Chapter 3 Steady state heat conduction

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```					  Chapter 3 Steady state heat conduction

Introduction
In this chapter, we will steady:
- The thermal resistance concept and its applications
- Spreading and contact thermal resistance
- Heat sink
- The heat flow path in a chip
- The heat flow across and along a PCB

1
3-1 Thermal resistance across plane walls
Thermal resistance of plates in series connection
Under the following conditions
- Perfect contact & constant thermal conductivity
Three plates are in contact: on the right-hand side of surface 3 there is a pool
of fluid and it is subjected to convection boundary condition. The left-hand side
of plate 1 is touched with a heat source with temperature T0. The heat
transmitted from the heat source to plate 1 must equal to that transmitted from
plate 3 to fluid. The thermal conductivities of the 3 plates are k1, k2 & k3,
respectively
0     1    2
3


Q        T0    T1    T2
T3                       h
k1 k2           T
& k1 A (T  T )  k2 A (T  T )  k3 A (T  T )  hA(T  T )
Q                                                   k3 3 
0   1           1   2           2   2
There exist 4 thermal resistances; 1 convection and 3 conduction thermal
L1            L2           L3

resistances.

2
3-1 Thermal resistance in plane walls
The temperature difference between each layer
&
QL1
T0  T1 
k1 A
&
QL2
T1  T2 
k2 A
&
QL3
T2  T3 
k3 A
Q&
T3  T 
         hA

& L1  L2  L3  1 )  Q R
T0  T  Q(                     &
i
k1 A k2 A k3 A hA
                                  i

The heat transfer rate is
T0  T
Q
&
L1   L     L   1
 2  3 
k1 A k2 A k3 A hA


3
3-1 Thermal resistance in plane walls
L1   L2   L3   1
The total thermal resistance is                 Rt              
k1 A k2 A k3 A hA
The thermal resistance diagram
T0       T1              T2                T

Q T3                                                  
Q
L1/k1A      L2/k2A    L3/k3A   1/h2A
The temperature at the interface of wall 2 & wall 3, or T3 can be calculated
by;
& T  T2  T0  T2  T  T  Q( L1  L2 )
Q 0                           &
 Ri L1  L2       2   0
k1 A k2 A
3
k1 A k2 A

The temperature at any surface can be calculated by using thermal
resistance concept. The above example is designed to simulate a chip with
3
layers of different materials from the junction (heat source) to the case
where
it is subjected to convection cooling.

4
Example 3-1 A chip subjected to convection cooling
A chip has surface area 5cm x 5cm. From the junction to the inner surface of
the case there is a protection layer with thickness 1mm and thermal
conductivity 4W/mK. The thickness of the case is also 1mm and thermal
conductivity 2W/mK. The ambient temperature and convection heat transfer
coefficient are 20oC and 20W/m2K. If the junction temperature, T1, is limited
to 100oC, what is power of chip? What is the case temperature ,T3 ?
T1             T2 T3                        tp/kpA   tc/kcA   1/h∞A
                                           
Q
Q     T1            T2       T3        T

The three thermal resistance are
tp            0.001                   t      0.001                         1
                0.1o C / W , c               0.05o C / W ,               2.2o C / W
kk A       4x0.05x0.05               kc A 2x0.05x0.05                 20x0.05x0.05

Q& T       80
 34W
 R 0.1 0.05  2.2
   i

Q & T3  T  34  T3  20  T  94.8o C
3
R3            2.2

5

•   Whenever heat flows from one layer to another of different cross-
sectional areas, the heat flow is not one-dimensional. In order to simplify
the calculation, the thermal resistance can be approximately estimated
using 1-D plus a correction factor; the spreading thermal resistance.
T
t                                 t1    t2
1         A                       Rt  R1D  Rsp              Rsp
T
A1
1   t                            k1 A1 k2 A2
2
2               2
T
Rsp is spreading thermal resistance, t, k and A are the thickness, thermal
3
conductivity and cross-sectional area of the two blocks, respectively. The
heat transfer rate across the two blocks

  T1  T3
Q
Rt
the thermal resistance diagram of heat flow across the two blocks
t1           t2
is                  ........     ...........Rsp
k1 A1       k2 A2
T1                                       T3

6
•   Rsp can be found from chart or empirical equations. A simple equation is
1
Rsp                 (oC/W)
 dk
d is the diameter of the smaller area k is thermal conductivity of the
larger area. For arbitrary shape of the smaller area, the equivalent
diameter is
1             4A
A  d  d 
2

4              

7
3-4 The thermal contact resistance
•   Contact thermal resistance is caused by imperfect
surface smoothness of the two surfaces in contact.
The enlarged view of an elemental area of the
surface
in contact is shown in figure.
•   Two rod in contact with perfect insulation in the
external surfaces, there exists a temperature drop at

thermal interface resistance, Rint,
the interface. The T
Q                                    Q          
Q
is
Rint
T
•   Thermal contact conductance, hc, is                  T
Q  h AT  R  1
defined asc        int
hc A

•   Units: Rint is (oC/W) and hc is (W/m2K) or (W/m2C)

8
• Thermal contact conductance is usually measured experimentally, it
depends on:
- the structure of surface roughness
- the hardness and elastic modulus of the materials in contact
- thermal conductivity of interstitial fluid in the gaps
- the applied pressure at the interface, etc.
• It contributes a significant portion to the total thermal resistance in
the electronic package.
• Use thermal interface material to reduce interface thermal
resistance

9
3-5 heat sink

•   Introduction
Convection thermal resistance is defined as 1/hA. In order to decrease the
thermal resistance to enhance the heat transfer rate. Two steps may be
considered
- Increase heat transfer coefficient—apply forced convection or increase v.
- Increase the heat transfer area →using heat sink

∞

10
-

11
•   Types of heat sink

12
•   Common types of fin
- Flat fin
- Pin fin

•   The fin heat transfer equations
Based on the following assumptions;
- Constant material and coolant properties
- No heat generation source inside the fins
- Constant convection heat transfer coefficient
- Uniform fin cross-sectional area, and
- Constant fin base temperature,
the fin heat transfer equation is derived

13
The fin heat transfer equations

                  dT
•   The heat conducted from the base to x is Q x     and Qx  kAc
                    dx
From the to x +Δx to the tip direction is Qx x
The difference between the two is equal to the
heat dissipated by convection                    W
 Q 
Qx      x x  hpx (T  T )

h = convection heat transfer coefficient
t
p = the perimeter of the fin = 2(t+w)
w =length of the fin
t = fin thickness                                               
                                                        Qconv
Qx x  Qx dQ
     hp(T  T )                      
Qx         
Qx x
dx      dx
The above equation becomes                                  x     x+Δx
d 2T              d 2T
kAc 2  hp(T  T )  2  a 2 (T  T )
dx                dx
1
hp 2
a(     )
kAc
14
Defining an excess temperature   T  T , the fin heat transfer equation
is         d 2
 a 2
dx 2
•    The solutions
The general solution is
  c1e ax  c 2 e  ax         (a)
c1 and c2 are the integration constants
At the base, x = 0, the temperature is equal to the base temperature Tb..
b  c1  c2 and b  Tb  T                             (1)
One more boundary condition is required
- Infinite long fin : If the fin is very long, at the tip x = L, TL  T
and c1 = 0. The temperature distribution is
  b e  ax  T  T  (Tb  T )e  ax
The heat transfer rate                                          Tb
T∞
         dT                                           1
Q  (kAc    ) x 0  kAcb (a)  b kAc a  (kAc hp) 2 b
dx
15
 dT 
- Insulated fin tip
 dx    0
  x L
From equation (a) c ae aL  c ae  aL  c e aL  c e  aL  0    (2)
1         2         1       2
Solving equation (1) and (2), the two integration constants are:

b e  aL                          b e aL
c1             aL
and      c1 
e e
aL
e aL  e  aL
Substituting c1 & c2 into equation (a), the temperature distribution is
cos ha( L  x)
  b                            Note :        T  T
cosh aL
1
         dT
The heat transfer rate is       Q  (kAc    ) x 0  (kAc hp) 2 b tanh aL
dx
When aL is large, tanhaL approaches to unity

16
- Convection tip boundary condition
The total fin surface area At including the tip area is At  Lp  Ac
Dividing the above equation by the perimeter, p, the corrected fin length
is                      A
Lc  L  c
P

the
If L is replaced by Lc, the tip surface area ofwt fin subjected to
wt      t
L  L
convection boundary condition is cconsidered              L     L
2( w  t )      2w      2
(1) For a rectangular fin
D
Lc  L 
4
(2) For a circular fin

t     t/2
L
c

17
The fin efficiency

•   Fin efficiency is defined as the ratio of actual heat transfer rate to the
maximum possible heat transfer rate
•   The surface temperature of the fin decreases from the base to the tip
direction. The degree of variation depends on the dimensions and the
thermal conductivity of the fin. If the thermal conductivity is very large, the
surface temperature may be constant and equals to Tb.
•   The heat transfer rate from the elemental area, pdx


dQ  hpdx (T  T )
•   The total heat transfer rate from the entire fin
L
Q fin   hp(T  Ta )dx

•   The maximum possible heat transfer rate of the entire fin
0
T = Tb = constant

Qmax  hpLb  hAfinb
•   The fin efficiency

Q fin
 
Qmax

18
The fin efficiency
•   Infinite long fin
1            1
(hpkAc ) b (hpkAc ) 2 b 1 kAc 12
2
1
                          (   ) 
hAfinb      hpLb      L hp     aL

•   Insulated fin tip        1
(hpkAc ) 2  b tanh aL tanh aL
                         
hAfin b           aL

•   Convection fin tip
tanh aLc

aLc

•   When aL is large than 3, the above will yield almost identical results.

19
Fin effectiveness of a single fin
•     Fin effectiveness is defined as the ratio of the rate of heat transfer from
the fin to the rate of heat transfer from the base area of the fin


Q fin hAfin (Tb  Ta ) Afin
 fin                              
Qnofin   hAb (Tb  Ta )   Ab    Base
fin
where Afin is the surface area of the fin and Ab is the base area.

•     For a very long fin
pL 1   pL kAc   kp
 fin                       ,,,,, ( Ab  Ac )
•
Ab aL Ab hpL    hAc                          single fin

- Smaller than unity, the fin reduces the heat transfer rate. It acts as an
insulator.
- Equal to unity, the fin has no enhancement in heat transfer.
- Larger than unity, the fin enhances heat transfer. For practical
consideration, it should be larger than 2

20
The overall effectiveness of fin array
•    The total rate of heat transfer of the fin array
                
Qtotal  Qunfin  Qfin  hAunfin (Tb  T ) hNAfin (Tb  T )
N is the number of fins and Afin is the
W
lateral surface area of a single and Aunfin is
the base area that is not occupied by fin

Qtotal  h( Aunfin  NAfin )(Tb  T )

•    The overall effectiveness of the fin array

Qtotal h( Aunfin   NAfin )(Tb  T ) ( Aunfin   NAfin )

o                                      
Qnofin         hAnofin (Tb  T )               Anofin
•    The thermal resistance of a fin array
Tb  T             1              1
R fin                                  

Qtotal   h( Aunfin   NAfin )  hNA fin

W
•    Number of fins N                     , t is the fin thickness & s is the spacing.
st
21
Example

•   1. A hot surface of 10cm long and 20cm wide at 100oC is to be cooled by
convection heat transfer. An engineer is asked to study the results by
attaching a heat sink of identical base area to the hot surface. The heat
sink has a number of equally spaced, rectangular fins of 5cm height and
2mm thick. The spacing between the fins is 4mm, assuming the tip of the
fin is under adiabatic boundary condition. The length of the fins is equal to
that of the hot surface (10cm). The thermal conductivity of the fin is
200W/mk and the convection heat transfer coefficient is 100 W/m2K & the
ambient air temperature is 20oC. Determine:
- The heat transfer rate of the hot surface? .
- The heat transfer rate by attaching the heat sink specified above
- The thermal resistance of the heat sink
•   Solutions
- The heat transfer rate of the hot surface

Q  hA(Ts  Ta )  100 x 0.2 x 0.1(100  20)  320W

22
•   The heat transfer rate of the fin array
- The fin efficiency
hp    100 x 2(0.1  0.002)
a                               22.6                                       100m
kAc     200 x0.002 x0.1                                                 m
L  0.05m                                                       t              50m
=2mm
  tanh aL / aL  0.83 /1.15  72%                                            m
- the heat transfer rate of the fin array                 200m
W     200                                   m
N               200 / 6  33 fins
t s 24
NAfin  NpL  33x 2(0.1  0.002) x0.05  0.336m 2
Aunfin  0.2 x0.1  33 x0.1x0.002  0.014m 2

Qtotal  ( Aunfin   Afin )(Tb  Ta )  100 x(0.014  0.72 x0.336)80  2047W
- The thermal resistance of the fin array

T   80
Rsin k        0.039o C / W
Q 2047

23
Thermal resistance across a semiconductor device

•   The junction to ambient thermal resistance
Tj               Tc          T

T j  T
R j 
Q&
R j  R jc  Rc
T T    T T      T T
Q & j   j   j c  Tc  T
R j R jc  Rc  R jc Rc

•   With heat sink, there must be an interface material in between, and therefore there
is an interface thermal resistance between the case and the base of the heat sink.
Tj            Tc          Tint         T

R j  R jc  Rint  Rs

24
Example: Interface thermal resistance
A device dissipates 50W, the maximum operation junction temperature is 85C
& the junction to case thermal resistance is 0.13C/W. The ambient temperature
is 50C. the junction to ambient thermal resistance at air velocity 5m/s without
heat sink is 4.7C/W. Determine whether a heat sink is required? If it is
required, a layer of interface material is necessary and the interface thermal
resistance is 0.1C/W. What is the maximum thermal resistance of the heat
sink? The heat dissipation to the board is neglected.
T  T
(a) Without heat sink.       Q  j       T j  T  QRj  50  50x4.7  144C
&
Rj

A heat sink is required, otherwise the junction T. is too high.
(b) Select the thermal resistance of the heat sink

R j  R jc  Rint  Rs
T j  T                 85  50
Rs              R jc  Tint           0.13  0.1  0.47C / W
Q&                     50

A heat sink with thermal resistance 0.47C/W is required

25
Heat flow in PCB

•   Construction and material properties
The main material is epoxy-glass,
- Good electrical insulator : A necessary requirement
- Poor heat conductor : ke = 0.26W/mK.
High thermal conductivity copper cladding, heat frame, or cores are used.
kc = 386W/mK. It is a multi-layer structure
W         L                  t
t   e
•   Temperature difference from center to the ends depends on the
c
thicknesses of each layer, the thermal conductivities, the length L, and
the heat flux on the board
•   The effective thermal conductivity
- The heat flow along the board
T          T                       T                    T
Qt  Qe  Qc  (kA
& & &               )e  (kA    )c  [(kA)e  (kA)c ]     keff ( Ae  Ac )
                   L           L                        L                     L
(kA)e  (kA)c (kA)e  (kA)c
keff                
- The effective thermal conductivity                 Ae  Ac              At

26
Example 3 Heat conduction along a PCB with copper cladding

•   Given : A = 10cm x 10cm, tc = 0.04mm, te = 0.16mm,
Kc = 380W/mK, ke = 0.26W/mK

•   Find (1) % of heat conducted along Cu and Ep
(2) keff
(kt )e  0.26 x0.16 x103  0.04 x103W / K
•   Solution
(kt )c  386 x0.04 x103  15.44 x103W / K
(kt )e  (kt )c  15.48 x103W / K
                      T           T                             wT
Qt  Qe  Qc  (ktw          )e  (ktw      )c  [(kt ) e  (kt ) c ]
L            L                               L
T
          (ktw        )e
Qe                  L                (kt )e         0.04 x103
                                                               0.26%

Qt [(kt )  (kt ) ]      wT (kt )e  (kt )c 15.48 x103
e        c
L
(kt )e  (kt )c 15.48 x103
keff                                  77.4W / mK
te  tc            0.2                                            27
Temperature distribution in PCB

•   Temperature distribution along the board
- Assumptions
a. uniform heat flux generation on the surface of the board
b. all heat from the devices on the board is transmitted along the board
to the heat sinks at the two ends
L                 L
HS                                     HS

- Equivalent uniform heat generation in the board
The heat from the surface area of the board may be considered a equivalent
heat generation in the board. Consider an elemental surface area


q
dA              t

g

  qdA  gtdA  g  q
Q              
t

28
•   The one-dimensional, steady state heat conduction equation is
d 2T    
g    d 2T   
q
2
       2        0
dx     keff  dx   tkeff
•   The boundary conditions
dT
x  0,     0                     TL                               TL
dx                                                   x
x   L, T  TL                                         0       L
•   The solution
dT      
q                    
q 2
Integration twice                   x  c1  T          x  C1 x  C2
dx    tkeff                2tkeff
qL2

Apply the boundary conditions            c1  0, c2  TL 

2tt keff
q
T  TL         ( L2  x 2 )
The solution is                          2tkeff

The maximum temperature is located at the center of the board, x = 0
2

qL
Tmax    TL 
2tkeff
29
Example 4: Heat conduction along a PCB with copper cladding

•   Given : Applying the identical conditions as given in example 3, A = 20cm x 10cm, tc
= 0.04mm, te = 0.16mm, Kc = 380W/mK, ke = 0.26W/mK, Uniform heat flux on the
surface of the board = 0.01W/cm2
•   Find the temperature difference between the center and the end of the board
Tmax       TL

10cm          10cm
•   Solution: From previous example the effective thermal conductivity is
keff  77.4W / mK
q  0.01W / cm2  0.01x104 W / m2  100W / m2

qL2
                      100 x0.12
Tmax    TL          Tmax  TL                   32.3o C
2tkeff               2 x0.0002 x77.4
If the two ends temperature is 20oC, the center temperature is 52.75oC

30
The following pages will not be taught

31
Example 5 temperature distribution along the board
Given: Identical conditions as given in example 3..

-L                      0                       +L
The two ends of the heat frame are at 20oC
Find the temperature at the center of the board using thermal resistance method
Dividing the board into 20 sections, each section has a surface area 1cm x10cm
& each section generates 0.01x10 = 0.1W. The thermal resistance for each
section is         L             0.01
Re                                  o
 6.5 C / W
keff At       77.4x(0.1x0.0002)

0.1W                    0.1W
center
0.1         0.1     0.1        0.1             0.1
W       W              0.1W        W
W                              W                                 0.1   0.1
W     W
TC        T1        T2     T3    T4      T5     T6       T7   T8     T9    TL

Consider only half of the board
32
•   The temperature drop across each section is
Tc  T1  Qc Rc  0.1x6.5  0.65C
&
T  T  2Q R  2x0.65  1.3C
1    2
&
c    c

T2  T3  3Qc Rc  3x0.65  1.95C
&
T3  T4  4Qc Rc  4x0.65  2.6o C
&
T  T  5Q R  5x0.65  3.25o C
4    5
&
c    c

T5  T6  4Qc Rc  6x0.65  3.9o C
&
T6  T7  4Qc Rc  7x0.65  4.55o C
&
T  T  4Q R  8x0.65  5.2o C
7    8
&
c   c

T8  T9  4Qc Rc  9x0.65  5.85o C
&
T9  TL  4Qc Rc  10x0.65  6.5o C
&

Adding the above equations, the maximum temperature difference is
Tc  TL  0.65 1.31.95  2.6  3.25  3.9  4.55  5.2  5.85  6.5)  35.73o C
The center temperature is 20 + 35.75 = 55.75oC
33
The following pages will not be taught

34
• Cupper cladding of equal width with epoxy layer, A=wt
wT           wT
Qt  [(kt)e  (kt)c ]
&                        keff tt
L             L
(kt)e  (kt)c
keff 
te  tc

• High thermal conductivity wires or rods with radius r
T               T
Qt  [(kA)e  (kA)c ]
&                             keff At
L                L
ke [(tw)e  n r 2 ]  (kn r 2 )c
keff 
(tw)e


n is number of wires o rods, te is the thickness of the epoxy board.

35
Example
•   Given: A = 10cm x 12cm, Power = 24W, copper heat frame has a thickness
1.2mm and thermal conductivity = 386W/mK,

Uniform heat flux on top
Epoxy, k = 0.26W/mK, t=
0.8mm
Epoxy adhesive, k = 1.8W/mK, t =
0.13mm
Copper frame, k = 386W/mk, t = 1.2mm

The two ends of the heat frame are at 20oC
•   Find the temperature at the center of the board
•   Dividing the board into 12 sections, each section generates 2W. The
thermal resistance diagram is shown in next page.

36
L          0.01
Re                             65o C / W
Example 15-8                      keff At 77.4x(0.1x0.002)

Thermal resistance diagram           1cm      1cm          1cm        1cm        1cm
1cm
2W       2W      2W      2W            2W
T 2W
h
R
e

Ra
+R
TL=
6 R 5 R 4 R 3 R 2 R
c1
c2       c2      c2       c2       c2                 20oC
1 R
c2
L               0.0008
Re  ( )epoxy                         3.077 o C / W
kA          0.26 x(0.01x0.1)
L                0.00013
Ra  ( ) adhesive                     0.072o C / W
kA            1.8 x(0.01x0.1)
L                0.0006
Rc1  ( )copper1                       0.002o C / W
kA           386 x(0.01x0.1)
L                    0.01
Rc 2  ( )copper 2                         0.216o C / W
kA            386 x(0.0012 x0.1)
37
Rv  Re  Ra  Rc1  3.151 C / W
o
•   The temperature at the center
Th  T6  2 Rv
T6  T5  2 Rc 2
T5  T4  4 Rc 2
T4  T3  6 Rc 2
T3  T2  8Rc 2
T2  T1  10 Rc 2
T1  T0  12 Rc 2
Th  T0  (2 Rv  2 Rc 2  4 Rc 2  6 Rc 2  8Rc 2  10 Rc 2  12 Rc 2 )
Th  T0  (2Rv  42Rc2 )
Th  20  (2x3.151  42x0.216)
 20  6.302  9.072  35.4o C

The center temperature is 35.4oC which is the maximum on board
38
3-2 Thermal resistance across a hollow cylinder
•   Assumptions
r
- One-dimensional                     ri
ro
- constant thermal conductivity
- No internal heat generation
•   Heat conduction equation and boundary conditions
d dT                 r = ri , T = Ti (1)
r    0          r = ro, T = To (2)
dr dr
dT        dT c1
•   Integrating twice       r     c1      T  c1 ln r  c2
dr        dr r
To  c1 ln ro  c2
•   Applying the two boundary conditions
Ti  c1 ln ri  c2
(To  Ti )              To  Ti
C1                   C2           ln ri  Ti
ro                     ro
• Solving the integration constants    ln( )                  ln( )
ri                     ri
39
The general solution

(To  Ti )        To  Ti              (To  Ti ) r
T  c1 ln r  c2             ln r          ln ri  Ti            ln( )  Ti
ro               ro                    ro       ri
ln( )            ln( )                 ln( )
ri               ri                    ri
•   The heat transfer rate
        dT               dT          T  Ti 1 Ti  To
Q   kA      k (2 rL)     2 rLk o      
dr               dr             ro r     R
ln( )
ri

•   The thermal resistance
ln(ro / ri )
R                          Ti              To
2 kL                         R

40

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