control volume and reynold transport

8/18/2019 control volume and reynold transport

1/43

Lecture 3

Fluid Kinematics: Velocity field,

Acceleration, Reynolds Transport

Theorem and its application


2/43

Aims

• Describing fluid flow as a field• How the flowing fluid interacts with the

environment (forces and energy)


3/43

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


4/43

Velocity field

u

v

dx

dy=


5/43

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.


6/43

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


7/43

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


8/43

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


9/43

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.


10/43

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

= = −′= + =


11/43

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


12/43

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


13/43

Example

Water flowing from an oscillating slit:

jviv yt uv

000 ))/(sin( +−= ω


14/43


15/43

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


16/43

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)


17/43

Example: acceleration

Velocity field is given by: ))(/( 0 j yi xlvv

−=

Find the acceleration and draw it on scheme


18/43

Example: acceleration


19/43

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


20/43

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


21/43

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:


22/43

• 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


23/43

The Reynolds Transport theorem

(for fixed nondeforming volume)

∫ ⋅=out CV

out dAb B nV ρ


24/43

∫ ⋅+∂∂

=CV

CV Sys dAbt

B

Dt

DBnV

ρ

General Reynoldstransport theorem forfixed control volume


25/43

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


26/43

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:


27/43

• 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


28/43

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

− + ⋅ =

⎡ ⎤⎛ ⎞− + − =⎢ ⎥⎜ ⎟⎝ ⎠⎢ ⎥⎣ ⎦

=

∫

∫


29/43

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


30/43

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


31/43

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

= − −

=


32/43

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


33/43

M t f M t E ti


34/43

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


35/43

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:


36/43

∫∫∫ ⋅+∂∂

=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


37/43


• 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


38/43

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


39/43

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

⎡ ⎤⎛ ⎞− + + ≥⎢ ⎥⎜ ⎟

⎝ ⎠⎣ ⎦



40/43


• 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


41/43

Example: Efficiency of a fan

Problem 4 20


42/43

Problem 4.20

• Determine local acceleration at points 1 and 2. Is theaverage convective acceleration between thesepoints negative, zero or positive?

Problems


43/43

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?

control volume and reynold transport

Documents