Heat Transfer in Microchannels
11.1 Introduction:
Applications
Cooling of microelectronics
Inkjet printer
Medical research
Micro-electro-mechanical systems (MEMS): Micro heat
exchangers, mixers, pumps, turbines, sensors and
actuators
1
11.1.1 Continuum and Thermodynamic
Equilibrium Hypothesis
Properties: (pressure, temperature, density, etc) are
macroscopic manifestation of molecular activity
Continuum: material having sufficiently large number
of molecules in a given volume to give unique values
for properties
Validity of continuum assumption: the molecularmean-free path, , is small relative to the characteristic
dimension of the system
Mean-free-path: average distance traveled by molecules
before colliding
2
Knudson number Kn:
Kn 

(1.2)
De
De = characteristic length
Gases: the criterion for the validity of the continuum
assumption is:
Kn  10 1
(1.3a)
Thermodynamic equilibrium: depends on collisions
frequency of molecules. The condition for thermodynamic
equilibrium is:
Kn  10 3
(1.3b)
3
At thermodynamic equilbirium:
fluid and an adjacent surface have the same velocity and temperature:
no-velocity slip
no-temperature jump
Continuity, Navier-Stokes equations, and energy equation
are valid as long as the continuum assuption is valid
No-velocity slip and no-temperature jump
are valid as long as thermodynamic equilibrium is justified
Microchannels: Channels where the continuum assumption
and/or thermodynamic equilibrium break down
4
9.1.2. Surface Forces. Examine ratio of surface to
volume for tube:
A
DL
4


V D 2 L / 4 D
(11.1)
For D = 1 m, A/V = 4 (1/m)
For D = 1 μm, A/V = 4 x 10 6 (1/m)
 Consequence:
(1) Surface forces may alter the nature of surface
boundary conditions
(2) For gas flow, increased pressure drop results in
large density changes. Compressibility becomes
important
5
9.1.3 Chapter Scope
 Classification
 Gases vs. liquids
 Surface boundary conditions
 Heat transfer in Couette flow
 Heat transfer in Poiseuille flow
9.2 Basic Consideration
9.2.1 Mean Free Path. For gases:

 
p
2
RT
(11.2)
6
Table 11.1
p = pressure
R = gas constant
T = temperature
μ = viscosity
values of  for common gases
R
gas
Air
Helium

  107
J/kg K kg/m3 kg/s m
287.0
1.1614
2077.1
0.1625
0.0808
Hydrogen 4124.3
Nitrogen
296.8 1.1233
Oxygen
259.8 1.2840
184.6
199.0
89.6
178.2
207.2

m
0.067
0.1943
0.1233
0.06577
0.07155
NOTE:
 Pressure drops along a channel   increases
 Kn increases
•  is very small, expressed in terms of the micrometer,
  10 m
6
7
11.2.2 Why Microchannels?
Nusselt number: fully developed
flow through tubes at uniform
surface temperature
hD
NuD 
 3.66
k
k
h  3.657
D
106
water
5
10
104
air
103
(6.57)
(11.3)
102
continuum
101
100
101
102
D( m)
103
104
Fig. 11.1
As D  h 
sink
Application:
Water cooled microchips
flow
microchip q
Fig. 11.2
8
11.2.3 Classification
Based on the 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)
Four important factors:
(1) Continuum
(2) Thermodynamic equilibrium
(3) Velocity slip
(4) Temperature jump
9
(1) Kn < 0.001: Macro-scale regime (previous chapters):
Continuum: valid
Thermodynamic equilibrium: valid
No velocity slip
No temperature jump
(2) 0.001 < Kn < 0.1: Slip flow regime:
Continuum: valid
Thermodynamic equilibrium: fails
• Velocity slip
• Temperature jump
Continuity, Navier-Stokes equations, and energy equations are valid
No-velocity slip and No-temperature jump conditions, conditions fail
Reformulate boundary conditions
10
(3) 0.1< Kn<10: Transition flow:
Continuity and thermodynamic equilibrium fail
Reformulate governing equations and boundary conditions
Analysis by statistical methods
(4) Kn>10: Free molecular flow: analysis by kinetic theory
of gases
11.2.4 Macro and Microchannels
Macrochannels: Continuum domain, no velocity slip,
no temperature jump
Microchannels: Temperature jump and velocity slip,
with or without failure of continuum assumption
11
Distinguishing factors:
(1) Two and three dimensional effects
(2) Axial conduction
(3) Viscous dissipation
(4) Compressibility
(5) Temperature dependent properties
(6) Slip velocity and temperature
(7) Dominant role of surface forces
12
11.2.5 Gases vs. Liquids
Macro convection:
 No distinction between gases and liquids
 Solutions for both are the same for the same geometry,
governing parameters (Re, Pr, Gr,…) and boundary
conditions
Micro convection:
 Flow and heat transfer of gases differ from liquids
Gas and liquid characteristics:
(1) Mean free path: liquid   gas
 Continuum assumption may hold for liquids but fail
for gases
13
 Typical MEMS applications: continuum assumption is
valid for liquids
(2) Knudsen number: used as criterion for thermodynamic
equilibrium and continuum for gases but not for liquids
(3) Onset of failure of thermodynamic equilibrium and
continuum: not well defined for liquids
(4) Surface forces: liquid forces are different from gas forces
(5) Boundary conditions: differ for liquids from gases
(6) Compressibility: liquids are almost incompressible while
gases are not
(7) Flow physics: liquid flow is not well known. Gas flow is
well known
14
(8) Analysis: more complex for liquids than gases
11.3 General Features
 Flow and heat transfer phenomena change as channel
size is reduced:
Rarefaction: Knudsen number effect
Compressibility: Effect of density change due to pressure
drop along channel
Viscous dissipation: Effect of large velocity gradient
Examine: Effect of channel size on:
 Velocity profile
 Flow rate
15
 Friction factor
 Transition Reynolds number
 Nusselt number
Consider:
Fully developed microchannel gas flow as the Knudsen
number increases from the continuum through the slip
flow domain
16
11.3.1 Flow Rate
Slip flow: increased velocity and
flow rate
Qe
1
Qt
(11.5)
(a) no-slip velocity
(b) slip velocity
Fig. 11.3
e = determined experimentally
t = from macrochannel theory or correlation equations
11.3.2 Friction Factor f
 Define friction coefficient C f
w
Cf 
2
( 1 / 2 )  um
(4.37a)
17
 w = wall shear stress
um = mean velocity
 Fully developed flow through channels: define friction
factor f
1 D p
f 
2
2 L  um
(11.6)
D = diameter
L = length
p = pressure drop
18
Macrochannels: fully developed laminar flow:
(1) f is independent of surface roughness
(2) Product of f and Reynolds number is constant for each
channel geometry:
f Re  Po
Po = Poiseuille number
(3) Po is independent of Reynolds number
Microchannels: compare experimental data, ( Po)e, with
theoretical value, ( Po) t , (macroscopic, continuum)
19
Po e
 Pot
 C*
(11.8)
Conclusion:
(1) C * departs from unity:
1  C *  1
(2) Unlike macrochannels, Po for fully developed flow
depends on the Re
(3) Conflicting findings due to: difficulties in measurements
of channel size, surface roughness, pressure distribution,
uncertainties in entrance effects, transition, and
determination of properties
20
11.3.3 Transition to turbulent flow
Macrochannels: smooth macrotubes
Re t 
uD

 2300
(6.1)
Microchannels: reported transition
300  Re t  16,000
Factors affecting the determination of Re t :
 Variation of fluid properties
 Measurements accuracy
 Surface roughness
21
11.3.4 Nusselt number. For fully developed conditions:
Macrochannel: Nusselt number is constant
Microchannels: In general, Nusselt number is not well
established:
 Nu varies along microchannels
 Nu depends on:
 Surface roughness
 Reynolds number
 Nature of gas
 Widely different reported results:
( Nu)e
0.21 
 100
( Nu)t
(11.9)
22
where:
( ) e = experimental
( ) t = macrochannel theory
Factors affecting the determination of Nue :
 Variation of fluid properties
 Measurements accuracy
23
11.4 Governing Equations
Slip flow regime: 0.001  Kn  0.1 :
Continuity , Navier - Stokes equations, and energy equation are valid
No - velocity slip and no - temperatur e jump conditions fail
Reformulat e boundary conditions
Factors to be considered:
 Compressibility
 Axial conduction
 Dissipation
11.4.1 Compressibility: Expressed in terms of Mach number
M=
fluid velocity
speed of sound
24
Macrochannels:
 Incompressible flow, M < 1
 Linear pressure drop
Microchannels:
 Compressible flow
 Non-linear pressure drop
 Decrease in Nusselt number
11.4.2 Axial Conduction
Macrochannels: neglect axial conduction for
Pe  Re D Pr  100
(6.30)
25
Pe = Peclet number
Microchannels: low Peclet numbers, axial conduction may
be important, it increases the Nusselt number
11.4.3 Dissipation
Microchannels: large velocity gradient, dissipation may
become important
11.5 Slip Velocity and Temperature Jump
Boundary Conditions
Slip velocity for gases:
2   u u( x ,0)
u( x ,0)  us 

u
n
(11.10)
26
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
Temperature jump for gases
2   T 2  T ( x ,0)
T ( x ,0)  Ts 
 T 1   Pr n
(11.11)
T(x,0) = fluid temperature at the boundary
T s = surface temperature
27
  c p / cv , specific heat ratio
 T = energy accommodating coefficient
NOTE
(1) Eq. (11.10) and (11.11) are valid for gases
(2) Eq. (11.10) and (11.11) are valid for Kn < 0.1
(3) σu and σT, are:
• Empirical factors
• They depend on the gas, geometry and surface
• Values range from zero (perfectly smooth) to unity
28
• Difficult to determine experimentally
• Values for various gases are approximately unity
29
11.4. 8 Analytic Solutions: Slip Flows
Two common flow types, extensive use in MEMS:
(1) Couette flow (shear driven): fluid is set in motion by
a moving surface
Examples:
100m
stator
rotor
stationary

2
movable
Fig. 11.4
Fig. 11.5
30
(2) Poiseuille flow (pressure driven): fluid is set in motion by
an axial pressure gradient
Examples:
Micro heat exchangers, mixers, microelectronic heat sinks
NOTE
 No pressure drop in Couette flow
 Signifiant pressure drop in Poiseuille flow
Boundary conditions: two types:
(1) Uniform surface temperature
(2) Uniform surface heat flux
31
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
32
(10) The accommodation coefficients are equal to unity,
 u   T  1.0
11.6.2 Couette Flow with Viscous Dissipation:
Parallel Plates with Surface Convection
y
ho T
 Infinitely large parallel plates
 Gas fills gap between plates
x
 Upper plate: moves with velocity us
us
u
H
Fig. 11.6
 Lower plate: stationary, insulated
 Convection at the upper plate
 Consider dissipation and slip conditions
33
Determine:
(1) Velocity distribution
(2) Mass flow rate
(3) Nusselt number
Find flow field and temperature distribution
Flow Field
 Normal velocity and all axial derivatives vanish
 Axial component of the Navier-Stokes equations, (2.9),
simplifies to
2
d u
dy 2
0
34
Boundary conditions: use (11.10), Set u  1
• Lower plate: n = y = 0 and us  0,
(11.10) gives
du( x ,0)
u( x ,0)  
dy
(g)
 Upper plate: n = H – y, (9.10) gives
du( x , H )
u( x , H )  us  
dy
Solution
y
u
1

(
 Kn )
u s 1  2 Kn H
(11.14)
35
Kn is the local Knudsen number
Kn 

H
(11.13)
NOTE
(1) Fluid velocity at the moving plate: set y = H in (11.14)
u( H ) 1  Kn

1
us
1  2 Kn
Effect of slip:
 Decrease fluid velocity at the moving plate
 Increase fluid velocity at the stationary plate
36
(2) Velocity distribution is linear
(3) Setting Kn = 0 in (11.14) gives the no-slip solution
y
u

us H
(k)
Mass Flow Rate m
m W
H
0
 u dy
(11.15)
W = channel width
Neglect variation of ρ along y, (11.14) into (11.15)
37
us
m   WH
2
 Flow rate is independent of the Knudsen number
 Compare with macrochannel flow rate mo
(k) into (11.15)
us
mo   WH
2
(11.17)
This is identical to (11.16), thus
m
1
mo
(11.18)
38
Nusselt Number
• Equivalent diameter for parallel plates, De = 2H
• Nusselt number
2 Hh
Nu 
k
(l)
Heat transfer coefficient h:
T ( H )
k
y
h
Tm  Ts
39
T ( H )
y
Nu  2 H
Tm  Ts
(11.19)
k = conductivity of fluid
T = fluid temperature
Ts = plate temperature
NOTE
(1) Fluid temperature at the moving plate, T (x,H), is not
equal to surface temperature
(2) h is defined in terms of surface temperature Ts
40
(3) Use temperature jump, (11.11), to determine Ts
(4) For the upper plate, n =H – y, eq. (11.11) gives
2  T ( x , H )
Ts  T ( x , H ) 
1   Pr
y
(11.20)
• Mean temperature Tm: defined in Section 6.6.2
mc pTm  W
H
0
 c p uT dy
(11.21)
• Neglect variation of cp and ρ along y, use (11.14)
. for u and (11.15) for m
41
2
Tm 
us H
H
0
uT dy
(11.22)
Determine temperature distribution:
 Use energy equation, (2.15)
 Apply above assumptions, note that axial derivatives
vanish, (2.15) gives
k
 2T
y
2
   0
(11.23)
42
(2.17) gives the dissipation function  which simplifies to
 u 
  
 y 
2
(11.24)
(9.24) into (9.23)
2
d T
dy
2

  du 
 
k  dy 
2
(11.25)
Boundary conditions
Lower plate:
dT (0)
0
dy
(m)
43
Upper plate:
dT ( H )
k
 ho (Ts  T )
dy
Use (920) to eliminate Ts
dT ( H )
2  T ( x , H )


k
 ho T ( x , H ) 
 T 
dy
1   Pr
n


(n)
Use velocity solution (9.14), solve for T

2
kH

H

2 Kn 2
2
T  y 


H   T
2
ho
2
  1 Pr
(11.26)
44
where

us

  
k  H (1  2 Kn ) 
2
(p)
Velocity solution (11.14), temperature solution (11.26)
giveTs , Tm and Nu
kH
Ts 
 T
ho
Tm 
(u)
1
2
2 Kn 2
1 2
2  kH
H


KnH



H   T (w)


1  2 Kn  4
3
ho
  1 Pr

45
2
Nu 
1
2 Kn
1 2 
 Kn  

1  2 Kn  4 3    1 Pr
(11.27)
Note the following regarding the Nusselt number
(1) It is independent of Biot number
(2) It is independent of the Reynolds number
(3) Unlike macrochannels, it depends on the fluid
(4) First two terms in the denominator of (11.27) represent
rarefaction (Knudsen number). The second term
represents effect of temperature jump
46
(5) Nusselt number for macrochannels, Nuo: set Kn = 0
in (11.27):
Nuo  8
(11.28)
Ratio of (11.27) and (11.28)
Nu
1  2 Kn

Nuo  8  8 Kn
1  3 Kn     1 Pr
(11.29)
NOTE: Ratio is less than unity
47
11.6.3 Fully Developed Poiseuille Channel
Flow: Uniform Surface Flux
 Pressure driven flow between parallel plates
 Fully developed velocity and temperature
 Inlet and outlet pressures are pi and po
• Uniform surface flux, q s
qs
y
Determine:
(1) Velocity distribution
(2) Pressure distribution
(3) Mass flow rate
(4) Nusselt number
H/2
H/2
x
qs
Fig. 11.7
48
Note: p  0
x
Major difference between macro and micro fully developed
slip flow:
Macrochannels: incompressible flow
(1) Parallel streamlines
(2) Zero lateral velocity component (v = 0)
(3) Invariant axial velocity ( u /  x  0)
(4) Linear axial pressure (dp / dx  constant)
49
Microchannels: compressibility and rarefaction change
above flow pattern:
(1) None of above conditions hold
(2) Large axial pressure drop  density changes
 compressible flow
(3) Rarefaction: pressure decreases   increases
 Kn increases with x
(4) Axial velocity varies with axial distance
(5) Lateral velocity v does not vanish
(6) Streamlines are not parallel
(7) Pressure gradient is not constant
50
Assumptions
(1) Steady state
(2) Laminar flow
(3) H / R  1
(4) Two-dimensional
(5) Slip flow regime (0.001 < Kn < 0.1)
(6) Ideal gas
(7) Constant viscosity, conductivity and specific heats
(8) Negligible lateral variation of density and pressure
(9)  u   T  1.0
51
(10) Negligible dissipation
Flow Field
Additional assumptions:
(11) Isothermal flow
(12) Negligible inertia forces:
u
v
 (u  v ) = 0
x
y
(13) The dominant viscous force is 
 2u
y 2
Navier-Stokes equations (2.9) simplify to:
p
 2u
 
0
2
x
y
52
Boundary conditions:
Symmetry at y = 0
u( x,0)
0
y
(e)
For the upper plate, n = H – y
u( x , H / 2)
u( x , H / 2)  
y
(f)
Solution to u
H 2 dp 
y2 
u
1  4 Kn( p)  4 2 
8 dx 
H 
53
For an ideal gas
Kn 

H

 
H
1
RT
2
p
(11.33)
Pressure Distribution p:
To determine p(x), must determine vertical component v:
start with continuity (2.2a)
 


   u    v     w   0
 t x
y
z
Apply above assumptions
54


  u    v   0
x
y
(h)
Use ideal gas to eliminate ρ:
p

RT
(11.31)
(11.31) into (h), assuming constant temperature


 pv     pu
y
x
(i)
(11.30) into (i)
55

H 2   dp
y2 
( pv ) 
 p (1  4 Kn( p)  4 2 )
y
8 x  dx
H 
(j)
Boundary conditions:
v (x,0)  0
(k)
v ( x, H / 2)  0
(l)
Multiply (j) by dy, integrate and using (k)

y
H 2   dp
y2 
d ( pv ) 
(1  4 Kn( p)  4 2 ) dy
p
8 x  dx 0
0
H 
y

(m)
56
Evaluate the integrals
H 3 1   dp
v
p
8  p x  dx

y 4 y 3  
1  4 Kn( p) 
3 
H
3
H  

(11.32)
Determination of p(x): Apply boundary condition (l) to (11.32)
  dp
p
x  dx

y 4 y 3  
0
1  4 Kn( p ) 

3
H 3 H  

y H / 2
Express Kn in terms of pressure. Equations (11.2) and
(11.13) give

 
1
Kn 
H

H
2
RT
p
(n)
(11.33)
57
Evaluate (n) at y = H/2, substitute (11.33) into (n) and
integrate
dp
p
dx
1
1 
 3  H 2 RT p   C


Integrate again (T is assumed constant)
1 2 
p 
2 RT p  Cx  D
6
H
(o)
Solve for p
p( x )  3

H
2 RT  18 RT
2
H
2
 6Cx  6 D
(p)
58
Pressure boundary conditions
p(0)  pi , p( L)  po
(q)
Apply (q) to (p)
1

C
( po  pi ) 
2RT ( po  pi )
6L
HL
2
2
pi 
D

2RT pi
6 H
2
Substitute into (p) and normalize by po
59
p( x )
3

2 RT 
po
Hpo
18 2  RT 
 1 
2
2
H
po 
6

2 RT (1 
2
po Hpo
pi2
pi  x
) 
po  L
pi2

2
po
6
2 RT
Hpo
(r)
pi
po
Introduce outlet Knudsen number Kno using (11.2) and
(11.13)
Kno 
 ( po )
H



H po
2
RTo
(11.34)
Substitute (11.34) into (r)
2


pi 
p( x )
 6 Kno  6 Kno 
 (1 

po
po 


pi  x
)  12 Kno (1  )
2
po  L
po
pi2
(11.35)
60
NOTE:
(1) Unlike macrochannel Poiseuille flow, pressure
variation along the channel is non-linear
(2) Knudsen number terms represent rarefaction effect
(3) The terms (pi/po)2 and [1- (pi/po)2](x/L) represent the
effect of compressibility
(4) Application of (11.35) to the limiting case of Kno =0
gives
p( x )

po
pi2
pi2 x
 (1  2 )
2
po
po L
(11.36)
This result represents the effect of compressibility alone
61
Mass Flow Rate
m  2W
H/2
0
 udy
(s)
W = channel width
(11.30) in (s)
WH 3
dp
1  6 Kn( p) 
m
12
dx
(t)
Density ρ :
p

RT
(11.37)
62
(11.33) gives Kn(p)
Kn( p ) 
 ( p)
H



Hp 2
RT
(11.33)
(11.33) and (11.37) in (t)
WH 3
m
12 RT

 dp
 
 p  6 H 2 RT  dx


(11.38)
(11.35) into (11.38) and let T=To
1 W H 3 po2
m
24  LRTo
 pi2

pi
 2  1  12 Kno 24(  1)
po
 po

(11.39)
63
Compare with no-slip, incompressible macrochannel case:

1 W H 3 po2  pi
mo 
 1

12  LRT  po

(11.40)

m 1  pi
   1  12 Kno 
mo 2  po

(11.41)
Taking the ratio
NOTE
(1) Microchannels flow rate is very sensitive to H
(2) (11.39) shows effect of rarefaction (slip) and
compressibility on m
64
(3) Since pi / po  1, (11.41) shows that neglecting
compressibility and rarefaction underestimates m
Nusselt Number
2 Hh
Nu 
k
(u)
For uniform surface flux q s
qs
h
Ts  Tm
Substitute into (g)
2 H qs
Nu 
k (Ts  Tm )
(v)
65
Plate temperature Ts: use (11.11)
2  T ( x , H / 2)
Ts  T ( x , H / 2) 
1   Pr
y
(11.42)
Mean temperature Tm:
H/2
Tm

 0
uT dy
(11.43)
H/2
0
udy
Need u(x,y) and T(x,y)
Velocity distribution: (11.30) gives u(x,y) for isothermal 66flow
Additional assumption:
(14) Isothermal axial velocity solution is applicable
(15) No dissipation,   0
(16) No axial conduction,  2T / x 2   2T / y 2
(17) Negligible effect of compressibility on the energy
equation
(18) Nearly parallel flow, v  0
Energy equation: equation (2.15) simplifies to
T
 2T
 c pu
k 2
x
y
(11.44)
67
Boundary conditions:
T ( x ,0)
0
y
T ( x , H / 2)
k
 qs
y
(w)
(x)
To solve (11.44), assume:
(19) Fully developed temperature
Solution: T(x,y) and Tm(x): Define
T ( x, H / 2)  T ( x, y )

T ( x, H / 2)  Tm ( x )
(11.45)
68
Fully developed temperature:
Thus

is independent of x
   ( y)
(11.46)

0
x
(11.47)
(11.45) and (11.46) give

 T ( x, H / 2)  T ( x, y ) 

0


x x  T ( x, H / 2)  Tm ( x ) 
69
Expanding and use (11.45)
dT ( x, H / 2) T
 dT ( x, H / 2) dTm ( x ) 

  ( y) 

 0 (11.48)

dx
x
dx
dx 

Determine:
T ( x , y ) dT ( x , H / 2)
dTm ( x )
,
and
x
dx
dx
Heat transfer coefficient h:
 T ( x , H / 2)
k
y
h
Tm ( x )  Ts ( x )
(y)
70
(11.42) gives Ts(x). (11.45) gives temperature gradient in (y)
T ( x, y)  T ( x, H / 2)  [T ( x, H / 2)  Tm ( x)]
Differentiate
T ( x, H / 2)
d ( H / 2)
 [T ( x, H / 2)  Tm ( x )]
y
dy
(z)
(z) into (y), use (11.42) for Ts(x)
k[T ( x, H / 2)  Tm ( x )] d ( H / 2)
h
Ts ( x )  Tm ( x )
dy
(11.49)
71
Newton’s law of cooling:
qs
h
Ts ( x )  Tm ( x )
Equate with (11.49)
qs
T ( x , H / 2)  Tm ( x )  
 constant
d ( H / 2)
dy
(11.50)
Differentiate
T ( x , H / 2) Tm ( x )

0
x
x
Combine this with (11.48)
72
dT ( x , H / 2) dTm ( x ) T


dx
dx
x
(11.51)
NOTE:
(11.51) replacesT with dTm in (11.44)
x
dx
Determine dTm :
dx
m
Conservation of energy for element:
qs
Tm 
Tm
dx
dTm
dx
dx
qs
Fig. 11.8
73
Conservation of energy for element:
dT


2qsWdx  mc pTm  mc p Tm  m dx
dx


Simplify
dTm 2Wqs

= constant
dx
mc p
(aa)
However
m  WH um
(bb)
(bb) into (aa)
74
dTm
2qs

= constant
dx
 c p um H
(11.52)
(11.52) into (11.51)
dT ( x , H / 2) dTm ( x ) T
2qs



dx
dx
x  c p um H
(11.53)
(11.53) into (11.44)
2qs u

2
kH um
y
 2T
(11.54)
75
Mean velocity:
um
2

H
H /2
0
udy
(cc)
(11.30) gives velocity u. (11.30) into (cc)
H 2 dp H / 2
um  
4 dx 0


y2 
1  4 Kn  4 2  dy
H 

Integrate
H 2 dp
1  6 Kn
um  
12 dx
(11.55)
76
Combining (11.30) and (11.55)
u
6 1
y2 

  Kn  2 
um 1  6 Kn  4
H 
(11.56)
(11.56) into (11.54)
12
qs  1
y2 

  Kn  2 
2
1  6 Kn kH  4
y
H 
 2T
(11.57)
Integrate twice
1 1
12qs
y4 
2
T ( x, y ) 
 f ( x) y  g( x)
 (  Kn) y 
2
(1  6 Kn)kH  2 4
12H 
(dd)
77
f(x) and g(x) are “constants” of integration
Boundary condition (w) gives
f (x )  0
Solution (dd) becomes
1 1
12qs
y4 
2
T ( x, y ) 
 (  Kn) y 
  g( x)
2
(1  6 Kn)kH  2 4
12 H 
(11.58)
NOTE:
(1)Boundary condition (x) is automatically satisfied
(2) g(x) is determine by formulating Tm using two methods
78
Method 1: Integrate (11.52)
x
2qs
dTm 
dx
 c p um H 0
T
mi

Tm

where
Tm (0)  Tmi
(11.59)
Evaluate the integrals
2qs
Tm ( x ) 
x  Tmi
 c p um H
(11.60)
79
Method 2: Use definition of Tm. Substitute (11.30) and
(11.58) into (11.43)
Tm ( x ) 

H /2
H dp

8 dx 0


1 1
12qs
y 2  
y4 
1  4 Kn  4 2  
 (  Kn) y 
  g ( x )dy
2

H   (1  6 Kn)kH  2 4
12 H 

H dp H / 2 
y2 

1  4 Kn  4 2 dy
8 dx 0
H 

2

Evaluate the integrals
3qs H
13 

2 13
Tm ( x ) 
( Kn)  Kn 
 g( x)

2 
40
560
k (1  6 Kn)
(11.61)
Equating (11.60) and (11.61) gives g(x)
g ( x )  Tmi
2qs
3qs H
13 

2 13

x
(
Kn
)

Kn


 c p um H
40
560
k (1  6 Kn) 2 
(11.62)
80
(11.58) into (11.42) gives Ts
3qs H  1
5
2 qs H
Ts ( x ) 
Kn   
Kn  g ( x )

k (1  6 Kn)  2
48    1 kPr
(11.63)
The Nusselt number is given in (v)
2 H qs
Nu 
k (Ts  Tm )
(v)
(11.61) and (11.63) into (v)
Nu 
2
3
5
1
13  
2 Kn
1

2 13
Kn


(
Kn
)

Kn




(1  6 Kn)  2
48 (1  6 Kn) 
40
560    1 Pr
(11.64)
81
NOTE:
(1) Kn in (11.64) depends of local pressure p
(2) Pressure varies with x, Kn varies with x
(3) Unlike macrochannels, Nu is not constant
(4) Unlike macrochannels, Nu depends on the fluid
(5) No-slip Nu for macrochannel
flow, Nuo: set Kn = 0 in (11.64)
0
8
Nu
6
140
Nuo 
 8.235
17
(11.65)
4
0
0.04
0.08
0.12
Kn
Fig. 11.9 Nusselt number for air
82
This agrees with Table 6.2
(6) Rarefaction and compressibility decrease the Nusselt
number
83
11.6.4 Fully Developed Poiseuille Channel
Flow: Uniform Surface Temperature
Repeat Section 11.6.3 with plates at uniform surface
temperature Ts
 Flow field: same for both cases:
y
Ts
(11.30) u(y)
(11.35)  p( x ) / po
H/2
H/2
x
(11.35)  m
Ts
Fig. 11.10
 Energy equation: (11.44) is modified to include axial
conduction
84
 Boundary conditions: different for the two cases
Nusselt number:
2 Hh
 2 H T ( x , H / 2 )
Nu 

k
Tm ( x )  Ts
y
Need T(x,y) and Tm(x)
Solution approach:
Solve the Graetz channel entrance problem and set
x   to obtain the fully developed solution
85
Axial conduction: can be neglected for:
Pe  PrRe  100
Microchannels:
Small Reynolds  Small Peclet number  Axial
conduction is important
Include axial conduction: modify energy equation (11.44)
T
 2T  2T
 c pu
 k( 2  2 )
x
x
y
(11.67a)
86
Boundary and inlet 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)
87
8
Axial velocity
Nu
1
u
6
y2 

  Kn  2 
um 1  6 Kn  4
H 
(11.56)
7
6
5
0
Solution
Pe = 0
1
5 8
0.04
0.08
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]
 Use method of separation of variables
• Specialize to fully developed:
set x  
Result: Fig. 11.11 shows Nu vs. Kn
88
NOTE
(1) Nu decreases as the Kn is increased
(2) No-slip solution overestimates microchannels Nu
(3) Axial conduction increases Nu
(4) Limiting case: no-slip (Kn = 0) and no axial conduction
( Pe   ) :
Nuo  7.5407
(11.73)
This agrees with Table 6.2
Heat Transfer Rate, q s:
Following Section 6.5
89
q s  m c p [Tm ( x )  Tmi ]
(6.14)
Tm (x) is given by
Tm ( x )  Ts  (Tmi  Ts ) exp [
Ph
x]
mc p
(6.13)
h , is determine numerically using (6.12)
1 x
h   h( x )dx
x 0
(6.12)
90
11.6.5 Fully Developed Poiseuille Flow in
Micro Tubes: Uniform Surface Flux
qs
r
r
ro
Consider:
 Poiseuille flow in
micro tube
 Uniform surface flux
z
qs
Fig. 11.12
 Fully developed velocity and temperature
• Inlet and outlet pressures are pi and po
91
Determine
(1) Velocity distribution
(2) Nusselt number
 Rarefaction and compressibility affect flow and heat
transfer
 Velocity slip and temperature jump
 Axial velocity variation
 Lateral velocity component
 Non-parallel stream lines
 Non-linear pressure
92
Assumptions
Apply the 19 assumptions of Poiseuille flow between
parallel plates (Sections 11.6.3)
Flow Field
 Follow analysis of Section 11.6.3
 Axial component of Navier-Stokes equations in
cylindrical coordinates:
1  v z
1 p
(r
)
r r
r
 z
(a)
v z ( r , z ) = axial velocity
93
Boundary conditions:
Assume symmetry and set σu = 1
v z (0, z )
r
0
v z ( ro , z )
u( ro , z )   
r
(b)
(c)
Solution
ro2 dp 
r2 
vz  
1  4 Kn  2 
4  dz 
ro 
(11.74)
94
Knudsen number
Kn 

(11.75)
2ro
Mean velocity vzm
v zm 
1
 ro2
ro
0
2 r v z dr
Use (11.74), integrate
ro2 dp
v zm  
(1  8 Kn )
8 dz
(11.76)
95
(11.74) and (11.76)
vz
1  4 Kn  ( r / ro ) 2
2
v zm
1  8 Kn
Solution to axial pressure
2


p( z )
pi 
pi
pi  z
 8 Kno  8 Kno 
 (1  )  16 Kno (1  )

po
po 
po  L
po


2
(11.78)
2
(11.76) and (11.78) give m
 ro4 po2
m
16  LRT
 pi2

pi
 1) (11.79a)
 2  1  16 Kno (
po
 po

96
For incompressible no-slip (macroscopic)
 ro4 po2 pi
mo 
(
 1)
8  LRT po
(11.79b)
Nusselt Number
 Follow Section 11.6.3
2ro h
Nu 
k
(d)
Heat transfer coefficient h:
qs
h
Ts  Tm
97
Substituting into (d)
2ro qs
Nu 
k (Ts  Tm )
(e)
Ts = tube surface temperature, obtained from temperature
jump condition (11.11)
2  T ( ro , z )
Ts  T ( ro , z ) 
1   Pr
r
Mean temperature:
(f)
ro
Tm

 0
v z T r dr
ro
0
v z rdr
(11.80)
98
Energy equation:
T k  T
 c pv z

(r )
 z r r r
(11.81)
Boundary conditions:
T (0, z )
0
r
(g)
T (ro , z )
k
 qs
r
(h)
Define
99
T (ro , z )  T (r , z )

T (ro , z )  Tm ( z )
(11.82)
Fully developed temperature:
Thus
   (r )
(11.83)

0
z
(11.84)
(11.82) and (11.84) give

 T (ro , z )  T (r , z ) 

0


 z  z  T (ro , z )  Tm ( z ) 
100
Expand (11.82)
 dT (ro , z ) dTm ( z ) 
dT (ro , z ) T

  (r ) 

0

dz
z
dz 
 dz
(11.85)
Determine:
T (r , z ) dT (ro , z )
dTm ( z )
and
,
z
dz
dz
Heat transfer coefficient h,:
T ( ro , z )
k
r
h
Tm ( z )  Ts ( z )
(i)
101
Rewrite (11.81)
T (r , z )  T (ro , z )  [T (ro , z )  Tm ( z )]
Differentiate and evaluating at r  ro
T (ro , z )
d (ro )
 [T (ro , z )  Tm ( z )]
r
dr
(j)
k[T (ro , z )  Tm ( z )] d (ro )
h
Ts ( z )  Tm ( z )
dr
(k)
(j) into (i)
102
Newton’s law of cooling h
qs
h
Ts ( z )  Tm ( z )
Equate with (k)
qs
T (ro , z )  Tm ( z )  
 constant
d (ro )
k
dr
(11.86)
Differentiate
T (ro , z ) Tm ( z )

0
z
z
103
Combine with (11.85)
dT (ro , z ) dTm ( z ) T


dz
dz
z
(11.87)
Will use (11.87) to replace T / z in (11.81) withdTm / dz.
qs
Conservation of energy to dx
m
Tm 
Tm
dx
dTm 



2 ro qs dz  mc pTm  mc p Tm 
dz 
dz


dTm
dx
dx
qs
Fig. 11.13
104
Simplify
dTm 2 roqs

dz
mc p
(l)
m   ro2 v z m
(m)
dTm
2qs

dz
 c p rov z m
(11.88)
However
(m) into (l)
(11.88) into (11.87)
dT (ro , z ) dTm ( z ) T
2qs



dz
dz
z  c p rov z m
(11.89)
105
(11.89) into (11.81)
2qs v z
 T
(r ) 
r
r r
kro v z m
(11.90)
(11.77) is used to eliminate v z / z zm in the above
qs
 T
4
(r
)
r r
1  8 Kn kro

r2 
1  4 Kn  2  r
ro 

(11.91)
Integrate
qs
T (r , z ) 
(1  8 Kn)k ro
4

2 1r 
(1  4 Kn) r 
  f ( z) y  g( z)
2
4 ro 

(n)
106
Condition (g) gives
f (z )  0
Solution (n) becomes
qs
T (r , z ) 
(1  8 Kn)k ro
4

2 1r 
(1  4 Kn) r 
  g( z)
2
4 ro 

(11.92)
Condition (h) is automatically satisfied
Determine g(z): Use two methods to determine Tm
Method 1: Integrate (11.88)
107
Tm
T
mi
where
z
2qs
dTm 
dz
 c p v z m ro 0

Tm (0)  Tmi
(11.93)
Evaluate the integral
2qs
Tm 
z  Tmi
 c pv z m ro
(11.94)
Method 2: Use definition of Tm in (11.80). Substitute (11.74)
and (11.92) into (11.80)
108
4 

1
4qs
r 2  
r
2
1  4 Kn  2  
 (  Kn) r 
  g ( z ) r dr
2

ro   (1  8 Kn)kro  4
16ro 

ro
2

r  rdr
1  4 Kn  2 
ro 
0 
ro 
Tm 
0

Integrate
Tm 
qs ro
7

2 14
16
Kn

Kn

 g(z)

2 
3
24 
k (1  8 Kn)
(11.95)
Equate (11.94) and (11.95), solve for g(z)
2qs
qsro
7

2 14
g ( z )  Tmi 
z
16Kn  Kn  

2
 c p rov zm k (1  8 Kn) 
3
24
(11.96)
109
Use (f) and (11.92) to determineTs (ro , z )
4qsro 
3
4 qsro
Ts (ro , z ) 
Kn   
Kn  g ( z )

k (1  8 Kn) 
16    1 kPr
(11.97)
Nusselt number: (11.95) and (11.97) into (e)
Nu 
2
4
3
1
14
7
4 1

( Kn 
)
16
Kn

Kn


Kn

2 
(1  8 Kn)
16 (1  8 Kn) 
3
24    1 Pr
2
(11.98)
110
Results: Fig. 11.14
 Fig. 11.14 gives Nu vs. Kn for air
• Rarefaction and compressibility 4.0
decrease the Nusselt number
 Nusselt number depends on Nu 3.0
the fluid
• Nu varies with distance
2.0
0
along channel
Fig. 11.14
• No-slip Nusselt number, Nuo,
is obtained by setting
Kn = 0 in (11.97)
48
Nuo 
 4.364
11
0.04
0.08
Kn
0.12
Nusselt number for air flow through
tubes at unifrorm surface heat flux
(11.99)
111
This agrees with (6.55) for macro tubes
11.9.6 Fully Developed Poiseuille Flow in
Micro Tubes: Uniform Surface Temperature
• Repeat Section 11.6.5
with the tube at surface
temperature Ts
 Apply same assumptions
r
r
ro
z
Fig. 11.15
Ts
 Boundary conditions are
different
Flow field solution is identical for the two cases
112
Nusselt number:
2ro h
 2ro T ( ro , z )
Nu 

k
Tm ( z )  Ts
r
• Determine T(r,z) and Tm(z)
 Follow the analysis of Section 11.6.4
 Solution is based on the limiting case of Graetz tube
entrance problem
 Axial conduction is taken into consideration
 Energy equation (11.81) is modified to include axial
conduction:
113
T k  T
 2T
 c pv z

(r
)k
 z r r r
z 2
(11.101a)
Boundary and inlet conditions
T ( r ,0)
0
z
(11.102a)
T ( ro , z )
2 2ro
Kn
  1 Pr
r
(11.103a)
T ( ro , z )  Ts 
T ( r ,0)  Ti
(11.104a)
T (r , )  Ts
(11.105a)
114
(11.76) gives axial velocity
vz
1  4 Kn  ( r / ro ) 2
2
v zm
1  8 Kn
(11.76)
 Solution by the method of separation of variables
 Solution is specialized for fully developed conditions at
large z
• Result for air shown in Fig. 11.16
 Neglecting axial conduction: set Pe  
 Axial conduction increases the Nu
115
• Limiting case: no slip and no axial conduction: at Kn = 0
and Pe  
4.5
Nuo  3.657
4.0
(11.72)

3.5
Nu
This agrees with (6.59)
• Limiting case: no slip with
axial conduction: at Kn =0
and Pe = 0:
3.0
2.5
2.0
0
0.04
0.08
0.12
Kn
Fig. 11.16 Nusselt number for flow through tubes
Nuo  4.175
at uniform surface temperature for air
116