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MECH 607 Micro Flows Fundamentals and Applications September 30,
2014American University of Beirut, Spring 2014 PS1
Problem Set 1
Problem 0This is more of a treasure hunt than a problem. You are
to conduct a relatively briefliterature review on the historical
evolution of the slip boundary condition since its in-ception by
Maxwell. Highlight the major achievement and challenges. You are to
includethe concept of Knudsen minimum in your discussion. Make sure
your include all thereferences you used.
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Problem 1: Pressure Driven Microchannel Flow
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Problem 2: Couette Flow
We consider the Couette flow problem in which the upper plate is
subject to harmonicdisplacement z = z0 sin(t). Assume h(x, y) = h0
is a constant and Ly >> Lx >> h0.(a) What are the
equation(s) (in dimensionless form) governing the Couette flow
under
consideration. Use the expression for the Stokes number =
(1/t0)h20
.
(b) Find the velocity distribution.(c) What is the total force
acting on the moving plate.(d) what are the damping (b) and spring
(k) coefficients if we model the upper platemotion by
mz + bz + kz = Fex
We consider Couette flow of a plate of dimensions 202m by 100m
by 5m as shown
202 m100 m
50 m
5 m4 m
5 m
Aluminum:
! = 2697 kg/m3E = 70 GPa
" = 0.3
2 m
-V cos(#t)
V cos(#t)
10 m
w
s
202 m100 m
50 m
5 m4 m
5 m
Aluminum:
! = 2697 kg/m3E = 70 GPa
" = 0.3
2 m
-V cos(#t)-V cos(#t)
V cos(#t)V cos(#t)
10 m
ww
s
in the Figure. Assume that the plate is thick enough to behave
as a rigid body. Theplate is actuated by a comb drive from both
sides. The plate is grounded while the statorcombs from both side
are subject to a similar magnitude but opposite sign voltage ( +and
-V0 cos( t)) so that when when is pulling the other is pushing and
vice versa. Theplate is suspended by four identical beams each of
50m length, 2m width, and 4mthick. The plate is separated from the
ground by h = 5m. On each side of the platethere are N comb fingers
each 5m thick and 25 length. Noting that the length of theplate is
L = (2N + 1)wf + 2Ns where N is the number of fingers connected to
the plate,wf is the finger width assumed equal for all fingers
(moving and stationary), and s isthe spacing between two
consecutive fingers, also assumed equal everywhere. Assumewf = 2m,
s = 2m, which leads to N = 25. The initial overlap between moving
andnon-moving fingers is l0 = 10m.(e) Estimate the damping
coefficient.
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Problem 3: Squeeze Film Damping
Consider squeeze film damping due to small amplitude step
vertical displacement (orvelocity impulse dh
dt= z0 (t) where is the Dirac delta function) of a plate
separated
initially from ground by h0. The Reynolds equation is
12(ph)
t=
[(1 + 6Kn)h3pp
]Assume the following:
h is only time dependent and is given by h(t) = h0 + h(t) where
h/h Lx >> h0.
(a) Simplify Reynolds equation and express the resulting
equation in terms of p = p/p0,
x = x/Lx, t = t/(Lx/U0), the squeeze number =12 (1/t0)L2x
p0h20. Specify the initial and
boundary conditions.(b) Find the pressure distribution, p(x).(c)
What is the total force acting on the upper plate F (t)?(d) Using
Laplace transform, what is the force in the frequency domain? What
is theLaplace transform of the force for a general time-dependent
motion of the upper plate?(e) What is the electric circuit analog
of the problem model? Comment on the nature ofthe force at low
frequencies and high frequencies.We consider squeeze film damping
of am RF switch, modeled as a plate of dimensions
150 m
25 m
25 m
100 m
20 m
5 m2 m
3 m
Aluminum:
! = 2697 kg/m3E = 70 GPa
" = 0.3
5 m
V cos(#t)
150 m
25 m
25 m
100 m
20 m
5 m2 m
3 m
Aluminum:
! = 2697 kg/m3E = 70 GPa
" = 0.3
5 m
V cos(#t)
150m (length) by 25m (width) by 5m (thickness). Assume that the
plate is thickenough to behave as a rigid body. The plate is
suspended by two 25m by 5m by 2mtethers as shown in the Figure. The
plate is separated from a rectangular 100m by 20melectrode by a gap
h0 = 3m. plate, tethers, and electrode are made of Aluminum and
are
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assumed to be perfect conductors. The plate-electrode perform as
a parallel plate capac-itor actuated by a voltage V = V0 cos( t).
The ambient pressure is atmospheric pressure.
Assume that the restoring force on the plate due to the tethers
is Ftether = EW H3/4L3 z
per tether, where E is the modulus of elasticity, W is the
tether width, H is the tetherthickness, L is the tether length, and
z is the vertical plate displacement from equilibrium.Also assume
that the electrostatic force applied on the plate is Felectrostatic
= 0AV
2/2h2
where 0 is the permittivity of free space, A is the area of the
electrode, V is the voltageand h is the gap between the the plate
and the electrode.
(a) Write down the balance of the various forces: electrostatic
force, restoring spring forcedue to tethers, and gas force related
to the inertia of the plate. Neglect weight of tethers.
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Solution of the Heat EquationWe consider the boundary-value
problem of heat conduction
2u
z2+g(z, t)
k=
1
u
tin 0 z 1, t > 0
k1uz
+ h1u = f1(t) at x = 0, t > 0
k2u
z+ h2u = f2(t) at x = 1, t > 0
u = F (z) in 0 z 1, t = 0
The general solution of the boundary-value heat conduction
equation is
u(z, t) =m=1
e2m tK(m, z)
[F (m) +
t0e
2m t
A(m, t
) dt]
where
A(m, t) =
kg(m, t
) +
[K(m, z)
k1
z=0
f1(t) +
K(m, z)
k2
z=1
f2(t)
]
F (m) = 10K(m, z
)F (z) dz
g(m, t) =
10K(m, z
) g(z, t) dz
If k1 = 0, replaceK(m,z)
k1
z=0
with 1h1
dK(m,z)dz
z=0
.
If k2 = 0, replaceK(m,z)
k2
z=1
with 1h2
dK(m,z)dz
z=1
.
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The kernel K and eigenvalues m have specific forms depending on
the type boundaryconditions, are listed in Table 1.
Table 1: Kernels and Eigenvalues for different boundary
conditions. With H1 = h1/k1and H2 = h2/k2.
B.C. at B.C. at Kernel K(m, z) Eigenvalues mx = 0 x = 1 positive
roots of:
3rd kind 3rd kind
2 m cos mx+H1 sin mx[(2m+H
21 )
(1+
H22
2m+H22
)+H1
]1/2 tan = (H1+H2)2H1H2H1 finite H2 finite
3rd kind 2nd kind
2[
2m+H21
(2m+H21 )+H1
]1/2cos m(1 x) tan = H1
H1 finite H2 = 0
2nd kind 3rd kind
2[
2m+H22
(2m+H22 )+H2
]1/2cos mx tan = H2
H1 = 0 H2 finite
3rd kind 1st kind
2[
2m+H21
(2m+H21 )+H1
]1/2sin m(1 x) cot = H1
H1 finite H2 =1st kind 3rd kind
2[
2m+H22
(2m+H22 )+H2
]1/2sin mx cot = H2
H1 = H2 finite2nd kind 2nd kind
2 cos mx
? sin = 0H1 = 0 H2 = 0
2nd kind 1st kind
2 cos mx cos = 0H1 = 0 H2 =1st kind 2nd kind
2 sin mx cos = 0
H1 = H2 = 01st kind 1st kind
2 sin mx sin = 0
H1 = H2 =
? For this particular case, replace
2 by 1 when = 0.
For example, ff the boundary conditions at z = 0 and z = 1 are
of the first type, thekernel and eigenvalues are
sin m = 0 with m > 0 m = mpiK(m, z) =
2 sin(mz) =
2 sin(mpi z)
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