Lecture 3
Final Version
Contents
• Control Volume and Control Surface
• Classification of Forces
• Reynolds Number
• Physical Meaning of Reynolds Number
• Streamlines
• Stream tube
• Laminar vs Turbulent Flow
• Other Non-Dimensional Numbers
• Conservation of Mass / Continuity
Equation
What Did We Do In Last Lecture?
Non-Newtonian Fluids
Newton’s Law of Viscosity
du
τ =µ
dy
Not Constant
Rheology
• The science that investigates deformation and
flow of matter and the properties and
molecular characteristics of fluids. …
• Listed various types of fluids
What Did We Do In Last Lecture?
Ideal Fluid vs. Real Fluid
Pressure
Pressure =
(Normal) Force Exerted
Area of Boundary
or
p=
• Variation of Pressure in a Fluid
• Hydrostatic Paradox
• Manometers
F
A
Note:
I have now put up
Four Example Sheets
on www.
Work through the questions!
Control Volume and Control Surface
•
Problem when dealing with flowing liquid is that one is
usually dealing with an endless stream of fluid.
•
Have to decide what part of stream shall constitute
element of system to be studied. What to do … ?
Example 1: Flow Over a Wing
Flow Speed V
Lift
CONTROL
VOLUME
(CV)
Drag (Resistance)
: CONTROL SURFACE
•
Alternatively body surface could be chosen as Control
Surface.
•
Usually it is convenient to assume that body is
stationary. In practice, however, body would usually
move through the fluid.
Continued ...
Example 1: Flow Past Ship
Upthrust
Free
Surface
Drag
V
CONTROL VOLUME (CV)
: CONTROL SURFACE
•
Again, alternatively body surface could be chosen as
Control Surface.
•
In this case 2 Fluids are involved - Water and Air
separated by a free surface.
Overall Goal:
Chose Control Surface and/or Control
Volume such that mathematical
description of problem becomes as
simple as possible!
Classification of Forces
•
We already mentioned several forces in fluid dynamics
and studied one (viscous force) in more detail ...
What are other forces that exist in a CV
and how can we classify them?
SURFACE FORCES
E.G.: Viscous Force
Pressure Force
Surface Tension
BODY FORCES
E.G.: Gravity
Electromagnetic
Force
Act ONLY on CS; usually
expressed in terms of stress
(= Force/Area)
Acts on ALL fluid elements
in CV
These forces can also be categorized as:
STATIC FORCES
E.G.: Gravity
Hydrostatic
Pressure
Surface Tension
Present
when
fluid
stationary - unaffected by
fluid motion.
DYNAMIC FORCES
E.G.: Viscous Force
Pressure
(Hydrodynamic
Component)
Absent in
Generated
Motion.
still fluid by
Fluid
REYNOLDS
NUMBER
Reynolds Number
One usually characterizes physical conditions present
in particular experiment/situation in terms of nondimensional numbers.
Example: Neutrally buoyant (i.e. density/weight such
that it neither sinks nor rises) sphere pulled through
water with velocity U
WATER
U
Sphere
Think: What might be the main forces
governing the dynamics?
•
•
•
Buoyancy force? … NO, of course not! We assumed
neutrally buoyant and, hence, excluded this force.
Viscous force? … YES, probably… No-slip on
surface… So, should make difference if pulled fast or
slow.
Inertial force? … YES, probably… When pulling
kinetic energy conveyed to fluid molecules. So, inertial
force associated with molecular motion will somehow
enter problem and compete with viscous forces.
Continued...
•So, expect that competition between Inertial Forces
and Viscous Forces governs problem.
‘competition’ is expressed by the ...
This
REYNOLDS NUMBER
The Reynolds number is a
DIMENSIONLESS QUANTITY
defined as
Re =
UL
ν
U
: Characteristic velocity of problem
(e.g. velocity of approaching fluid)
L
: Characteristic length scale of problem
(e.g. diameter or radius of sphere.
ν=
µ
: kinematic viscosity of fluid
ρ
Physical Meaning of Reynolds Number
•
The Reynolds number ...
Re =
UL
ν
… characterizes the RATIO of INERTIAL
FORCES and VISCOUS FORCES
Re =
Rate Of Change Of Fluid Momentum
Viscous Force
=
Inertia Force
Viscous Force
•
Note Re only gives you an order of magnitude estimate
for the force ratio (Re changes by factor 2 depending
on whether we use radius or diameter as characteristic
scale. Factor 2 usually not important).
•
However, a CRITICAL REYNOLDS NUMBER is
often stated to distinguish, for instance, between
Laminar and Turbulent flow in a particular flow
situation.If we want to calculate U or L from this value
then need to know if Re based on diameter or radius;
other wise we might end up with result out by factor 2 .
Example:
•
Ouestion (a): A sphere of diameter D=10 mm
moves with velocity V=2.3 m/s through water with
a kinematic viscosity of 1.007x10-6 m2/s. Find the
Reynolds number for the motion of the sphere.
Solution:
m
⋅ 0.01 m
V D
s
Re =
=
= 22,840
2
ν
m
1.007 × 10-6
s
2.3
Example:
•
Ouestion (a): A sphere of diameter D=10 mm
moves with velocity V=2.3 m/s through water with
a kinematic viscosity of 1.007x10-6 m2/s. Find the
Reynolds number for the motion of the sphere.
Solution:
m
⋅ 0.01 m
V D
s
Re =
=
= 22,840
2
ν
m
1.007 × 10-6
s
2.3
•
Ouestion (b): The same sphere now moves with
the same velocity through SAE50 oil. This has a
dynamic viscosity of 0.86 kg/ms and a density of
902 kg/m3. Find the Reynolds number ...
Solution:
Re =
V D
ν
=
V D
µ
ρ
=
ρV D
µ
902
=
kg
m
2
.
3
⋅
⋅ 0.01 m
3
s
m
= 24.12
kg
0.86
ms
Continued ...
•
Ouestion (c): A dust particle with a diameter of
D=0.01 mm sinks in air with a velocity V=0.001
m/s. The kinematic viscosity of air is 15.13x10-6
m2/s. Find the Reynolds number for the motion of
the particle.
Solution:
m
0.001 ⋅1.0 × 10 −5 m
V D
−4
s
Re =
=
=
6
.
6
×
10
2
ν
-6 m
15.13 × 10
s
Notes ...
• Q:
But we used diameter in the
calculations. What if we had used the
radius instead? Then all Reynolds
numbers would have halved!
• A:
Yes,… But it is important to
understand that Reynolds number only
gives you an order-of magnitude estimate
of ratio of inertial forces and viscous
forces! Had we used radius then high
values in our examples would have still
been high while low ones would have still
been low!
However ...
• Assume someone tells you that:
“The critical Reynolds number for laminarturbulent transition in my experiment was
Re=2350.”
•
If you want reproduce the experiment under
identical experimental conditions to verify this
result then you will, of course, need to know
exactly how the other person had defined the
Reynolds number! I.e. you need to know whether
the diameter or the radius was used to calculate
the value of 2350.
Streamlines
•
Let’s continue with studying fluids in motion. For
what lies ahead it is useful to introduce the concept of
Streamlines.
•
Consider flow past a circular cylinder. Assume that
flow has very low Reynolds number to ensure we have
laminar flow.
•
Imagine that you have small particles (e.g. Aluminium
dust) suspended in and transported with fluid.
•
Now take a short-duration photograph… What you
see on photo will look something like..
A
•
B
Note, in Points A and B flow comes to rest. These two
points are called STAGNATION POINTS.
Continued ...
•
If there are sufficient particles then the “dashes”
would join up to form STREAMLINES as
illustrated in the photograph below ...
A
B
•
Note, the surface of the cylinder is a special case of a
streamline.
•
Stagnation Points identified by red dots again and
labeled A and B.
Streamlines
Qualitative Definition of Streamlines
A streamline is an (imaginary)
continuous line drawn through the fluid
so that it has the direction of the
velocity vector at every point.
•
•
•
•
Note, definition implies that there can be no flow
across a streamline.
Definition also implies: Two streamlines cannot
intersect; except at position of zero velocity. (If two
streamlines would intersect where velocity not zero
then one would have the meaningless situation of
velocity with two different directions.
Patterns of streamlines are useful (particularly in 2D
flow) in providing a pictorial representation of a flow.
Compare/Read up on STREAKLINES and
PATHLINES.
FORMAL DEFINITION OF A
STREAMLINE
• In mathematical terms ...
• In 2D:
dy v
=
dx u
Flow Speed:
dx dy
=
u
v
or
U = u 2 + v2
dx dy dz
=
=
u
v
w
• In 3D:
Flow Speed:
U = u 2 + v 2 + w2
• Note that both the flow velocity,
U , and the flow speed
U , can change along a streamline.
Streamtube
A streamtube is formed by a closed
collection of individual streamlines as
illustrated in the sketch below ...
•
There is now flow across streamtube walls.
•
Streamtubes often used as control volumes. The
advantage that results from this choice being
that one does not have to consider flow across
walls of a streamtube but only flow at entry and
exit.
Laminar vs. Turbulent Flow
Laminar Flow
•
•
•
•
•
Fluid particles move along smooth paths in
laminas (layers).
One layer glides smoothly over an adjacent layer.
Tendencies to lateral or swirling motion are
strongly damped by viscosity.
Laminar flow governed by Newton’s law of
viscosity (or extension of it to three dimensional
[3D] flow.
Laminar flow is not stable in situations involving
combinations of low viscosity, high velocity, or
large flow passages (i.e. flow becomes unstable as
Re number becomes high enough.
Continued ...
Turbulent Flow
•
•
•
•
•
•
Most prevalent in engineering practice.
Fluid particles (or small molar masses) move very
irregular, swirling paths.
Exchange of momentum from one portion of fluid
to another one.
Turbulent swirls continuously range in size from
very small (say a few thousand fluid molecules) to
very large (a large swirl in a river or in an
atmospheric gust)
Turbulence sets up greater shear stresses and
causes more irreversible losses (Dissipation).
Calculation of shear stresses introduced by
turbulent flow is an extremely challenging
problem.
Generation of Turbulence by a Grid
Grid
Flow
Laminar Flow
Turbulent Flow
When Do We Encounter Low and High
Reynolds Numbers?
Low
Re Number
•
•
•
Laminar flow
Small organisms
swimming (e.g.
amoebae)
Small (micron-sized)
particles moving
through fluid (e.g. dust
particle sinking in air.
High
Re Number
•
•
•
•
Turbulent flow
Aircraft
Cars
… generally in most
engineering
applications
Re >> 1
Re << 1
Re ~ 107 − 108
for aircraft
•
•
Viscous forces >> Rate
of change of
momentum (i.e. inertia
force.
Also: Viscous force and
pressure comparable
(but you cannot
see/understand this
from our
considerations so far).
•
•
Viscous force << Rate
of change of
momentum (i.e. inertia
force).
Also: Viscous force <<
Pressure force and
inertia force (but you
cannot see/understand
this from our
considerations so far).
Other Non-Dimensional Numbers
•
There are many other non-dimensional numbers which
often represent force ratios and which are relavant in
various applications.
Examples:
Froude Number = Inertia/Gravitational Force
Free surface flows (ship on ocean). Water flow induces
inertia forces but gravity is important to the waves.
Weber Number = Inertia/Surface Tension
Bubble rising in viscous fluid displaces fluid and
deforms. Fluid displacement induces inertia effects and
deformation depends on surface tension
Drag Coefficient= Drag Force/Dynamic Force
Bond Number = Gravitational Force/Surface Tension
Grashof Number = Buoyancy Force/Viscous Force
•
But, not all non-dimensional numbers represent force
ratios!
Mach Number = Flow Speed/Speed of Sound
Nusselt Number =
Convection Heat Transfer/Conduction Heat Transfer
Conservation of Mass
•
For any fluid flow mass must be conserved:
Mass of
Fluid
Entering
Control Volume
Mass of Fluid
Entering per
Unit Time
•
Mass of
Fluid
Leaving
=
Mass of Fluid
Leaving per
Unit Time
+
Increase of
Mass of Fluid
in the CV per
Unit Time
For steady flow, i.e. when mass in CV remains
constant ...
Mass of Fluid
Entering per
Unit Time
=
Mass of Fluid
Leaving per
Unit Time
Continued...
•
Apply this principle to a stream tube ...
Mass of Fluid
Entering per Unit
Time at Section (1)
•
=
Mass of Fluid
Leaving per Unit
Time at Section (2)
How much does actually go in and out?
Mass in per Unit
Time at (1)
=
ρ1 δA1 u1
Mass out per Unit
Time at (2)
=
ρ 2 δA2 u2
ρ1 δA1 u1
=
ρ 2 δA2 u2
=
Constant
Equation of Continuity
for flow of compressible fluid through streamtube, u1
and u2 being velocities measured at right angles to
cross-sectional area at entry and exit.
Continued...
•
For flow of real fluid through pipe or other conduit,
velocity will vary over both cross-sectional area at
entry and exit. In this case one has to use mean
velocity at each cross-sectional area and Continuity
Equation becomes:
•
ρ1 A1 u1 = ρ 2 A2 u2 = m
•
If fluid is incompressible such that density at entry
and exit the same then ...
A1 u1 = A2 u2 = Q
!!!
The Continuity Equation is one of the major
tools of fluid mechanics. It provides a means
of calculating flow velocities at different
points in a system.
Example 1:
•
Apply continuity equation to determine relation
between flows into and out of junction sketched below.
A2 v2 Q2
A1
v1
Q1
A3 v3
Q3
Solution:
• Continuity requires
Total inflow to junction = Total outflow of junction
ρ1 Q1 = ρ 2 Q2 + ρ3 Q3
•
For incompressible fluid...
ρ1 = ρ 2 = ρ3
Q1 = Q2 + Q3
A1 v1 = A2 v2 + A3 v3
Continued...
A2 v2 Q2
A1
v1
Q1
•
A3 v3
Q3
In general, if we consider flow towards junction
as positive and flow away from junction as
negative then for steady flow ….
∑ρ Q = 0
Example 2:
•
Water flows from A to D and E through a series
pipeline. Given the pipe diameters, velocities and flow
rates below find the missing data...
Q1 = ?
v1 = ?
d1 = 50 mm
Q2 = ?
v2 = 2 m/s
d 2 = 75 mm
Q3 = 2Q4 = ?
v3 = 1.5 m/s
d3 = ?
1
Q4 = Q3 = ?
2
v4 = ?
Solution:
?
d 4 = 30 mm
Continued ...
•
Water flows from A to D and E through a series
pipeline. Given the pipe diameters, velocities and flow
rates below find the missing data...
Q1 = ?
v1 = ?
d1 = 50 mm
Q3 = 2Q4 = ?
Q2 = ?
v3 = 1.5 m/s
d3 = ?
v2 = 2 m/s
d 2 = 75 mm
1
Q4 = Q3 = ?
2
v4 = ?
d 4 = 30 mm
Solution:
• Calculate Q2 ...
2
3
⎛ 0.075 m ⎞
−3 m
Q2 = v2 A2 = 2 m/s π ⎜
⎟ = 8.839 × 10
2 ⎠
s
⎝
•
Mass conservation requires ...
Q1 = Q2
Continued ...
•
Water flows from A to D and E through a series
pipeline. Given the pipe diameters, velocities and flow
rates below find the missing data...
Q1 = ?
v1 = ?
d1 = 50 mm
Q2 = ?
v2 = 2 m/s
d 2 = 75 mm
Q3 = 2Q4 = ?
v3 = 1.5 m/s
d3 = ?
1
Q4 = Q3 = ?
2
v4 = ?
d 4 = 30 mm
Solution:
•
Then ...
Q1 = v1 A1
Q1 8.839 × 10−3 m3/s
m
= 4.5
v1 =
=
2
A1
s
⎛ 0.05 m ⎞
π⎜
⎟
⎝ 2 ⎠
Continued ...
Q2 = ?
Q1 = ?
v1 = ?
d1 = 50 mm
v2 = 2 m/s
d 2 = 75 mm
Q3 = 2Q4 = ?
v3 = 1.5 m/s
d3 = ?
1
Q4 = Q3 = ?
2
v4 = ?
•
Now ...
Q3 =
d 4 = 30 mm
1
3
Q2 = Q3 + Q4 = Q3 + Q3 = Q3
2
2
2
2
Q2 = 8.839 × 10−3 m3/s = 5.893 × 10−3 m3/s
3
3
Continued ...
Q2 = ?
Q1 = ?
v1 = ?
d1 = 50 mm
v2 = 2 m/s
d 2 = 75 mm
Q3 = 2Q4 = ?
v3 = 1.5 m/s
d3 = ?
1
Q4 = Q3 = ?
2
v4 = ?
•
Now ...
d 4 = 30 mm
1
3
Q2 = Q3 + Q4 = Q3 + Q3 = Q3
2
2
Q3 =
2
2
Q2 = 8.839 × 10−3 m3/s = 5.893 × 10−3 m3/s
3
3
Q4 =
1
1
Q3 = 5.893 × 10−3 m3/s = 2.946 × 10−3 m3/s
2
2
Continued ...
Q1 = ?
v1 = ?
d1 = 50 mm
Q3 = 2Q4 = ?
Q2 = ?
v3 = 1.5 m/s
d3 = ?
v2 = 2 m/s
d 2 = 75 mm
1
Q4 = Q3 = ?
2
v4 = ?
•
d 4 = 30 mm
Since we now know Q3 we get...
⎛d ⎞
Q3 = v3 A3 = v3 π ⎜ 3 ⎟
⎝ 2⎠
2
d3 =
4 Q3
π v3
4 ⋅ 5.893 × 10−3 m3/s
d3 =
= 0.707 m
π 1.5 m/s
Continued ...
Q1 = ?
v1 = ?
d1 = 50 mm
Q3 = 2Q4 = ?
Q2 = ?
v3 = 1.5 m/s
d3 = ?
v2 = 2 m/s
d 2 = 75 mm
1
Q4 = Q3 = ?
2
v4 = ?
•
d 4 = 30 mm
And finally...
Q
v4 = 4
A4
Q4 = v4 A4
v4 =
2.946 × 10−3 m3/s
0.03 m ⎞
π ⎛⎜
⎟
⎝ 2 ⎠
2
= 4.17
m
s