lecture 5 - drahmednagib.com · the navier–stokes equation isthe cornerstone of fluid mechanics....
TRANSCRIPT
REE 307Fluid Mechanics II
Lecture 5November 8, 2017
Dr./ Ahmed Mohamed Nagib Elmekawy
Zewail City for Science and Technology
DIFFERENTIAL ANALYSIS
OF FLUID FLOW
2
2
The fundamental differential equations of fluid motion are derived in
this chapter, and we show how to solve them analytically for some
simple flows. More complicated flows, such as the air flow induced by
a tornado shown here, cannot be solved exactly.
3
3
Objectives
• Understand how the differential equation of
conservation of mass and the differential linear
momentum equation are derived and applied
• Obtain analytical solutions of the equations of
motion for simple flow fields
4
4
LAGRANGIAN AND EULERIAN DESCRIPTIONS
Kinematics: The study of motion.
Fluid kinematics: The study of how fluids flow and how to describe fluidmotion.
With a small number of objects, such
as billiard balls on a pool table,
individual objects can be tracked.
In the Lagrangian description, one
must keep track of the position and
velocity of individual particles.
There are two distinct ways to describe motion: Lagrangian and Eulerian
Lagrangian description: To follow the path of individual objects.
This method requires us to track the position and velocity of each individual
fluid parcel (fluid particle) and take to be a parcel of fixedidentity.
5
• A more common method is Eulerian description of fluid motion.
• In the Eulerian description of fluid flow, a finite volume called a flow domain
or control volume is defined, through which fluid flows in and out.
• Instead of tracking individual fluid particles, we define field variables,
functions of space and time, within the control volume.
• The field variable at a particular location at a particular time is the value of
the variable for whichever fluid particle happens to occupy that location at
that time.
• For example, the pressure field is a scalar field variable. We define the
velocity field as a vector field variable.
Collectively, these (and other) field variables define the flow field. The
velocity field can be expanded in Cartesian coordinates as
6
6
In the Eulerian description, one
defines field variables, such as
the pressure field and the
velocity field, at any location
and instant in time.
7
• In the Eulerian description we
don’t really care what happens to
individual fluid particles; rather we
are concerned with the pressure,
velocity, acceleration, etc., of
whichever fluid particle happens
to be at the location of interest at
the time of interest.
• While there are many occasions in
which the Lagrangian description
is useful, the Eulerian description
is often more convenient for fluid
mechanics applications.
• Experimental measurements are
generally more suited to the
Eulerian description.
7
CONSERVATION OF MASS—THE CONTINUITYEQUATION
To derive a differential
conservation equation, we
imagine shrinking a control
volume to infinitesimal size.
8
The net rate of change of mass within the
control volume is equal to the rate at
which mass flows into the control volume
minus the rate at which mass flows out of
the control volume.
8
8
Derivation Using an Infinitesimal Control Volume
A small box-shaped control
volume centered at point P
is used for derivation of the
differential equation for
conservation of mass in
Cartesian coordinates; the
blue dots indicate the center
of each face.
At locations away from the center of the
box, we use a Taylor ser ies expansion
about the center of the box.
9
10
10
11
Conservation of Mass: Alternative forms
• Use product rule on divergence term
kz
jy
ix
kwjviuV
12
Continuity Equation in Cylindrical Coordinates
Velocity components and unit vectors in cylindrical coordinates: (a) two-
dimensional flow in the xy- or r𝜃-plane, (b) three-dimensional flow.
13
Conservation of Mass: Cylindrical coordinates
The divergence
operation in Cartesian
and cylindrical
coordinates.
15
Special C a s e s of the Continuity Equation
Special Case 1: Steady Compressible Flow
16
Special Case 2: Incompressible Flow
17
18
19
20
If the differential fluid
element is a material
element, it moves with the
flow and Newton’s second
law applies directly.
21
THE DIFFERENTIAL LINEAR MOMENTUM EQUATION
Derivation Using Newton’s Second Law
Acceleration Field
Newton’s second law applied to a fluid
particle; the acceleration vector (gray arrow)
is in the same direction as the force vector
(black arrow), but the velocity vector (red
arrow) may act in a different direction.
The equations of motion for fluid flow
(such as Newton’s second law) are
written for a fluid particle, which we
also call a material particle.
If we were to follow a particular fluid
particle as it moves around in the
flow, we would be employing the
Lagrangian description, and the
equations of motion would be directly
applicable.
For example, we would define the
particle’s location in space in terms
of a material position vector
(xparticle(t), yparticle(t), zparticle(t)).
22
Local
acce lera tion
Advective (convective)
acceleration
The components of the
acceleration vector in cartesian
coordinates:
Flow of water through the nozzle of a
garden hose illustrates that fluid
particles may accelerate, even in a
steady flow. In this example, the exit
speed of the water is much higher than
the water speed in the hose, implying
that fluid particles have accelerated
even though the flow is steady.
24
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 Va 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 ideal flow in a variable-area duct
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 and distributed over the volume of the fluid such as gravitational and electromagnetic.
• The force F acting on A may be resolved into two components, one normal and the other tangential to the area.
Positive components of the stress
tensor in Cartesian coordinates on the
positive (right, top, and front) faces of
an infinitesimal rectangular control
volume. The blue dots indicate the
center of each face. Positive
components on the negative (left,
bottom, and back) faces are in the
opposite direction of those shown here.
Body Forces
27
28
For fluids at rest, the only
stress on a fluid element is
the hydrostatic pressure,
which always acts inward
and normal to any surface.
𝜏𝑖𝑗 called the
viscous s t ress
tensor
Surface Forces
Sketch illustrating the surface forces acting in the x-
direction due to the appropriate stress tensor component
on each face of the differential control volume; the blue
dots indicate the center of each face.
30
If the differential fluid
element is a material
element, it moves with the
flow and Newton’s second
law applies directly.
31
Newtonian versus Non-Newtonian Fluids
Rheological behavior of fluids—shear
stress as a function of shear strain rate.
32
Rheology: The study of the deformation of flowing fluids.
Newtonian fluids: Fluids for which the shear stress is linearly proportional to the shear strain rate.
Newtonian fluids: Fluids for which the shear stress is not linearly related to the shear strain rate.
Viscoelastic: A fluid that returns (either fully or partially) to its original shape after the applied stress is released.
Some non-Newtonian fluids are called
shear thinning fluids or
pseudoplas t ic fluids, because the
more the fluid is sheared, the less
viscous it becomes.
Plastic fluids are those in which the
shear thinning effect is extreme.
In some fluids a finite stress called the
yield s t ress is required before the
fluid begins to flow at all; such fluids
are called Bingham plastic fluids.
Derivation of the Navier–Stokes Equation for Incompressible, Isothermal Flow
The incompressible flow
approximation implies constant
density, and the isothermal
approximation implies constant
viscosity.
33
The Laplacian operator, shown
here in both Cartesian and
cylindrical coordinates, appears
in the viscous term of the
incompressible Navier–Stokes
equation.
34
28
The Navier–Stokes equation is the
cornerstone of fluid mechanics.
The Navier–Stokes equation is an
unsteady, nonlinear, second order, partial
differential equation.
Equation 9–60 has four unknowns (three
velocity components and pressure), yet it
represents only three equations (three
components since it is a vector equation).
Obviously we need another equation to
make the problem solvable. The fourth
equation is the incompressible continuity
equation (Eq. 9–16).
35
Navier-Stokes Equations
)()]([)]([
)].3
22([)(
az
u
x
w
zy
u
x
v
y
Vx
u
xx
pg
z
uw
y
uv
x
uu
t
ux
)()]([)].3
22([
)]([)(
by
w
z
v
zV
y
v
y
y
u
x
v
xy
pg
z
vw
y
vv
x
vu
t
vy
)()].3
22([)]([
)]([)(
cVz
w
zy
w
z
v
y
z
u
x
w
xz
pg
z
ww
y
wv
x
wu
t
wz
x-momentum
y-momentum
z-momentum
Navier-Stokes Equations• For incompressible fluids, constant µ:
• Continuity equation: .V = 0
uz
u
y
u
x
u
z
w
y
v
x
u
xz
u
y
u
x
u
zx
w
yx
v
x
u
z
u
y
u
x
u
z
u
x
w
zy
u
x
v
yx
u
x
z
u
x
w
zy
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
az
u
y
u
x
u
x
pg
Dt
Dux
)()(2
2
2
2
2
2
bz
v
y
v
x
v
y
pg
Dt
Dvy
)()(2
2
2
2
2
2
cz
w
y
w
x
w
z
pg
Dt
Dwz
VpgDt
VD
2 Incompressible NSEwritten in vector form
Continuity and Navier–Stokes Equations in Cartesian Coordinates
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 be written 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:
)()( ax
pg
z
uw
y
uv
x
uu
t
ux
pgDt
VD
Euler equationswritten in vector form
)()( by
pg
z
vw
y
vv
x
vu
t
vy
)()( cz
pg
z
ww
y
wv
x
wu
t
wz
Continuity and Navier–Stokes Equations in Cylindrical Coordinates
Differential Analysis of Fluid Flow Problems
• 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 field
2. 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)
DIFFERENTIAL ANALYSIS OF FLUID FLOW PROBLEMS
•Calculating both the velocity and pressure fields for a flow of known geometry
and known boundary conditions
A general three-dimensional
incompressible flow field with
constant properties requires
four equations to solve for
four unknowns.
44
Exact Solutions of the NSE
1. Set up the problem and geometry, identifying all relevant dimensions and parameters
2. List all appropriate assumptions, approximations, simplifications, and boundary conditions
3. Simplify the differential equations as much as possible
4. Integrate the equations
5. Apply BCs to solve for constants of integration
6. Verify results
• Boundary conditions are critical to exact, approximate, and computational solutions.▪ BC’s used in analytical solutions are
• No-slip boundary condition
• Interface boundary condition
Procedure for solving continuity and NSE
32
Exact Solutions of the Continuity
and Navier–Stokes Equations
Procedure for solving the
incompressible continuity and
Navier–Stokes equations.
Boundary Conditions
A piston moving at speed VP in a cylinder.
A thin film of oil is sheared between the
piston and the cylinder; a magnified view of
the oil film is shown. The no-slip boundary
condition requires that the velocity of fluid
adjacent to a wall equal that of the wall.
46
33
At an interface between two
fluids, the velocity of the two
fluids must be equal. In addition,
the shear stress parallel to the
interface must be the same in
both fluids.
Along a horizontal free surface of
water and air, the water and air
velocities must be equal and the
shear stresses must match.
However, since air << water, a
good approximation is that the
shear stress at the water surface is
negligibly small.
47
34
Boundary conditions along a plane of
symmetry are defined so as to ensure
that the flow field on one side of the
symmetry plane is a mirror image of
that on the other side, as shown here
for a horizontal symmetry plane.
Other boundary conditions arise
depending on the problem setup.
For example, we often need to define
inlet boundary conditions at a
boundary of a flow domain where fluid
enters the domain.
Likewise, we define outlet boundary
conditions at an outflow.
Symmetry boundary conditions are
useful along an axis or plane of
symmetry.
For unsteady flow problems we also
need to define initial conditions (at
the starting time, usually t = 0).
48
Summary
• Introduction
• Conservation of mass-The continuity equation
Derivation Using an Infinitesimal Control Volume
Continuity Equation in Cylindrical Coordinates
Special Cases of the Continuity Equation
• The Navier-Stokes equation
Derivation of the Navier–Stokes Equation for Incompressible, Isothermal Flow
Continuity and Navier–Stokes Equations in CartesianCoordinates
Continuity and Navier–Stokes Equations in CylindricalCoordinates
• Differential analysis of fluid flow problems
Exact Solutions of the Continuity and Navier– Stokes Equations
49
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
• Assumptions1. Plates are infinite in x and z
2. Flow is steady, /t = 0
3. Parallel flow, v = 0
4. Incompressible, Newtonian, laminar, constant properties
5. No pressure gradient
6. 1D, 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
ggjgg y ,ˆ
Fully Developed Couette Flow
• Step 3: Simplify3 6
Note: these numbers referto 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
0
z
w
y
v
x
u
0
x
u
)()(2
2
2
2
2
2
z
u
y
u
x
ug
x
p
z
uw
y
uv
x
uu
t
ux
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
ygy
p
p = p(y)0
z
pyg
dy
dp
y
p
p = p(y)
)()(2
2
2
2
2
2
z
v
y
v
x
vg
y
p
z
vw
y
vv
x
vu
t
vy
)()(2
2
2
2
2
2
z
w
y
w
x
wg
z
p
z
ww
y
wv
x
wu
t
wz
Fully Developed Couette Flow
• Step 4: Integrate
y-momentum
X-momentum
integrate integrate
integrate
gdy
dp 3)( Cgyyp
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 an arbitrary constant (recall only P appears in NSE).• Let p = p0 at y = 0 (C3 renamed p0)
1. Hydrostatic pressure2. Pressure acts independently of flowgypyp 0)(
Fully Developed Couette Flow
• Step 6: Verify solution by back-substituting into differential equations• Given the solution (u,v,w)=(Vy/h, 0, 0)
• Continuity is satisfied0 + 0 + 0 = 0
• X-momentum is satisfied
)()(2
2
2
2
2
2
z
u
y
u
x
ug
x
p
z
uw
y
uv
x
uu
t
ux
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 the shear stress acting on the fluid yx
(Newton’s third law).
h
V
y
u
x
vyxxy
)(u=V/h
Parallel Plates (Poiseuille Flow)• Given: A steady, fully developed, laminar flow of a Newtonian fluid in a
rectangular channel of two parallel plates where the width of the channel is much larger than the height, h, between the 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 Important Assumptions
• The width is very large compared to the height of the plate.
• No entrance or exit effects.
• Fully developed flow.
• Steady flow
• THEREFORE…• Velocity can only be dependent on vertical location in the flow (u)
• v = w = 0
• The pressure drop is constant and in the x-direction only.
.in length a is where,p
Constantp
xLLx
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 we don’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 direction
2ax
y
ay
ay
gvvvv
•
2pt
Navier-Stokes EquationsIn Vector Form for incompressible flow:
z
y
x
gz
w
y
w
x
w
zz
ww
y
wv
x
wu
t
w
gz
v
y
v
x
v
yz
vw
y
vv
x
vu
t
v
gz
u
y
u
x
u
xz
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
gz
w
y
w
x
w
zz
ww
y
wv
x
wu
t
w
gz
v
y
v
x
v
yz
vw
y
vv
x
vu
t
v
gz
u
y
u
x
u
xz
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
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 yp is a function of x only
Rewriting (4), we get
(5)
u is a function of y only and therefore LHS is a function of y only
p is a function of x only and is a constant and therefore 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 an ordinary derivative.
(6)
Therefore, (5) becomes
(7)
L
p
x
p
y
u
12
2
x
p
L
p
L
p
dy
ud
2
2
1CyL
p
dy
du
integrate
Integrating again, we get
(8)
where C1 and C2 are constants to be determined using the boundary conditions given below:
Substituting the boundary conditions in (8), we get
Fixed plate
Fixed plate
Fluid flow direction
2ax
y
ayu at0
ayu at0
ay
ay
21
2
2CyC
y
L
pu
01 C2
2
2
a
L
pC
and
Therefore, (8) reduces to 22
2ya
L
pu
Parabolic velocity profile
}no-slip boundarycondition
Fixed plate
Moving plate at velocity U
Fluid flow direction
2ax
y
Steady, incompressible flow of Newtonian fluid in an infinite channel with one plate moving at uniform velocity - fully developed plane Couette-Poiseuille flow
where C1 and C2 are constants to be determined using the boundary conditions given below:
ay
ay
21
2
2CyC
y
L
pu
ayu at0
ayUu at
Substituting the boundary conditions in (8), we get
and
Therefore, (8) reduces toa
UC
21
22
2
2
a
L
pUC
yaa
Uya
L
pu
22
22
(8)
}no-slip boundarycondition
Parabolic velocity profile is super imposed on a linear velocity profile
Fixed plate
Moving plate at velocity U
Fluid flow direction
2ax
y
Steady, incompressible flow of Newtonian fluid in an infinite channel with one plate moving at uniform velocity - fully developed plane Couette-Poiseuille flow
ay
ay
yaa
Uya
L
pu
22
22
If zero pressure gradient is maintained in the flow direction, then ∆p = 0.
Therefore, we get
yaa
Uu
2Linear velocity profile
Points to remember: - Pressure gradient gives parabolic profile- Moving wall gives linear profile
Starting from plane Couette-Poiseuille flow
Let us get back to the velocity profile of fully developed plane Couette-Poiseuille flow
yaa
Uya
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
yy
yyLU
pau
1
2
11
2
22
U
uu
a
y
a
y
LU
pa
U
u1
2
11
2 2
22
R. Shanthini
15 March 2012
-1
0
1
-0.5 0 0.5 1 1.5
1
0.25
0
-0.25
-1
yyLU
pau
1
2
11
2
22
2/2 aLU
p
a
yy
U
uu
Plot the non-dimensionalized velocity profile of the fully developed plane Couette-Poiseuille flow
R. Shanthini
15 March 2012-1
0
1
-0.25 0.75
0
-0.5
yyLU
pau
1
2
11
2
22
2/2 aLU
p
a
yy
U
uu
BACK FLOW
Plot the non-dimensionalized velocity profile of the fully developed plane Couette-Poiseuille flow
R. Shanthini
15 March 2012
-1
0
1
-1 0 1
0
-0.25
-0.5
-1.5
yyLU
pau
1
2
11
2
22
2/2 aLU
p
a
yy
U
uu
Plot the non-dimensionalized velocity profile of the fully developed plane Couette-Poiseuille flow
yaa
Uya
L
pu
22
22
Let us determine the volumetric flow rate through one unit width fluid film along z-direction, using the velocity profile
Fixed plate
Moving plate at velocity U
Fluid flow direction2ax
y
ay
ay
a
a
a
a
a
a
yay
a
Uyya
L
pdyya
a
Uya
L
pudyQ
223222
23222
22322232
22
33
22
33 a
aa
Uaa
L
paa
a
Uaa
L
pQ
UaL
paQ
3
2 3
Determine the volumetric flow rate of the fully developed plane Couette-Poiseuille flow
-2
-1
0
1
2
3
4
-4 -2 0 2 4
Plot the non-dimensionalized volumetric flow rate of the fully 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
Laminar Flow in pipes
.oflinearaispressure..
0
zeidz
dp
z
p
rvv
vv
zz
r
Assumptions:
Now look at the z-momentum equation.
2
2
2
2
2
11
)(
dz
vv
rr
vr
rrdz
dp
z
vv
v
r
v
r
vv
t
v
zzz
zz
zzr
z
Steady Laminar pipe flow.
The above “assumptions” can be obtained from the single assumption 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
rrdz
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 the system (similar to what an electrical engineer might do in testing a linear system).
dz
dp
dr
dvr
dr
d
r
z
11
dz
dpr
dr
dvr
dr
d z
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
2C
dz
dpr
dr
dvr z
dz
dpr
dr
dvr z
2
2
dz
dpr
dr
dvz
2
Divide by r
2
2
4C
dz
dprvz
at pipe wall r = a velocity = 0 2
2
40 C
dz
dpa
dz
dpaC
4
2
2
Laminar flow in pipesSubstituting in the velocity equation gives:
)4
(44
2222
ra
dz
dp
dz
dpa
dz
dprvz
Volume flow rate is obtained from:
tioncross
zdAvQsec
ar
rzdrrvQ
02
ar
rdrrardz
dp
Q0
22
2
)(
dz
dpaQ
8
4
dz
dpDQ
128
4
Laminar flow in pipesAverage velocity, vav= Q/area
Maximum velocity at r = 0:
dz
dpDD
dz
dpDv avez
324/
128
224
,
)16
()4
(22
max,
D
dz
dpa
dz
dpvz
vavergae = ½ vmax
Shear stress:
)(8)4
()4
2)((
,
D
vD
dz
dpa
dz
dp
dr
dv avez
ar
z
Laminar flow in pipes
Coefficient of friction: f = 4 Cf
eave
ave
ave
wf
RDv
v
v
C1616
2
1 22
Drag Coefficient:
eRf
64