Transcript Document

Chapter 8:
Internal Incompressible Viscous Flow
Flows:
Laminar (some have analytic solutions)
Turbulent (no analytic solutions)
Incompressible:
For water  usually considered constant
For gas  usually considered constant
for M (~100m/s < 0.3)
Chapter 8: Internal Incompressible Viscous Flow
M2 = V2/c2
{c2 = kRT} M2 = V2/kRT
{p =  RT} M2 = V2/(kp/) = [2/k][1/2 V2/p]
M2 = 1.43 dynamic pressure / static pressure
M ~ 1.20 [dynamic pressure / static pressure]
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
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
Chapter 8:
Internal Incompressible Viscous Flow
At 200C the speed of sound is 343m/s;
If M=V/c=0.3, V=103m/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%
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
CONSERVATION OF MASS
& INCOMPRESSIBLE
•V = -(1/)D/Dt = 0
(5.1a)
Volume is not changing.
The density is not changing
as follow fluid particle.
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
Reynolds Number
~ ratio of inertial
to viscous forces
REYNOLDS
NUMBER
-- 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
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
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
Fully Developed Laminar Pipe/Duct Flow
LAMINAR Pipe Flow Re< 2300 (2100 for MYO)
LAMINAR Duct Flow Re<1500 (2000 for SMITS)
Uo = uavg
OUTSIDE BLUNDARY LAYER TREAT AS INVISCID, CAN USE B.E.
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
Le = 0.6D, Re = 10
Le = 140D, Re = 2300
Pressure gradient
balances wall
shear stress
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
20D < Le < 30D
104 < Re < 105
Pressure gradient
balances wall
shear stress
Parallel Plates - Re = UD/ = 140
Water Velocity = 0.5 m/s
FULLY DEVELOPED
LAMINAR PIPE &
DUCT FLOW
Circular Pipe – Re = UD/ = 195
Water Velocity = 2.4 m/s
Hydrogen Bubble Flow
Visualization
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
Le/D = 0.06 Re
LAMINAR FLOW – VELOCITY PROFILE
TURBULENT FLOW – VELOCITY PROFILE
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?
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).
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
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.
FULLY DEVELOPED LAMINAR FLOW
BETWEEN INFINITE PARALLEL PLATES
y=a
y=0
FULLY DEVELOPED LAMINAR FLOW
BETWEEN INFINITE PARALLEL PLATES
Assumptions: steady, incompressible,
no changes in z variables, v=w=0,
fully developed flow, no body forces
No Slip Condition: u = 0 at y = 0 and y = a
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).
= 0(3)
= 0(1)
= 0(4)
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, v=w=0,
FSx = 0
+
+
+
* Control volume not accelerating – see pg 131
=0
(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
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
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.
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
a
0
u = 0 at y = 0:
u = 0 at y = a:
c2 = 0
a = 1; dp/dx = 2
u = {(y/1)^2 -(y/1)}; channel height=1m
1.2
1
y (m)
0.8
Why velocity
negative?
a
0.6
0.4
0.2
0
-0.3
-0.25
-0.2
-0.15
u (m/s)
-0.1
-0.05
0
u(y) for fully developed laminar flow
between two infinite plates
y=a
y=0
negative
(next want to determine shear stress profile,yx)
yx = (du/dy)
SHEAR STRESS?
tau = [(y/1)-1/2]; a=1, dp/dx=1
1
0.9
0.8
0.7
y
0.6
0.5
0.4
0.3
Flow direction
0.2
0.1
0
-0.6
-0.4
-0.2
0
tau
0.2
0.4
0.6
tau = [(y/1)-1/2]; a=1, dp/dx=1
1
0.9
0.8
0.7
y
0.6
0.5
0.4
0.3
0.2
0.1
0
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
tau
y=a
-
dp/dx = negative
+
y=0
Does + and – shear stresses imply that
direction of shear force is different on
top and bottom plates?
Sign convention
for stresses
White
Positive stress is defined in the + x direction
because normal to surface is in the + direction
(next want to determine shear stress profile,yx)
+
Shear force
y=a
+
yx = (du/dy)
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
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 find some cells are dead.
What’s up?
very large shear
stresses at start-up
u(y) for fully developed laminar flow
between two infinite plates
y=a
y’ = a/2
y’=0
y=0
y’ = -a/2
y’ = y – a/2; y = y’ + a/2
(y’2 + ay’ + a2/4 –y’a – a2/2)/ a2 = (y’/a)2 – 1/4
(next want to determine volume flow rate, Q)
y=a
y=0
[y3/3 – ay2/2]oa = a3/3 – a3/2 = -a3/6
If dp/dx = const
(next want to determine average velocity)
A = la
= uavg
(next want to determine maximum velocity)
(a2/4)/a2 – (a/2)/a = -1/4
UPPER PLATE MOVING WITH CONSTANT SPEED U
Velocity distribution
UPPER PLATE MOVING WITH CONSTANT SPEED U
+
Boundary driven
Pressure driven
Shear stress distribution
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]
Average Velocity
l
Maximum Velocity
y=a
umax = a/2
y=0
EXAMPLE:
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
d yx/dy = g
yx = du/dy = gy + c1
du/dy = gy/ + c1/
u = gy2/(2) + yc1/ + c2
u = gy2/(2) + yc1/ + c2
u = gy2/(2) + yc1/ + c2 = gy2/(2) - ghy/ +U0
At y=h, u = gh2/(2) - gh2/ +U0
At y=h, u = -gh2/(2) +U0
FULLY DEVELOPED LAMINAR PIPE FLOW
r
APPROACH JUST
LIKE FOR DUCT FLOW
Note however, that
direction of positive
shear stress is opposite.
r
r
r
dFL
dFR
dFL = p2rdr
dFR = -(p + [dp/dx]dx) 2rdr
dFI = -rx2rdx
dFO = (rx + [d rx/dr]dr) 2(r + dr) dx
r
r
r
dFL
dFR
dFL = p2rdr
dFR = -(p + [dp/dx]dx)2rdr
dFL + dFR = -[dp/dx]dx2rdr
r
r
r
dFL
dFR
dFI = -rx2rdx
dFO = (rx + [d rx/dr]dr) 2(r + dr) dx
dFO+ dFI = -rx 2rdx + rx 2rdx + rx 2drdx +
[drx/dr)]dr2rdx + [drx/dr]dr 2dr dx
~0
dFO + dFI = rx 2drdx + [drx/dr]dr2rdx
r
r
r
dFL
dFR
dFL + dFR + dFI + dFO = 0
-[dp/dx]dx2rdr+rx 2drdx(r/r)+(drx/dr)dr2rdx = 0
[dp/dx] = rx/r + drx/dr = (1/r)d(rxr)/dr
dp/dx = (1/r)(d[rrx]/dr)
Because of spherical coordinates,
more complicated than for duct.
dp/dx = dxy/dy
dp/dx = (1/r)(d[rrx]/dr)
p is uniform at each
section, since F.D., so
not function of r or .
rx is at most a function
of r, because fully
developed, rx  f(x),
symmetry, rx  f().
dp/dx = constant = (1/r)(d[rrx]/dr)
dp/dx = constant = (1/r)(d[rrx]/dr)
d[r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 we you say about c1?
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
du/dr = r(dp/dx)/2
u = r2(dp/dx)/(4) + c2
u=0 at r=R, 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]
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
SHEAR STRESS PROFILE
FULLY DEVELOPED PIPE FLOW
= direction of shear stress on CV
FULLY DEVELOPED DUCT FLOW
- for flow to right
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)
VOLUME FLOW RATE – PIPE FLOW
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)
AVERAGE FLOW RATE – PIPE FLOW
Q = R4p/(8L)
uAVG = Q/A = Q/(R2)
= R4p/(R28L)
= R2p/(8L)
= -(R2/(8)) (dp/dx)
AVERAGE FLOW RATE – PIPE FLOW
uAVG = V = Q/A = Q/(R2) = R4p/(R28L)
uAVG = R2p/(8L) = -(R2/(8)) (dp/dx)
MAXIMUM FLOW RATE – PIPE FLOW
du/dr = (r/[2])p/x
At umax, du/dr = 0;
which occurs at r = 0
umax =
2
R (p/x)/(4)
MAXIMUM FLOW RATE – PIPE FLOW
FULLY DEVELOPED LAMINAR PIPE FLOW
u/umax = 1 –
u/umax
r/R
2
(r/R)
FULLY DEVELOPED LAMINAR PIPE FLOW
(r)/w
u/umax
Shear stress
CV exerts
r/R
THE END
END
L/D = 0.06 Re
Re = 2300
L = 140 D
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 = Uavg2[1-(y/r)2]
u = Uavg