 lgpp BA = sin2
2g
ppBA
=l
)(21 hHgpp ++=
hdHD 224
1
4
1 =
])(1[ 221D
dghpp +=
h = 1 cm
Papp 1.90])3010(1[01.08.91000827.0 221 =+=
1 Pa
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l224
1
4
1dHD = 2)(
D
dH l=
])(sin[ 2
D
dgpp aA ll += ])([sin 2
D
dg
ppaA +=
l
cm1=l pa= 0 pA
])10
8.0([sin8.910008.0
101
/1 2232
2
+=
s
m
m
kg
m
mN
118.0sin = o78.6=
2.3
1. resultant force
2.
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FR
===AAA
R ydAgdAgyghdAF sinsin
AyydA
c
A
=
x the first moment of the
area w.r.t. x-axis yc
centroid x y
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ApAghgAyF cccR === )(sin
hc
(xR, yR)
FR x moment
x
====
Ac
R
AAA
RR dAy
Ay
FdAygdAygydFyF 222 sinsin
Ay
dAy
yc
AR
=
2
xA
IdAy = 2 Ix x the second
moment of area or moment of inertia Ixc x
2cxcx AyII += c
c
xccR y
Ay
Iyy +=
0>Ay
I
c
xc center of pressure, yR
y xR
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Ay
I
Ay
xydA
xc
xy
c
AR ==
xy
A
IxydA=
Ixy y Ixyc
y
ccxycxy yAxII += AyI
xxc
xyc
cR +=
y Ixyc=
0 Why?
0)( =+=+== +++ xforxforxforxforA
xyc xydAxydAxydAxydAxydAI
1.FR= ghcA = pcA = ()x()
2.xR, yRxc, yc
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12
stop
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547.1160sin
10==
ocy
kNAghF cR 1230)4(4
1108.91000 2 ===
0=R
x 633.11)4(
4
155.11
)2(4
1
547.112
4
=
+=+=
Ay
Iyy
c
xc
cR
0866.0= cR yy
c
0)( = cRR
yyFM mkNmkNM == 107)0866.0)(1230(
pressure prism-
)])([(2
1
2bhghA
hgApF aveR === = ()
pa= ps= gage pressure
pa
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ghAApFFF sR 2
121 +=+=
3
2
2212211
hF
hFyFyFyF RR +=+=
yR
0.9
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NAghpF s 24400)6.0)(28.910009.050000()( 211 =+=+=
NAhh
gF 954)6.0)(2
6.0(8.910009.0
2
2122 ==
=
kNFFFR 4.2521 =+=
yo
)2.0()3.0( 21 FFyF oR += myo 296.0=
)3
1)()(
2
1()
2
1)(()
3
1)(()
2
1)(( 21 LLbgLLLbpLLbFLLbFLF st +=+=
N
LbgLLbpF st
2620
6
6.09.09.08.91600
2
9.06.05000)
3
1)()(
2
1()
2
1)((
=
+
=+=
Is Ft= F1+ F2? Why?
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2.4
dA
ApdFdrr
=
kFjFiFAdpF zRyRxRA
R
rrrrr
,,, ++==
inner product or dot product
=====A A
xx
AA
rxR dFpdAiAdpiFdiFF
rrrrrr
,
=====A A
yy
AA
ryR dFpdAjAdpjFdjFFrrrrrr
,
=====A A
zz
AA
rzR dFpdAkAdpkFdkFFrrrrrr
,
1. FR,x FR,y)
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x y2. FR,z FV)
zVgVgdghdApdAFF
VA
z
A
zVzR =====,
center of gravity
2FFH = WFFV += 1
BC ABC FH
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Fv 22 )()( VHR FFF +=
BC
O
HFCACAgCAghF =+= )1)()((21)1)()((2
BAabovefluidofweightBAghF ....)1)()((1 ==
))(1()(4
1.. 2 gBAABCofweightW ==
CBabovefluidofweightWFFV ....1 =+=
22 )()( VHR FFF +=
Ox1,x2 211 xFWx =
3
)(4
221
BABAxx =+
O
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2.5 buoyancy
buoyancy or buoyant force
Archimedes principle:
1. 2.
volumebodytoequivalentfluidofweight
surfaceaboveweightfluid
surfaceaboveweightfluidFFF VVB
=
==
)1(
)2()1()2(
)()()( 1212 volumebodygdAzzgdAppFbody
H
body
HB ===
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-
Bernoulli equation
inviscid or non-viscous
3.1 streamline
= Famrr
)()(
)()(
particleonforcegravitynetparticleonforcepressurenet
onacceleratiparticlemassparticle
+=
steadyfluidparticlestreamline
tangent
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s = s(t)
radius of curvature R = R(s)
dtsdV /rr
= dtVda /rr
=
nasadt
Vda ns
rrr
r+==
Vs
V
dt
ds
s
V
dt
dVas
=
== )( V = V(s) = V(s(t))
R
Van
2
=
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s
VVVamF ss
== )()( ynsV =
VgW =
sinsin VgWWs ==
(Why -?)
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p,
p + ps p - ps
infinitesimally small Taylor
expansion )2
( s
s
pps
Vs
pyns
s
p
ynpynppynppF sssps
=
=
=+= 2)()(
Vs
pgFWF psss )sin(
=+=
s
VV
s
pg
=
sin
potential flow AB
)1(3
3
x
aVV o +=
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AB = 0
s
VV
s
p
=
4
3
3
32
4
3
3
3
)1(3
)3
)(1(
x
a
x
a
V
x
aV
x
aV
x
VV
s
VV
o
oo
+=
+=
=
AB 4
3323 )/1(3
x
xaVa
x
p o +=
AB
]2
)/()[()(6
32
)(0
xaxaVdpdx
xpxp o
pp
gagep
xx
x
+=== =
=
=
=
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B 0 0
stagnation point
ds
Vd
s
p
ds
dzg
)(
2
1 2 =
dsdz /sin =
0)(2
1 2 =++ gdzVddp
1 2
0)2
1()
2
1( 1
2
2
2 =++++ gzVpgzVp
streamlinealongconstagzVp .2
1 2 =++
Bernoulli equation
1.2.3.4.
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12
222212112
1
2
1gzVpgzVp ++=++
oVV =1 12 zz = 0
2 =V
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221122
1
2
1oVVpp ==
ds
Vd
s
p
ds
dzg
)(
2
1 2 =
dx
Vd
x
p )(
2
1 2=
dxdx
Vddx
dx
dp =1
2
21
2
)(
2
1)( 2222112
2
1)(
2
1oVVVpp ==
pressure head
kinetic headpotential
head
230
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V2/2 gz p(gage)
1 0
2 0
3 0 0
2 12
23
3.2
centrifugal force
R
VV
R
mVFn
22 ==
coscos VgWWn ==
Vn
p
ysnn
p
yspysppysppF nnnpn
=
=
=+= 2)()(
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V
n
pgFWF pnnn )cos(
=+=
R
V
n
p
dn
dzg
2 =
dn
dz=cosQ
nr 1
2
0)()( 1
2
2
2
=++++ gzdnRV
pgzdnR
Vp
linenormalaalongconstagzdnR
Vp .
2
=++
(a)forced vortex(b)free
vortex(a) rCrV 1)( = (b)r
CrV 2)( =
oo prp =)(
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xy dz/dn = 0
rn = // R = rr
V
n
p
dn
dzg
2 =
(a) rCr
V
r
p 21
2
==
oo prrCrp += )(2
1)( 2221
(b) 32
2
2
r
C
r
V
r
p ==
oo
prr
Crp += )11
(2
1)(
22
2
2
orrrr
gzdnR
Vpgzdn
R
Vp == ++=++ )()(
22
(a) = drrrC
rpdrr
rCrp
o
o
o
22
1
22
1 )()(
oo prrCrp += )(2
1)( 2221
(b) = drrC
rpdrr
C
rpo
o 3
2
1
3
2
1
)()(
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oo
prr
Crp += )11
(2
1)(
22
2
2
0/ > rp
12
AB =R 12
21. =+ linenormalalongconstagzp
)(02 gagepp atm == 121221 )( =+= ghzzgpp
34 )(04 gagepp atm ==
4
int
2
43
int
2
3
43
)()( gzdzR
Vpgzdz
R
Vp
z
poany
z
poany
++=++
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== 4
3
4
3
2
34
2
343 )(
z
z
z
z
dzR
Vghdz
R
Vzzgp
3 CD 3
rigid-body rotation-
angular
velocity
cylindrical coordinates
zr e
z
pe
p
r
e
r
pp
rrrr
+
+
=
1
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0,0,22
==== zrrr aaerer
Va
rrrrr
)( keegpa zz
rrrrr==
gz
ppr
r
p
=
=
=
,0,2
r
z
gdzdrrdzz
pdr
r
pdp =
+
= 2
dp = 0
Crg
zg
r
dr
dz+== 2
22
2,
= gdzdrrdp 2
Cgzrzrp += 2
2
2),(
r
)( 2121 zzgpp =
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z
)(2
1 22
2
1
2
21 rrpp =
vortices
free vortex
forced vortex
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1 2
3 ~
3
2
2
2
2
3
22
3
2
2
2
1
2
1)(
)11
(2
1)(
Vr
Cprp
rAs
prr
Crp
o
o
==
+=
22
max22
1Vpp o =
2
2
2
22
22
2
1
2
1rgzprgzp +=+
-
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67
22
max
22
max2
2
1
2
1
2
1
VVp
VVpp
o
+=
+=
2
max1 Vpp o =
3.3
streamlineaalongconstapgzVp T==++ 2
2
1
1. p
(thermodynamic pressure
Why?
static pressurepiezometer
-
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68
ghgagep =)(1
2. gz hydrostatic pressure
3. V2/2 dynamic pressure
)(2
112
2hHgppV == (Why?)
2 stagnation point
p2stagnation pressure
2
122
1Vpp +=
-
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69
4. pTtotal
pressure
pitot-tube-
-
5/22/2018 _ _
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70
ps p1
ghgygxpghgygxp sg ++=++1
ghpp gs )(1 =
v21/2
ghv
g)(2
1
=
x y
-
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71
3.4
free jets
12 22
221
2
112
1
2
1gzVpgzVp ++=++
)(021 gagepp == hz =1 02=z 01V
2
2
1Vgh = ghV 2=
25 52
552
2
222
1
2
1gzVpgzVp ++=++
)(05 gagep = Hz =5
)(25 HhgV +=
34 42
443
2
332
1
2
1gzVpgzVp ++=++
)(04 gagep = 04=z l=3z 03V
)(3 l= hgp
-
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dj
dh vena contracta
90o
confined flows
conservation of mass
continuity equation
-
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73
0=+ SIO &&& { } { } { } 0.. =+ VCinmassofratetorageSmassofflowratenImassofflowrateutO &&&
mass flowratem& kg/s Qm =&
volume flowrateQm3/s AVQ =
222111 VAVA =
incompressible
2211 VAVA = 21 QQ =
-
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74
2
333
2
222
2
1112
1
2
1
2
1VpVpVp +=+=+
01V ).(03 jetfreep =
3
13
2
pV = 22212
2
1Vpp =
=== 321 3
31
11 /26.1
)27315(/29
/314.8
)1013(mkg
Kkmolekg
KkmolemkPa
kPa
RT
p=
+
+==
=== 321
smV /6926.1
1000323 =
=
smVAVAQ /00542.0)69()01.0(4
32
2233 ====
Pap
smVVAV
2963)67.7)(26.1(2
13000
/67.7/
2
2
2332
==
==
)(321 absoluteppp ===
321
-
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75
U h
SG SG < 1
22
221
2
112
1
2
1gzVpgzVp ++=++
2211 VAVAQ ==
])/(1[2
1)( 212
2
21221 AAVzzgpp +=
U ghSGgpghgzzgp = ll 2121 )(
ghSGzzgpp )1()( 1221 +=
])(1[2
1)1()( 2
1
22
2A
AVghSGghoil ==
)1(2
)/(1
)/(
2
122
2 SGg
AA
AQh
=
-
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76
h 12 z2 z1
p1 p2
cavitation
compressibility
saturation
pressure
condensationpressure
transient ~ 690 MPa
-
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77
siphoned tube
32
3332
2
2221
2
1112
1
2
1
2
1gzVpgzVpgzVp ++=++=++
mftz 57.4151 == Hz =2 mftz 52.153 ==
01V 031 == pp 32 VV = (Why?)
-
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78
smzzgVV /94.10)(2 3132 ===
221222121122
1)(
2
1
2
1VHzggzVgzVpp =++=
kPapCsat
o 71.115@ = kPap 29.9910171.12 ==
mH 6.8=
flowrate measurement
orifice plate
nozzle venturi
-
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79
22222
1112
1
2
1VpVp +=+ 2211 VAVAQ ==
])/(1[)(2
2
12
212
AAppAQ
=
Qactual Q
pQ
-
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80
Fluid Kinematics
fluid dynamics- F
=ma
fluid kinematics-
4.1 velocity field
ktzyxwjtzyxvitzyxuVrrr
),,,(),,,(),,,( ++=
zzrr ezrvezrvezrvV rrrr ),,(),,(),,( ++=
dtrdV AA /rr
=
),,,( tzyxVV
rr
= 2/1222
)( wvuVV ++==
r
-
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81
EulerianLagrangian
-
-
5/22/2018 _ _
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82
streamline
ktzyxwjtzyxvitzyxuVrrr
),,,(),,,(),,,( ++=
wdz
v
dy
u
dx
V
dr
===
-
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83
jyxviyxuVrrr
),(),( +=
u
v
dx
dy=
))(/( jyixVV orr
lr
=
x
y
xV
yV
u
v
dx
dy
o
o =
==)/(
)/(
l
l
= xdx
y
dy Cxy=
= constant on a streamline
= xy
= (x,y) stream function
-
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84
4.2 acceleration field
Eulerian
Lagrangian
steady-state
ar
),,,( tzyxVVrr
= )(),(),( tzztyytxx ===
z
Vw
y
Vv
x
Vu
t
V
dt
dz
z
V
dt
dy
y
V
dt
dx
x
V
t
V
dt
Vda
+
+
+
=
+
+
+
==
rrrrrrrrrr
t
V
r
local acceleration
z
V
wy
V
vx
V
u
+
+
rrr
convective
-
5/22/2018 _ _
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85
acceleration
gradient operator r
kz
jy
ix
rrrr
+
+
=
VVt
V
dt
Vda
rrrrr
r)( +
==
)()( += rrVtDt
Dordt
d
eulerian time-derivative operator
total differentiation
ktyjxzitVrrrr
23 ++=
kyxyztjtxytzi
jxtyktzxzjztkyiDt
VD
jxz
Vkty
y
Vjz
x
V
kyikt
wj
t
vi
t
u
t
V
tywxzvtu
rrr
rrrrrr
rr
rr
rr
rrrrrr
)2()3(3
))(()2)(())(3()3(
,2,
3
,,3
22
22
2
2
++++=
++++=
=
=
=
+=
+
+
=
===
kDt
Dwj
Dt
Dvi
Dt
Du
Dt
VD rrrr
Q ++=
zu
wy
u
vx
u
ut
u
Dt
Du
+
+
+
=
-
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86
z
vw
y
vv
x
vu
t
v
Dt
Dv
+
+
+
=
z
ww
y
wv
x
wu
t
w
Dt
Dw
+
+
+
=
)2
1()(L
xVxu o += Vo= 10 m/s L = 1 m
v = w = 0 u =u(x)
)2
1(22
)]2
1([2
L
x
L
V
L
V
L
xV
x
uu
Dt
Du ooo +=+=
=
22
/600)21(2
)( smL
V
Dt
Du oLx =+= =
60
-
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87
4.3 control volumesystem
1. conservation of mass
2. Newtons 2ndlaw
3. 1stlaw of thermodynamics
4.
-
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88
constMsys = 0=Dt
DMsys Why D/Dt ?
F
)()( momentumsytemDt
DVM
Dt
D
Dt
VDMaMF syssyssys ====
r
r
rr
Q W
0=+ sysEQW 0=+Dt
DE
Dt
DQ
Dt
DW sys
==sysM
sys VdeedmE
sys
gzVue ++=2
2
e total energy u
internal energy
Q T
T
Qds
==
sysM
sys VdssdmS
sys
S entropy
-
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89
1.
2.
-
5/22/2018 _ _
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90
4.4 Reynolds transporttheorem
nformulatio
VolumeControlTheoremTransportynolds
nformulatio
SystemRe
-
5/22/2018 _ _
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91
extensive propertyB
intensive propertyb
B = bm, b = dB/dm
B(extensive property) b(intensive property)
B = m b = 1
B = mv b = v
B = (1/2)mv2 b = (1/2)v2
B = E b = e
B = S b = s
== sys
ii
i
iV
sys VbdVbB
)(lim0
time rate of change
Dt
VbdD
Dt
DB syssys)(
=
(Why D/Dt ?)
-
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92
t
Vbd
t
Bcvcv
=
)( (Why /t ?)
B = m b = 1
Dt
VdD
Dt
Dm
Dt
DB syssyssys)(
==
t
Vd
t
m
t
Bcvcvcv
=
=
)(
-
5/22/2018 _ _
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93
0)(
=Dt
VdDsys
0
)(
= nVV rr
2
0
==in
SurfaceControlin
SurfaceControl
in dAnVbdAbVB
rr&
cos
-
5/22/2018 _ _
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97
0cos
-
5/22/2018 _ _
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98
finite control
volume analysis
1.integral C.V. analysis
2. infinitesimaldifferential C.V.
analysis
5.1 continuity
equation
-
5/22/2018 _ _
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99
t
+
=
+=
inininoutoutout
cv
CSCVsys
VAVAt
m
dAnVVdt
VdDt
D
rr
+
=
..int.. Vtheo
massofflow
ofratenet
VCcoincident
theofmasstheof
changeofratetime
systemcoincident
theofmasstheof
changeratetime
kg/s
0=DtDmsys
0=+
CSCV
dAnVVdt
rr
-
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100
0=
CV
Vdt
0=IO &&
0==
=
inout
inininoutoutout
CS
mm
VAVAdAnV
&&
rr
QAVm ==& mass
flowrate A
VA
dAnV
V A =
=
rr
volume flowrate
0=+
CSCV
dAnVVdt
rr
0
12 mm && = 1122 QQ = smAV
QQ
/0251.0 322
21
==
=
-
5/22/2018 _ _
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101
(1)
12 mm && = 111222 VAVA =
2
1
21 VV
=
RT
p=
sftRpsia
sftRpsia
Tp
VTpV
o
o
/219)453)(100(
)/1000)(540)(4.18(
21
2121
==
=
laminar(2)
-
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102
=
=
UAVA
VAVA
ininin
inininoutoutout
11
0
rdruVA
R
outoutout 20
22 = WHY?
02 110
22 = UArdruR
fully developed region
velocity
profileWhy?
])(1[ 2max2R
ruu =
0)42
(2 202
42
max = URR
rru
R
Uu
VUu ===2
,2 max2maxr
2
V Vcr
W
W:
-
5/22/2018 _ _
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103
Vcr:
W:
0=+
=
+
inininoutoutout
cv
CSCV
WAWAt
m
dAnWVdt
rr
WAWAm
WAWAmdAnW
infuel
infuel
CS
22111
222111 0
=
=+=&
&rr
-
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104
boundary layer theory
2)()(2)(
yy
U
yu=
bc
0=CS
dAnV rr
0=+++ dacdbcab AAAA
dAnVdAnVdAnVdAnV rrrrrrrr
==cdabbc AAA
bc dAnVdAnVdAnVm rrrrrr
&
)1()1(0
===
UUdyUdAdAnV
abab AA
rr
3
)1(2)1(])()(2[)(
0
2 ===
U
dyyy
UdAyudAnV
cdcd AA
rr
0)1(3
1)1(
3
2)1( >== UUUmbc&
-
5/22/2018 _ _
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105
5.2
{ }
++=
)( viscouspressure
forccessurfaceforcebody
systemtheof
momentumtheof
changeofratetime
=sys
sysFVdVDt
D rr
t
=CVcoincident
theofcontentssys FFr
+=
+
=
ininininoutoutoutout
CV
CSCVsys
VAVVAVVdVt
dAnVVVdVt
VdVDt
D
rrr
rrrrr)(
-
5/22/2018 _ _
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106
+
=
.... VCtheofout
momentumofflow
ofratenet
VCcoincident
theofmomentumtheof
changeofratetime
systemcoincident
theofmomentumtheof
changeratetime
=+
=
+
CVtheofcontentsininininoutoutoutout
CV
CSCV
FVAVVAVVdVt
dAnVVVdVt
rrr
rrrr)(
kg.m/s
=+CV
theofcontentsFSIO &&&
=
+
ii
F
smkg
CVthein
momentumof
ratetorageS
smkg
momentumof
ratenflowI
smkg
momentumof
rateutflowO
)/()/()/(
&&&
-
5/22/2018 _ _
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107
anchoring
force
kwiuVrrr
+=
= xFVAuVAu 111222
= zFVAwVAw 111222
AxFVAVVAV = 111121 cos
AzFVAVAV = 11121 )0(sin
)cos1()cos1( 12
1 == VmAVFAx &
sinsin 121 VmAVFAz &==
-
5/22/2018 _ _
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108
+
-
5/22/2018 _ _
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109
=CV
theofcontentsininininoutoutoutout FVAVVAV rr
2211
2,21,11122
212,21,1111222
)()()()(
)(
ApApWWFAppAppWWFmVmV
AApApApWWFVAwVAw
wnA
atmabsatmabswnA
atmabsabswnA
+=+=
++=
&&
mmm &&& == 21
0)1(
)(
2211
1
21
221121
>+++=
+++=
ApApWWA
AVm
ApApWWVVmF
wn
wnA
&
&
p1> p2 Why?
)(0 21 AApRWF atmznA +=
2,21,11122 )()( ApApWRmVmV absabswz += &&
-
5/22/2018 _ _
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110
2,21,121 )( ApApWRVVm absabswz +=&
Rz
221121 )( ApApWWVVmF wnA +++= &
Rz
U
+
kwjviuVrrrr
++=
-
5/22/2018 _ _
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111
2211
2,21,11122
212,21,1111222
111222
2
)()()()()(
0
ApApFVm
AppAppFmVmVAApApApFVAvVAv
F
FVAuVAu
Ay
atmabsatmabsAy
atmabsabsAy
Ax
Ax
++=
++= +++=
=
=
&
&&
FAy Why?
)(0 21 AApRF atmyAy +=
-
5/22/2018 _ _
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112
Ry Why?
2,21,12 ApApRVm absabsy ++= &
Rz
22112 ApApVmFAy = &
yR AyF
(1)(2)pressure drop
-
5/22/2018 _ _
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113
2211 ApWRApVAwVAw zinininoutoutout =
Rz
22
111111 )()( RwmwwAwVAw ininin === &
34
])(1[)2(2
2)(
22
1
222
0
1
0
2
2222
2
Rw
rdrR
rw
rdrwdAwwVAw
R
R
A
outoutout
=
=
==
2211
22
1
22
13
4ApWRApRwRw z =
11
2
121
3 A
R
A
Wwpp z++=
1.(1)uniform(2)
mw&(1)(2)
Why?
2.
-
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114
3.(1)(2)fully developed
region
viscous shear stress
lrrpp 2)( 221 =
r = R ()
A
rpp w
l221 =
wwall shear stress
R
wrR l2=
l
12 pp
dx
dp
-
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115
pressure gradient
Why?
boundary layer theory
2)()(2)(
yy
U
yu=
drag force
DFdAnVVdAnVV xAA
==+ 31
)()( rrrrrr
++=
00
)1()()1()( dyuudyUU
h
oo
)1()1(0
22 = dyuhUD o
h
-
5/22/2018 _ _
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116
+==h
o
CS
udydyUdAnV0 0
)1()1()(0)(
rr
)1()1(0
= dyuhUo
)1()(0
= dyuUuD o
momentum-integral
theory
)1(15
2)1()21)(2(
)1(])()(21)[)()(2()1()(
22
1
0
22
22
0
2
0
=+=
+=
oo
oo
UdU
dyyyyy
UdyuUuD
5.3
+
=
systemtheo
transferworkbyaddition
energyofratetimenet
systemtheo
transferheatbyaddition
energyofratetimenet
systemtheofenergy
storedtotaltheof
increaseofratetime
intint
-
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117
+= sysoutinsysoutsys
in WWQQVde
Dt
D)()( &&&&
sysinnetinnetsysWQVde
DtD )( && +=
e energy/mass
{ } { } { }energypotentialenergykineticenergyernalgzV
ue ++=++= int2
2(
t
CVcoincident
innet
innetsys
innet
innet WQWQ )()(
&&&& +=+
b = e
+
=
+=
+
=
ininoutout
CV
ininininoutoutoutout
CV
CSCVsys
memeVdet
VAeVAeVdet
dAnVeVdet
VdeDt
D
&&
rr
)(
CVinnet
innet
CSCV
WQdAnVeVde
t
)()( &&rr
+=+
W
usefulnoninnet
usefulinnet
innet WWW
+= &&&
pistonshaft
-
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118
flow work
flow work
pAVVFW stressnormalworkflow ==&
==CS
outoutoutininin
inworkflow
net dAnVpVApVApW )( rr
&
Why -?
+=+
CS
CV
inusefulnet
innet
CSCV
dAnVpWQdAnVeVdet
)()()( rr&&
rr
-
5/22/2018 _ _
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119
CV
inusefulnet
innet
CSCV
WQdAnVp
eVdet
)())(( &&rr
+=++
CV
inusefulnet
innet
CSCV
WQdAnVgzV
hVdet
)())(2
(2
&&rr
+=+++
CV
in
usefulnet
innet
ininoutout
CV
WQ
mgzV
hmgzV
hVdet
)(
)2
()2
(22
&&
&&
+=
+++++
puh += ( enthalpy)
+=+inusefulnet
innet WQSIO
&&&&&
=
+
)/(
int
inf
)/()/()/(
sJ
CVtheowork
usefulandheatof
ratelownettotal
sJ
CVthein
energyof
ratetorageS
sJ
energyof
ratenflowI
sJ
energyof
rateutflowO&
&&
-
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120
inusefulnet
innetinout
inout
inout WQzzgVV
hhm &&& +=+
+ )](2
)[(22
inusefulnet
innetinout
inoutinout wqzzg
VVhh +=+
+ )](
2)[(
22
kJ/kg
-
5/22/2018 _ _
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121
inusefulnet
innetinout
inoutinout wqzzg
VVhh +=+
+ )](
2)[(
22
kgkJw
kgkJ
kJ
J
sN
mkgmN
J
s
m
kg
kJ
VVhhw
out
inoutinout
inusefulnet
/797
/797
)1000](1[2
]1[)3060(
)33482550(
2)(
2
2
222
22
==
+=
+=
)(22
22
innetinoutin
inin
out
outout quugzVp
gzVp
++=++ ((
lossquuinnetinout = )(
((
kJ/kg
0.4 kW
-
5/22/2018 _ _
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122
losswzzgVV
hh
inusefulnetinout
inoutinout =+
+ )](2
)[(22
)2
()2
( 1
2
112
2
22 gzVp
gzVp
lossw
inusefulnet ++++=
kgmNsNmkg
smVlossw
inusefulnet /72
)]/()(1[2
)/12(
2 222
2 =
==
inusefulnet
inusefulnet
w
lossw
=
-
5/22/2018 _ _
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123
m
W
w in
usefulnet
inusefulnet
&
&
=
AVm =&
kgmN
smmmkg
skWmNkW
VD
W
w in
usefulnet
inusefulnet
/8.95
)/12](4/)6.0()[/23.1(
)]/()(1000)[4.0(
)4/( 2322
2
=
==
&
%2.758.95
72==
-
5/22/2018 _ _
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124
differential analysis
infinitesimal
6.2 fluid element
kinematics
translationlinear deformation
-
5/22/2018 _ _
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125
rotationangular
deformation
1. normal derivatives:z
w
y
v
x
u
,,
2. cross derivatives:y
w
x
w
z
v
x
v
z
u
y
u
,,,,,
normal derivatives
-
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126
cross derivatives
x 0xu t
))()(( tzyxx
uVinChange
=
x
u
t
tx
u
dt
Vd
V t
=
=
)][(lim)(1
0
yz 0,0
z
w
y
v
V
z
w
y
v
x
u
dt
Vd
V
rr=
+
+
=
)(1
-
5/22/2018 _ _
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127
volumetric dilation rate
Why?
t OA
ttOA
0lim
=
t
x
v
x
txxv
=
=
)/(tan
x
v
t
txv
tOA
=
=
)/([lim
0
0>x
v OA
t OB
-
5/22/2018 _ _
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128
tt
OB
0lim
=
tyux tyyu == )/(tan
y
u
t
tyu
tOB
=
=
)/([lim
0
0>y
u OB
x-y z
)(2
1
y
u
x
vz
= Why 1/2 ?
OBOA = x
v
y
u
=
y-z x
)(2
1
z
v
y
wx
=
x-z y
)(2
1
x
w
z
uy
=
kji zyx
rrrr
++=
-
5/22/2018 _ _
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129
curl
VVcurlrrrr
==2
1
2
1
wvu
zyx
kji
V
=
rrr
rr
(vorticity)
Vrrrr == 2
0== Vrrr
irrotational flow
velocity
potential
jyxixyVrrr
)(24 22 +=
0),(2,4 22 === wyxvxyu
0)(2
1=
=z
v
y
wx
0)(
2
1
=
x
w
z
uy
-
5/22/2018 _ _
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130
0)44(2
1)(
2
1==
= xxy
u
x
vz
y
u
x
vorz
=
= 0 OA OB
cross derivatives
OA
OB
+=
y
u
x
v
t
ty
ut
x
v
t tt
+
=
+
==
]
)()(
[limlim00
&
y
u
x
v
=
6.1
-
5/22/2018 _ _
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131
0=
+ CS CV
Vdt
dAnV rr
0=+ SIO &&&
])()()[(
])()()[(
),,(),,(),,(
),,(),,(),,(
dxdywdxdzvdydzu
dxdywdxdzvdydzu
VAVAdAnVIO
zyxzyxzyx
dzzyxzdyyxzydxx
inininoutoutout
CS
++
++=
==
+++
rr
&&
Taylors expansion
dxx
uuu zyxzydxx
++
)()()( ),,(),,(
dyy
vvv zyxzdyyx
++
)()()( ),,(),,(
dzz
www zyxdzzyx
++
)()()( ),,(),,(
-
5/22/2018 _ _
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132
dxdydzz
w
y
v
x
uIO ]
)()()([
+
+
= &&
dxdydzt
Vdt
Vdt
SCVCV
=== &
0])()()(
[ =
+
+
+
dxdydzz
w
y
v
x
udxdydz
t
0)()()(
=
+
+
+
z
w
y
v
x
u
t
0)( =+
Vt
rr
1. )(r
divergence
)(
)()()(
scalaraz
w
y
v
x
u
wkjviukz
jy
ix
V
+
+
=
++
+
+
= rrrrrrrr
2. )(r
gradient
-
5/22/2018 _ _
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133
)(
))(()(
vectorakz
pj
y
pi
x
p
pkz
jy
ix
p
rrr
rrrr
+
+
=
+
+
=
0)()()(
=
+
+
z
w
y
v
x
u
continuity equation
u = Ax, v = ?
0)()(
=
+
y
v
x
u
A
x
u
x
v=
=
)(
?)()(
xfAy
Whyxfdyy
vv
constx
+=
+
= =
jxfAyiAxVrrr
))(( +=
divergence theorem
0=
+ CS CV
Vdt
dAnV rr
0)()( == CS CV
VdVectordAnVectorrr
-
5/22/2018 _ _
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134
0)()( == CS CV
VdVdAnVrrrr
0)( =+ CV CV VdtVdV rr
0)]([ =+
VdVtCV
rr
0)( =+
Vt
rr
cylindrical polar coordinates
zzrr evevevtzrV rrrr
++= ),,,(
0)()(1)(1
=
+
+
z
vv
rr
rv
r
zr
-
5/22/2018 _ _
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135
fully developed region
r - direction
0)()(1)(1
=
+
+
z
vv
rr
rv
r
zr
fully-developed
symmetric in
0)(1
=
r
rv
r
r
r
CvCrv rr == ,
r = 0 0=C 0= rv
entrance region
stream function
0=
+
y
v
x
u
u v
xv
yu
=
=
,
-
5/22/2018 _ _
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136
( 0)]([)( =
+
xyyx
)
stream function udy vdx
1. u(x,y) v(x,y) (x,y)
2. streamlineaisconstayx .),( =
u
v
dx
dystreamlineaon =)(
(x,y)
(x+dx,y+dy)
0=+=
+
= udyvdxdyy
dxx
d
Cyx =),(
-
5/22/2018 _ _
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137
volume flow rate
+ d AC
ABC
ddxx
dyy
vdxudydq =
+
==
1 2
12 =q
2 1 q > 0
q < 0
13.0, == sAjAyiAxVrrr
-
5/22/2018 _ _
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138
y
Axu
==
)()( xfAxyxfdyy
constx+=+= =
f(x) v
CfAydx
dfAy
xv ===
= ,
)/
(3.03
m
smxy=
-
5/22/2018 _ _
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139
01)(1
=
+
v
rr
rv
r
r
(r,)
rv
rvr
=
=
,
1
axisymmetric
zzrr evevevzrV rrrr
++= ),,(
)0( =
=
v
0)()(1
=
+
z
v
r
rv
r
zr
-
5/22/2018 _ _
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140
0)()(
=
+
z
rv
r
rv zr (Why?)
(r,z)
rrv
zrv zr
=
= 1
,1
)(2 1221 =q (Why 2?)
6.3
=+
=
+
CVtheofcontentsininininoutoutoutout
CV
CSCV
FVAVVAVVdVt
dAnVVVdVt
rrr
rrrr)(
x
=
+=
=
+=
+
CVtheofcontents
CV
ininoutout
CV
ininininoutoutoutout
CSCV
FVdut
mumu
Vdut
VAuVAu
dAnVuVdut
&&
rr
)(
=+ CVtheofcontentsFSIO &&&
-
5/22/2018 _ _
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141
IO &&
mv& mw& x
x
-
5/22/2018 _ _
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142
[ ]{ }
[ ]{ }
[ ]{ }
dxdydzuwz
uvy
uux
dxdyuwdxdydzuwz
uw
dxdzuvdxdzdyuvy
uv
dydzuudydzdxuux
uuIO
)]()()([
)(
)(
)(
+
+
=
++
++
+= &&
dxdydzut
S )]([
=&
+= surfacexbodyxCV
theofcontentsx FFF ,,, dxdydzgF xbodyx =
,
+= viscousxpressurexsurfacex FFF ,,,
+
+
+
=
=
zzzyzx
yzyyyx
xzxyxx
zzzyzx
yzyyyx
xzxyxx
ij
p
p
p
zyyzzxxzyxxy === ,, Why?
-
5/22/2018 _ _
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143
[ ]{ }
[ ]{ }
[ ]{ }
dxdydzzyxx
p
dxdydzzyx
dxdydxdydzz
dxdzdxdzdyy
dydzdydzdxx
F
xzxyxx
xzxyxx
xzxzxz
xyxyxy
xxxxxxsurfacex
)]()()([
)]()()([
)(
)(
)(,
+
+
+
=
+
+
=
++
++
+=
-
5/22/2018 _ _
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144
SIO &&& +
?)()(
)(])()()(
[
)]()()([)(
Whyz
uw
y
uv
x
uu
t
u
z
uw
y
uv
x
uu
t
u
z
w
y
v
x
u
tu
uwz
uvy
uux
ut
+
+
+
=
+
+
+
+
+
+
+
=
+
+
+
x
)()(zyxx
pg
z
uw
y
uv
x
uu
t
u xzxyxxx
+
+
+
=
+
+
+
y
)()(zyxy
pg
z
vw
y
vv
x
vu
t
v yzyyyxy
+
+
+
=
+
+
+
z
)()(zyxz
pg
z
ww
y
wv
x
wu
t
w zzzyzxz
+
+
+=
+
+
+
ijpgVVt
V
Dt
VD
rrrrrrrrr
+=+
= ])([
6.4 inviscid flow
ijrr
-
5/22/2018 _ _
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145
x
pg
z
uw
y
uv
x
uu
t
ux
=
+
+
+
)(
y
pg
z
vw
y
vv
x
vu
t
vy
=
+
+
+
)(
z
pg
z
ww
y
wv
x
wu
t
wz
=
+
+
+
)(
pgVVt
V=+
rrrrrr
])([
Eulers equation
pgVV = rrrrr )(
-
5/22/2018 _ _
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146
zgkz
zj
y
zi
x
zgkjigkgg =
+
+
=++== rrrrrrrrr
)()100(
)()(2
1
)( VVVVVV
rrrrrrrrr
= What is )( Vrr
pzgVVVV =
rrrrrrrr )]()(
2
1[
)()(2
1 2VVzgV
p rrrrrr
=++
(inner product
sdVVsdzgsdVsdp rrrrrrrrrr
=++
)()(2
1 2
0)( = sdVV rrrr
)( VVrrr
Vr
sdr
kdzjdyidxsdrrrr
++=
dpdzz
pdy
y
pdx
x
psdp =
+
+
= )()()(rr
Eulers equation
0)(2
1 2 =++ gdzVddp
tconsgdzVddp
streamline
tan)(2
1 2 =++
ideal fluids
-
5/22/2018 _ _
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147
tconsgzVp
tan2
2
=++
1.2.3. 4.
irrotational
1.
2. velocitypotential
)(0)( directionanyindssdVsdVV == rrrrrrr
ds
tconsgdzVddp
spacein spotwoany
tan)(2
1
int
2 =++
-
5/22/2018 _ _
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148
2
2
221
2
11
22
gzVp
gzVp
++=++
velocity potential
0)(,0)(,0)( =
=
=
x
w
z
u
z
v
y
w
y
u
x
v
zw
yv
xu
=
=
=
,,
0)()( =
=
xyyxy
u
x
v
velocity potential
= rr
V
1.
-
5/22/2018 _ _
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149
2.
0)( 2 === rrrr
V
)()(2
scalarscalar = rr
Laplacian operator
02
2
2
2
2
2
=
+
+
zyx
potential flow
y
u
x
v
=
)()(yyxx
=
0
2
2
2
2
=
+
yx
-
5/22/2018 _ _
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150
u
v
dx
dyconstalong ==)(
(x,y)
(x+dx,y+dy) d
0=+=
+
= vdyudxdyy
dxx
d
v
u
dx
dyconstalong ==)(
-1
-
5/22/2018 _ _
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151
= rr
V
zv
rv
rv zr
=
=
=
,
1,
01
)(1
2
2
2
2
2 =
+
+
zrrr
rr
)/(2sin2 22 smr =
(1) 30 kPa(2)
?)(4cossin42sin2 22 Whyxyrr ===
-
5/22/2018 _ _
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152
yx
vxy
u 4,4 =
==
=
alirrotation
y
u
x
v==
,000
yv
xu
=
=
,
)(2 12 yfxudxconsty
+== =
)(2 22 xfyvdyconstx
+== =
Cyx += 22 22
(0,0) 0 22 22 yx =
2cos2
)sin(cos222
2
22222
r
ryx
=
==
2sin4,2cos41
rr
vrr
vr
=
==
=
=
=r
vr
vr1
,
)(2cos2 12
frdrvconst
r +== =
)(2cos2 22 rfrdrvconstr
+== =
Cr += 2cos2 2
-
5/22/2018 _ _
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153
(0,0) 0 2cos2 2r=
22
2222
22
)sin(cos22cos2
yx
rr
=
==
2
2
221
2
11
22gz
Vpgz
Vp++=++
)(2
2
2
2
112 VVpp +=
222222 16)2sin4()2cos4( rrrvvV r =+=+=
kPamkNsmkg
smmkgkPap 36
/)//.(1000
)/)(416(
2
)/(100030
22
223
2 =
+=
sin/Ar=
-
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154
cos/Ar=
6.5 viscous flow
normal stresses:
z
wp
y
vp
x
up
zz
yy
xx
+=
+=
+=
2
2
2
shear stresses
)(
)(
)(
z
u
x
w
y
w
z
v
x
v
y
u
zxxz
zyyz
yxxy
+
==
+
==
+
==
-
5/22/2018 _ _
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155
normal stresses:
z
vp
r
vv
rp
r
vp
zzz
r
rrr
+=
+
+=
+=
2
)1
(2
2
shear stresses
)(
)1
(
]
1
)([
r
v
z
v
v
rz
v
v
rr
v
rr
zrzrrz
zzz
rrr
+
==
+
==
+
==
x
)()(2
2
2
2
2
2
z
u
y
u
x
u
x
pg
z
uw
y
uv
x
uu
t
ux
+
+
+
=
+
+
+
y
)()(2
2
2
2
2
2
z
v
y
v
x
v
y
pg
z
vw
y
vv
x
vu
t
v
y
+
+
+
=
+
+
+
z
)()(2
2
2
2
2
2
z
w
y
w
x
w
z
pg
z
ww
y
wv
x
wu
t
wz
+
+
+
=
+
+
+
-Navier-Stokes
- u
-
5/22/2018 _ _
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156
v w p
(CFD)
-
r
)21
)(1
[
)(
2
2
22
2
22
2
z
vv
r
v
rr
v
r
vr
rrr
pg
z
vv
r
vv
r
v
r
vv
t
v
rrrrr
rz
rrr
r
+
+
+
=
+
+
+
)21
)(1
[1
)(
2
2
22
2
22 z
vv
r
v
rr
v
r
vr
rrr
p
rg
z
v
vr
vvv
r
v
r
v
vt
v
r
z
r
r
+
+
+
+
=
++
+
+
z
)1
)(1
[
)(
2
2
2
2
2 z
vv
rr
vr
rrz
pg
z
vv
v
r
v
r
vv
t
v
zzzz
zz
zzr
z
+
+
+
=
+
+
+
-
5/22/2018 _ _
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157
1.2.3. 4. 0=x (Why?)5.
kwjviuVrrrr
++=
4 3
u = u(x, y, z, t)
4 3 1
u = u(y)
-
5/22/2018 _ _
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158
x
)()(2
2
2
2
2
2
z
u
y
u
x
u
x
pg
z
uw
y
uv
x
uu
t
ux
+
+
+
=
+
+
+
1 4 4 3 5 4 3
y
)()(2
2
2
2
2
2
z
v
y
v
x
v
y
pg
z
vw
y
vv
x
vu
t
vy
+
+
+
=
+
+
+
1 4 4 3 -g 4 4 4
z
)()(2
2
2
2
2
2
z
w
y
w
x
w
z
pg
z
ww
y
wv
x
wu
t
wz
+
+
+
=
+
+
+
1 3 3 3 5 3 3 3
2
2
0x
u
x
p
+
=
gy
p
=0
z
p
=0 z
y
)(),( xfgyyxp +=
y
x
-
5/22/2018 _ _
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159
x
p
dy
ud
=
12
2
1)(1 Cyxp
dydu +=
21
2)(2
1)( CyCy
x
pyu ++
=
)(x
p
Why?
B.C #1 y = -h, u = 0
B.C #2 y = +h, u = 0
2
21 )(2
1,0 h
x
pCC
==
))((2
1)( 22 hy
x
pyu
=
)1)((3
2
)1())((2
1)1(
3
22
=
==
x
ph
dyhyx
pudyq
h
h
h
h
pressure drop
-
5/22/2018 _ _
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160
x
pppp rightleft
=
=
ll
)1(3
2 3
=
l
phq
l3)1(2
2ph
h
qVave
=
=
aveVxph
u 2
3)(2
2
max =
=
aveVu 2max=
)(),( xfgyyxp +=
opxxpxf
dxxdf
xp +==
)()(,)(
opxx
pgyyxp +
+= )(),(
po (0,0)
laminar flowReynolds
number ~ 1400
fluidofforceviscous
fluidofforceinertiahVave ==
)2(Re
~ 1400 turbulent
-
5/22/2018 _ _
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161
flow
Couette flow
U
Why?
21
2)(2
1)( CyCy
x
pyu ++
=
-
5/22/2018 _ _
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162
B.C #1 y = 0, u = 0
B.C #2 y = b, u = u
))((2
1)( 2 byy
x
p
b
yUyu
+=
dimensionless
)(2
)1)((
)1)()((2
)(
2
2
x
p
U
bP
b
y
b
yPb
y
b
y
b
y
x
p
U
b
b
y
U
yu
=
+=
=
P P
P = 0
byUyu =)(
-
5/22/2018 _ _
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163
1.2.3.4. 0=
y
(Why?)
5.kwjviuV rrrr ++=
4 3
v = v(x, y, z, t)
4 3 1
v = v(x)
x
)()(2
2
2
2
2
2
z
u
y
u
x
u
x
pg
z
uw
y
uv
x
uu
t
ux
+
+
+
=
+
+
+
u = 0 5 u = 0
y
)()(2
2
2
2
2
2
z
v
y
v
x
v
y
pg
z
vw
y
vv
x
vu
t
vy
+
+
+
=
+
+
+
1 4 4 3 -g 4 3
z
-
5/22/2018 _ _
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164
)()( 22
2
2
2
2
z
w
y
w
x
w
z
pg
z
ww
y
wv
x
wu
t
wz
+
+
+
=
+
+
+
w = 0 5 w = 0
x
p
=0 x
2
2
0dx
vdg
y
p +
=
zp
=0 z
0=x
p x
y
0=y
p
y
2
2
0dx
vdg +=
g
dx
vd=
2
2
1Cxg
dx
dv+=
v(x)
BC #1: x = h, 0== dxdv
xy
-
5/22/2018 _ _
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165
ghC = 1
22
2)( Cxghxgxv +=
BC #2: x = 0, v = 0
oVC = 2
oVxgh
xg
xv +=
2
2)(
)1)(
3(
)1()2
()1()(
3
0
2
0
=
+==
ghhV
dxVxgh
xg
dxxvq
o
h
o
h
-
Hagen-Poiseuille flow
-
-
5/22/2018 _ _
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166
1.2.3. 4. 0=
z (Why?)
5. zzrr evevevV
rrrr ++=
4 3
),,,( tzrvv zz =
3 4 1
)(rvv zz=
-
r
vr= v= 0
)21
)(1
[
)(
2
2
22
2
22
2
z
vv
r
v
rr
v
r
vr
rrr
pg
z
vv
r
vv
r
v
r
vv
t
v
rrrrr
rz
rrr
r
+
+
+
=
+
+
+
-g sin vr= v= 0
-
5/22/2018 _ _
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167
vr= v= 0
)21
)(1
[1
)(
2
2
22
2
22 z
vv
r
v
rr
v
r
vr
rrr
p
rg
z
vv
r
vvv
r
v
r
vv
t
v
r
z
r
r
+
+
+
+
=
++
+
+
-g cos vr= v= 0
z
1 4 3 4
)
1
)(
1
[
)(
2
2
2
2
2z
vv
rr
v
rrrz
p
g
z
vv
v
r
v
r
vv
t
v
zzz
z
zz
zzr
z
+
+
+
=
+
+
+
5 3 4
r
pg
= sin0
= p
rg
1cos0
)](1
[0r
vr
rrz
pz
+
=
)(
)(sin),,(
zfgyp
zfgrzrp
+=
+=
y
z
-
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168/169
168
z
zf
z
p
= )(
z
p
r
Z
)(1
)(1
z
p
r
vr
rr
z
=
1
2)(2
1Cr
z
p
r
vr z +
=
21
2 ln)(4
1)( CrCr
z
prvz ++
=
BC #1: r = 0, vz= finite
BC #2: r = R, vz= 0
2
21 )(4
1
,0 Rz
p
CC
==
))((4
1)( 22 rR
z
prvz
=
)(8
)2(4
0 z
pRdrrvQ
R
z
==
l
8
4pR
Q
=
Poiseuilles law
-
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l 8
2
2
pR
R
QVave
==
aveVpRzpRv 2
4)(
4
22
max === l
2
2
max
1R
r
v
vz =
aveVu2
3max=
laminar flowReynolds
number ~ 2100 ~ 2100