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8/18/2019 control volume and reynold transport
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Lecture 3
Fluid Kinematics: Velocity field,
Acceleration, Reynolds Transport
Theorem and its application
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Aims
• Describing fluid flow as a field• How the flowing fluid interacts with the
environment (forces and energy)
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Lecture plan
• Describing flow with the fields: Eulerian vs.
Lagrangian description.
• Flow analysis: Streamlines, Streaklines, Pathlines.
• How to perform calculations in the field description:
the Material Derivative• Reynold’s Transport Theorem
• Application of Reynolds transport theorem:
Continuity, Momentum and Energy conservation
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Velocity field
u
v
dx
dy=
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Velocity field: example
The calculated velocity field in a shipping channel is shown as the tide comes in
and goes out. The fluid speed is given by the length and color of the arrows. The
instantaneous flow direction is indicated by the direction that the velocity arrows
point.
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Velocity field is given by:
))(/( 0 j yi xlvv
−=
• Sketch the field in the first quadrant• find where velocity will be equal to v 0
Velocity field representation
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Eulerian and Lagrangian flow description
• Eulerian method – field concept is used, flow parameters (T,P, v etc.) are measured in every point in space vs. time
• Lagrangian method – an individual fluid particle is followed,parameters associated with this particle are followed in time
Example: Smoke coming
out of a chimney
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1D, 2D and 3D flow
Flow visualization of the complex
three-dimensional flow past a
model airfoil
The flow generated by an airplane is made visible
by flying a model Airbus airplane through two
plumes of smoke. The complex, unsteady, three-
dimensional swirling motion generated at the wingtips (called trailing vorticies) is clearly visible
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Flow types
• Steady flow – the velocity at any given point in space doesn’t vary withtime. Otherwise the flow is called unsteady
• Laminar flow – fluid particles follow well defined pathlines at any moment
in time, in turbulent flow pathlines are not defined.
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Const xy =
Streamlines
Velocity field is given by:
))(/( 0 j yi xlvv
−=
•draw the streamlines and
find there equation
u
v
dx
dy =
;
ln ln ; /
dy v y
dx u x
y x C y C x
= = −′= + =
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Streamlines, Streaklines, Pathlines• Streamline – line that
everywhere tangent tovelocity field
• Streakline – all particlesthat passed through acommon point
• Pathline – line traced by agiven particle as it flows
streamlines
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Imaging flow in a microfluidic channel
• Flow is seeded with fluorescent particles and imaged…(Project 5th semester Fall 2006)
Flow through a loosely packedmicrospheres bed Flow at a channel turn.Flow is disturbed by a microwire
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Example
Water flowing from an oscillating slit:
jviv yt uv
000 ))/(sin( +−= ω
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Material derivative
• Particle velocity
),( t r V A A
• Particle acceleration
t
z
z
V
t
y
y
V
t
x
x
V
t
V
dt
dV t r a A A A A A A A A A A
∂
∂
∂
∂+
∂
∂
∂
∂+
∂
∂
∂
∂+
∂
∂==),(
Dt
D
zw
yv
xu
t t r
VVVVV=
∂
∂+
∂
∂+
∂
∂+
∂
∂=),(a
• Material derivative )V ∇•+∂
∂=∂
∂+∂
∂+∂
∂+∂
∂= (t zw yv xut Dt
D
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Material derivative
• is the rate of changes for a given variable with time for agiven particle of fluid.
)V ∇•+
∂
∂=
∂
∂+
∂
∂+
∂
∂+
∂
∂= (
t z
w
y
v
x
u
t Dt
D
Unsteadiness of the flow(local acceleration)
Convective effect(convective acceleration)
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Example: acceleration
Velocity field is given by: ))(/( 0 j yi xlvv
−=
Find the acceleration and draw it on scheme
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Example: acceleration
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Control volume and system representation
• System – specific identifiable quantity of matter, that might
interact with the surrounding but always contains the same
mass
• Control volume – geometrical entity, a volume in space
through which fluid may flow
Governing laws of fluid motion are stated in terms in system, butcontrol volume approach is essential for practical applications
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Control volume and system representation
• Extensive property: B=mb (e.g. m (b=1), mv (b=v), mv2/2 etc)
intensive property
∫=
syssys bd B
V
ρ
∫
∫
∂
∂=
∂
∂∂
∂=
∂
∂
cv
cv
sys
sys
bd t t
B
bd
t t
B
V
V
ρ
ρ
0;0 <∂
∂=
∂
∂=
∂
∂=
∂
∂∫∫cv
cv
sys
sysd
t t
Bd
t t
BV V ρ ρ
Control volume: example
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The Reynolds Transport theorem (simplified)
II I CV SYS t t
CV SYS t
+−=+
=
)(:
:
δ
)()()()(
)()(
t t Bt t Bt t Bt t B
t Bt B
II I cvsys
cvsys
δ δ δ δ +++−+=+
=
t
t t B
t
t t B
t
t Bt t B
t
t Bt t B II I cvcv
syssys
δ
δ
δ
δ
δ
δ
δ
δ )()()()()()( ++
+−
−+=
−+
outflowinflow
Let’s consider an extensive property B:
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• For fixed control volume with one inlet, one outlet, velocity
normal to inlet/outlet
out incvsys B Bt
B Dt
DB +−∂
∂=
22221111 bV AbV A
t
B
Dt
DBcvsys ρ ρ +−
∂
∂=
• Can be easily generalized:
Th R ld T t th
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The Reynolds Transport theorem
(for fixed nondeforming volume)
∫ ⋅=out CV
out dAb B nV ρ
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∫ ⋅+∂∂
=CV
CV Sys dAbt
B
Dt
DBnV
ρ
General Reynoldstransport theorem forfixed control volume
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Application of Reynolds Transport Theorem
• We will apply it now to various properties:
– mass (continuity equation)
– momentum (Newton 2nd law)
– energy
∫ ⋅+∂∂
=CV
CV Sys dAbt
B
Dt
DBnV
ρ
C ti f
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Conservation of mass
• The amount of mass in the system should be conserved:
0=⋅+
∂
∂==
∫∫∫ CV CV sysSys
dAd t d Dt
D
Dt
DM
nV
ρ ρ ρ V V
Continuity equation
m Q AV ρ ρ = =
Mass flow rate through a section of control surface having area A:
Volume flow rate
A
V n d A
V A
ρ
ρ
⋅
=
∫
Average velocity:
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• For incompressible flow, the volume flowrate into a controlvolume equals the volume flowrate out of it.
• The overflow drain holes in a sink must be large enough toaccommodate the flowrate from the faucet if the drain hole atthe bottom of the sink is closed. Since the elevation head forthe flow through the overflow drain is not large, the velocitythere is relatively small. Thus, the area of the overflow drainholes must be larger than the faucet outlet area
E l
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Example
Incompressible laminar flow develops in a straight pipe of radius R. At section 1
velocity profile is uniform, at section 2 profile is axisymmetric and parabolic withmaximum value umax. Find relation between U and umax, what is average velocity
at section (2)?
2
1 1 2
2
1 max
0
max
0
1 2 0
2
A
R
AU V n dA
r AU u rdr
R
u U
ρ ρ
π
− + ⋅ =
⎡ ⎤⎛ ⎞− + − =⎢ ⎥⎜ ⎟⎝ ⎠⎢ ⎥⎣ ⎦
=
∫
∫
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A jet of fluid deflected by anobject puts a force on the object.This force is the result of thechange of momentum of the fluid
and can happen even though thespeed (magnitude of velocity)remains constant.
Newton second law and conservation of
momentum & momentum-of-momentum
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Newton second law and conservation of
momentum & momentum-of-momentum
∑∫ = syssys
F d
Dt
DV ρ V
Rate of change of themomentum of the
system
Sum of all externalforces acting on the
system
sys CV CS
D d d dA Dt t
ρ ρ ρ ∂= + ⋅∂∫ ∫ ∫
V V V V n V V On the other hand:
In an inertial coordinate system:
At a moment when system coincide with control volume:
sys contents of thecontrol volume
F F =∑ ∑
contentsof thecontrol volumeCV CS
d dA F t ρ ρ
∂
+ ⋅ =∂ ∑∫ ∫V V V n
V
Example: Linear momentum
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A=0.06 m2
V1=10 m/s
Example: Linear momentumDetermine anchoring forces required to keep the vane stationary vs angle Q.
Neglect gravity and viscosity.
x
CV CS
y
CV CS
u d u dA F
t
w d w dA F t
ρ ρ
ρ ρ
∂+ ⋅ =
∂
∂+ ⋅ =
∂
∑∫ ∫
∑∫ ∫
V n
V n
V
V
1 1 1 1 1 2
1 1 1 1 2
( ) cos ( )
0 ( ) sin ( )
Ax
Az
V V A V V A F
V A V V A F
ρ θρ
ρ θρ
− + =
− + =
2
1 1
2
1 1
(1 cos )
sin
Ax
Az
F V A
F V A
ρ θ
ρ θ
= − −
=
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Linear momentum: comments
• Linear momentum is a vector
• As normal vector points outwards, momentum flow inside aCV involves negative Vn product and moment flow outsideof a CV involves a positive Vn product.
• The time rate of change of the linear momentum of thecontents of a nondeforming CV is zero for steady flow
• Forces due to atmospheric pressure on the CV may needto be considered
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M t f M t E ti
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Moment-of-Momentum Equation
The net rate of flow of moment-of-momentum through a control surface equals the net
torque acting on the contents of the control volume.
Water enters the rotating arm of a lawn sprinkler along the axis of rotation with noangular momentum about the axis. Thus, with negligible frictional torque on the rotating
arm, the absolute velocity of the water exiting at the end of the arm must be in the radial
direction (i.e., with zero angular momentum also). Since the sprinkler arms are angled
"backwards", the arms must therefore rotate so that the circumferential velocity of the
exit nozzle (radius times angular velocity) equals the oppositely directed circumferentialwater velocity.
The Energy Equation
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The Energy Equation
Rate ofincrease of thetotal stored
energy of thesystem
Net rate ofenergyaddition by
heat transferinto the system
Net rate ofenergyaddition bywork transferinto the system
sys
sys
W Qd e Dt
D)( +=∫ V ρ
gz
V
ue ++= 2
2
Total stored energy per unit mass:
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∫∫∫ ⋅+∂∂
=CS CV sys
dAed et d e Dt
D
nV
ρ ρ ρ V V
Rate ofincrease of thetotal storedenergy of thesystem
rate ofincrease of thetotal storedenergy of thecontrol volume
Net rate ofenergy flowout of thecontrol volume
( )net net cvin inCV CS e d e dA Q W t ρ ρ
∂+ ⋅ = +
∂ ∫ ∫ V n V
Application of energy equation
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Application of energy equation
• Product V·n is non-zero only where liquid crosses the CS; ifwe have only one stream entering and leaving control volume:
2 2 2
2 2 2out in
CS out in
p V p V p V u gz dA u gz m u gz m ρ
ρ ρ ρ
⎛ ⎞ ⎛ ⎞ ⎛ ⎞+ + + ⋅ = + + + − + + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠ ⎝ ⎠∫ V n
• If no shaft power is applied and we assume flow steady
2 2
( ),2 2
net inout out in in
out in out in net net in in
Q p V p V gz gz u u q q
m ρ ρ + + = + + − − − =
lossavailable energy
Energy transfer
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Energy transfer
Work must be done on the device shown to turn it over because the system gainspotential energy as the heavy (dark) liquid is raised above the light (clear) liquid. This
potential energy is converted into kinetic energy which is either dissipated due to friction
as the fluid flows down the ramp or is converted into power by the turbine and then
dissipated by friction. The fluid finally becomes stationary again. The initial work done in
turning it over eventually results in a very slight increase in the system temperature
Second law of thermodynamics
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Second law of thermodynamics
• Let’s apply “stream line energy equation” to an infinitesimally thin volume
dU TdS pdV = −
2
2 net
in
p V m du d d gdz Qδ ρ
⎡ ⎤⎛ ⎞⎛ ⎞
+ + + =⎢ ⎥⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠⎣ ⎦
2
( )2 net
in
dp V d gdz Tds qδ ρ
⎡ ⎤⎛ ⎞
+ + = − −⎢ ⎥⎜ ⎟⎝ ⎠⎣ ⎦
1( )du Tds pd = −
For closed system in the absenceof additional work:
0dq
dS T
− ≥If we take into account Clausius inequality:
2
02
dp V d gdz
ρ
⎡ ⎤⎛ ⎞− + + ≥⎢ ⎥⎜ ⎟
⎝ ⎠⎣ ⎦
Application of energy equation
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Application of energy equation
• We can make the energy equation more concrete by noting:
– Work is usually transferred into liquid by rotating shaft:
shaft shaft T ω W =
– Or by normal stress acting on a free surface
CS
ˆ ˆ
ˆ
normal stress
normal stress
p A
p dA
δ σ δ δ ⋅ − ⋅
− ⋅∫
W = n V = V n
W = V n
• The energy equation is now:
2
( )2
net shaft cvin inCV CS
p V e d u gz dA Q W
t ρ ρ
ρ
⎛ ⎞∂+ + + + ⋅ = +⎜ ⎟
∂ ⎝ ⎠∫ ∫ V n
V
Example: Efficiency of a fan
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Example: Efficiency of a fan
Problem 4 20
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Problem 4.20
• Determine local acceleration at points 1 and 2. Is theaverage convective acceleration between thesepoints negative, zero or positive?
Problems
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Problems
• 5.102 Water flows steadily downthe inclined pump. Determine: – The pressure difference, p1-p2;
– The loss between sections 1 and 2
– The axial force exerted on the pipeby water
• 4.20. Determine localacceleration at points 1 and 2.Is the average convectiveacceleration between these
points negative, zero orpositive?
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