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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