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## Internal Incompressible Viscous Flow

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### Chapter 8: Internal Incompressible Viscous Flow Flows: Laminar (some have analytic solutions) Turbulent (no analytic solutions) Incompressible: – PowerPoint PPT presentation

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Title: Internal Incompressible Viscous Flow

1
Chapter 8 Internal Incompressible Viscous Flow
Flows Laminar (some have analytic solutions)
Turbulent (no analytic solutions) Incompressibl
e For water ? usually considered constant
For gas ? usually considered constant
for M (100m/s lt 0.3)
2
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3
Chapter 8 Internal Incompressible Viscous Flow
M2 V2/c2 c2 kRT M2 V2/kRT p ? RT M2
V2/(kp/?) 2/k1/2 ?V2/p M2 1.43
dynamic pressure / static pressure M 1.20
dynamic pressure / static pressure
4
What are static, dynamic and stagnation
pressures? The thermodynamic pressure, p, used
throughout this book chapters refers to the
static pressure (a bit of a misnomer). This is
the pressure experienced by a fluid particle as
it moves. The dynamic pressure is defined as ½
? V2. The stagnation pressure is obtained when
the fluid is decelerated to zero speed through
an isentropic process (no heat transfer, no
friction). For incompressible flow po p ½ ?
V2
5
Static pressure
atm. press. static pressure (what moving fluid
particle sees) Hand in steady wind Felt by
hand stag. press.
Stagnation pressure
for incompressible flow po p ½ ?V2
Dynamic pressure
6
Chapter 8 Internal Incompressible Viscous Flow
At 200C the speed of sound is 343m/s If
MV/c0.3, V103m/s ?p 1/2 ?(V22 - 0)
6400Pa 6 of 1 atm. p ?RT assume isothermal
(wrong) ?p/p ??/? 6 p/?k const assume
isentropic (right) ??/? 5
7
Chapter 8 Internal Incompressible Viscous Flow
• Compressibility requires work, may produce heat
and
• change temperature (note temperature changes
due
• to viscous dissipation usually not important)
• Need relatively high speeds (230 mph) for
• compressibility to be important
• Pressure drop in pipes usually not large enough
to
• make compressibility an issue

8
CONSERVATION OF MASS INCOMPRESSIBLE
?V -(1/?)D?/Dt 0
(5.1a)
Volume is not changing.
The density is not changing as follow fluid
particle.
9
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10
Chapter 8 Internal Incompressible Viscous Flow
Flows Laminar (some have analytic solutions)
Turbulent (no analytic solutions) Depends
of Reynolds number, Re I.F./V.F
11
REYNOLDS NUMBER
Reynolds Number ratio of inertial to viscous
forces -- hand waving argument --
Inertial Force Upstream Force on Front Fluid
Volume Face Inertial Force momentum flux
?f u l2 x u (mass flux x
velocity) Viscous Force Shear Stress Force on
Top Fluid Volume Face ? ?(du/dy)
?(u/kl) Viscous Force ? (u/kl)l2
?ul/k Re Inertial Force / Viscous Force Re
?f u l2 u / ?ul/k k?f ul/? Re ?f
ulc/? Where lc is a characteristic length.
?f
control volume
12
REYNOLDS NUMBER
Reynolds conducted many experiments using glass
tubes of 7,9, 15 and 27 mm diameter and water
temperatures from 4o to 44oC. He discovered that
transition from laminar to turbulent flow
occurred for a critical value of ?uD/? (or
uD/?), regardless of individual values of ? or u
or D or ?. Later this dimensionless number,
?uD/?, was called the Reynolds number in his
honor. Nakayama Boucher
13
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14
Chapter 8 Internal Incompressible Viscous Flow
Internal completely bounded - FMP
• Internal Flows can be Fully Developed Flows
• mean velocity profile not changing in x
• viscous forces are dominant - MYO

15
Fully Developed Laminar Pipe/Duct Flow
LAMINAR Pipe Flow Relt 2300 (2100 for
MYO) LAMINAR Duct Flow Relt1500 (2000 for SMITS)
Uo uavg
OUTSIDE BLUNDARY LAYER TREAT AS INVISCID, CAN USE
B.E.
16
As inviscid core accelerates, pressure must drop
Laminar Pipe Flow Entrance Length for Fully
Developed Flow L/D 0.06 Re L/D 0.03
Re, Smits
White
Pressure gradient balances wall shear stress
Le 0.6D, Re 10 Le 140D, Re 2300
17
As inviscid core accelerates, pressure must drop
Turbulent Pipe Flow Entrance Length for Fully
Developed Flow
Le/D 4.4 Re1/6 MYO
Note details of turbulence may take longer than
mean profile
White
Pressure gradient balances wall shear stress
20D lt Le lt 30D 104 lt Re lt 105
18
FULLY DEVELOPED LAMINAR PIPE DUCT FLOW
Parallel Plates - Re ?UD/? 140 Water Velocity
0.5 m/s
Circular Pipe Re UD/? 195 Water Velocity
2.4 m/s
Hydrogen Bubble Flow Visualization
19
2-D Duct Flow
a
b
c
Hydrogen Bubble Flow Visualization
Where taken a,b, or c?
Parallel Plates - Re 140 Water Velocity 0.5
m/s
20
Le/D 0.06 Re
21
LAMINAR FLOW VELOCITY PROFILE
TURBULENT FLOW VELOCITY PROFILE
22
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23
Upper plate moving at 2 mm/sec Re 0.03
(glycerin, h 20 mm)
Duct flow, umax 2 mm/sec Re 0.05(glycerin, h
40 mm)
VELOCITY 0 AT WALL NO SLIP CONDITION (DUST ON
FAN)
What happens if wall is made of water? Or what
happens to fluid particles next to no-slip layer?
24
No Slip Condition u 0 at y 0
Stokes (1851) On the effect of the internal
friction of fluids on the motion of pendulums
showed that no-slip condition led to
remarkable agreement with a wide range of
experiments including the capillary tube
experiments of Poiseuille (1940) and Hagen (1939).
25
VELOCITY 0 AT WALL NO SLIP CONDITION
Each air molecule at the table top makes about
1010 collisions per second. Equilibrium achieved
after about 10 collisions or 10-9 second, during
which molecule has traveled less than 1 micron
(10-4 cm). Laminar Boundary Layers - Rosenhead
26
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27
FULLY DEVELOPED LAMINAR FLOW BETWEEN INFINITE
PARALLEL PLATES
Perform force balance on differential control
volume to determine velocity profile, from which
will determine volume flow, shear stress,
pressure drop and maximum velocity.
28
FULLY DEVELOPED LAMINAR FLOW BETWEEN INFINITE
PARALLEL PLATES
ya
y0
29
FULLY DEVELOPED LAMINAR FLOW BETWEEN INFINITE
PARALLEL PLATES
in z variables, vw0, fully developed flow, no
body forces No Slip Condition u 0 at y 0
and y a
30
FULLY DEVELOPED LAMINAR FLOW BETWEEN INFINITE
PARALLEL PLATES
no changes in z variables, w 0
2-Dimensional, symmetry arguments v 0 du/dx
dv/dy 0 via Continuity, 2-Dim., Fully
Dev. du/dx 0 everywhere since fully
developed, therefore dv/dy 0 everywhere, but
since v 0 at surface, then v 0 everywhere
along y (and along x since fully developed).
31
0(4)
0(3)
0(1)
FSx FBx ?/?t (?cvu?dVol ) ?csu?V?dA Eq.
(4.17)
Assumptions (1) steady, incompressible, (3) no
body forces, (4) fully developed flow, no
changes in z variables, vw0,
FSx 0

0

Control volume not accelerating see pg 131
32
(Want to know what the velocity profile is.)

0

?p/?x d?xy/dy
?p/?x d?xy/dy constant
Since the pressure does not vary in the span-wise
or vertical direction, streamlines are straight
?p/?x dp/dx
33
Since the pressure does not vary in the span-wise
or vertical direction, streamlines are straight
?p/?x dp/dx
N.S.E. for incompressible flow with and constant
viscosity.
?(?v/?t u?v/?x v?v/?y w?v/?z) ?gy -
?p/?y ?(?2v/?x2 ?2v/?y2 ?2v/?z2
v 0 everywhere and always, gy 0 so left with
?p/?y 0
Eq 5.27b, pg 215
34
Important distinction because book integrates
?p/?x with respect to y and pulls ?p/?x out of
integral (pg 314), can only do that if dp/dx,
which is not a function of y.
35
integrate
?p/?x dp/dx d?xy/dy
(Want to know what the velocity profile is.)
For Newtonian fluid
substitute
integrate
USE 2 BOUNDARY CONDITIONS TO SOLVE FOR c1AND c2
36
a
0
c2 0
u 0 at y 0
u 0 at y a
37
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38
a 1 dp/dx 2?
Why velocity negative?
a
39
u(y) for fully developed laminar flow between
two infinite plates
y a
y 0
negative
40
(next want to determine shear stress profile,?yx)
?yx ?(du/dy)
41
SHEAR STRESS?
Flow direction
42
y a
-
dp/dx negative

y 0
Does and shear stresses imply that direction
of shear force is different on top and bottom
plates?
43
Sign convention for stresses
White
Positive stress is defined in the x direction
because normal to surface is in the direction
44
(next want to determine shear stress profile,?yx)
Shear force

?yx ?(du/dy)
y a
y 0

shear direction
For dp/dx negative ?yx on top is negative in
the x direction ?yx on bottom is positive in
the x direction
45
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46
Question? Given previous flow and ?wall 1
(N/m2) Set this experiment up and add cells that
are insensitive to shears less than ?wall Yet
Whats up?
47
very large shear stresses at start-up
48
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49
u(y) for fully developed laminar flow between
two infinite plates
y a/2
y a
y0
y 0
y -a/2
y y a/2 y y a/2
(y2 ay a2/4 ya a2/2)/ a2 (y/a)2
1/4
50
(next want to determine volume flow rate, Q)
ya
y0
y3/3 ay2/2oa a3/3 a3/2 -a3/6
If dp/dx const
51
(next want to determine average velocity)
A la
uavg
52
(next want to determine maximum velocity)
(a2/4)/a2 (a/2)/a -1/4
53
UPPER PLATE MOVING WITH CONSTANT SPEED U
54
Velocity distribution
55
UPPER PLATE MOVING WITH CONSTANT SPEED U

Pressure driven
Boundary driven
56
Shear stress distribution
57
Volume Flow Rate
Uy2/(2a) (1/(2?))(dp/dx)(y3/3) ay2/2 y
a Ua/2 (1/(2?))(dp/dx)(2a3 3a3)/6 Ua/2
(1/(12?))(dp/dx) a3
58
Average Velocity
l
59
Maximum Velocity
y a
umax a/2
y 0
60
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61
EXAMPLE
62
0
0
FSx FBx ?/?t (?cvu?dVol ) ?csu?V?dA Eq.
(4.17)
Assume (1) surface forces due to shear alone,
no pressure forces (patm on
either side along boundary) (2) steady flow and
(3) fully developed
Fsx FBx 0 Fs1 Fs2 - ?gdxdydz 0 F1 ?yx
(d?yx/dy)(dy/2)dxdz F2 ?yx -
(d?yx/dy)(dy/2)dxdz d?yx/dy ?g
63
d ?yx/dy ?g ?yx ?du/dy ?gy c1 du/dy
?gy/? c1/? u ?gy2/(2?) yc1/? c2
64
u ?gy2/(2?) yc1/? c2
u ?gy2/(2?) yc1/? c2 ?gy2/(2?) - ?ghy/?
U0
At yh, u ?gh2/(2?) - ?gh2/? U0 At yh, u
-?gh2/(2?) U0
65
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66
FULLY DEVELOPED LAMINAR PIPE FLOW
r
APPROACH JUST LIKE FOR DUCT FLOW Note however,
that direction of positive shear stress is
opposite.
67
r
r
r
dFL
dFR
dFL p2?rdr dFR -(p dp/dxdx) 2?rdr dFI
-?rx2?rdx dFO (?rx d ?rx/drdr) 2?(r
dr) dx
68
r
r
r
dFL
dFR
dFL p2?rdr dFR -(p dp/dxdx)2?rdr dFL
dFR -dp/dxdx2?rdr
69
r
r
r
dFL
dFR
dFI -?rx2?rdx dFO (?rx d ?rx/drdr) 2?(r
dr) dx dFO dFI -?rx 2?rdx ?rx 2?rdx
?rx 2?drdx d?rx/dr)dr2?rdx
d?rx/drdr 2?dr dx dFO dFI ?rx 2?drdx
d?rx/drdr2?rdx
0
70
r
r
r
dFL
dFR
dFL dFR dFI dFO 0 -dp/dxdx2?rdr?rx
2?drdx(r/r)(d?rx/dr)dr2?rdx 0 dp/dx
?rx/r d?rx/dr (1/r)d(?rxr)/dr
71
dp/dx (1/r)(dr?rx/dr) Because of spherical
coordinates, more complicated than for duct.
dp/dx d?xy/dy
72
dp/dx (1/r)(dr?rx/dr)
?rx is at most a function of r, because fully
developed, ?rx ? f(x), symmetry, ?rx ? f(?).
p is uniform at each section, since F.D., so
not function of r or ?.
dp/dx constant (1/r)(dr?rx/dr)
73
dp/dx constant (1/r)(dr?rx/dr) dr?rx/dr
rdp/dx integrating.. r?rx r2(dp/dx)/2 c1 ?rx
?du/dr ?rx ?du/dr r(dp/dx)/2 c1/r What
74
?rx ?du/dr r(dp/dx)/2 c1/r c1 0 or else
?rx ? ?rx ?du/dr r(dp/dx)/2
Shear forces on CV
For dp/dx negative, get negative shear stress on
CV but positive shear stress on fluid/wall
outside control volume
75
?du/dr r(dp/dx)/2 u r2(dp/dx)/(4?) c2 u0
at rR, so c2-R2(dp/dx)/(4?) u r2(dp/dx)/(4?)
- R2(dp/dx)/(4?) u r2 - R2 (dp/dx)/(4?) u
-R2(dp/dx)/(4?) 1 (r/R)2
76
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77
SHEAR STRESS PROFILE
?rx -?du/dr
?rx r(dp/dx)/2 TRUE FOR LAMINAR AND TURBULENT
FLOW
?du/dr r(dp/dx)/2 TRUE ONLY FOR LAMINAR FLOW
78
SHEAR STRESS PROFILE
FULLY DEVELOPED PIPE FLOW
direction of shear stress on CV
FULLY DEVELOPED DUCT FLOW
- for flow to right
79
VOLUME FLOW RATE PIPE FLOW
Q ?A V dA ?0R u2?rdr ?0R r2 - R2
(dp/dx)/(4?) 2?rdr Q (dp/dx)/(4?) r4/4 -
R2r2/2 ?0R (2?) (-?R4dp/dx)/(8?)
80
VOLUME FLOW RATE PIPE FLOW
81
VOLUME FLOW RATE PIPE FLOW as a function
of ?p/L
?p/?x constant (p2-p1)/L -?p/L
p2 p ?p
p1
L
Q (-?R4dp/dx)/(8?) ?R4?p/(8L?)
?D4(?p/(128L?)
82
AVERAGE FLOW RATE PIPE FLOW
Q ?R4?p/(8L?)
uAVG Q/A Q/(?R2) ?R4?p/(?R28L?)
R2?p/(8L?) -(R2/(8?)) (dp/dx)
83
AVERAGE FLOW RATE PIPE FLOW
uAVG V Q/A Q/(?R2) ?R4?p/(?R28L?) uAVG
R2?p/(8L?) -(R2/(8?)) (dp/dx)
84
MAXIMUM FLOW RATE PIPE FLOW
du/dr (r/2?)?p/?x At umax, du/dr 0
which occurs at r 0 umax R2(?p/?x)/(4?)
85
MAXIMUM FLOW RATE PIPE FLOW
86
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87
FULLY DEVELOPED LAMINAR PIPE FLOW
u/umax 1 (r/R)2
u/umax
r/R
88
FULLY DEVELOPED LAMINAR PIPE FLOW
?(r)/?w
u/umax
Shear stress CV exerts
r/R
89
THE END
90
END
91
L/D 0.06 Re Re 2300 L 140 D
92
Prandtl Tietjens
po
A
B
pA-po ½ ?Uavg2 what is ?p between A and B?
Flux of K.E. per unit volume u?½ ?u2(?r2)dr
at B
at A
u Uavg21-(y/r)2
u Uavg