3.155J/6.
152J Lecture 21:
Fluids Lab Testing
Prof. Akintunde I. Akinwande
Massachusetts Institute of Technology
5/2/2005
Slides prepared by Professor Martin Schmidt
Information
Take Home Quiz
Due Friday, May 6, Room 39-553
No Electronic Submissions
Course evaluations at the end of the
lecture on Wednesday, May 4.
Spring 2005 – A. I. Akinwande 3.155J/6.152J – Lecture 21 – Slide 2
1
Outline
Review of the Process and Testing
Fluidics
Solution of Navier-Stokes Equation
Solution of Diffusion Problem
Lab Report Guidance
References
Senturia, Microsystems Design, Kluwer, Sect.13.3
6.021 Web Site on Microfluidics Lab
Plummer, Chapter 7, p.382-384
Spring 2005 – A. I. Akinwande 3.155J/6.152J – Lecture 21 – Slide 3
Process Flow - Overview
Unexposed
Si SU-8 (100 µm) Surface treatment &
casting PDMS
photolithography
PDMS
UV light
Si
mask
removing elastomer from
Si master
PDMS
development
seal against glass after plasma
treatment and insert tubing
“master”
Si
tubing Our process
was changed
here
Spring 2005 – A. I. Akinwande 3.155J/6.152J – Lecture 21 – Slide 4
2
The Mixer
Width = 250µm, 500 µm
Depth = 100 µm
Inlet Length = 25 mm
Outlet Length = 35 mm
Images: Prof. D. Freeman
Spring 2005 – A. I. Akinwande 3.155J/6.152J – Lecture 21 – Slide 5
Packaging/Testing
Images: Prof. D. Freeman
Spring 2005 – A. I. Akinwande 3.155J/6.152J – Lecture 21 – Slide 6
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Experiment
Gravity feed of fluids
Requires ‘priming’ of channel
Particles for velocity measurement
We will attempt this
Dye for diffusion experiments
Food color
Measurements
Particle velocity
Diffusion
Spring 2005 – A. I. Akinwande 3.155J/6.152J – Lecture 21 – Slide 7
Navier-Stokes
The Navier-Stokes equation for
incompressible flow:
DU
ρm = η∇ 2 U − ∇P *
Dt
U = velocity P* = P − ρm g • r
P* = pressure (minus gravity body force)
ρm = fluid density (103 kg/m3 for water)
η = viscosity (10-3 Pa-s for water)
Spring 2005 – A. I. Akinwande 3.155J/6.152J – Lecture 21 – Slide 8
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Poiseuille Flow
Assume width (w) >> height (h)
Neglect entrance effects (L >> h)
L
w
Spring 2005 – A. I. Akinwande 3.155J/6.152J – Lecture 21 – Slide 9
Simplify to our problem
No time dependence DU
d/dt = 0 ρm = η∇ 2 U − ∇P *
Dt
Flow is constant in x-
direction (and 0 in z)
U = f(y)
Pressure is only a ∂2Ux K
function of x = −
A linear pressure drop
∂y 2 η
dP Stokes Flow
= −K (constant)
dx
Spring 2005 – A. I. Akinwande 3.155J/6.152J – Lecture 21 – Slide 10
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Poiseuille Flow
h
high
τw low
pressure τw pressure
Ux
Umax
Senturia, Figure 13.5
‘No-Slip’ Boundary conditions
Ux(y=0) = 0
Ux(y=h) = 0
Spring 2005 – A. I. Akinwande 3.155J/6.152J – Lecture 21 – Slide 11
Solution
∂2Ux K
= −
∂y 2 η
Solution is a quadratic polynomial
Ux = a + by + cy2
Using boundary conditions and
substitution
1
Ux = ⎡⎣ y ( h − y ) ⎤⎦ K
2η
Spring 2005 – A. I. Akinwande 3.155J/6.152J – Lecture 21 – Slide 12
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Parabolic Flow Profile
h Ux =
1
⎡ y ( h − y ) ⎤⎦ K
τw 2η ⎣
high low
pressure τw pressure
Ux
Umax
h2
Maximum velocity U max = K
8η
h Wh 3
Flow rate Q = W ∫ U x dy = K
0 12η
Average velocity Q h2 2
U= = K = U max
Wh 12η 3
Spring 2005 – A. I. Akinwande 3.155J/6.152J – Lecture 21 – Slide 13
Pressure drop over length
∆P 12ηL ∆P 12ηL
K= ∆P = Q R pois = =
L Wh 3 Q Wh 3
∆P = ρgH
H = height of water
g = gravity
Spring 2005 – A. I. Akinwande 3.155J/6.152J – Lecture 21 – Slide 14
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Flow Issues
Edge effects
Flow rate
Particle location in channel
Dimensions
Merging of channels
How to model
Spring 2005 – A. I. Akinwande 3.155J/6.152J – Lecture 21 – Slide 15
The Mixer – Mixing by diffusion
Width = 250µm, 500 µm
Depth = 100 µm
Inlet Length = 25 mm
Outlet Length = 35 mm
Images: Prof. D. Freeman
Spring 2005 – A. I. Akinwande 3.155J/6.152J – Lecture 21 – Slide 16
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Diffusion Image Sequence
Think of this axis as length or time
Spring 2005 – A. I. Akinwande 3.155J/6.152J – Lecture 21 – Slide 17
Imaging System Output
Spring 2005 – A. I. Akinwande 3.155J/6.152J – Lecture 21 – Slide 18
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Diffusion
Same problem as diffusion in an epi layer
As in the case of the design problem
Dopant
n - epi Concentration
n+ - silicon
Solution in Plummer, Chapter 7, p.382
Spring 2005 – A. I. Akinwande 3.155J/6.152J – Lecture 21 – Slide 19
Solution
Initial Conditions
C
Identical to Infinite
Source Problem:
C⎡ ⎛ x ⎞⎤
C ( x, t ) = ⎢1 − erf ⎜ ⎟⎥
2⎣ ⎝ 2 Dt ⎠ ⎦
Spring 2005 – A. I. Akinwande 3.155J/6.152J – Lecture 21 – Slide 20
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An ‘Intuitive’ way to look at it…
Think of the uniform
concentration as a
sum of dopant
‘pulses’
Each ‘pulse’ has a
Gaussian diffusion
profile
Dose = C ∆x
C n ⎡ ( x − x i )2 ⎤
Apply superposition C ( x, t ) =
2 πDt
∑ ∆x i exp ⎢ −
⎢⎣ 4Dt ⎥
⎥
since diffusion is i =1
⎦
linear
Spring 2005 – A. I. Akinwande 3.155J/6.152J – Lecture 21 – Slide 21
Solution
Taking the limit of ∆x
C n ⎡ ( x − x i )2 ⎤
C ( x, t ) =
2 πDt
∑ ∆x i exp ⎢ −
⎢⎣ 4Dt ⎥
⎥
i =1
⎦
C ⎡ ( x − α )2 ⎤
C ( x, t ) =
x
2 πDt 0
∫ exp ⎢⎢− 4Dt ⎥⎥dα
⎣ ⎦
(x − α) = η
2 Dt
C x 2 Dt
C ( x, t ) = ∫ exp ⎡⎣ −η2 ⎤⎦dη
π −x
Spring 2005 – A. I. Akinwande 3.155J/6.152J – Lecture 21 – Slide 22
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Solution
C
C ( x, t ) =
x 2 Dt
π ∫−x
exp ⎡⎣ −η2 ⎤⎦dη
C⎡ ⎛ x ⎞⎤
C ( x, t ) = ⎢1 − erf ⎜ ⎟⎥
2⎣ ⎝ 2 Dt ⎠ ⎦
erfc ( x ) = 1 − erf ( x )
C⎡ ⎛ x ⎞⎤
C ( x, t ) = ⎢ erfc ⎜ ⎟⎥
2⎣ ⎝ 2 Dt ⎠ ⎦
Spring 2005 – A. I. Akinwande 3.155J/6.152J – Lecture 21 – Slide 23
Error Function Solution
C⎡ ⎛ x ⎞⎤
C ( x, t ) = ⎢1 − erf ⎜ ⎟⎥
2⎣ ⎝ 2 Dt ⎠ ⎦
Spring 2005 – A. I. Akinwande 3.155J/6.152J – Lecture 21 – Slide 24
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Diffusion Issues
Fitting ideal curve to measured profiles
Scaling time to position
Choice of velocity
Non-ideal flow profiles
Spring 2005 – A. I. Akinwande 3.155J/6.152J – Lecture 21 – Slide 25
Fluids Lab Report
Follow the Letters format
Purpose: Characterization of a Liquid
Micromixer
Report Flow Velocity
Compare to calculated
Estimate errors
Extract an effective diffusion coefficient
Utilize ‘best estimate’ for flow velocity
Compare to expected (D ~ 2x10-6 cm2/s)
Identify relevant non-idealities
Spring 2005 – A. I. Akinwande 3.155J/6.152J – Lecture 21 – Slide 26
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