Viscous flow in pipesand channelsand channels.

Computational Fluid DynamicsComputational Fluid DynamicsLecture 6Lecture 6

Content

• Laminar and turbulent flowa a a d u bu e o• Entrance region• Flow in a pipe• Channels of non-circular cross-sectionChannels of non circular cross section• Circuit theory for fluidic channels• Computational fluid dynamics

General characteristic of Pipe flow

• pipe is completely filled with wateri d i i f i ll di t l th• main driving force is usually a pressure gradient along the

pipe, though gravity might be important as well

Pipe flow open-channel flow

Important facts from Fluid Dynamics

Re udρμ

=

• Reynolds number

μ

y– Can be interpreted as a ratio of inertia and viscous forces

Laminar or Turbulent flowd

well defined streakline one velocity component

• Reynolds number: can be interpreted as a ratio of inertia and viscous forcesRe udρ

μ=

well defined streakline, one velocity componentu=V i

Re 2100<Re 2100<

Re 4000>

velocity along the pipe is unsteady and accompanied by random component normal to pipe axis

u v w=V i + j + k

Laminar or Turbulent flow

• In this experiment water flows through a clear pipe with increasing speed. Dye is injected through a small diameter tube at the left portion of theDye is injected through a small diameter tube at the left portion of the screen. Initially, at low speed (Re <2100) the flow is laminar and the dye stream is stationary. As the speed (Re) increases, the transitional regime occurs and the dye stream becomes wavy (unsteady, oscillatory laminaroccurs and the dye stream becomes wavy (unsteady, oscillatory laminar flow). At still higher speeds (Re>4000) the flow becomes turbulent and the dye stream is dispersed randomly throughout the flow.

Entrance region and fully developed flow

• fluid typically enters pipe with nearly uniform velocityyp y p p y y• the length of entrance region depends on the Reynolds number

0.06Re for laminar flowelD=dimensionless

( )1/ 64.4 Re for turbulent flowe

DlD=

entrance length

Pressure and shear stress

no accelerationno acceleration, viscous forces balanced by pressure

pressure balanced byviscous forces and acceleration

Fully developed laminar flow

• we will derive equation for fully developed e de e equa o o u y de e opedlaminar flow in pipe using 3 approaches:

from 2nd Newton law directly applied– from 2nd Newton law directly applied– from Navier-Stokes equation– from dimensional analysis

2nd Newton’s law directly applied

2nd Newton’s law directly applied

2 21 1( ) 2 0

2p r p p r rl

pπ π τ π

τ− −Δ − =

Δ 2pl r

τΔ= , at / 2 stress is maximum

2 4 and

w

w w

Cr r Dr lp

D D

τ ττ ττ

= =

= Δ =

wall shear stress

doesn’t depend on radiusD D

2nd Newton’s law directly appliedfor Newtonian liquid:

dudr

τ μ= −2p rl

τ Δ⎛ ⎞= ⎜ ⎟⎝ ⎠

2du p rdr lμ

⎛ ⎞Δ= −⎜ ⎟

⎝ ⎠⎛ ⎞ 2

14pu r Clμ

⎛ ⎞Δ= − +⎜ ⎟

⎝ ⎠⎛ ⎞ 2

1

22

boundary condition: 0 at / 216

pu r D C Dlμ

⎛ ⎞Δ= = ⇒ = ⎜ ⎟

⎝ ⎠⎡ ⎤⎛ ⎞ ⎛ ⎞

22 2( ) 116pD ru r

l Dμ⎡ ⎤⎛ ⎞Δ ⎛ ⎞= −⎢ ⎥⎜ ⎟ ⎜ ⎟

⎝ ⎠⎢ ⎥⎝ ⎠ ⎣ ⎦

Flow rate: 4/ 2

( )2D D pQ dA d π Δ

∫ ∫0 ( )2128D pQ udA u r rdr

lππ

μΔ

= = =∫ ∫

2nd Newton’s law directly applied

• if gravity is present, it can be added to the pressure:g y p p

sin 2p glρ θ τΔ −

( ) 2sinl rp gl D

Vρ θ

=

Δ −( )

( ) 4

32sin

Vl

p gl DQ

μπ ρ θ

=

Δ −( )128

p gQ

lρ

μ=

Navier-Stokes equation applied0⋅ =

∂ ΔV∇ V

2p vt ρ

∂ Δ+ ⋅ = − + +

∂V g V∇V ∇V

1p u∂ ∂ ∂⎛ ⎞

in cylindrical coordinates:

1sinp ug rx r r r

ρ θ μ∂ ∂ ∂⎛ ⎞+ = ⎜ ⎟∂ ∂ ∂⎝ ⎠boundary conditions:

0; ( ) 0u u R∂= =

y

0

; ( )r∂

• The assumptions and the result are exactly the same as Navier-Stokes equation is drawn from 2nd Newton law

Creeping flow in microchannels• Let’s consider flow with very small Re numbers

2

0p v

⋅ =∂ Δ

+ ⋅ = − + +V g V∇

∇ V

V ∇V vt ρ+ + +

∂g V∇V ∇V

• DV/Dt part (non-linear) can be neglected leading to Stokes equation:Stokes equation:

2p v∇ = ∇FV +pρ

• equation is linear, and therefore is reversible. change of the velocity direction at the boundary will lead to change of the velocity direction in the whole domain

Dimensional analysis applied

( ), , ,p F V l D

D l

μΔ =

Δ ⎛ ⎞D p lV D

φμΔ ⎛ ⎞= ⎜ ⎟

⎝ ⎠

assuming pressure drop proportional to the length:

2

D p Cl p C VV D l D

μμΔ Δ

= ⇒ =

4( / 4 )C pDQ AVl

πμΔ

= =

Dimensional analysis of pipe flow• major loss in pipes: due to viscous flow in the straight

elements• minor loss: due to other pipe components (junctions etc.)

Major loss:Major loss:

( , , , , , )p F V D l ε μ ρΔ =roughness

• those 7 variables represent complete set of parameters for the problemset of parameters for the problem

p VD lρ εφ ⎛ ⎞Δ= ⎜ ⎟21

2

, ,V D D

φρ μ

= ⎜ ⎟⎝ ⎠

as pressure drop is proportional to length of the tube:

212

Re,p lV D D

εφρΔ ⎛ ⎞= ⎜ ⎟

⎝ ⎠

Dimensional analysis of pipe flow

p l εφΔ ⎛ ⎞ pDf Δ

friction factor

212

Re,p lV D D

εφρΔ ⎛ ⎞= ⎜ ⎟

⎝ ⎠ 212

pDfl VρΔ

=

2

Re, and2

l Vf p fD Dε ρφ ⎛ ⎞= Δ =⎜ ⎟

⎝ ⎠

• for fully developed laminar flow in a circular pipe:

64 / Ref = 64 / Ref =• for fully developed steady incompressible flow (from Bernoulli eq.):

2

2Lmajorp l Vh fg D gρ

Δ= =

g gρ

Non-circular ducts• Reynolds number based on hydraulic diameter:

4Re hVD ADρ= =

cross-sectionReh hD

Pμ= =

wetted perimeter

F i ti f t f i l d t• Friction factor for noncircular ducts:/ Ref C=for fully developed laminar flow:

2; w

wS p L fVττρ

Δ = Ρ =⎛ ⎞

2 2l l

2Vρ⎛ ⎞

⎜ ⎟⎝ ⎠

2 2

4 2 2h

l V l Vp f fS Dρ ρΡ

Δ = =

56 8hD a= 3

53 2hD a=

56.8ReDh

f = 53.2ReDh

f =

Non-circular ductsf f• Friction factor for noncircular ducts: / Ref C=

Moody chartFriction factor as a function of Reynolds number and relative roughness for round pipes

1 / 2.512.0 log3.7 Re

Df f

ε⎛ ⎞= − +⎜ ⎟⎜ ⎟

⎝ ⎠Colebrook formula

Equivalent circuit theory

p R QΔ = ×flow:

V R IV R I= ×electricity:

• channels connected in series

Equivalent circuit theory

• channels connected in parallelp

Complianceli (h d li it )• compliance (hydraulic capacitance):

Q – volume V/time I – charge/timeQ volume V/time I charge/timedqC =h d

dVC = −flow: electricity: CdUhydC

dpflow: electricity:

Equivalent circuits

Pipe networks

• Serial connection

1 2 3

1 2 3

A BL L L L

Q Q Qh h h h

−

= == + +

• Parallel connectionParallel connection

1 2 3

1 2 3

L L L

Q Q Q Qh h h= + +

= =1 2 3

COMPUTATIONAL FLUID DYNAMICS

Introduction• Computational fluid dynamics applications:

– Aerodynamics of aircraft and vehiclesHydrodynamics of ships– Hydrodynamics of ships

– Microfluidics and biosensors– Chemical process engineering

C b ti i d t bi– Combustion engines and turbines– Construction: External and internal environment– Electric and electronic engineering: heating and cooling of circuits

• Advantages of CFD approach– Reduction of time and costs

ff– Ability to do controlled experiment under difficult and hazardous condition

– Unlimited level of detailP ibili l l h i l– Possibility to couple several physical processes (momentum/mass/energy transfer, electrical/magnetic fields etc.)

Governing equations for fluid dynamics• Mass conservation:

mass change in a volume is equal to the net rate of flow∂ ( ) 0div u

tρ ρ∂+ =

∂( ) 0div u =Uncompressible fluid

• Moment conservation: Navier-Stokes equation

2 2 2

2 2 2

2 2 2

( ) ( )xu u u u p u u uu v w gt x y z x x y z

ρ ρ μ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂+ + + = − + + + +

∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂

∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂2 2 2

2 2 2

2 2 2

( ) ( )yv v v v p v v vu v w gt x y z y x y z

ρ ρ μ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂+ + + = − + + + +

∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂

∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂2 2 2

2 2 2( ) ( )zw w w w p w w wu v w gt x y z z x y z

ρ ρ μ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂+ + + = − + + + +

∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂

Discretization of the equations

• To obtain the solution the continuous non-linear equations are discretized and converted to algebraic equations.

• Discritization techniques:– Finite difference– Finite volume (finite element)– Boundary elements

Discretization Techniques

• Finite difference– Differential equations are converted to algebraic through the use of

Taylor series expansion

2 2u x u x⎛ ⎞∂ Δ ∂ Δ⎛ ⎞1, , 2

, ,

1, ,

...1! 2!

( )

i j i ji j i j

i j i j

u x u xu ux x

u uu O x

+

+

⎛ ⎞∂ Δ ∂ Δ⎛ ⎞= + + +⎜ ⎟⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠−∂⎛ ⎞ = + Δ⎜ ⎟∂ Δ⎝ ⎠ ,

( )i jx x⎜ ⎟∂ Δ⎝ ⎠

• How shall we represent the d d i ti ?second derivative?

Discretization Techniques• Finite element

– Similar to finite difference method continuous functions are replaced by– Similar to finite difference method, continuous functions are replaced by piecewise approximations valid on particular grid element

• Finite volume– Control Volume form of NSE is used on every grid elementControl Volume form of NSE is used on every grid element

• Boundary element method– Boundary is broken into discrete segments (panels), appropriate

singularities (sources, sinks, doublets, vortices) are distributed along the segments

CFD Methodologyfinite element method:1. Choose appropriate physical model2 Define the geometry2. Define the geometry.3. Define the mesh (grid): flow field is broken into set of

elements• Mesh could be structured (regular pattern) or unstructured.

Other types could be hybrid (several structured elements), moving (time dependent)

4. Define the boundary conditions5. Solving: conservation equations are (mass, momentum

and energy) are written for every element and solvedand energy) are written for every element and solved.6. Postprocessing: solution is visualized and hopefully

understood

CFD Methodology

• Common problems:C– Convergence issues

– Difficulties in obtaining the quality grid and managing the resourcesDifficulties in turbulent flow situations– Difficulties in turbulent flow situations

• Verification– Using other techniques

Flow in a 2D pipeC• Can be solved analytically:

Boundary condition (no slip) 0)()( =−= huhuBoundary condition (no slip) 0)()( =−= huhu

)(21 22 hy

xpu −∂∂

=μ

Velocity profile2 x∂μ

P t

Problem: Solve analytically and numerically and compare.

ParametersV=0.001m/s; h=0.2m; m=10-3 Pa*s

Questions:Q• What is the Reynolds number?• What is the calculated pressure drop in the pipe? What is the entrance pressure drop?p p p p• Is laminar flow fully developed?

S-cell Problem (home work)• Calculate velocity field in an S-

cell (3D fluidic cell, 50mm height). (6,15.5) (12 15 5)• What would flow field uniformity

across a 1mmx1mm array tt d i th iddl f th ll? (4.5,14)

( , ) (12,15.5)

spotted in the middle of the cell? ( , )

(12,14.5)(9,14.5)

(7.5,13)

Data:Flow rate: 100ul/minLiquid: water, T=298K (3,1)

(4.5,2.5)

Channel height (z): 50umChannel width: 1mm in/outlet;3mm chamberC t di 1 5

(0,1)(7.5,1.5)

(0,0)Curvature radius: 1.5mm

(6,0)

Problems• Ethanol solution of a dye (μ=1.197

mPa·s) is used to feed a fluidic lab-on-chip laser Dimension of the channelchip laser. Dimension of the channel are L=122mm, width w=300um, height h=10 um. Calculate pressure required

hi fl f Q 10 l/hto achieve flow rate of Q=10µl/h.

• 8.7 A soft drink with properties of 10 ºC water is sucked through p p ga 4mm diameter 0.25m long straw at a rate of 4 cm3/s. Is the flow at outlet laminar? Is it fully developed?

• Calculate total resistance of a microfluidic circuit shown. Assume that the pressure on all channels isthat the pressure on all channels is the same and equal Δp.