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 
 is the mean free path.
 Continuum: valid for:
 No-slip, no-temperature jump:

(1.2)
De
Kn  0.1
(1.3a)
Kn  0.001
(1.3b)
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
Air
Helium
11.2.1 Mean Free Path

Ideal gas:


 
RT
(11.2)
p 2
Properties if various gases; Table 11.1

J/kg K kg/m3
R
287.0
2077.1
Hydrogen 4124.3
Nitrogen
296.8
Oxygen
259.8

kg/s m  m
 107
1.1614
0.1625
184.6
199.0
0.08078
89.6
178.2
207.2
1.1233
1.2840
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
0.001 Kn  0.1 continuum, slip flow
0.1  Kn  10
transition flow
10  Kn
free molecular flow
(11.4)
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
1
Qt
(11.5)
11.3.2 Friction Factor

Friction coefficient C f
w
Cf 

(1 / 2)  u m2
Friction factor f
f 

1 D p
2 L  um2
(11.6)
Fully developed laminar flow in macrochannels:
f Re  Po


(4.37a)
(11.7)
Po is known as the Poiseuille number
Macrochannel theory does not predict Po. The following ratio is a measure of
prediction error
Poe
Pot
C*
(11.8)
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
Re t 
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
 100
( Nu) t
(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 u( x,0)
(11.10)
u( x,0)  u s 

u
n
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 2  T ( x,0)
 T 1   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 )  u s  

du ( x, H )
dy
(g)
(h)
Solution
u
1

( y  Kn)
u s 1  2 Kn H
(11.14)
Mass Flow Rate. The flow rate, m, for a channel of width W is
m W

H
0
 u dy
(11.15)
6
(11.14) into (11.15)
m   WH
us
2
(11.16)
us
2
(11.17)
Macrochannels flow m o
mo   WH
Thus
m
1
mo
(11.18)
Nusselt Number. Defined as
Nu 
Heat transfer coefficient h:
2 Hh
k
(l)
T ( H )
y
h
Tm  Ts
k
Substitute into (l)
T ( H )
y
Nu  2 H
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 Ts is given by the temperature jump condition:
Ts  T ( x, H ) 

2  T ( x, H )
1   Pr
y
Mean temperature Tm
2
Tm 
us H

(11.20)

H
uT dy
(11.22)
0
Temperature distribution: Energy equation simplifies to
k
 2T
y 2
   0
(11.23)
7
Dissipation function  :
 u 
   
 y 
2
(11.24)
(11.24) into (11.23)
d 2T
dy 2


  du 
2
 
k  dy 
(11.25)
Boundary conditions
dT (0)
0
dy
k
(m)
dT ( H )
 ho (Ts  T )
dy
Use (11.20)
k

dT ( H )
 ho
dy


2  T ( x, H )
 T 
T ( x, H )  1   Pr
n


(n)
Solution: Use (11.14) for u, substitute into (11.25), solve and use boundary
conditions (m) and (n)

kH H 2
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, Nuo : set Kn  0 in (11.27)
Nuo  8
Thus
(11.28)
8
Nu

Nuo
1  2 Kn
8
8 (1  2 Kn) Kn
1  Kn 
3
 1
Pr
(11.29)
11.6.3 Fully Developed Poiseuille Channel Flow: Uniform Surface Flux



Inlet and outlet pressures are p i and po
Surface heat flux: q s
Determine:
H/2
(1) Velocity distribution
(2) Pressure distribution
(3) Mass flow rate
(4) Nusselt number

H/2
x
qs
Poiseuille flow in microchannels differs
from macrochannels as follows:





qs
y
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.
(13) The dominant viscous force is 
 2u
y 2
.
Flow Field. Determine the axial velocity distribution.

Axial component of the Navier-Stokes equations



Boundary conditions
p
 2u
 2 0
x
y
u ( x,0)
0
y
u ( x, H / 2)
u ( x, H / 2)  
y
(c)
(e)
(f)
Solution
u
H 2 dp 
y2 
1  4 Kn( p )  4 2 
8 dx 
H 
(11.30)
Must determine pressure distribution and lateral velocity v. Continuity for compressible
flow:
9
Use ideal gas law in (h)

 u     v   0
x
y
(h)

 p v      pu 
y
x
(i)
(11.30) into (i)
2
2 


( pv)  H   p dp (1  4 Kn( p)  4 y 2 )
y
8 x  dx
H 
Boundary conditions on v
(j)
v( x,0)  0
(k)
v( x, H / 2)  0
(l)
Multiply (j) by dy, integrate
y
H 2   dp
d ( p v) 
p
8 x  dx

0


y
(1  4 Kn( p)  4
0
y2 
) dy
H 2 
(m)
Evaluate the integrals, solve for v, and use (l)
  dp 
y 4 y3 
1  4 Kn( p) 

p
x  dx 
H 3 H 3 

0

 y H / 2
(n)
Introduce Knudsen number
Kn 

H



H
2
RT
1
p
(11.33)
Evaluate (n) at y  H / 2, substitute (11.33) into (n) and integrate
1 2 
p 
6
H
2 RT p  Cx  D
(o)
where C and D are constants of integration. The solution to this quadratic equation is
p ( x)  3

H
2 RT  18 RT
2
H2
 6Cx  6 D
(p)
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
2


pi 
pi2
pi  x
p ( x)
 6 Kno  6 Kno 
)
  (1  2 )  12 Kno (1 
po
po 
po  L
po


Mass Flow Rate. The flow rate m for a channel of width W is
(11.35)
10
m  2W
H/ 2

 u dy
(s)
0
Use (11.30), (11.35) and the ideal gas law
WH 3 

m
p  6
12  RT 
H
 dp
RT 
2
 dx

(11.38)
Using (11.35) to formulate the pressure gradient, substituting into (11.38), assuming
constant temperature ( T  To ), and rearranging, gives

pi
1 W H 3 po2  pi2
m
 1)
 2  1  12 Kno (
24  LRTo  po
po

(11.39)
For macrochannel

1 W H 3 po2  pi
 1

12  LRTo  po 
(11.40)

m 1  pi
 
 1  12 Kno 
mo 2  po

(11.41)
mo 
Taking ratio
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 qs
k (Ts  Tm )
Ts is given by (11.11)
Ts  T ( x, H / 2) 
2  T ( x, H / 2)
1   Pr
y
(v)
(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
(16) Negligible axial conduction,  2T / x 2   2T / y 2
(17) Negligible effect of compressibility on the energy equation, u / x  v / y  0
(18) Nearly parallel flow, v  0


Energy equation: (2.15) simplifies to
T
 2T
 c pu
k 2
x
y
Boundary conditions
and
k
(11.44)
T ( x,0)
0
y
(w)
T ( x, H / 2)
 q s
y
(x)
Assume:
(19) Fully developed temperature. Define 

Fully developed temperature:
T ( x, H / 2)  T ( x, y )
T ( x, H / 2)  Tm ( x)
(11.45)
   ( y)
(11.46)

0
x
(11.47)
Thus
Equations (11.45) and (11.46) give
dT ( x, H / 2) T
 dT ( x, H / 2) dTm ( x) 

  ( y) 

0
dx
x
dx
dx 

(11.48)
The heat transfer coefficient h, is given by
T ( x, H / 2)
y
h
Tm ( x)  Ts ( x)
k
(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)
qs
Ts ( x)  Tm ( x)
(11.49)
12
q s
 constant
d ( H / 2)
dy
T ( x, H / 2)  Tm ( x)  
(11.50)
Combining this with (11.48), gives
dT ( x, H / 2) dTm ( x) T


dx
dx
x
(11.51)
Conservation of energy for the element in Fig. 11.8 gives
qs
dT


2q sWdx  mc pTm  mc p Tm  m dx 
dx


Simplify and eliminate m
m
dTm
2q s

= constant
dx
 c pum H
Tm 
Tm
(11.52)
dx
dTm
dx
dx
qs
(11.52) into (11.51)
2q s
dT ( x, H / 2) dTm ( x) T



dx
dx
x  c p u m H
Fig. 11.8
(11.53) into (11.44)
 2T
y 2

2q s u
kH u m
(11.54)
where u m is given by
2
um 
H

H /2
udy
(cc)
0
(11.30) into (cc)
um  
H 2 dp
1  6 Kn
12 dx
(11.55)
(11.30) and (11.55)
u
6 1
y2 


Kn



u m 1  6 Kn  4
H 2 
(11.56)
(11.56) into (11.54)
 2T
y 2

q s  1
y2 
12
  Kn  2 
1  6 Kn kH  4
H 
(11.57)
Integrating twice and use (w)
T ( x, y ) 
12q s
(1  6 Kn)kH
1 1
y4 
2
 (  Kn) y 
  g ( x)
12 H 2 
 2 4
(11.58)
13
To determine g(x), find Tm using two methods.
First method: Integrate (11.52)

Tm
2q s
dTm 
c p u m H
Tmi

x
dx
0
Tm (0)  Tmi
Evaluating the integrals
Tm ( x) 
(11.59)
2q s
x  Tmi
 c pum H
(11.60)
Second method: use definition in (11.43). (11.30) and (11.58) into (11.43)
Tm ( x) 
3q s H
13
13 

2
( Kn)  40 Kn  560   g ( x)

k (1  6 Kn) 
(11.61)
2
(11.60) and (11.61) give g(x)
g ( x)  Tmi 
2q s
3q s H
x
 c pum H
k (1  6 Kn) 2
13
13 

2
( Kn)  40 Kn  560 


(11.62)
Surface temperature Ts ( x, H / 2) :
3q s H  1
5  2 q s H
Ts ( x) 
Kn   
Kn  g ( x)

k (1  6Kn)  2
48    1 kPr
(11.63)
Nusselt number: (11.61) and (11.63) into (v)
Nu 
2
3
(1  6 Kn)
1
5
1
13
13   2 1


( Kn) 2 
Kn 
Kn
 Kn 


48 (1  6 Kn) 
40
560     1 Pr
2
9
NOTE:
(i) The Nusselt number is an implicit function of
x since Kn is a function p which is a function of
x.
(ii) Unlike macrochannels, the Nusselt number
depends on the fluid, as indicated by Pr and 
in (11.64).
(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, Nuo , is determined
by setting Kn  0 in (11.64)
(11.64)
8
7
Nu
6
5
4
0
0.04
0.08
0.12
Kn
Fig. 11.9 Nusselt number for air flow between
parallel plates at unifrorm surface
 = 1.4, Pr = 0.7,
heat flux for air,
u T 1
14
Nuo 
140
 8.235
17
(11.65)
(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

H/2
x
Fig. 11.10
2 Hh
 2 H T ( x, H / 2)

k
Tm ( x)  Ts
y
Ts
(11.66a)
Energy equation: Include axial conduction
 c pu

H/2
Temperature Distribution and Nusselt
Number. Newton’s law of cooling the
Nusselt number for this case is given by
Nu 
Ts
y
T
 2T  2T
k( 2  2 )
x
x
y
(11.67a)
Boundary conditions:
T ( x,0)
0
y
T ( x, H / 2)  Ts 
2 H
T ( x, H / 2)
Kn
  1 Pr
y
(11.68a)
(11.69a)
T (0, y)  Ti
(11.70a)
T (, y)  Ts
(11.71a)
Axial velocity is given by (11.56)
u
6 1
y2 

  Kn  2 
u m 1  6 Kn  4
H 
(11.56)
Dimensionless variables

2 um H
T  Ts
y
x
,   , Re 
, Pe  RePr
, 
H RePr
H

Ti  Ts
(11.72)
Use (11.56) and (11.72), into (11.66a)-(11.71a)
Nu  
2  ( , / 2)
m

(11.66)
15
2
2
6
( 1  Kn   2 )   1 2  2   2
1  6 Kn 4
 ( Pe) 

(11.67)
 ( ,0)
 0 

2 1
 ( ,1 / 2)
Kn
  1 Pr

(11.69)
 (0, )  1
(11.70)
 ( ,  )  0
(11.71)
 ( ,1 / 2)  

Solution: method of separation of variables

Result: Fig. 11.11.
8
NOTE:



Pe = 0
1
5 8
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  ) :
Nuo  7.5407

11
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.
qs
r
z
Determine the following:
(1) Velocity distribution
(2) Nusselt number
qs
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
p( z )
 8 Kno 
po
m
ro2 dp 
r2 
vz  
1  4 Kn  2 
4 dz 
ro 
(11.74)
1  4 Kn  (r / ro ) 2
vz
2
v zm
1  8Kn
(11.77)
2


pi 
pi2
pi  z
8
Kn


(
1

)

16
Kn
(
1

)

o
o


po 
p o  L
p o2



ro4 po2  pi2
pi
 2  1  16 Kno (  1)
16  LRTo  po
po


 ro4 p o2
mo 
8  LRT
(
pi
 1)
po
(11.78)
(11.79a)
(11.79b)
Nusselt Number. Define
Nu 
Nu 

2ro h
k
2 ro q s
k (Ts  Tm )
(d)
(e)
Results
T (r , z ) 
q s
(1  8 Kn) k ro

1 r4 
2
(
1

4
Kn
)
r


  g ( z)
4 ro2 

q s ro
14
7

2
16 Kn  3 Kn  24   g ( z )

k (1  8Kn) 
2q s
q sro
14
7

g ( z )  Tmi 
z
16 Kn 2  Kn  

2
 c p ro v z m
3
24 
k (1  8Kn) 
Tm 
Ts (ro , z ) 
Nu 
2
4q sro
k (1  8Kn)
3  4 q sro

 Kn  16     1 kPr Kn  g ( z )
2
4
1
14
7  4 1

( Kn  3 ) 
16 Kn 2  Kn   
Kn

2
(1  8 Kn)
16
3
24    1 Pr
(1  8 Kn) 
(11.92)
(11.95)
(11.96)
(11.97)
(11.98)

Nusselt number variation with Knudsen number for air, with   1.4 and Pr  0.7,
is plotted in Fig. 11.14.

No-slip Nusselt number, Nuo , is obtained by setting Kn  0 in (11.98)
Nuo 
48
 4.364
11
(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
2  u m ro
z
r
, R  , Re 
, Pe  RePr
, 
2ro RePr
ro

Ti  Ts
(11.106)
Nusselt number, energy equation, and boundary conditions in dimensionless form
Nu  
2  (1,  )
 m R
(11.100)
2
1  4 Kn  R 2  1 

(R  )  1 2  2
2(2  16 Kn)  R R
R
(2 Pe) 
 (0. )
0
R
4.5
(11
Pe = 0
1
5 
4.0
 (1,  )  
2 Kn  (1,  )
  1 Pr R
(11.101)
(11.103)
 ( R,0)  1
(11.104)
 ( R,  )  0
(11.105)
3.5
Nu
3.0
2.5
Solution: By separation of variables.
2.0
Results: Fig. 11.16.
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