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Picture credit: Sandia National Lab

Fluidic Dynamics

Dr. Thara SrinivasanLecture 17

Picture credit: A. Stroock et al., Microfluidic mixing on a chip

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Lecture Outline

• Reading• From S. Senturia, Microsystem Design, Chapter 13, “Fluids,”

p.317-334.

• Today’s Lecture• Basic Fluidic Concepts• Conservation of Mass → Continuity Equation• Newton’s Second Law → Navier-Stokes Equation• Incompressible Laminar Flow in Two Cases• Squeeze-Film Damping in MEMS

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5Viscosity

• Fluids deform continuously in presence of shear forces• For a Newtonian fluid,

• Shear stress = Viscosity × Shear strain• N/m2 [=] (N •s/m2) × (m/s)(1/m)• Centipoise = dyne s/cm2

• No-slip at boundaries

ηair = 1.8 × 10-5 N s/m2

ηwater = 8.91 × 10-4

ηlager = 1.45 × 10-3

ηhoney = 11.5

U

τw

τwh Ux

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Density

• Density of fluid depends on pressure and temperature…• For water, bulk modulus =• Thermal coefficient of expansion = • …but we can treat liquids as incompressible

• Gases are compressible, as in Ideal Gas Law

TMRPnRTPV

Wm

== ρ

3

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5Surface Tension

• Droplet on a surface

grh

ρθγ cos2=

∆P

2r

γγ

• Capillary wetting

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Lecture Outline

• Today’s Lecture• Basic Fluidic Concepts• Conservation of Mass → Continuity Equation• Newton’s Second Law → Navier-Stokes Equation• Incompressible Laminar Flow in Two Cases• Squeeze-Film Damping in MEMS

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5Conservation of Mass

• Control volume is region fixed in space through which fluid moves

rate of mass efflux

outflowofrateowinflofrateonaccumulatiofrate −=

dS

• Rate of accumulation

• Rate of mass efflux

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Conservation of Mass

0=∫∫ ⋅+∫∫∫ ∂∂

S mVm dSdV

tnUρρ

dS

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5Operators

• Gradient and divergence

zU

yU

xU

zyx

zU

yU

xU

zyx

zyx

zyx

∂∂

+∂∂

+∂∂

==⋅∇

∂∂

+∂∂

+∂∂

=∇

∂∂

+∂∂

+∂∂

=∇

UdivU

eee

kjiU

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Continuity Equation• Convert surface integral to volume integral using Divergence

Theorem

)( 0=∫∫∫

⋅∇+∂∂ dV

tV Uρρ )( 0=⋅∇+∂∂ Uρρ

t

• For differential control volume

Continuity Equation

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5Continuity Equation

• Material derivative measures time rate of change of a property for observer moving with fluid

)( 0=⋅∇+∇⋅+∂∂ UU ρρρ

t)( ρρρ

∇⋅+∂∂

= UtDt

D

∇⋅+∂∂

= UtDt

D

0=⋅∇+ UρρDtD 0=⋅∇+ Uρρ

DtD

For incompressible fluid

zU

yU

xU

t zyx ∂∂

+∂∂

+∂∂

+∂∂

=

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Lecture Outline

• Today’s Lecture• Basic Fluidic Concepts• Conservation of Mass → Continuity Equation• Newton’s Second Law → Navier-Stokes Equation• Incompressible Laminar Flow in Two Cases• Squeeze-Film Damping in MEMS

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5Newton’s Second Law for Fluidics• Newton’s 2nd Law:

• Time rate of change of momentum of a system equal to net force acting on system

Sum of forces acting on control

volume

Rate of momentum efflux

from control volume

Rate of accumulation of momentum in control volume

= +

( )∫∫ ⋅+∫∫∫==∑ SV dSdVdtd

dtd nUUUpF ρρ

Net momentum accumulation rate

Net momentum efflux rate

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Momentum Conservation

( ) ( )∫∫ ⋅+∫∫∫=∫∫∫+∫∫ +− SVVS dSdVdtddVdSP nUUUgτn ρρρ

• Momentum conservation, integral form

gravity forcepressure and shear stress forces

∑ =F

• Sum of forces acting on fluid

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5Navier-Stokes Equation

• Convert surface integrals to volume integrals

( )

∫∫∫ ⋅∇=∫∫ ⋅

∫∫∫ ∇−=∫∫−

∫∫∫

⋅∇∇+∇=∫∫

VS

VS

VS

dVdS

dVPdSP

dVdS

)()(

32

UUnUU

n

UUτ

ρρ

ηη

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Navier-Stokes Differential Form

( )U3

UU⋅∇∇+∇++−∇=

ηηρρ 2gPDtD

( )

( )dVdVt

dVP

VV

V

∫∫∫ ⋅∇+∫∫∫

∂∂

=∫∫∫

⋅∇∇+∇++∇−

UUU

U3

Ug

ρρ

ηηρ 2

dVDtD

V∫∫∫

Uρ

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5Incompressible Laminar Flow

( )U3

UgU ⋅∇∇+∇++−∇=ηηρρ 2P

DtD

• Incompressible fluid 0=⋅∇ U

UgU 2∇++−∇= ηρρ PDtD

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Cartesian Coordinates

UgU 2∇++−∇= ηρρ PDtD

∂∂+

∂∂+

∂∂++

∂∂−

=

∂∂+

∂∂+

∂∂+

∂∂

2

2

2

2

2

2

zU

yU

xUg

xP

zUU

yUU

xUU

tU

xxxx

x

xz

xy

xx

x

ηρ

ρ

x direction

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5Dimensional Analysis

ρη

ρUgU 2∇

++∇

−=P

DtD

• Each term has dimension L/t2 so ratio of any two gives dimensionless group• inertia/viscous = = Reynolds number• pressure /inertia = P/ρU2 = Euler number• flow v/sound v = = Mach number

• In geometrically similar systems, if dimensionless numbers are equal, systems are dynamically similar

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Two Laminar Flow Cases

U

τw

τwh

Ux

y

U

τwHigh P τw

Low P

Ux

y

Umax

h

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∂∂

+∂∂

++∂∂

−= 2

2

2

21yU

xU

xP

DtD xx

xx

ρη

ρgU

Couette Flow• Couette flow is steady viscous flow

between parallel plates, where top plate is moving parallel to bottom plate

• No-slip boundary conditions at platesxx

xx

yUUiU

iUU)(

0=

==⋅∇

hyatUUyatU

andyU

x

xx

====

=∂∂ 00

,02

2

=xU

U

τw

τwh Ux

y

U

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Couette Flow• Shear stress acting on plate due to motion, , is dissipative• Couette flow is analogous to resistor with power dissipation

corresponding to Joule heating

hA

UAR w ητ==Couette

2Couette RUP =

=∂∂

−==hy

xw y

UητU

τw

τwh

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5Poiseuille Flow

• Poiseuille flow is a pressure-driven flow between stationary parallel plates

• No-slip boundary conditions at plates

LP

xP ∆

−=∂∂

hyatUL

PyU

xx ,00,2

2

==∆

−=∂∂

η

=xU

=maxU

∂∂+

∂∂++

∂∂−= 2

2

2

21yU

xU

xP

DtD xx

xx η

ρgU

τwHigh P τw

Low P Ux

y

Umax

h

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Poiseuille Flow• Volumetric flow rate Q ~ [h3]• Shear stress on plates, , is

dissipative• Force balance

• Net force on fluid and plates is zero since they are not accelerating

• Fluid pressure force ∆PWh (+x) balanced by shear force at the walls of 2 WL (-x)

• Wall exerts shear force (-x), so fluid must exert equal and opposite force on walls, provided by external force

=∫= hxdyUWQ 0

LPh

yU

hy

x

2w

∆=∂∂−=

=

ητ

τw

Wτw

Uxmax3

22

12U

LPhU

WhQU

=∇=

=

η

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5Poiseuille Flow

• Lumped element model for Poiseuille flow

3pois

12Wh

LQPR η=

∆=

( )L

PrrU ox η4

22 ∆−= LPrQ o

ηπ32

4∆=

PerimeterArea4×

=hD ( )h

D DLUfP 2

21 ρ=∆

• Flow in channels of circular cross section

=DDf Re dimensionless constant

• Flow in channels of arbitrary cross section

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• Stokes, or creeping, flow• If Re << 1 , inertial term may

be neglected compared to viscous term

• Development length• Distance before flow assume

steady-state profile

Velocity Profiles

• Velocity profiles for a combination of pressure-driven (Poiseuille) and plate motion (Couette) flow

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5Lecture Outline

• Today’s Lecture• Basic Fluidic Concepts• Conservation of Mass → Continuity Equation• Newton’s Second Law → Navier-Stokes Equation• Incompressible Laminar Flow in Two Cases• Squeeze-Film Damping in MEMS

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Squeezed Film Damping

• Squeezed film damping in parallel plates• Gap h depends on x, y, and t • When upper plate moves

downward, Pair increases and air is squeezed out

• When upper plate moves upward, Pair decreases and air is sucked back in

• Viscous drag of air during flow opposes mechanical motion

moveable

fixed

F

h

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• Assumptions• Gap h << width of plates• Motion slow enough so gas moves under Stokes flow• No ∆P in normal direction• Lateral flow has Poiseuille like velocity profile• Gas obeys Ideal Gas Law• No change in T

Squeezed Film Damping

[ ]PPhKt

Phn ∇+⋅∇=

∂∂ 3)61()(12η

• Reynold’s equation• Navier-Stokes + continuity + Ideal Gas

Law• Knudsen number Kn is ratio of mean

free path of gas molecules λ to gap h• Kn < 0.01, continuum flow• Kn > 0.1, slip flow becomes important• 1 µm gap with room air, λ =

h

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Squeezed Film Damping

[ ]223

24)( Ph

tPh ∇=∂

∂η

• Approach• Small amplitude rigid

motion of upper plate, h(t) • Begin with non-linear partial

differential equation • Linearize about operating

point, average gap h0 and average pressure P0

• Boundary conditions• At t = 0, plate suddenly

displaced vertically amount z0 (velocity impulse)

• At t > 0+, v = 0• Pressure changes at edges

of plate are zero: dP/dh = 0 at y = 0, y = W

Wyand

PPp

dtdh

hp

WPh

tp

=∂

=

−∂∂

=∂∂

ξ

ξη

0

02

2

20

20 1

12

PPpandhhh ∂+=∂+= 00

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5General Solution

• Laplace transform of response to general time-dependent source z(s)

20

20

2

30

4

3

12,96),(

1)(

WPh

hLWbsszbsF c

cS η

πωπη

ω

==+

=

( ) )(1196)( 43

04

3

ssznh

LWsFoddn

nS

∑

+=

απη

plate velocity

1st term for small amplitude oscillation

damping constant

cutoff frequency

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• Squeeze number σd is a measure of relative importance of viscous forces to spring forces• ω < ωc : model reduces to linear resistive damping element

• ω > ωc : stiffness of gas increases since it does not have time to squeeze out

Squeeze Number

ωηωωπσ

020

22 12PhWnumbersqueeze

cd ==

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5Examples

• Analytical and numerical results of damping and spring forces vs. squeeze number for square plate

• Transient responses for two squeeze numbers, 20 and 60

Senturia group, MIT