Post on 08-Apr-2018
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N-S equations are 2nd order , nonlinear , non-homogeneous
partial differential equations . They are very difficult to solve .
Only for very limited number of cases , they can be solved in
closed form .
CASE STUDIES: Application of Equations of Motion
(N-S Equations)
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Flow of a Falling Film
The first example is the flow of a falling film . Consider the
Flow of a liquid at steady state along an inclined plate.
We are looking for (1) Volume flow rate
(2) Velocity profile perpendicular to the inclined plate
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Take a control
volume from
the system
Assumptions:(1) Constant temperature Constant density , viscosity .
(2) Laminar Flow
(3) Neglect entrance & exit effects .
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Start from the Navier-Stokes equation
For vz
:
vx=0 vz is not function of z vz is not function of y
0(steady state) vy=0 Assuming vz is not function of z
uniform pressure
Driving force-
Gravity only
GDE (Governing Differential Equation)
zzzzz
zz
yz
xz g
z
v
y
v
x
v
zz
vv
y
vv
x
vv
t
v
x
x
x
x
x
x
x
x!
x
x
x
x
x
x
x
x)(
2
2
2
2
2
2
RJ
02
2
!x
x@
z
zg
x
vR
0cos !x
x Fg
x
v
gz g
Fcos!g
gz
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2nd order , needs 2 Boundary Conditions.
B.C.
,2
,
!!
!xx!
xx!!
z
zz
xz
vx
x
v
x
vx
H
QX
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FVQ
F
VQ
cos1
cos
2
2
2
2
gx
v
g
x
v
z
z
!x
x
!xx
From GDE
,vz is a function of x only
Integrate
dx
dv
x
v zz !x
x@
1cos
1cxg
dx
dvz ! FVQ
From B.C.(1) c1=0
xgdx
dv! FV
Qcos
1
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Integrate again
From B.C. (2)
Velocity Profile
2
2co
2
cxgvz
! FV
Q
])([H
HFVQ
HFV
Q
FV
Q
HFVQ
HFVQ
xgv
gxgv
gc
cgv
z
z
z
!
!@
!@
!!
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When x=0 , vz=vz,max
For average velocity vz
For volume flow rate
0
3
00
co3
]3
co
|]3
co
]co
HFVQ
HHHFV
Q
HHFV
Q
HFHV
QHH
H
HH
!
!
!
!!
gv
g
xxg
dxx
gdxvv
z
zz
2
max,
cos2
1HFV
Q!@ gv
z
3cos3
1)( HFV
QH
wgwvQ z
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Flow Between Parallel Plates
Consider the flow of fluid between parallel plates .
Assumption :
(1) Constant density , viscosity
(2) Laminar flow
(3) Neglect entrance effect
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Start from the Navier-Stokes equation
For vz
:
vx is not function of x vz=0 vx is not function of z
steady state vy=0 vz is not function of x 0
Driving force-Pressure gradient
GDE
z
xxxxz
xy
xx
x gz
v
y
v
x
v
xz
vv
y
vv
x
vv
t
v
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x)(
2
2
2
2
2
2
RJ
xy
vx
x
x
!x
x J
R 2
2
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B.C. 1
!!
!x
x!!
x
x
yx
vy
y
vy
H
X
L
PP
y
vL
PP
xLx
L
)(1
1
0
0
!
x
x
!
x
x
Q
V
J
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2
20
0
1
10
)(
2
1
)(1
0)1.(.
)(1
cyL
PPv
y
L
PP
y
v
cCFromB
cy
L
PP
Y
V
Lx
Lx
LX
!
!
x
x
!
!
x
x
Q
Q
Q
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From B.C. (2)
L
PPyv
L
PPyL
PPv
L
PPc
cLPP
Lx
LL
x
L
L
)()(
2
1
)(21)(
21
)(
2
1
)(21
22
22
2
2
22
!
!
!
!
HQ
HQQ
HQ
HQ
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At y=0 , vx=vx,max
For average velocity
For volume flow rate
L
PP
v Lx 0max, 1 ! HQ
L
PPdyvv Lxx
0
2
03
1 !!
Q
H
H
H
L
PPwL 0
3
3
2
H
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Flow through a Circular Tube
Consider the fully developed flow of a fluid in a along tube of
Length L and radius R .
Some assumptions
(1) Constant density and viscosity
(2) Laminar Flow
(3) Neglect entrance effect
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Start from the Navier-Stokes eq in cylindrical coordinate
r
z L
Area=2rL Area=2(r+r)L
r
zzzzz
zzz
rz g
z
vv
rr
vr
rrzz
vv
v
r
v
r
vv
t
v
x
x
x
x
x
x
x
x
x
x!
x
x
x
x
x
x
x
x]
1)(
1[
2
2
2
2
2 UR
J
U
U
vr=0 vz not function of z
steady state v=0 vz not function of vz not function of z
gzr
vr
rr
gr
vr
rrz
z
z
xx
!x
x
xx
x
x
xx
xx
!
JR
RJ
)(1
)(1
0GDE , Driving force
gravity+pressure gradient
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B.C. (1) 0,0 !x
x!
r
vr
z
0, !!z
vR
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1
1
1
cgr
r
Lr
v
r
Integrate
gL
rr
vr
r
gz
Pr
r
vr
r
Lz
Lz
z
!
!
!
VL
VL
L
V
L
From B.C. (1) c1=0
2(
1 20
gr
r
L
PP
r
v
r
Lz
VL
!x
x
)2
(1
0 gr
L
PP
r
vLz
VL
!x
x
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Integrate again
From B.C.(2)
2
2
)
4
(1
cgrr
L
v Lz
!
V
L
)4
(1
)4
(1
0
0
0
grRL
c
cgrR
L
L
L
VL
VL
!
!
])(1)[4
)((
)4
(1
)4
(1
22
0
2
0
2
0
R
rRgr
L
PPv
grR
L
PPgr
r
L
PPv
Lz
LLz
!
!
LV
VL
VL
Velocity Profile
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At r=0 , vz=vz,max
For average velocity
For volume flow rate
Hagen-Poiseuille Law
related melt penetration into porous refractory
4((
2
0
max,LV
R
gL
PP
vL
z
!@
UT
T
8)(1
2
0
2
0 02
R
gL
PPd d
Rv L
R
zz
!!
)()(
4
LTU Rg !
LU
U
1
4
w
w R
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Creeping flow around a solid sphere
Consider the flow of an incompressible fluid about a solid sphere.The fluid approaches the sphere upward along the z-axis with a
uniform velocity v
Assumptions
(1) Const ,
(2) V is 0
(3) Very slow flow ( , Reynolds number < 1)
Application : Inclusions Removal
Rg
Dv
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Start from the Navier-Stokes equations in sphere coordinates(r, , )
For r-component
rrr
rrr
r
r
gv
rv
r
v
rv
rv
r
P
r
vvv
r
vv
r
v
r
vv
t
v
VJU
UU
L
JUUV
J
UU
JUJU
xx
xx
xx!
x
x
x
x
x
x
x
x
)sin
2cot222(
)sin
(
2222
2
22
Accelerationterms=0 due to very slow flow
0 v=0
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: known as nabla or del is the vector differential operator .
For Cartesian coordinate
kz
jy
ix
x
x
x
x
x!
2 known as Laplacian operator
2
2
2
2
2
22
zyx xx
xx
xx!
For spherical coordinate
!
xx
xx
xx!
sinJU H
JUH
UH
rrrr
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For the component
U
JUU
JUUJUUUU
VJU
U
UUL
U
U
JUUV
gv
rr
vv
rV
P
r
r
v
r
vvv
r
vv
r
v
r
vv
t
v
r
r
r
x
x
x
x
x
x
x
x
x
x
x
x
x
x
)sin
cos
sin(
1
)cot
sin(
Accelerationterms=0
V=0
Forcontinuity equation
0)(sin
1)sin(
sin
1)(
1!
x
x
x
x
x
x
x
xJ
VJ
VVV
vr
vr
vrrrt
r
const v=0
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G.D.E.
)si
2(
)cot222
(
)si(si
)(
222
2
222
2
2
2
!xx
xx
!x
x
xx
!xx
xx
UU
U
UU
U
VUUU
VUU
UUU
gvvv
gvv
vv
vv
3 eqns , 3 unknowns (P , vr , v)
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Solve the above equations , we obtain
U
U
UL
V
UL
X
U
U
sin]
os]22
os2
sin2
2
r
R
r
Rvv
r
R
r
Rvv
r
R
R
vgzPP
r
R
R
v
r
r
!
!
!
!
g
g
g
g
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The normal force acting on the solid surface is due to the pressure
at the solid surface , where r=R , z=Rcos
JUUUL
UV
UL
UV
T TddR
R
vgRPF
R
vgRPP
n
Rr
si]s2
3s[
cos2
3cos|
22
0 0 0
0
g
g
!
!@
!
g! RvgRFn TLVT 2
Buoyant force when Buoyant force due to
the fluid is stegnant fluid movement
Fn : Buoyant force on form drag
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As the surface , there is also shear stress acting tangentially , r
We are only interested in the z-component of the shear stress ,
rsin
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at the surface
g
g
g
!
!
!
!
Rv
ddRR
vF
R
v
t
Rrr
TL
JL
L
XT T
U
4
ii2
3
i2
3
|
2
0 0
2
Ft=frictio drag
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Total force acting upward
Stokes Law
Fs Fk
g! RvgRF TLVT 6
3
4 3
stegnant with fluid movement
R F
(inclusions rise fast)