Chapter 8:
Internal Incompressible Viscous Flow
Chapter 8:
Internal Incompressible Viscous Flow
Flows:
Laminar (some have analytic solutions)
Turbulent (no analytic solutions)
Depends of Reynolds number,
Re = Inertial Force/Viscous Force
Re = Lu/
cartoon approach
REYNOLDS
NUMBER
Reynolds
Number ~ ratio
of inertial =
to I.F./V.F.
viscous for
(Cartoon
approach)ces
(cartoon
approach)
-- hand waving argument -Inertial Force = (m) x (a)  ( l3) x (U/t)
U/t  U/(L/U)  U2/L
Inertial Force  ( l3 ) x (U2/L)
Inertial Force  ( L3 ) x (U2/L) = L2U2
fluid
element
f
L
U
REYNOLDS
NUMBER
Reynolds
Number ~ ratio
of inertial =
to I.F./V.F.
viscous for
(Cartoon
approach)ces
(cartoon
approach)
-- hand waving argument --
Viscous Force = () x (Area)  (dU/dy) x l2
dU/dy  U/l
ViscousForce  (U/l) x l2  ul  Ul
fluid
element
f
L
U
REYNOLDS
NUMBER
Reynolds
Number ~ ratio
of inertial =
to I.F./V.F.
viscous for
(Cartoon
approach)ces
(cartoon
approach)
-- hand waving argument -2
2
Inertial Force  L U
ViscousForce  UL
L
Define Re as LU/
where L & U are some
characteristic length scales
Re = Lu/
from N. S. E.
REYNOLDS NUMBER a la N.S.E.
a = Du/Dt = F
Eq. 5.27a:
(u/t + uu/x + vu/y + wu/z)
= - p/x + (2u/x2 + 2u/y2 + 2u/z2)
x-component
incompressible, constant , Newtonian,
ignore gravity, e-m, … forces,
REYNOLDS NUMBER a la N.S.E.
(u/t + uu/x + vu/y + wu/z)
= - p/x + (2u/x2 + 2u/y2 + 2u/z2)
Let u’ = u/U, v’ = v/U, w’ = w/U; x’ = x/L,
y’ = y/L, z’ = z/L; t’ = t/T and p’ = p/ (U2)
L and U are characteristic lengths and velocities and T=L/U
(Uu’)/(Tt’) + Uu’(Uu’)/(Lx’) +
Uv’(Uu’)/(Ly’) + Uw’(Uu’)/(Lz’)
= - (1/) p’U2 /Lx’ + (2(Uu’)/(Lx’)2 +
2(Uu’)/(Ly’)2 + 2(Uu’/(Lz’)2
REYNOLDS NUMBER a la N.S.E.
(Uu’)/(Tt’) + Uu’(Uu’)/(Lx’) +
Uv’(Uu’)/(Ly’) + Uw’(Uu’)/(Lz’)
= - (1/)  p’U2 /Lx’ + (2(Uu’)/(Lx’)2 +
2(Uu’)/(Ly’)2 + 2(Uu’/(Lz’)2
U/T = U/(L/U) = U2/L
{U2/L}[u’/t’+u’u’/x’+v’u’/y’+w’u’/z’]
= - {U2/L}p’/x’ +
{U/L2}(2u’/x’2 + 2u’/y’2 + 2u’/z’2)
REYNOLDS NUMBER a la N.S.E.
{U2/L}[u’/t’+u’u’/x’+v’u’/y’+w’u’/z’]
= - {U2/L}p’/x’ +
{U/L2}(2u’/x’2 + 2u’/y’2 + 2u’/z’2)
[u’/t’ + u’u’/x’ + v’u’/y’ + w’u’/z’]
= - p’/x’ +
{1/[UL]}(2u’/x’2 + 2u’/y’2 + 2u’/z’2)
1/ReL
REYNOLDS NUMBER a la N.S.E.
[u’/t’ + u’u’/x’ + v’u’/y’ + w’u’/z’]
= - p’/x’ +
{1/ReL}(2u’/x’2 + 2u’/y’2 + 2u’/z’2)
High Re # in some ways independent of viscosity.
However, near wall viscosity always important!
(why?)
REYNOLDS NUMBER a la N.S.E.
[u’/t’ + u’u’/x’ + v’u’/y’ + w’u’/z’]
= - p’/x’ +
{1/ReL}(2u’/x’2 + 2u’/y’2 + 2u’/z’2)
High Re # in some ways independent of viscosity.
However, near wall viscosity always important!
(because velocity gradients large)
REYNOLDS NUMBER a la N.S.E.
[u’/t’ + u’u’/x’ + v’u’/y’ + w’u’/z’]
= - p’/x’ +
{1/ReL}(2u’/x’2 + 2u’/y’2 + 2u’/z’2)
Two flows with the same geometry, same ReL
and satisfying the above equation (i.e. no body forces,
incompressible, constant visosity, Newtonian) will have
similar flow fields (dynamic similarity). Hence drag forces
measured in the lab can be extrapolated to full scale!
The principle of dynamic similarity makes
it possible to predict the performance of
full-scale aircraft from wind tunnel tests.
Drag coefficient is same for
dynamically similar flows
Drag coefficient =
Drag Force / (U2L2)
Lift coefficient =
Lift Force / (U2L2)
Reynolds Experiment (1883)
REYNOLDS NUMBER - EMPIRACLE
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 .
~ Nakayama & Boucher
Sommerfeld in a 1908 paper
first referred to uD/ as the
Reynolds number
REYNOLDS NUMBER - EMPIRACLE
Reynolds found that the
quality of the pipe inlet
affected transition – with a
smoother, bell-mouthed inlet
transition was delayed to
higher Reynolds numbers.
Laminar pipe flow is stable
to infinitesimal disturbances.
From Reynolds’ 1883 paper
Reynolds found
transition to occur
around Re = 13,000,
when experiment
repeated a hundred
years later (left)
transition was found
to be much less –
WHY?
Note: In pipe flow turbulence does not suddenly
at Retr appear throughout the pipe. It forms
turbulent slugs near the pipe entrance and grows
as it is passed through the pipe.
Pipe centerline: (a) fluctuating velocity; (b) mean velocity
u fluctuation
u mean
Re = 2550
----- from Triton
QUESTION: refer to data above, head not changing,
roughness not changing, viscosity not changing,
pipe diameter not changing – so why is flow rate?
Flow rate is reduced with appearance
of turbulent “slug”. Flow slows down
cause increased wall friction due to
turbulence. If near Recr then new Re
can be < Recr so no turbulent slugs near
entrance. After turbulent slug passes,
flow speeds up and Recr reoccurs and
pattern repeats.
breath
Fully Developed Flow
Chapter 8:
Internal Incompressible Viscous Flow
“viscous forces are dominant” - MYO
Internal Flows can be:
developing flows - velocity profile changing
fully developed - velocity profile not changing
Chapter 8:
Internal Incompressible Viscous Flow
V=0 AT WALL
D = 27 mm
Vavg = 6 cm/sec
ReD = 1600
V=0 AT WALL
Internal Flows can be:
developing flows - velocity profile changing
fully developed - velocity profile not changing
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}
{L/D = 0.06 Re, White}
{L/D = 0.13 Re,
Boussinesq 1891}
Pressure gradient
balances wall
shear stress
No acceleration
As “inviscid” core accelerates, pressure must drop
Pressure gradient
balances both
wall shear stress
and acceleration
Pressure gradient
balances wall
shear stress
No acceleration
Le = 140D, Re = 2300
{same trends for
turbulent flow}
As “inviscid” core accelerates, pressure must drop
Turbulent Pipe Flow
Entrance Length for
Fully Developed Flow
25-40 pipe diameter - Fox…
Le/D = 4.4 Re1/6 - MYO
20D < Le < 30D
104 < Re < 105 = White
Entrance length
much shorter now
in turbulent flow
Pressure gradient
balances wall
shear stress
No average
acceleration
LAMINAR Pipe Flow Re< 2300 (2100 for MYO)
PIPE
ReD = 1600
LAMINAR Duct Flow Re<1500 (2000 for SMITS)
DUCT FLOW: H = 0.2 cm, Uavg = 3.2 cm, ReH = 64
Fully Developed Laminar Pipe/Duct Flow
LAMINAR Pipe Flow Re< 2300 (2100 for MYO)
LAMINAR Duct Flow Re<1500 (2000 for SMITS)
Uo = V = Q/A
OUTSIDE BLUNDARY LAYER TREAT AS INVISCID, CAN USE B.E.
breath
Incompressible
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)
•For water  usually considered constant
•For gas  usually considered constant for M < 0.3
(~100m/s or 230 mph; / ~ 4%)
•Pressure drop in pipes “usually” not large enough to
make compressibility an issue (water hammer in an
exception).
Time Out
Static / Dynamic / Stagnation Pressures
What are static, dynamic and stagnation pressures?
The thermodynamic pressure, p, used throughout
this book refers to the static pressure. This is the
pressure experienced by a fluid particle as it moves
with the fluid.
static pressure
What are static, stagnation, and dynamic pressures?
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
What are static, dynamic and stagnation pressures?
The dynamic pressure is defined as ½  V2.
For incompressible flow: ½  V2 = po - p
breath
Laminar Flow – Theory
Fully Developed Flow
FULLY DEVELOPED LAMINAR FLOW
BETWEEN INFINITE PARALLEL PLATES
If gap between piston and cylinder is 0.005
mm or less than this flow can be modeled as
flow between infinite parallel plates.
(high pressure hydraulic system
like break system of car)
Want to know “stuff” like:
What’s pressure drop for specified flow & length?
What’s shear stress on bottom & top plates? Suppose plate moving?
What’s leakage flow rate of hydraulic oil between piston and cylinder…
need to know what u(y) is
FULLY DEVELOPED LAMINAR FLOW BETWEEN INFINITE PARALLEL PLATES
= 0(3)
= 0(1)
= 0(2)
FSx + FBx = /t (cvudVol )+ csuVdA
Eq. (4.17)
Assumptions: (1) steady, incompressible, (2) fully developed
flow (3) no body forces, (4) no changes in z variables, (5)
u = 0 at y = 0, y = a
FSx = surface forces
= pressure and shear forces
in x-direction
=0
+y
+x
FULLY DEVELOPED LAMINAR FLOW
BETWEEN INFINITE PARALLEL PLATES
Could use NSE directly, instead will derive velocity profile
using a differential control volume.
FULLY DEVELOPED LAMINAR FLOW
BETWEEN INFINITE PARALLEL PLATES
y=a
u = [a2/2](dp/dx)[(y/a)2 – (y/a)]
y=0
FSx = 0
y
x
+
+
+
=0
FULLY DEVELOPED LAMINAR FLOW BETWEEN INFINITE PARALLEL PLATES
FULLY DEVELOPED LAMINAR FLOW BETWEEN INFINITE PARALLEL PLATES
(Want to know what the velocity profile is.)
+
+
=0
- p/x + dxy/dy = 0
p/x = dp/dx = dxy/dy = constant
Left side is f(x) only [p(x)] = Right side f(y) only [u(y)]
Can only be true for all x and y if both sides equal a constant
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.
du/dx = 0 everywhere since fully developed,
therefore dv/dy = 0 everywhere,
but since v = 0 at boundary,
then v = 0 everywhere!
FULLY DEVELOPED LAMINAR FLOW BETWEEN INFINITE PARALLEL PLATES
Proof that p/y
= 0*
N.S.E. for incompressible flow with and constant viscosity.
v-component
(v/t + uv/x + vv/y + wv/z) =
gy - p/y + (2v/x2 + 2v/y2 + 2v/z2)
Eq 5.27b, pg 215
v = 0 everywhere and always, gy ~ 0 so left with:
p/y = 0; p = f(x) only!!!
FULLY DEVELOPED LAMINAR FLOW BETWEEN INFINITE PARALLEL PLATES
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.
FULLY DEVELOPED LAMINAR FLOW BETWEEN INFINITE PARALLEL PLATES
integrate
p/x = dp/dx = dxy/dy
yx = (dp/dx)y + c1
(Want to know what the velocity profile is.)
For Newtonian fluid*
substitute
integrate
USE 2 BOUNDARY CONDITIONS TO SOLVE FOR c1 AND c2
a
0
u = 0 at y = 0:
c2 = 0
u = 0 at y = a:
c1 = -1/2 (dp/dx)a
u = [1/(2)][dp/dx]y2 - [1/(2)] [dp/dx]ay
= [a2/(2)][dp/dx]{(y/a)2 – y/a}
FULLY DEVELOPED LAMINAR FLOW BETWEEN INFINITE PARALLEL PLATES
FULLY DEVELOPED LAMINAR FLOW BETWEEN INFINITE PARALLEL PLATES
u = [a2/(2)][dp/dx]{(y/a)2 – y/a}
u = {(y/1)^2 -(y/1)}; channel height=1m
a = 1; dp/dx = 2
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
FULLY DEVELOPED LAMINAR FLOW BETWEEN INFINITE PARALLEL PLATES
u(y) for fully developed laminar flow
between two infinite plates
y=a
y=0
negative
BREATH
Aside
Do you believe in the no slip condition?
NO SLIP
CONDITION
“Because of the no slip condition at the wall
we know that the velocity at the wall must
be zero along the entire length of the pipe.”
Pg 311
NO SLIP
CONDITION
“The fluid in direct contact with the solid
boundary has the same value as the boundary
itself; there is no slip at the boundary. This is
an experimental fact based on numerous
observations of fluid behavior.”
Pg 3
Parallel Plates - Re = UD/ = 140
Water Velocity = 0.5 m/s
Experimentally
found
(usually)
Circular Pipe – Re = UD/ = 195
Water Velocity = 2.4 m/s
Hydrogen Bubble Flow
Visualization
No slip condition
explains:
Why large particles are
easy to remove by
blowing but small
particles are not.
Why there is dust on a
fan blade.
Why it is difficult to all
the soap from a dish, just
by running water.
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)
Low Reynolds
number
High Reynolds
number
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
Neuman / Hagenbach (18581860) ~
Correct analytical solution to
laminar pipe flow
So you think you are comfortable
with the no-slip condition …
What happens to fluid particles
next to no-slip layer?
Surface Roughness
“It has been argued that the no-slip condition,
applicable when a viscous fluid flows over a solid surface,
may be an inevitable consequence of the fact that all
such surfaces are, in practice, rough on a microscopic
scale: the energy lost through viscous dissipation as a
fluid passes over and around these irregularities is
sufficient to ensure that it is effectively brought to rest.”
- On the No-Slip Boundary Condition, S.Richardson
Journal of Fluid Mechanics (1973), vol. 59, part 4, pp.707-719
No Slip Condition: u = 0 at y = 0
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
Bioluminescence on treated (lower)
and untreated (upper) surface
Bioluminescence on treated (upper)
and untreated (lower) surface
Flashlight
“It turns out –
although it is not at all self evident –
that in all circumstances where it has
been experimentally checked, the
velocity of a fluid is exactly zero at the
surface [with zero velocity]of a solid.”
The Feynman Lectures on Physics – 1964, Vol. II, 41-1
Slip Boundary conditions
for water flows in
hydrophobic nanoscale
geometries
J. H. Walther , R. L. Jaffe , T. Werder , and P. Koumoutsakos
Swiss Federal Institute of Technology, CH
Keywords: nanofluidics, slip condition , hydrophobic surfaces,
Abstract:
In a collaboration with experimental groups at NASA and ETH Zurich we conduct
computational studies towards the development of biosensors in aqueous
environments. Examples include arrays of carbon nanotubes that may operate
as artificial stereocillia (Noca et al., 2000) or as molecular sieves. Here we
present novel results assesing the validity of the no-slip boundary condition in
nanofluidics for prototypical geometries such as flow past a carbon nanotube
and flow between two graphite plates. The role of the geometry on the slip
length is investigated. The results show significant slip lengths (in disagreement
with the macroscale notion of no-slip at wall-fluid interfaces) and are consistent
with relevent experimental works of water flows over other hydrophobic
surfaces. First we report results from large scale non-equilibrium molecular
dynamics (NEMD) simulations of water flow past graphite surfaces in a setting
equivalent to a nanoscale planar Couette flow (Figure 1). A graphite surface is
known to be hydrophobic(Adamson:1997), and to exhibit physiochemical
similarity with carbon nanotubes in aqueous environments (Balavoine:1999).
The validity of the no-slip condition employed in macroscale Navier-Stokes
modeling has been questioned by experiments of water in hydrophobic
capillaries (Churaev:1984, Baudry:2001). In these experiments, the water is
found to exhibit a finite fluid velocity at the fluid-solid interface, with a slip
length of 28--30nm.The present NEMD simulations use the SPC/E water model
and the graphite-water interaction is modeled using a Lennard-Jones potential
calibrated to match the experimentally measured macroscopic contact angle of
water on graphite, cf. (Werder:2002). The average density profile in the channel
displays the well-known peaks in the vicinity of the interface and bulk properties
at the center of the channel cf. Figure 2a. Setting the upper walll in motion with
speeds of 50 to 100m/s drives the water and a linear velocity profiles is
established after 1-2ns as shown in Figure 2. The velocity profiles indicate a slip
length approximately Ls = 30nm in good agreement with the relevant
experimental values(Churaev:1984, Baudry:2001). In order to examine
geometry effects on the no-slip condition we conduct also simulations of flows
past carbon nanotubes (with diameters of 1 to 2 nm) whose axis is placed
perpedincular to the mean flow (Figure 1). In this case a slip length of 1nm is
observed. A systematic study is conducted where the effects of geometry and
driving mean velocity are assesed and a boundary condition for macroscale
simulations of water flows past hydrophobic surfaces is proposed.
Laminar Flow – Theory
Found u(y) / now yx(y)
(next want to determine shear stress profile,yx)
+ shear
direction
Shear force
+
y=a
yx = (du/dy)
y=0
+
For dp/dx = negative
yx in top ½ is negative & shear force is in the – x direction
yx in bottom ½ is positive & shear force is in the – x direction
FULLY DEVELOPED LAMINAR FLOW BETWEEN INFINITE PARALLEL PLATES
xy = a(dp/dx){y/a - 1/2}
negative
Shear force direction
y=a
Flow direction
dp/dx = negative
+
y=0
Shear force direction
Sign convention
for stresses
x
White
Positive stress is defined in the + x-direction Positive stress is defined in the – (x-direction)
as normal to surface is in the + z-direction as normal to surface is in the – (z-direction)
FULLY DEVELOPED LAMINAR FLOW BETWEEN INFINITE PARALLEL PLATES
xy = a(dp/dx){y/a - 1/2}
positive
tau = [(y/1)-1/2]; a=1, dp/dx=1
Shear force direction
1
0.9
0.8
0.7
y
a
0.6
0.5
y
0.4
0.3
Flow direction
0.2
0.1
0
-0.6
-0.4
-0.2
0
tau
0.2
0.4
Shear force direction
0.6
BREATH
FULLY DEVELOPED LAMINAR FLOW BETWEEN INFINITE PARALLEL PLATES
CHANGE OF VARIABLES
y=a
y’ = a/2
l
y’=0
l
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
Laminar Flow – Theory
Found u(y), yx(y); now Q, uavg, umax
FULLY DEVELOPED LAMINAR FLOW BETWEEN INFINITE PARALLEL PLATES
(volume flow rate, Q)
y=a
y=0
[y3/3 – ay2/2]oa = a3/3 – a3/2 = -a3/6
If dp/dx = const
FULLY DEVELOPED LAMINAR FLOW BETWEEN INFINITE PARALLEL PLATES
( average velocity)
A = la
= Uavg
FULLY DEVELOPED LAMINAR FLOW BETWEEN INFINITE PARALLEL PLATES
(maximum velocity)
(a2/4)/a2 – (a/2)/a = -1/4
BREATH
Laminar Flow – Theory
Upper Plate Moving
UPPER PLATE MOVING WITH CONSTANT SPEED U
Journal bearing (crankshaft inside car engine)
UPPER PLATE MOVING WITH CONSTANT SPEED U
UPPER PLATE MOVING WITH CONSTANT SPEED U
Velocity distribution
UPPER PLATE MOVING WITH CONSTANT SPEED U
+
Boundary driven
Pressure driven
UPPER PLATE MOVING WITH CONSTANT SPEED U
Shear stress distribution
UPPER PLATE MOVING WITH CONSTANT SPEED U
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]
UPPER PLATE MOVING WITH CONSTANT SPEED U
Volume Flow Rate
UPPER PLATE MOVING WITH CONSTANT SPEED U
Average Velocity
Area = al
l
UPPER PLATE MOVING WITH CONSTANT SPEED U
Maximum Velocity
y=a
umax = a/2
y=0
UPPER PLATE MOVING WITH CONSTANT SPEED U
very large shear
stresses at start-up
NOTE THAT STEADY FLOW FIELD IS
NOT ESTABLISHED INSTANTANEOUSLY
BREATH
Laminar Flow – Theory
Example
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
FBx = - gdxdydz
Fs1 – Fs2 - gdxdydz = 0
Fs1 = [yx + (dyx/dy)(dy/2)]dxdz
Fs2 = [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) - ghy/ +U0
ve locity profile
600
500
400
300
200
100
0
0
0.02
0.04
0.06
0.08
At y=h, u = gh2/(2) - gh2/ + U0
u = -gh2/(2) + U0
0.1
BREATH
Laminar Flow – Theory
Fully Developed Pipe Flow
Fully Developed Pipe Flow
CV
w
p1
V
l
w
p2
A = D2/4
 Fx  p1 A  p2 A   wDl  0
A p1  p2  D p1  p2 
w 

Dl
4l
Fully Developed Pipe Flow
CV
w
p1
V
l
(r) =
2
{r /4}{p
p
 r
2l
w
p2
A = r2/4
1-p2}/{2rl}
or  = (r/2)(dp/dx)
Eq 8.13a
p
 r
2l
p
w 
R
2l
R
 wr
2 w r


R
D
(r) on
control
volume
CV
+r
p1
w
V
+r
l
w
True for laminar and turbulent flow!!!
p2
 wr
2 w r


R
D
FULLY DEVELOPED LAMINAR PIPE FLOW
u/umax = 1 –
rx
=
du/dr
LAMINAR
2
(r/R)
rx
=
r(dp/dx)/2
u/umax
or
/w
r/R
LAMINAR AND
TURBULENT
p
 r
2l
only
laminar
du
  
dr
 p 
du  
rdr
 2 l 
p
 du  
 rdr
2l
p 2
u
r C
4l
u = 0, at y = R
p 2
C
R
4l
p 2
2
u
(R  r )
4l
p 2
2
u
(R  r )
4l
Eq. 8.12
CV
w
p1
V
l
w
p2
p 2
2
u
(R  r )
4l
Eq. 8.12
……
Eq. 8.12
Q = A V • dA
Q = 0R u2rdr
Q = 0R -[ R2 - r2] (dp/dx)/(4) 2rdr
Q = [(dp/dx)/(4)] (2)[ r4/4 - R2r2/2 ]0R
Q = (-R4dp/dx)/(8)
Eq. 8.13b
Eq. 8.12
umax = - (R2/(4)) (dp/dx)
Eq. 8.13e
V = uavg = Q/R2 = (-R4dp/dx)/(8R2)
V =uavg = -(R2/(8)) (dp/dx)
uavg = ½ umax
Eq. 8.13d
BREATH
Laminar Flow
1
2
f = {(p/L)D}/{ /2uavg } = ?
Laminar Flow
1
2
f = {(p/L)D}/{ /2uavg } = ?
uavg = -(R2/(8)) (dp/dx)
Eq. 8.13d
uavg = (R2/(8)) (p/L); p/L = uavg8/R2
f = [uavg8/R2] D/{1/2uavg2}
f = {64/D}/{uavg} = 64/{uavgD}
f = 64/ReD
THE END
Laminar Flow – Theory
Fully Developed Pipe Flow
Fox et al.’s development
FULLY DEVELOPED LAMINAR PIPE FLOW
APPROACH JUST
LIKE FOR DUCT FLOW
FULLY DEVELOPED LAMINAR PIPE
FLOW
r
r
r
r
dFL = p2rdr
dFR = -(p + [dp/dx]dx) 2rdr
dFI = -rx2rdx
dFO = (rx + [d rx/dr]dr) 2(r + dr) dx
FULLY DEVELOPED LAMINAR PIPE
FLOW
r
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
FULLY DEVELOPED LAMINAR PIPE FLOW
dp/dx = (1/r)(d[rrx]/dr)
p is uniform at each
section by symmetry.
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)
FULLY DEVELOPED LAMINAR PIPE FLOW
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?
FULLY DEVELOPED LAMINAR PIPE FLOW
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
dp/dx is
negative
For dp/dx negative, get negative shear stress on CV
FULLY DEVELOPED LAMINAR PIPE FLOW
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
SHEAR STRESS PROFILE
FULLY DEVELOPED PIPE FLOW
= direction of shear force on CV
FULLY DEVELOPED DUCT FLOW
- for flow to right
SHEAR STRESS PROFILE
rx = r(dp/dx)/2
TRUE FOR LAMINAR AND TURBULENT FLOW
du/dr = r(dp/dx)/2
TRUE ONLY FOR LAMINAR FLOW
FULLY DEVELOPED LAMINAR PIPE FLOW
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]
BREATH
FULLY DEVELOPED LAMINAR PIPE FLOW
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
FULLY DEVELOPED LAMINAR PIPE FLOW
VOLUME FLOW RATE
– 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)
FULLY DEVELOPED LAMINAR PIPE FLOW
AVERAGE FLOW RATE
Q = R4p/(8L)
uAVG = Q/A = Q/(R2)
= R4p/(R28L)
= R2p/(8L)
= -(R2/(8)) (dp/dx)
AVERAGE FLOW RATE
uAVG = V = Q/A = Q/(R2) = R4p/(R28L)
uAVG = R2p/(8L) = -(R2/(8)) (dp/dx)
FULLY DEVELOPED LAMINAR PIPE FLOW
MAXIMUM FLOW RATE
du/dr = (r/[2])p/x
At umax, du/dr = 0;
which occurs at r = 0
umax = R2(p/x)/(4)
MAXIMUM FLOW RATE
END