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8/18/2019 Lecture 2 0 Fluid Dynamics
1/35
Fluid Dynamics
Dr. Hamdy A. Kandil
Review: Fluid Kinematics -Acceleration Field
Consider a fluid particle and Newton's second law,
The acceleration of the particle is the time derivative of theparticle's velocity.
However, particle velocity at a point is the same as the fluidvelocity,
To take the time derivative of V, chain rule must be used because.
particle part icle particleF m a
par ticl e
par ticl e
dV a
dt
, , particle particle particle particleV V x t y t z t
particle particle particle
particle
dx dy dzV dt V V V a
t dt x dt y dt z dt
8/18/2019 Lecture 2 0 Fluid Dynamics
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Acceleration Field Since
In vector form, the acceleration can be written as
First term is called the local acceleration and is nonzero only forunsteady flows.
Second term is called the advective acceleration and accountsfor the effect of the fluid particle moving to a new location in theflow, where the velocity is different.
particle
V V V V
a u v wt x y z
, , particle particle particledx dy dz
u v wdt dt dt
V V t
V
dt
V d t z y xa
)(),,,(
then
k
z
j
y
i
x
k w jviuV
Where:
Acceleration Components
z
ww
y
wv
x
wu
t
wa
z
vw
y
vv
x
vu
t
va
z
uw
y
uv
x
uu
t
ua
z
y
x
The components of the acceleration are:
particle
V V V V a u v w
t x y z
Vector equation:
x- component
Y-component
z-component
Question: Give examples of steady flows with acceleration
Incompressible Steady idealflow in a variable-area duct
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Streamlines A Streamline is a curve thatis everywhere tangent to theinstantaneous local velocityvector.
Consider an arc length
must be parallel to thelocal velocity vector
Geometric arguments resultsin the equation for astreamline
dr dxi dyj dzk
dr
V ui vj wk
dr dx dy dz
V u v w
u
v
dx
dy
Properties of Streamlines A streamline can not have a sharp corner
Two streamlines can not intersect or meet at a sharp corner
No flow across a tube formed by streamlines (steam tube).
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Streaklines A Streakline is the locus of fluid
particles that have passedsequentially through aprescribed point in the flow.
Easy to generate inexperiments: dye in a waterflow, or smoke in an airflow.
Pathlines A Pathline is the actual pathtraveled (traced) by anindividual fluid particle oversome time period.
Same as the fluid particle'smaterial position vector
Particle location at time t:
Particle Image Velocimetry(PIV) is a modern experimentaltechnique to measure velocityfield over a plane in the flowfield.
, , par ticle particle par ticle x t y t z t
start
t
start
t
x x Vdt
For steady flow, streamlines, pathlines, andstreaklines are identical.
8/18/2019 Lecture 2 0 Fluid Dynamics
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Which type of lines?
Conservation of Mass Differential CV
Infinitesimal control volumeof dimensions dx, dy, dz Area of right
face = dy dz
Mass flow rate throughthe right face of thecontrol volume
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Conservation of Mass Differential CV
Now, sum up the mass flow rates into and out of the 6 faces of the CV
Plug into integral conservation of mass equation
Net mass flow rate into CV:
Net mass flow rate out of CV:
Rate of change of the control volume mass:
Conservation of Mass Differential CV
After substitution,
Dividing through by volume dxdydz
Or, if we apply the definition of the divergence of a vector
k z
j y
i x
k w jviuV
where
0)()()(
z
w
y
v
x
u
t
8/18/2019 Lecture 2 0 Fluid Dynamics
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Conservation of Mass: Alternative forms
Use product rule on divergence term
k z
j y
i x
k w jviuV
Conservation of Mass: Cylindrical coordinates
There are many problems which are simpler to solve if theequations are written in cylindrical-polar coordinates
Easiest way to convert from Cartesian is to use vector form anddefinition of divergence operator in cylindrical coordinates
8/18/2019 Lecture 2 0 Fluid Dynamics
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Conservation of Mass: Cylindricalcoordinates
Conservation of Mass: Special Cases Steady compressible flow
Cartesian
Cylindrical
0)()()(
z
w
y
v
x
u
8/18/2019 Lecture 2 0 Fluid Dynamics
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Conservation of Mass: Special Cases Incompressible flow
Cartesian
Cylindrical
= constant, and hence
0
z
w
y
v
x
u
k z
j y
i x
k w jviuV
Conservation of Mass In general, continuity equation cannot be used by itself to
solve for flow field, however it can be used to
1. Determine if a velocity field represents a flow.
2. Find missing velocity component
equation. continuity hesatisfy t torequired , w:Determine?
flow ibleincompressanFor
Example
222
w
z yz xyv
z y xu
),(2
3:Solution2
y xc z
xzw
8/18/2019 Lecture 2 0 Fluid Dynamics
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Conservation of MomentumTypes of forces:1. Surface forces: include all forces acting on the boundaries of a
medium though direct contact such as pressure, friction,…etc.
2. Body forces are developed without physical contact anddistributed over the volume of the fluid such as gravitationaland electromagnetic.
The force F acting on A may be resolved into two components,one normal and the other tangential to the area.
Stresses (forces per unit area)
Double subscript notation for stresses.• First subscript refers to the surface• Second subscript refers to the direction• Use for normal stresses and for tangential stresses
Surface ofconstant xSurface of
constant -x
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Momentum Conservation From Newton’s second law: Force = mass * acceleration
Consider a small element x y z as shown
The stresses at the center of the element are presented by thestress tensor:
x
y
z
The element experiences an acceleration
DVm ( )
Dt
as it is under the action of various forces:
normal stresses, shear stresses, and gravitational force.
V V V V x y z u v w
t x y z
zz zy zx
yz yy yx
xz xy xx
T
Surface forces in the x direction acting on afluid element.
Consider only the forces in the x direction acting on all six surfaces
zz zy zx
yz yy yx
xz xy xx
T
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Momentum Balance (cont.)
yxxx zx
Net force acting along the x-direction:
x x x x x y z x y z x y z g x y z
Normal stress Shear stressesBody force
z y xg y x z
z
y x z
z
z x y
y
z x y
y z y
x
x z y
x
xF
x zx
zx zx
zx
yx
yx
yx
yx xx
xx xx
xx x
)
2
()
2
()
2
(
)2
()2
()2
(
bxsx x F F F
)()( a z
uw
y
uv
x
uu
t
u
z y xg zx
yx xx x
The differential equation of motion in the x-direction is: xsxbx x x maF F maF
Similar equations can be obtained for the other two directions
Forces acting on the elementSurface forces
Body forces
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Equation of Motion (momentum eq.s)
Velocities stresses ----- Unknowns
fluid. aformotionof equationaldifferenti General
)()(
)()(
)()(
c z
w
w y
w
v x
w
ut
w
z y xg
b z
vw
y
vv
x
vu
t
v
z y xg
a z
uw
y
uv
x
uu
t
u
z y xg
zz yz xz z
zy yy xy
y
zx yx xx x
zszbz z z
ysyby x x
xsxbx x x
maF F maF
maF F maF
maF F maF
Conservation of Linear Momentum
Unfortunately, this equation is not very useful
10 unknowns
Stress tensor, : 6 independent components
Note:
Density
Velocity, V : 3 independent components
4 equations (continuity + 3 momentum) 6 more equations required to close problem!
zz zy zx
yz yy yx
xz xy xx
T
zy yz zx xz yx xy and ,
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Navier-Stokes Equations For Newtonian fluids the stress tensor components have some
definitions.
Substituting the stress tensor components into the equation ofmotion with produce the well-known Navier-Stokes equationsthat are suitable only for Newtonian Fluids:
Stress tensor components:
z
wV p
y
vV p
x
uV p
zz
yy
xx
2)3
2(
2)3
2(
2)3
2(
)(
)(
)(
y
w
z
v
z
u
x
w
y
u
x
v
zy yz
zx xz
yx xy
&
Navier-Stokes Equations
)()]([
)]([)].3
22([
a z
u
x
w
z
y
u
x
v
yV
x
u
x x
pg
Dt
Du x
)()]([
)].3
22([)]([
b y
w
z
v
z
V y
v
y y
u
x
v
x y
pg
Dt
Dv y
)()].3
22([
)]([)]([
cV z
w
z
yw
zv
y zu
xw
x z pg
Dt Dw z
x-momentum
y-momentum
z-momentum
8/18/2019 Lecture 2 0 Fluid Dynamics
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Navier-Stokes Equations For incompressible fluids, constant µ:
Continuity equation: .V = 0
u z
u
y
u
x
u
z
w
y
v
x
u
x z
u
y
u
x
u
z x
w
y x
v
x
u
z
u
y
u
x
u
z
u
x
w
z y
u
x
v
y x
u
x
z
u
x
w
z y
u
x
v
yV
x
u
x
2
2
2
2
2
2
2
2
2
2
2
2
2
22
2
2
2
2
2
2
2
2
)(
)()(
)()(
)]}[()][()]2[({
)]([)]([)].3
22([
Navier-Stokes Equations For incompressible flow with constant dynamic viscosity:
x- momentum
Y- momentum
z-momentum
In vector form, the three equations are given by:
)()(2
2
2
2
2
2
a z
u
y
u
x
u
x
pg
Dt
Du x
)()(2
2
2
2
2
2
b z
v
y
v
x
v
y
pg
Dt
Dv y
)()(2
2
2
2
2
2
c z
w
y
w
x
w
z
pg
Dt
Dw z
V pg Dt
V D
2 Incompressible NSEwritten in vector form
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Navier-Stokes Equation The Navier-Stokes equations for incompressible flow in vector
form:
This results in a closed system of equations!
4 equations (continuity and 3 momentum equations)
4 unknowns (U, V, W, p)
In addition to vector form, incompressible N-S equation can bewritten in several other forms including:
Cartesian coordinates
Cylindrical coordinates
Tensor notation
Incompressible NSEwritten in vector form
Euler Equations For inviscid flow (µ = 0) the momentum equations are given by:
x- momentum
Y- momentum
z-momentum
In vector form, the three equations are given by:
)()( a x
pg
z
uw
y
uv
x
uu
t
u x
pg Dt
V D
Euler equationswritten in vector form
)()( b y
pg
z
vw
y
vv
x
vu
t
v y
)()( c z
pg
z
ww
y
wv
x
wu
t
w z
8/18/2019 Lecture 2 0 Fluid Dynamics
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Differential Analysis of Fluid FlowProblems Now that we have a set of governing partial differential
equations, there are 2 problems we can solve
1. Calculate pressure (P ) for a known velocity field2. Calculate velocity (U, V, W ) and pressure (P ) for known
geometry, boundary conditions (BC), and initial conditions(IC)
There are about 80 known exact solutions to the NSE
Solutions can be classified by type or geometry, for example:
1. Couette shear flows
2. Steady duct/pipe flows (Poisseulle flow)
Exact Solutions of the NSE
1. Set up the problem and geometry, identifying all relevantdimensions and parameters
2. List all appropriate assumptions, approximations, simplifications,
and boundary conditions3. Simplify the differential equations as much as possible4. Integrate the equations5. Apply BCs to solve for constants of integration6. Verify results
Boundary conditions are critical to exact, approximate, andcomputational solutions.
BC’s used in analytical solutions are No-slip boundary condition
Interface boundary condition
Procedure for solving continuity and NSE
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boundary conditions
For a fluid in contact with asolid wall, the velocity of thefluid must equal that of the
wall
When two fluids meet at aninterface, the velocity andshear stress must be thesame on both sides
If surface tension effectsare negligible and thesurface is nearly flat
Interface boundary condition
No-slip boundary condition
Interface boundary condition Degenerate case of the interface BC occurs at the free surface
of a liquid. Same conditions hold
Since air
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Example exact solution
Fully Developed Couette Flow For the given geometry and BC’s, calculate the velocity and
pressure fields, and estimate the shear force per unit area
acting on the bottom plate
Step 1: Geometry, dimensions, and properties
Fully Developed Couette Flow
Step 2: Assumptions and BC’s
Assumptions
1. Plates are infinite in x and z
2. Flow is steady, /t = 03. Parallel flow, v = 0
4. Incompressible, Newtonian, laminar, constant properties
5. No pressure gradient
6. 2D, w = 0, /z = 0
7. Gravity acts in the -y direction,
Boundary conditions1. Bottom plate (y=0) – no slip condition: u=0, v=0, w=0
2. Top plate (y=h) : no slip condition: u=V, v=0, w=0
gg jgg y ,ˆ
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Fully Developed Couette Flow
Step 3: Simplify 3 6
Note: these numbers refer
to the assumptions on the
previous slide
This means the flow is “fully developed” or not changing in the direction of flow
Continuity
X-momentum
2 Cont. 3 6 5 7 Cont. 6
02
2
y
U
Fully Developed Couette Flow
Step 3: Simplify, cont.Y-momentum
2,3 3 3 3,6 3 33
Z-momentum
2,6 6 6 6 7 6 66
yg y
p
p = p(y)0 z
p yg
dy
dp
y
p
p = p(y)
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Fully Developed Couette Flow
Step 4: Integrate
y-momentum
X-momentumintegrate integrate
integrate
gdy
dp 3)( C gy y p
Fully Developed Couette Flow
Step 5: Apply BC’s
Y = 0, u = 0 = C1(0) + C2 C2 = 0
Y = h, u =V =C1h C1 = V/h
This gives
For pressure, no explicit BC, therefore C3 can remain anarbitrary constant (recall only P appears in NSE).
Let p = p0 at y = 0 (C3 renamed p0)
1. Hydrostatic pressure
2. Pressure acts independently of flowgy p y p 0)(
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Fully Developed Couette Flow
Step 6: Verify solution by back-substituting into differentialequations
Given the solution (u,v,w)=(Vy/h, 0, 0)
Continuity is satisfied
0 + 0 + 0 = 0
X-momentum is satisfied
Fully Developed Couette Flow
Finally, calculate shear force on bottom plate
Shear force per unit area acting on the wall
Note that w is equal and opposite to theshear stress acting on the fluid yx(Newton’s third law).
h
V
y
u
x
v yx xy
)(u=V/h
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Parallel Plates (Poiseuille Flow) Given: A steady, fully developed, laminar flow of a Newtonian
fluid in a rectangular channel of two parallel plates where thewidth of the channel is much larger than the height, h, betweenthe plates.
Find: The velocity profile and shear stress due to the flow.
Assumptions: Entrance Effects Neglected No-Slip Condition No vorticity/turbulence
Additional and Highlighted ImportantAssumptions
The width is very large compared to the height of the plate. No entrance or exit effects.
Fully developed flow. THEREFORE…
Velocity can only be dependent on vertical location in theflow (u )
v = w = 0 The pressure drop is constant and in the x-direction only.
.inlengthaiswhere, p
Constant p
x L L x
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Boundary Conditions No Slip Condition Applies
Therefore, at y = -a and y = +a, u=v=w = 0
The bounding walls in the z direction are often ignored. If wedon’t ignore them we also need: z = -W/2 and z = +W/2, u=v=w = 0, where W is the width of
the channel.
Fixed plate
Fixed plate
Fluid flow direction2ax
y
a y
a y
Incompressible Newtonian StressTensor
z
w
z
v
y
w
z
u
x
w
zv
yw
yv
yu
xv
z
u
x
w
y
u
x
v
x
u
2
2
2
τ
Now, we cancel terms out based
on our assumptions.This results in our new tensor:
000
00
00
y
u
y
u
T
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gvvvv
2 p
t
Navier-Stokes EquationsIn Vector Form for incompressible flow:
z
y
x
g z
w
y
w
x
w
z z
ww
y
wv
x
wu
t
w
g
z
v
y
v
x
v
y z
vw
y
vv
x
vu
t
v
g z
u
y
u
x
u
x z
uw
y
uv
x
uu
t
u
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
p
:component-z
p
:component-y
p
:component-x
Which we expand to component form:
z
y
x
g z
w y
w x
w z z
ww ywv
xwu
t w
g z
v
y
v
x
v
y z
vw
y
vv
x
vu
t
v
g
z
u
y
u
x
u
x z
uw
y
uv
x
uu
t
u
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
p
:component-z
p
:component-y
p
:component-x
Reducing Navier-Stokes
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Reducing Navier-Stokes This reduces to:
Including our constant pressure drop:
2
2
Pressure Modified
p0 y
u
x
2
2 p0
y
u
L
2
2 p0
: becomesequationtheTherefore b
dy
ud
L
yuuut
N-S equation therefore reduces to
02
2
xg y
u
x
p
0
yg y
p
0
zg z p
Ignoring gravitational effects, we get
02
2
y
u
x
p
0
y
p
0
z
pp is not a function of z
x - component:
y - component:
z - component:
x - component:
y - component:
z - component:
p is not a function of y
p is a functionof x only
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Rewriting (4), we get
(5)
u is a function of y only and therefore LHS is afunction of y only
p is a function of x only and is a constant andtherefore RHS is a function of x only
LHS = left hand side of the equationRHS = right hand side of the equation
function of (y) = function of (x)Therefore (5) gives,
x p
L p
x
p
y
u
12
2
= constant
It means = constant
That is, pressure gradient in the x -direction is a constant.
Rewriting (5), we get
where is the constant pressure gradient in the
x-direction
Since u is only a function of y, the partial derivative becomes anordinary derivative.
(6)
Therefore, (5) becomes
(7)
L
p
x
p
y
u
1
2
2
x
p
L
p
L
p
dy
ud
2
2
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Integrating (6), we get
(8)
where C1 and C2 are constants to bedetermined using the boundary
conditions given below:
Substituting the boundary conditions in (8), we get
Fixed plate
Fixed plate
Fluid flow direction
2ax
y
a yu at0a yu at0
a y
a y
21
2
2C yC
y
L
pu
01
C 2
2
2
a
L
pC
and
Therefore, (8) reduces to 222
ya L
pu
Parabolicvelocity profile
}no-slipboundarycondition
R. Shanthini15 March 2012
Fixed plate
Moving plateat velocity U
Fluid flow direction
2ax
y
Steady, incompressible flow of Newtonian fluid in an infinite channelwith one plate moving at uniform velocity- fully developed plane Couette-Poiseuille flow
where C1 and C2 are constants to bedetermined using the boundary conditionsgiven below:
a y
a y
21
2
2C yC
y
L
pu
a yu at0
a yU u at
Substituting the boundary conditions in (8), we get
and
Therefore, (8) reduces toa
U C
2
1
22
2
2
a
L
pU C
yaa
U ya
L
pu
22
22
(8)
}no-slipboundarycondition
Parabolic velocity profileis super imposed on alinear velocity profile
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R. Shanthini15 March 2012
Fixed plate
Moving plateat velocity U
Fluid flow direction
2ax
y
Steady, incompressible flow of Newtonian fluid in an infinite channelwith one plate moving at uniform velocity- fully developed plane Couette-Poiseuille flow
a y
a y
yaa
U ya
L
pu
22
22
If zero pressure gradient is maintained inthe flow direction, then ∆p = 0.
Therefore, we get
yaa
U u
2Linear velocity profile
Points to remember:- Pressure gradient gives parabolic profile- Moving wall gives linear profile
Starting from plane Coutte-Poiseuille flow
Let us get back to the velocity profile offully developed plane Couette-Poiseuille flow
yaa
U ya
L
pu
22
22
Let us non-dimensionalize it as follows:
a
yU
a
y
L
pau 1
21
2 2
22
Rearranging gives
Dividing by U gives
Introducing non-dimensional variables and , we get
a
y y
y y LU
pau
1
2
11
2
22
U
uu
a
y
a
y
LU
pa
U
u1
2
11
2 2
22
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R. Shanthini
15 March 2012-1
0
1
-0.5 0 0.5 1 1.5
1
0.25
0
-0.25
-1
y y LU
pau
1
2
11
2
22
2/2 a LU
p
a
y y
U
uu
Plot the non-dimensionalized velocity profile of thefully developed plane Couette-Poiseuille flow
R. Shanthini
15 March 2012-1
0
1
-0.25 0.75
0
-0.5
y y LU
pau
1
2
11
2
22
2/2 a LU
p
a y y
U
uu
BACK FLOW
Plot the non-dimensionalized velocity profile of thefully developed plane Couette-Poiseuille flow
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R. Shanthini
15 March 2012-1
0
1
-1 0 1
0
-0.25
-0.5
-1.5
y y LU
pau
1
2
11
2
22
2/2 a LU
p
a
y y
U
uu
Plot the non-dimensionalized velocity profile of thefully developed plane Couette-Poiseuille flow
yaa
U ya
L
pu
22
22
Let us determine the volumetric flow ratethrough one unit width fluid film along z -direction, using the velocity profile
Fixed plate
Moving plateat velocity U
Fluid flow direction
2a x
y
a y
a y
a
a
a
a
a
a
yay
a
U y ya
L
pdy ya
a
U ya
L
pudyQ
223222
23222
22322232
2
2
3
3
2
2
3
3 aaa
U aa L
paaa
U aa L
pQ
Ua L
paQ
3
2 3
Determine the volumetric flow rate of thefully developed plane Couette-Poiseuille flow
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-2
-1
0
1
2
3
4
-4 -2 0 2 4
Plot the non-dimensionalized volumetric flow rate of thefully developed plane Couette-Poiseuille flow
LU
pa
Ua
Q
2
3
21
Ua
Q
U
pa
2
Why the flow rate becomes zero?
Non-dimensionalizing Q, we get
Steady, incompressible flow of Newtonian fluid in a pipe- fully developed pipe Poisuille flow
Fixed pipe
z
r
Fluid flow direction 2a 2a
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Laminar Flow in pipes
.of linear ais pressure..
0
zeidz
dp
z
p
r vv
vv
z z
r
Assumptions:
Now look at the z -momentum equation.
2
2
2
2
2
11
)(
dz
vv
r r
vr
r r dz
dp
z
vv
v
r
v
r
vv
t
v
z z z
z z
z zr
z
Steady Laminar pipe flow.
The above “assumptions” can be obtained from the singleassumption of “fully developed” flow.
In fully developed pipe flow, all velocity components are assumed to
be unchanging along the axial direction, and axially symmetric i.e.:
Results from Mass/Momentum
Or rearranging, and substituting for the pressure gradient:
The removal of the indicated terms yields:
011
r
vr
r r dz
dp z
In a typical situation, we would have control over dp/dz . That is,we can induce a pressure gradient by altering the pressure at one
end of the pipe. We will therefore take it as the input to thesystem (similar to what an electrical engineer might do in testinga linear system).
dz
dp
dr
dvr
dr
d
r
z
11
dz
dpr
dr
dvr
dr
d z
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Laminar flow in pipes
But at r = 0 velocity = max or dvz/dr = 0 C1 = 0
Integrating once gives:
Integrating one more time gives:
1
2
2 C dz
dpr
dr
dv
r
z
dz
dpr
dr
dvr z
2
2
dz
dpr
dr
dv z
2
Divide by r
2
2
4C
dz
dpr v z
at pipe wall r = a velocity = 02
2
40 C
dz
dpa
dz
dpaC
4
2
2
Laminar flow in pipesSubstituting in the velocity equation gives:
)4
(44
2222
r a
dz
dp
dz
dpa
dz
dpr v z
Volume flow rate is obtained from:
tioncross
z dAvQsec
ar
r z dr rvQ
02
ar
r dr r ar dz
dp
Q0
22
2
)(
dz
dpaQ
8
4
dz
dp DQ
128
4
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Laminar flow in pipesAverage velocity, vav= Q/area
Maximum velocity at r = 0:
dzdp D D
dz
dp D
v ave z
324/128
224
,
)16
()4
(22
max,
D
dz
dpa
dz
dpv z
vavergae = ½ vmax
Shear stress:
)(8)4
()4
2)((
,
D
v D
dz
dpa
dz
dp
dr
dv ave zar
z
Laminar flow in pipes
Coefficient of friction: f = 4 Cf
eave
ave
ave
w f
R Dv
v
v
C 1616
2
1 22
Drag Coefficient:
e R f
64