thenotes to deal with thermodynamics and solution manual.... good document to refer and read for exam .. ... .... .... ............................ .............................

1. Lecture Aims
Conservation of mass
Conservation of
energy
Energy per unit mass
Rate of heat transfer
Power
Energy Equation
Head Form of the
Energy Equation
Power
A Simple Example
Bernoulli and Energy
Equations
A Second Simple
Example
1.1
Lecture 1
The Energy Equation
Derivation and a simple example of its use
CHE 2161/MEC2404 Mechanics of Fluids
Ravi Jagadeeshan
Department of Chemical Engineering
Monash University

2. Lecture Aims
Conservation of mass
Conservation of
energy
Energy per unit mass
Rate of heat transfer
Power
Energy Equation
Head Form of the
Energy Equation
Power
A Simple Example
Bernoulli and Energy
Equations
A Second Simple
Example
1.2
Aims of todays lecture
1 Derive the energy equation using control volume analysis
2 Simplify to the case of a one-dimensional inlet and outlet
3 Derive the head form of the energy equation
4 Compare Bernoulli and the energy equations
5 Solve some simple examples

3. Lecture Aims
Conservation of mass
Conservation of
energy
Energy per unit mass
Rate of heat transfer
Power
Energy Equation
Head Form of the
Energy Equation
Power
A Simple Example
Bernoulli and Energy
Equations
A Second Simple
Example
1.4
Conservation of mass
Mass flow rate
ρ1A1V1 = ρ2A2V2
Control Volume Application of the
First Law of Thermodynamics

4. Lecture Aims
Conservation of mass
Conservation of
energy
Energy per unit mass
Rate of heat transfer
Power
Energy Equation
Head Form of the
Energy Equation
Power
A Simple Example
Bernoulli and Energy
Equations
A Second Simple
Example
1.5
Conservation of energy
The rate of change of energy
within the control volume
=
The net rate of flow of energy into the volume
through the control surface
+
net time rate of energy addition
by heat transfer into the volume
−
net time rate of energy addition
by work out of the volume

5. Lecture Aims
Conservation of mass
Conservation of
energy
Energy per unit mass
Rate of heat transfer
Power
Energy Equation
Head Form of the
Energy Equation
Power
A Simple Example
Bernoulli and Energy
Equations
A Second Simple
Example
1.6
Conservation of energy
e = einternal + ekinetic + epotential
d
dt
Z
CV
e ρ dV
= −
Z
CS
e ρ (V · n) dA
+
dQ
dt
−
dW
dt
Q is positive when heat
is added to the system
W is positive when work
is done by the system

6. Lecture Aims
Conservation of mass
Conservation of
energy
Energy per unit mass
Rate of heat transfer
Power
Energy Equation
Head Form of the
Energy Equation
Power
A Simple Example
Bernoulli and Energy
Equations
A Second Simple
Example
1.7
Energy per unit mass
e = u +
1
2
V2
+ g z
u is the internal energy
(energy at the molecular scale – function of temperature)
1
2
V2
is the kinetic energy
g z is the potential energy

7. Lecture Aims
Conservation of mass
Conservation of
energy
Energy per unit mass
Rate of heat transfer
Power
Energy Equation
Head Form of the
Energy Equation
Power
A Simple Example
Bernoulli and Energy
Equations
A Second Simple
Example
1.8
Rate of heat transfer
dQ
dt
Represents all the ways in which energy is
exchanged between the control volume
contents and surroundings because of a
temperature difference
Mechanisms include, radiation, convection
and conduction
If the process is adiabatic, the heat transfer
rate is zero

8. Lecture Aims
Conservation of mass
Conservation of
energy
Energy per unit mass
Rate of heat transfer
Power
Energy Equation
Head Form of the
Energy Equation
Power
A Simple Example
Bernoulli and Energy
Equations
A Second Simple
Example
1.9
Rate of transfer of work
dW
dt
The rate of transfer of work is also
called the power
Work can be transferred across the
control surface by a moving shaft -
examples include turbines, fans, pumps
Ẇs
Work can also occur when a force
associated with a fluid normal stress
acts over a distance
Ẇp
pressure work (or
flow work)
Rate of work is equal to
force times displacement
per unit time Ẇp =
Z
CS
p (V · n) dA

9. Lecture Aims
Conservation of mass
Conservation of
energy
Energy per unit mass
Rate of heat transfer
Power
Energy Equation
Head Form of the
Energy Equation
Power
A Simple Example
Bernoulli and Energy
Equations
A Second Simple
Example
1.10
Steady State Energy Equation
d
dt
Z
CV
e ρ dV
= −
Z
CS
e ρ (V · n) dA
+
dQ
dt
−
dW
dt
0 = −
Z
CS
(u +
1
2
V2
+ g z) ρ (V · n) dA
+
dQ
dt
− Ẇs −
Z
CS
p (V · n) dA
Z
CS
(u +
1
2
V2
+ g z +
p
ρ
) ρ (V · n) dA =
dQ
dt
− Ẇs

10. Lecture Aims
Conservation of mass
Conservation of
energy
Energy per unit mass
Rate of heat transfer
Power
Energy Equation
Head Form of the
Energy Equation
Power
A Simple Example
Bernoulli and Energy
Equations
A Second Simple
Example
1.11
Steady State Energy Equation in One-Dimension
If the control volume has a one-dimensional inlet and outlet
Z
CS
(u +
1
2
V2
+ g z +
p
ρ
) ρ (V · n) dA
= (u2 +
1
2
V2
2 + g z2 +
p2
ρ2
) ṁ
−(u1 +
1
2
V2
1 + g z1 +
p1
ρ1
) ṁ
=
dQ
dt
− Ẇs

11. Lecture Aims
Conservation of mass
Conservation of
energy
Energy per unit mass
Rate of heat transfer
Power
Energy Equation
Head Form of the
Energy Equation
Power
A Simple Example
Bernoulli and Energy
Equations
A Second Simple
Example
1.12
Steady One-Dimensional Flow with Single Inlet and Outlet
q =
1
ṁ
dQ
dt
Energy in per unit mass
w =
1
ṁ
dẆs
dt
Energy out per unit mass
u1 +
1
2
V2
1 + g z1 +
p1
ρ1
!
+ q
= u2 +
1
2
V2
2 + g z2 +
p2
ρ2
!
+ w
First Law of Thermodynamics for a steady
one-dimensional flow with a single inlet and outlet
Each term has a unit of energy per unit mass

12. Lecture Aims
Conservation of mass
Conservation of
energy
Energy per unit mass
Rate of heat transfer
Power
Energy Equation
Head Form of the
Energy Equation
Power
A Simple Example
Bernoulli and Energy
Equations
A Second Simple
Example
1.13
Head Form of the Energy Equation
Dividing through out by g leads to the head form of
the energy equation


u1
g
+
1
2
V2
1
g
+ z1 +
p1
gρ1

 +
q
g
=


u2
g
+
1
2
V2
2
g
+ z2 +
p2
gρ2

 +
w
g
Each term has units of length (useful for pipe flow
calculations)

13. Lecture Aims
Conservation of mass
Conservation of
energy
Energy per unit mass
Rate of heat transfer
Power
Energy Equation
Head Form of the
Energy Equation
Power
A Simple Example
Bernoulli and Energy
Equations
A Second Simple
Example
1.14
Notational Formalities
For pipe flows, we assume the fluid is
incompressible (ρ1 = ρ2)
Introduce notation
hq =
q
g
hw =
w
g
Head form of the heat added
Head form of shaft work done
hw = hturbine − hpump

14. Lecture Aims
Conservation of mass
Conservation of
energy
Energy per unit mass
Rate of heat transfer
Power
Energy Equation
Head Form of the
Energy Equation
Power
A Simple Example
Bernoulli and Energy
Equations
A Second Simple
Example
1.15
Rearranging the Head Form of the Energy Equation
u1
g
+
1
2
V2
1
g
+ z1 +
p1
gρ1
+ hq =
u2
g
+
1
2
V2
2
g
+ z2 +
p2
gρ2
+ hw
1
2
V2
1
g
+ z1 +
p1
gρ1
!
=
1
2
V2
2
g
+ z2 +
p2
gρ2
!
+hturbine − hpump +
u2 − u1
g
− hq
#
1
2
V2
1
g
+ z1 +
p1
gρ1
!
+ hpump =
1
2
V2
2
g
+ z2 +
p2
gρ2
!
+hturbine + hfriction
1
2
V2
1
g
+ z1 +
p1
gρ1
+ hgain =
1
2
V2
2
g
+ z2 +
p2
gρ2
+ hloss

15. Lecture Aims
Conservation of mass
Conservation of
energy
Energy per unit mass
Rate of heat transfer
Power
Energy Equation
Head Form of the
Energy Equation
Power
A Simple Example
Bernoulli and Energy
Equations
A Second Simple
Example
1.16
Power
Power is energy per unit time
P = wṁ
=


w
g

 ρ Q g
= hw ρ Q g

16. Lecture Aims
Conservation of mass
Conservation of
energy
Energy per unit mass
Rate of heat transfer
Power
Energy Equation
Head Form of the
Energy Equation
Power
A Simple Example
Bernoulli and Energy
Equations
A Second Simple
Example
1.19
Relation between the Bernoulli and Energy Equations
Steady flow energy equation
u1 +
1
2
V2
1 + g z1 +
p1
ρ1
+ q =
u2 +
1
2
V2
2 + g z2 +
p2
ρ2
+ w
1
2
V2
1 + g z1 +
p1
ρ1
=
1
2
V2
2 + g z2 +
p2
ρ2
+ (u2 − u1 − q) + w
Bernoulli equation
1
2
V2
1 + g z1 +
p1
ρ1
=
1
2
V2
2 + g z2 +
p2
ρ2

17. Lecture Aims
Conservation of mass
Conservation of
energy
Energy per unit mass
Rate of heat transfer
Power
Energy Equation
Head Form of the
Energy Equation
Power
A Simple Example
Bernoulli and Energy
Equations
A Second Simple
Example
1.20
Relation between the Bernoulli and Energy Equations
List of assumptions in Bernoulli’s equation
1 Steady Flow
2 Incompressible Flow
3 Frictionless Flow
4 Flow along a single streamline
5 No shaft work between 1 and 2
6 No heat transfer between 1 and 2

18. Lecture Aims
Example of a
Hydroelectric Power
Plant
Viscous Flow in Pipes
and Ducts
General Character of
Pipe Flow
Entrance Region and
Fully-Developed Flow
Pressure and Shear
Stress
Flow in a Circular Pipe
Effect of Rough Walls
The Moody Chart
Example of a Pipe
Flow Calculation
2.1
Lecture 2
The Energy Equation
Viscous flow in pipes and ducts
CHE 2161/MEC2404 Mechanics of Fluids
Ravi Jagadeeshan
Department of Chemical Engineering
Monash University

19. Lecture Aims
Example of a
Hydroelectric Power
Plant
Viscous Flow in Pipes
and Ducts
General Character of
Pipe Flow
Entrance Region and
Fully-Developed Flow
Pressure and Shear
Stress
Flow in a Circular Pipe
Effect of Rough Walls
The Moody Chart
Example of a Pipe
Flow Calculation
2.2
Aims of todays lecture
1 Discuss the different flow regimes in pipe flow
2 Compare velocity profiles in fully-developed flow
3 Control volume analysis of flow in a pipe
4 Introduce the Darcy friction factor
5 Calculation of head loss in pipe flow
6 Discuss the role of rough walls and the Moody chart
7 Perform a simple pipe flow calculation

20. Lecture Aims
Example of a
Hydroelectric Power
Plant
Viscous Flow in Pipes
and Ducts
General Character of
Pipe Flow
Entrance Region and
Fully-Developed Flow
Pressure and Shear
Stress
Flow in a Circular Pipe
Effect of Rough Walls
The Moody Chart
Example of a Pipe
Flow Calculation
2.6
Viscous Flow in Pipes and Ducts
Numerous examples where it is important for daily
operations
Combination of analysis, experimental data and dimensional analysis
is needed for real world problems where viscous effects are important

21. Lecture Aims
Example of a
Hydroelectric Power
Plant
Viscous Flow in Pipes
and Ducts
General Character of
Pipe Flow
Entrance Region and
Fully-Developed Flow
Pressure and Shear
Stress
Flow in a Circular Pipe
Effect of Rough Walls
The Moody Chart
Example of a Pipe
Flow Calculation
2.7
General Character of Pipe Flow
• Flow in a pipe may be laminar, transitional or turbulent
• The Reynolds number
Re =
ρDV
µ
which is a ratio of inertial to viscous forces, characterizes
the different flow regimes

22. Lecture Aims
Example of a
Hydroelectric Power
Plant
Viscous Flow in Pipes
and Ducts
General Character of
Pipe Flow
Entrance Region and
Fully-Developed Flow
Pressure and Shear
Stress
Flow in a Circular Pipe
Effect of Rough Walls
The Moody Chart
Example of a Pipe
Flow Calculation
2.8
Entrance Region and Fully-Developed Flow
• Boundary layer develops where viscous effects are
important
• Viscous effects are unimportant in the inviscid core
• Initial velocity profile changes with distance until the end of
the entrance length `e
• Entrance length depends on the flow being laminar or
turbulent
`e
D
= 0.06Re (laminar)
`e
D
= 4.4Re
1
6 (turbulent)

23. Lecture Aims
Example of a
Hydroelectric Power
Plant
Viscous Flow in Pipes
and Ducts
General Character of
Pipe Flow
Entrance Region and
Fully-Developed Flow
Pressure and Shear
Stress
Flow in a Circular Pipe
Effect of Rough Walls
The Moody Chart
Example of a Pipe
Flow Calculation
2.9
Entrance Region and Fully-Developed Flow
Fully developed flow Turbulent profile
• Velocity profile and shear stress on the wall are constant
beyond the entrance length
• The flow is called fully-developed in this region
• The shape of the velocity profile depends on the flow
being laminar or turbulent

24. Lecture Aims
Example of a
Hydroelectric Power
Plant
Viscous Flow in Pipes
and Ducts
General Character of
Pipe Flow
Entrance Region and
Fully-Developed Flow
Pressure and Shear
Stress
Flow in a Circular Pipe
Effect of Rough Walls
The Moody Chart
Example of a Pipe
Flow Calculation
2.10
Pressure and Shear Stress
If there are no viscous forces, pressure is constant except for
hydrostatic variation
• Fluid accelerates or
decelerates in the
entrance
region–there is a
balance between
inertia, viscous and
pressure forces
• The magnitude of the
pressure gradient is
larger in the entrance
section than in the
fully-developed
region
• The pressure drop
4p = p2 − p1 causes
the fluid to flow in the
fully-developed
region
• The viscous force is just balanced by the pressure force, and the fluid
flows without acceleration

25. Lecture Aims
Example of a
Hydroelectric Power
Plant
Viscous Flow in Pipes
and Ducts
General Character of
Pipe Flow
Entrance Region and
Fully-Developed Flow
Pressure and Shear
Stress
Flow in a Circular Pipe
Effect of Rough Walls
The Moody Chart
Example of a Pipe
Flow Calculation
2.11
Flow in a Circular Pipe
We begin with a control volume analysis of the flow between sections 1 and 2
• The continuity equation is
Q1 = Q2 = constant
or V1 =
Q1
A1
= V2 =
Q2
A2
since the pipe is of constant
area
• The steady-flow energy
equation reduces to
1
2
V2
1 + gz1 +
p1
ρ1
=
1
2
V2
2 + gz2 +
p2
ρ2
+ ghloss
since there are no shaft work
or heat transfer effects
• The friction-head loss is therefore given by the expression
hloss = 4z +
4p
gρ

26. Lecture Aims
Example of a
Hydroelectric Power
Plant
Viscous Flow in Pipes
and Ducts
General Character of
Pipe Flow
Entrance Region and
Fully-Developed Flow
Pressure and Shear
Stress
Flow in a Circular Pipe
Effect of Rough Walls
The Moody Chart
Example of a Pipe
Flow Calculation
2.12
Flow in a Circular Pipe
We now apply the momentum balance to the control volume in
the figure, accounting for applied forces due to pressure,
gravity and shear
4p πR2
+ ρg (πR2
) 4 L sin φ =
−τw (2πR) 4 L = ṁ (V2 − V1) = 0
Since
4z = 4L sin φ
4z +
4p
gρ
=
2τw
gρ
4L
R
From our earlier expression for
hloss
hloss =
2τw
gρ
4L
R
We have not assumed laminar or turbulent flow

27. Lecture Aims
Example of a
Hydroelectric Power
Plant
Viscous Flow in Pipes
and Ducts
General Character of
Pipe Flow
Entrance Region and
Fully-Developed Flow
Pressure and Shear
Stress
Flow in a Circular Pipe
Effect of Rough Walls
The Moody Chart
Example of a Pipe
Flow Calculation
2.13
Flow in a Circular Pipe
• If we correlate τw with flow conditions, we have solved the
problem of head loss in pipe flow
• We can assume that τw depends on the following
variables
τw = F(ρ, V, µ, D, )
where is the wall roughness
• Dimensional analysis tells us that since there are 6
variables, and 3 reference dimensions, M, L, and T, there
will be 3 dimensionless groups
τw
ρV2
, Re,
D
• It is common to introduce a dimensionless parameter
called the Darcy friction factor,
f =
8τw
ρV2
• In terms of f, we expect to find a correlation, f = F(Re,
D
)

28. Lecture Aims
Example of a
Hydroelectric Power
Plant
Viscous Flow in Pipes
and Ducts
General Character of
Pipe Flow
Entrance Region and
Fully-Developed Flow
Pressure and Shear
Stress
Flow in a Circular Pipe
Effect of Rough Walls
The Moody Chart
Example of a Pipe
Flow Calculation
2.14
Flow in a Circular Pipe
hloss =
2τw
gρ
4L
R
=
8τw
ρV2
V2
4g
2 4 L
D
or
hloss = f
4L
D
V2
2g
Once we know, f = F(Re,
D
), we know hloss
In laminar flows,
f =
64
Re
can be derived analytically
Wall roughness does not affect laminar flows!
In turbulent flows, for a smooth walled pipe,
1
√
f
= 2.0 log(Re
√
f) − 0.8
cannot be derived analytically

29. Lecture Aims
Example of a
Hydroelectric Power
Plant
Viscous Flow in Pipes
and Ducts
General Character of
Pipe Flow
Entrance Region and
Fully-Developed Flow
Pressure and Shear
Stress
Flow in a Circular Pipe
Effect of Rough Walls
The Moody Chart
Example of a Pipe
Flow Calculation
2.15
Effect of Rough Walls
• Turbulent flows are strongly affected by wall roughness
• The structure and properties of the thin viscous sub-layer
near the wall are significantly influenced by even small
protuberances

30. Lecture Aims
Example of a
Hydroelectric Power
Plant
Viscous Flow in Pipes
and Ducts
General Character of
Pipe Flow
Entrance Region and
Fully-Developed Flow
Pressure and Shear
Stress
Flow in a Circular Pipe
Effect of Rough Walls
The Moody Chart
Example of a Pipe
Flow Calculation
2.16
Effect of Rough Walls
• Nikuradse simulated
roughness by gluing
uniform sand grains onto
the inner walls of the pipe
• After an onset point,
turbulent friction increases
with wall roughness
• For a given /D, friction
factor becomes constant at
high Reynolds numbers
• Colebrook devised a clever interpolation formula that
works for smooth and rough walls
1
√
f
= −2.0 log
/D
3.7
+
2.51
Re
√
f
• The Moody Chart is a plot of this formula

31. Lecture Aims
Example of a
Hydroelectric Power
Plant
Viscous Flow in Pipes
and Ducts
General Character of
Pipe Flow
Entrance Region and
Fully-Developed Flow
Pressure and Shear
Stress
Flow in a Circular Pipe
Effect of Rough Walls
The Moody Chart
Example of a Pipe
Flow Calculation
2.17
The Moody Chart
• No reliable friction factor for
2000 Re 4000
• Can be used for design calculations
for circular and non-circular pipe
flows, and open channel flows
• Roughness table for commercial
pipes
Most useful figure in Fluid Mechanics!

32. Lecture Aims
The Energy Grade
Line and The Hydraulic
Grade Line
Flow in a Pipe System
Minor Losses
Entrance Flow
Conditions
Exit Flow Conditions
Sudden Contraction or
Expansion
Gradual Expansion in
a Diffuser
Internal Structure of
Various Valves
Loss coefficients for
various pipe
components
Example of a pipe flow
calculation
3.1
Lecture 3
The Energy Equation
Minor losses due to pipe system components
CHE 2161/MEC2404 Mechanics of Fluids
Ravi Jagadeeshan
Department of Chemical Engineering
Monash University

33. Lecture Aims
The Energy Grade
Line and The Hydraulic
Grade Line
Flow in a Pipe System
Minor Losses
Entrance Flow
Conditions
Exit Flow Conditions
Sudden Contraction or
Expansion
Gradual Expansion in
a Diffuser
Internal Structure of
Various Valves
Loss coefficients for
various pipe
components
Example of a pipe flow
calculation
3.2
Aims of todays lecture
1 Discuss the energy grade line and the hydraulic grade line
2 Calculation of minor losses in the pipe flow system
3 Discuss entrance and exit flow conditions
4 Introduce losses in sudden contractions, expansions,
diffusers, valves and bends
5 Perform a simple pipe flow calculation with minor losses

34. Lecture Aims
The Energy Grade
Line and The Hydraulic
Grade Line
Flow in a Pipe System
Minor Losses
Entrance Flow
Conditions
Exit Flow Conditions
Sudden Contraction or
Expansion
Gradual Expansion in
a Diffuser
Internal Structure of
Various Valves
Loss coefficients for
various pipe
components
Example of a pipe flow
calculation
3.4
The Energy Grade Line and The Hydraulic Grade Line
Bernoulli Equation
p
γ
+
1
2
V2
g
+ z = constant = H
• According to the Bernoulli equation, the total
head remains constant along a stream line
• The elevation head, velocity head and pressure
head may vary along the streamline
• The energy grade line (EGL) is a line that
represents the total head available to the fluid

35. Lecture Aims
The Energy Grade
Line and The Hydraulic
Grade Line
Flow in a Pipe System
Minor Losses
Entrance Flow
Conditions
Exit Flow Conditions
Sudden Contraction or
Expansion
Gradual Expansion in
a Diffuser
Internal Structure of
Various Valves
Loss coefficients for
various pipe
components
Example of a pipe flow
calculation
3.5
The Energy Grade Line and The Hydraulic Grade Line
• The stagnation point at
the end of the Pitot
tube measures the total
head H of the flow
• The sum of the
pressure head and
elevation head is called
the piezometric head
p
γ
+ z
• The Hydraulic Grade Line (HGL) represents the
piezometric head along the flow - the HGL lies a
distance of one velocity head below the EGL

36. Lecture Aims
The Energy Grade
Line and The Hydraulic
Grade Line
Flow in a Pipe System
Minor Losses
Entrance Flow
Conditions
Exit Flow Conditions
Sudden Contraction or
Expansion
Gradual Expansion in
a Diffuser
Internal Structure of
Various Valves
Loss coefficients for
various pipe
components
Example of a pipe flow
calculation
3.6
The Energy Grade Line and The Hydraulic Grade Line
• The EGL is horizontal
and at the elevation of
the liquid in the tank
• A change in the fluid
velocity due to a
change in the pipe
diameter results in a
change in the elevation
of the HGL
• The distance from the
pipe to the HGL
indicates the pressure
within the pipe
• If the pipe lies below the HGL, the pressure within the pipe
is positive
• If the pipe lies above the HGL, the pressure is negative
• Regions of positive and negative pressure are readily
indicated when the pipe line and HGL are drawn together

37. Lecture Aims
The Energy Grade
Line and The Hydraulic
Grade Line
Flow in a Pipe System
Minor Losses
Entrance Flow
Conditions
Exit Flow Conditions
Sudden Contraction or
Expansion
Gradual Expansion in
a Diffuser
Internal Structure of
Various Valves
Loss coefficients for
various pipe
components
Example of a pipe flow
calculation
3.8
Flow in a Pipe System
hloss = f
V2
2g
!
4L
D
!
4p = f
ρV2
2
!
4L
D
!
• The friction factor is found from the Moody chart
f = F(Re,
D
)
• Its value depends on the flow being Laminar or
Turbulent
• Most pipe systems consist of additional
components such as valves, bends, tees etc.,
that add to the overall head loss of the system
• Such losses are termed Minor Losses

38. Lecture Aims
The Energy Grade
Line and The Hydraulic
Grade Line
Flow in a Pipe System
Minor Losses
Entrance Flow
Conditions
Exit Flow Conditions
Sudden Contraction or
Expansion
Gradual Expansion in
a Diffuser
Internal Structure of
Various Valves
Loss coefficients for
various pipe
components
Example of a pipe flow
calculation
3.9
Minor Losses
• Flow through a valve is a
common source of minor
head loss
• The head loss through the
valve may be a significant
portion of the resistance in
the system
• Theoretical analysis to predict
head loss through these
components is not yet
possible – head loss
information is based mostly
on experimental data
• Losses due to pipe system components are given in terms
of loss coefficients
hloss = Kl
V2
2g

39. Lecture Aims
The Energy Grade
Line and The Hydraulic
Grade Line
Flow in a Pipe System
Minor Losses
Entrance Flow
Conditions
Exit Flow Conditions
Sudden Contraction or
Expansion
Gradual Expansion in
a Diffuser
Internal Structure of
Various Valves
Loss coefficients for
various pipe
components
Example of a pipe flow
calculation
3.10
Minor Losses
• We expect a correlation
Kl = φ(geometry, Re)
• Usually the flow is
dominated by inertial
effects and viscous
effects are not as
important, so
Kl = φ(geometry) Flow into a pipe from a reservoir
• Changes in pipe diameter from one size to another leads
to losses that are not accounted for in the friction factor
calculation
• Extreme cases are flow into a pipe from a reservoir or out
of a pipe into a reservoir

40. Lecture Aims
The Energy Grade
Line and The Hydraulic
Grade Line
Flow in a Pipe System
Minor Losses
Entrance Flow
Conditions
Exit Flow Conditions
Sudden Contraction or
Expansion
Gradual Expansion in
a Diffuser
Internal Structure of
Various Valves
Loss coefficients for
various pipe
components
Example of a pipe flow
calculation
3.11
Entrance Flow Conditions
• A vena contracta region is
often developed at the
entrance to a pipe
because the fluid cannot
turn a sharp right corner
• The extra kinetic energy is
partially lost because of
viscous dissipation
• Figure on right shows
entrance loss coefficient
as a function of rounding
of the inlet edge
• The loss coefficient for a
square-edged entrance is
approximately Kl = 0.5

41. Lecture Aims
The Energy Grade
Line and The Hydraulic
Grade Line
Flow in a Pipe System
Minor Losses
Entrance Flow
Conditions
Exit Flow Conditions
Sudden Contraction or
Expansion
Gradual Expansion in
a Diffuser
Internal Structure of
Various Valves
Loss coefficients for
various pipe
components
Example of a pipe flow
calculation
3.12
Exit Flow Conditions
• A head loss is also produced when a fluid flows from a
pipe into a tank
• The entire kinetic energy of the exiting fluid is dissipated
through viscous effects
• The exit loss in this case is always Kl = 1.0

42. Lecture Aims
The Energy Grade
Line and The Hydraulic
Grade Line
Flow in a Pipe System
Minor Losses
Entrance Flow
Conditions
Exit Flow Conditions
Sudden Contraction or
Expansion
Gradual Expansion in
a Diffuser
Internal Structure of
Various Valves
Loss coefficients for
various pipe
components
Example of a pipe flow
calculation
3.13
Sudden Contraction or Expansion
• Losses also
occur because of
a change in pipe
diameter
• The sharp edged
entrance and exit
flows are limiting
cases of this
type of flow
Sudden Contraction
Sudden Expansion

43. Lecture Aims
The Energy Grade
Line and The Hydraulic
Grade Line
Flow in a Pipe System
Minor Losses
Entrance Flow
Conditions
Exit Flow Conditions
Sudden Contraction or
Expansion
Gradual Expansion in
a Diffuser
Internal Structure of
Various Valves
Loss coefficients for
various pipe
components
Example of a pipe flow
calculation
3.14
Gradual Expansion in a Diffuser
• For moderate to large angles, the conical diffuser is less
efficient than a sharp-edged expansion!
• It is difficult to efficiently decelerate a fluid!

44. Lecture Aims
The Energy Grade
Line and The Hydraulic
Grade Line
Flow in a Pipe System
Minor Losses
Entrance Flow
Conditions
Exit Flow Conditions
Sudden Contraction or
Expansion
Gradual Expansion in
a Diffuser
Internal Structure of
Various Valves
Loss coefficients for
various pipe
components
Example of a pipe flow
calculation
3.15
Internal Structure of Various Valves

45. Lecture Aims
The Energy Grade
Line and The Hydraulic
Grade Line
Flow in a Pipe System
Minor Losses
Entrance Flow
Conditions
Exit Flow Conditions
Sudden Contraction or
Expansion
Gradual Expansion in
a Diffuser
Internal Structure of
Various Valves
Loss coefficients for
various pipe
components
Example of a pipe flow
calculation
3.16
Loss coefficients for various pipe components
• A single pipe system
may have many minor
losses
• The total system loss is
calculated by summing
the major loss and all
the minor losses
• The length 4L in the
major loss calculation
includes the length in
bends etc

46. Lecture Aims
Energy grade line and
hydraulic grade line in
pipe flow
Three Types of Pipe
Flow Problems
Example of a Type 2
Problem
Example of a Type 3
Problem
Type 3 Problem with
Minor Losses
Multiple Pipe Systems
Parallel Pipe Systems
Three Reservoir Pipe
Junction
Water Distribution
System
4.1
Lecture 4
The Energy Equation
Three types of pipe flow problems and multiple pipe systems
CHE 2161/MEC2404 Mechanics of Fluids
Ravi Jagadeeshan
Department of Chemical Engineering
Monash University

47. Lecture Aims
Energy grade line and
hydraulic grade line in
pipe flow
Three Types of Pipe
Flow Problems
Example of a Type 2
Problem
Example of a Type 3
Problem
Type 3 Problem with
Minor Losses
Multiple Pipe Systems
Parallel Pipe Systems
Three Reservoir Pipe
Junction
Water Distribution
System
4.2
Aims of todays lecture
1 Discuss the energy grade line and the hydraulic grade line
in the presence of minor losses
2 Introduce three types of pipe flow problems
3 Solve examples of Type 2 and Type 3 problems
4 Example of Type 3 problem with minor losses
5 Calculate flows in multiple pipe systems
6 The three reservoir junction problem

48. Lecture Aims
Energy grade line and
hydraulic grade line in
pipe flow
Three Types of Pipe
Flow Problems
Example of a Type 2
Problem
Example of a Type 3
Problem
Type 3 Problem with
Minor Losses
Multiple Pipe Systems
Parallel Pipe Systems
Three Reservoir Pipe
Junction
Water Distribution
System
4.8
Three Types of Pipe Flow Problems
• The Moody chart can be used to solve most
problems involving friction losses in long pipes
• However, the Moody chart is a head loss chart,
i.e, given fluid properties, pipe dimensions and
flow rates, we can easily calculate hloss or 4p
• On the other hand, when we are required to
calculate the flow rate, or pipe dimensions, given
the friction factor f, then an iterative procedure is
required. This is because both D and V are
contained in the ordinate and the abscissa of the
chart
• Typical pipe flow problems can be classified as
belonging to three types

49. Lecture Aims
Energy grade line and
hydraulic grade line in
pipe flow
Three Types of Pipe
Flow Problems
Example of a Type 2
Problem
Example of a Type 3
Problem
Type 3 Problem with
Minor Losses
Multiple Pipe Systems
Parallel Pipe Systems
Three Reservoir Pipe
Junction
Water Distribution
System
4.9
Three Types of Pipe Flow Problems
Variable Type 1 Type 2 Type 3
a. Fluid
Density
Viscosity
Given
Given
Given
Given
Given
Given
b. Pipe
Diameter
Length
Roughness
Given
Given
Given
Given
Given
Given
Determine
Given
Given
c. Flow
Flow rate or
Average Velocity
Given Determine Given
d. Pressure
Pressure Drop or
head loss
Determine Given Given

50. Lecture Aims
Energy grade line and
hydraulic grade line in
pipe flow
Three Types of Pipe
Flow Problems
Example of a Type 2
Problem
Example of a Type 3
Problem
Type 3 Problem with
Minor Losses
Multiple Pipe Systems
Parallel Pipe Systems
Three Reservoir Pipe
Junction
Water Distribution
System
4.20
Multiple Pipe Systems
• The governing mechanisms for the flow in
multiple pipe systems are the same as for
single pipe systems
• Additional complexities arise because of the
number of unknowns involved

51. Lecture Aims
Energy grade line and
hydraulic grade line in
pipe flow
Three Types of Pipe
Flow Problems
Example of a Type 2
Problem
Example of a Type 3
Problem
Type 3 Problem with
Minor Losses
Multiple Pipe Systems
Parallel Pipe Systems
Three Reservoir Pipe
Junction
Water Distribution
System
4.21
Multiple Pipe Systems
• The simplest multiple pipe
systems can be classified into
series or parallel flows
• In a series pipe system, the flow
rate is the same in each pipe
Q = Q1 = Q2 = Q3 = . . .
and the head loss is the sum of
the head loss in each pipe
htotal = h1 + h2 + h3 + . . .
• In general the friction factors will be different for each pipe
because the Reynolds numbers and the relative
roughness will be different
• Type 1 is straightforward, Types 2 and 3 require iteration

52. Lecture Aims
Energy grade line and
hydraulic grade line in
pipe flow
Three Types of Pipe
Flow Problems
Example of a Type 2
Problem
Example of a Type 3
Problem
Type 3 Problem with
Minor Losses
Multiple Pipe Systems
Parallel Pipe Systems
Three Reservoir Pipe
Junction
Water Distribution
System
4.22
Parallel Pipe Systems
• In a parallel pipe system, the flow rate is the sum
of the flow rates in each of the pipes,
Q = Q1 + Q2 + Q3 + . . .
• However, the head loss experienced in each of
the pipes is the same,
hloss = h1 = h2 = h3 = . . .
• Again, the method of solution depends on what
is given and what is to be calculated

53. Lecture Aims
Energy grade line and
hydraulic grade line in
pipe flow
Three Types of Pipe
Flow Problems
Example of a Type 2
Problem
Example of a Type 3
Problem
Type 3 Problem with
Minor Losses
Multiple Pipe Systems
Parallel Pipe Systems
Three Reservoir Pipe
Junction
Water Distribution
System
4.23
Three Reservoir Pipe Junction
• Three or more pipes could meet at a junction
• If all the flows are considered positive toward the junction:
Q1 + Q2 + Q3 = 0
• This obviously implies that one or two of the flows must be
away from the junction
• The pressure must change through each pipe so as to
give the same static pressure pJ at the junction

54. Lecture Aims
Energy grade line and
hydraulic grade line in
pipe flow
Three Types of Pipe
Flow Problems
Example of a Type 2
Problem
Example of a Type 3
Problem
Type 3 Problem with
Minor Losses
Multiple Pipe Systems
Parallel Pipe Systems
Three Reservoir Pipe
Junction
Water Distribution
System
4.24
Three Reservoir Pipe Junction
We can write the energy equation from the free surface of each
reservoir to the junction point
1
2
V2
i
g
+ zi +
pi
γ
=
1
2
VJ
2
g
+ zJ +
pJ
γ
+ h(i)
loss ; i = 1, 2, 3
Note that p1 = p2 = p3 = V1 = V2 = V3 = VJ = 0
We then get 3 equations
z1 = zJ +
pJ
γ
+ h(1)
loss
z2 = zJ +
pJ
γ
+ h(2)
loss
z3 = zJ +
pJ
γ
+ h(3)
loss
If we define: hJ =
pJ
γ
+ zJ
z1 − hJ = h(1)
loss = f1
V1
2
2g
!
4L1
D1
z2 − hJ = h(2)
loss = f2
V2
2
2g
!
4L2
D2
z3 − hJ = h(3)
loss = f3
V3
2
2g
!
4L3
D3

55. Lecture Aims
Energy grade line and
hydraulic grade line in
pipe flow
Three Types of Pipe
Flow Problems
Example of a Type 2
Problem
Example of a Type 3
Problem
Type 3 Problem with
Minor Losses
Multiple Pipe Systems
Parallel Pipe Systems
Three Reservoir Pipe
Junction
Water Distribution
System
4.25
Three Reservoir Pipe Junction
The solution is an iterative procedure:
1 Guess the value of hJ and fi
2 Solve the set of equations for V1, V2, V3
3 Check if Q1 + Q2 + Q3 = 0
4 If not repeat the procedure

56. Lecture Aims
Energy grade line and
hydraulic grade line in
pipe flow
Three Types of Pipe
Flow Problems
Example of a Type 2
Problem
Example of a Type 3
Problem
Type 3 Problem with
Minor Losses
Multiple Pipe Systems
Parallel Pipe Systems
Three Reservoir Pipe
Junction
Water Distribution
System
4.26
Water Distribution System
Figure 9.1 Results of a Map Query
. Select a time period in which to query the map from the Map
Browser.
A complex multiple pipe system

57. CHE2161 Workshop Week 9
The Energy Equation

58. Example 1: Problem from 2012 S2 Exam
Consider a tank-pipe system, as shown in the figure. The pipe exits from the tank 4 m below the surface of the
water in the tank, has a diameter of 3 cm and length of 5 m. It is designed to deliver water into a reservoir,
which is 2 m below the exit of the pipe, at a flow rate of at least 11 m3/hr. The density of water can be taken to
be 998 kg/m3 and the viscosity to be 0.001 kg/m.s.

59. (a) Calculate the Reynolds number for the expected flow rate.
Re =
𝜌𝑣𝐷
𝜇
𝑄 = 11𝑚3
/ℎ
Q =
11
3600
= 3.06 × 10−3𝑚3/𝑠
𝑄 = 𝑣𝐴
=
3.06 × 10−3
𝜋 × 0.0152
𝑣 =
𝑄
𝐴
= 4.323𝑚𝑠−1
Re =
𝜌𝑣𝐷
𝜇
Re =
998 × 4.323 × 0.003
0.001
Re = 1.29 × 105
The first decision we need to make is in regard to the characteristic length we need to use in the calculation of
Reynolds number.
Since this is a pipe flow question, the pipe diameter should be used in calculating the Reynolds number.

60. (b) What is the head loss in the pipe at this flow rate?
𝑃1
𝜌𝑔
+
𝑣1
2
2𝑔
+ 𝑧1 =
𝑃2
𝜌𝑔
+
𝑣2
2
2𝑔
+ 𝑧2 + ℎ𝐿
𝑃1
𝜌𝑔
+
𝑣1
2
2𝑔
+ 𝑧1 =
𝑃2
𝜌𝑔
+
𝑣2
2
2𝑔
+ 𝑧2 + ℎ𝐿
ℎ𝐿 = 𝑧1 − 𝑧2 −
𝑣2
2
2𝑔
ℎ𝐿 = 𝑧1 − 𝑧2 −
𝑣2
2
2𝑔
ℎ𝐿 = 4 −
4.3232
2 × 9.81
= 3.046 𝑚~3.05 𝑚
To determine the head loss, we will need to use our energy equation:
Now we need to identify the points in our system between which we will conduct the analysis
Simplifying to make the head loss the subject of the equation:

61. (c) What is the friction factor for the pipe?
ℎ𝐿 = 𝑓
𝑙
𝐷
𝑣2
2𝑔
𝑓 =
ℎ𝐿 × 𝐷 × 2𝑔
𝑙 × 𝑣2
𝑓 =
3.05 × 0.03 × 2 × 9.81
5 × 4.3232
𝑓 = 0.0192
We will require our equation for the friction factor:
Rearranging

62. (d) What is the maximum roughness height allowable for the pipe?
𝑓
𝑅𝑒
𝜀
𝐷
𝑅𝑒 = 1.29 × 105
𝑓 = 0.0192
𝜀
𝐷
= 0.0004
𝜀 = 0.0004 × 𝐷 = 0.0004 × 0.03 = 0.000012 𝑚 = 0.012 𝑚𝑚
Here we will need to use the Moody chart, which allows the friction factor to be determined given the
Reynolds number and relative roughness.
In this case, we have the friction factor and Reynolds number, so we need to use these to determine the relative
roughness.
This can then be used to determine the actual roughness of the surface.

63. (e) If the Reynolds number is kept equal to its value at the
expected flow rate, and the pipe walls polished to a smooth
surface, what is the percent reduction in the friction factor and
the head loss?
𝑓
𝑅𝑒
𝜀
𝐷
Smooth pipe
𝑓 = 0.017
% 𝑟𝑒𝑑𝑢𝑐𝑡𝑖𝑜𝑛 𝑖𝑛 𝑓 =
0.019 − 0.017
0.019
= 10.5%
Here, we need to use the line on the Moody chart for a “smooth pipe”.
Using the previously calculated value for Reynolds number, we can read off the friction factor.

64. Example 2: Problem from 2011 S1 Exam
A pipe joins two reservoirs A and B. The elevation of the free surface in tank A is 10 m above that in tank B. The
pipe is 0.2 m diameter, 1000 m in length and has a friction factor of 0.032.

65. (a) What is the flow rate in the pipeline?
𝑃𝐴
𝜌𝑔
+
𝑣𝐴
2
2𝑔
+ 𝑧1 =
𝑃𝐵
𝜌𝑔
+
𝑣𝐵
2
2𝑔
+ 𝑧2 + ℎ𝐿
𝑃𝐴
𝜌𝑔
+
𝑣𝐴
2
2𝑔
+ 𝑧𝐴 =
𝑃𝐵
𝜌𝑔
+
𝑣𝐵
2
2𝑔
+ 𝑧𝐵 + ℎ𝐿
ℎ𝐿 = 𝑧𝐴 − 𝑧𝐵
ℎ𝐿 = 𝑓
𝑙
𝐷
𝑣2
2𝑔
= 10 𝑚
𝑣2 =
ℎ𝐿 × 𝐷 × 2𝑔
𝑓 × 𝑙
𝑣 =
ℎ𝐿 × 𝐷 × 2𝑔
𝑓 × 𝑙
=
10 × 0.2 × 2 × 9.81
0.032 × 1000
= 1.107 𝑚𝑠−1
𝑄 = 𝑣 × 𝐴
= 1.107 × 𝜋 × 0.12
= 0.0348 𝑚3/𝑠
For the first part of the question, we need to consider just a single pipe between the
reservoirs, and apply the energy equation between the reservoirs (points A and B).
ℎ𝐿 = 𝑓
𝑙
𝐷
𝑣2
2𝑔
How will we work out the flowrate?
Recall our equation for head loss.

66. (b) It is required to increase the flow to the downstream reservoir by
30%. This is to be done adding a second pipe of the same diameter that
connects at some point along the old pipe and runs down to the lower
reservoir. Assuming the diameter and the friction factor are the same as
the old pipe, how long should the new pipe be? Note that the length of
new pipe from the junction C to the lower reservoir can be assumed to
be the same as the length of the original pipe from C to the lower
reservoir.
𝑙2 = 𝑙3 𝐷2 = 𝐷3
𝑄 = 𝑄2 + 𝑄3
𝑄 = 1.3 × 0.03477 = 0.0452𝑚3/𝑠
𝑄2 = 𝑄3 =
𝑄
2
=
0.0452
2
= 0.0226 𝑚3/𝑠
The first thing to note is that the dimensions of pipes 2 and 3 are the same. That is:
We wish to increase the flow by 30%. Therefore:
Using conservation of mass and the fact that this is an incompressible fluid:
And given the identical nature of pipes 2 and 3:

67. 𝑃𝐴
𝜌𝑔
+
𝑣𝐴
2
2𝑔
+ 𝑧𝐴 =
𝑃𝐵
𝜌𝑔
+
𝑣𝐵
2
2𝑔
+ 𝑧𝐵 + ℎ𝐿
𝑃𝐴
𝜌𝑔
+
𝑣𝐴
2
2𝑔
+ 𝑧𝐴 =
𝑃𝐵
𝜌𝑔
+
𝑣𝐵
2
2𝑔
+ 𝑧𝐵 + ℎ𝐿
ℎ𝐿2 = ℎ𝐿3
ℎ𝐿 = 𝑧𝐴 − 𝑧𝐵
ℎ𝐿 = ℎ𝐿1 + ℎ𝐿2 or ℎ𝐿 = ℎ𝐿1 + ℎ𝐿3
In the length 1:
→ 𝑣 =
𝑄
𝐴
𝑄 = 𝑣𝐴 =
0.0452
𝜋 × 0.12 = 1.439 𝑚𝑠−1
In the length 2 (or 3):
→ 𝑣 =
𝑄
𝐴
𝑄 = 𝑣𝐴 =
0.0226
𝜋 × 0.12
= 0.7194 𝑚𝑠−1
Now we need to apply our energy equation between
points A and B
And so we have:
Important point: the head loss here needs to take into
account the head loss in the single pipe section (1) and
the parallel pipe section comprising (2) and (3).
For the parallel pipe section:
And therefore we can say:
To work out the head loss we will need the fluid
velocity in length (1) and length (2) or (3).

68. ℎ𝐿 = ℎ𝐿1 + ℎ𝐿2 = 𝑓
𝑙1
𝐷
𝑣1
2
2𝑔
+ 𝑓
(1000 − 𝑙1)
𝐷
𝑣2
2
2𝑔
ℎ𝐿 =
𝑓
2𝑔𝐷
𝑙1𝑣1
2
+ (1000 − 𝑙1)𝑣2
2
ℎ𝐿 =
𝑓
2𝑔𝐷
𝑙1𝑣1
2
+ 1000𝑣2
2
− 𝑙1𝑣2
2
ℎ𝐿 =
𝑓
2𝑔𝐷
𝑙1(𝑣1
2
− 𝑣2
2
) + 1000𝑣2
2
ℎ𝐿
2𝑔𝐷
𝑓
− 1000𝑣2
2
= 𝑙1(𝑣1
2
− 𝑣2
2
)
𝑙1 = ℎ𝐿
2𝑔𝐷
𝑓
− 1000𝑣2
2
÷ (𝑣1
2
− 𝑣2
2
)
𝑙1 = 456 𝑚
𝑙2 = 544 𝑚
𝑙2 = 𝑙3 = 1000 − 𝑙1
Note that:
So we can write:
Factorising:
Expanding inside the brackets
Collecting terms inside the brackets
Rearranging
Further rearranging to make l1 the subject:
= 10 ×
2 × 9.81 × 0.2
0.032
− 1000 × 0.71942 ÷ (1.4392 − 0.71942)

69. Example 3: Problem from 2011 S2 Exam
A parallel galvanized iron pipe system shown in the figure above delivers gasoline at 20°C with a total flow rate
of 0.04 m3/s. The pump is wide open and not running, with a loss coefficient of 1.5. Data ρ = 680 kg/m3, μ =
2.92 × 10-4 kg.m/s, ϵ = 0.15 mm.
Tips:
(i) Assume turbulent flow throughout.
(ii) Neglect losses in the bends.
(iii) Iterate once on the friction factor to solve the problem,
with an initial guess that f = 0.025 in both pipes.

70. (a) Assuming that the pressure drop is the same in both legs, show that,
hL1 = hL2 + hpump
where hL1 and hL2 are the major head losses in pipes 1 and 2, respectively,
and hpump is the minor loss in the pump.
𝑃𝑖𝑛
𝜌𝑔
+
𝑣𝑖𝑛
2
2𝑔
+ 𝑧𝑖𝑛 =
𝑃𝑜𝑢𝑡
𝜌𝑔
+
𝑣𝑜𝑢𝑡
2
2𝑔
+ 𝑧𝑜𝑢𝑡 + ℎ𝐿1
𝑃𝑖𝑛
𝜌𝑔
+
𝑣𝑖𝑛
2
2𝑔
+ 𝑧𝑖𝑛 =
𝑃𝑜𝑢𝑡
𝜌𝑔
+
𝑣𝑜𝑢𝑡
2
2𝑔
+ 𝑧𝑜𝑢𝑡 + ℎ𝐿2 + ℎ𝑝𝑢𝑚𝑝
Pipe I
ℎ𝐿1 =
𝑃𝑖𝑛 −𝑃𝑜𝑢𝑡
𝜌𝑔
Pipe II
ℎ𝐿2 + ℎ𝑝𝑢𝑚𝑝 =
𝑃𝑖𝑛 −𝑃𝑜𝑢𝑡
𝜌𝑔
∴ ℎ𝐿1 = ℎ𝐿2 + ℎ𝑝𝑢𝑚𝑝
In order to answer this question, let’s apply the
energy across each arm of the setup.

71. (b) Determine the flow rate in each pipe
ℎ𝐿1 = 𝑓1
𝑙1
𝐷1
𝑣1
2
2𝑔
ℎ𝐿2 = 𝑓2
𝑙2
𝐷2
𝑣2
2
2𝑔
ℎ𝑝𝑢𝑚𝑝 = 𝐾𝐿
𝑣2
2
2𝑔
ℎ𝐿1 = ℎ𝐿2 + ℎ𝑝𝑢𝑚𝑝
𝑓1
𝑙1
𝐷1
𝑣1
2
2𝑔
= 𝑓2
𝑙2
𝐷2
𝑣2
2
2𝑔
+ 𝐾𝐿
𝑣2
2
2𝑔
𝑓1
𝑙1
𝐷1
𝑣1
2
2𝑔
=
𝑣2
2
2𝑔
(𝑓2
𝑙2
𝐷2
+ 𝐾𝐿)
𝑄𝑖𝑛 = 𝑄1 +𝑄2
𝑄𝑖𝑛 = 0.04 𝑚3
𝑄𝑖𝑛 = 𝑣1𝐴1 + 𝑣2𝐴2
𝑄𝑖𝑛 = 𝑣1
𝜋𝐷1
2
4
+ 𝑣2
𝜋𝐷2
2
4
0.04 = 𝑣1
𝜋0.0752
4
+ 𝑣2
𝜋0.0252
4
𝐷1 = 0.075 𝑚 𝐷2 = 0.025 𝑚
81.487 = 9𝑣1 + 𝑣2
We have two unknowns here (Q1 and Q2), and so we will
require two equations to determine them.
First, consider the head loss relationship we just derived.
Now, consider conservation of mass (note that this is an
incompressible fluid).

72. 𝑓1
𝑙1
𝐷1
𝑣1
2
2𝑔
=
𝑣2
2
2𝑔
(𝑓2
𝑙2
𝐷2
+ 𝐾𝐿)
81.487 = 9𝑣1 + 𝑣2
Iterate with a first guess of f1 = f2 = 0.025
0.025
60
0.075
𝑣1
2
2 × 9.81
=
𝑣2
2
2 × 9.81
(0.025
55
0.025
+ 1.5)
1.0194𝑣1
2
= 2.8797𝑣2
2
𝑣1 = 1.6807𝑣2
81.487 = 9 × 1.6807𝑣2 +𝑣2
𝑣2 = 5.053 𝑚𝑠−1
𝑣1 = 8.492 𝑚𝑠−1
𝑣1 = 1.6807𝑣2
So now we have two equations:
(i)
(ii)
Turning our attention to Equation (ii):
𝑓1
𝑙1
𝐷1
𝑣1
2
2𝑔
=
𝑣2
2
2𝑔
(𝑓2
𝑙2
𝐷2
+ 𝐾𝐿)
Combining with Equation (i)
Solving, we obtain:

73. Arm #1
Re =
𝜌𝑣𝐷
𝜇
=
680 × 8.49 × 0.075
2.92 × 10−4
= 1.48 × 106
𝜀
𝐷
=
0.15 × 10−3
0.075
= 0.002
Arm #2
Re =
𝜌𝑣𝐷
𝜇
=
680 × 5.053 × 0.025
2.92 × 10−4 = 2.9 × 105
𝜀
𝐷
=
0.15 × 10−3
0.025
= 0.006
We now need to continue our iteration by checking the
guess of friction factor in each arm using the
determined fluid velocities.
𝑓
𝑅𝑒
𝜀
𝐷
Using the calculated Re and relative roughness, we can
now use the Moody Chart to determine the friction
factors in each arm.
𝑓1 = 0.023
𝑓2 = 0.032
𝑓1
𝑙1
𝐷1
𝑣1
2
2𝑔
= 𝑓2
𝑙2
𝐷2
𝑣2
2
2𝑔
+ 𝐾𝐿
𝑣2
2
2𝑔
→ 𝑣1 = 1.9768𝑣2
Now we return to the head loss equation:

74. 81.487 = 9𝑣1 + 𝑣2
𝑣1 = 1.9768𝑣2
81.487 = (9 × 1.9768𝑣2) + 𝑣2
𝑣2 = 4.34 𝑚𝑠−1
𝑣1 = 8.57 𝑚𝑠−1
→ 𝑄2= 𝑣2 × 𝐴2 = 4.34 × 𝜋 × 0.01252 = 0.00213 𝑚3/𝑠
→ 𝑄1= 𝑣1 × 𝐴1 = 8.57 × 𝜋 × 0.03752
= 0.0379 𝑚3
/𝑠
Arm #1
Re =
𝜌𝑣𝐷
𝜇
=
680 × 8.57 × 0.075
2.92 × 10−4
= 1.49 × 106
Arm #2
Re =
𝜌𝑣𝐷
𝜇
=
680 × 4.34 × 0.025
2.92 × 10−4
= 2.53 × 105
→ 𝑓1 = 0.023
→ 𝑓2 = 0.032
Therefore the solution has converged.
Finally, we should check the friction factors again to
confirm that the solution has converged.
Now we can use our two equations (from head loss and
conservation of mass) to determine values for the fluid
velocity in each arm.
Using the velocity we can determine the flowrate in arm 2:
Likewise, we can determine the velocity and flowrate in arm 1:

75. (c) Determine the overall pressure drop
ℎ𝐿1 =
𝑃𝑖𝑛 −𝑃𝑜𝑢𝑡
𝜌𝑔
ℎ𝐿2 + ℎ𝑝𝑢𝑚𝑝 =
𝑃𝑖𝑛 −𝑃𝑜𝑢𝑡
𝜌𝑔
= 𝑓1
𝑙1
𝐷1
𝑣1
2
2𝑔
→ ∆𝑃 =
𝜌 × 𝑓1 × 𝑙1 × 𝑣1
2
2 × 𝐷1
=
680 × 0.023 × 60 × 8.572
2 × 0.075
= 459 𝑘𝑃𝑎
= 𝑓2
𝑙2
𝐷2
𝑣2
2
2𝑔
+ 𝐾𝐿
𝑣2
2
2𝑔
→ ∆𝑃 =
𝜌 × 𝑓2 × 𝑙2 × 𝑣2
2
2 × 𝐷2
+ 𝐾𝐿
𝜌 × 𝑣2
2
2
∆𝑃 =
𝜌 × 𝑓2 × 𝑙2 × 𝑣2
2
2 × 𝐷2
+ 𝐾𝐿
𝜌 × 𝑣2
2
2
=
680 × 0.032 × 55 × 4.342
2 × 0.025
+ 1.5 ×
680 × 4.342
2
= 460 𝑘𝑃𝑎
To determine the pressure drop in each arm, we can return to the equations that we
derived via the energy equation. We can choose either (or both) arms.
Let’s check arm 2, which includes the pump.

76. Example 4: Problem from 2012 S1 Exam
Consider a reservoir feeding a pipe, as shown in the figure. The pipe diameter is 100mm and has
length 15m and feeds directly into the atmosphere at point C, which is 4m below the surface of the
reservoir (i.e. za − zc = 4.0m). The highest point on the pipe is B, which is 1.5m above the surface of
the reservoir (i.e. zb − za = 1.5m) and 5m along the pipe measured from the reservoir. Assume the
entrance to the pipe to be sharp and the value of friction factor f to be 0.32.

77. (a) Calculate the velocity of water leaving the pipe at point C.
𝑃𝐴
𝜌𝑔
+
𝑣𝐴
2
2𝑔
+ 𝑧𝐴 =
𝑃𝐶
𝜌𝑔
+
𝑣𝐶
2
2𝑔
+ 𝑧𝐶 + ℎ𝐿
First, let’s apply the energy equation between points A and C
𝑧𝐴 − 𝑧𝐶 =
𝑣𝐶
2
2𝑔
+ ℎ𝐿
Note that the head loss includes both:
(i) the major loss due to flow in the pipe; and
(ii) the minor loss due to the pipe entrance.

78. 𝑧𝐴 − 𝑧𝐶 =
𝑣𝐶
2
2𝑔
+ ℎ𝐿
𝑧𝐴 − 𝑧𝐶 =
𝑣𝐶
2
2𝑔
+
𝑣𝐶
2
2𝑔
(𝑓
𝑙
𝐷
+ 𝐾𝐿)
𝑧𝐴 − 𝑧𝐶 =
𝑣𝐶
2
2𝑔
(1 + 𝑓
𝑙
𝐷
+ 𝐾𝐿)
𝑣𝐶
2
=
2𝑔(𝑧𝐴−𝑧𝐶)
(1 + 𝑓
𝑙
𝐷
+ 𝐾𝐿
)
𝑣𝐶 =
2𝑔(𝑧𝐴−𝑧𝐶)
(1 + 𝑓
𝑙
𝐷
+ 𝐾𝐿)
=
2 × 9.81 × 4
(1 + 0.32 ×
15
0.1
+ 0.5)
= 1.26 𝑚𝑠−1
Recalling our equation from the previous page:
Incorporating the major loss from the pipe and the minor loss at the pipe entrance:
Factorising
Rearranging

79. (b) Calculate the pressure in the pipe at point B.
𝑃𝐴
𝜌𝑔
+
𝑣𝐴
2
2𝑔
+ 𝑧𝐴 =
𝑃𝐵
𝜌𝑔
+
𝑣𝐵
2
2𝑔
+ 𝑧𝐵 + ℎ𝐿
𝑃𝐵
𝜌𝑔
= 𝑧𝐴 − 𝑧𝐵 −
𝑣𝐵
2
2𝑔
− ℎ𝐿
𝑃𝐵
𝜌𝑔
= 𝑧𝐴 − 𝑧𝐵 −
𝑣𝐵
2
2𝑔
−
𝑣𝐵
2
2𝑔
(𝑓
𝑙
𝐷
+ 𝐾𝐿)
𝑃𝐵
𝜌𝑔
= −1.5 −
1.262
2 × 9.81
−
1.262
2 × 9.81
(0.32 ×
5
0.1
+ 0.5)
𝑃𝐵
𝜌𝑔
= −1.5 − 0.081 − 0.081 × (16.5)= −2.918
𝑃𝐵 = −2.918 × 9.81 × 1000 = −28620 𝑃𝑎 = −28.6 𝑘𝑃𝑎
Therefore the pressure is 28.6 kPa below the pressure at point A
𝑃𝐵
𝜌𝑔
= 𝑧𝐴 − 𝑧𝐵 −
𝑣𝐵
2
2𝑔
−
𝑣𝐵
2
2𝑔
(𝑓
𝑙
𝐷
+ 𝐾𝐿)
Again, we start with our energy equation.
Include the major loss due to pipe flow and
the minor loss at the pipe entrance:
Rearranging

80. Example 5: Problem from 2013 S1 Exam
Water flows in the pipe shown in the figure below, driven by pressurised air in the tank. (Neglect
minor losses in all calculations below and use the data for water provided in the aide memoire).
𝑃𝑖𝑝𝑒 𝑙𝑒𝑛𝑔𝑡ℎ 𝑙 = 30 𝑚 + 80 𝑚 + 60 𝑚 = 170 𝑚
Note that we are told to neglect the minor
losses due to the bends, pipe entrance etc.

81. (a) Estimate the gauge pressure required to provide a water flow rate Q = 60
m3/h?
𝑄 = 60𝑚3
/ℎ
𝑄 =
60
3600
= 0.0167𝑚3
/𝑠
𝑄 = 𝑣𝐴
𝑣 =
𝑄
𝐴
=
0.0167
𝜋 × 0.0252
= 8.4883 𝑚𝑠−1
Re =
𝜌𝑣𝐷
𝜇
Re =
1000 × 8.4883 × 0.05
1.2 × 10−3
Re = 3.5 × 105
We are going to need the fluid velocity at various points in the calculation, so
let’s start by calculating that.
To determine the friction factor we will need the Reynolds
number.
Firstly, make sure the units are correct.
Now use the volumetric flowrate to determine
the fluid velocity.

82. From the Moody chart with Re = 3.5 × 105
and a smooth pipe, we can read off:
𝑓 = 0.014
𝑃1
𝜌𝑔
+
𝑣1
2
2𝑔
+ 𝑧1 =
𝑃2
𝜌𝑔
+
𝑣2
2
2𝑔
+ 𝑧2 + ℎ𝐿
𝑃1
𝜌𝑔
= 𝑧2 − 𝑧1 +
𝑣2
2
2𝑔
+ ℎ𝐿
𝑃1
𝜌𝑔
= 𝑧2 − 𝑧1 +
𝑣2
2
2𝑔
+ 𝑓
𝑙
𝐷
𝑣2
2
2𝑔
𝑃1
𝜌𝑔
= 𝑧2 − 𝑧1 +
𝑣2
2
2𝑔
(1 + 𝑓
𝑙
𝐷
)
𝑃1
𝜌𝑔
= 70 +
8.48832
2 × 9.81
(1 + 0.014
170
0.05
) = 248.48
→ 𝑃1 = 248.48 × 𝜌 × 𝑔
𝑃1 = 248.48 × 1000 × 9.81
= 248.48 × 1000 × 9.81
= 2437545 𝑃𝑎
= 2438 𝑘𝑃𝑎
Now, return to the energy equation:
Rearranging
Including the major loss from pipe flow:
𝑃1
𝜌𝑔
= 𝑧2 − 𝑧1 +
𝑣2
2
2𝑔
(1 + 𝑓
𝑙
𝐷
)

83. (b) When a pressure p1 = 700 kPa is applied in the setup shown in the figure above, it is
found that gasoline (with density ρ = 679 kg/m3) flows out of the pipe with a flow rate Q
= 27 m3/h. What is the viscosity of gasoline?
𝑃1
𝜌𝑔
+
𝑣1
2
2𝑔
+ 𝑧1 =
𝑃2
𝜌𝑔
+
𝑣2
2
2𝑔
+ 𝑧2 + ℎ𝐿
𝑄 = 27 𝑚3
/ℎ
𝑄 =
27
3600
= 0.0075 𝑚3/𝑠
𝑄 = 𝑣𝐴
𝑣 =
𝑄
𝐴
=
0.0075
𝜋 × 0.0252
= 3.8197 𝑚𝑠−1
ℎ𝐿 = 𝑓
𝑙
𝐷
𝑣2
2
2𝑔
Again, we are going to need the fluid velocity at various
points in the calculation, so let’s start by calculating that.
Firstly, make sure the units are correct.
Now use the volumetric flowrate to determine
the fluid velocity.
Now let’s return to our energy equation:
The only head loss to be considered is the head
loss due to pipe flow.
ℎ𝐿 =
𝑃1
𝜌𝑔
+ 𝑧1 − 𝑧2 −
𝑣2
2
2𝑔
How can we get to viscosity? Reynolds number!
How can we get to Reynolds number? Friction factor!

84. ℎ𝐿 =
𝑃1
𝜌𝑔
+ 𝑧1 − 𝑧2 −
𝑣2
2
2𝑔
ℎ𝐿 =
700000
679 × 9.81
+ 10 − 80 −
3.81972
2 × 9.81
= 34.346 𝑚
ℎ𝐿 = 𝑓
𝑙
𝐷
𝑣2
2
2𝑔
→ 𝑓 = ℎ𝐿
𝐷
𝑙
2𝑔
𝑣2
2 = 34.346 ×
0.05
170
×
2 × 9.81
3.81972
→ 𝑓 = 0.0136
Using the Moody chart, for a smooth pipe with f =
0.0136, we can read off a value of Re = 4.66 × 105.
Re =
𝜌𝑣𝐷
𝜇
𝜇 =
679 × 3.8197 × 0.05
4.66 × 105
𝜇 = 2.78 × 10−4𝑁𝑠𝑚−2
→ 𝜇 =
𝜌𝑣𝐷
𝑅𝑒
So now we can evaluate our equation for head loss:
We also know:

85. CFD images by Professor Kunio Kuwahara, Institute of Space and Astronautical Science, Japan
(c) ISAS/JAXA, Courtesy of ISAS/JAXA
8.1 Boundary Layers – Introductory Concepts

86. No Slip Boundary Condition
Until now we have often approximated the flow near a solid boundary as having
the same velocity as the flow further away.
i.e. we have ignored viscous effects.
Real flows near a boundary satisfy the No Slip Boundary Condition: The flow
velocity exactly at the boundary must match the velocity of the boundary.
No slip boundary condition:
Velocity of flow at plate
boundary must be zero

87. Boundary Layer Theory: Ludwig Prandtl
Boundary Layers concept - Ludwig Prandtl 1904.
Prior to this theoretic studies ignored viscous effects – i.e. based on Euler equation (1755).
Prandtl split the flow into 2 regions:
• Boundary Layer region close to
surfaces where viscous effects are
important.
• Region outside the boundary layer
where viscous effects are small.

88. Boundary Layers
Boundary Layer (BL): a layer of fluid near a boundary
We’ve talked a lot about viscous forces.
•Viscous forces are important in a BL.
•Outside BL viscous forces are less important.
•I.e., the forces of interest relate to significant
velocity gradients in the BL, not outside the BL.
Viscous effects are significant in this region
BL
dy
du
A
F
m
t =
=

89. Flow Entering a Pipe

90. Wake: Downstream Effect of Viscous Forces
Credit: NASA

92. Credit: Cpl Craig Barrett
© Commonwealth of Australia, Department of Defence
http://images.airforce.gov.au/fotoweb/archives/5010-Air%20Force%20Images/DefenceImagery/2017/
AIA17_022/20170302raaf8185068_0669.jpg.info#c=%2Ffotoweb%2Farchives%2F5010-Air%2520Force%2520Images%2F

94. CFD images by Professor Kunio Kuwahara, Institute of Space and Astronautical Science, Japan
(c) ISAS/JAXA, Courtesy of ISAS/JAXA
8.2 Boundary Layer Analysis

95. Boundary Layers
An understanding of boundary layers for both external and internal flows allows
certain important flow behaviours to be studied:
• Viscous Flow in Pipes:
o Pressure drop – losses
• Flow over immersed Bodies:
o Lift Drag forces
o Flow separation

96. Right at the surface the flow is brought to rest.
Consider a uniform flow approaching a semi-infinite flat plate
Immediately adjacent to the plate flow is slowed by viscosity.
No Slip Boundary Condition
u(y=0) = 0
y
x

97. As flow moves along the plate, the area affected by viscosity will increase.
The boundary layer varies with both input flow and distance along the boundary
Aims of this section:
1.Define the boundary layer thickness: δ
2.Define the boundary layer displacement thickness: δ*
Boundary
Layer

98. Boundary Layer Thickness: d
Velocity goes from zero at the boundary surface u(y=0) = 0, to the free-stream velocity, U¥ at the
edge of the BL.
Difficult to define the exact point where BL ends.
Define edge of BL y = δ as point where velocity is 99% of free-stream value, u(y = δ) = 0.99U¥
Initially flow inside BL is laminar (smooth)
Boundary
Layer
Flow inside the BL is
slowed by viscous
forces
U¥
y = δ
y
x

99. A boundary layer effectively blocks or displaces the flow compared with an inviscid flow.
There is less flow within the boundary layer than there would have been had there been no
boundary layer, i.e., there is less flow compared with an inviscid (no boundary layer) flow.
The effect of the boundary layer is the same as having a blockage in an inviscid flow.
One way to quantify the boundary layer is the height of this blockage.
Define blockage height is the boundary layer displacement thickness d* .
U¥
y
x
d
Reduced mass
flow rate

100. Model the viscous boundary layer as an inviscid flow with a blockage of height d*, where d*
is the boundary layer displacement thickness.
U¥
y
x
d
U
d u
viscous d*
d
U
inviscid

101. Model the viscous boundary layer as an inviscid flow with a blockage of height
d*. where d* is the boundary layer displacement thickness.
Determine d* using the fact that the same mass rate must occur in both cases.
To determine mass flow rate, we need the velocity profile:
• Inviscid case u(y) = U (constant)
• Viscous case u(y) = ? – use an approximate function
U
d u
viscous d*
d
U
inviscid

102. Consider the mass flow rate/unit width through viscous and “inviscid boundary layers:
m = rudA =
0
d
ò ru bdy
0
d
ò , where b is width
mviscous = minviscid
ru(y) bdy
0
d
ò = rU bdy
d*
d
ò
u(y) dy
0
d
ò =U(d - d*
)
Ud*
=Ud - u(y) dy
0
d
ò =U 1dy
0
d
ò - u(y) dy
0
d
ò
Þ d*
= 1-
u(y)
U
æ
è
ç
ö
ø
÷dy
0
d
ò
U
d u
viscous
d*
d
U
inviscid

103. Boundary Layer Velocity Profile - Laminar
In order to determine the boundary layer displacement thickness, we need to know u(y).
Some reasonable velocity profiles for a Laminar** BL:
The velocity varies smoothly from 0 at the surface to U at the edge of the boundary layer,
i.e., y = d.
These equations approximate the flow profile inside the BL.
Outside the BL, the velocity is U.
**Laminar flow is smooth, we will consider turbulent BL later.
u(y) =Usin
p
2
y
d
æ
è
ç
ö
ø
÷ or u(y) =U 2
y
d
æ
è
ç
ö
ø
÷ -
y
d
æ
è
ç
ö
ø
÷
2
æ
è
ç
ç
ö
ø
÷
÷

104. d* - Laminar Boundary Layer
Calculate the boundary layer displacement thickness using the first velocity profile:
If we use the parabolic approximation, d* = 0.33d
d*
= 1-
u(y)
U
æ
è
ç
ö
ø
÷dy
0
d
ò , let u(y) =Usin
p
2
y
d
æ
è
ç
ö
ø
÷
d*
= 1-sin
p
2
y
d
æ
è
ç
ö
ø
÷
æ
è
ç
ö
ø
÷dy
0
d
ò = y+
2d
p
cos
p
2
y
d
æ
è
ç
ö
ø
÷
é
ë
ê
ù
û
ú
y=0
y=d
= d +0
( )- 0+
2d
p
æ
è
ç
ö
ø
÷ =d 1-
2
p
æ
è
ç
ö
ø
÷
d*
= 0.36d

105. Recall the physical meaning of the boundary layer displacement thickness:
A viscous boundary layer with boundary layer thickness δ is equivalent to
having an inviscid boundary flow with a blockage δ* high.
Where:
U
d u
viscous d*
d
U
inviscid
d*
= 0.36d

106. d* - Turbulent Boundary Layer
The velocity profiles in turbulent flows are different.
After we have explored these differences, we will return to define δ* for
turbulent flows.

107. CFD images by Professor Kunio Kuwahara, Institute of Space and Astronautical Science, Japan
(c) ISAS/JAXA, Courtesy of ISAS/JAXA
8.3 Boundary Layer Example

108. Exam Question 2008 S2
Air at 15 oC forms a boundary layer near a solid wall. The boundary layer
displacement thickness δ* represents the amount that the thickness of the body
must be increased so that an idealized (non-physical) inviscid flow mimics the
actual viscous boundary layer flow.

109. Use the figure above to help you answer the following questions, where d is the
boundary layer thickness, u is the viscous velocity profile as a function of y, and
U is the velocity outside the boundary layer. The velocity distribution in the
boundary layer can be approximated by:
with U = 30 m/s and d = 1 cm. The density of air at 15 oC is 1.22 kg/m3 and the
kinematic viscosity is 1.46x10-5 m2/s.
u
U
=1-exp -
2y
d
æ
è
ç
ö
ø
÷

110. a) In order to calculate d* what property is assumed to be the same in both the
viscous and idealized inviscid flows?
Answer: mass flow rate.

111. b) Given that the mass flow rate per unit width for a non-uniform velocity profile is
calculated using the integral, , show that
m = rudA
0
d
ò d*
= 1-
u(y)
U
æ
è
ç
ö
ø
÷dy
0
d
ò
ò
ò
ò
ò
ò
-
=
-
=
-
=
=
=
d
d
d
d
d
d
d
d
d
d
d
d
r
r
r
r
0
*
0
*
*
0
0
'
'
)
(
)
(
)
(
)
(
)
(
*
dy
U
y
u
dy
y
u
U
U
U
U
dy
y
u
Udy
dy
y
u
m
m inviscid
viscous
)
(
1
)
(
1
0
*
0
0
*
ò
ò
ò
÷
ø
ö
ç
è
æ
-
=
-
=
d
d
d
d
d
dy
U
y
u
dy
U
y
u
dy

112. c) What is the viscous shear stress at the wall?
Pa
U
y
dy
du
y
U
dy
du
y
U
y
u
09
.
0
)
1
)(
/
2
(
)
0
(
)
/
(
2
exp
)
/
2
(
/
)
2
exp
1
(
)
(
=
=
=
=
÷
ø
ö
ç
è
æ
-
=
÷
ø
ö
ç
è
æ
-
-
=
d
m
t
m
t
d
d
d

113. d) What is the viscous shear stress at y = 0.5 cm?
e) What is the shear stress outside the boundary layer?
Answer: zero. Velocity uniform. No relative motion between “layers.”
Pa
U
y
dy
du
03
.
0
))
1
)(exp(
/
2
(
)
5
.
0
(
)
/
(
=
-
=
=
=
d
m
t
m
t

114. CFD images by Professor Kunio Kuwahara, Institute of Space and Astronautical Science, Japan
(c) ISAS/JAXA, Courtesy of ISAS/JAXA
8.4 Viscous Friction Drag in a Boundary Layer

115. How to determine the Drag generated by the boundary layer:
• Apply the momentum equation to flow moving through a boundary layer control volume.
• Select a CV that has a streamline as its upper surface.
• Therefore there is no flow through upper or lower surfaces.
• Flow on this streamline enters BL at left end of CV.
• Before we can apply the momentum equation to find the force, we need to determine h.
For this step, we will use the continuity equation.
d
U
u
h y
x

116. δ
U
u
h
mass flow rate entering the CV = mass flow rate leaving the CV
Consider a width of flow, b, (which is normal to the page).
min = rUbh
mout = rb u(y)dy
0
d
ò
Þ rUbh = rb u(y)dy
0
d
ò
We could obviously solve for h, but it will
be more convenient to leave our work in
this form.
y
x

117. δ
U
u
h
Now apply the momentum equation:
Let’s consider the forces.
The drag force, D, is the force acting on the fluid in the CV, as shown.
Assume that the pressure is approximately constant.
D is the only force:
y
x
Fx
å = Vx rV ×n̂dA
CS
ò
Fx
å = -D

118. d
U
u
h
D
y
x
Fx
å = Vx rV ×n̂dA
CS
ò dA = bdy and Fx = -D
å
therefore
-D = br Vx
( ) 2
dy
out
ò - br Vx
( ) 2
dy
in
ò
-D = br u(y)
[ ]
2
dy
y=0
y=d
ò -br U2
dy
y=0
y=h
ò
-D = br u(y)
[ ]
2
dy
y=0
y=d
ò -brhU2
Recall: V·n is
positive for flow out and
negative for flow in.

119. d
U
u
h
D
-D = br u(y)
[ ]
2
dy
y=0
y=d
ò -brhU2
Also from continuity:
Substitute into the momentum equation:
rUbh = rb u(y)dy
0
d
ò
y
x
[ ]
( )
ò
ò
ò
=
=
=
=
=
=
-
=
-
=
-
d
d
d
r
r
r
y
y
y
y
y
y
dy
y
u
U
y
u
b
D
terms
Collecting
dy
y
u
Ub
dy
y
u
b
D
0
0
0
2
)
(
)
(
)
(
)
(

120. d
U
u
h
D
D = rb u U -u
( )
0
d
ò dy
Typically we present this as a non-dimensional drag co-efficient:
CDf =
D
1
2 rU2
bl
Skin Friction Drag Coefficient
y
x

121. CFD images by Professor Kunio Kuwahara, Institute of Space and Astronautical Science, Japan
(c) ISAS/JAXA, Courtesy of ISAS/JAXA
8.5 Skin Friction Drag Coefficient and BL
Thickness – Laminar Flow

122. Using an equation to approximate the velocity profile u(y), we can solve for the
skin friction coefficient (both are estimates for laminar flow).
Approximate velocity profiles for Laminar Boundary Layers:
is the Reynolds number based on the distance along the plate.
This accounts for the fact that the boundary layer height (δ) grows with x.
u
U
= sin(
p
2
y
d
)
u
U
= 2(
y
d
) - (
y
d
)2
CDf =
1.31
Rex
CDf =
1.46
Rex
Rex =
rUx
m
D = rb u U -u
( )
0
d
ò dy CDf =
D
1
2 rU2
bl

123. Variation of Laminar Boundary Layer along Plate
The boundary Layer grows along the plate.
If we assume a velocity profile apply the momentum continuity equations, we can
solve for d as a function of Re.
The proof requires a differential version of the momentum equations (3rd year fluids).
So we’ll just state without proof:
is the Reynolds number based on the distance along the plate.
U¥
y
x
δ
x
x Re
5
=
d
Rex =
rUx
m

124. Some numbers….
Air flows at 1.5 ms -1 past a semi-infinite flat plate. The flow is laminar.
What is the boundary layer thickness at x = 0.4 m along the plate?
d
x
=
5
Rex
where Rex =
rUx
m
r(air) =1.23kg / m3
, dynamic viscosity m(air) =1.8´10-5
Nsm-1
, U =1.5ms-1
and x = 0.4
d = 0.4´ 5´
1.23´1.5´ 0.4
1.8´10-5
æ
è
ç
ö
ø
÷
-1
2
d =10mm
U¥
y
x
δ

125. Repeat for laminar water flow:
d
x
=
5
Rex
where Rex =
rUx
m
r(water) = 998kg / m3
, m(water) =1.2´10-3
Nsm-1
, U =1.5ms-1
and x = 0.4
d = 0.4´5´
9.8´1.5´0.4
1.2´10-3
æ
è
ç
ö
ø
÷
-1
2
d = 2.7mm
U¥
y
x
δ

126. CFD images by Professor Kunio Kuwahara, Institute of Space and Astronautical Science, Japan
(c) ISAS/JAXA, Courtesy of ISAS/JAXA
8.6 Laminar and Turbulent Flows

127. For flow along a pipe, the characteristic length is the diameter of the pipe – the length
that most affects the flow.
For flow over a flat plate, the characteristic length is the distance along the plate, as this
is the length that most affects the flow.
At a critical values of Re, there will be a transition from laminar to turbulent flow (more
about this later). This value is different for different types of flows
Rex =
rVx
m
=
inertia forces
viscous forces
Credit: Willem van Aken
CSIRO
http://www.scienceimage.csiro.au/library/equipment
/i/6380/section-of-the-perth-kalgoorlie-water-supply-
pipeline-near-merredin-wa-1976-/
U¥
y
x
δ

128. Turbulent Boundary Layers
http://fdrc.iit.edu/research/images/
Turbulent flow is unsteady, fluctuating, chaotic, AND
contains eddies.
Turbulent flow is time averaged. It can be described
statistically, e.g., with average properties.

129. Transition to Turbulence
Laminar
Transition
(laminar or
turbulent)
Turbulent
m
r
=
Vx
Rex
y
x
Factors affecting transition:
•The Upstream velocity, U ® Re
•Distance along plate, x ® Re
•The Roughness of the surface
•The Turbulence upstream
•The Fluid properties
•The presence or absence of pressure gradients

130. Transition to Turbulence
Laminar
Transition
Turbulent
y
x
Typically on a flat plate transition occurs around Re ≈ 5x105.
But depending on other factors (e.g. surface roughness) transition can occur
between Re ≈ 2x105 to Re ≈ 3x106.
Transition for flow in a circular pipe occurs about Re ≈ 2100 to 4000.
Pathline of particle moving
through boundary layer
m
r
=
Vx
Rex

131. Boundary Layer Velocity Profile
Laminar
Transition
Turbulent
y
x
Turbulent boundary layers have different velocity profiles than Laminar
boundary layers.
Therefore there are also different shear stress and skin friction drag within
such boundary layers.
m
r
=
Vx
Rex

132. Turbulent Boundary
Layer
δ
Laminar Boundary
Layer
δ
Laminar Flow Turbulent Flow

133. Boundary Layer Velocity Profile
Turbulent BL profiles are flatter near the surface.
Therefore, the velocity gradient at the surface is higher.
du
dy y=0
laminar
( )
du
dy y=0
turbulent
( )
Boundary layer profiles over a flat plate
Turbulent BL
δ
Laminar BL
δ

134. Laminar flow (parabolic profile)
V
Turbulent flow
V

135. Turbulent d, d*, Average CDf
The turbulent boundary layer thickness and
displacement thickness are calculated using
an approximate velocity profile.
One uses the same method as with laminar
boundary profiles.
But an experimental/empirical method is
needed to find the Average Skin Friction Drag
Coefficient since the approximate velocity
profile at the surface is difficult to model.

136. Approximate equations - Turbulent BL
1. Boundary Layer Thickness:
d
x
=
0.37
Rex
1 5
2. Displacement Thickness: d*
=
d
8
3. Average Drag Coefficient: CDf =
0.07
Rex
1 5
These are approximate results.
Exact values also depend on other parameters such as the surface roughness
of the plate.

137. Friction Drag Coefficient for a Flat Plate
Note the difference in CDf for
Laminar and Turbulent BLs, and
the effect of surface roughness ε.
Note x = l is the length of the plate.
l
l
l
Re
33
.
1
,
Re
f
laminar
Df,
turbulent
Df,
=
÷
ø
ö
ç
è
æ
=
C
C
e
l
/
e

138. CFD images by Professor Kunio Kuwahara, Institute of Space and Astronautical Science, Japan
(c) ISAS/JAXA, Courtesy of ISAS/JAXA
8.7 Effect of Pressure Gradient

139. We are going to consider:
•The effect of a pressure gradient.
•What happens when the free-stream velocity varies due to a pressure gradient
(Bernoulli equation)?
•Laminar turbulent boundary layers behave differently.
•Positive negative pressure gradients have different effects.

140. Pressure Gradients
In region 1 the free stream velocity increases (approx. as Bernoulli’s) ® pressure decreases
Velocity is
increasing
Pressure is
decreasing
Velocity is
decreasing,
Pressure is
increasing
Region 1 Region 2 Region 3
¶p
¶x
0.0
“Favourable”
pressure gradient
“Adverse”
pressure gradient.
¶p
¶x
0.0

141. Recall the equations for boundary layer growth:
laminar turbulent
Therefore as velocity increases in a favourable pressure gradient, Rex increases
and the rate of boundary layer growth decreases. Thus:
The boundary layer will grow more slowly in a favourable pressure gradient.
Similarly:
The boundary layer will grow more quickly in an adverse pressure gradient.
d
x
=
5
Rex
;
d
x
=
0.37
Rex
1 5
where, Rex =
rUx
m
Effect of Pressure Gradient on BL Growth
¶p
¶x
0.0
¶p
¶x
0.0

142. Effect of Pressure Gradient on BL Growth
Region 1:
The free stream velocity is increasing the pressure is decreasing.
The boundary layer thickness increases more slowly.
Region 3:
The free stream velocity is decreasing the pressure is increasing.
The boundary layer thickness increases more rapidly.
Flow can separate.
Region 1 Region 2 Region 3
¶p
¶x
0.0
“Favourable”
pressure gradient
“Adverse”
pressure gradient.
¶p
¶x
0.0

143. Flow Separation in an Adverse Pressure Gradient
The descriptions in the above boxes refer to the growth of the boundary layer.
Flow is less likely to separate in a favourable (dp/dx 0) pressure gradient.
Flow more likely to separate in an adverse (dp/dx 0) pressure gradient.
flow reversal
slowly more rapidly
¶p
¶x
0.0
¶p
¶x
0.0

144. Flow Separation in an Adverse Pressure Gradient
y
d
1.0
0.8
0.6
0.4
0.2
0.0
U
• Consider first a boundary layer flow (RH velocity profile).
• As the free-stream velocity decreases, this alters the
velocity profile as shown (LH velocity profiles).
• Eventually, flow reversal occurs near the surface – flow
separation.
Decreasing
free stream
velocity
When flow near the surface
reverses (negative U), we say
the flow has separated.

145. Flow Separation in an Adverse Pressure Gradient

146. Turbulent Boundary Layer Separation
U
y
d
1.0
0.8
0.6
0.4
0.2
0.0
Turbulent
Laminar
Consider difference in velocity profile for turbulent laminar boundary layers.
This affects flow separation.
Velocity of a turbulent BL near the surface
is much larger than that in a laminar BL.
Thus in a turbulent BL flow reversal near
the surface will be more difficult to achieve
as larger changes in velocity there will be
required.
Thus a turbulent BL is more resistant to
flow separation than a Laminar BL

147. For example, flow around a cylinder experiences regions of favourable and adverse pressure
gradients.
Velocity increases from A-C reaching a maximum at C.
Therefore the pressure drops from A-C, reaching a minimum at C.
The pressure gradient is favourable from A-C, ¶p
¶q
0.0

148. Velocity decreases from C-D.
Therefore pressure increases from C-D,
Pressure gradient is adverse from C-D.
The flow separates at D due to the adverse pressure gradient.
The separated flow generates a relatively constant pressure between D F.
¶p
¶q
0.0

149. The point at which separation occurs depends
upon the Reynold’s number.
An idealized inviscid flow does not separate.
Very low Re flows are close to inviscid.

150. Full recovery of
pressure
Partial
recovery
Less
complete
recovery
Pressure Distribution on a Cylinder
The onset of an adverse pressure gradient
does not necessarily occur at 90 degrees
– complex flow, feedback from wake.

151. Separation affects the pressure drag (drag due to pressure difference).
Pressure drag directly depends on the pressure recovery i.e. pfront - prear.
Pressure Distribution on a Cylinder
Inviscid flow has full pressure recovery –
pressure drag is zero.
Turbulent flow separates later therefore
has improved pressure recovery
compared to laminar flow.
For this particular geometry pressure
drag is less for turbulent flow.

152. CFD images by Professor Kunio Kuwahara, Institute of Space and Astronautical Science, Japan
(c) ISAS/JAXA, Courtesy of ISAS/JAXA
8.8 Why Turbulent Flow Can Be Good

153. Effect of Separation on Wake Width
Different boundary layers
Different flow separation points
Different wake widths
Different drag

154. If turbulence is induced in the flow before the separation point at D, the flow will remain
attached further.
Thus, separation will be delayed until further along the “trailing” surface of the cylinder.
This will lead to a narrower wake and less drag.
Turbulent BL
δ
Laminar BL
δ

155. Surface roughness changes the BL
In this case, it causes a transition from
laminar to turbulent flow.
Separation is delayed for turbulent BL
(relative to a laminar BL).
This leads to a narrower wake and
thus reduced drag.
Turbulent / Laminar Boundary Layer Separation

156. Wing Vortex Generators
Induce turbulence to delay flow
separation and reduce drag.
NASA
https://www.google.com.au/search?q=wing+slats+nasasource=
lnmstbm=ischsa=Xved=0ahUKEwiXkPmjwpbTAhUJHJQKHVw_
D5YQ_AUIBigBbiw=1360bih=767#tbm=ischq=wing+vortex+
generators+nasaimgrc=RaEsPrHs1a4M3M:spf=646

157. CFD images by Professor Kunio Kuwahara, Institute of Space and Astronautical Science, Japan
(c) ISAS/JAXA, Courtesy of ISAS/JAXA
8.9 Types of Drag

158. Pressure Drag
Consider flow around a flat plate normal to the flow.
Flow separation causes a very large pressure decrease.
Majority of Drag is due to Pressure Drag.
Positive
(relative)
pressure
Negative
(relative)
pressure
-
-
-
-
-
-
-
-
+
+
+
+
+
+
+
+

159. Skin Friction Drag
Consider flow around a flat plate in-line with the flow.
Flow over plate surface is slowed down by viscous forces.
This loss of momentum is described as being caused by “skin friction”.
Majority of Drag is due to Skin Friction.

160. Automobile Drag Coefficient
Car drag is a combination of pressure skin friction drag
More streamlined shapes = delayed separation = decreased pressure drag

161. Some Drag Coefficients
From Figure 9.28
Munson

162. Some Drag Coefficients

163. Reduction in Drag
Different types of Drag:
• Pressure drag (due to flow separation)
• Skin Friction Drag
• Wave drag
Need to understand which types of drag
dominate in order to determine how to reduce
the total drag.
If the drag is due to flow separation, we can
often gain by streamlining the object

164. But if skin friction were the dominant source of drag, this approach would not help.
CD = 1.2 CD = 0.12
If the drag is due to flow separation, we can often gain by streamlining the object.
Drag = CD
1
2 rV2
A
( )= CD
1
2 rV2
LD
( )

165. Drag depends on (relative) velocity.
If we reduce relative velocity, we reduce the drag
E.g., Drafting a truck (both with a road speed of V), truck wake generates an
average velocity of Vinduced
( )
A
V
C
Drag 2
2
1
D r
=
Induced air speed
Vtruck relative to
air = V
Vinduced Vcar relative to air
= V - Vinduced

166. Monash University
http://www.flair.monash.edu.au/media/
Monash University
http://www.monash.edu/news/articles/
riding-for-olympic-glory

167. Frontal area Coefficient of Drag
0.5 m2
0.35 m2
0.35 m2
0.45 m2
Drag depends on area
If the area is reduced, the drag is reduced.
( )
A
V
C
Drag 2
2
1
D r
=

168. Property Laminar Turbulent
Reynolds No. Lower Higher
Composition Smooth, layered flow
parallel to surface.
Eddies, mixing, momentum
transfer across BL.
Skin Friction Drag (tw) Lower Higher
Velocity Profile Refer to diagram Refer to diagram
Thickness Small Larger grows quickly
Stability Less stable, separates
easily
More stable, resists separation
Summary: Laminar v.Turbulent

182. Workshop: Boundary Layers
CHE2161 Mechanics of Fluids

183. H2O
0.5 ms-1
Horizontal flat plate
𝑅𝑒 =
𝜌𝑈𝑥
𝜇
The transition from laminar to turbulent is not sharp, but occurs over a range.
Specifically, this transition occurs between 2 × 105 and 3 × 106
The value you will find in the text is Recrit = 5 × 105
𝑥crit
laminar
turbulent

184. H2O
0.5 ms-1
Horizontal flat plate
𝑅𝑒 =
𝜌𝑈𝑥
𝜇
So, if Recrit is 5 × 105, and knowing the density, free stream velocity and dynamic viscosity, we can evaluate
the distance x from the leading edge
This is water, so:
ρ = 1000 kg m-3
μ = 1.2 × 10-3 Pa.s
Rearranging the equation, we have:
𝑥 =
𝑅𝑒 × 𝜇
𝜌𝑈
𝑥 =
5 × 105 × 1.2 × 10−3𝜇
1000 × 0.5
= 1.2 𝑚

185. To answer this we need to use one of the formulae for boundary layer thickness.
Remember, we are at the transition from laminar to turbulent at this point, so there is probably justification to use
either the laminar or turbulent boundary layer thickness. Let’s choose laminar.
𝛿
𝑥
=
5
𝑅𝑒𝑥
𝑅𝑒𝑥 =
𝜌𝑈𝑥
𝜇
=
1000 × 0.5 × 1.2
1.2 × 10−3
= 5 × 105
𝛿
𝑥
=
5
𝑅𝑒𝑥
=
5
5 × 105
= 7.1 × 10−3
𝛿
𝑥
= 7.1 × 10−3 → 𝛿 = 7.1 × 10−3 × 1.2 = 8.5 × 10−3𝑚 = 8.5 𝑚𝑚

186. U ms-1
We know that for laminar flow the boundary layer thickness is given by:
𝛿
𝑥
=
5
𝑅𝑒
→ δ =
5𝑥
𝑅𝑒
=
5𝑥
𝜌𝑈𝑥
𝜇
= 5𝑥
𝜇
𝜌𝑈𝑥
= 5
𝑥
𝑥
𝜇
𝜌𝑈
= 5
𝜇
𝜌𝑈
𝑥 = 𝐶 𝑥 where 𝐶 = 5
𝜇
𝜌𝑈

187. We are told in the question that the boundary layer thickness is 12 mm at a distance of 1.3 m from the leading edge.
𝛿 = C 𝑥
𝐶 =
𝛿
𝑥
=
1.2 × 10−3
1.3
= 0.0105

188. 𝛿 = C 𝑥
𝛿 = 0.0105 𝑥
𝑥 𝛿
0.2 m
2.0 m
20 m
4.7 mm
15 mm
47 mm

189. U = 20 ms-1
2 m
×
×
2 m 2 m
𝑥crit
Case I Case II
laminar
laminar
turbulent

190. Outside In order to avoid the impact of viscous effects on the velocity profile near the plate
Why?
To assess whether the flow is laminar or turbulent we need to determine the Reynolds number 2 m along the plate.
𝑅𝑒 =
𝜌𝑈𝑥
𝜇
𝑅𝑒 =
𝜌𝑈𝑥
𝜇
=
1.23 × 20 × 2.0
1.8 × 10−5
= 2.7 × 106
2.7 × 106
5 × 105 ∴ turbulent‼

191. 𝛿
𝑥
=
0.37
𝑅𝑒𝑥
0.2
𝛿
𝑥
=
0.37
𝑅𝑒𝑥
0.2 → 𝛿 =
0.37𝑥
𝑅𝑒𝑥
0.2 =
0.37 × 2
2.75 × 106 0.2
= 0.038 𝑚 = 3.8 𝑐𝑚 = 38 𝑚𝑚
What do you think? At 38 mm? Further out? 38 mm
U = 20 ms-1
2 m
×
×

192. 𝑅𝑒 =
𝜌𝑈𝑥
𝜇
The first question we have is which length to
consider as the characteristic length? What do
you think?
The distance along the plate in the direction of
the flow is the characteristic length in relation to
the formation of the boundary layer.
𝑅𝑒 =
𝜌𝑈𝑥
𝜇
=
1000 × 3 × 2
1.2 × 10−3
= 5 × 106 ≫ 5 × 105
∴ 𝑡ℎ𝑒 𝑓𝑙𝑜𝑤 𝑖𝑠 𝑡𝑢𝑟𝑏𝑢𝑙𝑒𝑛𝑡

193. We have already established that the flow is turbulent at the edge of the plate.
However, is the flow in the boundary layer turbulent or laminar a distance of 0.4 m from the leading edge? Why
might this be relevant?
We need to know this because whether the flow is laminar or turbulent in the boundary layer will dictate the
thickness of the boundary layer.
𝑅𝑒 =
𝜌𝑈𝑥
𝜇
=
1000 × 3 × 0.4
1.2 × 10−3
= 1 × 106 5 × 105
∴ 𝑡ℎ𝑒 𝑓𝑙𝑜𝑤 𝑖𝑠 𝑡𝑢𝑟𝑏𝑢𝑙𝑒𝑛𝑡
𝛿
𝑥
=
0.37
𝑅𝑒𝑥
0.2
𝛿
𝑥
=
0.37
𝑅𝑒𝑥
0.2 → 𝛿 =
0.37𝑥
𝑅𝑒𝑥
0.2 =
0.37 × 0.4
1 × 106 0.2
= 9.3 × 10−3 𝑚 = 9.3𝑚𝑚

194. ሶ
𝑚 = න 𝜌𝑢 𝑦 𝑑𝐴
ሶ
𝑚 = 𝜌𝑈 𝛿 − 𝛿∗
× b
𝛿∗
=
𝛿
8
𝛿∗
= 0.34𝛿
Remember our continuity equation
The issue here is that we do not know the function u(y) in the boundary layer. So can we approximate this somehow?
By offsetting the flow inside the boundary by some distance δ*, we can treat the flow as inviscid (i.e., having a uniform
velocity profile with the free stream velocity).
δ*
ሶ
𝑚
δ
The distance δ* can be evaluated using relatively simple expressions:
For turbulent flow:
For laminar flow:

195. Firstly, recall that the flow is turbulent. Therefore:
𝛿∗
=
𝛿
8
= 𝜌𝑈 𝛿 −
𝛿
8
× b
= 𝜌𝑈
7
8
𝛿 × b
ሶ
𝑚 = 𝜌𝑈 𝛿 − 𝛿∗ × b
= 1000 × 3 ×
7
8
× 0.009338 × 10
= 245 𝑘𝑔/𝑠

196. ሶ
𝑚 = 𝜌𝑈
7
8
𝛿 × b
𝛿
𝑥
=
0.37
𝑅𝑒𝑥
0.2 𝑅𝑒 =
𝜌𝑈𝑥
𝜇
What do we need to
remember about
Reynolds number?

197. ሶ
𝑚 = 𝜌𝑈
7
8
𝛿 × b
ሶ
𝑚 = 𝜌𝑈
7
8
𝛿 × b
𝛿 =
ሶ
𝑚
𝜌𝑈 ൗ
7
8 𝑏
𝛿 =
245
1000 × 6 × ൗ
7
8 × 10
= 4.667 × 10−3 𝑚
𝛿
𝑥
=
0.37
𝑅𝑒𝑥
0.2
𝛿 =
0.37𝑥
𝑅𝑒𝑥
0.2
𝑅𝑒 =
𝜌𝑈𝑥
𝜇
=
0.37𝑥
ൗ
𝜌𝑈𝑥
𝜇
0.2
=
0.37𝑥
൘
𝜌0.2𝑈0.2𝑥0.2
𝜇0.2
=
0.37𝑥0.8𝜇0.2
𝜌0.2𝑈0.2
𝛿 =
0.37𝑥0.8
𝜇0.2
𝜌0.2𝑈0.2
𝑥0.8 =
𝛿𝜌0.2𝑈0.2
0.37𝜇0.2
𝑥 =
𝛿𝜌0.2
𝑈0.2
0.37𝜇0.2
1
0.8
𝑥 =
4.667 × 10−3 × 10000.2× 60.2
0.37 × (1.2 × 10−3)0.2
1
0.8
𝑥 = 0.1999 𝑚 = 0.2 𝑚

198. 𝑅𝑒 =
𝜌𝑈𝑥
𝜇
=
1.23 × 1 × 4
1.8 × 10−5
= 2.7 × 105 5 × 105 ∴ 𝑡ℎ𝑒 𝑓𝑙𝑜𝑤 𝑖𝑠 𝑙𝑎𝑚𝑖𝑛𝑎𝑟

199. 𝛿
𝑥
=
5
𝑅𝑒𝑥
𝛿
𝑥
=
5
𝑅𝑒𝑥
=
5
2.7 × 105
= 9.6 × 10−3
𝛿
𝑥
= 9.6 × 10−3
→ 𝛿 = 9.6 × 10−3
× 4 = 3.8 × 10−2
𝑚 = 3.8 𝑐𝑚
We determined the Reynolds number on the previous slide: Re = 2.7 × 105

200. δ
u(y)
δ
δ*
U

201. Is there flow across the streamline A-B?
Is there flow through the plate? NO!
NO!
QA = volumetric flowrate through the plane A
QB = volumetric flowrate through the plane B
𝑄𝐴 = 𝑈 × 𝑦𝐴 × 𝑏
𝑄𝐵 = 𝑈 × 𝛿𝐵 − 𝛿𝐵
∗
× 𝑏
𝑄𝐴 = 𝑄𝐵

202. Recall that this is laminar flow. So what is the equation for δ*?
𝛿∗ = 0.34𝛿
𝛿𝐵
∗
= 0.34𝛿𝐵
= 0.038 − 0.0129 = 0.025 𝑚
𝑄𝐴 = 𝑈 × 𝑦𝐴 × 𝑏
𝑄𝐵 = 𝑈 × 𝛿𝐵 − 𝛿𝐵
∗
× 𝑏
𝑄𝐴 = 𝑄𝐵
𝑈 × 𝑦𝐴 × 𝑏 = 𝑈 × 𝛿𝐵 − 𝛿𝐵
∗
× 𝑏
= 0.34 × 0.038 𝑚 = 0.0129 𝑚
𝑦𝐴 = 𝛿𝐵 − 𝛿𝐵
∗
Now, to determine yA we need to equate the flow through the plane at A with the plane at B

203. At any point on the streamline, the flow through the plane bounded by the streamline and the surface is constant.
𝑄𝐴 = 𝑄𝑥
At any point x we can determine the volumetric flowrate by using δ* to approximate flow in the boundary layer by
inviscid flow. Therefore for the plane bounded by the streamline and plate surface we can write:
𝑄𝑥 = 𝑈 × (𝑦𝑥 − 𝛿𝑥
∗
) × 𝑏
We can equate this with our expression for QA 𝑄𝐴 = 𝑈 × 𝑦𝐴 × 𝑏
𝑈 × 𝑦𝐴 × 𝑏 = 𝑈 × (𝑦𝑥 − 𝛿𝑥
∗
) × 𝑏
𝑦𝐴 = 𝑦𝑥 − 𝛿𝑥
∗
𝑦𝑥 = 𝑦𝐴 + 𝛿𝑥
∗
Now we simply need to determine δx* as a function of x.

204. 𝑦𝑥 = 𝑦𝐴 + 𝛿𝑥
∗
Now we simply need to determine δx* as a function of x.
𝛿𝑥
𝑥
=
5
𝑅𝑒
→ δ𝑥 =
5𝑥
𝑅𝑒
=
5𝑥
𝜌𝑈𝑥
𝜇
= 5𝑥
𝜇
𝜌𝑈𝑥
= 5
𝑥
𝑥
𝜇
𝜌𝑈
= 5
𝜇
𝜌𝑈
𝑥
Recall that this is laminar flow, so we have the following relationship between δ and x:
Now, we also know that for a boundary layer incorporating laminar flow:
𝛿𝑥
∗ = 0.34𝛿𝑥
Combining these we have:
𝛿𝑥
∗
= 0.34 × 5
𝜇
𝜌𝑈
𝑥 = 0.34 × 5
1.8 × 10−5
1.23 × 1
𝑥 = 0.0065 𝑥
Therefore we have: 𝑦𝑥 = 𝑦𝐴 + 0.0065 𝑥
Since yA = 0.025 m (calculated in part f) our final equation is: 𝑦𝑥 = 0.025 + 0.0065 𝑥

205. The first assumption we need to make is that the fluid is… INCOMPRESSIBLE!
This means that Qin = Qx

206. 𝑄0 = 𝑈 × 𝐴
= 0.6 × 0.32
= 0.054
𝑚3
𝑠

207. 𝑄𝑥 = 𝑈 × 𝐴(𝑥)

208. δ
u(y)
δ
u(y)
U
Actual duct
δ*
δ
U
Using the displacement thickness
If the dimension of the duct is d × d m2 (where d is a function of x), then the effective area can be given by:
𝐴𝑥 = 𝑑𝑥 − 2𝛿𝑥
∗ 2

209. If we want to maintain a constant velocity, we need to flare out the duct so as to maintain the cross-sectional area
as a constant. (We are told this in the question).
If the area at the inlet is 0.3 × 0.3 m2, and we wish to keep effective area constant throughout the duct, then we will
need to compensate for the displacement thickness as we move further into the duct from the inlet.
𝑑𝑥 = 0.3 + 2𝛿𝑥
∗
As the displacement thickness is a function of the boundary layer thickness, and the boundary layer thickness is a
function of the distance from the leading edge of the surface, then the displacement thickness is also a function
of the distance from the inlet to the duct.

210. 𝛿𝑥
𝑥
=
5
𝑅𝑒
→ δ𝑥 =
5𝑥
𝑅𝑒
=
5𝑥
𝜌𝑈𝑥
𝜇
= 5𝑥
𝜇
𝜌𝑈𝑥
= 5
𝑥
𝑥
𝜇
𝜌𝑈
= 5
𝜇
𝜌𝑈
𝑥
Now, we also know that for a boundary layer incorporating laminar flow:
𝛿𝑥
∗
= 0.34𝛿𝑥
Combining these we have:
𝛿𝑥
∗
= 0.34 × 5
1.8 × 10−5
1.23 × 0.6
𝑥 = 0.0084 𝑥
First, recall that we are to assume laminar flow in the boundary layer.

211. 𝑑𝑥 = 0.3 + 2𝛿𝑥
∗
First, recall the expression we derived at (e)
Secondly, recall the expression we derived at (f)
𝛿𝑥
∗ = 0.0084 𝑥
Combining these, we obtain:
𝑑𝑥 = 0.3 + 0.0168 𝑥

212. © State of Victoria
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5.1 Unsteady Continuity – Example

213. These slides include copyrighted material from the course textbook:
B.R. Munson, T.H. Okiishi, W.W. Huebsch and A.P. Rothmayer. Fundamentals of
Fluid Mechanics 7th Edition SI Version, John Wiley Sons, Inc., 2013.
These slides are for the use of Monash University students registered in this
course and are not to be further distributed.
The original development of this slide presentation was done by Dr Josie
Carberry and professors who preceded her in this course.
2

214. Review: Continuity Equation
As mass is conserved (can’t be created or destroyed) then the time rate of change of the
system mass = 0. The resulting equation is called the Continuity Equation:
D
Dt
r d
sys
ò =
¶
¶t
r
CV
ò d+ rV ×n̂
CS
ò dA
Time rate of
change of the
mass of the
coincident
system
=
Time rate of change
of the mass of the
contents of the
coincident control
volume
+
Net rate of flow of
mass through the
control surface
¶
¶t
r
CV
ò d+ rV ×n̂
CS
ò dA = 0
3

215. Review: Steady Continuity Equation
Outlet Surfaces, positive
where is the average velocity
(scalar) across the outlet surfaces
rV ×n̂
CS
ò dA = mout
å - min
å = 0
m = rV ×n̂
A
ò dA
V ×n̂
m
å out
= rV ×n̂
A,out
ò dA
m
å out
= rVA
( )out
V
Inlet Surfaces, negative
where is the average velocity
(scalar) across the inlet surfaces
V ×n̂
( )in
in
in
A
in
A
V
m
dA
n
V
m
r
r
=
×
-
=
å
ò
å
r
,
ˆ
V
4

216. Example – Conservation of Mass - Unsteady Flow
(Example 5.5 Munson)
Construction workers in a trench (10ft long, 5ft wide, 8ft deep).
The trench is near an intersection and subject to CO2 influx
from the traffic at a rate of 10ft3/min.
CO2 is heavier than air and will sink to the bottom displacing the
air the workers need to breathe.
Assume negligible mixing between the air and CO2
a) What is the time rate of change of the depth of CO2 in the
trench in ft/min?
b) How long for the level of CO2 to fully engulf the workers (i.e.,
to a 6ft depth)?
5

217. a) Time rate of change of the depth of CO2: Perform an analysis of the total mass
(air + CO2) or the individual masses. Easiest to consider just the mass of CO2.
Must use the full unsteady continuity equation:
¶
¶t
r
CV
ò d+ rV ×n̂
CS
ò dA = 0
The mass flux, or mass flow rate in
is a constant determined by QCO2.
Mass flow rate out (of CO2) is zero.
accumulation of
mass in CV
¶
¶t
rCO2
CV
ò d+ mout, CO2
- min, CO2
= 0
¶
¶t
rCO2
CV
ò d = min, CO2
Also d = 50 ft2
( )dh
and m = Qr

218. Time to accumulate a depth of 6 ft:
¶h
¶t
= 0.2 ft / min
( )
b) How long for the level of CO2 to fully engulf the workers (6ft depth)?
th=6 ft =
6 ft
( )
0.2 ft / min
( )
= 30min
¶h
¶t
50 ft2
( )rCO2
= Qout, CO2
rCO2
¶h
¶t
=
Qout, CO2
50 ft2
( )
=
10 ft3
/ min
( )
50 ft2
( )
= 0.2 ft / min
( )
7

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5.2 Moving Non-Deforming CV

220. Sometimes it makes sense to analyze a problem using a moving control volume
W is the fluid velocity seen by an observer moving with the control volume.
V is the velocity seen by a stationary observer in a fixed coordinate system.
VCV is the velocity of the control volume relative to the fixed coordinate system.
V = W + VCV
9

221. Moving Control Volume Example
A stationary nozzle sends a jet of water at 30 m/s onto a curved blade. The blade has a
turning angle of 60o.
The blade is attached to a cart which is moving at 10 m/s to the right. The cross sectional
area of the jet of water is constant throughout, and it has a value of 0.003 m2.
30 m/s
10 m/s
60 0
What is the velocity of
the water as it leaves
the blade?
10

222. Draw a control volume that is attached to the blade and write all velocities relative to this.
30 m/s
10 m/s
60 0
Flow is steady – doesn’t vary with time.
Velocity of CV also steady – doesn’t vary with time.
out
in m
m
= rAVout = rAVin
11

223. Change velocities from relative to ground to relative to moving CV.
rAVout = rAVin
Vin = 30 - 10m/s
r A are constant
Vout = 20 m/s (relative to moving CV)
10 m/s
60 0
Vin =
Vjet - Vblade
Vout (relative to the blade)
30 m/s
Need to find velocity
relative to ground
12

224. Change velocities from relative to ground to relative to moving CV.
20 m/s Velocity of the
water relative to the
blade.
Velocity of the blade.
10 m/s
20 m/s
10 m/s
60 0
Velocity of the water at exit
that a stationary observer
would see
26.5 m/s
13

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5.3 Linear Momentum Equation

226. The linear momentum equation is a restatement of Newton’s Second Law using the
Reynolds Transport Theorem.
Newton’s Second Law:
Time rate of change of
the linear momentum
of the system
=
Sum of external
forces acting on the
system
rd
Momentum is mass time velocity, therefore the momentum of a small particle of mass
is . Therefore Newton’s 2nd using a CV approach law becomes:
Vrd
D
Dt
Vr d = Fsys
å
sys
ò
Sum of forces on the
coincident control volume
Use Reynolds transport theorem
with Bsys = momentum
(1)
15

227. Reynolds Transport Theorem for Momentum
Reynolds transport theorem tells us the rate at which some property “B” changes within a CV:
where b = B/mass
Now let:
Bsys = system momentum (mass x velocity)
Thus: b = mV/m = velocity
DBsys
Dt
=
¶
¶t
rb
CV
ò d+ rbV ×n̂
CS
ò dA
Bsys = Vr d
sys
ò
D
Dt
Vr
sys
ò d =
¶
¶t
rV
CV
ò d+ VrV ×n̂
CS
ò dA
Also when a control volume is coincident with a system at an instant of time:
Fsys
å = Fcontents of the CV
å
(2)
(3) 16

228. Combining (2) and (3) into Newton’s 2nd Law (1):
Gives the Linear Momentum Equation:
D
Dt
Vr d = Fsys
å
sys
ò
¶
¶t
rV
CV
ò d+ VrV ×n̂
CS
ò dA = Fcontents of the CV
å
This is a vector equation – 3 equations in 3D space, e.g. in Cartesian coordinates:
V = uˆ
i + vĵ + wk̂
¶
¶t
ru
CV
ò d+ urV ×n̂
CS
ò dA = Fx
å
¶
¶t
rv
CV
ò d+ vrV ×n̂
CS
ò dA = Fy
å
¶
¶t
rw
CV
ò d+ wrV ×n̂
CS
ò dA = Fz
å 17

229. Linear Momentum Equation
is a scalar (component of velocity normal to the CV). Conceptually you
can think of it as the velocity that transports the momentum across the
control surface.
The sign of depends upon whether flow is moving in or out of the CV at
that point
The sign of depends on the coordinate system you have chosen and the
component of velocity (u,v,w) you are considering.
Consider the signs of and separately
¶
¶t
rV
CV
ò d+ VrV ×n̂
CS
ò dA = Fcontents of the CV
å
V ×n̂
V ×n̂
V
V ×n̂ V
18

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5.4 Forces on CV

231. The forces on the control volume are body forces and surface forces.
Body forces – we will only consider gravitational forces.
Surface forces – we will consider shear, pressure and resultant forces (e.g., drag).
You should always draw a diagram of the forces.
Linear Momentum Equation
¶
¶t
rV
CV
ò d+ VrV ×n̂
CS
ò dA = Fcontents of the CV
å
20

232. Forces on the Control Volume
Always draw a CV diagram showing the flow and the forces.
21

233. 22

234. 23

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5.5 Forces on Fluid and Anchoring Forces

236. Example (Munson example 5.10)
A horizontal jet of water exits a nozzle with a uniform speed of V1 strikes a vane, and it is
turned through and angle θ. The area of the flow remains constant along the vane.
Determine the anchoring force needed to hold the vane stationary if gravity and viscous
effects are negligible.
25

237. X
Z
Fanchor x Fanchor, z
¶
¶t
ru
CV
ò d+ urV ×n̂
CS
ò dA = Fx
å
26
These equations gives the resultant force on whatever is in the CV.
So, let’s start by thinking about a CV with just the fluid in the CV. And let’s consider the x-direction for now.
1. The force on the LHS of the equation is the sum of all the forces on the CV. Since there are no net
pressure or other forces to take account of, this is equal to Fx, the net resultant force on the fluid. The fluid
comes in purely in the x-direction and turns vertically upwards loosing x-momentum. Therefore we would
expect to calculate that it experiences a net force in the negative x-direction, i.e. a force slowing its
component in the x-direction.
¶
¶t
rw
CV
ò d+ wrV ×n̂
CS
ò dA = Fz
å

238. X
Z
Fanchor x Fanchor, z
¶
¶t
ru
CV
ò d+ urV ×n̂
CS
ò dA = Fx
å
27
2. Newton's third law tells us that if the fluid experiences a force in the x direction, Fx , due to the vane
surface, then the vane will experience and equal and opposite force. In this case, the slowing of the
incoming jet will push the vane to the right.
When we talk about an object experiencing a drag force this is what we are discussing – the object
experiences a net force because it has altered the momentum of the flow around it. The direction of drag
force on an object can be additionally confusing because of the reference frame: here we have a
stationary vane in a moving flow; often when we talk about drag on a car or a bike, we have a moving
vehicle in quasi-stationary air.

239. X
Z
Fanchor x Fanchor, z
¶
¶t
ru
CV
ò d+ urV ×n̂
CS
ò dA = Fx
å
28
3. Going back to Newton, the anchoring force to hold the vane is equal but opposite to the force on the
vane from the fluid. Now we see that the anchoring force is actually equal in both magnitude and direction
to the force on the fluid in the CV.
4. So if we had drawn the CV to include the vane, the resultant force we calculate from the equation
would be that to anchor the vane. This is exactly the same as the resultant force on the fluid so it doesn't
matter which CV we chose.

240. These equations give the force on the CV, if have CV around just the fluid: Fon fluid = Fx
The force the fluid exerts on the vane, Ffluid on vane,x (also called drag force) is of equal
magnitude but in the opposite direction: Ffluid on vane,x = - Fx
The anchoring force to hold the vane in place is again equal but opposite to the fluid force on
the vane : Fanchoring force, x = + Fx
X
Z
Fanchor x Fanchor, z
29
¶
¶t
ru
CV
ò d+ urV ×n̂
CS
ò dA = Fx
å
¶
¶t
rw
CV
ò d+ wrV ×n̂
CS
ò dA = Fz
å

241. © State of Victoria
HTTP://TRANSPORTSAFETY.VIC.GOV.AU/MARITIME-SAFETY/RECREATIONAL-VESSEL-OPERATORS/PERSONAL-WATERCRAFT-PWCS
5.6 Steady Flow Momentum Equation Example 1

242. Steady Momentum Equation
¶
¶t
rV
CV
ò d+ VrV ×n̂
CS
ò dA = Fcontents of the CV
å
0 in steady flow
VrV ×n̂
CS
ò dA = Fcontents of the CV
å
31