•
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

Newton’s Law of Viscosity
=
du dy
Not Constant
•
The science that investigates deformation and flow of matter and the properties and molecular characteristics of fluids. …
•
Listed various types of fluids

Pressure
=
(Normal) Force Exerted
Area of Boundary or p
=
F
A
•
Variation of Pressure in a Fluid
•
Hydrostatic Paradox
•
Manometers

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
Drag
Free
Surface 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
Act ONLY on CS; usually expressed in terms of stress
(= Force/Area)
BODY FORCES
E.G.: Gravity
Electromagnetic
Force
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 still fluid -
Generated by Fluid
Motion.

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. This
‘competition’ is expressed by the ...
The Reynolds number is a
U
DIMENSIONLESS QUANTITY defined as
=
U
ν
L
: 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
ρ
: kinematic viscosity of fluid

Physical Meaning of Reynolds Number
• The Reynolds number ...
Re
=
U
ν
L
… characterizes the RATIO of INERTIAL
FORCES and VISCOUS FORCES
Re
=
Rate Of Change Of Fluid
Viscous Force
Momentum
=
Inertia
Viscous
Force
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 which has a kinematic viscosity of 1.007x10 -6 m 2 /s.
Find the Reynolds number for the motion of the sphere.
Solution:
Re
=
V
ν
D
=
2 .
3 m s
⋅
0 .
01 m m
2
1.007
×
10
6 s
=
22 , 840
• 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/m 3 . Find the Reynolds number ...
Solution:
Re
=
V
ν
D
=
V
µ
D
ρ
=
ρ
V
µ
D
=
902 kg m
3
⋅
2 .
3 m s kg
⋅
0 .
01 m
0.86
m s
=
24 .
12

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 m 2 /s. Find the Reynolds number for the motion of the particle.
Solution:
Re
=
V
ν
D
=
0 .
001 m s
⋅
1 .
0
×
10
−
5 m
2
15.13
×
10
6 s m
=
6 .
6
×
10
−
4

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’ 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:
• In 3D: dy dx
= v u
Flow Speed:
Flow Speed: or u
= dx
= dy u v u
2 + v
2 dx u
= dy v
= dz w u
= u
2 + v
2 + w
2
• Note that both the flow velocity, u , can change along a streamline.
u , and the flow speed

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
• 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).
Re
>>
1
Re ~ 10
7 −
10
8 for aircraft
• 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
Leaving
Mass of Fluid
Entering per
Unit Time
=
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)
= ρ
1
δ
A
1 u
1
Mass out per Unit
Time at (1)
= ρ
2
δ
A
2 u
2
ρ
1
δ
A
1 u
1
= ρ
2
δ
A
2 u
2
= Constant
Equation of Continuity for flow of compressible fluid through streamtube, u
1 and u
2 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:
ρ
1
A
1 u
1
= ρ
2
A
2 u
2
=
• m
• If fluid is incompressible such that density at entry and exit the same then ...
A
1 u
1
=
A
2 u
2
=
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.
A
2 v
2
Q
2
A
1 v
1
Q
1
A
3 v
3
Q
3
Solution:
• Continuity requires
Total inflow to junction = Total outflow of junction
ρ
1
Q
1
= ρ
2
Q
2
+ ρ
3
Q
3
• For incompressible fluid...
ρ
1
= ρ
2
= ρ
3
Q
1
=
Q
2
+
Q
3
A
1 v
1
=
A
2 v
2
+
A
3 v
3

Continued...
A
2 v
2
Q
2
A
1 v
1
Q
1
A
3 v
3
Q
3
• 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...
Q
3
=
2 Q
4
=
?
Q
1
=
?
v
1 d
1
=
=
?
50 mm
Q
2
=
?
v
2 d
2
=
=
2 m/s
75 mm v
3 d
3
=
1 .
5
=
?
m/s v
4
=
?
Q
4
=
1
2
Q
3
=
?
d
4
=
30 mm
Solution:
?

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...
Q
3
=
2 Q
4
=
?
Q
1
=
?
v
1 d
1
=
=
?
50 mm
Q
2
=
?
v
2 d
2
=
=
2 m/s
75 mm v
3 d
3
=
1 .
5
=
?
m/s
Q
4
=
1
2
Q
3
=
?
v
4
=
?
d
4
=
30 mm
Solution:
• Calculate Q
2
...
Q
2
= v
2
A
2
=
2 m/s
π
⎝
⎜
⎛
0 .
075 m
2
⎠
⎟
⎞
2
=
8 .
839
×
10
−
3 m
3 s
• Mass conservation requires ...
Q
1
=
Q
2

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...
Q
3
=
2 Q
4
=
?
Q
1
=
?
v
1 d
1
=
=
?
50 mm
Q
2
=
?
v
2 d
2
=
=
2 m/s
75 mm v
3 d
3
=
1 .
5
=
?
m/s v
4
=
?
Q
4
=
1
2
Q
3
=
?
d
4
=
30 mm
Solution:
• Then ...
Q
1
= v
1
A
1 v
1
=
Q
1
A
1
=
8 .
839
×
10
−
3 m
3
/s
π
0 .
05 m
2
2
=
4 .
5 m s

Continued ...
Q
1
=
?
v
1 d
1
=
=
?
50 mm
Q
2
=
?
v
2 d
2
=
=
2 m/s
75 mm
Q
3
=
2 Q
4
=
?
v
3 d
3
=
1 .
5
=
?
m/s
• Now ...
v
4
=
?
Q
4
=
1
2
Q
3
=
?
d
4
=
30 mm
Q
2
=
Q
3
+
Q
4
=
Q
3
+
1
2
Q
3
=
3
2
Q
3
Q
3
=
2
3
Q
2
=
2
3
8 .
839
×
10
−
3 m
3
/s
=
5 .
893
×
10
−
3 m
3
/s

Continued ...
Q
1
=
?
v
1 d
1
=
=
?
50 mm
Q
2
=
?
v
2 d
2
=
=
2 m/s
75 mm
Q
3
=
2 Q
4
=
?
v
3 d
3
=
1 .
5
=
?
m/s
• Now ...
v
4
=
?
Q
4
=
1
2
Q
3
=
?
d
4
=
30 mm
Q
2
=
Q
3
+
Q
4
=
Q
3
+
1
2
Q
3
=
3
2
Q
3
Q
3
=
2
3
Q
2
=
2
3
8 .
839
×
10
−
3 m
3
/s
=
5 .
893
×
10
−
3 m
3
/s
Q
4
=
1
2
Q
3
=
1
2
5 .
893
×
10
−
3 m
3
/s
=
2 .
946
×
10
−
3 m
3
/s

Continued ...
Q
1
=
?
v
1 d
1
=
=
?
50 mm
Q
2
=
?
v
2 d
2
=
=
2 m/s
75 mm
Q
3
=
2 Q
4
=
?
v
3 d
3
=
1 .
5
=
?
m/s v
4
=
?
Q
4
=
1
2
Q
3
=
?
d
4
=
30 mm
• Since we now know Q
3 we get...
Q
3
= v
3
A
3
= v
3
π
⎝
⎜
⎛ d
3
2
⎠
⎟
⎞
2 d
3
=
4
π
Q
3 v
3 d
3
=
4
⋅
5 .
893
×
10
−
3
π
1 .
5 m/s m
3
/s
=
0 .
707 m

Continued ...
Q
1
=
?
v
1 d
1
=
=
?
50 mm
Q
2
=
?
v
2 d
2
=
=
2 m/s
75 mm
Q
3
=
2 Q
4
=
?
v
3 d
3
=
1 .
5
=
?
m/s v
4
=
?
Q
4
=
1
2
Q
3
=
?
d
4
=
30 mm
• And finally...
Q
4
= v
4
A
4 v
4
=
Q
4
A
4 v
4
=
2 .
946
×
10
−
3 m
3
/s
π
0 .
03 m
2
2
=
4 .
17 m s