Thermofluid principles are used in the design of these devices for both. “
performance” (i.e. functions) and “strength” (e.g. fluid-induced forces).
Thermofluid ...
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 microfabrication processes 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: y
Velocity profile, u(y)
The shear stress: τ = µ
uo θ
du ( y ) dy
(5.2)
where µ = dynamic viscosity of the fluid
u(y)
x
● Many fluid flow cases are characterized by Reynolds number: Re = in which ρ = mass density; V = velocity; L = characteristic length ● Laminar fluid flows occur at Re < 10-100 for compressible fluids, and Re< 1000 for incompressible fluids.
ρLV µ
(5.3)
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: 1
2
Reducer
V1
Q = V1A1 = V2A2 V2 To micro fluidic
1
Diameter d1 = 1000 µm
V1
2
V2
d2 = 20 µm
m3/s
(5.6)
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. 2
C’ 1
C D’
A’
A
V2
V1
D
dt V2
2
B’
B
V1dt
1
The force required to drive the fluid from 1-1 to 2-2, or the flow-induced forces to be:
r r ∑ F = m& (V2 − V1 )
(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. y o : 15 g n i n . ope x a M
1
Fy 2
3
5
Fx
4 Mx2,Vx2 Gas Flow
1 Closure plate
2 Dielectric base plate
3 Electrodes
4 Orifice
5 Silicon die
θ
V Vy Vx
Mx1,Vx1
x
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. y 1
Fy 2
3 5
Fx
4 Mx2,Vx2 Gas Flow
1 Closure plate
2 Dielectric base plate
3 Electrodes
4 Orifice
5 Silicon die
x
Mx1,Vx1
θ
V Vy Vx
The gas stream splits into two components, i.e. Mx1 induced by velocity Vx1 and Mx2 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
m& = ρQ = 0.0826 x (500 x10 −6 ) = 41.3 x10 −6
kg/s
y 1
Fy 2
3
Fx
4
5
Mx2,Vx2 Gas Flow 1 Closure plate
2 Dielectric base plate
3 Electrodes
4 Orifice
5 Silicon die
x
Mx1,Vx1
θ
V Vy Vx
Using Eq. (5.7), we have the following relations:
∑ F y = m& (V y 2 − V y ) and
∑ F x = ( M x1V x1 − M x 2 V x 2) − m& V x
(a) (b)
in which Vy2 = 0; Vx = V cosθ and Vy = V sinθ 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.
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:
( M x1V x1 − M x 2 V x 2) − m& V cos θ = 0 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:
M x1 − M x 2 = m& cos θ
M x1 + M x 2 = m&
(c) (d)
From which, we obtain the split mass flow rates to be: 41.3 x10 −6 m& (1 + cos 75 o ) = 26 x10 −6 M x1 = (1 + cos θ ) = 2 2
kg/s
m& 41.3 x10 −6 (1 − cos 75 o ) = 15.3 x10 −6 M x 2 = (1 − cos θ ) = 2 2
kg/s
A good design, of course, would desire Mx1 >> Mx2.
Laminar Fluid Flow in Circular Conduits - The Hagen-Poiseuille Equation τw
This equation relates the volumetric flow, Q and the corresponding pressure drop, ∆P.
r Velocity profile:
Shear stress profile: τ(r)
Vr(r)
x
r
a
x2 - x1 = ∆L x2
x1
π a4 ⎡ d ⎤ Q= P ρ gy − ( + ) ⎥ 8µ ⎢⎣ dx ⎦ where y = elevation of the tube from a reference plane. The pressure drop in the fluid over the tube length, L is: 8µLQ ∆P = π a4 NOTE: The pressure drop, ∆P ∝
(5.16)
(5.17)
1 meaning a reduction in half in the radius→24=16 a 4 times increase in pressure drop (pumping power)!!
The equivalent head loss in relation to Q is:
h f ,l =
128µLQ πρg d 4
(5.18)
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: dh =
4A p
(5.19)
where A = cross-sectional area of fluid flow p = wet perimeter.
w h1
h Rectangular conduit filled with fluid
4A 4( wh) 2 wh = = dh = p 2( w + h) w + h
Rectangular conduit filled with fluid up to h1
dh =
4 wh1 w + 2h1
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 spherical surface 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 forces of 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 itself exceed 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 tension Fs
Wet perimeter S
Coefficient of surface 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 volumes πa2∆P 2aL∆P 2 πaγ γL
γL
a
L
∆P =
γ a
a
∆P =
(5.24a)
2γ a
(5.24b)
Combining the above two cases for a volume in a microconduit: Tube wall
∆P =
3γ a
Fluid volume
Radius, a Tube diameter d Radius, a
L
∆P is the minimum pressure to be overcome for pumping this volume of liquid.
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:
∆P =
3γ 3x0.073 N/m2 or 876 Pa = = 876 a 250 x10 −6
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
Area, A d
Total amount of heat flow through the slab, Q during time period, t is: A(T a − T b )t Q (5.26) Q=k d where k = thermal conductivity of the solid with a 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: Q (T − T b ) =k a q= (5.27) At d 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: r v (5.28) q r , t = − k ∇T r , t
( )
In which
r r = position vector : ( x, y, z ) in Caartesian coordinate system qz
z
q(r,t)
q( x, y , z, t ) = q2x + q2y + q2z
qx
where
qy
x y
( )
Position vector: r: (x,y,z)
(5.29)
qx = − k x
∂T ( x, y , z, t ) ∂x
(5.30a)
qy = − k y
∂T ( x, y , z, t ) ∂y
(5.30b)
qz = − k z
∂T ( x, y , z, t ) ∂z
(5.30c)
The Heat Conduction Equation
r Q 1 ∂T (r , t ) v ∇ 2T ( r , t ) + = k α ∂t
(5.31)
where the Laplacian is defined as: ∂ ∂ ∂ + + ∇ = ∂ x2 ∂ y 2 ∂ z 2 2
2
2
2
∇ = 2
∂2 1 ∂ 1 ∂2 ∂2 + + + ∂ r 2 r ∂r r 2 ∂ θ 2 ∂ z 2
in Cartesian coordinate system, and
in cylindrical polar coordinate system
In the heat conduction equation , Eq. (5.31), the term Q = Q(r,t) is the heat generated by the solid material.
In MEMS and microsystems, electric resistance heating is commonplace. In such case, this amount of heat generation is equal to: Power P watts (W)
=
Current, I amperes (A)
2
χ
Resistance, R Ohms (Ω)
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:
α=
k ρC
(5.32)
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
Newton’s Cooling Law For heat flow in fluids
Ta A Fluid Ta > Tb
q
Fluid
Tb B
Heat flow from Point A to Point B is expressed as q-the heat flux (w/m2 or J/m2-s) in the expression: 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 dominating parameter 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.
Reynolds number:
Re =
Prandtl number
Pr =
Grashoff number
ρLV µ
Cpµ k
3 2 L ρ g Gr = 2 µ ( β∆t )
(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 surfaces being 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 of the solid structure and the surrounding fluid, the thermal boundary condition at the interface needs to be specifically defined.
Boundary condition at solid-fluid interface Boundary Layer SOLID: T(r,t)
FLUID: Tf
Boundary layer film resistance, 1/h
qs
qf Boundary surface
Normal line to the surface, n
position: rs
−k
r ∂T (r , t ) →
∂n
r rs
= h[T (rr s , t ) − T f ]
(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 in natural convection cases, which is common in microsystems.
Example 5.8 Show the differential equation and the appropriate initial and boundary conditions for 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 bottom surface 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 µm 1000 µm Cu film Si beam Support
100 µm
40 µm
Solution
h=0 T(x,t)
1000 µm
H e a t f l u x i n p u t, q
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 temperature function, T(x,t) in the beam with x being the coordinate in the thickness direction as shown below. The governing differential equation from the general form in Eq. (5-38) for the present Depth of Still air the beam case is:
or
∂ 2T ( x, t ) 1 ∂T ( x, t ) = 2 α ∂t ∂x
Length
Moving air at Tf = 20oC and h = 10-4 W/m2-oC
The initial condition is:
T ( x, t ) t =0 = 20 o C The boundary condition at the top of the beam, i.e. x = 0 is:
x Top face x= 0
Bottom face x = 40 µm
(5.39)
∂T ( x, t ) ∂x
=− x =0
q k
where the heat flux, q = I2R/A, with I = the current passing the thin copper film and R = the electric resistance of the copper film.
Still air h=0 T(x,t)
1000 µm
H e a t f l u x i n p u t,q
The boundary conditions at the bottom surface of the beam:
or
Depth of the beam
(a)
Length
Moving air at Tf = 20oC and h = 10-4 W/m2-oC
Bottom face x = 40 µm
∂T ( x, t ) ∂x
x = 40 x10−6 m
=0
(b) With moving surrounding air with Tf = 20oC and h = 10-4W/m2-oC: (b) We may derive the following boundary condition from Eq. (5.38).
x Top face x= 0
(a) With still surrounding air with h ≈ 0: From Eq. (5.42), we have
∂ ( x, t ) h h + T (x, t ) x = 40×10−6 = T f ∂x x = 40×10−6 k k in which k = thermal conductivity of the silicon beam
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:
Boundary conditions T1(x,t):
K1, α 1
T2(x,t):
K2, α 2
Ti(x,t):
Ki, αi
Boundary conditions
x
X = X1 X = X2 X = X3
X = Xi X = Xi+1
∂ 2Ti ( x, t ) ∂x
2
=
1 ∂Ti ( x, t ) αi ∂t
(5.40)
in which the layer designation, i = 1,2,3,…. with xi ≤ x ≤ xi +1 and t > 0, satisfying the following conditions: Prescribed initial conditions in xi ≤ x at t = 0, and Prescribed boundary conditions at x = 0 and x = xi+1 for t > 0. These conditions are: Ti(xi+1,t) = Ti+1(xi+1,t)
≤ xi +1
for i = 1,2,3,………. , and
∂ T i ( xi +1 , t ) ∂ T i +1 ( xi +1 , t ) for i = 1,2,3,………. = ki k i +1 ∂x ∂x
Example 5.9 The structure of a thermal actuator is made of a compound beam involving silicon and SiO2 as illustrated below. A thin copper film is deposited on the top of the SiO2 layer 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 the input temperature surface temperature 50oC. 1400 µm 1000 µm
50 µm
Cu film heater
Material 1
Silicon dioxide
2 µm
Silicon
Material 2
40 µm Support
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.
Solution: 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. Let T1(x,t) = temperature in SiO2 T2(x,t) = temperature in Si
SiO2
T1(x,t) Heat Flow
T2(x,t)
X = a = 2 µm X=0
From Eq. (5.47), we have the following DEs:
Thermally insulated boundary, q=0
Surface temperature, Ts = 50oC
Si
For SiO2 For Si
∂ 2 T 1 ( x, t ) 1 ∂ T 1 ( x, t ) = 2 ∂t ∂x α1 2 ∂ T 2 ( x, t ) 1 ∂ T 2 ( x, t ) = 2 ∂t ∂x α2
0≤ x≤a
(5.41a)
a≤ x≤b
(5.41b)
The initial conditions:
X = b = 42 µm
x o T 1 ( x, t ) t =0 = F 1 ( x) = 20 C
o T 2 ( x, t ) t =0 = F 2 ( x) = 20 C
The boundary conditions:
o T 1 ( x, t ) x =0 = 50 C
∂ T 2 ( x, t ) ∂x
=0 x =b = 42 µm
The compatibility conditions:
T 1 ( x , t ) x = a = 2 µm = T 2 ( x , t ) x = a = 2 µm
k1
∂ T 1 ( x, t ) ∂x
x = a = 2 µm
∂ ( x, t ) = k2 T 2 ∂x
x = a = 2 µm
The solution of this set of DEs and the associated conditions was carried out by using MathCad, a commercial software package, with graphical output: t = 600 µs
Temperature, oC
50
t = 100 µs
40 t = 50 µs
30
t = 1 µs
20
Time, t = 0
10 0
0
SiO2
2
4 Si
6
8 10 12 14 16 18 20 Depth of the Beam, x (µm)
30
The temperature variations in both layers at selected instances are plotted as shown in 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. ● Another major application of thermofluid engineering principle is in microfabrication such as chemical vapor deposition of thin films. ● Fluid-induced forces must be accounted for in the design of micro valves and pumps. Fluids also affect thermal behavior of matters. ● Thermal analysis in MEMS and microsystems involve conduction and convection heat transfer. ● 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 MEMS structure. This temperature field is used to assess the induced thermal stresses, strains and displacements. These thermally induced mechanical behavior is critical in the design of MEMS and microsystems. ● Thermofluids engineering principles for sub-mcrometer scale are radically different from those in macro-scale. Significant modifications of these principles and formulations are necessary.
End of
Chapter 5
