ME 189 Microsystems Design and Manufacture

Chapter 5

Thermofluid Engineering and Microsystems Design

There are many microscaled devices that involve “heat” or/and “fluid” flows.

Examples such as thermal force-actuated devices and valves and pumps in micro fluidic systems.

Thermofluid principles are used in the design of these devices for both“performance” (i.e. functions) and “strength” (e.g. fluid-induced forces).

Thermofluid principles are also used in the design of microfabricationprocesses such as chemical vapor deposition, oxidation, etc.

Characteristics of Moving Fluids

Fluids have volume but no shape.

Compressible fluids (gases) Incompressible fluids (liquids)

Fluids cannot withstand “normal” stresses, other than “hydrostatic pressures”

Shear stress is responsible for fluid flow. Shear stress is directly proportional to the velocity gradient in moving fluid:

x

yuo

θu(y)

The shear stress:dy

ydu )(µτ =

where µ = dynamic viscosity of the fluid

Velocity profile, u(y)

Many fluid flow cases are characterized by Reynolds number:µ

ρLV=Re

in which ρ = mass density; V = velocity; L = characteristic length

(5.2)

(5.3)

Laminar fluid flows occur at Re < 10-100 for compressible fluids, and Re< 1000 for incompressible fluids.

The Continuity Equation

It is often used to compute the volumetric flow, Q and the velocity, V of a moving fluid through conduits with variable cross-sectional areas.

In the situation such as illustrated below:

Diameterd1 = 1000 µm d2 = 20 µmV1 V2

V1 V2

To micro fluidic

Reducer

1

1 2

2

Q = V1A1 = V2A2 (5.6)m3/s

The Momentum Equation

This equation is derived on the basis of conservation of momentum. It is used to compute the fluid flow-induced forces on the interfacing solids.

It is used in assessing the strength of microvalves and pumps in a design process.

A A’

B B’

C

C’

D

D’

1

1

2

2

V1

V 2

V1dt

V 2dt

The force required to drive the fluid from 1-1 to 2-2, or the flow-induced forces to be:

)( 12 VVmFrr

& −∑ = (5.7)

Example 5.2 Assessing the flow-induced force in a micro valve.

A micromachined silicon valve utilizing electrostatic actuation is constructed. The valve unit has a similar configuration as that reported in [Ohnstein et.al. 1990] as illustrated below.

The thin closure plate is used as the valve with a dimension of 300 µm wide x 400 µm long x 4 µm thick. The plate is bent to open or close by electrostatic actuation to regulate the hydrogen gas flow. The maximum opening of the closure plate is 15-degree tilt from the horizontal closed position.

Determine the force induced by the flow of the gas at a velocity of 60 cm/min and a volumetric rate of 30000 cm3/min. Also, calculate the split of mass flow over the lower surface of the plate.

Gas Flow

1

2345

1 Closure plate 2 Dielectric base plate3 Electrodes 4 Orifice5 Silicon die

Fy

Fx x

y

Mx1,Vx1Mx2,Vx2

θ

VVx

Vy

Max. opening: 15o

Solution:

We look at the situation when the valve plate is at the maximum tilt angle of 15o, which leads to θ=75o in the following diagram.

Gas Flow

1

2345

1 Closure plate 2 Dielectric base plate3 Electrodes 4 Orifice5 Silicon die

Fy

Fx x

y

Mx1,Vx1Mx2,Vx2

θ

VVx

Vy

The gas stream splits into two components, i.e. Mx1 induced by velocity Vx1 andMx2 by velocity Vx2 . We designate Mx1 and Mx2 to be the respective components of the rate of mass flow of the gas, m&

The volumetric flow of the gas, Q = 30000 cm3/min or 500x10-6 m3/sec. The mass density of the gas, ρ = 0.0826 Kg/m3 [Janna 1993] with

66 103.41)10500(0826.0 −− === xxxQm ρ& kg/s

Gas Flow

1

2345

1 Closure plate 2 Dielectric base plate3 Electrodes 4 Orifice5 Silicon die

Fy

Fx x

y

Mx1,Vx1Mx2,Vx2

θ

VVx

Vy

Using Eq. (5.7), we have the following relations:

)( 2 VVmF yyy −=∑ & (a)

and VmVMVMF xxxxxx &−−=∑ )( 2211(b)

Thus, by substituting the values of θ = 75o and V = 60 cm/min or 10-2 m/sec into Eq. (a), we obtained the force Fy = 40x10-8 Kg-m/sec2, or 40x10-8 N.

in which Vy2 = 0; Vx = V cosθ and Vy = V sinθ

The horizontal force component, Fx on the plate exists only if the coefficient of friction between the gas and the contacting plate surface is known. However, we may reasonably assume a friction-less gas flow at that surface, which will then lead, according to Eq. (b), the following relationship:

0cos)( 2211 =−− θVmVMVM xxxx &

It is further reasonable to assume that Vx1 = Vx2 = V in a friction-less flow. Consequently, the split of mass flow at the lower surface of the plate can be obtained by solving the following simultaneous equations:

θcos21 mMM xx &=−

mMM xx &=+ 21

(c)(d)

From which, we obtain the split mass flow rates to be:

66

1 1026)75cos1(2103.41)cos1(

2−

−

=+=+= xxmM o

x θ& kg/s

66

2 103.15)75cos1(2103.41)cos1(

2−

−

=−=−= xxmM o

x θ& kg/s

A good design, of course, would desire Mx1 >> Mx2.

Laminar Fluid Flow in Circular Conduits- The Hagen-Poiseuille Equation

arx

x1 x2

x2 - x1 = ∆L

Vr(r)

Velocity profile:r

Shear stress profile: τ(r)

τw

This equation relates thevolumetric flow, Q and the corresponding pressure drop, ∆P.

⎥⎦⎤

⎢⎣⎡ +−= )(

8

4gyP

dxdaQ ρ

µπ (5.16)

where y = elevation of the tube from a reference plane.

The pressure drop in the fluid over the tube length, L is:

aLQP 4

8πµ

=∆ (5.17)

The equivalent head loss in relation to Q is:

dgLQ

h f 4,128πρ

µ=l (5.18)

NOTE: The pressure drop, 4

1a

P ∝∆ meaning a reduction in half in the radius→24=16times increase in pressure drop (pumping power)!!

Laminar Fluid Flow in Circular Conduits- The Hagen-Poiseuille Equation

For conduits with non-circular cross-sections.

In such cases, hydraulic diameter, dh is used in the Hagen-Poiseuille equations.

This diameter is defined as:

pA

d h4

=(5.19)

where A = cross-sectional area of fluid flowp = wet perimeter.

w

h h1

Rectangular conduit filled with fluid

Rectangular conduit filled with fluid up to h1

hwwh

hwwh

pA

d h +=

+==

2)(2

)(441

1

24

hwwhdh +

=

Incompressible Fluid Flow in Microconduits

Observation: Droplets of water on flat surfaces exhibit “spherical topography”and such phenomenon is possible only with “small” droplets.

Reason: It is the “surface tension” of the water that produces such sphericalsurface of droplets of liquids.

Surface Tension in Liquids

It is the cohesion forces of molecules that exist in all liquids. When a liquid is in contact with air or a solid, the inter- molecular forces

in the liquid bind the liquid molecules beneath the contacting surface, whereas no such force exist at the contacting surface.

Consequently, when the liquid is in contact with air, the inter-molecular forcesof the liquid tend to bond the liquid molecules together.

Since there is no force at the liquid/air contacting surface, the shape of the liquid at the interface becomes spherical.

In the case of larger sized droplets, the “weight” of the liquid droplet itselfexceed the inherited surface tension, and no droplet of spherical shape is possible.

Thus, surface tension is a dominant factor in “small” volume of liquids.

The surface tension of the small volume of fluids at the contacting surface of the conduits, and the friction at the interface result in radically different flow phenomena in microconduits.

Surface tension in small volume fluids presents obstacle to the flow,and extra pressure is required in pumping such flows – Capillary flow.

Magnitude of Surface tension in a liquid

Surface tensionFs

Wet perimeterS

Coefficient ofsurface tension, γ

The coefficient of surface tension, γ with a unit of N/m is a measure of the magnitude of the surface tension.

The γ - value for water can be obtained by the following empirical formula:

γ(T) = 0.07615 – 1.692 x 10-4T

where T is the temperature in oC and γ has units of N/m

(5.23)

Pressure change due to surface tension across liquid volumesπa2∆P

2 πaγa L

2aL∆P

γL γL

a

aP γ

=∆a

P γ2=∆

Combining the above two cases for a volume in a microconduit:

Radius, a

Radius, a Tube diameter

d

L

Fluid volume

Tube wall

aP γ3

=∆

∆P is the minimum pressure to be overcome for pumping this volume of liquid.

(5.24a) (5.24b)

Example 5.5

Determine the pressure required overcoming the surface tension of water in a small tube of 0.5 mm inside diameter. Assume that the water is at 20oC.

Solution:

We first determine the surface tension coefficient of water at 20oC from Eq. (5-23) to be γ = 0.073 N/m.

The tube has a radius of a = 250 µm = 250x10-6 m.

Following the expressions in Eqs. (5-24a,b), we have the pressure required to overcome the surface tension to be:

87610250073.033

6 ===∆−x

xa

P γ N/m2 or 876 Pa

Overview of Heat Conduction in Micro Structures

To assess temperature distribution (i.e. variations), T(r,t) in a micro structure subject heat flow, in which r = position vector, t = time.

The computed temperature distribution T(r,t) is used to assess the induced thermal stresses and strains (and displacements) in the structure such as In Eqs. (4.49) (4.51-4.57) in Chapter 4.

Fourier Law of Heat Conduction

Amount of heat flow, Q Q

d

Area, A

Total amount of heat flow through the slab, Qduring time period, t is:

dtTTAkQ ba )( −

= (5.26)

where k = thermal conductivity of the solid witha unit Btu/in-s-oF or W/m-oC

Thermal conductivity, k is a material property, which represents a material’s ability to conduct heat. It is normally a constant in normal range of temperature.

Heat flux, q, which is equal to:

dTTk

AtQq ba )( −

== (5.27)

is a more meaningful quantity in heat transfer analysis. It represents the “intensity”of heat flow. It has a unit of Btu/in2-s or W/m2.

The above is the Fourier law of heat conduction in simple one-dimensional case. For a more general case, it is expressed as:

( ) ( )trTktrq ,, vr∇−= (5.28)

In which systemcoordinateCaartesianinzyxvectorpositionr ),,(:=r

xy

z q(r,t)

qxqy

qz

Position vector: r: (x,y,z)

qqqtzyxq zyx222),,,( ++= (5.29)

xtzyxT

kq xx ∂∂

−=),,,(

ytzyxT

kq yy ∂∂

−=),,,(

ztzyxT

kq zz ∂∂

−=),,,(

where (5.30a)

(5.30b)

(5.30c)

The Heat Conduction Equation

ttrT

kQtrT

∂∂

=+∇),(1),(2

rv

α(5.31)

where the Laplacian is defined as:

zyx 2

2

2

2

2

22

∂∂+

∂∂+

∂∂=∇ in Cartesian coordinate system, and

zrrrr 2

2

2

2

22

22 11

∂∂+

∂∂+

∂∂

+∂∂=∇

θin cylindrical polar coordinate system

In the heat conduction equation , Eq. (5.31), the term Q = Q(r,t) is the heatgenerated by the solid material.

In MEMS and microsystems, electric resistance heating is commonplace.In such case, this amount of heat generation is equal to:

The power in the above expression has a unit of watt, which is equivalent to 1 Joule/sec. It is also equivalent to 1 N-m/sec in the SI units.

The constant α in Eq. (5-38) is called thermal diffusivity of the material with a unit of m2/sec. It has an important physical meaning of being a measure of how fast heat can conduct in solids (thermal inertia). Mathematically, it is equal to:

Ck

ρα =

in which ρ and C are the respective mass density and specific heat of the solid. The units for ρ is g/cm3, and the unit for C is J/g-oC.

Refer to Table 7.3 for the thermal physical properties of some common MEMS materials

Power Pwatts (W) = Current, I

amperes (A)

2χ Resistance, R

Ohms (Ω)

(5.32)

Newton’s Cooling Law For heat flow in fluids

Fluid

TaTb

q

Ta > Tb Fluid

Heat flow from Point A to Point B is expressed asq-the heat flux (w/m2 or J/m2-s) in the expression:

AB

q = h (Ta – Tb) (5.33)

where h = heat transfer coefficient, W/m2-oC

The magnitude of h depends on the properties of the fluid, but the dominatingparameter is the velocity of the fluid in motion (forced convection).

Heat convection also occur in fluid under no influence of external force, ie“natural (free) convection”. The h-value in forced convection is greater than that in natural convection.

Numerical values of h are determined by the values of the Nusselt number (Nu)from “dimensional analyses” in the following forms. The Nusselt number has an expression of Nu = hL/k, in which L = characteristic length and k = thermal conductivity of the fluid.

For forced convection:

Nu = α (Re)β(Pr)γ

For Natural convection:

Nu = α (Re)β(Pr) γ(Gr)δ

where α, β, γ and δ are constants determined by dimensional analyses with experiments.

µρLV

Re =

kC p µ

=Pr

)(2

23

tgLGr

∆=

βµρ

Reynolds number:

Prandtl number

Grashoff number

(5.3)

(5.34a)

(5.34b)

in which Cp is the specific of heat of fluids under constant pressure, β is the volumetric coefficient of thermal expansion, ∆t is the duration, and g is the gravitational acceleration.

Solid-Fluid Interaction

Modes of heat transfer:

Conduction in solids governed by Fourier law in Eq. (5.28)Convection in fluids governed by Newton’s cooling law in Eq. (5.33)

There are MEMS structures, e.g. thermally actuated beams with their surfacesbeing in contact with surrounding fluids.

At these interfaces the two modes of heat transfer take place with either:conduction to convection, or convection to conduction.

The situation is further complicated with the building of a “boundary layer” at the interface on the fluid side. Such boundary layer adds resistance to heat flow.Consequently, the temperature of the solid at the interface is not equal to that of the contacting fluids.

Because of both heat conduction and convection take place at the interface ofthe solid structure and the surrounding fluid, the thermal boundary conditionat the interface needs to be specifically defined.

Boundary condition at solid-fluid interface

SOLID:T(r,t)

FLUID:Tf

BoundaryLayer

Normal line tothe surface, nBoundary surface

position: rs

Boundary layer filmresistance, 1/hqs

qf

]),([),(TtTh

n

trTk fssr r −=∂

∂−

→rr

r

(5.35)

The thickness of the boundary layer relates to the velocity of the surrounding fluid.Thicker layers are produced with slow moving fluid, with extreme values innatural convection cases, which is common in microsystems.

Example 5.8

Show the differential equation and the appropriate initial and boundary conditionsfor a thermally actuated micro beam as illustrated below. A thin copper film is attached to the top surface of the silicon beam used as a resistant heater. The actuator is initially at 20oC. Consider two cases for the contacting air at the bottomsurface of the beam: (a) still air, (b) the air has a bulk temperature of 20oC but has a heat transfer coefficient of 10-4 W/m2-oC.

1200 µm1000 µm

Cu film

Si beam

Support

100 µm40 µm

Solution

We may consider the induced temperature field in the beam that will predominantly vary in the thickness of the beam. It is thus reasonable to assume a temperaturefunction, T(x,t) in the beam with x being the coordinate in the thickness direction as shown below.

T(x,t)

He a

t fl u

x i n

p ut, q

1000

µm

x

Still airh=0

or

Moving air atTf = 20oC and

h = 10-4 W/m2-oC

Top facex= 0

Bottom facex = 40 µm

Length

Depth ofthe beam

The governing differential equation from the general form in Eq. (5-38) for the present case is:

ttxT

xtxT

∂∂

=∂

∂ ),(1),(2

2

α(5.39)

The initial condition is:o

ttxT 20),( 0 == C

The boundary condition at the top of the beam, i.e. x = 0 is:

kq

xtxT

x−=

∂∂

=0

),(

where the heat flux, q = I2R/A, with I = thecurrent passing the thin copper film and R = the electric resistance of the copper film.

T(x,t)

Hea

t flu

x in

p ut, q

1000

µm

x

Still airh=0

or

Moving air atTf = 20oC and

h = 10-4 W/m2-oC

Top facex= 0

Bottom facex = 40 µm

Length

Depth ofthe beam

The boundary conditions at the bottom surface of the beam:

(a)

(b)

(a) With still surrounding air with h ≈ 0:From Eq. (5.42), we have

0),(61040

=∂

∂−= mxxx

txT

(b) With moving surrounding air with Tf = 20oC and h = 10-4W/m2-oC:We may derive the following boundarycondition from Eq. (5.38).

( ) ( ) fxx

TkhtxT

kh

xtx

=+∂

∂−

−×=

×=6

61040

1040

,,

in which k = thermal conductivity ofthe silicon beam

x

X = X1X = X2

X = X3

X = Xi

X = Xi+1

T1(x,t):

T2(x,t):

Ti(x,t):

K1, α1

K2, α2

Ki, αi

Boundary conditions

Boundary conditions

Heat Conduction in Multilayered Thin Films

Many MEMS devices are made of layers of dissimilar materials. Heat flow through these layers of dissimilar materials require special formulations.

The governing DE for a multi-layer solid is:

ttxT

xtxT i

i

i

∂∂

=∂

∂ ),(1),(2

2

α(5.40)

in which the layer designation, i = 1,2,3,….with and t > 0, satisfying the following conditions:

xxx ii 1+≤≤

Prescribed initial conditions in at t = 0, and

Prescribed boundary conditionsat x = 0 and x = xi+1 for t > 0.

These conditions are:

xxx ii 1+≤≤

Ti(xi+1,t) = Ti+1(xi+1,t) for i = 1,2,3,………. , and

xtxTkx

txTk iii

iii ∂

∂=

∂∂ ++

++ ),(),( 11

11 for i = 1,2,3,……….

Example 5.9

The structure of a thermal actuator is made of a compound beam involving siliconand SiO2 as illustrated below. A thin copper film is deposited on the top of the SiO2layer as the resistant heater.

This heater will provide a maximum temperature of 50oC at the top surface of the SiO2 layer. Determine the time required for the entire silicon beam to reach theinput temperature surface temperature 50oC.

1400 µm

1000 µm

Cu film heater

Silicon dioxide

Silicon

Support

50 µm

2 µm

40 µm

Given material properties are:Thermal conductivities: k1 = 1.4 w/m-oC for SiO2 and k2 = 157 w/m-oC for silicon.Thermal diffusivities: α1 = 0.62x10-6 m2/sec for SiO2 and α2 = 97.52x10-6 m2/sec for silicon.

Material 1

Material 2

Since heat will predominantly flow through the thickness of the compound beam due to the short distance of the passage, a one-dimensional heat conduction analysis along the thickness direction is justified.

Solution:

SiO2Si

Surf

ace

tem

pera

ture

, Ts =

50o C

Ther

mal

ly i n

sula

ted

b oun

dary

, q

= 0

x

X = 0

X = a = 2 µm X = b = 42 µm

T1(x,t)

T2(x,t)

Heat Flow

Let T1(x,t) = temperature in SiO2T2(x,t) = temperature in Si

ttxT

xtxT

∂∂

=∂

∂ ),(1),( 1

12

12

α

From Eq. (5.47), we have the following DEs:

bxa ≤≤

ax ≤≤0For SiO2

For Sit

txTx

txT∂

∂=

∂∂ ),(1),( 2

22

22

α

The initial conditions:o

t xFtxT 20)(),( 101 ===

ot xFtxT 20)(),( 202 ===C C

The boundary conditions:o

xtxT 50),( 01 ==

0),(

42

2 =∂

∂

== mbxxtxT

µC

The compatibility conditions:

maxmax txTtxT µµ 2221 ),(),(====

=maxmax x

txTkx

txTkµµ 2

22

2

11

),(),(

==== ∂∂

=∂

∂

(5.41a)

(5.41b)

The solution of this set of DEs and the associated conditions was carried outby using MathCad, a commercial software package, with graphical output:

0

10

20

30

40

50

20 4 6 8 10 12 14 16 18 20 30Depth of the Beam, x (µm)

Tem

pera

ture

, oC

Time, t = 0t = 1 µs

t = 50 µst = 100 µs

t = 600 µs

SiSiO2

The temperature variations in both layers at selected instances are plotted as shownin the graph above, from which we determined the time required for the silicon layer to reach the input temperature of 50oC is 600 micro seconds. This information will enable the design engineer to assess the sensitivity of the thermally actuated device.

SUMMARY Thermofluids engineering principles are used in the design of MEMS

microsystems such as micro valves and micro fluididcs. Many of these devices and systems are thermally actuated.

Fluid-induced forces must be accounted for in the design of microvalves and pumps. Fluids also affect thermal behavior of matters.

Thermal analysis in MEMS and microsystems involve conduction andconvection heat transfer.

Another major application of thermofluid engineering principle is in microfabrication such as chemical vapor deposition of thin films.

Fourier law governs heat conduction in solids, whereas Newton’s cooling law is used in convective heat transfer.

Heat conduction equation, with or without convective boundary conditions,is used to determine the temperature field (distribution) in the MEMSstructure. This temperature field is used to assess the induced thermal stresses, strains and displacements. These thermally induced mechanicalbehavior is critical in the design of MEMS and microsystems.

Thermofluids engineering principles for sub-mcrometer scale are radicallydifferent from those in macro-scale. Significant modifications of these principles and formulations are necessary.

End ofChapter 5