Lecture 3

Fluid Kinematics: Velocity fieldFluid Kinematics: Velocity field, Acceleration, Reynolds Transport

Theorem and its applicationTheorem and its application

Aims

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

environment (forces and energy)( gy)

Lecture plan

• Describing flow with the fields: Eulerian vs• Describing flow with the fields: Eulerian vs. Lagrangian description.Flow analysis: Streamlines Streaklines Pathlines• Flow analysis: Streamlines, Streaklines, Pathlines.

• How to perform calculations in the field description: h M i l D i ithe Material Derivative

• Reynold’s Transport Theorem• Application of Reynolds transport theorem:

Continuity, Momentum and Energy conservationy, gy

Velocity field• field representation of the flow: flow is represented as a

function of spatial coordinates (map)• example: velocity field

( , , , ) ( , , , ) ( , , , )V u x y z t i v x y z t j w x y z t k= + +rr r r

( , , , ) ( , , , ) ( , , , )y y j y

vdy=

udx

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.

Velocity field representationVelocity field is given by:

))(/( jyixlvvrrr

−= ))(/( 0 jyixlvv =• Sketch the field in the first quadrant

find where velocity will be equal to v• find where velocity will be equal to v0

Eulerian and Lagrangian flow description

• Eulerian method – field concept is used, flow parameters (T, P t ) d i i t i tiP, 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 timeparameters associated with this particle are followed in time

Example: Smoke coming out of a chimney

1D, 2D and 3D flow• In most of flow situations the flow is three-dimensional, though in many

situations it’s possible to reduce it to 2D or even 1D 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 wing tips (called trailing vorticies) is clearly visible

Flow types• Steady flow – the velocity at any given point in space doesn’t vary with

time Otherwise the flow is called unsteadytime. 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.

Streamlines• Streamline: line everywhere tangentional to

the elocit fieldthe velocity field

vdy=

udx=

Streamlines

Velocity field is given by:

))(/( jyixlvvrrr

= ))(/( 0 jyixlvv −=

•draw the streamlines and find there equation

uv

dxdy

=

;dy v ydx u x

= = −

Constxy =ln ln ; /dx u x

y x C y C x′= + =

Streamlines, Streaklines, Pathlines• Streamline – line that

everywhere tangent toeverywhere tangent to velocity field

• Streakline – all particles th t d th hthat passed through a common point

• Pathline – line traced by aPathline line traced by a given particle as it flows

streamlines

Micro Particle Image Velocimetry

•• Particle Image Velocimetry Particle Image Velocimetry (PIV): (PIV):

•• flow is seeded by small flow is seeded by small particlesparticlesparticles, particles, •• consecutive photographs of consecutive photographs of particles distribution are made particles distribution are made

I ti d i tI ti d i t•• Images are sectioned into Images are sectioned into interrogation regionsinterrogation regions•• Motion of particles within Motion of particles within interrogation region determined interrogation region determined by image crossby image cross--correlationcorrelation

Flow characterization on a microscale

– Particle Streak Velocimetry– Particle Image Velocimetry

(PIV)

Advantages:– Full-field technique– high spacial resolutionhigh spacial resolution– large range of velocities

covered (up to 8m/s)covered (up to 8m/s)– simplicity

Imaging flow in a microfluidic channel• Flow is seeded with fluorescent particles and imaged…

(Project 5th semester Fall 2006)

Flow through a loosely packed microspheres bed

Flow at a channel turn. Flow is disturbed by a microwire

t1t2=t+∆t

tt3t4

t5tn

t

Cross-correlation of 2 images is calculatedtn+1

∑∑= =

++⋅=Φq

j

p

injmigjifnm

1 1),(),(),(

t

j

t+∆tt

⊗

Example

Water flowing from an oscillating slit:

jvivytuvrrr

000 ))/(sin( +−= ω

Material derivative

)( trV• Particle velocity

),( trV AA

• Particle acceleration

tz

zV

ty

yV

tx

xV

tV

dtdVtra AAAAAAAA

AA ∂∂

∂∂

+∂∂

∂∂

+∂∂

∂∂

+∂∂

==),(

DtD

zw

yv

xu

ttr VVVVV

=∂∂

+∂∂

+∂∂

+∂∂

=),(a

• Material derivative )V ∇•+∂∂

=∂∂

+∂∂

+∂∂

+∂∂

= ( tz

wy

vx

utDt

D

Material derivative

• is the rate of changes for a given variable with time for a given particle of fluid.

)V ∇•+∂∂

=∂∂

+∂∂

+∂∂

+∂∂

= ( tz

wy

vx

utDt

D

Convective effect

Unsteadiness of the flow(local acceleration)

Convective effect(convective acceleration)

Example: acceleration

Velocity field is given by: ))(/( 0 jyixlvvrrr

−=

Find the acceleration and draw it on scheme

Example: accelerationExample: acceleration

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 spaceControl volume geometrical entity, a volume in space through which fluid may flow

Governing laws of fluid motion are stated in terms of the system, but control volume approach is essential for practical applications

Control volume and system representationControl volume: exampleControl volume and system representation

• Extensive property: B=mb (e.g. m (b=1), mv (b=v), mv2/2 etc)

intensive property∫=sys

sys bdB Vρsys

∫∂

=∂ sys bdB

Vρ

∫

∫∂

=∂

∂=

∂

cv

sys

bdB

bdtt

V

V

ρ

ρ

∫∂∂ cv

bdtt

Vρ

0;0 <∂

=∂

=∂

=∂

∫∫ cvsys dBdB

VV ρρ 0;0 <∂

=∂

=∂

=∂ ∫∫

cvsys

dtt

dtt

VV ρρ

The Reynolds Transport theorem (simplified)

CVSYSt =:IIICVSYStt

CVSYSt+−=+

=)(:

:δ

)()( tBtB cvsys =

Let’s consider an extensive property B:

initially, at time t:

)()()()(

)()(

ttBttBttBttB IIIcvsys

cvsys

δδδδ +++−+=+initially, at time t:

at time t+δt:

tttB

tttB

ttBttB

ttBttB IIIcvcvsyssys

δδ

δδ

δδ

δδ )()()()()()( +

++

−−+

=−+

tttt δδδδ

outflowinflowcvBt

∂∂

DB

• For fixed control volume with one inlet, one outlet, velocity outin

cvsys BBt

BDt

DB && +−∂∂

=

normal to inlet/outlet

bVAbVABDB cvsys +∂

22221111 bVAbVAtDtcvy ρρ +−

∂=

C b il li d• Can be easily generalized:

The Reynolds Transport theorem(for fixed nondeforming volume)

∫ ⋅=outCV

out dAbB nV )& ρ

∂BDB General Reynolds

∫ ⋅+∂

∂=

CV

CVSys dAbt

BDt

DBnV )ρ

General Reynolds transport theorem for fixed control volume

Application of Reynolds Transport Theorem

∫+∂

= CVSys dAbBDBnV )ρ∫ ⋅+

∂=

CV

dAbtDt

nVρ

• We will apply it now to various properties:– mass (continuity equation)– mass (continuity equation)– momentum (Newton 2nd law)– energy

Conservation of mass

• The amount of mass in the system should be conserved:

0=⋅+∂∂

== ∫∫∫CVCV

Sys dAdt

dDtD

DtDM

nV )ρρρ VV∂ CVCVsys tDtDt

C ti it tiContinuity equation

Mass flow rate through a section of control surface having area A:

m Q AVρ ρ= =&

g g

A

V n dAV

A

ρ ⋅=∫

r r

Average velocity:

Volume flow rate

Aρ

• For incompressible flow, the volume flowrate into a control volume equals the volume flowrate out of it.

• The overflow drain holes in a sink must be large enough to d t th fl t f th f t if th d i h l taccommodate the flowrate from the faucet if the drain hole at

the bottom of the sink is closed. Since the elevation head for the flow through the overflow drain is not large, the velocity there is relatively small Thus the area of the overflow drainthere is relatively small. Thus, the area of the overflow drain holes must be larger than the faucet outlet area

ExampleIncompressible laminar flow develops in a straight pipe of radius R At section 1Incompressible 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 with maximum value umax. Find relation between U and umax, what is average velocity at section (2)? a sec o ( )

∫r r

2

1 1 2

2

0A

R

AU V n dAρ ρ− + ⋅ =

⎡ ⎤⎛ ⎞

∫r

2

1 max0

1 2 0R rAU u rdr

Rπ

⎡ ⎤⎛ ⎞− + − =⎢ ⎥⎜ ⎟⎝ ⎠⎢ ⎥⎣ ⎦

∫

max 2u U=

Newton second law and conservation of momentum & momentum-of-momentum

A jet of fluid deflected by anA jet of fluid deflected by an object puts a force on the object. This force is the result of the change of momentum of the fluidchange of momentum of the fluid and can happen even though the speed (magnitude of velocity) remains constantremains constant.

Newton second law and conservation of momentum & momentum-of-momentum

In an inertial coordinate system:

∑∫ = syssys

FdDtD VρV

y

Rate of change of the momentum of the

Sum of all external forces acting on themomentum of the

systemforces acting on the system

At a moment when system coincide with control volume:∑ ∑

D d d dA∂+∫ ∫ ∫V V V V )V V

sys contentsofthecontrol volume

F F=∑ ∑

sys CV CS

d d dADt t

ρ ρ ρ= + ⋅∂∫ ∫ ∫V V V V nV VOn the other hand:

∂contentsofthecontrol volumeCV CS

d dA Ft

ρ ρ∂+ ⋅ =

∂ ∑∫ ∫V V V n)V

Example: Linear momentumDetermine anchoring forces required to keep the vane stationary vs angle Q. N l i d i i

A=0.06 m2

Neglect gravity and viscosity.

V1=10 m/s

xCV CS

u d u dA Ft

ρ ρ∂+ ⋅ =

∂ ∑∫ ∫ V n)V 1 1 1 1 1 2( ) cos ( )0 ( ) i ( )

AxV V A V V A FV A V V A F

ρ θρθ

− + =

+CV CS

yCV CS

w d w dA Ft

ρ ρ∂+ ⋅ =

∂ ∑∫ ∫ V n)V1 1 1 1 20 ( ) sin ( ) AzV A V V A Fρ θρ− + =

2 (1 cos )F V Aρ θ= − −1 12

1 1

(1 cos )

sinAx

Az

F V A

F V A

ρ θ

ρ θ

=

=

Linear momentum: comments

• Linear momentum is a vector• As normal vector points outwards, momentum flow inside a

CV involves negative V·n product and moment flow outside of a CV involves a positive V·n productof a CV involves a positive V n product.

• The time rate of change of the linear momentum of the contents of a nondeforming CV is zero for steady flowg y

• Forces due to atmospheric pressure on the CV may need to be considered

Example: Linear momentum – taking into account weight, pressure and change in speed

D t i th h i f i d t h ld i l i l l• Determine the anchoring force required to hold in place a conical nozzle attached to the end of the laboratorial sink facet. The water flow rate is 0.6 l/s, nozzle mass 0.1kg. The pressure at the section (1) is 464 kPa.

Example: Linear momentum – taking into account weight, pressure and change in speed

CV CS

w d w dAt

ρ ρ∂+ ⋅ =

∂ ∫ ∫ V n)V

1 1 2 2

CV CS

A n w

tF W p A W p A

∂

= − − − +

volume of a truncated cone:volume of a truncated cone:

( )2 21 2 1 2

112wV h D D D Dπ= + +

pressure distribution:

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 no g p gangular 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 circumferential water velocity.

The Energy Equation

f

WQdeD )( && +∫ Vρ

• The First Law of thermodynamics

syssys

WQdeDt

)( +=∫ Vρ

Rate of increase of the total stored

Net rate of energy

Net rate of energy

total stored energy of the system

addition by heat transfer into the system

gyaddition by work transfer into the system

2

2Ve u gz= + +)

Total stored energy per unit mass:2

g

∫∫∫∂D

∫∫∫ ⋅+∂∂

=CSCVsys

dAedet

deDtD nV )ρρρ VV

Rate of increase of the

rate of increase f h l

Net rate of increase of the total stored energy of the system

of the total stored energy of the control

l

energy flow out of the control volume

system volume

( )net net cve d e dA Q Wt

ρ ρ∂+ ⋅ = +

∂ ∫ ∫ V n) & &Vin inCV CSt∂ ∫ ∫

Power transfer due to normal and tangential stress

• Work transfer rate (i.e. power) can be transferred through a rotating shaft (e.g. turbines, propellers etc)through a rotating shaft (e.g. turbines, propellers etc)

shaft shaftW T w=& where Tshaft – torque and w – angular velocity

• or through the work of normal stressnormal stress

normal normalstress stress

W F V n A V pV n Aδ δ σ δ δ= ⋅ = ⋅ = − ⋅r r r rr r&on a single particle:

integrating: normalstress CS

W pV ndA= − ⋅∫r r&

CS

tangential stress: tangential tangentialstress stress

0W F Vδ δ= ⋅ =r r

&

Power transfer due to normal and tangential stress

• The first law of themrodynamics can be expressed now as:expressed now as:

shaft net in

netinCV CS CS

e d e dA Q W pV ndAt

ρ ρ∂+ ⋅ = + − ⋅

∂ ∫ ∫ ∫V nr r) & &V

• so, we can obtain the energy equation2p Ve d u gz dA Q Wρ ρ

⎛ ⎞∂+ + + + = +⎜ ⎟∫ ∫ V n)( & &V

so, we can obtain the energy equation

shaft net in2 net

inCV CS

e d u gz dA Q Wt

ρ ρρ

+ + + + ⋅ = +⎜ ⎟∂ ⎝ ⎠∫ ∫ V nV

Application of energy equation• Let’s consider a steady (in the

mean, still can be cyclical) flow and take a one streamand take a one stream

• Product V·n is non-zero only where liquid crosses the CS; ifwhere liquid crosses the CS; if we have only one stream entering and leaving control

l2p Vd dA Q W

⎛ ⎞∂+ + + + +⎜ ⎟∫ ∫ V n)( & &Vvolume:

2 2 2p V p V p VdA⎛ ⎞ ⎛ ⎞ ⎛ ⎞

+ + + + + + + + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟∫ V )( ( (& &

shaft net in2 net

inCV CS

pe d u gz dA Q Wt

ρ ρρ

+ + + + ⋅ = +⎜ ⎟∂ ⎝ ⎠∫ ∫ V nV

2 2 2out inCS out in

p p pu gz dA u gz m u gz mρρ ρ ρ

+ + + ⋅ = + + + − + + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠∫ V n

2 2V Vp p⎛ ⎞⎛ ⎞⎛ ⎞ ⎛ ⎞ − & &( ) shaft net in2

out inout in out in net

inout in

V Vp pm u u g z z Q Wρ ρ

⎛ ⎞⎛ ⎞⎛ ⎞ ⎛ ⎞ −− + − + + − = +⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎝ ⎠

( ( & &&

ph uρ

⎛ ⎞= + ⎜ ⎟

⎝ ⎠

( (

Energy transfer

Work must be done on the device shown to turn it over because the system gains y gpotential 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 p p ydissipated 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

Example: Temperature change at a water fall

• find the temperature change after a water gfall, cwater=4.19 kJ/kg·K

150m150m

( )2 2

2out in

out in out in netinout in

V Vp pm u u g z z Qρ ρ

⎛ ⎞⎛ ⎞⎛ ⎞ ⎛ ⎞ −− + − + + − =⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎝ ⎠

( ( &&

=0 =0 =0

Energy equation vs Bernoulli equation• Let’s return to our one-stream volume, steady flow (also no

shaft power)

( )2 2

2out in

out in out in neti

V Vp pm u u g z z Qρ ρ

⎛ ⎞⎛ ⎞⎛ ⎞ ⎛ ⎞ −− + − + + − =⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎝ ⎠

( ( &&

2 2 netQV V

&

2 inout inρ ρ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎝ ⎠

2 2

( ),2 2

netinout out in in

out in out in net netin in

Qp V p Vgz gz u u q q

mρ ρ+ + = + + − − − =( (

&

lossavailable energy

• Comparing with Bernoulli equation: 0out in netu u q− − =( (&• Comparing with Bernoulli equation: out in net

inq

i.e. steady incompressible flow should be also frictionless

Problems4 4 A l it fi ld i i b• 4.4. A velocity field is given by:

( 1)( 1)V xi x x y j= + − +r r r

where u and v are in m/s and x,y are in m. Plot the stream lines that passes through x y=0 Compare this streamline with

( 1)( 1)V xi x x y j+ +

lines that passes through x,y=0. Compare this streamline with the streakline through the origin.

• 4.10 Determine local acceleration at points 1 and 2. I th tiIs the average convective acceleration between these points negative, zero or p g ,positive?

Problems• 5.19 A converging elbow turns water

by 135º. The elbow flow volume is 0 2 3 b t ti 1 d 20.2m3 between sections 1 and 2, water flow rate is 0.4 m3/s, pressures at inlet and outlet 150kPa and 90 kPa, elbow mass 12 kg.

• 5.71 Water flows steadily down the inclined pump Determine:the inclined pump. Determine: – The pressure difference, p1-p2;– The loss between sections 1 and 2

Th i l f t d th i– The axial force exerted on the pipe by water