CHAPTER 11
CONVECTION IN MICROCHANNELS
11.1 Introduction
11.1.1 Continuum and Thermodynamic Hypothesis.
Previous chapters are based on two fundamental assumptions:
(1) Continuum: Navier-Stokes equations, and the energy equation are applicable
(2) Thermodynamic equilibrium: No-velocity slip and no-temperature jump at
boundaries.
Validity criterion: The Knudsen number:
Kn
(1.2)
De
is the mean free path.
Continuum: valid for:
Kn
0.1
(1.3a)
0.001
(1.3b)
No-slip, no-temperature jump:
Kn
11.1.2 Surface Forces
Surface area to volume ratio increases as channel size is decreases.
Surface forces become more important as channel size is reduced.
11.1.2 Chapter Scope
Classification
Gases vs. liquids
Rarefaction
Compressibility
Velocity slip and temperature jump
Analytic solutions: Couette and Poiseuille
flows
Table 11.1
gas
11.2 Basic Considerations
11.2.1 Mean Free Path
Ideal gas:
(11.2)
RT
p 2
Properties if various gases; Table 11.1
7
R
Air
Helium
J/kg K kg/m3
287.0
2077.1
Hydrogen 4124.3
Nitrogen
296.8
Oxygen
259.8
1.1614
0.1625
0.08078
1.1233
1.2840
10
kg/s m
m
184.6
199.0
89.6
178.2
207.2
0.067
0.1943
0.1233
0.06577
0.07155
2
11.2.2 Why Microchannels?
The heat transfer coefficient increases as channel size is decreased.
Examine fully developed flow through tubes and note the effect of diameter
h
3.657
k
D
(11.3)
11.2.3 Classification. Based on Knudsen number
Kn
0.001
continuum, no slip flow
(11.4)
0.001 Kn 0.1 continuum, slip flow
0.1 Kn 10
transition flow
10
Kn
free molecular flow
11.2.4 Macro and Microchannels
Macrochannels
Continuum and thermodynamic equilibrium model applies.
No-velocity slip and no-temperature jump.
Microchannels
Failure of macrochannel theory and correlation.
Distinguishing factors: two and three dimensional effects, axial conduction,
dissipation, temperature dependent properties, velocity slip and temperature
jump at the boundaries and the increasing dominant role of surface forces.
11.2.5 Gases vs. Liquids
Mean free paths of liquids are much smaller than those of gases.
Onset of failure of thermodynamic equilibrium and continuum is not well defined for
liquids.
Surface forces for liquids become more important.
Liquids are almost incompressible while gases are compressible.
11.3 General Features
Rarefaction: Knudsen number effect.
Compressibility: large channel pressure drop, changes in density (compressibility).
Dissipation: Increased viscous effects.
11.3.1 Flow Rate
No-velocity slip: Fig. 11.3a
Velocity slip: Fig. 11.3b
Flow rate Q : Macrochannel theory
underestimates flow rate:
(a)
(b)
Fig. 11.3
3
Qe
Qt
1
(11.5)
w
(4.37a)
11.3.2 Friction Factor
Friction coefficient C f
Cf
(1 / 2) u m2
Friction factor f
f
1D
2 L
p
(11.6)
u m2
Fully developed laminar flow in macrochannels:
f Re
(11.7)
Po
Po is known as the Poiseuille number
Macrochannel theory does not predict Po. The following ratio is a measure of
prediction error
Po
Po
e
C*
(11.8)
t
Po appears to depend on the Reynolds number.
Both increase and a decrease in C are reported.
11.3.3 Transition to Turbulent flow
Macrochannels
Ret
uD
2300
(6.1)
Microchannels: reported transition Reynolds numbers ranged from 300 to 16,000
11.3.4 Nusselt number
Macrochannels:
Fully developed laminar flow: constant Nusselt number, independent of
Reynolds number.
Microchannels: Macrochannel theory does not predict Nu. The following ratio is a
measure of reported departure from macrochannel prediction
0.21
( Nu ) e
( Nu ) t
100
(11.9)
11.4 Governing Equations
In the slip-flow domain, 0.001 Kn 0.1 , the continuity, Navier Stokes equations,
and energy equation are valid.
Important effects: Compressibility, axial conduction, and dissipation.
4
11.4.1 Compressibility
Compressibility affects pressure drop, Poiseuille number and Nusselt number.
11.4.2 Axial Conduction
Axial conduction is neglected in macrochannels for Peclet numbers greater than 100.
Microchannels typically operate at low Peclet numbers. Axial conduction may be
important.
Axial conduction increases the Nusselt number in the velocity-slip domain.
11.4.3. Dissipation
Dissipation becomes important when the Mach number is close to unity or larger.
11.5 Velocity Slip and Temperature Jump Boundary Conditions
In microchannels fluid velocity is not the same as surface velocity. The velocity slip
condition is
2
u ( x,0)
u
u ( x,0) u s
(11.10)
n
u
u(x,0) = fluid axial velocity at surface
u s surface axial velocity
x = axial coordinate
n = normal coordinate measured from the surface
u = tangential momentum accommodating coefficient
Gas temperature at a surface differs from surface temperature:
T ( x,0) Ts
2
T
T
2
1
T ( x,0)
Pr
n
(11.11)
T(x,0) = fluid temperature at the boundary
Ts = surface temperature
c p / cv , specific heat ratio
T
= energy accommodating coefficient
u
and
T
are assume equal to unity.
(11.10) and (11.11) are valid for gases.
11.6 Analytic Solutions: Slip Flows
Consider Couette and Poiseuille flows.
Applications: MEMS.
Thermal boundary conditions: Uniform surface temperature and uniform surface heat flux.
Examine the effects of rarefaction and compressibility.
5
11.6.1 Assumptions
(1) Steady state
(2) Laminar flow
(3) Two-dimensional
(4) Slip flow regime (0.001 < Kn < 0.1)
(5) Ideal gas
(6) Constant viscosity, conductivity, and specific heats
(7) Negligible lateral variation of density and pressure
(8) Negligible dissipation (unless otherwise stated)
(9) Negligible gravity
(10) The accommodation coefficients are assumed equal to unity,
u
T
1 .0 .
11.6.2 Couette Flow with Viscous Dissipation: Parallel Plates with Surface Convection
Stationary lower plate, moving upper plate.
Insulated lower plate, convection at the
upper plate.
y
ho T
us
Determine:
(1) Velocity distribution
(2) Mass flow rate
(3) Nusselt number
x
u
H
Fig. 11.6
Flow Field
x-component of the Navier-Stokes equations for compressible, constant viscosity
flow (2.9), simplifies to
d 2u
(11.12)
0
dy 2
Boundary conditions: apply (11.10)
du( x,0)
dy
u( x,0)
u ( x, H )
du( x, H )
dy
us
(g)
(h)
Solution
u
us
1
(y
1 2 Kn H
Kn)
(11.14)
Mass Flow Rate. The flow rate, m, for a channel of width W is
H
m W
u dy
0
(11.15)
6
(11.14) into (11.15)
m
us
2
WH
(11.16)
Macrochannels flow m o
mo
WH
us
2
(11.17)
Thus
m
mo
(11.18)
1
Nusselt Number. Defined as
2 Hh
k
Nu
(l)
Heat transfer coefficient h:
T (H )
y
Tm Ts
k
h
Substitute into (l)
Nu
T (H )
y
2H
Tm Ts
(11.19)
k thermal conductivity of fluid
T fluid temperature function (variable)
Tm fluid mean temperature
Ts
plate temperature
NOTE:
Fluid temperature at the surface, T ( x, H ), is not equal to surface temperature Ts .
Surface temperature is unknown in this example
Relation between T ( x, H ) and T s is given by the temperature jump condition:
Ts
2
1
T ( x, H )
T ( x, H )
Pr
y
(11.20)
uT dy
(11.22)
Mean temperature Tm
Tm
H
2
us H
0
Temperature distribution: Energy equation simplifies to
2
k
T
y2
0
(11.23)
7
Dissipation function
:
2
u
y
(11.24)
(11.24) into (11.23)
d 2T
du
k dy
dy 2
2
(11.25)
Boundary conditions
dT (0)
dy
k
dT ( H )
dy
(m)
0
ho (Ts T )
Use (11.20)
k
dT ( H )
dy
ho T ( x, H )
2
1
T ( x, H )
T
Pr
n
(n)
Solution: Use (11.14) for u, substitute into (11.25), solve and use boundary
conditions (m) and (n)
kH
H2
2 Kn 2
T
y2
H
T
(11.26)
2
ho
2
1 Pr
us
k H (1 2 Kn )
2
Nusselt number: Use (11.26) to formulate Ts ,
(11.19) Nu
(p)
dT ( H )
and Tm , substitute into
dy
8(1 2 Kn )
8
8 (1 2 Kn ) Kn
1
Kn
3
1
Pr
(11.27)
NOTE:
The Nusselt number is independent of Biot number.
The Nusselt number is independent of the Reynolds number. This is also the case
with macrochannel flows.
The Nusselt number depends on the fluid (Pr and ).
Nusselt number for macrochannel flow, Nu o : set Kn
Nu o
Thus
8
0 in (11.27)
(11.28)
8
Nu
Nu o
1 2 Kn
8 (1 2 Kn ) Kn
1
Pr
8
1
Kn
3
(11.29)
11.6.3 Fully Developed Poiseuille Channel Flow: Uniform Surface Flux
Inlet and outlet pressures are p i and p o
Surface heat flux: q s
Determine:
qs
y
H/2
(1) Velocity distribution
(2) Pressure distribution
(3) Mass flow rate
(4) Nusselt number
H/2
x
qs
Poiseuille flow in microchannels differs
from macrochannels as follows:
Fig. 11.7
Streamlines are not parallel.
Lateral velocity component v does not vanish.
Axial velocity changes with axial distance.
Axial pressure gradient is not linear.
Compressibility and rarefaction are important.
Assumptions. See Section 11.6.1. Additional assumptions:
(11) Isothermal flow.
(12) Negligible inertia forces.
2
(13) The dominant viscous force is
u
y2
.
Flow Field. Determine the axial velocity distribution.
Axial component of the Navier-Stokes equations
p
x
2
u
0
y2
(c)
Boundary conditions
u ( x,0)
y
0
(e)
u ( x, H / 2)
y
(f)
H 2 dp
y2
1 4 Kn ( p ) 4 2
8 dx
H
(11.30)
u ( x, H / 2)
Solution
u
Must determine pressure distribution and lateral velocity v. Continuity for compressible
flow:
9
u
x
v
y
(h)
0
Use ideal gas law in (h)
pv
y
(i)
pu
x
(11.30) into (i)
H2
dp
y2
p (1 4 Kn ( p ) 4 2 )
8
x dx
H
( p v)
y
(j)
Boundary conditions on v
v(x,0)
(k)
0
v( x, H / 2)
(l)
0
Multiply (j) by dy, integrate
y
H2
dp
p
8
x dx
d ( p v)
0
y
(1
4 Kn ( p) 4
0
y2
H2
)
dy
(m)
0
(n)
Evaluate the integrals, solve for v, and use (l)
x
p
4 y3
3 H3
dp
y
1 4 Kn( p)
dx
H
y H /2
Introduce Knudsen number
Kn
Evaluate (n) at y
H
H
2
1
p
RT
(11.33)
H / 2, substitute (11.33) into (n) and integrate
1 2
p
6
H
2 RT p
Cx
(o)
D
where C and D are constants of integration. The solution to this quadratic equation is
2
p( x)
3
2 RT
H
18 RT
H2
6Cx
(p)
6D
Boundary conditions on p
p (0)
pi , p ( L )
po
(q)
Use (q) to find C and D, substitute into (p) and use the definition of Knudsen number
p ( x)
po
6 Kn o
6 Kn o
pi
po
2
(1
pi2
2
po
) 12 Kn o (1
Mass Flow Rate. The flow rate m for a channel of width W is
pi x
)
po L
(11.35)
10
H/ 2
m
2W
(s)
u dy
0
Use (11.30), (11.35) and the ideal gas law
WH 3
p 6
12 RT
H
m
2
RT
dp
dx
(11.38)
Using (11.35) to formulate the pressure gradient, substituting into (11.38), assuming
constant temperature ( T To ), and rearranging, gives
m
1 W H 3 po2
24 LRTo
pi2
1 12 Kn o (
2
po
pi
po
1)
(11.39)
For macrochannel
mo
1 W H 3 po2
12 LRTo
pi
po
1
(11.40)
1 12 Kno
(11.41)
Taking ratio
m
mo
1
2
pi
po
NOTE:
m in microchannels is very sensitive to channel height H.
(11.39) shows the effect of rarefaction and compressibility.
Nusselt Number. Follow Section 11.6.2
Nu
2H qs
k (Ts Tm )
(v)
T s is given by (11.11)
Ts
T ( x, H / 2)
2
1
T ( x, H / 2)
Pr
y
(11.42)
Tm is given by
H /2
uT dy
Tm
0
H /2
udy
0
Temperature distribution. Solve the energy equation.
Additional assumptions:
(11.43)
11
(14) Axial velocity distribution is approximated by the solution to the isothermal case.
(15) Negligible dissipation,
0
2
T / y2
(16) Negligible axial conduction, 2T / x 2
(17) Negligible effect of compressibility on the energy equation, u / x
(18) Nearly parallel flow, v 0
Energy equation: (2.15) simplifies to
T
c pu
x
2
k
T
v/ y
0
(11.44)
y2
Boundary conditions
T ( x,0)
y
(w)
0
and
k
T ( x, H / 2)
y
(x)
qs
Assume:
(19) Fully developed temperature. Define
T ( x, H / 2) T ( x, y )
T ( x, H / 2) Tm ( x)
(11.45)
(y)
(11.46)
0
(11.47)
Fully developed temperature:
Thus
x
Equations (11.45) and (11.46) give
dT ( x, H / 2)
dx
T
x
( y)
dT ( x, H / 2)
dx
dTm ( x)
dx
0
(11.48)
The heat transfer coefficient h, is given by
T ( x, H / 2)
y
Tm ( x) Ts ( x)
k
h
(y)
Use (11.42) and (11.45) into (y)
h
k[T ( x, H / 2) Tm ( x)] d ( H / 2)
Ts ( x) Tm ( x)
dy
Newton’s law of cooling:
h
Equating the above with (11.49)
qs
Ts ( x) Tm ( x)
(11.49)
12
qs
d ( H / 2)
dy
T ( x, H / 2) Tm ( x)
(11.50)
constant
Combining this with (11.48), gives
dTm ( x)
dx
dT ( x, H / 2)
dx
T
x
(11.51)
Conservation of energy for the element in Fig. 11.8 gives
2q sWdx mc pTm
mc p Tm
qs
dTm
dx
dx
Simplify and eliminate m
dTm
dx
m
2q s
= constant
c pum H
Tm
Tm
(11.52)
dx
dTm
dx
dx
qs
(11.52) into (11.51)
dTm ( x)
dx
dT ( x, H / 2)
dx
T
x
Fig. 11.8
2q s
c pum H
(11.53) into (11.44)
2
2q s u
kH u m
T
y2
(11.54)
where u m is given by
um
2
H
H /2
(cc)
udy
0
(11.30) into (cc)
um
H 2 dp
1 6 Kn
12 dx
(11.55)
(11.30) and (11.55)
u
um
6
1
1 6 Kn 4
Kn
y2
(11.56)
H2
(11.56) into (11.54)
2
T
y2
qs 1
12
1 6 Kn kH 4
Kn
y2
(11.57)
H2
Integrating twice and use (w)
T ( x, y )
12q s
1 1
(
(1 6 Kn )kH 2 4
Kn) y 2
y4
12H 2
g ( x)
(11.58)
13
To determine g(x), find Tm using two methods.
First method: Integrate (11.52)
Tm
x
2q s
c pum H
dTm
Tmi
Tm (0)
dx
0
Tmi
(11.59)
2q s
x Tmi
c pum H
(11.60)
Evaluating the integrals
Tm ( x)
Second method: use definition in (11.43). (11.30) and (11.58) into (11.43)
Tm ( x)
3q s H
k (1 6 Kn)
2
13
Kn
40
( Kn) 2
13
560
(11.61)
g ( x)
(11.60) and (11.61) give g(x)
g ( x)
Tmi
2q s
x
c pum H
3q s H
k (1 6 Kn )
Surface temperature Ts ( x, H / 2) :
3q s H
1
Ts ( x )
Kn
k (1 6 Kn ) 2
2
5
48
( Kn ) 2
2
13
Kn
40
qs H
Kn
1 kPr
13
560
(11.62)
(11.63)
g ( x)
Nusselt number: (11.61) and (11.63) into (v)
Nu
2
3
1
Kn
(1 6 Kn) 2
5
48
1
( Kn) 2
(1 6 Kn)
(11.64)
13
Kn
40
(i) The Nusselt number is an implicit function of
x since Kn is a function p which is a function of
x.
(iii) The effect of temperature jump on the
Nusselt number is represented by the last term
in the denominator of (11.64).
(iv) The Nusselt for no-slip, Nu o , is determined
by setting Kn 0 in (11.64)
2
1
Kn
1 Pr
9
NOTE:
(ii) Unlike macrochannels, the Nusselt number
depends on the fluid, as indicated by Pr and
in (11.64).
13
560
8
7
Nu
6
5
4
0.04
0
0.08
0.12
Kn
Fig. 11.9 Nusselt number for air flow between
parallel plates at unifrorm surface
heat flux for air, = 1.4, Pr = 0.7,
u
T 1
14
Nu o
140
17
(11.65)
8.235
(v) Rarefaction and compressibility have the effect of decreasing the Nusselt number.
11.6.4 Fully Developed Poiseuille Channel Flow: Uniform Surface Temperature
Assumptions: same as the uniform flux
case.
The velocity, pressure, and mass flow
rate, are the same as for uniform flux.
Surface boundary condition is different.
Must determine temperature distribution
Ts
y
H/2
x
H/2
Fig. 11.10
Temperature Distribution and Nusselt
Number. Newton’s law of cooling the
Nusselt number for this case is given by
Nu
2 Hh
k
2H
T ( x, H / 2)
Tm ( x) Ts
y
Ts
(11.66a)
Energy equation: Include axial conduction
c pu
2
T
x
k(
T
2
2
y2
x
T
)
(11.67a)
Boundary conditions:
T ( x,0)
y
2
T ( x, H / 2) Ts
(11.68a)
0
H
T ( x, H / 2)
Kn
1 Pr
y
T (0, y )
(11.69a)
Ti
(11.70a)
T ( , y ) Ts
(11.71a)
Axial velocity is given by (11.56)
u
um
6
1
1 6 Kn 4
Kn
y2
(11.56)
H2
Dimensionless variables
T Ts
,
Ti Ts
y
, Re
H
x
,
H RePr
2 um H
, Pe
RePr
(11.72)
Use (11.56) and (11.72), into (11.66a)-(11.71a)
Nu
2
m
( , / 2)
(11.66)
15
6
(1
1 6 Kn 4
Kn
2
( Pe) 2
( ,0)
( ,1 / 2)
2
2
1
)
2
2
(11.67)
2
(11.68)
0
1
Kn
1 Pr
( ,1 / 2)
(11.69)
(0, ) 1
(11.70)
( , ) 0
(11.71)
Solution: method of separation of variables
8
Pe = 0
1
5 8
Result: Fig. 11.11.
NOTE:
7
Nu
The Nusselt number decreases as the Knudsen
number is increased.
Axial conduction
number.
increases
the
6
Nusselt
5
No-slip (Kn = 0) and negligible axial
conduction (Pe
):
Nu o
7.5407
0
0.08
0.04
0.12
Kn
Fig. 11.11 Nusselt number for flow between
parallel plates at uniform surface
temperature for air, Pr = 0.7,
,
1 .4 , u
T 1 [14]
(11.73)
11.6.5 Fully Developed Poiseuille Flow in Microtubes: Uniform Surface Flux
This problem is identical to
Poiseuille flow between
parallel plates at uniform flux
presented in Section 11.6.3.
qs
r
z
Determine the following:
(1) Velocity distribution
(2) Nusselt number
qs
Fig. 11.12
Assumptions. See Section 11.6.3.
Flow Field. Following the analysis of Section 11.6.3. Use cylindrical coordinates.
Results:
r
ro
16
ro2 dp
1 4 Kn
4 dz
vz
vz
v zm
p( z )
po
8Kn o
2
ro4 po2
16 LRTo
m
2
(11.77)
pi2
(1
pi z
)
po L
) 16 Kn o (1
2
po
pi2
pi
1
16
Kn
(
1)
o
po
po2
ro4 p o2
8 LRT
mo
(11.74)
ro2
1 4Kn (r / ro ) 2
1 8Kn
pi
po
8Kn o
r2
( pi
po
(11.78)
(11.79a)
1)
(11.79b)
Nusselt Number. Define
2ro h
k
Nu
Nu
(d)
2 ro q s
k (Ts Tm )
(e)
Results
T (r , z )
Tm
g ( z)
Tmi
q s ro
2
14
Kn
3
16Kn 2
1 r4
4 ro2
7
24
k (1 8Kn)
2q s
q s ro
z
16Kn 2
2
c p ro v z m
k (1 8Kn )
Ts (ro , z )
Nu
qs
(1 4 Kn ) r 2
(1 8 Kn ) k ro
4q s ro
3
Kn
k (1 8Kn )
16
4
(11.92)
g ( z)
(11.95)
g ( z)
14
Kn
3
7
24
(11.96)
q s ro
Kn g ( z )
1 kPr
(11.97)
2
4
( Kn 3 )
(1 8Kn )
16
1
(1 8Kn ) 2
16Kn
2
14
Kn
3
7
24
Nusselt number variation with Knudsen number for air, with
is plotted in Fig. 11.14.
No-slip Nusselt number, Nu o , is obtained by setting Kn
Nu o
48
11
4.364
4
1
Kn
1 Pr
(11.98)
1.4 and Pr
0.7,
0 in (11.98)
(11.99)
18
11.6.6 Fully Developed Poiseuille Flow in Microtubes: Uniform Surface Temperature
The uniform surface flux of Section 11.6.5
is repeated with the tube maintained at
uniform surface temperature Ts .
r
r
z
Temperature Distribution and Nusselt Number
Fig. 11.15
Ts
Same flow field as the uniform surface
flux case of Section 11.6.5
Follow the analysis of Section 11.6.4. and use the flow field of Section 11.6.5.
Dimensionless variables
T Ts
,
Ti Ts
z
, R
2ro RePr
2 um ro
r
, Re
ro
, Pe
RePr
(11.106)
Nusselt number, energy equation, and boundary conditions in dimensionless form
Nu
2
m
1 4 Kn R 2
2(2 16Kn)
(0. )
R
(1, )
2
0
Kn
1 Pr
(1, )
R
1
(R )
R R
R
(11.100)
(2 Pe) 2
(11.101)
2
4.5
(11.102)
(1, )
R
2
1
Pe = 0
1
5
4.0
(11.103)
3.5
(R,0) 1
(11.104)
(R, ) 0
(11.105)
Nu
3.0
Solution: By separation of variables.
2.5
Results: Fig. 11.16.
2.0
0
0.12
0.08
Kn
Fig. 11.16 Nusselt number for flow through tubes
at uniform surface temperature for air,
Pr = 0.7,
1.4, u T 1 , [14]
0.04
ro