Monash University Sunway campus is jointly owned by Monash University and the Jeffrey Cheah Foundation MEC 3451 Fluid Mechanics 2 School of Engineering Week 1 Semester 1, 2014
Dec 21, 2015
Monash University Sunway campus is jointly owned by Monash University and the Jeffrey Cheah Foundation
MEC 3451
Fluid Mechanics 2
School of Engineering
Week 1
Semester 1, 2014
Monash University Sunway campus is jointly owned by Monash University and the Jeffrey Cheah Foundation
2
Course Objectives
MEC 3451 Fluid Mechanics 2
o Derive the conservation equations governing fluid flows using finite control volume
and differential analysis
o Apply these governing equations to solve simple potential flow and viscous flow
problems
o Analyse internal flow, external flow and open channel flow problems
o Examine how boundary layers affect the behaviour of a fluid close to a surface
o Calculate lift and drag effects on a body
o Understand the concept of turbulence
o Appreciate compressibility effects in fluids and apply simple techniques to analyse
such flows
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Preliminaries
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MEC 3451 Fluid Mechanics 2
o Course delivery
• 33 Lectures (3 X 50 minute lectures per week)
• Practice Classes (1 X 3 hour session per week:
Check Allocate)
o Assessment
• Terminal examination (3 hours, 70%)
• Tests (30%)
o Course Text
• Munson, Young, and Okiishi, Fundamendal of
Fluid Mechanics, 6th Edition, John Wiley and
Sons
o Extra Help
• Moodle
• Tutors
• Consultation Hours
• Wednesday: 3PM – 5PM
• Thursday: 11AM – 12PM
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Lecture Topics
MEC 3451 Fluid Mechanics 2
Topic Textbook
(Munson et. al.)
Fluid Kinematics Chapter 4
Finite Control Volume Analysis Chapter 5
Differential Analysis Chapter 6
Similitude Chapter 7
Internal Flows Chapter 8
External Flows Chapter 9
Open Channel Flow Chapter 10
Compressible Flows Chapter 11
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Tutorials (Tutors: Mr. Ang & Mr. Ooi)
MEC 3451 Fluid Mechanics 2
o Practice classes (tutorials) held weekly starting Week 2
• Attendance at practice classes is compulsory
Test 1 (Assessment Task 1, 10%) o Approximately 1.5 to 2 hours
o 25th March, Tuesday (Week 4) • Materials covered in Week 1 to Week 3
• Approximately 1/3 of the marks on problems from tutorial sheets
• Approximately 1/10 of the marks on self-study topics
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MEC 3451 Fluid Mechanics 2
Test 2 (Assessment Task 2, 10%)
Test 3 (Assessment Task 3, 10%)
o Approximately 1.5 to 2 hours
o 15th April, Tuesday (Week 7) • Materials covered in Week 4 to Week 6
• Approximately 1/3 of the marks on problems from tutorial sheets
• Approximately 1/10 of the marks on self-study topics
o Approximately 1.5 to 2 hours
o 14th May, Wednesday (Week 10) • Materials covered in Week 7 to Week 9
• Approximately 1/3 of the marks on problems from tutorial sheets
• Approximately 1/10 of the marks on self-study topics
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Introduction:
Fluid Mechanics
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MEC 3451 Fluid Mechanics 2
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Fluid Mechanics
MEC 3451 Fluid Mechanics 2
o Study of the behaviour of fluids when
subject to applies forces
o Two subcategories
• Fluid statics: Behaviour of fluids at rest
• Fluid dynamics: Behaviour of fluids in motion
o Why study fluid mechanics?
• Fluids everywhere
✴ Everyday phenomenon
✴ Environmental flows
✴ Biological flows
✴ Medical devices
✴ Aerodynamics
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MEC 3451 Fluid Mechanics 2
o What is a fluid?
• Substance which continuously deform
(strained) when subject to a shear stress
• Solids, although deforming initially, do not
do so continuously
o Generally consists of liquids and
gases
• A liquid takes the shape of
the container it is in and
forms a free surface in the
presence of gravity
• Liquid is difficult to
compress
• A gas expands until
it encounters the walls
of the container and
fills the entire
available space
• Gases cannot form a
free surface
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Fluid Properties
MEC 3451 Fluid Mechanics 2
o Different fluids flow differently • This is because different fluids have different characteristics (for example
water, oil, honey, tar, air)
o Quantification of these fluids therefore requires the definition of
fluid properties • Density, specific volume, specific gravity
• Bulk modulus of compression
• Vapour pressure
• Surface tension
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Density, Specific Volume, Specific Gravity
MEC 3451 Fluid Mechanics 2
o Density • Mass 𝑚 per unit volume ∀
𝜌 ≡𝑚
∀
o Specific volume • Volume ∀ per unit mass 𝑚
∀ ≡∀
𝑚=1
𝜌
o Specific gravity (relative density) • Density relative to density of water at 4℃
SG ≡𝜌
𝜌H2O,40C
o Specific weight o Weight per unit volume
𝛾 ≡ 𝜌𝑔 To measure specific
gravity
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Viscosity
MEC 3451 Fluid Mechanics 2
o Recall definition of a fluid
• Substance which continuously deforms when
subject to a shear (tangential stress)
• Introduce concept of viscosity to describe the
‘fluidity’ of a fluid, i.e., how easily it flows
o Shear stress (force 𝐹 applied tangentially
to area 𝐴)
𝜏 =𝐹
𝐴∝𝑑𝑢
𝑑𝑦= 𝜇
𝑑𝑢
𝑑𝑦
(Shear Stress) = (Dynamic Viscosity) X (Rate of Strain)
• Constant of proportionality is the dynamic (or
absolute) viscosity
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MEC 3451 Fluid Mechanics 2
o Measure of a fluid’s resistance to deformation and hence flow
o Acts like friction between layers of fluid when they are forced to
move relative to each other
o Determine from slope of shear stress vs strain rate (deformation rate
or velocity gradient 𝑑𝑢 𝑑𝑦 )
• Linear for most common fluids (Newtonian)
• Non-Newtonian flows deal with deviations from linearity
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Hydrostatics
MEC 3451 Fluid Mechanics 2
o Pressure
• When fluid is at rest, the shear (tangential) stress is zero
• The only stress acting on the fluid is the pressure (force per unit area
acting normal to a surface)
• Scalar field
Pressure is the same at all points on a horizontal plane in a given fluid.
The length or the cross-sectional area of the tube has no effect on the height of the fluid column of a barometer.
𝑝𝐴1 = 𝑝𝐴2 = 𝑝𝐴3
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Hydrostatics
MEC 3451 Fluid Mechanics 2
o Variation with depth
• Newton’s Second Law
𝛿𝐹𝑧 = 𝜌𝑔 𝛿𝑧 𝐴 = 𝑝𝐴 − 𝑝 +𝑑𝑝
𝑑𝑧𝛿𝑧 𝐴
𝑑𝑝
𝑑𝑧= −𝜌𝑔
• Negative sign: pressure increases with depth
• Integrate between two elevations 𝑧1 and 𝑧2 to get
𝑝 = 𝑝0 + 𝜌𝑔ℎ
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Spatial Flow Field Variation (1, 2, 3-D Flow)
MEC 3451 Fluid Mechanics 2
o Flow can be exceedingly complex • A flow is either one-, two-, or three-dimensional depending on the number of
spatial components (𝑢𝑥 , 𝑢𝑦, 𝑢𝑧) in the velocity vector
𝑢 𝑥, 𝑦, 𝑧 = 𝑢𝑥𝐢 + 𝑢𝑦𝐣 + 𝑢𝑧𝐤
o However, it is possible to simplify the flow analysis • Flow between two flat plates
o For wide plates, negligible variation in the 𝑧 −direction
o Thin gaps, vertical velocity component 𝑢𝑦 is negligible
o Only one velocity component, i.e., 𝑢 = 𝑢(𝑦) needs to be considered
o Flow is one dimensional (although 𝑢𝑥 is a function of 𝑦)
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MEC 3451 Fluid Mechanics 2
o Flow over an infinitely long cylinder (into the plane) • Negligible variation in the 𝑧 −direction
• Both 𝑢𝑥 and 𝑢𝑦 important since flow circumnavigates cylinder
• Flow is two-dimensional
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Spatial Flow Field Variation (Uniform/Non-uniform Flow)
MEC 3451 Fluid Mechanics 2
o Uniform flow o A flow is uniform if the velocity does not vary along a streamline
• Flow between two plates
o An example of non-uniform flow is the flow over an aerofoil • The fluid accelerates on streamlines over the aerofoil and decelerates on
streamlines under the aerofoil to main flow conservation
• The velocity along a given streamline is therefore not constant
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Temporal Flow Field Variation (Steady/Unsteady Flow)
MEC 3451 Fluid Mechanics 2
o Steady flow • Flow invariant in time • All fluid properties at any spatial position in the flow do
not change with time 𝑑 𝑑𝑡 = 0 • Example of unsteady flow (𝑑 𝑑𝑡 ≠ 0)
o Flow over an oscillating plate
o Flow through a diffuser channel with moving walls
Steady flow
Unsteady flow
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Laminar and Turbulent Flow
MEC 3451 Fluid Mechanics 2
o Laminar flow • Flow is regular and highly ordered
o Each fluid layer moves smoothly and steadily with respect to the other layers (laminae) adjacent to it
o Deterministic system o Usually occur in viscous fluids where the velocity is low
o Turbulent flow • Flow is random and highly disordered
o Irregular and unsteady – characterised by velocity fluctuations o Chaotic movements of part of liquid in different directions superimposed on
main flow direction o All fluid properties at any spatial position in the flow field do not change with
time o System no longer deterministic
o Can only be described in term of statistical averages o Usually occurs in high velocity inviscid fluids
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Laminar and Turbulent Flow
MEC 3451 Fluid Mechanics 2
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Laminar and Turbulent Flow
MEC 3451 Fluid Mechanics 2
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Conservation Laws
MEC 3451 Fluid Mechanics 2
o Mass conservation (continuity equation)
o Mass cannot be created or destroyed
o Steady flow: 𝜕 𝜕𝑡 = 0
𝑚 in = 𝑚 out
𝜌𝑉in𝐴in = 𝜌𝑉out𝐴out
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Conservation Laws
MEC 3451 Fluid Mechanics 2
• Momentum conservation flow • Newton’s Second Law
Time rate of change of linear
momentum of a system
Sum of external forces
acting on the system =
𝑚 𝑉
out
− 𝑚 𝑉
in
= 𝐹sys
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Control Volume Analysis (Review):
Reynolds Transport Theorem
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Osborne Reynolds
(1842-1912)
MEC 3451 Fluid Mechanics 2
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MEC 3451 Fluid Mechanics 2
o Difficult to identify a fluid mass and track this for all times (Lagrangian description)
o Moreover, often not interested in a particular mass of fluid but rather the effect of
a flow in a structure or device
o Thus, helpful to formulate the fundamental equations of fluid flow for a finite
spatial region (geometric identity independent of mass), i.e., the control volume
(Eulerian description)
o Equations developed will be expressed in integral form
• Volume integrals are a convenient way to capture spatial variations in the fluid
properties
• These are related by the Reynolds Transport Theorem
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CV is arbitrarily chosen, however, selection of CV can either simplify or
complicate analysis.
– Clearly define all boundaries. Analysis is often simplified if control surface CS
is normal to flow direction.
– Clearly identify forces and torques of interest acting on the CV and CS.
– Clearly identify all fluxes crossing the CS.
Choosing a Control Volume
MEC 3451 Fluid Mechanics 2
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Let 𝐵 represents fluid parameters (mass, momentum, acceleration…..)
Let 𝑏 represents the amount of 𝐵 per unit mass (𝑚)
𝑏 =𝐵
𝑚
𝐵 = 𝑚 𝑏
Extensive Property (𝑩) Intensive Property (𝒃)
𝑚 1
1
2𝑚𝑉2
1
2𝑉2
Extensive and Intensive Property
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RTT - Physical Interpretation
The purpose of Reynolds transport theorem is to provide a link between control volume
ideas and system ideas.
A physical understanding of the concepts involved will show that it is a straightforward,
easy-to-use tool.
𝑑𝐵sys
𝑑𝑡 =
𝑑
𝑑𝑡 𝑏 𝜌 𝑑∀ + 𝑏 𝜌 𝐕 ∙ 𝐧 𝑑𝐴
CSCV
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𝑑𝐵sys
𝑑𝑡 =
𝑑
𝑑𝑡 𝑏 𝜌 𝑑∀ + 𝜌 𝑏 𝐕 ∙ 𝐧
CSCV
𝐴
• The time rate of change of an arbitrary extensive parameter of a system
• This may represent the rate of change of mass, momentum, energy, or
angular momentum of the system, depending on the choice of the
parameter 𝐵.
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𝑑𝐵sys
𝑑𝑡 =
𝑑
𝑑𝑡 𝑏 𝜌 𝑑∀ + 𝜌 𝑏 𝐕 ∙ 𝐧
CS
𝐴CV
This term represents the rate of change of 𝐵 within the control volume as the
fluid flow through it.
Because the system is moving and the
control volume is stationary, the time rate
of change of the amount of 𝐵 within the
control volume is not necessarily equal to
that of the system.
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𝑑𝐵sys
𝑑𝑡 =
𝑑
𝑑𝑡 𝑏 𝜌 𝑑∀ + 𝜌 𝑏 𝐕 ∙ 𝐧 𝐴
CSCV
This term represents
the net flowrate of the
parameter 𝐵 across the
entire control surface.
Over this portion of the control surface this
property is being carried out of the control
volume (𝑉 ∙ 𝑛 > 0)
Over this portion of the control surface
this property is being carried into the
control volume (𝑉 ∙ 𝑛 < 0)
Over the remainder of the control
surface there is no transport of 𝐵
across the surface since 𝑏𝑉 ∙ 𝑛 = 0,
because either 𝑉 = 0 or 𝑉 is parallel
to the surface at those locations.
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Reynolds Transport Theorem
Linear
Momentum
Equation
Moment (angular)
Momentum Equation
Continuity
Equation
The Energy Equation
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The general form of the continuity equation (conservation of mass) is obtained by
substituting the properties for mass into the Reynolds transport theorem
𝑑𝐵sys
𝑑𝑡 =
𝑑
𝑑𝑡 𝑏 𝜌 𝑑∀ + 𝑏 𝜌 𝐕 ∙ 𝐧 𝑑A
CSCV
Let 𝐵sys = 𝑚sys and 𝑏 = 𝑚sys/𝑚sys = 1 , resulting in
𝑑𝑚sys
𝑑𝑡=
𝑑
𝑑𝑡 𝜌 𝑑∀ + 𝜌 𝐕 ∙ 𝐧 𝑑A
CSCV
Control Volume Analysis:
Conservation of mass
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𝑑𝑚sys
𝑑𝑡=
𝑑
𝑑𝑡 𝜌 𝑑∀ + 𝜌 𝐕 ∙ 𝐧 𝑑A
CSCV
However, conservation of mass
𝑑𝑚sys
𝑑𝑡= 0
so the general, or integral form of the continuity equation is
𝑑
𝑑𝑡 𝜌 𝑑∀ + 𝜌 𝐕 ∙ 𝐧 𝑑A = 0
CSCV
This equation can be expressed in words as
The accumulation rateof mass in thecontrol volume
+The net flowrateof mass throughthe control surface
= 0
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For steady flow, the total amount of mass contained in CV is constant, that is,
total amount of mass entering must be equal to total amount of mass leaving,
𝑑
𝑑𝑡𝑚CV + 𝑚 𝑜 − 𝑚 𝑖
CSCS
= 0
For single-stream steady-flow systems,
𝑚 = 𝑚
outin
𝜌𝑖𝐴𝑖𝑉𝑖 in = 𝜌𝑖𝐴𝑖𝑉𝑖 out
𝑖𝑖
Conservation of mass:
Steady flow processes (𝝏 𝝏𝒕 → 𝟎 )
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For incompressible flows (𝜌 = constant),
The 𝑄𝑖 = 𝑉𝑖𝐴𝑖 is called the volume flow passing through the given cross
section. The volume flow 𝑄 = 𝑉𝐴 will have units of cubic meters per
second (m3/s).
Conservation of mass:
Incompressible flows (𝝆 = 𝐜𝐨𝐧𝐬𝐭𝐚𝐧𝐭)
𝐴𝑖𝑉𝑖 in = 𝐴𝑖𝑉𝑖 out
𝑖𝑖
𝑄in = 𝑄out𝑖𝑖
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Control Volume Analysis:
Linear Momentum Equation
𝑑𝐵sys
𝑑𝑡 =
𝑑
𝑑𝑡 𝑏 𝜌 𝑑∀ + 𝑏 𝜌 𝐕 ∙ 𝐧 𝑑𝐴
CSCV
Rate of changeof property 𝐵of system
=Rate of change of property 𝐵
in control volume
+Net outflowof property 𝐵
through control surface
The extensive property 𝐵𝑠𝑦𝑠 becomes the momentum of the system:
𝐵𝑠𝑦𝑠 = 𝑀𝑠𝑦𝑠
The intensive property of mass 𝑚 in the system is 𝑚𝑉, and so
𝑏 =𝑚𝑉
𝑚= 𝑉
MEC 3451 Fluid Mechanics 2
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𝑑 𝑀sys
𝑑𝑡=
𝑑
𝑑𝑡 𝑉𝜌 𝑑∀ + 𝑉𝜌 𝑉 ∙ 𝑛 𝑑𝐴
CSCV
Newton’s second law for a system of mass 𝑚 subjected to a force 𝐹 is expressed
as
𝐹 = 𝑚𝑎
𝐹 =𝑑 𝑚𝑉
𝑑𝑡
The law can also be formulated for a system composed of a group of particles.
𝐹 =𝑑 𝑀sys
𝑑𝑡
The 𝑀sys denotes the total momentum of all mass comprising the system.
MEC 3451 Fluid Mechanics 2
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Forces acting on CV consist of body forces that act throughout the entire body of
the CV (such as gravity, electric, and magnetic forces) and surface forces that
act on the control surface (such as pressure and viscous forces, and reaction
forces at points of contact).
Forces Acting on a CV
Body forces act on each volumetric
portion 𝑑𝑉 of the CV
Surface forces act on each portion
𝑑𝐴 of the CS
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The most common body force is
gravity, which exerts a downward
force on every differential element of
the CV
Total body force acting on CV
𝐹 𝑏𝑜𝑑𝑦 = 𝜌𝑔 𝑑∀ CV
= 𝑚CV𝑔
Body Force
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A surface force is defined as a force that requires physical contact, meaning that
the surface forces act at the control surface.
For example, 𝑝1𝐴1 acts at the control surface and requires contact between the
fluid outside the control volume and the fluid inside the control volume.
In addition to pressure, surface forces can be caused by shear stress, for
example the force 𝜏𝐴
Surface
Force
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If the flow crossing the control surface occurs through a series of inlet and outlet
ports, then:
𝐹 = 𝐹 S + 𝐹 B =𝑑
𝑑𝑡 𝑉𝜌 𝑑∀ + CV
𝑚 𝑜𝑉𝑜CS
− 𝑚 𝑖CS
𝑉𝑖
where the subscripts 𝑜 and 𝑖 refer to the outlet and inlet ports, respectively.
𝑑 𝑀sys
𝑑𝑡= 𝐹 = 𝐹 S + 𝐹 B =
𝑑
𝑑𝑡 𝑉𝜌 𝑑∀ + 𝑉𝜌 𝑉 ∙ 𝑛 𝑑𝐴
CSCV
Time rate of change of
momentum in control
volume
Net outflow rate of
momentum through
control surface
Sum of forces acting on
the matter in control
volume
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The first term on the right hand side of the equation represent the momentum
accumulation term. This term is zero when the momentum in each differential
volume is constant with time, that is, steady flow.
𝐹 = 𝐹 S + 𝐹 B =𝑑
𝑑𝑡 𝑉𝜌 𝑑∀ + CV
𝑚 𝑜𝑉𝑜CS
− 𝑚 𝑖CS
𝑉𝑖
Steady flow
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It is important to know that the momentum equation is a vector equation
(there is a direction associated with each term).
The 𝑚 (scalar) is the rate at which mass is passing across the control surface,
and 𝑉 (vector) is velocity evaluated at the control surface.
𝐹 = 𝐹 S + 𝐹 B =𝑑
𝑑𝑡 𝑉𝜌 𝑑∀ + CV
𝑚 𝑜𝑉𝑜CS
− 𝑚 𝑖CS
𝑉𝑖
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